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godesim Alternatives - Go Science and Data Analysis | LibHunt
Programming language: Go
License: BSD 3-clause "New" or "Revised" License
godesim alternatives and similar packages
Based on the "Science and Data Analysis" category.
Alternatively, view godesim alternatives based on common mentions on social networks and blogs.
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Problem Solving With Scientific Notation
Learning Objectives
• Solve application problems involving scientific notation
Solve application problems
Learning rules for exponents seems pointless without context, so let’s explore some examples of using scientific notation that involve real problems. First, let’s look at an example of how scientific
notation can be used to describe real measurements.
Think About It
Match each length in the table with the appropriate number of meters described in scientific notation below.
The height of a desk Diameter of water molecule Diameter of Sun at its equator
Distance from Earth to Neptune Diameter of Earth at the Equator Height of Mt. Everest (rounded)
Diameter of average human cell Diameter of a large grain of sand Distance a bullet travels in one second
Power of 10, units in meters Length from table above
Show Solution
One of the most important parts of solving a “real” problem is translating the words into appropriate mathematical terms, and recognizing when a well known formula may help. Here’s an example that
requires you to find the density of a cell, given its mass and volume. Cells aren’t visible to the naked eye, so their measurements, as described with scientific notation, involve negative exponents.
Human cells come in a wide variety of shapes and sizes. The mass of an average human cell is about [latex]2\times10^{-11}[/latex] grams^[1]Red blood cells are one of the smallest types of cells^[2],
clocking in at a volume of approximately [latex]10^{-6}\text{ meters }^3[/latex].^[3] Biologists have recently discovered how to use the density of some types of cells to indicate the presence of
disorders such as sickle cell anemia or leukemia. ^[4] Density is calculated as the ratio of [latex]\frac{\text{ mass }}{\text{ volume }}[/latex]. Calculate the density of an average human cell.
Show Solution
The following video provides an example of how to find the number of operations a computer can perform in a very short amount of time.
In the next example, you will use another well known formula, [latex]d=r\cdot{t}[/latex], to find how long it takes light to travel from the sun to the earth. Unlike the previous example, the
distance between the earth and the sun is massive, so the numbers you will work with have positive exponents.
The speed of light is [latex]3\times10^{8}\frac{\text{ meters }}{\text{ second }}[/latex]. If the sun is [latex]1.5\times10^{11}[/latex] meters from earth, how many seconds does it take for sunlight
to reach the earth? Write your answer in scientific notation.
Show Solution
In the following video we calculate how many miles the participants of the New York marathon ran combined, and compare that to the circumference of the earth.
Scientific notation was developed to assist mathematicians, scientists, and others when expressing and working with very large and very small numbers. Scientific notation follows a very specific
format in which a number is expressed as the product of a number greater than or equal to one and less than ten, and a power of 10. The format is written [latex]a\times10^{n}[/latex], where [latex]1\
leq{a}<10[/latex] and n is an integer. To multiply or divide numbers in scientific notation, you can use the commutative and associative properties to group the exponential terms together and apply
the rules of exponents. | {"url":"https://courses.lumenlearning.com/aacc-collegealgebrafoundations/chapter/read-problem-solving-with-scientific-notation/","timestamp":"2024-11-11T00:49:24Z","content_type":"text/html","content_length":"60498","record_id":"<urn:uuid:fb350187-911c-4a39-b2e7-13b244e9aaa0>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00378.warc.gz"} |
Code Golf - Posts tagged random
Posts tagged random
Challenge This is a simple randomness challenge: Given a non-negative integer $n$, and positive integer $m$, simulate rolling and summing the results of $n$ fair dice, each of which have $m$ sides...
Idea shamelessly stolen from caird and rak1507 Shuffle a subset of a list of unique, positive integers with uniform randomness, given the indices of that subset. For example, given the list $[A, B... | {"url":"https://codegolf.codidact.com/categories/50/tags/5333","timestamp":"2024-11-03T01:29:17Z","content_type":"text/html","content_length":"34282","record_id":"<urn:uuid:733dde52-48d3-42d0-b511-be78711d04a2>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00037.warc.gz"} |
Aperiodcast – MathsJam 2013!
You're reading: The Aperiodcast
Aperiodcast – MathsJam 2013!
We haven’t done one of these for absolutely ages. Since all three of us were at the big MathsJam conference a couple of weekends ago, we decided to introduce a local minimum into the fun curve by
sitting down and talking about how this site’s doing.
Actually, we ended up talking about the MathsJam baking competition for absolutely ages. When we got round to talking about the site, we mentioned:
Podcast: Play in new window | Download
Subscribe: Apple Podcasts | RSS | {"url":"https://aperiodical.com/2013/11/aperiodcast-mathsjam-2013/","timestamp":"2024-11-05T22:18:10Z","content_type":"text/html","content_length":"38782","record_id":"<urn:uuid:54ca1bb7-575e-4405-ad98-9d2a2c4fead3>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00650.warc.gz"} |
Which of the measurements has 3 significant figures?
Which of the measurements has 3 significant figures?
The quantities 0.456, 0.0456 and 0.00456 all contain 3 significant figures. In this case, you need to think in terms of exponential numbers. 0.0456 is 4.56 x 10-2 (only 3 significant figures) and
0.00456 is 4.56 x 10-3 (again, only three significant numbers).
How do you round significant figures on a calculator?
When rounding significant figures the standard rules of rounding numbers apply, except that non-significant digits to the left of the decimal are replaced with zeros. This calculator rounds down if
the next digit is less than 5 and rounds up when the next digit is greater than or equal to 5.
How many sig figs are in measurements?
The significant figures in a measurement consist of all the certain digits in that measurement plus one uncertain or estimated digit….Significant Figures.
Rule Examples
1. All nonzero digits in a measurement are significant A. 237 has three significant figures. B. 1.897 has four significant figures.
How many sig figs should be in a measurement?
Significant Figures Rules The rules below can be used to determine the number of significant figures reported in a measured number. Rule 1: All nonzero digits in a measurement are significant. 237
has three significant figures. 1.897 has four significant figures.
How many significant figures does 3.50 have?
How Many Significant Figures?
Number Scientific Notation Significant Figures
3500 3.5×103 2
300.00 3.0002 5
3.400 3.4×100 4
310 3.1×102 2
How many significant figures are there in 7500?
7500 has 2 e significant figures.
How do you calculate to 3 decimal places?
1. Round this number to 3 decimal places.
2. Count along the first 3 numbers to the right of the decimal point.
3. Count along the first 3 numbers to the right of the decimal point.
4. Count along the first 3 numbers to the right of the decimal point.
5. Look at the next number (the 4th number after the decimal place)
How do significant figures apply to measurements?
Significant digits (also called significant figures or “sig figs” for short) indicate the precision of a measurement. A number with more significant digits is more precise. For example, 8.00 cm is
more precise than 8.0 cm.
What is the rule for taking measurements?
Measurement Uncertainty
Rule Examples
1. All nonzero digits in a measurement are significant. 237 has three significant figures. 1.897 has four significant figures.
2. Zeros that appear between other nonzero digits (middle zeros) are always significant. 39,004 has five significant figures. 5.02 has three significant figures.
What is RND in Casio calculator?
Notes: It is advised to use the assignment symbol := in formulae with random functions; using the definition symbol = will result in the assignment of new random values during every calculation. The
maximum value for RND(Value) is around 2 billion, i.e. 2147483646 or 2.14E09.
How do you round to 3 significant figures in Excel?
For example, to round 2345678 down to 3 significant digits, you use the ROUNDDOWN function with the parameter -4, as follows: = ROUNDDOWN(2345678,-4). This rounds the number down to 2340000, with the
“234” portion as the significant digits.
What is 8 rounded to 3 significant figures?
Rounding to 3 significant figures is probably the most common way of rounding off. Rounding the number off to 3 significant fugures means you require 3 non-z…
What is the easiest way to identify significant figures?
Significant Figure Rules. Non-zero digits are always significant.
Uncertainty in Calculations. Measured quantities are often used in calculations.
Losing Significant Figures. Sometimes significant figures are ‘lost’ while performing calculations.
Rounding and Truncating Numbers.
Exact Numbers.
Accuracy and Precision.
How do you round to three significant digits?
look at the first non-zero digit if rounding to one significant figure.
look at the digit after the first non-zero digit if rounding to two significant figures.
draw a vertical line after the place value digit that is required.
look at the next digit.
What is the correct number of significant figures?
The zero to the left of a decimal value less than 1 is not significant.
All trailing zeros that are placeholders are not significant.
Zeros between non-zero numbers are significant.
All non-zero numbers are significant.
If a number has more numbers than the desired number of significant digits,the number is rounded. | {"url":"https://www.tag-challenge.com/2022/10/04/which-of-the-measurements-has-3-significant-figures/","timestamp":"2024-11-13T10:47:25Z","content_type":"text/html","content_length":"41385","record_id":"<urn:uuid:c495bf91-b854-42b5-b50a-93be02a42d9c>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00548.warc.gz"} |
Use VBA Union to Combine Ranges - wellsr.com
The VBA Union method in Excel is designed to combine ranges into a single range. You can use VBA Union to combine multiple ranges based on a common criteria, like all positive numbers, or even use it
to select a subset of a larger range.
This tutorial will introduce you to the VBA Union method and provide several examples to show you its limitations and teach you how to properly use it.
Basic VBA Union Macro
We’re going to start this tutorial with a basic Union example. Stick with us though because soon we’ll be discussing some very important limitations of the VBA Union method.
Sub BasicUnionDemo()
Dim rng1 As Range
Set rng1 = Union(Range("A1:C4"), Range("E1:F4"))
End Sub
Make powerful macros with our free VBA Developer Kit
Tutorials like this can be complicated. That’s why we created our free VBA Developer Kit and our Big Book of Excel VBA Macros to supplement this tutorial. Grab them below and you’ll be writing
powerful macros in no time.
Using VBA Union Method
When you run this macro, the ranges A1:C4 and E1:F4 are combined into one range, which we stored in variable rng1. Notice how we use the Set keyword to set our unified range to the variable rng1. You
can’t assign the combined range to the variable without the Set keyword.
After the union method is applied to the ranges, the macro selects the newly combined range, so you’re left with this:
It’s worthwhile to mention that the VBA Union method isn’t actually a global VBA method. It’s really a member of the Excel Type Library, so it technically should be entered like Application.Union
(...). Since we’re typically working directly in Excel when applying the Union method, we’re going to drop the Application and simply use the shorthand Union(...) notation here.
Working with the Combined Range
Selecting the combined range is just one of many things you can do with your newly created range object. You can iterate through each item in your combined range with a simple For Loop, like this:
Sub BasicUnionDemo2()
Dim rng1 As Range
Dim item As Range
Set rng1 = Union(Range("A1:C4"), Range("E1:F4"))
For Each item In rng1
Debug.Print item.Address
Next item
End Sub
When you run this macro, the address of each item in the range rng1 is printed to your immediate window, which you can open by pressing Ctrl+G from your VBA Editor.
Select Subset of a Range with VBA Union
One creative use for the VBA Union method is to use it to select a subset of cells in a range based on common criteria. For example, let’s say we wanted to store all the positive numbers in a column
to a single variable. How would you do that?
One way to do it is to iterate through each item in the column and apply the union method to each new positive number you encounter. There are simpler ways to do this, but we’re here to demonstrate
the VBA Union method.
To start, assume we have this dataset in our spreadsheet.
We’re going to loop through each row in this column and store each positive number in a shared range. To make things interesting, we’re actually going to use Union to store all zeroes in a range, all
positive numbers in a range, and all negative numbers in a range. That way, you can see the true power of organizing your data into separate ranges using the Union method. Doing it this way will also
highlight some of the limitations of the Union method.
Take a look at this macro:
Store numbers in different ranges based on value
Sub VBAUnionDemo()
Dim rngPOSITIVE As Range
Dim rngNEGATIVE As Range
Dim rngZERO As Range
Dim LastRow As Long
Dim i As Long
LastRow = Range("A" & Rows.Count).End(xlUp).Row
'categorize our ranges
For i = 1 To LastRow
If IsNumeric(Range("A" & i)) Then
If Range("A" & i) > 0 Then
If rngPOSITIVE Is Nothing Then
Set rngPOSITIVE = Range("A" & i)
Set rngPOSITIVE = Union(Range("A" & i), rngPOSITIVE)
End If
ElseIf Range("A" & i) < 0 Then
If rngNEGATIVE Is Nothing Then
Set rngNEGATIVE = Range("A" & i)
Set rngNEGATIVE = Union(Range("A" & i), rngNEGATIVE)
End If
Else 'equals zero
If rngZERO Is Nothing Then
Set rngZERO = Range("A" & i)
Set rngZERO = Union(Range("A" & i), rngZERO)
End If
End If
End If
Next i
'post-process our ranges
rngNEGATIVE.Font.Color = vbRed
rngZERO.Font.Italic = True
End Sub
In this example, we use the VBA IsNumeric function to check if a cell is a number. If it is, we then categorize it based on value (greater than 0, less than 0, equal to 0).
Again, there are definitely quicker ways to produce results like this, but we’re here to demonstrate how you can use the Union method in your own macros. Once you run this macro, your final column
will look like this:
Negative values will be red, cells equal to zero will be italicized, and all positive values will be selected.
VBA Union Limitations
Undefined (Nothing) parameters
The macro above highlights one of the primary limitations of the VBA Union method. Notice how we have an IF statement like this after testing the value of each cell:
If rngPOSITIVE Is Nothing Then
We have to perform this check because the Union method can’t combine ranges if one of the ranges doesn’t exist. In other words, until we define rngPOSITIVE the first time, we can’t include it in our
Union statement.
If we try to include a range equal to Nothing in a Union expression, we’ll get an “invalid procedure call or argument” error:
The first time you encounter a cell fitting your criteria, you have to add it to your range the traditional way, like this:
Set rngPOSITIVE = Range("A" & i)
After the range is defined the first time, you can add to the existing range with the Union command.
VBA Union on overlapping ranges
The second limitation deals with duplicates in a range. It’s important to point out that the VBA Union method is not the same as the mathematical Union operation. If the ranges you want to combine
intersect, VBA Union will list the intersecting ranges twice. Take the following macro, for example.
Sub UnionDemoIntersection()
Dim rng1 As Range
Dim item As Range
Set rng1 = Union(Range("A1:B3"), Range("B2:C4"))
For Each item In rng1
Debug.Print item.Address
Next item
End Sub
In this example, the two ranges overlap, which is obvious when you select the combined range:
Now take a look at you immediate window. You’ll notice that B2 and B3 are printed twice.
Because the intersecting ranges are included in your range twice, you’ll need to be careful when using the combined range in your macro.
Closing Thoughts
I use the VBA Union method often when I want to combine all the cells meeting a complex criteria into a common range. How do you plan on using the VBA Union method?
I hope you’ll take a minute to subscribe for more VBA tips. Simply fill out the form below and we’ll share our best time-saving VBA tips. | {"url":"https://wellsr.com/vba/2019/excel/use-vba-union-to-combine-ranges/","timestamp":"2024-11-12T22:14:31Z","content_type":"text/html","content_length":"38507","record_id":"<urn:uuid:5a8b2611-bc83-47b0-b733-6a2d1a9be81a>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00117.warc.gz"} |
Decode XORed Array | CodingDrills
Decode XORed Array
There is a hidden integer array arr that consists of n non-negative integers.
It was encoded into another integer array encoded of length n - 1, such that encoded[i] = arr[i] XOR arr[i + 1]. For example, if arr = [1,0,2,1], then encoded = [1,2,3].
You are given the encoded array. You are also given an integer first, that is the first element of arr, i.e., arr[0].
Return the original array arr. It can be proved that the answer exists and is unique.
Ada AI
I want to discuss a solution
What's wrong with my code?
How to use 'for loop' in javascript?
javascript (node 13.12.0) | {"url":"https://www.codingdrills.com/practice/decode-xored-array","timestamp":"2024-11-08T04:22:17Z","content_type":"text/html","content_length":"13503","record_id":"<urn:uuid:aaf1b08e-527c-4772-9fd5-19b9b6785046>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00652.warc.gz"} |
On the dynamics of a beam rotating at nonconstant speed
The response and its stability of a beam rotating at nonconstant angular speed are studied. The rotating speed is assumed to be the combination of a constant angular speed and a small periodic
perturbation. The axial and flexural deformations due to rotation are considered simultaneously. Thus, the rotating team at nonconstant speed yields a set of parametric excited partial differential
equations of motion. Extended Galerkin's method is employed for obtaining the discrete equations of motion. Then, the solution and the its stability are found by using the method of multiple scale.
名字 Proceedings of the ASME Design Engineering Technical Conference
卷 Part F168436-2
???event.eventtypes.event.conference??? ASME 1991 Design Technical Conferences, DETC 1991
國家/地區 United States
城市 Miami
期間 22/09/91 → 25/09/91
深入研究「On the dynamics of a beam rotating at nonconstant speed」主題。共同形成了獨特的指紋。 | {"url":"https://scholars.ncu.edu.tw/zh/publications/on-the-dynamics-of-a-beam-rotating-at-nonconstant-speed","timestamp":"2024-11-07T23:58:27Z","content_type":"text/html","content_length":"55120","record_id":"<urn:uuid:ff2907f4-3a16-4e74-a75f-603c81e033be>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00755.warc.gz"} |
DUALITY OF TIME - 2.11 Galileo Galilei
2.11 Galileo Galilei
Some years after its publication, the heliocentric model caused a strong controversy, especially after Tycho Brahe (1546-1601 AD) published his similar variation, followed by the advent of the
telescope. When the heliocentric model started to become popular, the church considered it formally heretical and the Pope banned all books and letters advocating it.
