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HOW DOES THE 3-CODE SMD RESISTOR VALUE WORK?
Thankfully for makers, SMD resistors (even down to 0603 sizes), have a three number code printed on them that can be used to identify its resistance. However, this number doesn’t tell us the
tolerance of that resistor, nor does it indicate its power, which can make it difficult to replace should a resistor break.
In most cases, the number code printed on a resistor is split up into two different numbers; the first two which indicate a decimal value, and a single number that indicates a multiplier. The
multiplier is not the number of time the number is multiplied, but the number of times that number is multiplied by 10. The best way to see how this works is to go through a few simple examples.
Example 1 - Decoding 123
In this case, the first to numbers are 1 and 2, so the decimal value is 12. The third number is 3, meaning that we need to multiply the number by 10 three times. So, 12 becomes 12 x 10 x 10 x 10
which is 12,000, or 12KΩ.
Example 2 - Decoding 56
In this case, the first two numbers are 5 and 6, so the decimal value is 56. As there is no third number, we don’t multiply the number at all (being 0 lots of 10). So, the value of this resistor
would be 56Ω.
Example 3 - Decoding 225
In this case, the first two numbers are 2 and 2, so the decimal value is 22. The third number is 5, meaning that we multiple the number by 10 five times. As such, the value of this resistor would be
22 x 10 x 10 x 10 x 10 x 10 which is 2,200,000, or 2.2MΩ. | {"url":"https://www.mitchelectronics.co.uk/resources/identifying-part-values/smd-resistor-identification","timestamp":"2024-11-12T15:57:34Z","content_type":"text/html","content_length":"6932","record_id":"<urn:uuid:e1eb4d86-9172-4f9c-8120-c17e9b2faca1>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00607.warc.gz"} |
Two Ways to Destroy the Economy: Hyperinflation and Central Planning
In the following video, Professor Engelhardt says that if he were a “Super Villain” and wanted to destroy an economy, there are two ways that seem obvious to him. The first is “Central Planning”,
aka. “Socialism” and the Second is Hyperinflation. Since he would be a lazy super villain, he says Central Planning is “too much work”, so he would choose hyperinflation.
He goes on to say that hyperinflation is difficult to define; even Economists like Murray Rothbard and Ludwig Von Mises used different definitions of plain inflation depending on the context.
Generally, inflation was originally defined as an “increase in the money supply in excess of the increase in the demand for money”. More recently, the term inflation has come to mean a “Big” increase
in prices. But the difficulty comes in determining how big is “Big”?
If we can’t even nail down the definition of inflation, how can we possibly nail down the meaning of hyperinflation… i.e. where does regular inflation end, and hyperinflation start?
Although it is difficult to say exactly… one definition of hyperinflation says that it begins at 50% PER MONTH. Personally, I would consider 50% PER YEAR close enough to be hyperinflation. In
addition to the supply of and demand for money, there are a couple of other factors that influence whether hyperinflation occurs or not. The first is “Velocity of Money” i.e. how quickly money
changes hands. The second is “Confidence in the Money”, when people lose confidence in their currency, they tend to want to unload it quickly and are willing to accept less value for each dollar.
The Three Phases of Hyperinflation
Phase 1- An increase in the supply of money paired with “Deflationary Expectations”
Typically, inflation is initially localized in a “handful of goods”, and people wonder why those goods are rising in price while other goods are still stable in price. So expectations are still that
prices are stable and “those prices will come back down”.
Phase 2- A further increase in the money supply paired with “Inflationary Expectations”
People realize that this inflation is not temporary, i.e. inflation is going to continue. People start spending faster because they no longer want to hold that money like they used to. Higher prices
are starting to spread throughout the rest of the economy.
Phase 3- A flight to real values
People totally lose confidence in the currency and begin seeking out other stores of value. This means that the demand for money is effectively zero. No one wants to hold money, they would rather
have virtually anything else. This raises the problem that if no one wants to hold money, who will be willing to accept it? Once no one wants the money, the economy begins to collapse, and it reverts
to a barter system which isn’t terribly efficient. So, if you want to destroy the economy, simply create hyperinflation by printing too much money.
Hyperinflation and the Destruction of Human Personality
One of the problems with hyperinflation is that it affects our habits of thought and action. Typically there are certain habits that lead to success, including:
• Having a “goal orientation”
• Choosing “delayed gratification”
• A “self-interest” combined with a desire to “serve others well”
• Desiring “thrift” i.e. don’t waste scarce resources, including time.
Hyperinflation (and Central Planning) destroys these normal incentives, in hyperinflation, we no longer have the long-term perspective; instead, we become survival-oriented and self-centered.
Professor Englehardt uses Venezuela as a current example of hyperinflation and its effects.
Recorded at the Mises Institute in Auburn, Alabama, on 28 July 2022.
• <a “https://inflationdata.com/articles/2008/02/24/what-is-core-inflation-and-why-doesnt-it-include-food-and-energy/”>What is Core Inflation?
The video originally appeared here.
Leave a Reply Cancel reply | {"url":"https://inflationdata.com/articles/2022/08/15/two-ways-to-destroy-the-economy-hyperinflation-and-central-planning/","timestamp":"2024-11-13T04:18:48Z","content_type":"text/html","content_length":"101730","record_id":"<urn:uuid:5293b442-7de5-4652-bd90-3b3543ecc8cb>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00824.warc.gz"} |
ASVAB Math Knowledge Practice Test 924537
Questions 5
Topics Pythagorean Theorem, Trapezoid, Triangle Classification, Triangle Geometry, Two Variables
The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2
+ b2): c2 = a2 + b2 or, solved for c, \(c = \sqrt{a + b}\)
A trapezoid is a quadrilateral with one set of parallel sides. The area of a trapezoid is one-half the sum of the lengths of the parallel sides multiplied by the height. In this diagram, that becomes
½(b + d)(h).
An isosceles triangle has two sides of equal length. An equilateral triangle has three sides of equal length. In a right triangle, two sides meet at a right angle.
A triangle is a three-sided polygon. It has three interior angles that add up to 180° (a + b + c = 180°). An exterior angle of a triangle is equal to the sum of the two interior angles that are
opposite (d = b + c). The perimeter of a triangle is equal to the sum of the lengths of its three sides, the height of a triangle is equal to the length from the base to the opposite vertex (angle)
and the area equals one-half triangle base x height: a = ½ base x height.
When solving an equation with two variables, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents,
Multiplication/Division, Addition/Subtraction.) | {"url":"https://www.asvabtestbank.com/math-knowledge/practice-test/924537/5","timestamp":"2024-11-04T13:41:04Z","content_type":"text/html","content_length":"11249","record_id":"<urn:uuid:50bc1396-70de-477c-950c-03e7dafec500>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00529.warc.gz"} |
Special Initial Conditions: A New Kind of Science | Online by Stephen Wolfram [Page 267]
In fact, it turns out that in any cellular automaton it is inevitable that initial conditions which consist just of a fixed block of cells repeated forever will lead to simple repetitive behavior.
For what happens is that each block in effect independently acts like a system of limited size. The right-hand neighbor of the rightmost cell in any particular block is the leftmost cell in the next
block, but since all the blocks are identical, this cell always has the same color as the leftmost cell in the block itself. And as a result, the block evolves just like one of the systems of limited
size that we discussed on page 255. So this means that given a block that is n cells wide, the repetition period that is obtained must be at most 2^n steps.
But if one wants a short repetition period, then there is a question of whether there is a block of any size which can produce it. The pictures on the next page show the blocks that are needed to get
repetition periods of up to ten steps in rule 30. It turns out that no block of any size gives a period of exactly two steps, but blocks can be found for all larger periods at least up to 15 steps.
But what about initial conditions that do not just consist of a single block repeated forever? It turns out that for rule 30, no other kind of initial conditions can ever yield repetitive behavior.
But for many rules—including a fair number of class 3 ones—the situation is different. And as one example the picture on the right below shows an initial condition for rule 126 that involves two
different blocks but which nevertheless yields repetitive behavior.
Rule 126 with a typical random initial condition, and with an initial condition that consists of a random sequence of the blocks | {"url":"https://www.wolframscience.com/nks/p267--special-initial-conditions--webview/","timestamp":"2024-11-04T15:31:35Z","content_type":"text/html","content_length":"83656","record_id":"<urn:uuid:5555ea7e-7dba-46e3-9041-4cf6e32e0d9f>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00488.warc.gz"} |
The Risk Bubble Chart – Part 1 | SumProduct are experts in Excel Training: Financial Modelling, Strategic Data Modelling, Model Auditing, Planning & Strategy, Training Courses, Tips & Online Knowledgebase
Charts and Dashboards: The Risk Bubble Chart – Part 1
Welcome back to our Charts and Dashboards blog series. This week, we’re going to look at how to make a Risk Bubble chart.
The Risk Bubble Chart
Have you ever wanted to organise your risks and consequences based on the likelihood of occurrence and the potential damage? Today, we will introduce you to a Risk Bubble chart which helps you to
keep track of the likelihood of any given problem, the consequences of the problem, and the size of said problem.
Here, this chart categorises types of risk in two dimensions, i.e. both the impact of the risk and the likelihood of it occurring. The size of the bubble shows how many categories meet the same
criteria, but this could be replaced by a monetary value impact, the number of customers affected, etc. And yes, I do appreciate that using bubbles (circles) may not be the best way to depict the
number of risk types due to the area of a circle being proportional to the square of its radius! The aim of this article is not to defend such practice, but to show how to create such a chart given
they are used for risk assessment / analysis.
A key element of this chart is the risk input table, which is automatically updated based upon your input(s) and the Risk Bubble chart. To assist us, there are a few key inputs we will be using here.
The first table we have is a table we shall name LU_Matrix (i.e. the “LookUp Matrix”), viz.
In this table, we will name the row headings (in grey) LU_Likelihood and the column / field headings (also in grey) LU_Consequences. With these inputs, we can create a risk table that changes colour
The next input we will need is the risk category table which determines the likelihood and consequence of a risk:
We name this table Risk_Category.
In this table, we can enter the Risk Category, Likelihood, and Consequence. These inputs will be used to make the Risk Bubble chart, but we will need to refine this data for it to be usable.
With these inputs available, the question now is: how do we make the Risk Bubble chart? Well, let’s get started with the risk table…
Risk Table
The first step is to construct the axis for this chart. For this task, we will need to employ the COUNTA, the SEQUENCE, and the INDEX functions. COUNTA as you guessed will “count” every cell that is
not blank. Therefore, we will count the number of Likelihood categories:
The output for this function will be five [5] as we have five [5] Likelihood categories in our example. As we generate a sequence of numbers, we will wrap up the COUNTA formula inside the SEQUENCE
This will generate an array from one [1] to five [5]:
However, we want this array to generate in reverse order, i.e., five [5] to one [1]. This can be achieved by setting the third argument of the SEQUENCE function to our number of Likelihood categories
and the fourth argument to negative one [-1]:
The third argument of the SEQUENCE function will specify the starting value and the fourth argument will specify the increment for each subsequent value. This will give us the following:
(If dynamic arrays are not available, alternative formulae can generate the same desired result).
Finally, we apply our INDEX function here to make the y-axis (vertical, or dependent, axis) of the chart:
The INDEX function will take the fifth item of the LU_Likelihood and put it on the top row, the fourth item put on the second row, and we repeat this process until we have the first item in the fifth
Then, we will go one cell below and one cell to the right of the word “Rare” here to enter our second, much simpler, formula:
After entering the formula in that cell (again, assuming you have Dynamic Arrays in your version of Excel), we will have the following visual:
Next, we apply some formatting:
We have formatted each of the Likelihood and Consequence values to make them look more like headings, whilst rotating the text in the y-axis. We have also resized all cells within the graph area to
dimensions of 100x100 pixels (a column width of 13.57 and a row height of 75).
Assuming the grid is positioned as in the graphic (above), in cell L89, we will enter the following formula, and then copy this throughout the entire table:
=INDEX(LU_Matrix[[Minimal]:[Catastrophic]], MATCH($K89,LU_Likelihood,0), MATCH(L$94,LU_Consequences, 0))
These MATCH functions will match the row position of the item on the left of the LU_Matrix and the column position of the item on the bottom of the LU_Matrix. Then the INDEX function will return the
item in the exact row and column that we match on the LU_Matrix. After entering the formula and populating the table we will have the following visual:
With the headings and shape of the chart in place, let’s take a break. Next week, we’ll look at making this a little more colourful.
That’s it for this week, come back next week for more Charts and Dashboards tips. | {"url":"https://www.sumproduct.com/blog/article/charts-and-dashboards-blogs/charts-and-dashboards-the-risk-bubble-chart-part-1","timestamp":"2024-11-08T15:48:26Z","content_type":"text/html","content_length":"42379","record_id":"<urn:uuid:77fa4714-d73c-479f-aa65-4f3820ff0a4f>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00508.warc.gz"} |
A chemical laboratory in Peru has discovered certain substances to lose weight in a short time to replace existing medications. The proportion of people in which the substance is effective is 0.7, while currently existing drugs are 60% effective a random sample of 100 people were given the new substance While the old drug was used in a sample of 150 What is the probability of observing a p1-p2 value less than 5%?
To begin with, we need to calculate the sample proportions for both the new substance and the old drug: For the new substance: Sample size (n1) = 100 Proportion of people for whom the substance is
effective (p1) = 0.7 Sample proportion (p̂1) = x1/n1, where x1 is the number of people in the sample for whom the substance was effective. We don't know the value of x1, but we can use the given
proportion to estimate it: x1 = p1 * n1 = 0.7 * 100 = 70 Sample proportion (p̂1) = x1/n1 = 70/100 = 0.7 For the old drug: Sample size (n2) = 150 Proportion of people for whom the drug is effective
(p2) = 0.6 Sample proportion (p̂2) = x2/n2, where x2 is the number of people in the sample for whom the drug was effective. We don't know the value of x2, but we can use the given proportion to
estimate it: x2 = p2 * n2 = 0.6 * 150 = 90 Sample proportion (p̂2) = x2/n2 = 90/150 = 0.6 The null hypothesis is that the proportion of people for whom the new substance is effective is equal to the
proportion for whom the old drug is effective, that is, H0: p1 = p2. The alternative hypothesis is that the new substance is more effective than the old drug, that is, Ha: p1 > p2. We can use a
two-sample z-test to compare the sample proportions and calculate the p-value: Test statistic: z = (p̂1 - p̂2) / sqrt(p̂(1-p̂)(1/n1 + 1/n2)) where p̂ = (x1 + x2) / (n1 + n2) is the pooled sample
proportion. z = (0.7 - 0.6) / sqrt(0.63 * 0.37 * (1/100 + 1/150)) = 2.22 (rounded to two decimal places) p-value = P(Z > 2.22) = 0.0139 (using a standard normal distribution table or a calculator)
The p-value is less than 5%, so we reject the null hypothesis at the 5% level of significance. We can conclude that there is strong evidence to suggest that the new substance is more effective than
the old drug for weight loss. | {"url":"https://math4finance.com/general/un-laboratorio-quimico-de-peru-ha-descubierto-ciertas-sustancias-para-reducir-de-peso-en-poco-tiempo-para-suplantar-a-los-medicamentos-ya-existentes-la-proporcion-de-personas-en-las-cuales-la-sustan","timestamp":"2024-11-08T14:28:55Z","content_type":"text/html","content_length":"32133","record_id":"<urn:uuid:f87a49e1-981c-4f84-8799-993575366720>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00461.warc.gz"} |
Essays on Dynamic Optimization for Markets and Networks
2023 Theses Doctoral
Essays on Dynamic Optimization for Markets and Networks
We study dynamic decision-making problems in networks and markets under uncertainty about future payoffs. This problem is difficult in general since 1) Although the current decision (potentially)
affects future decisions, the decision-maker does not have exact information on the future payoffs when he/she commits to the current decision; 2) The decision made at one part of the network usually
interacts with the decisions made at the other parts of the network, which makes the computation scales very fast with the network size and brings computational challenges in practice. In this
thesis, we propose computationally efficient methods to solve dynamic optimization problems on markets and networks, specify a general set of conditions under which the proposed methods give
theoretical guarantees on global near-optimality, and further provide numerical studies to verify the performance empirically. The proposed methods/algorithms have a general theme as “local
algorithms”, meaning that the decision at each node/agent on the network uses only partial information on the network.
In the first part of this thesis, we consider a network model with stochastic uncertainty about future payoffs. The network has a bounded degree, and each node takes a discrete decision at each
period, leading to a per-period payoff which is a sum of three parts: node rewards for individual node decisions, temporal interactions between individual node decisions from the current and previous
periods, and spatial interactions between decisions from pairs of neighboring nodes. The objective is to maximize the expected total payoffs over a finite horizon. We study a natural decentralized
algorithm (whose computational requirement is linear in the network size and planning horizon) and prove that our decentralized algorithm achieves global near-optimality when temporal and spatial
interactions are not dominant compared to the randomness in node rewards. Decentralized algorithms are parameterized by the locality parameter L: An L-local algorithm makes its decision at each node
v based on current and (simulated) future payoffs only up to L periods ahead, and only in an L-radius neighborhood around v. Given any permitted error ε > 0, we show that our proposed L-local
algorithm with L = O(log(1/ε)) has an average per-node-per- period optimality gap bounded above by ε, in networks where temporal and spatial interactions are not dominant. This constitutes the first
theoretical result establishing the global near-optimality of a local algorithm for network dynamic optimization.
In the second part of this thesis, we consider the previous three types of payoff functions under adversarial uncertainty about the future. In general, there are no performance guarantees for
arbitrary payoff functions. We consider an additional convexity structure in the individual node payoffs and interaction functions, which helps us leverage the tools in the broad Online Convex
Optimization literature. In this work, we study the setting where there is a trade-off between developing future predictions for a longer lookahead horizon, denoted as k versus increasing spatial
radius for decentralized computation, denoted as r. When deciding individual node decisions at each time, each node has access to predictions of local cost functions for the next k time steps in an
r-hop neighborhood. Our work proposes a novel online algorithm, Localized Predictive Control (LPC), which generalizes predictive control to multi-agent systems. We show that LPC achieves a
competitive ratio approaching to 1 exponentially fast in ρT and ρS in an adversarial setting, where ρT and ρS are constants in (0, 1) that increase with the relative strength of temporal and spatial
interaction costs, respectively. This is the first competitive ratio bound on decentralized predictive control for networked online convex optimization. Further, we show that the dependence on k and
r in our results is near-optimal by lower bounding the competitive ratio of any decentralized online algorithm.
In the third part of this work, we consider a general dynamic matching model for online competitive gaming platforms. Players arrive stochastically with a skill attribute, the Elo rating. The
distribution of Elo is known and i.i.d across players. However, the individual’s rating is only observed upon arrival. Matching two players with different skills incurs a match cost. The goal is
tominimize a weighted combination of waiting costs and matching costs in the system. We investigate a popular heuristic used in industry to trade-off between these two costs, the Bubble algorithm.
The algorithm places arriving players on the Elo line with a growing bubble around them. When two bubbles touch, the two players get matched. We show that, with the optimal bubble expansion rate, the
Bubble algorithm achieves a constant factor ratio against the offline optimal cost when the match cost (resp. waiting cost) is a power of Elo difference (resp. waiting time). We use players’ activity
logs data from a gaming start-up to validate our approach and further provide guidance on how to tune the Bubble expansion rate in practice.
• Gan_columbia_0054D_17841.pdf application/pdf 1.35 MB Download File
More About This Work
Academic Units
Thesis Advisors
Kanoria, Yashodhan
Ph.D., Columbia University
Published Here
June 7, 2023 | {"url":"https://academiccommons.columbia.edu/doi/10.7916/6eja-wd96","timestamp":"2024-11-09T06:56:39Z","content_type":"text/html","content_length":"30967","record_id":"<urn:uuid:dcc565f8-d65e-4b5c-8e66-eb689a00e810>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00555.warc.gz"} |
Exterior angle property of a Triangle
Jump to navigation Jump to search
To show that exterior angle of a triangle is equal to the sum of its interior opposite angles.
Estimated Time
30 minutes
Prerequisites/Instructions, prior preparations, if any
Knowledge about point, lines and angles, adjacent angles, alternate angles, corresponding angles, linear pair
Materials/ Resources needed
Digital : Click here to open the file
Non-digital : Worksheet, pencil, ruler, compass, protractor
Process (How to do the activity)
Download this geogebra file from this link.
1. Draw triangle ABC
2. Identify the angles of the triangle.
3. What is the sum of the angles of a triangle?
4. Extend one side, students should recognize the exterior angle formed.
5. Measure the exterior angle of a triangle
6. Identify interior opposite angles for the exterior angle of a triangle. Mark the angle in purple and green colour.
7. Move the sliders and observe the changes.
8. How are the two angles together related to the exterior angle?
9. Do you notice any relation between the exterior angle and the interior angles
Evaluation at the end of the activity
1. What would be the measure of exterior angle for each vertex of an equilateral triangle?
2. Does an exterior angle of a triangle is smaller than either of its interior opposite angles?
Go back to the page - click here | {"url":"https://www.karnatakaeducation.org.in/KOER/en/index.php/Exterior_angle_property_of_a_Triangle","timestamp":"2024-11-02T18:07:40Z","content_type":"text/html","content_length":"35561","record_id":"<urn:uuid:f75089b8-948e-461a-ac09-78d709071d2f>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00171.warc.gz"} |
Coverage estimation for Census 2021 in England and Wales
• Census 2021, as with any census, was subject to non-response and incorrect responses.
• The Office for National Statistics (ONS) developed on our successful approach to estimating coverage error from 2011 and built on the improved processing of Census 2021.
• We used logistic regression models rather than a stratified approach to dual system estimation, allowing us to account for the effect of many more characteristics on response.
• The results then went through a thorough quality assurance process, with adjustments made where necessary; the census count was 97% of the final estimate.
Nôl i'r tabl cynnwys
The Census of England and Wales provides an accurate, comprehensive and consistent picture of the England and Wales population, as laid out in Design for Census 2021. The key aim of the Census is to
produce high quality population counts, at subnational and national level, for demographic characteristics, which include:
• age
• sex
• ethnicity
• tenure
• accommodation type
• economic activity
Despite the best effort made by census data collection operations to count everyone, the complexity and size of the population results in census coverage errors. The most prevalent coverage error is
when a member of the target population is missed in the census (undercoverage). Less often, but still at a non-ignorable rate, a member of the target population is either duplicated or counted not at
their usual residence (overcoverage). The ONS uses a variety of statistical methods to estimate these coverage errors to produce corrected population totals for local authorities by the key
demographic characteristics. These estimates in general have higher accuracy than the raw census counts.
The Census Coverage Survey (CCS) was used to estimate the census coverage error. The data from this survey are then linked to the census data. A combination of capture-recapture, analysis of complex
survey data, and small area estimation methods are used. Finally, a bias adjustment process is used to adjust for issues which cannot be accounted for in the main design. These can occur as the
result of some of the statistical assumptions not being practically attainable in complex data collection exercises like the CCS and census. Variance estimation methods are used to assesses the
uncertainty around these estimates.
Census data collection covers both the general household population and managed residential accommodation, known as communal establishments. The general population itself is the population of
households and the population of individuals in these households. This methodology focuses on the 2021 Census coverage estimation for the general population, details of coverage estimation for
communal establishments will be published in early January 2023.
Nôl i'r tabl cynnwys
4. Census coverage estimation methodology
General population undercoverage estimation and adjustment
Once the coverage survey and census data are ready for estimation and matching complete, within sampled areas we know for each CCS record whether a census response exists or not. This can be used to
model the coverage probability given the set of observed census variables (or predictors) and their combinations (interactions).
Logistic regression is a powerful and well-understood tool for modelling probabilities. If modelling is done appropriately, it is possible to increase the precision of estimates thanks to using a
large sample. We can also reduce certain errors by simultaneously controlling for several important variables and some of their interactions. The approach uses the entire dataset to relate a
combination of demographic and other characteristics to the estimated probability that a member of the population with such characteristics will respond to census. Say, for example, a person who has
the characteristics:
• aged 32 years
• male
• white
• living in a rented purpose built flat
• looking for a job
• in a household size of two residents
• related to somebody in the household
• in hard-to-count 3
• in South West of England
• in a self-contained accommodation
• born in the UK
• not a student
• in an area that received access codes
• observed census return rate 0.965
This member of the population would have the estimated census response probability 0.95. While, a person with exactly the same characteristics except being female would have the estimated probability
The model predicts the probability for each combination of variables. Therefore, each observed census record gets a corresponding census response probability. This probability can be transformed to a
coverage weight by taking a reciprocal of the probability. In the example previously, we will have weights 0.95^-1 ≈ 1.053 and 0.953^-1 ≈ 1.049, respectively. If we observe 1000 individuals in the
census data like the first person in the above example, we can sum up the corresponding weights to estimate the population total for individuals with such characteristics: 0.95^-1× 1000 ≈ 1053.
Similarly, if we observe 1000 individuals in the census data as the second person, the corresponding estimate is 1049. Of course, we are never interested in such specific set of characteristics, but
rather something more useful, say, age-sex group by local authority. Since all census records are weighted, it is possible to produce the undercoverage adjusted total for any group of interest.
Mixed effects logistic regression was used both for household and person undercoverage estimation. It is similar to the model described above, but has the local authority as a "random effect". This
random effect allows the model to reflect the area specific variability in a more efficient way than having local authority as a fixed effect (like all the variables in the example shown). Without
the random effect the probabilities for these two persons would be 0.95 and 0.953, no matter what local authority within the region these two person were located in. However, with random effects,
those probabilities would be, say, 0.962 and 0.964 in local authority A, while 0.934 and 0.938 in local authority B. Mixed effects based estimation reflects local differences, but comes with the cost
of increased variability of estimates.
Undercoverage was estimated and corrected for both the household totals and person totals. The general approach was the same in both cases, though the actual models are different. In the case of the
household estimation, there was an additional adjustment for the distribution of household size. Model selection for two populations was run independently, but we tried to have as much consistency as
possible in terms of levels of variables and interactions used. The two populations are 'reconciled' by the adjustment process, further information on the adjustment process will be published in
winter 2022.
Figure 1a: Age-sex undercoverage probabilities (female)
England and Wales
Source: Office for National Statistics, Census 2021
Download this chart Figure 1a: Age-sex undercoverage probabilities (female)
Image .csv .xls
Figure 1b: Age-sex undercoverage probabilities (male)
England and Wales
Source: Office for National Statistics, Census 2021
Download this chart Figure 1b: Age-sex undercoverage probabilities (male)
Image .csv .xls
General population overcoverage estimation and adjustment
A similar approach to undercoverage estimation was used for overcoverage estimation at person level. Overcoverage occurs when a member of the census population is either enumerated:
• more than once
• in the wrong location
• despite not being a member of the target population (e.g. individuals born after census day)
• because of a completely fictitious census return
Where possible, data cleaning resolved erroneous records (Remove False Persons) and multiple responses at the same location (Remove Multiple Responses). As such, in overcoverage estimation we only
estimate for individuals enumerated more than once or enumerated in the wrong location.
Instead of modeling the coverage probability of those in the Census, overcoverage estimation was used to estimate the probability of correct enumeration in the census. Much like undercoverage
estimation, the linked census and census coverage survey allowed each linked record to have an outcome of 0 or 1, depending on if they were correctly enumerated or not. It is important here to assume
there is no overcoverage in the census coverage survey, as it is used as the correct location of census individuals. This is assumed due to the way the census and census coverage survey are designed,
where the time between the collection of them is designed to be large enough to optimise response rates but to reduce movement in the population. This linked outcome was used to model the probability
of correct enumeration, using a fixed effects, logistic regression model. Both numerical issues in the model fitting process and timescales meant that random effects were not included in this model.
Using the same example of characteristics, that person might have the estimated census correct enumeration probability 0.995. A person with exactly the same characteristics except being female, has
the estimated correct enumeration probability 0.9953.
In the same way, this overcoverage model then produces a correct enumeration probability for each combination of variables. Therefore, each observed census record gets the corresponding census
response probability and correct enumeration probability. The response probability can be transformed to the coverage weight by taking a reciprocal of the probability. However, for overcoverage
estimation, the aim is to down-weight the census estimate and therefore the undercoverage weights are multiplied by the correct enumeration probabilities. Where undercoverage error is estimated and
correcting for overcoverge error, we will have weights (0.995x0.95^-1 )≈ 1.047 and (0.995x0.953^-1 )≈ 1.044, respectively. If we observe 1000 individuals in the census data, we can sum up the
corresponding weights to estimate the population total for individuals with such characteristics: (0.995x0.95^-1) x 1000 ≈ 1047.37. Similarly, if we observe 1000 individuals in the census data as the
second person, the corresponding estimate is 1044.
Similarly, to 2011, matching of the census dataset to itself allowed for stronger estimates of the level of duplication, with high precision for each of 17 pre-specified groups within each region
across England and Wales. This method is outlined by Census to census matching strategy 2021. This census to census linkage exercise, enabled the estimated proportions of duplication across regions
and groups to be estimated with high precision. The estimated proportions of duplication found within each group in each region were then used to calibrate the estimated probabilities of correct
enumeration calculated by the model, to produce the final correct enumeration probabilities for each census record. Further information is available in The Proposed Duplication Calibration Method for
the 2021 Census of England and Wales.
The estimated level of overcoverage in the 2021 Census of England and Wales was 0.96%, compared to the 2011 Census of England and Wales where the estimated level of overcoverage was 0.6%.
Figure 2a: Age-sex correct enumeration probabilities (female)
England and Wales
Source: Office for National Statistics, Census 2021
Download this chart Figure 2a: Age-sex correct enumeration probabilities (female)
Image .csv .xls
Figure 2b: Age-sex correct enumeration probabilities (male)
England and Wales
Source: Office for National Statistics, Census 2021
Download this chart Figure 2b: Age-sex correct enumeration probabilities (male)
Image .csv .xls
Nôl i'r tabl cynnwys
Based on the experience of the previous censuses, some of the assumptions needed to produce the coverage adjusted population estimated with ignorable levels of bias may not be met in practice.
Therefore, the development of methods to adjust for certain biases is also a part of the coverage estimation. In addition, some ad-hoc adjustments may be implemented based on the quality assurance
results and availability of the data.
Producing coverage error corrected population size estimates using the Census Coverage Survey (CSS) and census data requires independence between these two data sources. Independence means that for
every member of the target population a chance of responding to the coverage survey does not depend on the member being census respondent or non-respondent. In practice, such independence is not
achievable and a dependence bias adjustment may be needed. In general, non-responders to census may be less likely to respond to the coverage survey, which would bias the estimates downwards -
leading to estimates that are too low. This is the type of bias for which correction was planned and prepared for in advance.
To do this, an alternative estimate was needed. Similar to previous censuses, an Alternative Household Estimate was calculated Alternative Household Estimate 2021. However, since the coverage
estimation in 2001 and 2011 Censuses used dual system, ratio, and synthetic approaches, while the coverage estimation in 2021 Census used the mixed-effects logistic regression approach. The way
adjustment was applied was very different this time aroundas outlined in Adjusting for the dependence bias in the Census 2021 coverage estimation.
There are two main challenges when using the Alternative Household Estimate to correct for the dependence bias. First, the alternative estimates are available at quite high level of aggregation
defined by local authority by hard-to-count index by accommodation type. The second challenge is that reliable alternative estimates are available for the household population only, whereas a
dependence bias adjustment is required both for the household and person populations.
There were several dependence bias adjustment methods designed and tested at the research stage for the 2021 Census. The approach chosen was the direct adjustment method with reweighting
(apportionment) based on the initial undercoverage probabilities.
In 2011, all local authorities were adjusted for dependence bias. However, in the 2021 only five were adjusted.
Another adjustment made was a single year of age adjustment for those aged zero to three years. Based on quality assurance, it was decided to adjust for these age groups across all local authorities
in England and Wales using administrative data. In addition, those aged 4 to 15 years were adjusted in Wales and North East based on the School Census.
Nôl i'r tabl cynnwys
6. Other quality assurance adjustments
Several other adjustments were made. There were 15 local authorities in undercoverage estimation were the random effect was forced to be 0 and only the fixed effects part of the model was used. These
was due to the fact that the Coverage survey in those areas was not of sufficient quality to reliably support the mixed effects logistic approach, while switching to the logistic regression for the
entire country would have had a negative effect for many other local authorities.
The estimated person coverage probabilities in several local authorities were constrained to the household coverage probabilities at the local authority by hard-to-count by accommodation type level.
The reason was not directly related to estimation. In this case, the combination of person and household level estimates meant that adjustment process might have experience difficulties. After a
careful consideration and assessing the impact, the decision to constrain the probabilities was made.
Nôl i'r tabl cynnwys
Variance estimation measures the variability of the estimates for the key domains of interest like person / household local authority total, local authority by age-sex group total, local authority be
tenure. This is outlined in Variance Estimation for 2021 Census Population Estimates. Similarly, to the 2011 Census, the bootstrap method is used. However, unlike the previous census, the bias
corrected percentile method was used to produce confidence intervals. This allowed reflecting non-symmetric distribution of the coverage error corrected estimates.
Nôl i'r tabl cynnwys | {"url":"https://cy.ons.gov.uk/peoplepopulationandcommunity/populationandmigration/populationestimates/methodologies/coverageestimationforcensus2021inenglandandwales","timestamp":"2024-11-08T05:58:49Z","content_type":"text/html","content_length":"83024","record_id":"<urn:uuid:30b9459d-7698-4b62-809a-b2118b9b66d7>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00765.warc.gz"} |
Chapter 2 - Force And Motion | Chapter Solution Class 9 - Flash Education
Publisher : Santra publication pvt. ltd.
Book Name : Madhyamik Physical Science And Environment
Class : 9 (Madhyamik)
Subject : Physical Science
Chapter Name : Force and Motion
Multiple Choice Question (MCQ)
[Each of Mark-1]
Question 1
A body whose momentum is constant must have constant
1. force
2. velocity
3. acceleration
4. all of these
A body whose momentum is constant must have a constant velocity. This is known as the law of conservation of momentum. In the absence of external forces, the total momentum of a system remains
constant. If the momentum of a body is constant, it means that there is no net external force acting on the body, and therefore the body will continue to move at a constant velocity.
Question 2
The motion of the rocket is based on the principle of conservation of
1. mass
2. kinetic energy
3. linear momentum
4. angular momentum
linear momentum
The motion of a rocket is based on the principle of conservation of linear momentum. According to Newton’s third law of motion, every action has an equal and opposite reaction. In a rocket, the force
of the gases expelled from the rocket engine creates an equal and opposite force in the opposite direction, which propels the rocket forward.
Question 3
A force of 250 N acts on a body and the momentum acquired is 125 kg m/s. What is the time for which the force acts on the body?
1. 0.2 s
2. 0.5 s
3. 0.4 s
4. 0.25 s
0.5 s
The formula for calculating the momentum acquired by an object due to an applied force is given by:
p = F × t
where p is the momentum acquired, F is the force applied, and t is the time for which the force acts.
We are given that a force of 250 N acts on a body and the momentum acquired is 125 kg m/s. Substituting these values into the above formula, we get:
125 = 250 × t
Solving for t, we get:
t = 125 / 250
t = 0.5 s
Question 4
A scooter of mass 120 kg is moving with a uniform velocity of 108 km/h. The force required to stop the vehicle in 10 s is
1. 720 N
2. 180 N
3. 360 N
4. 216 N
360 N
We can start by calculating the initial momentum of the scooter, which is given by:
p = m × v
where p is momentum, m is mass, and v is velocity.
Substituting the given values, we get:
p = 120× 108 ×{5 \over 18 }
To bring the scooter to a stop in 10 s, a net force must be applied in the opposite direction of motion. The magnitude of this force can be found using the equation:
F = Δp / t
where F is force, Δp is the change in momentum, and t is the time interval over which the change occurs.
Since the scooter comes to a stop, the final momentum is zero. Therefore, the change in momentum is equal to the initial momentum. Substituting the values, we get:
F = 36000 / 10 s = 3600 N
Question 5
A body is moving at a constant speed in a straight line path. A force is not required to
1. increase its speed
2. change the direction
3. decrease the momentum
4. keep it moving with uniform velocity
keep it moving with uniform velocity
A force is not required to keep a body moving at a constant speed in a straight line path. According to Newton’s first law of motion, a body will continue to move with uniform velocity in a straight
line path unless acted upon by an external force. In the absence of external forces, the momentum of the body will remain constant, and therefore, the body will continue to move with a constant speed
in a straight line path. Therefore, the correct answer is “keep it moving with uniform velocity”.
Question 6
A cricket player catches a ball of mass 0.1 kg moving with a speed of 10 m/s in 0.1 s. The force exerted by him is
1. 10 N
2. 5 N
3. 2 N
4. 1 N
10 N
F = Δp/t
F = m(v-u)\over t
F = 0.1 \times (0-10)\over 0.1
F = -10 N
The negative sign indicates that the force is in the opposite direction to the motion of the ball.
Question 7
What force will change the velocity of a body of mass 1 kg from 20 m/s to 30 m/s in 2 s?
1. 25 N
2. 5 N
3. 10 N
4. 2 N
5 N
F = Δp/t
F = m(v-u)\over t
F = 1 \times (30-20)\over 2
F = 1 \times (10)\over 2
F = 5 N
Question 8
A body of mass 2 kg is moving with a velocity of 8 m/s on a smooth surface. If it is to be brought to rest in 4 s, then the force to be applied is
1. 8 N
2. 4 N
3. 2 N
4. 1 N
4 N
F = Δp/t
F = m(v-u)\over t
F = 2 \times (0-8)\over 2
F = 1 \times (-8)\over 2
F = – 4 N
The negative sign indicates that the force is in the opposite direction to the motion of the ball.
Question 9
The rocket engines lift a rocket from the Earth because hot gases with high velocity
1. push it against the Earth
2. push is against the air
3. react against the rocket and push it up
4. heat up the air which lifts the rocket
react against the rocket and push it up
The rocket engines lift a rocket from the Earth because hot gases with high velocity react against the rocket and push it up. According to Newton’s third law of motion, every action has an equal and
opposite reaction. The rocket engines expel hot gases with high velocity in the opposite direction of the desired motion of the rocket. As a result, the rocket experiences a reaction force in the
opposite direction, which propels it upwards.
Question 10
A man is at rest in the middle of a pond on perfectly smooth ice. He can get himself to the shore by making use of Newton’s
1. first law
2. second law
3. third law
4. all the laws
third law
The man can get himself to the shore by making use of Newton’s third law of motion, which states that every action has an equal and opposite reaction. By applying a force on the ice, he can cause the
ice to react with an equal and opposite force, propelling him towards the shore.
Fill in the blanks
[Each of Mark-1]
1. To every ____ there is an ____ and ____ reaction.
2. The SI unit of momentum is ____.
3. In nature forces always occur in ____.
4. Newton’s first law holds good in the ____ frame.
5. ____ provides the measure of the inertia of a body.
6. A rocket works on the principle of conservation of ____
7. SI unit of force is ____
8. The ratio of the SI to the CGS unit of force is ____.
9. In any collision, the principle of conservation of ____ along holds
10. Action and reaction always act on two ____ bodies.
11. In collision and explosions, the total ____ remains constant, provided that no external ____ acts.
1. action, equal, opposite
2. kgm/s
3. pairs
4. inertial
5. mass
6. momentum
7. Newton (N)
8. 10^5
9. momentum
10. different
11. momentum
12. force
Answer in one word or in one sentence
[Each of Mark-1]
1. Which physical quantity corresponds to the rate of change of momentum ?
2. What is the SI unit of force?
3. State the relation between the momentum and force acting on a body.
4. Name the principle on which a rocket works.
5. What is the relationship between force and acceleration?
6. What is the relation between dyne and newton?
7. Name the physical quantity which makes it easier to accelerate a small car than a large car.
8. What is the force which produces an acceleration of 1 m/s2 in a body of mass 1 kg ?
9. What is the SI unit of inertia ?
10. Which law of motion is called the ‘law of inertia’?
11. Are action and reaction act on the same body or different bodies?
1. Force
2. Newton
3. Force (F) ∝ Rate of change of momentum (Δp\over t)
4. Conservation of momentum
5. F = ma, where F is the force and a is the acceleration
6. 1 N = 10^5 dynes
7. Mass
8. 1 newton
9. Kilogram
10. Newton’s first law
11. Different bodies
Short answer type questions
Question 1
State Newton’s first law of motion. Define force from this law.
Newton’s first law of motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. Force can be defined as a push or pull on an
object that can cause a change in its motion, according to this law.
Question 2
State Newton’s second law of motion and use it to define the unit of force in CGS system.
Newton’s second law of motion states that the acceleration of a body is directly proportional to the force acting on it and inversely proportional to its mass.
Mathematically, F = ma, where F is force, m is mass, and a is acceleration.
In the CGS system, the unit of force is defined as the dyne, which is the force required to impart an acceleration of 1 cm/s^2 to a body of mass 1 gram.
Question 3
‘Mass is a measure of inertia’-Explain.
Mass is a measure of inertia because it quantifies how much resistance an object has to changes in its state of motion. Inertia is the tendency of an object to maintain its state of motion. The more
massive an object is, the more difficult it is to change its state of motion.
Question 4
Establish the connection between Newton’s second law of motion with the first law of motion.
Newton’s second law of motion is directly connected to the first law of motion because it provides a quantitative relationship between force, mass, and acceleration. It states that a force is
required to change the state of motion of a body, which is consistent with the first law of motion. In fact, the first law can be seen as a special case of the second law, where the force acting on a
body is zero.
Question 5
Define newton. Establish its relation with dyne.
The newton is the SI unit of force, defined as the force required to impart an acceleration of 1 m/s^2 to a body of mass 1 kg.
1 Newton (N) = 1 kg m / s^2
= 1000 × 100 g cm / s^2
= 100000 dyne
= 10^5 dyne
Question 6
Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly.
When a speeding bus stops suddenly, passengers are thrown forward from their seats because of inertia. Inertia is the tendency of objects to keep moving in a straight line unless acted upon by an
external force. So when the bus suddenly stops, the passengers continue moving forward due to their inertia.
Question 7
State Newton’s third law of motion. Do action and reaction act on the same body?
Newton’s third law of motion states that for every action, there is an equal and opposite reaction. Action and reaction always act on two different bodies, not on the same body.
Question 8
What is the linear momentum of a body? Is it a scalar or a vector?
The linear momentum of a body is defined as the product of its mass and velocity. It is a vector quantity, as it has both magnitude and direction.
Question 9
If action and reaction are always equal and opposite, why do not they cancel each other?
Action and reaction do not cancel each other out because they act on different bodies. When a force is exerted on a body, it reacts by exerting an equal and opposite force on a different body, which
can result in motion or deformation of that body.
Question 10
State the basic conservation principle used in rocket motion.
The basic conservation principle used in rocket motion is the principle of conservation of momentum. The total momentum of the rocket and the ejected gases remains constant, and therefore, the rocket
can be propelled forward by expelling gases in the opposite direction to the desired motion.
Question 11
In a ‘Tug of war’ if both the parties exert a force of 7 dynes, what will be the tension of the string?
In a tug of war, the tension in the string will be equal to the force exerted by either party, which is 7 dynes each. The total tension in the string will be equal to the sum of these two forces,
which is 14 dynes.
Question 12
Explain why cleaning a garment from dust particles, it is suddenly set into motion.
When a garment is cleaned from dust particles, it is suddenly set into motion due to the sudden removal of the static friction force between the dust particles and the garment. The garment tends to
maintain its state of motion as per Newton’s first law of motion, which causes it to move suddenly.
Question 13
Explain why a gun receives a backward kick when a bullet is fired from it.
When a gun is fired, the bullet is accelerated in one direction, which results in a reactive force acting in the opposite direction on the gun. This causes the gun to experience a backward kick, as
per Newton’s third law of motion.
Question 14
Write down two differences between distance and displacement.
Distance is a scalar quantity that measures the total path covered by an object, whereas displacement is a vector quantity that measures the change in position of an object from its initial position
to its final position.
Question 15
Write down three differences between speed and velocity.
Speed is a scalar quantity that measures the rate of change of distance with respect to time, whereas velocity is a vector quantity that measures the rate of change of displacement with respect to
time. Speed is always positive, while velocity can be positive or negative, depending on the direction of motion. Speed does not indicate the direction of motion, while velocity does.
Question 16
Explain why in the unit of acceleration the unit of time comes twice.
Acceleration is defined as the rate of change of velocity with respect to time. Since velocity is a vector quantity that has both magnitude and direction, it requires two units of time in its
calculation to determine the change in velocity over a certain time interval. Therefore, the unit of acceleration is m/s^2, where the unit of time is squared.
Long answer type questions
[Each of Mark-3]
Question 1
State Newton’s laws of motion. Establish the relation F = ma.
Newton’s laws of motion are:
1. An object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
2. The rate of change of momentum of an object is directly proportional to the applied force and takes place in the direction in which the force acts.
3. For every action, there is an equal and opposite reaction.
Rate of change of momentum = Δp/t
= m(v-u)\over t [∵ Δp = m(v-u)]
= ma [∵ a = (v-u)\over t]
Now, according to Newton's second law of motion, the rate of change of momentum of an object is directly proportional to the applied force and takes place in the direction in which the force acts.
∴ F ∝ ma
or F = kma
where k is a constant. Here k = 1
Thus, F = ma
or, Force = mass × acceleration
Question 2
State Newton's second law of motion. Define dyne and newton. Establish a relationship between them.
Newton's second law of motion states that the force acting on an object is directly proportional to its mass and acceleration, and is given by the equation F = ma.
Dyne is a unit of force used in the CGS system of units. It is defined as the force required to impart an acceleration of 1 centimetre per second squared to a mass of 1 gram, and is denoted by the
symbol "dyn".
Newton is the SI unit of force, defined as the force required to impart an acceleration of 1 meter per second squared to a mass of 1 kilogram, and is denoted by the symbol "N".
1 Newton (N) = 1 kg m / s^2
= 1000 × 100 g cm / s^2
= 100000 dyne
= 10^5 dyne
∴ 1 Newton (N) = 10^5 dyne
Question 3
State and explain the principle of conservation of linear momentum and establish it from Newton's third law of motion.
The principle of conservation of linear momentum states that the total momentum of a system of objects remains constant if no external forces act on the system. This principle can be established from
Newton's third law of motion, which states that every action has an equal and opposite reaction. When two objects interact, they exert equal and opposite forces on each other, which means that the
total momentum of the system remains constant. The forces acting on each object may be different in magnitude, but their effects on the momentum cancel out each other. This principle of conservation
of linear momentum has wide applications in many areas of physics and has helped scientists to understand and explain many phenomena in the natural world.
Question 4
'All rest and motion are relative'- Explain. Define displacement, speed, velocity, acceleration and retardation.
"All rest and motion are relative" means that the state of motion of an object depends on the observer's frame of reference. In other words, an object may be considered to be at rest or in motion
depending on the reference point of the observer. For example, a person sitting in a moving train may appear to be at rest to the other passengers but may be moving relative to an observer standing
outside the train.
• Displacement is the change in position of an object from its initial position to its final position, measured in a specific direction. It is a vector quantity and is denoted by the symbol s.
• Speed is the rate of change of distance with respect to time. It is a scalar quantity and is denoted by the symbol v.
• Velocity is the rate of change of displacement with respect to time. It is a vector quantity and is denoted by the symbol v.
• Acceleration is the rate of change of velocity with respect to time. It is a vector quantity and is denoted by the symbol a.
• Retardation is the negative acceleration that opposes the motion of an object. It is also known as deceleration and is denoted by the symbol -a.
Question 5
Establish graphically :
1. v = u + at,
2. s = ut + 1⁄2 at^2,
3. v^2 = u^2 + 2as.
(i) Derivation of v = u + at
Change in velocity in time interval t ⇒ BE = BD - ED
If AE be drawn parallel to OD, then from the graph,
BD = BE + ED = BD + OA
or, v = BE + u
or, BE = v - u
Now, acceleration, a = Change\ in\ velocity\over time
= BE\over AE
= BE\over OD
Putting OD = t ⇒ a = BE\over t
⇒ BE = at = v - u
therefore,v = u + at
(ii) Derivation of s = ut + 1⁄2 at^2
In the figure, the distance travelled by the body is given by the area of the space between the velocity-time graph AB and the time axis OC, which is equal to the area of the figure OABD.
Thus, Distance travelled = Area of the trapezium OABD
But, Area of the figure OABD
= Area of rectangle OAED + Area of triangle ABE
= Area of rectangle OAED + area of triangle ABE
Now, find out the area of rectangle OAED and the area of triangle ABE.
(1) Area of rectangle OAED
= (OA)×(OC)
= (u)×(t)
(2) Area of triangle ABE
= 1\over 2 AE×BE
= 1\over 2 t×at
= 1\over 2 at^2
Distance travelled (s) is,
So, s = Area of rectangle OAED + Area of triangle ABE
= ut + 1\over 2 at^2
(iii) Derivation of v^2 = u^2 + 2as.
In the given figure, the distance travelled (s) by a body in time (t) is given by the area of the figure OABC which is a trapezium.
Distance travelld = Area of the trapezium OABC
So, Area of trapezium OABC
= 1\over 2 ×sum of parallel sides×height
= 1\over 2 (OA + CB) × OC
Now, (OA + CB) = u + v and (OC) = t.
Putting these values in the above relation, we get:
s = (u+v)\over 2 t ---- (1)
Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.
v = u + at
So, t = (v-u)\over a
Now, put this value of t in equation (1), we get:
s = (v+u)(v-u)\over 2a
On further simplification,
2as = v^2 – u^2
Finally the third equation of motion.
v^2 = u^2+2as
Numerical Problems
[Each of Mark-3]
Question 1
The speed-time graph of a particle moving along a fixed direction is shown in fig. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s. What is the average
speed of the particle over the intervals in (a) and (b)?
[Ans. : (a) 60 m, 6 m/s, (b) 24 m, 6 m/s.]
(a) t = 0 s to 10 s
Distance = Area under the graph
= 1\over 2 × 10 ×12
= 5 × 12 = 60 m
Average speed = total\ distance\over time
= 60\over 10
= 6 m/s
(b) beyond the scope (Should not be in class 9)
Question 2
The velocity-time graph of a body moving in a straight line is shown in the Fig. Find the displacement and distance travelled by the body in 6 seconds. [Ans. : 8 m, 16 m.]
(a) Displacement = Area under the graph
= 4 × 2 - 2 × 2 + 2 × 2
=8 - 4 + 4 = 8 m
(b) Distance = Area under the graph
= 4 × 2 + 2 × 2 + 2 × 2
= 8 + 8 + 8 = 24 m
Question 3
The velocity-time graph of the motion of a car is given. Find the distance travelled by the car in the first six seconds. What is the retardation of the car during the last two seconds? [Ans. 90 m,
15 m/s^2]
(a) Distance travelled in 6 s = Area under the graph from 0 to 6 s
= Area under the graph from 0 to 2 s + Area under the graph from 2 to 6 s
= 1\over 2 × 2 × 30 + (6 - 2) × 30
= 1\over 2 × 2 × 30 + 4 × 30
= 30 + 120 = 150 m
(b) Retardation (a) = (u - v)\over t
= (30 - 0)\over 2
= 30\over 2
= 15 m/s^2
Question 4
Calculate the acceleration of a body of mass 5 kg when a force of 15 Newton acts on it.
[Ans: 3 m/s^2]
Force (F) = 15 N
Mass (m) = 5 kg
Force = Mass × acceleration
or, acceleration = Force\over Mass
= 15\over 3 = 3 m/s^2
Question 5
The velocity of a body of mass 5 kg changes from 20 m/s to 10 m/s. Calculate the change of momentum of the body.
[Ans : 50 kg-m/s]
Mass (m) = 5 kg
Initial velocity (u) = 20 m/s
Final velocity (v) = 10 m/s
Change of momentum = m (v - u)
= 5 (10 - 20)
= 5 × (-10) = - 50 kg-m/s
The negative sign indicates that the momentum is in the opposite direction to the motion of the body.
Question 6
A body of mass 100 g is at rest. A force acts on it for 3 seconds and the body attains a velocity of 15 cm/ s. Calculate the magnitude of force.
[Ans: 500 dyne]
Mass (m) = 100 g
Initial velocity (u) = 0 cm/s
Final velocity (v) = 15 cm/s
Time (t) = 3 second
Force = m×(v - u)\over t
= 100 × (15 - 0)\over 3
= 100 × 5
= 500 dyne
Question 7
A body of mass 10 kg is moving with a velocity of 100 m/s. Find the force needed to stop it in 5 seconds.
[Ans: 200 N]
Mass (m) = 10 kg
Initial velocity (u) = 100 m/s
Final velocity (v) = 0 m/s
Time (t) = 5 seconds
Force = m×(v - u)\over t
= 10 × (0 - 100)\over 5
= 10 × (- 100)\over 5
= - 200 N
The negative sign indicates that the Force is in the opposite direction to the motion of the body. | {"url":"https://flasheducation.online/chapter-2-force-and-motion/","timestamp":"2024-11-06T05:03:40Z","content_type":"text/html","content_length":"216897","record_id":"<urn:uuid:cb43477f-71ed-4421-aba8-44fcff017ac6>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00636.warc.gz"} |
XGBoost Comparing Models with Box Plots
When evaluating the performance of XGBoost, it’s often useful to compare it against other popular machine learning models to gauge its relative effectiveness.
Box and whisker plots provide a clear and concise way to visualize the distribution of cross-validation scores for each model, making it easy to compare their performance and draw insights.
In this example, we’ll generate a synthetic multiclass classification dataset, train and evaluate XGBoost, Random Forest, and Support Vector Machine (SVM) models using cross-validation, and create a
box plot to compare the distribution of their cross-validation scores.
from sklearn.datasets import make_classification
from sklearn.model_selection import cross_val_score
from sklearn.ensemble import RandomForestClassifier
from sklearn.svm import SVC
import xgboost as xgb
import matplotlib.pyplot as plt
import numpy as np
# Generate a synthetic multiclass classification dataset
X, y = make_classification(n_samples=1000, n_classes=5, n_features=20, n_informative=10, random_state=42)
def evaluate_model(model):
scores = cross_val_score(model, X, y, cv=10, scoring='accuracy')
return scores
# Initialize models with default hyperparameters
models = {
'XGBoost': xgb.XGBClassifier(),
'Random Forest': RandomForestClassifier(),
'SVM': SVC()
# Evaluate each model and store the scores
results = {name: evaluate_model(model) for name, model in models.items()}
# Create a box and whisker plot
fig, ax = plt.subplots(figsize=(10, 6))
ax.boxplot(results.values(), labels=results.keys())
ax.set_title('Comparison of XGBoost and Other Models')
# Print the median score for each model
for name, scores in results.items():
print(f"{name}: Median accuracy = {np.median(scores):.4f}")
The resulting plot may look as follows:
In this example, we first generate a synthetic multiclass classification dataset using scikit-learn’s make_classification function. We then define a function called evaluate_model that takes a model
object, evaluates its performance using 10-fold cross-validation, and returns the scores.
Next, we initialize instances of XGBoost, Random Forest, and SVM models using their default hyperparameters. We evaluate each model using the evaluate_model function and store the cross-validation
scores in a dictionary called results.
To visualize the results, we create a box and whisker plot using matplotlib. Each box in the plot represents the distribution of cross-validation scores for a specific model. By comparing the boxes,
we can assess the relative performance and consistency of the models.
In this example, the box plot suggests that XGBoost and Random Forest have similar performance, with XGBoost having a slightly higher median accuracy and a smaller interquartile range, indicating
more consistent results. The SVM model appears to have lower accuracy and more variability compared to the other two models.
Finally, we print the median accuracy score for each model to provide a numerical summary of the results.
Using box and whisker plots to compare XGBoost against other models allows us to visually assess their relative performance and consistency, helping us make informed decisions when choosing a model
for our specific problem. | {"url":"https://xgboosting.com/xgboost-comparing-models-with-box-plots/","timestamp":"2024-11-13T04:28:23Z","content_type":"text/html","content_length":"11907","record_id":"<urn:uuid:b0d572af-0c7d-47cb-916e-51814d6aaaaf>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00309.warc.gz"} |
How to put margins on tables or arrays in R?
How to put margins on tables or arrays in R?, I’ll show you how to use the addmargins function in the R programming language to add margins to tables or arrays in this post.
Algorithm Classifications in Machine Learning – Data Science Tutorials
The addmargins function will be used in the tutorial’s two examples to annotate the margin values on table objects.
To be more specific, the instruction includes the following details:
Providing Examples of Data
The data listed below will serve as the foundation for this R tutorial.
data <- data.frame(x1 = c(LETTERS[1:3], "B", "B", "C"),x2 = letters[1:2])
x1 x2
1 A a
2 B b
3 C a
4 B b
5 B a
6 C b
Consider the previous table. It demonstrates how the sample data frame has two variables and six rows.
Boosting in Machine Learning:-A Brief Overview (datasciencetut.com)
Then, using these data, we can build a table object (a contingency table).
tab <- table(data)
x1 a b
A 1 0
B 1 2
C 1 1
Example 1:-Incorporate Sum Margins into the Contingency Table.
Addmargins() Function Utilization
I’ll show you how to add the sum of each row and column to a table’s margins using the addmargins function in Example 1.
Add Significance Level and Stars to Plot in R (datasciencetut.com)
Take into account the R code below:
tab_sum <- addmargins(tab, FUN = sum)
x1 a b sum
A 1 0 1
B 1 2 3
C 1 1 2
sum 3 3 6
As you can see, we have a row and column that contain the sum margins of our data that are annotated.
Hope you enjoyed it.
What is the best way to filter by row number in R? – Data Science Tutorials | {"url":"https://datasciencetut.com/how-to-put-margins-on-tables-or-arrays-in-r/","timestamp":"2024-11-06T22:13:32Z","content_type":"text/html","content_length":"112757","record_id":"<urn:uuid:79628abc-2614-4c68-80ac-b544f2bfa37b>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00282.warc.gz"} |
Add Fractions: Understanding The Basics
Add Fractions: Understanding the Basics
Learn how to add fractions with exceptional denominators, add and simplify fractions, or even add blended numbers. Get smooth-to-apprehend causes with actual-life examples. Master the skill of
including fractions these days. Adding fractions can be daunting for plenty of college students, however, it shouldn’t be. In this article, we’ll ruin down the basics of including fractions, such as
the way to upload bits with extraordinary denominators, add and simplify fractions, and even add blended numbers. With actual-existence examples and a conversational tone, you’ll be a pro at adding
fractions in no time.
Why Adding Fractions Matters
Add fractions are a fundamental quantity-associated concept used in our regular workouts. We use divisions while estimating elements for a recipe, running out a tip, or identifying the space we want
to tour. Adding portions permits us to consolidate at least two quantities to get them all out aggregate, making it a fundamental capability in technological know-how.
Adding Fractions with the Same Denominator
Add fractions with a comparable denominator is direct. To upload divisions, we want to have a shared aspect. The denominator addresses the all-out range of a balance that makes up a whole, at the
same time as the numerator addresses the number of components we manage. To upload divisions with similar denominators, we add the numerators and hold an equal denominator.
For example, we want to add 1/four and 2/four. Since the denominators are something very comparable, we can add the numerators and maintain a comparable denominator to get:
1/4 + 2/four = ¾
Adding Fractions with Different Denominators
Adding fractions with various denominators requires a couple of additional means. We need to music down a shared aspect to add divisions with extraordinary denominators. The most effective approach
for finding a shared element is to boom the denominators together.
For instance, we want to feature 1/four and 1/6. We can music down a shared issue employing duplicating the denominators together, much like this:
4 x 6 = 24
Then we need to trade every portion over completely to have a denominator of 24. To do that, we need to copy the numerator and denominator of each department with the aid of the same variety, to
provide us with the shared aspect of 24.
1/four x 6/6 = 6/24
1/6 x 4/4 = 4/24
Since we’ve got a similar denominator, we can upload the numerators:
6/24 + four/24 = 10/24
Even so, we will work on this element employing partitioning the numerator and denominator through their maximum essential popular variable, which for this example, is 2.
10/24 ÷ 2/2 = 5/12
Along these strains, 1/four + 1/6 = 5/12.
Adding and Simplifying Fractions
We can simplify add fractions with the aid of dividing the numerator and denominator by way of their maximum huge not unusual aspect. Simplifying fractions makes it less complicated to work with them
and decreases the chance of making a mistake.
For example, permits say we want to add 2/6 and 1/4. To upload those add fractions, we need to discover a commonplace denominator, that is 12:
2/6 x 2/2 = four/12
1/four x three/3 = 3/12
Now we will upload the numerators and simplify:
4/12 + three/12 = 7/12
7/12 is already simplified, so we don’t want additional simplification.
