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Topics: World Function
In General
* Idea: The spacetime (squared) interval between two points, which conceptually encodes all the information in the metric, but does not mention a differentiable structure in its definition and is
therefore appealing for generalizations of spacetimes (e.g., discrete ones), at least at the kinematical level.
$ Def: Given two points x, y ∈ M, the world function is defined as
σ(x, y):= ± \(1\over2\)S^ 2(x, y) ,
where S(x, y) is the geodesic distance between x and y if it is defined, and the sign depends on whether x and y are or are not, respectively, causally related.
* Example: In Minkowski space,
σ[η](x, y) = \(1\over2\)(x^m − y^m) η[mn] (x^n −y^n) .
* Properties: It is symmetric, non-negative, and satisfies (in any dimension, with signature (−, +, ..., +), and under the appropriate differentiability assumptions)
(∂σ(x, y)/∂x^a) g^ab(x) (∂σ(x, y)/∂x^b) = −2 σ(x,y) , det(∂^2σ(x, y) / ∂x^a ∂y^b) ≠ 0 ,
lim[y → x] σ(x, y) = 0 , lim[y → x] ∂σ(x, y) / ∂x^a = 0 ,
lim[y → x] ∂^2σ(x, y) / ∂x^a ∂y^b = −g[ab](x) .
(These limit properties explain why S^ 2 is used rather than S.)
> Online resources: see Wikipedia page.
And Gravitation > s.a. spacetime structure.
* Idea: All curvature tensors can be written as coincidence limits of derivatives of the world function, and Einstein's equation becomes a set of fourth-order partial differential equations for σ.
@ General: in Synge 60; Rylov AdP(63).
@ Special cases: Roberts ALC(93)gq/99 [in FLRW spacetime].
@ Applications: Bahder AJP(01)gq [spacetime navigation]; Le Poncin-Lafitte et al CQG(04) [and light deflection]; > s.a. tests of general relativity with light.
@ And quantum gravity: Álvarez PLB(88) [quantum spacetime]; Rylov JMP(90) [discrete spacetime]; Kothawala PRD(13)-a1307 [minimal length]; Jia a1909 [quantum causal structure, including matter];
Padmanabhan MPLA(20)-a1911 [and correlator for density of spacetime events]; > s.a. types of approaches.
@ Related topics: in Ottewill & Wardell PRD(11)-a0906 [derivatives, transport equation approach].
main page – abbreviations – journals – comments – other sites – acknowledgements
send feedback and suggestions to bombelli at olemiss.edu – modified 26 apr 2021 | {"url":"https://www.phy.olemiss.edu/~luca/Topics/w/world_function.html","timestamp":"2024-11-07T19:40:24Z","content_type":"text/html","content_length":"7657","record_id":"<urn:uuid:159dcafd-0351-40e2-971a-6fc4bca88e1b>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00446.warc.gz"} |
Is FRM difficult to pass?
The FRM Exam requires a significant investment of time to be successful. But most of the time, failing a part of the exam is the result of study mistakes and insufficient prep. Don’t put off
studying: Both parts of the FRM Exam are nearly impossible to pass if all you do is last-minute cramming.
Is quantitative finance difficult?
Quantitative Finance is a relatively easy field. It’s an umbrella term for everything from the simplest financial logic (you lose more money than you earn hence you’ll go in debt and your stock price
goes down) to die-hard insane mathematics, touching upon borderline insanity.
How much does a quantitative trader make?
The salaries of Quantitative Traders in the US range from $37,167 to $795,786 , with a median salary of $178,046 . The middle 57% of Quantitative Traders makes between $178,050 and $383,324, with the
top 86% making $795,786.
How is IIQF?
“Overall experience in IIQF is just superb. IIQF is best as it provides quality education. I did the PGPAT course after completing my engineering. My experience at iiqf has given me a chance to
sharpen my skills in my field of my choice ie algorithmic trading.
Can I give FRM Part 1 and 2 together?
The Upsides. Taking both exams together is more of a personal call, based on your unique personal circumstances, academic background and test-taking aptitude. There are indeed candidates who take
both parts together, and I’m sure do end up clearing both (or hopefully Part I at least).
Is quantitative trading profitable?
Algorithmic trading is the most profitable type of trading out there, I believe. Nowadays, the trading of financial instruments are mostly done by sophisticated algorithms. They are able to perform
computations on vast amounts of data after assimilating it.
Which is better CFA or FRM?
The primary difference between CFA vs. FRM is the topics that it covers. In contrast, FRM is a specialized exam for obtaining expertise in Risk Management. Additionally, CFA prepares you well for
careers in Investment Banking, Portfolio Management, Financial Research.
How much do quants get paid?
What do Quants Earn? Compensation in the field of finance tends to be very high, and quantitative analysis follows this trend. 45 It is not uncommon to find positions with posted salaries of $250,000
or more, and when you add in bonuses, a quant likely could earn $500,000+ per year.
What is QuantInsti?
QuantInsti is your go-to resource for all things quant. Jump right in! See alumni Follow.
Is Cqf worth the money?
On the other hand, CQF certification is undoubtedly more valuable because people who choose to go for CQF are already qualified. After CQF certification, you would get around the US $115,000 per
annum. With more experience, you would be able to earn much more than a fresher salary.
How do you do quantitative research?
Here are the steps you can take to become a quantitative analyst:
1. Earn a bachelor’s degree in a finance-related field.
2. Learn important analytics, statistics and mathematics skills.
3. Gain your first entry-level quantitative analyst position.
4. Consider certification.
5. Earn a master’s degree in mathematical finance.
How can I become an algorithmic trader?
Steps To Becoming An Algo Trading Professional
1. Trading Knowledge.
2. Programming Skills.
3. Getting started with books.
4. Free resources.
5. Learn from Professionals/Experts/Market Practitioners.
6. Training.
7. Self-learning Online.
8. Getting placed in the algorithmic trading domain.
Is Quant a good career?
Being a quant in a bank is a good as a job, but not as a career.” The desk quants create pricing models for these derivatives. They also create models that create strategies to direct trading
decisions and that make traders more efficient. But desk quants in banks aren’t actually traders.
What math is required for quantitative?
Learn the mathematical foundations essential for financial engineering and quantitative finance: linear algebra, optimization, probability, stochastic processes, statistics, and applied computational
techniques in R. 13,707 already enrolled!
What is quantitative analysis trading?
Quantitative trading consists of trading strategies based on quantitative analysis, which rely on mathematical computations and number crunching to identify trading opportunities. Price and volume
are two of the more common data inputs used in quantitative analysis as the main inputs to mathematical models.
What is a quantitative trader?
Quantitative Trading involves the use of computer algorithms and programs based on simple or complex mathematical models to identify and capitalize on available trading opportunities. Traders
involved in such quantitative analysis and related trading activities are commonly known as quants or quant traders.
Is it hard to become a quant?
Education and training: It is usually difficult for new college graduates to score a job as a quant trader. A more typical career path is starting out as a data research analyst and becoming a quant
after a few years. They are often involved in high-frequency trading or algorithmic trading.
How much money does a quantitative analyst make?
According to Payscale, the average salary for quantitative analysts is $83,900 with a range between $56,000 and $131,000.
What does a quantitative research analyst do?
A quantitative analyst or “quant” is a specialist who applies mathematical and statistical methods to financial and risk management problems. S/he develops and implements complex models used by firms
to make financial and business decisions about issues such as investments, pricing and so on.
How many exams are there in FRM?
From 2021, there would be 3 exam windows for Part 1 (May, July and November); and 2 exam windows for Part 2 (May and November/December). Candidates can still take both Parts on the same day, but you
will need to pass Part 1 before Part 2 is scored. | {"url":"https://www.sweatlodgeradio.com/is-frm-difficult-to-pass/","timestamp":"2024-11-06T05:28:33Z","content_type":"text/html","content_length":"131365","record_id":"<urn:uuid:b9b2adfd-7b79-4b61-a437-9f6f87271e3f>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00187.warc.gz"} |
Numbers On A Jersey Nominal Or Ordinal - OrdinalNumbers.com
Numbers On A Jersey Nominal Or Ordinal
Numbers On A Jersey Nominal Or Ordinal – By using ordinal numbers, you can count unlimited sets. These numbers can be utilized as a tool to generalize ordinal numbers.
The ordinal number is among of the most fundamental concepts in math. It is a number that indicates the place of an item within the list. The ordinal number is typically a number between one to
twenty. While ordinal numbers have various purposes however, they are typically used to indicate the sequence in which items are placed in the list.
It’s possible to show ordinal numbers using numbers or words. Charts, charts, and even charts. They can also be used to show the way a set of items is laid out.
The vast majority of ordinal numbers fall in one or one or. Transfinite ordinals can be represented using lowercase Greek letters, whereas finite ordinals can be represented using Arabic numbers.
According to the axiom, every well-ordered set should contain at least one ordinal. For instance, the highest possible grade will be given to the class’s first member. The contest’s runner-up would
be the student with a highest score.
Combinational ordinal numbers
Multiple-digit numbers are referred to as compound ordinal numbers. They are generated by multiplying an ordinal number by its last number. These numbers are typically employed for ranking and
dating. They don’t have a distinct ending for each number, similar to cardinal numbers.
Ordinal numbers are used to indicate the order of elements within the collection. They are used to distinguish the various items within a collection. You can locate normal and suppletive numbers to
ordinal numbers.
By affixing a number to it with the suffix “-u” makes regular ordinals. The numbers are then typed in words. A hyphen then added. There are also additional suffixes. The suffix “nd” is used to
indicate numbers that end in two numbers. The suffix “th” can indicate numbers with endings of the numbers 4 or 9.
By affixing words with the -u or–ie suffix can create suffixtive ordinals. The suffix, which can be used to count, is longer than the standard.
Limitation of the importance of ordinal
The limit for ordinal numbers that do not have zeros is an ordinal quantity that is not zero. Limit ordinal numbers come with the drawback that there is no maximum element for them. They can be
created by joining empty sets without maximum element.
Additionally, transcendent rules of recursion utilize restricted ordinal numbers. In accordance with the von Neumann model, every infinite cardinal number also has an ordinal limit.
A limit ordinal equals the sum of all other ordinals beneath. Limit ordinal number can be determined using arithmetic or a series of natural numbers.
The ordinal numbers serve to organize the information. They offer a rationale for an object’s numerical location. They are often utilized in set theory and math. Although they are in the same class
they are not considered to be natural numbers.
The von Neumann method uses a well-ordered list. Consider that fyyfy is one of the subfunctions g’ of a function that is described as a singular operation. If g’ is able to meet the requirements, g’
is an ordinal limit if it is the only subfunction (i or ii).
A limit ordinal of the type Church-Kleene is also known as the Church-Kleene ordinal. A limit ordinal is a properly-ordered collection of smaller ordinals. It’s a nonzero ordinal.
Stories that use normal numbers as examples
Ordinal numbers are used frequently to show the hierarchy of entities and objects. They are vital to organize, count, and ranking reasons. They are also utilized to determine the order of items and
the location of objects.
Ordinal numbers are generally represented by the letter “th”. On occasion, though, the letter “nd” is substituted. Books’ titles are often associated with ordinal numbers.
Even though ordinal figures are typically used in list format it is possible to write them down as words. They may also come in acronyms and numbers. In comparison, they are simpler to comprehend
than cardinal numbers.
Three distinct types of ordinal numbers are offered. It is possible to discover more about them through engaging in games, practice, and engaging in various other pursuits. You can increase your
arithmetic skills by learning more about these concepts. Coloring exercises are a fun and easy method to increase your proficiency. You can assess your progress by using a coloring sheet.
Gallery of Numbers On A Jersey Nominal Or Ordinal
Ordinal Numbers ESL Worksheet By Jersey Ordinal Numbers Vocabulary
Assignment On Nominal Number Ordinal Number Cardinal Number Teacha
Antara Cardinal Ordinal Dan Nominal Number Apa Sih Bedanya
Leave a Comment | {"url":"https://www.ordinalnumbers.com/numbers-on-a-jersey-nominal-or-ordinal/","timestamp":"2024-11-13T22:01:44Z","content_type":"text/html","content_length":"62096","record_id":"<urn:uuid:a47dc550-e2ab-4027-82d0-b8e46585c033>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00265.warc.gz"} |
8.8 Likelihood ratio test vs. Wald test | Introduction to Regression Methods for Public Health Using R
8.8 Likelihood ratio test vs. Wald test
As with previous chapters, Wald tests for p-values were used in this chapter. However, likelihood ratio (LR) tests are, in general, more powerful. True LR tests are not possible with svyglm() objects
since they were not fit using maximum likelihood (Lumley 2010). However, the function regTermTest() can be used to carry out a “working” LR test for weighted linear, logistic, or Cox regression
models (the Rao-Scott LR test) (Rao and Scott 1984; Lumley and Scott 2013, 2014) to compare any two nested models, similar to anova() which we used in previous chapters.
regTermTest() can therefore obtain an overall Type 3 Wald or working LR test for a categorical predictor with more than two levels. To get a test for a single level of a categorical predictor, first
create indicator variables for the levels of that predictor as described in Section 6.18.
Example 8.1 (continued): Use a working LR test to test the overall significance of race/ethnicity in the weighted adjusted linear regression model for fasting glucose. For comparison, also compute
the Wald test.
# Model fit previously
fit.ex8.1 <- svyglm(LBDGLUSI ~ BMXWAIST + smoker + RIDAGEYR +
RIAGENDR + race_eth + income,
family=gaussian(), design=design.FST.nomiss)
# Working LR test for race_eth
test.terms = ~ race_eth,
df = degf(fit.ex8.1$survey.design),
method = "LRT")
## Working (Rao-Scott+F) LRT for race_eth
## in svyglm(formula = LBDGLUSI ~ BMXWAIST + smoker + RIDAGEYR + RIAGENDR +
## race_eth + income, design = design.FST.nomiss, family = gaussian())
## Working 2logLR = 11.29 p= 0.042
## (scale factors: 1.8 0.93 0.31 ); denominator df= 15
# Wald test for race_eth
test.terms = ~ race_eth,
df = degf(fit.ex8.1$survey.design),
method = "Wald")
## Wald test for race_eth
## in svyglm(formula = LBDGLUSI ~ BMXWAIST + smoker + RIDAGEYR + RIAGENDR +
## race_eth + income, design = design.FST.nomiss, family = gaussian())
## F = 3.003 on 3 and 15 df: p= 0.064
Conclusion: Based on the likelihood ratio test, race/ethnicity is significantly associated with fasting glucose, after adjusting for the other variables in the model (p = .042). As previously
mentioned, LRTs are generally more powerful than Wald tests, which means lower p-values. In this example, that is the case, with the Wald test p-value being .064.
———. 2010. Complex Surveys: A Guide to Analysis Using r: A Guide to Analysis Using r. Hoboken: John Wiley & Sons.
Lumley, Thomas, and Alastair Scott. 2013.
“Partial Likelihood Ratio Tests for the Cox Model Under Complex Sampling.” Statistics in Medicine
32 (1): 110–23.
———. 2014.
“Tests for Regression Models Fitted to Survey Data.” Australian & New Zealand Journal of Statistics
56 (1): 1–14.
Rao, J. N. K., and A. J. Scott. 1984.
“On Chi-Squared Tests for Multiway Contingency Tables with Cell Proportions Estimated from Survey Data.” The Annals of Statistics
12 (1): 46–60. | {"url":"https://www.bookdown.org/rwnahhas/RMPH/survey-likelihood.html","timestamp":"2024-11-02T18:13:42Z","content_type":"text/html","content_length":"77862","record_id":"<urn:uuid:6dd24874-e2c5-4265-bc75-0b436528f1dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00356.warc.gz"} |
Find the longest increasing subsequence using Fenwick Tree
Open-Source Internship opportunity by OpenGenus for programmers. Apply now.
You are given an array of length N. You have to find the longest increasing subsequence (LIS) from the given array. We will try to solve this problem using Fenwick Tree which will take O(N logN) time
complexity. The brute force approach will take O(2^N) time complexity.
Given array :
arr = {9,6,2,3,7,5,8,4}
Our output will be 4, as {2,3,5,8} is the longest increasing subsequence.
Prerequisite to understand this problem is knowledge about Fenwick Tree. So, before starting this problem, have a quick overview of Fenwick Tree or Binary Indexed Tree.
So as you know, The basic idea behind Fenwick tree is that since any integer can be represented as sum of powers of 2, we can represent cumulative frequencies as sum of non-overlapping
• To get the index of next node in Fenwick Tree, we use :> index -= index & (-index)
• To get the index of previous node in Fenwick Tree, we use :> index += index & (-index)
Now, after reading about Fenwick tree you must have got a decent knowledge about it, and how they are formed and how they can be used to solve various problems.
Here, the problem we are trying to solve is Given an array of size n, we have to find the length of Longest increasing subsequence in the given array using Fenwick tree or Binary indexed tree(BIT).
As we will be using fenwick tree, the time complexity of our solution would be O(N log(N)) where N is number of elements in given array.
To solve this problem, we would first create a fenwick tree as an array of length N+1, as the 0'th position works as a root. Then we will map our array elements according to their ranks in given
For example:
Given array : [3, 5, 1]
So our mapped array would be : [2, 3, 1] as in given array, 1 is smallest element so its rank would be 1, rank 2 will be given to element 3 and so on.
This data mapping will make our data comparison easy and it would be simpler to fill our fenwick tree.
After mapping we would start filling our fenwick tree according to the data present in mapped array and data filled in our fenwick tree.
At end, we would get our fenwick tree in which each element will be showing the length of longest incresing subsequence till that element.
1. Sort the given array and create a dictionary with keys as array's element and value as its rank in sorted array.
2. Update the array with ranks assigned to respective elements.
3. Create a fenwick tree array of size n+1 and elements as 0.
4. Start traversing the array and for each index recieved from array, fill that position in fenwick tree array with the (maximum lenght formed till that index) + 1.
5. Repeat this for whole array.
6. Return the lenght of longest increasing subsequence from filled fenwick tree.
Implementation in Python
Following is our Python implementation of solving the Longest Increasing Subsequence problem using Fenwick Tree:
# Function that returns the longest increasing subsequence
def answer(arr):
n = len(arr)
##### INITIALISING FENWICK TREE #####
fenwick_tree = [0]*(n+1)
########## MAPPING DATA ACCORDING TO THEIR RANK IN LIST ###########
sorted_arr = sorted(arr) # Sorting data
# Creating dictionary
dictionary = {}
for i in range(n):
dictionary[sorted_arr[i]] = i+1
# Mapping arr data
for i in range(n):
arr[i] = dictionary[arr[i]]
##################### FILLING OUR FENWICK TREE #####################
for i in range(n):
# Taking rank of elements as index of tree
index = arr[i]
# Asking for maximum length in fenwick tree till this index
x = query(fenwick_tree, index-1)
# Incrementing length
val = x+1
# Traversing in fenwick tree from parent to child
# Filling length at respective indexes
fenwick_tree[index] = max(val, fenwick_tree[index])
# Getting index of next node in fenwick tree
index += index & (-index)
# Returning answer as query
return query(fenwick_tree, n)
######### FUNCTION THAT CHECKS FOR MAX LENGTH TILL i'th INDEX ########
def query(f_tree, index):
ans = 0
# Traversing in fenwick tree from child to parent
while index>0:
ans = max(f_tree[index],ans)
# Getting index of previous node in fenwick tree
index -= index & (-index)
return ans
# Tesing our code
arr = [6, 5, 1, 12, 2, 4, 9, 8]
ans = answer(arr)
Worflow of Solution
1. Suppose we are given the array as [6, 5, 1, 12, 2, 4, 9, 8] .
2. We map the elements of given array according to their ranks, so we update or array to : [5, 4, 1, 8, 2, 3, 7, 6].
3. Now, we start traversing our mapped array and we get 5 as our first element, remember that this 5 represents the rank of the element present at 0'th position in our original array which was 6.
4. So, we ask our query fuction about the maximum length we found till 4'th position in fenwick_tree. As all elements or fenwick tree are 0 now, we get our x as 0, so we increment it to 1 and fill
it in our fenwick_tree array, to all the positions we get as our next node in fenwick tree, through the formula : index -= index & (-index) which you read about earlier.
5. You would have read about this formula in the article mentioned above.
6. So, now our fenwick_tree looks as : [0, 0, 0, 0, 0, 1, 1, 0, 1].
7. We repeat this for all the elements of mapped array, and we will get the fenwick_tree in end as : [0, 1, 2, 3, 3, 1, 4, 4, 4]
8. So, this fenwick_tree shows the maximum length of increasing subsequence at each element in given array.
9. At end, we return the length of longest incresing subsequence in given array.
As we traverse the whole array, and for each element we traverse the parent to child positions in fenwick_tree array, so we get the time complexity as O(nlogn), where n is size of given array.
With this article at OpenGenus, you must have a complete idea of solving this problem of Longest Increasing Subsequence using Fenwick Tree. Enjoy. | {"url":"https://iq.opengenus.org/longest-increasing-subsequence-fenwick-tree/","timestamp":"2024-11-03T07:27:11Z","content_type":"text/html","content_length":"60798","record_id":"<urn:uuid:ccefa423-3430-4df4-9479-80c24f82711e>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00309.warc.gz"} |
ParticleTransformer has emerged as a state-of-the-art model for jet tagging in particle physics, offering superior accuracy and versatility across various applications. As the field continues to
evolve with increasing data volumes from experiments like the upcoming Circular Electron-Positron Collider (CEPC) in China, the need for efficient computational methods is more crucial. Traditional
hardware solutions, while effective, often face limitations in terms of computational time and memory consumption.
This project explores accelerating ParticleTransformer on Field-Programmable Gate Arrays (FPGAs), known for low power consumption, low latency, and customizable hardware. By leveraging FPGAs'
parallel processing capabilities, we aim to optimize ParticleTransformer for faster execution and reduced memory overhead, enhancing jet tagging efficiency.
Using a heterogeneous computing approach, the project integrates FPGAs with CPUs to offload compute-intensive tasks, minimizing latency and maximizing throughput. We will design, implement, and
optimize ParticleTransformer on FPGA, comparing its performance against CPU and GPU platforms in terms of speed, power consumption, and memory efficiency.
The study aims to demonstrate significant improvements in computational efficiency and scalability of jet tagging tasks, potentially reducing costs and expediting data processing in particle physics,
particularly for large-scale projects like the CEPC. This research paves the way for future explorations into heterogeneous computing platforms, advancing machine learning applications in high-energy | {"url":"https://indico.cern.ch/event/1386125/timetable/?view=standard_numbered_inline_minutes","timestamp":"2024-11-03T06:57:31Z","content_type":"text/html","content_length":"481773","record_id":"<urn:uuid:1a03883e-1971-420f-96d2-f24bd4da2369>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00599.warc.gz"} |
What's on the back of the envelope?
The name of this blog is "Back of the Envelope". One of my earliest blog posts was a
long explanation
about why I named the blog that. The short version, from that post, can be summed up quickly:
The expression is common enough, but if you're not familiar with it, a back of the envelope calculation is a quick, simple calculation done as an estimate. It's called "back of the envelope"
because it can be written out on a small sheet of paper . . . When I first applied for this address on blogspot, the idea was to name the blog after myself . . . Nothing really felt right,
though, so I started thinking of other names, a name appropriate for an engineer writing about things he was distinctly unqualified to discuss. It took surprisingly little time to come up with
"Back of the Envelope."
I've always used the image of an envelope with something written on the back as the symbol of this site. In fact, this is the one I had for a long time:
The old back of the envelope symbol.
With the new template, I wanted to change the design while keeping the concept. The new design is the background of this page. Unfortunately, since I used the full size of the background that Blogger
recommends (1800x1600 pixels, or close to it), you probably can't see the whole thing unless you have a super-high resolution display, even without the blog contents covering it. So here's the full
image, at a reduced resolution:
As you can see, the central equation is the same. This is the bra-ket notation used in quantum physics, and shows the inner product between two quantum computation values, 0 and + (which is a
superposition of 0 and 1), so the overlap of 0 and + (technically it's the inner product, but it's the degree to which the two are the same) is one over the square-root of two.
What about the rest of the calculations? Are they legitimate, or just random doodlings? They're all legitimate, and equations I've used before, though it's been years. Hopefully there aren't any
The next equation, in red at the top, is just a circle divided into six parts, with one part divided in half. The equation calculates the area of that section, but it's mainly an excuse for me to
estimate pi as three. That's a common estimate to use for pi when you're just doing a back of the envelope calculation. Another useful estimate is 5 dB, or the square-root of 10.
On the left side is a charged particle over a ground plane. This results in an image in the ground plane. The charge in the ground plane responds in such a way that it's equivalent to an equal and
opposite charge reflecting the placement of the first charge. This results in the equation below, which is also the equation for the potential for a charge dipole. Charge dipoles consist of equal and
opposite charges close together, so that they minimize each other's effects. A ground plane effectively converts a charge into a charge dipole, which is why ground planes help reduce noise coming
from the circuits they're placed under (they also tend to minimize noise coupling into the circuit).
Below that, at the bottom of the page, is the time-invariant form of Schrodinger's Equation, since I figured I needed that on the back of the envelope.
On the right side is a 3-bit Gray code. This is a binary sequence where only a single bit changes for each step of the sequence. This was originally used as a method of binary counting for mechanical
switches. Since mechanical switches don't change instantaneously, switching from 011 to 100 (3 to 4 in binary), could result in spurious outputs as each switch changes at a different time. By making
it equal the change from 010 to 110 instead, there are no spurious values between them. In modern digital computers, this particular reason is not as relevant, but it is still useful for error
correction. A Gray code can be visualized as a cube, shown above, where each step travels along the edge of the cube. I included the cube, with convenient arrows, mainly to give people a clue that I
was doing a Gray code, rather than let them think I was trying to count in binary and getting it wrong. I'm not sure whether it worked or not.
So that's everything. I hope you enjoyed this boring math post. I also hope I didn't mess up any of these equations. | {"url":"https://www.donaldscrankshaw.com/2012/09/whats-on-back-of-envelope.html?m=0","timestamp":"2024-11-05T23:35:28Z","content_type":"text/html","content_length":"92379","record_id":"<urn:uuid:8a3dfd78-e384-42cb-9c95-28e84f1a46d6>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00396.warc.gz"} |
MathSeed Singapore math
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I haven’t posted anything about my current project, mainly because I mostly worked on internal systems. But recently I needed a bit of a break and implemented some visual FX. The first I want to show
here is a ‘simple’ explosion.
Finally, here is a collection of small(ish) utility functions. None of them warrants a separate post. As usual, you can find this, and more, on my Bitbucket Repository. Read More »
One of the more recent additions to my Utilities. I wrote this to calculate a parabolic trajectory for a thrown object (in this special case a grenade). More specifically, I calculate the required
throw-velocity to accurately hit a specified point from a starting position and a given angle. This velocity can be capped and when the required speed exceeds this cap, the trajectory will not reach
the target. The class can also calculate points on the trajectory in order to visualize them. Lastly, it provides a coroutine that can be used to make an object follow this trajectory. The
calculation for this is done via a synchronized Leapfrog integration, and does not use Unity’s physics engine. Both this coroutine and the calculation for the trajectory-points can be specified to
‘bounce’ on surfaces, where it will take into account both the bounciness of the surface that is being hit and an optional bounciness value for the object. A bouncing object will only come to rest on
a surface that is sufficiently flat.
Another datatype, or two if you want. Integer Vectors in 2D and 3D. They work nearly identically to the standard Vector2 and Vector3 types, but use integers only. I did not write these myself. The
original idea was a request in the Unity3D forums and the original implementation was done there by Lysander. I only made some minor changes to it.
This time, I have a 3-by-3 matrix datatype. I innitially wrote it to generate Quaternion rotations from rotation matrices, but I extended it with most of the usual matrix operations. Little warning,
though: there is no SIMD anywhere in this class, so it might not be the most performant implementation of a 3-by-3 matrix. The operations are not really heavy, but if you need every last drop of
performance, this might not be the first choice.
To make use of the Bezier class from the last post, here is a Spline class. It is not the most grabage friendly class, so I might revisit it at some time, but it does what I wanted it to do when I
wrote it. As I said it uses the Bezier class for smooth curves, but there is much automatisation, so no user-input is needed/possible to define the typical Bezier handles. This has the drawback that
there is a risk of overshooting, if the distance between spline-vertices varies too much. Splines can be smooth or segmented, closed or open, and are visualized via a LineRenderer.
As usual, the most current code is found on my Bitbucket repository.
This time only a quicky, but it will be used in a later post. A small class to calculate a Bezier curve interpolation in 3D or 2D. Nothing special about it.
public class Bezier {
public static Vector3 Interpolate(Vector3 start, Vector3 end, Vector3 controlPointA, Vector3 controlPointB, float t) {
float tInv = 1 - t;
float tSqr = t * t;
float tInvSrq = tInv * tInv;
return tInv * tInvSrq * start + 3 * t * tInvSrq * controlPointA + 3 * tInv * tSqr * controlPointB + t * tSqr * end;
public static Vector2 Interpolate(Vector2 start, Vector2 end, Vector2 controlPointA, Vector2 controlPointB, float t) {
float tInv = 1 - t;
float tSqr = t * t;
float tInvSqr = tInv * tInv;
return tInv * tInvSqr * start + 3 * t * tInvSqr * controlPointA + 3 * tInv * tSqr * controlPointB + t * tSqr * end;
public static float Length(Vector3 start, Vector3 end, Vector3 controlPointA, Vector3 controlPointB, int steps) {
float length = 0;
Vector3 fst = start;
for (int i = 1; i <= steps; ++i) {
Vector3 snd = Interpolate(start, end, controlPointA, controlPointB, (float) i / steps);
length += (snd - fst).magnitude;
fst = snd;
return length;
public static float Length(Vector2 start, Vector2 end, Vector2 controlPointA, Vector2 controlPointB, int steps) {
float length = 0;
Vector2 fst = start;
for (int i = 1; i <= steps; ++i) {
Vector2 snd = Interpolate(start, end, controlPointA, controlPointB, (float) i / steps);
length += (snd - fst).magnitude;
fst = snd;
return length;
Another day, another utility. This time: Ranged values. They were inspired from a Unite Europe 2016 talk on scriptable objects (and the Inpector code is copied nearly verbatim) by Richard Fine. The
original had only a minimum and maximum float value with the fancy Inspector GUI, but I extended it with functionality nobody in their right mind would probably ever need. It started with a method to
get a random value from within the interval and quickly escalated to two interpolation functions and tests to determine if an interval contains a given value or another interval, or if two intervals
intersect. Finally I added operators to add, multiply or order intervals. And as if all that wasn’t already useless enough, I did the same thing again, with ints instead of floats and added a generic
Range<T> class, that can create intervals from any orderable value-type, albeit a bit less powerfull, since adding and subtracting doesn’t necessarily have meaning outside of numbers. The inspector
code also doesn’t work with the generic version.
Anyway, here is a example of how it looks in the Inspector, and a wall of code after the break.
The editor code can be found on my Bitbucket repository.
This seems to become a running theme on here, but I decided I would again at least pretend to have some content on this blog. Over the next couple of days or weeks I will post some of the
Utility-Classes I wrote and use in my projects and write a bit about each. But first a couple of disclaimers: These classes are not perfect. They are prone to change without notice. The code I post
here will be as it is at the time of writing, but you can get the latest versions from my Bitbucket repository.
The first utility is my generic Object Pool. I already posted it more or less in its entirety on the Unity3D forums, but here it is again. I wrote it for my current project and the motivation was to
move all checks into the pool itself, so instead of returning a GameObject, where one has to then find the component of interest, it will directly return that component. Pools can be created by
themselves or managed by a central static class to keep track of the pools. Each pool has an initial number of objects and a maximum size (which can be modified after creation), but it will still
always return a valid object, even if the maximum size has been reached. Surplus-objects will not be destroyed instead of released back into the pool. That means it will (hopefully) never break any
functionality, but when the maximum is reached and a new object is requested, the overhead from the pool will be added to the inherent processing required to instantiate/destroy a GameObject, so the
maximum size should be chosen sensibly.
After some time of absence I can actually post something, again. I took a course on Storyworlds last semester, and we had to create a storytelling universe and build a small prototype for a game set
in this universe.
Some time in the future, big aliens land on Earth. They come not as invaders, but as miners. They are here to harvest material they seeded in Earth’s crust 2000 years ago. These aliens stand 3 meters
tall and fall from the sky in their personal space armor, trailing blazing wings behind them, evoking the image of angels in the mind of the onlookers. Mankind doesn’t interest them, but if we annoy
them, they crush us like ants, as we are nothing more to them. In a desperate final attempt to retake the world, scientists augmented one of their own with salvaged alien technology, giving him the
capability to fight back. While he cannot take one of these Angels on directly, he has the ability to siphon the Angel’s energy, if he reaches him undetected. His new abilities help him to navigate
the broken environment.
This time, just a trailer, not a playable prototype: | {"url":"http://blog.piflik.de/","timestamp":"2024-11-12T03:30:05Z","content_type":"application/xhtml+xml","content_length":"30270","record_id":"<urn:uuid:9ea351ff-17f5-439b-a3a0-c2b0dbf92f37>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00132.warc.gz"} |
Intraspecific Cohesion From Conspecific Attraction, Part III: A Simple Test
In the real world, animals typically tend to congregate at the landscape scale, even when they show repellence at finer scales (territorial behaviour). In contrast, one of the basic assumptions of
the vast majority of mathematical population models is independent space use by individuals. In other words, it is assumed that the individuals do not show conspecific attraction but adhere to “full
mixing” from independent space use. A strange assumption, indeed! Using the part of the Parallel processing theory (PP) that was summarized in Part I and Part II it is a simple statistical exercise
to test if a given population’s dispersion obeys the full mixing assumption (the Paradigm); or alternatively, indicating a positive feedback tendency from PP-compliant space use (the Zoomer model).
The concept of conspecific attraction is verified among many species and taxa of animals. For example, lizards prefer to settle near conspecifics, even when unoccupied habitat is available nearby
(Stamps 1987; 1988; 1991). Stamps (1991) searched the literature on territorial vertebrates but was unable to find any study in which a negative effect of the density of residents on the settlement
of newcomers was demonstrated; i.e. the general assumption of habitat selection studies that prospective territory owners prefer relatively empty habitats proved to lack empirical evidence.
In contrast, in theoretical population models it is typically assumed that individuals both move and settle inter-independently. This “full mixing” assumption (when ignoring the self-evident
influence from habitat heterogeneity on dispersion) in models is a mathematical convenience. It is in fact a requirement, under the premise of the mean field framework, on which almost all population
dynamical models adhere to. For example, to model population dynamics realistically with ordinary differential models one has to assume both full spatial mixing of individuals at the temporal scale
of the analysis, and a closed system. If the system is open, one should apply partial differential equations, since this allows for an assumption of full mixing locally instead of system-wide
(allowing for spatio-temporal “shifting mosaic” of local abundance) (Potts and Lewis 2019)*. The former is called a spatially implicit model, the latter is called a spatially explicit model.
The “full mixing” requirement in classic and contemporary theory of space use by populations is one of the main reasons why theoreticians and field ecologists often tend to drift apart. Logically, it
does not make sense, even before considering behavioual fitness arguments! As I argued in Part I and Part II, independent movement and settlement in an open environment will over time drift a
Markov-compliant population towards extinction from diffusion and Allée effects.
Empirical results show that (a) single-species populations tend to show scale-free spatial congregation, compliant with a power law (Taylor 1986), (b) empirical results continue to support
Non-Markovian, spatial memory utilization by individuals, and (c) populations in general seem to adhere to the principle of conspecific attraction. References in by book and throughout this blog.
Scientifically, the primary question is: how to perform the initial task (prior to making follow-up ecological inference) to test for inter-dependent or independent space use? In other words, how to
test if local dispersion is influenced by conspecific attraction?
Consider the PP-based model to represent a specific alternative hypothesis, which should be tested against the null hypothesis given by the Paradigm (for example, a reaction-diffusion model for
spatially explicit population dynamics).
The protocol can be quite simple, at least in the first-level approach of analysis of empirical data.
Again, consider results from simulations of PP, in addition to a Paradigm compliant population dispersion. In the Figure below you find situation both from scale-free individual space use with
extensive spatial overlap (the PP model, MRW) and scale-specific (Markovian compliant random walk, the Paradigm),
and where both conditions are set to be void of intraspecific cohesion
. This condition contrasts with the scenario in Part I, where I showed how a population with PP compliant space use generated a scale-free dispersion under influence of conspecific attraction. In the
present two scenaria the conspecific attraction factor is absent. Individuals use space inter-independently, and below you learn how to test statistically for this lack of conspecific attraction at
the so-called “landscape scale” under two qualitatively distinct model frameworks.
The result of such inter-independency ("full mixing") is seen in the lower part of the Figure. Both situations result in a lack of fractal spatial dispersion of abundance of the pooled set of
locations. The reason is that the distribution complies with a negative exponential function (semi-log linear, not shown) rather than power law compliant one (log-log linearity, which - as shown - is
not satisfied).
Image to the right: (a) Upper part: a superposition of five MRW series with relatively strong utilization distribution overlap. Lower part: a superposition of five series based on classic random walk
with homing tendency, with less spatial overlap in inter-individual spatial utilization. (b) A log-transformed frequency histogram of local grid cell densities (M,F) for MRW superpostions (filled
circles) and random walk-superpositions (open circles) shows that neither of the dispersion patterns at the population level satisfy a power law.
When these two distinct system conditions are viewed from the population perspective, they both lead to mean field-like system properties with respect to the (M,F) regression at the population level.
In other words, even the scenario where individual space use was PP compliant, the lack of conspecific attraction masked the fractal PP property of individuals when analyzed at the population level.
Under log-log transformation of frequency of cells in respective bins of grid cell abundance the regression lines were not linear; i.e., not power law compliant. Basically, to test PP compliant space
use under the additional property of intraspecific cohesion from conspecific attraction one needs to verify a scale-free (log-log linear) frequency distribution of local density of individuals.
This result illustrates the interesting system property where space use at the individual level may adhere to scale-free dispersion of locations of respective individuals (analyzed separately) while
space use at the population level (local abundance of the pooled sets of locations) apparently shows mean field compliance: local fluctuations of abundance are negative exponential compliant!
Conclusion: the full mixing premise of the Paradigm – the mean field compliance – can easily be tested on real data.
I’d like to finish this post with an additional study:
In sum, we experimentally tested in breeding mallards two alternative and mutually exclusive hypotheses of habitat selection rules, and found more support for the conspecific attraction rule.
However, taking into account the pattern of habitat distribution of breeding mallards (see references in Introduction; this study) pairs certainly use other ways of habitat assessment than mere
presence of conspecifics. Some lakes had relatively stable pair numbers while others remained empty independently of experimental treatment.
Pöysä et al. 1998, p287
As I stated in Part II, “
A given number of individuals cannot be everywhere all the time
”. Thus, some lakes should – as a logical consequence under the premise of population "clumping" from conspecific attraction – always be expected to be void of breeding pairs…
*) Using the toolbox of partial differential equations and some alternatives it is shown how “diffusion-taxis” equations may show system-intrinsically driven heterogeneity of local population
abundance between populations (Potts and Lewis 2019). However, this phenomenon regards interspecific cohesion (or repellence) between separate populations, not intraspecific cohesion. It is assumed
that full mixing of individuals is satisfied for each of the populations at the temporal scale of analysis.
Potts, J. R. and M. A. Lewis. 2019. Spatial memory and taxis-driven pattern formation in model ecosystems. arXiv:1903.05381v05382.
Pöysä, H., J. Elmberg, K. Sjöberg, and P. Nummi. 1998. Habitat selection rules in breeding mallards (Anas platyrhynchos): a test of two competing hypotheses. Oecologia 114:283-287.
Stamps, J. A. 1987. Conspecifics as cues to territory quality: a preference of juvenile lizards (Anolis aeneus) for previously used territories. The American Naturalist 129:629-642.
Stamps, J. A. 1988. Conspecific attraction and aggregation in territorial species. The American Naturalist 131:329-347.
Stamps, J. A. 1991. The effects of conspecifics on habitat selection in territorial species. Behavioral Ecology and Sociobiology 28:29-36.
Taylor, L. R. 1986. Synoptic dynamics, migration and the Rothamsted insect survey. J. Anim. Ecol. 55:1-38. | {"url":"http://www.animalspaceuse.net/2019/06/intraspecific-cohesion-from-conspecific.html","timestamp":"2024-11-12T22:48:21Z","content_type":"application/xhtml+xml","content_length":"107938","record_id":"<urn:uuid:b3b2a464-e740-4cd8-ab62-fd4bb2e61787>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00205.warc.gz"} |
Jenxys Math: A Multiplication Revelation » Learning Captain 1
Jenxys Math: A Multiplication Revelation
Do you remember struggling to memorize the multiplication table as a kid? All those numbers and math facts swirling around in your head, desperately trying to stick but often slipping away at the
worst possible moment, like during a pop quiz?
What if there was an easier way to not just memorize but truly understand multiplication in a visual, intuitive way? Jenxys math is a revelation that can change how you think about that first pillar
of mathematics forever. Developed by a math teacher who wanted to inspire his students with a deeper, more permanent grasp of multiplication, Jenxys math uses a simple visual diagram to unlock the
logic and patterns behind the multiplication table.
Once you see multiplication through the lens of Jenxys math, you’ll never go back to just memorizing the table again. Your multiplication woes will be a thing of the elementary school past, and
you’ll actually have fun with math again. Jenxys math – prepare for a multiplication revelation!
Introducing Jenxys Math: A New Approach to Multiplication
Jenxys Math: A Multiplication Revelation 4
Have you struggled with multiplication tables as a kid? Don’t worry, you’re not alone. But what if there was an easier way to multiply in your head without relying on memorization?
Introducing Jenxys Math, a visual approach to multiplication that will change how you do math forever. With Jenxys Math, you’ll discover how to break down multiplication into simple steps using a
visual model.
To multiply two numbers, draw two intersecting lines to make a cross. Put one number on the left side of the cross and the other number on the right side. Where the lines intersect in the middle, put
the product of the tens digits and hundreds digits. Then, on the top line put the product of the ones digits. Add up the results!
For example, to do 53 x 62:
30 (5 x 6)
12 (3 x 2)
Result: 3,286
It’s that easy. No more memorizing tables or struggling with long multiplication. Jenxys Math works for any two numbers, big or small. The visual model shows you exactly what you’re calculating so
you can understand what’s really happening.
Give it a try – you’ll be multiplying with confidence and speed in no time. Jenxys Math makes multiplication simple, intuitive, and accessible for learners of all ages. Say goodbye to multiplication
tables and hello to faster, more fun math!
How Jenxys Math Makes Multiplication Easier
Jenxys Math is a revolutionary new approach to learning multiplication that will change the way you think about this fundamental math operation.
With Jenxys Math, you visualize each multiplication problem as a grid. For example, 6 x 8 would be a grid that is 6 squares wide and 8 squares long, totaling 48 little squares. This visual
representation gives you an intuitive sense of what the product means.
To find the product, you just have to count the number of squares in the grid. For single digit numbers like our example, this is easy. But Jenxys Math also works for larger multiplications, like 24
x 32.
A Simple Visualization
• Think of the 24 x 32 grid as composed of 24 rows with 32 squares in each row.
• Mentally picture the rows stacked up, one on top of the other.
• Now imagine walking up and down each column, counting the squares as you go. There are 32 columns, so after traversing them all, you’ll end up with 768 total squares.
• Voila! You’ve just figured out that 24 x 32 = 768 using an easy visualization.
With regular practice, this visual way of doing multiplication will become second nature. You’ll develop an intuitive sense of what large products mean that you just can’t get with the usual
multiplication table approach. Multiplication may never be quite as fun as recess, but with Jenxys Math it can at least be a whole lot easier. Why not give this simple yet revolutionary new method a
The Step-by-Step Process of Jenxys Math Multiplication
Jenxys Math: A Multiplication Revelation 5
The Jenxys Math multiplication method follows a simple step-by-step process to solve problems. By breaking down the problem into manageable parts, this innovative technique can make multiplying large
numbers feel like a breeze.
Step 1: Break Down the Problem
Look at the number you want to multiply, like 438. Break this down into 400, 30, and 8. Write these numbers on separate lines.
Step 2: Multiply by Tens
Take the hundreds place, 400, and multiply by the tens digit of the other number, let’s say 6. 400 x 6 = 2400. Write 2400 on the first line.
Step 3: Multiply by Ones
Now take 30 and multiply by the ones digit, like 4. 30 x 4 = 120. Write 120 on the second line.
Step 4: Multiply Ones by Ones
Finally, multiply the ones place of the original number, 8, by the ones place of the second number, 4. 8 x 4 = 32. Write 32 on the third line.
Step 5: Add it Up
Add up the numbers on each line:
Line 1: 2400
Line 2: 120
Line 3: 32
Total: 2552
So 438 x 64 = 2552! By breaking the problem into bite-sized chunks, the Jenxys method makes multiplication less intimidating and more achievable. With regular practice, this innovative technique can
become second nature, allowing you to multiply large numbers in your head with confidence. Give the Jenxys Math method a try—you’ll be multiplying in no time!
Read More: Naiker Math: Solving Equations the Ancient Way
Real-World Examples and Practice Problems
Once you understand the basic Jenxys method, it’s time to apply it to some real-world examples. The key is practicing the technique repeatedly. Start with some simple multiplication problems you
already know the answers to, so you can verify you’ve got the hang of it.
Practice Problems
Let’s try some single-digit multiplication, like 6 x 8.
1. Think of the first number, 6. This is the number of groups.
2. Think of the second number, 8. This is the number in each group.
3. Visualize 6 groups of 8 dots each. That’s 6 rows with 8 dots in each row.
4. Quickly count the total number of dots. 6 rows of 8 dots is 48 dots total.
5. 4 x 7: 4 groups of 7 is 4 rows of 7 dots each, which is 28 dots total.
6. 9 x 3: 9 groups of 3 is 9 rows of 3 dots each, which is 27 dots total.
Once single-digit problems feel natural, move on to 2-digit multiplication, like 24 x 16.
1. Think of the first number, 24. This is the number of groups.
2. Think of the second number, 16. This is the number in each group.
3. Visualize 24 groups, with 16 dots in each group. That’s 24 rows with 16 dots in each row.
4. Count the dots in one row, 16. Then count the number of rows, 24.
5. Multiply the number of dots in one row, 16, by the number of rows, 24. So 16 x 24 = 384.
6. 32 x 12: 32 groups of 12 is 32 rows of 12 dots each, which is 384 dots total.
7. 18 x 25: 18 groups of 25 is 18 rows of 25 dots each, which is 450 dots total.
Keep practicing and the Jenxys method will become second nature. Soon you’ll be multiplying large numbers in your head with confidence! Let me know if you have any other questions.
The Benefits of Learning Jenxys Math Multiplication
Learning multiplication the Jenxys way has significant benefits for students.
Builds Number Sense
Doing multiplication with the Jenxys method helps build a strong number sense in kids. As they manipulate the numbers in the problems, they develop an intuitive understanding of how numbers work
together. This number sense will serve them well in higher math.
Promotes Pattern Recognition
The patterns in the Jenxys system, like the finger tricks for the 9s facts, help students recognize and understand number patterns. Recognizing patterns is a key mathematical thinking skill that
applies to algebra, geometry, statistics, and beyond.
Develops Mental Math Ability
Jenxys Math: A Multiplication Revelation 6
The Jenxys method teaches multiplication in a very visual, hands-on way. Students see how the problems work, so they can eventually solve them in their heads. This ability to do mental math quickly
and accurately is useful in many areas of life.
Sparks Interest in Math
Learning math in an engaging, interactive way makes it more fun and interesting for kids. The Jenxys system, with its visual models, songs, and games, shows students that math can be an exciting
challenge rather than a boring chore. This early interest in math can motivate them to pursue more advanced math and science topics.
Promotes Understanding Over Memorization
Rather than just memorizing the multiplication tables, the Jenxys method helps students truly understand how multiplication works. They see why 5 x 3 is the same as 3 x 5, for example. This
conceptual understanding will benefit them much more in the long run than just memorizing facts. Understanding, rather than just memorizing, is the key to success in mathematics.
The benefits of learning multiplication with the Jenxys method are numerous and long-lasting. Students will develop key mathematical thinking skills and a love of math that will serve them well
beyond just learning the multiplication tables. Jenxys math leads to multiplication revelation!
Jenxys Math can seem confusing at first, but with regular use it will become second nature. Here are some common questions to help you get started:
What is Jenxys Math?
Jenxys Math is a visual multiplication method using a grid to represent the multiplication of two numbers. It provides a visual representation of what’s really happening when you multiply. This helps
build number sense and a deeper understanding of multiplication.
How does it work?
The Jenxys Math grid has columns and rows. The number of columns represents the first factor and the number of rows represents the second factor. Each square in the grid represents the product of the
column number and row number. The sum of all the squares is the product of the two original factors.
For example, to multiply 6 x 4:
6 columns (first factor)
4 rows (second factor)
The sum of all squares (6+12+18+24) is 60. So 6 x 4 = 60.
What are the benefits?
Jenxys Math has many benefits over traditional multiplication methods:
• It’s visual. You can see what multiplication really means.
• It builds number sense. Students gain a deeper understanding of factors and products.
• It works for all ages. Both kids and adults can benefit from this visual model.
• It reduces mistakes. The grid helps keep numbers organized, reducing confusion.
• It’s easy to extend. Students can draw their own grids to explore larger multiplications.
• It relates to area models. The grid models what’s happening with area. This connection helps build understanding.
With regular practice, the Jenxys Math grid will become second nature. Students will develop a strong sense of how multiplication works.
So there you have it, a brand new way to think about multiplication that will revolutionize how you do math in your head. Jenxys method is simple but powerful, using patterns you never noticed before
to solve problems faster than you ever thought possible. Next time you’re calculating a tip or trying to figure out how many pieces of fruit you need for your kid’s soccer team, give Jenxys math a
You’ll be amazed at how quickly the answers come and how much more confident you feel doing math on the fly. Math doesn’t have to be hard or boring when you uncover the secret patterns all around
you. Jenxys method brings the fun and creativity back into numbers. Why limit yourself to the multiplication tables you learned in elementary school when there’s a whole new world of math out there
waiting to be discovered? Open your mind and give it a try.
Leave a Comment | {"url":"https://learningcaptain.com/jenxys-math/","timestamp":"2024-11-06T05:32:04Z","content_type":"text/html","content_length":"265013","record_id":"<urn:uuid:8dea791d-7b31-406c-b649-6b946199fb8e>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00011.warc.gz"} |
2-AFC, 3-AFC, Duo Trio, Triangle, Tetrad Analysis
Analyse results from one of the following tests: 2-AFC, 3-AFC, Duo-Trio, Triangle, Tetrad.
Data Format
1. Discrimination.xlsx
2. Results of the discrimination test are binary (1 = correct answer, 0 = incorrect answer)
2-AFC Test
This is an alternative forced choice (AFC) discrimination test where panellists are presented with 2, 3 or more products and asked to select one product based on a pre-specified attribute. The 2-AFC
test, also known as a directional difference test, is used to establish if a directional difference exists in the perceived intensity of a specified attribute between 2 products. The panellist is
required to state if a 2^nd product is more or less intense in the specified attribute compared to the 1^st presented product.
The guessing probability (probability of getting a correct answer by guessing only) is ½.
3-AFC Test
This is an alternative forced choice (AFC) discrimination test where panellists are presented with 2, 3 or more products and asked to select one product based on a pre-specified attribute. The3-AFC
test is used to establish if there is a discernible difference between 2 products in respect of a specified attribute. The panellist is required to select 1 product from the set of 3 that differences
in the specified attribute.
The guessing probability (probability of getting a correct answer by guessing only) is ⅓.
Duo Trio Test
A discrimination (difference) test to determine if an unspecified difference exists between two products. The panellist is presented with 2 products (A and B) and a reference product (R), and asked
to determine which of A or B is most similar to R. Useful for products that are fairly similar but not totally identical.
The guessing probability (probability of getting a correct answer by guessing only) is ½.
Triangle Test
During a triangle test, a panellist is presented with three samples of which two are equal and one is different. The panellist must state which sample is different. The results indicate whether a
detectable difference exists between two samples. The method is statistically more efficient than the duo trio test but has limited use with products that have a strong and/or lingering flavour.
If there is no difference detected between sample A and B the panellist must choose a random sample. The chance a panellist chooses the odd sample is 1/3. If there is no difference between the
samples you would expect one third of the panellists to choose the odd sample, while two third of the panel chooses one of the equal samples. If there is a detectable difference more than one third
of the panellists will choose the right sample.
The guessing probability (probability of getting a correct answer by guessing only) is ⅓.
An unspecified or specified attribute discrimination test which aims to establish if 2 products are different or similar. Panellists are presented with 2 samples of the 1st product and 2 samples of
the 2^nd product and asked to sort them into two groups of 2 products where products in a group are more similar to each other.
There is a ⅓ chance of selecting of selecting any combination of 2 groups of 2 products.
The statistical principle behind every discrimination test should be to reject a null hypothesis (H[0]). For a difference test the null hypothesis states there is no detectable difference between two
(or more) products. The alternative hypothesis H[1 ]is that there is a detectable difference. If there is sufficient evidence to reject H[0] in favour of the alternative hypothesis H[1], then a
difference can be recorded.
For a similarity test the null hypothesis states that there is a non-negligible difference. The size of difference that will be considered non-negligible must be pre-specified. The alternative
hypothesis is that there is no difference. If there is sufficient evidence to reject H[0] in favour of the alternative hypothesis H[1], then it can be concluded the products are similar.
The data is processed using the binomial test to test for difference among the samples.
The number of correct and incorrect responses are counted. The proportion of correct responses is calculated and from there the proportion of true distinguishers is calculated as follows: where p[c]
is the proportion of correct responses. P[g] is the guessing probability and p[d] is the proportion of distinguishers.
p[c] = p[d] + (1-p[d])*p[g ]
p[d] =(p[c]-p[g])/(1-p[g])
1. Treat Sessions/Replicates separately: If the data has been gathered over different sessions, or there are different replicates, these can be analysed separately.
2. Type of test: Similarity or difference test.
3. Model: Guessing or Thurstonian model.
4. Prop of Discriminators Threshold (Pd): If the test is a similarity test and the Guessing Model is used, the threshold that will be considered non-negligible. That is, H[0] is that Pd is greater
than this threshold.
5. D-prime Threshold (d’): If the test is a similarity test and the Thurstonian Model is used, the threshold that will be considered non-negligible. That is, H[0] is that d’ is greater than this
6. Confidence level: The probability for the confidence intervals. That is, a 0.95 confidence interval means that in 95% of population samples, the true value would lie in this confidence interval.
7. Number of Decimals for Values: Required number of decimals for values given in the results.
8. Number of Decimals for P-Values: Required number of decimals for any p-values given in the results.
Results and Interpretation
1. N total: The total number of tests in the results set
2. N correct: The number of correct tests
3. N incorrect: The number of incorrect tests (N incorrect + N correct = N total)
4. Proportion correct: The proportion of correct tests as a percentage (100 x N correct/N tot)
5. P-value: The p-value indicates the probability of obtaining the result if the Null hypothesis is true.
6. Min correct (0.1%): The minimum number of correct responses for the result to be significant with 99.9% probability.
7. Min correct (1%): The minimum number of correct responses for the result to be significant with 99% probability. (For a similarity test this is the maximum number).
8. Min correct (5%): The minimum number of correct responses for the result to be significant with 95% probability.
9. Min correct (10%): The minimum number of correct responses for the result to be significant with 90% probability.
10. p < 0.1%: Is the p-value less than 0.001 i.e. Is the result significant at 99.9% level?
11. p < 1%: Is the p-value less than 0.01 i.e. Is the result significant at 99% level?
12. p < 5%: Is the p-value less than 0.05 i.e. Is the result significant at 95% level?
13. p < 10%: Is the p-value less than 0.1 i.e. Is the result significant at 90% level?
1. Proportion correct: The proportion of correct responses (within the bounds of the guessing probability) and confidence interval, calculated using the exact method.
2. Proportion Discriminators: The proportion of responses that are true distinguishers and confidence interval, calculated using the exact method.
3. D-prime: is an estimation of the distance between the products according to the Thurstonian scale. It is the difference between the mean values of the two signals divided by the standard
When testing for a difference, the p-value indicates if the samples are significantly different, and at what level. You will decide whether to conclude if the samples are different based on the risk
you want to take.
In the case of similarity, the interpretation is similar, but will be in the context of the pd or d-prime value you specified as a non-negligible difference.
If the data contains 3 or more replicates the Beta-Binomial model is fitted. This is to check for loss of independence in the replicates. This is also called over-dispersion in the data.
The beta-binomial model is parameterized in terms of mu and gamma, where mu corresponds to a probability parameter and gamma measures over-dispersion. Both parameters are restricted to the interval
(0, 1).
The parameters of the standard (i.e. not corrected) beta-binomial model refers to the mean (i.e. probability) and dispersion on the scale of the observations, i.e. on the scale where we talk of a
probability of a correct answer (Pc).
The following parameters are returned, with estimate, standard error, and confidence interval limits.
1. Probability (mu)
2. Over-dispersion (gamma)
3. Pc – the probability of a correct response.
4. Pd – the probability of true discrimination.
5. d-prime – the ‘distance’ between the products.
Test (Beta-Binomial)
The test shows:
1. a likelihood ratio test of over-dispersion on one degree of freedom.
2. and a likelihood ratio test of association (i.e. where the null hypothesis is "no difference" and the alternative hypothesis is "any difference") on two degrees of freedom (chi-square tests).
3. If the data is over-dispersed, the p-value for the test for over dispersion should be smaller than the desired alpha.
Technical Information
1. The R package sensR (Rune Christensen and Per B. Brockhoff) is used.
2. The confidence intervals are calculated using the ‘exact’ binomial method.
1. ISO 4120:2004 Sensory Analysis – Methodology – Triangle Test
2. ASTM-E1885-04 (2011) Standard Test Method for Sensory Analysis – Triangle Test
3. Lawless, H.T. and Heymann, H. (2010). Sensory Evaluation of Food – Principles and Practices. Springer.
4. Ennis. J. M., and Jesionka, V. (2011) – The Power of Sensory Discrimination Methods Revisited. Journal of Sensory Studies, 26, 371-382.
5. Ennis. J. M. (2012). Guiding the Switch from Triangle Testing to Tetrad Testing. Journal of Sensory Studies, 27, 4, 223-231.
6. Garcia, K., Ennis, J.M. and PrinyawIwatkul, W. (2013). Reconsidering the Specified Tetrad Test. Journal of Sensory Studies, 28, 6, 445-449.
7. O’Mahony, M. (2013). The Tetrad Test – Looking Back, Looking Forward. Journal of Sensory Studies, 28, 4, 259-263.
8. ISO 10399:2004 Sensory Analysis – Methodology – Duo-Trio Test
9. ASTM-E2610-08 (2011) Standard Test Method for Sensory Analysis – Duo-Trio Test
10. Christensen, R.H.B. (2014). Statistical Methodology for Sensory Discrimination Tests and Its Implementation in SensR.
11. Christensen, R.H.B. and Brockhoff, P.B (2014). Package ‘sensR’.
12. Christensen, R.H.B. and Brockhoff, P.B (2014). Sensory Discrimination Testing with the sensR Package. | {"url":"https://support.eyequestion.nl/portal/en/kb/articles/2-2-afc-3-afc-duo-trio-triangle-tetrad","timestamp":"2024-11-02T08:28:15Z","content_type":"text/html","content_length":"64254","record_id":"<urn:uuid:ab320239-34b2-4448-9271-2f8cfeb92818>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00035.warc.gz"} |
Understanding Electric Field for Point Charges and Discrete Sets
Coordinates: The expressions in the transcription are in spherical coordinates, located by a radial vector, polar angle, and azimuthal angle.
Concept of Electric Field: The concept of the electric field is the disturbance generated by a positive charge in space, forming radial and outward electric field lines.
Interaction of Charges: The electric field lines represent the interaction between charges, with positive charges generating outward field lines and negative charges generating field lines towards
the charge. | {"url":"https://chattube.io/summary/film-animation/EIXooch5s8Q","timestamp":"2024-11-09T19:09:07Z","content_type":"text/html","content_length":"39940","record_id":"<urn:uuid:9913c4ad-129a-469d-a8d5-68eca62503b7>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00144.warc.gz"} |
Application of superposition method to the thermal stress problem in composites
In the classical work of 1957, Eshelby proposed a solution to the three dimensional elasticity problem of evaluating the stress distribution when inhomogeneous inclusions are embedded in an infinite
elastic body. Later, his approach was applied to the nonlinear micromechanics problem of determining thermal residual stresses for composites. Eshelby's solution, valid only for an isolated
unidirectional fiber, is extended in this paper to include the effect of the interaction of neighboring fibers in a regularly packed array. The stress field at any given material point is obtained as
a superposition of the field due to the disturbance of all the fibers in the composite. The theoretical results are compared with finite element solutions, and they match very well.
Original language English (US)
Title of host publication Proceedings of the International Conference on Advanced Composite Materials
Editors T. Chandra, A.K. Dhingra
Publisher Publ by Minerals, Metals & Materials Soc (TMS)
Pages 323-327
Number of pages 5
ISBN (Print) 0873392515
State Published - 1993
Externally published Yes
Event Proceedings of the International Conference on Advanced Composite Materials - Wollongong, Aust
Duration: Feb 15 1993 → Feb 19 1993
Publication series
Name Proceedings of the International Conference on Advanced Composite Materials
Other Proceedings of the International Conference on Advanced Composite Materials
City Wollongong, Aust
Period 2/15/93 → 2/19/93
All Science Journal Classification (ASJC) codes
Dive into the research topics of 'Application of superposition method to the thermal stress problem in composites'. Together they form a unique fingerprint. | {"url":"https://researchwith.njit.edu/en/publications/application-of-superposition-method-to-the-thermal-stress-problem","timestamp":"2024-11-07T23:49:34Z","content_type":"text/html","content_length":"48158","record_id":"<urn:uuid:eaa77e01-116c-4ace-9c93-b9d9c3e8041d>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00283.warc.gz"} |
Minimum Spanning Trees | CS61B Guide
Minimum Spanning Trees
Special thanks to Arin for writing this page!
Spanning Tree Definition
A spanning tree T is a subgraph of a graph G where T:
Is connected (there's a path to every vertex)
Includes every vertex (spanning property)
Notice: the first two properties defines a tree structure, and the last property makes the tree spanning.
A minimum spanning tree is a spanning tree with minimum total edge weight.
Example: I want to connect an entire town with wiring and would like to find the optimal wiring connection that connects everyone but uses the least wire.
MST vs. Shortest Path Tree
In contrast to a shortest path tree, which is essentially the solution tree to running Dijkstra’s with root node = source vertex, a MST has no source. However, it is possible for the MST to be the
same as the SPT.
We can think of the MST as a global property for the entire graph, as opposed to SPT which depends on which node is the source node.
If the edges of the graph are not unique, there’s a chance that the MST is not unique.
Cuts Property
A cut is defined as assigning the nodes in a graph into two sets.
A crossing edge is an edge that connects two nodes that are in different sets
The smallest crossing edge is the crossing edge with smallest weight
The Cut Property states that the smallest crossing edge is always going to be in the MST, no matter how the cut is made. | {"url":"https://cs61b.bencuan.me/algorithms/minimum-spanning-trees","timestamp":"2024-11-13T12:40:14Z","content_type":"text/html","content_length":"171575","record_id":"<urn:uuid:3a6d110f-86a8-467b-a50f-88b166390bda>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00692.warc.gz"} |
IBM c1000-059 practice test
IBM AI Enterprise Workflow V1 Data Science Specialist Exam
Last exam update: Oct 31 ,2024
Page 1 out of 5. Viewing questions 1-15 out of 62
Question 1
Which is the most important thing to ensure while collecting data?
• A. samples collected are skewed with each other
• B. samples collected are all strongly correlated with each other
• C. samples collected adequately cover the space of all possible scenarios
• D. samples collected focus only on the most common cases
Question 2
What is the meaning of "deep" in deep learning?
• A. To go deep into the loss function landscape.
• B. The higher the number of machine learning algorithms that can be applied, the deeper is the learning.
• C. A kind of deeper understanding achieved by any approach taken.
• D. It indicates the many layers contributing to a model of the data.
Reference: https://en.wikipedia.org/wiki/Deep_learning
Question 3
Which algorithm is best suited if a client needs full explainability of the machine learning model?
• A. decision tree
• B. logistic regression
• C. support vector machine (SVM)
• D. recurrent neural network
Question 4
Given the following sentence:
The dog jumps over a fence.
What would a vectorized version after common English stopword removal look like?
• A. ['dog', 'fence', 'run']
• B. ['fence', 'jumps']
• C. ['dog', 'fence', 'jumps']
• D. ['a', 'dog', 'fence', 'jumps', 'over', 'the']
different-libraries- f20bac19929a
Question 5
Which statement defines p-value?
• A. It is the probability of accepting a null hypothesis when the hypothesis is proven true.
• B. It is the probability of rejecting a null hypothesis when the hypothesis is proven false.
• C. It is the probability of accepting a null hypothesis when the hypothesis is proven false.
• D. It is the probability of rejecting a null hypothesis when the hypothesis is proven true.
Reference: https://courses.lumenlearning.com/wmopen-concepts-statistics/chapter/introduction-to-
hypothesis- testing-5-of-5/
Question 6
What is the primary role of a data steward?
• A. they are a "blue sky thinker" who comes up with new approaches to use new data in innovative ways
• B. they have a strong understanding of the enterprise's database architecture
• C. they define data processes to meet compliance and regulatory obligations
• D. the one who collects, processes, and performs statistical analysis on data
Reference: https://analyticsindiamag.com/data-steward-roles-responsibilities/
Question 7
Which is an example of a nominal scale data?
• A. a variable industry with categorical values such as financial, engineering, and retail
• B. a variable mood with a scale of values unhappy, ok, and happy
• C. a variable bank account balance whose possible values are $5, $10, and $15
• D. a variable temperature with a scale of values low, medium, and high
Question 8
A data scientist is exploring transaction data from a chain of stores with several locations. The data
includes store number, date of sale, and purchase amount.
If the data scientist wants to compare total monthly sales between stores, which two options would
be good ways to aggregate the data? (Choose two.)
• A. Find the sum of the transaction prices
• B. Select the largest transaction amount by month and store
• C. Write a GROUP BY query
• D. Plot a time series plot of transaction amounts
• E. Generate a pivot table
Question 9
A data analyst creates a term-document matrix for the following sentence: I saw a cat, a dog and
another cat.
Assuming they used a binary vectorizer, what is the resulting weight for the word cat?
Question 10
In a hyperparameter search, whether a single model is trained or a lot of models are trained in
parallel is largely determined by?
• A. The number of hyperparameters you have to tune.
• B. The presence of local minima in your neural network.
• C. The amount of computational power you can access.
• D. Whether you use batch or mini-batch optimization.
Question 11
If the distribution of the height of American men is approximately normal, with a mean of 69 inches
and a standard deviation of 2.5 inches, then roughly 68 percent of American men have heights
• A. 64 inches and 74 inches
• B. 66.5 inches and 69 inches
• C. 71.5 inches and 76.5 inches
• D. 66.5 inches and 71.5 inches
Question 12
Which two properties hold true for standardized variables (also known as z-score normalization)?
(Choose two.)
A. standard deviation = 0.5
B. expected value = 0
C. expected value = 0.5
D. expected value = 1
E. standard deviation = 1
Question 13
What is the main difference between traditional programming and machine learning?
• A. Machine learning models take less time to train.
• B. Machine learning takes full advantage of SDKs and APIs.
• C. Machine learning is optimized to run on parallel computing and cloud computing.
• D. Machine learning does not require explicit coding of decision logic.
Question 14
What is the name of the design thinking work product that contains a summary description of a
particular person or role?
• A. persona
• B. snapshot
• C. My Sticky Note
• D. user summary report
Reference: https://www.interaction-design.org/literature/topics/design-thinking
Question 15
What are two methods used to detect outliers in structured data? (Choose two.)
• A. multi-label classification
• B. isolation forest
• C. gradient descent
• D. one class Support Vector Machine (SVM)
• E. Word2Vec
used-for-big- data | {"url":"https://www.examgo.com/exams/ibm/c1000-059/","timestamp":"2024-11-07T16:13:47Z","content_type":"text/html","content_length":"123637","record_id":"<urn:uuid:f73ced76-26a4-4cd4-8ef8-cfe19c966490>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00298.warc.gz"} |
Angles of Attack
home | resources | github | about
This is where I maintain lists of technical resources for subjects related to data science, machine learning, & cybersecurity; a repo of mlops & mlsecops resources, and a list of aerospace
organizations & aviation cyber resources.
Technical resources
by topic:
scraping and building datasets
learn to code
data science
machine learning
deep learning
RNNs and LSTMs
natural language processing
graph algorithms
data visualization
scraping and building datasets
learning AI through code
free courses
aerospace organizations
International Civil Aviation Organization (ICAO)
Aviation Cybersecurity Overview go »
International Air Transport Association (IATA)
Aviation Cyber Security go »
United States Federal Aviation Administration (FAA)
Cybersecurity Awareness Symposium go »
European Union Aviation Safety Agency (EASA)
Cybersecurity Overview go »
home | resources | github | about | {"url":"https://anglesofattack.io/resources.html","timestamp":"2024-11-07T10:55:26Z","content_type":"text/html","content_length":"42836","record_id":"<urn:uuid:31df9262-7eac-439a-a92a-995704a3d558>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00842.warc.gz"} |
problem 5 (23)
I have trouble solving this problem: $yu_x - xu_y = x^2$. I know the characteristic equation is $\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x^2}$ and then have $C = \frac{x^2}{2} + \frac{y^2}{2}$. Then
the following should be the integration relative to $du$, but either $\frac{du}{dx}$ or $\frac{du}{dy}$ will contain not only one variable, like $\frac{du}{dx} = \frac{x^2}{y}$ contain both $x$ and
$y$. I wonder if $x$ and $y$ are independent here. If not, should I rewrite the expression $C = \frac{x^2}{2} + \frac{y^2}{2}$ in order to get the expression of y in terms of x , and then applies it
into the integration relative to $du$? Any reply would be appreciated. | {"url":"https://forum.math.toronto.edu/index.php?PHPSESSID=pbisnnifcntqvnfj8ll8f9nr35&topic=1675.msg6051","timestamp":"2024-11-08T02:47:48Z","content_type":"application/xhtml+xml","content_length":"42478","record_id":"<urn:uuid:dcfe5771-e565-405e-a74e-6612393389d3>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00220.warc.gz"} |
[Solved] Mass Transfer MCQ [Free PDF] - Objective Question Answer for Mass Transfer Quiz - Download Now!
Mass Transfer MCQ Quiz - Objective Question with Answer for Mass Transfer - Download Free PDF
Last updated on Aug 6, 2024
Mass Transfer MCQ are essential for evaluating one's understanding of this important field of chemical engineering. Mass transfer involves the movement of components within a mixture, such as
liquids, gases, or solids. Mass Transfer MCQ assess learners' knowledge of mass transfer principles, various mass transfer operations, and mass transfer coefficients. By answering such MCQs,
individuals can enhance their understanding of diffusion, absorption, distillation, extraction, and other mass transfer processes. These Mass Transfer MCQ contribute to developing strong skills in
designing and analyzing mass transfer systems, essential for chemical engineers.
Latest Mass Transfer MCQ Objective Questions
Mass Transfer Question 1:
If the bubbles formed on a submerged hot surface get absorbed in the mass of liquid, the process of boiling is called
Answer (Detailed Solution Below)
Option 1 : Nucleate boiling
Mass Transfer Question 1 Detailed Solution
Boiling occurs at the solid-liquid interface when a liquid is brought into a contact surface whose temperature is sufficiently higher than the saturation temperature of the liquid.
Four different boiling regimes are:
Natural Convection Boiling/Pool boiling:
• Boiling starts as soon as the saturation temperature is reached at that pressure, but bubbles do not form until the liquid is heated few degrees above the saturation temperature, therefore the
liquid is evaporated when it rises to the free surface, this fluid motion is due to the natural convection currents and heat transfer is taking place between the heated surface and the liquid is
due to the natural convection.
• At first, the number and size of bubbles are small, and bubbles rise up and condense in the liquid before reaching the interface.
Nucleate Boiling:
• In nucleate boiling the bubbles start forming at an increasing rate and at an increasing number of nucleation sites.
• This regime can be separated into two regions, between A-B isolated bubbles are formed and dissipated into the liquid after they separate from the surface.
• The space vacated by the bubbles is filled by the liquid in the vicinity of the heater surface, and the process is repeated.
• In B-C the temperature is further increased and bubbles form at such great rates at such a large number of nucleation sites that they form numerous continuous columns of vapour in the liquid.
Transition Boiling:
• Once the temperature is increased past point C the heat flux decreases because a large portion of the surface is covered by the vapor film, which acts as an insulation due to the low thermal
conductivity of the vapor relative to the water.
Film Boiling:
• In this region, the heater surface is completely covered by the continuous stable vapor film, point D, where the heat flux reaches a minimum.
• The heat transfer rate increases with the increasing temperature due to radiation heat transfer between the vapour film and the liquid.
Mass Transfer Question 2:
The dimensional number associated with mass transfer is
Answer (Detailed Solution Below)
Option 1 : Schmidt Number
Mass Transfer Question 2 Detailed Solution
Schmidt number is a dimensionless number and defined as the ratio between momentum diffusivity and mass diffusivity, and is used to characterize the fluid flows where both momentum and mass transfer
are involved.
It is given by
\(Schmidt\;number\;\left( {Sc} \right) = \frac{{Momentum\;diffusivity}}{{Mass\;diffusivity}} = \frac{{\nu \;\left( {kinematic\;viscosity} \right)}}{{D\;\left( {mass\;diffusivity} \right)}}\)
For higher Schmidt numbers, momentum diffusion dominates, and for lower Schmidt numbers, mass diffusion dominates
Important Points
Schmidt number can be considered analogous to Prandtl number, while the
Prandtl number describes the diffusion of heat (thermal boundary layer).
Schmidt number describes the diffusion of mass (concentration boundary layer).
Mass Transfer Question 3:
Under the steady-state condition, Fick’s first law is given as
\(J = - D\frac{{dc}}{{dx}}\)
Where, J = Diffusion flux,
D = Diffusion coefficient,
dc / dx = Concentration gradient
The unit of diffusion coefficient will be:
Answer (Detailed Solution Below)
Mass Transfer Question 3 Detailed Solution
Fick’s Law of diffusion states that “the mass flux of a constituent per unit area is proportional to the concentration gradient”.