It was Galileo Galilei (1564-1642 AD) who took the challenge to defend this controversial model, but he was met with strong opposition from astronomers and theologians which later led to his
Galileo was an Italian polymath interested in astronomy, physics, philosophy, and mathematics. He studied gravity and free fall, velocity and inertia, projectile motion and and pendulums, and the
principle of relativity, in addition to many other related applications. He contributed in transforming Europe from natural philosophy to modern science.
One of Galileo’s greatest contributions was to recognize that the role of science was not to explain “why” things happened as they do in nature, but only to describe them, which greatly simplified
the work of scientists, and liberated them from the influence of theologians. Subsequently, this led Galileo himself to describe natural phenomena using mathematical equations, supported with
experimentation to verify their validity. This marked a major deviation from the qualitative science of Aristotelian philosophy and Christian theology.
Based on these ideas Galileo was able to develop the mechanics of falling bodies from the earlier ideas of the theory of impetus that tried to explain projectile motion against gravity. By dropping
balls of the same material, but with different masses, from the Leaning Tower of Pisa, he showed that all compact bodies fell at the same rate. Galileo then proposed that a falling body would fall
with a uniform acceleration, as long as the resistance of the medium through which it was falling remained negligible, which allowed him to derive the correct kinematic law that the distance traveled
during a uniform acceleration is proportional to the square of the elapsed time:
However, as it was the case with Copernicus, Galileo’s discoveries had been also clearly stated by many Muslim scholars more than five centuries before, and they even quoted and developed older
theories in this regard. For example, we find Hibatullah ibn Malaka al-Baghdadi (1080–1164), an Islamic philosopher and physician of Jewish descent from Baghdad, originally known by his Hebrew birth
name Baruch ben Malka and was given the name of Nathanel by his pupil Isaac ben Ezra before his conversion from Judaism to Islam towards the end of his life. In one of his anti-Aristotelian
philosophical works Kitab al-Mutabar (The Book of What Has Been Established by Personal Reflection), he proposed an explanation of the acceleration of falling bodies by the accumulation of successive
increments of power with successive increments of velocity Crombie (1959). In this and other books and treatises, he described the same laws of motion that were later presented by Newton, except that
they were not formulated in mathematical equations.
Nonetheless, by the 17th century, the Copernican and Galilean heliocentric models started to replace the classical ancient worldview, at least by knowledgeable researchers. Between the years
1609-1619, the scientist Johannes Kepler (1571-1630 AD) formulated his three mathematical statements that accurately described the revolution of the planets around the Sun. In 1687, in his major book
Philosophiae Naturalis Principia Mathematica, Isaac Newton provided his famous theory of gravity, which supported the Copernican model and explained how bodies more generally move in space and time,
as we shall discuss in section 16. | {"url":"https://www.smonad.com/time/book.php?id=68","timestamp":"2024-11-10T15:29:00Z","content_type":"text/html","content_length":"34088","record_id":"<urn:uuid:8d5a7080-3bb5-4c03-bbd5-918e1e56eaab>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00141.warc.gz"} |
Registration of Complex Data
We have developed a number of image registration methods with a particular focus on image time-series. We have developed (1) a method for the longitudinal atlas-building from diffusion tensor images
[1], (2) a generalization of least-squares line fitting to the space of images (termed geodesic regression) which allows for compact representation of image-time series by their initial momenta and
hence for simplified subsequent statistical analysis [2,3,4], and (3) have developed a registration method which can account for image appearance changes (as for example caused by a traumatic brain
injury or a brain tumor) by jointly estimating a global space deformation and an overlaid geometric model change affecting image appearance [5].
A driving problem for the development of our methods has been brain imaging. However, our recent work has been motivated by the need of the pediatric airway group register airways. Airways show
spatial variations on a variety of levels: (1) within-subject variation during the normal breathing cycle; (2) change within a subject over time; (3) change between subjects over time; (4)
across-subject variation during the normal breathing cycle; (5) presence of pathology (e.g., stenosis) and articulated structures (such as the vocal fold and the epiglottis); and (6) variation in
extracted information (for example for segmentation). Hence, the pediatric airway project is an ideal candidate to advance the state of the art in registration methods for temporal data for improved
analysis for a clinically relevant problem.
Geodesic Regression for Image Time Series
Registration of image-time series has so far been accomplished (1) by concatenating registrations between image pairs, (2) by solving a joint estimation problem resulting in piecewise geodesic paths
between image pairs, (3) by kernel based local averaging or (4) by augmenting the joint estimation with additional temporal irregularity penalties. We proposed a generative model extending least
squares linear regression to the space of images by using a second-order dynamic formulation for image registration [2]. Unlike previous approaches, the formulation allows for a compact
representation of an approximation to the full spatio-temporal trajectory through its initial values. The method also opens up possibilities to design image-based approximation algorithms. The
resulting optimization problem is solved using an adjoint method. Key to the formulation is to be able to write the image registration problem in initial value form. In the scalar-valued case (for
linear regression) this amount to recasting the least square estimation for the line model y=mx+c into the second order dynamical system form d^2/dt^2 y = 0, y(0)=c, dy/dt(0)=c, where the initial
conditions are simply the y-intercept and the slope. For the image-case we have an initial image and its initial momentum. In the optimization all images along the geodesic exert forces, which
influence the initial condition for the geodesic. In the scalar-valued case this amounts to having a least squares solution, which can be interpreted as a physical system under force and momentum
balance. We have also developed an approximate variant of this algorithm, which greatly simplifies the computation [3]. Furthermore, we have proposed a related approach which jointly captures spatial
deformations and changes in image intensity within a regression framework [4]. The most exciting part about the overall solution approach is that it can be extended to general manifold data, which we
will exploit in our future work on pediatric airways.
Geometric Metamorphosis
Geometric Metamorphosis. An image is explained by a global deformation (via v) and a geometric model deformation (via v^\\tau ). Corresponding structures in the source and target guide the estimation
of v and v^\\tau addresses additional appearance differences at the pathology. To avoid faulty evaluation of image similarities, a suitable image composition method is required which discards
non-matchable image information.
Standard image registration methods do not account for changes in image appearance. Hence, metamorphosis approaches have been developed which jointly estimate a space deformation and a change in
image appearance to construct a spatio-temporal trajectory smoothly transforming a source to a target image. For standard metamorphosis, geometric changes are not explicitly modeled. We proposed a
geometric metamorphosis formulation [5], which explains changes in image appearance by a global deformation, a deformation of a geometric model, and an image composition model. This work is motivated
by the clinical challenge of predicting the long-term effects of traumatic brain injuries based on time-series images. This work is also applicable to the quantification of tumor progression (e.g.,
estimating its infiltrating and displacing components) and predicting chronic blood perfusion changes after stroke.
[1] G. Hart, Y. Shi, M. Sanchez, M. Styner, and M. Niethammer, “DTI Longitudinal Atlas Construction as an Average of Growth Models,” MICCAI International Workshop on Spatio-Temporal Image Analysis
for Longitudinal and Time-Series Data, 2010. [2] M. Niethammer, Y. Huang, and F.-X. Vialard, “Geodesic regression for image time-series,” in MICCAI 2011. [3] Y. Hong, M. Sanchez, M. Styner, and M.
Niethammer, “Simple Geodesic Regression for Image Time-Series,” Workshop on Biomedical Image Registration (WBIR), 2012. [4] Y. Hong, S. Joshi, M. Sanchez, M. Styner, and M. Niethammer, “Metamorphic
Geodesic Regression,” Proceedings of the International Conference on Medical Image Computing and Computer Assisted Intervention (MICCAI), 2012. [5] M. Niethammer, G. Hart, D. Pace, P. M. Vespa, A.
Irimia, J. D. Van Horn, S. Aylward, “Geometric Metamorphosis,” in Medical Image Computing and Computer-Assisted Intervention (MICCA), 2011, pp. 639-646. | {"url":"https://cismm.web.unc.edu/core-projects/biomedical-image-analysis/registration-of-complex-data/","timestamp":"2024-11-10T02:24:19Z","content_type":"text/html","content_length":"65268","record_id":"<urn:uuid:cfc1dc8e-6a87-4dfa-a0b6-1828953fadaa>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00046.warc.gz"} |
AS183/KH0201 - Ascher Sum Fieldmarks
Fieldmark Fieldmark XRay
Pendant Sum
Group Group
Ascher Databook Notes:
1. This is a small piece of cord wound around the pendant for 0.3 cm. between the 2 long knots.
Notes: 2. AS182-AS186 are associated in that they are tied together. For a comparison of them, see AS182.
3. By spacing, the khipu contains 2 pairs of pendants. Each pendant in the second pair is W with a DB subsidiary.
4. The sum of the pendants in the first pair equals the sum of the pendants in the second pair. Each pair sums to 100.
5. Multiplication by 2 is suggested by the first 4 values. P1 is a W cord with value 85+5; P2 is DB with value 6+4; P3 is W with value 45; and P3s1 is DB with value 3+2. Note that 2
(3+2) = 6+4. If you double 45 and keep the tens position and units position separated, you get 2 (4 tens & 5 ones) = 8 tens + 2(5 ones) = 85+5. | {"url":"https://khipufieldguide.com/notebook/fieldmarks/catalog/AS183.html","timestamp":"2024-11-11T18:04:29Z","content_type":"application/xhtml+xml","content_length":"30683","record_id":"<urn:uuid:55b1ce61-6dce-40c3-9277-f8d4c7d3a03e>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00281.warc.gz"} |
VITEEE Important Topics for Exam Preparation for 2024
VITEEE is a university-level examination, and thousands of students appear yearly. This is a challenging exam. Therefore, the candidates must have a very high level of preparation to beat the
competition and crack the exam to get admission to their dream college. In this article, we will give you complete information about VITEEE Important Topics 2024.
VITEEE Important Topics for Preparation
As mentioned, understanding the important chapters for VITEEE 2024 can help aspirants focus their preparation more effectively. Here’s a breakdown of the important chapters for each subject:
1. Optics: This includes laws of reflection and refraction, lenses, optical instruments, and wave optics.
2. Electrostatics: Topics like Coulomb’s law, electric field and potential due to point charges, Gauss’s law, capacitance, and electric potential energy fall under this category.
1. Atomic Structure: Focus on concepts related to the structure of atoms, Bohr’s model, quantum numbers, electronic configuration, and periodic properties.
2. Thermodynamics: Understanding the laws of thermodynamics, heat, work, internal energy, enthalpy, entropy, Gibbs free energy, and various thermodynamic processes is crucial.
1. 3D Geometry: Topics such as direction cosines and ratios, distance between two points, equation of lines and planes in space, and intersection of lines and planes are important.
2. Vectors: Understanding vector algebra, scalar and vector products, and their applications in geometry and physics is essential.
3. Integration: Focus on definite and indefinite integrals, properties of integrals, integration by substitution, by parts, and by partial fractions.
Focusing on these important chapters can help aspirants prioritize their study schedule and allocate more time to topics that carry more weightage in the VITEEE exam. However, it’s also important to
have a balanced preparation and cover all topics to ensure comprehensive understanding and better performance on the exam.
Important topics of Chemistry
Topics No. of Questions Topics Approx questions
P-Block Elements 5 The Solid State 2-3
Basic Concepts 5-6 Alcohols, Phenols, and Ethers 3-4(only approx phenol weightage of questions 2-3)
Equilibrium 4 Chemical Kinetics 2-3
Aldehydes, Ketones and Carboxylic Acids 4 Hydrogen and the s-Block Element 2-3
Chemical Bonding 3-4 Polymers 2-3
Coordination Compounds 2-3 States of Matter 2-3
Thermodynamics and Thermochemistry 3-4 The d and f Block Elements 3-4
Alkanes, Alkenes and Alkynes 3-4 Solutions 2-3
Haloalkanes and Haloarenes 2-3
Important topics of Physics
Topics Approx. questions Topics Approx. questions
Mechanical Properties 5 Atomic Study 3-4
Current Electricity 4 Oscillation 3-4
Dual Nature of Matter & Radiation 3-4 Gravitation 3-4
Magnetism & Moving Charges 3-4 Kinematics 2-3
Thermodynamics 3-4 Alternating Current 2-3
Rigid Body Dynamics 3-4 Waves 2-3
Work, Energy, and Power 3-4 Electrical Fields 2-3
Planar Motion 3-4 Capacitance & Electrical Potential 2-3
Ray Optics 3-4 Semiconductors 2-3
Wave Optics 3-4 Laws of motion 2-3
Wave Optics 3-4
Important topics of Mathematics
1. Applications of Matrices and Determinants
2. Complex Numbers
3. Coordinate Geometry
4. Trigonometry(Trigonometric equations,Inverse trigonometric functions)
5. Vector Algebra
6. Analytical Geometry of Three Dimensions
7. Differential Calculus (Derivatives of different Functions, Tangents and Normals, Maxima and Minima, Rolle’s Theorem, Mean Value Theorem, and Intermediate Value Theorem)
8. Integral Calculus and its Applications (Definite and Indefinite Integrals)
9. Differential Equations(Solution of Differential Equations, Linear first-order differential equations)
10. Probability Distributions(Conditional Probability, Baye’s Theorem, Independent Events, Mean and Variance distribution)
11. Statistics(Measures of dispersion, Mean and Variance)
Important topics of Biology with weightage
1. Taxonomy
2. Cell and Molecular Biology-5
3. Reproduction-5
4. Genetics and Evolution-(3-4)
5. Human health and diseases
6. Plant physiology
7. Biochemistry
8. Human physiology
9. Biotechnology and its applications
10. Biodiversity, ecology, and environment- (2-3)
Important topics of English
It includes topics such as Grammar, Comprehension, and Vocabulary that would generally be prepared by the candidate during their school preparations.
1. Comprehensive passage
2. Grammar(Question tags, Phrasal verbs, Determiners, Prepositions, Modals, Adjectives, Agreement, Time and Tense, Parallel construction, Relative pronouns, Voice, Transformation)
3. spotting errors
4. Vocabulary(Synonyms, Antonyms, Odd Word, One Word, Jumbled letters, Homophones, Spelling)
5. Idioms/Phrases
6. Compositions(Rearrangement, Paragraph Unity, Linkers /Connectives)
Logical Reasoning
1. Analogy
2. Classification
3. Series Completion
4. Logical Deduction – Reading Passage
5. Chart Logic
1. Pattern Perception
2. Figure Formation and Analysis
3. Paper Cutting
4. Figure Matrix
5. Rule Detection
Candidates are advised to analyze the important topics before commencing the preparation for the exam.
As you know, every competitive examination needs a diligent analysis of its syllabus. In the VITEEE examination too, we advise you to focus on the syllabus and prepare a strategy accordingly.
Additionally, it is important to follow a credible and reliable set of study material for your preparation. Do not build a stack of books and create unnecessary pressure. Instead, keep relevant books
and revise them repeatedly.
Lastly, appear for as many mock tests as possible. It helps you to understand your weak and strong areas.
Maintain consistency and follow a healthy routine.
Good luck!
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1. VITEE Preparation Tips | {"url":"https://learndunia.com/viteee-important-topics/","timestamp":"2024-11-05T09:13:07Z","content_type":"text/html","content_length":"127682","record_id":"<urn:uuid:72eefed2-58bd-498d-b423-f97bbc75706b>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00506.warc.gz"} |
Tube in a Sphere
A cylindrical tube is inserted in a sphere. You can produce partial tubes and spheres in various ways. The volume of the remaining part of the sphere is , which is the volume of a sphere of radius ,
where is half the length of the tube. | {"url":"https://www.wolframcloud.com/objects/demonstrations/TubeInASphere-source.nb","timestamp":"2024-11-05T09:10:14Z","content_type":"text/html","content_length":"203988","record_id":"<urn:uuid:4c668a82-99e3-4fee-afa1-0a01ab5d5846>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00043.warc.gz"} |
Boson Sampling Tests Quantum Computing
Written by Mike James
Wednesday, 02 January 2013
Despite all the hype, we still don't know if we can build a quantum computer that is worth anything. Now we have something that might provide a shortcut to the test of a full machine - boson
We have made a lot of progress with quantum computing, but there is still the nagging doubt at the back of every researchers mind that there might be a "no-go" principle in operation. The problem
with building a quantum computer is keeping the quantum effects operating properly as the device grows larger. Roughly speaking you can build a quantum computer that works with a few bits, but
extending things to enough bits to do some thing that goes beyond what can be achieved using a classical computer is a tough practical problem.
This is fine as long as it is a tough practical problem and there is no basic theoretical result which limits what a quantum computer can do. If it is just practical then we need to push on and
refine our technology until we reach the point where a quantum computer can beat a classical computer. If there is a "no-go" theorem then we might as well give up.
To be clear about this - no quantum computer to date has computed anything that isn't well within the ability of a classical computer.
The quantum computers of today have done things like factoring two-digit numbers in a time that can be easily beaten by a classical computer.
If you can increase the number of bits that the quantum computer works with then they could in the same time factor numbers that would take a classical computer the rest of the life of the universe
to factor.
You can see that it would not be unreasonable for there to be a deep principle that says something like - "no quantum computer can ever compute something that is beyond the reach of a classical
At the moment no one knows if there is such a theorem although Scott Arronson, a well-known MIT computer scientist, has offered a prize of $100,000 for any proof that quantum computers are
impossible. He also is responsible for thinking up a test that might be easy enough to complete that would at least prove that there is no such "no-go" theorem.
Image courtesy of Alisha Toft
His experiment is quite simple. You set up a quantum state with n bosons (integer spin particles such as a photon of light) in particular configurations. You then allow the system to evolve and
interact with gadgets such as beam splitters and phase shifters and observe the final state. In an experimental setup this corresponds to having n light sources interact with standard optical gadgets
and then see where the photons exit the machine. This is fairly easy to set up and there are lots of quantum optics labs doing this sort of thing every day.