Adding Mixed Numbers
Mixed numbers are an aggregate of a whole variety and a fraction. To add mixed numbers, we want to convert them into mistaken fractions after which upload them.
For instance, we want to feature 2 1/four and three 2/three. To add those blended numbers, we first convert them to incorrect fractions:
2 1/four = nine/4
three 2/three = eleven/three
Now we can add the mistaken fractions by using locating a commonplace denominator:
nine/4 x 3/three = 27/12
eleven/three x 4/4 = forty four/12
Now we can add the numerators:
27/12 + forty four/12 = 71/12
Finally, we can simplify the fraction:
71/12 ÷ three/3 = 23 2/three
Therefore, 2 1/four + three 2/three = 23 2/three. Great post to read about Solar System Project Ideas.
Tips for Add Fractions
Here are a few recommendations to keep in thoughts when adding fractions:
• Always discover a not-unusual denominator before adding fractions with exclusive denominators.
• Simplify fractions earlier than adding them to lessen the chance of creating a mistake.
• Always double-take a look at your work to make sure your solution is accurate.
How do you upload fractions with one-of-a-kind denominators?
To upload fractions with unique denominators, discover a common denominator by using multiplying the denominators collectively. Convert each bit to have the common denominator after which upload the
numerators. Simplify the fraction if feasible.
How do you add and sum fractions?
To upload and sum fractions, find a common denominator, and convert every bit to have the common denominator, after which add the numerators. Simplify the fraction if viable.
How do you add and simplify fractions?
To upload and simplify fractions, discover a common denominator, convert each bit to have the commonplace denominator, add the numerators, and then simplify the fraction by dividing both the
numerator and denominator via their most tremendous commonplace component.
How can you upload 3 10 and a couple of five?
To upload three 10 and a pair of five, convert each combined wide variety to an unsuitable fraction, discover a commonplace denominator, and add the numerators, after which simplify the fraction if
Table: Add fractions
Type of Fractions Steps to Add
Same denominator Add the numerators and keep the same denominator.
Different denominators Find a common denominator, convert each fraction to have the common denominator, add the numerators, and simplify the fraction if possible.
Mixed numbers Convert each mixed number to an improper fraction, find a common denominator, add the numerators, and simplify the fraction if possible.
Jasper Bruxner is a passionate and versatile blogger with a keen eye for trends and a knack for crafting engaging content. As the founder of WendyWaldman, he has established himself as a trusted
resource in a diverse range of niches, including food, tech, health, travel, business, lifestyle, and news. He tends to share the latest tech news, trends, and updates with the community built around
Wendywaldman. His expertise and engaging writing style have attracted a loyal following, making him a respected voice in the online community. | {"url":"https://wendywaldman.com/add-fractions-understanding-the-basics/","timestamp":"2024-11-10T05:32:45Z","content_type":"text/html","content_length":"119294","record_id":"<urn:uuid:1b3de303-c446-4c03-993a-d4dafca55fd0>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00072.warc.gz"} |
Validated example of LLM acceleration
Theory of Fast Cross Entropy Loss
As a start of LLM acceleration project, I learnt Fast Cross Entropy Loss in unsloth, it shows a better realize solution of cross entropy loss than default pytorch code.
As default, pytorch will use log_softmax to realize CE loss, it’s common for all solution, but, the target has only one valid value in the training process of LLM and other element are 0, this is the
different part between LLM training and common CE, that’s why LLM acceleration solution works.
So, we can update the calculate process of CE as below
\[\begin{align*}\label{2} & CT(x, y)_i = -y_i(log(softmax(x))) \\ & =-y_i(log(p_i)) \\ & =-(log(\frac{exp(x_y)}{sum(exp(x))})) \\ & =logsumexp(x) - x_y \\ \end{align*}\]
After optimizaiton, this can reduce time complexity from $O(4n)$ to $O(2n)$, which can reduce the cost of time and GPU memory for the best suitation.
This is all the theory of solution
Taking finetuning Gemma2 as example, I tried to compare the result of new CE loss and default one, but I found some thing that I have never know before.
(1) smoothed_loss
There is a smoothed_loss which is used in CE Loss except traditional log_softmax, which is the first noise during my validation.
But I also realize it throught new method.
(2) A trick that subtracting the max-logit to make softmax more stable
After I realize above code and fine-tune Gemma2, I found it can reduce fine-tune time by 4.8%, but loss of new code didn’t reduce by training step, so, I ask help in Machine Learning subreddit and
get what I want.
After applying subtracting max-logit to my code, it works, there almost zero differences for loss between default pytorch code and my new code, the loss as below:
This can prove that My new code have the same result compare to default official code.
default offical key code:
1 # time complexity 2n for subtracting max-logit
2 log_probs = -nn.functional.log_softmax(logits, dim=-1) # time complexity 4n
4 nll_loss = log_probs.gather(dim=-1, index=labels) # 1n
5 smoothed_loss = log_probs.sum(dim=-1, keepdim=True, dtype=torch.float32) # 1n
My new code:
1 logitsmax = logits.max(dim = -1)[0].unsqueeze(-1) # 1n
2 logsumexp = ((logits - logitsmax.repeat(1, 1, logits.shape[-1])).exp().sum(dim = -1).log()).unsqueeze(-1) + logitsmax # 3n
4 nll_loss = logsumexp - logits.gather(dim=-1, index=labels) # 0
5 smoothed_loss = (logits.shape[-1] * logsumexp - logits.sum(dim = -1, keepdim=True, dtype = torch.float32)) # 1n
PS: In theory, time complexity reduce from $8n$ to $5n$, I’m not 100% confirm this value right and it’s hard to prove that.
In terms of time cost in finetune Gemma2, I only sample 100 sample(my fault) from yahma/alpaca-cleaned and run 60 steps two times for both official code and new code.
offical CE Loss code New code differences
362.5s (average) 352.5s 2.8%
I validated the new method to calculate CE Loss and reduce time cost by 2.8%.
In additional, using Triton to realize should have better performance.
This is just a example for LLM acceleration, more similiar thing can be done to accelerate the process. | {"url":"https://informal.top/posts/validated-example/","timestamp":"2024-11-09T00:40:41Z","content_type":"text/html","content_length":"28641","record_id":"<urn:uuid:e7472122-d6f3-497e-900e-92ac11ad6614>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00448.warc.gz"} |
Question: Problems Related to Markov Chains 1 Problenm Consider a memory system which only allows you to do… | Assignment Writing Service: EssayNICE
Show transcribed image text Problems Related to Markov Chains 1 Problenm Consider a memory system which only allows you to do sequential search. For example a read/write tape drive. If you want to
look for a file you have to search sequentially looking at the first file, then the second file and so on until you find the file. A reasonable strategy would be place the most recently retrieved
file at the front (imagine that the tape system can magically do this). This way the files that are accessed more often will be "at the front" and require less searching time in the long run.
Consider the case with only 3 files A, B, and C. ordered A and then B followed by C, then Xo = Λ BC. Enumerate the state space. 2. If Xo = Λ BC, list all possible states of X1. 3. If p,1,pB, and
PC, = 1-PA-pB are the probabilities with which files A, B, and C are accessed, respectively, determine the one-step state transition matrix. 4. If pA-06, pB = 0.10, PC 0.3, determine the steady state
probability for the file order ABC. 5. In general show that the steady state probability of the state ABC is given by PAPB PB PC
Problems Related to Markov Chains 1 Problenm Consider a memory system which only allows you to do sequential search. For example a read/write tape drive. If you want to look for a file you have to
search sequentially looking at the first file, then the second file and so on until you find the file. A reasonable strategy would be place the most recently retrieved file at the front (imagine that
the tape system can magically do this). This way the files that are accessed more often will be "at the front" and require less searching time in the long run. Consider the case with only 3 files A,
B, and C. ordered A and then B followed by C, then Xo = Λ BC. Enumerate the state space. 2. If Xo = Λ BC, list all possible states of X1. 3. If p,1,pB, and PC, = 1-PA-pB are the probabilities with
which files A, B, and C are accessed, respectively, determine the one-step state transition matrix. 4. If pA-06, pB = 0.10, PC 0.3, determine the steady state probability for the file order ABC. 5.
In general show that the steady state probability of the state ABC is given by PAPB PB PC | {"url":"https://essaynice.com/question-problems-related-to-markov-chains-1-problenm-consider-a-memory-system-which-only-allows-you-to-do/","timestamp":"2024-11-02T21:55:59Z","content_type":"text/html","content_length":"302448","record_id":"<urn:uuid:6bf71583-be0b-4bd9-9be5-80d2e94fa5cc>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00526.warc.gz"} |
ECMJ: First models of climate change
Let's develop a first simple model of climate change. These notes are based on this lecture.
We first need to recall the fundamental concept of differential equation and see how to solve them in Julia.
Differential equations
What is a scientific model? A matematical description of (some aspects of) reality.
What's the language with which such a description can be formulated? It's the language of equations, namely equalities among formulas that impose constraints on what can happen.
What else can we say about how these equations usually look like? Usually, when modeling reality, we are interested in knowing what happens after some time, given some initial conditions. Our
equations will likely be functions of a time variable $t$. Moreover, it is typically the case that, rather than being able to describe with an equation how the system look like at a certain time $t$,
we only have hypotheses about how the system changes across time: for example, we can rarely predict, a-priori, where a moving mass will be located in space, but we can provide equations that
approximate how the position of the mass will change from an instant to the next one.
The equations describing how a function changes are called differential equations.
A simple example of ordinary (one-variable) differential equation is $D_t y = a - bt$. In Julia we can define such differential equation using the ModelingToolkit library:
using ModelingToolkit
@variables t y(t)
@parameters a b
D = Differential(t)
equation = [
D(y) ~ a - b*t
@named system = ODESystem(equation)
Model system with 1 equations
States (1):
Parameters (2):
The @named macro essentially rewrites the last line as
system = ODESystem(equation; name=:example_system)
We can then solve the system with a compatible Julia library of numerical solvers:
using DifferentialEquations
init_condition = [y => 0.0]
time_span = (0.0, 10.0)
parameters = [a => 2.0, b => .5]
problem = ODEProblem(system, init_condition, time_span, parameters)
solution = solve(problem)
retcode: Success
Interpolation: specialized 4th order "free" interpolation, specialized 2nd
order "free" stiffness-aware interpolation
t: 7-element Vector{Float64}:
u: 7-element Vector{Vector{Float64}}:
Let's see the solution:
using Plots
plot(solution, ylim=(-5,5), framestyle=:origin)
The simplest climate model
Let's consider arguably the simplest nontrivial climate model:
\[ \begin{align*} & \color{brown}{\text{change in heat content}} \newline &= \color{orange}{\text{incoming solar radiation (energy from the Sun's rays)}} \newline & \phantom{=} - \color{blue}{\text
{outgoing thermal radiation (i.e. blackbody cooling to space)}} \newline & \phantom{=} + \color{grey}{\text{human-caused greenhouse effect (trapped outgoing radiation)}}. \end{align*} \]
Each of the above terms should be intended as an average over the entire globe.
Incoming radiation ☀️
Let's address the first term, $\color{orange}{\text{incoming solar radiation}}$.
The amount of energy that Earth receives from the Sun, i.e. Earth's insolation, is
S = 1368 # measured in W/m^2
Part of it is reflected back to space. The measure of such effect is called albedo, and in Earth's case can be estimated to be
α = .3
Thus, out of the energy S that is received by the Sun, the fraction that it is retained is 1-α.
There's another important correction that we should apply to the above measure of insolation.
The above number S is estimated for a square meter of surface that lies perpendicularly to the sunrays. However, the Earth's surface is spherical. What's the correction which would apply? A disc of
radius $R$ has area $\pi R^2$ while a sphere of the same radius has area $4 \pi R^4$, so there's a factor $\frac 14$ that should multiplay S.
We then have that the incoming solar radiation on the r.h.s. of our starting equation is
in_radiation = S*(1-α)/4;
Earth without outgoing radiation 🔥
Let's start by considering only the $\color{orange}{\text{incoming solar radiation}}$ and ignoring the other terms: $\begin{align*} & \color{brown}{\text{change in heat content}} = \color{orange}{\
text{incoming solar radiation}} \end{align*}$
We want to model the Earth's temperature in this simple case:
@variables temp(t)
1-element Vector{Symbolics.Num}:
The equation above is about the heat content of the planet. 🤔 However, when we enter a swimming pool we don't care about the amount of heat stored in it, we care about the water's temperature, don't
we? Recall that the temperature is given by the heat content divided by the heat capacity $C$, in other words
\[ \color{brown}{\text{change in heat content}} = (\color{red}{\text{change in temperature}}) \cdot C. \]
Therefore $\begin{align*} & \color{red}{\text{change in temperature}} = \frac 1C (\color{orange}{\text{incoming solar radiation}} ). \end{align*}$
Hence, in order to estimate how much the incoming radiation will influence the temperature of the Earth, we need to have an estimate the heat capacity of the atmosphere and upper-ocean:
@parameters C = 51. # J/m^2/°C
1-element Vector{Symbolics.Num}:
We thus have the system
@named sys_only_in = ODESystem([ D(temp) ~ in_radiation / C ])
Model sys_only_in with 1 equations
States (1):
Parameters (1):
C [defaults to 51.0]
In order to define an ODEProblem, we also need an initial condition. Let's assume that the time 0 of our equation corresponds to the year 1850, before the start of the Second Industrial Revolution;
at the time, it is estimated that
temp₀ = 14.0 # °C in 1850
init_condition = [temp => temp₀]
1-element Vector{Pair{Symbolics.Num, Float64}}:
temp(t) => 14.0
We can now solve system:
prob_only_in = ODEProblem(sys_only_in, init_condition, (0, 170))
sol_only_in = solve(prob_only_in)
retcode: Success
Interpolation: specialized 4th order "free" interpolation, specialized 2nd
order "free" stiffness-aware interpolation
t: 4-element Vector{Float64}:
u: 4-element Vector{Vector{Float64}}:
How does the solution look like? Let's plot it:
legend = false,
xlabel = "Years since 1850",
ylabel = "°C")
hline!( [temp₀], ls=:dash)
annotate!( 80, temp₀,
text("Preindustrial Temperature = $(temp₀)°C",:bottom))
title!("Earth's temperature without outgoing radiation")
Earth with outgoing radiation 🥶
Let's move on and consider the $\color{blue}{\text{outgoing thermal radiation}}$, i.e. the black-body radiation which allows Earth to dissipate some of the heat. It is a complicated phenomenon, which
we simplify by linearizing it and assuming that, together with the incoming radiation, the overall contribution to $D(\text{temp}(t))$ is equal to $B(\text{temp₀-temp(t)})$ for some $B$:
@parameters B = 1.3 # "climate feedback parameter" in W/m^2/°C
@named sys_in_and_out = ODESystem([
D(temp) ~ B*(temp₀-temp)/C
Model sys_in_and_out with 1 equations
States (1):
Parameters (2):
B [defaults to 1.3]
C [defaults to 51.0]
This time Earth can cool down. Let's solve the problem assuming that the initial temperature is higher, e.g. 20 °C, and plot the solution:
init_condition = [temp => 20]
prob_in_and_out = ODEProblem(sys_in_and_out, init_condition, (0, 170))
sol_in_and_out = solve(prob_in_and_out)
legend = false,
xlabel = "Years after start",
ylabel = "°C")
hline!( [temp₀], ls=:dash)
annotate!( 80, temp₀,
text("Initial temperature = $(temp₀)°C",:bottom))
title!("Earth's temperature with outgoing radiation")
😲 Wow! Unfortunately we are still ignoring one term of our initial equation...
The infamous Greenhouse effect ⛽
The greenhouse effect can be modeled by the equation
\[ \begin{align*} \color{grey}{\text{human-caused greenhouse effect (trapped outgoing radiation)}} \newline = \text{forcing_coefficient} \cdot \ln \left(\frac {\text{CO}_2}{\text{CO}^{(\text{0})}_2}\
right), \end{align*} \]
where $\text{CO}^{(\text{0})}_2$ is the initial value of the CO₂ level in the atmosphere.
Empirical data shows that
forcing_coef = 5.0 # W/m^2
CO₂⁽⁰⁾ = 280. # pre-industrial CO₂ concentration, in ppm
Our prediction of the Earth's temperature will depend on the amount of CO₂ emission; we thus need an hypothesis on the latter. We can extrapolate such hypothesis from past data.
Let's look at data from the Mauna Loa Observatory (more info here):
using CSV, DataFrames
CO2_historical_data_url = "https://scrippsco2.ucsd.edu/assets/data/atmospheric/stations/in_situ_co2/monthly/monthly_in_situ_co2_mlo.csv"
CO2_historical_data = CSV.read(download(CO2_historical_data_url), DataFrame, header=55, skipto=58);
first(CO2_historical_data, 11)
11 rows × 10 columns (omitted printing of 2 columns)
Yr Mn Date Date CO2 seasonally fit seasonally
Int64 Int64 Int64 Float64 Float64 Float64 Float64 Float64
1 1958 1 21200 1958.04 -99.99 -99.99 -99.99 -99.99
2 1958 2 21231 1958.13 -99.99 -99.99 -99.99 -99.99
3 1958 3 21259 1958.2 315.71 314.44 316.2 314.91
4 1958 4 21290 1958.29 317.45 315.16 317.3 314.99
5 1958 5 21320 1958.37 317.51 314.7 317.87 315.07
6 1958 6 21351 1958.45 -99.99 -99.99 317.26 315.15
7 1958 7 21381 1958.54 315.87 315.2 315.86 315.22
8 1958 8 21412 1958.62 314.93 316.21 313.98 315.29
9 1958 9 21443 1958.71 313.21 316.1 312.44 315.35
10 1958 10 21473 1958.79 -99.99 -99.99 312.43 315.41
11 1958 11 21504 1958.87 313.33 315.21 313.61 315.46
We plot the data after removing missing values (here represented by entries $-99.99$):
validrowsmask = CO2_historical_data[:, " CO2"] .> 0
plot( CO2_historical_data[validrowsmask, " Date"],
CO2_historical_data[validrowsmask, " CO2"],
label="Mauna Loa CO₂ data (Keeling curve)")
ylabel!("CO₂ (ppm)")
Looking at the data, we guess that the CO₂ in the atmosphere follows the following function (given the current human behavior):
CO₂(t) = CO₂⁽⁰⁾ * ( 1 + ((t-1850)/220)^3 )
CO₂ (generic function with 1 method)
We verify how much our guess overlaps with the data:
years = 1850:2022
plot!( years, CO₂.(years), lw=3,
label="Cubic Fit", legend=:topleft)
title!("CO₂ observations and fit")
Our guess seems quite good. Let's use it to put the GHE together with the radiation absorbed by the sun, in a complete toy model.
A climate toy model 🌍
Putting everything together we obtain the system
greenhouse_effect(CO₂) = forcing_coef * log(CO₂/CO₂⁽⁰⁾)
@named sys_climate = ODESystem([
D(temp) ~ (1/C)*( B*(temp₀-temp) + greenhouse_effect(CO₂(t)) )
Model sys_climate with 1 equations
States (1):
Parameters (2):
B [defaults to 1.3]
C [defaults to 51.0]
How fast will the system heat up?
using LaTeXStrings # used in `plot` in `label` property
init_condition = [temp => temp₀]
years = (1850, 2022)
params = [C => 51, B => 1.3]
prob_climate = ODEProblem(sys_climate, init_condition, years, params)
sol_climate = solve(prob_climate)
xlabel = "Years after start",
ylabel = "°C",
ylim = (10, 20),
label = L"B=1.3, C=51")
hline!( [temp₀], ls=:dash, label=nothing)
annotate!( 1900, temp₀-.8,
text("Initial temperature = $(temp₀)°C",:bottom))
title!("Earth's temperature in our toy climate model")
The variables C (heat capacity) and B (feedback parameter) are parameters of ModelingToolkit, meaning that we can try out different values for them, and add some information to the plot:
parameters2 = [C => 20, B => 0.6]
prob_climate2 = ODEProblem(sys_climate, init_condition, years, parameters2)
sol_climate2 = solve(prob_climate2)
xlabel = "Years after start",
ylabel = "°C",
ylim = (10, 20),
label = L"B=0.6, C=20")
hline!( [16], ls=:dash, label=nothing)
annotate!( 1930, 16,
text("Paris Agreement threshold (2°C warming)", :bottom))
Testing our model 📈
Similarly to what we have done to check our guess for the CO₂ evolution, we can test our overall model against historical data from NASA:
T_url = "https://data.giss.nasa.gov/gistemp/graphs/graph_data/Global_Mean_Estimates_based_on_Land_and_Ocean_Data/graph.txt";
T_df = CSV.read(download(T_url), DataFrame, header=3, skipto=5, delim=" ", ignorerepeated=true);
T_df[1:6, :]
6 rows × 3 columns
Year No_Smoothing Lowess(5)
Int64 Float64 Float64
1 1880 -0.17 -0.1
2 1881 -0.09 -0.13
3 1882 -0.11 -0.17
4 1883 -0.18 -0.2
5 1884 -0.28 -0.24
6 1885 -0.33 -0.26
Let's check our solution sol_climate:
plot(sol_climate[t], sol_climate[temp], lw=2, label="Temperatures predicted by the model", legend=:topleft)
ylabel!("Temperature (°C)")
plot!( T_df[:,1], T_df[:,2] .+ 14.15, color=:black, label="Observations by NASA", legend=:topleft) | {"url":"https://natema.github.io/ECMJ-GSSI-2022/notebooks/simple_climate","timestamp":"2024-11-09T20:33:46Z","content_type":"text/html","content_length":"387310","record_id":"<urn:uuid:09856467-6499-4d04-8e8d-21bdf031b637>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00499.warc.gz"} |
US 26627
Download all exemplars
This annotated exemplar is intended for teacher use only. Annotated exemplars are extracts of student evidence, with commentary, that explain key parts of a standard. These help teachers make
assessment judgements at the grade boundaries.
Download all exemplars and commentary [PDF, 6.5 MB]
Using 266 unit standards for the Literacy and Numeracy co-requisite (external link) - NCEA.education
Meets Performance Criteria 1.1 and 1.2
26627 Learner 1 (PDF | 345 KB)
This sample of learner evidence contributes to a portfolio of naturally occurring evidence generated over an acceptable period of time to meet the requirements of Guidance Information (GI) 2 and GI3.
The evidence reflects skills described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for Adult Numeracy (GI4).
Solving this real world problem (1) contributes evidence towards Outcome 1. The learner has made a reasonable estimation of the tree height (2) and taken appropriate and accurate measurements of
length (3) and angle (4) (using a clinometer app on an iphone). The learner has then used these measurements in calculations to find the height of the tree (5). Note: using the clinometer app is
acceptable, see GI5 and Performance Criteria (PC) 1.2, because it still requires the learner to take a measurement. The app must be calibrated to zero, positioned correctly at eye level, and read
The measuring tool and units used are appropriate to the problem and context (PC1.1). Although useful formula are provided, the learner has independently chosen the measurements to take, formulae to
use, and calculations to make to reach an acceptable solution (as required by GI7, PC1.1 and PC1.2). The signed attestation by the supervisor (6) provides the information necessary to verify that
these PCs have been met.
This sample provides acceptable evidence for four of the seven range items required to meet Outcome 1: length, angle, estimation (2) and conversion (7).
Meets Performance Criteria 1.1 and 1.2
26627 Learner 2 (PDF | 293 KB)
This sample of learner evidence contributes to a portfolio of naturally occurring evidence generated over an acceptable period of time to meet the requirements of Guidance Information (GIs) 2 and 3.
The evidence reflects skills described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for Adult Numeracy (GI4).
The problem (1) has a useful, real life purpose, which meets GI1 and the intent of the standard.
Taking measurements of length and using them in calculations to solve a capacity problem contributes evidence towards Outcome 1. The learner has measured the room’s dimensions (2) and used them to
calculate the total area to be painted (3) to solve the problem of the amount of paint required (4). The learner has independently selected an effective method to solve the problem, see GI7 and
Performance Criteria (PC) 1.2, and checked the reasonableness of the answer (5). The accuracy of the measurements taken has been attested to by the assessor (6) to meet PC1.1.
This sample provides acceptable evidence for solving a problem involving both length and capacity (two of the possible range items of Outcome 1).
Meets Performance Criteria 1.1 and 1.2
26627 Learner 3 (PDF | 634 KB)
This sample of learner evidence contributes to of a portfolio of naturally occurring evidence generated within the context of an agricultural training course and over an acceptable time period (as
required by Guidance Information (GI) 2 and GI3. The evidence reflects skills described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for Adult Numeracy
Taking measurements of length and making calculations to solve the problem of how much grass seed is needed (1) contributes evidence towards Outcome 1. The learner has used an informal measurement of
length by pacing out the dimensions of the area to be sown. The assessor has recorded what the learner actually did (2) and how the learner checked the accuracy of the informal measurements by
measuring the length of his pace (3).
Using informal measurement (such as paces) can be acceptable depending on the context and the level of accuracy required for the problem. In this instance the assessor has noted that a high degree of
accuracy is not required (4). The learner has calculated the total area using sensible rounding (6), and the amount of grass seed required (7) to solve the problem. The assessor has signed and dated
the observation sheet (5).
This sample provides acceptable evidence of three of the seven range items required to meet Outcome 1: solving problems involving length and mass and conversion within the metric system (8).
Meets Performance Criteria 1.1 and 1.2
26627 Learner 4 (PDF | 282 KB)
This sample of learner work contributes to a portfolio of evidence generated over an acceptable period of time, see Guidance Information (GI) 3. The evidence presented is naturally occurring, from an
assessment of standard 5236 (GI 2). The evidence meets (or exceeds) the level of demand described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for Adult
Numeracy (GI4).
The problem to be solved (1) is applied and relevant to the learner’s work and training context. The NZ Building Code regulations (Resource A) has been provided for reference.
The learner has taken appropriate and accurate measurements of length (2) and used them in calculations (3) to find the angle of the ramp (4). This solves the problem of whether the ramp meets the
building code (5), and contributes evidence towards Outcome 1.
The measuring tool and the units used are appropriate to the problem and context (PC1.1). The signed attestation by the supervisor (6) provides the information necessary to verify that PC1.1 and
PC1.2 have been met.
This sample provides acceptable evidence for two of the seven range items required to meet Outcome 1: solving problems involving length and angle.
Meets Performance Criteria 1.1 and 1.2
26627 Learner 5 (PDF | 682 KB)
This sample of learner work contributes to a portfolio of evidence generated over an acceptable period of time GI3. The evidence presented is naturally occurring, from an assessment of standard 5236
as part of a Trades Block course (GI 2). The evidence meets (or exceeds) the level of demand described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for
Adult Numeracy (GI4).
The problem to be solved (1) is applied and relevant to the learner’s work and training context. The NZ Building Code regulations (Resource A) has been provided for reference.
As required by Outcome 1, the learner has taken an appropriate and accurate measurement of length (2) and used it in calculations (3) to find the length of the ramp (4) and the horizontal distance
from the landing (5). This solves the problem of building a ramp that meets the building code (1). The measuring tool and the units used are appropriate to the problem and context (PC1.1) and the
signed attestation by the supervisor (6) provides the information necessary to verify that PC1.1 and PC1.2 have been met.
This sample provides acceptable evidence for two of the seven range items required to meet Outcome 1: solving problems involving length and angle.
Meets Performance Criteria 1.1 and 1.2
26627 Learner 6 (PDF | 452 KB)
This sample of learner evidence contributes to a portfolio of naturally occurring evidence generated within the context of a learning programme and over an acceptable period of time to meet the
requirements of Guidance Information (GI) 2 and GI3. The evidence reflects skills described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for Adult
Numeracy (GI4). The activity meets the intent of the standard that the problems posed are in a real context and relevant to learners and/or everyday life.
The learner has taken measurements of mass (1) and capacity (2) and used them in calculations (3) (4) to solve problems (5) (6) contributing evidence towards Outcome 1. The learner has selected and
used appropriate and effective methods to reach a reasonable solution, see GI7 and Performance Criteria (PC) 1.2.
By completing and signing the attestation on the learner work (7) the assessor has verified that the learner has taken accurate measurements without assistance, using appropriate measuring tools and
units of measurement (PC1.1).
This sample provides acceptable evidence of four of the seven range items required to meet Outcome 1: solving problems involving mass and capacity, as well as estimation (8) and conversion within the
metric system (9).
Meets Performance Criteria 1.1 and 1.2
26627 Learner 7 (PDF | 195 KB)
This sample of learner evidence contributes to a portfolio of naturally occurring evidence generated within the context of a foundation learning programme and over an acceptable period of time to
meet the requirements of Guidance Information (GIs) 2 and 3. The evidence reflects skills described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for
Adult Numeracy (GI4). The activity meets the intent of the standard that the problems posed are in a real context and relevant to learners and/or everyday life.
Measurements of mass (1) and time (2) have been taken and used in calculations (3) to solve the problems posed (4), thereby contributing evidence towards Outcome 1. The learner has independently
chosen effective methods to use to reach a solution, see GI7 and Performance Criteria (PC) 1.2.
The assessor has verified (5) that the learner has taken measurements (to an acceptable degree of accuracy), using appropriate measuring tools and units of measurement (PC1.1), and has judged the
solutions to be reasonable (6).
This sample provides acceptable evidence for three of the seven range items required to meet Outcome 1: solving problems involving mass and time, as well as conversion within the metric system (7).
Meets Requirements of Guidance Information 2, 3 and 4
26627 Learner 8 (PDF | 3.7 MB)
This sample of learner evidence contributes to a portfolio of naturally occurring evidence generated over an acceptable period of time to meet the requirements of Guidance Information (GI) 2 and GI3.
The evidence reflects skills described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for Adult Numeracy (GI4).
This sample provides acceptable evidence for location as the learner has shown an understanding of location in terms of direction and distance in marking the sections of the ride on the map (1) in
order to determine the finish point for the ride (2).
Meets Requirements of Guidance Information 2, 3 and 4
26627 Learner 9 (PDF | 732 KB)
This sample of learner evidence contributes to a portfolio of naturally occurring evidence generated over an acceptable period of time to meet the requirements of Explanatory Notes (GIs) 2 and 3. The
evidence reflects skills described by step 5 of the Measure and Interpret Shape and Space strand of the Learning Progressions for Adult Numeracy (GI4).
This sample provides acceptable evidence for location as the learner has described the distance and direction of each section of the course (1). Directions have been described using bearings and the
8 point compass. | {"url":"https://www2.nzqa.govt.nz/ncea/subjects/litnum/26622-26627/exemplars/us-26627/","timestamp":"2024-11-09T23:02:07Z","content_type":"text/html","content_length":"75472","record_id":"<urn:uuid:17ed2b96-b76b-4911-8b70-3d326add9cfd>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00884.warc.gz"} |
Plotting Discrete-Time Signals - Neil Robertson
Plotting Discrete-Time Signals
A discrete-time sinusoid can have frequency up to just shy of half the sample frequency. But if you try to plot the sinusoid, the result is not always recognizable. For example, if you plot a 9 Hz
sinusoid sampled at 100 Hz, you get the result shown in the top of Figure 1, which looks like a sine. But if you plot a 35 Hz sinusoid sampled at 100 Hz, you get the bottom graph, which does not look
like a sine when you connect the dots. We typically want the plot of a sampled sinusoid to resemble its continuous-time version. To achieve this, we need to interpolate (I discussed interpolation in
my last post [1]).
In fact, we can design a general-purpose interpolator for plotting any sampled signal, as long as the signal’s bandwidth is less than some defined fraction of the Nyquist frequency f[N] = f[s]/2. A
reasonable upper frequency for many applications is 0.8*f[N], or 0.4*f[s]. A sinewave at 0.4*fs has 1/0.4 = 2.5 samples per cycle, or 5 samples every two cycles. If we interpolate it by 8, we will
then have 20 samples/cycle. When we then connect these samples with straight lines, the plot looks to the eye like a smooth sinewave, as we’ll see.
The frequency response of an example interpolate-by-8 filter is shown in Figure 2. The filter is a linear-phase FIR lowpass, with passband of 0.4*fs and stopband of 0.6 *fs. We’ll use it in a Matlab
function interp_by_8.m to interpolate some example signals. The Matlab function’s code is listed at the end of this article.
Figure 1. Sinusoids sampled at 100 Hz. Top: f[0 ]= 9 Hz Bottom: f[0] = 35 Hz.
Figure 2. Interpolation-by-8 filter frequency response.
Example 1. Sinusoid
In this example, we’ll plot a 39 Hz sinusoid originally sampled at 100 Hz. Here is the Matlab code to create the signal and interpolate it to a sample rate of 800 Hz:
fs= 100; % Hz sample frequency
Ts= 1/fs; % s sample time
N= 32; % number of samples
n= 0:N-1; % time index of x
f0= 39; % Hz frequency of sinusoid
x= sin(2*pi*f0*n*Ts); % sinusoid
x_interp= interp_by_8(x); % interpolated sinusoid
M= length(x_interp);
m= 0:M-1; % time index of x_interp
axis([0 32 -1 1]),xlabel('n')
axis([0 210 -1 1]),xlabel('m')
The original and interpolated signals are plotted in Figure 3. The interpolated signal shows the original sample times by plotting every 8^th sample as an orange dot. The interpolated version
includes an initial transient response of length equal to that of the interpolation filter (121 samples). The transient is due to the discontinuity at the start of the sine signal. The interpolator
output is valid starting at m = 121.
This example poses the same display problem that comes up in digital oscilloscopes [2]. A scope might have a sinusoidal input at f[0] = 390 MHz, sampled at f[s] = 1000 MHz. The scope uses
interpolation to display the signal.
Figure 3. Top: Sinusoid with f[0] = 39 Hz and f[s] = 100 Hz.
Bottom: Sinusoid interpolated by 8, with orange dot at every 8^th sample.
Example 2. Filtered Pulse
In this example, x is a pulse signal filtered by a 3^rd order Butterworth filter. The filter has a -3 dB frequency of 0.3*f[s]. The following Matlab code creates the pulse and then interpolates it by
8. Figure 4 shows x and the interpolated version. As before, the interpolated signal shows every 8^th sample plotted as an orange dot.
% 3rd order butterworth filter coeffs, fc/fs= 0.3
% [b,a]= butter(3,2*.3)
b= .2569*[1 3 3 1];
a = [1 0.5772 0.4218 0.0563];
u= [0 0 0 0 ones(1,12) zeros(1,10)]; % rectangular pulse
x= filter(b,a,u); % filtered pulse signal
x_interp= interp_by_8(x); % interpolated signal
N= length(x);
n= 0:N-1; % time index of x
M= length(x_interp);
m= 0:M-1; % time index of x_interp
Figure 4. Top: Filtered pulse.
Bottom: Filtered pulse after interpolation by 8, with orange dot at every 8^th sample.
Matlab Function interp_by_8.m
This Matlab function interpolates-by-eight a signal with bandwidth up to 0.4*fs. Note that we don’t need to know the sample frequency of the signal x to perform the interpolation. The only input to
interp_by_8 is the vector x itself. See my last post [1] for a discussion of interpolation.
% funtion interp_by_8.m 8/29/19 Neil Robertson
% interpolation by 8 using 121-tap FIR filter
% passband = 0 to 0.4*fs stopband = 0.6 fs to 4fs
function x_interp= interp_by_8(x)
b1= [-32 -12 -11 -6 2 12 22 31 37 38 33 20 2 -20 -43 -62 -74 -74 ...
-62 -36 0 42 84 119 139 138 114 67 1 -74 -148 -207 -241 -239 ...
-196 -114 -1 129 257 361 420 418 345 202 1 -234 -473 -676 ...
-805 -822 -700 -426 -1 554 1205 1901 2586 3197 3681 3991];
b= [b1 4098 fliplr(b1)]/2^15; % interpolation filter coeffs
N= length(x);
x_up= zeros(1,8*N);
x_up(1:8:end)= 8*x; % upsampled signal
x_interp= conv(x_up,b); % interpolated signal
The FIR interpolation filter’s coefficients are plotted in Figure 5. In order to attenuate the images in the upsampled signal x_up, the filter’s stopband must start at 0.6 f[s], where f[s] is the
sample frequency of the interpolator’s input x. This is shown in figure 6, where the images of x_up’s spectrum are in orange. The filter’s frequency response magnitude is also plotted. The response
in dB is plotted in Figure 2. Accuracy of interpolation depends on the filter’s passband flatness and stopband attenuation.
The filter coefficients were computed using the Parks-McClellan algorithm [3] as follows:
Ntaps= 121; % number of taps
f= [0 .4 .6 4]/4; % frequency vector relative to 4*fs
a= [1 1 0 0]; % amplitude goal vector
b= firpm(Ntaps-1,f,a); % coefficients
b= round(b*2^15)/2^15; % quantized coefficients
The frequency vector f has passband of 0 to 0.4*f[s] and stopband of 0.6*f[s] to 4*f[s]. For example, if f[s] = 100 Hz (8*f[s] = 800 Hz), the passband is 0 to 40 Hz and the stopband is 60 to 400 Hz.
Figure 5. Coefficients of 121-tap interpolate-by-8 filter.
Figure 6. Top: Example spectrum of x_up occupying up to 0.4*f[s], showing images beginning at 0.6 f[s].
Bottom: Interpolation filter response (linear amplitude scale)
1. Robertson, Neil, “Interpolation Basics”, DSP Related website, Aug, 2019, https://www.dsprelated.com/showarticle/1293.php
2. AN1494, “Advantages and Disadvantages of Using DSP Filtering on Oscilloscope Waveforms”, Agilent (Keysight), 2004 http://literature.cdn.keysight.com/litweb/pdf/5989-1145EN.pdf
3. The Mathworks website https://www.mathworks.com/help/signal/ref/firpm.html
Neil Robertson September, 2019
[ - ]
Comment by ●September 16, 2019
Hi Neil. Thanks for your blog and thanks for the Reference link to the interesting Keysight oscilloscope application note. Oscilloscopes sure have come a long way since the 300 kHz bandwidth
Dumont scope I first touched in the late 1960s.
[ - ]
Comment by ●September 16, 2019
Hi Rick,
I noticed that Dumont was based in New Jersey, my home state. Did you use the scope with an HP audio generator?
[ - ]
Comment by ●September 20, 2019
Hi Neil. Ha ha. No, I didn't use an HP audio oscillator.
For you young folks out there, Neil is poking a little geriatric fun at me. His reference to an HP audio oscillator is, I think, referring to the famous audio oscillators that were the *very* first
product of the Hewlett Packard company. Hewlett & Packard built their first audio oscillator in a garage in Palo Alta California in 1939. That garage is now listed on the National Register of
Historic Places.
[ - ]
Comment by ●September 20, 2019
Actually, I used an HP audio oscillator in EE lab courses in the 1970's.
It looked like the example at this site:
[ - ]
Comment by ●September 22, 2019
Hi Neil. That's a neat photo. It's been a long time since I last saw "banana plugs" used as input connectors to a piece of equipment.
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Overloading unary operators in C#
This is the second entry in a series of posts about overloading operators in C#.
Previous entries:
-Introduction to operator overloading in C#
This time we're gonna see how to overload unary operators and some use cases.
Remember that in C# the unary operators are:
• Addition (+)
• Subtraction (-)
• Logical negation (!)
• Increment and decrement (++ and --)
• Bitwise complement (~)
• true and false
Basics of operator overloading
The C# language has a simple structure for the overload of operators, basically you define a static method on a type whose return type and parameters are the type itself. For example:
public static <TypeName> operator <OperatorSymbol>(<Type>Name> typeName)
where TypeName is the enclosing type, and OperatorSymbol is the operator to overload (+, -, and so on). You will understand it better when we get to the concrete examples.
For giving concrete examples and an easier comprehension, I'll be using a Temperature class on which we will overload the operators and apply our custom logic.
namespace UnaryOperators
public enum TemperatureType
Celsius = 1,
public class Temperature
public double Value { get; set; }
public TemperatureType Type { get; set; }
public Temperature(double value = 0, TemperatureType type = TemperatureType.Celsius)
Value = value;
Type = type;
We'll see how to use the operators on the Temperature object but modifying its underlying Value property.
The addition (+) operator
The addition operator by default just returns the current value of a numeric type in its unary form. We will change this behavior so it returns the absolute value of the temperature if it's lesser
than 0 and return the current temperature value otherwise.
public static Temperature operator + (Temperature tempA)
if (tempA.Value < 0) tempA.Value = System.Math.Abs(tempA.Value);
return tempA;
As you can see, the method receives a Temperature object and returns a Temperature object which, should be the rule since you will apply the operator on a Temperature object. When we get to the
binary operators we'll see how to apply operands between our custom type and any other type.
The subtraction (-) operator
The subtraction operator by default returns the negative value of a numeric type. We will override this behavior so we can modify the underlying value by applying the operator to the Temperature
public static Temperature operator - (Temperature temp)
temp.Value = - temp.Value;
return temp;
The increment (++) and decrement (--) operators
These are very straightforward, we will modify their behavior so they increment or decrement the temperature by 10 units.
public static Temperature operator ++(Temperature temp)
temp.Value += 10;
return temp;
public static Temperature operator --(Temperature temp)
temp.Value -= 10;
return temp;
True and False operators
I don't really see a need to overload these operators, since you can use boolean flags instead. But one particular and very specific use these operators have is when you need to overload the OR (|)
and AND (&) operators, since you need to implement the True and False for them to work.
Also, True and False work in pairs so if you overload one you must overload the other.
public static bool operator true (Temperature temp)
if (temp.Type == TemperatureType.Celsius) return temp.Value < 40;
if (temp.Type == TemperatureType.Farentheit) return temp.Value < 104;
return temp.Value < 313.5;
public static bool operator false (Temperature temp)
if (temp.Type == TemperatureType.Celsius) return temp.Value >= 40;
if (temp.Type == TemperatureType.Farentheit) return temp.Value >= 104;
return temp.Value >= 313.5;
Say you'd like to check if a Temperature is at a high level and can cause dehydration .As you see in the code, I set the 40 grades celsius temperature as the threshold, so in practice you could use
it like this:
var temperature = new Temperature(41, TemperatureType.Celsius);
if (temperature)
Console.WriteLine($"Outside temp is {temperature.Value}. Looks like a sunny day");
Console.WriteLine($"Outside temp is {temperature.Value}. It's dangerous to be outside with this heat");
// Prints Outside temp is 41. It's dangerous to be outside with this heat
Logical negation (!) and bitwise (~) operators
I couldn't find concrete use cases for these operators, but i'll explain the bitwise operator since the logical negation is very basic.
The bitwise operator works on numeric types by changing the zeros and ones that compose the number to their opposite value. Let's see an example:
If you have the decimal 78, in binary it would be 1001110.
When you apply the bitwise operation, all 1's are converted to 0's and viceversa; thus our binary number would be 0110001 which corresponds to the decimal 49.
This entry covered the basics of the operator overloading in C#, and how they can be used to change the behavior of some operations on our custom types. On the next post I'll be covering binary
operators, which are a more interesting and useful aspect of the C# language.
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Time Complexity
The computational complexity that describes the amount of time it takes to run an algorithm. Time complexity is commonly estimated by counting the number of elementary operations performed by the
algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Thus, the amount of time taken and the number of elementary operations performed by the algorithm are
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Operators in C LanguageOperators in C Language
Operators in C Language
An operator is a symbol that tells the compiler to perform specific mathematical or logical functions. C language is rich in built-in operators and provides the following types of operators −
• Arithmetic Operators
• Relational Operators
• Logical Operators
• Bitwise Operators
• Assignment Operators
• Misc Operators
We will, in this chapter, look into the way each operator works.
Arithmetic Operators
The following table shows all the arithmetic operators supported by the C language. Assume variable A holds 10 and variable B holds 20 then −
Operator Description Example
+ Adds two operands. A + B = 30
− Subtracts second operand from the first. A − B = -10
* Multiplies both operands. A * B = 200
/ Divides numerator by de-numerator. B / A = 2
% Modulus Operator and remainder of after an integer division. B % A = 0
++ Increment operator increases the integer value by one. A++ = 11
— Decrement operator decreases the integer value by one. A– = 9
Relational Operators
The following table shows all the relational operators supported by C. Assume variable A holds 10 and variable B holds 20 then −
Operator Description Example
== Checks if the values of two operands are equal or not. If yes, then the condition becomes true. (A == B) is not true.
!= Checks if the values of two operands are equal or not. If the values are not equal, then the condition becomes true. (A != B) is true.
> Checks if the value of left operand is greater than the value of right operand. If yes, then the condition becomes true. (A > B) is not true.
< Checks if the value of left operand is less than the value of right operand. If yes, then the condition becomes true. (A < B) is true.
>= Checks if the value of left operand is greater than or equal to the value of right operand. If yes, then the condition becomes true. (A >= B) is not true.
<= Checks if the value of left operand is less than or equal to the value of right operand. If yes, then the condition becomes true. (A <= B) is true.
Logical Operators
Following table shows all the logical operators supported by C language. Assume variable A holds 1 and variable B holds 0, then −
Operator Description Example
&& Called Logical AND operator. If both the operands are non-zero, then the condition becomes true. (A && B) is false.
|| Called Logical OR Operator. If any of the two operands is non-zero, then the condition becomes true. (A || B) is true.
! Called Logical NOT Operator. It is used to reverse the logical state of its operand. If a condition is true, then Logical NOT operator will make it false. !(A && B) is true.
Bitwise Operators
Bitwise operator works on bits and perform bit-by-bit operation. The truth tables for &, |, and ^ is as follows −
p q p & q p | q p ^ q
Assume A = 60 and B = 13 in binary format, they will be as follows −
A = 0011 1100
B = 0000 1101
A&B = 0000 1100
A|B = 0011 1101
A^B = 0011 0001
~A = 1100 0011
The following table lists the bitwise operators supported by C. Assume variable ‘A’ holds 60 and variable ‘B’ holds 13, then −
Operator Description Example
& Binary AND Operator copies a bit to the result if it exists in both operands. (A & B) = 12, i.e., 0000 1100
| Binary OR Operator copies a bit if it exists in either operand. (A | B) = 61, i.e., 0011 1101
^ Binary XOR Operator copies the bit if it is set in one operand but not both. (A ^ B) = 49, i.e., 0011 0001
~ Binary One’s Complement Operator is unary and has the effect of ‘flipping’ bits. (~A ) = ~(60), i.e,. -0111101
<< Binary Left Shift Operator. The left operands value is moved left by the number of bits specified by the right operand. A << 2 = 240 i.e., 1111 0000
>> Binary Right Shift Operator. The left operands value is moved right by the number of bits specified by the right operand. A >> 2 = 15 i.e., 0000 1111
Assignment Operators
The following table lists the assignment operators supported by the C language −
Operator Description Example
= Simple assignment operator. Assigns values from right side operands to left side operand C = A + B will assign the value of A + B to C
+= Add AND assignment operator. It adds the right operand to the left operand and assign the result to the left operand. C += A is equivalent to C = C + A
-= Subtract AND assignment operator. It subtracts the right operand from the left operand and assigns the result to the left operand. C -= A is equivalent to C = C – A
*= Multiply AND assignment operator. It multiplies the right operand with the left operand and assigns the result to the left operand. C *= A is equivalent to C = C * A
/= Divide AND assignment operator. It divides the left operand with the right operand and assigns the result to the left operand. C /= A is equivalent to C = C / A
%= Modulus AND assignment operator. It takes modulus using two operands and assigns the result to the left operand. C %= A is equivalent to C = C % A
<<= Left shift AND assignment operator. C <<= 2 is same as C = C << 2
>>= Right shift AND assignment operator. C >>= 2 is same as C = C >> 2
&= Bitwise AND assignment operator. C &= 2 is same as C = C & 2
^= Bitwise exclusive OR and assignment operator. C ^= 2 is same as C = C ^ 2
|= Bitwise inclusive OR and assignment operator. C |= 2 is same as C = C | 2
Misc Operators ↦ sizeof & ternary
Besides the operators discussed above, there are a few other important operators including sizeof and ? : supported by the C Language.
Operator Description Example
sizeof() Returns the size of a variable. sizeof(a), where a is integer, will return 4.
& Returns the address of a variable. &a; returns the actual address of the variable.
* Pointer to a variable. *a;
? : Conditional Expression. If Condition is true ? then value X : otherwise value Y
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Installation of MATLAB Software - The Engineering KnowledgeInstallation of MATLAB Software - The Engineering Knowledge
INTRODUCTION TO MATLAB. MATLAB is software that is used to perform data manipulation calculations visualization math and programming. This software has features to handle basic and high-level
operations in the simple and easiest way. It is mostly used for the solution of differential equations and optimization of different operations.
Through the use of this software, we can solve larger systems of equations effectively and so it is used as the finest option to solve differential equations. We can also get data visualization and
symbolic features from this software. Through this software, we can solve easy equations effectively but it has features to resolve high-level operation that has complications. In this post, we will
cover different parameters for MATLAB and related problems. So let’s get started INTRODUCTION TO MATLAB
Common Uses of MATLAB
• MATLAB is the finest option to solve linear algebra and matrix computation. It is a very famous solution for industrial solution engineers used to solve different research-based projects and math
• With that this software can also be used for automatic control solutions, statistics, and digital signal processing.
• The common uses of this software for industries and scientific-based projects are listed below
□ Signal Processing
□ Communication System
□ Control Systems
□ Fuzzy Logic
□ Wavelets
□ Generic Algorithms
□ Optimization
□ Neural Networks
How to Use MATLAB
• First of all, you must have MATLAB installed on your computer. Then press on the MATLAB icon to start it.
• When you press on the Icon MATLAB screen will be shown. In the below figure, you can see the interface. The MATLAB desktop is also called IDE or Integrated Development Environment.
• In the below screen, there is a start button in the lower left-side corner that will show the MTALB tools, shortcuts, and documentation
How to Use MATLAB as Calculator
□ To use MATLAB as a calculator we solve simple equations in the software. Let solve an expression
□ we have 2+5*2
□ to solve this equation write in prompt command(>>) as below
□ 2+5*2 ans=12
MATLAB Mathematical Functions
• In the below table, you can see the standard MATLAB Functions
Let’s solve an example with these functions
y= e-a sin(x) + 10√y
for a=5, x=2, and y=8 is computed by
>a=5; x=2; y=8;
>y=exp (-a)*sin(x) +10*sqrt(y)
• Let’s solve another example
To find the value of sin (pi/4) and e10
MATLAB Variables, Expression, and Statements
• Variable is one word having no space and numbers of characters up to 31. These variables are sensitive to the case and their names start with a letter. There is no use of punctuation characters
• MATLAB statements are written as
>>Variables= Expression
• Expressions consist of function operators and variable names. When the evaluation is done value is given to the variable and shown. If the variable name and =sign are not added the answer of the
variable is automatically generated and the result will be assigned to ans
Precedence Rules
• Expressions are solved from left to right through the use of exponential operation with a high value of precedence, after that multiplication and division with equal precedence then addition at
last subtraction with equal precedence.
• There is the use of parentheses to change this sequence for which case these rules of precedence are functioned in every set of parentheses initiating with the inner set and preceding to forward.
• The value that was added recently given to the variable that was used is given here. Let’s suppose if we write x at the prompt that output will be
X =
• The numerical value can be seen in another format that is mentioned below
e= 1/3
e = 0.3333
• format long (long decimal format)
e = 0.33333333333333
format short e (short exponential format)
e =
3.3333 e-01
>>format long e (long exponential format) e =
3.33333333333333 e-04
>format ( default format)
e =
Matrices in MATLAB
• There are different ways to enter the matrices in MATLAB. Which are listed below
• Write Explicit elements lists
• Loading of data from outer data files
• Through the use of built-in functions produce the matrices
• Make matrices through the use of functions and then fave in files
• The operation of MATLAB is based on a single type of object like a rectangular numerical matrix with complex entries.
• All variable denotes the matrices. If we have to store matrix 1 2 3 in variable x through the use of MATLAB we have to write the following code
• For row, separation semicolons are used and for elements, separation space or commas are used. So above shown matrix can be stored through the use of the below command
>X = [1,2,3;4,5,6;7,8,9];
• For this purpose, the statement can also use
X = [ 1 2 3 4 5 6 7 8 9 ]
• The existence of a
• semicolons at the command end suppress the results
• Here transpose of matrix b is taken and stored in y through the below command
>>Y = X’
Y =
• Matrices operations performed in the MATLAB are seen here
Matrix Operations In MATLAB
>>X=[1 2 3 4; 5 6 7 8]
□ You can assigned to new matrix to selected part of matrix
• To choose the elements of the main diagonal
Changing and deleting matrix elements
>>X= 1:5
>>X([1 3])=0
MATLAB Matrix Manipulating
>>X=[1 2 4: 5 7 8]
To take transpose
>fliplr (X)
Matrix Operation
>>X= [1 3 4; 5 7 8]
> 3+X
Element-by-element operations vs. Matrix Operations
□ Certain types of measure taken for Matrix multiplication and has differenece from basic element to element multiplication.
□ It not like the addition operation where two matrices added just addtion of elements.
□ So we have certain froup of operators which differentiate the matrix operation from elementy by element opeation. For perform the matrix operation use * for multiplication, / for division , ^
for exponentiation.
□ While for side by side opeation we uses .* for multiplcation, ./ for division and .^ for exponentation.
>>X=[2 3 4; 5 6 7; 2 1 0];
>>Y=[1 2 3; 5 6 2; 0 0 5];
Plotting in MATLAB
• MATLAB software comes in different graphic tools. Through the use of less commands, we can plot the given data into a graphical representation
• Graphical representation of data is very simple and easy in MATLAB.
• Here we will discuss the data plotting in the MATLAB
• For plotting the graph command of Matlab is “plot(x,y)”. Let’s suppose we have two vectors x= (1; 2; 3; 4; 5; 6) and y= (3; -1; 2; 4; 5; 1)
>> x= [1 2 3 4 5 6];
>> y= [3 -1 2 4 5 1];>>plot(x,y)
The result is given here
• Now we solve this below expression
Now try the following:
>>x = 0: pi/100: 2*pi; >>y= sin (x);
plot (x,y)
Axis Labels and Adding Titles
• Through the use of Matlab, we can add titles and labels axis. For example for the previous example we add X- and Y-axis labels
>> label (‘x = 0:2\pi’); >>ylabel (‘sine of x’);
>>title (‘Plot of the Sine Function’)
MATLAB multiple Data Set in a Single Plot
• Through the use of a single-cell plot many (x,y) pairs of expressions are created. For example, this expression draws three plots for functions x: y1=2 cos(x), y2=cos(x), and y3=0.5*cos (x) in
the interval 0 ≤ x ≤ 2∏
• The output of our code is shown below
Introduction to MATLAB Programming
• For the creation of file in MATLAB press on New in the file menu in the menu bar then you will see the new file tab
• When you write the code and have to save the file the called m-file will be saved in bin folder of MATLAB by default function
• These files is known as M-Files since they have ”.m” in the title. There are two types of M-files Script files and Function files
Here sample M-File exist
z= x+y
What is the Function Files
• For extension of Matlab functions files used. We can make new functions according to our issue that will have a similar status like other MATLAB functions.
• Variables in the function file are through default local. Though we can make a variable globel according to requirements
• Let solve example
• Function X=prod(c,d)
X= c*d
• Resolve this file having filename prod.m after that write MATLAB prompt
>>prod(2,4) ans=
• To get the input from users syntax rule is applied
>>variable name = input (‘variable’);
• Let us have another example
>>a=input (‘ Enter a number ‘);
• And for the command prompt, if we function this code it will be shown like:>> Enter a number
• if we write 1 at the cursor will have a result like x=1
MATLAB Control FLOW
• MATLAB has below control flow structures
• If- Else –End
• For loops
• While loops
• Switch-Case constructions
What is Script Files
• The script file has a series of normal MATLAB statements. if the file has a name like rkg.m then MATLAB command>>rkg will result in statements in the file to be operated.
• The script file has global variables and will vary the value of variables of the same title in conditions of the current MATLAB section
• Larger matrix solutions are done by script files
What is Load Function MATLAB
• The load function of MATLAB reads binary files consisting of matrices produced by previous MATLAB sessions or reads text files comprising numeric data.
• A text file must be configured in the rectangular table having numbers, parted by blanks having one row in one line and an equal number of elements in every row.
That is all about the MATLAB Software all details has explained if you have any queries ask here
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mirror symmetry
So you are both saying “A mirror symmetry is a (any) equivalence between two Calabi-Yau A ∞-categories.” ?
I never said that nonsense. E.g. even in the most classical case, one has two equivalences simultaneously: A(X) and B(Y) and A(Y) and B(X).
No, Domenico, I was talking about my full agreement with the quotation long entry 21.
we’re here for that! :)
Yes, indeed. Three of us now. The rest is lost in the landscape, literally. ;-)
Not all the rest of course. For instance Distler-Freed-Moore is all about identifying the right notion of string background.
There is a huge amount of formalization of concepts that still needs to be done.
we’re here for that! :)
By the way: even though I am disagreeing, I am delighted that we are discussing this.
It is kind of weird that this kind of discussion we are having here isn’t present in the literature, resolved or not. I think there is so much confusion in high energy physics and string theory
literature due to the fact that such conceptual basics are never properly discussed. There is a huge amount of formalization of concepts that still needs to be done.
I agree fully with 15
is that 25?
So you are both saying “A mirror symmetry is a (any) equivalence between two Calabi-Yau $A_\infty$-categories.” ?
Hopefully I can eventually manage to convince you that this is not a good idea… ;-)
I agree fully with 15
is that 25?
More about supersymmetry: there is a famous non-renormalization theorem for both twisted models which says that there are sufficiently supersymmetry-related cancellations that there are no quantum
corrections from renormalization. This also enables the business of path integral localization which in most examples needs supersymmetry to work. I think, that these localization effects are true
for general compact Kahler manifolds, not necessary Calabi-Yau manifolds, and it is also OK for B-model on Calabi-Yau orbifolds, and also in the case of A-model for any closed symplectic manifold.
I agree fully with 15, it is a bit more complicated to articulate concerns from 16 on.
Another angle to look at it is in terms of quantization:
“geometric background” in the sense that makes things like mirror symmetry a “remarkable” equivalence in your terms is classical background . In that: it defines a classical sigma-model QFT. The
“remarkable” fact then is that two inequivalent classical sigma-models become equivalent after quantization.
That this does not happend for 1dQFT is by the way a famous theorem by Connes, if I recall correctly: there are non-isometric but isospectral Riemannian manifolds, but if one takes on top of the
spectrum of the Laplace operator (= Hamiltonian of the quantized 1 d sigma model that they induce) their full spectral triple into account, then that serves to distinguish these and one can “hear the
shape of the drum” after all.
So I think it makes not much sense to talk about mirror symmetry in such a context.
that’s why in my version I didn’t say “mirror symmetry is”, but “a mirror symmetry is”. it would be just a synonym for “equivalence” in the context of Calabi-Yau categories (or of TCFTs, which is the
same). even in the duality invariant notion of geometry, I think equivalences would be relevant: we do not only want to say that two “geometries” are equivalent, but also how they are equivalent.
As you know, I am thinking that there should be one which is such that the input datum is a differential cocycle in an $\infty$-connected $(\infty,1)$-topos
and you know I agree with and enjoy a lot this point of view. but I wonder whether forcing to the extreme this generality one would not be able to consider a given Calabi-Yau category as an example.
and then involved the sigma-model in this particular case would just be the identity :-) what I mean is that there must be social agreement on what a geometric input is and what is not. so one cannot
say whether differential cocycles can encompass all kind of geometric inputs unless there’s agreement on what geometric inputs are. and the only way I can see to comprehend all possible inputs is to
include all Calabi-Yau categories.
but now I want to write down a few more specific questions/remarks on “classic” mirror symmetry.. give me a couple of hours to think to what I want to write and I’ll be back :)
I agree, but what is a geometric background? where does the notion of geometric ends?
Right, one can in fact take the stance that the SCFT defines the geoemtry. This is the point of view of spectral geometry: a spectral triple is just a 1d SQFT and the idea of spectral triples is
really to define geometry as whatever the corresponding 1d SQFT sees it. The case of 2d SCFTs is really a straightforward generalization/categorification of this. (Yan Soibelman is in the process of
writing that up for our book…)
So but if we take this point of view, then the statements about dualities vanish. Because then we have adopted a “duality invariant” notion of geometry, where I mean dulity in the strict sense of
perturbative string theory. This is a good point of view for many purposes, but it removes the subtlety that “mirror symmetry” is interested in , by definition.
So I think it makes not much sense to talk about mirror symmetry in such a context.
