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Department of Mathematics
When men started to do counting, Mathematics began from that day itself we can’t come to conclusion the day from which Mathematics came into use by humans. Thus, Mathematics is a very old subject as
well as a very important subject in this world.
In this college, the Department of Mathematics was introduced during 2003-2004 and 24 students were joined in First Batch. In this department Physics, Statistics, Operation Research, and Graph Theory
are Allied subjects for the first year, second year, and Final year students respectively.
The Enthusiastic students of the Mathematics Department are participating in various competitions with other college students they are encouraged by awarding Scholarships by Trusts.
They are not only enlightened in Mathematics subjects but also with other required information which is required for their whole life to uplift them.
There are Eight teaching faculty educate and intellectualize the students in Mathematics Department
In accordance with the valuable guidelines being given by our Management and by the Principal of this college a good standard of Education as being provided to our students in Mathematics there by
the following students have got university Rank.
• Imparting of quality mathematics education and the inculcating of the spirit of research through innovative teaching and research methodologies.
• To achieve high standards of excellence in generating and propagating knowledge in Mathematics. Department is committed to providing an education that combines rigorous academics with joy of
• To provide an environment where students can learn, become competent users of mathematics, and understand the use of mathematics in other disciplines.
• Mission
• To be a leading Mathematics Department in the country.
• To emerge as a global center of learning, academic excellence, and innovative research.
• Mathematics principals Equations formulas are proved one and this subject is base for the various scientifically inventions, Hence we have to bring this information to the society.
• Create awareness to the students for the benefits of learning these subjects.
• To make it understand to the students that the job opportunities available by learning this subjects.
• We should add interest of learnings Mathematics, subtract Hesitation, Multiply Motivation, Divide Fear, then only the number of students will be increased remarkably.
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How Aleo executes Decentralized Private Computation
In the realm of blockchain and decentralized technologies, the quest for privacy-preserving computation stands at the forefront of innovation. This blog post delves into the intricate world of
Zero-Knowledge Proofs (ZKPs) and their application in advanced frameworks like ZEXE (Zero-Knowledge Execution) and its subsequent evolution into Aleo. We will explore the technical nuances,
challenges, and solutions that these frameworks offer, including code snippets to illustrate key concepts.
The Foundation: Zero-Knowledge Proofs in Cryptocurrencies
Zero-Knowledge Proofs have been a cornerstone in the privacy aspect of cryptocurrencies. Initially, their implementation, as seen in Bitcoin, was basic, focusing primarily on transactional privacy.
However, as the demand for complex computations (like those needed in smart contracts) grew, the limitations of traditional ZKP implementations became apparent.
The Limitation of Fixed Circuits
Traditional ZKP systems are designed for specific computations, known as fixed circuits. This design restricts their application to pre-defined tasks, making them unsuitable for the dynamic and
varied nature of smart contract computations.
Code Snippet: Basic ZKP Implementation
# Simplified ZKP for a fixed computation (e.g., a specific hash function)
def prove_knowledge_of_preimage(hash_value, preimage):
assert hash(preimage) == hash_value
# The proof generation process would go here
# For simplicity, we're returning the preimage as 'proof'
return preimage
def verify_proof(hash_value, proof):
return hash(proof) == hash_value
Addressing Arbitrary Computations
To enable ZKPs for arbitrary computations, two main approaches have been proposed: universal circuits and proof recursion.
Universal Circuits
Universal circuits offer the flexibility to plug in any program into a fixed circuit. However, they are computationally intensive and impractical for large-scale applications due to significant
Proof Recursion
Proof recursion, on the other hand, involves verifying a proof of program execution rather than executing the program within the ZKP system. This method leverages the succinctness of ZKPs, making it
more suitable for decentralized networks.
Code Snippet: Proof Recursion Concept
# Hypothetical proof recursion function
def verify_program_proof(program_proof, verification_key):
# The verifier checks the proof against the verification key
# This is a simplified representation
return check_proof_validity(program_proof, verification_key)
From ZEXE to Aleo: A Paradigm Shift
ZEXE introduced a framework for private payments on decentralized ledgers using ZKPs. It emphasized privacy in both function and data within transactions. However, its model, involving separate
predicates for the birth and death of records, was complex.
Aleo simplified this by combining these predicates into a single program proof, streamlining the development process.
Function Privacy vs. Public State in Aleo
Aleo adopts a pragmatic approach to function privacy. It recognizes the inherent compromises in applications interacting with public state, like Uniswap contracts, and opts for a balanced privacy
Code Snippet: Aleo's Simplified Model
# Aleo's model: Single program proof for a transaction
def verify_transaction(transaction, program_proof, verification_key):
# Verify the entire transaction against a single program proof
return verify_program_proof(program_proof, verification_key)
The Future: Balancing Privacy, Efficiency, and Functionality
The ongoing development in decentralized private computation is about finding the right balance between privacy, efficiency, and functionality. Frameworks like Aleo represent a step towards this
balance, offering a spectrum of privacy options to cater to diverse application needs.
Potential Code Evolution: Hybrid Public-Private State
Looking ahead, we might see hybrid models where public and private states coexist, offering flexibility and enhanced privacy.
Hypothetical Code Snippet: Hybrid State Model
def execute_transaction(transaction, private_state, public_state):
# Process transaction using both private and public states
private_result = process_private_state(transaction, private_state)
public_result = process_public_state(transaction, public_state)
return combine_results(private_result, public_result)
The journey towards decentralized private computation is marked by continuous innovation and adaptation. As we progress, the focus will likely be on developing frameworks that offer a range of
privacy options, tailored to the diverse needs of blockchain applications. This evolution promises a future where secure, private, and efficient digital transactions become the norm, reshaping our
interaction with blockchain technology and its applications.
More Blog Posts
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sunk cost calculator
Calculating sunk costs is an essential aspect of financial analysis, helping businesses make informed decisions by considering irreversible expenses. In this article, we’ll provide you with a
comprehensive guide on how to use a sunk cost calculator, including the formula, examples, and frequently asked questions.
How to Use
To utilize the sunk cost calculator, follow these simple steps:
1. Input the initial cost of the project or investment.
2. Enter the salvage value, if any.
3. Specify the useful life of the project or investment.
4. Click the “Calculate” button to obtain the sunk cost.
The sunk cost formula is straightforward:
• Initial Cost: The total expenditure on the project or investment.
• Salvage Value: The estimated residual value of the asset at the end of its useful life.
Example Solve
Let’s consider an example:
• Initial Cost: $50,000
• Salvage Value: $10,000
\text{Sunk Cost} = $50,000 – $10,000 = $40,000
In this case, the sunk cost is $40,000.
Frequently Asked Questions
What is a sunk cost?
A sunk cost is an expenditure that has already occurred and cannot be recovered. It should not influence future decision-making.
How does the calculator help in decision-making?
The calculator assists in determining the sunk cost of a project, aiding businesses in evaluating the financial implications of continuing or discontinuing an investment.
Can the salvage value be zero?
Yes, the salvage value can be zero, indicating that there is no residual value for the asset at the end of its useful life.
Understanding sunk costs is crucial for effective financial management. By using the sunk cost calculator and considering the derived value, businesses can make strategic decisions that align with
their long-term goals.
Calculate sunk costs effortlessly with our online sunk cost calculator. Make informed decisions by understanding the financial implications of your investments. Try it now!
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15.1 Paths
When a Racket procedure takes a filesystem path as an argument, the path can be provided either as a string or as an instance of the path datatype. If a string is provided, it is converted to a path
using string->path. Beware that some paths may not be representable as strings; see Unix Path Representation and Windows Path Representation for more information. A Racket procedure that generates a
filesystem path always generates a path value.
By default, paths are created and manipulated for the current platform, but procedures that merely manipulate paths (without using the filesystem) can manipulate paths using conventions for other
supported platforms. The bytes->path procedure accepts an optional argument that indicates the platform for the path, either 'unix or 'windows. For other functions, such as build-path or
simplify-path, the behavior is sensitive to the kind of path that is supplied. Unless otherwise specified, a procedure that requires a path accepts only paths for the current platform.
Two path values are equal? when they are use the same convention type and when their byte-string representations are equal?. A path string (or byte string) cannot be empty, and it cannot contain a
nul character or byte. When an empty string or a string containing nul is provided as a path to any procedure except absolute-path?, relative-path?, or complete-path?, the exn:fail:contract exception
is raised.
Most Racket primitives that accept paths first cleanse the path before using it. Procedures that build paths or merely check the form of a path do not cleanse paths, with the exceptions of
cleanse-path, expand-user-path, and simplify-path. For more information about path cleansing and other platform-specific details, see Unix and Mac OS Paths and Windows Paths.
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Summation notation calculator
Author Message
T50 Posted: Friday 18th of Sep 18:27
Heya guys! Is someone here know about summation notation calculator? I have this set of questions about it that I just can’t understand. Our class was tasked to solve it and know how we
came up with the answer . Our Math professor will select random people to solve the problem as well as show solutions to class so I require detailed explanation about summation notation
calculator. I tried answering some of the questions but I guess I got it completely wrong . Please help me because it’s urgent and the due date is close already and I haven’t yet
understood how to solve this.
kfir Posted: Sunday 20th of Sep 09:39
Algebra Master is the latest hot favourite of summation notation calculator students. I know a couple of teachers who actually ask their students to use a copy of this software at their
residence .
From: egypt
daujk_vv7 Posted: Sunday 20th of Sep 12:11
Algebra Master really is a great piece of math software. I remember having difficulties with converting fractions, greatest common factor and matrices. By typing in the problem from
homework and merely clicking Solve would give step by step solution to the algebra problem. It has been of great help through several Algebra 2, Intermediate algebra and Remedial
Algebra. I seriously recommend the program.
From: I
dunno, I've
lost it.
nemosbortiv Posted: Monday 21st of Sep 19:43
It looks great . How could I acquire that program ? Could you give me a link that could lead me to more details regarding that software?
From: United
DoniilT Posted: Tuesday 22nd of Sep 19:48
It is simple to access this program. Click here for details . You are guaranteed satisfaction. If not you get a refund. So what is there to lose anyway? Cheers and good luck.
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sing Fourier Transform
Figure 1. Different image patterns: (a) circle, (c) annulus, (e) square, (g) horizontal double slit, (i) vertical double slit and (k) two dirac deltas symmetric with the y-axis and their
corresponding Fourier transform (b, d, f, h, j and l, respectively)
The circle and the annulus have a similar Fourier Transform, which are both circular in nature. The square has a square-ish FT which is indicative of its shape. More shapes (or apertures in this
case) have unique FTs. Next are the FTs of the slits. Again, both FTs are indicative of its original image. Last is the FT of two dots symmetric along the y-axis. We can see that it has a somewhat
sinusoidal pattern along the x-direction.
Note that we may be exchanging terms FT and FFT. FT means Fourier Transform while FFT means Fast Fourier Transform. The main difference is well, the latter is, uhm, fast. Yeah, I’m not joking! FFT
uses the 2^x where x is a number to optimize FT, making it fast. Most programs use FFT to save time instead of directly evaluating FT by itself. :)
For the next part of the activity, we were asked to find the FT of two dirac deltas, two circles and two squares. Figure 2 shows the results. The first column shows the original images, the second
column shows the 3 Dimensional look of the original image, then we display is FFT in 3D and 2D. The dots show a fan-like FT with sinusoidal ridges. As for the circles, it is similar to the FT of that
of in Figure 1b. Conversely, the FFT of the squares is similar to that of Figure 1f. This is again because certain apertures have a unique corresponding FFT pattern.[S::S]
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28 CIs for comparing two odds or proportions | Scientific Research and Methodology
28 CIs for comparing two odds or proportions
So far, you have learnt to ask a RQ, design a study, classify and summarise the data, and form confidence intervals. In this chapter, you will learn to:
• identify situations where comparing two qualitative variables is appropriate.
• form confidence intervals for the difference between two proportions.
• form confidence intervals for odds ratios using software output.
• determine whether the conditions for using the confidence intervals apply in a given situation.
28.1 Introduction: eating habits
Mann and Blotnicky (2017) examined the relationship between where university students usually ate, and where the student lived, for students from two Canadian east coast universities. The researchers
cross-classified the \(n = 183\) students (the units of analysis) according to two qualitative variables:
• Where they lived: with their parents, or not with their parents;
• Where they ate most meals: off-campus or on-campus.
Both variables are qualitative, so means are not appropriate for summarising the data. The data can be compiled into a two-way table of counts (Table 28.1), also called a contingency table. Both
qualitative variables have two levels, so this is a \(2\times 2\) table. Every cell in the \(2\times 2\) table contains different students, so the comparison is between individuals.
The study has one sample of students, classified according to two variables (i.e., each student is placed into one of the four cells in the \(2\times 2\) table).
The table can be constructed with either variable as the rows or the columns. However, software commonly compares rows, so it makes sense to place the groups to be compared (i.e., the explanatory
variable) in the rows of the table.
TABLE 28.1: Where university students live and eat.
Most off-campus Most on-campus
Living with parents \(52\) \(\phantom{0}\phantom{0}2\)
Not living with parents \(105\) \(\phantom{0}24\)
The proportion of students who eat most meals off-campus can be compared between those who live with their parents and those who do not live with their parents. Then, the parameter is the difference
between the population proportions in each group, and the RQ could be written as:
Among university students, what is the difference between the proportion of students eating most meals off-campus, comparing those who do and do not live with their parents?
Alternatively, the odds of students who eat most meals off-campus can be compared between those who live with their parents and those who do not live with their parents. Then, the parameter is the
odds ratio (OR); specifically, the odds ratio of eating most meals off-campus, comparing those living with parents to those not living with parents. Using the OR, the RQ could be written as:
Among university students, what is the odds ratio of students eating most meals off-campus, comparing those who do and do not live with their parents?
Take care defining the odds ratios! Recall (Sect. 12.5): software usually compares Row 1 to Row 2, and Column 1 to Column 2 (that is, the last row is usually the reference level). For this reason,
defining your OR in the same way makes sense.
28.2 Summarising data
With two qualitative variables, an appropriate numerical summary includes the odds and proportions (or percentages) for the outcome for both comparison groups, and the sample sizes (Table 28.2).
To compare the proportions, define the sample proportion of students eating most meals off-campus as \(\hat{p}\), and write \(\hat{p}_P\) for the proportion living with parents and \(\hat{p}_N\) for
the proportion not living with parents. Then, \[ \hat{p}_P = \frac{52}{52 + 2} = 0.962963 \quad\text{and}\quad \hat{p}_N = \frac{105}{105 + 24} = 0.8139535. \] The difference between the two
proportions is \[ \hat{p}_P - \hat{p}_N = 0.9630 - 0.8140 = 0.1490, \] (as in the software output: Fig. 28.1), or \(14.9\)%. By this definition, the difference is how much greater the proportion
eating most meals off-campus is for students living with their parents, compared to students not living with their parents.
Be clear about how differences are defined! Differences could be computed as:
• the proportion eating most meals off-campus for those living with their parents, minus the proportion not living with their parents. This measures how much greater the proportion is for those
living with their parents; or
• the proportion eating most meals off-campus for those not living with their parents, minus the proportion living with their parents. This measures how much greater the proportion is for those not
living with their parents.
Either is fine, provided you are consistent, and clear about how the difference are computed. The meaning of any conclusions will be the same.
To compare the odds, first see that the odds of eating most meals off-campus is:
• \(52 \div 2 = 26\) for students living with their parents (Row 1 of Table 28.1).
• \(105\div 24 = 4.375\) for students not living with their parents (Row 2 of Table 28.1).
(Notice the numbers in the second column are always on the bottom of the fraction.) So the odds ratio (OR) of eating most meals off-campus (the first column), comparing students living with parents
to students not living with parents (second column), is \(26 \div 4.375 = 5.943\) (as in the software output: Fig. 28.1).
The numerical summary (Table 28.2) shows the proportion and odds of eating most meals off-campus, comparing students living at home and those not living at home.
The odds ratio can be interpreted in either of these ways (i.e., both are correct):
• The odds compare Row 1 counts to Row 2 counts, for both columns. The odds ratio then compares the Column 1 odds to the Column 2 odds.
• The odds compare Column 1 counts to Column 2 counts. The odds ratio then compares the Row 1 odds to the Row 2 odds.
Odds and odds ratios are computed with the first row and first column values on the top of the fraction. In this case, both of the above approaches produces an OR of \(5.943\). Since the explanatory
variable is usually in the rows, the first is usually the most useful.
An appropriate graph is a side-by-side bar chart (Fig. 28.1, left panel) or a stacked bar chart. The side-by-side bar is a good display for comparing the odds. For instance, in the two left-most bars
in Fig. 28.1 (left panel), the first bar is \(26\) times as high as the second bar (and \(26\) is the odds); in the two right-most bars, the first bar is \(4.375\) times as high as the second bar
(and \(4.375\) is the odds). A stacked bar chart would be a good visual display for comparing the proportions.
TABLE 28.2: The odds and proportion of university students eating most meals off-campus.
Odds of having most meals off-campus Proportion having most meals off-campus Sample size
Living with parents \(26.000\) \(\phantom{-}0.963\) \(\phantom{0}54\)
Not living with parents \(\phantom{-}4.375\) \(\phantom{-}0.814\) \(129\)
\(\phantom{-}5.943\) \(\phantom{-}0.149\)
Each sample of students comprises different students, giving different proportions and odds of having most meals off-campus for both groups (living with, and not living with, parents). Hence, the
difference between the two proportions, and the odds ratio, will vary between samples. This means that both the difference between the two proportions, and the odds ratio, have sampling
28.3 CIs for the difference between two proportions
The sample proportions for each group will vary from sample to sample, and the difference between the sample proportions will be different for each sample. Hence, the difference between the sample
proportions has a sampling distribution and standard error. Under certain conditions (Sect. 28.5), this sampling distribution has a normal distribution.
Definition 28.1 (Sampling distribution for the difference between two sample proportions) The sampling distribution of the difference between two sample proportions \(\hat{p}_A\) and \(\hat{p}_B\) is
(when the appropriate conditions are met; Sect. 28.5) described by:
• an approximate normal distribution,
• centred around a sampling mean whose value is \({p_{A}} - {p_{B}}\), the difference between the population proportions,
• with a standard deviation, called the standard error of the difference between the proportions, of \(\displaystyle\text{s.e.}(\hat{p}_A - \hat{p}_B)\).
The standard error for the difference between the proportions is found using \[$$\text{s.e.}(\hat{p}_A - \hat{p}_B) = \sqrt{ \text{s.e.}(\hat{p}_A)^2 + \text{s.e.}(\hat{p}_B)^2}, \tag{28.1}$$\]
though this value will often be given (e.g., on computer output).
For the student-eating data, the standard error of the sample proportions for each group are computed using Eq. (23.4) as \[\begin{align*} \text{s.e.}(\hat{p}_L) &= \sqrt{\frac{0.962963\times ( 1 -
0.962963)}{54}} = 0.025700; \text{and}\\ \text{s.e.}(\hat{p}_N) &= \sqrt{\frac{0.8139535\times (1 - 0.8139535)}{129}} = 0.034262. \end{align*}\] The standard error of the difference between the
proportions is \[ \text{s.e.}(\hat{p}_P - \hat{p}_N) = \sqrt{ \text{s.e.}(\hat{p}_P)^2 + \text{s.e.}(\hat{p}_N)^2} = \sqrt{ 0.025700^2 + 0.034262^2 } = 0.042830. \]
Thus, the differences between the sample proportions will have:
• an approximate normal distribution,
• centred around the sampling mean whose value is \(p_P - p_N\),
• with a standard deviation, called the standard error of the difference, of \(\text{s.e.}(\hat{p}_P - \hat{p}_N) = 0.04282954\).
The sampling distribution describes how the values of \(\hat{p}_P - \hat{p}_N\) vary from sample to sample. Then, finding a \(95\)% CI for the difference between the proportions is similar to the
process used previously, since the sampling distribution has an approximate normal distribution: \[ \text{statistic} \pm \big(\text{multiplier} \times\text{s.e.}(\text{statistic})\big). \] When the
statistic is \(\hat{p}_P - \hat{p}_N\), the approximate \(95\)% CI is \[ (\hat{p}_P - \hat{p}_N) \pm \big(2 \times \text{s.e.}(\hat{p}_P - \hat{p}_N)\big). \] So, in this case, the approximate \(95\)
% CI is \[ 0.1490 \pm (2 \times 0.042830), \] or \(0.149 \pm 0.0857\) after rounding (i.e., from \(0.0633\) to \(0.235\)). This approximate CI is very similar to the (exact) CI from software (Fig.
28.1). We write:
The difference between the proportions of students eating most meals at home is \(0.1490\), higher for those living with their parents (\(0.963\); \(n = 52\)) that those not living with their
parents (\(0.814\); \(n = 129\)), with the \(95\)% confidence interval from \(0.0633\) to \(0.235\).
The plausible values for the difference between the two population proportions are between \(0.063\) to \(0.235\), larger for those living with parents.
Giving the CI alone is insufficient; the direction in which the differences were calculated must be given, so readers know which group had the higher proportion.
28.4 CIs for odds ratios
A CI can be formed for the difference between the two proportions, and a CI can also be formed for the odds ratio. Every sample of students is likely to be different, and hence the odds of students
eating off campus will vary from sample to sample (in both groups). Hence, the OR varies also from sample to sample. That is, sampling variation exists, so the odds ratio has a sampling distribution.
However, the sampling distribution of the sample OR does not have a normal distribution^5. Fortunately, a simple transformation of the sample OR does have a normal distribution, though we omit the
details. For this reason, we will only use software output for finding the CI for the odds ratio, and not discuss the sampling distribution directly. In other words, we will rely on software to find
CIs for odds ratios.
Software (Fig. 28.1, right panel) gives the sample OR as \(5.94\), and the (exact) \(95\)% CI as \(1.35\) to \(26.1\). The value of the OR is the same as computed manually.
We write:
The odds of students eating most meals off-campus is \(5.94\) higher for students living with their parents (odds: \(26.0\); \(n = 54\)) than for students not living with their parents (odds: \
(4.38\); \(n = 129\)), with the \(95\)% confidence interval from \(1.35\) to \(26.1\).
There is a \(95\)% chance that this CI straddles the population OR. Notice that the meaning of the OR is explained in the conclusions: the odds of eating most meals off-campus, and comparing students
living with parents to not living with parents.
The CI for an OR is not symmetrical, like the others we have seen^6; that is, the sample OR of \(5.94\) is not in the centre of the confidence interval.
Interpreting and explaining ORs can be challenging, so care is needed!
28.5 Statistical validity conditions
As usual, these results hold under certain conditions. The CIs computed above are statistically valid if
• All expected counts in the table are at least five.
Some books may give other (but similar) conditions. The units of analysis are also assumed to be independent (e.g., from a simple random sample). If the statistical validity conditions are not met, a
confidence interval based on the non-parametric Fisher's method may be used (Fisher 1962).
Importantly, this condition is based on the expected counts, not the observed counts. The expected counts are the counts expected if there was no relationship between the two variables in the two-way
table. If there was no relationship between the two variables for the student-meals data (Table 28.1), students living with or not with their parents would have a similar percentage of meals eaten
off-campus. That is, the two conditional probabilities would be the same. The overall percentage of students eating meals off-campus is \(157/183\times 100 = 85.79\)% (from Table 28.1). If there was
no relationship between the two variables, this percentage would be the same for students living with or not with their parents. In other words, we would expect \(85.79\)% of the \(54\) students who
do live with their parents to eat most meals off-campus (which is \(46.33\)), and we would expect \(85.79\)% of the \(129\) students who do not live with their parents to eat most meals off-campus
(which is \(110.67\)). This statistical validity condition is explained further in Sect. 35.3.1.
Usually, you do not have to compute these expected values, as software can produce the expected counts (see Fig. 28.2). However, a quick check for the statistical validity is to compute the value of
the smallest expected value, using \[ \frac{(\text{Smallest row total})\times(\text{Smallest column total})}{\text{Overall total}}. \] If this value is greater than five, the CIs are statistically
Example 28.1 (Statistical validity) For the students-eating data, software can be used to compute the expected counts (Fig. 28.2) None are less than five, and so the conclusion is statistically
valid. (One observed count is less than five, but this is not relevant to checking for statistical validity.)
In Table 28.1, the smallest row total is \(54\) and the smallest column total is \(26\). Then, \[ \frac{54\times 26}{183} = 7.67, \] which is larger than five, so the CIs are statistically valid.
(The value of \(7.67\) is also the smallest expected count in Fig. 28.2.)
28.6 Example: turtle nests
The hatching success of loggerhead turtles on Mediterranean beaches is often compromised by fungi and bacteria. Candan, Katılmış, and Ergin (2021) compared the odds of a nest being infected, between
nests relocated due to the risk of tidal inundation, and non-relocated nests (Table 28.3). Note that the explanatory variable (whether the nest is relocated) is in the rows. The researchers were
interested in knowing:
For Mediterranean loggerhead turtles, what is the difference between the proportion of infected nests, comparing natural to relocated nests?
The parameter here is the difference between the proportions infected, comparing natural to relocated nests.
Alternatively, the researchers could have asked:
For Mediterranean loggerhead turtles, what are the odds of infections comparing natural to relocated nests?
The parameter is the odds ratio of non-infection, comparing natural to relocated nests. The odds ratio can be defined in other ways also, but this definition is consistent with how software computes
odds given Table 28.3 (i.e., first row to second row; first column to second column).
TABLE 28.3: Non-infected and infected
turtle nests.
Non-infected Infected
Natural \(29\) \(10\)
Relocated \(14\) \(\phantom{0}8\)
The data are summarised graphically (Fig. 28.3) and numerically (Table 28.4). From the software output (Fig. 28.4), the \(95\)% CI for the difference between the proportions is from \(-0.1361\) to \
(0.3505\). The negative value is not a negative proportion; it is a negative difference between two proportions. Specifically, it means that the population proportion of infected nests is larger for
relocated nests than natural nests by \(0.1361\). Write:
The difference between proportion of infected nests is \(0.107\) (\(95\)% CI: \(-0.136\) to \(0.527\)), comparing natural nests (proportion: \(0.744\); \(n = 39\)) to relocated nests (\(0.636\);
\(n = 22\)).
In addition, from the software output (Fig. 28.4) the \(95\)% CI for the odds ratio is from \(0.537\) to \(5.12\). Write:
The OR of a non-infected nest, comparing natural nests (odds: \(2.90\); \(n = 39\)) to relocated nests (odds: \(1.75\); \(n = 22\)), is \(1.66\) with a \(95\)% CI from \(0.537\) to \(5.12\).
The smallest expected count is \(6.49\) (Fig. 28.4), so the CIs are statistically valid. (Alternatively, since the smallest row and column counts are \(22\) and \(18\) respectively, we see that \(22\
times18/61 = 6.49\), which is greater than five.)
TABLE 28.4: The odds and proportion of non-infected nests.
Odds non-infected Proportion non-infected Sample size
Natural \(2.900\) \(0.744\) \(39\)
Relocated \(1.750\) \(0.636\) \(22\)
\(1.657\) \(0.107\)
28.7 Chapter summary
To compare a two-level qualitative variable between two groups, a confidence interval can be formed for the difference between two proportions, or for an odds ratio.
To compute a confidence interval (CI) for the difference between two proportions, compute the difference between the two sample proportions, \(\hat{p}_A - \hat{p}_B\), and identify the sample sizes \
(n_A\) and \(n_B\). Then compute the standard error, which quantifies how much the value of \(\hat{p}_A - \hat{p}_B\) varies across all possible samples: \[ \text{s.e.}(\hat{p}_A - \hat{p}_B) = \sqrt
{ \text{s.e}(\hat{p}_A) + \text{s.e.}(\hat{p}_B)}, \] where \(\text{s.e.}(\hat{p}_A)\) and \(\text{s.e.}(\hat{p}_B)\) are the standard errors of Groups \(A\) and \(B\) (Eq. (23.4)). The margin of
error is (multiplier\({}\times{}\)standard error), where the multiplier is \(2\) for an approximate \(95\)% CI (using the \(68\)--\(95\)--\(99.7\) rule). Then the CI is: \[ (\hat{p}_A - \hat{p}_B) \
pm \left( \text{multiplier}\times\text{standard error} \right). \] Software is used to compute a confidence interval (CI) for the odds ratio, as the sampling distribution does not have a normal
distribution. The statistical validity conditions should be checked: all expected counts should exceed five.
28.8 Quick review questions
Egbue, Long, and Samaranayake (2017) studied the adoption of electric vehicles (EVs) by a certain group of professional Americans (Table 28.5). Software output is shown in Fig. 28.5.
1. What percentage of people without post-graduate study would buy an EV in the next \(10\) years? (do not add the percentage symbol)
2. What are the odds that a person without post-graduate study would buy an EV in the next \(10\) years?
3. Using the output, what is the OR of buying an electric vehicle in the next \(10\) years, comparing those without post-grad study to those with post-grad study?
4. True or false: The CI means that the sample OR is likely to be between \(0.68\) and \(4.28\).
5. The negative proportion in output suggests the statistical validity conditions are not met.
TABLE 28.5: Responses to 'Would you
purchase an electric vehicle in the next
\(10\) years?' by education.
Yes No
No post-grad \(24\) \(\phantom{0}8\)
Post-grad study \(51\) \(29\)
1. The number without post-grad study: \(24 + 8 = 32\). The percentage of people without post-grad study who would buy an EV in the next \(10\) years is \(24/32 = 0.75\), or \(75\)%.
2. The people with post-grad study are in the bottom row. The odds of people without post-grad study who would buy an EV in the next \(10\) years is \(24/8 = 3\).
3. The odds of people without post-grad study who would buy an electric vehicle is \(24/8 = 3\).
The odds of people with post-grad study who would buy an electric vehicle is \(51/29 = 1.7586\).
So the OR is \(3/1.7586 = 1.706\).
4. Not at all. We know exactly what the sample OR is (in this sample, it is \(1.706\)... exactly). CIs always give an interval in which the population parameter is likely to be within.
5. The CI is statistically valid if all the expected counts exceed \(5\). So we don't really know for sure from the given information. But the observed counts are all reasonably large, so it is very
probably statistically valid.
28.9 Exercises
Answers to odd-numbered exercises are available in App. E.
Exercise 28.1 Suppose the sample odds ratio has a value of one. What will be value of the difference between the sample proportions? Explain.
Exercise 28.2 Suppose the sample odds ratio has a value smaller than one. Will the difference between the sample proportions be a positive or a negative value? Explain.
Exercise 28.3 [Dataset: CarCrashes] Wang et al. (2020) recorded information about car crashes in a rural, mountainous county in western China (Table 28.6).
1. Compute the proportion of crashes involving a pedestrian in 2011.
2. Compute the proportion of crashes involving a pedestrian in 2015.
3. Compute the difference between the proportion of crashes involving a pedestrian, comparing 2011 to 2015.
4. Compute the difference between the proportions.
5. Compute an approximate \(95\)% CI for the difference between the proportions.
6. Use the output (Fig. 28.6) to write down a \(95\)% CI for the difference between the proportions.
7. Interpret what this CI means.
8. Compute the odds of crashes involving a pedestrian in 2011.
9. Compute the odds of crashes involving a pedestrian in 2015.
10. Compute the odds ratio of crashes involving a pedestrian, comparing 2011 to 2015.
11. Interpret what this odds ratio means.
12. Write down the CI for the odds ratio.
13. Construct an appropriate numerical summary table for the data.
14. Sketch a suitable graph to display the data.
15. Determine if the CIs are statistically valid.
TABLE 28.6: Responses to
'Would you purchase an
electric vehicle in the
next \(10\) years?' by
pedestrians vehicles
2011 \(15\) \(35\)
2015 \(37\) \(85\)
Exercise 28.4 [Dataset: ScarHeight] Wallace et al. (2017) compared the heights of scars from burns received in Western Australia (Table 28.7). Software was used to analyse the data (Fig. 28.7).
1. Compute the proportion of men having a smooth scar (that is, height is \(0\)).
2. Compute the proportion of women having a smooth scar (that is, height is \(0\)).
3. Compute the difference between the proportions of men and women having a smooth scar.
4. Compute the standard error for the difference between the proportions.
5. Compute the approximate \(95\) CI for the difference between the proportions.
6. Write down the \(95\)% CI for the difference between the proportions, using the software output.
7. Interpret what this CI means.
8. Compute the odds of having a smooth scar (that is, height is \(0\)) for men.
9. Compute the odds of having a smooth scar (that is, height is \(0\)) for women.
10. Compute the odds ratio of having a smooth scar, comparing men to women.
11. Interpret what this odds ratio means.
12. Write down the CI for the odds ratio.
13. Construct an appropriate numerical summary table for the data.
14. Sketch a suitable graph to display the data.
15. Determine if the CIs are statistically valid.
TABLE 28.7: Heights of scars for men
and women.
Men Women
Smooth \(216\) \(\phantom{0}99\)
0mm to 1mm \(115\) \(\phantom{0}62\)
Exercise 28.5 Sketch the sampling distribution for the difference between the proportions of students eating most meals off-campus, for those living with parents minus those not living with parents.
What is the sampling distribution for the equivalent odds ratio?
Exercise 28.6 Sketch the sampling distribution for the difference between the proportion of non-infected turtle nests, for natural nests minus relocated nests (in Sect. 28.6). What is the sampling
distribution for the equivalent odds ratio?
Exercise 28.7 [Dataset: EarInfection] A study of ear infections in Sydney swimmers (Smyth 2010) recorded whether people reported an ear infection or not, and where they usually swam. Use Fig. 28.8 to
answers these questions.
1. Compute the standard error for the difference between the proportions of people not reporting ear infections, comparing non-beach to beach swimmers.
2. Compute an approximate \(95\)% CI for the difference between the proportions.
3. Write down the \(95\)% CI for the difference between the proportions.
4. Interpret the CI.
5. Confirm that the odds ratio in the output is correct.
6. Use the software output to write down a \(95\)% CI for the odds ratio.
7. Interpret the CI.
8. Are the CIs statistically valid?
9. Construct the summary table for the data.
Exercise 28.8 [Dataset: EmeraldAug] The Southern Oscillation Index (SOI) is a standardised measure of the air pressure difference between Tahiti and Darwin, and is related to rainfall in some parts
of the world (Stone, Hammer, and Marcussen 1996), and especially Queensland (Stone and Auliciems 1992).
The rainfall at Emerald (Queensland) was recorded for Augusts between 1889 to 2002 inclusive (P. K. Dunn and Smyth 2018), where the monthly average SOI was positive, and when the SOI was non-positive
(that is, zero or negative), as shown in Table 28.8.
Use the software output in Fig. 28.9 to answer these questions.
1. Compute the standard error for the difference between the proportions of wet August, comparing months with a positive SOI to months with a non-positive SOI.
2. Compute an approximate \(95\)% CI for the difference between the proportions.
3. Write down the \(95\)% CI for the difference between the proportions.
4. Interpret the CI.
5. Confirm that the odds ratio in the output is correct.
6. Use the software output to write down a \(95\)% CI for the odds ratio.
7. Interpret the CI.
8. Are the CIs statistically valid?
9. Construct the summary table for the data.
TABLE 28.8: The SOI, and whether rainfall
was recorded in Augusts between 1889 and
2002 inclusive.
Positive SOI Non-positive SOI
Rain \(53\) \(40\)
No rain \(\phantom{0}7\) \(14\)
Exercise 28.9 [Dataset: Turbines] A study of turbine failures (Myers, Montgomery, and Vining 2002; Nelson 1982) ran \(73\) turbines for around \(1\,800\), and found that seven developed fissures
(small cracks). They also ran a different set of \(42\) turbines for about \(3\,000\), and found that nine developed fissures.
1. Construct the two-way table for the data.
2. Compute the standard error for the difference between the proportions.
3. Compute an approximate \(95\)% CI for the difference between the proportions.
4. Write down the \(95\)% CI for the difference between the proportions.
5. Interpret the CI.
6. Use the software output (Fig. 28.10, left panel) to write down a \(95\)% CI for the odds ratio.
7. Interpret the CI.
8. Are the CIs statistically valid?
Exercise 28.10 [Dataset: HatSunglasses] B. Dexter et al. (2019) recorded the number of people at the foot of the Goodwill Bridge, Brisbane, who wore hats between \(11\):\(30\)am to \(12\):\(30\)pm.
Of the \(386\) males observed, \(79\) wore hats; of the \(366\) females observed, \(22\) wore hats. The software output is shown in Fig. 28.10 (right panel).
1. Construct the two-way table for the data.
2. Compute the standard error for the difference between the proportions.
3. Compute an approximate \(95\)% CI for the difference between the proportions.
4. Write down the \(95\)% CI for the difference between the proportions.
5. Interpret the CI.
6. Use the software output to write down a \(95\)% CI for the odds ratio.
7. Interpret the CI.
8. Are the CIs statistically valid?
Exercise 28.11 [Dataset: PetBirds] Kohlmeier et al. (1992) examined people with lung cancer, and a matched set of controls who did not have lung cancer, and recorded the number in each group that
kept pet birds. One RQ of the study was:
What is the odds ratio of keeping a pet bird, comparing people with lung cancer (cases) compared to people without lung cancer (controls)?
The data, compiled in a \(2\times2\) contingency table, are given in Table 28.9.
1. Construct a numerical summary table.
2. Sketch a graphical summary.
3. Use the software output (Fig. 28.11, left panel) to find a \(95\)% CI, making sure to describe the odds ratio carefully.
4. Use the software output to find a \(95\)% CI for the difference between the proportions.
5. Are the CIs statistically valid?
TABLE 28.9: The pet bird data.
Adults with lung cancer Adults without lung cancer
Did not keep pet birds \(141\) \(328\)
Kept pet birds \(\phantom{0}98\) \(101\)
Exercise 28.12 [Dataset: B12Diet] Gammon et al. (2012) examined B12 deficiencies in 'predominantly overweight/obese women of South Asian origin living in Auckland', some of whom were on a vegetarian
diet and some of whom were on a non-vegetarian diet. One RQ was:
What is the odds ratio of these women being B12 deficient, comparing vegetarians to non-vegetarians?
The data appear in Table 28.10, and the software output in Fig. 28.11 (right panel).
1. Construct a numerical summary table.
2. Sketch a graphical summary.
3. Use the software output to find a \(95\)% CI, making to describe the odds ratio carefully.
4. Is the CI statistically valid?
TABLE 28.10: The number of vegetarian and
non-vegetarian women who are (and are not) B12
B12 deficient Not B12 deficient
Vegetarians \(\phantom{0}8\) \(\phantom{0}26\)
Non-vegetarians \(\phantom{0}8\) \(\phantom{0}82\)
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Random Masters - Download Randomly Generated Mathematics Worksheets for Primary and High School in Excel or PDF Format
Worksheets in this category help students to read, write and understand common fractions and mixed numbers, and to be able to apply the four operations to solve problems involving fractions.
Worksheet Name Files
Fractions: Equivalent Fractions
Fractions: Comparing Using >, <, =
Fractions: Arranging in Order
Adding Fractions: Like Denominators
Adding Fractions: Related Denominators
Adding Fractions: Mixed Denominators
Subtracting Fractions: Like Denominators
Subtracting Fractions: Related Denominators
Subtracting Fractions: Mixed Denominators
Multiplying Fractions
Dividing Fractions
Fraction Review: Four Operations on Fractions
Converting Mixed Numbers to Improper Fractions
Converting Improper Fractions to Mixed Numbers
Adding and Subtracting Mixed Numbers
Multiplying and Dividing Mixed Numbers
Four Operations on Mixed Numbers
Fraction of a Quantity
|
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please help me!!!! … - QuestionCove
Ask your own question, for FREE!
Mathematics 75 Online
OpenStudy (anonymous):
please help me!!!! 15 minutes at 40 kph!
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OpenStudy (anonymous):
help me!! pleas
OpenStudy (anonymous):
How far/long?
OpenStudy (anonymous):
how far will i travel at these speeds for these times? 15 minutes at 40 kph?
OpenStudy (anonymous):
Since its 40 kilometers per HOUR, then you think of how long an hour is. 60 minutes. How long is 15 minutes compared to 60? One fourth. Divide 40 by 4 and you get the distance traveled in 15 minutes
while going 40kph.
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OpenStudy (anonymous):
thank you!
OpenStudy (anonymous):
No problem ^_^
OpenStudy (anonymous):
45 minutes at 100 kph so do you divide 45 by 100?
OpenStudy (anonymous):
Not exactly. You find what 45 minutes is of an hour. 45/60 simplifies to 3/4. What is 3 fourths of 100? That's the distance traveled in 45 minutes at 100kph. The way to do it boils down to a few
simple steps. 1.Find what part the time is of the perhour part like I did above. 45 minutes to 60 minutes because the speed is a perhour rate. 2.Once you have that, find that portion of the speed. 3/
4 of 100 is 75 as calculated by the fraction we got.
OpenStudy (anonymous):
so you divide 100 by 4 and multiply it by 3
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OpenStudy (anonymous):
umm i have this question sorry to bother u again 80 kph in 12 minutes so 12/60 simplify it
OpenStudy (anonymous):
No XD Find what the ratio between the rate/speed and the time actually traveled. Like I did above, you said you traveled 45 minutes, right? Your speed was 100 kilometers per hour (this is what KPH
means). This means that for every 60 minutes you travel, you'll cover 100 kilometers. But we didn't go the full 100 kilometers. We only went 45. We need to find out what the ratio for 45 out of 60 is
so we can find out how to fix 100 kph. 45/60 simplifies to 3/4. This means that if you traveled 45 minutes at 100kph, you would have traveled 3/4 of an hour. What is 3/4 of 100? We know its 75. This
means that we went 75 kilometers per 45 minutes. Does this make sense? Once you get this concept down the other problems should come quite easily.
OpenStudy (anonymous):
OpenStudy (anonymous):
If you have any other questions about this, be sure to message me ^_^
OpenStudy (anonymous):
i have about 20 to do :( i only get it wen u just tell wat to divide or watever this is too long
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OpenStudy (anonymous):
yea ino u just divide 100 by 4 and multiply by 3 and i got 75 that's the same way
OpenStudy (anonymous):
I explained the problem you gave me in detail. I'll replace the numbers with variables so you can plug in the numbers in each of the new problems and solve them easily.
OpenStudy (anonymous):
OpenStudy (anonymous):
how do u simplify??????
OpenStudy (anonymous):
If you traveled X minutes and your speed was Y kilometers per hour (this is what KPH means). This means that for every X minutes you travel, you'll cover Y kilometers. But we didn't go the full Y
kilometers. We only went N. We need to find out what the ratio for X out of 60 (the minute values, 60 is there because it is the number of minutes in an hour) is so we can find out how to fix Y kph.
Simplify as much as you can X/60. This means that if you traveled X minutes at Y kph, you would have traveled (insert simplified fraction here) of an hour. To find how far you traveled in X minutes,
take Y and the fraction. Using the fraction, find out what portion of Y it is. (Like I did above) 3 fourths of 100 is 75. When all is worked out you should get the distance traveled in X minutes.
Does this make more sense?
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OpenStudy (anonymous):
You simplify by finding the number you can divide both the numerator and denominator by that gives you the smallest answer. As I showed above, 45/60 was able to simplify to 3/4 because both the
numbers were divisible by 15. If you can find the largest number to divide both the top and the bottom by to get a whole number without a decimal so that you can no longer divide the top and bottom
anymore, you;v simplified the fraction. ALWAYS make sure you divide the top and bottom by the SAME number at the SAME time. If you can't divide one, you can't divide the other.
OpenStudy (anonymous):
Since your question is answered, could you move this to Closed ^_^? Click the Close button at the top of this post.
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vexpf_(3mvec) [opensolaris man page]
vexpf_(3mvec) [opensolaris man page]
vexp_(3MVEC) Vector Math Library Functions vexp_(3MVEC)
vexp_, vexpf_ - vector exponential functions
cc [ flag... ] file... -lmvec [ library... ]
void vexp_(int *n, double * restrict x, int *stridex,
double * restrict y, int *stridey);
void vexpf_(int *n, float * restrict x, int *stridex,
float * restrict y, int *stridey);
These functions evaluate the function exp(x) for an entire vector of values at once. The first parameter specifies the number of values to
compute. Subsequent parameters specify the argument and result vectors. Each vector is described by a pointer to the first element and a
stride, which is the increment between successive elements.
Specifically, vexp_(n, x, sx, y, sy) computes y[i * *sy] = exp(x[i * *sx]) for each i = 0, 1, ..., *n - 1. The vexpf_() function performs
the same computation for single precision data.
These functions are not guaranteed to deliver results that are identical to the results of the exp(3M) functions given the same arguments.
Non-exceptional results, however, are accurate to within a unit in the last place.
The element count *n must be greater than zero. The strides for the argument and result arrays can be arbitrary integers, but the arrays
themselves must not be the same or overlap. A zero stride effectively collapses an entire vector into a single element. A negative stride
causes a vector to be accessed in descending memory order, but note that the corresponding pointer must still point to the first element of
the vector to be used; if the stride is negative, this will be the highest-addressed element in memory. This convention differs from the
Level 1 BLAS, in which array parameters always refer to the lowest-addressed element in memory even when negative increments are used.
These functions assume that the default round-to-nearest rounding direction mode is in effect. On x86, these functions also assume that the
default round-to-64-bit rounding precision mode is in effect. The result of calling a vector function with a non-default rounding mode in
effect is undefined.
On SPARC, the vexpf_() function delivers +0 rather than a subnormal result for arguments in the range -103.2789 <= x <= -87.3365. Other-
wise, these functions handle special cases and exceptions in the same way as the exp() functions when c99 MATHERREXCEPT conventions are in
effect. See exp(3M) for the results for special cases.
An application wanting to check for exceptions should call feclearexcept(FE_ALL_EXCEPT) before calling these functions. On return, if
fetestexcept(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW | FE_UNDERFLOW) is non-zero, an exception has been raised. The application can then
examine the result or argument vectors for exceptional values. Some vector functions can raise the inexact exception even if all elements
of the argument array are such that the numerical results are exact.
See attributes(5) for descriptions of the following attributes:
| ATTRIBUTE TYPE | ATTRIBUTE VALUE |
|Interface Stability |Committed |
|MT-Level |MT-Safe |
exp(3M), feclearexcept(3M), fetestexcept(3M), attributes(5)
SunOS 5.11 14 Dec 2007 vexp_(3MVEC)
Check Out this Related Man Page
vexp_(3MVEC) Vector Math Library Functions vexp_(3MVEC)
vexp_, vexpf_ - vector exponential functions
cc [ flag... ] file... -lmvec [ library... ]
void vexp_(int *n, double * restrict x, int *stridex,
double * restrict y, int *stridey);
void vexpf_(int *n, float * restrict x, int *stridex,
float * restrict y, int *stridey);
These functions evaluate the function exp(x) for an entire vector of values at once. The first parameter specifies the number of values to
compute. Subsequent parameters specify the argument and result vectors. Each vector is described by a pointer to the first element and a
stride, which is the increment between successive elements.
Specifically, vexp_(n, x, sx, y, sy) computes y[i * *sy] = exp(x[i * *sx]) for each i = 0, 1, ..., *n - 1. The vexpf_() function performs
the same computation for single precision data.
These functions are not guaranteed to deliver results that are identical to the results of the exp(3M) functions given the same arguments.
Non-exceptional results, however, are accurate to within a unit in the last place.
The element count *n must be greater than zero. The strides for the argument and result arrays can be arbitrary integers, but the arrays
themselves must not be the same or overlap. A zero stride effectively collapses an entire vector into a single element. A negative stride
causes a vector to be accessed in descending memory order, but note that the corresponding pointer must still point to the first element of
the vector to be used; if the stride is negative, this will be the highest-addressed element in memory. This convention differs from the
Level 1 BLAS, in which array parameters always refer to the lowest-addressed element in memory even when negative increments are used.
These functions assume that the default round-to-nearest rounding direction mode is in effect. On x86, these functions also assume that the
default round-to-64-bit rounding precision mode is in effect. The result of calling a vector function with a non-default rounding mode in
effect is undefined.
On SPARC, the vexpf_() function delivers +0 rather than a subnormal result for arguments in the range -103.2789 <= x <= -87.3365. Other-
wise, these functions handle special cases and exceptions in the same way as the exp() functions when c99 MATHERREXCEPT conventions are in
effect. See exp(3M) for the results for special cases.
An application wanting to check for exceptions should call feclearexcept(FE_ALL_EXCEPT) before calling these functions. On return, if
fetestexcept(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW | FE_UNDERFLOW) is non-zero, an exception has been raised. The application can then
examine the result or argument vectors for exceptional values. Some vector functions can raise the inexact exception even if all elements
of the argument array are such that the numerical results are exact.
See attributes(5) for descriptions of the following attributes:
| ATTRIBUTE TYPE | ATTRIBUTE VALUE |
|Interface Stability |Committed |
|MT-Level |MT-Safe |
exp(3M), feclearexcept(3M), fetestexcept(3M), attributes(5)
SunOS 5.11 14 Dec 2007 vexp_(3MVEC)
|
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Spreadsheet-like Dataflow Programming in TypeScript
The reactive library for the spreadsheet driven development
import { cell, formula, swap, deref } from '@snapview/sunrise'
const inc = (a) => a + 1
// Source Cell with initial value 1
const x = cell<number>(1)
// Formula Cell created by incrementing the source cell
const y = formula(inc, x)
// Formula Cell without any value, but with side effect
const printCell = formula(console.log, y)
// Swapping the value of initial source cell
swap(inc, x)
deref(x) // 2
deref(y) // 3, this value is already printed to console because of printCell
Sunrise provides a spreadsheet-like computing environment consisting of source cells and formula cells and introduces the Cell interface to represent both
All Cells
• Contain values
• Implement the Dereferencable interface and their value can be extracted via the deref function
• Implement the Destroyable interface and can be destroyed via the destroy function
• Implement the Subscribable interface and can be subscribed via the subscribe function
and unsubscribed via the unsubscribe function
Source Cells
The source cell is just a container including a value of arbitrary type. To construct the source cell the cell function should be used
const a = cell<number>(1) // a cell of number
const b = cell<string>('hello') // a cell of string
const c = cell({ x: 1, y: 2 }) // an object cell
const d = cell<string | undefined>(undefined) // a cell that can be either string or undefined
To receive the current value of a cell the deref function is used. Unlike many other reactive libraries in Sunrise this is considered to be a totally valid operation. A cell is not a stream or any
other magic thing, it's just a box with a value inside
deref(a) // 1
deref(b) // 'hello'
deref(c) // { x: 1, y: 2 }
deref(d) // undefined
There are two ways to change a value inside a cell reset and swap. reset just sets a new value to and swap accepts a function from old value to new value, applies it and swap the cell to the new
const a = cell<number>(1)
reset(2, a)
deref(a) // 2
swap((x) => x + 1)
deref(a) // 3
reset and swap are async operations, the new value will be set not immediately, but they implement the Software Transaction Memory and they are always consistent.
In case you don't need a cell anymore, the cell can be destroyed with the destroy function. Be careful because destroying the cell will also destroy all the dependent cells as well. After the
destruction, any operation on a cell is illegal, and throw the OperationOnDestroyedCellError
const x = cell<number>(1)
const y = formula((a) => a + 1, x)
destroy(x) // both x and y are destroyed now
Formula Cells
A formula cell is a sort of materialized view of a function. You can look at it as a cell with a formula inside in some table processor program. To create a formula cell you need a formula (function)
and an arbitrary number of source cells as an input
const a = cell<number>(1)
const b = formula((x) => x + 1, a) // now b is always an increment of a
deref(b) // 2
reset(5, a)
deref(b) // 6
You can also use simple values as input to formula instead of cells. This might be quite handy when you don't know if the input is a cell or just a value
const x = cell<number>(1)
const y = cell<number>(2)
const z: number = 3
const sum = formula((a, b, c) => a + b + c, x, y, z)
deref(sum) // 6
reset(5, x)
deref(sum) // 10
Predefined formula cells
There are quite some formula cells predefined for faster cell generations
Object's field
To extract one field from an object you can use the field function
const x = cell({ a: 1, b: 2 })
const fld = field('a', x)
deref(fld) // 1
swap((x) => ({ ...x, a: 2 }), x)
deref(fld) // 2
An element of an array
To extract an element from an array by index you can use the byIndex function. The type of the result is Cell<T | undefined> because it's not guaranteed that the element is presented
const x = cell(['a', 'b', 'c'])
const el = byIndex(1, x)
deref(el) // 'b'
swap((x) => ['z', ...x], x)
deref(el) // 'a'
Convert to boolean
To check that an element is truthy you can use the toBool function.
const x = cell(1)
deref(toBool(x)) // true
const y = cell<string | undefined>(undefined)
deref(toBool(y)) // false
const x = cell<boolean>(true)
deref(not(x)) // false
const y = cell(1)
deref(not(y)) // false
In some cases, it's useful to have both the old cell's value and the new one. For this purpose, history can be used. It serves a tuple with the old and new
values inside. Be aware, initially, the old value is undefined
const x = cell<number>(1)
const hist = history(x)
deref(hist) // [1, undefined]
reset(2, x)
deref(hist) // [2, 1]
Download Details:
Author: snapview
Download Link: Download The Source Code
Official Website: https://github.com/snapview/sunrise
License: MIT
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Topography and Data Mining Based Methods for Improving Satellite Precipitation in Mountainous Areas of China
Department of Hydraulic Engineering, Tsinghua University, Beijing 100084, China
State Key Lab of Hydroscience and Engineering, Tsinghua University, Beijing 100084, China
Author to whom correspondence should be addressed.
Submission received: 16 February 2015 / Revised: 14 July 2015 / Accepted: 14 July 2015 / Published: 24 July 2015
Topography is a significant factor influencing the spatial distribution of precipitation. This study developed a new methodology to evaluate and calibrate the Tropical Rainfall Measuring Mission
Multi-satellite Precipitation Analysis (TMPA) products by merging geographic and topographic information. In the proposed method, firstly, the consistency rule was introduced to evaluate the fitness
of satellite rainfall with measurements on the grids with and without ground gauges. Secondly, in order to improve the consistency rate of satellite rainfall, genetic programming was introduced to
mine the relationship between the gauge rainfall and location, elevation and TMPA rainfall. The proof experiment and analysis for the mean annual satellite precipitation from 2001–2012, 3B43 (V7) of
TMPA rainfall product, was carried out in eight mountainous areas of China. The result shows that the proposed method is significant and efficient both for the assessment and improvement of satellite
precipitation. It is found that the satellite rainfall consistency rates in the gauged and ungauged grids are different in the study area. In addition, the mined correlation of
location-elevation-TMPA rainfall can noticeably improve the satellite precipitation, both in the context of the new criterion of the consistency rate and the existing criteria such as Bias and RMSD.
The proposed method is also efficient for correcting the monthly and mean monthly rainfall of 3B43 and 3B42RT.
1. Introduction
Precipitation is an important factor in water cycle systems, providing critical information for land-surface hydrological processes, climatological research, and water resource management.
Measurements of precipitation include traditional ground gauge stations, and more recently, satellite-based remote sensing monitoring. Although there are potential uncertainties caused by various
factors, such as systematic and random errors [
], and difficulties in capturing solid precipitation [
], gauge station rainfall has been widely accepted in terms of both accuracy and effectiveness due to its direct measurement. However, the sparse gauge networks, especially in rough terrains and
mountainous areas, hinder the application of gauge rainfall to basins/regional scales [
Fortunately, satellites provide an unprecedented possibility for retrieving global precipitation with satisfactory spatial and temporal resolutions over large scales [
]. Numerous satellite-based precipitation products in a variety of spatiotemporal scales have been publicly introduced despite widely diverse levels of accuracy [
]. Among them, the Tropical Rainfall Measuring Mission Multi-satellite Precipitation Analysis (TMPA) is one of the most widely used products [
]. TMPA provides two kinds of products, near-real-time products and post-real-time products. The near-real-time products provide valuable data 6–9 h after observation for real-time hydrological and
water-related disaster forecasts at the expense of accuracy [
]. Post-real-time products incorporate the global gauge dataset aiming at providing high quality rainfall data for research purposes with two months’ latency of observation, and so called research
products. The latest version is TMPA V7 issued in 2012 including 3B42RT for near-real-time and 3B43 for post-real-time. The 3B43 has been adjusted according to the gauge dataset produced by the
Global Precipitation Climatology Center [
Because of the indirect measurement and lack of reliable microwave data or high quality algorithms in areas characterized by cold land temperatures, snow-cover and ice-cover [
], uncertainties are inevitable for satellite rainfall products. Normally, satellite rainfall over grid cells with gauges is extracted and compared with gauge data, and the accuracy is evaluated by
statistical metrics (e.g
, mean bias and root mean squared error) [
]. Many efforts have been taken to investigate the errors in satellite rainfall measurements at regional, national and global scales, with some consistent findings. In general, research products of
TMPA fit the gauge rainfall data better than near-real-time products because of the gauge-merging procedures [
In addition to the error evaluation, a number of correction methods for satellite rainfall have been developed, in which the satellite-gauge rainfall bias over grid cells with gauges was considered
as critical information to derive the areal bias map by interpolation [
]. Some studies have indicated that the topography factors exert strange and complex controls on precipitation both globally and regionally [
], especially in the mountainous areas [
]. It was also found that the satellite-gauge difference varies with elevation [
]. Many approaches attempted to overcome this problem by deriving a linear regression model for correcting satellite rainfall with geographic/topographic information [
], or by stochastic model to adjust biases on rainfall frequency and intensity [
]. The linear rain-elevation relation was the common assumption [
However, the higher accuracy of satellite rainfall over gauged grid cells does not necessarily imply a higher accuracy over non-gauged grid cells; it could vary significantly with regions and
seasons. For instance, the TMPA product tends to overestimate rainfall in areas with lower actual rainfall but underestimate it in areas with higher actual rainfall [
]. It is also shown that higher error occurs at a higher temporal resolution, in winter, and in mountainous areas [
]. Moreover, the relationship between topography and rainfall is regionally dependent and still remains unclear so far. The non-linear relationship has been increasingly recognized in the
hydrological community [
Current studies of evaluation and correction of satellite precipitation products mainly focus on grid cells with gauge measurements. Few studies examine grid cells without gauges even though a
majority of grid cells do not have gauge measurements because of the sparse gauge network, especially in mountainous areas. In practice, the post-real-time satellite rainfall products (such as TMPA
3B43) incorporated global gauge network data to remove bias, so that the grid cells with gauges may have a higher accuracy, whereas grid cells without gauges have not been bias-corrected but are
considered to have the same level of accuracy as those grids with gauges. There is a large uncertainty and over-optimistic.
This study aimed to develop a comprehensive method to improve the current evaluation and correction methods for the satellite precipitation products. The methodology was described in details in
Section 2
following the introduction. In addition, eight mountainous areas of China were selected for testing the effectiveness of the proposed methods taking TMPA (V7) as a case study in
Section 3
. The discussion focus on the suitability of the methods to other satellite precipitation products and in deferent time scales, which was put in the front of the section of Conclusion.
2. Methodology
2.1. The Gauge-Elevation-Consistency (GEC) Rule for Assessment
The consistency rule is a general principle in various researches and public management, such as hydrology stationarity assumption. The consistency rule in this study is defined as that the rainfall
in a closer region should have the similar rainfall characteristics. The definition has two contents: (1) the satellite precipitation should have the same value as the ground gauged precipitation in
the same grid cells as ground gauges in; (2) the satellite precipitation should have the similar value as the ground gauged precipitation in the grid cells closer to the ground gauges.
The mathematical expression of the consistency rule in this study is expressed as follows.
$Δ P = P g − P s → 0 , ∀ L s = L g$
$Δ P = P u − P s → 0 , ∀ L s = L u$
$P g = P u , ∀ L g − L u < D$
means the rainfall and
means the location of the gauges or satellite grid cells. The subscript
stands for gauge station, while
represent the satellite grids with gauges (gauged gird cells) and without gauges (ungauged grid cells).
is a close region (see detail expression below).
The Equation (1) is the first content of the definition, which has been used numerously in comparing the difference between satellite precipitation and ground precipitation. The Equation (2) is the
second content of the definition, which just extends the Equation (1) from the grid cells with gauges to the cells without gauges. The Equation (3) is the bridge of ground gauges, gauged and ungauged
satellite cells. The Equation (3) has also been used frequently, like the Thiessen polygon for areal rainfall interpolation.
The closer region (i.e., closer grid cells) in this study is the areas which have the same elevation, the same slope aspect, and similar location as the ground gauges. The same elevation refers to
the relationship of rainfall and altitude; the same slope aspect refers to relationship of rainfall and vapor sources direction; the closer location is to limit the ungauged grid cells in a tolerant
distance from gauged grid cells. In the tolerant range, all ungauged grid cells’ ground precipitation is the same as the closest gauge’s precipitation. Taken the definition above, the error
assessment of satellite precipitation can be expanded from grid cells with gauge to grid cells without gauge.
For grid cells with gauge measurements, the corresponding grid values of the satellite rainfall are compared with ground measured precipitation according to the Equation (1). The assessment can be
quantitatively expressed by the statistic parameters of Mean Bias and Root Mean Square Deviation as follows.
$Bias = ∑ i = 1 M ( P S i − P G i ) ∑ i = 1 M P G i × 100 %$
$R M S D = 1 M ∑ i = 1 M ( P S i − P G i ) 2$
is the gauged grid cell rank number and
is the total number of ground gauges,
is the satellite rainfall,
is the ground gauge measurement.
For grid cells without gauge measurements, the corresponding grid values of the satellite rainfall will be compared with the ground measured rainfall according to Equations (2) and (3). The
assessment is quantitatively expressed by a newly proposed statistic criterion, Consistency Rate (CR), as follow.
$c o u n t j = { 1 , if P S j ∈ D 0 , if P S j ∉ D$
$n = ∑ j = 1 N c o u n t j$
is the ungauged grid cell rank number.
is the Rainfall-Elevation Mask (REM) for qualification the ungauged cells precipitation,
, the exact closer region in Equation (3).
is the number of satellite ungauged grid cells within the REM having comparable rainfall as gauges, and
is the total number of ungauged grid cells.
The REM can be derived based on the rainfall measurement and elevation of the gauges. Assuming there are
≥ 2) ground rainfall gauges in the same slope of a mountain. Sort the
gauges in ascending (or descending) order of elevation. Set every
= 2, …,
) sequential gauges as a group
= 1, …,
+ 1). For each
, there are a lowest and highest elevation (
$E k l o w , E k h i g h$
), as well as a minimum and maximum gauge rainfall (
$P G k l o w , P G k h i g h$
). The rectangle space of (
$E k l o w , P G k l o w ; E k h i g h , P G k h i g h$
) forms a closer region,
. All the closer regions consist of the whole rainfall-elevation mask
. The mask physically denotes the possible or reasonable rainfall range for each elevation.
Figure 1
shows an example of deriving the rainfall-elevation mask (
= 3,
= 5).
Figure 1. The derivation of rainfall-elevation mask (l = 3, M = 5). The red solid lines are the up and low limit of rainfall and the black dashed lines are the up and low limit of elevation of each
sub-mask. The whole pink region is the final rainfall-elevation mask (REM).
A larger CR means that a larger proportion of satellite grid cells have comparable rainfall as gauge measurements within the same elevation range. In other words, satellite precipitation tends to
have higher accuracy. Therefore, the CR value can quantifiably measure the consistency between the gauge and satellite rainfall over grid cells without gauges.
2.2. The Location-Elevation-TMPA (LET) Correlation for Improvement
Considering the significant influence of topographic and geographical features on rainfall, topography and geography information is involved in improving the satellite rainfall in this study. The
gauge data are assumed to be the actual rainfall values. For the grid cells with gauges, the relationships between the actual rainfall and topographic/geographic information as well as TMPA rainfall
were investigated. Compared with the original TMPA rainfall, the final predicted rainfall incorporates topographic and geographic information. In other words, TMPA rainfall was corrected by the
topographic and geographic information.
Given the unclear influence of topographic and geographic factors on rainfall, Genetic Programming (GP) was used as a tool to mine the relationship between rainfall and related factors. In the
present study, the real rainfall (gauge measurements) was used as the target of GP, and the inputs include geographical location (north latitude and east longitude), elevation, and TMPA rainfall. The
method is expressed below:
$P A = f ( X , Y , E , P S )$
is the actual rainfall,
are the locations (longitude and latitude, respectively),
is the elevation, and
is the rainfall from TMPA. Elevation is considered as the main variable that influences the spatial distribution of precipitation in mountainous areas, geography information is regarded as the factor
of regional and local climate patterns, and the TMPA data is involved to make full use of the satellite information.
There are three steps for mining the robust and explicit formula of Equation (9).
Step (1), testing calibration. Exclude part of ground gauges and then put the remaining gauges into the mining dataset together with the same grid cells of satellite for data mining. The information
of the dataset includes location of the gauges, elevation and satellite precipitation in the same location as gauges. The problem of GP is as follows.
$Min R 2 = ( M ∑ i = 1 M ( P G i • P A i ) − ∑ i = 1 M P G i • ∑ i = 1 M P A i ) 2 / [ M ∑ i = 1 M P G i 2 − ( ∑ i = 1 M P G i ) 2 ] [ M ∑ i = 1 M P A i 2 − ( ∑ i = 1 M P A i ) 2 ] Min C V ( R M S D
) = R M S D P G ¯ × 100 % = M ∑ i = 1 M P G i ( P G i − P A i ) 2 M × 100 %$
is the coefficient of correlation between gauge measurements (
) and modeled actual rainfall (
is coefficient of variation of the
, which is calculated by normalizing
by the mean value of the measurements. The target of the GP problem is to minimize the
between actual rainfall from Equation (9) and gauge measurements.
Step (2), cross-validation. Predict the satellite precipitation in the grid cells excluded gauges located by the mined explicit formula of Equation (9) and assess the fitness of the mined explicit
Repeat step (1) and (2) until all the gauges are included and excluded at least once to validate the effectiveness of LET method.
Step (3), final calibration. If the cross-validation process indicates that LET is valid, put all gauges into mining dataset together with the same grid cells of satellite for data mining. The mind
formula is the final explicit Equation (9), which can be used to predict and adjust satellite precipitation in the grid cells both with and without ground gauges.