The whole point of the boson sampling setup is that as soon as you have more than a few starting photons the number of interactions grows so fast that you can't do the sums classically. In fact, the
problem is in the complexity class #P and it is #P-complete. A boson sampling machine can beat a classical computer when it reaches, say, 20 photons. Unfortunately at the moment the number of photons
actually used in the machines is around three or four. Currently the researchers are investigating how it all works and what can go wrong.
At least four groups have published papers showing that the idea works - for small numbers of bosons. They are all of the opinion that boson sampling is easy and they see no problems in scaling up
their work. The only real question is - if it really is easy why only four photons? Perhaps the coming year will reveal if there is, or is not, a "no-go" theorem but it is worth noting that there are
a few possible ways that things can go wrong.
The first is that boson sampling isn't a universal computational machine - perhaps a specialized machine is allowed to be faster than a classical computer. The "no-go" theorem could be of the form
that no quantum Turing machine can deliver results beyond a classical Turing machine. There is also the small matter of proving that a classical machine cannot do the boson sampling calculation as
fast as the "real thing".
TypeScript Improves Never-Initialized Variables Checks
Microsoft has announced TypeScript 5.7 in beta, with improvements including stronger checks for variables that have never been initialized before use, and path rewriting for relative paths.
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Flutter Forked As Flock
One of developers who worked on the Flutter team at Google has created an open-source form of the framework. Matt Carroll says Flock will be "Flutter+", will remain constantly up to date with
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Last Updated ( Wednesday, 02 January 2013 ) | {"url":"https://www.i-programmer.info/news/112-theory/5271-boson-samplling-tests-quantum-computing.html","timestamp":"2024-11-13T14:17:17Z","content_type":"text/html","content_length":"36882","record_id":"<urn:uuid:f8342ead-1f16-44dc-82e4-a6dd002845f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00318.warc.gz"} |
MathFiction: Barr’s Problem (Julian Hawthorne)
Contributed by Vijay Fafat
A cute, tall-tale about one Professor Brooks - presumably one of mathematics - his past student, Barr, and his 19-year old niece, Susan Wayne. The two youngsters are in love with each other but the
Professor had not yet consented to their union because, as he tells Barr:
(quoted from Barr’s Problem)
“You are not, at present, worth a decent girl’s acceptance. [,,,] You are a babe in swaddling clothes. You have never once kicked out a leg on your own independent account. Susan agreed to tolerate
you on the theory that you might turn out to he, hereafter, less of a prig and a poke than you appear now. I have yet to be satisfied that she has any grounds for her expectation. I even question
your love for her. It is only her pretty outside that has attracted you.”
Barr, of course, remonstrates vehemently, professing a soulful love for Susan. So the professor decides to put him to a test. He first asks Barr if he believes in the fourth dimension of space, an
idea which Barr dismisses as “a mathematical lark which nobody believes”. So the professor shows him a small coil of rope with a knot in it and claims that a woman had taken a perfectly normal,
unknotted loop of rope and given it a shake in the fourth dimension to end up with a knot, something impossible to do in three dimensions without cutting the rope. Barr does not know what to believe
but tells the professor that while his disbelief in the fourth dimension might be wavering following the professor’s claim, his love for Susan remained just as strong, not subject to any wavering. So
then the professor goes through an elaborate theory of dimensional scales, invariance of measurement ratios if everything were to change size by the same factor, how a movement in the fourth
dimension would shield one from any global change of scale in 3 dimensions and the like. Barr (and likely, the reader) has no idea where all this mathematical talk is leading but plays along.
Finally, the Professor demonstrates the reality of the fourth dimension by making his niece disappear in the higher spatial dimension, from where she emerges to be just 6 inches tall... Magic realism
at its best.
Barr loses his nerve and hems and haws about how he loves the miniature Susan but cannot really marry her and so on. Susan also feels the Professor has played an unfair trick on Barr. So then the
Professors unravels his magic. It was all an exercise in hypnotism to test their love and to fortify it against unexpected occurrences in the future. The fourth dimension, if one exists, was not to
be tested that evening. Geometry was the magician’s head-fake in this instance.
A bit hurried in the end but still worth a good smile. | {"url":"https://kasmana.people.charleston.edu/MATHFICT/mfview.php?callnumber=mf1504","timestamp":"2024-11-09T13:30:39Z","content_type":"text/html","content_length":"10823","record_id":"<urn:uuid:208d8e0d-084d-4f27-ac81-f25f476434cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00548.warc.gz"} |
Cosmological Consequences of Massive Photons
We all learned in our junior high science classes that photons are massless. This statement has resulted in a lot of confusion for laypeople. In our junior high science classes we were also taught
that photons possess energy and, thanks to Einstein’s special relativity theory, energy is equivalent to mass.
I find that no matter what audience I address, all the attendees know that E = mc^2. It may be the only physics equation they know, but they understand that it implies mass can be converted into
energy, and energy into mass.
The resolution of massless photons and the mass equivalence of photon energy is that the conversion of energy into matter requires that energy packets (photons) be accelerated to a very high
velocity. (Note the c^2 term in Einstein’s special relativity equation.)
Whenever physicists state that a photon is a massless particle, they mean that it has a “zero rest mass.” In fact, the textbook mass of any particle is its rest mass. When a particle is at rest, its
relativistic mass possesses a minimum value—namely, the “rest mass.”
Is the Photon Rest Mass Exactly Zero?
Physicists believe the photon rest mass is exactly zero. However, they do not know that for certain. It is impossible to make any observation or do any experiment that would prove that the photon
rest mass equals exactly zero.
The best that physicists and astronomers can do is to place an upper limit on a possible positive photon rest mass. Experiments and observations done so far establish that the photon rest mass, if it
is nonzero, must be very tiny indeed.
What If the Photon Had a Nonzero Rest Mass?
Even if the photon rest mass is very tiny instead of exactly zero, serious consequences for both particle physics and cosmology could ensue. For starters, the theory of quantum electrodynamics would
be in big trouble. Nobel laureate Richard Feynman called quantum electrodynamics the “jewel of physics.”^1 Quantum electrodynamics provides a complete integration, or unification, of classical
electromagnetism with quantum mechanics and special relativity. It describes how light and matter interact in both the classical and quantum realms. If photons possess a nonzero rest mass, charge
conservation would no longer be guaranteed, and the gauge invariance that is crucial for quantum electrodynamics would be lost. At least one Nobel Prize in physics would need to be rescinded.
If photons possess a nonzero rest mass, not all photons would travel at the same velocity. The velocity of light would be a function of frequency. Astronomers would face difficulties in integrating
their measurements at radio wavelengths with those at optical and X-ray wavelengths. Also, these different velocities would disturb cosmic distance scales and yield a different picture of the
geometry of the universe.
Another consequence of a nonzero photon rest mass is that the electrostatic force would be weaker over large distances compared to small distances. Such variations would imply that the magnetic
fields of galaxies and galaxy clusters are weaker than what astronomers think. Astronomers’ galactic dynamics models would need to be revised. Such revisions would also imply adjustments in the
values of cosmic density parameters, which form the foundational basis of all cosmic creation models.
A zero rest mass for the photon implies that a photon can be polarized in only two directions—the two that are orthogonal to the photon’s direction of motion. A nonzero photon rest mass means that
there would be a third polarization direction—one along the photon’s direction of motion. Since our models of the cosmic hyperinflation event that occurred when the universe was younger than 10^-33
seconds critically depend upon determining the polarization levels of the cosmic microwave background radiation (the radiation left over from the cosmic creation event—see figure 1), a nonzero photon
rest mass would give a much different picture of the early history of the universe, with serious consequences for the universe’s present properties.
Figure 1: Planck Satellite Map of the Cosmic Microwave Background (CMB) Radiation. Polarization measures of the CMB radiation reveal what kind of early inflation event the universe experienced.
A nonzero rest mass for the photon affects the cosmic microwave background radiation in another way. It would affect the spectral behavior of the cosmic microwave background dipole anisotropy (see
figure 2). The distortion would increase with wavelength and would lead to different conclusions about the Great Attractor and the Monster Attractor (both are dense concentrations of galaxy
clusters), which astronomers have deduced are pulling our Milky Way Galaxy in a direction that explains the cosmic microwave background dipole anisotropy.
Figure 2: Map of the Cosmic Microwave Background Dipole Anisotropy
Upper Limits to the Photon Mass
The consequences of a nonzero rest mass for the photon, especially for cosmology and particle physics, are so devastating that most physicists and astronomers are persuaded that photons really are
massless. However, these consequences have not stopped theoreticians from proposing alternatives to the standard cosmic creation models and the standard particle creation models based on nonzero
photon rest masses. Thus, a major effort in both physics and astronomy is to develop observations and experiments to place evermore stringent upper bounds on the photon mass. I will discuss these
efforts in my next blog post.
Featured image: World’s largest photon collecting machine—Five-hundred-meter Aperture Spherical Telescope (FAST). Image credit: www.news.cn/Xinhua | {"url":"https://reasons.org/explore/blogs/todays-new-reason-to-believe/cosmological-consequences-of-massive-photons","timestamp":"2024-11-07T01:08:09Z","content_type":"text/html","content_length":"129545","record_id":"<urn:uuid:265bbaff-d37a-47a1-8fef-7ac80c3ec839>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00848.warc.gz"} |
Dijkstra's Rallying Cry for Generalization
Submitted by egdaylight on
Dijkstra tackled the problem of allowing multiple users share the university's X8 computer and its peripheral hardware devices (EWD 51). To solve this problem, Dijkstra first reformulated it in terms
of as many symmetries as he could find, thereby obtaining a more general problem description. Afterwards, he introduced case distinctions in a controlled manner, gradually refining the general
problem into an X8-specific problem (by resorting to flip-flops, compact storage techniques, and what we today call caching).
Starting with Dijkstra's generalizations, here are four ways in which he transformed the initial problem into a more symmetric one:
1. Dijkstra removed the distinction between a user's program and a peripheral hardware device by viewing both as a process (— each of which shares the memory of the X8).
2. Dijkstra removed the distinction between a non-sequential and a sequential process. A non-sequential process can be redefined in terms of two or more sequential processes (that communicate with
each other via the shared memory). Therefore, Dijkstra only considered sequential processes.
3. Dijkstra removed the distinction between terminating and cyclic processes. A terminating process can be redefined as a cyclic process. Therefore, Dijkstra only considered cyclic processes.
4. The number of processes can fluctuate: processes can suddenly appear and disappear. A fixed number is, after all, a special case of a changing number; whence Dijkstra's preference for the latter.
Dijkstra subsequently used the word “machine” instead of “sequential cyclic process” in the rest of his exposition. The general problem thus amounted to: allowing a changing number of machines to
communicate with each other in an orderly fashion by means of a shared memory.
Afterwards, Dijkstra introduced the notions “semaphore”, “V operation”, “P operation”, “atomic action”, etc. — concepts that I take for granted here — and illustrated how they can be used to
synchronize two or more “communicating machines”. (Today we would rather use the words: “communicating processes”.)
After symmetry and machine independence comes asymmetry. In Dijkstra's words:
Once we take the limited hardware facilities of the university's computer system into account we need to distinguish between `concrete machines' and `abstract machines'. [My paraphrased
translation from Dutch, EWD 51]
Using modern terminology:
• Dijkstra's “concrete machine” corresponds to a hardware task, and
• Dijkstra's “abstract machine” corresponds to a software task.
Note thus that a “concrete machine” is not a refinement of an “abstract machine”. A “concrete machine” and an “abstract machine” are both refinements of a “machine”. (The concrete machine typically
resembled a peripheral hardware device of the X8 computer.)
Dijkstra did not want to distinguish between slow abstract machines and fast abstract machines. Each abstract machine, Dijkstra emphasized, is either working or halting — and that is thus the only
distinction we can make about any two abstract machines. (I believe Dijkstra made the same generalization for concrete machines in one of his later writings.)
Dijkstra then focused on the semaphore that stands in between a concrete machine and an abstract machine. The semaphore has an integer value t which we will store in the shared memory at a fixed
memory location — a location that is known to both the concrete and the abstract machine. Again, in the interest of generality, Dijkstra noted that t can be positive, zero, or even negative. For as
long as there is no practical reason to constrain t's range, Dijkstra stressed, we should keep it as broad as possible.
There is a complication, however. An integer t stored in memory is a passive entity and a semaphore has to be active in that it should inform its pending machines (if any) whenever its integer
becomes positive. Therefore, Dijkstra said, we will add a flip-flop to the semaphore. Dijkstra denoted the value of the flip-flop — which can be either 0 or 1 — by T.
Let change_t and change_T denote change of the semaphore's integer t and flip-flop T, respectively. Dijkstra remarked that there is two ways to ensure that change_t and change_T occur in an orderly
fashion, so that the semaphore's t and T values remain in sync with each other. One way (A) is to define the combination of change_t and change_T as one atomic action. Another way (B) is to retain
change_t and change_T as separate actions but to guarantee that any temporal discrepancy between t and T does not cause any problems.
Whenever the concrete machine modifies the semaphore we will go for (A), Dijkstra remarked. The choice for (A) is natural here because the concrete machine (hardware) will have a higher priority than
the abstract machine (software) when demanding control of the shared memory. Whenever the abstract machine modifies the semaphore we will go for (B).
The asymmetry between concrete and abstract machines thus gradually entered the picture, leading Dijkstra to introduce two case distinctions:
1. The concrete machine invokes the P operation and the abstract machine invokes the V operation.
2. The concrete machine invokes the V operation and the abstract machine invokes the P operation.
I will briefly discuss the first case distinction.
The P operation, implemented on the concrete machine (hardware), amounts to:
“if T=1 then begin t:=t-1; if t≤0 then T:=0 end”.
The V operation, implemented on the abstract machine (software), amounts to:
if T=0 then
begin if 0<t
then T:=1 end”
Dijkstra explained why these code snippets do the job, an explanation that I omit here.
“In practice”, Dijkstra continued, “the X8 forces us to deal with more complexities”. In the interest of reducing the communication between the abstract machine on the X8 and the concrete machine on
a peripheral device, we will store the flip-flop T in the peripheral device. This design choice makes the query T=1 and the assignment T:=0 less costly. On the other hand, it makes the query T=0 more
costly. Therefore, we will also cache a copy of T in the main memory, denoted by ST.
These considerations led Dijkstra to present refined code snippets of the P and V operations. The P operation on the concrete machine becomes:
“if T=1 then begin t:=t-1; if t≤0 then ST:=T:=0 end”.
The V operation on the abstract machine becomes:
if ST=0 then
begin if 0<t then
begin ST:=1;
Dijkstra took more machine-specific concerns into account: he placed t and ST in the same word in main memory. This storage compaction led to refined code snippets (not shown here).
In summary, then, Dijkstra first reasoned machine independently and as symmetrically as possible. He gradually took machine-specific concerns into account, breaking symmetries only when compelled to
do so. At the end of the day, he had a solution that was both simple and practical. Contrast this, then, with my previous observation that Dijkstra was not perceived as an engineer by some of his | {"url":"https://www.dijkstrascry.com/node/113","timestamp":"2024-11-13T12:20:26Z","content_type":"text/html","content_length":"37635","record_id":"<urn:uuid:7421d157-ac3e-4fa1-bb4d-59c763be25d2>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00038.warc.gz"} |
Chris Adolph :: 503
POLS/CSSS 503 Spring 2014
Advanced QuantitativePolitical Methodology Class meets: Tuesdays 4:30-7:20 pm Electrical Engineering 031
Offered every Spring at the University of Washington by various instructors TA: Carolina Johnson (UW Political Science)
Section meets: F 1:30-3:20 pm Savery 117
Lectures Click on lecture titles to view slides or the buttons to download them as PDFs.
Topic 1
Introduction to the Course and to R
R code and data for the GDP example. R code and data from the fertility example. You’ll find detailed instructions for downloading, installing, and learning my recommended software for quantitative
social science here. Focus on steps 1.1 and 1.3 for now, and then, optionally, step 1.2. (Note: These recommendations may seem dated, as many students prefer to use RStudio as an integrated design
environment in combination with RMarkdown. You are free to follow that model, which minimizes start-up costs. I still prefer a combination of Emacs, the plain R console, and Latex/XeLatex for my own
productivity, with occasional use of Adobe Illustrator for graphics touch-up.)
Topic 2
Review of Matrix Algebra for Regressionand Regression and Graphics in R
We will work through Kevin Quinn’s matrix algebra review. R code and csv data for an example of how the base graphics package can create scatterplots and perform linear regression.
Topic 3
Linear Regression in Matrix Form andProperties and Assumptions of Linear Regression
You may find useful three review lectures on basic probability theory, discrete distributions, and continuous distributions.
Topic 4
Inference and Interpretation of Linear Regression
Example code for estimating a linear regression, extracting confidence intervals for the parameters, and plotting fitted values with a confidence envelope.
Topic 5
Specification and Fitting in Linear Regression
Topic 6
Outliers and Robust Regression Techniques
Student Assignments
Problem Set 1
Due Tuesday, 15 April, in class
Data for problem 1 in comma-separated variable format.
Problem Set 2
Due Friday, 25 April, in section
Data for problem 1 in comma-separated variable format.
Problem Set 3
Due Tuesday, 6 May, in class
Five R script templates for simulation of the performance of linear regression with different kinds of data: when the Gauss-Markov assumptions apply; when there is an omitted variable; when there is
selection on the response variable; when there is heteroskedasticity; and when there is autocorrelation in the response variable.
Problem Set 4
Due Tuesday, 20 May, in class
Data for problem 2 in comma-separated variable format.