I think what we really need – and we have been discussing this at some length already elsewhere, of course – is a full formalization of the notion “$\sigma$-model”. As you know, I am thinking that
there should be one which is such that the input datum is a differential cocycle in an $\infty$-connected $(\infty,1)$-topos. Not clear yet if that works out fully, but somethin like that I expect
should count as “geometric input”.
I agree, but what is a geometric background? where does the notion of geometric ends? can we really say that noncommutative geometry is geometry and $A_\infty$-category theory is not? as a matter of
personal taste I would restrict mirror symmetry to the switch of A-model and B-model for pairs of Calabi-Yau manifolds, but this is too narrow and has been overcome by the current use of the term
mirror symmetry, which now refers to a plethora of different examples. so, even if I’m not satisfied with it, I am unable to see a better definition for what is mirror symmetry than the one I
suggested above, and would consider $\sigma$-model dualities as remarkable examples. but I see this is an extreme point of view, and that’s why I didn’t edit Mirror symmetry along these lines.
so, what is mirror symmetry? it seems to me that, as Zoran points out, the modern use of the term is something more vast than it was at the beginning, which we could nPOV state as: a mirror
symmetry is an equivalence between Calabi-Yau A ∞-categories.
I would think that would not be the right way to say it.
I think the nature of all these “dualities”, T-duality, mirror symmetry, U-duality etc. is that they describe invariances of the sigma-model construction(functor?)
$\sigma-model : geometric background data \to 2dQFT$
Here “geometric data” may be quite a bit more sophisticated than just the naive “manifold with extra structure”, it may be some non-commutative geometry etc. but it is still “geometric data” of
Then what the “dualities” assert is that there are certain systematic operations on the collection of geometric data under which $\sigma-model$ is invariant, up to equivalence.
This is what mirror symmetry is an example of.
A second step then is to notice that parts of the category of 2dQFTs of sorts is equivalent to a category of certain categories, which transports this statement to a statement about an invariant of a
$\sigma-model : geometric background data \to CertainCategories$
But it is still important, I think, to read it is a statement about such an assignment, not just as a statement that certain categories are equivalent. It is the fact that these categories that are
equivalent are induced from geometric data, and from different geometric data, that makes clear in which sense their equivalence is “remarkable” in your sense.
I created a stub for Hodge numbers
Thanks, I have edited it slightly.
I notice that we are lacking an entry on complex manifold…
so, what is mirror symmetry? it seems to me that, as Zoran points out, the modern use of the term is something more vast than it was at the beginning, which we could nPOV state as: a mirror symmetry
is an equivalence between Calabi-Yau $A_\infty$-categories. after this we could add: it derives its name from the following phenomenon, observed for a quantity of Calabi-Yau manifolds: etc. etc.
the one above would at least be a definition, and examples of mirror symmetries in literature would be from this point of view examples of “reamarkable” mirror symmetries, where “remarkable” is
something undefined with a matter of personal taste in it (where personal can involve a large part of the community at a certain time), depending on how a priori urelated were the two Calab-Yau
categories which turn out to be equivalent in a given example. to make an analogy, one could say that there’s a trivial isomorphism between the group of automorphisms of the set $\{a,b,c\}$ and the
group of automorphism of the set $\{1,2,3\}$, but there is a remarkable isomorphism between the the group of automorphisms of the set $\{a,b,c\}$ and the quotient of the free group on two generators
$\rho,\sigma$ modulo the subgroup generated by $\{\rho^3,\sigma^2, (\sigma\rho)^2\}$.
fine. I’ve now edited it. I’ll try to recover the remark on $A_\infty$-categories as a model for stable $(\infty,1)$-categories at some point, during future revisions of that entry.
Domenico, feel free to polish. In expanding these paragraphs, I had tried to preserve as much as possible of the previous paragraphs, which maybe led to some repetition. Please, if you have the time,
try to make a well-flowing piece out of it.
in the Lab entry on Mirror symmetry, at the point here homological mirror symmetry is introduced, I find exposition abit confused: namely, apparently we first say that the TCFT formulation is
equivalent to the categorical one, then we say the categorical version is an almost established conjecture, and finally we say that the categorical version is equivalent to the TCFT version. I would
rephrase that part as follows:
“This categorical formulation was introduced by Maxim Kontsevich in 1994 under the name homological mirror symmetry. The equivalence of the categorical expression of mirror symmetry to the SCFT
formulation has been proven by Maxim Kontsevich and independently by Kevin Costello, who showed how the datum of a topological conformal field theory is equivalent to the datum of a Calabi-Yau
A-infinity-category (see TCFT).”
in this I would omit the digression on $(\infty,1)$-categories, not because it is not correct, but because it seems to me to break the sentence, making it harder to follow the argument.
There are some earlier results for abelian varieties and for toric varieties, e.g.
V. Golyshev, V. Lunts, D. Orlov, Mirror symmetry for abelian varieties. J. Algebraic Geom. 10 (2001), no. 3, 433–496, math.AG/9812003
Okay, let’s change the statement about the involution.
We also need to say something about Landau-Ginzburg model, eventually.
Meanwhile, I created a section Complete proofs that is meant to list articles in which (homological) mirror symmetry is completely established on certain (types of) spaces.
Currently this contains the Efimov article and the complete proofs that he cites.
I disagree with the statement that we have really an involution, even not conjecturally. For example, to some Calabi-Yau varieties one associates a rather different mirror which is not a variety but
sort of deformation of the corresponding category in noncommutative world, which has no underlying variety in commutative sense. So locally one has another Calabi Yau variety or sometimes a bit
different thing. Next there is that business about families and considering this only in the large volume limit. So it is not really a picture of internal involution set-theoretically.
I know of this paper, Efimov is the most talented undergraduate student at Moscow University, about to graduate with Orlov and to go to graduate school at Harvard with Seidel.
I am scanning the arXiv for recent statements of results. This here seems to have a good deal of such:
• Alexander I. Efimov, Homological mirror symmetry for curves of higher genus
Abstract. Katzarkov has proposed a generalization of Kontsevich’s mirror symmetry conjecture, covering some varieties of general type. Seidel has proved a version of this conjecture in the
simplest case of the genus two curve. Basing on the paper of Seidel, we prove the conjecture (in the same version) for curves of genus $g\geq 3,$ relating the Fukaya category of a genus $g$ curve
to the category of Landau-Ginzburg branes on a certain singular surface. We also prove a kind of reconstruction theorem for hypersurface singularities. Namely, formal type of hypersurface
singularity (i.e. a formal power series up to a formal change of variables) can be reconstructed, with some technical assumptions, from its D$(\Z/2)$-G category of Landau-Ginzburg branes. The
precise statement is Theorem 1.2.
A survey of established results as of 2008 I find in
• M. Ballard, Meet homological mirror symmetry.
See in particular page 30 and following.
True, maybe you could add caveats, as indicated.
I basically agree, but I would say “there is a conjectural involution” at the very beginning. also it is not clear to me in which sense the categorical version of mirror symmetry is now pretty much
I created a stub for Hodge numbers
triggered by discussion with Zoran and Domenico, I have tried to expand the Idea-section at mirror symmetry. Check.
Good job
Well, I just introduced nLab (and category theory) to a room of physicists at the March Meeting of the American Physical Society and there was a good amount of interest. I included the URL so
hopefully we'll get some contributions. Even if we don't, I'm trying to add as much to the physics stuff as I can. I've just been very busy lately with a gazillion things to do. But I'll squeeze time
in for it eventually. Plus I have so many problems in my own research - and now apparently the research of other quantum physicists - that could benefit from a categorical approach that I think it
will begin to expand a bit. So have faith! The physicists are coming! (And, no, that's not a threat... :-)
I have added two standard references and also linked quantization to a unpolished short post of mine at mathlight on quantization. I do not have enough spare time now to write polished version for
cafe now though. Hopefully in few weeks.
Edit: 1. how does one get rid of "Possibly related posts: (automatically generated)" at wordpress ? they link idiotic blogs as possibly related
1. (I wrote 2 and stupid software shows 1) I suspect that the field theory version of the geometric quantization can not sense the singularities of the corresponding D-modules. Thus, unlike the
(finite degree of freedom) QM, I do not expect much use of geometric quantization method. BV of course supposedly works.
One way to attract people looking into this is to have blog posts
That is for random people outside of the circle of some 30-40 nlab regulars. I am more disappointed that the 30-40 nlab regulars, do not seem to be that interested in physics contributions. I value
nlab work far above the ncafe discussions. You can see from your own posts that those (edit: entries at cafe) which had the deepest technical content (e.g. your post on Bousfield localization)
usually got less response than those mild ones where everybody can step in. Thanks, anyway, I could do some guest post occasionally.
The "3000 and One Things to Think About" post did a good job in my opinion in giving an overview of pages (nLab and otherwise) that are sufficiently advanced to deserve a closer look - and every
reader could pick the topics of interest her-/himself. Once there is enough material on mathematical physics in the n-lab, you, Urs, could write a similar post ("1001 aspects of mathematical physics"
or something like that).
I still feel the mathematical physics is still badly underrrepresented in nlab. I feel if I write something related to physics it will get much less response and comments from other nlabizants
than if I write any math nonsense.
One way to attract people looking into this is to have blog posts on the corresponding topic, with maybe some pointers to the nLab. I can forward a guest post of yours, if you want to write one.
Of course also on the blog the interest in mathematical physics is badly underrepresented. But there are a few regulars who are interested. One would need to lure them to get more involved with the
Lab, too.
I still feel the mathematical physics is still badly underrrepresented in nlab. I feel if I write something related to physics it will get much less response and comments from other nlabizants than
if I write any math nonsense.
One thing which would be very practical in nlab is to have pages listing all the various actions and couplings which are in standard usage in various falvours of string and M-theory, together with
sketch of argument for the shape of corresponding actions. These things are hard to find systematized in the literature, especially from descent mathematical viewpoint. The 1997 two IAS volumes on
strings for mathematicians are obsolete as they have almost no D-brane mathematics. Zwiebach has lots of those cases listed, but it is low level, without deeper explanations; still it is the main
resource to start. Old books focus on classical cases and sectors.
Urs, the 3 links you put in a comment do not work any more. Reason: you do not have published version but show version now for your publicly available private web. These links you will easily
I noticed that symplectic geometry is in a mess. In addition to other problems, there was some strange and rambling paragraph at the end which looks like somebody's informal journal club
presentation. I moved it entirely to equivariant localization and elimination of nodes. I found that the source of the text is one cafe discussion.
Thans, Zoran.
I added some links.
mirror symmetry (needs more well chosen references, I am runnning out of time and will be busy next few days; there are hundereds of references available so we should choose important and/or well
written ones)
so here is the questio/remak I was promising.. it concerns something very basic, but it seems it has somehow gone lost in the modern take on mirror symmetry: Hodge numbers (for instance they are
mentioned only once, and in the introduction, in the recent survey by Ballard mentioned in #15).
at the very beginning of the mirror symmetry story, one had an $n$-dimensional Calabi-Yau manifold, computed its Hodge numbers and organized them into the shape of a square with $h^{0,0}$ as the
bottom vertex and $h^{n,n}$ as the top vertex (Hodge diamond). this can be seen as a morphism
$Calabi-Yau manifolds \stackrel{Hodge diamond}{\to} numerology$
and it was remarked that for suitable pairs of Calabi-Yau manifolds, the numerologies one obtained were related by a symmetry of the Hodge diamond. the physical interpretation of this is very simple:
very roughtly, one has two TCFTs attached to a Calabi-Yau manifold, namely the A-model and the B-model, and the Hodge numbers appear as dimensions of suitable eigenspaces for $\mathfrak{u}(1)\times\
so the above arrow is refined as
$Calabi-Yau manifolds\stackrel{A/B model}{\to}TCFT \stackrel{Hodge diamond}{\to} numerology$
in the categorical approach, this becomes
$Calabi-Yau manifolds\stackrel{A/B model}{\to}Calabi-Yau category \stackrel{Hodge diamond}{\to} numerology$
so, what is the arrow $Calabi-Yau category \stackrel{Hodge diamond}{\to} numerology$?
a similar numerological example is Candelas-de la Ossa-Green-Parkes formula (and all the mathematics it generated..). that can be sketched as
$Calabi-Yau manifolds\stackrel{GW potential/ Yukawa coupling}{\to} numerology$
so in categorical terms one should have something like
$Calabi-Yau manifolds\stackrel{A/B model}{\to}Calabi-Yau category \stackrel{?}{\to} numerology$
where “?” is presumibly Kevin Costello’s GW-potential associated to a TCFT. but in neither of Costello’s papers on TCFTs one can find any occurrence of “Yukawa”. so the question is: where have the
dear old basics of mirror symmetry gone in the categorical refomulation? I’m sure Maxim Kontsevich had this extremely clear in his mind when he formulated the homological mirror conjecture, but I’m
quite surprised these basics seem to have disappeared from the categorical treatment of mirror symmetry. are we still talking of the same thing? surely yes (at least I hope so), but I’d like to find
this written out in more evidence somewhere..
Domenico, it is not only Hodge diamond it is also correspondence between the variation of Hodge structures of type A and of type B. In categorical framework there is a suitable version of a more
general “noncommutative Hodge structures” which tell you again more than Hodge diamond. See again Katzarkov, Kontsevich, Pantev arxiv/0806.0107 for recent state of the art.
even in the most classical case, one has two equivalences simultaneously: A(X) and B(Y) and A(Y) and B(X).
but this can be asked only when both models are available for X and Y, and in the modern usage this is not the case: for instance one says that the mirror of the A-model on $\mathbb{P}^n$ is the
B-model on $w:(\mathbb{C}^*)^n\to \mathbb{C}$, where $w(z_1,\dots,z_n)=z_1+\cdots+z_n+q/(z_1\cdots z_n)$, where $q\in \mathbb{C}^*$. (this is the first mirror symmetry example in Katzarkov,
Kontsevich and Pantev).
it is not only Hodge diamond it is also correspondence between the variation of Hodge structures of type A and of type B
sure. waht I was pointing out is that even something much poorer such as the Hodge diamond seems to be lost in the abstract nonsense of the categorical formulation. for instance in Katzarkov,
Kontsevich and Pantev Hodge structures are described in a setting that though noncommutative is still geometric, and there is nothing such as a Hodge structure of a Calabi-Yau category (by teh way
there is no occurrence of “Calabi-Yau category” in that paper). so what? I don’t know. it could mean that the real framework for mirror symmetry are not Calabi-Yau categories, but rather “Calabi-Yau
categories with a fixed Hodge structure”? by the way, does every Calabi-Yau category admit an Hodge strucure (whatever this means)? clearly “geometric” Calabi-Yau categories do have natural Hodge
structures, so this reinforces Urs point of view that one is interested only in Calabi-Yau categories with a geometric origin. as far as concerns me, I’m now in a worse position than a few hours ago,
not only I do not now what is mirror symmetry, but neither I know what it is about. :)
#36 is sadly true..
I added that not only F(X) = D(Y) but simultaneously oine requires F(Y)=D(X). Previously just one half of the statement was there.
but this can be asked only when both models are available for X and Y, and in the modern usage this is not the case
It is not easy to show full mirror symmetry.
It is not true that the Hodge diamond is lost in modern formulation; I mean did you ever attend a homological mirror symmetry conference ? There is almost no talk without derived category formulation
and without several examples full of Hodge diamond data illustrating various aspects.
“geometric” Calabi-Yau categories do have natural Hodge structures, so this reinforces Urs point of view that one is interested only in Calabi-Yau categories with a geometric origin
One thing is the apperance of Calabi-Yau categories and their relation to TFTs and another (though related) thing is the mirror symmetry. It is not clear what you mean by “geometric origin”.
Restricting to commutative varieties is certainly not the proper scape as mirror partner of a variety is often not commutative; families of deformations can be studied using deformation theory and
they contain many interesting members which are not geometric in naive sense. Everything is geometric eventually at sufficiently abstract level. It is not good to try to make generalizations with so
little experience, this is a huge subject, and making easy souding generalizations leads to easy failures. It is best to go along the program which I proposed few months in nlab without any success:
to build in nlab the expositions of separate notions like equivariant localization, path integral localization, Picard-Fuchs equations, variations of Hodge structure, the language of smooth
A-infinity categories, Gromov-Witten invariants, Picard-Lefschetz theory, Maslov index, quantum D-module, Floer homology etc.
In Kontsevich’s Homological algebra of mirror symmetry the homological mirror conjecture is only one way, see at page 18, but I agree that for Calabi-Yau manifolds one should ask for both ways.
Hodge structures are described in a setting that though noncommutative is still geometric
Domenico, the complex parameter in the study of monodromy is a complex parameter. This is just one dimension. This is not about the underlying space, but about the business of meromorphic connection.
In typical applications in Katzarkov et al. one takes something like cyclic homology of very abstract category and put on top of it such structure.
It is not true that the Hodge diamond is lost in modern formulation
I think that if in a paper from 2008 whose title is “Meet homological mirror symmetry” the word “Hodge” appears once, then one is entitled to say that Hodge diamond is lost.. ;) by that I didn’t mean
that who really works in mirror symmetry is not concerned with Hodge structures, but that a basic question like what is mirror symmetry about? seems not to have a clear answer (since I can remember
Urs writing somewhere having got no clear answer to What is string theory? I can’t complain too much :) )
Young people learn hi fashionable techniques in particular areas. String theorists of today do not know the papers of classics from 1980s. You picked a paper from a new knowledgeable young specialist
so no wonder you face such peculiarieties. Read instad Orlov, Kontsevich, Gross, Seidel, Fukaya…
one takes something like cyclic homology of very abstract category and put on top of it such structure.
and this very abstract category is what has to be thought as some (derived) category of (quasi-coherent) sheves over a noncommutative space, right?
derived) category of (quasi-coherent) sheves over a noncommutative space
something like (derived) noncommutative deformation of a complex projective variety, I guess
Read instad Orlov, Kontsevich, Gross, Seidel, Fukaya…
but I do want to read these! but I need a framework where pieces of the jigsaw puzzle goes in, otherwise I’m lost. can you choose for me a few selected papers to read to get a basic but at the same
time neat and complete picture of what is mirror symmetry about?
There is no complete picture. Every picture or formalism captures just some aspects.
I think for the very beginning introduction the recent book might be the best
• Paul Aspinwall, Tom Bridgeland, Alastair Craw, Michael R. Douglas, Mark Gross, Dirichlet branes and mirror symmetry, Amer. Math. Soc. Clay Math. Institute 2009.
which I do not have but had browsed it online few pages as googlebooks or whatever. Orlov’s lectures on derived categories and mirror symmetry are very clearly written:
• A. N. Kapustin, D. O. Orlov, Lectures on mirror symmetry, derived categories, and D-branes, Uspehi Mat. Nauk 59 (2004), no. 5(359), 101–134; translation in Russian Math. Surveys 59 (2004), no. 5,
907–940, math.AG/0308173
but have very little excursion into physics.
well, actually in Kasputin and Orlov the only appearance of Hodge is in the physics introduction at the beginning of the lecture :)
just joking. now I’ll look more carefully at these references before coming back here. thanks a lot!
If somebody gets a file of Aspinwall et al. above I would like to have it :)
OK, so most of this discussion is waaaaay over my head. However, I spent some time over the last couple of years working with a student on developing a link - primarily grounded in conceptual, i.e.
physical, reasoning - between symmetry and entropy. This is based on a certain conceptual interpretation of entropy. We have a very tentative, but rather weak mathematical result, but one that is
based on fairly solid physical arguments. So my question to you guys is, at a higher level like what’s being discussed in this thread, could any of you envision a link between symmetry and entropy?
could any of you envision a link between symmetry and entropy?
Sure, it’s not deep: if your system has symmetry, then its entropy is invariant under the symmetry operation.
This has nothing special to do with mirror symmetry, but of course it applies there, too: if you want to compute the entropy of an SCFT and find it too hard, you can equivalently compute the entropy
of an equivalent mirror SCFT, if that turns out to be easier. Because they are, well, equivalent.
This is done for instance in this article here:
• Aspinwall, Maloney, Simons, Black Hole Entropy, Marginal Stability and Mirror Symmetry (pdf)
The authors want to compute entropy of a type IIB SCFT, find that too hard, invoke the mirror symmetry (-conjecture, for their purpose) and instead compute with the mirror dual type IIA SCFT. Or at
least argue that there is such a computation.
@domenico_fiorenza: I think that Kontsevich's original 1994 paper "Homological algebra of mirror symmetry" is (still!) the best place to start. This is where the homological mirror symmetry proposal
first appeared.
I added some words, and some references, regarding mirror symmetry beyond the Calabi-Yau case.
@Kevin: thanks for the reference. I actually knew that, but looking back to it after your suggestion has been a good idea: now I more clearly see which is the question I’m interested in. namely,
Kontsevich writes on page 18 We expect that the equivalence of derived categories will imply numerical predictions. and this is the statement I’d like to see worked out in detail. any reference?
(Costello? Kasputin-Orlov? others?)
@domenico_fiorenza: As far as I know, that statement has not yet been rigorously established in any cases. This is something I’ve been thinking about…
Try looking at Costello’s paper “The Gromov-Witten potential associated to a TCFT”. Also look at Katzarkov-Kontsevich-Pantev.
Edit to: mirror symmetry by Urs Schreiber at 2018-04-01 01:33:24 UTC.
Author comments:
adde pointer to textbookby Ibanez-Uranga | {"url":"https://nforum.ncatlab.org/discussion/937/mirror-symmetry/?Focus=6540","timestamp":"2024-11-03T09:51:09Z","content_type":"application/xhtml+xml","content_length":"164927","record_id":"<urn:uuid:5cfba3b0-e6eb-4731-8050-5ad1f514f3ae>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00600.warc.gz"} |
Calculating Relationships with Correlation Matrices · Advanced SQL · SILOTA
1. Summarizing Data
2. Calculating Relationships with Correlation Matrices
Making Correlation Coefficient Matrices to understand relationships in SQL
A scatter X-Y plot is a straightforward way to visualize the dependency between two variables. However, at times you want to understand how more than two variables are related. The correlation
In this recipe, we'll look for correlations betwen two or more variables and visualize it as a matrix. You can use the correlation matrix to figure out what activities are correlated, to plan future
activities. For example, within your customer support department, how are net promoter scores (NPS) related to support wait times.
The Correlation Coefficient math
One of the most commonly used correlation formula is Pearson’s. For a sample,
The SQL to correlate email subject lengths and open rates
Using Mailchimp's email campaign data – we are interested in finding if there's a relationship between the length of the subject and open/click rates.
1. Calculating correlation by hand
We are going to be using Common Table Expressions (CTEs) to organize our intermediate results:
with table_mean as
select avg(char_length(subject_line)) as mean_subject_length,
avg(report_summary_open_rate) as mean_open_rate
from mailchimp.campaigns
table_corrected as
select char_length(subject_line) - mean_subject_length as mean_subject_length_corrected,
report_summary_open_rate - mean_open_rate as mean_open_rate_corrected
from table_mean, mailchimp.campaigns
select sum(mean_subject_length_corrected * mean_open_rate_corrected) / sqrt(sum(mean_subject_length_corrected * mean_subject_length_corrected) * sum(mean_open_rate_corrected * mean_open_rate_corrected)) as r
from table_corrected;
This is a direct translation of the math equation. Fortunately, we don't have to repeat this each time, we can simply use the in-built corr function to calculate the correlation for us.
2. Calculating pairwise correlation using corr
The correlation can be calculated as follows:
select corr(char_length(subject_line), report_summary_open_rate) as r;
For more than two variables, we are going to repeat the correlation calculation pairwise between the variables and organize the results in the follow format. In step 3, it'll be clear why we use this
row col coeff
subject_length subject_length (always 1)
subject_length open_rate coefficient value
subject_length click_rate coefficient value
open_rate open_rate (always 1)
open_rate click_rate coefficient value
click_rate click_rate (always 1)
where subject_length is char_length(subject_line).
select 'subject_length' as row,
'subject_length' as col,
corr(subject_length, subject_length) as coeff
from mailchimp.campaigns
select 'subject_length' as row,
'open_rate' as col,
corr(subject_length, open_rate) as coeff
from mailchimp.campaigns
select 'subject_length' as row,
'click_rate' as col,
corr(subject_length, click_rate) as coeff
from mailchimp.campaigns
select 'open_rate' as row,
'open_rate' as col,
corr(open_rate, open_rate) as coeff
from mailchimp.campaigns
select 'open_rate' as row,
'click_rate' as col,
corr(open_rate, click_rate) as coeff
from mailchimp.campaigns
select 'click_rate' as row,
'click_rate' as col,
corr(click_rate, click_rate) as coeff
from mailchimp.campaigns
3. Pivoting the table to get a matrix
In this step, we'll be building a manual pivot table using the values from the col column. After the pivot, the values of the col column will become new columns in the resulting table.
select row,
sum(case when col='subject_length' then coeff else 0 end) as subject_length,
sum(case when col='open_rate' then coeff else 0 end) as open_rate,
sum(case when col='click_rate' then coeff else 0 end) as click_rate
from (
// ... query as before
group by row
order by row DESC
and our table will look like this:
row subject_length open_rate click_rate
subject_length 1 -0.31595114872448 -0.102740332427165
open_rate 0 1 0.798087505055931
click_rate 0 0 1
which matches the values we get from Excel or Google sheets. Our calculations show that open and click rates are negatively correlated with subject lengths and there's a strong positive relation
between open and click rates (which is obvious because they are not independent variables – you need to open an email in order to click it.)
👋 No fuss, just SQL We are open sourcing everything from the experience working with our agency clients. They spend thousands of dollars to get this level of detailed analysis – which you can now get
for free. We send one update every week. Join 400+ data analysts who are leveling up with our recipes. 👊
No spam, ever! Unsubscribe any time. See past emails here. | {"url":"http://www.silota.com/docs/recipes/sql-correlation-coefficient-matrix-summary.html","timestamp":"2024-11-05T15:59:48Z","content_type":"text/html","content_length":"80979","record_id":"<urn:uuid:9cbbc81f-8d8f-42b2-840d-611982d6e6f8>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00485.warc.gz"} |
Finding an Unknown Using the Similarity between Polygons
Question Video: Finding an Unknown Using the Similarity between Polygons Mathematics • First Year of Secondary School
Given that the two polygons are similar, find the value of π ₯.
Video Transcript
Given that the two polygons are similar, find the value of π ₯.
When dealing with any question involving similar polygons, we know that the corresponding angles are congruent or the same and the corresponding sides are proportional. One pair of corresponding
sides are π ½π and π π . A second pair of corresponding sides are π Άπ ½ and π π . As the corresponding sides are proportional, we know that their ratios are the same. Two π ₯ plus six
over seven π ₯ minus seven must be equal to 24 over 28.
In order to find the value of π ₯, we could cross multiply at this stage. However, it is easier to simplify our fractions first. We can factor the numerator and denominator of the left-hand
fraction. Two π ₯ plus six becomes two multiplied by π ₯ plus three. And seven π ₯ minus seven becomes seven multiplied by π ₯ minus one. Dividing the numerator and denominator of our right-hand
fraction by four gives us six over seven as 24 divided by four is six and 28 divided by four is equal to seven.
The denominators on both sides are divisible by seven. And the numerators are divisible by two. This leaves us with a simplified equation of π ₯ plus three over π ₯ minus one is equal to three. We
can multiply both sides of this equation by π ₯ minus one. Distributing the parentheses gives us π ₯ plus three is equal to three π ₯ minus three. We can then subtract π ₯ and add three to both
sides of the equation. This gives us six is equal to two π ₯. Finally, dividing both sides by two gives us π ₯ is equal to three.
If the two polygons are similar, the value of π ₯ is three. We can check this by substituting three back into the expressions for the smaller shape. Two multiplied by three is equal to six, and
adding six to this gives us 12. Seven multiplied by three is equal to 21 and subtracting seven gives us 14. It is therefore clear that our two polygons are similar with a scale factor of two as 12
multiplied by two is 24 and 14 multiplied by two is 28. | {"url":"https://www.nagwa.com/en/videos/915169734858/","timestamp":"2024-11-06T10:24:57Z","content_type":"text/html","content_length":"249464","record_id":"<urn:uuid:03184183-b9da-4611-81c0-40e108b16a40>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00339.warc.gz"} |
56 Comment(s)
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This brings the skills to real world experiences and shows how learning can allow for essential development in young people. Education is a different experience for everyone.
The article mentions that education isn't just what we think of when we think of the word education. It can be many different experiences, because education can also be things like appreciation of
others and new experiences and how that appreciation can help us grow.
Bringing the skills to real world experiences is important for students to learn. Also taking the skills learned and using them properly.
This brought some knowledge in how to bring the skills onto the real world to help understand when it is more applicable for your students.
I think that the bringing all of the skills into the real world will help to understand when it is set up for a student to learn from those experiences
You must help each other to gain knowledge and you must be inquisitive. Bringing skills into the real world makes it more applicable for students. Change is a part of the real world but is
facilitated by the social innovators of the world.
Education plus is more than just learning but the knowledge you hold and what could be expected from one, through this many teachers are able to help their students by working with them to help them
understand and remember what's being taught.
Education plus is more than just learning but the knowledge you hold and what could be expected from one, through this many teachers are able to help their students by working with them to help them
understand and remember what's being taught.
Education is more than just was is being taught in your classroom currently. You have to expect other things.
Education plus is important to talk about where learners are at and how they got the skills and real world application.
I think that Education Plus is important to the discussion of deep learning because it teaches us how to educate people on how to make a difference in the world. It gives us the tools we need to help
others succeed.
Education Plus talks about where learners who are educated come to acquire new things that are apart of real world problem solving.
Education plus, helps understand what the teaching should be like, understand others in the outside world. Changes happens all the time. I love this article.
I love that the article uses appreciation as a determining factor of what it means to be educated. We as people need to be better at appreciating opportunities, perspectives, opinions, cultures, and
Education is a broad base of many different people and things, teachers today must be aware of the outside world in the present as well as what may come in the future.
A teacher should not only be prepared for what is going on in today's world, but also for the possibilities of what may happen in tomorrow's world as well.
Education is more than a teacher and a student.
One of the things that stood out to me from this article is where it states that being educated means recognizing, appreciating, assessing, reinforcing, and cultivating. I believe that this is
important because being educated comes more than just learning but also passing that knowledge to someone else.
The digital revolution could be the key that unlocks true educational innovation, but it is at risk if teachers don't see the power in the tools (or the tools don't deliver as promised.) | {"url":"https://socinn.tiged.org/innovation/gallery/view/4529077","timestamp":"2024-11-03T07:14:14Z","content_type":"text/html","content_length":"84858","record_id":"<urn:uuid:ec10cd99-579c-43dd-8b30-3fb92ab3d7e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00841.warc.gz"} |
ShearLab is a MATLAB library developed for processing two- and threedimensional data with a certain class of basis functions named shearlets. Such shearlet systems are particularly well adapted to
represent anisotropic features (such as curves) that are often crucial in multidimensional data. The resulting representation has proven well-suited for image processing tasks such as inpainting,
denoising or image separation. On this website we provide the full MATLAB code, a framework for numerical tests as well as general information on shearlets.
Similiar to wavelet systems, shearlet systems are constructed by modifying generator functions. For wavelet systems, these functions are isotropically scaled and translated. While this is enough to
provide an optimally sparse representation for an interesting class of 1D functions it fails to do so in higher dimensions. To compensate this shortcoming, the direction of the generator functions
has to be varied. In shearlet theory, this is accomplished by shearing and anisotropic scaling.
Various desirable properties of such shearlet systems have been mathematically proven in recent years. In particular, it has been shown that there are compactly supported generator functions that
form a frame for L^2(R^2) and provide optimally sparse representations of cartoon-like functions up to a logarithmic factor. The latter is a key finding in shearlet theory because it is often assumed
that most natural signals can in fact be modelled as a cartoon-like function.
However, for a successful application of such a theory a good implementation is necessary. In contrast to rotation, shearing does not disrupt the integer grid and thus makes a unified treatment of
the continuous and digital shearlet theory possible. This unity allows for a rather faithful implementation of the digital shearlet transform in MATLAB.
The implementation of ShearLab has been thoroughly tested in various numerical experiments. The detailed results and their analysis can be found in the corresponding publications, the full code for
the experiments in the software section.
We invite you to explore the site, test the ShearLab library yourself and contact us if you have any questions or remarks! | {"url":"http://shearlab.math.lmu.de/","timestamp":"2024-11-07T05:50:39Z","content_type":"application/xhtml+xml","content_length":"15783","record_id":"<urn:uuid:9b6321f7-c277-40f5-83db-9cb6a5756d07>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00347.warc.gz"} |
A macroscopic crowd motion model of gradient flow type
Published Paper
Inserted: 18 sep 2009
Last Updated: 22 jul 2011
Journal: M3AS
Volume: 20
Number: 10
Year: 2010
revised version, with Section 6 on extensions and limits of the model
A simple model to handle the flow of people in emergency evacuation situations is considered: at every point $x$, the velocity $U(x)$ that individuals at $x$ would like to realize is given. Yet, the
incompressibility constraint prevents this velocity field to be realized and the actual velocity is the projection of the desired one onto the set of admissible velocities. Instead of looking at a
microscopic setting (where individuals are represented by rigid discs), here the macroscopic approach is investigated, where the unknwon is the evolution of a density $\rho(t,x)$. If a gradient
structure is given, say $U=-\nabla D$ where $D$ is, for instance, the distance to the exit door, the problem is presented as a Gradient Flow in the Wasserstein space of probability measures. The
functional which gives the Gradient Flow is neither finitely valued (since it takes into account the constraints on the density), nor geodesically convex, which requires for an ad-hoc study of the
convergence of a discrete scheme.
Keywords: Wasserstein distance, Gradient Flow, crowd motion, continuity equation | {"url":"https://cvgmt.sns.it/paper/269/","timestamp":"2024-11-04T12:28:48Z","content_type":"text/html","content_length":"9168","record_id":"<urn:uuid:9cf45347-ba7a-4b8c-b008-7de75d725b84>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00412.warc.gz"} |
Calculation of the unsteady pressure distribution on an oscillating cascade with thick profiles and steady turning angle
A method based on Martensen's theory of vorticity layers on a surface in steady flow is derived to calculate the unsteady pressure distribution on a harmonically oscillating plane cascade with an
arbitrary profile shape in incompressible flow. After linearizing the basic equations of the initial boundary value problem, which is generally nonlinear, a Fredholm integral equation is developed
for the pressure distribution on oscillating blades. Among other characteristic parameters, this equation describes the interaction of steady and unsteady flow. Pressure distributions and aerodynamic
coefficients are given for a special case in order to examine the influence of parameters.
Ph.D. Thesis
Pub Date:
November 1981
□ Cascade Flow;
□ Computational Fluid Dynamics;
□ Pressure Distribution;
□ Rotor Aerodynamics;
□ Unsteady State;
□ Aerodynamic Coefficients;
□ Fredholm Equations;
□ Harmonic Oscillation;
□ Incompressible Flow;
□ Vorticity Equations;
□ Fluid Mechanics and Heat Transfer | {"url":"https://ui.adsabs.harvard.edu/abs/1981PhDT........86C/abstract","timestamp":"2024-11-10T19:13:08Z","content_type":"text/html","content_length":"34893","record_id":"<urn:uuid:992b98d3-47f4-4d7c-9f92-14542048a1ee>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00485.warc.gz"} |
1.7: Errors in Measurement
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Scientific measurements are of no value (or at least, they're not really scientific) unless they are given with some statement of the errors they contain. If a poll reports that one candidate leads
another by 5%, that may be politically useful for the winning candidate to point out. But all respectable polls are scientific, and report errors. If the error in measurement is plus or minus 10%,
which indicates anything from the candidate leading by 15% to trailing by 5%, the poll really does not reliably tell who is in the lead. If the poll had an error of 1%, the leading candidate could
make a more scientific case that for being in the lead (by a 4% to 6% margin, or 5% +/-1%).
Luckily, we have an easy method for indicating the error in a measurement, and it is suggested in the results we gave for an air-pollution experiment, in which the mass of smoke collected was given
as 3.42 × 10^–2 g and the volume of the balloon was given as 1.021 926 4 × 10^7 cm^3.
"Significant Figures"
The volume was calculated from a balloon diameter of 106 inches. There is something strange about the second quantity, though. It contains a number which was copied directly from the display of an
electronic calculator and has Too Many Digits to represent the error of measurement properly.
The reliability (more specifically, the random error or precision) of a quantity derived from a measurement is customarily indicated by the number of significant figures (or significant digits) it
contains. For example, the three significant digits in the quantity 3.42 × 10^–2 g tell us that a balance was used on which we could distinguish 3.42 × 10^–2 g from 3.43 × 10^–2 or 3.41 × 10^–2 g.
There might be some question about the last digit, but those to the left of it are taken as completely reliable. Another way to indicate the same thing is (3.42±0.01) × 10^–2 g. Our measurement is
somewhere between 3.41 × 10^–2 and 3.43 × 10^–2 g.
To illustrate how the number of significant digits indicates the error, suppose we had a measurement reported as 3.42 g. From the measurement, we would assume an error of 0.01 g, and the percent
error is
\[ \text{Percent Error } = \dfrac{\text{Error}}{\text{Value}} \times 100\%\label{1} \]
\[ \text{Percent Error} = \dfrac{0.01 \text{ g}}{3.42 \text{ g}} \times 100\% = 0.29\% \nonumber \]
If the correctly recorded measurement had more digits, like 3.4275 g, the measurement itself would indicate that a more expensive balance was used to give better precision (a smaller percent error).
In this case, the error is 0.0001 g, and the percent error is:
\[ \text{Percent Error} = \dfrac{0.0001 \text{ g}}{3.4265 \text{ g}} \times 100\% = 0.0029\% \nonumber \]
So we record measurements to the proper number of digits as a rudimentary method^[1] for indicating their approximate error.
Figure \(\PageIndex{1}\) The level of a liquid in a graduated cylinder. The saucer-shaped surface of a liquid in a tube is called a meniscus.
Illustration of a part of a graduated cylinder filled with a liquid. The liquid surface has a downwards curvature that gives off a reading of 38 centimeter cubed.
As another example of choosing an appropriate number of significant digits, let us read the volume of liquid in a graduated cylinder.
The bottom of the meniscus lies between graduations corresponding to 38 and 39 cm^3. We can estimate that it is at 38.5 cm^3, but the last digit might be off a bit-perhaps it looks like 38.4 or 38.6
cm^3 to you. Since the third digit is in question, we should use three significant figures. The volume would be recorded as 38.5 cm^3, indicating 0.1 cm^3 error. Laboratory equipment is often
calibrated similarly to this graduated cylinder—you should estimate to the nearest tenth of the smallest graduation.
Counting Significant Digits
In some ordinary numbers, for example, 0.001 23, zeros serve merely to locate the decimal point. They do not indicate the reliability of the measurement and therefore are not significant. Another
advantage of scientific notation is that we can assume that all digits are significant. Thus if 0.001 23 is written as 1.23 × 10^–3, only the 1, 2, and 3, which indicate the reliability of the
measurement, are written. The decimal point is located by the power of 10.
If the rule expressed in the previous paragraph is applied to the volume of air collected in our pollution experiment, 1.021 926 4 × 10^7 cm^3, we find that the volume has eight significant digits.
This implies that it was determined to ±1 cm^3 out of about 10 million cm^3, a reliability which corresponds to locating a grasshopper exactly at some point along the road from Philadelphia to New
York City. For experiments as crude as ours, this is not likely. Let us see just how good the measurement was.
You will recall that we calculated the volume from the diameter of the balloon, 106 in. The three significant figures imply that this might have been as large as 107 in or as small as 105 in. We can
repeat the calculation with each of these quantities to see how far off the volume would be:
\( & \text{ }r\text{ = }\dfrac{\text{1}}{\text{2}}\text{ }\times \text{ 107 in = 53}\text{.5 in }\times \text{ }\dfrac{\text{1 cm}}{\text{0}\text{.3937 in}}\text{ } \\ & \text{ = 135}\text{.890 27
cm} \\ & \text{ }V\text{ = }\dfrac{\text{4}}{\text{3}}\text{ }\times \text{ 3}\text{.141 59 }\times \text{ (135}\text{.890 27 cm)}^{\text{3}}\text{ } \\ & \text{ = 10 511 225 cm}^{\text{3}}\text{ =
1}\text{.051 122 5 }\times \text{ 10}^{\text{7}}\text{ cm}^{\text{3}} \\ & \text{or }V\text{ = }\dfrac{\text{4}}{\text{3}}\text{ }\times \text{ 3}\text{.141 59 }\times \text{ }\left( \dfrac{\text{1}}
{\text{2}}\text{ }\times \text{ 105 in }\times \text{ }\dfrac{\text{1 cm}}{\text{0}\text{.3937 in}} \right)^{\text{3}}\text{ } \\ & \text{ = 9 932 759 cm}^{\text{3}}\text{ = 0}\text{.993 275 9 }\
times \text{ 10}^{\text{7}}\text{ cm}^{\text{3}} \\ \)
That is, the volume is between 0.99 × 10^7 and 1.05 × 10^7 cm^3 or (1.02 ± 0.03) × 10^7 cm^3. We should round our result to three significant figures, for example, 1.02 cm³, because the last digit,
namely 2, is in question.
So the conclusion is simple: The calculated result cannot be more precise than the measurement from which it was made. To show this in an alternative way, the error in 106 in would be about 1 in, so
the percent error is
\[ \text{Percent Error} = \dfrac{1 \text{ in}}{106 \text{ in}} \times 100\% = 0.9\% \nonumber \]
If the result is reported as 10 219 264 cm^3, the implied error is 1 part in 10 219 264 and the percent error is:
\[ \text{Percent Error} = \dfrac{1 \text{ cm}^{\text{3}}}{10219264 \text{ cm}^\text{3}} \times 100\% = 0.000009\% \nonumber \]
Clearly, we should not be able to increase the precision of our data by merely manipulating the data mathematically, so we need simple rules for rounding numbers in calculated quantities.
If the result is reported properly as 1.02 × 10^7 cm^3, the percent error is
\[ \text{Percent Error} = \dfrac{0.01 \text{ cm}^{\text{3}}}{1.02 \text{ cm}^\text{3}} \times 100\% = 1\% \nonumber \],
approximately the same as the original measurement.
1. All digits to be rounded are removed together, not one at a time.
2. If the left-most digit to be removed is less than five, the last digit retained is not altered.
3. If the left-most digit to be removed is greater than five, the last digit retained is increased by one.
4. If the left-most digit to be removed is five and at least one of the other digits to be removed is nonzero. the last digit retained is increased by one.
5. If the left-most digit to he removed is five and all other digits to he removed are zero, the last digit retained is not altered if it is even, but is increased by one if it is odd.
Application of the Rules for Rounding Numbers can be illustrated by an example.
Round each of the numbers below to three significant figures.
1. 34.7449
2. 34.864
3. 34.754
4. 34.250
5. 34.35
a) Apply rules 1 and 2:
449 are digits to be removed. Arrow points to 7 as the last digit retained and that is less than five. The final results shows 34.7.
(Note that a different result would be obtained if the digits were incorrectly rounded one at a time from the right.)
Apply rules 1 and 3: 34.8
64 →
c) Apply rules 1 and 4: 34.754 → 34.8
d) Apply rules 1 and 5: 34.250 → 34.2
e) Apply rule 5: 34.35 → 34.4
To how many significant figures should we round our air-pollution results? We have already done a calculation involving multiplication and division to obtain the volume of our gas-collection balloon.
It involved the following quantities and numbers:
a calculation involving multiplication and division to obtain the volume of our gas-collection balloon
│ \( 106 \text{ in}\) │ A measured quantity with three significant figures │
│\( 0.3937 \text{ in/cm}\) │A conversion factor given with four significant figures^(1) │
│\(3.141 59\) │π used with six significant figures (we could obtain more if we wanted) │
│\(\tfrac{\text{4}}{\text{3}} \text{ and } \tfrac{\text{1}}{\text{2}}\)│Numbers^(2) with an infinite number of significant figures since the integers in these fractions are exact by definition│
1. some conversion factors like 1 mg = 0.001 g are exact, and have an infinite number of significant figures).
2. Note that pure numbers are not measurements or quantities, and so they are unlimited in the number of significant figures (numbers have zero error).
The result of the calculation contained three significant figures — the same as the least-reliable number. This illustrates the general rule that for multiplication and division the number of
significant figures in the result is the same as in the least-reliable measurement. Defined numbers such as π, ½ or 100 cm/1m are assumed to have an infinite number of significant figures.
In the case of addition and subtraction, a different rule applies. Suppose, for example, that we weighed a smoke-collection filter on a relatively inaccurate balance that could only be read to the
nearest 0.01 g. After collecting a sample, the filter was reweighed on a single-pan balance to determine the mass of smoke particles.
• Final mass: 2.3745 g (colored digits are in question)
• Initial mass: –2.32 g
• Mass of smoke: 0.0545 g
Since the initial weighing could have been anywhere from 2.31 to 2.33 g, all three figures in the final result are in question. (It must be between 0.0445 and 0.0645 g). Thus there is but one
significant digit, and the result is 0.05g. The rule here is that the result of addition or subtraction cannot contain more digits to the right than there are in any of the numbers added or
subtracted. Note that subtraction can drastically reduce the number of significant digits when this rule is applied.
Rounding numbers is especially important if you use an electronic calculator, since these machines usually display a large number of digits, most of which are meaningless. The best procedure is to
carry all digits to the end of the calculation (your calculator will not mind the extra work!) and then round appropriately. Answers to subsequent calculations in this book will be rounded according
to the rules given. You may wish to go back to previous examples and round their answers correctly as well.
Evaluate the following expressions, rounding the answer to the appropriate number of significant figures.
1. \(32.61 \text{g} + 8.446 \text{g} + 7.0 \text{g} \)
2. \(0.136 \text{cm}^3 \times 10.685 \text{g cm}^{-3}\)
1. \(32.61 \text{g} + 8.446 \text{g} + 7.0 \text{g} = 48.056 \text{g} = 48.1 \text{g (7.0 has only one figure to the right of the decimal point.)}\)
2. \(0.136 \text{cm}^3 \times 10.685 \text{g cm}^{-3} = 1.453 \text{g} = 1.45 \text{g (0.136 has only three significant figures.)}\)
Precision vs. Accuracy
When we suggested filling a surplus weather balloon to measure how much gas was pumped through our air-pollution collector, we mentioned that this would be a rather crude way to determine volume. For
one thing, it would not be all that simple to measure the diameter of an 8- or 9-ft sphere reliably. Using a yardstick, we would be lucky to have successive measurements agree within half an inch or
so. It was for this reason that the result was reported to the nearest inch. The degree to which repeated measurements of the same quantity yield the same result is called precision. Repetition of a
highly precise measurement would yield almost identical results, whereas low precision implies numbers would differ by a significant percentage from each other.
A highly precise measurement of the diameter of our balloon could be achieved, but it would probably not be worthwhile. We have assumed a spherical shape, but this is almost certainly not exactly
correct. No matter how precisely we determine the diameter, our measurement of gas volume will be influenced by deviations from the assumed shape. When one or more of our assumptions about a
measuring instrument are wrong, the accuracy of a result will be affected. An obvious example would be a foot rule divided into 11 equal inches. Measurements employing this instrument might agree
very precisely, but they would not be very accurate. The following data on the mass of the smoke, measured repeatedly with two balances and a scale, show the difference between accuracy and
precision. If he mass has been determined to be exactly 0.03420 g by an independent measurement, the first balance is both accurate (the correct value is obtained) and precise (the range of
measurements is small); the second balance is precise, but not accurate; the scale is neither accurate nor precise. Note that we are comparing the three to one another; relative to a balance that had
a range of 0.00001 g, all three would be imprecise.
Table \(\PageIndex{1}\): Accurate vs. Precise
│ │Mass, Balance A │Mass, Balance B │Mass, Scale│
│ │0.0342 │0.0362 │0.0380 │
│ │0.0341 │0.0361 │0.0370 │
│ │0.0342 │0.0363 │0.0390 │
│ │0.0343 │0.0362 │0.0400 │
│average│0.0342 │0.0362 │0.0385 │
│range │0.001 │0.001 │0.013 │
The following figures help to understand the difference between precision (small expected difference between multiple measurements) and accuracy (difference between the result and a known value).
Figure \(\PageIndex{1}\): (left) High accuracy and Low Precision. (middle left) Low accuracy and High precision. (middle right) High accuracy and Low Precision. (middle left) Low accuracy and Low
precision(Public Domain; DarkEvil)
An important point of a different kind is illustrated in the last two paragraphs. A great many common words have been adopted into the language of science. Usually such an adoption is accompanied by
an unambiguous scientific definition which does not appear in a normal dictionary. Precision and accuracy are many times treated as synonyms, but in science each has a slightly different meaning.
Another example is quantity, which we have defined in terms of “number × unit.” Other English words like bulk, size, amount, and so forth, may be synonymous with quantity in everyday speech, but not
in science. As you encounter other words like this, try to learn and use the scientific definition as soon as possible, and avoid confusing it with the other meanings you already know.
Even granting the crudeness of the measurements we have just described, they would be adequate to demonstrate whether or not an air-pollution problem existed. The next step would be to find a chemist
or public health official who was an expert in assessing air quality, present your data, and convince that person to lend his or her skill and authority to your contention that something was wrong.
Such a person would have available equipment whose precision and accuracy were adequate for highly reliable measurements and would be able to make authoritative public statements about the extent of
the air-pollution problem. | {"url":"https://chem.libretexts.org/Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.07%3A_Errors_in_Measurement","timestamp":"2024-11-07T22:11:54Z","content_type":"text/html","content_length":"157316","record_id":"<urn:uuid:1006fb7c-ebb2-4d50-9287-4d2dc55da0d6>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00550.warc.gz"} |
Spaceship Math Addition A: 1+2, 2+1, 1+3, 3+1
Spaceship Math is a difficult addition method that is used in Math Class at school. The students are allowed to do their own homework, but they need to follow a chart of instructions. This edition
works on a line by line basis. When it is completed in the classroom, it is then tested and added to another. So the student can’t see how many things they have done before they start the second
Spaceship Math Addition A
Worksheet 1
In addition to Spaceship Math, there are other ways to get your math work done and to find out what you are doing wrong. You can use the online form to find out what you are doing wrong. A regular
teaching guide can also be a good addition to your study time.
Sometimes, the instructions given by Spaceship Math can be confusing to a student. This can be because of all the different keys on the board. As long as you remember the key that you use, you will
have no trouble understanding the work.
Worksheet 2
However, you need to remember that you can’t take one approach to Math and apply it to other methods of addition. It all depends on how you want to approach your work. Here are some ways that you can
use to solve problems:
Take the first two problems that you solved to the times’ tables that they come from. You will know that there are times tables for different areas of the table. Find the times table for the item
that you want to purchase so that you can use the information to find the times table of whatever you are looking for.
Worksheet 3
Go to the Math Worksheet on your computer. You will see the row of numbers from zero to nine, with the row indicated by the number that is right next to it. Start working on the item that is on the
left side of the calculation formula.
Try to find a formula that matches the item that you are working on. It will be easy for you to do this if you remember the name of the formula that you are using. You should make notes when you find
the formula, and you should also make a graph that you will use to see the differences between the formulas and the item.
Worksheet 4
Spaceship Math may be difficult, but it can be a great tool to learn the numbers. You will be able to find out what is wrong with your approach to problems and if you have made any mistakes.
Gallery for Spaceship Math Addition A: 1+2, 2+1, 1+3, 3+1
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Assessment of the Mechanical Integrity of a Pipeline with Metal Loss, Using Finite Element Method
Monday, April 11, 2016
Exhibit Hall E (George R. Brown )
Following the guidelines proposed by the ASME B31G and API579 standards, for the assessment of systems that compromise its mechanical integrity, given the existence of defects associated with the
presence of corrosion damage, the probability of failure of a pipeline, which transports hazardous materials associated with the Oil & Gas industry, and is exposed to pitting corrosión damage was
calculated when it overcomes its yield stress. For this, a methodology that combines the corresponding structural analysis of the system and subsequent reliability analysis was proposed. Through the
first, the calculation of the von Mises stress present in areas where the pitting defects was made; for this, the corrosion growth model to be used in the present work was defined by implementing the
correlation proposed by the Southwest Research Institute (SwRI) to find out the corrosion rate growth in annual terms. Subsequently the pipeline section where the study would be carried out was
defined, based on the results obtained in a previous study, which analyzed information related to corrosion damage present on the pipeline, collected through the In Line Inspection (ILI) technique.
This allowed to restrict the test area to a total length of 80 cm and 1/8 section of pipe, lastly simulating the system and obtaining the loads associated with the von Mises stress affecting
corrosion defined areas. For reliability analysis it was necessary to find the probability distribution function associated with the von Mises stress on the system studied; defining an experimental
design which sought to evaluate different values of the random variables defined on the model proposed by SwRI, thus simulating the system a sufficient number of times in order to find the
distribution of von Mises stress for this case study. With the distribution function defined, system reliability analysis was carried out through Monte Carlo simulations; thus, it was established
that the limit state function would have a conventional structure comparing resistance with loads. Thus, the resistance was set as the yield stress of the material, while the load would be the von
Mises stress, and thus the random variable from which random numbers governed by the probability distribution found previously were generated. As mentioned previously, it was defined that the present
system fails when the von Mises stress exceeds the yield stress of the material; finding that for the time range evaluated (year one, five, ten and fifteen), the probability of failure is behaved as
expected, increasing year by year if no maintenance is not performed to the pipeline section evaluated.
Extended Abstract: File Not Uploaded | {"url":"https://aiche.confex.com/aiche/s16/webprogram/Paper441629.html","timestamp":"2024-11-10T05:27:03Z","content_type":"text/html","content_length":"8732","record_id":"<urn:uuid:1d6cff7d-a04e-44c6-ad94-1e4d75d59601>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00355.warc.gz"} |
Cipher (Cryptography)
A cipher, in the context of cryptography, is an algorithm used to transform a message into another string of symbols in order to hide its meaning from a third party.
A cipher operates by replacing each symbol in the original string with a word in some alphabet.
The specific nature of the replacement depends on the specifics of the cipher.
In the context of cryptography, the plaintext is the message being transformed into an encrypted string.
In the context of cryptography, the ciphertext is the ciphered string after the plaintext has been processed by the cipher.
In the context of cryptography, a key is a piece of information which:
is used to convert plaintext to ciphertext
allows the ciphertext to be transformed back into the original plaintext.
These two processes may use different keys.
Also see
The word cipher can also be found in its less common spelling: cypher.
The word ultimately derives from the Arabic صِفْر (ṣifr), meaning zero or empty. | {"url":"https://proofwiki.org/wiki/Definition:Cipher_(Cryptography)","timestamp":"2024-11-04T17:48:37Z","content_type":"text/html","content_length":"46080","record_id":"<urn:uuid:85629784-5a99-42b2-ae4b-a17a50de47b7>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00744.warc.gz"} |
FM Synthesis is an old-school way of generating musical instrument sounds, initially popularised by the Adlib and SoundBlaster PC sound cards in the late ’80s (and, of course, in piano keyboards).
Here’s an example of what FM Synth music sounded like in games. Ahh the nostalgia.
A friend who is a school music teacher found that his students all use the same identical samples for instruments for their creations. So I created YouSynth, a web app that allows you to create any
instrument you like using a basic form of FM synthesis, and download that instrument as WAV file you can use anywhere, as well as play around with it using an attached MIDI keyboard. Please check it
So as to not leave out the maths teachers, I thought I’d write an article about how the maths for FM synthesis works! I think it’s fascinating, hopefully you might too. My dream is that maybe a
maths teacher somewhere would use this as an interesting demonstration of applied maths to pique their students’ interest :)
To start with, here’s the gist of it - for each sample, the value is:
carrierFrequency * time * 2 * pi
sin(modulatorFrequency * time * 2 * pi) * modulatorEnvelope
) * carrierEnvelope
Now let’s break that down.
Carrier frequency
The carrier frequency is the fundamental frequency of the note. Eg for A4, it’s 440 Hz. For Middle C, aka C4, it’s ~261.6 Hz.
For each note you go up (including sharps), the frequency is multiplied by 2^(1/12). The 1/12 is because there are 12 freqencies in each octave when including the sharps. The 2^ is because
frequencies double with each octave. Eg A4 is 440 Hz, and A5 is 880 Hz.
When working with MIDI, each note gets a number representation: C4=60, C#4=61, D4=62, etc. To convert from a midi note to a frequency, the formula is: 440 * 2 ^ ((midiNote - 69) / 12).
The time in the above formula is in seconds since the note started playing. Since you’d typically be generating samples at a rate of 44100 or 48000 Hz, to convert from the sample number to the
time, this formula applies: time = sample / sampleRate.
The 2 * pi is necessary because sin repeats its output every multiple of 2 * pi on its input. An interesting aside: Credible mathematicians consider that tau (2 * pi) should be taught to students
instead of pi, because it is so common that we need to double pi before using it, so why not just use the double as the famous constant, then? See the Tau manifesto.
Modulator frequency
The modulator is the waveform that ‘modulates’ the fundamental frequency. Think of it as the whammy bar on a guitar being wiggled up and down quickly.
Typically the modulator frequency is a whole-number multiple or fraction of the fundamental frequency. Eg for a fundamental of 440 Hz, the following modulator frequencies all sound ‘nice’: 110
(440/4), 146.7 (440/3), 220 (440/2), 440, 880, 1320, etc.
The envelopes control the amplitude/volume of the carrier and modulator over time. From initially zero, quickly up to 100%, then down to a sustained volume of perhaps 50%, where it remains while the
piano key is held, then when the key is released, it gradually returns to 0.
A common strategy is the ADSR envelope.
During the attack stage: amplitude = time / attackDuration.
During decay stage: amplitude = 1 - (time - attackDuration) / decayDuration * (1 - sustainAmplitude).
During sustain stage: amplitude = sustainAmplitude.
During release stage: amplitude = sustainAmplitude - releasingTime / releaseDuration.
Other waves
To make more interesting sounds, other waveforms besides sine waves can be used. Some common ones are square, triangle, and sawtooth. Here are their formulae which repeat every multiple of 1 on the
• Sine = sin(x * 2 * pi)
• Square = 4 * floor(x) - 2 * floor(2 * x) + 1
• Triangle = 2 * abs(2 * (x + 0.25 - floor(x + 0.75))) - 1
• Sawtooth = 2 * (x - floor(x + 0.5))
So there you have it, the maths behind basic FM Synthesis. Thanks for reading, hope you found this fascinating, at least a tiny bit, God bless!
Photo by Vackground on Unsplash | {"url":"http://www.splinter.com.au/","timestamp":"2024-11-10T15:44:27Z","content_type":"text/html","content_length":"13555","record_id":"<urn:uuid:62bb17b4-b10a-4e72-b0ee-b89e842af33f>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00601.warc.gz"} |
The Neural Network Input-Process-Output Mechanism -- Visual Studio Magazine
Neural Network Lab
The Neural Network Input-Process-Output Mechanism
Understanding the feed-forward mechanism is required in order to create a neural network that solves difficult practical problems such as predicting the result of a football game or the movement of a
stock price.
An artificial neural network models biological synapses and neurons and can be used to make predictions for complex data sets. Neural networks and their associated algorithms are among the most
interesting of all machine-learning techniques. In this article I'll explain the feed-forward mechanism, which is the most fundamental aspect of neural networks. To get a feel for where I'm headed,
take a look at the demo program in Figure 1 and also a diagram of the demo neural network's architecture in Figure 2.
[Click on image for larger view.] Figure 1. Neural network feed-forward demo.
If you examine both figures you'll see that, in essence, a neural network accepts some numeric inputs (2.0, 3.0 and 4.0 in this example), does some processing and produces some numeric outputs (0.93
and 0.62 here). This input-process-output mechanism is called neural network feed-forward. Understanding the feed-forward mechanism is required in order to create a neural network that solves
difficult practical problems such as predicting the result of a football game or the movement of a stock price.
[Click on image for larger view.] Figure 2. Neural network architecture.
If you're new to neural networks, your initial impression is likely something along the lines of, "This looks fairly complicated." And you'd be correct. However, I think the demo program and its
explanation presented in this article will give you a solid foundation for understanding neural networks. This article assumes you have expert-level programming skills with a C-family language. I
coded the demo program using C#, but you shouldn't have too much trouble refactoring my code to another language such as Visual Basic or Python.