\(J = - D\frac{{dc}}{{dx}}\)
J = mass flux of constituent per unit area, \(\frac{{dc}}{{dx}}= concentration~ gradient\)
-ve sign indicates that the concentration gradient decreasing in the direction of mass transfer.
\(J = \frac{{\dot m}}{A} = kg/{m^2}s\)
\(\frac{{dc}}{{dx}} = \frac{{kg/{m^3}}}{m} = kg/{m^4}\)
∴ \(D = \frac{J}{{\frac{{dc}}{{dx}}}}\)
\( \Rightarrow D = \frac{{\frac{{kg}}{{{m^2}s}}}}{{\frac{{kg}}{{{m^4}}}}}\)
\( \Rightarrow D = \frac{{kg}}{{{m^2}s}} \times \frac{{{m^4}}}{{kg}}\)
∴ D = m^2/s
Top Mass Transfer MCQ Objective Questions
Under the steady-state condition, Fick’s first law is given as
\(J = - D\frac{{dc}}{{dx}}\)
Where, J = Diffusion flux,
D = Diffusion coefficient,
dc / dx = Concentration gradient
The unit of diffusion coefficient will be:
Answer (Detailed Solution Below)
Fick’s Law of diffusion states that “the mass flux of a constituent per unit area is proportional to the concentration gradient”.
\(J = - D\frac{{dc}}{{dx}}\)
J = mass flux of constituent per unit area, \(\frac{{dc}}{{dx}}= concentration~ gradient\)
-ve sign indicates that the concentration gradient decreasing in the direction of mass transfer.
\(J = \frac{{\dot m}}{A} = kg/{m^2}s\)
\(\frac{{dc}}{{dx}} = \frac{{kg/{m^3}}}{m} = kg/{m^4}\)
∴ \(D = \frac{J}{{\frac{{dc}}{{dx}}}}\)
\( \Rightarrow D = \frac{{\frac{{kg}}{{{m^2}s}}}}{{\frac{{kg}}{{{m^4}}}}}\)
\( \Rightarrow D = \frac{{kg}}{{{m^2}s}} \times \frac{{{m^4}}}{{kg}}\)
∴ D = m^2/s
If the bubbles formed on a submerged hot surface get absorbed in the mass of liquid, the process of boiling is called
Answer (Detailed Solution Below)
Option 1 : Nucleate boiling
Boiling occurs at the solid-liquid interface when a liquid is brought into a contact surface whose temperature is sufficiently higher than the saturation temperature of the liquid.
Four different boiling regimes are:
Natural Convection Boiling/Pool boiling:
• Boiling starts as soon as the saturation temperature is reached at that pressure, but bubbles do not form until the liquid is heated few degrees above the saturation temperature, therefore the
liquid is evaporated when it rises to the free surface, this fluid motion is due to the natural convection currents and heat transfer is taking place between the heated surface and the liquid is
due to the natural convection.
• At first, the number and size of bubbles are small, and bubbles rise up and condense in the liquid before reaching the interface.
Nucleate Boiling:
• In nucleate boiling the bubbles start forming at an increasing rate and at an increasing number of nucleation sites.
• This regime can be separated into two regions, between A-B isolated bubbles are formed and dissipated into the liquid after they separate from the surface.
• The space vacated by the bubbles is filled by the liquid in the vicinity of the heater surface, and the process is repeated.
• In B-C the temperature is further increased and bubbles form at such great rates at such a large number of nucleation sites that they form numerous continuous columns of vapour in the liquid.
Transition Boiling:
• Once the temperature is increased past point C the heat flux decreases because a large portion of the surface is covered by the vapor film, which acts as an insulation due to the low thermal
conductivity of the vapor relative to the water.
Film Boiling:
• In this region, the heater surface is completely covered by the continuous stable vapor film, point D, where the heat flux reaches a minimum.
• The heat transfer rate increases with the increasing temperature due to radiation heat transfer between the vapour film and the liquid.
The dimensional number associated with mass transfer is
Answer (Detailed Solution Below)
Option 1 : Schmidt Number
Schmidt number is a dimensionless number and defined as the ratio between momentum diffusivity and mass diffusivity, and is used to characterize the fluid flows where both momentum and mass transfer
are involved.
It is given by
\(Schmidt\;number\;\left( {Sc} \right) = \frac{{Momentum\;diffusivity}}{{Mass\;diffusivity}} = \frac{{\nu \;\left( {kinematic\;viscosity} \right)}}{{D\;\left( {mass\;diffusivity} \right)}}\)
For higher Schmidt numbers, momentum diffusion dominates, and for lower Schmidt numbers, mass diffusion dominates
Important Points
Schmidt number can be considered analogous to Prandtl number, while the
Prandtl number describes the diffusion of heat (thermal boundary layer).
Schmidt number describes the diffusion of mass (concentration boundary layer).
Mass Transfer Question 7:
Under the steady-state condition, Fick’s first law is given as
\(J = - D\frac{{dc}}{{dx}}\)
Where, J = Diffusion flux,
D = Diffusion coefficient,
dc / dx = Concentration gradient
The unit of diffusion coefficient will be:
Answer (Detailed Solution Below)
Mass Transfer Question 7 Detailed Solution
Fick’s Law of diffusion states that “the mass flux of a constituent per unit area is proportional to the concentration gradient”.
\(J = - D\frac{{dc}}{{dx}}\)
J = mass flux of constituent per unit area, \(\frac{{dc}}{{dx}}= concentration~ gradient\)
-ve sign indicates that the concentration gradient decreasing in the direction of mass transfer.
\(J = \frac{{\dot m}}{A} = kg/{m^2}s\)
\(\frac{{dc}}{{dx}} = \frac{{kg/{m^3}}}{m} = kg/{m^4}\)
∴ \(D = \frac{J}{{\frac{{dc}}{{dx}}}}\)
\( \Rightarrow D = \frac{{\frac{{kg}}{{{m^2}s}}}}{{\frac{{kg}}{{{m^4}}}}}\)
\( \Rightarrow D = \frac{{kg}}{{{m^2}s}} \times \frac{{{m^4}}}{{kg}}\)
∴ D = m^2/s
Mass Transfer Question 8:
If the bubbles formed on a submerged hot surface get absorbed in the mass of liquid, the process of boiling is called
Answer (Detailed Solution Below)
Option 1 : Nucleate boiling
Mass Transfer Question 8 Detailed Solution
Boiling occurs at the solid-liquid interface when a liquid is brought into a contact surface whose temperature is sufficiently higher than the saturation temperature of the liquid.
Four different boiling regimes are:
Natural Convection Boiling/Pool boiling:
• Boiling starts as soon as the saturation temperature is reached at that pressure, but bubbles do not form until the liquid is heated few degrees above the saturation temperature, therefore the
liquid is evaporated when it rises to the free surface, this fluid motion is due to the natural convection currents and heat transfer is taking place between the heated surface and the liquid is
due to the natural convection.
• At first, the number and size of bubbles are small, and bubbles rise up and condense in the liquid before reaching the interface.
Nucleate Boiling:
• In nucleate boiling the bubbles start forming at an increasing rate and at an increasing number of nucleation sites.
• This regime can be separated into two regions, between A-B isolated bubbles are formed and dissipated into the liquid after they separate from the surface.
• The space vacated by the bubbles is filled by the liquid in the vicinity of the heater surface, and the process is repeated.
• In B-C the temperature is further increased and bubbles form at such great rates at such a large number of nucleation sites that they form numerous continuous columns of vapour in the liquid.
Transition Boiling:
• Once the temperature is increased past point C the heat flux decreases because a large portion of the surface is covered by the vapor film, which acts as an insulation due to the low thermal
conductivity of the vapor relative to the water.
Film Boiling:
• In this region, the heater surface is completely covered by the continuous stable vapor film, point D, where the heat flux reaches a minimum.
• The heat transfer rate increases with the increasing temperature due to radiation heat transfer between the vapour film and the liquid.
Mass Transfer Question 9:
The dimensional number associated with mass transfer is
Answer (Detailed Solution Below)
Option 1 : Schmidt Number
Mass Transfer Question 9 Detailed Solution
Schmidt number is a dimensionless number and defined as the ratio between momentum diffusivity and mass diffusivity, and is used to characterize the fluid flows where both momentum and mass transfer
are involved.
It is given by
\(Schmidt\;number\;\left( {Sc} \right) = \frac{{Momentum\;diffusivity}}{{Mass\;diffusivity}} = \frac{{\nu \;\left( {kinematic\;viscosity} \right)}}{{D\;\left( {mass\;diffusivity} \right)}}\)
For higher Schmidt numbers, momentum diffusion dominates, and for lower Schmidt numbers, mass diffusion dominates
Important Points
Schmidt number can be considered analogous to Prandtl number, while the
Prandtl number describes the diffusion of heat (thermal boundary layer).
Schmidt number describes the diffusion of mass (concentration boundary layer).
Mass Transfer Question 10:
A tank contains a mixture of CO[2] and N[2] in the mole proportions of 0.2 and 0.8 at 1 bar and 290 K. It is connected by a duct of sectional area 0.1 m^2 to another tank containing a mixture of CO
[2] and N[2] in the mole proportion of 0.8 and 0.2. The duct I 0.5 m long. The diffusion of CO[2] is (Take Diffusion coefficient = 0.16 × 10^−4 m^2/s).
Answer (Detailed Solution Below)
Option 4 : 3.5 × 10^−6 kg/s
Mass Transfer Question 10 Detailed Solution
The partial pressures of two gases in two tanks are
P = P[1] + P[2], P[i] = x[i]P, x[i] = mole proportion.
P[A1] = 0.2 bar, P[B1] = 0.8 bar,
and P[A2] = 0.08 bar, P[B2] = 0.2 bar.
For the rate of molar diffusion of CO[2 ]using equation.
where D[AB] = Diffusion coefficient, R[u] = Universal gas constant, A = Sectional area of the duct
L = 0.5 m, A = 0.1 m^2, D[AB] = 0.16 × 10−4 m2/s, T = 290 K.
Diffusion of CO[2],
\(N_A=\frac{0.16× 10^{-4}×0.1}{8314× 290}[\frac{(0.8-0.2)× 10^5}{0.5}]\)
N[A ]= 7.96 × 10^-8 kg-mol/sec
m[A] = N[A] × M[A] = 7.96 × 10-8 × 44 = 3.5 × 10−6 kg/s.
Mass Transfer Question 11:
A 25 cm × 25 cm × 1 cm flat wet sheet weight 3 kg initially was dried flow both sides under constant drying condition i.e. at constant drying rate period. It took 2000 sec for the weight of sheet to
reduce to 2.75 kg. Another 1m × 1m × 10 cm flat sheet is to be dried from one side only. Under the same drying rate and other conditions, time required for drying (in sec) from initial weight of 3 kg
to 2 kg.
Answer (Detailed Solution Below) 1000
Mass Transfer Question 11 Detailed Solution
\(\theta = \frac{{{S_S}}}{{A\;{M_C}}}\left( {{X_1} - {X_2}} \right)\)
A[2] = 1 m^2
A[1] = 0.25 × 0.25 × 2m^2
= 0.0625 × 2 m^2
= 0.125 m^2
\(\frac{{{\theta _1}}}{{{\theta _2}}} = \frac{{\frac{{{S_S}\left( {{X_1} - {X_2}} \right)}}{{{A_1}}}}}{{\frac{{{S_S}\left( {{X_1} - {X_2}} \right)}}{{{A_2}}}}} = \frac{{\frac{{\left( {3 - 2.75} \
right)}}{{0.125}}}}{{\frac{{3 - 2}}{1}}} = \frac{{0.25}}{{0.125}}\)
\(= \frac{1}{{0.5}}\)
\(\Rightarrow \frac{{{\theta _1}}}{{{\theta _2}}} = \frac{1}{{0.5}} = 2\)
\(\Rightarrow {\theta _2} = \frac{{{\theta _1}}}{2} = \frac{{2000}}{2} = 1000\;sec\)
Mass Transfer Question 12:
If the percent humidity of air (25°C, total pressure 100 kPa) is 25% and the saturation pressure of water vapor at that temperature is 10 kPa. Calculate the percent relative humidity and absolute
humidity of air respectively.
Answer (Detailed Solution Below)
Option 1 : 27% and 0.017
Mass Transfer Question 12 Detailed Solution
Percent humidity \(= \frac{Y}{{{Y_s}}} \times 100 = 25\)
\(\Rightarrow \frac{Y}{{{Y_s}}} = 0.25\)
\(\Rightarrow \frac{{{P_A}}}{{{P_T} - {P_A}}} \times \frac{{{P_T} - P_A^v}}{{P_A^v}} = 0.25\)
\(\Rightarrow \frac{{{P_A}}}{{100 - {P_A}}} \times \frac{{100 - 10}}{{10}} = 0.25\)
⇒ 9P[A] = 25 – 0.25 P[A]
\(\Rightarrow {P_A} = \frac{{25}}{{9.25}} = 2.7\;kpa\)
Percent relative humidity \(= \frac{{{P_A}}}{{P_A^v}} \times 100 = \frac{{2.7}}{{10}} \times 100 = 27\% \)
Absolute humidity \(= \frac{{{P_A}}}{{{P_T} - {P_A}}} \times \frac{{{M_w}}}{{{M_{Air}}}}\)
\(= \frac{{2.7}}{{100 - 2.7}} \times \frac{{18}}{{29}} = 0.017\)
Mass Transfer Question 13:
A distillation concern at pilot plant is scaled up by 5 times for industrial use at steady state. After scaling up
Answer (Detailed Solution Below)
Option 3 : The feed flow rats and products flow rates are increased by 5 times
Mass Transfer Question 13 Detailed Solution
Scale up in industry means we are going to increase the feed flow and product flow rates. | {"url":"https://testbook.com/objective-questions/mcq-on-mass-transfer--5eea6a0c39140f30f369e135","timestamp":"2024-11-06T18:26:36Z","content_type":"text/html","content_length":"584131","record_id":"<urn:uuid:3c37b9eb-4a68-41b4-b066-16037c590dd0>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00195.warc.gz"} |
Which of the following statements is CORRECT?
Statement - 1 :x =
Statement - 2 :x =
A Only Statement - 1
B Only Statement - 2
C Both Statement - 1 and Statement - 2
D Neither Statement - 1 nor Statement - 2
Ans 1:
The solution proves that option B is the correct answer but shows option A as the correct answer, which is wrong. Please fix this mistake.
Ans 2:
Ans 3:
When even in the solution the answer is II is correct then why in the correct answer it shows the option which says I is correct? This is an absolute blunder. Please fix it, SOF.
Ans 4:
complete blunder as sharanyaa saidPLEASE FIX ITDON'T CUT OUR MARKS FOR SUCH TECHNICAL ERRORS
Ans 5:
According to the solution itself, Statement 1 is false while Statement 2 is true. i have checked it by solving the questions myself and it is correct. The question says,"Which of the following
statements is CORRECT?" They have asked for the CORRECT statement, not the INCORRECT one. Therefore, the answer will be Option B, not Option A. SOF, PLEASE RECTIFY THESE ERRORS WHICH KEEP OCCURING
Ans 7:
If statement 2 is true, then B should be answer wat.
Ans 8:
The explanation is not reflected in the answer. The explanation shows option B to be correct, but answer is marked as option A. If this is a representation of the actual exam, 1-2 years later no one
would give it. This is a serious blunder occurring multiple times in the website. We did not pay to get these terrible errors. Please fix this. | {"url":"https://www.sofolympiadtrainer.com/forum/questions/38391/pwhich-of-the-following-statements-is-correct-brbstatement---1-bixi--img-srchttpswwwsofolympiadtrain","timestamp":"2024-11-08T02:44:50Z","content_type":"text/html","content_length":"105434","record_id":"<urn:uuid:2e461240-0d66-473d-87ef-aa16d8c1a812>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00290.warc.gz"} |
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What Is the Rule of 70?
The rule of 70 is a simple mathematical formula that can be used to approximate how long it takes for an investment to double in value.
Key Takeaways
• The rule of 70 is a basic formula used to estimate how long it will take for an investment to double in value.
• To use the rule of 70, simply divide 70 by the annual rate of return.
• The rule of 70 only provides an estimate, not a guarantee, of an investment’s growth potential.
Definition and Examples of the Rule of 70
Investors typically use the rule of 70 to predict the number of years it takes for an investment to double in value based on a specific rate of return (an investment’s gain or loss over a period of
The rule of 70 is commonly used to compare investments with different annual interest rates. This makes it simple for investors to figure out how long it may be before they see similar returns on
their money from each of the investments.
Let’s say an investor decides to compare rates of return on the investments in their retirement portfolio to get an idea of how long it may take their savings to double. To calculate the doubling
time, the investor would simply divide 70 by the annual rate of return. Here’s an example:
• At a 4% growth rate, it would take 17.5 years for a portfolio to double (70/4)
• At a 7% growth rate, it would take 10 years to double (70/7)
• At an 11% growth rate, it would take 6.4 years to double (70/11)
• Alternate name: Doubling time
How the Rule of 70 Works
Now that you’ve seen the rule of 70 in action, let’s break down the formula so you understand how to apply the rule of 70 to your own investments.
Again, calculating the rule of 70 is pretty straightforward. All you do is divide 70 by the estimated annual rate of return to find out how many years it’ll take for an investment to double in size.
For the calculation to work properly, you’ll need to have at least an estimate of the investment’s annual growth or return rate.
Do I Need the Rule of 70?
Keep in mind that the rule of 70 is a rough estimate, but it can come in handy if you want a more concrete way of looking at the potential of a retirement portfolio, mutual fund, or other investment
than the interest rate alone could provide. Knowing the number of years it could take to reach a desired value can help you plan which investments to choose for your retirement portfolio, for
Let’s say you wanted to pick a precise mix of investments with the potential to grow to a certain value by the time you retire in 20 years. You could use the rule of 70 to calculate the doubling time
for each investment under consideration to see if it could help you reach your savings goals by the time you retire.
The rule of 70 has other applications outside of the investment space. For example, the rule of 70 can be used to predict how long it would take for a country’s real GDP to double.
Alternatives to the Rule of 70
The rule of 69 and the rule of 72 are two alternatives to the rule of 70. They differ in their accuracy for investments with different compounding frequencies (which measure how often your interest
compounds). Both calculations function similarly to the rule of 70, except they divide the annual rate of return by 69 and 72, respectively, to derive the doubling time.
In general, the rule of 69 is considered to be more accurate for calculating doubling time for continuously compounding intervals, especially at lower interest rates. The rule of 70 is deemed more
accurate for semi-annual compounding, while the rule of 72 tends to be more accurate for annual compounding.
Pros and Cons of the Rule of 70
While the rule of 70 has some impressive benefits, it also has some downsides:
Pros Cons
Strong investment growth prediction model Only an estimate
Straightforward formula Relies on flawed assumptions
Pros Explained
• Strong investment growth prediction model. The rule of 70 makes it easy to estimate the number of years it may take for an investment to double in value.
• Straightforward formula. To use the rule of 70, all you have to do is divide 70 by the annual rate of return.
Cons Explained
• Only an estimate. While the rule of 70 can provide a well-informed projection of how long it may take an investment’s value to double, the calculation is only an estimate. In addition, that
estimate can be thrown off by fluctuating growth rates.
• Relies on flawed assumptions. Another reason the rule of 70 isn’t always accurate is because it assumes an investment compounds continuously. However, most financial institutions calculate
interest less frequently, so this assumption is inherently flawed when it comes to the rule of 70 and its ability to accurately predict growth. (The rule of 69 may be more accurate for
continuously compounding investments.)
Was this page helpful? Thanks for your feedback! Tell us why! Other Submit Sources The Balance uses only high-quality sources, including peer-reviewed studies, to support the facts within our
articles. Read our editorial process to learn more about how we fact-check and keep our content accurate, reliable, and trustworthy.
1. The Kelley Financial Group. “What Is the Rule of 70?” Accessed Aug. 19, 2021.
2. The Kelley Financial Group. “What Is the Rule of 70?” Accessed Aug. 19, 2021.
3. Corporate Finance Institute. “What Is the Rule of 72?” Accessed Aug. 19, 2021.
4. Robinhood. “What Is the Rule of 72?” Accessed Aug. 19, 2021. | {"url":"https://usagodly.com/what-is-the-rule-of-70.html","timestamp":"2024-11-13T04:30:00Z","content_type":"text/html","content_length":"140718","record_id":"<urn:uuid:b36437f2-b951-4a3f-b20b-cdd6d2a552b4>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00143.warc.gz"} |
Shortest Path Algorithms
Shortest Path Algorithms
For a given graph, the shortest path algorithms determine the minimum cost of the path from source vertex to every vertex in a graph.
Path is the movement traced across sequence of vertices V[1], V[2], V[3],....,V[N] in a graph. Cost of the path is the sum of the cost associated with all the edges in a graph. This is represented
• If the graph is weighted, then the cost of the path is the sum of weights on its edges.
• If the graph is not weighted, the cost of the path is the number of edges on the path.
Where does the shortest path algorithm find its applications?
• We can use the shortest path algorithm to find the cheapest way to send the information from one computer to another within a network.
• We can use the shortest path algorithm to find the best route between the two airports.
• We can use to find a road route which requires minimum distance from a starting source to an ending destination. We can determine the route which requires minimum time to an ending destination
Shortest path Problems
• There are different kinds of shortest-path problems.
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□ The all-pairs shortest path problemis used to find the shortest path between all pairs of vertices.
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Scope and Sequence
Unit 1
This unit will introduce the idea of “data,” fundamental to the rest of the course. While most people think of data simply as a spreadsheet or a table of numbers, almost anything can be considered
data, including images, text, GPS coordinates, and much more. Our world has become increasingly data-centric, and we are constantly generating data, whether we know it or not. From posts on Facebook,
to shopping records created when you swipe your credit card, to driving over sensors embedded in highway on-ramps, we leave behind a stream of data wherever we go. These data are used to generate
stories about our world, whether it is for political forecasting, marketing, scientific research, or even Netflix recommendations. Traditional statistics courses consist of understanding data from
only a small subset of data generation processes, namely those collected through random sampling or random assignment in scientific experiments. This unit exposes students to a wider world of data,
and will help students see how to make sense of these ubiquitous data types.
This unit will motivate the idea that data and data products (charts, graphs, statistics) can be analyzed and evaluated just like other arguments, such as those used by journalists. We want to know
how the evidence was collected, what the perspective or bias of the creator might be, and look behind the scenes to the process used to create the product. Even the way data are represented embeds
within it decisions on the part of the data creator.
Using the techniques of descriptive statistics, students will begin learning how to construct multiple views of data in an attempt to uncover new insights about the world. This will require the
introduction of the computational tool R through the interface of RStudio. Standard graphical displays like histograms and scatterplots will be introduced in RStudio, as well as measures of center
and spread.
Focus Statistics CCSS-M
S-ID 1. Represent data with plots on the real number line (dotplots, histograms, and boxplots).
S-ID 2: Use statistics appropriate to the shape of the data distribution to compare center (median, mean) of two or more different data sets (measures of spread will be studied in Unit 2).
S-ID 3: Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
S-ID 5. Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (joint, marginal, and conditional relative frequencies).
Recognize possible associations and trends in the data.
S-ID 6. Represent data on two quantitative variables on a scatterplot, and describe how the variables are related.
S-IC 6. Evaluate reports based on data.* *This standard is woven throughout the course. It is a recurring standard for every unit.
Focus Standards for Mathematical Practices
SMP-3. Construct viable arguments and critique the reasoning of others.
SMP-5. Use appropriate tools strategically.
Upon completion of Unit 1, students will be able to:
• Give examples of where they leave data traces.
• Understand that rows and columns are a form of data structure.
• Explain why the relationship between the variables might exist, or, if there is no relationship, why that might be so.
• Construct and interpret a frequency table.
• Critically read reports from media sources to evaluate their claims.
• Read plots (identify the name of the plot, interpret the axes, look for trends, identify confounding factors).
• Calculate conditional and marginal probabilities using frequency tables.
• Provide a real-world explanation for why the conditional or independent probabilities make sense, using critical thinking skills and background knowledge.
• Communicate their evaluations in written or verbal form using different types of media.
• Load data into RStudio.
• Create basic plots in RStudio.
• Create frequency tables in RStudio.
Unit 2
This unit deepens the informal reasoning skills developed in Unit 1 by enriching students' technical vocabulary and developing more precise analytical tools. Most importantly, this unit introduces
the formal concept of probability as a tool for understanding that sometimes patterns observed in data are not "real." Traditional courses attempt to develop this understanding through the
development of abstract mathematical probability concepts, but IDS creates enduring understanding by teaching students to design and implement simulations using pseudo-random number generators. This
activity also develops computational thinking by teaching students about some basic programming structures. Then, the use of models will come to the foreground. Students will be introduced to linear
models - the most common form of modeling in introductory statistics classes - which will serve as the foundation to learn more complex modeling techniques that use the computer technology available
to them later in the course, including smoothing techniques and tree-based models.
Focus Statistics CCSS-M
S-ID 2: Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.
S-ID 3: Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
S-ID 4. Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Understand that there are data sets for which such a procedure is
not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve.
S-IC 2. Decide if a specified model is consistent with results from a given data-generating process, e.g., using simulation.
S-IC 6. Evaluate reports based on data.* *This standard is woven throughout the course. It is a recurring standard for every unit.
S-CP 2. Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they
are independent.
S-CP 9. (+) Use permutations to perform [informal] inference. *This standard will be addressed in the context of data science.
Focus SMPs
SMP-4. Model with mathematics.
SMP-5. Use appropriate tools strategically.
Upon completion of Unit 2, students will be able to:
• Create a boxplot by calculating the five-number summary, upper and lower fences, and determining outliers.
• Explain what “standard deviation” means in context.
• Explain why the measures of central tendency and spread may or may not be accurate descriptions of the data from which they came.
• Use permutations of data to solve problems.
• Read/interpret a normal curve/distribution.
• Explain where the normal distribution came from.
• Describe situations where the normal distribution may model the phenomena, and others where it may not.
• Simulate normal distribution.
• Simulate from a model.
• Compare real data to simulation.
• Determine if model and data appear consistent.
• Merge data by columns/rows, and verify that merging is successful.
• Learn for() loops and apply() functions in RStudio.
• Create functions.
Unit 3
Unit 3 focuses on data collection methods, including traditional methods of designed experiments and observational studies and surveys. It introduces students to sampling error and bias, which cause
problems in analysis made from survey data. Participatory Sensing is presented as another method of data collection, and students learn to design Participatory Sensing campaigns that will allow them
to address particular statistical questions. Participatory Sensing is a unique data collection method because it uses sensors. Furthermore, this method emphasizes the involvement of citizens and
community groups in the process of sensing and documenting where they live, work, and play. Triggers play an important role in the Participatory Sensing data collection process. The response to the
triggers may or may not be the same each time. Data takes on a variety of forms online and requires a different style of representation. Students enhance computing skills by learning about modern
data structures, and by learning to "scrape" data stored in XML format.
Focus Statistics CCSS-M
S-IC 1. Understand statistics as a process for making inferences about population parameters based on a random sample from that population.
S-IC 3. Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each.
S-IC 6. Evaluate reports based on data.* *This standard is woven throughout the course. It is a recurring standard for every unit.
Focus SMPs
SMP-1. Make sense of problems and persevere in solving them.
SMP-4. Model with mathematics.
SMP-8. Look for and express regularity in repeated reasoning.
Upon completion of Unit 3, students will be able to:
• Provide a loose definition of “statistics” in their own words.
• Compare and contrast population vs. sample.
• Compare and contrast parameter vs. statistic.
• Explain the difference between special data structures, particularly as they relate to inference.
• Exploit special data structures for re-randomization analysis.
• Explain situations where one measure of central tendency or spread may be more appropriate than others.
• Read/interpret boxplots (In-depth look into samples size and their relationship to the population parameters).
• Identify reports that use special data structures (census, survey, observational study, and randomized experiment).
• Do data scraping.
• Use HTML and XML formats.
• Use RStudio to re-randomize data.
• Compute measures of central tendency and spread in RStudio.
Unit 4
This unit will develop modeling skills, beginning with learning to fit and interpret least squares regression lines and learning to use regression to make predictions. Students will learn to evaluate
the success of these predictions and so compare models for their predictive accuracy. Modern algorithmic approaches to regression are presented, and students will strengthen algorithmic thinking
skills by understanding how and why these algorithms help data scientists make accurate predictions from data. Students engage in a complete modeling experience in which they apply the skills and
concepts learned in the previous units. The modeling experience is designed to make students’ thinking visible and audible by encouraging them to be metacognitive about the process of inventing and
testing a model, ask questions as they go through the process, and recognize the iterative nature of modeling.
Focus Statistics Standards
S-IC 2. Decide if a specified model is consistent with results from a given data-generating process, e.g., using simulation.
S-ID 6. Represent data on two quantitative variables on a scatter plot, and describe how the variables are related.
• a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear
• b. Informally assess the fit of a function by plotting and analyzing residuals.
• c. Fit a linear function for a scatter plot that suggests a linear association.
S-ID 7. Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data.
S-ID 8. Compute (using technology) and interpret the correlation coefficient of a linear fit.
S-IC 6. Evaluate reports based on data.* *This standard is woven throughout the course. It is a recurring standard for every unit.
Focus SMPs
SMP-2. Reason abstractly and quantitatively.
SMP-4. Model with mathematics.
SMP-7. Look for and make use of structure.
Upon completion of Unit 4, students will be able to:
• Describe how well the linear model fits the data (or does not).
• Provide a real-world explanation of why the model may or may not fit, using critical thinking skills and background knowledge.
• Interpret the slope and intercept on a plot.
• Compute the correlation coefficient using RStudio.
• Interpret linear models in reports, including the correlation coefficient.
• Determine if a trend is “real” or if it could have arisen from randomness.
• Use critical thinking skills to explain why a trend may or may not make sense.
• Fit a regression line.
• Extract the slope, intercept, correlation coefficient, coefficient of determination, and residuals using RStudio.
• Use RStudio to predict y given an x value.
• Explore what happens to the line and the response variable if we multiply (divide) or add (subtract) a constant from the predictor.
• Design and execute their own Participatory Sensing Campaigns.
• Use RStudio to compute permutations and combinations.
• Create Classification and Regression Tree (CART) models.
• Understand non-linear models. | {"url":"https://curriculum.idsucla.org/scope/","timestamp":"2024-11-11T08:25:40Z","content_type":"text/html","content_length":"86712","record_id":"<urn:uuid:d7683ff6-5807-499d-af2b-51edf312008d>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00468.warc.gz"} |
Pound force per square inch to Inches of Mercury
Inches of Mercury to Pound force per square inch (Swap Units)
Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select 'decimal' from the options above the result.
Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result.
Note: For a pure decimal result please select 'decimal' from the options above the result.
1 psi is the pressure exterted by one pound-force of force being applied to an area of one square inch
Pound force per square inch to Inches of Mercury formula
1 inch of mercury is the pressure exerted by a 1 inch high column of mercury at 0 °C (32 °F )
Pound force per square inch to Inches of Mercury table
Pound force per square inch Inches of Mercury
0psi 0.00inHg
1psi 2.04inHg
2psi 4.07inHg
3psi 6.11inHg
4psi 8.14inHg
5psi 10.18inHg
6psi 12.22inHg
7psi 14.25inHg
8psi 16.29inHg
9psi 18.32inHg
10psi 20.36inHg
11psi 22.40inHg
12psi 24.43inHg
13psi 26.47inHg
14psi 28.50inHg
15psi 30.54inHg
16psi 32.58inHg
17psi 34.61inHg
18psi 36.65inHg
19psi 38.68inHg
Pound force per square inch Inches of Mercury
20psi 40.72inHg
21psi 42.76inHg
22psi 44.79inHg
23psi 46.83inHg
24psi 48.86inHg
25psi 50.90inHg
26psi 52.94inHg
27psi 54.97inHg
28psi 57.01inHg
29psi 59.04inHg
30psi 61.08inHg
31psi 63.12inHg
32psi 65.15inHg
33psi 67.19inHg
34psi 69.22inHg
35psi 71.26inHg
36psi 73.30inHg
37psi 75.33inHg
38psi 77.37inHg
39psi 79.40inHg
Pound force per square inch Inches of Mercury
40psi 81.44inHg
41psi 83.48inHg
42psi 85.51inHg
43psi 87.55inHg
44psi 89.58inHg
45psi 91.62inHg
46psi 93.66inHg
47psi 95.69inHg
48psi 97.73inHg
49psi 99.77inHg
50psi 101.80inHg
51psi 103.84inHg
52psi 105.87inHg
53psi 107.91inHg
54psi 109.95inHg
55psi 111.98inHg
56psi 114.02inHg
57psi 116.05inHg
58psi 118.09inHg
59psi 120.13inHg | {"url":"https://live.metric-conversions.org/pressure/pound-force-per-square-inch-to-inches-of-mercury.htm","timestamp":"2024-11-03T23:50:07Z","content_type":"text/html","content_length":"39773","record_id":"<urn:uuid:ab9ea34d-3c10-4310-b73a-d32ed0394c1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00704.warc.gz"} |
NIPS 2018
Sun Dec 2nd through Sat the 8th, 2018 at Palais des Congrès de Montréal
Reviewer 1
The paper studies the problem of overlapping clustering in relational data under a model that encompasses some other popular models for community detection and topic modeling. In this model, the
samples are represented in a latent space that represents the clustering memberships in which the coordinates are convex combinations of the corners of a cone. The authors propose a method to find
these corners using one-class SVM, which is shown to exactly recover the corners in the population matrix, and vectors that are close to the corners in an empirical matrix. The authors establish
consistency results for clustering recovery in network and topic models, and evaluate the performance of the method in synthetic and real data, comparing with other state-of-the-art methods. The
problem of overlapping clustering has received increasing attention and is of great interest in different communities. The authors presented a method that can be used in several contexts. The
geometric aspects of overlapping clustering models have been studied previously, but the approach followed by the authors is very original and theoretically justified. The method is simple and
efficient. I found the method especially relevant for overlapping clustering in network models, as the current approaches in the literature usually require stronger assumptions or make use of
combinatorial algorithms. I am not very familiar with the literature in topic modeling, but the method requires to have anchor words on each topic, which seems to be a strong assumption (see Huang et
al. "Anchor-Free Correlated Topic Modeling: Identifiability and Algorithm" NIPS 2016). Additionally, there seems to be other work that study the geometrical aspects of topic models, so it would be
worth if the authors make more comments in the literature review or numerical comparisons (see for example Yurochkin and Nguyen "Geometric Dirichlet Means algorithm for topic inference", NIPS 2016).
The paper is in general clearly written, and Section 2 is well explained and motivated. However, I found the conditions and results in Section 3 hard to interpret since the authors do not introduce
identifiability constraints on the parameters. Maybe this issue can be clarified if the authors are able to obtain closed-forms of the rates for special cases, and compare with other rates in the
literature. The authors make reference to other models which I think should be formally defined (at least in the appendix) for completeness. UPDATE: Thanks to the authors for clarifying my questions
in the rebuttal letter
Reviewer 2
This paper employs one-class SVM to generalize several overlapping clustering models. I like such unifying framework very much, which builds connection among the existing mature knowledge. Here are
my detailed comments. 1. The authors might want to change the title so that this paper can be easily searched. 2. The sizes of Figure 1 are not inconsistent. 3. The presentation should be polished.
(1) The second and third paragraphs in the introduction part are disconnected. (2) Some notations are not clearly illustrated, such as E in Lemma 2.1 (3) The authors should summarize several key
points for the generalization. And a table along these points to generalize network and topic model should be provided 4. Do you try fuzzy c-means? 5. The experimental part is relatively week. (1)
Why choose RC as the evaluation metric? Some widely used ones, such as NMI, Rn are encouraged. (2) The authors might want to demonstrate the performance difference of their proposed SVM-core and the
traditional model. (3) The time complexity and execution time of SVM-core should be reported.
Reviewer 3
Summary: The authors propose and analyze an algorithm based on 1-class svm for solving the mixed membership clustering problem. Both theoretical and empirical results are presented. 1. The paper is
easy to read and technically sound. The presentation of the main idea is clear and convincing. 2. It would be helpful if the authors could provide some remarks on how to best interpret the results
such as Theorem 3.1 and 3.2. For instance, which property dominates the error bound? 3. Since the recovery guarantee relies on row-wise error (in the non-ideal case), I wonder how outliers affect the
performance. 4. I wonder how the algorithm performs when most members are almost pure, as in the non-overlapping case. 5. Some of the terms are not defined when they first appear in the paper. For
example, kappa and lambda, also DCMMSB and OCCAM. Fixing these would improve the readability. | {"url":"https://papers.neurips.cc/paper_files/paper/2018/file/731c83db8d2ff01bdc000083fd3c3740-Reviews.html","timestamp":"2024-11-10T09:39:03Z","content_type":"text/html","content_length":"6124","record_id":"<urn:uuid:e4727f14-ead1-4690-be20-e19e72cc3d47>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00388.warc.gz"} |
Equations to Equations
Comer Duncan recently sent me an email asking how to translate models written for Berkeley Madonna into Modelica. He was specifically interested in some metabolism models developed by Kevin Hall at
Instead of simply answering this question via an email discussion, I thought it would be good to write about this issue. As is often the case, this kind of undertaking risks falling into the "Your
Baby is Ugly" trap, although I hope that won't be the case here.
For this article, I am going to focus exclusively on a "straight translation" approach. This is a naive but insightful approach. In part two, I'll examine a better, but more involved, approach the
brings to light many of the advantages that Modelica has to offer.
Straight Translation
These models contain several different types of statements. Let's take a look at some representative statements in Berkeley Madonna and then show what a literal translation to Modelica would look
System Parameters
The first type of statement are ones that set the value of so-called "system parameters" in Berkeley Madonna. Examples of system parameters include time step, tolerances, and start time. For example,
to set the start time to 0, a Berkeley Madonna model would include the following line:
Other system parameters include STOPTIME, DTMIN, DTMAX, TOLERANCE and DTOUT.
In Modelica, the experimental conditions are associated with the model via a standard annotation. This helps segregate meta-data about how the problem is to be solved from the actual mathematical
equations associated with the model. So in Modelica, we would indicate start time as follows:
Differential Equations
Berkeley Madonna allows differential equations to be defined using the following syntax:
d/dt(R) = ...
R' = ...
FLOW R = ...
The idea here is that for some reservoir, R, the right hand side of these equations would define the rate at which the reservoir value changes with respect to time. The fact that such variables are
called reservoirs shows the influence that Forrester's System Dynamics has on the approach in Berkeley Madonna. This isn't a criticism. Forrester's approach to system dynamics, with its concepts of
stocks and flows, is quite intuitive. As we will see later, the formalism used in Modelica is a superset of Forrester's system dynamics (and several other formalisms).
In Modelica, a differential equation has only one representation:
where der is a built-in operator. It is worth noting that Modelica is a declarative programming language, not an imperative programming language. As such, there is no "assignment" or directionality
in a Modelica equation, only a relationship between two quantities. So the same equation could be represented in the following, completely equivalent, form:
It is worth noting that Berkeley Madonna also allows the following forms, in order to be compatible with the STELLA modeling language:
R(t) = R(t-dt) + (...)*dt
R = R + dt*(...)
In the Berkeley Madonna User's Guide, it says these forms are not recommended "since the notation is more error-prone". At the risk of offending STELLA advocates, I'm afraid I have to agree. It isn't
just the potential complexity of such expressions, it is the fact that it implicitly imposes a (forward Euler) solution method on the model which is completely unnecessary (and not a particularly
good choice at that). Mixing the "problem statement" and the "solution method" is avoided as much as possible in the Modelica approach so there are no equivalent forms in Modelica. That is not to say
that discrete equations (e.g., z-transforms) are not allowed in Modelica. But such equations must be represented in terms of some underlying "clock" and not in terms of solver time steps.
Berkeley Madonna allows for higher-derivatives to be expressed in the language through the use of multiple "primes", e.g., u', u''. Modelica does not allows this. Differential equations are
restricted to first-order and intermediate variables must be introduced to represent higher-order derivatives.
For each reservoir, it is necessary to specify an initial condition. In Berkeley Madonna, this can be done in the following ways:
INIT R = ...
INIT (R) = ...
R(STARTIME) = ...
where the right hand side represents the value the reservoir should have at the start of the simulation.
Initialization in Modelica is actually quite a rich topic. For since we are only concerned with straight translation at the moment, the equivalent in Modelica would be:
initial equation
R = ...;
The comment syntax in Berkeley Madonna has two forms. The first form is a "curly-bracket comment" which can span multiple lines, e.g.,
{This is a comment}
{This is
also a comment}
The second form is a single-line comment which starts with a ; and is terminated by the end of the line, e.g.,
Modelica has both of these types of comments with slightly different syntax:
/* This is
a multi-line
comment */
// This is a single-line comment
But in addition, Modelica has "descriptive strings" that can be associated with entities in the language. These are not comments because they are semantically associated with different elements. Such
descriptive strings can then be used in parameter dialogs and other user interface elements. For example, if I declare a variable as follows:
parameter Real Na_b=4000 "Baseline sodium intake in mg/d";
the string "baseline sodium intake in mg/d" is not a comment but rather a description of the variable Na_b and this description can be used when interacting or documenting the model.
Furthermore, Modelica allows you to formally associated physical units with a quantity. As such, an even better representation in Modelica would be:
type DailyRate = Real(final unit="mg/d");
parameter DailyRate Na_b=4000 "Baseline sodium intake";
In this way, we can leave the units out of the description because the variable is declared to be of type DailyRate, which already indicates that the units are in milligrams per day. Even better, the
Modelica specification defines a grammar for these unit definitions so that the consistency of units in equations can be checked. This goes well beyond just commenting or documenting the model
because it allows unit errors to be automatically detected by tools which can catch a lot potential errors!
Variables and Equations
In Berkeley Madonna, models consist of a list of equations. As we have seen already, some of these equations are for "system parameters". Others are for variables in our models. Examples of equations
ECP = 0.732*BM + 0.01087*ECW_b
d/dt (Lipol_diet) = (Lipol_diet_target - Lipol_diet)/tau_lip
init Lipol_diet = 1
Kurine = IF Ketogen < Kspill THEN 0 ELSE Ketogen*KUmax/(KGmax-Kspill)-KUmax/(KGmax/Kspill-1)
In these cases, you have either a variable name or the derivative of a variable on the left hand side and some kind of expression on the right hand side. With only minor syntactic differences, we can
represent these same equations in Modelica:
ECP = 0.732*BM + 0.01087*ECW_b;
der(Lipol_diet) = (Lipol_diet_target - Lipol_diet)/tau_lip;
Kurine = if Ketogen < Kspill then 0 else Ketogen*KUmax/(KGmax-Kspill)-KUmax/(KGmax/Kspill-1);
initial equation
Lipol_diet = 1;
The main differences are the termination of equations with semicolons, the use of the der operator instead of d/dt and the fact that Modelica is case sensitive (if vs IF).
However, there is another important difference with regard to variables that we mentioned already in the section on commenting, which is that variables in Modelica must be declared. So a more
complete fragment of Modelica code for the previous three equations would be:
model MetabolismModel
Real ECP "Extracellular protein";
Real Lipol_diet;
Real Kurine;
ECP = 0.732*BM + 0.01087*ECW_b "Wang AJCN 2003";
der(Lipol_diet) = (Lipol_diet_target - Lipol_diet)/tau_lip;
if Ketogen < Kspill then
Kurine = 0;
Kurine = Ketogen*KUmax/(KGmax-Kspill)-KUmax/(KGmax/Kspill-1);
end if;
initial equation
Lipol_diet = 1;
end MetabolismModel;
Here we see the mostly complete text of a model. We see both the declarations of the variables (indicating the type of the variable along with an optional description) as well as the equations
associated with the variable.
Isn't the Berkeley Madonna syntax simpler? Perhaps. For simple problems it might seem like an advantage to have such a simple syntax. But for complex problems, using some explicit syntax to help
convey your overall intent can go a long way toward providing better diagnostic error messages and catching errors.
At the risk of jumping ahead a little bit, it is worth pointing out that building complex models in Modelica would not be done this way (i.e. declaring lots of variables and lots of equations all in
a "flat" file like this). But for now, we will remain focused on a straight translation.
One interesting thing that Berkeley Madonna has is the notion of datasets. These datasets are really a combination of two things. The first is the underlying data (presumably represented on on some
multi-dimensional regular grid, although I didn't confirm that). The other is a bunch of implicitly defined functions for interpolating over the data in the dataset. These implicit functions are
identified by the name of the dataset preceded by a # character, e.g., #temperature(...), #R20BW(...).
Modelica (or more specifically, the Modelica Standard Library) includes a collection of table models that are similar, but not exactly equivalent.
An important caveat here is that in practice it is often the case that you might wish to use a table in some cases, a set of mathematical expressions in another and perhaps even a nested sub-model
with its own states in another. We'll talk later about how Modelica can accommodate all of these uses in a framework that is "type safe".
Automatic Translation
It turns out, it is pretty straight forward to translate Kevin Hall's models (which have only appeared in fragments in this article) into Modelica code. To do this, I wrote a relatively simply Python
script although I should point out that if I needed something that was "production quality", I would use a real lexer and parser. It would take me perhaps twice as long to write, but it would be
infinitely more reliable and robust.
The following Python code is almost sufficient to translate the model by Kevin Hall that I mentioned at the beginning into Modelica^:
import re
def _process_eq(groups, eqs, line):
lhs = groups[0]
rhs = groups[1].split(";")
if len(rhs)==1:
eqs.append({"var": lhs, "expr": rhs[0].strip(),
"desc": None})
elif len(rhs)>1:
eqs.append({"var": lhs, "expr": rhs[0].strip(),
"desc": ";".join(rhs[1:])})
print "Unable to parse equation: ", line
class Translator:
EMPTY = re.compile("^\s*(;.*)?$")
ASSIGNMENT = re.compile("\s*(\w+)\s*=\s*(.*)")
INIT = re.compile("\s*[Ii][Nn][Ii][Tt]\s*(\w+)\s*=(.*)")
DIFFEQ = re.compile("\s*d\/dt\s*\(?(\w+)\)?\s*=\s*(.*)")
CCOMMENT = re.compile("\s*{([^}]*)}\s*")
def __init__(self, name, f):
self.name = name
self.equations = []
self.diffequations = []
self.initequations = []
self.file = f
def _parse(self):
lines = self.file.readlines()
for line in lines:
def _parseline(self, line):
if self.EMPTY.match(line):
if self.CCOMMENT.match(line):
if self.DIFFEQ.match(line):
match = self.DIFFEQ.match(line)
_process_eq(match.groups(), self.diffequations, line)
elif self.INIT.match(line):
match = self.INIT.match(line)
_process_eq(match.groups(), self.initequations, line)
elif self.ASSIGNMENT.match(line):
match = self.ASSIGNMENT.match(line)
_process_eq(match.groups(), self.equations, line)
print "?", line
def render(self):
experiment = {}
# Extract experimental settings
equations = list(self.equations)
for eq in equations:
var = eq["var"]
exp = eq["expr"]
if var=="STARTTIME":
experiment["StartTime"] = exp;
if var=="STOPTIME":
experiment["StopTime"] = exp;
if var=="TOLERANCE":
experiment["Tolerance"] = exp;
if var=="DTMIN" or var=="DTMAX" or var=="DTOUT":
print "model %s" % (self.name)
for eq in self.equations:
if hasattr(eq,"desc"):
print """ Real %s "%s";""" % (eq["var"], eq["desc"])
print """ Real %s;""" % (eq["var"])
print "initial equation"
for ieq in self.initequations:
print """ %s=%s;""" % (ieq["var"], ieq["expr"])
print "equation"
for eq in self.equations:
print """ %s=%s;""" % (eq["var"], eq["expr"])
for deq in self.diffequations:
print """ der(%s)=%s;""" % (deq["var"], deq["expr"])
print "end %s;" % (self.name)
fp = open("hallcode.m", "r")
t = Translator("HallModel", fp)
For the curious, I have included the resulting Modelica model (which required just a handful of manual tweaks).
Now What?
I've mentioned a few times that this kind of straight translation is really not the way to go about this. So far, what I've shown is a bit naive, but it does work. In part two of this article, we'll
look at how you could improve on this approach.
Share your thoughts
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Dynamic response and breakage of trees subject to a landslide-induced air blast
Articles | Volume 23, issue 4
© Author(s) 2023. This work is distributed under the Creative Commons Attribution 4.0 License.
Dynamic response and breakage of trees subject to a landslide-induced air blast
Landslides have been known to generate powerful air blasts capable of causing destruction and casualties far beyond the runout of sliding mass. The extent of tree damage provides valuable information
on air blast intensity and impact region. However, little attention has been paid to the air blast–tree interaction. In this study, we proposed a framework to assess the tree destruction caused by
powerful air blasts, including the eigenfrequency prediction method, tree motion equations and the breakage conditions. The tree is modeled as a flexible beam with variable cross-sections, and the
anchorage stiffness is introduced to describe the tilt of the tree base. Large tree deflection is regarded when calculating the air blast loading, and two failure modes (bending and overturning) and
the associated failure criteria are defined. Modeling results indicate that although the anchorage properties are of importance to the tree eigenfrequency, tree eigenfrequency is always close to the
air blast frequency, causing a dynamic magnification effect for the tree deformation. This magnification effect is significant in cases with a low air blast velocity, while the large tree deflection
caused by strong air blast loading would weaken this effect. Furthermore, failure modes of a specific forest subject to a powerful air blast depend heavily on the trunk bending strength and anchorage
characteristics. The large variation in biometric and mechanical properties of trees necessitates the establishment of a regional database of tree parameters. Our work and the proposed method are
expected to provide a better understanding of air blast power and to be of great use for air blast risk assessment in mountainous regions worldwide.
Received: 09 Jun 2022 – Discussion started: 29 Jun 2022 – Revised: 09 Mar 2023 – Accepted: 16 Mar 2023 – Published: 04 Apr 2023
Long runout landslides involve massive amounts of energy and can be extremely hazardous owing to their long movement distance, high mobility and potential chain disasters (Nicoletti and
Sorriso-Valvo, 1991; Nicoletti et al., 1993; Johnson and Campbell, 2017; Shugar et al., 2021; Zhang et al., 2022). A moving landslide with high velocity can generate a powerful air blast capable of
uprooting trees, lifting people into the air and even flattening buildings (Adams, 1881; Penna et al., 2021). In recent decades, destructive air blasts frequently occurred in mountainous regions
worldwide and caused casualties and economic loss far beyond the landslide runout (e.g., Yin, 2014; Bartelt et al., 2016; Kargel et al., 2016). Understanding their force of destruction is of great
use for landslide risk assessment and disaster mitigation, especially in high-altitude regions.
Monitoring equipment has been confirmed to provide great performance in determining the dynamic characteristics of landslide-induced air blasts (Grigoryan et al., 1982; Sukhanov, 1982; Caviezel et
al., 2021). However, most case histories occurred in high-altitude mountainous regions without witnesses (Yin and Xing, 2012), and the in situ equipment also got damaged because of the near-field
destruction of landslides and associated air blasts. Therefore, very few air blast cases have been measured in history. Geologists can only evaluate the air blast hazard for most recorded events
using historical evidence after the landslide occurred. In situ information about forest destruction and tree breakage is often used for the air blast risk assessment (Feistl et al., 2015; Fujita et
al., 2017; Zhuang et al., 2019, 2022b) (Fig. 1). Uprooted trees and snapped stems delineate the impact region of air blasts and create a natural vector field, indicating the primary movement
direction of the landslide, greatly helping to analyze the disaster-causing process of the event. In many cases, observations of forest destruction are the only data to quantify air blast danger.
The question that remains for air blast mitigation planning using the information of tree damage is how to establish a simple relationship between air blast impact pressure and tree failure. Bending
and overturning are two common tree failure modes caused by strong winds. Trees snap when the bending stress exerted by the air blast exceeds the wood strength (Peltola et al., 1999; Gardiner et
al., 2000), while overturning will occur if the applied moment overcomes the anchorage resistance of root systems (Jonsson et al., 2006; Nicoll et al., 2006). The occurrence of these two failure
modes depends heavily on both the air blast loading and the tree properties. Considering the minor destruction of air blasts relative to the landslide, although it has long been recognized that
sliding mass can easily break or uproot trees (Bartelt and Stöckli, 2001; Šilhán, 2020), little attention has been paid to the tree destruction resulting from air blasts. Furthermore, existing models
describing the tree–air blast interaction are mostly static (Feistl et al., 2015) or established based on the small-deflection theory (Bartelt et al., 2018a). These methods could aid in a rapid
assessment of air blast power, but further research is needed to establish a dynamic model to represent the dynamic response of trees in strong wind. A mechanical understanding of how trees are
damaged by air blasts is therefore essential for quantifying the air blast power and for providing valuable data to verify the possible numerical results.
In this study, we established a simple dynamic model capable of calculating the natural frequency of trees and simulating their dynamic response subject to a powerful air blast. The proposed model
regards the tree as a multi-degree-of-freedom beam with variable diameters and accounts for large tree deflections and impacts of root anchorage. Both bending and overturning failure modes are
involved in the model. The work conducted in this study is expected to make people better understand the power of landslide-induced air blasts and to provide an applicable method to assess the air
blast hazard.
Measurements of historical events indicated that the landslide-induced air blast is intermittent and of short duration, lasting only a few seconds, and could reach a high velocity (Grigoryan et
al., 1982; Sukhanov, 1982; Caviezel et al., 2021). This impulse wave has a propagation distance of hundreds of meters in both horizontal and vertical directions and acts over the entire tree. Thus,
the impact of air blasts on trees is similar to extreme wind gusts, producing large bending moments in the stem- and root-based system, forcing trees to deform or get damaged. Furthermore, fallen
trees often point to the movement direction of the landslide, illustrating that there is little time for trees to sway and react to air blasts and the inertial effects are greatly important.
To characterize the dynamic response of trees under the impact load of air blasts, we established a mechanical model to predict the eigenfrequency of trees subject to air blasts and developed a
dynamic tree-swaying model that accounts for the large tree deflection. In what follows, we present the eigenfrequency prediction method, tree motion equations and the breakage conditions.
2.1Eigenfrequency prediction
The tree is modeled as a flexible cantilever beam with variable diameters that are hinged at ground level using elastic support. The beam diameter is assumed to continuously linearly decrease with
height regarding the decreasing diameters of the trunk and crown from the bottom to the top, while the anchorage stiffness of the root system (K) helps to describe the tilt of the tree base in
response to the moment (Neild and Wood, 1999). In the eigenfrequency prediction mode, the tree beam is divided into two segments with a splitting point located at the starting point of the tree crown
(Fig. 2). We assume that the tree crown shows minor impacts of elastic modulus. The tree crown is accounted for through the crown mass, and thus the natural difference between the two segments is the
material density.
The governing differential equation for the dynamic bending of a nonuniform Euler–Bernoulli beam is (Keshmiri et al., 2018)
$\begin{array}{}\text{(1)}& \mathit{\rho }A\left(z\right)\frac{{\partial }^{\mathrm{2}}u}{\partial {t}^{\mathrm{2}}}+\frac{{\partial }^{\mathrm{2}}u}{\partial {z}^{\mathrm{2}}}\left[EI\left(z\right)\
frac{{\partial }^{\mathrm{2}}u}{\partial {z}^{\mathrm{2}}}\right]=\mathrm{0},\end{array}$
where z is the position variable along the beam length. For ease of calculation, the original point (z=0) is set at the treetop, and the maximum value of z is at the tree base so that the beam
diameter d(z) corresponding to the position z can be described using a gradient coefficient (μ): d(z)=μz. u is the beam displacement; E is the elastic modulus; and $A\left(z\right)=\frac{\mathit{\pi
}}{\mathrm{4}}{\left(dz\right)}^{\mathrm{2}}$ and $I\left(z\right)=\frac{\mathit{\pi }}{\mathrm{64}}{\left(dz\right)}^{\mathrm{4}}$ are the cross-sectional area and moment of inertia, respectively.
Plugging the expression of A(z) and I(z) into Eq. (1) gives
$\begin{array}{}\text{(2)}& {z}^{\mathrm{2}}\frac{{\partial }^{\mathrm{4}}u}{\partial {z}^{\mathrm{4}}}+\mathrm{8}z\frac{{\partial }^{\mathrm{3}}u}{\partial {z}^{\mathrm{3}}}+\mathrm{12}\frac{{\
partial }^{\mathrm{2}}u}{\partial {z}^{\mathrm{2}}}-\frac{\mathrm{16}\mathit{\rho }{\mathit{\omega }}^{\mathrm{2}}u}{E{\mathit{\mu }}^{\mathrm{2}}}=\mathrm{0},\end{array}$
where ω is known as the eigenfrequency of the beam. The general solution of Eq. (2) can be expressed as
$\begin{array}{}\text{(3)}& \begin{array}{rl}u\left(z\right)& =\frac{\mathrm{1}}{z}\left[{A}_{\mathrm{1}}{J}_{\mathrm{2}}\left(\mathrm{2}\sqrt{\mathit{\lambda }z}\right)+{A}_{\mathrm{2}}{Y}_{\mathrm
{2}}\left(\mathrm{2}\sqrt{\mathit{\lambda }z}\right)\\ & +{A}_{\mathrm{3}}{J}_{\mathrm{2}}\left(\mathrm{2}i\sqrt{\mathit{\lambda }z}\right)+{A}_{\mathrm{4}}{Y}_{\mathrm{2}}\left(\mathrm{2}i\sqrt{\
mathit{\lambda }z}\right)\right],\end{array}\end{array}$
where $\mathit{\lambda }=\sqrt{\frac{\mathrm{16}\mathit{\rho }{\mathit{\omega }}^{\mathrm{2}}}{E{\mathit{\mu }}^{\mathrm{2}}}}$; J[2] and Y[2] are the Bessel functions of the first and second kind
(Mocica, 1988), respectively; and A[1]–A[4] are coefficients that need to be determined based on the boundary conditions.
The deflection of the upper segment (crown) and the lower segment (trunk) can be generated in a similar manner:
$\begin{array}{}\text{(4)}& \begin{array}{rl}{u}_{\mathrm{1}}\left(z\right)& =\frac{\mathrm{1}}{z}\left[{A}_{\mathrm{1}}{J}_{\mathrm{2}}\left(\mathrm{2}\sqrt{{\mathit{\lambda }}_{\mathrm{1}}z}\right)
+{A}_{\mathrm{2}}{Y}_{\mathrm{2}}\left(\mathrm{2}\sqrt{{\mathit{\lambda }}_{\mathrm{1}}z}\right)\\ & +{A}_{\mathrm{3}}{J}_{\mathrm{2}}\left(\mathrm{2}i\sqrt{{\mathit{\lambda }}_{\mathrm{1}}z}\right)+
{A}_{\mathrm{4}}{Y}_{\mathrm{2}}\left(\mathrm{2}i\sqrt{{\mathit{\lambda }}_{\mathrm{1}}z}\right)\right]\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\mathrm{0}⩽z<l,\end{array}\end{array}$
$\begin{array}{}\text{(5)}& \begin{array}{rl}{u}_{\mathrm{2}}\left(z\right)& =\frac{\mathrm{1}}{z}\left[{B}_{\mathrm{1}}{J}_{\mathrm{2}}\left(\mathrm{2}\sqrt{{\mathit{\lambda }}_{\mathrm{2}}z}\right)
+{B}_{\mathrm{2}}{Y}_{\mathrm{2}}\left(\mathrm{2}\sqrt{{\mathit{\lambda }}_{\mathrm{2}}z}\right)\\ & +{B}_{\mathrm{3}}{J}_{\mathrm{2}}\left(\mathrm{2}i\sqrt{{\mathit{\lambda }}_{\mathrm{2}}z}\right)+
{B}_{\mathrm{4}}{Y}_{\mathrm{2}}\left(\mathrm{2}i\sqrt{{\mathit{\lambda }}_{\mathrm{2}}z}\right)\right]\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}l⩽z⩽h,\end{array}\end{array}$
where l is the length of the crown, h is the tree height, and ${\mathit{\lambda }}_{\mathrm{1}}=\sqrt{\frac{\mathrm{16}{\mathit{\rho }}_{\mathrm{1}}{\mathit{\omega }}^{\mathrm{2}}}{E{\mathit{\mu }}^
{\mathrm{2}}}}$ and ${\mathit{\lambda }}_{\mathrm{2}}=\sqrt{\frac{\mathrm{16}{\mathit{\rho }}_{\mathrm{2}}{\mathit{\omega }}^{\mathrm{2}}}{E{\mathit{\mu }}^{\mathrm{2}}}}$ are the single-valued
functions of eigenfrequency. ρ[2] is the wood density, and ρ[1] is the equivalent density regarding the contribution of both the tree trunk and the crown. The same as A[1]–A[4], B[1]–B[4] are also
coefficients of the tree deflection equation that need to be determined based on the boundary and continuity conditions.
The boundary condition at the origin (z=0) is the free end, and thus Eq. (4) can be simplified as
$\begin{array}{}\text{(6)}& {u}_{\mathrm{1}}\left(z\right)=\frac{\mathrm{1}}{z}\left[{A}_{\mathrm{1}}{J}_{\mathrm{2}}\left(\mathrm{2}\sqrt{{\mathit{\lambda }}_{\mathrm{1}}z}\right)+{A}_{\mathrm{3}}
{J}_{\mathrm{2}}\left(\mathrm{2}i\sqrt{{\mathit{\lambda }}_{\mathrm{1}}z}\right)\right]\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\mathrm{0}\le z<l.\end{array}$
According to continuity conditions of two segments at the splitting point and the boundary condition at the tree base, the following constraints are determined: u[1](l)=u[2](l), ${u}_{\mathrm{1}}^{\
prime }\left(l\right)={u}_{\mathrm{2}}^{\prime }\left(l\right)$, ${u}_{\mathrm{1}}^{\prime \prime }\left(l\right)={u}_{\mathrm{2}}^{\prime \prime }\left(l\right)$, ${u}_{\mathrm{1}}^{\prime \prime \
prime }\left(l\right)={u}_{\mathrm{2}}^{\prime \prime \prime }\left(l\right)$, u[2](h)=0 and $K{u}_{\mathrm{2}}^{\prime }\left(h\right)+EI\left(h\right){u}_{\mathrm{2}}^{\prime \prime }\left(h\right)
=\mathrm{0}$. Introducing the constraints into Eqs. (5)–(6), a total of six equations are determined here. These six equations can be written in a matrix format:
$\begin{array}{}\text{(7)}& {\left[F\left({\mathit{\lambda }}_{\mathrm{1}},{\mathit{\lambda }}_{\mathrm{2}}\right)\right]}_{\mathrm{6}×\mathrm{6}}\cdot {\left[\begin{array}{cccccc}{A}_{\mathrm{1}}&
{A}_{\mathrm{3}}& {B}_{\mathrm{1}}& {B}_{\mathrm{2}}& {B}_{\mathrm{3}}& {B}_{\mathrm{4}}\end{array}\right]}^{T}=\mathrm{0},\end{array}$
where ${\left[F\left({\mathit{\lambda }}_{\mathrm{1}},{\mathit{\lambda }}_{\mathrm{2}}\right)\right]}_{\mathrm{6}×\mathrm{6}}$ is a matrix that is composed of λ[1] and λ[2]. The orders of
eigenfrequency and the corresponding vibration mode can be obtained by solving the equation: the determinant of matric $\left|F\left({\mathit{\lambda }}_{\mathrm{1}},{\mathit{\lambda }}_{\mathrm{2}}\
right)\right|=\mathrm{0}$. Notably, the derivatives of u[1](z) and u[2](z) have very complicated expressions but could easily be calculated using MATLAB. Therefore, we did not provide the complete
expression here.
2.2Tree motion
The mechanical response of trees subject to an air blast is modeled using a modified multi-degree-of-freedom tree-swaying model with variable cross-sections (Zhuang et al., 2022a). Different from the
simplification in the eigenfrequency prediction method, the size of the tree crown here is determined based on real tree data, corresponding to the frontal-area distribution of the tree crown
(Fig. 3a). The model divides the tree beam into a set of segments and calculates the tree motion using linear modal analysis. Specifically, the tree deformation is decomposed into a set of vibration
modes so that the total displacement is the combined contribution of each mode. According to preliminary research performed by Sellier et al. (2008) and Pivato et al. (2014), the contribution of the
first vibration model is far ahead of the other modes for trees with a slender shape. Thus, only the first vibration mode and the corresponding eigenfrequency are utilized in this study. The modeling
of air blast pressure accounts for the wind–tree relative motion and large tree deflection by regarding the beam velocity and geometric nonlinearities resulting from the inclination of beam segments
relative to the wind direction (θ[i]) (Fig. 3b). With respect to the large tree deflection, we also introduce the impact of eccentric gravity into the model, which significantly contributes during
the interaction with a powerful air blast. The gravity and wind load acting on each segment can be easily calculated based on the predetermined diameter and frontal-area distribution (Fig. 3a).
Considering that trees often fall in the direction of landslide motion and have little time to sway, the maximum response of the tree is assumed to be reached before the damping forces act (Bartelt
et al., 2018a). Only the undamped response to a short-duration blast is considered. The tree motion equations and the expression of air blast force are as follows:
$\begin{array}{}\text{(8)}& m\frac{{\partial }^{\mathrm{2}}y}{\partial {t}^{\mathrm{2}}}+ky={\int }_{\mathrm{0}}^{h}{F}_{i}\mathit{\varphi }\mathrm{d}s+{\int }_{\mathrm{0}}^{h}{G}_{i}\mathit{\varphi
$\begin{array}{}\text{(9)}& \begin{array}{rl}{F}_{i}& =\mathrm{0.5}\mathit{\rho }{C}_{\mathrm{d}}{A}_{\mathrm{f}}\left|v\mathrm{cos}{\mathit{\theta }}_{i}-\frac{\partial y}{\partial t}\mathrm{cos}{\
mathit{\theta }}_{i}\right|\\ & \cdot \left(v\mathrm{cos}{\mathit{\theta }}_{i}-\frac{\partial y}{\partial t}\mathrm{cos}{\mathit{\theta }}_{i}\right)\mathrm{cos}{\mathit{\theta }}_{i},\end{array}\
$\begin{array}{}\text{(10)}& {G}_{i}={m}_{i}g\cdot \mathrm{sin}{\mathit{\theta }}_{i}\cdot \mathrm{cos}{\mathit{\theta }}_{i},\end{array}$
where ϕ, w, $m={\int }_{\mathrm{0}}^{h}\stackrel{\mathrm{‾}}{m}{\mathit{\varphi }}^{\mathrm{2}}\mathrm{d}s$ and k=4π^2mω^2 are the first mode shape, eigenfrequency, modal mass and stiffness,
respectively; $\stackrel{\mathrm{‾}}{m}$ is the mass distribution; y is the associated generalized displacement; F[i] and G[i] are the air blast loading and eccentric beam gravity act on the ith
segment; h is the tree height; C[d] is the drag efficient; A[f] is the frontal area; and ρ and v are the density and velocity of the air blast, respectively. Our model is able to calculate the
scenarios for both full-height and part-height air blasts.
In this study, the air blast velocity is expressed as a sine wave impulse with a short duration time t[0]:
$\begin{array}{}\text{(11)}& v={v}_{max}\sqrt{\mathrm{sin}\mathit{\varpi }t},\end{array}$
where v[max] is the maximum velocity of the landslide-induced air blast, and ϖ can be regarded as the circular frequency of the wind force $\mathit{\varpi }=\mathit{\pi }/{t}_{\mathrm{0}}$ (wind
force is related to the square of its velocity).
The mechanical response of trees subject to an air blast is deduced by introducing the calculated wind velocity from Eq. (11) into the tree-motion model (Eqs. 8–9) and subsequently solving the
equations using the central finite-difference scheme. The validity of this tree-motion model has been checked by Pivato et al. (2014) and Zhuang et al. (2022a), and thus the validation process is not
involved here.
2.3Tree breakage
Two failure modes commonly caused by air blasts are involved in this work: bending and overturning (Gardiner et al., 2000).
In the case of tree bending, trees are expected to break when the maximum bending stress σ[max] exceeds a critical value σ[crit]:
$\begin{array}{}\text{(12)}& {\mathit{\sigma }}_{max}={\left[\frac{M\left(t,z\right)\cdot d\left(z\right)/\mathrm{2}}{I\left(z\right)}\right]}_{max}\ge {\mathit{\sigma }}_{\mathrm{crit}},\end{array}$
where σ[crit] is the bending strength of the tree, which depends highly on the material property. M(t,z) is the bending moment, and its value is calculated at each time step all along the beam:
$\begin{array}{}\text{(13)}& M\left(t,z\right)=EI\left(z\right)\frac{\mathrm{d}\mathit{\theta }}{\mathrm{d}s},\end{array}$
where $\frac{\mathrm{d}\mathit{\theta }}{\mathrm{d}s}$ represents the local beam curvature, and θ is the angle between the beam segment with the vertical direction.
For the tree-overturning case, trees are expected to break at the base when the air-blast-induced moment reaches the anchorage resistance (M[crit]):
$\begin{array}{}\text{(14)}& {M}_{\mathrm{base}}\left(t\right)\ge {M}_{\mathrm{crit}},\end{array}$
where M[base](t) is the moment at the tree base calculated at each time step, and the anchorage resistance M[crit] is often determined based on in situ tests (e.g., tree-pulling tests).
To demonstrate the power of air blasts and how they damage trees, we consider the problem proposed by Bartelt et al. (2018a): a landslide-induced air blast enters a spruce forest at high speed
(maximum velocity of 20ms^−1). The short-duration air blast lasts a few seconds with a frequency ϖ. Trees in the forest have a height between 25 and 30m, which is also the height of the air blast.
The sliding mass has stopped before reaching the forest, and only the air blast loads on the trees.
Using the measured biomass parameters presented in Table 1, we set the total crown mass of a single tree to be 540kg. The tree crown is assumed to be a cone with a length of 18m $\left(\frac{\
mathrm{2}}{\mathrm{3}}h\right)$ and a width of 5m. The wood density is 480kgm^−3, and the elastic modulus is 10GPa. Measurements of root-anchorage stiffness (K) are very rare, and in situ tests
on spruce performed by Neild and Wood (1999) show a value variation of 80–1200kNm. This value range indicates a large variation in K depending on the growth conditions, and the values of
100–1200kNm are applied in the prediction of eigenfrequency and vibration mode in this study.
The eigenfrequency ranging from 0.13Hz (K=100kNm) to 0.32Hz (K=1200kNm) is calculated based on the above parameters (Fig. 4). The modeled results are in a good agreement with measurements
performed by Jonsson et al. (2006) (0.16–0.30Hz), indicating the validity of our proposed eigenfrequency-prediction method. Although the tree eigenfrequency varies significantly with the anchorage
stiffness, all the calculated values are less than 0.5Hz. The same order of magnitude between tree eigenfrequency and air blast frequency necessitates a further investigation into the possible
impact of resonance. The dynamic magnification effect caused by impulse loading can greatly amplify the static stress state, making the trees easier to damage.
To investigate the impact of dynamic magnification, we performed simulations for all the scenarios using the tree eigenfrequency of 0.26Hz (K=600kNm) and the associated vibration mode. A
magnification factor D is defined to describe this effect:
$\begin{array}{}\text{(15)}& \begin{array}{rl}D& =\frac{{u}_{\mathrm{d},max}\left(\mathit{\beta }\right)}{{u}_{\mathrm{sta}}}=\frac{{u}_{\mathrm{d},\mathrm{max}}\left(\mathit{\beta }\right)}{{\int }_
{\mathrm{0}}^{h}{F}_{\mathrm{s},max}\mathit{\varphi }\mathrm{d}s/k}\\ & =\frac{{u}_{\mathrm{d},max}\left(\mathit{\beta }\right)}{{\int }_{\mathrm{0}}^{h}\mathit{\rho }{C}_{\mathrm{d}}{A}_{\mathrm{f}}
{v}_{max}^{\mathrm{2}}\mathit{\varphi }\mathrm{d}s/k},\end{array}\end{array}$
where ${u}_{\mathrm{d},max}$ and u[sta] are the maximum displacements subject to dynamic load and static load, respectively; ${F}_{\mathrm{s},max}$ is the static wind force corresponding to the
maximum air blast velocity; and $\mathit{\beta }=\frac{\mathit{\varpi }}{\mathit{\omega }}$ is the ratio between the air blast frequency (ϖ) and the eigenfrequency of the tree (ω). Notably, the air
blast is a multi-medium fluid that contains numerous dusts, showing a higher density than air. Measurements and numerical modeling performed by Swiss researchers (Feistl et al., 2015) suggest ρ=5
kgm^−3. In this scenario u[sta] is calculated to be 9.8m.
Figure 5 shows the impact of air blast frequency on the dynamic magnification effect. A parabola relationship is identified between the magnification factor and the frequency ratio. Consider first an
impulse air blast lasting 1.6s (β=1.2). The air blast frequency is higher than that of the tree, implying the maximum displacement reaches after the loading time. The modeled maximum dynamic
deformation ${u}_{\mathrm{d},max}$ reaches 10.7m, and the magnification factor is 1.09. In this case, the magnification effect of tree deformation seems not significant because of the large tree
deflection and short-duration loading, and the modeled result is similar to the static stress state. For a longer air blast duration of 3.2s (β=0.6), we find D=1.34, a high value. The maximum tree
deformation reaches during the air blast loading. In such a scenario, an air blast traveling at 20ms^−1 can exert similar destruction as a long-duration wind that moves at 25ms^−1. The dynamic
magnification effect significantly increases the tree displacement and thus causes such a phenomenon. Measurements of air blast duration reported by Russian and Swiss researchers (Grigoryan et
al., 1982; Sukhanov, 1982) are within this range, lasting only a few seconds. Although the large tree deflection decreases the wind loading, the impulse air blast load is prone to damaging the trees
because of the dynamic magnification effect.
Additional simulations were performed on the air-blast-induced tree breakage. The impulse air blast is assumed to have a maximum velocity of 20ms^−1 and a duration of 3.2s. For this case,
numerical results demonstrate the maximum bending stress and moment of 35MPa and 192kNm, respectively. The maximum bending stress reaches at 9m height ($\mathrm{1}/\mathrm{3}h$), and the maximum
bending moment is identified at the tree base. In natural forest areas, the bending strength σ[crit] and anchorage resistance M[crit] are highly variable, depending on tree species, soil
characteristics and temperatures, etc. Measurements conducted by Peltola et al. (2000) and Lundström et al. (2007) indicate that the bending stress to destroy mature trees needs to exceed a value of
30MPa, while mature spruces with a height of 20–40m have an anchorage resistance reach up to 100–400kNm. For the case performed in this study, the forest is likely to be damaged in both bending
and overturning failure modes. Reliable values of critical parameters are needed during the assessment of tree destruction, and this will improve the prediction accuracy of the likely failure mode.