Problem Set 5 (Optional)
Due Friday, 6 June, in section
Data for problems 1. Data for problem 2. Data for problem 3. (All data in comma-separated variable format.)
Final Paper
Due Monday, 9 June, at 3:00 PM, in my Gowen mailbox
See the syllabus for paper requirements, and see my guidelines and recommendations for quantitative research papers. | {"url":"https://faculty.washington.edu/cadolph/index.php?page=20&photo=3&uwphoto=1&banner=6","timestamp":"2024-11-08T08:36:47Z","content_type":"application/xhtml+xml","content_length":"20925","record_id":"<urn:uuid:f8e3d0c6-3088-464b-86bd-5fedad77bd17>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00639.warc.gz"} |
6 x 6 magic squares
6 x 6 magic squares requires JavaScript
A mouseclick on any number of the green-bordered square ("number-pool") moves this number into the first empty field of the red-bordered square ("magic area"); a click on any number of the magic area
brings the number back to its original place in the pool.
Aim: A magic square of size 6 x 6 is to be constructed, (with additional properties: nine of the 2x2 subsquares have equal sums and the inner 4x4 subsquare is pandiagonal).
When "show" or "quick" is activated, a backtracking algorithm will continue the search for a solution; interruption can be caused by clicking the option "mouse". When a solution is found, the
algorithm will stop, and in that case, a search for a next solution may be continued with the corresponding button. The blue-bordered square always contains the last found solution. The delay (in
milliseconds) regulates the frequency of image output, if "show" is activated.
© H.B. Meyer 6 x 6 magic squares and patterns general 6x6 magic squares
magic squares and cubes 4 x 4 magic squares 5 x 5 magic squares nuclei for magic squares | {"url":"http://hbmeyer.de/backtrack/mag6en.htm","timestamp":"2024-11-07T23:42:29Z","content_type":"text/html","content_length":"26248","record_id":"<urn:uuid:3fac769a-7236-4354-8cb1-fbeab5305d21>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00281.warc.gz"} |
Development of a Tool to Analyze the Economic Viability of Energy Communities
Issue Renew. Energy Environ. Sustain.
Volume 6, 2021
Achieving Zero Carbon Emission by 2030
Article Number 28
Number of page(s) 5
DOI https://doi.org/10.1051/rees/2021028
Published online 26 August 2021
Renew. Energy Environ. Sustain.
, 28 (2021)
Research Article
Development of a Tool to Analyze the Economic Viability of Energy Communities
University of Applied Sciences Upper Austria, Energy Research Group ASIC, Wels, Austria
^* e-mail: fernando.carreras@fh-wels.at
Received: 28 June 2021
Received in final form: 22 July 2021
Accepted: 4 August 2021
Energy Communities (EC) are an instrument to improve the efficiency and autarky of Smart Grids by increasing the local consume of the energy locally produced. Energetic (energy flows, CO[2]
emissions) and economic (operative costs, acquisition and maintenance of technologies) aspects of all components of the EC must be evaluated to quantify the participation of the EC to achieve the
proposed goal. Effective analysis of EC must account for numerous complexities and uncertainties, requiring advanced computational tools. The main contribution of this paper is the introduction of a
software package to analyze the viability of ECs focused on the particularities imposed by the new Austrian law for renewable energies, which optimizes the energy flows between all participants. The
results of the test case show more than a 14.2% reduction of global cost. At the same time, all participants achieve better results operating inside of the EC than alone. The range of cost reductions
varies between 2.75% and 51%. The spread of these reductions opens a question about a fair and optimal way to set trade prices inside of the EC for future works.
© F. Carreras and G. Steinmaurer, Published by EDP Sciences, 2021
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1 Introduction
The Austrian government set recently as goal to cover 100% of the Austrian electricity consumption in 2030 from renewable energies [1]. Energy Communities (EC) are an instrument to achieve this goal
by increasing the local consume of the energy locally produced and hence to increase the efficiency and autarky of grids. Energetic (energy flows, CO[2] emissions) and economical (operative costs,
acquisition and maintenance of technologies) aspects of an EC must be evaluated to quantify the participation of the EC to achieve the proposed goal.
We programmed a software tool based on Mixed Integer Linear Programming (MILP) to evaluate the economic and energetic viability of a potential EC. Different software tools are already programmed to
regulate and control micro and smart grids [2–6] but this new development covers the gap of the particularities imposed by [1] on ECs. In this sense, the software takes into account not only the
different roles and parameters of the participants of the EC (producers and/or consumers, private/industry, constant/dynamic prices) and the installed technologies (photovoltaic, solar heat, heat
pumps, electric batteries, heat water puffers, ab- und adsorption chillers, etc.) but also the particularities of [1].
Most of the usual tools above mentioned, make use of data reduction methods (by instance, by defining one pattern working and weekend day for each month and extrapolating the results along the month)
or mean aggregation of time steps of energy loads and generation (considering then one-hour data instead 15-minutes data). Although these methods speed up the computations, our tool is based on a
screen moving strategy, which analyses the EC by generating optimisation strategies each 15 minutes for the next 48 hours. This procedure introduce several advantages: it allows considering dynamic
buy and selling prices of energy inside and outside of the EC, storage devices can be continuously used, and a higher time resolution provides a more realistic performance of the EC.
The first part of this paper describes the formulation to model the EC and its energy flows as an MILP problem. In a second section, the environment and data for the experimentation are introduced.
The third part, presents the results of a practical case. The paper closes with a discussion and plans for future development.
2 EC modelling
An EC is considered in its general form as the composition of three levels, corresponding to one kind of energy: electricity, heal and cool. At each level belong all energy sources, which do not
require any transformation (photovoltaic, wind, geothermal) and storage technologies. Transformation operators represent sector coupling technologies (heat pump, absorption chiller). Each participant
can be connected to a public grid and to the grid of the community, to share energy. Members of the community are only allow to sell self-generated energy. For this reason, each participant is
modelled mean two nodes. The first node bundles the public grid, loads and energy flows from the second node. This one connects the generators of renewable energies with storage components, community
exchange. Both nodes are connected, however in an only one-way (from second to first node) kind, because only energy from renewable sources are allowed to be feed in to the public grid and energy
from the public grid does not flow into the community grid. A schema of this construction is illustrated in Figure 1.
In a close way to [5], a mathematical description of the optimization problem is based on balance equations (1), (2) for each node; capacity constrains (3), (4), (5) and storage continuity
constraints (6). The objective function to minimize can be in terms of costs and/or emissions defined.$X k , i n ( 1 ) ( t ) − X k , o u t ( 1 ) ( t ) = L k ( t )$(1)where $X k , i n ( n ) ( t )$
refers to the input flow of energy k at time t for the node n, $X k , o u t ( n ) ( t )$ refers to the output flow of energy k at time t for the node n, L[k](t) is the load of the k energy time t.$G
k , g ( t ) + S k , i n ( t ) − S k , o u t ( t ) + X k , i n ( 2 ) ( t ) − X k , o u t ( 2 ) ( t ) = 0$(2)where G[k,g](t) refers to the provided k energy by generator/source g at time t, S[k,in](t
) and S[k,out](t) are the input and output flows of energy k at time t.$G k , g ( t ) ≤ G max , k , g ( t )$(3) $X k ( 1 ) ( t ) ≤ X max , k ( 1 ) ( t )$(4) $X k ( 2 ) ( t ) ≤ X max , k ( 2 ) ( t )$
(5) $S O C k ( t ) =S O C k ( t − 1 ) + η k , c h ⋅ S k , i n ( t ) − S k , o u t ( t ) / η k , d i s = 0$(6)where SOC[k](t) is the state of charge of storage k at time t. Finally η[k,ch] and η[k,
dis] are the charging and discharging efficiencies of the storage system.
The evaluation of the community comprises two stages. The first stage regards the individual-case of each participant alone with the technologies to be installed but any energy exchange with the
participants of the EC takes place. At this stage, the energy flows $X k , i n ( n ) ( t )$ and $X k , o u t ( n ) ( t )$ are limited to connections of the considered participant. With the
information of the equations (1)–(6) one can define the optimization problem for the i participant (7):$min x ‾ i ∈ Ω ( x ‾ i ) f i ( x ‾ i ) { A e q , i ⋅ x ‾ i = b e q , i x ‾ l b , i ≤ x ‾ i ≤ x ‾
u b , i$(7)
This optimization problem is solved by using MILP methods for each participant and its solution (costs and/or emissions) provides the reference values for the third stage. For EC with N participants:
$V = { v 1 , v 2 , … , v N }$(8)with these values will be defined a new optimization problem (second stage), which incorporates these values as a new constraint b[ineq]:$min x ‾ ∈ Ω ( x ‾ ) f ( x ‾ )
A e q ⋅ x ‾ = b e q A i n e q ⋅ x ‾ ≤ b i n e q x ‾ l b ≤ x ‾ ≤ x ‾ u b$(9)
This second optimization considers the whole EC and the new constraint b[ineq] guaranties that each participant of the EC profits by the cooperation inside of the community.
Relevant parameters to analyse the community are the time to analyse, the time resolution and the time forecast. These parameters depend on the available data measurements. In practical cases, a
whole optimization (both stages) is carried out once each 15 minutes. It provides a control strategy for a time horizon of 24 hours. In addition, to avoid seasonal effects, the time to analyse used
to be one or more whole years. Additional parameters to take into account are the sell and buy prices of the public and community grids.
Fig. 1
Scheme of a two-participants EC.
3 Environment and data
The new law for renewable energies in Austria encourages municipalities to cooperate to achieve the objectives. Aspects of this cooperation are divers. From one part, public institutions should
provide roof surfaces from public buildings (schools, town halls, sport halls, and gymnasiums) to install new PV panels. On the other hand, installations of technologies to produce renewable and
sustainable energies are promoted and financial supported.
Considered Participants are private citizens, the municipality and industrial partners. Prices are different for small and big consumer. Some of the participants are consumers and at the same time
energy producers. In a first approach, we have considered an EC with only electricity exchange (heating and cooling loads are carried out by Power to Heat (P2H) and Power to Cool (P2C) technologies
and hence they are integrated as electric loads).
Based on realistic values of energy loads [7] and productions [8] in central Europe, we evaluate the economic and energetic viability of an EC composed by 10 participants placed in a medium size
rural town under the constraints and rules [1] pointed out at previously. The analysis was carried out with a 15-minutes time resolution and it covers one year.
Table 1 summarizes the parameters of the EC, the planned dimension of PV to be installed, the currently electricity consumption per year and the type of consumer. Electricity prices for private
participants (P) and for industrial participants (I) are showed in Table 2. Sell price to feed the public grid is the same for private and industrial participants and it is regulated by the
Abwicklungsstelle für Ökostrom AG [9]. The current value of the compensation is 7.67 €Cent/kWh but it will be reduced for new contracts to 7.06 €Cent/kWh. The trade prices inside of the EC are only
one €Cent/kWh lower (buy price) and one €Cent/kWh higher (sell price) that the respectively prices of the public grid. These prices are set with the only purpose to show, how a one Cent variation is
able to produce a reduction of the global costs inside of the EC. With these prices is guaranteed that the producers sell electricity into the EC first and they feed the public grid later.
Table 1
Description of the participants of the Energy Community.
Table 2
Buy and sell prices for each consumer type.
4 Results
Table 3 shows a summary of the yearly results. Three situations are considered for each participant: the current yearly costs without PV, the costs with PV but without collaboration with the
community and the costs inside of the community.
The installation of PVs supposes for the participants a costs reduction of a 19.5%. With respect to the current situation, the EC achieves a global costs reduction of 41.6%. The collaboration inside
of an EC supposes a win-win situation for all participants: electricity producers are getting an advantageous price by selling electricity into the community and all consumers are paying less by
buying electricity from other participants.
Figure 2 shows on the positive part of the diagram the monthly consumption of electricity of the EC (bought from grid, self-consumed and bought inside of the EC) and on the negative part of the
diagram, the sold electricity (sold to the EC and sold to grid). Seasonal effects of electricity production are clearly visible. The EC always buys electricity from the public grid because of the
nature of PV production. The EC must always sell electricity to the public grid because the consumption of industrial participants sinks on weekends.
As the objective of an EC is to increase the local consumption of locally produced energy as much as possible, a first analysis of the results suggest a potential improvement of the use of the
produced energy for the suggested EC. From one part, the use of energy storage technologies could increase the local autarky of the EC. However, the investment costs of such technologies (for the
currently subventions policy in Austria these technologies are not the first priority) compared with the potential benefits should be investigated. Another possible explanation for the excess could
be an oversized PV production.
An EC can be considered from the point of view of grid operators as an instrument to stabilize the grid at regional and national levels. For this reason, the implementation of storage technologies
and its consequences must be considered not only at the level of the EC.
Fig. 2
Monthly distribution of the electricity consumption.
5 Conclusion and future work
In this paper, a new software tool for analysis of EC focused on the restrictions of the new Austrian law for renewable energies was presented. It was shown that this tool is able to manage
heterogeneous configurations of participants attending to the played roles and their nature as consumer. The tool provides not only economic results but also analyses of the energy flows, which help
to identify possible improvements by the configuration of the EC.
The tool was tested with realistic parameters for the Austrian case and with a difference of only 0.01 €/kWh between the trade prices of public and community grids, the cooperation between all the
participants achieves a global costs reduction of a 14.2%.
ECs seem to be a promising way to achieve the objectives set by the new Austrian law of renewable energies. For this reason, future extensions of the software tool plan to extend the integration of
transformation technologies for sector coupling and the incorporation of new participation forms such as mobility (Vehicle to Grid or Vehicle to Community).
The software tool is able also to handle dynamic buy and sell prices for each participant. This feature was not used at the current experimentation because it has to do with an open question: how can
one establish fair trade prices inside of the community? As the participants of the EC are involved with different investments, the prices should be set in a way that these economic efforts must be
somehow compensated. Future research is here needed.
This paper and the research behind it would not have been possible without the exceptional financial support of the Government of Upper Austria through the project FTI-OÖ Förderung Energieforschung
FH OÖ − Methodenentwicklung für Energieflussoptimierung.
Cite this article as: Fernando Carreras, Gerald Steinmaurer, Development of a Tool to Analyze the Economic Viability of Energy Communities, Renew. Energy Environ. Sustain. 6, 28 (2021)
All Tables
Table 1
Description of the participants of the Energy Community.
Table 2
Buy and sell prices for each consumer type.
All Figures
Fig. 1
Scheme of a two-participants EC.
In the text
Fig. 2
Monthly distribution of the electricity consumption.
In the text
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Teaching Statistics in Integration with Psychology
Marie Wiberg
Umeå University, Sweden
Journal of Statistics Education Volume 17, Number 1 (2009), jse.amstat.org/v17n1/wiberg.html
Copyright © 2009 by Marie Wiberg, all rights reserved. This text may be freely shared among individuals, but it may not be republished in any medium without express written consent from the author
and advance notification of the editor.
Key Words: Student-centered learning, Research problems, Course revision.
The aim was to revise a statistics course in order to get the students motivated to learn statistics and to integrate statistics more throughout a psychology course. Further, we wish to make students
become more interested in statistics and to help them see the importance of using statistics in psychology research. To achieve this goal, several changes were made in the course. The theoretical
framework to motivate teaching method changes was taken from the statistics education literature together with the ideas of student-centered learning and Kolb’s learning circle. One of the changes
was to give the students research problems in the beginning of the course that were used throughout the course and which they should be able to solve at the end of the course. Other changes were to
create a course webpage and to use more computer-based assignments instead of assignments with calculators. The students’ test results and their answers on the Survey of Attitudes Toward Statistics,
SATS, (Schau, Stevens, Dauphinee, & Del Vecchio, 1995) together with course evaluations showed that by changing the course structure and the teaching, students performed better, and were more
positive towards statistics even though statistics was not their major.
1. Background
Teaching statistics with statistics students can be relatively easy compared to teaching statistics with students who are not primarily interested in statistics. Teaching statistics integrated with
another subject might even be more difficult. This paper focuses on how to motivate students in another subject area to study statistics. In particular, the focus is on students who learn statistics
integrated with a psychology course. Hogg (1991) points out that "students frequently view statistics as the worst course taken in college". Further, Gal and Ginsburg (1994) noted that the fact that
students tend to have negative feelings about statistics is recognized by many teachers in statistics. Peterson (1991) noted that most students have to take at least an introductory statistics course
in college but they tend to remember the pain more than the substance. Using different teaching methods the ambition was to revise a course and change students’ negative attitudes toward statistics.
The statistics course of interest is integrated in an undergraduate second semester psychology course. The course runs through the whole semester and is part of three out of four psychology modules
offered during the semester. The statistics part in each module consists of a few lectures in the beginning of each module. At the end of each module, one or two exam questions in their written exam
focus on statistics. The students need at least 50 percent correct on the statistics questions in order to pass the statistics taught in each particular module. The department of psychology is
responsible for the course and decides the contents of the course. The structure of the statistics part and the detailed contents of the lectures are decided by the statistics teacher. The statistics
part is intended to help the students to become better psychology researchers. For details about the course outline, objectives and a thorough description see Appendix A. Note, there are no
statistical prerequisites for the course although one semester of psychology is mandatory. This makes this course comparable to an elementary introductory statistical course.
During the first semester I taught the course, I noted frustration and lack of motivation among the students since they failed to see the links between psychology and statistics. The course
evaluations and the students’ test results showed that something was wrong with the course. This led me to rethink the course and search for possible solutions on how to motivate students to learn
statistics and change the students’ attitudes toward statistics.
1.1. Aim
The aim was to revise a statistics course in order to get the students motivated to learn statistics and to integrate statistics more throughout a psychology course. A further objective was to find a
possible solution for how statistics should be taught to psychology students in a way which makes them become interested in statistics and to help them see the importance of using statistics in
psychology research.