3. Case Study and Results
3.1. Data
The data used in this paper are mainly the satellite precipitation products and ground gauges precipitation. The study area was focused on the eight mountainous areas of China, including Himalaya
(the part in China), Kunlun, Tianshan (the part in China), Qilian, Qinling, Taihang, Changbai, and Wuyi. The location and basic information were shown in
Figure 2
Table 1
The Himalaya, Kunlun and Qilian are located in the Qinghai–Tibetan High Plateau. The Himalaya tops at 8848 m and blocks the wet monsoon winds from south, drier and colder winds from the north [
]. Kunlun tops at 7576 m controlled by the Westerlies in summer and Mongolian–Siberian High in winter with a cold and arid climate [
]. Qilian has the peak of 5820 m and the climate varies from arid to semiarid with large temporal–spatial precipitation difference [
]. Tianshan is also located in the west of China and is kept away from the moisture, resulting in an arid continental climate [
]. Taihang and Qinling are located in the central part of mainland of China with the peak of 3059 m and 3747 m respectively. The Taihang serves as an important geographical boundary with the Loess
Plateau to its west and the North Plain to its east [
]. The Qinling acts as a significant climate boundary in China, blocking cold, dry airflow from the north in winter and humid, warm airflow from the south in summer [
]. Influenced by the East Asian summer monsoon, Taihang and Qinling have the typical continental monsoon climate with wet summers and dry winters. Changbai and Wuyi are located in northeast and
southeast of China, respectively, with relatively low altitude. The Changbai stretches along the boundary between China and North Korea and has a temperate continental climate with long cold winter
and short summer [
]. Located in southeast China near the sea, the Wuyi has a warm, humid subtropical climate characterized as monsoon patterns. The hot season (April to September) has the greatest precipitation
brought by summer monsoon and typhoons [
Figure 2. The location of the studied mountainous areas of China. The red lines are the boundaries of the mountainous areas, and the blue dots are the gauge stations.
Table 1. Basic information for the studied mountainous areas.
Region Area (10^3 km^2) Mean Elevation ^a(m) Peak Elevation (m) Gauges Information
Gauges Numbers Gauges Altitudes (m) Mean Annual Rainfall (2001–2012) (mm)
Himalaya 1054.7 4592 8848 33 2328–4900 467
Kunlun 786.7 2897 7576 15 887–3504 102
Tianshan 392.2 1712 7125 19 35–2458 180
Qilian 337.6 2954 5820 23 1139–3367 230
Qinling 129.5 921 3747 13 249–2065 770
Taihang 223.2 1012 3059 22 63–2208 498
Changbai 631.9 334 2667 48 4–775 663
Wuyi 366.2 386 2154 43 3–1654 1589
^a The elevation is from the DEM of the Shuttle Radar Topography Mission with a spatial resolution of 90 m.
The satellite precipitation (2001–2012) comes from the last version (V7) of TMPA 3B43, which was adjusted by global gauge dataset and provides monthly precipitation with the spatial resolution of
0.25° and global coverage of 50°S–50°N data for research (
). The ground gauge precipitation (2001–2012) comes from the last version of China Daily Ground Climate Dataset, which was produced by National Meteorological Information Center of China (NMIC) and
provides daily precipitation (
). This dataset has been applied in numerous studies as an actual rainfall reference [
]. The publicly available NMIC dataset comprises daily climate observations from 824 meteorological stations covering almost the entire China. This research used 216 stations within the eight
mountainous regions (see
Figure 2
). Following other researches [
], the precipitation is converted into the mean annual data (average of 2001–2012) in each grid cells of satellite and each gauge of ground measurement. The monthly 3B43 rainfall and near-real-time
3-h 3B42RT rainfall were also discussed in
Section 4
The digital elevation model (DEM) comes from the Geological Survey Earth Resources Observation and Science Center (
) of the United States. The original spatial resolution of the DEM data over China is 90 m.
3.2. Assessment the Uncertainty of Satellite Precipitation
3.2.1. Grid Cells with Gauges
For grid cells with gauge measurements, the mean annual rainfall (2001–2012) from TMPA 3B43 (V7) was compared with that from the gauge stations over the eight mountainous regions, as shown in
Table 2
Figure 3
Table 2. Comparison of mean annual rainfall (2001–2012) from gauge measurements and the original 3B43.
Table 2. Comparison of mean annual rainfall (2001–2012) from gauge measurements and the
original 3B43.
Region Altitude of Gauges (m) Gauge (mm) 3B43 (mm) Bias (%) RMSD (mm)
Himalaya 2328–4900 453 667 47.2 272
Higher Mountains Kunlun 887–3504 106 131 23.6 76
Tianshan 35–2458 175 200 14.3 54
Qilian 1139–3367 233 264 13.3 65
Qinling 249–2065 776 791 1.9 46
Lower Mountains Taihang 63–2208 502 542 8.0 49
Changbai 4–775 674 757 12.3 103
Wuyi 3–1654 1560 1654 6.0 161
Average -- 560 626 15.8 103
It shows that the 3B43 products tended to overestimate rainfall by 15.8% on mean annual scale. In the higher mountains with peaks above 5800 m, including Himalaya, Kunlun, Tianshan and Qilian, the
mean overestimation bias is 24.5% in the altitude from 35 m to 4900 m on gauged grid cells. In the other mountains with peaks lower than 3800 m, including Qinling, Taihang, Changbai and Wuyi, the
mean overestimation bias is only 7.0% in the altitude from 3 m to 2208 m on gauged grid cells.
The result presented in the study is mostly consistent with previous reports, such as the study in east and northeast China during the period between 2005 and 2007 [
], in Gangjiang river basin originated from Wuyi Mountains [
], in Jinghe river basin originated from Qinling Mountains [
]. However, it varied in the Himalaya Mountains in literatures, both underestimation in Nepal and Pakistan [
] and significant overestimation over the Tibetan Plateau [
] was reported.
Figure 3. Validation of the mean annual rainfall (2001–2012) from gauge measurements and the original 3B43.
Despite the merging of the global gauge network, TMPA 3B43 (V7) still shows a noteworthy error in most mountainous areas over China, especially in the higher mountain areas.
3.2.2. Grid Cells without Gauges
In order to refer to the closer region for the grid cells without gauges, each mountain was divided into two regions along the hillsides according to the watershed and vapor transportation direction.
The three connected gauges (
= 3) rainfall-elevation mask was derived hillside by hillside, as shown in
Figure 4
Taking the filter by mask
, the Consistency Rate (CR) can be calculated obviously. The CR of gauged grid cells was also calculated to make a comparison as well as to test the effectiveness of CR in evaluating the rainfall
accuracy. The result is shown in
Table 3
. For grid cells with gauges, the CR showed an average of 57.9% in higher mountains (including Himalaya, Kunlun, Tianshan, and Qinling) and 66.9% in other mountains (including Qinling, Taihang,
Changbai, and Wuyi). This result is consistent with that from traditional assessment in
Table 2
There is a significant difference between the CR values for gauged and ungauged grid cells, and the difference can reach up to 17% in Qinling. In some cases, such as in Qilian and Wuyi, the CR was
relatively low for grid cells without gauges despite of the high CR for gauged grid cells. Thus, satellite rainfall accuracy in grid cells with gauges might fail to represent the accuracy of grid
cells without gauges.
Figure 4. The mean annual (2001–2012) rainfall-elevation mask and rainfall filter (the pink space, l = 3). The green/black forks stand for TMPA rainfall grid cells with gauges located in/out of the
masks. The grey circles are the cells without gauges. Kunlun has only one mask because of lacking of enough gauges on another hillside. (a,b) Himalaya, (c) Kunlun, (d,e) Tianshan, (f,g) Qilian, (h,i)
Qinling, (j,k) Taihang, (l,m) Changbai, (n,o) Wuyi.
A common issue in the current evaluation is the mismatch in spatial scale between the satellite rainfall (always 0.25° × 0.25° grid) and the gauge measurements (point scale). Before the potential
improvement of the spatial resolution of satellite rainfall products, some endeavors for the mismatch issue were attempted by averaging the gauge rainfall within the satellite grid cells [
], or obtaining grid gauge rainfall by an interpolation technique [
] if a denser gauge network is available. Given the sparse available gauges, the corresponding grid cell values were separately extracted for comparison with gauge data over grid cells with gauges,
which was the common method to conduct accurate evaluations [
Table 3. Consistency rate (%) of mean annual rainfall (2001–2012) of the original 3B43.
Region CR in the Whole Region CR in the Hillside Face to Vapor Transportation CR in the Hillside Back to Vapor Transportation
Gauged Grids Ungauged Grids Gauged Grids Ungauged Grids Gauged Grids Ungauged Grids
Himalaya 51.5 57.4 100.0 84.7 42.9 47.7
Higher Mountains Kunlun 57.1 60.0 57.1 60.0 -- --
Tianshan 57.9 63.5 55.6 74.4 60.0 53.4
Qilian 65.2 58.7 50.0 51.4 88.9 69.5
Qinling 92.3 75.0 100.0 73.1 87.5 77.6
Lower Mountains Taihang 63.6 64.6 42.9 55.0 73.3 68.2
Changbai 60.4 67.0 58.1 67.0 64.7 67.5
Wuyi 51.2 33.8 50.0 36.4 51.9 32.5
Average 62.4 60.4 64.2 62.8 67.0 59.5
3.3. Improvement the Robust of Satellite Precipitation
3.3.1. Testing Calibration and Cross Validation
The LET method was validated by cross-validation process before being applied to the satellite rainfall. For each mountain, 3–6 gauges were excluded when searching for the best fit through GP, and
then the mined relation was applied on the excluded gauge for validating the relation. The procedure is then repeated until all the gauges have been excluded only once. Given the number of available
gauges (see
Table 1
), the cross-validation process is repeated 4–8 times for the eight areas.
Figure 5
shows the comparison between corrected rainfall and measurements.
of original 3B43 rainfall ranged within 46~272 mm (see
Table 2
). However after the correction by the LET method,
was reduced to 31~97 mm. Corrected rainfall fitted gauge data well in
Figure 5
. Therefore, the relationships between the actual rainfall and relevant factors obtained from GP were capable of deriving reliable actual rainfall over grid cells with gauges.
3.3.2. Final Calibration and Correction of TMPA
Put all the ground gauges into the mining dataset to correct the mean annual rainfall of 3B43 of the period 2001–2012 by LET. The mined final equations are listed in the
. The error parameters are listed in
Table 4
. It indicates that the accuracy of 3B43 rainfall was significantly improved in gauge grid cells after the calibration, compared with the original 3B43 accuracy in
Table 2
. The average bias was reduced from 15.8% to −4.5%; meanwhile, the average
was reduced from 103 mm to 47 mm. Compared with
Table 3
, the average CR was improved from 62.4% to 76.6% and from 60.0% to 64.2% for the grid cells with and without gauges, respectively.
Figure 5. Cross-Validation of the Location-Elevation-TMPA (LET) method using mean annual rainfall (2001–2012) from gauge measurements and corrected 3B43.
Table 4. Comparison of mean annual rainfall (2001–2012) from gauges and corrected 3B43.
Region Mean of Gauges (mm) Mean of gauged grids (mm) Bias (%) RMSD (mm) CR of Gauged Grids (%) CR of Ungauged Grids (%)
Himalaya 453 422 −6.8 92 84.8 78.2
Kunlun 106 76 −28.2 49 64.3 63.7
Tianshan 175 171 −2.2 31 84.2 66.7
Qilian 233 235 1.0 26 82.6 60.4
Qinling 776 777 0.2 27 76.9 71.6
Taihang 502 503 0.3 19 72.7 77.0
Changbai 674 674 0.0 44 75.0 55.1
Wuyi 1560 1561 0.0 89 72.1 41.1
Average 560 552 −4.5 47 76.6 64.2
Figure 6
shows the rainfall-elevation scatters from the original 3B43, corrected 3B43 and gauged data. Scatter for 3B43 was plotted every 200 m in bins based on the elevation range of each mountain. It can be
found that the rainfall presented an obvious tendency
elevation in most cases. The rainfall presents a positive correlation to elevation in Wuyi, Changbai and Qilian, as well as Tianshan and Kunlun in the range of elevation <4000 m. The
rainfall-elevation relation did not show an obvious trend in Qinling (
Figure 6
e). However, it shows a negative relation between rainfall and elevation in Taihang (
Figure 6
f), and in Himalaya when elevation <4500 m (
Figure 6
a). The similar feature was also reported in the Himalaya north foreland to the Tibetan Plateau [
] and south foreland to Nepal [
], and also in Awash River in Ethiopia [
]. Rainfall-elevation scatters from corrected satellite rainfall were much closer to that from gauge data, especially in Himalaya (see
Figure 6
Figure 6. The original (black squares), corrected (black circles) 3B43 and gauged (red triangles) mean annual rainfalls (2001–2012) versus elevation in study areas. The error bars denote the lower 5%
and upper 95% rainfall value within each elevation range. (a) Himalaya, (b) Kunlun, (c) Tianshan, (d) Qilian, (e) Qinling, (f) Taihang, (g) Changbai, (h) Wuyi.
It should be noted that the LET method relies on gauge data to fit the relationship. The mined relation may only be valid within areas that have similar rainfall error patterns as the gauge grid
cells do. To extend the relation to where exceeding to the gauge elevation range, more attention should be paid.
4. Discussion
4.1. The Sensitive of CR to l of Rainfall-Elevation Mask
As described in
Section 2.1
is the number of the gauges for the sequential gauge group used to derive the rainfall-elevation mask. The
values were calculated with different
as shown in
Table 5
. The
increases with the increasing of
, which is reasonable since a larger
contributes a larger overlap for rainfall-elevation mask. However, sensitivity of
decreases when
surpasses 3, and
= 2 might be too strict,
= 3 might be a good option for evaluating the accuracy of satellite rainfall without gauges.
Table 5. Consistency rate (%) changes with l for
mean annual rainfall (2001–2012) of original 3B43.
Region l = 2 l = 3 l = 4 l = 5
Himalaya 42.0 57.4 61.4 62.8
Higher Mountains Kunlun 31.9 60.0 69.4 75.9
Tianshan 29.8 63.5 75.1 83.0
Qilian 36.9 58.7 71.0 77.8
Qinling 40.5 75.0 89.7 94.0
Lower Mountains Taihang 44.0 64.6 82.5 85.9
Changbai 38.2 67.0 79.8 84.0
Wuyi 12.9 33.8 44.0 53.6
Average 34.5 60.0 71.6 77.1
4.2. The Suitability of LET for Monthly Precipitation of TMPA 3B43 (V7)
To test the effectiveness of LET on monthly scale, Kunlun was taken as a case study. The rainfall of every July and every month were corrected from the dataset of 2001–2012, and the
before and after correction were listed in
Table 6
. Comparison between satellite rainfall and measurements were shown in
Figure 7
. For both every July and every month cases, the corrected rainfall has higher
and lower
than originals, indicating the valuable effectiveness of LET method on monthly scale.
Table 6. Errors of original and corrected monthly
rainfall (2001–2012) of 3B43 in Kunlun.
Time Scale Original Corrected
R^2 CV(RMSD) (%) R^2 CV(RMSD) (%)
Every July 0.55 72.9 0.73 57.1
Every month 0.53 124.9 0.61 109.2
Figure 7. Comparison between 3B43 rainfall (original and corrected) and gauge rainfall on every July (a) and every month (b) in Kunlun during 2001–2012.
4.3. The Effectivity for the of TMPA 3B42RT (V7)
4.3.1. Effective for Assessment
In this section, the evaluation and correction methods were applied on TMPA 3B42RT (V7) to test the effectiveness of the proposed methods on near-real-time products which is free from bias-correction
procedure. The statistical parameters and scatter comparison showing the fitness between rainfall from 3B42RT and gauges were shown in
Table 7
Figure 8
Table 7. Comparison of mean annual rainfall (2001–2012) from gauge measurements and original 3B42RT.
Table 7. Comparison of mean annual rainfall (2001–2012) from gauge measurements and original 3B42RT.
Region Mean of Gauge (mm) Mean of Gauged Grids (mm) Bias (%) RMSD (mm) CR of Gauged Grids (%) CR of Ungauged Grids (%)
Himalaya 453 1457 221.6 1050 0.0 1.9
Kunlun 106 360 239.6 332 14.3 44.0
Tianshan 175 771 340.6 660 0.0 2.3
Qilian 233 454 94.8 276 30.4 47.8
Qinling 776 835 7.6 81 76.9 58.6
Taihang 502 653 30.1 162 22.7 16.5
Changbai 674 676 0.3 104 77.1 67.3
Wuyi 1560 1562 0.1 313 34.9 22.5
Average 560 846 116.8 372 32.0 32.6
Figure 8. Validation of the mean annual rainfall (2001–2012) from gauge measurements and original 3B42RT.
The result showed that the TMPA 3B42RT products tended to overestimate rainfall in most study areas too. In the higher mountains with peak above 5800 m, including Himalaya, Kunlun, Tianshan and
Qilian, the mean overestimation bias is 224.0%. In the others with peak lower than 3800 m, including Qinling, Taihang, Changbai and Wuyi, the mean overestimation bias is 9.5%. For the grid cells with
gauges, CR is averagely 32.0%, 11.2% in higher mountains, and 52.9% in the other mountains. For the grid cells without gauges, CR is averagely 32.6%, 24.0% in higher mountains, and 41.2% in other
mountains. Comparison with the result of 3B43 in
Table 2
Table 3
, the 3B42RT presents a larger uncertainty judged by both traditional criteria of bias and
and the new criterion of CR. The similar result was also found in western and northern China from 2008 to 2011 [
]. This is obviously reasonable since 3B42RT has not been adjusted by ground gauged data.
4.3.2. Effective for Correction
The finesses between LET-corrected 3B42RT rainfall and measurements in gauged grid cells are listed in
Table 8
Figure 9
shows the comparison between corrected 3B42RT and measurements. It is clear that the accuracy of 3B42RT rainfall was significantly improved in gauge grid cells after the calibration, compared with
the original 3B42RT accuracy in
Table 7
. Average Bias was reduced from 116.8% to 8.7% and average
was reduced from 372 to 72 mm, the average CR was improved significantly from 32.0% to 77.1% in the grid cells with gauges and from 32.6% to 62.7% in the grid cells without gauges. Compared with that
of 3B43 (
Table 3
Table 4
), the improvement for 3B42RT is almost doubled. This is reasonable since 3B43 had been corrected by GPCC dataset.
Table 8. Comparison of mean annual rainfall (2001–2012) from gauge measurements and corrected 3B42RT.
Table 8. Comparison of mean annual rainfall (2001–2012) from gauge measurements and corrected 3B42RT.
Region Mean of Gauges (mm) Mean of Gauged Grids (mm) Bias (%) RMSD (mm) CR of Gauged Grids (%) CR of Ungauged Grids (%)
Himalaya 453 454 0.2 95 87.9 76.1
Kunlun 106 117 10.8 35 64.3 78.8
Tianshan 175 290 65.2 130 42.1 46.8
Qilian 233 231 −0.6 38 78.3 61.1
Qinling 776 770 −0.8 41 100.0 62.1
Taihang 502 496 −1.2 30 86.4 71.8
Changbai 674 656 −2.7 60 83.3 50.5
Wuyi 1560 1544 −1.0 143 74.4 54.3
Average 560 570 8.7 72 77.1 62.7
Figure 9. Validation of the mean annual rainfall (2001–2012) from gauge measurements and corrected 3B42RT.
5. Conclusions
Based on gauge data and TMPA data on the eight mountainous regions of China, methods were proposed for evaluating and calibrating satellite rainfall. The main findings are as follows.
Most evaluation and calibration approaches for satellite rainfall are primarily conducted on grid cells with gauges. However, the evaluation method based on the context of consistency rule shows that
a higher accuracy of satellite rainfall in gauged grid cells does not absolutely indicate the same accuracy in ungauged grid cells. The consistency rule should be paid enough attention to the grid
cells both on the gauged and ungauged satellite grid cells.
Many literatures reported that calibration for satellite rainfall in mountainous regions is unsatisfactory, especially at higher elevations, which is attributed to the complex topography and
geography of higher mountains. This study demonstrated that the topographic and geographic information is valuable for correcting satellite rainfall. The location-elevation-TMPA (LET) method based on
GP data mining presents great potential to overcome the difficulty of unknown pattern of the relationship and is valid both for near-real-time and research products, and both on monthly and annual
It should be mentioned that the basic assumption of the evaluation and correction methods is the reliability of the gauge rainfall. Gauges may suffer from various uncertainty caused by the many
reasons, such as difficulty of capturing solid precipitation. Meanwhile, satellite rainfall also tends to have large errors due to low signal-to-noise ratio in high-altitude and cold area. However,
reliable methods to fix this problem are still missing with the current techniques. Results from the proposed methods should be double checked in these areas. Moreover, particular attention should
also be paid when applying the relationship from gauged cells to areas with elevation exceeding the gauge elevation range.
This work was financially supported by the National Natural Science Foundation of China via Grant 51279076 and 91125018, the National Science and Technology Support Program Project of China
(2013BAB05B03) and the Public Non-profile Project of the Ministry of Water Resources of China (Grant No. 201301082, 201401031). We greatly thank the editors and reviewers for providing thorough and
constructive comments to improve the manuscript.
Author Contributions
Zhongjing Wang and Ting Xia designed this study, conducted the analysis, and wrote the manuscript; Hang Zheng provided important suggestions, and improved the manuscript.
Conflicts of Interest
The authors declare no conflicts of interest.
Table A1. Relationship between actual rainfall and related factors (P[T]: TMPA rainfall; E: elevation; X: longitude; Y: latitude).
Table A1. Relationship between actual rainfall and related factors (P[T]: TMPA rainfall; E: elevation; X: longitude; Y: latitude).
Region Relationships for 3B42RT Relationships for 3B43
Himalaya $P a = { exp [ exp ( 1 4 X 1 2 ) − 2 E − X 3 2 ] } 1 4 ( P T − E 1 2 + 5.67 P T − 9.35 X + E X 1 $P a = ( P T − 2 Y − X ) 1 4 exp ( 0.25 ( X + Y ) 1 2 ) Y ( E − X 1 2 E − 1 2 − 2 P T ) 1 8 −
2 ) 1 2 − X 2 Y$ 4 X − Y$
Kunlun $P a = E exp ( 3 Y 1 2 ) X − 4 [ 1 + E 4 Y 3 exp ( 3 Y 1 2 ) + X Y P T + Y 1 2 + Y ] − 1$ $P a = Y + X − 3 2 exp [ 6.13 E ( T + 3 X ) X + Y 3 2 ( X 2 − 1.35 E ) 1 16 P T 1 8 X − 1 ]$
Tianshan $P a = exp { [ exp ( P T X ) E + P T + ( E − 9.88 ) E 2 P T 3 exp ( E Y ) E + P T 3 exp ( P T X $P a = P T ( X − Y ) [ ( P T + 3 X ) ( Y − X ) exp ( E T − X ) + 1.69 P T − Y ( Y − X ) + Y ]
) + X P T 3 E + E ] 1 16 Y 3 2 X − 1 }$ − 1 − P T exp ( E P T ) − Y + P T 1 2$
Qilian $P a = ( P T X ) 1 2 ( 4 P T + E ) 1 4 ( X − 2 Y ) − E Y − 1228.5 ( P T E ) 2 − Y$ $P a = E 1 2 P T { ( 3 Y − X ) [ P T 3 X 0.005 E 2 ( E − X ) + Y ] + 0.62 E } − 1 2$
Qinling $P a = 1 + exp [ 0.5 exp ( 0.5 P T 1 4 ) − 0.5 X 1 2 ] P T 1 2 E 1 2 X + 2 X + T + T 1 2 − ( 1 Y $P a = ( E + 2 X ) X ( P T − E ) 2 + ( 2 X + P T ) E ( 2 P T − E ) ( P T − E ) + P T − Y$
2 − 1 X ) exp ( X 1 2 )$
Taihang $P a = X − 1.51 Y { ( 1.41 Y 1 2 P T − E ) 1 2 [ X Y 8 + X 3 2 ( X − 2.35 ) 1 2 Y 15 2 + P T E 4 $P a = ( X − Y − 1.5 ) [ ( X Y E ) 1 2 + X 33 16 Y E 1 16 − Y − X + P T − 1.2 ] ( X − E Y X Y
] 3 2 1.19 Y 61 4 + X }$ − E ) − 1$
Changbai $P a = ( P T − 2 Y + X P T E − 2 + E 3 2 + X ) 2 [ Y − ( P T − Y ) ( E 2 + X Y P T + E Y ) Y E 2 $P a = P T − Y − E Y 1 2 − ( 0.731 − X − Y − E ) P T − 2 ( Y − E ) 2 ( 0.80 P T − E ) ( Y − E
+ X Y 2 P T + E Y 2 − Y P T 2 ] − 1$ )$
Wuyi $P a = X + Y + [ P T + P T 9 ( P T E − 1 + E ) 3 Y 9 E 3 ( E 2 + Y + X ) 1 2 ] 1 16 P T 5 16 E 1 $P a = P T + ( 3 E + P T + Y + P T 1 2 − X ) 1 2 − ( 2 E + P T + Y ) { X E P T + E − X + exp
16 X 7 8$ [ ( X − Y ) 1 2 − E 1 2 ] } − 1$
© 2015 by the authors; licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution license (http://
Share and Cite
MDPI and ACS Style
Xia, T.; Wang, Z.-J.; Zheng, H. Topography and Data Mining Based Methods for Improving Satellite Precipitation in Mountainous Areas of China. Atmosphere 2015, 6, 983-1005. https://doi.org/10.3390/
AMA Style
Xia T, Wang Z-J, Zheng H. Topography and Data Mining Based Methods for Improving Satellite Precipitation in Mountainous Areas of China. Atmosphere. 2015; 6(8):983-1005. https://doi.org/10.3390/
Chicago/Turabian Style
Xia, Ting, Zhong-Jing Wang, and Hang Zheng. 2015. "Topography and Data Mining Based Methods for Improving Satellite Precipitation in Mountainous Areas of China" Atmosphere 6, no. 8: 983-1005. https:/
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Probabilistic Analyses of Combinatorial Optimization Problems on Random Shortest Path Metrics
Simple heuristics for combinatorial optimization problems often show a remarkable performance in practice. Worst-case analysis often falls short of explaining this performance. Because of this,
‘beyond worst-case analysis’ of algorithms has recently gained a lot of attention, including probabilistic analysis of algorithms.
The instances of many combinatorial optimization problems are essentially a discrete metric space. Probabilistic analysis for such metric optimization problems has nevertheless mostly been conducted
on instances drawn from Euclidean space, which provides a structure that is usually heavily exploited in the analysis. However, most instances from practice are not Euclidean. Little work has been
done on metric instances drawn from other, more realistic, distributions. Some initial results have been obtained by Bringmann et al. (Algorithmica, 2015), who have used random shortest path metrics
generated from complete graphs to analyse heuristics.
In this thesis we look at several variations of the random shortest path metrics, and perform probabilistic analyses for some simple heuristics for several combinatorial optimization problems on
these random metric spaces. A random shortest path metric is constructed by drawing independent random edge weights for each edge in a graph and setting the distance between every pair of vertices to
the length of a shortest path between them, with respect to the drawn weights.
We provide some basic properties of the distances between vertices in random shortest path metrics. Using these properties, we perform several probabilistic analyses. For random shortest path metrics
generated from (dense) Erdős-Rényi random graphs we show that the greedy heuristic for the minimum-distance perfect matching problem, the nearest neighbor and insertion heuristics for the traveling
salesman problem, and a trivial heuristic for the k-median problem all achieve a constant expected approximation ratio. Additionally, we show a polynomial upper bound for the expected number of
iterations of the 2-opt heuristic for the traveling salesman problem in this model.
For random shortest path metrics generated from sparse graphs we show that the greedy heuristic for the minimum-distance perfect matching problem, and the nearest neighbor and insertion heuristics
for the traveling salesman problem all achieve a constant expected approximation ratio. Additionally, we show that the 2-opt heuristic for the traveling salesman problem also achieves a constant
expected approximation ratio in this model.
For random shortest path metrics generated from complete graphs we analyse a simple greedy heuristic for the facility location problem: opening the κ cheapest facilities (with κ only depending on the
facility opening costs). If the facility opening costs are such that κ is not too large, then we show that this heuristic is asymptotically optimal. For large values of κ we provide a closed-form
expression as upper bound for the expected approximation ratio and we evaluate this expression for the special case where all facility opening costs are equal.
Moreover, we show in this model that a simple 2-approximation algorithm for the Steiner tree problem is asymptotically optimal as long as the number of terminals is not too large. We also present
some numerical results that imply that the 2-opt heuristic for the traveling salesman problem seems to perform rather poorly in this model.
Dive into the research topics of 'Probabilistic Analyses of Combinatorial Optimization Problems on Random Shortest Path Metrics'. Together they form a unique fingerprint.
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Given F(x) = {(4 - X)?, What Is The Value Of F(16)?Record Your Answer And Fill In The Bubbles On Your
Step-by-step explanation:
Put 16 where x is and do the arithmetic.
f(16) = 1/3(4 -16)^2 = 1/3(-12)^2 = 1/3(144)
f(16) = 48
The required simplest form of 8mm; 24cm is 1mm; 3cm.
We have to determine, the simplest form of 8 mm: 24 cm.
To obtain the simplest form of the equation calculation must be done in a single unit following all the steps given below.
The ratio is just a way to compare two quantities while the proportion is an equation that shows that two ratios are equivalent.
Here, 8 mm: 24 cm.
[tex]8mm \ ; 24cm[/tex]
Divided by 8 both sides,
[tex]= \dfrac{8mm}{24cm}\\\\= \dfrac{1mm}{3cm}[/tex]
Hence, The required simplest form of 8mm; 24cm is 1mm; 3cm.
To know more about the Time period click the link given below.
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Surface Area
Calculate the surface areas of the given basic solid shapes using standard formulae.
This is level 4; Find the surface area of a variety of cylinders. The diagrams are not to scale.
Find the surface area of this solid cylinder if the radius of the circular top is 43cm and its height is 39cm. Give your answer to the nearest square centimetre.
☐ cm^2 ☐ ✓ ✗
Find the surface area of a solid cylinder if the diameter of the circular end is 70cm and its length is 42cm. Give your answer to the nearest square centimetre.
☐ cm^2 ☐ ✓ ✗
Find the surface area of this cylinder.
Give your answer to three significant figures.
☐ cm^2 ☐ ✓ ✗
A cylindrical tube is used to store circular salted potatoe crisps. All of its external surface area except one of the circular ends is painted red. Calculate the area of the painted region if the
length of the tube is 31cm and the diameter of the tube is 7cm. Give your answer to the nearest square centimetre.
☐ cm^2 ☐ ✓ ✗
A lumberjack needed to calculate the surface area of a log. He estimated that the log was a cylinder with a length of 12 metres and diameter 70 centimetres.
What did he calculate the surface area to be? Give your answer to the nearest square metre.
☐ cm^2 ☐ ✓ ✗
A cylindrical telegraph pole was ten times as tall as it was wide.
Calculate the curved surface area of the pole if it's radius was 30cm. Give your answer in square metres to nearest square metre.
☐ cm^2 ☐ ✓ ✗
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Description of Levels
Level 1 - Find the surface area of shapes made up of cubes.
Level 2 - Find the surface area of a variety of cuboids.
Level 3 - Find the surface area of a variety of prisms.
Level 4 - Find the surface area of a variety of cylinders.
Level 5 - Find the surface area of a variety of cones.
Level 6 - Find the surface area of a variety of pyramids.
Level 7 - Find the surface area of a variety of spheres.
Level 8 - Find the surface area of composite shapes.
Level 9 - Mixed, more challenging questions involving surface area.
Volume - Find the volume of basic solid shapes.
Surface Area = Volume - Can you find the ten cuboids that have numerically equal volumes and surface areas? A challenge in using technology.
Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers).
More on 3D Shapes including lesson Starters, visual aids, investigations and self-marking exercises.
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Surface Area Formulae
Cube: \(6s^2\) where \(s\) is the length of one edge.
Cuboid: \(2(lw + lh + wh)\) where \(l\) is the length, \(w\) is the width and \(h\) is the height of the cuboid.
Cylinder: \(2\pi rh + 2\pi r^2\) where \(h\) is the height (or length) of the cylinder and \(r\) is the radius of the circular end.
Cone: \(\pi r(r+l)\) where \(l\) is the distance from the apex to the rim of the circle (sloping height) of the cone and \(r\) is the radius of the circular base.
Cone: \(\pi r(r+\sqrt{h^2+r^2})\) where \(h\) is the height of the cone and \(r\) is the radius of the circular base.
Square based pyramid: \(s^2+2s\sqrt{\frac{s^2}{4}+h^2}\) where \(h\) is the height of the pyramid and \(s\) is the length of a side of the square base.
Rectangular based pyramid: \(lw+l\sqrt{\frac{w^2}{4}+h^2}+w\sqrt{\frac{l^2}{4}+h^2}\) where \(h\) is the height of the pyramid, \(l\) is the length of the base and \(w\) is the width of the base.
Sphere: \(4\pi r^2\) where \(r\) is the radius of the sphere.
Prism: Double the area of the cross section added to the product of the length and the perimeter of the cross section.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a
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Search results for: multiple linear regression
Commenced in January 2007
Search results for: multiple linear regression
3583 Research on the Problems of Housing Prices in Qingdao from a Macro Perspective
Authors: Liu Zhiyuan, Sun Zongdi, Liu Zhiyuan, Sun Zongdi
Qingdao is a seaside city. Taking into account the characteristics of Qingdao, this article established a multiple linear regression model to analyze the impact of macroeconomic factors on housing
prices. We used stepwise regression method to make multiple linear regression analysis, and made statistical analysis of F test values and T test values. According to the analysis results, the model
is continuously optimized. Finally, this article obtained the multiple linear regression equation and the influencing factors, and the reliability of the model was verified by F test and T test.
Keywords: Housing prices, multiple linear regression model, macroeconomic factors, Qingdao City.
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3582 A Comparison of the Sum of Squares in Linear and Partial Linear Regression Models
Authors: Dursun Aydın
In this paper, estimation of the linear regression model is made by ordinary least squares method and the partially linear regression model is estimated by penalized least squares method using
smoothing spline. Then, it is investigated that differences and similarity in the sum of squares related for linear regression and partial linear regression models (semi-parametric regression
models). It is denoted that the sum of squares in linear regression is reduced to sum of squares in partial linear regression models. Furthermore, we indicated that various sums of squares in the
linear regression are similar to different deviance statements in partial linear regression. In addition to, coefficient of the determination derived in linear regression model is easily generalized
to coefficient of the determination of the partial linear regression model. For this aim, it is made two different applications. A simulated and a real data set are considered to prove the claim
mentioned here. In this way, this study is supported with a simulation and a real data example.
Keywords: Partial Linear Regression Model, Linear RegressionModel, Residuals, Deviance, Smoothing Spline.
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3581 Internet Purchases in European Union Countries: Multiple Linear Regression Approach
Authors: Ksenija Dumičić, Anita Čeh Časni, Irena Palić
This paper examines economic and Information and Communication Technology (ICT) development influence on recently increasing Internet purchases by individuals for European Union member states. After
a growing trend for Internet purchases in EU27 was noticed, all possible regression analysis was applied using nine independent variables in 2011. Finally, two linear regression models were studied
in detail. Conducted simple linear regression analysis confirmed the research hypothesis that the Internet purchases in analyzed EU countries is positively correlated with statistically significant
variable Gross Domestic Product per capita (GDPpc). Also, analyzed multiple linear regression model with four regressors, showing ICT development level, indicates that ICT development is crucial for
explaining the Internet purchases by individuals, confirming the research hypothesis.
Keywords: European Union, Internet purchases, multiple linear regression model, outlier
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3580 Multi-Linear Regression Based Prediction of Mass Transfer by Multiple Plunging Jets
The paper aims to compare the performance of vertical and inclined multiple plunging jets and to model and predict their mass transfer capacity by multi-linear regression based approach. The multiple
vertical plunging jets have jet impact angle of θ = 90^O; whereas, multiple inclined plunging jets have jet impact angle of θ = 60^O. The results of the study suggests that mass transfer is higher
for multiple jets, and inclined multiple plunging jets have up to 1.6 times higher mass transfer than vertical multiple plunging jets under similar conditions. The derived relationship, based on
multi-linear regression approach, has successfully predicted the volumetric mass transfer coefficient (K[L]a) from operational parameters of multiple plunging jets with a correlation coefficient of
0.973, root mean square error of 0.002 and coefficient of determination of 0.946. The results suggests that predicted overall mass transfer coefficient is in good agreement with actual experimental
values; thereby, suggesting the utility of derived relationship based on multi-linear regression based approach and can be successfully employed in modeling mass transfer by multiple plunging jets.
Keywords: Mass transfer, multiple plunging jets, multi-linear regression.
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3579 Relationship between Sums of Squares in Linear Regression and Semi-parametric Regression
Authors: Dursun Aydın, Bilgin Senel
In this paper, the sum of squares in linear regression is reduced to sum of squares in semi-parametric regression. We indicated that different sums of squares in the linear regression are similar to
various deviance statements in semi-parametric regression. In addition to, coefficient of the determination derived in linear regression model is easily generalized to coefficient of the
determination of the semi-parametric regression model. Then, it is made an application in order to support the theory of the linear regression and semi-parametric regression. In this way, study is
supported with a simulated data example.
Keywords: Semi-parametric regression, Penalized LeastSquares, Residuals, Deviance, Smoothing Spline.
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3578 The Relative Efficiency of Parameter Estimation in Linear Weighted Regression
Authors: Baoguang Tian, Nan Chen
A new relative efficiency in linear model in reference is instructed into the linear weighted regression, and its upper and lower bound are proposed. In the linear weighted regression model, for the
best linear unbiased estimation of mean matrix respect to the least-squares estimation, two new relative efficiencies are given, and their upper and lower bounds are also studied.
Keywords: Linear weighted regression, Relative efficiency, Mean matrix, Trace.
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3577 Clustering Protein Sequences with Tailored General Regression Model Technique
Authors: G. Lavanya Devi, Allam Appa Rao, A. Damodaram, GR Sridhar, G. Jaya Suma
Cluster analysis divides data into groups that are meaningful, useful, or both. Analysis of biological data is creating a new generation of epidemiologic, prognostic, diagnostic and treatment
modalities. Clustering of protein sequences is one of the current research topics in the field of computer science. Linear relation is valuable in rule discovery for a given data, such as if value X
goes up 1, value Y will go down 3", etc. The classical linear regression models the linear relation of two sequences perfectly. However, if we need to cluster a large repository of protein sequences
into groups where sequences have strong linear relationship with each other, it is prohibitively expensive to compare sequences one by one. In this paper, we propose a new technique named General
Regression Model Technique Clustering Algorithm (GRMTCA) to benignly handle the problem of linear sequences clustering. GRMT gives a measure, GR*, to tell the degree of linearity of multiple
sequences without having to compare each pair of them.
Keywords: Clustering, General Regression Model, Protein Sequences, Similarity Measure.
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3576 Comparison of Polynomial and Radial Basis Kernel Functions based SVR and MLR in Modeling Mass Transfer by Vertical and Inclined Multiple Plunging Jets
Presently various computational techniques are used in modeling and analyzing environmental engineering data. In the present study, an intra-comparison of polynomial and radial basis kernel functions
based on Support Vector Regression and, in turn, an inter-comparison with Multi Linear Regression has been attempted in modeling mass transfer capacity of vertical (θ = 90O) and inclined (θ multiple
plunging jets (varying from 1 to 16 numbers). The data set used in this study consists of four input parameters with a total of eighty eight cases, forty four each for vertical and inclined multiple
plunging jets. For testing, tenfold cross validation was used. Correlation coefficient values of 0.971 and 0.981 along with corresponding root mean square error values of 0.0025 and 0.0020 were
achieved by using polynomial and radial basis kernel functions based Support Vector Regression respectively. An intra-comparison suggests improved performance by radial basis function in comparison
to polynomial kernel based Support Vector Regression. Further, an inter-comparison with Multi Linear Regression (correlation coefficient = 0.973 and root mean square error = 0.0024) reveals that
radial basis kernel functions based Support Vector Regression performs better in modeling and estimating mass transfer by multiple plunging jets.
Keywords: Mass transfer, multiple plunging jets, polynomial and radial basis kernel functions, Support Vector Regression.
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3575 On the outlier Detection in Nonlinear Regression
Authors: Hossein Riazoshams, Midi Habshah, Jr., Mohamad Bakri Adam
The detection of outliers is very essential because of their responsibility for producing huge interpretative problem in linear as well as in nonlinear regression analysis. Much work has been
accomplished on the identification of outlier in linear regression, but not in nonlinear regression. In this article we propose several outlier detection techniques for nonlinear regression. The main
idea is to use the linear approximation of a nonlinear model and consider the gradient as the design matrix. Subsequently, the detection techniques are formulated. Six detection measures are
developed that combined with three estimation techniques such as the Least-Squares, M and MM-estimators. The study shows that among the six measures, only the studentized residual and Cook Distance
which combined with the MM estimator, consistently capable of identifying the correct outliers.
Keywords: Nonlinear Regression, outliers, Gradient, LeastSquare, M-estimate, MM-estimate.
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3574 Economic Dispatch Fuzzy Linear Regression and Optimization
Authors: A. K. Al-Othman
This study presents a new approach based on Tanaka's fuzzy linear regression (FLP) algorithm to solve well-known power system economic load dispatch problem (ELD). Tanaka's fuzzy linear regression
(FLP) formulation will be employed to compute the optimal solution of optimization problem after linearization. The unknowns are expressed as fuzzy numbers with a triangular membership function that
has middle and spread value reflected on the unknowns. The proposed fuzzy model is formulated as a linear optimization problem, where the objective is to minimize the sum of the spread of the
unknowns, subject to double inequality constraints. Linear programming technique is employed to obtain the middle and the symmetric spread for every unknown (power generation level). Simulation
results of the proposed approach will be compared with those reported in literature.
Keywords: Economic Dispatch, Fuzzy Linear Regression (FLP)and Optimization.
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3573 Climate Change in Albania and Its Effect on Cereal Yield
This study is focused on analyzing climate change in Albania and its potential effects on cereal yields. Initially, monthly temperature and rainfalls in Albania were studied for the period 1960-2021.
Climacteric variables are important variables when trying to model cereal yield behavior, especially when significant changes in weather conditions are observed. For this purpose, in the second part
of the study, linear and nonlinear models explaining cereal yield are constructed for the same period, 1960-2021. The multiple linear regression analysis and lasso regression method are applied to
the data between cereal yield and each independent variable: average temperature, average rainfall, fertilizer consumption, arable land, land under cereal production, and nitrous oxide emissions. In
our regression model, heteroscedasticity is not observed, data follow a normal distribution, and there is a low correlation between factors, so we do not have the problem of multicollinearity.
Machine learning methods, such as Random Forest (RF), are used to predict cereal yield responses to climacteric and other variables. RF showed high accuracy compared to the other statistical models
in the prediction of cereal yield. We found that changes in average temperature negatively affect cereal yield. The coefficients of fertilizer consumption, arable land, and land under cereal
production are positively affecting production. Our results show that the RF method is an effective and versatile machine-learning method for cereal yield prediction compared to the other two
methods: multiple linear regression and lasso regression method.
Keywords: Cereal yield, climate change, machine learning, multiple regression model, random forest.
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3572 A Multiple Linear Regression Model to Predict the Price of Cement in Nigeria
Authors: Kenneth M. Oba
This study investigated factors affecting the price of cement in Nigeria, and developed a mathematical model that can predict future cement prices. Cement is key in the Nigerian construction
industry. The changes in price caused by certain factors could affect economic and infrastructural development; hence there is need for proper proactive planning. Secondary data were collected from
published information on cement between 2014 and 2019. In addition, questionnaires were sent to some domestic cement retailers in Port Harcourt in Nigeria, to obtain the actual prices of cement
between the same periods. The study revealed that the most critical factors affecting the price of cement in Nigeria are inflation rate, population growth rate, and Gross Domestic Product (GDP)
growth rate. With the use of data from United Nations, International Monetary Fund, and Central Bank of Nigeria databases, amongst others, a Multiple Linear Regression model was formulated. The model
was used to predict the price of cement for 2020-2025. The model was then tested with 95% confidence level, using a two-tailed t-test and an F-test, resulting in an R^2 of 0.8428 and R^2 (adj.) of
0.6069. The results of the tests and the correlation factors confirm the model to be fit and adequate. This study will equip researchers and stakeholders in the construction industry with information
for planning, monitoring, and management of present and future construction projects that involve the use of cement.
Keywords: Cement price, multiple linear regression model, Nigerian Construction Industry, price prediction.
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3571 Studding of Number of Dataset on Precision of Estimated Saturated Hydraulic Conductivity
Authors: M. Siosemarde, M. Byzedi
Saturated hydraulic conductivity of Soil is an important property in processes involving water and solute flow in soils. Saturated hydraulic conductivity of soil is difficult to measure and can be
highly variable, requiring a large number of replicate samples. In this study, 60 sets of soil samples were collected at Saqhez region of Kurdistan province-IRAN. The statistics such as Correlation
Coefficient (R), Root Mean Square Error (RMSE), Mean Bias Error (MBE) and Mean Absolute Error (MAE) were used to evaluation the multiple linear regression models varied with number of dataset. In
this study the multiple linear regression models were evaluated when only percentage of sand, silt, and clay content (SSC) were used as inputs, and when SSC and bulk density, Bd, (SSC+Bd) were used
as inputs. The R, RMSE, MBE and MAE values of the 50 dataset for method (SSC), were calculated 0.925, 15.29, -1.03 and 12.51 and for method (SSC+Bd), were calculated 0.927, 15.28,-1.11 and 12.92,
respectively, for relationship obtained from multiple linear regressions on data. Also the R, RMSE, MBE and MAE values of the 10 dataset for method (SSC), were calculated 0.725, 19.62, - 9.87 and
18.91 and for method (SSC+Bd), were calculated 0.618, 24.69, -17.37 and 22.16, respectively, which shows when number of dataset increase, precision of estimated saturated hydraulic conductivity,
Keywords: dataset, precision, saturated hydraulic conductivity, soil and statistics.
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3570 A Fuzzy Linear Regression Model Based on Dissemblance Index
Authors: Shih-Pin Chen, Shih-Syuan You
Fuzzy regression models are useful for investigating the relationship between explanatory variables and responses in fuzzy environments. To overcome the deficiencies of previous models and increase
the explanatory power of fuzzy data, the graded mean integration (GMI) representation is applied to determine representative crisp regression coefficients. A fuzzy regression model is constructed
based on the modified dissemblance index (MDI), which can precisely measure the actual total error. Compared with previous studies based on the proposed MDI and distance criterion, the results from
commonly used test examples show that the proposed fuzzy linear regression model has higher explanatory power and forecasting accuracy.
Keywords: Dissemblance index, fuzzy linear regression, graded mean integration, mathematical programming.
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3569 Two New Relative Efficiencies of Linear Weighted Regression
Authors: Shuimiao Wan, Chao Yuan, Baoguang Tian
In statistics parameter theory, usually the parameter estimations have two kinds, one is the least-square estimation (LSE), and the other is the best linear unbiased estimation (BLUE). Due to the
determining theorem of minimum variance unbiased estimator (MVUE), the parameter estimation of BLUE in linear model is most ideal. But since the calculations are complicated or the covariance is not
given, people are hardly to get the solution. Therefore, people prefer to use LSE rather than BLUE. And this substitution will take some losses. To quantize the losses, many scholars have presented
many kinds of different relative efficiencies in different views. For the linear weighted regression model, this paper discusses the relative efficiencies of LSE of β to BLUE of β. It also defines
two new relative efficiencies and gives their lower bounds.
Keywords: Linear weighted regression, Relative efficiency, Lower bound, Parameter estimation.
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3568 Fuzzy Logic Approach to Robust Regression Models of Uncertain Medical Categories
Authors: Arkady Bolotin
Dichotomization of the outcome by a single cut-off point is an important part of various medical studies. Usually the relationship between the resulted dichotomized dependent variable and explanatory
variables is analyzed with linear regression, probit regression or logistic regression. However, in many real-life situations, a certain cut-off point dividing the outcome into two groups is unknown
and can be specified only approximately, i.e. surrounded by some (small) uncertainty. It means that in order to have any practical meaning the regression model must be robust to this uncertainty. In
this paper, we show that neither the beta in the linear regression model, nor its significance level is robust to the small variations in the dichotomization cut-off point. As an alternative robust
approach to the problem of uncertain medical categories, we propose to use the linear regression model with the fuzzy membership function as a dependent variable. This fuzzy membership function
denotes to what degree the value of the underlying (continuous) outcome falls below or above the dichotomization cut-off point. In the paper, we demonstrate that the linear regression model of the
fuzzy dependent variable can be insensitive against the uncertainty in the cut-off point location. In the paper we present the modeling results from the real study of low hemoglobin levels in
infants. We systematically test the robustness of the binomial regression model and the linear regression model with the fuzzy dependent variable by changing the boundary for the category Anemia and
show that the behavior of the latter model persists over a quite wide interval.
Keywords: Categorization, Uncertain medical categories, Binomial regression model, Fuzzy dependent variable, Robustness.
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3567 Orthogonal Regression for Nonparametric Estimation of Errors-in-Variables Models
Authors: Anastasiia Yu. Timofeeva
Two new algorithms for nonparametric estimation of errors-in-variables models are proposed. The first algorithm is based on penalized regression spline. The spline is represented as a
piecewise-linear function and for each linear portion orthogonal regression is estimated. This algorithm is iterative. The second algorithm involves locally weighted regression estimation. When the
independent variable is measured with error such estimation is a complex nonlinear optimization problem. The simulation results have shown the advantage of the second algorithm under the assumption
that true smoothing parameters values are known. Nevertheless the use of some indexes of fit to smoothing parameters selection gives the similar results and has an oversmoothing effect.
Keywords: Grade point average, orthogonal regression, penalized regression spline, locally weighted regression.
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3566 Density Estimation using Generalized Linear Model and a Linear Combination of Gaussians
Authors: Aly Farag, Ayman El-Baz, Refaat Mohamed
In this paper we present a novel approach for density estimation. The proposed approach is based on using the logistic regression model to get initial density estimation for the given empirical
density. The empirical data does not exactly follow the logistic regression model, so, there will be a deviation between the empirical density and the density estimated using logistic regression
model. This deviation may be positive and/or negative. In this paper we use a linear combination of Gaussian (LCG) with positive and negative components as a model for this deviation. Also, we will
use the expectation maximization (EM) algorithm to estimate the parameters of LCG. Experiments on real images demonstrate the accuracy of our approach.
Keywords: Logistic regression model, Expectationmaximization, Segmentation.
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3565 Analyzing of Public Transport Trip Generation in Developing Countries; A Case Study in Yogyakarta, Indonesia
Authors: S. Priyanto, E.P Friandi
Yogyakarta, as the capital city of Yogyakarta Province, has important roles in various sectors that require good provision of public transportation system. Ideally, a good transportation system
should be able to accommodate the amount of travel demand. This research attempts to develop a trip generation model to predict the number of public transport passenger in Yogyakarta city. The model
is built by using multiple linear regression analysis, which establishes relationship between trip number and socioeconomic attributes. The data consist of primary and secondary data. Primary data
was collected by conducting household surveys which randomly selected. The resulted model is further applied to evaluate the existing TransJogja, a new Bus Rapid Transit system serves Yogyakarta and
surrounding cities, shelters.
Keywords: Multiple linear regression, shelter evaluation, travel demand, trip generation.
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3564 Delay-independent Stabilization of Linear Systems with Multiple Time-delays
Authors: Ping He, Heng-You Lan, Gong-Quan Tan
The multidelays linear control systems described by difference differential equations are often studied in modern control theory. In this paper, the delay-independent stabilization algebraic criteria
and the theorem of delay-independent stabilization for linear systems with multiple time-delays are established by using the Lyapunov functional and the Riccati algebra matrix equation in the matrix
theory. An illustrative example and the simulation result, show that the approach to linear systems with multiple time-delays is effective.
Keywords: Linear system, Delay-independent stabilization, Lyapunovfunctional, Riccati algebra matrix equation.
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3563 A Cost Optimization Model for the Construction of Bored Piles
Authors: Kenneth M. Oba
Adequate management, control, and optimization of cost is an essential element for a successful construction project. A multiple linear regression optimization model was formulated to address the
problem of costs associated with pile construction operations. A total of 32 PVC-reinforced concrete piles with diameter of 300 mm, 5.4 m long, were studied during the construction. The soil upon
which the piles were installed was mostly silty sand, and completely submerged in water at Bonny, Nigeria. The piles are friction piles installed by boring method, using a piling auger. The volumes
of soil removed, the weight of reinforcement cage installed, and volumes of fresh concrete poured into the PVC void were determined. The cost of constructing each pile based on the calculated
quantities was determined. A model was derived and subjected to statistical tests using Statistical Package for the Social Sciences (SPSS) software. The model turned out to be adequate, fit, and have
a high predictive accuracy with an R2 value of 0.833.
Keywords: Cost optimization modelling, multiple linear models, pile construction, regression models.
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3562 Computational Aspects of Regression Analysis of Interval Data
Authors: Michal Cerny
We consider linear regression models where both input data (the values of independent variables) and output data (the observations of the dependent variable) are interval-censored. We introduce a
possibilistic generalization of the least squares estimator, so called OLS-set for the interval model. This set captures the impact of the loss of information on the OLS estimator caused by interval
censoring and provides a tool for quantification of this effect. We study complexity-theoretic properties of the OLS-set. We also deal with restricted versions of the general interval linear
regression model, in particular the crisp input – interval output model. We give an argument that natural descriptions of the OLS-set in the crisp input – interval output cannot be computed in
polynomial time. Then we derive easily computable approximations for the OLS-set which can be used instead of the exact description. We illustrate the approach by an example.
Keywords: Linear regression, interval-censored data, computational complexity.
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3561 Empirical Statistical Modeling of Rainfall Prediction over Myanmar
Authors: Wint Thida Zaw, Thinn Thu Naing
One of the essential sectors of Myanmar economy is agriculture which is sensitive to climate variation. The most important climatic element which impacts on agriculture sector is rainfall. Thus
rainfall prediction becomes an important issue in agriculture country. Multi variables polynomial regression (MPR) provides an effective way to describe complex nonlinear input output relationships
so that an outcome variable can be predicted from the other or others. In this paper, the modeling of monthly rainfall prediction over Myanmar is described in detail by applying the polynomial
regression equation. The proposed model results are compared to the results produced by multiple linear regression model (MLR). Experiments indicate that the prediction model based on MPR has higher
accuracy than using MLR.
Keywords: Polynomial Regression, Rainfall Forecasting, Statistical forecasting.
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3560 Quantitative Structure Activity Relationship and Insilco Docking of Substituted 1,3,4-Oxadiazole Derivatives as Potential Glucosamine-6-Phosphate Synthase Inhibitors
Authors: Suman Bala, Sunil Kamboj, Vipin Saini
Quantitative Structure Activity Relationship (QSAR) analysis has been developed to relate antifungal activity of novel substituted 1,3,4-oxadiazole against
Candida albicans
Aspergillus niger
using computer assisted multiple regression analysis. The study has shown the better relationship between antifungal activities with respect to various descriptors established by multiple regression
analysis. The analysis has shown statistically significant correlation with R
values 0.932 and 0.782 against
Candida albicans
Aspergillus niger
respectively. These derivatives were further subjected to molecular docking studies to investigate the interactions between the target compounds and amino acid residues present in the active site of
glucosamine-6-phosphate synthase. All the synthesized compounds have better docking score as compared to standard fluconazole. Our results could be used for the further design as well as development
of optimal and potential antifungal agents.
Keywords: 1, 3, 4-Oxadiazole, QSAR, Multiple linear regression, Docking, Glucosamine-6-Phosphate Synthase.
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3559 Artificial Neural Network based Modeling of Evaporation Losses in Reservoirs
Authors: Surinder Deswal, Mahesh Pal
An Artificial Neural Network based modeling technique has been used to study the influence of different combinations of meteorological parameters on evaporation from a reservoir. The data set used is
taken from an earlier reported study. Several input combination were tried so as to find out the importance of different input parameters in predicting the evaporation. The prediction accuracy of
Artificial Neural Network has also been compared with the accuracy of linear regression for predicting evaporation. The comparison demonstrated superior performance of Artificial Neural Network over
linear regression approach. The findings of the study also revealed the requirement of all input parameters considered together, instead of individual parameters taken one at a time as reported in
earlier studies, in predicting the evaporation. The highest correlation coefficient (0.960) along with lowest root mean square error (0.865) was obtained with the input combination of air
temperature, wind speed, sunshine hours and mean relative humidity. A graph between the actual and predicted values of evaporation suggests that most of the values lie within a scatter of ±15% with
all input parameters. The findings of this study suggest the usefulness of ANN technique in predicting the evaporation losses from reservoirs.
Keywords: Artificial neural network, evaporation losses, multiple linear regression, modeling.
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3558 A Hybrid Model of ARIMA and Multiple Polynomial Regression for Uncertainties Modeling of a Serial Production Line
Authors: Amir Azizi, Amir Yazid b. Ali, Loh Wei Ping, Mohsen Mohammadzadeh
Uncertainties of a serial production line affect on the production throughput. The uncertainties cannot be prevented in a real production line. However the uncertain conditions can be controlled by a
robust prediction model. Thus, a hybrid model including autoregressive integrated moving average (ARIMA) and multiple polynomial regression, is proposed to model the nonlinear relationship of
production uncertainties with throughput. The uncertainties under consideration of this study are demand, breaktime, scrap, and lead-time. The nonlinear relationship of production uncertainties with
throughput are examined in the form of quadratic and cubic regression models, where the adjusted R-squared for quadratic and cubic regressions was 98.3% and 98.2%. We optimized the multiple quadratic
regression (MQR) by considering the time series trend of the uncertainties using ARIMA model. Finally the hybrid model of ARIMA and MQR is formulated by better adjusted R-squared, which is 98.9%.
Keywords: ARIMA, multiple polynomial regression, production throughput, uncertainties
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3557 Optimization of Slider Crank Mechanism Using Design of Experiments and Multi-Linear Regression
Authors: Galal Elkobrosy, Amr M. Abdelrazek, Bassuny M. Elsouhily, Mohamed E. Khidr
Crank shaft length, connecting rod length, crank angle, engine rpm, cylinder bore, mass of piston and compression ratio are the inputs that can control the performance of the slider crank mechanism
and then its efficiency. Several combinations of these seven inputs are used and compared. The throughput engine torque predicted by the simulation is analyzed through two different regression
models, with and without interaction terms, developed according to multi-linear regression using LU decomposition to solve system of algebraic equations. These models are validated. A regression
model in seven inputs including their interaction terms lowered the polynomial degree from 3^rd degree to 1^st degree and suggested valid predictions and stable explanations.
Keywords: Design of experiments, regression analysis, SI Engine, statistical modeling.
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3556 Non-Methane Hydrocarbons Emission during the Photocopying Process
Authors: Kiurski S. Jelena, Aksentijević M. Snežana, Kecić S. Vesna, Oros B. Ivana
Prosperity of electronic equipment in photocopying environment not only has improved work efficiency, but also has changed indoor air quality. Considering the number of photocopying employed, indoor
air quality might be worse than in general office environments. Determining the contribution from any type of equipment to indoor air pollution is a complex matter. Non-methane hydrocarbons are known
to have an important role on air quality due to their high reactivity. The presence of hazardous pollutants in indoor air has been detected in one photocopying shop in Novi Sad, Serbia. Air samples
were collected and analyzed for five days, during 8-hr working time in three time intervals, whereas three different sampling points were determined. Using multiple linear regression model and
software package STATISTICA 10 the concentrations of occupational hazards and microclimates parameters were mutually correlated. Based on the obtained multiple coefficients of determination (0.3751,
0.2389 and 0.1975), a weak positive correlation between the observed variables was determined. Small values of parameter F indicated that there was no statistically significant difference between the
concentration levels of nonmethane hydrocarbons and microclimates parameters. The results showed that variable could be presented by the general regression model: y = b0 + b1xi1+ b2xi2. Obtained
regression equations allow to measure the quantitative agreement between the variables and thus obtain more accurate knowledge of their mutual relations.
Keywords: Indoor air quality, multiple regression analysis, nonmethane hydrocarbons, photocopying process.
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3555 Aircraft Gas Turbine Engines Technical Condition Identification System
Authors: A. M. Pashayev, C. Ardil, D. D. Askerov, R. A. Sadiqov, P. S. Abdullayev
In this paper is shown that the probability-statistic methods application, especially at the early stage of the aviation gas turbine engine (GTE) technical condition diagnosing, when the flight
information has property of the fuzzy, limitation and uncertainty is unfounded. Hence is considered the efficiency of application of new technology Soft Computing at these diagnosing stages with the
using of the Fuzzy Logic and Neural Networks methods. Training with high accuracy of fuzzy multiple linear and non-linear models (fuzzy regression equations) which received on the statistical fuzzy
data basis is made. Thus for GTE technical condition more adequate model making are analysed dynamics of skewness and kurtosis coefficients' changes. Researches of skewness and kurtosis coefficients
values- changes show that, distributions of GTE work parameters have fuzzy character. Hence consideration of fuzzy skewness and kurtosis coefficients is expedient. Investigation of the basic
characteristics changes- dynamics of GTE work parameters allows to draw conclusion on necessity of the Fuzzy Statistical Analysis at preliminary identification of the engines' technical condition.
Researches of correlation coefficients values- changes shows also on their fuzzy character. Therefore for models choice the application of the Fuzzy Correlation Analysis results is offered. For
checking of models adequacy is considered the Fuzzy Multiple Correlation Coefficient of Fuzzy Multiple Regression. At the information sufficiency is offered to use recurrent algorithm of aviation GTE
technical condition identification (Hard Computing technology is used) on measurements of input and output parameters of the multiple linear and non-linear generalised models at presence of noise
measured (the new recursive Least Squares Method (LSM)). The developed GTE condition monitoring system provides stage-bystage estimation of engine technical conditions. As application of the given
technique the estimation of the new operating aviation engine temperature condition was made.
Keywords: Gas turbine engines, neural networks, fuzzy logic, fuzzy statistics.
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3554 Harmonics Elimination in Multilevel Inverter Using Linear Fuzzy Regression
Authors: A. K. Al-Othman, H. A. Al-Mekhaizim
Multilevel inverters supplied from equal and constant dc sources almost don-t exist in practical applications. The variation of the dc sources affects the values of the switching angles required for
each specific harmonic profile, as well as increases the difficulty of the harmonic elimination-s equations. This paper presents an extremely fast optimal solution of harmonic elimination of
multilevel inverters with non-equal dc sources using Tanaka's fuzzy linear regression formulation. A set of mathematical equations describing the general output waveform of the multilevel inverter
with nonequal dc sources is formulated. Fuzzy linear regression is then employed to compute the optimal solution set of switching angles.
Keywords: Multilevel converters, harmonics, pulse widthmodulation (PWM), optimal control.
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Molly took part in a math quiz and scored 55 points on 30 questions. Of the questions, there were 3-point word problems, 2-point calculation - DocumenTVMolly took part in a math quiz and scored 55 points on 30 questions. Of the questions, there were 3-point word problems, 2-point calculation
Molly took part in a math quiz and scored 55 points on 30 questions. Of the questions, there were 3-point word problems, 2-point calculation
Molly took part in a math quiz and scored 55 points on 30 questions. Of the questions, there were 3-point word problems, 2-point calculation questions, and 1-point multiple-choice questions. She
answered 5 more 2-point questions than 1-point questions correctly. Find how many of each type of question she answered correctly.
in progress 0
Mathematics 3 years 2021-08-27T08:12:46+00:00 2021-08-27T08:12:46+00:00 1 Answers 5 views 0
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Rahul P.
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About Rahul P.
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The Quote of the Week: Lyons on Single Event Probabilities
"Given that a repeated series of trials is required, frequentists are unable to assign probabilities to single events. Thus, with regard to whether it was raining in Manchester yesterday, there
is no way of creating a large number of `yesterdays' in order to determine the probability. Frequentists would say that, even though they might not know, in actual fact it either was raining or
it wasn't, and so this is not a matter for assigning a probability. And the same remains true even if we replace `Manchester' by `the Sahara Desert'.
Another example would be the unwillingness of a frequentist to assign a probability to the statement that `the first astronaut to set foot on Mars will return to Earth alive.' This does not mean
it is an uninteresting question, especially if you have been chosen to be on the first manned-mission to Mars, but then, don't ask a frequentist to assess the probability."
Louis Lyons, "
Bayes and Frequentism: a Particle Physicist's Perspective
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Fundamentals of
This book was written for advanced undergraduates, graduate students, and mature
scientists in mathematics,computer science,engineering,and all disciplines in which
numerical methods are used . At the heart of most scientific computer codes lie matrix
computations,soitis important to understand how to perform such computation seffi-
ciently and accurately. This book meets that need by providing a detailed introduction
to the fundamental ideas of numerical linear algebra .
The prerequisites are a first course in linear algebra and some experience with
computer programming. For the understanding of some of the examples, especially
in the second half of the book,the student will find it helpful to have had a first course
in differential equations .
There are several other excellent books on this subject,including those by Demmel
[15],GolubandVanLoan[33],andTrefethenandBau[71]. Students who are new to
this material often find those books quite difficult to read. The purpose of this book
is to provide a gentler, more gradual introduction to the subject that is nevertheless
mathematically solid . The strong positive student response to the first edition has
assured me that my first attempt was successful and encouraged me to produce this
updated and extended edition.
The first edition was aimed mainly at the undergraduate level. As it turned out,
the book also found a great deal of use as a graduate text. I have therefore added
new material to make the book more attractive at the graduate level. These additions
are detailed below. However, the text remains suitable for undergraduate use, as
the elementary material has been kept largely intact, and more elementary exercises
have been added. The instructor can control the level of difficulty by deciding which
sections to cover and how far to push into each section. Numerous advanced topics
are developed in exercises at the ends of the sections.
The book contains many exercises, ranging from easy to moderately difficult.
Some are interspersed with the textual material and others are collected at the end
of each section. Those that are interspersed with the text are meant to be worked
immediately by the reader. This is my way of getting students actively involved in
the learning process. In order to get something out, you have to put something in.
Many of the exercises at the ends of sections are lengthy and may appear intimidating
at first. However,the persistent student will find that s/he can make it through them
with the help of the ample hints and advice that are given. I encourage every student
to work as many of the exercises as possible.
Nearly all numbered items in this book, including theorems, lemmas, numbered
equations,examples,and exercises,share a single numbering scheme. For example,
the first numbered item in Section 1.3 is Theorem 1.3.1. The next two numbered
items are displayed equations,which are numbered(1.3.2) and (1.3.3),respectively.