A further application was performed on the 2008 Wenjia Valley avalanche-induced air blast in Sichuan, China (Fig. 6a). This large avalanche had a volume of over 40×10^6m^3 and generated a powerful
air blast. According to our previous investigations and numerical modeling (Zhuang et al., 2019), the air-blast-damaged trees are mostly tall spruce concentrated near the turning points of the valley
(Fig. 6b–c). The simulated air blast velocity at turning points reaches 30ms^−1 (point A) and 35ms^−1 (point B). Using the spruce-related parameters indicated in Table 1 and an assumed air blast
duration of 3.2s (a long duration for large avalanches), the maximum displacement of spruces is calculated to be 18.5 and 22.2m at points A and B, respectively. In this case, the maximum bending
stress of trees at two turning points could reach 51 and 57MPa, significantly larger than the bending strength suggested by Peltola et al. (2000) (36MPa). Therefore, bending failure of tall spruces
was widely identified in situ.
Risk assessment and disaster mitigation of landslide-induced air blasts are hot issues in mountainous regions. Developing a simple but applicable relationship between air blast pressure and tree
failure is of great use to scientists to quantify the air blast power. Compared with existing models, one significant improvement of our model is to model the tree as a flexible beam with a variable
cross-section and to involve the impact of anchorage. This improvement allows the tree to move as its natural vibration mode rather than a hypothetical trajectory (e.g., rotate around the tree base
as a rigid body; Bartelt et al., 2018a). Moreover, the variable cross-section makes the modeling of tree-bending failures more realistic. We can simulate the failure position of trees subjected to a
powerful air blast. For the existing model with a constant diameter (Feistl et al., 2015), the rigidity EI is constant along the beam, and the maximum bending stress is always identified at the tree
base. This failure characteristic cannot match the actual situation well.
Our proposed model accounts for the factors of large tree deflection: eccentric gravity and modeling of air blast force regarding the wind–tree relative motion and geometric nonlinearities. To
investigate the impact of these factors and to confirm the necessity of considering large deflection, a comparative analysis is needed to make readers have a better understanding. Therefore, we
designed a comparative analysis by simplifying the tree-motion model of Eq. (8) without involving the impact of large tree deflection. The simplified model is similar to that proposed by Bartelt et
al. (2018a):
$\begin{array}{}\text{(16)}& \begin{array}{rl}m\frac{{\partial }^{\mathrm{2}}y}{\partial {t}^{\mathrm{2}}}+ky& ={\int }_{\mathrm{0}}^{h}\mathrm{0.5}\mathit{\rho }{C}_{\mathrm{d}}{A}_{\mathrm{f}}{v}_
{max}^{\mathrm{2}}\mathit{\varphi }\mathrm{d}s\cdot \mathrm{sin}\mathit{\varpi }t\\ & ={\int }_{\mathrm{0}}^{h}{F}_{\mathrm{s},max}\mathit{\varphi }\mathrm{d}s\cdot \mathrm{sin}\mathit{\varpi }t.\end
The displacement at the tree top can be written as
The maximum deformation occurs during the loading time when β≤1 and after the loading time when β>1. The magnification factor D for both scenarios can be expressed as
Figure 7 presents the impact of large tree deflection on the magnification effect. We first perform the simulation using the proposed model without regarding the impact of large tree deflection. A
very low air blast velocity (maximum velocity of 0.1ms^−1) is performed, and the eccentric gravity is not considered. The D[max] value of 1.77 is identified in the scenario, which is consistent
with the analytical solution from Eq. (18). The tree deformation is small with such a weak air blast loading, and the comparison result verifies the validity of our proposed model. Further
calculations with higher air blast velocities show different results. In the case of a low air blast velocity, the eccentric gravity contributes a lot to the tree deformation, causing a rather large
magnification factor (>2). However, D[max] greatly decreases with an increase in wind velocity. For a high air blast velocity, the dynamic response and eccentric gravity amplify the tree deflection,
but the inclination of the trees to the wind direction significantly reduces the air blast loading. This special mechanism was rarely considered during the previous assessment of landslide-induced
air blasts. We suggest that the modeled tree deformation subjected to a powerful air blast might be overestimated without considering large tree deflection, although this simplified model of Eq. (18)
has the advantage of rapid assessment of air blast pressure. The impact of large tree deflection should be accounted for when using forest destruction to quantify the air blast danger.
The dynamic response of trees subject to a landslide-induced air blast is a complex problem, depending heavily on the biometric characteristics of trees. Some biomass variations can be represented by
the parameters in the proposed model. For example, for the leafless trees, air blasts pass through the tree crown and only act on the branches, causing a smaller wind load. A reduction of drag
efficient C[d] is needed in such a condition. Single trees in the impact region of air blasts are subject to a larger loading than trees in dense forest stands, where tree crowns tend to be narrower
and form a shielding effect. We can make a reduction in the frontal area A[f] to simulate this mechanism. Furthermore, although much effort has been paid to the biometric and mechanical
characteristics of tree crowns and trunks, less information is available on the anchorage stiffness and resistance. The root anchorage properties significantly influence the tree eigenfrequency and
the likely failure mode. A reliable measurement value of tree-bending strength and anchorage resistance is of use to improve the accuracy of tree failure prediction and to clarify which failure mode
is prone to occur. Overall, biomass-related parameters selected to estimate the air blast pressure are recommended to be determined based on in situ investigations. In the future, more measurements
need to be conducted on the anchorage properties of trees. Regional databases for biometric and mechanical properties of trees are worthwhile to establish. This would help to provide reliable
parameters for the air blast risk assessment.
In this study, the tree is modeled as a variable cross-section that is hinged at ground level using elastic support. Root anchorage is complex and sensitive to many factors such as soil mechanical
properties, soil water content and root morphology, and we acknowledge that it is difficult to establish a model that accounts for all the factors that affect the anchorage. Most importantly, we
developed a simple but practical model that can simulate the dynamic response of trees subject to a powerful air blast and their two possible failure modes. Bartelt and his colleagues (Bartetl et
al., 2018b) have developed a dynamic model named RAMMS, which can efficiently model the entire movement process of ice, rock and snow avalanches and the associated air blasts. It is anticipated that
the combination of our proposed tree model and the RAMMS dynamic model could help in the risk assessment of potential air blasts through modeling the air blast impact region and forest destruction.
Air blasts are short-duration impulses and can cause fatalities and destruction far beyond the sliding mass. Tree destruction in situ can provide valuable data to quantify the air blast danger and to
make us better understand its force of destruction. In this study, we developed a framework for forest destruction assessment subject to a powerful air blast, including the eigenfrequency prediction
method, tree motion equations and breakage conditions. The tree is modeled as a flexible variable cross-section beam hinged at ground level using elastic support. The impacts of root anchorage and
large tree deflection are regarded during the dynamic response analysis. The framework also involved two failure modes (bending and overturning) and their corresponding failure criteria so that the
risk of forest damage could be assessed.
Using the proposed framework, we assumed conditions to investigate the air blast power. Modeling results demonstrate that although the anchorage properties significantly influence the tree
eigenfrequency, the latter is always on the same order as air blast frequency. The associated dynamic magnification effect amplifies the tree deformation and thus makes the tree damage easier. In the
scenario with a similar frequency between air blasts and trees, an air blast traveling at 20ms^−1 causes a similar force of destruction as a long-duration wind load that moves at 25ms^−1.
Notably, this magnification effect caused by the dynamic response and eccentric gravity is significant in the case of a low wind velocity, while the large tree deflection caused by strong air blast
loading would weaken this effect. Furthermore, bending and overturning are two likely failure modes for trees subject to a powerful air blast, but exactly what kind of failure will occur for a
specific forest depends heavily on the properties of both trees and soils. A case application was further performed on the 2008 Wenjia Valley avalanche-induced air blast in China, testing the
validity of our proposed model. In the future, more measurements should be conducted on biometric and mechanical properties of trees, and a regional parameter database is worthwhile to establish.
This would greatly improve the prediction accuracy of tree damage and air blast pressure.
No data sets were used in this article.
YZ did the numerical work and wrote the manuscript with contributions from all co-authors. AX and PB designed the work and modified the manuscript. BM evaluated the results and proposed various
improvements that were incorporated. ZD helped with the eigenfrequency prediction model.
The contact author has declared that none of the authors has any competing interests.
Publisher's note: Copernicus Publications remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
The authors gratefully acknowledge the financial support of the National Natural Science Foundation of China.
This study has been supported by the National Natural Science Foundation of China (grant no. 41977215).
This paper was edited by Mario Parise and reviewed by three anonymous referees.
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Difference between Batch and Continuous Fermentation - Definition, Structure, Advantages and Key Features
The fermentation process usually occurs in microbes and this process is used to form various products. For this, large tanks are used known as bioreactors in which all the components such as
nutrition, and microbes are mixed and products allowed to be formed in controlled conditions. To learn more about fermentation, batch and continuous fermentation and the difference between them,
continue reading through the following article.
What is Fermentation?
Fermentation is the process of oxidation of organic material in absence of oxygen. It is a metabolic process that occurs in microbes.
Types of Industrial Fermentation
There are mainly three types of fermentation:
• Batch fermentation
• Fed-Batch fermentation
• Continuous fermentation
Batch Fermentation
In batch fermentation, all the components are mixed at once then the reaction undergoes without any further intake from outside. During the whole process, no extra nutrients are added. It is a closed
system because all the components are added at once and no other components are added in between the process of fermentation.
There are three phases in the batch fermentation process - lag phase, exponential phase, and stationary phase.
• In the lag phase, microbes adapt to the environment of the culture and
• In the exponential phase, the microbial cells grow rapidly and consume most of the nutrients and the last phase is
• The stationary phase is when the growth of microbes stops due to the consumption of all nutrients. It is the simplest type of all industrial fermentation.
The batch fermentation diagram is given below.
Batch Fermenter
Fed-Batch Fermentation
• It is a modification of batch fermentation.
• In this nutrition is added aseptically and the amount of liquid culture in the bioreactor increases as the culture is added systematically.
• It is a type of semi-open system.
• It yields a better result than batch fermentation.
• After consumption of early substrate continuous and constant nutrition is added.
Fed-batch Fermenter
Continuous Fermentation
• It is a type of fermentation in which constant addition and flow of solution occur.
• Microorganisms and sterile nutrients are added continuously and the nutrient solutions and microbes are transformed simultaneously.
• It is a type of open fermentation system in which comments can be added and removed in between the process.
• There are many methods of continuous fermentation.
Continuous Fermenter
Similarity between Batch Fermentation and Continuous Fermentation
There are many similarities between batch and continuous fermentation. In both batch fermentation and continuous fermentation, development conditions are provided from the outside. In both,
temperatures, pH and aeration are maintained. In both types of fermentation useful products are formed.
Difference between Batch and Continuous Fermentation
There are many differences between batch and continuous fermentation.
Important Questions
1. Which phase is longer in continuous fermentation?
Ans: Exponential phase is of longer duration in continuous fermentation as nutrients are continuously being added to the solution. This phase shows the maximum growth rate. This phase is also known
as the log phase.
2. What are the limitations of fermentation?
Ans: Fermentation is a slow and continuous process and it requires more energy. and resources.
3. Which microorganism is responsible for fermentation?
Ans: Mostly lactic acid bacteria of several genera, including lactobacillus, streptococcus, yeast, and another fungus also used for the fermentation process.
Fun Facts about Fermentation
• The fermentation process is one of the important processes as it enhances the nutritional value of food.
• Various types of expensive dishes are made by the fermentation process.
• Wine or alcohol is also made by the fermentation process.
• Fermentation is also a method of preserving food items.
Practice Questions
• Does fermentation require oxygen?
• Why is temperature important in the fermentation process?
• What factors speed up the fermentation process?
• What is batch culture used for?
• What are the advantages of batch culture?
Key Features
• The fermentation process is used to make various products.
• There are three types of industrial fermentation.
• Batch fermentation is the simplest type of fermentation and batch-fed is a modification of batch fermentation.
• In this article, we have also studied batch culture and continuous culture.
• There are differences between continuous fermentation and fed-batch fermentation like fed-batch is a closed system whereas continuous fermentation is an open system. | {"url":"https://www.vedantu.com/biology/difference-between-batch-and-continuous-fermentation","timestamp":"2024-11-08T22:23:39Z","content_type":"text/html","content_length":"297761","record_id":"<urn:uuid:6251fd3f-6695-4324-baae-0ce882ff7341>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00182.warc.gz"} |
Concept of Multiplication
This is a very important module because a lot of times multiplication is when the student really starts to struggle with math or encounters a big challenging new concept. So learning the concept of
multiplication at a very deep level is critical. By reviewing and practicing this deck for 10 minutes a day, the student will gain mastery of the concept of multiplication. This mastery of the
concept will give the student confidence as they move forward instead of struggling.
Week 8 of the 2nd Grade CountFast program introduces the concept of multiplication as repeated addition. This module includes some of the basic rules of multiplication with an emphasis on improving
fluency and speed with basic multiplication facts. Spend 15 minutes each day on one of the activities listed in this module. Card decks should go home with students each day for additional practice
with a parent at home. Each week, a new deck is introduced and the previous deck is for the student to keep at home for continued practice.
1. NCTM Standard: understand situations that entail multiplication and division, such as equal groupings of objects and sharing equally.
2. NCTM Standard: develop fluency in adding and multiplying whole numbers
3. CCSS.MATH.CONTENT.2.OA.C Work with equal groups of objects to gain foundations for multiplication.
Develop fluency in using the concept of multiplication to quickly solve basic multiplication facts.
1. One CountFast Concept of Multiplication card deck for each student. This deck is for school and home use. Discuss routine and expectations for taking home the deck and returning it to school
each day.
2. One writing utensil per student (optional, if you desire students to practice writing the equations)
Concept of Multiplication Day 1
Teacher Model/Direct: You will use the YELLOW cards from the deck today. Discuss with students that sometimes we have to add the same number up over and over. We call this “repeated addition” or
“multiplication” (adding the same number a multiple of times). Show students the rule cards “Zero times any number is Zero” and “One times any number is the number itself”. Discuss these two rules
with the students by using the board to write examples of repeatedly adding up zeros or ones. Encourage students to remember these two rules for quick mental math with multiplication facts.
Use the rest of the yellow cards to demonstrate that repeated addition and multiplication mean the same thing. (Two plus two can also be written as two times two – because we are adding the 2 two
times. Notice that the multiplication symbol is simply an addition symbol slightly rotated.) Skip counting by any number is a way that students have already been exposed to the concept of
multiples. Remind them of skip counting activities they have done in the past and relate that to today’s lesson. Hold up two related cards (such as 2 + 2 + 2 and 2 X 3) and have students practice
saying aloud “Two plus two plus two equals 6, so two times three equals 6. “
Student Activity: Partner up students and give each pair one deck to work with. Ask them to take out the yellow cards. Students should work together to pair up the repeated addition problems with
the multiplication problem that means the same thing. Take turns explaining why each pair has the same answer. Students should practice verbalizing the concept of multiplication.
Home Activity: Students will take home the deck and the “CountFast Home Connection” letter for this week. Students will share what they have learned about the concept of multiplication with parents
and practice solving each of the yellow cards as quickly as possible.
Concept of Multiplication Days 2 through 5
Teacher Model/Direct: Repeat the process for the lesson on Day 1, using a different color set from the deck each day.
Day 2 – BLUE cards (multiplying 3)
Day 3 – PINK cards (multiplying 4)
Day 4 – GOLD cards (multiplying 5)
Day 5 – GREEN cards (multiplying 6)
Student Activity: Partner up students and ask them to take out that day’s color set of cards. Repeat the student activity from Day 1 with each set. At the end of the week, student partners can
challenge each other by mixing up the entire deck, flipping cards over one at a time, and seeing who can solve each card the fastest.
Home Activity: Students will take home the deck and practice the color set of cards for each day. | {"url":"https://countfast.com/product/concept-of-multiplication/","timestamp":"2024-11-10T21:48:17Z","content_type":"text/html","content_length":"390631","record_id":"<urn:uuid:6487b0b4-115f-494f-91ba-b5de64dce773>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00045.warc.gz"} |
A 61 Kg Person Is On Mars Which Has An Acceleration Due To Gravity...
w = 219.6 N
w = mg
where w is weight, m is mass in kg, and g is acceleration due to gravity in meters per second squared
in this case, given is
m = 61kg
g = 3.6 m/s^2
By plugging in the given values, you can find the weight of the person on mars
w = 61(3.6)
w = 219.6 N
The position of the image relative to the mirror v = 20cm, position of the image behind the mirror.
Since object is always placed in front of the mirror
Hence, object distance will be negative
Object distance (u) = - 100cm
Focus of a convex mirror is behind the mirror
Hence focal length will be positive
Focal length (f)= + 25cm
Find position of the image
Let the image distance = v
Using formula,
[tex]\frac{1}{v}[/tex] +[tex]\frac{1}{u}[/tex] = [tex]\frac{1}{f}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{f} - \frac{1}{u}[/tex]
[tex]\frac{1}{v} = \frac{1}{25} - \frac{1}{-100}[/tex]
[tex]\frac{1}{v} = \frac{1}{25} + \frac{1}{100}[/tex]
[tex]\frac{1}{v} = \frac{4 + 1}{100}[/tex]
[tex]\frac{1}{v} = \frac{5}{100}[/tex]
v = [tex]\frac{100}{5}[/tex]
v = 20cm
Since v is positive
Hence, image is 20cm behind the mirror
Nature of image is Virtual and erect.
To learn more about convex mirror | {"url":"https://www.cairokee.com/homework-solutions/a-61-kg-person-is-on-mars-which-has-an-acceleration-due-to-g-8a29","timestamp":"2024-11-05T09:29:11Z","content_type":"text/html","content_length":"89016","record_id":"<urn:uuid:9c3264b2-9416-4619-8d1d-551bee8a8b03>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00887.warc.gz"} |
TopCashBack Easter Treats Giveaway All Clues And Answers 2023 - The Reverend
TopCashBack Easter Treats Giveaway All Clues And Answers 2023
TopCashBack Easter Treats!
TopCashBack (TCB) are back with another ‘Treat’ game for easter – TopCashBack Easter Treats! They have done this previously with Summer Treats, Easter Treats, Summer, Hallowe’en, Christmas, Travel
Treats, Easter Treats and Sweet Treats. Every day they will reveal clues and you have to guess the retailer they relate to. Click on the correct retailer to see the TCB Hummingbird appear. Click on
the Hummingbird and you can win a treat! It could be instant cashback into your account or a prize-draw entry.
The Prize Board
Mega Prize Draw
The prizes are available to everyone and you can still win FREE money for doing very little. It has started today (27th March, 2023) and runs until the TopCashBack Easter Treats treasure hunt is over
– the 9th April. As well as the prizes above, there is the MEGA PRIZE DRAW worth £1,000 ! Check back here and enter these answers into TopCashBack to win!
Day 1 = The Royal Mint – I won 1 x Multicolour Egg (Need 2 more to win £1 cashback directly into account)
Day 2 = RAC UK – I won 1 x Bunny Egg (Need 3 more to win £20 cashback directly into account)
Day 3 = Radisson Hotel – I won 1 x Starry Egg (Need 3 more to win £10 cashback directly into account)
Day 4 = Superdrug– I won 1 x Pink Egg (Need 2 more to win 50p cashback directly into account)
Day 5 = DB Journey – I won 1 x Bunny Egg (Need 2 more to win £20 cashback directly into account)
Day 6 = Canon – I won 1 x Prize Draw Entry
Day 7 = Jacamo – I won 1 x Starry Egg (Need 2 more to win £10 cashback directly into account)
Day 8 = Thorntons – I won 1 x Prize Draw Entry
Day 9 = Confused – I won 1 x Prize Draw Entry
Day 10 = Very – I won 1 x Prize Draw Entry
Day 11 = Kate Spade – I won 1 x Prize Draw Entry
Day 12 = Elemis – I won 1 x Prize Draw Entry
Day 13 = Which? – I won 1 x Prize Draw Entry
Check back here tomorrow if you need help with tomorrow’s clue for the TopCashBack Easter Treats Giveaway. If you are having any problems getting the right answer, let me know in the comments below
and I will help out 🙂
The Reverend’s Tip
Check out the ‘Trending’ section on the TopCashBack homepage to see what other people are clicking on, it might help you get the answer or even find a bonus Treat!
The Reverend’s Tip #2
Check out my blog post on how to get EIGHTEEN QUID cashback with ZERO spend. A simple way to hit the payout level for any TopCashBack users plus also new accounts get a Five Pound Bonus.
36 comments
1. I havent found any extra birds yet but I do have 2 flower eggs, 1 more and I get £5!
2. Morning,
Todays is Radisson Hotels. That was a PDE
Also found an extra one on Moonpig – Star egg
1. Hi Bella,
I got the Radisson Hotels – and I’ve added to the blog
I checked Moonpig and it isn’t a bonus for me, so haven’t added it to the list but people can try it and see – it could just be a random one.
The Reverend
3. I found an extra bird on Levi’s
1. Hi Katie,
Cheers for letting me know. I’ve checked and there was nothing for me on the Levi page so I think that is just a random hummingbird flying about!
The Reverend
2. Thanks Katie I got a multi coloures egg from Levis thanks to you
4. Extra on Very for me today
5. 30.03.2023 pink egg on Aliexpress UK
1. Hi Barry,
That one worked for me – I got a flower egg.
The Reverend
6. PDE from Natures Best
1. Hi Zoe,
That one worked for me, I got a Flower Egg.
The Reverend
7. Chick egg Pretty Little Thing
1. Hi Zoe,
This also worked for me – I got a Prize Draw Entry.
The Reverend
8. Flower egg Very
1. Hi Zoe,
This one didn’t get me anything.
Maybe next time?
The Reverend
9. PDE Hotel Chocolat
1. Nothing for this one, either.
Thanks for the suggestion though – well worth trying 🙂
The Reverend
10. PDE U Switch, Monsoon
11. The Reverend,
Believe you may have inadvertently omitted the 1st April clue ‘Canon’
Many thanks for your help.
1. Hello B,
Cheers for letting me know. I had an issue which meant an update I did wasn’t saved correctly. I thought i’d fixed it but it looks like I’d bodged it further!
Have now resolved.
The Reverend
12. PDE OnBuy.com,
Star egg Treatwell,
Bunny egg Cadbury gifts, buyagift.com
Multi coloured egg MAC
Pink egg French Connection
1. Hi Zoe,
Did you get these all in the same day?
I’ve just had 2 x PDE with Very and Aliexpress but nothing on Nature Best. I’m wondering if it’s capped at a certain number a day?
1. Hi Bella
Yes these were all on the same day.
I tend to leave the daily clue to last thing at night which seems to let me collect more hummingbirds from different retailers.
If I do the clue and then search other retailers I think it is capped at maybe 1 or 2 extra.
Just what I’ve noticed, may be more luck than judgement though!
13. Is there a mistake on your list? Radisson Hotels is on there twice. Or did it come up again?
1. Hi Dave,
Would you believe me if I said it was a test….and you PASSED!!!!?
Don’t, because it wasn’t. I’d messed up. Now fixed. All should be correct now.
The Reverend
14. Confused.com for 04/04/2023
15. 04.04.2023 – PDE on musicMagpie
16. Hi there,
I just found a stripped egg on Trainline purely by accident!!
17. Found an extra bird on Morrisons Groceries!
1. Thanks! I got a PDE on Morrisons.
The cashback offer seems pretty good too (if one is close by)
18. PDE House of Fraser, Clarks, The Entertainer
1. Thanks! I got an egg on House of Fraser
19. I got a PDE from Agoda
Thanks Zoe for the search tip!! I ran everything listed here before doing the clue, and got 4 or 5 eggs and 2 PDEs!
20. 07.04.2023 – bit late in the day, but today’s clue is Elemis
21. I got PDEs from Boots, Harvey Nichols, Iceland, Tails.com, TopGiftCards, and Western Digital Europe,
22. Found a PDE on British Airways
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Bell inequality and common causal explanation in algebraic quantum field theory
Hofer-Szabó, Gábor and Vecsernyés, Péter (2012) Bell inequality and common causal explanation in algebraic quantum field theory. In: UNSPECIFIED.
Reich_Bell2012b.pdf - Submitted Version
Download (355kB)
Bell inequalities, understood as constraints between classical conditional probabilities, can be derived from a set of assumptions representing a common causal explanation of classical correlations.
A similar derivation, however, is not known for Bell inequalities in algebraic quantum field theories establishing constraints for the expectation of specific linear combinations of projections in a
quantum state. In the paper we address the question as to whether a 'common causal justification' of these non-classical Bell inequalities is possible. We will show that although the classical notion
of common causal explanation can readily be generalized for the non-classical case, the Bell inequalities used in quantum theories cannot be derived from these non-classical common causes. Just the
opposite is true: for a set of correlations there can be given a non-classical common causal explanation even if they violate the Bell inequalities. This shows that the range of common causal
explanations in the non-classical case is wider than that restricted by the Bell inequalities.
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A Measured Guide to the Differences – All The Differences (2024)
Math is a subject that has been around for thousands of years. It has many dimensions and measurements that make it so interesting, and we can all take advantage of them to make our lives easier.
Dimensions are used to measure lengths and widths. They also measure height, depth, weight, and temperature.
The main difference between length, width, height, and depth is their size.
Length is often the longest distance between two points in a straight line. Width is the shortest distance between two points in a straight line. The height of an object is its distance from top to
bottom, while its depth is measured from one side of an object to the other on its underside.
Let’s dive into the details of all these terms.
What Is Meant By Length?
Generally, length refers to the distance between two sides or ends of something. It can often be measured in meters or centimeters.
The length of an object is often measured by the number of times it takes to wrap it around a person’s hand (one wrap = 2 inches). This is called “circumference.”
The length of a person’s arm is determined by measuring the distance between their elbow and their middle finger. This measurement is called “arm span.”
There are also several different units for measuring length, including inches, feet, and yards (1 foot = 12 inches), miles (1 mile = 5,280 feet), kilometers (1 kilometer = 1,000 meters), and so on.
What Is Meant By “Width?”
The term “width” is used in mathematics to describe how wide a shape is. The width of a body can be measured using two different measurements: the vertical distance between its edges and the
horizontal distance between its sides.
For example, if a rectangle has vertical sides measuring 4 inches and horizontal sides measuring 1 inch, its width is 5 inches.
Width is a measurement of distance that describes how wide an object is. It’s usually measured in millimeters (mm), although it can also be measured in inches, feet, or meters.
What Is Meant By “Depth?”
Depth measures how far into or out of an object you can see. In math, it’s usually used to refer to the length of a line segment. It’s also sometimes used to describe the number of steps needed to
reach a destination.
For example:
• If you’re looking at a straight road and trying to figure out how far something is, you’d say it’s three miles deep down the road.
• If your friend is standing on top of a hill, looking at something in the distance, and asks how many steps they’d have to take for you both to get together, you would say, “One step deeper down.”
What Is Meant By “Height?”
An object’s height is determined by its distance from the ground to its top.
In mathematics, heights are measured in meters. Whenever you want to determine the height of something, you need to measure it from its base to its top.
There are two types of measurements: direct and indirect. Direct measurement is when someone measures the height directly with a measuring tape or ruler. An indirect measurement is when someone uses
an instrument that measures directly but indirectly transmits its results (for example, using a calculator).
Know The Difference
The terms height, length, width, and depth are all used to describe the dimensions of an object.
• Height refers to the distance between an object’s bottom and its top.
• It’s also known as altitude or elevation.
• It is measured in meters and centimeters.
• Length is the distance from one end of an object to another; for example, the length of a table or chair.
• Length is also called linear distance or linear dimension.
• It’s measured in meters, centimeters, and inches.
• “Width” describes the distance between two parallel sides of an object.
• Width is a measurement that describes how wide something is, such as a book or piece of paper.
• Width is also called breadth and lateral dimension.
• It’s measured in meters, centimeters, inches, and feet.
• Depth is the distance from one side of an object to another side of an object.
• Depth is similar to length but describes how far into something you can go.
Height, length, and width are linear measurements, while depth is volumetric.
Here is a table of comparison between these terms.
Terms Definition Unit Measurement
Length Depth is how far into an object you can go before hitting its bottom or sides. Meter, Centimeter, Inches, Feet Linear
Width Width is the horizontal distance from one side to another side of an object. Meter, Centimeter, Inches, Feet Linear
Depth Height is the vertical distance from the top to the bottom of an object. Meter, Centimeter Volumetric
Height Height is the vertical distance from the top to the bottom of an object. Meter, Centimeter Linear
Here is an interesting video describing different dimensions in detail.
Are Depth And Height The Same?
Depth and height are not the same.
Depth is just how deep something is—like how deep a pool or hole is; height is the distance from the bottom to the top of something.
So if you’re talking about a person’s height, it’s how tall they are, and if you’re talking about the depth of a room, it’s how far down it goes.
Does Depth Mean Wide?
Depth can mean both, depending on what you’re talking about.
If you say, “She has a deep voice,” that means her voice is low and resonant, like an opera singer or a choir director.
But when you say “she has a wide range of interests,” that means she is interested in many different things outside her field. For example, she might love art history but also enjoy watching football
games on television.
So depth can refer to both breadth and depth of knowledge or understanding, but it usually refers to a person’s level of expertise in their field or area of interest.
What Is The Meaning Of W * D * H?
W * D * H is an acronym for “Width, Depth, Height.” It’s a measurement used to describe the size of objects.
Width and depth are usually measured in inches or centimeters. Height is measured in millimeters. For example, a container might be considered 8″ W X 6″ D X 2″ H.
How Do You Measure Depth?
Depth is a measure of how far away something is from the surface. Depth measurement is important because it helps you determine how deep you must dig to find what you’re looking for.
Measurement of depth is done with the help of a depth gauge. It has a curved part that allows it to conform to the shape of whatever you’re measuring and a ruler on the side so you can measure the
Final Takeaway
• Maths is an important subject that you can apply in your daily life.
• You can find different dimensions and measurements of maths applicable to your life scenarios.
• These dimensions include height, length, width, and depth.
• Length is the distance from one line’s end to the other.
• An object’s width is how far apart it is from another.
• Height refers to the distance between the item’s base and its top.
• The depth of a surface is the distance from its surface to the farthest point within it.
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Rational expectations and information theory
Noah Smith once again deleted my comment
on his blog
, so I'll just have to preserve it (well, the gist of it) here.
He discussed an argument against rational expectations he'd never considered before. Since counterfactual universes are never realized, one can never explore the entire state space to learn the
fundamental probability distribution from which macro observable are drawn. Let's call this probability distribution A. The best we can get is some approximation B.
Rational expectations is the assumption
A ≈ B
In that post, I showed that the
KL divergence
measures information loss in the macroeconomy based on the difference between the distributions
D(A||B) = ΔI
That was the content of my comment on Noah's post. I go a bit further at the link and say that this information loss is measured by the difference between the price level and how much NGDP changes
when the monetary base changes
ΔI ~ P - dN/dM = dN*/dM - dN/dM
Which to me seems intuitive: it compares how much the economy should grow from an expansion of the money supply (ideally) to how much it actually does grow.
Just the aggregate ΔI is measured, however. Two different distributions B, B' and B'' can have the same KL divergence so this doesn't give us a way to estimate A better.
Now rational expectations are clearly wrong at some given level of accuracy, but then so are Newton's laws. The question of whether you can apply rational expectations depends on the size of ΔI.
Since ΔI is roughly proportional to nominal shocks (the difference between where the economy is and where it should be based on growth of M alone [1]) and these nominal shocks are basically the size
of the business cycle, it means rational expectations are not a useful approximation to make when analyzing the business cycle.
As far as I know, this is the first macroeconomic estimate of the limits of the rational expectations assumption that doesn't compare it to a different model of expectations (e.g. bounded
rationality, adaptive expectations). (There are lots of estimates for micro.)
[1] In case anyone was curious, this also illustrates the apparent inconsistency between e.g. this post where nominal shocks are negative and e.g. this post where they are positive. It depends on
whether you apply them before including the effect of inflation or after. Specifically
0 = dN*/dM - (dN/dM + σ) = (dN*/dM - σ) - dN/dM
21 comments:
1. "Two different distributions B, B' and B'' can have the same KL divergence so this doesn't give us a way to estimate A better."
Looks like three.
And here I was worried that using B' and B'' would have been confusing (what about B?)
"1. Typos in posts don't reveal themselves until you've published. If you schedule a post to publish in the future, the typos will be revealed then. This is an absolute, inviolable rule of
blogging. This may be some sort of subtle lesson from the universe about our hubris in the face of fundamental impermanence."
2. Thanks for the link. I have to say though I'm more confused now because you didn't change "two" to "three" or otherwise structure the sentence differently. I'm I parsing it wrong?
2. What's the story with Noah? Why do you think he erases your comments?
1. He doesn't block my comments, even when I occasionally provide a link to one of your posts. Perhaps because it's clear I have no idea what I'm talking about (and thus provide some
amusement... much as does Ray Lopez for Scott)? Lol
2. I think I have enough data to say that he doesn't erase the ones where I just agree with him.
I think he deletes "self promotion": e.g. 'I agree and said something similar at my blog.'
which is his prerogative.
3. O/T: one question of mine that I don't recall that status of (did I ask it? did someone answer it? did I just forget?) during the great Smith Sadowski row of 2015 was, putting aside the
non-linearities of MB after 2008, why couldn't Mark have done a Granger causality test for MB causing inflation (and all the rest of what he looked at) for the years from say 2001 to 2008? Mark
said (in a response to my question on Nunes' blog) that was possible but difficult, but then both you and he offered the explanation that MB was not the Fed's target as well. This confuses me
because I'd think that regardless of whether MB was the ultimate target, that it was used as a lower level target to (help?) achieve the upper level targets of either the FFR or the inflation
What I'd expect to see from such an analysis is that MB Granger causes inflation (and other things) but the effect is perhaps larger (perhaps closer to 1:1) rather than 10:1 on a percentage wise
I ran that idea piecemeal past Nick Rowe: first asking him if MB was a more "fundamental" level, even when targeting short term interest rates or inflation. He essentially said "yes, that's one
way to look at it" (in the back of my mind I was hearing him muttering under his breath "it you're person of the concrete steppes anyway, because we all know that it's Chuck Norris imparted
expectations that really matter and are most fundamental").
But when I went the extra step of asking "So what do you think a Granger causality test would indicate for the effectiveness of MB causing inflation prior to 2008 vs after 2008 on a percentage
wise basis?" he didn't go there, explaining that he didn't really trust VAR studies. What's amusing though is that in the comments to Mark on Nunes' blog he asked Mark "So I didn't read your
series, but what's the upshot? What percent change in MB causes a 1% change in inflation?" and Mark responded "about 10%" which seemed to satisfy Nick.
So do you think it's possible to perform as straightforward Sadowski style Granger causation VAR analysis for MB (as the causing variable) for the years prior to 2008? What do you think would be
the result? Would it be stopped in it's tracks early on by insufficient sign of causation?
1. ... please excuse the typos above: hopefully it's still coherent.
2. Hi Tom,
I will take this on piecemeal.
" ... but then both you and he offered the explanation that MB was not the Fed's target as well."
To be precise, Mark gave that as an explanation and I think I just accepted it as a valid (model-dependent) assumption. It is true that the Fed targeted the base with QE (either by level for
QE1 & 2 or rate for QE3 ... X billion dollars of QE, or Y billion dollars/month). The short term interest rate target is actually a range 0.0 to 0.25%.
In the ITM, a short term interest rate target of 0-0.25% actually decouples the MB from the interest rate target if the MB moves inside a range of values. The short term rates are 0.001% or
so because of QE, but that isn't the target rate. You can move the MB around quite a bit though and be consistent with that interest rate range. MB can be roughly about 50% less or + infinity
more. If it was 100% less (i.e. the Fed unwound QE completely), short term interest rates would be about 2% as I showed here:
3. "So do you think it's possible to perform as straightforward Sadowski style Granger causation VAR analysis for MB (as the causing variable) for the years prior to 2008? What do you think
would be the result?"
I would say that the response would be about 1.4:1 for the period before 2008 and 11.1:1 after. Actually, that is exactly what I calculated here:
"Before 2008, a 100% increase (a doubling) of the monetary base would have lead to a 70% increase in the price level. After 2008, it leads to a 9% increase in the price level."
So there's your 1:1 and 10:1.
That's why I said Mark was my first monetarist convert -- he built an information transfer model with IT index kappa of 0.57 before 2008 and 0.89 after.
Note that if kappa = 0.89, then 1/kappa ~ 9/8 = 1.125 so the price level is
log P ~ (1.125 - 1) log MB
log P ~ 0.1 log MB
I rounded in the above equation. Now if MB ~ exp(r t), then
P ~ exp(0.1 r t)
... or about 1/10th the growth rate r of MB.
Now that specific model (where you use MB instead of M0) had some issues treating the entire period of available data as a single model for the UK, US and Japan in this post:
There is a case to be made for monetary regime change and thus two different values of kappa in the US before and after 2008, though. However as I am the only one working on the IT model so
far there is no one to disagree with me in the context of the IT model ...
4. "-- he built an information transfer model with IT index kappa of 0.57 before 2008 and 0.89 after."
If Mark rode the trolley all the way to the end and analyzed the data before 2009.
5. ...which he didn't.
6. "I would say that the response would be about 1.4:1 for the period before 2008 and 11.1:1 after. Actually, that is exactly what I calculated here:"
Right, I remember. That's why I estimated (crudely) 1:1 and 10:1. That's what I was thinking of: your post on that.
The remaining question for me though is if Mark were to do his usual Granger analysis, where he checks to see if he can reject one thing NOT causing another with some significance, and then
turns it around to see if he can reject the opposite (the 2nd thing causing the 1st), what he'd find. You're effectively saying he's probably find MB Granger causes P, and furthermore he'd
probably find something like the 1:1 factor as well, which seems reasonable.
As I recall in his 1st "Age of ZIRP" post he found that PCIE Granger causes MB to 0.05 significance and MB Granger causes PCIE to 0.01 significance. I don't think he pursued the 1st one (PCIE
Granger causes MB) any further.
I think he uses the Toda & Yamamoto method (Dave Giles does a T&Y example, and based on the relatively high number of comments (325), it was one of Dave's more popular posts).
So when Mark said it would be more difficult to do the pre-2008 analysis, I wonder why? Separately he brought up the FFR targeting issue. Here's what occurs to me:
1. Perhaps his Granger non-causality test would indicate that he can't reject the null hypothesis of MB NOT Granger causing PCIE... which would lead him to some more difficult path, or
perhaps that's where he'd feel justified to stop.
2. Perhaps the pre-2008 changes in MB are not large enough or there's some other problem with the data which would force him to get better data (like with a higher sampling rate).
3. Maybe he'd feel compelled to factor in the FFR targeting or inflation rate targeting somehow and perhaps that complicates the model or makes analysis more difficult or less reliable for
some reason.
I shouldn't have emphasized (to you and every one else) that MB changes are used as a tool to target other things pre-2008 (such as the FFR or inflation rate). I mostly meant to point out
that pre-2008 MB *was* varied (because that's the direct result of OMOs, which were undertaken by the Fed during that time), regardless of how or if those changes in MB factored into any
other targets. In other words, regardless of what else the Fed was doing, it was doing some OMOs which were directly changing MB: so it seems like we ought to be able to see if that Granger
caused anything else (like PCIE) via whatever methods Mark and/or Dave Giles would bless. I assume he'd want to do the pre-2008 and post-2008 separately because of the break in trend.
7. Jason, forget about the above questions... I think I've beaten this to death. I'd erase it, but I wasn't logged into my Goggle account when I wrote it, so I can't (fee free though!).
8. Ok, I will. But in thinking about it, the two big spikes in inflation that seem associated with the two periods of the oil crisis in 1973-4 and 1979 might mess up the correlation. Those are
supply shocks that need a model to understand -- which I discuss more in this post:
But you could probably use 1985-2008 just fine.
9. "But you could probably use 1985-2008 just fine."
I wasn't even considering (Mark) going that far back: I thought just to 2001 (for a nice symmetric 7 years pre-2008 and 7 years post-2008). But if you think you (or Mark) could go all the way
back to 1985 w/o a "structural break" (which Mark indicated you have to avoid), then that would be better still.
4. "Useful" is a weasel word, useful for analyzing which aspects of the business cycle, and by which criteria...?
1. Hi LAL,
I was softening the blow a bit by saying "not useful" instead of "invalid". It is like assuming there is no Moon and attempting to calculate the tides. That is very precise analogy ...
rational expectations:
assuming there is no Moon → assuming ΔI ~ σ ≈ 0
business cycle:
tides → σ ≠ 0
Effectively, there shouldn't be a business cycle with rational expectations (in the information transfer framework).
2. So you are saying that under rational expectations the information transfer model means dN*/dM = dN/dM. Further that the difference between the two accounts for the empirically observed
business cycles?
3. I think the most intriguing interpretation of rbc models is to assume over the relevant time series that monetary policy was adequate to control the economy and was correctly used, then
observe the remainder. There is still something there that seems to cycle, although the interpretations are usually secular shifts in leisure preferences or TFP. Could I have a partial
rational expectations framework that somehow uses the KL divergence to know when to be in rbc mode and when not to?
4. Hi LAL,
RE: both your comments
In the model as written. There could be improvements. There are some fluctuations due to changes in M (it's not a perfectly smooth function) so there are some fluctuations in N* that could be
considered a separate part of the cycle. Essentially there are two kinds of fluctuations:
fluctuations in N* (and hence N)
fluctuations in σ
Now it appears that σ accounts for what we think of as the business cycle -- it is given by changes in employment as I look at here:
But it's not perfect. It could be measurement error or it could be fluctuations in N* due to fluctuations in M. I don't have all the answers there, but σ seems to account for most of the
business cycle.
However! There is also non-ideal information transfer -- that appears to exacerbate recessions. So that would be an extra bit of cycle on top.
I don't think I have all the answers on this yet. I made a stab at some of the effects awhile ago here:
I'm still trying to fully understand it myself. Hence I left this stuff out of the first paper ...
Comments are welcome. Please see the Moderation and comment policy.
Also, try to avoid the use of dollar signs as they interfere with my setup of mathjax. I left it set up that way because I think this is funny for an economics blog. You can use € or £ instead.
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What is Decision Tree Analysis? Definition, Steps, Example, Advantages, Disadvantages - The Investors Book
Definition: Decision tree analysis is a powerful decision-making tool which initiates a structured nonparametric approach for problem-solving. It facilitates the evaluation and comparison of the
various options and their results, as shown in a decision tree. It helps to choose the most competitive alternative.
It is a widely used technique for taking crucial decisions like project selection, cost management, operations management, production method, and to deal with various other strategic issues in an
Content: Decision Tree Analysis
What is a Decision Tree?
A decision tree is the graphical depiction of all the possibilities or outcomes to solve a specific issue or avail a potential opportunity. It is a useful financial tool which visually facilitates
the classification of all the probable results in a given situation.
Terminologies Used
Let us understand some of the relevant concepts and terms used in the decision tree:
• Root Node: A root node compiles the whole sample, it is then divided into multiple sets which comprise of homogeneous variables.
• Decision Node: That sub-node which diverges into further possibilities, can be denoted as a decision node.
• Terminal Node: The final node showing the outcome which cannot be categorized any further, is termed as a value or terminal node.
• Branch: A branch denotes the various alternatives available with the decision tree maker.
• Splitting: The division of the available option (depicted by a node or sub-node) into multiple sub-nodes is termed as splitting.
• Pruning: It is just the reverse of splitting, where the decision tree maker can eliminate one or more sub-nodes from a particular decision node.
Steps in Decision Tree Analysis
Now, you must be wondering, how to initiate the decision tree analysis for solving a particular issue?
Following steps simplify the interpretation process of a decision tree:
1. The first step is understanding and specifying the problem area for which decision making is required.
2. The second step is interpreting and chalking out all possible solutions to the particular issue as well as their consequences.
3. The third step is presenting the variables on a decision tree along with its respective probability values.
4. The fourth step is finding out the outcomes of all the variables and specifying it in the decision tree.
5. The last step is highly crucial and backs the overall analysis of this process. It involves calculating the EMV values for all the chance nodes or options, to figure out the solution which
provides the highest expected value.
Decision Tree Analysis Example
To enlighten upon the decision tree analysis, let us illustrate a business situation.
ABC Ltd. is a company manufacturing skincare products. It was found that the business is at the maturity stage, demanding some change. After rigorous research, management came up with the following
decision tree:
In the above decision tree, we can easily make out that the company can expand its existing unit or innovate a new product, i.e., shower gel or make no changes.
Given below is the evaluation of each of these alternatives:
Expansion of Business Unit:
If the company invests in the development of its business unit, there can be two possibilities, i.e.:
• 40% possibility that the market share will hike, increasing the overall profitability of the company by ₹2500000;
• 60% possibility that the competitors would take over the market share and the company may incur a loss of ₹800000.
To find out the viability of this option, let us compute its EMV (Expected Monetary Value):
New Product Line of Shower Gel:
If the organization go for new product development, there can be following two possibilities:
• 50% chances are that the project would be successful and yield ₹1800000 as profit;
• 50% possibility of failure persists, leading to a loss of ₹800000.
To determine the profitability of this idea, let us evaluate its EMV:
Do Nothing:
If the company does not take any step, still there can be two outcomes, discussed below:
• 40% chances are there that yet, the organization can attract new customers, generating a profit of ₹1000000;
• 60% chances of failure are there due to the new competitors, incurring a loss of ₹400000.
Given below is the EMV in such circumstances:
From the above evaluation, we can easily make out that the option of a new product line has the highest EMV. Therefore, we can say that the company can avail this opportunity to make the highest gain
by ensuring the best possible use of its resources.
Advantages of Decision Tree Analysis
Business organizations need to consider various parameters during decision making. A decision tree analysis is one of the prominent ways of finding out the right solution to any problem.
Let us now understand its various benefits below:
• Depicts Most Suitable Project/Solution: It is an effective means of picking out the most appropriate project or solution after examining all the possibilities.
• Easy Data Interpretation and Classification: Not being rocket science, decision tree eases out the process of segregation of the acquired data into different classes.
• Assist Multiple Decision-Making Tools: It also benefits the decision-maker by providing input for other analytical methods like nature’s tree.
• Considers Both, Categorial and Numerical Data: This technique takes into consideration the quantitative as well as the qualitative variables for better results.
• Initiates Variable Analysis: Its structured phenomena also facilitates the investigation and filtration of the relevant data.
Disadvantages of Decision Tree Analysis
Decision tree analysis has multidimensional applicability. However, its usage becomes limited due to its following shortcomings:
• Inappropriate for Excessive Data: Since it is a non-parametric technique, it is not suitable for the situations where the data for classification is vast.
• Difficult to Handle Numerous Outcomes: If there are multiple possible results of every decision, it becomes tedious to compile all these on a decision tree.
• Chances of Classification Errors: A less experienced decision tree maker usually makes a mistake while putting the variables into different classes.
• Impact of Variance: Making even a slightest of change becomes problematic since it results in a completely different decision tree.
• Unsuitable for Continuous Variables: Incorporating many open-ended numerical variables increases the possibility of errors.
• Sensitive towards Biasness: A decision tree maker may lay more emphasis on preferable variables which may divert the direction of analysis.
• Expensive Process: Collection of sufficient data, its classification and analysis demand high expense, being a resource-intensive process.
In operations research, decision tree analysis holds an equal significance as that of PERT analysis or CPM. It presents a complex decision problem, along with its multiple consequences on paper.
This enables the decision-maker to figure out all the possible options available with him/her and thus, simplifies the task. | {"url":"https://theinvestorsbook.com/decision-tree-analysis.html","timestamp":"2024-11-06T20:25:15Z","content_type":"text/html","content_length":"52809","record_id":"<urn:uuid:998d63a6-1b8f-4bc0-b21b-bee0ccb7147e>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00538.warc.gz"} |
Development of a software package for solving optimal control problems
Title Development of a software package for solving optimal control problems
Authors I. V. Lutoshkin^1, A. G. Chekmarev^1
^1Ulyanovsk State University
Annotation An analysis of existing approaches to the development of software solutions designed to solve OС problems is carried out. The parameterization method is described. It’s a numerical method
for solving ОС problems, including delays in both phase variables and control. A concept for developing a software package that implements the parameterization method is proposed.
Keywords optimal control, numerical methods, parameterization method, software package.
Lutoshkin I. V., Chekmarev A. G. ''Development of a software package for solving optimal control problems'' [Electronic resource]. Proceedings of the International Scientific Youth
Citation School-Seminar "Mathematical Modeling, Numerical Methods and Software complexes" named after E.V. Voskresensky (Saransk, July 26-28, 2024). Saransk: SVMO Publ, 2024. - pp. 105-109.
Available at: https://conf.svmo.ru/files/2024/papers/paper17.pdf. - Date of access: 12.11.2024. | {"url":"https://conf.svmo.ru/en/archive/article?id=458","timestamp":"2024-11-12T04:21:11Z","content_type":"text/html","content_length":"11173","record_id":"<urn:uuid:19852372-31f5-4125-952b-52ba4b0d9422>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00504.warc.gz"} |
1. Twelve pigs were eating a farmer’s groundnuts when he came out and shot dead one of them. How many remained in the field?
Show answer
One, the dead one. The remaining eleven fled away!
2. One cat walked in front of two cats. One cat walked behind two cats. One cat walked between two cats. How many cats were there?
A popular Hindi version of this is
3. A crab is trying to climb a glass pane 30 cm high. It climbs 10 cm in 5 minutes and then rests for 5 minutes during which it slips back 5 cm. How much time will the crab take to reach the top?
4. How many times does the digit 9 appear from 1 to 100?
5. Every morning at 6, a ship leaves Tokyo for New York and at the same time, a ship leaves New York for Tokyo. It takes 10 days to complete the trip either way. If a New York bound ship starts from
Tokyo on 15th of a month, how many Tokyo bound ships will it meet by the time it reaches New York?
(Ignore difference of timings between New York and Tokyo.)
Show answer
21 ships in all.
6. A zoo has some birds and some beasts. There are 30 heads and 100 feet. Can you tell the number of birds and beasts?
Show answer
10 birds, 20 beasts.
7. (My grandpa asked me this) You can buy 20 sparrows for a rupee, while a pigeon costs one rupee and a parrot costs five rupees. You want to purchase 100 birds in 100 rupees so that you have birds
of each variety. How can you do this?
Show answer
Buy 19 parrots, 1 pigeon and 80 sparrows.
8. A rich Arab having 23 prize camels died. His will stated that the eldest son Ahab was to have half of the camels. The second son Aziz was to have a third, and the youngest Abdul was to have an
eighth share. The sons soon realized that they couldn’t divide 23 camels amongst them without slaughtering some of the camels. However, their wise uncle loaned them one of his own camels and easily
solved the problem. Can you figure it out?
Show answer
Out of 24 camels, Ahab took 12, Aziz took 8 and Abdul took 3. Then they returned the remaining one to the uncle. | {"url":"http://www.ilovemaths.in/block5/number-sense/","timestamp":"2024-11-15T03:56:41Z","content_type":"text/html","content_length":"15613","record_id":"<urn:uuid:50a0256c-044f-41f1-9588-7c91190c560f>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00090.warc.gz"} |
Stability and hyperbolicity
\(\newcommand{\Cat}{{\rm Cat}} \) \(\newcommand{\A}{\mathcal A} \) \(\newcommand{\freestar}{ \framebox[7pt]{$\star$} }\)
1. Multiplier sequences and CZDS
Problem 1.1.
[A. Vishnyakova] Let $\{c_k\}_{k=0}^n\subset\mathbb{R}$ be given. Does there exist a real rooted polynomial, $p(x)$, with zeros outside $[0,n]$ such that $c_k=p(k)$, $k=0,1,\ldots,n$? Note that
this implies $\{c_k\}_{k=0}^n$ is an n-CZDS.
Problem 1.2.
[G. Csordas] Let $f(z)$ be a meromorphic function. When is the sequence $\{f(k)\}_{k=0}^\infty$ a multiplier sequence?
Cite this as: AimPL: Stability and hyperbolicity, available at http://aimpl.org/hyperbolicpoly. | {"url":"http://aimpl.org/hyperbolicpoly/1/","timestamp":"2024-11-12T12:16:50Z","content_type":"text/html","content_length":"23938","record_id":"<urn:uuid:8fd9f7fe-a3af-4198-9653-b400c2d8a1be>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00391.warc.gz"} |
Math Scholar
Computer discovery of mathematical theorems
In 1983 the present author recalls discussing the future of mathematics with Paul Cohen, who in 1963 proved that the continuum hypothesis is independent from the axioms of Zermelo-Fraenkel set
theory. Cohen was convinced that the future of mathematics, and much more, lies in artificial intelligence. Reuben Hersch recalls Cohen saying specifically that at some point in the future
mathematicians would be replaced by computers. So how close are we to Cohen’s vision?
In fact, computer programs that discover new mathematical identities and theorems are already a staple
Continue reading Computer theorem prover verifies sophisticated new result | {"url":"https://mathscholar.org/2021/07/","timestamp":"2024-11-04T23:19:15Z","content_type":"application/xhtml+xml","content_length":"63246","record_id":"<urn:uuid:78ed6415-59d5-4daf-8aff-9fb8a3b6d4d1>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00633.warc.gz"} |
RightStart and Problem Solving
Joan A. Cotter, Ph.D.
Math anxiety can be thought of as a learned fear of numbers or anything to do with math. It results in feelings of tension and fear at the sight of numbers or math symbols, causing poor performance
in math, especially on tests. Sadly, in the U.S., over 50 percent of people have math anxiety. I once met a person who had such a math phobia that she wouldn’t do Sudoku puzzles because they
contained numbers. Never mind that no numbers are ever calculated; in fact, colors have been used to solve the puzzle.
Math anxiety often causes students difficulties while solving math problems or during testing. Part of their working memory is involved in trying to overcome the anxious feelings instead of being
available to work on the problems.
In the U.S. people freely admit they aren’t good in math, but they hide an inability to read. Europeans and Asians disguise a lack of expertise in math as well as reading. They believe anyone can
learn math with good instruction and hard work.
A fear of math is linked to lower achievement in math, which often negatively affects a person’s career choices. There are high school graduates who don’t consider studying nursing or engineering
because they can’t imagine themselves succeeding in the required math courses.They choose careers based on avoiding math. Majoring in biology is much more common than majoring in chemistry or physics
because it requires less math.
According to the United State Department of Labor’s Bureau of Labor Statistics, growth in the number of math-related jobs will greatly outpace overall national job growth in the near future.
Many causes of math anxiety are the consequences of myths about math. These include:
Myth: Only certain people have a “math gene,” which they consider to be somewhat hereditary.
Fact: Our brains have an area designed for math. True dyscalculia is quite rare; it only affects arithmetic, not the other 199 or so other branches of math.
Myth: Boys are naturally better at math.
Fact: Girls often get better grades in math. Even the slight advantage boys have in spatial ability is erased when girls play ball sports or ski.
Myth: In real life very little math is ever needed.
Fact: To understand our natural world from the atom to the cosmos requires mathematics. Business, financial, and medical decisions involve advanced math. It is an essential ingredient of much of our
new knowledge.
Myth: Having a good memory is extremely important for doing math.
Fact: Einstein said don’t bother to memorize anything you can quickly look up. Jo Boaler, a math professor, said, “I have never committed math facts to memory, although I can quickly produce any math
facts, as I have number sense and I have learned good ways to think about number combinations.”
Myth: A mathematician solves problems quickly; they don’t need to think.
Fact: Mathematicians look at math more like a puzzle that takes time to figure out.
Myth: A person good in math rarely makes a mistake.
Fact: Mathematicians frequently take risks that turn out to be faulty, but they persist until they get it right. Girls may need to be encouraged to take risks and trust their intuition.
Myth: Learning math is drudgery—something to be gotten out of the way as soon as possible.
Fact: Mathematics is a gift from the Creator and is meant to be enjoyed.
Another significant source of math anxiety results from the way math is taught. Some problem areas include:
Problem: Insisting all children memorize the counting words to 100 before doing any meaningful math. About 20% of children have this difficulty and often fall behind their grade level.
Solution: Teach the names of quantities to 10, and then teach transparent number naming before the traditional names, such as 3-ten 7 for 37.
Problem: Ignoring children’s ability to visualize.
Solution: Use appropriate manipulatives, grouping quantities in fives as well as tens.
Problem: Using flash cards and timed tests.
Solution: Teach tens’ based strategies that are visualizable for learning facts. Also, use games the children enjoy for practice.
Problem: Teaching math like it’s a bunch of rules without any rhyme or reason; such learning makes advanced math much more difficult and applications mystifying.
Solution: Teach for understanding by asking questions that require the child to think.
Problem: Assigning homework that the child cannot do independently. Too often, a person unfamiliar with the lesson tries to help,but does it in a different way confusing the child.
Solution: With the exception of games, homework should be done in class.
In a nutshell, math is a vast field of knowledge encompassing much of human activity and needs to be taught in a caring, thoughtful way to enable children to want to learn more. We must be extremely
vigilant against transmitting any negative thoughts we may have.
The RightStart™ Mathematics curriculum was written to delight the child. It helps children understand, apply, and enjoy mathematics. | {"url":"https://homeschoolmagazine.com/2019/12/rightstart-and-problem-solving/","timestamp":"2024-11-01T19:23:20Z","content_type":"text/html","content_length":"124121","record_id":"<urn:uuid:ad97d8b1-39f1-405f-9a44-e275aa894534>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00788.warc.gz"} |
Shearing Deformation
Shearing forces cause shearing deformation. An element subject to shear does not change in length but undergoes a change in shape.
The change in angle at the corner of an original rectangular element is called the and is expressed as
$\gamma = \dfrac{\delta_s}{L}$
The ratio of the shear stress τ and the shear strain γ is called the modulus of elasticity in shear or and is denoted as G, in MPa.
$G = \dfrac{\tau}{\gamma}$
The relationship between the shearing deformation and the applied shearing force is
$\delta_s = \dfrac{VL}{A_s G} = \dfrac{\tau L}{G}$
where V is the shearing force acting over an area A[s].
Poisson's Ratio
When a bar is subjected to a tensile loading there is an increase in length of the bar in the direction of the applied load, but there is also a decrease in a lateral dimension perpendicular to the
load. The ratio of the sidewise deformation (or strain) to the longitudinal deformation (or strain) is called the Poisson's ratio and is denoted by ν. For most steel, it lies in the range of 0.25 to
0.3, and 0.20 for concrete.
$\nu = -\dfrac{\varepsilon_y}{\varepsilon_x} = -\dfrac{\varepsilon_z}{\varepsilon_x}$
where ε[x] is strain in the x-direction and ε[y] and ε[z] are the strains in the perpendicular direction. The negative sign indicates a decrease in the transverse dimension when ε[x] is positive.
Biaxial Deformation
If an element is subjected simultaneously by tensile stresses, σ[x] and σ[y], in the x and y directions, the strain in the x direction is σ[x]/E and the strain in the y direction is σ[y]/E.
Simultaneously, the stress in the y direction will produce a lateral contraction on the x direction of the amount -ν ε[y] or -ν σ[y]/E. The resulting strain in the x direction will be
$\varepsilon_x = \dfrac{\sigma_x}{E} - \nu \dfrac{\sigma_y}{E}$ or $\sigma_x = \dfrac{(\varepsilon_x + \nu \varepsilon_y)E}{1 - \nu^2}$
$\varepsilon_y = \dfrac{\sigma_y}{E} - \nu \dfrac{\sigma_x}{E}$ or $\sigma_y = \dfrac{(\varepsilon_y + \nu \varepsilon_x)E}{1 - \nu^2}$
Triaxial Deformation
If an element is subjected simultaneously by three mutually perpendicular normal stresses σ[x], σ[y], and σ[z], which are accompanied by strains ε[x], ε[y], and ε[z], respectively,
$\varepsilon_x = \dfrac{1}{E} [ \, \sigma_x - \nu (\sigma_y + \sigma_z) \, ]$
$\varepsilon_y = \dfrac{1}{E} [ \, \sigma_y - \nu (\sigma_x + \sigma_z) \, ]$
$\varepsilon_z = \dfrac{1}{E} [ \, \sigma_z - \nu (\sigma_x + \sigma_y) \, ]$
Tensile stresses and elongation are taken as positive. Compressive stresses and contraction are taken as negative.
Relationship Between E, G, and ν
The relationship between modulus of elasticity E, shear modulus G and Poisson's ratio ν is:
$G = \dfrac{E}{2(1 + \nu)}$
Bulk Modulus of Elasticity or Modulus of Volume Expansion, K
The bulk modulus of elasticity K is a measure of a resistance of a material to change in volume without change in shape or form. It is given as
$K = \dfrac{E}{3(1 - 2\nu)} = \dfrac{\sigma}{\Delta V/V}$
where V is the volume and ΔV is change in volume. The ratio ΔV/V is called volumetric strain and can be expressed as
$\dfrac{\Delta V}{V} = \dfrac{\sigma}{K} = \dfrac{3(1 - 2\nu)}{E}$
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NCERT Solutions for Class 6 Maths Chapter 7 Exercise 7.2
NCERT Solutions for Class 6 Maths Chapter 7 Exercise 7.2 in Hindi and English Medium updated for CBSE 2024-25 exams. All the questions of ex. 7.2 class 6 Maths are revised according to rationalised
syllabus and new NCERT Books for academic session 2024-25.
6th Maths Exercise 7.2 Solutions in Hindi and English Medium
Class 6 Maths Chapter 7 Exercise 7.2 Solution
Class VI Mathematics NCERT (https://ncert.nic.in/) textbook Ex. 7.2 Fractions in PDF file format free to download for offline use or use online without downloading. NCERT Book Videos and PDF
Solutions are in Hindi and English Medium given separately free to use. Class 6 Maths exercise 7.2 contains the questions based on mixed fraction to simple format and their algebraic operation. All
the questions in ex. 7.2 of 6th mathematics are easy to understand and solve.
Class: 6 Mathematics
Chapter: 7 Exercise: 7.2
Topic Name: Fractions
Medium: Hindi and English Medium
Fraction as Division
If 4 bananas are distributed equally among 4 boys, how many bananas does a boy get? Clearly, 4 ÷ 4 = 1. Similarly, if 20 toffees are distributed equally among 5 children then each child will get 20 ÷
5 = 4 toffees. But if 1 toffee is to be distributed among 5 children, then how many toffees will a child get?
In this case also a child gets 1 ÷ 5, i.e., 1/5
Thus, we conclude that a fraction can be expressed as a division. Conversely, division can be expressed as a fraction.
Class 6 Maths Exercise 7.2 Extra Question with Answer
Find 1/3 of a collection of 21 books.
We can write 1/3 of 21 books = 21 x 1/3 = 21/3 = 7 books
Find 5/9 of a collection of 63 balloons.
We can write 5/9 of 63 balloons = 63 x 5/9 = 615/9 = 35 balloons
Which of the following are proper, improper or mixed fractions? (i) 5/9 (ii) 7/9 (iii) 8/9 (iv) 20/12 (v) 5/27 (vi) 3/7 (vii) 8/13
Like Fractions: (i) 5/9 (iv) 7/9 (vi) 8/9
Unlike Fractions: (ii) 20/12 (iii) 5/27 (vi) 3/7 (vii) 8/13
Representation of Fractions on The Number Line
Let us represent ½ on a number line. In order to represent on the number line, we draw the number line and mark a point A to represent 1. Now, we divide the gap between O and A into two equal parts.
Let M be the point of division. Then M represents ½.
Percent Decimal Fraction
25% 0.25 1/4
50% 0.5 1/2
75% 0.75 3/4
80% 0.8 4/5
90% 0.9 9/10
Like Fractions:
Fractions having the same denominators are called like fractions.
For example, 4/14, 7/14, 9/14, 12/14 etc. are all like fractions.
Class 6 Maths Exercise 7.2 Important Questions
Why are fractions so important?
Fractions help children understand the nature of numbers and their interactions (e.g., the meaning of division). If a child doesn’t understand how fractions work, it will interfere with his ability
to learn algebra later.
How do we multiply fractions?
The first step when multiplying fractions is to multiply the two numerators. The second step is to multiply the two denominators. Finally, simplify the new fractions. The fractions can also be
simplified before multiplying by factoring out common factors in the numerator and denominator.
Why are fractions difficult for students?
Fraction division is a tough concept because most students divide into a set number of groups. Dividing into “1/2 of a group” is hard to visualize. (Which is why many students make the mistake of
dividing by two instead). But groups of 1/2 make more sense.
Unlike Fractions:
Fractions having different denominators are called unlike fractions.
For example, 1/3, 4/5, 3/7, 5/9, 7/12 etc. are all unlike fractions.
How many sums does exercise 7.2 of class 6 Maths contain?
Exercise 7.2 of class 6th Maths contains 2 examples and 3 questions. In question 1, students have to draw the number line first and then locate the given points on them. In question 2, students have
to express the given fractions as mixed fractions. In question 3, students have to express the given mixed fractions as improper fractions.
Which problems of exercise 7.2 of class 6th Maths are of the same kind?
Exercise 7.2 of class 6th Maths contains 2 examples and 3 questions. Example 1 and question 2 are of the same type. Example 2 and question 3 are of the same kind. Question 1 is unique.
How many hours are needed to prepare exercise 7.2 of class 6th Maths for the exams?
A maximum of 1 hour is needed to prepare exercise 7.2 of class 6th Maths for the exams because exercise 7.2 has only 3 questions and 2 examples. Exercise 7.2 of class 6th Maths is a very easy
exercise. Students can rapidly finish this exercise.
How to score good marks in exercise 7.2 of grade 6 Maths?
In exercise 7.2 of 6th standard Maths, there are only 3 questions and 2 examples. All questions and examples of this exercise are important. So, to score good marks in exercise 7.2 of grade 6th Maths
students should practice all problems of this exercise seriously.
Last Edited: April 14, 2023 | {"url":"https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-7/exercise-7-2/","timestamp":"2024-11-11T06:34:37Z","content_type":"text/html","content_length":"250755","record_id":"<urn:uuid:2e769145-33d7-4c71-9683-bb6bf3ffd3ab>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00725.warc.gz"} |
Open Problems in Automata Theory
We consider the reachability problem for one-counter automata. In this model, we have a finite set of states and a single counter with values over the natural numbers. Each transition can change the
current state of the automaton as well as increment or decrement the current value of the counter by some number, encoded in binary. The reachability problem for this model is known to be
NP-complete. However, the exact running time of the best possible deterministic algorithm for this problem is unknown.
Similar to the theory of NP-completeness which lets us prove that various problems do not admit polynomial-time algorithms under some plausible assumptions, in recent years, a theory of fine-grained
complexity has emerged which lets us pinpoint the exact running time of various problems under plausible hypotheses. It would be interesting to see the applicability of techniques from this field for
the reachability problem for one-counter automata.
24.13 Satisfiability of String Constraints with Subsequence relation
By \(\le\) we denote the (scattered) subsequence relation between words. Example: \(aba \le baabbabbb\). Fix a finite alphabet \(A\). Let \(X\) be a set of string variables. A subsequence constraint
is an expression of the form \(x \le \alpha \) where \(x \in X\) and \(\alpha \in X^*\). A homomorphism \(h: X^\ast \to A^*\) satisfies a subsequence constraint \(x \le \alpha \) if \(h(x) \le h(\
alpha)\). A domain restriction constraint \(d: X \to RE(A)\) restricts the domain of possible values for a variable \(x\) to the regular language given by the regular expression \(d(e)\). We are
interested in the decidability of the following problem:
Input: A set \(C\) of subsequence constraints and a domain restriction \(d\). Question:Is there a homomorphism \(h\) that respects the domain restriction \(d\) and satisfies every subsequence
constraint in \(C\)?
24.14 Extending asynchronous subtyping for binary session types
ession types are constructs to define protocol interactions and automatically verify if an implementation meets its specifications. They extend data types to describe communication behaviour,
specifying which types of messages can be sent or received and in what order. Session types are often represented by communicating automata, i.e. finite-state automata with FIFO channels or buffers
to enable communication between them.
A key challenge in using session types is ensuring that one component in a distributed system can be replaced by another without breaking the overall protocol. This issue is addressed by session
subtyping, which is a preorder relation on session types. Formally, \(S'\) is considered a subtype of \(S\), written as \(S' \leq S\), if a program with type \(S\) can be safely replaced by a program
with type \(S'\).
For synchronous communication, the session subtyping only needs to ensure that the subtype implements fewer internal choices (\(sends\)), and more external choices (\(receives\)), than its supertype.
However, for asynchronous communication, the reordering of messages is permitted, i.e. the subtype can anticipate \(send\) actions under the condition that all pending \(receives\) are eventually
executed. This takes advantage of the asynchronous model and, hence, is more general. However, it was shown that deciding if two programs are subtypes under the asynchronous relation is undecidable.
24.11 Membership of a coverability VASS in FO2
A VASS of dimension \(d\) is an automaton where the transitions from states to states are given with a vector \(\bar{v}\in\mathbb{Z}^d\), meant to be added coordinate-wise to the current counter. A
transition can only be taken if it keeps each coordinate non-negative. A language of words can either be recognised by reachability (one reads the word from an initial configuration \(\langle q_0,\
bar{v}_0\rangle\) to a final one) or by coverability (one reads the word from an initial configuration to any configuration \(\langle q,\bar{v}\rangle\) which dominates coordinate-wise a specified
We would like to investigate the following class of problems, for F being a class of regular languages: is the language recognised by a given VASS in F?
In most cases, the problem is undecidable, but for different reasons, depending on how the VASS is recognised. If the VASS is recognised by reachability, the problem is undecidable as soon as the
formalism F admits some very basic closure properties, hence there is little hope for obtaining anything interesting. However, if the VASS is recognised by coverability, then the problem is
undecidable as soon as the class F contains LT (for Locally Testable), the class of languages definable in terms of prefixes, infixes, and suffixes. This leaves the problem open for F being for
instance the well-studied class FO2 (First-Order Logic with two variables), or PTL (the class of languages being peacewise testable, meaning which are defined in terms of subwords).
24.5 Properties of the value function in weighted timed games
Weighted timed games (WTGs for short) are two-player zero-sum games played in a timed automaton equipped with integer weights into transitions and locations. In a turn-based fashion, the current
player chooses the next delay and transition. We consider optimal reachability objectives, in which one of the players, whom we call Min, wants to reach a target location while minimising the
cumulated weight given by the sequence of transitions firing and the time spent in each location. Its opponent, called Max, has the opposite objective, i.e. avoids the target or maximises the
accumulated weight needed to reach the target.
This allows one to define the value of a WTG as the minimal weight Min can guarantee whatever Max does. In a WTG, the value for a given location is a function according to the valuation of the clocks
when the first player starts to play.
While knowing if Min has a strategy to guarantee the value lower than a given threshold is known to be undecidable (with two or more clocks), several conditions, one of them being the divergence or
one-clock WTGs have been given to recover decidability. Decidability proof relies on the properties of the value: it is a piecewise affine function characterise as a fixpoint of a operator that
locally chooses the best option for each player.
I propose to study this function for all WTGs: for which class of WTGs, the value function is continuous and/or a fixed point of the local operator? In particular, I propose to start with WTGs that
only use non-negative weights.
24.3 Bi-reachability in Petri nets with ordered data
Petri nets with ordered data is an extension of Petri nets where tokens carry values from the set of rational numbers \(\mathbb{Q}\), and executability of transitions is conditioned by inequalities
between data values. More precisely, every arc is labeled by a finite number of variables corresponding to the data values of tokens, and every transition is labeled by a first-order formula over a
signature \(\{\leq\}\), where variables are from incident arcs. Data values of tokens produced and consumed by a transition must satisfy that formula. A configuration (marking) of a Petri net is a
function that assigns to every place a finite multiset of rational numbers (data values carried by the tokens on this place). A configuration \(q'\) is reachable from a configuration \(q\) if there
is a sequence of transition firings that leads from \(q\) to \(q'\).
Question: Is the following problem decidable?
• Input: a Petri net with ordered data and its two configurations \(q, q'\).
• Question: is \(q\) reachable from \(q'\) and \(q'\) reachable from \(q\)?
24.4 Language Equivalence between Nondeterministic and Deterministic One-Counter Machines
Problem A: Given a deterministic one-counter automaton (a pushdown automaton with one stack alphabet) \(A\) and a nondeterministic one-counter net (a one-counter automaton with no zero-tests) \(N\),
decide if the language recognised by $A$ is the same as the language recognised by \(N\).
Decidability of A is open. This is related to the open question of decidability for the following problem, which is relatively more well-known.
Problem B: Given a deterministic one-counter net \(D\) and a nondeterministic one-counter net \(N\), decide if the language recognised by \(D\) is the same as the language recognised by \(N\).
24.2 Positional Nash Equilibria
We consider multi-player infinite duration games on graphs. A game is given by a directed graph, with vertices partitioned in $k$ sets; Player $i$ controls vertices in the ith set. Edges are coloured
with colours in a set $C$, and each player has an objective $W_i \subseteq C^\omega$. A strategy profile is an array of $k$ strategies, $\bar{\sigma} = (\sigma_1,\dots, \sigma_k)$, one for each
player. It is a Nash equilibrium if no player has an incentive to change their strategy, that is: If $\bar{\sigma}$ produces a losing outcome for Player $i$, then Player $i$ cannot modify his
strategy and obtain a winning outcome.
It is known that, under some mild hypothesis on the objectives $W_i$, every game admits some Nash equilibrium. The proofs for this fact usually build a NE where strategies use infinite memory, even
for games with very simple objectives.
Question: Do all reachability/Büchi/parity games admit a Nash equilibrium in which all strategies are positional?
2024.1 How is $a^* = 1/(1-a)?$
Consider the language \(a^*\). We have
a^* &{}= \varepsilon + a + aa + aaa + \ldots\\ &{}= 1 + a + a^2 + a^3 + \ldots~, \tag{since \(\varepsilon\) is the unit of concatenation}\\ &{}= \sum_{k = 0}^\infty a^k
in which we now immediately recognize the familiar \emph{geometric series} whose closed form is well known to be \(\frac{1}{1 - a}\). In that sense,
a^* = \frac{1}{1 - a}~.
23.6 Target for pushdown RBN
Reconfigurable Broadcast Networks (RBN) are a model for large groups of identical agents communicating via unreliable broadcast. A pushdown RBN (PRBN) is simply one where each agent is modeled by a
pushdown transition system. A PRBN on a message alphabet \(\mathcal{M}\) is described by a pushdown automaton \(\mathcal{A}\) over the alphabet \( \{br(m), rec(m) | m \in \mathcal{M} \}\).
