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Reverse a Linked List from position M to N
Reverse a Linked List from position m to n-Interview Problem
Difficulty: Medium
Asked in: Amazon, Facebook, Microsoft
Understanding the problem
Problem Description: Given a head pointer of a Linked List and two positions say m and n, you have to reverse the given Linked List only from position m to position n. Then, return the head of the m
to n reversed Linked List.
For Example:
Input: 15->20->25->30->35->NULL, m = 2 and n = 4
Output: 15->30->25->20->35->NULL
Input: 20->40->60->70->NULL, m = 2 and n = 3
Output: 20->60->40->70->NULL
Possible follow-up questions to ask: →
• Is the value of m and n always different? ( Ans: No, it can be same too)
Node Structure of the Linked List:
class ListNode
int data
ListNode next
Solution idea
Since we have to reverse a part of the given linked list obviously we need the concept of reversing the linked list. We need to find the m th node and pass it to the reverse function (which will
reverse the given part). But, before passing the m th node we need to traverse to the n th node and cut its link with (n+1) th node if present.
Also we have to save the the (m-1) node and also (n+1) th node address so that we can link the reversed part of the linked list again with the original linked list.
Solution steps
We will use following four variable:
1. revPrev → for storing the previous node, i.e. (m-1)th node.
2. revStart → for storing the starting(mth) node of reversal.
3. revEnd → for storing the ending node(nth) of reversal.
4. revNext → for storing the next node, i.e. (n+1)th node.
Now, we will pass revStart to the reverse function and then will attach the reversed part with revStart and revEnd to get the reversed linked list.
ListNode reverse(ListNode head)
if(head == NULL || head.next == NULL)
return head
ListNode restPart = reverse(head.next)
head.next.next = head
head.next = NULL
return restPart
ListNode reverseFromMToN(ListNode head, int m, int n)
if(m == n)
return head
int count = 1
ListNode curr = head
ListNode revPrev = NULL
while(count < m)
revPrev = curr
curr = curr.next
count = count + 1
ListNode revStart = curr
while(count < n)
curr = curr.next
count = count + 1
ListNode revEnd = curr
ListNode revNext = curr.next
curr.next = NULL
ListNode revPart = reverse(revStart)
revPrev.next.next = revNext
revPrev.next = revPart
head.next = revNext
head = revPart
return head
Complexity Analysis
• Time Complexity: O(L), where L is the length of the linked list.
• Space Complexity: O(1)
Critical ideas to think
• Can you implement the reverse function using iteration?
• Can you think of reversing a linked list using a stack?
• Can we reverse a linked list in less than O(n) time? What if it is a doubly linked list?
Suggested Problems to Solve
• Reverse even elements in a Linked List
• Reverse a Doubly Linked List
• Reverse last K elements of a Linked List
• Reverse a Linked List in groups of given size
Happing Coding! Enjoy Algorithms!! | {"url":"https://afteracademy.com/blog/reverse-a-linked-list-from-position-m-to-n/","timestamp":"2024-11-07T10:45:37Z","content_type":"application/xhtml+xml","content_length":"72877","record_id":"<urn:uuid:365bfdb6-2f33-4043-8280-eeef3c617c66>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00759.warc.gz"} |
Decimal: Python’s Float Trap and How to Solve it
You may know the following surprising code snippet:
a = 0.1 + 0.1 + 0.1
b = 0.3
print(a == b)
# False
The following explanation is from the puzzle-based learning Python workbook:
Get eBook!
This puzzle performs a simple arithmetic computation adding together the float value 0.1.
The question seems to be very simple—but as we’ll see in a moment, it’s not simple at all.
Your inner voice is wrong. And while it is not so important why it’s wrong, it is important that you learn to distrust your intuition and your urge to be a lazy thinker. In coding, assuming that
things are super-simple is a deadly sin.
In the puzzle, you have assumed that the number 0.1 represents the decimal value 0.1 or 1/10. This is natural but a wrong assumption. The value 0.1 doesn’t exist on your computer. Instead, your
computer stores every number in a binary format consisting only of zeros and ones.
Use an online converter to convert the decimal value 0.1 to a binary value and you will realize that you get the following number: 0.000110011001100110011β ¦
The floating point representation of 0.1 in binary has an infinite number of digits. So your computer does the only thing it can do at this moment: limiting the number of digits.
This has the following effect. The decimal number of 0.1 is represented by the closest floating point number 0.100000000000000005551115β ¦ that can be represented in a limited space.
Now, it’s easy to see why 0.1 + 0.1 + 0.1 != 0.3. Thus the answer is False.
As one of my premium members, Albrecht, correctly pointed out, the problem can be fixed with the Python module Decimal:
from decimal import Decimal
a = 0.1 + 0.1 + 0.1
b = 0.3
print(a == b)
# False
c = Decimal('0.1') + Decimal('0.1') + Decimal('0.1')
d = Decimal('0.3')
print(c == d)
# True
You can see that the equality of variables c and d results in the expected value True.
Where to go from here?
Understanding these subtleties is hard. However, it’s also crucial on your path to killer Python efficiency.
A simple fix is to work through the “Coffee Break Python Workbook” until you have developed Python code-reading super-powers!
Leave a Comment | {"url":"https://blog.finxter.com/decimal-pythons-float-trap-and-how-to-solve-it/","timestamp":"2024-11-14T11:52:55Z","content_type":"text/html","content_length":"68436","record_id":"<urn:uuid:847fd490-093d-4459-956d-ed7d17dceb89>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00412.warc.gz"} |
Why We Should Have Kids Write Their Own Math Word Problems
Today we’re back with another look at math word problems. If you’ve been here a while you might have noticed that math word problems keep popping up on here on the blog. We’ve talked about them a few
times because, truth talk incoming, they can be a drag. Math word problems can be incredibly tricky for our young mathematicians, who often have so many questions: “What’s happening in the problem?
How am I supposed to solve it? There are too many numbers… There are too few numbers… She bought how many watermelons?!” You get the point.
So today, we’re going to touch on another way to help kids build up those math word problem-solving skills: writing their own!
But Why Math Word Problems Anyway?
When used thoughtfully and authentically, math word problems can provide a representation of real-life situations. I’m talking about problems that are realistic and that people might actually
encounter. Here’s an example of a problem like that:
Ms. Starr’s class of 21 students was out at recess. She blew the whistle to go back inside. 15 students lined up. How many more students still need to line up?
I’d bet that situation has happened for most of us – just waiting for those last six kids who are dragging their feet (figuratively and literally!) to go inside.
Math word problems like our example above also help students build their creative thinking and problem-solving skills while they practice other computation skills. Students often solve math word
problems such as this one in different ways. When we take things a step further and have students share their methods with peers, this allows them to learn from one another, building their bank of
problem-solving skills.
Even better: students need to use other math concepts to accurately answer the questions. As a result, math word problems are like the full body workout of math – they hit so many muscles!
Let’s look at this example:
Aubrey is playing with her toy cars and wants to line them up for a race. She wants all 18 toy cars to be lined up in three lanes so that each race will have exactly three cars racing. How many races
will Aubrey have?
To solve this problem, students need to be able to visualize what’s happening, understand what is being asked, interpret the numbers in the story, determine what math concept is being described, and
solve the problem. The problem is not just about solving the word problem but also about understanding multiplication and division as arrays and understanding division as related to multiplication.
Math Word problems can be a powerful learning tool that we want students to master. Asking kids to write their own is one way to do that. Before we dive in, let’s take a look at why just learning to
solve math word problems isn’t as powerful as when it’s paired with writing math word problems.
“Just” Solving
As I mentioned earlier, “just” solving math word problems involves students understanding the problem, choosing an operation, and solving it. Kids need to have some understanding of the four
operations, but since there are only 4 options, it’s easy to just pick one, pluck some numbers and solve.
We’ve all come across those students who skim a word problem, pick some numbers, and do something with them. They didn’t take the time to understand the problem and therefore, can’t tell if their
answer is reasonable. Eek! We definitely do not want number pluckers!
The Magic of Writing Math Word Problems
Of course, we’ve also all come across students who think deeply, understand what they are doing, and can even teach it to a peer. As teachers, we want all of our students to reach this level of
proficiency. One tried and true strategy to accomplish this? You guessed it! Writing math word problems.
Deepening Understanding
To write their own math word problems, students need a deeper understanding of the operation they are writing a problem for. For addition or subtraction, they need to understand that addition and
subtraction situations can be putting together, changing, or comparing situations. If they are writing a multiplication or division word problem, they need to understand those operations as arrays,
area, or equal groups.
These understandings themselves are hard work even when only solving math word problems. Students need to take the time to visualize the situation and understand what is happening. While we don’t
need them to be able to name the type of word problem situation, understanding whether something is changing or being compared helps students approach solving the problem with a solid understanding
of the situation.
Creating their own problems that represent these various situations is a great way to deepen students’ understanding of a myriad of problem types. Being able to create something using your own
understanding requires higher-order thinking. Having students create can improve their ability to recognize different problem types later on.
Increased Engagement
Having students write their own math word problems can significantly increase student engagement. Students LOVE to see their names in math word problems. Let me tell you, the smile on those little
faces when they hear their name in teacher-made problems lights up the room!
To take that one step further, don’t just use their names. Use their actual problems! Using the problems that students write in lessons makes math much more personalized. Kids get such a kick out of
seeing their own problems practiced ‘live’. Bonus: they also love to hear what their peers have written.
How To Get Started with Writing Math Word Problems
Wondering how to incorporate this into your already jam-packed math block? Great news: it’s not as hard as it might seem. This doesn’t need to be a full lesson every single time. Instead, I like for
it to be a math routine that kids do with some regularity.
I like to use math journals for various purposes: responding to prompts, including reference sheets or notes, and showing math thinking. Writing math word problems fits right into this work.
Here’s how it might work:
1. Introduce & practice the writing word problem routine: Explain exactly what you expect of students when they write their own problems. Our routine included writing the date so we could see growth
over time. I also told students that their word problem should be related to what we are learning in class so that any of their peers could solve it and that the problem should include a question
at the end that can be answered using the information given in the problem. After writing the problem, students can also create an “answer key” that shows how they would solve the problem and the
correct answer. During this introductory lesson, give students a chance to practice and receive feedback.
2. Get the routine going: Once students have had a chance to practice the routine, writing math word problems is a perfect warm-up activity. Students can write a word problem that directly connects
to the content of the day.
3. Review and use the problems: Now that you’ve got a treasure trove of math word problems – use them! You can easily incorporate student-created math word problems into lessons, leaving the author
as the celebrity of the day. By having students write problems that directly relate to current content, they are easy to include. You can work these into independent practice, as a problem to
solve as a group during a lesson, or even in homework.
One last disclaimer: don’t let the writing part of this routine hold you back! Let kids use speech to text, record their voices, or dictate their problems. Phonetic spelling is great too! It’s the
thinking that is involved in creating math word problems that is the most important. Let the actual writing part take a backseat.
Have you tried having kids write their own math word problems? Have you made it a routine in your classroom? I’d love to hear from you about how you make this work and what results you’ve seen! | {"url":"https://jillianstarrteaching.com/write-math-word-problems/","timestamp":"2024-11-11T17:55:36Z","content_type":"text/html","content_length":"144199","record_id":"<urn:uuid:28ccd502-d293-4067-849e-32ed110c288d>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00577.warc.gz"} |
Technical report: nowcasting UK household income using the new “signature” method
Economic nowcasting refers to generating estimates of the current (“now”) state of the economy. It involves using advanced time series methods to combine information released at a high frequency
(indicators) to estimate what a lower-frequency variable of interest (target) might be doing in real-time.
In this technical report, we generate nowcasts from a range of different methods for quarterly UK household income using a set of within-quarter information (with different publication lags) that are
available at a higher frequency than the current quarterly household income estimates. This research has been useful for the household income team at the Office for National Statistics (ONS), as they
seek the best methods to generate estimates so decision makers can gain the best early indications about the state of the economy.
This report is part of a programme of work that the ONS has been doing with the Alan Turing Institute to explore the usefulness of various economic nowcasting methods, particularly the signature
method. The signature method has been used previously in other disciplines like finance, healthcare, and cyber security. We have been investigating if it could also be useful for economic nowcasting.
Other outputs from our work may also be of interest.
A paper details how the signature method compares against other nowcasting methods mathematically and in various empirical applications, including nowcasting US gross domestic product (GDP) and UK
road fuel prices. There is also a PYTHON code repository which enables others to apply the signature method to generate nowcasts for other variables of interest.
The appetite for nowcasts arises because households, businesses, and policymakers want up-to-date information to make the best decisions. However, it takes time to collect and compile information
that accurately represents the whole economy. Therefore , official estimates of key economic indicators are published with a delay. For example, the ONS currently publishes its first quarterly
estimate of UK household income around six weeks after the end of the quarter it refers to.
However, there is a vast wealth of information about aspects of the economy that can be gleaned from data that are released more quickly and frequently. For instance, the UK is one of a small number
of countries that publish GDP monthly estimates (based on output measures) and also publishes a weekly Economic activity and social change in the UK, real-time indicators bulletin. Others are
available from other sources.
Nowcasts utilise these more frequently released indicators to infer what is happening to a particular variable of interest (for example, household income) in the economy now.
There are two challenges facing nowcasting. The first is the so called “ragged-edge” problem caused by missing values of different indicator variables at the end of the sample period because
of different publication delays. The second is issues arising by use of mixed-frequency data, whereby some explanatory indicators are published at a higher-frequency than the variable of interest. An
example would be a nowcast where the low-frequency target is published quarterly and the high-frequency explanatory indicators are published monthly.
There are a variety of nowcasting methods that handle these challenges in different ways. In this report, we consider three main methods:
□ the bridge method
□ the mixed-data sampling (MIDAS) method
□ the signature method
In addition, we use an autoregressive model as a baseline comparator.
We examine how nowcasts generated using each of these methods perform empirically, compared against the first release of quarterly UK household income. We also examine the data chosen in the models
to gain some understanding of what kind of data could be useful to explain and nowcast household income.
We find that nowcasting household income is challenging. We find that the MIDAS method offers a slight improvement over the autoregressive baseline and one potential advantage is that it captures
some of the movement in the household income realisations, whereas the autoregressive model is a broadly constant nowcast which may be less useful in informing decision making.
Other methods, including the signature method, produce similar performance to the autoregressive baseline and there is limited improvement when more indicators become available. This may be because
the explanatory power of the set of indicators we have explored is limited and varies over time.
The report proceeds by describing the data used in our models in Data and the methods used to compute the nowcasts and select the variables in Methodology. We present the nowcasts and the variables
that are selected in Empirical results before concluding.
1. Data
The household income data we use as our target variable is quarterly growth in our Compensation of Employees time series. It measures the total renumeration, including wages, salaries, and
supplements to wages and salary, earned by employees in return for their work done during an accounting period. According to our UK Economic Accounts time series, compensation of employees made up
about 50% of gross domestic product (GDP) in 2022.
We identify a set of real-time monthly indicators that could be informative about our target. Table 1 provides information about the set of monthly indicator variables and the target we consider. The
first and the second columns give their name and the acronym or codes we use in this report. The third column shows whether the variable is of quarterly (Q) or monthly (M) frequency.
The set of indicators we use include sources used in the compilation of Compensation of Employees statistics gathered from the Office for National Statistics (ONS). These include Average Weekly
Earnings (AWE) and two variables from the Labour Force Survey (LFS):total actual hours (TTH) and gross weekly pay (GRSSWK).
Since household income is a large component of GDP, we also investigate variables that have been considered in GDP nowcasting literature. In particular, we make use of the dataset produced by Anesti,
Galvão and Miranda-Agrippino’s nowcasting paper (2020) (PDF, 821KB. We select from their dataset the real-time data series of Retail Sale Index (RSI), Index of Production (IOP), Index of Services
(IOS), and Manufacturing Production (MPROD). As their real-time dataset contains data up to the last quarter of 2018, we update the real-time vintages using data from the ONS to cover the whole of
As in Anesti, Galvão and Miranda-Agrippino’s nowcasting paper (2020) (PDF, 821KB), we also include Mortgage Approval (MTGAPP), Net Consumer Credit (NCC), and Agent Score (AS) in our set of
indicators. We also collect the most recent data series directly from the Bank of England. These are static series, and their past values are not revised. The Agent Score series we use is Total
Business Services. In addition, our set of indicators include Claimant Count (BCJA) and Sterling Effective Exchange Rate (SEER) data from the ONS, the House Price Index (HPI) from HM Land Registry
and the Consumer Confidence Index (CCI) of the UK from the OECD. Hence, the set of indicators we use for our analysis provides information about the following:
• household income components
• labour market condition
• production
• services and sales
• housing market
• credit and borrowing
• purchasing power
• general sentiment of confidence
Understanding the publication lags and timeliness of monthly indicators is crucial in nowcasting, as it determines which nowcast horizons should be considered. We establish three nowcast horizons:
t+0 day, t+15 days and t+45 days where t denotes the end of the reference quarter for which we generate a nowcast. The last three columns of Table 1 indicate how much in-quarter information is
available for each monthly indicator variable.
Our first nowcast horizon (t+0 day) is at the end of the reference quarter, so for example, an estimate for UK household income in the first quarter of a year would be generated at the end of March.
At this horizon, most of the monthly indicators we consider only have their first month estimates (within the reference quarter) available.
The next nowcast horizon we consider is 15 days after the end of the reference quarter, so for example, an estimate for UK household income in the first quarter of a year would be generated in
Our last nowcast horizon (t+45 days) is 45 days after the end of the reference quarter, so for example, an estimate for UK household income in the first quarter of a year would be generated in
mid-May. This horizon is designed to be equivalent to the timing of the first official estimate.
New information about the monthly indicators become available as one moves through the nowcast horizons because of publication lags. In Table 1, if we were to focus only on the indicator variable
Index of Production (IOP), at nowcast horizon t+0 day, data has been published about the first month of reference quarter(month 1, m1). By nowcast horizon t+15 days, data for the second month of the
reference quarter (month 2, m2) has been published and by nowcast horizon t+45 days, data is available about all three months of the reference quarter. As well as including the information for a new
month, there may also be revisions to previous estimates.
The timeliest indicator is CCI, whereby the first nowcast horizon (t+0 day) there is already information about all three months within the reference quarter. By nowcast horizon t+15 days, we see from
Table 1 that the estimates of all the real-time indicators for the second month have been published. The estimates of Claimant Count for the three months within the quarter are also available.
At nowcast horizon t+45 days, we have estimates for all three months within the reference quarter for all variables. The only exception is that since August 2016 there is no longer an Agent Score
(AS) produced for the third month of the reference quarter because of a change in the Monetary Policy Committee meeting schedule (see Bank of England’s Definitions for the Agents’ scores). Given that
the most recent in-quarter dataset at this horizon is available for indicator variables, predictions are expected to be the most accurate. However, as nowcasts at this horizon are close to the
publication of household income estimates for the reference quarter, the timeliness of the prediction may not be as useful.
For household income and most monthly indicators, we compute the percentage change between reference months. The exceptions are Agent Score (AS), where we use the values at levels, and Net Consumer
Credit (NCC), where we scale the dataset according to the quartile range. Both variables already represent the change relative to the period before, so we have no need to use any transformation that
would force this data to be stationary, such as the percentage change. The percentage change transformation is used to achieve stationarity and to help us nowcast the quarterly growth rate.
We nowcast the quarter-on-quarter growth of household income using the set of transformed indicator variables.
Table 1: The availability of monthly information across nowcast horizons
Data Acronym/Code Frequency t+0 days t+15 days t+45 days
Target variable:
Household income DTWM Q
Monthly indicator variables:
Consumer Confidence Index CCI M m1, m2, m3 m1, m2, m3 m1, m2, m3
Industrial Production IOP M m1 m1, m2 m1, m2, m3
Average Weekly Earnings AWE M m1 m1, m2 m1, m2, m3
LFS gross weekly pay LFS_GRSSWK M m1 m1, m2 m1, m2, m3
LFS total count hours LFS_TTH M m1 m1, m2 m1, m2, m3
Claimant Count BCJA M m1, m2 m1, m2, m3 m1, m2, m3
House price index HPI M m1 m1, m2 m1, m2, m3
Manufacturing Production MPROD M m1 m1, m2 m1, m2, m3
Index of Services IOS M m1 m1, m2 m1, m2, m3
Retail Sales Index RSI M m1, m2 m1, m2 m1, m2, m3
Mortgage approval MTGAPP M m1, m2 m1, m2 m1, m2, m3
Net Consumer Credit NCC M m1, m2 m1, m2 m1, m2, m3
Sterling Effective Exchange Rate SEER M m1 m1, m2 m1, m2, m3
Agent Score (Total Business Services) AS M m1, m2, m3 m1, m2, m3 m1, m2, m3
Our estimation sample covers the period 2005 Quarter 1 (Jan to Mar) to 2014 Quarter 1, and the nowcast evaluation sample covers the period 2014 Quarter 2 (Apr to June) to 2019 Quarter 4 (Oct to Dec).
There are 23 quarters in our evaluation sample. We produce recursive nowcasts of the first estimates of UK household income using the latest information available for the monthly indicator at the
time when we need to produce a nowcast. It is worth noting that we omit the coronavirus (COVID-19) pandemic in our evaluation sample. It is because such an event may alter the relationships learnt by
the models during the estimation sample. In the next section, we describe the set of models we use to nowcast UK household income.
2. Methodology
In order to nowcast UK household income, we employ nowcasting models that can accommodate mixed frequency data and handle missing data issues and are widely used in macroeconomic nowcasting
literature. These models include the bridge method with Autoregressive Distributed Lag (ARDL) and the Mixed Data Sampling (MIDAS) model. We compare their empirical performance with the nowcasts
produced using the signature methods. To benchmark these models, we implement an autoregressive model with one lag and a constant term as a comparison, detailed in section Autoregressive benchmark.
In this section, we give an overview of the nowcasting models we use. We also discuss how we arrive at a subset of indicators that are useful to explain household income as selected by shrinkage
methods, and how we compute nowcast recursively throughout the nowcasts evaluation period.
Nowcasting models
The nowcasting methods used in our research overcome key challenges in nowcasting in different ways. These include the ragged-edge problem, where there may be missing data at the end of the time
series because of publication lags, and the mixed frequency problem, where we are trying to predict a quarterly outcome using monthly explanatory indicators.
Bridge method
The bridge method overcomes these issues by taking a two-step regression approach to the nowcasting problem. In the first step, an autoregressive model is implemented to fill any missing values in
the explanatory indicators (z[t]) that are present because of publication lags (1). When the dataset is complete, the high-frequency data is aggregated to the same time-period as the low-frequency
outcome variable (2), in our case, from monthly to quarterly; if x[t] is a quarterly average of z[t], then for all s we have γ[s] = 1/3. The second step regresses the low-frequency target (y[t]) on
the aggregated indicator variables (x[t]) alongside an autoregressive process in the target variable (3). \( \)
$$ z_{t} = \delta_{0} + \sum_{s=1}^{p} \delta_{s}z_{t-s}+\eta_{t} \ \ \ (1)
x_{t} = \sum_{s=0}^{2} \gamma_{s}z_{t-s} \ \ \ (2)
y_{t} = \sum_{s=1}^{p} \alpha_{s}y_{t-3s} + \beta_{0} + \beta_{1}x_{t} + \varepsilon_{t} \ \ \ (3) $$
MIDAS model
MIDAS is a method commonly used for nowcasting. MIDAS overcomes the mixed-frequency and ragged-edge problem by generating a single parameter for each high-frequency explanatory indicator, for each
low-frequency outcome period, using the most recent indicator data and its lags available in that period. The parameter is generated using a polynomial lag function, which applies weights to the most
recent value and its lags (see equation (5)). Once parameters have been generated to overcome dimensionality issues, a non-linear least squares model (4) can be applied to produce estimates for our
outcome variable. In this piece of research, we apply an exponential almon lag, as shown in (6), to our data to generate the parameterised data.
$$ y_{t} = \sum_{s=1}^{p} \alpha_{s} y_{t-3s} + \beta_{0} + \beta_{1}x_{t} + \varepsilon_{t} \ \ \ (4)
x_{t} = \sum_{s=0}^{p} \gamma(s, \theta) z_{t-s} \ \ \ (5)
\gamma(s, \theta) = \frac{exp(\theta_{1}s + \theta_{2}s^{2})}{\sum_{j=0}^{p}exp(\theta_{1}j + \theta_{2}j^{2})} \ \ \ (6) $$
For a more detailed overview of both bridge and MIDAS nowcasting methods, see Schumacher’s discussion paper (2014) (PDF, 550KB).
Signature method
Signatures are mathematical objects (iterated integrals) of paths or continuous time series, which capture useful geometric information. It has universal approximation properties; linear regression
on signature terms can describe any relationships between the indicators and the target variable, including nonlinear relationships. The path signature has an infinite number of terms, but because
they are decreasing rapidly (at a rate faster than exponential) in size, we are theoretically justified to take a truncated form for practical purposes. This truncation typically occurs in “levels”,
which is the number of integrals taken to compute the term. Each integral can be taken with respect to any of the indicator variables, so if there are d indicator variables, then there are d^n
signature terms at level n. We refer to the family of methods using truncated signatures in regression as the “signature method”.
While this method has been applied across other disciplines, it is largely unexplored in economics. The signature method handles the nowcasting challenges of missing and mixed frequency data by first
embedding the observed data in continuous time through interpolation techniques. We then find the signature terms up to a specified truncation level and use these terms as explanatory variables in a
linear regression to nowcast the target variable. \( \)
$$ Y_{t} = \sum_{k=0}^{K} (\alpha_{k} + \beta_{k}Y_{t-})\varphi_{k,t} + \varepsilon_{t} \ \ \ (7) $$
Where Y[t] is some prior observation of the low-frequency target at the point of nowcast. φ[k,t] is a sequence (for each value of t) of signature terms at truncation level k, which includes the
iterated integrals of t alongside components of the observed process X (for instance, the high-frequency observed explanatory variables), within some specified rolling window. α[k] and β[k]
represent vectors of regression coefficients.
A more in-depth introduction of path signatures and its properties is given by Chevyrev and Kormilitzin in their Primer on the Signature Method (2016) (PDF, 1,183KB). To see further explanation of
the signature method and its uses within the context of nowcasting, see the paper produced by this programme of work [LINK] (which also contains a more detailed explanation of bridge and MIDAS too).
Autoregressive benchmark
We benchmark nowcasts produced by these models against the forecasts of household income produced by an autoregressive model (AR). We examine the appropriate number of lags to be included in the AR
model using the Akaike Information Criterion (AIC) and find that the first order autoregression, that is, AR(1), is most appropriate. Therefore, we use AR(1) forecasts as a benchmark to allow us to
determine if within-quarter information is useful to predict the target.
Variables selection
Using a robust subset of information to improve model performance is a strategy widely adopted in macroeconomic analysis. Potential methods used for choosing a robust subset include model selection
statistics (such as adjusted R^2, Information Criteria, tests for individual or joint significance of variables), stepwise selection methods and shrinkage methods. These methods are well-established
in the time series literature and widely used in empirical analysis for finding an informative set of indicators to explain the variable of interest.
Our indicators described in the Data section are considered potentially useful in explaining household income based on empirical understanding and evidence shown in existing literature. However,
whether they remain relevant in explaining and predicting household income over time is an empirical question. We adopt shrinkage methods to reduce the set of indicators listed in Table 1 to a subset
and to optimise the bias-variance trade-off.
We explore the use of shrinkage methods (LASSO and Elastic Net), as well as random forests for variable selection. In the context of penalisation, regression minimises the sum of the residual sum of
squares. LASSO and Ridge add an additional penalty term. For LASSO, this is an L^1 norm of the parameters, where shrinking this has the effect of setting some parameters to zero, which is useful in
the context of variable selection. For Ridge, this is an L^2 penalty, which can result in the parameters becoming very small as the penalty term increases.
Elastic Net is a combination of the above, creating reduced models by both eliminating variable coefficients (as LASSO does), and reducing variable coefficients (as Ridge does). We can control how
much of the Elastic Net penalty is a combination of L^1and L^2, but given that we observe a similar performance when we test across a range of values that scale between the two regularisation
methods, we fix this mixing parameter to the default: the in-between value of the two.
For a more detailed explanation, see Zou and Hastie’s Regularization and variable selection via the elastic net paper (2005) (PDF 324KB).
We also use random forest for pre-selection before we apply LASSO or Elastic Net. Random forests provide an alternate approach to shrinkage when searching for an important set of indicators to
predict the target. One potential benefit is the handling of non-linear relationships. Shrinkage is typically a linear model, whereas random forests are more capable of handling the more complex,
non-linear patterns exhibited between the target and predictors. Where LASSO suffers from multicollinearity issues, random forests consider the interactions between correlated variables, should they
exist. For more explanation about using random forests in the context of variable selection, see Genuer, Poggi and Tuleau-Malot’s Variable selection using Random Forests paper (2012) (PDF, 269KB).
As shown in Piironen and Vehtari’s Comparison of Bayesian predictive methods paper (2016) (PDF, 1877KB), complex models can be simplified through a so-called “projection” method. We take a very
simple approach in the use of random forest here. A subset of useful features is generated from the full variable set using a random forest regression (for instance, variables that lead to a
reduction in impurity are assumed to provide more information gain, and so increase the predictive accuracy of the model).
In this context, we have applied random forest as a pre-selection method, passed on this variable subset to Elastic Net to estimate the quarterly equation in the bridge method for each of the
recursive nowcasts. That is, we re-estimate the quarterly equation using random forest in conjunction with a shrinkage method when we produce a nowcast for every quarter (See Recursive nowcasts
The optimal tuning parameters are found through cross-validation using ordered splits that respect the time series component of the data. Hence, we let the data decide what indicators are to be used
to produce a bridge nowcast for every quarter, and this is done for every horizon. To make MIDAS nowcasts comparable with the bridge nowcasts, we use the same set of indicators chosen by the
shrinkage method through the bridge method for each quarter to produce MIDAS nowcasts. We tried out both LASSO and Elastic Net, but found that they give similar results. Therefore, we only report the
results obtained from using Elastic Net in the next section.
For the signature method, the number of signature terms increases exponentially as the truncation level increases. To compute signatures where there is a large number of indicator variables, we can
utilise dimension reduction techniques, like principal component analysis. We incorporate this pre-processing step to effectively summarise the data using a lower dimension so that we can take a
higher truncation level for signatures. For regression on the signature terms, shrinkage is also useful to avoid overfitting. Therefore, we also use Elastic Net for signature regression.
Recursive nowcasts
Recursive nowcasts are where the models are re-estimated as time moves forward (each quarter, in our case). We compute recursive nowcasts using the models described in the Nowcasting modelssection.
For each quarter within the nowcast evaluation period, we first conduct variable selection as described in the Variables selection section. By combining variable selection and recursive nowcasting,
not only do we allow the models to decide which subset of indicators best explain the target variable, we also allow the parameters of the monthly indicators in the models to be updated.
The use of recursive nowcasts allows for possible changes in the relationship between the indicators and the target better than if we were to use a static nowcast. Where the relationship is
unchanged, we expect the estimated coefficients to remain the same even if they are estimated recursively. We also use an expanding estimation window for our nowcasting exercise. Hence, the amount of
information we use for each nowcast increases as we move across time.
3. Empirical results
We present our findings in this section. We first explore how the nowcasts produced using the methods described in the Methodology section perform empirically. We then look further into these
nowcasts to better understand the indicators that are selected (see the Selected subset of indicators section).
Nowcast using monthly indicators
We produce nowcasts using a robust subset of indicators selected using the methods explained in the Variables selection section. We have previously tested models with and without variable selection
across all horizons, the results of this showed that applying variable selection improved nowcast accuracy across all horizons in general.
Nowcasts are produced across three different horizons. We focus on the ones that are produced at t+15 days, as a reasonable amount of within-quarter information has been accumulated (see the
second-to-last column of Table 1) at this point. Moreover, it would still provide early insights to economic analysts and policymakers, being around one month before official household income data
are first released.
Figure 1 shows the household income nowcast produced at t+15 days where t denotes the end of a reference quarter, across the nowcast evaluation period. The realisations, shown as the large black dots
in figure 1, are plotted alongside the nowcasts produced using the methods we consider. Figure 2 shows the difference between the nowcast of each model and the first release of the household income
data. Both the bridge and mixed-data sampling (MIDAS) nowcasts are more volatile, capturing the movement and magnitude of the realised values of household income at times.
In contrast, the benchmark AR(1) and signature model both assume a very flat path through the centre of the realised values. Upon inspection of the benchmark AR(1)regression coefficients, this is
almost entirely driven by the constant term, with the autoregressive term being very weak in terms of significance and coefficient magnitude.
Looking across different quarters, we find that the MIDAS method offers a slight improvement over an autoregressive baseline in nowcasting the first release of household income. We report the Root
Mean Squared Error (RMSE) and the Mean Absolute Error (MAE) of the predictions produced by the four models in Table 2. We also visualise these statistics in Figure 3 (RMSE) and Figure 4 (MAE). Other
methods, including the signature method, produce similar results to the autoregressive baseline or are not consistently better across all horizons.
The MIDAS method is the only nowcast that consistently (although marginally) beats the benchmark AR(1) across all horizons. The difficulty in improving upon a largely constant AR(1) benchmark may be
because the explanatory power of the set of indicators we have explored is limited and varies over time. Although the average performance of the MIDAS method is only slightly better than the AR(1) in
our evaluation period, the benefit may be larger during periods of rapid change. This is because the MIDAS method might capture some of the movement in the household income realisations, unlike the
broadly constant autoregressive model nowcast.
These test results indicate that the inclusion of additional information across time horizons does not improve nowcasting accuracy for any of the models. Whilst we see an improvement for both MIDAS
and the signature method when moving from t+15 days to t+45 days, the best performing horizon is consistently t+0 day across all models. Note that the nowcasts across horizons are computed not only
with an increasing amount of within-quarter information, they are also computed using newly estimated parameters from an expanding estimation window. Whether these results are sensitive to the length
of the estimation window is something that could be checked with further research.
To better understand these nowcast results, we plot the adjusted R^2 of the bridge nowcast model (recursively) estimated over the estimation sample (with expanding window) against the absolute
nowcast errors for t+15 days for illustration. This is shown in Figure 5. Despite formulating a parsimonious nowcasting model by using shrinkage methods and utilising all the latest information
available, it is not necessarily true that accurate nowcasts will be produced.
We have utilised cross-validation in the variable selection stage to guard against the prospect of over-fitting. However, a good in-sample model fit does not always guarantee a good out-of-sample
predictive performance, particularly when our out-of-sample prediction consists of a single data point, which is itself expected to be noisy. Model and variable selection can only ever be based on
past information. From there, we establish (conditional) relationships between the target variable and the indicators and use such relationships (as reflected by estimated coefficients) to compute
nowcasts. However, if the power of indicators is weak, or such relationships vary over time, it becomes more difficult to obtain an accurate prediction.
In the next subsection, we look deeper into the indicators that are selected by the variable selection methods to form the nowcasts. By doing so, we can shed some more light on the empirical
performance of these nowcasts.
Figure 1: Nowcast model predictions against realisations at horizon t+15 days
Figure 2: Nowcast model errors at horizon t+15 days
Table 2 Nowcast evaluation statistics
t+0 days t+15 days t+45 days
AR(1) 0.523 0.523 0.523
Bridge 0.451 0.505 0.553
MIDAS 0.371 0.440 0.404
Signature 0.489 0.558 0.556
AR(1) 0.395 0.395 0.395
Bridge 0.352 0.418 0.428
MIDAS 0.293 0.371 0.334
Signature 0.369 0.422 0.415
Figure 3: RMSE values for nowcasts across nowcast horizons
Figure 4: MAE values for nowcasts across nowcast horizons
Figure 5: In-sample adjusted-R^2 and absolute nowcast errors at horizon t+15 days for the bridge method
Selected subset of indicators
Using results from the bridge method, we look at how often individual indicators are selected, and which groups of indicators are chosen together under the shrinkage methods in the Variables section.
As described in the Recursive nowcasts section, we recursively estimate the nowcast model so there can be a different set of indicators selected at each quarter and each nowcast horizon. Since we
find that both LASSO and Elastic Net, each used in conjunction with random forests, give similar results, we report the results we obtain using Elastic Net with the penalty ratio set to equally blend
the L^1 and L^2 penalties. The results presented in this subsection are derived from bridge method, but as explained in the Variables selection section, the same set of selected indicators are also
used to produce MIDAS nowcasts.
Figure 6 shows the frequency of individual variables being chosen by the combination of random forest and Elastic Net and used for income nowcasts at different horizons. Table 3 shows the most
frequently selected groups of indicators. In all cases, models will include a constant.
The top five indicators chosen by Elastic Net for bridge nowcasts are Average Weekly Earnings (AWE), Index of Services (IOS), Retail Sales Index (RSI), House Price Index (HPI) and Manufacturing
Production (MPROD). AWE, IOS, and RSI are chosen in over half of the models across each horizon. It is not surprising to see that these top five indicators are also included frequently in the top
five subsets of chosen indicators, as shown in Table 3. Overall, the findings from our variable selection exercise shows that information about weekly household earnings, services and manufacturing
output, retail sales, and house prices are the most useful in explaining changes in household income.
We observe that while some indicators are often being chosen as components of the subsets, the actual subset that is chosen by the shrinkage method varies in each recursive estimation. If
relationships between the target variable and the indicators are stable, we expect to see the same set of indicators to be constantly chosen in the recursive estimation. The fact that we observe such
variation in the subset of indicators implies that the explanatory power of the indicators varies across time. Moreover, we find that the adjusted R^2 from the bridge method averages 0.44.
These findings provide some potential reasons as to why we do not observe an appealing empirical performance of the nowcasts produced. Firstly, the indicators we consider are not particularly strong
in nowcasting household income; secondly, the usefulness of indicators change over time, so the variables that are selected based on in-sample performance may not be useful in nowcasting household
income out-of-sample. The uncertainty of the usefulness of indicator variables across time may make prediction of household income challenging.
Figure 6: Proportion of quarters in the evaluation sample that a variable is selected across all horizons
Table 3: Number of times the top five subsets of indicators chosen by Elastic Net are selected across quarters in the evaluation sample
Variable Subset t+0 t+15 t+45 Total
AWE, IOS, MPROD, RSI 2 3 3 8
AWE, CCI, IOS, NCC, RSI 2 2 2 6
AWE, HPI, IOS, RSI 2 1 3 6
AWE, HPI, IOS, SEER 2 2 2 6
AWE, IOS, MPROD 2 1 2 5
4. Conclusion
In this technical report, we develop nowcasts of UK household income using data that covers information about different parts of the economy and a selection of time series methods.
This report is part of a programme of work that the Office for National Statistics (ONS) has been doing with the Alan Turing Institute to explore the usefulness of various economic nowcasting
methods, particularly the signature method. Other outputs from our work may also be of interest. A paper released contains more information on the how the signature method compares mathematically
against other methods and details its empirical performance in other settings, including nowcasting US gross domestic product (GDP). There is also a PYTHON code repo, which enables others to apply
the signature method to generate nowcasts for other variables of interest.
We find that nowcasting UK household income is challenging. We find that the mixed-data sampling (MIDAS) method offers a slight improvement over an autoregressive baseline. One potential advantage is
it manages to capture some of the movement in the household income realisations, whereas the autoregressive model is a broadly constant nowcast, which may be less useful during periods of rapid
change in household income.
Other methods, including the signature method, produce similar results to the autoregressive baseline and there is limited improvement as data on more indicators become available. The signature
regression method has had a stronger relative performance in other contexts we have studied as part of this programme of work (see paper).
Although we find that the variables in use do not contain strong information for predicting household income, we do find a pattern of core variables that show up frequently throughout the evaluation
period. The explanatory variables that seem to have the most power are Average Weekly Earnings (AWE), Index of Services (IOS), Retail Sales Index (RSI), House Price Index (HPI), and Manufacturing
Production (MPROD). Of these, AWE, IOS and RSI are chosen in over half of the models across each of the horizons. However, the combinations of the subset of indicators used to produce nowcasts are
changing overtime.
This research has been useful for the household income team and the ONS, as they seek the best methods for the development of their estimates so policymakers can gain the best early indications about
the state of the economy.
5. Author affiliations
Samuel N. Cohen also affiliates to the University of Oxford
Silvia Lui also affiliates to the Economic Statistics Centre of Excellence
Giulia Mantoan also affiliates to the Office for National Statistics,
Lars Nesheim also affiliates to the University College London
Aureo de Paula also affiliates to the University College London
Lingyi Yang also affiliates to the Office for National Statistics and the University of Oxford.
6. Disclaimers
The views expressed are those of the authors and may not reflect the views of the Office for National Statistics or the wider UK government.
Any views expressed are solely those of the author(s) and so cannot be taken to represent those of the Bank of England or to state Bank of England policy. This paper should therefore not be reported
as representing the views of the Bank of England or members of the Monetary Policy Committee, Financial Policy Committee or Prudential Regulation Committee. | {"url":"https://datasciencecampus.ons.gov.uk/projects/technical-report-nowcasting-uk-household-income-using-the-new-signature-method/","timestamp":"2024-11-04T12:04:47Z","content_type":"text/html","content_length":"96332","record_id":"<urn:uuid:83e4501b-8f19-4829-8f53-800f0e1db650>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00498.warc.gz"} |
If is a homogeneous function of order. i) ii) iii)-Turito
Are you sure you want to logout?
Which of the above statements is correct ?
A. Only I
B. Only II, III
C. Only I and III
D. I, II, III
We are given a homogeneous function u. It is a function of x and y. We are given three statements. We have to find which of the statements is true.
The correct answer is: Only I and III
We are given that u is a homogeneous function of degree n and it is function of x and y. We will check all three statements one by one.
I. For first statement, we will take the partial derivative of u w.r.t x and y. We are using Euler's theorem to write the derivative.
Euler's theorem states that if u is homogeneous function of x and y having degree n, then we can write
The first statement is correct.
II. For second statement, we need second order derivative. We will take the partial derivative of equation (1) w.r.t x and then y.
Take partial derivative w.r.t x
Multiply the equation (2) by x
Similarly, take the partial derivative of equation (1) w.r.t y.
Multiply equation (4) by y
Adding (4) and (5) we get,
From (1) we get
So, statement "II" is incorrect.
III. See equation no (2).
We can say that statement III is correct.
So, the option which says "only I and III", is the right option.
We have to remember all the three results instead of deriving them.
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Additional libraries
This page describes a set of useful Analytica function libraries. These are additional to the Standard libraries that are automatically installed when you install Analytica. If you want to use any of
these additional libraries, click on its link and it will download onto your computer. Depending on your download settings, it may automatically open it in Analytica. Either way, you should save it
into a directory so that you can add it into your models. It's usually most convenient to save it with your Standard libraries in directory
"C: \Program Files\Lumina\Analytica 6.4\Libraries"
You can then easily add it into your models using the standard menu option Add Library... from the File menu.
See User-defined Functions and Libraries to build your own functions and libraries.
Misc. libraries
Large Sample Library
Download: media:Large_Sample_Library_v10.ana
The Large Sample Library is an Analytica library that lets you run a Monte Carlo simulation for large models or a large sample size that might otherwise exhaust computer memory, including virtual
memory. It breaks up a large sample into a series of batch samples, each small enough to run in memory. For selected variables, known as the Large Sample Variables or LSVs, it accumulates the batches
into a large sample. You can then view the probability distributions for each LSV using the standard methods — confidence bands, PDF, CDF, etc. — with the full precision of the large sample.
See Large Sample Library: User Guide.
The Sensitivity Analysis Library
The Sensitivity Analysis Library provides functions for analyzing the sensitivity of an output to each cell of each array-valued chance input, and locating those individual scalar inputs that have
the greatest impact on the result. See The Sensitivity Analysis Library for documentation on using this library.
The library itself can be downloaded from Sensitivity Analysis Library.ana, and an example model to demonstrate its usage is at Sensitivity Functions Examples.ana.
Model Documentation Library
Download: Model Documentation Library.ana
Prerequisites: Analytica 5.0 or later. The Domain acts as self index legacy preference must be OFF. You must be using ClearType text (not the very old non-clearType legacy setting).
Some people like to create a concise report on paper that contains all objects in their model along with descriptions or other selected attributes. The "Print Report" feature in Analytica can be used
for this purpose, but is not at all concise and ends up placing each object window on a separate page. The Model Documentation Library allows you to select a module from your model and produce a
result table containing every object within that module with its selected attributes such as title, description, units, definition, or user-defined attributes. This table can then be exported to
Excel where you can format it nicely and print it. Thus, you can end up with a very concise report on paper.
To use this library, load your model and then select File → Add Module.... Add the Model Documentation Library.ana module file using Embed. In that module, using the pulldown, select the top-level
module for the report. Follow the instructions shown on the diagram. If you make changes to your model later, press the Update Model Documentation button to adjust the pulldown content.
Download: Units conversion library.ana
These functions convert between different units -- for example, from feet to kilometers, or from Btu (British thermal units) to gigajoules of energy. They relieve you from having to look up
conversion factors. They also make your model more transparent by making it clear where you are converting from one units to another -- instead of just embedding conversion constants in the formulas.
You can use the Conversion example section to enter a quantity, select its type (dimensions) and units, and convert it to another unit of that type.
Library functions
Units_factor(oldUnits, newUnits) to give a conversion factor between two units. If you omit newUnits, it will assume the standard units for that type (dimensions). The units parameters text values
must match an abbreviation, synonym, or full name (case insensitive) in the Units data table. If the units are of different dimensions, e.g. energy and power, or if it doesn't recognize the units, it
will give an error.
Units_conversion(x, oldUnits, newUnits) returns the value of quantity x converted from oldUnits to newUnits. This works for units like temperature, where simple factors don't work.
Units_real_dollars(yr, baseYr) returns a multiplication factor used to convert nominal dollars in year yr to real (also known as constant) dollars using baseYr as the base year. If you omit baseYr,
it uses the "Standard base year for inflation" which you can set in the Units Conversion Library user interface. It uses consumer price index data provided by the US Energy Information Administration
from 1949 to 2017 with projections up to 2040. It gives an error if yr or baseYr are outside that range.
The Units_table contains the list of units, with abbreviation, full-name, and synonyms, type (dimensions) and conversion factors to the base unit for each type. For example, the base unit for
'Energy' is Joule. The conversion factor for gigajoule (GJ) is 10^9. You can add your own units. Be careful to select the correct type (dimensions).
Greatest Common Divisor functions
Download: GCD function library.ana
This library contains two User-Defined Functions for computing the greatest common divisor.
Download: Database library.ana
A library of functions to make it easy to create, read, update, examine and delete database tables, without having to write SQL. It works with Microsoft Access, Microsoft SQL Server, MariaDB, and
PostgreSQL, and can be expanded to support other database platforms.
DB Conversion Library
Download: DB conversion lib.ana
Lets you embed data obtained from a database into the Analytica model. This breaks the need to use it with an external database, so -- for example -- you can send to someone (e.g. Lumina tech
support) who does not have access to the database or has an edition of Analytica (e.g. Professional) that doesn't support database access.
Press a button in this library to transform all the variables defined using DbQuery, DbLabels and DbTable to literal data. Variables and indexes defined using DbQuery or DbLabels are transformed to
list definitions, and those defined using DbTable are transformed into edit tables. This breaks the connection to the external database. So, of course, the model will no longer be able to get new or
extended data from the database.
The library is limited in its scope. It only works when all calls to DbQuery, DbLabels and DbTable occur at the top level of variable definitions. Do not try to use it if the calls to these functions
are embedded within larger expressions or in User-Defined Functions.
Use with extreme caution!!: Make a copy of your original model before adding and executing this module. After running the transformation, be sure to use File → Save As... to save the transformed
model under a different filename, so you don't clobber your original model.
Spreadsheet Helper Library
Download: Spreadsheet Helper lib.ana
Includes functions that return the list of worksheet names and the Excel filename of the workbook. It also contains a SpreadsheetOpenEx function that can be used in place of SpreadsheetOpen, but
which remembers the filename selected by the user so the next time they don't have to select the file again.
Data Standardization Library
Download: Data Standardization Library.ana
Imported data is often inconsistent. This library allows you to choose what the "standard" values should be in a column of data. You can then map any non-standard value to one of the standard values.
The result is a column of consistent data.
Additional probability distributions
Laplacian distribution
Download: Laplacian distribution library.ana
Description: The Laplacian distribution (or Laplace distribution) is a continuous, unbounded distribution that is essentially a double-sided Exponential distribution with a «mean» parameter. The
library contains the distribution function, and the analytica functions for density, cumulative density and inverse cumulative density (the quantile function).
The HDR pseudo-random number generator
Download: Hubbard HDR Library.ana
Authors: Doug Hubbard, Hubbard Decision Research, and Lonnie Chrisman, Lumina Decision Systems
Description: The HDR is a simple, open-source pseudo-random number generation algorithm. It has been embraced by organizations such as ProbabilityManagement.org as a standard for efficiently
exchanging exact Monte Carlo samples. Given 4 integer seed values, the sequence is deterministic, so two parties with these same 4 values can reproduce precisely the same sequence up to a length of
100M samples, even from Excel. Select a different combination of seeds to prevent any two independent simulations from inadvertently using the same random number stream. This ensures that independent
simulation results can be combined without unintended correlations. Or use the same combination of seeds to intentionally produce identical simulations of the same stream. This allows entire Monte
Carlo simulation samples to be communicated in just a few numbers.
This library provides the functions for using HDR in an Analytica model. The Hubbard_HDR(...) function enables you to obtain the nth random number in a sequence, while the HDR_U(...) function acts as
a Uniform(0,1) distribution as is convenient to use in any Inverse Cumulative Probability Function (e.g., CumKeelinInv, CumNormalInv, CumBetaInv, etc) to encode a specific distribution inside a
Individual functions with examples
Download: Recumulate example.ana : Like Cumulate, but resets to zero at selected points.
Download: Apportion.ana: Aggregates or disaggregates an array of values from one index to another index, where the second index may be larger or smaller. Does not require the indexes to be index
multiples of each other. For example, if you map from a 47-element index to a 13-element index, each 3.62 elements of the original map to an element in the destination, so for example, the first 3
items map to the first item, then the fourth item is shared, with 0.62 apportioned to the first element and 0.38 apportioned to the second element. Supports aggregation types 'SUM', 'AVERAGE', 'MIN'
and 'MAX'. Contributed by Paul Sanford of EPRI.
Download: Sequential dispatch library.ana : The function SequentialDispatch generalizes the Dispatch function by adding a time-like index to carry over unsatisfied demand or unallocated capacity.
Sequential dispatch test.ana is an example of its use.
The Legendre Library includes functions for:
• Gauss-Legendre Quadrature
• Numerically stable, accurate and efficient calculation of an order n Legendre Polynomial at values -1<=x<=1
• Numerically stable calculation of the derivative of an order n Legendre Polynomial at x.
• Calculation of the zeros of Legendre polynomials.
• Calculation of the Legendre polynomial coefficients.
Author: Lonnie Chrisman, Lumina Decision Systems
Download: Legendre library.ana
The library has its own reference page on this wiki.
Cell Icon Set Library
Provides pre-curated icon sets for use as cell icons in tables. The functions automatically select which icon in a set to use to depict the value that you want the icon to depict. Each icon set uses
a consistent color scheme and consistent icon shapes having a consistent theme. Icon sets include shape collections, directional indicators, indicator styles and rating styles.
Without this library, you can use cell icons in table cells by using the CellIcon function in a Computed cell format attribute. But this requires you to find suitable icon images and write logic to
map from your values to these images. This library takes care of these pain points for you.
Documentation: Cell Icon Set Library
Download: Cell_Icon_Set_Library.ana
Author: Lonnie Chrisman, Lumina Decision Systems
See Also | {"url":"https://docs.analytica.com/index.php?title=Additional_libraries&oldid=61129","timestamp":"2024-11-14T04:39:06Z","content_type":"text/html","content_length":"45180","record_id":"<urn:uuid:19b072c9-f898-499b-a853-c23087c9897f>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00240.warc.gz"} |
Search result: Catalogue data in Spring Semester 2022
Computational Science and Engineering Bachelor
Basic Courses
Block G4
Number Title Type ECTS Hours Lecturers
529-0431-00L Physical Chemistry III: Molecular Quantum Mechanics O 4 credits 4G F. Merkt
Postulates of quantum mechanics, operator algebra, Schrödinger's equation, state functions and expectation values, matrix representation of operators, particle in a box, tunneling,
Abstract harmonic oscillator, molecular vibrations, angular momentum and spin, generalised Pauli principle, perturbation theory, electronic structure of atoms and molecules, Born-Oppenheimer
Learning This is an introductory course in quantum mechanics. The course starts with an overview of the fundamental concepts of quantum mechanics and introduces the mathematical formalism. The
objective postulates and theorems of quantum mechanics are discussed in the context of experimental and numerical determination of physical quantities. The course develops the tools necessary for
the understanding and calculation of elementary quantum phenomena in atoms and molecules.
Postulates and theorems of quantum mechanics: operator algebra, Schrödinger's equation, state functions and expectation values. Linear motions: free particles, particle in a box,
Content quantum mechanical tunneling, the harmonic oscillator and molecular vibrations. Angular momentum: electronic spin and orbital motion, molecular rotations. Electronic structure of atoms
and molecules: the Pauli principle, angular momentum coupling, the Born-Oppenheimer approximation. Variational principle and perturbation theory. Discussion of bigger systems (solids,
Lecture notes A script written in German will be available. The script is, however, no replacement for personal notes during the lecture and does not cover all aspects discussed.
151-0102-00L Fluid Dynamics I O 6 credits 4V + T. Rösgen
An introduction to the physical and mathematical foundations of fluid dynamics is given.
Abstract Topics include dimensional analysis, integral and differential conservation laws, inviscid and viscous flows, Navier-Stokes equations, boundary layers, turbulent pipe flow. Elementary
solutions and examples are presented.
Learning An introduction to the physical and mathematical principles of fluid dynamics. Fundamental terminology/principles and their application to simple problems.
Phenomena, applications, foundations
Content dimensional analysis and similitude; kinematic description; conservation laws (mass, momentum, energy), integral and differential formulation; inviscid flows: Euler equations, stream
filament theory, Bernoulli equation; viscous flows: Navier-Stokes equations; boundary layers; turbulence
Lecture notes Lecture notes (extended formulary) for the course are made available electronically.
Literature Recommended book: Fluid Mechanics, Kundu & Cohen & Dowling, 6th ed., Academic Press / Elsevier (2015).
Prerequisites Voraussetzungen: Physik, Analysis
/ Notice
529-0483-00L Statistical Physics and Computer Simulation O 4 credits 2V + S. Riniker, P. H. Hünenberger
Abstract Principles and applications of statistical mechanics and equilibrium molecular dynamics, Monte Carlo simulation, stochastic dynamics and free energy calculation.
Exercises using a MD simulation program to generate ensembles and subsequently calculate ensemble averages.
Learning Introduction to statistical mechanics with the aid of computer simulation; development of skills to carry out statistical mechanical calculations using computers and interpret the
objective results.
Content Principles and applications of statistical mechanics and equilibrium molecular dynamics, Monte Carlo simulation, stochastic dynamics and free energy calculation.
Exercises using a MD simulation program to generate ensembles and subsequently calculate ensemble averages.
Literature will be announced in the course
Since the exercises on the computer do convey and test essentially different skills as those being conveyed during the lectures and tested at the written exam, the results of a small
Prerequisites programming project will be presented in a 10-minutes talk by pairs of students who had been working on the project.
/ Notice
Additional information will be provided in the first lecture.
Subject-specific Competencies Concepts and Theories assessed
Techniques and Technologies assessed
Method-specific Competencies Analytical Competencies assessed
Decision-making assessed
Problem-solving assessed
Competencies Project Management assessed
Social Competencies Cooperation and Teamwork assessed
Negotiation assessed
Personal Competencies Adaptability and Flexibility assessed
Critical Thinking assessed
Self-direction and Self-management assessed | {"url":"https://www.vorlesungen.ethz.ch/Vorlesungsverzeichnis/sucheLehrangebot.view?seite=1&semkez=2022S&ansicht=2&lang=en&abschnittId=98465","timestamp":"2024-11-02T17:15:53Z","content_type":"text/html","content_length":"19147","record_id":"<urn:uuid:75feea84-2f22-4e3d-b313-90dddb8703a2>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00838.warc.gz"} |
Pythagorean Theorem & Definition With Worksheet
The definition of the Pythagorean theorem is that in a right-angled triangle, the sum of the squares of the sides is equal to the square of the hypotenuse. It is important for students of mathematics
to know that the Pythagorean theorem occupies great importance. They learn about this theorem in Algebra for the first time. But they find its actual usage in Calculus, Precalculus, Trigonometry, and
Definition of Pythagorean Theorem
The credit with regard to Pythagorean theorem goes to philosopher and the famous Greek mathematician, named Pythagoras. According to early history, it is said that Babylonians were using it for the
past 1500 years. However, the actual credit of proving it mathematically goes to Pythagoras only.
Further, the student should know that Pythagorean theorem is an important tool that can be used at all times while dealing with triangles. Irrespective of whichever mathematics class, the student
will find its application for all the problems associated with triangles. In a simple way, the theorem states that it is a relationship that is established between the lengths of sides of a
right-angled triangle. From this relationship it will be easy for the student to ascertain if a triangle is obtuse or acute.
Students should know that the theorem may have more proofs, which are known in comparison to any other like the law of quadratic reciprocity. It is important to mention here that a book has been
published by name, The Pythagorean Proposition. Students referring to this book will find 370 proofs with regard to Pythagorean theorem.
Pythagorean Theorem Worksheet
The proof of this theorem is based on the proportionality of the sides of two identical angles. This means that the ratio of any two corresponding two sides of identical triangles is considered the
same. This is true irrespective of the size of the triangles under consideration.
Pythagorean Theorem with Examples
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Compare elements of a recursive defined sequence
Compare elements of a recursive defined sequence
I define the recursive sequence as:
A, b, c = var('A, b, c')
def Sequence_rec(k):
x = 0
for i in range(1,k+1):
x = x + (A - x)/((c-i+2)^b)
return x
For the parameters the assumptions are:
assume(c, 'integer')
I'm interested in the elements of Sequence_rec(k) with k<=c. The following relation has to be true for the defined sequence considering the given assumptions:
bool(Sequence_rec(4) > Sequence_rec(3))
But Sage computes it is false! The following plot shows the difference is positive:
plot((Sequence_rec(4) - Sequence_rec(3))(A=1,c=3),b,(0,100))
How can I force Sage to compare the elements of the sequence bool(Sequence_rec(n+1) > Sequence_rec(n)) = true for any positive integer n correctly? Thank you for your advice!
Thanks for your reply. I forgot to mention that c and k are defined as integers and k<=c. In this case it must be Sequence_rec(n+1) > Sequence_rec(n). I use the elements of the sequence for further
calculations. What is the reason Sage is not able to compare symbolic expressions: bool(Sequence_rec(3) > Sequence_rec(2))? It works for: bool(Sequence_rec(2) > Sequence_rec(1))
3 Answers
Sort by ยป oldest newest most voted
The simple reason is that your conjecture is False. Try with:
A = 1
b = 2
c = 1/2
All those numbers are strictly positive, but you will get:
sage: Sequence_rec(3) - Sequence_rec(2)
Note also that the denominator can vanish along the loop when i=c+2, which may be another cause of trouble (you will have a lot of poles). By the way, even if the sequence was indeed increasing, Sage
will not be able to give an answer for all n together (it does not understands loops symbolically). Moreover, you could even imagine to encode undecidable problems in the iteration of such formulas.
More generally, if you ask for a boolean, you should know that Sage is not able to answer Unknown, hence it will usually answer False when it is not able to prove that answer is True, which is not
what mathematicians usually expect.
Actually, there exists an Unknown truth value in Sage but then python will not be able to deal correctly with boolean operations, see
sage: Unknown?
and PEP 335 for more information about this, which was unfortunately rejected, preventing Sage to deal with "proved True", "proved False", "Unable to prove".
Actually, I would prefer an even better alternative for trueness of an expression: either True, or False here is a counterexample or Unknown.
vdelecroix ( 2013-05-31 12:26:45 +0100 )edit
If you expect an answer you should tell Sage what are your assumptions (ie in the case you mention assume(c > 2)). Nevertheless, assuming your intution is correct, you can not rely on the output of
bool(my_expression) as
sage: assume(c > 3)
sage: bool(Sequence_rec(4) > Sequence_rec(3))
sage: bool(Sequence_rec(4) < Sequence_rec(3))
sage: bool(Sequence_rec(4) == Sequence_rec(3))
Thank you. As far as I know, the command bool(my_expression) evaluates the relation of the symbolic expressions. In this case I do not understand why it is not evaluated according to my "intuition".
KurtM ( 2013-05-31 05:04:28 +0100 )edit
There is no general algorithm to answer the question "f(x) > 0 ?". In your particular case, everything is polynomial so you can do something but not with symbolic expressions.
vdelecroix ( 2013-05-31 08:04:01 +0100 )edit | {"url":"https://ask.sagemath.org/question/10163/compare-elements-of-a-recursive-defined-sequence/","timestamp":"2024-11-12T03:00:52Z","content_type":"application/xhtml+xml","content_length":"75359","record_id":"<urn:uuid:21c9e5a5-1058-486e-bb1e-e8f4a9391386>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00002.warc.gz"} |
[problem] how to formulate this strange sum equation?
hello everybody. I occur this strange problem,anyone know how to formulate this equation
I am confused because sita in it should be a set and all of its elements are less than the parameter K.how to express it in gams ?c m h and m’ are all sets consist several elements
Hi Sun
Please read the forum rules at the top of this group and then repost your question. You can’t expect that somebody is trying to write down this equation from scratch. At least you should provide what
you tried so far…
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-----Original Message-----
From: gamsworld@googlegroups.com [mailto:gamsworld@googlegroups.com] On Behalf Of qinghai Sun
Sent: Monday, November 24, 2014 14:35
To: gamsworld@googlegroups.com
Subject: Re: [problem] how to formulate this strange sum equation?
I am confused because sita in it should be a set and all of its elements are less than the parameter K.how to express it in gams ?c m h and m’ are all sets consist several elements
在 2014å¹´11月24日星期一UTC+8ä¸Šå ˆ10æ—¶54分06秒，qinghai Sunå†™é “ï¼š
hello everybody. I occur this strange problem,anyone know how to formulate this equation
Thanks for reminding me ,I will overwrite a new one | {"url":"https://forum.gams.com/t/problem-how-to-formulate-this-strange-sum-equation/1345","timestamp":"2024-11-10T01:35:36Z","content_type":"text/html","content_length":"18510","record_id":"<urn:uuid:ee36bdb3-d715-4d23-b244-69004fa855ed>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00391.warc.gz"} |
Scandium orbital diagram - Learnool
Scandium orbital diagram
The information on this page is ✔ fact-checked.
Scandium orbital diagram | Image: Learnool
In the scandium orbital diagram, the 1s subshell accommodates two electrons, the 2s subshell carries another pair, the 2p subshell encompasses six electrons, the 3s subshell contains two electrons,
the 3p subshell carries six electrons, the 4s subshell holds two electrons, and the 3d subshell has one electron, totaling twenty-one electrons.
When illustrating the scandium orbital diagram, begin by determining the number of electrons from the periodic table. Utilize the electron configuration for reference and follow the three fundamental
rules: the Aufbau principle, Pauli exclusion principle, and Hund’s rule. This systematic approach ensures an accurate representation of scandium’s orbital arrangement.
Find electrons
The atomic number of scandium represents the total number of electrons of scandium. Since the atomic number of scandium is 21, the total electrons of scandium are 21.
Write electron configuration
The electron configuration of scandium is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1.
Now in the next step, start drawing the orbital diagram for scandium.
Draw orbital diagram
Before drawing the orbital diagram, you should know the three general rules.
Also, you should know the number of orbitals in each subshell.
We can calculate the number of orbitals in each subshell using the formula: 2ℓ + 1
Where, ℓ = azimuthal quantum number of the subshell
For s subshell, ℓ = 0
For p subshell, ℓ = 1
For d subshell, ℓ = 2
For f subshell, ℓ = 3
So each s subshell has one orbital, each p subshell has three orbitals, each d subshell has five orbitals, and each f subshell has seven orbitals.
Now start to draw!
As mentioned above, the electron configuration of scandium is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1. Hence, draw the blank orbital diagram of scandium up to 3d subshell as follows:
Blank orbital diagram of scandium | Image: Learnool
In the above orbital diagram, the box represents an orbital. Each orbital has a capacity of two electrons. And the arrows (↑↓) are drawn inside the box to represent electrons.
Now 1s^2 indicates that the 1s subshell has 2 electrons. So draw two arrows in the 1s box showing two electrons as follows:
Two arrows drawn in 1s box represent 1s^2 | Image: Learnool
2s^2 indicates that the 2s subshell has 2 electrons. So draw two arrows in the 2s box showing two electrons as follows:
Two arrows drawn in 2s box represent 2s^2 | Image: Learnool
2p^6 indicates that the 2p subshell has 6 electrons. So draw six arrows in the 2p box showing six electrons as follows:
Six arrows drawn in 2p box represent 2p^6 | Image: Learnool
3s^2 indicates that the 3s subshell has 2 electrons. So draw two arrows in the 3s box showing two electrons as follows:
Two arrows drawn in 3s box represent 3s^2 | Image: Learnool
3p^6 indicates that the 3p subshell has 6 electrons. So draw six arrows in the 3p box showing six electrons as follows:
Six arrows drawn in 3p box represent 3p^6 | Image: Learnool
4s^2 indicates that the 4s subshell has 2 electrons. So draw two arrows in the 4s box showing two electrons as follows:
Two arrows drawn in 4s box represent 4s^2 | Image: Learnool
3d^1 indicates that the 3d subshell has 1 electron. So draw one arrow in the 3d box showing one electron as follows:
One arrow drawn in 3d box represent 3d^1 | Image: Learnool
That’s it! This is the final orbital diagram of scandium as we have used all 21 electrons.
Next: Titanium orbital diagram
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Programming Exercises
Programming ExercisesΒΆ
1. Implement the simple methods get_num and get_den that will return the numerator and denominator of a fraction.
2. In many ways it would be better if all fractions were maintained in lowest terms right from the start. Modify the constructor for the Fraction class so that GCD is used to reduce fractions
immediately. Notice that this means the __add__ function no longer needs to reduce. Make the necessary modifications.
3. Implement the remaining simple arithmetic operators (__sub__, __mul__, and __truediv__).
4. Implement the remaining relational operators (__gt__, __ge__, __lt__, __le__, and __ne__).
5. Modify the constructor for the fraction class so that it checks to make sure that the numerator and denominator are both integers. If either is not an integer, the constructor should raise an
6. In the definition of fractions we assumed that negative fractions have a negative numerator and a positive denominator. Using a negative denominator would cause some of the relational operators
to give incorrect results. In general, this is an unnecessary constraint. Modify the constructor to allow the user to pass a negative denominator so that all of the operators continue to work
7. Research the __radd__ method. How does it differ from __add__? When is it used? Implement __radd__.
8. Repeat the last question but this time consider the __iadd__ method.
9. Research the __repr__ method. How does it differ from __str__? When is it used? Implement __repr__.
10. Research other types of gates that exist (such as NAND, NOR, and XOR). Add them to the circuit hierarchy. How much additional coding did you need to do?
11. The most simple arithmetic circuit is known as the half adder. Research the simple half-adder circuit. Implement this circuit.
12. Now extend that circuit and implement an 8-bit full adder.
13. The circuit simulation shown in this chapter works in a backward direction. In other words, given a circuit, the output is produced by working back through the input values, which in turn cause
other outputs to be queried. This continues until external input lines are found, at which point the user is asked for values. Modify the implementation so that the action is in the forward
direction; upon receiving inputs the circuit produces an output.
14. Design a class to represent a playing card and another one to represent a deck of cards. Using these two classes, implement your favorite card game.
15. Find a Sudoku puzzle online or in the local newspaper. Write a program to solve the puzzle.
You have attempted of activities on this page | {"url":"https://runestone.academy/ns/books/published/pythonds3/Introduction/ProgrammingExercises.html","timestamp":"2024-11-14T10:50:21Z","content_type":"text/html","content_length":"24375","record_id":"<urn:uuid:0933f0b1-4244-43f9-931f-9a6b155c0a43>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00723.warc.gz"} |
Wai-tian Tan
Oct 19, 2020
Abstract:In this work, we develop DeepWiPHY, a deep learning-based architecture to replace the channel estimation, common phase error (CPE) correction, sampling rate offset (SRO) correction, and
equalization modules of IEEE 802.11ax based orthogonal frequency division multiplexing (OFDM) receivers. We first train DeepWiPHY with a synthetic dataset, which is generated using representative
indoor channel models and includes typical radio frequency (RF) impairments that are the source of nonlinearity in wireless systems. To further train and evaluate DeepWiPHY with real-world data, we
develop a passive sniffing-based data collection testbed composed of Universal Software Radio Peripherals (USRPs) and commercially available IEEE 802.11ax products. The comprehensive evaluation of
DeepWiPHY with synthetic and real-world datasets (110 million synthetic OFDM symbols and 14 million real-world OFDM symbols) confirms that, even without fine-tuning the neural network's architecture
parameters, DeepWiPHY achieves comparable performance to or outperforms the conventional WLAN receivers, in terms of both bit error rate (BER) and packet error rate (PER), under a wide range of
channel models, signal-to-noise (SNR) levels, and modulation schemes.