These are followed by the first exercise of the section,which bears the number 1.3.4.
Thus each item has a unique number: the only item in the book that has the number
1.3.4 is Exercise1.3.4. Although this schemeis unusual,I believe that most readers
will find it perfectly natural ,once they have gotten used to it. Its big advantage is that
it makes things easy to find: The reader who has located Exercises 1.4.15 and 1.4.25
but is looking for Example 1.4.20,knows for sure that this example lies some where
between the two exercises.
There are a couple of exceptions to the scheme. For technical reasons related
to the type setting, tables and figures (the so-called floating bodies) are numbered
separately by chapter. For example,the third figure of Chapter1 is Figure1.3.
New Features of the Second Edition
Use of MATLAB
By now MATLAB is firmly established as the most widely used vehicle for teaching
matrix computations. MATLAB is an easy to use, very high-level language that
allows the student to perform much more elaborate computational experiments than
before. MATLAB is also widely used in industry. I have therefore added many
examples and exercises that make use of MATLAB. This book is not, however, an
introduction to MATLAB,nor is it a MATLAB manual. For those purposes there are
other book savailable,for example,the MATLAB Guide by Higham and Higham[40].
However, MATLAB’s extensive help facilities are good enough that the reader may
feel noneed for a supplementary text. In an effort to make it easier for the student to
use MATLAB with this book,I have included an index of MATLAB terms ,separate
from the ordinary index.
I used to make my students write and debug their own Fortran programs. I have
left the Fortran exercises from the first edition largely intact. I hope a few students
will choose to work through some of these worth while projects.
More Applications
In order to help the student better understand the importance of the subject matter of
this book,I have included more examples and exercises on applications ( solved using
MATLAB ),mostly at the beginnings of chapters. I have chosen very simple applica-
tions: electrical circuits, mass-spring systems, simple partial differential equations.
In my opinion the simplest examples are the ones from which we can learn the most.
Earlier Introduction of the Singular Value Decomposition (SVD)
The SVD is one of the most important tools in numerical linear algebra. In the first
edition it was placed in the final chapter of the book, because it is impossible to
discuss methods for computing the SVD until after eigenvalue problems have been
discussed. I have since decided that the SVD needs to be introduced sooner, so
that the student can find out earlier about its properties and uses . With the help
of MATLAB, the student can experiment with the SVD without knowing anything
about how it is computed. ThereforeI have addeda brief chapteron the SVD in the
middle of the book.
New Material on Iterative Methods
The biggest addition to the book is a chapter on iterative methods for solving large,
sparse systems of linear equations . The main focus of the chapter is the powerful
conjugate-gradient method for solving symmetric, positive definite systems. How-
ever, the classical iterations are also discussed, and so are preconditioners. Krylov
subspace methods for solving indefinite and non symmetric problems are surveyed
There are also two new sections on methods for solving large, sparse eigenvalue
problems. The discussion includes the popular implicitly-restarted Arnoldi and
Jacobi-David son methods.
I hope that these additions in particular will make the book more attractive as a
graduate text.
Other New Features
To make the book more versatile,a number of other topics have been added,including:
•a backward error analysis of Gaussian elimination , including a discussion of
the modern component wise error analysis.
•a discussion of reorthogonalization,apractical means of obtaining numerically
or thonormalvectors.
•a discussion of how to update the decomposition when arow or columnis
added to or deleted from the data matrix, as happens in signal processing and
data analysis applications.
•a section introducing new methods for the symmetric eigen value problem that
have been developed since the first edition was published.
A few topics have been deleted on the grounds that they are either obsolete or too
specialized. I have also taken the opportunity to correct several vexing errors from
the first edition. I can only hope that I have not introduced too many new ones.
I am greatly indebted to the authors of some of the early works in this field. These
[24], G. W. Stewart [67], C. L. Lawson and R . J. Hanson[48], B. N. Parlett [54], A.
Guide [64], and the LINPACK Users’ Guide [18]. All of them influenced me
profoundly. Bytheway,everyone of these books is still worth reading today. Special
thanks goto CleveMoler for inventing MATLAB,which has changed everything .
Most of the first edition was written while I was on leave, at the University of
Bielefeld, Germany. I am pleased to thank once again my host and longtime friend,
Ludwig Elsner. During that stay I received financial support from the Fulbright
commission. A big chunk of the second edition was also written in Germany,at the
Technical University of Chemnitz. I thank my host (and another longtime friend),
VolkerMehrmann. On that visit I received financial support from Sonder for schungs-
bereich 393, TU Chemnitz. I am also indebted to my home institution, Washington
State University,forits support of my work on both editions.
I thank once again professors Dale Olesky, Kemble Yates, and Tjalling Ypma,
who class-tested a preliminary version of the first edition. Since publication of the
first edition, numerous people have sent me corrections, feedback, and comments.
These include A. Cline, L. Dieci, E. Jessup, D. Koya, D. D. Olesky, B. N. Parlett,
A. C. Raines, A. Witt, and K. Wright. Finally, I thank the many students who have
helped me learn how to teachthis material over the years.
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Biohealth science's connection to quantitative sciences | Department of Chemistry
Redefining quantitative and biohealth sciences
Faculty and researchers in the College of Science are interpreting and advancing biohealth sciences in innovative new ways by applying the natural sciences, such as mathematics, statistics and
chemistry. In recent times, researches in biology and medicine have been guided by biomolecular analysis technologies, mathematics and computations, and scientists are using these tools to address a
spectrum of biological questions about diseases, from how they spread to risk factors.
In the last few years, our College has experienced an impressive spurt of transdisciplinary research in the quantitative and biohealth sciences. Ongoing studies and research advances range from
analyzing genetic data on epidemics and inventing disease-detecting biosensors to developing statistical methods to better understand neuron connectivity and the transmission of signals in the brain.
Through collaborative research across our campus, our faculty are paving the way for innovative biohealth science research which broadens the training of students across scientific disciplines.
Biological systems and mathematical models
Connections between biology and the mathematical sciences are fueling innovation and expansion in those disciplines. Statistician Sharmodeep Bhattacharyya explains how interpreting data from various
experimental sources can generate new insights and solutions in the areas of neuroscience and genomics.
“Statistical methods, with their inherent objective of analyzing the uncertainty of a system help identify key interesting factors in the deluge of interesting data," said Bhattacharyya. "Such jobs
can range from identifying a key set of genes affecting a disease for a specific group of people (like in precision medicine) or identifying the interaction between key regions of the brain for
people who have a set of genes that causes a neurological disease."
Bhattacharyya has developed new statistical methods to analyze Electro-Cortico Graph (ECoG) array data from human and rat brains to identify connections involving speech and hearing.
Mathematician Vrushali Bokil’s research demonstrates how mathematical modeling, analysis and numerical simulations can illuminate insights in complex biological systems and how the health sciences,
in turn, can spark new mathematical ideas. She collaborates with a mix of biologists and mathematicians across the country as well as in the UK, France and Germany on a project funded by NIMBioS (the
National Institute of Mathematical and Biological Synthesis).
The project will allow Bokil and her colleagues to generate novel mathematical and statistical methods involving multiple hosts and multiple pathogens and that operate across a range of
spatiotemporal scales, and to analyze the effects of climate change and human activities on the emergence of new plant viruses. Bokil points to the increasing use of mathematics to model complicated
biological systems.
“It is exciting to be at the interface of biology and math,” said Bokil. “I write down a system of equations that models the physical or biological system. While the mathematical modeling and
numerical simulations are fascinating in and of themselves, the added value of feeding back into biological applications is very rewarding.”
Benjamin Dalziel, an assistant professor in Integrative Biology, is part of a growing breed of biologists who are turning the biological sciences into a more quantitative field. Dalziel is a
population biologist who uses mathematical tools to answer questions about the spread of infectious diseases, such as influenza and measles in populations and cities.
Dalziel, who also has an appointment in the mathematics department, maps hotspots of pathogen activity and diversification, and develops mathematical models to explain the patterns he finds. A
current project explores whether there are systematic differences among cities with respect to their epidemic risk.
“I find the connections between mathematical modeling and biology very interesting. After developing a model, we ask, 'Is this happening in nature and how do we test it?' And if nature is doing
something different, 'What did we get wrong with the model?' Sometimes there is a lot you have to do with the model besides [reviewing the] data to understand its behavior and to get it to interface
with the real world," said Dalziel, who is developing a new mathematics course specifically for the life sciences.
Innovative disease imaging
A major application of analytical chemistry and its quantitative aspects to biology involves the creation of tools that directly aid in the diagnosis of cancer, heart disease, strokes and other
serious ailments.
Chemistry assistant professor Sean M. Burrows runs a busy lab comprising undergraduates and doctoral students and their research is focused on innovating technologies to visualize biomarkers of
disease. They pioneer novel, colorful fluorescent biosensor designs—analytical devices that relate biological molecules to a fluorescent signal—for visualizing and quantifying microRNAs, which are
small non-coding RNA molecules that have a role in a plethora of gene regulatory events.
MicroRNAs hold great potential to yield information about the beginning stages of a disease and cell/tissue activity. Burrows and his team are trying to develop highly efficient fluorescent
technologies for basic research and clinical use.
“Basically the idea is to design an imaging technology that will give us more information on the molecular interactions within the cell,” explains Borrows. “[For example], can we create an instrument
that greatly advances the information content in terms of the numbers of colors we can look at in a cell? With the current technology, you could see one or two colors from the cell. But if we can
look at 10 or more different colors, that will tell us much more about a biological mechanism," adds Burrows.
In an exciting breakthrough, the Burrows group designed a more efficient fluorescent biosensor for better signal interpretation from microRNA biosensors. The innovation has attracted significant
attention in the field and was favorably reviewed in an article on the field of emerging microRNA biosensors in Analytical Chemistry.
However, existing imaging technology to learn about the underlying details of cellular mechanisms, such as the super resolution microscopy, is expensive. Burrows is keen to develop a cheaper
alternative that can be used in a regular microscope.
“We can then open the door for more researchers to get more information from the cells they are interested in studying. This, in turn, will enable more transformative breakthroughs to understand
disease progression and ultimately find cures.”
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Test Bank for A Concise Introduction to Logic 13th Edition
Product details:
• ISBN-10 : 1305958098
• ISBN-13 : 978-1305958098
• Author: Patrick Hurley
Over a million students have learned to be more discerning at constructing and evaluating arguments with the help of A CONCISE INTRODUCTION TO LOGIC, 13th Edition. The text’s clear, friendly,
thorough presentation has made it the most widely used logic text in North America. The book shows you how the content connects to real-life problems and gives you everything you need to do well in
your logic course. Doing well in logic improves your skills in ways that translate to other courses you take, your everyday life, and your future career. The accompanying technological resources
offered through MindTap, a highly robust online platform, include self-grading interactive exercises, a new digital activity that allows you to apply the skills you learn to a real-world problem, and
videos to reinforce what you learn in the book and hear in class.
Table contents:
Part I: INFORMAL LOGIC.
1. Basic Concepts.
Arguments, Premises, and Conclusions. Exercise. Recognizing Arguments. Exercise. Deduction and Induction. Exercise. Validity, Truth, Soundness, Strength, Cogency. Exercise. Argument Forms: Proving
Invalidity. Exercise. Extended Arguments. Exercise.
2. Language: Meaning and Definition.
Varieties of Meaning. Exercise. The Intension and Extension of Terms. Exercise. Definitions and Their Purposes. Exercise. Definitional Techniques. Exercise. Criteria for Lexical Definitions.
3. Informal Fallacies.
Fallacies in General. Exercise. Fallacies of Relevance. Exercise. Fallacies of Weak Induction. Exercise. Fallacies of Presumption, Ambiguity, and Illicit Transference. Exercise. Fallacies in Ordinary
Language. Exercise.
Part II: FORMAL LOGIC.
4. Categorical Propositions.
The Components of Categorical Propositions. Exercise. Quality, Quantity, and Distribution. Exercise. Venn Diagrams and the Modern Square of Opposition. Exercise. Conversion, Obversion, and
Contraposition. Exercise. The Traditional Square of Opposition. Exercise. Venn Diagrams and the Traditional Standpoint. Exercise. Translating Ordinary Language Statements into Categorical Form.
5. Categorical Syllogisms.
Standard Form, Mood, and Figure. Exercise. Venn Diagrams. Exercise. Rules and Fallacies. Exercise. Reducing the Number of Terms. Exercise. Ordinary Language Arguments. Exercise. Enthymemes. Exercise.
Sorites. Exercise.
6. Propositional Logic.
Symbols and Translation. Exercise. Truth Functions. Exercise. Truth Tables for Propositions. Exercise. Truth Tables for Arguments. Exercise. Indirect Truth Tables. Exercise. Argument Forms and
Fallacies. Exercise.
7. Natural Deduction in Propositional Logic.
Rules of Implication I. Exercise. Rules of Implication II. Exercise. Rules of Replacement I. Exercise. Rules of Replacement II. Exercise. Conditional Proof. Exercise. Indirect Proof. Exercise.
Proving Logical Truths. Exercise.
8. Predicate Logic
Symbols and Translation. Exercise. Using the Rules of Inference. Exercise. Quantifier Negation Rule. Exercise. Conditional and Indirect Proof. Exercise. Proving Invalidity. Exercise. Relational
Predicates and Overlapping Quantifiers. Exercise. Identity. Exercise.
Part III: INDUCTIVE LOGIC.
9. Analogy and Legal and Moral Reasoning.
Analogical Reasoning. Legal Reasoning. Moral Reasoning. Exercise.
10. Causality and Mill’s Methods.
“Cause” and Necessary and Sufficient Conditions. Mill’s Five Methods. Mill’s Methods and Science. Exercise.
11. Probability.
Theories of Probability. The Probability Calculus. Exercise.
12. Statistical Reasoning.
Evaluating Statistics. Samples. The Meaning of “Average.” Dispersion. Graphs and Pictograms. Percentages. Exercise.
13. Hypothetical/Scientific Reasoning.
The Hypothetical Method. Hypothetical Reasoning: Four Examples from Science. The Proof of Hypotheses. The Tentative Acceptance of Hypotheses. Exercise.
14. Science and Superstition.
Distinguishing Between Science and Superstition. Evidentiary Support. Objectivity. Integrity. Concluding Remarks. Exercise.
Answers to Selected Exercises.
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Calculus answer tests
There are two answer tests for dealing with calculus problems.
This test is a general differentiation test: it is passed if the arguments are algebraically equivalent, but gives feedback if it looks like the student has integrated instead of differentiated. The
first argument is the student's answer. The second argument is the model answer. The answer test option must be the variable with respect to which differentiation is assumed to take place.
There are edge cases, particularly with \(e^x\) where differentiation is indistinguishable from integration. You may need to use the "quiet" option in these cases.
This test is designed for a general indefinite integration question: it is passed if both the arguments are indefinite integrals of the same expression. The first argument is the student's answer.
The second argument is the model answer. The answer test option needs to be the variable with respect to which integration is assumed to take place, or a list (see below).
Getting this test to work in a general setting is a very difficult challenge. In particular, the test assumes that the constant of integration is expressed in a form similar to \(+c\), although which
variable used is not important.
The Int test has various additional options.
The question author must supply these options in the form of a list [var, opt1, ...]. The first argument of this list must be the variable with respect to which integration is taking place.
If one of the opt? is exactly the token NOCONST then the test will condone a lack of constant of integration. That is, if a student has missed off a constant of integration, or the answers differ by
a numerical constant, then full marks will be awarded. Weird constants (e.g. \(+c^2\)) will still be flagged up.
If one of the opt? is exactly the token FORMAL then the test will condone the formal derivative of the student's answer matching that of the teacher. This is useful in examples such as \(\log(|x+3|)/
2\) vs \(\log(|2x+6|)/2\) where effectively the constant of integration differs by a numerical constant. Note, if you use the FORMAL option then by definition you will accept a missing constant of
The answer test architecture only passes in the answer to the test. The question is not available at that point; however, the answer test has to infer exactly which expression, including the
algebraic form, the teacher has set in the question. This includes stripping off constants of integration and constants of integration may occur in a number of ways, e.g. in logarithms. In many cases
simply differentiating the teacher's answer is fine, in which case the question author need not worry. Where this does not work, the question author will need to supply the expression from the
question in the right form as an option to the answer test. This is done simply by adding it to the list of options.
[x, x*exp(5*x+7)]
The test cannot cope with some situations. Please contact the developers when you find some of these. This test is already rather overloaded, so please don't expect every request to be accommodated!
This test, in particular, has a lot of test cases which really document what the test does in detail.
The issue of \( \int \frac{1}{x} dx = \log(x)+c\) vs \( \int \frac{1}{x} dx = \log(|x|)+c\) is a particular challenge. What mark would you give a student who integrated \[ \int \frac{1}{x} dx = \log
(k\times abs(x))?\] If the teacher uses \(|..|\) in their answer then the student is also expected to use the absolute value. The test is currently defined in such a way that if the teacher uses \( \
log(|x|)+c \) in their answer, then they would expect the student to do so. If they don't use the absolute value function, then they don't expect students to but will accept this in an answer.
For example, if the teacher's answer is \( \log(x)+c \) (i.e. no absolute value) then all the following are considered to be correct. \[ \log(x)+c,\ \log(|x|)+c,\ \log(k\,x),\ \log(k|x|),\ \log(|k, x
|) \]
If the teacher's answer is \( \log(|x|)+c \) (i.e. with absolute value) then all the following are considered to be correct. \[ \log(|x|)+c,\ \log(k|x|),\ \log(|k, x|)\ \] Now, the following are
rejected as incorrect, as the studnet should have used \(|..|\) \[\log(x)+c,\ \log(k\,x)\]
Note that STACK sets the value of Maxima's logabs:true, which is not the default in Maxima. This has the effect of adding the absolute value funtion when integrate is used.
In the case of partial fractions where there are more than one term of the form \(\log(x-a)\) then we insist the student is at least consistent. If the teacher has any \(\log(|x-a|)\) then the
student must use \(|...|\) in all of them. If the teacher has no \(\log(|x-a|)\) (i.e. just things like \(\log(x-a)\)) then the student must have all or none.
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Addition Practice: 5 Super Easy Ways to Enrich Adding 3 Numbers for 1st GradersAddition Practice: 5 Super Easy Ways to Enrich Adding 3 Numbers for 1st Graders - Teaching Perks
Addition practice in 1st grade, specifically adding three numbers, can be a fun and challenging activity for first-grade students.
To help students stay engaged and challenged, it’s important to gradually increase the difficulty of problems they’re working on. This blog post outlines five super easy ways to enrich addition
practice and adding 3 numbers for first-graders.
Addition Practice With Larger Numbers
To keep students engaged and challenged in their addition practice, it’s important to gradually increase the difficulty of problems they’re working on. One way to do this is by using larger numbers
when adding 3 numbers. This strategy requires students to not only use their basic addition skills but also their understanding of place value and mental math.
By using bigger numbers, students will be required to break down the numbers into smaller parts and add them together in a way that makes sense. This can help them develop stronger problem-solving
skills and become more confident in their ability to tackle challenging addition practice.
For example, 24 + 38 + 15 = ?
To solve this problem, students must first break down each number into its place value parts. For example, 24 can be broken down into 20 + 4, 38 can be broken down into 30 + 8, and 15 can be left as
is. Next, students can add the numbers with the same place value parts, starting from the largest place value and moving down. For example, they can add 20 + 30 to get 50, then add 4 + 8 to get 12
Now they are left with 12 + 15 which can be broken down into 10 + 10 and 5 + 2. When they add the tens and ones they should get 77. See the image above to see this broken down.
Another addition practice strategy would be to let students use or draw base ten blocks. They can then move and manipulate the blocks to solve.
By using larger numbers and three addends, students are required to use their place value knowledge and mental math skills to solve the problem. This can help them develop stronger problem-solving
skills and become more confident in their ability to tackle challenging math problems.
Use Mental Math to Enrich Adding 3 Numbers
Mental math is an important addition practice skill that students can develop and improve through practice. One way to do this is by challenging them to solve addition problems with three addends
without using any writing tools, such as a pencil or paper.
This forces students to rely on their mental math skills and enhances their ability to perform arithmetic operations quickly and accurately. Additionally, solving problems mentally can help students
see connections between numbers and develop a deeper understanding of mathematical concepts.
Encourage your students to start with smaller numbers and gradually increase the difficulty as they become more comfortable with mental math. This strategy can be both fun and challenging for
students and help them become more confident in their math abilities.
Let’s look at this example: 14+12+9
To begin, students can add the first two numbers: 14 + 12 by making a ten. They would give 2 from the 12 to the 14 making it 16 + 10 = 26.
Next, they can add the third number, 9, to 26. One way to do this is to break the 9 into two parts: 4 and 5. Then, they can add 4 to 26 to get 30, and then add the remaining 5 to get 35. Therefore,
the answer to the problem 14 + 12 + 9 is 35.
By completing addition practice with three addends mentally, students can develop their mental math skills and become more confident in their ability to perform arithmetic operations without relying
on a pencil or paper. This strategy encourages them to think more critically and creatively.
Addition Practice With Missing Addends For Adding 3 Numbers
Adding missing addends to addition problems with three addends can be a great way to challenge your students’ math skills and encourage them to think more critically.
One way to do this is by providing problems with larger numbers and missing addends, such as 8 + __ + 6 = 20. To begin, students can add the three known addends: 8 + 6 = 14. Next, they can subtract
this sum from the total sum of 20 to find the missing addend: 20 – 16 = 4. Therefore, the missing addend is 4.
By adding missing addends to addition problems with three addends, students are required to use their place value knowledge and mental math skills to figure out what number is missing. This can help
them develop stronger problem-solving skills and become more confident in their ability to tackle challenging math problems.
Create a 3 Addend Problem To Match a Sum
Providing students with an addition practice challenge where they must find a solution using a certain number is a great way to encourage them to think creatively and use their problem-solving
One way to do this is by challenging students to create an addition with 3 addends problem that equals a given number. For instance, you can ask them to create an addition problem with three addends
that equals 15.
You could also ask them how many different possibilities they can come up with for that specific number.
This requires students to think critically and apply their knowledge of addition to come up with a solution. By creating their own addition problems, students can develop their ability to think
abstractly. This challenge can be a fun and engaging addition practice activity while also encouraging creativity and critical thinking.
Mix Addition Practice of Adding 3 Numbers with Subtraction
Mixing addition and subtraction in 3 addend problems can provide a new level of challenge for students, while also helping them to practice their addition and subtraction skills together.
For example, you can give students problems like 10 + 8 + 3 – 6. In order to solve this type of problem, students must add the first three addends (10 + 8 + 3) and then subtract the fourth addend (6)
from the sum.
This requires them to use both addition and subtraction skills in the same problem, while also developing their ability to use inverse operations to check their work. Encouraging students to practice
with a mix of addition and subtraction problems can help them become more confident in their math abilities and provide a deeper understanding of mathematical concepts.
Enrichment For Adding 3 Numbers Conclusion
In conclusion, mastering the skill of adding three numbers is an essential building block for a child’s mathematical development. With the help of these tips and tricks, students can gain confidence
and fluency in this area. It is important to keep in mind that every child learns differently, and differentiation is key in ensuring that all students are given the opportunity to succeed.
If you want differentiated small group plans for adding 3 numbers that not only provides enrichment lessons but also lessons for your on level and struggling learners, check out these 1st grade math
Addition practice for adding 3 numbers in 1st grade.
This packet comes with five different lessons to teach this skill, along with differentiated student work mats, answer keys, independent practice pages, and games. Teachers can also ask the higher
order thinking questions provided to deepen understanding. In addition, differentiated word problems are provided to help students apply this skill to real-world situations. With these resources,
students can build their confidence and proficiency in adding three numbers.
Want more strategies for adding 3 numbers? Check out this post!
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1999 U.S. Mint Proof Set Coin Value (Errors List & "S" Mint Mark Worth)
The popularity of U.S. mint Proof set coins has continued to rise steadily, especially among new collectors. This is majorly because they allow collectors to collect valuable coins in a single
Although each proof set represents a celebration in history, Proof sets from 1999 are sought after as they consist of five coins from the 50 State Quarters Program.
We’ve packed this post with relevant information for collectors and enthusiasts interested in learning about this unique coin collection. Here, We’ll explore the 1999 U.S. Mint Proof Set Value,
revealing notable facts in its history, variety, grading, and errors.
1999 U.S. Mint Proof Set Value Chart
Mint Mark Proof
1999-S U.S. Mint Proof Set $12.2
1999-S U.S. Mint Quarter Proof Set $5.1
1999-S U.S. Mint Silver Proof Set $92
1999-S Mint Mark 9-Piece Proof Set Value
A quick peek into the 1999-S mint mark U.S. mint proof set, and you’ll find nine distinct coins. The collection comprises five quarters of the 50 State Quarters Program and four presidential coins,
all in different denominations.
Coins in this Proof set possessed the S mint mark, a telltale that the San Francisco Mint produced them. The total mintage of this coin is 2,543,401 pieces, making it the highest quantity proof set
produced in 1999.
In addition to this, you will be interested in knowing that this proof set was the first to come in two separate cases. One of the cases contained the four presidential coins, and the other housed
the five State quarters for the year.
It all began early in the 19th century when the U.S. Mint began producing mint coins. These coins were tagged special, and their minting required greater attention to detail than regular ones.
Although the U.S. Mints initially only made proof versions of a few selected coins to serve as a collector’s item and not for commercial purposes, as proof coins became popular, collectors began
demanding proofs of each previously struck coin.
In response to this request, the U.S. Mint made proof coin sets (the penny, nickel, quarter, Half dollar, and one dollar). Each coin set was sealed in a clear plastic case, packaged, and sold only to
At the turn of 1999, the U.S. Mint launched the 50 States Quarters, a program designed to include additional five new quarters to the Proof set each year. The program was slated to run for ten years,
and five different Quarters would be included each year, representing each of the American States in the order they joined the Union.
With this move announced, many collectors couldn’t wait to get their hands on one, and when the set was finally released, the U.S. Mint recorded an impressive profit compared to the previous year,
with over 2.5 million dollars in sales.
This was shocking because the set sold at nearly double the price of the previous ones due to the addition of the five new quarters (Connecticut quarter, New Jersey quarter, Georgia quarter,
Pennsylvania quarter, and Delaware quarter).
In May 2002, the mint decided to sell off some unsold proof sets it had in storage; this included the 1999 Proof set. These sets sold for an initial price of $94.95, and Collectors could buy as much
as they wanted as individual orders were not restricted to any specified amount. However, the total number of orders could not exceed the available quantity of 150,000.
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As time passed, an announcement from the U.S. Mint halted the sales of the Proof sets not too long after production began due to damage from storing the coins.
The Obverse
The obverse of the four presidential coins in the 1999 U.S. mint Proof set features relief portraits of some of the nation’s past presidents.
On the Lincoln Cent, you’ll find a right-facing bust of Abraham Lincoln taking up the cent’s center. “IN GOD WE TRUST” is written curvedly above his head, close to the coin’s rim, while “LIBERTY”
appears on the left, near the back of his neck.
On the Jefferson Nickel, Thomas Jefferson’s close-up profile faces the left direction. Next to it is the phrase, “IN GOD WE TRUST,” which you can spot near the coin’s left rim.
Franklin Roosevelt’s head also appears facing left on the Roosevelt dime. Next to it is the boldly written “LIBERTY” inscription, which is seen on the coin’s left rim.
The last former president featured in this set is John F. Kennedy. On the obverse of the Kennedy half dollar Proof coin, his portrait occupies the center. Also, on close examination, you would
observe that the top of his head touching the “BER” in the spaced-out “LIBERTY” curved along the coin’s top half rim.
Finally, beside his neck, you’ll notice the words “IN GOD” and “WE TRUST”, with the former on the left and the latter on the right.
The Reverse
On the reverse of each quarter coin in the 1999 U.S. Mint proof set, you can see unique elements that symbolize one of the first five states in the Union.
For instance, On the Delaware Quarter, there’s an image of Caesar Rodney riding a horse, indicating his journey to Philadelphia in 1776. This journey is particularly important because of its purpose,
which was to sign the Declaration of Independence
The Pennsylvania Quarter is even more appealing because it features a striking image of the Commonwealth statue located at the state’s Capitol dome. It also features what looks like the state’s
outline behind the statue.
Moving further, another interesting feature is the 1776 depiction of George Washington crossing the Delaware River with his troops, this t the center of the New Jersey Quarter’s reverse.
The Georgia Quarter’s more simplistic reverse design features the Peach, the state’s symbol, and its motto and state outline. Finally, the Connecticut Quarter dedicates its reverse to the Charter Oak
located in the state, symbolizing the country’s independence.
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The Edges
In the 1999 U.S. mint Proof set, only the Lincoln Cent and Jefferson Nickel have plain edges. All the others, including the five new quarters and two presidential coins, have reeded edges.
The five quarters in this Proof set are made of 75% copper and 25% nickel bonded to a 100% copper core; they weigh 2.27 grams and have a diameter of 17.9mm. The details of presidential coins vary,
For one, the Roosevelt dime and Kennedy half dollar comprise copper and nickel in similar amounts as the quarters. They weigh 2.27 grams and 11.34 grams, respectively.
The Lincoln Cent is the only coin having zinc metal in its composition. It’s also the lightest presidential coin in the set, weighing 2.4 grams and 19.05mm in diameter. The Jefferson nickel is a
slightly heavier coin weighing 5 grams. It has a diameter of 21.2mm and is composed of 75% copper and 15% nickel.
The S mint mark 9-piece Proof set has a value that fluctuates depending on market conditions. It is currently worth $12.2 and has maintained this value since February 1, 2021, when it rose from $8.
1999-S Mint Mark 5-Piece Quarter Proof Set Value
The 1999 S mint mark 5-piece quarter Proof set contains only the first five quarter coins for the year. The coins bear the “S” mint mark to indicate they were minted at the San Francisco Mint. Also,
with a total mintage of 1,169,958, this proof set has the second-highest quantity among the three proof sets of 1999.
The 5-piece quarter Proof set is the least favorable of the three set types among collectors, with a slightly lesser value than the 9-piece set earlier discussed. However, according to the current
market price, it is worth $5.1, almost five times its face value of $1.25. It returned to this price on February 1, 2021, after an initial drop to $5 in December 2019.
1999 S Mint Mark 9-Piece Silver Proof Set Value
Although coins in all the 1999 proof set types have a cameo or deep cameo finish that many collectors find appealing, collectors consider the silver proof set more valuable. The 1999 silver proof set
was so widely accepted that it sold about 804,565 pieces, a notable increase from the previous year.
As collectors’ interest in the State Quarters grew, the demand for the silver proof set increased way more than the quantity available.
This increase caused a significant spike in the market value of the 1999 silver proof set. As a result, it ranks the highest in terms of value among the three set types, with a current market price
of $92. A quick look at the coin chart history shows that it rose to $110 in 2021.
1999 U.S. Mint Proof Set Grading
Proof coins usually come in higher grades because they have not gone through many hands over the years, unlike regular coins used for commercial purposes. Therefore, each coin in the 1999 proof set
is graded between 60 to 70 on the coin grading scale.
This video shows how the proof coins are graded according to PCGS.
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Rare 1999 U.S. Mint Proof Set Error Lists
An error coin has visible dents or markings resulting from a poorly-executed striking process. Although proof sets consisting of only flawed coins are impossible to find, finding one quarter, cent,
or nickel with an error is possible.
Let’s look at some possible errors contained in the 1999 proof set.
1999 U.S. Mint Proof Set Die Crack Error
A coin die is a tool used to impress designs onto blank coins. Typically, this tool strikes several thousand coins before becoming too worn out. Through regular use, the dies leave visible jagged
line markings and cracks on the coin’s obverse or reverse.
The value of coins in the 1999 proof set having the die crack error will increase depending on the size and position of the crack. However, because proof coins are minted with more care and
precision, cracks will often appear faint and sometimes unnoticeable to the naked eye.
1999 U.S. Mint Proof Set Strike-Through Error
A strike-through error is one fascinating error that commands a high price in the coin market. It occurs when a foreign object sticks to the die immediately before striking the coin, resulting in a
raised or indented spot on its surface. The strike-through error can appear on any 1999-proof coin, but a known occurrence was on a Connecticut silver quarter.
A flawless 1999 U.S. Mint Silver proof quarter in excellent condition, especially one with the highest grade a proof coin can receive: a PF70, will sell for about $45. On the other hand, you can find
one with a strike-through error at a much higher price due to its rarity.
A tiny error might only add a few dollars to its value, while if the error is apparent, it could increase the coin’s value by hundreds or a few thousand dollars.
1999 U.S. Mint Proof Set Close “AM” Error
You’ll find the close AM error on the reverse of a few 1999-proof pennies. It occurs when a proof die specially made for close-collar striking is mistakenly used to produce a penny. This error makes
coins marked with it highly desirable by collectors.
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You can spot coins having this error by looking at the first two letters, “AM”, in the word “AMERICA ”. Usually, they’ll appear too close to each other, unlike error-free coins with relatively wider
spacing between their “AM”. In terms of value, coins with this error are worth significantly more than those without it.
A 1999 close “AM” proof penny has a current market price of over $200, a great value when compared to error-free ones that are only worth $5.
1999 U.S. Mint Proof Set FAQs
How Can I Tell If a 1999 U.S. Mint Proof Set is Fake?
You can start by looking at its packaging. The 1999 U.S. Mint proof set comes in a cardstock box containing two cases. For the regular 9-piece Proof set, inspect each case to be sure it has blue
inserts, while for the silver proof set, confirm that the case containing the presidential coins has a red insert.
Also, remember that an original 1999 U.S. Mint proof set comes with a Certificate of Authenticity. You can find this vital piece of paper inside the box. Finally, we advise you to purchase your Proof
set from credible online and local coin vendors to reduce your chances of getting fake ones.
How Much Silver is in a 1999 Silver Proof Set?
Coins in the 1999 silver proof set have a 90% silver content with a total silver weight of 1.33826 ounces. This high content can fetch you good money if you’re going by the current silver spot price
of $25.16 per ounce.
When Can I Get More Info About The 1999 U.S. Mint Proof Set?
You can browse the U.S. Mint’s official website to learn more about the 1999 U.S. Mint proof set. Additionally, you’ll learn about the history, value, and varieties of the Proof set on the online
platforms of some authorized coin services, such as the Professional Coin Grading Service (PCGS) and the Numismatic Guarantee Company (NGC).
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Z: **Archived from Fall 2007**: American Economic History: Economics 113
(A) One-Sentence Identifications: Briefly state the importance of each of the following people/places/things/concepts for American economic history (40 pts):
Henry Ford; Crash of 1929; New Deal; Bretton Woods; Gold Standard; General Motors; Jim Crow; National Industrial Recovery Act; National Labor Relations Act; Civil Rights Act of 1964; Plessy v.
Ferguson; John Maynard Keynes; Milton Friedman; Mass Production; World War II; Federal Reserve; Morrill Land Grant Colleges; Gilded Age; Wage Discrimination; Sharecropping.
(B) One-Paragraph Discussions: Write one paragraph (three to six sentences) answering each of the following questions (40 pts):
1. How was America different from other more-or-less equally developed countries in sending its children to high school in the first half of the twentieth century?
2. What role did the stock market crash of 1929 play in bringing on the Great Depression of 1929-1941?
3. Why did the distribution of income among white male Americans become so much more equal between the 1920s and the 1960s?
4. Chicago-school economists tend to argue that large-scale persistent discrimination has small economic effects unless it is supported and enforced by the government. Do you think this is accurate?
5. What role did the United States play in putting western Europe on the right economic track (or a right economic track) in the years after World War II?
(C) Essay: Write an essay on one of the two following topics (40 pts):
1. What, in your view, are the similarities and differences in economic discrimination against women and against African-Americans since 1865? To what extent is it helpful and insightful to view
these two phenomena as analogous? To what extent is it misleading and destructive?
2. The Great Depression and the political reaction to it changed the American economy. How did these change the American economy? To what extent were the changes permanent?
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Yuejiang Li
Sep 11, 2023
Abstract:According to mass media theory, the dissemination of messages and the evolution of opinions in social networks follow a two-step process. First, opinion leaders receive the message from the
message sources, and then they transmit their opinions to normal agents. However, most opinion models only consider the evolution of opinions within a single network, which fails to capture the
two-step process accurately. To address this limitation, we propose a unified framework called the Two-Step Model, which analyzes the communication process among message sources, opinion leaders, and
normal agents. In this study, we examine the steady-state opinions and stability of the Two-Step Model. Our findings reveal that several factors, such as message distribution, initial opinion, level
of stubbornness, and preference coefficient, influence the sample mean and variance of steady-state opinions. Notably, normal agents' opinions tend to be influenced by opinion leaders in the two-step
process. We also conduct numerical and social experiments to validate the accuracy of the Two-Step Model, which outperforms other models on average. Our results provide valuable insights into the
factors that shape social opinions and can guide the development of effective strategies for opinion guidance in social networks.
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CLASS 11th PHYSICS SYLLABUS 2018-19 & COURSE STRUCTURE WITH MARKS DISTRIBUTION
CLASS 11th PHYSICS SYLLABUS 2018-19
UNIT CHAPTER MARKS DOWNLOAD
Chapter–1: Physical World
1. Physical World and Measurement 20
Chapter–2: Units and Measurements
Chapter–3: Motion in a Straight Line
2. Kinematics
Chapter–4: Motion in a Plane
3. Laws of Motion Chapter–5: Laws of Motion
4. Work, Energy and Power Chapter–6: Work, Energy and Power 17
5. Motion of System of Particles and Rigid Body Chapter–7: System of Particles and Rotational Motion
6. Gravitation Chapter–8: Gravitation
Chapter–9: Mechanical Properties of Solids
7. Properties of Bulk Matter Chapter–10: Mechanical Properties of Fluids 16
Chapter–11: Thermal Properties of Matter
8. Thermodynamics Chapter–12: Thermodynamics
9. Behaviour of Perfect Gases and Kinetic Theory
of Chapter–13: Kinetic Theory
Chapter–14: Oscillations and waves
10. Mechanical waves and Ray Optics 17
Chapter–15: Ray optics
Max Marks: 70
CBSE CLASS XI : MATHEMATICS COURSE STRUCTURE FOR 2018-2019 | SYLLABUS | MARKS DISTRIBUTION | CHAPTERS | BOOKS- CLICK HERE
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Mathematics - Textbook for class XI - NCERT Publication
Mathematics exemplar problems for class XI, NCERT publication. Guidelines for Mathematics Laboratory in Schools, class XI - CBSE Publication
Unit I: Physical World and Measurement Chapter–1: Physical World
Physics-scope and excitement; nature of physical laws; Physics, technology and society.
Chapter–2: Units and Measurements
Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. Length, mass and time measurements; accuracy and precision of measuring instruments; errors in
measurement; significant figures. Dimensions of physical quantities, dimensional analysis and its applications.
Unit II: Kinematics
Chapter–3: Motion in a Straight Line
Frame of reference, Motion in a straight line: Position-time graph, speed and velocity.
Elementary concepts of differentiation and integration for describing motion, uniform and non-uniform motion, average speed and instantaneous velocity, uniformly accelerated motion, velocity - time
and position-time graphs.
Relations for uniformly accelerated motion (graphical treatment).
Chapter–4: Motion in a Plane
Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of
vectors, relative velocity, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors.
Motion in a plane, cases of uniform velocity and uniform acceleration-projectile motion, uniform circular motion.
Unit III: Laws of Motion
Chapter–5: Laws of Motion
Intuitive concept of force, Inertia, Newton's first law of motion; momentum and Newton's second law of motion; impulse; Newton's third law of motion.
Law of conservation of linear momentum and its applications.
Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication.
Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).
Unit IV: Work, Energy and Power
Chapter–6: Work, Engery and Power
Work done by a constant force and a variable force; kinetic energy, work-energy theorem, power.
Notion of potential energy, potential energy of a spring, conservative forces: conservation of mechanical energy (kinetic and potential energies); non-conservative forces: motion in a vertical
circle; elastic and inelastic collisions in one and two dimensions.
Unit V: Motion of System of Particles and Rigid Body
Chapter–7: System of Particles and Rotational Motion
Centre of mass of a two-particle system, momentum conservation and centre of mass motion. Centre of mass of a rigid body; centre of mass of a uniform rod.
Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications.
Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions.
Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation). Statement of parallel and perpendicular axes theorems and their applications.
Unit VI: Gravitation
Chapter–8: Gravitation
Kepler's laws of planetary motion, universal law of gravitation.
Acceleration due to gravity and its variation with altitude and depth.
Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite, Geo-stationary satellites.
Unit VII: Properties of Bulk Matter
Chapter–9: Mechanical Properties of Solids
Elastic behaviour, Stress-strain relationship, Hooke's law, Young's modulus, bulk modulus, shear modulus of rigidity, Poisson's ratio; elastic energy.
Chapter–10: Mechanical Properties of Fluids
Pressure due to a fluid column; Pascal's law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure.
Viscosity, Stokes' law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli's theorem and its applications.
Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise.
Chapter–11: Thermal Properties of Matter
Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv - calorimetry; change of state - latent heat
Heat transfer-conduction, convection and radiation, thermal conductivity, qualitative ideas of Blackbody radiation, Wein's displacement Law, Stefan's law, Green house effect.
Unit VIII: Thermodynamics
Chapter–12: Thermodynamics
Thermal equilibrium and definition of temperature (zeroth law of thermodynamics), heat, work and internal energy. First law of thermodynamics, isothermal and adiabatic processes.
Second law of thermodynamics: reversible and irreversible processes, Heat engine and refrigerator.
Unit IX: Behaviour of Perfect Gases and Kinetic Theory of Gases
Chapter–13: Kinetic Theory
Equation of state of a perfect gas, work done in compressing a gas.
Kinetic theory of gases - assumptions, concept of pressure. Kinetic interpretation of temperature; rms speed of gas molecules; degrees of freedom, law of equi-partition of energy (statement only) and
application to specific heat capacities of gases; concept of mean free path, Avogadro's number.
Unit X: Mechanical Waves and Ray Optics
Chapter–14: Oscillations and Waves
Periodic motion - time period, frequency, displacement as a function of time, periodic functions.
Simple harmonic motion (S.H.M) and its equation; phase; oscillations of a loaded spring-restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation
of expression for its time period.
Free, forced and damped oscillations (qualitative ideas only), resonance.
Wave motion: Transverse and longitudinal waves, speed of wave motion, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings
and organ pipes, fundamental mode and harmonics, Beats, Doppler effect.
Chapter–15: RAY OPTICS
Ray Optics: Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and its applications, optical fibres, refraction at spherical surfaces, lenses, thin
lens formula, lensmaker's formula, magnification, power of a lens, combination of thin lenses in contact, refraction and dispersion of light through a prism.
Scattering of light - blue colour of sky and reddish apprearance of the sun at sunrise and sunset.
Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.
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Duality in the space of sections - Equivalences between Calabi–Yau manifolds and roofs of proje
Let us recall here the geometry of the roof of typre 𝐴^𝐺
𝑋 𝐺(2, 𝑉[5]) 𝐺(3, 𝑉[5]) 𝑌
𝑓[1] 𝑓[2]
𝑝 𝑞
The notation is the following:
◦ 𝑉[5] is a five-dimensional vector space and 𝐹 =𝐹(2,3, 𝑉[5]).
◦ 𝑝 and𝑞 are the natural projections from𝐹 to the two Grassman-nians.
◦ The flag variety 𝐹 has Picard group generated by the pullbacks of the hyperplane bundles of the two Grassmannians 𝐺(2, 𝑉[5]) and 𝐺(3, 𝑉[5]). We denote the pullbacks ofO𝐺(2,𝑉[5])(1) andO𝐺(3,𝑉[5])(1)
by O (1,0) and O (0,1) respectively. In this notation 𝑀 is the zero locus of a section 𝑠 ∈𝐻^0(𝐹 ,O (1,1)).
◦ One has that 𝑝[∗]O (1,1) = Q^∨
2(2) and 𝑞[∗]O (1,1) = U[3](2), where we call U𝑖 the universal bundle of a Grassmannian 𝐺(𝑖, 𝑉[5]) and Q𝑖 its universal quotient bundle. The varieties 𝑋 and 𝑌 are, respectively, the zero loci of the
sections 𝑝[∗]𝑠 and 𝑞[∗]𝑠 of Q^∨
2(2) and U[3](2),
◦ 𝑓[1] is a fibration over 𝐺(2, 𝑉[5]) with fiber isomorphic to P^1, for points outside the subvariety 𝑋 whereas the fibers are isomorphic toP^2 for points on 𝑋. Similarly 𝑓[2] is a map onto𝐺(3, 𝑉[5])
whose fibers are P^1 outside𝑌 and P^2 over𝑌.
The rest of this chapter is focused on proving that the general Calabi–
Yau pair associated to the roof of type 𝐴[4], in the sense of Definition 4.1.8 is non birationally equivalent.
Lemma 7.1.1. Let𝑋 be the zero locus of a regular section 𝑠[2]∈ 𝐻^0(𝐺(2, 𝑉[5]),Q^∨[2](2)).
Then 𝑠[2] is uniquely determined by 𝑋 up to scalar multiplication. Simi-larly, if𝑌 is the zero locus of a regular section𝑠[3]ofU[3](2)on𝐺(3, 𝑉[5]), 𝑠[3]is uniquely determined by𝑌.
Proof. We will prove the result for 𝐺(2, 𝑉[5]), the proof for the case of 𝐺(3, 𝑉[5]) is identical. Let us suppose 𝑋 is the zero locus of two sections 𝑠[2] ande𝑠[2]. Then, the Koszul resolutions with
respect to these two sections can be extended to the diagram:
· · · Q[2](−2) I𝑋 0
where the existence of the arrow 𝛽 is given by the following claim:
Claim. The map
Hom(Q[2](−2),Q[2](−2)) −→Hom(Q[2](−2),I𝑋) (7.1.3) is surjective.
This can be verified by proving surjectivity of the following map:
𝐻^0(Q[2]^∨⊗ Q[2]) −→ 𝐻^0(Q^∨[2](2) ⊗ I𝑋) (7.1.4) This can be achieved by tensoring the Koszul resolution ofI𝑋 byQ^∨
In fact, by the identities detQ[2]=O (1) and Q[2]' ∧^2Q^∨
2(1) one has the exact sequence
0−→ Q[2]^∨(−3) −→ Q^∨[2] ⊗ Q[2]^∨(−1) −→ Q^∨[2] ⊗ Q[2] −→ Q (2) ⊗ I𝑋 −→0 (7.1.5) where by the Borel-Weil-Bott theorem one finds:
𝐻^•(𝐺(2, 𝑉[5]),Q^∨[2](−3))=0, 𝐻^•(𝐺(2, 𝑉[5]),Q[2]^∨⊗ Q^∨[2](−1))=C[−2],
𝐻^•(𝐺(2, 𝑉[5]),Q[2]^∨⊗ Q[2]) =C[0].
which proves our claim.
In particular, if two sections define the same 𝑋, then the identity of the ideal sheaf lifts to an automorphism of Q[2](−2). However, since Ext^•(Q[2],Q[2]) = C[0], the only possible automorphisms of
Q[2](−2) are scalar multiples of the identity. That implies that the sections differ by
multiplication with a nonzero constant.
Corollary 7.1.2. Let 𝑋 = 𝑍(𝑠[2]) ⊂ 𝐺(2, 𝑉[5]). Then there exists a unique hyperplane section 𝑀 of 𝐹 such that the fiber 𝑝|^−^1
𝑀(𝑥) is isomorphic to P^2for𝑥 ∈ 𝑋 and is isomorphic toP^1for𝑥 ∈ 𝐺(2, 𝑉[5]) \𝑋. Similarly for 𝑌 = 𝑍(𝑠[3]) ⊂ 𝐺(3, 𝑉[5]) there exists a unique hyperplane section𝑀 of 𝐹 such that the fiber𝑞|^−1
𝑀(𝑥)is isomorphic toP^2for𝑥 ∈𝑌and is isomorphic toP^1for𝑥 ∈𝐺(3, 𝑉[5]) \𝑌.
Proof. We consider only the case 𝑋 =𝑍(𝑠[2]) ⊂𝐺(2,5) the other being completely analogous. Since𝐹 is the projectivization of a vector bundle over 𝐺(2, 𝑉[5]), then the pushforward 𝑝[∗] defines a
natural isomorphism
𝐻^0(𝐹 ,O (1,1))= 𝐻^0(𝐺(2, 𝑉[5]),Q^∨[2](2)).
Hence𝑠[2]= 𝑝[∗](𝑠)for a unique𝑠 ∈ 𝐻^0(𝐹 ,O (1,1)). We define𝑀 =𝑍(𝑠) which satisfies the assertion by the discussion above. The uniqueness of 𝑀 follows from Lemma 7.1.1. Indeed, for any hyperplane
𝑀 = 𝑍(𝑠˜), the fibers 𝑝|^−[˜]^1
𝑀(𝑥) are isomorphic to P^2 exactly for 𝑥 ∈ 𝑍(𝑝[∗]𝑠˜), but 𝑍(𝑝[∗]𝑠˜) = 𝑋 only if 𝑝[∗]𝑠˜ is proportional to 𝑠[2] which means that ˜𝑠 is proportional to 𝑠 and this proves uniqueness.
Let us consider an isomorphism
𝑓 :𝐺(2, 𝑉[5]) −→𝐺(3, 𝑉[5]). (7.1.7) Every such isomorphism is induced by a linear isomorphism𝑇[𝑓] :𝑉[5] −→
5 in the following way:
𝑓 =𝐷◦𝜙[2]:𝐺(2, 𝑉[5]) −→𝐺(3, 𝑉[5]). (7.1.8)
where 𝐷 is the canonical isomorphism
𝐷 :𝐺(𝑖, 𝑉[5]) −→ 𝐺(5−𝑖, 𝑉^∨
5) (7.1.9)
and 𝜙[𝑖] is the induced action of 𝑇[𝑓] on the Grassmannian:
𝜙[𝑖] :𝐺(𝑖, 𝑉[5]) −→𝐺(𝑖, 𝑉^∨
5) (7.1.10)
Similarly, we consider dual maps 𝑓^∨ : 𝐺(3, 𝑉^∨
5) −→ 𝐺(2, 𝑉^∨
5), ex-pressed as 𝑓^∨=𝜙^∨
2 ◦𝐷^∨.
Note that above maps 𝑓, 𝐷, 𝜙[2], 𝜙[3] are restrictions of linear maps between the Plücker spaces of the corresponding Grassmannians. By abuse of notation we shall use the same name for their linear
We can now introduce the following notion of duality.
Definition 7.1.3. Given an isomorphism 𝑓 : 𝐺(2, 𝑉[5]) −→ 𝐺(3, 𝑉[5]), we say𝑋 ⊂ 𝐺(2, 𝑉[5])is 𝑓-dualto𝑌 ⊂ 𝐺(2, 𝑉[5])if(𝑋 , 𝑓(𝑌))is a Calabi–Yau pair associated to the roof of type 𝐴^𝐺
4, in the sense of Definition 4.1.8.
Let us start by defining 𝑃=P(∧^2𝑉[5]) ×P(∧^2𝑉^∨
5), where ∧^2𝑉[5] is identi-fied with ∧^3𝑉^∨
5 by means of 𝐷. In that case 𝐹 is a linear section of 𝑃 (in its Segre embedding) by a codimension 25 linear space.
Remark 7.1.4. Recall that (Wey03, Proposition 3.1.9) the equations of 𝐹 in 𝑃 are described by the following sections 𝑠[𝑥]∗⊗𝑦 ∈𝐻^0(𝑃,O (1,1))
𝑠[𝑥]∗⊗𝑦(𝛼, 𝜔) =𝜔(𝑥^∗) ∧𝛼∧𝑦 (7.1.11) for 𝜔 ∈Λ^2𝑉^∨
5 = Λ^3𝑉[5], 𝛼∈Λ^2𝑉[5] and for every 𝑥^∗⊗ 𝑦 ∈𝑉^∨
5 ⊗𝑉[5].
In other words, we have
𝑠[𝑥]∗⊗𝑦(𝛼, 𝜔) =0 for ( [𝛼],[𝜔]) ∈𝐹(2,3, 𝑉[5]) ⊂ P(Λ^2𝑉[5]) ×P(Λ^3𝑉[5]). This defines a 25 dimensional subspace𝐻^0(𝑃,I𝐹(1,1)) ⊂𝐻^0(𝑃,O (1,1)) spanned by linearly independent sections corresponding to
𝑥^∗ = 𝑒^∗
𝑖, 𝑦=𝑒[𝑗] for 𝑖, 𝑗 ∈ {1. . .5} and a chosen basis {𝑒[𝑖]} for 𝑉[5].
Now, for every 𝑓 as in (7.1.7) we define the following function:
𝑃 𝑃
(𝑥 , 𝑦) ( (𝑓^∨)^−1(𝑦), 𝑓(𝑥))
which induces the following map at the level of sections:
𝐻^0(𝑃,O𝑃(1,1)) 𝐻^0(𝑃,O𝑃(1,1))
𝑠 𝑠◦𝜄[𝑓]
Note that 𝜄[𝑓] is a linear extension of an automorphism of the flag variety 𝐹 ⊂ 𝑃. It is constructed in such a way that we have that 𝑋 is defined by a section 𝑝[∗](𝑠) ∈ 𝐻^0(𝐺(2, 𝑉[5]),Q^∨
2(2)) if and only if 𝑓(𝑋) is defined by 𝑞[∗](
e𝜄[𝑓](𝑠)) ∈𝐻^0(𝐺(3, 𝑉[5]),U^∨
Our aim is to interpret 𝑓-duality in the setting above as explicitly as possible. For that we will identify 𝐻^0(𝐹 ,O (1,1)) with a subspaceH𝐹
of sections in 𝐻^0(𝑃,O (1,1)) invariant under our transformations. The following lemmas will be useful in the proof of non-birationality of general Calabi–Yau pairs.
Lemma 7.1.5. The space𝐻^0(𝑃,O (1,1))decomposes as𝐻^0(I𝐹|𝑃(1,1))⊕
3, which is the representation space of the product of represen-tations of weights𝜔[2]and𝜔[3]of𝐺 𝐿(𝑉[5]). By the Littlewood-Richardson rule, this space decomposes in the following way, and the
decomposi-tion is 𝐺 𝐿(𝑉[5])-invariant: Moreover, again by the Borel–Weil–Bott theorem, one has 𝑉[𝜔]
2+𝜔[3] = 𝐻^0(𝐹 ,O (1,1)) from which we get a surjection:
𝐻^0(𝑃,O (1,1)) −→𝐻^0(𝐹 ,O (1,1)) (7.1.15) from which the claim follows once we set H𝐹 :=𝑉[𝜔]
Alternatively, one can proceed in the following way: it is well known that Aut(𝐹) ' 𝐺 𝐿(𝑉[5])o Z/2. Moreover, the action of Aut(𝐹) on 𝐹 is linear and extends to an action of Aut(𝐹) on 𝑃 compatible
withe𝜄[𝑓]. It follows that 𝐻^0(I𝐹|𝑃(1,1)) is invariant under e𝜄[𝑓] since it is clearly invariant under Aut(𝐹). Furthermore the dual action of Aut(𝐹) on 𝑃^∨ preserves the dual flag variety, hence 𝐻^0
(I𝐹^∨|𝑃^∨(1,1)) is invariant under the dual action ofe𝜄[𝑓]. We can define H𝐹 = 𝐻^0(I𝐹^∨|𝑃^∨(1,1))^⊥. The latter space is invariant under Aut(𝐹), so it is also invariant under e𝜄[𝑓] and the map H𝐹 → 𝐻
^0(𝐹 ,O (1,1)) defined by restriction is an
Note that, by construction, the action of e𝜄[𝑓] on 𝐻^0(𝐹 ,O (1,1)) cor-responds to the action e𝜄[𝑓] on H𝐹. It means that we can think of 𝐻^0(𝐹 ,O (1,1)) equipped with the action induced by e𝜄[𝑓] as a
subset of 𝐻^0(𝑃,O (1,1)) invariant under the action of e𝜄[𝑓] on 𝐻^0(𝑃,O (1,1)) .
Remark 7.1.6. Note that, by applying the procedure of Remark 7.1.4 to describe the equations of the dual flag 𝐹^∨ with respect to the dual basis of 𝑉[5], one can find explicit equations defining H𝐹
in terms of matrices in 𝐻^0(𝑃,O (1,1)) ' 𝑀[10][×][10]. In particular, in our choice of basis, Equation 7.1.11 provides explicit linear conditions on the entries of 10×10 matrices to be elements of
H𝐹. This will be useful in the proof of Theorem 7.2.6.
Lemma 7.1.7. The variety 𝑋 is 𝑓-dual to𝑌 if and only if there exists a constant𝜆 ∈ C^∗such that sections𝑠[𝑋] ∈ H𝐹, 𝑠[𝑌] ∈ H𝐹 defining 𝑋 and𝑌 respectively satisfye𝜄[𝑓](𝑠[𝑌]) =𝜆 𝑠[𝑋].
Proof. By definition, 𝑋 is 𝑓-dual to 𝑌 if there exists a section ˆ𝑠 ∈ 𝐻^0(𝐹 ,O (1,1)) such that 𝑝[∗]𝑠ˆ defines 𝑋 while 𝑞[∗]𝑠ˆ defines 𝑓(𝑌). By Lemma 7.1.5 there then exists a unique section 𝑠 ∈ H𝐹
such that
𝑠 = 𝑠|𝐹. Now, by definition of e𝜄[𝑓], since 𝑞[∗]𝑠 defines 𝑓(𝑌) we have 𝑝[∗](
e𝜄[𝑓])^−1(𝑠) defines 𝑌. Furthermore by Lemma 7.1.5 we know that (e𝜄[𝑓])^−1(𝑠) ∈ H𝑓. We conclude from Lemmas 7.1.1 and 7.1.5 that up to multiplication by constants 𝑠=𝑠[𝑋] and (
e𝜄[𝑓])^−1(𝑠) =𝑠[𝑌]. From now on, let us fix a basis of𝑉[5] inducing a dual basis on 𝑉^∨
5, and natural bases on ∧^2𝑉[5] and ∧^2𝑉^∨
5 which are dual to each other.
A section 𝑠 ∈𝐻^0(𝑃,O𝑃(1,1)) is represented by a 10×10 matrix 𝑆 in the following way
𝑠: (𝑥 , 𝑦) 𝑦^𝑇𝑆𝑥 (7.1.16)
where 𝑥 and 𝑦 are expansions of 𝑥 and 𝑦 in the chosen bases of ∧^2𝑉[5] and ∧^2𝑉^∨
5. Once fixed our bases, 𝜙[2]is represented by a 10×10 invert-ible matrix 𝑀[𝑓], which is the second exterior power of the invertible matrix associated to𝑇[𝑓].
We can now describe very explicitly the 𝑓-duality in terms of matrices using the following.
Lemma 7.1.8. If𝑆is the matrix associated to𝑠 ∈ 𝐻^0(𝑃,O𝑃(1,1))then the matrix associated toe𝜄[𝑓](𝑠)is𝑀^−1
𝑓 𝑆^𝑇𝑀[𝑓].
Proof. On a pair(𝑥 , 𝑦), the map𝜄[𝑓] acts via𝜄[𝑓](𝑥 , 𝑦) = ( (𝜙^∨
2)^−1(𝑦), 𝜙[2](𝑥)).
Furthermore, in our choice of basis 𝜙[2](𝑥) = 𝑀[𝑓]𝑥 and (𝜙^∨
2)^−1(𝑦) = (𝑀^𝑇
𝑓)^−1𝑦. This yields:
e𝜄[𝑓](𝑠) (𝑥 , 𝑦)=𝑠◦𝜄[𝑓](𝑥 , 𝑦) =(𝑀[𝑓]𝑥)^𝑇𝑆(𝑀^𝑇
𝑓)^−1𝑦 =𝑦^𝑇𝑀^−1
𝑓 𝑆^𝑇𝑀[𝑓]𝑥 (7.1.17)
Remark7.1.9. In (OR17, sec. 5), it is proven that[𝑣] ∈P(𝔤𝔩(𝑉))defines a section𝑠[𝑣]of∧^2𝑉(1), whose projection to𝐻^0(𝐺(2, 𝑉[5]),∧^2Q[2](2))cuts out the threefold 𝑋[[][𝑣][]]. Then 𝑠[𝑣] corresponds to
a 10×10 matrix 𝑆 that we defined in (7.1.16). Hence, from Lemmas 7.1.7 and 7.1.8 follows
that 𝑋[[][𝑣][]] and 𝑋[[]
𝑣^𝑇] are 𝐷-dual. This means that our duality relation on X[25] between 𝑋 and𝑌 given by the condition of (𝑋 , 𝑌) being a Calabi–
Yau pair associated to the roof of type 𝐴^𝐺
4 is equivalent to the duality notion defined in (OR17, Section 5), extending the duality defined on X[25].
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Tides in binary star systems
When neutron stars emit gravitational waves
Scientists at the Max Planck Institute for Gravitational Physics in Potsdam develop an accurate model for the detection and interpretation of gravitational waves emitted by neutron stars in binary
systems. This model contains, for the first time, a realistic description of how neutron stars are deformed just before they collide. As the deformation depends on the exotic physics in neutron star
interiors and directly influences the gravitational waves more detailed information about the science contained in the expected signals is now available. This will enable more robust measurements
leading to an improved understanding of the properties of the densest objects in our universe.
The first discovery of gravitational waves from merging black holes announced earlier this year has initiated the use of gravitational waves as unique probes of the most violent astrophysical
processes. A highly anticipated source of gravitational waves are collisions of neutron stars. Neutron stars are among the most fascinating objects in the universe: they have up to two times the mass
of our Sun contained in a diameter of less than 20 kilometers. The nature of such extremely dense matter has remained a mystery for decades. If we could probe the interiors of neutron stars, we could
understand the unknown physics of these extreme celestial bodies. Gravitational wave astronomy will allow us to do so, as neutron stars in binaries emit waves in spacetime when they merge. These
gravitational waves carry unique information about the neutron stars. However, such signals of astrophysical origin are weak compared to the instrumental noise of current detectors. Nevertheless, the
extraction of a signal from the noisy data and its analysis becomes possible with accurate theoretical models of the plausible signals that these systems emit. In particular, the Effective One Body
model for binary black holes developed at the Max Planck Institute for Gravitational Physics in Potsdam and the University of Maryland was instrumental in assessing the highest detection confidence
and maximizing the science gains from the recent discovery of gravitational waves with LIGO detectors.
Fig. 1: Tidal forces deform a neutron star (left) orbiting another compact object - a second neutron star or a black hole.
© T. Hinderer/AEI
The present work extends this Effective One Body model to include the imprint of the rich neutron star physics on the waves. When a neutron star orbits another compact object - a second neutron star
or a black hole - it is deformed due to tidal forces (figure 1). This effect is reminiscent of what happens here on Earth when the moon’s gravity raises the ocean tides. Similarly, the neutron star
deforms in response to its companion. This has been the focus of several past studies. The present work significantly improves the modeling of tidal effects by taking into account that internal
oscillations in the neutron star will arise when the companion’s tidal force varies at a frequency that is close to a characteristic frequency of the star itself. This is analogous to oscillations of
a bridge excited by a band marching at a pace that matches the bridge’s characteristic frequency. The characteristic frequency of neutron stars is in the kHz range and is approached just before the
neutron star and its companion merge. In this final stage of the collision the neutron star orbits its companion in less than a millisecond at about half of the speed of light. Both the amount of
tidal deformation and the characteristic frequency of a neutron star depend sensitively on the microphysical properties of the neutron star matter. Any tidal response of the star leaves a distinct
imprint on the gravitational waves emitted by the binary. Thus, gravitational waves will reveal unique information about the exotic interior of the neutron stars.
“Our detailed model more accurately predicts the waveforms and thus tells us what to look for in the data”, says Dr Andrea Taracchini, co-author of the study and scientist in the Astrophysical and
Cosmological Relativity division at AEI. “We tested our model against results from numerical relativity simulations produced by our collaborators in the US and Japan. The model shows a better
agreement with the numerical results than models which neglect the characteristic frequency." "This means that our model is capturing genuine physical effects”, says Dr Tanja Hinderer, main author of
the paper, scientist at the University of Maryland and long-term visitor at the AEI. Numerical simulations provide the most realistic predictions for gravitational waves, however, they are too
expensive to deliver enough waveforms for the detectors. The newly developed analytical model not only enables generating arbitrarily many waveforms, but also explains physical characteristics of the
The search and analysis of gravitational waves requires detailed knowledge of an enormous number of different waveforms. Various different parameter combinations must be calculated: different
compositions of the binary system, different mass ratios, spins, and models of neutron star matter. The new analytical model allows to calculate thousands of waveforms in a short time. The extraction
of science from the gravitational wave data is then performed using these templates.
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SSM (State-Space Model)
State-Space Model
Mathematical frameworks used to model dynamic systems by describing their states in space and how these states evolve over time under the influence of inputs, disturbances, and noise.
State-space models are integral to control theory and signal processing, representing systems as a set of input, output, and state variables related by first-order differential equations. In these
models, the system's current state is described by a set of state variables, and the evolution of these states is determined by linear or nonlinear equations. SSMs are particularly powerful for
dealing with multi-variable systems where the interactions between variables may be complex and hidden. They are used extensively for system identification, time series analysis, forecasting, and
control systems design, allowing for the accommodation of noise and other uncertainties in the modeling process.
The concept of state-space in control engineering was primarily developed in the late 1950s and early 1960s. It became a fundamental aspect of modern control theory, particularly through the work on
optimal control and the development of the Kalman filter in the early 1960s. The state-space approach provided a unified framework that was applicable to both continuous and discrete systems, marking
a significant shift from classical control methods that focused on transfer functions.
Rudolf E. Kalman was particularly influential in the development of state-space models through his work on the Kalman filter, which efficiently estimates the state of a linear dynamic system from a
series of incomplete and noisy measurements. His contributions laid foundational principles for the use of SSMs in various applications, including aerospace and economics. Other notable contributors
include Pierre Simon Laplace and Andrey Markov, who developed early concepts related to state estimation and stochastic processes, respectively.
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Weighted Geometric Mean
Weighted Geometric Mean Selected
for Viewperf Composite Numbers
by Bill Licea-Kane
At its February 1995 meeting in Salt Lake City, a subcommittee within the OPC project group was given the task of recommending a method for deriving a single composite metric for each viewset running
under the Viewperf benchmark. Composite numbers had been discussed by the OPC group for more than a year.