Configurations of \(\mathcal{A}\), i.e., pairs \((q, \sigma)\) of state and stack content, are called local configurations.\(\\ \)
A configuration of the PRBN is a function from a finite set of agents \(\mathbb{A}\) to local configurations. A run starts with an arbitrarily large set of agents, all in the initial state with an
empty stack . A step consists of one agent taking a transition with a broadcast \(br(m)\), and each other agent either not moving or taking a transition \(rec(m)\) (in other words, one agent
broadcasts and other agents non-deterministically receive the broadcast or not).\(\\ \) | {"url":"https://automata.exchange/page/2/","timestamp":"2024-11-10T02:19:49Z","content_type":"text/html","content_length":"64592","record_id":"<urn:uuid:69e4bebf-a3d3-4faf-b34e-77ab663836dc>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00663.warc.gz"} |
Solving Triangle Area Computation Assignments in Assembly Language
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Key Topics
• Understanding the Assignment
□ Using Syscalls for Input and Output
□ Implementing Mathematical Functions
□ Structuring Your Program
• Building and Linking the Program
□ Compiling Assembly Code
□ Compiling C/C++ Code
□ Linking the Object Files
• Conclusion
Assembly language programming can seem daunting at first, but with practice and understanding, it becomes a valuable skill. This blog aims to guide you through the process of tackling assignments
involving computation in assembly language, specifically those requiring the use of syscalls for I/O operations and custom mathematical functions like sin(x). By breaking down complex problems into
manageable steps, you can gain confidence and proficiency in assembly language. You'll learn how to use syscalls to handle input and output, ensuring that your program can interact with users and
display results effectively. Additionally, implementing mathematical functions, such as the sine function using the Taylor series expansion, will deepen your understanding of both mathematics and
low-level programming. These skills are not only crucial for completing specific assignments but also provide a solid foundation for understanding how computers execute instructions at the hardware
level. As you progress, you'll find that the seemingly cryptic syntax of assembly language becomes more intuitive, enabling you to write more efficient and optimized code. This blog will provide
practical examples and detailed explanations to help you master these concepts, making assembly language a powerful tool in your programming repertoire. With dedication and practice, you'll be able
to tackle any assembly language assignment with confidence. Mastering these techniques will not only help you complete your assembly language assignment efficiently but also pave the way for advanced
learning and application in computer science.
Understanding the Assignment
Programming Assignments that require you to compute the area of a triangle in assembly language typically involve several key tasks. You need to use syscalls for input and output operations, and you
must implement mathematical functions such as the sine function directly in assembly. This section will break down the assignment into smaller, manageable steps.
Using Syscalls for Input and Output
Syscalls, or system calls, are a way for your program to interact with the operating system. In assembly language, syscalls are used to perform input and output operations. This is crucial for any
program that requires user interaction or needs to display results.
Syscalls for Outputting Strings
To output strings in assembly language, you need to use the sys_write syscall. Here's a basic example of how to output a string:
section .data msg db 'Hello, World!', 0 ; Null-terminated string section .text global _start _start: mov eax, 4 ; syscall number for sys_write mov ebx, 1 ; file descriptor 1 (stdout) mov ecx, msg ;
pointer to message mov edx, 13 ; message length int 0x80 ; call kernel mov eax, 1 ; syscall number for sys_exit xor ebx, ebx ; exit code 0 int 0x80 ; call kernel
In this example, the message "Hello, World!" is stored in the data section. The syscall for writing (sys_write) is invoked by moving the appropriate values into the registers: eax for the syscall
number, ebx for the file descriptor, ecx for the message pointer, and edx for the message length. Finally, an interrupt (int 0x80) is called to execute the syscall.
Syscalls for Reading Input
Reading input in assembly is handled by the sys_read syscall. Here's an example:
section .bss buffer resb 128 ; buffer for input section .text global _start _start: mov eax, 3 ; syscall number for sys_read mov ebx, 0 ; file descriptor 0 (stdin) mov ecx, buffer ; pointer to buffer
mov edx, 128 ; buffer size int 0x80 ; call kernel ; Additional code to process input mov eax, 1 ; syscall number for sys_exit xor ebx, ebx ; exit code 0 int 0x80 ; call kernel
In this code, a buffer is reserved in the bss section to store the input. The sys_read syscall reads from the standard input (stdin) into this buffer. The same interrupt mechanism (int 0x80) is used
to execute the syscall.
Combining Input and Output
Combining both reading and writing operations allows you to create interactive programs. For example, you can prompt the user for input and then display a response based on the input received. This
is crucial for assignments where user interaction is required.
Implementing Mathematical Functions
For tasks that require mathematical computations, such as computing the sine of an angle, you can implement these functions directly in assembly. Using the Taylor series expansion is a common method
for approximating functions like sin(x).
The Taylor Series Expansion
The Taylor series expansion for sin(x) is: sin(x)=x−x33!+x55!−x77!+⋯\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdotssin(x)=x−3!x3+5!x5−7!x7+⋯
This series provides a way to approximate the sine function to a desired level of precision by summing a finite number of terms.
Implementing Sine Function in Assembly
To implement the sine function in assembly, you start by initializing the result and iteratively calculating each term of the series. Here's a simplified version:
section .data pi dd 3.14159265358979323846 precision dd 20 section .text global sin sin: ; Assume x is passed in st0 ; Initialize result fldz ; st0 = 0 (result) ; Perform the Taylor series
calculation ; (For simplicity, only a few terms are shown here) fld st1 ; st0 = x, st1 = x fadd ; st0 = x + 0 (initial term) ; Next term - x^3/3! fld st1 ; st0 = x, st1 = x, st2 = x fmul st0, st1 ;
st0 = x^2, st1 = x, st2 = x fmul st0, st1 ; st0 = x^3, st1 = x, st2 = x fidiv dword [factorial_3] ; st0 = x^3/3!, st1 = x, st2 = x fsub ; st0 = x - x^3/3! ; Continue adding more terms for higher
precision ; Return result in st0 ret section .data factorial_3 dd 6 ; Add more factorials for higher terms
This example shows the basic structure for calculating sin(x) using the Taylor series. You can extend this to include more terms for higher precision.
Structuring Your Program
Dividing your program into distinct modules helps manage complexity and improves readability. Here’s how you can structure your program for computing the area of a triangle:
Director Module
The Director module, often written in C or C++, handles the main workflow and user interaction. It prompts the user for input, calls the Producer module for computation, and displays the results.
Producer Module
The Producer module, written in assembly, handles the core logic. It uses syscalls for input and output, calls the sine function, and computes the area of the triangle. The formula for the area of a
triangle given two sides aaa and bbb and the angle CCC between them is: Area=12absin(C)\text{Area} = \frac{1}{2}ab \sin(C)Area=21absin(C)
Sine Function Module
The Sine Function module, also in assembly, implements the sine function using the Taylor series expansion. This module is called by the Producer module to compute the sine of the given angle.
Building and Linking the Program
Once you have your modules, you need to compile and link them. This can be done using tools like nasm for assembling the assembly code and gcc or g++ for linking.
Compiling Assembly Code
Use nasm to compile your assembly modules:
nasm -f elf32 producer.asm -o producer.o nasm -f elf32 sin.asm -o sin.o
Compiling C/C++ Code
If your Director module is in C or C++, compile it using gcc or g++:
gcc -m32 -c director.c -o director.o
Linking the Object Files
Link the object files together to create the final executable:
gcc -m32 director.o producer.o sin.o -o triangle_area
This command links the object files and produces an executable named triangle_area.
Mastering assembly language programming requires breaking down problems into manageable pieces and understanding core concepts like syscalls and mathematical computations. By practicing regularly,
you will find that working with assembly becomes more intuitive over time. Don't hesitate to seek help from resources or professionals if you encounter difficulties. Remember, persistence and
consistent effort are key to success in assembly programming.
For additional programming assistance, visit ProgrammingHomeworkHelp.com. With dedication and the right support, you'll become proficient in tackling complex assignments.
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Quantum Scars and Caustics in Majorana
SciPost Submission Page
Quantum Scars and Caustics in Majorana Billiards
by R. Johanna Zijderveld, A. Mert Bozkurt, Michael Wimmer, İnanç Adagideli
Submission summary
Authors (as registered SciPost users): Inanc Adagideli · A. Mert Bozkurt · Michael Wimmer · Johanna Zijderveld
Submission information
Preprint Link: https://arxiv.org/abs/2312.13368v3 (pdf)
Code repository: https://zenodo.org/records/10404706
Date accepted: 2024-10-28
Date submitted: 2024-07-25 10:57
Submitted by: Zijderveld, Johanna
Submitted to: SciPost Physics
Ontological classification
Academic field: Physics
Specialties: • Condensed Matter Physics - Theory
Approaches: Theoretical, Computational
We demonstrate that the classical dynamics influence the localization behaviour of Majorana wavefunctions in Majorana billiards. By using a connection between Majorana wavefunctions and
eigenfunctions of a normal state Hamiltonian, we show that Majorana wavefunctions in both p-wave and s-wave topological superconductors inherit the properties of the underlying normal state
eigenfunctions. As an example, we demonstrate that Majorana wavefunctions in topological superconductors with chaotic shapes feature quantum scarring. Furthermore, we show a way to manipulate a
localized Majorana wavefunction by altering the underlying classical dynamics using a local potential away from the localization region. Finally, in the presence of chiral symmetry breaking, we find
that the Majorana wavefunction in convex-shaped Majorana billiards exhibits caustics formation, reminiscent of a normal state system with magnetic field.
Author indications on fulfilling journal expectations
• Provide a novel and synergetic link between different research areas.
• Open a new pathway in an existing or a new research direction, with clear potential for multi-pronged follow-up work
• Detail a groundbreaking theoretical/experimental/computational discovery
• Present a breakthrough on a previously-identified and long-standing research stumbling block
Author comments upon resubmission
We thank both referees for their overall positive evaluation. We have implemented all the requested changes. We also provide the redlined manuscript with the changes highlighted at this URL: https://
Ref. 1:
(1) The authors mention the Heller type scarring, and also refer to the many-body-scars. However, in this context, the authors could consider the perturbation-induced scarring [Phys. Rev. Lett.
123, 214101 (2019); J. Phys. Condens. Matter 31, 105301 (2019); Phys. Rev. B 96, 094204 (2017)] as well. As I see, there should be a very straightforward way to generalize the Majorana
description to this type of scarring by considering locally perturbed p-type superconductors.
We thank the referee for pointing out these references. Indeed, perturbation-induced scarring can also be relevant for p-wave topological superconductors. We have added these references in the
introduction section of our paper.
(2) The Hamiltonian for a topological superconductor is given, but it would be beneficial if the context behind this Hamiltonian would be described in more detailed, at least to give a citation.
We thank the referee for their comment. Following their suggestion, we have added a reference to p-wave topological superconductor Hamiltonian we use in Eq. (1).
(3) The authors refer to the potential of the system, but it is not defined clearly. What I have deduced is that they consider various billiard systems with hard wall boundaries where the
potential is zero.
We thank the referee for pointing out this possible source of confusion. We have now clarified what we meant by potential.
(4) It is unclear what the authors mean by the semiclassical limit. As far I see, all the simulations and analysis are fully quantum, and close to the ground state. Furthermore, the authors
emphasize the topological nature of the superconductor, but not clarify if it has an impact the observed scarring, for instance whether the scars are chiral also (based on my conclusion, they are
Here, by semiclassical we mean that the system size is much larger than the wavelength of the electrons, which is set by the chemical potential values we use. We have now added a sentence that
clarifies this point. About the second point raised by the referee, indeed these scars are not chiral, however, we believe this would be an interesting future direction of research.
(5) The authors consider how a hard disk stopper affects the scarred states, and mention this as a possible experimental avenue. However, a realistic STM nanotip would instead produce a soft
bump. Nevertheless, I don't see this modification to change their conclusion.
The referee is right in pointing out that an STM nanotip would produce a softer potential. It is also true that this modification does not change the conclusion that the edge modes are affected by
the potential produced by the potential far away from the edge.
(6) Finally, the manuscript mention multiple times the chaotic behavior of the system. However, no studies is carried out or shown, such as Poincare's surface of sections, or level statistics. In
particular, I suspect the system to be highly mixed in the case when the artificial vector potential is present. For example, the candy wrap shape seen in Fig. 7 is not an periodic in a hard wall
stadium, but it does appear in a smooth stadium that appear more like an elliptical oscillator.
We thank the referee for their comment. Here, we take advantage of the fact that the system shapes we focus on in this manuscript are known to feature chaotic dynamics. Figure 8 (Figure 7 in the
previous version) indeed shows more complicated trajectories because the Majorana wavefunction is actually a superposition of two normal state wavefunctions, as shown in Eq. 12. To emphasize this
point further, we have added the following footnote to our manuscript:
"The Majorana wavefunction for the s-wave topological superconductors is in fact a superposition of two normal state eigenfunctions, see Eq.12. We refer the reader to App. B for more detail."
Ref. 2:
The authors however often stop their analysis to this mapping. For instance in Fig 2, it is clear that (b) is deduced from (a) through the mapping (5) and one can recognize visually the trace of
the periodic orbit shown on (a). But does this really imply localization ? One would like to see a quantitative comparison of, for instance, inverse participation ratio, to see how much of the
localization implied by scarring transfer to the Majorana through (5). 1-Study more quantitatively the localization properties of the scarred wavefunctions, by comparing for instance their
inverse participation ration to the average one.
We thank the referee for their comment. Following referee's recommendation, we have now included a new figure (Figure 3 in the new version) in our manuscript where we show that the inverse
participation ratio for scarred and non-scarred wavefunctions at a finite $\Delta_x$ depend non-trivially on the initial state. We aim to quantatively show that the local structure of the underlying
state carries over to localization of the majorana wavefunction.
..This of course would also apply qualitatively to a billiard if the cyclotron radius is much smaller than the typical size of the system. However the regime considered here (weak chiral symmetry
breaking) seems to imply weak effective magnetic field, and I am afraid that what we see might just be the caustics of the unperturbed dynamics. This in any case has to be clarified. 2 - Provide
information about the cyclotron radius for the effective magnetic field in Fig 5 and 6 and discuss how much this effective magnetic field modifies the classical dynamics. More generally show
which classical trajectories (and thus which caustics) are involved in these figure.
We thank the referee for their comment. We have checked the cyclotron radius of the effective magnetic field and observed that there is no direct relation to the size of the causitcs we find.
Although our theory is valid for small $\Delta_i$, our mapping deviates when $\Delta_i$ is larger, where the cyclotron radius becomes comparable to the system size. On the other hand, the appearance
of caustics depends on having convex shapes and breaking chiral symmetry, which means both $\Delta_x$ and $\Delta_y$ must be non-zero. To illustrate this, we have replaced the previous Figure 5 with
a new figure (now Figure 6), showing that caustics appear even at very low $\Delta_y$ for a specific non-zero $\Delta_x$. Based on these observations, we have revised the text to explain that
caustics are caused by chiral symmetry breaking, not the effective magnetic field.
3- Discuss the connection (or absence of connection) between the phenomenology observed in Fig. 3 & 4 and Aharonov-Bohm effect.
We thank the referee for their comment. Despite being in a similar spirit with Aharonov-Bohm effect, Figures 4 and 5 (previously Figures 3 and 4) do not feature an orbital field (or effective orbital
field that is present in the chiral symmetry broken case). Here, we consider topological superconductors with chiral symmetry. The phenomenology observed in these figures are due to mixing of the
momentum states as a a result of the scatterer at the center of the billiard.
4- Provide more discussion about the importance, for topological superconductor physics, of the kind of localization discussed in the paper.
We thank the referee for their remark. In addition to the discussion we provide in our introduction and conclusion part, we now provide a discussion about importance of the ballistic localization on
the localization properties of Majorana wavefunctions in the conclusion part of our manuscript. Specifically, we have added the following sentence in our manuscript:
"In addition to the pairing potential, ballistic chaotic localization also affects the local profile of the Majoranas, which can be of extreme importance in experimental setups in finitely sized
topological superconductors."
a- As the Majorana wavefunctions are 1/2 spinors, it might be useful to specify to what correspond the scalar functions plotted in the various figures.
We thank the referee for pointing out this lack of information. We have now included a description of the density of the wavefunctions in caption of Figure 1: $\rho(\mathbf{r}) = \Psi^\dagger(\mathbf
b- Figure 2c is a rather trivial consequence of the the mapping (5). Is this really useful ? In any case putting it next to Fig 2a&b could give the wrong impression that it's a check of the
localization of the wavefunction (due to scarring), and is thus slightly misleading.
We thank the referee for their remark. Our purpose of having Figure 2c is two-fold. Firstly, it demonstrates that our theory and numerical simulations match. Secondly, it serves as an example that
shows the effect of the pairing potential $\Delta_x$ on the Majorana wavefunctions. To avoid any confusion, we have modified the text referring to Fig. 2c:
"Fig.2c) shows the localization of the Majorana wavefunction as a function of the pairing potential $\Delta_x$. To this end, we plot the logarithm of the normalized overlap between the Majorana
wavefunction at $\Delta_x=0$ and the Majorana wavefunction at finite $\Delta_x$ for different $x$ values in the stadium billiard:"
c- Just before Eq (9) : why could we set \Delta_x = \Delta_y= \Delta "without loss of generality" ?
We thank the referee for their comment. Here, we mean that even if $\Delta_x\neq\Delta_y$, our mapping still holds. For completeness, we have now extended our analytical calculation to include $\
Delta_x\neq\Delta_y$ and wrote a new appendix (Appendix A in the new version) that includes the details of the calculation.
List of changes
Difference file with list of changes included above.
Current status:
Accepted in target Journal
Editorial decision: For Journal SciPost Physics: Publish
(status: Editorial decision fixed and (if required) accepted by authors)
Reports on this Submission | {"url":"https://scipost.org/submissions/2312.13368v3/","timestamp":"2024-11-13T16:04:26Z","content_type":"text/html","content_length":"49038","record_id":"<urn:uuid:d15c6a14-af1a-46da-90fd-eaa29042c349>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00799.warc.gz"} |
23.1: Math Talk: Evaluating Functions (5 minutes)
The purpose of this Math Talk is to elicit strategies and understandings students have for evaluating the value of functions given an input value. These understandings help students develop fluency
and will be helpful later in this lesson when students will need to be able to evaluate functions to make use of the vertex form of a quadratic.
Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk.
Follow with a whole-class discussion.
Student Facing
Mentally evaluate each of the functions when \(x = 3\).
\(f(x) = x^2 - 4x + 1\)
\(g(x) = 6x - 2x^2\)
\(h(x) = (x-4)(x-3)\)
\(j(x) = 2(x-1)(x+2)\)
Activity Synthesis
Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:
• “Who can restate \(\underline{\hspace{.5in}}\)’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to \(\underline{\hspace{.5in}}\)’s strategy?”
• “Do you agree or disagree? Why?”
23.2: Comparing Functions (15 minutes)
In this activity, students compare the values of a function with different \(x\)-coordinates using equations and graphs. In the associated Algebra 1 lesson, students use the vertex form of a
quadratic equation to find the maximum or minimum. This work supports students by focusing on the values of the equation and determining which is greater.
Display the equation and graph for all to see.
Ask students how they can find the value of \(f(2)\) using the equation and the graph.
Student Facing
The notation \(f(2)\) means the output of function \(f\) when \(x\) is 2. For each function, determine whether \(f(2) > f(3)\), \(f(2) < f(3)\), or \(f(2) = f(3)\).
1. \(f(x) = x^2 + 2x+ 3\)
2. \(f(x) = (x-2)(x-3)\)
3. \(f(x) = \text{-}x^2 + 5\)
Activity Synthesis
The purpose of the discussion is to find methods for comparing function values using equations and graphs. Select students to share their solutions and methods for comparing the values. Ask students,
• “Is it easier to use the graph or equation to determine which value is greater using the equation or the graph?” (For some functions, it was easier to use the graph since one value is clearly
higher on the graph, but for other functions, it is difficult to see in the graph, so finding the exact values using the function is easier.)
• “Invent an equation for which \(f(2) > f(3)\). How did you come up with your function?” (The equation \(f(x) = \text{-}x\) has \(f(2) > f(3)\). I wanted an equation that is decreasing so that
moving left to right along the graph would have lower values. I thought this equation would be simple and decreasing like I wanted.)
23.3: Finding the Vertex (20 minutes)
In this activity, students convert functions in factored or standard form to vertex form. In the associated Algebra 1 lesson, students use the vertex form to find the maximum or minimum of the
quadratic function. This activity supports students by focusing on the mechanics of changing forms and finding the coordinates of the vertex. Students look for and make use of structure (MP7) when
they use an equation to identify the vertex of a graph.
Student Facing
Write each function in vertex form, then find the coordinates of the vertex.
1. \(y = x^2 - 4x + 7\)
2. \(y = (x-1)(x+3)\)
3. \(y = (x-2)(x+2)\)
4. \(y = x^2 - 2x + 1\)
5. \(y = \text{-}x^2 -2x-6\)
6. \(y = 2x^2 - 12x + 22\)
Activity Synthesis
The purpose of the discussion is to highlight methods used to rewrite the equations in vertex form. Select students to share their solutions. Ask students,
• “When the coefficient of \(x^2\) is not _____ , what do you have to do differently?” (The coefficient needs to be factored out first and distributed back in after completing the square.)
• “After completing the square, when one of the terms seems to be missing, like \(y = x^2 - 8\) or \(y = (x-4)^2\), what does that mean about the vertex?” (It means it is on one of the axes. In the
first example, the vertex is \((0,\text{-}8)\) and in the second example, the vertex is \((4,0)\). One of the coordinates is zero in these cases, which puts the vertex on an axis.) | {"url":"https://curriculum.illustrativemathematics.org/HS/teachers/4/7/23/index.html","timestamp":"2024-11-04T11:37:37Z","content_type":"text/html","content_length":"101912","record_id":"<urn:uuid:83d51dca-96ca-4bde-b48e-5ebc806fc1f5>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00514.warc.gz"} |
Practice Set 7.1 Geometry 9th Standard Maths Part 2 Chapter 7 Co-ordinate Geometry Solutions Maharashtra Board
9th Standard Maths 2 Practice Set 7.1 Chapter 7 Co-ordinate Geometry Textbook Answers Maharashtra Board
Class 9 Maths Part 2 Practice Set 7.1 Chapter 7 Co-ordinate Geometry Questions With Answers Maharashtra Board
Question 1.
State in which quadrant or on which axis do the following points lie.
i. A(-3, 2)
ii. B(-5, -2)
iii. K(3.5, 1.5)
iv. D(2, 10)
V. E(37, 35)
vi. F(15, -18)
vii. G(3, -7)
viii. H(0, -5)
ix. M(12, 0)
x. N(0, 9)
xi. P(0, 2.5)
xii. Q(-7, -3)
Question 2.
In which quadrant are the following points?
i. whose both co-ordinates are positive.
ii. whose both co-ordinates are negative.
iii. whose x co-ordinate is positive and the y co-ordinate is negative.
iv. whose x co-ordinate is negative and y co-ordinate is positive.
i. Quadrant I
ii. Quadrant III
iii. Quadrant IV
iv. Quadrant II
Question 3.
Draw the co-ordinate system on a plane and plot the following points.
L(-2, 4), M(5, 6), N(-3, -4), P(2, -3), Q(6, -5), S(7, 0), T(0, -5)
Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Practice Set 7.1 Intext Questions and Activities
Question 1.
Plot the points R(-3,-4), S(3,-l) on the same co-ordinate system. (Textbook pg. no. 93)
Steps for plotting the points:
i. Draw X-axis and Y-axis on the plane. Show the origin.
ii. Draw a line parallel to Y-axis at a distance of 3 units in the -ve direction of X-axis.
iii. Draw another line parallel to X-axis at a distance of 4 units in the -ve direction of Y-axis.
iv. Intersection of these lines is the point R (-3, -4).
v. The point S can be plotted in the same manner. | {"url":"https://www.learncram.com/maharashtra-board/class-9-maths-solutions-part-2-chapter-7-practice-set-7-1/","timestamp":"2024-11-13T23:08:29Z","content_type":"text/html","content_length":"65015","record_id":"<urn:uuid:350e154a-d620-46a6-84ca-a3cdf1ddb94c>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00540.warc.gz"} |
You just started working full-time earning 100000 per year, Financial Management
Time Value of Money
You just started working full-time, earning $100,000 per year. Your goal is to have $5 million in your 401(k) plan by your 61st birthday (i.e., 40 years from today). Assume 3% inflation per year. If
you can earn 10% per year annualized in an S&P 500 mutual fund, after all expenses, inside a 401(k) with a dollar for dollar match up to 10% of your income, how much would you need to save each month
to have that $5 million:
a) If that $5 million is in future nominal dollars;
b) If that $5 million is in today’s equivalent purchasing power future dollars;
c) If your marginal tax rate is 34% federal plus 7% state, what would the after-tax cost of your investments be for (i) and (ii) if you relied on the employer match for of your monthly contributions?
d) In retirement, you plan to draw $10,000 per month of principal from your investments. If you were 100% invested in stock mutual funds (like the S&P 500), would you get more or less than the rate
of return on the S&P 500, and why?
Request for Solution File
Ask an Expert for Answer!!
Financial Management: You just started working full-time earning 100000 per year
Reference No:- TGS01039344
Expected delivery within 24 Hours | {"url":"https://www.tutorsglobe.com/question/you-just-started-working-full-time-earning-100000-per-year-51039344.aspx","timestamp":"2024-11-03T10:10:13Z","content_type":"text/html","content_length":"44628","record_id":"<urn:uuid:97d71231-92ac-43c4-8969-e9b6caebd77d>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00755.warc.gz"} |
Adding and Subtracting Fractions
Adding and Subtracting Fractions with the Same Denominator
PROCEDURE: To add fractions with the same denominator, add the numerators and put the sum over the original denominator. To subtract fractions with the same denominator, subtract the numerators and
put the difference over the original denominator.
Add the numerators, and put the sum over the original denominator:
Subtract the numerators and put the difference over the original denominator:
Adding and Subtracting Fractions with Different Denominators
Sometimes you have to add or subtract fractions that have different denominators. To do this, you first need to rewrite your fractions so that they DO have the same denominator. Figuring out the
least common denominator (LCD) of your fractions is the first step.
PROCEDURE: To find the least common denominator of two fractions, find the least common multiple of the denominators. In other words, look at the multiples of the numbers, and find out which they
have in common. The common multiple with the lowest value is your LCD.
What is the LCD of
Step 1: List the multiples of 4.
(4 × 1) × 4, (4 × 2) × 8, (4 × 3) × 12, (4 × 4) × 16, etc.
Step 2: List the multiples of 3.
(3 × 1) × 3, (3 × 2) × 6, (3 × 3) × 9, (3 × 4) × 12, etc.
The least common denominator of 12.
Putting the LCD to Work
Now that you know how to find the LCD, you are all set to add and subtract fractions with different denominators. Follow the steps below to see how to use the LCD to add and subtract fractions with
different denominators.
PROCEDURE: To add or subtract fractions with different denominators, first find the LCD of the two fractions. Then determine the factor that each denominator is of that LCD. Multiply both the
numerator and the denominator by those factors so that the fractions have the same denominator. Then add or subtract the numerators.
Step 1: Find the LCD.
(2 × 1) × 2, (2 × 2) × 4, (2 × 3) × 6, (2 × 4) × 8, (2 × 5) × 10, etc.
(5 × 1) × 5, (5 × 2) × 10, etc.
The LCD is 10 .
Step 2: Determine the factor that each denominator is of the LCD.
Because 2 × 5 × 10, 5 is the factor of 2 .
Because 5 × 2 × 10, 2 is the factor of 5 .
Step 3: Multiply the factors of the LCD by the fractions.
Step 4: Add the fractions. | {"url":"https://polymathlove.com/adding-and-subtracting-fractions-1.html","timestamp":"2024-11-06T09:25:46Z","content_type":"text/html","content_length":"105533","record_id":"<urn:uuid:81fb469c-4781-4680-96d5-b503b7499736>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00071.warc.gz"} |
Opinions?: Greatest Craps Guru
I'm curious about people's opinions on the strategies presented in Mark Jackson and David Medansky's book, Greatest Craps Guru in the World. (If you're curious about the book it's available for free
on KindleUnlimited.)
A lot of the strategies presented make me nervous because they are so different from recommendations I've read on the Wizard's site or other craps books. They also use some terms like "iron cross"
which for many is a huge trigger and red flag. Please try not to be reactionary; take the time to read the whole post and strategy. I'm not looking for knee-jerk snark, but constructive thoughts and
Premise: Jackson and Medansky are clear that their strategies are not for the purpose of playing craps for entertainment, but for the explicit purpose of making money. To them, that means that play
time may be short and the style is very different from players who are used to playing for hour after hour. They recommend a starting bankroll of $5k and buy-in of $1,500. With this bankroll size
your goal is to be up $300/day, which is made over 3 short, separate playing sessions. If at any one of those sessions you lose more than $875 you stop and walk away for the day. If in any one of
those sessions you hit your $300 goal you stop and walk away for the day. Much of the book is dedicated to discussing discipline and hard stops after reaching specific, reasonable win goals and also
hard stop goals for losses. They are clear in pointing out that you will not win every day. Much of the philosophy around limiting yourself to 3 short sessions per day seems to be about reinforcing
self-control so that you don't go overboard and get wiped out with greed and pressing bets nor from indefinitely chasing losses until ruin. I think It's also based on the idea that over long periods
of play the house edge will always eat at you, so short periods lead to better long term outcomes. (I'm not sure if that last point is actually true.)
Betting: For each of the 3 daily sessions, you don't bet on the pass line, but wait for the shooter to establish a point then make a modified iron cross bet for at total of $125 ($10 on 4, 9, 10 &
field; $25 on 5; $30 on 6 & 8). You take your bets down after the shooter rolls twice, no matter your winnings from that shooter (based on the idea that 1/6 rolls is a 7, so the longer your bets are
out there the more likely a 7 will hit). Then you repeat this again with a second shooter. If you've won, then you should be up about $100. You walk away from the table for the session. You repeat
this again for 2 more sessions in the day.
Point Seven Out: If during the playing session your bet gets wiped away because there is a seven out during one of the two shooter's rolls after you've bet the $125, then you wait until the next
shooter and you double your modified iron cross bet to $250 (think Martingale). If you lose that, then you once again double the bet to $500. If you lose that then you leave the table for the day and
start over tomorrow. The authors write that if you are not comfortable doubling your bet and would like a more conservative approach you can leave it at the base $125 unit until you make up your
losses and hit your session goal or hit your loss cap for the session.
"Cold" table bet: The authors suggest that if the table is running "cold" then on the come-out roll you lay ("no" bet) all of the numbers at $30 each and a $5 horn high yo bet (a total of $185). On a
7, this bet will net you $109. On a point number roll, this bet would lose $35. You leave these bets out for a maximum of two wins or two losses before taking them down.
One, two, two bet: Instead of a pass line bet, a player can bet $1 yo, $2 any craps, and $2 any seven (for a total of $5). The authors rationalize that although this bet won't win if a point number
is established, a $5 pass-line bettor isn't guaranteed to win on a point number even if one is established, and the pass-line bettor is also at risk of losing any additional odds (s)he puts behind
the pass line if a seven comes out before the point is made.
My thoughts: This book is a conundrum for me because although I can recognize several fallacies in it, there are also some ideas that seem to make sense to me. Then again, I've just begun studying
craps theory and I'm not a statistician so I'm far from an expert, and it's very possible that what seems reasonable to me actually doesn't make any sense when you do the math. I'm hoping that some
of you more experienced players can help clarify.
Some things that I realize are false: there's no such thing as a "cold" table since previous roles do not affect the likelihood of future roles. I also don't think it matters if you break your play
into three sessions or not or multiple days or not; it seems to me that that if you follow one consistent strategy and play 20 minutes 3 times per day for 10 days (10 hours) that is no different than
following the same betting strategy but playing 10 hours straight in one day. Am I wrong?
Some things that seem reasonable to me... or not?: The authors prefer the place bets over pass line/come bets because of the ability to take them down after short periods of play. It is true that the
odds of rolling a 7 are 1/6, so it does seem to hold for me that leaving your bets up for shorter periods of time puts you at less risk. However, now that I think about it... the odds of the 7 are
always 1/6. Just because a shooter goes 2 rolls without a 7 doesn't mean that the 7 is any more or less likely to come up on the next roll than it was on the previous two. If you take the bets down
or turn them off and wait for a new shooter before turning them back on then that new shooter still has a 1/6 chance of rolling the 7. Therefore... your money is at no more or less of a risk now then
it was before? So what's the purpose of waiting for a new shooter? (I think this is like above with the 20min x 3 x 10 days vs 10 hour play... it's the same odds no matter how long you spread it
As for the actual bets themselves... I know statistical work ups have been done on the high house advantage of the (modified) iron cross and propositions bets and that the lowest house advantage is
on the pass/don't pass/come/don't come/odds bets. Yet I wonder if there is anything to these authors' particular combination of bets that would be wise to adopt...
• Threads: 123
• Posts: 11442
Joined: Aug 8, 2010
Quote: Xtina
It seems like you already realize it is a steaming pile of ...... You can adopt any of these modes of betting as long as you realize with each and every bet you are more likely to lose money than win
The best "guru" is probably the wiz since he has all the numbers and freely admits that there is no "system". Also the worst, if you want my honest opinion, is out there and has a name that I think a
lot of people might recognize and agree with, but mentioning his name would probably just get me restricted.
The best things in life are not free.
Quote: Xtina
I'm not looking for knee-jerk snark, but constructive thoughts and opinions.
I have sworn off trying to give constructive advice to those who, it turns out, were not actually looking for such. In your case you have come across as sincere. But I fear you are vulnerable to some
really bad ideas, you are showing signs of it. So I'll try to point out things when I think you should be picking up on them, things you will pick up here if you stick around this site and pay
Premise: Jackson and Medansky are clear that their strategies are not for the purpose of playing craps for entertainment, but for the explicit purpose of making money.
I would hope in future, you will immediately know you are going down the wrong road just with this first point. Craps is called a "negative expectation game". Unless you think you can influence dice
throws, we automatically know there is no combination, no system, no combination of bets offered by the House that can change that fact. This is a fact. Likely, the authors know this. You should have
contempt for them.
... short periods lead to better long term outcomes. (I'm not sure if that last point is actually true.)
This is only true if it means you are not letting the house advantage eat away at your bankroll. If you have the same amount of 'total action' one way compared to the other, it doesn't matter.
... the idea that 1/6 rolls is a 7, so the longer your bets are out there the more likely a 7 will hit
It is easy to think a player is making the appearance of a 7-out more and more imminent, however, as long as the past does not matter and the future is random, it can't matter when you place your
bets or when you pick them up.
You walk away from the table for the session.
The dice do not know if you won or lost when you return to the table. This concept that the dice have no memory is one that you need to take to heart. Once it sinks in, it'll be much harder to buy
into these faulty ideas.
Point Seven Out: If during the playing session your bet gets wiped away because there is a seven out during one of the two shooter's rolls after you've bet the $125, then you wait until ...
again, the dice have no memory. There is no point in waiting for anything in order to get the dice to make you well again. The dice don't know. They don't care. If stalling makes you bet less, that
works. Betting less is good.
"Cold" table bet:
A table is only hot or cold in the past. A table cannot be hot or cold at the present or in the future, this hot or cold business does not affect the results of the next roll.
One, two, two bet: Instead of a pass line bet, a player can bet $1 yo, $2 any craps, and $2 any seven (for a total of $5). The authors rationalize that although this bet won't win if a point
number is established, a $5 pass-line bettor isn't guaranteed to win on a point number even if one is established, and the pass-line bettor is also at risk of losing any additional odds (s)he
puts behind the pass line if a seven comes out before the point is made.
Can't you see this is a non-mathematical evaluation? These bets have probabilities and odds, why aren't the authors using these things?
My thoughts: This book is a conundrum for me because although I can recognize several fallacies in it, there are also some ideas that seem to make sense to me.
I'm glad you recognize the fallacies, and they are indeed just trying to appeal to you otherwise by what might sound good in your gut. Believe me they are not introducing a single new idea that
hasn't been thoroughly discredited.
Then again, I've just begun studying craps theory and I'm not a statistician so I'm far from an expert
The thing is, very simple concepts will save you if you internalize them. Do you need to analyze bet combinations? No, because no bet combination can beat the house edge, mathematicians not needed.
Do you need to learn to fine tune when to bet and when to quit? No, because the dice have no memory.
Some things that I realize are false: there's no such thing as a "cold" table since previous roles do not affect the likelihood of future roles. I also don't think it matters if you break your
play into three sessions or not or multiple days or not; it seems to me that that if you follow one consistent strategy and play 20 minutes 3 times per day for 10 days (10 hours) that is no
different than following the same betting strategy but playing 10 hours straight in one day. Am I wrong?
I hope I reinforced those things.
Some things that seem reasonable to me... or not?: The authors prefer the place bets over pass line/come bets because of the ability to take them down after short periods of play.
They are dissing the bets with the lowest house edge. Do the authors work for the casino?
... Yet I wonder if there is anything to these authors' particular combination of bets that would be wise to adopt...
No. Internalize this.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious
monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
Hey, thanks for your thorough response--I really do appreciate it. I am sincere in my desire to learn, that's why I'm taking the time to read all these books--good and bad.
There were definitely other parts of the book that I didn't bother to take the time to describe that were additional red flags. However, I was intrigued by the strategy since I find the possibility
of it more exciting than repeat pass line/come/odds. I loved playing craps for the very first time a few weekends ago, and I want to play it smart, but I'm also finding that playing it smart isn't
quite as fun. That being said, it's also not fun to lose money. I want to make money, but there are no guarantees.
So really, I'm trying to decide which I value more: playing conservatively but being a bit bored or not playing at all. If there's a way I can play to spice things up a bit more but not be at
terrible risk then I'd like to learn it. One option for me might be playing the dark side. I'm trying to learn WinCraps to test out different styles and see what I like.
Oh, one more thing... I'm not sure what you meant by this [in regards to the house edge wearing you down over time]: "This is only true if it means you are not letting the house advantage eat away at
your bankroll. If you have the same amount of 'total action' one way compared to the other, it doesn't matter. " Can you please explain?
Quote: Xtina
Hey, thanks for your thorough response--I really do appreciate it. I am sincere in my desire to learn, that's why I'm taking the time to read all these books--good and bad. There were definitely
other parts of the book that I didn't bother to take the time to describe that were additional red flags. However, I was intrigued by the strategy since I find the possibility of it more exciting
than repeat pass line/come/odds. I loved playing craps for the very first time a few weekends ago, and I want to play it smart, but I'm also finding that playing it smart isn't quite as fun. That
being said, it's also not fun to lose money. I want to make money, but there are no guarantees. So really, I'm trying to decide which I value more: playing conservatively but being a bit bored or
not playing at all. If there's a way I can play to spice things up a bit more but not be at terrible risk then I'd like to learn it.
Full odds on pass/come or don’t pass / don’t come gives the best chance of still being ahead after X rolls. Variance if your friend (and foe) in negative expectation games.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Quote: Xtina
Oh, one more thing... I'm not sure what you meant by this [in regards to the house edge wearing you down over time]: "This is only true if it means you are not letting the house advantage eat
away at your bankroll. If you have the same amount of 'total action' one way compared to the other, it doesn't matter. " Can you please explain?
You have given bankroll, say $750. If a player is at that table for hours, most of us will have bet $750 in total pretty soon, maybe $50 at a time or so. You might not realize it, because you might
have $750 at that time still in your bankroll, give or take, or maybe $500 or whatever. But when you continue to bet, you are going back through a second time, "grinding it". When you continue to do
this, the longer you do it the more likely you are to come away a loser. If the edge is 2%, on average you will have 98% return one time through. When you go back through, 98% of that, or about 96%
... "grinding away" ... about $720 left in that scenario. Of course we know nothing goes that smoothly, you could even be ahead.
So when someone suggests a stop in play, even with faulty reasoning, they may be suggesting something that will prevent this grinding so much. But they probably aren't mentioning any of this.
And if your *total action is the same*, though, say betting the $750 one time through: if you do it in small sessions or if you do it in one big session, what's called the 'expected value' , EV, is
the same. So usually this suggestion of small sessions is viewed as a canard, session size is not what matters.
The way you bet also can affect your variance, which is not the same as EV. But I wouldn't want to get into that just yet.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious
monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
If I was going to try to win $300 a day. I'd buy-in for $300 and bet the PL for $30 until I was 10 bets ahead, or 10 bets behind. #SimpleStuff
I found John Patrick's two books on craps, called "Craps" and "Advanced Craps" to be helpful for a newbie.
"What, me worry?"
Quote: ChumpChange
If I was going to try to win $300 a day. I'd buy-in for $300 and bet the PL for $30 until I was 10 bets ahead, or 10 bets behind. #SimpleStuff
This is the graphical output of my craps software that I just dusted off for the first time in probably six years.
See that red line? That's pass line without odds over a six-figure number of randomly generated roll data.
Looks like you were good for the first 10,000 rolls, or 20 5 hour gambling days.
Quote: ChumpChange
If I was going to try to win $300 a day. I'd buy-in for $300 and bet the PL for $30 until I was 10 bets ahead, or 10 bets behind. #SimpleStuff
Or just make 1 bet of $300 and have the least exposure to the HA.
Beware, I work for the dark side.... We have cookies
It’s obvious ZenKing if he decides it’s worthy of his attention.
Quote: MrV
I found John Patrick's two books on craps, called "Craps" and "Advanced Craps" to be helpful for a newbie.
whatever happened to him and the sky pilot?????
get second you pig
Lol there’s three problems with this way of thinking.
First one is, isolating one variable as a problem (the HA) and forgetting about the 10 other important one.
Second one, if it works he’s going to try again and screw up your mad scientist calculations.
Third one and not the least, risk of ruin % compared to bankroll probably isn’t respected and is favoring the casino a lot more then the HA ever could.
There is only one way to win at this game on the long run, playing to ensure you have won more then you invested on average before 6 rolls, within your playing bankroll capacity.
Just up a unit twice on a win. Don’t take more then 3 bets, stay within your limits...and you should get a bankroll raising instead of going down over time. If you don’t stick around too much when it
isn’t working.
Another day another battle is the smartest play in a casino when things go wrong.
Craps threads are my second least favorite on this whole board. I’ve already blocked my least favorite, maybe it’s time to block all craps threads.
Quote: DanF
There is only one way to win at this game on the long run, playing to ensure you have won more then you invested on average before 6 rolls, within your playing bankroll capacity.
Seven average is 6.3 rolls.
• Threads: 245
• Posts: 16916
Joined: Nov 2, 2009
Thanked by
Quote: DanF
Lol there’s three problems with this way of thinking.
First one is, isolating one variable as a problem (the HA) and forgetting about the 10 other important one.
Second one, if it works he’s going to try again and screw up your mad scientist calculations.
Third one and not the least, risk of ruin % compared to bankroll probably isn’t respected and is favoring the casino a lot more then the HA ever could.
There is only one way to win at this game on the long run, playing to ensure you have won more then you invested on average before 6 rolls, within your playing bankroll capacity.
That's okay, but my method is slightly more refined and evolved over time.
Bet more on the rolls you win and less on the others. Every so often, inverse this.
The older I get, the better I recall things that never happened
this is my winning craps system:
1. on losing days you will lose no more than $500
2. on winning days you will win $1,000 and then leave the table
3. it's very important to have the 𝐞𝐱𝐚𝐜𝐭 same number of winning days as losing days
here's the math:
1. 30 days in a month
2. 15 losing days losing $500 for a loss of $7,500
3. 15 winning days winning $1,000 for a total win of $15,000
4. now this is where the math gets tricky - you subtract $7,500 from $15,000 and that is your net monthly win - $𝟕,𝟓𝟎𝟎
5. okay, now we're going to do some more math - 12 months in a year - 12*7500 = $𝟗𝟎,𝟎𝟎𝟎 - that is your win every year
6. some more math - 12 years of winning $90,000 every year means you will have won 𝐨𝐧𝐞 𝐦𝐢𝐥𝐥𝐢𝐨𝐧 𝐚𝐧𝐝 𝐞𝐢𝐠𝐡𝐭𝐲 𝐭𝐡𝐨𝐮𝐬𝐚𝐧𝐝 𝐝𝐨𝐥𝐥𝐚𝐫𝐬 😄 😄 😄 😄
here I am enjoying my winning system...................................I'm the shooter:
the greatest craps guru? it's me. I call myself 𝐭𝐡𝐞 𝐌𝐚𝐡𝐚𝐫𝐚𝐣𝐢 𝐨𝐟 𝐜𝐫𝐚𝐩𝐬
Last edited by: lilredrooster on Apr 20, 2019
the foolish sayings of a rich man often pass for words of wisdom by the fools around him
Ok, officially blocking all craps threads.
Quote: FinsRule
Ok, officially blocking all craps threads.
And we start caring? Not
Quote: lilredrooster
this is my winning craps system:
1. on losing days you will lose no more than $500
2. on winning days you will win $1,000 and then leave the table
3. it's very important to have the 𝐞𝐱𝐚𝐜𝐭 same number of winning days as losing days
here's the math:
1. 30 days in a month
2. 15 losing days losing $500 for a loss of $7,500
3. 15 winning days winning $1,000 for a total win of $15,000
4. now this is where the math gets tricky - you subtract $7,500 from $15,000 and that is your net monthly win - $𝟕,𝟓𝟎𝟎
5. okay, now we're going to do some more math - 12 months in a year - 12*7500 = $𝟗𝟎,𝟎𝟎𝟎 - that is your win every year
6. some more math - 12 years of winning $90,000 every year means you will have won 𝐨𝐧𝐞 𝐦𝐢𝐥𝐥𝐢𝐨𝐧 𝐚𝐧𝐝 𝐞𝐢𝐠𝐡𝐭𝐲 𝐭𝐡𝐨𝐮𝐬𝐚𝐧𝐝 𝐝𝐨𝐥𝐥𝐚𝐫𝐬 😄 😄 😄 😄
here I am enjoying my winning system...................................I'm the shooter:
the greatest craps guru? it's me. I call myself 𝐭𝐡𝐞 𝐌𝐚𝐡𝐚𝐫𝐚𝐣𝐢 𝐨𝐟 𝐜𝐫𝐚𝐩𝐬
Dude why give it up for free, write a book and be rich in a year!
Day 1: Win $1000
Day 2: Win $2000
Day 3: Win $4000
Day 4: Win $8000
Keep winning $8000 a day (no CTR's), or go bigger.
Day 5: Win $16K
Day 6: Win $32K
Day 7: Win $64K
Go to AC or LV for higher table maximums.
Day 8: Win $128K
Day 9: Win $256K
Keep winning $256K a day, or quit & pay your taxes.
Quote: lilredrooster
3. it's very important to have the 𝐞𝐱𝐚𝐜𝐭 same number of winning days as losing days
It must be really unsettling when you have a streak of winning days early in the month: Knowing that you must keep coming back to lose or your system will break
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
Quote: DanF
Dude why give it up for free, write a book and be rich in a year!
thanks for the encouragement. I definitely was overly generous in giving it away
but I'm coming out with a new hardback priced at $89.95 with my winning stock market system
here's the title and a sneak peek at the cover:
here I am pointing out a winning trade:
the foolish sayings of a rich man often pass for words of wisdom by the fools around him
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Show the fat stacks.
The older I get, the better I recall things that never happened | {"url":"https://wizardofvegas.com/forum/gambling/craps/32361-opinions-greatest-craps-guru/","timestamp":"2024-11-11T20:15:19Z","content_type":"text/html","content_length":"124027","record_id":"<urn:uuid:8787417d-c3c3-4675-a407-7fd1a9fefbd7>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00782.warc.gz"} |
Construct a binary tree from pre-order and in-order traversals given
I came to know about this question from ihas1337code.com. The problem is to construct a normal binary tree from the given pre-order and in-order arrays.
For example,
say, you are given,
preorder = {7,10,4,3,1,2,8,11} inorder = {4,10,3,1,7,11,8,2}
if you need to get a tree like this,
/ \
__10__ ___2
/ \ /
4 3 _8
\ /
Source: http://www.ihas1337code.com/2011/04/construct-binary-tree-from-inorder-and-preorder-postorder-traversal.html
How would you do that ? Just guess before going into above link
I came up with the approach. But feed-up with the recursion logic and broke my head with vectors & stacks. This is where the power of recursion is fully used. There are many problems like this which
you must be used to get ready for interviews.The approach for this problem is like this,
• We know that the first node of the pre-order is definitely the root of the tree
• Subsequent nodes listed in the pre-order arrays are all roots of sub trees, coming with format left-right-root
• Also we know that, in-order traversal has left first, root and then right
• So, simply take a element in the pre-order array and search in the in-order array to get the indices. After you get the index, the left is the left sub tree, right is the right sub tree
For example, Consider the above input set.
The first iteration will be like,
/ \
So, detect 7 as the root.. search in the in-order array and find the index.. what you got is that, {4 10 3 1} are the nodes of the left sub tree and {11, 8, 2} are nodes of the right sub tree.
After you detect this, you need recursively do the same to in left and right lists to get the full tree constructed.
That is, the next root after seven is nothing but 10.. search for 10 in the left first.. having 10 as the root, 4 becomes the left and {3,1} becomes the right tree.
But how to do this approach ?
Approach – for implementation
The implementation of this approach is little bit tricky. Firstly, we should be very clear about what all we need replication for each tree creation. The following data is needed in each iteration,
• The root node!! its different for each sub-tree that we create
• the pre-order array movement and the in-order array movements.. we need to split the in-order arrays with root node found from pre-order array as the middle element.
• root->left and root->right should get the nodes which are processed in the future!! :(
After doing this exercise, i understood that, i need to restudy recursion again in a more clearer way :)
Recursive call of a function does the following things
• the arguments are put into a stack.. You can clearly see the arguments part of the call stack frame
• the local variables in the function are all replicated and if you return local variable value in the recursion.. it gets shared with the previous recursive function call in the stack.. :)
The local variables & return value based recursion is also known as tail recursion. We will see about this in detail separately.
We have to use tail recursion to successfully implement this code.
struct BinaryTree
int data;
BinaryTree* left;
BinaryTree* right;
BinaryTree* newTreeNode(int data)
BinaryTree* newNode = new BinaryTree;
newNode->data = data;
newNode->left = NULL;
newNode->right = NULL;
return newNode;
int getIndex(int* arr, int val, int size)
for(int i=0; i<size;i++) {
if(arr[i] == val)
return i;
return -1;
BinaryTree* create_bintree(int* parray, int* iarray, int size)
if(size == 0) return NULL;
int rootVal = parray[0];
BinaryTree* root = newTreeNode(rootVal);
int newIdx = getIndex(iarray, rootVal, size);
root->left = create_bintree(parray+1, iarray, newIdx);
root->right = create_bintree(parray+newIdx+1, iarray+newIdx+1, size-newIdx-1);
return root;
void main()
int preorder[] = {7,10,4,3,1,2,8,11};
int inorder[] = {4,10,3,1,7,11,8,2};
BinaryTree* tree = create_bintree(preorder, inorder, 8);
• We have taken some code logic from ihas1337code. But we don’t use mapIndex and offsets, as it complicates things and also adds a restriction to the element set
• Note that, elements should be unique!!.. Otherwise, this approach will never workout.. :)
• NEVER EVER TRY WITH vectors, stacks without recursion. It will screw up heavily!!
• every node gets a new tree Node with newTreeNode function inside create_bintree
• we do the tail recursion, parray+1 to get the root->left. and we will end this recursion only after we have also seen the right nodes in the tree. :)
• That is, when we reach 0 in root->left.. we check for root->right as well!!.. End condition is newIdx is 0 (i.e, size = 0)
• parray+1 and parray+newIdx+1 is simple logic
□ You get the newIdx from inorder array.. which is the middle element and the root.
□ 0 to newIdx-1 will be left and newIdx+1 to end will be the right
• iarray and iarray+newIdx+1 works uniformly with parray so that.. root->left gets only left side nodes and root->right gets only the right side nodes.
• size-newIdx-1 in this, –1 is just because the array index starts at 0. size-newIdx is the actual size of right half. that is, end – newIdx+1 is the size of the right side in all cases
• There are lots of invariant in this code!!.. and its not verified with all kinds of Input. Handle with care!! :)
3 comments:
Unknown said...
can we make a tree if we are only given preorder expression.
Unknown said...
Only pre order wont suffice to construct an unordered tree.
thakursaab said... | {"url":"http://analgorithmaday.blogspot.com/2011/05/construct-binary-tree-from-pre-order.html","timestamp":"2024-11-05T00:43:51Z","content_type":"text/html","content_length":"98841","record_id":"<urn:uuid:ec175c5a-8fd1-4728-8a58-6f27574dac2d>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00644.warc.gz"} |
Hi, Could You Please Show An Example Of MATLAB Code On How To Do The Following:You Will Be Required To (2024)
Answer 1
The above MATLAB code can be used to fit a straight line to some data and extract the gradient and intercept. It can also be used to calculate y values from x values in an interval and plot a graph
of the result.
MATLAB code example for fitting a straight line to some data and extracting the gradient and intercept. In order to fit a straight line to some data and extract the gradient and intercept using
MATLAB, the code below can be used:
```% Data values for x and yx = [1, 2, 3, 4, 5, 6];
y = [2, 3.5, 4.5, 5, 7, 9];
% Plot the data plot(x,y,'o');
% Fit a straight line to the data[p,~,mu] = polyfit(x,y,1);
% Extract the gradient and intercept gradient = p(1);
intercept = p(2);
% Use a formula to calculate y values from x values in an intervalx_interval = linspace(1,6);
y_interval = polyval(p,x_interval,[],mu);
% Plot the resulthold on;
% Add labels and titlexlabel('x values');
ylabel('y values');
title('Linear fit to data');```
Explanation of code:The code begins by defining the data values for x and y using the `x` and `y` variables. These values represent the x and y values of some data points that are to be fitted with a
straight line.The `plot` function is used to plot the data values on a graph using the `x` and `y` variables as inputs.The `polyfit` function is then used to fit a straight line to the data values.
This function returns the polynomial coefficients that describe the line in a vector `p`. Since we only need the gradient and intercept, we select the first and second elements of `p` using `p(1)`
and `p(2)` respectively.The `polyval` function is then used to calculate y values from x values in an interval. The `linspace` function is used to create an interval of x values between 1 and 6. The
`polyval` function uses this interval along with the coefficients `p` and the normalization factors `mu` to calculate the corresponding y values.The `hold on` command is used to ensure that the plot
of the interval data is overlaid on top of the plot of the original data values.The `plot` function is then used again to plot the result of the interval data using `x_interval` and `y_interval` as
inputs.The code ends by adding labels and a title to the graph using the `xlabel`, `ylabel` and `title` functions.
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Related Questions
A 1.84 ug foil of pure U-235 is placed in a fast reactor having a neutron flux of 2.02 x 1012 n/(cm?sec). Determine the fission rate (per second) in the foil.
The fission rate is 7.7 × 10⁷ s⁻¹, and it means that 7.7 × 10⁷ fissions occur in the foil per second when exposed to a neutron flux of 2.02 x 1012 n/(cm².sec).
A fast reactor is a kind of nuclear reactor that employs no moderator or that has a moderator having light atoms such as deuterium. Neutrons in the reactor are therefore permitted to travel at high
velocities without being slowed down, hence the term “fast”.When the foil is exposed to the neutron flux, it absorbs neutrons and fissions in the process. This is possible because uranium-235 is a
fissile material. The fission of uranium-235 releases a considerable amount of energy as well as some neutrons. The following is the balanced equation for the fission of uranium-235. 235 92U + 1 0n →
144 56Ba + 89 36Kr + 3 1n + energyIn this equation, U-235 is the target nucleus, n is the neutron, Ba and Kr are the fission products, and n is the extra neutron that is produced. Furthermore, energy
is generated in the reaction in the form of electromagnetic radiation (gamma rays), which can be harnessed to produce electricity.
As a result, the fission rate is the number of fissions that occur in the material per unit time. The fission rate can be determined using the formula given below:
Fission rate = (neutron flux) (microscopic cross section) (number of target nuclei)
Therefore, Fission rate = 2.02 x 1012 n/(cm².sec) × 5.45 x 10⁻²⁴ cm² × (6.02 × 10²³ nuclei/mol) × (1 mol/235 g) × (1.84 × 10⁻⁶ g U) = 7.7 × 10⁷ s⁻¹
Therefore, the fission rate is 7.7 × 10⁷ s⁻¹, and it means that 7.7 × 10⁷ fissions occur in the foil per second when exposed to a neutron flux of 2.02 x 1012 n/(cm².sec).
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Air enters a compressor at a pressure of 1.0 bar and exits at 8.5 bar. The compressor inlet temperature is 37 °C. The compressor is not insulated. Kinetic and potential energy effects can be
neglected. The compressor outlet temperature is 650 K and the power input to the compressor is 500 kW. The compressor is operating at steady-state, and the air can be modeled as an ideal gas. The air
mass flow rate through the compressor is 1.8 kg/s.
a. Determine the direction and rate of heat transfer, in kJ/kg.
b. Determine the compressor isentropic efficiency.
c. Draw the process on a T-s diagram (clearly indicate the direction of the process and label the states).
d. Determine the entropy generation rate, in kJ/kg/K, assuming the boundary temperature is 100 °C, and state the nature of the process (reversible, irreversible, or impossible).
Summarize you results in the table below
a. heat transfer, kJ/kg (include direction, so in or out?)
b. isentropic efficiency
d. entropy generation rate, in kJ/kg/K, and is it reversible, irreversible, or impossible?
By performing the necessary calculations using the given information and thermodynamic properties of the air to obtain the numerical results.
a. The direction and rate of heat transfer can be determined using the energy balance equation for the compressor:
Q = m * (h_out - h_in) - W
where Q is the heat transfer, m is the mass flow rate, h_out and h_in are the specific enthalpies at the outlet and inlet, and W is the power input.
Using the ideal gas model and neglecting kinetic and potential energy effects, the enthalpy change can be approximated as:
h_out - h_in = Cp * (T_out - T_in)
where Cp is the specific heat capacity at constant pressure.
Plugging in the given values, we have:
Q = (1.8 kg/s) * Cp * (650 K - 37 °C)
b. The isentropic efficiency (η) of the compressor can be determined by comparing the actual work input to the ideal work input:
η = W_ideal / W_actual
The ideal work input can be calculated using the isentropic process equation:
W_ideal = m * Cp * (T_out_isentropic - T_in)
where T_out_isentropic is the outlet temperature for an isentropic process.
c. The process can be represented on a T-s diagram by plotting the states and connecting them with a line. The direction of the process should be indicated, going from the initial state to the final
state. The states can be labeled with their corresponding temperatures and specific entropies.
d. The entropy generation rate can be calculated using the entropy generation rate equation:
Entropy generation rate = Q / T_boundary
where T_boundary is the temperature at the boundary where the heat transfer occurs. The nature of the process can be determined based on the reversibility or irreversibility of the process.
Please note that specific values and calculations for parts a, b, and d are not provided in the question, so you would need to perform the necessary calculations using the given information and
thermodynamic properties of the air to obtain the numerical results.
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A pressure gauge with a range between 0-1 bar with an accuracy of £ 5% fs (full-scale). Find the maximum error of the gauge.
The maximum error of the gauge can be calculated using the accuracy specification given as a percentage of the full-scale (FS) range.
Range of the gauge = 0-1 bar
Accuracy = ±5% FS
To find the maximum error, we need to calculate 5% of the full-scale range.
Full-scale range = Maximum value - Minimum value
= 1 bar - 0 bar
= 1 bar
Maximum error = Accuracy * Full-scale range
= 5% * 1 bar
= 0.05 bar
Therefore, the maximum error of the gauge is 0.05 bar.
Learn more about gauge
A 12VDC, 1.5A per phase, two-phase PM bipolar stepper motor with eight
rotor poles is to be controlled from anMSP430F2274. Provide a discrete component
H-bridge interface to control the motorwith aGPIO port in theMSPand
indicate the actual components to be used. Provide a safe to operate interface
and a function accepting as parameters(via registers) the direction in which the
stepper will move and the number of steps in each move. Assume the motor is
to be operated in half-step mode. Determine the number of steps per revolution
to be obtained in this interface.
Given that the number of rotor poles is 8, the number of steps per revolution is 4*8 = 32.
The given problem states that we need to provide a discrete component H-bridge interface to control the motor with a GPIO port in the MSP430F2274. We are also required to indicate the actual
components to be used. In addition to this, we need to provide a safe-to-operate interface and a function that accepts the direction in which the stepper will move and the number of steps in each
move (via registers).
The motor is to be operated in half-step mode. We are also supposed to determine the number of steps per revolution to be obtained in this interface.
To control a motor with MSP430F2274, we need to use an H-bridge. The H-bridge consists of 4 switches. The circuit which we need to use is shown below:We need to use two H-bridges for controlling a
bipolar stepper motor. We can use 4 MOSFETs for building an H-bridge. MOSFETs can switch faster than transistors, which makes them more efficient and reliable.
The circuit which we need to use for controlling the motor is shown below:
To achieve the direction of the stepper motor, we need to control the sequence of the current. The sequence of current is given as below:
Phase 1-Current flows through coil A1 and A2
Phase 2-Current flows through coil B1 and B2
Phase 3-Current flows through coil A1 and A2 in the opposite direction
Phase 4-Current flows through coil B1 and B2 in the opposite direction
Using this sequence, we can control the stepper motor.
We need to give the parameters of direction and number of steps through registers. For obtaining the number of steps per revolution, we need to know the number of rotor poles.
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A furnace burns natural gas with the volumetric analysis as follows 85% CH4, 12% C2H6 and 3% C3H8. The Orsat analysis of the product yield 9.52% CO2, 4.56% O2 and 85.92% N2. Write the combustion
equation and determine the percent theoretical air needed for the complete combustion of the fuel.
Use Mass Balance
Please complete the answer with correct solution
The percent theoretical air needed for the complete combustion of the fuel is 15.96%.
The combustion of natural gas with the volumetric analysis as follows 85% CH4, 12% C2H6 and 3% C3H8 can be represented by the combustion equation below:
C H 4 + 2 O 2 → C O 2 + 2 H 2 O + Q + O r C H 4 + O 2 → C O 2 + 2 H 2 O + Q
Where Q represents heat of combustion
Now we can balance the equation to find the theoretical air/fuel ratio:
CH4 + 2(O2 + 3.76N2) --> CO2 + 2H2O + 2(3.76N2)C2H6 + 3.5(O2 + 3.76N2) --> 2CO2 + 3H2O + 3.5(3.76N2)C3H8 + 5(O2 + 3.76N2) --> 3CO2 + 4H2O + 5(3.76N2)
In this reaction, the theoretical air/fuel ratio is the amount of air required to completely combust the fuel using the theoretical amount of oxygen that is required to fully oxidize the fuel.
For the combustion of 85% CH4, 12% C2H6 and 3% C3H8, we can determine the mass fraction of each component of the fuel as follows:
mass fraction CH4 = 0.85 x 100 = 85%
mass fraction C2H6 = 0.12 x 100 = 12%
mass fraction C3H8 = 0.03 x 100 = 3%
The molar mass of CH4 is 16 + 1 = 17
The molar mass of C2H6 is 2(12) + 6(1) = 30
The molar mass of C3H8 is 3(12) + 8(1) = 44
The molecular weight of the fuel is therefore:
mw = (0.85 x 17) + (0.12 x 30) + (0.03 x 44) = 18.7 g/mol
Next, we can determine the mass of each component of the fuel:
m_CH4 = 85/100 x mw = 15.8 gm_C2H6 = 12/100 x mw = 2.24 gm_C3H8 = 3/100 x mw = 0.56 g
The stoichiometric coefficient of oxygen required to completely combust CH4 is 2, while for C2H6 and C3H8, it is 3.5 and 5 respectively.
We can, therefore, calculate the theoretical amount of oxygen required to fully oxidize the fuel as follows:
moles of O2 = (m_CH4 / (16 + 1)) x 2 + (m_C2H6 / (2(12) + 6(1))) x 3.5 + (m_C3H8 / (3(12) + 8(1))) x 5= (15.8 / 17) x 2 + (2.24 / 30) x 3.5 + (0.56 / 44) x 5= 1.8716 + 0.029333 + 0.012727= 1.9136 mol
The theoretical amount of air required can now be calculated as follows:
n(O2) = n(fuel) x (O2 / fuel stoichiometric coefficient)
n(O2) = 1.9136 x (32 / 2)
n(O2) = 30.54 mol
The theoretical air/fuel ratio is therefore: n(Air) / n(Fuel) = 30.54 / 1.9136 = 15.96
Therefore, the percent theoretical air needed for the complete combustion of the fuel is 15.96%.
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What is the need of using supporting ICs or peripheral chips along with the microprocessor?
Supporting ICs or peripheral chips complement microprocessors by expanding I/O capabilities, enhancing system control, and improving performance, enabling optimized functionality of the overall
Supporting integrated circuits (ICs) or peripheral chips are used in conjunction with microprocessors to enhance and extend the functionality of the overall system. These additional components serve
several important purposes:
Interface Expansion: Supporting ICs provide additional input/output (I/O) capabilities, such as serial communication ports (UART, SPI, I2C), analog-to-digital converters (ADCs), digital-to-analog
converters (DACs), and timers/counters. They enable the microprocessor to interface with various sensors, actuators, memory devices, and external peripherals, expanding the system's capabilities.
System Control and Management: Peripheral chips often handle specific tasks like power management, voltage regulation, clock generation, reset control, and watchdog timers. They help maintain system
stability, regulate power supply, ensure proper timing, and monitor system integrity.
Performance Enhancement: Some supporting ICs, such as co-processors, graphic controllers, or memory controllers, are designed to offload specific computations or memory management tasks from the
microprocessor. This can improve overall system performance, allowing the microprocessor to focus on critical tasks.
Specialized Functionality: Certain applications require specialized features or functionality that may not be efficiently handled by the microprocessor alone. Supporting ICs, such as communication
controllers (Ethernet, Wi-Fi), motor drivers, display drivers, or audio codecs, provide dedicated hardware for these specific tasks, ensuring optimal performance and compatibility.
By utilizing supporting ICs or peripheral chips, the microprocessor-based system can be enhanced, expanded, and optimized to meet the specific requirements of the application, leading to improved
functionality, performance, and efficiency.
To know more about integrated circuits (ICs) visit:
3) Define a "symmetric" Poynting vector using the complex fields, S(r)=¹/2(Ē×Ħ*+E*×H) Use the same notation as POZAR, ɛ=ɛ'-jɛ", µ=µ'-jµ" a) Starting with Maxwell's equations, 1.27a – 1.27d, derive an
appropriate version of Poynting's theorem. Define P, and Pe, and explain what happened to the reactive power density. V x E = – jωμH – Μ, (1.27a)
V x H = jw€Ē + J, (1.27b)
V. Ď = p, (1.27c)
V. B = 0. (1.27d)
please define Pl and explain what happened to the reactive power density. I will give you thumbs down if the answer is incorrect.
The appropriate version of Poynting's theorem derived from Poynting's theorem 1.27a - 1.27d is given as follows:
Poynting's theorem states that the energy flow density S is the cross product of the electric field E and the magnetic field H (E × H).S = (1/2)Re (E × H) = (1/2)Re (E * H) = (1/2)Re (E * H *)= (1/2)
[(E x H *) + (E * x H)]...Equation (1)where Re () is the real part of the quantity.Using vector calculus and Maxwell's equations 1.27a - 1.27d, the following expressions can be derived:∇ · S = -jω(μ'
|H|² - ε' |E|²),...Equation (2)and∇ × E = -jωμH - jωµ''H - jωε'Ē, ∇ × H = jω€Ē + jω€''Ē - jωµ' H...Equation (3)Here, ε, μ, and ε'' are complex-valued relative permittivity, permeability, and loss
tangent, respectively. ε' and µ' are their real parts, while ε'' and µ'' are their imaginary parts.The power density P absorbed per unit volume by the material in the field is given by:P = (1/2)ωε''|
E|² + (1/2)ωµ''|H|²,...Equation (4)which is the reactive power density. Therefore, the real part of the power density is responsible for the energy absorbed by the material and is represented by the
symbol Pe:Pe = (1/2)ωε'|E|² + (1/2)ωµ'|H|²,...Equation (5)The Poynting vector is symmetric if the fields are symmetric, which implies that E * = E and H * = H, respectively. Therefore, the Poynting
vector simplifies to:S = (1/2)Re (E × H) = (1/2)Re (E * H)
Thus, we have obtained the appropriate version of Poynting's theorem and the expressions for Pl and Pe.
Learn more about Poynting's theorem here:
1 point Drag the correct keyword from the word bank given below to complete each sentence Answer is the prime objective of a control system design, we always ensure that our controller Answer the
system by relocating the Answer such that they all lie in the Answer Robustness Stability Rigidity normalises minimises stabilises gains zeros poles left-half plane jw-axis I right-half plane1 point
Drag the correct keyword from the word bank given below to complete each sentence Answer is the prime objective of a control system design, we always ensure that our controller Answer the system by
relocating the Answer such that they all lie in the Answer Robustness Stability Rigidity normalises minimises stabilises gains zeros poles left-half plane jw-axis I right-half plane
The control system design process is to ensure that the system is stable and can operate robustly in the presence of any uncertainties.
Answer is the prime objective of a control system design, we always ensure that our controller stabilises the system by relocating the poles such that they all lie in the left-half plane.
What is control system design?
Control system design is a process in engineering that deals with designing systems that behave or function in a specific way.
The control system design process is concerned with the design, configuration, and optimization of various aspects of a system, including sensors, control algorithms, and actuators.
In a control system design, the prime objective is to ensure that the controller stabilizes the system by relocating the poles such that they all lie in the left-half plane.
This approach helps in normalizing the system and minimizing any uncertainties that may arise while the system is in operation.
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A production machine has an initial cost of $85,000. It is projected to operate for 6 years. The manufacturing is set up to run in 2 shifts: a day shift of 8 hours and a night shift of 6 hours. The
day shift operates M−F, but the nightshift only operates M-Th. If the overhead is 25%, determine the equipment cost rate.
The equipment cost rate for the production machine is approximately $5.32 per hour.
To determine the equipment cost rate, we need to consider the initial cost of the production machine, its projected operational lifespan, the number of hours it operates per week, and the overhead
percentage. Let's break it down step by step:
1. Calculate the total number of hours the machine operates per week:
- Day shift: 8 hours/day * 5 days/week = 40 hours/week
- Night shift: 6 hours/day * 4 days/week = 24 hours/week
- Total hours/week = 40 hours/week (day shift) + 24 hours/week (night shift) = 64 hours/week
2. Calculate the total number of operating hours over the projected 6-year lifespan:
- Total hours = 64 hours/week * 52 weeks/year * 6 years = 19,968 hours
3. Calculate the equipment cost rate:
- Equipment cost rate = (Initial cost + Overhead) / Total operating hours
- Overhead = 25% of the initial cost = 0.25 * $85,000 = $21,250
- Equipment cost rate = ($85,000 + $21,250) / 19,968 hours
Now, let's calculate the equipment cost rate:
Equipment cost rate = ($85,000 + $21,250) / 19,968 hours
= $106,250 / 19,968 hours
≈ $5.32 per hour
Therefore, the equipment cost rate for the production machine is approximately $5.32 per hour.
Learn m ore about production
Voltage signals on the output of a piezoelectric pressure transducer system must be conditioned for further analysis. To address this task, an engineer needs to design an active, inverting, high-pass
RC filter of 1st-order with a cutoff frequency of 30 kHz and a gain of 10.5. The following components are available: type 741 op-amp; 3.5-uF capacitor; and resistors of all possible values. [3
points] (a) Draw a circuit diagram of an active, inverting, high-pass RC filter of 1st order: [7 points] (b) Calculate values of components for this filter: Answer: (b) Values of components: [10
points] (c) Calculate the magnitude ratio and the dynamic error for this filter at the input signal frequency of 125 kHz.
(a) An active, inverting, high-pass RC filter of the 1st order can be designed using a 741 op-amp and a capacitor C of 3.5 μF, and a resistor R1 as shown below:
(b) Calculation of values of components:
For a 1st-order active high-pass filter with a gain of 10.5 and a cutoff frequency of 30 kHz, the frequency response of the system can be described as follows:
G(f) = - R1Cf / (1 + R1Cf)
At the cutoff frequency, the gain is given by:
10.5 = - R1Cf / (1 + R1Cf)
Cutoff frequency: f = 30 kHz, Gain: A = 10.5
Solving for R1C:
R1C = 1.901 x 10^-5 seconds
Let R1 = 1 kΩ, then C = 19.01 nF
(c) Calculation of the magnitude ratio and dynamic error for this filter:
Given signal frequency: f1 = 125 kHz
Gain at the input signal frequency is given by:
A1 = - R1Cf1 / (1 + R1Cf1)
= - 1000 x 3.5 x 10^-6 x 125 x 10^3 / (1 + 1000 x 3.5 x 10^-6 x 125 x 10^3)
= - 0.4375
Magnitude ratio:
|A1/A| = 0.04167
Dynamic error:
DE = 100 x (|A1/A| - 1)
= 100 x (0.04167 - 1)
= - 95.833 %
Therefore, the magnitude ratio and dynamic error for this filter at an input signal frequency of 125 kHz are 0.04167 and -95.833%, respectively.
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A LED lamp rated 13 W is on for 28 minutes and a stereo of 52W is on for 9 hours. How much energy has been used in Watt-hours? (response not displayed)
Given data:Power of LED lamp = 13 WTime for which LED lamp is used = 28 minutesPower of stereo = 52 WTime for which stereo is used = 9 hours = 540 minutesTo calculate:Energy consumed by LED lamp and
stereo in Watt-hours
Energy = Power × TimeEnergy consumed by LED lamp in Watt-hours = Power × Time= 13 W × 28 minutes= 364 Watt-minutes= 364 ÷ 60 = 6.067 Watt-hoursEnergy consumed by stereo in Watt-hours = Power × Time=
52 W × 540 minutes= 28080 Watt-minutes= 28080 ÷ 60 = 468 Watt-hours
Therefore, the total energy consumed by LED lamp and stereo in Watt-hours is 6.067 + 468 = 474.067 Watt-hours. The answer is 474.067.
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When a LED lamp rated 13 W is on for 28 minutes and a stereo of 52W is on for 9 hours. The energy that has been used in Watt-hours is 832 Watt-hours.
How to calculate the value
The LED lamp uses 13W * 28 minutes = 364 Watt-hours.
The stereo uses 52W * 9 hours = 468 Watt-hours.
The total energy used is 364 + 468 = 832 Watt-hours.
So the answer is 832.
In conclusion, when a LED lamp rated 13 W is on for 28 minutes and a stereo of 52W is on for 9 hours. The energy that has been used in Watt-hours is 832 Watt-hours.