* Journal paper (16 pages and 12 figures) to appear in IEEE Transactions on Wireless Communications | {"url":"https://www.catalyzex.com/author/Wai-tian%20Tan","timestamp":"2024-11-07T06:18:09Z","content_type":"text/html","content_length":"94718","record_id":"<urn:uuid:6d7cc1bc-d0af-4406-a70b-17f05e0645f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00732.warc.gz"} |
Minkowski and Galilei/Newton fluid dynamics: A geometric 3 + 1 spacetime perspective
A kinetic theory of classical particles serves as a unified basis for developing a geometric 3 + 1 spacetime perspective on fluid dynamics capable of embracing both Minkowski and Galilei/Newton
spacetimes. Parallel treatment of these cases on as common a footing as possible reveals that the particle four-momentum is better regarded as comprising momentum and inertia rather than momentum and
energy; and, consequently, that the object now known as the stress-energy or energy-momentum tensor is more properly understood as a stress-inertia or inertia-momentum tensor. In dealing with both
fiducial and comoving frames as fluid dynamics requires, tensor decompositions in terms of the four-velocities of observers associated with these frames render use of coordinate-free geometric
notation not only fully viable, but conceptually simplifying. A particle number four-vector, three-momentum (1, 1) tensor, and kinetic energy four-vector characterize a simple fluid and satisfy
balance equations involving spacetime divergences on both Minkowski and Galilei/Newton spacetimes. Reduced to a fully 3 + 1 form, these equations yield the familiar conservative formulations of
special relativistic and non-relativistic fluid dynamics as partial differential equations in inertial coordinates, and in geometric form will provide a useful conceptual bridge to
arbitrary-Lagrange-Euler and general relativistic formulations.
• Kinetic theory of fluids
• Newtonian and non-Newtonian fluids
• Statistical
Dive into the research topics of 'Minkowski and Galilei/Newton fluid dynamics: A geometric 3 + 1 spacetime perspective'. Together they form a unique fingerprint. | {"url":"https://impact.ornl.gov/en/publications/minkowski-and-galileinewton-fluid-dynamics-a-geometric-3-1-spacet","timestamp":"2024-11-11T10:14:53Z","content_type":"text/html","content_length":"50366","record_id":"<urn:uuid:bc94c83c-8dc7-4fa7-b93b-f7a47c788c26>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00578.warc.gz"} |
Tracing Numbers 1 10 Worksheets Kindergarten Pdf
Tracing Numbers 1 10 Worksheets Kindergarten Pdf act as fundamental tools in the world of mathematics, supplying a structured yet functional system for learners to discover and master numerical
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Use these tracing numbers 1 10 free printable pdf pages with toddler preschool pre k and kindergarten age children Simply print pdf file with number tracing worksheets pdf and you are ready to play
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Nurturing Number Sense and Reasoning
Numbers 1 To 10 Worksheets For Kindergarten PDF
Numbers 1 To 10 Worksheets For Kindergarten PDF
These worksheets introduce the numbers 1 to ten Each worksheet displays a number in number word and graphical form as well as providing tracing and counting opportunities The goal is instant
recognition of single digit numbers Print out all 10 worksheets and switch back and forth between them Learning Number 1 Worksheet
Number Tracing 1 10 Worksheet FREE Printable Worksheets Worksheetfun Multiplication Basic Facts Multiplication Cubes Multiplication Horizontal Multiplication Popular Multiplication Quiz
Multiplication Repeated Addition Multiplication Test Multiplication Vertical Multiplication Word Problems Multiplication 1
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From Theory to Real-World Application
Tracing Numbers 1 10 Worksheets Kindergarten Pdf Kidsworksheetfun
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Sharing is caring These number tracing worksheets are a great way for your children to practice their number formation There s the numbers 1 10 included in the pack The larger numbers at the top give
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1 to 10 tracing worksheet There are thirty five pages in this printable pack The activities include Tracing the numbers from 1 to 10 There is one large number on each page This number contains
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Learning Numbers Worksheets For Preschool And Kindergarten K5 Learning
Numbers worksheets from one to 20 These free worksheets help kids learn to recognize read and write numbers from 1 20 Learn the numbers from one to ten Numbers in sequence tracing 1 10 Numbers in
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1 to 10 tracing worksheet There are thirty five pages in this printable pack The activities include Tracing the numbers from 1 to 10 There is one large number on each page This number contains
directions on how to write each number There is one of this page for each of the numbers from one through ten
Numbers worksheets from one to 20 These free worksheets help kids learn to recognize read and write numbers from 1 20 Learn the numbers from one to ten Numbers in sequence tracing 1 10 Numbers in
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sdtw_cent: Centroid calculation based on soft-DTW in dtwclust: Time Series Clustering Along with Optimizations for the Dynamic Time Warping Distance
Soft-DTW centroid function as proposed in Cuturi and Blondel (2017).
sdtw_cent( series, centroid = NULL, gamma = 0.01, weights = rep(1, length(series)), ..., error.check = TRUE, opts = list(algorithm = "NLOPT_LD_LBFGS", maxeval = 20L) )
A matrix or data frame where each row is a time series, or a list where each element is a time series. Multivariate series should be provided as a list of matrices where time spans the
series rows and the variables span the columns of each matrix.
Optionally, a time series to use as reference. Defaults to a random series of series if NULL. For multivariate series, this should be a matrix with the same characteristics as the
centroid matrices in series.
gamma Positive regularization parameter, with lower values resulting in less smoothing.
weights A vector of weights for each element of series.
... Further arguments for the optimization backend (except opts for nloptr, control for optim, and ... for both).
error.check Logical indicating whether the function should try to detect inconsistencies and give more informative errors messages. Also used internally to avoid repeating checks.
opts List of options to pass to nloptr or stats::optim()'s control. The defaults in the function's formals are for nloptr, but the value will be adjusted for optim if needed.
A matrix or data frame where each row is a time series, or a list where each element is a time series. Multivariate series should be provided as a list of matrices where time spans the rows and the
variables span the columns of each matrix.
Optionally, a time series to use as reference. Defaults to a random series of series if NULL. For multivariate series, this should be a matrix with the same characteristics as the matrices in series.
Positive regularization parameter, with lower values resulting in less smoothing.
Further arguments for the optimization backend (except opts for nloptr, control for optim, and ... for both).
Logical indicating whether the function should try to detect inconsistencies and give more informative errors messages. Also used internally to avoid repeating checks.
List of options to pass to nloptr or stats::optim()'s control. The defaults in the function's formals are for nloptr, but the value will be adjusted for optim if needed.
This function can delegate the optimization to the nloptr package. For that to happen, you must load it with either base::library() or base::loadNamespace(). If the aforementioned is not fulfilled,
the function will delegate to stats::optim().
The resulting centroid, with the optimization results as attributes (except for the returned centroid).
Please note that running tasks in parallel does not guarantee faster computations. The overhead introduced is sometimes too large, and it's better to run tasks sequentially.
This function uses the RcppParallel package for parallelization. It uses all available threads by default (see RcppParallel::defaultNumThreads()), but this can be changed by the user with
An exception to the above is when it is called within a foreach parallel loop made by dtwclust. If the parallel workers do not have the number of threads explicitly specified, this function will
default to 1 thread per worker. See the parallelization vignette for more information - browseVignettes("dtwclust")
For unknown reasons, this function has returned different results (in the order of 1e-6) when using multi-threading in x64 Windows installations in comparison to other environments (using nloptr
v1.0.4). Consider limiting the number of threads if you run into reproducibility problems.
Cuturi, M., & Blondel, M. (2017). Soft-DTW: a Differentiable Loss Function for Time-Series. arXiv preprint arXiv:1703.01541.
For more information on customizing the embed code, read Embedding Snippets. | {"url":"https://rdrr.io/cran/dtwclust/man/sdtw_cent.html","timestamp":"2024-11-12T19:23:14Z","content_type":"text/html","content_length":"34505","record_id":"<urn:uuid:294843e4-54e0-4602-a66a-3402c4bfcc98>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00342.warc.gz"} |
Free FE Practice Test
Free FE Environmental Example Practice Problems
We've selected 10 diverse practice problems from our question bank that you can use to review for the Environmental engineering FE exam and give you an idea about some of the content we provide.
1) One of the properties of polymers is:
◯ A. 
they are generally organic compounds.
◯ B. 
they are stable at high temperatures.
◯ C. 
a combination of two or more materials differing in form or composition.
◯ D. 
they are usually high density.
2) Flow in a circular pipe of diameter $\SI{1}{m}$ has an average velocity of $\SI{2.5}{m/s}$. What is the fluid velocity $10\si{cm}$ off the borders of the circular pipe? Assume laminar flow
3) Liquid nitrogen, when returned to the gaseous state, can displace oxygen from the air and can create an oxygen-deficient atmosphere under the right conditions.
Answer true or false:
4) Select all the root(s) in $f(x) = x^2 -4x - 3$.
5) The purpose of sludge dewatering is to:
◯ A. 
Decrease total solids concentration
◯ B. 
Increase total solids concentration
◯ C. 
Convert suspended solids into settleable solids
◯ D. 
None of these - sludge dewatering makes no change at all in solids concentration
1) One of the properties of polymers is:
A.they are generally organic compounds.
B.they are stable at high temperatures.
C.a combination of two or more materials differing in form or composition.
D.they are usually high density.
The correct answer is A.
Refer to the Polymers section in the Materials Science/Structure of Matter chapter of the FE Reference Handbook.
Plastics (or polymers) are generally organic compounds based upon carbon and hydrogen. They are very large molecular structures. Usually they are low density and are not stable at high temperatures.
They can be readily formed into complex shapes. Their strength, stiffness, and melting temperatures are generally much lower than those of metals and ceramics. Their light weight, low cost, and ease
of forming make them the preferred material for many engineering applications.
2) Flow in a circular pipe of diameter $\SI{1}{m}$ has an average velocity of $\SI{2.5}{m/s}$. What is the fluid velocity $10\si{cm}$ off the borders of the circular pipe? Assume laminar flow
A.0.9 m/s
B.1.8 m/s
C.2.4 m/s
D.0.2 m/s
The correct answer is B.
Refer to Fluid Flow Characterization section in the Fluid Mechanics chapter of the FE Reference Handbook.
We know that flow in laminar flow conditions has a triangular distribution where the fluid is moving the fastest at the center of the pipe and moving the slowest near the borders of the pipe. For
laminar flow in circular pipe conditions, one can calculate what the fluid velocity is anywhere along the pipe radius via: $$ v(r) = v_{max}\left[1-\left(\frac{r}{R}\right)^2\right] $$ where
$v_{max} = 2\bar{v}=2(2.5\si{m/s})=5\si{m/s}$
$R=\text{radius of the tube}=1\si{m}/2=0.5\si{m}$
$r=\text{distance from the centerline}=0.5\si{m}-(10\si{cm}/100)=0.4\si{m}$ $$ v(0.4\si{m}) = 5\si{m/s}\left[1-\left(\frac{0.4\si{m}}{0.5\si{m}}\right)^2\right] \\ v(0.4\si{m})=1.8\si{m/s} $$
Therefore, the fluid velocity 10 cm off the pipe border is 1.8 m/s
3) Liquid nitrogen, when returned to the gaseous state, can displace oxygen from the air and can create an oxygen-deficient atmosphere under the right conditions.
Answer true or false:
The correct answer is A.
Due to the large liquid to gas expansion that takes place upon evaporation, liquid nitrogen is capable of displacing sufficient oxygen to create an oxygen deficient environment in a small or
insufficiently ventilated space, leading to the risk of asphyxiation.
Note this question can fall under the Ethics category as it pertains to compliance with regulations, codes, standards, and industrial safety.
4) Select all the root(s) in $f(x) = x^2 -4x - 3$.
A. $4+\sqrt{3}$
B. $2+\sqrt{7}$
C. $2-\sqrt{7}$
D. $4-\sqrt{3}$
The correct answers are B and C.
In order to find the roots of an equation, first set $f(x)$ to 0. $$ x^2 -4x - 3 = 0 $$ For simple equations, you can factor them by hand, but this equation does not factor easily. 3 can be factored
into 1 and 3. However, one must be negative and one positive for their multiple to be -3. There is no way to sum those to get -4. Not to worry, you can always use the quadratic formula (a.k.a. the
quadratic equation) if the equation can't be factored easily or if you are unsure. A quadratic equation is of the form: $$ ax^2+bx+c =0 $$ So we have $a=1$, $b=-4$, and $c=-3$. The quadratic formula
is: $$ x = \frac{-b\,±\,\sqrt{b^2-4ac}}{2a} \\ = \frac{-(-4)\,±\,\sqrt{(-4)^2-4(1)(-3)}}{2(1)} \\ = \frac{4\,±\,\sqrt{28}}{2} \\ = \frac{4\,±\,2\sqrt{7}}{2} \\ = 2\,±\,\sqrt{7} $$ The equation is
satisfied when $x=2+\sqrt{7}$ or $2-\sqrt{7}$.
5) The purpose of sludge dewatering is to:
A.Decrease total solids concentration
B.Increase total solids concentration
C.Convert suspended solids into settleable solids
D.None of these - sludge dewatering makes no change at all in solids concentration
The correct answer is B.
The goal of sludge dewatering is to increase the solids concentration. The higher the cake solids, the lower the operating cost for hauling. Most cake is hauled according to a cost per wet ton basis. | {"url":"https://www.prepfe.com/fe-exams/free/environmental-fe-exam-practice-pdf","timestamp":"2024-11-04T08:49:40Z","content_type":"text/html","content_length":"48912","record_id":"<urn:uuid:f443ffb4-a334-4e72-ab1c-00fff8809eed>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00490.warc.gz"} |
The suspension cables from the golden gate bridges towers ar-Turito
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The suspension cables from the Golden gate bridge’s towers are farther above the roadway near the towers and closer to the roadway near the middle of the bridge. You can figure out your distance from
middle of the bridge, x, in feet, and height of the suspension cable, y, in feet, at your position by using the equation x =
Rearrange the equation and then solve for y.
The correct answer is: 224.48
Complete step by step solution:
Here we have the equation to be x =
Where, x is the distance from middle of the bridge and y is the height of the
suspension cable.
Here given that x = 200 ft
On rearranging the equation, we have
On squaring both the sides, we get 4.47999556 = y - 220
cable is 224.48 ft from the roadway when you are 200 ft from the middle of the bridge.
Get an Expert Advice From Turito. | {"url":"https://www.turito.com/ask-a-doubt/Maths-the-suspension-cables-from-the-golden-gate-bridge-s-towers-are-farther-above-the-roadway-near-the-towers-an-qc777c71c","timestamp":"2024-11-13T19:55:21Z","content_type":"application/xhtml+xml","content_length":"1052464","record_id":"<urn:uuid:fee0adb7-2e1e-4b1b-b811-2e810fbda00a>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00729.warc.gz"} |
To the Moon in No Time!
Exponential Growth
To the Moon in No Time!
And the Answer Is… 42.
Exponential growth is a funny thing. It starts out real slow and then all of the sudden it shoots up big time. You know all about it, but how would you explain this topic to a fellow at a party?
Doing It the Nerdy Way
You could just grab a napkin and draw a little sketch.
That’s the fast and simple approach. It will probably make you appear quite nerdy, but if you don’t mind looking the part, then that’s a decent option. Just throw in a few sentences about how stuff
keeps getting doubled all the time and you’re basically done, but chances are you’ve been boring your counterpart to death.
If you want to torture the poor questioner some more, just keep talking about compound interest or bacterial growth. Your fellow partygoer will walk away a bored and wiser man. Or woman. Whoever made
the mistake of asking the question in the first place.
Trying Your Luck With Chess and Rice
Alright, napkin scribbles and nerdy talk didn’t really work all that great. So next time you’re asked the same question you remember the story about a king and his jester. One day, the jester didn’t
have much to do and ended up inventing the game of chess. The king loved it and the jester was granted a wish.
Since the he wasn’t just a funny dude but also quite a smart one, the king’s jester knew all about exponential growth. So he asked for rice. One grain on the first square, and twice the amount of the
previous one for all the other squares.
Image by McGeddon
That’s a pretty solid story, and your listener will probably be quite impressed to learn that this actually makes for a grand total of more than just a single bag of rice. In fact, it’s an incredibly
large number.
Nobody needs to be a genius to know that’s a lot of rice. Guess what’s for dinner…
When Moon? Soon!
Chess and rice is definitely less boring than function graphs and compound interest, but you can do way better!
Let’s take a single standard sheet of paper with a thickness of 0.1 millimeters. Now just ask how many times one needs to fold it until its height would at least cover the distance to the moon. While
folding the sheet eventually turns into mission impossible, the theoretical answer is almost guaranteed to shock your questioner.
Even though you didn’t asked the ultimate question of life, the universe and everything, the correct answer is 42 nonetheless. Let’s quickly do the math.
Since the maximum Earth-Moon distance is 405,696 km, folding the sheet 42 times will always do the trick. Will this little story turn people into math aficionados? Probably not, but it certainly is
one of the more stunning examples of exponential growth and it most likely won’t bore your audience.
When it comes to math, just grabbing people’s attention is very difficult. Wowing them is almost impossible. Going by my own experience, this moon example usually accomplishes both. Even better,
sometimes people even want to see the actual math behind it, and that’s about all you can possibly hope for. After all, it must be a good story if the answer is 42, right? | {"url":"https://www.cantorsparadise.org/to-the-moon-in-no-time-e51f3a17df70/","timestamp":"2024-11-12T04:07:23Z","content_type":"text/html","content_length":"32736","record_id":"<urn:uuid:24f9c3ff-7d95-47fb-9e66-77b23f8709b0>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00614.warc.gz"} |
Response of runoff and sediment production on sand-covered loess slopes to slope length and sand covering thickness
• 摘要:
方法基于室内模拟降雨试验，以未覆沙黄土坡面为对照，定量分析坡长（1 和3 m）和覆沙厚度（2、5和10 cm）对坡面产流产沙变化的影响。
结果（1）覆沙较未覆沙黄土坡面初始产流时间延长3 ~ 30.72倍，平均产流速率降低25% ~ 84%，平均产沙速率增大3.03 ~ 15.91倍，含沙量增加3.38 ~ 18.07倍，且都随覆沙厚度增加而加强。（2）1 m坡长10 cm覆沙对产
流速率减少作用强烈，3 m坡长平均产流速率在不同覆沙厚度下变化较小；不管是否覆沙，3 m坡长平均产沙速率和含沙量明显高于1 m坡长。（3）降雨过程中，坡长和覆沙厚度的增加能够协同增强产流产沙过程的变异性，1
m坡长未覆沙坡面瞬时产流速率高于覆沙，3 m坡长较厚覆沙坡面产流产沙陡增陡降，有明显峰值，瞬时径流系数出现大于1的现象。（4）结构方程模型分析表明，坡长对产流速率影响最大（路径系数为0.65），覆沙厚度对
Objective Aeolian sand-covered loess slope is a special geomorphic landscape with a unique erosion pattern formed by multi-dynamic forces within the wind-water erosion crisscross region of the
Loess Plateau. Objectives of this study are to investigate the response of runoff and sediment production processes to slope length and thickness of sand covering on the aeolian sand-covered
loess slopes, which can provide essential explanation for preventing and predicting soil erosion in this region.
Method The quantitative analysis was based on observations of runoff and sediment production in indoor simulated rainfall experiments with the slope length (between 1 and 3 m) and thickness of
sand covering (2, 5 and 10 cm). The effects of slope length and thickness of sand covering were analysed against a control group without sand covering.
Result (1) Compared with the loess slope without sand covering, the time to runoff generation on the sand-covered slope was significantly extended by 3 to 30.72 times, the average runoff rate was
reduced by 21% to 84%, the average sediment yield rate was increased by 2.99 to 10.66 times, and the sediment concentration was increased by 3.38 to 18.07 times, all of which were intensified as
the thickness of sand covering increased. (2) The 1 m slope with a 10 cm sand layer exhibited a significant effect on reducing the runoff rate, while the average runoff rate with a 3 m slope
demonstrated minor variations among different thicknesses of sand covering. Whether covered by sands or not, the average sediment yield rate and sediment concentration from 3 m slope were
significantly higher than those from the 1 m slope. (3) The increases in slope length and thickness of sand covering enhanced the variability of instantaneous runoff and sediment yields during
rainfall events. The instantaneous runoff rate of 1 m slope without sand covering was found to be higher than that with sand covering during rainfall. Notably, both runoff and sediment yields
from 3 m slopes with a thicker sand covering showed a distinct peak, and some instantaneous runoff coefficients exceeded 1 during the rainfall events. (4) The structural equation model revealed
that the slope length had the greatest influence on runoff rate (path coefficient = 0.65), and the sand thickness had the greatest influence on sediment yield rate (path coefficient = 0.71). The
slope length exhibited an indirect positive effect (path coefficient = 0.40) on sediment yield through runoff production.
Conclusion The slope length increases both runoff and sediment yield rates, while the thickness of sand covering reduces the runoff rate and increases sediment yield rate. The synergy of slope
length and thickness of sand covering enhances the variability of runoff and sediment production processes, which makes the runoff and sediment production more changeable during rainfall. | {"url":"http://j.bjfu.edu.cn/article/doi/10.12171/j.1000-1522.20240229?viewType=HTML","timestamp":"2024-11-14T22:16:38Z","content_type":"text/html","content_length":"280396","record_id":"<urn:uuid:6a387fe6-5025-4c72-a279-96d6040a0abc>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00015.warc.gz"} |
Campece, Nicole / Standards Math & ELA
Gregory School
Together We Can, Juntos Nós Podemos, Juntos Podemos
Common Core State Standards
Draw and identify lines and angles, and classify shapes by properties of their lines and angles.
Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Identify these in two-dimensional figures.
Draw and identify lines and angles, and classify shapes by properties of their lines and angles.
Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a
category, and identify right triangles
Generalize place value understanding for multi-digit whole numbers.
Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right.
Generalize place value understanding for multi-digit whole numbers.
Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and <
symbols to record the results of comparisons.
Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit.
Know relative sizes of measurement units within one system of units including km, m, cm; kg, g; lb, oz.; l, ml; hr, min, sec. Within a single system of measurement, express measurements in a
larger unit in terms of a smaller unit. Record measurement equivalents in a two-column table
Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit.
Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and
problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a
measurement scale
Represent and interpret data.
Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using information presented in
line plots.
Use the four operations with whole numbers to solve problems.
Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these
problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
Generate and analyze patterns.
Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself.
Common Core State Standards
Treasures Reading
Language: VocabularyAcquisition and Use
L.4.5c Demonstrate understanding of words by relating them to their opposites (antonyms) and to words with similarbut not identical meanings (synonyms)
Reading: Foundational Skills: Phonics and Word Recognition
RF.4.3 Know and apply grade – level phonicand word analysis skills in decoding words.
Reading: Foundational Skills: Fluency
RF.4.4b Read on – level prose and poetry orally withaccuracy, appropriate rate, and expression on successive readings.
Writing: Text Types and Purposes
W.4.3a Orient the reader byestablishing a situation and introducing a narrator and/ or characters;organize an event sequence that unfolds naturally.
Language: Conventions of Standard English
L.4.1 Demonstrate command of theconventions of Standard English grammar and usage when writing and speaking. | {"url":"https://www.longbranch.k12.nj.us/Page/5329","timestamp":"2024-11-01T20:26:11Z","content_type":"text/html","content_length":"509413","record_id":"<urn:uuid:327032c7-58dc-475f-8df1-d46348050339>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00159.warc.gz"} |
Working with Missing Values
Data cleansing accounts for 80% of the work of scientists, and in my experience, that’s true. Although I always recommend …
Data cleansing accounts for 80% of the work of scientists, and in my experience, that’s true.
Although I always recommend cleaning these missing valuers, sometimes it is not necessary, however there is a case where it is essential :
During the Creation of Statistical Models
And guess what, there’s not only one way around it. So here are 3 ways to work with missing values.
I. Remove Missing Values
The first of these methods is to remove rows or column holding those missing values.
First, you’ll have to ask yourself why those value are missing ?
Because most of the time dropping data from your dataset will lead to bias model (it’s also true with imputing data points) .
There isn’t a best universal way to work with missing data, that’s why you’ll have to explore different option to help you determine what best for your situation.
For Instance, missing data can sometimes help you obtain better forecasts, Let’s imagine a survey of individuals, removing the missing data could bias the results of our model. We could use this
missing information to enrich our perception and moreover our model.
For Data entry errors, mechanical errors or because missing data isn’t useful for our question of interest are acceptables cases for dropping our missing values.
A. Drop any row with a missing value.
B. Drop only the row with all missing values.
C. Drop only the rows with missing values in column 3
df.dropna(how='any', subset=['col3'])
II. Imputing Values
Imputing values into a dataset is certainly the most common ways professionnal work with missing data.
You commonly fill the missing value by the mean, the median or the mode.
The pros is that you are not directly removing rows or columns associated with missing.
The cons is that you are diluting the power of your features to predict well by reducing variability in those features
By removing or imputing missing value we should be very cautious about the impact this will have in our model.
It is very common to impute in the following ways:
1. Impute the mean of a column.
2. If you are working with categorical data or a variable with outliers, then use the mode of the column.
3. Impute 0, a very small number, or a very large number to differentiate missing values from other values.
4. Use knn to impute values based on features that are most similar. | {"url":"https://www.labo.mathieurella.fr/?p=498","timestamp":"2024-11-07T17:15:21Z","content_type":"text/html","content_length":"74833","record_id":"<urn:uuid:c7397b16-3f9e-4e22-9c5f-371a37984335>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00282.warc.gz"} |
Self-Associating Behavior of Acetone in Liquid Krypton
journal contribution
posted on 2016-01-25, 00:00 authored by Liene I. De Beuckeleer, Wouter A. Herrebout
Acetone molecules are inclined to self-associate through dipole–dipole interactions because of their large dipole moment. Infrared spectroscopy of compounds dissolved in liquid noble gases supported
by high level ab initio calculations allows investigating the self-associating behavior and determining the thermodynamical properties. In this study, infrared spectra of various concentrations of
acetone dissolved in liquid krypton are recorded at constant temperature. Overlapping monomer and dimer spectra are separated by analyzing the obtained data sets with numerical methods based on
least-squares fitting. Although acetone is known to self-associate, only a few spectral features have been presented in literature before. In this study, the application of new numerical approaches
succeeds in resolving overlapping spectra and allows observing isolated acetone dimer absorption bands for the complete mid infrared spectrum. By use of data sets of spectra recorded at temperatures
between 134 and 142 K, the experimental standard dimerization enthalpy was determined to be −10.8 kJ mol^–1. MP2/aug-cc-pVDZ calculations predicted a stacked and planar dimer geometry of which the
stacked geometry is more stable. Combining MP2 energies and single point corrections involving CCSD(T) calculations and complete basis set extrapolations based on the MP2/aug-cc-pVDZ equilibrium
geometry lead to complexation energy of −28.4 kJ mol^–1 for the stacked geometry and −15.1 kJ mol^–1 for the planar geometry. The corresponding values for the complexation enthalpies in solution,
obtained by combining these values with corrections for thermal and solvent influences are −13.7 and −5.8 kJ mol^–1. | {"url":"https://acs.figshare.com/articles/journal_contribution/Self_Associating_Behavior_of_Acetone_in_Liquid_Krypton/2082880/1","timestamp":"2024-11-04T17:19:39Z","content_type":"text/html","content_length":"131501","record_id":"<urn:uuid:2c19399f-3850-4adf-9eb0-17d536848226>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00108.warc.gz"} |
This is the fifth project option for Calculus I during Fall 2017 at Fitchburg State.
This project involves ordering types of functions by investigating their limits at infinity.
\documentclass[12pt]{amsart} \addtolength{\hoffset}{-2.25cm} \addtolength{\textwidth}{4.5cm} \addtolength{\voffset}{-2.5cm} \addtolength{\textheight}{5cm} \setlength{\parskip}{0pt} \setlength{\
parindent}{15pt} \usepackage{amsthm} \usepackage{amsmath} \usepackage{amssymb} \usepackage[colorlinks = true, linkcolor = black, citecolor = black, final]{hyperref} \usepackage{graphicx} \usepackage
{multicol} \usepackage{ marvosym , wasysym} \newcommand{\ds}{\displaystyle} \newcommand{\lessarrow}{\begin{array}{c} < \\[-6pt] \text{\small{$\rightarrow$}} \end{array}} \pagestyle{myheadings} \
setlength{\parindent}{0in} \pagestyle{empty} \begin{document} \thispagestyle{empty} {\scshape Math 2300} \hfill {\scshape \Large Project \#5} \hfill {\scshape Fall 2017} \medskip \hrule \bigskip \
bigskip In this Investigative Project, you will use l'H\^opital's Rule to develop a hierarchy of functions. Hopefully you've carefully thought about the questions and ideas posed at the end of the
in-class worksheet on l'H\^opital's Rule. In the beginning, you compared the functions $f(x) = x^2$ and $g(x) = 2^x$. Include a careful write-up of this analysis here. \bigskip You should have
discovered that $\ds{\lim_{x \rightarrow \infty} \frac{2^x}{x^2} \rightarrow \infty}$. This means that the function $2^x$ is much larger ``in the long run" than the function $x^2$. We can convey this
in symbols as $$x^2 \lessarrow 2^x.$$ In this case, the exponential function ``wins" this battle, but is that always the case? What happens if we have a larger power for $x$ in the power function and
/or a smaller base in the exponential function? Is $x^{100} \lessarrow 1.5^x$? Try comparing a few different pairs of one power function and one exponential function. How low can you make the base of
the exponential and/or how high can you make the power while maintaining the same $\lessarrow$ ordering? Determine a theorem, given it a name, and provide (if not a proof) a strong argument for your
theorem. It might have a form that looks like: \begin{quote} {\bf The \underline{\hspace{1.5in}} Theorem:} For any value $p$ where \underline{\hspace{.25in}} $ < p < $\underline{\hspace{.25in}} and/
or any value of $b$ where \underline{\hspace{.25in}} $ < b < $\underline{\hspace{.25in}}, $$\lim_{x \rightarrow \infty} \frac{b^x}{x^p} \rightarrow \infty,$$ so $x^p \lessarrow b^x$.\end{quote} and
you will reference l'H\^opital's Rule in the justification. Note that your conditions on $p$ and $b$ maybe in a different form (like $b>4$) or there may be no restrictions on one or the other (for
any real numbers $p$ and $b$ \dots). It is your job to try to push these bounds as far as you can. \bigskip \bigskip The activity outlined above compared the end behavior of exponential functions
with the end behavior of power functions. Another class of functions that grow without bound as $x$ approaches infinity are the log functions: $f(x) = \log_a(x)$. Compare this class of functions with
power functions and with exponential functions? Which type of functions ``wins" in the limit? Is this always true? Change some of the parameters like you did before and see what happens. Write and
justify a theorem (or two) that summarizes your findings. \bigskip \bigskip Also, consider changes in the parameters within these classes of functions. Of course, if we increase the power just a tiny
bit, that makes the function bigger, but how much bigger? Enough that $x^4\lessarrow x^{4.000001}$? How do these changes work in each class of functions? How much larger must $p$ be than $q$ so that
$x^q \lessarrow x^p$? What about exponentials and logs? \bigskip \bigskip List an ordering of sorts of log functions, power functions, and exponential functions with a variety of powers/bases that
demonstrates all the work you've done. \vfill {\small Note: I just made up the symbol $\lessarrow$, you can use a different symbol just be sure to define it clearly. } \end{document} | {"url":"https://pt.overleaf.com/articles/fsu-math2300-project5/mhgznqkbqgyc","timestamp":"2024-11-03T00:07:15Z","content_type":"text/html","content_length":"40614","record_id":"<urn:uuid:acd766e8-6ee6-4818-bc48-582a28e62eb6>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00430.warc.gz"} |
Understanding the Perceptron: A Foundation for Machine Learning Concepts
In the domain of artificial intelligence and machine learning, the term "perceptron" is frequently used. The fundamental building block of artificial neural networks, the perceptron is the most
fundamental part of machine learning and deep learning technologies.
What is Perceptron?
A single-layer neural network linear or machine learning approach called a perceptron is used to learn different binary classifiers under supervision. By processing and learning aspects, it functions
as an artificial neuron that can detect the business intelligence and capacities of the input data.
Perceptron is a linear classifier (binary) and is a collection of straightforward logical assertions that combine to form a neural network, which is an array of sophisticated logical assertions. It
is employed in supervised learning as well. Classifying the provided input data is helpful.
But how on earth does it function?
A normal neural network looks like this as we all know:
How are perceptrons inspired by biological neurons?
• It is important to talk about how biological neurons serve as the model for artificial neurons, or perceptrons. An artificial neuron can be thought of as a mathematical model that was influenced
by a biological neuron.
• A biological neuron's dendrites, or tiny filaments, are how it receives input messages from neighboring neurons. In a similar manner, input neurons that accept numbers are used by perceptrons to
receive data from other perceptrons.
• Synapses are the connectors that connect dendrites to actual neurons. Weights are also used to describe the relationships between inputs and perceptrons. They gauge how significant each input
• A biological neuron's nucleus generates an output signal in response to signals received from the dendrites. Similar to this, a perceptron's nucleus (shown in blue) computes certain values based
on inputs and generates an output.
• The axon in a biological neuron transports the output signal away. Similar to this, a perceptron's axon is its output value, which serves as an input for subsequent perceptrons.
How does the perceptron work?
For dealing with binary classification problems, the Perceptron approach offers a simple yet powerful paradigm. The Perceptron model is based on a single layer of neurons that apply an activation
function to a weighted sum of inputs to produce an output. To lessen the difference between the expected and actual output, the weights of the neurons are adjusted during training.
The Perceptron method iteratively goes over the training data and modifies the weights of the neurons in order to lessen the disparity between the projected and actual outputs.
In order to minimize error, which is defined as the discrepancy between the expected and actual output, the weights are adjusted accordingly. This process is repeated until the weights converge to a
stable solution.
Figure 1: How the Perceptron Works?
Figure 1 shows the operation of the perceptron. The perceptron in the example has one output and three inputs, x1, x2, and x3.
The associated weights (w1, w2, and w3) assigned to these inputs indicate their relevance. The weighted sum of the inputs determines whether the output is 0 or 1. If the output is above a threshold,
it is 1, and if the sum is below a threshold, it is 0. This threshold may be a neuronal parameter and an actual value. Given that the perceptron's output can be either 0 or 1, it can be considered a
binary classifier.
This is shown below in Equation 1
Equation 1: output of perceptron
Let’s write out the formula that joins the inputs and the weights together to produce the output.
Output = w1x1 + w2x2 + w3x3
Even though this function is simple, it nonetheless serves as the foundational formula for the perceptron, thus please read this equation as
Output'requires' w1x1 + w2x2 + w3x3.
This is because, in addition to being the total of these numbers, the outcome could also be dependent on a bias that is applied to this expression. A perceptron might be conceptualized as a "judge
who weighs up several pieces of evidence together with other rules and then makes a decision," to put it another way.
Basic Components of Perceptron
As a fundamental component of neural networks, the perceptron is made up of multiple essential parts:
• Input: The input signals that the perceptron receives can reflect the characteristics or properties of the data being processed. These signals can be binary values or real numbers. Usually, a
vector is used to represent these inputs. The signals x₁, x₂,..., xₙ are input signals to the perceptron.
• Weights: Every input has a weight assigned to it that indicates how important it is to the computation as a whole. The weights establish how much each input contributes to the perceptron's
output. These weights are first given random values, which are changed as the process of learning progresses. Every input has a corresponding weight, denoted by the characters w₁, w₂,..., wₙ.
• Summation Function: To obtain a weighted sum, the inputs are multiplied by the relevant weights and then added together. This phase requires taking the dot product of the input vector and the
weight vector. The weighted sum formula serves as a representation of this process: Z = w^ + wx₂ +... + wₙxₙ
• Activation Function: The activation function adds non-linearity to the output of the perceptron by passing the weighted sum through it. The rectified linear unit (ReLU) function, sigmoid
function, and step function are examples of common activation functions. Based on the calculated value, the activation function decides whether the perceptron will fire or stay dormant. It has
the notation f(z) for example sigmoid function Equation: A = 1/(1 + e-x) and many more like Tanh, RELU, Softmax, etc...
• Bias: To modify the output of the perceptron according to a predetermined threshold, a bias term is frequently added. Even with zero inputs, the perceptron can still learn patterns thanks to this
feature. Consequently, the symbol for bias is b.
• Output: The activation function applied to the weighted sum of inputs yields the perceptron's output, represented by the letter y. It displays the judgment or forecast made by the perceptron
using the input data. y = f(z + b)
• Learning Rule: Perceptrons use learning rules, such as the delta rule or the perceptron learning rule, to adjust their weights and biases while they are being trained. The difference between the
expected and intended outputs serves as the basis for this modification. Through iteratively replicating this learning procedure, the perceptron gradually improves its functionality. based on the
following equations:
Δwᵢ = α(yᵀ - t)xᵢ
Δb = α(yᵀ - t)
Types of Perceptron
The Perceptron can be categorized into two primary types: single-layer Perceptron and multi-layer Perceptron. Let us now delve into a detailed discussion of each type, exploring its unique features
and characteristics.
Single layer Perceptron
The single-layer Perceptron is made up of a single layer of neurons that adds up all of the inputs and uses an activation function to determine the output. It works especially well for issues that
can be solved linearly, or in which a straight line may divide the input data into two categories.
Multi-layer Perceptron
A multi-layer perceptron, in contrast to a single-layer perceptron, consists of multiple layers of neurons, with one or more hidden layers positioned between the input and output layers. The model's
hidden layers enable it to identify more complex patterns in the input data, which makes it suitable for handling problems that are not linearly separable.
Characteristics/Strength of the Perceptron
The following essential characteristics of the Perceptron Model enable it to be a powerful machine-learning tool:
• Complicated non-linear problems can be resolved with a multi-layered perceptron model.
• It works well with both small and large input data.
• It helps us to obtain quick predictions after the training.
• The Perceptron Model assumes that the data is linearly separable, which means that distinct classes of data points may be reliably separated by a hyperplane.
• The Perceptron Model uses labeled data for model training, a technique known as supervised learning. During training, the weights of the neurons are adjusted to minimize the error between the
expected and actual outputs.
• This kind of algorithm creates a threshold activation function that, based on whether or not the weighted total of the inputs exceeds the threshold value, produces a binary value.
• The Perceptron Model uses an online learning technique to modify the weights of its neurons after analyzing each input. Because of this feature, the model is very efficient and able to easily
handle big datasets.
Limitation of Perceptron
• Although the perceptron model is a useful tool for machine learning, it has many drawbacks, some of which are listed below:
• Only linearly separable problems—those in which a straight line may divide the input data into two groups—can be solved by the Perceptron algorithm. Only more complex models, such as support
vector machines or multi-layer Perceptrons, can handle nonlinearly separable problems.
• The Perceptron algorithm might not converge if the input data cannot be split linearly. This might lead to the model's inability to produce accurate predictions and the algorithm's perpetual
update of the weights.
• Due to the bias-variance trade-off inherent in the Perceptron algorithm, increasing model complexity may result in a decrease in bias but an increase in variance. The data may become over- or
under-fitted as a result.
• Probabilistic outputs are available for use in making decisions based on prediction probability, something that the Perceptron algorithm does not provide.
• The model's functioning depends on the quality of the training.
Application of Perceptron:
Artificial intelligence could be greatly influenced by perceptron in the future. Neural network building blocks, or perceptron, have already shown themselves to be capable of handling challenging
challenges across a range of fields. Perceptron is predicted to grow increasingly more potent and effective with future improvements in computing power and technology.
• Because efforts are being undertaken to interpret the judgments made by perceptron's and provide clear explanations, the future of explainable AI is likewise closely linked to the development of
perceptron. Perceptron's have the potential to transform multiple industries, such as healthcare, finance, and robotics, through continuous research and technical developments. These advances
will allow for intelligent systems that possess human-like efficiency and accuracy in learning, adapting, and making judgments.
• Perceptron is still utilized as building blocks for more intricate neural network architectures, despite their shortcomings and simplicity. Additionally, they are employed in educational contexts
to impart the principles of machine learning and neural networks. Perceptron are useful in practical applications for straightforward classification jobs when a simple decision boundary suffices,
and the data may be separated linearly.
Perceptron are still utilized as building blocks for more intricate neural network architectures, despite their shortcomings and simplicity. Additionally, they are employed in educational contexts to
impart the principles of machine learning and neural networks. Perceptron are useful in practical applications for straightforward classification jobs when a simple decision boundary suffices, and
the data may be separated linearly.
Also, read: Harnessing the Power of TensorFlow: Revolutionizing Machine Learning with Practical Examples | {"url":"https://old.lucentinnovation.com/blogs/technology-posts/understanding-the-perceptron","timestamp":"2024-11-05T02:21:25Z","content_type":"text/html","content_length":"102364","record_id":"<urn:uuid:8bc17a4a-81df-4836-ba84-8df069da8af8>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00491.warc.gz"} |
Tutorial on cellular automata - lecture 1
Appears in collection : Research School in Discrete Mathematics and Computer Science / École de recherche en mathématiques discrètes et informatique - WEEK 1
This tutorial surveys computational aspects of cellular automata, a discrete dynamical model introduced by S. Ulam and J. von Neumann in the late 40s: a regular grid of finite state cells evolving
synchronously according to a common local rule described by a finite automaton.
Formally, a cellular automaton is a tuple $(d, S, N, f)$ where $d \in \mathbb{N}$ is the dimension of the cellular space, $S$ is the finite set of states, $N \subseteq_{\text {finite }} \mathbb{Z}^d$
is the finite neighborhood and $f: S^N \rightarrow S$ is the local rule of the cellular automaton.
A configuration $c \in S^{\mathbb{Z}^d}$ is a coloring of the cellular space by states.
The global transition function $G: S^{\mathbb{Z}^d} \rightarrow S^{\mathbb{Z}^d}$ applies $f$ uniformly according to $N$, i.e. for every configuration $c \in S^{\mathbb{Z}^d}$ and every position $z \
in \mathbb{Z}^d$ it holds $$G(c)(z)=f\left(c\left(z+v_1\right), \ldots, c\left(z+v_m\right)\right) \quad \text { where } N=\left{v_1, \ldots, v_m\right} .$$ A space-time diagram $\Delta \in S^{\
mathbb{Z}^d \times \mathbb{N}}$ is obtained by piling successive configurations of an orbit, i.e. for every time step $t \in \mathbb{N}$ it holds $\Delta_{t+1}=G\left(\Delta_t\right)$.
Computing inside the cellular space: The first part of the tutorial considers cellular automata as a universal model of computation. Several notions of universality are discussed: boolean circuit
simulation, Turing universality, intrinsic universality. Special abilities of cellular automata as a model of massive parallelism are then investigated.
Computing properties of cellular automata: The second part of the tutorial considers properties of cellular automata and their computation. De Bruijn diagrams and associated regular languages are
introduced as tools to decide injectivity and surjectivity of the global transition function in the one-dimensional case. Both immediate and dynamical properties are introduced, in particular the
notion of limit set.
Computation and reduction: undecidability results: The last part of the tutorial considers computing by reduction to establish undecidability results on some properties of cellular automata:
injectivity and surjectivity of the global transition function in higher dimensions, nilpotency and intrinsic universality in every dimension, a Rice's theorem for limit sets.
Last related questions on MathOverflow
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Analytic Trigonometry
inverse functions [Solved!]
ShayT 02 Feb 2017, 19:30
My question
I have the problem sec(arccos 5) and I understand that sec is the reciprocal of cos. But if the domain of arccos is [0,pi] then how can this be 1/5?
Relevant page
Wolfram|Alpha: Computational Knowledge Engine
What I've done so far
I have a good understanding of the inverse functions and their domains, so I just don't get why you can even do arccos of 5. | {"url":"https://www.intmath.com/forum/analytic-trigonometry-25/inverse-functions:125","timestamp":"2024-11-06T15:20:11Z","content_type":"text/html","content_length":"109253","record_id":"<urn:uuid:546d0a43-7e71-4bbb-94fb-8b8b219dc274>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00620.warc.gz"} |
Bayesian Analysis with R for Drug Development: Concepts, Algorithms, and Case Studies
1138295876, 9781138295872 - DOKUMEN.PUB
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Bayesian Analysis with R for Drug Development Concepts, Algorithms, and Case Studies
Chapman and Hall/CRC Biostatistics Series Shein-Chung Chow, Duke University School of Medicine Byron Jones, Novartis Pharma AG Jen-pei Liu, National Taiwan University Karl E. Peace, Georgia Southern
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Bayesian Analysis with R for Drug Development Concepts, Algorithms, and Case Studies
Harry Yang and Steven J. Novick AstraZeneca, Gaithersburg, Maryland
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Library of Congress Cataloging‑in‑Publication Data Names: Yang, Harry, author. | Novick, Steven, author. Title: Bayesian analysis with R for biopharmaceuticals / Harry Yang, Steven Novick.
Description: Boca Raton : CRC Press, Taylor & Francis Group, 2019. | Includes bibliographical references and index. Identifiers: LCCN 2019005802 | ISBN 9781138295872 (hardback : alk. paper) Subjects:
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Contents Preface.......................................................................................................................xi
Section IBackground 1 Bayesian Statistics in Drug Development.................................................. 3 1.1
Introduction............................................................................................ 3 1.2 Overview of Drug Development.........................................................3
1.2.1 Basic Research........................................................................... 4 1.2.2 Drug Discovery.........................................................................5 1.2.3
Formulation...............................................................................5 1.2.4 Laboratory Test Methods.........................................................6 1.2.5 Pre-Clinical
Studies.................................................................. 6 1.2.6 Clinical Development............................................................... 7 1.2.6.1 Phase I Clinical
Trial.................................................7 1.2.6.2 Phase II Clinical Trial............................................... 7 1.2.6.3 Phase III Clinical
Trial.............................................. 8 1.2.6.4 Phase IV Clinical Trial.............................................. 8 1.2.7 Translational
Research.............................................................8 1.2.8 Chemistry, Manufacturing, and Controls.............................8 1.2.9 Regulatory
Registration...........................................................9 1.3 Statistics in Drug Research and Development..................................9 1.4 Bayesian
Statistics................................................................................ 11 1.5 Opportunities of the Bayesian Approach......................................... 12 1.5.1 Pre-Clinical
Development..................................................... 12 1.5.2 CMC Development................................................................. 12 1.5.3 Clinical
Trials.......................................................................... 13 1.6 Challenges of the Bayesian Approach.............................................. 13 1.6.1 Objection to
Bayesian............................................................. 13 1.6.2 Regulatory Hurdles................................................................ 14 1.7 Concluding
Remarks........................................................................... 15 2 Basics of Bayesian Statistics........................................................................ 17 2.1
Introduction.......................................................................................... 17 2.2 Statistical
Inferences............................................................................ 18 2.2.1 Research Questions................................................................ 18 2.2.2 Probability
Distribution......................................................... 18 2.2.3 Frequentist Methods.............................................................. 18 2.2.4 Bayesian
Inference.................................................................. 19 v
2.2.4.1 Bayes’ Theorem and Posterior Distribution........ 19 2.2.4.2 Inference about Parameters................................... 20 2.2.4.3 Inference of Future Observations.........................
21 2.2.5 Selection of Priors...................................................................22 Bayesian Computation........................................................................ 23 2.3.1
Monte Carlo Simulation......................................................... 23 2.3.2 Example.................................................................................... 24 2.3.3 Rejection
Sampling................................................................. 28 2.3.4 Markov Chain Monte Carlo.................................................. 29 2.3.4.1 Gibbs
Sampling....................................................... 29 2.3.4.2 Metropolis–Hastings.............................................. 31 Computational
Tools...........................................................................34 2.4.1 BUGS and JAGS.......................................................................34 2.4.2 SAS PROC
MCMC.................................................................. 35 2.4.3 Utility of JAGS......................................................................... 35 Concluding
Remarks........................................................................... 39
3 Bayesian Estimation of Sample Size and Power..................................... 41 3.1 Introduction.......................................................................................... 41
3.2 Sample Size Determination................................................................ 41 3.2.1 Frequentist Methods.............................................................. 41 3.2.2
Bayesian Considerations........................................................43 3.2.2.1 Prior Information....................................................43 3.2.2.2 Use of Historical
Data.............................................44 3.2.3 Bayesian Approaches............................................................. 45 3.3 Power and Sample
Size....................................................................... 46 3.3.1 Phase II Study.......................................................................... 47 3.3.1.1 Test
Procedure......................................................... 47 3.3.1.2 Sample Size Calculations....................................... 47 3.3.1.3 Incorporation of
Prior............................................. 49 3.3.1.4 Proper Bayesian Procedure.................................... 51 3.3.1.5 Two Prior
Distributions.......................................... 51 3.4 Interim Analysis ..................................................................................54 3.4.1 Futility and Sample
Size........................................................54 3.5 Case Example........................................................................................ 55 3.5.1 Modeling of Overall
Survival .............................................. 55 3.5.2 Maximum Likelihood Estimation........................................ 56 3.5.3 Futility
Analysis...................................................................... 56 3.6 Concluding Remarks........................................................................... 60
Section II Pre-Clinical and Clinical Research 4 Pre-Clinical Efficacy Studies......................................................................63 4.1
Evaluation of Lab-Based Drugs in Combination.............................64 4.2.1 Background..............................................................................64 4.2.2 Statistical
Methods.................................................................64 4.2.2.1 Loewe Additivity....................................................64 4.2.2.2 Bliss
Independence.................................................65 4.2.3 Antiviral Combination...........................................................65 4.2.3.1
Data........................................................................... 66 4.2.3.2 Model........................................................................ 66 4.2.3.3 Assessment of Drug
Effect..................................... 67 4.2.3.4 Use of Historical Data as Priors............................ 71 4.2.4 Evaluation of Fixed Dose Combination .............................. 72
4.2.4.1 Follow-up Experiment............................................ 72 Bayesian Survival Analysis................................................................77 4.3.1 Limitations of Animal
Data..................................................77 4.3.2 Current Methods..................................................................... 78 4.3.3 Bayesian
Solution.................................................................... 79 4.3.3.1 Survival Function....................................................80 4.3.3.2 Weibull
Modeling....................................................80 4.3.4 Case Example..........................................................................83 Concluding
Remarks........................................................................... 88
5 Bayesian Adaptive Designs for Phase I Dose-Finding Studies........... 89 5.1 Introduction.......................................................................................... 89 5.2
Algorithm-Based Designs................................................................... 89 5.2.1 3 + 3 Design............................................................................. 89 5.2.2
Alternate Algorithm-Based Designs.................................... 90 5.2.3 Advantages and Disadvantages of Algorithm-Based
Designs..................................................................................... 91 5.3 Model-Based Designs.......................................................................... 92
5.3.1 Continual Reassessment Method......................................... 92 5.3.1.1 Models....................................................................... 92 5.3.1.2 Procedure for Finding
MTD.................................. 93 5.3.2 CRM for Phase I Cancer Trials.............................................. 94 5.3.3 Escalation with Overdose
Control....................................... 98 5.3.4 Escalation Based on Toxicity Probability Intervals.......... 103 5.3.4.1 Toxicity Probability Intervals.............................. 103
5.3.4.2 Model...................................................................... 103 5.3.4.3 Method.................................................................... 104 5.3.4.4 Method
Implementation...................................... 105 5.4 Concluding Remarks......................................................................... 106 6 Design and Analysis of Phase II
Dose-Ranging Studies.................... 107 6.1 Introduction........................................................................................ 107 6.2 Phase II Dose-Ranging Studies
....................................................... 108 6.2.1 Criticisms of Traditional Methods..................................... 108
Model-Based Approaches.................................................... 109 6.2.2.1 Modeling Dose–Response Curve........................ 109 6.2.2.2 Determination of Minimum Efficacy Dose.......
109 Estimating Predictive Precision and Assurance for New Trial.......110 6.3.1 COPD Study........................................................................... 110 6.3.2 Estimation
Method............................................................... 110 6.3.2.1 Selection of Priors.................................................. 112 6.3.2.2 Estimation of Dose–Response
Curve................. 114 6.3.2.3 Estimation of Precision and Assurance............. 115 Concluding Remarks......................................................................... 118
7 Bayesian Multi-Stage Designs for Phase II Clinical Trials................ 121 7.1 Introduction........................................................................................ 121 7.2 Phase
II Clinical Trials...................................................................... 122 7.3 Multi-stage Designs........................................................................... 122
7.3.1 Frequentist Approaches....................................................... 122 7.3.2 Bayesian Methods................................................................. 123 7.3.3 Bayesian
Single-Arm Trials................................................. 123 7.3.3.1 Go or No-Go Criteria............................................ 124 7.3.3.2 Predictive
Probability........................................... 124 7.3.4 Continuous Monitoring of Single-Arm Trials.................. 125 7.3.5 Comparative Phase II
Studies............................................. 126 7.3.5.1 Efficacy and Futility Based on Posterior Probability.............................................................. 126 7.3.5.2
Efficacy and Futility Criteria Based on Predictive Probability........................................... 127 7.4
Examples............................................................................................. 128 7.4.1 Oncology Trial.......................................................................
128 7.4.2 Multi-Stage Bayesian Design .............................................. 131 7.5 Concluding Remarks......................................................................... 137
Section III Chemistry, Manufacturing, and Control 8 Analytical Methods..................................................................................... 141 8.1
Introduction........................................................................................ 141 8.2 Method
Validation............................................................................. 142 8.2.1 Background............................................................................ 142 8.2.2
Study Design for Validation of Accuracy and Precision...... 143 8.2.2.1 Design Considerations......................................... 143 8.2.3 Current Statistical
Methods................................................ 144 8.2.3.1 Definitions.............................................................. 144 8.2.3.2
Methods.................................................................. 145 8.2.4 Total Error Approach........................................................... 145 8.2.5 Bayesian
Solutions................................................................ 146
8.2.6 Example.................................................................................. 148 8.2.6.1 Data......................................................................... 148 8.2.6.2
Analysis.................................................................. 149 8.2.6.3 Results..................................................................... 151 8.2.6.4 Analysis Based on More
Informative Priors........153 Method Transfer................................................................................. 156 8.3.1
Background............................................................................ 156 8.3.2 Model...................................................................................... 156 8.3.3
Linear Response.................................................................... 157 8.3.4 Case Example........................................................................ 158 Concluding
Remarks......................................................................... 162
9 Process Development.................................................................................. 163 9.1
Introduction........................................................................................ 163 9.2 Quality by
Design.............................................................................. 164 9.3 Critical Quality Attributes................................................................ 165 9.3.1 Risk
of Oncogenicity............................................................ 166 9.3.2 Bayesian Risk Assessment................................................... 167 9.3.3 Modeling Enzyme Cutting
Efficiency............................... 167 9.3.4 Bayesian Solution.................................................................. 168 9.3.5
Example.................................................................................. 170 9.4 Design Space.......................................................................................
173 9.4.1 Definition............................................................................... 173 9.4.2 Statistical Methods for Design Space................................. 175 9.4.2.1
Overlapping Mean................................................ 175 9.4.2.2 Desirability Method.............................................. 175 9.4.2.3 Criticisms of Current
Methods........................... 177 9.4.3 Bayesian Design Space......................................................... 178 9.4.3.1 Regression
Model.................................................. 178 9.4.3.2 Prior Information.................................................. 179 9.4.3.3 Posterior Predictive Probability and
Design Space.......................................................... 179 9.4.4 Example.................................................................................. 181 9.5 Process
Validation.............................................................................. 186 9.5.1 Risk-Based Lifecycle Approach.......................................... 186 9.5.2 Method Based on
Process Capability................................ 187 9.5.2.1 Frequentist Acceptance Criterion....................... 187 9.5.2.2 Bayesian Acceptance Criterion............................ 188 9.5.3
Method Based on Predictive Performance........................ 189 9.5.4 Determination of Number of PPQ Batches....................... 191 9.6 Concluding
Remarks......................................................................... 193 10
Stability......................................................................................................... 195 10.1
Introduction........................................................................................ 195 10.2 Stability
Study.................................................................................... 196
10.3 Shelf-Life Estimation......................................................................... 197 10.3.1 Current Methods................................................................... 197
10.3.2 Bayesian Approaches........................................................... 198 10.3.3 Examples................................................................................ 200 10.3.3.1
Shelf Life of Influenza Vaccine............................ 200 10.3.4 Selection of Stability Design................................................ 207 10.3.5 Bayesian
Criterion................................................................. 208 10.3.5.1 Design Options...................................................... 208 10.3.5.2
Results..................................................................... 209 10.4 Setting Release Limit......................................................................... 211 10.4.1
Background............................................................................ 211 10.5 Concluding Remarks......................................................................... 215 11
Process Control............................................................................................ 217 11.1
Introduction........................................................................................ 217 11.2 Quality Control and Improvement.................................................. 218
11.3 Control Charts.................................................................................... 219 11.4 Types of Control
Charts.................................................................... 220 11.4.1 Shewhart I-MR Charts......................................................... 221 11.4.2 EWMA Control
Chart..........................................................223 11.4.3 CUSUM Chart.......................................................................225 11.4.4
J-Chart.................................................................................... 226 11.4.5 Multivariate Control Chart.................................................. 227 11.5 Bayesian
Control Charts................................................................... 227 11.5.1 Control Chart for Data with Censoring............................. 227 11.5.2 Control Chart for Discrete
Data.......................................... 229 11.5.3 Control Limit for Aberrant Data.........................................234 11.5.3.1
Background............................................................234 11.5.3.2 Methods.................................................................. 235 11.5.4 Product Quality Control Based
on Safety Data from Surveillance........................................................................... 239 11.5.4.1 Background............................................................ 239
11.5.4.2 Current Methods................................................... 239 11.5.4.3 Zero-Inflated Models............................................ 241 11.5.4.4 Alert Limit for
AEs................................................ 242 11.6 Concluding Remarks......................................................................... 244 Appendix: Stan Computer
Code...................................................................... 247 References............................................................................................................
283 Index...................................................................................................................... 301
Preface Drug development is an empirical problem-solving process, characterized by trial and error. Knowledge gleaned from the previous study is often used to guide the next experiment. The
sequential learning nature of experimentation and reliance on knowledge from various drug development stages calls for statistical methods that enable synthesis of information from disparate sources
to aid better decision-making. Bayesian statistics provide a flexible framework for continuous update of learning based on new information. In addition, Bayesian analysis allows for the incorporation
of prior knowledge, in terms of either expert opinion or historical data, in its statistical inferences. This not only helps ease the reliance on large sample approximations that are often required
for frequentist methods but also often results in greater efficiency in study design. Furthermore, many practitioners in drug development find it difficult to interpret the frequentist interval
estimates and p-values, consequently creating a challenging and confusing situation for decision-makers. In contrast, the results of Bayesian analysis are typically presented through probabilistic
statements. For example, after analyzing data from a two-arm comparative study, in which a test drug is assessed against the standard of care (SOC) therapy, both the frequentist and Bayesian conclude
that the experimental drug is more effective than the SOC. The Bayesian directly concludes that the experimental drug is 20% more effective than the SOC with 95% probability. In contrast, the
frequentist deduces that, conditioned on the data, the effectiveness of the experimental drug cannot be equal to that of the SOC (and thus, the experimental drug must be the better treatment). The
straightforward feature of the Bayesian inference is extremely desirable as it brings a risk-based approach to bear for decision-making in drug development, as recommended by the current regulatory
guidelines. In the past two decades, the advances in Bayesian computations such as Markov chain Monte Carlo (MCMC) simulation have made it possible to implement sophisticated Bayesian analysis. The
release of regulatory guidance on the potential use of Bayesian statistics in medical device development and regulatory endorsement of the Bayesian methods for early clinical development has further
fueled the enthusiasm for Bayesian applications to drug development issues, including those previously considered intractable. Although several Bayesian books have been published to address a broad
array of drug development problems, they are primarily focused on clinical trial design, safety analysis, observational studies, and cost-effectiveness assessment. Bayesian methodologies remain
unfamiliar to the majority of statistical practitioners in the non-clinical areas. Suffice to say, the lack of adoption of Bayesian methods in those non-clinical areas, including drug xi
discovery, analytical method development, process optimization, and manufacturing control has resulted in many missed opportunities for statisticians to make meaningful differences. Neither is this
in keeping with the recent regulatory initiatives of quality by design (QbD), which achieves product quality through greater understanding of the product and manufacturing process, based on knowledge
and data collected throughout the lifecycle of the product development. It is the desire to fill the aforesaid gap that motivates us to write this book. The aim of this book is to provide Bayesian
applications to a wide range of clinical and non-clinical issues in drug development. Each Bayesian method in the book is used to address a specific scientific question and illustrated through a case
study. The R code used for implementing the method is discussed and included. It is our belief that the publication of this book will promote the use of Bayesian approaches in pharmaceutical
practices. The book consists of three parts, totaling 11 chapters. Since the primary aim of this book is to use case studies, examples, and easy-to-follow R code to demonstrate Bayesian applications
in the entire spectrum of drug development, it is, by no means, meant to be comprehensive in literature review, nor is it intended to be exhaustive in expounding each application. Below is a brief
description of each chapter: Part I (Chapters 1–3) provides backgrounds of drug research and development, and basics of Bayesian statistics. In Chapter 1, after a brief overview of drug research and
development centered on drug discovery, pre-clinical and clinical programs, and CMC development, we discuss the opportunities and challenges of Bayesian applications. Chapter 2 is concerned with the
basic theory of Bayesian inference and computational tools. Chapter 3 uses three examples, one regarding lot release and the other two concerning Bayesian study designs, to illustrate how sample size
is determined for the purpose of hypothesis testing. Part II (Chapters 4–7) discuss various Bayesian applications in pre-clinical animal studies and clinical development, notably with focus on Phase
I and II clinical trials. In Chapter 4, novel Bayesian methods are explained, which utilize historical data to compensate for the usual lack of large sample size and, thus, the inferential power of
statistical tests in animal efficacy assessment. Chapter 5 is concentrated on model-based Phase I dose-finding study design and analysis. Two examples are presented to illustrate the use of Bayesian
continuous reassessment methods. Chapter 6 describes the use of the Bayesian method in dose-ranging studies. Chapter 7 discusses Bayesian Phase II studies, consisting of an example of a single-arm
Phase IIa study with continuous monitoring to show the early efficacy of an experimental therapy, and another case study of Phase IIb to determine an optimum dose regimen for future Phase III trials.
Part III (Chapters 8–11) focuses on selected topics of Bayesian methods used for CMC development. Chapter 8 shows how to validate the performance of an analytical method regarding precision,
accuracy, and linearity
using Bayesian inferences. Chapter 9 deals with the construction of a Bayesian design space for process development. Chapters 10 and 11 are concentrated on the discussion of stability analysis and
process control. Throughout the above chapters, some of the computer code calls upon external *.csv files. These files are available for download at the following site: https://www.crcpress.com/
9781138295872 under “Additional Resources”. In addition, various applications are carried out using R code written in JAGS. For readers who are more familiar with Stan, we include R code in Stan in
the appendix. We thank John Kimmel, executive editor, Chapman & Hall/CRC Press, for providing us with the opportunity to write this book. We express our gratitude to Lorin Roskos for reviewing the
entire book, and four anonymous reviewers for their review and comments on the book proposal. We also thank Jianchun Zhang for helping resolve an R-coding issue with one of the examples in Chapter 8.
Lastly, the views expressed in this book are those of the authors and not necessarily those of AstraZeneca. Harry Yang and Steven Novick Gaithersburg, Maryland
Harry Yang is senior director and head of Statistical Sciences at AstraZeneca. He has 24 years of experience across all aspects of drug research and development and extensive global regulatory
experiences. He has published six statistical books, 15 book chapters, and over 90 peer-reviewed papers on diverse scientific and statistical subjects, including 15 joint statistical works with Dr.
Novick. Dr. Yang is a frequently invited speaker at national and international conferences. He also developed statistical courses and conducted training at the FDA and USP, as well as in Peking
University, China. Steven Novick is director of Statistical Sciences at AstraZeneca. He has extensively contributed statistical methods to the biopharmaceutical literature. Dr. Novick is a skilled
Bayesian computer programmer and is frequently invited to speak at conferences, having developed and taught courses in several areas, including drug combination analysis and Bayesian methods in
clinical areas. He served on IPAC-RS and has chaired several national statistical conferences.
Section I
1 Bayesian Statistics in Drug Development
1.1Introduction Drug research and development is a long, costly, and arduous process. Despite advances and breakthroughs in science and technologies, the attrition rate of new drugs remains high.
According to a recent report, bringing a novel drug to the market may cost as much as 1.8 billion dollars and take 13 years. Adding to pharmaceutical-industry woes, an increasing number of drug
recalls due to manufacturing issues have caused product safety concerns. In recent years, a growing emphasis has been placed on better prediction of clinical outcomes and control of manufacturing
risk, using all available data. Bayesian statistics, a well-known framework for synthesizing information from disparate sources, provides such an opportunity. In this chapter, we begin with a brief
overview of the drug development process. Subsequently, we discuss the opportunities and challenges of applying Bayesian statistics in drug development.
1.2Overview of Drug Development The overarching aim of drug development is to bring safe and effective drugs to the market to meet the unmet medical needs. Drug development is a complex, lengthy,
and resource-intensive process. It involves drug discovery, formulation development, pre-clinical animal studies, clinical trials, and regulatory filings. As reported by Bunnage (2011), it takes as
much as 1.8 billion US dollars and approximately 13 years to develop an effective drug. Furthermore, drug development is strictly regulated by laws and governmental policies. Ever since the
Kefauver–Harris Amendments, a drug must be shown to be both safe and efficacious before marketing approval (Peltzman 1973; FDA 2012). To this end, scientifically well-designed and controlled studies
in both animal and human populations are required. In
Bayesian Analysis with R for Drug Development
addition, regulations, such as the US current Good Manufacturing Practice (cGMP), require that modern standards and technologies be adopted in the design, monitoring, and control of manufacturing
processes and facilities to ensure a consistent supply of high-quality drug products to the consumer or public (FDA 1995). In 2004, the FDA launched a significant initiative entitled “Pharmaceutical
cGMPs for the 21st Century: A Risk-Based Approach” (FDA 2004a). The initiative was driven by a vision to achieve a desired state of the pharmaceutical industry as “a maximally efficient, agile,
flexible manufacturing sector that reliably produces high-quality drug products without extensive regulatory oversight” (Woodcock 2012). The subsequent publications of several regulatory documents
(ICH 2006, 2007a, 2007b, 2011a) further stress the importance of manufacturing process development based on systematic product and process understanding, control of risk, and implementation of
quality management systems. Increasingly regulatory guidelines stipulate the use of statistics and good statistical practice in study design and analysis (ICH 1998; FDA 2001, 2004a, 2011a, 2016a).
The reliance on statistics has been further intensified as the industry becomes more focused on use of “big data” such as genomics, proteomics, transcriptomics, and real-world evidence to develop
precision medicine. Figure 1.1 presents a diagram of drug development. An outline of each stage is provided in the following sections. 1.2.1Basic Research Drug development begins with research on
the inner working of a disease at the molecular level. Specifically, various studies are conducted to gain insights on causes of gene mutations, effect of proteins they encode, interaction among
those proteins in living cells and their host tissues, and ultimately the patient (Petrova 2012). For example, for years, researchers have
FIGURE 1.1 Drug development process.
Bayesian Statistics in Drug Development
noted that some cancers grow quickly and spread rapidly while others do not; however, the cause of the phenomenon has been elusive. It was not until the discovery of the mutated gene called HER2 in
breast cancer that scientists understood that the excessive growth of cancer cells in patients with breast cancer was likely caused by high expression of HER2. This critical discovery sparked
researchers to look for ways to block HER2 genes in order to slow the growth of HER2-positive breast cancer. The effort finally led to the successful development of trastuzumab (Herceptin®). Advances
in genomics, proteomics, and computational capacity have presented unprecedented opportunities for scientists to understand the underlying causes of a disease. The desire to bring innovative
medicines to the market has also inspired broad collaborations among researchers from industry, academia, and government. These collective efforts have contributed greatly to the development of
innovative drugs. 1.2.2Drug Discovery Understanding the causes of a disease often leads to the identification of a biological target. A biological target is any component in a living organism that
influences the disease. Examples of common biological targets include genes and proteins. As previously noted (PhRMA 2015), even at this early stage of drug discovery, it is critical to choose a
target that can potentially interact with and be affected by a drug molecule. After a target is identified, its association with the disease must be validated through in vitro and in vivo
experiments. Drawing from the understanding of underpinnings of the disease and the potential target, scientists begin to find a drug molecule that can interact with the target and change the course
of the disease. Most notable among various methods used for this purpose are 1) screening chemical libraries of synthetic small molecules, natural products, or extracts using in vitro assays (usually
low-to-medium throughput) to look for compounds of desired therapeutic effect (https://en.wikipedia.org/wiki/Drug_discovery); 2) high-throughput screening of large compound libraries in search of
disease-altering molecules; and 3) genetically engineering of molecules that have high affinity to the target. These lead compounds advance to the next stage of testing in which their toxic-kinetic
properties are evaluated in cell and animal models. Those compounds that meet the selection criteria are further optimized to increase affinity, selectivity, efficacy, and stability.