In May 1995, the OPC project group decided to adopt a weighted geometric mean as the single composite metric for each viewset.
What is a Weighted Geometric Mean?
Above is the formula for determining a weighted geometric mean, where "n" is the number of individual tests in a viewset, and "w" is the weight of each individual test, expressed as a number between
0.0 and 1.0. (A test with a weight of "10.0%" is a "w" of 0.10. Note the sum of the weights of the individual tests must equal 1.00.)
The weighted geometric mean of CDRS-03, for example, is expressed by the following formula:
wgm-cdrs-03 =
The same formula for the weighted geometric mean as expressed in a Microsoft Excel expression:
Why the Weighted Geometric Mean?
The OPC subcommittee that recommended a method for determining composite numbers started with the description for assigning weights that is provided to each creator of a viewset: "Assign a weight to
each path based on the percentage of time in each path..."
Given this description, the weighted geometric mean of each viewset is the correct composite metric. This composite metric is a derived quantity that is exactly as if you ran the viewset tests for
100 seconds, where test 1 was run for 100 × weight[1] seconds, test 2 for 100 × weight[2] seconds, and so on.
The end result would be the number of frames rendered/total time which will equal frames/second. It also has the desirable property of "bigger is better"; that is, the higher the number, the better
the performance.
Why Not Weighted Harmonic Mean?
Since the results of Viewperf are expressed as "frames/second," the subcommittee was asked why we did not choose the weighted harmonic mean. The weighted harmonic mean would have been the correct
composite if the description published for Viewperf read as follows: "Assign a weight to each path based on the percentage of operations in each path..."
Given this description, the weighted harmonic mean would be as if you ran the viewset tests for 100 frames, where 100 × weight[1] frames were drawn with test1, the next 100 × weight[2] frames were
drawn by test2, and so on. The 100 frames divided by the total time would be the weighted harmonic mean.
Since the weights for the view sets were selected on percentage of time, not percentage of operations, we chose the weighted geometric mean over the weighted harmonic mean.
What About Weighted Arithmetic Mean?
The weighted arithmetic mean is correct for calculating grades at the end of a school term. It is not correct for the situation we face here.
Consider for a moment a trivial example, where there are two tests, equally weighted in a viewset:
Test 1 Test2 Weighted Arithmetic Mean
50% 50%
System A 1.0 100.0 50.5
System B 1.1 100.0 50.55
System C 1.0 110.0 55.5
System B is 10-percent faster at Test1 than System A. System C is 10-percent faster at Test2 than System A. But look at the weighted arithmetic means. System B's weighted arithmetic mean is only
.1-percent higher than System A's, while System C's weighted arithmetic mean is 10-percent higher. Even normalization doesn't help here.
Why Not Normalized Weighted Geometric Mean?
Here the OPC project group departs company from the nearly universal practice in benchmarking of normalizing test results. SPECint92, PLBsurf93 and Xmark93, for example, are all normalized results
based on a variety of "reference" systems.
Since our weights were percentage of time and since the results from Viewperf are expressed in frames/sec, we were not obligated to normalize. Normalization introduces many issues of its own,
starting with something as simple as how to select a reference system.
We invite readers to select two different systems whose results are published in this newsletter and to use each one as the reference system. You will discover quickly that the normalized weighted
geometric means change only in absolute magnitude. If the weighted geometric mean of System B is 10-percent higher than System A, for example, the normalized weighted geometric mean of System B will
be 10-percent higher than System A, no matter what reference system you choose.
Is There a Disadvantage to Weighted Geometric Mean?
As with any composite, the weighted geometric mean can act as a "filter" for results; this introduces the danger that important information might be lost and inappropriate conclusions could be drawn.
So, proper use of these composites is important. Use the composite as an additional piece of information. But also take a look at each individual test result in a viewset.
Please don't rely exclusively on any synthetic benchmark such as Viewperf. In the end, isn't actual application performance on an actual computer system w hat you are really attempting to find?
Bill Licea-Kane is responsible for graphics performance measurement within Digital Equipment Corp.'s Computer Systems Performance Group. He serves on all three GPC subcommittees and is chair of the
PLB group. He can be reached by phone at 603-881-2804 or by e-mail at wwlk@perfit.zko.dec.com.
[GPC Home] [OPC Project] [PLB Project] [XPC Project]
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Re: [tlaplus] Re: Why strong fairness implies weak fairness?
That should certainly be: if F is stronger than (implies) G, then G => H is stronger than F => H.
For any properties F, G, and H, if F is stronger than (implies) G, then G => F is stronger than F => H.
On Friday, December 27, 2019 at 10:58:40 PM UTC-8, Shiyao MA wrote:
In the pluscal manual, it is stated that,
Strong fairness of (action) A is stronger than (implies) weak fairness of A. In other words, if SFvars (A) is true of a behavior σ, then WFvars (A) is also true of σ.
Strong fairness is a relaxation of weak fairness as it only requires an action to be *indefinitely* enabled instead of *continuously* enabled, so why will SF(A) => WF(A) ?
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The line y = k intersects the graph of y = 3x^2 in points A and B. Points C and D are on the x-axis and ABCD is a rectangle. If the area of ABCD = 128/9, what is the value of k? Calculators not
allowed. Show your method clearly!
The above question is designed for students in their second year of algebra or precalculus. It can also be used for practice for the Algebra 2 questions on the SAT, although it is somewhat above that
level. Many students, including the more advanced, tend to struggle with problems like these because they don't have that much experience with coordinate geometry questions. These types of problems
are critical for their later development. Once the students have done a few of these they do not find them so formidable. A useful pedagogical tool is to let them try it, review the method clearly,
then erase the board and call on students to recall each step. Tell them you will do this, encouraging them to take good notes and pay careful attention! When using it on an assessment, make it a
bonus the first time, then make it count. Some of my readers may recall a similar parabola problem a few months ago.
Since there has been some discussion about the complications in the previous post on percents, how do we as educators deal with adversity and transform it into a teachable moment. We're explaining a
difficult problem that students are struggling with, so we try to explain it again, but to no avail. What do experienced educators do?
(A) Abandon the problem - it was simply too hard or they're not ready for it yet. Perhaps assign it for extra credit?
(B) Make the question simpler by removing some of the complexity. Consider a scaffolding approach?
(C) Re-think what prerequisite skills were needed?
In this case, let's re-work the problem as follows:
In the senior class, there are 20% more girls than boys. If there are 180 girls, how many more girls than boys are there among the seniors?
What do you think the results will be now that we've concretized the problem?
Do you think some students will make the classic error of taking 20% of 180 and subtract to obtain 144 boys? You betcha!! There's no getting around the issue of recognizing the correct BASE for the
20% in my opinion. Whether you consider the boys to be 100% and the girls to be 120% or you let x = number of boys and 1.2x = number of girls, the central issue is recognizing that you cannot take
20% of the girls! Sometimes students need to just use algebra as a tool - it's a great one! By the way, one student used algebra with ratios to solve this:
Boys = x
Girls = 1.2x
Difference = 0.2x
Therefore, the difference/girls = 0.2x/1.2x = 1/6 and 180 divided by 6 equals 30!
How would you have reworded the question to make it more accessible?
Thanks to Joanne Jacobs for this interesting article about how an engineer, a native of Ghana and now living in Detroit, decided to give up his full-time job to dedicate himself to writing a series
of math texts to help his own children (who were struggling) and now others. After reading the full article I must say I want to see more details and samples of what he's written. According to the
article, his materials adhere to his state's standards and apparently blend the traditional algorithms and skill development with conceptual understanding. Gee, does this sound familiar! The fact
that other parents and school administrators are now taking notice and want copies is fascinating. Textbook publishers pour tons of money into developing a new math series and along comes one
individual who decided these materials were simply not working for his children. Is there a message here?
For now, I'll just leave the title as the problem to be discussed. Please consider how middle and high school students would approach this. How many might incorrectly guess that 60% of the seniors
are girls and 40% are boys. If the problem were rephrased as a multiple-choice question, this type of error would occur frequently from my experience. What causes the confusion? Is it just the
wording of the question or is there also an underlying issue regarding conceptual understanding of percents? Could it be that the question is asking for a percent and hasn't provided any actual
numbers of students? What are the most effective instructional methods and strategies to help students overcome these issues? Certainly, algebraic methods would be a direct approach, but what
foundation skills and concepts should middle schoolers develop even before setting up algebraic expressions?
Lines L and M are perpendicular. Line L contains (0,0). Lines L and M intersect at A(5,2). Find x.
(1) Would students in geometry be more likely to consider some or all of the following: similar triangles, altitudes on hypotenuse theorems, areas, Pythagorean approaches? If this were presented as a
coordinate problem in Algebra 2, what would be the most likely approach? Would some students use the distance formula?
(2) As always, it is our role as educators to present these kinds of challenges and to encourage students to think more deeply. Making connections between algebra and geometry happens naturally for
some, but certainly not for all! We must enable this dialogue via the classroomm environment we establish and the kinds of questions we ask. It is important to first decide the goal. What concepts
are we trying to develop? Slope? Ratios of corresponding parts? How many students never make the connection between the two!
(3) The answer is 29/5 or 5.8. Think of at least two methods!
What do you think would be the results of giving the following Algebra 1 problems to your students before, during, and after the course?
Do you believe that either or both of these could be or have been SAT questions?
Do students normally have exposure to these kinds of problems in their regular assignments?
Do these kinds of questions require a deeper conceptual understanding of algebra?
1. Given: x^2 - 9 = 0
Which of the following must be true?
I. x = 3
II. x = -3
III. x^2 = 9
(A) I only (B) I, II only (C) I, III only (D) I, II, III
(E) none of the preceding answers is correct
2. Given: (a - b) (a^2 - b^2) = 0
Which of the following must be true?
I. a = b
II. a = -b
III. a^2 = b^2
(A) I only (B) I, II only (C) I, III only (D) III only (E) I, II, III
(a) AB = 8, BC = 6 in rectangle ABCD. Find the lengths of all segments shown in the diagram above.
Specifically: BD, CF,DE, BE, FE, CE
Comments: This is another in a series of rectangle investigations. To deepen student understanding of triangle relationships and to provide considerable practice with these ideas, the question asks
for more than just one result. Students should be encouraged to first draw ALL of the triangles in the diagram separately and recognize why they are all similar! Using ratios, students should be able
to find all the segments efficiently. One could also demonstrate the altitude on hypotenuse theorems as well!
(b) In case, students need a bit more of a challenge, have them derive expressions for all of the above segments given that AB = b and BC = a. To make life easier, assume b > a. This should keep your
stronger students rather busy! This algebraic connection is powerful stuff. We want our students to appreciate that algebra is the language of generalization.
More practice for students...
As indicated many many times on this blog, there is no substitute for experience!
The keys to success here are:
(1) Careful reading (do students often miss the key word!)
(2) Knowledge of facts (why is zero the most important number in life!)
(3) Knowledge of strategies, methods (multiplication principle, organized lists, counting by groups, etc)
If these problems are helpful for students, let me know...
Identity Crisis...
I imagined the following dialogue taking place between a traditionalist (T) and a reformist (R) after both had finished reading the posts I've written on this blog for the past 6-7 months:
T: He is definitely a radical reformer. He uses phrases like investigations, explorations, discovery-learning, problem-solving, working in groups and encourages the use of the calculator for some
activities. He is more concerned with conceptual understanding than with content and skills that all students must know. He actually believes that young children can think profoundly about concepts
before they have completely mastered their skills. He talks 'less is more.' He has no documented research base for any of his wild ideas, and pretends that decades of classroom instruction are just
as legitimate as a carefully developed research study. In what accredited journal of education research has he published?
R: Nonsense. He's one of your kind. He preaches strong foundation in basic skills, automaticity of basic facts, facility with percents, decimals and fractions and generally does not promote the use
of the calculator in the lower grades. Other than some esoteric mortgage activity he wrote, most of his math challenges have little to do with the real world, focusing instead on number theory and
combinatorial math, topics which are above the heads of most middle schoolers. His geometry problems are very traditional. He is often critical of calculator use. He focuses on standards and
curriculum, even suggesting we need a nationalized math curriculum (and we know who will determine that!). He doesn't even applaud the efforts of his own national math teacher organization. He covers
up his 'back to basics' approach by pretending he is a centrist. We all know one cannot be a centrist here. Either you're pregnant or you're not! He has no documented research base for any of his
wild ideas, and pretends that decades of classroom instruction are just as legitimate as a carefully developed research study. In what accredited journal of education research has he published?
I know the regular readers of this blog know who I am and I know where their heart lies, but what can be done to move math education into a zone of reality that is sorely needed for our children. All
these wonderful ideas but there is still so much confusion out there. How far have we come in the past dozen years or so to change the reality of a curriculum that is 'one inch deep and one mile
wide'? Your thoughts are welcome...
Ok, here's a fairly typical SAT-type of question that requires application of fundamental ideas in geometry. Some students 'see' a way to find the value of x in less than 15 seconds. These students
have strong conceptual ability and are confident of their knowledge and reasoning. They are not thrown by a question that is somewhat different from the textbook problems they've seen.
Some of the issues as I see them are:
(1) How do we raise the knowledge/experiential base and confidence of those students who cannot seem to find a solution path and inevitably give up?
(2) How do we extend the thinking of those talented students who solve the problem in short order and just sit there complacently?
Perhaps, beyond these considerations and the instructional strategies employed is the bigger issue of PROVIDING FREQUENT CHALLENGES FOR OUR STUDENTS THAT GO BEYOND NORMAL TEXTBOOK FARE.
Where does one normally find these types of challenges in school curricula? Embedded in a natural way in the regular set of textbook exercises that can be routinely assigned? OR are they labeled as
Standardized Test Practice at the end of a section or in a separate part of the chapter or text? OR are they found primarily in ancillary materials provided by the publisher? Can you guess where I
think they should be and if they should be labeled? Should they be stand-alone multiple-choice questions or more open-ended with several parts that go beyond the 'answer?'
Before I suggest an activity based on this innocent-looking problem, I invite readers to consider a variety of methods of solution (so far I've observed about 6-7 'different' approaches) and how one
might go beyond this standardized test question to deepen student reasoning. To remove the element of surprise and focus attention on process, the 'answer' is 90... (sorry!).
Have fun but if you 'see' the answer in 15 seconds or less, don't stop there!
If interested in purchasing my NEW 2012 Math Challenge Problem/Quiz book, click on BUY NOW at top of right sidebar. 175 problems divided into 35 quizzes with answers at back. Suitable for SAT/Math
Contest/Math I/II Subject Tests and Daily/Weekly Problems of the Day. Includes both multiple choice and constructed response items.
Price is $9.95. Secured pdf will be emailed when purchase is verified. DON'T FORGET TO SEND ME AN EMAIL (dmarain "at gmail dot com") FIRST SO THAT I CAN SEND THE ATTACHMENT!
I can't believe I'm sinking to these depths but it is summer and this may be the time for silliness. Two of my students shared the first couple (I'm doing it from memory so they're not exactly the
originals) and the third I saw somewhere in my travels across the web (I forget the site to give proper attribution):
1. Who designed King Arthur's RoundTable?
Sir Cumference, of course. (ugh, groan,...)
2. What did pi say to i?
Get real!
3. Number two said to number three: "Boy, number 6 is really bad news. He's always in trouble."
Number three replied, "What do you expect. He's a product of our times!"
Ok, let me have it!
Of course, there would be more harmony in the universe if this had been the 13th Carnival but enjoy the 12th Carnival nevertheless over at Vedic's Math Forum.
Vedic has selected an interesting assortment of posts covering a broad range of math. For example, Knot Homology in Algebraic Topology, Key Concepts in PerCents, Math Mnemonics, Observing Objects
Near the Speed of Light, Sudoku and Graph Theory, An Interesting Way to Calculate the Square Root of 2, a discussion about finding Next Gen Math Teachers and a couple of recent posts from this blog.
Enjoy! A great way to beat the summer doldrums!
The next Carnival (#13!) will be hosted on 7-27 over at Polymathematics.
Two of the opposite vertices of square PQRS have coordinates P(-1,-1) and R(4,2). (a) SAT-Type: Find the area of PQRS.
SAT Level of difficulty: 4-5 (i.e., moderately difficult to difficult).
Note: For standardized tests, in particular, students are encouraged to learn the special formula for the area of a square in terms of its diagonal.
Now for a more significant challenge that can be used to extend and enrich. Students can work individually or in small groups:
(b) Explain why there is only one possible square with the given pair of opposite vertices. Use theorems to justify your reasoning. Would this also be true if a pair of consecutive vertices were
given? How many rectangles, in general, are determined if 2 vertices are given (opposite or consecutive)?
(c) Determine the coordinates of Q and S, the other pair of vertices of the square.
Note: There are many many approaches here. Students often get hung up on the distance formula leading to messy algebra (with 2 variables). There are simpler coordinate methods. You may want to
provide a toolkit for students here: Graph paper, Geometer's Sketchpad, etc. Students who estimate or 'guess' the coordinates must verify (PROVE) that these vertices do in fact form a square.
Students who quickly 'solve' the problem should be encouraged to find more than one method. This is the only way they will expand their thinking!
And now another coordinate problem that can be solved by a variety of methods...
In right triangle PQR, with right angle at Q, the coordinates of the vertices are:
Determine the value of the area of the triangle. Assume p and r are positive.
Notes: Students should again be encouraged to try both synthetic (Euclidean) and analytical (algebraic, coordinates) methods.
General Comment: Students often forget how powerful slopes can be when solving geometry problems by coordinate methods!
Sorry for the silly title but when the temperature approaches 100, I start becoming delusional!
Algebra teachers know that the equations of horizontal and vertical lines (the coordinate axes in particular) are stumbling blocks for students and creative educators and desperate students often
resort to clever mnemonics and other memory aids to recall these. I invite readers to share their favorite. The teachers in my department (former department that is -- be kind, I'm adjusting to
retirement) became enamored of HOY-VUX. I'm not sure who originated it but this person deserves credit! The name is silly (reminiscent of horcruxes from Harry Potter), the students laugh at it, but
when the test comes around, they write it at the top of their paper. Here's how it works:
HOY: Horizontal, slope 0, Y=...)
VUX: Vertical, Undefined slope, X=...)
Now I know that others out there have their favorite ways of teaching this so PLEASE SHARE!
Believe it or not, the above was not the intent of this post but it's probably more interesting than the technical stuff to follow! This discussion is intended for Algebra 2 students and beyond. A
full treatment requires some vector analysis but I will avoid that for now. Unlike most of my offerings, I did not set this up as a worksheet but you'll get the idea. You may want to bookmark this
and save it for when 3-dimensional graphing comes up in the curriculum.
Start with a horizontal number line: <------------------|---------------->x
Ask students to plot x = 3 on the line. No ambiguity here, right!?!
Thus, the 1-dimensional graph of x = 3 is a POINT! Easy, so far.
Now draw both coordinate axes. Plot the point at 3 on the x-axis, ask a student for both of its coordinates and ask if it satisfies the equation 1x + 0y = 3.
Confirm this: 1(3) + 0(0) = 3.
Students generally treat x = 3 as an exceptional case of the equation of a line, but having both variables may help them see it isn't that special (other than its slope of course!).
Ask students to verify that (3,1), (3,2), (3,-1), (3,-2) all satisfy the equation 1x + 0y = 3.
Have them plot these points.
Ask the class (I didn't feel like writing this in the form of a worksheet today) to verify that (3,k) satisfies this equation for any real value of k. Students need to understand this significance of
the zero coefficient of y.
This should help them to recognize that the graph of x = 3 is a vertical line. Don't get me wrong. Understanding this does not necessarily lead to getting it right on a test! They still need survival
gear (mnemonics) for that! HOY-VUX to the rescue!
BUT THERE'S MUCH MORE TO THIS!
Point out that in the equation 1x + 0y = 3, we see that the resulting line is PERPENDICULAR to the axis with the non-zero coefficient (x-axis here) and PARALLEL to the axis whose coefficient is zero
(the y-axis in this case). We're not proving anything here or explaining why this is true, just making an observation that we will generalize later.
Ok, so where's the plane in all of this?
In 3-dimensional space, we examine the equation 1x + 0y + 0z = 3. We can still graph the point corresponding to 3 on the x-axis, the vertical line graphed above (y can be chosen arbitrarily) and now
z can be any real number. Corresponding to each variable whose coefficient is zero, the graph will now be a plane PARALLEL to that axis and PERPENDICULAR to the axis whose coefficient is not zero.
Thus, our graph is now a plane parallel to the y- and z-axes (therefore parallel to the yz-plane determined by these axes) and perpendicular to the x-axis. Of course, software like Mathematica,
Derive, or even freeware available on the web will help students visualize this better. Cardboard or Styrofoam models are also highly effective here. The more the students construct these models and
label the axes and planes, the better they will be able to make sense of all this.
So what is the graph of x = 3? All of the above!
Now have your students analyze the equation y = 3 following this model!
Of course, July 7th, 2077 or 7-7-77 will be fun too for those around to enjoy it 70 years from now!
Now you all know the story of the inveterate gambler who waited until the 7th day of the 7th month to bet $777 on horse #7 in the 7th race at 7-1 odds. Of course, the horse finished 7th! Sorry, I
couldn't resist telling this groaner...
If the difference of 2 numbers is less than the sum of the 2 numbers, which of the following must be true?
(A) Exactly one of the numbers is positive
(B) At least one of the numbers is positive
(C) Both numbers are positive
(D) At least one of the numbers is negative
The answer given in the original source was (B). Do you agree? See notes below for further discussion of the wording of the question (before you react!).
This SAT-type question was posted about a year ago in my discussion group, MathShare (which is still extant but possibly being phased out). On that forum, I discussed how students struggled with the
subtleties of logic in their analyses of the problem. That online discussion led to a meaningful debate (involving some exceptionally thoughtful educators) about how and when logical thinking needs
to be developed in our students. All agreed that critical thinking and logical reasoning must begin when children enter school, long before the formalism of an axiomatic approach. Do you believe this
is currently happening in most elementary schools? What materials are being used by those districts or teachers who are infusing critical thinking and logic? If we move toward a more standardized
curriculum nationally, how important is this? I'm sure you know how I feel!
Other Notes about the problem above:
(1) Is the question ambiguous or flawed because the term difference fails to specify in what order the numbers are subtracted? Should the domain of numbers be specified (would integers be better?).
On the SAT, it is understood that the domain is always real numbers.
(2) Why do you think so many students (these were strong SAT prep students) struggled with this and had great difficulty accepting that (B) was the correct choice? Do you think phrases like at least
one and exactly one are problematic for many students?
(3) What methods do you think were used by students? There were several approaches as I recall.
(4) Do you think students should be encouraged to use an algebraic approach here rather than plugging in numbers and testing various cases?
(5) Would restating the question in its contrapositive form make it easier to grasp? (how many students remember this from geometry?!?)
(6) Would this question lead to a richer discussion if it were open-ended, i.e., no choices given?
(7) Could this question be given to middle-schoolers after they have learned the rules of integers or do you believe they do not have sufficient maturity to handle the logic?
Your thoughts....
How many even 4-digit positive integers greater than 6000 are multiples of 5?
Students who have had many experiences with problems like this have a huge advantage on Math Contests and SATs. To level the playing field, you might want to consider giving your middle and secondary
students a daily SAT/Math Contest Problem of the Day like the one above. Questions like this require:
(a) Careful reading skills (encourage underlining or circling of keywords
(b) Knowledge of the Fundamental Principle of Counting (most often termed the Multiplication Principle)
(c) Clear thinking
(d) Careful reasoning
The answer is 399 (pls correct this if you feel I erred!). The process is equally important. Students should be encouraged to list a few examples (preferably the first 2-3) of numbers satisfying the
conditions. The challenge is to recognize HOW MANY conditions are subtly embedded in the dozen or so words in the question ! Some students will prefer to make an organized list and count by grouping,
which is fine, but, as they develop, they should recognize that is the basis for the Multiplication Principle (and later on, permutations).
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Implementation of Shi’s non-degeranate Vuong test
Yves Croissant
The original Vuong test
Vuong (1989) proposed a test for non-nested model. He considers two competing models, \(F_\beta = \left\{f(y|z; \beta); \beta \in \beta\right\}\) and \(G_\gamma = \left\{g(y|z; \gamma); \gamma \in \
Gamma\right\}\). Denoting \(h(y | z)\) the true conditional density. The distance of \(F_\beta\) from the true model is measured by the minimum KLIC:
\[ D_f = \mbox{E}^0\left[\ln h(y\mid z)\right] - \mbox{E}^0\left[\ln f(y\mid z; \beta_*)\right] \]
where \(\mbox{E}^0\) is the expected value using the true joint distribution of \((y, X)\) and \(\beta_*\) is the pseudo-true value of \(\beta\). As the true model is unobserved, denoting \(\theta^\
top = (\beta ^ \top, \gamma ^ \top)\), we consider the difference of the KLIC distance to the true model of model \(G_\gamma\) and model \(F_\beta\):
\[ \Lambda(\theta) = D_g - D_f = \mbox{E}^0\left[\ln f(y\mid z; \beta_*)\right]- \mbox{E}^0\left[\ln g(y\mid z; \gamma_*)\right] = \mbox{E}^0\left[\ln \frac{f(y\mid z; \beta_*)}{g(y\mid z; \gamma_*)}
\right] \]
The null hypothesis is that the distance of the two models to the true models are equal or, equivalently, that: \(\Lambda=0\). The alternative hypothesis is either \(\Lambda>0\), which means that \
(F_\beta\) is better than \(G_\gamma\) or \(\Lambda<0\), which means that \(G_\gamma\) is better than \(F_\beta\). Denoting, for a given random sample of size \(N\), \(\hat{\beta}\) and \(\hat{\
gamma}\) the maximum likelihood estimators of the two models and \(\ln L_f(\hat{\beta})\) and \(\ln L_g(\hat{\gamma})\) the maximum value of the log-likelihood functions of respectively \(F_\beta\)
and \(G\gamma\), \(\Lambda\) can be consistently estimated by:
\[ \hat{\Lambda}_N = \frac{1}{N} \sum_{n = 1} ^ N \left(\ln f(y_n \mid x_n, \hat{\beta}) - \ln g(y_n \mid x_n, \hat{\gamma})\right) = \frac{1}{N} \left(\ln L_f(\hat{\beta}) - \ln L_g(\hat{\gamma})\
right) \]
which is the likelihood ratio divided by the sample size. Note that the statistic of the standard likelihood ratio test, suitable for nested models is \(2 \left(\ln L^f(\hat{\beta}) - \ln L^g(\hat{\
gamma})\right)\), which is \(2 N \hat{\Lambda}_N\).
The variance of \(\Lambda\) is:
\[ \omega^2_* = \mbox{V}^o \left[\ln \frac{f(y \mid x; \beta_*)}{g(y \mid x; \gamma_*)}\right] \]
which can be consistently estimated by:
\[ \hat{\omega}_N^2 = \frac{1}{N} \sum_{n = 1} ^ N \left(\ln f(y_n \mid x_n, \hat{\beta}) - \ln g(y_n \mid x,_n \hat{\gamma})\right) ^ 2 - \hat{\Lambda}_N ^ 2 \]
Three different cases should be considered:
• when the two models are nested, \(\omega^2_*\) is necessarily 0,
• when the two models are overlapping (which means than the models coincides for some values of the parameters), \(\omega^2_*\) may be equal to 0 or not,
• when the two models are stricly non-nested, \(\omega^2_*\) is necessarely strictly positive.
The distribution of the statistic depends on whether \(\omega^2_*\) is zero or positive. If \(\omega^2_*\) is positive, the statistic is \(\hat{T}_N = \sqrt{N}\frac{\hat{\Lambda}_N}{\hat{\omega}_N}\)
and, under the null hypothesis that the two models are equivalent, follows a standard normal distribution. This is the case for two strictly non-nested models.
On the contrary, if \(\omega^2_* = 0\), the distribution is much more complicated. We need to define two matrices: \(A\) contains the expected values of the second derivates of \(\Lambda\):
\[ A(\theta_*) = \mbox{E}^0\left[\frac{\partial^2 \Lambda}{\partial \theta \partial \theta ^ \top}\right] = \mbox{E}^0\left[\begin{array}{cc} \frac{\partial^2 \ln f}{\partial \beta \partial \beta ^ \
top} & 0 \\ 0 & -\frac{\partial^2 \ln g}{\partial \beta \partial \beta ^ \top} \end{array}\right] = \left[ \begin{array}{cc} A_f(\beta_*) & 0 \\ 0 & - A_g(\gamma_*) \end{array} \right] \]
and \(B\) the variance of its first derivatives:
\[ B(\theta_*) = \mbox{E}^0\left[\frac{\partial \Lambda}{\partial \theta}\frac{\partial \Lambda}{\partial \theta ^ \top}\right]= \mbox{E}^0\left[ \left(\frac{\partial \ln f}{\partial \beta}, - \frac
{\partial \ln g}{\partial \gamma} \right) \left(\frac{\partial \ln f}{\partial \beta ^ \top}, - \frac{\partial \ln g}{\partial \gamma ^ \top} \right) \right] = \mbox{E}^0\left[ \begin{array}{cc} \
frac{\partial \ln f}{\partial \beta} \frac{\partial \ln f}{\partial \beta^\top} & - \frac{\partial \ln f}{\partial \beta} \frac{\partial \ln g}{\partial \gamma ^ \top} \\ - \frac{\partial \ln g}{\
partial \gamma} \frac{\partial \ln f}{\partial \beta^\top} & \frac{\partial \ln g}{\partial \gamma} \frac{\partial \ln g}{\partial \gamma^\top} \end{array} \right] \]
\[ B(\theta_*) = \left[ \begin{array}{cc} B_f(\beta_*) & - B_{fg}(\beta_*, \gamma_*) \\ - B_{gf}(\beta_*, \gamma_*) & B_g(\gamma_*) \end{array} \right] \]
\[ W(\theta_*) = B(\theta_*) \left[-A(\theta_*)\right] ^ {-1}= \left[ \begin{array}{cc} -B_f(\beta_*) A^{-1}_f(\beta_*) & - B_{fg}(\beta_*, \gamma_*) A^{-1}_g(\gamma_*) \\ B_{gf}(\gamma_*, \beta_*) A
^{-1}_f(\beta_*) & B_g(\gamma_*) A^{-1}_g(\gamma_*) \end{array} \right] \]
Denote \(\lambda_*\) the eigen values of \(W\). When \(\omega_*^2 = 0\) (which is always the case for nested models), the statistic is the one used in the standard likelihood ratio test: \(2 (\ln L_f
- \ln L_g) = 2 N \hat{\Lambda}_N\) which, under the null, follows a weighted \(\chi ^ 2\) distribution with weights equal to \(\lambda_*\). The Vuong test can be seen in this context as a more robust
version of the standard likelihood ratio test, because it doesn’t assume, under the null, that the larger model is correctly specified.
Note that, if the larger model is correctly specified, the information matrix equality implies that \(B_f(\theta_*)-A_f(\theta_*)\). In this case, the two matrices on the diagonal of \(W\) reduce to
\(-I_{K_f}\) and \(I_{K_g}\), the trace of \(W\) to \(K_g - K_f\) and the distribution of the statistic under the null reduce to a \(\chi^2\) with \(K_g - K_f\) degrees of freedom.
The \(W\) matrice can be consistently estimate by computing the first and the second derivatives of the likelihood functions of the two models for \(\hat{\theta}\). For example,
\[ \hat{A}_f(\hat{\beta}) = \frac{1}{N} \sum_{n= 1} ^ N \frac{\partial^2 \ln f}{\partial \beta \partial \beta ^ \top}(\hat{\beta}, x_n, y_n) \]
\[ \hat{B}_{fg}(\hat{\theta})= \frac{1}{N} \sum_{n=1}^N \frac{\partial \ln f}{\partial \beta}(\hat{\beta}, x_n, y_n) \frac{\partial \ln g}{\partial \gamma^\top}(\hat{\gamma}, x_n, y_n) \]
For the overlapping case, the test should be performed in two steps:
• the first step consists on testing whether \(\omega_*^*\) is 0 or not. This hypothesis is based on the statistic \(N \hat{\omega} ^ 2\) which, under the null (\(\omega_*^2=0\)) follows a weighted
\(\chi ^ 2\) distributions with weights equal to \(\lambda_* ^ 2\). If the null hypothesis is not rejected, the test stops at this step and the conclusion is that the two models are equivalent,
• if the null hypothesis is reject, the second step consists on applying the test for non-nested models previously described.
The non-degenerate Vuong test
Shi (2015) proposed a non-degenerate version of the Vuong (1989) test. She shows that the Vuong test has size distortion, leading to subsequent overrejection. The cause of this problem is that the
distribution of \(\hat{\Lambda}\) is discontinuous in the \(\omega^2\) parameter (namely a normal distribution if \(\omega^2 > 0\) and a distribution related to a weight \(\chi^2\) distribution if \
(\omega^2 = 0\)). Especially in small samples, it may be difficult to distinguish a positive versus a zero value of \(\omega ^ 2\) because of sampling error. To solve this problem, using local
asymptotic theory, Shi (2015) showed that, rewriting the Vuong statistic as:
\[ \hat{T} = \frac{N \hat{\Lambda}_N}{\sqrt{N \hat{\omega} ^ 2_N}} \]
the asymptotic distribution of the numerator and of the square of the denominator of the Vuong statistic is the same as:
\[ \left( \begin{array}{cc} N \hat{\Lambda}_N \\ N \hat{\omega} ^ 2 _ N \end{array} \right) \rightarrow^d \left( \begin{array}{cc} J_\Lambda \\ J_\omega \end{array} \right) = \left( \begin{array}{cc}
\sigma z_\omega - z_\theta ^ \top V z_\theta / 2 \\ \sigma ^ 2 - 2 \sigma \rho_* ^ \top V z_\theta + z_\theta ^ \top V ^ 2 z_\theta \end{array} \right) \]
\[ \left(\begin{array}{c}z_\omega \\ z_\theta \end{array}\right) \sim N \left(0, \left(\begin{array}{cc} 1 & \rho_* ^ \top \\ \rho_* & I \end{array}\right) \right), \]
\(\rho_*\) is a vector of length \(K_f + K_g\), \(\sigma\) a positive scalar and V is the diagonal matrix containing the eigen values of \(B ^ {\frac{1}{2}} A ^ {-1} B ^ {\frac{1}{2}}\).
Based on this result, Shi (2015) showed:
• that the expected value of the numerator is \(-\mbox{trace}(V) / 2\), the classical Vuong statistic is therefore biased and this bias can be severe in small samples and when the degree of
parametrization of the two models are very different,
• that the denominator, being random, can take values close to zero with a significant probability, which can generate fat tails in the distribution of the statistic.
Shi (2015) therefore proposed to modify the numerator of the Vuong statistic:
\[\hat{\Lambda}^{\mbox{mod}}_N = \hat{\Lambda}_N + \frac{\mbox{tr}(V)}{2 N}\]
and to add a constant to the denominator, so that:
\[ \left(\hat{\omega}^{\mbox{mod}}(c)\right) ^ 2 = \hat{\omega} ^ 2 + c \; \mbox{tr}(V) ^ 2 / N \]
The non-degenarate Vuong test is then:
\[ T_N^{\mbox{mod}} = \frac{\hat{\Lambda}^{\mbox{mod}}_N}{\hat{\omega}^{\mbox{mod}}}= \sqrt{N}\frac{\hat{\Lambda}_N + \mbox{tr}(V) / 2N}{\sqrt{\hat{\omega} ^ 2 + c \;\mbox{tr}(V) ^ 2 / N}} \]
The distribution of the modified Vuong statistic can be estimated by simulations: drawing in the distribution of \((z_\omega, z_\theta^\top)\), we compute for every draw \(J_\Lambda\), \(J_\omega\)
and \(J_\Lambda / \sqrt{J_\omega}\). As \(\sigma\) and \(\rho_*\) can’t be estimated consistently, the supremum other these parameters are taken, and Shi (2015) indicates that \(\rho_*\) should be in
this case a vector where all the elements are zero except for the one that coincides with the highest absolute value of \(V\) which is set to 1.
The Shi test is then computed as follow:
1. start with a given size for the test, say \(\alpha = 0.05\),
2. for a given value of \(c\), choose \(\sigma\) which maximize the simulated critical value for \(c\) and \(\alpha\),
3. adjust \(c\) so that this critical value equals the normal critical value, up to a small disperency (say 0.1); for example, if the size is 5%, the target is \(v_{1 - \alpha / 2} = 1.96 + 0.1 =
4. compute \(\hat{T}_N^{\mbox{mod}}\) for the given values of \(c\) and \(\sigma\) ; if \(\hat{T}_N^{\mbox{mod}} > v_{1 - \alpha / 2}\), reject the null hypothesis at the \(\alpha\) level,
5. to get a p-value, if \(\hat{T}_N^{\mbox{mod}} > v_{1 - \alpha / 2}\) increase \(\alpha\) and repeat the previous steps until a new value of \(\alpha\) is obtained so that \(\hat{T}_N^{\mbox{mod}}
= v_{1 - \alpha^* / 2}\), \(\alpha^*\) being the p-value of the test.
Shi (2015) provides an example of simulations of non-nested linear models that shows that the distribution of the Vuong statistic can be very different from a standard normal. The data generating
process used for the simulations is:
\[ y = 1 + \sum_{k = 1} ^ {K_f} z^f_k + \sum_{k = 1} ^ {K_g} z^g_k + \epsilon \]
where \(z^f\) is the set of \(K_f\) covariates that are used in the first model and \(z^g\) the set of \(K_g\) covariates used in the second model and \(\epsilon \sim N(0, 1 - a ^ 2)\). \(z^f_k \sim
N(0, a / \sqrt{K_f})\) and \(z^g_k \sim N(0, a / \sqrt{K_g})\), so that the explained variance explained by the two competing models is the same (equal to \(a ^ 2\)) and the null hypothesis of the
Vuong test is true. The vuong_sim unables to simulate values of the Vuong test. As in Shi (2015), we use a very different degree of parametrization for the two models, with \(K_f = 15\) and \(K_G = 1
## [1] 1.5588591 2.5625929 -1.0904569 0.7207765 2.7263322 1.1310495
## [1] 1.119354
## [1] 0.207
We can see that the the mean of the statistic for the 1000 replications is far away from 0, which means that the numerator of the Vuong statistic is seriously biased. 20.7% of the values of the
statistic are greater than the critical value so that the Vuong test will lead in such context a noticeable overrejection. The empirical pdf is shown in the following figure, along with the normal
Implementation of the non-degenarate Vuong test
The micsr package provides a ndvuong function that implements the classical Vuong test. It has a nest argument (that is FALSE by default but can be set to TRUE to get the nested version of the Vuong
test). This package also provide a llcont generic which returns a vector of length \(N\) containing the contribution of every observation to the log-likelihood.
The ndvuong package provides the ndvuong function. As for the vuongtest function, the two main arguments are two fitted models (say model1 and model2). The \(\hat{\Lambda}_n\) vector is obtained
using llcont(model1) - llcont(model2). The relevant matrices \(A_i\) and \(B_i\) are computed from the fitted models using the estfun and the meat functions from the sandwich package. More precisely,
\(A^{-1}\) is bdiag(-bread(model1), bread(model2) and \(B\) is crossprod(estfun(model1), - estfun(model2)) / N, where N is the sample size. Therefore, the ndvuong function can be used with any models
for which a llcont, a estfun and a bread method is available.
Voter turnout
The first application is the example used in Shi (2015) and is used to compare our R program with Shi’s stata’s program. Coate and Conlin (2004) used several models of electoral participation, using
data concerning referenda about alcool sales regulation in Texas. Three models are estimated: the prefered group-utilitarian model, a “simple, but plausible, alternative: the intensity model” and a
reduced form model estimated by the seemingly unrelated residuals method. They are provided in the ndvuong package as turnout, a list of three fitted models. The results of the Shi test are given
below. We first compute the Shi statistic for an error level of 5%. We therefore set the size argument to 0.05 (this is actually the default value) and the pval argument to FALSE.
## Non-degenerate Vuong test for non-nested models
## data: turnout$group-turnout$intens
## z = 1.7759, size = 0.050000, vuong_stat = 2.084528, constant =
## 0.381107, crit-value = 2.059963, sum e.v. = -10.997224, vuong_p.value =
## 0.018556
## alternative hypothesis: different models
The Shi statistic is 1.776, which is smaller that the critical value 2.06. Therefore, based on the Shi test, we can’t reject the hypothesis that the two competing models are equivalent at the 5%
level. The value of the constant \(c\) is also reported, as is the sum of the eigen values of the \(V\) matrix (sum e.v.). The classical Vuong statistic is also reported (2.085) and is greater than
the 5% normal critical value (the p-value is 0.019). Therefore, the classical Vuong test and the non-degenerate version lead to opposite conclusions at the 5% level.
To get only the classical Vuong test, the nd argument can be set to FALSE:
## Vuong test for non-nested models
## data: turnout$group-turnout$intens
## z = 2.0845, p-value = 0.01856
## alternative hypothesis: different models
To get the p-value of the non-degenerate Vuong test, the pval argument should be set to TRUE.
## Non-degenerate Vuong test for non-nested models
## data: turnout$group-turnout$intens
## z = 1.8125, vuong_stat = 2.084528, constant = 0.000000, sum e.v. =
## -10.997224, vuong_p.value = 0.018556, p-value = 0.0864
## alternative hypothesis: different models
The results indicate that the p-value is 0.086, which confirms that the Shi test concludes that the two model are equivalent at the 5% level.
Transport mode choice (nested models)
The third example concerns transport mode choice in Canada. The dataset, provided by the mlogit package is called ModeCanada and has been used extensively in the transport demand litterature (see in
particular ???; ???; and ???). The following example is from (???). The raw data set is first transformed to make it suitable for the estimation of discrete choice models. The sample is restricted to
the individuals for which 4 transport modes are available (bus, air, train and car).
## Le chargement a nécessité le package : dfidx
## Attachement du package : 'dfidx'
## L'objet suivant est masqué depuis 'package:stats':
## filter
## Attachement du package : 'mlogit'
## Les objets suivants sont masqués depuis 'package:micsr':
## rg, scoretest, stdev
MC <- mlogit.data(ModeCanada, subset = noalt == 4, chid.var = "case",
alt.var = "alt", drop.index = TRUE)
We first estimate the simplest discrete choice model, which is the multinomial logit model. The bus share being negligible, the choice set is restricted to the three other modes and the reference
mode is set to car.
ml <- mlogit(choice ~ freq + cost + ivt + ovt | urban + income, MC,
reflevel = 'car', alt.subset = c("car", "train", "air"))
This model relies on the hypothesis that the unobserved component of the utility functions for the different modes are independent and identical Gumbell variables. (???) proposed the heteroscedastic
logit for which the errors follow a general Gumbell distributions with a supplementary scale parameter to be estimated. As the overall scale of utility is not identified, the scale parameter of the
reference alternative (car) is set to one.
hl <- mlogit(choice ~ freq + cost + ivt + ovt | urban + income, MC,
reflevel = 'car', alt.subset = c("car", "train", "air"),
heterosc = TRUE)
## Estimate Std. Error z-value Pr(>|z|)
## (Intercept):train 0.678393435 0.332762598 2.038671 4.148288e-02
## (Intercept):air 0.656754399 0.468163091 1.402832 1.606668e-01
## freq 0.063924677 0.004916769 13.001360 0.000000e+00
## cost -0.026961457 0.004283139 -6.294789 3.078178e-10
## ivt -0.009680773 0.001053874 -9.185892 0.000000e+00
## ovt -0.032165526 0.003593007 -8.952258 0.000000e+00
## urban:train 0.797131578 0.120739176 6.602096 4.053868e-11
## urban:air 0.445472634 0.082160945 5.421951 5.895197e-08
## income:train -0.012597857 0.003994180 -3.154053 1.610196e-03
## income:air 0.018859983 0.003215926 5.864558 4.503294e-09
## sp.train 1.237182865 0.110460959 11.200182 0.000000e+00
## sp.air 0.540323852 0.111835294 4.831425 1.355592e-06
The two supplementary coefficients are sp.train and sp.air. The student statistics reported are irrelevant because they test the hypothesis that these parameters are 0, as the relevant hypothesis of
homoscedasticity is that both of them equal one. The heteroscedastic logit being nested in the multinomial logit model, we can first use the three classical tests: the Wald test (based on the
unconstrained model hl), the score test (based on the constrained model ml) and the likelihood ratio model (based on the comparison of both models).
To perform the Wald test, we use lmtest::waldtest, for which a special method is provided by the mlogit package. The arguments are the unconstrained model (hl) and the update that should be used in
order to get the constrained model (heterosc = FALSE). To compute the scoretest, we use mlogit::scoretest, for which the arguments are the constrained model (ml) and the update that should be used in
order to get the unconstrained model (heterosc = TRUE). Finally, the likelihood ratio test is performed using lmtest::lrtest.
## Wald test
## data: homoscedasticity
## chisq = 25.196, df = 2, p-value = 3.38e-06
## score test
## data: heterosc = TRUE
## chisq = 9.4883, df = 2, p-value = 0.008703
## alternative hypothesis: heteroscedastic model
## Likelihood ratio test
## Model 1: choice ~ freq + cost + ivt + ovt | urban + income
## Model 2: choice ~ freq + cost + ivt + ovt | urban + income
## #Df LogLik Df Chisq Pr(>Chisq)
## 1 12 -1838.1
## 2 10 -1841.6 -2 6.8882 0.03193 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The three statistics are \(\chi ^2\) with two degrees of freedom under the null hypothesis of homoscedascticity. The three tests reject the null hypothesis at the 5% level, and even at the 1% level
for the Wald and the score test. These three tests relies on the hypothesis that, under the null, the constrained model is the true model. We can get rid of this hypothesis using a Vuong test. Note
the use of the nested argument that is set to TRUE:
## Non-degenerate Vuong test for nested models
## data: hl-ml
## z = 0.4554, vuong_stat = 6.888241, vuong_p.value = 0.047927, p-value =
## 0.211
## alternative hypothesis: different models
The homoscedasticity hypothesis is still rejected at the 5% level for the classical Vuong test (the p-value is 0.048), but it is not using the non-degenerate Vuong test (p-value of 0.211).
Transport mode choice (overlapping models)
We consider finally another dataset from mlogit called RiskyTransport, that has been used by (???) and concerns the choice of one mode (among water-taxi, ferry, hovercraft and helicopter) for trips
from Sierra Leone’s international airport to downtown Freetown.
RT <- mlogit.data(RiskyTransport, shape = "long", choice = "choice",
chid.var = "chid", alt.var = "mode", id.var = "id")
We estimate models with only one covariate, the generalized cost of the mode. We estimate three models: the basic multinomial logit model, the heteroscedastic model and a nested model where
alternatives are grouped in two nests according to the fact that they are fast or slow modes.
ml <- mlogit(choice ~ cost, data = RT)
hl <- mlogit(choice ~ cost , data = RT, heterosc = TRUE)
nl <- mlogit(formula = choice ~ cost, data = RT,
nests = list(fast = c("Helicopter", "Hovercraft"),
slow = c("WaterTaxi", "Ferry")),
un.nest.el = TRUE)
modelsummary::msummary(list(multinomial = ml, heteroscedastic = hl, nested = nl))
multinomial heteroscedastic nested
(Intercept) × WaterTaxi 1.754 1.955 1.581
(0.159) (0.224) (0.127)
(Intercept) × Ferry 1.445 0.417 1.359
(0.168) (0.401) (0.118)
(Intercept) × Hovercraft 0.844 −3.666 0.461
(0.167) (2.511) (0.148)
cost −0.009 −0.022 −0.007
(0.001) (0.004) (0.001)
sp.WaterTaxi 0.745
sp.Ferry 2.840
sp.Hovercraft 4.551
iv 0.554
Num.Obs. 7172 7172 7172
AIC 3344.2 3300.7 3330.4
RMSE 0.63 0.63 0.63
mcfadden’s r2 0.150521340495322 0.163130381343293 0.154549418820887
Compared to ne multinomial model, the heteroscedastic model has 3 supplementary coefficients (the scale parameters for 3 modes, the one for the reference mode being set to 1) and the nested logit
model has one supplementary parameter which is the nest elasticity (iv in the table). Both models reduce to the multinomial logit model if:
• sp.WaterTaxi = sp.Ferry = sp.Hovercraft = 1 for the heteroscedastic model,
• iv = 1 for the nested logit model.
Therefore, the two models are over-lapping, as they reduce to the same model (the multinomial logit model) for some values of the parameters.
The first step of the test is the variance test. It can be performed using ndvuong by setting the argument vartest to TRUE:
## variance test
## data: nl-hl
## w2 = 0.047327, p-value < 2.2e-16
## alternative hypothesis: positive variance
The null hypothesis that \(\omega^2 = 0\) is rejected. We can then proceed to the second step, which is the test for non-nested models.
## Non-degenerate Vuong test for non-nested models
## data: hl-nl
## z = 1.7298, vuong_stat = 1.829208, constant = 0.000000, sum e.v. =
## -1.832021, vuong_p.value = 0.033684, p-value = 0.0975
## alternative hypothesis: different models
The classical and the non-degenerate Vuong tests both conclude that the two models are equivalent at the 5% level, but that the heteroscedastic model is better than the nested logit model at the 10%
Coate, Stephen, and Michael Conlin. 2004. “A Group Rule-Utilitarian Approach to Voter Turnout: Theory and Evidence.” American Economic Review 94 (5): 1476–1504.
Shi, Xiaoxia. 2015. “A Nondegenerate Vuong Test.” Quantitative Economics, 85–121.
Vuong, Quang H. 1989. “Likelihood Ratio Tests for Selection and Non-Nested Hypotheses.” Econometrica 57 (2): 397–33.
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Anthropometry for Paraplegics
Thank you very much to all those who replied to the request I posted (twice
somehow) for help with estimating anthropometric parameters for paraplegics.
While I am grateful for the assistance, I don't think we've got to the heart of
the question yet. I wish to be able to calculate parameters for a subjects who
clearly do not fit normative tables. Surely the same problem has been
approached by every researcher who has performed kinetic analyses. How do
people studying the kinetics of weightlifting derive these parameters? The
standard tables must be nearly bad for these people as they would be for
paraplegics. We need a method to CALCULATE values for any given population. The
geometric methods described below will assist, but they cannot provide the
whole solution. I would welcome a reply from people from people doing all kinds
of kinetic research to know how they get parameters for any non-standard
Thank you once again to all those who replied before. I have included all your
responses below.
Peter Sinclair
The University of Sydney
From: SMTP%"IVMEMOL@HDETUD2.TUDELFT.NL" 27-JAN-1993 18:56:05.01
Maybe I can help tomorrow; together with Douglas Hobson I analyzed
anthropometric data from 122 people with cerebral palsy. He brought
the dataset with him from Memphis Tennessee when he vistited our
lab for his sabattical.
We used regression equations of Clauser, CE, JT McConville and JW Young
Weight, volume and center of mass of segments of human body
Wright-Patterson Air Force Base Ohio(1969) AMRL-TR-69-70
for example Center of Mass of Tibia:
0.309 * tibial height - 0.558 * knee breadth +5,786 cm
The data about the CP-sample is published as:
Hobson, DA and JFM Molenbroek
Anthropometry and Design for the disabled: Experiences with seating
design for cerebral palsy population
Applied Ergonomics 21(1990)1,43-54
With regards
Johan FM Molenbroek
Delft University of Technology
The Netherlands
From: SMTP%"GA4020@SIUCVMB.SIU.EDU" 28-JAN-1993 02:20:32.14
-although not directly designed for paraplegics, you might use part of
the Hanavan body model to predict the location of segmental mass
centers. The input data are radii of the distal and proximal ends of
the segment and the segment weight (which of course you have to derive
from some other means). The output will also include the moment of
inertia around the mass center.
The reference is: Hanavan, E.P. (1964) A mathematical model of the
human body. AMRL Technical Documentary Report 64-102, Wright-Patterson
Air Force Base.
-The hanavan body model is a set of equations that predict the moment of
inertia and location of the center of mass of the frustum of a right
circular cone (the base of the cone with the tip cut off). The basic
assumption of the model is that the segment has a constant density
throughout the length of the segment. The application of this model to
a human body segment also assumes the segment to be shaped like the
frustum of a cone. Namely, for any of the extremity segments, the wider
proximal end is the base of the cone and the narrower distal end is the
top of the cone.
-The input parameters to the model are: segment length, segment weight,
segment mass, radius of proximal end, and radius of distal end.
We obtain these values from:
segment length - position data of segment in space
segment mass - we use the values from Dempster predicting each segment
mass from total body mass
segment weight - mass * accel of gravity
radii - we measure the proximal and distal circumferences of each
on each subject and calculate radii.
-The equations are as follows. The variable names come directly from
hanavan. R is the proximal radius, RR the distal radius (R > RR).
SL, SM, SW are segment length, mass, and weight.
Eta is location of mass center expressed as a ratio of the length from
proximal end to mass center and the segment length (eg Eta = .5 is mass
center at mid point, Eta < .5 is mass center closer to proximal end).
Seg In is the moment of inertia in SI units.
Delta = (3 * SW) / (SL * (R^2 + R * RR + RR^2) * 3.14159)
Mu = RR / R
Sigma = 1 + Mu + Mu^2
Eta = (1 + 2 * Mu + 3 * Mu^2) / (4 * Sigma)
Aa = (9 / (20 * 3.14159)) * ((1 + Mu^2 + Mu^3 + Mu^4) / Sigma^2)
Bb = (3 / 80) * ((1 +4 * Mu + 10 * Mu^2 + 4 * Mu ^3 + Mu^4) /
Seg In = (Aa * SM^2) / (Delta * SL) + (Bb * SM * SL^2)
Here is some test data for you to check the equations:
Total body mass = 83.4
SL = 0.353
SM = 8.34 ( this was a sub's thigh = 0.10 * body mass)
SW = 81.8154
R = 0.1003
RR = 0.0653
Delta = 10614.2246
Mu = 0.6508
Sigma = 2.0743
Eta = 0.4305 (notice how similar to other predictions of CM location)
Aa = 0.0625
Bb = 0.0795
Seg In = 0.0837 (a very reasonable number)
Good luck with this, Peter. If you have trouble let me know.
Paul DeVita
email: ga4020@siucvmb.siu.edu
From: SMTP%"GA4020@SIUCVMB.SIU.EDU" 30-JAN-1993 03:17:16.44
From: SMTP%"blacknl@tuns.ca" 28-JAN-1993 05:51:50.48
I am doing some research work in structural and functional anthropometry
of wheelchair mobile paraplegics. I attempted to perform some
work using the methods decribed by Jensen (1979), J. of Biomechanics.
However, I found it difficult to get subjects to volunteer for the
slides necessary for his method. Please keep me posted of your
John Kozey
ps I am using the account of Nancy Black for this letter. Please respond
to her account.
From: SMTP%"MICHEL@physocc.lan.mcgill.ca" 28-JAN-1993 09:10:21.51
Our laboratory is also currently facing the problem of getting
good anthropometrical estimates of paraplegics and paraparetics.
We are working presently on using Hatze's equations even
though the density of the segments will have errors.
We would appreciate it very much if you could send us the replies
you will get with your query so that we could improve our kinetic
Thank you for your time.
Michel Ladouceur, M.Sc.
Human Gait Laboratory
School of P. & O. T.
McGill University
Montreal, Canada
e-mail: michel@physocc.lan.mcgill.can
From: SMTP%"steiner@clio.rz.uni-duesseldorf.de" 28-JAN-1993 21:11:09.93
I had intended to post a similar call for help, but know I am lazy enough
to ask you to summarize the answers you get and either post them on BIOMCH-L
or send them to me.
We are working on FES for paraplegics and need the data to feed our dynamical
simulation program.
Thank you for your kind help
Rene Steiner
Neurologisches Therapiecentrum
Hohensandweg 37
D-4000 Duesseldorf 13
From: SMTP%"E_DOW@uvmvax.uvm.edu" 2-FEB-1993 04:05:21.83
I wonder why you want to do this? There is some school of thought that
anthropometrics is not everything it has been cracked up to be! It
certainly helps in defining the range of values for design purposes
but is dangerous if you expect to design for the "average" person. I
find it interesting to think in terms of an "average para's" legs.
Of what use would it be considering that the legs are non-functional?
I think you might be hard pressed to find the data you are looking for
considering how different every para is....even more so than able-bodied
Good luck!..
Jerry Weisman
Vermont Rehab Engineering Center
University of Vermont
From: SMTP%"marko@robo.fer.uni-lj.si" 4-FEB-1993 01:59:59.57
All anthropometric variables you want: mass, location of mass
centre can be foud in Winter DA, Biomechanics of human movement,
John Willey&Sons, New York 1979.
I modelled shank and thigh as truncated cone with bigger
and smaller radius. This is not published yet, but:
from body mass and hight and according to Winter body density,
segment densities and segment masses are determined. From mass and
density segment volume can be found. You can also express mass
location in terms of r1 and r2 of cone. And you can express volume
in terms of r1 and r2. Both nonlinear equations include two radius
r1 and r2 and can be solved with appropriate numeric method
(Newton-Raphson). Works.
I don't have experience with CT scans, but for body density
non CT scan measurements would only count.
With best regards,
Marko Munih
Marko Munih, M. Sc. Faculty of El. & Comp. Eng.
Teaching Assistant Trzaska 25, 61000 Ljubljana, Slovenia
marko@robo.fer.uni-lj.si tel.: +386 1 265 161
fax.: +386 1 264 990
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How do I use Mathematica to solve this problem?
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How do I use Mathematica to solve this problem?
Consider a large Ferris wheel above a lake. The wheel is 30 meters in radius and its center stands 80 meters above the lake level. At t = 0, a stunt person stands on the top of the Ferris wheel
(theta=0) which is rotating at a constant angular velocity w = 0.2 rad/s. At t = 0, a rescue boat is 150 m from the vertical center line of the Ferris wheel and travels toward the base of the wheel
at a constant speed of 10 m/s.(In other words, if the center of the wheel has coordinates (0, 80) and the initial coordinates of the person are (0, 110), the initial position of the front of the boat
is (150, 0)). Assume the person has no initial velocity other than that of the rotating wheel; assume also that there are no sources of friction in this problem. Assume further that the boat is one
meter in length and the long axis of the boat is moving directly toward the Ferris wheel. The Ferris wheel is rotating toward the incoming boat.Your program will allow you to determine when should
the stunt person step off the Ferris wheel to safely land in the boat as it speeds by. At what angle (with respect to the vertical) should the person step off to accomplish this?
This is my code. Whenever I test values of theta, I do not get the correct answer. Here are some relevant equations for this problem:
My code:
Clear[g, \[Omega], R, H, h, Vb, \[Theta], Pxo, Pyo, Vxo, Vyo, Ta, Tw, \ Ttotal, Bx, P, nterms]
g = 9.8; \[Omega] = .2; R = 30; H = 80; h = .01; Vb = 10;
\[Theta][n_] := n*h;
Pxo[n_] := Px[n] = R*Sin[\[Theta][n]];
Pyo[n_] := Pyo[n] = H + R*Cos[\[Theta][n]];
Vxo[n_] := Vxo[n] = \[Omega]*R*Cos[\[Theta][n]];
Vyo[n_] := Vyo[n] = -\[Omega]*R*Sin[\[Theta][n]];
Ta[n_] := (Vyo[n] + Sqrt[Vyo[n]^2 + 2*g*Pyo[n]])/g;
Tw[n_] := \[Theta][n]/\[Omega];
Ttotal[n_] := Tw[n] + Ta[n];
Bx[n_] := Bx[n] = 150 - (Vb*Ttotal[n]);
P[n_] := P[n] = Vxo[n]*Ttotal[n] + Pxo[n];
nterms = Catch[Do[If[(Bx[n] - P[n]) < 0, Throw[n]], {n, 1000}]; Throw[0]];
Print["Jump at ", (\[Theta][nterms]) (180/\[Pi]), \[Degree]]
1 Reply
I'd say your solution looks like 'traditional programming.' No doubt it can be easily troubleshooted to give roughly the right answer.
There is a better way. For a system this simple we should be able to write down everything we know, and then let the algebra sort itself out. Doing this by hand we'd skip easy simplifications. But
you have a computer, so put things in full form with intelligible variable names.
This problem is really a lot more interesting with different rotation rates. I initially solved for a rotation rate of $2\pi/10$ which has three solutions. Notably the possibility of multiple
solutions makes a lot of the assumptions in the linked pdf $extremely \ dubious!$
See attached, solution for Jump time k and splash time t. (error in notebook, conversion of time to radians to degrees is flawed, should use 36 roots / pi, yeilding 110.2 degrees.)
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Math, Grade 6, Putting Math to Work, Project Rubric & Relevant Criteria
Project Work
Work Time
Project Work
Work on your project with your project group.
Part of your work today may involve planning, making calculations, and/or finding needed information.
When your presentation is finished, it must contain:
• A written explanation of the mathematics in your project
• Accurate representations (such as graphs, tables, and/or diagrams) of the mathematics in your project
• Use the project rubric to evaluate your project in its current state to make sure you are on the right track.
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How much is 1 Lac?
How much is 1 Lac?
As per the Indian numbering system, one lakh is a unit which is equal to 100,000 (one hundred thousand) in the international unit system.
How can I write 1 lakh in English?
A lakh (/læk, lɑːk/; abbreviated L; sometimes written lac) is a unit in the Indian numbering system equal to one hundred thousand (100,000; scientific notation: 105). In the Indian 2,2,3 convention
of digit grouping, it is written as 1,00,000.
How many dollars is 5 lakhs?
Convert Indian Rupee to US Dollar
INR USD
1,000 INR 13.6289 USD
5,000 INR 68.1445 USD
10,000 INR 136.289 USD
50,000 INR 681.445 USD
How do you write 5 lakhs in a Cheque?
For 5,00,000 Rs write it as Five Lakh Rupees Only. For Rs write it as Ten Lakh Rupees Only. Which one is correct Lakh or Lac ? In general, you can use any spelling, but when you are writing a cheque
or any financial related document then write as Lakh.
How do you write 5 lakhs in numbers?
Note: We found that some people call it 5 lakhs or 5 lac, but the correct way to say it is 5 lakh (lakh without the trailing “s”).
How can I write 3 lakh in English?
Note: We found that some people call it 3 lakhs or 3 lac, but the correct way to say it is 3 lakh (lakh without the trailing “s”).
How can I make 1 billion?
1,(one billion, short scale; one thousand million or milliard, yard, long scale) is the natural number following and preceding 1, One billion can also be written as b or bn. In standard form, it is
written as 1 × 109.
How many zeros does 80 Lacs have?
How do you write 80 lakhs in numbers?
1,80,00,000 As you can see, 1 crore and 80 lakh is the same as 18 million.
What does 1 crore weigh?
In One Crore Indian Rupees, there will be 5,000 notes or 50 numbers of 100 x Rs. 2000 packs. Now, for 1 Crore = 5000 x 1.006 grams i.e. 5 Kg and 30 grams.
How many millions is equal to 1 billion?
1000 millions
What is the period in a number?
When a number is written in standard form, each group of digits separated by a comma is called a period . The number 5,has four periods.
How do you write 12 lakhs in numbers?
Note: We found that some people call it 12 lakhs or 12 lac, but the correct way to say it is 12 lakh (lakh without the trailing “s”).
How many lakhs are there in a million?
ten lakhs
Do you spell out numbers in an essay?
It is generally best to write out numbers from zero to one hundred in nontechnical writing. In scientific and technical writing, the prevailing style is to write out numbers under ten.
When should you write out numbers in an essay?
Writing Small and Large Numbers A simple rule for using numbers in writing is that small numbers ranging from one to ten (or one to nine, depending on the style guide) should generally be spelled
out. Larger numbers (i.e., above ten) are written as numerals.
How much is 1.3 billion in millions?
0.8 × 1000 = 800 million
Billion to Million
Billion Million
1 Billion 1000 Million
1.2 Billion 1200 Million
1.3 Billion 1300 Million
How do you write 70 lakhs in numbers?
1,70,00,000 As you can see, 1 crore and 70 lakh is the same as 17 million.
How do you write 4 lakhs in numbers?
Note: We found that some people call it 4 lakhs or 4 lac, but the correct way to say it is 4 lakh (lakh without the trailing “s”).
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Lumber Calculator
Lumber: How to Calculate Board Feet Needed and Cost
Understanding how to calculate board feet is essential for anyone involved in construction, woodworking, or carpentry. Whether you’re planning a DIY project, managing a construction site, or running
a lumber business, accurately determining the amount of lumber needed and its cost ensures efficiency and budget adherence. This comprehensive guide delves into the fundamentals of board feet
calculations, explores the board feet formula, provides step-by-step methods for calculating board feet and cost, offers detailed example problems, and highlights practical applications to enhance
your proficiency in lumber management.
Understanding Lumber Calculations
Lumber calculations involve determining the volume of wood required for a project, typically measured in board feet. A board foot is a standard unit of measure in the lumber industry, representing a
volume of 144 cubic inches (1 foot long by 1 foot wide by 1 inch thick). Mastering lumber calculations ensures that you purchase the right amount of material, avoid waste, and stay within budget
At the core of lumber calculations are the measurements of the wood’s dimensions: length, width, and thickness. By accurately measuring these dimensions and applying the board feet formula, you can
determine the total volume of lumber needed for your project. Additionally, understanding how to calculate the cost based on board feet helps in effective budgeting and financial planning.
The Board Feet Formula
The board feet formula is a mathematical equation used to calculate the volume of lumber in board feet. It takes into account the dimensions of the wood pieces to provide an accurate measurement of
the total volume required.
Board Feet (BF) = (Length (ft) × Width (in) × Thickness (in)) / 12
• Length (ft) = The length of the lumber in feet.
• Width (in) = The width of the lumber in inches.
• Thickness (in) = The thickness of the lumber in inches.
This formula calculates the volume of lumber in board feet by converting the measurements into consistent units and applying the appropriate mathematical operations. Understanding and applying this
formula is crucial for accurate lumber estimation and cost calculation.
How to Calculate Board Feet
Calculating board feet involves a straightforward application of the board feet formula. Follow these steps to accurately determine the amount of lumber needed for your project:
1. Measure the Lumber: Determine the length, width, and thickness of each piece of lumber you intend to use.
2. Convert Units if Necessary: Ensure all measurements are in the correct units as per the board feet formula (length in feet, width and thickness in inches).