2. Method
2.1. Participants and procedure
The course was first given as a traditional course to 20 students and background material was collected. The revised course was taken by 24 students and the same material as when the traditional
course was given was collected. The two groups are assumed to be similar since they had the same average admission points. Further, the groups had similar average ages and about 1/5 were male
students and 4/5 were female students.
2.2. Instruments
The collected materials included test score results, course evaluations, teacher reflections and students’ beliefs about statistics collected using the affect subscale, i.e. students’ feelings
concerning statistics statements from the Survey of Attitudes Toward Statistics, SATS (Schau, Stevens, Dauphinee, & Del Vecchio, 1995). The SATS with 28 items was used which have items on a 7-point
Likert-type scale which ranged from strongly disagree to strongly agree, with higher ratings indicating a more positive attitude. There are four subscales within the instrument; a 6-item Affect
subscale, a 6-item Cognitive Competence subscale, a 9-item Value subscale and a 7-item Difficulty subscale. The SATS was chosen since it correlates highly with the Attitude Toward Statistics, ATS (
Wise, 1985) and the Statistics Attitude Survey (Roberts & Bilderback, 1980) see e.g. Schau et al (1995) or Carmona (online). Especially the affect scale, which is suppose to measure students’
feelings concerning statistics and the value subscale, which is suppose to measure students’ attitudes about the usefulness, relevance, and worth of statistics in personal and professional life, were
of interest. The choice of SATS was also due to the fact that the SAS and the ATS have been criticized for not measuring what they intend to measure (Gal & Ginsburg, 1994). A choice was made to only
use the SATS post test, as opposed to use both SATS pre and post tests. The reason behind this choice was our main interest to examine attitude differences between groups which receive different
teaching methods and there were no explicit interest in examining attitude differences within individuals.
In order to decide what and which changes should be made in the course in order to facilitate the learning of statistics for the students the statistics education and pedagogy literature were
examined. The literature search included searching the databases Science Direct, Educational Resources Information Center (ERIC) and Academic Search Elite. The course changes made were motivated by
the material found in articles from the statistics education literature and from the pedagogy literature.
3. Theoretical frameworks
3.1. Teaching methods in statistics education
There has been substantial research over the past years about how one should teach statistics at different levels and for different students. Cobb & Moore (1997, p. 814) states clearly that
"Statistics should be taught as statistics." By that statement they emphasize that Statistics is a field on its own and should be treated as such and not as a subfield of something else. In that
sense it is especially important to use the opportunity as a statistician to be allowed to teach statistics for e.g. psychologists in their own course. A statistician can contribute to a large extent
although it helps to have some knowledge of the students’ major area of study.
Cobb (1991) noted that most courses can be improved by emphasizing the use of empirical data and concepts linked to the data, at the expense of giving the students less theory and fewer "recipes" to
follow. Cobb also emphasizes using descriptive statistics such as graphs throughout the teaching. Cobb & Moore (1997) suggested using data in the statistical course in order to introduce statistical
ideas and applications. Smith (1998) stated that students learn statistics by doing statistics. This includes; collecting data, performing analyses and communicating the results. Smith’s students’
results on the final exam and the students’ course evaluations showed that the students appreciated this method. Hogg (1991) also emphasized that students benefit in their learning if they also
collect data and not just work with other people’s data. Summing up, in statistics education it is a common belief that students learn by working with real data which they are involved in collecting.
Teaching statistics to non statisticians has been explored in different contexts before. Roback (2003) e.g. sketched how to develop a course for a mixed audience. Roback’s first aim of the course was
that students should develop a "statistical literacy", and an "…understanding of basic elements of statistics that can help in critically evaluating data-driven results…" in the students’ field of
interest and in their lives. This goal is in line with the ideas of data-driven learning in statistics. In this light, when teaching statistics to non statisticians it is important to focus on real
problems which can appear in the industrial or business world. Romero, Ferrer, Capilla, Zunica, Balasch and Serra (1995) also emphasized that the focus should be more on the student than on the
A problem with traditional introductory statistics courses is usually the lack of motivation among the non statistician students. In the past, researchers (see e.g. Gal & Garfield, 1997; Hoerl, Hahn
and Doganaksoy 1997) have emphasised the importance of enthusiasm and of interest for students together with teaching statistical literacy and thinking as desired in introductory statistical courses.
Gal & Ginsburg (1994) showed that the students’ motivations, expectations, attitudes and their impression of the discipline play a role in their success in statistical courses. Varhoof, Sotos,
Onhena, Verschaffel, van Dooren, & Noortgate (2006) also noted a positive attitude between the students’ results and their attitudes.
3.2. Suggestions from pedagogy literature
The pedagogy literature search revealed that there are a number of theories on how to increase motivation among students and how to change the student perspective and focus in a course (see e.g.
Heikkilä & Lanka, 2006). Here, student-centered learning (e.g. Stuart, 1997) was chosen since it was in line with the statistics education literature together with the idea of a learning circle as
discussed in Kolb (1984).
3.2.1. Student centered learning
The main idea with student-centered learning is to involve the students in the learning process in order to make learning more meaningful. Further, the idea is to relate the topics taught to the
students to their lives, their interests and to let the students engage in the creating and the understanding of knowledge (McCombs & Whistler, 1997). One of the ideas is that if the students are
involved in the learning process they want to learn more and not just memorize given facts. Student-centered learning is the first step to let students own the material that they learn. It can also
be adapted to the needs of each individual student (Stuart, 1997). Students learn more efficiently if they are involved in their learning and they share the learning process with the teacher instead
of the teacher being the only source of information. Students should be treated as creators together with the teacher in the learning process (McCombs & Whistler, 1997).
3.2.2. Kolb’s learning circle
In order to describe the structure in the learning process one can e.g. use Kolb’s learning circle as shown in Figure 1. Kolb (1984) created a model of learning using the four elements; concrete
experience, observation and reflection, the formation of abstract concepts and testing in new situations.
Figure 1. A model of Kolb’s learning circle.
The idea is to present a concrete problem for the students, and use this problem to teach general principles and theories to the students. In the first phase, some problem is posed or action is
performed. In the next phase, the students try to understand this concept in a specific setting or in similar settings. In the third phase, the students should try to generalize the problem so that
in the fourth and last phase they can test hypotheses and use their knowledge in new situations. The overall idea is to make the students understand that they have to know some theories in order to
solve problems. Here, the circle has been described as going from phase 1 to phase 4 but it is important to point out that Kolb & Fry (1975) argued that you can start anywhere in the learning cycle
and it can be viewed as a continuous spiral. The interested reader can read more about Kolb’s learning circle in Kolb (1984) or Kolb & Frye (1975).
3.3. Reflections upon the suggestions from the literature
The ideas from the statistics education literature were combined with the ideas from the pedagogy literature. In particular, when teaching statistics to non statisticians it is important to emphasize
data and concepts at the expense of less theory (Cobb, 1991; Cobb & Moore, 1997) and to have hands on problems (Smith, 1998) or real-life problems (Roback, 2003) which the students have been part of
collecting/creating (Hogg, 1991). To use real data and focus on the students’ learning instead of the lecturing (Romero, Ferrer, Capilla, Zunica, Balasch, Serra, Alcover, 1995) will engage the
students more actively and can thus be viewed as student-centered learning (Stuart, 1997; McCombs & Whistler, 1997). The same goes for the ideas of Smith & College (1998) who stated that one learns
statistics by performing statistics. In other words, by letting students take responsibility over their learning process and do a lot of data analysis they will become better in statistics.
The combined ideas from the statistics education and pedagogy literature resulted in constructing and using small research problems and more extensively using the psychology labs which were already
in the course. Descriptions of the changes in the course are given in more details in the next section. In order to get a structured way to work with the students in their learning process Kolb’s
(1984) learning circle was used.
The problems when teaching a course integrated with the students’ major subject area are similar to the problems Roback (2003) encountered when teaching a course to a mixed audience. The aim is,
however, the same: the students should develop an understanding of basic elements of statistics which can help them in critically evaluating data-driven results in their field of interest.
Finally, a problem with traditional statistics courses is usually the lack of motivation among the students. It is important both for the students’ learning and their performance that the students
are motivated, interested and feel enthusiastic about learning the subject (Gal & Garfield, 1997; Gal & Ginsburg, 1997; Hoerl, Hahn and Doganaksoy, 1997). These ideas about student motivation can be
combined with the pedagogical literature on how to increase motivation among students and how to change the student perspective and focus in a course by introducing an area differently or changing
teaching methods, the assessments or focus in a course (see e.g. Heikkilä & Lanka, 2006). The research problems mentioned and the use of real life examples, were therefore important when motivating
students to learn statistics.
4. Implemented changes in the course
The implemented changes in the courses were on two levels; course administrative changes and teaching method changes. From a teacher’s perspective, the course administrative changes were necessary
and needed in order to make the teaching method changes. Note that the course objectives were stated by the department of psychology and hence could not be changed, however the teaching methods,
assignments, course materials and assessments were possible to change.
4.1. Course administrative changes
In the planning phase, some changes were made in the course schedule so that statistics would be more evenly spread out during each module in order to give the students time to reflect on their
learning. The book Statistical Methods for Psychology by Howell (2006), which was already used in a higher level course, was added. This might seem like a small change but since the students only had
handouts before, it was a huge difference for them to have a course book that dealt with statistical methods in the area of psychology. This was especially important for students who wanted to learn
more and wanted to see the usefulness of statistics in psychology. By having a course book that discusses research problems in their own field they could more easily accept and understand why they
had to learn statistics.
Another administrative change was to create a webpage for the course where figures and overheads shown in the lectures could be displayed together with handouts, computer assignments and extra
assignments. One of the great advantages with the webpage was that students could choose whether to do the computer assignments at the scheduled hours or on their own on their home computers. The
last administrative change was to add computer assignments to each module. The first time I taught the course, many lectures and assessment questions were focused on statistics with calculators and
there was only one computer assignment in the last module. To mainly use calculators when teaching statistics instead of computers is something that was common a few years ago. Today, many
statistical problems that were previously solved using a calculator are now solved with the help of computers and using computer software reflects the actual practice of researchers in many different
To change the assessment, or part of the assessment, is a way to achieve changed student behaviour (Biggs, 1999). Therefore, the assignments were somewhat changed in the course. Today’s psychology
researchers have access to user-friendly statistical software and might need sophisticated statistical methods to evaluate their results. So, the focus in the course shifted from teaching students to
merely doing problems using calculators to also teaching them how to use computers as a tool. If a computer is used to solve a statistical problem, it is important for the students to know what
questions should be answered before using a statistical method. It is also important to know how the data material from the research question can best be structured in order to be analysed. Most
importantly, are the assumptions in a model satisfied? What do we do if the assumptions in the model are not satisfied? Many computer programs will give an analysis regardless of whether the
assumptions are satisfied; hence the researcher needs to know which method and what assumptions should be satisfied in order to obtain a meaningful analysis.
4.2. Revision of teaching methods
The most important change involved the more data driven and concepts based teaching found in the statistics education literature as well as models from the pedagogy literature: Kolb’s learning circle
and the idea of student-centered learning including changing the structure of the lectures. Instead of teaching statistical methods and then applying them to psychological problems and research
questions, the order was swapped. The students were presented a psychology research article containing statistics which they would encounter during the course. The new order included giving the
students small research problems in psychology. One example is given below and five more examples are given in Appendix B. Note that the level of information and structure of the research problems
A researcher is interested in determining if the decoration of a café affects people’s abilities to relax, in terms of staying longer and drinking more coffee with their friends. The researcher has
access to 8 test subjects. Design a suitable experiment.
Depending on the design of experiment the students chose, different data were given to the students and they could perform a different analysis, or the students were asked to perform a similar
experiment. After completing the course the students should be able to solve similar problems. The problems were also used to demonstrate and motivate the importance of learning statistics for
psychologists. To give the students a small psychology research problem was both an idea of trying to give the students a concrete experience as pointed out in the statistics education literature (
Cobb, 1991; Cobb & Moore, 1997; Smith, 1998), and phase 1 in Kolb’s learning circle and to let the students own the material.
In order to solve the problems, the students needed to search for knowledge. In this sense, the psychology research questions helped create a learning situation that is student-centered. In the
second phase in Kolb’s learning circle the teacher gave the students tools to help them solve the initial problem. This was done mainly in form of lectures and computer assignments with tutoring.
When the students achieve the specific knowledge they can move to the next phase in the circle, hence try to generalize the problem. In general, the students used the research problems and examples
in order to figure out the generalization of the problem.
The last phase described in Kolb’s learning circle was not strictly part of the statistics integrated portion in each course module but can be said to be a part of each module anyway. In each of the
four modules, the students had to perform psychology experiments (see examples of psychology labs in Appendix C) and write lab reports. These experiments were also part of the traditional course. The
last phase states that the students should test hypotheses and use their knowledge in new situations. This is exactly what the students do in the experiments and when writing the lab reports. For
example, if the students have learnt inference theory and testing of hypotheses in the statistics parts of the course module, the psychology experiment could include writing a lab report where they
draw conclusions about statistical hypotheses in the psychology experiment they have performed. By letting students work with real problems they feel that they have to understand some basic concepts
in order to solve the overall problem. The changes made in the course were mainly motivated from the literature but the main idea was to involve the students more so they would feel they were part of
the learning process.
5. Evaluation of changes
In order to evaluate the revised course four sources of information were used; student course evaluations test score results from the written exams, the SATS questionnaire and the teacher’s
reflections. The results from the revised course were compared with the results from the when the course was taught traditionally.
5.1. Student perspective: course evaluations
At the end of the semester students filled out a course evaluation for the whole statistics course, i.e. including all the three modules which involved statistics. A copy of this form is given in
Appendix D.
5.1.1. Traditional course
Fourteen students out of the 20 students who participated in the traditional course expressed some kind of negative general impression about the course or were sceptical of statistics in general.
Eight out of these fourteen students reported that they had learned psychology and statistics separately during the semester, in other words they failed to see the links between psychology and
statistics. Hence, the students did not see the connection between the disciplines. The most confusing part was why they needed to learn statistics at all and how they would use their statistics
knowledge in the psychology field. Six students expressed dissatisfaction with the statistics part of their education. Three students claimed that they wanted to learn psychology and were forced to
learn statistics at the same time but they did not see why they should learn it.
Fifteen students wanted to change the structure of the course, either to spread the statistics part over each module or compress all statistics into four weeks. Nine students mentioned that the
course material was not enough and the connection to psychology was lacking. All students left blank the questions on what they did not want to change in the course.
Only three students answered the question about what was the most interesting part of the course. They thought the most interesting part was when psychology examples were used or when the
applicability to psychology of the statistics tools was evident. The least interesting part was learning to calculate statistics as mentioned by four students. Finally, the structure of the course,
i.e. the students got two or three days of statistics in the beginning of each module and were supposed to study by themselves for the final exam for that module four weeks later, was disliked by 15
5.1.2. Revised course
In the revised course, the general impressions of the students were much more positive than the students who took the traditional course. Nineteen of the 24 students had in general a positive
impression about the course. Four students wrote that they had started the semester with a dislike of statistics but now wanted to learn more statistics. Seven students especially expressed that
statistics was interesting. No students had a negative general impression although three students were neutral about this question.
Five students wanted to change the textbook since they thought it was somewhat advanced. The rest of the students left this question blank. Fourteen students did not want to change the fact that
there was an extensive course webpage, which included all materials taught in class, assignments and handouts. Eleven students mentioned that they had used the webpage extensively and it had helped
them in their learning process.
Twelve students thought that the most interesting part was to analyze psychology materials. These students felt that they could actually use statistics in their psychology assignments and psychology
experimental lab reports. Fourteen students were satisfied with the textbook, especially students who planned on taking the advanced psychology course. Five students wrote that they hoped they would
have more statistics when they were going to take psychology in advanced levels. The least interesting part was when purely statistics was taught as mentioned by three students. Four students
suggested that the psychology labs should be used even more in the statistics part, maybe as a way to assess part of the statistics taught in the course.
The structure/schedule question revealed that twenty students were satisfied with the structured although four students would have preferred just a statistics course instead of an integrated course.
The students who were positive said that it gave them time to reflect when the statistics part was spread over the semester.
In addition to the course evaluations a short group discussion was held at the end of the course which aimed on discussing the students’ feelings and impressions about how it has been to be part of a
problem-driven and student-centered course. Many students said that they felt challenged in the course and the fact that the problems were connected to their area of interest made them more engaged
in the course than they had thought they would be before they started the course. Some students said that they gradually started to like Statistics as they were working with interesting problems.
Several students also said that they had come to value Statistics once they realised how it could be used in their field of study. The only negative aspect which was raised was that some students
felt there was much more work in this kind of course.
5.2. SATS questionnaire
In order to examine the students’ attitudes towards statistics the student were given the SATS questionnaire (Schau, Stevens, Dauphinee, & Del Vecchio, 1995). The students answered the items on a
scale where 1 indicates strongly disagree, 4 indicates neither disagree nor agree and 7 indicates strongly agree. The instrument was distributed at the end of the semester to both the students who
were taught the traditional course and the students in the revised course. In Table 1 the results from SATS are shown for the traditional and the revised course. High values indicate a more positive
The Affect component which is intended to measure students’ positive and negative feelings concerning statistics showed a substantial large difference between the two student groups. In the
traditional course the median value was 3.33, which should be compared with a median value of 4.08 in the revised course. Further, the Cognitive competence component, which measures the attitudes
about intellectual knowledge and skills when applied to statistics, also showed a statistically significant higher median value for the students who took the revised course as compared with the
traditional course. The Value component, which measures the attitudes about usefulness, relevance, and worth of statistics in personal and professional life, also had statistically significant higher
values for students in the revised course. Note the Difficulty component, which measure attitudes about the difficulty of statistics as a subject, had similar values in both groups. The internal
consistency estimated with coefficient alpha revealed that most components in the SATS scale had reasonable high internal consistency in both courses.