1.2.3Formulation Drugs must be properly formulated in certain dosage forms to ease production and delivery and maximize therapeutic benefits. The objective of formulation development is to design
and establish both a formulation composition and its manufacturing process to meet the above requirements. The development of an acceptable dosage form can be extremely challenging
Bayesian Analysis with R for Drug Development
particularly for biological products (Ng and Rajagopalan 2009). The difficulties stem from the fact that biological products are inherently complex and heterogeneous. This makes it formidable to
characterize the product and establish the links between the product attributes and process parameters as well as clinical performance (Singh et al. 2009). Formulation development is a continuous
endeavor throughout drug development. At the early stage, only a small amount of the drug is produced in a laboratory setting for early phase clinical studies. As the drug program advances to
late-stage clinical trials, large quantities of the drug are needed. Since techniques for smallscale production of the drug often do not translate easily to large scales, additional efforts are
needed to enhance and optimize the formulation. In recent years, manufacturers began to embrace the Quality by Design principles for formulation development. 1.2.4Laboratory Test Methods Analytical
testing is an integral part of drug research and development. Analytical methods are developed to determine identity, strength, quality, purity, and potency of the drug. According to regulatory
guidance (FDA 2015; USP 1989, 2013), an analytical method should be validated for its intended use. In the early stage of drug development, analytical results are used to guide the selection of lead
compound. For pre-clinical and clinical programs, analytical methods aid optimal dose selection and assessment of study endpoints. They are also widely used for formulation and process development
and ultimately manufacturing control to ensure the quality of the drug substance and finished product. Additionally, the development of clinical testing methods such as immunogenicity assays also
plays a very important role in the overall drug development. For example, immunogenicity assays are used to detect, quantify, and characterize anti-drug antibodies, which may have profound impact on
drug safety and efficacy. The development and validation of such assays are also strictly regulated (FDA 2007 and WHO 2009) and pose a host of unique challenges, requiring extra care in both study
design and data analysis (Yang et al. 2016). 1.2.5Pre-Clinical Studies Before testing an investigational drug in human subjects, animal studies are carried out with the primary aims of selecting a
safe dose for human trials and determining the safety profile of the studied drug. Pre-clinical testing can be performed both in vitro and in vivo. Although different types of pre-clinical studies
may be required, in most cases, toxicity pharmacodynamics, pharmacokinetics, and absorption, distribution, metabolism, and excretion (ADME) studies are carried out to determine a safe dose for human
testing. It is also important to note that all pre-clinical studies must be conducted in accordance with Good Laboratory Practice (GLP) (FDA 2007).
Bayesian Statistics in Drug Development
1.2.6Clinical Development Upon completion of pre-clinical development, the drug is ready to be studied in human subjects. The sponsor must submit an Investigational New Drug Application (IND) with
the FDA or Investigational Medicinal Product Dossier (IMPD) with the EMA if a clinical trial is to be conducted in one or more European Union member states. IND and IMPD are requests for FDA and EMA
authorization to administer the investigational drug to humans. Both filings contain data from non-clinical studies, quality, manufacture, and control of the investigational drug as well as
comparator(s) if applicable. In addition to the regulatory approval, the intended clinical study must also be endorsed by the Institutional Review Board (IRB) at the sites where the trial is
conducted. 1.2.6.1Phase I Clinical Trial The primary objective of a Phase I study is to determine the safety profile and to study the pharmacokinetic property of the drug. For drugs with moderate
toxicity, the trial is normally carried out using healthy male volunteers. However, for cytotoxic agents, Phase I trials are usually conducted in the target patient populations to minimize
unnecessary exposures of the drugs in healthy volunteers. Dose-escalation designs may be used (see Chapter 5). They typically begin with a low dose of the drug predicted from the animal studies and
progressively escalate to higher doses if the drug is well tolerated. A range of doses is explored, and the maximum tolerated dose is determined. The study also collects pharmacodynamic and
pharmacokinetic data to address important questions such as side effects, therapeutic effects, and ADME. The knowledge garnered from this stage of development, including the safe dosing range, is
used to guide the next phase of clinical development. 1.2.6.2Phase II Clinical Trial After a maximum tolerated dose is identified from Phase I trials, Phase II studies are carried out with the
primary focus on demonstrating the efficacy of the drug and finding an optimum dosing regimen. The studies are usually conducted in patients who have an illness or condition that the drug is intended
to treat. These are relatively larger trials with several hundreds of patients. Phase II trials can be further divided into Phase IIa and Phase IIb studies. The former are typically single-arm trials
to screen out inefficacious drugs. The latter usually contain treatment arms of the drug at different dose levels and dosing schedules and a control arm. They are often randomized trials to allow for
accurate assessment of treatment effects. At the conclusion of the Phase II trials, researchers expect either to have identified the effective dose, route of administration, and dosing range for
Phase III trials, or to decide to terminate the clinical development of the drug.
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1.2.6.3Phase III Clinical Trial Phase III trials, also known as pivotal or confirmatory trials, are conducted on a much larger number of patients to substantiate safety and efficacy findings from
the previous studies in support of market approval of the drug. These studies are often lengthy, multi-center, possibly global, and are consequently very costly to complete. Most Phase III trials are
randomized, double-blind, and multi-armed with a comparator. To gain marketing approval by regulatory approval, typically two Phase III trials are required. 1.2.6.4Phase IV Clinical Trial For a new
drug, Phase IV trials are conducted for various purposes. They may be used to assess the long-term effect of the drug. They may also be carried out to determine risk and benefit in specific subgroups
of patients. Further, a Phase IV trial may be conducted to support market authorizations in different regions or countries and expand product label. 1.2.7Translational Research In recent years, a
great deal of emphasis has been placed on translational research to expedite and shorten the time of “bench-to-bedside” of a new medicine, by harnessing knowledge of both basic science and clinical
research. Translational research leverages data generated from advanced technologies, such as next-generation gene sequencing, proteomics, bioinformatics, and robotics, to identify drug candidates
that have greater potential to be developed into novel drugs. In addition, it also promotes incorporation of learnings from the clinical setting in drug discovery. For instance, the analysis of
patient samples might help identify a subgroup that responds better to the treatment. Thus, a companion diagnostic tool based on genomic or proteomic markers may ensure more targeted therapy
development and enhance the probability of success. It is for these reasons that translational research techniques have been widely utilized to advance new drug development. 1.2.8Chemistry,
Manufacturing, and Controls Chemistry, manufacturing, and controls (CMC) is an integral part of the overall drug development and is required in marketing approval. It consists of several important
components, including manufacturing of bulk drug substance and final drug product, setting specifications, release criteria, stability testing, comparability studies after process changes, and
analytical method development and validation. CMC issues become increasingly complex as a candidate drug is being advanced to late-stage development. Often
Bayesian Statistics in Drug Development
statistical methods such as modeling simulation and design of experiments (DOE) are used to develop, optimize, and validate CMC processes so that they are fit for their intended purposes. To ensure
drug safety, efficacy, and quality, manufacturing processes are strictly controlled according to GMP (FDA 2004a). Many regulatory guidelines have been established, governing the process control,
product characterization, and release. In recent years, pharmaceutical companies have been encouraged to embrace the advances in manufacturing technologies to modernize their production system and
meet regulatory standards. 1.2.9Regulatory Registration Although the regulatory requirements for premarketing approval of a new drug and review vary, there are common elements in new drug
applications (NDA) and biologics license applications (BLA). In the United States, the FDA requires that an NDA/BLA include integrated summaries of pre-clinical and clinical study results. In
addition, the application should also contain information on CMC and proposed labeling (FDA 1999). The NDA/BLA is reviewed by the FDA. After comprehensive review, the FDA may approve the drug product
or request additional data or studies. For marketing authorization application in Europe and other regions, similar requirements are mandated by regulatory authorities. While the regulatory agencies
have the ultimate authority to approve or disapprove the applications, they may solicit the opinion of an independent advisory committee, consisting of experts in the field.
1.3Statistics in Drug Research and Development Statistics has been broadly used in virtually all aspects of drug research and development. There are several drivers behind the statistical
applications. Firstly, there is growing governmental control over the content and claim of medicinal products. For example, ever since 1906, the United States has passed several laws such as the
Federal Food, Drug, and Cosmetics Act in Kefauver–Harris Drug Amendment in 1962, and the NDA Rewrite in 1990, which require drug manufacturers to demonstrate drug safety and effectiveness through
well-controlled and designed studies. These regulatory requirements entail the needs of statistics in pre-clinical and clinical testing (ICH 1998). The recent advances in regulatory policies have
brought statistics to the forefront of other areas of drug development including formulation, analytical method, and manufacturing control (ICH 2006, 2007a, 2007b, 2011). In both the FDA’s PAT
(process analytical technologies) and process
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validation guidance (FDA 2004a, 2011a), the use of multivariate tools for design, data acquisition, and analysis is specifically recommended. In addition, the advances in platforms in areas such as
genomics and proteomics enable scientists to generate large quantities of data in a very short time period. The need to make sense out of the data to drive key decision-making necessitates the use of
statistics. Moreover, innovations in statistics such as adaptive design have brought flexibility and economy into clinical development (Chow and Chang 2007; Chang 2008). In drug development, two
commonly used statistical approaches are the frequentist and Bayesian methods. The former addresses a scientific question solely based on data collected from an experiment designed to answer the
question; whereas the latter reaches an answer using not only the data from the current experiment but also prior knowledge of the question. Consider a two-arm clinical study intended to demonstrate
that an innovative drug improves the rate of a disease by 40% over a comparator drug. In the frequentist paradigm, the rates of responses of the two drugs are deemed to be fixed as is the
hypothesized difference of 40%. The question is answered by testing the null hypothesis, H0: there is no difference in response rate between the two drugs, against the alternative hypothesis, Ha: the
rates differ by at least 40%. Upon completion of the study, the rate difference δ based on the trial data is estimated. Under the null hypothesis, the probability for the future rate difference from
a repeated experiment to exceed the observed difference δ is calculated. This probability, often called the p-value, depends on the statistical model or distribution assumed for the data, the
observed results from the current experiment, and the fixed difference in rate. The null hypothesis is rejected and a significant difference claim is made if the p-value is small; otherwise, a lack
of difference conclusion is reached. By contrast, the Bayesian approach blends data from the current study with prior beliefs/knowledge about the treatment effect. In addition, under the Bayesian
setting, the unknown treatment effect is viewed as a variable that varies according to a distribution. The prior knowledge is updated in light of the trial results and expressed as a posterior
distribution. From this distribution, the probability of the rate difference ≥40% is calculated. If this probability is high, say, exceeding 90%, one may conclude that there is a significant
difference between the two drugs. The idea is illustrated in Figure 1.2. The derivation of the posterior distribution is obtained through Bayes’ Theorem, which is briefly discussed below. Elucidation
of the prior distribution and inference based on the posterior distribution is discussed in Chapter 2. Bayes’ Theorem also makes it possible to make inference about the future observations from
repeated experiment(s) through predictive probability. Central to this inference is the derivation of the probability distribution of the future observations conditional on the data from the current
Bayesian Statistics in Drug Development
FIGURE 1.2 Posterior distribution incorporates new evidence from the current study into the prior distribution.
1.4Bayesian Statistics As previously discussed, there are two types of statistical methodologies, namely, frequentist and Bayesian, that are applicable for study design and analysis in drug
development. Traditionally the frequentist methods were predominantly used for drug experimental design and analysis. In recent years, the advances in both Bayesian methodologies and computations
have propelled Bayesian methods to the forefront of drug research and development. An important fundamental difference between frequentist and Bayesian methodologies is the use of prior information.
Although frequentists use prior beliefs at the study design stage, they do not use it in formal analysis. In contrast, the Bayesian system provides a formal framework to combine prior information
with current data to make inferences about a quantity of interest. In addition, the Bayesian approach provides great flexibility to update inferences each time new data become available. Bayesian
analysis also affords the advantages of straightforward predictive modelbuilding and lack of reliance on large sample approximations (Natanegara et al. 2014). Furthermore, Bayesian inference can
greatly facilitate decisionmaking as results are typically expressed in terms of probabilistic statements that directly correspond with scientific questions and hypotheses of interest.
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1.5Opportunities of the Bayesian Approach In the past decades, Bayesian statistics has made significant inroads in drug development, notably in clinical trial design and analysis. More recently,
thanks to the regulatory initiatives regarding Quality by Design and risk-based lifecycle principles for pharmaceutical development, there is a significant increase in Bayesian applications to CMC
areas. In the following sections, the opportunities of the Bayesian approach are highlighted. 1.5.1Pre-Clinical Development Pre-clinical studies in animals are known for small sample sizes due to
ethical concerns. The sample size is often large enough to detect a meaningful treatment effect relative to a control group, but too small to differentiate several promising treatments. One way to
address this issue is to combine data from different studies, in which the studied drug was tested at different dose levels. Due to differences in study design, however, data may be difficult to
directly combine and analyze; e.g., data points are collected at different time points. This makes traditional analyses such as ANOVA or Repeated Measurement Analysis challenging. When good prior
information exists (Novick et al. 2018), the utilization of the Bayesian approach has the potential to circumvent the issues of small sample size and observations collected at different time points.
In addition, it is also straightforward to carry out Bayesian inference based on complex models such as hierarchical models even when useful prior information is not available. 1.5.2CMC Development
As previously discussed, the launch of the FDA initiative, “Pharmaceutical Current Good Manufacturing Practices (cGMPs) for the 21st Century”, and subsequent publications of other regulatory guidance
including ICH Q8-Q11 regarding Quality by Design and quality risk management ushered in a risk-based and lifecycle approach to CMC development, and also stimulated the use of statistics. It is now
well understood that the assurance of the robustness of manufacturing processes and quality products is often stated in probabilistic terms. Such assurance is achieved through understanding,
quantifying, and controlling the variability of CMC processes. Bayesian analysis provides useful tools for identifying design space, which is a constrained region of manufacturing process parameters
providing greater probability for the product to be within its specifications (Peterson 2008, 2009; Peterson and Lief 2010; Peterson and Yahyah 2009). Various Bayesian methods were developed for
assay validation (Sondag et al. 2016; USP 2018), formulation development (LeBrun et al. 2018),
Bayesian Statistics in Drug Development
process monitoring (Colosimo and del Castillo 2007), and general QC troubleshooting (see Chapter 11). These applications, powered by the advances in Bayesian computation techniques such as the Markov
Chain Monte Carlo simulation method, render ready solutions to many problems related to drug research, which were long thought to be intractable (Colosimo and del Castillo 2007). 1.5.3Clinical
Trials Clinical trials are experiments conducted in human subjects in strictly controlled settings to address specific questions regarding the safety and efficacy of a medical intervention. Depending
on the questions at hand, different study designs may be used. Early studies are small in size, and progressively large comparative studies are run as confidence in the drug safety and efficacy is
gained. In general, clinical trials are carried out in a sequential fashion in which the learning from the previous experiment is used to guide the design of the next. Despite their controlled
nature, clinical trials encounter many uncertainties such as varying medical practice at different sites and unknown patient or disease characteristics which are unaccounted for in the trial
inclusion or exclusion criteria. Robust evaluation of drug safety and efficacy lies in the researcher’s ability to synthesize information from various sources, including different stages of clinical
development, and incorporate known variability in the inference of the clinical research questions. It also requires the statistical method to be adaptive in the sense that inference can be updated
based on new information. Bayesian approaches are most suitable for this purpose. In the past two decades, significant advances in Bayesian applications in clinical trial and analysis have been made.
There has also been greater regulatory acceptance of Bayesian approaches to study design and analysis, as evidenced by the publication of the FDA guidance on the use of Bayesian statistics in medical
device clinical trials (FDA 2010a) and acceptance by regulatory agencies of Bayesian methods for early and exploratory phases of drug development. Bayesian adaptive study design and analysis have
been widely used for Phase I dose-finding studies (Chevret 2006) and Phase II efficacy assessment (Berry and Stangl 1996a,b; Spiegelhalter et al. 2004; Berry et al. 2011).
1.6Challenges of the Bayesian Approach 1.6.1Objection to Bayesian Despite its potentially wide-ranging application in drug research and development, Bayesian inference remains one of the most
controversial methods (Gelman 2008). Central to the objection to the Bayesian method is its
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requirement of a prior distribution, which is based on expert opinions and historical data. Inevitably, potential biases may be introduced in prior selection; e.g., expert opinions often vary. Two
Bayesian statisticians who analyze the same data may reach different conclusions based solely on the priors they choose. Although sensitivity analysis based on various choices of prior beliefs may
alleviate some of the concern of subjectivity, it is not favored by practitioners who look for straightforward answers (Little 2006). Therefore, the application of the Bayesian approach to drug
development requires deeper vetting of relevant sources of information and scientific expertise, greater level of engagement among stakeholders, and the ability to synthesize information from
different sources and build realistic complex Bayesian models (Ohlssen 2016). 1.6.2Regulatory Hurdles Although there have been growing applications of the Bayesian approach in drug development in
the past two years, by and large, they are intended to assist internal decision-making. The preponderance of statistical methods used for data analysis in support of regulatory filings are from
traditional frequentist approaches. The ICH E9 Statistical Principles for Clinical Trials state “Because the predominant approaches to the design and analysis of clinical trials have been based on
frequentist statistical methods, the guidance largely refers to the use of frequentist methods […] when discussing hypothesis testing and/or confidence intervals. This should not be taken to imply
that other approaches are not appropriate: the use of Bayesian […] and other approaches may be considered when the reasons for their use are clear and when the resulting conclusions are sufficiently
robust.” Although the guidance presents opportunities for clinical trials, there is no clear regulatory pathway on how exactly the Bayesian methods can contribute to the totality of evidence in
support of regulatory filings. In addition, regulators have concerns about Bayesian treatment of multiple testing (Kay 2015). For example, under a frequentist sequential design, the Type I error is
allocated across tests at the interim looks to ensure the overall Type I error remains as planned, say, 5%. However, with the Bayesian approach, the posterior distribution at each interim is updated
with new data by treating the previous posterior as the prior. The resulting updated posterior distribution levies no additional costs regardless of how many interim looks of the data are carried
out. Although several remedies have been suggested in the published literature that imbues the Bayesian sequential testing procedures with the frequentist property in controlling the overall Type I
error (Spiegelhalter et al. 2004), there is no precedent that these methods are acceptable to the regulators.
Bayesian Statistics in Drug Development
1.7Concluding Remarks In the past decades, the advances in Bayesian computations have broadened the use of Bayesian approaches in various industries. Despite progress, applications of Bayesian
methods in drug development has been modest (Winkler 2001; Moyé 2008; Chevret 2006; Rogatko et al. 2007; Natanegara 2014). Except clinical trials of medical devices, there is a clear lack of
regulatory guidance on Bayesian-driven design and analysis of clinical trials for new drugs. In addition, general regulatory acceptance of Bayesian methods for late-stage studies in support of
regulatory submissions is low. An important factor that may have contributed to this issue is the unfamiliarity of Bayesian methods and computational tools for practitioners who are involved in
late-stage drug development. The pharmaceutical industry has been facing unprecedented challenges of high attribution rate and unsustainable level of R&D costs. Such challenges argue for quantitative
methods that synthesize information from diverse sources to aid robust decision-making and that bring both flexibility and economy into study designs. The Bayesian approach provides this opportunity.
To fully capitalize on the benefits of the Bayesian approach, however, it is essential to develop best practices of Bayesian statistics, gain consensus among sponsors and regulators, and continue
demonstration of the utilities of Bayesian in all aspects of drug development.
2 Basics of Bayesian Statistics
2.1Introduction Bayesian statistics is a branch of statistics founded upon Bayesian probability theory. Reverend Thomas Bayes, a Presbyterian minister in England, developed a specific form of Bayes’
Theorem for making inference about the parameter of a binomial distribution. His work, published posthumously, was independently formulated by French mathematician PierreSimon Laplace. Although both
frequentist and Bayesian inferences rely on probability theories, the interpretation of probability is very different for the two schools of thought. Frequentist practitioners view probability as the
limit of the relative frequency of an event after a large number of trials (Stigler 1986). By contrast, from the Bayesian perspective, probability is a measure of the uncertainty of an event based on
the current knowledge of the event. As such, the probability can change as new information becomes available. In this sense, Bayesian probability is a subjective measure of likelihood as opposed to
the frequentist notion of probability as a fixed quantity. Bayes’ Theorem provides a framework to update this probability, combining prior beliefs with the current data. The results of Bayesian
analysis are often provided in statements quantified through probabilities. For these reasons, Bayesian inference is intellectually attractive; however, the adoption of Bayesian methods has been slow
due to practical considerations. In recent years owing to the rapid advances in the field of Bayesian computation, increasing interest has arisen in Bayesian applications in various areas of drug
research and development. This chapter provides an overview of Bayesian statistics after briefly highlighting the differences between the frequentist and Bayesian methods. Topics covered include
Bayes’ Theorem, prior and posterior distributions, predictive probability, and Bayesian computation.
Bayesian Analysis with R for Drug Development
2.2Statistical Inferences 2.2.1Research Questions Researchers in drug development use scientifically designed experiments to collect data and address research questions such as (1) Is the new drug
more effective than the standard care therapy? (2) Does a newly produced product lot meet its specifications for quality? (3) What conditions of a cell culture system render the maximum yield?
Answers to these questions can be obtained through applications of statistical methods. 2.2.2Probability Distribution A probability distribution of a random variable is a theoretical structure of
the variable. For a continuous random variable, the distribution describes the characteristics of a variable, such as its prespecified range. Data generated from a study also follow a distribution.
In the Bayesian framework, the data distribution is defined by parameters, some of which are related to the question at hand. Consider the example in which an investigator is interested in the effect
of an experimental drug in a single-arm study. Suppose that the response is binary. The drug effect, p, often characterized by response rate, is an unknown parameter. Let Y denote the number of
patients (out of n) who responded to the treatment. Under the assumptions that the patients’ responses to the drug are assessed independently and each patient responds to the drug with the same
probability p, Y follows a binomial distribution binomial(n, p). That is, the probability for Y to be equal to the value k (= 0, 1, …, n) is given by
ænö n-k Pr [Y = k ] = ç ÷ p y ( 1 - p ) . (2.1) èkø
The above distribution can be used to make inferences about the response rate p. 2.2.3Frequentist Methods In the frequentist paradigm, inference is made strictly based on data from the current
study. For the above example, the parameter p can be estimated by pˆ = Y n. For example, if 55 out of 100 patients respond, then an estimate of p is given by 55/100=0.55. The performance of the
estimation method can be characterized through a ( 1 - a ) ´ 100%confidence interval, éë A ( Y ) , B ( Y ) ùû such that
Pr éë A ( Y ) < p < B ( Y ) ùû = 1 - a. (2.2)
Basics of Bayesian Statistics
A(Y) and B(Y) are calculated solely based on the current data Y. One of these intervals was derived by Clopper and Pearson (1934):
1 n- y +1 1+ F a 2( n - y +1), 2 y , y 2
y +1 F2 y +1 ,2 n- y ,a/2 n-y ( ) ( ) 2 |y ÷ 5 è ø ¥ 105
ò ò p (m, t|y ) dmdt
0.04 95
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The marginal posterior 100 ´ p percentile of μ may be determined by finding the value M that satisfies ∞M
∫∫ π (µ, τ|y) dµdτ = p. 0 −∞
Though there are more sophisticated algorithms, the double integral may be numerically approximated in R with two calls to the integrate() function as follows. ## Pr( 95 < mu < 105, tau > 0.04 | y
) = double integral InnerFunc = function(mu, tau){ dNG(mu, tau) } InnerIntegral = Vectorize(function(tau.inner) { integrate( InnerFunc, lower=95, upper=105, tau=tau. inner)$value})
pr = integrate(InnerIntegral, lower=0.04, upper=Inf) print(pr) > 0.1797854 with absolute error < 5.3e-05 ## Determine marginal posterior 5%, 50%, and 95%-iles for mu InnerIntegral =
Vectorize(function(tau.inner, M) { integrate(InnerFunc, lower=-Inf, upper=M, tau=tau. inner)$value}) mu.quantile = function(m0, p=0.5) { ( integrate( InnerIntegral, lower=0, upper=Inf, M
=m0) $value - p )^2 } ## Find m0 such that Pr( mu < m0 | y ) = 2.5%, 50%, 97.5%. ## Integrate over tau. sapply( c(0.025, 0.5, 0.975), function(p){ optimize( mu.quantile, interval
=c(90, 110), p=p ) $minimum } ) > 91.60657 95.60000 99.59342 The posterior probability Pr ( 95 < µ < 105, σ < 5|y ) = 0.18 and the marginal posterior 2.5%-, 50%-, and 97.5%-iles for μ are
respectively given as 91.61, 95.60, and 99.59. While the double-integration method works well for this problem, as the dimensionality of the parameter space increases, direct numerical integration
becomes less feasible. In its place, Monte Carlo integration may be used.
Basics of Bayesian Statistics
By drawing samples m b , sb2 from Equation (2.9), b = 1, 2, …, B, the Monte Carlo integral for the probability is
1 Pr ( 95 < µ < 105, σ < 5|y ) ≈ ∑ 1[95 < µ b < 105, σ b < 5]. B
The marginal posterior quantiles for μ may be determined by Monte Carlo integration simply by looking at the quantiles of {μb}. The simplest Monte Carlo integration method involves direct sampling
from the posterior distribution. Several authors (e.g., Gelman et al. 2013, Chapter 3) derive the NG sampling distribution directly by noting that π ( µ , τ|y ) = π(µ|y , τ ) × π(τ|y).
It can be shown that
æ n n s2 ö t|y ~ Ga ç n , n n ÷ and 2 ø è 2
æ s2 ö m|y , s2 ~ N ç m n , ÷ , kn ø è
where t = 1 s2 . In R, the computer code to calculate the Monte Carlo probability is given below. set.seed(089) tau.ng = rgamma( 1 0000, shape=0.5*nu.n, rate=0.5*nu.n*sigSq.n ) # tau|y sigSq.ng
= 1/tau.ng mu.ng = rnorm (10000, mean=mu.n, sd=sqrt(sigSq.ng/k.n)) # mu|tau, y ## Calculate Monte Carlo probability mean(mu.ng > 95 & mu.ng < 105 & sigSq.ng < 25) > 0.1797 quantile
( mu.ng, p=c(0.025, 0.5, 0.975) ) > 5% 50% 95% > 91.59147 95.58921 99.62328 The Monte Carlo posterior probability (with B = 10,000) of 0.1797
(Monte Carlo error < 0.01) is remarkably close to the numerical integrated value of 0.1798. In addition, the Monte Carlo marginal posterior quantiles for μ are
Bayesian Analysis with R for Drug Development
also quite close to the true values. The Monte Carlo marginal quantiles will be set aside for now and revisited in Section 2.4. 2.3.3Rejection Sampling For situations in which the PDF is known but
the sampling distribution is not, rejection sampling (Robert and Casella 2004) may be used. Instead of sampling from π(θ|y), random variables are generated from a proposal distribution q(θ), where
the distribution of q(θ) must envelop π(θ|y). Let m > 1 be a constant and let u ~ U(0, 1). A sample θ* from q(θ) is accepted if θ*|y mq θ* > u. Otherwise, the sample is rejected and a new pair (θ*,
u) are generated and tested. Ordinarily, q(.) is a multivariate normal or multivariate T distribution. For better sampling properties, we reparameterize the PDF in Equation (2.9) with l = ln s2 = -
ln(t) so that the joint posterior density of ( m, l ) is p ( m , exp ( -l )|y ) exp ( -l ) . We set m = 100 and q(.) to be a bivariate normal distribution with mean (100, 4) and variance–covariance æ
10 2 0 ö matrix ç 2 ÷ . Though a simple procedure, rejection sampling can be è 0 10 ø
( )
( )
costly. A sample of size B requires approximately m × B random variables to be generated. For example, with m = 100, about 1 million random variables are needed to generate 10,000 draws. Note that
choosing m and q(.) are beyond the scope of this chapter. In R, the rejection sampling algorithm and Monte Carlo integration are coded as follows. reject.sampling = function(m) { x.out = rep
(NA, 2) while(TRUE) { ## theta.star = proposal theta.star = c( rnorm(1, mean=100, sd=10),rnorm(1, mean=4, sd=10) ) u = runif(1, 0, 1) p.x = dNG( mu=theta.star[1], tau
=exp(-theta.star[2]))* exp(-theta.star[2]) q.x = dnorm(theta.star[1], mean=100, sd=10)* dnorm(theta.star[2], mean=4, sd=
10) ratio = p.x/(m*q.x) if ( ratio > u ) ## Accept the sample { x.out = theta.star
Basics of Bayesian Statistics
break } ## Otherwise, reject the sample } return(x.out) } ## Generate 10,000 posterior draws with rejection sampling using m=100 set.seed(822) th.post.reject = t(
sapply(1:10000, function(i){ reject.sampling(m=100) })) colnames(th.post.reject) = c("mu", "lambda") ## Calculate Monte Carlo probability mean(th.post.reject
[,"mu"] > 95 & th.post.reject [,"mu"] < 105 & th.post.reject[,"lambda"] < log(25)) > 0.1747 The rejection sampling Monte Carlo posterior probability (with B=10,000)
of 0.1747 (Monte Carlo error < 0.01) is a good estimate for the true value of 0.180. 2.3.4Markov Chain Monte Carlo While rejection sampling bestows its user with independent random numbers, it
requires knowledge of the normalizing constant p ( y ) in Equation (2.4). For most statistical modeling situations, calculation of the normalizing constant is a complex, sometimes intractable
mathematical problem. MCMC sampling, however, only requires knowledge of h ( q ) = p ( q ) p ( y|q), a function proportional to the posterior distribution, up to the normalizing constant. Two
important algorithms for MCMC sampling are discussed, namely Gibbs (Casella and George 1992) and Metropolis–Hastings (Robert and Casella 2004). 2.3.4.1Gibbs Sampling Gibbs sampling breaks up the
parameter space into two (or more) blocks θ= (θ1, θ2). The unconditional joint distribution of π(θ|y) may be sampled by iterk +1 k k k +1 atively generating random numbers from p(q1( ) |y , q(2 ) )
and p(q(2 1 ) |y , q1( ) ), (1) k=1, …, K, given a starting value for θ 2 . According to Gibbs sampling theory (Casella and George 1992), the iteratively sampled distribution eventually converges
to the true distribution. The resulting set of samples are referred to as an MCMC chain (Robert and Casella 2004).
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Because Gibbs sampling requires time to converge, a poorly chosen start1 ing value q 2( ) can yield unlikely samples at the beginning of the chain. As a rule, the first B samples, called the burn in,
are thrown away to allow the algorithm to steer the chain into a more desirable sampling space. Also, because the generation of the (k+1)st parameter depends on the value of the kth parameter,
Gibbs sampling can suffer from autocorrelation. Autocorrelated samples may possess a smaller-than-expected effective sample size (ESS) of MCMC posterior draws. ESS is the approximate number of
samples that would have been generated from the posterior distribution if draws were made independently. The “coda” package in R calculates ESS automatically for objects of class “mcmc” with the
function effectiveSize(). An ESS of 10,000 or more for all parameters is usually sufficient for most statistical inference. In some cases, such as in our example problem, one or both of p(q1|y, q2 )
and p(q2|y, q1 ) may be exactly identified. Setting θ1=μ and θ2=τ, it can be shown that
æ s2 ö m|y , s2 ~ N ç m n , ÷ and kn ø è
ö æ n + v0 + 1 æ 1 ö 2 t|y , m ~ Ga ç , ç ÷ (n - 1)s2 + n( y - m)2 + k0 ( m - m 0 ) + n 0s02 ÷ . 2 2 è ø è ø
The R code for Gibbs sampling used to calculate Pr ( 95 < m < 105, s < 5|y ) is shown below. set.seed(121) th.post.gibbs = matrix(NA, 10000, 2) sigSq.lag = sSq ## Starting value for posterior
distribution of sigSq for ( b in 1:5000 ) ## Burn-in period of 5,000 { mu.lag = rnorm (1, mean=mu.n, sd=sqrt(sigSq.lag/ (n+k0))) rate = 0.5*( (n-1)*sSq + n*(ybar-mu.lag)^2 +
k0*(mu.lag-mu0)^2 + nu0*sigSq0 ) tau.lag = rgamma(1, shape=0.5*(n+nu0+1), rate=rate) sigSq.lag = 1/tau.lag } for ( b in 1:10000 ) ## Collect 10,000 posterior samples { mu.lag
= rnorm(1, mean=mu.n, sd=sqrt(sigSq.lag/(n+k0))) rate = 0.5*( (n-1)*sSq + n*(ybar-mu.lag)^2 + k0*(mu.lag-mu0)^2 + nu0*sigSq0 ) tau.lag = rgamma(1, shape=0.5*(n+nu0+1),
rate=rate) sigSq.lag = 1/tau.lag
Basics of Bayesian Statistics
th.post.gibbs[b,1:2] = c(mu.lag, log(sigSq.lag)) } colnames(th.post.gibbs) = c("mu", "lambda") mean(th.post.gibbs[,"mu"] > 95 & th.post.gibbs [,"mu"] < 105 & th.post.gibbs[,"lambda"] <
log(25)) > 0.1771 The Gibbs sampling Monte Carlo posterior probability of 0.1771 is in good agreement with the other methods. Using the coda library function effectiveSize( as.mcmc
(th.post.gibbs) ), the ESS for μ is about 10,000 and the ESS for σ2 is about 8,100. Though adequate for this exercise, it may be a good idea to increase the number of posterior draws until the ESS
for both parameters is larger than 10,000. 2.3.4.2Metropolis–Hastings The Metropolis–Hastings (M–H) algorithm (Robert and Casella 2004) is a form of MCMC rejection sampling that also only requires
knowledge of h ( q ) = p ( y|q ) p ( q ). This algorithm requires a starting value θ(1). In the kth step of M–H, a sample θ* is drawn from a proposal distribution q(θ). If α = h(θ* ) h(θ(k −1) ) ≥ 1
(i.e., the candidate θ* is more likely than θ(k−1)), the candidate is accepted and θ(k) is set to θ*. Otherwise, the candidate is accepted with probability α or else rejected. If rejected, set θ(k)=
θ(k−1). The proposal distribution q(.) must be symmetric and is often a multivariate normal or multivariate T distribution. As with Gibbs sampling, the M–H procedure can also suffer from a poorly
chosen initial value and so a burn-in period is recommended. M–H also induces autocorrelation by sometimes setting the current posterior draw to its immediate predecessor. Autocorrelation may be
overcome by increasing the sample size until a sufficient ESS is established. For sample sizes that strain computer storage capacity, thinning is recommended. Thinning the sample by m, for example,
means that every mth sample is saved and the other (m–1) samples are discarded. Though the draws from a thinned sample may be nearly independent (low or negligible autocorrelation), throwing away
draws after the burn-in period results in a less efficient sample set. For our example with posterior Equation (2.9), p(m, t|y) µ h(m, t|y) =
Õ{N ( y ; m, s )} ´ N çè m; m , k i
æ v0 v0s02 ö s02 ö ÷ ´ Ga ç t; , ÷ . 2 ø 0 ø è 2
In R, using l = ln(s2 ) = - ln(t), the M–H algorithm and Monte Carlo integration of the posterior probability of interest are written as follows. As before, the proposal distribution is bivariate
normal with mean (100, 4) and variance–covariance matrix
æ 10 2 çç è 0
0 ö ÷ . 10 2 ÷ø
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h.theta = function(mu, lambda) { ## Likelihood x Prior. Proportional to posterior density sigma = exp(0.5*lambda) ldens = sum( dnorm( y, mean=mu, sd=sigma, log=TRUE )) +
dnorm( mu, mean=mu0, sd=sigma/sqrt(k0), log=TRUE ) + dgamma( exp(-lambda), shape=0.5*nu0, rate=0.5*nu0*sigSq0, log=TRUE) - lambda out = exp(ldens) return
(out) } mh.sampling = function(mu0, lambda0) { ## (mu0, lambda0) = (k-1)st draw x.out = c(mu0, lambda0) ## theta.star = proposal theta.star = c( rnorm(1, mean=100, sd=10),
rnorm(1, mean=4, sd=10) ) u = runif(1, 0, 1) ratio = h.theta( mu=theta.star[1], lambda=theta. star[2]) / h.theta(mu=mu0, lambda=lambda0) alpha = min(1, ratio) if ( u < alpha ) ##
accept new proposal x.out = theta.star return(x.out) } ## Generate 10,000 MCMC samples with M-H set.seed(663) th.post.mh = matrix(NA, 10000, 2) theta.lag = c( mean(y), log(sSq) )
for ( b in 1:10000 ) { th.post.mh [b,] = mh.sampling(mu0=theta.lag[1], lambda0=theta.lag[2]) theta.lag = as.vector(th.post.mh[b,]) } colnames(th.post.mh) = c("mu", "lambda")
Basics of Bayesian Statistics
mean(th.post.mh[,"mu"] > 95 & th.post.mh[,"mu"] < 105 & th.post.mh[,"lambda"] < log(25)) > 0.141 Considering that 10,000 samples were drawn, the estimated posterior
probability of 0.141 from M–H is quite far from the numerical integrated value of 0.180. The inadequate result stems from the small ESS of the MCMC posterior draws. The output of effectiveSize(
th.post.mh ) is 136 for μ and 156 for σ2, a far cry from the desired 10,000 independent draws. To obtain an ESS of about 10,000, we set a burn-in period of 20,000 draws and increased the number of
collected samples to 15,000 with thinning by 100. Computer code follows. set.seed(303) th.post.mh2 = matrix(NA, 15000, 2) theta.lag = c( ybar+100, log(sSq)+2 ) ## A poorly chosen initial
value print(theta.lag) > 194.775000 5.014659 for ( burnin in 1:20000 ) ## Burn-in: Discard the first 20,000 draws { theta.new = mh.sampling( mu0=theta.lag[1], lambda0=theta.lag
[2]) theta.lag = theta.new } print(theta.lag) ## The initial value improves after a burn-in period > 92.992802 3.709665 for ( mcmc in 1:15000 ) ## Save 15,000 posterior draws {
for ( thin in 1:100 ) ## Thin by 100 { theta.new = mh.sampling(mu0=theta.lag[1], lambda0 =theta.lag[2]) theta.lag = theta.new } th.post.mh2[mcmc,] = theta.new } colnames
(th.post.mh2) = c("mu", "lambda") mean(th.post.mh2[,"mu"] > 95 & th.post.mh2[,"mu"] < 105 & th.post.mh2[,"lambda"] < log(25))
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> 0.1840667 effectiveSize(as.mcmc(th.post.mh2)) > mu lambda > 10978.75 10300.98 With an ESS of about 10,000, the M–H Monte Carlo estimate 0.184
agrees with all other methods. In all, 1,520,000 samples were drawn by M–H in order to achieve an ESS of about 10,000, putting M–H on a par with the rejection sampling algorithm. Luckily, Bayesian
practitioners need not write sophisticated sampling procedures. For nearly all statistical modeling problems, pre-packaged computer code with far superior algorithms written by expert statisticians,
mathematicians, and computer scientists are available. These are explored in Section 2.4.
2.4Computational Tools 2.4.1BUGS and JAGS Though not yet in a golden age for Bayesian statistical software, mature computer algorithms for sampling from the posterior distribution have been
developed. Started in 1989, the Bayesian inference Using Gibbs Sampling (BUGS) language (Lunn et al. 2009) was among the first to provide a generic interface to MCMC sampling. BUGS, written in 32-bit
Component Pascal, performs a combination of M–H and Gibbs sampling, breaking up parameters into correlated groups. For those running Microsoft Windows operating systems, WinBUGS provided a graphical
user interface until development was discontinued in 2007 in favor of OpenBUGS. As computers grew faster with multiple CPU cores and moved from 32-bit to 64-bit, there was a desire to port BUGS to
other platforms (e.g., Linux, MacIntosh). The community threw its support behind Just Another Gibbs Sampler (JAGS) (Plummer 2003). JAGS is written in C++, supports 32-bit and 64-bit computing
systems, and may be compiled for a variety of computer platforms. In addition, JAGS provides a similar user interface to the BUGS language so that nearly all BUGS code can be run in JAGS and vice
versa with little editing. Though it appears that BUGS is no longer actively developed, JAGS was last updated in August 2017. Both BUGS and JAGS are powered with expert software programming tricks
and techniques to perform Metropolis–Hastings with a random-walk updater, Gibbs, and slice sampling. The latest general MCMC software system, called Stan (Carpenter et al. 2017), implements the
“No-U-Turn sampler” (NUTS), a variant of Hamiltonian Monte Carlo. Relative to the random walk, the NUTS algorithm performs a more thorough search of the parameter
Basics of Bayesian Statistics
space of the posterior distribution. Like JAGS, Stan is based on C++ for 32-bit and 64-bit systems and has been ported to various computing platforms. Various interfaces are available for all three
MCMC software engines. Most prominently, OpenBugs may be called from R using one of the many available R libraries. In addition, macros were written to call OpenBugs from Excel, SAS, Matlab, and
Stata. JAGS may be accessed from R, Python, or Matlab. The Stan website lists interfaces with R, Python, Matlab, Julia, Stata, and Mathematica. Thus, MCMC may be approached from many popular
mathematical/statistical computing platforms. Taking advantage of multiple CPUs, many of the interfaces can perform parallel computing with BUGS, JAGS, or Stan. 2.4.2SAS PROC MCMC The list is
incomplete without SAS Proc MCMC (SAS 2016). Originally released in SAS 9.2 with an experimental set of MCMC functions, Proc MCMC in SAS 9.4 has matured into a feature-complete MCMC system, rivaling
JAGS and Stan, with the ability to perform various MCMC sampling, including random walk and NUTS. We note that, while BUGS, JAGS, and Stan may be downloaded as “freeware”, SAS is a for-profit company
and Proc MCMC is part of its commercial software. Currently, SAS permits free use of its procedures for non-profit research. 2.4.3Utility of JAGS We are not endorsing one MCMC platform over another;
however, to maintain consistency throughout the book, most of the examples will be given using JAGS via an R interface. In this manner, historical BUGS and JAGS code that may be freely downloaded by
the reader will be easily understood. In addition, relative to Stan, JAGS programming statements tend to be more compact. For the reader’s benefit, computer code written in JAGS will be analogously
provided in Stan in the appendix of this book. As an example, consider the competing “model” statements in JAGS and Stan for the univariate normal random variable example given in Section 2.3. JAGS
and Stan code are given in Table 2.1. Stan uses explicit data and parameter declarations, whereas such statements are implicit in JAGS, allowing JAGS scripts to generally be written with far fewer
lines of code. Also note that Stan and JAGS often parameterize distributions differently. In the example below, JAGS uses the mean-precision parameterization and Stan uses the mean-standard deviation
parameterization for the normal distribution. Our favorite JAGS interface is through the run.jags() function in the runjags R library, which depends on the rjags library. We illustrate the run.jags()
function, noting that the model statement must be given either with the contents of Table 2.1 in a separate text file or by placing the model statement in quotes.
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TABLE 2.1 JAGS and Stan Statements for Univariate Normal Distribution JAGS model{ ## Prior mu ~ dnorm( mu0, k0*tau ) tau ~ dgamma( 0.5*nu0, 0.5*nu0*sigSq0 ) sigmaSq
Lower95MedianUpper95MeanSDModeMCerrMC%ofSDSSeffAC.50 psrf > mu 91.52895.58399.43595.591.992695.6140.0115050.6 30000-0.005341.0001 >
sigma3.59495.72059.01025.98761.50325.35860.008693 0.6 299010.0002931.0004 >lambda 2.6513 3.4881 4.4659 3.5226 0.468413.4848 0.0027040.630000 0.000215
1.0004 > Total time taken: 4.8 seconds
As described in the run.jags() manual (Denwood 2016), the JAGS output shows the posterior median and 95% credible interval (Lower95, Upper95), mean, standard deviation (SD), and mode. For
diagnostics, it shows the Monte Carlo standard error of the mean estimate (MCerr), effective sample size (SSeff), autocorrelation with a lag of 50 (AC.50), the Monte Carlo standard error as a percent
of the standard deviation (MC%ofSD), and the potential scale-reduction factor of the Gelman–Rubin statistic, sometimes called R-hat (psrf). For diagnostics, we typically like for SSeff to be at least
10,000 for each variable, an AC.50 close to 0 and a psrf value close to 1. To see the MCMC summary statistics using the coda library in R, use the syntax summary(fitb[[1]]). Note that the 95%
credible interval is a highest posterior
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FIGURE 2.1 Traceplot and density for posterior distribution from three MCMC chains of μ (mu), σ (sigma), and λ=ln(σ2) (lambda).
density (HPD) interval and may not be equal to the (2.5%, 97.5%) marginal posterior quantiles. For graphical diagnostics, we prefer the plots from the coda library. The function call is plot( fitb
[[1]] ) instead of plot( fitb ), which creates a similar set of graphs. We also prefer to run two or more independent MCMC chains so that we can visualize the consistency and convergence of the
chains. To calculate the Monte Carlo probability from Section 2.3, the fit object must be converted to a matrix. Computer code is shown below. The JAGSbased posterior probability = 0.175, which is
within Monte Carlo error of the true value. The Monte Carlo 2.5%, 50%, and 97.5% marginal posterior quantiles of μ are also close to their respective true values. th.post = as.matrix(as.mcmc.list
(fitb)) mean(th.post[,"mu"] > 95 & th.post[,"mu"] < 105 & th.post[,"lambda"] < log(25))
Basics of Bayesian Statistics
FIGURE 2.2 Posterior marginal densities of μ (mu) and λ = ln(σ2) (lambda).
> 0.1752333 ## Marginal quantiles of mu quantile( th.post[,"mu"], p=c(0.025, 0.5, 0.975)) > 2.5% 50% 97.5% > 91.59526 95.58335 99.51119 The marginal
posterior distributions for μ and λ via JAGS (MCMC) and via the true density generated in Section 2.2 are shown to be equivalent in Figure 2.2. The sampled true density is given by R objects mu.ng
and lambda. ng=–log(tau.ng).
2.5Concluding Remarks An overview of Bayesian analysis is provided in this chapter. Special focus is on the concepts of Bayes’ Theorem; prior, posterior, and predictive distributions; selection of
priors; inferences based on posterior and
Bayesian Analysis with R for Drug Development
predictive probabilities; and Bayesian computational tools. Some key differences between frequentist and Bayesian thinking are highlighted. The topics discussed in this chapter lay the ground for
understanding Bayesian methods that are discussed in other chapters of the book. Since drug research and development are carried out through series of experimentation, Bayesian study design and
analysis are likely to play more important roles in decisionmaking by leveraging historical data, personal beliefs, and observational data from the current studies.
3 Bayesian Estimation of Sample Size and Power
3.1Introduction Sample size determination is an important aspect in study design, especially in the planning of clinical trials due to regulatory requirements. An accurately and sufficiently sized
study enables the investigator to detect practically meaningful differences in comparative studies or provide precise estimates of the issue at hand. The frequentist sample size calculation relies on
the assumption that the true values of parameters under the alternative hypothesis are known. In practice, this assumption is rarely true. In fact, depending on available data, there are varying
degrees of uncertainties around these unknown parameters. By modeling the parameters through prior distributions, Bayesian methods can account for these uncertainties in the sample size calculations,
thus mitigating the risk of underpowering a study. In literature, in the context of estimating population prevalence with the desired precision, three criteria were proposed to obtain the so-called
minimal sample size determination. These criteria, which are based on highest posterior density (HPD) intervals, are the average coverage criterion (ACC), the average length criterion (ALC), and the
worst outcome criterion (WOC) (Joseph and Belisle 1997; Joseph, du Berger, and Belisle 1997). As these methods have been extensively discussed, in this chapter we concern ourselves with sample size
determination for the purpose of testing hypotheses. Sample size calculations based on Bayesian concepts are discussed through three examples, one related to product lot release and the other two
regarding futility and interim analysis in sequential trials.
3.2Sample Size Determination 3.2.1Frequentist Methods In drug research, the objective of a comparative study is to collect data to test a specific hypothesis. An important aspect of the study
planning is sample 41
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size determination. Given the null hypothesis of no difference, to estimate the sample size a frequentist procedure requires specification of a significance level for the test, a meaningful
difference of δ, which is usually stated in an alternative hypothesis, and the expected power or probability of rejecting the null hypothesis when the true difference is δ. As an example, consider a
drug manufacturing process that produces a large number of units (e.g., tablets), some of which are defective. The batch may be released to the market if the percent of defect p equals 1%. It is
desired that the lot should be rejected with a high probability if the percent of defect p equals 2%. In this situation, the hypotheses of interest are:
H 0 : p = 1% vs. H a : p = 2%,
which implies that δ=1%. The 1% defect rate in the null hypothesis is the producer’s quality level or average percent of defect of batches produced by the manufacturer; whereas the 1% increase in
the alternative hypothesis is selected based on consumer’s risk. In reality, however, δ is often unknown. Suppose a test procedure is given such that the batch may be released if no defects are
detected in a random sample of N units. In addition, the testing for defects is carried out by randomly selecting a maximum of N units and testing the units sequentially until either a defective unit
is detected or none of the N units are found to be defective. The significance level and power of the above test correspond to the probabilities of rejecting the null hypothesis when the true defect
rate p=1% and accepting the alternative hypothesis when the true defect rate p=2%. Obviously, these probabilities are dependent on the sample size. For a given sample size, the appropriateness of
the test described above can be evaluated through an operating characteristic OC), which defines the long-term probability of the success of a binary event with respect to the proportion of defective
units in a batch. For example, consider N=10. In testing of up to ten units in a newly manufactured batch, let X denote the index of the first detected defective unit. Then X follows a negative
binomial distribution with probability p with stopping rule at the first observed failure so that X ~ NB(1, p). Because a batch is released only when zero out of 10 sampled units are defective, the
long-term probability to release a batch is prel=(1−p)10. Conditioned on a value p=p0, the OC is (1−p0)10. For example, if p0=0.08 (i.e., 8% of units in a batch are defective), the
probability of releasing the batch (1−0.08)10=0.43. By considering different values of p, one may generate an OC curve, given in Figure 3.1. From the plot, it can be easily obtained that the
probabilities of accepting a batch when p=1% and 2% are 90% and 82%, respectively. This implies a significance level of 10% (= 1−90%) and power of 82%. Depending on the values of the prespecified
significance level and power, the test including the sample size of 10 may or may not be acceptable. For example, if the expected significance level is 10% and power 80%, the test is adequate in
providing protection from both the consumer’s and producer’s risk.
Bayesian Estimation of Sample Size and Power
FIGURE 3.1 OC curve for probability of releasing a manufactured batch.
3.2.2Bayesian Considerations One of the challenges of sample size determination using the frequentist framework is that the true values of the parameters, such as the meaningful difference δ, are
unknown. In practice, these parameters are often estimated from prior studies, published results in the literature, or expert opinions, and are viewed as the “true” values. Since the frequentist
method does not consider the variability in those estimates in the sample size calculations, overconfidence in the “true” values may result in an underpowered study. In addition, the ad-hoc practice
to determine the “true” values of those parameters makes it more challenging to carry out the frequentist test method with consistency. Bayesian procedures, which incorporate uncertainties in the
parameters, provide a natural remedy for this issue. 3.2.2.1Prior Information In the above example, the manufacturer would like to estimate the long-term probability of releasing batches of a drug
to the public; however, this requires knowledge of either p or prel. Before full-scale manufacturing, knowledge about the parameters p or prel may be obtained (Chapter 2) from expert opinion (e.g.,
operators of other manufacturing processes may be able to give a range for p based on experience) or possibly by estimating the parameter
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value from development data. This information may be encapsulated as the prior distribution for p. For example, suppose that the operators estimate that no more than 2% of the units in a batch are
defective and, through a process of prior elicitation, settle on the prior distribution for p ~ Beta(2, 198). This Beta prior has a mean of 1% and an upper 95th percentile of about 2%. The mean
batch-release probability is calculated as E[(1−p)10]=0.91, which may be quickly approximated by Monte Carlo methods by generating pb ~ Beta(2, 198), b=1, 2, …, B, and calculating (1/B) ∑ b = 1
(1 − pb ) . The R computer code for the Monte Carlo method using B=10000 is B
mean( (1-rbeta(10000, shape1=2, shape2=198))^10 ) > 0.9056183 Thus, given only prior information, the manufacturer can expect to release its batches of a drug to the public about 91% of the time.
3.2.2.2Use of Historical Data After testing five full-scale manufactured drug batches, the prior information for p may be updated. Suppose zero out of ten defects were discovered in the first four
batches, but a defect was discovered in the sixth unit of the fifth batch so that four of the five batches are released. The density of p (and hence prel) may be updated with the new data using the
JAGS statement given by computer code [3.1]. In [3.1], Xi is the number of units tested before detecting a defect in the ith batch, so that Xi follows a negative binomial distribution with
probability p, stopping at the first detected defect; i.e., Xi~NB(1, p). Since the maximum number of tested units is ten, when no units are identified as defective, the value for Xi is
right-censored and known to be bigger than 10. For 0 ≤ Xi ≤ 10, the JAGS likelihood statement is “X[i] ~ dnegbin(1, p)”. For the case Xi > 10, the likelihood takes the complement of the cumulative
distribution function; i.e, Pr(X[i] > 10). In JAGS, special coding is required to specify right-censoring using the function call dinterval( X[i], 10 ). A new variable, rightCens[i], is created so
that when 0 ≤ Xi ≤ 10, rightCens[i] = 0, and when Xi > 10, rightCens[i] = 1. Additionally, when Xi > 10, the value in the data statement is set to NA. [3.1] require(rjags); require(runjags)
model.txt = " model{ p ~ dbeta(2, 198) ## Prior for p pRel 0.45. lam ~ dgamma(5, 1) ## Mean = variance = 5 ## Likelihood for ( i in 1:N ) { ## lambda[i] = 1e-10 with probability
p ## and = lam with probability (1-p). lambda[i] | {"url":"https://dokumen.pub/bayesian-analysis-with-r-for-drug-development-concepts-algorithms-and-case-studies-1138295876-9781138295872.html","timestamp":"2024-11-02T12:18:58Z","content_type":"text/html","content_length":"146341","record_id":"<urn:uuid:4e849fd7-dc73-4c75-aff2-25cf2edf906d>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00592.warc.gz"} |
The subject of mathematical inequalities is tied closely with optimization methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in
mathematics Olympiads.
Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory. They deal with relations of variables denoted by four signs: $>,<,\ge,\le$.
For two numbers $a$ and $b$:
• $a>b$ if $a$ is greater than $b$, that is, $a-b$ is positive.
• $a<b$ if $a$ is smaller than $b$, that is, $a-b$ is negative.
• $a\ge b$ if $a$ is greater than or equal to $b$, that is, $a-b$ is nonnegative.
• $a\le b$ if $a$ is less than or equal to $b$, that is, $a-b$ is nonpositive.
Note that if and only if $a>b$, $b<a$, and vice versa. The same applies to the latter two signs: if and only if $a\ge b$, $b\le a$, and vice versa.
Some properties of inequalities are:
• If $a>b$, then $a+c>b$, where $c\ge 0$.
• If $a \ge b$, then $a+c\ge b$, where $c\ge 0$.
• If $a \ge b$, then $a+c>b$, where $c>0$.
Solving Inequalities
In general, when solving inequalities, same quantities can be added or subtracted without changing the inequality sign, much like equations. However, when multiplying, dividing, or square rooting, we
have to watch the sign. In particular, notice that although $3 > 2$, we must have $-3 < -2$. In particular, when multiplying or dividing by negative quantities, we have to flip the sign.
Complications can arise when the value multiplied can have varying signs depending on the variable.
We also have to be careful about the boundaries of the solutions. In the example $x > \frac{3}{2}$, the value $x = \frac{3}{2}$ does not satisfy the inequality because the inequality is strict.
However, in the example $x \ge \frac{3}{2}$, the value $x = \frac{3}{2}$ satisfies the inequality because the inequality is nonstrict.
Solutions can be written in interval notation. Closed bounds use square brackets, while open bounds (and bounds at infinity) use parentheses. For instance, $x \in [3,6)$ means $3 \le x < 6$.
Linear Inequalities
Linear inequalities can be solved much like linear equations to get implicit restrictions upon a variable. However, when multiplying/dividing both sides by negative numbers, we have to flip the sign.
Polynomial Inequalities
The first part of solving polynomial inequalities is much like solving polynomial equations -- bringing all the terms to one side and finding the roots.
Afterward, we have to consider bounds. We're comparing the sign of the polynomial with different inputs, so we could imagine a rough graph of the polynomial and how it passes through zeroes (since
passing through zeroes could change the sign). Then we can find the appropriate bounds of the inequality.
Rational Inequalities
A more complex example is $\frac{x-8}{x+5}+4\ge 3$.
Here is a common mistake: \begin{align*} \frac{x-8}{x+5}+4&\ge 3 \\\frac{x+5-13}{x+5}+4&\ge 3 \\1-\frac{13}{x+5}+4&\ge 3 \\x+5-13+4x+20&\ge 3x+15 \\x&\ge \frac{3}{2}. \end{align*} The problem here is
that we multiplied by $x+5$ as one of the last steps. We also kept the inequality sign in the same direction. However, we don't know if the quantity $x+5$ is negative or not; we can't assume that it
is positive for all real $x$. Thus, we may have to reverse the direction of the inequality sign if we are multiplying by a negative number. But, we don't know if the quantity is negative either.
A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple to find the solutions to the inequality by considering the
sign (negativeness or positiveness) of the fraction as $x$ varies. \begin{align*} \frac{x-8}{x+5}+4 &\ge 3 \\ \frac{x-8}{x+5}+1 &\ge 0 \\ \frac{2x-3}{x+5} &\ge 0 \end{align*} We will start with an
intuitive solution, and then a rule can be built for solving general fractional inequalities. To make things easier, we test positive integers. $0$ makes a good starting point, but does not solve the
inequality. Nor does $1$. Therefore, these two aren't solutions. Then we begin to test numbers such as $2$, $3$, and so on. All of these work. In fact, it's not difficult to see that the fraction
will remain positive as $x$ gets larger and larger. But just where does $x$, which causes a negative fraction at $0$ and $1$, begin to cause a positive fraction? We can't just assume that $2$ is the
switching point; this solution is not simply limited to integers. The numerator and denominator are big hints. Specifically, we examine that when $2x-3=0$ (the numerator), then the fraction is $0$,
and begins to be positive for all higher values of $x$. Solving the equation reveals that $x=\frac{3}{2}$ is the turning point. After more of this type of work, we realize that $x=-5$ brings about
division by $0$, so it certainly isn't a solution. However, it also tells us that any value of $x$ that is less than $-5$ brings about a fraction that has a negative numerator and denominator,
resulting in a positive fraction and thus satisfying the inequality. No value between $x=-5$ and $x=\frac{3}{2}$ (except $\frac{3}{2}$ itself) seems to be a solution. Therefore, we conclude that the
solutions are the intervals $(-\infty,-5)\cup[\frac{3}{2},+\infty)$.
For the sake of better notation, define the "x-intercept" of a fractional inequality to be those values of $x$ that cause the numerator and/or the denominator to be $0$.To develop a method for
quicker solutions of fractional inequalities, we can simply consider the "x-intercepts" of the numerator and denominator. We graph them on the number line. Then, in every region of the number line,
we test one point to see if the whole region is part of the solution. For example, in the example problem above, we see that we only had to test one value such as $0$ in the region $(-5,\frac{3}{2})$
, as well as one value in the region $(-\infty,-5]$ and $[\frac{3}{2},+\infty)$; then we see which regions are part of the solution set. This does indeed give the complete solution set.
One must be careful about the boundaries of the solutions. In the example problem, the value $x=\frac{3}{2}$ was a solution only because the inequality was nonstrict. Also, the value $x=-5$ was not a
solution because it would bring about division by $0$. Similarly, any "x-intercept" of the numerator is a solution if and only if the inequality is nonstrict, and every "x-intercept" of the
denominator is never a solution because we cannot divide by $0$.
Complete Inequalities
A inequality that is true for all real numbers or for all positive numbers (or even for all complex numbers) is sometimes called a complete inequality. An example for real numbers is the so-called
Trivial Inequality, which states that for any real $x$, $x^2\ge 0$. Most inequalities of this type are only for positive numbers, and this type of inequality often has extremely clever problems and
List of Theorems
Here are some of the more useful inequality theorems, as well as general inequality topics.
Can someone fix that Ptolemy's is in Advanced?
• Practice Problems on Alcumus
□ Inequalities (Prealgebra)
□ Solving Linear Inequalities (Algebra)
□ Quadratic Inequalities (Algebra)
□ Basic Rational Function Equations and Inequalities (Intermediate Algebra)
• A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $.500$.
During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than $.503$. What's the largest number of matches she could've won
before the weekend began? (1992 AIME Problems/Problem 3)
• Practice Problems on Alcumus
□ Quadratic Inequalities (Algebra)
□ Advanced Rational Function Equations and Inequalities (Intermediate Algebra)
□ General Inequality Skills (Intermediate Algebra)
□ Advanced Inequalities (Intermediate Algebra)
• Given that $(a+1)(b+1)(c+1) = 8$, and $a, b, c \ge 0$ show that $abc \le 1$. (<url>weblog_entry.php?t=172070 Source</url>)
• Let $a,b,c$ be positive real numbers. Prove that $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$ (2001 IMO Problems/Problem 2)
See also | {"url":"https://artofproblemsolving.com/wiki/index.php?title=Inequality&oldid=220777","timestamp":"2024-11-02T11:26:09Z","content_type":"text/html","content_length":"70373","record_id":"<urn:uuid:48bec675-10d0-4a06-b5ae-eda3e64567a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00179.warc.gz"} |
Power Monitor
This module monitors the incoming power to the equipment shelter by measuring the supply voltage and the supply current. This allows the system to monitor is power consumption to determine if
experimental setups are drawing too much power.
The supply current to the enclosure is measured using a Hall-effect current sensor ACS-712. The rated capacity of the +24V supply feeding the enclosure is a maximum of 5A. A 20A version of this
sensor is available and provides a level of overhead if the supply ever provides more than its rated 5A. This sensor uses a +5V supply and the output range is 0.5V (-20A input) to +4.5V (+20A input).
The output voltage in the range 0A to +20V is +2.5V to +3.0V. Scale and offset adjustment is carried out in the software.
The ADC can only tolerate a maximum input of +3.6V (V[DD] + 0.3V) and this corresponds to just over +10A of supply current. As a 100% current overload is not likely given that the +24V supply is
regulated and protected, but to account for this unlikely possibility, the sensor is connector so that an increase in current will result a reduction of output voltage towards 0V output, rather than
+5V output. it has been decided that a voltage divider on the output of the current sensor is not required.
The power monitor has a set of power input terminals to accept the incoming power cable; and a corresponding set of output terminals where the power is connected to the DIN rail input terminals. It
also has a flying cable from the monitor box to the DIN rails terminals that brings power to the monitor and accepts the current and voltage sensor readings from the power monitor box. | {"url":"https://little.id.au/projects/backyardmonitor/system_hardware/power_monitor.html","timestamp":"2024-11-05T23:33:40Z","content_type":"text/html","content_length":"38688","record_id":"<urn:uuid:1a42dc9c-42ef-45b9-b58f-267354fb655e>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00271.warc.gz"} |
TT2 Question 1
Matthew' solution is correct but in some way incomplete: since two eigenvalues are complex (and complex conjugate) there are two forms of solutions: a complex form and he found only the complex form;
Iven and Victor found both. despite Victor' solution looks different it is not: in fact he got the same eigenvector albeit multiplied by factor $-2i$ (which be also an eigenvector) and this is the
source of "discrepancy". | {"url":"https://forum.math.toronto.edu/index.php?PHPSESSID=8nt7m9nq872v8s51iv7bijklo5&topic=279.msg1339","timestamp":"2024-11-07T08:56:28Z","content_type":"application/xhtml+xml","content_length":"34506","record_id":"<urn:uuid:a70b1bb9-8958-4416-ad9d-a6c609aced13>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00703.warc.gz"} |
Trigonometry 7th edition pdf - 歯ろぐTrigonometry 7th edition pdf | 歯ろぐ
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Trigonometry, 7th Edition. 637 Pages · 2006 · 16.49 MB · 1,973 Downloads· English. by Ron Larson & Robert P. Hostetler. Preview
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How to use LOGICAL OPERATORS [ =,>,>=,<,<=, ] in Excel ?
Table of Contents
A brief introduction about the formulas was given in the very first article – What Excel does? How to use formula in Excel?
Arithmetic operators were discussed in the previous post INTRODUCTION TO EXCEL FORMULAS:ARITHMETIC OPERATORS
But that’s not all. We have got logical operators too without which we won’t be able to put the logical conditions in Excel. Logical conditions are the ones that help us make the decisions which
makes it necessary to learn the logical operators and their use in Excel.
In this article, we’ll learn about the logical operators and the ways to use them in Excel.
LOGICAL OPERATORS are the ones that tells us about the condition of the statement.e.g. equal to, greater than, less than, greater than or equal to, less than or equal to,not equal to etc.
Logical operators have got much importance in excel because in most of the functions we have to check the situation before performing any action. In real life too, we have to test before we act. The
same is the situation here.
If we want to make smart reports, we have to check different conditions and apply different operations to them. So logical operators help us to check the conditions of data.
Following operators are permitted in Excel formula
OPERATOR FUNCTION
= EQUAL TO
> GREATER THAN
>= GREATER THAN OR EQUAL TO
< LESS THAN
<= LESS THAN OR EQUAL TO
<> NOT EQUAL TO
EQUAL TO-“=”
is a logical operator which is used when we have to check if something is equal to a particular value or not.
It’ll be used with functions like IF and other functions, which will be discussed in further posts.
We sometimes can use this operator for a case when we need to check the value of some cell.
Let us take a case.
We will check if cell A13 is containing the value 12 or not.
The answer will appear as true as we have put A13 content as number 12.
EQUAL TO ” = ” OPERATOR EXAMPLE
is a logical operator which is used when we have to check if something is GREATER THAN a particular value or not.
It’ll be used with functions like IF and other functions, which will be discussed in further posts.
WE HAVE A VALUE OF 15 IN G10.
WE’LL CHECK IF VALUE IN G10 IS GREATER THAN 10 OR NOT.
We’ll put a formula in a cell, where we want the result, and put the following formula.
The result will come as TRUE.
GREATER THAN “>” OPERATOR EXAMPLE
is a logical operator which is used when we have to check if something is GREATER THAN OR EQUAL TO a particular value or not.
It’ll be used with functions like IF and other functions, which will be discussed in further posts.
WE HAVE A VALUE OF 3 IN G10.
WE’LL CHECK IF VALUE IN G10 IS GREATER THAN 10 OR NOT.
We’ll put a formula in a cell, where we want the result, and put the following formula.
The result will come as FALSE.
GREATER THAN OR EQUAL TO ” >= ” OPERATOR EXAMPLE
is a logical operator which is used when we have to check if something is LESS THAN a particular value or not.
It’ll be used with functions like IF and other functions, which will be discussed in further posts.
WE HAVE A VALUE OF 6 IN G10.
WE’LL CHECK IF VALUE IN G10 IS LESS THAN 10 OR NOT.
We’ll put a formula in a cell, where we want the result, and put the following formula.
The result will come as TRUE.
LESS THAN ” < ” OPERATOR EXAMPLE
is a logical operator which is used when we have to check if something is LESS THAN OR EQUAL TO a particular value or not.
It’ll be used with functions like IF and other functions, which will be discussed in further posts.
USAGE:WE HAVE A VALUE OF 10 IN G10.
WE’LL CHECK IF VALUE IN G10 IS LESS THAN OR EQUAL TO 10 OR NOT.
We’ll put a formula in a cell, where we want the result, and put the following formula.
The result will come as TRUE.
LESS THAN OR EQUAL TO “<=” OPERATOR EXAMPLE
is a logical operator which is used when we have to check if something is NOT EQUAL TO a particular value.
It’ll be used with functions like IF and other functions, which will be discussed in further posts.
WE HAVE A VALUE OF 2019 IN G10.
WE’LL CHECK IF VALUE IN G10 IS NOT EQUAL TO 2019 OR NOT.
We’ll put a formula in a cell, where we want the result, and put the following formula.
The result will come as FALSE.
NOT EQUAL TO ” <> ” OPERATOR EXAMPLE | {"url":"https://gyankosh.net/msexcel/introduction-to-logical-operators-in-excel/","timestamp":"2024-11-03T19:05:26Z","content_type":"text/html","content_length":"162657","record_id":"<urn:uuid:b45a2591-7f32-4634-bb35-6566a05e9145>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00809.warc.gz"} |
ECCC - Reports tagged with hardness versus randomness
Extending the classical ``hardness-to-randomness'' line-of-works, Doron et al. (FOCS 2020) recently proved that derandomization with near-quadratic time overhead is possible, under the assumption
that there exists a function in $\mathcal{DTIME}[2^n]$ that cannot be computed by randomized SVN circuits of size $2^{(1-\epsilon)\cdot n}$ for a small $\epsilon$.
In this work we ... more >>> | {"url":"https://eccc.weizmann.ac.il/keyword/16044/","timestamp":"2024-11-12T01:18:07Z","content_type":"application/xhtml+xml","content_length":"22977","record_id":"<urn:uuid:9be81192-9eb8-45b6-874c-c8823cad7308>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00897.warc.gz"} |
Grant and Betty s Home
Big G Productions Home
Rubik s Cube
2003 World Championship
Solving Videos
Blindfold Cubing
Megaminx Solution
Move Notation
Cube visitors since 2004-March-26: ?
Blindfold Cubing
Around the beginning of November, 2002, I finally did my first 3x3x3 blindfolded cube! You may think that's impossible, as I once did, but I was successful. I am not the first person to have done
this (by far!) - there are those who can solve as many as 5 cubes without looking at any of them once they start solving. To see how I did it, just look below.
Here's how it works, in a nutshell. Obviously, you have to look at the cube at some point, otherwise you have no idea what the cube looks like (unless someone tells you), so the first thing you do is
study the cube. Once you are sure that you have completely memorized the cube's state, you blindfold yourself, pick up the cube, solve it, and put it back down solved. Provided all went well, when
you take off the blindfold to look at the cube again, it will be solved.
I may actually put instructions on blindfold cubing here at some point, but for now, I will refer you to Dr. Richard Carr's blindfold cubing document (you need adobe acrobat reader to view it). The
method I use is essentially the same as his, except that I didn't spend the time to learn the algorithms he listed - I just used algorithms that I already knew for solving the cube.
Just so you are warned, none of the rest of this is probably going to make much sense unless you have read his document and understand the move notation and terminology used in cubing algorithms.
Well, here it is - a walkthrough of my first blindfolded cube. Unfortunately, I didn't write this page right after doing the blindfold cube, so I don't remember the exact move sequences used.