The demo can be characterized as a fully connected three-input, four-hidden, two-output neural network. Unfortunately, neural network terminology varies quite a bit. The neural network shown in
Figure 2 is most often called a two-layer network (rather than a three-layer network, as you might have guessed) because the input layer doesn't really do any processing. I suggest this by showing
the input nodes using a different shape (square inside circle) than the hidden and output nodes (circle only).
Each node-to-node arrow in Figure 2 represents a numeric constant called a weight. For example, assuming nodes are zero-indexed starting from the top of the diagram, the weight from input node 2 to
hidden node 3 has value 1.2. Each hidden and output node, but not any input node, has an additional arrow that represents a numeric constant called a bias. For example, the bias for hidden node 0 has
value -2.0. Because of space limitations, Figure 2 shows only one of the six bias arrows.
For a fully connected neural network, with numInputs input nodes, numHidden hidden nodes and numOutput output nodes, there will be (numInput * numHidden) + (numHidden * numOutput) weight values. And
there will be (numHidden + numOutput) bias values. As you'll see shortly, biases are really just a special type of weights, so for brevity, weights and biases are usually collectively referred to as
simply weights. For the three-input, four-hidden, two-output demo neural network, there are a total of (3 * 4) + (4 * 2) + (4 + 2) = 20 + 6 = 26 weights.
The demo neural network is deterministic in the sense that for a given set of input values and a given set of weights and bias values, the output values will always be the same. So, a neural network
is really just a form of a function.
Computing the Hidden-Layer Nodes
Computing neural network output occurs in three phases. The first phase is to deal with the raw input values. The second phase is to compute the values for the hidden-layer nodes. The third phase is
to compute the values for the output-layer nodes.
In this example, the demo does no processing of input, and simply copies raw input into the neural network input-layer nodes. In some situations a neural network will normalize or encode raw data in
some way.
Each hidden-layer node is computed independently. Notice that each hidden node has three weight arrows pointing into it, one from each input node. Additionally, there's a single bias arrow into each
hidden node. Understanding hidden node computation is best explained using a concrete example. In Figure 2, hidden node 0 is at the top of the diagram. The first step is to sum each input times each
input's associated weight: (2.0)(0.1) + (3.0)(0.5) + (4.0)(0.9) = 5.3. Next the bias value is added: 5.3 + (-2.0) = 3.3. The third step is to feed the result of step two to an activation function.
I'll describe this in more detail shortly, but for now: 1.0 / (1.0 + Exp(-3.3)) = 0.96. The values for hidden nodes 1, 2 and 3 are computed similarly and are 0.55, 1.00 and 0.73, as shown in Figure 1
. These values now serve as inputs for the output layer.
Computing the Output-Layer Nodes
The output-layer nodes are computed in the same way as the hidden-layer nodes, except that the values computed into the hidden-layer nodes are now used as inputs. Notice there are a lot of inputs and
outputs in a neural network, and you should not underestimate the difficulty of keeping track of them.
The sum part of the computation for output-layer node 0 (the topmost output node in the diagram) is: (0.96)(1.3) + (0.55)(1.5) + (1.00)(1.7) + (0.73)(1.9) = 5.16. Adding the bias value: 5.16 + (-2.5)
= 2.66. Applying the activation function: 1.0 / (1.0 + Exp(-2.66)) = 0.9349 or 0.93 rounded. The value for output-layer node 1 is computed similarly, and is 0.6196 or 0.62 rounded.
The output-layer node values are copied as is to the neural network outputs. In some cases, a neural network will perform some final processing such as normalization.
Neural network literature tends to be aimed more at researchers than software developers, so you'll see a lot of equations with Greek letters such as the one in the lower-right corner of Figure 2.
Don't let these equations intimidate you. The Greek letter capital phi is just an abbreviation for the activation function (many other Greek letters, such as kappa and rho, are used here too). The
capital sigma just means "add up some terms." The lowercase x and w represent inputs and weights. And the lowercase b is the bias. So the equation is just a concise way of saying: "Multiply each
input times its weight, add them up, then add the bias, and then apply the activation function to that sum."
The Bias As a Special Weight
Something that often confuses newcomers to neural networks is that the vast majority of articles and online references treat the biases as weights. Take a close look at the topmost hidden node in
Figure 2. The preliminary sum is the product of three input and three weights, and then the bias value is added: (2.0)(0.1) + (3.0)(0.5) + (4.0)(0.9) + (-2.0) = 3.30. But suppose there was a dummy
input layer node with a value of 1.0 that was added to the neural network as input x3. If each hidden-node bias is associated with the dummy input value, you get the same result: (2.0)(0.1) + (3.0)
(0.5) + (4.0)(0.9) + (1.0)(-2.0) = 3.30.
Treating biases as special weights that are associated with dummy inputs that have constant value 1.0 simplifies writing research-related articles, but with regard to actual implementation I find the
practice confusing, error-prone and hack-ish. I always treat neural network biases as biases rather than as special weights with dummy inputs.
Activation Functions
The activation function used in the demo neural network is called a log-sigmoid function. There are many other possible activation functions that have names like hyperbolic tangent, Heaviside and
Gaussian. It turns out that choosing activation functions is extremely important and surprisingly tricky when constructing practical neural networks. I'll discuss activation functions in detail in a
future article.
The log-sigmoid function in the demo is implemented like so:
private static double LogSigmoid(double z)
if (z < -20.0) return 0.0;
else if (z > 20.0) return 1.0;
else return 1.0 / (1.0 + Math.Exp(-z));
The method accepts a type double input parameter z. The return value is type double with value between 0.0 and 1.0 inclusive. In the early days of neural networks, programs could easily generate
arithmetic overflow when computing the value of the Exp function, which gets very small or very large, very quickly. For example, Exp(-20.0) is approximately 0.0000000020611536224386. Even though
modern compilers are less susceptible to overflow problems, it's somewhat traditional to specify threshold values such as the -20.0 and +20.0 used here.
Overall Program Structure
The overall program structure and Main method of the demo program (with some minor edits and WriteLine statements removed) is presented in Listing 1. I used Visual Studio 2012 to create a C# console
application named NeuralNetworkFeedForward. The program has no significant Microsoft .NET Framework dependencies, and any version of Visual Studio should work. I renamed file Program.cs to the more
descriptive FeedForwardPrograms.cs and Visual Studio automatically renamed class Program, too. At the top of the template-generated code, I removed all references to namespaces except the System
Listing 1. Feed-Forward demo program structure.
using System;
namespace NeuralNetworkFeedForward
class FeedForwardProgram
static void Main(string[] args)
Console.WriteLine("\nBegin neural network feed-forward demo\n");
Console.WriteLine("Creating a 3-input, 4-hidden, 2-output NN");
Console.WriteLine("Using log-sigmoid function");
const int numInput = 3;
const int numHidden = 4;
const int numOutput = 2;
NeuralNetwork nn = new NeuralNetwork(numInput, numHidden, numOutput);
const int numWeights = (numInput * numHidden) +
(numHidden * numOutput) + numHidden + numOutput;
double[] weights = new double[numWeights] {
0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2,
-2.0, -6.0, -1.0, -7.0,
1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0,
-2.5, -5.0 };
Console.WriteLine("\nWeights and biases are:");
ShowVector(weights, 2);
Console.WriteLine("Loading neural network weights and biases");
Console.WriteLine("\nSetting neural network inputs:");
double[] xValues = new double[] { 2.0, 3.0, 4.0 };
ShowVector(xValues, 2);
Console.WriteLine("Loading inputs and computing outputs\n");
double[] yValues = nn.ComputeOutputs(xValues);
Console.WriteLine("\nNeural network outputs are:");
ShowVector(yValues, 4);
Console.WriteLine("\nEnd neural network demo\n");
catch (Exception ex)
} // Main
public static void ShowVector(double[] vector, int decimals) { . . }
public static void ShowMatrix(double[][] matrix, int numRows) { . . }
} // Program
public class NeuralNetwork
private int numInput;
private int numHidden;
private int numOutput;
private double[] inputs;
private double[][] ihWeights; // input-to-hidden
private double[] ihBiases;
private double[][] hoWeights; // hidden-to-output
private double[] hoBiases;
private double[] outputs;
public NeuralNetwork(int numInput, int numHidden,
int numOutput) { . . }
private static double[][] MakeMatrix(int rows, int cols) { . . }
public void SetWeights(double[] weights) { . . }
public double[] ComputeOutputs(double[] xValues) { . . }
private static double LogSigmoid(double z) { . . }
} // Class
} // ns
The heart of the program is the definition of a NeuralNetwork class. That class has a constructor, which calls helper MakeMatrix; public methods SetWeights and ComputeOutputs; and private method
LogSigmoid, which is used by ComputeOutputs.
The class containing the Main method has two utility methods, ShowVector and ShowMatrix. The neural network is instantiated using values for the number of input-, hidden- and output-layer nodes:
const int numInput = 3;
const int numHidden = 4;
const int numOutput = 2;
NeuralNetwork nn = new NeuralNetwork(numInput,
numHidden, numOutput);
Next, 26 arbitrary weights and bias values are assigned to an array:
const int numWeights = (numInput * numHidden) +
(numHidden * numOutput) +
numHidden + numOutput;
double[] weights = new double[numWeights] {
0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2,
-2.0, -6.0, -1.0, -7.0,
1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0,
-2.5, -5.0 };
The weights and bias values are stored so that the first 12 values are the input-to-hidden weights, the next four values are the input-to-hidden biases, the next eight values are the hidden-to-output
weights, and the last two values are the hidden-to-output biases. Assuming an implicit ordering isn't a very robust strategy, you might want to create four separate arrays instead.
After the weights are created, they're copied into the neural network object, and then three arbitrary inputs are created:
double[] xValues = new double[] { 2.0, 3.0, 4.0 };
Method ComputeOutputs copies the three input values into the neural network, then computes the outputs using the feed-forward mechanism, and returns the two output values as an array:
double[] yValues = nn.ComputeOutputs(xValues);
Console.WriteLine("\nNeural network outputs are:");
ShowVector(yValues, 4);
Console.WriteLine("\nEnd neural network demo\n");
The Neural Network Data Fields and Constructor
The NeuralNetwork class has three data fields that define the architecture:
private int numInput;
private int numHidden;
private int numOutput;
The class has four arrays and two matrices to hold the inputs, weights, biases and outputs:
private double[] inputs;
private double[][] ihWeights;
private double[] ihBiases;
private double[][] hoWeights;
private double[] hoBiases;
private double[] outputs;
In matrix ihWeights[i,j], index i is the 0-based index of the input-layer node and index j is the index of the hidden-layer node. For example, ihWeights[2,1] is the weight from input node 2 to hidden
node 1. Similarly, matrix hoWeights[3,0] is the weight from hidden node 3 to output node 0. C# supports a true two-dimensional array type, and you may want to use it instead of implementing a matrix
as an array of arrays.
The constructor first copies its input arguments into the associated member fields:
this.numInput = numInput;
this.numHidden = numHidden;
this.numOutput = numOutput;
Then the constructor allocates the inputs, outputs, weights and biases arrays, and matrices. I tend to leave out the .this qualifier for member fields:
inputs = new double[numInput];
ihWeights = MakeMatrix(numInput, numHidden);
ihBiases = new double[numHidden];
hoWeights = MakeMatrix(numHidden, numOutput);
hoBiases = new double[numOutput];
outputs = new double[numOutput];
Helper method MakeMatrix is just a convenience to keep the constructor code a bit cleaner and is defined as:
private static double[][] MakeMatrix(int rows, int cols)
double[][] result = new double[rows][];
for (int i = 0; i < rows; ++i)
result[i] = new double[cols];
return result;
Setting the Weights and Biases
Class NeuralNetwork method SetWeights transfers a set of weights and bias values stored in a linear array into the class matrices and arrays. The code for method SetWeights is presented in Listing 2.
Listing 2. The SetWeights method.
public void SetWeights(double[] weights)
int numWeights = (numInput * numHidden) +
(numHidden * numOutput) + numHidden + numOutput;
if (weights.Length != numWeights)
throw new Exception("Bad weights array");
int k = 0; // Points into weights param
for (int i = 0; i < numInput; ++i)
for (int j = 0; j < numHidden; ++j)
ihWeights[i][j] = weights[k++];
for (int i = 0; i < numHidden; ++i)
ihBiases[i] = weights[k++];
for (int i = 0; i < numHidden; ++i)
for (int j = 0; j < numOutput; ++j)
hoWeights[i][j] = weights[k++];
for (int i = 0; i < numOutput; ++i)
hoBiases[i] = weights[k++];
Neural networks that solve practical problems train the network by finding a set of weights and bias values that best correspond to training data, and will often implement a method GetWeights to
allow the calling program to fetch the best weights and bias values.
Computing the Outputs
Method ComputeOutputs implements the feed-forward mechanism and is presented in Listing 3. Much of the code consists of display messages to show intermediate values of the computations. In most
situations you'll likely want to comment out the display statements.
Listing 3. Method ComputeOutputs implements the feed-forward mechanism.
public double[] ComputeOutputs(double[] xValues)
if (xValues.Length != numInput)
throw new Exception("Bad inputs");
double[] ihSums = new double[this.numHidden]; // Scratch
double[] ihOutputs = new double[this.numHidden];
double[] hoSums = new double[this.numOutput];
for (int i = 0; i < xValues.Length; ++i) // xValues to inputs
this.inputs[i] = xValues[i];
Console.WriteLine("input-to-hidden weights:");
FeedForwardProgram.ShowMatrix(this.ihWeights, -1);
for (int j = 0; j < numHidden; ++j) // Input-to-hidden weighted sums
for (int i = 0; i < numInput; ++i)
ihSums[j] += this.inputs[i] * ihWeights[i][j];
Console.WriteLine("input-to-hidden sums before adding i-h biases:");
FeedForwardProgram.ShowVector(ihSums, 2);
Console.WriteLine("input-to-hidden biases:");
FeedForwardProgram.ShowVector(this.ihBiases, 2);
for (int i = 0; i < numHidden; ++i) // Add biases
ihSums[i] += ihBiases[i];
Console.WriteLine("input-to-hidden sums after adding i-h biases:");
FeedForwardProgram.ShowVector(ihSums, 2);
for (int i = 0; i < numHidden; ++i) // Input-to-hidden output
ihOutputs[i] = LogSigmoid(ihSums[i]);
Console.WriteLine("input-to-hidden outputs after log-sigmoid activation:");
FeedForwardProgram.ShowVector(ihOutputs, 2);
Console.WriteLine("hidden-to-output weights:");
FeedForwardProgram.ShowMatrix(hoWeights, -1);
for (int j = 0; j < numOutput; ++j) // Hidden-to-output weighted sums
for (int i = 0; i < numHidden; ++i)
hoSums[j] += ihOutputs[i] * hoWeights[i][j];
Console.WriteLine("hidden-to-output sums before adding h-o biases:");
FeedForwardProgram.ShowVector(hoSums, 2);
Console.WriteLine("hidden-to-output biases:");
FeedForwardProgram.ShowVector(this.hoBiases, 2);
for (int i = 0; i < numOutput; ++i) // Add biases
hoSums[i] += hoBiases[i];
Console.WriteLine("hidden-to-output sums after adding h-o biases:");
FeedForwardProgram.ShowVector(hoSums, 2);
for (int i = 0; i < numOutput; ++i) // Hidden-to-output result
this.outputs[i] = LogSigmoid(hoSums[i]);
double[] result = new double[numOutput]; // Copy to this.outputs
this.outputs.CopyTo(result, 0);
return result;
Method ComputOutputs uses three scratch arrays for computations:
double[] ihSums = new double[this.numHidden]; // Scratch
double[] ihOutputs = new double[this.numHidden];
double[] hoSums = new double[this.numOutput];
An alternative design is to place these scratch arrays as class data members instead of method local variables. In the demo program, method ComputeOutputs is only called once. But a neural network
that solves a practical problem will likely call ComputeOutputs many thousands of times; so, depending on compiler optimization, there may be a significant penalty associated with thousands of array
instantiations. However, if the scratch arrays are declared as class members, you'll have to remember to zero each one out in ComputeOutputs, which also has a performance cost.
Method ComputeOutputs uses the LogSigmoid activation method for both input-to-hidden and hidden-to-output computations. In some cases different activation functions are used for the two layers. You
may want to consider passing the activation function in as an input parameter using a delegate.
Wrapping Up
For completeness, utility display methods ShowVector and ShowMatrix are presented in Listing 4.
Listing 4. Utility display methods.
public static void ShowVector(double[] vector, int decimals)
for (int i = 0; i < vector.Length; ++i)
if (i > 0 && i % 12 == 0) // max of 12 values per row
if (vector[i] >= 0.0) Console.Write(" ");
Console.Write(vector[i].ToString("F" + decimals) + " "); // 2 decimals
public static void ShowMatrix(double[][] matrix, int numRows)
int ct = 0;
if (numRows == -1) numRows = int.MaxValue; // if numRows == -1, show all rows
for (int i = 0; i < matrix.Length && ct < numRows; ++i)
for (int j = 0; j < matrix[0].Length; ++j)
if (matrix[i][j] >= 0.0) Console.Write(" ");
Console.Write(matrix[i][j].ToString("F2") + " ");
In order to fully understand the neural network feed-forward mechanism, I recommend experimenting by modifying the input values and the values of the weights and biases. If you're a bit more
ambitious, you might want to change the demo neural network's architecture by modifying the number of nodes in the input, hidden or output layers.
The demo neural network is fully connected. An advanced but little-explored technique is to create a partially connected neural network by virtually severing the weight arrows between some of a
neural network's nodes. Notice that with the design presented in this article you can easily accomplish this by setting some weight values to 0.
Some complex neural networks, in addition to sending the output from one layer to the next, may send their output backward to one or more nodes in a previous layer. As far as I've been able to
determine, neural networks with a feedback mechanism are almost completely unexplored.
It's possible to create neural networks with two hidden layers. The design presented here can be extended to support multi-hidden-layer neural networks. Again, this is a little-explored topic. | {"url":"https://visualstudiomagazine.com/Articles/2013/05/01/Neural-Network-Feed-Forward.aspx?m=1","timestamp":"2024-11-14T21:08:56Z","content_type":"text/html","content_length":"148014","record_id":"<urn:uuid:f8405eff-307d-47f0-a3f7-2b6ee7e00762>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00014.warc.gz"} |
[Solved] CECS328 Programming assignment 1 | Assignment Chef
Implement linearSearch(a,key) and binarySearch(a,key) functions.
Part A. In this part we will calculate the average-case running time of each function.
1. Request the user to enter a positive integer, and call it n. (n = 10^5)
2. Generate n random integers between -1000 to 1000 and save them in array a.
3. Sort a (you can use any sorting algorithm you want.)
4. Pick a random number in a and save it in variable called key.
5. Call each function separately to search for the key in the given array.
6. To calculate the average-running time, you need to have a timer to save the total runtime when repeating step 4 and 5 for 100
(Note1: Do not forget to divide the runtime by the number of the times you run step 4-5)
(Note2: Remember to choose a different random number each time you go back to step 4.)
Part B. In this part we will calculate the worst-case running time of each function.
1. Repeat steps 1 to 3 in part A.
2. Now to have the worst-case scenario, set the value of the key to 5000 to make sure it does not exist in the array.
3. Run each function ONLY once to calculate the worst-case running time when n = 10^5.
4. Calculate how much time your machine takes to run one single step using your binary search function. (Hint: look at HW4)
5. Now using the previous step, write a code to calculate the worst-case running time for each algorithm when n=10^7 (You do not use a timer for this step. Just a simple calculation!).
There are no reviews yet. | {"url":"https://assignmentchef.com/product/solved-cecs328-programming-assignment-1-2/","timestamp":"2024-11-11T16:40:32Z","content_type":"text/html","content_length":"235060","record_id":"<urn:uuid:1d44ff82-5cc6-46d4-836e-6fde5866b883>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00024.warc.gz"} |
Sudoku | Brilliant Math & Science Wiki
Sudoku is a logic-based puzzle. It is a type of constraint satisfaction problem, where the solver is given a finite number of objects (the numerals 1-9) and a set of conditions stating how the
objects must be placed in relation to one another. The puzzle consists of a \(9\times 9\) grid further divided into nine \(3×3\) sub-grids (also called boxes, blocks, regions, or sub-squares).
The objective is to place the digits within the grid so that each column, row, and sub-grid contains all of the digits from \(1\) to \(9\). Each puzzle is published with some of the boxes already
filled in, and those constraints help define the problem's difficulty level.
People sometimes say there is no math required to solve Sudoku. What they really mean is, there is no arithmetic involved in Sudoku. Mathematical thinking in the forms of deductive reasoning and
algorithms are the basic tools for solving Sudoku puzzles.
Strategies for Solving Sudoku
There are numerous ways to solve a Sudoku puzzle. Backtracking is the most commonly used and easy to understand, though probably not the most efficient. The constraints given at the beginning of the
puzzle narrow down the available numerals that can be placed in the empty boxes. For each blank, the solver can build a candidate bank, a list of all the numbers that meet the known constraints. When
the candidate bank has been narrowed to one for a particular space, that numeral can be added to the grid. Each (correct) answer adds further constraints for the remaining spaces in the same row,
column, or box, helping to narrow the candidate banks for the remaining questions.
When backtracking, it helps to keep all the dimensions of the puzzle in mind and to look horizontally and vertically for clues, as well as within each \(3\times 3\) box.
In an easy-to-solve Sudoku, careful observation and backtracking are usually all it takes to solve the puzzle. In more advanced puzzles, it may be necessary to guess and check (which is why Sudoku is
best practiced with a pencil) or to find a more sophisticated way to process the clues.
Donald Knuth developed a backtracking algorithm called Algorithm X and a technique called Dancing Links (DLX) to utilize it. This algorithm provides a more efficient means for solving Sudoku puzzles.
Solve the Sudoku below:
Using the rules and goals of a Sudoku puzzle, there are many thousands of possible arrangements of digits that would satisfy the conditions for a proper Sudoku solution.
Given a proper Sudoku solution, how many distinct (unique) arrangements of the digits in the grid could be presented by simply swapping the digits in individual cells?
For example, changing all the 9s to 1s and 1s to 9s would produce a different arrangement in the grid, but would still be a legitimate correct solution.
Solve the Sudoku and find the sum of the numbers that are there in the middle of each box.
Sudoku puzzles have several qualities that make them useful in cognition studies. Sudoku is a sustained task that requires concentration and strategizing, but the rules are relatively simple and can
be learned quickly by most test subjects. Sudoku has been used as a performance measure in studies on aging [1], distractions [2], multitasking [3], and other brain-related subjects.
A World Sudoku Championship has been held annually since 2006 [4]. The venue rotates among countries in Asia, Europe, and North America.
Although the format previously discussed, the \(9\times 9\) grid with \(3\times 3\) regions, is the most common, many other variations exist. They often incorporate new shapes into the puzzle design,
expand the grid, or add extra constraints to the puzzle.
This game is very similar to Sudoku but uses the letters A, B, C, and D.
Which letter would go into the red box with the question mark?
\[x + y + z + w = K\] \[x + y + z + w \neq K\] Multiple solutions
The following are the rules of the \(4\times 4\) Kenken above:
• Each row contains exactly one of each digit 1 through 4.
• Each column contains exactly one of each digit 1 through 4.
• Each colored group of cells is a cage containing digits which achieve the result \((K)\) using the specified operation \((+\) or \(\times)\).
First, solve the puzzle, where \(K\) is a strictly positive integer.
If there is a unique value of \(K\), determine whether or not \(x + y + z + w = K\).
Otherwise, i.e. if you believe there are more than one solutions with different values of \(K\) for the given puzzle, choose "Multiple solutions."
Note: Unlike standard Sudoku, KenKen can have cage(s) with repeated digits.
Hint: Observe the structure of this puzzle carefully. What can be said about those tetrominoes and the numbers in each row and column?
All sums must be the same All but the center circle must be the same The center circle sum has multiple values that work
Here are the rules of sum Sudoku:
• Each and every column, row, and \(2\times2\) box contains distinct numbers from 1 to 4.
• The white circle marked at four adjacent cells indicate that the sum of upper-left and bottom-right cells is the sum of upper-right and bottom-left cells.
What can we conclude about the circle sums in the puzzle?
The setup in the top-left box is valid since \(2 + 3 = 1 + 4 = 5\). However, the four cell values at the center are invalid since \(4 + 2 \neq 3 + 1\).
Write a single positive digit \((\leq 6)\) into each empty square in the figure below, such that every number \((1, 2, 3, 4, 5, 6)\) appears in each row and in each column. The number in a circle
shows the product of the four numbers around it.
If you are finished, find the minimum value of the product of the numbers in the pink squares!
Note: Here is a sample, with digits \(1, 2, 3, 4\):
[1] Pieramico V, Esposito R, Sensi F, et al. Combination training in aging individuals modifies functional connectivity and cognition, and is potentially affected by dopamine-related genes. PLoS ONE.
[2] Strigo IA, Simmons AN, Matthews SC, Craig AD. The relationship between amygdala activation and passive exposure time to an aversive cue during a continuous performance task. PLoS ONE. 2010;5
[3] Shih SI. A null relationship between media multitasking and well-being. PLoS ONE. 2013;8(5):e64508.
[4] World Sudoku Championships http://www.worldpuzzle.org/championships/wsc/ Accessed February 16, 2016.
[5] Image from https://commons.wikimedia.org/wiki/File:WorldSudokuChampionship2015SofiaBulgaria02.JPG under Creative Commons licensing for reuse and modification.
[6] Image from https://commons.wikimedia.org/wiki/File:WorldSudokuChampionship2015SofiaBulgaria05.JPG under Creative Commons licensing for reuse and modification.
[7] Image from https://commons.wikimedia.org/wiki/File:25by25sudoku.png under Creative Commons licensing for reuse and modification.
[8] Image from https://commons.wikimedia.org/wiki/File:OceansHypersudoku18Solution.svg under Creative Commons licensing for reuse and modification.
[9] Image from https://commons.wikimedia.org/wiki/File:Sudoku3Dsol.gif under Creative Commons licensing for reuse and modification. | {"url":"https://brilliant.org/wiki/sudoku/?subtopic=puzzles&chapter=grid-puzzles","timestamp":"2024-11-05T17:10:51Z","content_type":"text/html","content_length":"69627","record_id":"<urn:uuid:83f368fa-608f-44c4-b452-9c421b4f155b>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00673.warc.gz"} |
The total angular momentum of light has received attention for its application in a variety of phenomena such as optical communication, optical forces, and sensing. However, the quantum behavior
including the commutation relations has been relatively less explored. Here, we derive the correct commutation relation for the total angular momentum of light using both relativistic and
non-relativistic approaches. An important outcome of our work is the proof that the widely assumed quantum commutation relation for the total observable angular momentum of light is fundamentally
incorrect. Our work will motivate experiments and lead to new insights on the quantum behavior of the angular momentum of light.
Last updated on 08/08/2024 | {"url":"https://electrodynamics.org/publications/what-are-quantum-commutation-relations-total-angular-momentum-light","timestamp":"2024-11-04T20:27:27Z","content_type":"text/html","content_length":"27846","record_id":"<urn:uuid:e9ac9ec8-3407-427b-bb83-3fb241fc52f9>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00831.warc.gz"} |
Total Resistance Calculator - Parallel
Total Resistance Calculator – Parallel
This calculator determines the total effective resistance of any number of resistors in parallel.
Enter the resistor values separated by commas.
Parallel Resistance Formula
R[total] = (1/R[1] + 1/R[2] + 1/R[3 ]+…. + 1/R[n])^-1
Example Calculation
Consider four resistors in parallel: 1 Ω, 6 Ω, 15 Ω and 100 Ω. The total effective resistance is 0.80 Ω.
If you have two resistors – one large and the other small in parallel, the effective resistance is closer to the smaller value. Take for instance 1 Ω and 100 Ω in parallel. The effective resistance
is 0.99 ≈ 1 Ω.
Related Calculator
Total Resistance Calculator – gives the effective resistance for a combination of series and parallel resistors. | {"url":"https://3roam.com/total-resistance-calculator-parallel/","timestamp":"2024-11-14T08:05:19Z","content_type":"text/html","content_length":"190483","record_id":"<urn:uuid:515cbeb5-eadc-49ce-9e8f-ea22fcb30a23>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00808.warc.gz"} |
EPSRC Reference: EP/N035178/1
Title: Extended Gyrokinetic Modelling for Dynamic Flows
Principal Investigator: McMillan, Dr BF
Other Investigators:
Researcher Co-Investigators:
Project Partners: Culham Centre for Fusion Energy Swiss Federal Inst of Technology (EPFL)
Department: Physics
Organisation: University of Warwick
Scheme: Standard Research
Starts: 01 October 2016 Ends: 29 February 2020 Value (£): 328,689
EPSRC Research Topic Classifications: Energy - Nuclear Plasmas - Laser & Fusion
EPSRC Industrial Sector Classifications:
Related Grants:
│Panel Date │Panel Name │Outcome │
Panel History: ├───────────┼─────────────────────────────────────────────────────────┼─────────┤
│12 May 2016│EPSRC Physical Sciences Materials and Physics - May 2016 │Announced│
Summary on Grant Application Form
Plasma physics is the study of large collections of charged particles and their interactions. One key application of plasma physics is in magnetically confined nuclear fusion, where nuclei, typically
of Hydrogen, are confined by a magnetic field, so that they can reach the high temperatures necessary to fuse together as they collide. A large research effort is underway to develop fusion reactors
that exploit the energy release from fusion to produce electricity.
In a tokamak, which is a specific kind of fusion reactor design, the vacuum chamber containing the fusion fuel can be conceptually separated into a 'closed field region' where particles cannot escape
along magnetic field lines, and an 'open field region' where they can stream along the field lines and hit the wall.
The quite sharp boundary between the open and closed field region is particularly interesting and important, because a strong additional 'transport barrier', called a pedestal, can develop there. A
large difference in temperature and density is sustained across the narrow pedestal, allowing the core plasma to reach higher pressure, and leading to a major increase in fusion power. How the
pedestal is formed, and when it breaks down, are questions of vital importance to fusion reactor operation, but these issues are quite poorly understood at present.
This proposal seeks to answer basic questions about the pedestal, and similar structures that develop in other laboratory and space plasmas. This is an investigation of the fundamental properties of
magnetised plasmas. How do such structures evolve, and how does this interact with the plasma turbulence? What plasma instabilities develop in these plasmas?
To answer these questions, we need models of how the particles individually and collectively respond to electromagnetic fields, and for the hot plasmas of interest we usually need to keep track of
the motion of particles, rather that just look at the overall fluid motion. In magnetised plasmas, the basic motion of plasma particles is a circular orbit, or gyration, perpendicular to the magnetic
field, as well as a parallel motion along the magnetic field line. This can be formalized mathematically using a framework known as 'gyrokinetics', which has become the dominant way to understand the
transport of hot plasma in tokamaks.
A new gyrokinetic formalism has been developed by the proposer which is designed to be more accurate in regions with large amplitude structures like the tokamak pedestal. It is based on a rethinking
of the assumptions usually made, so that both short wavelength fluctuations and more global physics may be handled in a unified way. We relax the requirement that perturbations be small amplitude but
instead require that the 'vorticity', which measures how rapidly blobs of plasma rotate, is relatively limited.
This proposal will develop and exploit this mathematical framework to solve a range of fundamental physics problems in magnetised plasmas with large perturbations. A computer code to embody this
plasma model will be further developed, and this will require the development of new algorithms. This code will then be deployed to understand both fundamental physics problems of magnetised plasmas,
as well as the specific applications to structured regions of tokamaks.
As well as computational work, physical understanding of these plasmas requires us to develop a deeper understanding of the mathematical framework. We will use limiting cases to determine how the
physics relates to simpler formalisms, and determine the underlying conservation laws to tie the turbulent dynamics to the large scale physics of momentum and energy transport.
Key Findings
This information can now be found on Gateway to Research (GtR) http://gtr.rcuk.ac.uk
Potential use in non-academic contexts
This information can now be found on Gateway to Research (GtR) http://gtr.rcuk.ac.uk
Description This information can now be found on Gateway to Research (GtR) http://gtr.rcuk.ac.uk
Date Materialised
Sectors submitted by the Researcher
This information can now be found on Gateway to Research (GtR) http://gtr.rcuk.ac.uk
Project URL:
Further Information:
Organisation Website: http://www.warwick.ac.uk | {"url":"https://gow.epsrc.ukri.org/NGBOViewGrant.aspx?GrantRef=EP/N035178/1","timestamp":"2024-11-04T14:43:08Z","content_type":"application/xhtml+xml","content_length":"27797","record_id":"<urn:uuid:38fd02fc-79a0-45ae-8de7-2ffcc751d945>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00142.warc.gz"} |
Decreasing sample size:
A decreases precision
B decreases bias
C increases precision
D decreases precision and...
Decreasing sample size: A decreases precision B decreases bias C increases precision D decreases precision and...
Decreasing sample size:
A decreases precision
B decreases bias
C increases precision
D decreases precision and bias
Sample size is the number of items selected to get estimates for the overall population. The larger the sample size, the better the prediction / precision and vice versa. Bias is when the items do
not have equal chance of being selected that is the sample is not randomly selected. It can happen with any sample irrespective of the size. Hence the sample size does not affect bias.
Decreasing sample size:
A decreases precision
B decreases bias
C increases precision
D decreases precision and bias | {"url":"https://justaaa.com/statistics-and-probability/376303-decreasing-sample-size-a-decreases-precision-b","timestamp":"2024-11-06T11:31:04Z","content_type":"text/html","content_length":"39092","record_id":"<urn:uuid:f94db487-7cc7-432e-8845-717d5886051e>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00478.warc.gz"} |
Question ID - 156012 | SaraNextGen Top Answer
The stiffened cross-section of a long slender uniform structural member is idealized as shown in the figure below. The lumped areas at $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ have equal
cross-sectional area of $3 \mathrm{~cm}^{2}$. The webs $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and $\mathrm{DA}$ are each $5 \mathrm{~mm}$ thick. The structural member is subjected to a twisting
moment of $10 \mathrm{kNm}$. The magnitudes of the shear flow in the webs, $q_{A B}, q_{B C}, q_{C D}$, and $q_{D A}$ in $\mathrm{kN} / \mathrm{m}$ are, respectively
(A) 20,20,20,20
(B) 0,0,50,50
(C) 40,40,0,0
(D) 50 , $, 50$ | {"url":"https://www.saranextgen.com/homeworkhelp/doubts.php?id=156012","timestamp":"2024-11-06T20:22:18Z","content_type":"text/html","content_length":"15427","record_id":"<urn:uuid:4fe83ba1-f1d8-4f1b-b43b-2da4ee791a1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00246.warc.gz"} |
Math 410. Introduction to Quantum Computing, Spring 2024
Instructor: Chi-Kwong Li
Meeting time, format, office hours.
• TR 3:30 - 4:50 p.m. Boswell 40.
• Office hours: Wednesday TWT 9:30 - 10:30 am.m. or by appointment.
Quantum information science is a rapidly growing area. Quantum cryptography is in commercial use, and the construction of practical quantum computer still require a lot of research from different
disciplines including mathematics, physics, computer science, chemistry, engineering, material science, etc.
In this course, an introduction of the subject will be given based on the first 11 chapters of the book.
and some complementary notes, which will be put on blackboard.
We will cover topics including: basic linear algebra background, the mathematical framework for quantum mechanics, qubits and quantum key distributions, quantum gates and quantum circuits in quantum
computing, quantum integral transforms, quantum algorithms of Deutsch, Joza, Grover and Shor, decoherence, quantum error correction, DiVinzenzo criteria, physical realizations. Current research
problems will be mentioned. Here are some other useful reference books.
• Nielsen and Chuang, Quantum Computation and Quantum Information Science, Cambridge.
• Watrous, The Theory of Quantum Information, Cambridge.
• Yanofsky and Mannucci, Quantum Computing for Computer Scientists, Cambridge.
• The Functional Analysis of Quantum Information Theory a collection of notes based on lectures by Gilles Pisier, K. R. Parthasarathy, Vern Paulsen and Andreas Winter. https://arxiv.org/pdf/
Problem sets
• There will be 10-12 problems sets assigned weekly. Texfiles of the problems will be put on the blackboard site. Pdf files of the solutions will be uploaded to the blackboard after the due dates.
• Challenging problems will be assigned from time to time, extra-credits will be given to successful (or partially successful) attempts.
• Help will be provided during office hours or group homework sessions.
• You have to use LaTex to typset mathematical document, an excellent skill to acquire. You may go to overleaf to use the online program for the typesetting.
Assessment will be based on the homeworks sets
%: 0 - 60 - 65 - 70 - 75 - 80 - 83 - 87 - 90 - 93 - 100
F D C- C C+ B- B B+ A- A | {"url":"https://cklixx.people.wm.edu/teaching/m410.html","timestamp":"2024-11-04T20:13:16Z","content_type":"text/html","content_length":"3581","record_id":"<urn:uuid:49f56b2c-597f-4b14-994d-57758c78229f>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00660.warc.gz"} |
July 2016 – Good Math/Bad Math
Sometimes, I think that I’m being punished.
I’ve written about Cantor crankery so many times. In fact, it’s one of the largest categories in this blog’s index! I’m pretty sure that I’ve covered pretty much every anti-Cantor argument out there.
And yet, not a week goes by when another idiot doesn’t pester me with their “new” refutation of Cantor. The “new” argument is always, without exception, a variation on one of the same old boring
But I haven’t written any Cantor stuff in quite a while, and yet another one landed in my mailbox this morning. So, what the heck. Let’s go down the rabbit-hole once again.
We’ll start with a quick refresher.
The argument that the cranks hate is called Cantor’s diagonalization. Cantor’s diagonalization as argument that according to the axioms of set theory, the cardinality (size) of the set of real
numbers is strictly larger than the cardinality of the set of natural numbers.
The argument is based on the set theoretic definition of cardinality. In set theory, two sets are the same size if and only if there exists a one-to-one mapping between the two sets. If you try to
create a mapping between set A and set B, and in every possible mapping, every A is mapped onto a unique B, but there are leftover Bs that no element of A maps to, then the cardinality of B is larger
than the cardinality of A.
When you’re talking about finite sets, this is really easy to follow. If I is the set {1, 2, 3}, and B is the set {4, 5, 6, 7}, then it’s pretty obvious that there’s no one to one mapping from A to
B: there are more elements in B than there are in A. You can easily show this by enumerating every possible mapping of elements of A onto elements of B, and then showing that in every one, there’s an
element of B that isn’t mapped to by an element of A.
With infinite sets, it gets complicated. Intuitively, you’d think that the set of even natural numbers is smaller than the set of all natural numbers: after all, the set of evens is a strict subset
of the naturals. But your intuition is wrong: there’s a very easy one to one mapping from the naturals to the evens: {n → 2n }. So the set of even natural numbers is the same size as the set of all
natural numbers.
To show that one infinite set has a larger cardinality than another infinite set, you need to do something slightly tricky. You need to show that no matter what mapping you choose between the two
sets, some elements of the larger one will be left out.
In the classic Cantor argument, what he does is show you how, given any purported mapping between the natural numbers and the real numbers, to find a real number which is not included in the mapping.
So no matter what mapping you choose, Cantor will show you how to find real numbers that aren’t in the mapping. That means that every possible mapping between the naturals and the reals will omit
members of the reals – which means that the set of real numbers has a larger cardinality than the set of naturals.
Cantor’s argument has stood since it was first presented in 1891, despite the best efforts of people to refute it. It is an uncomfortable argument. It violates our intuitions in a deep way. Infinity
is infinity. There’s nothing bigger than infinity. What does it even mean to be bigger than infinity? That’s a non-sequiter, isn’t it?
What it means to be bigger than infinity is exactly what I described above. It means that if you have a two infinitely large sets of objects, and there’s no possible way to map from one to the other
without missing elements, then one is bigger than the other.
There are legitimate ways to dispute Cantor. The simplest one is to reject set theory. The diagonalization is an implication of the basic axioms of set theory. If you reject set theory as a basis,
and start from some other foundational axioms, you can construct a version of mathematics where Cantor’s proof doesn’t work. But if you do that, you lose a lot of other things.
You can also argue that “cardinality” is a bad abstraction. That is, that the definition of cardinality as size is meaningless. Again, you lose a lot of other things.
If you accept the axioms of set theory, and you don’t dispute the definition of cardinality, then you’re pretty much stuck.
Ok, background out of the way. Let’s look at today’s crackpot. (I’ve reformatted his text somewhat; he sent this to me as plain-text email, which looks awful in my wordpress display theme, so I’ve
rendered it into formatted HTML. Any errors introduced are, of course, my fault, and I’ll correct them if and when they’re pointed out to me.)
We have been told that it is not possible to put the natural numbers into a one to one with the real numbers. Well, this is not true. And the argument, to show this, is so simple that I am
absolutely surprised that this argument does not appear on the internet.
We accept that the set of real numbers is unlistable, so to place them into a one to one with the natural numbers we will need to make the natural numbers unlistable as well. We can do this by
mirroring them to the real numbers.
Given any real number (between 0 and 1) it is possible to extract a natural number of any length that you want from that real number.
Ex: From π-3 we can extract the natural numbers 1, 14, 141, 1415, 14159 etc…
We can form a set that associates the extracted number with the real number that it was extracted from.
Ex: 1 → 0.14159265…
Then we can take another real number (in any arbitrary order) and extract a natural number from it that is not in our set.
Ex: 1 → 0.14159266… since 1 is already in our set we must extract the next natural number 14.
Since 14 is not in our set we can add the pair 14 → 0.1415926l6… to our set.
We can do the same thing with some other real number 0.14159267… since 1 and 14 is already in our set we will need to extract a 3 digit natural number, 141, and place it in our set. And so on.
So our set would look something like this…
A) 1 → 0.14159265…
B) 14 → 0.14159266…
C) 141 → 0.14159267…
D) 1410 → 0.141
E) 14101 → 0.141013456789…
F) 5 → 0.567895…
G) 55 → 0.5567891…
H) 555 → 0.555067891…
… …
Since the real numbers are infinite in length (some terminate in an infinite string of zero’s) then we can always extract a natural number that is not in the set of pairs since all the natural
numbers in the set of pairs are finite in length. Even if we mutate the diagonal of the real numbers, we will get a real number not on the list of real numbers, but we can still find a natural
number, that is not on the list as well, to correspond with that real number.
Therefore it is not possible for the set of real numbers to have a larger cardinality than the set of natural numbers!
This is a somewhat clever variation on a standard argument.
Over and over and over again, we see arguments based on finite prefixes of real numbers. The problem with them is that they’re based on finite prefixes. The set of all finite prefixes of the real
numbers is that there’s an obvious correspondence between the natural numbers and the finite prefixes – but that still doesn’t mean that there are no real numbers that aren’t in the list.
In this argument, every finite prefix of π corresponds to a natural number. But π itself does not. In fact, every real number that actually requires an infinite number of digits has no corresponding
natural number.
This piece of it is, essentially, the same thing as John Gabriel’s crankery.
But there’s a subtler and deeper problem. This “refutation” of Cantor contains the conclusion as an implicit premise. That is, it’s actually using the assumption that there’s a one-to-one mapping
between the naturals and the reals to prove the conclusion that there’s a one-to-one mapping between the naturals and the reals.
If you look at his procedure for generating the mapping, it requires an enumeration of the real numbers. You need to take successive reals, and for each one in the sequence, you produce a mapping
from a natural number to that real. If you can’t enumerate the real numbers as a list, the procedure doesn’t work.
If you can produce a sequence of the real numbers, then you don’t need this procedure: you’ve already got your mapping. 0 to the first real, 1 to the second real, 2 to the third read, 3 to the fourth
real, and so on.
So, once again: sorry Charlie: your argument doesn’t work. There’s no Fields medal for you today.
One final note. Like so many other bits of bad math, this is a classic example of what happens when you try to do math with prose. There’s a reason that mathematicians have developed formal
notations, formal language, detailed logical inference, and meticulous citation. It’s because in prose, it’s easy to be sloppy. You can accidentally introduce an implicit premise without meaning to.
If you need to carefully state every premise, and cite evidence of its truth, it’s a whole lot harder to make this kind of mistake.
That’s the real problem with this whole argument. It’s built on the fact that the premise “you can enumerate the real numbers” is built in. If you wrote it in careful formal mathematics, you wouldn’t
be able to get away with that. | {"url":"http://www.goodmath.org/blog/2016/07/","timestamp":"2024-11-10T22:22:05Z","content_type":"text/html","content_length":"71164","record_id":"<urn:uuid:c38fade2-5014-4d34-8f62-5b5ee4a87edb>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00765.warc.gz"} |
What is the Speed of Light? in context of light speed distance
30 Aug 2024
What is the Speed of Light?
The speed of light, denoted by c, is a fundamental constant in physics that represents the maximum velocity at which information or matter can travel in the universe. It is a crucial concept in
various branches of physics, including electromagnetism, special relativity, and quantum mechanics.
Definition and Formula
The speed of light is defined as the distance traveled by light per unit time. In vacuum, it is denoted by c and has a value of approximately 299,792,458 meters per second (m/s). The formula for the
speed of light in vacuum is:
c = λf
where λ (lambda) is the wavelength of light and f is its frequency.
Context: Light Speed Distance
The concept of light speed distance is crucial in understanding the nature of space and time. According to special relativity, as an object approaches the speed of light, its mass increases and time
appears to slow down relative to an observer at rest. This phenomenon is known as time dilation.
The formula for the Lorentz factor, which describes this effect, is:
γ = 1 / sqrt(1 - (v^2/c^2))
where v is the velocity of the object and c is the speed of light.
The speed of light has far-reaching implications in various areas of physics. For instance, it sets a fundamental limit on the maximum speed at which any object or information can travel. This has
significant consequences for our understanding of space-time and the behavior of particles at high energies.
In addition, the speed of light is used as a reference point in many physical measurements, such as determining the distance to celestial objects using astronomical methods like parallax and radar
The speed of light is a fundamental constant that plays a central role in our understanding of the universe. Its value has been precisely measured through various experiments and theoretical
calculations. The concept of light speed distance highlights the importance of considering the effects of relativity when dealing with high-speed phenomena. Further research into the properties of
light and its interactions with matter will continue to refine our understanding of this fundamental constant.
• Einstein, A. (1905). Does the inertia of a body depend upon its energy content? Annalen der Physik, 18(13), 639-641.
• Jackson, J. D. (1999). Classical Electrodynamics. John Wiley & Sons.
• Peskin, M. E., & Schroeder, D. V. (1995). An Introduction to Quantum Field Theory. Addison-Wesley Publishing Company.
Note: The references provided are a selection of classic and influential works in the field of physics that relate to the topic of light speed distance.
Related articles for ‘light speed distance ‘ :
• Reading: **What is the Speed of Light? in context of light speed distance **
Calculators for ‘light speed distance ‘ | {"url":"https://blog.truegeometry.com/tutorials/education/7d774e6dcfaf513becad54710d9e31fd/JSON_TO_ARTCL_What_is_the_Speed_of_Light_in_context_of_light_speed_distance_.html","timestamp":"2024-11-06T08:23:49Z","content_type":"text/html","content_length":"16742","record_id":"<urn:uuid:4523f61e-fbb0-4fb7-ab89-4f9207e24a6a>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00598.warc.gz"} |
Midpoint of Line Segment Calculation
What is the coordinate of the midpoint of the line segment with endpoints F(-408) and G(-416)?
Write your answer as an integer or a decimal, or mixed number in simplest form.
Final answer:
The midpoint formula is used to find the coordinate of the midpoint of a line segment.
To find the coordinate of the midpoint of FG, we can use the midpoint formula. The formula for the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is: midpoint = ((x1 + x2)/2, (y1 +
Since we don't have the coordinates of F and G, we can't calculate the midpoint. Please provide the coordinates of F and G so that we can help you find the midpoint of FG. | {"url":"https://airdocsolutions.com/geography/midpoint-of-line-segment-calculation.html","timestamp":"2024-11-09T07:24:05Z","content_type":"text/html","content_length":"20669","record_id":"<urn:uuid:12000125-59b8-460a-bead-616463439fde>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00707.warc.gz"} |
Australian Racing Bet Types & Calculators
But based on the calculated Tennis Trading Articles percentage probability, the example helps you to understand the odds. With the example, you can read out the percentage probability of a bet in the
future. The simple rule of thumb with decimal betting odds is the bigger the number, the larger the return will be. For example, decimal odds of 3.75 will result in larger winnings than decimal odds
of 1.75. When something is just as likely to happen as not, it is given even odds. These are presented as 1/1 in fractional odds or 2.0 in decimal odds.
If bettors choose to use one, it’s just to see the value in certain wagers. And so, the sportsbook will continue to raise the line until the other side starts receiving action (from 10.5 to 11 to
11.5, etc). While receiving equal money isn’t always possible, it’s always the goal because if that’s the case, the sportsbook will come out in the positive no matter the final result.
If all of the individual bets in a parlay hit, then the payout is much bigger. A parlay is treated as one big bet, even though it is made up of several individual bets. As such, every individual bet
within the parlay has to hit in order for you to win. For example, if your parlay has 10 over/under picks, and you get 9 of them right — you’ll still lose the bet. Our tricast bet calculator gives
you an excellent way to play around with the selections in your tricast bet.
• Use our free bet calculator to work out exactly how much profit you stand to win from your selections.
• If you think the combined score for both teams will be 104 points or less, you would bet the UNDER.
• Enter a value and the calculator will automatically convert it to the other formats.
• If you live in a state with several online sportsbooks to choose from, you’re missing out on value and optimized betting strategy if you’re not using an odds calculator.
Rest assured, your bet will payout as though they are still +120 underdogs. Pretend you and a friend are going to bet on the result of a million-coin flips. Everything you need to know about placing
a bet, from layout to the user experience. Any odds in which the first number is bigger than the second are odds against, while any odds in which the first number is smaller than the second are odds
on. Odds-on events are considered more likely to happen than not by bookmakers, and vice versa for events that odds against. Now you know odds are set out with two numbers separated by a forward
slash, you can use them to work out the probability of an event happening.
What Do Odds Of +200 Mean?
However, each sportsbook has a different policy on occurrences like that. So, be sure to find out how yours handles such situations before you place a wager. You can mess around with the bet slip,
and add a ton of events to see what your potential winnings could be. If you want to remove any events, just click the “X” next to the ones you don’t want on your bet slip. If you don’t want to bet
any of the games you selected, there’s usually a “clear all” option at either the top or bottom of your bet slip. The minus in front of the New England Patriots odds means they are favourites and the
calculation is different.
Using Betting Odds To Calculate Winnings
Most often, each way bets are paid out at ¼ odds, but occasionally bookmakers may offer slightly better odds on races and this can be reflected by altering the each way price. For a fractional bet
enter the first number of the odds in the first box and the second in the second. The next stage of the process is to enter the details of your bet and you can enter your odds in one of two formats,
either as a fraction (such as 11/2) or as a decimal (3.5). Now over 100 bets, we are forecast to make a loss of £19.
How Does A Yankee Payout?
Unlike a coin toss, sports betting odds are subjective, and therefore if you outsmart the bookmaker, you’re likely to make money. Our parlay calculator will combine up to 12 games and calculate your
payout based on your bet amount and the odds for each game. To Convert Odds to Different Odds Formats,input either decimal odds, American odds, fractional odds, or implied probability into the “Enter
Your Odds” section of the converter above.
A photograph determines the result when the finish is close and the result cannot be determined by the naked eye. When that is the cases the result is declared as a dead heat because the horses
cannot be separated at the line. The total stake is apportioned to the backed horse and the unbacked horse in a dead heat. The bet calculator will take dead heat rules into account when calculating
the return from a bet. In the event of a dead heat, a button must be clicked in the bet calculator and the return is adjusted accordingly.
In sports betting, each team is assigned odds — assigned by a sportsbook — that represent the likelihood of its winning the game. In a betting line between two teams, the team expected to win is
called the favorite. The team expected to lose the game is called the underdog.
If the real probability is 2/1 (3.0 in decimal, 200 in American) then the bookie will subtract their 5% margin and the real odds given will be 19/10 (2.90, 190). Every game has a linked odds for home
winner, draw, away winner and under 2.5 goals. Setting these odds, the system will filter and search the games in our database. Then the bet calculator will offer the statistics for any kind of bet.
To calculate the implied probability you need to convert the odds into a percentage. | {"url":"https://ordispremieresnations.ca/australian-racing-bet-types-calculators/","timestamp":"2024-11-06T05:47:41Z","content_type":"text/html","content_length":"32886","record_id":"<urn:uuid:97014e9b-5386-4db3-b435-8f7a0e3af38a>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00558.warc.gz"} |
Solve the math problem
400 students are asked if they live in an apartment and have a pet:
Apartment: 120
Both: 30
Pet: 90
The probability that a randomly selected student not living in an apartment has a pet is
Solve the math problem 400 students are asked if they live in an apartment and have a pet: Apartment: 120 Both: 30 Pet: 90 The probability that a randomly selected student not living in an apartment
has a pet is
1495 views
Answer to a math question Solve the math problem 400 students are asked if they live in an apartment and have a pet: Apartment: 120 Both: 30 Pet: 90 The probability that a randomly selected student
not living in an apartment has a pet is
101 Answers
To find the probability that a randomly selected student not living in an apartment has a pet, you can use conditional probability. The notation for this probability is \( P(\text{Pet | Not
Apartment}) \), and it is calculated as the number of students with both a pet and not living in an apartment divided by the total number of students not living in an apartment. Given: - Students
living in an apartment with a pet: 30 - Students not living in an apartment with a pet: \(90 - 30 = 60\) (since there are 90 students with a pet and 30 of them live in an apartment) Now, calculate
the probability: \[ P(\text{Pet | Not Apartment}) = \frac{\text{Number of students with a pet and not living in an apartment}}{\text{Total number of students not living in an apartment}} \] \[ P(\
text{Pet | Not Apartment}) = \frac{60}{400 - 120} \] \[ P(\text{Pet | Not Apartment}) = \frac{60}{280} \] Now, simplify the fraction if possible. \frac{60}{280}=\frac{3}{14} So, the probability that
a randomly selected student not living in an apartment has a pet is \( \frac{3}{14} \)
Frequently asked questions (FAQs)
What is the average number of hours spent on social media per day by a group of 50 people?
What is the measure of angle A in a triangle ABC if the angle bisector of angle A splits BC into segments of lengths 2x-3 and 3x+4?
Math question: Find the slope-intercept equation of a line passing through the points (2, 5) and (-3, 8). | {"url":"https://math-master.org/general/solve-the-math-problem-400-students-are-asked-if-they-live-in-an-apartment-and-have-a-pet-apartment-120-both-30-pet-90-the-probability-that-a-randomly-selected-student-not-living-in-an-apa","timestamp":"2024-11-07T17:29:31Z","content_type":"text/html","content_length":"244687","record_id":"<urn:uuid:561e666c-7bce-46e6-bf46-d0efe8bdf643>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00194.warc.gz"} |
Things that Count
« | | Part 1 Origins »
Welcome to things-that-count.net. This website describes a collection of antique calculators (“collection Calculant”^1) and uses it to help develop a historical account of the way humans developed
the need and capacity to calculate, the things they used to help them, and how human societies (and even human brains) evolved with those developments.
The account given here is of a 37,000 year story, one in which humans came to count, record their counting and do simple arithmetic sums. Over time these capabilities became essential ingredients in
what became increasingly complex societies. More citizens needed to be able to manipulate numbers in their daily lives and, over time, a range of aides of various sorts were developed to assist. At
the beginning, the aides were very simple - for example, marks scribed on bones and patterns arranged with pebbles. Later, primitive devices and tables were developed and sold. Over time, much more
elaborate mechanical devices were developed to help in this task. Many of these devices, where they survive, now represent little more than mechanical fossils. Unused and largely forgotten their
remains are scattered across history from earliest human pre-history to our present moment.
The need for calculation, however has prospered. As societies have become more complex, transactions in them depending on arithmetic (the familiar tasks of counting, adding, subtracting, multiplying
and dividing), as well as more complex mathematics, have intensified. Yet over much of this period, for many people in these societies, doing even the simplest arithmetic tasks has been neither easy
nor, for some, comprehensible. For this reason finding ways to do these tasks faster and more accurately, and to spread the ability across more people, has been a preoccupation in many societies. It
is the approaches that have been taken to aid achieving these simple goals (rather than the development of complex mathematics) which is the primary focus of this website.
Early “calculators” were not things. Rather they were people who were employed to calculate. Over time these people were first aided, but later replaced by calculating devices. These devices became
very widely used across many countries. There is evidence we may now be passing the heyday of such stand-alone calculators. This is because, increasingly since the advent of electronic computing, the
aides to calculation have begun to appear in virtual form as apps in phones, tablets and laptops. The end of calculators, seen as devices, in this sense is looming.
One might imagine that a history of calculators would consist simply of the progressive discovery and invention of ever more effective and sophisticated calculating devices. Indeed many such accounts
do focus on this with loving details of the minutae of mechanical invention. But to focus simply on that is to oversimplify and lose much of what is potentially interesting. Across human history many
weird things were indeed devised for doing simple calculations. But the development of these begs a series of questions: When and why were they made, how were they used, and why at times were they
forgotten for centuries or even millennia?
The objects in “collection Calculant” described here, which are used to help illustrate answers to these sort of questions, are drawn from across some 4,000 years of history. Each of them was created
with a belief that it could assist people in thinking about (and with) numbers. They range from little metal coin-like disks to the earliest electronic pocket calculators - representing a sort of
‘vanishing point’ for all that had come before. The collection and the history it illustrates in a sense form a duet - the two voices each telling part of the story. The history has shaped what has
been collected, and the collection has helped shape how the story is told.
1. Initial observations.
As most people know, the spread of electronic personal calculators of the 1970s was followed quickly by the first personal computers. Before long, computer chips began to be embodied in an ever
expanding array of converging devices. In turn, ever greater computing power spread across the planet. However sophisticated these modern computers appeared on the outside, and whatever the diversity
of functions they performed, at heart they achieved most of this by doing a few things extremely fast. (Central to the things they did were logic operations such as “if”, “and”, “or” and “not”, and
the arithmetic operations of addition and subtraction - from which multiplication and division can also be derived). On top of this were layers of sophisticated programming, memory and input and
Prior calculating technologies had to rely on slower mechanical processes. This meant they were much more limited in speed, flexibility and adaptability. Nevertheless they too were designed to
facilitate the same fundamental arithmetic and logical operations. The technologies of mathematics are in this sense much simpler than the elaborate analytic structures which make up mathematical
analysis. And for this reason, it is not necessary to consider all of mathematics in order to follow much of the history of how the technologies to aid mathematical reasoning developed. Just
considering the history of the development of aids to calculation can tell a great deal. As already noted, it is that which is dealt with here.
The calculational devices that were developed show an unmistakeable progression in complexity, sophistication and style from the earliest to the latest. Corresponding to this it is possible to
construct histories of calculational aids as some sort of evolution based on solving technical problems with consequent improvements in design building one upon the other. But, as already noted, it
is also important to understand why they were invented and used.
Fortunately in order to understand what has shaped the development of these calculational aids we can largely avoid talking much about mathematics. This is lucky because mathematics is by now a huge
field of knowledge. So you are entitled to relief that in this site we will avoid much of mathematics. We need not touch, for example, on calculus, set and group theory, the mathematics of infinite
dimensional vector spaces that make the modern formulation of quantum mechanics possible, and tensors which Einstein used to express his wonderfully neat equations for the shape of space-time.^2 It
will be sufficient to note that many modern challenges - from the prediction of climate under the stress of global warming, to the simulation of a nuclear reactor accident, to the deconstruction of
DNA - could not occur without enormous numbers of calculations which in the end are constituted out of additions and subtractions (and multiplications and divisions) and can only be carried out in
workable times with the use of ever faster calculating devices.
Even keeping our attention restricted to the basic arithmetic operations, it turns out we will still encounter some of the curly issues that we would have to think about if we were focussing on the
whole evolving field of mathematical thought. Of course history of mathematics is itself a field of scholastic study which can be developed from many perspectives. These include those from the
mainstream of history and philosophy of science^3 through to the sociology of science.^4 Even though this discussion here focuses on only a tiny “arithmetic core” of mathematics it will still be
useful to take some account of this literature and its insights. In particular, whether concerning ourselves with the evolution of the simple areas of mathematics or the more obstruse areas, one
question is always raised: what led to this particular development happening as and when it did?
2. Did increases in the power of mathematics lead the development of calculators? Was it the other way round?
It might be assumed that arithmetic, and more broadly, mathematics, developed through a process that was entirely internal to itself. For example, this development might have been propelled forward
because people could ask questions which arise within mathematics, but require the invention of new mathematics to answer them.