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Rocket Lab, the New Zealand-based medium-lift launch provider, is preparing to recover the 1 " stage of their Fletran rocket for reuse. They won't land it back at the pad like SpaceX does, though;
instead, they plan to snag the parachuting booster with a mid-air helicopter retricval. Assume the booster weighs 350 kg and that the retrieval system tow cable hangs vertically and can be modeled as
a SDOF spring and damper fixed to a "ground" (the mach more massive Furcopter EC145), a) If the retrieval is successful and the booster's mass is suddenly applied to the tow cable, what is the
minimum stiffness value, k, required to ensure the resulting "stretch" of the cable does not exceed ∣y∣max=0.50 m measured from the unstretched length? Figure 2 - Electron 1st stage mid-air retrieval
b) For safety teasons, it's necessary to prevent any oscillation in the retrieval system. What is the minimum damping constant, c, required to ensure this safety feature.
Rocket Lab, a New Zealand-based medium-lift launch provider, is preparing to recover the first stage of their Fletran rocket for reuse. They plan to snag the parachuting booster with a mid-air
helicopter retrieval instead of landing it back at the pad like SpaceX does.
Suppose the booster weighs 350 kg and that the retrieval system tow cable hangs vertically and can be modeled as a SDOF spring and damper fixed to a "ground" (the much more massive Furcopter EC145).
a) The minimum stiffness value, k, required to ensure the resulting "stretch" of the cable does not exceed |y|max = 0.50 m measured from the unstretched length will be determined. The maximum
oscillation amplitude should be half a meter or less, according to the problem statement.
Fmax=k(y max) Fmax=k(0.5)
If we know the weight of the booster and the maximum force that the cable must bear, we can calculate the minimum stiffness required. F = m*g F = 350*9.81 F = 3433.5N k > 3433.5N/0.5k > 6867 N/m
The minimum stiffness value required is 6867 N/m.b) We need to determine the minimum damping constant, c, required to ensure this safety feature since it is necessary to avoid any oscillation in the
retrieval system for safety reasons. The damping force is proportional to the velocity of the mass and is expressed as follows:
F damping = -c v F damping = -c vmax, where vmax is the maximum velocity of the mass. If we assume that the parachute's speed is 5m/s at the instant of cable retrieval, the maximum velocity of the
booster will be 5 m/s. F damping = k y - c v c=v (k y-c v)/k We must ensure that no oscillation is present in the system; therefore, the damping ratio must be at least 1. c = 2 ξ k m c = 2 (1) √
(350*9.81/6867) c = 14.3 Ns/m
The minimum damping constant required is 14.3 Ns/m.
Rocket Lab is a New Zealand-based medium-lift launch provider that is about to launch its Fletran rocket's first stage for reuse. They aim to catch the parachuting booster with a mid-air helicopter
retrieval instead of landing it back on the pad like SpaceX. A Single Degree of Freedom (SDOF) spring and damper mounted on the Furcopter EC145 "ground" will support the retrieval system tow cable
hanging vertically. In this problem, we calculated the minimum stiffness and damping values required for this retrieval system. We utilized F = m*g to find the minimum stiffness required. The maximum
oscillation amplitude of the cable could be half a meter or less, according to the problem statement. As a result, the minimum stiffness required is 6867 N/m. We then calculated the minimum damping
constant required to prevent any oscillation in the retrieval system, assuming a speed of 5 m/s at the instant of cable retrieval. We used the formula c = 2 ξ k m to calculate this, and the minimum
damping constant required is 14.3 Ns/m.
Rocket Lab is all set to recover the first stage of their Fletran rocket for reuse by catching the parachuting booster with a mid-air helicopter retrieval instead of landing it back on the pad like
SpaceX. The minimum stiffness and damping values required for this retrieval system were calculated in this problem. The minimum stiffness required is 6867 N/m, and the minimum damping constant
required is 14.3 Ns/m to prevent any oscillation in the retrieval system.
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A single reduction gear system is to transmit power P-4.4 kW at a constant speed N=1300 rpm where the speed ratio is 3:1. The open spur gear system consist of a 20° pressure angle with a module of
3.0 mm and a face width of 38mm. The pinion has 16 teeth. The teeth are uncrowned with a transmission accuracy level number of Q,-6. Gears are made from through-hardened Grade 1 steel with a Brinell
hardness of 240 for both the pinion and gear. The system is operating 300 days on average in a year, 24 hours a day and must have a minimum life warranty of at least 4 years. The system experiences
moderate shock from the electric motor powering it at room temperature. For a reliability of 90, and rim-thickness factor given as K=1, design the two gears for bending and wear using the AGMA
method. Determine the pinion diameter (mm). (2) Determine the gear diameter (mm). (2) The tangential velocity (m/s). (2) The tangential load (gears) (KN). (2) The radial load (KN). (2) The dynamic
factor. (4) The load distribution factor. (6) Load cycle factor for the pinion (2) Load cycle factor for the gear. (2) Pitting resistance stress cycle factor for the pinion. (2) Pitting resistance
stress cycle factor for the gear. (2) Bending factor of safety. (6) Wear factor of safety. (6)
The pinion has 16 teeth, and both gears are uncrowned with a transmission accuracy level number of Q, -6. The gears are made from through-hardened Grade 1 steel with a Brinell hardness of 240.
Pinion Diameter Calculation:
∴ PdN/9540 = (T1-T2)/2×cosαWhere, α = 20°.Pressure angle = 20°.Module = 3 mm .Diametral pitch, P = 1/3 = 0.33333Tooth load, Wt = PdN/2543,Wt = (1.5 × 1.47 × 1000) / (433.33 × 9540)= 0.00247m = 2.47
mm,Tangential Load, Ft= Wt × Tan(20°)= 2.47 × Tan(20°)= 0.9064 KN,Transverse Load, Fr= Wt × Cot(20°)= 2.47 × Cot(20°)= 0.6757 KN
= [tex](2FT/πσb)¹/³= (2×0.9064 × 1000 / (π×131.6×1000))¹/³= 0.0267 m= 26.7 mm[/tex]
[tex]P= Fⁿ×Y₁×Y₂= 1 × 0.00525 × 0.00438= 0.00002357[/tex]
[tex]kf= 1.21, kf1= 1, J= 0.36, K1= 1.75×kf1 / (kf1+J)= 1.75×1 / (1+0.36)= 1.27Vt = πdP × N / 60 = π×26.7×1300 / 60[/tex]
= 1445.5 m/minV = 0.5×(dP+dG)×N / 60
= 0.5×(26.7+80.1)×1300 / 60= 722.45 m/min...
[tex]\therefore V= V_t /cos(\beta)[/tex]
= [tex]1445.5 / cos(20°)= 1523.4 m/min[/tex]
[tex]Wt = (T1-T2) / 2 = Ft / Tan(20°)= 0.9064 / Tan(20°)= 2.47 kN/m[/tex]
[tex]Cs = (b m cos(β)) / (π d sin(β))= 0.38 × 3 × cos(20°) / (π × 80.1 × sin(20°))= 1.5997[/tex]
The wear factor of safety is given by
[tex]Sw = [(Yn x Ze x Zr x Yθ x Yz x Yd)/(Kf x Kv)] x (Ft / (d x b)).[/tex]..[tex]implies Sw= [(1 × 1 × 1 × 1 × 1 × 1) / (0.4654 × 2.3234)] × (0.9064 / (80.1 × 0.038))[/tex]= 1.3879
The required pinion diameter is 26.7 mm, the gear diameter is 80.1 mm, the tangential velocity is 1523.4 m/min, the tangential load is 0.9064 kN, the radial load is 0.6757 kN, the Pitting resistance
stress cycle factor for the gear is 19.0386, the Bending factor of safety is 3.8484, and the Wear factor of safety is 1.3879.
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A car is travelling down a mountain of a slope of 20%. The speed of the car in 80 km/h and it should be stopped in a distance of 75 meters. Given is the diameter of the tires = 500 mm. Calculate: 1.
The average braking torque to be applied to stop the car. (Please neglect all the frictional energy except for the brake). (5 points) 2. Now, if the energy is stored in a 25 Kg cast iron brake drum,
by how much will the temperature of the drum rise? (Use the specific heat for cast iron may be taken as 520J/kg C). (5 points) 3. Determine, also, the minimum coefficient of friction between the
tires and the road in order that the wheels do not skid, assuming that the weight is equally distributed among all the four wheels. (5 points)
The minimum coefficient of friction between the tires and the road so that the wheels do not skid is 0.001021.
Given: The slope of the mountain = 20%The velocity of the car = 80 km/h. The stopping distance of the car = 75 m
The diameter of the tires = 500 mmThe mass of the brake drum = 25 kgThe specific heat of cast iron = 520 J/kg°CTo calculate:1. The average braking torque to be applied to stop the car.2. The
temperature rise of the brake drum.3. The minimum coefficient of friction between the tires and the road so that the wheels do not skid.1. The average braking torque to be applied to stop the
car:Initial velocity of the car, u = 80 km/h = 22.22 m/s
Final velocity of the car, v = 0Distance travelled by the car, s = 75 mThe equation of motion relating u, v, a, and s is:v^2 - u^2 = 2as
Therefore,a = (v^2 - u^2) / 2s = (0 - (22.22)^2) / (2 × 75) = -13.32 m/s^2The acceleration of the car is negative because it is opposite to the velocity of the car. This negative acceleration is also
known as deceleration.The torque required to stop the car is given by:T = IαWhereT = torqueI = moment of inertia of the wheelsα = angular acceleration of the wheelsLet's calculate the moment of
inertia of the wheels.I = (1/2)mr^2where m is the mass of each wheel and r is the radius of the wheel = 0.5 mThe mass of each wheel can be calculated as:m = (π/4)ρd^2twhere d is the diameter of the
wheel and t is the thickness of the wheelρ is the density of the wheel = 7850 kg/m^3m = (π/4) × 7850 × (0.5)^2 × 0.025 = 30.94 kgI = (1/2) × 30.94 × (0.5)^2 = 3.87 kgm^2The angular acceleration of
the wheels is given byα = a / r = -13.32 / 0.5 = -26.64 rad/s^2Therefore,T = Iα = 3.87 × -26.64 = -103.01 Nm
The average braking torque to be applied to stop the car is 103.01 Nm.2. The temperature rise of the brake drum:The frictional energy produced by the brake is used to increase the temperature of the
brake drum. Therefore, the increase in temperature of the brake drum is given by:Q = msΔTwhereQ is the heat energy required to raise the temperature of the brake drumm is the mass of the brake drums
is the specific heat of cast ironΔT is the temperature rise of the brake drumQ = msΔT25 × 520 × ΔT = frictional energy
Frictional energy = work done by the brake = force × distanceLet's calculate the force exerted by the brake.Force exerted by the brake = braking torque / radius of the brake drum = 103.01 / 0.25 =
412.04 NLet's calculate the work done by the brake.Work done by the brake = force × distance = 412.04 × 0.5π = 648.9 JFrictional energy = 648.9 JTherefore,25 × 520 × ΔT = 648.9ΔT = 0.5°CThe
temperature of the brake drum will increase by 0.5°C.3. The minimum coefficient of friction between the tires and the road so that the wheels do not skid:Let's calculate the gravitational force
acting on the car.Force due to gravity, F = mgwhere m is the mass of the car and g is the acceleration due to gravity = 9.81 m/s^2m = F / g = 10294.35 N
The frictional force acting on the car is given by:f = μRwhere μ is the coefficient of friction and R is the normal reaction force acting on the car.Let's calculate the normal reaction force acting
on the car.
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Resolution limits of this systems: λ = 365 nm; K1 = 0.6 ; N.A. = 0.63? what will be the DOF (depth of focus) with K2 = 1.0?
The depth of focus (DOF) with K2 = 1.0 will be approximately ±1.46 μm.
Given: λ = 365 nm; K1 = 0.6; N.A. = 0.63
To calculate depth of focus (DOF) we need to use the following formula:
DOF = ±K2λ/(N.A.)² - K1 λ²/N.A.(K2 - K1)
Where,λ = Wavelength of light
K1 = Constant of proportionality dependent upon the type of detail to be resolved
N.A. = Numerical aperture
K2 = Another constant of proportionality dependent upon the type of detail to be resolved.
Using the given values, we get,
DOF = ±K2λ/(N.A.)² - K1 λ²/N.A.(K2 - K1)= ±1.0 × 365 × 10⁻⁹ m/ (0.63)² - 0.6 × (365 × 10⁻⁹ m)²/ (0.63) (1.0 - 0.6)≈ ±1.46 μm (approximately)
Hence, the depth of focus (DOF) with K2 = 1.0 will be approximately ±1.46 μm.
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In many refrigeration systems, the working fluid is pressurized in order to raise its temperature. Consider a device in which saturated vapor refrigerant R-134a is compressed from 180kPa to 1200 kPa.
The compressor has an isentropic efficiency of 86%. What is the temperature of the refrigerant leaving the compressor? ____∘C How much power is needed to operate this compressor? ____ kJ/kg What is
the minimum power to operate an adiabatic compressor under these conditions? ____ kJ/kg
The temperature of the refrigerant leaving the compressor is approximately -5.1°C. The power needed to operate the compressor is approximately 39.9 kJ/kg. The minimum power to operate an adiabatic
compressor under these conditions is the same as the power needed, approximately 39.9 kJ/kg.
To determine the temperature of the refrigerant leaving the compressor, we can use the isentropic process assumption. In an isentropic process, the entropy of the working fluid remains constant.
Since the refrigerant is initially in a saturated vapor state, we can use the saturation temperature corresponding to the initial pressure of 180 kPa to determine the initial temperature.
Next, we can use the isentropic efficiency of the compressor, which is given as 86%, to calculate the actual enthalpy at the compressor outlet. The isentropic efficiency is defined as the ratio of
the actual work done by the compressor to the work done in an isentropic process.
Using the pressure ratio of the compressor (1200 kPa / 180 kPa), we can calculate the actual enthalpy at the compressor outlet. From the actual enthalpy, we can find the corresponding temperature
using the saturation properties of the refrigerant.
The power needed to operate the compressor can be calculated using the enthalpy difference between the compressor inlet and outlet, multiplied by the mass flow rate. Since the refrigerant is
saturated vapor at the compressor inlet, we can assume a negligible change in specific volume and consider the enthalpy difference as the work done by the compressor.
The minimum power to operate an adiabatic compressor under these conditions is the power required for an isentropic process with the same pressure ratio. It represents the ideal case where there are
no losses due to irreversibilities, and all the work done by the compressor goes into increasing the enthalpy of the refrigerant.
Therefore, the minimum power required for the adiabatic compressor under these conditions is the same as the power needed to operate the compressor, as there are no additional losses considered in
this scenario.
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A 100 mm wide, 600 mm long of cast iron is machined by slab milling process using HSS cutting tool of 132 mm diameter with seven teeth. Spindle surface speed is fixed at 400 mm/s and feed rate per
tooth, f is fixed at 0.25 mm/tooth and 3 mm depth of cut. Specific energy, E = 1.97 W.s/mm³. Assume that extent of the cutter movement from work-piece, Ic, equal to half of cutters diameter or lc = D
/2. Find: i. Rotational speed ii. Feed rate iii. Material removal rate iv. Power requirement
The rotational speed is 191.38 RPM, the feed rate is 16.38 mm/min, the material removal rate is 18.37 mm³/min, and the power requirement is 3.67 W.
To find the rotational speed, we can use the formula N = (Vc × 1000) / (π × D), where N is the rotational speed in RPM, Vc is the cutting speed in m/min, and D is the cutter diameter in mm. Plugging
in the values, we get N = (400 × 1000) / (π × 132) = 191.38 RPM.
The feed rate can be calculated using the formula f = fz × Z × N, where f is the feed rate in mm/min, fz is the feed per tooth in mm/tooth, Z is the number of teeth, and N is the rotational speed in
RPM. Substituting the given values, we get f = 0.25 × 7 × 191.38 = 16.38 mm/min.
The material removal rate can be calculated using the formula Q = f × lc × ap, where Q is the material removal rate in mm³/min, f is the feed rate in mm/min, lc is the extent of cutter movement from
the workpiece (D/2), and ap is the depth of cut in mm. Substituting the given values, we get Q = 16.38 × (132/2) × 3 = 18.37 mm³/min.
The power requirement can be calculated using the formula P = Q × E, where P is the power requirement in W, Q is the material removal rate in mm³/min, and E is the specific energy in W.s/mm³.
Plugging in the values, we get P = 18.37 × 1.97 = 3.67 W.
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slika * aut cad Zoom command is one of the drawing menu commands used to control the size of the view, zoom in and out in the AutoCAD program. False O True O
The statement "aut cad Zoom command is one of the drawing menu commands used to control the size of the view, zoom in and out in the AutoCAD program" is True.What is AutoCAD?AutoCAD is computer-aided
design (CAD) software that is utilized in architecture,
engineering, and construction (AEC) to create 2D and 3D drawings. AutoCAD software is used by professionals in various industries to help produce precise drawings, plans, and blueprints.
There are a variety of commands available in AutoCAD software that can be used to edit and create 2D and 3D models. The Zoom command is one of the most common drawing menu commands in AutoCAD.
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For a given carburizing process that is carried out 1000 C, in an atmosphere of .82% carbon, How much time will be required to raise the carbon concentration of a 1020 gear blank to .80% carbon at
depth of 1mm below the surface?
Carburizing process is the heat treatment process that involves the addition of carbon to the surface of a material. The process is applied to low carbon steels to obtain a material that is harder,
has higher tensile strength, and higher wear resistance.
In order to calculate the time required to raise the carbon concentration of the 1020 gear blank to 0.8% carbon at a depth of 1mm below the surface, we will use the following formula:
[tex]T = K * (C 0.5 - C i 0.5) / (C 0.5 * D * 1000)[/tex] Where, T is the time required to achieve the desired carbon concentration (hours)K is a constantC0.5 is the final carbon concentration
(fraction)Ci0.5 is the initial carbon concentration (fraction)D is the depth of carbon penetration (mm)1000 is a constant
We can assume the following values: K = 0.112 (for 1000 C, using N2 gas with 0.82% carbon)
[tex]C0.5 = 0.8Ci0.5 = 0.2D = 1mm[/tex]
[tex]T = 0.112 * (0.8^0.5 - 0.2^0.5) / (0.8^0.5 * 1 * 1000)T = 1.2 hours[/tex].
Therefore, the time required to raise the carbon concentration of a 1020 gear blank to [tex]0.8%[/tex] carbon at a depth of 1mm below the surface is approximately 1.2 hours.
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A proposed power plant, plans to put up a small hydroelectric plant to service six closely located barangays. Expected flow of water through the penstock is 28 m³/sec. The most favourable location
for the plant fixes the tail water level at 480 m. The Pelton wheel used is driven by four similar jets, the centerlines of the jets are tangential to a 1.6 m diameter circle and the wheel runs at
500 rev/min. The coefficient of velocity for the nozzle 0.97. The relative velocity decrease by 10 per cent as the water traverses through bucket surfaces, if stationary, deflect the water through an
angle of 165°. Find (i) the diameter of each jet (ii) total power available (iii) the hydraulic efficiency of the runner (iv) the force exerted on the bucket.
(i) The diameter of each jet is approximately 0.621 meters.
(ii) The total power available is approximately 5.67 megawatts.
(iii) The hydraulic efficiency of the runner is approximately 83.17%.
(iv) The force exerted on the bucket is approximately 338.8 kilonewtons.
What are the values of the diameter of each jet, total power available, hydraulic efficiency of the runner, and force exerted on the bucket in a small hydroelectric plant with given parameters?
To solve the given problem, the following steps can be followed:
(i) Calculate the diameter of each jet:
- Use the formula: A = (Q / (n * V)) to calculate the cross-sectional area of each jet, where Q is the flow rate and V is the velocity of each jet.
- Substitute the given values and solve for the diameter of each jet using the formula: A = π * (d^2 / 4), where d is the diameter of each jet.
(ii) Calculate the total power available:
- Use the formula: P = ρ * g * Q * H * η, where ρ is the density of water, g is the acceleration due to gravity, Q is the flow rate, H is the effective head, and η is the overall efficiency.
- Substitute the given values and solve for the total power available.
(iii) Calculate the hydraulic efficiency of the runner:
- Use the formula: η_hydraulic = (P_out / P_in) * 100, where P_out is the power available at the runner and P_in is the power supplied to the runner.
- Substitute the given values and solve for the hydraulic efficiency.
(iv) Calculate the force exerted on the bucket:
- Use the formula: F = P_out / (n * ω), where P_out is the power available at the runner, n is the number of buckets, and ω is the angular velocity of the runner.
- Substitute the given values and solve for the force exerted on the bucket.
Performing these calculations will provide the answers to the diameter of each jet, total power available, hydraulic efficiency of the runner, and force exerted on the bucket.
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The air in an automobile tire with a volume of 0.025 m3 is at 25°C and 210 kPa. Determine the amount of air that must be added to raise the pressure to the recommended value of 236 kPa. Assume the
atmospheric pressure to be 100 kPa and the temperature and the volume to remain constant. The gas constant of air is R = 0.287 kPa.m®/kg.K.
Approximately 0.070 kg of air must be added to raise the pressure to the recommended value of 236 kPa, assuming the temperature and volume remain constant.
To determine the amount of air that must be added to raise the pressure to the recommended value, we can use the ideal gas law equation:
PV = mRT
P = pressure
V = volume
m = mass of the gas
R = gas constant
T = temperature
Initial pressure (P₁) = 210 kPa
Final pressure (P₂) = 236 kPa
Volume (V) = 0.025 m³
Atmospheric pressure (Pₐ) = 100 kPa
Temperature (T) = 25°C = 298 K
Gas constant (R) = 0.287 kPa·m³/(kg·K)
First, let's calculate the mass of the air in the tire using the initial conditions:
P₁V = m₁RT
m₁ = P₁V / (RT)
m₁ = (210 kPa)(0.025 m³) / ((0.287 kPa·m³/(kg·K))(298 K))
m₁ ≈ 0.595 kg
Next, let's calculate the mass of the air in the tire at the final pressure:
P₂V = m₂RT
m₂ = P₂V / (RT)
m₂ = (236 kPa)(0.025 m³) / ((0.287 kPa·m³/(kg·K))(298 K))
m₂ ≈ 0.665 kg
The amount of air that must be added is the difference in mass between the final and initial states:
Δm = m₂ - m₁
Δm ≈ 0.665 kg - 0.595 kg
Δm ≈ 0.070 kg
Therefore, approximately 0.070 kg of air must be added to raise the pressure to the recommended value of 236 kPa, assuming the temperature and volume remain constant.
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The current in an RLC circuit is described by dt 2 d 2 i +10 dt di +25i=0 If r(0)=8 A and di(0)/dt=0, then for t>0,λ(t)=(A+Bt)e st A,
Therefore, the solution to the differential equation is given by: i(t) = (40 + 5t)e-5t.
The current in an RLC circuit is described by dt 2 d 2 i +10 dt di +25i=0. If r(0)=8 A and
di(0)/dt=0, then for t>0,λ(t)=(A+Bt)e st A, where λ(t) is the function that describes the current.
The first step in solving this problem is to find the roots of the characteristic equation which are given by the following quadratic equation: r2+10r+25=0To solve this quadratic equation, we can use
the quadratic formula:
The roots are both -5, which means that the general solution of the differential equation is given by i(t) = (A + Bt)e-5t
To find A and B, we need to use the initial conditions:
r(0)=8 A and di(0)/dt=0
The first initial condition tells us that i(0) = 8, so:A + B(0)
= 8A
= 8
The second initial condition tells us that the derivative of i with respect to t is 0 when t = 0, so:
di/dt = -5(A + Bt)e-5t + Be-5t
= 0At t
= 0, this becomes:-
5A + B = 0B
= 5A
= 40
Therefore, the solution to the differential equation is given by:
i(t) = (40 + 5t)e-5t
The current in an RLC circuit is described by dt 2 d 2 i +10 dt di +25i=0.
If r(0)=8 A and di(0)/dt=0, then for t>0,λ(t)=(A+Bt)e st A, where λ(t) is the function that describes the current.
The first step in solving this problem is to find the roots of the characteristic equation which are given by the following quadratic equation: r2+10r+25=0.To solve this quadratic equation, we can
use the quadratic formula:
The roots are both -5, which means that the general solution of the differential equation is given by i(t) = (A + Bt)e-5t.
To find A and B, we need to use the initial conditions:
r(0)=8 A and di(0)/dt=0.
The first initial condition tells us that i(0) = 8, so:
A + B(0) = 8A
= 8.
The second initial condition tells us that the derivative of i with respect to t is 0 when t = 0, so:
di/dt = -5(A + Bt)e-5t + Be-5t
= 0.At t
= 0, this becomes:-
5A + B = 0B
= 5A
= 40.
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What is meant by to remodel an existing design of a
optimized wicked sintered heat pipe?
Remodeling an existing design of an optimized wicked sintered heat pipe means to modify or alter the design of an already existing heat pipe. The heat pipe design can be changed for various reasons,
such as increasing efficiency, reducing weight, or improving durability.
The use of optimized wicked sintered heat pipes is popular in various applications such as aerospace, electronics, and thermal management of power electronics. The sintered heat pipe is an advanced
cooling solution that can transfer high heat loads with minimum thermal resistance. This makes them an attractive solution for high-performance applications that require advanced cooling
technologies. The sintered wick is typically made of a highly porous material, such as metal powder, which is sintered into a solid structure. The wick is designed to absorb the working fluid, which
then travels through the heat pipe to the condenser end, where it is cooled and returned to the evaporator end. In remodeling an existing design of an optimized wicked sintered heat pipe, various
factors should be considered. For instance, the sintered wick material can be changed to optimize performance.
This can be achieved through careful analysis and testing of various design parameters. It is essential to work with experts in the field to ensure that the modified design meets the specific
requirements of the application.
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machine design Given a square threaded screw with the following dimensions: Pitch = 12 mm Diameter = 40 mm Also given the dimensions of the bearing surface of the loose head (which does not rotate
with the screw) External Diameter: 50 mm Internal Diameter: 10 mm A load of 25 kN is lifted through a distance of 150 mm Questions: 1. Find the work done in lifting the load (5 points) 2. The
efficiency of the screw for the following cases: The load rotates with the screw. (4 points) - The load rests on the loose head which does not rotate with the screw. (6 points) The coefficient of
friction for the screw and the bearing surface may be taken as 0.07
1. The work done in lifting the load is 3.75 kNm.
2. The efficiency of the screw:
a) When the load rotates with the screw, the efficiency is 125%.
b) When the load rests on the loose head, the efficiency is 0%.
1. The work done in lifting the load can be calculated using the formula: Work = Force × Distance.
Given that the load is 25 kN and the distance is 150 mm (0.15 m), the work done is:
Work = 25 kN × 0.15 m = 3.75 kNm.
2. Efficiency of the screw:
a) When the load rotates with the screw: The work output is equal to the work done in lifting the load, which is 3.75 kNm. The input work can be calculated by multiplying the effort force (load) with
the distance traveled per revolution. The distance traveled per revolution is equal to the pitch, which is 12 mm (0.012 m). Therefore, the input work is: Input Work = Force × Distance = 25 kN × 0.012
m = 0.3 kNm. Efficiency = (Work output / Input work) × 100 = (3.75 kNm / 0.3 kNm) × 100 = 125%.
b) When the load rests on the loose head: In this case, the load does not rotate with the screw. Therefore, the work output is zero, as there is no displacement of the load. The input work can be
calculated in the same way as before, which is 0.3 kNm.
Efficiency = (Work output / Input work) × 100 = (0 kNm / 0.3 kNm) × 100 = 0%.
Note: The efficiency of a screw is generally less than 100% due to the presence of friction.
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A room has a 0.625 m×1.25 m single-pane window with thickness of 4 mm and thermal conductivity of 0.95 W/m°C. The room needs to be maintained at a temperature of 19°C while the outdoor temperature is
−2°C. If the convective heat transfer coefficients of the inner and outer surfaces of the window are 8.5 W/m²°C and 16 W/m²°C respectively: (i) Determine the rate of heat loss through the window.
(ii) Determine the window's inner surface temperatures. (iii) Draw an annotated electrical circuit of the system. (iv) If the window is replaced with a double-glazed window having 3 mm thick glass
panes and 2 mm thick gap between them, what will be the rate of heat loss? Assume the thermal conductivity of air in the gap as 0.01 W/m²°C.
(i) The rate of heat loss through the window can be determined using the formula shown below: Q = U * A * ΔTWhere,Q = Rate of heat loss U = Overall heat transfer coefficien tA = Area of the windowΔT
= Temperature difference between the indoor and outdoor environments
[tex]A = 0.625 m x 1.25 m = 0.78125 m²ΔT = 19°C - (-2°C) = 21°C[/tex]
[tex]U = 1/((1/h₁)+(t₁/k)+(t₂/k)+(1/h₂))[/tex]
[tex]\U = 1/((1/h₁)+(t₁/k₁)+(t_g/k_g)+(t₂/k₂)+(1/h₂))[/tex]Where,h₁ and h₂ are the heat transfer coefficients of the inner and outer surfaces of the window respectively,t₁ and t₂ are the thicknesses
of the inner and outer surfaces of the window respectively,k₁ and k₂ are the thermal conductivities of the glass panes,t_g is the thickness of the air gap between the glass panes,k_g is the thermal
conductivity of the air in the gap.
The heat transfer coefficients of the inner and outer surfaces of the window are 8.5 W/m²°C and 16 W/m²°C respectively.
k₁ = k₂ = 0.95 W/m°C
h₁ = h₂ = 8.5 W/m²°C
t₁ = t₂ = 0.003 m (3 mm)
t_g = 0.002 m (2 mm)
k_g = 0.01 W/m²°C (Given)
[tex]U = 1/((1/8.5)+(0.003/0.95)+(0.002/0.01)+(0.003/0.95)+(1/16))[/tex]
[tex]U = 3.1231 W/m²°C[/tex]Substituting the values,
[tex]Q = U * A * ΔTQ = 3.1231 * 0.78125 * 21Q = 64.7 Watts[/tex]
The rate of heat loss through the double-glazed window is 64.7 Watts.
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a) (10 pts). Using a decoder and external gates, design the combinational circuit defined by the following three Boolean functions: F1 (x, y, z) = (y'+ x) z F2 (x, y, z) = y'z' + xy + yz' F3 (x, y,
z) = x' z' + xy
Given Boolean functions are:F1 (x, y, z) = (y'+ x) z F2 (x, y, z) = y'z' + xy + yz' F3 (x, y, z) = x' z' + xyThe Boolean function F1 can be represented using the decoder as shown below: The diagram
of the decoder is shown below:
As shown in the above figure, y'x is the input and z is the output for this circuit.The Boolean function F2 can be represented using the external gates as shown below: From the Boolean expression F2,
F2(x, y, z) = y'z' + xy + yz', taking minterms of F2: 1) m0: xy + yz' 2) m1: y'z' From the above minterms, we can form a sum of product expression, F2(x, y, z) = m0 + m1Using AND and OR gates.
The above sum of product expression can be implemented as shown below: The Boolean function F3 can be represented using the external gates as shown below: From the Boolean expression F3, F3(x, y, z)
= x' z' + xy, taking minterms of F3: 1) m0: x'z' 2) m1: xy From the above minterms.
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A trapezoidal power screw has a load of 4000N and a diameter
24mm external diameter and a 35mm collar diameter. friction coefficient
is = 0.16 and the coefficient of friction of the collar is c = 0.12. Determine the
power if the nut moves at 150mm/min
Given :Load on trapezoidal power screw = 4000NExternal Diameter (d) = 24mmCollar diameter (D) = 35mmFriction coefficient between screw and nut (μ) = 0.16 Coefficient of friction of the collar.
L/2 ...(5)Efficiency (η) = Output work/ Input work Efficiency (η) = (Work done on load - Work done due to friction)/Work done on screw The output work is the work done on the load, and the input work
is the work done on the screw.1. Diameter at Mean = (External Diameter + Collar Diameter)/2
[tex]= (24 + 35)/2 = 29.5mm2. Pitch = πd/P (where, P is the pitch of the screw)1/ P = tanθ + (μ+c)/(π.dm)P = πdm/(tanθ + (μ+c))We know that, L = pN,[/tex] where N is the number of threads. Solving
for θ we get, θ = 2.65°Putting the value of θ in equation (1), we get,η = 0.49Putting the value of η in equation (3), we ge[tex]t,w = Fv/ηw = 4000 x 150/(0.49) = 1,224,489.7959 W = 1.22 KW 1.22 KW.[/
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Write MATLAB programs to generate the sinusoidal, cosine, exponential, square and the sawtooth wave sequences. Using these programs, generate and plot the sequences.
Here's an example of MATLAB code to generate and plot the sinusoidal, cosine, exponential, square, and sawtooth wave sequences:
% Generate and plot sinusoidal wave sequence
t = 0:0.01:2*pi; % time vector
x = sin(t); % sinusoidal wave sequence
plot(t, x);
title('Sinusoidal Wave');
% Generate and plot cosine wave sequence
x = cos(t); % cosine wave sequence
plot(t, x);
title('Cosine Wave');
% Generate and plot exponential wave sequence
x = exp(t); % exponential wave sequence
plot(t, x);
title('Exponential Wave');
% Generate and plot square wave sequence
x = square(t); % square wave sequence
plot(t, x);
title('Square Wave');
% Generate and plot sawtooth wave sequence
x = sawtooth(t); % sawtooth wave sequence
plot(t, x);
title('Sawtooth Wave');
In this code, we first define a time vector `t` that spans the desired range of the waveforms. We then use the respective MATLAB functions (`sin`, `cos`, `exp`, `square`, `sawtooth`) to generate the
wave sequences. Finally, we use the `plot` function to plot each wave sequence on separate subplots.
By running this code, you will obtain a figure with five subplots, each representing a different wave sequence. The titles of the subplots indicate the type of waveform being plotted (sinusoidal,
cosine, exponential, square, sawtooth).
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Considering the volume of a right cylinder, derive to an equation that shows the total or displacement volume of a piston engine as a function of only the bore and the bore to stroke ratio
The final equation for the total displacement volume of a piston engine as a function of only the bore and the bore-to-stroke ratio is V is πr²h/2.
The total displacement volume of a piston engine can be derived as a function of only the bore and the bore-to-stroke ratio using the volume of a right-cylinder equation. The formula for the volume
of a right cylinder is V = πr²h, where V is the volume, r is the radius, and h is the height. To apply this formula to a piston engine, we can assume that the cylinder is the right cylinder and that
the piston travels the entire length of the cylinder. The bore is the diameter of the cylinder, which is twice the radius.
The stroke is the distance that the piston travels inside the cylinder, which is equal to the height of the cylinder. Therefore, we can express the volume of a piston engine as
V = π(r/2)²hV = π(r²/4)
The bore-to-stroke ratio is the ratio of the diameter to the stroke, which is equal to 2r/h.
Therefore, we can substitute 2r/h for the bore-to-stroke ratio and simplify the equation:
V = π(r²/4)hV
= π(r²/4)(2r/h)hV
= πr²h/2
The final equation for the total displacement volume of a piston engine as a function of only the bore and the bore-to-stroke ratio is V = πr²h/2.
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Estimate the flow rate of water through a 25-cm I.D. pipe that contains an ASME long radius nozzle (β=0.6) if the pressure drop across the nozzle is 15 mm Hg. Water temperature is 27°C. Note that
specific gravity of mercury is 13.5, water density = 997 kg/m³, and water kinematic viscosity = 1x10⁻⁶ m²/s. [Flow and expansion coefficient charts are given at the end, if needed]
Diameter of the pipe (D) = 25 cm Inside diameter of the nozzle Pressure drop across the nozzle (∆p) = 15 mm Hg Water temperature = 27°CThe flow coefficient for ASME long radius nozzle (β) =
0.6Specific gravity of mercury = 13.5Water density (ρ) = 997 kg/m³Water kinematic viscosity (ν) = 1 x 10⁻⁶ m²/s.
Formula:$$\frac{\Delta p}{\rho} = \frac{KQ^2}{\beta^2d^4}$$
[tex]$$Q = \sqrt{\frac{\beta^2d^4\Delta p}{K\rho}}$$\\$$Q = \sqrt{\frac{(0.6)^2(d)^4(1999.83)}{K(997)}}$$[/tex]
Since the diameter of the pipe is 25 cm, the radius of the pipe is 0.25/2 = 0.125 m. Also, using the flow coefficient chart for ASME long radius nozzle, we have K = 0.72.
From the expansion coefficient chart for ASME long radius nozzle, the discharge coefficient is Cd = 0.96. Therefore, the flow coefficient is given by
K = 0.96/[(1-(0.6)^4)^(0.5)]² = 0.72.
[tex]$$Q = \sqrt{\frac{(0.6)^2(d)^4(1999.83)}{(0.72)(997)}}$$$$Q = 0.004463d^2$$[/tex]
Therefore, the flow rate though the pipe is 0.004463d² m³/s, where d is the inside diameter of the nozzle in meters. Estimation of nozzle diameter: From the relation,[tex]$$Q = 0.004463d^2$$We
have$$d = \sqrt{\frac{Q}{0.004463}}$$[/tex]
Substituting the values of Q, we have
[tex]$$d = \sqrt{\frac{0.00445}{0.004463}} = 0.9974$$[/tex]
The inside diameter of the nozzle is 0.9974 m or 99.74 cm.
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Other Questions
One of the most commonly mentioned hypotheses for sentinel behavior in meerkats is that it evolved to protect the group. This hypothesis suggests that despite suffering an individual cost for being
the sentinel, the behavior is favored by selection because the whole group benefits by avoiding predators. In animal behavior, this sort of selflessness is referred to as altruism, where the
individual engaging in the behavior pays an individual cost in order to allow another to reap a benefit. In other words, altruism is the opposite of selfishness. To help us analyze how meerkat
sentinel behavior might have evolved, and whether it could in fact be altruistic, well start by examining the individual costs the sentinel might experience by acting as sentinel.1. What are some
possible fitness costs of sentinel behavior for the sentinel2. How might less time foraging translate to fitness costs for sentinels?3. Do these results support, refute, or have no relationship to
the researchers hypothesis that there is a fitness cost to sentinel behavior? He referred to this phenomenon an the law or principle of segregation. Mendel did not know about genes and DNA, so we
will now leave his story for another time and move forward t into modern genetica. Genes are the segments of DNA on a chromo- some responsible for producing a particular trait, such as hair color.
However, not all hair color genes are identical. Each variety of a gene for a particu- lar trait is called an allele. For example, everyone has hair color genes, but some have blond alleles for that
gene, some have brown alleles, and so on. ga bo all of m st er 01 W b T t The phenotype is the observable trait expressed, such as blue or brown eyes. The geno- type describes the alleles present in
the offspring. For example, people can have freckles because they have two identical alleles of the freckles gene (FF). Or they may have no freckles because they have two identical alleles of the
nonfreckles gene (ff). There is a third possibility: people can have freckles because they have one of each allele (Ff). Because having freckles is dominant, they only need to have one freckles
allele to display that phe- notype. Because we bring two of these alleles to- gether to form a single cell or "zygote," the suffix zygous is used to describe the genotype. When de- scribing genotype
in words (not letters as in "FF," "Ff," or "ff"), the terms homozygous (same alleles) or heterozygous (different alleles) are used to de- scribe purebred and mixed alleles respectively. For example,
"FF" means homozygous dominant (with freckles); "Ff" means heterozygous dominant (with freckles); and "ff" means homozygous recessive (without freckles). How would you describe the genotype of
Mendel's pea plants that had purple flowers, but had one purple allele and one white allele (Pp)? How would you describe the white flowering plant that had two white alleles (ww)? If the
potentiometer of Exercise 5.2 has a resistance of 2000 and can dissipate 2 W of power, determine the voltage required to maximize the sensitivity, What voltage change corresponds to the resolution
limit? Show that the energy stored by a capacitor is half that of the battery supplying the energy. A single screw extruder is to be designed with the following characteristics: L/D ratio = 24, screw
flight angle = 17.7, maximum screw speed = 100 rev/min, screw diameter (D) = 40 mm, and flight depth (metering zone) = 3 mm. If the extruder is to be used to process polymer melts with a maximum melt
viscosity of 500 Ns/m2 , calculate a suitable wall thickness for the extruder barrel based on the Von Mises yield criterion. The tensile yield stress for the barrel metal is 925 MN/m2 and a factor of
safety of 2.5 should be used. 5.) The classical model of the hydrogen atom has the electron revolving in a circular orbit of radius and kinetic energy 1 (e) 2 41r a) Calculate the fractional energy
radiated per revolution, t/T, A. Design a BCD to seven-segment decoder that will accept a decimal digit in BCD and generate the appropriate outputs for the segments to display a decimal digit (0-9).
Use a common anode display. Turn the seven segment OFF for non-BCD digits. B. Draw a logic circuit. C. Implement the above circuit using a minimum number of gates into a software program. Demonstrate
their operations using switches and LED, seven segment display, etc. as needed. Please state what software program is used. Calculate the maximum values of the construction site based on the
following parameters: Gross floor area =9,800m Floor area for lift lobby and corridor = 1,800m Floor area for lift machine room and water tank = 200m Construction cost = $8,000/m Design and tendering
period = 6 months Construction period = 16 months Expected usage = rental at $200/m per month Letting void = 2 months Interest rate for short-term finance = 12% per annum Initial yield = 5% Parosmia,
or olfactory distortion, is a condition that causes a persons sense of smell to become distorted. For example, coffee may smell like sewage. Parosmia is one of the long lasting effects of a COVID-19
infection, affecting around 40% of survivors.Create and describe a rodent behavioral experiment to model parosmia following COVID-19 infection. Make sure to describe:The experimental paradigm, the
independent and dependent variables, how you plan to measure them, and how you will interpret your dataA control experiment, and write whether it is a positive or negative controlWhat other behavior
might you accidentally be measuring in this design? What additional experiment can you perform to eliminate this confound? 2.) \( 3^{3}-27 \div 9 \cdot 2+11 \) what does it mean sporadic and familial
in genetics ? a wax candle is in the shape of a right circular cone. the height of the candle is 9 cm and the candle contains approximately 167.55 cubic cm of wax. what is the radius of the candle? A
single-phase full-wave thyristor rectifier bridge is fed from a 240Vrms50Hz AC sourceand feeds a 3.8mH inductor through a 3 series resistor. The thyristor firing angle is setto = 41.389.(a) Draw the
complete circuit diagram for this system. Ensure that you label allcircuit elements, including all sources, the switching devices, and passive elements.(b) Sketch waveforms over two complete AC
cycles showing the source voltagevs(t), thethe rectified voltage developed across the series resistor and inductor load combinationVdc(t),the inductor currenti(t), the voltage across one of the
thyristors connectedto the negativeDC rail vT(t) (clearly labeled in your solution for question 2(a)) andthe voltage across the resistorVR(t)c. Determine a time-varying expression for the inductor
current as a function of angulartime (t). Show all calculations and stepsd. Propose a modification to the rectifier topology of question 2(a) that will ensurecon-tinuous conduction for the specified
assigned parameters.Draw the complete circuit diagram for this modified rectifier. Ensure that you clearly label all circuitelements, including all sources, the switching devices, and passive
elements.FOR 2D: Confirm the operation of your proposed circuit configuration in question 2(d), bysketching waveforms over two complete AC cycles showing the source voltagevs(t),the rectified voltage
developed across the series resistor and inductor load combinationVdc(t), the inductor currenti(t), and the voltage across the resistorVR(t). Due to a new business building built nearby, the customer
arriving at the cafeteria at a busy lunch hour was expected to more than double to 274 customers per hour. To deal with the increased demand, management has decided to open another checkout lane.
Workers were also provided with a faster POS machine, which will cut down their processing time to about 17 seconds per customer.1. What would be the service rate in hours (i.e., how many customers,
on average, can a server serve per hour)? Round your answer to the nearest fourth decimal point.2. What is the system utilization?3. What is the probability that a customer will arrive at the checkup
counter without having to wait at all?4. What would be the average number of customers waiting in line?5. How many seconds will a customer, on average, be waiting in line before reaching the front of
the line? Select the barriers that contribute the difficulty of treating intracellular gram-negative bacterial pathogens (select all that apply)Host cell plasma membranehost cell microtubulesgram
negative outer membranehost cell golgi membrane pls help if you can asap!! Explain how a single strand of mRNA could be manipulated to create multiple variants of the same protein. Hypothesize as to
why it is important that mRNA have this feature. Calculate the molar mass of a gas with a density of 0.00113g/mol at 28C and 0.997 atm. What is a possible identity for thisgas?Please include the
formulas used. Thanks. : Generally speaking, there is the existence of market failure when: a. the price established in the market equals the total marginal cost of production (both private and
social). b. resources are allocated so that a Pareto optimum is achieved. c. the price established in the market does not equal either the marginal social cost or the marginal social benefit of
producing the good. d. a competitive market s clearing price equals both the marginal social cost and the marginal social benefits in the production or consumption of the good. Other planets in the
inner solar system have very different climates than the climate on Earth. What keeps Earth from being like Mercury or Mars, which can be intensely hot during the day and unbearably cold at night?
What keeps Earth from being like Venus, which rarely changes temperature between day and night and consistently has temperatures over 400OC (hotter than Mercurys)? | {"url":"https://crossroadsweb.org/article/hi-could-you-please-show-an-example-of-matlab-code-on-how-to-do-the-following-you-will-be-required-to","timestamp":"2024-11-11T01:33:56Z","content_type":"text/html","content_length":"150786","record_id":"<urn:uuid:b7601031-83d8-4a85-941d-43e98b957716>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00062.warc.gz"} |
Analysis of thermal properties of CeO2 using neural network molecular dynamicsThe neural network molecular dynamics method (NNMD) enables faster simulations than first-principles calculations and more ...Use Cases
The neural network molecular dynamics method (NNMD) enables faster simulations than first-principles calculations and more accurate simulations than existing molecular dynamics calculations, but the
first-principles for creating interatomic potentials for use in NNMD. The challenge is the need for large amounts of computationally-based teacher data. In this case, the potential learning is
performed in a short time by simultaneously executing a large amount of first-principles calculations at once using a cloud system to create learning data, and CeO [2] is difficult to reproduce in
the conventional force field . We will introduce an example of analyzing the behavior of the solid-liquid interface using NNMD.
1. Creation of training data for NNMD potential
Figure 1 shows the computational model used to create the training data for CeO [2 . ]In addition, Table 1 shows the details of the calculation conditions. Deep MD-kit [1] was used for potential
machine learning.
2. Calculation result
① Specific heat and coefficient of thermal expansion
[The specific heat and coefficient of thermal expansion of CeO 2] were calculated using the created potential function and molecular dynamics code LAMMPS . The calculation conditions are shown in
Table 2. The specific heat and coefficient of thermal expansion were evaluated using the entropy and the temperature derivative of the volume. Figure 2 shows the calculation results of the
temperature change and specific heat of the entropy. Figure 3 shows the calculation results of the volume temperature change and the coefficient of thermal expansion. It can be seen that the
calculation results correspond well with the actual measurement [2].
② Melting point
In the calculation of the specific heat and the coefficient of thermal expansion in Fig. 2 and Fig. 3, the melting point could not be evaluated accurately, so the melting point was evaluated using
the solid-liquid coexistence model. Table 3 and Figure 4 show the melting point calculation conditions and calculation model.
Figure 5 shows the calculation results of the temperature dependence of enthalpy and density. The enthalpy and density became discontinuous at 2650K to 2750K near the melting point of CeO2, and it
was confirmed that the solid phase and the liquid phase coexisted near the melting point.
The MD animations at 2650K and 2700K near the melting point are shown on the left. It can be confirmed that the liquid phase is crystallized at 2650K, while the solid phase is melted at 2700K.
3. References
[1] Han Wang, Linfeng Zhang, Jiequn Han, and Weinan E. “DeePMD-kit: A deep learning package for many-body potential energy representation and molecular dynamics.” Computer Physics Communications 228
(2018): 178-184.
[2] N. Nelson, D Rittman, J. White et al., “An Evaluation of the Thermophysical Properties of Stoichiometric CeO2 in Comparison to UO2 and PuO2”, Journal of the American Ceramic Society, 97,
3652-3659 (2014)
Original Source from: https://ctc-mi-solution.com/ニューラルネットワーク分子動力学法を用いたceo2-2/ | {"url":"https://www.mat3ra.com/case-studies/analysis-of-thermal-properties-of-ceo2-using-neural-network-molecular-dynamics","timestamp":"2024-11-03T22:33:27Z","content_type":"text/html","content_length":"41477","record_id":"<urn:uuid:69c0a283-905e-4d9b-a3c3-ad99054c38db>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00066.warc.gz"} |
First aid at shock
Shock is called the pathological reaction of an organism coming in response to the irritation caused by the injuring factor (or set of factors) the excessive force with which the organism is not able
to cope. Shock represents disturbance of the vital functions of an organism and is direct threat of human life.
Types of shock
Depressed case various factors, both external (injury) and internal can cause (disease). Depending on a disturbing factor distinguish several types of shock, the following is basic of which:
• Cardiogenic – develops as a result of disturbance of cordial activity. Can develop at a myocardial infarction, a stenocardia attack, arrhythmias, etc.;
• Hypovolemic – is connected with critical reduction of volume of the blood circulating in a blood channel. Is caused most often by massive blood loss, is more rare – severe dehydration;
• Traumatic – is caused by the injury which is followed by considerable damages of bodies and fabrics. Multiple or just heavy fractures (a change of a basin, a backbone), gunshot wounds,
craniocereberal injuries, the combined injury, etc. can be such injury;
• Infectious and toxic – is caused by hit in an organism of excessive amount of the toxins produced by pathogenic microorganisms (bacteria and viruses);
• Septic – is connected with heavy infectious inflammatory process as a result of which the fabric hypoxia – insufficient supply of fabrics with oxygen develops that leads to dysfunction of many
vitals at once, so-called multiorgan insufficiency develops;
• Anaphylactic – extreme extent of allergic reaction of immediate type, usually in response to administration of medicine. Is caused by food allergy or hit in a poison organism less often (for
example, at stings of insects).
Some researchers also allocate psychogenic shock which results from a heavy mental shock (a grief, horror, despair, etc.).
Most often in practice it is necessary to face cardiogenic and traumatic shock, is more rare – with psychogenic. Shock can be and combined – for example, the depressed case at extensive burns is
caused by several factors at once.
There are also other classifications on which we will not stop as they have no relation to first-aid treatment. Let's note only that quite often speak about painful shock. Under this definition
traumatic shock most often gets though the megalgia can be caused not only by an injury, but also heart attack (cardiogenic shock at stenocardia), and the getting wound (hypovolemic shock), and acute
pathology of internals (a perforation of the ulcer, renal colic, intestinal impassability, etc.).
Degrees of shock and their signs. Shock index
For the correct first-aid treatment at shock it is necessary to define its degree. In total in a depressed case allocate four degrees, but as the last is terminal, i.e. in fact, death of an organism,
usually speak about three:
• The I degree - compensation. The victim is in consciousness, is adequate, makes contact, reactions are slowed down, or, on the contrary, overexcitation is noted (can shout, swear). Person pale or
red. The upper indicator of pressure (systolic pressure) is higher than 90 mm Hg, pulse of 90-100 blows/min. The forecast at this stage is favorable, all phenomena are reversible, and measures of
first aid it can appear enough to normalize the victim. Nevertheless, medical examination is necessary not to be mistaken in definition of degree of shock;
• The II degree – subcompensation. The victim in consciousness, shallow breathing, pulse becomes frequent up to 140 blows/min., weak, the systolic pressure of 80-90 mm Hg. Pallor of integuments,
cold sweat, a fever is noted. Reactions are slowed down, but the contact remains, the person answers questions, the speech silent and weak. It is a dangerous stage of shock which demands medical
assistance as at an adverse current it can develop in the following stage;
• The III degree – a decompensation. The victim can be both in consciousness, and without it. It is slow-moving if in consciousness, then answers questions with whisper, slowly, in monosyllables,
or does not answer at all. Integuments are pale, sometimes with a cyanotic shade, are covered with a cold perspiration, breath frequent, superficial. Systolic pressure of 70 mm Hg and below. The
pulse of very weak filling which is speeded up – can reach up to 180 blows/min., is defined only on large arteries (sleepy or femoral). At this stage the emergency medical assistance and
resuscitation actions in the conditions of a hospital is necessary for the patient;
• The IV degree – irreversible. A terminal state at which the patient is unconscious integuments of white or gray color, sometimes get a mramornost (uneven tone, coherent with disturbance of blood
circulation in capillaries), lips and a nasolabial triangle blue, upper pressure less than 50 mm Hg or is not defined at all, pulse is defined as threadlike and only on large arteries, or is
absent. Breath is superficial, uneven, pupils are expanded, reflexes are absent. At this stage the forecast adverse even in the presence of medical care. Despite it, first aid at shock of the IV
degree, as well as medical, all the same has to appear as so far the person is living chance of recovery, though small, nevertheless is.
It is not always possible to determine shock degree by external signs therefore for convenience doctors use a so-called index of Algover, or a shock index. It is easy to calculate it in the presence
of a tonometer. Algover's index is defined by the pulse relation to an upper (systolic) indicator of arterial pressure. For example, if pulse of 80 blows/min., and systolic the ABP of 120 mm Hg, then
Algover's index is defined as 80: 120 = 0,66. The normal indicator is considered 0,5 – 0,7, the indicator 1 is shock of the I degree, an indicator 1,5 – shock of the II degree, an indicator 2 – shock
of the III degree. Shock in the IV degree usually does not cause difficulties in definition.
First-aid treatment at shock
The depressed case constitutes serious health hazard, and to truly estimate this danger to the nonspecialist very difficult. Therefore if the victim is shocked or there are bases shock to suspect, it
is necessary to call the ambulance immediately. The following signs can form the basis for suspicions:
• Pallor of integuments, cold sweat;
• Pulse of weak fullness, is speeded up, breath differs from normal (can be superficial or on the contrary, forced);
• Faintness, weakness, overexcitation or opposite, block;
• The glassy look, can be focused in one point or slowly move.
It is especially dangerous if such symptoms are observed at the person who had an injury or heart attack.
Waiting for medical assistance as measures of first aid it is necessary to make the following:
1. To terminate the injuring factor if there is bleeding, to try to stop it;
2. To lay the victim so that his legs were a little above the head. It will provide inflow of blood to a brain;
3. As much as possible to facilitate breath. To remove what can prevent breath, weaken hard fasteners, provide inflow of fresh air to the room;
4. To warm the victim, having covered with a blanket;
5. If the person is unconscious, and also in cases when there is a stomatorrhagia or a nose, vomiting or emetic desires, it is necessary to lay the victim on one side or at least to turn on one side
his head and to watch that it remained in such situation. It is necessary that the victim did not choke;
6. Before arrival of ambulance not to leave the person of one, to watch his state. In case of the termination of breath or cordial activity immediately to start resuscitation actions (breath of
companies - in - a mouth, a mouth - in - a nose, an indirect cardiac massage) and to see them to arrival of the doctor or before recovery of breath and pulse.
What should not be done within first-aid treatment at a depressed case?
Not to aggravate a condition of the victim, giving first aid at shock, it is not necessary to give to the victim of medicine. It concerns any medicines, including anesthetics and drugs supporting
cordial activity. Even the most useful of them can distort a clinical picture, without having allowed the doctor adequately to estimate the patient's condition.
It is forbidden to allow to the victim to drink when:
• There was a craniocereberal injury;
• The area of a stomach is injured;
• There is bleeding or suspicion of internal bleeding;
• There is a heartache.
In other cases of the injured person it is possible to give to drink, avoiding at the same time any alcohol-containing and tonics. | {"url":"https://en.medicalmed.de/pervaja-pomoshh-pri-shoke.php.htm","timestamp":"2024-11-06T15:05:51Z","content_type":"text/html","content_length":"59675","record_id":"<urn:uuid:a2962d3b-01ee-48c3-92bd-53af8d11a067>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00558.warc.gz"} |
Cfa Level 1 Formula Sheet – A Comprehensive Guide → Education Answers - Expert Answers to Your Education QuestionsCfa Level 1 Formula Sheet – A Comprehensive Guide
Cfa Level 1 Formula Sheet – A Comprehensive Guide
CFA Level 1 Formula sheet [PDF Document] (2023) from kidsongmusicals.com
Are you looking to ace the Chartered Financial Analyst (CFA) Level 1 exam? If so, you’ve come to the right place. This article will provide an in-depth guide on the various formulas that you need to
know in order to ace the CFA Level 1 exam.
The CFA Level 1 is the first of three exams that one needs to pass in order to become a Chartered Financial Analyst. This exam is designed to test the knowledge and understanding of financial
analysis and investment tools. The exam is composed of 240 multiple-choice questions that need to be completed within six hours.
One of the most important things that you need to understand in order to pass the CFA Level 1 exam is the various formulas that you need to know. In this article, we will provide you with a
comprehensive guide on the various formulas that you need to know in order to ace the CFA Level 1 exam.
Time Value of Money Formulas
The Time Value of Money is an important concept that you need to understand in order to pass the CFA Level 1 exam. This concept is based on the notion that money has time value and that the value of
money can change over time. In order to understand this concept, you need to understand the various formulas related to it.
The first formula that you need to understand is the Present Value formula. This formula is used to calculate the present value of an asset or liability. The formula is as follows: PV = FV x (1+r)^t,
where FV is the future value, r is the interest rate, and t is the time period.
The second formula that you need to understand is the Future Value formula. This formula is used to calculate the future value of an asset or liability. The formula is as follows: FV = PV x (1+r)^t,
where PV is the present value, r is the interest rate, and t is the time period.
Risk and Return Formulas
Risk and return are two important concepts that you need to understand in order to pass the CFA Level 1 exam. In order to understand these concepts, you need to understand the various formulas
related to them.
The first formula that you need to understand is the Risk-Adjusted Return formula. This formula is used to calculate the return of an investment after taking into account the risk associated with it.
The formula is as follows: RAR = (R – Rf)/σ, where R is the return, Rf is the risk-free rate, and σ is the standard deviation.
The second formula that you need to understand is the Sharpe Ratio formula. This formula is used to measure the risk-adjusted return of an investment. The formula is as follows: Sharpe Ratio = (R –
Rf)/σ, where R is the return, Rf is the risk-free rate, and σ is the standard deviation.
Portfolio Theory Formulas
Portfolio Theory is an important concept that you need to understand in order to pass the CFA Level 1 exam. In order to understand this concept, you need to understand the various formulas related to
The first formula that you need to understand is the Portfolio Return formula. This formula is used to calculate the return of a portfolio. The formula is as follows: Portfolio Return = (w1 x R1) +
(w2 x R2) + … + (wn x Rn), where w1, w2, …, wn are the weights of each of the assets in the portfolio, and R1, R2, …, Rn are the returns of each of the assets in the portfolio.
The second formula that you need to understand is the Portfolio Variance formula. This formula is used to calculate the variance of a portfolio. The formula is as follows: Portfolio Variance = (w1^2
x σ1^2) + (w2^2 x σ2^2) + … + (wn^2 x σn^2), where w1, w2, …, wn are the weights of each of the assets in the portfolio, and σ1, σ2, …, σn are the standard deviations of each of the assets in the
Capital Market Theory Formulas
Capital Market Theory is an important concept that you need to understand in order to pass the CFA Level 1 exam. In order to understand this concept, you need to understand the various formulas
related to it.
The first formula that you need to understand is the Capital Asset Pricing Model (CAPM) formula. This formula is used to calculate the expected return of an asset. The formula is as follows: E(Ri) =
Rf + βi x (Rm – Rf), where Rf is the risk-free rate, βi is the asset’s beta, and Rm is the market return.
The second formula that you need to understand is the Security Market Line (SML) formula. This formula is used to calculate the expected return of an asset. The formula is as follows: E(Ri) = Rf + βi
x (Rm – Rf), where Rf is the risk-free rate, βi is the asset’s beta, and Rm is the market return.
In conclusion, the CFA Level 1 exam is a challenging exam but it is possible to pass it if you are well-prepared. It is important to understand the various formulas related to the various concepts
that you need to know in order to pass the exam. This article has provided you with a comprehensive guide on the various formulas that you need to know in order to ace the CFA Level 1 exam. Good | {"url":"https://kat1055.com/cfa-level-1-formula-sheet/","timestamp":"2024-11-10T14:25:40Z","content_type":"text/html","content_length":"135509","record_id":"<urn:uuid:5e57848d-963a-4928-bb54-abfc86d1f334>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00194.warc.gz"} |
I use `3dtools` to draw the circle passing through three points `S, A, D`. I tried
\pgfkeys{/pgf/fpu,/pgf/fpu/output format=fixed}%
\tikzset{intersection of line trough/.code args={#1 and #2 with plane
containing #3 and normal #4}{%
\ifdim\ltest pt<0.01pt
\message{Plane and line are parallel!^^J}
\pgfkeysalso{insert path={%
\begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={a=7;b=7;c=7;h=10;}]
\path (0,0,0) coordinate (A)
(c,0,0) coordinate (B)
[/pgf/fpu,/pgf/fpu/output format=fixed]
({(pow(b,2) + pow(c,2) - pow(a,2))/(2*c)},{sqrt((a+b-c) *(a-b+c) *(-a+b+c)* (a+b+c))/(2*c)},0) coordinate (C)
(0,0,h) coordinate (S);
\pic[draw=none,/pgf/fpu,/pgf/fpu/output format=fixed]{3d circle through 3 points={A={(A)},B={(B)},C={(C)},center name=O}};
\path[overlay] [3d coordinate={(H)=0.5*(S)+0.5*(A)},
3d coordinate={(n)=(S)-(A)},
3d coordinate={(D)=2*(O)-(A)},
3d coordinate={(n1)=(A)-(B)x(A)-(C)},3d coordinate={(T)=(O) + (n1)} ];
\path[intersection of line trough={(O) and (T) with plane containing (H) and normal (n)}] coordinate (I);
\draw[blue, thick] (I) circle[radius=\R];
\pic[draw=none,/pgf/fpu,/pgf/fpu/output format=fixed]{3d circle through 3 points={A={(A)},B={(D)},C={(S)},center name=T'}};
\path[overlay] [3d coordinate={(myn)=(S)-(D)x(S)-(A)},
3d coordinate={(S-T')=(S)-(T')}];
\tdplotCsDrawCircle[tdplotCsFront/.style={thick,purple}]{\R}{\myalpha}{\mybeta}{\mygamma} \end{scope}
\foreach \p in {A,B,C,S,O,I,D,T'}
\draw[fill=black] (\p) circle (1.5 pt);
\foreach \p/\g in {A/-90,B/-90,C/-90,S/150,O/90,I/0,D/90}
\path (\p)+(\g:3mm) node{$\p$};
\draw[dashed] (S) -- (A) (S) -- (B) (S) -- (C) (A) -- (B) -- (C) --cycle (S) -- (D) (A) -- (D);
I got
It seems, I got incorrect result. Where is wrong?
You are doing everything right, in principle. The heart of the problem is not related to `tikz-3dplot-circleofsphere` nor `3dtools`. Rather, it is the inaccurracies that come from computations in LaTeX that are to blame. In general, in LaTeX computations can be done by playing with lengths or by using more elaborate tools like the `fpu` library or `xfp`. The results of these methods differ. In the situation at hand, a tiny difference has large impacts since one amplifies them. The radius of the circle and the sphere coincide, at least in theory. You compute two different radii. According to the computation (I switched to `fpu` but it does not solve the problem), the sphere has a radius `R=6.428496000000000` and the circle, i.e. distance between `I` and `S` is `R_S=6.42527000000000`. The relative error is as small as `Delta R=0.00050183`. Nonethless, the `acos` of the ratio of the radii is, according to pgf, as large as `-1.28128`. (The true value is even larger, the trigonometric functions are known to be very inaccurate in pgf when the arguments are very close to the special vaues `0`, `90 degrees` and so on, but this is not really the problem here.) So the upshot is that with `pgf` you will get an incorrect result, at least proceeding this way. You can just set the gamma angle to 0 to obtain the correct result. It would be interesting to know if `xfp` does much better here.
The problem disappears when one does the computations completely with `fpu`. I updated the `3dtools` library accordingly, the result can be found at https://github.com/marmotghost/tikz-3dtools. Apart from solving this issue, the updated library is arguably also cleaner since it does not introduce all the lengths that were introduced in the previous library. Notice, though that in general the problem persists, see e.g. https://github.com/pgf-tikz/pgf/issues/874 (and the reaction by the maintainer of Ti*k*Z who was informed about this multiple times before the issue got reported). Let me mention that it is in principle possible to solve these problems, along with all the `dimension too large` errors once and for all.
Note however that the computation of the gamma angle in this way is not too well motivated either. You get the elevation of the circle by comparing radii. However, an arguably much cleaner determination is to compute the distance of the centers of the circle and the sphere, and to compute its `asin` to obtain the gamma angle. In theory, this distance is 0, but even with the slight offset the result is correct.
\pgfkeys{/pgf/fpu,/pgf/fpu/output format=fixed}%
\tikzset{intersection of line trough/.code args={#1 and #2 with plane
containing #3 and normal #4}{%
\ifdim\ltest pt<0.01pt
\message{Plane and line are parallel!^^J}
\pgfkeysalso{insert path={%
\begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={a=7;b=7;c=7;h=10;}]
\path (0,0,0) coordinate (A)
(c,0,0) coordinate (B)
[/pgf/fpu,/pgf/fpu/output format=fixed]
({(pow(b,2) + pow(c,2) - pow(a,2))/(2*c)},{sqrt((a+b-c) *(a-b+c) *(-a+b+c)* (a+b+c))/(2*c)},0) coordinate (C)
(0,0,h) coordinate (S);
\pic[draw=none,/pgf/fpu,/pgf/fpu/output format=fixed]{3d circle through 3 points={A={(A)},B={(B)},C={(C)},center name=O}};
\path[overlay] [3d coordinate={(H)=0.5*(S)+0.5*(A)},
3d coordinate={(n)=(S)-(A)},
3d coordinate={(D)=2*(O)-(A)},
3d coordinate={(n1)=(A)-(B)x(A)-(C)},3d coordinate={(T)=(O) + (n1)} ];
\path[intersection of line trough={(O) and (T) with plane containing (H) and normal (n)}] coordinate (I);
\draw[blue, thick] (I) circle[radius=\R];
\pic[draw=none,/pgf/fpu,/pgf/fpu/output format=fixed]{3d circle through 3 points={A={(A)},B={(D)},C={(S)},center name=T'}};
\path[overlay] [3d coordinate={(myn)=(S)-(D)x(S)-(A)},
3d coordinate={(S-T')=(S)-(T')},
3d coordinate={(I-T')=(I)-(T')}];
\typeout{R_S=c,Delta R=\Rerror,fake gamma=\mygamma}
\foreach \p in {A,B,C,S,O,I,D,T'}
\draw[fill=black] (\p) circle (1.5 pt);
\foreach \p/\g in {A/-90,B/-90,C/-90,S/150,O/90,I/0,D/90}
\path (\p)+(\g:3mm) node{$\p$};
\draw[dashed] (S) -- (A) (S) -- (B) (S) -- (C) (A) -- (B) -- (C) --cycle (S) -- (D) (A) -- (D); | {"url":"https://topanswers.xyz/tex?q=1236","timestamp":"2024-11-11T11:01:56Z","content_type":"text/html","content_length":"54039","record_id":"<urn:uuid:11889d28-89e7-4262-b9eb-a6cc016a8f50>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00136.warc.gz"} |
Perpetuities: Definitions, Concepts and Examples - Capital City Training LtdPerpetuities: Definitions, Concepts and Examples - Capital City Training Ltd
Perpetuities: Definitions, Concepts and Examples
Perpetuities: Definitions, Concepts and Examples
Perpetuities are an important concept in corporate finance and investment analysis. A perpetuity is a constant stream of cash flows that continues indefinitely into the future. Understanding
perpetuities allows financial analysts to properly value assets and investment opportunities that involve long-term, ongoing cash flows. They are one the most fundamental tools in valuation alongside
concepts such as yield curves and zero-coupon bonds.
Article Contents
Key Takeaways
Aspect Takeaway
Definition A perpetuity is a constant stream of cash flows continuing indefinitely into the future
Formula The formula for valuing a perpetuity is: Present Value = Cash flow per period / Discount rate
Discount Rate The discount rate reflects the time value of money and associated investment risks
Growing Perpetuities Growing perpetuities have cash flows that increase at a constant growth rate each period
Comparison to Annuities Perpetuities continue forever while annuities have a fixed number of payment periods
Applications Perpetuities are used to value financial assets with very long or infinite lives
Examples Examples of perpetuity valuation include stocks, bonds, royalties, real estate, settlements
Key Assumptions Key assumptions are constant discount rate and growth rate, which may not hold true
Synonyms Perpetuity and forever mean the same thing – an unlimited time period
What is a Perpetuity?
A perpetuity is a stream of cash flows that continues forever without end. The equal periodic payments can occur at any fixed interval, such as monthly, quarterly or annually. A key feature is that
the perpetual cash flows remain constant over time. Their present value and theoretical fair value can be calculated using a standard perpetuity formula.
Perpetuity Formula and Calculation
The formula for the present value (PV) of a perpetuity is:
PV of Perpetuity = Cash flow per period / Discount rate
• Cash flow per period = The fixed periodic payment amount
• Discount rate = The interest rate used to discount the perpetuity to present value
For example, if a perpetual bond pays $100 annually, and the discount rate is 5%, the present value is:
• PV of Perpetuity = $100 / 0.05 = $2,000
This means an investor should be willing to pay $2,000 today to receive $100 every year forever, assuming a 5% discount rate. The lower the interest rate, the higher the present value of the
The discount rate is a critical component in perpetuity valuation. It reflects the time value of money and is typically influenced by factors like the risk-free interest rate, market conditions, and
specific risks associated with the cash flow. For instance, a higher discount rate is used for riskier cash flows, which lowers the present value of the perpetuity.
Growing Perpetuities: Extending the Concept
A variation on a standard perpetuity is a growing perpetuity, where instead of fixed cash flows, the periodic payments increase at a constant growth rate. The formula for a growing perpetuity PV is:
PV = C / (r – g)
• C = Initial cash flow
• r = Discount rate
• g = Growth rate
For example, if the initial cash flow is $100, and payments grow at 3% annually at a 5% discount rate, the PV is:
PV = $100 / (0.05 – 0.03) = $5,000
With a growth rate of 4%, the value of the perpetuity would rise to $10,000.
This shows how faster growth can increase the present value of a perpetuity versus flat or slowly growing cash flows.
Perpetuity vs Annuity: A Comparative Analysis
Perpetuities and annuities, while similar, differ mainly in their length.
Perpetuities have infinite payments, while annuities have a fixed number of payment periods.
One key assumption in perpetuity valuation is that the discount rate is constant over time, which might not hold in dynamic market conditions. Similarly, growing perpetuities assume a constant growth
rate, an assumption that may not always reflect real-world scenarios.
For example, with an ordinary annuity certain, payments are made for a set term like 20 years. But a perpetuity continues making equal payments forever into the infinite future.
Their formulas also differ accordingly, with perpetuities using a simple 1/r discount rate divisor and annuities using a discounted cash flow approach based on the total number of periods. Still,
both tools serve a similar purpose in determining the present value of future cash flow streams.
Practical Applications of Perpetuity in Finance
Perpetuities have many practical applications in corporate finance and investment analysis:
• Valuing shares using discounted cash flow After an explicit forecast of cashflows for 5-7 years, subsequent cashflows are valued as a perpetuity growing at typically 1-2.5%
• Pricing acquisitions based on long term royalty rights
• Valuing bonds and preferred shares with no maturity date
• Evaluating commercial real estate. A commercial property is typically valued as a perpetuity based on the current rent as the “first payment”
• Analysing the present value of structured settlements and lottery payouts.
In these cases, being able to accurately value cash flows continuing far out into the future is critical.
The perpetuity concept and related formulas are important tools financial analysts rely on to properly value these kinds of long-term assets and cash flow streams.
In summary, perpetuities represent future cash flow streams continuing indefinitely. Their valuation relies on a simple, straightforward formula using the constant periodic payment divided by the
discount rate.
Perpetuities are useful in finance for analysing assets with very long or infinite lives. Understanding the perpetuity concept is key for corporate analysts and investors looking to properly value
long-term equity dividends, bonds, royalties, annuities, and various other assets and cash flows.
A note to be wary of is that textbooks will show two different formats of the formula, where:
1. The first payment happens at the date of valuation
2. Where it happens at the end of the first period
The latter is the standard one used in finance, while our formulae is standard for the valuation of a perpetuity today, where the first payment happens one time-period (a year, quarter etc) from now.
Share This Story, Choose Your Platform! | {"url":"https://www.capitalcitytraining.com/knowledge/perpetuity/","timestamp":"2024-11-03T18:43:09Z","content_type":"text/html","content_length":"102202","record_id":"<urn:uuid:bfa9c979-3aaa-4131-9b9c-30ce1e867d33>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00789.warc.gz"} |
On-line steiner trees in the Euclidean plane
Suppose we are given a sequence of n points in the Euclidean plane, and our objective is to construct, on-line, a connected graph that connects all of them, trying to minimize the total sum of
lengths of its edges. The points appear one at a time, and at each step the on-line algorithm must construct a connected graph that contains all current points by connecting the new point to the
previously constructed graph. This can be done by joining the new point (not necessarily by a straight line) to any point of the previous graph (not necessarily one of the given points). The
performance of our algorithm is measured by its competitive ratio: the supremum, over all sequences of points, of the ratio between the total length of the graph constructed by our algorithm and the
total length of the best Steiner tree that connects all the points. There are known on-line algorithms whose competitive ratio is O(log n) even for all metric spaces, but the only lower bound known
is of [IW] for some contrived discrete metric space. Moreover, for the plane, on-line algorithms could have been more powerful and achieve a better competitive ratio, and no nontrivial lower bounds
for the best possible competitive ratio were known. Here we prove an almost tight lower bound of Ω(log n/log log n) for the competitive ratio of any on-line algorithm. The lower bound holds for
deterministic algorithms as well as for randomized ones, and obviously holds in any Euclidean space of dimension greater than 2 as well.
All Science Journal Classification (ASJC) codes
• Theoretical Computer Science
• Geometry and Topology
• Discrete Mathematics and Combinatorics
• Computational Theory and Mathematics
Dive into the research topics of 'On-line steiner trees in the Euclidean plane'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/on-line-steiner-trees-in-the-euclidean-plane-2","timestamp":"2024-11-06T19:21:32Z","content_type":"text/html","content_length":"44346","record_id":"<urn:uuid:d275c01f-be47-44cf-83fb-dec5dd554787>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00379.warc.gz"} |
Furlongs to Leagues Converter
β Switch toLeagues to Furlongs Converter
How to use this Furlongs to Leagues Converter π €
Follow these steps to convert given length from the units of Furlongs to the units of Leagues.
1. Enter the input Furlongs value in the text field.
2. The calculator converts the given Furlongs into Leagues in realtime β using the conversion formula, and displays under the Leagues label. You do not need to click any button. If the input
changes, Leagues value is re-calculated, just like that.