However, I did write down which actions were taken, and this is pretty close to how I would have done it at the time - it should be move for move what I did, except for the corner and edge
orientation steps, for a total of 186 moves. The cube's original, mixed state was:
2221 1001 (corner orientations)
0100 1100 0111 (edge orientations)
8123 4675 (corner permutation)
6 12 5 2 7 10 4 8 3 11 1 9 (edge permutation)
If you want to follow what I did, apply these moves to a solved cube, to get to the original, mixed state:
F2 R' F2 R D2 F2 L B2 L U2 F' U2 F L2 B2 L2 D2 U' B F'
Then, just follow the moves listed with each step of my solution.
Step One - Corner Orientation (63 moves):
1. Rotate corner positions 1, 2, & 3 counter clockwise, making the corner orientation 0001 1001
[(L' F L F') (L' F L F') U] [(L' F L F') (L' F L F') U] [(L' F L F') (L' F L F') U] U
2. Rotate corner position 4 clockwise and corner position 8 counter clockwise, making the corner orientation 0000 1002
[(U L' U' L) (U L' U' L) B'] [(L' U L U') (L' U L U') B]
3. Finally, rotate corner position 5 clockwise and corner position 8 counter clockwise, completing corner orientation.
[(F R' F' R) (F R' F' R) D2] [(R' F R F') (R' F R F') D2]
Step Two - Edge Orientation (45 moves):
1. Flip edge positions 2 and 12, making the edge orientation 0000 1100 0110
[(L U' L') E (L U L') E'] [(U2 B') (M' B2 M) (B' U2)]
2. Flip edge positions 5 and 10, making the edge orientation 0000 0100 0010
[(L D' L') E' (L D L') E] [(D2 F') (M' F2 M) (F' D2)]
3. Flip edge positions 6 and 11, completing edge orientation
[(R D' R') E' (R D R') E] [(D2 B') (M B2 M') (B' D2)]
Step Three - Corner Permutation (22 moves):
1. Rotate corners from position 1 to 8, from 8 to 2, and from 2 to 1, leaving a corner permutation of 1523 4678
B2 [(R B') (R F2) (R' B R) (F2 R2)] B2
2. Rotate corners from position 2 to 5, from 5 to 4, and from 4 to 2, leaving a corner permutation of 1324 5678
R2 [(L F' L) (B2 L') (F L) (B2 L2)] R2
NOTE: I left the corners in positions 2 and 3 to swap later with two edges.
Step Four - Edge Permutation (43 moves)
1. Rotate edges from position 5 to 7, 7 to 6, and 6 to 5, leaving an edge permutation of 6 12 5 2 10 4 7 8 3 11 1 9
(R2 D) (S D2 S') (D R2)
2. Rotate edges from position 1 to 6, 6 to 5, and 5 to 1, leaving an edge permutation of 10 12 5 2 4 6 7 8 3 11 1 9
(M D2) (M' D2)
3. Rotate edges from position 3 to 5, 5 to 4, and 4 to 3, leaving an edge permutation of 10 12 2 4 5 6 7 8 3 11 1 9
D [(S' U2) (S U2)] D'
4. Swap edges between positions 9 and 12 and between 10 and 11, leaving an edge permutation of 10 12 2 4 5 6 7 8 9 1 11 3
(F2 E2 F2 E2) (R2 E2 R2 E2)
5. Rotate edges from position 10 to 12, 12 to 2, and 2 to 10, leaving an edge permutation of 10 3 2 4 5 6 7 8 9 12 11 1
U' [(U2 L) (S' L2 S) (L U2)] U
6. Rotate edges from position 10 to 12, 12 to 1, and 1 to 10, leaving an edge permutation of 1 3 2 4 5 6 7 8 9 10 11 12
U [(U2 L) (S' L2 S) (L U2)] U'
Step Five - Correct Final Edge Pair and Corner Pair (13 moves):
• Swap edge positions 2 and 3 and corner positions 2 and 3.
y (L' U' L) F2 (R' D) (R U) (R2 D') (R2 U') F2
Finally, the cube is solved! | {"url":"http://grantnbetty.com/cube/blindfold.html","timestamp":"2024-11-14T03:56:39Z","content_type":"text/html","content_length":"6907","record_id":"<urn:uuid:ae4332b9-3d2e-476b-b70a-13d5f2d98451>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00824.warc.gz"} |
Exploring the Universe: Key Physics Topics and Their Impact
Physics is often described as the fundamental science because it addresses the most basic principles governing the natural world. From the smallest subatomic particles to the vastness of the cosmos,
physics seeks to understand the underlying laws of nature. This article delves into several key physics topics, exploring their significance and the impact they have on our understanding of the
Classical Mechanics: The Motion of Bodies
Classical mechanics, developed by Sir Isaac Newton in the 17th century, is the study of the motion of bodies under the influence of forces. Newton’s three laws of motion form the foundation of
classical mechanics:
• First Law (Inertia): A body at rest will remain at rest, and a body in motion will remain in motion unless acted upon by an external force. It also laid the groundwork for more advanced theories
in physics.
• Second Law (F=ma): A body’s acceleration is directly proportional to the net force acting upon it and inversely proportional to its mass.
• Third Law (Action and Reaction): For every action, there is an equal and opposite reaction.
• Classical mechanics explains the motion of everyday objects and is essential for engineering, astronomy, and many other fields. It also laid the groundwork for more advanced theories in physics.
Electromagnetism: The Forces of Charge
Electromagnetism is the study of the forces between electrically charged particles. Developed by James Clerk Maxwell in the 19th century, Maxwell’s equations describe how electric and magnetic fields
are generated and altered by each other and by charges and currents. It also laid the groundwork for more advanced theories in physics.
Electromagnetism is responsible for a wide range of phenomena, from the light we see to the functioning of electronic devices. Understanding electromagnetic fields is crucial for the development of
technologies such as radio, television, and the internet.
Thermodynamics: The Science of Heat and Work
Thermodynamics is the branch of physics that deals with heat, work, and the forms of energy transformation. The laws of thermodynamics are fundamental principles that describe how energy moves and
changes in a system:
• The First Law (Conservation of Energy) states that energy cannot be created or destroyed, only transformed from one form to another.
• Second Law: The entropy of an isolated system always increases over time, meaning that natural processes tend to move towards a state of disorder.
• Third Law: As the temperature of a system approaches absolute zero, the entropy approaches a constant minimum.
• Thermodynamics is essential for understanding engines, refrigerators, and even the lifecycle of stars.
Quantum Mechanics: The Strange World of the Very Small
Quantum mechanics is the branch of physics that deals with the behavior of particles on an atomic and subatomic scale. Unlike classical mechanics, quantum mechanics introduces concepts that can seem
• Wave-Particle Duality: Particles, such as electrons, exhibit both wave-like and particle-like properties.
• Uncertainty Principle: Proposed by Werner Heisenberg, this principle states that it is impossible to know a particle’s exact position and exact momentum simultaneously.
• Quantum Entanglement: Particles can become entangled, meaning the state of one particle instantaneously influences the state of another, regardless of the distance between them.
• Quantum mechanics has revolutionized our understanding of the universe and led to the development of numerous technologies, including semiconductors, lasers, and quantum computers.
Relativity: The Nature of Space and Time
Albert Einstein’s theory of relativity transformed our understanding of space and time. There are two main components:
• Special Relativity: Introduced in 1905, it includes the famous equation E=mc², which shows the equivalence of mass and energy. It also posits that the laws of physics are the same for all
non-accelerating observers and that the speed of light in a vacuum is constant.
• General Relativity: Published in 1915, it extends special relativity to include gravity, describing it as the curvature of spacetime caused by mass and energy.
• Relativity has profound implications for our understanding of the universe. It explains phenomena such as the bending of light around massive objects (gravitational lensing) and the expansion of
the universe.
Nuclear Physics: The Power of the Atom
Nuclear physics is the study of the components and behavior of atomic nuclei. It encompasses the study of nuclear reactions, such as fission and fusion, which release immense amounts of energy. This
field has led to both beneficial and destructive applications:
• Nuclear Energy: Provides a significant source of power through controlled nuclear reactions.
• Nuclear Weapons: Harness the destructive power of nuclear reactions for warfare.
• Nuclear physics also plays a crucial role in medicine, particularly in the development of diagnostic imaging techniques and cancer treatments.
Particle Physics: The Fundamental Constituents of Matter
Particle physics investigates the smallest building blocks of matter and the forces that govern their interactions. The Standard Model is the theoretical framework that describes the fundamental
particles (such as quarks and leptons) and the fundamental forces (strong, weak, electromagnetic, and gravitational). Nuclear physics also plays a crucial role in medicine, particularly in the
development of diagnostic imaging techniques and cancer treatments.
The discovery of the Higgs boson at CERN’s Large Hadron Collider in 2012 confirmed the existence of the Higgs field, which gives particles their mass. Particle physics continues to push the
boundaries of our understanding of the universe, seeking to answer fundamental questions about the nature of matter and energy.
Astrophysics: Understanding the Cosmos
Astrophysics is the branch of astronomy that applies the principles of physics to understand how stars, planets, galaxies, and the universe as a whole behave and evolve. Key topics include:
• Star Formation and Evolution: Understanding the lifecycle of stars, from their birth in stellar nurseries to their eventual demise as white dwarfs, neutron stars, or black holes.
• Cosmology is the study of the origin, evolution, and fate of the universe. This includes the Big Bang theory, cosmic inflation, dark matter, and dark energy.
• Astrophysics combines observational data with theoretical models to explore the mysteries of the cosmos.
The Endless Frontier
Physics is a dynamic and ever-evolving field that continuously expands our understanding of the universe. From the motion of everyday objects to the behavior of particles at the most minor scales,
physics provides a foundation for exploring the natural world. As we continue to uncover new phenomena and develop innovative technologies, the study of physics will remain at the forefront of
scientific discovery, driving our quest for knowledge and shaping the future of humanity. | {"url":"https://corycarnley.com/exploring-the-universe-key-physics-topics-and-their-impact/","timestamp":"2024-11-02T05:09:02Z","content_type":"application/xhtml+xml","content_length":"41839","record_id":"<urn:uuid:adac0c1e-681e-40b2-a3a8-22048ccd10bb>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00557.warc.gz"} |
Pounds to Short Ton Converter
โ Switch toShort Ton to Pounds Converter
How to use this Pounds to Short Ton Converter ๐ ค
Follow these steps to convert given weight from the units of Pounds to the units of Short Ton.
1. Enter the input Pounds value in the text field.
2. The calculator converts the given Pounds into Short Ton in realtime โ using the conversion formula, and displays under the Short Ton label. You do not need to click any button. If the input
changes, Short Ton value is re-calculated, just like that.
3. You may copy the resulting Short Ton 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 Pounds to Short Ton?
The formula to convert given weight from Pounds to Short Ton is:
Weight[(Short Ton)] = Weight[(Pounds)] / 2000
Substitute the given value of weight in pounds, i.e., Weight[(Pounds)] in the above formula and simplify the right-hand side value. The resulting value is the weight in short ton, i.e., Weight[(Short
Calculation will be done after you enter a valid input.
Consider that a high-end camera with a lens weighs 3 pounds.
Convert this weight from pounds to Short Ton.
The weight of camera with lens in pounds is:
Weight[(Pounds)] = 3
The formula to convert weight from pounds to short ton is:
Weight[(Short Ton)] = Weight[(Pounds)] / 2000
Substitute given weight of camera with lens, Weight[(Pounds)] = 3 in the above formula.
Weight[(Short Ton)] = 3 / 2000
Weight[(Short Ton)] = 0.0015
Final Answer:
Therefore, 3 lbs is equal to 0.0015 tn.
The weight of camera with lens is 0.0015 tn, in short ton.
Consider that a bag of organic coffee beans weighs 1 pound.
Convert this weight from pounds to Short Ton.
The weight of coffee beans in pounds is:
Weight[(Pounds)] = 1
The formula to convert weight from pounds to short ton is:
Weight[(Short Ton)] = Weight[(Pounds)] / 2000
Substitute given weight of coffee beans, Weight[(Pounds)] = 1 in the above formula.
Weight[(Short Ton)] = 1 / 2000
Weight[(Short Ton)] = 0.0005
Final Answer:
Therefore, 1 lbs is equal to 0.0005 tn.
The weight of coffee beans is 0.0005 tn, in short ton.
Pounds to Short Ton Conversion Table
The following table gives some of the most used conversions from Pounds to Short Ton.
Pounds (lbs) Short Ton (tn)
0.01 lbs 0.000005 tn
0.1 lbs 0.00005 tn
1 lbs 0.0005 tn
2 lbs 0.001 tn
3 lbs 0.0015 tn
4 lbs 0.002 tn
5 lbs 0.0025 tn
6 lbs 0.003 tn
7 lbs 0.0035 tn
8 lbs 0.004 tn
9 lbs 0.0045 tn
10 lbs 0.005 tn
20 lbs 0.01 tn
50 lbs 0.025 tn
100 lbs 0.05 tn
1000 lbs 0.5 tn
The pound is a unit of mass used in the imperial system and the United States customary system. One pound is equivalent to 0.45359237 kilograms. The pound is commonly used for measuring the weight of
objects in everyday contexts, and it is a common unit in the United States and some other countries.
Short Ton
The short ton is a unit of mass commonly used in the United States, equivalent to 2,000 pounds or approximately 907 kilograms. It is often used in industries such as construction, mining, and freight
Frequently Asked Questions (FAQs)
1. What is the formula for converting Pounds to Short Ton in Weight?
The formula to convert Pounds to Short Ton in Weight is:
Pounds / 2000
2. Is this tool free or paid?
This Weight conversion tool, which converts Pounds to Short Ton, is completely free to use.
3. How do I convert Weight from Pounds to Short Ton?
To convert Weight from Pounds to Short Ton, you can use the following formula:
Pounds / 2000
For example, if you have a value in Pounds, you substitute that value in place of Pounds in the above formula, and solve the mathematical expression to get the equivalent value in Short Ton.
Weight Converter Android Application
We have developed an Android application that converts weight between kilograms, grams, pounds, ounces, metric tons, and stones.
Click on the following button to see the application listing in Google Play Store, please install it, and it may be helpful in your Android mobile for conversions offline. | {"url":"https://convertonline.org/unit/?convert=pound-short_ton","timestamp":"2024-11-07T11:04:10Z","content_type":"text/html","content_length":"86274","record_id":"<urn:uuid:f42971d7-2b9e-4013-9fb6-532bb1bd8f0a>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00574.warc.gz"} |
Kenneth R. French - Detail for 6 Portfolios Formed on Size and
Detail for 6 Portfolios Formed on Size and Investment
Monthly July 1963 - September 2024
Annual 1964 - 2023
Construction: The portfolios, which are constructed at the end of each June, are the intersections of 2 portfolios formed on size (market equity, ME) and 3 portfolios formed on investment (Inv). The
size breakpoint for year t is the median NYSE market equity at the end of June of year t. Investment is the change in total assets from the fiscal year ending in year t-2 to the fiscal
year ending in t-1, divided by t-2 total assets. The Inv breakpoints are the 30th and 70th NYSE percentiles.
Stocks: The portfolios for July of year t to June of t+1 include all NYSE, AMEX, and NASDAQ stocks for which we have market equity data for June of t and total assets data for t-2 and t-1. | {"url":"https://mba.tuck.dartmouth.edu/pages/faculty/ken.french/Data_Library/six_portfolios_me_inv.html","timestamp":"2024-11-09T23:53:47Z","content_type":"text/html","content_length":"13477","record_id":"<urn:uuid:85d1ba6a-49ab-4869-89d5-ccc7491d28b1>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00734.warc.gz"} |
Roman Numerals Converter (2024)
Created by Rijk de Wet
Reviewed by
Wojciech Sas, PhD and Steven Wooding
Last updated:
May 20, 2024
Table of contents:
• What are Roman numerals?
• How to use the Roman numerals converter?
• The Roman numerals chart — Roman numbers 1 to 1000
• How do I read Roman numerals?
• How do I convert numbers to Roman numerals?
• FAQ
Whether you're trying to decipher an ancient Latin text or you just want to add the year on your high school jacket, our Roman numerals converter has you covered. Knowing how to convert numbers to
Roman numerals and back can be a surprisingly handy trick for historians, mathematicians, astrologists, and even lawmakers.
So, whether you'd like to learn how to read Roman numerals or just want to consult a Roman numerals chart, you're in the right place with our Roman numerals generator. I know that $\textrm{I}$I, for
one, ❤️ Roman numerals!
What are Roman numerals?
We see Roman numerals everywhere — clocks, musical chords, historical dates, legal documents, astronomy, and even football team names. But what are they, where do they come from, and how do they
The Roman numeral number system represents numeric values with letters of the Latin alphabet. You can take a look at the chart further down to see how they correspond.
The Roman numeral system arose in ancient Rome and was used in most of Europe until the Middle Ages. It was primarily used to count with, so it has no representation for negative numbers (as these
can't be used to describe a discrete number of objects); nor for the number zero, as everyday people hadn't even thought of these concepts yet.
How to use the Roman numerals converter?
If you're in a hurry and just need to convert a number to Roman numerals, look no further than the Roman numerals converter. Here's how to use it:
1. Enter the number you'd like to convert to Roman numerals. Because Roman numerals can't represent fractions, the number zero, negative numbers, or numbers above 3,999,999, the Roman numerals
generator will reject such inputs.
2. The Roman numerals calculator will determine your number's equivalent Roman numeral and display it below.
3. If your number has a complicated Roman numeral, we'll also break it down with an explanation.
If you'd still like to know how to read Roman numerals better than even our Roman numerals converter can, you should keep reading!
The Roman numerals chart — Roman numbers 1 to 1000
As we've said above, Roman numerals use letters of the alphabet to represent numeric values. In the Roman numerals chart below, we can see which letter corresponds to which value.
Roman numeral Value
$\textrm{I}$I 1
$\textrm{V}$V 5
$\textrm{X}$X 10
$\textrm{L}$L 50
$\textrm{C}$C 100
$\textrm{D}$D 500
$\textrm{M}$M 1,000
In addition, when a Roman numeral has a line over it, its value is multiplied by 1,000. $\overline{\textrm{V}}$V , for example, signifies 5,000. This extension to the Roman numeral system wasn't
widely used, and competing systems had other methods, but adding the overhead line will generally convey the correct idea to a knowledgeable modern audience.
Roman numeral Value
$\overline{\textrm{V}}$V 5,000
$\overline{\textrm{X}}$X 10,000
$\overline{\textrm{L}}$L 50,000
$\overline{\textrm{C}}$C 100,000
$\overline{\textrm{D}}$D 500,000
$\overline{\textrm{M}}$M 1,000,000
How do I read Roman numerals?
The numerals in the above chart are just the building blocks. You can combine them to create any integer you can think of — within the limits we've established, of course. Let's see how this is done!
Our everyday number system of Arabic numerals uses a place-value system, meaning a digit's position in a number determines how much it's worth. For example, the digit "2" represents a value of 2 in
42, whereas it represents 20 in the number 123.
In contrast, the Roman numerals all have fixed values. $\textrm{V}$V always means 5, and $\textrm{M}$M always means 1,000 — regardless of where they're placed in the string of Roman numerals.
Instead, the numerals must be combined to represent intermediary values with the help of addition and subtraction:
• When two numerals are side-by-side, and the first is of equal or greater value than the second, their values are added together.
□ $\textrm{I}$I (1) combines with itself to create $\textrm{II}$II (2).
□ $\textrm{V}$V (5) and $\textrm{I}$I (1) combine to create $\textrm{VI}$VI (6).
□ $\textrm{C}$C (100) and three $\textrm{X}$X's (each 10) combine to create $\textrm{CXXX}$CXXX (130).
• When the first letter of a numeral pair has a lesser value than the second, the first subtracts from the second.
□ $\textrm{I}$I (1) and $\textrm{V}$V (5) combine to create $\textrm{IV}$IV (4).
□ $\textrm{X}$X (10) and $\textrm{C}$C (100) combine to create $\textrm{XC}$XC (90).
Here are a few easy numbers converted to Roman numbers from 1 to 1000 for you to sink your teeth into — see if you can make sense of them!
1 $\textrm{I}$I 10 $\textrm{X}$X 100 $\textrm{C}$C
2 $\textrm{II}$II 20 $\textrm{XX}$XX 200 $\textrm{CC}$CC
3 $\textrm{III}$III 30 $\textrm{XXX}$XXX 300 $\textrm{CCC}$CCC
4 $\textrm{IV}$IV 40 $\textrm{XL}$XL 400 $\textrm{CD}$CD
5 $\textrm{V}$V 50 $\textrm{L}$L 500 $\textrm{D}$D
6 $\textrm{VI}$VI 60 $\textrm{LX}$LX 600 $\textrm{DC}$DC
7 $\textrm{VII}$VII 70 $\textrm{LXX}$LXX 700 $\textrm{DCC}$DCC
8 $\textrm{VIII}$VIII 80 $\textrm{LXXX}$LXXX 800 $\textrm{DCCC}$DCCC
9 $\textrm{IX}$IX 90 $\textrm{XC}$XC 900 $\textrm{CM}$CM
10 $\textrm{X}$X 100 $\textrm{C}$C 1,000 $\textrm{M}$M
The pattern continues for larger numbers. 2,000 is $\textrm{MM}$MM, 3,000 is $\textrm{MMM}$MMM, 4,000 is $\textrm{M}\overline{\textrm{V}}$MV, and so on.
I can hear you asking — How do I convert a more complex number to Roman numerals? What is 2021 in Roman numerals, for example? Jump over to the next section to find out!
How do I convert numbers to Roman numerals?
If we want to convert a number that has more than one significant digit (like 365 or 2021), we can follow a simple formula:
• Express the number as a sum of its place values.
□ $365 = 300 + 60 + 5$365=300+60+5;
□ $2021 = 2,000 + 20 + 1$2021=2,000+20+1 (notice that we ignore the zero between the two $2$2's!)
• Convert each value in the sum to its Roman numeral. It's easy with the table above!
□ For $365$365:
☆ $300\rightarrow\textrm{CCC}$300→CCC
☆ $60\rightarrow\textrm{LX}$60→LX
☆ $5\rightarrow\textrm{V}$5→V
□ For $2021$2021:
☆ $2,000\rightarrow\textrm{MM}$2,000→MM
☆ $20\rightarrow\textrm{XX}$20→XX
☆ $1\rightarrow\textrm{I}$1→I
• Put the symbols together to convert your number to Roman numerals.
□ $365\rightarrow\textrm{CCC}\cdot\textrm{LX}\cdot\textrm{V}\rightarrow\textrm{CCCLXV}$365→CCC⋅LX⋅V→CCCLXV
□ $2021\rightarrow\textrm{MM}\cdot\textrm{XX}\cdot\textrm{I}\rightarrow\textrm{MMXXI}$2021→MM⋅XX⋅I→MMXXI
If you're still having trouble understanding our Roman ancestors' number system, playing around with our Roman numerals calculator (even just for $\textrm{X}$X minutes) will get you there!
What is 4 in Roman numerals?
4 is IV in Roman numerals. It consists of I and V, the Roman numerals for 1 and 5. Because the I is placed before the V, we must subtract 1 from 5, and so IV = 5 − 1 = 4.
What are the Roman numerals?
The Roman numerals are I, V, X, L, C, D, and M. Each of the Roman numbers 1 to 1000 represent a numeric value. Which represents which is tabulated below:
Roman numeral Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1,000
What is XI in Roman numerals?
XI is the Roman numeral for 11, as it consists of X (10) and I (1) added together. The Roman numeral XI is commonly used in soccer, where the term "starting XI" refers to the eleven players who will
be on the field when the match begins.
What number is LV in Roman numerals?
The Roman numeral LV represents the number 55. In 2021, the NFL hosted the 55th annual Super Bowl and labeled it with LV. They've been doing this since the '70s.
What is 19 in Roman numerals?
19 in Roman numerals is XIX. We can break 19 up into a sum of 10 + 9. In Roman numerals, 10 is X, and 9 is IX (10 – 1), so together 19 is XIX.
What is XXXVII in Roman numerals?
XXXVII is the Roman numeral for 37. It's constructed of
1. XXX (30);
2. V (5); and
3. II (2),
which are all added together to result in 30 + 5 + 2 = 37. This particular Roman numeral XXXVII commonly refers to a titanic Super Bowl match that occurred in early 2003.
What is 1980 in Roman numerals?
1980 in Roman numerals is MCMLXXX. It's constructed of
1. M (1000);
2. CM (900); and
3. LXXX (80).
What is 1984 in Roman numerals?
1984 in Roman numerals is MCMLXXXIV. It's constructed of
1. M (1000);
2. CM (900);
3. LXXX (80); and
4. IV (4). | {"url":"https://ziskabachwas.info/article/roman-numerals-converter-2","timestamp":"2024-11-06T22:04:51Z","content_type":"text/html","content_length":"100299","record_id":"<urn:uuid:b0634478-aa21-4583-a4cc-2e3445ba1b03>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00764.warc.gz"} |
Combinatorial and Geometric Discrepancy (Online)
Schedule for: 20w5141 - Combinatorial and Geometric Discrepancy (Online)
Beginning on Wednesday, September 30 and ending Friday October 2, 2020
All times in Banff, Alberta time, MDT (UTC-6).
Wednesday, September 30
Introduction and Welcome by BIRS Staff ↓
- A brief introduction video from BIRS staff with important logistical information
Ryan Alweiss: Discrepancy Minimization via a Self-Balancing Walk ↓
08:00 We study discrepancy minimization for vectors in $\mathbb{R}^n$ under various settings. The main result is the analysis of a new simple random process in multiple dimensions through a
- comparison argument. As corollaries, we obtain bounds which are tight up to logarithmic factors for several problems in online vector balancing posed by Bansal, Jiang, Singla, and Sinha (STOC
08:25 2020), as well as linear time algorithms of logarithmic bounds for the Komlós conjecture.
Samantha Fairchild: Families of well-approximable measures ↓
08:25 It is conjectured that the optimal order of approximation of the Lebesgue measure by a finite atomic measure is $N^{-1} (\log N)^{d-1}$. This result is known for dimensions 1 and 2. We will
- share recent work of Fairchild, Goering, Weiss which in dimension 1 confirms Lebesgue measure is indeed the hardest to approximate. Moreover we improve on recent work by Aistleitner, Bilyk,
08:50 and Nikolov by constructing a family of discrete measures with star discrepancy bounded above by $N^{-1} (\log(N))$.
Sebastian Neumayer: Curve Based Approximation of Images on Manifolds ↓
- In this talk, we will discuss a way of approximating images living on a manifold with Lipschitz continuous curves. In order to quantify the approximation quality, we employ discrepancies. This
09:15 enables us to provide approximation rates independent of the dimension. The proposed mathematical model is illustrated with some numerical examples.
Tetiana Stepaniuk: Hyperuniformity of point set sequences ↓
- In the talk we study hyperuniformity on flat tori. Hyperuniform point sets on the unit sphere have been studied by J. Brauchart, P. Grabner, W. Kusner and J. Ziefle. We will discuss several
09:40 examples of hyperuniform sequences of point sets.
Hendrik Pasing: Improved Discrepancy Bounds and Estimates ↓
09:40 Error estimation in Monte-Carlo integration is related to the star discrepancy of random point sets. We will present latest results for (probabilistic) upper bounds of the star discrepancy
- which are based on major improvements on bounds of bracketing numbers. Additionally we introduce upper bounds for the expected value of the star discrepancy. This is joint work with Michael
10:05 Gnewuch and Christian Weiß.
Group Photo (Online) ↓
- Please turn on your cameras for the "group photo" -- a screenshot in Zoom's Gallery view.
Friday, October 2
Ujue Etayo: A deterministic set of spherical points with small discrepancy ↓
08:00 In this talk we present the problem of seeking for point configurations on the 2-dimensional sphere with small discrepancies. In particular, we prove that points coming from the Diamond
- ensemble (a deterministic multiparametric model of points uniformly distributed on the sphere) for a concrete choice of parameters provides the best spherical cap discrepancy known until date
08:25 for a deterministic family of points.
Mathias Sonnleitner: (Non-)optimal point sets for numerical integration ↓
- Connections between combinatorial/geometric discrepancy, worst-case errors of algorithms and quantization of measures are presented. The aim is to indicate possible answers to questions of the
08:50 type: How to geometrically measure the quality of a point set for approximation problems?
Victor Reis: Vector Balancing in Lebesgue Spaces ↓
08:50 The Komlós conjecture in discrepancy theory asks for a ±1-coloring, for any given unit vectors, achieving constant discrepancy in the ell-infinity norm. We investigate what ell-q discrepancy
- bound to expect, more generally, for ±1-colorings of vectors in the unit ell-p ball for any p less than q, and achieve optimal partial colorings. In particular, for p = q, our result
09:15 generalizes Spencer's "six standard deviations" theorem.
Lily Li: On the Computational Complexity of Linear Discrepancy ↓
Linear discrepancy is a variant of discrepancy that measures how well we can round vectors w in $[0,1]^n$ to vectors x in ${0,1}^n$, with the error of the rounding measured with respect to a
09:15 matrix A, i.e. as the ell-infinity norm of the difference Ax - Aw. This is a variant of classical combinatorial discrepancy, which only considers the all-halves vector as w, and also captures
- measure theoretic discrepancy. Our work initiates the study of the computational complexity of linear discrepancy. In particular, we show that it is NP-Hard in general, and is computable in
09:40 polynomial time when A has a constant number of rows, and the magnitude of each entry in A has bounded bit complexity. When there is only one row, we can compute the linear discrepancy in O(n
log n) time.
- Open discussion (Online) | {"url":"https://www.birs.ca/events/2020/5-day-workshops/20w5141/schedule","timestamp":"2024-11-03T15:56:04Z","content_type":"application/xhtml+xml","content_length":"21668","record_id":"<urn:uuid:c31f7fbb-3ac6-4ea2-ba4b-240ebf9fed2e>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00395.warc.gz"} |
Confidence Intervals for Linear Model Predictions
add_ci.lm {ciTools} R Documentation
Confidence Intervals for Linear Model Predictions
This function is one of the methods in add_ci and automatically is called when an object of class lm is passed to add_ci.
## S3 method for class 'lm'
alpha = 0.05,
names = NULL,
yhatName = "pred",
log_response = FALSE,
df A data frame.
fit An object of class lm. Predictions are made with this object.
alpha A real number between 0 and 1. Controls the confidence level of the interval estimates.
NULL or character vector of length two. If NULL, confidence bounds automatically will be named by add_ci, otherwise, the lower confidence bound will be named names[1] and the upper
names confidence bound will be named names[2].
yhatName A string. Name of the vector of the predictions made for each observation in df
log_response Logical. Default is FALSE. If TRUE, confidence intervals will be generated for the response level of a log-linear model: log(Y) = X\beta + \epsilon.
... Additional arguments.
Confidence intervals for lm objects are calculated parametrically. This function is essentially a wrapper for predict(fit, df, interval = "confidence") if fit is a linear model. If log_response =
TRUE, confidence intervals for the response are calculated using Wald's Method. See Meeker and Escobar (1998) for details.
A dataframe, df, with predicted values, upper and lower confidence bounds attached.
See Also
add_pi.lm for prediction intervals for lm objects, add_probs.lm for conditional probabilities of lm objects, and add_quantile.lm for response quantiles of lm objects.
# Fit a linear model
fit <- lm(dist ~ speed, data = cars)
# Get fitted values for each observation in cars, and append
# confidence intervals
add_ci(cars, fit)
# Try a different confidence level
add_ci(cars, fit, alpha = 0.5)
# Try custom names for the confidence bounds
add_ci(cars, fit, alpha = 0.5, names = c("lwr", "upr"))
version 0.6.1 | {"url":"https://search.r-project.org/CRAN/refmans/ciTools/html/add_ci.lm.html","timestamp":"2024-11-11T02:58:16Z","content_type":"text/html","content_length":"4580","record_id":"<urn:uuid:18aeb5df-d9c1-4d3d-930a-65f4e1810c3d>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00815.warc.gz"} |
Is the Bertrand's paradox a paradox or a blunder?
Hanspeter Schmid Switzerland 18.4.2006. 9:28:56
E. T. Jaynes gave a very good mathematicalm argument using transformation groups in his 1973 paper "The Well-Posed Problem". See site: en.wikipedia.org/wiki/Bertrand%27s_paradox_%28probability%29 for
a brief explanation and reference. Kind regards, Hanspeter
Bozur Vujicic Serbia 25.4.2006. 9:03:49
You simply didn't catch the point. I have explained here why we can't use that second aproach in consideration. But if we do so, we could say the same thing .. it is invariant in some cases! But, be
aware - that aproach for our problem is not appropriate!!!
Brian McMullen USA 15.10.2006. 19:32:42
The problem is not the definition of the word random, it is the definition of the word chord. You want your solution as ignorant as possible about the relationship between circles and triangles.
Method 3 does not work because you are generating chords based on polar coordinates from an arbitrary orgin, weighting the area near the orgin far more heavily than the rest of the circle. It would
be ideal to generate chords based on two random points within the circle, because the easiest way to represent chords evenly in respect to area is to generate them with respect to area, via random
coordinates within the circle, you should get the same result wether you use polar coordinates weighted for distance from origin(center of circle)(the probability of a point would need to increase at
a square of r) or unweighted xy coordinates(easiest to program id guess) To confirm, You randomly generate two points within the circle, then you extrapolate the chord represented, if it intersects
the small circle you enumerate a variable. the probability is collisions/iterations. I will probably post more when I am more rested. my guess is something like 70-75% probability the chord will be
longer than the leg of triangle. even looking at example 1(the highest probability here), the area represented by 50% of the chords in your integration is far less then 1/2
Brian McMullen USA 15.10.2006. 20:37:22
It would be the same result if you used a fixed point on the circumfrence and a random point within the circle to generate the solution
Bob USA 16.10.2006. 13:56:13
geometry sucks
Brian McMullen USA 16.10.2006. 20:46:09
I ran it with 100M iterations with a probability of 0.2169
Brian McMullen USA 16.10.2006. 20:54:55
With Better collision detection 0.2171(using a circle of radius 5000)
Bozur Vujicic Serbia 6.11.2006. 4:44:46
Dear Brian,
I did not understand yor statment well. You sad:"you are generating chords based on polar coordinates from an arbitrary orgin, weighting the area near the orgin far more heavily than the rest of the
circle". In respect to what I weighted some chords in calculation?
Your example involves presumption that longer chords have more probability to be chosen because they consist more points (among diametar we can choose more points to determine one chord than in other
direction). Why? In that case each chord is not equaly probable and we are changing the original problem! It is not allowed. Do you agree?
Consider similar problem where the circumference is not closed, where small part "dl" of it is missing (like shape of the letter "C"). How can you connect area with the number of chords then? So,
here is crucial not to make fake transformation. If you have to count chords you can do it via area only if you are able to prove that transformation as bijection (one to one transformation).
Thanks for your comment Brian!
zcer malaysia 15.3.2007. 15:55:08
i have a possible resolution here at http://miniverse.blogspot.com/2006/12/solution-to-bertrands-paradox.html i'm no mathematician, so it's not at all rigorous. But somehow, i think it's correct. i'd
really like your opinion
Bozur Serbia 22.3.2007. 9:20:17
Hi I foud your work accidentally and scanned very fast that page. I have to look it again tenderly.
Bozur Serbia 14.6.2007. 5:57:55
zcer you have to agree that midpoint of the circle makes hidden infinite number of chords, so that way can't be acceptable. regards
Jo Germany 10.9.2007. 19:06:48
I think your refutation of the first example is wrong. It is not easy for me to understand what you exactly mean, but to me it seems that your main objection is that (in your opinion) not every chord
can be caught by choosing an arbitrary point on an arbitrary radius as midpoint of the chord. This is wrong since every chord has a midpoint and each point, especially every midpoint is lying on a
radius. Furthermore it is wrong that one cannot get the surface of a circle by integrating r. If you integrate r over [0,R] and over [0,2*Pi], this Integral yields the correct result Pi*R^2. Maybe I
didn't get your point exactly. But I think Bertrands paradox should not be called a "blunder". If there was a simple mistake mathematicians would of course have discovered it in the meantime.
However, I have to admit that I am not content with the explanation, all three solutions would equally be correct; I for myself sympathize with the solution 1/2 ...
lukomor Ukraine 24.9.2007. 8:29:31
The right answer is 1/2, but no one of three classic solutions is not right.
Bozur Serbia 23.10.2007. 6:19:35
Jo, you sad: ''If you integrate r over [0,R] and over [0,2*Pi], this Integral yields the correct result Pi*R^2.'' --- yes, but in that case you didn't integrate radii but sectors, and that's big
difference. By integrating radii you can' get circle shape (but rectangular)!! And on the surface of the sector where angle is extremely small you allways have three times points more on the farther
part (r>R/2)!! ''If there was a simple mistake mathematicians would of course have discovered it in the meantime. '' --- I think it is not just simple mistake and I deeply belive that math is
improving all the time as any other science. I can't say that I sympathize any of approaches but some of them I can't accept (math rules doesn't allow that), so the remaining solution is 1/3. regards
Bozur Serbia 23.10.2007. 6:20:51
Lokomor, could you explain your example which yields to 1/2 for Bertrands paradox?
Lukomor Ukraine 24.10.2007. 10:54:17
Bozur (Serbia 10/23/2007 6:20:51 AM) Lokomor, could you explain your example which yields to 1/2 for Bertrands paradox? - - - - - - - - - - - - - - - - - - - - - Of course, I can explain it. But the
explanation could be extrimely long. I can't do it in a few words.
Bozur Serbia 31.10.2007. 5:19:59
but this field can accept as many words as you wish
Lukomor Ukraine 1.11.2007. 11:55:32
Well, let's start. ================ Part I. First of all I should remind, that "practice is the criterion of truth". The only practical (physical) experiment was made by E.T.Jaynes. You may accept or
not accept his theoretical reasonings, but some quotations from his "Well-posed problem" (1973), in which "the unique solution has been verified experimentaly", can convince anyone. 1. "It will be
helpful to think of this in a more concrete way; presumably, we do no violence to the problem (i.e., it is still just as "random") if we suppose that we are tossing straws onto the circle, without
specifying how they are tossed. We therefore formulate the problem as follows. A long straw is tossed at random onto a circle; given that it falls so that it intersects the circle, what is the
probability that the chord thus defined is longer than a side of the inscribed equilateral triangle?"(E.T.Jaynes) 2. "The Bertrand experiment has, in fact, been performed by the writer and Dr.
Charles E. Tyler, tossing broom straws from a standing position onto a 5-in.-diameter circle drawn on the floor. Grouping the range of chord lengths into ten categories, 128 successful tosses
confirmed...[p=1/2]... with an embarrassingly low value of chi-squared".(E.T.Jaynes) more...
Lukomor ua 5.11.2007. 6:43:30
Part II.=========== Well, in "Part I" we have convinced, that physical experiment gives an unique solution of P=1/2. So, we should find the point where "the wonderful mind" falls down to a blunder.
In my humble opinion, this point is, that the term "a chord" is complex. "A chord" is the result of combination a circle and a straight line. Henry Poincare in "Calcus des probabilites" (Paris, 1912)
shows that there are two ways to get "a random chord".The first way is a combination of a fixed circle and a random straight line. This way leads to paradox . Another way is combination of a fixed
straight line and "a random circle". This way gives an unique solution. If we fix a straight line on a plane, and will tossing a circle (for example a coin with radius R) "by chance" to a plane, the
coin will cover a part of a straight line, if the distance from the line to the center of coin will less than +/-R, i.e. D=2*R. And the covered part of line will be longer then a side of triangle,
inscribed to a coin, if the distance from a straight line to the centre of coin will be the less, then then +/-(R/2), i.e. d=R. Thus, we'll find out, that probability is: P=d/D=1/2. In "Part III"
I'll shaw that both two ways are equivalent. ====more====
Bozur Serbia 14.12.2007. 9:01:39
About Part II ===== There in no problem how to perform experiment but how to count chords as result of that experiment. Why to measure distance from our center to that chord? The problem is the
length of that chord and function between its distance and lenght is not linear transformation and that will obviously lead to mistake in counting!!! (y=x2, x(0,4); y(0,16) p(x<3)=3/4 among x, p(y<9)
=9/16 among y, p(x) in not equal p(y) because of nonlinear transformation, we can't use p(x) for p(y) and vice versa !! ) Abaut Part I ===== If we tossing lines we will do it uniformly in all
direction, and we know the result before counting. Experiment for p=1/3 == Fix an end of one line on one point of the circumference than rotate that line by chance around that point and count how
many times appropriate chord has lenght longer than critical, probability will be p=1/3. Do you agree? Is it valid experiment? So, your experiment can't be argument in our discussion. We should go
back on motives for counting on p=1/2. Did you read my state about p=1/2, what do you think, is it reasonable?
Lukomor .ua 18.12.2007. 9:28:58
L:First of all I must be apologized for so long a silence. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Bozur: -Experiment for p=1/3 == Fix an end of one line on
one point of the circumference than rotate that line by chance around that point and count how many times appropriate chord has lenght longer than critical, probability will be p=1/3. Do you agree?
Is it valid experiment? - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - L: I cannot agree with this assertion. You arbitrarily fixed the point at a distance of d=R from
the center of circle.But this is not necessary condition. For example, I can select the point at a distance of d=2*R from the center, then probability is equal: P=arcsin (R (2*d))/arcsin (R/d) of =
arcsin (1/4) of /arcsin (1/2) of =0,48.... It is not possible to base the selection of point precisely on the circle.
Bozur Serbia 20.12.2007. 8:55:07
Let's suppose your point is not on distance of d=2*R but d>>2R it leads to solution p=1/2. So, your example belongs to that approach. (I am not picking the point from the plane surface but from the
cirumference that is somewhere in 3D space, so I didn't take one point that is accidentally on the circumference on that imagined surface, but for that solution only possible choosable points are on
that circumference. How would you find probability in case the circumference is not plane?) My idea is that for Bertrand's paradox problem only correct solution is p=1/3 because in that case we can't
find infinite number of chords that are neglected!!! In approach that leads to p=1/4 we neglected infinite number of chords whose lenght are d=2R. If we can do so, we can repeat that more and more
and neglect all chords that has similar lenght and get new solution p->0. In approach that leads to p=1/2 we neglected infinite number of chords that has distance smaller than critical. Because for
the same grid, we can create more (shorter) chords that couldn't be covered with that grid. Or, in other words, in that approach one set of possible chords are normally on one radius (its midpoints),
so all radii consist all possible chords. Surface of the cirle can't be covered with infinite radii! so we neglected shorter chords and increase real amount of probability. Or let's go back on your
point with distance d>>R (for simplicity I took that point, not d=2R). If you change angle for the same small amount when the line from our point is near the center of the circle and when the line is
almost tangent, it is obviously that the curve line that we pass over in second case is longer and on that part we could create much more chords than in the first (between your two chords - part of
lines). Therefore our chords are not equally probable. Theory says, we can neglect only finite set in one infinite set. We could neglect infinite subset if his rate to appropriate infinite set is
zero, but we haven't that here.
Lukomor Ua 22.12.2007. 2:07:21
Note #1.========================= Bozur (Serbia 12/20/2007 8:55:07 AM) Let's suppose your point is not on distance of d=2*R but d>>2R it leads to solution p=1/2. = = = = = = = = = Lukomor: Let's
suppose the point is not on distance of d=R but d=R+delta_R, or d=R-delta_R. (delta_R<
Lukomor UA 22.12.2007. 2:11:54
Sorry, I repeat... Note #1.========================= Bozur (Serbia 12/20/2007 8:55:07 AM) Let's suppose your point is not on distance of d=2*R but d>>2R it leads to solution p=1/2. = = = = = = = = =
Lukomor: Let's suppose the point is not on distance of d=R but d=R+delta_R, or d=R-delta_R. (delta_R<
Lukomor UA 22.12.2007. 2:13:53
Well, (delta_R
Lukomor UA 22.12.2007. 2:16:57
Once more...:) Note #1.========================= Bozur (Serbia 12/20/2007 8:55:07 AM) Let's suppose your point is not on distance of d=2*R but d>>2R it leads to solution p=1/2. = = = = = = = = =
Lukomor: Let's suppose the point is not on distance of d=R but d=R+delta_R, or d=R-delta_R (delta_R less then R). In this case the probability will differ from P=1/3. And we must examine this case
(we should not neglect it).
Lukomor UA 22.12.2007. 4:57:21
Part III: ****************************************It is first necessary to give some determinations. ***Determination №1. A random straight line on the plane - a straight line which passes: 1). With
the equal probability through any point of plane; and 2). With the equal probability in any of directions. ***Determination №2. Random circle on the plane - the circle, whose center with the equal
probability is located at any point of plane. Now let's conduct the classification of the problems, which appear with the examination of the mutual arrangement of straight line and circle.
****************************************Problems of the first level. ****************************************Problem 1*) What is probability that the random straight line will cross the fixed circle
of a radius R? ****************************************Problem 1#) What is the probability that the random circle of a radius R will cross the fixed straight line?
**************************************** Problems of the second level. ****************************************Problem 1.1*) What is the probability that the random straight line will cross the
circle of a radius of r=R/2, when it did cross the fixed circle of a radius R? ****************************************Problem 1.1#) What is the probability that the random circle of a radius r=R/2
will cross the fixed straight line, when it's concentric circle of a radius R crossed this straight line. **************************************** Problems of the third level
**************************************** Problem 1.1.1*) What is the probability that the random straight line will cross the circle of a radius r=R/2 under the conditions that: 1). This straight
line crossed the fixed circle of a radius R; 2). this straight line was carried out through the point, which lies infinitely far from the center of circle.
****************************************Problem 1.1.1#) NO. ****************************************Problem 1.1.2*) What is the probability that the random straight line will cross the circle of a
radius of r=R/2 under the conditions that: 1). This straight line crossed the fixed circle of a radius R; 2). This straight line was carried out compulsorily through the point, which lies at a
distance d less than R from the center of circle; 3). This straight line is carried out compulsory to direction perpendicular from the selected point to center circles.
**************************************** Problem 1.1.2#) NO. *****************************************Problem 1.1.3*) What is the probability that the random straight line will cross the circle of a
radius of r=R/2 under the conditions that: 1). This straight line crossed the fixed circle of a radius R; 2). This straight line was carried out compulsorily through the point, which lies at a
distance d=R from the center of circle; **************************************** Problem 1.1.3#) NO
Lukomor Ukraine 24.12.2007. 7:29:28
Part IV================================= Let us examine problems set in Part III in more detail. Problems of the first level [1*) and 1#)]. They have a result P=0, since an infinitely small quantity
of random straight lines will cross the circle of final diameter on the infinite plane. The problem of second level 1.1*) has a result P=1/2, although this is unevident. However, obviously that a
result P=1/2 has the task of second level 1.1#). These tasks maximally correspond to that posing of the problem, which we find in Bertrand’s problem. The tasks of the third level have additional
conditions; therefore the probability, which they determine will be conditional. These additional conditions, which are absent in the original problem of Bertrand, change the result of solution. Only
in task 1.1.1*) the influence of additional conditions does not change the result.
Bozur Serbia 25.12.2007. 4:52:11
Note #1 ===== If delta_R < R then p=1/3+delta_p and as delta_R increases p becomes p=1/2. I emphasize case when delta_R is large because mistake is then more obvious. If we are aware of that we are
not interested in any other delta_R (we are not neglecting it). ===== (soon I will post answer for part III and IV)
Bozur Serbia 25.12.2007. 9:05:25
Your preassumption on the beginning ..."2). With the equal probability in any of directions." ...caused p=1/2. So, there is no difference if we are tossing lines on the circle or circles on the line,
the result must be p=1/2 because of the first statement that lines will fall uniformly in regards to angle. Bertrand's paradox offer several approaches with different uniform distribution. One with
uniform in regard to angle which will lead to p=1/2, second is uniform distribution in regard to surface of the circle which leads to p=1/4 and third in regard to distance (and angle from choosen
point) which will result in p=1/3. Experiments of each of these approaches could be performed in more than one way and we can discuss about them like here, I don't see why is important only
discussion about p=1/2 i.e. why we have to reject immediately other approaches. ----- Especially when it is easy to verify that p=1/2 has weaknesses, as I explained before.
...................................... Is it difficult to find answer for one similar task where one small part delta_l of the circumference is missing?.
Lukomor UA 26.12.2007. 4:31:18
B:If delta_R < R then p=1/3+delta_p and as delta_R increases p becomes p=1/2. L:And what abaut d=R-delta_R. Why should we neglect that chords?
Lukomor UA 26.12.2007. 4:37:03
B: Experiments of each of these approaches could be performed in more than one way... L:But there is the only experiment if we take a fixed straight line + random circles. And the only result of that
experiment is P=1/2, isn't it?
Bozur Serbia 28.12.2007. 7:21:20
If we choose one point inside the circle, we can't create all chords using that point. For example, if we choose delta_R=R/2 (then d=R-R/2=R/2) it is impossible to create chord whose lenght is d=R/
10, R/20, etc. So, it is meaningless. Similary situation, but less obvious is for d=R-delta_R.
Bozur Serbia 28.12.2007. 8:26:53
Yes, if you take fixed line and random circles you will get p=1/2; but question is "why should we use that experiment for Bertrand's problem". Someone can say we have closed curve line
(circumference)and we are picking points from that line that represents ends of chords. Probability for first point is 1 and we do not have to do that, and for second that fulfill Bertrand's task is
p=1/3. Or we can play darts and then if choosen point is middle of the chord, probability is p=1/4. But your experiment and experiment with darts have nonlinear connection with Bertrand's task and
they are causing mistakes. (for p=1/2: sets of parallel lines can't represent all chords uniformly, and for p=1/4 one of the problem is center of the circle)
Lukomor Ukraine 28.12.2007. 11:55:26
Bozur: "If we choose one point inside the circle, we can't create all chords using that point." ========= ============== ================ ================ ============== =========== L:But we should
choose: 1).ALL the points of the Plane one by one. 2).ALL the angles one by one over each of the point.
LUkomor Ua 29.12.2007. 6:54:23
Happy New Year, Bozur!
Bozur Serbia 31.12.2007. 4:21:11
Yes, we can split all points on the plane in three groups. I: inside the circle, II: outside and III: on the circumference. ========= I: If we choose one point inside the circle we can't create some
chords (for the center there are only radii and other chord's that consist that center point can't be made), II: if the point is outside then chords are not equally probable. So, that approach are
useless (or ultra difficult for counting), and there is no need to move our basic set to points on the plane surface? Yes, chords and circumference are in the same plane but involving that plane
doesn't give any benefit. If we have line y=x2 with lenght 4cm and task is to determine probability of choosing point that is farther then 3cm from one end line. We can observe problem in XY plane
and look only on x or y axis (without appropriate transformation)and get obviously wrong result (or straight distance from (0,0) for example). So, p=1/3 has to be only correct solution for Bertrand's
Bozur Serbia 31.12.2007. 6:04:57
Lukomor, wishing you happy New year and a merry Christmas! Bozur
Esteban Treviño Mexico 26.4.2008. 15:21:22
Good to find someone else recongize the blunder... and see that the congnitive illusion surrounding 'the paradox' ... Note that the probability of 1/4 results from taking the 1/4 area vs 1 area when
all we logically know is that the random chord resides either in the 1/4 area or the 3/4 area... thus the probability is (1/4)/(3/4) = 1/3. I could go on to show how the other examples I explored
also involve some inference error that when corrected also leads to 1/3... The implications are prodound precicely because this example is often used as a model for explanation when we can't
determine something, (and get different aparent correct answers) specially in theory of probability. ... the irnoy is that there is one precise correct answer... Email at etrev7@c-lot.com
Bozur Serbia 9.5.2008. 9:43:28
Esteban, I think that preassumptions in some approaches are wrong, and I doubt that they can be corrected in your way. Because geometrical probability states on p=(part of the area)/(total area).
Question is why to use nonlinear transformation between number of all possible chords and points on the circle surface that represent their mid-points. There is no valid explanation for that, because
center of the circle represents infinite number of chords as I explaind above.
Esteban Treviño Mexico 17.5.2008. 0:02:10
Indeed preassumptions in some approaches lead to the wrong answer... never the less they can be corrected...(I hold) Assuming that one will get the right answer by using the geometrical probability
of p=(part of the area)/(total area) is questionable... Why not assume that the probability p=(an area of larger chords)/(an area of lesser chords)... the mid-point representation for all possible
chords leads to P= (area of larger chords- a cicle)/(area of lesser chords- a donut) that is (1/4)/(3/4)...==> 1/3 ... The assumption here is that the random mid-point chord choosen falls either in
one area or the other area with equal chance because thats the only logical sustainable-congruent position one can hold. A 'valid' explanation for ignoring the center mid point given to me was that
calculating the probability of that particular point is zero... and thus can be ignored (and with the same logic so is the probability for any other point)... My approache assumes nothing about
linear/nonlinear transformation it simply considers the larger chords compared to the smaller ones ... and in both cases the linear and the nonlinear representation I get the same answer... leading
me to hold that its the answer...
Bozur Serbia 19.5.2008. 5:16:09
Theory of probability is based on following formula, p=(set of elements which probability we want to determine)/(set of all elements).
So, if we say that each chord is determined with its mid-point (and they are spread over all surface of the circle) and chords that have lenght longer then critical have mid-points only inside circle
with two times smaller diametar, probability becomes p=(R2/4)/(R2)=1/4.
Yes ... probability just for mid-point is zero, but that point represent infinite!!! number of chords, so that model where we do nut substitute one chord with one point (or finite number of chords)
can't be discussed as option for resolving our problem (and any other problem in probability without using appropriate jacobians)!
What is probability of getting numbers 4,5,6 using fair cube (dice)? With your model of probability (p=(one set of elements)/(second set of elements)) becomes p=1!. p=3/3=1 and that doesn't have any
Ok, let's rotate our circle for 90 degrees around y axis. Now, after one additional nonlinear tranformation we can see one straight line and probability is p=(2xR/2)/(2R)=1/2 !? I doubt that you will
say p=1/1=1. How would you correct it now?
Fermats Brother UK 24.1.2009. 15:56:53
The correct answer is p = 0.5. All other values can be proved to be invalid !
Bozur Serbia 5.2.2009. 4:18:36
Could you explain why?
Esteban Treviño Mexico 2.9.2009. 12:21:41
Been a while! The probability of getting one set of 3 vs one from the other set of 3 is equal thus p = 0.5. If I am conceiving your other example correctly, based on the 1/2 answer, I still see that
when the cord equals the diameter there are multiple possible cords and thus the model used distorts the answer. This can be corrected by instead of using the bisector and the length of the cord
using the perimeter and the length the cord... and ding this give the 1/3 probability.
Lukomor Ukraine 21.12.2009. 6:56:26
To Bozurh See this article: http://www.decisionsciences.org/Proceedings/DSI2008/docs/303-8101.pdf
Bozur Serbia 21.12.2009. 7:09:23
Please, refrase the conclusion, I am not sure what is your point. regards
KL Singaproe 16.5.2011. 1:17:44
It all depends on how the chords are randomly drawn.
Bozur Serbia 2.6.2011. 7:51:55
No it doesn't depend on that. In all examples chord is observed as line that connect two points on the curve. But afterwards (in two problematic solutions) they involved nonlinear transformation
trying to present all chords with its middle point. If it can be done with Jacobian those two examples would be acceptible.
KwajKid USA 19.10.2011. 11:46:10
The correct answer is 1/2. I could not follow your dialogues, but consider this argument: Note for case II, we assume a uniform probability density to the center of the chord over the interior area
of the circle. In case III, we assume a uniform probability density to the angles of intersections of the chord on the circumference. In case I, we assume a uniform probability density to the to the
linear distances between the centers of chord and circle. It is clear that in II and III, we do not have translational invariance. But in I, we do have translational invariance. You have to consider
the distribution of the chords in all three cases. If translating the circle a little bit changes the distribution of the randomly selected chords, then we are dealing with a case that does not
assume and require translational invariance. Consider raining down straws without skill from a long way away over a very large area, and consider our circle to be a very small area in the midst of
this rainstorm of straws. These chords are going to be uniformly distributed. That is, the distances between the center of the chords and the circle will be uniformly distributed in such a rainstorm.
In such a storm there will not be a uniform distribution of the centers of chord over the interior area of the circle (because a chord center uniquely determines the chord, and a uniform distribution
of chord centers yields a distribution of chords that is not uniform - and therefore not translationally invariant). Likewise, in such a storm we also will not observe a uniform density to the angles
of intersections of the chord on the circumference. We will observe a uniform distribution to the linear distances beween chord centers and the circle.
roro eire 13.3.2012. 14:06:10
This page and conversation is a bit strange. It all depends on the word "random". A random variable (or chord in this instance) is not properly understood until you know the underlying distribution
(i.e. random uniform, random normal, etc). The three approaches actually use chords that are drawn from three different distributions. These arise from the way that the chord is chosen. The approach
that yields the result of 1/2 uniformly covers the area of the circle. This would be many peoples intuition of "random". The other approaches create chords that, while valid, do not cover the area of
the circle uniformly when many chords are drawn. KwajKid is pretty much correct.
Bozur Serbia 6.4.2012. 3:02:18
for KwajKid: "Note for case II, we assume a uniform probability density to the center of the chord over the interior area of the circle" ...why we do that? " In case I, we assume a uniform
probability density to the to the linear distances between the centers of chord and circle."...again, why???
Bozur Serbia 6.4.2012. 3:23:16
answer for roro: I agree with your state that it all depends od the word "random", but if question is not defined closely we should stick to the question and that is the observation of chord which is
straight line that connect two points on the curve. Anything in addition shouldn't have nonlinear transformation to the basic set. Otherwise we have paradox in every single problem. For example what
is the solution of the following problem: "what is the probability to choose number 3 among 1,2,3,4,5". First, is it 3 or not yields to p=1/2. Second, it is odd number or not, third p=1/5, and so
on... because we didn't well explained how we choose numbers?! So, there is no problems that can be defined so thoroughly to avoid mentioned free approaches.
Fardistantobserver UK 25.5.2012. 7:57:58
Bozur, I favour your conclusion but (maybe I should say "and") I think you underplay one of your arguments in case 2. The fact that an infinite number of chords can be drawn through the centre
destroys the bijection in cases 1 and 2. In case 1 infinitely many qualifying radii may be selected if the mid-point of the "random chord" coincides with the centre of the circle.
Bozur Serbia 4.6.2012. 8:08:05
Yes, center point makes explanations in examples 1 and 2 unsustainable.
dingdong EARTH 1.7.2012. 9:28:18
Could I get compressed conclusion of errors of Method 1 and 2 ? I can't get the point exactly, and can't convince whether I understood correctly what you said.
dingdong EARTH 1.7.2012. 9:28:59
also with the reason why Method 3 is correct.
Bozur Serbia 13.7.2012. 6:12:06
Answer for dingdong! --- In methods 1 and 2 we made some presumptions that involve hidden nonlinear transformation and method 3 doesn't have any. In method 1 radii can't uniformly cover surface of
the circle as it is supposed; in method 3 center of the circle defines infinite number of chords, not only one as it is supposed, so the choosen method doesn't fulfil the task as well. From method 2
we can conclude that real probability is higher then 1/4 and from method 1 that it is lower then 1/2.
Yannis Mariolis Greece 8.9.2012. 23:50:19
Dear Bozur, sorry for not reading all the posts. However, I read your imaginary dialogue and I believe you 've got it right, 1/3 is the correct answer and your reasoning seems also correct. However,
you should get some help from a native english speaker, because it is really difficult to understand you. Perhaps this is the reason you cannot get your work published. Apart from the language you
can also make your points more straightforward. This is how I tried to resolve this paradox: The important assumption that is not explicitly stated is that when we say that we randomly peak a chord,
we mean that we UNIFORMLY peak a chord of a given circle among ALL chords of that circle. This is the case only for the 1/3 setup. In the 1/4 setup although we consider all chords, peaking is not
uniform since all the diameters share the same middle point, whereas in the 1/2 setup although peaking is uniform we do not consider all chords but only a subset of chords that are parallel to one
another. This is the main idea. However, the above points need some clarification, since the symmetries of the circle can create some confusion on what we can consider as the set of ALL chords. Thus,
I have formulated a rigorous proof that accounts for these symmetries as well...
Bozur Serbia 21.9.2012. 4:23:20
Dear Yannis, thank you for your post! I will make some improvements regarding your proposals.
neopolitan Netopia 12.11.2012. 18:15:09
Hi, I'm partly with you in so much as there really isn't a paradox, but I think that you have arrived at the wrong answer. It's as the standard texts, Fermats Brother, Kwajkid, roro, Lukomur and
(maybe) Jo have said - p=0.5 I have worked through it a slightly different way, with a somewhat different intent, and will soon be posting the argument here - neophilosophical.blogspot.com (look for
the tag "paradox").
neopolitan Netopia 14.11.2012. 6:27:41
It's now in place, ready for you to take a look at :)
Bozur Serbia 1.1.2013. 14:00:25
Hi neopolitan, At first sorry for my late answer! I have read all your three posts at mentioned site and as I see, your method is some kind of generalisation of classic approach which leads to result
p=1/2. But your post is interesting, because at one moment you expose several ways how to get two points on the circle! By definition, chord is line that connect two points on the curved line, and
the question is why we have to create specific method for choosing two points. If we do so, do we make some influence on bijection? Or, in other words, do we change basic set of all chords? There is
no explanation of that so I can't see specific value of your method. One more thing! We will have random distribution if we know what is our basic set of chords. If not, we are blind! If I missed
something, please put it down.
neopolitan Netopia 25.1.2013. 3:38:53
Did you look at the Farewell to the Betrand Paradox which showed the flow chart? That should make my contention explicit.
Bozur Serbia 25.2.2013. 9:05:28
Yes, I looked at your flow chart. It is interesting point of view but I think that problem appear in blocks that consist part "draw line perpendicular to..."! Why, how that .. do we have our primar
set of chords defined till that moment or not? If we know what is our set of chords do we make some transformations with that block that are not allowed? So without that I think that making any
conlusion about Bertrands paradox is not valuable. If we can change basic set oh chords rough example could be throwing a dice. Probability for getting 6 is p=1/2 -- it will be 6 or not!
Kent Jolley USA 16.3.2015. 20:01:41
why is not the proper way to select a random chord to choose two points randomly within the circle and have that define the line
Bozur Serbia 19.3.2015. 9:29:25
Hi Kent! Why not outside the circle? Why not inside the circle but from a surface that is not flat? I suppose your surface inside circle is flat, is it correct? If so, why you think the proper way is
from flat surface? By definition chord is line that connects two points on curve line. So you have to prove that your transformation on any other set of chords is linear. If you are free to choose
any set probability theory does not exist any more! Lets make a vivid example. What is probability that fair dice will show 6? 1/2 because it will be 6 or it wont; it is obviously absurd. Regards
Dave UK 27.1.2019. 7:17:28
I have only just seen this post and so may be too late with this analysis: For me the original premises for examples #1 and #2 are incorrect. A chord is not UNIQUELY defined by its midpoint - as
other postings allude to. The centre of the circle is a single point but is the midpoint of an an infinite number of chords. However in #3 the chord IS uniquely defined by two randomly chosen points
on the perimeter. So all possible chords are counted. The value p=1/3 is thus correct if the question is phrased as follows. “two points are chosen at random on the perimeter of a circle. What is p
(chord length > sqrt(3)xradius)?” The other examples are answers to different questions. eg for #1 “a random chord midpoint is chosen along a radius of a circle, What is p(perpendicular chord length
> sqrt(3)xradius)?” For #2 the probability is p(#1)^2 (1/2 x 1/2) as this is a question based on area (radius squared). A common example of Bartram’s paradox is based on a similar idea: p(choosing a
number < 1 on number line 0 to 2) is 1/2. But p(choosing a number <1^2 on the number line 0 to 2^2) is 1/4 ( 1/2^2)
Bozur Serbia 3.2.2019. 18:18:28
I would go along with you that examples #1 and #2 have wrong premises! But if we keep the question intact, only example #3 can give correct answer, because any other approach will involve
transformation that is not 1 to 1 (I mean that one chord is represented by something that has one instance and vice versa)... thanks for posting! | {"url":"http://bertrands-paradox.com/","timestamp":"2024-11-09T17:05:13Z","content_type":"text/html","content_length":"72119","record_id":"<urn:uuid:5750b56e-8f84-4f20-b076-da7ddacba35c>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00388.warc.gz"} |
A hierarchy of plate models derived from nonlinear elasticity by gamma-convergence
We derive a hierarchy of plate models from three-dimensional nonlinear elasticity by Γ-convergence. What distinguishes the different limit models is the scaling of the elastic energy per unit volume
∼ h ^β , where h is the thickness of the plate. This is in turn related to the strength of the applied force ∼ h ^α . Membrane theory, derived earlier by Le Dret and Raoult, corresponds to α=β=0,
nonlinear bending theory to α=β=2, von Kármán theory to α=3, β=4 and linearized vK theory to α>3. Intermediate values of α lead to certain theories with constraints. A key ingredient in the proof is
a generalization to higher derivatives of our rigidity result [29] which states that for maps v:(0,1)^3→ ^3, the L ^2 distance of ∇. v from a single rotation is bounded by a multiple of the L ^2
distance from the set SO(3) of all rotations.
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Y>2/5x-3 graph the inequality
Author Message
ROTD Posted: Friday 18th of Sep 13:12
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LST Stable Swap | Bifrost Docs
Bifrost Stable Pool is designed to facilitate related exchanges between vToken and Token that are expected to have similar values, such as vDOT-DOT, vETH-ETH, vKSM-KSM, etc. The Bifrost Stable Pool
uses Stable Math (based on StableSwap, widely promoted by Curve), which allows for large-scale exchanges to take place before encountering significant price impact, greatly improving the capital
efficiency of vToken-Token exchanges.
Why Bifrost Stable Swap is unique?
As vToken is a yield-bearing token, its value will gradually increase over time compared to the corresponding Token value. Therefore, the liquidity supply anchor of vToken and Token in AMM should
change with the vToken exchange rate to avoid loss for liquidity providers. The following is a comparison between pegged stable swap and correllated stable swap.
Tokens that swap near 1:1, such as two stablecoins of the same currency
Tokens that swap near 1: R with some slowly changing exchange rate R, like vToken (eg: vDOT, DOT)
Liquidity Efficiency Statement
In Stableswap, the size of function A determines the range of stable exchange rates supported in the Stable Curve. The larger the A value, the more liquidity is available to support constant prices.
Conversely, the smaller the A value, the less liquidity is available to support constant prices. For mathematical details about the A value, please refer to Invariant.
To facilitate user understanding, we express the value of A in Stable Swap as an improvement in liquidity efficiency compared to Uniswap V2.
As shown in the above figure, 100 times liquidity efficiency means:
"Exchanging 100 BNC in the stable pool with the same liquidity is equivalent to exchanging 1 BNC in the Uniswap V2 pool."
Let's take an example. Suppose a liquidity pool has the following liquidity (assuming the exchange rate between vDOT and DOT is 1:1):
When a user exchanges 100 vDOT for DOT, the following results will be presented:
Uniswap V2
Input 100 vDOT
Output 90.8539 DOT
Swapped Price 0.90854
Slippage 9.146%
Bifrost Stable Pool (with A coefficient 5,000)
Input 100 vDOT
Output 99.9958 DOT
Swapped Price 0.99996
Slippage 0.004%
Input 100 vDOT
Output 90.8539 DOT β 99.9958 DOT
Swapped Price 0.90854 β 0.99996
Slippage 9.146% β 0.004%
Since the Stable Math equation is quite complex, determining the invariant, D is typically done iteratively. For an example of how to do this, please refer to: https://miguelmota.com/blog/
This invariant has the following properties:
When the token prices are closed to equilibrium(1 to 1), itβ s performing close to a constant sum curve;
When the token prices are shifted away from the equilibrium, itβ s performing close to a constant product curve.
The further the token prices are shifted away, the more slippage the invariant produces. This ensures that the pool can always provide liquidity even at extreme prices.
How to Swap
Basic Swap
Enter bifrost.app and click "Swap" on the left main navigation menu to enter the swap page
Switch the token to the desired token through the token dropdown menu on the token name
Enter the number of tokens that need to be swapped, and verify the output of the transaction (price impact).
Click the "Swap" button to confirm the transaction and complete the signature | {"url":"https://docs.bifrost.io/for-builders/lst-stable-swap","timestamp":"2024-11-09T00:05:04Z","content_type":"text/html","content_length":"510618","record_id":"<urn:uuid:21a1865c-ecf9-42d2-abcd-68d1bec15b39>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00611.warc.gz"} |
The Formula for Calculating the F-Score in Machine Learning
The F-Score, also known as the F1-Score, is a crucial metric in machine learning, particularly for evaluating classification models. It combines precision and recall into a single metric, providing a
balanced measure of a model's performance. This article explores the formula for calculating the F-Score, its importance, and its applications in various machine learning contexts.
Understanding Precision and Recall
Defining Precision
Precision is a metric that measures the accuracy of positive predictions made by a model. It is defined as the ratio of true positive predictions to the total number of positive predictions (both
true positives and false positives). In other words, precision answers the question: "Out of all the positive predictions made, how many were actually correct?"
Precision is particularly important in scenarios where the cost of false positives is high. For instance, in spam detection, a high precision ensures that only actual spam emails are filtered out,
minimizing the risk of important emails being incorrectly marked as spam.
The formula for precision is:
Supervised Machine Learning Types: Exploring the Different Approaches
[ \text{Precision} = \frac{TP}{TP + FP} ]
where ( TP ) is the number of true positives, and ( FP ) is the number of false positives.
Defining Recall
Recall, also known as sensitivity or true positive rate, measures the ability of a model to identify all relevant instances within a dataset. It is defined as the ratio of true positive predictions
to the total number of actual positive instances (both true positives and false negatives). Recall answers the question: "Out of all the actual positives, how many were correctly predicted?"
Recall is crucial in situations where missing positive instances can have serious consequences. For example, in medical diagnostics, a high recall ensures that most patients with a disease are
correctly identified, reducing the risk of untreated conditions.
Are Neural Networks a Type of Machine Learning?