Table 1. Median (MD), mean (M), standard deviation (SD) and coefficient alpha (α) of the attitude components from the SATS scale.
│ │Traditional course │ Revised course │
│Component │ MD │ M │ SD │ α │ MD │ M │ SD │ α │
│Affect │3.33│3.23│1.04│0.85│4.08│4.15│1.20│0.83│
│Cognitive competence │4.00│3.94│0.89│0.79│4.67│4.72│0.87│0.74│
│Value │4.11│4.13│0.70│0.63│4.94│4.85│0.83│0.71│
│Difficulty │3.43│3.41│0.55│0.51│3.43│3.41│0.64│0.65│
5.3. Exam results
After each completed module written exam was given to the students. For each module the students needed to have at least 50 percent of the statistical questions correctly answered. Summing the
maximum score on the items in each module yielded a maximum total score of 20 for the whole semester. In order to compare the students’ exam results before and after the change the average score
among the students in each group was calculated. The exams are assumed to be of equal difficulty. The result showed that the 20 students who were given the traditional course earned an average score
of 12.98 (SD = 2.20). This result is in line with results obtained from previous years when the course was given by other lecturers. In the revised course the 24 students had an average score of
15.63 (SD = 2.57), hence a large (and a statistically significant) improvement. This course has been given again since these changes were made and the reported results are in line with other times
the revised course has been given.
5.4. Teacher perspective: reflections
The administrative changes in the course were necessary in order to change the rest of the course. The schedule change yielded more time for the students to reflect on what they learnt. The course
book was a very good tool both for extra exercises and for discussions. The course webpage was a convenient way to give extra information or more material to the students.
The teaching method changes were stimulating, not just because it made teaching more interesting, but mainly because of the positive feedback from the students. The students performed better in the
course, in terms of better understanding of key concepts which was evident when discussing statistical problems and their answers on the exam. The data-driven statistical learning seemed to increase
the motivation and hence gave positive results. In the revised course the students had to take greater responsibility for their own learning and be more active in what they needed to learn. The
workload was about the same as for the students in the previous semester, but instead of calculator assignments there were more computer assignments.
In the future, the student-centered learning should be even more stressed, maybe in terms of letting the students come up with some research problems that they wish to solve. Further, it could also
be effective to have the statistics assessment as part of the students’ lab reports. In that way the students get an even closer connection to real psychology research problems and it is easy to
assess whether the students can use the statistics they learned in their psychology research. Finally, changing a course takes a lot of time in planning and finding new materials. Since this course
was given every year for some years some of the materials could be reused.
6. Discussion
The aim was to revise a statistics course in order to get the students motivated to learn statistics and to integrate statistics more throughout a psychology course. Further, we wished to make
students more interested in statistics and to help them see the importance of using statistics in psychology research. The rationale for revising the course was that students felt they were forced to
learn statistics. The evaluation of the changes included students’ course evaluations, average test score results, the SATS questionnaire and teacher reflections. As a teacher I observed that the
students came to the course with low expectations and they had very limited pre-knowledge. A possible solution to the problem was to use the idea of data-driven problems suggested from the statistics
education literature (Cobb, 1991; Cobb & Moore, 1997; Smith, 1998), together with student-centered learning and Kolb’s learning circle in order to increase motivation. These ideas were realised
through introducing a psychology related research problem in the beginning of the course and letting the students find the necessary tools to solve the problem. The idea was to give the students
tools to understand the statistical concept in specific situations and help them generalise them in order to apply them in similar settings. Later, when the students had completed the statistics part
they would be able to generalize the statistics concepts taught in order to use them when solving research problems in their psychology experiments.
What was evident in the revised course was that most of the students had a more positive attitude towards statistics than in previous courses although they perceived the same level of difficulty of
statistics as a subject. The students were much more satisfied in the revised course as seen in the course evaluations. This is important to notice, since students whose primary study area is
something other than statistics tend to give the lowest ratings in course evaluations compared with students taking courses in line with their majors (i.e. pure statistical students). However, I have
not investigated whether the same result is obtained in a regular statistics course if the same teaching methods are used.
So what has been learned? To integrate and develop a new course is time consuming but doing it by using great pedagogic and statistics education ideas makes it easier when one can see that it works
in practice. It is a learning experience for both students and teachers which do becomes smoother over time, which also Rinaman (1998) noted. The involvement by students (Smith, 1998; Cobb & Moore,
1997) is important both from a student learning perspective and a teacher perspective.
By putting in a real effort I believe that we can change the common students’ view summarised by Hogg (1991) "students frequently view statistics as the worst course taken in college" into thinking
that it could be the most useful course they ever taken in college. At least that is my vision.
Appendix A
COURSE DESCRIPTION AND LECTURE OUTLINES
Course description: Integrated within three out of the four modules in the full semester psychology course, with emphasis on quantitative methods, corresponding to four credits, is an orientation
about research methods and statistical procedures. This orientation includes descriptive Statistics, hypothesis testing, statistical inference, methods to estimate reliability and validity.
Integrated with the three modules are also experimental design, t-tests (module 1), nonparametric tests (module 2) correlation and linear regression analysis (module 3). Details are given below.
Goals of the course: To obtain an introduction to research methods and statistical procedures. An overall aim is after finishing the course to be able to use basic statistical theory in psychology
Tools: Calculators and the software SPSS are used to analyze data.
Webpage: General information is given about the course. Especially, presentations from class, study materials, old exams, assignments and test results can be found on the webpage. Students are
encouraged to visit the webpage regularly.
Text: Howell, D. C. (2006). Statistical methods in Psychology, 6^th edition. New York: Thomson Learning.
Student activity and achieving the goals of the course: Students need to be active and benefit from working through the assignments. Emphasis is on understanding statistical concept, knowing when to
use them, how to use them and when not to use them. Reading the text is important. Computer assignments, laboratory work and tests during each module are used to evaluate student learning.
Module 1: Descriptive statistics, experimental design and hypothesis tests
- Introduction to Statistics
- Descriptive Statistics (Graphs, measuring center, measuring spread)
- Elementary probability theory
- Random variables
- Normal distribution (properties, z-scores, model)
- Sampling distributions
- Hypothesis test of mean(s)
- t distribution
- One sample t-tests
- Two samples t-tests
- Paired samples t-tests
Module 2: Nonparametric tests
- Goodness of fit - chi square tests
- Test of independence
- Kruskal-Wallis test, Mann-Whitney Wilcoxon test
Module 3: Observational design, correlation & regression
- Reliability and validity
- Correlation (linearity, graphs)
- Linear regression (including model assumptions, dummy variables, residual plots, outliers, least squares regression line)
Appendix B
EXAMPLES OF RESEARCH PROBLEMS
1. Is it possible to train the memory in order to remember spontaneous observable events? The researcher has eight volunteers and wants to compare two programs. The researcher randomly divides them
into two groups; one with training and one without training. A week later the researcher shows the eight volunteers a movie with a bank robbery. The volunteers are questioned about the robbery.
Those who have trained their memory had the following correctly answered items: 20, 25, 24, and 23. The other four had: 14, 22, and 18, 19. Is the program effective?
2. Examine the number of reversals for 30 students when they examine the Necker (1832) cube. In other words, the number of times a person shifts between seeing one kind of cube to see the other kind
of cube. Design an experiment and examine if there is a difference between men and women.
3. Use the data obtained from your experiment in problem 2 and compare it to the historic mean result of 16 reversals per minute (Orbach, Ehrlich, & Heath, 1963).
4. Assume that we want to examine if coffee helps students to be more alert during a lecture. Design and perform such an experiment!
5. Use two different psychology tests, which are intended to measure the same or similar concept. Distribute them among volunteers. Examine them using regression. What can we learn?
Appendix C
EXAMPLES OF PSYCHOLOGY LABS
1. Neisser (1964) conducted an experiment about recognizing the letter Z. Neisser showed that the letter Z was detected much faster when displayed in a context of rounded letters (O,S,Q) as compared
with a context of angular letters (K,E,L). Neisser showed when using a similar background to the letter Z it took twice the time to find it compared to when a dissimilar background was used.
Neisser concluded that searching a visual array is primarily concerned with distinctive features. Design a replicate of this experiment using at least 30 persons. Use valid statistical tools to
evaluate your experiment.
2. Risk perception: Estimate the intensity of the smell of iso-amylacetat and butanol. The smells are recorded using the Borg CR 10 scale developed by Borg (1982; 1998). To learn about odor
perception read e.g. Dalton (1996). Perform the experiment and draw valid conclusions.
3. Noise: Develop a noise experiment and try it on several individuals. The experimental design can be a between group design with one control and one experimental group or a comparison of different
levels of noise etc… Give the different groups some kind of psychological test (e.g. memory test, perception test etc.). Design and conduct an experiment and make relevant analysis.
4. Environmental stress: Conduct an experiment in a stressful environment and perform several tasks at the same time (e.g. listen to someone reading a text at the same time as they count backwards
and perform Raven’s (1938) test). Afterwards the individuals should fill in a survey about perceived stress. Perform this experiment, analyze the obtained data and draw valid conclusions.
5. Survey construction: Construct a survey about psychological environment, e.g. environmental behavior. Choose a sample of individuals and have them fill in your survey. Analyze your survey and
discuss reliability and validity concepts.
Appendix D
COURSE VALUATION
Volume 17, Number 2, of the Journal of Statistics Education contains a Letter to the Editor concerning this article.
Biggs, J. (1999). Teaching for Quality Learning at University. Buckingham: Open University Press.
Borg, G. (1982). Psychophysical bases of perceived exertion. Medicine & Science in Sports & Exercise. 14(5): 377-381.
Borg, G. (1998). Borg's Perceived Exertion and Pain Scales. Champaign, IL: Human Kinetics.
Carmona, J. (online) Mathematical background and attitudes toward statistics in a sample of undergraduate students. [Online ] (http://www.stat.auckland.ac.nz/~iase/publications/11/Carmona.doc)
Cobb, G. W. (1991). Teaching Statistics: More Data, Less Lecturing. Amstat News, December, 1991, pp. 1,4.
Cobb, G. W. & Moore, D. (1997). Mathematics, Statistics, and Teaching. The American Mathematical Monthly, 104, pp. 801-823.
Dalton, P. (1996). Odor perception and beliefs about risk. Chemical Senses, 21, 4, pp. 447-458.
Gal, I and Garfield, J. B. (eds.) (1997). The Assessment Challenge in Statistics Education. Amsterdam: IOS press.
Gal, I. & Ginsburg, L. (1994). The Role of Beliefs and Attitudes in Learning Statistics: Towards an Assessment Framework. Journal of Statistics Education, 2, 2 [Online] (http://jse.amstat.org/v2n2/
Hoerl, R., Hahn, G. and Doganaksoy, N. (1997). Discussion: Let’s stop Squandering Our Most Strategic Weapon, International Statistical Review, 65, 2, pp 147-153.
Hogg, R. V. (1991). Statistical Education: Improvements are Badly Needed. The American Statistician, 45, 342-343.
Heikkilä, A. & Lanka, K. (2006). Studying in Higher Education: Students’ approaches to Self-learning, Self-regulation and Cognitive Strategies. Studies in Higher Education, 31, 1, pp 99-117.
Howell, D. C. (2006). Statistical methods for psychology (6th ed). New York: Thomson learning.
Kolb, D. A. (1984). Experimental Learning: Experience as the source of learning and development. New Jersey: Prentice Hall.
Kolb, D. A. & Fry, R. (1975). Toward an Applied Theory of Experimental Learning; in C. Cooper (ed.) Theories of Group Process, London: John Wiley.
McCombs, B. & Whitler, J. S. (1997). The Learner-Centered Classroom and School: Strategies for Increasing Student Motivation and Achievement. San Francisco: Josey-Bass Publishers.
Necker, L. A. (1832). Observations on some remarkable Optical Phenomena seen in Switzerland; and on an Optical Phenomenon which occurs on viewing a Figure of a Crystal or geometrical Solid, The
London and Edinburgh Philosophical Magazine and Journal of Science 1 (1832) (5), pp. 329–337.
Neisser, U. (1964). Visual search. Scientific American, 210(6):94-102.
Orbach, J., Ehrlich, D., & Heath, H. (1963). Reversibility of the Necker cube: I. An examination of the concept of ‘‘satiation of orientation. Perceptual and Motor Skills, 17, 439–458.
Peterson, I. (1991). Pick a Sample. Science News, 140, 56-58.
Raven, J.C. (1938). Progressive matrices: A perceptual test of intelligence. London: H.K. Lewis.
Roback, P. J. (2003). Teaching an Advanced Methods Course to a Mixed Audience. Journal of Statistics Education, 11, 2, [Online] (http://jse.amstat.org/v11n2/roback.html)
Roberts, D. M. & Bilderback, E. W. (1980). "Reliability and Validity of a Statistics Attitude Survey", Educational and Psychological Measurement, 40, 235-238.
Rinaman, W. C. (1998). Revising a Basic Statistics Course. Journal of Statistics Education, 41, 3, [Online] (http://jse.amstat.org/v6n2/rinaman.html)
Romero, R. Ferrer, A., Capilla, C., Zunica, L. Balasch, S., Serra, V. Alcover, R. (1995). Teaching Statistics to Engineers: An Innovative Pedagogical Experience. Journal of Statistics Education, 41,
3, [Online] (http://jse.amstat.org/v3n1/romero.html)
Schau, C., Stevens, J., Dauphinee, T. L., & Del Vecchio, A. (1995). The development and validation of the Survey of Attitudes Toward Statistics. Educational and Psychological Measurement, 55,
Smith, G. & Colleger, P. (1998). Learning Statistics by doing Statistics. Journal of Statistics Education, 6, 3. [Online] (http://jse.amstat.org/v6n3/smith.html)
Stuart, A. (1997). Student-Centered Learning. Learning, 26, 1, pp. 53-56.
Varhoof, S, Sotos, A.E. C., Onghena, P., Verschaffel, L., van Dooren, W., & van den Noortgate, W. (2006). Attitudes Toward Statistics and Their Relationship with Short- and Long-Term Exam Results.
Journal of Statistics Education, 14, 3. [Online] (http://jse.amstat.org/v14n3/vanhoof.html)
Wise, S. L. (1985). Attitudes Toward Statistics. Educational and Psychological Measurement, 45, pp 401-405.
Marie Wiberg
Department of Statistics
Umeå University
901 87 Umeå
+ 46 90 786 95 24
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environment where scientists interact effectively. Because mathematical reasoning provides valuable insight within other fields of science, the interactions that we have in mind extend beyond
mathematics to other disciplines within the Faculty of Science (i.e., physics, computer science, chemistry and biology) as well as Geoscience, Healthcare, and Economics, all of which play important
roles in the research profile of the Mathematical Institute. Along the same lines: we strive to break through the traditional boundary between “fundamental” and “applied” mathematics. Formally, the
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from the vertex to that same base. Find the area of each of these triangles. 5th grade 6th grade 7th grade. Web apply the formula a = 1/2 * base * height; Area of a triangle #1. A is the area of the
triangle. Area of a triangle (basic) calculate the area of each of the nine triangles shown. This array of 5th grade printable worksheets on area of triangles comprises problems in three different
formats, with integer dimensions offered in two levels.
Web to find the area of any triangle we require two measures: Area of a triangle #1. Find the area of each of these triangles. Web grade 5 math worksheets on finding the area of triangles, including
triangles which are not right triangles. Calculate the areas of triangles on these printable worksheets.
Xyz with yz = 9mm, xy = 13mm and angle xyz = 121°. Web area of triangles worksheets. Web the corbettmaths practice questions on the area of a triangle using sine. Finding the area of a triangle.
This array of 5th grade printable worksheets on area of triangles comprises problems in three different formats, with integer dimensions offered in two levels. The area of the triangle is 20cm2, nind
x. H, the height is the perpendicular distance from the base to the opposite vertex.
You may have to remind pupils how to measure the area of a triangle (½ base x height) » as above » worksheet Free pdf worksheets from k5 learning's online reading and math program. Show the pupils
area c and d and ask them how they might measure the area of these shapes.
Nevertheless, It Sparks A Nerdy Satisfaction.
Web to find the area of a triangle, you can use the following formula: Find the area of each triangle. Area of a triangle #1. They should know about breaking the shape into two;
Web Apply The Formula A = 1/2 * Base * Height;
(total for question 6 is 3 marks) the area of the triangle is 100m2 work out the value of x. A= x 7 cm x 4 cm1 2. Multiply the base and height and divide by two, to calculate the area. Web the area
of the triangle is 70cm2 work out the value of x.
These Sheets Are Graded From Easiest To Hardest, And Each Sheet Comes Complete With Answers.
The area of the triangle is 20cm2, nind x. This array of 5th grade printable worksheets on area of triangles comprises problems in three different formats, with integer dimensions offered in two
levels. A=\frac {1} {2}bh a = 21bh. 7 diverse types to calculate area of a triangle.
Focusing On Finding The Area Of Triangles, This Set Of Worksheets Features Triangles Whose Dimensions Are Given As Integers, Decimals And Fractions Involving Conversion To Specified Units As Well.