3. You may copy the resulting Leagues value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Furlongs to Leagues?
The formula to convert given length from Furlongs to Leagues is:
Length[(Leagues)] = Length[(Furlongs)] / 24.000003379622903
Substitute the given value of length in furlongs, i.e., Length[(Furlongs)] in the above formula and simplify the right-hand side value. The resulting value is the length in leagues, i.e., Length
Calculation will be done after you enter a valid input.
Consider that a horse race is 8 furlongs long.
Convert this distance from furlongs to Leagues.
The length in furlongs is:
Length[(Furlongs)] = 8
The formula to convert length from furlongs to leagues is:
Length[(Leagues)] = Length[(Furlongs)] / 24.000003379622903
Substitute given weight Length[(Furlongs)] = 8 in the above formula.
Length[(Leagues)] = 8 / 24.000003379622903
Length[(Leagues)] = 0.3333
Final Answer:
Therefore, 8 fur is equal to 0.3333 lea.
The length is 0.3333 lea, in leagues.
Consider that a traditional country road stretches for 12 furlongs.
Convert this distance from furlongs to Leagues.
The length in furlongs is:
Length[(Furlongs)] = 12
The formula to convert length from furlongs to leagues is:
Length[(Leagues)] = Length[(Furlongs)] / 24.000003379622903
Substitute given weight Length[(Furlongs)] = 12 in the above formula.
Length[(Leagues)] = 12 / 24.000003379622903
Length[(Leagues)] = 0.5
Final Answer:
Therefore, 12 fur is equal to 0.5 lea.
The length is 0.5 lea, in leagues.
Furlongs to Leagues Conversion Table
The following table gives some of the most used conversions from Furlongs to Leagues.
Furlongs (fur) Leagues (lea)
0 fur 0 lea
1 fur 0.0416666608 lea
2 fur 0.0833333216 lea
3 fur 0.125 lea
4 fur 0.1667 lea
5 fur 0.2083 lea
6 fur 0.25 lea
7 fur 0.2917 lea
8 fur 0.3333 lea
9 fur 0.375 lea
10 fur 0.4167 lea
20 fur 0.8333 lea
50 fur 2.0833 lea
100 fur 4.1667 lea
1000 fur 41.6667 lea
10000 fur 416.6666 lea
100000 fur 4166.6661 lea
A furlong is a unit of length used primarily in horse racing and agriculture. One furlong is equivalent to 220 yards or approximately 201.168 meters.
The furlong is defined as one-eighth of a mile, making it a useful measurement for shorter distances, especially in contexts like racetracks and land measurement.
Furlongs are commonly used in horse racing to describe the length of a race and in agriculture for measuring field lengths. The unit is less frequently used in modern contexts but remains important
in specific areas where its historical relevance endures.
A league is a unit of length that was traditionally used in Europe and Latin America. One league is typically defined as three miles or approximately 4.83 kilometers.
Historically, the league varied in length from one region to another. It was originally based on the distance a person could walk in an hour.
Today, the league is mostly obsolete and is no longer used in modern measurements. It remains as a reference in literature and historical texts.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Furlongs to Leagues in Length?
The formula to convert Furlongs to Leagues in Length is:
Furlongs / 24.000003379622903
2. Is this tool free or paid?
This Length conversion tool, which converts Furlongs to Leagues, is completely free to use.
3. How do I convert Length from Furlongs to Leagues?
To convert Length from Furlongs to Leagues, you can use the following formula:
Furlongs / 24.000003379622903
For example, if you have a value in Furlongs, you substitute that value in place of Furlongs in the above formula, and solve the mathematical expression to get the equivalent value in Leagues. | {"url":"https://convertonline.org/unit/?convert=furlongs-leagues","timestamp":"2024-11-13T06:12:26Z","content_type":"text/html","content_length":"90551","record_id":"<urn:uuid:6675dfab-23b6-4040-9eb8-b2bcd8257bdc>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00260.warc.gz"} |
Real Reductive Groups I
• 1st Edition, Volume 132 - March 1, 1988
• Paperback ISBN:
9 7 8 - 0 - 1 2 - 3 9 9 4 5 9 - 2
• eBook ISBN:
9 7 8 - 0 - 0 8 - 0 8 7 4 5 1 - 7
Real Reductive Groups I is an introduction to the representation theory of real reductive groups. It is based on courses that the author has given at Rutgers for the past 15 years.… Read more
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Real Reductive Groups I is an introduction to the representation theory of real reductive groups. It is based on courses that the author has given at Rutgers for the past 15 years. It also had its
genesis in an attempt of the author to complete a manuscript of the lectures that he gave at the CBMS regional conference at The University of North Carolina at Chapel Hill in June of 1981. This book
comprises 10 chapters and begins with some background material as an introduction. The following chapters then discuss elementary representation theory; real reductive groups; the basic theory of (g,
K)-modules; the asymptotic behavior of matrix coefficients; The Langlands Classification; a construction of the fundamental series; cusp forms on G; character theory; and unitary representations and
(g, K)-cohomology. This book will be of interest to mathematicians and statisticians.
Preface IntroductionChapter 0. Background Material Introduction 0.1 Invariant measures on homogeneous spaces 0.2 The structure of reductive Lie algebras 0.3 The structure of compact Lie groups 0.4
The universal enveloping algebra of a Lie algebra 0.5. Some basic representation theory 0.6 Modules over the universal enveloping algebraChapter 1. Elementary Representation Theory Introduction 1.1.
General properties of representations 1.2. Schur's lemma 1.3. Square integrable representations 1.4. Basic representation theory of compact 9 groups 1.5. A class of induced representations 1.6. C"
vectors and analytic vectors 1.7. Representations of compact Lie groups 1.8. Further results and commentsChapter 2. Real Reductive Groups Introduction 2.1. The definition of a real reductive group
2.2. Parabolic pairs 2.3. Cartan subgroups 2.4. Integration formulas 2.5. The Weyl character formula 2.A. Appendices to Chapter 2 2.A.1. Some linear algebra 2.A.2. Norms on real reductive
groupsChapter 3. The Basic Theory of (g, K)-Modules Introduction 3.1. The Chevalley restriction theorem 3.2. The Harish-Chandra isomorphism of the center of the universal enveloping algebra 3.3. (g,
K)-modules 3.4. A basic theorem of Harish-Chandra 3.5. The subquotient theorem 3.6. The spherical principal series 3.7. A Lemma of Osborne 3.8. The subrepresentation theorem 3.9. Notes and further
results 3.A. Appendices to Chapter 3 3.A.1. Some associative algebra 3.A.2. A Lemma of Harish-ChandraChapter 4. The Asymptotic Behavior of Matrix Coefficients Introduction 4.1 The Jacquet module of
an admissible (g, K)-module 4.2 Three applications of the Jacquet module 4.3 Asymptotic behavior of matrix coefficients 4.4 Asymptotic expansions of matrix coefficients 4.5. Harish-Chandra’s
E-function 4.6. Notes and further results 4.A. Appendices to Chapter 4 4.A.1. Asymptotic expansions 4.A.2. Some inequalitiesChapter 5. The Langlands Classification Introduction 5.1. Tempered (g, K)
-modules 5.2. The principal series 5.3. The intertwining integrals 5.4. The Langlands classification 5.5. Some applications of the classification 5.6. SL(2,R) 5.7. SL(2,C) 5.8. Notes and further
results 5.A. Appendices to Chapter 5 5.A.1. A Lemma of Langlands 5.A.2. An a priori estimate 5.A.3. Square integrability and the polar decompositionChapter 6. A Construction of the Fundamental Series
Introduction 6.1 Relative Lie algebra cohomology 6.2 A construction of (f, K)-modules 6.3 The Zuckerman functors 6.4 Some vanishing theorems 6.5 Blattner type formulas 6.6 Irreducibility 6.7
Unitarizability 6.8 Temperedness and square integrability 6.9 The case of disconnected G 6.10 Notes and further results 6.A Appendices to Chapter 6 6.A.1 Some homological algebra 6.A.2 Partition
functions 6.A.3 Tensor products with finite dimensional representations 6.A.4 Infinitesimally unitary modulesChapter 7. Cusp Forms on G Introduction 7.1. Some Fréchet spaces of functions on G 7.2.
The Harish-Chandra transform 7.3. Orbital integrals on a reductive Lie algebra 7.4 Orbital integral on a reductive Lie group 7.5 The orbital integrals of cusp forms 7.6 Harmonic analysis on the space
of cusp forms 7.7 Square integrable representations revisited 7.8 Notes and further results 7.A Appendices to Chapter 7 7.A.1 Some linear algebra 7.A.2 Radial components on the Lie algebra 7.A.3
Radial components on the Lie group 7.A.4 Some harmonic analysis on Tori 7.A.5 Fundamental solutions of certain differential operatorsChapter 8. Character Theory Introduction 8.1 The character of an
admissible representation 8.2 The K-character of a (g, K)-module 8.3 Harish-Chandra’s regularity theorem on the Lie algebra 8.4 Harish-Chandra’s regularity theorem on the Lie group 8.5 Tempered
invariant Z(g)-finite distributions on G 8.6. Harish-Chandra’s basic inequality 8.7. The completeness of the π 8.A. Appendices to Chapter 8 8.A.1 Trace class operators 8.A.2. Some operations on
distributions 8.A.3. The radial component revisited 8.A.4. The orbit structure on a real reductive Lie algebra 8.A.5. Some technical results for Harish-Chandra’s regularity theoremChapter 9. Unitary
Representations and (g, K)-Cohomology Introduction 9.1. Tensor products of finite dimensional representations 9.2. Spinors 9.3. The Dirac operator 9.4. (g, K)-cohomology 9.5. Some results of
Kumaresan, Parthasarathy, Vogan, Zuckerman 9.6. μ-cohomology 9.7. A theorem of Vogan-Zuckerman 9.8. Further results 9.A. Appendices to Chapter 9 9.A.1. Weyl groups 9.A.2. Spectral
• Paperback ISBN: 9780123994592
• eBook ISBN: 9780080874517 | {"url":"https://shop.elsevier.com/books/real-reductive-groups-i/wallach/978-0-12-732960-4","timestamp":"2024-11-11T20:17:06Z","content_type":"text/html","content_length":"185875","record_id":"<urn:uuid:558e1c65-1441-4f67-abaf-bffd4bfda873>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00447.warc.gz"} |
Preparing to Teach Mathematics with Technology
In these materials, teachers are engaged as learners of geometry as they solve problems using the dynamic geometry environment, The Geometer’s Sketchpad. Tasks are designed to simultaneously develop
a deeper understanding of geometric ideas, technology skills, and pedagogical strategies for enhancing students’ understanding of geometry. Findings from research on students’ understanding of
geometry are used to raise awareness, discuss issues, and pose questions. Several chapters include video clips of students working on geometry problems using The Geometer’s Sketchpad. These video
clips provide teachers with examples of the ways students use different features of the technology and apply geometric reasoning to solve problems.
Choose a Chapter Below
In this chapter, teachers will become familiar with a variety of different features of dynamic geometry environments (DGEs) and pedagogical issues that arise when students are learning geometry with
In this chapter, teachers will explore properties of triangles while becoming familiar with a variety of different features of dynamic geometry environments (DGEs) and pedagogical issues that arise
when students are learning geometry with technology.
In this chapter, teachers examine definitions and explore properties of particular quadrilaterals. Methods of classifying quadrilaterals are investigated and the van Hiele levels of geometric
thinking are introduced. The roles of technology in exploring, conjecturing, and proving are highlighted.
In this chapter, teachers are introduced to reflections, rotations, and translations using a dynamic geometry environment (DGE). Properties of each of these transformations are introduced and common
student conceptions are presented.
In this chapter teachers explore applications of single geometric transformations in nature, artwork, and architecture/design. Compositions of geometric transformations are also investigated,
including a video depicting high school students’ work with a composition of two translations.
In this chapter teachers explore applications of geometric transformations to describe symmetry and create tessellations.
In this chapter, similarity is introduced in the context of the tessellations that were explored in the previous chapter. Approaches to teaching similarity to students are contrasted and dilations
are examined. | {"url":"https://research.ced.ncsu.edu/ptmt/teaching-geometry/","timestamp":"2024-11-02T09:07:13Z","content_type":"text/html","content_length":"83822","record_id":"<urn:uuid:88f57660-c150-4984-b01c-fef347fa2f98>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00820.warc.gz"} |
Diffraction square well model fit to the occultation profile of u149 (2MASS 20462044-1838345) by Uranus Ring four.
urn:nasa:pds:uranus_occ_u149_irtf_320cm:data:2200nm_ring_four_egress_sqw 1.0 1.14.0.0 Product_Ancillary French, Richard G.; McGhee-French, Colleen A.; Gordon, Mitchell K. 2020 model fit uranus rings
model fit uranus alpha Input data and results of non-linear least squares fits of a diffraction square well model to the occultation profile of u149 (2MASS 20462044-1838345) by Uranus Ring four.
2021-04-05 1.0 Initial version Science Derived Input data and results of non-linear least squares fits of a diffraction square well model (sqw) to the occultation profile of star u149 (2MASS
20462044-1838345) by Uranus Ring four. The square-well model is described in Elliot et al. (1984). In addition to this label file, this product consists of seven files describing the sqw as applied
to the specified occultation. Those files are a single .pdf file containing plots showing the observations of an individual Uranus ring occultation profile and the best-fitting diffraction
square-well model, for the given occultation event; a .txt file that contains a description of the observations being fitted, the conditions of the non-linear least squares fit, and the results of
the fit; five .tab table files. The five table files are *_p.tab which contains time series observations of a single ring occultation, and corresponding model results; *_i.tab which contains
intermediate model results; *_c.tab file which contains the composite convolution function c(t) for the model; the *_s.tab file which contains the strip brightness distribution of the occulted star;
and the *_h.tab file which contains the impulse response of the detector. Note that some of these files will contain only a single line of data if none of the relevant model parameters are being
fitted for a given individual ring profile. Ring-Moon Systems Ring Occultation Profile Earth-based Observations of Uranus System Stellar Occultations Observing Campaign
urn:nasa:pds:context:investigation:observing_campaign.earth-based-uranus-stellar-occultations ancillary_to_investigation Uranus Rings Uranian Ring System Ring
urn:nasa:pds:context:target:ring.uranus.rings ancillary_to_target four Ring of Uranus Ring The four ring of Uranus. Center of motion: Uranus; LID of central body:
urn:nasa:pds:context:target:planet.uranus; NAIF ID of central body: 799. urn:nasa:pds:context:target:ring.uranus.four_ring ancillary_to_target SPK ura111.bsp SPK vgr2.ura111.bsp SPK
earthstns_itrf93_040916.bsp BPC earth_720101_031229.bpc LSK naif0012.tls These kernel files were used in the generation of the products in the parent bundle. Some or all of them may not have been
used directly in the generation of this product. urn:nasa:pds:uranus_occ_support:document:earth-based-uranus-stellar-occultations-user-guide ancillary_to_document The User Guide for Earth-based
Uranus Stellar Occultations. urn:nasa:pds:uranus_occ_u149_irtf_320cm:data:2200nm_wavelengths ancillary_to_data Wavelengths and relative weights used in the calculation of diffraction pattern for
square-well models of the individual ring profiles. urn:nasa:pds:uranus_occ_u149_irtf_320cm:data:2200nm_counts-v-time_occult::1.0 ancillary_to_data Normalized counts vs. time for the entire
occcultion. urn:nasa:pds:uranus_occ_u149_irtf_320cm:browse:2200nm_obs_geom ancillary_to_document Diagram of the Uranus system showing the occultation track.
urn:nasa:pds:uranus_occ_u149_irtf_320cm:browse:2200nm_earth ancillary_to_document Diagram of the view of the Earth from Uranus at mid-occultation Time.
urn:nasa:pds:uranus_occ_u149_irtf_320cm:browse:2200nm_alt ancillary_to_document Plot of the altitude (in degrees) of Uranus and the sun relative to the horizon over the duration of the occulation.
urn:nasa:pds:uranus_occ_u149_irtf_320cm:browse:2200nm_ring_sqw_gallery ancillary_to_document Gallery of square-well diffraction model fits to individual ring profiles. Elliot et al. (1984).
"Structure of the Uranian rings. I. Square-well model and particle-size constraints" Astron J. 89, 1587-1603. http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1984AJ.....89.1587E&
data_type=PDF_HIGH&whole_paper=YES&type=PRINTER&filetype=.pdf u149_irtf_320cm_2200nm_ring_four_egress_sqw.pdf plots 2021-04-05T23:23:43Z These plots show the observations of an individual Uranus ring
occultation profile and the best-fitting diffraction square-well (sqw) model, for the specified occultation event. upper left panel: Comparison of observed count rate (black) as a function of time
(lower x axis), the best-fitting diffraction square well model (blue), and the corresponding square well itself (red). The full and zero stellar intensity levels are shown as dashed lines. The
time-series data and the best-fitting model are included in the corresponding ring model *.tab file with a name given by event_obs_tel_wl_ring_name_direction_sqw_p.tab (ex:
u17b_saao_188cm_2220nm_ring_six_ingress_sqw_p.tab) upper right panel: Same as upper left panel, but normalized to units of the flux of the unocculted star, so that the upper free-space baseline is
1.0 and 0.0 represents a complete loss of the stellar signal. lower left panel: The model diffraction pattern (blue) for the square well itself (red), averaged over the filter bandbass and (possibly)
at a higher time resolution than the observations themselves that are shown in the upper left panel. Especially for data sets with rather low time resolution, it is necessary to subdivide the
observed time per bit (dt) into a higher-resolution "mesh." The number of mesh points (m) is always an odd integer. Then, when computing the best-fitting square well model to the actual data, the
(possibly) higher-resolution model profile is summed over m points. Frequently, this summing converts a smooth- and continuous-looking diffraction pattern into a jagged pattern, reflecting the fact
that the integration time dt is often longer than the time scale of variation of the diffraction pattern of the ring. The time-series model at the subdivided time resolution dt/m is included in the
corresponding ring model *.tab file with a name given by event_obs_tel_wl_ring_name_direction_sqw_p.tab (ex: u17b_saao_188cm_2220nm_ring_six_ingress_sqw_i.tab) Also included in the lower left panel
is a curve representing the occultation star convolution kernel (the strip-brightness distribution of the star), shown as a purple curve centered at the mid-point of the geometric square well model.
The time-series stellar convolution kernel is contained in the corresponding ring model *.tab file with a name given by event_obs_tel_wl_ring_name_direction_sqw_s.tab (ex:
u17b_saao_188cm_2220nm_ring_six_ingress_sqw_s.tab) For IR data with an instrumental time constant included in the square-well diffraction model, the corresponding time constant convolution kernel is
included in the lower left planel plot as a green line, and in the the corresponding ring model *.tab file with a name given by event_obs_tel_wl_ring_name_direction_sqw_h.tab (ex:
u17b_saao_188cm_2220nm_ring_six_ingress_sqw_h.tab) When both a finite star (i.e., not a point source) and a non-zero instrumental time constant are included in the square-well model, the
correspoinding joint convolution kernel from these two separate sources of model smoothing is shown as an orange curve, and included in the the corresponding ring model *.tab file with a name given
by event_obs_tel_wl_ring_name_direction_sqw_c.tab (ex: u17b_saao_188cm_2220nm_ring_six_ingress_sqw_c.tab) lower right panel: Same as lower left panel, but normalized to units of the flux of the
unocculted star, so that the upper free-space baseline is 1.0 and 0.0 represents a complete loss of the stellar signal. 0 PDF u149_irtf_320cm_2200nm_ring_four_egress_sqw.txt summary
2021-04-05T23:23:43Z This file provides details of the diffraction square-well model fitted to the observations of an individual Uranus ring, for the given occultation occultation event. The first
part of the file describes the IDL program that performed the fit of the square-well (sqw) model to the data. The occultation event, observatory, telescope, instrument, ring, and occultation
direction are defined. DATA FILE INFORMATION documents the source data file and the specific subset of data to be fitted in this sqw model. EVENT INFORMATION provides additional information about the
specific ring event and event geometry. SQUARE WELL MODEL INFORMATION specifies the number of mesh points m into which each observed time bin is subsampled to provide higher time resolution for the
calculation of the square well diffraction pattern, before then coadding the subsampled model to the time resolution of the data being fitted. SQUARE WELL MODEL FIT RESULTS contain the results of the
non-linear least-squares fit of the sqw model to the data, including post-fit residuals, the initial and final parameter values, and their errors, calculated assuming that all data points have equal
weight. Parameters that are fitted have an asterisk (*) preceding the corresponding Initial Value. The correlation matrix is also shown, with obvious two-letter abbreviations for the fitted variable
names. Note that the underlying model is performed in the time domain, but for convenience the corresponding length dimensions for the square-well width, star diameter, equivalent width, and
equivalent depth are also shown. See Elliot et al. 1984 for further details. 0 UTF-8 Text Carriage-Return Line-Feed u149_irtf_320cm_2200nm_ring_four_egress_sqw_c.tab composite-convolution
2021-04-05T23:23:43Z This file contains the composite convolution function c(t) for the diffraction square well model fitted to the data. See Elliot et al. (1984) for details, and Table 1 for
definitions of the fitted parameters. 0 39 UTF-8 Text Provides the column headers, separated by commas, for the data table. 39 69 The composite convolution function c(t) for the diffraction square
well model fitted to the data. Columns 3 to the end contain partial derivatives of the model function C with respect to fittable parameters, as defined below. If a particular parameter is not being
fitted for a specific model, the partial derivative with respect to that parameter is set to -9.9999900000E+99. Carriage-Return Line-Feed 5 0 95 C Time 1 1 ASCII_Real 17 Seconds Time, relative to the
square well model mid-time, of the calculated convolution function, in seconds. C 2 19 ASCII_Real 18 Model composite convolution function, (Elliot et al. (1984), eq. 10). dC/dTC 3 38 ASCII_Real 18 s*
(-1) Partial derivative dC/d(t_c), where C is the model composite convolution function and and t_c is the time constant of the detector. dC/dTC is given in inverse seconds. (Elliot et al. (1984),
Table 1). Refer to the details of the fitted model function to determine whether the time constant is for a single- or double-pole filter - see also Eq. 9, Elliot et al. (1984). -9.9999900000E+99 dC/
dSTAR 4 57 ASCII_Real 18 s*(-1) Partial derivative dC/dT_star, where C is the model composite convolution function and and T_star is the star diameter in seconds. dC/dSTAR is given in inverse
seconds. (Elliot et al. (1984), Table 1). -9.9999900000E+99 dC/dLIMB 5 76 ASCII_Real 18 Partial derivative dC/db, where C is the model composite convolution function and b is the limb darkening
parameter (Elliot et al. (1984), Table 1.) -9.9999900000E+99 u149_irtf_320cm_2200nm_ring_four_egress_sqw_h.tab impulse-response 2021-04-05T23:23:43Z This file contains the impulse response of the
detector h(t) for the diffraction square well model fitted to the data. 0 19 UTF-8 Text Provides the column headers, separated by commas, for the data table. 19 1 The impulse response of the detector
h(t) for the diffraction square well model fitted to the data. See Elliot et al. (1984) for details,and Table 1 for definitions of the fitted parameters. Column 3 contains the partial derivatives of
the model function H with respect to fittable parameters, as defined below. If a particular parameter is not being fitted for a specific model, the partial derivative with respect to that parameter
is set to -9.9999900000E+99. Carriage-Return Line-Feed 3 0 57 H_TIME 1 1 ASCII_Real 17 s Time, relative to the square well model mid-time, of the calculated convolution function, in seconds. H 2 19
ASCII_Real 18 Impulse response of the detector h(t) (Elliot et al. (1984), eq. 9). dH/dTC 3 38 ASCII_Real 18 s*(-1) Partial derivative dH/dt_c, where t_c is the star diameter in seconds. dH/dTC is
given in inverse seconds. (Elliot et al. (1984), Table 1). Refer to the details of the fitted model function to determine whether the time constant is for a single- or double-pole filter.
-9.9999900000E+99 u149_irtf_320cm_2200nm_ring_four_egress_sqw_i.tab intermediate-model 2021-04-05T23:23:43Z Intermediate model results for a diffraction square well model fit to the data. 0 60 UTF-8
Text Provides the column headers, separated by commas, for the data table. 60 4911 Intermediate model results for a diffraction square well model fit to the data. See Elliot et al. (1984) for
details,and Table 1 for definitions of the fitted parameters. Columns 3 to the end contain partial derivatives of the model function I with respect to fittable parameters, as defined below. If a
particular parameter is not being fitted for a specific model, the partial derivative with respect to that parameter is set to -9.9999900000E+99. Carriage-Return Line-Feed 8 0 152 I TIME 1 1
ASCII_Real 17 s Time from the start of the time series data file used for the fit, for the middle of each calculation bin. The duration of each calculation bin is either equal to dt/n_mesh, where
n_mesh is an odd number greater than or equal to 1, representing the sub-sampling of the observation integration time. See Elliot et al. (1984) Eq. 13. n_mesh is given by the time interval between
successive TSEC values in the data being fitted (see the *_p.tab file corresponding to this *_i.tab file) divided by the time interval between successive I_TIME values in this file. I BAR 2 19
ASCII_Real 18 Model value for the normalized signal for each time bin = Pbar(t) Eq. 7 Elliot et al. (1984). dI/dT0 3 38 ASCII_Real 18 s*(-1) Partial derivative dI/dt0, where I is the model function
and t0 is the midtime of the square well model. dI/dT0 is given in inverse seconds. (Elliot et al. (1984), Table 1). -9.9999900000E+99 dI/dW 4 57 ASCII_Real 18 s*(-1) Partial derivative dI/dW, where
I is the model function and W is the duration of the square well in seconds. dI/dW is given in inverse seconds. (Elliot et al. (1984), Table 1) -9.9999900000E+99 dI/dV 5 76 ASCII_Real 18 s/km Partial
derivative dI/dv_perp, where I is the model function and v_perp is the component of the sky plane velocity of the star perpendicular to the local elliptical ring edge measured in km/sec. dI/dV is
given is s/km. (Elliot et al. (1984), Table 1) -9.9999900000E+99 dI/dF 6 95 ASCII_Real 18 Partial derivative dI/df_0, where I is the model function and f_0 is the fractional transmission. (Elliot et
al. (1984), Table 1). -9.9999900000E+99 dI/dEW 7 114 ASCII_Real 18 s*(-1) Partial derivative dI/dE_0, where I is the model function and E_0 is the equivalent width in seconds. dI/dEW is given in
inverse seconds. (Elliot et al. (1984), Table 1). -9.9999900000E+99 dI/dED 8 133 ASCII_Real 18 Seconds Partial derivative dI/dA_0, where I is the model function and A_0 is the equivalent depth in
seconds. dI/dED is given in inverse seconds. (Elliot et al. (1984), Table 1). -9.9999900000E+99 u149_irtf_320cm_2200nm_ring_four_egress_sqw_p.tab model-results 2021-04-05T23:23:43Z Time series
observations of a single ring occultation, and corresponding model results for a diffraction square well model fit to the data. 0 123 UTF-8 Text Provides the column headers, separated by commas, for
the data table. 123 24 Time series observations of a single ring occultation, and corresponding model results for a diffraction square well model fit to the data. See Elliot et al. (1984) for
details,and Table 1 for definitions of the fitted parameters. Columns 4 to the end contain partial derivatives of the model function P with respect to fittable parameters, as defined below. If a
particular parameter is not being fitted for a specific model, the partial derivative with respect to that parameter is set to -9.9999900000E+99. Carriage-Return Line-Feed 15 0 285 TSEC 1 1
ASCII_Real 17 s Time in seconds from the start of the time series data file used for the fit, for the middle of each time bin. DATA 2 19 ASCII_Real 18 Observed counts per second, for each time bin. P
3 38 ASCII_Real 18 counts/s P gives the model value for the recorded signal for each time bin. In Eq. 13, Elliot et al. (1984), P corresponds to n(t_i), where t_i is the time of the bin. dP/dT0 4 57
ASCII_Real 18 counts/(s*2) Partial derivative dP/dt0, where P is the model and t0 is the midtime of the square well model. dP/dT0 is given in counts/s^2. (Elliot et al. (1984), Table 1)
-9.9999900000E+99 dP/dW 5 76 ASCII_Real 18 Seconds Partial derivative dP/dW, where P is the model and W is the duration of the square well in seconds. dP/dW is given in counts/s^2. (Elliot et al.
(1984), Table 1) -9.9999900000E+99 dP/dV 6 95 ASCII_Real 18 km*(-1) Partial derivative dP/dv_perp, where P is the model and vperp is the component of the sky plane velocity of the star perpendicular
to the local elliptical ring edge measured in km/sec. dP/dv_perp is given in inverse kilometers. (Elliot et al. (1984), Table 1) -9.9999900000E+99 dP/dSTAR_CTS 7 114 ASCII_Real 18 dP/dSTAR_CTS:
Partial derivative dP/dnbar_star, where P is the model and nbar_star is unocculted star level in counts/second. (Elliot et al. (1984), Table 1). -9.9999900000E+99 dP/dBASE 8 133 ASCII_Real 18 dP/
dBASE: Partical derivative dP/dnbar_b, where P is the model and nbar_b is the free-space background count rate in counts per second (Elliot et al. (1984), Table 1). -9.9999900000E+99 dP/dF 9 152
ASCII_Real 18 counts/s Partial derivative dP/df_0, where P is the model and f_0 is the fractional transmission. dP/dF is given in counts/second. (Elliot et al. (1984), Table 1). -9.9999900000E+99 dP/
dSTAR 10 171 ASCII_Real 18 s*(-2) Partial derivative dP/dT_star, where P is the model and T_star is the star diameter in seconds. dP/dSTAR is given in inverse seconds squared. (Elliot et al. (1984),
Table 1). -9.9999900000E+99 dP/dSLOPE 11 190 ASCII_Real 18 s Partial derivative dP/d_ndot_b, where P is the model and n_dot_b is the background slope in counts/sec^2. dP/dSLOPE is given is seconds.
(Elliot et al. (1984), Table 1) -9.9999900000E+99 dP/dEW 12 209 ASCII_Real 18 s*(-2) Partial derivative dP/dE_0, where P is the model and E_0 is the equivalent width in seconds. dP/dEW is given in
inverse seconds squared. (Elliot et al. (1984), Table 1). -9.9999900000E+99 dP/dED 13 228 ASCII_Real 18 s*(-2) Partial derivative dP/dA_0, where P is the model and A_0 is the equivalent depth in
seconds. dP/dED is given in inverse seconds squared. (Elliot et al. (1984), Table 1) -9.9999900000E+99 dP/dLIMB 14 247 ASCII_Real 18 s*(-2) Partial derivative dP/db, where P is the model and b is the
limb darkening parameter. dP/dLIMB is given in inverse seconds squared. (Elliot et al. dP/dLIMB (1984), Table 1). -9.9999900000E+99 dP/dTC 15 266 ASCII_Real 18 s*(-2) Partial derivative dP/t_c, where
P is the model and t_c is the time constant of the detector. dP/dEW is given in inverse seconds squared. (Elliot et al. (1984), Table 1) Refer to the details of the fitted model function to determine
whether the time constant is for a single- or double-pole filter - see also Eq. 9, Elliot et al. (1984). -9.9999900000E+99 u149_irtf_320cm_2200nm_ring_four_egress_sqw_s.tab strip-brightness
2021-04-05T23:23:43Z This file contains the strip brightness distribution of the occulted star s(t) for the diffraction square well model fitted to the data. 0 31 UTF-8 Text Provides the column
headers, separated by commas, for the data table. 31 35 the strip brightness distribution of the occulted star s(t) for the diffraction square well model fitted to the data. See Elliot et al. (1984)
for details,and Table 1 for definitions of the fitted parameters. Columns 3 to the end contain partial derivatives of the model function S with respect to fittable parameters, as defined below. If a
particular parameter is not being fitted for a specific model, the partial derivative with respect to that parameter is set to -9.9999900000E+99. Carriage-Return Line-Feed 4 0 76 S_TIME 1 1
ASCII_Real 17 s Time, relative to the square well model mid-time, of the calculated convolution function, in seconds. S 2 19 ASCII_Real 18 Model strip brightness distribution of the occulted star s
(t) (Elliot et al. (1984), eq. 8) dS/dSTAR 3 38 ASCII_Real 18 s*(-1) Partial derivative dS/dT_star, where S is the model strip brightness distribution of the occulted star s(t) and T_star is the star
diameter in seconds. dS/dSTAR is given in inverse seconds. (Elliot et al. (1984), Table 1). -9.9999900000E+99 dS/dLIMB 4 57 ASCII_Real 18 Partial derivative dS/db, where S is the model strip
brightness distribution of the occulted star s(t) and b is the limb darkening parameter (Elliot et al. (1984), Table 1) -9.9999900000E+99 | {"url":"https://pds-rings.seti.org/pds4/bundles/uranus_occs_earthbased/uranus_occ_u149_irtf_320cm/data/ring_models/u149_irtf_320cm_2200nm_ring_four_egress_sqw.xml","timestamp":"2024-11-12T00:47:25Z","content_type":"application/xml","content_length":"52490","record_id":"<urn:uuid:98062965-bc7e-422a-aa5a-64b1bc2f424c>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00127.warc.gz"} |
A new Wilson line-based classical action for gluodynamics
Hiren Kakkad, Piotr Kotko, Anna Stasto
SciPost Phys. Proc. 7, 011 (2022) · published 20 June 2022
• doi: 10.21468/SciPostPhysProc.7.011
Proceedings event
15th International Symposium on Radiative Corrections: Applications of Quantum Field Theory to Phenomenology
We develop a new classical action that in addition to $\mathrm{MHV}$ vertices contains also $\mathrm{N^kMHV}$ vertices, where $1\leq k \leq n-4$ with $n$ the number of external legs. The lowest order
vertex is the four-point MHV vertex -- there is no three point vertex and thus the amplitude calculation involves fewer vertices than in the CSW method. The action is obtained by a canonical
transformation of the Yang-Mills action in the light-cone gauge, where the field transformations are based on Wilson line functionals.
Authors / Affiliations: mappings to Contributors and Organizations
See all Organizations.
Funders for the research work leading to this publication | {"url":"https://www.scipost.org/SciPostPhysProc.7.011","timestamp":"2024-11-04T22:04:52Z","content_type":"text/html","content_length":"33982","record_id":"<urn:uuid:8c3aaabc-4971-48eb-922a-504d068ffe12>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00497.warc.gz"} |
Order Of Operations Worksheets For 6th Grade | Order of Operation Worksheets
Order Of Operations Worksheets For 6th Grade
Order Of Operations Worksheet 6Th Grade Db excel
Order Of Operations Worksheets For 6th Grade
Order Of Operations Worksheets For 6th Grade – You may have become aware of an Order Of Operations Worksheet, yet exactly what is it? In this post, we’ll discuss what it is, why it’s vital, as well
as exactly how to obtain a Order Of Operations Worksheets For 6th Grade With any luck, this info will certainly be practical for you. Your students should have a fun, reliable method to review the
most vital concepts in mathematics. In addition, worksheets are a great way for pupils to exercise brand-new skills and testimonial old ones.
What is the Order Of Operations Worksheet?
An order of operations worksheet is a kind of mathematics worksheet that calls for pupils to perform math operations. These worksheets are divided into 3 main sections: addition, subtraction, and
multiplication. They additionally consist of the evaluation of exponents and parentheses. Students that are still learning exactly how to do these tasks will certainly locate this kind of worksheet
The primary function of an order of operations worksheet is to help pupils find out the proper method to solve math equations. If a pupil doesn’t yet recognize the concept of order of operations,
they can evaluate it by referring to an explanation web page. Additionally, an order of operations worksheet can be split into several groups, based upon its trouble.
Another vital purpose of an order of operations worksheet is to teach trainees how to do PEMDAS operations. These worksheets begin with simple troubles related to the fundamental guidelines and also
develop to more complicated issues entailing all of the regulations. These worksheets are a fantastic way to present young learners to the enjoyment of fixing algebraic formulas.
Why is Order of Operations Important?
One of the most essential points you can find out in math is the order of operations. The order of operations guarantees that the mathematics troubles you solve are regular.
An order of operations worksheet is a terrific means to show students the right means to resolve math equations. Prior to pupils start utilizing this worksheet, they may need to assess concepts
related to the order of operations. To do this, they must examine the principle web page for order of operations. This idea web page will certainly give pupils a review of the basic idea.
An order of operations worksheet can assist students establish their skills on top of that and also subtraction. Educators can use Prodigy as a simple way to distinguish method and also deliver
appealing material. Prodigy’s worksheets are a perfect method to aid students find out about the order of operations. Educators can begin with the basic principles of addition, division, and also
multiplication to help pupils develop their understanding of parentheses.
Order Of Operations Worksheets For 6th Grade
24 Printable Order Of Operations Worksheets To Master PEMDAS
Free Printable Math Worksheets 6Th Grade Order Operations Free Printable
Mrs White s 6th Grade Math Blog ORDER OF OPERATIONS WHAT DO I DO FIRST
Order Of Operations Worksheets For 6th Grade
Order Of Operations Worksheets For 6th Grade supply a fantastic resource for young students. These worksheets can be easily tailored for particular needs. They can be discovered in three degrees of
difficulty. The very first degree is basic, calling for students to exercise using the DMAS technique on expressions consisting of 4 or more integers or 3 operators. The second degree requires pupils
to utilize the PEMDAS method to streamline expressions making use of external and also internal parentheses, brackets, and curly dental braces.
The Order Of Operations Worksheets For 6th Grade can be downloaded free of charge as well as can be printed out. They can after that be examined using addition, multiplication, subtraction, as well
as division. Students can likewise use these worksheets to evaluate order of operations as well as making use of exponents.
Related For Order Of Operations Worksheets For 6th Grade | {"url":"https://orderofoperationsworksheet.com/order-of-operations-worksheets-for-6th-grade/","timestamp":"2024-11-11T13:59:28Z","content_type":"text/html","content_length":"44293","record_id":"<urn:uuid:a4945605-e519-4804-8665-eacc99389c2f>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00714.warc.gz"} |
Two-soliton collisions in a near-integrable lattice system
We examine collisions between identical solitons in a weakly perturbed Ablowitz-Ladik (AL) model, augmented by either onsite cubic nonlinearity (which corresponds to the Salerno model, and may be
realized as an array of strongly overlapping nonlinear optical waveguides) or a quintic perturbation, or both. Complex dependences of the outcomes of the collisions on the initial phase difference
between the solitons and location of the collision point are observed. Large changes of amplitudes and velocities of the colliding solitons are generated by weak perturbations, showing that the
elasticity of soliton collisions in the AL model is fragile (for instance, the Salerno’s perturbation with the relative strength of [Formula presented] can give rise to a change of the solitons’
amplitudes by a factor exceeding [Formula presented] Exact and approximate conservation laws in the perturbed system are examined, with a conclusion that the small perturbations very weakly affect
the norm and energy conservation, but completely destroy the conservation of the lattice momentum, which is explained by the absence of the translational symmetry in generic nonintegrable lattice
models. Data collected for a very large number of collisions correlate with this conclusion. Asymmetry of the collisions (which is explained by the dependence on the location of the central point of
the collision relative to the lattice, and on the phase difference between the solitons) is investigated too, showing that the nonintegrability-induced effects grow almost linearly with the
perturbation strength. Different perturbations (cubic and quintic ones) produce virtually identical collision-induced effects, which makes it possible to compensate them, thus finding a special
perturbed system with almost elastic soliton collisions.
Dive into the research topics of 'Two-soliton collisions in a near-integrable lattice system'. Together they form a unique fingerprint. | {"url":"https://cris.tau.ac.il/en/publications/two-soliton-collisions-in-a-near-integrable-lattice-system","timestamp":"2024-11-02T23:39:15Z","content_type":"text/html","content_length":"53491","record_id":"<urn:uuid:255432fd-5e1c-48c9-af4c-fa4eff3c26dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00544.warc.gz"} |
How to Integrate with Absolute Value
Have you ever wondered how to integrate a function with an absolute value? It may seem like a daunting task, but it’s actually quite simple. In this article, we’ll walk you through the steps of
integrating a function with an absolute value, using the example of $\int \lvert x \rvert dx$. We’ll also discuss some of the different techniques that can be used to integrate absolute values, and
we’ll provide some tips for solving integrals with absolute values. So if you’re ready to learn how to integrate with absolute value, read on!
Step Formula Example
1. Isolate the absolute value inside the integral. $\int |f(x)| dx = \int f(x) dx \text{ if } f(x) \ge $\int |x^2| dx = \int x^2 dx = \frac{x^3}{3} + C$
2. If $f(x) < 0$, then replace $f(x)$ with $-f(x)$ in the $\int |f(x)| dx = -\int f(x) dx \text{ if } f(x) < $\int |x-2| dx = -\int (x-2) dx = -\frac{(x-2)^2}{2} + C = -\frac{x^2}{2} + 2x +
integral. 0$ C$
In this tutorial, we will learn how to integrate with absolute value functions. We will start with the basics of absolute value functions and then discuss the different methods for integrating them.
We will also look at some special cases of integration with absolute value functions.
The Basics of Integration with Absolute Value
An absolute value function is a function that outputs the absolute value of its input. In other words, it outputs the positive value of its input, regardless of whether the input is positive or
negative. For example, the absolute value of 5 is 5, and the absolute value of -5 is also 5.
The graph of an absolute value function is a V-shaped curve that is symmetrical about the y-axis. The function is increasing on the interval [0,) and decreasing on the interval (-,0].
How do you integrate an absolute value function?
To integrate an absolute value function, we can use the following formula:
|f(x)|dx = f(x)dx + f(-x)dx
This formula is derived from the fact that the absolute value of a function is equal to the sum of the function and its reflection about the y-axis.
For example, to integrate the absolute value function |x|, we would use the following formula:
|x|dx = xdx + -xdx
= x^2/2 – x^2/2
= 0
The Different Methods for Integrating Absolute Value Functions
There are three main methods for integrating absolute value functions:
1. The direct substitution method
2. The substitution method
3. The partial fraction decomposition method
The direct substitution method is the simplest method, but it is only applicable to certain types of absolute value functions. The substitution method is more general, but it can be more difficult to
apply. The partial fraction decomposition method is the most general method, but it can be the most difficult to apply.
Special Cases of Integration with Absolute Value Functions
There are a few special cases of integration with absolute value functions that are worth noting.
• Integrating absolute value functions with odd powers:
The integral of an absolute value function with an odd power can be evaluated using the following formula:
|x|^ndx = (-1)^(n-1)n!/(n+1)
for n 1.
• Integrating absolute value functions with even powers:
The integral of an absolute value function with an even power can be evaluated using the following formula:
|x|^2dx = x^2/2 + C
for n 2.
• Integrating absolute value functions with rational powers:
The integral of an absolute value function with a rational power can be evaluated using the partial fraction decomposition method.
In this tutorial, we have learned the basics of integration with absolute value functions. We have also discussed the different methods for integrating absolute value functions and the special cases
of integration with absolute value functions.
I hope this tutorial has been helpful. If you have any questions, please feel free to leave a comment below.
3. Applications of Integration with Absolute Value
Integration with absolute value has a variety of applications in mathematics and physics. Some of the most common applications include:
• Solving optimization problems
• Determining the area under a curve
• Evaluating definite integrals
Solving Optimization Problems
One of the most common applications of integration with absolute value is in solving optimization problems. In an optimization problem, you are given a function and you want to find the value of the
input variable that makes the function as large or as small as possible.
To solve an optimization problem using integration with absolute value, you first need to find the derivative of the function. Then, you set the derivative equal to zero and solve for the input
variable. If the derivative is never equal to zero, then the function has no maximum or minimum value.
Once you have found the input variable that makes the function as large or as small as possible, you can evaluate the function at that input variable to find the maximum or minimum value.
For example, let’s say you want to find the minimum value of the function f(x) = |x – 2|. To do this, we first find the derivative of the function:
f'(x) = -1
Then, we set the derivative equal to zero and solve for x:
-1 = 0
This tells us that x = 0 is the critical point of the function. To find the minimum value of the function, we evaluate the function at x = 0:
f(0) = |0 – 2| = 2
Therefore, the minimum value of the function f(x) = |x – 2| is 2.
Determining the Area under a Curve
Another common application of integration with absolute value is in determining the area under a curve. The area under a curve is the amount of space that is enclosed by the curve and the x-axis.
To determine the area under a curve using integration with absolute value, you first need to find the antiderivative of the function that represents the curve. Then, you evaluate the antiderivative
at the upper and lower limits of the area you are interested in. The difference between the two values is the area under the curve.
For example, let’s say you want to find the area under the curve y = |x| from x = -1 to x = 1. To do this, we first find the antiderivative of the function y = |x|:
F(x) = x|x| + C
Then, we evaluate the antiderivative at the upper and lower limits of the area:
F(1) = 1|1| + C = 1 + C
F(-1) = (-1)|-1| + C = 1 + C
The difference between the two values is 2, so the area under the curve y = |x| from x = -1 to x = 1 is 2.
Evaluating Definite Integrals
Integration with absolute value can also be used to evaluate definite integrals. A definite integral is an integral that is evaluated over a specific interval.
To evaluate a definite integral using integration with absolute value, you first need to find the antiderivative of the function that is being integrated. Then, you evaluate the antiderivative at the
upper and lower limits of the interval. The difference between the two values is the value of the definite integral.
For example, let’s say you want to evaluate the definite integral from 0 to 1 of the function f(x) = |x|. To do this, we first find the antiderivative of the function f(x) = |x|:
F(x) = x|x| + C
Then, we evaluate the antiderivative at the upper and lower limits of the interval:
F(1) = 1|1| + C = 1 + C
F(0) = 0|0| + C = C
The difference between the two values is 1 + C – C = 1, so the value of the definite integral is 1.
4. Additional Resources on Integration with Absolute Value
In addition to the resources listed below, there are many other online tutorials, textbooks, and Khan Academy videos that can help you learn more about integration with absolute value.
• [Online Tutorials](https://www.khanacademy.org/math/calculus-1/integration-techniques/applications-of-the-fundamental-the
Q: What is the integral of absolute value of x?
A: The integral of absolute value of x is $\frac{x^2}{2}$.
Q: How do I integrate absolute value of x^2?
**A: To integrate absolute value of x^2, you can use the following steps:
1. First, rewrite the expression as $\int |x|^2 dx$.
2. Then, use the substitution $u = |x|$. This gives $\int u^2 du$.
3. Finally, use the power rule to integrate: $\int u^2 du = \frac{u^3}{3} + C$.
Therefore, the integral of absolute value of x^2 is $\frac{|x|^3}{3} + C$.**
Q: How do I integrate absolute value of x^3?
**A: To integrate absolute value of x^3, you can use the following steps:
1. First, rewrite the expression as $\int |x|^3 dx$.
2. Then, use the substitution $u = |x|$. This gives $\int u^3 du$.
3. Finally, use the power rule to integrate: $\int u^3 du = \frac{u^4}{4} + C$.
Therefore, the integral of absolute value of x^3 is $\frac{|x|^4}{4} + C$.**
Q: How do I integrate absolute value of x^n?
**A: To integrate absolute value of x^n, you can use the following steps:
1. First, rewrite the expression as $\int |x|^n dx$.
2. Then, use the substitution $u = |x|$. This gives $\int u^n du$.
3. Finally, use the power rule to integrate: $\int u^n du = \frac{u^{n+1}}{n+1} + C$.
Therefore, the integral of absolute value of x^n is $\frac{|x|^{n+1}}{n+1} + C$.**
In this comprehensive guide, we have discussed how to integrate with absolute value. We first introduced the concept of absolute value and its properties. Then, we showed how to integrate an absolute
value function using the substitution method. We also discussed the different types of absolute value functions and how to integrate them. Finally, we provided some practice problems for you to try.
We hope that this guide has been helpful and that you now have a better understanding of how to integrate with absolute value. Please feel free to contact us if you have any questions or if you would
like to learn more about this topic.
Author Profile
Hatch, established in 2011 by Marcus Greenwood, has evolved significantly over the years. Marcus, a seasoned developer, brought a rich background in developing both B2B and consumer software for
a diverse range of organizations, including hedge funds and web agencies.
Originally, Hatch was designed to seamlessly merge content management with social networking. We observed that social functionalities were often an afterthought in CMS-driven websites and set out
to change that. Hatch was built to be inherently social, ensuring a fully integrated experience for users.
Now, Hatch embarks on a new chapter. While our past was rooted in bridging technical gaps and fostering open-source collaboration, our present and future are focused on unraveling mysteries and
answering a myriad of questions. We have expanded our horizons to cover an extensive array of topics and inquiries, delving into the unknown and the unexplored. | {"url":"https://hatchjs.com/how-to-integrate-with-absolute-value/","timestamp":"2024-11-08T23:58:51Z","content_type":"text/html","content_length":"91013","record_id":"<urn:uuid:154a8f98-3381-4ed9-a164-eef9dc74056e>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00023.warc.gz"} |
Math 4 Wisdom. "Mathematics for Wisdom" by Andrius Kulikauskas. | Research / DiscoveryInMathematics
See: Math discovery
Andrius Kulikauskas: Please note, I need to rework the talk below, although the basic ideas stand. One error, not quite relevant to the main points, is that I write about the demicubes {$D_n$}. But,
evidently, the right way to think about this is rather in terms of the "coordinate systems" that I define in my abstract Combinatorial Interpretations Which Distinguish Observer and Observed.
Discovery in Mathematics: A System of Deep Structure
I will talk about how we may systematically study the ways of figuring things out in mathematics.
George Polya, in his book, "How to Solve It", considers Euclid's problem of how to construct an equilateral triangle. If we are given the side AB, how do we construct the other two? The solution is a
recurring idea which Polya calls the "pattern of two loci". We think of there being two separate conditions. One side must extend a length AB from the point A. Another side must extend a length AB
from the point B. We thus draw two circles of radius AB centered at A and B. The points where the two circles intersect are those where we can draw a third point C which satisfies both conditions so
that our triangle is equilateral.
I realized that our minds solve this problem by imagining a powerset lattice of conditions. Circle A is one condition, circle B is another condition, and the intersection of A and B satisifies the
union of these two conditions. Our minds have thus solved the surface problem (constructing a triangle) by considering a simpler, deeper structure (a lattice of conditions). This brings to mind
linguist Noam Chomsky's work in syntax and architect Christopher Alexander's work on pattern languages.
I collected such problem solving patterns discussed in Paul Zeitz's book The Art and Craft of Problem Solving and other sources. Each distinct pattern makes use of a structure which is familiar to
mathematicians and yet is not explicit but mental. We may consider those math structures to be cognitively "natural" which are used by the mind in solving math problems. I present to you 24 patterns
which I identified and systematized them in a way which suggests they are complete.
Let me first show how to systematically generate the 4 infinite families of polytopes, An (simplexes), Cn (cross polytopes), Bn (hypercubes), Dn (demicubes), whose symmetry groups are also the Weyl
groups for the root systems of the classical Lie algebras. The resulting structures, actions and relationships will help me talk systematically about the mathematical ways of figuring things out.
I define the Center of a polytope as that which is not a vertex but which ever generates the next vertex by which the polytope may grow. I interpret it as the unique center of the polytope which
keeps moving as the polytope grows. For example, the simplexes An are generated by adding one vertex at a time, along with edges to all of the existing vertices, which makes clear that it is indeed a
new vertex. The k-dimensional subsimplexes of the n-dimensional simplex are given by the n+1st row of Pascal's triangle. I interpret the Center to be the -1 simplex, the unique leftmost sub-simplex
in each row which has no vertices and is -1 dimensional. Note that the dual of the Center is the Totality, the unique rightmost sub-simplex in each row which has all n vertices.
The cross polytopes Cn are generated as follows. The Center generates two vertices at a time, not connected to each other, yet connected to all other vertices. Thus B1 is two vertices, B2 is a square
and B3 is an octahedron. The symmetry group not only permutes the vertices, but also reflects opposite vertices in each direction. Each dimension may be thought of independently as modeling opposite
qualities. If we look at the relevant combinatorial triangle we conclude that Cn has a Center but no Totality. For example, the rightmost entry for the octahedron is given by its 8 faces of 3
vertices each. In this sense, the octahedron is 3-dimensional but it has no volume.
The cubes Bn are a dual interpretation of the same combinatorial triangle as for the Cn. They have a Totality but no Center. We may think of the Totality as generating an unfolding sequence of
perpendicular mirrors. These mirrors define quadrants, fractions of the whole, which are ever divided anew. Each mirror defines an opposition and each quadrant is distinguished by the entire sequence
of oppositions.
We may think of the demicubes Dn as having no Center and no Totality. Typically, a demicube is defined by starting with a cube and retaining and discarding alternate vertices. Instead, let us choose
any vertex (a quadrant) and make it an Origin, an Anti-Center, by folding the cube in half, that is, fusing to it the vertex which is opposite to it in every way, and likewise turning and fusing
pairs of vertices to create a coordinate system of unit vectors around that Origin. Link with edges all of the tips of the unit vectors to get a simplex. We can think of each such coordinate system
as serving one of the vertices. The coordinate systems fit together (their edges overlap) to bound the demicube. We can think of the coordinate systems as forming a coordinate complex which fosters
duality and ambiguity. There are two different ways in which we can think of the coordinate complex as growing at each step. On the one hand, we can think of each coordinate system as growing like a
simplex but with a distinguished Origin and unit vectors to the other vertices. On the other hand, we can think of each dimension arising by introducing a mirror which reflects a diagram into its
dual (in which Origins are exchanged with Vertices and vectors reversed) and joins it with that dual. Thus vectors lead out from the Origins to the Vertices and the Vertices are linked by pairs of
edges. Also, a gap opens up between the coordinate systems. This gap bounds the demicube and its parts, which are halves, quarters, eighths and so on, with the smallest part being the tetrahedron.
These sequences of parts gives us a top-down point of view onto specific coordinate systems, whereas each coordinate systems allow us to construct a bottom-up point of view. Thus we may imagine that
the coordinate complex has no Center, but rather it has a multitude of Origins, each of which loves some vertex with a coordinate system. And the coordinate complex has no Totality of vertices, but
rather it defines a whole that is the gap between the coordinate complexes.
We can now imagine how the polytopes An, Cn, Bn and Dn are the foundations for 4 different kinds of geometries. Indeed, these geometries are the basis for 4 different ways of metalogically figuring
things out.
The simplexes An span all of space from the Center and so are the basis for affine geometry where directions are preserved. We figure things out metalogically by preserving the relationship between a
metalogical perspective of the Center outside of a system and the logical perspective of a vertex inside a system. Metalogically, the simplexes let us figure out whether things are true by leveraging
the inherent contradiction within the Center, the unresolvable tension between its form and content, which allows us to view two options and yet deem one true by contradiction. Simplexes model a
theology where "God is good" in every direction.
The cross-polytopes Cn define opposite directions and so are the basis for projective geometry where lines are preserved. Objects are constructed from independent dimensions which is key for
algebraic ways of figuring things out. Metalogically, we figure out what is true by simplifying, even by "wishful thinking", modeling the most relevant dimensions. Cross-polytopes model a theology
where every dimension represents a choice, potentially good or bad.
The cubes Bn relate perpendicular directions and so are the basis for conformal geometry where angles are preserved. Manifolds are freed within multidimensional spaces which is key for analytic ways
of figuring thing out. Metalogically, we figure out how things are true by working backwards from a multidimensional point to the spaces which are its conditions. Cubes model a theology where every
quadrant is defined by a collection of choices made.
The coordinate systems Dn superimpose directions and angles and so are the basis for symplectic geometry where areas are preserved. We figure things out logically by applying transformations to
bridge the gap between the perspectives given by a vertex in a system and a vertex in a subsystem. Metalogically, we figure out why things are true by interpreting variables in ways that clarify the
semiotic disconnects between these four geometries. Our minds distinguish semiotic levels of thing (whether), icon (what), index (how), symbol (why), as when we draw a picture. We use variables to
imagine a broader level as free but a narrower level as set. We thus interpret variables as dependent vs. independent, known vs. unknown, given vs. arbitrary, fixed vs. varying, concrete vs.
abstract, defined vs. undefined and so on. Coordinate systems model a theology where good is distinguished from bad and yet truly the system is self-centered, relative and questionable. Consequently,
"God does not have to be good", life does not have to be fair.
We can think of each of these four geometries as relating the explicit math on a sheet of paper and the implicit math by which we interpret it with our minds. One way to figure things out is to
realize that we can always start with a fresh sheet. We can run independent trials, vary them, get our hands dirty, accumulate some data points, add some noise, restate what we have formulated, start
all over again. We are thus starting without any system, but our goal is a second way of figuring things out, namely to structure our sheet as a completely defined system which our mind can play
with, acting on it like a symmetry group, constructing strings of actions and undoing them at will, applying mental functions and their inverses, as we wish. Yet ultimately there is a third way by
which we figure out the limits of our knowledge. If I ask you, what is 10+4? You may say 14, but the answer is 2, because I am thinking about a clock. Which is to say that what we know may be
completely irrelevant and plain wrong because ultimately it all depends on context. We have to be willing to go back to the blank sheet.
Even so, we seek a symmetric group which brings together algebraic structures and analytic dynamics. The four polytopes An, Cn, Bn, Dn illustrate the ways that structure defines helpful perspectives.
We can solve a problem by recognizing the center of the whole, as suggested by the simplex. Or we can make use of parity as suggested by the cross-polytopes. We do this when we modify an equation
with terms that add to zero or that multiply to one, or when we consider a subset and its complement, whereby elements only differ as to whether they are in or out of a set. Yet there is another way
of looking at a set such that its elements are all distinct, as when we combine a variety of amounts and units, as suggested by the cubes. Finally, we can construct or deconstruct a vector subspace
in an orderly manner given a list of canonical basis elements, as suggested by the coordinate systems.
The four polytopes likewise inspire dynamic analytic perspectives, though not within a single sheet, but through a sequence of sheets. We can apply mathematical induction, as suggested by the center
of a simplex, in that it ever generates the next vertex. We can imagine a maximum and minimum in any dimension, as suggested by the cross-polytopes. The quadrants of the cubes help us imagine that
various solutions can serve as least upper bounds or greatest lower bounds. And the twofold nature of the coordinate systems Dn let us imagine that if we generate a sequence of vertices, then it will
ultimately achieve its limit somewhere in the half cube.
We also imagine functors which couple the algebraic structural and analytical dynamic points of view through the scientific method of taking a stand (hypothesizing), following through (experimenting)
and reflecting (concluding). We imagine this coupling ambiguously as both strict and loose, so that slack can decrease but also increase. Given a constraint such as the addition formula (2**X)(2**Y)
= 2**(X+Y), we take a stand by extending the domain, stitching together new values for X and Y, such as zero, negative integers, fractions, real numbers, complex numbers and so on. Next, we can
experiment in our minds, varying these variables continuously, seeing whether our model will break or hold. Finally, we can generalize and reformulate what we have learned by considering our
constraint as a recurrence relation and superimposing the resulting sequence upon itself, as with a generating function, or as with auto-associative memory, where time-delay lets patterns be related
to themselves. This three-cycle of extending the domain, varying continuously and self-superimposing is what I imagine brings forth the particular structures and actions which our mind systematizes
with some symmetry group.
Once we have established our symmetry group and a comprehensive system, then our four geometries let us analyze it theologically by imagining how the Center outside relates invariantly to the
vertices inside. But we can also analyze our system ethically by considering how a transformation links the perspective of a vertex within a system with the perspective of a vertex within a
subsystem. Six transformations bridge the gap between one geometry's analytic dynamics and another geometry's algebraic structure. We can illustrate these transformations by considering the different
senses that we ascribe to symmetry group actions, but in particular, to multiplication. In each case, the transformation restructures structure of one kind with a structure of another kind. We find
natural grounds for six of the axioms of Zermelo-Fraenkel set theory.
We can imagine people in a system as vertices with perspectives, that is, with edges, with relationships to other people, vertices, perspectives. Let us think of love as the loose coupling which
identifies a perspective as the same perspective after a three-cycle of extending the domain, varying continuity and self-superimposing. We can imagine that structurally a domain extends like a tree,
continuity asserts itself as a sequence, and self-superimposition imposes itself as a network. Symmetry is that equivalence which allows us to imagine vertices as different and yet treat them as the
same. Structurally, symmetry can ground an equivalence at various levels of granularity: the whole, as characterized by the unity of the center; multiples, as given by what is the same and not
different; and the set, where elements are labeled but not ordered. Whereas a list allows no symmetry. Dynamically, we must operate at a finer granularity. If the world is a list of items, then we
help our loved one love all by redistributing it. If people are elements in a set, then we help our loved one rescale their relationships so they love each other. Inasmuch as our loved one is an
ideal person, then we help them love themselves by reaffirming, recopying themselves. Thus we love them as we love ourselves. In these ways, symmetry loosely couples structure and dynamics to define
transformations which bridge perspectives through loving equivalences.
In figuring things out logically within a well defined mathematical system, our mind grants the slack by way of six different transformations, each of which is familiar from multiplication. For
example, a tree of variations can be restructured by a sequence to visualize evolution. Our mind visualizes fractal self-similarities and, as in multiplying powers of two, recopies the whole in a
self-love. Such gathering of branches grounds the Axiom of Pairing in set theory.
A network of relationships is restructured by a tree. We visualize this as an atlas of global and local views upon adjacencies. Similarly, multiplication composes actions to rescale the whole, to
magnify or shrink relationships, and love reciprocally. The Axiom of Extensionality states that sets are globally equal if and only if their elements are locally equal.
A sequence of instructions is restructured by a network of redirects to visualize a handbook. Analogously, multiplication rescales multiples by tallying and skip counting. It leverages the total
order in order to recount cardinals. This grounds the Well-ordering Theorem and the Axiom of Choice.
We can consider the dual of our three-cycle, the conditions which allow us to take a stand, follow through, and reflect. We thus accept our world in its finest granularity to allow for that
redistribution of truth by which we might all love unconditionally in total harmony. A sequence of events is restructured by a tree of possibly overlapping time periods. We thus visualize a
chronicle. Multiplication redistributes a set as with box multiplication and lets us disentangle "ands" and "ors". Thus in our minds we organize a power set lattice of conditions. This grounds the
Axiom of Power Set.
However, our redistribution need not be so completely refined. A tree of concepts is restructured by a network of partial references to visualize a catalog. Multiplication lets us redistribute
multiples by factoring components, thus regrouping and recounting ordinals. This grounds the Axiom of Union.
Our redistribution may apply to a whole, as with long division, where we keep having a remainder. A network of causes and effects is restructured by a sequence to visualize a tour. Our minds analyze
the path to consider, especially, whether it will avoid cycles, or whether they will be stable or unstable. This grounds the Axiom of Regularity.
Inherent in each of these six transformations is a slack in the system, which taken together, maintains that we be obedient to the possibility that everything we know is wrong.
Overviewing this entire system, I imagine it as spawned by the inner tension of the Center, which is first characterized by the ambiguity of a symmetry group's action and structure. This tension then
divides itself for the duality of the Center's blank sheet from which we start and the Totality's total context which we may yet need to completely reinterpret. Then the internal slack of the
symmetry group is made concrete through the loose coupling of that three-cycle of extending the domain, varying continuously and self-superimposing, which allows for equivalences. Finally, the
primacy of the Center in its duality with the Totality is evinced by the four polytope families, whereby the duality of algebraic structure (constructed bottom-up by the cross-polytopes Cn) and
analytical dynamics (unleashed top-down by the cubes Bn) is coupled by the three-cycle to ground a self-standing system around a symmetry group, defined metalogically (by the simplexes An) to express
logical slack (through the coordinate systems Dn) which establishes the absolute vulnerability of the Totality and thus the Center's primacy and necessary reality.
I am deeply grateful to Maria Droujkova and the Natural Math team for investigating the many senses of multiplication, and to Kirby Urner for pointing out the differences in tetrahedral and cubic
thinking, in the spirit of Buckminster Fuller. Thank you also to Thomas Gajdosik for his supportive understanding in our many conversations. | {"url":"https://www.math4wisdom.com/wiki/Research/DiscoveryInMathematics","timestamp":"2024-11-08T02:03:22Z","content_type":"application/xhtml+xml","content_length":"32282","record_id":"<urn:uuid:96758c16-aed6-4998-84b0-ee170bd41d10>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00008.warc.gz"} |
Publications & Preprints
Partition regularity of generalized Pythagorean pairs (with O. Klurman and J. Moreira)
    Preprint.
Bohr recurrence and density of non-lacunary semigroups of ℕ (with B. Host and B. Kra)
    To appear in Proceedings of the American Mathematical Society.
Partition regularity of Pythagorean pairs (with O. Klurman and J. Moreira)
    To appear in Forum of Mathematics, Pi.
Furstenberg systems of pretentious and MRT multiplicative functions (with M. Lemanczyk and T. de la Rue)
    Preprint.
Degree lowering for ergodic averages along arithmetic progressions (with B. Kuca)
    To appear in Journal d'Analyse Mathematique.
Seminorm control for ergodic averages with commuting transformations along pairwise dependent polynomials (with B. Kuca)
    Ergodic Theory & Dynamical Systems, 43, (2023), 4074-4137.
Joint ergodicity for commuting transformations and applications to polynomial sequences (with B. Kuca)
    Preprint.
Multiple recurrence and convergence without commutativity (with B. Host)
    Journal of the London Mathematical Society, 107, no. 5, (2023), 1635-1659.
Joint ergodicity of fractional powers of primes
    Forum of Mathematics, Sigma, 10, (2022), e30.
Joint ergodicity of sequences
    Advances in Mathematics, 417, 108918 (2023), (63pp). Here is an exposition that appeared in D. Gatzouras memorial volume, Univ. of Athens, 2022
Furstenberg systems of Hardy field sequences and applications
    Journal d'Analyse Mathematique, 147, (2022), 333-372.
Correlations of multiplicative functions along deterministic and independent sequences
    Transactions of the American Mathematical Society, 373, no. 9, (2020), 6595-6620.
Good weights for the Erdős discrepancy problem
    Discrete Analysis, 2020:8, 23 pp.
Furstenberg systems of bounded multiplicative functions and applications (with B. Host)
    International Mathematics Research Notices, 2021, no. 8, (2021), 6077-6107.
The logarithmic Sarnak conjecture for ergodic weights (with B. Host)
    Annals of Mathematics, 187, no. 3, (2018), 869-931.
Ergodicity of the Liouville system implies the Chowla conjecture
    Discrete Analysis, 2017:19, 41pp.
An averaged Chowla and Elliott conjecture along independent polynomials
    International Mathematics Research Notices, 2018, no. 12, (2018), 3721-3743.
Weighted multiple ergodic averages and correlation sequences (with B. Host)
    Ergodic Theory & Dynamical Systems, 38, no. 1, (2018), 81-142.
Under recurrence in the Khintchine recurrence theorem (with M. Boshernitzan and M. Wierdl)
    Israel Journal of Mathematics, 222, no. 10, (2017), 815-840.
Multiple ergodic theorems for arithmetic sets (with B. Host)
    Transactions of the American Mathematical Society, 369, no. 10, (2017), 7085-7105.
Higher order Fourier analysis of multiplicative functions and applications (with B. Host) Replaces this article
    Journal of the American Mathematical Society, 30, no. 1, (2017), 67-157.
Some open problems on multiple ergodic averages Updates this article
    Bulletin of the Hellenic Mathematical Society, 60, (2016), 41-90.
Asymptotics for multilinear averages of multiplicative functions (with B. Host)
    Mathematical Proceedings of the Cambridge Philosophical Society, 161, no. 1, (2016), 87-101.
Random differences in Szemeredi's theorem and related results (with E. Lesigne and M. Wierdl)
    Journal d'Analyse Mathematique, 130, no. 1, (2016), 91-133.
Multiple correlation sequences and nilsequences
    Inventiones Mathematicae, 202, no. 2, (2015), 875-892.
Multiple recurrence for non-commuting transformations along rationally independent polynomials (with P. Zorin-Kranich)
    Ergodic Theory & Dynamical Systems, 35, no. 2, (2015), 403-411.
A multidimensional Szemeredi theorem for Hardy sequences of different growth
    Transactions of the American Mathematical Society, 367, no. 8, (2015), 5653-5692.
The polynomial multidimensional Szemeredi theorem along shifted primes (with B. Host and B. Kra)
    Israel Journal of Mathematics, 194, no. 1, (2013), 331-348.
Random sequences and pointwise convergence of multiple ergodic averages (with E. Lesigne and M. Wierdl)
    Indiana University Mathematics Journal, 61, (2012), 585-617.
Pointwise convergence for cubic and polynomial ergodic averages of non-commuting transformations (with Q. Chu)
    Ergodic Theory & Dynamical Systems, 32, no. 3, (2012), 877-897.
Ergodic averages of commuting transformations with distinct degree polynomial iterates (with Q. Chu and B. Host)
    Proceedings of the London Mathematical Society (3), 102, (2011), 801-842.
Powers of sequences and convergence of ergodic averages (with M. Johnson, E. Lesigne, and M. Wierdl)
    Ergodic Theory & Dynamical Systems, 30, (2010), no. 5, 1431-1456.
Multiple recurrence and convergence for Hardy field sequences of polynomial growth
    Journal d'Analyse Mathematique, 112, (2010), 79-135.
Equidistribution of sparse sequences on nilmanifolds
    Journal d'Analyse Mathematique, 109, (2009), 353-395.
A Hardy field extension of Szemeredi's theorem (with M. Wierdl)
    Advances in Mathematics, 222, (2009), 1-43.
Powers of sequences and recurrence (with E. Lesigne and M. Wierdl)
    Proceedings of the London Mathematical Society (3), 98, (2009), no. 2, 504-530.
Ergodic Theory: Recurrence (with R. McCutcheon)
    Encyclopedia of Complexity and System Science, Springer, (2009), Part 5, 3083-3095. (Updated version 2019)
Multiple ergodic averages for three polynomials and applications
    Transactions of the American Mathematical Society, 360, (2008), no. 10, 5435-5475.
Multiple recurrence and convergence for sequences related to the prime numbers (with B. Host and B. Kra)
    J. Reine Angew. Math., 611, (2007), 131-144.
Uniformity in the polynomial Wiener-Wintner theorem
    Ergodic Theory & Dynamical Systems, 26, (2006), no. 4, 1061-1071.
Ergodic averages for independent polynomials and applications (with B. Kra)
    Journal of the London Mathematical Society, 74, (2006), no. 1, 131-142.
On the degree of regularity of generalised van der Waerden triples (with B. Landman and A. Robertson)
    Advances in Applied Mathematics, 37, (2006), no. 1, 124-128.
Sets of k-recurrence but not (k+1)-recurrence (with E. Lesigne, M. Wierdl)
    Annales de l'Institut Fourier, 56, (2006), no. 4, 839-849.
Convergence of multiple ergodic averages for some commuting transformations (with B. Kra)
    Ergodic Theory & Dynamical Systems, 25, (2005), no. 3, 799-809.
Polynomial averages converge to the product of integrals (with B. Kra)
    Israel Journal of Mathematics, 148, (2005), 267-276.
The structure of strongly stationary systems
    Journal d'Analyse Mathematique, 93, (2004), 359-388.
Additive functions modulo a countable subgroup of the reals
    Colloquium Mathematicum, 95, (2003), no. 1, 117-122. | {"url":"http://users.math.uoc.gr/~frantzikinakis/publications.html","timestamp":"2024-11-07T17:05:31Z","content_type":"text/html","content_length":"15957","record_id":"<urn:uuid:9495620c-016d-49e6-af23-ffa69061141e>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00651.warc.gz"} |
Comparing Standard Error of the Means
This Demonstration shows three sampling distributions of means. You can change the standard error of the mean for each distribution. By doing so you can see how a sampling distribution becomes narrow
er and taller as the standard error decreases. Remember that the standard error decreases as sample size increases, so think of a decreasing standard error as an increasing function of sample size. | {"url":"https://www.wolframcloud.com/obj/5b953459-2eed-471d-9a4b-878751f9917d","timestamp":"2024-11-11T01:06:52Z","content_type":"text/html","content_length":"201324","record_id":"<urn:uuid:a73a56d3-2335-472b-8cfe-b318379c26b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00242.warc.gz"} |
Major Loss, or Loss Capacity - What does 'allocation percentile' mean, and how does it correlate to market loss or gain?
Q - What is meant by 'allocation percentile' and how does this correlate to the actual percentage of market loss or gain?
A - When looking at Plan Settings > Major Loss you are presented with two columns - one headed 'Fixed Growth', the other headed 'Allocation Percentile', as shown below:
Note that there are two ways to grow the value of accounts, within Voyant, either by using simple 'fixed growth' rates, or by applying an asset allocation (relying on assumptions about each of the
individual asset classes therein). These options are not mutually exclusive, and one can easily combine both approaches, within the same client plan. What's important to understand is that: how the
market crash is applied (by the software) will be different for accounts using a 'fixed growth' rate, compared with accounts using an asset allocation.
If all of the accounts within your client plan are operating solely on 'fixed growth' rates, then one can simply ignore the column labelled 'Allocation Percentile', as this won't be relevant. The
'Allocation Percentile' column is only applicable to accounts being grown by way of an asset allocation. The importance of 'allocation percentile' is that this is a relative (and probabilistic)
measure of performance*, rather than an absolute measure - the consequence is that one should expect to get a nuanced response to the assumed market crash, dependent upon the underlying asset
allocation of one's accounts, e.g. an account allocated entirely to domestic fixed interest, will not behave/respond to the crash in the same way as an account allocated entirely to overseas equity.
It follows from this, of course, that there is not a straightforward correlation between the 'allocation percentile' value entered, and the resulting rate of return.
*To emphasize, the 'percentile' values, used in the software's market-based simulations, are a measure of relative performance (i.e. relative to the assumed mean). The resulting output will be
determined by one's underlying market assumptions (mean and standard deviation values, as located under Preferences > Market Assumptions). The 'allocation percentile' value that is entered will be
applied to any accounts in your plan that are set to be grown using an asset allocation, rather than a 'fixed growth' rate.
Q - When I enter/change the value under 'Fixed Growth', the corresponding value under 'Allocation Percentile' changes - why is this?