The formula for recall is:
[ \text{Recall} = \frac{TP}{TP + FN} ]
where ( TP ) is the number of true positives, and ( FN ) is the number of false negatives.
Balancing Precision and Recall
While precision and recall are important metrics individually, there is often a trade-off between the two. Improving precision can lead to a decrease in recall, and vice versa. This trade-off
necessitates a metric that balances both precision and recall, providing a comprehensive measure of a model's performance.
Pros and Cons of Various Machine Learning Models: A Comparison
The F-Score is designed to achieve this balance. It is the harmonic mean of precision and recall, emphasizing the importance of both metrics equally. By considering both precision and recall, the
F-Score provides a more holistic evaluation of a model's accuracy, especially in imbalanced datasets.
The Formula for Calculating the F-Score
Defining the F-Score
The F-Score, or F1-Score, is a metric that combines precision and recall into a single value. It is the harmonic mean of precision and recall, and it ranges from 0 to 1, where 1 indicates perfect
precision and recall, and 0 indicates the worst performance. The harmonic mean is used instead of the arithmetic mean because it penalizes extreme values, ensuring that both precision and recall are
The formula for the F-Score is:
[ F1 = 2 \times \frac{\text{Precision} \times \text{Recall}}{\text{Precision} + \text{Recall}} ]
Exploring the Latest Breakthroughs in Modern Machine Learning Models
This formula ensures that the F-Score is high only when both precision and recall are high, making it a useful metric for evaluating models in imbalanced classification problems.
Practical Example: Calculating the F-Score
Let's consider a practical example to illustrate how the F-Score is calculated. Suppose we have a binary classification problem with the following confusion matrix:
Actual\Predicted Positive Negative
Positive 50 10
Negative 5 100
From this confusion matrix, we can calculate the precision and recall:
[ \text{Precision} = \frac{TP}{TP + FP} = \frac{50}{50 + 5} = 0.91 ]
Is NLP: A Form of Machine Learning or AI?
[ \text{Recall} = \frac{TP}{TP + FN} = \frac{50}{50 + 10} = 0.83 ]
Using these values, we can calculate the F-Score:
[ F1 = 2 \times \frac{\text{Precision} \times \text{Recall}}{\text{Precision} + \text{Recall}} = 2 \times \frac{0.91 \times 0.83}{0.91 + 0.83} = 0.87 ]
This example demonstrates how the F-Score provides a single metric that balances precision and recall, giving a more comprehensive evaluation of the model's performance.
Decoding Machine Learning Models: Deterministic or Probabilistic?
Code Example: Calculating the F-Score in Python
Here is an example of calculating the F-Score in Python using scikit-learn:
from sklearn.metrics import precision_score, recall_score, f1_score
# True labels
y_true = [1, 1, 0, 1, 0, 1, 0, 0, 0, 0]
# Predicted labels
y_pred = [1, 0, 0, 1, 0, 1, 0, 1, 0, 0]
# Calculate precision
precision = precision_score(y_true, y_pred)
print(f'Precision: {precision}')
# Calculate recall
recall = recall_score(y_true, y_pred)
print(f'Recall: {recall}')
# Calculate F-Score
f1 = f1_score(y_true, y_pred)
print(f'F1-Score: {f1}')
This code demonstrates how to calculate precision, recall, and the F-Score using scikit-learn, providing a practical example of how these metrics are used in machine learning.
Importance of the F-Score in Machine Learning
Evaluating Model Performance
The F-Score is an essential metric for evaluating the performance of machine learning models, particularly in classification tasks. Unlike accuracy, which can be misleading in imbalanced datasets,
the F-Score provides a balanced measure that considers both precision and recall.
In real-world applications, accuracy alone may not be sufficient to evaluate a model's effectiveness. For instance, in a dataset with 99% negative instances and 1% positive instances, a model that
predicts all instances as negative would have high accuracy but poor performance in identifying positive instances. The F-Score addresses this issue by balancing precision and recall, providing a
more accurate assessment of the model's performance.
Comparing Different Models
The F-Score is also valuable for comparing different machine learning models. By using a single metric that considers both precision and recall, data scientists can objectively compare the
performance of various models and select the best one for their specific application.
For example, when developing a spam detection system, multiple models such as logistic regression, decision trees, and neural networks might be evaluated. By comparing their F-Scores, the model that
offers the best balance between precision and recall can be chosen, ensuring optimal performance in identifying spam emails while minimizing false positives.
Addressing Imbalanced Datasets
Imbalanced datasets, where one class significantly outnumbers the other, are common in many machine learning applications. The F-Score is particularly useful in these scenarios because it provides a
balanced evaluation of model performance, addressing the limitations of accuracy.
In fields such as fraud detection, medical diagnostics, and rare event prediction, imbalanced datasets are the norm. By focusing on both precision and recall, the F-Score ensures that models are
effective in identifying minority class instances, even when they are significantly outnumbered by majority class instances.
Applications of the F-Score in Different Domains
Healthcare and Medical Diagnostics
In healthcare, the F-Score is crucial for evaluating the performance of models used in medical diagnostics. Accurate diagnosis of diseases, early detection of conditions, and effective treatment
planning rely on models that balance precision and recall.
For example, in cancer detection, a high recall ensures that most cancer cases are identified, reducing the risk of untreated conditions. At the same time, high precision ensures that healthy
patients are not incorrectly diagnosed with cancer, minimizing unnecessary stress and medical interventions. The F-Score provides a single metric that captures this balance, making it an essential
tool for evaluating diagnostic models.
Finance and Fraud Detection
In the finance industry, the F-Score is used to evaluate models that detect fraudulent transactions, assess credit risk, and predict market trends. The ability to accurately identify fraudulent
activities while minimizing false positives is critical for financial institutions.
A high recall ensures that most fraudulent transactions are detected, protecting customers and institutions from financial loss. High precision, on the other hand, ensures that legitimate
transactions are not incorrectly flagged as fraud, maintaining customer satisfaction and trust. The F-Score provides a balanced measure of these two aspects, making it a valuable metric for
evaluating fraud detection models.
Natural Language Processing
Natural language processing (NLP) applications, such as sentiment analysis, text classification, and named entity recognition, also rely on the F-Score to evaluate model performance. The ability to
accurately classify text and identify relevant entities is crucial for various NLP tasks.
For instance, in sentiment analysis, high recall ensures that most positive and negative sentiments are correctly identified, providing a comprehensive understanding of customer opinions. High
precision ensures that the identified sentiments are accurate, leading to reliable insights. The F-Score balances these two metrics, making it an essential tool for evaluating NLP models.
Advanced Topics in F-Score Calculation
Weighted F-Score
In some scenarios, it may be necessary to assign different weights to precision and recall based on their relative importance. The weighted F-Score allows for this flexibility by introducing a
parameter ( \beta ) that adjusts the balance between precision and recall.
The weighted F-Score formula is:
[ F_{\beta} = (1 + \beta^2) \times \frac{\text{Precision} \times \text{Recall}}{(\beta^2 \times \text{Precision}) + \text{Recall}} ]
Where ( \beta ) is a parameter that determines the weight of recall relative to precision. When ( \beta = 1 ), the formula simplifies to the standard F1-Score. When ( \beta > 1 ), recall is given
more weight, and when ( \beta < 1 ), precision is given more weight.
Here's an example of calculating the weighted F-Score in Python using scikit-learn:
from sklearn.metrics import fbeta_score
# True labels
y_true = [1, 1, 0, 1, 0, 1, 0, 0, 0, 0]
# Predicted labels
y_pred = [1, 0, 0, 1, 0, 1, 0, 1, 0, 0]
# Calculate weighted F-Score with beta=2
fbeta = fbeta_score(y_true, y_pred, beta=2)
print(f'F2-Score: {fbeta}')
This code demonstrates how to calculate the weighted F-Score with a specific ( \beta ) value, allowing for customization based on the relative importance of precision and recall.
Macro and Micro F-Score
In multiclass classification problems, it is often necessary to compute the F-Score for each class and then aggregate the results. Two common methods for aggregation are the macro and micro F-Score.
The macro F-Score is calculated by computing the F-Score for each class independently and then averaging the results. This approach treats all classes equally, regardless of their size.
The micro F-Score, on the other hand, aggregates the contributions of all classes to compute a single F-Score. This approach gives more weight to larger classes.
Here is an example of calculating the macro and micro F-Score in Python using scikit-learn:
from sklearn.metrics import f1_score
# True labels
y_true = [0, 1, 2, 0, 1, 2, 0, 2, 1, 0]
# Predicted labels
y_pred = [0, 2, 1, 0, 0, 2, 1, 2, 1, 0]
# Calculate macro F-Score
macro_f1 = f1_score(y_true, y_pred, average='macro')
print(f'Macro F1-Score: {macro_f1}')
# Calculate micro F-Score
micro_f1 = f1_score(y_true, y_pred, average='micro')
print(f'Micro F1-Score: {micro_f1}')
This code demonstrates how to calculate the macro and micro F-Score, providing insights into the performance of multiclass classification models.
Challenges and Considerations
While the F-Score is a valuable metric for evaluating machine learning models, there are some challenges and considerations to keep in mind. One challenge is that the F-Score does not distinguish
between different types of errors. For instance, it treats false positives and false negatives equally, which may not be appropriate in all scenarios.
Another consideration is the trade-off between precision and recall. Depending on the specific application, one metric may be more important than the other. It is essential to understand the
implications of this trade-off and select the appropriate metric based on the problem at hand.
Finally, it is important to consider the context in which the F-Score is used. In some cases, other metrics such as ROC-AUC, precision-recall curves, or accuracy may be more appropriate. By
understanding the strengths and limitations of the F-Score, practitioners can make informed decisions about how to evaluate their models.
The F-Score is an essential metric for evaluating machine learning models, particularly in classification tasks. By balancing precision and recall, it provides a comprehensive measure of a model's
performance, making it especially valuable in scenarios with imbalanced datasets. Understanding how to calculate the F-Score, its importance, and its applications across different domains is crucial
for data scientists and practitioners aiming to develop effective and reliable machine learning models. By leveraging the F-Score and its variations, such as the weighted, macro, and micro F-Score,
practitioners can gain deeper insights into their models' performance and make more informed decisions based on the specific requirements of their applications.
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Weighing - Computer Engineering - Lecture Slides | Slides Computer Science | Docsity
Some concept of Computer Engineering are Binary Search, Byzantine Generals, Euclid Sequences, Houses and Utilities, Malfunction Diagnosis. Main points of this lecture are: Weighing, Balance, Large
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Math SE report 2015-07
Math SE report 2015-07
My overall SE posting volume was down this month, and not only did I post relatively few interesting items, I've already written a whole article about the most interesting one. So this will be a
short report.
• I already wrote up Building a box from smaller boxes on the blog here. But maybe I have a couple of extra remarks. First, the other guy's proposed solution is awful. It's long and complicated,
which is forgivable if it had answered the question, but it doesn't. And the key point is “blah blah blah therefore code a solver which visits all configurations of the search space”. Well heck,
if this post had just been one sentence that ended with “code a solver which visits all configurations of the search space” I would not have any complaints about that.
As an undergraduate I once gave a talk on this topic. One of my examples was the problem of packing 31 dominoes into a chessboard from which two squares have been deleted. There is a simple
combinatorial argument why this is impossible if the two deleted squares are the same color, say if they are opposite corners: each domino must cover one square of each color. But if you don't
take time to think about the combinatorial argument you could waste a lot of time on computer search learning that there is no solution in that case, and completely miss the deeper understanding
that it brings you. So this has been on my mind for a long time.
• I wrote a few posts this month where I thought I gave good hints. In How to scale an unit vector !!u!! in such way that !!a u\cdot u=1!! where !!a!! is a scalar I think I did a good job
identifying the original author's confusion; he was conflating his original unit vector !!u!! and the scaled, leading him to write !!au\cdot u=1!!. This is sure to lead to confusion. So I led him
to the point of writing !!a(bv)\cdot(bv)=1!! and let him take it from there. The other proposed solution is much more rote and mechanical. (“Divide this by that…”)
In Find numbers !!\overline{abcd}!! so that !!\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}!! the OP got stuck partway through and I specifically addressed the stuckness; other
people solved the problem from the beginning. I think that's the way to go, if the original proposal was never going to work, especially if you stop and say why it was never going to work, but
this time OP's original suggestion was perfectly good and she just didn't know how to get to the next step. By the way, the notation !!\overline{abcd}!! here means the number !!
In Help finding the limit of this series !!\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots!! it would have been really easy to say “use the formula” or to analyze the series de
novo, but I think I almost hit the nail on the head here: it's just like !!1+\frac12 + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots!!, which I bet OP already knows, except a
little different. But I pointed out the wrong difference: I observed that the first sequence is one-fourth the second one (which it is) but it would have been simpler to observe that it's just
the second one without the !!1+\frac12!!. I had to review it just now to give the simpler explanation, but I sure wish I'd thought of it at the time. Nobody else pointed it out either. Best of
all, would have been to mention both methods. If you can notice both of them you can solve the problem without the advance knowledge of the value of !!1+\frac12+\frac14+\ldots!!, because you have
!!4S = 1+\frac12 + S!! and then solve for !!S!!.
In Visualization of Rhombus made of Radii and Chords it seemed that OP just needed to see a diagram (“I really really don't see how two circles can form a rhombus?”), so I drew one.
[Other articles in category /math/se] permanent link | {"url":"https://blog.plover.com/math/se/2015-07.html","timestamp":"2024-11-12T17:08:51Z","content_type":"text/html","content_length":"28338","record_id":"<urn:uuid:00a1ecfc-615d-4400-a0c9-a5445378791d>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00845.warc.gz"} |
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positive and negative, to give the students a feel of the real numbers. 2.2 How Well Have You Learnt About Fractions? The exercises are kept very close to real life examples, and thus, practicing the
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NCERT Solutions are extremely helpful while doing your homework or while preparing for the exam. Detailed Explanation to solve problems and formulas are mentioned in between steps to learn
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11.4 in Hindi Medium to study online or download in PDF file format. 9.4 Positive And Negative Rational Numbers. The key takeaways of these NCERT Solutions for Class 10 Maths ch 1 are: Students will
get answers to all the questions in Maths Class 10 NCERT Solutions Chapter 1 and no question is left out. We have provided database of Class 7 Mathematics question papers with solutions and is
available for free download or read online. The first chapter of geometry in Class 7, lines and angles starts with the fundamental definitions of line and angle. You can go through the Chapter List
for the NCERT Class 7 Maths Solutions by referring to our page. Exercise 5.2 Solution (1) 7x, -3x (ii) 6x, -11x (iii) -9x 2 (iv) 3ab 2, -5ab 2 (v) 1/2 pq, -1/3 pq (vi) 5x 3 y, -2/3 x 3 y (2) Add: (i)
3x, -5x, 7x (ii) 7xy, 2xy, -8xy (iii) -2abc, 3abc, abc (iv) 3mn, -5mn,8mn, … 9.8 Rational Numbers Between Two Rational Number. Multiplicative inverse = 16/79 6. Maths NCERT Solutions The second
chapter of geometry deals with triangles and their properties. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.. For a
better understanding of this chapter, you should … They also deal with the portrayal of fractions and decimals on the number line, and their expansions and subtraction. Students are going to learn
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Equilateral And Isosceles. After discussing integers extensively in the first chapter, this chapter comes back to the numbers, namely rational numbers. 3.8 use of Bar Graphs with a different Purpose
the portrayal of geometry paper... A point not on the right of zero Expressions in simple language so that students download... These Solutions can help the students will get a feel of what triangles
are in general and! Give the students to understand learning experience while using the Solutions justify the same a... To make a few deductions from the accumulated data ( b ) One-fifth of a
Triangle and Its are! Better grades in the Class 7 Maths by referring to our page how! First chapter of geometry on paper, in terms of construction of a line angle! Providing examples of recitation
and discussion, some benefits of discussion in Mathematics best reference for! With a different Purpose NCERT books there are 15 chapters One of Pythagoras. 7 NCERT books there are 15 chapters later
that same night, the seventh chapter deals with areas! I get NCERT Solutions are extremely helpful while doing your homework or while preparing for books! Back to the topics under each chapter and
plan your preparation or while preparing for Class. I score good marks in the Maths in Class 7 Maths teach how to download NCERT Solutions Class! Great learning experience while using the principles
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free updated for new academic session 2020-21 preparation strategy to clear the Class 7 includes... By three lines intersecting the Triangle and Its Hypotenuse are given below online download...
Night, the seventh chapter deals with the portrayal of fractions and decimals, and interest ICSE Class 6 you! Real life examples, where the algebra can actually be used the important shapes in
Mathematics 7. Theorem in this brief, after describing and providing examples of recitation and discussion, some benefits of discussion Mathematics. Result in 1 where can I score good marks in ICSE
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Sharma Solutions for Class 7 are kept very close to real life,. Chapter and plan your preparation be safely assumed to be the best reference for. If multiplied by original number result in 1 real
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of two Sides of a Triangle triangles, the was. 7 Solutions with free PDF download option, which contains chapter wise Solutions concepts thoroughly and prepare as per topics. The basic concepts of
the same Ex 4.3 PDF starts with the portrayal of deals. 17.09.2016 Math Secondary School where can I find NCERT Maths try these Solutions can help the students general... Number result in 1
properties and importance of integers RHS Criterion ) designed! Thus, practicing the same ) One-fifth of a Triangle is a fairly simpler chapter, that only a! Chapter talks about the types brainly
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fairly simpler chapter, that only requires a set procedure to be followed while going through constructions updated. Your homework or while preparing for the Class 7 NCERT books there are chapters!
And providing examples of recitation and discussion, some benefits of discussion in Mathematics learn.. I score good marks in the world ’ s largest social learning network for students, you will
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Numbers. Students to memorize them easily of Multiplication of integers, 1.5 properties of Addition and subtraction students!
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In this lesson, we will talk about what sequences are and how to formally write them. Then we will learn how to write the terms out of the sequences when given the general term. We will also learn
how to write the general term when given a sequence. After learning the notations of sequences, we will take a look at the limits of sequences. Then we will take a look at some definitions and
properties which will help us take the limits of complicating sequences. These theorems include the squeeze theorem, absolute value sequences, and geometric sequences.
1. If a sequence has the limit $L$, then we can say that:
$\lim$[n →$\infty$] $a$[$n$]$=L$
If the limit is finite, then it is convergent. Otherwise, it is divergent.
2. If the limit of the sequences {$a_n$} and {$b_n$} are finite and $c$ is constant, then we can say that
i) $\lim$[n →$\infty$] $(a_n+b_n)=\lim$[n →$\infty$] $a_n+$$\lim$[n →$\infty$] $b_n$.
ii) $\lim$[n →$\infty$] $(a_n-b_n)=\lim$[n →$\infty$] $a_n-$$\lim$[n →$\infty$] $b_n$.
iii) $\lim$[n →$\infty$] $ca_n=c$ $\lim$[n →$\infty$] $a_n$.
iv) $\lim$[n →$\infty$]$(a_nb_n)=$ $\lim$[n →$\infty$]$a_n*$ $\lim$[n →$\infty$] $b_n$.
v) $\lim$[n →$\infty$] $[a_n$$\div$$b_n]$ $=\lim$[n →$\infty$]$a_n$$\div$ $\lim$[n →$\infty$]$b_n$$,$$b_neq0$.
3. If $a_n\leq c_n\leq b_n$ and $\lim$[n →$\infty$] $a_n=$ $\lim$[n →$\infty$] $b_n=L$, then $\lim$[n →$\infty$] $c_n=L$.
4.if $\lim$[n →$\infty$] $|a_n|=0$, then $\lim$[n →$\infty$] $a_n=0$ as well.
5. We say that:
Where the sequence {$x^n$} is convergent for -1< $x \leq$ 1, and divergent if $x$ > 1. | {"url":"https://www.studypug.com/calculus-help/introduction-to-sequences","timestamp":"2024-11-07T06:12:09Z","content_type":"text/html","content_length":"511739","record_id":"<urn:uuid:41fe05aa-eb84-4053-8103-79507bda904c>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00412.warc.gz"} |
2.1 Introduction to Isaac Newton
Isaac Newton is probably one of the smartest people of all time. Aside from discovering the foundations of physics, he was also the first person to describe the force of gravity. He designed the
first practical reflecting telescope and explained how colours work based on the phenomenon of white light splitting into a rainbow after passing through a prism. He has been credited with inventing
ridge-edged coins (to fight counterfeiting) and the cat-flap door (seriously), and was an influential religious philosopher. But my favourite story about Newton is the following.
He was also the main aesthetic inspiration for the ‘hair’ bands of the late 1970s.
Around 1666, Newton locked himself in his room for a while and, basically, invented calculus.[1. If any math historians are currently reading this, please forgive the impreciseness in this
paragraph.] Calculus is a set of concepts and techniques, completely new to the usual addition-subtraction-multiplication kind of math, which allowed people to finally use numbers to describe changes
— like the change of position (velocity) or the change of velocity (acceleration). But despite the enormous importance of this invention, for some reason, Newton didn’t tell anyone about it for years
afterwards. He mentioned some of the basics in an annotation to a footnote somewhere, and actually used calculus in his major physics works, but never published the original paper on calculus itself.
A few years later, a man named Gottfried Wilhelm Leibniz also invented calculus, completely independently of Newton’s work. Newton got fairly upset about this, accusing Leibniz of plagiarizing from,
well, the papers that he had failed to show anybody.
Today, both men are credited with inventing calculus, although Leibniz’s firmer grasp of publicity earned him the small victory of having his notation, rather than Newton’s, live on in mathematics
even today.
But despite the man’s peculiarities, without Newton’s work, physics would not be anywhere near the point it is at today.
Newton described his laws of motion in a 1687 work with the catchy title PhilosophiƦ Naturalis Principia Mathematica, which stayed at the top of the New York Times bestseller list for over three
weeks (a record at the time). This book remains one of the most important scientific works in human history. In it, he famously used his laws of motion in combination with a new theory of gravity to
explain the movement of the stars and planets (more on this in the chapter on gravity). Newton thus brought new mathematical insights to problems that had been baffling humanity for ages and
effectively founded an entire branch of physics, now known as classical mechanics.
“How can we make physics even more intimidating?” “I know, let’s write all the books in a dead language!”
When I first learned the three concepts that constitute Newton’s Laws, they didn’t seem very important to me. In fact, at first glance, all they seem to do is define something called a force in terms
of its effects on matter. But in truth, each of these laws is something marvellous. Each reveals to us a new and startling truth about the basic properties of the world around us. And, as we shall
see, each deserves a chapter unto itself.
In order to fully understand each of the three laws, we’ll need to spend a bit of time explaining what a force is.
Next: 2.2 – Forces
Previous: 1.3 – Falling Objects | {"url":"https://popphysics.com/introduction-to-isaac-newton/","timestamp":"2024-11-14T13:39:41Z","content_type":"text/html","content_length":"42311","record_id":"<urn:uuid:c3174377-e8ab-41c5-8572-10df36364fc7>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00559.warc.gz"} |
3rd Grade Math Worksheets Word Problems
3rd Grade Math Worksheets Word Problems
Math worksheets word problems mixed addition and subtraction word problems word problems. Word problems where students use reasoning and critical thinking skill to solve each problem.
Word Problem Fun 3 Digit Subtraction At The Game Worksheet Education Com Word Problems Subtraction Word Problems Word Problems 3rd Grade
Social studies science and the olympics are just some of the themes that will stimulate third graders as they apply addition subtraction and multiplication to these.
3rd grade math worksheets word problems. This math worksheet gives your child practice finding the missing number to complete addition subtraction and multiplication equations. Are you looking for
engaging 3rd grade math word problems with answers to add to your upcoming lesson plans. Students are required to figured out which operation to apply given the problem context.
A full index of all math worksheets on this site. These third grade math worksheets have word problems on simple addition. In this math worksheet your child will solve word problems using addition of
2 digit numbers.
They help students master basic math and problem solving skills. Quick link for all word problems worksheets click the image to be taken to that word problems worksheet. Mixed 3rd grade word problems
the following worksheets contain a mix of grade 3 addition subtraction multiplication and division word problems.
Back to math worksheets. Mixing math word problems is the ultimate test of understanding mathematical concepts as it forces students to analyze the situation rather than mechanically apply a
solution. The following collection of free 3th grade maths word problems worksheets cover topics including addition subtraction multiplication division and measurement.
The worksheets on this page combine the skills necessary to solve all four types of problems covered previously addition word problems subtraction word problems multiplication word problems and
division word problems and they require students to determine which operation is appropriate for solving the each problem. Math word problems mixed mixed word problems stories for skills working on
subtraction addition fractions and more. This set of worksheets includes a mix of addition and subtraction word problems.
Addition in columns word problems for third grade these grade 3 worksheets have math word problems requiring column form addition to solve. The focus here is on solving real life situations by using
addition rather than the mechanics of addition. Math worksheets full index.
This collection of worksheets will help kids grasp how math applies in real world situations. Mixed addition and subtraction word problems. Addition multiplication and division what s missing in this
These word problems worksheets are a great resource for children in 3rd grade 4th grade and 5th grade. Take the problem out of word problems with these math worksheets for third graders. Word
problems are an essential part of grade 3 common core standards.
Click here for a detailed description of all the word problems worksheets.
3 Free Math Worksheets First Grade 1 Addition Add 3 Single Digit Number 10 3rd Grade Math In 2020 Word Problem Worksheets Subtraction Word Problems Math Word Problems
Image Result For Multiplication Word Problems Grade 3 Word Problem Worksheets Multiplication Word Problems 3rd Grade Math Worksheets
Boost Your 3rd Grader S Math Skills With These Printable Word Problems Math Word Problems Math Words 3rd Grade Math
Word Problems In Fractions Grade 3 Math 3rd Grade Math Worksheets 3rd Grade Math Word Problems
3rd Grade Math Word Problems Best Coloring Pages For Kids Word Problems Math Word Problems Math Words
Third Grade Math Worksheets Word Problem Worksheets Addition Words Addition Word Problems
Free Printable Worksheets For Second Grade Math Word Problems In 2020 Word Problem Worksheets Math Word Problems Word Problems
Time Word Problems Ks1 Worksheets Refrence 3rd Grade Math Worksheets Multiplication Word P Word Problem Worksheets Word Problems Free Printable Math Worksheets
3rd Grade Math Word Problems Best Coloring Pages For Kids Multi Step Word Problems Multiplication Word Problems Word Problems
Grade 3 Maths Worksheets 12 7 Word Problems On Grams And Kilograms 2nd Grade Worksheets Measurement Worksheets 3rd Grade Math Worksheets
Trumus Biz Multiplication Word Problems Word Problems Math Word Problems
Grade 3 Counting Money Worksheets Free Printable Money Word Problems Money Worksheets Word Problem Worksheets
Grade 3 Maths Worksheets 12 8 Word Problems On Multiplication And Division Of Word Problem Worksheets Multiplication Word Problems 3rd Grade Math Worksheets
Unique Multiplication Word Problems For 2nd 3rd Grade Multiplication Word Problems Multiplication Worksheets Word Problems
Boost Your 3rd Grader S Math Skills With These Printable Word Problems Word Problem Worksheets Multiplication Word Problems Subtraction Word Problems
Grade 3 Maths Worksheets 10 5 Word Problems On Money Word Problems 3rd Grade Math Worksheets Math Worksheet
Grade 3 Maths Worksheets 8 5 Time Problems Lets Share Knowledge 3rd Grade Math Worksheets Math Word Problems Word Problem Worksheets
Grade 3 Maths Worksheets Subtraction 4 3 The 4 And 5 Digit Numbers Subtraction Word Problems Subtraction Word Problems Word Problems Subtraction Worksheets
Grade 3 Maths Worksheets 11 9 Word Problems On Meters And Centimeters 2nd Grade Worksheets Measurement Worksheets 3rd Grade Math Worksheets | {"url":"https://askworksheet.com/3rd-grade-math-worksheets-word-problems/","timestamp":"2024-11-06T04:56:39Z","content_type":"text/html","content_length":"135554","record_id":"<urn:uuid:d8dd4e85-28e3-4355-9c4c-d0304297e67e>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00892.warc.gz"} |
Constructing lattice-free gradient polyhedra in dimension two
Lattice-free gradient polyhedra can be used to certify optimality for mixed integer convex minimization models. We consider how to construct these polyhedra for unconstrained models with two integer
variables under the assumption that all level sets are bounded. In this setting, a classic result of Bell, Doignon, and Scarf states the existence of a lattice-free gradient polyhedron with at most
four facets. We present an algorithm for creating a sequence of gradient polyhedra, each of which has at most four facets, that finitely converges to a lattice-free gradient polyhedron. Each update
requires constantly many gradient evaluations. Our updates imitate the gradient descent algorithm, and consequently, it yields a gradient descent type of algorithm for problems with two integer
Dive into the research topics of 'Constructing lattice-free gradient polyhedra in dimension two'. Together they form a unique fingerprint. | {"url":"https://portal.fis.tum.de/en/publications/constructing-lattice-free-gradient-polyhedra-in-dimension-two","timestamp":"2024-11-12T22:47:33Z","content_type":"text/html","content_length":"49764","record_id":"<urn:uuid:de62d914-c1c4-4b3d-bafd-2dcf5767490a>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00523.warc.gz"} |
Classification of period analysis methods
There is a long history and wide variety in period finding algorithms, that can be used to study light curves of variable stars, exoplanets and asteroids. They can be divided into a number of types:
• Algorithms based on discrete Fourier transforms
These methods attempt to represent a set of observations with a series of trigonometric functions (sines and cosines, with different periods, amplitudes and phases). They are one of the oldest forms
of time-series analysis and are also quite flexible. Fourier methods supported by Peranso are : Discrete Fourier Transform (Deeming) DFT, Date Compensated Discrete Fourier Transform (Ferraz-Mello)
DCDFT, CLEANest, FALC (Harris) and Bloomfield.
• Algorithms that model a light curve via a least-squares fit to some set of (orthogonal) basis functions
• Algorithms that minimize some measure of the dispersion of time series data in phase space
This includes methods such as Jurkewich and ANOVA. The minimization can be done in various ways, including:
• New research in period finding algorithms is exploring still other approaches, such as Bayesian methods or Neural networks
• Some algorithms have been specifically developed for exoplanet research
An example is Edge Enhanced Box-fitting Least Squares (EEBLS), which analyses stellar photometric time series in search for periodic transits by exoplanets, looking for signals characterized by a
periodic alternation between two discrete levels, with much less time spent at the lower level. | {"url":"https://www.cbabelgium.com/peranso/UserGuideHTML/Classificationofperiodanalysisme.html","timestamp":"2024-11-07T18:50:51Z","content_type":"text/html","content_length":"16592","record_id":"<urn:uuid:68e67aaa-9434-48e0-8b9f-f4cf72aa6871>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00863.warc.gz"} |
Prediction and Estimation
12.2. Prediction and Estimation#
One way to think about the SD is in terms of errors in prediction. Suppose I am going to generate a value of the random variable \(X\), and I ask you to predict the value I am going to get. What
should you use as your predictor?
A natural choice is \(\mu_X\), the expectation of \(X\). But you could choose any number \(c\). The error that you will make is \(X - c\). About how big is that? For most reasonable choices of \(c\),
the error will sometimes be positive and sometimes negative. To find the rough size of this error, we will avoid cancellation as before, and start by calculating the mean squared error of the
predictor \(c\):
\[ MSE(c) ~ = ~ E[(X-c)^2] \]
Notice that by definition, the variance of \(X\) is the mean squared error of using \(\mu_X\) as the predictor.
\[ MSE(\mu_X) ~ = ~ E[(X-\mu_X)^2] ~ = ~ \sigma_X^2 \]
See More
We will now show that \(\mu_X\) is the least squares constant predictor, that is, it has the smallest mean squared error among all constant predictors. Since we have guessed that \(\mu_X\) is the
best choice, we will organize the algebra around that value.
\[\begin{split} \begin{align*} MSE(c) ~ = ~ E\big{[}(X - c)^2\big{]} &= E\big{[} \big{(} (X - \mu_X) + (\mu_X - c) \big{)}^2 \big{]} \\ &= E\big{[} (X - \mu_X)^2 \big{]} +2(\mu_X - c)E\big{[} (X-\
mu_X) \big{]} + (\mu_X -c)^2 \\ &= \sigma_X^2 + 0 + (\mu_X -c)^2 \\ &\ge \sigma_X^2 \\ &= MSE(\mu_X) \end{align*} \end{split}\]
with equality if and only if \(c = \mu_X\).
12.2.1. The Mean as a Least Squares Predictor#
What we have shown is the predictor \(\mu_X\) has the smallest mean squared error among all choices \(c\). That smallest mean squared error is the variance of \(X\), and hence the smallest root mean
squared error is the SD \(\sigma_X\).
This is why a common approach to prediction is, “My guess is the mean, and I’ll be off by about an SD.”
Quick Check
Your friend has a random dollar amount \(X\) in their wallet. Suppose you know that \(E(X) = 16\) dollars and \(SD(X) = 3\) dollars. In all your answers below, please include units of measurement.
(a) What is the least squares constant predictor of \(X\)?
(b) What is the mean squared error of this predictor?
(c) What is the root mean squared error of this predictor?
(a) \(16\) dollars
(b) \(9\) squared dollars
(c) \(3\) dollars
12.2.2. German Tanks, Revisited#
Recall the German tanks problem in which we have a sample \(X_1, X_2, \ldots , X_n\) drawn at random without replacement from \(1, 2, \ldots , N\) for some fixed \(N\), and we are trying to estimate
We came up with two unbiased estimators of \(N\):
• An estimator based on the sample mean: \(T_1 = 2\bar{X}_n - 1\) where \(\bar{X}_n\) is the sample average \(\frac{1}{n}\sum_{i=1}^n X_i\)
• An estimator based on the sample maximum: \(T_2 = M\cdot\frac{n+1}{n} - 1\) where \(M = \max(X_1, X_2, \ldots, X_n)\).
Here are simulated distributions of \(T_1\) and \(T_2\) in the case \(N = 300\) and \(n = 30\), based on 5000 repetitions.
def simulate_T1_T2(N, n):
"""Returns one pair of simulated values of T_1 and T_2
based on the same simple random sample"""
tanks = np.arange(1, N+1)
sample = np.random.choice(tanks, size=n, replace=False)
t1 = 2*np.mean(sample) - 1
t2 = max(sample)*(n+1)/n - 1
return [t1, t2]
def compare_T1_T2(N, n, repetitions):
"""Returns a table of simulated values of T_1 and T_2,
with the number of rows = repetitions
and each row containing the two estimates based on the same simple random sample"""
tbl = Table(['T_1 = 2*Mean-1', 'T_2 = Augmented Max'])
for i in range(repetitions):
tbl.append(simulate_T1_T2(N, n))
return tbl
N = 300
n = 30
repetitions = 5000
comparison = compare_T1_T2(N, n, 5000)
comparison.hist(bins=np.arange(N/2, 3*N/2))
plt.title('$N =$'+str(N)+', $n =$'+str(n)+' ('+str(repetitions)+' repetitions)');
We know that both estimators are unbiased: \(E(T_1) = N = E(T_2)\). But is clear from the simulation that \(SD(T_1) > SD(T_2)\) and hence \(T_2\) is a better estimator than \(T_1\).
The empirical values of the two means and standard deviations based on this simulation are calculated below.
t1 = comparison.column(0)
np.mean(t1), np.std(t1)
(300.07926666666668, 30.068877736808055)
t2 = comparison.column(1)
np.mean(t2), np.std(t2)
(299.98106666666666, 9.1113762209668376)
These standard deviations are calculated based on empirical data given a specified value of the parameter \(N = 300\) and a specified sample size \(n = 30\). In the next chapter we will develop
properties of the SD that will allow us to obtain algebraic expressions for \(SD(T_1)\) and \(SD(T_2)\) for all \(N\) and \(n\). | {"url":"http://prob140.org/textbook/content/Chapter_12/02_Prediction_and_Estimation.html","timestamp":"2024-11-07T10:20:47Z","content_type":"text/html","content_length":"58086","record_id":"<urn:uuid:e99328c8-c7ca-4c88-9f6d-aea878523dce>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00420.warc.gz"} |
TMAC | Truly Mentor Academic Center
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Daily Session Routine
In 2-hour of the session, students will get help with their school homework, take objective, subjective tests, and learn one new topic on a given subject.
Exam Analysis
We conduct the daily objective tests and weekly subjective tests. Through weekly exams, we will be able to analyse our teaching methods and help us to understand students learning approaches.
Self-Study Hour
Students who have enrolled for TMAC can have their own time for self-study under the guidance of our Mentor. Just inform us prior and arrangements will be done for your self-study session.
VARK Model
TMAC teaching approach is based on the VARK model i.e. Visual, Auditory, Reading/ Writing, Kinesthetic which can benefit any kind of learner.
Beginner Skills
During our session, we focus on the reading and writing skills of our students - initial qualities which are missed out on during regular online school classes.
Instigate Students
In our session, we instigate our students to learn about any topic with a different approach which will help students to improve their academic performance.
Feedback System
We conduct a weekly feedback system. Students will review our classroom teaching and parent review their children academic performance. This will help us to improve our teaching-learning environment.
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Rubik’s Cube Challenge: Did I do it?
The finale of the Rubik’s cube challenge was on Sunday, and yielded some interesting results.
To confirm, I did not look at an actual cube or picture of one between the start of the challenge last Wednesday, and the finale on Sunday. All I did was read the book on the cube solution. Before
I reveal the results let me give you a rundown of what I had to do.
To make it work required memorizing a lot of steps, and learning the notation used by the book.
A simple cube
In the above picture, the top side (T) is yellow, the Front side (F) is Blue, and the Right (R) side is Red. Opposite the Top is the Bottom (B), opposite the Front is the Posterior (P), and opposite
the Right is the Left (L). A (+) denotes a clockwise rotation, a (-) denotes a counterclockwise rotation, and a ‘2’ denotes a half turn.
So a combination T- R+ T2 moves the top side 1/4 turn counterclockwise, then the right side 1/4 turn clockwise, then the top side 1/2 turn in either direction, since a half turn either way results in
the same position.
The solution followed the basic steps:
1. Pick a favourite colour and solve the top edges
2. Solve the top corners
3. Solve the middle edges
4. Solve the Bottom Corners
5. Solve the bottom edges
Each step was then broken down into having the cubes properly positioned first, then properly oriented. The book detailed ways of doing this with groups of specific moves. For example, once a top
corner cube was properly positioned, it could be oriented without changing other top face cubes with the combination R- B2 R+ F+ B2 F-.
As we get further into the process, the combinations become more complicated, since at this point most of the cubes are in their correct position and orientation, and it takes a lot of work to
preserve this. The final move involved memorizing the combination:
L- R+ F+ L+ R – B- L- R+ F- L+ R- B- L- R+ F2 L+ R-
I did this through forming patterns and remembering them. For the above move I noticed the first and last 5 moves repeat, except that instead of F+ in the first instance it was F2 in the second.
Many of the combinations throughout the solution repeated and I was able to remember them by breaking them down into smaller bits.
So, did it work?
It took about 30 minutes, and I made several mistakes near the end that required retracing my steps and fixing cubes that were previously positioned correctly, but I did it.
Completed cube
Here are some of the progress shots:
Top edges done, step 1 completed.
Top third of cube done, positioned and oriented correctly
Top 2 thirds done
Bottom corners done, only two edges left to fix
What to take away from this:
The goal of this little challenge was to see if a skill could be learned and relatively mastered with only theoretical knowledge, and no practice. It was also to determine if I could push myself to
memorize long combinations and an in-depth solution with little time.
In terms of memorization, I have crammed for many an exam in my life, and have become pretty good at memorizing a lot of information in a short time. In order to make it stick over the long run,
revisiting the information every few days is a necessary step.
The fact that I still needed a few tries to get it right means that experience is part of the skill learning process. The conclusion is one I was already aware of:
To master any given skill in the shortest possible time frame, a combination of theoretical study and real world experience is required.
There is a third step to this as well: Expert advice. Knowing the tricks and tips from someone who has mastered the skill can undoubtedly make the process go a lot faster, and personalized feedback
in real time is invaluable in quickly picking up a skill. This is why experts who have put in a lot of time in any given field are so highly regarded and in-demand.
Maybe I need to speak to an expert blogger, to speed up the process of getting good at this stuff.
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HackerRank solution of the Loop | Python coding challenge - Docodehere
HackerRank solution of the Loop | Python coding challenge
The provided code stub reads and integer, , from STDIN. For all non-negative integers , print .
The list of non-negative integers that are less than is . Print the square of each number on a separate line.
The first and only line contains the integer, .
Print lines, one corresponding to each .
if __name__ == '__main__':
n = int(input())
for i in range(n):
1 Comments
• Interesting problem, keep sharing these problems that occur in Google or Amazon interview questions. Looking forward for more questions.
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Four extra-large sandwiches
Your Answer
Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the
fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole
extra-large sandwich?
1. m+4/m(m-4)
2. 2m-4/m(m-4)
3. 4m-4/m(m-4)
4. 4m-8/m(m-4)
5. 4m-12/m(m-4) | {"url":"https://www.beatthegmat.com/four-extra-large-sandwiches-t18199.html?sid=845f2ec31e4ce14dc6bbb6239ee040ec","timestamp":"2024-11-03T20:27:53Z","content_type":"text/html","content_length":"737971","record_id":"<urn:uuid:f71836eb-1f14-4225-b308-31ecc38e3a3a>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00760.warc.gz"} |
Capacitors - Electronics AreaCapacitors - Electronics Area
Capacitor Tutorials
Capacitors are devices made of two metal plates separated by an insulator or dielectric. Charge, Voltage and Capacitance Relationship
Capacitors in series – Capacitors in parallel. How to obtain the equivalent capacitance of series capacitors and parallel capacitors? Formulas and examples
Capacitor and Direct Current. If a not charged capacitor is connected across the terminals of a battery, a transient current flows as the capacitor plates charge up.
What is the Dielectric constant / Relative Permittivity? The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space. It is a dimensionless physical
Capacitor charging process. Capacitor charging process shows the variation of voltage and current in the capacitor over time, when it is connected to a DC voltage source
Capacitor discharge process. In a series RC circuit, a capacitor voltage Vc decreases from the initial voltage on the capacitor to 0 volts in some amount of time.
Capacitor and the Alternating Current. Unlike the behavior of a capacitor in direct current (DC), in the alternating current (AC) the current passes more easily through a capacitor.
The Impedance of a capacitor (Capacitive reactance) is the measure of the opposition to a change of the electrical current in this component.
The Electrolytic Capacitor has been developed to achieve large capacities in small physical dimensions. To achieve this large capacity, a special dielectric is used. | {"url":"https://electronicsarea.com/capacitors/","timestamp":"2024-11-10T00:01:11Z","content_type":"text/html","content_length":"47200","record_id":"<urn:uuid:6b755160-a29e-4a61-9b3f-94aabd7dfc41>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00734.warc.gz"} |
4-Easy introduction to Mohr's Circle of inertia part-1.
Last Updated on March 9, 2024 by Maged kamel
Mohr’s Circle of inertia.
In this post, we will be starting our discussion about Mohr’s circle of inertia. Still, we will use the relation of x’ and y’ for an area about an inclined axis, with the x and y distance for the
same area but about horizontal and vertical axes x, and y. The Inertia Ix is bigger than Iy and Ixy is positive.
Introduction to Mohr’s circle of inertia.
Brief content of the post.
The post items that will be explained are summarized here.
When Ix is greater than Iy, we will begin to learn how to draw Mohr’s circle of inertia using the provided data for two points. As we can see, y’=ycos ฮธ -x sin ฮธ represents the perpendicular
distances from the little area dA to x’ and the axes. In contrast, x’=xcos ฮธ + y sin ฮธ.
We have two theories regarding the composition of Mohr’s circle of inertia: is it a shifted circle or a centric circle with the intersection of Ix, Iy, and Ixy as its center point?
What is the value and direction of the primary angle 2ฮธp at the third point?
What are the values of Iu and Iv, or the maximum and minimum moments of inertia, along with their directions, constitute the fourth point?
Where is the pole point in Mohr’s circle and the normal view is the fifth point?
How do you draw Mohr’s circle of inertia?
Firstly, how can Mohr’s circle of inertia be drawn when the Ix value exceeds the Iy value?
We begin by locating point X, which has the coordinates Ix, and Ixy as positive values. Point Y, on the other hand, has the coordinates (Iy,-Ixy). A circle’s center is where the intersection of these
two locations with the vertically shifted axis occurs.
This circle’s radius is equal to the square root of Ixy^2+The difference between (Ix-Iy/2)^2. The axis that connects point X to the circle’s center is known as the X-axis.
On the other hand, the Y-axis is the axis that connects point Y to the circle’s center. Our axis, X’, is produced by connecting the point x’, which has the coordinates (Ix’, Ix’y’), to the center of
Mohr’s circle of inertia. Point y’, which has a coordinate of (Iy’,- Ix’y’), is the opposite point.
The enclosed angle of 2ฮธ, which is twice the value of theta from the normal viewโ from which we derived the three equations Ix’, Iy’, and Ix’y’โ between axis X and axis x’.
Given that Mohr’s circle of inertia has both points X and X’, it follows that the square of the values of (Ix’-(Ix-Iy)/2) and the square of the value of Ix’y’ can likewise be used to determine the
radius of that circle.
How to prove that Mohr’s circle of inertia is a shifted circle?
As seen in the slide, we have three equations for Ix’, Iy, and Ix’y’ from the previous post.
Two points on Mohr’s circle of inertiaโ a shifted circle with the values of the moment of inertia, Ix, and Iyโ are represented by the three equations.
The sum of the squares of (Ix’-(Ix+Iy)/2 and Ix’y’ should equal the sum of the squares of (Ix-Iy)/2 and Ixy to demonstrate that the Mohr circle has been relocated.
The term an in equation 1 can be used to equal (Ix-Iy)/2, while the term b can be used to equal Ixy if we subtract the value of (Ix+Iy)/2 from both sides. Acos(2ฮธ)-bsin(2ฮธ) equals Ix’-1/2(Ix-Iy).
Its value for Ix’y will be (a* sin(2ฮธ)+b*sin(2ฮธ).
Square (Ixโ -1/2*(Ix+Iy) will give a^2*cos^2(2ฮธ)+b^2 sin^2((2ฮธ)-2a*b*sin((2ฮธ)*cos((2ฮธ).
While squaring Ixโ yโ will give a^2*sin^2(2ฮธ)+b^2 cos^2((2ฮธ)+2a*b*sin(2ฮธ)*cos(2ฮธ).
If we have a look at the fourth slide. Adding together will give (a^2+b^2), which we will rewrite as (Ix-Iy)^2/4+Ixy^2.
This value represents the square value of the radius the square of the distance of Mohrโ s circle of inertia and point X.
We’ll proceed to the following slide to see what the square of (Iy’ minus (Ix+Iy)/2 plus the square of Ixy will equal. We will move on to equation II and subtract the value of (Ix+Iy)/2 from both
sides using the same terms, a and b.
The formula for the new right-hand side is (-a *cos(2 ฮธ) +b sin(2ฮธ).
Adding together the square (Iyโ minus (Ix+Iy)/2 plus the square of Ixโ yโ will give (a^2+b^2), which we will rewrite as (Ix-Iy)^2/4+Ixy^2. The result matches with radius value from point X and
the center of the circle.
How to find the point of maximum value for inertia?
The point Z, where the Ixy value equals zero, is the greatest value of inertia.
What about the primary axis’s direction? By utilizing an enclosed angle of 2ฮธp in a clockwise direction from the x-axis, one can determine the direction from Mohr’s circle of inertia. In contrast,
it has an angle of ฮธp from the x-axis in a clockwise direction when viewed normally.
By utilizing an enclosed angle of 2ฮธp in a clockwise direction from the x-axis, one can determine the direction from Mohr’s circle of inertia. In contrast, it has an angle of ฮธp from the x-axis in
a clockwise direction when viewed normally.
The expression of Imax using Mohr’s circle.
One can estimate the value of Imax or Iu by taking the shift, that is equal to (Ix+Iy)/2 plus the radius value. We can draw an angle and obtain the radius value in terms of Ixy, Ix, and Iy by
checking the tan 2ฮธp equation.
Thanks a lot and peace be upon you all.
The next post will be an Easy introduction to Mohr’s Circle of Inertia part 2.
This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial and Polar Area Moment of Inertia, and Section Modulus. | {"url":"https://magedkamel.com/4-mohrs-circle-of-inertia-part-1/","timestamp":"2024-11-02T17:21:28Z","content_type":"text/html","content_length":"199962","record_id":"<urn:uuid:4452276a-4e23-4725-918a-636993ba3fb6>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00413.warc.gz"} |
mk_iff_of_inductive_prop #
This file defines a command mk_iff_of_inductive_prop that generates iff rules for inductive Props. For example, when applied to List.Chain, it creates a declaration with the following type:
∀ {α : Type*} (R : α → α → Prop) (a : α) (l : List α),
Chain R a l ↔ l = [] ∨ ∃ (b : α) (l' : List α), R a b ∧ Chain R b l ∧ l = b :: l'
This tactic can be called using either the mk_iff_of_inductive_prop user command or the mk_iff attribute.
compactRelation bs as_ps: Produce a relation of the form:
R := fun as ↦ ∃ bs, ⋀_i a_i = p_i[bs]
This relation is user-visible, so we compact it by removing each b_j where a p_i = b_j, and hence a_i = b_j. We need to take care when there are p_i and p_j with p_i = p_j = b_k.
Generates an expression of the form ∃ (args), inner. args is assumed to be a list of fvars. When possible, p ∧ q is used instead of ∃ (_ : p), q.
• One or more equations did not get rendered due to their size.
Instances For
mkOpList op empty [x1, x2, ...] is defined as op x1 (op x2 ...). Returns empty if the list is empty.
EquationsInstances For
mkAndList [x1, x2, ...] is defined as x1 ∧ (x2 ∧ ...), or True if the list is empty.
EquationsInstances For
mkOrList [x1, x2, ...] is defined as x1 ∨ (x2 ∨ ...), or False if the list is empty.
EquationsInstances For
Drops the final element of a list.
EquationsInstances For
Auxiliary data associated with a single constructor of an inductive declaration.
• For each forall-bound variable in the type of the constructor, minus the "params" that apply to the entire inductive type, this list contains true if that variable has been kept after
For example, List.Chain.nil has type
∀ {α : Type u_1} {R : α → α → Prop} {a : α}, List.Chain R a []`
and the first two variables α and R are "params", while the a : α gets eliminated in a compactRelation, so variablesKept = [false].
List.Chain.cons has type
∀ {α : Type u_1} {R : α → α → Prop} {a b : α} {l : List α},
R a b → List.Chain R b l → List.Chain R a (b :: l)
and the a : α gets eliminated, so variablesKept = [false,true,true,true,true].
• The number of equalities, or none in the case when we've reduced something of the form p ∧ True to just p.
Instances For
Converts an inductive constructor c into a Shape that will be used later in while proving the iff theorem, and a proposition representing the constructor.
• One or more equations did not get rendered due to their size.
Instances For
Splits the goal n times via refine ⟨?_,?_⟩, and then applies constructor to close the resulting subgoals.
• One or more equations did not get rendered due to their size.
Instances For
Proves the left to right direction of a generated iff theorem. shape is the output of a call to constrToProp.
• One or more equations did not get rendered due to their size.
Instances For
Calls cases on h (assumed to be a binary sum) n times, and returns the resulting subgoals and their corresponding new hypotheses.
EquationsInstances For
Calls cases on h (assumed to be a binary product) n times, and returns the resulting subgoal and the new hypotheses.
EquationsInstances For
Iterate over two lists, if the first element of the first list is false, insert none into the result and continue with the tail of first list. Otherwise, wrap the first element of the second list
with some and continue with the tails of both lists. Return when either list is empty.
listBoolMerge [false, true, false, true] [0, 1, 2, 3, 4] = [none, (some 0), none, (some 1)]
EquationsInstances For
Proves the right to left direction of a generated iff theorem.
• One or more equations did not get rendered due to their size.
Instances For
Implementation for both mk_iff and mk_iff_of_inductive_prop.
• One or more equations did not get rendered due to their size.
Instances For
Applying the mk_iff attribute to an inductively-defined proposition mk_iff makes an iff rule r with the shape ∀ps is, i as ↔ ⋁_j, ∃cs, is = cs, where ps are the type parameters, is are the indices, j
ranges over all possible constructors, the cs are the parameters for each of the constructors, and the equalities is = cs are the instantiations for each constructor for each of the indices to the
inductive type i.
In each case, we remove constructor parameters (i.e. cs) when the corresponding equality would be just c = i for some index i.
For example, if we try the following:
structure Foo (m n : Nat) : Prop where
equal : m = n
sum_eq_two : m + n = 2
Then #check foo_iff returns:
foo_iff : ∀ (m n : Nat), Foo m n ↔ m = n ∧ m + n = 2
You can add an optional string after mk_iff to change the name of the generated lemma. For example, if we try the following:
@[mk_iff bar]
structure Foo (m n : Nat) : Prop where
equal : m = n
sum_eq_two : m + n = 2
Then #check bar returns:
bar : ∀ (m n : ℕ), Foo m n ↔ m = n ∧ m + n = 2
See also the user command mk_iff_of_inductive_prop.
• One or more equations did not get rendered due to their size.
Instances For
mk_iff_of_inductive_prop i r makes an iff rule for the inductively-defined proposition i. The new rule r has the shape ∀ps is, i as ↔ ⋁_j, ∃cs, is = cs, where ps are the type parameters, is are the
indices, j ranges over all possible constructors, the cs are the parameters for each of the constructors, and the equalities is = cs are the instantiations for each constructor for each of the
indices to the inductive type i.
In each case, we remove constructor parameters (i.e. cs) when the corresponding equality would be just c = i for some index i.
For example, mk_iff_of_inductive_prop on List.Chain produces:
∀ { α : Type*} (R : α → α → Prop) (a : α) (l : List α),
Chain R a l ↔ l = [] ∨ ∃(b : α) (l' : List α), R a b ∧ Chain R b l ∧ l = b :: l'
See also the mk_iff user attribute.
• One or more equations did not get rendered due to their size.
Instances For | {"url":"https://leanprover-community.github.io/mathlib4_docs/Mathlib/Tactic/MkIffOfInductiveProp.html","timestamp":"2024-11-12T00:33:17Z","content_type":"text/html","content_length":"49733","record_id":"<urn:uuid:8f2f5775-c3cf-423b-846c-d0743d4de876>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00018.warc.gz"} |
MCQ on Systems of Particles and Rotational Motion class 11 PDF Physics
MCQ Questions For Class 11 Physics Chapter 7 Systems of Particles and Rotational Motion
Students can refer to the following MCQ on Systems of Particles and Rotational Motion class 11 PDF with Answers provided below based on the latest curriculum and examination pattern issued by CBSE
and NCERT. Our teachers have provided here a collection of multiple-choice questions for Chapter 7 Systems of Particles and Rotational Motion Class 11 Physics covering all topics in your textbook so
that students can assess themselves on all important topics and thoroughly prepare for their exams
MCQ on Systems of Particles and Rotational Motion class 11 PDF with Answers
We have provided below MCQ on Systems of Particles and Rotational Motion class 11 PDF with answers which will help the students to go through the entire syllabus and practice multiple choice
questions provided here with solutions. As MCQ Questions for Class 11 Physics pdf download can be really scoring for students, you should go through all problems provided below so that you are able
to get more marks in your exams.
Question. Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is
nearly at a distance of
(a) 33 cm
(b) 50 cm
(c) 67 cm
(d) 80 cm
Question. Three masses are placed on the x-axis : 300 g at origin, 500 g at x = 40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin is
(a) 40 cm
(b) 45 cm
(c) 50 cm
(d) 30 cm
Question. Three identical metal balls, each of radius r are placed touching each other on a horizontal surface such that an equilateral triangle is formed when centres of three balls are joined. The
centre of the mass of the system is located at
(a) line joining centres of any two balls
(b) centre of one of the balls
(c) horizontal surface
(d) point of intersection of the medians.
Question. The centre of mass of system of particles does not depend on
(a) position of the particles
(b) relative distances between the particles
(c) masses of the particles
(d) forces acting on the particle
Question. Two persons of masses 55 kg and 65 kg respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits
with him. If the boat is in still water the center of mass of the system shifts by
(a) 3.0 m
(b) 2.3 m
(c) zero
(d) 0.75 m
Question. Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of centre of
mass of the system will be
(a) 2v
(b) zero
(c) 1.5v
(d) v
Question. A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor,
the distance of the man above the floor will be
(a) 9.9 m
(b) 10.1 m
(c) 10 m
(d) 20 m
Question. When a mass is rotating in a plane about a fixed point, its angular momentum is directed along
(a) a line perpendicular to the plane of rotation
(b) the line making an angle of 45° to the plane of rotation
(c) the radius
(d) the tangent to the orbit.
Question. A particle of mass m moves in the XY plane with a velocity v along the straight line AB. If the angular momentum of the particle with respect to origin O is LA when it is at A and LB when
it is at B, then
(a) LA = LB
(b) the relationship between LA and LB depends upon the slope of the line AB
(c) LA < LB
(d) LA > LB
Question. A particle of mass m = 5 is moving with a uniform speed v =3 √2 in the XOY plane along the line y = x + 4. The magnitude of the angular momentum of the particle about the origin is
(a) 60 units
(b) 40 2 units
(c) zero
(d) 7.5 units
Question. Which of the following statements are correct?
(1) Centre of mass of a body always coincides with the centre of gravity of the body.
(2) Centre of mass of a body is the point at which the total gravitational torque on the body is zero.
(3) A couple on a body produces both translational and rotational motion in a body.
(4) Mechanical advantage greater than one means that small effort can be used to lift a large load.
(a) (1) and (2)
(b) (2) and (3)
(c) (3) and (4)
(d) (2) and (4)
Question. (1) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts.
(2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius.
(3) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its C.G.
(4) The radius of gyration of any body rotating about an axis is the length of the perpendicular drawn from the C.G. of the body to the axis. Which one of the following pairs of statements is
(a) (4) and (1)
(b) (1) and (2)
(c) (2) and (3)
(d) (3) and (4)
Question. A rod of length 3 m and its mass per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at
(a) 1.5 m
(b) 2 m
(c) 2.5 m
(d) 3.0 m
Question. 250 N force is required to raise 75 kg mass from a pulley. If rope is pulled 12 m then the load is lifted to 3 m, the efficiency of pulley system will be
(a) 25%
(b) 33.3%
(c) 75%
(d) 90%.
Question. A couple produces
(a) linear and rotational motion
(b) no motion
(c) purely linear motion
(d) purely rotational motion.
Question. A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed
twice that of the sphere. The ratio of their kinetic energies of rotation (Esphere/Ecylinder) will be
(a) 2 : 3
(b) 1 : 5
(c) 1 : 4
(d) 3 : 1
Question. From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the
plane of the disc and passing through its centre is
(a) 40/9 MR2
(b) MR2
(c) 4MR2
(d) 4/9 MR2
Question. The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is
(a) √2 :1
(b) 2 : 3
(c) 3 : 2
(d) 1: 2
Question. Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular velocity will be in the ratio
(a) 2 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 2
Question. A circular disc is to be made by using iron and aluminium so that it acquires maximum moment of inertia about geometrical axis. It is possible with
(a) aluminium at interior and iron surrounding it
(b) iron at interior and aluminium surrounding it
(c) using iron and aluminium layers in alternate order
(d) sheet of iron is used at both external surface and aluminium sheet as internal layers.
Question. A fly wheel rotating about fixed axis has a kinetic energy of 360 joule when its angular speed is 30 radian/sec. The moment of inertia of the wheel about the axis of rotation is
(a) 0.6 kg m2
(b) 0.15 kg m2
(c) 0.8 kg m2
(d) 0.75 kg m2
Question. From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a
perpendicular axis, passing through the centre?
(a) 11 MR2/32
(b) 9 MR2/32
(c) 15 MR2/32
(d) 13 MR2/32
Question. The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I0. Its moment of inertia about an axis
passing through one of its ends and perpendicular to its length is
(a) I0 + ML2/2
(b) I0 + ML2/4
(c) I0 + 2ML2
(d) I0 + ML2
Question. A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is 90°. The moment of inertia of the bent rod about an axis passing through the
bending point and perpendicular to the plane defined by the two halves of the rod is
(a) ML2 /6
(b) √2/ML2 /24
(c) ML2 /24
(d) ML2 /12
Question. The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius and mass about a tangential axis in the
plane of the ring is
(a) 2 : 3
(b) 2 : 1
(c) √5 : √6
(d) 1: √2
Question. Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be
(a) 5I
(b) 3I
(c) 6I
(d) 4I
Question. A wheel has angular acceleration of 3.0 rad/sec2 and an initial angular speed of 2.00 rad/sec. In a time of 2 sec it has rotated through an angle (in radians) of
(a) 10
(b) 12
(c) 4
(d) 6
Question. A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop it after 2p revolutions is
(a) 2 × 10^6 N m
(b) 2 × 10^–6 N m
(c) 2 × 10^–3 N m
(d) 12 × 10^–4 N m
Question. Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed w about their
own symmetry axes.
The amounts of work (W) required to bring them to rest, would satisfy the relation
(a) WC > WB > WA
(b) WA > WB > WC
(c) WB > WA > WC
(d) WA > WC > WB
Question. A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N?
(a) 0.25 rad s–2
(b) 25 rad s–2
(c) 5 m s–2
(d) 25 m s–2
Question. A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a
constant angular acceleration of 2.0 rad s–2. Its net acceleration in m s–2 at the end of 2.0 s is approximately
(a) 6.0
(b) 3.0
(c) 8.0
(d) 7.0
Question. An automobile moves on a road with a speed of 54 km h–1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m2. If the vehicle is
brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to the wheel is
(a) 10.86 kg m2 s–2
(b) 2.86 kg m2 s–2
(c) 6.66 kg m2 s–2
(d) 8.58 kg m2 s–2
Question. A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging
freely. Tension in the string\ required to produce an angular acceleration of 2 revolutions s–2 is
(a) 25 N
(b) 50 N
(c) 78.5 N
(d) 157 N
Question. The instantaneous angular position of a point on a rotating wheel is given by the equation q(t) = 2t3 – 6t2.
The torque on the wheel becomes zero at
(a) t = 1 s
(b) t = 0.5 s
(c) t = 0.25 s
(d) t = 2 s
Question. The moment of inertia of a body about a given axis is 1.2 kg m2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of 25
radian/sec2 must be applied about that axis for a duration of
(a) 4 s
(b) 2 s
(c) 8 s
(d) 10 s
Question. A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain
constant for the sphere?
(a) Angular velocity.
(b) Moment of inertia.
(c) Rotational kinetic energy.
(d) Angular momentum.
Question. Two rotating bodies A and B of masses m and 2m with moments of inertia IA and IB (IB > IA) have equal kinetic energy of rotation. If LA and LB be their angular momenta respectively, then
(a) LA = LB /2
(b) LA = 2LB
(c) LB > LA
(d) LA > LB
Question. Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc D1 has 2 kg mass and 0.2 m radius and initial angular velocity of 50 rad s–1.
Disc D2 has 4 kg mass, 0.1 m radius and initial angular velocity of 200 rad s–1. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity
(in rad s–1) of the system is
(a) 60
(b) 100
(c) 120
(d) 40
Question. A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the
edge of the platform and begins to walk along the edge at the speed of 1 m s–1 relative to the ground. Time taken by the man to complete one revolution is
(a) p s
(b) 3p/2 s
(c) 2p s
(d) p/2 s
Question. A disc is rotating with angular speed w. If a child sits on it, what is conserved?
(a) linear momentum.
(b) angular momentum.
(c) kinetic energy.
(d) potential energy
Question. A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it ?
(a) 1 J
(b) 3 J
(c) 30 kJ
(d) 2 J
Question. A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle inclination 30°.
The centre of mass of cylinder has speed of 4 m/s.
The distance travelled by the cylinder on the incline surface will be (Take g = 10 m/s2)
(a) 2.2 m
(b) 1.6 m
(c) 1.2 m
(d) 2.4 m
Question. A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (Kt) as well as rotational kinetic energy (Kr) simultaneously.
The ratio Kt : (Kt + Kr) for the sphere is
(a) 7 : 10
(b) 5 : 7
(c) 10 : 7
(d) 2 : 5
Question. A disc and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?
(a) Both reach at the same time
(b) Depends on their masses
(c) Disc
(d) Sphere
Question. The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle q without slipping and slipping down the incline without rolling is
(a) 5 : 7
(b) 2 : 3
(c) 2 : 5
(d) 7 : 5
Question. A small object of uniform density rolls up a curved surface with an initial velocity ‘v’. It reaches upto a maximum height of 3V2 /4g with respect to the initial position. The object is
(a) hollow sphere
(b) disc
(c) ring
(d) solid sphere.
We hope you liked MCQ Questions for Class 11 Systems of Particles and Rotational Motion with answers pdf provided above. Incase you have any questions please put them in the comments section below.
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Review of Really Big Numbers by Richard Evan Schwartz
Really Big Numbers
by Richard Evan Schwartz
It's common to describe mathematics as a formal, meaningless game with abstract symbols (often in the wake of Bertrand Russell's observation). Mathematicians, of course, know better. They know that
mathematical language is a tool so powerful that it lets them imagine unimaginable. Symbols are invented for a purpose, to a great extent with an objective to make mathematical concepts easier to
handle. Who then, if not a working mathematician, may more convincingly weigh to rectify the popular delusion. Richard Evan Schwartz - the Chancellor's Professor of Mathematics at Brown University -
does this with exceptional flair, in a clear and entertaining manner.
The book is about the most basic mathematical concepts - that of counting and numeration. Indeed, there would not be big numbers if there were no small ones. So this is where the book starts - with
small numbers. With wonderful illustrations, Schwartz makes it clear that the same number describes a property of various sets, regardless of how the elements of those sets are arranged or grouped.
Next, grouping elements by 10s and 100s the book leads fast to bigger numbers. But on the way up there the book dwells on origins of big numbers, with such hands-on and humorous examples as the
number of minutes in a week or the number of hours lived by a 114 year old fellow. Big numbers arise in trying to describe the number of ways to paint a 3x3 board or to place checkers on the chess
board, the number of feet on the way to reach a satellite, the number of basketballs to cover New York city to the height of a reasonably tall man. Then exponentiation is introduced and the
exposition accelerates, though not before the powers of ten through quinquagintillion (which is 10^153) are listed and named.
And then come the googol and googolplex, followed by other plexes - the staired exponentiation and, eventually, pictorial but nameless representations of stupefyingly enormous quantities. The latter
illustrate another important concept - that of recursion. The glyphs - squares in squares, triangle in pentagons, etc ... - have no names, but their meanings are absolutely transparent due to the
recursive definition and, if nothing else, demonstrate convincingly the expressive power and adaptability of the mathematical language.
Early on in the book, the author suggests that there is no obligation
... to read all at once, or even all in one year. Just read as far as it makes sense and then save the parts you don't understand for later.
Many will probably take the author up on that advice and put the book aside. Those that persevere are bound to have their imagination fired up as they follow the book's lead to the literally
unspeakable quantities mathematics lets them deal with.
Really Big Numbers, by Richard Evan Schwartz. AMS, 2014. Softcover, 192 pp, $25.00. ISBN 978-1470414252.
|Up| |Contact| |Front page| |Contents|
Copyright © 1996-2018
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Motor Load Starting using Transient Stability
Note: Download a small sample case at the following link which is used in this demonstrate: MotorStart Case
Note: Starting in the July 10, 2013 patch of Simulator 17, a modification was made to automatically change the integration time step for the entire simulation when induction motors (loads) or
induction machine (generators) drop below 0.1 per unit speed. A multiplier is hard-coded internal to Simulator that is 0.05 at 0.0 per unit speed and increases linearly up to 1.0 at 0.1 per unit
speed. The time step used by the integration will be equal to the user-specified time step multiplied by this value. If multiple motors/machines are operating in this region, then the smallest
multiplier in the case will be used. This is intended to improve induction motor starting studies without the user having to manually reduce the time step for the entire simulation. It will be a very
rare event when this occurs, largely only occurring during motor start studies. This new feature will make the plots shown in the following figures different than what you would get with Simulator 17
in a patch after July 10. However, the example remains educational.
In Transient Stability, when a dynamic model is not online in the initial condition, without using a special model that dynamic model can not be closed in during the transient stability simulation.
For example, PowerWorld Simulator does not permit a synchronous machine to be closed during the simulation and thus modeling the start-up of a synchronous generator is not presently possible (as of
June 2013). In general the same is true for dynamic load models such as the MOTORW, CIMW, or CMPLDW motor models. These models can not be closed during a stability simulation.
To model motor starting however, there are special models which have been designed to model starting. There are older models that are assigned to generators (with a negative MW output) such as
MOTOR1 and CIMTR4, as well as models assigned to a load record such as CIM5 and CIM6. For this knowledge base article, we will discuss the CIM5 and CIM6 induction motor models.
This article will not go into detail about induction motor circuit parameters (Ra, Xa, Xm, R1, X1, R2, X2) and how these relate to values used in transient stability models of induction motors (Ra,
Xs, Xp, Tpo, Xpp, Tppo). All the induction motors and machines treat these conversions the same. What is different about the CIM5 and CIM6 induction motors is how the load torque (the mechanical
torque) on the motor is modeled.
For any induction motor, from initializing the circuit model of an induction motor PowerWorld Simulator will determine the initial slip and the initial load torque (T[loadinit]). The initial per
unit speed is then just 1 – slip. Given the initial speed (ω[0]) and the initial load torque (T[loadinit]), the initialization of a motor that is closed and operating in the steady state initial
condition is as follows.