Suppose we know about addition and that 2+2 =4. Then it is possible to ask what number added to 4 would give 2. Answering that involves inventing the idea of a negative number. This leads to progress
through ‘completing mathematics’ (i.e. seeking to answer all the questions that arise in mathematics which cannot yet be answered.) That must be part of the story of how mathematics develops. Yet the
literature on the history of mathematics tells us this cannot be all.
The idea of ‘mathematics’, and doing it, are themselves inventions. The question of when mathematics might be useful will have different answers in different cultures. Different societies may
identify different sorts of issues as interesting or important (and only some of these will be usefully tackled with mathematics). Also different groups of people in those societies will be educated
in what is known in mathematics. Finally, different groups of people, or organisations, may have influence in framing the questions that mathematicians are encouraged (and resourced) to explore.
But the same is true of invention. At different times and in different cultures there have been quite different views taken on the value of change, and thus invention. At some points in history the
mainstream view has been that the crucial task is to preserve the known truth (for example, as discovered by some earlier civilisation - notably the ancient Greeks, or as stated in a holy book). At
other times or places much greater value has been placed on inventing new knowledge. Even when invention is in good standing there can be a big question of who is to be permitted to do it. And even
if invention is applauded it may be still true that this may only be in certain areas considered appropriate or important. This is as true in mathematics as in other areas of human activity.
In short, a lot of factors can shape what is seen as “mathematics”, what it is to be used for, and by whom. As an illustration it is worth remembering that astrology has until relatively recently
been considered both a legitimate area of human knowledge and a key impetus for mathematical development. Thus E. G. Taylor writes of the understandings in England in the late sixteenth century: “The
dictum that mathematics was evil for long cut at the very roots of the mathematical arts and practices. How were those to be answered for whom mathematics meant astronomy, astronomy meant astrology,
astrology meant demonology, and demonology was demonstrably evil?”^5 Indeed, it was noted that when the first mathematical Chairs were established at Oxford University, parents kept their sons from
attending let they be ‘smutted with the Black Art’.^6 However, despite these negative connotations, practioners of “the dark arts” played a strong role in developing and refining instruments and
methodologies for recording and predicting the movement of “star signs” as they moved across the celestial sphere.
One of the key features of the contemporary world is its high level of interconnection. In such a world it is easy to imagine that developments in “mathematics” which happen in one place will be
known and built on almost simultaneously in another. Yet that is a very modern concept. In most of history the movement of information across space and time has been slow and very imperfect. So what
at what one time has been discovered in one place may well have been forgotten a generation or two later, and unheard of in many other places. For this reason, amongst others already mentioned, talk
of the evolution of mathematics as if it had a definite timetable, and a single direction is likely to be very misleading.
History of course relies on evidence. We can only know where and when innovations have occurred when evidence of them can be uncovered. Even the partial picture thus uncovered reveals a patchwork of
developments in different directions. That is certainly a shadow of the whole complex pattern of discovery, invention, forgetting, and re-discovery which will have been shaped at different times by
particular needs and constraints of different cultures, values, political structures, religions, and practices. In short, understanding the evolution of calculating machines is assisted by
investigating it in the context of the evolution of mathematical thinking. But that is no simple picture. The history of developments in calculators and mathematics has been embroidered and shaped by
the the social, political and economic circumstances in which they emerged. At times, mathematical developments have shaped developments in calculators, and and other times, the opposite has been
3. What is a calculator?
“Calculator” could be taken to mean a variety of things. It could be calculation ‘app.’ on a smart phone, a stand alone elctronic calculator from the 1970s, or the motorised and before that
hand-cranked mechanical devices that preceded the electronic machines. In earlier times it could simply mean someone who calculates. It is difficult to see where the line should be drawn in this
regress all the way back to the abstract manipulation of ‘numbers’.
In this discussion, “calculator” is used as shorthand for “calculating technology”. In particular it is taken to mean any physically embodied methodology, however basic, used to assist the
performance of arithmetic operations (including counting). Thus a set of stones laid out to show what the result is if two are added to three (to give five), or if in three identical rows of five
what the outcome is of multiplying five by three (to give fifteen) will be regarded as a simple calculator. So too, will the fingers of the hand, when used for similar purpose, and even the marking
of marks on a medium (such as sand, clay or papyrus) to achieve a similar result.
This approach is certainly not that taken in all the literature. Ernest Martin in his widely cited book The Calculating Machines (Die Rechenmaschinen) is at pains to argue of the abacus (as well as
slide rules, and similar devices), that “it is erroneous to term this instrument a machine because it lacks the characteristics of a machine”.^7 In deference to this what is referred to here is
“calculators” (and sometimes “calculating technologies or “calculating devices”). Where the phrase “calculating machine” is used it will be in the sense used by Martin, referring to something with
more than just a basic mechanism which would widely be understood to be a machine. But with that caveat, the term “calculator” will be used here very broadly.
The decision to apparently stretch the concept of calculator so far reflects a well known observation within the History and Philsophy of Science and Technology that in the end, technique and
technology, or science and technology, are not completely distinct categories. Technologies embody knowledge, the development of technologies can press forward the boundaries of knowledge, and
technological development is central to discovery in science. As Mayr says in one of many essays on the subject, “If we can make out boundaries at all between what we call science and technology,
they are usually arbitrary.”^8 Indeed, as will be described later, the mental image that mathematics is the work of mathematicians (‘thinkers’) whilst calculators are the work of artisans (‘practical
working people’) is an attempt at a distinction that falls over historically, sociologically, and philosophically.
4. A discussion in three parts.
This is a work in progress, which in part is why it is formed as a website. So please regard it as a first draft (for which there may never be a final version). For this reason, corrections,
additional insights, or links to other resources I should know about will be much appreciated.
A word also about the way I have constructed the historical account. In keeping with the analysis I have contributed to elsewhere (in a book by Joseph Camilleri and myself^9), human development, will
roughly be divided into a set of semi-distinct (but overlapping) epochs, preceded by a “pre-Modern Era” spanning the enormous time period from the birth of the first modern homo-sapiens to the
beginning of the “Modern Period”. This beginning is set as beginning (somewhat earlier than is conventional) in the middle of the sixteenth century, with the “Early Modern Period” continuing from the
mid-sixteenth to late eighteenth century, and the “Late Modern Period” stretching forward into the twentieth century, and terminating around the two world wars. From thereon the world is regarded by
Joseph Camilleri and myself as entering a period of transition^10 (but there is not much need to focus on that here).
Thus the historical account is broken into three parts. The first part looks at the relationship between the evolution of calculating and calculators in the pre-Modern period. That forms a backdrop
but only one object in the collection is of an appropriate age. Apart from that object (which is some 4,000 years old) the objects in this “collection Calculant” are drawn from the Modern Period (the
earliest of these objects being from the early sixteenth century), and the Late Modern Period (from 1800) when mechanical calculation began to gain greater use in the broader society.
« | | Part 1 Origins »
^1 “Calculant” in Latin means literally “they calculate”. (More precisely it is the third-person plural present active indicative of the Latin verb calculo ( calculare, calculavi) meaning “they
calculate, they compute”). (↑)
^2 See for example http://mathworld.wolfram.com/EinsteinFieldEquations.html or for more explanation http://physics.gmu.edu/~joe/PHYS428/Topic9.pdf (both viewed 26 Dec 2011) (↑)
^3 See for example, Eleanor Robson, Jacqueline A. Stedall, The Oxford handbook of the history of mathematics, Oxford University Press, UK, 2009 (↑)
^4 See for example, Sal Restivo, Mathematics in Society and History: Sociological Inquiries, Kluwer Academic Publishers, Netherlands, 1992 (↑)
^5 E.G.R. Taylor, The Mathematical Practitioners of Tudor & Stuart England 1485–1714, Cambridge University Press for the Institute of Navigation, 1970, p. 4. (↑)
^6 John Aubrey quoted in Taylor, ibid, p. 8. (↑)
^7 Ernest Martin, The Calculating Machines (Die Rechenmaschinen), 1925, Translated and reprinted by Peggy Aldrich Kidwell and Michael R. Williams for the Charles Babbage Institute, Reprint Series for
the History of Computing, Vol 16, MIT Press, Cambridge, Mass, 1992, p. 1. (↑)
^8 Otto Mayr, “The science-technology relationship”, in Barry Barnes and David Edge (eds), Science in Context, The MIT Press, Cambridge USA, 1982, p.157. (↑)
^9 Joseph Camilleri and Jim Falk, Worlds in Transition: Evolving Governance Across a Stressed Planet, Edward Elgar, UK, 2009 (↑)
^10 ibid, pp. 132–45. (↑)
« | | Part 1 Origins »
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How to Use Exponential Moving Average (EMA) In Trading?
Exponential Moving Average (EMA) is a commonly used technical indicator in trading that helps to identify trends and potential entry or exit points. It is similar to a simple moving average (SMA) but
places greater emphasis on recent price data. The EMA gives more weight to recent prices, making it more responsive to changes in the market.
To calculate the EMA, you first need a set period of time (such as 9, 20, or 50), or you can choose your own timeframe depending on your trading strategy. Then, you need a starting point, which is
usually the SMA for the chosen period.
To calculate the EMA for subsequent periods, you can use the following formula:
EMA = (Price - EMA(previous day)) * (2 / (1 + n)) + EMA(previous day)
• Price is the current closing price of the asset.
• EMA(previous day) is the EMA value of the previous day.
• n is the chosen period or timeframe.
This formula calculates the weighting multiplier (2 / (1 + n)), and by applying it to the difference between the current price and the previous day's EMA, you can gradually calculate the EMA for each
Traders use EMA in various ways, but one common usage is to identify bullish or bearish trends. When the price is above the EMA line, it suggests a bullish trend, whereas when the price is below the
EMA line, it indicates a bearish trend. Traders may use this information to make buy or sell decisions accordingly.
EMA can also help traders identify potential entry or exit points. For example, when the price crosses above the EMA line, it may signal a bullish reversal and an opportunity to buy. Conversely, when
the price crosses below the EMA line, it may indicate a bearish reversal and a time to sell.
Additionally, the EMA can be used in conjunction with other technical indicators to create trading strategies. For instance, some traders use the crossover of the short-term EMA (e.g., 9-day EMA)
with the long-term EMA (e.g., 20-day EMA) as a confirmation signal for entering or exiting trades.
It's important to note that while EMA is a useful tool, it should not be solely relied upon for trading decisions. Traders should consider using it in conjunction with other technical indicators,
fundamental analysis, and risk management strategies to enhance their trading strategies and reduce the risk of false signals.
How to interpret EMA crossover signals?
EMA crossover signals are used by traders to identify potential changes in market trends. When the shorter-term EMA (exponential moving average) crosses above the longer-term EMA, it generates a
bullish signal, indicating a possible uptrend. Conversely, when the shorter-term EMA crosses below the longer-term EMA, it generates a bearish signal, indicating a possible downtrend.
Here are some steps to interpret EMA crossover signals:
1. Identify the two EMAs: Determine the two EMAs you want to use for the crossover analysis. Common options include the 20-day and 50-day EMAs or the 50-day and 200-day EMAs.
2. Look for a crossover: Keep a close eye on the chart to see if the shorter-term EMA crosses above or below the longer-term EMA.
3. Confirm the crossover: Don't solely rely on the crossover. Validate the signal by observing other technical indicators or performing additional analysis. Look for supporting factors such as
volume patterns, price action, or other indicators aligning with the crossover signal.
4. Determine the strength of the crossover: Consider the angle or slope of the EMAs during the crossover. A steep cross may indicate a more significant trend change, while a shallow cross might
suggest a weaker signal.
5. Identify the market trend: If the shorter-term EMA crosses above the longer-term EMA, it generates a bullish signal, indicating a potential uptrend. Conversely, if the shorter-term EMA crosses
below the longer-term EMA, it generates a bearish signal, indicating a potential downtrend.
6. Take other factors into account: Consider other factors that might influence the interpretation of the EMA crossover signal, such as overall market conditions, fundamental analysis, or news
Remember that EMA crossovers are not foolproof and can sometimes generate false signals. Therefore, it is essential to use them in conjunction with other indicators and analysis techniques to make
more informed trading decisions.
How to use EMA to calculate moving average envelopes?
To use the EMA (Exponential Moving Average) to calculate moving average envelopes, follow these steps:
1. Determine the period for the moving average envelopes. This is the number of periods you want to calculate the average for. For example, if you want to calculate the average for the past 20
periods, the period would be 20.
2. Calculate the EMA. The formula for EMA is: EMA = (Close - EMA(previous day)) * (2 / (period + 1)) + EMA(previous day)
Start by calculating the EMA for the first period using the closing price of that period. Then, for each subsequent day, use the formula above with the closing price of that day and the EMA from the
previous day.
1. Determine the percentage for the moving average envelopes. This is the percentage above and below the EMA that you want the envelopes to be set at. For example, if you want the envelopes to be 5%
above and below the EMA, the percentage would be 5.
2. Calculate the upper and lower envelopes. To calculate the upper envelope, multiply the EMA by (1 + percentage/100). To calculate the lower envelope, multiply the EMA by (1 - percentage/100).
For example, if the EMA is 100 and the percentage is 5, the upper envelope would be 100 * (1 + 5/100) = 105, and the lower envelope would be 100 * (1 - 5/100) = 95.
These upper and lower envelopes can be used for various purposes in technical analysis, such as determining potential overbought or oversold conditions or identifying support and resistance levels.
What are some common misconceptions about using EMA in trading?
1. EMA guarantees accurate predictions: While EMA is a widely used indicator in trading, it is important to note that it does not guarantee accurate predictions. EMA is based on historical price
data and is designed to provide a smooth average of recent price movements. However, it cannot accurately predict future price movements or guarantee profitable trades.
2. EMA is a foolproof trend indicator: EMA is often used to determine the direction of a trend in technical analysis. However, it is important to remember that trends can be complex and influenced
by various factors. Relying solely on EMA for trend analysis may lead to false signals or misinterpretations of market conditions.
3. EMA is always superior to SMA (Simple Moving Average): The Exponential Moving Average (EMA) is often preferred over the Simple Moving Average (SMA) due to its weighting formula, which places more
significance on recent price data. While EMA may provide a more timely response to price changes, this does not mean it is always superior to SMA. The choice between EMA and SMA depends on the
trader's preferences, trading strategy, and the specific market conditions.
4. EMA can accurately predict market reversals: Some traders mistakenly believe that EMA can reliably predict market reversals. While EMA can indicate potential reversals based on changes in the
slope or crossover of different EMA lines, it is not a foolproof method. False signals and failed reversals can occur, and it is crucial to corroborate EMA indications with other technical
analysis tools and indicators.
5. EMA works well in all market conditions: EMA performs differently depending on the market conditions and asset being traded. It may work well in trending markets but may generate false signals or
provide choppy results in sideways or ranging markets. Traders should consider adapting their EMA parameters or using additional tools to suit different market conditions and assets.
6. Using EMA alone is enough for successful trading: Relying solely on EMA or any single indicator for trading decisions is not recommended. Successful trading requires a comprehensive approach that
considers multiple factors, including fundamental analysis, market sentiment, and risk management. EMA can be a useful tool within a broader trading strategy, but it should not be the sole basis
for trades. | {"url":"https://topminisite.com/blog/how-to-use-exponential-moving-average-ema-in","timestamp":"2024-11-03T22:24:01Z","content_type":"text/html","content_length":"255457","record_id":"<urn:uuid:f9aee0a8-094f-4766-9138-7ce45ededaf5>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00037.warc.gz"} |
Boolean Algebra
Boolean Algebra Examples No2. Find the Boolean algebra expression for the following system. The output of the system is given as Q = (A.B) + (A+B), but the. The meaning of BOOLEAN ALGEBRA is a system
of algebra in which there are only two possible values for a variable (often expressed as true and false or as 1. Draw a logic circuit to generate F. Page 4. Simplification of Boolean functions.
Using the theorems of Boolean Algebra, the algebraic forms of functions can. Buy Logic and Boolean algebra on pro-dvijenie24.ru ✓ FREE SHIPPING on qualified orders. PyEDA has an extensive library for
the creation and analysis of Boolean functions. This document describes how to explore Boolean algebra using PyEDA.
Boolean Algebra. Definition: A Boolean Algebra is a math construct (B,+,., ', 0,1) where B is a non-empty set, + and. are binary operations in B, ' is a. The basic postulate of Boolean algebra is the
existence of a two-valued Boolean variable which can take any of the two distinct values 0 and 1. It. Boolean Algebra expression simplifier & solver. Detailed steps, Logic circuits, KMap, Truth
table, & Quizes. All in one boolean expression calculator. An algebraic system is the combination of a set and one or more operations. A Boolean algebra is defined by the set B ⊇ B≡ {0,1} and by two
operations, denoted. Boolean Algebra. 1. Boolean Functions. Boolean Functions. Definitions 1. A Boolean variable is a variable that may take on values only from the set. The basic operations on bits
were discussed extensively by George Boole and the rules governing them are eponymously named “Boolean algebra.”. All arithmetic operations performed with Boolean quantities have but one of two
possible outcomes: either 1 or 0. With this text, he offers an elementary treatment that employs Boolean algebra as a simple medium for introducing important concepts of modern algebra. Boolean
algebra is the theoretical foundation for digital systems. Boolean algebra formalizes the rules of logic. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by
millions of students & professionals. For math Boolean algebra. Natural. In the world of Boolean algebra, there are only two possible values for any quantity and for any arithmetic operation: 1 or 0.
This is known as the Boolean algebra duality principle. The order of operations for Boolean algebra, from highest to lowest priority is NOT, then AND, then OR. Boolean algebra is the category of
algebra in which the variable's values are the truth values, true and false, ordinarily denoted 1 and 0 respectively. A Boolean algebra is a mathematical structure that is similar to a Boolean ring,
but that is defined using the meet and join operators instead of the usual. Named after George Boole (–), an English mathematician, educator, philosopher and logician. The calculator will try to
simplify/minify the given boolean expression, with steps when possible. Applies commutative law, distributive law. Boolean algebra is a branch of mathematics that deals with operations on logical
values and incorporates the principles of logic into algebra. A Boolean algebra or Boolean lattice is a complemented distributive lattice. This type of algebraic structure captures essential
properties of both set. It is a method of representing expressions using only two values (True and False typically) and was first proposed by George Boole in Boolean algebra is a mathematical system
that describes the arithmetic of a two-state system and is the mathematical language of digital electronics.
Boolean algebra can be applied to any system in which each variable has two states. This chapter closes with sample problems solved by. Boolean algebra. The Wolfram Language represents Boolean
expressions in symbolic form, so they can not only be evaluated, but also be symbolically manipulated and. Using Boolean Algebra to describe truth tables. Truth tables can be represented using
Boolean algebra. For each row where Q = 1 there is a Boolean expression. Boolean Algebra (Binary Logic). Theorem. A + 0 = A. A + 1 = 1. A A A. A * 0 = 0. A * 1 = A. A*A A. A + A = A. A + A' = 1. A *
A = A. A * A' = 0. A + B = B + A. Two Boolean values: true and false. Used to control flow of control in programs. C has no Boolean data type--we substitute boolean-valued integer variables.
Boolean Algebra. ▫ the algebra of propositions. ▫ Basis for computation in binary computer systems. □ Constants/Truth Values. ▫ False (0) or True (1). Boolean algebra which was formulated by George
Boole, an English mathematician () described propositions whose outcome would be either true or false. The boolean logic itself is not necessarily part of the boolean algebra, just the truth values.
If A and B are truth values, so are A∨B, A∧B and ¬A.
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Elements of Geometry;
Elements of Geometry;: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle and the Geometry of Solids; to which are Added, Elements of Plane and Spherical
John Playfair
Popular passages
If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.
The diameter is the greatest straight line in a circle; and of all others, that which is nearer to the centre is always greater than one more remote; and the greater is nearer to the centre than the
less. Let ABCD be a circle, of which...
IF from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it ; if the rectangle contained by the whole line which cuts the circle, and the
part of it without the circle be equal to the square of the line which meets it, the line which meets shall touch the circle.
THE greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA : the side AC
is likewise greater than the side AB. For, if it be not greater, AC must...
If then the sides of it, BE, ED are equal to one another, it is a square, and what was required is now done: But if they are not equal, produce one of them BE to F, and make EF equal to ED, and
bisect BF in G : and from the centre G, at the distance GB, or GF, describe the semicircle...
IN a right angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another. Let ABC be a right
angled triangle, having the right angle BAC ; and from the point A let AD be drawn perpendicular to the base BC : the triangles ABD, ADC are similar to the whole triangle ABC, and to one another.
If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced...
When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is
called a perpendicular to it.
Bibliographic information | {"url":"https://books.google.com.jm/books?id=maUBAAAAYAAJ&vq=%22is+equal+to+the+fquare+of+the+ftraight+line+which+is+made+up+of+the+half+and+the+part+produced.%22&source=gbs_navlinks_s","timestamp":"2024-11-04T17:22:09Z","content_type":"text/html","content_length":"52124","record_id":"<urn:uuid:565bfcf3-bd67-403f-81b1-26f9dc0e0dde>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00249.warc.gz"} |
SubtractRect function (winuser.h) - Win32 apps
SubtractRect function (winuser.h)
The SubtractRect function determines the coordinates of a rectangle formed by subtracting one rectangle from another.
BOOL SubtractRect(
[out] LPRECT lprcDst,
[in] const RECT *lprcSrc1,
[in] const RECT *lprcSrc2
[out] lprcDst
A pointer to a RECT structure that receives the coordinates of the rectangle determined by subtracting the rectangle pointed to by lprcSrc2 from the rectangle pointed to by lprcSrc1.
[in] lprcSrc1
A pointer to a RECT structure from which the function subtracts the rectangle pointed to by lprcSrc2.
[in] lprcSrc2
A pointer to a RECT structure that the function subtracts from the rectangle pointed to by lprcSrc1.
Return value
If the resulting rectangle is empty, the return value is zero.
If the resulting rectangle is not empty, the return value is nonzero.
The function only subtracts the rectangle specified by lprcSrc2 from the rectangle specified by lprcSrc1 when the rectangles intersect completely in either the x- or y-direction. For example, if *
lprcSrc1 has the coordinates (10,10,100,100) and *lprcSrc2 has the coordinates (50,50,150,150), the function sets the coordinates of the rectangle pointed to by lprcDst to (10,10,100,100). If *
lprcSrc1 has the coordinates (10,10,100,100) and *lprcSrc2 has the coordinates (50,10,150,150), however, the function sets the coordinates of the rectangle pointed to by lprcDst to (10,10,50,100). In
other words, the resulting rectangle is the bounding box of the geometric difference.
Because applications can use rectangles for different purposes, the rectangle functions do not use an explicit unit of measure. Instead, all rectangle coordinates and dimensions are given in signed,
logical values. The mapping mode and the function in which the rectangle is used determine the units of measure.
Requirement Value
Minimum supported client Windows 2000 Professional [desktop apps only]
Minimum supported server Windows 2000 Server [desktop apps only]
Target Platform Windows
Header winuser.h (include Windows.h)
Library User32.lib
DLL User32.dll
See also | {"url":"https://learn.microsoft.com/en-us/windows/win32/api/winuser/nf-winuser-subtractrect?redirectedfrom=MSDN","timestamp":"2024-11-10T22:31:03Z","content_type":"text/html","content_length":"49578","record_id":"<urn:uuid:efd412bd-6c3e-4b99-9fbf-c1eb30fcffed>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00730.warc.gz"} |
Gearing Ration
Miss Dazy Registered Posts: 34 Regular contributor ⭐
I am also having truble with the gearing ratios...If i am given the long term loans of £800, and share capital of £100, are they the only figures i need?
i.e. 800 x 100 / 100 + 800 = 800
or should there be more figures that i should be putting in?
There is only the profit figure next to the share capital and there is no other long term debt.
• there are two ratios for gearing
1 debt/equity = (800/100)x 100 = 800%
2 debt/(equity + Debt) = [800/(800+100)] x100 = 89%
Most of the time the second formula is used because it indicates how much of the company's assets are owned by non-equity holders.
There is an obligation to pay interest charges on the borrowed amount.
Should there be a down turn in business, there may be a risk of non payment of interest charges and a posibility that equity holders may lose all or a substancial value of the equity if the
lenders force the company into liquidation for the non payment of interest payment.
PS net profit is part of equity. you will have to add the net profit and share capital to arrive at Equity value.
• Hiya,
Gearing ratio = long term loan/long term loan + equity
Hope this helps
• gearing ratio is long term loan/shareholders funds + long term loan x 100
so in your case it is 800/800 + 100 x 100
= 88.9%
• gearing ratios?
i have my exam next week and haven't heard or been taught about gearing ratios? Which module i it for?
• I think it comes around in ECR?, PEV or PCR, MAC and DFS.
Often it's called slightly different depending on the unit though.
If you do auditing or CMCC it would be in there as well, but those are skilltests.
• If you start from the formula rather than what it shows it can be a chore to understand whereas if you start from the information you want and work back to the formula it might make more sense.
If my house is worth £200,000 and I still owe £120,000 on the mortgage I have long term liabilities representing 60% of my capital.
If my car is worth £2,500 and I owe £1,000 then 40% of the capital is borrowed
You can go into book after book in your university/college library and see gearing in the context of the subject the book has been written - and there are a lot of ratios but if you use it for
what you want to use it, then the ratio drops into the analysis rather than the analysis looking for a formula. | {"url":"https://forums.aat.org.uk/Forum/discussion/28103/gearing-ration","timestamp":"2024-11-04T14:11:44Z","content_type":"text/html","content_length":"300168","record_id":"<urn:uuid:dbf7ac61-3004-48b6-8ee2-00a8563c0199>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00794.warc.gz"} |
What Is An Example Of Interpolation?
Examples of Interpolation Let's suppose a gardener planted a tomato plant and she measured and kept track of the growth of the tomato plant every other day. This gardener is a very curious person,
and she would like to estimate how tall her plant was on the fourth day. Linear Interpolation Method: The Linear Interp... Cubic Spline Interpolation Method: The cube S... Shape-Preservation Method:
The Shape-pres... Nearest Neighbor Method: In this method the v...
Interpolation is a useful Mathematical and Statistical tool that is used to estimate values between any two given points. In this article, you will learn about this tool, the formula for
Interpolation and how to use it.
Interpolation can be defined as the process of finding a value between two points on a line or curve.
Now to help us remember what it means, we should think of the first part of the word, which is 'inter,' and which means 'enter,' and that reminds us to look 'inside' the data we originally had.
Interpolation is a tool which is not only useful in Statistics, but is a tool that is also useful in the field of science, business or any time whenever there is a need to predict values that fall
within any two existing data points.
Examples of Interpolation
Here's an example which will illustrate the concept of Interpolation and give you a better understanding of the concept of Interpolation. Let’s suppose a gardener planted a tomato plant and she
measured and kept track of the growth of the tomato plant every other day. This gardener is a very curious person, and she would like to estimate how tall her plant was on the fourth day.
Her table of observations basically looked like the table given below:
Height (mm)
Based on the given chart, it's not too difficult to figure out whether the plant was probably 6 mm tall on the fourth day and this is because this disciplined tomato plant grew in a linear pattern;
that is there was a linear relationship between the number of days measured and the plant's growth. Linear pattern basically means that the points created a straight line. We could estimate it by
plotting the given data on a graph.
(Image to be added soon)
But what if the plant does not grow with a convenient linear pattern? What if its growth looked more like that in the picture given below?
(Image to be added soon)
What do you think the gardener will do in order to make an estimation based on the above curve? Well, that is where the Interpolation formula comes into picture.
Formula of Interpolation
The Interpolation formula can be written as -
y- y1= ((y2-y1)/ (x2- x1))* (x2- x1)
Now , if we go back to the tomato plant example, the first set of values for day three are given as (3,4), the second set of values for day five are given as (5,8), and the value for x is 4 since we
want to find the height of the tomato plant, y, on the fourth day. After substituting these given values into the formula, we can easily calculate the estimated height of the plant on the fourth day.
y – y1 =( (y2- y1) / ( x2- x1))* (x- x1)
Putting the values we have been given,
—What is an interpolation? It’s taking from the Composition, without using any of the Sound recording. Best example: taking the melody. The fact that it’s bypassing Sound recording also means it
only uses one CR, which makes it legally less imposing.
— ❀ anna ❀ 8 makes one butt ☹☻ (@SteamPoweredDM) Feb 28, 2022
y – 4 = ((8- 4) / ( 5- 3))* (x- 3)
y – 4 =4/2 (x-3)
y – 4 = 2(x-3)
y – 4 = 2(4-3)
y= 2(1) +4
y = 6
Types of Interpolation Methods
There are various different types of Interpolation Methods. Here they are:
Types of Interpolation
Linear Interpolation Method
The Linear Interpolation Method applies a distinct linear polynomial between each pair of the given data points for the curves, or within the sets of three points for surfaces.
Nearest Neighbor Method
In this method the value of an interpolated point is inserted to the value of the most adjacent data point. Therefore, the nearest neighbor method does not produce any new data points.
Cubic Spline Interpolation Method
The cube Spline method fits a different cubic polynomial between each pair of the given data points for the curves, or between sets of three points for surfaces.
Shape-Preservation Method
The Shape-preservation method is also known as Piecewise cubic Hermite Interpolation (PCHIP). This method preserves the monotonicity and the shape of the given data. It is for curves only.
Thin-plate Spline Method
The Thin-plate Spline method basically consists of smooth surfaces that also extrapolate well. This method is only for surfaces.
Biharmonic Interpolation Method
The Biharmonic method is generally applied to the surfaces only.
Why is the concept of Interpolation Important?
The concept of Interpolation is used to simplify complicated functions by sampling any given data points and interpolating these data points using a simpler function.
Commonly Polynomials are used for the process of Interpolation because they are much easier to evaluate, differentiate, and integrate and are known as polynomial Interpolation.
Drawbacks of Interpolation Method
While Interpolation is known to solve a lot of Mathematical and Statistical problems, it does have certain drawbacks and criticisms. One such drawback is that although the method of Interpolation is
simple and has been known to Mathematicians and people in general, for a long time, it has been known to lack the necessary accuracy and precision.
In the ancient Greek and Babylonian civilizations, the method of Interpolation was crudely used for prediction purposes. They would determine various factors such as the right time for sowing seeds
(in farming practices), calculate astronomical points in space and time to determine celestial events up in the sky, and plan strategies for monsoons, crop yield, growth and movement.
Today, the same methods are being used in the modern-day problems of the world. People use these methods of Interpolation for the fairly unpredictable stock markets, in solving data related to
security analysis, for determining volatility of the highly unpredictable public-traded shares and bonds, and this overpowering mass of data makes the employment of Interpolation unreasonable as it
can lead to many faulty predictions.
More often than not, the use of Interpolation in regression analysis, in this way leads to the yielding of an “error term”, that is obtaining a set of values that do not represent the factual
relationship between the variables most crucial for successful prediction. Interpolation must be employed for simple predictions such as determining the interest rate or value of any variable for
which the data point is missing. | {"url":"https://askandanswer.info/what-is-an-example-of-interpolation/","timestamp":"2024-11-08T14:16:57Z","content_type":"text/html","content_length":"58073","record_id":"<urn:uuid:bde8b888-3055-4361-bd54-6e017c351baf>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00527.warc.gz"} |
The FREQUENCY function in Google Sheets returns an array of frequencies of values in a dataset. It is commonly used to create a histogram. The output is an array with the same number of elements as
the classes parameter (second argument).
Use the FREQUENCY formula with the syntax shown below, it has 2 required parameters:
=FREQUENCY(data, classes)
1. data (required):
The array or range of values to be counted.
2. classes (required):
The array or range of intervals (bins) to group the values into.
Here are a few example use cases that explain how to use the FREQUENCY formula in Google Sheets.
Count the number of values within a specific range
By using the FREQUENCY function, you can easily count the number of values within a specific range. For example, if you have a list of grades and you want to know how many students scored between
60-69, 70-79, and 80-89, you can use the FREQUENCY function to group the grades into these ranges and count the number of values in each range.
Create a histogram
The FREQUENCY function is commonly used to create histograms. A histogram is a chart that shows the distribution of values in a dataset. By using the FREQUENCY function to group the values into
intervals (bins), you can create a histogram that shows how many values fall into each interval.
Calculate the mode
The mode is the value that appears most frequently in a dataset. By using the FREQUENCY function, you can easily calculate the mode. The mode is the value that corresponds to the highest frequency in
the array returned by the FREQUENCY function.
Common Mistakes
FREQUENCY not working? Here are some common mistakes people make when using the FREQUENCY Google Sheets Formula:
Incorrect data range
One common mistake is providing an incorrect data range that does not match the number of classes specified. Double-check that the data range and classes match.
Incorrect class range
Another common mistake is providing an incorrect class range that does not include all of the values in the data range. Double-check that the class range includes all the values in the data range.
Non-numeric data
If your data contains non-numeric values, the FREQUENCY function will return an error. Make sure your data is numeric before using this function.
Incorrect array formula syntax
If you forget to enter the formula as an array formula (by pressing Ctrl+Shift+Enter), the FREQUENCY function will not work properly. Make sure to enter the formula correctly.
Undefined classes
If you do not specify the number of classes, the FREQUENCY function will return an error. Make sure to specify the number of classes.
Related Formulas
The following functions are similar to FREQUENCY or are often used with it in a formula:
• COUNTIF
The COUNTIF formula counts the number of cells within a specified range that meet a certain criterion. This formula is commonly used to count cells that meet a specific condition or criteria.
• SUMIF
The SUMIF formula is used to add up values in a range that meet a specific criterion. It can be used to sum values based on text, numbers, or dates. The formula is most commonly used in financial
analysis, budgeting, and data analysis.
• AVERAGEIF
The AVERAGEIF function calculates the average of a range of cells that meet a specified criteria. It is commonly used when working with large datasets to quickly calculate the average of a subset
of data. The function takes a range of cells to evaluate (criteria_range), a string or value to compare against (criterion), and an optional range of cells to average (average_range). If the
average_range is not specified, the function will use the same range as the criteria_range.
• MEDIAN
The MEDIAN function returns the median (middle) value of a set of numbers. It is commonly used to find the middle value in a range of data points. If the number of data points is even, it returns
the average of the two middle values. This function can be useful in statistical analysis and data visualization.
Learn More
You can learn more about the FREQUENCY Google Sheets function on Google Support. | {"url":"https://checksheet.app/google-sheets-formulas/frequency/","timestamp":"2024-11-10T04:55:55Z","content_type":"text/html","content_length":"46101","record_id":"<urn:uuid:5c98f925-175c-4e42-b520-e58e2c22cbb1>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00143.warc.gz"} |
Table Calculations
Of Table Calculations
So now we know a little more about table calculations, typically where are they used and more importantly, how do they differ from normal and level-of-detail calculations?
Lets tackle this in stages:
How do Table Calculations Differ from the other Calculations
Well it is simple really, with exception of the First(), Index(), Last() and Size() calculations, all table calcs are aggregate which means the value they are working with, first needs to be
aggregated such as Sum(Sales), Count(Product) or {Include [Region] : Sum(Sales)} and secondly, these calculations take place inside Tableau, directly on the data that is visible (including scrolling)
to you.
As a result, table calcs are used in connection with normal and LOD calcs rather than in place of them.
So How do Table Calculations Actually Work?
As data is returned to Tableau and materialised (by you) into the working window, table calculations are generated at this time, which means that table calculations are dependant on the filters that
have been applied to your view.
Once you materialise your data, Tableau is simply running your selected calculation against teh defined scope but crucially, only against the data in your working window, all other data that is not
materialised into the view will bear no relevance on the output of the expression used, even if you have defined the scope at the expression-level in the expressions editor.
And Finally, How and Where Will I Use Them?
How breaks down to two important criteria:
• Type of calculation / purpose
• Scope
Calculation Type
Is down to what you are trying to achieve, with row numbering calculations simply written as FunctionName() so the Index function for example is written simply as Index()
Whereas the remaining calculations (see the table above) all enclose an aggregate calculation on number in some form for example the running_sum() calculation for Sales would be Running_Sum(Sum
Arguably the most important part about table calculations is that of scope.
The scope of a calculation arguably detracts from the simplicity of Tableau, as no other function within the software requires you to consider beyond the drag-and-drop interface, exactly how Tableau
needs to address the data.
Normally, when cutting data, it is enough to use case and/or if statements to determine which data is including in the calculation or, in the case of LOD calcs, what additional fields need to be
considered (or removed); this is still true of table calculations as those calcs are wrapped inside table calcs, however, when you are writing your table calculation, you will be presented with the
"Default Table Calculation" indicator:
Normal Calculation Table Calculation
1. Indicates the partition and direction of the scope
2. Click to open the scope dialogue box to hard-set the scope
Upon first creating your calculation, Tableau will default the table calc to Table (across) - do not worry if you have no data running across your table, i this instance, Tableau will switch to Table
(down) in this instance.
title Applying the scope from the calculation editor
Be careful when setting the scope from the calculation editor:
• Defining the scope in the calculation editor affects the scope for all table calculations that form the calculation:
Looking at this table:
Assume you want to compare the total for furniture against the overall total: 779,103 / 2,938,103, using this calculation:
Code Block
Window_Sum(Sum(Sales),First(),Last()) / Total(Sum(Sales)
Will mean the scope is set for the whole calculation, so regardless as to how you define your scope, you will always return 100%
But, by breaking down the calculation into two separate calcs and then using a third calc to perform the calculation, means that whilst Tableau will not provide the ability to set the scope from
the third calculation window, you can set the scope from editor for each calculation and, more importantly, when applying the value to the chart:
but by breaking the calcs down, we can see how we want to define the scope:
And so now we create the individual calculations, the combination calculation and then set the scope - note the delta symbol () on the pill indicates the calculation is a table calculation
1. Right-mouse select the pill that holds the complete calculation
2. Select Edit Table Calculation
3. Select your first nested calculation
4. Adjust the scope of the first table calculation to Sub-Category - this tells Tableau that you are only interested in calculating the total for each Category
5. Switch the Nested Calculations to the Total calculation and leave the scope at Table (down)
• Similar to defining the scope of LOD calcs, setting the scope from the calculation editor will permanently fix the calculation to the items. Whilst this is not as problematic as an LOD calc in
that the scope can be adjusted on-the-fly for each viz, it will mean that the chart will break if the items are not present, but only until they are either added or the scope is adjusted
About: Use
So you have thoroughly explained how table calculations work, although I am still unsure as to when and where I shall use them?
Where you use them will be down to your requirements, however it is possible that you may have already used them such as:
Such as when testing for change in the case of A/B testing - variant vs control or, period over period eg:
Code Block
Lookup comparisons -
) / Lookup(Sum(Sales),-1)
Will enable you to compare the current row to the previous row
% Total You can easily use a level-of-detail calculation to calculate total: {Sum(Sales)} however, what if you want your total to be a sub-total?
Rank The classic Top n.
Running_Sum Without which, generating cumulatives would be impossible
Many of the more complex charts require the table calculations for positioning in order hang data from them - the measure waterfall I built could not work without table calculations.
And performing calculations when using an OLAP source cannot be done without them (unless you use MDX which then only creates specific outputs):
OLAP sources (cubes), or any source of pre-aggregated data held within a rigid structure is often limited to simply retrieving data from an intersection of one-to-many members such that if-then-else
logic calculations are usually unavailable, such the it then becomes necessary to use a lookup calculation to step in. | {"url":"https://datawonders.atlassian.net/wiki/pages/diffpagesbyversion.action?pageId=108462167&selectedPageVersions=20&selectedPageVersions=21","timestamp":"2024-11-13T10:37:39Z","content_type":"text/html","content_length":"73763","record_id":"<urn:uuid:be747ae0-1fe3-44b4-89b2-25dc9605b005>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00242.warc.gz"} |
Referring to Cells in Calculations
You can refer to columns and rows using a form of cell notation that identifies the intersection of a row and a column as ®, c).
Syntax: How to Use Cell Notation for Rows and Columns in a RECAP Expression
A row and column can be addressed in an expression by the notation:
Is a required constant.
Is the row number.
Is the column number. Use an asterisk (*) to indicate the current column.
Example: Referring to Columns Using Cell Notation in a RECAP Expression
In this request, two RECAP expressions derive VARIANCEs (EVAR and WVAR) by subtracting values in four columns (1, 2, 3, 4) in row three (PROFIT). These values are identified using cell notation
SUM E_ACTUAL E_BUDGET W_ACTUAL W_BUDGET
3000 AS 'SALES' OVER
3100 AS 'COST' OVER
BAR OVER
RECAP PROFIT/I5C = R1 - R2; OVER
" " OVER
RECAP EVAR(1)/I5C = E(3,1) - E(3,2);
AS 'EAST--VARIANCE' OVER
RECAP WVAR(3)/I5C = E(3,3) - E(3,4);
AS 'WEST--VARIANCE'
The output is shown as follows.
E_ACTUAL E_BUDGET W_ACTUAL W_BUDGET
-------- -------- -------- --------
SALES 6,000 4,934 7,222 7,056
COST 4,650 3,760 5,697 5,410
------ ------ ------ ------
PROFIT 1,350 1,174 1,525 1,646
EAST--VARIANCE 176
WEST--VARIANCE -121
Note: In addition to illustrating cell notation, this example demonstrates the use of column numbering. Notice that the display of the EAST and WEST VARIANCEs in columns 1 and 3, respectively, are
controlled by the numbers in parentheses in the request: EVAR (1) and WVAR (3). | {"url":"https://ecl.informationbuilders.com/focus/topic/shell_7706/FOCUS_CreatingReports/source/topic207.htm","timestamp":"2024-11-15T01:26:10Z","content_type":"text/html","content_length":"4999","record_id":"<urn:uuid:8dbd12cf-461c-4359-a0e7-730fde94da63>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00642.warc.gz"} |
Zero Polynomial: Definition, Degree & Solved Examples
What is zero polynomial?
Any polynomial in which all the variables have a coefficient equal to zero is known as a zero polynomial. For example \(0, 0x, 0x^2 \) and so on.
What is the degree of a zero polynomial?
The degree of a zero polynomial is undefined, therefore it is considered as negative (usually -1 or -).
Is 7 a zero polynomial?
No, a zero degree polynomial simply means 0. It can be nothing other than 0.
What is a zero degree polynomial?
A polynomial in which the degrees of all the variables is 0 is called a zero degree polynomial. For example- \(6x^0\), \(-9a^0\). Therefore, any integer, positive or negative is a zero degree
What are the zeros of a polynomial?
Zeros of a polynomial are nothing but the roots of the polynomial. Zeros or roots of a polynomial are those values of the variable (x) which make the polynomial equal to 0. For example- For the
polynomial x2+7x-18, the zero or the root will be 2, because (2)2+72-18=4+14-18=0. | {"url":"https://testbook.com/maths/zero-polynomial","timestamp":"2024-11-11T16:10:25Z","content_type":"text/html","content_length":"868539","record_id":"<urn:uuid:3f2afe13-3c32-4feb-9c67-58dc82f39a01>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00676.warc.gz"} |
maximum stable set problem
When addressing the maximum stable set problem on a graph G = (V,E), rank inequalities prescribe that, for any subgraph G[U] induced by U ⊆ V , at most as many vertices as the stability number of G
[U] can be part of a stable set of G. These inequalities are very general, as many of … Read more | {"url":"https://optimization-online.org/tag/maximum-stable-set-problem/","timestamp":"2024-11-12T21:58:19Z","content_type":"text/html","content_length":"83214","record_id":"<urn:uuid:ff4aae7a-82ef-48e8-a01c-2710768c4b23>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00025.warc.gz"} |
Merge Sort in C# - Code Maze
Merge Sort in C#
Merge sort in C# is one of the algorithms that we can use to sort elements. Merge Sort is known to be efficient as it uses the “divide and conquer” strategy that we are going to discuss as we
implement it.
Let’s start.
What is Merge Sort?
Merge Sort achieves its purpose using a two-step process:
Divide: merge sort starts by dividing the input array into two halves. The algorithm recursively calls itself for each of those halves until there are no half-arrays to divide during the sorting
Conquer: the algorithm sorts and merges the sub-arrays in this step to return an array whose values are sorted.
Generally, we use these high-level steps when sorting an array or a list with a merge sort:
Step 1: Check if the array has one element. If it does, it means all the elements are sorted.
Step 2: Use recursion to divide the array into two halves until we can't divide it anymore.
Step 3: Merge the arrays into a new array whose values are sorted.
How Does Merge Sort Algorithm Work?
Let’s say we want to sort an array that has eight elements:
int[] array = { 73, 57, 49, 99, 133, 20, 1, 34 };
Dividing Arrays
Using the merge sort algorithm, let’s start subdividing the array into equal halves. We can see here that in the first iteration, the array is split into two equal arrays of half the size (4):
73, 57, 49, 99 133, 20, 1, 34
Now, we divide these arrays into halves without changing the sequence of the elements in the original array:
73, 57 49, 99 133, 20 1, 34
We are going to subdivide the arrays further into arrays that have one element each (we achieve the atomic value in this step):
After subdividing the arrays, we are going to now start merging them while comparing the elements and placing them in the right order.
Merging and Sorting
This algorithm requires an extra array to store the sorted elements. The algorithm starts by traversing the subarrays that hold 73 and 57. 57 is less than 73, so we swap their positions while merging
them into a new array that has two elements:
57, 73
Next, we traverse the subarrays that hold 49 and 99 while comparing the elements at each position. Since 49 is less than 99, the merged array of size 2 becomes:
57, 73 49, 99
The same process applies here as the algorithm starts traversing the subarrays while comparing the values they hold in each position. 20 is less than 133, so the algorithm swaps their positions when
merging them and the array becomes:
57, 73 49, 99 20, 133
1 and 34 are already in the correct order so we don’t swap their positions when combining them back into an array:
57, 73 49, 99 20, 133 1, 34
Next Merge Iteration
In the next iteration of the merging process, the arrays should have four elements each, sorted in the correct order.
The algorithm starts traversing the two subarrays of size two while comparing the values at each position. In the first position, 49 is less than 57. We move 49 to the first position followed by 57.
Now we move on to 73 and since it is less than 99 they don’t need to swap positions.
So, the new merged array of size 4 is:
49, 57, 73, 99
We repeat the same process for the right half of the array.
The algorithm compares the elements on each subarray while merging them into a new array. 1 is less than 20 so it becomes the first element in the merged array followed by 20.
The algorithm compares 133 with 34 and adds 34 first into the new array before adding 133. So, the new array after the merging process is:
1, 20, 34, 133
In the next iteration of the merging process, the array should have eight sorted elements (same as the original array).
The algorithm starts traversing the subarrays while comparing the values at each position while adding them to a new array that has sorted elements. For example, at position 1 in the subarrays, 1 is
less than 49, so we store 1 as the first element in the merged array:
49, 57, 73, 99 1, 20, 34, 133
The algorithm iterates through the arrays while placing them in the correct order in the merged array which becomes:
1, 20, 34, 49, 57, 73, 99, 133
Let’s implement the merge sort algorithm in C#.
How to Implement Merge Sort in C#?
We are going to define a method SortArray() as our entry point into the sorting algorithm. The method takes three parameters int[] array, int left and int right:
public int[] SortArray(int[] array, int left, int right)
if (left < right)
int middle = left + (right - left) / 2;
SortArray(array, left, middle);
SortArray(array, middle + 1, right);
MergeArray(array, left, middle, right);
return array;
First, SortArray() uses the left and right integer values to define the index of the element in the middle of the array.
The method recursively calls itself to subdivide the right and left subarrays. The merging process commences after each array has one element. Let’s write a method that implements the merge sort
public void MergeArray(int[] array, int left, int middle, int right)
var leftArrayLength = middle - left + 1;
var rightArrayLength = right - middle;
var leftTempArray = new int[leftArrayLength];
var rightTempArray = new int[rightArrayLength];
int i, j;
for (i = 0; i < leftArrayLength; ++i)
leftTempArray[i] = array[left + i];
for (j = 0; j < rightArrayLength; ++j)
rightTempArray[j] = array[middle + 1 + j];
i = 0;
j = 0;
int k = left;
while (i < leftArrayLength && j < rightArrayLength)
if (leftTempArray[i] <= rightTempArray[j])
array[k++] = leftTempArray[i++];
array[k++] = rightTempArray[j++];
while (i < leftArrayLength)
array[k++] = leftTempArray[i++];
while (j < rightArrayLength)
array[k++] = rightTempArray[j++];
The first thing to note is that the MergeArray() method takes four parameters. The leftArrayLength and rightArrayLength variables help us define temporary arrays to hold values during the sorting
var leftArrayLength = middle - left + 1;
var rightArrayLength = right - middle;
var leftTempArray = new int[leftArrayLength];
var rightTempArray = new int[rightArrayLength];
We copy data into those temporary arrays using two loops as the next step:
for (i = 0; i < leftArrayLength; ++i)
leftTempArray[i] = array[left + i];
for (j = 0; j < rightArrayLength; ++j)
rightTempArray[j] = array[middle + 1 + j];
We then proceed to compare the elements in the leftTempArray[i] and rightTempArray[j] objects and swap their positions if the element in the leftTempArray[i] is less than or equal to the element in
the rightTempArray[j] object while storing them in the array[k] position in the merged array:
while (i < leftArrayLength && j < rightArrayLength)
if (leftTempArray[i] <= rightTempArray[j])
array[k++] = leftTempArray[i++];
array[k++] = rightTempArray[j++];
The process completes by copying any remaining elements from the leftTempArray[i] and the rightTempArray[j] objects into the merged array:
while (i < leftArrayLength)
array[k++] = leftTempArray[i++];
while (j < rightArrayLength)
array[k++] = rightTempArray[j++];
Finally, we can verify that the method sorts an array accurately:
var array = new int[] { 73, 57, 49, 99, 133, 20, 1 };
var expected = new int[] { 1, 20, 49, 57, 73, 99, 133 };
var sortFunction = new Merge();
var sortedArray = sortFunction.SortArray(array, 0, array.Length - 1);
CollectionAssert.AreEqual(sortedArray, expected);
Time and Space Complexity of Merge Sort
The space complexity of the merge sort algorithm is O(N) because the merging process results in the creation of temporary arrays as we can see in the implementation step.
The time complexity of the merge sort algorithm is O(N*log N).
Let’s see why.
Assuming the length of the array is N, every time we divide it into equal halves, we can represent that process as a logarithmic function log N. In each step, we find the middle of any subarray,
which takes O(1) time.
As we merge the subarrays, the algorithm takes O(N) which we determine through the length of the array we are sorting.
Therefore, the total time that the merge sort algorithm takes is N(log N + 1), which we can convert to O(N*log N).
The best, average, and worst-case complexity of the merge sort algorithm is O(N*log N).
The best-case complexity scenario occurs when the algorithm sorts an array whose values are sorted. In this case, the number of comparisons the algorithm makes is minimal.
In a scenario where we have a mix of values that are in the correct order and those that are not, the algorithm encounters an average-case complexity scenario.
On the other hand, the worst-case complexity scenario occurs when assuming we have a list of numbers that are in descending order and we need to sort them in ascending order. The algorithm would
encounter a worst-case complexity as the length of that array grows to N.
Advantages and Disadvantages of Merge Sort Algorithm
The merge sort algorithm is faster when sorting large arrays than other sorting algorithms such as bubble sort.
It has consistent execution times as all the cases take O(N*log N) time.
However, merge sort requires O(N) space to run, which makes it less efficient than bubble sort and selection sort which use constant space O(1). Besides that, it does not perform well as bubble sort
for smaller arrays and lists.
Merge Sort in C# Performance
Let’s verify the merge sort algorithm has a time complexity of O(N*log N) by measuring the time it takes for the algorithm to sort an array.
First, let’s write a method to generate a set of random numbers that are going to be added into the array:
public static int[] GenerateRandomNumber(int size)
var array = new int[size];
var rand = new Random();
var maxNum = 10000;
for (int i = 0; i < size; i++)
array[i] = rand.Next(maxNum + 1);
return array;
The GenerateRandomNumber method takes an integer as its sole input. Using the inbuilt Random class, we generate integer values (less than 10000) that we’re going to put into the array.
Let’s write a method that generates a sorted array:
public static int[] GenerateSortedNumber(int size)
var array = new int[size];
for (int i = 0; i < size; i++)
array[i] = i;
return array;
Next, we are going to create a collection that holds different arrays that have random and sorted values:
public IEnumerable<object[]> ArrayData()
yield return new object[] { GenerateRandomNumber(200), 0, 199, "Small Unsorted" };
yield return new object[] { GenerateRandomNumber(2000), 0, 1999, "Medium Unsorted" };
yield return new object[] { GenerateRandomNumber(20000), 0, 19999, "Large Unsorted" };
yield return new object[] { GenerateSortedNumber(200), 0, 199, "Small Sorted" };
yield return new object[] { GenerateSortedNumber(2000), 0, 1999, "Medium Sorted" };
yield return new object[] { GenerateSortedNumber(20000), 0, 19999, "Large Sorted" };
Each object entry has four values. An integer array such as GenerateRandomNumber(200) , the index of the first value in the array (0), the index of the last element in the array (199), and a string
object storing the name of that array (“Small Unsorted”).
The array objects have different sizes (to simulate time complexity scenarios) and hold random numbers that are added by the GenerateRandomNumber method. The GenerateSortedNumber method creates
arrays that have values that are sorted.
Let’s assess the sample best, average, and worst-case complexity performance results of the algorithm:
| Method | array | left | right | arrayName | Mean | Error | StdDev |
|---------- |------------- |----- |------ |---------------- |------------:|-----------:|-----------:|
| SortArray | Int32[200] | 0 | 199 | Small Sorted | 16.95 μs | 0.577 μs | 1.702 μs |
| SortArray | Int32[200] | 0 | 199 | Small Unsorted | 17.02 μs | 0.653 μs | 1.925 μs |
| SortArray | Int32[2000] | 0 | 1999 | Medium Sorted | 204.07 μs | 8.722 μs | 25.717 μs |
| SortArray | Int32[2000] | 0 | 1999 | Medium Unsorted | 203.08 μs | 8.809 μs | 25.972 μs |
| SortArray | Int32[20000] | 0 | 19999 | Large Sorted | 2,352.66 μs | 102.065 μs | 300.940 μs |
| SortArray | Int32[20000] | 0 | 19999 | Large Unsorted | 2,530.11 μs | 111.928 μs | 330.023 μs |
We can see that the time merge sort takes to sort an array increases as the size of the array grows.
The algorithm sorts a large array in 2,530.11 μs (~0.00253 seconds) but sorts a small array in 17.02 μs (1.702e-5 seconds).
Besides that, we can see that the algorithm performs slightly better when sorting ordered arrays than when sorting random elements. The difference can be seen when sorting larger arrays.
Merge sort in C# is efficient when sorting large lists and arrays but needs extra space O(N) to achieve its goals. It uses the “divide and conquer” strategy, which makes it have a time complexity of
O(N*log N). If you want to learn how to implement other sorting algorithms, check out these bubble sort and selection sort articles. | {"url":"https://code-maze.com/csharp-merge-sort/","timestamp":"2024-11-05T08:56:20Z","content_type":"text/html","content_length":"122955","record_id":"<urn:uuid:dd68c766-a65a-44fc-ab87-214ca83369c3>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00395.warc.gz"} |
Taoism and Freemasonry
Matéo Simoita
The Masonic approach draws its reflection from all traditions; we propose a reflection on the application of Taoist principles to Masonic symbolism. (see the slideshow which presents a reminder of
the history of Taoism)
It might seem like a challenge to want to use Taoist concepts that appeared several hundred years BC to explain the operation of a Masonic lodge, but, if we accept the famous adage of "Everything is
in everything! "the exercise should not leave indifferent. This is of course a first approach to popularization that could be more detailed if the request is expressed.
There are several possible entries in the understanding of Taoist philosophy; we suggest two of them:
-From Chinese energetics,
-Using the five elements.
We could define energy as a "setting in motion" of a dynamism affecting the living, whether in a collective or individual form.
The principle of Taoism could, in a simplified way, be presented as follows:
Total Yang energy, immaterial, fecundates total Yin energy, material...
This initial fertilization of total yin energy produces a succession of six energy forms:
-The small yin energy (Tae Yin)
-Average energy (Chao yin)
-The Great Yin Energy (Tsue Yin)
-The little yang energy (Chao yang)
-The average yang energy (Yang ming)
-The great yang energy (Tae Yang)
From this great yang energy there is the possibility to recreate the total yang energy (see attached diagram).
In this evolution of the energy cycle, we can notice three key moments:
-The "fertilization" of the total yin energy by the total yang energy.
-The mutation of the great yin energy into small yang energy
-The transition from great yang energy to total yang energy
If we compare with the Masonic "logic", we can see that there is a similarity between the total yang energy and the Light and between the total yin energy and the Darkness.
When we proclaim "Ordo Ab Chao", which can also be stated as "Light transcends matter", the Taoist doctrine affirms that the total yang energy transforms and fecundates the total yin energy.
This logic of relations between energies naturally applies to the living world. Each organ has its own energy.
There are positive energies which are earlier "helping" and negative or "perverse" energies which can disturb the functioning of the system. The energy can also be of external origin and affect our
organism; for example cold, heat, trauma are energies but also anger, fear, joy.
All the subtlety of the Taoist approach has been to describe the changes in the functioning of the organism and therefore of the human being according to the variations of all these energies and to
imagine a logic that explains the expected and foreseeable consequences of these interactions.
Classically one describes a global energy and six "functional" energies, each of which can be in the yang form or in the yin form; schematically let us recall that the yang form is characterized by
an activation whereas the yin form corresponds to a certain passivity. Naming these energies in French is necessarily simplistic because the Chinese terms have a wide variety of nuances, but the
objective being to be understood, the stakes are high and Taoist specialists will excuse us.
These six main types of energy are:
-Distribution energy ensures the distribution and "feeding" of the different components of the system.
-The energy of fulfilment is manifested in the realisation of intimate processes.
-The energy of preparation includes the capacity for evaluation and anticipation prior to the commitment.
-The energy of movement has an impact on what we could call the physical body and in particular its exteriorization;
-The energy of separation affects all processes that seek to separate the essential from the non-essential;
-The energy of resistance is dedicated to preserving the essential, i.e. the germ of the renewal of life and thus in a way the memory of the code.
These six functional energies, if they work in synergy, allow the system to fulfil its role, to protect itself and to ensure their durability. As soon as one fails, a fragility appears and the whole
becomes vulnerable.
The complexity of the phenomenon of life is such that Taoism refers to other energies, but it is necessary here to be concise and to know how to limit oneself.
If we consider the lodge as a living organism, it is possible to use this modeling of an energetic functioning. One can thus analyze from them the global functioning and that of the different
Globally, we could say that the lodge is a Yin entity, to which we could give the adjective feminine insofar as it corresponds to a maieutic that gives "birth" to initiates who will have their own
existence. In addition to this function of initiation of laymen, the lodge has another reason for being that of strengthening the fraternity between its members and extending it to society. A lodge
that is experiencing internal disorders is functionally unable to perform these functions.
The general orientation of the Lodge towards mysticism or towards the production of societal reforms, proceeds in a certain way from the same dynamism, that of transformation; in fact one could say
that the Lodge is destined to create the initiate and to "shape" him/her towards sanctification and towards the exercise of a mission by becoming an "architect" (the one who creates).
To ensure its role the lodge has a college of officers; using the Taoist energy model, each officer could be assigned a functional energy:
To the venerable and the secretary, the energy of movement (the secretary could be the yin form of this energy and the venerable the yang form).
-to supervisors, the energy of preparation
-to the speaker and the roofer, the energy of resistance (the speaker is in the yang form and the roofer in the yin form)
-to the Master of Ceremonies, the energy of fulfillment
-to the expert, the energy of the separation for what comes from the outside and for verification from the inside
-to the hospitaller and the treasurer, the energy of distribution (the treasurer in the yin form and the hospitaller in the yang form)
When we apply Taoist logic to the functioning of the lodge we highlight the weak intervention of the energy of separation whereas it is a fundamental dynamism in the living world because it allows to
get rid of all that is impure. In lodge, the expert to whom one could attribute this role, does not have prerogatives so affirmed except when it is a question of admitting or not a visitor; this
quasi-absence of internal self-regulation could moreover explain quite a few problems!
The balance of the system is achieved if each function plays its role; as an example:
if the function of resistance is no longer ensured, the lodge loses its "soul" by losing its history and specificity,
if the hospitaller does not play his role of distribution in particular of fraternity and solidarity, the link between the members of the lodge will dissolve in an aseptic formalism!