The worksheet includes worked examples to support and great range of questions and answers. Finding the area of a triangle. Area of a triangle (basic) calculate the area of each of the nine triangles
shown. 5th grade 6th grade 7th grade. | {"url":"https://cosicova.org/eng/finding-the-area-of-a-triangle-worksheet.html","timestamp":"2024-11-04T10:46:13Z","content_type":"text/html","content_length":"26527","record_id":"<urn:uuid:1cb4641d-1e01-4aee-8c47-5252e5c001f8>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00309.warc.gz"} |
Write the equilibrium equation
Write the equilibrium equations on which the following \(K_{\mathrm{sp}}\) expressions are based. (a) \(\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}\) (b) \(\left[\mathrm{Pb}^
{2+}\right]\left[\mathrm{CrO}_{4}^{2-}\right]\) (c) \(\left[\mathrm{Mn}^{4+}\right]\left[\mathrm{O}^{2-}\right]^{2}\) (d) \(\left[\mathrm{Al}^{3+}\right]^{2}\left[\mathrm{~S}^{2-}\right]^{3}\)
Short Answer
Expert verified
Based on the provided step-by-step solution, write a short answer explaining the process and the results: To find the equilibrium equations for various solubility products (Ksp), we first need to
identify the dissociation equations for each ionic compound in water. We then derive the corresponding equilibrium equation using the given Ksp expression. The process involves determining the
dissociated ions and their corresponding coefficients in the equilibrium equation. By following this method, we identified the equilibrium equations for Hg2(Cl)2, PbCrO4, MnO2, and Al2S3.
Step by step solution
(a) Write the dissociation equation for Hg2(Cl)2
In order to find the equilibrium equation, we will first write the dissociation equation for Hg2(Cl)2. The dissociation equation for Hg2(Cl)2 in water is: Hg2(Cl)2(s) <=> Hg2^2+(aq) + 2 Cl^-(aq)
(a) Write the equilibrium equation for Hg2(Cl)2 based on given Ksp expression
By utilizing the given Ksp expression, we can derive its corresponding equilibrium equation as follows: Ksp = [Hg2^2+] [Cl^-]^2
(b) Write the dissociation equation for PbCrO4
Now we will write the dissociation equation for PbCrO4. The dissociation equation for PbCrO4 in water is: PbCrO4(s) <=> Pb^2+(aq) + CrO4^2-(aq)
(b) Write the equilibrium equation for PbCrO4 based on given Ksp expression
By utilizing the given Ksp expression, we can derive its corresponding equilibrium equation as follows: Ksp = [Pb^2+] [CrO4^2-]
(c) Write the dissociation equation for MnO2
We will write the dissociation equation for MnO2. The dissociation equation for MnO2 in water is: MnO2(s) <=> Mn^4+(aq) + 2 O^2-(aq)
(c) Write the equilibrium equation for MnO2 based on given Ksp expression
By utilizing the given Ksp expression, we can derive its corresponding equilibrium equation as follows: Ksp = [Mn^4+] [O^2-]^2
(d) Write the dissociation equation for Al2S3
We will now write the dissociation equation for Al2S3. The dissociation equation for Al2S3 in water is: Al2S3(s) <=> 2 Al^3+(aq) + 3 S^2-(aq)
(d) Write the equilibrium equation for Al2S3 based on given Ksp expression
By utilizing the given Ksp expression, we can derive its corresponding equilibrium equation as follows: Ksp = [Al^3+]^2 [S^2-]^3
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ksp Expressions
The abbreviation **Ksp** stands for the solubility product constant. This is a special type of equilibrium constant unique to sparingly soluble salts. It represents the product of the concentrations
of the ions in solution at equilibrium, each raised to the power of their coefficients in the dissociation equation. Ksp expressions are crucial in predicting how much of a compound can dissolve in
water.For example, in the Ksp expression for **Hg2(Cl)2**: \[K_{sp} = [\text{Hg}_2^{2+}][\text{Cl}^-]^2\]at equilibrium, the concentration of **Hg2^2+** and squared concentration of **Cl^-** are
multiplied to give the Ksp value. By knowing Ksp values, you can determine the point at which a solution becomes saturated and begins to precipitate. This helps chemists gauge the solubility of
compounds in different conditions.
Dissociation Equations
**Dissociation equations** represent the process where a solid ionic compound dissolves in water to form its ions. This is a chemical reaction where bonds in the solid break, freeing ions into the
solution.For **PbCrO4**, the dissociation equation is: \[\text{PbCrO}_4(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{CrO}_4^{2-}(aq)\]This shows that solid lead chromate dissociates into lead
ions and chromate ions when dissolved in water.Understanding dissociation is key to writing equilibrium equations, which relate closely to Ksp expressions, and are used to understand how compounds
behave in aqueous solutions.The dissociation process impacts not just solubility, but also the ionic strength and electrical conductivity of the solution.
Equilibrium Constants
The **equilibrium constant** is a value that expresses the ratio of concentrations of products to reactants at equilibrium. For solubility, these are the Ksp values. They are dimensionless numbers
which provide insight into the extent of a reaction.Each reaction has a unique equilibrium constant that is determined experimentally. For the dissociation of **MnO2**, the equilibrium equation is:\
[K_{sp} = [\text{Mn}^{4+}][\text{O}^{2-}]^2\]The equilibrium constant tells you how far the reaction proceeds before equilibrium is reached.Larger Ksp values mean higher solubility, whereas smaller
values indicate limited solubility. By comparing Ksp, one can predict which of different compounds is more likely to remain dissolved in solution.
Chemical Solubility
**Chemical solubility** is the extent to which a substance can dissolve in a solvent to form a homogeneous solution.It is directly linked to Ksp because the solubility product constant helps predict
how much solute will dissolve at a given temperature.For instance, in the case of **Al2S3**:\[\text{Al}_2\text{S}_3(s) \rightleftharpoons 2 \text{Al}^{3+}(aq) + 3 \text{S}^{2-}(aq)\]The Ksp value
calculates the maximum amount of dissolved ions in a saturated solution.Various factors affect solubility, such as temperature, pressure, and the presence of other ions in solution. This knowledge is
used in fields like pharmacology, environmental science, and many industries where dissolving and precipitating substances are critical. | {"url":"https://www.vaia.com/en-us/textbooks/chemistry/chemistry-principles-and-reactions-8-edition/chapter-15/problem-7-write-the-equilibrium-equations-on-which-the-follo/","timestamp":"2024-11-12T00:11:03Z","content_type":"text/html","content_length":"266451","record_id":"<urn:uuid:3ad733a0-6adf-4dc1-9fee-931b0003be7b>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00895.warc.gz"} |
Binary Search In Python - CopyAssignment
In the previous post, the binary search in python was discussed but not in enough detail. Therefore, in this post, the binary search in python shall be discussed in depth.
Binary search in python involves looking for a specific element in a list or an array. It is referred to as a binary search because it involves checking if the middle element of the array is the
element being looked for. If not, then the list is split into two. We will take one of the new subarrays created as an array and do to it what we previously did to the original array. This is dome
until we find the intended element.
NOTE: The list to be searched must be sorted.
Recursive Binary Search in Python
In this method, some key concepts need to be understood.
First, the length of the array will keep changing, and to keep track of this, we use the beginning point of the respective array. This simply means that whenever we have an array, we take the middle
position with respect to the beginning value and the length of the array. When we split an array to get a subarray, we get a new beginning position and a new middle position.
The function we are going to define will take four parameters. The first the list that acts as a dictionary to Binary Search from. The second and the third are the beginning position and the length
of the array in question. These two will be used to get the middle value. The algorithm will involve function recalling hence we cannot hardcode these two since they need to be dynamic. If we hard
code them, every time we recall the function, their values will be the same and we will get the same middle value making the function be stuck in a never-ending loop.
With all this key information, let us now look at the first Binary Search in Python algorithm.
# Method one
def binarySearch (values, l, r, element):
# l is the starting reference to find the middle.
# It will keep changing time we split the array to get subarrays.
# Hence it is not hard coded into the algorithm.
# It won't make much importance to the user since the user needs to know the first middle point is in reference with the first index which is zero.
# Therefore when calling the function, l will be 0.
# R is the lenght of the list.
# Python's len() function gives the accurate lenght but indexing starts at one value less.
# Hence the lenght will be one value less.
# Check if the list has not been exhausted.
if r >= l:
# We are checking i
mid = l + (r - l) // 2
# We start at the middle of the list and check if the element
if values[mid] == element:
return "Element " + str(element) + " is at " + str(mid + 1)
# We check if the element is on the left side of the split array
elif values[mid] > element:
return binarySearch(values, l, mid-1, element)
# Otherwise, the element is in the right side of the split array.
return binarySearch(values, mid + 1, r, element)
# If we fail to find the element in the list we return an absent statement.
# Element is not present in the array
return "Element " + str(element) + " is not in the list"
Loop Binary Search in Python
This method is a little simpler as compared to the first one but is quite complex in its calculation.
Unlike the previous Binary Search, we do not have any recursive function. This means we do not recall the function and hence we can hard code some values.
In place of the recursive function, we implement a loop that will increment these values. A while loop is used since we are not iterating over the list. If we were iterating over the list then the
binary search will change to a linear search since we are not splitting the list as well.
The while loop will run as long as the subarrays have not been exhausted. Let us look at the second Binary Search in Python algorithm
def binarySearch2(values, element):
# The function takes two parameters.
# Values is the list that acts as a dictionary to search the element from.
# The element is the value being searched for.
start = 0
# This is the beginning value of the array in question.
# The value will change depending on the array.
length = len(values)-1
# This is the length of the array in question.
# This number will vary on whether the list is split or not.
while start <= length and element >= values[start] and element <=values[length]:
#This means that the search loop will run as long as these three conditions are met.
# The first is as long as the list has not been exhausted. As long as the start is smaller than or equal to the lenght
# The second is that the element is larger than the start value on the list.
# The last one is that the element is smaller than the last element on the list.
# Find the middle point.
mid = start + int(((float(length-start)/(values[length]-values[start]))*(element-values[start])))
# This calculation simply adds the previous mid to an incremented value.
# Compare the middle value on the list to the element being searched for.
if values[mid] == element:
# If we find the element, we print out its value position on the list.
return "Element " + str(element) + " is at " + str(mid + 1)
if values[mid] < element:
#If we fail to get the element in this current loop, we increment and move to the next loop.
start = mid + 1
#If we fail to get the number completely, we print out an absent statement.
return "Element " + str(element) + " has not been found."
Thank you for reading through my post. Please leave a comment or any query below.
You can check out other posts below:
Sorting and Searching in Python
Download chrome driver using Selenium
Guess the number game in Python | {"url":"https://copyassignment.com/binary-search-in-python-program/","timestamp":"2024-11-04T04:15:25Z","content_type":"text/html","content_length":"69656","record_id":"<urn:uuid:ff60de1d-a586-4175-91af-7cdc80fba784>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00403.warc.gz"} |
Example i.i.d. model
Note that you can also use the standard way of specifying priors component-wise on individual variance components, we show this below.
Prior options
Before going into the details on how to use the software, we give you a short introduction to the HD prior. We also refer to Fuglstad et al. (2020) for details.
We use the penalized complexity (PC) prior (Simpson et al. 2017) to induce shrinkage. This can make a robust prior that stabilizes the inference. We do not go into details on the PC prior here, but
the following priors are available in makemyprior:
Consider a random intercept model \(y_{i,j} = a_i + \varepsilon_{i,j}\) for \(i,j = 1, \dots, 10\), where \(a_i \overset{\text{iid}}{\sim} N(0, \sigma_{\mathrm{a}}^2)\) is a group effect and \(\
varepsilon_i \overset{\text{iid}}{\sim} N(0, \sigma_{\varepsilon}^2)\) is a residual effect. We define the variance proportion \(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} = \frac{\sigma_{\
mathrm{a}}^2}{\sigma_{\mathrm{a}}^2 + \sigma_{\varepsilon}^2}\). Then we denote the different PC prior distributions as: * \(\sigma_{\mathrm{*}} \sim \mathrm{PC}_{\mathrm{0}}(U, \alpha)\), with \(\
mathrm{Prob}(\sigma_{\mathrm{*}} > U) = \alpha\), and shrinkage towards \(\sigma_{\mathrm{*}} = 0\). * \(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} \sim \mathrm{PC}_{\mathrm{0}}(m)\) with \(\
mathrm{Prob}(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} > m) = 0.5\) so that \(m\) defines the median, and shrinkage towards \(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} = 0\), i.e.,
the base model is a model with only \(\pmb{\varepsilon}\). * \(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} \sim \mathrm{PC}_{\mathrm{1}}(m)\) with \(\mathrm{Prob}(\omega_{\frac{\mathrm{a}}{\
mathrm{a+\varepsilon}}} > m) = 0.5\) so that \(m\) defines the median, and shrinkage towards \(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} = 1\), i.e., the base model is a model with only \(\
pmb{a}\). * \(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} \sim \mathrm{PC}_{\mathrm{M}}(m, c)\) with \(\mathrm{Prob}(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} > m) = 0.5\) and \(\
mathrm{Prob}(\mathrm{logit}(1/4) < \mathrm{logit}(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}}) - \mathrm{logit}(m) < \mathrm{logit}(3/4)) = c\) so that \(m\) defines the median, and \(c\) says
something about how concentrated the distribution is around the median. The shrinkage is towards \(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}} = m\), i.e., the base model is a combination of
the effects \(\pmb{a}\) and \(\pmb{\varepsilon}\).
Note that \(\mathrm{PC}_{\mathrm{1}}(m)\) on \(\omega_{\frac{\mathrm{a}}{\mathrm{a+\varepsilon}}}\) is equivalent to \(\mathrm{PC}_{\mathrm{0}}(1-m)\) on \(1-\omega_{\frac{\mathrm{a}}{\mathrm{a+\
varepsilon}}} = \omega_{\frac{\mathrm{\varepsilon}}{\mathrm{a+\varepsilon}}}\).
The priors listed above are denoted pc, pc0. pc1, and pcM in makemyprior.
Model description
Consider the hierarchical model for the \(n = m \cdot p\) observations \(y_{i,j}\), \(i = 1, \ldots p\) and \(j = 1, \ldots, m\), given by \[\begin{align*} y_{i,j}|\eta_{i,j}, \sigma_{\varepsilon}^2
&\sim N(\eta_{i,j}, \sigma_{\varepsilon}^2), \\ \eta_{i,j} &= \mu + x_i \beta + a_i + b_j, \end{align*}\] where \(\mu\) is an intercept, \(x_i\) is a covariate with coefficient \(\beta\), and \(a_1,
a_2, \ldots, a_p \overset{\text{iid}}{\sim} N(0, \sigma_\mathrm{a}^2)\) and \(b_1, b_2, \ldots, b_m \overset{\text{iid}}{\sim} N(0, \sigma_\mathrm{b}^2)\) are random effects. The residuals \(\
varepsilon_1, \varepsilon_2, \dots, \varepsilon_n \sim N(0, \sigma_{\varepsilon}^2)\).
Make data and linear predictor
First we specify our model by making a formula object (see ?mc):
Then we put our data in a list (a data.frame can also be used). We simulate the data here.
Make prior
Then we make the prior object using the function make_prior. It needs the arguments we made above, formula and data, a likelihood family (Gaussian likelihood is the default), and optional priors for
intercept and covariate coefficients (both have a Gaussian distribution with \(0\) mean and a standard deviation of \(1000\)). Note that the observations y are not used to create the prior, but is
included in the prior object as all the information about the inference is stored there.
prior <- make_prior(formula, data, family = "gaussian",
intercept_prior = c(0, 1000),
covariate_prior = list(x = c(0, 100)))
#> Warning: Did not find a tree, using default tree structure instead.
This gives the default prior, which is a prior where all model effects are assigned an equal amount of variance through a symmetric Dirichlet distribution. The default prior on the total variance
depends on the likelihood. See Section default settings for details on default settings.
We print details about the prior, plot the prior to see how the distributions look, and plot the prior tree structure:
#> Model: y ~ x + mc(a) + mc(b)
#> Tree structure: a_b_eps = (a,b,eps)
#> Weight priors:
#> (w[a/a_b_eps], w[b/a_b_eps]) ~ Dirichlet(3)
#> Total variance priors:
#> V[a_b_eps] ~ Jeffreys'
#> Covariate priors: intercept ~ N(0, 1000^2), x ~ N(0, 100^2)
plot_prior(prior) # or plot(prior)
Now we can use a graphical interface to choose our prior. We do not show this in the vignette, but it can be opened with the following command:
The output (which we store in new_prior) is of the same class as the output from make_prior, and can be used directly for inference.
With the following command, we specify this prior:
\[$$\omega_{\frac{\mathrm{a}}{\mathrm{a+b}}} \sim \mathrm{PC}_{\mathrm{M}}(0.7, 0.5),\, \omega_{\frac{\mathrm{a+b}}{\mathrm{a+b} + \varepsilon}} \sim \mathrm{PC}_{\mathrm{0}}(0.25),\,\text{and}\, \
sigma_{\mathrm{a+b} + \varepsilon} \sim \mathrm{PC}_{\mathrm{0}}(3, 0.05). \label{eq:software:examplemodel_prior}$$\]
new_prior <- make_prior(
formula, data,
prior = list(
tree = "s1 = (a, b); s2 = (s1, eps)",
w = list(s1 = list(prior = "pcM", param = c(0.7, 0.5)),
s2 = list(prior = "pc1", param = 0.75)),
V = list(s2 = list(prior = "pc0", param = c(3, 0.05)))
covariate_prior = list(x = c(0, 100))
#> Model: y ~ x + mc(a) + mc(b)
#> Tree structure: a_b = (a,b); eps_a_b = (eps,a_b)
#> Weight priors:
#> w[a/a_b] ~ PCM(0.7, 0.5)
#> w[eps/eps_a_b] ~ PC0(0.25)
#> Total variance priors:
#> sqrt(V)[eps_a_b] ~ PC0(3, 0.05)
#> Covariate priors: intercept ~ N(0, 1000^2), x ~ N(0, 100^2)
We can carry out inference with Stan (Carpenter et al. 2017) and INLA (Rue, Martino, and Chopin 2009). Note that we in this vignette do not run the inference, as it takes time and will slow down the
compilation of the vignette and thus the package download, but the code is included below and the user can carry out the inference with that.