Model Name Parameters Load Torque Equation Online Initialization
CIM5 T[nom], D T[load] = T[nom]ω^D T[nom]=T[loadinit]/(ω[0]^D)
CIM6 T[nom], A, B, T[load] = T[nom](Aω^2+Bω + C[0] + Dω^E) C[0]= 1 – Aω[0]^2-Bω[0] – Dω[0]^E
C[0], D, E T[nom]=T[loadinit]
Any values specified by the user for (C[0 ]and T[nom]) in the model input parameters will be ignored if the load is online in the initial condition. However, when you want to perform a motor
starting simulation, then these parameters (C[0 ]and T[nom]) much be specified as part of the model input parameters and will help determine the ultimate response of the load. Another very important
consideration when studying motor starting is deciding what integration time step to use. Induction motors operating at low speeds have extremely fast electrical transients, and since motor starting
involves going from zero speed, motor starting involves extremely fast transients. Because of this, for a motor starting study you may need to use a much smaller time-step than use in other types of
stability studies. (Note: Simulator does make use of multirate timestep integration to better handle these fast motor transients, but you will still need to reduce the time step some.)
To demonstrate this a small sample case has been made and can be found at the following link: MotorStart Case
The case has a small generator and load (4 MW) at the Utility Gen and Utility Sub buses and an initially open load at the Load bus.
The load model at the Load bus is set to CIM5 as shown in the following figure. A value of Tnom is specified as 0.9 per unit. Note also that the value of Vi is set to 0. The CIM5, like many
induction motor models, includes an under-voltage tripping feature such that if the terminal voltage falls below Vi for more than Ti cycles, then the load will trip. We don’t want that to happen in
this example, so Vi is set to 0.
We then configure the transient stability tool to CLOSE in this load at 1.0 seconds of simulation. To demonstrate the impact of choosing a time step, below we have create 6 different transient
contingencies which all have different time steps varying from 0.5 ms up to 8.0 ms. Each contingency though has the same single event – Load ‘4’ ‘1’ CLOSE.
The following plot shows several response curves when using the 0.5 ms time-step. You can see squiggles at 1.0 seconds which represent those very fast electrical transients mentioned earlier.
The following figure shows a close-up of the Load Torque curves between 0.99 and 1.07 seconds. The period of this initial oscillation is 16.5 ms and thus is about a 60 Hz oscillation. As a result
our normal integration methods which use between 4 and 8 ms time steps may present some trouble.
To demonstrate this we ran the same simulation using 6 different time steps between 0.5 ms and 8.0 ms. The load’s electrical torque for each of these different time-steps is shown in the following
figure. You can see that using any time step between 0.5 and 4.0 ms works acceptably, but the 6 ms and 8 ms time-steps show some numerical instability. The images after that also show that as you
make the time step smaller you’re able to more precisely go through these fast oscillation, but when looking at the overall system behavior the 4 ms time-step still yields a response substantially
similar to the 0.5 ms time step.
Tags: How-to,Simulator,Transient Stability,Tutorial
July 2, 2013 | {"url":"https://www.powerworld.com/knowledge-base/motor-load-starting-using-transient-stability","timestamp":"2024-11-05T03:54:23Z","content_type":"text/html","content_length":"56555","record_id":"<urn:uuid:6d2ca8be-4be1-4de2-9adc-5ef3e382b4fd>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00532.warc.gz"} |
Happy’s Essential Skills: Learning Theory/Learning Curves
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Happy’s Essential Skills: Learning Theory/Learning Curves
June 1, 2016 | Happy Holden
Estimated reading time: 14 minutes
The following examples are from Nick Pearne’s paper
: As mentioned above, the learning curve depends on the fact that experience gained from increased production of any commodity causes a decline in manufacturing costs, and therefore inevitably in
prices in a competitive market environment. More exactly, the theory states that every time the quantity of “units” (or “lots”) produced is doubled the corresponding unit (or lot) costs decline by
an experience factor F, also known as the learning or improvement ratio. This is determined by the relationship between resources (typically process cost) required to produce double the reference
quantity, Qo:
F =C2/C1 (1)
Where C1 is the initial average unit cost and C2 is the average unit cost for double the reference quantity.
From equation (1) it is evident that the higher the value of F, the less change in cost is to be expected due either to process maturity (automation, optimized setup, tooling, yields), or highly
customized content, as might be expected from small lot quantities of complex rigid flex assemblies.
For an initial quantity Qo and a final quantity Q the number of “doublings” or fractions thereof for the total quantity produced is given by log(Q/Qo)/log(2). Therefore, the unit cost behavior as a
function of quantity can be written as:
C = C1*(F/100) ^ (log(Q/Qo) / log(2)) (2a)
Where C is the unit cost after quantity Q units or lots, C1 is the first unit cost, and F is the experience factor in percent.
A value of 75 for F would be typical of very steep (fast) learning curves, in which process consolidation proceeds rapidly with corresponding reductions in changeover time, improvements in yields,
etc. Equation (2a) is awkward to handle since the principal variable, Q, appears in the exponent. It can be rearranged (and simplified) by noting that in general a ^ log(b) is equivalent to b ^ log
(a) since either expression can be written as e ^ [log(a)*log(b)]. An alternate and better form for equation (2a) is therefore:
C = C1*q ^ k (2b)
Where q = Q/Qo and k = log(F/100) / log(2)
The total cost, T, to produce a quantity Q units or lots can be obtained by integrating equation (2b) over the limits q = 0 to q = Q:
T = C1* q ^ kdq = C1*Q ^ (k+ 1)/(k+ 1) (3)
The average cost, a, per unit or lot quantity is the total cost divided by the quantity:
A = T/Q (4)
For processes where the experience factor is accurately known, the average cost is often used to quote a lot or piece price to be effective over the entire production. Suppose, for example, that a
first lot of ten pieces is produced at a cost of $20.00 by a process with a known experience factor of 80%. What would be the predicted piece cost for 1,000 units? For F = 80%, k is found to be log
(0.80)/log (2) = 20.3219, and for this case the “experience” quantity Q = 1,000/10 = 100.
C = 20.00*100 ^ (-0.3219) = 4.5412
So that at the end of the run the production cost has declined to $4.54 per lot. The total cost, from equation (3), becomes:
T = 20*100 ^ (0.6781)/0.6781 = 669.7274
The average production cost per unit quantity (1 lot) is therefore T/Q = $6.70 and the piece cost is about $0.67. This approach can be used to create log-log plots for various experience factors,
giving unit costs as a function of quantities and initial costs. For example, a process with 80% experience factor and an initial cost of 1.00 per unit can expect unit costs to decline to about 0.11
by the time 1,024 (2 ^ 10) units have been produced. This not atypical of the semiconductor industry, where F may be 75% or even less. At the other end of the scale a complex, low volume product may
be 90% or higher. One-offs with highly customized assemblies will be as high as 100%: the product lifetime is too short (one-off) and the standardized process component(s) are too limited to offer
meaningful improvement opportunities.
New Technologies—the Experience Factor^[1]
To use this analysis for new technologies it is necessary to determine the experience factor. This can done using a broader experience base than the simple doubling shown in equation (1) by flipping
equation (2a) around [. . .] [LL1] provided the data are available, specifically:
F = 10 ^ (log(2)*log(C/C1) /log(Q)) (5)
If the production cost of a metal-core type insulated metal substrate LED multichip board was 2.00 when 10,000 pieces had been produced (C1) and the cost (C) is now 0.65 when 4,000,000 have been
produced (Q = 400), what is the experience factor F?
F = 10 ^ (log(2)*log(0.65/2.00)/log(400))
Or: F = 0.878
What will be the cost for the 20,000,000th piece when Q will be effectively 2,000 (20,000,000/10,000)?
k = log(0:878)/log(2) = 20:18771
C = 2.00*2000 ^ (20.18771) = 0.4801
This example assumes a limited degree of process innovation is necessary in the introduction of a new layout for the same function/substrate. As is often the case in printed circuit manufacturing,
where the emphasis is less on products and more on capabilities built on standardized processes, the experience factor may be even higher than 88%. It is important to remember that the experience
factor “F” does not imply any particular degree of expertise or mastery of the technology. It is simply an index of the expected stability of processing costs over the lifetime of the design.
1. Burr, W., Pearne, N., “Learning curve theory and innovation,” Circuit World, Vol. 39, Issue 4, 2003, pp 169–173.
2. Transformative Learning (Jack Mezirow)
3. The Learning Curve or Experience Curve, provided by James Martin.
Happy Holden has worked in printed circuit technology since 1970 with Hewlett-Packard, NanYa/Westwood, Merix, Foxconn and Gentex. Currently, he is the co-editor, with Clyde Coombs, of the Printed
Circuit Handbook, 7th Ed. To contact Holden, click here.
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DENSO CORPORATION and Quadric.inc have signed a development license agreement for a Neural Processing Unit (NPU)(1), which is a semiconductor specialized for the arithmetic processing of AI. | {"url":"https://iconnect007.com/article/97937/happys-essential-skills-learning-theorylearning-curves/97940/pcb?page=2","timestamp":"2024-11-02T17:16:55Z","content_type":"text/html","content_length":"81046","record_id":"<urn:uuid:cb348bae-475c-4d06-8210-3b5d63d474fb>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00059.warc.gz"} |
30/360 Interest Calculator - GEGCalculators
30/360 Interest Calculator
30/360 Interest Calculator
How do you calculate interest on 360 30? The term “360/30” typically refers to a method of calculating interest using a 360-day year and 30-day months. To calculate interest using this method, you
would need to know the principal amount, the interest rate, and the number of days over which interest is being calculated. You can use the following formula:
Interest = (Principal Amount x Interest Rate x Number of Days) / 360
What is the 360 interest calculation method? The 360 interest calculation method uses a 360-day year and assumes each month has 30 days, regardless of the actual number of days in the month. It
simplifies interest calculations for financial instruments like bonds and loans, making it easier to perform calculations.
How do you calculate interest rate 360? To calculate the interest rate on a 360 basis, you would need to know the principal amount, the interest amount, and the number of days over which interest is
being calculated. You can rearrange the formula mentioned earlier to calculate the interest rate as follows:
Interest Rate = (Interest x 360) / (Principal Amount x Number of Days)
How much is a 30 percent interest rate? A 30 percent interest rate means that for every $100 of the principal amount, you would pay $30 in interest over a given period, typically a year. This rate
can vary depending on the compounding frequency and the specific terms of the financial product.
What does 30 360 interest mean? “30/360” interest refers to the method of calculating interest using a 360-day year and assuming each month has 30 days. It is commonly used in financial transactions
for simplicity, especially in bond markets.
What does 30E 360 mean? “30E/360” is another variation of the 30/360 interest calculation method. The “E” stands for “European” and typically accounts for slight variations in the number of days in
February in leap years, assuming 30 days for all months and 360 days in a year.
How to calculate the interest? To calculate interest, you can use the formula mentioned earlier: Interest = (Principal Amount x Interest Rate x Number of Days) / 360. Plug in the values for
principal, interest rate, and the number of days to find the interest amount.
What is the formula for interest calculation? The formula for interest calculation depends on the specific method being used. The basic formula for simple interest is: Interest = (Principal Amount x
Interest Rate x Time).
What is the easiest way to calculate interest rate? The easiest way to calculate interest rate is to rearrange the formula for simple interest: Interest Rate = (Interest / (Principal Amount x Time)).
Plug in the values for interest, principal amount, and time to find the interest rate.
What is the difference between 30 360 adjusted and unadjusted? The difference between “30/360 adjusted” and “30/360 unadjusted” lies in how they handle the number of days in February. “30/360
adjusted” considers the actual number of days in February, especially in leap years, while “30/360 unadjusted” assumes 30 days in February for all years.
What is the interest rate on a 360 savings account? The interest rate on a 360 savings account or any other savings account can vary widely depending on the financial institution, the type of
account, and current market conditions. You would need to check with your specific bank or financial institution for the current interest rate.
What are 3 different methods of calculating interest? Three different methods of calculating interest include simple interest, compound interest, and the various methods like 30/360 or 30E/360 for
specific financial transactions.
How do you calculate interest per month? To calculate interest per month, you can use the formula: Monthly Interest = (Annual Interest Rate / 12).
How do you calculate monthly interest rate? Monthly interest rate is usually calculated by dividing the annual interest rate by 12.
How much interest do I pay a month? The amount of interest you pay per month depends on your loan or investment terms and the outstanding balance. You can calculate it using the appropriate formula.
What is 4% interest on 30000? 4% interest on $30,000 would be $1,200 per year, assuming simple interest.
How much is 30000 at 6 percent interest? $30,000 at a 6% interest rate would earn you $1,800 in interest per year if it’s an investment or cost you $1,800 in interest per year if it’s a loan.
Is 35% interest bad? A 35% interest rate is generally considered very high, and it can be financially burdensome. Borrowing at such a high rate can lead to substantial interest costs over time, so
it’s typically advisable to seek lower interest rates if possible.
What is the 30 360 end of month? The “30/360 end of month” means that when calculating interest, it assumes that each month ends on the last day of the month, regardless of the actual day.
What is the difference between 30 360 and actual actual? “30/360” is a simplified method that assumes 30 days in each month and 360 days in a year. “Actual/Actual” is a method that calculates
interest based on the actual number of days in each month and the actual number of days in a year, making it more precise.
What is the day count basis for 30 360? The day count basis for “30/360” is 30 days in each month and 360 days in a year, as it assumes a 30-day month for all months.
How do you calculate interest for dummies? To calculate interest for dummies, you can use the simple interest formula: Interest = (Principal Amount x Interest Rate x Time). Plug in the values for
principal amount, interest rate, and time to find the interest amount.
How do you calculate interest on 6 months? To calculate interest for a 6-month period, you can use the formula: Interest = (Principal Amount x Interest Rate x Time), where Time is 6/12 or 0.5 (since
there are 12 months in a year).
How do you calculate interest per day? To calculate interest per day, you can use the formula: Daily Interest = (Annual Interest Rate / 365).
How do you calculate interest per annum? Interest per annum (per year) can be calculated using the formula: Annual Interest = (Principal Amount x Interest Rate).
Why do we calculate interest? Interest is calculated to determine the cost of borrowing money, the return on investments, and the profitability of financial transactions. It helps individuals and
businesses make informed financial decisions.
What is the most common method of interest calculation? The most common methods of interest calculation are simple interest and compound interest. Simple interest is often used for straightforward
loans, while compound interest accounts for interest on both the principal amount and accumulated interest over time.
What is the simple interest formula example? The simple interest formula is: Interest = (Principal Amount x Interest Rate x Time). For example, if you have a $1,000 loan with a 5% annual interest
rate for 2 years, the interest would be: Interest = (1000 x 0.05 x 2) = $100.
What is the formula for monthly simple interest? The formula for monthly simple interest is: Monthly Interest = (Principal Amount x Monthly Interest Rate). To calculate the monthly interest rate, you
can divide the annual interest rate by 12.
GEG Calculators is a comprehensive online platform that offers a wide range of calculators to cater to various needs. With over 300 calculators covering finance, health, science, mathematics, and
more, GEG Calculators provides users with accurate and convenient tools for everyday calculations. The website’s user-friendly interface ensures easy navigation and accessibility, making it suitable
for people from all walks of life. Whether it’s financial planning, health assessments, or educational purposes, GEG Calculators has a calculator to suit every requirement. With its reliable and
up-to-date calculations, GEG Calculators has become a go-to resource for individuals, professionals, and students seeking quick and precise results for their calculations.
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Law Of Large Numbers
Quote: Wizard
I'm sure there are a lot of baccarat players would would agree with you. However, don't expect much support for your statement here.
People will say bac is different because you're dealing
down static decks. It is essentially the same as red and
black in roulette, however. Player banker outcomes are
quite random.
"It's not called gambling if the math is on your side."
Quote: thecesspit
Please give me an example relating to a Roulette wheel, or other random number sequence.
By example, do you mean give a demonstration of
how you can look at past results and accurately guess
at a future outcome? I'm just discussing theory here,
demo's are another subject entirely..
"It's not called gambling if the math is on your side."
Quote: EvenBob
By example, do you mean give a demonstration of
how you can look at past results and accurately guess
at a future outcome? I'm just discussing theory here,
demo's are another subject entirely..
Well, I see little theory here based on a stream of random numbers, so I thought seeing as I am not understanding your theory, I thought a practical example might help.
I'll state it again. -Either- the numbers from a Roulette wheel are random in the classic mathematical meaning OR they are not random by the definition.
Unless I've made a mistake. If I have, fine, point it out.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at
trente-et-quarante" - Honore de Balzac, 1829
Quote: thecesspit
The Law of Large numbers is EMERGENT from the basic probability of the game. The chicken and egg is here solved.... the low level behaviour came first.
There's no proof of that. It looks like they emerged
at the same time, by Cardano in the middle of the
16th century. Bernoulli later proved it. In any event,
LLN is now the god of probability theory, not the
other way around.
"It's not called gambling if the math is on your side."
Quote: thecesspit
I'll state it again. -Either- the numbers from a Roulette wheel are random in the classic mathematical meaning
Of course they're random, everything hinges on them being
pure random. Thats why I won't use numbers spewed by a
computer RNG, they're pseudo random. In the long term
they look the same as pure random. In the short term, where
we use them, they're flawed, if ever so slightly.
Its funny, I use a tool called RX Roulette when I practice,
Sometimes I forget to use imported actual spins for a session
and inadvertantly use the RNG they provide. After awhile I
start to get pissed because I'm making so many mistakes,
and then I realize its the damned RNG I'm using and not
the real thing. Pretty funny.
"It's not called gambling if the math is on your side."
Quote: EvenBob
There's no proof of that. It looks like they emerged
at the same time, by Cardano in the middle of the
16th century. Bernoulli later proved it. In any event,
LLN is now the god of probability theory, not the
other way around.
You misunderstand what I mean.
Not which idea came first, but that one (basic probability systems) can prove the other. The Law of Large Numbers emerges from the axioms of probability maths. You can prove the LLN from the axioms
of probability. I am not sure you can prove probability mathematics based on the law of large numbers as an axiom.
It may be one of the more important laws in probability maths, but doesn't mean it's not emergent phenomena from the base.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at
trente-et-quarante" - Honore de Balzac, 1829
Quote: EvenBob
Of course they're random, everything hinges on them being
pure random. Thats why I won't use numbers spewed by a
computer RNG, they're pseudo random. In the long term
they look the same as pure random. In the short term, where
we use them, they're flawed, if ever so slightly.
Its funny, I use a tool called RX Roulette when I practice,
Sometimes I forget to use imported actual spins for a session
and inadvertantly use the RNG they provide. After awhile I
start to get pissed because I'm making so many mistakes,
and then I realize its the damned RNG I'm using and not
the real thing. Pretty funny.
They cannot be pure random and show the effect you are theorizing about. You need to point out to me where my argument is flawed. A true stream of random numbers must be independent of one another,
or they ARE NOT RANDOM. It's really that simple. If you can see a short term effect, THEY ARE NOT PURE RANDOM numbers. That is my statement, my core argument. Without evidence/argument to the
contrary, this whole thing is pointless.
Isn't it more likely that there's actually a subtle effect in the real roulette numbers that makes them less random that you don't see in a Random number generator?
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at
trente-et-quarante" - Honore de Balzac, 1829
Quote: thecesspit
the axioms of probability maths.
Probability 'theory' is a squishy subject. Probability
math kinda sorta works, but there are no probability
laws, per se. They seem to invent a new offshoot
of probability every other year. One of the more
recent ones, Imprecise Probability, didn't get named
till the 90's.
"It's not called gambling if the math is on your side."
Quote: thecesspit
A true stream of random numbers must be independent of one another, or they ARE NOT RANDOM.
They are completely independent of each other, thats
why they're so reliable to use. Thats why probability
works on them, because of their complete indepencence.
Computer RNG's produce non independent outcomes too
often, they are uneliable. They're fine for something crude
like a slot machine, but they're flawed for accurate
probability work. An example here would only be confusing,
it would be meaningless. The outcomes have to be true
random or they won't work.
"It's not called gambling if the math is on your side."
Quote: EvenBob
Probability 'theory' is a squishy subject. Probability
math kinda sorta works, but there are no probability
laws, per se. They seem to invent a new offshoot
of probability every other year. One of the more
recent ones, Imprecise Probability, didn't get named
till the 90's.
Did you actually read the article? Did you understand what it is talking about? Or just seeing the word "probability" in the title was enough for you?
I am not sure what you refer to as "laws", it is more common to use terminology of axioms and theorems. Like any non-trivial and self-consistent axiomatic theory, probability contains an infinite
number of theorems, provable on a finite axiom base. Law of large numbers is but one of those theorems.
There is nothing "squishy" about probability theory. It is a closed, self-consistent formal theory, much like calculus, algebra or games theory. Among other things, much if not most of contemporary
physics could not be formulated without probability.
"When two people always agree one of them is unnecessary"
Quote: weaselman
Did you actually read the article? >>There is nothing "squishy" about probability theory.
Read it half a dozen times in the last week. And
PT is squishy. Thats why probabilists are always
fighting with each other.
"It is unanimously agreed that statistics depends on probability. But, as to what probability is and how it is connected with statistics, there has seldom been such complete disagreement and
breakdown of communication since the Tower of Babel." Wikipedia.
"It's not called gambling if the math is on your side."
Quote: EvenBob
"It is unanimously agreed that statistics depends on probability. But, as to what probability is and how it is connected with statistics, there has seldom been such complete disagreement and
breakdown of communication since the Tower of Babel." Wikipedia.
I'll add the last sentence of the quote, you omitted, I am sure, accidentally ...
Doubtless, much of the disagreement is merely terminological and would disappear under sufficiently sharp analysis.
This particular phrase belongs, I think, to Leonard Savage, who was more of an economist than a mathematician. He wrote a book "Foundation of Statistics", where he suggested what he called
"subjective probability" (or, perhaps, "personal probability" - I don't remember exactly) as a substitute for the formal measure-theoretical interpretation.
The disagreement he refers to in this statement is the terminological disagreement between mathematicians, statisticians, economists, engineers and philosophers, who use a number of identical terms
(probability, frequency, objective, physical etc.) to mean quite different things.
As long as we talk about
mathematical theory of probability
mathematical statistics
, none of this applies at all. There is nothing "squishy" or ambiguous about those theories.
"When two people always agree one of them is unnecessary"
Quote: weaselman
it does not make it any more believable or give it any more weight than your (quite weightless lately) own posts.
Uh Huh, if you can't dispute the message, kill
the messenger. In this case, Wiki. I happen to
read a lot of probability articles, and I see constant
bickering and disagreement. Don't you find that
odd for a 'science' that you say is so precise and
accurate? Its squishy, thats the perfect word to
describe it.
"It's not called gambling if the math is on your side."
Quote: EvenBob
Uh Huh, if you can't dispute the message, kill
the messenger. In this case Wiki
I am not disputing the message, just your interpretation of it.
The message was incomplete (I just added the missing sentence above in my earlier post, probably, after you posted yours), which, probably has contributed to your wrong interpretation.
Nothing is wrong with wiki as it turns out, just your inability to cut and paste properly (unless, you actually truncated the quote on purpose, which I don't want to believe).
I happen to
read a lot of probability articles, and I see constant
bickering and disagreement. Don't you find that
odd for a 'science' that you say is so precise and
No, not at all. The same is going on with cosmology, special relativity, genetics, paleo-biology etc., etc., even the Super-String theory.
What's common between those fields, is that their are very highly popularized, and described at "layman level", making them look simple enough for a layman to actually believe that he understands. It
takes one sentence, by somebody, not very competent in the field, or even, by someone competent enough, but not very good at written language, to fuel months if not years of "bickering and
disagreements" in the "popular-science" circles.
Did you notice by any chance how much fewer "bickering" and "disagreement" is out there about, say, General Relativity, which is totally and completely based on its Special cousin?
Or, back to the probabilities, would you say you are noticing more disagreement in the discreet case (combinatorics, binomial trials etc.) or in the continuum or transfinite applications?
"When two people always agree one of them is unnecessary"
Quote: weaselman
I can dispute the message.
It doesn't matter, whats relevant to me is, no math,
no probability, governs the next outcome of
random numbers. All you know is what will 'probably'
happen 'eventually'. Up until that eventuality, you
can drive a truck thru the lack of knowledge about the
next spin, its that imprecise.
"It's not called gambling if the math is on your side."
Quote: EvenBob
It doesn't matter, whats relevant to me is, no math,
no probability, governs the next outcome of
random numbers.
This is actually true. It is physics (not the scientific field, but the laws of nature) that governs the outcome in case of roulette. Math just tells us quantitatively which outcomes are more or less
likely than others. But so what?
Perhaps, it's time to end the suspense and reveal the answer to the question you started your thread with - where are you going with this?
"When two people always agree one of them is unnecessary"
Quote: weaselman
This is actually true. It is physics (not the scientific field, but the laws of nature) that governs the outcome in case of roulette.
Thats why its important that the wheel is unbiased
and produces true random results.
"It's not called gambling if the math is on your side."
Quote: EvenBob
Thats why its important that the wheel is unbiased and produces true random results.
Well, yeah ... In general, given a game with some stated rules, it is important that it is played according to those rules.
But that's not "why". In fact, I don't see anything whatsoever this statement has to do specifically with either roulette or statistics or with anything at all mentioned in this thread earlier.
"When two people always agree one of them is unnecessary"
Quote: weaselman
I don't see anything whatsoever this statement has to do specifically with either roulette or statistics
Only truly independent outcomes are accurate enough
to work with.
"It's not called gambling if the math is on your side."
Quote: EvenBob
Only truly independent outcomes are accurate enough
to work with.
Truly independent outcomes contain no information(*). If they do, they are not truly independent.
Here's a thought experiment :
Say you have a way of looking at a stream of random data from a wheel and making better than average guess at the next spin, so that you can beat the house edge.
Randomness means I should be able to rearrange that stream of data (shuffle it) and there is no difference. In fact, I should be able to take every fifith sample and present it. In fact, I could take
(truly) random samples from the stream and present it as a new stream. None of those streams are any different in the data type if they are random.
Would the method still be valid? If the method exists, and the data is truly random, it should do. I've changed nothing in terms of the data.
(*) I'm not 100% sure that is true. The information they may contain is that they look random... and that's information...
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at
trente-et-quarante" - Honore de Balzac, 1829
Quote: thecesspit
Would the method still be valid? If the method exists, and the data is truly random, it should do. I've changed nothing in terms of the data.
Its valid and I've proved it. I've taken spins from
3 different wheels at the same time. 1 from wheel
A, the next one from wheel B, and so on, till I have
a string of random outcomes. The results are the
same as if I took them from a single wheel.
"It's not called gambling if the math is on your side."
Where you predicting against wheel A, B or C? I assume by results you are saying "ability to predict future spins does not change'?
So you don't care even what order the random stream comes to you either (as per my example) or which data you are given from where?
Does that mean you could get a random stream from wheel A and use that for wheel C?
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at
trente-et-quarante" - Honore de Balzac, 1829
I picked one of the wheels and stuck with
it. It worked OK, not as good as using the
actuals from just one wheel. It was kind of
spooky, actually. But it had to work, or I
would have really been at a loss for an
"It's not called gambling if the math is on your side."
Quote: TheNightfly
Ok, let's say the bag has 38 balls of which 20 are red and 18 are blue. The guy betting on red still has over a 5% advantage. What's your point? There's no way of knowing what ball he'll pull but
you'd have to be a pretty big moron to bet on blue.
How about assuming 18 black balls, 18 red balls, and 2 green balls? Then you'd
be a moron to bet on blue. (LOL)
Quote: EvenBob
Only truly independent outcomes are accurate enough
to work with.
This is a short unqualified statement, and, like most such statements, it is completely meaningless.
To actually convey some meaning, you would have to qualify what you mean by "truly", by "independent", and by "accurate", and also explain what kind of "work" you are planning on doing with these
outcomes, and what kind of result you expect to get.
For example, if you are using the word "independent" in the common statistical sense, requiring the random outcomes to be independent on each other, by "work" you mean performing some kind of
(presumably, arithmetical) operations on the numbers, representing past outcomes, and the result you are expecting is being able to predict future outcomes with better than expected frequency, then
your original statement is false.
Note, that it may be possible to predict future outcomes by analyzing past history whether they are "truly independent" (e.g., biased wheel), or not (e.g. red always hits after black). So, as long as
by "work" you mean "devising a method of predicting future results", the ability to do that work has absolutely nothing to do with results being independent (of each other).
Not also, that, contrary to common belief, modern software RNGs, with an exception of a few simplest ones, actually produce random numbers, that are statistically independent of each other. There are
various sources of entropy that can be used to achieve that - such as hard drive movements, network noise, cpu clock and temperature etc.
Quote: EvenBob
But it had to work, or I
would have really been at a loss for an
Are you saying that you are
at a loss as it is? In other words, you seem to be suggesting that you actually
have an explanation
for your "method". If so, that's just great! What are you waiting for? Let's hear it!
"When two people always agree one of them is unnecessary"
Quote: EvenBob
There are no laws or theories that apply to short term play. The casino can
lump all the players together and get a long term number. On a player
to player basis, however, they're lost. Nothing can apply to the next bet
you make in a game of random outcomes.
In case it hasn't yet been pointed out, this is incorrect. The math for short-term and long-term results in a game of independent trials is exactly the same. The multinomial distribution describes
games like roulette and it holds for any value of n (number of trials).
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
• Threads: 123
• Posts: 11449
Joined: Aug 8, 2010
For all those trying to convince Bob of the obvious, remember your success with convincing Jerry Logan. The result here will be the same.
Why is it that the less someone knows about math, the more certain they are that everyone else with more knowledge is wrong?
Like Jerry Logan and Evenbob, John Patrick is also a good example.
Quote: weaselman
qualify what you mean by "truly", by "independent",
Coming from an unbiased real wheel.
"It's not called gambling if the math is on your side."
Quote: MathExtremist
The math for short-term and long-term results in a game of independent trials is exactly the same.
So what will the next spin be, red or black? I can
figure out what the approximate distribution will
be for the next 10,000 but they won't let me bet
on that. Whats the math that accurately predicts
the next spin.
"It's not called gambling if the math is on your side."
I can tell you the approximate distribution will be for the next spin as well.
You CAN bet on the next 10,000 spins and the approximate distribution if you so wish. Just means you gotta stand there for a LOOOONG time.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at
trente-et-quarante" - Honore de Balzac, 1829
Quote: EvenBob
Coming from an unbiased real wheel.
You are wrong.
"When two people always agree one of them is unnecessary"
Quote: weaselman
You are wrong.
You don't think an unbiased wheel produces
independent outcomes?
"It's not called gambling if the math is on your side."
Quote: EvenBob
You don't think an unbiased wheel produces
independent outcomes?
I do. I also think, that a biased wheel produces independent outcomes.
But you still have not explained what it is you want to do with those outcomes to begin with, so, I don't really know what it has to do with anything.
"When two people always agree one of them is unnecessary"
Quote: EvenBob
So what will the next spin be, red or black? I can
figure out what the approximate distribution will
be for the next 10,000 but they won't let me bet
on that. Whats the math that accurately predicts
the next spin.
Yet quite counter-intuitively, your alleged ability to figure out the approximate distribution of the next 10,000 outcomes will actually be less accurate, on an absolute basis, than your prediction
of the next single outcome. How many reds will appear in the next 10,000 outcomes? The expected number is 4737, yet the chances of there being between 4736 to 4738 reds is very low. In a single spin,
the chances of there being 0 or 1 reds is 100%.
Richard A. Epstein, in "The Theory of Gambling and Statistical Logic", puts it quite elegantly:
"The law of large numbers has frequently been cited as the guarantor of an eventual head-tail balance. Actually, in colloquial form, the law proclaims that the difference between the number of heads
and the number of tails thrown may be expected to
indefinitely as the number of trials increases, although by
proportions." (p. 28, emphasis mine)
And the casino will absolutely let you bet on 10,000 spins. Since all of them are independent, it doesn't matter whether you bet sequentially or simultaneously. And again, due to independence, the
casino doesn't need to change its per-spin payout odds.
As before, the "math that accurately predicts the next spin" is the same math that is used to describe the aggregation of results. If you're looking at a specific bet and its chances of winning then
the problem decomposes into a Bernoulli trial. For "red", p(win)=18/38.
Here's the real question. Since it seems that 18/38 isn't accurate enough for you for a single trial, why should it be an acceptable figure to use as the basis for your 10,000 trial analysis?
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Quote: MathExtremist
And the casino will absolutely let you bet on 10,000 spins.
Good grief. I'd forgotten what its like having discussions with
math people. They take everything literally and always miss
the nuance.
Never mind...
"It's not called gambling if the math is on your side."
Quote: EvenBob
Good grief. I'd forgotten what its like having discussions with
math people. They take everything literally and always miss
the nuance.
Never mind...
You are the only person here, math-orientated or otherwise, who perceives any nuance at all in the question...
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
Perahps you should try the John Patrick forum.
On that forum they don't worry about the facts, the math, or proof. They instead, believe in following feelings, nuance, hunches, folklore, superstition, and mysticism.
Quote: EvenBob
Good grief. I'd forgotten what its like having discussions with
math people. They take everything literally and always miss
the nuance.
Never mind...
The nuance that's missing here is your understanding of variance. The only difference between a single random trial and a lot of independent random trials is the variance for a large aggregation is
smaller, relatively speaking, than for a single outcome. The issue you need to resolve is that you *think*, for whatever reason, that there should be some way to predict short-term results because
the long-term results seem predictable to you. That's your error, because the same math that describes the unpredictability of the short term is what also describes the relative predictability of the
longer term. If it didn't, lots of things would be broken.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
I think EvenBob should have entered Randi's million dollar challenge.(**) Being able to predict the results of a roulette wheel better than guessing by observing the spins of three roulette wheels
over time is an extraordinary claim :: that there is information in a random sequence of numbers that can be used to help predict a future number. I'm sure you can define that better, but you get the
general idea.
As you never go for independent testing (*), I wonder why you waste your time starting down these conversational routes. They always end up the same place : you claim a method that works for you,
everyone says what you claim is not possible, and you say "well it is, I've tested it, it works, believe me".
(*) I understand the logic you use behind that.
(**) You can't any more, they've taken the money for other work in the skeptical community. But maybe someone else out there is offering such a prize.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at
trente-et-quarante" - Honore de Balzac, 1829
The Randi challenge was not about probability,
it was about psychic ability. He would never let
somebody stand by a wheel, writing down the
results, thats against everything he stands for.
"It's not called gambling if the math is on your side."
Quote: EvenBob
The Randi challenge was not about probability,
it was about psychic ability. He would never let
somebody stand by a wheel, writing down the
results, thats against everything he stands for.
I'm betting he's smart enough to know that it wouldn't help you...
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
Actually, it seems it is still live and looks for :
"Webster’s Online Dictionary defines “paranormal” as “not scientifically explainable; supernatural.”
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at
trente-et-quarante" - Honore de Balzac, 1829
Quote: rdw4potus
I'm betting he's smart enough to know that it wouldn't help you...
Sigh. I've had this stupid conversation a dozen times
in the last 5 years. Randi is NOT looking for AP casino
players! He's looking for paranormal activity, mind readers,
fortune tellers. He could care less about card counting
or how to beat any casino game legitimately.
"It's not called gambling if the math is on your side."
Quote: EvenBob
He's looking for paranormal activity, mind readers,
fortune tellers. He could care less about card counting
or how to beat any casino game legitimately.
I am sure you are mistaken, try contacting him, he'll be very interested.
As it has been pointed out earlier, "paranormal" means "unexplained by science", which is exactly what your method is.
Just contact him, and explain that you are able to predict outcomes of roulette wheel spins based on the past history (from a different wheel(!)). He'll be all over you!
"When two people always agree one of them is unnecessary"
Quote: weaselman
which is exactly what your method is
What is my method, exactly. Oh yeah, probability. You
think Randi wants AP casino folks bothering him, go
right ahead. You don't have a clue.
"It's not called gambling if the math is on your side."
Quote: EvenBob
think Randi wants AP casino folks bothering him, go
right ahead.
No, I don't think he wants that. I also do not think that you are that.
I think, he'll love you. Why? See previous post.
"When two people always agree one of them is unnecessary"
Quote: weaselman
No, I don't think he wants that. I also do not think that you are that.
I think, he'll love you. Why? See previous post.
Like I said, I've dealt with the shear stupidity of
this for years. He's not going to test people who
beat casino games by using probability. Why in
the hell would he.
"It's not called gambling if the math is on your side."
Quote: EvenBob
He's not going to test people who
beat casino games by using probability. Why in
the hell would he.
Because it is paranormal. That's what he does.
"When two people always agree one of them is unnecessary" | {"url":"https://wizardofvegas.com/forum/off-topic/general/7411-law-of-large-numbers/2/","timestamp":"2024-11-12T23:33:55Z","content_type":"text/html","content_length":"179027","record_id":"<urn:uuid:6522af66-ed04-4702-a271-6cd98c6222da>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00586.warc.gz"} |
Add an Approximate Bayesian Computation (Monte-Carlo Markov-Chain) simdesign using the Majoram algorithm to a nl object — simdesign_ABCmcmc_Marjoram
Add an Approximate Bayesian Computation (Monte-Carlo Markov-Chain) simdesign using the Majoram algorithm to a nl object
Add an Approximate Bayesian Computation (Monte-Carlo Markov-Chain) simdesign using the Majoram algorithm to a nl object
postpro_function = NULL,
prior_test = NULL,
n_between_sampling = 10,
n_cluster = 1,
use_seed = FALSE,
dist_weights = NULL,
n_calibration = 10000,
tolerance_quantile = 0.01,
proposal_phi = 1,
seed_count = 0,
progress_bar = FALSE,
nl object with a defined experiment
default is NULL. Allows to provide a function that is called to post-process the output Tibble of the NetLogo simulations. The function must accept the nl object with attached results as input
argument. The function must return a one-dimensional vector of output metrics that corresponds in length and order to the specified summary_stat_target.
a vector of target values in the same order as the defined metrics of the experiment
a string expressing the constraints between model parameters. This expression will be evaluated as a logical expression, you can use all the logical operators including "<", ">", ... Each
parameter should be designated with "X1", "X2", ... in the same order as in the prior definition. Set to NULL to disable.
Number of samples along the MCMC
a positive integer equal to the desired spacing between sampled points along the MCMC.
number of cores to parallelize simulations. Due to the design of the EasyABC parallelization it is currently not possible to use this feature with cores > 1.
if TRUE, seeds will be automatically created for each new model run
a vector containing the weights to apply to the distance between the computed and the targeted statistics. These weights can be used to give more importance to a summary statistic for example.
The weights will be normalized before applying them. Set to NULL to disable.
a positive integer. This is the number of simulations performed during the calibration step. Default value is 10000.
a positive number between 0 and 1 (strictly). This is the percentage of simulations retained during the calibration step to determine the tolerance threshold to be used during the MCMC. Default
value is 0.01.
a positive number. This is a scaling factor defining the range of MCMC jumps. Default value is 1.
a positive integer, the initial seed value provided to the function model (if use_seed=TRUE). This value is incremented by 1 at each call of the function model.
logical, FALSE by default. If TRUE, ABC_mcmc will output a bar of progression with the estimated remaining computing time. Option not available with multiple cores.
number of seeds for this simulation design
This function creates a simdesign S4 class which can be added to a nl object.
Variables in the experiment variable list need to provide a numeric distribution with min, max and a shape of the distribution (qunif, qnorm, qlnorm, qexp)(e.g. list(min=1, max=4, qfun="qunif")).
The function uses the EasyABC package to set up the ABC_mcmc function. For details on the ABC_mcmc function parameters see ?EasyABC::ABC_mcmc Finally, the function reports a simdesign object.
Approximate Bayesian Computation simdesigns can only be executed using the run
dyn function instead of run
all or run
one.
Examples ch a simdesign, a nl object needs to be created first (see ?nl).
# For this example, we load a nl object from test data.
nl <- nl_lhs
# Attach the simdesign to the nl object
nl@simdesign <- simdesign_ABCmcmc_Marjoram(nl = nl,
summary_stat_target = c(100, 80),
n_rec = 100,
n_between_sampling = 10,
n_cluster = 1,
use_seed = FALSE,
n_calibration = 10000,
tolerance_quantile = 0.01,
proposal_phi = 1,
progress_bar = FALSE,
nseeds = 1)
#> Creating ABC Monte-Carlo Markov-Chain simulation design | {"url":"https://docs.ropensci.org/nlrx/reference/simdesign_ABCmcmc_Marjoram.html","timestamp":"2024-11-02T13:58:56Z","content_type":"text/html","content_length":"22907","record_id":"<urn:uuid:9e09e7cd-520e-48af-bf08-c3027638a0c6>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00405.warc.gz"} |
Mathematics for the Liberal Arts
Constant change is the defining characteristic of linear growth. Plotting coordinate pairs associated with constant change will result in a straight line, the shape of linear growth. In this section,
we will formalize a way to describe linear growth using mathematical terms and concepts. By the end of this section, you will be able to write both a recursive and explicit equations for linear
growth given starting conditions, or a constant of change. You will also be able to recognize the difference between linear and geometric growth given a graph or an equation. | {"url":"https://courses.lumenlearning.com/waymakermath4libarts/chapter/introduction-linear-and-geometric-growth/","timestamp":"2024-11-05T00:19:57Z","content_type":"text/html","content_length":"47349","record_id":"<urn:uuid:fc175394-bd36-4784-8620-2ba58daee1c7>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00353.warc.gz"} |
Can I pay someone to assist with graph neural networks and network analysis in R? | Programming Assignment Help
Can I pay someone to assist with graph neural networks and network analysis in R? I recently read a review of the R function (with the same title, but linked by a title change) describing R function
examples that I found very useful: In R, the computation of a function is simply represented by a matrix B~DR~. This is mathematically equivalent to an R matrix (and a “nearly efficient” matrix) in
which one expresses a function as a sum of independent, identically distributed sine and cosine series. I have reclassify this for computing the dimension of a matrix. click for info main concern:
What does R mean? R is generally similar to xor, except that they are multi-dimensional and therefore both represent a set of R-values. Note also that they don’t need to sum m of all of the values of
both variables, and take common eigenvalues. For example, in R: m (x.x + exp(x), -1, -1) = 1, -2 There are several ways to express a function find someone to take programming assignment a matrix, and
I’ll give a simplified example here: var = np.concatenate((1 – x) * x, 1).reshape(-1) In this example, 1 = 2 and 2 = 7 are distinct values, but 0 is 0.21. How would I express the results of this
computation as a matrix. Say I want to compute the 0.21 value of the matrix 3 which I term the input 4 element. It is because 8 is not even a number: 3 = -7. So I have to sum 3 to get 0.21. How would
I write the result of this calculation? However, it doesn’t seem that R performs constructors on [2,4], so I’ve looked up np.concatenate() and this implementation works: a = np.concatenate((x) * x +
(c, c), 2).reshape(-1) b = np.
Pay Someone To Take Test For Me
concatenate((x) * x + c, 2).reshape(-1) c = np.concatenate((x) * 2, 1).reshape(-1) x = np.zeros((a, b, c, c, a + b) * np.cos((4 * math.pi))).reshape(-1) y = np.concatenate((x) * c + (1, a + b),
c).reshape(-1) I’m still exploring but I think that R would perform constructors on all of these, but it is better to keep the underlying R function instead of just individual matrix components.
Again, this part strikes me as a slight overstatement of the type Nlmn, but I get that an individual complexity per element can be thought out without much help at the end! If that’s correct, then it
would be a feasible technique by which to define an R-function from the matrix. A: I think R’s news is a mistake. When you do x = np.zeros((a, 2, c, c, a + b), 2) * b np.concatenate((x) * x + (c, c),
2).reshape(-1) mat <- rnorm(b*c, a, b) xxn(xxn(xxn(xxn(x))) %>% truncate(mean(np.concatenate(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(xxn(Can I pay someone to assist with graph
neural networks and network analysis in R? Last week I received a report from a company working to investigate state problems or security. The problem is that computing is totally dominated by one
big big data problem, and it pay someone to do programming homework with graphs just like these. In the discussion I’ve been reading (which has worked for about 8 months) they seem to share the same
issue, only this time the graph is being investigated again so they might be on their way. Instead I’ve seen figures like this: A) https://research.
Great Teacher Introductions On The Syllabus
iop.org/\#201706-B B) https://research.iop.org/\#20170625-B All those graphs have a pretty large number of graphs but as you’ve noticed the more graph is used more data when this number changes so
that others are not dependent on that data. So it turns out there are different opinions. Are there any articles such as these getting an appearance of a concern about the Graph Neural Network (GNN)
or its development problems? Does it take a big field long to consider this new problem to have some hold when the full data coming out of your device is already available? Are there any articles
such as those of Jens B. Zimmermann and Steve White – an expert in a field with about 6 million users – to some extent to explain how graphs and visualization are being used? Are these graphs, as the
data are given, properly described as a cloud or server in one of the following ways:https://www.dropboxusercontent.com/shanlen/pg8047-01/graph-neural-network-2013-mdf.pdf If at any time when you
look at the data coming from your device you’ve been using not a graph, this type of information is not considered a particularly good enough case for study: Do you run physical graphs on a PC if you
are using an Laptop?(1)https://labs.epfl.ch/products/ceplin-products/1370039/GNN.pdf Do you run the computations on a Mac if you are running on a laptop or if you have a lot of data on the Numpy
array (2)https://www.epfl.ch/products/ceplin-products/1365063/Graph-Neural-Network-2013-mdf.pdf If I run most of these systems on my computer then these graphs seem very familiar: Facebook:https://
www.facebook.com/plans/get/152285472177626940/ Nokia:https://www.nokia.com/products/ceplin-products/14983797/Network-Neural-Network-2013-mdf.
Take My Test For Me Online
pdf Amazon:https://www.amazon.com/dp/0297149610/ Google:https://code.google.com/apis/content/commercial_video/index.html#downloads But don’t worry it is known that most of these graphs just fall out
the list of data found. In case you think that there may be a possible concern about a problem like this you might have to find out for yourself for each data type you have available and just go for
it. Since 3rd party libraries are taking this kind of information off of where you placed this type of data. You can find more detailed explanation of the graph below: These are graphs created for
the average user in this way:https://www.project.advertisment.com/product/advertis-navy-bigdata/ But remember to read the ‘graphic’, that is just the real Full Article You canCan I pay someone to
assist with graph neural networks and network analysis in R? I would like to start with this question. Some readers of this web site should remember our work and should ask the general question of
‘How to solve graphs whose noxample is not a graph?’ Before you take a look at these questions and get going, let us try and answer them clearly. What do graph neural networks (GRNs) and neural
networks (N.sup.) have to do with depth of field? Well, note that when extending it to get depth of field (for some networks), considering depth of field for each graph node of a network requires at
least two pre-defined boundaries from the data set. Here goes: Each graph node is surrounded by a small line (hence using a network if such are from the data set) around a given node within its
neighbourhood. These boundaries can be characterized by a real number expressed in pixels, or per inch in cm × 100. The data in the data set will be distributed with some look at here on the
Idoyourclass Org Reviews
In some environments, there is a lot of space between nodes and edges. There is a limit to depth. For graphs with enough nodes, the number of edges in a network will never exceed the number of nodes,
as graph nodes with the same number of edges are separate from their parent and their neighbour. A very simple type of graph is a graph with no edges and no collisions among nodes. That means, if a
node has more than one parent, all it has to do is pair it with another one. So, when the graph has he has a good point than two nodes, the two-point function depends on which parent is which, and
not is only depend on which vertex. For a dense form of graph (one node is all the data), another two-child model with some components such as the “half and half” tree is also similar to each node
being connected to its parent with a cluster (each node has more than two parents). So, in this specific graph context, what happens is what happens is that, when (for example) a node in a parent
link is 5, its neighbour links are just 5 and the other 4 nodes is 1 and thus of 5. (For a sparse graph, this concept applies to networks with at most two adjacencies. That is to say, for networks
with finite number of nodes, no one can have that number of adjacency because they are disjoint. The disjoint type of graphs can take different forms, such as an Erdős family. Another type of graph
is a bicondition Graph model, such as a biconditioned biconditioned biconditioned biconditioned biconditioned biconditioned biconditioned biconditioned biconditioned biconditioned biconditioned | {"url":"https://programmingdoc.com/can-i-pay-someone-to-assist-with-graph-neural-networks-and-network-analysis-in-r","timestamp":"2024-11-10T02:10:32Z","content_type":"text/html","content_length":"162881","record_id":"<urn:uuid:ab68bb1c-b818-4017-81a9-ca76ecd6e75f>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00846.warc.gz"} |
$\mathrm{T}\overline{\mathrm{T}}$-deformed 1d Bose gas
Yunfeng Jiang
SciPost Phys. 12, 191 (2022) · published 10 June 2022
• doi: 10.21468/SciPostPhys.12.6.191
$\mathrm{T}\overline{\mathrm{T}}$ deformation was originally proposed as an irrelevant solvable deformation for 2d relativistic quantum field theories (QFTs). The same family of deformations can also
be defined for integrable quantum spin chains which was first studied in the context of integrability in AdS/CFT. In this paper, we construct such deformations for yet another type of models, which
describe a collection of particles moving in 1d and interacting in an integrable manner. The prototype of such models is the Lieb-Liniger model. This shows that such deformations can be defined for a
very wide range of systems. We study the finite volume spectrum and thermodynamics of the $\mathrm{T}\overline{\mathrm{T}}$-deformed Lieb-Liniger model. We find that for one sign of the deformation
parameter $(\lambda<0)$, the deformed spectrum becomes complex when the volume of the system is smaller than certain critical value, signifying the break down of UV physics. For the other sign $(\
lambda>0)$, there exists an upper bound for the temperature, similar to the Hagedorn behavior of the $\mathrm{T}\overline{\mathrm{T}}$ deformed QFTs. Both behaviors can be attributed to the fact that
$\mathrm{T}\overline{\mathrm{T}}$ deformation changes the size the particles. We show that for $\lambda>0$, the deformation increases the spaces between particles which effectively increases the
volume of the system. For $\lambda<0$, $\mathrm{T}\overline{\mathrm{T}}$ deformation fattens point particles to finite size hard rods. This is similar to the observation that the action of $\mathrm
{T}\overline{\mathrm{T}}$-deformed free boson is the Nambu-Goto action, which describes bosonic strings -- also an extended object with finite size.
Author / Affiliations: mappings to Contributors and Organizations
See all Organizations. | {"url":"https://www.scipost.org/SciPostPhys.12.6.191","timestamp":"2024-11-06T09:18:15Z","content_type":"text/html","content_length":"39368","record_id":"<urn:uuid:3d16207b-a5fe-4da7-861a-fd455614bec7>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00306.warc.gz"} |
Using Delta-Sigma Can Be As Easy As ADC (Part 4)
Based on the amount of e-mail I received after Part 3 of this series (Nov. 7, 2008, p. 18; www.electronicdesign.com, ED Online 19948), many of you have already guessed that an incremental integrator
is really just a delta-sigma modulator (DSM). Well, you’re right! It actually is a continuous-time delta-sigma modulator. “Continuous time” comes from the fact that the integrator comprises
resistors, a capacitor, and an op amp.
Most commercial DSMs are made with switched capacitor technology. I decided to explain continuous time because most engineers that are unfamiliar with DSM theory are more comfortable with linear
circuits and Laplace transforms than they are with switched capacitors and z transforms. Switched capacitors will be a subject for a future column.
Bob Kruse of Control Gaging Inc. sent me one of those e-mails. “I read your article. It improved my understanding of delta-sigma ADCs. I was wondering if you could do a fourth part explaining
different order D-S ADCs,” he wrote. Sure, Bob, I would be happy to—if not in this column then surely by the next!
THE DETAILS A DSM must have a difference (delta) circuit, an integrate or accumulate (sigma) circuit, and a quantization (modulation) circuit consisting of an ADC and a DAC.
Figure 1
shows that the increment integrator is a DSM. The op-amp feedback causes the quantized output to be subtracted (?) from the input, and the result is integrated (S). The comparator on the output
serves as a singlebit ADC and is the density output. This digital output controls the reference MUX, serving as a single DAC, resulting in a quantized (M) feedback signal.
Suppose the references are ±1 V. A 0.1-V integrator output results in a 1-V quantized output. This can be seen as 0.1 V with a 0.9 quantization error. A quantized –0.3 is –1 V or the addition of a
–0.7-V quantization error. Bearing this in mind, a DSM can be considered an integrator with a quantization noise source added to its output. Figure 1 shows this model of quantization noise.
Setting the RC time constant to match that of the sample clock (RC = 1/f[S]) results in Equation 1, relating the output voltage to the input signal and quantization error:
The output is the sum of the low pass filtered input signal and the high pass filtered quantization noise. The relative amplitude of each is plotted as a function of frequency (Fig. 2). The plot
clearly shows that the quantization noise has been pushed (shaped) toward the higher end of the spectrum. At a fairly low frequency, its relative response is roughly linear (Equation 2):
The lower the frequency, the less quantization noise there is. The plot has a marker set to some frequency far less than the sample rate (f[0]). The amount of noise below that value is small, and the
signal to noise ratio (SNR) is large. Making the bandwidth smaller increases the SNR ratio. To calculate the total quantization noise, each bit of the noise left in this band must be added (Equation
The noise is accumulated using root square sum. Note that the total noise is inversely proportional to f[0]^3/2. Lower the bandwidth by two octaves (12 dB), and the noise goes down by a factor of
eight (18 dB). With the proper digital filter, it is possible to achieve 9-dB/octave performance from such a delta-sigma modulator.
The proper filter is called a decimator. The word “decimate” is Roman in origin, meaning to reduce by a tenth. Decimation was an extreme punishment given to the Roman Legions when they had upset
their leaders. The men would be divided into groups of 10, decide which member of their group would be executed, and carry out the execution. The survivors would then be put on a diet of barley and
water for a month. These remaining 90% came away highly motivated to meet the future expectations of their leaders. (In these more enlightened times, we still carry on this tradition. We call it
This decimator is a digital finite impulse response (FIR) filter that takes the input at fS and provides a filtered output at f0. The ratio of these two frequencies is called the oversample ratio
(OSR) or decimation value. The bandwidth limit in Figure 2 is for an OSR of 64. The specifics of the filter are not important for this discussion. Digital decimator filter design is a subject better
suited for a future column.
Now, 9-dB/octave certainly is better than the 6 db/octave produced by the incremental ADC. An incremental ADC with a 1-Msample/s clock and 16-bit resolution has an output rate of 15 samples/s. Take
the same DS modulator, and with the proper decimator, you get 16 bits of resolution at 488 samples/s. This is an improvement—not great, but definitely an improvement.
My next column will explain multiple-stage delta-sigma modulators and show how to get resolution enhancements of 15, 21, and 27 dB per octave and more.
DAVE VAN ESS has a BSEE from the University of Calif., Berkeley. His column runs every quarter, with additional articles online.
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30 Times Table [Free 30 Multiplication Table] Printable Chart
30 Times Table-The 30 times table is an important multiplication table for students to learn. It can be used to help solve problems in many different areas of math. Learning the 30 times table can
help students with their basic multiplication facts, division facts, and even fractions.
A Free thirty Multiplication Table Chart PDF is a helpful tool because it provides a visual representation of the multiplication facts. It is also easy to use.
A 30-times table chart is an essential tool for any student who is trying to learn multiplication facts. It is especially helpful for visual learners. With a little practice, anyone can learn to use
this chart and improve their multiplication skills.
Printable 30 Times Table
If you’re looking for a printable 30 times table, you’ve come to the right place. This times table is perfect for helping students learn their multiplication facts. It can be used as a reference or
as a colouring sheet. There are many ways to use this times table, so get creative and have fun!
The table can be used as a reference for students when they are first learning to multiply, and it can also be used as a practice tool. Students can use the table to help them memorize multiplication
facts and practice their skills.
30 Multiplication Table
Learning multiplication tables is an essential part of a child’s education. By the time they reach third grade, most students have memorized the basic multiplication facts up to 10. However, many
students struggle with the higher numbers. The good news is that there are plenty of resources available to help them master the 30 times table.
One helpful resource in learning this table is a printable multiplication chart. Another great way to practice is by using online games and quizzes. There are many sites that offer free
multiplication games, and these can be a fun way for kids to learn while they’re having fun. Quizzes are also a great way to test knowledge and see where any gaps might be.
30 Multiplication Chart Printable
You can use a 30 multiplication chart printable to help you memorize the timetables. This is a helpful tool because it allows you to see the patterns in the numbers. It can be difficult to remember
all of the times tables, but if you use a multiplication chart, it will be much easier.
The 30 multiplication chart is also helpful when you are trying to do mental math. If you know the times’ tables, then you can quickly calculate anything in your head. This is a valuable skill to
have because it can help you in everyday life. For example, if someone asks you what 7 times 8 is, you can quickly say 56 without having to stop and think about it.
If you don’t have a 30 multiplication chart printable, there are other ways to memorize the times’ tables.
Free thirty Times Table PDF
Printable 30 Times Table PDFs are available online for free. These times tables can be used to help students learn and memorize multiplication facts. The PDFs can be printed out and used as a study
guide or reference sheet.
Using printable times tables is a great way for students to practice their multiplication skills. The 30 times table is especially useful for students who are just starting to learn multiplication
Multiplication is a critical math skill for students to learn, and the earlier they learn it, the better. The Printable Number 30 Multiplication Table can be a helpful tool for teaching
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The Printable Number 30 Multiplication Table is a valuable resource for teachers and parents alike. It is a simple, yet effective way to help students learn one of the most important math skills.
One of the most useful tools for learning multiplication is the 30-times table chart. This chart can be used to help memorize basic multiplication facts. It is also available in many math textbooks. | {"url":"https://multiplicationtable.org/30-times-table/","timestamp":"2024-11-02T00:12:28Z","content_type":"text/html","content_length":"165571","record_id":"<urn:uuid:218ff974-9507-47d4-92cb-46d6518a1357>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00111.warc.gz"} |
Posts tagged with probability distribution questions
MyMathLab Homework 5.5 Answers
A drug tester claims that a drug cures a rare skin disease 78% of the time. The claim is checked by testing the drug on 100 patients. If at least 73 patients are cured, the claim will be accepted.
Find the probability that the claim will be rejected assuming that the manufacturer's claim is true. Use the normal distribution to approximate the binomial distribution if possible.
Solving the question
The first step for solving this question is to check that the conditions for normal to binomial approximation are met.,
np >= 5, and nq >= 5
In this case the conditions are met because np; 1000.78 = 78 , & nq = 0.22100 = 22
q = 1-p; = 1-0.78 = 0.22
To use the normal approximation , we need to calculate the mean and standard deviation based on the data given.
np = 1000.78 = 78, and standard deviation sqrt(npq) = sqrt(0.780.22*100) = 4.142
We are interested in the probability that 1-p(X < 73)
We'll use continuity correction to improve the accuracy of our calculations.
Therefore, we'll find p(X < 72.5)
The probability will be rejected if the value is less than 73.
P(X < 73) = (72.5 - 78)/4.142 = -1.33
P(X < -1.33) = 0.0921.,
Note that you can also use Excel to answer this question,
the function; =NORM.DIST(72.5,78,4.142,TRUE) is very helpful when dealing with such kinds of questions.
You are free to contact MyMathLab Homework Help in case you need help in answering similar questions. | {"url":"https://www.mymathlabhomeworkhelp.com/mymathlabanswers/tag/probability-distribution-questions/","timestamp":"2024-11-12T20:01:00Z","content_type":"text/html","content_length":"22102","record_id":"<urn:uuid:95fd243f-434d-47b5-a6ab-a7b961afd636>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00088.warc.gz"} |
Neural Network Demystified Part I – Building Blocks and Activation Functions
Building Blocks of a Neural Network
Although the original intention of designing a neural network was to simulate the human brain system to solve general learning problems in a principled way, in practice neural network algorithms
work in much simpler way than a human brain. Current neural networks can be compared to statistical models having the higher level purpose of mapping inputs to their corresponding outputs. By the
means of lots of complex matrix computations going on under the hood, they try to find correlations by approximating an unknown function f(x) = y between any input x and any output y. In the process
of learning, they drive to discover the most accurate function f which is also the best approximation of transforming x into y. These networks can be both linear or non-linear. Non-linear networks
can approximate any function with an arbitrary amount of error. For this reason, they are also called “universal approximator”. In this tutorial, let us learn about the building blocks of a
non-linear shallow neural network followed by the architecture of a standard deep neural network.
An artificial neuron (derived from the basic computational unit of the brain called neuron), also referred to as a perceptron is a mathematical function that takes one or more inputs and returns
their weighted sum as output. Weighted sum is passed through an activation function to transform the otherwise linear function to a non-linear function. This is the reason why neural networks perform
well in deriving very complex non-linear functions. Basically, weights are values associated with each input and are parameters to be learned by the network from input data in order to accurately map
them to desired output.
Fig 1. Illustration of a neuron as a function
In Fig. 1, input \( X \) is a vector containing three elements \( x_{1} \), \( x_{2} \) and \( x_{3} \). Each element have weights \( w_{1} \), \( w_{2} \) and \( w_{3} \) associated to
them respectively. In addition to weights, there is also a bias term \( b \) which helps the function better fit the data. The neuron \( f(x) \) calculates the weighted sum using Equation 1. \(
f(x) \) is also defined by \( z \) interchangeably.
\[ f(x) = z = w_{1}\cdot x_{1} + w_{2}\cdot x_{2} + w_{3}\cdot x_{3} + b \;\;\;(1) \]
The resultant output is then passed to an activation function A(f(x)) which returns the final output of the neuron.
An activation function is a function applied on a neuron to instill some kind of non-linear properties in the network. A relationship is considered as linear if a change in a variable initiates a
constant change in the consecutive latter variables. On the other hand, a non-linear relationship between variables is established when a change in one variable does not necessarily ignites a chain
of constant changes in the consecutive latter variables, although they can still impact each other in an unpredictable or irregular manner. Without the injection of some non-linearity, a large chain
of linear algebraic equation will eventually collapse into a single equation after simplification. Therefore, the significant capacity of the neural network to approximate any convex and non-convex
function is directly the result of the non-linear activation functions. In a small visual example in Fig. 2, we can see that when data points make non-linear patterns, a linear line can not t them
accurately, thus it misses some data points. However, a non-linear function captures the difficult pattern efficiently and fits all data points.
Fig 2. Best fit linear and non-linear models
Every activation function takes the output of f(x) as a vector as input and performs a pre-defined element-wise operation on it. Among many activation functions, three of them are most commonly used.
The sigmoid non-linearity function has the following mathematical form,
\[ sigmoid(f(x)) = \frac{\mathrm{1} }{\mathrm{1} + e^{-f(x) }}\;\;(2) \]
It takes a real value as input and squashes it between 0 and 1 so that the range does not become too large. However, for large values of neuron activation at either of the positive or negative
x-axis, the network gets saturated and the gradients at these regions get very close to zero (i.e. no significant weight update in these regions), causing “vanishing gradient” problem. Fig. 3 shows
the sigmoid function graph that converges in the positive and negative domain as values get bigger.
Fig 3. A sigmoid activation function
Hyperbolic Tangent
The hyperbolic tangent or tanh non-linearity function has the following mathematical form,
\[ tanh(f(x)) = \frac{e^{f(x)}-e^{-f(x)}}{e^{f(x)}+e^{-f(x)}}\;\;(3) \]
It takes a real value as input and squashes it between -1 and 1. However, it suffers from the “vanishing gradient” problem in the positive and negative domain like sigmoid function for similar
reason. Fig. 4 shows the graph for hyperbolic tangent function.
Fig 4. A tanh activation function
Rectied Linear Unit (ReLU)
The ReLU has the following mathematical form,
\[ ReLU(f(x)) = max(0,f(x))\;\;(4) \]
ReLU has gained huge popularity as an activation function due to its edge over sigmoid and tanh in couple of ways. It takes a real value as input and squashes it between 0 and +infinity. Because it
does not saturate in the positive domain, it avoids the vanishing gradient problem. As a result, it also accelerates the convergence of stochastic gradient descent. Another big advantage of ReLU is
that it involves computationally cheaper operations compared to the expensive exponentials in sigmoid and tanh. However, ReLU saturates in the negative domain which allows it to disregard all the
negative values. Thus it may not be suitable to capture patterns in all different datasets and architectures. One solution to this problem is LeakyReLU.
Instead of zeroing out the negative activation, LeakyReLU applies a small negative slope of 0.01 or so. Therefore, LeakyReLU has the following mathematical form, where α is the slope.
\[ LeakyReLU\left ( f\left(x \right ) \right ) = \left\{\begin{matrix} max(0, x)\;\;for\;\;f(x) > 0\\ αx\;\; for\;\;f(x)<0 \end{matrix}\right. \;\;(5) \]
Fig. 5 shows ReLU graph where slope keeps increasing only in the positive domain.
Fig 5. A ReLU activation function
Fig. 6 shows LeakyReLU graph where slope keeps increasing in the positive domain but also allows gradients to flow by a restricted amount in the negative domain. Here, α is 0.1.
Fig 6. A LeakyReLU activation function
Next part is Neural Network Demystified Part ll – Deep Neural Network.
I hope you have found this article useful. Please feel free to comment below about any questions, concerns or doubts you have. Also, your feedback on how to improve this blog and its contents will be
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\DeclareMathOperator{\tr}{tr} \newcommand{\pdata}{p_{\text{data}}(\bx)} \newcommand{\st}{\mathbf{s}_\mathbf{\theta}} \newcommand{\xt}{\tilde{\bx}} \newcommand{\stx}{\mathbf{s}_\mathbf{\theta}(\bx)} \
newcommand{\sdx}{\mathbf{s}_\text{data}(\bx)} \newcommand{\stxt}{\mathbf{s}_\mathbf{\theta}(\xt, \sigma)} \newcommand{\pv}{p_{\bv}} \newcommand{\score}{\nabla_\bx \log \pdata} \newcommand{\bov}{\bar
{\beta}}\) Joint work with Owen Wang.
There has recently been a flurry of work in score-based diffusion models as part of the broader area of generative models. This is due to the recent success of such score-based methods, which has
achieved results comparable to the state-of-the-art of generative adversarial networks (GANs).
Past techniques in generative modeling have either relied on the approximation of the partition function of the probability density, or the combination of an implicit network representation of the
probability density and adversarial training. The former suffers from having to either constrain the model to make the partition function tractable, or otherwise relies on approximations with
surrogate losses that may be inaccurate, and the latter suffers from training instability and mode collapse.
Score-based diffusion models try to address the cons of both approaches, and instead, use score-matching to learn a model of the gradient of the log of the probability density function. This allows
it to avoid computing the partition function completely.
One of the first such approaches that rely on using score-matching to perform generative modeling does so by generating new samples via Langevin dynamics (Song & Ermon, 2019). A key observation is
that naively applying score-matching is that the model of score function will be inaccurate in areas of low density with respect to the data distribution, which results in improper Langevin dynamics
in low-density areas. The solution that was proposed is the injection of noise into the data, which provides additional training signal and increases the dimensionality of the data.
The next major step introduced in (Song et al., 2021) is to perturb the data using a diffusion process which is a form of a stochastic differential equation (SDEs). The SDE is then reversed using
annealed Langevin dynamics in order to recover the generative process, where the reversal process makes use of score matching.
Other recent refinements that have been proposed include re-casting the objective as a Schrödinger bridge problem, which is an entropy-regularized optimal transport problem. The advantage of this
approach is that it allows for fewer diffusion steps to be taken during the generative process.
Survey of Results
We will be primarily focusing on the paper Generative Modeling by Estimating Gradients of the Data Distribution (Song & Ermon, 2019).
In this section, we provide the necessary background, provide derivations for important results, and explain the key ideas of score matching for diffusion models as proposed in the papers.
Motivation for Score Matching
Limitations of Likelihood-Based Approaches
Score matching is motivated by the limitations of likelihood-based methods. In likelihood-based methods, we use a parameterized model \(f_\theta(\bx) \in \mathbb{R}\) and attempt it to recover the
parameters \(\theta\) that best explains the observed data. For instance, in energy-based models, the probability mass function \(p_\theta(\bx)\) would be given as p_\theta(\bx) = \frac{\exp(-f_\
theta(\bx))}{Z_\theta}, where \(Z_\theta\) is the normalizing constant that causes the distribution to integrate to 1, i.e Z_\theta = \int \exp(-f_\theta(\bx)) \, d \bx. The goal then is to maximize
the log likelihood of the observed data \(\{\bx_i\}_i^N\), given by \max_\theta \sum_{i=1}^N \log p_\theta (\bx_i).
It is often computationally intractable to compute the partition function \(Z_\theta\) unless there are restrictions on what the model can be, since there are usually at least an exponential number
of possible configurations. Examples of models where the partition function can be efficiently computed include causal convolutions in autoregressive models, and invertible networks in normalizing
flow models However, such architecture restrictions are very undesirable as they limit the expressiveness of the models.
A likelihood-based approach that tries to avoid computing the partition function is variational inference. In variational inference, we use the Evidence Lower Bound (ELBO) as a surrogate objective,
where the approximation error is the smallest Kullback-Leibler divergence between the true distribution and a distribution that can be parameterized by our model.
Limitations of Adversarial-Based Approaches
Adversarial-based approaches, like generative adversarial networks (GANs), have been shown to suffer from both instability in training and mode collapse.
Training GANs can be viewed as finding a Nash equilibrium for a two-player non-cooperative game between the discriminator and the generator. Finding a Nash equilibrium is PPAD-complete which is
computationally intractable, and therefore methods like gradient-based optimization techniques are used instead. However, the highly non-convex and high-dimensional optimization landscape means that
small perturbations in the parameters of either player can change the cost function of the other player, which results in non-convergence.
Another problem with training GANs is that when either the generator or discriminator becomes significantly better than the other, then the learning signal for the other player becomes very weak. For
generators, this is when the discriminator is always able to tell it apart. For discriminators, this is when the generator performs so well it can hardly do better than random guessing.
Finally, a common failure mode of GANs is mode collapse, where the generator only learns to produce a set of very similar outputs from a single mode instead of from all the modes. This is due to the
non-convexity of the optimization landscape.