3. Apply the Board Feet Formula: Substitute the measurements into the formula: BF = (Length × Width × Thickness) / 12.
4. Calculate the Board Feet: Perform the multiplication and division to find the board feet for each piece.
5. Sum the Board Feet: Add up the board feet of all individual pieces to find the total board feet needed for your project.
By following these steps, you can accurately calculate the total board feet required, ensuring you purchase the right amount of lumber for your project.
Calculating Lumber Cost
Once you’ve determined the total board feet needed, the next step is to calculate the cost of the lumber. This involves knowing the price per board foot and multiplying it by the total board feet
Total Cost = Board Feet (BF) × Price per Board Foot
• Board Feet (BF) = The total board feet calculated for your project.
• Price per Board Foot = The cost of one board foot of the lumber you’re purchasing.
Understanding how to calculate lumber cost allows you to budget effectively, compare prices from different suppliers, and make informed purchasing decisions.
Common Lumber Calculation Examples
To reinforce your understanding of lumber calculations, let’s explore some common examples and scenarios where precise measurements are essential.
1. Calculating Board Feet for a Deck
Suppose you are building a deck that requires 50 boards. Each board is 10 feet long, 6 inches wide, and 2 inches thick. How many board feet are needed?
1. Identify the measurements: Length = 10 ft, Width = 6 in, Thickness = 2 in.
2. Apply the board feet formula: BF = (10 × 6 × 2) / 12 = 120 / 12 = 10 BF per board.
3. Calculate total board feet: 10 BF × 50 boards = 500 BF.
4. Result: 500 board feet are needed.
2. Determining Lumber Cost for a Shed
Imagine you need to build a shed that requires 300 board feet of lumber. The price per board foot is $3.50. What is the total cost?
1. Identify the total board feet: 300 BF.
2. Identify the price per board foot: $3.50.
3. Apply the cost formula: Total Cost = 300 × 3.50 = $1,050.
4. Result: The total cost is $1,050.
3. Calculating Board Feet for a Furniture Project
You’re crafting a custom table that requires the following lumber pieces:
• Top: 5 ft long, 12 in wide, 1 in thick.
• Legs (4): 3 ft long, 4 in wide, 4 in thick.
Calculate the total board feet needed.
1. Calculate board feet for the top: BF = (5 × 12 × 1) / 12 = 60 / 12 = 5 BF.
2. Calculate board feet for one leg: BF = (3 × 4 × 4) / 12 = 48 / 12 = 4 BF.
3. Calculate board feet for four legs: 4 BF × 4 legs = 16 BF.
4. Calculate total board feet: 5 BF + 16 BF = 21 BF.
5. Result: 21 board feet are needed.
Practical Applications of Lumber Calculations
Accurate lumber calculations are vital in numerous fields. Understanding how to calculate board feet and cost can enhance efficiency and precision in multiple contexts:
1. Construction and Building Projects
Builders and contractors use lumber calculations to estimate the amount of wood needed for structures like decks, sheds, and framing. Accurate estimates prevent over-purchasing and minimize waste.
2. Carpentry and Woodworking
Carpenters and woodworkers calculate board feet to determine the materials required for furniture, cabinetry, and other wood products. Precise calculations ensure cost-effective and high-quality
3. Lumber Retail and Wholesale
Lumber retailers and wholesalers use board feet calculations to price their products accurately, manage inventory, and meet customer demands efficiently.
4. DIY Home Improvement
Homeowners engaged in DIY projects use lumber calculations to plan and budget their projects effectively, ensuring they purchase the right amount of materials without unnecessary expenditure.
5. Landscaping and Outdoor Structures
Landscapers and outdoor structure builders calculate board feet for projects like pergolas, fences, and garden structures, ensuring durability and aesthetic appeal.
6. Industrial Applications
Industries that utilize wood products for manufacturing or packaging rely on accurate lumber calculations to optimize production processes and control costs.
7. Educational and Training Purposes
Educational institutions and training programs teach lumber calculations to students and apprentices, equipping them with essential skills for their careers in construction and woodworking.
Additional Example Problems
Problem 1: Finding Board Feet for a Roof Framework
Question: A roof framework requires 25 pieces of lumber, each 12 feet long, 5 inches wide, and 2 inches thick. How many board feet are needed?
1. Identify the measurements: Length = 12 ft, Width = 5 in, Thickness = 2 in.
2. Apply the board feet formula: BF = (12 × 5 × 2) / 12 = 120 / 12 = 10 BF per piece.
3. Calculate total board feet: 10 BF × 25 pieces = 250 BF.
4. Result: 250 board feet are needed.
Problem 2: Calculating Lumber Cost for a Pergola
Question: You need 150 board feet of lumber for a pergola project. If the price per board foot is $4.25, what is the total cost?
1. Identify the total board feet: 150 BF.
2. Identify the price per board foot: $4.25.
3. Apply the cost formula: Total Cost = 150 × 4.25 = $637.50.
4. Result: The total cost is $637.50.
Problem 3: Determining Board Feet for a Custom Bookshelf
Question: A custom bookshelf requires 3 shelves, each measuring 6 feet long, 10 inches wide, and 1.5 inches thick. Calculate the total board feet needed.
1. Identify the measurements: Length = 6 ft, Width = 10 in, Thickness = 1.5 in.
2. Apply the board feet formula for one shelf: BF = (6 × 10 × 1.5) / 12 = 90 / 12 = 7.5 BF.
3. Calculate total board feet: 7.5 BF × 3 shelves = 22.5 BF.
4. Result: 22.5 board feet are needed.
Problem 4: Calculating Cost for Multiple Lumber Pieces
Question: You need 400 board feet of lumber for a construction project. The lumber costs $3.75 per board foot. What is the total cost?
1. Identify the total board feet: 400 BF.
2. Identify the price per board foot: $3.75.
3. Apply the cost formula: Total Cost = 400 × 3.75 = $1,500.
4. Result: The total cost is $1,500.
Tips for Effective Lumber Calculations
• Double-Check Measurements: Ensure all measurements are accurate to prevent over-purchasing or material shortages.
• Understand Lumber Grades: Different grades affect the price per board foot. Choose the appropriate grade for your project needs.
• Factor in Waste: Include an additional percentage of lumber to account for waste, cuts, and errors.
• Compare Prices: Shop around and compare prices from different suppliers to get the best deal on lumber.
• Use Reliable Tools: Utilize online board feet calculators and measurement tools to enhance accuracy.
• Plan Ahead: Thoroughly plan your project to determine the exact amount of lumber needed, reducing unnecessary costs.
• Stay Consistent with Units: Ensure all measurements are in the same units before performing calculations to maintain consistency.
• Consult Building Codes: Adhere to local building codes and regulations regarding lumber usage and dimensions.
Mastering lumber calculations is a crucial skill for anyone involved in construction, woodworking, or carpentry. Whether you’re undertaking a DIY project, managing a construction site, or running a
lumber business, accurately determining board feet needed and cost ensures efficiency, budget adherence, and material optimization. By understanding the board feet formula, practicing with detailed
example problems, and applying practical tips, you can enhance your proficiency in lumber management. Leveraging reliable calculation tools and maintaining a consistent approach to measurements will
empower you to execute projects with confidence and precision, minimizing waste and maximizing cost-effectiveness.
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Limitations of the Central Restrict Theorem – Watts Up With That? - DAILY WOWLimitations of the Central Restrict Theorem – Watts Up With That?
Guest Essay by Kip Hansen — 17 December 2022
The Central Limit Theorem is particularly good and valuable especially when have many measurements that have slightly different results. Say, for instance, you wanted to know very precisely the
length of a particular stainless-steel rod. You measure it and get 502 mm. You expected 500 mm. So you measure it again: 498 mm. And again and again: 499, 501. You check the conditions:
temperature the same each time? You get a better, more precise ruler. Measure again: 499.5 and again 500.2 and again 499.9 — one hundred times you measure. You can’t seem to get exactly the same
result. Now you can use the Central Limit Theory (hereafter CLT) to good result. Throw your 108 measurements into a distribution chart or CLT calculator and you’ll see your central value very darned
close to 500 mm and you’ll have an idea of the variation in measurements.
While the Law of Large Numbers is based on repeating the same experiment, or measurement, many times, thus could be depended on in this exact instance, the CLT only requires a largish population
(overall data set) and the taking of the means of many samples of that data set.
It would take another post (possibly a book) to explain the all the benefits and limitations of the Central Limit Theory (CLT), but I will use a few examples to introduce that topic.
Example 1:
You take 100 measurements of the diameter of ball bearings produced by a machine on the same day. You can calculate the mean and can estimate a variance in the data. But you want a better idea, so
you realize that you have 100 measurements from each Friday for the past year. 50 data sets of 100 measurements, which if sampled would give you fifty samples out of 306 possible daily samples of
the total 3,060 measurements if you had 100 samples for every work day (six days a week, 51 weeks).
The central limit theory is about probability. It will tell you what the most likely (probable) mean diameter is of all your ball bearings produced on that machine. But, if you are presented with
only the mean and the SD, and not the full distribution, it will tell you very little about how many ball bearings are within specification and thus have value to the company. The CLT can not tell
you how many or what percentage of the ball bearings would have been within the specifications (if measured when produced) and how many outside spec (and thus useless). Oh, the Standard Deviation
will not tell you either — it is not a measurement or quantity, it is a creature of probability.
Example 2:
The Khan Academy gives a fine example of the limitations of the Central Limit Theorem (albeit, not intentionally) in the following example (watch the YouTube if you like, about ten minutes) :
The image is the distribution diagram for our oddly loaded die (one of a pair of dice). It is loaded to come up 1 or 6, or 3 or 4, but never 2 or 5. But twice more likely to come 1 or 6 than 3 or
4. The image shows a diagram of expected distribution of the results of many rolls with the ratios of two 1s, one 3, one 4, and two 6s. Taking the means of random samples of this distribution out of
1000 rolls (technically, “the sampling distribution for the sample mean”), say samples of twenty rolls repeatedly, will eventually lead to a “normal distribution” with a fairly clearly visible
(calculable) mean and SD.
Here, relying on the Central Limit Theorem, we return a mean of ≈3.5 (with some standard deviation).(We take “the mean of this sampling distribution” – the mean of means, an average of averages).
Now, if we take a fair die (one not loaded) and do the same thing, we will get the same mean of 3.5 (with some standard deviation).
Note: These distributions of frequencies of the sampled means are from 1000 random rolls (in Excel, using fx=RANDBETWEEN(1,6) – that for the loaded die was modified as required) and sampled every 25
rolls. Had we sampled a data set of 10,000 random rolls, the central limit would narrow and the mean of the sampled means — 3.5 —would become more distinct.
The Central Limit Theorem works exactly as claimed. If one collects enough samples (randomly selected data) from a population (or dataset…) and finds the means of those samples, the means will tend
towards a standard or normal distribution – as we see in the charts above – the values of the means tend towards the (in this case known) true mean. In man-on-the-street language, the means are
clumping in the center around the value of the mean at 3.5, making the characteristic “hump” of a Normal Distribution. Remember, this resulting mean is really the “mean of the sampled means”.
So, our fair die and our loaded die both produce approximate normal distributions when testing a 1000 random roll data set and sampling means. The distribution of the mean would improve – get closer
to the known mean – if we had ten or one hundred times more of the random rolls and equally larger number of samples. Both the fair and loaded die have the same mean (though slightly different
variance or deviation). I say “known mean” because we can, in this case, know the mean by straight-forward calculation, we have all the data points of the population and know the mean of the
real-world distribution of the dies themselves.
In this setting, this is a true but almost totally useless result. Any high school math nerd could have just looked at the dies, maybe made a few rolls with each, and told you the same: the range
of values is 1 through 6; the width of the range is 5; the mean of the range is 2.5 + 1 = 3.5. There is nothing more to discover by using the Central Limit Theorem against a data base of 1000 rolls
of the one die – though it will also tell you the approximate Standard Deviation – which is also almost entirely useless.
Why do I say useless? Because context is important. Dice are used for games involving chance (well, more properly, probability) in which it is assumed that the sides of the dice that land facing up
do so randomly. Further, each roll of a die or pair of dice is totally independent of any previous rolls.
Impermissible Values
As with all averages of every type, the means are just numbers. They may or not have physically sensible meanings.
One simple example is that a single die will never ever come up at the mean value of 3.5. The mean is correct but is not a possible (permissible) value for the roll of one die – never in a million
Our loaded die can only roll: 1, 3, 4 or 6. Our fair die can only roll 1, 2, 3, 4, 5 or 6. There just is no 3.5.
This is so basic and so universal that many will object to it as nonsense. But there are many physical metrics that have impermissible values. The classic and tired old cliché is the average number
of children being 2.4. And we all know why, there are no “.4” children in any family – children come in whole numbers only.
However, if for some reason you want or need an approximate, statistically-derived mean for your intended purpose, then using the principles of the CLT is your ticket. Remember, to get a true mean
of a set of values, one must add all the values together divide by the number of values.
The Central Limit Theorem method does not reduce uncertainty:
There is a common pretense (def: “Something imagined or pretended“) used often in science today, which treats a data set (all the measurements) as a sample, then take samples of the sample, use a CLT
calculator, and call the result a truer mean than the mean of the actual measurements. Not only “truer”, but more precise. However, while the CLT value achieved may have small standard deviations,
that fact is not the same as more accuracy of the measurements or less uncertainty regarding what the actual mean of the data set would be. If the data set is made up of uncertain measurements, then
the true mean will be uncertain to the same degree.
Distribution of Values May be More Important
The Central Limit Theory-provided mean would be of no use whatever when considering the use of this loaded die in gambling. Why? … because the gambler wants to know how many times in a dozen
die-rolls he can expect to get a “6”, or if rolling a pair of loaded dice, maybe a “7” or “11”. How much of an edge over the other gamblers does he gain if he introduces the loaded dice into the
game when it’s his roll?
(BTW: I was once a semi-professional stage magician, and I assure you, introducing a pair of loaded dice is easy on stage or in a street game with all its distractions but nearly impossible in a
Let’s see this in frequency distributions of rolls of our dice, rolling just one die, fair and loaded (1000 simulated random rolls in Excel):
And if we are using a pair of fair or loaded dice (many games use two dice):
On the left, fair dice return more sevens than any other value. You can see this is tending towards the mean (of two dice) as expected. Two 1’s or two 6’s are rare for fair dice … as there is only
a single unique combination each for the combined values of 2 and 12. Lots of ways to get a 7.
Our loaded dice return even more 7’s. In fact, over twice as many 7’s as any other number, almost 1-in-3 rolls. Also, the loaded dice have a much better chance of rolling 2 or 12, five times
better than with fair dice. The loaded dice don’t ever return 3 or 11.
Now here we see that if we depended on the statistical (CLT) central value of the means of rolls to prove the dice were fair (which, remember is 3.5 for both fair and loaded dice) we have made a
fatal error. The house (the casino itself) expects the distribution on the left from a pair of fair dice and thus the sets the rules to give the house a small percentage in its favor.
The gambler needs the actual distribution probability of the values of the rolls to make betting decisions.
If there are any dicing gamblers reading, please explain to non-gamblers in comments what an advantage this would be.
Finding and Using Means Isn’t Always What You Want
This insistence on using means produced approximately using the Central Limit Theorem (and its returned Standard Deviations) can create non-physical and useless results when misapplied. The CLT
means could have misled us into believing that the loaded dice were fair, as they share a common mean with fair dice. But the CLT is a tool of probability and not a pragmatic tool that we can use to
predict values of measurements in the real world. The CLT does not predict or provide values – it only provides estimated means and estimated deviations from that mean and these are just numbers.
Our Khan academy teacher, almost in the hushed tones of a description of an extra-normal phenomenon, points out that taking random same-sized samples from a data set (population of collected
measurements, for instance) will also produce a Normal Distribution of the sampled sums! The triviality of this fact should be apparent – if the “sums divided by the [same] number of components”
(the means of the samples) are normally distributed then the sums of the samples must need also be normally distributed (basic algebra).
In the Real World
Whether considering gambling with dice – loaded and fair – or evaluating the usability of ball bearing from the machinery we are evaluating – we may well find the estimated means and deviations
obtained by applying the CLT are not always what we need and might even mislead us.
If we need to know which, and how many, of our ball bearings will fit the bearing races of a tractor manufacturing customer, we will need some analysis system and quality assurance tool closer to
If our gambler is going to bet his money on the throw of a pair of specially-prepared loaded dice, he needs the full potential distribution, not of the means, but the probability distribution of the
Averages or Means: One number to rule them all
Averages seem to be the sweetheart of data analysts of all stripes. Oddly enough, even when they have a complete data set like daily high tides for the year, which they could just look at visually,
they want to find the mean.
The mean water level, which happens to be 27.15 ft (rounded) does not tell us much. The Mean High Water tells us more, but not nearly as much as the simple graph of the data points. For those
unfamiliar with astronomic tides, most tides are on a ≈13 hour cycle, with a Higher High Tide (MHHW) and a less-high High Tide (MHW). That explains what seems to be two traces above.
Note: the data points are actually a time series of a small part of a cycle, we are pulling out the set of the two higher points and the two lower points in a graph like this. One can see the
usefulness of a different plotting above each visually revealing more data than the other.
When launching my sailboat at a boat ramp near the station, the graph of actual high tide’s data points shows me that I need to catch the higher of the two high tides (Higher High Water), which
sometimes gives me more than an extra two feet of water (over the mean) under the keel. If I used the mean and attempted to launch on the lower of the two high tides (High Water), I could find
myself with a whole foot less water than I expected and if I had arrived with the boat expecting to pull it out with the boat trailer at the wrong point of the tide cycle, I could find five feet less
water than at the MHHW. Far easier to put the boat in or take it out at the highest of the tides.
With this view of the tides for a month, we can see that each of the two higher tides themselves have a little harmonic cycle, up and down.
Here we have the distribution of values of the high tides. Doesn’t tell us very much – almost nothing about the tides that is numerically useful – unless of course, one only wants the means, which
would be just as easily eye-ball guessed from the charts above or this chart — we would get a vaguely useful “around 29 feet.”
In this case, we have all the data points for the high tides at this station for the month, and could just calculate the mean directly and exactly (within the limits of the measurements) if we needed
that – which I doubt would be the case. But at least we would have a true precise mean (plus the measurement uncertainty, of course) but I think we would find that in many practical senses, it is
useless – in practice, we need the whole cycle and its values and its timing.
Why One Number?
Finding means (averages) gives a one-number result. Which is oh-so–much easier to look at and easier to understand than all that messy, confusing data!
In a previous post on a related topic, one commenter suggested we could use the CLT to find “the 2021 average maximum daily temperature at some fixed spot.” When asked why one would want do to so,
the commenter replied “To tell if it is warmer regarding max temps than say 2020 or 1920, obviously.” [I particularly liked the ‘obviously’.] Now, any physicists reading here? Why does the
requested single number — “2021 average maximum daily temperature” — not tell us much of anything that resembles “if it is warmer regarding max temps than say 2020 or 1920”? If we also had a
similar single number for the “1920 average maximum daily temperature” at the same fixed spot, we would only know if our number for 2021 was higher or lower than the number for 1920. We would not
know if “it was warmer” (in regards to anything).
At the most basic level, the “average maximum daily temperature” is not a measurement of temperature or warmness at all, but rather, as the same commenter admitted, is “just a number”.
If that isn’t clear to you (and, admittedly, the relationship between temperature and “warmness” and “heat content of the air” can be tricky), you’ll have to wait for a future essay on the topic.
It might be possible to tell if there is some temperature gradient at the fixed place using a fuller temperature record for that place…but comparing one single number with another single number does
not do that.
And that is the major limitation of the Central Limit Theorem
The CLT is terrific at producing an approximate mean value of some population of data/measurements without having to directly calculate it from a full set of measurements. It gives one a SINGLE
NUMBER from a messy collection of hundreds, thousands, millions of data points. It allows one to pretend that the single number (and its variation, as SDs) faithfully represents the whole data set/
population-of-measurements. However, that is not true – it only gives the approximate mean, which is an average, and because it is an average (an estimated mean) it carries all of the limitations
and disadvantages of all other types of averages.
The CLT is a model, a method, that will produce a Mean Value from ANY large enough set of numbers – the numbers do not need to be about anything real, they can be entirely random with no validity
about anything. The CLT method pops out the estimated mean, closer and closer to a single value whenever more and more samples from the larger population are supplied it. Even when dealing with
scientific measurements, the CLT will discover a mean (that looks very precise when “the uncertainty of the mean” is attached) just as easily from sloppy measurements, from fraudulent measurements,
from copy-and-pasted findings, from “just-plain-made-up” findings, from “I generated my finding using a random number generator” findings and from findings with so much uncertainty as to hardly be
called measurements at all.
Bottom Lines:
1. Using the CLT is useful if one has a large data set (many data points) and wishes, for some reason, to find an approximate mean of the data set, then using the principles of the Central Limit
Theorem; finding the means of multiple samples from the data set, making a distribution diagram, and with enough samples, by finding the mean of the means, the CLT will point to the approximate mean,
and give an idea of the variance in the data.
2. Since the result will be a mean, an average, and an approximate mean at that, then all the caveats and cautions that apply to the use of averages apply to the result.
3. The mean found through use of the CLT cannot and will not be less uncertain than the uncertainty of the actual mean of original uncertain measurements themselves. However, it is almost
universally claimed that “the uncertainty of the mean” (really the SD or some such) thus found is many times smaller than the uncertainty of the actual mean of the original measurements (or data
points) of the data set.
This claim is a so generally accepted and firmly held as a Statisticians’ Article of Faith that many commenting below will deride the idea of its falseness and present voluminous “proofs” from their
statistical manuals to show that they such methods do reduce uncertainty.
4. When doing science and evaluating data sets, the urge to seek a “single number” to represent the large, messy, complex and complicated data sets is irresistible to many – and can lead to serious
misunderstandings and even comical errors.
5. It is almost always better to do much more nuanced evaluation of a data set than simply finding and substituting a single number — such as a mean and then pretending that that single number can
stand in for the real data.
# # # # #
Author’s Comment:
One Number to Rule Them All as a principal, go-to-first approach in science has been disastrous for reliability and trustworthiness of scientific research.
Substituting statistically-derived single numbers for actual data, even when the data itself is available and easily accessible, has been and is an endemic malpractice of today’s science.
I blame the ease of “computation without prior thought” – we all too often are looking for The Easy Way. We throw data sets at our computers filled with analysis models and statistical software
which are often barely understood and way, way too often without real thought as to the caveats, limitations and consequences of varying methodologies.
I am not the first or only one to recognize this – maybe one of the last – but the poor practices continue and doubting the validity of these practices draws criticism and attacks.
I could be wrong now, but I don’t think so! (h/t Randy Newman)
# # # # #
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ACP Atmospheric Chemistry and Physics ACP Atmos. Chem. Phys. 1680-7324 Copernicus Publications Göttingen, Germany 10.5194/acp-4-413-2004 Optimizing CO<sub>2</sub> observing networks in the presence
of model error: results from TransCom 3 Rayner R. J ^1 CSIRO Atmospheric Research, Melbourne, Australia 03 03 2004 4 2 413 421 Copyright: © 2004 R. J Rayner 2004 This work is licensed under the
Creative Commons Attribution-NonCommercial-ShareAlike 2.5 Generic License. To view a copy of this licence, visit https://creativecommons.org/licenses/by-nc-sa/2.5/ This article is available from
https://acp.copernicus.org/articles/4/413/2004/acp-4-413-2004.html The full text article is available as a PDF file from https://acp.copernicus.org/articles/4/413/2004/acp-4-413-2004.pdf
We use a genetic algorithm to construct optimal observing networks of atmospheric concentration for inverse determination of net sources. Optimal networks are those that produce a minimum in average
posterior uncertainty plus a term representing the divergence among source estimates for different transport models. The addition of this last term modifies the choice of observing sites, leading to
larger networks than would be chosen under the traditional estimated variance metric. Model-model differences behave like sub-grid heterogeneity and optimal networks try to average over some of this.
The optimization does not, however, necessarily reject apparently difficult sites to model. Although the results are so conditioned on the experimental set-up that the specific networks chosen are
unlikely to be the best choices in the real world, the counter-intuitive behaviour of the optimization suggests the model error contribution should be taken into account when designing observing
networks. Finally we compare the flux and total uncertainty estimates from the optimal network with those from the 3 control case. The 3 control case performs well under the chosen
uncertainty metric and the flux estimates are close to those from the optimal case. Thus the 3 findings would have been similar if minimizing the total uncertainty guided their network choice.
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Construct a triangle with sides 3 cm, 4 cm and 5 cm. - WorkSheets Buddy
Construct a triangle with sides 3 cm, 4 cm and 5 cm.
Construct a triangle with sides 3 cm, 4 cm and 5 cm. Draw its circumcircle and measure its radius.
Using a ruler and compasses only:
(i) Construe a triangle ABC with the following data:
Base AB = 6 cm, AC = 5.2 cm and ∠CAB = 60°.
(ii) In the same diagram, draw a circle which passes through the points A, B and C. and mark its centre O.
More Solutions:
Leave a Comment
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Total Return On Investment Formula
Total returns can be calculated as a dollar amount, or as a percentage. In other words, you can say that a stock's total return was $8 per share over a certain. Total return is a metric that
represents all returns on an investment, including capital gains and other financial rewards. To calculate ROI, you first add income received — interest or dividends — to the ending investment value.
Then, you divide this number by the beginning. Total Returns · Equity Multiple · Annualized Rate of Return (ARR) · Internal Rate of Return (IRR) · Return on Investment (ROI). The calculator covers
four different methods of calculating ROI: net income, capital gain, total return, and annualized return. ROI Calculator - Free Download.
ROI Calculation for these Examples ; Total net returns, , ; Basic ROI, ROI = / = %, ROI = / = % ; Annualized return ROI (r). Expanded ROI formula · Total revenue: The total income generated from the
investment · Total costs: All costs associated with the investment, including initial. ROI = Net Income / Cost of Investment ; ROI = Investment Gain / Investment Base ; ROI Formula: = [(Ending Value
/ Beginning Value) ^ (1 / # of Years)] – 1. The formula for ROI looks at the benefit received from an investment, divided by the initial investment cost. Ques. What is considered a good ROI? Ans. A
good. ROI = (net profit / total investment) x The net profit equals the difference between the net benefit and the net cost related to making the investment. The. Annualised return can be calculated
with the following formula: End Value – Beginning Value/Beginning Value * * (1/holding period of the investment) For. The ROI formula is the ratio between the net profit earned on an investment and
the cost of the investment, expressed as a percentage. The ROI ratio is a. ROI Calculation for these Examples ; Total net returns, , ; Basic ROI, ROI = / = %, ROI = / = % ; Annualized return ROI (r).
As evident, ROI is a value metric used to calculate an investment's success or to compare the productivity of different investments in delivering a return on. The calculation for return on investment
is also straightforward— it's determined by dividing the profit earned by the cost of the investment itself. If you. Basic formula to calculate return on investment. Working out the ROI is pretty
simple, so you don't need an ROI calculator or any other special tool to do it.
“ROI is calculated by dividing a company's net income by its total assets.” ROI formula: Net income. Total assets. X Traditionally, ROI is calculated by dividing the net income from an investment by
the original cost of the investment, the result of which is expressed as a. The basic ROI formula is: Net Profit / Total Investment * = ROI. Let's apply the formula with the help of an example. You
are a house flipper. You purchased. Total investment returns are measured for stand-alone, one-off projects. ROI is a calculation of the monetary value of an investment versus its cost. The ROI
formula is: (profit minus cost) / cost. **) The return of an investment including re-invested dividends and other values that have been separated from the share is usually called Total Return. All.
To calculate the total return over the period, divide the ending value by the beginning value and then subtract one. Return on investment (ROI) is a financial ratio expressed as a percentage, used as
a metric to evaluate investments and rank them compared to other investment. The calculator covers four different methods of calculating ROI: net income, capital gain, total return, and annualized
return. ROI Calculator - Free Download.
Return on Investment estimates the loss and gain generated on the amount of money invested. ROI (Return on Investment) is generally expressed in the percentage. Free return on investment (ROI)
calculator that returns total ROI rate and annualized ROI using either actual dates of investment or simply investment. Total rate of return The total return calculation effectively shows you the
percentage return that your original investment generated. Note that this. Return rate – For many investors, this is what matters most. · Starting amount – Sometimes called the principal, this is the
amount apparent at the inception of. calculator to ensure you have the information you need to evaluate the investment. total value that's still due on the mortgage to determine the ROI. The.
The formula for ROI looks at the benefit received from an investment, divided by the initial investment cost. Ques. What is considered a good ROI? Ans. A good.
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Hypothesis Testing Problem
Hi there!
I have a problem with stating the right Hypothesis Test for the following problem (chi-squared test):
“A study was conducted to determine whether the standard deviation of monthly maintenance costs of a Pepper III aircraft is $300.”
The correct statement is:
• H0: pop. variance = 300^2
• Ha: pop. variance =/ 300^2
However, when I read the problem, I was sure that Ha: pop. variance = 300^2. The solution says that the wording is a little tricky, but it does not explain how I am supposed to interpret it… Could
you guys explain me the right way to interpret the wording of the problem? Thank you so much!
The null hypothesis always includes the equal sign; the alternative hypothesis does not. If you’re testing whether the standard deviation of something is equal to a given value, then H0 is σ = k and
Ha is σ ≠ k.
Of course! I didn’t think about that! Thanks s2000magician
You’re welcome.
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A funnel is made up of a partial cone and a cylinder as shown in the figure. The maximum amount of liquid that can be in the funnel at any given time is 16.59375π cubic centimeters. Given this information, what is the volume of the partial cone that makes up the top part of the funnel?
1. Home
2. General
3. A funnel is made up of a partial cone and a cylinder as shown in the figure. The maximum amount of l...
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Expressing any given point on plane with one unique number
• I
• Thread starter Mashiro
• Start date
TL;DR Summary
Question on if exists a fractal of a line such that after infinite iterations it could cover any given point on a plane.
Currently, as far as I know, the two main ways to express any given point on a plane is through either cartesian plane or polar coordinates. Both of which requires an ordered pair of two numbers to
express a point. However, I wonder if there exists such a system that could express any given point on a plane using only one number.
Intuitionally, I think of lines. I know there exists fractals of a line that could theoretically fill a plane after infinite iterations. Therefore, I believe we can construct a system of expressing a
plane based on such fractal:
1. The fractal must pass through all points, therefore arbitrarily define any point as origin, denoted as zero.
2. Based on zero, define a positive direction.
3. To express any given point, simple trace from zero, alone the fractal. The distance from your point to the origin would be the unique number describing the point on the plane.
However, some problems are also raised:
1. Does there exist such rule so that a fractal could pass through every single point on a plane? No matter rational or not.
2. If the theory is correct, due to the fact that set of irrational numbers is a higher level of infinity compared to the set of rational numbers, there is a 100% chance of meeting an "irrational
point" by choosing arbitrarily. Is this going to be problematic?
If everything about this theory works out, could we apply the same method to a higher dimension (for instance 3-dimensional), and express any given point in a three dimensional space with two
numbers? Or perhaps even one.
I am currently a sophomore and my knowledge about mathematics is basic. I might have made stupid mistakes anywhere above. Please point them out to me if you spot any.
Staff Emeritus
Science Advisor
2023 Award
You can cover a certain square ##S\subseteq \mathbb{R}^2## of the plane with a single line, a
Hilbert curve
, or a
Peano curve
. Hence if our curve ##\gamma \, : \,[0,1] \longrightarrow S## hits your point ##p\in S## for the first time (more than once is possible), say ##\gamma (t_0)=p,## then ##t_0## can be thought of as a
representation of ##p## by a single number.
The disadvantage is, that different squares lead to different curves lead to different representations.
fresh_42 said:
You can cover a certain square ##S\subseteq \mathbb{R}^2## of the plane with a single line, a
Hilbert curve
, or a
Peano curve
. Hence if our curve ##\gamma \, : \,[0,1] \longrightarrow S## hits your point ##p\in S## for the first time (more than once is possible), say ##\gamma (t_0)=p,## then ##t_0## can be thought of
as a representation of ##p## by a single number.
The disadvantage is, that different squares lead to different curves lead to different representations.
Thank you so much! I was learning polar coordinates today in class and could not help thinking about this question. However, I still don't understand why it is possible to hit a same point more than
Staff Emeritus
Science Advisor
2023 Award
Mashiro said:
Thank you so much! I was learning polar coordinates today in class and could not help thinking about this question. However, I still don't understand why it is possible to hit a same point more
than once.
A surjective mapping looks easier to create than a bijective one so I wrote this passage a bit as an insurance. Let me look it up.
Cantor wrote in a letter to Dedkind in 1874 that it was impossible and that proof was almost unnecessary.
Three years later in 1877, Cantor wrote Dedekind a letter that he now believed that it is possible and added a sketch of proof.
At least I have been faster than Cantor. It is possible without crossings, but it can last until your point is covered.
"At least I have been faster than Cantor." - Ha! :-D
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Fall 2005
CS 583 - Fall 2005
Data Mining and Text Mining
Course Objective
This course has three objectives. First, to provide students with a sound basis in data mining tasks and techniques. Second, to ensure that students are able to read, and critically evaluate data
mining research papers. Third, to ensue that students are able to implement and to use some of the important data mining and text mining algorithms.
Think and Ask!
If you have questions about any topic or assignment, DO ASK me or even your classmates for help, I am here to make the course undersdood. DO NOT delay your questions. There is no such thing as a
stupid question. The only obstacle to learning is laziness.
General Information
• Instructor: Bing Liu
□ Email: Bing Liu
□ Tel: (312) 355 1318
□ Office: SEO 931
• Course Call Number: 22887
• Lecture times:
□ 11:00am-12:15pm, Tuesday & Thursday
• Room: 319 SH
• Office hours: 2:00pm-3:30pm, Tuesday & Thursday (or by appointment)
• Final Exam: 40%
□ Time and date: Thurs, Dec 8 10:30 - 12:30pm 319SH
• Midterm: 25%
• Projects: 35%
• Knowledge of probability and algorithms
• Knowledge of C or C++ for assignments
Teaching materials
• Text: Reading materials will be provided one-two days before the class
• Reference books
□ Data mining: Concepts and Techniques, by Jiawei Han and Micheline Kamber, Morgan Kaufmann Publishers, ISBN 1-55860-489-8.
□ Principles of Data Mining, by David Hand, Heikki Mannila, Padhraic Smyth, The MIT Press, ISBN 0-262-08290-X.
□ Introduction to Data Mining, by Pang-Ning Tan, Michael Steinbach, and Vipin Kumar, Pearson/Addison Wesley, ISBN 0-321-32136-7.
□ Machine Learning, by Tom M. Mitchell, McGraw-Hill, ISBN 0-07-042807-7
□ Modern Information Retrieval, by Ricardo Baeza-Yates and Berthier Ribeiro-Neto, Addison Wesley, ISBN 0-201-39829-X
• Data mining resource site: KDnuggets Directory
Topics (subject to change)
Introduction Slides
1. Data pre-processing Slides
□ Data cleaning
□ Data transformation
□ Data reduction
□ Discretization
2. Association rule mining Slides
□ Basic concepts
□ Apriori Algorithm
□ Mining association rule with multiple minimum supports
□ Mining class association rules
□ Summary
3. Supervised learning (Classification) Slides
□ Basic concepts
□ Decision trees
□ Classifier evaluation
□ Rule induction
□ Classification based on association rules
□ Naive-Bayesian learning
□ Naive-Bayesian learning for text classification
□ Support vector machines
□ K-nearest neighbor
□ Summary
4. Unsupervised learning (Clustering) Slides
□ Basic concepts
□ K-means algorithm
□ Representation of clusters
□ Hierarchical clustering
□ Distance functions
□ Data standardization
□ Handling mixed attributes
□ Which clustering algorithm to use?
□ Cluster evaluation
□ Discovering holes and data regions
□ Summary
5. Introduction to information retrieval Slides
□ Basic text processing and representation
□ Cosine similarity
□ Relevance feedback and Rocchio algorithm
6. Post-processing: Are all the data mining results interesting? Slides
□ Objective interestingness
□ Subjective interestingness
7. Partially supervised learning Slides
□ Semi-supervised learning
☆ Learning from labeled and unlabeled examples using EM
☆ Learning from labeled and unlabeled examples using co-training
□ Learning from positive and unlabeled examples
8. Link analysis and Web search Slides
□ Social network analysis: centrality and prestige
□ Citation analysis: co-citation and bibliographic coupling
□ The PageRank algoithm (of Google)
□ The HITS algorithm: authorities and hubs
□ Mining communities on the Web
9. Introduction to Web content mining Slides
□ Structured data extraction
□ Opinion extraction and analysis
□ Information integration
10. Summary
Projects - graded (you will demo your programs to me)
• Each group consists of 3 students, and will work on two assignments
1. One standard algorithm implementtion, and
2. One research project.
• Deadlines: Implementation - Nov 15 2005, Research - Nov 29 2005
Rules and Policies
• Statute of limitations: No grading questions or complaints, no matter how justified, will be listened to one week after the item in question has been returned.
• Cheating: Cheating will not be tolerated. All work you submitted must be entirely your own. Any suspicious similarities between students' work (this includes, exams and program) will be recorded
and brought to the attention of the Dean. The MINIMUM penalty for any student found cheating will be to receive a 0 for the item in question, and dropping your final course grade one letter. The
MAXIMUM penalty will be expulsion from the University.
• MOSS: Sharing code with your classmates is not acceptable!!! All programs will be screened using the Moss (Measure of Software Similarity.) system.
• Late assignments: Late assignments will not, in general, be accepted. They will never be accepted if the student has not made special arrangements with me at least one day before the assignment
is due. If a late assignment is accepted it is subject to a reduction in score as a late penalty.
Back to Home Page By Bing Liu, Aug 20 2005
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Lesson 8
Relate Quotients to Familiar Products
Warm-up: Number Talk: Multiplication and Division (10 minutes)
The purpose of this Number Talk is to elicit strategies and understandings students have for multiplying and dividing within 100. These understandings help students develop fluency and identify
division facts that are related to known products.
When students use the relationship between multiplication and division to find division facts they don’t know, they are looking for and making use of structure (MP7).
• Display one expression.
• “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
• Record answers and strategy.
• Keep expressions and work displayed.
• Repeat with each expression.
Student Facing
Find the value of each expression mentally.
• \(4 \times 10\)
• \(40 \div 4\)
• \(40 \div 10\)
• \(60 \div 6\)
Activity Synthesis
• “How do the first 3 expressions show that multiplication and division are related?” (The two division expressions both have one of the factors missing. In the first expression, the 10 is missing.
In the second, the 4 is missing.)
Activity 1: Card Sort: Multiplication (20 minutes)
The purpose of this activity is for students to check-in on their progress towards fluent multiplication within 100. Students work in groups of 2 to sort products into groups they know right away,
can find quickly, or don’t know yet. The launch provides time for a class discussion about what it means to know a fact quickly. Students identify five products with which they’d like to be more
proficient, share their strategies, and practice finding the products they choose. The cards from this activity will be used in the next activity.
MLR8 Discussion Supports. Synthesis: Display a sentence frame to support whole-class discussion: “The next time I multiply _____ and _____, I will . . . .”
Advances: Listening, Speaking
Representation: Internalize Comprehension. To support working memory, provide students with sticky notes or mini whiteboards.
Supports accessibility for: Memory, Organization
Required Materials
Materials to Copy
• Card Sort: Multiplication
• Card Sort: Multiplication Recording Sheet
Required Preparation
• Create a set of cards from the blackline master for each group of 2.
• The Multiplication Fact sort cards from this activity will be used again in the next activity.
• Groups of 2
• “Today we’re going to revisit the multiplication facts to see how many you’ve learned so far. Remember, though, you have the rest of the year to learn them.”
• “We all know what it means to know a product right away, but what does it mean to know a product quickly?” (We can figure it out in a couple of seconds with a strategy. We can figure it out in
less than 5 seconds.)
• Discuss as a class and come to an agreement about what it means to find a product quickly.
• Consider asking:
□ “Does anyone want to add on to what _____ says it means to find a product quickly?”
□ “Does anyone have different ideas about what it means to find a product quickly?”
□ “Based on this discussion, does anyone want to revise their ideas about what it means to find a product quickly?”
• Give each group one set of pre-cut cards and a sort table.
• “Take some time to quiz each other on multiplication facts. As you quiz your partner, use the table to sort the expressions into three groups that show if they know it right away, they can find
it quickly, or they don’t know it yet.”
• 7–10 minutes: partner work time
• “Choose 5 multiplication facts that you don’t know yet and write down the expressions. These are the products you will practice finding.”
• 1 minute: independent work time
• “Now, share the products you want to practice with your partner and have them help you think of some strategies you could use to find the products quickly.”
• “After you have some strategies, take some time to practice finding the products you chose.”
• 5–7 minutes partner work time
Student Facing
Quiz your partner on their multiplication facts. Sort your partner’s facts into one of these columns:
1. know it right away
2. can find it quickly
3. don’t know it yet
Multiplication expressions I’m going to practice:
Advancing Student Thinking
If students don’t yet have a strategy for one of the facts they’ve chosen to practice, consider asking:
• “What have you tried so far to find this product?”
• “Could you check in with another group to see if they could suggest a strategy for finding this product?”
Activity Synthesis
• “What were some useful strategies for finding products you didn’t know yet?” (Thinking of a product I already know and using that product to find the one I didn’t know yet. Using products of 2,
5, and 10 to figure out other products.)
Activity 2: If I Know, Then I Know (15 minutes)
The purpose of this activity is for students to identify division facts that are related to multiplication facts that they know. Students complete “If I know, then I know” statements using their
multiplication fact cards from the previous activity. Give students time, if needed, to determine the product before generating the related division equation. Some students may generate 4 related
division equations for each product by moving the quotient to the left side of the equal sign. If this comes up, recognize that this is possible, but keep the emphasis on generating two related
division facts, one for each of the factors as the unknown number.
When students use the relationship between multiplication and division identify two division facts from a multiplication fact, they look for and make use of structure (MP7).
Required Preparation
• Each group of 2 needs a set of cards from the previous activity.
• Groups of 2
• “Read the first statement of the activity. Talk with your partner about how you could finish the statement.”
• 1 minute: partner discussion
• Share responses.
• “Now, you’re going to take turns drawing a card and using the fact you chose to complete an ‘If I know, then I know’ statement with the multiplication fact that you drew and the related division
facts. Take some time to figure out the multiplication fact together if you need to. After every turn, record the multiplication equation and related division equations in the table.”
• 7–10 minutes: partner work time
Student Facing
If I know \(4 \times 5 = 20\), then I know _____.
1. Set the multiplication fact cards in a stack face down.
2. Take turns drawing a multiplication fact card.
3. Use the multiplication fact on the card to record a multiplication equation in the “If I know . . .” column.
4. Then, record related division equations in the “Then I know . . .” column.
│If I know . . . , │then I know . . . │
Activity Synthesis
• “How many division equations were you able to come up with for each multiplication equation? Explain your reasoning.” (2, because I could come up with 1 equation where the quotient was 1 factor
and 1 equation where it was the other factor. 4, because I could come up with 1 for each of the factors being the quotient, and I could have the quotient on the right or the left of the equal
Lesson Synthesis
“Today we thought about multiplication facts that we know and worked on some that we don’t know yet. How did this help you with finding division facts?” (We could use a multiplication fact to find
related division facts. We realized that if we know a multiplication fact, then there are some division facts that we know too.)
Cool-down: Multiplication and Division Facts (5 minutes)
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Simplify algebraic expression in matlab
simplify algebraic expression in matlab Related topics: how to solve polynomial on ti 89
adding and subtracting integers enrichment
topics in algebra solutions
completing the square on a parabolic equation
compare rational expression and polynomial division
Factor Online Polynomials Program
prentice hall math oklahoma pre algerbra answers
system by elimination calculator
technical mathmetic
algebra properties
intermediate algebra,11
Author Message Author Message
drunsdix42 Posted: Thursday 04th of Jan 09:53 Osx Hi Fi Zame Posted: Monday 08th of Jan 07:15
I am in a real problem . Somebody assist me please. I It looks helpful. How could I get that software? Could
get a lot of dilemma with system of equations, graphing you give me a link that could lead me to more
and subtracting exponents and especially with simplify information regarding that software?
Reg.: 24.09.2002 algebraic expression in matlab. I need to show some Reg.: 09.02.2007
rapid progress in my math. I read there are various
Software Tools available online which can help you in
algebra. I can pay some cash too for an effective and
inexpensive software which helps me with my studies.
Any hint is greatly appreciated. Thanks. thicxolmed01 Posted: Wednesday 10th of Jan 09:22
Jahm Xjardx Posted: Friday 05th of Jan 16:16 I would recommend trying out Algebrator. It not only
assists you with your math problems, but also displays
I find these routine queries on almost every forum I visit. all the required steps in detail so that you can improve
Please don’t misunderstand me. It’s just as we Reg.: 16.05.2004 the understanding of the subject.
enter high school , things change suddenly . Studies
Reg.: 07.08.2005 become complex all of a sudden. As a result, students
encounter trouble in doing their homework. simplify
algebraic expression in matlab in itself is a quite complex
subject. There is a program named as Algebrator Admilal`Leker Posted: Friday 12th of Jan 09:48
which can assist you in this situation.
Finding the program is as effortless , as kid’s play.
fveingal Posted: Saturday 06th of Jan 12:24 You can go to:
Hello there. Algebrator is really amazing ! It’s been Reg.: 10.07.2002 > for further details and access the program. I am
months since I used this program and it worked like confident you will be happy with it just as I was. Also , it
magic! Algebra problems that I used to spend solving offers you a money back promise if you aren’t
Reg.: 11.07.2001 for hours just take me 4-5 minutes to answer now. pleased .
Just enter the problem in the program and it will take
care of the solving and the best thing is that it shows
the whole solution so you don’t have to figure out
how did the software come to that answer.
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Kantorovich Initiative Seminar: Ziv Goldfeld
Gromov-Wasserstein Alignment: Statistical and Computational Advancements via Duality
The Gromov-Wasserstein (GW) distance quantifies dissimilarity between metric measure (mm) spaces and provides a natural alignment between them. As such, it serves as a figure of merit for
applications involving alignment of heterogeneous datasets, including object matching, single-cell genomics, and matching language models. While various heuristic methods for approximately evaluating
the GW distance from data have been developed, formal guarantees for such approaches—both statistical and computational—remained elusive. This work closes these gaps for the quadratic GW distance
between Euclidean mm spaces of different dimensions. At the core of our proofs is a novel dual representation of the GW problem as an infimum of certain optimal transportation problems. The dual form
enables deriving, for the first time, sharp empirical convergence rates for the GW distance by providing matching upper and lower bounds. For computational tractability, we consider the entropically
regularized GW distance. We derive bounds on the entropic approximation gap, establish sufficient conditions for convexity of the objective, and devise efficient algorithms with global convergence
guarantees. These advancements facilitate principled estimation and inference methods for GW alignment problems, that are efficiently computable via the said algorithms.
Event Type
Scientific, Seminar
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Adding Integers Calculator - Calculator Hub
Adding Integers Calculator
Integer addition has never been this simple before. To easily add any two integers, use this online adding integers calculator.
<iframe src="https://calculatorhub.org/?cff-form=114" style="width:100%;height:100%;"></iframe>
What is Integer?
Often referred to as directed numbers, integers are numbers that may be thought of as both positive and negative whole numbers.
There are an endless number of them, some of which are given below: I = {. . .,−3,−2,−1, 0, 1, 2, 3, . . .}. The standard symbol for describing integers is ‘I’.
There is an equal and opposite negative number for every positive integer. In actual life, integers are used to demonstrate that some values can be negative.
A temperature that is 25° below zero is represented by the symbol -25°, and one that is 17° above it is represented by the symbol +17°, or more simply by the number 17°.
How to Use Adding Integers Calculator?
• This integer addition calculator is designed in such a way that even a young child can use it successfully.
• You can compute the addition of two integers in just two steps.
• Enter the first integer in the first input box, followed by the second in the second input box.
• The sum of the two integers will be instantly returned by the calculator.
Adding Integers Formula
The formulas listed below are used for adding integers. Four formulas illustrate the four situations that could arise when adding integers.
How to Add Integers?
Listening to the term “addition” may cause you to smile. Because everyone knows that addition is one of the simplest mathematical terms to grasp. However, we should remind you that we are discussing
integer addition here.
Due to the fact that integers can be both positive and negative, you must exercise caution while adding integers if even a single number has the negative sign (-). since even one negative sign (-)
can alter the outcome.
For you to fully grasp the idea of the addition of integers even with negative numbers, we have solved every type of case below.
1. Add Integers +9 & +27?
Solution :
Let’s start with a simple example. Because both numbers are positive, there is no need to think twice about performing the addition. Simply follow the standard addition procedure.
We’ll use the above formula where both values are positive.
∴ (+) + (+) = (+)
∴ (+9) + (+27) = +36 or 36
The addition of integers +9 and +27 is +36.
2. Add integers -25 & -13?
Solution :
Since both of the integers in this problem are negative(-), we will proceed in the same manner as for the preceding example, but we must implement the second formula provided above.
∴ (-) + (-) = (-)
∴ (-25) + (-13) = -38
As both the integers have negative sign the answer will also get a negative sign(-)
The addition of integers -25 and -13 is -38.
3. Add integers +125 & -85?
Solution :
Both of the integers in this problem have opposite signs; one is positive and the other is negative. Because the positive integer is greater than the negative integer in this situation (125>85), you
will use the third formula.
∴ (+) + (-) = (+)
∴ (+125) + (-85) = (+)
Important Note : Since the signs of the integers in the first two examples were identical, we did not pay much care when adding the numbers. But we need to exercise caution as the integers’ signs
diverge from one another. Here, a single minus (-) sign can convert an addition into a subtraction.
Since the operation is positive and the second integer has a negative sign, (+) (-) = (-). This forces us to execute a two-number subtraction, but the outcome will always bear the sign of the larger
∴ (+125) + (-85) = +40
As the larger number has positive sign the result will also get the positive sign.
Trick : When adding two integers, if the signs of the integers differ, simply subtract the smaller number from the larger number. When you have the result, give it the sign of the larger number.
The addition of integers +125 and -85 is +40.
4. Add integers +60 & -120?
Solution :
Even in this problem, the signs of the integers are different. We will use the fourth formula because the larger number has a negative sign.
∴ (+) + (-) = (-)
∴ (+60) + (-120) = (-)
Because the operation is positive and the second integer is negative, (+) (-) = (-). Even in this example, we must perform a two-number subtraction. Simply subtract the smaller number from the larger
number and apply the sign of the larger number to the result value.
∴ (+60) + (-120) = (-60)
Because the larger number has a negative sign, the answer will have a negative sign as well.
The addition of integers +60 and -120 is -60.
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Exploiting the Diffie-Hellman bug in socat
A few days ago socat, a popular networking tool, issued a curious sounding security advisory:
"In the OpenSSL address implementation the hard coded 1024 bit DH p parameter was not prime. The effective cryptographic strength of a key exchange using these parameters was weaker than the one one
could get by using a prime p. Moreover, since there is no indication of how these parameters were chosen, the existence of a trapdoor that makes possible for an eavesdropper to recover the shared
secret from a key exchange that uses them cannot be ruled out."
More background information on this vulnerability can be found on Ars Technica and Hacker News, in this post I want to focus on building an exploit.
The patch shows that
p =
It has been discovered that
p = 271 * 13597 *
The last factor, let's denote it F3889, is a 1002-bit non-prime integer, whose factors remain unknown. By trial division we know that its smallest factors are larger than $2^{40}$. If we want to go
further, we'll need to deploy a proper factorization algorithm. In that case I'll choose Lenstra elliptic curve factorization, whose running time depends on the size of smallest factor of F3889
rather than the size of the number itself. If the smallest factor is smaller than $2^{250}$, the Lenstra's algorithm would recover it before too long. The other factors can then be recovered by
either Lenstra's or the general number field sieve algorithm.
On the other hand if F3889 is a product of two 500-bit primes, chances are we might never be able to factor it (without spending a million of dollars or so). This is very unlikely, if $p$ was indeed
randomly generated. Thus, it's reasonable to assume that anyone determined and lucky enough can factor F3889 completely. It'll be a fun project, let me know if you want to work on it :).
But, you might ask, why do we care so much about the factors of $p$? It seems to have nothing to do with solving the discrete log problem (DLP) on $Z_p$, doesn't it? The answer is yes, knowing the
factors of $p$, and if they are small enough, allows one to solve DLP on $Z_p$, thanks to the Chinese Remainder Theorem (CRT).
As I wrote before on this blog, CRT is one of the most powerful cryptanalysis tools ever. I've seen countless systems broken because of it. Whenever analyzing or designing a new system, ask yourself
if you can break it using this simple trick, and you'll be surprised that most of the times the answer is yes. Pohlig and Hellman probably asked this question themselves, and eventually figured out
that if the order of a group is a product of small primes (i.e., a smooth number), one can solve DLP in the subgroups, which is easier, and combine the results using CRT to obtain the discrete log in
the original group.
Let's look at an example. Suppose that we want to solve for $x$, given $g$ and $h = g^x \pmod{n}$, where the order of group $Z_n$ is $\phi(n) = q_1 * q_2$ with $q_1$ and $q_2$ are prime. We can break
this problem into three smaller sub-problems:
1/ Find $x_1$ such that $h^{q_1} = (g^{q_1})^{x_1} \pmod{n}$
2/ Find $x_2$ such that $h^{q_2} = (g^{q_2})^{x_2} \pmod{n}$
3/ We can prove that $x = x_1 \pmod{q_2}$ and $x = x_2 \pmod{q_1}$, and thus can figure out $x$ with CRT.
Note that 1/ and 2/ are themselves DLP, but they are in subgroups of order $q_1$ and $q_2$, respectively. Hence, the Pohlig-Hellman algorithm reduces DLP in group of order $q_1 * q_2$ to DLP in group
of order $q_1$ or $q_2$. If $q_1$ or $q_2$ or small, we can brute-force for $x_1$ or $x_2$. If they are larger, we can deploy the Pollard's rho or the index calculus algorithm. It has been estimated
that an academic team can break discrete log if $q_1$ or $q_2$ is a 768-bit prime and that a nation-state can break a 1024-bit prime.
I hope that it's clearer now why we need to factor $p$. We want to calculate $\phi(p)$ and its factors, which we need to deploy the Pohlig-Hellman attack. Is it surprising that the factors of the
order of the group determines how hard DLP is on that group?
If the largest factor of $p$ is smaller than 2^800 it's reasonable to assume that anyone knowing this factor can solve DLP on the multiplicative group $Z_p$. That is, they can find $x$ given $g$ and
$h$, where $g^x = h \pmod{p}$; this in turn allows them recovering the shared secret just by passively eavesdropping on the Diffie-Hellman key exchange. Note that if the larger factors of $p$ are not
safe prime, i.e., not in the form of $2 * q + 1$ where $q$ is also a prime, the computation cost is even smaller.
In summary you can exploit this bug by following these steps:
1/ Using Lenstra elliptic curve factorization and the general number field sieve to factor $p$ completely, and
2/ Using Pohlig-Hellman and CRT to reduce DLP on the multiplicative group $Z_p$ to the multiplicative group $Z_{p'}$ where $p'$ is the largest factor of $\phi(p)$, and
3/ Using Pollar's rho or index calculus to solve DLP on $Z_{p'}$, and
4/ Sniffing socat traffic and recovering the DH shared secret, and
5/ Profit!
If this is a backdoor, it's trivial for its creators to exploit it, because they can skip 1/ and pre-compute most of 2/ and 3/. Even if this is not a backdoor, I suspect that it doesn't cost more
than \$10K on AWS or Google Cloud Platform to perform step 1/, 2/ and 3/. If we're so lucky that F3889 is a product of 3 primes, 2 of which are 250-bit, step 1/ might cost less than \$2K, and
pre-computation for step 2/ and step 3/ might cost even less.
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Great introduction, and that is the reason why we just need to care about DLP on cyclic groups of large prime orders. And to avoid most attacks (which are very similar to methods mentioned in your
post), we must use strong primes.
As far as I know (please correct me if I am wrong), to generate a strong prime, i.e. a prime $p$ that has the form $kq + 1$, where $q$ is also large prime, we use the properties of arithmetic
progresses, which is stated as Dirichlet's theorem: "Let $r, m$ are integers such that $\gcd(r,m)=1$, then there exists infinitely many primes of the form $r + km$ (that is all the set of prime $p$
such that $p \equiv r \mod m$. Denote this set as $\pi_{r,m}$. Then $\pi_{r,m}$ approximates $\frac{1}{\phi(m)}\frac{x}{\log x}$, where $\phi(m)$ is the Euler's totient function." For reference, see
From this, if we want to find a 100 bit strong prime, we can start with a 80 bit prime $q$, and choose random $k$, and then test whether $kq + 1$ is prime. The theorem of Dirichlet above give us the
estimation of success probability. Also, the next question might be how we can test a prime number in an efficient way? As far as I know, people prefer using elliptic curve primality proving. The
construction of suitable curves for testing is difficult, and Morain-Atkin have to use complex multiplication method.
But these things cannot defend the very efficient attack, the index calculus. Amazingly, the analogue generally fails for ECDLP, though it is again true for curves of higher genus (hyperelliptic
curves)! These things mentioned in this comment can be found in the book of L. C. Washington about elliptic curves.
Awesome post! I've been looking into it myself. Some people also told me about pollard's p-1 and p+1 factoring algorithm that might work.
The other thing is that the new 2048bits prime they used might have a "bad" order. If you can factor it in primes small enough, there is Pohlig-Hellman attack as well.
Last thing: this is not only vulnerable to passive Pohlig-Hellman attacks, but also to more active attacks like small subgroup attacks where you would send a different generator of each of the
smaller subgroups. 1) Since the subgroup of the generator they used (2) for the new prime is unknown, it must be that they're not verifying that publickey sent to their program are indeed in the
right subgroup 2) since we can generate our own generators, we can build smaller base for our discrete log problems and it should be faster than Pohlig-Hellman. Significantly faster? I don't know
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This comment has been removed by the author.
> I haven't seen anyone using any other values for k rather than 2 though
You can't use something else than 2, otherwise you get an even number > 2 => not a prime
Thanks you for your waiting :D. Unfortunately, I know very little on the topic of complex multiplication. I just mention a little bit about the definition here. Though it is used to construct curves
with given number of points (very important in ECDLP, right?), I really don't know how it works.
Let us consider an elliptic curve over $\mathbb{Q}$, denoted $E(\mathbb{Q})$. As we know, it is an abelian group, which has the ring of endomorphism. Recall that an endomorphism is a polynomial map
from $E(\overline{\mathbb{Q}})$ to itself, that transform $\infty$ to $\infty$. Then it is obvious to see that the multiplication by an integer $n$, defined by $P$ to $nP$ is an endomorphism. For
ordinary curves, $End(E(\overline{\mathbb{Q}}))$ is isomorphic to $\mathbb{Z}$, the ring of integers. For curves with complex multiplication, this ring is strictly larger than $\mathbb{Z}$.
For example, consider the curve $y^2=x^3+x$, it has the complex multiplication $\phi(x,y)=(-x,iy)$, where $i^2+1=0$ (Recall that this curve defined over $\mathbb{F}_p$, where $p > 3$ and $p \equiv 3\
mod 4$, it is a supersingular curve. In the finite fields case, the Frobenius map $(x,y) \rightarrow (x^p,y^p)$ is also an endomorphism, the endomorphism ring is strictly larger than $\mathbb{Z}$ in
this situation).
A very important application of this curve is the following theorem, that has a profound meaning: Let $F/\mathbb{Q}(i)$ be the Galois extension of $\mathbb{Q}(i)$ of finite degree, and suppose that
$Gal(F/\mathbb{Q}(i))$ is abelian. Then there is an integer $n \ge 1$ such that $F \subset \mathbb{Q}(i)(E[n])$, where $E[n]$ is the set of $n$-torsion group, and $\mathbb{Q}(i)(E[n])$ is the
extension field of $\mathbb{Q}(i)$ obtained by adding $x$ and $y$ coordinates of elements in $E[n]$.
The content of this comment can be found in the book of Silverman and Tate "Rational points on elliptic curves" (Chapter VI). For applications of complex multiplication in cryptography, if you are
curious (me too), why don't we establish a small program for understanding about this?
P.S: TEX really works in the comments, when we see the full post mode.
Thanks you for your comment. I really appreciate your ideas about creating a group to discuss on ECC. It is a very good way to learn. I will think about this and send email to you later.
For the question, I do think that the addition law on elliptic curves is naturally defined in the algebraic viewpoint. In Chapter XI, the author develops the theory of divisors, which allows us to
define addition law for curves of higher genus. The addition law on these groups is not "point plus point" as elliptic curves. We have to work with reduced divisors. And these groups are called
Jacobian varieties. For an elliptic curve, we have a bijective map between "points" on its Jacobian variety and its "actual" points. Hence, we can define the addition law based on the addition law on
the variety. It is an interesting story (but a little bit long), and it cannot be fully covered in this comment.
> how the hell did mathematicians discover that points on elliptic curve form a group
I think they were just trying to figure out how to find rational points on a curve. At first there were easy points to find, then how to find new ones? A good way was to trace a line between two of
them, and you get a new point. And by reversing it on the y axis you would get a new point again (this is because of y^2). Later on they figured out that it was forming a group (apparently the proof
is pretty easy so I guess it is not so difficult to imagine). There is a video of Bonneh on the subject: https://www.youtube.com/watch?v=4M8_Oo7lpiA
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How do specific surface area and particle size distribution change when granular media dissolve?
Fig. 1. Incremental change in specific surface during dissolution for a granular medium with two particle sizes having a size ratio of β, given that the number fraction of smaller particles is p: (a)
shows the curve that distinguishes the regions where the change is positive or negative, with a dotted line at p=1 for comparison; (b) shows the same curve, with the quantity (1−p) on the y-axis.
The filled circles in (b) indicate the starting points of the systems considered in Fig. 2, and the arrows indicate the direction of the path taken.
Fig. 2. Predictions of the evolution in scaled specific surface area (S=s(ϵ)∕s(0)) during uniform dissolution of a bimodal particle size distribution. The left plot shows the influence of the number
fraction of the smaller particles for a fixed initial size ratio of β=10. The right plot shows the influence of that initial size ratio for a fixed number fraction p=0.999 of the smaller particles.
The initial behavior, whether increasing, decreasing, or remaining relatively constant, is indicated by the diagram in Fig. 1.
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Find the sum of the following series if the number of terms is 183 and the series is 4+(-4)+4+(-4).........? | Socratic
Find the sum of the following series if the number of terms is 183 and the series is 4+(-4)+4+(-4).........?
1 Answer
We can solve this using inspection.
We see that the series is:
$4 - 4 + 4 - 4 + 4 - 4 + 4. \ldots$
We see that the first two terms give $0$. Working on the first four, six, and eight terms we get the same answer. Adding the first three, five and seven terms give us $4$.
We can say that if the series has an even number of items, it sums up to $0$, and if it has an odd number of terms, it sums up to $4$.
$183$ is odd, and so the series sums up to $4$.
Impact of this question
1605 views around the world
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game theory – Skeptical Sports Analysis
Going into the final round of ESPN’s Stat Geek Smackdown, I found myself 4 points behind leader Stephen Ilardi, with only 7 points left on the table: 5 for picking the final series correctly, and a
bonus 2 for also picking the correct number of games. The bottom line being, the only way I could win is if the two of us picked opposite sides. Thus, with Miami being a clear (though not
insurmountable) favorite in the Finals, I picked Dallas. As noted in the ESPN write-up”
“The Heat,” says Morris, “have a better record, home-court advantage, a better MOV [margin of victory], better SRS [simple rating system], more star power, more championship experience, and had a
tougher road to the Finals. Plus Miami’s poor early-season performance can be fairly discounted, and it has important players back from injury. Thus, my model heavily favors Miami in five or six
But I’m sure Ilardi knows all this, so, since I’m playing to win, I’ll take Dallas. Of course, I’m gambling that Ilardi will play it safe and stick with Miami himself since I’m the only person
close enough to catch him. If he assumes I will switch, he could also switch to Dallas and sew this thing up right now. Game-theoretically, there’s a mixed-strategy Nash equilibrium solution to
the situation, but without knowing any more about the guy, I have to assume he’ll play it like most people would. If he’s tricky enough to level me, congrats.
Since I actually bothered to work out the equilibrium solution, I thought some of you might be interested in seeing it. Also, the situation is well-suited to illustrate a couple of practical points
about how and when you should incorporate game-theoretic strategies in real life (or at least in real games).
Some Game Theory Basics
Certainly many of my readers are intimately familiar with game theory already (some probably much more than I am), but for those who are less so, I thought I should explain what a “mixed-strategy
Nash equilibrium solution” is, before getting into the details on the Smackdown version (really, it’s not as complicated as it sounds).
A set of strategies and outcomes for a game is an “equilibrium” (often called a “Nash equilibrium”) if no player has any reason to deviate from it. One of the most basic and most famous examples is
the “prisoner’s dilemma” (I won’t get into the details, but if you’re not familiar with it already, you can read more at the link): the incentive structure of that game sets up an equilibrium where
both prisoners rat on each other, even though it would be better for them overall if they both kept quiet. “Rat/Rat” is an equilibrium because an individual deviating from it will only hurt
themselves. Bother prisoners staying silent is NOT an equilibrium, because either can improve their situation by switching strategies (note that games can also have multiple equilibriums, such as
the “Which Side of the Road To Drive On” game: both “everybody drives on the left” and “everybody drives on the right” are perfectly good solutions).
But many games aren’t so simple. Take “Rock-Paper-Scissors”: If you pick “rock,” your opponent should pick “paper,” and if he picks “paper,” you should take “scissors,” and if you take “scissors,”
he should take “rock,” etc, etc—at no point does the cycle stop with everyone happy. Such games have equilibriums as well, but they involve “mixed” (as opposed to “pure”) strategies (trivia note:
John Nash didn’t actually discover or invent the equilibrium named after him: his main contribution was proving that at least one existed for every game, using his own proposed definitions for
“strategy,” “game,” etc). Of course, the equilibrium solution to R-P-S is for each player to pick completely at random.
If you play the equilibrium strategy, it is impossible for opponents to gain any edge on you, and there is nothing they can do to improve their chances—even if they know exactly what you are going to
do. Thus, such a strategy is often called “unexploitable.” The downside, however, is that you will also fail to punish your opponents for any “exploitable” strategies they may employ: For example,
they can pick “rock” every time, and will win just as often.