The five Taoist elements applied to the Masonic approach
Reminder: In a global system where the living is under the combined influences of Earth and Heaven, if the energies can be considered as a celestial influence, the five elements include the earthly
These five symbolic elements speak to us much better than the energies because we find them in our rituals, the air can very well be assimilated to the symbol of Wood.
Things become more complicated when we understand that to each of the five elements we attach a dynamism, an organ, a function, a quality, an orientation, etc.
The attached diagrams visualize the relationships between the five elements: relations of generation and control.
According to the Taoist tradition, the harmony of the whole is the consequence of the coherence between the 5 elements; if an element is failing, the system is weakened and can cause consequences.
As an example, if we restrict ourselves to the primary meanings, a flood for example corresponds to an excess of the Water element which will first affect the Fire element by weakening it; the Earth
element whose function is to control the Water element will be exhausted.
This example can be found for the entire contents of the Water element.
In relation to the functioning of the Lodge, we will find these interactions between the five elements from the contents that are close to them:
Wood is the element of movement but also that of spirituality; located in the East, it corresponds to the venerable
Fire, an element of beauty, of the imaginary; located in the South, it corresponds to the second supervisor at the REAA.
The earth, the element of food but also of social ties and therefore of worries; it is located in the centre ;
Metal, element of judgement; located in the west, it corresponds to the roofer.
Water, an element of memory and history; it is located in the North
Disturbances to the system of the five elements can occur:
This simplified adaptation of Taoist notions to the Masonic Lodge can allow a better understanding of the logic of the Masonic approach and the imperatives to which we must submit if we want
everything to work in harmony:
The interdependence of functions and dynamics implies the absolute necessity of consensus to make a Lodge function properly and to find solutions when a problem arises;
Each function has its specificity; none takes precedence over the others; and this is true both in the officers' college and for each member of the lodge; likewise for the training of apprentices.
The Tao is the Way; harmony is to stay on the Way! All this is easily found in the Masonic approach!
This slideshow covers the history of Taoism, its logic and its various applications.
Taoist" reading of the Masonic dress (as defined by the Lodge meeting)
Without modifying the ritual used (which would be possible) it is possible to refer in the lodge to certain Taoist principles and in particular :
the principle of the transformation of the organism (of the subject) according to the energy cycle,
the humility of the initiate before the great laws of nature (of Heaven and Earth),
protection against the aggression of external and internal perverse energies!
If we consider that the purpose of Masonic dress is, on the one hand, to allow a "resourcing" of the initiate in the search for Truth in an ethical dimension and, on the other hand, to encourage the
strengthening of fraternal ties, these Taoist principles can encourage brothers and sisters to take better advantage of the conduct of their lodges. | {"url":"https://www.idealmaconnique.com/post/taoism-and-freemasonry","timestamp":"2024-11-07T02:40:07Z","content_type":"text/html","content_length":"1050487","record_id":"<urn:uuid:0aea1e85-6dd5-417a-a542-12c2a034d6f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00012.warc.gz"} |
Using the wave equation
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1. Question 1 of 6
Correct 1 / 1 Points
Incorrect / 1 Points
Remember to convert from cm to metres first! The formula is v = f λ
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2. Question 2 of 6
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3. Question 3 of 6
□ The picture below shows the frequency rating of a microwave. You can check the back of your own microwave for this. The frequency is 2450 MHz, or 2.45 * 10^9Hz.
Ori tried to measure the wavelength of microwaves. He put a piece of chocolate in a non-spinning microwave for a short time and noticed two very melted spots. He interpreted these to be the
peaks of the microwave. The length between them was 12cm.
Using the frequency and wavelength provided, calculate the speed of light. (All electromagnetic waves, including microwaves, are a type of ‘light’ and therefore travel at the same speed)
Round your answer to two significant figures.
Answer: [ ] * 10^8 m/s
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4. Question 4 of 6
The time between the peaks is the period. To convert that to frequency, you take its inverse, so 1/10s. Then just apply the v = f λ formula
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5. Question 5 of 6
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Set frequency to 1, and speed to 2. Plug into the wave speed equation.
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6. Question 6 of 6
□ In liquid C, a specific wave has a wavelength of 0.01 m and a speed of 2m/s. The wave then moves to liquid D, in which its wavelength changes to 0.04m. Calculate its new speed. (Assume that
the frequency stays constant).
Answer: [ ] m/s
(Not sure how to do this? See hint)
Correct 1 / 1 Points
Incorrect / 1 Points
First calculate the wave speed, v, in the first liquid. Then plug that speed into the equation for the second liquid.
[Back] [Hint] [Check] [Next] | {"url":"https://sciencemesoftly.com/quizzes/using-the-wave-equation/","timestamp":"2024-11-05T22:53:12Z","content_type":"text/html","content_length":"130571","record_id":"<urn:uuid:b5518f27-6f10-42c6-b65a-da14f0eb2501>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00317.warc.gz"} |
Zachary Colburn
Bioi provides functions to perform connected component labeling on 1, 2, or 3 dimensional arrays, single linkage clustering, and identification of the nearest neighboring point in a second data set.
Determine the minimum separation distance between two sets of points
Dual channel PALM or iPALM data can represent the localizations of two distinct fluorophore-labeled proteins. To determine the minimum distance separating each localization of protein A from protein
B, the following code could be implemented. Each row in the output of find_min_dists corresponds to the point denoted by the same row in mOne. “dist” is the distance to the nearest point in mTwo,
whose row number in mTwo is given in the “index” column.
Single linkage clustering
PALM/iPALM localizations in very close proximity may belong to the same superstructure, for example a single focal adhesion. By linking localizations separated by less than some empirically
determined distance, these different super structures can be identified. Below, 2D PALM data is simulated and all points falling within a distance of 0.8 are linked.
Clusters can be easily visualized using ggplot2.
Label connected components (i.e. find contiguous blobs)
Following background subtraction and thresholding, distinct cellular structures (focal adhesions for example) can be identified in fluorescent microscopy images. Array elements that are connected
horizontally/vertically or diagonally are labeled with the same group number. Each contiguous object is labeled with a different group number. The function find_blobs can be used to implement this
functionality in 1, 2, or 3 dimensions. Below, this operations is performed on a matrix structure. | {"url":"https://cran.uib.no/web/packages/Bioi/vignettes/Bioi.html","timestamp":"2024-11-03T06:29:35Z","content_type":"text/html","content_length":"18918","record_id":"<urn:uuid:7303bced-ed87-426d-a38c-082feee36f0d>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00055.warc.gz"} |
Journée MathStic «Combinatoire et Probabilités» au LIPN
We will first introduce translation surfaces, which are Riemann surfaces built from gluing polygons in the plane via translations. The torus is the most basic example of a translation surface. The
closed geodesics we count are called saddle connections, and are found by following geodesics which start and end at a marked point. In the case of the torus, the saddle connections correspond to
pairs of integers (a,b) which are coprime to each other. We will present probabilistic results counting saddle connections with length conditions, as well as counting pairs of saddle connections with
various pairing conditions. We will finish with highlighting the open questions and difficulties of counting triples of closed geodesics. | {"url":"https://indico.in2p3.fr/event/25439/","timestamp":"2024-11-11T13:22:04Z","content_type":"text/html","content_length":"115575","record_id":"<urn:uuid:73006888-4b03-47c3-b647-f8154f384a90>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00058.warc.gz"} |
What is a Logic Puzzle?
Answer: SONGBOOKS
First, solve the logic puzzles. There are nine of them, and in each puzzle the name of the genre is given, perhaps with rules, but in some of the puzzles you have to look up the rules separately. A
full solution to all nine puzzles is in the appendix.
Then, there are several observations to make simultaneously. There are several colored regions in each grid. However, only three of the puzzles (LITS, Simple Loop, Star Battle) tell you how to
extract from the colored regions. Thankfully, one of these puzzles is a shading puzzle, one is a loop drawing puzzle, and one is an object placement puzzle. Because all nine puzzles fall into one of
these three categories, it makes sense to apply the same type of extraction to the other puzzles of each category.
The key idea of this puzzle is that we are creating a purist/rebel alignment chart of these nine logic puzzles. The flavortext clues this with the words "rebel" and "alignment", as well as providing
the two axes, and the title of the puzzle loosely hints towards this as well, because purist/rebel alignment charts are questioning what a title is. See the sandwich alignment chart for an example of
The flavor states that the speaker is "down" for "any grid shape", implying that the three puzzles with the weird grid shapes (Nurikabe, Simple Loop, Tents) are the bottom row of the grid. Similarly,
they think not including the rules is "right", so the puzzles that only state the name of the genre (Akari, Masyu, Nurikabe) are on the right column. We can then infer that the top row consists of
square grids while the middle row is rectangular grids, and that the left column consists of rules featuring extraction while the middle column does not have extraction. Thus, we can place the nine
genres as follows:
Rules Purist Rules Neutral Rules Rebel
Grid Purist Star Battle Uso-One Masyu
Grid Neutral LITS Yajilin Akari
Grid Rebel Simple Loop Tents Nurikabe
Now, we can examine the ??? puzzle at the bottom. There is a 3x3 grid provided that maps colors to locations, and a 9x9 grid split into 3x3 sections. Because we have our nine puzzles each in one of
the locations in a 3x3 grid, we can place the extracted digits from each puzzle into corresponding locations in the bottom 9x9 grid.
Upon doing this, the puzzle looks like
. This looks remarkably like a Sudoku; indeed, it solves logically as one. Taking the green cells in each box and summing them up, then converting these numbers to letters via A1Z26 yields the
Author's Notes
So this puzzle is... a bit of a marathon. I tried to make the feeders easy while still being interesting, although I think the logic is still a little tricky if you're not already familiar with the
genres. Six out of the nine puzzles were nerfed from the original version of this puzzle. If you're not logicked out, here are these original puzzles: LITS Simple Loop Star Battle Tents Uso-One
Appendix: Logical Walkthrough
Throughout these solutions, we use RxCy to refer to a cell in a rectangular grid; R1C1 is the top left cell.
The initial break-in of this puzzle is the outside border. Observe that there can be at most six light bulbs among the outside border, because every line in this image can contain at most one light
However, consider each of the clues on the outside border (R1C1, R1C9, etc.) Each of these places a bound upon the minimum number of light bulbs around it. In this image, each of the clues has at
least one of the pictured light bulbs around it:
But this makes a minimum of six light bulbs! Therefore, this minimum must be attained. This means that the outside border only has two options to take, that can be thought of as a clockwise circle or
a counterclockwise circle:
There are two key takeaways from this. First, we know that no other cell on the outside border can have a light bulb (otherwise there would be more than six). Second, consider the 2-clues on the
border. They cannot have both of their adjacent border cells have light bulbs, so the one remaining cell must have a light bulb.
The next observation is to consider the cell R4C6. We claim this cell cannot have a light bulb. Indeed, if it contains a light bulb, then the 2-clue in R5C7 is resolved, but this will form a
contradiction with the 1-clue in R8C7, which has no place it can go.
Thus R4C6 cannot have a light bulb, which places light bulbs in R3C5 and R5C5. Because the light bulb in R5C5 sees R9C5, we now know which of the two border options was correct.
The puzzle is pretty easy to just guess from here (and indeed, it may be faster in this puzzle to just assume both of the cases of the border and see which one works), but let's finish the puzzle
logically. R2C6 must have a light bulb because it's the only way to resolve the 1-clue in R2C7, but also, let's consider what happens if R4C7 does not have a light bulb. If this is the case, then
R5C8 and R6C7 must both have light bulbs to resolve the 2-clue in R5C7, but if this happens then there's no way to resolve the 1-clue in R8C7.
Therefore there must be a light bulb in R4C7. This resolves the 1-clue in R5C10, and the rest of the puzzle is finished off quickly.
There are two "halves" of this puzzle that are connected at the end. Let's first try to resolve the left half. First, consider the two cross-shaped regions. Both of these need to contain a
T-tetromino; there is no other way to resolve them. However, this means that the two T-tetrominos cannot touch. It turns out that if R4C2 or R4C3 are used, then this will always occur, so therefore
these cells cannot be used, and the tetrominos are resolved:
Applying the no 2x2 rule leaves only one option for the bottom left region, which is an I-tetromino. Afterwards, there are only two ways to resolve the region next to it, but one of these two ways
makes an I-tetromino that touches the other region, so that region is also resolved. Finally, considering the top region, R1C1-3 cannot be used because there are only three cells that do not connect
to any other usable cell. This leaves only two options for that region, but one of them forms a 2x2 (R45C45), so that region is resolved also.
Now, we can turn our attention to the right half. The first step is to consider the two top regions, especially the one that has five cells in it. There are only two ways to resolve that region, and
both of these ways form an L-tetromino with cells R1-3C10 shaded, so we can mark those three cells as shaded. Afterwards, the region next to it also must form an L-tetromino regardless, so it cannot
touch those three cells, which leaves only one option available.
Now, consider the region below the filled-in region (the one in the middle-right). In particular, we claim R3C12 cannot be shaded. If this were to happen, the only way to resolve that region that
does not form a 2x2 will make an L-tetromino that touches the one we just filled in. After this, R4C13 must be shaded because all shaded cells are connected in the end, and furthermore there is only
one way to resolve that entire region without forming a 2x2 or a touching L-tetromino.
Actually, we can now resolve the region to the left of that T-tetromino now as well. The cell R6C11 is shaded because not shading it leaves only four cells in that region which form a 2x2, which in
turn shades R5C11. Afterwards, we must avoid a T-tetromino because this touches one, but also, if we were to form an L-tetromino in this region (with cells R4C10 and R4C11), then this would touch the
above five-cell region which we established at the beginning of the puzzle must contain an L-tetromino. This leaves only one option for this region, forming an S-tetromino.
Consider the region below and left of this S-tetromino region. There are now only two options for where the tetromino can go. However, if it goes to the left, there is no way to place it that does
not form a 2x2 tetromino; either you take both cells R4C9 and R5C9, or you form a 2x2 in that region itself. Therefore that region is resolved fully.
Now, we need to consider how to connect the two sides to satisfy connectivity. If we do not use the L-tetromino at the top, then we could connect via the bottom, but this requires two I-tetrominos to
touch. So, we need to use the L-tetromino, and if we were to use R1C9, then no two cells from opposite sides are close enough. This forces R3C9 to be shaded, after which there is still only one way
to resolve the connection that does not form a 2x2 or force two L-tetrominos to touch each other.
We can start this puzzle off by drawing some "trivial" lines. If a black pearl is too close to the edge, then there must be a line coming out of it in the opposite direction. In addition, a white
pearl on the edge must be passed parallel to the edge so it does not go into the edge of the grid.
We can apply white pearl rules a few times on the border cells now, as well as on R9C6. In addition, the two adjacent black pearls at the bottom left must not be connected directly, because it isn't
possible to let both of them continue to go straight. Then, the bottom one of this pair must also turn left because if it turned right it would cross the white pearls and intersect at R9C6, which is
a problem.
R9C5 must go up and down because of the adjacent parallel line, and R6C6 cannot go down; if it did then the white pearl on R9C6 doesn't have an adjacent turn. However, the key observation to make at
this point concerns R3C5. Basically, a white pearl can never be at a corner, so this black pearl is very restricted in how it goes. If it goes up and left, then R2C4 is at a corner and cannot be
resolved; similar issues happen with two of the other directions, meaning it must go down and right.
We apply a few more direct deductions from here; R2C6 and R4C4 have their directions resolved by adjacent parallel lines, R5C5 goes straight, which in turn resolves R6C6 to go right. This now forces
R7C7 to go down, and now R8C6 must turn left (it needs to turn due to the white pearl and now it can't turn right). R11C6 must go left, because two to the right of it is a used cell. Finally, R2C3
and R2C6 cannot be horizontal; if they did then they would connect to R2C6 and R2C4 would not satisfy the white pearl rule. That's a lot of deductions, but all of them follow directly from the basic
Masyu rules.
We can make some connectivity deductions now. R3C1 must go right and then down to escape and not form a smaller loop. R5C4 goes left by the white pearl next to it, and connects to R3C3. R1C4 goes
right and connects to R2C5. R11C4 must go left to avoid a self-loop, which resolves a bit of R10C5's path. In fact, applying the black pearl at R8C3 and continuing similar deductions, along with the
white pearl at R5C5 causing R6C5 to turn left, resolves the entire left hand of the puzzle.
Now, R7C7 must go right because if it goes left then it cannot escape, but the key observation right now involves R4C8. If it goes down and left, this forms a smaller loop, which is bad. If it goes
down and right, this causes an issue with the white pearl at R5C9. This means that it simply cannot go down, so it has to go up. Now, R3C7 has to go down, because it is connected to R3C6 by the left
hand side of the puzzle. This resolves a bit more of the puzzle by similar direct deductions as the rest of the puzzle.
The top part of this puzzle can be resolved as well; in particular, R2C7 must connect to R1C8, because if either of them connected to R2C8, then the other would be trapped. This makes R2C8 connect to
R3C10, and to avoid forming multiple loops, both of these need to start going down.
The white pearl on R7C10 must go up and down to avoid trapping R6C10. It then turns left to connect up to R7C9. From here, there is only one way to resolve the white pearl on R9C9. To finish off the
puzzle, just observe that R10C8 must go right to have any hope of not trapping everything, and the puzzle resolves.
This absolutely cursed creation is actually a lot easier than it seems at first glance. There are several cells that must be immediately shaded to not connect two different clues together.
Now, the central 5 only has one way to escape, which then shades another cell to avoid touching the 4. This 4 in turn only has one way to escape, which shades yet another cell in relation with the
upper 2. This 2 is resolved now, and both the 4 and the 5 can be shaded a bit more; in fact, using shaded cell connectivity with the shaded cell to the left of the 2, the 4 can be resolved entirely.
Now, we can apply the "no internal point" rule (this topology's equivalent of the no 2x2 rule) to resolve the top-left 3. Furthermore, two other cells must be unshaded by this rule, one above the
rightmost 3 and one a bit below it. That 3 must connect to the upper unshaded cell, since there's no other way to resolve it that doesn't touch the 10.
The 10 can now actually be fully resolved; the left unresolved cell on the top row (R1C9 if you squint) must be unshaded, forcing a corridor of unshaded cells on the right column, and in order for
the shaded cells to escape they go left afterwards, which unshades another cell by the internal point rule, making a region of exactly size 10.
The next step involves resolving the 5-clue. There is an unshaded cell near it by the internal point rule (R4C2) so this the 5, from R4C4, either goes left twice or down and then left, to avoid the
3. However, in the latter scenario, that unshaded cell connects to the 4, but it can only do that in a region of size at least 5, which is a contradiction. This resolves the section. Afterwards, the
4 is immediately resolved by applying the internal point rule.
The central 3 is resolved immediately from this image. In addition, the other 3 is quickly resolved; connectivity forces R8C1 to be shaded and R9C2 to be unshaded. This leaves only one unshaded
region left, although the internal point rule gives us a few unshaded cells in it.
This 8 clue is very sparse; most attempts to resolve it end up forming a shaded internal point somewhere. One logical way to resolve it is as follows: First, among the two rightmost cells on the top
row, at least one must be unshaded to not break the 2x2 rule. Then, among the bottom four, at least one must be unshaded for a similar reason. Finally, there must be at least one cell to deal with
connectivity (the three left cells, the 8 clue, and the right region must all be connected, and aren't yet). However, this totals to eight cells including the five unshaded cells already in the grid,
which means that we cannot waste any unshaded cells. This only leaves one way to resolve it because many ways (e.g. including the bottom right corner cell) would waste an unshaded cell.
Simple Loop
Simple Loops like this one usually flow very smoothly. The most basic deduction in this genre is that a cell only has two adjacent cells that it can go to, because that cell is immediately resolved.
There are some more cells that can be resolved trivially (bottom left), but the key deduction involves the right region; especially the region's R5C1. This cell cannot go left and right, because it
would form a small loop. Therefore, it must go up. This resolves a whole bunch of information on the right subgrid, because right's R4C2 has only two adjacent nodes it can be connected to, and this
path repeats. In fact, the entire right subgrid resolves in a snakelike pattern.
We can resolve some more based on this; on the top subgrid, we need to not form a small loop, which forces R5C4 and R5C5 to go up. In addition, right subgrid's R5C1 has to go left now, which resolves
a lot more in a snakelike pattern again.
To avoid forming multiple loops, left subgrid's R3C2 and R3C3 both go up, and the left grid is fully resolved. We can continue making basic deductions, both two-adjacency deductions and multiple
loops, to resolve the whole puzzle.
Star Battle
In this genre, a dot between some cells means there must be a star within those cells. When there is a dot between two adjacent cells, the cells adjacent to this pair cannot have a star.
In Star Battles, you can start the puzzle by resolving smaller regions; in particular the top left region is quite restricted. If either R2C1 or R2C2 has a star, then no other cell in that region can
contain a star. After this, there must be a star within R1C1-2 and there must be a star placed in R3C1.
After this, we can consider what happens on the top right region. If there are two stars within the top four cells in that region, then that makes three stars on the top row, which is bad. So, one of
the lower two cells has a star, which blocks out R2C9 and R3C9 from having stars. Now we can examine R4C9. If R4C9 does not contain a star, then that region now only has a single 2x2 square where
both stars need to go, which is a problem because only one star can go in a 2x2 square. Thus, R4C9 has a star, which resolves a star on R2C10 as well.
This places a star between R1C7-8, which forces a star in R3C7. Now, we can examine the other upper region. It cannot contain a star in Row 1, because there are already two known stars there.
Similarly, it cannot contain one in Row 3. So there must be one star in Row 2 and one star in Row 4, which means that R4C3 has a star.
This next step is quite tricky. Consider where stars are going in Rows 5 and 6. Within the red region, two stars go in Rows 5 and 6, because there are no other cells. Within the blue region, at least
one star goes in Rows 5 and 6 (in fact, exactly one star). Within the brown region, at least one star goes in Rows 5 and 6. However, this makes four stars, which is the number of stars that can go in
those two rows. This means that either R7C6 or R7C7 must contain a star in the brown region, which means the cells above and below this pair don't have a star. Similarly, one of the bottom three
cells in the blue region contains a star, so there cannot be a star in R8C5.
Now, consider the bottom-middle region. There must be at most one star within R9-10C5-6, so there must also be one star in R7-8C8. This actually resolves the pair to the left of it and means that
R7C6 has a star.
We can do more with the red region. There must be one star in the left three (still yet to be marked as wrong) cells, and one star in the right two. This forms a pair that resolves R7C4 as not a
star, and R8C4 to have a star. There is also now a pair in R10C4-5.
At this point we should consider the left two columns. There are two stars contributed to it by the top left region, and the red region must also have a star in the left two columns. So, the bottom
left region must have at most one star in these two columns, which means that R10C3 must have a star. Because this is the second star in column 3, this resolves a lot of information immediately;
R10C5, R6C4 both contain stars, which then places stars in R2C5 and R5C6 as well. Also, there is a star in R5C1 since that's the only way to get two stars in Row 5, which then places a star in R1C2.
Now, let's consider row 9. Almost every cell from it does not have a star; the only four that are still possible are all in the gray region, so they indeed come from there. This places a pair in
R9C7-8 and another in R9C9-10. This forces a star into R7C8.
After this, the rest of the puzzle is straightforward. Only R6C10 can contain the second star in Column 10, which resolves the gray pairs and the pair from Row 1. Finally, R8C2 contains a star.
This is kind of a weird puzzle, but it's not too bad. The first observation is that there are some cells we know cannot contain tents, because there are no trees directly adjacent to them. Most
importantly, R3C2 and R3C3 do not contain tents, so the 3-right clue is very restricted. There must be a tent in R3C1, since only one of R3C4-5 and only one of R3C6-7 can contain a tent. In addition,
because R1C5 does not have a tent, there is only one way to place the tents in the top row to satisfy the 2-right clue: One on R1C2 and R1C4. These three tents can be connected to their only adjacent
Now, R3C4 cannot have a tent because the tree in R2C4 is already used. This resolves row 3 to have tents on R3C5 and C7. In addition, the 1 down-left clue is satisfied, which marks quite a few cells
as not having tents.
Consider the 2 down-left clue. There are four possible locations for tents that split into two pairs: R4C3, R5C3, R6C2, R7C1. But the tree in R5C1 restricts this; its tent is either in R5C2 or R6C1,
and in either case R6C2 cannot also contain a tent. This places a tent in R7C1, which then places one in R5C2 and R4C3.
The next step involves the 3 up-left clue. There is already one tent along that line, and there are only two more trees that are close enough to have their tent there. So this means R6C5's tree
connects to a tent on R5C6, and there's a tent in one of R7C6 and R8C6. But similar to the earlier deduction, the tree on R9C4 means there cannot be a tent on R8C6, so this resolves both of these
trees. Finally, the 1-right clue resolves the final tent on R8C2.
In this solution, we use an O on top of a clue to mark that it must be telling the truth, while an X marks that it must be lying.
In Uso-One, some clues are known to be liars. For instance, a 3 on the edge cannot be a truth teller because if it were true then that cell would be separated from the rest; there is a similar issue
for 2s in the corner. This immediately resolves the liar in two regions. Then, because R3C5 is a 3-clue and R3C6 cannot be shaded, we get three shaded cells immediately. Also, note that some clues
are the only one in their region, which makes them liars by default.
Consider the region containing the 2,0,2. If the top 2 is true, then the 0 is lying, and vice versa. Similarly, if the bottom 2 is true, then the 0 is lying, and vice versa. Thus, if the 0 were true
then both 2s would be lying, so this isn't the case and both 2s are true. We can also apply connectivity of unshaded cells to get R3C1 to be unshaded.
Now, consider the 2 and the 0 in the 2,0,3 region. Exactly one of these must be true, since if the 2 is true then R6C4 is shaded and if the 0 is true then it's unshaded. Thus, the 3-clue in R7C3 is
always true. This shades some cells which then resolves that the 0 is true and the 2 is lying.
A similar deduction can be made on the top right region. The 2 in R3C7 is either lying or telling the truth. If it is true, then the 2 in R3C8 cannot be satisfied, so one of those two is a lie, which
means the other two clues in the region are definitely true. Then, to avoid breaking connectivity, R1C9 and R2C10 cannot both be shaded, which means R2C8 and R3C9 are shaded. This implies that R3C7
is lying, and we can resolve a bit more based on the fact that the 0-clue nearby is definitely lying, as well as using the 2-clue from R2C6.
There is a rather subtle step now; consider the region with 3,3,0. If both 3s are true, then they are both resolved, but this breaks connectivity, separating out the top right region from the rest.
Thus, the 0 must be true, and one of the 3s is lying. This, in turn, means that R7C6 must be lying. Much of the bottom right gets resolved using the three true clues there.
The 3s are resolved too, with R5C7 being true. Because the 2 in R9C9 being true would break connectivity, it's false, which, along with the 0 being wrong, resolves that section. Finally, the bottom
left is also easily resolvable, since R9C3 and R6C1 must not be satisfied.
We can resolve quite a bit of this puzzle immediately by some Yajilin counting deductions. First, the right column can be entirely resolved. This is because on an edge, no two shaded cells can be one
cell apart from each other (or adjacent), so they have to be at least two apart, which resolves it to be in Rows 2, 5, and 8. In addition, Column 1 is similarly resolved; there must be a shaded cell
in Row 1, and then there cannot be one in Columns 4 or 6. In order to force three in that column, they must be in Columns 3 and 7. We can draw a bit of the loop based on these shaded cells.
In fact, the 5-left clue can actually be fully resolved. Because we must put five shaded cells within 10 cells, one in each pair is shaded, which means R7C8 and R7C10 are shaded. Furthermore,
consider what happens if, say, both R7C3 and R7C5 are shaded. Then R8C4 is a problem, because it has three directions it's forced to go out of. Similarly, if R7C4 and R7C6 are both shaded, then R8C5
is broken. Thus, it must be the case that both R7C3 and R7C6 are shaded.
From here, the 3-right in R3C6 is also resolved, since there are only five cells that can be shaded. After doing that (and resolving R3C8, which has to go up and down), the 3-up is also immediately
resolved, which has more basic deductions we can make from there.
Now, we can do some entrance-counting deductions. Consider the bottom right region. It currently has three entrances (R4C10, R4C12, R6C12), and one possible-entrance (R4C8). If R4C8 goes left, then
there is an odd number of entrances in the bottom right, which means there will definitely be one that is left unpaired. This is a problem, so R4C8 goes right, and we can resolve a bit more because
the cells adjacent to the shaded ones are immediately resolved. By a similar token, R2C8 must go left to satisfy parity, which resolves the top right, and to avoid a smaller loop, the right side is
resolved entirely.
The 3-right in R2C1 is actually resolved; by the same deduction as the 5-left, we can't have two cells that are one space apart that are both shaded, and R2C2 can't be shaded because if it were then
R1C2 is trapped, so R2C3 and R2C6 are shaded. Also, the 1-down is resolved too, since R7C5 can't have a shaded cell, so R5C5 must have one.
One final entrance-counting deduction solves the puzzle. Consider R6C4. If it goes up, then there are an odd number of entrances to the bottom unresolved section, which is bad. Thus it can't go up,
but this means that R5C4 must go both up and left. The rest of the puzzle resolves immediately from there.
Unlike the others, this puzzle is computer-generated; it's also very straightforward, and only requires simple deductions, either "This number can only go in one place" or "Only one number can go
There are a lot of clues that can immediately be filled in. 7 in Box 5 must be in R4C4, and after this, 7 must be in R2C5 in Box 2. 9s are forced in R4C6 and R2C8 in their respective boxes. The 7 and
9 in Row 2 let us resolve that row; R2C4 has a 2 and R2C9 a 1.
The 1 now places a 1 in R4C7 and R7C8. Also, 6 is forced into R4C9 in box 6. The rest of the box resolves immediately, with the 2 and 5 only having one spot each.
All of the 8s can now be resolved; Box 3 has only one spot, then Box 2, and similarly Box 4 only has one spot and Box 5 only have one spot each. Also, Box 3 gets resolved, with the 3 being placed,
and then the 5, with the 7 going in the final spot.
The remaining 1s can also get resolved because Box 2 only has one spot for a 1. Box 2 also gets resolved by the 4 going to R4C1. After placing a 6 in R3C4, the 6s can be entirely resolved in the
We can now place 3s in Box 1, 4, 5, 8, and then 7, resolving the 3s. Then, we can resolve the remaining 5s in Boxes 4, 8, and 9. This now resolves a 7 in Box 9, which gives all of the 7s, and from
there the rest of the puzzle resolves the remaining digits straightforwardly. | {"url":"https://puzzlehunt.club.cc.cmu.edu/protected/solutions/21011_sol/index.html","timestamp":"2024-11-07T06:25:38Z","content_type":"text/html","content_length":"45196","record_id":"<urn:uuid:28fc1ddc-6bc8-4716-ad81-ed83301097d7>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00289.warc.gz"} |
From enigmas in physics to a structural version of idealism
Quote:Our first-person perspective is primary, the external world emergent, argues physicist and member of the Austrian Academy of Sciences, Dr. Markus Müller, in his presentation during Essentia
Foundation’s 2020 online work conference. A follow-up interview with Dr. Müller, expanding on the topics of his presentation, is also available. [I'm posting the interview in the next post - Sci]
Dr. Müller is Group Leader at the Institute for Quantum Optics and Quantum Information in Vienna, Austria, and a member of the Austrian Academy of Sciences. He is also a Visiting Fellow at the
Perimeter Institute for Theoretical Physics, in Waterloo, Canada. Prior to these appointments, Dr. Müller has been an Assistant Professor and Canada Research Chair in the Foundations of Physics
at the University of Western Ontario in London, Canada. He received his doctorate from the Berlin University of Technology in 2007, with a thesis titled Quantum Kolmogorov Complexity and the
Quantum Turing Machine, under the supervision of Prof. Ruedi Seiler. Dr. Müller has over 50 technical publications to his name, and is a member of Essentia Foundation‘s Academic Advisory Board.
'Historically, we may regard materialism as a system of dogma set up to combat orthodox dogma...Accordingly we find that, as ancient orthodoxies disintegrate, materialism more and more gives way to
- Bertrand Russell
(This post was last modified: 2021-01-29, 01:36 AM by Sciborg_S_Patel.)
An interview with physicist Dr. Markus Müller
Quote:An obvious way out—for which I argue—is to give up the container view altogether, at least fundamentally. For some reason, this seems to be psychologically difficult. Hence, if the quantum
puzzles were the only ones we encounter in physics or philosophy, then one might well choose to be a Bohmian or an Everettian and enjoy the resulting psychological comfort. But as I argue in my
work, there are many more challenges to this container view than just quantum mechanics; for example, the puzzles and thought experiments that I have mentioned in my talk, which arguably render
the container view methodologically inadequate. I believe that there is something important about our world and our place in it that we have so far failed to grasp.
Quote:Given the puzzles of quantum mechanics, the many-worlds view that you have mentioned in your question aims at telling a coherent story about the quantum world in terms of this container or
stage play view. Does this Everettian interpretation succeed in doing so? Yes, absolutely! But the problem is that you can make every worldview consistent with modern physics if you stretch it
far enough. Do you dislike the idea of many worlds and would like to hold on to a single-world classical picture? Then pick Bohmian mechanics! Do you prefer to abandon any notion of randomness
altogether? Then pick ‘t Hooft’s superdeterministic cellular-automaton interpretation! Pick whatever you like—the experimental predictions will be identical, and nobody can prove you wrong.
Given this situation, I think that the only reliable way to understand what we can really learn from QM—what it tells us about the world—is to disregard interpretations and look at actual
scientific practice. There, we find that quantum states are nothing but our calculational tool to determine probabilities of measurement outcomes—and all we ever see are these outcomes. And it
turns out that these probabilities have surprising properties. For example, they violate Bell’s inequalities. The simplest logical conclusion to me is to see this as a hint that the world is,
first, fundamentally probabilistic in some sense, and second, that we cannot consistently regard the outcomes of our measurements as predetermined in any way that would deserve this name.
'Historically, we may regard materialism as a system of dogma set up to combat orthodox dogma...Accordingly we find that, as ancient orthodoxies disintegrate, materialism more and more gives way to
- Bertrand Russell
Law without law: from observer states to physics via algorithmic information theory
Markus P. Mueller
Quote:According to our current conception of physics, any valid physical theory is supposed to describe the objective evolution of a unique external world. However, this condition is challenged
by quantum theory, which suggests that physical systems should not always be understood as having objective properties which are simply revealed by measurement. Furthermore, as argued below,
several other conceptual puzzles in the foundations of physics and related fields point to limitations of our current perspective and motivate the exploration of an alternative: to start with the
first-person (the observer) rather than the third-person perspective (the world). In this work, I propose a rigorous approach of this kind on the basis of algorithmic information theory. It is
based on a single postulate: that universal induction determines the chances of what any observer sees next. That is, instead of a world or physical laws, it is the local state of the observer
alone that determines those probabilities. Surprisingly, despite its solipsistic foundation, I show that the resulting theory recovers many features of our established physical worldview: it
predicts that it appears to observers as if there was an external world that evolves according to simple, computable, probabilistic laws. In contrast to the standard view, objective reality is
not assumed on this approach but rather provably emerges as an asymptotic statistical phenomenon. The resulting theory dissolves puzzles like cosmology's Boltzmann brain problem, makes concrete
predictions for thought experiments like the computer simulation of agents, and suggests novel phenomena such as "probabilistic zombies" governed by observer-dependent probabilistic chances. It
also suggests that some basic phenomena of quantum theory (Bell inequality violation and no-signalling) might be understood as consequences of this framework.
'Historically, we may regard materialism as a system of dogma set up to combat orthodox dogma...Accordingly we find that, as ancient orthodoxies disintegrate, materialism more and more gives way to
- Bertrand Russell
The following 2 users Like Sciborg_S_Patel's post:2 users Like Sciborg_S_Patel's post
• stephenw, Brian
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• stephenw, Brian | {"url":"https://psiencequest.net/forums/thread-from-enigmas-in-physics-to-a-structural-version-of-idealism","timestamp":"2024-11-02T10:53:39Z","content_type":"application/xhtml+xml","content_length":"53178","record_id":"<urn:uuid:79f17c33-c96d-4385-8d6e-17b1bf5f899f>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00066.warc.gz"} |
495 Femtometer/Second Squared to Feet/Minute/Second
Femtometer/Second Squared [fm/s2] Output
495 femtometer/second squared in meter/second squared is equal to 4.95e-13
495 femtometer/second squared in attometer/second squared is equal to 495000
495 femtometer/second squared in centimeter/second squared is equal to 4.95e-11
495 femtometer/second squared in decimeter/second squared is equal to 4.95e-12
495 femtometer/second squared in dekameter/second squared is equal to 4.95e-14
495 femtometer/second squared in hectometer/second squared is equal to 4.95e-15
495 femtometer/second squared in kilometer/second squared is equal to 4.95e-16
495 femtometer/second squared in micrometer/second squared is equal to 4.95e-7
495 femtometer/second squared in millimeter/second squared is equal to 4.95e-10
495 femtometer/second squared in nanometer/second squared is equal to 0.000495
495 femtometer/second squared in picometer/second squared is equal to 0.495
495 femtometer/second squared in meter/hour squared is equal to 0.0000064152
495 femtometer/second squared in millimeter/hour squared is equal to 0.0064152
495 femtometer/second squared in centimeter/hour squared is equal to 0.00064152
495 femtometer/second squared in kilometer/hour squared is equal to 6.4152e-9
495 femtometer/second squared in meter/minute squared is equal to 1.782e-9
495 femtometer/second squared in millimeter/minute squared is equal to 0.000001782
495 femtometer/second squared in centimeter/minute squared is equal to 1.782e-7
495 femtometer/second squared in kilometer/minute squared is equal to 1.782e-12
495 femtometer/second squared in kilometer/hour/second is equal to 1.782e-12
495 femtometer/second squared in inch/hour/minute is equal to 0.0000042094488188976
495 femtometer/second squared in inch/hour/second is equal to 7.0157480314961e-8
495 femtometer/second squared in inch/minute/second is equal to 1.1692913385827e-9
495 femtometer/second squared in inch/hour squared is equal to 0.00025256692913386
495 femtometer/second squared in inch/minute squared is equal to 7.0157480314961e-8
495 femtometer/second squared in inch/second squared is equal to 1.9488188976378e-11
495 femtometer/second squared in feet/hour/minute is equal to 3.507874015748e-7
495 femtometer/second squared in feet/hour/second is equal to 5.8464566929134e-9
495 femtometer/second squared in feet/minute/second is equal to 9.744094488189e-11
495 femtometer/second squared in feet/hour squared is equal to 0.000021047244094488
495 femtometer/second squared in feet/minute squared is equal to 5.8464566929134e-9
495 femtometer/second squared in feet/second squared is equal to 1.6240157480315e-12
495 femtometer/second squared in knot/hour is equal to 3.463930899e-9
495 femtometer/second squared in knot/minute is equal to 5.773218165e-11
495 femtometer/second squared in knot/second is equal to 9.622030275e-13
495 femtometer/second squared in knot/millisecond is equal to 9.622030275e-16
495 femtometer/second squared in mile/hour/minute is equal to 6.6437007874016e-11
495 femtometer/second squared in mile/hour/second is equal to 1.1072834645669e-12
495 femtometer/second squared in mile/hour squared is equal to 3.9862204724409e-9
495 femtometer/second squared in mile/minute squared is equal to 1.1072834645669e-12
495 femtometer/second squared in mile/second squared is equal to 3.0757874015748e-16
495 femtometer/second squared in yard/second squared is equal to 5.4133858267717e-13
495 femtometer/second squared in gal is equal to 4.95e-11
495 femtometer/second squared in galileo is equal to 4.95e-11
495 femtometer/second squared in centigal is equal to 4.95e-9
495 femtometer/second squared in decigal is equal to 4.95e-10
495 femtometer/second squared in g-unit is equal to 5.0475952542407e-14
495 femtometer/second squared in gn is equal to 5.0475952542407e-14
495 femtometer/second squared in gravity is equal to 5.0475952542407e-14
495 femtometer/second squared in milligal is equal to 4.95e-8
495 femtometer/second squared in kilogal is equal to 4.95e-14 | {"url":"https://hextobinary.com/unit/acceleration/from/fms2/to/ftmins/495","timestamp":"2024-11-04T18:33:56Z","content_type":"text/html","content_length":"97974","record_id":"<urn:uuid:c9778fcc-3858-4ae2-a3a6-c71fc0a45396>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00319.warc.gz"} |
Solutions to Stochastic Processes Ch.7
《随机过程-第二版》（英文电子版） Sheldon M. Ross 答案整理，此书作为随机过程经典教材没有习题讲解，所以将自己的学习过程记录下来，部分习题解答参考了网络，由于来源很多，出处很多也不明确，无法一一注明，
Solutions to Stochastic Processes Sheldon M. Ross Second Edition(pdf)
Since there is no official solution manual for this book, I
handcrafted the solutions by myself. Some solutions were referred from web, most copyright of which are implicit, can’t be listed clearly. Many thanks to those authors! Hope these solutions be
helpful, but No Correctness or Accuracy Guaranteed. Comments are welcomed. Excerpts and links may be used, provided that full and clear credit is given.
7.1 Consider the following model for the flow of water in and out of a dam. Suppose that, during day \(n, Y_n\) units of water flow into the dam from outside sources such as rainfall and river flow.
At the end of each day, water is released from the dam according to the following rule: If the water content of the dam is greater than \(a\), then the amount of \(a\) is released. If it is less than
or equal to \(a\), then the total contents of the dam are released. The capacity of the dam is \(C\), and once at capacity any additional water that attempts to enter the dam is assumed lost. Thus,
for instance, if the water level at the beginning of day \(n\) is \(x\), then the level at the end of the day (before any water is released) is \(min(x + Y_n, C)\). Let \(S_n\) denote the amount of
water in the dam immediately after the water been released at the end of day \(n\). Assuming that the \(Y_n, n \geq 1\), are independent and identically distributed, show that \(\{S_n, n \geq 1\}\)
is a random walk with reflecting barriers at 0 and \(C-a\).
7.2 Let \(X_1, \dots, X_n\) be equally likely to be any of the \(n!\) permutations of \((1,2,\dots, n)\). Argue that,
$$P\{\sum_{j=1}^njX_j \leq a\} = P\{\sum_{j=1}^njX_j\geq n(n+1)^2/2 -a\}$$
P\{\sum_{j=1}^njX_j \leq a \} &= P\{nS_n – \sum_{i=1}^{n-1}S_i \leq a\} \\
&= P\{\sum_{i=1}^nS_i \geq (n+1)S_n – a\} \\
&= P\{\sum_{j=1}^njX_j\geq n(n+1)^2/2 -a\} \end{align}$$
7.3 For the simple random walk compute the expected number of visits to state \(k\).
Suppose \(p \geq 1/2\), and starting at state \(i\). When \(i \leq k\),
$$p_{ik} = 1 \\
f_{kk} = 2 -2p\\
E = 1 + \frac{1}{2p – 1}$$
When \(i > k\)
$$p_{ik} = (\frac{1-p}{p})^{i – k}\\
f_{kk} = 2 – 2p\\
E = p_{ik}(1 + \frac{1}{2p-1})$$
7.4 Let \(X_1, X_2, \dots, X_n\) be exchangeable. Compute \(E[X_1|X_{(1)}, X_{(2)}, \dots, X_{(n)}]\), where \(X_{(1)} \leq X_{(2)} \leq \dots \leq X_{(n)}\) are the \(X_i\) in ordered arrangement.
Since \(X_i\) is exchangeable, \(X_1\) can be any of \(X_{(i)}\) with equal probability. Thus,
$$E[X_1|X_{(1)}, X_{(2)}, \dots, X_{(n)}] = \frac{1}{n}\sum_{i=1}^nX_{(i)}$$
7.6 An ordinary deck of cards is randomly shuffled and then the cards are exposed one at a time. At some time before all the cards have been exposed you must say “next”, and if the next card exposed
is a spade then you win and if not then you lose. For any strategy, show that at the moment you call “next” the conditional probability that you win is equal to the conditional probability that the
last card is spade. Conclude from this that the probability of winning is 1/4 for all strategies.
Let \(X_n\) indicate if the nth card is a spade and \(Z_n\) be the proportion of spades in the remaining cards after the \(n\) card. Thus \(E|Z_n| < \infty\) and
$$E[Z_{n+1}|Z_1, \dots , Z_n] = \frac{(52 -n)Z_n – 1}{52 -n-1}Z_n + \frac{(52-n)Z_n}{52-n-1}(1 – Z_n) = Z_n$$
Hence \(Z_n\) is a martingale.
Note that \(X_52 = Z_51\). Thus
E[X_{n+1}|X_1, \dots, X_n]&= E[X_{n+1}|Z_1, \dots, Z_n] = Z_n\\
&= E[Z_51|Z_1, \dots, Z_n] = E[X_52|X_1, \dots, X_n]
Finally, let \(N\) be the stopping time corresponding to saying “next” for a given strategy.
P\{\text{Win}\} &= E[X_{N+1}] = E[E[X_{N+1}|N]] \\
&= E[Z_N] = E[Z_1] = 1/4
7.7 Argue that the random walk for which \(X_i\) only assumes the values \(0, \pm 1, \dots, \pm M\) and \(E[X_i] = 0\) is null recurrent.
7.8 Let \(S_n, n \geq 0\) denote a random walk for which
$$\mu = E[S_{n+1} – S_n] \neq 0$$
Let, for \(A >0, B > 0\),
$$N = min\{n: S_n \geq A \text{ or } S_n \leq -B\}$$
Show that \(E[N] < \infty\). (Hint: Argue that there exists a value \(k\) such that \(P\{S_k > A +B\} > 0\). Then show that \(E[N] \leq kE[G]\), where \(G\) is an appropriately defined geometric
random variable.)
Suppose \(\mu > 0\), and let \(k > (A+B)/\mu\), then
k\mu – A -B &= E[S_k – A – B] \\
&= E[S_k – A – B|S_k > A + B]P\{S_k > A+B\} \\&+ E[S_k – A – B|S_k \leq A + B]P\{S_k \leq A+B\}\\
&\leq E[S_k – A – B|S_k > A + B]P\{S_k > A+B\}
\end{align}$$ Thus, there exists \(k > (A+B)/\mu, p = P\{S_k > A +B\} > 0\). Let \(Y_i = \sum_{j = ik+1}^{(i+1)k} X_j\), then \(P\{Y_i > A + B\} = p\). And it’s obviously that if any of \(Y_i\)
exceeds \(A+B\), \(S_N\) occurs. Hence, \(E[N] \leq k/p\)
7.10 In the insurance ruin problem of Section 7.4 explain why the company will eventually be ruined with probability 1 if \(E[Y] \geq cE[X]\).
7.11 In the ruin problem of Section 7.4 let \(F\) denote the interarrival distribution of claims and let \(G\) be the distribution of the size of a claim. Show that \(p(A)\), the probability that a
company starting with \(A\) units of assets is ever ruined, satifies
$$p(A) = \int_0^{\infty}\int_0^{A + ct}p(A + ct -x)dG(x)dF(t) + \int_0^{\infty}\bar{G}(A+ct)dF(t)$$
Condition on the first claim, then
p(A) &= P\{\text{ruined at first claim}\} + P\{\text{ruined after first claim}\} \\
&= \int_0^{\infty}\bar{G}(A+ct)dF(t) + \int_0^{\infty}\int_0^{A + ct}p(A + ct -x)dG(x)dF(t)
7.12 For a random walk with \(\mu = E[X] > 0\) argue that, with probability 1,
$$\frac{u(t)}{t} \to \frac{1}{\mu} \quad \text{as } t \to \infty$$
where \(u(t)\) equals the number of \(n\) for which \(0 \leq S_n \leq t\).
7.13 Let \(S_n = \sum_{i=1}^n X_i\) be a random walk and let \(\lambda_i, i > 0\), denote the probability that a ladder height equals \(i\) — that is, \(\lambda_i = P\){first positive value of \(S_n
\) equals \(i\)}.
(a) Show that if
$$P\{X_i = j\} = \left\{\begin{array}{ll}
q \quad j = -1 \\
\alpha_j \quad j \geq 1 \\ \end{array}\right. \\
q + \sum_{j=1}^{\infty} \alpha_j = 1$$
then \(\lambda_i\) satisfies
$$\lambda_i = \alpha_i + q(\lambda_{i+1} + \lambda_1\lambda_i) \quad i > 0$$
(b) If \(P\{X_i = j\} = 1/5, j = -2,-1,0,1,2\), show that
$$\lambda_1 = \frac{1+\sqrt{5}}{3+\sqrt{5}} \quad \lambda_2 = \frac{2}{3+\sqrt{5}} $$
7.14 Let \(S_n, n\geq 0\), denote a random walk in which \(X_i\) has distribution \(F\). Let \(G(t,s)\) denote the probability that the first value of \(S_n\) that exceeds \(t\) is less than or equal
to \(t+s\). That is,
$$G(t,s) = P\{\text{first sum exceeding } t \text{ is } \leq t+s\}$$
Show that
$$G(t, s) = F(t + s) – F(t) + \int_{-\infty}^t G(t-y, s)dF(y)$$
\(S_n|X_1\) is distributed as \(X_1 + S_{n-1}\). Thus if \(A\)={first sum exceeding \(t\) is \(\leq t + s\)},
G(t,s) &\equiv P\{A\} = E[P\{A|X_1\}] \\
&= F(t+s) – F(t) + \int_{-\infty}^t G(t-y, s)dF(y) | {"url":"http://www.charmpeach.com/stochastic-processes/solutions-to-stochastic-processes-ch-7/977/","timestamp":"2024-11-07T12:49:47Z","content_type":"text/html","content_length":"88047","record_id":"<urn:uuid:dc043513-c2c7-4bed-bde0-b1d3822b2522>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00653.warc.gz"} |
[11:30am] Nitin Nitsure, Bhaskaracharya Pratishthana, Pune
Date Tuesday, 26 March, 11.30 am
Venue Room 215
11:00am Host Sudarshan Gurjar
speaker Nitin Nitsure
Affiliation Bhaskaracharya Pratishthana
Title Galois descent and Galois cohomology: examples and calculations.
We will begin by recalling the calculus of Galois descent, and its translation into Galois cohomology classes for twisted forms. After that, we will look at common examples,
Abstract involving Galois twisted forms of algebraic groups, Brauer Severi varieties, and vector bundles (Hilbert 90). The final example will be the Galois twisted from over reals of
a non-separated affine line with doubled origin. This twisted form is an algebraic space which is not a scheme.
[2:30pm] Sudarshan Gurjar, IIT Bombay
Topology and
Related Topics
Date Tuesday 26 March, 2:30 pm
Venue Ramanujan Hall
2:00pm Host Rekha Santhanam
speaker Sudarshan Gurjar
Affiliation IIT Bombay
Title Vector bundles and Characteristic Classes
This is the second talk in the series of three talks. We will give an introduction to the characteristic classes of a vector bundle. Characteristic classes are
Abstract invariants of a vector bundle taking values in the singular cohomology of the base and satisfying the obvious functoriality property with respect to pullback. They are
the measure of the non-triviality of the vector bundle.
The background assumed will depend on the audience present.
[4:00pm] R. V. Gurjar, IIT Bombay
Date Tuesday, 26 March 2024, 4-5 pm
Venue Room 215
4:00pm Host Tony J. Puthenpurakal
speaker R. V. Gurjar
Affiliation IIT Bombay
Title Some more results about Brieskorn-Pham singularities-II
We will discuss the following results. (1) Classification of 3-dimensional factorial B-P singularities by U.Storch. (2) Classification of affine B-P 3-folds which admit a
non-trivial action of the additive group G_a (equivalently, a non-zero locally nilpotent derivation on their coordinate ring) by M. Chitayat. (3) Classification of affine
B-P 3-folds which are rational (i.e. their function field is a purely transcendental extension of the ground field) by M Chitayat. | {"url":"https://www.math.iitb.ac.in/webcal/day.php?date=20240326","timestamp":"2024-11-11T04:35:29Z","content_type":"text/html","content_length":"27384","record_id":"<urn:uuid:e2dc878a-48e2-4e6d-b38d-779c57a5c772>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00171.warc.gz"} |
How to use dimensionality reduction techniques in machine learning?
Dimensionality reduction techniques are useful in machine learning when the original dataset has a large number of features, which can increase the complexity of the model and result in longer
training times, more memory usage, and potential overfitting. Here are some common methods for using dimensionality reduction techniques in machine learning:
1. Principal Component Analysis (PCA): PCA is a statistical method that transforms a larger set of variables into a smaller set of uncorrelated variables, known as principal components. These
components explain the maximum amount of variance in the data, making them an effective tool for dimensionality reduction.
2. Linear Discriminant Analysis (LDA): LDA is a supervised method that identifies linear combinations of features that best separate the classes or groups in the dataset. This technique is commonly
used for classification problems where the features have high correlation with each other.
3. t-SNE: t-SNE is a nonlinear method that reduces the dimensions of the dataset while preserving the local structure of the data. This technique is commonly used for visualizing high-dimensional
data in two or three dimensions.
4. Autoencoders: Autoencoders are a neural network-based approach to dimensionality reduction that can learn a compressed representation of the input data. They work by mapping the input data to a
lower-dimensional space and then reconstructing the original data from this latent space.
In summary, dimensionality reduction techniques can help simplify complex datasets and improve the performance of machine learning models. The choice of technique depends on the characteristics of
the dataset and the specific task at hand. | {"url":"https://devhubby.com/thread/how-to-use-dimensionality-reduction-techniques-in","timestamp":"2024-11-08T16:11:22Z","content_type":"text/html","content_length":"114042","record_id":"<urn:uuid:fc1ba1a3-04be-40c6-b5a0-070721c056f0>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00192.warc.gz"} |
The Relationship Between Mathematics & Physics
Understanding the relationship can eliminate much confusion.
In mathematics you are free to construct anything you like as long as it's consistent. So for example, in mathematics n-dimensional spaces really exist. They are well defined mathematical objects.
Physics uses mathematical objects to model reality - with varying levels of success. The fact that physics may use a mathematical object to model reality does not mean that object now exists outside
of mathematics. It doesn't. It simply means the object is useful in modeling reality.
So for example, do n-dimensional spaces "really" exist? The question is meaningless. They exist in mathematics, and they are useful in modeling reality. But to ask if they "really" exist is a
meaningless question.
Heck, it's even conceivable that something other that mathematics will ultimately be better at modeling reality. Very unlikely, but conceivable.
Content written and posted by Ken Abbott | {"url":"http://www.math-math.com/2018/07/the-relationship-between-mathematics.html","timestamp":"2024-11-12T20:26:18Z","content_type":"text/html","content_length":"32787","record_id":"<urn:uuid:094f30a7-9b18-47af-bb3c-8aed75e01efa>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00440.warc.gz"} |
Math Expression: Learn about raphing linear equations
Graphs: Graphing Linear Equations
Lesson Objective
This lesson shows you the steps on plotting (i.e. graphing) the graph of linear equations.
Why Learn This?
If you are given a simple equation and have been asked to visualize it, you will very likely able to do it mentally.
Now, if the equation gets more complex, you will soon find it difficult to visualize it mentally. This is where the coordinate plane is here to help.
With several simple steps, you will be able to draw out any equation. As a start, we will use the equation for straight lines.
Note that, straight line equation is also known as linear equation.
Study Tips
As you watch the math video below, try to remember the following:
Tip #1
You need to draw a x-y table to help you to record the coordinates of your points in orderly manner.
Tip #2
Take note on how the coordinates of each point are calculated.
Tip #3
In this lesson, only few points are used to plot the equation. For more complicated equations, you may need to get more points to have a complete picture of the equation.
Now, watch the following math video to know more.
Math Video
Math Video Transcript
Welcome to a new lesson. After mastering on how to read and write coordinates, we can now use it to visualize the graph a linear equation.
Let's say you are given the equation y = 2x + 1. This equation, by itself, may not makes much sense.
But if we were to use the Coordinate plane to plot a graph of this equation, you get to actually see how this equation looks like.
To do so, first, you have to draw out a table like this. This table is useful to store the coordinates of the point that you are going to calculate soon.
Now, let the variable x to take up the first row of this table.
Similarly, let the variable y take up the second row. With this, you can see that, these empty column here will used to store coordinates of one point.
Now let's start filling this table. Let me put the value, 1 in the box right here. By doing so, you will have a point on the plane with it's x-coordinate = 1.
But you have a problem. Though you have the x-coordinate as 1, you still do not know what is the y coordinate.
This is where you will have to use this equation, to find out the y coordinate of this point when x = 1.
Now, substitute x=1 right here, solve for y by first multiplying 2 with 1. You will get 2. 2 plus 1 gives you 3. Let's put this new value y = 3 into the table.
It now means that, when this point's x coordinate = 1, the value of y coordinate for this equation is 3.
With this newly calculated point, let us now adjust this point to the correct coordinates (1,3).
Alright, you now have your first point from y = 2x + 1.
To continue further, we will do the same thing for x = 2, x = 3 and x = 4.
For x =2 , we will now solve this equation. Substitute x=2. multiply 2 with 2 gives 4 .Add 4 with 1. You will get y = 5. Let's put this value into the table. Now, you have a new coordinates (2,5).
This point is located right over here.
For x =3 , we will now solve this equation. Substitute x =3..Multiply 2 with 3 gives 6. Add 6 with 1. You'll get y = 7. Let's put this value into the table. Now, you have a new coordinates (3,7).
This point is located right over here.
For x =4 , we will now solve this equation. Substitute x =4. Multiply 2 with 4 gives 8. Add 8 with 1. You'll get y = 9. Let's put this value into the table. Now, you have a new coordinates (4,9).
This point is located right over here.
You see that these points forms a pattern here.
Now, when we connect these points together, we will get a straight line. This line is the graph of y = 2x + 1.
So, what you have learn showed you how draw the graph of a linear equation.
Alright, that is all for this lesson. You can move on to the practice questions to test your understanding.
Practice Questions & More
Multiple Choice Questions (MCQ)
Now, let's try some MCQ questions to understand this lesson better.
You can start by going through the series of questions on graphing linear equations or pick your choice of question below.
• Question 1 on using a linear equation to calculate y-coordinates
• Question 2 on the basics of graphing linear equations
Site-Search and Q&A Library
Please feel free to visit the Q&A Library. You can read the Q&As listed in any of the available categories such as Algebra, Graphs, Exponents and more. Also, you can submit math question, share or
give comments there. | {"url":"https://www.mathexpression.com/graphing.html","timestamp":"2024-11-08T21:31:01Z","content_type":"text/html","content_length":"32742","record_id":"<urn:uuid:13cbeb3a-c7b0-4624-bc35-fe2369a2db72>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00624.warc.gz"} |
A Quadrafilar Helical Antenna for Low Elevation GPS Applications
This article presents a low cost implementation of a quadrafilar helical antenna for Global Positioning System (GPS) applications. The antenna is designed to produce a right-hand circularly polarized
(RHCP), elevated omnidirectional radiation pattern with constant azimuth gain at low elevation angles. This type of antenna is ideal for satellite communications (SATCOM) ground terminals, which
operate at low elevation angles with respect to the satellite position. The antenna is designed to operate in a typical SATCOM receive band from 1525 to 1559 MHz and is specifically intended for use
in differential GPS (DGPS) network applications.
Senglee Foo
CAL Corp.
Ottawa, Ontario, Canada
DGPS technologies are being used extensively for various commercial services related to geophysical surveying and mapping, geographic information management, construction and farming. For such
applications at remote sites, these services rely on a constant link to a SATCOM satellite to receive the precise DGPS information. Antennas with low elevation radiation patterns allow these
receivers to operate at low tracking angles relative to the satellite with a substantial improvement in the signal-to-noise ratio compared to a typical zenith pointing antenna. A RHCP, elevated
omnidirectional radiation pattern is achieved by using a quadrafilar helix with a long element length (longer than one wavelength).
The Antenna Configuration
The concept of a resonant, fractional-turn, quadrafilar helix (also known as a volute antenna) was first introduced by Kilgus.^1 It was shown that this antenna can produce a highly directive
circularly polarized radiation pattern by feeding a proper complex excitation to the four helices. It was later shown^2 that a shaped-conical pattern also can be achieved by using helices of an
integral number of turns and varying pitches. Since then, several derivatives of the concept have evolved. This article extends this antenna concept to produce an elevated omnidirectional radiation
pattern with dielectric-loaded four-arm helices.