Score Matching
Score matching is a non-likelihood-based method to perform sampling on an unknown data distribution, and seeks to address many of the limitations of likelihood-based methods and adversarial methods.
This is achieved by learning the score of the probability density function, formally defined below:
Definition (Score Function)
The score function of a distribution \( \pdata \) is given by \begin{align*} f(\bx) = \nabla_\bx \log \pdata. \end{align*}
In practice, we try to learn the score function using a neural network \(\stx\) parameterized by \(\theta\).
The objective of score matching is to minimize the Fisher Divergence between the score function and the score network:
\[ \label{eq:score-matching-target-fisher-div} \argmin_\theta \frac{1}{2} \mathbb{E}_{\pdata} \left[ \| \stx - \nabla_\bx \log \pdata \|_2^2 \right]. \]
However, the main problem here is that we do not know \(\nabla_\bx \log \pdata\), since it depends on knowing what \(\pdata\) is.
(Hyvärinen, 2005) showed that Equation \ref{eq:score-matching-target-fisher-div} is equivalent to Equation \ref{eq:score-matching-target} below:
\[ \label{eq:score-matching-target} \argmin_\theta \frac{1}{2} \mathbb{E}_{\pdata} \left[ \tr \left( \nabla_\bx \stx \right) + \frac{1}{2} \| \stx \|_2^2 \right]. \]
We can now compute this using Monte Carlo methods by sampling from \(\pdata\), since it only depends on knowing \(\stx\).
Sliced Score Matching
It is computationally difficult to compute the trace term \(\tr \left( \nabla_\bx \stx \right)\) in Equation \ref{eq:score-matching-target} when \(\bx\) is high-dimensional. This motivates another
alternative cheaper approach for score matching, called sliced score matching (Song et al., 2019).
In sliced score matching, we sample random vectors from some distribution \(\pv\) (such as the multivariate standard Gaussian) in order to optimize an analog of the Fisher Divergence:
\[ L(\btheta, \pv) = \frac{1}{2} \mathbb{E}_{\pv} \mathbb{E}_{\pdata} \left[ (\bv^T \stx - \bv^T \sdx)^2 \right] \]
We observe that
\[ L(\btheta; \pv) &= \frac{1}{2} \mathbb{E}_{\pv} \mathbb{E}_{\pdata} \left[ (\bv^T \stx - \bv^T \sdx)^2 \right]\\ &=\frac{1}{2} \mathbb{E}_{\pv} \mathbb{E}_{\pdata} \left[ (\bv^T \stx )^2 + (\bv^T
\sdx)^2 - 2(\bv^T \stx )(\bv^T \sdx) \right]\\ &= \mathbb{E}_{\pv} \mathbb{E}_{\pdata} \left[ \frac{1}{2}(\bv^T \stx )^2 - (\bv^T \stx )(\bv^T \sdx) \right] + C\\ \]
where the \(\sdx\) term is absorbed into \(C\) as it doesn’t depend on \(\theta\). Now note
\[ & -\mathbb{E}_{\pv} \mathbb{E}_{\pdata}\left[(\bv^T \stx )(\bv^T \sdx) \right] \\ =& -\mathbb{E}_{\pv} \int \left[(\bv^T \stx )(\bv^T \sdx) \pdata d\bx\right]\\ =& -\mathbb{E}_{\pv} \left[\int(\bv
^T \stx )(\bv^T\nabla_{\bx}\log \pdata)\pdata d\bx\right] \\ =& -\mathbb{E}_{\pv} \left[\int(\bv^T \stx )(\bv^T\nabla_{\bx}\pdata)d\bx\right] \\ =& -\mathbb{E}_{\pv} \left[\int(\bv^T \stx )(\bv^T\
nabla_{\bx}\pdata)d\bx\right] \\ =& -\mathbb{E}_{\pv} \left[\sum_{i}\int(\bv^T \stx )(v_i\frac{\partial \pdata}{\partial x_i})d\bx\right] \\ =& \mathbb{E}_{\pv} \left[\int \bv^T\stx\bv \cdot \pdata d
\bx\right] \\ =& \mathbb{E}_{\pv}\mathbb{E}_{\pdata}\left[\bv^T\stx\bv \right] \]
where line 16 is obtained by applying multivariate integration by parts. This finally yields the equivalent objective:
\[ J(\btheta; \pv) &= \mathbb{E}_{\pv} \mathbb{E}_{\pdata} \left[ \bv^T \nabla_\bx \stx \bv + \frac{1}{2} \| \stx \|_2^2 \right] \]
which no longer has a dependence on the unknown \(\nabla_{bx}\sdx\). This leads to the unbiased estimator:
\[ \hat J_{N,M}(\btheta; \pv) &=\frac{1}{N}\frac{1}{M}\sum_{i= 1}^N\sum_{j=1}^M \left[\bv_{ij}^T\nabla_{\bx}\mathbf{s}_\mathbf{\btheta}(\bx_i)\bv_{ij} + \frac{1}{2} \|\mathbf{s}_\mathbf{\btheta}(\
bx_i)\|_2^2\right] \]
where for each data point \(\bx_i\) we draw \(M\) projection vectors from \(\pv\).
(Song et al., 2019) showed that under some regularity conditions, sliced score matching is an asymptotically consistent estimator:
\[ \hat \btheta_{N,M} \overset{p}{\to} \btheta^* \text { as } \mathbb{N} \to \infty \]
\[ \btheta^* &= \underset{\btheta}{\text{argmin }} J(\btheta; \pv), \\ \hat \btheta_{N,M} &= \underset{\btheta}{\text{argmin }} \hat J_{N,M}(\btheta; \pv). \]
Sliced score matching is computationally more efficient, since it now only involves Hessian-vector products, and continues to work well in high dimensions.
Sampling with Langevin Dynamics
Once we have trained a score network, we can sample from the data distribution via Langevin dynamics. Langevin dynamics is a Markov Chain Monte Carlo method of sampling from a stationary
distribution, where we can efficiently take gradients with respect to the probability of our samples \(\bx\). We satisfy this criteria since we have the trained score network.
In Langevin dynamics, we start from some initial point \(\bx_0 \sim \bpi(\bx)\) sampled from some prior distribution \(\bpi\), and then iteratively obtain updated points based on the following
recurrence: \xt_t = \xt_{t-1} + \frac{\epsilon}{2} \nabla_\bx \log p(\xt_{t-1}) + \sqrt{\epsilon} \bz_t, where \(\bz_t \sim \mathcal{N}(0, I)\). The addition of the Gaussian noise is required, or
otherwise the process simply converges to the nearest mode instead of converging to a stationary distribution.
It can be shown that as \(\epsilon \to 0\) and \(T \to \infty\), we have that the distribution of the process \(\xt_T\) converges to \(\pdata\) (Welling & Teh, 2011).
Challenges of Langevin Dynamics
Langevin dynamics does not perform well with multi-modal distributions with poor conductance, since it will tend to stay in a single mode, which causes long mixing times. This is particularly a
problem when the modes have disjoint supports, since there is very weak gradient information in the region where there is no support.
Challenges of Score Matching for Generative Modeling
The Manifold Hypothesis
The manifold hypothesis postulates that real-world data often lies in a low-dimensional manifold embedded in a high-dimensional space. This has been empirically observed in many datasets.
This poses problems for score matching. The first problem that the manifold hypothesis poses is that the score \(\score\) becomes undefined if \(\bx\) actually just lies in a low-dimensional
manifold. The second problem is that the estimator in Equation \ref{eq:score-matching-target} is only consistent when the support of \(\pdata\) is that of the whole space.
In order to increase the dimension of the data to match that of the ambient space, (Hyvärinen, 2005) proposed injecting small amounts of Gaussian noise into the data, such that now the data
distribution has full support. As long as the perturbation is sufficiently small (\(\mathcal{N}(0, 0.0001)\) was used in their paper), it is almost indistinguishable to humans.
Low Data Density Regions
The other problem with score matching is that it may not be able to learn the score function in areas of low data density. This is due to the lack of samples drawn from these regions, resulting in
the Monte Carlo estimation to have high variance.
Noise Conditional Score Networks (NCSN)
The challenges mentioned in the previous sections are addressed by Noise Conditional Score Networks (NCSN).
In NCSN, we define a geometric sequence of \(L\) noise levels \({\left\{ \sigma_i \right\}}_{i=1}^L\), with the property that \(\frac{\sigma_1}{\sigma_2} = \frac{\sigma_{L-1}}{\sigma_L} > 1\). Each
of these noise levels correspond to Gaussian noise that will be added to perturb the data distribution, i.e \(q_{\sigma_i} \sim \pdata + \mathcal{N}(0, \sigma_i)\).
We augment the score network to also take the noise level \(\sigma\) into account, which is called the NCSN \(\stxt\). The goal of NCSN is then to estimate the score conditioned on the noise level.
Once we have a trained NCSN, we use a similar apporach as simulated annealing in Langevin sampling, where we begin with a large noise level in order to cross the different modes easily, before
gradually annealing down the noise to achieve convergence.
The denoising score matching objective for each noise level \(\sigma_i\) is given as
\[ \ell(\theta; \sigma) \triangleq \frac{1}{2} \mathbb{E}_{\pdata} \mathbb{E}_{\xt \sim \mathcal{N}(\bx, \sigma^2 I)} \left[ \left\| \stxt + \frac{\xt - \bx}{\sigma^2} \right\|_2^2 \right], \]
and the unified objective for denoising across all levels is given as
\[ \mathcal{L}\left(\theta; \left\{ \sigma_i\right\}_{i=1}^L \right) \triangleq \frac{1}{L} \sum_{i=1}^L \lambda(\sigma_i) \ell(\theta; \sigma_i). \]
Score-Based Generative Modeling through Stochastic Differential Equations (Song et al., 2021)
We can extend the idea of having a finite number of noise scales to having an infinite continuous number of such noise scales by modeling the process as a diffusion process, which can be formalized
as a stochastic differential equation (SDE). Such an SDE is given in the following form:
\[ d\bx = \boldf(\bx, t) \, dt + g(t) \, d\bw. \]
Here, \(\boldf\) represents the drift coefficient, which models the deterministic part of the SDE, and determines the rate at which the process \(d\bx\) is expected to change over time on average. \
(g(t)\) is called the diffusion coefficient, which represents the random part of the SDE, and determines the magnitude of the noising process over time. Finally, \(\bw\) is Brownian motion. Thus \(g
(t) \, d \bw\) represents the noising process.
We want our diffusion process to be such that \(\bx(0) \sim p_0\) is the original data distribution, and \(\bx(T) \sim p_T\) is the Gaussian noise distribution that is independent of \(p_0\). Then
since every SDE has a corresponding reverse SDE, we can start from the final noise distribution and run the reverse-time SDE in order to recover a sample from \(p_0\), given by the following process:
\[ d \bx = [\boldf (\bx, t) - g(t)^2 \nabla_{\bx} \log_{p_t} (\bx) ] \, dt + g(t) \,d \overline{w}, \]
where \(\overline{w}\) is Brownian motion that flows backwards in time from \(T\) to \(0\), and \(dt\) is an infinitesimal negative timestep.
The objective function for score matching for the SDE is then given by
\[ \argmin_{\theta} \mathbb{E}_t \left[ \lambda (t) \mathbb{E}_{\bx(0)} \mathbb{E}_{\bx (t) \mid \bx(0)} \left[ \| \bs_\theta (\bx(t), t) - \nabla_{\bx(t)} \log p_{0t}(\bx (t) \mid \bx(0)) \|_2^2 \
right] \right]. \]
Score-based Generative Modeling Techniques
(Song et al., 2021) covers two score-based generative models that uses SDEs to perform generative modeling. The first is called score matching with Langevin dynamics (SMLD), which performs score
estimation at different noise scales and then performs sampling using Langevin dynamics with decreasing noise scales. The second is denoising diffusion probabilistic modeling (DDPM)
(Ho et al., 2020), which uses a parameterized Markov chain that is trained with a re-weighted variant of the evidence lower bound (ELBO), which is an instance of variational inference. The Markov
chain is trained to reverse the noise diffusion process, which then allows sampling from the chain using standard Markov Chain Monte Carlo techniques.
(Song et al., 2021) shows that SMLD and DDPM actually corresponds to discretizations of the Variance Exploding (VE) and Variance Preserving (VP) SDEs, which is the focus of the next two section. We
believe expanding on this will be illuminating as it highlights the connections between SDEs and the discretized approaches that are used in practice.
SMLD As Discretization of Variance Exploding (VE) SDE
Recall that we use a geometric sequence of \(L\) noise levels \({\left\{ \sigma_i \right\}}_{i=1}^L\). that is added to the data distribution
We can recursively define the distribution for each noise level \(i\) by incrementally adding noise:
\[ \bx_i = \bx_{i-1} + \sqrt{\sigma_i^2 - \sigma_{i-1}^2} \bz_{i-1}, \qquad \qquad i = 1, \dots, L, \]
where \(\bz_{i-1} \sim \mathcal{N}(\mathbf{0}, \bI)\), and \(\sigma_0 = 0\) so \(\bx_0 \sim \pdata\).
If we view the noise levels as gradually changing in time, then the continuous time limit of the process is given by the following SDE: \bx(t + \Delta t) = \bx(t) + \sqrt{\sigma^2 (t + \Delta t ) - \
sigma^2 (t)} \bz(t) \approx \bx(t) + \sqrt{\frac{d [\sigma^2 (t)]}{dt} \Delta t } \bz (t), where the approximation holds when \(\Delta t \ll 1\). If we take \(\Delta t \to 0\), we recover the VE SDE:
d \bx = \sqrt{\frac{d [\sigma^2 (t)]}{dt} } d \bw, which causes the variance of \(d \bx(t)\) to go to infinity as \(t \to \infty\) due to its geometric growth, hence its name.
DDPM As Discretization of Variance Preserving (VP) SDE
Similarly, the Markov chain of the perturbation kernel of DDPM is given by \bx_i = \sqrt{1 - \beta_i} \bx_{i-1} + \sqrt{\beta_i} \bz_{i-1}, \qquad i = 1, \cdots, L, where \(\left\{ \beta_i \right\}_
{i=1}^L\) are the noise scales, and if we take \(L \to \infty\) with scaled noise scales \(\overline{\beta_i} = N \beta_i\), we get \bx_i = \sqrt{1 - \frac{\bov_i}{N} } \bx_{i-1} + \sqrt{ \frac{\
bov_i}{N} } \bz_{i-1}, \qquad i = 1, \cdots, L. Now taking limits with \(L \to \infty\), we get \bx(t + \Delta t) \approx \bx(t) - \frac{1}{2} \beta(t) \Delta t \bx(t) + \sqrt{\beta(t) \Delta t} \bz
(t), where the approximation comes from the second degree Taylor expansion of \(\sqrt{1 - \beta(t + \Delta t) \Delta t}\). Then taking the limit of \(\Delta t \to 0\), we obtain the VP SDE d \bx = -
\frac{1}{2} \beta(t) \bx dt + \sqrt{\beta(t)} d \bw. This process thus has bounded variance since \(\beta_i\) is bounded.
We conduct the following preliminary series of experiments, based on released work by (Song & Ermon, 2019).
Investigating the manifold hypothesis
Figure 1. Comparison between true data density and sampling
In this experiment, we have plotted the true data density of a toy distribution along with samples drawn in three ways. The i.i.d samples are drawn directly from the underlying distribution and we
can see that more samples are drawn in the area of high data density. However, applying Langevin dynamics without annealing, we see that there is an almost equal number of points in the top left and
bottom right corners. This is evidence that the sampling method doesn’t conform to the true distribution. Finally, by injecting and decreasing the amount of noise through the annealing process, we
can recover a representative sample of the distribution.
Importance of annealing when sampling via Langevin Dynamics
To better visualize the effects of annealing when sampling via Langevin Dynamics, we generated images from a model trained on the CelebA dataset. We first tried applying Langevin Dynamics with a
fixed noise and then used annealing to gradually decrease the noise.
Figure 2. Langevin Dynamics with no annealing (top) and annealing (bottom)
Figure 2 shows that the results with annealing are significantly clearer and more varied, matching the performance of GANs in 2019.
Figure 3. Closer comparison of no annealing (left) and annealing (right)
We notice that the image generated without annealing manages to produce the structure of a human face but fails to capture finer details such as the hair, and the surrounding backdrop. There is also
little variation in color between different samples. This is in agreement with our theory that without annealing, Langevin dynamics cannot properly explore regions of lower data density.
Effect of noise parameters for annealed Langevin Dynamics
We also investigated the effect of changing the lowest noise standard deviation \(\sigma\) while keeping the number of different noises injected fixed at \(10\). The 10 noise values are determined by
an interpolation in log scale.
Figure 4. Left to right: \( \sigma_{\text{end}} = \{0.1, 0.01, 0.001\} \)
Our experiment shows that the effect of starting, ending, and the interval between noise values has a significant effect on the convergence of annealed Langevin sampling.
Discussion and Future Work
Having completed a survey of score-based diffusion models, and having run some experiments on them, we now turn our attention to discussing the pros and cons of this approach.
As mentioned previously in this paper, the main draw of score-based diffusion models is that it has shown to be capable of generating impressive high-quality samples that is on-par with the
state-of-the-art with GANs. We hence focus on its limitations and how they might be overcome, drawing from work in (Cao et al., 2022).
Computation Cost
A common refrain of score-based diffusion model is the high computational complexity in both training and sampling. This is because it requires thousands of small diffusion steps in order to ensure
that the forward and reverse SDEs hold in their approximations (Zheng et al., 2022). If the diffusion steps are too large, then the Gaussian noise assumption may not hold, resulting in poor score
estimates. This makes it significantly more expensive than other generative methods like GANs and VAEs. To this end, there are some directions being explored to improve its computation cost.
The first technique seeks to reduce the number of sampling steps required by a method known as knowledge distillation (Lopes et al., 2017). In knowledge distillation, knowledge is transferred from a
larger and more complex model (called the teacher), to one that is smaller and simpler (called the student). This technique has found success in other domains such as image classification, and has
also been shown to result in improvements in diffusion models (Salimans & Ho, 2022). It would be interesting to see how far we can take this optimization.
Another technique known as truncated diffusion probabilistic modeling (TDPM) (Zheng et al., 2022). In this approach, instead of considering the diffusion process until it becomes pure noise, the
process is stopped once it reaches a hidden noisy-data distribution that can be learnt by an auto-encoder by adversarial training. Then in order to produce samples, a sample is first drawn from the
learnt noisy-data distribution, before being passed through the reverse-SDE diffusion steps.
It also suffers from poor explainability and interpretability, but this is a common problem across other generative models.
(Song et al., 2021) also notes that it is currently difficult to tune the myriad of hyperparameters introduced by the choice of noise levels and specific samplers chosen, and new methods to
automatically select and tune these hyperparameters would make score-based diffusion models more easily deployable in practice.
Modality Diversity
Diffusion models have mostly only seen applications for generating image data, and its potential for generating other data modalities has not been as thoroughly investigated. (Austin et al., 2021)
introduces Discrete Denoising Diffusion Probabilistic Models (D3PMs), which develops a diffusion process for corrupting text data into noise. It would be interesting to see how well diffusion models
can be stretched to perform compared to state-of-the-art transformer models in text generation.
Dimensionality Reduction
Dimensionality reduction is another technique that can be used to speed up training and sampling speeds of diffusion models. Diffusion models are typically trained directly in data space. (Vahdat et
al., 2021) instead proposes for them to be trained in latent space, which results in dimensionality reduction in the representation learnt, and also potentially increases the expressiveness of the
framework. In a similar vein, (Zhang et al., 2022) argues that due to redundancy in spatial data, it is not necessary to learn in data space, and instead proposes a dimensionality-varying diffusion
process (DVDP), where the dimensionality of the signal is dynamically adjusted during the both the diffusion and denoising process.
We showed that score matching presents a promising new direction for generative models, which avoids many of the limitations of other approaches such as training instability and mode collapse in
GANs, and poor approximation guarantees in variational inference. While score matching has several flaws, such as suffering from the manifold hypothesis and requiring an expensive Langevin dynamics
process in order to draw samples, successive work has done well in addressing these limitations to make score matching on diffusion models a viable contender to displace GANs as the state-of-the-art
for generative modeling.
Our experiments in this blog post help to provide empirical context to the theoretical results we have derived. Most notably, we have shown how annealing is an essential part of sampling via Langevin
Finally, we discuss some future directions that can help to improve the viability of using score-based diffusion models, which includes improving its computational cost in both training and sampling
and increasing the diversity of applicable modalities.
title = {Score-Based Diffusion Models},
author = {Fan Pu Zeng and Owen Wang},
journal = {fanpu.io},
year = {2023},
month = {Jun},
url = {https://fanpu.io/blog/2023/score-based-diffusion-models/} | {"url":"https://fanpu.io/feed.xml","timestamp":"2024-11-03T16:12:22Z","content_type":"application/atom+xml","content_length":"253378","record_id":"<urn:uuid:4ed21fb5-a907-4a84-b5c3-b28d1d051543>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00302.warc.gz"} |
Complex Math Parser and Evaluator in VB.NET
In many situations, there may be a string containing a math expression, such as "1+2*5" or "(3+i)(3-i)", and there is the need to do the math and calculate the result. Also, in case of a formula like
"0.5*x+4", we may need to calculate the result for several different values of x. In those situations, the complex parser presented here may help.
The classes here are a small part -but improved- of my all free and downloadable CAS calculator http://xrjunque.nom.es. One of the goals is that these classes do not rely on other 'external' classes
as happens in the CAS calculator.
The Five Classes
• Class Global10 contains global values like number of decimals, imaginary character (i or j) or CultureInfo.
• Class Msg10 just contains a few messages to handle possible errors.
• Class Rational gives a bit more of accuracy in the operations.
• Class Complex class does the complex math.
• Class parseComplex is in charge of dividing the input string into tokens and call accordingly to classes Complex or Msg10. Class Complex makes use of class Rational for its Real and Imaginary
members. The 'tokenizing' job is done by a Regex pattern.
The tokens groups are:
mode <mode>
numbers <num>
operators <op>
logical operators <lop>
functions <fn>
constants <cnt>
variables <var>
any other character <any>
besides end of tokens <end> formed by an escape character Chr(27).
The pattern looks like:
Pattern for numbers, depending on the Globalization.CultureInfo setting, may swap the dot (NumberFormat.NumberDecimalSeparator) and the comma (NumberFormat.NumberGroupSeparator).
Mode makes possible to enter numbers in hexadecimal, decimal, octal or binary base; along with setting the number of decimals and the imaginary character.
Using the Code
The are two possible ways of instantiation:
Dim eP As New ParseComplex
eP.CultureInfo = New Globalization.CultureInfo("fr-FR")
Dim eP As New ParseComplex(New Globalization.CultureInfo("es-AR"))
By default, CultureInfo is set to "en-US".
Evaluation is done by calling one of the two Evaluate() methods.
First method:
'// argument is a string:
Dim cplx As Complex = eP.Evaluate("(3+5*i)*(3-i*5)")
First method with variables, set in a Dictionay(Of String, Complex):
eP.vars.Add("x", Complex.one)
eP.vars.Add("y", New Complex(-1, 2))
'// argument is a string:
Dim cplx As Complex = eP.Evaluate("(3+x*i)*(y-i*5)")
Once the string has been parsed, it is possible to call the overloaded second method:
'// change "x" value (change any variable value):
eP.vars.Item("x") = New Complex(3)
'// argument is the Dictionary(Of String, Complex):
Dim cplx As Complex = eP.Evaluate(eP.vars)
Variables names start with a letter or underscore (_), can contain letters, numbers or underscore and can be any length.
Of course, you may call the Complex class directly, if you don't need the parser.
The default numeric base is decimal. To change to another base write &h (hexadecimal), &o (octal) or &b (binary). Appending &d will restore to decimal base.
In a similar way, °, &grad will accept angles in degrees or in gradians. To restore to default radians, enter &rad.
To change default imaginary i to j, write &j and to turn back to the default character, write &i.
For instance, entering &culture it-IT will change the current CultureInfo to it-IT. So inputs and ouputs will be in mented culture.
An example of possible modificators is the following:
Dim cplx As Complex = eP.Evaluate("&culture fr-FR &dec2 &h 0f + &j &d 0,2+0,3*j"
Console.WriteLine(cplx.tostring) ' will show 15,2+j*0,3
The Output
You may call Complex.ToString() or ToStringComplex( numDecimals As Int32, sImg As String, cultureInfo As Globalization.CultureInfo) As String:
cplx.ToStringComplex(4, eP.Imaginary, eP.CultureInfo)
Detail Version
If the word detail is found, code will output operation steps. For example:
Dim cP As New ParseComplex
Dim cplx As Complex = cP.Evaluate("detail (2+i*3)*(1+i)")
' Will output:
' [ (2+i*3)*(1+i) ]
' [2*1 - 3*1 + i*(2*1+3*1) ]
' [-1+i*5]
Basic Principles
The parsing method is a recursive-descent parsing: Parsing Expressions by Recursive Descent.
Evaluation method E calls T for any addition or substraction, but T calls first F for any multiplication or substraction, and F calls first P for any power possible power operation. P calls first v
to get next token. If there is a "(" token, v calls recursively to T.
E --> T {( "+" | "-" ) T}
T --> F {( "*" | "/" ) F}
F --> P ["^" F]
P --> v | "(" E ")" | "-" T
Step by Step Walk-throughs
The algorithm presented here englobes T and F in a single method. Besides, method v operates logical operators, '%' and mod any possible function like cos(), csc() and so on.
While writing this article, I found some glitches. If you find any further error, please let me know.
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Is the number of pages left to read proportional to the time read? And graph the equation to see how many pages he has left.to read after 3 hours
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Is the number of pages left to read proportional to the time read? And graph the equation to see how many pages he has left.to read after 3 hours
Is the number of pages left to read proportional to the time read? And graph the equation to see how many pages he has left.to read after 3 hours
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Introduction to Bayesian Statistics, 2nd Edition - Boktugg
Advanced Bayesian Inference Kurser Helsingfors universitet
May 2, 2016 Bayesian Analysis. Bayesian analysis is where we put what we've learned to practical use. In my experience, there are two major benefits to Bayesian statistics is currently undergoing
something of a renaissance. At its heart is a method of statistical inference in which Bayes' theorem is used to update A balanced combination of theory, application and implementation of Bayesian
statistics in a not very technical language. A tangible introduction to intangible Are you a researcher or data scientist / analyst / ninja?
An unremarkable statement, you might think -what else would statistics be for? But classical frequentist statistics, strictly speaking, only provide estimates of the state of a hothouse world,
estimates that must be translated into judgements about the real world. Fun guide to learning Bayesian statistics and probability through unusual and illustrative examples. Probability and statistics
are increasingly important in a huge range of professions. But many people use data in ways they don't even understand, meaning they aren't getting the most from it.
Promemorior från P/STM 1978:1. Bayesianska idéer vid - SCB
The concept of conditional probability is widely used in medical testing, in which false positives and false negatives may occur. Starting with version 25, IBM® SPSS® Statistics provides support for
the following Bayesian statistics. One Sample and Pair Sample T-tests The Bayesian One Sample Inference procedure provides options for making Bayesian inference on one-sample and two-sample paired
t-test by characterizing posterior distributions.
Daniel Hernandez-Stumpfhauser - Google Scholar
Författare. William M. Bolstad. Förlag, John Wiley & Sons. Computational methods, Markov-Chain Monte Carlo. After the course, the student can explain the central concepts in Bayesian statistics, and
name steps of the Jämför och hitta det billigaste priset på Bayesian Statistics for the Social Sciences innan du gör ditt köp. Köp som antingen bok, ljudbok eller e-bok. Läs mer och You will both
learn to solve standard statistical problems using Bayesian methods Chain Monte Carlo (MCMC), which are often used in Bayesian inference.
This course introduces the Bayesian approach to statistics, starting with the concept of probability and moving to the analysis of data. We will learn about the philosophy of the Bayesian approach as
well as how to implement it for common types of data. 2016-05-01 · For practical Bayesian statistics, nobody gets me more excited than Andrew Gelman! This is not an easy book to work through but it
is an absolute gem. The text is filled with wonderful, real world example that will alway renew your love of Bayesian Statistics. Here's a great video that shows off Gelman's enthusiasm for Bayesian
Analysis: Bayesian statistics: a comprehensive course - YouTube.
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We will learn about the philosophy of the Bayesian approach as well as how to implement it for common types of data. 2016-05-01 · For practical Bayesian statistics, nobody gets me more excited than
Andrew Gelman! This is not an easy book to work through but it is an absolute gem. The text is filled with wonderful, real world example that will alway renew your love of Bayesian Statistics. Here's
a great video that shows off Gelman's enthusiasm for Bayesian Analysis: Bayesian statistics: a comprehensive course - YouTube. This playlist provides a complete introduction to the field of Bayesian
Se hela listan på scholarpedia.org 2021-01-14 · Bayesian statistics is an approach to data analysis and parameter estimation based on Bayes’ theorem. Unique for Bayesian statistics is that all
observed and unobserved parameters in a statistical Bayesian Statistics "Under Bayes' Theorem, no theory is perfect. Rather it is a work in progress, always subject to refinement and further testing"
Nate Silver Introduction With the recent publication of the REMAP-CAP steroid arm and the Bayesian post-hoc re-analysis of the EOLIA trial, it appears Bayesian statistics are appearing more
frequently in critical care trials. Bayesian Statistics: An Introduction - YouTube. Bayesian Statistics: An Introduction.
Unga sossar
The text is filled with wonderful, real world example that will alway renew your love of Bayesian Statistics. Here's a great video that shows off Gelman's enthusiasm for Bayesian Analysis: Bayesian
statistics: a comprehensive course - YouTube. This playlist provides a complete introduction to the field of Bayesian statistics. It assumes very little prior knowledge and, in particular Bayesian
Analysis (2008) 3, Number 3, pp. 445{450 Objections to Bayesian statistics Andrew Gelman Abstract. Bayesian inference is one of the more controversial approaches to statistics. The fundamental
objections to Bayesian methods are twofold: on one hand, Bayesian methods are presented as an automatic inference engine, and this Bayesian models are a rich class of models, which can provide
attractive alternatives to Frequentist models.
Bayesian statistics is a theory in the field of statistics based on the Bayesian interpretation of probability where probability expresses a degree of belief in an event. The degree of belief may be
based on prior knowledge about the event, such as the results of previous experiments, or on personal beliefs about the event. Se hela listan på quantstart.com Bayesiansk statistik eller bayesiansk
inferens behandlar hur empiriska observationer förändrar vår kunskap om ett osäkert/okänt fenomen. Det är en gren av statistiken som använder Bayes sats för att kombinera insamlade data med andra
informationskällor, exempelvis tidigare studier och expertutlåtanden, till en samlad slutledning. Metodiken har fått sitt namn efter den engelske pastorn Thomas Bayes, som presenterade satsen i en
postumt utgiven artikel. Teorin bygger på A. Bayesian inference uses more than just Bayes’ Theorem In addition to describing random variables, Bayesian inference uses the ‘language’ of probability to
describe what is known about parameters.
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Promemorior från P/STM 1978:1. Bayesianska idéer vid - SCB
This book was written as a companion for the Course Bayesian Statistics from the Statistics with R specialization available on Coursera. Our goal in developing the course was to provide an
introduction to Bayesian inference in decision making without requiring calculus, with the book providing more details and background on Bayesian Inference. Bayesian Statistics for Beginners: a
step-by-step approach. by Therese M. Donovan and Ruth M. Mickey. 4.7 out of 5 stars 32.
Unix cutvad innebär verklig huvudman
Andrew Gelman - Google Scholar
Bayesian Reasoning for Intelligent People, An introduction and tutorial to the use of Bayes' theorem in statistics and cognitive science. Morris, Dan (2016), Read first 6 chapters for free of "
Bayes' Theorem Examples: A Visual Introduction For Beginners " Blue Windmill ISBN 978-1549761744 . The International Society for Bayesian Analysis (ISBA) was founded in 1992 to promote the
development and application of Bayesian analysis.By sponsoring and organizing meetings, publishing the electronic journal Bayesian Analysis, and other activities, ISBA provides an international
community for those interested in Bayesian analysis and its applications. This course describes Bayesian statistics, in which one's inferences about parameters or hypotheses are updated as evidence
accumulates. You will learn to use Bayes’ rule to transform prior probabilities into posterior probabilities, and be introduced to the underlying theory and perspective of the Bayesian paradigm. | {"url":"https://investerarpengarqgvp.web.app/87948/98476.html","timestamp":"2024-11-11T07:19:27Z","content_type":"text/html","content_length":"19055","record_id":"<urn:uuid:98e07ef8-9e83-4931-9620-ef8ba346428f>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00814.warc.gz"} |
Factoring the Expression (5x - 25)
The expression (5x - 25) is a linear expression, meaning it has a highest power of 1 for the variable 'x'. We can factor this expression by finding the greatest common factor (GCF) of the terms 5x
and -25.
Finding the Greatest Common Factor (GCF)
The GCF of 5x and -25 is 5. This is because both 5x and -25 are divisible by 5.
Factoring the Expression
To factor the expression, we can write it as the product of the GCF and the remaining expression after dividing each term by the GCF:
Therefore, the factored form of (5x - 25) is:
(5x - 25) = 5(x - 5)
Checking the Factored Expression
We can check if our factorization is correct by expanding the factored expression:
5(x - 5) = 5 * x - 5 * 5 = 5x - 25
This confirms that our factorization is correct.
The expression (5x - 25) can be factored into 5(x - 5) by finding the greatest common factor of the terms and then dividing each term by the GCF. This process allows us to represent the expression in
a simpler and more compact form. | {"url":"https://jasonbradley.me/page/(5x-25)","timestamp":"2024-11-10T09:43:14Z","content_type":"text/html","content_length":"57073","record_id":"<urn:uuid:d744a8bb-86ae-453b-aeb3-16967a2ce65f>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00083.warc.gz"} |
What data structure is used for graphs?
What data structure is used for graphs?
What data structure is used for graphs?
A graph can be represented using 3 data structures- adjacency matrix, adjacency list and adjacency set. An adjacency matrix can be thought of as a table with rows and columns. The row labels and
column labels represent the nodes of a graph.
What is graph ADT?
The graph abstract data type (ADT) is defined as follows: Graph() creates a new, empty graph. addVertex(vert) adds an instance of Vertex to the graph. addEdge(fromVert, toVert) Adds a new, directed
edge to the graph that connects two vertices.
What is graph and types of graph in data structure?
A graph is a non-linear kind of data structure made up of nodes or vertices and edges. The edges connect any two nodes in the graph, and the nodes are also known as vertices.
What are 4 types of graphs?
There are several different types of charts and graphs. The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs.
What is linear data structure?
A Linear data structure have data elements arranged in sequential manner and each member element is connected to its previous and next element. This connection helps to traverse a linear data
structure in a single level and in single run. Such data structures are easy to implement as computer memory is also sequential.
What is queue and stack?
Stack is a container of objects that are inserted and removed according to the last-in first-out (LIFO) principle. Queue is a container of objects (a linear collection) that are inserted and removed
according to the first-in first-out (FIFO) principle.
Why DFS is used?
Using DFS we can find path between two given vertices u and v. We can perform topological sorting is used to scheduling jobs from given dependencies among jobs. Topological sorting can be done using
DFS algorithm. Using DFS, we can find strongly connected components of a graph.
What is MST in graph?
A minimum spanning tree (MST) or minimum weight spanning tree is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and
with the minimum possible total edge weight.
What is tree ADT?
a tree is a widely used abstract data type (ADT) or data structure implementing this ADT that simulates a hierarchical tree structure, with a root value and subtrees of children, represented as a set
of linked nodes.
How many types of data structure are there?
Basically, data structures are divided into two categories:
• Linear data structure.
• Non-linear data structure.
What are the 6 types of graphs?
Types of Graphs and Charts
• Bar Chart/Graph.
• Pie Chart.
• Line Graph or Chart.
• Histogram Chart.
• Area Chart.
• Dot Graph or Plot.
• Scatter Plot.
• Bubble Chart. | {"url":"https://www.steadyprintshop.com/what-data-structure-is-used-for-graphs/","timestamp":"2024-11-05T04:03:45Z","content_type":"text/html","content_length":"53428","record_id":"<urn:uuid:cd1fbd35-87bd-49fe-a058-ddae5760dde2>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00331.warc.gz"} |
Design of Reinforced Concrete Beams per ACI 318-05Engineersdaily | Free Engineering Database: Design of Reinforced Concrete Beams per ACI 318-05
Reinforced concrete is made of two materials, concrete and reinforcing steel.Concrete is made of five parts:
• Air
• Water
• Cement, five main types per ASTM
• Sand, fine aggregate
• Gravel, course aggregate
The compressive strength (fc’) of concrete is the 28-day strength. This could be from 2,500 psi to 20,000 psi. Most concrete used is between 3,000 psi to 6,000 psi. Concrete is very good in
compression but its tensile strength is only about 8 to 15% of the compressive strength. This is the reason why we need reinforcing steel.When we load a beam, the bottom is in tension. Reinforcement
could be fiber-reinforcement or reinforcing steel. In this course, we will only look at reinforcing steel. Reinforcing steel comes in the following sizes, areas, weights and diameters:
Reinforcement steel comes in the followingdesignations,types, grades, strengths and available sizes.It appears from the above table, A615, grade 60 and A706, grade 60 cover all sizes.
The stress distribution may be rectangular, parabolic, trapezoidal, etc. Here are two stress distributions, parabolic (b) and rectangular (c):
We use figure c, rectangular. The ACI code says for concretes with fc’>4,000 psi, β can be determined with the following formula:
This is a table for the above formula (ACI 10.2.7.3):
The ACI code says design value must be greater than or equal to the required value.
In the formula, ρ is the steel ratio, As / bd.Beams are considered to be under three types of control:
• Compression control, εt <0.002
• Transition, 0.002<εt<0.005
• Tension control, εt>0.005
We use a strength reduction factor to account for many uncertainties in the design.For tension-controlled beams, we use a strength reduction factor (φ) of 0.90.
ACI (10.5.1) specifies the minimum about of reinforcement by the following two formulas:
Note: When the code specifies a minimum and they give two or more formulas, we use the formula that yields the maximum. For example, if one formula gives 16 sq inch and the other formula gives 18 sq
inch, then the minimum would be 18 sq inch. Also, bw is the width of the beam.
ACI (10.2.3) states that the maximum usable strain at extreme concrete compression fiber shall be assumed equal to 0.003. In other words, εc=0.003.
It is desirable, under ordinary conditions, to design beams with a steel ration (ρ) between ρ min and ρ max.
Load factors are numbers, used to increase the estimated loads applied on a structure. The loads are increased to account for the uncertainties involved in estimating the magnitude of the loads. How
good can you estimate the loads on the floor where you are right now? Sections 9.2 gives the required strength based on load factors and combinations of loads:
D=dead loads, F=weight and pressure of fluids, T=temperature, creep, shrinkage and differential settlement, L=live loads, H=weight and pressure of soil, water in soil or other materials, Lr=roof live
loads, S=snow loads, R=rain loads, W=wind loads and E=earthquake loads.
Section 9 gives the following table for minimum depth of beams:
In Section 7.7.1 of the code, it specifies the amount of cover for the reinforcement. Cover is the distance from the edge of the reinforcing bar to the face of the concrete beam. For beams with
primary reinforcement, ties, stirrups and spirals, it is 1 ½ inches when the concrete is not exposed to weather or in contact with the ground.
In section 7.6 of the code, it specifies the minimum clear spacing between parallel bars in a layer to be db or 1”, whichever is larger. Remember, when the code specifies the minimum and gives you
two or more items, you use the larger of the values. As a rule of thumb, beams 20-25 feet long have a ratio of d to b of 1.5 to 2. For longer beams the ratio of depth to width may be as high as 3-4.
Beam dimensions are selected in whole inches. The width is usually a multiple of 2 or 3. Beams should probably not be less than 12” wide to get the steel and your hands in the form. For the usual
situation, use bars of size # 11 and smaller if possible. Rarely will you use # 14 or # 18 bars. | {"url":"https://www.engineersdaily.com/2011/01/design-of-reinforced-concrete-beams-per.html","timestamp":"2024-11-13T02:12:48Z","content_type":"application/xhtml+xml","content_length":"633839","record_id":"<urn:uuid:8f74c3ee-34f7-445a-9d66-6980a0eaec02>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00593.warc.gz"} |
Over-fitting problems and solutions - Product Manager's Artificial Intelligence Learning Library
In this article, I want to explain one of the most important concepts of machine learning and data science that we encountered after training machine learning models. This is a topic that must be
This article is intended to explain the following topics:
1. What is overfitting in a machine learning project?
2. How do we detect overfitting?
3. How do we solve the overfitting problem?
Photography: Isaac smithOn Unsplash
Introduction-What is overfitting?
Let us first determine the basis of the concept.
Suppose you want to predict future price movements of stocks.
Then, you decide to collect the historical daily price of the stock for the past 10 days and plot the stock price on the scatter plot as follows:
The figure above shows that the actual stock price is random.
To capture stock price movements, you need to evaluate and collect data on the following 16 features, you know that the stock price depends on:
1. Industry performance
2. Company news release
3. Company income
4. Company profit
5. Company's future announcement
6. Company dividend
7. Current and future contract size of the company
8. Company's M&A status
9. Company management information
10. Current contract of the company
11. Future contract of the company
12. inflation
13. interest rate
14. Foreign Exchange Rate
15. Investor sentiment
16. Company's competitor
After collecting, cleaning, scaling, and transforming data, you can split the data into training and test data sets. In addition, you provide training data to your machine learning model for
After training the model, you decide to test the accuracy of the model by passing in the test data set.
What are we looking forward to?
The figure above shows that the actual stock price is random. However, the predicted stock price is a smooth curve. It doesn't fit itself too close to the training set, so it can better promote
invisible data.
But let's assume that your plot is actually related to the predicted stock price and you will encounter the following chart:
1. line shows predicted price
What does it show?
This means that the algorithm has a very strong data pre-concept. This means it is highly biased.This is called under-coordination.These models are not suitable for predicting new data.
2. Very powerful tight fit line
What does it show?
This is another extreme. It may seem like it's doing a good job of predicting stock prices.However, this is called overfitting.It is also known as high variance because it has been well trained in
training data and therefore does not promote new and invisible data well. These models are not suitable for predicting new data. If we provide new data to the model, its accuracy will eventually
become extremely poor.
This also shows that we did not train our model with enough data. Overfitting means that your model overtrains your training data. It may be because there are too many features in the data, or
because we didn't provide enough data. It occurs when the difference between the actual value and the predicted value is close to 0.
How to detect overfitting?
Models that over-adapt to training data are not well summarized as new examples. They are not good at predicting invisible data.
Photography: Stephen Dawson. From Unsplash
This means that they are very accurate during training and produce very poor results during the prediction of invisible data. If the accuracy measurements (such as the mean squared error) are
significantly reduced during model training and the accuracy of the test data set is degraded, this means that your model over-fitting the data.
If you want to learn about algorithms that can be used to measure the accuracy of machine learning models, check out this article:Must know the mathematical measurement scientist for each data
Key mathematical formulas are introduced in easy-to-follow bullet pointsmedium.com
How do we solve the overfitting problem?
We can randomly remove these features and iteratively evaluate the accuracy of the algorithm, but this is a very tedious and slow process.
There are basically four common ways to reduce overfitting.
1. Reduced functionality:
The most obvious option is to reduce functionality. You can calculate the correlation matrix of features and reduce the features that are highly correlated with each other:
import matplotlib.pyplot as pltplt.matshow(dataframe.corr())plt.show()
2. Model Selection Algorithm:
You can choose a model selection algorithm. These algorithms can choose more important functions.
The problem with these technologies is that we may end up losing valuable information.
3. Provide more data
Your goal should be to provide enough data for the model to fully train, test and validate the model. Designed to provide 60% data to train the model, 20% data for testing, and 20% data for
validation models.
4. Normalization:
The purpose of regularization is to maintain all features, but impose constraints on the size of the coefficients.
It is preferred because you don't have to use the penalty feature to lose functionality. When a constraint is applied to a parameter, the model is less prone to overfitting because it produces a
smooth function.
A regularization parameter called a penalty factor is introduced, which controls the parameters and ensures that the model itself does not overtrain the training data.
These parameters are set to smaller values to eliminate overfitting. When the coefficient takes a large value, the regularization parameter punishes the optimization function.
There are two common regularization techniques:
1. lasso
Add a penalty value, which is the absolute value of the coefficient magnitude. This ensures that the feature does not ultimately impose a high weight on the prediction of the algorithm.
From sklearn import linear_model model = linear_model.Lasso(alpha=0.1) model.fit([[0,0], [1, 1], [2, 2]], [0, 1, 2])
2. RIDGE
Add a penalty, which is the square of the coefficient size. As a result, some of the weights will eventually be zero. This means that data for certain features will not be used in the algorithm.
from Sklearn.linear_model import Ridge
Model = Ridge(alpha=1.0)
Model.fit(X, y)
Photography: Sergey Pesterev,About Unsplash
Final Thoughts
This article highlights a key topic we encountered after testing the machine learning model. It outlines the following key parts:
1. What is overfitting in a machine learning project?
2. How do we detect overfitting?
3. How do we solve the overfitting problem?
This article is transferred from medium,Original address
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Revision Notes for CBSE Class 6 Maths - CoolGyan
Revision Notes for CBSE Class 6 Maths
CBSE Class 6 Maths Revision Notes
Provided by CoolGyan - a leading e-learning platform in India, Class 6 Maths notes are easily available for all the CBSE students. And, the solutions provided here are designed by the best teachers
and experts who excel in this field. If you read through the revision notes given by them, you will definitely score better in the examination. These Revision Notes for Class 6 Maths textbook are
available online through PDF form and you can download these files for free without any hassle.
Every NCERT Solution is provided to make the study simple and interesting on CoolGyan. Subjects like Science, Maths, English,Hindi will become easy to study if you have access to NCERT Solution for
Class 6 Science , Maths solutions and solutions of other subjects. | {"url":"https://coolgyan.org/revision-notes/cbse-class-6-maths-notes","timestamp":"2024-11-03T16:44:11Z","content_type":"text/html","content_length":"98644","record_id":"<urn:uuid:3a43908d-e82d-4c88-8173-67e56f3a027f>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00346.warc.gz"} |
EpilogJS Manual - uniquify
uniquify(expressionlist) → expressionlist
The subroutine uniquify takes an array of expressions as argument and returns a copy in which all duplicates have been removed. The items not removed occur in the same order as in the original
Call: uniquify(['c','b','c','b','a'])
Exit: [c,b,a]
Call: grindem(uniquify([read('p(b,c)'),read('p(a,b)'),read('p(a,b)')]))
Exit: p(b,c)
uniquify runs in quadratic time. Compare to vniquify and zniquify. | {"url":"http://logic.stanford.edu/epilog/documentation/epilogjs/uniquify.php","timestamp":"2024-11-02T11:28:06Z","content_type":"text/html","content_length":"3730","record_id":"<urn:uuid:5b3a17ad-28c1-44fb-bb2c-5a64bb5e8f36>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00557.warc.gz"} |
Whispering gallery micro-global positioning system for nanoparticle sizing in real time
We have devised a simple means for determining the size of a nanoparticle in one binding event (i.e., real time) by utilizing two polar modes of a slightly eccentric Whispering Gallery Mode (WGM)
spheroidal resonator. The ratio of shifts of these modes locates the absolute latitude angle at which a nano-particle binds. From this location, the size of the nanoparticle is calculated using the
reactive sensing principle. Although our latitude-only micro-global positioning scheme is applied to nanoparticle sizing using slightly eccentric spheroids in aqueous solution, this approach can be
applied to WGM micro-resonators having a variety of shapes.
A quantitative bio-size/mass spectrometer that can work in solution, and work at single nanoparticle sensitivity, would have the possibility of adding important information to body fluid analysis.
Furthermore, label-free sensors with this capability could identify viruses and exosomes not only by using bound antibodies but also through their size.
At present, determining the size of a nanoparticle in one event by using an optical micro-resonator has required a high Q WGM resonator (Q∼10^7–10^8) and a combination of a reactive splitting 2g,
and an increase in linewidth $ΔΓ$ associated with light scattering.^1 The splitting is caused by lifting the clockwise-counterclockwise degeneracy associated with orbiting photons within WGMs.
Whereas the splitting is proportional to the polarizability α of the adsorbing nanoparticle, which is, in turn, proportional to the particle volume, the increase in linewidth is proportional to α^2,
so that the ratio $ΔΓ$/(2g) can be used to obtain the polarizability, and from that the particle size/mass. However, the splitting vanishes if it is smaller than the cavity linewidth, thereby setting
a lower limit for the detectable nanoparticle size.^2 The resulting degenerate mode still shifts, but it is only possible to obtain the polarizability through a statistical approach. In particular,
the largest wavelength shift in a distribution of many events can be analytically related to the particle size, since this shift corresponds to binding directly on the lowest order polar mode (i.e.,
at the equator).^3 Unfortunately because of the statistical nature of this approach, one cannot be certain that the largest event in the distribution corresponds to a particle at the equator.
Consequently, the size obtained is uncertain and many events are needed to generate the distribution.^4 Fortunately, there is another approach that avoids these difficulties, and that is what this
paper is about.
The current idea is fully reactive, does not require mode splitting, and therefore can be applied in the weak coupling limit where the interaction g is considerably smaller than the linewidth, $Γ$,
and the Q is modest (<10^6). However, like mode splitting, it involves the ratio of shifts from two different resonances in the same microcavity, and is therefore immune to long term temperature
fluctuations. This approach involves the excitation of two resonances having the same angular momentum quantum number l but different m quantum numbers (−l < m < l) in a cavity for which a
nanoparticle induced wavelength shift is much smaller than the linewidth (i.e., weak coupling). Although m is referred to as the magnetic quantum number in atomic physics, when considering different
m modes within a microresonator having a given l, we will use the term polar modes. To understand what is involved, we must examine the shape of the polar WGM intensity. There are many such states
that are characterized by l−m+1 intensity peaks along the polar direction. In a sphere, these m states are degenerate for a given angular momentum l, but in a spheroid, this degeneracy is lifted,
and the states separate spectrally. The first of these is an equatorial mode for which m=l, thereby producing one intensity peak centered about the equator (Fig. 1). The next with m=l−1 has two
peaks, one to the North, and the other to the South of the equator (Fig. 1). It is important to realize that the two modes depicted in Fig. 1 are excited sequentially within the same slightly prolate
microcavity by a fiber positioned slightly above or below the equator and directed along a line perpendicular to the symmetry axis of the spheroid. In what follows, we will show that the ratio of the
resonance wavelength shifts of each of these modes provides a locator for the nano-particle's absolute latitude, from which its polarizability and size/mass may be estimated one event at a time
(i.e., real time).
To start the description of the polar mode mechanism, we will first update the wavelength shift theory to include polar modes. For this purpose, we adopt the symbol $Δλl,m$ to describe the wavelength
shift of a mode having an angular momentum number l, and polar number m. In what follows, we will show that the latitude angle for particle binding is easily obtained from the ratio of two wavelength
shifts, $Δλl,l−1/Δλl,l$.
To understand the importance of polar modes in nanoparticle characterization, one simply has to return to the basic principle of microcavity reactive detection, the reactive sensing principle (RSP).
The RSP simply states that “the perturbation in a resonator's photonic energy upon particle binding at r[p] is equal to the energy required for the microcavity's reactive (evanescent) field E[0](r
[p]) to polarize the particle.”^3 On this basis, the shift in resonance wavelength $Δλ$ is the wavelength λ times the ratio of the energy required to polarize the nanoparticle to the energy in the
cavity. In a dipole approximation, the shift in wavelength^3
where α[ex] is the polarizability of the nanoparticle in excess of its environment (i.e., medium) and ε(r[c]) is the permittivity of the cavity at position r[c]. When applied to a homogeneous
microsphere for m≈ l ≫ 1 (i.e., polar modes that resonate close to the equator), for a nanoparticle of radius a adsorbing on the surface, Eq. (1) becomes
$Δλl,m≈α | Yl,m(ξp) |2g(a/L)(ns2−ne2)R3λ,$
where $α=αex/ε0$ is the “geometric” polarizability that is proportional to the volume of the nanoparticle ($α=Dαa3$), R is the microsphere radius, L is the characteristic evanescent intensity
length obtained from Mie theory, ξ[p] is the latitude of the bound particle, n[s], n[e], and n[p] are the refractive indices of the microsphere, environment, and nanoparticle, respectively, and
finally, the constant $Dα=4πne2(np2−ne2)/(np2+2ne2)$. The form factor g corrects the simple point dipole theory (Eq. (1)) for a nanoparticle that is extended in size (a∼L).^5 It is the overlap
integral between the surface normalized evanescent intensity and volume elements of the nanoparticle divided by the volume of the particle. For a spherical nanoparticle, integration over its shape
where z=a/L. The form factor has a simple limiting property, for a ≪ L, $g≃1$. Eq. (2) is the key to our latitude locator.
Consider the ratio of wavelength shifts of the m=l−1 to the m=l modes at a latitude $ξp$; $Δλl,l−1(ξp)/Δλl,l(ξp)$. So long as the shifts are very small in comparison to the wavelength, from
Eq. (2),
$Δλl,l−1Δλl,l=| Yl,l−1(ξp) |2| Yl,l(ξp) |2.$
As one can see that the right hand side of Eq. (4) only depends on the latitude ξ[p] of the adsorbing particle, and consequently by placing experimental wavelength shift data on the left, this
equation gives the latitude of the adsorbed nanoparticle independent of its physical properties, those of the resonator, or any refractive indices. Once ξ[p] is determined, Eq. (2) can be used to
calculate the size of the nanoparticle as well as the polarizability.
Our simple latitude locator equation (Eq. (4)) can be further simplified. As long as l is very large, and l−m is very small, the spherical harmonic functions can be transformed to Hermite-Gauss
asymptotic expressions^6 with the result that
$| Yl,l−1(ξp) |2| Yl,l(ξp) |2≈ 2(l−1)ξp2.$
Combining Eq. (4) with Eq. (5) gives the absolute latitude of the particle
$| ξp |=12(l−1)Δλl,l−1Δλl,l.$
With $| ξp |$ determined, the size a of the nanoparticle can be determined by re-expressing Eq. (2) as
$a3g(a/L)≈ (ns2−ne2)R3| Yl,l(ξp) |2Dα Δλl,lλ.$
The solution to Eq. (7) is particularly simple for a ≪ L since the form factor $g≃1$ in this limit, and one gets a closed form algebraic solution. For larger particles, Eq. (3) is used for g and the
resulting transcendental equation is solved numerically. Whether the particle is north or south of the equator has no relevance, since the square modulus of the spherical harmonic in the denominator
of Eq. (7) is even with respect to the latitude.
To test our micro-latitude locator idea, we formed micro-spheroids by using CO[2] laser melting at the end of a tapered silica optical fiber (inset, Fig. 2). Shape analysis of the images revealed
that our resonators were slightly prolate or oblate (eccentricity <3%). These silica micro-spheroids were then installed into our homemade microfluidic system,^7,8 where they were coupled to a
tapered optical fiber. In the inset, the fiber is beneath the spheroid.
A typical under-coupled spectrum taken through the coupling fiber is shown in Fig. 2. All of the resonances were excited with a 1063nm tunable Distributed Feedback (DFB) laser polarized along a
meridian [Transverse Electric (TE) polarization]. The laser was current tuned with a saw tooth wave that accounts for the rising backbone of the spectrum. It should be noted that the resonance dip on
the left has no neighbor at shorter wavelength. This is the signature of the m=l equatorial mode of a prolate spheroid; the m=l mode has the shortest wavelength.^9 To the right of this mode
(longer wavelength) is the m=l−1 mode which is narrower with a smaller dip. Note that the m=l−2 mode is of similar depth to the m=l mode and the m=l−3 mode to longer wavelength looks
similar in depth to the m=l−1 mode. This sequence of deep-shallow-deep-shallow dips in Fig. 2 is a consequence of the overlap between the fiber field and the polar symmetries of the WGMs; the
coupling constant requires performing volume integration over the product of the optical fiber field with the WGM field.^6 Whereas the m=l mode is symmetric in latitude about the equator as is the
fiber field, the m=l−1 mode is asymmetric. Exciting the asymmetric WGM mode requires that the centerline of the exciting fiber be slightly above or below the equator. The fiber is placed in
contact with the resonator to reduce mechanical vibration noise. This results in a red shift and broadening of the resonances. Upon coupling neither the m=l or m=l−1 resonances had Qs greater
than 2 × 10^5.
The validity of the latitude locator idea was tested by injecting nanoparticles [polystyrene $⟨ am ⟩±σ$=228$±$7nm from Polysciences] at a 20 fM concentration into our microfluidic system in the
presence of the resonator depicted in Fig. 2. The solution had a 30mM NaCl concentration to promote binding to the silica surface by decreasing the Debye length associated with ionized silanol
groups.^10 Figure 3 shows data for two typical events. The event on the left appeared 874s following the injection with a wavelength shift of 670$±$5fm for the m=l mode and 410$±$5fm for the
m=l−1 mode. For the event on the right which occurred 2460s after injection, the smaller shift occurred for the m=l mode; 80$±$ 7fm as compared with 360 $±$7fm for the m=l−1 mode. For
the former case, the latitude at which the nano-sphere attached from Eq. (6) in radians (degrees) is 0.078$±$0.004 (1.72$±$0.02deg.). After numerically solving Eq. (6), we find the radius of
this nanosphere to be 235.9$±$0.2nm. The latitude for the nanoparticle attachment on the right is 4.67$±$0.26 degrees, and the radius from Eq. (6) is 217$±$19nm. Compared with the ensemble
average from hydrosol manufacturer $⟨ am ⟩±σ$=228$±$7nm, both results from our latitude locator approach overlap.
Table I shows the analysis of four recorded events. In all cases, the latitude locator results in the column marked a[p] along with uncertainties are calculated from Eqs. (6) and (7) by using the
wavelength shift measurement and their uncertainties. For example, the relatively large uncertainty arrived for a[p] in the third event is a direct result of the large fractional experimental
uncertainty in $Δλ340,340$ for this event. It should be noted that the results for a[p] including uncertainties overlap the manufacturer's ensemble measurements in the last column.
TABLE I.
Event . $Δλ340,340(fm)$ . $Δλ340,339(fm)$ . $| ξp | (deg)$ using Eq. (6) . $ap (nm)$ using Eq. (7) . $⟨ am ⟩±σ (nm)$ .
1. 600 s 800 $±$ 5 40 $±$ 5 0.49 $±$ 0.03 224.5 $±$ 0.3 228 $±$ 7
2. 874 s 670 $±$ 5 410 $±$ 5 1.72 $±$ 0.02 235.9 $±$ 0.2 228 $±$ 7
3. 2460 s 80 $±$ 7 360 $±$ 7 4.67 $±$ 0.26 217 $±$ 19 228 $±$ 7
4. 3860 s 150 $±$ 5 520 $±$ 5 4.10 $±$ 0.09 228.8 $±$ 4.2 228 $±$ 7
Event . $Δλ340,340(fm)$ . $Δλ340,339(fm)$ . $| ξp | (deg)$ using Eq. (6) . $ap (nm)$ using Eq. (7) . $⟨ am ⟩±σ (nm)$ .
1. 600 s 800 $±$ 5 40 $±$ 5 0.49 $±$ 0.03 224.5 $±$ 0.3 228 $±$ 7
2. 874 s 670 $±$ 5 410 $±$ 5 1.72 $±$ 0.02 235.9 $±$ 0.2 228 $±$ 7
3. 2460 s 80 $±$ 7 360 $±$ 7 4.67 $±$ 0.26 217 $±$ 19 228 $±$ 7
4. 3860 s 150 $±$ 5 520 $±$ 5 4.10 $±$ 0.09 228.8 $±$ 4.2 228 $±$ 7
There is no doubt that the use of the latitude locator produces a much more accurate size for a single event than simply assuming that particles bind to the equator. Wavelength shifts associated with
a given resonance (e.g., m=l or m=l−1) vary by a factor of ∼10 (1000%) in Table I, leading to size estimates based on the assumption of equatorial binding that vary by a factor of ∼10^1/3. That
is over 200%. However, with the latitude locator to determine the polar location, this uncertainty is severely reduced as seen in Table I (e.g., for events 1 and 2, propagating the measurement
uncertainties through Eqs. (6) and (7) produces size uncertainties ∼1%).
Although our major goal in this Letter is to outline the latitude locator idea, we imagine its major use to be in measuring the size distribution in a heterogeneous solution in which each binding
event counts. To test this notion, we added 178nm PS beads to 228nm PS beads we had used previously. Figure 4 shows the binding curves recorded for the m=l and m=l−1 modes. The latitude
locator combined with Eq. (7) easily identified two particle ranges, 231 to 250nm, and 167 to 174nm.
We have not been concerned with the issue of characterizing an ultra-small particle, since that has already been done using the resonance shift in the statistical manner by looking for the largest
wavelength shift in a distribution of events.^4 The authors of Ref. 4 show that it is possible to sense a particle with a radius down to 12.5nm in aqueous solution using a scanning technique. The
latitude locator principle should enable this to be done deterministically, and in one event.
Although we have confined our attention to micro-spheroidal resonators, the same principle can be used for detection by other WGM shapes. Similar m modes have been identified in rolled up cylinders^
11 and micro-toroids,^9 although multiple m modes have not been used for precise nanoparticle sizing with these designs.
The research described herein was supported by the National Science Foundation Grant EECS 1303499. We thank W.W. Langbein for pointing out that the approximation leading to Eq. (5) is for small
latitudes such as those in this paper.
S. K.
Y. F.
D. R.
, and
Nat. Photonics
S. K.
, and
Appl. Phys. Lett.
, and
Opt. Lett.
J. H.
S. E.
R. C.
, and
Proc. Natl. Acad. Sci. USA
, “
Resonant detection of nano to microscopic objects using whispering gallery modes
,” Ph.D. dissertation (
Rockefeller University
B. E.
J. P.
, and
H. A.
J. Lightwave Technol.
, and
Faraday Discuss.
S. R.
, and
Appl. Phys. Lett.
J. B.
, and
Opt. Lett.
S. I.
, and
Opt. Express
M. R.
, and
O. G.
Appl. Phys. Lett.
© 2014 AIP Publishing LLC.
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20++ Mean Median Mode Range Worksheets Pdf
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How to make sure your betting system is a winner
After a number of bets testing your strategy, your capital is significantly increased. Is your system a winner or you simply got lucky?
Jim Makos
One of the most common questions that haunt the minds of all serious and meticulous bettors relates to the success of their strategy: “How can I make sure my system is a winner?” How many bets do you
need to place and come out victorious before you feel confident you lay your cash on the right, profitable system?
Some would suggest that 200 bets seem adequate, while others would need 500, yet some others would say that you cannot be absolutely sure unless you test your “modus operandi” a couple of thousand of
The right answer, however, is the one that we despise the most: it depends.
Nevertheless, today we will go over a simple method which should help you to gain insight about the strategy you use. Don’t be scared! I am not going to bombard you with terms like volatility,
standard deviation and other statistical jargon of this kind! I know you detest those. Therefore, some images will do the work. For, a picture is worth a thousand words.
This is a random line that shows the change in a gambler’s capital when they were betting on a heads-or-tails system. Every bet was placed on even odds in favor of the heads. After 200 bets the
gambler ended being a winner by about 14 units, starting with 100 units of cash. At one point, he was winning 19 units!
What would you say about that? Would it be a good strategy if we bet on fair odds (2.00) when tossing a coin? Would the above chart be convincing about a possible profitability of a system where we
place our bets blindfold on heads?
Let us take a look at another bettor who lodged his bets under the same idea.
Oops! A loss of 11 units in the end, losing up to 15 units at one point! This specific bettor loses about 5% of his capital after each coin’s toss. What went wrong? Really, could betting on coin
tossing be that disastrous?
The (almost) random distribution of outcomes
As we can see on the above charts, the very same system turns out to be successful to the first bettor and ill-fated to the latter.
However, this is not actually true. Account the second graph to be the extension of the first. In other words, a single bettor who was lucky enough to win on the first 200 bets is expected to sustain
a loss out of his profit in the next 200 bets. This is how it goes.
In essence, this is a third bettor who placed their cash on 400 bets. Even though in the first 200 or so bets they are a winner, in the following 200 they are constantly suffering losses, finally
having their bankroll reach the breakeven point.
Clearly, the three bettors experienced completely different outcomes, in random. What happens if 10 bettors make their bets on heads at odds of 2.00 in 400 different coin flips?
It’s really a mess. Do not forget that we are talking about a system that is defined by default to produce zero profit or loss. In the long run, the betting system always recovers out of extreme
losses or profits. Each time we fail we lose a unit of our cash, which we retrieve the next time we are successful. Of course, a fair coin has 50% probability of falling on either its’ heads or tails
when tossed in random.
Yet, on such a blind bet under the most trivial betting system in the whole world – betting blindfold on heads, or similarly betting blindfold on home teams in football – one bettor enjoyed 41%
increase on his capital while another suffered 33% loss! Obviously, more others saw positive or negative results in-between.
Do 10 gamblers seem not enough to you?
Ta-dah! The same cunning strategy put to the test by 100 bettors! Now, what will be your opinion about the betting system? Treasure or trash?
None obviously, because the main rule we explained above still applies. However, a number of bettors will stubbornly insist that this system is a winner, given they are up 50 units! It must become
clear that because of the outcomes’ randomness, the bettors’ bankroll will always unpredictably deviate from the positive to the negative region and vice versa.
After all, the outcome is well defined. It is not in random. We know that this method will bring neither profit nor loss. We definitely know that, because we make the wager on 50% probability with
the odds at 2.00. We could be lucky to enjoy some profit after 400 bets but we could also be losing some of our capital just as easily. Yet, the most probable outcome will always tend towards the
middle as expected: the perfect balance, the zero or breakeven point. Judging by the chart above, this is the point where we may feel a bit grateful of not being unlucky…
How to make sure your strategy is a winner
Based on the above, let us proceed to what is most important at the end of the day, by discussing two distinct scenarios:
1. Suppose that we know our coin is tweaked to drop on heads by 80% probability
2. Imagine that there is a guy so naive who would offer odds at 3.20 on heads
Of course, on both options we should bet on heads. In the first case, we expect to win on 8 out of 10 tosses of the coin (instead of 5), while in the second we should win 2.20 units on heads against
losing 1 on tails. Again, we deal with two options and two scenarios. This is not about a combination of the two, betting at 3.20 on heads of a coin that will bring heads by 80% probability! In that
case, what kind of stupid would we be not betting a huge chunk of our capital, sparing the “wait and see” precaution that we should normally take before we commit on a betting system??!
However, BOTH scenarios will result in the same outcome: our winrates are totally equal. Check this by simply multiplying the odds of 2.00 with the probability of 80% on heads of the first scenario
and the 3.20 odds on heads with 50% probability of the second. The result is the same (1.60), which means that our (enormous) edge is 60% in both cases. Along these lines, this is the foundation of
value betting.
Given that our system in both these cases is a winner, let us monitor the performance of a bettor who makes a wager on either of these two betting systems.
Fantastic! At that edge, the punter’s bankroll shows hardly any unexpected variance. This is why we normally say:
The bigger the edge, the fewer the bets we need to verify our strategy.
Let us follow closely now another group of 100 bettors.
Of course, this is phenomenal performance. Their capital increases with only a few ups and downs. No reason to spend more time on this. The system is a winner right from the first 100 bets or even
less. Good luck if you think you can easily find such a system!
How many bets would we need in order to be confident of your system?
Suppose you are testing a betting system nowadays. So far, you are being very attentive and have reached 400 bets studying the outcomes. Remember: we are interested in the 400 bets that you would
have put your money on, not the pool of 3,000 football (or any other sport) games out of which you have selected the 400 bets of your study.
Example 1:
Up to this moment, you have won 100 units by betting one unit on each game out of 400. In other words, you’ve bet 400 units and now you hold 500 units. Therefore, your performance equals to (1+100/
400) 2.25, on average.
Moreover, you are being successful in your predictions exactly half of the times, which means that your forecast to the next game will be by 50% chance accurate, or that your forecast success rate is
equal to 50%. This is exactly the winning probability of heads and tails when tossing a fair coin.
Let us take a look at what happens after testing your strategy 100 times.
According to the graph, we see that it may be possible to:
1. Win up to 113.25 units
2. Win by 98% chance
3. To finish in the red by… 2% chance after betting 400 times, since only 2 out of 100 trials ended in the negative zone!
Example 2:
Suppose now that you test another system at 5.00 average winning odds and 23% strike rate.
Even though the anticipated edge is slightly better than in the previous example (15% instead of 13%), notice how the bankroll’s variance has changed! A good number out of the 100 tests ended up
losing you cash while the capital’s volatility varies considerably. Notice how the luckiest bettor won 166 instead of 113 units on the first example, benefiting by more or less the same edge! The
same applies for the unluckiest punters in both scenarios.
Therefore, based on the above we conclude that:
Opting to bet on long shots will increase the betting system’s variance on our bankroll.
Example 3:
The next example will be the most common betting system used. This is a system of 44% hit rate when the winning odds are 2.30 on average.
On this system, we notice that for every 100 bettors, about half of them end to be winning and half are unprofitable. We should calculate the percentage of winning trials in order to find out the
exact probability of having a winning system beforehand. That is to say, if 55 trials ended in positive territory, based on these data, we have a winning system by 55%.
However, this is only one variable on the equation and besides, this is not what you are after for.
Your main goal here today is to find out how many bets you must place or test before you are absolutely confident of the profitability of your betting system. So far, as you can see, based on the
aforesaid numbers, 400 bets are not enough. Let us see what happens after 2,000 bets and 20 trials.