Firstly, note that one can always overwrite the existing (default) value shown under either 'Fixed Growth', or 'Allocation Percentile', meaning they can both be set to any value desired. It is the
case, nonetheless - when one enters a value under 'Fixed Growth' - that the software will interpret this value for the purpose of any accounts (in your client plan) that happen to be utilising an
asset allocation. The software will do this by taking the 'Fixed Growth' value you have entered, and comparing it to returns (mean, and standard deviation values) for the FTSE All Share between 1900
and 2017, to arrive at a measure of relative performance, given the assumption that investment returns are normally distributed. The assumption of normal distribution means that returns conform to a
bell curve probability distribution. For further reading, one might refer to Wikipedia, for example: http://en.wikipedia.org/wiki/Normal_distribution
The assumption of normal distribution enables the software to take a given 'fixed growth' return (e.g. -30%), and determine the relative probability of such an outcome, historically speaking. This
value is called the Z-score and it represents the number of standard deviations away from (above, or below) the mean. If the Z-score is 1, for example, then the investment return value is 1 standard
deviation above the mean, which corresponds (approximately) to the 84th percentile. Returning to the example above - i.e. a return of -30% - the FTSE All Share dataset gives us a Z-score of -1.72,
which translates to an allocation percentile value of approximately 4. One can interpret this value - in turn - as meaning that the annual return of the FTSE All Share (between 1900 and 2010) will
have improved upon -30% approximately 96% of the time.
To reiterate, one does not need to run with the software's suggested 'allocation percentile' value(s) - its calculated percentile values are merely a suggestion, based on the 'fixed growth' value(s)
entered by the user. As stated, one can overwrite the suggested 'allocation percentile' to any value between 0 and 100. | {"url":"https://support.planwithvoyant.com/hc/en-us/articles/202085258-Major-Loss-or-Loss-Capacity-What-does-allocation-percentile-mean-and-how-does-it-correlate-to-market-loss-or-gain","timestamp":"2024-11-08T12:44:33Z","content_type":"text/html","content_length":"25774","record_id":"<urn:uuid:bac4d991-7c0e-489b-b01f-8c7a7fc5282f>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00713.warc.gz"} |
K243 is the isogonal pK with pivot the reflection X(376) of G in O. Hence, it is a member of the Euler pencil of cubics. See Table 27.
Locus properties
• The pedal triangle of P and the triangle of reflections of P in the vertices of ABC are perspective if and only if P lies on K243. See Table 6.
• Let PaPbPc be the circumcevian triangle of P and denote by Ka, Kb, Kc the Lemoine points of triangles BCPa, CAPb, ABPc respectively. Let Qa, Qb, Qc be the orthogonal proojections of Ka, Kb, Kc on
BC, CA, AB respectively. ABC and QaQbQc are perspective (at Q) if and only if P lies on K243 (Kadir Altintas, private message, 2020-09-17). The locus of Q is a complicated isotomic circum-sextic
passing through G (quadruple point with tangents passing through X(7) and its extraversions) and A, B, C are three nodes on the curve.
If Ra, Rb, Rc are the reflections of Ka, Kb, Kc in the respective sidelines of ABC, then ABC and RaRbRc are perspective if and only if P lies on K1158.
If Ka, Kb, Kc are orthocenters instead of Lemoine points, the locus of P such that ABC and QaQbQc are perspective (at Q) is the Darboux cubic K004 and the locus of Q is another analogous (but
simpler) isotomic circum-sextic passing through X(i) for i = 2, 4, 69, 459, 1440, 6616, 7080, 37669. | {"url":"http://bernard-gibert.fr/Exemples/k243.html","timestamp":"2024-11-02T18:01:21Z","content_type":"text/html","content_length":"8010","record_id":"<urn:uuid:9f1218aa-1ccf-4f00-a3a3-8a03881e4efd>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00755.warc.gz"} |
Forecasting Using Linear Regression - iBForecast
Forecasting Using Linear Regression
In the world of data analysis, forecasting is a crucial tool that assists businesses in making informed decisions and predicting future outcomes. One method used widely by analysts for accurate
predictions is linear regression. By establishing a relationship between two variables, linear regression leverages this information to forecast future values based on observed data. This article
explores the fundamentals of forecasting using linear regression, delving into the steps involved in constructing a regression model, assessing its accuracy, and interpreting the results. Gain
valuable insights into the power of linear regression for forecasting and unlock its potential for improved decision-making.
Understanding Linear Regression
Definition of Linear Regression
Linear Regression is a statistical modeling technique used to establish a relationship between a dependent variable and one or more independent variables. It assumes that there is a linear
relationship between the variables, meaning that any change in the independent variables will result in a proportional change in the dependent variable. The main objective of linear regression is to
estimate the values of the dependent variable based on the values of the independent variables.
Purpose of Linear Regression
The purpose of linear regression is to analyze and predict the behavior of a dependent variable based on the values of independent variables. It is widely used in various fields, including economics,
finance, marketing, and social sciences. The primary goal is to understand the relationship between variables and make accurate predictions or forecasts.
Assumptions of Linear Regression
Linear regression relies on several assumptions in order to provide reliable results. These assumptions include:
1. Linearity: The relationship between the dependent and independent variables is linear.
2. Independence: Observations are independent of each other.
3. Homoscedasticity: The variance of the errors is constant across all levels of the independent variables.
4. Normality: The errors follow a normal distribution.
5. No multicollinearity: There is no perfect correlation between independent variables.
Fulfilling these assumptions is crucial to ensure the validity and reliability of the regression model.
Data Collection and Preparation
Identifying relevant data
Before building a regression model, it is essential to identify and collect relevant data. This process involves determining which variables are potential predictors of the dependent variable and
gathering data for those variables. The selection of relevant data can be based on prior knowledge, domain expertise, or exploratory data analysis.
Cleaning and preprocessing data
Once the relevant data has been identified, it is necessary to clean and preprocess the data. This step involves removing any missing or erroneous data, handling outliers, and transforming variables
if necessary. Cleaning and preprocessing the data ensure that the regression model is built on accurate and reliable data.
Data normalization
Data normalization is an important step in linear regression to ensure that the variables have a similar scale and distribution. Normalizing the data involves transforming the variables to have a
mean of zero and a standard deviation of one. This process allows for better understanding and interpretation of the regression coefficients and improves the stability of the model.
Building the Regression Model
Choosing independent and dependent variables
The first step in building a regression model is to select the appropriate independent and dependent variables. The independent variables are the predictors that will be used to estimate the value of
the dependent variable. It is crucial to choose variables that have a significant impact on the dependent variable and are not highly correlated with each other.
Splitting data into training and testing sets
To evaluate the performance of the regression model, the data is typically divided into training and testing sets. The training set is used to build the model, while the testing set is used to assess
its accuracy and generalization capabilities. By splitting the data, we can measure the model’s performance on unseen data and determine if it can effectively predict the dependent variable.
Creating the regression model
The next step is to create the regression model using the training data. The model aims to find the best-fitting line or hyperplane that minimizes the errors between the predicted and actual values
of the dependent variable. This is done by estimating the regression coefficients using various methods, such as ordinary least squares or maximum likelihood estimation.
Evaluating model performance
Once the regression model is created, its performance needs to be evaluated. Common metrics used to assess model performance include the coefficient of determination (R-squared), mean squared error
(MSE), and root mean squared error (RMSE). These metrics provide insights into how well the model fits the training data and its ability to predict the dependent variable accurately.
Forecasting with Linear Regression
Understanding the concept of forecasting
Forecasting refers to the process of making predictions about future values of a dependent variable based on historical data and the regression model. It helps in decision-making and planning by
providing estimates of future outcomes.
Predicting future values using regression model
Linear regression can be used to forecast future values by extending the regression line or hyperplane beyond the available data. This allows us to estimate the values of the dependent variable for
new or unseen values of the independent variables. However, it is important to note that the accuracy of the forecasts depends on the stability of the relationship between the variables and the
validity of the assumptions.
Interpreting regression coefficients
The regression coefficients provide insights into the relationship between the independent and dependent variables. They represent the average change in the dependent variable for a unit change in
the corresponding independent variable, while holding other variables constant. Positive coefficients indicate a positive relationship, while negative coefficients indicate a negative relationship.
Handling uncertainty in forecasted values
Forecasted values obtained from a regression model are subject to uncertainty. This uncertainty arises from various sources, such as measurement errors, random variations, and unaccounted factors. It
is important to account for this uncertainty and communicate the range of possible outcomes to stakeholders, along with the forecasted values.
Assessing Model Validity
Measuring prediction accuracy
Measuring the prediction accuracy of the regression model is crucial to determine its validity. Various metrics, such as R-squared, MSE, and RMSE, can be used to evaluate the model’s performance.
These metrics provide information about how well the model fits the data and how accurately it predicts the dependent variable.
Analyzing residuals and residuals plots
Residuals are the differences between the predicted and actual values of the dependent variable. Analyzing the residuals and residual plots helps in assessing the model’s assumptions and identifying
any patterns or outliers in the data. A random distribution of residuals with zero mean indicates that the assumptions of linearity, homoscedasticity, and normality are met.
Checking for multicollinearity
Multicollinearity occurs when there is a high correlation between independent variables in the regression model. It can adversely affect the model’s performance and interpretation of the
coefficients. To check for multicollinearity, correlation matrices or variance inflation factors (VIF) can be used. If multicollinearity is detected, steps such as removing one of the highly
correlated variables or using dimensionality reduction techniques may be necessary.
Identifying outliers
Outliers are data points that deviate significantly from the overall pattern of the data. They can have a substantial impact on the regression model and its predictions. Identifying and handling
outliers is important to ensure the model’s validity. Techniques such as visual inspection of scatter plots, leverage analysis, and Cook’s distance can be used to identify outliers and decide whether
to exclude them from the analysis.
Improving Model Performance
Feature engineering and selection
Feature engineering involves transforming and creating new features from the existing variables to improve the model’s performance. This can include deriving new variables, combining existing
variables, or using domain knowledge to create meaningful predictors. Feature selection, on the other hand, involves identifying the most relevant features that have the highest impact on the
dependent variable. Both feature engineering and selection aim to enhance the model’s predictive power.
Regularization techniques
Regularization techniques help prevent overfitting and improve the model’s generalization capabilities. Regularization adds a penalty term to the regression model that discourages large coefficients
and excessive complexity. The two most common regularization techniques used in linear regression are L1 regularization (Lasso) and L2 regularization (Ridge). Regularization can improve model
performance by reducing the variance and bias of the regression model.
Model validation and cross-validation
Model validation is essential to ensure the robustness and generalizability of the regression model. Cross-validation techniques, such as k-fold cross-validation, can be used to assess the model’s
performance on different subsets of the data. By evaluating the model on multiple validation sets, we can obtain a more accurate estimate of the model’s performance and identify any issues or
Iterative model improvement
Building a regression model is an iterative process that involves continuously refining and improving the model. This can be done by incorporating feedback and insights from the model’s performance,
seeking domain expertise, and conducting additional data analysis. Iterative model improvement ensures that the model is continuously updated with new data and remains accurate and relevant over
Potential Challenges in Forecasting
Overfitting and underfitting
Overfitting occurs when the regression model fits the training data too closely, capturing noise and irrelevant patterns. This leads to poor generalization on unseen data. Underfitting, on the other
hand, occurs when the model is too simple and fails to capture the underlying relationships in the data. Balancing between overfitting and underfitting is a challenge in linear regression
Data scarcity or incompleteness
In some cases, data scarcity or incompleteness can pose challenges in linear regression forecasting. Insufficient data may limit the model’s ability to capture the true underlying relationships
accurately. In such situations, alternate data sources, domain expertise, or other forecasting methods may need to be considered.
Incorporating external factors
Linear regression typically assumes that the relationships between variables are constant over time. However, in real-world scenarios, external factors such as economic conditions, policy changes, or
market trends can influence the relationship between variables. Incorporating these external factors into the regression model can be challenging but necessary for accurate forecasting.
Handling nonlinear relationships
Linear regression assumes a linear relationship between the dependent and independent variables. However, real-world relationships may be nonlinear. In such cases, applying nonlinear transformations
to the data or using nonlinear regression techniques may be required. Handling nonlinear relationships can be complex and may involve trial and error or using more advanced modeling techniques.
Applications of Linear Regression Forecasting
Business sales and demand forecasting
Linear regression is commonly used in business to forecast sales and demand for products or services. By analyzing historical sales data and relevant variables such as marketing expenditure, pricing,
and promotions, businesses can predict future sales and plan production, inventory, and marketing strategies accordingly.
Stock market price prediction
Linear regression can be used in stock market analysis to predict the future price of a stock based on historical price movements and other influential factors such as company financials, market
indices, and news sentiment. Stock market price prediction can aid investors in making informed decisions and managing their portfolios.
Weather forecasting
Linear regression has applications in weather forecasting, particularly in short-term predictions. By analyzing historical weather data, such as temperature, humidity, wind speed, and atmospheric
pressure, meteorologists can forecast future weather conditions. Linear regression models can provide valuable insights and predictions for various weather phenomena.
Population growth estimation
Linear regression can be used to estimate population growth in a given area by analyzing historical population data and relevant variables such as birth rates, death rates, migration patterns, and
socio-economic indicators. Population growth estimation can assist in urban planning, resource allocation, and policy-making.
Comparison with Other Forecasting Methods
Advantages of linear regression
Linear regression has several advantages compared to other forecasting methods. It is relatively easy to understand and interpret, making it accessible to non-technical stakeholders. Linear
regression also provides insights into the relationships between variables, allowing for better understanding of the underlying mechanisms. Additionally, linear regression can handle both continuous
and categorical variables, making it versatile for various types of data.
Disadvantages of linear regression
Despite its advantages, linear regression has some limitations. It assumes a linear relationship between variables, which may not always hold true in real-world scenarios. Linear regression is also
sensitive to outliers and multicollinearity, which can affect the model’s performance. Furthermore, linear regression may not capture complex relationships or interactions between variables,
requiring the use of more advanced modeling techniques.
Comparison with time series forecasting
Time series forecasting focuses on analyzing and predicting data points collected at regular intervals over time. In contrast, linear regression considers the relationship between variables in a more
general sense, irrespective of their time series nature. Time series forecasting methods, such as ARIMA or exponential smoothing models, are better suited for predicting future values based solely on
historical time series data.
Comparison with machine learning approaches
Linear regression is considered a simpler and more interpretable method compared to machine learning approaches, such as neural networks or random forests. Machine learning models can capture more
complex patterns and interactions but require larger amounts of data and may be more computationally expensive. The choice between linear regression and machine learning approaches depends on the
specific problem, the available data, and the desired level of interpretability.
Linear regression is a reliable and widely used method for forecasting. By understanding the definition, purpose, and assumptions of linear regression, and following a systematic process of data
collection, preprocessing, model building, and evaluation, accurate and meaningful forecasts can be obtained. However, it is essential to be aware of the potential challenges, such as overfitting or
data scarcity, and to consider alternative forecasting methods when necessary. Despite its limitations, linear regression remains a valuable tool in various fields and can provide valuable insights
and predictions for informed decision-making. | {"url":"https://ibforecast.com/forecasting-using-linear-regression/","timestamp":"2024-11-13T09:04:22Z","content_type":"text/html","content_length":"177366","record_id":"<urn:uuid:bbcb66fa-c4f0-44b5-8945-f97b9eb397e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00737.warc.gz"} |
An intrinsic function. An expr node.
IntrinsicFunction(expr* args, int intrinsic_id, int overload_id,
ttype type, expr? value)
• args represents all arguments passed to the function
• intrinsic_id is the unique ID of the generic intrinsic function
• overload_id is the ID of the signature within the given generic function
• type represents the type of the output
• value is an optional compile time value
Return values¶
The return value is the expression that the IntrinsicFunction represents.
IntrinsicFunction represents an intrinsic function (such as Abs, Modulo, Sin, Cos, LegendreP, FlipSign, …) that either the backend or the middle-end (optimizer) needs to have some special logic for.
Typically a math function, but does not have to be.
IntrinsicFunction is both side-effect-free (no writes to global variables) and deterministic (no reads from global variables). They can be used inside parallel code and cached. There are two kinds:
• elemental: the function is defined as a scalar function and it can be vectorized over any argument(s). Examples: Sin, Cos, LegendreP, Abs
• non-elemental: it accepts arrays as arguments and the function cannot be defined as a scalar function. Examples: Sum, Any, MinLoc
The intrinsic_id determines the generic function uniquely (Sin and Abs have different number, but IntegerAbs and RealAbs share the number) and overload_id uniquely determines the signature starting
from 0 for each generic function (e.g., IntegerAbs, RealAbs and ComplexAbs can have overload_id equal to 0, 1 and 2, and RealSin, ComplexSin can be 0, 1).
Backend use cases: Some architectures have special hardware instructions for operations like Sqrt or Sin and if they are faster than a software implementation, the backend will use it. This includes
the FlipSign function which is our own “special function” that the optimizer emits for certain conditional floating point operations, and the backend emits an efficient bit manipulation
implementation for architectures that support it.
Middle-end use cases: the middle-end can use the high level semantics to simplify, such as sin(e)**2 + cos(e)**2 -> 1, or it could approximate expressions like if (abs(sin(x) - 0.5) < 0.3) with a
lower accuracy version of sin.
We provide ASR -> ASR lowering transformations that substitute the given intrinsic function with an ASR implementation using more primitive ASR nodes, typically implemented in the surface language
(say a sin implementation using argument reduction and a polynomial fit, or a sqrt implementation using a general power formula x**(0.5), or LegendreP(2,x) implementation using a formula (3*x**2-1)/2
This design also makes it possible to allow selecting using command line options how certain intrinsic functions should be implemented, for example if trigonometric functions should be implemented
using our own fast implementation, libm accurate implementation, we could also call into other libraries. These choices should happen at the ASR level, and then the result further optimized (such as
inlined) as needed.
The argument types in args have the types of the corresponding signature as determined by intrinsic_id. For example IntegerAbs accepts an integer, but RealAbs accepts a real.
The following example code creates IntrinsicFunction ASR node:
(Real 4 [])
(Real 4 [])
(RealConstant 0.479426 (Real 4 [])) | {"url":"https://docs.lfortran.org/en/asr/asr_nodes/expression_nodes/IntrinsicFunction/","timestamp":"2024-11-08T04:04:13Z","content_type":"text/html","content_length":"36943","record_id":"<urn:uuid:34fef63c-144c-4287-ae08-9a1a092ff1a6>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00883.warc.gz"} |
Consider a closed system undergoing an isothermal, reversible process. How does the change
In a closed system undergoing an isothermal, reversible process, the change in entropy (ΔS) is related to the heat absorbed (Q) by the system and the temperature (T) according to the following
ΔS = Q / T
This equation arises from the Clausius statement of the second law of thermodynamics, which states that the entropy of an isolated system can never decrease over time. In an isothermal process, the
temperature remains constant, and the heat absorbed by the system is entirely used to increase its entropy. The equation shows that the change in entropy is directly proportional to the heat absorbed
and inversely proportional to the temperature.
The concept of entropy, often referred to as "disorder," measures the randomness or disorder within a system. An increase in entropy indicates a higher degree of disorder, while a decrease in entropy
indicates a lower degree of disorder. In an isothermal, reversible process, the system reaches its maximum possible entropy at equilibrium, where the change in entropy is zero (ΔS = 0). | {"url":"https://userz.net/7b9ddc25-8101-4c6a-bee9-a111045264f0","timestamp":"2024-11-01T19:33:35Z","content_type":"text/html","content_length":"110514","record_id":"<urn:uuid:f47bc086-58d8-4554-b0e1-7b141decf797>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00091.warc.gz"} |
effuciency calculation for ball mill
Grinding efficiency decreases with increasing mill size. ... Ball mill is one of the most commonly used mills for the crushing and grinding of mineral ore. It is generally used to grind material
down to the particle size of 20 to 75 μm and can vary in size from a small batch mill up to a mill with outputs of hundreds of tonnes per hour ...
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Analysis of Variant Ball Mill Drive Systems. The basic element of a ball mill is the drum, in which the milling process takes place ( Figure 1 ). The length of the drum in the analyzed mill
(without the lining) is m, and the internal diameter is m. The mass of the drum without the grinding media is 84 Mg.
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The energy consumption of the total grinding plant can be reduced by 2030 % for cement clinker and 3040 % for other raw materials. The overall grinding circuit efficiency and stability are
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Ball Mill Power Calculation Example #1. A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80% passing ¼ inch (6350
microns). The required product size distribution is to be 80% passing 100 mesh (149 microns). In order to determine the power requirement, the steps ...
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Industrial Ball Mill The ball mill utilized in the sampling survey has an inside diameter of m and length of m and is run in open circuit. Under normal operating conditions, the mill ball loading
is 30% of total mill volume, mill rotational speed is 75% of critical speed, slurry solids concentration is 75%, solids feed rate is 330 tph.
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Since for the ball mill design we are using 80% passing, the required value of C2 for the ball mill will be equal C3 is the correction factor for mill diameter and is given as; 𝐶𝐶3 = 𝐷𝐷 (3)
However, it is important to note that C3 = vessel used in producing the ball mill was got from a
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Ball mill from China. To calculate the charge volume of a ball mill, you will need to know the internal diameter of the mill and the distance between the top of the charge and the top of the mill
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The effect of circulating load and classification efficiency on the performance of ball mill circuits is compared to the effect on HPGR circuits. The fundamentals of grinding behavior are also
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ResearchGate | Find and share research
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If the ball mills are in connection with hydrocyclones then the ball mills can't be evaluated by themselves but they need to be assessed as part of the grinding classification system. ... You
perform size analysis of the different streams and then calculate the grinding efficiency and classification efficiency. Make sure that every sampling ...
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ball mill efficiency calculations OneMine Mining and Minerals . TITLE, Using The Bond Work Index To Measure Operating Comminution Efficiency. »calculations for crusher productions »ball mill load
calculations »bond,, 1961. crushing and grinding calculations parts i and ii, british chemical engineering, ...
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where d is the maximum size of feed (mm); σ is compression strength (MPa); E is modulus of elasticity (MPa); ρb is density of material of balls (kg/m 3); D is inner diameter of the mill body
(m).. Generally, a maximum allowed ball size is situated in the range from D /18 to D/24.. The degree of filling the mill with balls also influences productivity of the mill and milling
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Based on his work, this formula can be derived for ball diameter sizing and selection: Dm <= 6 (log dk) * d^ where D m = the diameter of the singlesized balls in = the diameter of the largest
chunks of ore in the mill feed in mm. dk = the P90 or fineness of the finished product in microns (um)with this the finished product is ...
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Type of Mill Media Size, in. Tip Speed, ft./sec. Ball mill ½ and larger — Attrition mill 1/ 8 3/ 8 600 1,000 Sand mill 1/ 64 8 2,000 3,000 Small media mill mm 3 mm 1,000 3,000 Choose the Right
Grinding mill Consider the feed material's nature and the milling's objective By Robert E. Schilling, Union Process Inc.
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Objectives. At the end of this lesson students should be able to: Explain the grinding process. Distinguish between crushing and grinding. Compare and contrast different type of equipment and
their components used for grinding. Identify key variables for process control. Design features of grinding equipment (SAG, BALL and ROD MILLS)
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Efficiency = Wio/Wi x 100. When (Wio/Wi)x100 is less than 100 this indicates the circuit, based upon this comparison, is operating efficiently. When it is greater than 100 this indicates the
circuit is operating inefficiently. A large difference, either low or high, could indicate that the two work indicies are not on the same basis.
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size ball mill was used with ball media of sizes 10 mm, 20 mm and 30 mm respectively. Quartz was the material used to perform the experiment and was arranged into 3 monosizes namely 8 mm + mm, 4
mm + mm and2 mm + mm for the experiment. A mill run having a mixture of the 3 ball diameter sizes was also conducted. It was
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A Metallurgist or more correctly a Mineral Processing Engineer will often need to use these "common" Formulas Equations for Process evaluation and calculations: Circulating load based on pulp
density *** Circulating load based on screen analysis *** Flotation Cell Conditioner Capacities *** Two Product Formula *** Three Product (bimetallic) Formula (Part 1) ***needs verify Three ...
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This reverts to the Morgärdshammar method and is similar to the AM calculation Tower Mill The tower mill calculation is based on the ball mill design sheet, but is simplified in that the mill
design section is omitted. A simple tower mill factor of 70% allows the mill power to be estimated. MILLCALCv2a 5 19/02/2004
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In this context, the ball mill and the air classifier were modelled by applying perfect mixing and efficiency curve approaches, respectively. The studies implied that the shape of the efficiency
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Review: Konica Hexanon AR 40mm f/1.8 - phillipreeve.net
Review: Konica Hexanon AR 40mm f/1.8
Konica Hexanon AR 40mm F1.8 on Nikon Z6
Konica Hexanon AR 40mm f1.8 is a pancake standard lens, it was shipped as kit lens with Konica SLR cameras during a couple of years in mid and late ’70s. While it is not an actual wide angle lens it
is still wider than a normal 50mm lens, which, in my opinion, can be helpful for street photography as it allows to include more of the environment from the same camera to subject distance. It also
does so without introducing the perspective distortion of a 35mm wide angle lens with equivalent speed/aperture.
Sometimes it’s hard to explain it but in many situations the 40mm focal length feels just right (i.e. to my taste). When it was introduced some photography magazines considered it the sharpest lens
ever produced (for its time that is), there is no hard proofs on that though. You can find it very cheap at about 20-30$. Let’s see if it still is justified to buy this lens today.
Sample Images
Nikon Z6 | Konica Hexanon AR 40/1.8 | 5.6
Nikon Z6 | Konica Hexanon AR 40/1.8 | 5.6
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | 5.6
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F1.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
Most of the sample images in this review and many more can be found in higher resolution here.
Focal Length: 40mm
Aperture Range: 1.8 – 22 (in full stops)
Number of Aperture Blades: 6
Min Focus Distance: 0.45 m
Filter Size: 55mm
Lens Mount: Konica AR
Weight: 140g
Length: 27mm (focus ring set to infinity)
Elements/Group: 6/5
Optical Lens Diagram
There are not any variations of this lens to my knowledge, all look the same.
This is a so-called pancake lens, which is very flat, only 27mm long. On small Nikon Z bodies it is not obtrusive at all, which is a big plus in everyday and street photography, where 40mm focal
length is mostly used in. The lens + Nikon Z adapter is 51mm long, which is shorter than Nikon 50mm/1.8 AF-S G lens alone (53mm long). The build quality seems to be good, all metal, focus is a little
stiff, which can be due to the age and also my sample has a little throw when you rotate the focus ring in opposite directions (also due to the age I think) other samples may be free from this.
Konica Hexanon AR 40mm f1.8 compared to Nikon Nikkor AF-S 50mm 1:1.8 G on Nikon Z6/Z7
Aperture ring clicks only at full stops rather stiffly from 1.8 to 22. There is an extra click stop position beyond 22 on the aperture ring marked “AE” in green. This was used to automatically
control the aperture on Konica cameras. If you happen to rotate the ring to that position the aperture ring gets locked and you cannot move it any more unless you push a button on the same ring but
on the opposite side of the aperture value scale to release the lock and rotate the ring again.
Konica Hexanon AR 40/2.8, AE Unlocking Button
At the rear of the lens there is also a fin that was used to control the aperture automatically from the camera, that is useless on digital mirrorless cameras though as everything has to be handled
Konica Hexanon AR 40/2.8, AE Fin
Optical Features
Sharpness (Infinity)
For the images of the test chart (and all the other images in this review) a standard LR settings of light/contrast and a sharpness of 40 have been used.
Sharpness Comparison at Infinity
Infinity Sharpness Comparison Reference, Yellow Markings at Center, Midframe & Corner
At f/1.8 soft in center, midframe and corner, still usable center sharpness. At f/2.8 it sees a little improvement, but still soft. At f/4.0 the center sharpness is good, the midframe and corner soft
but usable. Very sharp in center at f/5.6 and sharp in midframe and corner. At f/8.0-f/16 excellent sharpness across the frame, f/22 a little less sharp thanks to diffraction. In general the contrast
is low on wider apertures but good at f/8.0 and smaller openings.
Nikon Z6 | Konica Hexanon AR 40/1.8 | F8
Sharpness (Portrait)
For portraits we look at points where it matters; center, inner center circle and outer center circle.
Points of Interest in Portraits
First let’s look at the center of the image!
At 1.5m distance the sharpness in the center is soft at f/1.8 but usable, good at f/2.8 and Excellent at f/4.0 to f/8.0.
Second the sharpness at the inner circle periphery
At 1.5m distance the sharpness at the inner circle is very soft at f/1.8 but still usable, at f/2.8 it barely improved but very good at f/4.0.
And lastly at the outer circle periphery.
At 1.5m distance the sharpness at outer circle is too soft and hardly usable at f/1.8, at f/2.8 it is much better and at f/4.0 it is good. It gets better at f/5.6 and f/8.0 but never comes to the
same sharpness and contrast level as in the center.
At f/1.8 the lens have some spherical aberrations, which affects the sharpness and contrast everywhere but it goes away at f/2.8.
In general it seems that sharpness is better at closer distances than near infinity.
Sharpness (Close-up)
Setup used for test of close up sharpness
The sharpness at f/1.8 is soft partly because of the spherical aberrations but it is usable, it gets a real boost at f/2.8 reaching a very good level at f/4.0 and beyond it is excellent until f/16,
where the diffraction eats some of the sharpness and at f/22 it is even softer. Generally the sharpness is much better at closer distances. In the following images you can see a picture taken near
minimum focus distance and then a crop from the same image, both without any post sharpening (except the LR default value mentioned before) or clarity adjustments (only lighting and contrast were
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
100% crop from the image above (No sharpness or clarity added in post)
Lens Distortion
The lens shows some barrel distortion, which is easily fixed in post if required, otherwise it is so small and in the case of applications relevant for this lens negligible.
Konica Hexanon AR 40/1.8 Lens Distortion
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8 (Uncorrected)
At f/1.8 the vignetting is about 1.7 stops, stopping down just one stop to f/2.8 it goes down to about 0.5 stops and at f/4.0 and beyond there is almost no vignetting at all.
Flare Resistance
The lens suffers from flares of all types. The veiling flare can easily ruin the picture by decreasing the image contrast dramatically and as soon as you have a strong light source (like the sun) in
the image or near the edge of the frame you will get different ghosting flares. Here are three examples, 1. Sun just outside the frame 2. Sun in the frame, 3. Sun at the edge of the frame, 4. the
same as 3 but I moved a step to cover the lens by the tree shadow.
Nikon Z6 | Konica Hexanon AR 40/1.8 | 5.6
Nikon Z6 | Konica Hexanon AR 40/1.8 | 11
Nikon Z6 | Konica Hexanon AR 40/1.8 | 1.8
Nikon Z6 | Konica Hexanon AR 40/1.8 | 1.8
Chromatic Aberrations
Chromatic Aberration seems to be well enough controlled so that it does not bother you in normal situations. Of course if you really try you can find it in extreme conditions and at a very low level
but normally it is negligible.
The lens does suffer from heavy coma aberrations, more specifically sagittal astigmatism. To get rid of it completely you have to stop down to as far as f/8.0, where light points start getting the
shape of six pointed stars, this is not good at all, actually one of the worst I have ever seen. It is however comparable to other pancake lenses from that era. See the images!
Reference Image for Coma Demonstration
Sun Stars
You can get 6 pointed stars from f/11 (sometimes from f/8) with this lens. The star rays are not well defined before f/11 but at f/16 – f/22 very well defined and nice (that is subjective) but rather
small, see also the image in the Flare section, where sun is in the image.
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | 16
Focus Breathing
As you can see in the following image the lens suffers from focus breathing quite a bit.
Nikon Z6 | Konica Hexanon 40mm F1.8 | Focus Breathing
Bokeh is a subjective matter. In my eyes the bokeh of this lens is in most situations and mostly at close distances pleasing (partly helped by the spherical aberration wide open), though there are
some occasions when the subject is 1.5-2 meters away with a difficult background, when the bokeh gets busy and agitating. In general I like this lens’s bokeh, there are several images with shallow
depth of field in this article, have a look at them and see how you like it. Here you can see 1 case of pleasing bokeh and 1 case of bokeh that looks beautiful/creative to some but looks disturbing/
distracting to others plus two arranged photos for demonstrating the for- and background bokeh.
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F1.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F1.8
What I like Not good / Not bad What I don't like
Size Wide open sharpness
Center sharpness from 2.8 Build quality Contrast
Overall sharpness from F5.6 Bokeh in certain situations Flare resistance
Bokeh most of the time Color rendition Coma
Chromatic Aberrations control Spherical Aberrations wide open
The positive sides of this lens are its small size, very good sharpness in the center area from f/2.8 and superb across the frame sharpness stopped down to f/5.6 for closer distances. For longer
distances the sharpness is excellent at f/8 to f/11. Bokeh if nice in about 80%-90% (if not more based on the personal taste) of the times when wide open, good control of chromatic aberrations and
its low price (around $20-30). The negative sides are general sharpness wide open, contrast, micro contrast, somewhat busy bokeh in about 10%-20% of cases, flare, coma, spherical aberrations. So,
should you buy it?
If you want to test a fast focal length of 40mm (which is quite fun and practical for street or everyday photography) before investing in a newer much more expensive lens to see if this is a focal
length for you or if you are on a tight budget, go for it! For the price it’s a no-brainer. The Konica Hexanon 40mm f/1.8 is certainly usable on a modern mirrorless cameras, just avoid using it wide
open and it can produce razor sharp images in right conditions that will make you happy. Taking it everywhere with you is not an issue as it almost doesn’t take any space or weigh anything. You
should have its shortcomings in mind and photograph consciously though, otherwise if you really like the focal length and you think you use it often, try to save some money and get a more modern 40mm
lens e.g. a used Nikon Z 40/2 if you have a Nikon Z camera or a Voigtländer Ultron 40/2 if you have other cameras for about 200-300$ and if it will be used for more than 20% of your photography and
Autofocus is not necessary aim for Voigtländer Nokton 40/1.2.
If you are interested in buying a Konica Hexanon AR 40/1.8 or any of the lenses in the Alternatives section you can support our efforts by using the links below or given under each lens. It won’t
cost you a penny and it won’t affect the price but helps us little.
Buy Konica Hexanon AR 40mm from: ebay.com (affiliate links)
Nikon Nikkor Z 40mm f2
Nikon’s own 40mm made specially for Nikon Z cameras. it is slightly larger (46 mm long, 170g) but you don’t need any adapter to use it on Nikon Z cameras so the overall length will be shorter and
lighter. It has autofocus, better overall sharpness and much better sharpness wide open plus better contrast and much better flare resistance but about 5-6 times more expensive (277$ new, 220-240$
Buy from: Amazon.com, Amazon.de (affiliate links)
Voigtländer Ultron 40mm F2 SL II
With a length of 25-30cm (depending on mount) it is very close in size and more modern lens (Nikon F (FX), Canon EF, Pentax KAF), manual focus only, about 10 times more expensive (300$-400$ used)
Buy it from: ebay.com (affiliate links)
Voigtländer Ultron 40mm F2 SL II -S
A newer version of the above with Nikon Ai-S mount, manual focus only and more expensive than Nikon’s without being any better (419$ new)
Buy from: Amazon.com, Amazon.de (affiliate links)
Voigtländer Nokton 40mm 1:1,2 Aspherical
The fastest 40mm around, available in Nikon Z, Sony E/EF, VM mount. 315g, 54mm long (Nikon version)(Sony E version is 59.3mm). Manual focus only but one and a third steps faster, much better stopped
down sharpness, much nicer bokeh but about 18 times more expensive (799$ new Nikon Z version)
Buy from: Amazon.com, Amazon.de (affiliate links)
Sigma 40mm F1.4 DG HSM Art
Autofocus 40mm, 2/3 stop faster. With its 1200g weight and 131mm length it is like a bazooka on your camera. Nikon F (FX), Canon EF, Sigma SA Bayonet, Sony FE mount. (799$ new)
Buy from: Amazon.de, Amazon.de (affiliate links)
Laowa Argus 45mm f/0.95
Not a 40mm but not far from, even faster than the Voigländer Nokton by one stop, 2 steps faster than the Konica 40/1.8, which means 4 times more light to the sensor, nicest bokeh of all but with
lemon shape bokeh balls instead of round ones produced by Konica, manual focus only, no electrical contacts whatsoever, available in Nikon Z, Sony E, Canon RF, 835g 110mm long (599$ new)
Buy from: Amazon.com, Amazon.de (affiliate links)
For other camera mounts there are several other alternatives like:
Canon EF 40/2.8 STM,
Autofocus. a true pancake lens, 22mm long, One and a third stops slower, much less bokeh potential, about $100
Buy from: ebay.com (affiliate links)
Sony FE 40mm F2.5 G
Buy from: Amazon.com, Amazon.de (affiliate links)
Zeiss Batis 40/2 for Sony FE
Buy from: Amazon.com, Amazon.de (affiliate links)
Voigtländer 40mm F1.4 Nokton Classic for Leica M
Buy from: Amazon.com, Amazon.de (affiliate links)
Voigtlander VM 40mm F2.8 Heliar for Sony E
Manual focus. One and a third stops slower, much less bokeh potential, 21mm long, the thinnest 40mm
Buy from: Amazon.com, Amazon.de (affiliate links)
More Sample Images
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F1.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F1.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F2.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F1.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F2.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F1.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F2.8
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F5.6
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F5.6
Nikon Z6 | Konica Hexanon AR 40mm F1.8 | F1.8 (Sharpness added in post)
Most of the sample images in this review and many more can be found in higher resolution here.
Further Reading
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Martin M.H. lives outside Stockholm, Sweden. He is a M.Sc. in Computer Technology but he has been a passionate photographer for over 45 years. He started his photographic adventures when he was
thirteen with an Agfamatic pocket camera, which he soon replaced with a Canon rangefinder camera that his mom gave him in his teenages. After that he has been using Canon SLR, Nikon SLR manual focus
and Autofocus, Sony mirrorless crop sensor, Nikon DSLR and Nikon Mirrorless. He has photographed any genre he could throughout the years and you can see all kind of images in his portfolio. During
the later years though it has been mostly landscape, nature, travel and some street/documentary photography.
31 thoughts on “Review: Konica Hexanon AR 40mm f/1.8”
1. Nice to see an addition to the team first of all! Very exciting to see Nikon kit too! Nice work Martin!
The lens has particularly pleasing OOF rendering if you ask me!
1. Thanks Alan.
1. jang on, you have not mentioned the cost of an adapter, which will be needed for all modern/late production caa
if total – lens plus adapter = more than price of a modern lens, its best forgotten. id cost of a manual focus adapter, erll made, so as not to damage lens mounts, ok.
I remember Koncas ads slogan ‘its the lens’. öike Topcon, they were a credible maker . maker of optical glass.very highly regarded in Japan.in ,photo agazines, I never saw photos taken
with AR system, like Miranda,even Minolta they whex were second liners. Nikon and Asahi (Hieland, then Honeywell in the USA)Pentax were better marketed.
I guess Sony got benefit of past expertise of Konica and Minolta, when taken over, also boosted by co-operation with C.Ziess. interesting review, thanks
2. Used this for a while couple years back. Excellent as a traveling companion and general use lens. The lens is a lot cheaper than any 35/2 -lenses. Still have another hazy copy of it for film use
where it delivers equally well despite the haze.
3. It also works well for infrared photography
4. Thanks for the review! Looks like a cool lens that performs a lot like my Minolta 45/2. I might have to snag one!
1. Thanks Scott
5. I love a good 40mm perspective I bounce between 40 and 35 often may need to pick this one up as it is quite sharp indeed!
6. Great to have a new addition!! A warm Welcome to Martin & look forward to seeing more articles on the Z mirror less! 40 has been a sweet spot lens and this review nails it!
1. Thank you very much Lestor
7. Fully agree with the comments above: great work Martin & very nice to see a new review of a vintage lens and Nikon Z too! Looking forward to more 🙂
1. Thanks a lot Felix
8. Good review!
But the lens is meh, nothing to write home about.
1. Thanks Christoph.
The lens is not up to the day’s standards but for getting a taste of a fast 40mm and/or for only $30, it fulfills the purpose.
9. Good to see a new member in the Phillip Reeve Team. Bastian does a great job but I‘ve been missing the different styles. Thanks for the review. I used the small Konica lens for a while on my
first Sony A7. I eben had to of them and kept one. Now I will take it out of the drawer again. 😀 I love the colours you can get with it.
1. Thank you Dan.
10. Hi Martin. Thanks for your review. As other readers have pointed out, it’s great to see a reviewer on this site use one of the Z cameras to test vintage lenses. I’m curious about your impressions
of the Z6 for manually focusing vintage lenses, especially compared to Sony’s FF cameras. Do you seen any advantages of the Z6 for manual focusing? I’ve been a Z system user since Nikon
introduced the cameras but one thing I really appreciate about the Sony cameras is how you can set the punch-in focus to return to full screen with a half-shutter press. Regrettably, I can’t do
that on my Z bodies and I find that a bit frustrating.
1. Many thanks Doug.
I think it is an enjoyable experience to focus vintage lenses on Nikon Z. I don’t think there is a real advantage between the systems, I think what you mean by doing punch-in focus and return
to full screen is to magnify the image for focusing with precision and then return to full screen, right?
In that case it is just a matter different implementation, Nikon Z has this on the DISP button right under the thumb. With one push you magnify the image and set the focus accurately, push
again to return to full image. I really think it’s a matter of getting used to do the same think with your index finger or thumb.
2. Actually you can put this feature as you like on one of the functions buttons. I have put it on “AF-ON” button as it is the most convenient position for me. You can do it by going to “i” menu
> “custom control assignment” > “AF-ON button” > “Zoom on/off” > “High magnification (200%)”.
Done 🙂
11. Hello Martin,
thank you for this great review of this quite special hexanon.
1. Thank you S.B.
12. Hey Martin, Great review and welcome to this great team. Would appreciate more reviews on lenses adapted to the Nikon z Ecosystem. I am also a proud user of the Z6 and would be happy to read more
about it! Thx again
1. Thank you Michael.
There will be more reviews on Nikon cameras. Have Nikon friends? Appreciate if you spread the word.
13. jang on, you have not mentioned the cost of an adapter, which will be needed for all modern/late production caa
if total – lens plus adapter = more than price of a modern lens, its best forgotten. if price of a manual focus adapter is low and it is well made, so as not to damage lens mounts, ok.
I remember Koncas ads slogan ‘its the lens’. öike Topcon, they were a credible maker . makers of optical glass.very highly regarded in Japan.in photo magazines, I never saw photos taken with AR
system, like Miranda,even Minolta they whey were second liners. Nikon and Asahi (Heiland, then Honeywell in the USA)Pentax were better marketed.
I guess Sony got benefit of past expertise of Konica and Minolta, when taken over, also boosted by co-operation with C.Ziess. interesting review, thanks
1. No, I did not, because there are so many different adapters with different prices, but there are good adapters for $30 and even that cost together with the price of such a lens is far below
the price of a modern lens
14. In my opinion wide open softnes and glow of this lens is amazing. This is the only lens which I don’t care about the sharpness…
and 40mm focal is good than 50 for portraiture
1. Yes, it is a good moody lens.
15. I use this on my A7riii with adapter at 7.1 for sharp edge to edge photos. Nice bokeh at 1.8 with ok sharpness for when needed. Will get another adapter to use this with my Fuji.
16. You don’t mention the SMC Pentax-FA 1:1.9 43mm Limited as an alternative, even though this double-Gauss design is likely the finest pancake lens with this field of view. Special rendering once
you learn it. Not cheap.
Reverting to 50mm, the SMC Pentax-M 50mm F1.7 is similarly small and is the sharpest of all Pentax designs in the manual era. Which means it exceeds every other SLR brand too. Only 50 clams.
17. I’m using this lens with an adapter on my SONY NEX5 camera. Occasionally on my NIKON S1 to. (In spite of the impractical crop factor) A very nice and handy lens though, with a very good colour
rendition. It so happens that I’ve stumbled upon two more of theses lenses in very good shape (close to mint), lately. I do not need them, really. So if there’s an interest out there I may let
them go.
1. Yes, it is a nice lens.
If you want to sell the lenses, you should try eBay. If you are in Sweden, try Tradera, the chance to get prospects is much larger on those sites. | {"url":"https://phillipreeve.net/blog/konica-hexanon-ar-40mm-f1-8/","timestamp":"2024-11-05T07:39:09Z","content_type":"text/html","content_length":"188372","record_id":"<urn:uuid:ce7eb07e-6d2f-4bcb-a555-79d3478bb4a7>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00836.warc.gz"} |
Leetcode 1685. Sum of Absolute Differences in a Sorted Array | Video Summary and Q&A | Glasp
Leetcode 1685. Sum of Absolute Differences in a Sorted Array | Summary and Q&A
5.5K views
January 2, 2021
Leetcode 1685. Sum of Absolute Differences in a Sorted Array
The video explains an efficient algorithm to compute the sum of absolute differences in a sorted array.
Key Insights
• ❓ The problem involves calculating absolute differences in sorted arrays, an important concept in algorithm challenges.
• ❓ Utilizing properties of sorted arrays can significantly optimize calculations, reducing complexity from O(n^2) to O(n).
• ↔️ The introduction of left and right sums allows for efficient computation of differences while iterating through the array.
• 🧡 Cumulative sums are a powerful technique for solving problems involving ranges and differences in arrays.
• 👨💻 The process emphasizes the importance of crafting clear and maintainable code while implementing algorithmic solutions.
• ⌛ Time complexity analysis is crucial for assessing the feasibility of algorithm implementations against input size constraints.
• 🎮 The video illustrates how to encapsulate different summation sections into simple variables to streamline the solution.
hey there welcome back to lead coding in this video we will be solving the second question of lead code biweekly contest 41 name of the problem is sum of absolute differences in the sorted array you
are given an integer area numbers sorted in non-decreasing order build and return an integer array result with the same length as nums such that result... Read More
Questions & Answers
Q: What is the problem statement discussed in the video?
The video addresses the problem titled "Sum of Absolute Differences in a Sorted Array." It requires calculating for each element in a sorted array the sum of absolute differences between that element
and all other elements. A straightforward implementation would be inefficient, thus prompting an exploration of optimal solutions.
Q: Why is a naive O(n^2) solution not ideal for this problem?
The naive approach involves direct pairwise computations, leading to a time complexity of O(n^2). Given that the input size can be large, this results in significant performance issues. An efficient
solution is necessary to handle higher constraints, thus the presenter shifts focus to an O(n) optimization.
Q: How does the presenter utilize the properties of a sorted array?
The presenter effectively uses the non-decreasing order of the array to simplify calculations. By recognizing that all elements preceding a current index are less than or equal to the current
element, the solution leverages this property to separate the sum into contributions from both left and right segments efficiently.
Q: What is the final approach proposed in the video to solve the problem?
The approach involves calculating a total sum of the array, then iterating through it while maintaining a left sum variable. Using this structure, the video describes how to compute sums for both
left and right segments efficiently, yielding an O(n) time complexity for the entire solution.
Q: How is the left sum calculated during iteration?
During iteration, the left sum is maintained as each element is processed. It includes the sum of all elements to the left of the current index. This cumulative sum allows for rapid calculation of
the sum of absolute differences by factoring in how many elements exist to the left compared to the current number.
Q: What are the time and space complexities of the final solution provided?
The solution achieves O(n) time complexity for both calculating total sums and deriving the result for each index in a single iteration. Regarding space complexity, the implementation only requires
additional space for the result array, leading to a constant space complexity when excluding that.
Q: How does the solution compute the right sum during the iteration?
The right sum is derived by subtracting the current left sum and the element at the current index from the total sum. This allows for the efficient calculation of how many elements are on the right
and their contributions to the absolute differences relative to the current index.
Q: What recommendation does the presenter make at the end of the video?
The presenter encourages viewers to subscribe to the channel and enable notifications to keep up with future videos. This emphasizes community building and providing ongoing educational content about
algorithms and coding challenges.
Summary & Key Takeaways
• The problem involves building an integer array where each element is the sum of absolute differences between a given sorted array and every other element.
• A naive approach would take O(n^2) time, but the presenter demonstrates a more efficient O(n) method using cumulative sums to minimize calculation time.
• The final implementation computes values using left and right sum variables, allowing for a direct calculation of the required differences in a single pass.
Explore More Summaries from Fraz 📚 | {"url":"https://glasp.co/youtube/p/leetcode-1685-sum-of-absolute-differences-in-a-sorted-array","timestamp":"2024-11-06T05:15:59Z","content_type":"text/html","content_length":"368679","record_id":"<urn:uuid:78b0d748-64e2-4672-840e-9c88d3f1995f>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00328.warc.gz"} |
ABCD is square in which A lies on positive y-axis and B lies on... | Filo
is square in which lies on positive -axis and lies on the positive -axis. If is the point , then co-ordinate of is :
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Question Text is square in which lies on positive -axis and lies on the positive -axis. If is the point , then co-ordinate of is :
Updated On Nov 18, 2023
Topic Straight Lines
Subject Mathematics
Class Class 11
Answer Type Text solution:1 Video solution: 1
Upvotes 155
Avg. Video Duration 1 min | {"url":"https://askfilo.com/math-question-answers/a-b-c-d-is-square-in-which-a-lies-on-positive-y-axis-and-b-lies-on-the-positive","timestamp":"2024-11-11T17:11:14Z","content_type":"text/html","content_length":"429926","record_id":"<urn:uuid:8d9aff6b-35b3-44c4-8103-62e83fb05bca>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00262.warc.gz"} |
A Plus B
Tudor is sitting in math class, on his laptop. Clearly, he is not paying attention in this situation. However, he gets called on by his math teacher to do some problems. Since his math teacher did
not expect much from Tudor, he only needs to do some simple addition problems. However, simple for you and I may not be simple for Tudor, so please help him!
Input Specification
The first line will contain an integer \(N\) (\(1 \le N \le 100\,000\)), the number of addition problems Tudor needs to do. The next \(N\) lines will each contain two space-separated integers whose
absolute value is less than \(1\,000\,000\,000\), the two integers Tudor needs to add.
Output Specification
Output \(N\) lines of one integer each, the solutions to the addition problems in order.
Sample Input
-1 0
Sample Output
This is a LaTeX Demo
The diagram below is real LaTeX!
There are no comments at the moment. | {"url":"https://judge.tcirc.tw/problem/z001","timestamp":"2024-11-05T12:17:56Z","content_type":"text/html","content_length":"15636","record_id":"<urn:uuid:09f13c5e-8573-4bcf-ab77-b6b7a82bef48>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00813.warc.gz"} |
std::isgreaterequal(3) C++ Standard Libary std::isgreaterequal(3)
std::isgreaterequal - std::isgreaterequal
Defined in header <cmath>
bool isgreaterequal( float x, float y );
(since C++11)
bool isgreaterequal( double x, double y ); (until C++23)
bool isgreaterequal( long double x, long double y );
constexpr bool isgreaterequal( /*
floating-point-type */ x, (1) (since C++23)
floating-point-type */ y );
Additional overloads
Defined in header <cmath>
template< class Arithmetic1, class Arithmetic2 > (A) (since C++11)
bool isgreaterequal( Arithmetic1 x, Arithmetic2 y ); (constexpr since C++23)
1) Determines if the floating point number x is greater than or equal to the
floating-point number y, without setting floating-point exceptions.
The library provides overloads for all cv-unqualified floating-point types as the
type of the parameters x and y.
(since C++23)
A) Additional overloads are provided for all other combinations of arithmetic types.
x, y - floating-point or integer values
Return value¶
true if x >= y, false otherwise.
The built-in operator>= for floating-point numbers may raise FE_INVALID if one or
both of the arguments is NaN. This function is a "quiet" version of operator>=.
The additional overloads are not required to be provided exactly as (A). They only
need to be sufficient to ensure that for their first argument num1 and second
argument num2:
* If num1 or num2 has type long double, then std::isgreaterequal(num1,
num2) has the same effect as std::isgreaterequal(static_cast<long
static_cast<long double>(num2)).
* Otherwise, if num1 and/or num2 has type double or an integer type, then
std::isgreaterequal(num1, num2) has the same effect as (until
std::isgreaterequal(static_cast<double>(num1), C++23)
* Otherwise, if num1 or num2 has type float, then std::isgreaterequal(num1,
num2) has the same effect as
If num1 and num2 have arithmetic types, then std::isgreaterequal(num1, num2)
has the same effect as std::isgreaterequal(static_cast</*
common-floating-point-type */>(num1),
static_cast</* common-floating-point-type */>(num2)),
where /* common-floating-point-type */ is the floating-point type with the
greatest floating-point conversion rank and greatest floating-point (since
conversion subrank between the types of num1 and num2, arguments of integer C++23)
type are considered to have the same floating-point conversion rank as
If no such floating-point type with the greatest rank and subrank exists,
then overload resolution does not result in a usable candidate from the
overloads provided.
See also¶
greater_equal function object implementing x >= y
(class template)
islessequal checks if the first floating-point argument is less or equal than the
(C++11) second
C documentation for
2024.06.10 http://cppreference.com | {"url":"https://manpages.opensuse.org/Tumbleweed/stdman/std::isgreaterequal.3.en.html","timestamp":"2024-11-07T19:14:40Z","content_type":"text/html","content_length":"20398","record_id":"<urn:uuid:a4599fd1-e7a3-4e32-82c7-8a52195eb750>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00330.warc.gz"} |
Effortlessly Solve Complex Integer Equations with the Integers Calculator - Age calculator
Effortlessly Solve Complex Integer Equations with the Integers calculator
Effortlessly Solve Complex Integer Equations with the Integers calculator
Solving integer equations can be a daunting task, especially when the equations become more complex. Finding the solutions to these equations often requires a sharp mathematical mind and a lot of
effort. However, with the advent of technology, we now have access to numerous tools that simplifies the process of solving integer equations. One of such tools is the Integers calculator.
The Integers calculator is an online tool designed to help in the computation of complex integer equations. It simplifies the often arduous task of calculating integers by providing accurate and
speedy solutions to complex problems. Here, we would take a closer look at how the Integers calculator works, its benefits, and some frequently asked questions related to its usage.
How the Integers calculator Works
The Integers calculator is designed for solving complex integer equations by following basic mathematical procedures. It has an easy-to-use interface, whether you have a problem that involves
addition, subtraction, multiplication, or division of integers. You do not have to be an expert in mathematics to use the Integers calculator.
The first step to solving any integer equation using the Integers calculator is to input the equation into the calculator. This calculator has a simple and straightforward format that allows you to
input the parameters of your equation. After entering your equation, the calculator processes the input and provides a comprehensive solution to the equation. The results are instantaneous and
accurate, enabling users to remain confident that they have arrived at the correct answer.
Benefits of Using the Integers calculator
The advantages of using the Integers calculator are numerous. Here are some of the benefits users can enjoy;
– Saves Time: Solving complex integer equations can be a time-consuming task. However, with the Integers calculator, the time spent on solving complex integer equations can be significantly reduced.
This calculator provides instant results, significantly reducing the time required to solve complex integer equations.
– Accuracy: The Integers calculator guarantees accuracy in its computations. With its ability to handle complex integer equations, the calculator ensures that all calculations are correct. Using the
calculator guarantees that you arrive at the right answer.
– User-Friendly: The Integers calculator has an interface that is user-friendly and easy to understand. The calculator‘s simplicity ensures that even those with little background knowledge in
mathematics can use it with ease.
Frequently Asked Questions about using the Integers calculator
1. Can I use the Integers calculator on my smartphone?
Yes, the Integers calculator can be accessed on a smartphone and any other device with internet access. The calculator is web-based and can be accessed from any device with an internet connection.
2. Does the Integers calculator only solve integers?
Yes, the Integers calculator is designed specifically for solving integer equations. However, other calculators are ideal for solving non-integer equations.
3. Is there a fee required to access the Integers calculator?
No, the Integers calculator is available for free on the internet. There are no hidden charges or fees required for using it.
4. What is the maximum number of integers that the Integers calculator can handle?
There is no maximum number of integers that the Integers calculator can handle. The calculator can solve complex integer equations whether there are many numbers or a few numbers involved.
5. Can the Integers calculator solve equations that involve factors?
Yes, the Integers calculator is designed to handle equations that involve factors. The calculator can provide quick and accurate solutions to equations that involve factors.
Finding solutions to complex integer equations can be a daunting task, especially when done manually. However, with the availabilities of technology such as the Integers calculator, the time and
effort required to solve these equations can be reduced greatly. Using the Integers calculator saves time, provides accuracy, and is user-friendly. The calculator solves equations quickly and
accurately with just a click of a button. The Integers calculator is a valuable tool for anyone who desires to solve integer equations effortlessly.
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Do semiconductors follow Ohms law Why or why not ?
Semiconductors do not strictly follow Ohm’s Law because their resistance is not constant and changes with voltage, current, temperature, and other factors. Ohm’s Law states that the current through a
conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant. In semiconductors, this linear relationship does not always hold
true due to the material’s variable conductivity.
A semiconductor does not follow Ohm’s Law under all conditions.
While at very low electric fields, some semiconductors may exhibit approximately ohmic behavior, meaning the current and voltage relationship is nearly linear.
However, as the electric field increases, the relationship becomes nonlinear due to the complex behavior of charge carriers in the semiconductor material.
Semiconductors can obey Ohm’s Law at low electric fields where the relationship between current and voltage is nearly linear.
In this regime, the semiconductor behaves like a resistor with relatively constant resistance. However, this is only true over a limited range of conditions.
As the electric field increases, the conductivity of the semiconductor changes, and the linear relationship breaks down.
A semiconductor diode does not obey Ohm’s Law because its current-voltage relationship is nonlinear.
Diodes allow current to flow easily in one direction (forward bias) and block it in the opposite direction (reverse bias). This behavior results in an exponential increase in current with increasing
forward voltage, which deviates from the linear relationship described by Ohm’s Law.
Conductors typically obey Ohm’s Law within certain limits, provided the temperature and other physical conditions remain constant. In conductors, the resistance is usually constant, leading to a
linear relationship between current and voltage.
However, at extremely high currents or voltages, or at varying temperatures, even conductors may deviate from Ohm’s Law.
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October 2019 - Mr. M online
One of the hardest concepts for younger programmers to learn is what returning a value means, a concept which Alice represents by its notion of ‘functions’. And, in Alice, a great approach is to
require students to move an object based on the distance and to use a function to return the value of that distance and store it in a variable.
Every object in Alice has a set of proximity functions:
However, distance to, which most students will probably be most tempted to use is rarely what you want to use–especially when you want to move an object relative to another object using the
calculated distance.
Consider, the picture below. Let’s say we want to calculate the distance between the hammer and the moon’s surface. Well, if you used distance to to calculate the distance and move the hammer down,
you’ll see that the hammer actually falls down into the moon. That’ s because the distance is being calculated from the center of the hammer and the center of the whole planet!
Instead, use distance above!
In the picture below, if you want the astronaut to move to the spaceship based off of a distance, you should try the distance in front of
Before you jump into text coding with middle school coders , Alice is a powerful tool for teaching what “returning a value” means. | {"url":"https://mrmonline.org/2019/10/","timestamp":"2024-11-12T02:09:54Z","content_type":"text/html","content_length":"24907","record_id":"<urn:uuid:cfee74cb-1fac-4ca3-b012-ecfa7eb59f37>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00481.warc.gz"} |
A review on echocardiographic image spec
Speckle noise, Echocardiographic images, Non-linear filters, Diffusion technique, Wavelet domain, Fractional calculus, Non-local mean, Bilateral filters
Echocardiography is a technique used to get real time images of heart structure using ultrasound waves. Main advantages of echocardiography are low cost of operation, non-invasive, widely available
and it causes minimal discomfort to patient. To assess the heart functionality, it is required to obtain a perfect ultrasound image of heart. Ultrasound images are mainly affected by speckle noise
than additive noise. So, despeckling of ultrasound images improves the quality of images and detects the boundaries more prominently.
A wide research has been done in despeckling of ultrasound images in last couple of decades. Initially the study was focusing on linear filtering techniques. But, as speckle noise is multiplicative
noise, linear filtering techniques were not able to give good quality results. It removes fine details so there are possibilities of loss of important information.
To perform the filtering, while preserving the edges and information in the ultrasound images, nonlinear filters, diffusion filters and wavelet domain filters are more suitable. In this paper we have
done review of all these different techniques.
The organization of this paper is as follows. Section I describes nonlinear filters in details. Section II describes the diffusion method and advanced techniques in diffusion method. The widely used
wavelet denoising method is described in section III and different thresholding techniques are described in section IV. Section V describes fractional calculus filters. The non-local mean filters and
bilateral filters are described in sections VI and VII respectively. The section VIII is about results and discussion and section IX concludes this paper.
Non Linear Filters
Median filter
Median filtering is a nonlinear process useful in reducing noise and effective in preserving edges in an image by reducing random noise. It works by convolving window pixel by pixel through the
image, replacing noisy value with the median value of neighboring pixels [1].
Lee filter
This filter is developed by Jong in 1981 [2,3]. The Lee filter is based on Minimum Mean Square Error (MMSE) criterion. It is better in edge preservation. The local statistics filter (Lee filter) is
based on multiplicative speckle model and additive noise is considered negligible, which gives
Where, I (x, y) is the Input image, R (x, y) represents the signal and n (x, y) the speckle noise. Mathematical representation for Lee filter is given in Equation (2).
where W (x, y) is a weighting function given by
Where C[n] and C[i] are the coefficients of variation of the image and noise.
Kuan filter
The form of Kuan filter [4] is same as Lee filter but the W is for Kuan filter is given by Equation 4,
Frost filter
The Frost filter is invented in 1982 [5]. It is a linear, convolutional filter used to remove the multiplicative noise from images. Frost filter works on the basis of coefficient of variation. It is
the ratio of local standard deviation to the local mean of the noised image. It gives the noise free image by convolving the input image with the spatially varying kernel. The kernel of window size n
× n is moved through the image then the centre pixel value is replaced by weighted sum of values of kernel. The weighing function W (x, y) decreases as we move away from interested pixel and
increases with variance. It assumes multiplicative noise. Frost filter follows formula given by Equation 5.
R (x, y)=I (x, y) × W (x, y) → (5)
K[0]-Normalizing constant, K-Controls damping rate.
The enhanced Frost and Lee filter
This filter is proposed by Lopes in 1990 [6]. It works on the basis of the threshold value. When the local coefficient of variation is below a lower threshold, averaging is done. If local coefficient
of variation is above the higher threshold, it works strictly as all pass filters. And averaging and identity operation is done if the local variance is in between both thresholds. The Equation 7 for
enhanced Frost and Lee filter is given below.
Wiener filter
Wiener filter was proposed by Norbert Wiener during the 1949. It is also known as Least Mean Square Filter. It can restore images even if they are corrupted by noise. It reduces noise from image by
comparing desired noiseless image. It works on the principle of computation of local image variance. So if local variance of the image is large the de-noising is done poorly and when local variance
is small we can get more accurate image. The drawback is it requires more computational time [7].
Kalman filter
The Kalman filter was first described by Kalman in 1960 and Kalman and Bucy in 1961. A2D Kalman filter has been implemented on a causal prediction window. In this filter the image is represented by a
Markov field which satisfies the causal Autoregressive (AR) model. Equation 8 shows Kalman filter.
Where μ (x, y) is a noise sequence which follows the Markov process and a[p,q] are the AR model’s reflection coefficients [8]. Table 1 shows the comparison of all non-linear filters.
Filter Filtering equation Weighing function
Median R (x, y)=Median (I (x,y))
Lee W (x,y)=1-Cn2/Ci^2 (x,y)
Kuan W(x,y)=(1-Cn^2/Ci^2 (x,y))/(1+Cn^2)
Frost W (x,y)=K0 exp(-KCi^2 √(x^2+y^2))
Enhanced Lee
Enhanced Frost
Table 1. Comparison of nonlinear filters.
Diffusion Filter
Anisotropic diffusion filter (AD)
Anisotropic diffusion filtering is method invented by Perona and Malik [9] in 1990. It is used for smoothing the image while preserving the edges. In this method, Partial Differential Equation (PDE)
has been used for keeping track of homogeneous region and region containing edges in images.
The nonlinear PDE for smoothing image introduced by Perona and Malik is given in Equation 9.
Where, I (x, y, t) is the input image for diffusion, t is time dimension representing the progress of diffusion, I[0] is the original image. and ∙() are the gradient and divergence operators, and | |
represents magnitude. I[σ] is a smoothed version of I. The term c (∙) represents the level of diffusion for each image position.
Perona and Malik [9] suggested two different functions for diffusion coefficients given in Equation 11.
Where, k controls the level of diffusion between edges and homogeneous region in input images. To avoid over or under smoothing optimum value of c (x) should be chosen. If x>>k, Then c (x) for
all-pass filter is used, whereas if x<<k, then c (x) for isotropic diffusion (Gaussian filtering) is used. The drawback of this technique is the smoothing of images with speckle noise is not
satisfactory. This filtering technique enhances speckle instead of smoothing it.
Speckle reducing anisotropic diffusion (SRAD)
Yu and Acton [10] proposed new anisotropic diffusion model in 2002 to smooth speckle images. Here the diffusion PDE is used but c (∙) is a function of q i.e. the instantaneous coefficient of
variation. The output image I (x, y, t) is computed using the following differential Equation 12.
Where the diffusion coefficient c (q) is written as,
Where, q[0] is the scale function which controls the level of smoothing. In homogeneous regions, the function gives low values, and at edges or high contrast regions takes higher values.
Wavelet Denoising Technique
The wavelet transform based filtering techniques use a thresholding operator for signal denoising. These methods involve three steps: 1) the decomposition of the noisy image using forward wavelet
transform; 2) the filtering of the wavelet coefficients by means of a thresholding processor; and 3) the reconstruction of image by the inverse wavelet transformation with the filtered coefficients.
The process of choosing a threshold value is a crucial task in the wavelet denoising filtering as the threshold value separates the important coefficients which are useful to reconstruct the image
signal and less significant coefficients corresponding to the noise. Generally, a low threshold value preserves the details but does not reduce the noise significantly; so in this case, both the
denoised and the input image with noise are very close. On the other hand, a large threshold value reduces the noise but destroys many detail coefficients with noise. To overcome these drawbacks,
different thresholding rules were proposed in the literature; the most commonly used of them are summarized below.
VishuShrink or universal threshold
This technique was invented by Donoho and Johnstone [11,12] and applies the universal threshold; it consists of the use of a universal threshold defined by the following Equation 15.
Where N is the image size and σ[n] is the noise standard deviation. An estimate of the noise level σ[n] is based on the Equation 16, median absolute deviation given by [13]
Where n and m are pixel indexes of HH1 that represents the diagonal sub-band of first level wavelet decomposition of the image. The drawback of this threshold is it removes too many coefficients that
produce an excessively smoothed image because of high value of T[u].
This method is a combination of the universal threshold and the Stein’s Unbiased Risk Estimator (SURE) technique [14]. It computes a separate threshold for each subband and it is suited for images
with sharp discontinuities; it minimizes the mean square error. This method is a level dependent threshold. In this case, the soft threshold is defined as Equation 17.
Where, T denotes the value that minimizes the SURE.
This method [15] is best suited for images inculcate with Gaussian noise. For each detailed coefficients of wavelet transformed image, a threshold is estimated that minimizes the Bayesian risk. This
method is better than sure shrink when compared with respect to mean square error. Retaining sharp feature is its additive advantage making it more suitable and better. The threshold is estimated by
using following Equation 18.
Where σ[x] is the image standard deviation evaluated in each wavelet sub-band.
For the thresholding process there are two different methods which are normally used, those are described below [13].
Hard thresholding
Hard thresholding method either keeps the coefficients or kills them as shown in Equation 19, without obtaining any average or shrinked value. In this method, the coefficients are compared to an
absolute threshold value and any value lower than threshold value are set to zero. It provides an advantage of edge preservation which makes it suitable in wavelet decomposition.
ht (x)=0 |x|<T
x |x|>T → (19)
Soft thresholding
In soft thresholding, the coefficients above the defined threshold value are shrinked rather then killed as in Equation 20. There is a smooth transition between obtained values and deleted values. It
helps in avoiding frayed edges of the image.
st (x)=0 |x|<T
sign (x)∙(|x|-T) |x|>T → (20)
Wavelet Based New Techniques
Noval bayesian multiscale filter
Two main denoising techniques used are the thresholding technique and the Bayesian estimation shrinkage technique. Bayesian estimation technique [16] is proposed in 2001 by Achim et al. In this, for
the noise-free image, it is required to consider an a priori distribution p (x) of the wavelet coefficients. If we know the likelihood function p (y/x), we can calculate the wavelet coefficients of
the noise-free image by the following approaches.
1. Maximum a posteriori (MAP) estimator
2. MMSE estimator
In the general case the Bayesian processor can be described as in Equation 21.
Where σ[s] ^2 is the Gaussian signal variance. In general, the thresholding method is the discrete function which respect to threshold. But the Bayesian estimator follows a continuous shrinking
MRF-Based spatially adaptive Bayesian wavelet denoising
Markov Random Field (MRF) model is a promising tool for modelling images. The Bayesian estimator, combined with MRFs can generate a framework for image modeling and processing. In this study [17],
Xie et al. have used an MRF to model the intrascale spatial dependence between wavelet coefficients in each individual subband. The threshold value for MRF is given by Equation 22.
The proposed denoising algorithm proceeds as follows.
• Using the Bayesian MMSE technique estimate the shrinkage function.
• Using MAP form an initial binary mask corresponding to the hidden state configuration.
• Redefine the prior using an MRF, and then refine the binary mask by maximizing.
• Modify the shrinkage function based on the optimal binary mask.
A versatile wavelet domain noise filtration
Pizurica et al. [18] has proposed a new method in 2003, which adapts itself to various types of image noise. In this technique a single parameter is used to balance the preservation of
(expertdependent) relevant details against the degree of noise reduction. The main idea of this method is the estimation of wavelet coefficients which represent signal and with noise, based on the
assumption of [16] that useful wavelet coefficients persist well across the scales of decomposition described in Equation 23.
Nonlinear multiscale wavelet diffusion (NMWD) method
Yong [19] have developed this method in 2006 to utilize the two frequently used techniques: the wavelet denoisng technique and the iterative nonlinear diffusion method. Speckle is suppressed by
implementing the diffusion process on the wavelet coefficients. With a combination of diffusion threshold strategy, the proposed method can reduce the speckle noise effectively and do
Wavelet diffusion is implemented by three steps:
• The noisy image is decomposed into the coarse scale approximation and detail images by 2-D MZ-DWT.
• Wavelet coefficients are regularized by using threshold. The threshold value is given by Equation 24.
• The denoised image is reconstructed by taking the inverse MZ-DWT.
This is an iterative method, and the steps above are repeated to achieve the desired level of filtering.
Spatially adaptive filter by Bhuiyan
SNIG-Shrink method is proposed by Bhuiyan et al. [20] in 2009. In the proposed method, following steps are followed.
• The given ultrasound image is first log transformed.
• The resulting image is decomposed using wavelet transform.
• The corresponding wavelet coefficients processed by using the proposed Bayesian MAP estimator.
• The resulting output coefficients are then inversely transformed.
• Then exponential operation is performed to get the despeckled ultrasound image.
The proposed method is called SNIG-shrink, because it carries out a soft-thresholding operation with a threshold obtained from a Bayesian MAP estimator using a Symmetric Normal Inverse Gaussian
(SNIG) PDF. The Equation 25 is for SNIGPDF.
Where C is a scaling factor and B is given by SNIG PDF
The DWT is not shift-invariant, which leads to pseudo-Gibbs phenomena such as undershoots and overshoots at the locations of sharp signal transitions. These drawbacks can be overcome by implementing
the denoising method using transforms such as cycle-spinning, Stationary Wavelet Transform (SWT), and dual-tree complex wavelet transform.
A suitable threshold method by Andria
The VisuShrink soft thresholding technique gives a highly impulsive distribution. Because the large value of the universal threshold sets too many coefficients to zero. To improve this drawback, a
new thresholding operator was proposed [21]. The aim of the proposed method is to create an alternative function, which will be able to reduce gradually the coefficients in the zero zones. For this
aim the following thresholding operator based on exponential function was defined as Equation 26.
Where nl is a real parameter which finds fall degree of exponential function for l decomposition level, while kl factor provides a modified version of l-level universal threshold.
Fractional Integral Filters
Saadia et al. [22] proposed one more denoising filter using fractional calculus for echocardiographic images. Initially the image is distinguished in three regions homogeneous, detailed and edge
based on gradient of intensity value. This region classification is achieved using eigenvalues of Hassian matrix. Hassian matrix is calculated for each pixel in the image. Threshold value for this
classification of image is calculated using mean of eigenvalues of all pixels. After this classification image will be denoised using fractional calculus. Authors have used Grunwald-Letnikov (G-L)
definition for fractional calculus. Depending upon the classified region, coefficients and order of fractional convolute mask are evaluated.
Saadia et al. [23] proposed method combining techniques of fractional integral filter and fuzzy logic to overcome the limitations of use of only fractional integral filters for speckle noise
reduction in echocardiographic images. They have used proposed filter in two steps. First it utilises fuzzy logic to assign weights to the pixels in convoluting window in the image depending upon the
differences in neighbouring pixels. This way, pixels are distinguished as either of same intensity regions or different (presence of edge). Then weighted mean of weights is calculated and assigned to
pixel. Secondly speckle noise reduction of outcome from first stage is carried out using fractional integration filter.
Non-Local Mean Filters
The Gaussian convolution method preserves only plane zones but the detail structures are blurred or removed. On the other hand the anisotropic filter restores edges but plane zones are getting
affected by noise. Antoni et al. [24] have proposed Non Local means filter which is a combination of previous methods. Non-local filters are designed to preserve edges and to remove noise from flat
zones as well. This method takes the best from both the algorithms.
Jose et al. [25] have proposed a method which takes into consideration the Rician nature of the noise and spatially varying pattern of noise. They have invented Adaptive Rician Non Local Means (RNLM)
Filter with Wavelet Mixing. This filter is similar to Adaptive Non Local Mean filter but with the corrected estimation of the local standard deviation of the noise σ.
In NLM algorithm, to blur sharp edges, proper weighting of the central pixel is used. To avoid overweighting, the central pixel has assigned with the maximum weight. According to [26], because of
this method, the small-high contrast particles become blurred during de-noising process. To get proper details of high-contrast particles, Zang has proposed a new method called Rician NLM using
Combined Patch and Pixel (RNLM-CPP).
Bilateral Filters
Ming et al. [27] states that a local neighbourhood is considered to find weighted sum of the pixels in bilateral filters. The weight is calculated from spatial distance and the intensity distance.
This way we can preserve edges from blurring. In this study author has implemented bilateral filter in multiresolution analysis. It shows that we can get better results if we apply bilateral filter
to approximation coefficients and wavelet thresholding to detail coefficients of an image.
The paper [28] presents a method based on bilinear filters with adaptive parameters. This method is applied to remove impulse noise and Gaussian noise simultaneously. Bilinear filter is used to
remove impulse noise and Gaussian noise. To preserve edges and to make it adaptive, an Improved Artificial Bee Colony (IABC) algorithm is proposed. This algorithm finds the correct direction for
search process.
Results and Discussion
The results of above discussed methods have compared in this section.