The Smackdown Game
The situation between Ilardi and I going into our final Smackdown picks is just such a game: If Ilardi picked Miami, I should take Dallas, but if I picked Dallas, he should take Dallas, in which case
I should take Miami, etc. When you find yourself in one of these “loops,” generally it means that the equilibrium solution is a mixed strategy.
Again, the equilibrium solution is the set of strategies where neither of us has any incentive to deviate. While finding such a thing may sound difficult in theory, for 2-player games it’s actually
pretty simple intuitively, and only requires basic algebra to compute.
First, you start with one player, and find their “break-even” point: that is, the strategy their opponent would have to employ for them to be indifferent between their own strategic options. In this
case, this meant: How often would I have to pick Miami for Miami and Dallas to be equally good options for Ilardi, and vice versa.
So let’s formalize it a bit: “EV” is the function “Expected Value.” Let’s call Ilardi or I picking Miami “iM” and “bM,” and Ilardi or I picking Dallas “iD” and “bD,” respectively. Ilardi will be
indifferent between picking Miami and Dallas when the following is true:
Let’s say “WM” = the odds of the Heat winning the series. So now we need to find EV(iM) in terms of bM and WM. If Ilardi picks Miami, he wins every time I pick Miami, and every time Miami wins when
I pick Dallas. Thus his expected value for picking Miami is as follows:
When he picks Dallas, he wins every time I don’t pick Miami, and every time Miami loses when I do:
Setting these two equations equal to each other, the point of indifference can be expressed as follows:
Solving for bM, we get:
What this tells us is MY equilibrium strategy. In other words, if I pick Miami exactly as often as we expect Miami to lose, it doesn’t matter whether Ilardi picks Miami or Dallas, he will win just
as often either way.
Now, to find HIS equilibrium strategy, we repeat the process to find the point where I would be indifferent between picking Miami or Dallas:
In other words, if Ilardi picks Miami exactly as often as they are expected to win, it doesn’t matter which team I pick.
Note the elegance of the solution: Ilardi should pick each team exactly as often as they are expected to win, and I should pick each team exactly as often as they are expected to lose. There are
actually a lot of theorems and such that you’d learn in a Game Theory class that make identifying that kind of situation much easier, but I’m pretty rusty on that stuff myself.
So how often would each of us win in the equilibrium solution? To find this, we can just solve any of the EV equations above, substituting the opposing player’s optimal strategy for the variable
representing the same. So let’s use the EV(iM) equation, substituting (1-WM) anywhere bM appears:
$EV(iEq)=1 - WM +WM^2$
Here’s a graph of the function:
Obviously, it doesn’t matter which team is favored: Ilardi’s edge is weakest when the series is a tossup, where he should win 75% of the time. The bigger a favorite one team is, the bigger the
leader’s advantage.
Now let’s Assume Miami was expected to win 63% of the time (approximately the consensus), the Nash Equilibrium strategy would give Ilardi a 76.7% chance of winning, which is obviously considerably
better than the 63% chance that he ended up with by choosing Miami to my Dallas—so the actual picks were a favorable outcome for me. Of course, that’s not to say his decision was wrong from his
perspective: Either of us could have other preferences that come into play—for example, we might intrinsically value picking the Finals correctly, or someone in my spot (though probably not me) might
care more about securing their 2nd-place finish than about having a chance to overtake the leader, or Ilardi might want to avoid looking bad if he “outsmarted himself” by picking Dallas while I
played straight-up and stuck with Miami.
But even assuming we both wanted to maximize our chances of winning the competition, picking Miami may still have been Ilardi’s best strategy given when he knew at the time, and it would have been a
fairly common outcome if we had both played game-theoretically anyway. Which brings me to the main purpose for this post:
A Little Meta-Strategy
In reality, neither of us played our equilibrium strategies. I believed Ilardi would pick Miami more than 63% of the time, and thus the correct choice for me was to pick Dallas. Assuming Ilardi
believed I would pick Dallas less than 63% of the time, his best choice was to pick Miami. Indeed, it might seem almost foolhardy to actually play a mixed strategy: what are the chances that your
opponent ever actually makes a certain choice exactly 37% of the time? Whatever your estimation, you should go with whichever gives you the better expected value, right?
This is a conundrum that should be familiar to any serious poker players out there. E.g., at the end of the hand, you will frequently find yourself in an “is he bluffing or not?” (or “should I bluff
or not?”) situation. You can work out the game-theoretically optimal calling (or bluffing) rate and then roll a die in your head. But really, what are the chances that your opponent is bluffing
exactly the correct percentage of the time? To maximize your expected value, you gauge your opponent’s chances of bluffing, and if you have the correct pot odds, you call or fold (or raise) as
So why would you ever play the game-theoretical strategy, rather than just making your best guess about what your opponent is doing and responding to that? There are a couple of answers to this.
First, in a repeating game, there can be strategic advantages to having your opponent know (or at least believe) that you are playing such a strategy. But the slightly trickier—and for most people,
more important—answer is that your estimation might be wrong: playing the “unexploitable” strategy is a defensive maneuver that ensures your opponent isn’t outsmarting you.
The key is that playing any “exploiting” strategy opens you up to be exploited yourself. Think again of Rock-Paper-Scissors: If you’re pretty sure your opponent is playing “rock” too often, you can
try to exploit them by playing “paper” instead of randomizing—but this opens you up for the deadly “scissors” counterattack. And if your opponent is a step ahead of you (or a level above you), he
may have anticipated (or even set up) your new strategy, and has already prepared to take advantage.
Though it may be a bit of an oversimplification, I think a good meta-strategy for this kind of situation—where you have an equilibrium or “unexploitable” strategy available, but are tempted to play
an exploiting but vulnerable strategy instead—is to step back and ask yourself the following question: For this particular spot, if you get into a leveling contest with your opponent, who is more
likely to win? If you believe you are, then, by all means, exploit away. But if you’re unsure about his approach, and there’s a decent chance he may anticipate yours—that is, if he’s more likely to
be inside your head than you are to be inside his—your best choice may be to avoid the leveling game altogether. There’s no shame in falling back on the “unexploitable” solution, confident that he
can’t possibly gain an advantage on you.
Back in Smackdown-land: Given the consensus view of the series, again, the equilibrium strategy would have given Ilardi about a 77% chance of winning. And he could have announced this strategy to
the world—it wouldn’t matter, as there’s nothing I could have done about it. As noted above, when the actual picks came out, his new probability (63%) was significantly lower. Of course, we
shouldn’t read too much into this: it’s only a single result, and doesn’t prove that either one of us had an advantage. On the other hand, I did make that pick in part because I felt that Ilardi was
unlikely to “outlevel” me. To be clear, this was not based on any specific assessment about Ilardi personally, but based my general beliefs about people’s tendencies in that kind of situation.
Was I right? The outcome and reasoning given in the final “picking game” has given me no reason to believe otherwise, though I think that the reciprocal lack of information this time around was a
major part of that advantage. If Ilardi and I find ourselves in a similar spot in the future (perhaps in next year’s Smackdown), I’d guess the considerations on both sides would be quite different.
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Investigation of torsion angle measurement method in spatial cables of suspension bridges
Spatial cables in suspension bridges may twist between two adjacent suspension points due to transverse tension. To determine the spatial torsion angle of the cables, a scale-reduced suspender bridge
model was fabricated and tested in this study. The cable torsion angles of the bridge model under different loading cases were obtained and presented. Theoretical methods (e.g., the coordinate method
and the angle method) to calculate the spatial cable torsion angle in the suspender bridge were developed by analyzing the deforming characteristics of the suspender cables. The feasibility and
predictive accuracy of the proposed methods were assessed using the experimental results. Results show that the proposed coordinate and angle methods produced similar torsion angles for the main
cables, showing a maximum relative error of 10.4 %. Compared to the coordinate method's complex data transformation and measurement procedures, the angle method is more convenient, efficient, and
easier to calculate the torsion angle of the main cable. Despite the capability of calculating the spatial cable torsion angles, the coordinate method also exhibits advanced capability in determining
the geometric curves of the main cable under different spatial angles.
• A scale-reduced suspender bridge was tested for torsion measurement
• Theoretical methods to calculate cable torsion angle were developed
• Method to determine geometric curves of main cable was proposed
1. Introduction
The main cable is vital for a suspender bridge in transferring dead and live loads from bridge decks to the main tower. The main cable in the suspender bridges is usually made of high-strength steel
wires, which are very easy to deform under constructing and serving loads [1-3]. The incompatible deformation of steel wires would change the curve line of the main cable, bringing unexpected cable
torsions and stress redistributions among the wires. Along with the main cable's twisting, the cable claps' lateral dip angle would also change. Suppose the cable clamp angle exceeds the allowable
angle of the sling. In that case, the clamp ear plates connecting with the slings may squeeze each other and cause bending deformation to the slings, seriously affecting the service life of the
anchoring structure of the sling [4, 5]. In order to avoid torsion stresses in cable clips and prevent torsion damage to the bridge members, the torsion deformation of the main cable should be
calculated accurately during the design of the suspender bridge. Previous studies on the torsion of main cables mainly focused on the tow and cable formation process, and numerical studies on the
torsion of main cables caused by the uneven internal force after the completion of cable erection are also found in the literature [5-7]. With the gradual deep understanding of the cable torsion in
suspender bridges, scholars worldwide devote themselves to exploring effective predicting methods for calculating the cable torsion angles under various loading cases. Unfortunately, due to the
comprehensive impacts of many variables, a theoretical predictive method for spatial cables in the suspender bridge is currently unavailable.
To partially cover this knowledge gap, this paper investigates the calculation method for predicting the cable torsion angle during erecting of the slings. A scale-reduced suspender bridge model was
fabricated and tested to obtain the cable angle. Based on the deformation characteristics of the main cable, a coordinate method and angle method to calculate the spatial cable torsion angle in
suspender bridges are developed. The feasibility and predictive accuracy of the proposed methods are assessed based on the experimental data. The present study's outcomes can serve as a potential
method for designing spatial cables in suspender bridges.
2. Description of suspender bridge model
The experimental program selects a real twin-tower three-span suspension bridge located in China as the prototype bridge. Based on the configuration and geometric size of the prototype bridge, a
scale-reduced suspender bridge model was designed and constructed. As shown in Fig. 1, the total length of the model bridge is 28.39 m, with a span arrangement of 17.33 m in the middle and 5.53 m on
each side. The overall height of the main tower is 4.7 m for the PM1 tower and 4.8 m for the PM2 tower. No sling was set between the main girder and the main cable in the side spans. In the middle
span, there are seven vertical slings with the same adjacent intervals numbered by 1), 2),… 7) from east to west between the main cable and girder. The main cable comprises 19 steel strands and is
arranged in a plane layout. Each steel strand comprises seven steel wire bundles with a nominal diameter of 1.8 mm and an elastic modulus of 2.0×10^5 MPa. The unstressed length of the main cable in
the mid-span is 29.28 m, while that in the side span has a cable length of 3 m (the sum of the net length and extends into anchorage length). All the slings are fabricated using steel wire ropes with
a material elastic modulus of 1.95×10^5 MPa. The Q345qD steel (with a nominal yield strength of 345 MPa) was adopted for the main towers, saddles, and anchorage blocks. In order to simulate the
geometric boundaries above the towers and allow predetermined deformation of the cable, the circular arc of the main saddle along the longitudinal direction of the bridge is set in both the façade
and plane. There are two main parallel cables in the selected prototype bridge. Only a single cable was fabricated in the model bridge to ease model construction and save costs.
Fig. 1Layout of the suspension bridge model (UNIT: m)
3. Cable torsion angle measurement method and apparatus
Under the transverse tension of slings, the main cable in suspender bridges may convert from a planar state to a spatial state, accompanied by the constant changing of the cable linearity and
deflection angle. In order to characterize the torsion angle of the main cable, the following two predictive methods are developed to calculate the variations in cable angles.
3.1. Coordinate method
The coordinate method calculates the cable linearity and torsion angle based on the geometry relationships among the coordinates of key points. In order to obtain the critical coordinates, several
key points (e.g., the sling lifting point, the midpoint between adjacent slings, and the saddle point) on the cable curve should be designated. At the location of each key point, a measurement device
with two displacement observation targets is preset horizontally to characterize the displacement caused by loads. The position changes on the observation targets can be measured by the total
stations. As shown in Fig. 2, the measurement device fastened to the main cable is not allowed to slide during testing, and the direction and length of the target plates on the device can be flexibly
adjusted. The measurement points on the target plate center are labeled as points $A$ and $B$ (see Fig. 2).
Fig. 2Measurement device and target plates
In the measurement system, the $X$, $Y$, and $Z$ axis are along the bridge’s lateral, longitudinal, and vertical directions. Before applying the testing load to the bridge model, the total station
can determine the initial coordinates of points $A\left({x}_{A},{y}_{A},{z}_{A}\right)$ and $B\left({x}_{B},{y}_{B},{z}_{B}\right)$. After the load is applied to the bridge model, the measurement
points move to the new positions ${A}^{"}\left({x}_{{A}^{"}},{y}_{{A}^{"}},{z}_{{A}^{"}}\right)$ and ${B}^{"}\left({x}_{{B}^{"}},{y}_{{B}^{"}},{z}_{{B}^{"}}\right)$. By using the measured
coordinates, the direction vector of a straight line, $S$, of points $A$ and $B$ can be expressed as:
$\left\{\begin{array}{l}{S}_{1}=\left({m}_{1},{n}_{1},{p}_{1}\right)=\left({x}_{B}-{x}_{A},{y}_{B}-{y}_{A},{z}_{B}-{z}_{A}\right),\\ {S}_{2}=\left({m}_{2},{n}_{2},{p}_{2}\right)=\left({x"}_{B}-{x"}_
where ${S}_{1}$ is the direction vector before loading; ${S}_{2}$ is the direction vector after loading.
According to the cosine formula of the angle, the angle $\theta$ between vectors ${S}_{1}$ and ${S}_{2}$ can be determined by Eq. (2):
$\left\{\begin{array}{l}\mathrm{c}\mathrm{o}\mathrm{s}\theta =\frac{{m}_{1}{m}_{2}+{n}_{1}{n}_{2}+{p}_{1}{p}_{2}}{\sqrt{{{m}_{1}}^{2}+{{n}_{1}}^{2}+{{p}_{1}}^{2}}\cdot \sqrt{{{m}_{2}}^{2}+{{n}_{2}}^
{2}+{{p}_{2}}^{2}}},\\ \theta =\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{{m}_{1}{m}_{2}+{n}_{1}{n}_{2}+{p}_{1}{p}_{2}}{\sqrt{{{m}_{1}}^{2}+{{n}_{1}}^{2}+{{p}_{1}}^{2}}\
cdot \sqrt{{{m}_{2}}^{2}+{{n}_{2}}^{2}+{{p}_{2}}^{2}}}\right).\end{array}\right\$
Analyzing the geometric relationships between the target plates makes it easy to know that the achieved $\theta$ is the cable torsion angle caused by loads. Normally, the torsion direction of the
main cable is identical to the transverse force of the slings (positive torsion). However, during the construction of the suspenders, anti-torsion may occur at the local position of the main cable (
negative torsion). The following conditions can be used to distinguish the twisting direction of the cable: (i) for ${m}_{1}{m}_{2}+{n}_{1}{n}_{2}+{p}_{1}{p}_{2}\ge$ 0, it is a positive torsion and
designated as "+"; and (ii) for ${m}_{1}{m}_{2}+{n}_{1}{n}_{2}+{p}_{1}{p}_{2}<$ 0, it is a negative torsion and designated as "–". It should be noted that as the main cable is constantly located at
the midpoint of strait line $AB$ (${A}^{"}{B}^{"}$), the introduced coordinate method can also be used to determine the cable curve at any spatial angle before and after deformation.
3.2. Angle method
The angle method determines cable torsion angle by deriving the direction vector from measured angles. Fig. 3 shows the inclinometers for angle measurement and the arrangement of the inclinometers.
As can be seen, in order to precede the angle measurement, two inclinometers are fixed to the main cable at the measurement point. The L-shaped component hinged on the block can swing freely along
the transverse direction, and its longitudinal and vertical directions are fixed. During the angle measurements, the position of the carrier axis changes, and the L-shaped component swing in the same
plane, contributing to a constant right triangle between the L-shaped component and the carrier axis. By installing the inclinometer on the horizontal level of the L-shaped component, the angle
between the projection of the carrier axis and the horizontal axis can be identified. With the help of this measurement system, the vertical angle $\alpha$and horizontal angle $\beta$ of the carrier
axis on the cable can be read from the inclinometers.
Fig. 3Configuration and arrangements of inclinometers
Fig. 4Angle method calculation diagram
In order to obtain the cable torsion angle, it is necessary to know the vector and plane angle of the carrier axis. Fig. 4 shows the geometrical relationship between the vectors. As can be seen, the
length of the carrier axis is $L$, and the carrier axis before and after loading is $MA$ and $NB$, respectively. The vertical angle was measured before and after loading ${\alpha }_{1}$ and ${\alpha
}_{2}$, respectively. The projection of the carrier axis in the $xoy$ horizontal plane is ${M}^{""}{A}^{""}$ and ${N}^{""}{B}^{""}$, respectively. The angle between ${M}^{""}{A}^{""}$and $x$ is ${\
beta }_{1}$, while the angle between ${N}^{""}{B}^{""}$ and $x$ is ${\beta }_{2}$. When the arrow of the inclinometer points upward, the ${\alpha }_{1}$ and ${\alpha }_{2}$ in the vertical plane and
the ${\beta }_{1}$ and ${\beta }_{2}$ in the horizontal plane are defined as positive, marked as "+".
Assume that the coordinate of point $M$ is $\left({x}_{M},{y}_{M},{z}_{M}\right)$, and the coordinate of point $A$ can be obtained by geometric relationship:
$\left\{\begin{array}{l}{x}_{A}={x}_{M}+L\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{1}\mathrm{c}\mathrm{o}\mathrm{s}{\beta }_{1},\\ {y}_{A}={y}_{M}+L\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{1}\mathrm
{s}\mathrm{i}\mathrm{n}{\beta }_{1},\\ {z}_{A}={z}_{M}+L\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_{1}.\end{array}\right\$
The change in points $A$ and $M$ is the direction vector ${S}_{1}$ of carrier axis $MA$:
$\left\{\begin{array}{l}{S}_{1}=\stackrel{\to }{MA}=\left({x}_{A}-{x}_{M},{y}_{A}-{y}_{M},{z}_{A}-{z}_{M}\right),\\ {S}_{1}=\left(L\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{1}\mathrm{c}\mathrm{o}\
mathrm{s}{\beta }_{1},L\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\beta }_{1},L\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_{1}\right).\end{array}\right\$
Similarly, the direction vector ${S}_{2}$ of carrier axis $NB$ can be obtained:
$\left\{\begin{array}{l}{S}_{2}=\stackrel{\to }{NB}=\left({x}_{B}-{x}_{N},{y}_{B}-{y}_{N},{z}_{B}-{z}_{N}\right),\\ {S}_{2}=\left(L\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{2}\mathrm{c}\mathrm{o}\
mathrm{s}{\beta }_{2},L\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{2}\mathrm{s}\mathrm{i}\mathrm{n}{\beta }_{2},L\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_{2}\right).\end{array}\right\$
The vector length of ${S}_{1}$ and ${S}_{2}$ is the axis length of carrier $L$, and the main cable torsion angle determined by ${S}_{1}$ and ${S}_{2}$ is:
$\left\{\begin{array}{l}\mathrm{c}\mathrm{o}\mathrm{s}\theta =\frac{{S}_{1}\cdot {S}_{2}}{\left|{S}_{1}\right|\left|{S}_{2}\right|}={k}_{1}{k}_{2}+{v}_{1}{v}_{2}+{w}_{1}{w}_{2},\\ \theta =\mathrm{a}\
where: ${k}_{1}=\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{1}\mathrm{c}\mathrm{o}\mathrm{s}{\beta }_{1}$, ${v}_{1}=\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\beta }_{1}$
, ${w}_{1}=\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_{1}$, ${k}_{2}=\mathrm{c}\mathrm{o}\mathrm{s}{\alpha }_{2}\mathrm{c}\mathrm{o}\mathrm{s}{\beta }_{2}$, ${v}_{2}=\mathrm{c}\mathrm{o}\mathrm{s}{\
alpha }_{2}\mathrm{s}\mathrm{i}\mathrm{n}{\beta }_{2}$, ${w}_{2}=\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_{2}$.
4. Model test and discussions
The experimental tests applied symmetrical loading to the bridge model from the middle to both sides. In order to examine the torsion angle of the main cable under different loading cases,
experimental results were obtained from the bridge model when the sling and plumb line possessed an inclination angle of 12° in transverse direction under 20 % and 80 % of the design load,
respectively, were analyzed. The sling transverse tensioning force triggers certain torsion in the spatial cable between the two suspension points, the midpoints between two adjacent slings, and that
between the side suspender and main tower are selected as the research object. For the convenience of discussion, the selected key points from east to west are numbered 1, 2, …, and 8. Table 1
presents a summary of the experimental data measured from the model.
With the help of the above-introduced calculation procedures, the torsion angles of the main cable in the test model caused by the loads are achieved and summarized in Table 2. As can be seen, the
coordinate method and angle method produced similar cable torsion angle results for the tested bridge model, showing a maximum relative error of 10.4 %. It is noticed that the difference between the
calculated torsion angles for the two methods is comparatively large for the key points near side spans. This can be ascribed to the fact that under the transverse tensioning effects of the slings,
the cable torsion angle is a spatial angle rather than a plane angle, and the changes in cable linearity near the towers are more significant than that near the mid-span. Table 2 also shows that the
coordinate method produced torsion angle is larger than the (${\alpha }_{2}-{\alpha }_{1}$). This also confirms that the main cable under the three-dimensional suspender force is spatially twisted.
Table 1Measured data before and after loading
Before loading After loading
Key point number
$X$ (m) $Y$ (m) $Z$ (m) ${\alpha }_{1}$[](°) ${\beta }_{1}$[](°) $X$ (m) $Y$ (m) $Z$ (m) ${\alpha }_{2}$[](°) ${\beta }_{2}$[](°)
1-1 –0.223 0.587 36.011 –0.139 0.521 35.923
1 16.5 6.3 36.0 21.4
1-2 0.406 0.685 36.264 0.337 0.777 36.343
2-1 0.172 3.401 34.121 0.194 3.440 34.147
2 3.8 4.5 7.4 6.0
2-2 0.834 3.287 34.086 0.848 3.302 34.076
3-1 0.399 5.458 33.005 0.423 5.481 33.005
3 –9.5 2.6 –4.8 4.0
3-2 1.049 5.448 33.131 1.080 5.453 33.081
4-1 0.511 7.618 32.469 0.535 7.619 32.429
4 7.2 2.4 14.5 3.5
4-2 1.135 7.634 32.746 1.192 7.629 32.623
5-1 0.487 10.047 32.636 0.525 10.036 32.575
5 –2.3 2.0 1.4 3.0
5-2 1.150 10.065 32.686 1.190 10.062 32.582
6-1 0.412 11.754 33.054 0.443 11.743 32.988
6 10.0 2.8 16.8 3.7
6-2 1.070 11.804 33.079 1.094 11.767 33.091
7-1 0.197 13.913 34.095 0.209 13.819 34.187
7 –7.8 5.8 4.3 9.5
7-2 0.852 13.910 34.218 0.870 13.904 34.209
8-1 –0.212 16.623 36.223 –0.191 16.529 36.315
8 1.6 9.0 17.4 21.6
8-2 0.459 16.676 36.296 0.448 16.727 36.208
Table 2Calculated torsion angle of main cable
Key point Angle by coordinate method $\ ${\alpha }_{2}-{\alpha }_{1} ${\beta }_{2}-{\beta }_{1} Angle by angle method $\ Angle difference by the two Relative error by the two
number theta$ (°) $ (°) $ (°) theta$ (°) methods (°) methods (%)
1 23.3 19.5 15.1 23.7 0.4 1.7
2 3.8 3.6 1.5 3.9 0.1 2.6
3 4.6 4.7 1.4 4.9 0.3 6.5
4 7.6 7.3 1.1 7.4 –0.2 –2.6
5 3.8 3.7 1.0 3.8 0.0 0.0
6 7.2 6.8 0.9 6.9 –0.3 –4.2
7 11.5 12.1 3.7 12.7 1.2 10.4
8 19.8 15.8 12.6 20.1 0.3 1.5
5. Conclusions
This paper investigates the calculation method for predicting the cable torsion angle during erecting the slings. A scale-reduced suspender bridge model was fabricated and tested to obtain the cable
torsion angle. Based on the deformation characteristics of the main cable, the coordinate method and angle method to calculate the spatial cable torsion angle in suspender bridges are developed. The
main conclusion drawn from this study includes:
1) Under the transverse tensioning effects of the slings, the cable torsion angle is a spatial angle rather than a plane angle, and the changes in cable linearity near the towers are more significant
than that near the mid-span.
2) The proposed coordinate and angle methods produced similar torsion angles for the main cables, showing a maximum relative error of 10.4 %. Despite the capability of calculating the spatial cable
torsion angles, the coordinate method also exhibits advanced capability in determining the geometric curves of the main cable under different spatial angles.
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About this article
Materials and measurements in engineering
suspension bridge
spatial cable
torsion angle
calculation method
The authors express their sincere gratitude for the support provided by the Changsha University of Science and Technology.
Data Availability
The datasets generated during and/or analyzed during the current study are available from the corresponding author on reasonable request.
Conflict of interest
The authors declare that they have no conflict of interest.
Copyright © 2023 Guangqing Xiao, et al.
This is an open access article distributed under the
Creative Commons Attribution License
, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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How many values of c satisfy the conclusion of the Mean Value Theorem for f(x) = x^3 + 1 on the interval [-1,1]? | Socratic
How many values of c satisfy the conclusion of the Mean Value Theorem for f(x) = x^3 + 1 on the interval [-1,1]?
1 Answer
There are two values of $c$ that work.
The derivative of $f$ is $f ' \left(x\right) = 3 {x}^{2}$. The slope of the secant line is $\frac{f \left(1\right) - f \left(- 1\right)}{1 - \left(- 1\right)} = \frac{2 - 0}{2} = 1$. Setting $f ' \
left(x\right) = 1$ and solving for $x$ results in $x = \setminus \pm \frac{1}{\sqrt{3}}$. These are the two "$c$" values, and they are in the interval $\left[- 1 , 1\right]$.
Here's a picture for the given situation:
The red line is the secant line between $\left(- 1 , f \left(- 1\right)\right) = \left(- 1 , 0\right)$ and $\left(1 , f \left(1\right)\right) = \left(1 , 2\right)$. The green line is the tangent line
to $f$ at $\left(- \frac{1}{\sqrt{3}} , f \left(- \frac{1}{\sqrt{3}}\right)\right) \setminus \approx \left(- 0.577 , 0.808\right)$ and the gold line is the tangent line to $f$ at $\left(\frac{1}{\
sqrt{3}} , f \left(\frac{1}{\sqrt{3}}\right)\right) \setminus \approx \left(0.577 , 1.192\right)$. All these lines are parallel with a slope of 1.
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Reducing Data Center Energy Consumption via Coordinated Cooling and Load Management
Reducing Data Center Energy Consumption via Coordinated Cooling and Load Management
Luca Parolini, Bruno Sinopoli, Bruce H. Krogh
Dept. of Electrical and Computer Engineering
Carnegie Mellon University
This paper presents a unified approach to data center energy management based on a modeling framework that characterizes the influence of key decision variables on computational performance, thermal
generation, and power consumption. Temperature dynamics are modeled by a network of interconnected components reflecting the spatial distribution of servers, computer room air conditioning (CRAC)
units, and non-computational components in the data center. A second network models the distribution of the computational load among the servers. Server power states influence both networks.
Formulating the control problem as a Markov decision process (MDP), the coordinated cooling and load management strategy minimizes the integrated weighted sum of power consumption and computational
performance. Simulation results for a small example illustrate the potential for a coordinated control strategy to achieve better energy management than traditional schemes that control the
computational and cooling subsystems separately. These results suggest several directions for further research.
C.0GeneralSystem architecture[Data center, Energy management, Energy efficiency, Thermal modeling, Optimization, Optimal control, Cyber-physical systems]
Data servers account for nearly 1.5% of total electricity consumption in the U.S. at a cost of approximately $4.5 billion per year [16]. Data center peak load exceeded 7 GW in 2006, and without
intervention is destined to reach 12 GW by 2011, about the power generated by 25 baseload power plants. To address this emerging problem, this paper advocates a coordinated approach to data center
energy management.
In most data centers today, the cooling and the computational subsystems are controlled independently. Computer room air conditioning (CRAC) units operate at levels necessary to keep the hottest
regions in the data center below critical temperature limits. Computational tasks are allocated to achieve the desired performance with server power states that minimize overall server energy
consumption. Some load management strategies consider the influence of the server loads on the temperature distribution, but CRAC control remains thermostatic. Coordinating cooling and load
management requires a unified modeling framework that represents the interactions of these two subsystems. We present the elements of such a framework and formulate the coordinated control problem as
a constrained Markov decision process (CMDP). Simulation results for a small system illustrate the potential benefits of this approach.
The following section reviews previous work on energy management strategies in data centers. Section 3 proposes an integrated thermal and computational model. Temperature dynamics are modeled by a
network of interconnected components reflecting the spatial distribution of servers, CRAC units, and non-computational components in the data center. A second network models the distribution of the
computational load among the servers. The server power states influence both networks. In Sec. 4, the energy management problem is formulated as a CMDP. Decision variables include the CRAC set
points, the allocation of tasks to servers, and server power states. The dynamic MDP strategy determines these control variables as a function of the system state to minimize the integral of the
weighted sum of power consumption and computational performance. Section 5 presents simulation results for a simple system, comparing the optimal MDP strategy to a conventional decoupled control of
the cooling and computational subsystems. The final section identifies several directions for further research.
Figure 1: A layered data center model: a computational network of server nodes (rectangles); and a thermal network of server nodes (rectangles), CRAC nodes (circles), and environment nodes
(diamonds). Server power states influence both networks.
2 Previous Work
Early work on energy minimization for data centers focused on computational fluid dynamic (CFD) models to analyze and design server racks and data center configurations to optimize the delivery of
cold air, and thereby minimize cooling costs [13,12]. Subsequent papers focused on the development of optimal load-balancing policies, at both the server and rack levels. Constraints on these
policies were either the minimum allowed computational performance or the maximum allowed peak power consumption [5,3,4,18,6,15].
Some recent papers have considered policies to minimize the total data center energy consumption by distributing the workload in a temperature-aware manner [11,2]. Research on load distribution based
on the thermal state of the data center, led to the development of fast and accurate estimation techniques to determine the temperature distribution in a data center, based on sparsely distributed
temperature measurements [10,14]. Approaches to the data center energy management problem based on queueing theory and Markov decision processes can be found in [9,17].
3 Integrated thermal - computational model
As illustrated in Fig. 1, we model a data center as two coupled networks: a computational network and a thermal network. The computational network describes the relationship between the streams of
data incoming from external users and generated by the data center itself and the computational load assigned to each server. The computational network is composed of server nodes that interact
through the exchange of workloads. We call jobs the computational requests that enter and leave the data center and tasks the computational requests exchanged within the data center. A single job can
be broken into multiple tasks and similarly, multiple tasks can be merged into a single job. A server's workload is then defined as the number of tasks arriving at the node per unit of time. The data
center networks interact with the external world by exchanging jobs at the computational network level and through the consumption of electrical power at the thermal network level.
The thermal network describes the thermal energy exchange at the physical level of the data center and accounts for the power consumed by the CRAC units, IT devices and non-computational devices. The
thermal network presents three types of nodes: server, CRAC and environment nodes. Environment nodes represent IT devices other than servers, non-computational devices and external environmental
effects (e.g. weather). Each node in the thermal network has an input temperature
Following the thermal model proposed in [14], we assume that the input temperature
Each computational server node is associated to a unique thermal server node. We call server node the combination of them. The coupling between the two parts of a server node is given by the power
state 7], we assume that each server node has a finite number of possible power states denoted by the set
We model the computational server node as a G/M/1 queue. The task inter-arrival distribution is given by a combination of the inter-arrival jobs at the data center and server interaction in the
computational network, while the service time is exponentially distributed with parameter 2, show that the thermal part of server node can be modeled as a first-order linear time-invariant (LTI)
system defined by the following differential equation:
Figure 2: Empirical modeling of thermal server nodes. Left: Server node power consumption. Right: measured
A server node can represent either a single server or a set of servers. In the second case it is necessary to identify aggregate definitions of the node power state, power consumption, input and
output temperatures.
CRAC nodes reduce the input temperatures of other thermal nodes. Their inputs are: the input air temperature
Environment nodes model all other subsystems that influence the data center thermal dynamics, including IT devices that cannot be directly controlled, non-computational devices, and the external
environment. The environment node output temperature is determined by a function
There are three sets of controllable variables in our proposed data center model: the computational network workload exchange, the server node power states and the CRAC node reference temperatures.
Standard scheduling algorithms, in this framework, become controllers that try to optimize either the power consumption of server nodes or the power consumption of the whole data center, relying only
on the information contained at the computational network level. Such an approach leads to sub-optimal solutions since the information given by the thermal network and consequently the possibility to
act on a greater number of variables are discarded. Even thermal aware scheduling algorithms are sub-optimal since they do not take advantage of the possibility to control the CRAC reference
4 CMDP Formulation
In order to synthesize data center energy management strategies, we use the model developed in the previous section to build a constrained Markov decision process (CMDP) [1]. The theory of CMDPs
provides a powerful framework for tackling dynamic decision-making problems under uncertainty. Clearly our control problem can be naturally cast as a dynamic decision making problem under
uncertainty. For example, we don't know when a task will arrive to a server node, or when it will be completed. CMDP techniques are the key to the synthesis of a controller when a system can be
described by a finite set of states and a finite set of actions. In these cases the stochastic dynamic control problem can be solved using standard linear programming techniques.
For the sake of clarity, we show how to formulate the data center energy management problem as a finite CMDP using a particular instantiation of the data center model. Generalization to an arbitrary
case is straightforward, but would require the introduction of more notation. Here we assume that the data center is composed of
In order to formulate our optimization problem as a finite CMDP we have to identify: a finite set states, a finite set actions from which the controller can choose at each step transition
probabilities representing the probability of moving from a state immediate costs for each time step. The total cost over a given time horizon is the sum of the cost incurred at each time step.
For this example, we assume jobs arrive only through the third server node, called the scheduler, and there is no workload exchange between server nodes one and two, as illustrated in Fig. 3.
Jobs arrive at the scheduler one at a time at the beginning of every time slot
In this example, there are no differences between jobs and tasks and the controller action on the computational network workload exchange reduces to the choice of the mean value of
The scheduler does not participate in thermal energy exchange between server nodes and CRAC node and its power consumption
CRAC node is assumed to have no dynamics and hence, its output temperature coincides with
In order to reach the goal of a finite set of states we have to discretize the set of admissible values for
We model the uncertainty on
The selected cost function is:
1 Optimization constraints
The constraints of the optimization problem are the maximum and minimum input temperature allowed by the two server nodes:
Another constraint we add imposes that each server node has to be set to its most conservative power state if no tasks will be sent to it and that its mean task execution rate has to be greater than
or equal to its task arrival rate.
The cost function we want minimize is:
where we denote with discount factor.
This particular cost function can be called a discounted energy^ cost since we are weighting the energy spent at the present time more than the one that will be spent in the far future.
5 Simulation Results
In order to solve the CMDP problem we use the Markov Decision Process Toolbox for MATLAB developed by the decision team of the Biometry and Artificial Intelligence Unit of INRA Toulouse, [8].
We compare our control algorithm against a typical hierarchical greedy scheduling algorithm that first chooses the best values for 6) and then it selects the best value for
In this simulation we assume that server node 1 consumes more power than server node 2 for the same value of the task execution rate, while its thermal effect on the input temperature of the CRAC
node is reduced with respect to the effects of server node 2. Finally we assume that the power consumption of the data center is mainly driven by the power of the CRAC node.
Figure 4: Reference temperature, input and output temperature of each node.
Figure: Optimal controller case.
Figure: Greedy controller case.
Figure 5: Tasks execution rate and scheduler switching probability for both the optimal and greedy controller case. Black vertical lines are set correspondently to the instants where
Figure: Task execution rate.
Figure: Switching probability
As shown in Fig. 4(a) and 4(b), using mainly server node one instead of the more power efficient server node two, the optimal controller is able to maintain a smaller difference between the input and
output temperature of the CRAC node if compared to the greedy controller. Hence, as shown in Fig. 7(a), the optimal controller reaches a lower data center power consumption profile than the greedy
controller and consequently a lower overall energy consumption, Fig. 7(b). As depicted in Fig. 6 energy minimization is obtained without heavily affecting the computational characteristics of the
data center.
This paper presents a coordinated cooling and load management strategy to reduce data center energy consumption. This strategy is based on a holistic and modular approach to data center modeling that
represents explicitly the coupling between the thermal and computational subsystems. This leads to a constrained Markov decision process (CMDP) formulation of the energy management problem for which
optimal strategies can be computed using linear programming. Simulation results for a small example illustrate the potential for a coordinated control strategy to achieve better energy management
than traditional schemes that control the computational and cooling subsystems separately. We are currently instrumenting a small data center at Carnegie Mellon to test our algorithms in a realistic
There are several directions for further research. The CMDP formulation will be intractable for detailed models of real systems due to the exponential growth of the state space. This problem can be
addressed by using hierarchical models. In our modeling framework, the nodes in each network can represent aggregate sets of devices, each of which can be modeled by lower-level thermal and
computational networks. We are currently developing the details of such hierarchical models. Hierarchical models will lead naturally to hierarchical and distributed control strategies. Initial steps
in this direction can be found in [15].
Figure 7: Comparison of energy and power consumption with optimal and greedy controller.
Figure: Power consumption.
Figure: Energy consumption.
Another issue regards the modeling and allocation of jobs and tasks in the computational network. Our current model assumes homogeneous and independent tasks and simple FIFO tasks queues at each
server. In real systems, there can be a complex mixture of tasks and servers work on more than one task at a time. There may also be dynamic exchanges and reallocation of tasks that are in progress.
We are currently investigating methods for incorporating and exploiting these features.
Finally, our model has several parameters that need to be determined for specific applications. This requires the development of effective system identification algorithms, which need to be
implemented on line since the parameter values can be time varying.
E. Altman.
Constrained Markov Decision Processes.
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C.-H. Lien, Y.-W. Bai, M.-B. Lin, P.-A. Chen.
The saving of energy in web server clusters by utilizing dynamic sever management.
IEEE Int. Conf. on Networks, Nov. 2004.
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Web server power estimation, modeling and management.
In IEEE, editor, ICON '06. 14th IEEE Int. Conf. on Networks, 2006.
W. Felter, K. Rajamani, T. Keller, and C. Rusu.
A performance-conserving approach for reducing peak power consumption in server systems.
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In ATEC, 2005.
Patel C. D., Bash C. E., Sharma R. K, Beitelmal A, Friedrich R. J.
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Patel, C.D., Sharma, R.K, Bash, C.E., Beitelmal.
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Sensor-based fast thermal evaluation model for energy efficient high-performance datacenters.
Oct. 2006.
R. Raghavendra, P. Ranganathan, V. Talwar, Z. Wang, X. Zhu.
No "power" struggles: coordinated multi-level power management for the data center.
ASPLOS, 2008.
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Report to congress on server and data center energy efficiency.
Technical report, ENERGY STAR Program, Aug. 2007.
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Managing server energy and operational costs in hosting centers.
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An energy-efficient management mechanism for large-scale server clusters.
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Reducing Data Center Energy Consumption via Coordinated Cooling and Load Management
This document was generated using the LaTeX2HTML translator Version 2002-2-1 (1.70)
Copyright © 1993, 1994, 1995, 1996, Nikos Drakos, Computer Based Learning Unit, University of Leeds.
Copyright © 1997, 1998, 1999, Ross Moore, Mathematics Department, Macquarie University, Sydney.
When 5) represents the expected energy spent by the discretized system.
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PKMDS for Web
Find PKMDS for Web on... the web! https://pkmds.app/
GitHub Repo: https://github.com/codemonkey85/PKMDS-Blazor
Issue tracker: https://github.com/codemonkey85/PKMDS-Blazor/issues
Hello all. Some of you might remember me and / or PKMDS, but most likely not. But I have returned from years of quiet meditation (work and family) to bring PKMDS back to a modern generation of
Introducing: PKMDS for Web!
Built as an ASP.NET Core web app using Blazor WebAssembly for UI, and leveraging PKHeX.Core as the logical foundation, PKMDS for Web is intended to be a save editor for all Pokémon games, with
support for all modern browsers (including mobile). Development will probably be very slow, since this is just a fun side-project for me. Contributions and suggestions are welcome - feel free to
create an issue and / or pull request on the GitHub repo.
Special thanks to @Kaphotics and everyone who contributed to PKHeX over the years. I'm standing on your collective shoulders (although feel free to steal as much as you want if you ever intend to
make a web-based PKHeX
If you have any issues, you may report them here, or feel free to create an issue on the GitHub repository.
EDIT: Please note, the app is under development and is super unfinished.
This web app is amazing! I've tried unsuccesfully to get the mobile app of PkHex working, but this appears to work great!
• 1
• 2 weeks later...
Currently getting a 403 error trying to reach the app, is this offline intentionally?
• 3 weeks later...
How to import saved pokemons?
Hi, I've been using this web app and it's amazing. I'm using it on my pokemon scarlet save file for a while but after the teal mask and indigo disk update, I can't edit it anymore because it won't
load and there's this unhandled error message at the bottom. Do you have any thoughts why? Thank you.
Wow, I had no idea people were using this app! I checked on this topic several times and hadn't seen any activity. Then for whatever reason I never received any notifications that there were new
posts. Sorry about the delayed response folks!
On 12/13/2023 at 5:40 PM, oddgo said:
Currently getting a 403 error trying to reach the app, is this offline intentionally?
I had moved hosting from Azure Static Web Apps to GitHub Pages, so I assume that's where you ran into the error.
On 1/8/2024 at 2:56 PM, Gray0001 said:
How to import saved pokemons?
How to import saved Pokémon from where? Once you make changes to a save file, you can export the file, and the exported copy will have your changes.
On 1/13/2024 at 8:31 PM, Lucent said:
Hi, I've been using this web app and it's amazing. I'm using it on my pokemon scarlet save file for a while but after the teal mask and indigo disk update, I can't edit it anymore because it
won't load and there's this unhandled error message at the bottom. Do you have any thoughts why? Thank you.
Can you please DM me your save file so I can take a look? Or, you can try again yourself - I just updated the app to the latest version of PKHeX.Core.
On 1/18/2024 at 11:26 PM, codemonkey85 said:
Can you please DM me your save file so I can take a look? Or, you can try again yourself - I just updated the app to the latest version of PKHeX.Core.
Sure, but I don't think I can DM you. Not allowed to send messages to your acc mate.
• 2 weeks later...
hello I use pkmds-web to increase the levels of my pokemon on hearthgold, and when I launch my game (on desmume emulator) it resets. helpppppp
• 2 weeks later...
I was wondering if it would be possible to have a way of copying the box and party data, like pkhex.
It may just be me being stupid, but I can't find a way to transfer my violet save file to my android. Can anyone help?
Is it possible to get a save file into a calculator? Like https://hzla.github.io/Dynamic-Calc/?data=8f199f3f40194ecc4b8e&dmgGen=4&gen=7&switchIn=4&types=5 ?
I finally found a way to use a save editor (the app) but i would really like to use it for my nuzlocks
• 4 weeks later...
On 2/5/2024 at 5:44 AM, furtimax said:
hello I use pkmds-web to increase the levels of my pokemon on hearthgold, and when I launch my game (on desmume emulator) it resets. helpppppp
I’m not sure what this means. Do you mean the changes are not appearing? Or do you mean your save file is being erased?
On 2/16/2024 at 10:11 PM, Gideonion said:
I was wondering if it would be possible to have a way of copying the box and party data, like pkhex.
What do you mean by copying the box and party data? Do you mean copying a Pokémon from one slot into another?
Congratulations on this amazing development. I loved it and I think is really really useful form some o us. I just want to suggest to add a drag and drop function for moving pokemon into de boxes.
I'll explain. Many of us use PokeHex to transport pokemon from game to game in emulators, what I propose is a Drag and drop function that convert let's say for example .pk1 to .pk8 automaticaly on
drag and drop. Since you can already export single pokemons, it would be extemely usefull. Thanks.
• 4 weeks later...
On 3/17/2024 at 9:16 PM, codemonkey85 said:
I’m not sure what this means. Do you mean the changes are not appearing? Or do you mean your save file is being erased?
What do you mean by copying the box and party data? Do you mean copying a Pokémon from one slot into another?
I assume they mean being able to export an entire box's showdown data just like it can be done with a party but I am not sure
• 3 weeks later...
On 3/17/2024 at 12:46 PM, codemonkey85 said:
What do you mean by copying the box and party data? Do you mean copying a Pokémon from one slot into another?
I think they mean transferring Pokemon from one save file to another because there is an option to export Pokemon from the save file but not one to import.
1 minute ago, VR2308 said:
I think they mean transferring Pokemon from one save file to another because there is an option to export Pokemon from the save file but not one to import.
I also tried "making" a new Pokemon but there was no option to insert the PID into a blank slot.
Hello there!
Such a great work for mobile user like me! I really appeciate it.
Got 1 question, Is there anyway to edit my pokemon status to FATEFUL ENCOUNTER in pkmds? I don’t see any options or checkbox whatsoever. Can you tell me how can I flag FATEFUL ENCOUNTER? or is it
just impossible for web version?
Thanks to everyone for your feedback!
On 3/18/2024 at 10:31 AM, AllanBrito said:
I just want to suggest to add a drag and drop function for moving pokemon into de boxes. I'll explain. Many of us use PokeHex to transport pokemon from game to game in emulators, what I propose
is a Drag and drop function that convert let's say for example .pk1 to .pk8 automaticaly on drag and drop. Since you can already export single pokemons, it would be extemely usefull. Thanks.
I can add drag and drop to move Pokémon around in boxes / the party within one game (I was already planning this), but I don't know if or how I can support transferring between games yet. Will have
to think about that one.
On 4/28/2024 at 8:51 PM, VR2308 said:
I also tried "making" a new Pokemon but there was no option to insert the PID into a blank slot.
I've added this to the 'Issues' section in the repo. It will get attention when I have time.
11 hours ago, PAPADOG said:
Got 1 question, Is there anyway to edit my pokemon status to FATEFUL ENCOUNTER in pkmds? I don’t see any options or checkbox whatsoever. Can you tell me how can I flag FATEFUL ENCOUNTER? or is it
just impossible for web version?
There is not yet a way to do this in the web app. I've added this to the 'Issues' section in the repo. It will get attention when I have time.
• 4 weeks later...
Hello, when I open the save file on the web app, it won't load, but when I open the game file itself it shows like where the things would be, any suggestions to fix this?
thanks for making this, cant add a salac berry as my held item or a jirachi. Is that intentional?
Is there a way to import individual pokemon which have been previously exported?
On 5/31/2024 at 11:35 AM, manic404 said:
thanks for making this, cant add a salac berry as my held item or a jirachi. Is that intentional?
It is not intentional. I'll need more info, such as which game you are working with.
On 6/2/2024 at 2:09 PM, Trillick said:
Is there a way to import individual pokemon which have been previously exported?
Not yet! I have a massive to-do list, and that's on there.
[S:Quick update: I've made some architectural changes to the app, so now the first time you visit the page, it will run off the server using a persistent connection. The next time you visit, it will
have silently downloaded an offline version, so it will run directly on your browser. The end result hopefully will be invisible to users, but it should mean that it starts a bit faster!:S]
This change has been reverted, as PKMDS for Web is now able to do everything in WASM, and to eliminate server costs to me.
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xtended Kalman
Perform simultaneous localization and mapping using extended Kalman filter
Since R2021b
The ekfSLAM object performs simultaneous localization and mapping (SLAM) using an extended Kalman filter (EKF). It takes in observed landmarks from the environment and compares them with known
landmarks to find associations and new landmarks. Use the associations to correct the state and state covariance. The new landmarks are augmented in the state vector.
slamObj = ekfSLAM creates an EKF SLAM object with default properties.
slamObj = ekfSLAM(Name,Value) sets properties using one or more name-value pair arguments in addition to any combination of input arguments from previous syntaxes. Any unspecified properties have
default values.
slamObj = ekfSLAM('MaxNumLandmark',N,Name,Value) specifies an upper bound on the number of landmarks N allowed in the state vector when generating code. This limit on the number of landmarks applies
only when generating code.
slamObj = ekfSLAM('MaxNumLandmark',N,‘MaxNumPoseStored’,M,Name,Value) specifies the maximum size of the pose history M along with the maximum number of landmarks N in the state vector while
generating code. These limits apply only when generating code.
You cannot change the value of the properties State, StateCovariance, StateTransitionFcn, and MaxNumLandmark after the object is created. Set the value of these properties as a default or while
creating the object.
State — State vector
[0; 0; 0] (default) | M-element column vector
State vector, specified as an M-element column vector.
Data Types: single | double
StateCovariance — State estimation error covariance
eye(3) (default) | M-by-M matrix
State estimation error covariance, specified as an M-by-M matrix. M is the number of states in the state vector.
Data Types: single | double
StateTransitionFcn — State transition function
nav.algs.velocityMotionModel (default) | function handle
State transition function, specified as a function handle. This function calculates the state vector at time step k from the state vector at time step k-1. The function can take additional input
parameters, such as control inputs or time step size.
The function also calculates the Jacobians with respect to the current pose and controller input. If not specified, the Jacobians are computed using numerical differencing at each call to the predict
function. This computation can increase processing time and numerical inaccuracy.
The function considers nonadditive process noise, and should have this signature:
[pose(k),jacPose,jacControl] = StateTransitionFcn(pose(k-1),controlInput,parameters)
• pose(k) is the estimated pose at time k.
• jacPose is the Jacobian of StateTransitionFcn with respect to pose(k-1).
• jacControl is the Jacobian of StateTransitionFcn with respect to controlInput.
• controlInput is the input for propagating the state.
• parameters are any additional arguments required by the state transition function.
Data Types: function_handle
MeasurementFcn — Measurement function
nav.algs.rangeBearingMeasurement (default) | function handle
Measurement function, specified as a function handle. This function calculates an N-element measurement vector for an M-element state vector.
The function also calculates the Jacobians with respect to the current pose and landmark position. If not specified, the Jacobians are computed using numerical differencing at each call to the
correct function. This computation can increase processing time and numerical inaccuracy.
The function considers additive measurement noise, and should have this signature:
[measurements(k),jacPose,jacLandmarks] = MeasurementFcn(pose(k),landmarks)
• pose(k) is the estimated pose at time k.
• measurements(k) is the estimated measurement at time k.
• landmarks are the positions of the landmarks.
• jacPose is the Jacobian of MeasurementFcn with respect to pose(k).
• jacLandmarks is the Jacobian of MeasurementFcn with respect to landmarks.
Data Types: function_handle
InverseMeasurementFcn — Inverse measurement function
nav.algs.rangeBearingInverseMeasurement (default) | function handle
Inverse measurement function, specified as a function handle. This function calculates the landmark position as an M-element state vector for an N-element measurement vector.
The function also calculates the Jacobians with respect to the current pose and measurement. If not specified, the Jacobians are computed using numerical differencing at each call to the correct
function. This computation can increase processing time and numerical inaccuracy.
The function should have this signature:
[landmarks(k),jacPose,jacMeasurements] = InverseMeasurementFcn(pose(k),measurements)
• pose(k) is the estimated pose at time k.
• landmarks(k) is the landmark position at time k.
• measurements are the observed landmarks at time k.
• jacPose is the Jacobian of InverseMeasurementFcn with respect to pose(k).
• jacMeasurements is the Jacobian of InverseMeasurementFcn with respect to measurements.
Data Types: function_handle
DataAssociationFcn — Data association function
nav.algs.associateMaxLikelihood (default) | function handle
Data association function, specified as a function handle. This function associates the measurements with the landmarks already available in the state vector. The function may take additional input
The function should have this signature:
[associations,newLandmarks] = DataAssociationFcn(knownLandmarks,knownLandmarksCovariance,observedLandmarks,observedLandmarksCovariance,parameters)
• knownLandmarks are known landmarks in the map.
• knownLandmarksCovariance is the covariance of knownLandmarks.
• observedLandmarks are the observed landmarks in the environment.
• observedLandmarksCovariance is the covariance of observedLandmarks.
• parameters are any additional arguments required.
• associations is a list of associations from knownLandmarks to observedLandmarks.
• newLandmarks are the indices of observedLandmarks that qualify as new landmarks.
Data Types: function_handle
ProcessNoise — Process noise covariance
eye(2) (default) | W-by-W matrix
Process noise covariance, specified as a W-by-W matrix. W is the number of process noise terms.
Data Types: single | double
MaxAssociationRange — Maximum range for landmarks to be checked for association
inf (default) | positive integer
Maximum range for the landmarks to be checked for association, specified as a positive integer.
Data Types: single | double
MaxNumLandmark — Maximum number of landmarks in state vector
inf (default) | positive integer
Maximum number of landmarks in the state vector, specified as a positive integer.
Data Types: single | double
MaxNumPoseStored — Maximum size of pose history
inf (default) | positive integer
Maximum size of pose history, specified as a positive integer.
Data Types: single | double
Object Functions
copy Create deep copy of EKF SLAM object
correct Correct state and state error covariance
landmarkInfo Retrieve landmark information
poseHistory Retrieve corrected and predicted pose history
predict Predict state and state error covariance
removeLandmark Remove landmark from state vector
reset Reset state and state estimation error covariance
Perform Landmark SLAM Using Extended Kalman Filter
Load a race track data set that contains the initial vehicle state, initial vehicle state covariance, process noise covariance, control input, time step size, measurement, measurement covariance, and
validation gate values.
load("racetrackDataset.mat","initialState","initialStateCovariance", ...
"processNoise","controllerInputs","timeStep", ...
Create an ekfSLAM object with initial state, initial state covariance, and process noise.
ekfSlamObj = ekfSLAM("State",initialState, ...
"StateCovariance",initialStateCovariance, ...
Initialize a variable to store the pose.
storedPose = nan(size(controllerInputs,1)+1,3);
storedPose(1,:) = ekfSlamObj.State(1:3);
Predict the state using the control input and time step size for the state transition function. Then, correct the state using the data of the observed landmarks, measurement covariance, and
validation gate for the data association function.
for count = 1:size(controllerInputs,1)
% Predict the state
% Get the landmarks in the environment
observedLandmarks = measurements{count};
% Correct the state
if ~isempty(observedLandmarks)
correct(ekfSlamObj,observedLandmarks, ...
% Log the estimated pose
storedPose(count+1,:) = ekfSlamObj.State(1:3);
Visualize the created map.
fig = figure;
figAx = axes(fig);
axis equal
grid minor
hold on
landmarks = reshape(ekfSlamObj.State(4:end),2,[])';
legend("Robot trajectory","Landmarks","Start","End")
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
Version History
Introduced in R2021b
See Also
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A329694 - OEIS
The Motzkin step set is U=(1,1), H=(1,0) and D=(1,-1). An excursion is a path starting at (0,0), ending at (n,0) and never crossing the x-axis, i.e., staying at nonnegative altitude.
G.f.: (1+t)*(1-2t^3-sqrt(1-4t^3-4t^4))/(2t^4).
a(4)=3 since we have the following 3 excursions of length 4: UHDH, HUHD and HUDH.
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National Mathematics Day Significance 2023 Archives - Universe Public School
Mathematics Day, on December 22, celebrates the birth anniversary of India’s famed Mathematician Srinivasa Ramanujan. Mathematicians consider the genius of Ramanujan to be comparable to that of the
eighteenth and nineteenth century’s Euler and Jacobi. He made significant advancements in the partition function and is highly acclaimed for his work in number theory. Since 2012, December 22 has
been designated as National Mathematics Day in India, with several educational activities taking place at colleges and universities around the nation on this day. In 2017, the day’s significance was
enhanced by the opening of the Ramanujan Math Park in Kuppam, in Chittoor, Andhra Pradesh. Mathematics lovers like Sri Ramanujan are all around the world and some even support others in enhancing
their knowledge on the subject.
The Theme of National Mathematics Day 2023
The Theme of National Mathematics Day 2023 is Mathematics for Everyone.
National Mathematics Day History
• In India, the idea for Mathematics Day was inspired by the well-known Mathematician Srinivasa Ramanujan, whose contributions have influenced people all across the world. Ramanujan was born in
1887 to an Lyengar Brahmin family in Erode, Tamil Nadu. He had received little formal education, yet by the age of twelve, he had mastered trigonometry and had developed several theorems of his
• Ramanujan ran away from home at the age of 14 and joined Pachaiyappa’s college in Madras. Ramanujan was unable to earn a Fellow of Arts degree because, like his peers there, he was unable to
excel in other subjects and could only succeed in mathematics. Ramanujan chose to conduct independent research in Mathematics despite his extreme poverty.
• The aspiring Mathematician quickly caught the attention of Chennai’s mathematicians. In 1912, the founder of the Indian Mathematical Society, Ramaswamy Iyer, helped him secure a clerkship at the
Madras Port Trust. Ramanujan subsequently started submitting his work to British Mathematicians. A Cambridge-based Mathematician named GH Hardy was so pleased by Ramanujan theorems that he
invited him to London in 1913.
• In 1914, Ramanujan traveled to Britain, where Hardy helped him enroll in Trinity College, Cambridge. Ramanujan was well on his way to success after being elected as a member of the London
Mathematical Society in 1917. He also became a fellow of the Royal Society in 1918, making him one of the youngest holders of this esteemed title.
Significance of National Mathematics Day 2023
Let’s take a closer look at National Mathematics Day 2023.
• On National Mathematics Day 2023, read about Trigonometry. Ramanujan developed the theorems of trigonometry on his own. It is among the most significant mathematical inventions.
• On National Mathematics Day 2023 try to Ramanujan biopic movie. The life of Ramanujan is faithfully portrayed in the film.
• As Ramanujan shows, each student has unique strengths and shortcomings. Try our hardest at all times, but don’t forget to help and motivate a student who does particularly well in a particular
• National Mathematics Day is observed to pay tribute to and celebrate the exceptional mathematician Srivasa Ramanujan, who taught himself more mathematics after dropping out of college due to his
struggles in other subjects.
Mathematics Day Timeline
• 1887- Ramanujan was born: Born into a poor Iyengar Brahmin family in Erode, Tamil Nadu, Ramanujan went on to become a talented mathematician who made a lasting contribution to the subject.
• 1918- A High Achievement: Ramanujan is made a fellow of the Royal Society, making him one of the youngest recipients in history. Immediately following his election to the British London
Mathematical Society membership.
• 2012- Mathematics Day is Recognized: In honor of Ramanujan’s accomplishments, former Indian Prime Minister Manmohan Singh proclaims December 22, the day of his birth, to be National Mathematics
• 2019- The Royal Society Honors Ramanujan: The prestigious Royal Society- The United Kingdom’s National Academy of Sciences- tweets a special message in honor of the fellow.
How to Celebrate Mathematics Day
• Read Up About Trignometry: Around the age of twelve, Srinivasa Ramanujan began writing stories. He had mastered trigonometry’s bewildering logic and was independently deriving theorems. Not
everyone feels the need to celebrate mathematics. It’s still an important subject. Why not study up on trigonometry or attempt to learn the topic yourself? It’s one of the most significant areas
in mathematics, and students ought to concentrate on it. Great Trignometry skills allow students to work out complex angles and dimensions in relatively little time.
• Watch the Movie About Ramanujan: The brilliant Mathematician inspired the field of Mathematics and many students across India. If you’re unfamiliar with him, you may see his amazing success story
from the comfort of your own home. Take a look at Dev Patel’s “The Man Who Knew Infinity.” it’s a great biopic of Ramanujan inspiring journey.
• Encourage Other Students Strengths: One crucial lesson to be learned from Srinivasa Ramanujan’s incredible life story and achievements in mathematics is that he persisted even in the face of
appalling performance in other areas like English, philosophy, and Sanskrit. This demonstrates that every student has unique talents and shortcomings. While it’s crucial to constantly try our
hardest, it’s also critical to support and encourage a student who shines in a particular topic. Who knows, maybe that compliment will help nurture their will to perform better and even pursue
their interests to amazing heights.
Why We Love Mathematics Day
• Mathematics is a Universal Language: Whether you enjoy it or not, mathematics is an essential part of the world, and without it, we couldn’t make any sense of it at all. Mathematics is a
methodical application of matter, makes our lives orderly, and prevents chaos. Certain qualities that are nurtured by Mathematics are the power of reasoning, abstract or spatial thinking,
creativity, problem-solving ability, critical thinking, and even more effective communication skills. A day worthy of celebration is one to honor this field.
• It Inspires Us to Educate Ourselves: The main goal of Mathematics Day is to honor and celebrate the outstanding mathematician Srinivasa Ramanujan, who dropped out of college because he didn’t
perform well in other areas and instead taught himself mathematics. Ramanujan studied mathematics on his own, living in extreme poverty and almost going hungry, without getting a formal degree.
Despite his difficult circumstances, his dedication and hard work helped him become one of the most well-known mathematicians of today. Hard work, and a little bit of luck, really can lead us to
fulfill our dreams.
• Practically Every Career Uses Math: The most fundamental tasks performed by mathematicians and scientists, including testing hypotheses, obviously depend on mathematical concepts. While
scientific careers famously involve math, they are not the only careers to do so. Basic arithmetic comprehension is even necessary for operating a cash register. To maintain track of the pieces
on the assembly line and, occasionally, to manage fabrication software that uses geometric features to produce their goods, workers in factories need to be proficient in mental math. Really, and
job requires math because you must know how to interpret your paycheck and balance your budget.
Events and Activities Related to National Mathematics Day
Different educational institutions organize several events for National Mathematics Day. Some run quiz contests with both general and math questions about the same subject. Universe Public School
encourages students to play games and puzzles like Suduko, number Jigsaw, Kakuro, word search, and other similar activities to add a little pleasure to their day. This allows children to enjoy math
in a slightly different way. Schools celebrate this Ramanujan day by holding a variety of events, from plays to skits, in addition to math-related activities. Students perform a whole skit to depict
the life of this great Mathematician and how he became one. In addition, light shows his contribution to the field of numbers so that students can become more knowledgeable about it and interested in
related fields.
A proper mathematics fair is held on the day so that all the students can participate in something related to math and then, later on, can look at each other’s projects to learn something new. Some
schools have taken a very different approach to celebrating this day, asking their students to come up with interesting math and physics projects beforehand.
Some interesting methods to celebrate this day include arithmetic workshops and talks, math arts and crafts, movie screenings, scavenger hunts, storytelling, math and technology expos, and
math-themed dress-up days.
When is National Mathematics Day 2023 observed?
National Mathematics Day 2023 is observed on 22nd December.
In which year National Mathematics Day was first observed?
National Mathematics Day was first observed in 2012.
What is the theme of National Mathematics Day 2023?
“Mathematics for Everyone” is the theme for National Mathematics Day 2023.
Who is the father of National Mathematics Day?
Since Mathematics Day is commemorated on his birthday, December 22, to honor his contributions to mathematics, the renowned Indian mathematician Srinivasa Ramanujan is regarded as the founder of the
Who is the father of Mathematics?
Archimedes, an Ancient Greek Mathematician, engineer, Astronomer, Physicist, and inventor is considered the father of Mathematics for his significant contribution to the development of Mathematics.
How do we celebrate Math Day?
Mathematical Day is observed in a variety of ways not only in India but also globally. In India, NASI hosts a workshop on Ramanujan and mathematics applications to commemorate National Mathematics
Day. The workshop is attended by popular lecturers and experts in the field of Mathematics from across the country.
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Editorial - AtCoder Regular Contest 127
Instead of strings, let us construct ternary numbers. Here, numbers are already padded with zeros, so being the lexicographically largest is equivalent to being the largest number.
We need \(N\) numbers that begin with \(2\). Thus, it is necessary that \(2 \times 3^{(L-1)} +N-1 \leq t\). Actually, we can achieve \(2 \times 3^{(L-1)} +N-1 = t\).
We begin with \(2 \times 3^{(L-1)},2 \times 3^{(L-1)}+1,\cdots,2 \times 3^{(L-1)}+N-1\). Let us convert each of these numbers as follows:
• in ternary, replace 0 with 1, 1 with 2, 2 with 0.
This results in numbers that all begin with 0 and are pairwise distinct. Similarly, perform the following conversion, too:
• in ternary, replace 0 with 2, 1 with 0, 2 with 1.
Now we can print these \(3N\) numbers that satisfy the conditions.
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How do you calculate velocity?
Determine the object’s original velocity by dividing the time it took for the object to travel a given distance by the total distance. In the equation V = d/t, V is the velocity, d is the distance
and t is the time.
What are the 3 formulas for motion?
• First Equation of Motion : v = u + a t.
• Second Equation of Motion : s = u t + 1 2 a t 2.
• Third Equation of Motion : v 2 = u 2 + 2 a s.
What are the 4 formulas of motion?
The equations are as follows: v=u+at,s=(u+v2)t,v2=u2+2as,s=ut+12at2,s=vtโ 12at2.
What are the 4 variables of motion?
Motion is described using the following variables, position, velocity, acceleration, and, implicitly, time.
How do you explain motion in physics?
motion, in physics, change with time of the position or orientation of a body. Motion along a line or a curve is called translation. Motion that changes the orientation of a body is called rotation.
How do you calculate motion?
Newton’s second law, which states that the force F acting on a body is equal to the mass m of the body multiplied by the acceleration a of its centre of mass, F = ma, is the basic equation of motion
in classical mechanics.
Is physics easy or hard?
Students and researchers alike have long understood that physics is challenging. But only now have scientists managed to prove it. It turns out that one of the most common goals in physicsโ finding
an equation that describes how a system changes over timeโ is defined as “hard” by computer theory.
Is velocity a speed?
Why is it incorrect to use the terms speed and velocity interchangeably? The reason is simple. Speed is the time rate at which an object is moving along a path, while velocity is the rate and
direction of an object’s movement.
Can velocity be negative?
Velocity: The velocity of an object is the change in position (displacement) over a time interval. Velocity includes both speed and direction, thus velocity can be either positive or negative while
speed can only be positive.