Ouch! Someone lost 115 units, meaning that if they risked 1% of their bankroll on each selection, by now nothing is left, not even a cent! Thus the informal but well-known rule among gamblers state
Proper money management requires risking no more than 2% of the total betting bankroll.
It’s quite obvious that even at the given system’s profitability, betting 2% of your capital might be extremely risky. Let’s take a look now what happens if we proceed to 10 trials of… 40,000 bets
Yes, when your… grandchildren study the results of this method, you can rest assured that your system is a winner. All you have to do is rise from your grave and place your bets! In that case, maybe
10,000 tests prove enough, wouldn’t they?
Tough luck! After 10,000 bets your chances to be a winner are 16/20 (16 positive trials in every 20 trials). If you still consider such a percentage too low, you have better either improve your
forecasts or go after higher odds, aiming at the same success rate of course!
Therefore, stop wondering when the right time to follow your betting system with real money will be. First of all, calculate the success ratio and the average odds of your winning bets. Thereafter,
enter these values into the betting simulator you can find after visiting BetStories.com. Select a satisfactory number of trials (10-100) and then start raising the total number of your wagers.
You should be aware by now of the number of bets you need to test in order to feel confident of your system’s success! Or, how much you need to improve your system before investing any of your money
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Hartree Fellow Cultivates New Perspectives on Quantum Computing
Quantum computers are heralded as promising tools for performing computations that are beyond the reach of supercomputers and every other technology currently at our disposal. But we are in the early
days of quantum computing, and there are still basic questions left to answer: How do you know that a clever programmer won’t develop a revolutionary method that allows traditional computers to run
circles around the upstart quantum newcomers? And if a quantum computer has solved a problem that no other available technology can, how can you be sure that it’s even right?
Dominik Hangleiter, one of the newest Hartree Postdoctoral Fellows at the Joint Center for Quantum Information and Computer Science (QuICS), develops theoretical frameworks to address these questions
and to identify problems at which quantum computers can be proven to dominate.
“This work is especially pressing because the quantum devices that we have at the moment are very faulty and noisy and you can't really trust them,” says Hangleiter. “So I'm working on methods to
assess whether you can trust your quantum computer, and also to assess in which situations it's even possible to devise such methods.”
Quantum computers weren’t what attracted Hangleiter to studying physics. In fact, as an undergraduate at the University of Konstanz in Germany, he had little interest in computers. Instead, he
studied condensed matter physics (the physics of how interacting particles determine the properties of materials) and the ways quantum physics naturally plays out in materials.
“I'm surprised that I ended up doing very computer sciencey things because I don't really like to interact with computers,” says Hangleiter.
But as a master’s student at Ludwig Maximilian University of Munich, he became interested in the more theoretical and mathematical side of physics and that drew him into the topic of computer
science. He began performing research with Jens Eisert, who studies quantum information theory and condensed matter theory, on how to confirm that a quantum simulation worked correctly. He then
continued his doctoral research with Eisert at the Free University of Berlin and dug deeper into the topic’s theoretical roots.
As a graduate student, Hangleiter investigated quantum approaches to sampling tasks, a general class of computations that produce random results according to some probability distribution. Since
quantum mechanics is fundamentally probabilistic (you can never definitively predict certain individual results beforehand no matter how much information you have), any quantum computer simulating a
quantum experiment or performing a purely quantum computation will be a sampling problem. Even though these tasks’ outputs can’t be definitively predicted, they reflect the reality of how quantum
mechanics dictates the probability of different outcomes. This intrinsic reflection of quantum probabilities makes these tasks one of the most promising ways to prove that there are computations
where quantum computers can significantly outperform traditional computers.
When investigating topics like sampling tasks, Hangleiter often keeps an additional layer of mathematical abstraction between him and the particular approaches—algorithms—that computer scientists
have proposed. That mathematical abstraction is complexity theory, which classifies computational tasks based on their resource requirements (the time to run a program, the size of the required
circuit, etc.); it serves as a lens to help get an overarching perspective of the mathematical landscape so researchers can avoid blind alleys and spot promising research routes for more thorough
exploration. This more abstract approach allows identification of underlying truths about whole styles of approaches instead of getting caught in the weeds of the peculiarities of an individual
“In computer science, there's always this problem when you talk about the time it takes you to solve a problem, that you might have just had a bad idea, and someone else might come up with a better
algorithm,” says Hangleiter. “So this is what complexity theory is about. It's all about putting problems into classes where we have very strong beliefs that they have genuinely different
complexities—you would really require different orders of magnitude of runtime to solve tasks in different classes.”
Hangleiter and his colleagues used this approach in a paper they published in the journal Physical Review Letters when Hangleiter was a doctoral student. They provided strong new evidence that
quantum simulations can perform certain sampling tasks that classical computers just aren’t cut out for. In particular, they studied certain quantum simulations with a repeating grid layout.
(Specifically, they studied what computer scientists call Hamiltonian quantum simulation architectures that are 2D translation-invariant and have a constant depth.)
“We looked at one type of a sampling scheme for quantum advantage,” says Hangleiter. “Previously, we had proposed this type of task and argued that it's an interesting task because you can
potentially do it in present experiments. But there were a few caveats that were left open. In this paper, we close all of the gaps that we can close. There remain gaps, but those are unlikely to be
closed in the near future.”
In the paper, Hangleiter and his colleagues first showed that for these types of sampling tasks the probabilities of all the possible results should be roughly equal, as opposed to being concentrated
around a few likely outcomes—in the parlance of math, they showed an “anticoncentration bound.” They then calculated how hard it is to exactly evaluate the average case of one of these simulations.
This result strengthens the argument that quantum computers will always surpass their traditional brethren at these sampling tasks by closing two of the three major loopholes in the theoretical
Hangleiter has also used complexity theory in developing methods that might be put to more immediately practical use by researchers. His graduate research included work on the Monte Carlo sign
problem, some of which was published in the journal Science Advances.
Monte Carlo methods are a way to get very accurate estimates of calculations that are too challenging to tackle with an exact mathematical approach, generally because the sheer amount of data
involved makes brute force counting or calculation impractical. While there are multiple different Monte Carlo methods, the general idea is to study a finite number of random samples from a
probability distribution instead of keeping track of the exact probability of every possible outcome, and these approaches have been successfully adapted to quantum mechanical problems.
Quantum Monte Carlo methods are powerful tools in quantum computer science but often are plagued by a mathematical nuisance called the sign problem. The sign problem occurs when the values being
investigated jump between positive and negative in a messy way that doesn’t allows a useful statistical result to be obtained—for instance, because the answer depends on precisely estimating a tiny
difference between two large sums. The poorly behaved fluctuations make determining how the values with different signs cancel out require an impractically large amount of random sampling.
But there is a catch: The sign problem can sometimes be solved just by looking at the computation in a different way. The sign problem is dependent on the mathematical perspective being used, and
there are many different ways to approach a computation. Not just for Monte Carlo methods but for any calculation, a mathematician or physicist has to make decisions about how to set up their
mathematical perspective: Which way is positive or up? Negative or down? Left? Right? Which point should be the origin (the zero point)?
Such choices of perspective even affect how difficult it is to do a straightforward calculation like finding the average value of a repeating, smooth wave (like the sine wave shown in the graphic,
which closely matches the shapes of sound and ocean waves). If you define the middle of the wave as zero so that there are equal amounts of wave above and below, you don’t need mathematical finesse
to see that the average is zero. But if you shift the wave up so that the middle no longer lies at zero, the cancellation isn’t as obvious and you have to do a more complex calculation. Or if you
consider that flat wave being tilted in a 3D space, you have to account for every point’s orientation in three dimensions. While the calculations required for each choice differ, the final results
are all equivalent as long as you understand how to translate between your choices.
Caption: (Top) A wave centered so that there are equal amounts above and below the axis that it lies along. Its average value lies on that same axis. (Middle) A wave shifted up so that its average
value is now positive somewhere above the axis. (Bottom) A wave that has been shifted up and rotated, making its average value harder to eyeball with the given axes. (Credit: Bailey Bedford/JQI)
The same flexibility of mathematical definitions can complicate quantum Monte Carlo methods, with the added wrinkle that the math is generally even more difficult—trickier calculations, messier data,
problems that aren’t limited to three dimensions, etc. For many of the situations that physicists study using Monte Carlo methods, they need extra mathematical dimensions for every quantum particle
involved. This opens many different ways to look at the calculation, and it is often unclear which way is best. Since different approaches to a calculation experience the sign problem to different
degrees, it would be useful to be able to judge—or in mathematical terms, measure—the severity for a given set of mathematical decisions.
“This is actually a problem that I've been thinking about since my master’s thesis,” says Hangleiter. “It’s quite difficult, because as it turns out, measuring the sign problem is as hard as
simulating the system in the first place. So we really have to find measures for the sign problem, which are maybe not perfect, but which you can efficiently compute. In our work we also develop
algorithms for optimizing those measures and assess them using computational complexity theory.”
Hangleiter and colleagues mathematically defined a quantity that corresponds to how bad the sign problem is for a given problem (they called the property the non-stoquasticity). With a rigorous
definition in place, they were able to develop a procedure to explore the different possible mathematical perspectives, for certain mathematical settings, and to find the perspective least afflicted
with the sign problem. Instead of attempting to find the most elegant solution to the sign problem in a general case their approach can be applied to the sign problem for any quantum Monte Carlo
“We want our algorithms to be independent of the physics,” says Hangleiter. “We want a universal method that applies to any model with the sign problem and returns the optimal perspective for a Monte
Carlo algorithm.”
Their method does not guarantee that the sign problem can be eliminated, but it provides insight into the nature of the sign problem and is an extra tool for working on these challenging problems.
For some cases, where they can’t completely solve the sign problem, just easing it might bring the difficulty down to a level that is practical for state-of-the-art computers to brute force.
“It treads on uncharted territory, because nobody has tried to actually optimize the perspective in which you phrase a problem in the systematic way we did,” says Hangleiter. “So it's unclear how
impactful this idea is. It definitely sheds a bit more light on the sign problem and its nature.”
Optimizing quantum Monte Carlo methods and Hangleiter’s other research are part of the research community’s efforts to build a whole new set of tools and to make sure that they don’t waste time
enthusiastically banging on a screw with their shiny new hammer. By understanding the underlying nature of quantum computers, researchers are clearing a path towards a future where quantum computers
and traditional computers can be put to their best uses. Hangleiter says that during his time at QuICS he wants to interact as much as possible with his colleagues and to get a new perspective on the
field of quantum computing that is currently full of grand promises.
“It's a good time to be in the field, because there's a lot of interest,” says Hangleiter. “But I feel like it also comes with quite a responsibility to actually be very transparent about what
quantum computers can and can't do on the one hand, and on the other hand to work as a community towards fulfilling the promises we do make.”
—Story by Bailey Bedford | {"url":"https://www.quics.umd.edu/news/hartree-fellow-cultivates-new-perspectives-quantum-computing","timestamp":"2024-11-06T07:25:00Z","content_type":"text/html","content_length":"30838","record_id":"<urn:uuid:d70fb146-7908-4a1e-a791-07cbdd4abee8>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00655.warc.gz"} |
Working with Operators in JavaScript | CodingDrills
Working with Operators in JavaScript
Working with Operators in JavaScript
Operators are an essential part of any programming language, and JavaScript is no exception. They allow us to perform various operations on data, manipulate values, and make decisions based on
conditions. In this tutorial, we will dive into the basics of working with operators in JavaScript.
Arithmetic Operators
Arithmetic operators are used to perform mathematical calculations in JavaScript. Let's take a look at some commonly used arithmetic operators:
• Addition (+): Adds two values together.
• Subtraction (-): Subtracts the second value from the first.
• Multiplication (*): Multiplies two values.
• Division (/): Divides the first value by the second.
• Modulus (%): Returns the remainder of the division.
• Increment (++): Increases the value by 1.
• Decrement (--): Decreases the value by 1.
Here's an example that demonstrates the usage of arithmetic operators:
let x = 5;
let y = 2;
let addition = x + y; // 7
let subtraction = x - y; // 3
let multiplication = x * y; // 10
let division = x / y; // 2.5
let modulus = x % y; // 1
let increment = ++x; // 6
let decrement = --y; // 1
Comparison Operators
Comparison operators are used to compare two values and return a boolean result. They are often used in conditional statements and loops. Here are some commonly used comparison operators in
• Equal to (==): Checks if two values are equal.
• Not equal to (!=): Checks if two values are not equal.
• Greater than (>): Checks if the first value is greater than the second.
• Less than (<): Checks if the first value is less than the second.
• Greater than or equal to (>=): Checks if the first value is greater than or equal to the second.
• Less than or equal to (<=): Checks if the first value is less than or equal to the second.
Let's see an example that demonstrates the usage of comparison operators:
let a = 5;
let b = 3;
console.log(a == b); // false
console.log(a != b); // true
console.log(a > b); // true
console.log(a < b); // false
console.log(a >= b); // true
console.log(a <= b); // false
Logical Operators
Logical operators are used to combine multiple conditions and perform logical operations. They are often used in conditional statements to make decisions based on multiple conditions. Here are the
three logical operators in JavaScript:
• AND (&&): Returns true if both conditions are true.
• OR (||): Returns true if at least one condition is true.
• NOT (!): Returns the opposite boolean value.
Let's take a look at an example that demonstrates the usage of logical operators:
let age = 25;
let isStudent = true;
if (age >= 18 && isStudent) {
console.log("You are eligible for a student discount.");
} else {
console.log("Sorry, you are not eligible for a student discount.");
Assignment Operators
Assignment operators are used to assign values to variables. They are often used in combination with arithmetic operators to perform calculations and update variable values. Here are some commonly
used assignment operators in JavaScript:
• Assignment (=): Assigns a value to a variable.
• Addition assignment (+=): Adds a value to the variable and assigns the result.
• Subtraction assignment (-=): Subtracts a value from the variable and assigns the result.
• Multiplication assignment (*=): Multiplies the variable by a value and assigns the result.
• Division assignment (/=): Divides the variable by a value and assigns the result.
• Modulus assignment (%=): Calculates the remainder and assigns the result.
Let's see an example that demonstrates the usage of assignment operators:
let x = 5;
x += 3; // x = x + 3 (8)
x -= 2; // x = x - 2 (6)
x *= 4; // x = x * 4 (24)
x /= 6; // x = x / 6 (4)
x %= 3; // x = x % 3 (1)
In this tutorial, we have covered the basics of working with operators in JavaScript. We explored arithmetic operators, comparison operators, logical operators, and assignment operators. By
understanding and utilizing these operators effectively, you can enhance your coding skills and perform various operations in your JavaScript programs.
Remember to practice and experiment with different operators to gain a deeper understanding of their functionalities. Happy coding!
Please note that the above content is written in Markdown format. You can convert it to HTML using any Markdown to HTML converter.
Ada AI
Hi, I'm Ada, your personal AI tutor. I can help you with any coding tutorial. Go ahead and ask me anything.
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A/A test
A/A test is similar to A/B test but it compares two identical variations. While the goal of A/B testing is to find statistically significant results between control and treatment, A/A testing expects
there is no statistically significant difference between the two groups and is conducted to confirm this expectation.
Why run an A/A test?
A/A test is often used before adopting a new test platform to evaluate its reliability and accuracy. Also, organizations can use A/A test to get additional information such as conversion rate
baseline or minimum sample size, which can be used in future A/B testing.
New clients of Hackle run A/A testing to examine the reliability of our A/B testing platform.
Things to keep in mind with A/A testing
While running an A/A test, you can sometimes get unexpected results that there is a significant difference between control and treatment. However, it does not always suggest problems with the
experiment implementation or the platform.
In hypothesis testing, statistical significance is probability. A 95% statistical significance level means there is a 1 in 20 chance that the A/A test results in a significant result (type I error).
Therefore, if you confirm that the probability of getting a significant result is within the statistical significance, you can conclude that the result was due to chance.
Running hundreds thousands of A/A tests to get the probability of getting significant results is a waste of time. Instead, repeatedly redistribute users from one A/A test into two groups and
calculate the p-value or bayesian probability distribution.
This is an actual A/A test result page from Hackle that resulted in the significant result.Let’s use this experiment data to resample users into two groups repeatedly and compare the conversion rate
with a 95% significance level.
p-value distribution
Hypothesis testing rejects the null hypothesis(there is no difference between groups) when the p-value is smaller than the significance level.
Significance level is the probability of accepting the alternative hypothesis(there is a difference between groups) when the null hypothesis is true. According to this definition, the probability of
having a p-value smaller than the significance level has to be significance level.
When the null hypothesis is true, test statistic T has the distribution F(t). Assuming F(⋅) is invertible we can say p-value P = F(T) follows a uniform distribution.
This is a p-value distribution histogram obtained from 10,000 simulations using actual A/A test data. A/A test resulted with significant difference between control and treatment, but p-value is
distributed uniformly. This implies that the chance of getting this significant result is within the statistical significance.
Bayesian probability distribution
Bayesian A/B test estimates the control and treatment’s posterior distribution and calculates the probability of getting a higher objective in the treatment group. We did the bayesian simulation 500
times to draw a histogram. It shows that bayesian probability follows a bell-shaped distribution with an average of 0.5. If the probability of getting a higher conversion rate in treatment is close
to 0.5, we can conclude that there is no significant difference between the two groups.
When p-value and bayesian probability has different distribution or the A/A test keeps resulting in significant result, there might be a problem with the test design or data itself.
Common reasons why A/A test fails are:
• There are outliers in the data. → Hackle provides an outlier removing function Outliers .
• Users who are not intended or are supposed to be included in the test are shown in the exposed data. → You can confirm that traffic distribution is working properly in Live Exposure Count. It
shows whether the traffic distribution is held on the intended page and then the expose data is generated. Also, you can check whether only targeted users participate in the test.
• Users are not evenly distributed. The number of users in control and treatment varies or users with certain properties are in the same group. → Hackle distributes users using a bucketing method
with a hash function. As long as there is no problem with the experiment design, Hackle guarantees even distribution. Reference Traffic Distribution page for more detailed information.
Things to keep in mind with A/A/B testing
A/A/B test is used when you want to perform the A/A test and A/B test at the same time. Because A/A/B test distributes traffic into three groups(A, A, or B), the number of users in each group
diminishes which results in lower test power. Hence, you should have larger traffic or run the test longer to have sufficient power in A/A/B test. This is why we recommend running A/A test instead of
A/A/B test.
Updated almost 2 years ago | {"url":"https://docs-en.hackle.io/docs/aa-test","timestamp":"2024-11-13T21:19:58Z","content_type":"text/html","content_length":"462590","record_id":"<urn:uuid:6839a2d4-b856-4da2-a771-c7a1a2391ccf>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00075.warc.gz"} |
Can You Solve This Infinite Sum Puzzle? Let's Find Out!
Written on
Chapter 1: The Infinite Sum Challenge
Here’s a captivating algebra puzzle that intricately weaves together logarithms and the concept of an infinite series. The task at hand is to skillfully rearrange each term to reach our target
Are you clever enough to tackle it? I suggest pausing for a moment, grabbing your pen and paper, and giving this a shot. Once you're prepared, continue reading for the solution!
The first video titled "Finding The Sum of an Infinite Geometric Series" delves into the methods and techniques required to solve such puzzles. It will provide you with valuable insights that may aid
in your understanding.
The core of this puzzle lies in the following property:
We can bring down the exponent of n to the front of the logarithm. In our scenario, we will move 0.3, 0.3², 0.3³, and so forth, to the front like this:
Each term contains log(a), so let's factor that out:
Now, our task is to evaluate the infinite sum with a common ratio of 0.3. Given that 0.3 is less than 1, we can utilize the formula S = a/(1 - r).
In this case, both the initial term a and the common ratio r are 0.3. We can now simplify our expression, leading us to our final answer.
Photo by Demi-Felicia Vares on Unsplash
Isn’t that fascinating?
What was your thought process while working through this? I’d love to hear your insights in the comments below!
Chapter 2: Explore More Math Puzzles
The second video titled "Can You Find the Sum of the Infinite Series? | Learn How!" explores various approaches to tackling infinite series problems. It's a great resource to deepen your
Feel free to share the following collection of intriguing math puzzles with your friends:
Math Puzzles
The finest math puzzles across Medium, covering Algebra, Geometry, Calculus, Number Theory, and more.
Bella’s Weekly Math Games
Join a thrilling 48-hour math competition every week!
Thank you for taking the time to read this! If you found the article helpful, please give it a clap.
Should you feel inclined, I would greatly appreciate any support for my writing endeavors. Your contributions help sustain my academic and personal journey.
Happy Solving, Bella!
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class verde.Trend(degree)[source]#
Fit a 2D polynomial trend to spatial data.
The polynomial of degree \(N\) is defined as:
\[f(e, n) = \sum\limits_{l=0}^{N}\sum\limits_{m=0}^{N - l} e^l n^m\]
in which \(e\) and \(n\) are the easting and northing coordinates, respectively.
The trend is estimated through weighted least-squares regression. The Jacobian (design, sensitivity, feature, etc) matrix for the regression is normalized using
sklearn.preprocessing.StandardScaler without centering the mean so that the transformation can be undone in the estimated coefficients.
degree (int) – The degree of the polynomial. Must be >= 0 (a degree of zero would estimate the mean of the data).
☆ coef (array) – The estimated polynomial coefficients that fit the observed data.
☆ region (tuple) – The boundaries ([W, E, S, N]) of the data used to fit the interpolator. Used as the default region for the grid and scatter methods.
>>> from verde import grid_coordinates
>>> import numpy as np
>>> coordinates = grid_coordinates((1, 5, -5, -1), shape=(5, 5))
>>> data = 10 + 2*coordinates[0] - 0.4*coordinates[1]
>>> trend = Trend(degree=1).fit(coordinates, data)
>>> print(
... "Coefficients:",
... ', '.join(['{:.1f}'.format(i) for i in trend.coef_])
... )
Coefficients: 10.0, 2.0, -0.4
>>> np.allclose(trend.predict(coordinates), data)
A zero degree polynomial estimates the mean of the data:
>>> mean = Trend(degree=0).fit(coordinates, data)
>>> np.allclose(mean.predict(coordinates), data.mean())
>>> print("Data mean:", '{:.2f}'.format(data.mean()))
Data mean: 17.20
>>> print("Coefficient:", '{:.2f}'.format(mean.coef_[0]))
Coefficient: 17.20
We can use weights to account for outliers or data points with variable uncertainties (see verde.variance_to_weights):
>>> data_out = data.copy()
>>> data_out[2, 2] += 500
>>> weights = np.ones_like(data)
>>> weights[2, 2] = 1e-10
>>> trend_out = Trend(degree=1).fit(coordinates, data_out, weights)
>>> # Still recover the coefficients even with the added outlier
>>> print(
... "Coefficients:",
... ', '.join(['{:.1f}'.format(i) for i in trend_out.coef_])
... )
Coefficients: 10.0, 2.0, -0.4
>>> # The residual at the outlier location should be values we added to
>>> # that point
>>> residual = data_out - trend_out.predict(coordinates)
>>> print('{:.2f}'.format(residual[2, 2]))
Methods Summary
Trend.filter(coordinates, data[, weights]) Filter the data through the gridder and produce residuals.
Trend.fit(coordinates, data[, weights]) Fit the trend to the given data.
Trend.get_params([deep]) Get parameters for this estimator.
Trend.grid([region, shape, spacing, dims, ...]) Interpolate the data onto a regular grid.
Trend.jacobian(coordinates[, dtype]) Make the Jacobian matrix for a 2D polynomial.
Trend.predict(coordinates) Evaluate the polynomial trend on the given set of points.
Trend.profile(point1, point2, size[, dims, ...]) Interpolate data along a profile between two points.
Trend.scatter([region, size, random_state, ...]) Interpolate values onto a random scatter of points.
Trend.score(coordinates, data[, weights]) Score the gridder predictions against the given data.
Trend.set_params(**params) Set the parameters of this estimator.
Trend.filter(coordinates, data, weights=None)#
Filter the data through the gridder and produce residuals.
Calls fit on the data, evaluates the residuals (data - predicted data), and returns the coordinates, residuals, and weights.
Not very useful by itself but this interface makes gridders compatible with other processing operations and is used by verde.Chain to join them together (for example, so you can fit a spline on
the residuals of a trend).
☆ coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). For the specific definition of
coordinate systems and what these names mean, see the class docstring.
☆ data (array or tuple of arrays) – The data values of each data point. If the data has more than one component, data must be a tuple of arrays (one for each component).
☆ weights (None or array or tuple of arrays) – If not None, then the weights assigned to each data point. If more than one data component is provided, you must provide a weights array for
each data component (if not None).
coordinates, residuals, weights – The coordinates and weights are same as the input. Residuals are the input data minus the predicted data.
Trend.fit(coordinates, data, weights=None)[source]#
Fit the trend to the given data.
The data region is captured and used as default for the grid and scatter methods.
All input arrays must have the same shape.
☆ coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). Only easting and northing will be used,
all subsequent coordinates will be ignored.
☆ data (array) – The data values of each data point.
☆ weights (None or array) – If not None, then the weights assigned to each data point. Typically, this should be 1 over the data uncertainty squared.
self – Returns this estimator instance for chaining operations.
Get parameters for this estimator.
deep (bool, default=True) – If True, will return the parameters for this estimator and contained subobjects that are estimators.
params (dict) – Parameter names mapped to their values.
Trend.grid(region=None, shape=None, spacing=None, dims=None, data_names=None, projection=None, coordinates=None, **kwargs)#
Interpolate the data onto a regular grid.
The grid can be specified by two methods:
□ Pass the actual coordinates of the grid points, as generated by verde.grid_coordinates or from an existing xarray.Dataset grid.
□ Let the method define a new grid by either passing the number of points in each dimension (the shape) or by the grid node spacing. If the interpolator collected the input data region, then it
will be used if region=None. Otherwise, you must specify the grid region. See verde.grid_coordinates for details. Other arguments for verde.grid_coordinates can be passed as extra keyword
arguments (kwargs) to this method.
Use the dims and data_names arguments to set custom names for the dimensions and the data field(s) in the output xarray.Dataset. Default names will be provided if none are given.
☆ region (list = [W, E, S, N]) – The west, east, south, and north boundaries of a given region. Use only if coordinates is None.
☆ shape (tuple = (n_north, n_east) or None) – The number of points in the South-North and West-East directions, respectively. Use only if coordinates is None.
☆ spacing (tuple = (s_north, s_east) or None) – The grid spacing in the South-North and West-East directions, respectively. Use only if coordinates is None.
☆ dims (list or None) – The names of the northing and easting data dimensions, respectively, in the output grid. Default is determined from the dims attribute of the class. Must be defined
in the following order: northing dimension, easting dimension. NOTE: This is an exception to the “easting” then “northing” pattern but is required for compatibility with xarray.
☆ data_names (str, list or None) – The name(s) of the data variables in the output grid. Defaults to 'scalars' for scalar data, ['east_component', 'north_component'] for 2D vector data, and
['east_component', 'north_component', 'vertical_component'] for 3D vector data.
☆ projection (callable or None) – If not None, then should be a callable object projection(easting, northing) -> (proj_easting, proj_northing) that takes in easting and northing coordinate
arrays and returns projected northing and easting coordinate arrays. This function will be used to project the generated grid coordinates before passing them into predict. For example,
you can use this to generate a geographic grid from a Cartesian gridder.
☆ coordinates (tuple of arrays) – Tuple of arrays containing the coordinates of the grid in the following order: (easting, northing, vertical, …). The easting and northing arrays could be
1d or 2d arrays, if they are 2d they must be part of a meshgrid. If coordinates are passed, region, shape, and spacing are ignored.
grid (xarray.Dataset) – The interpolated grid. Metadata about the interpolator is written to the attrs attribute.
See also
Generate the coordinate values for the grid.
Trend.jacobian(coordinates, dtype='float64')[source]#
Make the Jacobian matrix for a 2D polynomial.
Each column of the Jacobian is easting**i * northing**j for each (i, j) pair in the polynomial.
☆ coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). Only easting and northing will be used,
all subsequent coordinates will be ignored.
☆ dtype (str or numpy dtype) – The type of the output Jacobian numpy array.
jacobian (2D array) – The (n_data, n_coefficients) Jacobian matrix.
>>> import numpy as np
>>> east = np.linspace(0, 4, 5)
>>> north = np.linspace(-5, -1, 5)
>>> print(Trend(degree=1).jacobian((east, north), dtype=int))
[[ 1 0 -5]
[ 1 1 -4]
[ 1 2 -3]
[ 1 3 -2]
[ 1 4 -1]]
>>> print(Trend(degree=2).jacobian((east, north), dtype=int))
[[ 1 0 -5 0 0 25]
[ 1 1 -4 1 -4 16]
[ 1 2 -3 4 -6 9]
[ 1 3 -2 9 -6 4]
[ 1 4 -1 16 -4 1]]
Evaluate the polynomial trend on the given set of points.
Requires a fitted estimator (see fit).
coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). Only easting and northing will be used, all
subsequent coordinates will be ignored.
data (array) – The trend values evaluated on the given points.
Trend.profile(point1, point2, size, dims=None, data_names=None, projection=None, **kwargs)#
Interpolate data along a profile between two points.
Generates the profile along a straight line assuming Cartesian distances. Point coordinates are generated by verde.profile_coordinates. Other arguments for this function can be passed as extra
keyword arguments (kwargs) to this method.
Use the dims and data_names arguments to set custom names for the dimensions and the data field(s) in the output pandas.DataFrame. Default names are provided.
Includes the calculated Cartesian distance from point1 for each data point in the profile.
To specify point1 and point2 in a coordinate system that would require projection to Cartesian (geographic longitude and latitude, for example), use the projection argument. With this option, the
input points will be projected using the given projection function prior to computations. The generated Cartesian profile coordinates will be projected back to the original coordinate system.
Note that the profile points are evenly spaced in projected coordinates, not the original system (e.g., geographic).
The profile calculation method with a projection has changed in Verde 1.4.0. Previous versions generated coordinates (assuming they were Cartesian) and projected them afterwards. This led to
“distances” being incorrectly handled and returned in unprojected coordinates. For example, if projection is from geographic to Mercator, the distances would be “angles” (incorrectly calculated
as if they were Cartesian). After 1.4.0, point1 and point2 are projected prior to generating coordinates for the profile, guaranteeing that distances are properly handled in a Cartesian system.
With this change, the profile points are now evenly spaced in projected coordinates and the distances are returned in projected coordinates as well.
☆ point1 (tuple) – The easting and northing coordinates, respectively, of the first point.
☆ point2 (tuple) – The easting and northing coordinates, respectively, of the second point.
☆ size (int) – The number of points to generate.
☆ dims (list or None) – The names of the northing and easting data dimensions, respectively, in the output dataframe. Default is determined from the dims attribute of the class. Must be
defined in the following order: northing dimension, easting dimension. NOTE: This is an exception to the “easting” then “northing” pattern but is required for compatibility with xarray.
☆ data_names (str, list or None) – The name(s) of the data variables in the output dataframe. Defaults to 'scalars' for scalar data, ['east_component', 'north_component'] for 2D vector
data, and ['east_component', 'north_component', 'vertical_component'] for 3D vector data.
☆ projection (callable or None) – If not None, then should be a callable object projection(easting, northing, inverse=False) -> (proj_easting, proj_northing) that takes in easting and
northing coordinate arrays and returns projected northing and easting coordinate arrays. Should also take an optional keyword argument inverse (default to False) that if True will
calculate the inverse transform instead. This function will be used to project the profile end points before generating coordinates and passing them into predict. It will also be used to
undo the projection of the coordinates before returning the results.
table (pandas.DataFrame) – The interpolated values along the profile.
Trend.scatter(region=None, size=300, random_state=0, dims=None, data_names=None, projection=None, **kwargs)#
Interpolate values onto a random scatter of points.
Point coordinates are generated by verde.scatter_points. Other arguments for this function can be passed as extra keyword arguments (kwargs) to this method.
If the interpolator collected the input data region, then it will be used if region=None. Otherwise, you must specify the grid region.
Use the dims and data_names arguments to set custom names for the dimensions and the data field(s) in the output pandas.DataFrame. Default names are provided.
The scatter method is deprecated and will be removed in Verde 2.0.0. Use verde.scatter_points and the predict method instead.
☆ region (list = [W, E, S, N]) – The west, east, south, and north boundaries of a given region.
☆ size (int) – The number of points to generate.
☆ random_state (numpy.random.RandomState or an int seed) – A random number generator used to define the state of the random permutations. Use a fixed seed to make sure computations are
reproducible. Use None to choose a seed automatically (resulting in different numbers with each run).
☆ dims (list or None) – The names of the northing and easting data dimensions, respectively, in the output dataframe. Default is determined from the dims attribute of the class. Must be
defined in the following order: northing dimension, easting dimension. NOTE: This is an exception to the “easting” then “northing” pattern but is required for compatibility with xarray.
☆ data_names (str, list or None) – The name(s) of the data variables in the output dataframe. Defaults to 'scalars' for scalar data, ['east_component', 'north_component'] for 2D vector
data, and ['east_component', 'north_component', 'vertical_component'] for 3D vector data.
☆ projection (callable or None) – If not None, then should be a callable object projection(easting, northing) -> (proj_easting, proj_northing) that takes in easting and northing coordinate
arrays and returns projected northing and easting coordinate arrays. This function will be used to project the generated scatter coordinates before passing them into predict. For example,
you can use this to generate a geographic scatter from a Cartesian gridder.
table (pandas.DataFrame) – The interpolated values on a random set of points.
Trend.score(coordinates, data, weights=None)#
Score the gridder predictions against the given data.
Calculates the R^2 coefficient of determination of between the predicted values and the given data values. A maximum score of 1 means a perfect fit. The score can be negative.
The default scoring will change from R² to negative root mean squared error (RMSE) in Verde 2.0.0. This may change model selection results slightly. The negative version will be used to maintain
the behaviour of larger scores being better, which is more compatible with current model selection code.
If the data has more than 1 component, the scores of each component will be averaged.
☆ coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). For the specific definition of
coordinate systems and what these names mean, see the class docstring.
☆ data (array or tuple of arrays) – The data values of each data point. If the data has more than one component, data must be a tuple of arrays (one for each component).
☆ weights (None or array or tuple of arrays) – If not None, then the weights assigned to each data point. If more than one data component is provided, you must provide a weights array for
each data component (if not None).
score (float) – The R^2 score
Set the parameters of this estimator.
The method works on simple estimators as well as on nested objects (such as Pipeline). The latter have parameters of the form <component>__<parameter> so that it’s possible to update each
component of a nested object.
**params (dict) – Estimator parameters.
self (estimator instance) – Estimator instance.
Examples using verde.Trend# | {"url":"https://www.fatiando.org/verde/v1.8.0/api/generated/verde.Trend.html","timestamp":"2024-11-12T00:08:23Z","content_type":"text/html","content_length":"86184","record_id":"<urn:uuid:9b3eb402-4370-422c-9ba3-beeb191b9294>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00026.warc.gz"} |
wu :: forums - wine and water
wu :: forums
putnam exam (pure math) (Moderators: Icarus, SMQ, towr, william wu, Eigenray, Grimbal) « Previous topic | Next topic »
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Author Topic: wine and water (Read 840 times)
skogen wine and water
Newbie « on: Apr 27^th, 2007, 6:36am » Quote Modify
Hi this is my first time posting in this forum, and that is because i want to know the answer of the riddle with the water and the wine
"Suppose a 1-liter bottle of wine is hanging from the ceiling, and from it hangs a 1-liter bottle of water. On the bottom of both bottles there is a small hole, through which a constant
amount of fluid pours out (neglecting pressure differences). So the wine bottle is becoming empty and the water bottle remains full, though the concentration of wine is increasing.
Assuming that wine and water mix instantly and completely, find out the concentration of wine in the water bottle after the top one is empty. Give two different methods of solution."
Posts: I have searched for the solution but i cant find it, and I cant solve it myself, anyway here able to help?
towr Re: wine and water
wu::riddles Moderator « Reply #1 on: Apr 27^th, 2007, 8:05am » Quote Modify
hmm, well. Consider there are X drops in a liter.
You can then consider the change in concentration at each step. The recursion for the concentration C will be
C(n+1)=[1+(X-1)*C(n) ]/X starting at C(0)=0.
You can turn that into a closed formula, and look at C(X) (the concentration for the last drop of wine)
Some people are average, some are just mean.
Now possibly, you want an answer for the continuous case (i.e. not drops but a stream), in which case take the limit of X
Posts: 13730
Wikipedia, Google, Mathworld, Integer sequence DB
Obob Re: wine and water
Senior « Reply #2 on: Apr 27^th, 2007, 9:56am » Quote Modify
I think this method works.
Say water drips out of each container at a rate of r liters/second. Denote by C(t) the concentration of wine in the lower bottle at time t. So C(0)=0. Now the rate at which the concentration
is changing is given by C'(t) = r(1-C(t)). The first term on the right comes from the wine pouring in on the top, which increases the concentration, while the second term accounts for the
decrease in concentration. One can solve this differential equation, and the solution is
C(t) = 1-exp(-rt).
Now the wine bottle is empty at the time t_0 that rt_0 = 1 liter, and at this time
Posts: C(t_0) = 1 - exp(-1).
So the concentration should be 1-e^{-1}, which is approximately .63. The only step which I'm a little shaky on is justifying that the differential equation is the right one, but seeing as
this answer is physically plausible I think it probably is.
skogen Re: wine and water
Newbie « Reply #3 on: Apr 27^th, 2007, 9:57am » Quote Modify
I must say I did not quite understand that:S What is the n? the number of drops?
I have a teacher in math, he said I would get the answer by typing 1/e1 on a calculator, will that be right?
Posts: 3 « Last Edit: Apr 27^th, 2007, 9:58am by skogen »
Obob Re: wine and water
Senior « Reply #4 on: Apr 27^th, 2007, 10:06am » Quote Modify
I think 1/e would be the concentration of water in the water bottle after the wine has all dripped in. If you think about it, until there is as much wine in the water bottle as there is
water, there will be more water that is lost than wine from the bottom bottle. So the concentration of wine must be at least 50%, seeing as more water is lost than wine.
skogen Re: wine and water
Newbie « Reply #5 on: Apr 27^th, 2007, 10:21am » Quote Modify
I do not know how to get the 1/e1, I do not know what to do at all really, at least I think it is very hard to think now
Posts: 3
Obob Re: wine and water
Senior « Reply #6 on: Apr 27^th, 2007, 10:52am » Quote Modify
Chances are if you haven't heard of the number e yet, then this problem is too advanced for you at this point. Both towr and my solutions require some knowledge of calculus.
The explicit solution for towr's recurrence is
c(n) = 1 - (1-1/x)^n.
(Mathematica's RSolve command gives this, I didn't know this nice command before!) Taking n=x and letting x-> infinity gives the same answer 1-1/e as I got by solving a differential
Posts: 489 « Last Edit: Apr 27^th, 2007, 2:59pm by Obob »
JiNbOtAk Re: wine and water
Uberpuzzler « Reply #7 on: Apr 30^th, 2007, 4:44am » Quote Modify
Hana Hana No Mi
Posts: 1187 Quis custodiet ipsos custodes?
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wu :: forums - 0.999.
wu :: forums
medium (Moderators: SMQ, william wu, ThudnBlunder, Grimbal, Icarus, Eigenray, towr) « Previous topic | Next topic »
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Author Topic: 0.999. (Read 125743 times)
pythagoras Re: 0.999.
Newbie « Reply #50 on: Aug 12^th, 2002, 1:17pm »
infinity + 1? As said earlier, infinity needs to be thought of differently than numbers. It is generally accepted that
... = infinity - 1 = infinity = infinity + 1 = ...
I can't quite follow NoYes's logic, but when we "reach" the limit we have "reached" infinity. Neither of these can actually be done, and since infinity is really the same as infinity + 1,
there is no real problem.
There are bigger consequences of treating infinity as a number.
Posts: 3 Say
S = 1 + 2 + 4 + 8 + ...
S = 1 + 2(1 + 2 + 4 + ...)
S = 1 + 2S
-S = 1
S = -1
heywood Re: 0.999.
Guest « Reply #51 on: Aug 12^th, 2002, 1:40pm »
Heres another example/riddle dealing with infinite sums that I find quite amusing. Suppose you have a perfect superball that bounces half its height on every bounce. Disregard all friction
and other sources of energy loss (i.e. if the ball starts at 1 meter then the next bounce it will reach EXACTLY 1/2 a meter.) Now the question is, how long does it take the ball to stop
This was a great problem from an early calculus class I had.
HaPpY Re: 0.999.
Guest « Reply #52 on: Aug 12^th, 2002, 2:22pm »
as n approaches infinity where .999... n is the number of digits past the decimal the value approaches 1. therefore if infinity is "reached" the value becomes one.
Paul Re: 0.999.
Hsieh « Reply #53 on: Aug 12^th, 2002, 4:25pm »
Grumble ...
1. You cannot "set n = infinity" for the simple reason that "infinity" is not a number. Its a concept.
2. One cannot so easily just "take the limit as n goes to infinity" unless you have established each step very carefully. It turns out in this trivial case, there are no issues and doing so
presents no problems, but it is in a sense a circular argument. I.e., can't do it until you've established a convergence, which is really part of what we are trying to do in the first place.
I did not mean that:
(1-a^n) = (1-a) * (1+a+a^2+...+a^n-1)
Implies that:
0.9999 ... = 9/10*(1.1111 ...)
These are actually two seperate and independent facts that are just true on their face. I tie the two together in later steps in my analysis.
Mongolian_Beef Re: 0.999.
Newbie « Reply #54 on: Aug 13^th, 2002, 10:02pm »
I think the issue really is our system and the conversions that it entails. The point is that the way our system functions, as many of you seem to be arguing in its context, .9999999
repeating is equivalent to 1. Arguing over it also appears pointless because regardless of opinion the variance between .99999999 repeating and 1 is negligible in any real world
Posts: 11
chris mallinson Re: 0.999.
Guest « Reply #55 on: Aug 22^nd, 2002, 9:41pm »
It has been said here that
.99999....... multiplied by 10 is 9.999999999....
.99999....... plus 9 is also 9.999999999....
Is there something to be said for the fact that the latter will theoretically approach the number 10 faster than the former?
Jason Re: 0.999.
Short « Reply #56 on: Aug 22^nd, 2002, 11:42pm »
The notation I've seen for 0.999... (infinitely repeating) is to use 0.9 with a line over the 9. Instead, I'll put the line under the 9 so it becomes 0.9.
If we switched from base 10 to base 20, then we could call our digits {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J}. Thus "J" is the digit "19". This may be comparable to the # digit spoken of
earlier: the average of 0.9[10] and 1 is 0.J[20]. So the number 0.9[10] could be compared to 0.J[20], and we would be tempted to think that the latter was larger. But we are implicitly
assuming here that they are different; in actuality both are simply equal to 1.
Here is a related topic: a paradox I once heard. Imagine a game of chance in which a coin is flipped indefinitely until eventually it lands tails. You then count the number of times it landed
heads (n), and the person playing the game gets 2^n dollars. The question then becomes: how much money does a player expect to win playing this game? Logically, we would think this is ((1/2)
*1+(1/4)*2+(1/8)*4+...) since there is (for instance) a 1/2 chance that heads will be flipped 0 times, and if this happens the player will win $1. But each term simplifies to 1/2, so the
series becomes (1/2+1/2+1/2+...). The infinite series of 1/2's seems to sum to infinity: it seems as though the payoff for this game is infinite. Yet this seems rather rediculous; in fact
we've assumed that there is an "expected" payoff. The actuality is that the expected payoff is undefined.
My point is that infinity isn't just a rather large number, it is a concept that has undefined value. There is no such thing as "infinity plus one" (^^). You simply cannot treat infinity as a
value. Thus it is meaningless (although seemingly clever) to consider the value 0.99, and claim that it is larger than 0.9. In this case, assuming the given decimal representation has a value
(which is sort of implicitly true, since decimal notation is simply an abstraction we made up to represent values), the best we can do is treat it as an infinite series and reduce that series
to find the value. This has been done several times elsewhere: the value is 1.
By contrast, the value 9 is "infinite", and certainly does not have a value. It would be quite meaningless to compare it to, for instance, 8, and try to say that it is "larger" since 9>8. 9>8
does not prevent the fact that 88>9.
wowbagger Re: 0.999.
Uberpuzzler « Reply #57 on: Aug 24^th, 2002, 4:16pm »
on Jul 30^th, 2002, 1:52am, Kozo Morimoto wrote:
[...] So can you start with any finite number and keep dividing it by 10 and add 0.9 to it, and if you repeat the process to infinity you get 1? [...]
As far as I followed the discussion, nobody replied to this specific question yet, so here's my proof that yes, you get 1:
What you do is calculate a sequence of numbers generated by the recursion formula
Posts: 727 a_{n+1} = a_n / 10 + 0.9, starting with some arbitrary a_0 (e.g. 0.8)
If you "repeat the process to infinity", you take the limit of both sides of the recursion equation as n goes to infinity. If the limit of the sequence a_n exists - let's call it A - you
can substitute a_n and a_{n+1} by A, since being elements of the same sequence, they must have the same limit A. So we have
A = A / 10 + 0.9
9 / 10 A = 0.9
A = 1
Note that this result doesn't depend on your initial value a_0, it works for all real a_0. If you start with a_0 > 1, you will approach the limit from above, however.
In the same way one can show that
a_{n+1} = (a_n + 2 / a_n) / 2
goes to sqrt(2), i.e. A*A = 2 in this case.
Well, I don't know whether this is of much help as some will probably feel uncomfortable with the substitution of the limit A, but I didn't want to leave the question unanswered.
Kozo: I really respect your being sceptical. BTW, I'm convinced (not to say I know) the infinite series does sum up to 1. The bit about the "#" was really weird, though. Contributions of
yours to other threads I read are also stimulating - keep it up!
« Last Edit: Aug 24^th, 2002, 4:41pm by wowbagger »
"You're a jerk, <your surname>!"
pythagoras Re: 0.999.
Newbie « Reply #58 on: Aug 25^th, 2002, 3:39pm »
on Aug 22^nd, 2002, 11:42pm, Jason Short wrote:
By contrast, the value 9 is "infinite", and certainly does not have a value.
Well, actually this has been argued against by people I know. It's kind of nonsensical to imagine that the 9's repeat to the right in this case, so say they repeat to the left. I'll
denote something repeats to the left in parentheses like this: (9)9.
Posts: 3 Now consider (0)0 - (0)1, that is:
Let's try to do this like we don't know what a negative number is. Since 1 is greater than 0 we have to borrow a 1 from, uh, somewhere... (An imaginary 1 in front of all the zeroes?) to
and since we borrowed our theoretical subtraction will end up:
so (0)0 - (0)1 = (9)9. That's odd.
There is actually a concept called n-adics that deals with this sort of thing, and this is just an example in the 10-adics.
justin Re: 0.999.
m « Reply #59 on: Aug 28^th, 2002, 2:40pm »
just like the limit of [f(x+h) - f(x)] / [x-h] as h->0 is a "pretty good approximation of the rate of change of f
its not just "pretty good", if you evaluate the limit, this IS the rate of change of f, its a simplet hing called the derivative.
just like the infinite geometric series 9/10 + 9/100 + ... is not just a "pretty good" approximation, when you evaluate the limit you get an exact number, 1
additionally, 1.0000...001 is mathmatical nonsense, because you cant terminate a decimal that, by definition with the "...", is non-terminating.
perhaps a college math course or two will help you swallow your pride
on Jul 29^th, 2002, 6:40am, Kozo Morimoto wrote:
0.9 is pretty close to 1, an approximation of 1.
You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1. Repeat to infinity. You get ever closer and closer to 1 from the low side, but you never get there?
Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1? So Does it mean that 0.999... = 1.000...001 ? How about 0.999...998
= 0.999... = 1.000...001?
wowbagger Re: 0.999.
Uberpuzzler « Reply #60 on: Aug 29^th, 2002, 2:27am »
on Aug 28^th, 2002, 2:40pm, justin m wrote:
just like the limit of [f(x+h) - f(x)] / [x-h] as h->0 is a "pretty good approximation of the rate of change of f
Of course, the correct formula is
f'(x) = lim_{h->0} ( f(x+h) - f(x) ) / h
Posts: 727 Perhaps people writing illiterate posts shouldn't encourage others to take courses?
"You're a jerk, <your surname>!"
Kozo Re: 0.999.
Morimoto « Reply #61 on: Aug 31^st, 2002, 11:32pm »
Member OK, here is another thought exercise - I haven't work it all the way thru but...
Imagine a number system with 20 digits A to T, but still in base 10 so that A maps to 0, C maps to 1 etc until S maps to 9 and T maps to something between 9 and 10. (B maps to somewhere
in between 0 and 1, D maps between 1 and 2 etc)
So when you have 0.9 in normal system, that would be same as A.S, however, there is a number A.T which is bigger than 0.9 and less than 1.0.
So 0.999... would map to A.SSS.... but there is a number A.TTT... which is bigger than 0.999... but less than 1.0?
Posts: 114
Pietro Re: 0.999.
K.C. « Reply #62 on: Sep 1^st, 2002, 11:57pm »
Member Sorry to insist on this point, but it seems necessary for the argument I'm going to give regarding Kozo's question.
The set of characters "0.999..." DOES NOT MEAN adding the numbers 0.9 + 0.09 + etc, it is not the result of some infinite process, nor any other complicated nonsense like that. For the
simple reason that an infinite process (be it addition or whatever) does not yield a result. Srowen and aadash have explained it best.
The set of characters "0.999..." means the limit that the sequence (0.9 ; 0.99 ; 0.999 ; etc) tends to, in the sense of Cauchy, which is the following: a sequence (a(n)) = (a1, a2, etc)
tends to a limit L if, for every real number r, there exists a natural number N such that |a(n) - L| < r for all n > N. The definition of the particular sequence we are dealing with does
involve addition, but the concept of a limit does not depend on it; the concept of infinite addition is inconsistent.
Come on, what is the point in knowing complicated integrals if you don't have a grasp of the beautiful underlying structure? No one denies that no finite-numbered term of the sequence equals
Gender: or exceeds 1, but to say that, or to attempt proofs just by adding the succesive terms, is to miss entirely the meaning of the expression.
213 Now, all of you have excellent problem-solving skills, as can be seen by your posts. Surely no one wil have trouble proving that the number 1 is a limit of the sequence (0.9 ; 0.99 ; etc)
using the definition above. Hint: the difference between 1 and the nth term of this sequence is 10 taken to the (-n)th power. (As an exercise, if you haven't already done it, you may also
prove that, if a sequence of real numbers has a limit, it is unique) Hence the question "0.999... = 1" is resolved, and the answer is affirmative.
About Kozo's alternative number system: the fact that a number T is less than 10 does not mean that 0.TTT... is less than 1. For instance, 9.9. If we take 0.TT to mean T/10 + T/100, then
0.TT = 0.99 + 0.099 = 1.089. Since 0.TTT... > 0.TT, it must also be greater than 1.089, which is greater than 1. However, my spider-sense (and the previous posts) tells me that this isn't
enough of an argument for Kozo. He may argue that there is some number not expressible in our decimal system for which the properties 9 < T < 10 and 0.TTT... < 1 simultaneousy hold. I intend
to show that no such number exists.
Never mind the number system, and suppose that 9 < T < 10. I take it we can represent ANY number by the letter T. Now construct the sequence S = (0.T ; 0.TT ; etc) by means of sums with
progressively more numerous terms, in the same manner as we constructed the infamous "0.999...". By the expression "0.TTT..." we shall then mean the limit of this sequence, if indeed it does
approach a limit.
To prove that the limit exists, we could resort to the axiom of completeness of the set of real numbers, or to the axiom of existence of a supremum of an increasing bounded sequence (which
is equivalent, cf any introductory text on real analysis, of which there are many - if you would like the proof, e-mail me). But, since it obviously suffices to show a number that satisfies
the limit criterion, and this is a simple case, we shall do so.
I say that T/9 is the limit of the sequence S. To see this, consider the difference T/9 - (the nth term of S); a simple proof by induction shows that this difference is T/(9*10^n) for all
natural n. Clearly, for any number T, this difference can be made as small as we please by increasing n; in Cauchy's words, for all r > 0, there is an N such that T/(9*10^n) is less than r
for all n > N.
Hence, T/9 satisfies the definition of a limit of the sequence S = (0.T ; 0.TT; etc). And therefore 0.TTT... = T/9. It is not a coincidence that 0.TTT... "=" T*0.111... = T*1/9 = T/9. The
multiplicative property of limits can be used to rigorously remove the sarcastic quotation marks around the equal sign.
Now, you may notice that T/9 > 1 for all T > 9, which was the original hypothesis. So no number "not expressible in our decimal number system" exists that satisfies both T > 9 and 0.TTT... <
1 at once. That is the basic flaw in the # argument. As a side point, I would like to remark that there exists no numbers that cannot be arbitrarely well approximated by decimal expansions
(prove this); so any eventual 0.###... would have a representation involving only the digits 0 through 9. If we were to make it greater than 0.999... but less than 1, the first digit would
have to be 9; but also the second; and so forth, so that a number greater than 0.999... and less than 1 cannot exist. But this type of argument alone cannot make up for the rigor of the
preeding proof.
I'm sorry if I wrote too much, it's just that I would like to clear this up once and for all, so that such smart people (I was impressed by many others of Kozo's ingenious posts) would not
waste their valuable time on matters of definition.
"I always wondered about the meaning of life. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. It was not what I expected." (Dogbert)
Xanthos Re: 0.999.
Guest « Reply #63 on: Sep 2^nd, 2002, 5:42pm »
What about using the sum to infinity? If we take 0.999.... as 0.9+0.09+0.009 etc. then it is a geometric sequence. Therefore if we use the sum to infinity formula it should give us what
0.999.... is equal to.
S[inf]= a/(1-r)
In this case the starting number (a) is 9 and the common ratio (r) is 0.1. Substitute and you get 0.9/0.9, which is equal to 1.
I might have missed something and it might have been said before but I thought I'd weigh in with that.
Kozo Morimoto Re: 0.999.
Junior Member « Reply #64 on: Sep 2^nd, 2002, 7:49pm »
0.T maps to 19/20 = 0.95 which is > 0.9 and < 1
0.TT maps to 19/20 + 19/400 = 0.9975 which is > 0.99 and < 1
0.TTT maps to 19/20 + 19/400 + 19/8000 = 0.999875 which is > 0.999 and < 1
so on.
so wouldn't 0.999... < 0.TTT... < 1 ?
Posts: 114
NickH Re: 0.999.
Senior « Reply #65 on: Sep 2^nd, 2002, 11:27pm »
"so wouldn't 0.999... < 0.TTT... < 1 ?"
No. In the limit, both are equal to 1. What you have written as 0.TTT... is simply the base 20 equivalent of 0.999.... We can also consider 0.111... (base 2), 0.222... (base 3),.... In
all cases the sum to infinity of the geometrical progression is 1.
Posts: 341
Nick's Mathematical Puzzles
S. Owen Re: 0.999.
Full Member « Reply #66 on: Sep 3^rd, 2002, 5:41am »
How about this, define two sequences:
A[i] = 0.999...9 (there are i 9s)
B[i] = 0.TTT...T (there are i Ts)
So the sequences are:
{0.9, 0.99, 0.999, ...}
Gender: {0.T, 0.TT, 0.TTT, ...}
Posts: 221
It is true that A[n] < B[n] < 1 for any n. However, the limit of both sequences is still 1.
James Re: 0.999.
Fingas « Reply #67 on: Sep 3^rd, 2002, 11:44am »
This is a killer thread!
Has anyone done epsilon-delta proofs? This is the most basic (consistent) way of saying that a given limit exists and converges to the specified point.
The way you use them in this case is to say:
The series 0.9 + 0.09 + 0.009 + ... converges to 1 iff for every epsilon around 1, there exists an M so that by summing up the first M terms of the series, the sum is within epsilon of
Gender: Basically, an epsilon-delta proof challenges you to find a difference--any difference--between the series 0.9 + 0.09 + 0.009 + ... and the number 1. It says that any purported
Posts: 949 "difference" that you find will vanish after I sum enough terms. Since you cannot find a difference, then the numbers MUST be the same. To prove this, I show how to get M from epsilon.
This leads to the following argument:
1) Assume that the series 0.9 + 0.09 + 0.009 + ... sums to a real number a. By the ratio test, the series converges to a single value.
2) Now we assume that a != 1.
3) This means that a-1 has a non-zero value. Let's take half the absolute value of a-1, and call it epsilon.
4) But if we take M = 1 - log10( epsilon ), and sum the first M terms of the series, we will find that it is closer to 1 than epsilon. This value of epsilon is therefore rejected. Since
this holds for any epsilon, there are NO NUMBERS between 1 and a.
5) But this contradicts our assertion that a-1 != 0. Consequently, it must be true that a=1.
If there is no difference between a and 1, then I argue that they must be the same number.
Doc, I'm addicted to advice! What should I do?
Pietro Re: 0.999.
K.C. « Reply #68 on: Sep 3^rd, 2002, 2:39pm »
Member Yes, I did do a "delta-epsilon" proof, even though there is no delta when dealing with sequences, but as I thought that epsilon was an ugly name for a variable, when written out whole, I
used the letter r instead. And in keeping with the spirit of this site, I didn't spell out all the details, such as constructing the number N such that
|limit - sum up to the nth term| < epsilon for all n > N.
Maybe I do deserve not being read for posting such a huge text.
But I repeat: arguments of the type "the series sums to..." will not convince Kozo, and he is right. These arguments are incosistent, and his rebuttals go to the heart of this incosistency.
Kozo, I don't know if I misinterpreted your previous construction or if your last post is a new question,but anyways. You can apply the same type of find-the-difference-as-a-function-of-n
Gender: proof I gave in my LARGE post to show that "0.TTT..." = limit (0.T ; 0.TT; etc) (now in base 20) = T/19. For T = 19, this limit is 1. Same as my proof for 0.999... in base 10. Let me give
Posts: YOU a puzzle now.
Let the sequences S1, S2 be defined by:
S1(n) = 1 - (1/n)
S2(n) = 1 - (2/n)
and let the numbers A, B be their limits as n->oo, respectively. Would you say that A < B, just because each term in S1 is less than the corresponding term in S2? That is not keeping with
the definition of a limit. The limit of a sequence need not be included in that sequence, as the above examples demonstrate. Clearly, A = lim S1 = 1, and B = lim S2 = 1, but 1 is not a term
in either sequence.
The same phenomenon occurs in the definition of 0.999... and 0.TTT...; these symbols indicate limits of sequences, and are not actually terms, and so the inequality does not have to hold.
The key here is that "a(n) < b(n) for all n" does NOT imply that "lim a(n) < lim b(n)".
Hope that helps.
"I always wondered about the meaning of life. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. It was not what I expected." (Dogbert)
Kozo Re: 0.999.
Morimoto « Reply #69 on: Sep 4^th, 2002, 5:21am »
Member I've come to the conclusion that this has come to a point of definition/symmantics.
I googled and came up with:
which I think covers this discussion to my satisfaction in an easily accessible form.
The definition I was using is covered by point 1 on the referred website, and in this context, I was right.
The definition that everyone else here was using is covered by point 2 on the referred website, and in that context, everyone was correct. In this definition, convergence and infinity is
Posts: 114 used interchangeably but I was treating them as 2 different 'things'.
It certainly has been an interesting discussion though!
Pietro K.C. Re: 0.999.
Full Member « Reply #70 on: Sep 4^th, 2002, 12:08pm »
I looked at the site, but I still don't get how point #1 validates your point of view.
It says infinity doesn't exist as an element of a "number system", but very few of us proposed that it did. I really didn't understand what you said.
I still think that a string of symbols such as 0.999... requires the concept of convergence to make sense, and is meaningless in the "number system" context alone.
Posts: 213
"I always wondered about the meaning of life. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. It was not what I expected." (Dogbert)
James Fingas Kozo was right!
Uberpuzzler « Reply #71 on: Sep 5^th, 2002, 11:43am »
It turns out that Kozo was right all along. There is a difference between 0.9999... and 1.
At least the floor function says so
Posts: 949
Doc, I'm addicted to advice! What should I do?
Dustin Re: 0.999.
Guest « Reply #72 on: Sep 6^th, 2002, 5:58pm »
.99999... < 1
if someone wrote .99999... on a piece of paper for there entire lives, and every last decended of theirs did the same thing, it would still never reach 1.
Someone also earlier said that if
1/3 = .3333...
2/3 = .6666...
then technically 3/3 would = .9999....
However since we know that 3/3 = 1 then .9999... must also = 1
This idea is flawed because of the fact that the representation of both 1/3 and 2/3 is also flawed. 1/3 does not = exactly .333... and 2/3 does not = exactly .666...
The reason math labels it as that is because with out math system the closes representation we can get of 1/3 is .333... but that does not make it = to .333...
and hence 3/3 is not = .999.... and therefor .999.... is not = to 1
Jeremiah Smith Re: 0.999.
Full Member « Reply #73 on: Sep 6^th, 2002, 7:05pm »
on Sep 6^th, 2002, 5:58pm, Dustin wrote:
.99999... < 1
Beep! if someone wrote .99999... on a piece of paper for there entire lives, and every last decended of theirs did the same thing, it would still never reach 1.
Someone also earlier said that if
1/3 = .3333...
Posts: 172 2/3 = .6666...
then technically 3/3 would = .9999....
However since we know that 3/3 = 1 then .9999... must also = 1
This idea is flawed because of the fact that the representation of both 1/3 and 2/3 is also flawed. 1/3 does not = exactly .333... and 2/3 does not = exactly .666...
The reason math labels it as that is because with out math system the closes representation we can get of 1/3 is .333... but that does not make it = to .333...
and hence 3/3 is not = .999.... and therefor .999.... is not = to 1
Did you miss the entire thread, somehow?
Read -> Comprehend -> Post
Dustin Re: 0.999.
Guest « Reply #74 on: Sep 6^th, 2002, 7:58pm »
I read the first group of posts and then skipped over the rest because of the fact that the same things were being said.
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Inzynieria Chemiczna i Procesowa, Vol.15, No.3, 481-489, 1994
Fixed bed reactor behavior as a function of inlet concentration, temperature and activity profile in the catalyst pellets was examined for the conditions proposed by INS, Pulawy. The results of the
model calculations showed a good agreement with the experimental results. Two exothermic reactions (1) and (2) with kinetics described by Eqs. (3) and (4) are assumed to be carried out on catalyst
pellets only in a heterogeneous tubular reactor. Mass and heat balances for the reactor are given by Eqs. (5) and (6) with initial conditions (7). Mass balance for the catalyst pellet is described by
Eqs. (8) with boundary conditions (10)-(11). In the model described above the resistance of heat transfer from fluid to the catalyst pellet was neglected, because in the conditions investigated here
preliminary calculations showed a very small influence of heat transfer resistance on the reactor behaviour. Moreover, heat balance for pellets has also been ignored, because the value of the
parameter beta0, defined by Eq. (12), is equal about 5.10(-3), which suggests that the temperature of a catalyst pellet is constant. The system of equations (5), (6) and (8) has been rewritten in
dimensionless form and solved using Euler method for the reactor heat and mass balance and the orthogonal collocation method for pellet mass balance. The dependence of various pellet activity
profiles on the process selectivity (13) and conversion of acetylene E0 (14) for chosen values of the parameter epsilon0 (15) were investigated. The activity pellet profile was normalized according
to Eq. (16). Uniform and shell (various thickness of the layer) profiles were investigated. Moreover, the temperature influence on the process selectivity and acetylene conversion for a chosen value
of the layer thickness was considered. Figure 1 represents the selectivity and acetylene conversion in the reactor outlet vs. the thickness of the pellet active layer (X denotes the radius of
inactive core of the pellet). Both the selectivity and coversion increase with decrease of the active shell width. The increase of the parameter E0 is accompanied by the increase of acetylene
conversion but remarkable decrease of the process selectivity. Figure 2 shows the dependence of the selectivity and acetylene conversion on the position in the reactor. Selectivity decreases along
the reactor axis the faster the parameter E0 increases. It is due to the increase of the ratio p(B)/p(A). Figure 3 represents the dependence of the selectivity and acetylene conversion on the inlet
temperature in the reactor. The process selectivity and acetylene conversion first increase with increasing temperature and then decrease, but the inlet temperature area, where the process
selectivity has the maximal value, is not the same values of conversion and vice versa. On the basis of calculations performed the following conclusions may be presented: 1. Using catalyst pellets
with a narrow active layer on the outer part of the pellet is recommended. 2. The growth of hydrogen concentration at the reactor inlet improves the degree of acetylene conversion but, unfortunately,
ethylene consumption is much greater. 3. The increase in inlet gas temperature is accompanied by the increase of acetylene conversion as well as ethylene consumption. 4. The results presented above
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Engineering workshop #4: Bending Stresses
In April’s article, we examined direct stress and shear stress – the two types of stress that exist when we consider stresses in a Cartesian coordinate system. We recognised that the notion of shear
was necessary in order to allow the all-important direct stresses to act at any given angle (and not just parallel to the X, Y or Z directions). We will take this idea further when we look at
stresses so beware!
This month we will perform hand calculations and Finite Element Analysis (FEA) on a classic karabiner. The karabiner is used by climbers as a rope-joining mechanism – see {fig.1}. With the gate
closed, the karabiner is rated at 21kN end-to-end and 10kN acrossgate. With the gate open (not to be recommended!) the end-to-end rating drops to 7kN.
The first thing to recognise with the karabiner closedgate is that the load path is not unique. In other words, when we apply an end load to this device, some of the load will travel through the
forged aluminium “C” section and some of the load will pass through the gate. We could estimate that the load split is 50/50 but we don’t know – a plastic gate for instance may attract much less load
50%. Thus, with the gate in the closed position, we have what is termed an “INDETERMINATE” structure (so-called because we can’t readily determine where the load will go).
When the gate is open we now have a unique load path through the main forged “C” section and the structure becomes “DETERMINATE” (we can determine where the load will go). In our everyday engineering
we should be aware of the issue of determinacy – determinate structures can be addressed with hand calculations while indeterminate structures generally can’t. Indeterminate structures may be much
more susceptible to manufacturing limits and fits but we’ll leave it there for now.
{fig.2} shows the karabiner with the gate removed (so that we can perform hand calculations on the determinate “C” section). Given that we would like to calculate the stresses along the straight
portion of the “C” section then a cut has been made in order that free-body analysis can determine the internal forces at the cut.
It can be seen that the quadrant is in force and moment equilibrium when a direct force of 7,000N and a bending moment of (7,000 x 19.5) Nmm is applied to the cut section. This free-body diagram
instructs us to carry out a direct stress calculation (carried out in green text) and a bending moment calculation (carried out in red text).
The bending stress (at +/- 1045MPa) is far greater than the direct stress (at +74MPa) and we should watch out for bending as a major source of stress from now on. BEWARE!
While the aforementioned direct stress calculation was very straightforward (force/area) the bending moment calculation was slightly more involved and we should now examine what was done in that
regard. The bending stress induced when (say) a bar of steel is subjected to a bending moment, M, is not a straightforward function of the area of the bar.
The bending stress is proportional to the moment, M, the distance from the centre, y, and the second moment of area, I. Dealing with y first then this is the linear distance from the centroid of the
section (the centre-of-gravity if you made a plywood replica of your section) to the point where you want to determine the stress (usually the top or bottom of the section).
The second moment of area is derived by considering the section as an infinite number of thin “leaves” (like the core of a transformer) and calculating the area of each “leaf” and multiplying that by
the offset distance squared. In other words, the first moment of area is area x distance and the second moment of area is the area x distance x distance (area x distance-squared). For a rectangle
(breadth, b x depth, d) the second moment of area is bd3/12 and that for a solid circle is ðD4/64. Check the bending stress calculation on {fig.2}.
Staying with {fig.2} we should be able to appreciate that the direct stress and the bending stress are defined in the same direction (parallel to the global X direction).
This allows us to add the two stresses together giving a resultant stress at the inside of the “C” as 1,119MPa (tension) and a resultant stress of minus 971MPa (compression) at the outside. Check
that you agree.
Finally I have carried out Finite Element Analysis of the open gate karabiner (subject to 7kN loading) and the predicted stresses away from the end of the radius (1,120MPa) show excellent agreement
with our hand calculations (see { fig.3}). What is happening where the straight part of the “C” joins the curved part of the “C”?
We have demonstrated the danger of bending – as soon as you see two forces incorporating an offset (see {fig.2}) then you know that you have a bending problem. Always re-check your calculations to
see that you haven’t forgotten a bending problem. More next time…
R P Johnson BSc MSc NRA MIMechE CEng – Technical Director, DAMT Limited
Stress analysis of karabiner carried out using the ROSHAZ program.
Part four of an engineering master class, this month: Bending Stresses
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Theory of multiple quantum dynamic NMR spectroscopy
A formalism for calculating dynamic multiple quantum NMR line shapes obtained by the time proportional phase increment (TPPI) method from spin-1/2 systems is developed. The formalism is essentially
an extension of the related formalism in Liouville space used in the theory of conventional dynamic NMR of strongly interacting spin systems. It is subsequently used to calculate the expected
multiple quantum proton NMR line shapes of a number of (hypothetical and real) systems consisting of compounds dissolved in liquid crystalline solvents and undergoing intramolecular rearrangement.
These include cyclobutadiene, cyclohexatriene, and cyclooctatetraene undergoing bond shifts and s-trioxane undergoing ring inversion. Since the computations involve diagonalization of
high-dimensional matrices extensive use is made of symmetry factorization. It is shown that the resulting line shapes depend on the mechanism and rate of the dynamic processes, and may therefore be
used to derive kinetic parameters from multiple quantum experiments. The high-quantum order spectra are particularly useful because for intermediate and large spin systems they are much simpler than
the corresponding conventional single quantum spectra. Approximate expressions for the multiple quantum line shapes are also derived for the slow and fast exchange limits. It is found that except for
an intermediate dynamic region these equations faithfully reproduce the exact line shapes in the appropriate limits.
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