A conventional volute antenna consists of two to four helices of short element length (a multiple of a quarter-wavelength). Typically, these designs form one-quarter-, one-half- or full-turn
quadrafilar helical antennas. To obtain the elevated beam peak angle in this design, the overall element length of each helix is slightly longer than a conventional volute antenna helix and has an
effective length of approximately one-and-a-quarter wavelength. One side of the helices is connected directly to the feeds while the other terminals are left open. One major advantage of this type of
antenna is that it allows the setting of the beam pointing angle by selecting an appropriate overall helix length and pitch angles. The actual length and pitch angle of the helices are adjusted to
compensate for the loading of the dielectric tubing. Figure 1 shows the physical configuration of the antenna. The design of the quadrafilar helix is based on the concept of four helices operating in
a quadrature phase separation, that is, Df = ± (m – 1)p /4, where m is the helix number. The sign of the phase difference between the helices and the winding direction of the helices are related
intimately. One direction of the phase difference produces cardioid-shaped radiation patterns over the front hemisphere; the other direction produces the desired elevated omnidirectional patterns.
The sign of the phase difference between helices also depends on the winding direction of the helices. To obtain an elevated omnidirectional pattern, the sign of the phase difference and the
direction of the helix winding occur in the opposite direction. For instance, a Df = +90° phase difference between the helices in the clockwise direction produces a RHCP omnidirectional pattern if
the helices are wound in a left-hand sense. Similarly, the same RHCP radiation pattern also can be obtained using right-hand-wound helices by implementing the phase difference between the helices in
the exact opposite direction (Df = –90°). Figure 2 shows the design. The four helices are wound on thin cylindrical polyvinylchloride (PVC) dielectric tubing. In this case, the helices are wound in a
left-hand sense and the feed network is designed to produce a +90° phase difference in the clockwise direction. The feed network is printed on a thin, low loss, woven-glass polytetraflouroethylene
laminate. The feed input of each helix is soldered directly to the circuit side of the microstrip feed network. This microstrip feed network provides a simple means of achieving a low cost balun/
combiner. This circuit is designed and optimized using the Series IV microwave CAD program. Amplitude and phase balances of this circuit are critical. If all the helices are impedance matched to the
50 W microstrip line, an amplitude balance of less than ± 0.25 dB and phase balance of less than 10° can be achieved. In practice, this precision can be achieved by tuning each helix element with all
other helices in place and terminated to a 50 W load. By doing so, the effect of mutual coupling between the helices is included in the matching process. It usually takes a few iterations to achieve
the match condition for all ports.
Impedance Matching
Each of the helix elements is matched to its corresponding 50 W microstrip feed line by fine-tuning the helix element length and shaping the first quarter turn of the helix winding during
prototyping. The pitch angle of this first quarter turn is designed to act as a direct impedance transformer between the helix winding and the 50 W microstrip transmission line. The shape of this
matching section is critical to obtaining a good input SWR. The coordinate of this matching section must be measured and reproduced precisely using a mandrel for volume production. Using this
technique, an SWR of less than 1.3 can be achieved over a five percent frequency bandwidth.
The EM Model and Simulation
The design was modeled and optimized using moment-method-based electromagnetic (EM) codes. The computer modeling does not include the dielectric loading, which will be corrected by using a
measurement method. During the design, two software tools were used for the simulation. One tool is a moment method using composite wire and plate structures (WIPL-4) developed at the University of
Belgrade, Yugoslavia. The other tool is a moment method using a thin wire model (NEC-2) developed by Lawrence Livermore Laboratory, CA. Figure 3 shows the electrical model. The two models are
identical except for the modeling of the finite ground plane. In the WIPL model, the ground plane is modeled as a single metal plate; the NEC model approximates the ground plane using a wire mesh
model. Both methods are based on the solution of the electric field integral equation method. However, the WIPL code uses high order polynomial basis functions to approximate the current distribution
in the wires and plates; the NEC model uses sinusoidal basis functions. A total of 300 and 472 unknowns are present in the WIPL and NEC models, respectively. These two models are run on a 90 MHz 486
DOS platform. The total run time is 51 s for the WIPL-4 simulation and 59 s for the NEC simulation. Figure 4 shows a comparison of the co-polar radiation patterns (principal plane) generated by the
two models. As shown, the results are closely related. The peak gain between the two models is within 0.5 dB. The pattern difference within the main beam is relatively small. The larger pattern
difference at low gain angle, in the null area, is a direct result of the differences in the modeling of the ground plane. The wire mesh approximation in the NEC model cannot approximate the surface
current distribution on the ground plane as accurately as the WIPL model, especially in the excitation regions. Figure 5 shows the contour plots of the co-polar and cross-polar radiation patterns of
the antenna generated by the NEC model. In this case, the antenna is designed to provide an elevation beam peak approximately 30° from the horizon and produce an omnidirectional pattern with a
minimum of 3.4 dBiC gain in the azimuth plane at the elevation beam peak angle. This antenna exhibits a relatively constant gain in the azimuth direction. Typically, the left-hand cross-polar
component within the 3 dB beamwidth is below –16 dB, which produces an axial ratio of better than 2.77 dB. These characteristics also allow the antenna to be used for low elevation communications
without requiring azimuth tracking.
Dielectric Loading and Measurement Results
The effect of dielectric loading on a helical antenna has been investigated extensively.^3,4 The loading of the dielectric shell alters the phase velocity of the RF current along the helix windings.
As a result, it also severely alters the radiation patterns of the antenna. To correct for this effect, the pitch angle and overall helix length are reduced slightly to compensate. Figure 6 shows the
measured radiation patterns of the compensated and uncompensated antennas compared to the NEC result. In this case, the dielectric cylinder has a wall thickness of less than a few millimeters with a
dielectric constant of less than 4. In the uncompensated case, the dielectric loading causes the radiation pattern to peak at a lower elevation angle and suffer a significant loss in the peak gain. A
reduction in the helix length and pitch angle effectively shortens the electrical path length of the helices and compensates for the phase error. By using measurement results, an appropriate change
in the pitch angle and helix length can be determined. Note that the null at the zenith is also partially filled in the compensated case. Figure 7 shows the measured return loss at the input of a
typical quadrafilar antenna. This type of antenna can offer a relatively wide frequency bandwidth. In this case, an SWR of 1.2 is achieved over a 10 percent frequency bandwidth.
The design and low cost implementation of a quadrafilar helical antenna have been presented. The design uses four electrically long (greater than one wavelength) helical elements. It has been shown
that, with proper phase excitations at the four input ports, an elevated omnidirectional beam pattern with a beam peak at the desired elevation angle can be obtained. The design of a simple
microstrip power combiner including the quadrature phase distribution was shown to provide the proper balun/power-combining network. It was also shown that the effect of the dielectric loading is
significant and can be compensated by varying the pitch angles and helix length.
The Series IV microwave CAD program is a product of HPEEsof, Santa Rosa, CA.
1. C.C. Kilgus, "Resonant Quadrafilar Helix," IEEE Transactions on Antennas and Propagations, Vol. AP-17, May 1969, pp. 349–451.
2. C.C. Kilgus, "Shaped-conical Radiation Pattern Performance of the Backfire Quadrafilar Helix," IEEE Transactions on Antennas and Propagations, Vol. AP-23, May 1975, pp. 392–397.
3. S. Foo, P. Wood and P. Cowles, "A Low Cost, Lightweight, Aeronautical SATCOM Antenna," Microwave Journal, Vol. 40, No. 1, January 1997, pp. 166–173.
4. D.E. Baker, "Design of a Broadband Impedance-matching Section for Peripherally Fed Helical Antennas," Antenna Applications Symposium Proceedings, University of Illinois, September 1980. | {"url":"https://www.microwavejournal.com/articles/2229-a-quadrafilar-helical-antenna-for-low-elevation-gps-applications","timestamp":"2024-11-14T18:02:21Z","content_type":"text/html","content_length":"71300","record_id":"<urn:uuid:af2b5425-168e-4691-9993-ee27196c07e7>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00778.warc.gz"} |
How do you solve (x-6)^2 = 25? | HIX Tutor
How do you solve # (x-6)^2 = 25#?
Answer 1
Rewriting #(x-6)^2=25# as #(x-6)^2-5^2=0#
we see the left side is the difference of squares with factors: #((x-6)+5)*((x-6)-5) = 0#
Either #{: (,(x-6)+5 = 0, ," or ",,(x-6)-5=0), (rArr,x-1=0, , , ,x-11=0), (rArr,x=1, ," or ", ,x=11) :}#
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Answer 2
To solve the equation (x-6)^2 = 25:
1. Expand the left side of the equation: (x-6)^2 = (x-6)(x-6) = x^2 - 12x + 36.
2. Set the expanded expression equal to 25: x^2 - 12x + 36 = 25.
3. Rearrange the equation by subtracting 25 from both sides: x^2 - 12x + 36 - 25 = 0, which simplifies to x^2 - 12x + 11 = 0.
4. Factor the quadratic equation: (x - 11)(x - 1) = 0.
5. Set each factor equal to zero and solve for x:
□ x - 11 = 0 => x = 11
□ x - 1 = 0 => x = 1.
So, the solutions to the equation are x = 11 and x = 1.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Texture Classification with Wavelet Image Scattering
This example shows how to classify textures using wavelet image scattering. In addition to Wavelet Toolbox™, this example also requires Parallel Computing Toolbox™ and Image Processing Toolbox™.
In a digital image, texture provides information about the spatial arrangement of color or pixel intensities. Particular spatial arrangements of color or pixel intensities correspond to different
appearances and consistencies of the physical material being imaged. Texture classification and segmentation of images has a number of important application areas. A particularly important example is
biomedical image analysis where normal and pathologic states are often characterized by morphological and histological characteristics which manifest as differences in texture [4].
Wavelet Image Scattering
For classification problems, it is often useful to map the data into some alternative representation which discards irrelevant information while retaining the discriminative properties of each class.
Wavelet image scattering constructs low-variance representations of images which are insensitive to translations and small deformations. Because translations and small deformations in the image do
not affect class membership, scattering transform coefficients provide features from which you can build robust classification models.
Wavelet scattering works by cascading the image through a series of wavelet transforms, nonlinearities, and averaging [1][3][5]. The result of this deep feature extraction is that images in the same
class are moved closer to each other in the scattering transform representation, while images belonging to different classes are moved farther apart.
This example uses a publicly available texture database, the KTH-TIPS (Textures under varying Illumination, Pose, and Scale) image database [6]. The KTH-TIPS dataset used in this example is the
grayscale version. There are 810 images in total with 10 textures and 81 images per texture. The majority of images are 200-by-200 in size. This example assumes you have downloaded the KTH-TIPS
grayscale dataset and untarred it so that the 10 texture classes are contained in separate subfolders of a common folder. Each subfolder is named for the class of textures it contains. Untarring the
downloaded kth_tips_grey_200x200.tar file is sufficient to provide a top-level folder KTH_TIPS and the required subfolder structure.
Use the imageDatastore to read the data. Set the location property of the imageDatastore to the folder containing the KTH-TIPS database that you have access to.
location = fullfile(tempdir,'kth_tips_grey_200x200','KTH_TIPS');
Imds = imageDatastore(location,'IncludeSubFolders',true,'FileExtensions','.png','LabelSource','foldernames');
Randomly select and visualize 20 images from the dataset.
numImages = 810;
perm = randperm(numImages,20);
for np = 1:20
im = imread(Imds.Files{perm(np)});
colormap gray; axis off;
Texture Classification
This example uses MATLAB®'s parallel processing capability through the tall array interface. Start the parallel pool if one is not currently running.
if isempty(gcp)
Starting parallel pool (parpool) using the 'local' profile ...
Connected to the parallel pool (number of workers: 6).
For reproducibility, set the random number generator. Shuffle the files of the KTH-TIPS dataset and split the 810 images into two randomly selected sets, one for training and one held-out set for
testing. Use approximately 80% of the images for building a predictive model from the scattering transform and use the remainder for testing the model.
Imds = imageDatastore(location,'IncludeSubFolders',true,'FileExtensions','.png','LabelSource','foldernames');
Imds = shuffle(Imds);
[trainImds,testImds] = splitEachLabel(Imds,0.8);
We now have two datasets. The training set consists of 650 images, with 65 images per texture. The testing set consists of 160 images, with 16 images per texture. To verify, count the labels in each
ans=10×2 table
Label Count
______________ _____
aluminium_foil 65
brown_bread 65
corduroy 65
cotton 65
cracker 65
linen 65
orange_peel 65
sandpaper 65
sponge 65
styrofoam 65
ans=10×2 table
Label Count
______________ _____
aluminium_foil 16
brown_bread 16
corduroy 16
cotton 16
cracker 16
linen 16
orange_peel 16
sandpaper 16
sponge 16
styrofoam 16
Create tall arrays for the resized images.
Ttrain = tall(trainImds);
Ttest = tall(testImds);
Create a scattering framework for an image input size of 200-by-200 with an InvarianceScale of 150. The invariance scale hyperparameter is the only one we set in this example. For the other
hyperparameters of the scattering transform, use the default values.
sn = waveletScattering2('ImageSize',[200 200],'InvarianceScale',150);
To extract features for classification for each the training and test sets, use the helperScatImages_mean function. The code for helperScatImages_mean is at the end of this example.
helperScatImages_mean resizes the images to a common 200-by-200 size and uses the scattering framework, sn, to obtain the feature matrix. In this case, each feature matrix is 391-by-7-by-7. There are
391 scattering paths and each scattering coefficient image is 7-by-7. Finally, helperScatImages_mean obtains the mean along the 2nd and 3rd dimensions to obtain a 391 element feature vector for each
image. This is a significant reduction in data from 40,000 elements down to 391.
trainfeatures = cellfun(@(x)helperScatImages_mean(sn,x),Ttrain,'Uni',0);
testfeatures = cellfun(@(x)helperScatImages_mean(sn,x),Ttest,'Uni',0);
Using tall's gather capability, gather all the training and test feature vectors and concatenate them into matrices.
Trainf = gather(trainfeatures);
Evaluating tall expression using the Parallel Pool 'local':
- Pass 1 of 1: Completed in 1 min 39 sec
Evaluation completed in 1 min 39 sec
trainfeatures = cat(2,Trainf{:});
Testf = gather(testfeatures);
Evaluating tall expression using the Parallel Pool 'local':
- Pass 1 of 1: Completed in 23 sec
Evaluation completed in 23 sec
testfeatures = cat(2,Testf{:});
The previous code results in two matrices with row dimensions 391 and column dimension equal to the number of images in the training and test sets, respectively. So each column is a feature vector.
PCA Model and Prediction
This example constructs a simple classifier based on the principal components of the scattering feature vectors for each class. The classifier is implemented in the functions helperPCAModel and
helperPCAClassifier. The function helperPCAModel determines the principal components for each digit class based on the scattering features. The code for helperPCAModel is at the end of this example.
The function helperPCAClassifier classifies the held-out test data by finding the closest match (best projection) between the principal components of each test feature vector with the training set
and assigning the class accordingly. The code for helperPCAClassifier is at the end of this example.
model = helperPCAModel(trainfeatures,30,trainImds.Labels);
predlabels = helperPCAClassifier(testfeatures,model);
After constructing the model and classifying the test set, determine the accuracy of the test set classification.
accuracy = sum(testImds.Labels == predlabels)./numel(testImds.Labels)*100
We have achieved 99.375% correct classification, or a 0.625% error rate for the 160 images in the test set. A plot of the confusion matrix shows that our simple model misclassified one texture.
In this example, we used wavelet image scattering to create low-variance representations of textures for classification. Using the scattering transform and a simple principal components classifier,
we achieved 99.375% correct classification on a held-out test set. This result is comparable to state-of-the-art performance on the KTH-TIPS database.[2]
[1] Bruna, J., and S. Mallat. "Invariant Scattering Convolution Networks." IEEE Transactions on Pattern Analysis and Machine Intelligence. Vol. 35, Number 8, 2013, pp. 1872–1886.
[2] Hayman, E., B. Caputo, M. Fritz, and J. O. Eklundh. “On the Significance of Real-World Conditions for Material Classification.” In Computer Vision - ECCV 2004, edited by Tomás Pajdla and Jiří
Matas, 3024:253–66. Berlin, Heidelberg: Springer Berlin Heidelberg, 2004. https://doi.org/10.1007/978-3-540-24673-2_21.
[3] Mallat, S. "Group Invariant Scattering." Communications in Pure and Applied Mathematics. Vol. 65, Number 10, 2012, pp. 1331–1398.
[4] Pujol, O., and P. Radeva. “Supervised Texture Classification for Intravascular Tissue Characterization.” In Handbook of Biomedical Image Analysis, edited by Jasjit S. Suri, David L. Wilson, and
Swamy Laxminarayan, 57–109. Boston, MA: Springer US, 2005. https://doi.org/10.1007/0-306-48606-7_2.
[5] Sifre, L., and S. Mallat. "Rotation, scaling and deformation invariant scattering for texture discrimination." 2013 IEEE Conference on Computer Vision and Pattern Recognition. 2013, pp 1233–1240.
[6] KTH-TIPS image databases homepage. https://www.csc.kth.se/cvap/databases/kth-tips/
Appendix — Supporting Functions
function features = helperScatImages_mean(sf,x)
x = imresize(x,[200 200]);
smat = featureMatrix(sf,x);
features = mean(mean(smat,2),3);
function model = helperPCAModel(features,M,Labels)
% This function is only to support wavelet image scattering examples in
% Wavelet Toolbox. It may change or be removed in a future release.
% model = helperPCAModel(features,M,Labels)
% Copyright 2018 MathWorks
% Initialize structure array to hold the affine model
model = struct('Dim',[],'mu',[],'U',[],'Labels',categorical([]),'s',[]);
model.Dim = M;
% Obtain the number of classes
LabelCategories = categories(Labels);
Nclasses = numel(categories(Labels));
for kk = 1:Nclasses
Class = LabelCategories{kk};
% Find indices corresponding to each class
idxClass = Labels == Class;
% Extract feature vectors for each class
tmpFeatures = features(:,idxClass);
% Determine the mean for each class
model.mu{kk} = mean(tmpFeatures,2);
[model.U{kk},model.S{kk}] = scatPCA(tmpFeatures);
if size(model.U{kk},2) > M
model.U{kk} = model.U{kk}(:,1:M);
model.S{kk} = model.S{kk}(1:M);
model.Labels(kk) = Class;
function [u,s,v] = scatPCA(x,M)
% Calculate the principal components of x along the second dimension.
if nargin > 1 && M > 0
% If M is non-zero, calculate the first M principal components.
[u,s,v] = svds(x-sig_mean(x),M);
s = abs(diag(s)/sqrt(size(x,2)-1)).^2;
% Otherwise, calculate all the principal components.
% Each row is an observation, i.e. the number of scattering paths
% Each column is a class observation
[u,d] = eig(cov(x'));
[s,ind] = sort(diag(d),'descend');
u = u(:,ind);
function labels = helperPCAClassifier(features,model)
% This function is only to support wavelet image scattering examples in
% Wavelet Toolbox. It may change or be removed in a future release.
% model is a structure array with fields, M, mu, v, and Labels
% features is the matrix of test data which is Ns-by-L, Ns is the number of
% scattering paths and L is the number of test examples. Each column of
% features is a test example.
% Copyright 2018 MathWorks
labelIdx = determineClass(features,model);
labels = model.Labels(labelIdx);
% Returns as column vector to agree with imageDatastore Labels
labels = labels(:);
function labelIdx = determineClass(features,model)
% Determine number of classes
Nclasses = numel(model.Labels);
% Initialize error matrix
errMatrix = Inf(Nclasses,size(features,2));
for nc = 1:Nclasses
% class centroid
mu = model.mu{nc};
u = model.U{nc};
% 1-by-L
errMatrix(nc,:) = projectionError(features,mu,u);
% Determine minimum along class dimension
[~,labelIdx] = min(errMatrix,[],1);
function totalerr = projectionError(features,mu,u)
Npc = size(u,2);
L = size(features,2);
% Subtract class mean: Ns-by-L minus Ns-by-1
s = features-mu;
% 1-by-L
normSqX = sum(abs(s).^2,1)';
err = Inf(Npc+1,L);
err(1,:) = normSqX;
err(2:end,:) = -abs(u'*s).^2;
% 1-by-L
totalerr = sqrt(sum(err,1));
See Also
Related Examples
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A car slows from 22m/s to 3m/s with a constant acceleration of –2.1m/s2. How long does this require? | Socratic
A car slows from 22m/s to 3m/s with a constant acceleration of –2.1m/s2. How long does this require?
1 Answer
$v = u + a t$
Use above equation to find time ($t$)
#t = (v - u)/a = ("3 m/s "-" 22 m/s")/(-"2.1 m/s"^2) = "9.05 s"#
Impact of this question
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The writing bot Question. Read and write activity needs to be captured. - FcukTheCode
The writing bot Question. Read and write activity needs to be captured.
A scientist has created a writing bot, which will read from one book and write into another. Both books may have different dimensions.i.e. number of lines on each page and number of pages itself.
Read and Write also happen at different speeds.
The bot first reads from the first book fully, then processes the format to write into the second book(done instantaneously) and finally starts writing into the second book.
Your task is to identify, after a specified interval of time, if the bot is reading or writing.
For each of these activities how much read and write activity has happened needs to be captured in terms of pages and number of pages and number of lines on the current page.
Input has 7 lines
->The first line contains an integer denoting the number of pages in the first book.
->The second line contains an integer denoting the number of lines per page of the first book.
->The third line contains an integer denoting the number of pages in the second book.
->The fourth line contains an integer denoting the number of lines per page of the second book.
->The fifth line contains an integer denoting the reading speed in lines/second.
->The sixth line contains an integer denoting the writing speed in lines/second.
->The seventh line contains integers denoting the time in seconds at which the results are to be processed.
On one time, print three items.
Current activity performed (READ OR WRITE),
Page number, Line number.
All three items should be delimited by space.
Np1=int(input("Enter the No: of pages in the First Book: ")) #the number of pages in the first book.
Nl1=int(input("Enter the No: of lines per page of the First Book: ")) #the number of lines per page of the first book.
Np2=int(input("Enter the No: of pages in the Second Book: ")) #the number of pages in the second book.
Nl2=int(input("Enter the No: of lines per page of the Second Book: ")) #the number of lines per page of the second book.
Rs=int(input("Enter the reading speed in lines per second: ")) #the reading speed in lines/second.
Ws=int(input("Enter the writing speed in lines per second: ")) #the writing speed in lines/second.
Tot=int(input("Enter the time in seconds at which the results are to be processed: ")) #the time in seconds at which the results are to be processed.
if (Tot-secs)>=0:
print("WRITE {} {}".format(Pnumb,Remlines))
print("READ {} {}".format(Pnumb,Remlines))
Enter the No: of pages in the First Book: 100
Enter the No: of lines per page of the First Book: 10
Enter the No: of pages in the Second Book: 500
Enter the No: of lines per page of the Second Book: 6
Enter the reading speed in lines per second: 8
Enter the writing speed in lines per second: 4
Enter the time in seconds at which the results are to be processed: 145
WRITE 13 2 (current activity, page number, line number)
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Operators :: Intro CS Textbook
Video Script
Boolean Operators (Slide 6)
Let’s take a look at what the Boolean and operator does out in this Venn diagram here we’re going to kind of use these to showcase where the Boolean statements that we’re checking out with the
Boolean operator where these statements are actually evaluated to be true. So if we assume that we are two facts here, A and B are both true. So let’s say this inner circle here, this left circle
here represents a and this right circle here represents B and the square on the outside represents everything else. So when a and b are not, so if A and B are both true, then the statement evaluates
to true. So the left hand side and the right hand side of the operator both must be true in order for the whole statement to be true. If A is false, or B is true. false, then the whole thing would be
false. Now if we introduce a third fact, see, we kind of get the same results, right. So similar kind of story with the Venn diagram. Here we have a, b, and c. But notice that this little square here
where we were actually filled in over here is no longer filled in. And that’s because all three facts must be true for the whole statement to evaluate to true. So a has to be true, B has to be true.
And C has to be true in order for the whole statement to evaluate true.
So we’ll have similar representation here for the OR operator where the English word is the representation in Python of the OR operator the double bar symbol, so this is just the two vertical bars
from your keyboard that is the OR operator in Java, C sharp and other programming languages and then a capital V is great. To be the OR operator for Boolean algebra. So let’s take a look at how or
operates. So if we have the same statement, as we have before the same two premises A and B both being true initially, and the circles are kind of the representation of the same thing here, but the
OR operator will evaluate to be true if either side of the operator are true. So if the left hand side is true, or the right hand side is true, the whole thing will be true. So if A is true, or B is
true, so left hand side and the right hand side, so if A or B is true, then the whole statement is also true. And so nothing on the outside will actually be filled in quite yet. And the similar kind
of story is for a third fact or fifth, that one has to be true for the whole statement to be true. So if any of them are true, the whole thing is true, but things kind of get tricky when we’re in
introduce this next operator the exclusive OR exclusive or you won’t really find normally in programming languages. The Exclusive OR can be simulated using and or and the next operator that we’ll be
covering here in just a second.
But the exclusive OR works in a little bit of a different way than the regular or the exclusive OR operates very similar to what we would expect the normal or operator to be in the sense that if A is
true, or B is true, the statement is true, but notice that a and b is now false. If a and b are both true, the statement is false. So the exclusive OR operator is expecting one or the other. So that
means the left side or the right side must be true, but not both. That is where the exclusive portion of the exclusive OR or the X or operator actually comes out. If I introduce a third fact things
get to be a little bit more difficult to understand because now I would expect it to be kind of similar pattern as we have up here just to be, or on my just with my two facts, everything in the
middle would be white, but everything on the outside would be red. But you notice when when we have all three to be true, all three can be true and the exclusive OR would still be true. Now let’s
take a look at why that would be the case.
XOr White Board Example
So let’s take a look at the example that we saw on the slides. We have our facts a, and our XOR operator. So we have a XOR B XOR. See, now if we kind of do our substitutions here now we could
substitute ones and zeros here are true and false. So let’s go ahead and substitute our Boolean values here for our variables. So if A is true So we have a XOR B, which is also true. And C, which is
also true. Now, just like most of your math problems, even when you’re just doing multiplication and division, you’re always going to evaluate your statement from the left to the right. So we need to
first evaluates true x or true now XOR Exclusive OR exclusive right, the left or the right can be true, but not both. Since the left hand side of my operation is true, and the right hand side of my
operator is true, this portion of the statement evaluates to false. Then all we need to do then is keep on working the rest of our statements. So we still have one XOR left. So we have false x or
true? Now exclusive or one side or the other must be true but not both. So false x or true is actually true because both sides aren’t true. So this statement you say a false x or true, evaluates to
true. So the whole thing true x our true, false XOR true, ends up being true.
Not (Slide 6)
So our last Boolean operator here is the NOT operator. The NOT operator acts pretty much like negation, as you would expect, like multiplying something by negative one not something is the opposite
of what it actually is in Python as the previous Boolean operators the and and the OR operator, the NOT operator in Python is very English light not but Python is kind of weird. You will also see the
exclamation point In some operations, but it doesn’t mean the traditional knots or negation operator as in many other programming languages. So, again, you’ll see not in Python, the exclamation
point, this is going to be things like Java. And the weird sideways l here, this is going to be your Boolean algebra. So let’s take a look at what the Boolean operator not actually looks like. As I
mentioned, the NOT operator is a negation, so not something as the opposite. So not a or not true is false. So not a if I write a here, so everything in the circle is a so when A is actually true, so
everything inside of the circle then everything on the outside of a is actually true because it’s negated and similar idea for B when B is true, everything But B is true. So the whole statement is
evaluated as such and similar idea if I introduce a third fact. So if I have three facts A, B and C, not B means that everything but B is true, just like in this example here. | {"url":"https://textbooks.cs.ksu.edu/cs-zero/i-concepts/03-bits-and-boolean-algebra/02-operators/","timestamp":"2024-11-12T13:06:43Z","content_type":"text/html","content_length":"73277","record_id":"<urn:uuid:c553809e-fe03-40c4-8aa6-dc1335ddc02b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00745.warc.gz"} |
INF4480 Project III: Spectral estimation solved
Figure 1: Georeferenced sidescan sonar image (or mosaic)
The exercises in this project will be related to sonar data collected by the HUGIN autonomous underwater vehicle. Figure 1 shows the recorded sonar data after pulse compression (matched filtering)
and beamforming (by delay-and-sum). The range profile
from each ping has been rotated and georeferenced relative to each other. This is called
a sidescan sonar image. The yellow lines indicates the particular ping to be used in this
The sonar geometry is shown in Figure 2. We will consider a single ping (or pulse) with
recorded timeseries from one horisontal receiver array with Nh = 32 hydrophones (receivers).
Figure 2: Sonar geometry
The data file sonardata4.mat at the course web page, is a matlab mat-file that contains
the recorded timeseries from a single ping of sonar data as described above. The variable data contains a complex matrix of size [Nt
, Nh], where Nh = 32 is the number of
hydrophones, and Nt
is the number of time samples. The sampling frequency is stored
in variable fs (in Hertz). The start time for the recordings is zero.
The transmitted signal is a Linear Frequency Modulated (LFM) pulse of pulse length Tp,
with signal bandwidth of B, as follows
sT x(t) =
exp(j2παt2/2) −Tp/2 ≤ t ≤ Tp/2
0 |t| > Tp/2
where α is the chirp rate related to the signal bandwidth as
α = B/Tp (2)
The programmed signal bandwidth was B = 30 kHz, and the center frequency of the
signal was fc = 100 kHz. Note that the received timeseries are basebanded at reception
(the carrier frequency has been removed). We have therefore taken out the carrier (center
frequency) from the signal model.
2 INF4480
1 Spectral estimation
Exercise 1 A
Create an array out of the following selection:
• Channel 14
• Start sample M = 1200
• Number of samples N = 1024
Assume that the sequence is WSS within the data window.
Implement, calculate and plot the following spectral estimates:
• The periodogram
• The modified periodogram with a window of choice
• The Welch method with segment size L = 256 and overlap D = L/2
• The multitaper spectral estimator using matlab’s function pmtm. Choose the order
of the method (should be larger than 3).
See the lecture notes for details about the methods. Matlab source code must be written
in the presentation. Do not use matlab’s periodogram and pwelch.
The plots should have frequency in kHz on the x-axis, and Power spectral density in
dB on the y-axis. Remember proper normalisation (such that the y-values from all the
spectral estimators can be compared).
1. Does bias reduction help? Explain (from data).
2. Does variance reduction help? Explain (from data).
The transmitted (ideal) signal has a flat frequency response within the bandwidth of 30
kHz. There should be no signal outside the frequency band.
1. Estimate the true bandwidth (from guessing when the signal energy has dropped
below 6 dB from the passband)
2. What is wrong with the spectrum (compared to the ideal spectrum)?
Exercise 1 B
Create an array out of the following selection:
• Channel 9
• Start sample M = 7000
• Number of samples N = 2048
Assume that the sequence is WSS within the data window.
Calculate and plot the following spectral estimates:
3 INF4480
• The periodogram
• The modified periodogram with a window of choice
• The Welch method with segment size L = 256 and overlap D = L/2
• The multitaper spectral estimator using matlab’s function pmtm
This part of the signal is dominated by additive noise and interference (lines from other
1. Does bias reduction help? Explain (from data).
2. Does variance reduction help? Explain (from data).
3. What is the concequence of variance reduction?
2 Spectrogram analysis
Exercise 2 A
Create an array out of the following selection:
• Channel 14
• Start sample M = 400
• Number of samples N = 8192
The sequence has time-varying statistics within the data window.
Implement the short-time Fourier transform (STFT) method (or spectrogram).
• Choose the modified periodogram with a window (taper) of choice.
• Choose a segment length of L = 64, 128, 256, 512
• Zeropad each segment with a factor 4
Follow the method description in the lecture notes. Display the STFT as colorcoded image
in dB, with frequency in kHz on the y-axis and time in ms on the x-axis. Show at least 50
dB dynamic range. Matlab code has to be written in the presentation. Compare the four
1. What is the concequence of choosing segment length?
2. The chips seen around the time-frequency representation (especially from 100 ms
to 200 ms, are from other interfering sensors. What is the length of the chips (in
ms)? Are the frequencies constant from chip to chip?
4 INF4480 | {"url":"https://www.programmingmag.com/answers/project-iii-spectral-estimation-solved/","timestamp":"2024-11-13T01:11:14Z","content_type":"text/html","content_length":"100542","record_id":"<urn:uuid:807bf354-2414-408a-bccd-3fc5ab58ff40>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00364.warc.gz"} |
Quantum Amplitudes, Classical Ignorance & Quantum Information Processing
Date: Saturday, June 30, 2018 - 10:30
Venue: Martin Wood Lecture Theatre, Clarendon Laboratory
A major theme in current theoretical physics is understanding the implications of quantum mechanics for the dynamics of composite systems. When sub-systems interact, they naturally become correlated.
In the 1980s it was realised that by exploiting the correlations between quantum sub-systems it should be possible to (i) break current cryptographic systems by rapidly decomposing large numbers into
their prime factors, and (ii) exchange confidential information in the secure knowledge that messages have not been intercepted. As we have become ever more dependent on the internet these explosive
implications of quantum mechanics for cryptography have driven efforts to build quantum communication channels and quantum computers.
On June 30, to mark the completion of the Beecroft Building, the members of the Rudolf Peierls Centre for Theoretical Physics, who occupy the building above ground, and the members of the Quantum
Information Technology hub (NQIT), who are installing kit in many of its subterranean laboratories, will join forces to explain the basic principles underlying the quantum dynamics of composite
systems and to describe the challenge of implementing quantum computation and cryptography practically.
In the afternoon there will be three talks from NQIT, which is led by Oxford University. A general introduction to the programme will be followed by two talks about using ion traps and photonics to
build a quantum computer. There will also have a chance to see the new labs via exclusive video and an opportunity to questions NQIT researchers.
You will have an opportunity to explore the Beecroft Building (above ground) between 09.00-10.30.
Quantum Systems from Group up
The first talk will review the modern formulation of the basic ideas of quantum mechanics. We start by explaining what quantum amplitudes are, how they lead to the idea of a quantum state and how
these states evolve in time. We then discuss what happens when a measurement is made before describing correlated ('entangled') systems. Applying these ideas to two-state systems ('qubits') we point
out that the complexity of computing the evolution of an N qubit system grows like exp(N).
The second talk will review how to deal with quantum systems that are coupled to the outside world, as in reality all systems are. We first introduce density operators and explain how quantum states
give rise to them. We then turn to measures of entanglement that can be computed from a density operator, and show that entanglement grows with time. Finally, we show how the interaction with the
environment gives rise to the phenomenon of decoherence.
Networked Quantum Information Technologies (NQIT)
Quantum logic with trapped-ion qubits | {"url":"https://saturdaytheory.web.ox.ac.uk/quantum-amplitudes-classical-ignorance-quantum-information-processing","timestamp":"2024-11-10T00:09:43Z","content_type":"text/html","content_length":"110349","record_id":"<urn:uuid:50dd28e1-1e8e-4860-9e74-6e05941a947c>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00500.warc.gz"} |
spin-statistics theorem
Bologna is good change for bad universities and disaster for good, as every democracy-like change in a structured society is: detrimental for elite and good for the masses.
From my experience, this is quite right.
There has been endless discussion about this in the press here. One simple fact seemed to be implicit in each and every contribution that I have read, but was never made explicit:
nowadays there are simply two different kind of people going to universities, for two different kinds of reasons: one group wants to enter academia, the other group wants a high-level professional
education. The whole problem is that these two groups are not being distinguished.
Bologna is good change for bad universities and disaster for good, as every democracy-like change in a structured society is: detrimental for elite and good for the masses.
I would be curious to know what kind of arguments we should be using with undergraduates.
My outdated answer is: When I was a student, theoretical QFT was an optional class on the way to the Diplom, so the professors would simply say “we will do it our way and anyone who does not like it
does not have to attend” (which means that there were introductory classes in QFT starting with the definition of a $C^*$ algebra).
There has been a big reform recently, the Bologna process, which changed that. I do not know how they do it now.
Hmmm. Interesting. I’ve heard of the Halvorson/Müger paper but have never read it. I would be curious to know what kind of arguments we should be using with undergraduates. Personally, I prefer
correctness over simplicity, but, on the other hand, I also prefer that the student have some level of actual understanding and not simply be able to parrot what I say. It’s a tough line to straddle
(I sincerely hope people appreciate how hard it can be to teach undergraduates well - sometimes I feel as if undergraduate education gets treated as the “ugly duckling” as it were).
I like the Heisenberg anecdote, by the way. I went hunting for Weyl’s grave in Zürich in January but wasn’t able to find it (I found Pauli’s, Hopf’s, James Joyce’s, and Fritz Zwicky’s).
Yes, I mean the chapter II.4 Spin-Statistics Connection, subchapter “The price of perversity”: The problem with the kind of argument presented here is that there are several no-go theorems that say
that the objects manipulated here cannot be made precise in certain ways, one of the best review papers in AQFT, Hans Halvorson, Michael Müger, “Algebraic Quantum Field Theory” (reference is on the
AQFT page) lists some of these.
This leads some people working in AQFT to the conclusion that arguments like this should be avoided at all costs, even at an undergraduate level.
This reminds me of an anecdote about Heisenberg: First, he intended to study mathematics, but when he told the math professor whom he consulted in Göttingen about this (I forgot his name) that he had
read “Raum, Zeit, Materie” by Weyl, the professor said: “Then you have been spoilt for mathematics anyway”. Heisenberg then went to Arnold Sommerfeld and decided to study physics instead…
@Tim: Are you referring to Zee’s conceptual description? Wasn’t sure.
Sure, why not. Some people would say that handwaving of this kind does more harm than good, but I’m not one of them. (Where is the argument for spin 1/2 ? :-)
I’m not quite happy with the situation in the Haag-Kastler either, as far as I understand it, I think there should be an argument that is easier…
There’s a really nice discussion of this from a conceptual POV in Tony Zee’s Quantum Field Theory in a Nutshell.
Added an idea paragraph and the reference to the Guido/Longo paper.
stub for spin-statistics theorem. Just recording a first few references so far.
The whole problem is that these two groups are not being distinguished.
I think that is generally true, but, oddly enough, students in my department tend to be undecided until sometimes their senior year. One former student actually went into industry, disliked it, and
returned to academia to get his PhD.
We've been having this sort of discussion on campus all year in relation to our revised core curriculum and strategic plan. One thing we realized that we do really well is, in general, provide a
well-rounded, classical education to future professionals, meaning, as a college, the majority of our students end up as professionals (this is not entirely true for physics, though) but have gotten
a classical education grounded in philosophy, theology, science, art, etc.
I guess what I'm saying is that it is possible to do both and do both well. The problem is that most institutions don't take the time to try. Instead they focus on one or the other. In the
professional category, they tend to (at least in this country) cater to student desires a little too much (i.e. it's like they're selling a product).
But, back on the original topic, I'm always looking for new and creative - but correct! - ways to teach complex things to undergraduates. Like I said somewhere before, undergraduate education tends
to be like the "ugly duckling" in math and physics (at least in this country, anyway) but it's only doing the field a disservice to treat it as such.
I would like to mention that, after my question at mathoverflow here about the PCT and the spin-statistics theorem did not get an answer (was worth a try anyway), I asked Professor Rehren from the
Göttinger AQFT group the same question, and he confirmed that the references cited on the nLab are pretty much the state of the art. (He also mentions some papers about CFT that I don’t know, I’ll
have a look).
Thanks, Tim!
By the way, I just read in a new QM text, that Feynman once challenged the physics/mathematics community to develop an elementary proof of the spin-statistics theorem. In 2002 the editor of the
American Journal of Physics brought up Feynman’s challenge again as having gone unfulfilled. Apparently this is still the case.
“Elementary” is in the eye of the beholder. I think Feynman originally said (in response to a challenge) that he would would prepare a “freshman lecture” to explain the theorem, but eventually gave
up. (“Freshman lecture” meaning something on the order of his famous lecture notes on physics.)
“Elementary” is in the eye of the beholder.
My concept of “simple” in this case is:
a) axioms used: Can I understand the physical motivation of those?
c) how easy is it to spot when and where and how which of the axioms are used in the proof?
For example, the Reeh-Schlieder theorem needs heavy math machinery like the SNAG-theorem and the edge-of-the-wedge theorem, but on the other hand it is very easy to see which axioms are used, and
that e.g. the causality axiom is not (which is a little bit disappointing for anyone who thinks that the Reeh-Schlieder theorem violates causality and that we should change the causality axiom to
dodge it).
…the editor of the American Journal of Physics brought up Feynman’s challenge again as having gone unfulfilled.
Ok, but is there any money in it?
@Tim: What happened to point b)? ;)
Ok, but is there any money in it?
Probably not, but if my nanotech startup takes off and I get rich (as opposed to simply delusional), I’ll sponsor a prize for it. :-)
finally polished up this bibitem
• Raymond F. Streater, Arthur S. Wightman, PCT, Spin and Statistics, and All That, Princeton University Press (1989, 2000) [ISBN:9780691070629, jstor:j.ctt1cx3vcq]
and added pointer to this one:
• Franco Strocchi, §4.2 in: An Introduction to Non-Perturbative Foundations of Quantum Field Theory, Oxford University Press (2013) [doi:10.1093/acprof:oso/9780199671571.001.0001]
diff, v13, current
Added a reference to a recent arXiv paper on spin-statistics in TFT
Arun Debray
diff, v16, current
added (here) a bunch of references treating the spin-statistics theorem for non-relativistic particles via the topology of their configuration space of points.
Will put that list now into its on entry spin-statistics via configuration spaces – references, to be !include-ed here and at configuration space of points
diff, v19, current
added pointer to:
• Cameron Krulewski, Lukas Müller, Luuk Stehouwer: A Higher Spin Statistics Theorem for Invertible Quantum Field Theories [arXiv:2408.03981]
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cement plant ball mills critical speed formulae
W = power consumption expressed in kWh/short to (HPhr/short ton = kWh/short ton) Wi = work index, which is a factor relative to the kwh/short ton required to reduce a given material from
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Custom Functions/Events
Hi there,
So I have finally started learning about metatables and its usage.
I was wondering,
How would I acheive something like this-
Which basically should return me a table of only basepart class items in workspace.
I know I could use “IsA”, but I want a fresh & more advanced way.
local Bricks = Workspace:GetOnlyBaseParts()
--Or even better:
local Target = Workspace:GetItemsOfClass(ClassName)
Or even something cool like this-
local ItemsToRemove = Workspace:GetPartsToRemove(classname,name(optional))
Or even:
local Parts = Workspace:GetPartsOfClass(classname)
local PaintParts = Parts:Paint(Color3)
1 Like
How could I make something like those?
Custom functions are really vast and nice, & an opportunity to let us be creative a bit and do some stuff
1 Like
You can use a wrapper.
2 Likes
What is a wrapper? I am new to this topic
I understood how to make simple custom functions such as ‘VanishObject’,
But how could I make a function that returns a table?(custom function)
Assuming you know of the __index and __newindex metamethods, you can use it to return the instance instead of a table.
local realWorkspace = workspace
local workspace = {}
workspace.__index = realWorkspace
function workspace:PrintHi()
function workspace.new()
local newWorkspace = {}
-- inside the newWorkspace table, you add your properties and events, not functions
return setmetatable(newWorkspace, workspace)
Everything is explained well in the tutorial so you should check that out. You’ll notice you won’t be able to call native workspace methods natively so you can use this:
1 Like
Oh, that’s nice.
So if I wanted to make a custom function that returns me only the given class(name) , for e.g:
local Detect = Workspace:GetOnlyModels()
How would I do that?
(The examples you sent me, talk about simple functions that return instances, not tables
Make a function that iterates through the workspace and adds all of the models to a table then returns it.
Alright, so far I have this
local realWorkspace = workspace
local functions = {}
functions.__index = realWorkspace
function functions:GetItems(class:string)
local items = {}
for _,Item in pairs(game.Workspace:GetDescendants()) do
if Item:IsA(class) then
return items
function functions.new()
local newFunctions = {}
return setmetatable(newFunctions, functions)
Okay, so you need to now call functions.new(), and with the object returned from that, call :GetItems on it and it should work.
2 Likes
local realWorkspace = workspace
local functions = {}
functions.__index = realWorkspace
function functions:GetItems(class:string)
local items = {}
for _,Item in pairs(game.Workspace:GetDescendants()) do
if Item:IsA(class) then
return items
function functions.new()
local newFunctions = {}
return setmetatable(newFunctions, functions)
local result = functions.new()
Gave an error
Oh I see the issue, you will have to make the __index metamethod a function that returns a function inside of functions if it exists, else returns the real instance’s function. So like
local functions = {}
-- class methods go in the functions table
function functions:__index(key)
if functions[key] then
return functions[key]
return workspace[key]
function functions.new()
local object = {}
-- properties and events here
return setmetatable(object, functions)
local realWorkspace = workspace
local functions = {}
functions.__index = realWorkspace
function functions:GetItems(class:string)
if functions[class] then
return functions[class]
return workspace[class]
function functions.new()
local object = {}
return setmetatable(object, functions)
local result = functions.new()
error again
This has to be a function that returns functions[index] if functions[index] exists, or workspace[index]
local realWorkspace = workspace
local functions = {}
functions.__index = functions
function functions:GetItems(class:string)
if functions[class] then
return functions[class]
return workspace[class]
function functions.new()
local object = {}
return setmetatable(object, functions)
local result = functions.new()
it doesnt print
nor does this one -
local realWorkspace = workspace
local functions = {}
functions.__index = functions
function functions:GetItems(class:string)
local items = {}
for _,Item in pairs(game.Workspace:GetDescendants()) do
if Item:IsA(class) then
return items
function functions.new()
local object = {}
return setmetatable(object, functions)
local result = functions.new()
This still is not a function. It has to be a function that checks if functions[index] exists where index is the second argument passed to the __index function. If it does exist, return functions
[index], else return workspace[index]
1 Like
Well, that’s a huge confusion right here…
Never thought it’d be that confusing & difficult to understand that…
Yeah, object wrappers are difficult to understand but one day it just clicked for me so I’ll try explaining it as best as I can. So, __index is a metamethod that can either be a function or another
If it is a function it will call that function, if it is another object, it will look in this other object for the index
1- newObject = object.new() - newObject is a table that contains properties/events, and whose metatable has a __index metamethod that is a function.
2- newObject.nonexistentkey - Since no entry in newObject exists with the index nonexistentkey, it gets the __index metamethod of the newObject object, and calls that __index metamethod if
object.nonexistentkey does not exist. In our case it does not exist, so it passes the table that was attempted to be indexed as the first argument (in our case it’s newObject) and as the second
argument it’s the key that was used to attempt to index newObject. We do not need to call __index with parentheses because indexing newObject with a nonexistent key calls this already.
3- In our __index function, we attempt to index functions, which is the table with class methods that extend the actual workspace methods, with nonexistentkey. This does not invoke any metamethods.
If it exists return it.
4- nonexistentkey does not exist in the functions table, so we index the real instance with nonexistentkey, this will either error properly, or returns the real instance property.
In the end, imagine it like this,
This line:
local myProperty = newObject.nonexistentkey
Is the same same as doing
local myProperty = getmetatable(newObject).__index(newObject, 'nonexistentkey')
So, in the end, the process is like this, where you attempt each step and if one of the steps is successful, use the result from the successful attempt, if it errors, do nothing, if nil is returned
but it did not error, proceed:
1- look in newObject for nonexistentkey
2- look in newObject’s metatable for nonexistentkey
3- look in realWorkspace for nonexistentkey
4- no options left, return nil.
2 Likes
Hey again,
So I watched AlvinBlox's tutorial on metatables, and now I understand how to use __index and __newindex .
But unfortunately, I still am confused how would I use these in my case?
Put the metamethods in your functions table, then use the functions table as your metatable.
function functions:__index(key)
if functions[key] then
return functions[key]
return workspace[key]
return setmetatable(newTable, functions)
This function is defined as a metamethod, it confuses me
Why isnt it defined with a name? (Like regular funtions), will there be a difference if I were to name it function functions:Test(key) | {"url":"https://devforum.roblox.com/t/custom-functionsevents/1817054","timestamp":"2024-11-04T09:17:05Z","content_type":"text/html","content_length":"75357","record_id":"<urn:uuid:d12783b6-b843-439f-88d2-454fbda95241>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00248.warc.gz"} |
Model info
Source code
• [ View in CSDMS GitHub repository]
Model citations
Nr. of publications: 2
Total citations: 54
h-index: 1
m-quotient: 0.08
Link to this page
Other models by author
The model gFlex implements multiple (user-selectable) solution methods to solve for flexure and isostasy due to surface loading in both one (line loads) and two (point loads) dimensions. It works for
elastic lithospheric plates of both constant and spatially variable elastic thickness and allows the user to select the solution method.
This page contains some information on gFlex, but its main documentation is in the README.md file, displayed at the gFlex GitHub repository at https://github.com/umn-earth-surface/gFlex.
Solution methods
An analytical approach to solving the flexure equations is computed by the superimposition of analytical solutions for flexure with a constant elastic thickness. Current implementations perform this
superposition in the spatial domain. This works both on uniform grids and arbitrary meshes.
The good:
• The analytical solution method for an arbitrary mesh is useful for coupling Flexure with finite element models such as CHILD without requiring any regridding.
The bad:
• Analytical solutions by superposition are an N^2 problem, making this method become increasingly problematic for larger grids and numbers of nodes.
• Analytical solutions are computed based on sets of point loads at nodes or centers of cells, so they will fail and show too much isostatic response if the cells become larger than a modest
fraction of a flexural wavelength; this is because at too large of a grid size, the approximation of summing immediately adjacent loads breaks down.
• Analytical solutions work only with approximations of constant elastic thickness.
Analytical solutions using spectral techniques are not yet implemented.
For the numerical implementation, gFlex computes a finite difference solution to the flexure equations for a lithospheric plate of nonuniform (or uniform, if one so desires) elastic thickness via a
thin plate assumption. It uses the UMFPACK direct solvers to compute the solutions via a lower-upper decomposition of a coefficient matrix. The coefficient matrix in the 1D case is a pentadiagonal
sparse matrix that is trivial to generate. In the 2D case, 2D grid is reordered into a 1D vector. UMFPACK solution routines are then able to copmute solutions to flexure in around a second or less,
though sometimes up to a minute for very large grids. Iterative solution methods may also be used, but are not tested.
Some advantages of using a numerical (rather than analytical) solution are that:
• This method permits spatial variability in lithospheric elastic thickness. This allows the use of real maps of elastic thickness in models or synthetic maps of elastic thickness variability to
test hypotheses.
• This rapid solution once the coefficient matrix is built makes this method a good choice for numerical models that require frequent updating of flexural deformations of the lithosphere.
• The rapid solution technique likewise allows efficient calculations of mixed sediment and/or water loading, or water loading with onlap and offlap, such that a constant fill density cannot be
assumed and solutions must be produced iteratively.
• This model will not break down when grid cell sizes are increased greatly, as will happen for superposition of analytical solutions if the grid cells become too large relative to a flexural
wavelength for the point load assumption to hold true.
Key physical parameters and equations
The largest component of gFlex is a solution to the PDE for lithospheric flexure in 2 dimensions. Our finite difference solutiosn follow van Wees and Cloetingh (1994):
<math> \frac{\partial^2}{\partial{x_i^2}} \left( D(x,y) \frac{\partial^2 w}{\partial{x_i^2}} \right) + \Delta \rho g w = q(x,y) </math>
• Flexure was developed first in MATLAB (Spring / early Summer 2010) and then in Python with Numpy, Scipy, and Matplotlib (translated October 2010).
• As of October 2011, Flexure became IRF- and CMT-compliant, and it was coupled to the landscape evolution model CHILD for the Fall 2011 CSDMS meeting. (Abstract and presentation are here, though I
haven't dared to watch myself.)
• Current work is being done to improve boundary condition handling and the speed of finite difference solutions to constant elastic thickness problems (Early 2012).
• The project was revived in fall 2014m renamed gFlex, and brought to completion in late winter 2015.
Nr. of publications: 2
Total citations: 54
h-index: 1
m-quotient: 0.08
Version 0.9 (development) and higher
Instructions for the current version of gFlex are available from the README.md file at: https://github.com/umn-earth-surface/gFlex
Version 0.1
(Modified from instructional emails)
This is the old version of Flexure".
You need to have python, numpy, scipy, and matplotlib installed to use flexure.
The version 0.1 structure (prone to change) consists of two main files.
• flexcalc.py is a python module which contains all of the functions needed to execute flexure.
• flexit.py is the frontend that, via optparse, gives you the ability to specify inputs and outputs and run flexural solutions via an interactive command-line interface.
In addition to these files, version 0.1 comes with some basic test loads and elastic thickness maps.
Trial run
For the basic functionality on the first runthrough, you navigate to the directory with the flexit.py and type something in like:
python flexit.py -vcrp --dx=20000 --Te=Te_sample/bigrange_Te_test.txt --q0=q0_sample/top_bottom_bars.txt
You select which of the sample Te and q0 files you want. Andy_output.png is the flexural response (variable=w) output from:
python flexit.py -vcrp --dx=20000 --Te=Te_sample/bigrange_Te_test.txt --q0=q0_EW_bar.txt
The flags are explained in the help file (python flexit.py -h), as are other options for running the code. Basically you will want to run "-c" the first time, but not again unless you are going to
redo the coefficient matrix (i.e., use a different pattern of elastic thickness). This calculation can take a long time for large grids, so you will want to store these files. When running without
making a coefficient matrix, unless you're using the default coefficient matrix name (as we do above), you will have to specify its location with "--coeff-file=NAME". For example:
python flexit.py -vrp --Te=Te_sample/bigrange_Te_test.txt --q0=q0_sample/top_bottom_bars.txt --coeff_file=coeffs.txt
If all else fails
Have you typed:
python flexit.py -h
? This gives you all of the in-program help information.
Feel free to email Andy Wickert for anything related to this model (see contact info in box at top).
Input Files
The coefficient matrix for the 2D finite difference solution with variable elastic thickness requires a map of elastic thicknesses in *.txt / ASCII format. This elastic thickness map must be two
cells wider on each side than the map of loads; this is because the finite difference solution must "look" two cells in every direction. It also requires the specification of several parameters,
• Young's modulus (defaults to 10^11 Pa)
• Poisson's ratio (defaults to 0.25)
• Mantle density (defaults to 3300 kg/m^3)
• Density of infilling material (defaults to 0 kg/m^3)
This outputs an ASCII sparse matrix file (Matrix Market *.mtx format).
The flexural solution requires the ASCII file for the sparse coefficient matrix generated above and an imposed array of loads (also ASCII), along with the specification of input and output file
Output Files
The coefficient matrix creator writes a *.mtx sparse matrix ASCII file that is used in the direct solution. This matrix is characteristic to a given pattern of elastic thickness, and therefore can be
reused if elastic thickness does not change.
The real solver outputs an ASCII grid of deflections due to the load. This is the output that is of scientific interest and/or useful to plug into other modules (e.g., for flexural subsidence). | {"url":"https://csdms.colorado.edu/csdms_wiki/index.php?title=Model:GFlex&oldid=300005","timestamp":"2024-11-06T10:16:51Z","content_type":"text/html","content_length":"69817","record_id":"<urn:uuid:f1eff204-120c-45cf-aecb-d24b191442a0>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00460.warc.gz"} |
Multiplication Charts For Kids | Multiplication Chart Printable
Multiplication Charts For Kids
Multiplication Table Education Chart Poster Posters Multiplication
Multiplication Charts For Kids
Multiplication Charts For Kids – A Multiplication Chart is a helpful tool for youngsters to find out how to increase, split, as well as discover the smallest number. There are numerous uses for a
Multiplication Chart.
What is Multiplication Chart Printable?
A multiplication chart can be utilized to help youngsters learn their multiplication truths. Multiplication charts come in many forms, from complete page times tables to single web page ones. While
individual tables serve for providing chunks of info, a complete page chart makes it easier to assess facts that have actually currently been grasped.
The multiplication chart will usually feature a top row as well as a left column. When you desire to find the product of 2 numbers, choose the very first number from the left column and also the 2nd
number from the top row.
Multiplication charts are valuable learning tools for both adults and children. Multiplication Charts For Kids are readily available on the Internet and also can be printed out and laminated flooring
for durability.
Why Do We Use a Multiplication Chart?
A multiplication chart is a diagram that shows how to increase 2 numbers. It commonly includes a left column and also a top row. Each row has a number representing the item of the two numbers. You
choose the first number in the left column, relocate down the column, and then select the 2nd number from the top row. The product will be the square where the numbers meet.
Multiplication charts are valuable for several reasons, including helping youngsters learn just how to divide as well as streamline portions. Multiplication charts can likewise be practical as desk
resources since they serve as a constant tip of the trainee’s development.
Multiplication charts are additionally beneficial for assisting students memorize their times tables. They help them discover the numbers by minimizing the variety of actions needed to finish each
procedure. One method for memorizing these tables is to concentrate on a single row or column at once, and after that move onto the following one. At some point, the whole chart will be committed to
memory. Just like any ability, memorizing multiplication tables takes some time and also practice.
Multiplication Charts For Kids
Free And Printable Multiplication Charts Activity Shelter
Multiplication Facts Classroom Math Chart Kids Chart Paper Etsy
10 LAMINATED Educational Math Posters For Kids Multiplication Chart
Multiplication Charts For Kids
If you’re seeking Multiplication Charts For Kids, you’ve involved the ideal area. Multiplication charts are readily available in different styles, including full size, half size, and a range of
charming designs. Some are vertical, while others feature a straight style. You can likewise find worksheet printables that include multiplication formulas as well as math truths.
Multiplication charts and tables are crucial tools for kids’s education. You can download and also publish them to make use of as a teaching aid in your kid’s homeschool or class. You can also
laminate them for durability. These charts are great for usage in homeschool mathematics binders or as class posters. They’re specifically valuable for kids in the second, 3rd, as well as fourth
A Multiplication Charts For Kids is a helpful tool to enhance math realities as well as can help a kid learn multiplication quickly. It’s additionally a great tool for avoid counting as well as
learning the times tables.
Related For Multiplication Charts For Kids | {"url":"https://multiplicationchart-printable.com/multiplication-charts-for-kids/","timestamp":"2024-11-07T04:21:37Z","content_type":"text/html","content_length":"41162","record_id":"<urn:uuid:d839bb44-2999-404f-bb24-73d840201e2e>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00215.warc.gz"} |
Robert Langlands
Writing for NRC Handelsblad, Robbert Dijkgraaf, IAS Director and Leon Levy Professor, likens IAS Professor and Abel Prize Laureate Robert Langlands’s program, “a deep connection between two
completely different parts of mathematics: on the one hand, numbers and their relations, on the other hand, geometrical patterns and their symmetries,” to a mathematical Rosetta Stone. | {"url":"https://www.ias.edu/idea-tags/robert-langlands","timestamp":"2024-11-03T07:25:10Z","content_type":"text/html","content_length":"77044","record_id":"<urn:uuid:08f9c1b6-27e0-4825-8190-20af9b0594c8>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00550.warc.gz"} |
Area Calculat
Area Calculator
Calculated Area:
15.00 m²
Area Calculations in Building and Construction
Area calculations play a crucial role in the construction industry, serving as the foundation for various project phases, from planning and design to material estimation and cost management. Accurate
area measurements are essential for ensuring the successful execution of construction projects, helping professionals to optimize resources, meet regulatory requirements, and achieve desired
outcomes. This guide provides an in-depth overview of area calculations for common geometric shapes, their practical applications in construction, and key considerations for effective use.
Understanding Area Calculations
Area is defined as the amount of space enclosed within a two-dimensional boundary, expressed in square units. In construction, area measurements are typically represented in square meters (m²), which
is the standard unit of measurement for most calculations. Calculating the area of a shape involves applying specific mathematical formulas based on the geometry of the shape. The most common shapes
encountered in construction include rectangles, squares, triangles, and circles.
1. Rectangles and Squares
• Formula:
For a rectangle:
Area = Length × Width
For a square (a special case of a rectangle):
Area = Side × Side
• Use Cases:
Rectangular and square areas are ubiquitous in construction. They represent spaces such as rooms, plots, foundations, walls, floors, and roofs. Calculating these areas is essential for
determining the quantity of materials like tiles, paint, flooring, and concrete needed for construction. For instance, knowing the area of a floor helps in estimating the number of tiles required
or the amount of paint needed to cover a wall.
2. Triangles
• Formula:
Area = ½ × Base × Height
• Use Cases:
Triangular areas are common in construction, especially in roof designs, gable ends, trusses, and certain architectural elements. Calculating the area of triangles helps in designing and
estimating the materials for roof coverings, structural supports, and decorative elements. Triangular calculations are also useful in land surveying where land parcels may not conform to standard
rectangular shapes.