What is the formula for final velocity?
Final velocity (v) squared equals initial velocity (u) squared plus two times acceleration (a) times displacement (s). Solving for v, final velocity (v) equals the square root of initial velocity (u)
squared plus two times acceleration (a) times displacement (s).
What is the first equation of motion?
The first equation of motion is v = u +at. Here, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time. The velocity-time relation gives the first equation of
motion and can be used to find acceleration.
What is motion and example?
What is Motion? The free movement of a body with respect to time is known as motion. For example- the fan, the dust falling from the carpet, the water that flows from the tap, a ball rolling around,
a moving car etc. Even the universe is in continual motion.
Why should we study motion?
Force and motion are important parts of everyday life. As students study this unit, they will learn how these physical factors impact their lives and work. The lessons and activities will help
students become aware of factors like friction, gravity, and magnetic force.
What are the concepts of motion?
Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed and frame of reference to an observer and measuring the change in position of the body relative to
that frame with change in time.
What is velocity vs acceleration?
Velocity is the rate of displacement of an object. It is measured in m/s. Acceleration is the rate of change of velocity of an object.
How do you solve for acceleration?
Summary. According to Newton’s second law of motion, the acceleration of an object equals the net force acting on it divided by its mass, or a = F m . This equation for acceleration can be used to
calculate the acceleration of an object when its mass and the net force acting on it are known.
How do you find kinetic energy?
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the
kinetic energy has units of kilograms-meters squared per second squared.
What are Newton’s 1st 2nd and 3rd laws of motion?
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two
objects interact, they apply forces to each other of equal magnitude and opposite direction.
What is the 5 description of motion?
Mathematical terms used to describe motion include displacement, distance, velocity, acceleration, speed, and time.
How do you solve a motion question?
Is physics harder than calculus?
Physics is absolutely harder than calculus. Calculus is an intermediate level of mathematics that is usually taught during the first two years of most STEM majors. Physics on the other hand is a very
advanced and difficult and highly researched field.
Is physics harder than chemistry?
Physics is considered comparatively harder than chemistry and various other disciplines such as psychology, geology, biology, astronomy, computer science, and biochemistry. It is deemed difficult
compared to other fields because the variety of abstract concepts and the level of maths in physics is incomparable.
Is physics harder than biology?
Beginning university students in the sciences usually consider biology to be much easier than physics or chemistry. From their experience in high school, physics has math and formulae that must be
understood to be applied correctly, but the study of biology relies mainly on memorization.
Is time a vector or scalar?
For example, displacement, velocity, and acceleration are vector quantities, while speed (the magnitude of velocity), time, and mass are scalars.
Is velocity a scalar or vector?
Speed is a scalar quantity โ it is the rate of change in the distance travelled by an object, while velocity is a vector quantity โ it is the speed of an object in a particular direction.
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On distance dominator packing coloring in graphs
Let $G$ be a graph and let $S=(s_1, s_2, \ldots, s_k)$ be a non-decreasing sequence of positive integers. An {\em $S$-packing coloring} of $G$ is a mapping $c: V(G) \rightarrow \{1,2, \ldots, k\}$
with the following property: if $c(u)=c(v)=i$, then $d(u, v) > s_i$ for any $i \in \{1, 2, \ldots, k\}$. In particular, if $S=(1,2,3, \ldots, k)$, then $S$-packing coloring of $G$ is well known under
the name \textit{packing coloring}. %The smallest integer $k$ such that there exists an packing coloring of $G$, is called the {\em packing chromatic number} of $G$ and is denoted by $\chi_{\rho}(G)
Next, let $r$ be a positive integer and $u, v \in V(G)$. A vertex $u$ \textit{$r$-distance dominates} a vertex $v$ if $d_G(u,v) \leq r$.
In this paper, we present a new concept of a coloring, namely \textit{distance dominator packing coloring}, defined as follows. A coloring $c$ is a \textit{distance dominator packing coloring} of $G$
if it is a packing coloring of $G$ and for each $x \in V(G)$ there exists $i \in \{1,2,3, \ldots\}$ such that $x$ $i$-distance dominates each vertex from the color class of color $i$. The smallest
integer $k$ such that there exists a distance dominator packing coloring of $G$ using $k$ colors, is the \textit{distance dominator packing chromatic number} of $G$, denoted by $\chi_{\rho}^d(G)$.
In this paper, we provide some lower and upper bounds on the distance dominator packing chromatic number, characterize graphs $G$ with $\chi_{\rho}^d(G) \in \{2,3\}$, and provide the exact values of
$\chi_{\rho}^d(G)$ when $G$ is a complete graph, a star, a wheel, a cycle or a path. In addition, we consider the relation between $\chi_\rho(G)$ and $\chi_{\rho}^d(G)$ for a graph $G$.
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Main [14]. To get a distinctive viewpoint, the readers may perhaps consult Reference [15]. two. | m-entrepreneurship.com
Main [14]. To get a distinctive viewpoint, the readers may perhaps consult Reference [15]. two. Graph Coverings and Conjugacy Classes of a Finitely Generated Group Let rel( x1 , x2 , . . . , xr ) be
the relation defining the finitely presented group f p = x1 , x2 , . . . , xr |rel( x1 , x2 , . . . , xr ) on r letters (or generators). We are serious about the conjugacy classes (cc) of subgroups
of f p with respect towards the nature of the relation rel. In a nutshell, 1 observes that the cardinality structure d ( f p) of conjugacy classes of subgroups of index d of f p is all of the closer
to that on the no cost group Fr-1 on r – 1 generators as the option of rel consists of far more non local structure. To arrive at this statement, we experiment on protein foldings, musical types and
poems. The former case was very first explored in [3]. Let X and X be two graphs. A graph epimorphism (an onto or surjective homomor phism) : X X is known as a ML-SA1 site covering projection if, for
every vertex v of X, maps the neighborhood of v bijectively onto the neighborhood of v. The graph X is referred to as a base graph (or possibly a quotient graph) and X is named the covering graph.
The conjugacy classes of subgroups of index d within the fundamental group of a base graph X are in one-to-one correspondence with the connected d-fold coverings of X, as it has been known for some
time [16,17]. Graph coverings and group actions are closely related. Let us get started from an enumeration of integer partitions of d that satisfy:Sci 2021, 3,3 ofl1 2l2 . . . dld = d, a well-known
trouble in analytic quantity theory [18,19]. The number of such partitions is p(d) = [1, 2, 3, five, 7, 11, 15, 22 ] when d = [1, 2, 3, four, five, 6, 7, eight ]. The number of d-fold coverings of a
graph X from the first Betti number r is ([17], p. 41), Iso( X; d) =l1 2l2 …dld =d(l1 !2l2 l2 ! . . . dld ld !)r-1 .Yet another interpretation of Iso( X; d) is found in ([20], Euqation (12)). Taking
a set of mixed quantum states Icosabutate In Vitro comprising r 1 subsystems, Iso( X; d) corresponds to the steady dimension of degree d regional unitary invariants. For two subsystems, r = 1 and
such a stable dimension is Iso( X; d) = p(d). A table for Iso( X, d) with modest d’s is in ([17], Table three.1, p. 82) or ([20], Table 1). Then, 1 desires a theorem derived by Hall in 1949 [21]
concerning the quantity Nd,r of subgroups of index d in Fr Nd,r = d(d!)r-1 -d -1 i =[(d – i)!]r-1 Ni,rto establish that the number Isoc( X; d) of connected d-fold coverings of a graph X (alias the
number of conjugacy classes of subgroups in the fundamental group of X) is as follows ([17], Theorem 3.2, p. 84): Isoc( X; d) = 1 dm|dNm,r d l| md l (r -1) m 1 , mlwhere denotes the number-theoretic
M ius function. Table 1 offers the values of Isoc( X; d) for smaller values of r and d ([17], Table 3.two).Table 1. The quantity Isoc( X; d) for smaller values of initially Betti number r (alias the
number of generators on the absolutely free group Fr ) and index d. As a result, the columns correspond to the quantity of conjugacy classes of subgroups of index d within the absolutely free group
of rank r. r 1 2 three four 5 d=1 1 1 1 1 1 d=2 1 three 7 15 31 d=3 1 7 41 235 1361 d=4 1 26 604 14,120 334,576 d=5 1 97 13,753 1,712,845 207,009,649 d=6 1 624 504,243 371,515,454 268,530,771,271 d=7
1 4163 24,824,785 127,635,996,839 644,969,015,852,The finitely presented groups G = f p may possibly be characterized when it comes to a very first Betti number r. To get a group G, r will be the
rank (the number of generators) from the abelian quotient G/[ G, G ]. To some extent, a group f p whose 1st Betti numb.
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Mathematical modeling and stochastic H∞ identification of the dynamics of the MF-influenced oxidation of hexane
This paper presents a mathematical model explicitly reflecting the magnetic-field-induced transitions in a biologically significant process: n-hexane oxidation. The range of the magnetic field
strength is found (0.05-0.3 T) with the trend indicating significant magnetic-field-induced change in the rates of reactions involving hexane (up to 50% at 0.2 T). The equations describing the
effects of the magnetic field on the photoinduced free radical reaction of oxidation involving a lipid-modeling substance, hexane, are obtained on the basis of chemical kinetics and data from a batch
experiment. The magnetic-field-induced changes in n-hexane oxidation are validated using the identification technique based on the real time input-output data in a separately conducted flow-through
• Lipid-modeling substance
• Magnetic-field-induced transitions
• Mathematical modeling
• Oxidation
• Stochastic H∞ dentification
ASJC Scopus subject areas
• Statistics and Probability
• Modeling and Simulation
• General Biochemistry, Genetics and Molecular Biology
• General Immunology and Microbiology
• General Agricultural and Biological Sciences
• Applied Mathematics
Dive into the research topics of 'Mathematical modeling and stochastic H∞ identification of the dynamics of the MF-influenced oxidation of hexane'. Together they form a unique fingerprint.
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What is 24 VDC Filtered, Regulated, and Peak Voltage?
Power Converter AC (Alternating Current)
A power supply doesn't supply power; a power supply converts power. This is true of a DC power supply, and this is true of the utility power supplying power to the power supply.
At the utility power generating station, they take the power that has been converted to steam, the power that has come from a dam across a river, the power that comes from wind, and then turn
generators, creating the utility power. The power they generate is in the thousands of volts, then converted to higher voltage and finally down to the 120 Volts AC (Alternating Current) or 230 Volts
AC we think of as utility power.
Like plug-in power supply or a power supply panel, the utility power company didn't create the power, they just converted it.
Peak-to-Peak Voltage
Here, the voltage goes from 10 Volts Positive Peak to 10 Volts Negative Peak, or 20 Volts Peak-to-Peak
AC Equivalent Voltage or RMS Voltage
Pictured here is 7.07 V RMS. Whether it's AC Sinusoidal voltage or rectified, pulsating voltage, RMS voltage is 70.7 % of the Peak Sinusoidal voltage.
The reason that it's called AC is that current goes one direction, and then the other direction, and then it goes the first direction again. The current does this double reversal 50 or 60 times a
second (50 Hz or 60 Hz).
The 120 Volts AC or 230 Volts AC is not true voltage, the voltage is the heating value of what a steady DC voltage would be; it's the voltage in RMS. RMS or Root-Mean-Square is a mathematical way of
measuring the heating value from varying voltage.
In other words, the AC RMS voltage is the equivalent of the DC voltage that would light a light bulb. If you have a battery of 120 volts, the 120 Volts AC RMS would light the light bulb exactly a
brightly as 120 Volts DC of the battery; if you have a battery of 230 volts, the 230 Volts AC RMS would light the light bulb exactly a brightly as 230 Volts DC of the battery.
The trouble is that the AC voltage isn't constant. The AC voltage goes gradually from negative to positive and back to negative. (OK, for us humans 50 Hz or 60 HZ is extremely fast, but for a
computer, 50 Hz or 60 Hz is very slow motion, so this alternating voltage is slowly changing back and forth.)
The voltage starts out at a negative peak, quite a bit more than the 120 volts or 230 volts, goes up to zero volts, continues until it reaches a positive peak of quite a bit more than 120 volts or
230 volts, and then goes back past zero volts to the negative peak.
It does this 50 or 60 times a second, and the heating value (somewhat similar to averaging the voltage) is the same as a 120 Volt or 230 Volt battery.
Peak Voltage
Many times, rectification is performed using diodes, like a full wave bridge rectifier (four diodes arranged inside a small black box).
In this case, the negative peaks on the AC power have been reversed so that along with the normally positive peaks, they are also positive.
Reversing the plus and minus leads on the Full Wave Bridge Rectifier will produce negative peaks rather than positive peaks.
The positive and negative peaks in voltage, like I said, is higher than the RMS voltage. For 120 Volts AC RMS voltage, the positive and negative peaks in voltage is 169 Volts, for 230 Volts AC RMS,
the peaks are 325 Volts.
The voltage is at the peak voltages only momentarily; and most the time, the voltage is moving between the peaks, so the instantaneous voltages are usually lower.
To find peak-to-peak voltage, measure the total voltage between the positive peaks and the negative peaks. For 120 Volts AC RMS, the peak-to-peak voltage is 388 Volts, and for 230 Volts AC RMS, the
peak-to-peak voltage is 650 Volts.
A Power Supply Converts in Stages
There are three power supply output types: "Rectified", "Filtered", and "Regulated".
Pictured here is AC power (50Hz or 60 Hz) that has been rectified into pulsating DC. The voltage is pulsing at a rate of 100 to 120 times a second.
The rectified voltage is 10 Volts Peak. If it wasn't rectified, the voltage would be 20 Volts Peak-to-Peak.
A DC voltmeter may read 7.07 Volts (RMS Voltage), or the DC voltmeter may read 10 Volts (Peak Voltage). As "Ripple Voltage", an AC meter may read 7.07 volts, 10 volts, or some other voltage. You
can't count on exact AC voltage readings, but if the AC meter shows significant voltage, at least you can count on there being significant voltage.
- This is the cheapest to build power supplies. Using diodes, the transformer-reduced AC voltage is changed from an AC RMS voltage (where the voltage is alternating
positive-negative-positive-negative) to a voltage that pulses positive-positive-positive-positive . . . or negative-negative-negative-negative . . .
Basically, if it's a
24 Volt rectified power supply
, it's still the 24 Volts RMS that's described earlier, with a peak voltage of about 34 Volts.
Because there's almost nothing else added to it, there's a "Ripple Voltage" of 34 Volts Peak-to-Peak; the voltage is continually changing between zero volts and 34 Volts. If you use a DC Voltmeter on
this, most DC Voltmeters will show 24 Volts. If you use an AC Voltmeter on this, the AC Voltmeter will show something like 10 Volts AC to 34 Volts AC, depending on the AC Voltmeter.
As a contrast, an AC Voltmeter will show about Zero volts when measuring the AC Voltage on a battery because there's no ripple voltage.
Here, the power at the peak voltage charges up a capacitor, and the capacitor provides a more continuous power. The capacitor is used to filter out most of the ripple from the rectified power.
A DC voltmeter will show close to 10 Volts, an AC voltmeter will show less than 2 volts (ripple voltage).
- This is a rectified power supply that has added to it capacitors, resistors, and sometimes a few other components that can be used to "Filter Out" some or all of the AC "Ripple Voltage". The actual
voltage on the output can go up down like a nominal power supply. This varying can be from changes in the 120 Volts AC or 230 Volts AC input, changes in current use by the equipment being supplied
power, or when the power supply is on "Battery Backup".
Here, the filtered power is held to a constant voltage. To keep it constant, the regulated voltage is always less than the rectified voltage.
A DC voltmeter will show a constant voltage, here somewhere around 8 Volts (unvarying). An AC voltmeter will show zero volts.
- This is a rectified and filtered power supply that also has added circuitry to keep the voltage exactly the same, even when utility power voltage and changes, or the load current changes.
Nowadays, some power supplies inside fire alarm systems are using these regulated power supplies. Watch out, though. Some of the regulated power supplies may say "Regulated", but when the supply goes
to emergency battery backup power, the power supply might become "Nominal Voltage" because the batteries are nominal voltage.
Battery Voltage - Nominal Voltage
Most power supplies for fire alarm systems are 24 Volt DC power supplies, but that's 24 volts nominal. They use transformers to reduce the 120 Volts AC or 230 Volts AC to what can be used in the rest
of the power supply. The actual voltage is usually tied to the battery voltage, which is nominal.
A 12 Volt Nominal power supply is not 12 volts, the 12 volts only refers to the name (Nominal), not to the actual voltage. The same is true of a 24 Volt Nominal power supply commonly used in fire
alarm systems. The 24 volts only refers to the name, not to the actual voltage.
The label on batteries has only a little to do with what voltage can be expected from the battery. If the battery is a 24 Volt battery, the voltage of the battery can be as low as 20 Volts for a
nearly dead battery or as high as 27.2 Volts for a fully charged battery. The battery's name may be 24 Volts, but the voltage is anywhere between 20 Volts and 27.2 Volts.
Nominal means "In Name Only". A fire alarm system may have a power supply that is named 24 Volts, but in reality, the actual voltage of the power supply is almost always somewhere between 20 Volts
and 27 Volts.
Douglas Krantz
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[QSMS Monthly Seminar] Closed orbits and Beatty sequences
Date: 6월24일 금요일, 2021년
Place: 129동
Title: Closed orbits and Beatty sequences
Speaker: 강정수 교수
A long-standing open question in Hamiltonian dynamics asks whether every strictly convex hypersurface in R^{2n} carries at least n closed orbits. This was answered affirmatively in the non-degenerate
case by Long and Zhu in 2002. The aim of this talk is to outline their proof and to highlight its connection to partitioning the set positive integers.
Title: On the $\tilde{H}$-cobordism group of $S^1 \times S^2$'s
Speaker: 이동수 박사
Kawauchi defined a group structure on the set of homology S1 × S2’s under an equivalence relation called $\tidle{H}$-cobordism. This group receives a homomorphism from the knot concordance group,
given by the operation of zero-surgery. In this talk, we review knot concordance invariants derived from knot Floer complex, and apply them to show that the kernel of the zero-surgery homomorphism
contains an infinite rank subgroup generated by topologically slice knots.
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How to get the offset value of the first of the last so many times a condition occurred?
I'm essentially trying to get the offset value of the first "peak" that occurred out of the last so many "peaks" that occurred of a moving average, but I'm having a hard time figuring out how to go
about that. I feel like there's a simple solution to this that I just can't seem to pinpoint. Please refer to the code below:
input length = 12;
def avg = MovingAverage(averageType.SIMPLE, close, length);
def avgpeak = avg < avg[1] and avg[1] > avg[2];
def avgpeakprice = if avgpeak then avg[1] else 0;
What I want to do is to make an average out of the "
"s. So let's say I want an average price of the last 5 "
"s. In my mind, this means I need to somehow get the offset value of the first of the last 5 "
"s that occurred so I can use the offset value to add up all of the "
"s of the last 5 "
"s using the Sum function. I'd then divide the sum of the last 5 "
"s by 5 to get the average I'm looking for:
def peakpriceaverage = sum(avgpeakprice, <offset value of the first of the last 5 avgpeaks>) / 5;
I feel like I'm overthinking or overcomplicating something and I'll honestly be surprised if there isn't someone out there who can find a simple solution to my issue. I'm fairly certain I'm going to
have a big "DUH" moment as soon as someone answers my question.
I'm essentially trying to get the offset value of the first "peak" that occurred out of the last so many "peaks" that occurred of a moving average, but I'm having a hard time figuring out how to
go about that. I feel like there's a simple solution to this that I just can't seem to pinpoint. Please refer to the code below:
input length = 12;
def avg = MovingAverage(averageType.SIMPLE, close, length);
def avgpeak = avg < avg[1] and avg[1] > avg[2];
def avgpeakprice = if avgpeak then avg[1] else 0;
What I want to do is to make an average out of the "
"s. So let's say I want an average price of the last 5 "
"s. In my mind, this means I need to somehow get the offset value of the first of the last 5 "
"s that occurred so I can use the offset value to add up all of the "
"s of the last 5 "
"s using the Sum function. I'd then divide the sum of the last 5 "
"s by 5 to get the average I'm looking for:
def peakpriceaverage = sum(avgpeakprice, <offset value of the first of the last 5 avgpeaks>) / 5;
I feel like I'm overthinking or overcomplicating something and I'll honestly be surprised if there isn't someone out there who can find a simple solution to my issue. I'm fairly certain I'm going
to have a big "DUH" moment as soon as someone answers my question.
like many people with an idea, you are jumping around with your thoughts and guessing at functions to use, before you define the tasks to do.
you didn't explain what you want to do with these numbers. what do you want to see? a label? a line?
what you want, determines how and when to collect the data.
asking these 3 things will help steer your project.
..where - on a chart
..what - ? a line ? a label ?
..when - a line spanning the last 5 peaks? the last 5 bars ?
if you want an average of several values, then your first paragraph is irrelevent. it doesn't matter where the first in a group is. you just need a way to look at previous bars until you have found
the desired quantity of signals. then add the values as you find them. then divide the total by the quantity to get an average.
sum() looks at many bars, so you might read from 50 bars and divide by 5, which would be wrong. might be able to use an if then to supply data to sum(), but the length will always be different, and i
don't think thinkscript will allow a variable for sum length.
avgavg = sum( (if peak then avg alse 0), length ) / 5
what i did,
define peaks on an average
count them
create a reverse count
if the reverse count <= the peak quantity then thet are the last 5 peaks, process them.
add up the 5 peak values then /5, to get an average
draw a horizontal line at the average level
# avg_of_prev_x_peaks_01_rev
def na = double.nan;
def bn = barnumber();
def lastbn = HighestAll(If(IsNaN(close), 0, bn));
def lastbar = if (bn == lastbn) then 1 else 0;
#def lastbar = !isnan(close[0]) and isnan(close[-1]);
def price = close;
input peak_qtys = 5;
input avg1_len = 12;
#input ma1_type = AverageType.EXPONENTIAL;
input avg1_type = AverageType.simple;
def avg1 = MovingAverage(avg1_type, price, avg1_len);
input show_avg = yes;
plot z1 = if show_avg then avg1 else na;
# avg peaks , ref is middle bar , 1 bar on each side
#def avgpeak = avg1[1] < avg1[0] and avg1[0] > avg1[-1];
def avgpeak = if lastbar then 0 else (avg1[1] < avg1[0] and avg1[0] > avg1[-1]);
# ref 3rd bar , low, high, low , bar after peak
#def avgpeak = avg1[2] < avg1[1] and avg1[1] > avg1[0];
def t = 0;
# count the peaks
def avgpeakcnt = if bn == 1 then 1
else if avgpeak[t] then avgpeakcnt[1] + 1
else avgpeakcnt[1];
def maxcnt = highestall(avgpeakcnt);
def revcnt = maxcnt - avgpeakcnt + 1;
# define the group of bars that include x peaks
def ispeaks = (avgpeak[t] and revcnt <= peak_qtys);
def peakrange = (revcnt <= peak_qtys);
def avgtotal = if bn == 1 then 0
else if lastbar then avgtotal[1]
else if ispeaks then round(avgtotal[1] + avg1, 2)
else avgtotal[1];
def avgavg = round(avgtotal / peak_qtys, 2);
def avgavg2 = highestall(if lastbar then avgavg else 0);
plot za = if peakrange then avgavg2 else na;
addlabel(1, " ", color.black);
addlabel(0, peak_qtys + " peaks", color.yellow);
addlabel(1, avgavg2 + " average of the last " + peak_qtys + " peaks, on an average", color.yellow);
addlabel(1, " ", color.black);
input test1_peaks = yes;
addchartbubble(test1_peaks and avgpeak[t], high,
, (if peakrange then color.yellow else color.gray), yes);
input test2_avg = no;
#addchartbubble(test2_avg and (revcnt <= peak_qtys), low,
addchartbubble(test2_avg, low,
revcnt + "\n" +
peak_qtys + "\n" +
avgavg2 + "\n" +
ispeaks + "\n" +
, (if ispeaks then color.yellow else color.gray), no);
#, color.yellow, no);
# peak data
input test1_peak_cnts = no;
addchartbubble(test1_peak_cnts and avgpeak[t], avg1,
avg1 + "\n" +
avgpeak[t] + "\n" +
avgtotal + "\n" +
avgavg + "\n\n" +
avgpeakcnt + "\n" +
maxcnt + "\n" +
, (if ispeaks then color.yellow else color.gray), yes);
MRNA 15min
find 5 peaks on an average line
blue line - average of those 5 peaks
Join useThinkScript to post your question to a community of 21,000+ developers and traders.
like many people with an idea, you are jumping around with your thoughts and guessing at functions to use, before you define the tasks to do.
you didn't explain what you want to do with these numbers. what do you want to see? a label? a line?
what you want, determines how and when to collect the data.
asking these 3 things will help steer your project.
..where - on a chart
..what - ? a line ? a label ?
..when - a line spanning the last 5 peaks? the last 5 bars ?
if you want an average of several values, then your first paragraph is irrelevent. it doesn't matter where the first in a group is. you just need a way to look at previous bars until you have
found the desired quantity of signals. then add the values as you find them. then divide the total by the quantity to get an average.
sum() looks at many bars, so you might read from 50 bars and divide by 5, which would be wrong. might be able to use an if then to supply data to sum(), but the length will always be different,
and i don't think thinkscript will allow a variable for sum length.
avgavg = sum( (if peak then avg alse 0), length ) / 5
what i did,
define peaks on an average
count them
create a reverse count
if the reverse count <= the peak quantity then thet are the last 5 peaks, process them.
add up the 5 peak values then /5, to get an average
draw a horizontal line at the average level
# avg_of_prev_x_peaks_01_rev
def na = double.nan;
def bn = barnumber();
def lastbn = HighestAll(If(IsNaN(close), 0, bn));
def lastbar = if (bn == lastbn) then 1 else 0;
#def lastbar = !isnan(close[0]) and isnan(close[-1]);
def price = close;
input peak_qtys = 5;
input avg1_len = 12;
#input ma1_type = AverageType.EXPONENTIAL;
input avg1_type = AverageType.simple;
def avg1 = MovingAverage(avg1_type, price, avg1_len);
input show_avg = yes;
plot z1 = if show_avg then avg1 else na;
# avg peaks , ref is middle bar , 1 bar on each side
#def avgpeak = avg1[1] < avg1[0] and avg1[0] > avg1[-1];
def avgpeak = if lastbar then 0 else (avg1[1] < avg1[0] and avg1[0] > avg1[-1]);
# ref 3rd bar , low, high, low , bar after peak
#def avgpeak = avg1[2] < avg1[1] and avg1[1] > avg1[0];
def t = 0;
# count the peaks
def avgpeakcnt = if bn == 1 then 1
else if avgpeak[t] then avgpeakcnt[1] + 1
else avgpeakcnt[1];
def maxcnt = highestall(avgpeakcnt);
def revcnt = maxcnt - avgpeakcnt + 1;
# define the group of bars that include x peaks
def ispeaks = (avgpeak[t] and revcnt <= peak_qtys);
def peakrange = (revcnt <= peak_qtys);
def avgtotal = if bn == 1 then 0
else if lastbar then avgtotal[1]
else if ispeaks then round(avgtotal[1] + avg1, 2)
else avgtotal[1];
def avgavg = round(avgtotal / peak_qtys, 2);
def avgavg2 = highestall(if lastbar then avgavg else 0);
plot za = if peakrange then avgavg2 else na;
addlabel(1, " ", color.black);
addlabel(0, peak_qtys + " peaks", color.yellow);
addlabel(1, avgavg2 + " average of the last " + peak_qtys + " peaks, on an average", color.yellow);
addlabel(1, " ", color.black);
input test1_peaks = yes;
addchartbubble(test1_peaks and avgpeak[t], high,
, (if peakrange then color.yellow else color.gray), yes);
input test2_avg = no;
#addchartbubble(test2_avg and (revcnt <= peak_qtys), low,
addchartbubble(test2_avg, low,
revcnt + "\n" +
peak_qtys + "\n" +
avgavg2 + "\n" +
ispeaks + "\n" +
, (if ispeaks then color.yellow else color.gray), no);
#, color.yellow, no);
# peak data
input test1_peak_cnts = no;
addchartbubble(test1_peak_cnts and avgpeak[t], avg1,
avg1 + "\n" +
avgpeak[t] + "\n" +
avgtotal + "\n" +
avgavg + "\n\n" +
avgpeakcnt + "\n" +
maxcnt + "\n" +
, (if ispeaks then color.yellow else color.gray), yes);
MRNA 15min
find 5 peaks on an average line
blue line - average of those 5 peaks
I need to inspect the code you provided more closely at a later time, but I think you may have answered my question and found a solution to my issue as the last paragraph before the code provided
essentially describes what I was going for.
That's essentially what I was trying to do was to a plot a line that defines the average price of X amount of previous peaks defined by the user. The same would be done for previous troughs. This way
you'd essentially be able to define a price range between the average of peaks and the average of troughs to more safely trade within, using the trough average as a fair price estimate to buy and the
average peak as a fair price estimate to sell. How it's displayed doesn't matter so much. What matters is what the price of each average is. My preference would be to have two horizonal lines like
the one provided in your screenshot that together defines the range to trade within.
That's essentially what I was trying to do was to a plot a line that defines the average price of X amount of previous peaks defined by the user. The same would be done for previous troughs. This
way you'd essentially be able to define a price range between the average of peaks and the average of troughs to more safely trade within, using the trough average as a fair price estimate to buy
and the average peak as a fair price estimate to sell. How it's displayed doesn't matter so much. What matters is what the price of each average is. My preference would be to have two horizonal
lines like the one provided in your screenshot that together defines the range to trade within.
find 2 average price levels, from a group of peaks and a group of valleys.
choose how many peaks/valleys to find (default is 5)
choose the data source, average line (default) or highs and lows,
choose to include last bar or not
find peaks and valleys on the chosen data,
..peaks are based on the current bar > (1 bar before and 1 bar after)
..valleys are based on the current bar < (1 bar before and 1 bar after)
..2 labels display the prices and data type
dots are plotted at data points. yellow for peaks, cyan for valleys.
# avg_of_prev_x_peaksvalleys_02
#How to get the offset value of the first of the last so many times a condition occurred?
def na = double.nan;
def bn = barnumber();
def lastbn = HighestAll(If(IsNaN(close), 0, bn));
def lastbar = if (bn == lastbn) then 1 else 0;
#def lastbar = !isnan(close[0]) and isnan(close[-1]);
input qty_peaks_valleys = 5;
input price_data = { default average , high_low };
input last_bar_can_be_peakvalley = yes;
# data - average
def price = close;
input avg1_len = 12;
#input ma1_type = AverageType.EXPONENTIAL;
input avg1_type = AverageType.simple;
def avg1 = MovingAverage(avg1_type, price, avg1_len);
input show_avg_line = yes;
plot z1 = if show_avg_line then avg1 else na;
# choose price levels to analyze, avg, close,...
#def peakdata = avg1;
#def valleydata = avg1;
#def peakdata = high;
#def valleydata = low;
#input price_data = { default average , high_low };
def peakdata;
def valleydata;
def datatype;
if price_data == price_data.average then {
peakdata = avg1;
valleydata = avg1;
datatype = 1;
} else if price_data == price_data.high_low then {
peakdata = high;
valleydata = low;
datatype = 2;
} else {
peakdata = na;
valleydata = na;
datatype = 0;
def t = 0;
# peak data
def peak = (peakdata[1] < peakdata[0] and peakdata[0] > peakdata[-1]);
def peaklast = if !last_bar_can_be_peakvalley then 0
else if lastbar and peakdata > peakdata[1] then 1
else 0;
def peak2 = if lastbar then peaklast else peak;
# count the peaks
def avgpeakcnt = if bn == 1 then 1
else if peak2[t] then avgpeakcnt[1] + 1
else avgpeakcnt[1];
def pkmaxcnt = highestall(avgpeakcnt);
def pkrevcnt = pkmaxcnt - avgpeakcnt + 1;
# define the group of bars that include x peaks
def ispeaks = (peak2[t] and pkrevcnt <= qty_peaks_valleys);
def peakrange = (pkrevcnt <= qty_peaks_valleys);
def pkavgtotal = if bn == 1 then 0
# else if lastbar then avgtotal[1]
else if lastbar and peaklast then pkavgtotal[1] + peakdata
else if ispeaks then round(pkavgtotal[1] + peakdata, 2)
else pkavgtotal[1];
def pkavgavg = round(pkavgtotal / qty_peaks_valleys, 2);
def pkavgavg2 = highestall(if lastbar then pkavgavg else 0);
plot zpk = if peakrange then pkavgavg2 else na;
addlabel(1, " ", color.black);
#addlabel(0, qty_peaks_valleys + " peaks", color.yellow);
addlabel(1, pkavgavg2 + " average of the last " + qty_peaks_valleys + " peaks, on " + (if datatype == 1 then "an average" else "highs"), color.yellow);
addlabel(1, " ", color.black);
# valley data
def valley = (valleydata[1] > valleydata[0] and valleydata[0] < valleydata[-1]);
def valleylast = if !last_bar_can_be_peakvalley then 0
else if lastbar and valleydata < valleydata[1] then 1
else 0;
def valley2 = if lastbar then valleylast else valley;
#def t = 0;
# count the peaks
def avgvalleycnt = if bn == 1 then 1
else if valley2[t] then avgvalleycnt[1] + 1
else avgvalleycnt[1];
def valmaxcnt = highestall(avgvalleycnt);
def valrevcnt = valmaxcnt - avgvalleycnt + 1;
# define the group of bars that include x peaks
def isvalleys = (valley2[t] and valrevcnt <= qty_peaks_valleys);
def valleyrange = (valrevcnt <= qty_peaks_valleys);
def valavgtotal = if bn == 1 then 0
# else if lastbar then avgtotal[1]
else if lastbar and valleylast then valavgtotal[1] + valleydata
else if isvalleys then round(valavgtotal[1] + valleydata, 2)
else valavgtotal[1];
def valavgavg = round(valavgtotal / qty_peaks_valleys, 2);
def valavgavg2 = highestall(if lastbar then valavgavg else 0);
plot zval = if valleyrange then valavgavg2 else na;
addlabel(1, " ", color.black);
#addlabel(0, qty_peaks_valleys + " valleys", color.yellow);
addlabel(1, valavgavg2 + " average of the last " + qty_peaks_valleys + " valleys, on " + (if datatype == 1 then "an average" else "lows"), color.yellow);
addlabel(1, " ", color.black);
input show_peak_dots = yes;
plot zpd = if peak2[t] and peakrange then peakdata else na;
input show_valley_dots = yes;
plot zvd = if valley2[t] and valleyrange then valleydata else na;
input test1_peaks = no;
addchartbubble(test1_peaks and peak2[t], peakdata,
pkrevcnt + "\n" +
, (if peakrange then color.yellow else color.gray), yes);
input test2_peakdata = no;
#addchartbubble(test2_avg and (revcnt <= peak_qtys), low,
addchartbubble(test2_peakdata, peakdata,
pkrevcnt + "\n" +
qty_peaks_valleys + "\n" +
pkavgavg2 + "\n" +
ispeaks + "\n" +
, (if ispeaks then color.yellow else color.gray), no);
#, color.yellow, no);
# peak data
input test1_peak_cnts = no;
addchartbubble(test1_peak_cnts and peak2[t], peakdata,
peakdata + "\n" +
peak2[t] + "\n" +
pkavgtotal + "\n" +
pkavgavg + "\n\n" +
avgpeakcnt + "\n" +
pkmaxcnt + "\n" +
, (if ispeaks then color.yellow else color.gray), yes);
input test1_valleys = no;
addchartbubble(test1_valleys and valley2[t], valleydata,
valrevcnt + "\n" +
, (if valleyrange then color.cyan else color.gray), no);
input test2_valleydata = no;
#addchartbubble(test2_avg and (revcnt <= peak_qtys), low,
addchartbubble(test2_valleydata, valleydata,
valrevcnt + "\n" +
qty_peaks_valleys + "\n" +
valavgavg2 + "\n" +
isvalleys + "\n" +
zval + "\n" +
valleydata + "\n" +
valley2[t] + "\n"
, (if isvalleys then color.cyan else color.gray), no);
#, color.yellow, no);
# peak data
input test1_valley_cnts = no;
addchartbubble(test1_valley_cnts and valley2[t], valleydata,
valleydata + "\n" +
valley2[t] + "\n" +
valavgtotal + "\n" +
valavgavg + "\n\n" +
avgvalleycnt + "\n" +
valmaxcnt + "\n" +
, (if isvalleys then color.cyan else color.gray), yes);
PM 15min
5 peaks , valleys.
data from an average line
include last bar , yes
PM 15min
5 peaks , valleys
data from high, low
Last edited:
find 2 average price levels, from a group of peaks and a group of valleys.
choose how many peaks/valleys to find (default is 5)
choose the data source, average line (default) or highs and lows,
choose to include last bar or not
find peaks and valleys on the chosen data,
..peaks are based on the current bar > (1 bar before and 1 bar after)
..valleys are based on the current bar < (1 bar before and 1 bar after)
..2 labels display the prices and data type
dots are plotted at data points. yellow for peaks, cyan for valleys.
# avg_of_prev_x_peaksvalleys_02
#How to get the offset value of the first of the last so many times a condition occurred?
def na = double.nan;
def bn = barnumber();
def lastbn = HighestAll(If(IsNaN(close), 0, bn));
def lastbar = if (bn == lastbn) then 1 else 0;
#def lastbar = !isnan(close[0]) and isnan(close[-1]);
input qty_peaks_valleys = 5;
input price_data = { default average , high_low };
input last_bar_can_be_peakvalley = yes;
# data - average
def price = close;
input avg1_len = 12;
#input ma1_type = AverageType.EXPONENTIAL;
input avg1_type = AverageType.simple;
def avg1 = MovingAverage(avg1_type, price, avg1_len);
input show_avg_line = yes;
plot z1 = if show_avg_line then avg1 else na;
# choose price levels to analyze, avg, close,...
#def peakdata = avg1;
#def valleydata = avg1;
#def peakdata = high;
#def valleydata = low;
#input price_data = { default average , high_low };
def peakdata;
def valleydata;
def datatype;
if price_data == price_data.average then {
peakdata = avg1;
valleydata = avg1;
datatype = 1;
} else if price_data == price_data.high_low then {
peakdata = high;
valleydata = low;
datatype = 2;
} else {
peakdata = na;
valleydata = na;
datatype = 0;
def t = 0;
# peak data
def peak = (peakdata[1] < peakdata[0] and peakdata[0] > peakdata[-1]);
def peaklast = if !last_bar_can_be_peakvalley then 0
else if lastbar and peakdata > peakdata[1] then 1
else 0;
def peak2 = if lastbar then peaklast else peak;
# count the peaks
def avgpeakcnt = if bn == 1 then 1
else if peak2[t] then avgpeakcnt[1] + 1
else avgpeakcnt[1];
def pkmaxcnt = highestall(avgpeakcnt);
def pkrevcnt = pkmaxcnt - avgpeakcnt + 1;
# define the group of bars that include x peaks
def ispeaks = (peak2[t] and pkrevcnt <= qty_peaks_valleys);
def peakrange = (pkrevcnt <= qty_peaks_valleys);
def pkavgtotal = if bn == 1 then 0
# else if lastbar then avgtotal[1]
else if lastbar and peaklast then pkavgtotal[1] + peakdata
else if ispeaks then round(pkavgtotal[1] + peakdata, 2)
else pkavgtotal[1];
def pkavgavg = round(pkavgtotal / qty_peaks_valleys, 2);
def pkavgavg2 = highestall(if lastbar then pkavgavg else 0);
plot zpk = if peakrange then pkavgavg2 else na;
addlabel(1, " ", color.black);
#addlabel(0, qty_peaks_valleys + " peaks", color.yellow);
addlabel(1, pkavgavg2 + " average of the last " + qty_peaks_valleys + " peaks, on " + (if datatype == 1 then "an average" else "highs"), color.yellow);
addlabel(1, " ", color.black);
# valley data
def valley = (valleydata[1] > valleydata[0] and valleydata[0] < valleydata[-1]);
def valleylast = if !last_bar_can_be_peakvalley then 0
else if lastbar and valleydata < valleydata[1] then 1
else 0;
def valley2 = if lastbar then valleylast else valley;
#def t = 0;
# count the peaks
def avgvalleycnt = if bn == 1 then 1
else if valley2[t] then avgvalleycnt[1] + 1
else avgvalleycnt[1];
def valmaxcnt = highestall(avgvalleycnt);
def valrevcnt = valmaxcnt - avgvalleycnt + 1;
# define the group of bars that include x peaks
def isvalleys = (valley2[t] and valrevcnt <= qty_peaks_valleys);
def valleyrange = (valrevcnt <= qty_peaks_valleys);
def valavgtotal = if bn == 1 then 0
# else if lastbar then avgtotal[1]
else if lastbar and valleylast then valavgtotal[1] + valleydata
else if isvalleys then round(valavgtotal[1] + valleydata, 2)
else valavgtotal[1];
def valavgavg = round(valavgtotal / qty_peaks_valleys, 2);
def valavgavg2 = highestall(if lastbar then valavgavg else 0);
plot zval = if valleyrange then valavgavg2 else na;
addlabel(1, " ", color.black);
#addlabel(0, qty_peaks_valleys + " valleys", color.yellow);
addlabel(1, valavgavg2 + " average of the last " + qty_peaks_valleys + " valleys, on " + (if datatype == 1 then "an average" else "lows"), color.yellow);
addlabel(1, " ", color.black);
input show_peak_dots = yes;
plot zpd = if peak2[t] and peakrange then peakdata else na;
input show_valley_dots = yes;
plot zvd = if valley2[t] and valleyrange then valleydata else na;
input test1_peaks = no;
addchartbubble(test1_peaks and peak2[t], peakdata,
pkrevcnt + "\n" +
, (if peakrange then color.yellow else color.gray), yes);
input test2_peakdata = no;
#addchartbubble(test2_avg and (revcnt <= peak_qtys), low,
addchartbubble(test2_peakdata, peakdata,
pkrevcnt + "\n" +
qty_peaks_valleys + "\n" +
pkavgavg2 + "\n" +
ispeaks + "\n" +
, (if ispeaks then color.yellow else color.gray), no);
#, color.yellow, no);
# peak data
input test1_peak_cnts = no;
addchartbubble(test1_peak_cnts and peak2[t], peakdata,
peakdata + "\n" +
peak2[t] + "\n" +
pkavgtotal + "\n" +
pkavgavg + "\n\n" +
avgpeakcnt + "\n" +
pkmaxcnt + "\n" +
, (if ispeaks then color.yellow else color.gray), yes);
input test1_valleys = no;
addchartbubble(test1_valleys and valley2[t], valleydata,
valrevcnt + "\n" +
, (if valleyrange then color.cyan else color.gray), no);
input test2_valleydata = no;
#addchartbubble(test2_avg and (revcnt <= peak_qtys), low,
addchartbubble(test2_valleydata, valleydata,
valrevcnt + "\n" +
qty_peaks_valleys + "\n" +
valavgavg2 + "\n" +
isvalleys + "\n" +
zval + "\n" +
valleydata + "\n" +
valley2[t] + "\n"
, (if isvalleys then color.cyan else color.gray), no);
#, color.yellow, no);
# peak data
input test1_valley_cnts = no;
addchartbubble(test1_valley_cnts and valley2[t], valleydata,
valleydata + "\n" +
valley2[t] + "\n" +
valavgtotal + "\n" +
valavgavg + "\n\n" +
avgvalleycnt + "\n" +
valmaxcnt + "\n" +
, (if isvalleys then color.cyan else color.gray), yes);
PM 15min
5 peaks , valleys.
data from an average line
PM 15min
5 peaks , valleys
data from high, low
Precisely what I was going for, can't thank you enough! Well done!
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Publications | Junggi's Home
1. P. Narayan, Euihun Joung and J. Yoon, “Gravitational Edge Mode in Asymptotically AdS2: JT Gravity Revisited,” JHEP 2405 (2024) 244 arXiv:2304.06088 [hep-th].
2. K. Lee and J. Yoon, “TT Deformation of N = (1,1) Off-Shell Supersymmetry and Partially Broken Supersymmetry,” Phys. Rev. D110 (2024) 025001 arXiv:2306.08030 [hep-th].
3. A. Jevicki, D. Mukherjee and J. Yoon, “Emergent factorization of Hilbert space at large N and black hole,” (2024) arXiv:2404.07862 [hep-th].
4. K. Lee, A. Sivakumar and J. Yoon, “Gravitational Edge Mode in N = 1 Jackiw-Teitelboim Supergravity,” (2024) arXiv:2403.17182 [hep-th].
5. P. Chen, M. Sasaki, D. Yeom and J. Yoon, “Tunneling between Multiple Histories as a Solution to the Information Loss Paradox,” Entropy 2023, 25 (2023) 1663 arXiv:2206.10251 [gr-qc].
6. S. Ryu and J. Yoon, “Unitarity of Symplectic Fermion in α-vacua with Negative Central Charge,” Phy. Rev. Lett. 130 (2023) 241602 arXiv:2208.12169 [hep-th].
7. K. Alkalaev, E. Joung and J. Yoon, “Color decorations of Jackiw-Teitelboim gravity,” JHEP 2208 (2022) 286 arXiv:2204.10214 [hep-th].
8. K. Alkalaev, E. Joung and J. Yoon, “Schwarzian for colored Jackiw-Teitelboim gravity,” JHEP 2209 (2022) 160 arXiv:2204.09010 [hep-th].
9. P. Chen, M. Sasaki, D. Yeom and J. Yoon, “Resolving information loss paradox with Euclidean path integral,” (2022) arXiv:2205.08320 [gr-qc].
10. D. Bak, C. Kim, S. Yi and J. Yoon, “Python’s Lunches in Jackiw-Teitelboim gravity with matter,” JHEP 2022 (2022) 175 arXiv:2112.04224 [hep-th].
11. J. Yoon, “A Bound on Chaos from Stability,” JHEP 2111 (2021) 097 arXiv:1905.08815 [hep-th].
12. P. Chen, M. Sasaki, D. Yeom and J. Yoon, “Solving information loss paradox via Euclidean path integral,” (2021) arXiv:2111.01005 [hep-th].
13. A. Jevicki, X. Liu, J. Yoon and J. Zheng, “Dynamical Symmetry and the Thermofield State at Large N,” Universe 2022, 8(2) (2022) 114 arXiv:2109.13381 [hep-th].
14. B. Kang and J. Yoon, “Sign-Problem-Free Variant of Complex Sachdev-Ye- Kitaev Model,” Phys. Rev. B105 (2022) 045117 arXiv:2107.13572 [cond-mat.str-el].
15. K. Lee, P. Yi and J. Yoon, "TTbar-deformed Fermionic Theories Revisited," 2107 (2021) 217 arXiv:2104.09529 [hep-th].
16. D. Bak, C. Kim, S. Yi and J. Yoon, “Unitarity of Entanglement and Islands in Two-Sided Janus Black Holes,” JHEP 2101 (2021) 155 arXiv:2006.11717 [hep-th].
17. Y. Qi, S. Sin and J. Yoon, “Quantum Correction to Chaos in Schwarzian Theory,” JHEP 1911 (2019) arXiv:1906.00996 [hep-th].
18. V. Jahnke, K. Kim and J. Yoon, "On the Chaos Bound in Rotating Black Holes" JHEP 1905 (2019) 037 arXiv:1903.09086 [hep-th]
19. P. Narayan and J. Yoon, "Chaos in Three-dimensional Higher Spin Gravity" JHEP 1907 (2019) 046 arXiv:1903.08761
20. R.d.M. Koch, A. Jevicki, K. Suzuki and J. Yoon, "AdS Maps and Diagrams of Bi-local Holography," JHEP 1903 (2019) 133 arXiv:1810.02332 [hep-th].
21. T. Nosaka, D. Rosa and J. Yoon, “Thouless time for mass-deformed SYK,” JHEP 1809 (2018) 041 arXiv:1804.09934 [hep-th].
22. P. Narayan and J. Yoon, “Supersymmetric SYK Model with Global Symmetry,” JHEP 1808 (2018) 159 arXiv:1712.02647 [hep-th].
23. J. Yoon, “SYK Models and SYK-like Tensor Models with Global Symmetry,” JHEP 1710 (2017) 183 arXiv:1707.01740 [hep-th].
24. J. Yoon, “Supersymmetric SYK Model: Bi-local Collective Superfield / Super- matrix Formulation,” JHEP 1710 (2017) 172 arXiv:1706.05914 [hep-th].
25. P. Narayan and J. Yoon, “SYK-like Tensor Models on the Lattice,” JHEP 1708 (2017) 083 arXiv:1705.01554 [hep-th].
26. S. Chaudhuri, V. I. Giraldo-Rivera, A. Joseph, R. Loganayagam and J. Yoon, “Abelian Tensor Models on the Lattice,” submitted to Phys. Rev D (2017) arXiv:1705.01930 [hep-th].
27. A. Jevicki, K. Suzuki and J. Yoon, “Bi-Local Holography in the SYK Model,” JHEP 1607 (2016) 007 arXiv:1603.06246 [hep-th].
28. A. Jevicki and J. Yoon, “SN Orbifolds and String Interactions,” J. Phys. A49 (2016) 205401 arXiv:1511.07878 [hep-th].
29. A. Jevicki and J. Yoon, “Bulk from Bi-locals in Thermo Field CFT,” JHEP 1602 (2016) 090 arXiv:1503.08484 [hep-th].
30. R. d. M. Koch, A. Jevicki, J. P. Rodrigues and J. Yoon, “Canonical Formu- lation of O(N) Vector/Higher Spin Correspondence,” J. Phys. A48 (2015) 105403 arXiv:1408.4800 [hep-th] .
31. R. d. M. Koch, A. Jevicki, J. P. Rodrigues and J. Yoon, “Holography as a Gauge Phenomenon in Higher Spin Duality,” JHEP 1501 (2015) 055 arXiv:1408.1255 [hep-th].
32. A. Jevicki, K. Jin and J. Yoon, “1/N and loop corrections in higher spin AdS4/CFT3 duality,” Phys. Rev. D89 (2014) 085039 arXiv:1401.3318 [hep-th].
33. A. Jevicki and J. Yoon, “Field Theory of Primaries in WN Minimal Models” JHEP 1311 (2013) 060 arXiv:1302.3851 [hep-th].
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Drs. Huang, Kovalsky, and Marzuola, UNC - Mathematical and Statistical Analysis of Compressible Data on Compressive Network - Department of Mathematics
Drs. Huang, Kovalsky, and Marzuola, UNC – Mathematical and Statistical Analysis of Compressible Data on Compressive Network
February 2, 2023 @ 3:30 pm - 4:30 pm
Mode: In-Person
Title: Mathematical and Statistical Analysis of Compressible Data on Compressive Network
We present an overview of the FRG project on the “Mathematical and Statistical Analysis of Compressible Data on Compressive Networks”.
Compressible features of data include the low-rank, low-dimension, sparsity, and features from the classification/categorization/clustering process.
Discovering such compressible features is a major challenge in data analysis, which we will address using hierarchical decompositions derived from
spectral, statistical, and algebraic geometric analysis of data. We also study how to construct optimally defined compressive networks, specifically tailored
to the discovered compressible features, to enable an accurate and efficient extraction and manipulation of sparse representations in complex and
high dimensional systems in an inherently interpretable manner. Sample ongoing projects include the accurate and efficient representation of layer
potential using algebraic variety, spectral flow and fast computation of the eigensystems, frequency domain based statistical analysis, fast algorithms
for high dimensional truncated multivariate Gaussian expectations, and recursive tree algorithms for orthogonal matrix generation and matrix-vector
multiplications in rigid body Brownian dynamics simulations.
Related Events
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Dr. Calculus - PCB Heaven
Dr. Calculus technical calculators
Welcome to the Dr. Calculus pages.
What you are about to find here are sophisticated, on-line, real time calculators for specific jobs. They are designed in order to save you from the time consuming common calculations. Those
calculators are not to solve "riddles" and impossible mathematic equations. They are simple in operation (as simple as can be) and each one does one specific task. These tasks are some calculations
that an amateur (and not only) project designer will need to do as for example, you may need to solve several times the equation for the calculation of oscillation of a 555. Dr. calculus will do
those calculations for you saving you precious time and keeping your results error free.
As you understand, those calculators are far from complete. I would need more than my life to script a complete list of calculators. The list is from time to time enriched with new scripts. In case
you badly need a calculator, you may ask from us (with email of course) and you will have it the soonest possible.
All calculators are copyrighted. You may NOT copy the scripts without our permission. This would be against the international law for copyrights.
Follow this link to go back to the Dr. Calculus Index...
Select the tolerance:
E6 - 20%
E12 - 10%
E24 - 5%
E48 - 2%
E96 - 1%
E192 - 0.5%, 0.25% or higher tolerance
Enter the resistor value and select the measurement:
Previous resistor value:
Best match
Next resistor value
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» Cofiber Sequence «
Monday, December 10th, 2012 | Author: Konrad Voelkel
I want to explain a particularly easy example of a motivic cellular decomposition: That of $n$-dimensional projective space. The discussion started with cohomology (part 1), continued with bundles
and cycles (part 2) and in this part 3, we discuss motivic stuff.
Continue reading «Invariants of projective space III: Motives»
Category: English, Mathematics | Comments off
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Matrix summation method
From Encyclopedia of Mathematics
One of the methods for summing series and sequences using an infinite matrix. Employing an infinite matrix
If the series on the right-hand side converges for all
then the sequence
then this series is said to be summable to the sum
A matrix summation method for series can be also defined directly by transforming the series (1) into a sequence
Less often used are matrix summation methods defined by a transformation of a series (1) into a series
or by a transformation of a sequence
which use matrices
The matrix of a summation method all entries of which are non-negative is called a positive matrix. Among the matrix summation methods one finds, for example, the Voronoi summation method, the Cesàro
summation methods, the Euler summation method, the Riesz summation method Hausdorff summation method, and others (see also Summation methods).
[1] G.H. Hardy, "Divergent series" , Clarendon Press (1949)
[2] R.G. Cooke, "Infinite matrices and sequence spaces" , Macmillan (1950)
[3] G.P. Kangro, "Theory of summability of sequences and series" J. Soviet Math. , 5 : 1 (1976) pp. 1–45 Itogi Nauk. i Tekhn. Mat. Anal. , 12 (1974) pp. 5–70
[4] S.A. Baron, "Introduction to the theory of summability of series" , Tartu (1966) (In Russian)
How to Cite This Entry:
Matrix summation method. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Matrix_summation_method&oldid=12057
This article was adapted from an original article by I.I. Volkov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098.
See original article
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You Should Try Quordle If You're Too Good at Wordle
Within seconds of finishing his first Wordle game, my 12-year-old son opened a new tab and googled “infinite Wordle.” There are, of course, many ways to play Wordle over and over again. (Here’s one;
here’s another.) What’s better, though, is playing multiple Wordle-style puzzles at the same time.
(If you’re out of the loop: in Wordle, you have six chances to guess a five-letter word. Green squares mean a guessed letter is correct; yellow means that letter is in the word but not in that
position. That’s it, that’s the whole game.)
Dordle is the first simultaneous Wordle variant I was introduced to. It’s a double Wordle: each of your guesses is entered for both puzzles at the same time, and your goal is to solve both in seven
guesses or less. I’ve been playing the daily Dordle since puzzle #0002, but today I learned I’ve been missing out on Quordle, which allows you to play four puzzles at once. In just nine guesses.
How do you win at Dordle and Quordle?
First of all, both of these games feature a “daily” puzzle that is the same for everyone, and a “free” or “practice” puzzle with a different solution every time. So, yes, the truly addicted can play
four puzzles at a time all day long. If nothing else, you can take advantage of practice games as you’re learning the ropes.
Now, on to the strategy. No matter how many puzzles you’re solving, I like to think of guesses as answering one (or more) of three questions:
1. What letters are in the solution?
2. I know some yellows; where are they in the solution?
3. Is this word the solution?
It’s a mistake to play the game with #3 as your only strategy. Pretty quickly you’ll discover exploratory guesses are important, and you’ll pick starter words that answer question #1 efficiently. (My
mnemonic: ETAOIN SHRDLU, pronounced “Edwin Shirdloo” as if it were a name, is a list of the arguably most common letters in English. My starters always use letters from this list.)
Simultaneous Wordles require lots of #1 and #2 guesses. You need to constantly ask yourself: what information can I gather with this guess? The same guess can do double duty for different puzzles at
the same time: maybe you combine a yellow from one word (#2) with some brand-new letters (#1) and a yellow from another word (#2 again).
You definitely don’t want to do a #3 guess until you’re pretty sure of the answer, since your attempted solution on one puzzle is a guess that will probably be useless for the other(s).
How to solve four Wordle puzzles at once
Alright, let’s look at this in action. The puzzle I’m solving (shown in the image up top) is a “practice” puzzle, so you don’t have to worry about spoilers.
We start with TRASH, and get hits on three of the puzzles. Next up, to start making progress on the upper right puzzle, I choose a word that uses common letters but doesn’t repeat any of the ones we
just tried: CLINK.
Still only one yellow in that upper right puzzle, but we’re in a really good position now on the bottom two puzzles, which now each have four letters confirmed—some of them are even greens. Three on
the top left isn’t shabby either. As we put together future guesses (#2's and #3's for the puzzles where we’re close), let’s keep feeding in new letters (#1) to help with the top right.
We can probably solve one already: the lower left has to be CHAS_, giving us either CHASE or CHASM. I get cocky and go with CHASE, which is wrong, but at least gets an E into play. So then I solve
with CHASM, and then I see what I’ve learned from introducing that E.
The lower right is _ _ A_E with an S, a T, and a K in there somewhere, so it’s either STAKE or SKATE. Rather than use one of those as my next pick, I want to knock out a couple possibilities. I still
don’t know where that L goes in the top right puzzle, and I’d like to get another unknown letter or two into the mix. It also wouldn’t hurt to stick an A in there somewhere to help us out on the top
left, where we know there’s an A but we don’t know where.
I settle on ATOLL, which gives valuable clues for all three. We now know that the top right puzzle has an O, and we know several places that the L cannot be. We know that the top left must be either
_AIS_ or _ _ISA, and my gut is saying DAISY. We have also confirmed that the lower right must be SKATE and not STAKE.
So I guess SKATE and then DAISY, which reveals a Y at the end of the top right puzzle. We don’t have many letters to go on, but the fact that we’ve guessed so many, with so few hits, suggests there
might be at least one double letter. It can’t be a double L, since there are no Ls in the second and fourth spots, so I consider doubling up a letter I haven’t guessed yet. What could fit in LO_ _Y?
All I can think of is LOBBY—yup, that’s it. Four puzzles solved in eight guesses, with one to spare. The little squares won’t necessarily fit in a tweet, so you also get a little shareable graphic
that looks like this:
Strategy for Dordle is the same, except you’ll be done sooner. I think simultaneous puzzle-solving is a good learning tool for regular Wordle, since it encourages thoughtfulness in information
gathering rather than just trying to guess the correct answer. Give it a try and see how you fare! (Meanwhile, I need to try Octordle.)
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ACO Seminar
The ACO Seminar (2016–2017)
Sep. 22, 3:30pm, Wean 8220
Misha Lavrov, CMU
Improving lower and upper bounds
(1) A version of the Hales-Jewett theorem says that for any t there exists a dimension on such that any two-coloring of the grid [t]^n contains t points on a line that are the same color.
Generalizations of this problem lead to the Graham-Rothschild parameter sets theorem; the well-known Graham's number is an upper bound on the dimension required for the first nontrivial instance of
the parameter sets theorem.
We prove tighter bounds on the Hales-Jewett number for t=4 as well as on the problem for which Graham's number is an upper bound (the latter is joint work with John Mackey and Mitchell Lee).
(2) An n-vertex graph is considered to be distance-uniform (for a distance d and a tolerance ε>0) if, for any vertex v, at least (1-ε)n of the vertices in the graph are at distance exactly d from v.
Random graphs provide easy examples of distance-uniform graphs, with d = O(log n); Alon, Demaine, Hajiaghayi, and Leighton conjectured that this is true for all distance-uniform graphs.
We provide examples of distance-uniform graphs with much larger d, and for sufficiently small ε, prove matching upper bounds on d (joint work with Po-Shen Loh).
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Homework Physics 645 Fall 2007
Homework 1 due 9/21/07
Chapter 1 (Cushman-Rosin)
problems 3, 4
1-3) The land sea breeze is a lighrt wind blowing from the sea as a result of the temperature difference between the land and the sea. As this temperature difference reverses from day to night, the
daytime sea breeze turns into a nighttime land breeze. If you were developing a numerical model of the sea-land breeze, should you include the effects of the earths rotation? Justify!
1-4) The Great Red Spot of Jupiter, centered at 22 degrees south and spaning 12 degrees in latitude and 25 degrees in longitude, exhibitswind speeds of 100m/sec. The planets radius is 71,400 km and
the rotation rate is 1.763 x 10^-4 sec^-1. Is the Great Red Spot influenced by the planet's rotation? Justify!
Homework 2 due 9/28/07
Chapter 2(Cushman-Rosin)
(handed out int class)
problems 2, 3 , 4 (Cushman-Rosin)
problem 1.2 (from Acheson, handed out in class)
Homework 3 due 10/5/07
problem 1.7, 2.1, (moved to next week 2.3) (from Acheson, handed out in class)
Homework 4 due 10/12/07
problem 2.3 (from Acheson, handed out in class) +
problems 3-3 (Cushman-Rosin)
Homework 5 due 11/21/07
problems 4-1, 4-8, (Cushman-Rosin)
problems 5-1, 5-2, 5-5 (Cushman-Rosin)
note: I like 2.27 from Vallis ;)
Last updated 9 November, 2007
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Regularisation – Introduction
In other words, bias in a model is high when it does not perform well on the training data itself, and variance is high when the model does not perform well on the test data. Please note that a model
failing to fit on the test data means that the model results on the test data varies a lot as the training data changes. This may be because the model coefficients do not have high reliability.
You also saw that there is a trade-off between bias and variance with respect to model complexity. As seen in the video, a simple model would usually have high bias and low variance, whereas a
complex model would have low bias and high variance. In either case, the total error would be high.
What we need is lowest total error, i.e., low bias and low variance, such that the model identifies all the patterns that it should and is also able to perform well with unseen data.
For this, we need to manage model complexity: It should neither be too high, which would lead to overfitting, nor too low, which would lead to a model with high bias (a biased model) that does not
even identify necessary patterns in the data.
Another point to keep in mind is that the model coefficients that we obtain from an ordinary-least-squares (OLS) model can be quite unreliable if among all the predictors that we used to build our
model, only a few are related significantly to the response variable.
So, what is regularisation and how does it help solve this problem?
Regularisation helps with managing model complexity by essentially shrinking the model coefficient estimates towards 0. This discourages the model from becoming too complex, thus avoiding the risk of
Let’s try and understand this a bit better now.
We know that when building an OLS model, we want to estimate the coefficients for which the cost/loss, i.e., RSS, is minimum. Optimising this cost function results in model coefficients with the
least possible bias, although the model may have overfitted and hence have high variance.
In case of overfitting, we know that we need to manage the model’s complexity by primarily taking care of the magnitudes of the coefficients. The more extreme values of the coefficients are (high
positive or negative values of the coefficients), the more complex the model is and, hence, the higher are the chances of overfitting. Let’s try and understand in the forthcoming video.
When we use regularisation, we add a penalty term to the model’s cost function.
Here, the cost function would be Cost = RSS + Penalty.
Adding this penalty term in the cost function helps suppress or shrink the magnitude of the model coefficients towards 0. This discourages the creation of a more complex model, thereby preventing the
risk of overfitting.
When we add this penalty and try to get the model parameters that optimise this updated cost function (RSS + Penalty), the coefficients that we get given the training data may not be the best (maybe
more biased). Although with this minor compromise in terms of bias, the variance of the model may see a marked reduction. Essentially, with regularisation, we compromise by allowing a little bias for
a significant gain in variance.
As we saw in the video, when we perform regularisation, it has a smoothening effect on the model fit; in other words, when regularisation is used, the curve smoothens out and the fit is close to what
we want it to be.
Note that we also need to remember two points about the model coefficients that we obtain from OLS:
• These coefficients can be highly unstable – this can happen when only a few of the predictors that we have considered to build our model are related significantly to the response variable and the
rest are not very helpful, hence random variables.
• There may be a large variability in the model coefficients due to these unrelated random variables such that even a small change in the training data may lead to a large variance in the model
coefficients. Such model coefficients are no longer reliable, since we may get different coefficient values each time we retrain the model.
Multicollinearity, i.e., the presence of highly correlated predictors, may be another reason for the variability of model coefficients. Regularisation helps here as well.
So, to summarise,we use regularisation because we want our models to work well with unseen data, without missing out on identifying underlying patterns in the data. For this, we are willing to make a
compromise by allowing a little bias for a significant reduction in variance. We also understood that the more extreme the values of the model coefficients are, the higher are the chances of model
overfitting. Regularisation prevents this by shrinking the coefficients towards 0. In the next two segments, we will discuss the two techniques of regularisation: Ridge and Lasso and understand how
the penalty term helps with the shrinkage.
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Random Page
A game in which the number of points are fixed: for example, chess and checkers.
Since the number of points are fixed, any time one player gains points, another player directly loses those points --- or the potential of gaining those points. The total change in points for each
move is zero, and the sum of all changes at the end of the game is therefore zero.
In any zero sum game, it is impossible for all players to win and equally impossible for all players to lose --- ties are the result of stalemates or other cyclical game loops.
See also NonZeroSumGame.
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How do you calculate total time?
Table of contents:
How do you calculate total time?
To get the total decimal hours we use the formula: h = hours + (minutes / 60) + (seconds / 3600)....11.
How do I sum time duration in Excel?
HOW TO ADD TIME IN EXCEL
1. Step 1: Enter your hours and minutes in a hh:mm format in the column cells.
2. Step 2: Change the Format of your total cell to: [h]: mm.
3. Step 3: In your Total cell enter the Excel formula " =SUM( " and then select the cells with the hours in it.
4. Step 4: Click Enter. The total sum of your hours should now show up!
What are the steps for subtracting time?
To subtract time, subtract the minutes then subtract the hours. Since we can't have negative minutes, add 60 to the minutes and subtract 1 from the hours (60 minutes = 1 hour).
Can Excel add time?
Tip: You can also add up times by using the AutoSum function to sum numbers. Select cell B4, and then on the Home tab, choose AutoSum. The formula will look like this: =SUM(B2:B3). Press Enter to get
the same result, 16 hours and 15 minutes.