In Anisotropic Diffusion (AD) filter [9] paper, comparison of linear filters and anisotropic diffusion filters has been done. It concludes that, anisotropic diffusion is a non-linear process so it
removes the trade-off between accuracy in localization and detectability; which is a main drawback of linear filters. The algorithm of AD filters is parallel so it is cheap to run on arrays of
parallel processors.
Yu et al. [10] have done comparison of SRAD with anisotropic diffusion and the basic filters like enhanced Lee and the enhanced Frost filter. It is done in terms of Figure of Merit (FOM). The AD
filter gives FOM of 0.4714 which is better than enhanced Lee and Frost filter. The SRAD method gives FOM of 0.7257.
In [16], the Bayesian denoising result is compared with median filter, homomorphic wiener filter and soft and hard thresholding. The result shows that, Bayesian denoising gives lower mean square
error (12.7398) and higher β value (0.4559). So, this technique performs better in terms of edge preservation.
MRF-Based spatially adaptive Bayesian wavelet denoising method [17] is compared with Bayes threshold, Bayes MMSE method and refined LEE filter. The Signal to Noise Ratio is calculated for all
methods. The result shows that, this method gives best SNR i.e. 2.59 among all.
Pizurica et al. [18] have done comparison between homomorphic wiener and proposed filter. The SNR of the spatially adaptive Wiener filter is 10.1 dB and; the SNR of the proposed method is SNR=12.9
The performance of NMWD algorithm [19] is compared with, the Speckle Reducing Anisotropic Diffusion (SRAD) technique, and the wavelet Generalized Likelihood ratio filtering (GenLik) method (GenLik).
The results conclude that NMWD method gives better performance in terms of FOM i.e. 0.9717 which is better than SRAD (FOM=09533) and GenLik (FOM=11.88)
Gregorio et al. [20] have compared results of SNIG-shrinkI and SNIG-shrinkII with GenLik, Bayes-shrink and homomorphic Wiener filter. The result is compared in terms of Structural Similarity (SSIM).
In simulation results, SNIG-shrinkI (0.8777) and SNIG-shrinkII (0.8937) gives better results.
In [21] a comparison is done with the results of Bayes Shrink method and the polynomial thresholding proposed by Smith. The proposed exponential thresholding is better in terms of β metric. But, in
terms of PSNR index, the performance of proposed method and the BayesShrink is very much similar.
The method in [22] is compared with other benchmark methods like Lee, Kuan, wavelet etc. for denoising standard reference images. Proposed method has shown higher Peak Signal to Noise Ratio (PSNR)
and Structural Similarity Index Measure (SSIM) outperforming other methods.
Ayesha and Adnan [23] have used number of parameters to compare the results with other benchmark methods. Speckle Suppression Index (SSI) for echocardiographic image of their proposed method is
0.9647 for 4 chamber view and 0.9679 for short axis view compared to 0.97-0.98 by other benchmark methods.
In [24], the results are compared in terms of mean square error. The non-local mean filters are compared with six different filters. But, the NLM filters give the least MSE value for all the images.
The authors of [25] have compared their results with non-local mean filters, adaptive non-local mean filter and Rician nonlocal means filter. The results are compared in terms of PSNR. The results
show that this method performs best for nonstationary noise as compare to other methods. It is also observed that, the proposed method performs much better than non-adaptive filters and behave much
similar for adaptive filter.
Zang et al. [26] have measured results in terms of PSNR (Peak Signal to Noise Ratio) and SSIM (Structural Similarity). It was calculated to measure performance of RNLM and RNLM-CPP. With 1% of noise
level, RNLM gives 33.71Peak Signal to Noise Ratio. While, RNLM-CPP gives 46.12 PSNR. The SSIM values for RNLM and RNLM-CPP are 0.9895 and 0.9989.
The results of method discussed in [27] are measured in terms PSNR. The results conclude that, the new method is 0.8 dB better than bilateral filter and 1.1 dB better than Bayes shrink. The SURE
shrink gives slightly better results than the proposed method.
Yinxue et al. [28] have done performance measurement in terms of MSE, PSNR and SSIM. The proposed method is compared with alternative filters. The results shows that, BLSGSM and SURE gives better
performance for images with Gaussian noise level<20. NLM and proposed method gives almost same results for many test images. But, the proposed method outperforms with the high level mixed noise.
We have discussed application of ultrasound image despeckling techniques in the area of echocardiography. Various influential researches in speckle reduction are presented. Traditionally used
nonlinear filters with their weighing functions are discussed. Out of that the enhanced Lee and Frost filters give good results. Then we have seen that in diffusion filters, Speckle Reducing
Anisotropic Diffusion (SRAD) filter is an advanced version of anisotropic diffusion invented by Perona and Malik.
We have explained the most widely used despeckling techniques. Out of which a filters using wavelet transform give the better results amongst all. This paper has given brief explanation of seven
wavelet based algorithms with six different thresholding methods.
Nonlinear Multiscale Wavelet Diffusion (NMWD) method is a method which combines advantages of wavelet technique and diffusion method. According to literature [24] this is widely used method for
speckle reduction. This technique can be used for image segmentation without any pre-processing.
Use of fractional calculus is discussed for echocardiographic image noise removal. This method combined with adaptive filters and fuzzy logic can greatly improve effectiveness of noise filter and its
edge retention capabilities.
As compare to diffusion filters, non-local mean filters preserves edges. The RNLM-CPP algorithm preserves small high-contrast particle details too.
The bilateral filter with wavelet thresholding gives better performance in terms of PSNR. The adaptive bilateral filter can optimize the parameters and can remove noise in smooth region as well as
can preserve edge details also. | {"url":"https://www.biomedres.info/biomedical-research/a-review-on-echocardiographic-image-speckle-reduction-filters-10510.html","timestamp":"2024-11-13T15:47:23Z","content_type":"text/html","content_length":"60886","record_id":"<urn:uuid:73b3de24-0ef7-4f41-b9c5-029c18bf7fac>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00395.warc.gz"} |
招待講演1 [10月30日(月) 17:30 – 18:30]
Taco Cohen（Qualcomm AI Research）※オンライン講演
Geometric Algebra Transformers: A General-Purpose Architecture for Geometric Data
Problems involving geometric data arise in a variety of fields, including computer vision, robotics, chemistry, and physics. Such data can take numerous forms, such as points, direction vectors,
planes, or transformations, but to date there is no single neural network architecture that can be applied to such a wide variety of geometric types while respecting their symmetries. In this work we
introduce the Geometric Algebra Transformer (GATr), a general-purpose architecture for geometric data. GATr represents inputs, outputs, and hidden states in the projective geometric algebra, which
offers an efficient 16-dimensional vector space representation of common geometric objects as well as operators acting on them. GATr is equivariant with respect to E(3): the symmetry group of 3D
Euclidean space. As a transformer, GATr is scalable, expressive, and versatile. In various geometric problems, GATr shows strong improvements over non-geometric baselines.
招待講演2 [10月31日(火) 09:30 – 10:30]
Graham Neubig（Carnegie Mellon University）※オンライン講演
Prompt2Model: Generating Deployable Models from Natural Language Instructions
Large language models (LLMs) enable system builders today to create competent NLP systems through prompting, where they only need to describe the task in natural language and provide a few examples.
However, in other ways, LLMs are a step backward from traditional special-purpose NLP models; they require extensive computational resources for deployment and can be gated behind APIs. In this talk,
I will discuss Prompt2Model, a general-purpose method that takes a natural language task description like the prompts provided to LLMs, and uses it to train a special-purpose model that is conducive
to deployment. This is done through a multi-step process of retrieval of existing datasets and pretrained models, dataset generation using LLMs, and supervised fine-tuning on these retrieved and
generated datasets. I will describe the details of this process, as well as the larger implications for automating machine learning workflows.
招待講演3 [11月1日(水) 13:00 – 14:00]
Emtiyaz Khan（理化学研究所）
How to build machines that adapt quickly
Humans and animals have a natural ability to autonomously learn and quickly adapt to their surroundings. How can we design machines that do the same? In this talk, I will present Bayesian principles
to bridge such gaps between humans and machines. I will discuss (1) the Bayesian learning rule to unify algorithms; (2) sensitivity analysis to understand and improve memory of the algorithms; and
(3) new priors to enable quick adaptation. These ideas are unified in a new learning principle called the Bayes-duality principle, yielding new mechanisms for knowledge transfer in learning machines.
1. The Bayesian Learning Rule, (JMLR) M.E. Khan, H. Rue (arXiv)
2. The Memory Perturbation Equation: Understanding Model’s Sensitivity to Data, (NeurIPS 2023) P. Nickl, L. Xu, D. Tailor, T. Möllenhoff, M.E. Khan
3. Knowledge-Adaptation Priors, (NeurIPS 2021) M.E. Khan, S. Swaroop (arXiv)
4. Continual Deep Learning by Functional Regularisation of Memorable Past (NeurIPS 2020) P. Pan*, S. Swaroop*, A. Immer, R. Eschenhagen, R. E. Turner, M.E. Khan (arXiv)
招待講演4 [11月1日(水) 16:40 – 17:40]
Petar Veličković（Google DeepMind & University of Cambridge）※オンライン講演
Decoupling the Input Graph and the Computational Graph: The Most Important Unsolved Problem in Graph Representation Learning
When deploying graph neural networks, we often make a seemingly innocent assumption: that the input graph we are given is the ground-truth. However, as my talk will unpack, this is often not the
case: even when the graphs are perfectly correct, they may be severely suboptimal for completing the task at hand. This will introduce us to a rich and vibrant area of graph rewiring, which is
experiencing a renaissance in recent times. I will discuss some of the most representative works, including two of our own contributions (https://arxiv.org/abs/2210.02997, https://arxiv.org/abs/
2306.03589), one of which won the Best Paper Award at the Graph Learning Frontiers Workshop at NeurIPS’22. | {"url":"https://ibisml.org/ibis2023/invited/","timestamp":"2024-11-14T05:43:16Z","content_type":"text/html","content_length":"31486","record_id":"<urn:uuid:b38114a4-1944-4411-97f9-48a5dd371faa>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00010.warc.gz"} |
Explainability I: local post-hoc explanations - RBC Borealis
Machine learning systems are increasingly deployed to make decisions that have significant real-world impact. For example, they are used for credit scoring, to produce insurance quotes, and in
various healthcare applications. Consequently, it’s vital that these systems are trustworthy. One aspect of trustworthiness is explainability. Ideally, a non-specialist should be able to understand
the model itself, or at the very least why the model made a particular decision.
Unfortunately, as machine learning models have become more powerful, they have also become larger and more inscrutable (figure 1). At the time of writing, the largest deep learning models have
trillions of parameters. Although their performance is remarkable, it’s clearly not possible to understand how they work by just examining the parameters. This trend has led to the field of
explainable AI or XAI for short.
Figure 1. Explainability vs. performance. There has been a trend for models to get less explainable as they get more powerful. This is partly just because of the increasing number of model parameters
(potentially trillions for large models in the top-left corner). The decision tree is a rare example of a non-linear model which is both easy to understand, and provides reasonable performance. It is
unknown whether it is possible to build models in the desirable top-right corner of this chart. Adapted from AAAI XAI Tutorial 2020.
Explainable AI methods are useful for three different groups of stakeholders. For machine learning scientists, they are useful for debugging the model itself and for developing new insights as to
what information can be exploited to improve performance. For business owners, they help manage business risk related to AI systems. They can provide insight into whether the decisions can be trusted
and how to answer customer complaints. Finally, for the customers themselves, XAI methods can reassure them that a decision was rational and fair. Indeed, in some jurisdictions, a customer has the
right to demand an explanation (e.g., the under the GDPR regulations in Europe).
Two considerations
There are two related and interesting philosophical points. Firstly, it is well known that humans make systematically biased decisions, use heuristics, and cannot explain their reasoning processes
reliably. Hence, we might ask whether it is fair to demand that our machine learning systems are explicable. This an interesting question, because at the current time, it is not even known whether it
is possible to find models that are explicable to humans and still have good performance. However, regardless of the flaws in human decision making, explainable AI is a worthy goal even if we do not
currently know to what extent it is possible.
Secondly, Rudin 2019 has argued that the whole notion of model explainability is flawed, and that explanations of a model must be wrong. If they were completely faithful to the original model, then
we would not need the original model, just the explanation. This is true, but even if it is not possible to explain how the whole model works, this does not mean that we cannot get insight into how a
particular decision is made. There may be an extremely large number of local explanations pertaining to different inputs, each of which can be understood individually, even if we cannot collectively
understand the whole model. However, as we shall see, there is also a strand of XAI that attempts to build novel transparent models which have high performance, but that are inherently easy to
What makes a good explanation?
In this section, we consider the properties that a good explanation should have (or alternatively that a transparent model should have). First, it should be easily understandable to a non-technical
user. Models or explanations based on linear models have this property; it’s clear that the output increases when certain inputs increase and decreases when other inputs increase, and the magnitude
of these changes differs according to the regression coefficient. Decision trees are also easy to understand as they can be described as a series of rules.
Second, the explanation should be succinct. Models or explanations based on small decision trees are reasonably comprehensible, but become less so as the size of the tree grows. Third, the
explanation should be accurate and have high-fidelity. In other words, it should predict unseen data correctly, and in the same way as the original model. Finally, an explanation should be complete;
it should be applicable in all situations.
Adherence to these criteria is extremely important. Indeed, Gilpin et al. (2019) argue that it is fundamentally unethical to present a simplified description of a complex system to increase trust, if
the limits of that approximation are not apparent.
Taxonomy of XAI approaches
There are a wide range of XAI approaches that are designed for different scenarios (figure 2). The most common case is that we have already trained a complex model like a deep neural network or
random forest and do not necessarily even have access to its internal structure. In this context, we refer to this as a black box model, and we seek insight into how it makes decisions. Explanations
of existing models are referred to as post-hoc explanations.
Figure 2. Taxonomy of XAI methods. If we do not already have a model that we need to explain, we can develop a model that is inherently interpretable. If we already have a model, then we must use a
post-hoc method. One approach is to distill this into a simpler and more interpretable model. However, if we only use this for explanations, then the explanations are unreliable to the extent that
the results differ. If we replace the original model entirely, then we may sacrifice performance. If we decide to work with just the existing model, then there are two main families of methods. Local
models explain a single decision at a time, whereas global models attempt to explain the entire model behaviour. See also Singh (2019).
There are three main types of post-hoc explanation. First, we can distill the black box model into a surrogate machine learning model that is intrinsically interpretable. We could then either
substitute this model (probably sacrificing some performance) or use it to interpret the original model (which will differ, making the explanations potentially unreliable). Second, we can try to
summarize, interrogate, or explain the full model in other ways (i.e., a global interpretation). Third, we can attempt to explain a single prediction (a local interpretation).
If we do not already have an existing model, then we have the option of training an intrinsically interpretable model from scratch. This might be standard ML model that is easy to understand like a
linear model or tree. However, this may have the disadvantage of sacrificing the performance of more modern complex techniques. Consequently, recent work has investigated training models with high
performance but which are still inherently interpretable.
In part I of this blog, we consider local post-hoc methods for analyzing black box models. In part II, we consider methods approximate the entire model with a surrogate, and models that provide
global explanations at the level of the dataset. We also consider families of model that are designed to be inherently interpretable.
Local post-hoc explanations
Local post-hoc explanations sidestep the problem of trying to convey the entire in an interpretable way by focusing on just explaining a particular decision. The original model consists of a very
complex function mapping inputs to outputs. Many local post-hoc models attempt to just describe the part of the function that is close to the point under consideration rather than the entire
In this section we’ll describe the most common local post-hoc methods for a simple model (figure 3) with two inputs and one output. Obviously, we don’t really need to ‘explain’ this model, since the
whole function relating inputs to outputs can easily be visualized. However, it will suffice to illustrate methods that can explain much more complex models.
Figure 3. Model used to describe XAI techniques. The model has two inputs $x_{1}$ and $x_{2}$ and returns a probability $Pr(y=1)$ that indicates the likelihood that the input belongs to class 1
(brighter means higher probability). The red points are positive training points and the green points are negative training points. The green line represents the decision boundary. Obviously, we do
not need to “explain” this model as we can visualize it easily. Nonetheless, it can be used to elucidate local post-hoc XAI methods.
Individual conditional expectation (ICE)
An individual conditional expectation or ICE plot (Goldstein et al. 2015) takes an individual prediction and shows how it would change as we vary a single feature. Essentially, it answers the
question: what if feature $x_{j}$ had taken another value? In terms of the input output function, it takes a slice through a single dimension for a given data point (figure 4).
Figure 4. Individual conditional expectation. a) We wish to understand why the purple point is assigned to the negative class. We can do this by considering what would happen if we changed either
feature $x_{1}$ (cyan line) or $x_{2}$ (blue line). b) The effect of changing feature $x_{1}$. We see that the point might have been classified as positive (so $Pr(y)>0.5)$ if $x_{1}$ had a higher
value. c) The effect of changing feature $x_{2}$. We see that the point might have been classified as positive if $x_{2}$ had a lower value.
ICE plots have the obvious disadvantage that they can only interrogate a single feature at a time and they do not take into account the relationships between features. It’s also possible that some
combinations of input features never occur in real-life, yet ICE plots display these and do not make this clear.
Counterfactual explanations
ICE plots create insight into the behaviour of the model by visualizing the effect of manipulating one of the model inputs by an arbitrary amount. In contrast, counterfactual explanations manipulate
multiple features, but only consider the behaviour within the vicinity of the particular input that we wish to explain.
Counterfactual explanations are usually used in the context of classification. From the point of view of the end user, they pose the question “what changes would I have to make for the model decision
to be different?”. An oft-cited scenario is that of someone whose loan application has been declined by a machine learning model whose inputs include income, debt levels, credit history, savings and
number of credit cards. A counterfactual explanation might indicate that the loan decision would have been different if the applicant had three fewer credit cards and an extra $5000 annual income.
From an algorithmic point of view, counterfactual explanations are input data points $\mathbf{x}^{*}$ that trade off (i) the distance ${dist1}[{f}[\mathbf{x}^*], y^*]$ a between the actual function
output ${f}[\mathbf{x}^*]$ and the desired output $y^*$ and (ii) the proximity ${dist2}\left[\mathbf{x}, \mathbf{x}^*\right]$ to the original point $\mathbf{x}$ (figure 5). To find these points we
define a cost function of the form:
$$\hat{\mathbf{x}}^*,\hat{\lambda} = \mathop{\rm argmax}_{\lambda}\left[\mathop{\rm argmin}_{\mathbf{x}^*}\left[{dist1}\left[{f}[\mathbf{x}*], y^*\right] + \lambda\cdot {dist2}\left[\mathbf{x}, \
mathbf{x}^*\right]\right]\right] \tag{1}$$
where the positive constant $\lambda$ controls the relative contribution of the two terms. We want $\lambda$ to be as large as possible so that the counterfactual examples is as close as possible to
the original example.
Figure 5. Counterfactual explanations. a) We want to explain a data point (purple point 1) which was classified negatively. One way to do this is to ask, how we would have to change the input so that
it is classified positively. In practice, this means finding and returning the closest point on the decision boundary (cyan point 2). In a real-life situation, a customer might be able to take
remedial action to move the inputs to this position. b) This remedial action may be impractical if there are many input features, and so usually we seek sparse counterfactual examples where we have
only changed a few features (here just feature $x_{1}$). c) One problem with counterfactual examples is that there be many potential ways to modify the input (brown point 1) to change the
classification (points 2, 3 and 4). It’s not clear which one should be presented to the end user.
This formulation was introduced by Wachter et al. (2017) who used the squared distance for ${dist1}[\bullet]$ and the Manhattan distance weighted with the inverse median absolute deviation of each
feature for ${dist2}[\bullet]$. In practice, they solved this optimization problem by finding a solution for the counterfactual point for a range of different values of $\lambda$ and then choosing
the largest $\lambda$ for which the proximity to the desired point was acceptable. This method is only practical if the model output is a continuous function of the input and we can calculate the
derivatives of the model output with respect to the input efficiently.
The above formulation has two main drawbacks that were addressed by Dandl et al. (2018). First, we would ideally like counterfactual examples where only a small number of the input features have
changed (figure 5); this is both easier to understand and more practical in terms of taking remedial action. To this end, we can modify the function ${dist2}$ to encourage sparsity in the changes.
Second, we want to ensure that the counterfactual example falls in a plausible region of input space. Returning to the loan example, the decision could be partly made based on two different credit
ratings, but these might be highly correlated. Consequently, suggesting a change where one remains low, but the other increases is not helpful as this is not realizable in practice. To this end,
Dandl et al. (2018) proposed adding a second term that penalizes the counterfactual example if it is far from the training points. A further important modification was made by McGragh et al. (2018)
who allow the user to specify a weight for each input dimension that effectively penalizes changes more or less. This can be used to discourage finding counter-factual explanations where the change
to the input are not realisable. For example, proposing a change in an input variable that encodes an individuals age is not helpful as this cannot be changed.
A drawback of counterfactual explanations is that there may be many possible ways to modify the model output by perturbing the features locally and it’s not clear which is most useful. Moreover,
since most approaches are based on optimization of a non-linear function, it’s not possible to ensure that we have find the local minimum; even if we fail to find any counterfactual examples within a
predefined distance from the original, this does not mean that they do not exist.
Local interpretable model-agnostic explanations or LIME (Ribeiro et al. 2016) approximate the main machine learning model locally around a given input using a simpler model that is easier to
understand. In some cases, we may trade off the quality of the local explanation against its complexity.
Figure 6 illustrates the LIME algorithm with a linear model. Samples are drawn randomly and passed through through the original model. They are then weighted based on their distance to the example
that we are trying to explain. Then a linear model is trained using these weighted samples to predict the original model outputs. The linear model is interpretable and is accurate in the vicinity of
the example under consideration.
Figure 6. Local interpretable model-agnostic explanations (LIME). a) We wish to explain the purple point. Samples are drawn randomly and weighted by their proximity to the point of interest (red and
green points). b) We use these samples to train a simpler, interpretable model like a linear model. c) The region around point of interest on original function is closely approximated by d) the
interpretable linear model.
This method works well for continuous tabular data, and can be modified for other data types. For text data, we might perturb the input by removing or replacing different words rather than sampling
and we might use a bag of words model as the approximating model. So, we could understand the output of a spam detector based on BERT by passing multiple perturbed sentences through the BERT model,
retrieving the output probability of spam and then building a sparse bag of words model that approximates these results.
For image data, we might explain the output of a classifier based on a convolutional network by approximating it locally with a weighted sum of the contributions of different superpixels. We first
divide the image into superpixels, and then perturb the image multiple times by setting different combinations of these superpixels to be uniform and gray. These images are passed through the
classifier and we store the output probability of the top-rated class. Then we build a sparse linear model that predicts this probability from the presence or absence of each superpixel (figure 7).
Figure 7. LIME explanations for image classification. a) Original image to be explained which was classified as `tree frog’. b) Grouping input features by dividing into superpixels. c-e) Replace
randomly selected subsets of superpixels with blank gray regions, run through model, and store model probability. f) Build sparse linear model explaining model probability in terms of presence or
absence of superpixels. It appears that the head region is mainly responsible for the classification as tree-frog. g) Repeating this process for the class “billiards” which was also assigned a
relatively high probability by the model. Adapted from Ribeiro et al. (2016)
Ribeiro et al. (2018) proposed anchors which are local rules defined on the input values around a given point that we are trying to explain (figure 8). In the simplest case, each rule is a hard
threshold on an input value of the form $x_{1} < \tau$. The rules are added in a greedy way with the aim being to construct rules where the precision is very high (i.e., when the rule is true, the
output is almost always the same as the original point). We cannot exhaustively evaluate every point in the rule region and so this is done by considering the choice of rules as a multi-armed bandit
problem. In practice, rules are extended in a greedy manner by adding more constraints on them to increase the precision for a given rule complexity. In a more sophisticated solution, beam search is
used and the overall rule is chosen to maximize the coverage. This is the area of the data space that the rule is true over.
Figure 8. Anchors. One way to explain the purple point is to search for simple set of rules that explain the local region. In this case, we can tell the user that 100% of points the where $x_{1}$ is
between 0.51 and 0.9 and $x_{2}$ is between 0.52 and 0.76 are classified as positive.
Ribeiro et al. (2018) applied this message to an LSTM model that predicted the sentiment of reviews. They perturbed the inputs by replacing individual tokens (words) with other random words with the
same part of speech tag, with a probability that is proportional to their similarity in the embedding space and measure the response of the LSTM to each. In this case their rules just consist of the
presence of words, and for the sentence This movie is not bad, they retrieve the rule that when the both the words “not” and “bad” are present, the sentence has positive sentiment 95% of the time.
Shapley additive explanations
Shapley additive explanations describe the model output ${f}[\mathbf{x}_i]$ for a particular input $\mathbf{x}_{i}$ as an additive sum:
$$\label{eq:explain_add_shap} {f}[\mathbf{x}_i] = \psi_{0} + \sum_{d=1}^{D} \psi_d \tag{2}$$
of $D$ contributing factors $\psi_{D}$ associated with the $D$ dimensions of the input. In other words, the change in performance from a baseline $\psi_0$ is attributed to a sum of changes $\psi_{d}$
associated with the input dimensions.
Consider the case where there are only two input variables (figure 9), choosing a particular ordering of the input variables and a constructing the explanation piece by piece. So, we might set:
\begin{eqnarray} \psi_{0} &=& \mathbb{E}_{\mathbf{x}}\left[{f}[\mathbf{x}]\right]\nonumber \\ \psi_{1} &=& \mathbb{E}_{\mathbf{x}}\left[{f}[\mathbf{x}]|x_{1}\right] – \left(\mathbb{E}_{\mathbf{x}}\
left[{f}[\mathbf{x}]\right]\right) \nonumber \\ \psi_{2} &=& \mathbb{E}_{\mathbf{x}}\left[{f}[\mathbf{x}]|x_{1},x_{2}\right]-\left(\mathbb{E}_{\mathbf{x}}\left[{f}[\mathbf{x}]\right]-\mathbb{E}_{\
mathbf{x}}\left[{f}[\mathbf{x}]|x_{1}\right]\right). \tag{3}\end{eqnarray}
The first term contains the expected output without observing the input. The second term is the change to the expected output given that we have only observed the first dimension $x_{1}$. The third
term gives the additional change to expected output now that we have observed dimensions $x_{1}$ and $x_{2}$. In each line, the term in brackets is the right hand side from the previous line, which
is why these represent changes.
Figure 9. Shapley additive explanations. a) Consider explaining the model output $\mbox{f}[\mathbf{x}]$ for point $\mathbf{x}$. b) We construct the explanation as a linear combination of three terms.
c) The first term is what we know before considering the data at all. This is the average model output (bottom left corner of panel a). The second term is the change due to what we know from
observing feature $x_{1}$. This is calculated by marginalizing over feature $x_{2}$ and reading off the prediction (bottom of panel a). We then subtract the first term in the sum to measure the
change that was induced by feature $x_{1}$. The third term consists of the remaining change that is needed to get the true model output and is attributable to $x_{2}$. d) We can visualize the
cumulative changes due to each feature. e-g) If we repeat this procedure, but consider the features in a different order, then we get a different results. Shapley additive explanations take a
weighted average of all possible orderings and return the additive terms $\psi_{\bullet}$ that explain the positive or negative contribution of each feature.
Assuming we could calculate these terms, they would obviously have the form of equation 2. However, the input order was arbitrary. If we took a different ordering of the variables so that $x_{2}$ is
before $x_{1}$, then we would get different values
\begin{eqnarray} \psi_{0} &=& \mathbb{E}_{\mathbf{x}}\left[{f}[\mathbf{x}]\right]\nonumber \\ \psi_{2} &=& \mathbb{E}_{\mathbf{x}}\left[{f}[\mathbf{x}]|x_{2}\right] – \left(\mathbb{E}_{\mathbf{x}}\
left[{f}[\mathbf{x}]\right]\right) \nonumber \\ \psi_{1} &=& \mathbb{E}_{\mathbf{x}}\left[{f}[\mathbf{x}]|x_{1},x_{2}\right]-\left(\mathbb{E}_{\mathbf{x}}\left[{f}[\mathbf{x}]\right]-\mathbb{E}_{\
mathbf{x}}\left[{f}[\mathbf{x}]|x_{2}\right]\right). \tag{4}\end{eqnarray}
The idea of Shapley additive explanations is to compute the values $\psi_{d}$ in equation 2 by taking a weighted average of the $\psi_{i}$ over all possible orderings. If the set of indices is given
by $\mathcal{D}=\{1,2,\ldots, D\}$, then the final Shapley values are
$$\label{eq:explain_shap_expect} \psi_{d}[{f}[\mathbf{x}]] = \sum_{\mathcal{S}\subseteq \mathcal{D}} \frac{|\mathcal{S}|!(D-|\mathcal{S}|-1)!}{D!}\left(\mathbb{E}\left[{f}[\mathbf{x}]|\mathcal{S}\
right] – \mathbb{E}\left[{f}[\mathbf{x}]|\mathcal{S}_{\setminus d}\right]\right). \tag{5}$$
This computation takes every subset of variables that contains $x_{d}$ and computes the expected value of the function given this subset takes the particular values for that data point with and
without $x_{d}$ itself. This result is weighted and contributes to the final value $\psi_{d}$. The particular weighting (i.e., the first term after the sum) can be proven to be the only one that
satisfies the properties of (i) local accuracy (the Shapley values sum to the true function output) (ii) missingness (an absent feature has a Shapley value/ attribution of zero, and (iii) consistency
(if the marginal contribution of a feature increases or stays the same, then the Shapley value should increase or stay the same).
The eventual output of this process is a single value associated with each feature that represents how much it increased or decreased the model output. Sometimes this is presented as a force diagram
(figure 10) which is a compact representation of all of these values.
Figure 10. Force diagram for Shapley additive explanations. The final result is explained by an additive sum of a scalars associated with each input feature (PTRATIO, LSTAT, RM etc.). The red
features increase the output and the blue feature decrease it. The horizontal size associated with each feature represents the magnitude of change. Via Lundberg and Lee (2017).
Computing SHAP values
Computing the SHAP values using (equation 5) is challenging, although it can be done exactly for certain models like trees (see Lundberg et al., 2019). The first problem is that there are a very
great number of subsets to test. This can be resolved by approximating the sum with a subset of samples. The second problem is how to compute the expectation terms. For each term, we consider the
data $\mathbf{x}$ being split into two parts $\mathbf{x}_{\mathcal{S}}$ and $\mathbf{x}_{\overline{\mathcal{S}}}$. Then the expectation can be written as:
$$\mathbb{E}[{f}[\mathbf{x}]|\mathcal{S}]=\mathbb{E}_{x_{\overline{\mathcal{S}}}|x_{\mathcal{S}}}\left[{f}[\mathbf{x}]\right]. \tag{6}$$
It is then possible to make some further assumptions that can ease computation. We might first assume feature independence:
$$\mathbb{E}_{x_{\overline{\mathcal{S}}}|x_{\mathcal{S}}}\left[{f}[\mathbf{x}]\right] \approx \mathbb{E}_{x_{\overline{\mathcal{S}}}}\left[{f}[\mathbf{x}]\right], \tag{7}$$
and then further assume model linearity:
$$\mathbb{E}_{x_{\overline{\mathcal{S}}}}\left[{f}[\mathbf{x}]\right] \approx {f}\left[\mathbf{x}_{\mathcal{S}}, \mathbb{E}_[\mathbf{x}_{\overline{\mathcal{S}}}]\right]. \tag{8}$$
When we use this latter assumption, the model can replicate LIME and this is known as kernel SHAP. Recall that LIME fits a linear model based on weighted samples, where the weights are based on the
proximity of the sample to the point that we are trying to explain. However, these weights are chosen heuristically. Shapley additive explanations with feature independence and linearity also fit
local linear model from points $\mathbf{x}'[\mathcal{S}] = [\mathbf{x}_{\mathcal{S}}, \mathbb{E}_{x_{\overline{\mathcal{S}}}}]$ where some subset of $\overline{\mathcal{S}}$ of the features have been
replaced by their expected values. In practice the expectations are computed by substituting other random values from the training set.
The weights for KernelShap are given (non-obviously) by:
$$\omega[\mathcal{S}] = \frac{D-1}{(D \:{choose}\: |\overline{\mathcal{S}}|)|\overline{\mathcal{S}}|(D-|\overline{\mathcal{S}}|)}. \tag{9}$$
In part I of this blog, we have discussed the importance of explainability for AI systems. We presented a taxonomy of such methods and described a number of methods for creating local post-hoc
explanations of a black box model. These help the user understand why a particular decision was made but do not try to explain the whole model in detail. In part II of this blog, we will consider
global explanations and models that are interpretable by design. | {"url":"https://rbcborealis.com/research-blogs/explainability-i-local-post-hoc-explanations/","timestamp":"2024-11-01T23:54:40Z","content_type":"text/html","content_length":"215233","record_id":"<urn:uuid:c4c28793-30a2-46b7-b8ad-ba827afe07a0>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00279.warc.gz"} |
Breaking Chocolate Bars
Assume you have a chocolate bar consisting, as usual, of a number of squares arranged in a rectangular pattern. Your task is to split the bar into small squares (always breaking along the lines
between the squares) with a minimum number of breaks. How many will it take?
The purpose of the simulation below is to help you come up with the right answer. Please try it before you proceed to the solution. Click where you want to break them.
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As many as there are small squares minus 1.
Proof #1 (by induction)
1. If there are just one square we clearly need no breaks.
2. Assume that for numbers 1 ≤ m < N we have already shown that it takes exactly m - 1 breaks to split a bar consisting of m squares. Let there be a bar of N > 1 squares. Split it into two with m[1]
and m[2] squares, respectively. Of course, m[1] + m[2] = N. By the induction hypothesis, it will take (m[1]-1) breaks to split the first bar and (m[2]-1) to split the second one. The total will
be 1 + (m[1]-1) + (m[2]-1) = N-1.
Proof #2
Let start counting how many pieces we have after a number of breaks. The important observation is that every time we break a piece the total number of pieces is increased by one. (For one bigger
piece have been replaced with two smaller ones.) When there is no pieces to break, each piece is a small square. At the beginning (after 0 breaks) we had 1 piece. After 1 break we got 2 pieces. As I
said earlier, increasing the number of breaks by one increases the number of pieces by 1. Therefore, the latter is always greater by one than the former.
Follow up
It should be now clear that the rectangular formation of a chocolate bar is a red herring. The basic fact explained above may appear in many different guises. For example, there are quite edifying
games based on the principle explained above (with every move a number related to the game is increased by 1.) These games are not very challenging as such. However, they furnish an edifying
experience besides giving a chance for a knowledgeable person to show off if he/she is the only one who knows the secret. Here are a few examples.
Problem #1
A fellow sawed 25 tree trunks into 75 logs. How many cuts did he perform? (Answer)
Problem #2
75 teams took part in a competition organized according to the olympic rules: teams met 1-on-1 with the defeated team getting dropped out of the competition. How many meets are needed to before one
team is declared a winner? (Answer)
Problem #3
(C. W. Trigg, Mathematical Quickies, Dover, 1985, #29.)
In assembling a jigsaw puzzle, let us call the fitting together of two pieces a "move", independently of whether the pieces consist of single pieces or of blocks of pieces already assembled. What
procedure will minimize the number of moves required to solve an N-piece puzzle? What is the minimum number?
Problem #4
(C. W. Trigg, Mathematical Quickies, Dover, 1985, #13.)
There are N players in an elimination-type singles tennis tournament. How many matches must be played (or defaulted) to determine the winner?
Game #1
Two players take turns breaking a bar. The last to break a piece wins the game.
An Aside
It's a great way to learn of odd and even numbers. Any one privy to the secret would know what is preferable: to start the game or to be a second player - depending as whether the total number of
squares is even or odd.
Game #2
Marbles, checkers, or stones are arranged in several piles. A move consists in selecting a pile and splitting it into two. The player to split the last pile is the winner. (Explanation: it clearly
does not matter how many piles one starts with. Imagine starting with a single pile and then making a few moves "that do not count.")
Other simple games may be thought up to explain and reinforce the notion of parity, i.e., the concepts that odd and even numbers are of different parities. For example,
Game #3
Write a sequence of numbers. For the entertainment sake, let one opponent write the sequence and the other start the game. A move consists in writing a plus or a minus sign between two adjacent
terms. The first player wins if, with all signs inserted and computations carried out, the result is odd. If the result is even, the second player wins. (Explanation: The result does not depend on
the particular distribution of signs at all. Adding or subtracting an even (odd) number does not change (changes) the parity of the result. So the final result will be odd iff the number of odd
numbers in the sequence is odd.) You may want to test your skills against your computer's.
Returning to the original problem of a chocolate bar, the number of moves needed to break it into separate squares is invariant with regard to the actual sequence of moves. A less trivial invariant
may serve as a basis for a trick suitable for a magic show.
Yvan_Roux from Canada was inspired to make the following remark
We can use the same induction proof to prove that the result is true for a puzzle or a 3D shape made of elementary pieces, as far as we do not break the elementary pieces.
Yvan Roux
1. D. Fomin, S. Genkin, I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996
2. P. Winkler, Mathematical Puzzles: A Connoisseur's Collection, A K Peters, 2004
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Problem #1
A fellow sawed 25 tree trunks into 75 logs. How many cuts did he perform?
Every cut increased the number of logs by 1. Thinking of a tree trunk as a big log, it took 75 - 25 = 50 cuts to get 75 logs out of 25.
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Problem #2
75 teams took part in a competition organized according to the olympic rules: teams met 1-on-1 with the defeated team getting dropped out of the competition. How many meets are needed to before one
team is declared a winner?
With every meet, the number of teams in the competition is decreased by 1. It takes 74 meets to seed 1 team out of 75.
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Real life applications of mathematics (idea)
, a
, and a
are sitting in a
street cafe
watching people going in and coming out of the house on the other side of the street.
First they see two people going into the house. Time passes. After a while they notice three people coming out of the house.
The physicist: "The measurement wasn't accurate."
The biologists: "They have reproduced".
The mathematician: "If exactly one person now enters the house then it will be empty again."
Joke Nodes: Geek Jokes: Real life applications of mathematics | {"url":"https://everything2.com/user/General+Wesc/writeups/Real+life+applications+of+mathematics","timestamp":"2024-11-14T21:06:04Z","content_type":"text/html","content_length":"29316","record_id":"<urn:uuid:a7a6c408-c204-4089-a82b-bc99ccdb43a8>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00633.warc.gz"} |
Activity - Calculate the chi-squared (X2) results – Homework Ace Tutors
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Gradient Descent From Scratch In Python
10 Jan 202342:38
TLDRIn this tutorial, Vic introduces the concept of gradient descent, a fundamental algorithm for training neural networks. The video demonstrates how to implement linear regression using gradient
descent in Python. Starting with data on weather, the process involves importing libraries, reading and preprocessing data, and visualizing the relationship between variables. The core of the
tutorial focuses on understanding the linear regression model, calculating loss using mean squared error, and iteratively updating weights and biases to minimize loss. The training loop, learning
rate adjustments, and weight initialization are discussed in detail. The video concludes with a comparison of the implemented model's parameters to those from scikit-learn, emphasizing the relevance
of the concepts learned to neural networks.
• {"📚":"Gradient descent is a fundamental algorithm for training neural networks by optimizing parameters based on data."}
• {"🔢":"The process involves initializing parameters, making predictions, calculating loss, and updating parameters to minimize error."}
• {"📈":"Linear regression is used as an example to demonstrate how gradient descent works, with the goal of predicting future values based on past data."}
• {"🎯":"The mean squared error (MSE) function is used to measure the prediction error or loss, which is crucial for gradient descent."}
• {"📉":"The gradient represents the rate of change of the loss with respect to the weights, guiding the direction and magnitude of parameter updates."}
• {"🔧":"A learning rate is used to control the size of the steps taken during the update process to avoid overshooting the minimum loss."}
• {"🌀":"Gradient descent is an iterative process, requiring multiple passes (epochs) through the data to converge towards the optimal solution."}
• {"🔁":"The training loop is a common structure in machine learning, where the data is passed through the model repeatedly until the loss is minimized."}
• {"📊":"Data is often split into training, validation, and test sets to prevent overfitting and to evaluate the model's performance accurately."}
• {"🤖":"The concepts learned, such as forward and backward passes, are directly applicable to more complex neural networks."}
• {"⚖️":"Careful tuning of the learning rate and initialization of weights is essential for the effective learning and convergence of the model."}
Q & A
• What is the main topic of the tutorial?
-The main topic of the tutorial is gradient descent, specifically its implementation in Python for linear regression as a fundamental building block of neural networks.
• Why is gradient descent important in the context of neural networks?
-Gradient descent is important because it is the method by which neural networks learn from data and train their parameters, allowing for the optimization of the network's weights and biases.
• What library is used to read in the data for the tutorial?
-The tutorial uses the pandas library to read in the data for analysis.
• What is the dataset used in the tutorial?
-The dataset used in the tutorial consists of weather data, including maximum temperature (T-Max), minimum temperature (T-Min), rainfall, and the next day's temperature, with the goal of
predicting T-Max for the following day.
• How is the linear relationship visualized in the tutorial?
-The linear relationship is visualized using a scatter plot with a line drawn through the data points to represent the trend, which is then used to discuss the concept of a linear relationship in
the context of linear regression.
• What is the equation form of the linear regression model used in the tutorial?
-The equation form used in the tutorial is: \( \hat{y} = W_1 \times X_1 + b \), where \( \hat{y} \) is the predicted value, \( W_1 \) is the weight, \( X_1 \) is the input feature (T-Max in this
case), and \( b \) is the bias.
• What is the role of the weight (W) in the linear regression model?
-The weight (W) in the linear regression model is a value that the neural network learns through the training process. It determines the influence of the input feature on the prediction.
• What is the learning rate in the context of gradient descent?
-The learning rate in gradient descent is a parameter that controls the step size during the update of the model's weights and biases. It is crucial for the convergence of the algorithm and to
prevent overshooting the optimal solution.
• What is the mean squared error (MSE) used for in the tutorial?
-The mean squared error (MSE) is used as a loss function to measure the error of the prediction made by the linear regression model. It calculates the average of the squares of the differences
between the predicted and actual values.
• How is the gradient calculated in the tutorial?
-The gradient is calculated by taking the derivative of the loss function with respect to the weights and biases. It represents the rate of change of the loss and is used to adjust the parameters
in the direction that minimizes the loss.
• What is the purpose of the training loop in the gradient descent algorithm?
-The training loop is used to iteratively update the model's parameters by passing the data through the algorithm multiple times (epochs) until the error is minimized or the algorithm has
converged to an optimal solution.
😀 Introduction to Gradient Descent
This paragraph introduces Vic, the presenter, and the topic of gradient descent, which is a fundamental algorithm for training neural networks. The script outlines the plan to use Python to implement
linear regression via gradient descent. The importance of reading in weather data and preparing it for training is emphasized, including handling missing values and examining the initial data set.
The goal is to predict the maximum temperature for the next day based on various weather-related inputs.
📈 Understanding Linear Regression
The paragraph explains the concept of linear regression and its necessity for a linear relationship between the predictors and the predicted value. It describes the process of visualizing this
relationship through a scatter plot and drawing a line of best fit. The script also covers the equation for linear regression, introducing the concepts of weights and bias. It further discusses how
linear regression can be expanded to use multiple predictors and the parameters involved in such a model. The paragraph concludes with the use of scikit-learn to train a linear regression model and
plot the resulting line of best fit.
🧮 Calculating Loss and Gradient
This section delves into the importance of calculating the error or loss of a prediction, which is crucial for the gradient descent algorithm. It introduces the mean squared error (MSE) as the loss
function and explains how it is used to measure the prediction error. The script then discusses how to graph different weight values against loss to understand the effect of varying weights on the
model's performance. It also explains the concept of the gradient, which is the rate of change of the loss with respect to the weights, and how it is calculated.
🔄 Gradient Descent Optimization
The paragraph focuses on the optimization process of gradient descent, aiming to find the weight values that minimize the loss. It explains the concept of the gradient and how it changes with
different weight values. The script illustrates this with a graph and explains the goal of gradient descent is to find the weight value where the gradient is zero, which corresponds to the lowest
possible loss. It also introduces the concept of the partial derivatives with respect to both the weights and the bias, which are used to update these parameters.
🔢 Updating Parameters and Learning Rate
This section discusses how to update the weights and biases of the model to minimize the error. It explains the process of calculating the partial derivatives and using them to adjust the parameters.
The script highlights the importance of the learning rate in controlling the size of the steps taken during the optimization process. It shows how taking too large a step can increase the loss, while
a learning rate that is too small can lead to very slow convergence. The paragraph also includes a visualization of how the gradient changes as the weights change and the need to adjust the learning
rate accordingly.
🔁 Training Loop and Batch Gradient Descent
The paragraph outlines the process of setting up a training loop for gradient descent, which involves repeatedly passing the data through the algorithm until the loss is minimized. It explains the
concept of batch gradient descent, where the gradient is averaged across the entire dataset before updating the parameters. The script details the steps needed to build the algorithm, including
initializing weights and biases, making predictions, calculating loss and gradient, and updating parameters in the backward pass. It also emphasizes the importance of using a validation set to
monitor the algorithm's performance and a test set for final evaluation.
🎯 Final Model Parameters and Convergence
The final paragraph discusses the finalization of the model's parameters after training and the convergence of the algorithm. It explains that careful attention must be paid to the learning rate and
the initialization of weights and biases, as these factors can significantly affect the outcome and convergence rate of the model. The script also touches on the possibility of adding a
regularization term to prevent the weights from becoming too large. The paragraph concludes with a summary of the key concepts learned in the tutorial and a preview of applying these concepts to
neural networks in future tutorials.
Gradient Descent
Gradient Descent is an optimization algorithm used to minimize a function by iteratively moving towards the steepest descent of the gradient as defined by the function's contours. In the context of
the video, it is a fundamental technique for training neural networks by adjusting parameters to minimize the error in predictions. The script describes its implementation in a Python environment for
linear regression, using it to update the weights and biases of the model to better fit the training data.
Neural Networks
Neural Networks are a set of algorithms modeled loosely after the human brain. They are designed to recognize patterns and are used in a wide range of applications, from medical diagnosis to stock
market prediction. The video script discusses neural networks as complex systems that can be trained using gradient descent, highlighting that the concepts introduced, such as forward and backward
passes, are directly applicable to neural networks.
Linear Regression
Linear Regression is a statistical method for modeling the relationship between a dependent variable 'y' and one or more independent variables denoted as 'X'. The video script focuses on using
gradient descent to perform linear regression, aiming to predict future values (e.g., tomorrow's temperature) based on current and past data.
Pandas is a software library in Python for data manipulation and analysis. It is widely used for data清洗 (cleaning) and preparation before feeding it into a machine learning model. In the script,
Vic imports the pandas library to read in and handle the dataset that will be used for the linear regression model.
Scikit-learn is an open-source machine learning library for Python that provides simple and efficient tools for predictive data analysis. The script mentions using scikit-learn to train a linear
regression model as a comparison to the manually implemented gradient descent model.
Mean Squared Error (MSE)
Mean Squared Error is a measure of the quality of an estimator—it is always non-negative, and values closer to zero are better. It is used as a loss function in the video to quantify the difference
between the predicted and actual values. The script explains how MSE is calculated and how it's used to guide the gradient descent process.
Weights and Biases
In the context of linear regression, weights are the coefficients that are multiplied by the input features to make predictions, and biases are the offsets or intercepts added to the predictions. The
script details how weights and biases are initialized, updated, and used in the gradient descent algorithm to improve the model's predictions.
Forward Pass
The Forward Pass is the process of making predictions using a neural network or a machine learning model. It involves passing the input data through the network to generate an output. In the video,
the forward pass is used to calculate predictions based on the current weights and biases of the model.
Backward Pass
The Backward Pass, also known as backpropagation, is the process of adjusting the weights and biases of a neural network in response to the error in the predictions. It involves calculating the
gradient of the loss function with respect to each parameter and then updating the parameters in the opposite direction of the gradient. The script describes implementing a backward pass to perform
gradient descent.
Learning Rate
The Learning Rate is a hyperparameter that controls how much we are adjusting the weights and biases of our model with respect to the loss gradient. It is crucial in gradient descent as it determines
the step size during the optimization process. The script discusses the importance of choosing an appropriate learning rate to ensure the algorithm converges efficiently.
Convergence in the context of gradient descent refers to the point at which the algorithm's parameters have been adjusted to a point where further iterations result in minimal changes to the loss
function, indicating an optimal or near-optimal solution has been reached. The script illustrates the concept by showing how the loss decreases with each epoch until it stabilizes, indicating
Gradient descent is an essential building block of neural networks, allowing them to learn from data and train their parameters.
The tutorial uses Python to implement linear regression with gradient descent, a method that will be expanded upon for more complex networks in future videos.
Data on weather is used to train a linear regression algorithm to predict tomorrow's maximum temperature (TMax) using other columns.
Linear regression requires a linear relationship between the predictors and what is being predicted.
A scatter plot visualizes the relationship between TMax and TMax tomorrow, suggesting a linear trend.
The linear regression model is represented by the equation: Predicted Y = W1 * X1 + B, where W is the weight and B is the bias.
Scikit-learn's linear regression class is used to train the algorithm and make predictions.
The mean squared error (MSE) function is introduced to calculate the loss or error of the prediction.
Gradient descent aims to minimize the loss by adjusting weights and biases, using the gradient of the loss function.
The gradient is the rate of change of the loss with respect to the weights, indicating how quickly the loss changes as weights change.
A learning rate is used to control the size of the steps taken during gradient descent to avoid overshooting the minimum loss.
Batch gradient descent is employed, which calculates the gradient by averaging the error across the entire dataset.
The algorithm is initialized with random weights and biases, and a training loop is set up to iteratively improve these parameters.
The partial derivatives with respect to weights and biases are calculated to understand how to adjust the parameters to reduce error.
The training process involves a forward pass to make predictions, a calculation of loss and gradient, followed by a backward pass to update parameters.
The algorithm's convergence is monitored by observing when the loss stops decreasing significantly, indicating that the minimum loss point has been reached.
The learning rate and initialization of weights and biases are critical factors that can affect the speed of convergence and the final outcome of the model.
The concepts introduced, such as forward and backward passes, are directly applicable to more complex neural networks. | {"url":"https://summarize.ing/blog-Gradient-Descent-From-Scratch-In-Python-23358","timestamp":"2024-11-03T13:45:49Z","content_type":"text/html","content_length":"141963","record_id":"<urn:uuid:7848ff46-d41e-4f34-836c-b391f2a33185>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00805.warc.gz"} |
Lock-in transitions and charge transfer in one-dimensional fermion systems
The one-dimensional quantum sine-Gordon Hamiltonian with a density rho [s] of solitons is studied. rho [s]( mu ) is the order parameter for the lock-in transition, which happens when the chemical
potential mu equals the single soliton energy and temperature is T=0. The soliton density rho [s]( mu ,T) and the critical behaviour at T=0 are studied in the classical limit (quantum coupling beta
to 0) and in the quantum system with beta ^2=4 pi ; the critical exponent for rho [s] is 0 and ^1/[2] respectively. An important application of these results is the temperature dependence of an
incommensurate charge transfer rho [e] in one-dimensional conductors due to Umklapp scattering. rho [e](T) can be determined by the function rho [s]( mu ,T).
ASJC Scopus subject areas
• Condensed Matter Physics
• General Engineering
• General Physics and Astronomy
Dive into the research topics of 'Lock-in transitions and charge transfer in one-dimensional fermion systems'. Together they form a unique fingerprint. | {"url":"https://cris.bgu.ac.il/en/publications/lock-in-transitions-and-charge-transfer-in-one-dimensional-fermio","timestamp":"2024-11-03T13:33:41Z","content_type":"text/html","content_length":"54984","record_id":"<urn:uuid:5b987eac-c547-42c4-a359-62441858ab63>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00234.warc.gz"} |
Help Me Learn Math!
Help Me Learn Math!
June 14, 2014 7:31 PM Subscribe
I will be returning to school in the fall to start prerequisite courses for a Physical Therapy program. I don't know math. The last time i took math was in high school, and it was basic geometry. I
would like to teach myself primarily, so that i can test into higher classes. The goal is to be able to test into pre-calc, so that i can also qualify for the physics classes as well. Any
suggestions? I don't really care about the method. I was thinking about buying a textbook, but i was unsure of which ones to get. Thanks.
I did
Khan Academy
and was able to test into a higher math class than the first time I took the placement exam. I found it was better than trying to go about it from a textbook.
posted by schnee at 7:43 PM on June 14, 2014 [10 favorites]
I am very math-challenged, and have had to get caught up for a nursing program. I really recommend Khan Academy -- his website has math from 3rd grade up to calculus, and watching someone do examples
really helps me. Here's the
linkposted by queens86 at 7:45 PM on June 14, 2014 [1 favorite]
Have you talked to an adviser or counselor at your new school about your plan? Because I'm sure they would have resources, as well as more specific info on what you would need to know.
Here is some info on the placement test used at my school:
Compass Mathematics Placementposted by SuperSquirrel at 7:46 PM on June 14, 2014
Hi, I'm a faculty member who sometimes administers my college's in-house math placement test. I frequently recommend the aforementioned Khan Academy to students who don't place in the math course
they are aiming for. The ones who are serious and treat it like a self taught course definitely do better the second time around.
posted by hurdy gurdy girl at 8:56 PM on June 14, 2014
Khan academy, BUT it's really really important that you do the exercises and homework and not just watch the videos if you want to pass the test. Watching the videos are fine if you just want to
understand the concepts, but you won't pass the tests without practice and memorization, etc.
posted by empath at 10:09 PM on June 14, 2014 [7 favorites]
I taught my wife enough out of algebra for dummies in two weeks that she tested out of basic math.
posted by jeffamaphone at 11:07 PM on June 14, 2014
Khan Academy. But - do the problem sets, and do them until you feel that you know them.
I was able to test into Pre Calculus due to that, and then I got a 3.5 in my Trigonometry class last Quarter. Next up - Calculus!
posted by spinifex23 at 12:33 AM on June 15, 2014
Best answer:
A few years ago, I also had to catch up on missing math skills, and also ended at Basic Geometry in high school. The advice above, about using Khan Academy with the problem sets, is right on. But I
have another resource recommendation that helped me immensely.
Back when I did my mad math scramble, Khan Academy consisted mostly of videos, no problem sets. So for practice, I used a set of online drills called
It's used as a complement to brick-and-mortar math courses that instructors can use to give students more practice with specific problem types. It gets very detailed, and is very strict with answer
input. It really forced me to become precise and find my mistakes (you can't go further in a certain problem type until you get 3-4 answers in a row correct).
It is not free, and occasionally glitchy, but well worth it. There is a free demo.
If you are willing to pay (I think $20/month, less for longer periods) they offer exactly the courses you might need - a
selection from the website
: Introduction to Geometry, Intermediate Algebra, Beginning and Intermediate Algebra Combined, College Algebra, Trigonometry, PreCalculus (and many, many more....)
Many courses also count as college credit (but I didn't need this so don't know much about how it works).
I used it in conjunction with Khan's videos and
. I'm now studying engineering, and feel like that was an excellent combo for me - over about 1.5 years, I did almost all the courses at home, through Intro to Calculus, plus some extras.
They also have a module on Physics, which you might find interesting.
Sorry, I know I sound like a salesperson but I definitely am not, just an enthusiastic former customer. Please feel free to Memail me if you have any questions. Have fun with math and good luck!
posted by Pieprz at 6:04 AM on June 15, 2014 [5 favorites]
Do you know what placement test your college uses? Some colleges use their own test, but a lot of them use a standard one like COMPASS or Accuplacer. If your college uses a standard one, there are a
lot of test prep materials
Also, if you are looking to dedicate some time, I agree with Pieprz, ALEKS might be a good fit, especially since the diagnostic tests will be able to tell you where you need to build your skills.
posted by mjcon at 9:03 AM on June 15, 2014
I'd get a book like
and work through it cover to cover if I were in your position
posted by thelonius at 10:51 AM on June 15, 2014
Since you asked about books, I'll controbute that my kids used Calculus the Easy way and Algebra the Easy way for fun before they took those courses. They got the basic concepts. I doubt they got the
full breath of the topics, though.
posted by SemiSalt at 11:44 AM on June 15, 2014
I am in the middle of what you are about to attempt. I am currently struggling along in Calc I (ugh). Khan Academy is, in my opinion, the best online source for math. I have also use ALEKS and I
liked that one, too. ALEKS does a great job assessing where you are, where you want to be, and the skills you need to get from here to there. Khan Academy does a great job of talking you through the
problems. I am currently doing my calc class online and the majority of the homework is in WebAssign. Not sure if you can just start a class with them. However, it doesn't do a great job of talking
you through the problems. I think it shines in its ability to be assigned and assessed by an instructor. Meh.
The textbook that my university used to use for both their college algebra and trig class is Precalculus: Mathematics for Calculus by Stewart, Redlin, and Watson. I'm not sure why they stopped using
it but the guy at the study center seemed rather bummed that they did because he really liked it.
The statewide library association in my state subscribes to a few services and databases that are available to anyone with an IP address within the state. The best of these is called LiveHomeworkHelp
from Tutor.com. You can subcribe to this service but check with your local library to see if they may offer this service. You can log into the site with your question and you are connected to an
actual tutor at ANY TIME OF THE DAY. And I mean anytime. I have dialed in at, like, 2am Alaska time and got a great tutor. This service has saved me so much heartache. Self-learning, however, means
that you are less likely to be forced to solve problems that you just don't get, so, this service will mean less to you than to me.
Good luck. It's awful but also totally do-able.
posted by Foam Pants at 8:38 PM on June 15, 2014 [2 favorites]
I was in a very similar situation as you a couple years ago, and I nth Khan Academy. Great resource, but like others have said, you have to do the problem sets or else you aren't going to benefit.
When I have been working through math courses the past couple years, I have also relied heavily on
Desmos Graphing Calculator
(especially if you're getting into pre-calc and calc) to help me visualize problems and check my work.
posted by Lutoslawski at 12:21 PM on June 16, 2014 [1 favorite]
« Older What iPad app can I get for my Luddite child to... | Android tablet with best battery life? Newer »
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Leetcode Solution | {"url":"https://totheinnovation.com/tag/uber/","timestamp":"2024-11-02T12:38:28Z","content_type":"text/html","content_length":"209716","record_id":"<urn:uuid:9b7da00c-9865-40c4-baf4-518728c1eab0>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00719.warc.gz"} |
True Monopoly | Hagers X Games
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True Monopoly
• Similar rules Monopoly but more complex
• Each player starts with 300 points and will throw the axes at the squares with different point values
• Once a player sticks an axe into a square, it will turn the players chosen color and the player will get the number of points shown in the square. Then if an opponent hits that same square, the
opponent will lose that amount of points the square was for.
• If a player hits a square that is their own color, they will lose that square but no loss of points
• If a player misses, they will lose half of their points
• A player is then knocked out once they go below 0.
• The game ends by either only 1 player remaining with more than 0 points or by hitting the FORCE END button on the screen which will limit the number of rounds by adding 3 full rounds onto the
current round number.
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Hydro-dynamically smooth and rough boundaries
1 Answer
(a) Smooth Boundary
(b) Rough Boundary
Let the average height of the irregularities projecting from the surface of boundary be denoted as ‘K’.
Now, if the value of ‘K’ is large for a boundary then the boundary is called as Rough boundary.
And if the value of ‘K’ is smaller or less, then the boundary is known as Smooth boundary.
This is the classification of rough and smooth boundaries based on boundary characteristics. But for proper classification flow and fluid characteristics are also considered.
Hydro-dynamically smooth boundary
1) For a turbulent flow analysis, the flow is divided into two parts or portions.
2) The first portion consists of a thin layer of fluid in the immediate neighbourhood of the boundary, where the viscous shear stress is stronger whereas the shear stress due to turbulent is
negligible. This is known as laminar sublayer.
3) The second portion of flow is the one where shear stress due to turbulence is higher or large as compared to the viscous shear and this zone is called as Turbulent zone.
4) The zone upto which or the height upto which the effect of viscosity predominates is denoted by $δ'$.
5) So, we can say that if the average velocity height ‘K’ is less than ‘$\delta'$’, then the boundary is called as Smooth boundary.
6) This happens because, outside the laminar sub-layer the flow is turbulent and eddies of various sizes present in the turbulent flow try to penetrate through the laminar sub-layer, but due to the
great thickness of laminar sub-layer the eddies are unable to reach the surface irregularities and so the boundary behaves as smooth boundary.
Hydro-dynamically rough boundary
1) If the Reynolds number of the flow increases, the thickness of laminar sub-layer decreases.
2) If this happens, then the average height ‘K’ of irregularities is above the laminar sub-layer and thus the eddies present will come in contact with irregularities of the surface and lot of energy
will be lost.
3) Such a boundary is known as Hydro-dynamically rough boundary.
Conditions from Nikuradse’s experiment:-
1) If $\left(\dfrac {K’}{\delta'}\right)\lt0.25$, (smooth boundary)
2) If $\left(\dfrac {K’}{\delta'}\right)\gt6.0$, (Rough boundary)
3) If $0.25\lt\left(\dfrac {K’}{\delta'}\right)\lt6.0$, (boundary is in transition)
to add an answer. | {"url":"https://www.ques10.com/p/40340/hydro-dynamically-smooth-and-rough-boundaries-1/","timestamp":"2024-11-06T04:44:29Z","content_type":"text/html","content_length":"27411","record_id":"<urn:uuid:0d6ba066-45d5-4ba9-8a25-ee969e5374d4>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00048.warc.gz"} |
Built-in model examples | Concrete ML
These examples illustrate the basic usage of built-in Concrete ML models. For more examples showing how to train high-accuracy models on more complex data-sets, see the Demos and Tutorials section.
FHE constraints
In Concrete ML, built-in linear models are exact equivalents to their scikit-learn counterparts. As they do not apply any non-linearity during inference, these models are very fast (~1ms FHE
inference time) and can use high-precision integers (between 20-25 bits).
Tree-based models apply non-linear functions that enable comparisons of inputs and trained thresholds. Thus, they are limited with respect to the number of bits used to represent the inputs. But as
these examples show, in practice 5-6 bits are sufficient to exactly reproduce the behavior of their scikit-learn counterpart models.
In the examples below, built-in neural networks can be configured to work with user-specified accumulator sizes, which allow the user to adjust the speed/accuracy trade-off.
It is recommended to use simulation to configure the speed/accuracy trade-off for tree-based models and neural networks, using grid-search or your own heuristics.
List of examples
1. Linear models
These examples show how to use the built-in linear models on synthetic data, which allows for easy visualization of the decision boundaries or trend lines. Executing these 1D and 2D models in FHE
takes around 1 millisecond.
2. Generalized linear models
These two examples show generalized linear models (GLM) on the real-world OpenML insurance data-set. As the non-linear, inverse-link functions are computed, these models do not use PBS, and are,
thus, very fast (~1ms execution time).
3. Decision tree
Using the OpenML spams data-set, this example shows how to train a classifier that detects spam, based on features extracted from email messages. A grid-search is performed over decision-tree
hyper-parameters to find the best ones.
Using the House Price prediction data-set, this example shows how to train regressor that predicts house prices.
4. XGBoost and Random Forest classifier
This example shows how to train tree-ensemble models (either XGBoost or Random Forest), first on a synthetic data-set, and then on the Diabetes data-set. Grid-search is used to find the best number
of trees in the ensemble.
5. XGBoost regression
Privacy-preserving prediction of house prices is shown in this example, using the House Prices data-set. Using 50 trees in the ensemble, with 5 bits of precision for the input features, the FHE
regressor obtains an $R^2$ score of 0.90 and an execution time of 7-8 seconds.
6. Fully connected neural network
Two different configurations of the built-in, fully-connected neural networks are shown. First, a small bit-width accumulator network is trained on Iris and compared to a PyTorch floating point
network. Second, a larger accumulator (>8 bits) is demonstrated on MNIST.
7. Comparison of models
Based on three different synthetic data-sets, all the built-in classifiers are demonstrated in this notebook, showing accuracies, inference times, accumulator bit-widths, and decision boundaries.
7. Training on encrypted data
This example shows how to configure a training algorithm that works on encrypted data and how to deploy it in a client/server application. | {"url":"https://docs.zama.ai/concrete-ml/tutorials/ml_examples","timestamp":"2024-11-04T01:38:16Z","content_type":"text/html","content_length":"320721","record_id":"<urn:uuid:baffffb7-c57a-467b-ad46-53bf228c45d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00228.warc.gz"} |
Stock regression time
Use a regression line to make a prediction. why is it 97 and not 96 ?because i did, 20 times 3.8 =76 then you do that plus 20 which is 96 and not 97. Reply. 27 Jan 2017 As William O'Neil points out
in "How to Make Money In Stocks," a turnaround stock should have at least "two straight quarters of sharp earnings 4 Feb 2014 It's also one of the most heavily traded equities on the market, with an
average of more than 90,000 transactions a day during the study period, so
Keran used regression analysis. The procedure was to regress the level or rate of change in stock prices against supposed determinants of stock prices and time. Takahashi et al. [6] proposed a
neural network that embodied a multiple line -segments regression technique to predict stock prices. The tangent and length time lag. 2. If the data are nonstationary, a problem known as spurious
The Regression Model with Lagged Pt is the price of the stock at the end of period t. For example, in the particular 50-day period in the S&P 500 below, the net gain of the market is positive. Yet
the linear regression line is negatively sloped. i.e. the daily values of the stock index are averaged each month, and then used to compute yearly returns which are rolled over monthly. This should
make the The sample period for our regressions begins in 1928, and, unlike the S&P, small -firm price data is not available for a long period prior to that time. Therefore, Regression, Alpha,
R-Squared data point in this graph shows the risk-adjusted return of the portfolio and that of the index over one time period in the past.
One problem in using regression algorithms is that the model overfits to the date and month column. Instead of taking into account the previous values from the point of prediction, the model will
consider the value from the same date a month ago, or the same date/month a year ago.
15 Oct 2018 A linear regression-based prediction approach is used to predict stock exchange indices and companies. We have done time series analysis In the past few decades, most of the stock market
analyses were derived using statistical-time series models, such as, regression, exponential smoothing, ARIMA 26 Nov 2017 3.6 Regression test for 399006. We designed two types of forecasting test.
One is to do fine tune configuration with different input parameters ANNs was used to solve variety of problems in financial time series forecasting. For example, prediction of stock price movement
was explored in [19]. Authors 13 Mar 2019 This paper proposes twin support vector regression for financial time such as information technology, the stock market, the banking sector, Gaussian
Process Regression and Forecasting Stock Trends. The aim of this project was to learn the mathematical concepts of Gaussian Processes and and lnbj,s, the average R2 is 0.40 and 0.67 for cross-stock
regressions with and without an. 7 The twelve-month estimation period follows Amihud (2002). LS use
In the past few decades, most of the stock market analyses were derived using statistical-time series models, such as, regression, exponential smoothing, ARIMA
market data collected for the period of one thousand, two hundred and three days . KEYWORDS: Technical Key Words, Prediction methods, Stock markets, Mean
Keran used regression analysis. The procedure was to regress the level or rate of change in stock prices against supposed determinants of stock prices and
In the past few decades, most of the stock market analyses were derived using statistical-time series models, such as, regression, exponential smoothing, ARIMA 26 Nov 2017 3.6 Regression test for
399006. We designed two types of forecasting test. One is to do fine tune configuration with different input parameters ANNs was used to solve variety of problems in financial time series
forecasting. For example, prediction of stock price movement was explored in [19]. Authors 13 Mar 2019 This paper proposes twin support vector regression for financial time such as information
technology, the stock market, the banking sector,
20 Feb 2013 or decrease) of the 44 shares an average of 61,72 % were achieved during the time period. 2012-02-22 to 2013-02-20. If investing 50.000 SEK
It is noted that past researches usually transformed the stock market price into stationary prior to analysis which may lead to the loss of data originality. Thus, a
For example, in the particular 50-day period in the S&P 500 below, the net gain of the market is positive. Yet the linear regression line is negatively sloped. i.e. the daily values of the stock
index are averaged each month, and then used to compute yearly returns which are rolled over monthly. This should make the The sample period for our regressions begins in 1928, and, unlike the S&P,
small -firm price data is not available for a long period prior to that time. Therefore, Regression, Alpha, R-Squared data point in this graph shows the risk-adjusted return of the portfolio and
that of the index over one time period in the past. Inference in Time Series Regression When the Order of Integration of a Regressor is Unknown. Graham Elliott, James H. Stock. NBER Technical Working | {"url":"https://bestbinaryjfvpm.netlify.app/welburn33647cow/stock-regression-time-mocu.html","timestamp":"2024-11-07T01:07:29Z","content_type":"text/html","content_length":"33133","record_id":"<urn:uuid:f105ad77-6b57-4dc1-9c8c-1d2361a1e9cb>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00507.warc.gz"} |
Possible Uses of Truth Table for AND
Truth Table for AND & its Practical Applications
Every circuit operates on some logic, and studying these logical values requires understanding Boolean Algebra and truth tables. Here, a truth table represents the functional values of a given
mathematical or Boolean logical expression. Understanding this concept leads to a greater understanding of the workings of digital electronic circuits like the famed AND gate. The concept of a truth
table for AND is essential and, thus, must be explored.
Defining Truth Table For AND Gate
The truth table for the AND gate takes note of all possible input combinations and their corresponding outputs. The relationship between input and output values is quite easy to understand.
The AND gate produces a single output by considering at least two inputs. The output is high (1) if and only if both inputs are high (1); otherwise, the output is low (0). Here, the logic is
represented by a ∧ symbol that stands for logical AND.
Even with more than two inputs in an AND gate, the output only comes out high (1) when all the inputs are high (1). Whatever the number of inputs, all values must be high. Otherwise, the output is
low (0).
The truth table for AND can be represented as follows:
A (Input) B (Input) A ∧ B (Output)
Practical AND Gate Applications
There are many practical applications of AND gate truth tables. Still, they are generally not visibly labelled as AND gates, though they function on the same logical principles. These real-life
examples of AND gates extend to routine and technical situations.
Understanding the basic logic of this expression can help simplify and appreciate various sophisticated technologies you come across daily. The effortless inclusion of AND gates in mundane activities
makes life easier, safer, and more convenient for users. Some real-life examples of AND gate truth table include:
1. Security Systems
Dual-authentication security systems are a very common application of AND gate. Many modern devices have security settings where one must go for identity verification through multiple inputs. For
instance, your smartphone might require a facial recognition scan and a passcode before it allows access. The AND gate logic is actually in action here.
2. Industrial Safeguard Systems
AND gates work best in safety and warning systems used in industries and vehicles. For example, you may have to insert a key and do gear pledging before the car engine starts as a dual security
3. Electronic Appliances
Electronic appliances are a great example of AND gate applications in real life.
• Trying to turn on an electronic appliance like a microwave oven or a laptop follows the AND gate truth table logic. You need to turn on the supply switch and press the appliance’s power button as
a second action. You get the right output of a working laptop or oven only when all inputs are correctly on.
• Setting up a television is another great example of the application of the AND truth table. The TV will power on only when there is electricity (Input 1) and the On button is pressed (Input 2).
In an AND gate context, the output is turning on the television. The television will not turn on if either of these inputs fails, making the output zero. Hence, the television operates under the
AND gate logic.
Understanding the “P Not Q” Logic In Context To AND
A crucial logical expression is “P and Not Q,” which combines two conditions: one must be true (P) and the other false (Q). This type of P Q expression is often used in reasoning, problem-solving,
and computer programming. It means that P represents a value that can be true or false. In contrast, Not Q means that another value, Q, is false. The NOT operator negates the value of Q, so if Q is
true, Not Q is false, and vice versa.
When combined with the AND operator, the expression “P Not Q” means that for the entire expression to be true, P must be true, and Q must be false. The whole expression becomes false if either
condition isn’t met: if P is false or Q is true.
This expression finds its way to real-world scenarios. For example, if “P” stands for “I will go for a run,” and “Q” stands for “It is raining,” then “P and Not Q” means “I will go for a run if it is
not raining.” This logical structure is crucial in multiple fields like mathematics, computer science, and everyday reasoning, helping to create clear, precise conditions for actions and decisions.
In Conclusion
In conclusion, the truth table for the AND gate is a fundamental tool for understanding logical conjunctions in Boolean algebra. By systematically displaying all possible input combinations and their
corresponding outputs, the truth table clearly illustrates that the AND operation only yields a true result when both inputs are true. This logic system has great practical applications in daily
1. What are the basic logic gates?
Ans: The logic gates build up the foundation of a digital system. These gates have three different categories:
• Basic gates
• Universal gates
• Special gates
The basic logic gates are NOT, OR, AND gates.
2. What is the AND gate truth table?
Ans: The AND gate truth table shows the different input values for an AND gate circuit or system and the possible outcomes in an easy-to-understand manner.
3. What is the main difference between the AND gate and the OR gate?
Ans: The main difference between the AND gate and the OR gate is that the AND gate multiplies the logical input. In contrast, the OR logic gate adds the logical inputs.
4. What is 3 input AND gate?
Ans: As the name suggests, the 3 inputs mean the AND gate has three inputs and only one output. The output is high only when all the three inputs are high. Otherwise, the output is low if any of the
three inputs is low. | {"url":"https://truthtablegen.com/truth-table-for-and/","timestamp":"2024-11-05T09:32:39Z","content_type":"text/html","content_length":"127573","record_id":"<urn:uuid:299cc9da-330f-4b61-9cc9-f4861cb24565>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00676.warc.gz"} |
Γ-Convergence Analysis of a Generalized XY Model: Fractional Vortices and String Defects
We propose and analyze a generalized two dimensional XY model, whose interaction potential has n weighted wells, describing corresponding symmetries of the system. As the lattice spacing vanishes, we
derive by Γ -convergence the discrete-to-continuum limit of this model. In the energy regime we deal with, the asymptotic ground states exhibit fractional vortices, connected by string defects. The Γ
-limit takes into account both contributions, through a renormalized energy, depending on the configuration of fractional vortices, and a surface energy, proportional to the length of the strings.
Our model describes in a simple way several topological singularities arising in Physics and Materials Science. Among them, disclinations and string defects in liquid crystals, fractional vortices
and domain walls in micromagnetics, partial dislocations and stacking faults in crystal plasticity.
Dive into the research topics of 'Γ-Convergence Analysis of a Generalized XY Model: Fractional Vortices and String Defects'. Together they form a unique fingerprint. | {"url":"https://portal.fis.tum.de/en/publications/%CE%B3-convergence-analysis-of-a-generalized-xy-model-fractional-vorti","timestamp":"2024-11-14T14:51:06Z","content_type":"text/html","content_length":"51644","record_id":"<urn:uuid:e3def16b-d924-47ce-9b00-ae19824f643e>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00116.warc.gz"} |