3. Circles
• Formula:
Area = π × Radius²
• Use Cases:
Circular areas are often encountered in the construction of round columns, circular windows, domes, and roundabouts. Accurate calculation of circular areas is vital for determining the quantity
of materials like steel and concrete for columns, the glass needed for circular windows, or the land area required for a circular driveway or garden feature.
Practical Applications of Area Calculations in Construction
Area calculations are not merely academic exercises; they have real-world applications that are critical to the success of construction projects. Here are some common scenarios where area
calculations are indispensable:
1. Material Estimation
One of the primary applications of area calculations is in material estimation. Knowing the area of surfaces allows construction professionals to accurately estimate the quantity of materials needed,
reducing waste and ensuring that sufficient supplies are available. For example:
• Flooring: The area of floors is used to determine the number of tiles, carpet, or wooden planks required.
• Painting: The area of walls is essential for estimating the amount of paint needed, taking into account factors such as the number of coats and type of paint.
• Concrete: The area of a slab or foundation helps in calculating the volume of concrete required when combined with thickness measurements.
2. Cost Estimation and Budgeting
Accurate area calculations contribute to precise cost estimation and budgeting. By knowing the area of various construction elements, contractors can better estimate labor costs, material expenses,
and other project-related costs. This ensures that the project stays within budget and helps in making informed decisions about cost-saving measures.
3. Land and Site Planning
In site planning, area calculations are used to determine the size and layout of plots, building footprints, and open spaces. This is especially important in urban planning, where maximizing the use
of limited land is critical. Area measurements also play a role in complying with zoning regulations, which often dictate minimum and maximum building sizes and open space requirements.
4. Structural Design
Engineers use area calculations to determine the load-bearing capacity of structural elements. For example, the area of a beam or column's cross-section is crucial for calculating its ability to
support weight. Similarly, the area of a wall influences its ability to withstand lateral forces, such as wind or seismic activity.
5. Interior Design and Space Planning
In interior design, area calculations are used to plan the layout of furniture, fixtures, and fittings within a space. Knowing the area of rooms helps designers create functional and aesthetically
pleasing layouts that optimize the use of available space. This is particularly important in small or irregularly shaped rooms, where efficient space utilization is key.
Advanced Considerations in Area Calculations
While basic area calculations for common shapes are relatively straightforward, the construction industry often requires more advanced techniques for irregular shapes, complex structures, and
large-scale projects. Understanding these advanced considerations can enhance the accuracy and applicability of area measurements in construction.
1. Composite Shapes
In many construction projects, areas are not perfect rectangles, squares, triangles, or circles. Instead, they may be composite shapes, composed of multiple simple geometric forms combined together.
To calculate the area of such shapes, one must:
• Decompose the Shape: Break down the composite shape into simpler shapes (e.g., divide an L-shaped room into two rectangles).
• Calculate Individual Areas: Calculate the area of each component shape using the appropriate formula.
• Sum the Areas: Add together the areas of all the component shapes to get the total area.
2. Irregular Shapes
When dealing with irregular shapes that cannot be easily decomposed into standard geometric forms, more advanced methods such as numerical integration, grid approximation, or software tools may be
required. For example:
• Grid Method: Overlay a grid of known dimensions on the irregular shape and count the number of full and partial grid squares within the shape. Estimate the total area based on the proportion of
squares that are filled.
• Software Tools: Utilize CAD (Computer-Aided Design) software or specialized area calculation tools to compute the area of irregular shapes with high precision.
3. Scaling and Proportions in Drawings
In architectural and engineering drawings, areas are often represented on a scale different from real life. When calculating area from scaled drawings:
• Understand the Scale: Ensure you know the scale of the drawing (e.g., 1:100 means 1 unit on the drawing equals 100 units in reality).
• Apply the Scale Factor: Convert the measurements from the drawing scale to the real-world scale before calculating the area.
4. Accuracy in Large-Scale Projects
For large-scale construction projects, even small errors in area calculations can lead to significant cost overruns or material shortages. In such cases, it's crucial to:
• Double-Check Measurements: Verify all measurements using high-precision tools and techniques.
• Consult with Experts: Involve structural engineers, surveyors, or architects to validate area calculations for critical components.
• Use Professional Tools: Employ professional-grade software and tools designed for large-scale construction planning to ensure accuracy.
Area calculations are fundamental to the construction industry, providing the basis for material estimation, cost planning, site design, and more. By mastering the formulas for common shapes and
understanding their practical applications, construction professionals can ensure the accuracy and efficiency of their projects. Whether calculating the area of a simple room or a complex roof
design, precise area measurements are essential for achieving the desired outcomes in construction.
Understanding advanced techniques and considerations, such as composite shapes, irregular forms, and scaling, further enhances the reliability of area calculations in complex construction scenarios.
Accurate area measurements not only contribute to project success but also optimize resource use, reduce waste, and ensure that construction projects are completed on time and within budget. | {"url":"https://constructcalc.com/area.php","timestamp":"2024-11-09T03:10:57Z","content_type":"text/html","content_length":"25265","record_id":"<urn:uuid:6b145c12-d55c-4b8c-b55c-6b69529aba50>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00123.warc.gz"} |
Total time calculator
This calculator is useful for fast time addition, you just enter time entries in hh mm ss format and the calculator gives you the sum of there time entries
The calculator uses the following format:
• three numbers separated by spaces are interpreted as hh mm ss, f.e. 1 0 24 - means 1 hour 0 minutes 24 seconds
• two numbers separated by spaces are interpreted as mm ss, f.e. 12 24 - means 12 minutes 24 seconds
• one number is interpreted as ss, f.e. 5 means 5 seconds
Using this format, you can quickly get the sum of your time entries. But it also displays the reminder after dividing the total sum by 24 hours, that is, for 51 hours the reminder will be 3 (51 -
Similar calculators
PLANETCALC, Total time calculator | {"url":"https://planetcalc.com/65/","timestamp":"2024-11-14T16:39:37Z","content_type":"text/html","content_length":"30609","record_id":"<urn:uuid:89309378-b570-4941-95d2-770e199b2520>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00057.warc.gz"} |
Multiplying Binomials Calculator
Last updated:
Multiplying Binomials Calculator
Welcome to Omni's multiplying binomials calculator, where we'll learn just what the name suggests: how to multiply binomials. It is a special case of polynomial multiplication, which we covered in
the multiplying polynomials calculator, but it's so common in applications and coursebooks that it deserved its very own calculator. So for today, binomials are the only polynomials we need (well,
sort of...), and if you're not too sure what they are, we give the binomial definition in math just in case.
Binomials, polynomials
They call mathematics the language of the universe for a reason: it describes the rules that govern the world. Sure, physicists do something similar, but in the end, physics is just applying
mathematics to specific scenarios. *grabs popcorn*
Anyway, the way we present those rules is through formulas. These vary in length and difficulty from that for the area of a rectangle, through the mass moment of inertia, to some crazy equations only
a handful of people understand (or claim to understand). All of these have one important thing in common: variables.
Variables represent objects (usually numbers) that we don't know or don't want to specify. For instance, the formula for gravitational pull involves the variable $m$ for mass, but since we don't give
the number straight away, we can use the equation for the Earth, the Moon, or any other object. The variables make the whole thing universal.
Polynomials are expressions that contain variables only in non-negative integer powers. In other words, a polynomial cannot involve variables inside roots, logarithms, trigonometric functions, or any
other fancy mathematical tool. However, they can involve many variables.
Below, we list a few examples of polynomials:
• $x + 2y$
• $a^2 + 2ab + b$
• $n^3 - 0.7n + \frac{3}{8}$
• $1 + 3 + x^5 - x^7 + 19x^9$
Note how the exponents can differ; they can even repeat, as long as they aren't negative or fractional. For the latter, you can refer to our fraction exponent calculator.
A binomial is a polynomial with two terms. As such, among the ones above, only the first expression is a binomial. However, note that although the binomial definition in math is fairly general,
Omni's multiplying binomials calculator deals only with so-called linear binomials. To be precise, we'll have expressions of the form $ax + b$, meaning with only one variable $x$ and in the first
power. Such binomials are most common in applications and coursebooks and are more than enough to explain the concept.
Now that we've come to know our enemy, we're ready to fight them! Let's learn how to multiply binomials.
How to multiply binomials?
For those who like word-based explanations, there is a method of multiplying binomials called the FOIL method. If you're interested, feel free to check out Omni's FOIL calculator. Here, however, we
focus on formulas.
When multiplying binomials (or any polynomials, for that matter), the basic rule is: multiply every term of the first expression by every term of the second. We don't want to go into too much
generality, so let's just take two binomials (for simplicity, we'll use the same notation our multiplying binomials calculator uses): $a_1x + a_0$ and $b_1x + b_0$. If we apply the rule mentioned, we
$(a_1x + a_0)(b_1x + b_0) \\[1em] = (a_1x\times b_1x) + (a_1x \times b_0) \\[1em] \ \ \ \ + (a_0 \times b_1x) + (a_0 \times b_0)$
Recall that $x \times x = x^2$ and that we can take the numbers in front of the variables (because multiplication is commutative). Therefore:
$(a_1x + a_0)(b_1x + b_0) \\[1em] = (a_1x\times b_1x) + (a_1x \times b_0) \\[1em] \ \ \ \ + (a_0 \times b_1x) + (a_0 \times b_0) \\[1em] = (a_1\times b_1)x^2 + (a_1\times b_0)x \\[1em] \ \ \ \ (a_0\
times b_1)x + (a_0 \times b_0)$
Next, we group together the two summands with $x$ (i.e., use the fact that multiplication is associative), and get:
$(a_1x + a_0)(b_1x + b_0) \\[1em] = (a_1b_1)x^2 + (a_1b_0)x \\[1em] \ \ \ \ + (a_0b_1)x + (a_0b_0) \\[1em] = (a_1b_1)x^2 + (a_1b_0 + a_0b_1)x \\[1em] \ \ \ \ + (a_0b_0)$
Voilà! We got the formula for multiplying binomials. Let us just mention here that it's not too difficult to go from binomials to polynomials in general; there's just a bit more work to do.
If you've multiplied binomials and want to learn how to undo this operation, check our factoring trinomials calculator.
We've seen the binomial definition in math, we've learned how to multiply binomials, so there's only one thing left to do: see an example.
Example: using the multiplying binomials calculator
Let's find the product of $3x - 2$ and $x + 5$. Before we get our hands dirty (well, not really), we'll show you how to get the answer using the multiplying binomials calculator.
At the top of our tool, we see a symbolic expression representing the problem at hand:
$(a_1x + a_0)(b_1 x + b_0) \\[1em] = c_2 x^2 + c_1 x + c_0$
The same notation is used underneath in corresponding sections. Looking back at our example, we input:
$a_1 = 3, a_0 = -2, b_1 = 1, b_0 = 5$
Note how we have $b_1 = 1$ even though there was no $1$ in our binomial. That is because by convention, if the number in front of the variable is $1$, we don't write it. Also, observe how some c's
get calculated even before we input all four entries. For instance, it's enough to have $a_1$ and $b_1$ to see what $c_2$ is because the formula in the above section needs no other values. Lastly,
note how the multiplying binomials calculator also gives a step-by-step solution once you give it all the necessary numbers.
But before we reveal the answer, let's try to arrive at it ourselves. The task shouldn't be too difficult: we simply apply the formula from the above section and do the simple addition and
$(3x -2)(x + 5) \\[1em] = (3\times1)x^2 \\[1em] \ \ + (3\times5 + (-2) \times 1)x \\[1em] \ \ + ((-2)\times5) \\[1em] = 3x^2 + 13x - 10$
And just like that, we're done! A piece of cake, wouldn't you say?
Make sure to check out other Omni tools dedicated to polynomials in the algebra calculators section, for example, the polynomial division calculator. | {"url":"https://www.omnicalculator.com/math/multiplying-binomials","timestamp":"2024-11-07T05:54:23Z","content_type":"text/html","content_length":"717489","record_id":"<urn:uuid:86e2d5da-cd7a-418b-8138-b0df355e58c1>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00220.warc.gz"} |
Troubleshooting Servers
This guide contains instructions for troubleshooting high server CPU usage and memory leaks from the game server using various diagnostics tools.
We'll go over how to interact with Metaplay-managed cloud environments, such as connecting to the Kubernetes cluster and retrieving files from the cloud. This page also covers .NET Performance
Counters, which are a great starting point to find possible problems. We'll also cover capturing and analyzing CPU captures for the pods and troubleshooting memory looks using tools like
dotnet-cgdump and dotnet-dump.
To troubleshoot the pods running in the cloud, you'll need to interact with the Kubernetes cluster using the kubectl tool.
# Login to Metaplay cloud npx @metaplay/metaplay-auth@latest login # Get a kubeconfig against the specified project and environment npx @metaplay/metaplay-auth@latest get-kubeconfig <organization>-
<project>-<environment> --output my-kubeconfig # Use the generated kubeconfig with kubectl export KUBECONFIG=$(pwd)/my-kubeconfig # On windows powershell $env:KUBECONFIG="$(pwd)\my-kubeconfig"
The generated my-kubeconfig uses the metaplay-auth login to authenticate to the cluster, so the authentication will be valid for as long as the metaplay-auth session lasts.
This is how to list the server pods in an environment:
If the kubeconfig was generated with metaplay-auth older than v1.3.0, the kubectl requires explict namespace argument. The argument is in the form of kubectl -n <namespace> .., for example kubectl -n
idler-develop get pods. On later metaplay-auth versions the namespace can be omitted as the generated kubeconfig defaults to the target environment namespace.
You can run the following command to start a Kubernetes ephemeral diagnostics container against one of the server pods:
kubectl debug <pod-name> -it --profile=general --image metaplay/diagnostics:latest --target shard-server # For example: kubectl debug all-0 -it --profile=general --image metaplay/diagnostics:latest
--target shard-server
This gives you a shell which can access the running server process. The metaplay/diagnostics image contains various diagnostics tools to help debug the server's CPU or memory consumption, including
the .NET diagnostics tools, Linux perf, curl, and others.
The container automatically detects which user is running the target server process: app for chiseled base images and root for classic base images. A shell is opened for that user so that the .NET
diagnostics tools work without further tricks. The shell starts in /tmp directory as all users can write files there.
Starting from Release 28, Metaplay uses the chiseled .NET base images, which are distroless and contain no shell, so an ephemeral container needs to be used. The distroless images are considered much
safer as their attack surface is substantially smaller than that of a full OS image.
Depending on infrastructure version, the target container name may be metaplay-server instead of shard-server. If kubectl debug gives an error message of a missing target container, try again with
--target metaplay-server.
Run the following command to copy a file from a pod to your local machine:
kubectl cp <pod-name>:<path-to-file> ./<filename> # For example: kubectl cp all-0:/tmp/some-diagnostic-file ./some-diagnostic-file # To copy a file from the `kubectl debug` debug container: kubectl
cp <pod-name>:<path-to-file> ./<filename> --container <debugger-container-id> # For example: kubectl cp all-0:/tmp/some-diagnostic-file ./some-diagnostic-file --container debugger-qqw2k
The file system on the containers is ephemeral and gets wiped out if the container is restarted. If you perform diagnostics operations that generate any files you'd like to keep, you should retrieve
them immediately to your local machine to avoid accidentally losing them.
If the source file does not exist, kubectl cp does not always print an error message and instead silently completes. Due to ephemerality of the source filesystem, always copy files to a new file name
or delete the destination file first to avoid using data from an earlier kubectl cp call.
Taking heap dumps of the game server binary can take a long time, during which the game server is completely unresponsible. This causes the Kubernetes liveness probes to fail which leads to the
container getting killed by Kubernetes after enough probe failures (around 30sec by default).
The metaplay-gameserver Helm chart v0.6.1 introduces sdk.useHealthProbeProxy which, when enabled, causes the health probes to get sent via the entrypoint binary running in the container.
If the sdk.useHealthProbeProxy is enabled, you can override the health probe proxy to always return true for the liveness probe with the following which prevents Kubernetes from killing the pod due
to the liveness probes failing during a heap dump operation:
The override is applied for 30 minutes after which is returns to normal behavior, i.e., forwarding the health probes to the game server process. You can explicitly remove the proxy override with the
Debugging crashed servers can be tricky in Kubernetes as by default the container restarts also cause the contents of the file system to be lost. This means that crash dumps written to the disk are
also lost as the container restarts after a crash.
Metaplay configures a volume mount to be mapped onto the game server containers at /diagnostics and core dumps are written there in case of server crashes. These volumes share the lifecycle of the
pod, as opposed to the container. Thus, the core dumps are retained over server restarts and can be retrieved for debugging purposes.
kubectl cp <pod-name>:/diagnostics/<file> ./core-dump # For example: kubectl cp all-0:/diagnostics/<file> ./core-dump
To analyze the heap dump, you can use the interactive analyzer:
See .NET Guide to dotnet-dump on how to use dotnet-dump to analyze the core dump.
You can use the /diagnostics directory for your own purposes as well, if you need storage that is slightly more persistent than the regular container file system.
Note that re-scheduling the pod, either due to deploying a new version of the server, or for example due to a failed EC2 node, the volumes are lost and any crash dumps along with them.
The .NET performance counters are a good troubleshooting starting point and give a good overview of the health of the running pod.
# Find the PID of the running server dotnet-counters ps # Monitor PID 100 dotnet-counters monitor -p 100
Here are some good overall indicators to check and see if the following counters are within reasonable limits:
You can also take a look at the Dotnet Diagnostic Tools CLI Design page from the .NET Diagnostics repository for an overview of the .NET troubleshooting tools available.
You can use the dotnet-trace command to collect a CPU profile from the game server. Here we're collecting it from a 30-second interval.
# Find the PID of the running server dotnet-trace ps # Collect from PID 100 dotnet-trace collect -p 100 --duration 00:00:00:30
This will output a file called trace.nettrace. To retrieve the file to your local machine, use the following:
# Note: File is written to the debug container's filesystem kubectl -n <namespace> cp <pod-name>:<path>/trace.nettrace ./trace.nettrace --container <debugger-container-id> # For example: kubectl -n
idler-develop cp all-0:/tmp/trace.nettrace ./trace.nettrace --container debugger-qqw2k
If you're using Visual Studio (recommended for Windows users), just drag the file into your IDE.
You can also use Speedscope. Run dotnet-trace convert <tracefile> --format Speedscope to convert the image to Speedscope format, which generates a JSON file that you can open in https://
Note that by default, Speedscope only shows a single thread. You can switch threads from the top-center widget. There are a few different views available. Left Heavy view is good for getting overall
CPU usage, and the Time Order view is good for analyzing short-term spikes.
You can check out the Dotnet Docs if you want to dive a little deeper into the Dotnet Trace tools.
The recommended way for tracing memory leaks is to start with a Load Testing run on your local machine:
Collecting a memory dump with either dotnet-gcdump or dotnet-dump can take a long time, and the process is completely frozen during the operation. This will pause the game for all the players.
If the heap is large enough (generally multiple gigabytes), the operation can take long enough for the Kubernetes health checks to consider the container unhealthy and restart it. Please see Health
Probe Overrides on how to override the health probes temporarily to prevent this from happening.
For a more in-depth guide, see the .NET Guide to Debugging Memory Leaks.
In general, if you have access to a Windows machine, you should start with dotnet-gcdump:
See .NET Diagnostics Tools: dump vs. gcdump for a more detailed comparison between the two and detailed instructions on using each.
In the cloud, the diagnostic docker image comes with the tool pre-installed. To install the tool locally, you can run the following command on your machine:
# Find the PID of the running server dotnet-gcdump ps # Use tool on PID 100 dotnet-gcdump collect -p 100
This will create a file named something like 20240325_095122_28876.gcdump. To retrieve the file from the Kubernetes pod, run the following:
# Note: File is written to the debug container's filesystem kubectl cp <pod-name>:<path>/<filename> ./<filename> --container <debugger-container-id> # For example: kubectl cp all-0:/tmp/
20240325_095122_28876.gcdump ./20240325_095122_28876.gcdump --container debugger-qqw2k
To analyze the heap dump, you can drag the file into Visual Studio to open it or open the file in PerfView.
The dotnet-dump tool can be used to collect and analyze full memory dumps of a running process. Analyzing the dump must happen on the same OS where the dump originated, so a Linux machine is required
to analyze dumps from docker images running in Kubernetes clusters.
In the cloud, the docker images have the tool pre-installed. Alternatively, you can run the following command to install the tool in locally in your machine:
# Find the PID of the running server dotnet-dump ps # Use tool on PID 100 dotnet-dump collect -p 100
This will create a file named something like dump_20240325_095609.dmp. To retrieve the file from the Kubernetes pod, run the following:
kubectl cp <pod-name>:<path>/xxxxx.dmp ./xxxxx.dmp # For example: kubectl cp all-0:/tmp/xxxxx.dmp ./xxxxx.dmp
To analyze the heap dump, you can use the interactive analyzer.
Here are some useful commands for a good starting point:
These are some resources you can use to learn more about these tools and how to use them effectively:
The LLDB debugger can be used to dig deeper into the memory heap dumps. You can use it to dump aggregate amounts of memory used by types and trace the object graph to understand which objects are
referenced by whom.
Take a look at the following articles for more information:
This slide presentation by Pavel Klimiankou also has some useful insights about using Perfview, LTTng and LLDB: https://www.slideshare.net/pashaklimenkov/troubleshooting-net-core-on-linux | {"url":"https://docs.metaplay.io/game-server-programming/how-to-guides/troubleshooting-servers.html","timestamp":"2024-11-15T00:22:14Z","content_type":"text/html","content_length":"175091","record_id":"<urn:uuid:68b15e29-3697-4018-88b1-e6e5f60ffe16>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00399.warc.gz"} |
Precise zero energy bound for supersymmetry
Usually we can shift the energy $$E$$ by any amount $$\delta$$ to redefine the lowest energy as $$E + \delta.$$
However, in supersymmetry, there is a precise $$E=0$$ must be true, so that the supercharge $$Q$$ annihilates some state $$|\psi_{min}\rangle$$ to give the minimal energy $$Q |\psi_{min}\rangle =0$$
and also $$H|\psi_{min}\rangle =Q^2|\psi_{min}\rangle=0.$$
This raises the question that we have a precise zero energy bound for supersymmetry theory.
Does it mean that we cannot shift the energy $$E$$ to $$E + \delta$$ in supersymmetry theory? What is the deep reason behind it?
This post imported from StackExchange Physics at 2020-12-13 12:43 (UTC), posted by SE-user annie marie heart
I will offer two arguments to try to exhibit what the problem with ground states with non zero energy in a supersymmetric theory is.
The case of gauged supersymmetry: Energy backreacts on the geometry.
If the energy of the ground states of a supersymmetric theory were non exactly zero the underlying geometry of the background would be allowed to be curved in arbitrary ways. The latter is imposible
in a supersymmetric theory because supersymmetry enforce at least an spin structure of the underlying background geometry or possibly a trivial canonical bundle or a special holonomy (as in the
Calabi-Yau, $$G_{2}$$ or $$Spin(7)$$ cases) on the target manifold.
Supersymmetry in quantum mechanics: The hamiltonian in a $$(0+1)$$-supersymmetric theory can be schematically written (see Supersymmetry and Morse theory) as $$H=\frac{1}{2}(Q_{1}^{2}+Q_{2}^{2}).$$
If for some ground state $$\psi$$$$H\psi eq 0,$$ it would follow that $$\psi$$ wouldn't be annihilated by the (positive definite) squares of the supercharges; then $$\psi$$ wouldn't preserve some
It is illustrative to notice that in the case of a Riemannian manifold $$\mathcal{M}$$ the hamiltonian coincides with the laplacian of $$\mathcal{M}$$ (see chapter 2 in Supersymmetry and Morse theory
); if the laplacian was not zero, then you can make an analogy with electrodynamics, you can't be possibly describing the vaccum of the theory, sources must be present.
This post imported from StackExchange Physics at 2020-12-13 12:43 (UTC), posted by SE-user Ramiro Hum-Sah | {"url":"https://www.physicsoverflow.org/43993/precise-zero-energy-bound-for-supersymmetry","timestamp":"2024-11-04T12:08:20Z","content_type":"text/html","content_length":"116136","record_id":"<urn:uuid:604bf939-9eff-46da-b9d4-3c90600c6754>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00060.warc.gz"} |
Fine-Structure Contamination: Observations and a Model of a Simple Two-Wave Case
1. Introduction
Eulerian records of ocean quantities (e.g., shear) typically suffer from “fine-structure contamination,” where existing vertical structure is swept past fixed-depth sensors by internal wave motions.
Perceived time variability (in the absence of horizontal advection) has intrinsic and advective terms:
Frequency spectra of these contaminated records can be difficult to interpret since Doppler-shifted frequencies, indicative of the
term, are present as well as those associated with the intrinsic
This Doppler shifting can be reduced by examining quantities in a semi-Lagrangian or isopycnal-following frame (
Pinkel and Anderson 1997
). The resultant spectra are much more indicative of intrinsic processes (
Anderson 1993
). For example,
Kunze et al. (1990)
found that observed shear frequency spectra could be almost entirely explained by vertically advected near-inertial motions.
Typically, the intrinsic and advective motions have broad and overlapping frequency ranges. A common “narrowband” situation occurs when near-inertial shear layers are advected by tidal displacements.
But, in midlatitudes the nearness of the inertial and tidal frequencies makes even this situation difficult to interpret.
This paper seeks to provide insight into fine-structure contamination by demonstrating the mechanism in a narrowband (two wave) situation. Measurements of velocity, shear, and isopycnal displacement
at 6.5°S are presented, where the Coriolis frequency (f = 1/4.4 cpd) is well below the dominant (K[1], diurnal) tidal frequency. This spectral separation allows the tidal heaving of the near-inertial
shear layers to be seen clearly. The mechanism is evident in both depth–time maps and Eulerian frequency spectra, which show sharp peaks at f ± K[1]. The time series and spectra are well modeled by a
simple kinematic two-wave model. As expected, the shifted peaks are absent when the shear is examined in the isopycnal-following frame.
The theoretical framework and the two-wave model are described in the next section, followed by a brief description of the data collection and processing in section 3. Depth–time maps and frequency
spectra of shear are presented and compared to model predictions in section 4. Conclusions follow.
2. Theory
a. Semi-Lagrangian frame
Consider a motionless, stratified ocean. The density profile,
, may be inverted to yield the at-rest depth,
of any isopycnal surface of density
The coordinate
defines the semi-Lagrangian (s-L) or isopycnal-following frame, and, in the at-rest ocean, is identical to the Eulerian depth
Both coordinates are taken positive downward.
When some process (say, the tide) vertically displaces the isopycnals, their instantaneous depth is given by
ρ, t
) is the instantaneous upward displacement of the isopycnal with density
Now, an Eulerian measurement of shear (say) at a particular depth records the shear on whatever isopycnal is presently at that depth. This is the essence of fine-structure contamination.
Equation (3)
is the mathematical equivalent of this statement, and relates the s-L and Eulerian frames. Transformation to the s-L frame removes the variability associated with vertical structure that is swept
past fixed-depth sensors by isopycnal displacements.
b. Two-wave tidal heaving model
Within the above framework, consider the following two-wave model of tidally heaved inertial shear.
The rotary velocity field,
of a near-inertial wave (frequency
) is given in the isopycnal-following frame by
the real and imaginary parts of
yield the zonal and meridional velocity components of a counterclockwise-polarized (Southern Hemisphere) inertial wave with upward phase propagation (
Fig. 1
, gray lines).
A sinusoidal internal tide (solid lines) of amplitude
displaces the isopycnals at the diurnal frequency,
Since the internal tide is typically very low mode, the depth dependence is neglected. This assumption is valid over a limited depth range away from the surface and bottom, provided the inertial
wavelength is less than that of the internal tide. In this limit, there is no strain (∂
= 0), and therefore
t, z
) =
t, ζ
The plane-wave solution,
, is distorted by tidal heaving when viewed in the Eulerian reference frame, as seen by substitution of
If the advection is small compared to an inertial wavelength (
/2 ≪ 1), then
Substitution of
Doppler shifting has introduced, in addition to the original wave, constituents at the frequencies
. The ratio of the spectral amplitude, Φ, of the Doppler-shifted peaks (last two terms) to that of the original (first term) is then
The Eulerian shear of the inertial wave is given by
, ∂
= 1 + ∂
For small strain, ∂
≪ 1, then ∂
≈ ∂
and Eulerian and semi-Lagrangian shear differ only via advection and not by straining. The analysis for shear is thus identical to that for velocity.
3. Data
Data for this study was collected during October/November 1998 aboard the Indonesian Vessel Baruna Jaya IV, as part of the ARLINDO Microstructure experiment. The ship repeated 36-km legs centered at
6.5°S, 128°E, in the central Banda Sea. The goal of the study was to quantify the mixing occurring in the Indonesian Throughflow. These results are described in Alford et al. (1999).
The shear field was dominated by a downward-propagating near-inertial wave of amplitude S[max] ≈ 0.02 s^−1, which was responsible for three-fourths of the observed mixing (Alford and Gregg 2001).
Specifically, dissipation rate and diapycnal diffusivity were both coherent at the 95% confidence level with inertial-band shear and Froude number.
Diurnal tidal currents (0.11 m s^−1) and displacements (9 m) were larger than their semidiurnal counterparts (0.09 m s^−1 and 4 m). Both constituents were very low wavenumber (as often observed),
resulting in very weak tidal shear.
a. Isopycnal displacement data
Isopycnal displacement data were obtained from the Modular Microstructure Profiler (MMP), a loosely tethered vehicle equipped with microstructure and Sea-Bird CTD sensors. Profiles, spaced 20 minutes
apart, were made to 300-m depth in 4-hour bursts separated by 2-hour gaps. To form a continuous time series, these data were interpolated onto a 5-m, 3-h grid, with each temporal bin containing 2
hours of data and a 1-hour gap. Isopycnal displacement was computed by first selecting a set of evenly spaced isopycnal surfaces, based on the cruise-mean density profile. Then, the instantaneous
depth of each was computed from each density profile by linear interpolation.
b. Velocity and shear data
Velocity and shear records were obtained using an RDI 150-KHz broadband acoustic Doppler current profiler mounted on the ship's bottom. Low scattering strength limited the device's useful range to
about 160 m (quantitative reliability begins to fail sooner, at 120 m). The vertical resolution was 4 m. Shear estimates were computed by differencing over 8 m. Finally, velocity and shear were
averaged over the same 3 hours, and interpolated onto the same 5-m grid, as the isopycnal displacement data.
Semi-Lagrangian records of shear and velocity were computed from the Eulerian records by interpolating onto the instantaneous depth of each isopycnal surface obtained from MMP.
4. Results
a. Depth–time maps
The basic tidal heaving mechanism is visible in the Eulerian map of zonal shear (Fig. 2a). Isopycnal depths z(ρ, t), with mean spacing 10 m, are plotted in black. The downgoing near-inertial wave is
evident as broad upward-sloping shear bands with wavelength 2πm^−1[o] ≈ 70 m, and period 2πω^−1[o] ≈ 2πf^−1 = 4.4 days.^^4
The near-inertial shear layers are advected by the (primarily diurnal) isopycnal displacement field. The s-L shear field (Fig. 2b), which has most of the advection removed, does not show this
distortion. Consequently, the inertial bands are much straighter when viewed in the s-L frame. Residual deviations of s-L shear from inertial phase lines are due either to errors in interpolation and
isopycnal depth, or to intrinsic diapycnal motions.
Figure 2a shows fine-structure contamination. To make this clear, a single representative Eulerian time series from 93-m depth (a horizontal slice through Fig. 2a) is low-pass filtered with a cutoff
1.5 cpd (to focus on heaving by the diurnal tide alone), and plotted in Fig. 2c. The time series clearly contains frequency components other than inertial. Good agreement is seen with the two-wave
model (Fig. 2c, black line) when observed wave parameters and a visual best-fit phase (2πm^−1[o] = 70 m, η[o] = 9 m, ϕ[o] = π/2) are used.^^5
A time series of s-L shear at the same depth (that is, shear across the isopycnal whose mean depth is 93 m) is much more sinusoidal (Fig. 2d, black line), indicating that the effects of advection are
much reduced. The model (black line) is, of course, perfectly sinusoidal in the s-L frame [Eq. (4)].
b. Frequency spectra
The frequency content of the data are examined via rotary spectral analysis (Gonella 1972). Spectra of rotary velocity u + iυ, rotary shear u[z] + iυ[z], and isopycnal displacement are computed by
demeaning, detrending, and Hanning and Fourier transforming each time series between 50 and 120 m and over yeardays 296.5–307.4. The transformed series are then averaged together to form one spectral
estimate. For shear and velocity, spectra are computed for both Eulerian and semi-Lagrangian quantities.
The resulting spectra are plotted vs frequency in Fig. 3. The rotary spectra (Figs. 3a,c) are plotted vs negative/positive frequency, representing clockwise/counterclockwise motions. (The
displacement spectrum is plotted vs positive frequency, since the positive/negative portions are redundant for a real time series.) Note that the plotting scale is linear in the abscissa but
logarithmic in the ordinate. Confidence limits are indicated. The spectra are plotted as stairs to indicate the limited frequency resolution, Δf = 1/(10.9 days) = 0.09 cpd. Eulerian and
semi-Lagrangian spectra are plotted in red and blue, respectively.
A strong inertial peak is present in the spectrum of all three quantities. Its presence at +f in the rotary spectra (Figs. 3a,c) indicates its counterclockwise sense of rotation, consistent with
Southern Hemisphere dynamics. It especially dominates the shear records, containing 68% of the total variance.
The internal tide is evident in the velocity and displacement fields. The K[1] tide is nearly completely counterclockwise polarized, while the M[2] constituent^^6 is much more nearly even.
The absence of corresponding peaks at the K[1] and M[2] frequencies in the Eulerian shear spectrum (Fig. 3c, red line) confirms the earlier assertion that the tidal motions are low mode. Present
instead are peaks at f ± K[1], whose magnitudes are accurately predicted by the two-wave model [Eq. (9), black circles]. A cursory inspection of the spectrum would lead to the erroneous conclusion
that the K[1] tide contains strong shear. The presence of the shifted peaks in the Eulerian shear spectrum, which has no discernible tidal constituents, is instead a direct consequence of diurnal
tidal heaving of near-inertial shear.
This conclusion is supported by the semi-Lagrangian spectra (blue lines): the peaks at f ± K[1] are greatly attenuated. [The effect is especially visible in the shear spectra (Fig. 3c), where there
are no peaks at K[1], but can also be seen in those of velocity (Fig. 3a).] The loss of these peaks in the s-L spectra is due to the absence of advective frequencies in the moving reference frame.
Results for M[2] are less clear. A peak at f + M[2] is also close to the predicted magnitude for heaving by the M[2] tide (black circle), but a corresponding peak is not present at f − M[2]. Though
the data are not conclusive, a peak at f + M[2] but not at f − M[2] is suggestive of a freely propagating wave resulting from nonlinear interaction between f and M[2], as argued by Mihaly et al.
(1998). A proper treatment of this issue is hampered by the weaker M[2] signal and the possibility of influence from the 3-h bin length, which yields only four points per M[2] tidal cycle.
5. Conclusions
Data and a model are presented that illustrate a simple, two-wave case of the “fine-structure contamination” problem. Depth–time maps and spectra of velocity, shear and isopycnal displacement are
examined from a low-latitude site, where the Coriolis frequency is much less than the diurnal tidal frequency. As a result of this spectral separation, diurnal tidal heaving of the near-inertial
motions is clearly identifiable in depth–time maps and Eulerian frequency spectra, which show “contamination” peaks at f ± K[1]. A simple two-wave model produces good agreement with observed time
series and accurately predicts the magnitude of the shifted peaks. The effect is much reduced by transforming to an isopycnal-following frame, where advection is minimized.
No new physics are introduced in this paper. It is emphasized that while nonlinear interactions between the internal tides and the near-inertial wave are not ruled out, they are not necessary to
explain the observations: a purely kinematic effect is responsible for the observed distortion of the near-inertial shear layers. These observations are a reminder (hardly needed to most
oceanographers) that even the simplest ocean situations can contain subtle complications that require careful interpretation.
This work was supported by M.A.'s startup funding at the Applied Physics Laboratory. The data collection and initial analysis were supported by NSF Grant OCE9729288. I am grateful to Mike Gregg for
data, guidance, and support. Conversations with Chris Garrett, Eric Kunze, Steve Mihaly, and Dave Winkel were helpful.
• Alford, M., and M. Gregg, 2001: Near-inertial mixing: Modulation of shear, strain and microstructure at low latitude. J. Geophys. Res., in press.
• Alford, M., M. Gregg, and M. Ilyas, 1999: Diapycnal mixing in the Banda Sea: Results of the first microstructure measurements in the Indonesian Throughflow. Geophys. Res. Lett, 26 (17) 2741–2744.
• Anderson, S. P., 1993: Shear, strain and thermohaline vertical fine structure in the upper ocean. Ph.D. thesis, University of California, San Diego, 143 pp.
• Gonella, J., 1972: A rotary-component method for analysing meteorological and oceanographic vector time series. Deep-Sea Res, 19 , 833–846.
• Kunze, E., M. G. Briscoe, and A. J. Williams III, 1990: Observations of shear and vertical stability from a neutrally buoyant float. J. Geophys. Res, 95 , (C10). 18127–18142.
• Mihaly, S. F., R. Thomson, and A. B. Rabinovich, 1998: Evidence for nonlinear interaction between internal waves of inertial and semidiurnal frequency. Geophys. Res. Lett, 25 (8) 1205–1208.
• Pinkel, R., and S. Anderson, 1997: Shear, strain and Richardson number variations in the thermocline. Part I: Statistical description. J. Phys. Oceanogr, 27 , 264–281.
Fig. 1.
Schematic diagram of a downward-propagating near-inertial wave in the presence of diurnal isopycnal displacements. Locations where the velocity of the near-inertial wave exceeds 90% of its maximum
are shaded in gray. The vertical Eulerian coordinate (z, dotted lines) differs from the semi-Lagrangian depth coordinate, ζ, by diurnal isopycnal heaving of amplitude η[o] [Eq. (5), solid lines]
Citation: Journal of Physical Oceanography 31, 9; 10.1175/1520-0485(2001)031<2645:FSCOAA>2.0.CO;2
Fig. 2.
Eulerian (a) and semi-Lagrangian (b) depth–time maps of zonal 8-m shear. Isopycnal depths whose mean spacing is 10 m are overplotted in (a). Single-depth (93 m) Eulerian time series (c) of observed
(red) and model [Eq. (6)] (black) shear. Semi-Lagrangian (d) observed (blue) and model (black) time series of shear from the isopycnal whose mean depth is 93 m
Citation: Journal of Physical Oceanography 31, 9; 10.1175/1520-0485(2001)031<2645:FSCOAA>2.0.CO;2
Fig. 3.
Frequency spectra of rotary velocity (a), isopycnal displacement (b), and rotary shear (c). Eulerian spectra are plotted in red, and semi-Lagrangian spectra in blue. Counterclockwise/clockwise
motions appear in (a) and (c) at positive/negative frequencies. Displacement is real, so its spectrum is plotted vs positive frequency only. Vertical dotted lines indicate inertial, tidal, and
inertial plus/minus tidal frequencies, indicated at the top of (a). In (c), the predicted amplitudes from the two-wave heaving model [Eq. (9)] are shown as black circles for the K[1] and M[2]
Citation: Journal of Physical Oceanography 31, 9; 10.1175/1520-0485(2001)031<2645:FSCOAA>2.0.CO;2
A similar derivation of the model, with some algebra errors, may be found in Anderson (1993).
Throughout this paper, all frequencies are taken in radian units, but plotting is done in cyclic units.
The internal wave equations are specified in terms of z, not ζ. Equation (4) is a valid solution because tidal vertical accelerations associated with the displacements are miniscule, and do not
affect the dynamics.
Alford and Gregg (2001) found 2πm^−1[o] ≈ 100 m when they accounted for WKB refraction. Here, the lower 2πm^−1[o] ≈ 70 m reflects the reduction of the wave scale near the stratification peak near
100-m depth.
Since m[o]η[o]/2 = 0.35 renders (7) marginally valid, a numerical evaluation of (6) is used rather than (7). The two differ somewhat in phase, but are nearly identical in magnitude.
The spectral resolution is not sufficient to distinguish between 2 cpd and M[2]. Peaks near 2 cpd are interpreted as M[2] tidal, though they could be harmonics of the diurnal tide. Likewise,
frequency shifting of the near-inertial wave by the mean flow and background vorticity, as examined by Alford and Gregg (2001), is too small to be resolved spectrally. | {"url":"https://journals.ametsoc.org/view/journals/phoc/31/9/1520-0485_2001_031_2645_fscoaa_2.0.co_2.xml","timestamp":"2024-11-07T09:50:19Z","content_type":"text/html","content_length":"451002","record_id":"<urn:uuid:87bbc3a2-f06d-4b20-8771-f64bb7052dc7>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00036.warc.gz"} |
What are some examples of deductive reasoning?
What are some examples of deductive reasoning?
Examples of deductive logic:
• All men are mortal. Joe is a man. Therefore Joe is mortal.
• Bachelors are unmarried men. Bill is unmarried. Therefore, Bill is a bachelor.
• To get a Bachelor’s degree at Utah Sate University, a student must have 120 credits. Sally has more than 130 credits.
What activities require deductive reasoning?
1 Guess the Coin. Give students clues about which coin you are thinking of as students narrow down the possibilities using deductive reasoning.
2 Detective Work.
3 Classmate Claims.
4 Dystopian Literature Propaganda.
Is deductive reasoning based on examples?
Deductive reasoning is a type of deduction used in science and in life. It is when you take two true statements, or premises, to form a conclusion. For example, A is equal to B. Given those two
statements, you can conclude A is equal to C using deductive reasoning.
How can deductive reasoning be used in daily life?
The following is a formula often used in deduction: If A = B and B = C, then in most cases A = C. So, for example, if traffic gets bad starting at 5 p.m. and you leave the office at 5 p.m., it can be
deductively reasoned that you’ll experience traffic on your way home.
Which is the best example of deductive reasoning?
For example, “All men are mortal. Harold is a man. Therefore, Harold is mortal.” For deductive reasoning to be sound, the hypothesis must be correct. It is assumed that the premises, “All men are
mortal” and “Harold is a man” are true.
What does deductive reasoning mean?
Deductive reasoning, or deductive logic, is a type of argument used in both academia and everyday life. Also known as deduction, the process involves following one or more factual statements (i.e.
premises) through to their logical conclusion.
Is deductive reasoning always true?
Deductive reasoning goes in the same direction as that of the conditionals, and links premises with conclusions. If all premises are true, the terms are clear, and the rules of deductive logic are
followed, then the conclusion reached is necessarily true. In deductive reasoning there is no uncertainty.
What is a valid deductive argument?
A deductive argument is said to be valid if and only if it takes a form that makes it impossible for the premises to be true and the conclusion nevertheless to be false. Otherwise, a deductive
argument is said to be invalid.
What is the difference between inductive and deductive arguments?
Deductive reasoning is sometimes described as a “top-down” form of logic, while inductive reasoning is considered “bottom-up.”. A deductive argument is one in which true premises guarantee a true
conclusion. In other words, it is impossible for the premises to be true but the conclusion false.
What is the difference between inductive and deductive?
The main difference between inductive and deductive approaches to research is that whilst a deductive approach is aimed and testing theory, an inductive approach is concerned with the generation of
new theory emerging from the data.
What are non – deductive arguments?
Definition: A non-deductive argument is an argument for which the premises are offered to provide probable – but not conclusive – support for its conclusions. In a good non-deductive argument, if the
premises are all true, you would rightly expect the conclusion to be true also,…
What is deductive logic?
Deductive reasoning, also deductive logic, is the process of reasoning from one or more statements (premises) to reach a logically certain conclusion. Deductive reasoning goes in the same direction
as that of the conditionals, and links premises with conclusions. | {"url":"https://www.pursuantmedia.com/2020/06/23/what-are-some-examples-of-deductive-reasoning/","timestamp":"2024-11-08T04:11:56Z","content_type":"text/html","content_length":"59708","record_id":"<urn:uuid:bc002e1b-8b69-40a4-bb1d-a40202251d85>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00103.warc.gz"} |
The Three Little Pigs | R-bloggersThe Three Little Pigs
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Jesse, you asked me if I was in the meth business or the money business. Neither. I’m in the empire business (Walter White in Breaking Bad)
The game of pig has simple rules but complex strategies. It was described for the first time in 1945 by a magician called John Scarne. Playing the pig game is easy: each turn, a player repeatedly
rolls a die until either a 1 is rolled or the player decides to hold:
• If the player rolls a 1, they score nothing and it becomes the next player’s turn
• If the player rolls any other number, it is added to their turn total and the player’s turn continues
• If a player chooses to hold, their turn total is added to their score, and it becomes the next player’s turn
The first player who reach at least 100 points is the winner. For example: you obtain a 3 and then decide to roll again, obtaining a 1. Your score is zero in this turn. Next player gets the sequence
3-4-6 and decides to hold, obtaining a score of 13 points in this turn.
Despite its simplicity, the pig game has a very complex and against-intuition optimal strategy. It was calculated in 2004 by Todd W. Neller and Clifton Presser from Gettysburg College of
Pennsilvania with the help of computers.
To illustrate the game, I simulated three players (pigs) playing the pig game with three different strategies:
• The Coward pig, who only rolls the die a small number of times in every turn
• The Risky pig, who rolls the die a more times than the coward one
• The Ambitious pig, who tries to obtain in every turn more points than two others
I simulated several scenarios.
• Some favorable scenarios for Coward pig:
In first scenario, the Coward pig rolls the die between 1 and 5 times each round and wins if the Risky pig asumes an excessive level of risk (rolling each time between 10 and 15 times). Trying to
obtain more than the Coward is a bad option for the Ambitious pig. Simulating this scenario 100 times gives victory to Coward a 51% of times (25% to Risky and 24% to Ambitious).
Second scenario puts closer Coward and Risky pigs (first one rolls the die between 4 and 7 times each round and second one between 6 and 9 times). Coward wins 54% of times (34% Risky and only 12%
Being coward seems to be a good strategy when you play against a reckless or when you are just a bit more conservative than a Risky one.
• Some favorable scenarios for Risky pig:
Rolling the die between 4 and 6 times each round seems to be a good option, even more when you are playing against a extremely conservative player who rolls no more than 3 times each time. Simulating
100 times these previous scenarios gives victory to Risky pig a 58% of times in first the case in which Coward rolls allways 1 and Risky 6 times each round (0% for Coward and only 42% form Ambitious)
and 66% of times in the second one (only 5% to Coward and 29% to Ambitious).
Being Risky is a good strategy when you play against a chicken.
• Some favorable scenarios for Ambitious pig:
The Ambitious pig wins when two others turn into extremely coward and risky pigs as can be seen in the first scenario in which Ambitious wins 65% of the times (31% for Coward and 4% for Risky).
Ambitious pig also wins when two others get closer and hit the die a small number of times (2 rolls the Coward and 4 rolls the Risky). In this scenario the Ambitious wins 58% of times (5% for Coward
and 37% for Risky). By the way, these two scenarios sound very unreal.
Being ambitious seems to be dangerous but works well when you play against a crazy and a chicken or against very conservative players.
From my point of view, this is a good example to experiment with simulations, game strategies and xkcd style graphics.
The code:
#Number of hits for Coward
#Number of hits for Risky
game=data.frame(ROUND=0, part.p1=0, part.p2=0, part.p3=0, Coward=0, Risky=0, Ambitious=0)
while(max(game$Coward)<100 & max(game$Risky)<100 & max(game$Ambitious)<100)
#Coward Little Pig
p1=sample(1:6,sample(CowardLower:CowardUpper,1), replace=TRUE)
#Risky Little Pig
p2=sample(1:6,sample(RiskyLower:RiskyUpper,1), replace=TRUE)
#Ambitious Little Pig
repeat {
if (p3==1|s3>max(s1,s2)) break
panel.background = element_rect(fill="darkolivegreen1"),
panel.border = element_rect(colour="black", fill=NA),
axis.line = element_line(size = 0.5, colour = "black"),
axis.ticks = element_line(colour="black"),
panel.grid.major = element_line(colour="white", linetype = 1),
panel.grid.minor = element_blank(),
axis.text.y = element_text(colour="black"),
axis.text.x = element_text(colour="black"),
text = element_text(size=25, family="xkcd"),
legend.key = element_blank(),
legend.position = c(.2,.75),
legend.background = element_blank(),
plot.title = element_text(size = 50)
ggplot(game, mapping=aes(x=game$ROUND, y=game$Coward)) +
geom_line(color="red", size=1.5) +
geom_line(aes(x=game$ROUND, y=game$Risky), color="blue", size=1.5) +
geom_line(aes(x=game$ROUND, y=game$Ambitious), color="green4", size=1.5) +
geom_point(aes(x=game$ROUND, y=game$Coward, colour="c1"), size=5.5) +
geom_point(aes(x=game$ROUND, y=game$Risky, colour="c2"), size=5.5) +
geom_point(aes(x=game$ROUND, y=game$Ambitious, colour="c3"), size=5.5) +
ggtitle("THE THREE LITTLE PIGS") +
xlab("ROUND") + ylab("SCORING") +
geom_text(aes(max(game$ROUND), max(max(game$Coward, game$Risky, game$Ambitious)), hjust=1.2, family="xkcd", label="WINNER!"), size=10)+
geom_hline(yintercept=100, linetype=2, size=1)+
scale_y_continuous(breaks=seq(0, max(max(game$Coward, game$Risky, game$Ambitious))+10, 10))+
scale_x_continuous(breaks=seq(0, max(game$ROUND), 1))+
labels = c(paste("Coward: ", CowardLower, "-", CowardUpper, " hits", sep = ""), paste("Risky: ", RiskyLower, "-", RiskyUpper, " hits", sep = ""), "Ambitious"),
breaks = c("c1", "c2", "c3"),
values = c("red", "blue", "green4"))+ opts | {"url":"https://www.r-bloggers.com/2014/06/the-three-little-pigs/","timestamp":"2024-11-06T21:03:00Z","content_type":"text/html","content_length":"102665","record_id":"<urn:uuid:5b85c072-c572-42b0-8ec6-fb6ad39c3680>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00046.warc.gz"} |
▷ INEQUALITIES | What is an inequality and how is it resolved?
Inequalities are algebraic expressions. They have two members between which one of these signs appears: <, = <,>,> =. Solution of an inequality is called any value of the unknown that makes the
inequality true.
Sometimes, the statements that give rise to an algebraic expression do not say “is equal to”, but “is greater than” or “is less than”. These statements give rise to expressions like these:
Inequalities usually have infinite solutions (there is only one equal number, but there are infinite numbers less than another). If the two members of an inequality are multiplied or divided by a
negative number, the inequality changes direction.
Solving inequalities
To solve an equation, we followed a series of steps: remove parentheses, remove denominators, pass x to one member and numbers to the other … All of them are valid, exactly the same, for
inequalities, except one:
If the two members of an inequality are multiplied or divided by a negative number, the inequality changes direction.
Ejemplo de inecuación:
1. Solve:
Ejercicios de inecuaciones
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The number 26 has 4 factors and is a composite number. To calculate the factors of 26 we divide,...
El número 25 tiene 3 divisores y es compuesto. Para calcularlos dividimos siempre que podamos...
The number 24 has 8 factors and is a composite number. To calculate the factors of 24 we divide,...
The number 23 has 2 factors and is a prime number. To calculate the factors of 23 we divide,...
The number 22 has 4 factors and is a composite number. To calculate the factors of 22 we divide,...
The number 21 has 4 factors and is a composite number. To calculate the factors of 21 we divide,...
The number 20 has 6 factors and is a composite number. To calculate the factors of 20 we divide,...
The number 19 has 2 factors and is a prime number. To calculate the factors of 19 we divide,...
Los divisores de 18 son 6 y es compuesto. Para calcularlos dividimos siempre que podamos entre...
The number 17 has 2 factors and is a prime number. To calculate the factors of 17 we divide,...
Factors of 26
Factors 25
Factors of 24
Factors of 23
Factors of 22
Factors of 21
Factors of 20
Factors of 19
Factors of 18
Factors of 17 | {"url":"https://aulaprende.com/en/inequalities/page/2/?et_blog","timestamp":"2024-11-02T18:46:15Z","content_type":"text/html","content_length":"313426","record_id":"<urn:uuid:4fa8889a-0557-4ea4-a3d2-fd5b2b2cac8a>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00053.warc.gz"} |
Measure Your Carbon Footprint - Sustainable BuildMeasure Your Carbon Footprint
Measure Your Carbon Footprint
It’s a case where smaller really is more beautiful.
So how do you go about calculating your carbon footprint, when virtually everything you do creates carbon? You need to go through every little thing, and be thorough and honest, until you reach a
final total. That includes what you spend on heating and running the appliances in your home, your car, every facet.
Your Home
For many of us, it’s gas that brings water and heat into the home. Look at the kilowatt hours (kwh) on each of the four quarterly bills for the last year, and add them up. To find out your total
emissions, multiply the number of kilowatt hours by 0.19, then divide by the amount of adults in the house to discover the figure. If you’re not bothered about being completely exact, calculate
10,000 kwh for a small house, 20,500 for a medium-sized house, and 28,000 for a large house
For electricity you should also total the kilowatt hours used over a year, then multiply that figure by 0.43, and finally divide by the number of adults in the house. If the bills have vanished, try
a rough total of 1,650 kwh for a small house, 3,300 kwh for a medium sized house, or 5,000 kwh for a large house.
Your Car
Figuring the carbon footprint for you car is a lot trickier, and any figure is going to be a guesstimate. So much depends on how far you drive, as well as how you drive, on top of the type of vehicle
you own.
As a very general rule, any car will emit its own weight in carbon dioxide for every 6000 miles that you drive. So, first work out how many miles you drive each year. Divide this by 6000 to find out
how many times its own weight of carbon dioxide your car produces in a year. Then you need to estimate the weight of your car, in kilograms.
A small car like a Fiat or a Renault Clio or a Golf will weigh about 1100 kg. A larger family car like a BMW or a large Rover or Mazda might weight around 1800 kg. And if you have a MPV or a Range
Rover or a big, high vehicle like that that can take around 7 people, then your vehicle possible weighs around 2800 kg. If you want to make your calculation as accurate as possible, find the actual
weight in your handbook or from your car manufacturer or from the internet. Search for the model name, the word weight and the word specification and hunt around. Some cars are heavier than you
expect – a mini clubman S, for example, has a gross weight of 1690 kg.
Take the figure you have for the weight of your car and multiply it by the number you got by dividing your total annual mileage by 6000. This will give you an idea of the number of kilograms of
carbon dioxide your car produces each year. Remember though, that this doesn’t take into account the carbon dioxide that was produced when it was manufactured. Your carbon output will also be higher
in any car if you drive it very fast, or do a lot of stop-start journeys in heavy traffic.
Other Items
There are, of course, many other things in your life – your shopping, for instance, and it’s almost impossible to come up with a carbon total for that over the course of a year. Much of our food
comes from abroad, meaning it has “food miles” attached to it – but did you buy strawberries 10 months ago? How would you remember?
Still, there are a few things you can calculate. Take your holiday, for instance. Where did you go? Did you fly there? If so, you can come up with a figure for the CO2 emissions on the flight. Maybe
you went to Turkey, a very popular destination. If you did, that’s 1275 kg of carbon – per person each way. If you take two foreign holidays a year and fly, the figures soon add up. It can be very
instructive to sit down and calculate your carbon footprint, and the chances are that you’ll be surprised at just how large it is. That knowledge can be an important first step in lowering it.
If maths isn’t your strongest point, don’t worry. There are a number of web sites where you can simply enter the figures and they’ll make the calculations for you. However you do it, it’s the
results, and the determination to change things, that matter. | {"url":"https://sustainablebuild.co.uk/measuringyourcarbonfootprint/","timestamp":"2024-11-02T12:27:38Z","content_type":"text/html","content_length":"126971","record_id":"<urn:uuid:35cc5ca1-680f-436d-998a-b5dafa338acc>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00239.warc.gz"} |
Friday Riddle
nudgie Member Posts: 1,478 Member
RIDDLE: COUNTING TO 50
Henry and Gretchen are going to play a game. Henry explains, "You and I will take turns saying numbers. The first person will say a number between 1 and 10. Then the other person will say a number
that is at least 1 higher than that number, and at most 10 higher. We will keep going back and forth in this way until one of us says the number 50. That person wins. I'll start."
"Not so fast!" says Gretchen. "I want to win, so I will start."
What number should Gretchen say to start?
• Guessing
I say 50 - ok I win. LOL I know it has to be between 1 and 10 but I cheated
I say 50 - ok I win. LOL I know it has to be between 1 and 10 but I cheated
i thought that
at first.....even posted it, and went back in and changed it after re-reading....
• Let's play.
For those of you who think Gretchen should start with some number other than 6, I'll be Henry to your Gretchen. If you start with 1-5, as Henry I say 6. If Gretchen starts with 7-10, Henry says
17. It's your turn.
Discussion Boards
• 6 CSN Information
• 121.8K Cancer specific | {"url":"https://csn.cancer.org/discussion/179224/friday-riddle","timestamp":"2024-11-13T22:44:20Z","content_type":"text/html","content_length":"290288","record_id":"<urn:uuid:6babbd69-ea99-4471-9a66-9c4d16daae55>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00658.warc.gz"} |
GED Test Tips: Answering Fill-in-the-Blank Questions - dummies
Fill-in-the-blank questions on the GED test require that you fill in the answer without the benefit of four answer choices to choose from. Often, they involve some calculation, using the information
provided in the question.
Here's an example: Demitri wanted to buy a new television set. His old one had a diagonal measurement of 32 inches, but he wanted to buy a 50-inch diagonal set. The new television set would be
inches wider, measured diagonally.
To answer this question, you have to find the difference between the two TV sets. The new set would be 50 – 32 = 18 inches wider, measured diagonally.
Now try another: Carol found a part-time job to augment her scholarship. She was paid $13.45 an hour and was promised a 15-percent raise after three months. Business had been very poor during that
period, and the owner of the business called Carol in to explain that he could afford only an 11-percent raise but would reassess the raise in the next quarter depending on how business was. With
this raise, Carol's new hourly rate would be
Carol's new salary would be calculated at the rate of $13.45 times 11 percent, or
(to the nearest penny). If you want to calculate the amount of an 11-percent raise, you can multiply by 111 percent
About This Article
This article is from the book:
This article can be found in the category: | {"url":"https://www.dummies.com/article/academics-the-arts/study-skills-test-prep/ged/ged-test-tips-answering-fill-blank-questions-241407/","timestamp":"2024-11-11T03:59:15Z","content_type":"text/html","content_length":"76239","record_id":"<urn:uuid:2cd591c3-55df-4ead-92ef-208cedb15186>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00101.warc.gz"} |
ACO Seminar
We discuss two recent methods in which an object with a certain property is sought. In both, using of a straightforward random object would succeed with only exponentially small probability. The new
randomized algorithms run efficiently and also give new proofs of the existence of the desired object. In both cases there is a potentially broad use of the methodology.
(i) Consider an instance of K-SAT in which each clause overlaps (has a variable in common, regardless of the negation symbol) with at most D others. Lovasz showed that when eD ≤ 2^K (regardless of
the number of variables) the conjunction of the clauses was satisfiable. The new approach due to Moser is to start with a random true-false assignment. In a WHILE loop, if any clause is not satisfied
we "fix it" by a random reassignment. The analysis of the algorithm (due to Don Knuth and others) is unusual, connecting the running of the algorithm with certain Tetris patterns, and leading to some
algebraic combinatorics. [These results apply in a quite general setting with underlying independent "coin flips" and bad events (the clause not being satisfied) that depend on only a few of the coin
(ii) No Outliers. Given n vectors r_j in n-space with all coefficients in [-1,+1] one wants a vector X=(x_1,...,x_n) with all x_i=+1 or -1 so that all dot products (X dot r_j) are at most K sqrt(n)
in absolute value, K an absolute constant. A random X would make (X dot r_j) Gaussian but there would be outliers. The existence of such an X was first shown by the speaker. The new approach, due to
Lovett and Meka, is to begin with X=(0,...,0) and let it float in kind of restricted Brownian Motion until all the coordinates hit the boundary. | {"url":"https://aco.math.cmu.edu/abs-12-02/apr12.html","timestamp":"2024-11-05T07:03:21Z","content_type":"text/html","content_length":"3219","record_id":"<urn:uuid:82249b35-7c51-435f-9058-45fafec7ba74>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00398.warc.gz"} |