How do I add 30 minutes in Excel?
Add or Subtract Time
1. Enter a time into cell A1.
2. To add 2 hours and 30 minutes to this time, enter the formula shown below. ...
3. Select cell B1.
4. Right click, and then click Format Cells (or press CTRL + 1).
5. In the Category list, select Time, and select a Time format.
6. Click OK.
7. Enter a time into cell A1 (use 2 colons to include seconds).
How do I calculate time between dates in Excel?
In a new cell, type in =DATEDIF(A1,B1,”Y”). The “Y” signifies that you'd like the information reported in years. This will give you the number of years between the two dates. To find the number of
months or days between two dates, type into a new cell: =DATEDIF(A1,B1,”M”) for months or =DATEDIF(A1,B1,”D”) for days.
Why is Datedif not in Excel?
DATEDIF is not a standard function and hence not part of functions library and so no documentation. Microsoft doesn't promote to use this function as it gives incorrect results in few circumstances.
But if you know the arguments, you may use it and it will work and in most of the cases will give correct results.
How do I convert hours to working days in Excel?
FLOOR Function Now, for example say you have in Cell A1 then to break it down to number of days (as your requirement is to calculate number of days based on 8 hours a day), you'll have to use =FLOOR
(A1*24/8,1) which is same as =FLOOR(A1*3,1) . This will give result as 1 day .
What is the Datedif function in Excel?
The Excel DATEDIF function returns the difference between two date values in years, months, or days. The DATEDIF (Date + Dif) function is a "compatibility" function that comes from Lotus 1-2-3. ...
end_date - End date in Excel date serial number format. unit - The time unit to use (years, months, or days).
What is Edate in Excel?
The Excel EDATE function returns a date on the same day of the month, n months in the past or future. You can use EDATE to calculate expiration dates, maturity dates, and other due dates. Use a
positive value for months to get a date in the future, and a negative value for dates in the past.
How do I enable Datedif in Excel?
Using The DateDif Function in Microsoft Excel
1. Have you ever needed to work out the difference between two dates? ...
2. There are a couple of ways to insert a function into an Excel Worksheet.
3. This will bring up the Insert Function Window.
4. Search for the Function you want to use, and select go.
How do I calculate months and days in Excel?
Calculate elapsed year, month and days Select a blank cell which will place the calculated result, enter this formula =DATEDIF(A2,B2,"Y") & " Years, " & DATEDIF(A2,B2,"YM") & " Months, " & DATEDIF
(A2,B2,"MD") & " Days", press Enter key to get the result.
How do I manually calculate my service length?
Part 1: Calculate Length of Service to Year & Month Unit in Excel
1. Step 1: In E2, enter the formula =DATEDIF(C2,D2,”y”)&” years “&DATEDIF(C2,D2,”ym”)&” months”.
2. Comments:
3. Step 2: Click Enter to get result. ...
4. Step 3: Drag the fill handle to fill cells for other employees in this table.
How do you calculate years months and days between two dates?
Use DATEDIF to find the total years. In this example, the start date is in cell D17, and the end date is in E17. In the formula, the “y” returns the number of full years between the two days.
How much time should be between dates?
The first several dates should be spaced close together in an effort to keep the momentum going. The second date should not take place more than two weeks after the first date. If the first date went
exceptionally well, the best thing you can do is lock in a second date soon after.
How do you calculate years experience?
How to calculate work experience?
1. Step 1: First, consider the Date of Joining (i.e) DOJ.
2. Step 2: Then, consider the Last Working Date (i.e) LWD.
3. Step 3: Calculate the difference between Date of Joining and Last Working Date.
4. Step 4: Minus the two dates.
5. Step 5: Hence, the difference is mathematically proved.
How many days is 3 years from now?
Years to Days Conversion Table
Years Days
How can I calculate my age?
In some cultures, age is expressed by counting years with or without including the current year. For example, one person is twenty years old is the same as one person is in the twenty-first
year of his/her life.
How is work experience calculated in hours?
How to calculate your work experience. Your entries will be calculated automatically by multiplying the number of weeks worked by the number of hours worked per week. For this calculation, a
working week comprises 35-60 hours and a working year comprises 40-50 weeks.
Year How do you calculate hours and minutes for payroll?
You do this by dividing the minutes worked by 60. You then have the hours and minutes in numerical form, which you can multiply by the wage rate. For example, if your employee works 38 hours
and 27 minutes this week, you divide 27 by 60. This gives you 0.
How do I calculate my work hours per month?
Working hours per day multiplied by working days per week multiplied by 52 weeks in the year divided by 12 months in the year equals the average number of working hours per month.
How is work experience calculated?
Work experience can be calculated by adding up the number of weeks of full-time (or equivalent) paid work, for example: 30 hours/37.
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An Evaluation Strategy for Commercial Precision Marketing Based on Artificial Neural Network
2.1 EIS construction
Web 2.0 offers a consumer-centered model of information dissemination. The maturation of Web 2.0 allows consumers to acquire useful information via the Internet, and share information with other
consumers. The rise of Internet tools, such as Weibo, WeChat, and Taobao, has brought fundamental changes to consumer behavior pattern and media marketing model. With the growing accuracy of search
websites, great progress has been made in the precision placement and benefit-sharing mechanism of commercial advertisements. Meanwhile, media marketing has become increasingly in-depth and precise.
Against this background, the core functions of we media have extended from releasing personal information and sharing group information to the comparison and discussion of news and commercial
advertisements. In addition, the information release has also changed from the traditional merchant-to-consumer model into the modern merchant-to-consumer plus consumer-to-consumer model. Considering
the changing model of information release, the consumption process of attention-interest-desire-memory-action (AIDMA) was adjusted into attention-interest-search-action-share (AISAS). As shown in
Figure 1, two Internet-based actions, namely, the search for the information on products of interest, and the post-purchase information sharing with other consumers, are the key links in the
consumption process.
Figure 1. The consumption process in the improved AIDMA model
To improve the accuracy of commercial marketing strategy in the context of big data, the first step is to make scientific and reasonable evaluation of commercial precision marketing. Based on the
improved AIDMA model, this paper sets up a three-layer hierarchical EIS for the effect of commercial precision marketing. Taking the different phases of the consumption process as primary indices,
the established EIS is comprehensive and concise, and composed of both quantitative and qualitative indices.
Layer 1 (goal): B={effect of commercial precision marketing};
Layer 2 (primary indices): B={B1, B2, B3, B4, B5}={attention, interest, search, action, share};
Layer 3 (secondary indices):
B1={B11, B12}={online ad impression, brand recognition};
B2={B21, B22}={page view, product favorite ratio};
B3={B31, B32, B33, B34}={keyword search times, keyword search index, related word search times, related word search index};
B4={B41, B42, B43, B44, B45}={conversion rate, growth rate of online orders, growth rate of purchasers, growth rate of consumption frequency meeting merchant standard, growth rate of consumption
B5={B51, B52, B53, B54}={number of topics posted, number of topic replies, number of product reviews, number of product reposts}.
In the attention phase B1, brand recognition B12 is positively correlated with the effect of commercial precision marketing. Let N be the total number of consumers, and N[I] be the number of
consumers that effectively recognize a brand. Then, the value of B12 can be calculated by:
$B_{12}=\frac{N_{I}}{N}$ (1)
After a product catches the attention in B1, its favorite ratio (the ratio of those adding the product to the favorite list to those attracted by the product) in interest phase B2 is positively
correlated with the effect of commercial precision marketing. Let N[A] be the number of consumers interested in the product, and N[F] be the number of consumers adding the product to the favorite
list. Then, the value of B22 can be calculated by:
$B_{22}=\frac{N_{F}}{N_{A}}$ (2)
Every potential consumer will search for more information about the product of interest online. In the search phase B3, the keyword search times and related word search times can be denoted as N[SK]
and N[SR], respectively. Then, B32 and B34 can be respectively calculated by:
$B_{32}=\frac{B_{31}}{N_{S K}}$ (3)
$B_{34}=\frac{B_{33}}{N_{S R}}$ (4)
In phase B3, the consumers gain more understanding about the product, and receive advertisements released by the merchant. In the following action phase B4, the consumers are very likely to directly
purchase the product by clicking the link or place an order for the product. Let N[B] be the number of consumers that purchase or order the product, and N[VT] be the page view of the product. Then,
the value of B41 can be calculated by:
$B_{41}=\frac{N_{B}}{N_{V T}}$ (5)
Let O[i], P[i], F[i], and C[i] be the growth rates of online orders, purchasers, consumption frequency meeting merchant standard, and consumption amount of the i-th month after the commercial
precision marketing, respectively. Then, the values of B42-45 in that month can be respectively calculated by:
$B_{42}=\frac{O_{i-1}}{O_{i}}$ (6)
$B_{43}=\frac{P_{i-1}}{P_{i}}$ (7)
$B_{44}=\frac{F_{i-1}}{F_{i}}$ (8)
$B_{45}=\frac{C_{i-1}}{C_{i}}$ (9)
2.2 PCA-based index screening
The PCA can reduce the dimensionality of multidimensional indices into a few comprehensive principal indices. Each principal index can reflect the actual information of the object, and reduce the
overlap with the information reflected by other indices. Therefore, the PCA can simplify the evaluation of commercial precision marketing. In this paper, the PCA is introduced to screen the
evaluation indices for the effect of commercial precision marketing in the following steps:
Step 1. Collect sample data and initialize the set B of evaluation indices, which differ in dimensionality and value range. To minimize the impact of the difference on evaluation result, normalize
the positive indies by:
$B_{i j}^{\prime}=\frac{B_{i j}-B_{i j-\min }}{B_{i j-\max }-B_{i j-\min }}$ (10)
Normalize the negative indices by:
$B_{i j}^{\prime \prime}=\frac{B_{i j-\max }-B_{i j}}{B_{i j-\max }-B_{i j-\min }}$ (11)
Complete the translation by:
$B_{i j}^{\prime \prime}=B_{i j}^{\prime}+10^{-3}$ (12)
Step 2. Compute the correlation coefficient r[jk] between two indices B[ij] and B[ik] on the same layer by
$r_{j k}=\frac{\sum_{j=1}^{n} \sum_{k=1}^{n}\left(B_{i j}-\hat{B}_{i j}\right)\left(B_{i k}-\hat{B}_{i j}\right)}{\sqrt{\sum_{j=1}^{n}\left(B_{i j}-\hat{B}_{i j}\right)^{2} \sum_{k=1}^{n}\left(B_{i
k}-\hat{B}_{i j}\right)^{2}}}, j \neq k$ (13)
On this basis, formulate a correlation coefficient matrix for the indies on the same layer.
Step 3. Solve the covariance matrix BB^T/n of B, and obtain m eigenvalues λ[r] and corresponding eigenvectors b[r] of the matrix.
Step 4. Determine the principal components l corresponding to the eigenvalues, whose cumulative contribution rate is greater than 85%:
$\frac{\sum_{r=1}^{l} \lambda_{r}}{\sum_{r=1}^{m} \lambda_{r}} \geq 85 \%$ (14)
The l principal indices explain most information of all indices. Then, compute the contribution rate of each principal index by:
$\omega_{r}=\frac{\lambda_{r}}{\sum_{r=1}^{m} \lambda_{r}}$ (15)
Step 5. Let N be the index samples. Determine the entropy of each index by:
$e_{r}=-\frac{1}{\ln N} \sum_{r=1}^{m} \omega_{r} \ln \omega_{r}$ (16)
Step 6. Compute the coefficient of variation of each index:
$c_{r}=1-e_{r}$ (17)
Step 7. Compute the weight of each index (Table 1):
$W_{r}=\frac{c_{r}}{\sum_{r=1}^{m} c_{r}}$ (18)
Step 8. Determine the composite score of commercial precision marketing by:
$B_{i r}^{*}=\sum_{r=1}^{m} W_{r} B_{i r}^{\prime}$ (19)
Table 1. The weight of each secondary index
│Index│Weight │Index│Weight │
│B11 │0.0622 │B42 │0.1068 │
│B12 │0.1370 │B43 │0.0316 │
│B21 │0.0216 │B44 │0.0154 │
│B22 │0.1468 │B45 │0.0049 │
│B31 │0.0693 │B51 │0.0054 │
│B32 │0.1650 │B52 │0.2031 │
│B33 │0.0891 │B53 │0.6390 │
│B34 │0.0301 │B54 │0.6561 │
│B41 │0.1135 │ │ │
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This number is a composite.
There are 80 four-digit primes which are concatenations of two-digit primes. [De Geest]
(10^80 - 80)/80 is prime. [Luhn]
80^(80+1) + (80+1)^80 is prime.
The largest natural number n such that all prime factors of n and n+1 are smaller or equal to the prime digit 5. [Capelle]
The only two-digit integer n that yields a prime in the form n^(n+1)+(n+1)^n. [Silva]
The smallest integer n such that both n and n+1 are products of at least 4 primes. [Post]
Theoretically, the fewest prime factors in the numbers spanning a non-trivial 8-tuple. [Merickel]
Printed from the PrimePages <t5k.org> © G. L. Honaker and Chris K. Caldwell
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[Solved] A normal distribution can be described co | SolutionInn
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A normal distribution can be described completely by what? Average variance through time Variance and cumulative probability Mean and standard deviation Geometric and arithmetic average
A normal distribution can be described completely by what? Average variance through time Variance and cumulative probability Mean and standard deviation Geometric and arithmetic average None of the
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The course on Networking aims to provide a general and basic introduction to the art of “networking”, that tries to unravel the operation and behavior of networks, both man-made (infrastructures such
as the Internet, power grids and transportation networks) as well as networks appearing in nature (such as the human brain, biological networks and social human interactions). The course on
Networking will introduce concepts of Network Science, that basically studies the interplay between, on the one hand, the processes - also called functions or services - on the network and, on the
other hand, the underlying topology or structure, that can also change over time as an evolving organism. Network Science combines many disciplines such as graph and network theory, probability
theory, physical processes, control theory and algorithms.
After this course, students are expected to represent/abstract real-world infrastructural network (e.g. a communication system) as a complex network, understand the basic methods to analyze
properties of networks and dynamic processes on networks. Students will also understand why processes on networks and design of networks are so complex. Finally, students may appreciate the
fascinatingly rich structure and behavior of networks and may realize that much in the theory of networks still lies open to be discovered.
Course Outline
• PART 1: Network topology or graph
□ Basics of networking and introduction to Network Science
□ Graph theory: what is a network? Representation of a graph, overview of the relatively new theory of complex networks, called Network Science.
□ Graph metrics: important characterizers of a network (network metrics)
□ Graph models
□ Examples of real-world networks (airline transportation, the web and Internet, social networks, brain networks, etc.) and applications of network science
• PART 2: Network function or process and service
□ Electrical networks: the power of the Laplacian and the effective resistance matrix
□ network robustness (failure, cascading effects,...)
□ traffic management and scheduling
□ overlay networking and new aspects of networking such as interdependent networks
• PART 3: Applications and examples of networks Some classes will be taught by a guest lecturer. Ranging from year to year, a selection among the following will be covered:
□ Electrical networks (smart grids)
□ Networks on Chip (NoC)
□ Optical networks
□ Computer Networks (the Internet)
□ Mobile communication networks
□ Sensor networks
□ Biological networks
□ Social networks
□ Transportation networks
Course announcements
• TUDelft code: EE4C06
• 5 EC
• The course material consists of the slides posted on Brightspace after each class. The slides will contain references to the literature providing an extra depth. In particular, references to
Graph Spectra for Complex Networks, Second edition, 2023 will be added in the slides.
• The examination is an open book examination.
• Due to the large number of interested students (> 250 over the last years), the examination consists of 20 multiple choice questions in Brightspace on a computer.
• There are precisely two examination possibilities per year.
• All other announcements are mentioned on Brightspace.
Last modified: 5-09-2023
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The Square Root Function in Python (Summary) – Real Python
The Square Root Function in Python (Summary)
Knowing how to use sqrt() is only half the battle. Understanding when to use it is the other. Now you’ve seen some of both, so go and apply your newfound knowledge about the Python square root
In this course, you’ve covered:
• A brief introduction to square roots
• The ins and outs of the Python square root function, sqrt()
• A practical application of sqrt() using a real-world example
Congratulations, you made it to the end of the course! What’s your #1 takeaway or favorite thing you learned? How are you going to put your newfound skills to use? Leave a comment in the discussion
section and let us know.
Why do you force one of the operands in a division to be a float? Isn’t that redundant in Python 3.x?
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BEC condenstate thermal statistics vs. coherence
BEC condenstate thermal statistics vs. coherence
• Thread starter sam_bell
• Start date
In summary: This is because it would be overwhelmed by the number of particles in the system. In order to calculate the density operator, one would need to use a cutoff point, say after a certain
number of particles.
Hi. I'm reading an introductory section on the Bose-Einstein condensation of a non-interacting, spinless boson gas. I'm confused by the claim that the ground state is in a coherent state with
eigenvalue sqrt(N0) exp(i theta), where N0 is the expected number of particles in the ground state. The justification is that the commutator [a0/sqrt(V), a0*/sqrt(V)] = 1/V goes to zero in the
thermodynamic limit V = volume goes to infinity (a0 annihilation operator for ground state). Therefore a0 acts like a complex number and so the ground state must be in a coherent state. Huh? Who
asked you to divide by V anyway? Totally opaque. And doesn't statistical mechanics say the system is in an ensemble of definite particle eigenstates with probability exp(-beta*mu*N)/Z (i.e. NOT a
coherent superposition of definite particle states)?? Does someone understand this better?
The expectation value of a_0^*a_0 is N, so in the thermodynamic limit you have to divide by V so as to obtain something finite, i.e. N/V.
The vanishing of the commutator (and of the commutators with all other well defined operators) means that by Schur's theorem a/\sqrt{V} is represented as a pure number iff the representation of the
operator algebra is irreducible. However, no one forces you to use an irreducible representation. You can also work with a state of the BEC with the number of particles being sharp. However,
calculations become somewhat more involved. Nevertheless, this is the correct description for finite systems.
And finally, no, statistical mechanics does not tell you that particle number has to be sharp as N is an operator in that expression you write down and not a number.
OK, you've given me quite a bit to chew on there. I might have a follow-up later, but thanks for the head-start.
I understand that exp(beta*mu*N)/Z could be viewed as an operator and therefore that <N> = Tr[ N exp(beta*mu*N)/Z ] could be calculated in any basis (say coherent states). But this doesn't tell you
about the distribution of particle number. In the grand canonical ensemble the density operator rho = Sum[ exp(beta*mu*n)/Z |n><n|, n=0..+inf ], i.e. diagonal only in the basis of definite particle
states. If as claimed the ground state is actually in a coherent state with eigenvalue alpha, then shouldn't we have rho = |alpha><alpha| (i.e. in a pure state)? In an experiment I imagine it would
be difficult to measure anything but <N>. Is this a case of trying to apply statistical mechanics to rigidly? Or am I interpreting wrong?
Last edited:
sam_bell said:
I understand that exp(beta*mu*N)/Z could be viewed as an operator and therefore that <N> = Tr[ N exp(beta*mu*N)/Z ] could be calculated in any basis (say coherent states). But this doesn't tell
you about the distribution of particle number. In the grand canonical ensemble the density operator rho = Sum[ exp(beta*mu*n)/Z |n><n|, n=0..+inf ], i.e. diagonal only in the basis of definite
particle states. If as claimed the ground state is actually in a coherent state with eigenvalue alpha, then shouldn't we have rho = |alpha><alpha| (i.e. in a pure state)? In an experiment I
imagine it would be difficult to measure anything but <N>. Is this a case of trying to apply statistical mechanics to rigidly? Or am I interpreting wrong?
One of the problems here is that the density operator generally does not exist in the thermodynamical limit of infinite sample size.
FAQ: BEC condenstate thermal statistics vs. coherence
1. What is BEC (Bose-Einstein condensate)?
BEC is a state of matter that occurs at extremely low temperatures, close to absolute zero. It is a phenomenon observed in certain materials, where a large number of particles behave as a single
quantum entity, exhibiting wave-like properties.
2. What is the difference between thermal statistics and coherence in BEC?
Thermal statistics refers to the distribution of particles within the BEC, which follows the Bose-Einstein distribution. Coherence, on the other hand, refers to the phase relationship between
particles within the BEC, which is maintained due to the wave-like nature of the particles.
3. How does BEC behavior differ from that of a gas?
In a gas, particles move independently and randomly, while in a BEC, particles exhibit collective behavior due to coherence. Additionally, the energy distribution in a BEC follows the Bose-Einstein
distribution, while in a gas it follows the Maxwell-Boltzmann distribution.
4. What are the applications of BEC in science and technology?
BEC has potential applications in fields such as quantum computing, precision measurements, and creating new states of matter. It has also been used in creating atom lasers and studying
5. How is BEC created in a laboratory setting?
BEC is created by cooling a gas of atoms to extremely low temperatures, typically using laser cooling techniques. The atoms are then confined in a magnetic trap and further cooled using evaporative
cooling, until they reach the critical temperature for BEC formation.
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Excel Formula: Count Days Between Cells
In this tutorial, we will learn how to count the number of days between two cells in Excel using a Python formula. This can be achieved by using the DATEDIF function, which calculates the difference
in days between two dates. The formula takes the start date and end date as arguments, and returns the number of days between them.
To count the days between cell L1 and M1, we can use the following formula in Excel:
This formula calculates the difference in days between the dates specified in cells L1 and M1. The unit of measurement is set to "d" to indicate days.
For example, if cell L1 contains the date 01/01/2022 and cell M1 contains the date 01/10/2022, the formula would return the value 9, representing the number of days between the two dates.
It's important to note that the DATEDIF function does not include the end date in the calculation. If you want to include the end date, you can add 1 to the formula:
=DATEDIF(L1, M1, "d") + 1
This modified formula would return 10 for the first example.
In conclusion, the formula =DATEDIF(L1, M1, "d") allows us to easily count the number of days between two cells in Excel using Python. It provides a convenient way to calculate date differences and
can be customized to include or exclude the end date based on your requirements.
An Excel formula
Formula Explanation
The formula uses the DATEDIF function to calculate the number of days between two dates. In this case, the dates are specified in cells L1 and M1.
Step-by-step explanation
1. The DATEDIF function takes three arguments: the start date, the end date, and the unit of measurement.
2. In this formula, L1 is the start date and M1 is the end date.
3. The unit of measurement is specified as "d" to calculate the difference in days.
4. The DATEDIF function returns the number of days between the two dates.
For example, if cell L1 contains the date 01/01/2022 and cell M1 contains the date 01/10/2022, the formula =DATEDIF(L1, M1, "d") would return the value 9, which represents the number of days between
the two dates.
Similarly, if cell L1 contains the date 01/01/2022 and cell M1 contains the date 01/01/2023, the formula would return the value 365, as there are 365 days between the two dates.
Note that the DATEDIF function does not include the end date in the calculation. If you want to include the end date, you can add 1 to the formula:
=DATEDIF(L1, M1, "d") + 1
This modified formula would return 10 for the first example and 366 for the second example.
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Applying the Power Rule - Knowunity
Applying the Power Rule: AP Calculus Study Guide
Welcome, mathletes! It’s time to flex those calculus muscles as we dive into the dynamic world of differentiation. Today, we’re going to unravel one of the biggest tricks in the calculus book – the
Power Rule. Think of it as the superhero of differentiation, always ready to save the day! 🦸♂️📚
The Power Rule: Math's Secret Weapon
The Power Rule is the VIP pass to quickly finding derivatives without all the hassle of the limit definition. Here's the drill: If you have a function ( f(x) = x^n ), where ( n ) is a constant (it
doesn’t change, just like your love for calculus), then the derivative ( f'(x) ) is given by:
[ f'(x) = n \cdot x^{(n-1)} ]
So, it's kind of like reducing power – literally! The Power Rule snips off a bit of power from ( x ) and makes life a whole lot easier. Kind of like having a cheat code for your calculus problems 🎮📉.
Rocking the Power Rule: Practice Problems
Alright, math ninjas, it's practice time! Let's see the Power Rule in action. 🥋💡
1. Given ( f(x) = x^4 ), find ( f'(x) ).
2. Given ( f(x) = \frac{1}{x^5} ), find ( f'(x) ).
3. Given ( f(x) = \sqrt{x} ), find ( f'(x) ).
4. Given ( f(x) = x^6 + 2x^4 - 10 ), find ( f'(x) ).
Remember, the Power Rule is your go-to tool, but sometimes you may need to tweak the function a bit to use it effectively, especially with fractions and square roots. Let's uncover these mysteries
one by one.
Insights Before the Reveal
• Tip number one: The Power Rule loves simplicity. Rewriting functions to make them power-friendly can be a game-changer.
• Pro-tip number two: Constants don’t change, and their derivatives are as zero as the chances of a snowstorm in the Sahara. 🌵❄️
Power Rule In Action: Solutions
Let's reveal the magic curtain and see those derivatives pop out:
1. For ( f(x) = x^4 ):
[ f'(x) = 4 \cdot x^{(4-1)} = 4x^3 ]
2. For ( f(x) = \frac{1}{x^5} ), rewrite it as ( f(x) = x^{-5} ):
[ f'(x) = -5 \cdot x^{(-5-1)} = -5x^{-6} = -\frac{5}{x^6} ]
3. For ( f(x) = \sqrt{x} ), rewrite it as ( f(x) = x^{\frac{1}{2}} ):
[ f'(x) = \frac{1}{2} \cdot x^{(\frac{1}{2}-1)} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} ]
4. For ( f(x) = x^6 + 2x^4 - 10 ):
[ f'(x) = 6x^5 + 8x^3 - 0 ] (Note: the derivative of the constant (-10) is just zero).
Summary and Cheer
High fives all around! 🎉 You’ve now got the Power Rule under your belt, and you’re ready to take on more of calculus’ greatest hits. Keep practicing, and remember: With great power (rule) comes great
responsibility... to ace those exams.
Key Terms to Remember
• Differentiate: Finding the derivative of a function to measure the rate of change.
• Inverse Function (f^{-1}(x) = \sqrt{x}): A function that reverses the effect of another function.
• Inverse Functions: Paired functions that undo each other's effect.
• Power Rule: The key to breaking down the derivative of a function raised to a constant power.
• Product Rule: Used to differentiate products of two functions.
Fun Fact
The Power Rule is like the magic beans of calculus – a small trick with giant benefits. Plant these rules in your mind, and watch your mathematical skills grow as high as a beanstalk! 🌱✨
Until next time, keep mathematically flexing those brains and stay curious, my friends! 🌟
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How to Pass Study.com Math 105 : Precalculus Algebra Exam
Welcome to the one and only guide for passing your study.com Math 105: Precalculus Algebra exam! So if you are preparing for this course, you will be set for a mathematical ride where the outcome
would prepare you on how to approach different higher mathematics courses plus real-life problems. This course is intended to study mathematical modeling deeply, derivatives, and solutions of
different equations and inequalities with symbols and graphs. From the purely mathematical lover to the student merely enduring, there’s something interesting in this class.
Think of study.com Math 105: Precalculus Algebra as your tool kit for how you want to read and perceive mathematics. You will be able to write and solve equations that describe the situations you
meet in everyday life, compare the values of functions, and even take a glimpse at the derivatives. It concerns going beyond simple analytics and learning a systematic way to think that will be
valuable in every future assignment. Therefore, sit tight, put on your seat belt, and let me help you to defeat the precalculus algebra and the world with it!
A Sneak Peek into Math 105📖
Embarking on study.com Math 105: Precalculus Algebra tells you that you are going to study a range of mathematical ideas that will help pave the way for other math classes. This class comprises
numerous subjects each of which is central to achieving mastery in precalculus algebra.
Starting with Mathematical Modeling, here are some of the courses you are going to learn. This entails the use of symbols on a pattern, which best represents an occurrence in the real world. The
topics include how to write and assess expressions associated with geometric figures, learn about linear models, also use systems of linear equations to get the breakeven points and market
equilibrium. These are not just theoretical ideas; they are the strategies that are used in different decision-making processes in different areas including commerce and engineering.
Then, Linear Equations & Inequalities present the next topic in the course. By the end of the combined course, you will be able to recognize and plot linear equations, comprehend the concepts of
slope and intercepts, as well as solve systems of equations. This segment is elementary because linear equations are encountered throughout mathematics, as well as in other spheres of human life and
Further on, Quadratic Functions and Rational Expressions & Functions are included in what is to be covered. These sections are important for the proctored final exam, specifically the Speaking
Section. Quadratic equations will be factored in, the process of completing the square will be viewed, and solving quadratic inequalities will also be developed. For rational functions, you will
learn operations on rational expressions, their graphs, and asymptotes.
The content of the book is divided into chapters and each chapter progresses from the previous one, hence a clear understanding of precalculus algebra. It is pertinent to spend time doing problems
since this is an excellent way of revising and grasping the concepts in the book. self-paced will enable you to take more time on sections that you need to practice as you prepare for the final exam
of the course.
How to Study Weekly for Math 105📝
Are you ready to ace your study.com Math 105: Precalculus Algebra? Below is a suggested plan of study for a week to ensure that you do not lag behind and are well-prepared for the final exam. The
reader of this guide will be guided towards successful abstractions that lay a strong base and use regular exercises for creating different concepts
Week 1: Introduction & Mathematical Modeling
• Lessons to Cover: Introduction to algebraic expressions and linear models.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice writing and solving algebraic expressions.
□ External Resource: Khan Academy – Algebra Basics
Week 2: Linear Equations & Inequalities
• Lessons to Cover: Linear equations, slope-intercept form, and graphing inequalities.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Solve practice problems on graphing linear equations.
□ External Resource: PatrickJMT – Linear Equations
Week 3: Quadratic Functions
• Lessons to Cover: Factoring quadratics, quadratic formula, and graphing parabolas.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice solving quadratic equations using different methods.
□ External Resource: Professor Leonard – Quadratics
Week 4: Rational Expressions & Functions
• Lessons to Cover: Multiplying, dividing, adding, and subtracting rational expressions.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice simplifying rational expressions and solving rational equations.
□ External Resource: Mathispower4u – Rational Expressions
Week 5: Polynomial Functions
• Lessons to Cover: Adding, subtracting, and multiplying polynomials; polynomial long division.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice polynomial operations and solve practice problems.
□ External Resource: Khan Academy – Polynomial Operations
Week 6: Geometry Basics
• Lessons to Cover: Area and circumference of a circle, distance formula, and midpoint formula.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Solve problems involving geometric figures and formulas.
□ External Resource: Math Antics – Geometry
Week 7: Functions Overview
• Lessons to Cover: Domain and range, power and radical functions, points of discontinuity.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice identifying domain and range and graphing functions.
□ External Resource: PatrickJMT – Functions
Week 8: Function Operations
• Lessons to Cover: Adding, subtracting, multiplying, and dividing functions; inverse functions.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice performing operations on functions and finding inverses.
□ External Resource: Khan Academy – Function Operations
Week 9: Graph Symmetry
• Lessons to Cover: Symmetry in graphs, even and odd functions.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice identifying symmetry and classifying functions as even or odd.
□ External Resource: Mathispower4u – Graph Symmetry
Week 10: Exponential and Logarithmic Functions
• Lessons to Cover: Exponential growth and decay, properties of logarithms, graphing logarithmic functions.
• Activities:
Week 11: Essentials of Trigonometry
• Lessons to Cover: Trigonometric functions, identities, and equations.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice solving trigonometric equations and proving identities.
□ External Resource: Khan Academy – Trigonometry
Week 12: Introduction to the Derivative
• Lessons to Cover: Rate of change, formal definition of the derivative, graphing derivatives.
• Activities:
□ Watch video lessons and take notes.
□ Complete quizzes for each lesson.
□ Practice calculating and graphing derivatives.
□ External Resource: Paul’s Online Math Notes – Derivatives
This, of course, will help you prepare well for the final exam, and if you stick to the study plan above, you’ll definitely be ready for the test. Repetition is important, do not rush through the
material and if there is a hard topic that was discussed just go over it again and do not be shy to use an extra resource.
Supplement Your Math 105 Learning with Free Resources📂
Below is a list of the best and actually free resources on the Internet that will make it easy to understand the intricate concepts in study.com Math 105: Precalculus Algebra, and perform great on
the final exam.
Khan Academy
YouTube Channels and Playlists
1. PatrickJMT: For clear and concise explanations on various math topics.
2. Mathispower4u: A great resource for step-by-step tutorials on numerous algebra and precalculus topics.
3. Professor Leonard: Excellent in-depth lectures that cover a wide range of math topics.
These resources are perfect for supplementing your study.com materials, providing different perspectives, explanations, and plenty of practice problems. Make sure to take advantage of them!
Focus Topics for Success in Math 105🔑
Explaining the major terms used in study.com Math: Precalculus Algebra is helpful in understanding the subject, especially in the final exam in Math 105. Below are four big subjects that are clearly
described with examples and the possibilities to compare to enhance your knowledge.
1. Quadratic Functions
Quadratic functions are a cornerstone of this course. They are typically written in the form y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c. Understanding how to factor these functions, use the quadratic
formula, and complete the square is essential.
2. Rational Expressions
Rational expressions involve fractions where the numerator and/or the denominator are polynomials. You need to know how to perform operations on these expressions and solve rational equations.
3. Exponential and Logarithmic Functions
Exponential and logarithmic functions are vital for modeling growth and decay processes.
4. Trigonometric Functions
Trigonometric functions relate angles to side lengths in right triangles. Understanding their properties and identities is crucial.
These key concepts are not only heavily featured in the course but are also critical for solving complex problems efficiently. Mastering these areas will give you a significant advantage on the final
Top Questions about Math 105 Answered❓
While going through study.com Math 105: Precalculus Algebra, you may have these few questions. Here are some frequently asked questions that might help clear up any confusion:
Q: What should I do to make preparation for the proctored final exam efficient?
A: If you want to follow a good week-by-week study schedule, you can use the following guidelines: Each day should be devoted to understanding a particular concept and the Migliorini Chapter Recaps
can be used for self-testing. Take the quizzes at the end of each lesson, make use of the additional external resources, and revisit the topics that were covered frequently.
Q: Is it allowed to use a calculator while taking proctored exams?
A: Yes, the use of a non-graphing, scientific calculator will be allowed. Study.com will similarly allow the use of the Desmos scientific calculator during the exam. Nevertheless, any computer,
calculator especially graphing calculators, and any other programs are prohibited.
Q: What should I do if I find some topic to be a bit difficult for me to understand?
A: If you are having a problem comprehending a certain topic, go back to the video lessons and jot down notes in detail. The Use of external resources mentioned above, for other explanations and the
practice problems mentioned above. Of course, do not forget to contact study.com for further support.
Q: To what extent are the quizzes in this course essential?
A: It is common knowledge that quizzes are very important as they contribute to a large percentage of the final grade (points 100 out of 300). They assist in re-stating and checking up on what has
been understood and being practiced. Here, one gets to attempt each quiz three times and the best score will be considered.
Q: What will I do if I fail the final exam in the initial attempt?
A: In case you are not satisfied with the score you got on the proctored final exam, you have to wait for 3 days to take the test again. You can take this exam twice, after which you can no longer
try to improve your score. For this reason, ensure you use the study guide and take your time when seeking the exam.
Q: Are there any specific strategies for solving quadratic equations?
A: Yes, there are several strategies including factoring, using the quadratic formula, and completing the square. Practice each method to understand their applications and choose the most suitable
one depending on the equation.
In Math 105: Precalculus Algebra, it will be evident that this course provides you with a core foundation for progression mathematically and solving real-life applications. Through a particular
schedule, using resources from the Internet, and studying the main topics of a course, you are to be ready for the more severe proctored final exam.
Remember, consistency is key. Always solve exercises, go through the information that is difficult to understand, and do not shy away from asking a tutor for more assistance. This makes it easy for
you not only to learn but to also apply this knowledge as seen from the video lessons, quizzes, and even the proctored exam offered.
Come to this course with confidence and open mind as each topic connects to the next to provide the overall picture of precalculus algebra. Good luck, and enjoy your mathematical journey!
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Crystal Prison Zone
I have two problems with statistics in psychological science. They are:
1. Everybody speaks in categorical yes/no answers (statistical significance) rather than continuous, probabilistic answers (probably yes, probably no, not enough data to tell).
2. There's a lot of bullshit going around. The life cycle of the bullshit is extended by publication bias (running many trials and just reporting the ones that work) and p-hacking (torturing the
data until it gives you significance).
Meta-analysis is often suggested as one solution to these problems. If you average together everybody's answers, maybe you get closer to the true answer. Maybe you can winnow out truth from bullshit
when looking at all the data instead of the tally of X significant results and Y nonsignificant results.
That's a nice thought, but publication bias and p-hacking make it possible that the meta-analysis just reports the degree of bias in the literature rather than the true effect. So how do we account
for bias in our estimates?
Bayesian Spike-and-Slab Shrinkage Estimates
One very simple approach would be to consider some sort of "bullshit factor". Suppose you believe, as John Ioannidis does, that half of published research findings are false. If that's all you know,
then for any published result you believe that there's a 50% chance that there's an effect such as the authors report it (p(H1) = .5) and a 50% chance that the finding is false (p(H0) = .5). Just to
be clear, I'm using H0 to refer to the null hypothesis, H1 to refer to the alternative hypothesis.
How might we summarize our beliefs if we wanted to estimate the effect with a single number? Let's say the authors report d = 0.60. We halfway believe in them, but we still halfway believe in the
null. So on average, our belief in the true effect size delta is
delta = (d | H0) * probability(H0) + (d | H1) * probability(H1)
delta = (0) * (0.5) + (0.6) * (0.5) = 0.3
So we've applied some shrinkage or regularization to our estimate. Because we believe that half of everything is crap, we're able to improve our estimates by adjusting our estimates accordingly.
This is roughly a Bayesian spike-and-slab regularization model: the spike refers to our belief that delta is exactly zero, while the slab is the diffuse alternative hypothesis describing likely
non-zero effects. As we believe more in the null, the spike rises and the slab shrinks; as we believe more in the alternative, the spike lowers and the slab rises. By averaging across the spike and
the slab, we get a single value that describes our belief.
Bayesian Spike-and-Slab system. As evidence accumulates for a positive effect, the "spike" of belief in the null diminishes and the "slab" of belief in the alternative soaks up more probability.
Moreover, the "slab" begins to take shape around the true effect.
So that's one really crude way of adjusting for meta-analytic bias as a Bayesian: just assume half of everything is crap and shrink your effect sizes accordingly. Every time a psychologist comes to
you claiming that he can make you 40% more productive, estimate instead that it's probably more like 20%.
But what if you wanted to be more specific? Wouldn't it be better to shrink preposterous claims more than sensible claims? And wouldn't it be better to shrink fishy findings with small sample sizes
and a lot of p = .041s moreso than a strong finding with a good sample size and p < .001?
Bayesian Meta-Analytic Thinking by Guan & Vandekerckhove
This is exactly the approach given in a recent paper by
Guan and Vandekerckhove
. For each meta-analysis or paper, you do the following steps:
1. Ask yourself how plausible the null hypothesis is relative to a reasonable alternative hypothesis. For something like "violent media make people more aggressive," you might be on the fence and
assign 1:1 odds. For something goofy like "wobbly chairs make people think their relationships are unstable" you might assign 20:1 odds in favor of the null.
2. Ask yourself how plausible the various forms of publication bias are. The models they present are:
1. M1: There is no publication bias. Every study is published.
2. M2: There is absolute publication bias. Null results are never published.
3. M3: There is flat probabilistic publication bias. All significant results are published, but only some percentage of null results are ever published.
4. M4: There is tapered probabilistic publication bias: everything p < .05 gets published, but the chances of publishing get worse the farther you get from p < .05 (e.g. p = .07 gets published
more than p = .81).
3. Look at the results and see which models of publication bias look likely. If there's even a single null result, you can scratch off M2, which says null results are never published. Roughly
speaking, if the p-curve looks good, M1 starts looking pretty likely. If the p-curve is flat or bent the wrong way, M3 and M4 start looking pretty likely.
4. Update your beliefs according to the evidence. If the evidence looks sound, belief in the unbiased model (M1) will rise and belief in the biased models (M2, M3, M4) will drop. If the evidence
looks biased, belief in the publication bias models will rise and belief in the unbiased model will drop. If the evidence supports the hypothesis, belief in the alternative (H1) will rise and
belief in the null (H0) will drop. Note that, under each publication bias model, you can still have evidence for or against the effect.
5. Average the effect size across all the scenarios, weighting by the probability of each scenario.
If you want to look at the formula for this weighted average, it's:
delta = (d | M1, H1) * p(M1, H1) + (d | M1, H0)*p(M1, H0) + (d | M2, H1)*p(M2, H1) + (d | M2, H0)*p(M2, H0) + (d | M3, H1)*p(M3, H1) + (d | M3, H0)*p(M3, H0) + (d | M4, H1)*p(M4, H1) + (d | M4,
H0)*p(M4, H0)
(d | Mx, H0) is "effect size d given that publication bias model X is true and there is no effect." We can go through and set all these to zero, because when the null is true, delta is zero.
(d | Mx, H1) is "effect size d given that pubication bias model X is true and there is a true effect." Each bias model makes a different guess at the underlying true effect. (d | M1, H1) is just the
naive estimate. It assumes there's no pub bias, so it doesn't adjust at all. However, M2, M3, and M4 say there is pub bias, so they estimate delta as being smaller. Thus, (d | M2, H1), (d | M3, H1),
and (d | M4, H1) are shrunk-down effect size estimates.
p(M1, H1) through p(M4, H0) reflect our beliefs in each (pub-bias x H0/H1) combo. If the evidence is strong and unbiased, p(M1, H1) will be high. If the evidence is fishy, p(M1, H1) will be low and
we'll assign more belief to skeptical models like p(M3, H1), which says the effect size is overestimated, or even p(M3, H0), which says that the null is true.
Then to get our estimate, we make our weighted average. If the evidence looks good, p(M1, H1) will be large, and we'll shrink d very little according to publication bias and remaining belief in the
null hypothesis. If the evidence is suspect, values like p(M3, H0) will be large, so we'll end up giving more weight to the possibility that d is overestimated or even zero.
So at the end of the day, we have a process that:
1. Takes into account how believable the hypothesis is before seeing data, gaining strength from our priors. Extraordinary claims require extraordinary evidence, while less wild claims require less
2. Takes into account how likely publication bias is in psychology, gaining further strength from our priors. Data from a pre-registered prospective meta-analysis is more trustworthy than a look
backwards at the prestige journals. We could take that into account by putting low probability in pub bias models in the pre-registered case, but higher probability in the latter case.
3. Uses the available data to update beliefs about the hypothesis and publication bias both, improving our beliefs through data. If the data look unbiased, we trust it more. If the data looks like
it's been through hell, we trust it less.
4. Provides a weighted average estimate of the effect size given our updated beliefs. It thereby shrinks estimates a lot when the data are flimsy and there's strong evidence of bias, but shrinks
estimates less when the data are strong and there's little evidence of bias.
It's a very nuanced and rational system. Bayesian systems usually are.
That's enough for one post. I'll write a follow-up post explaining some of the implications of this method, as well as the challenges of implementing it.
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An introduction to Slot Attention
In this blog post, I’ll cover the basic slot attention mechanism (Locatello et al., 2020) and go over some intuition as to why it works.
The problem we tackle is learning representations for particular regions of images, like a representation for a box, sphere or background of the scene.
Intuition, k-means clustering
Hard K-means clustering:
K-means clustering in the most simple form has the following steps:
Setup: Randomly initialize clusters by assigning each vector \(\mathbf{x}_n\) to one of \(N\) clusters
1. Compute cluster means \(\boldsymbol{\mu}_n\) by averaging all points assigned to cluster \(n\)
2. Reassign each each point to the cluster corresponding to it closest mean \(\boldsymbol{\mu}_k\).
3. If some no new assignments were made, we’ve converged and can stop.
While the example I gave above was for 2D Points, this algorithm has been used for images as well, for segmenting regions of the image by the pixel values in some colour space.
When this algorithm converges, we expect to have \(N\) centroids, where each one should have a high affinity to the points closest to it.
Soft K-Means clustering
The above algorithm doesn’t take into account that a certain point may share characteristics of more than one cluster. Hard K-Means is not expressive enough to capture this sort of relation, so we
use Soft K-Means instead.
Setup: Randomly initialize clusters by assigning each vector \(\mathbf{x}_n\) to one of \(N\) clusters. We also define a distance measure \(\phi_n(i)\), that measures the “closeness” of \(\mathbf{x}
_n\) to cluster \(\boldsymbol{\mu}_i\).
Repeat: Iterate the following
1. Updating \(\phi\):
\[\phi_n(i)=\frac{\exp \left\{-\frac{1}{\beta}\left\|\mathbf{x}_n-\boldsymbol{\mu}_i\right\|^2\right\}}{\sum_j \exp \left\{-\frac{1}{\beta}\left\|\mathbf{x}_n-\boldsymbol{\mu}_j\right\|^2\right
\}}, \text { for } i=1, \ldots, N\]
2. Update \(\boldsymbol{\mu}_i\) : For each \(i\), update \(\boldsymbol{\mu}_k\) with the weighted average
\[\boldsymbol{\mu}_i=\frac{\sum_n \mathbf{x}_n \phi_n(i)}{\sum_n \phi_n(i)}\]
Now we have a more expressive model, but notice a few things:
• The model is dependent on the data that it has fit to. Eg: Say you run this on image A and get some clusters. In order to recieve new clusters, you need to rerun the algorithm.
• Highly dependent on initialization. For some cluster initializations, you can get a very poor clustering performance
• The cluster mean might not be the best representation of the vectors inside the cluster itself
This motivates the question, can we use parameters and highly expressive neural networks to perform much better than K-Means?
Slot Attention
How to compute slots:
The slot attention operation works as follows:
Setup: Initialize embedding weights for key, query and value projection \(q( \cdot ), k(\cdot), v(\cdot)\) (Vaswani et al., 2023). Also initialize \(N_\text{slots}\) slots of embedding dimension \(D
\), basically a matrix of shape \(N_{\text{slots}} \times D\). These can be sampled from an isotropic normal with learned mean \(\boldsymbol{\mu} \in \mathbb{R}^{D}\)
Repeat T times:
1. Given image of shape \(B\times 3 \times H \times W\), use a CNN encoder to encode this into a feature map of dimension \(B\times D \times H \times W\). Then flatten this into a sequence of tokens
of shape \(N_{\text{data}} \times D\), call this \(\text{inputs}\).
2. Compute softmax weights over the data embeddings:
\[\text{Softmax}(\frac{1}{\sqrt{D}} k(\text{inputs}) q(\text{slots})^T, \text{axis='slots'})\]
Unpacking this, we have \(\text{inputs} \in \mathbb{R}^{N_{data} \times D}\) and \(\text{slots} \in \mathbb{R}^{N_{slots} \times D}\) (excluding batch dimension). Therefore, our softmax weights
are of shape \((N_{\text{data}}, N_{\text{slots}})\) and the softmax is done across the \(N_{\text{slots}}\) axis. This is different from regular attention, which is done over the \(N_{\text
{data}}\) axis, as this promotes “competition” across the slots. I’ll explain the intuition for that later.
3. Compute slot updates as using a weighted average, the shape of which is \((N_{\text{slots}},D)\). Note that is the same shape as our slots.
\[\text{Softmax}(\frac{1}{\sqrt{D}} k(\text{inputs}) q(\text{slots})^T, \text{axis='slots'})^T v(\text{inputs})\]
4. Update the previous slots using a small GRU network and the slot updates as the input.
As you can see above, each slot update is a linear combination of \(\text{inputs}\). The coefficients of input embedding vector is determined by the softmax matrix, very similar to the regular
attention mechanism we all know and love.
To see why the slots enforce competition, we need to take a look at the softmax matrix in more detail. Denote \(I_i\) as the \(i^{th}\) input vector, and \(S_j\) as the \(j^{th}\) slot vector.
\[k(\text{inputs}) q(\text{slots})^T = \begin{pmatrix} I_1\cdot S_1 & \cdots & I_1 \cdot S_{N_{\text{slots}}} \\ \vdots & \ddots & \vdots \\ I_{N_{\text{data}}} \cdot S_1 & \cdots & I_{N_{\text
{data}}}\cdot S_{N_{\text{slots}}} \end{pmatrix}\]
For which direction to take the softmax in, we have two options, either row wise (on the data axis) or column wise (on the slot axis):
• If we normalize across the data axis, each row will sum up to one. So when we right multiply by \(v(\text{inputs})\), each slot update will be a convex linear combination of the input vectors.
This is exactly what is used in the regular attention mechanism. However, each slot is unrestricted in what parts of the input sequence it can attend to. For example, the first embedding could
have a softmax weight of \(1.0\).
• If we normalize across the slot dimension, each column will sum up to one. Now when we right multiply by \(v(\text{inputs})\), each slot update won’t be a convex linear combination anymore.
However, now we are constraining the attention weights for each embedding across all slots. For example, if slot \(S\) has a high attention weight \(\approx 1\) for embedding \(I\), then it must
be the case that the other slots have attention weights \(\approx 0\) for \(I\). This promotes “competition” for input vectors among the slots, as only a few slots will be able to have a high
weight for any given input vector due to the properties of softmax.
Ultimately, if a slot has high coefficients for a set of input embedding vectors, it should be representative of those input embedding vectors.
Therefore, I view this as a more expressive version of the K-Means operation we covered earlier, as the goal of both is to compute embeddings for distinct regions of the input images, and both do so
via linear combinations of the input data. I believe expressive comes from the high degree of nonlinearity within the \(q,k,v\) projections and \(\text{Softmax}\).
What does this actually do?
So now we run slot attention on a given image, and receive \(N_{\text{slots}}\) slots. What do we do now?
Well, we need some sort of signal to update these weights, and quantify how good of a representation we learned. A simple answer (and what is used in the original paper) is to simply reconstruct the
original image.
So assume each slot has learned to have high affinity (high inner product) with a particular region of the image. Eg: One slot binded to a sphere, another to a cube. Then, if we reconstruct each slot
into an image, we should get a set of images for each object the slot binds to.
This is exactly what is done in the paper. Here is the sequence of steps to decode:
1. For each slot of shape \((D, )\), use add two new spatial dimensions and repeat, getting a shape of \((D, \hat{H}, \hat{W})\)
2. Run this through a series of transpose convolutions, to get images of shape \((4, H, W)\).
3. The first 3 channels are RGB respectively, the last is an alpha channel. So for each slot \(i \in \{1, \dots, K\}\), we split each feature map into \(C_i \in \mathbb{R}^{3\times H \times W}\) and
\(\alpha_i \in \mathbb{R}^{3 \times H \times W}\) (we simply took our alpha channel and repeated it 3 times, so the shape lines up with \(C_i\)).
4. Finally we compose these by to get predicted image \(\hat{Y}\).
\[\hat{Y} = \sum_{i=1}^K \alpha_i \odot C_i\]
Our loss function is the simple MSE loss between original image \(Y\) and our predicted image \(\hat{Y}\):
\[L(\hat{Y}, Y) = \left | \left | \hat{Y} - Y \right | \right |^2_2\]
Here’s a visual of the pipeline I presented above for this image with 7 objects and 7 slots. While the decoded image isn’t perfect, notice how each slot representation, when decoded, has roughly
learned to represent a specific object in the scene.
Shortcomings and future directions:
To be added later…
Attention Is All You Need
Object-Centric Learning with Slot Attention
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Author Message
qtebed Posted: Thursday 21st of Mar 16:37
Hey dudes, I am about halfway through the semester, and getting a bit worried about my course work. I just don’t seem to grasp the stuff I am learning, especially things to do
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but if anyone knows about other ways of mastering topics like parallel lines or reducing fractions effectively, please let me know Thanks
Registered: 16.10.2002
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kfir Posted: Saturday 23rd of Mar 08:27
I really don’t know why God made math , but you will be happy to know that a group of people also came up with Algebrator! Yes, Algebrator is a program that can help you crack
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Registered: 07.05.2006
From: egypt
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Outafnymintjo Posted: Saturday 23rd of Mar 15:11
You must go through Algebrator. I had always thought math to be a difficult subject but this program made it easy to learn . You can type in the question and it gives you the
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Registered: 22.07.2002
From: Japan...SUSHI
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erx Posted: Sunday 24th of Mar 21:51
solving a triangle, adding exponents and cramer’s rule were a nightmare for me until I found Algebrator, which is truly the best algebra program that I have come across. I have
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Registered: 26.10.2001
From: PL/DE/ES/GB/HU
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Ccotticas Posted: Tuesday 26th of Mar 21:04
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Registered: 10.07.2003
From: Tacoma, WA
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CHS` Posted: Thursday 28th of Mar 10:47
Here you go, click on this link – https://softmath.com/ordering-algebra.html. I personally think it’s a really good software and the fact that they even offer an unconditional
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April 20: Kate Stange
Time: 3 pm Eastern/noon Pacific
Title: The geometry of number theory, through Mobius transformations
Abstract: Mobius transformations beautifully illustrate the geometry of complex numbers. My own interest arose when playing with some questions from number theory. I’ll show you some of the hidden
geometry of number theory through Schmidt arrangements, which are fractal circle packings built of Mobius transformations. Along the way, I’ll share my own story of computer-driven mathematical
experimentation and illustration.
Bio: Katherine Stange is a number theorist, cyclist and mom at the University of Colorado, Boulder. Mathematically speaking, she likes to dig around in the mud and get messy, with examples and
computer data. She enjoys number theory’s diverse tools in the face of elementary problems, loves to explore mathematics visually and computationally, and is attracted to problems involving geometry.
She also enjoys algorithms and works on post-quantum cryptography. She did her Bachelor of Mathematics at the University of Waterloo, and her PhD at Brown University.
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Category: Fibonacci Numbers
In the realm of understanding, where concepts intertwine like threads in a grand web, the idea of recursion stands as a pillar, echoing the patterns found in nature, mathematics, and Archives
the very fabric of reality. This exploration delves into the profound interconnectedness of recursion, the eternal now, unraveling time, fractal understanding, and the cause and
effect, woven together in a feedback loop that mirrors the infinite complexity of the Fibonacci sequence. Categories
Figure 1. Recursion and the Eternal Now: This figure presents a journey through recursion, the eternal now, unraveling time, and fractal understanding, all linked by cause and effect
within an ongoing feedback loop. Inspired by Fibonacci recursion, it shows how each concept fluidly transitions into the next, reflecting universal patterns. This visual exploration
invites us on a discovery where time and consciousness blend, unveiling the interconnected weave of reality.
The Foundation: Recursion and Fibonacci
Recursion, at its core, is a method where the solution to a problem depends on solutions to smaller instances of the same problem. It's a principle beautifully exemplified by the
Fibonacci sequence, where each number is the sum of the two preceding ones. This sequence is not just a mathematical curiosity but a pattern that recurs throughout nature, from the
spirals of shells to the branching of trees.
In our conceptual framework, recursion represents the starting point, the seed from which our understanding of the eternal now begins. It's the process of looking inward, of finding
within each moment the seeds of past and future, each now a reflection of the infinite recursion that defines our existence.
The Eternal Now: Pure Awareness
At the heart of our exploration lies the eternal now, a state of pure awareness unbound by the linear constraints of time. It's a concept that transcends the ticking of clocks and the
pages of calendars, representing a moment of clarity and presence that is both fleeting and everlasting.
This eternal now is not a static point but a gateway to deeper understanding, a lens through which we view the dance of interconnection that binds all things. It's in this eternal
moment that we find the essence of recursion, the echo of the Fibonacci sequence in the rhythm of our thoughts and the patterns of our consciousness.
Unraveling Time: The Dance of Interconnection
From the eternal now, we move to the unraveling of time, a process that reveals the interconnected nature of our experiences and perceptions. Time, in this context, is not a line but a
web, a complex weave of causes and effects that defy simple linear explanation.
This unraveling is a journey through the fractal complexity of reality, where each moment contains within it the seeds of countless others. It's a recognition that every event, every
thought, and every action is part of a greater whole, a symphony of interconnection that resonates with the patterns of recursion and the Fibonacci sequence.
Fractal Understanding: Beyond Cause and Effect
As we delve deeper into the web of time, we arrive at fractal understanding, a perspective that transcends the traditional linear chain of cause and effect. In this realm, the
simplicity of one-to-one relationships gives way to a more fluid, dynamic understanding of how things relate.
Fractal understanding is about recognizing the self-similar patterns that recur at different scales, from the microcosm to the macrocosm. It's about seeing the Fibonacci sequence not
just in a series of numbers but in the very structure of our thoughts and the universe itself.
The Feedback Loop: A Return to Recursion
Completing our conceptual journey is the feedback loop, a path that leads from the complexities of fractal understanding back to the simplicity of recursion. This loop is a reminder
that in every end is a beginning, in every complexity, a simplicity waiting to be discovered.
The feedback loop is the breath of the cosmos, the rhythm of existence that dances to the tune of the Fibonacci sequence. It's a cycle of renewal and understanding that brings us back
to the eternal now, enriched by our journey through the unraveling of time and the depths of fractal understanding.
Conclusion: The Web of Understanding
In this exploration of recursion, the eternal now, and the fractal nature of reality, we find a reflection of the infinite complexity and beauty of the universe. Like the Fibonacci
sequence, our understanding unfolds in ever-expanding spirals, each turn revealing new vistas of awareness and connection.
The journey through these concepts is not just an intellectual exercise but a meditation on the nature of existence itself. It's a reminder that in the patterns of the universe, from
the smallest shell to the vastness of the cosmos, lies a message of interconnection and eternal renewal, a symphony of patterns waiting to be discovered and understood.
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The Fibonacci sequence is a marvel that has captivated mathematicians throughout history. But its allure extends beyond math, influencing fields like art, biology, architecture, music,
botany, and finance.
Is this truly the world's most mesmerizing number sequence, or are we stretching our imagination to find patterns where none lie?
Join us as we delve deep into this mathematical wonder that has fascinated minds across disciplines and eras.
1. Does Architectural design reflect the golden ratio?
The golden ratio is often touted as appearing in many architectural marvels, both ancient and contemporary. Examples span from the Great Pyramid of Giza and the Parthenon to modern
landmarks like the Eiffel Tower, Toronto’s CN Tower, and the United Nations Secretariat building.
It's debated whether all these structures consciously employ the golden ratio in their designs. The Parthenon, with its intricate layout, leaves us guessing about its architects'
intentions regarding the golden ratio.
However, there seems to be a deliberate use of the golden ratio in the design of Toronto’s CN Tower. The proportion between its total height (553.33 meters) and the height to its
observation deck (342 meters) is strikingly close to 1.618, the golden ratio.
The CN Tower is a communications tower built in 1976. It was the world’s tallest free-standing structure at the time.
2. Is the spiral of the Nautilus shell based on the golden ratio?
Indeed, there's more nuance than popular narratives suggest. The classic "golden spiral" is crafted using consecutive golden rectangles, resulting in a spiral that expands by the
golden ratio for every quarter turn. However, the Nautilus shell's spiral isn't an exact match to this golden spiral.
The spiral constructed from a Golden Rectangle is NOT a Nautilus Spiral.
Here's the crux of the matter: The golden ratio can shape spirals in multiple ways.
For instance, a spiral that widens according to the golden ratio after every 180-degree twist more closely mirrors the spirals found in numerous Nautilus shells.
A spiral expanding by the golden ratio at every 180-degree turn is a closer match to some Nautilus shells for the first few rotations
I trust you grasp the distinction.
Here's another intriguing insight: The nautilus shell expands according to the number 108 (refer to "
The Number 108
"). This number resonates with the pentagram, which fundamentally operates on the principles of the golden ratio.
Thus, it's evident that the Nautilus spiral can indeed reflect proportions approaching Phi.
3. Did renaissance artists use the golden ratio in their paintings?
Absolutely. Leonardo Da Vinci's "Last Supper" prominently features golden ratios.
Inspired by him, Salvador Dalí, in 1955, crafted "The Sacrament of the Last Supper" adhering to golden ratio dimensions.
Similarly, in Michelangelo’s "Creation of Adam," the iconic touch between God’s and Adam’s fingers occurs exactly at the golden ratio point of the width within their frame.
These instances highlight just a fraction of the Renaissance artists' fascination with the golden ratio in their masterpieces.
4. Are the spirals seen in nature based on the golden ratio?
Spirals in nature are often linked to the golden ratio, and while it's not universally true, there are noteworthy instances.
Sunflower seeds, for instance, form interconnecting spirals based on Fibonacci numbers. This Fibonacci-driven pattern maximizes the number of seeds on a seed head.
Beyond sunflowers, the golden ratio influences the growth of leaves, branches, and petals, optimizing sunlight absorption as new leaves emerge.
Nature frequently showcases logarithmic spirals: from spiral galaxies and ram horns to hurricanes and whirlpools. The golden spiral is a subset of these logarithmic spirals, and while
it may be present in the examples listed, not all natural logarithmic spirals are golden spirals.
5. Is there a new algorithm based on the golden ratio that can predict spiritual experience?
It seems that there might be a connection between mathematics and spiritual experience. Sacco's 2017 study tentatively suggested that the dynamic effects observed at age 18 could
potentially predict an elevation in spiritual experiences. This observation aligns with the predictions of the FLCM. This offers a tentative link between the relationship of the golden
ratio and spiritual experience.
However, the dynamical effects of age 11 and 30 did not support the hypothesis of increased spiritual experience as predicted. This result required an alternative explanation.
It should be evident that children at age 11 may not be able to communicate the complexities of some spiritual experiences. And it’s entirely reasonable that spiritual experiences
could be linked with the instability that is less likely at age 30 when personality becomes stabilized.
In the real world, there are many compelling examples of the golden ratio in natural phenomena. But not all phenomena in nature are based on the golden ratio.
You can find plenty of examples where the golden ratio is not found in nature as might be expected. But you can’t deny the power of data and evidence to prove the inherent power of the
golden ratio in reality.
The critical question you need to ask yourself is: Why does the golden ratio show up so commonly in nature?
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The Golden Ratio: A Principle of Energy Flow
The golden ratio, seen in structures as vast as galaxies or as intricate as DNA, has long been the symbol of ideal harmony. Duke University's Adrian Bejan ties this unique ratio to a
universal law of nature's design. Through his work on the constructal law, Bejan reveals how nature shapes itself to ease flow. The essence of the golden ratio, he suggests, lies in
achieving maximal efficiency with minimal energy (Bejan, 2009).
This ratio allows structures to scale infinitely without changing their core shape, resulting in its recurrence throughout nature. By choosing the path of least resistance, the golden
ratio epitomizes energy-efficient flow. This is evident in falcons, which harness the golden spiral for an energy-saving approach to their prey (Tucker, 2000), and in plants, where
Fibonacci spirals display energy optimization in phyllotaxis (Li et al., 2007).
The Purpose of Life: Flow
The Constructal Law intertwines modern science with ancient spiritual writings, hinting at a unified purpose for all existence: to facilitate energy flow. Energy naturally seeks
equilibrium, flowing from concentrations, like the sun, to the expansive universe.
Our individual purpose is deciphering our unique energy flow. Despite being made of the same elements, each person's energy interplays differently. Our brains might share a similar
number of nerves, but the connections, molded by experiences, are distinct, crafting our personalities.
When our energies align, we enter a "flow" state, a concept explored by Csikszentmihalyi (1990) and Maslow (1964). In this state:
Causes of Flow:
- Clear goals
- Immediate performance feedback
- Confidence in handling challenges
Characteristics of Flow:
- Absolute focus
- Control and openness
- Enhanced learning and positivity
Consequences of Flow:
- Transcendence of self-awareness
- Altered time perception
- Intrinsic reward from the activity
Discovering your energy's optimal flow—whether through passions or talents—uncovers your unique life purpose. Like fingerprints, our energy patterns are singular. Aligning with this
flow often leads to peak performance, happiness, and fulfillment, underscoring our individual roles in the universe.
Bejan, A. (2009). The golden ratio predicted: Vision, cognition and locomotion as a single design in nature. International Journal of Design & Nature and Ecodynamics, 4(2), 97-104.
Csikszentmihalyi, M. (1990). Flow: The Psychology of Optimal Experience. Harper and Row, New York, NY.
Li, C., Ji, A., & Cao, Z. (2007). Stressed Fibonacci spiral patterns of definite chirality. Applied Physics Letters, 90(16), 164102.
Maslow, A. H. (1964). Religions, values, and peak-experiences. Columbus, OH: Ohio State University Press. (Original work published 1940)
Tucker, V. A. (2000). The deep fovea, sideways vision and spiral flight paths in raptors. Journal of Experimental Biology, 203(24), 3745-3754.
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The Sacred Significance of "108"
The number "108" holds profound significance across various spiritual traditions. In Hinduism and Buddhism, it represents a path of balance with 108 virtues to nurture and 108
defilements to shun. This sense of equilibrium extends to their malas (or rosaries), which feature 108 beads. In Islamic traditions, "108" is sometimes invoked in place of God's name.
Even in tai chi, a practice rooted in fluid motion and balance, there are 108 movements. The recurring presence of this number across diverse spiritual practices suggests its
importance goes beyond mere coincidence.
108 and Growth
The number 108 has intriguing ties to the cosmos: the average distance from the Sun and Moon to Earth is 108 times their respective diameters. Beyond this celestial connection, the
natural world showcases 108 in the growth of a nautilus shell. As the nautilus matures, each new chamber of its shell expands to be 1.08 times larger than the previous one, crafting a
mesmerizing logarithmic spiral on the shell's exterior.
In the BBC Two documentary "The Code," Professor Marcus du Sautoy from the University of Oxford unveils the elegant spiral shell of the nautilus, which grows at a constant rate of
1.08. He delves into how spirals and the wonders of math manifest across nature.
108 and The Number 5
The number 108 is related to the number 5. The clue is related to the angle of the Pentagon, being 108˚.
108 and The Fibonacci Sequence
The number 108 has a profound significance when looked through the lens of decimal parity, an ancient numerical system used by cultures like Egypt and India to discern the essence of
Decimal parity simplifies numbers to single digits. Take 361: 3 + 6 + 1 = 10, and further, 1 + 0 = 1. Thus, 361's decimal parity is 1.
Consider the Fibonacci Sequence's first 24 numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657.
Applying decimal parity to this sequence yields a recurring pattern of 24 digits: 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1. The sum? A familiar 108.
Intriguingly, this pattern mirrors the 1.08 growth constant the nautilus adopts for its spiral shell. The connection between nature's designs and ancient numerical systems is truly
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