First, we look at inference with Stan. We must start by compiling the Stan-code:
Then we can do the inference:
We can look at the graphs of the posterior:
plot_posterior_stan(posterior1, param = "prior", prior = TRUE) # on the scale of the prior, together with the prior
plot_posterior_stan(posterior1, param = "variance") # on variance scale
plot_fixed_posterior(posterior1) # fixed effects
We can also sample from the prior and compare on variance scale:
Inference with INLA is carried out in a similar way:
And we can look at some posterior diagnostics. Note that we can only look at the posteriors on variance/precision/standard deviation scale when doing inference with INLA.
See vignette("plotting", package = "makemyprior") for more details on functions for plotting.
Default settings
• If no prior is specified (neither tree structure nor priors), the prior will be a joint prior where all latent components (including a possible residual effect) get an equal amount of variance in
the prior.
• The prior on the total variance (top nodes) varies with likelihood:
□ Jeffreys’ prior for Gaussian likelihood for a tree structure with one tree, \(\mathrm{PC}_{\mathrm{0}}(3, 0.05)\) otherwise.
□ \(\mathrm{PC}_{\mathrm{0}}(1.6, 0.05)\) for binomial likelihood.
□ \(\mathrm{PC}_{\mathrm{0}}(1.6, 0.05)\) for Poisson likelihood.
• The default prior on individual variance (singletons) varies with likelihood:
□ \(\mathrm{PC}_{\mathrm{0}}(3, 0.05)\) for Gaussian likelihood.
□ \(\mathrm{PC}_{\mathrm{0}}(1.6, 0.05)\) for binomial likelihood.
□ \(\mathrm{PC}_{\mathrm{0}}(1.6, 0.05)\) for Poisson likelihood.
• The default prior on a variance proportion (split node) is a Dirichlet prior assigning equal amount of variance to each of the model components involved in the split.
for details on default settings.
Additional examples
We include some additional examples on how to create various prior distributions. We still use the same model and data, and change the joint prior on the variances. We do not run inference. Note that
the values of the priors are NOT based on knowledge about the model, but chosen to show the different options of the package. See vignette("wheat_breeding", package = "makemyprior"), vignette
("latin_square", package = "makemyprior"), and vignette("neonatal_mortality", package = "makemyprior") for examples where we discuss how expert knowledge can be used to set the priors.
prior2 <- make_prior(formula = formula, data = data,
prior = list(tree = "(a); (b); (eps)",
V = list(
a = list(prior = "pc", param = c(1, 0.05)),
b = list(prior = "pc", param = c(2, 0.05)),
eps = list(prior = "pc", param = c(3, 0.05))
prior3 <- make_prior(formula = formula, data = data,
prior = list(tree = "s1 = (a, b); (eps)",
V = list(
s1 = list(prior = "pc", param = c(3, 0.05)),
eps = list(prior = "pc", param = c(3, 0.05))),
w = list(
s1 = list(prior = "pcM", param = c(0.5, 0.8))
#> R version 4.3.2 (2023-10-31)
#> Platform: aarch64-apple-darwin20 (64-bit)
#> Running under: macOS Sonoma 14.1.1
#> Matrix products: default
#> BLAS: /Library/Frameworks/R.framework/Versions/4.3-arm64/Resources/lib/libRblas.0.dylib
#> LAPACK: /Library/Frameworks/R.framework/Versions/4.3-arm64/Resources/lib/libRlapack.dylib; LAPACK version 3.11.0
#> locale:
#> [1] C/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
#> time zone: Europe/Oslo
#> tzcode source: internal
#> attached base packages:
#> [1] stats graphics grDevices utils datasets methods base
#> other attached packages:
#> [1] makemyprior_1.2.2
#> loaded via a namespace (and not attached):
#> [1] Matrix_1.6-1.1 gtable_0.3.4 jsonlite_1.8.8 highr_0.10
#> [5] dplyr_1.1.4 compiler_4.3.2 promises_1.2.1 tidyselect_1.2.0
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#> [13] scales_1.3.0 yaml_2.3.8 fastmap_1.1.1 mime_0.12
#> [17] lattice_0.21-9 ggplot2_3.4.4 R6_2.5.1 labeling_0.4.3
#> [21] shinyjs_2.1.0 generics_0.1.3 knitr_1.45 htmlwidgets_1.6.4
#> [25] visNetwork_2.1.2 MASS_7.3-60 tibble_3.2.1 munsell_0.5.0
#> [29] shiny_1.8.0 bslib_0.6.1 pillar_1.9.0 rlang_1.1.3
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#> [37] sass_0.4.8 cli_3.6.2 withr_3.0.0 magrittr_2.0.3 | {"url":"https://cran.wustl.edu/web/packages/makemyprior/vignettes/make_prior.html","timestamp":"2024-11-05T10:24:00Z","content_type":"text/html","content_length":"1048959","record_id":"<urn:uuid:1e9e3b40-49ba-4dca-ad11-396c77b6bde1>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00708.warc.gz"} |
One vs Rest
In the previous segment, you learnt about the One vs One classifier method. Now, in this segment, you will learn about the One vs Rest classification technique in detail. Similar to One vs One, this
technique involves three steps, each with its own function. In the coming video, Ankit will provide a detailed explanation of all three steps.
As explained by Ankit, the One vs Rest technique involves three steps.
• Data sets creation
• Model training
• Model prediction
In the first step, data set creation, as explained in the video, multiple data sets are created, which have these features.
• All the data points will be present in each data set.
• Each created data set will have two target variables only.
In the video, Ankit took a data set having 1,000 rows, which has ‘n’ target variables. The target variables are represented in different colours. As shown in the image below, multiple data sets are
created, each with two target variables only, i.e., they are represented as 1 and 0. 1 represents a particular colour point and the rest are marked as 0. It is important to note that unlike the One
vs One technique, where each subset has fewer than 10,000 rows, in this technique, all the data sets will have 10,000 rows.
In the second step, which is model training, for each data set, a model is built and trained. Since the data set has ‘n’ target variables, we will create ‘n’ data sets in total. We will, thus, build
n models (one model for each data set). In the image below, you can see that each yellow box represents a model built on a particular data set.
Similar to the One vs One method, it is important to note that the same algorithm is to be used for all the classifiers. If you are using logistic regression for one model, then you will have to use
logistic regression for all the models. You will learn about the implementation/application part in the segment on Python coding.
In the third step, which is model prediction, test samples are passed to each of these ‘n’ models, and the models predict/classify the test samples accordingly. In the image below, you can see that a
test sample represented with green is passed to all the n models. For each model, the output is a probability score for that respective target variable (represented as 1).
For example, let us consider the model built on dataset1. The model will give a probability score for the test sample to be classified under the blue class. Similarly, the model built on dataset2
will give a probability score for the test sample to be classified under the red class.
After all the n models have performed all the classification, the test sample gets classified under the target variable that has the highest probability.
In the end, Ankit summarises the One vs Rest method diagrammatically. The image below summarises all the three steps executed in the One vs Rest method in a single slide. You can see that Step 1
indicates data set creation, Step 2 indicates model training and the final step indicates model prediction.
The image below is quite useful if you want to revise the concept as it explains the concept both diagrammatically and conceptually. You can see that in Step 3 (model prediction), the models classify
the points based on the data sets on which they are trained.
For example, classifier 1 in the image gives the probability score for the test sample to be classified under the class 1 category. It is similar for the other data sets as well. In the final box,
you can see the probability table, where for each class, there is a specific probability score. The test sample gets classified under the target variable with the maximum probability score as written
in the box. | {"url":"https://www.internetknowledgehub.com/one-vs-rest/","timestamp":"2024-11-09T22:55:06Z","content_type":"text/html","content_length":"81592","record_id":"<urn:uuid:f8673db1-1548-449a-a075-f7a2cf061a0c>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00299.warc.gz"} |
A bit confused about time as the fourth dimension
Hi. I’m under the impression that w/ relativity “space-time” is four space. And four space means that every axis is orthogonal from all the rest. But, if I travel really quickly, the flow of time
changes for me relative to others’ frames of reference. But, doesn’t that imply that the two aren’t really orthogonal?
From the traveler’s frame of reference, the flow of time doesn’t change, so they are orthogonal (I guess). But it’s that it is from another frame of reference that the change in time is perceived —
in the same way that the length of a moving object shortens relative to an observer in a different frame of reference.
So, that should clear up my confusion, right? To the person moving, traveling doesn’t affect the flow of time, just like moving sideways doesn’t move one forward or back. That’s why we can think of
space-time as four dimensions.
Is this roughly correct?
More or less. An even better analogy is to consider rotated coordinate systems on a plane. I can set down a set of coordinates and say that “north is this way and east is this way”. You can sit in
the same place and set down another set of coordinates that don’t necessarily coincide with mine, even if your north is at right angles to your east. If I then say that a particular tree is due east
of us by 100 yards, you won’t agree to that it’s due north, since your “east” is different from mine.
Now replace “north” with “time-direction” and “east” with “space-direction” in the above paragraph, and that’s roughly how special relativity works.
The way I heard it is, the 4th dimensional axis isn’t merely time (t), it’s the speed of light multiplied by time (ct).
The thinking is that you’re always moving exactly at the speed of light © relative to everything else. If you’re “stationary”, you’re actually travelling at c in the ct direction. If you’re moving in
the x direction, you’re actually travelling at c on a vector somewhere between the x direction and the ct direction. Since the ct component of your vector is shorter than c when you’re moving in any
of the x-y-z coordinates, you’re actually moving more slowly through time, which is why time dilation occurs.
I’m not really certain how that whole shortening-in-the-direction-of-motion thing fits in with this model, however.
Yeah, I knew that the changing coordinates was involved, in this case it’s not a rotation. I think what screwed me up was confusing the observer w/ the actor. Thanks for the help!
There is an excellent discussion of of this notion in Brian Greene’s The Fabric of the Cosmos. I don’t have it handy so I can’t give you the page number. But I have been reading this stuff as a
layman for many years and Greene’s description was the first time I saw this concept described, and what a great job of explaining it.
Think of the fourth dimension as this:
Suppose I told you i would meet you in the Empire State Building. With the given information, you can narrow the location of our meeting to an x,y coordinate. But you realize now that the Empire
State Building consists of 102 floors. You now need a z coordinate to find our meeting place. The room number and the exact floor of the Empire State Building now gives you the precise location. But
I never told you what time I was going to meet you there. That is how time is the fourth dimension. A bit less confusing I hope.
For those of you who presume that I stole this, you are right. I took it from a Science Channel special on physics. Can’t remember the episode or anything.
Oooh! Oooh! Ordering this right now, thanks!
Yes, that is a pretty clear explanation. What may also help to grasp the concept is an understanding that time is used to measure movement relative to other movement. If you compare it to the example
above, time doesn’t become a factor if you factor out one of the two people meeting.
Just as you cannot detect movement if you only have one object, you cannot measure movement relative to other movement if you only have one object moving. This is why any time measuring device always
adds some form of movement to the movement that is being measured.
Actually, a number of years ago I was researching this very topic, and came across something (lord knows where at this date) that stated explicitly that Einstein’s relativity did away with the notion
of time as a fourth dimension.
It puzzled me at the time, and for all I know a serious physicist might regard it as wrong, but I think the reasoning was that to regard time as a fourth dimension, you need to think of it in the
same terms as the three space dimensions. That is, two one-inch measures, brought together, will match.
However, due to relativistic time dilation for different bodies under different velocities, the same can not be said. If you compare the measure of one second in both time frames, they will not be
Thus time can not be seen as a fourth dimension in the same way as space dimensions.
This is how I recall the argument going, but it’s been years and my thoughts on it are hazy.
Two different things are going on here: “covariance” and “Lorentz metrics”
First of all, when you bring two one-inch measures together, you’re also implicitly rotating them in space to make them point the same direction, and rotating them in spacetime to make them both
travel at the same velocity. Relativity states how things transform as you change your point of view – how they vary together, thus “covariance”. In fact, what you’ve stated is almost exactly
backwards: covariance means that we can compare two lengths by putting them in the same frame of reference.
To look at it another way, if the argument about time we correct, then by relativistic spatial transformations the same would go for spatial measurements.
Now, if we ask how intervals (spatial or temporal) are actually measured, we work in analogy from 3-d space: the Pythagorean theorem. If we pick a rectangular coordinate system, then points (x[sub]1
[/sub],y[sub]1[/sub],z[sub]1[/sub]) and (x[sub]2[/sub],y[sub]2[/sub],z[sub]2[/sub]) are separated by a length whose square is
(x[sub]1[/sub]-x[sub]2[/sub])[sup]2[/sup] + (y[sub]1[/sub]-y[sub]2[/sub])[sup]2[/sup] + (z[sub]1[/sub]-z[sub]2[/sub])[sup]2[/sup]
When we add time, though, it turns out that the proper formula for the square of the interval between (x[sub]1[/sub],y[sub]1[/sub],z[sub]1[/sub],t[sub]1[/sub]) and (x[sub]2[/sub],y[sub]2[/sub],z[sub]
2[/sub],t[sub]2[/sub]) is
(x[sub]1[/sub]-x[sub]2[/sub])[sup]2[/sup] + (y[sub]1[/sub]-y[sub]2[/sub])[sup]2[/sup] + (z[sub]1[/sub]-z[sub]2[/sub])[sup]2[/sup] - c[sup]2/sup[sup]2[/sup]
The c[sup]2[/sup] is a conversion factor – nothing more, nothing less. Imagine if measurements in the x-direction were in meters and those in the y-direction were in kilometers. Then to get a
distance-squared, we’d have to multiply the displacement in the y-direction by 1/1000 (and then square that) to convert to meters. Similarly, t is measured in seconds, so we have to multiply by c (in
meters per second) to convert to meters.
That minus sign, though, is a different matter. That means that the metric (way of measuring distances) is “Lorentz”, which is basically saying there’s one sign in it that’s different from the
others. That means that temporal displacements are different than spatial ones in that the squares of their intervals are negative rather than positive. Still, as far as the mathematics goes, the two
are united in a single 4-d framework.
I cerftainly would never argue with you on a point of mathematics.
My post was an attempt to illustrate the thrust of what I was remembering. Transposition actually played no part in the argument as I recall it. I was merely trying to reconstruct my own thinking on
how it made sense.
The main gist was that because of the relativistic effects that bring about phenomena like the “twin paradox”, we must understand time as a purely local phenomenon. As it varies with frame of
reference, it can not be used to understand the universe globally in the same way the three spatial dimensions can, and thus does not qualify as a dimension.
I myself actually prefer to think of time as a fourth dimesion, which greatly weakens my ability to argue the other side, but I thought I was duty-bound to at least bring it up.
I like to think, for example, that if space were two dimensions instead of three, and we had a square that grew from a point to a certain size at a constant rate, then abruptly shrunk again at the
same rate, it would form an octahedron in 3-dimensional space-time. We could think of ourselves as we change over time as a 3-dimensional cross-section of some 4-dimenstional shape.
The problem with this is that the twin paradox is inherently a GR problem. For the twins to come together again, at least one must accelerate, which removes it from the SR realm. At that point, the
methods of measuring spacetime intervals I was describing before become methods of measuring lengths of tangent vectors to a curved spacetime. In that sense, then, all measurements must be done
locally, and things may depend on how we move two objects together to be compared. It’s not time-dilation that causes the problem, it’s something differential geometers call “parallel transport” or
“the Levi-Civita connection”. Even so, spacetime still is locally composed of four dimensions – three of space and one of time.
If you have this book in hardcover, the discussion is from pp. 47-50. If you have a different edition than mine, you can check the index under
speed of light:
combined motion through space and time and
It is not mathematical but conceptual. Good for the layperson. | {"url":"https://boards.straightdope.com/t/a-bit-confused-about-time-as-the-fourth-dimension/327301","timestamp":"2024-11-03T00:32:29Z","content_type":"text/html","content_length":"57707","record_id":"<urn:uuid:03fbe16f-801c-4204-b9a0-78cfe5eb6734>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00433.warc.gz"} |
Understanding Sabermetrics by Gabriel B. Costa, Michael R. Huber, John T. Saccoma Read Free Book Online
Let’s start with Ruth. From the table above, we see that the Babe had 8398 AB and 2062 BB for a total of 10,460 PA. Hence, if x is the number of additional AB, the following proportion preserves the
AB to PA ratio:
To solve this equation for x , we merely “cross multiply” and isolate the unknown quantity to obtain x = 1638. So Ruth would get an additional 1638 AB. This implies that he would also receive an
additional 402 BB, because 8398 AB + 1638 (additional AB) + 2062 BB + 402 (additional BB) = 12,500 PA.
So, if Ruth was just as good as he always was for these extra 1638 AB, then his projected HR total would be:
We note that the term
is nothing more than a prorating of the 714 statistic. But we are assuming that Ruth would be 5 percent less the hitter he was during the rest of his career. Therefore, the
term should be multiplied by 0.95, giving the true projected HR figure as:
Note that the left-hand side of this last equation has a “714” in both terms. If we factor out the 714 (and recall that the number “1” is always an understood coefficient of any term), we see that
In other words, if we multiply 714 by the coefficient 1.185, we get the projected cumulative HR total for Babe Ruth.
We call this the equivalence coefficient , because it gives us a “reasonable” estimate of the desired cumulative HR total, given our defined “equivalent” scenario.
With regard to Ted Williams, his additional AB compute to 2197, while his additional BB come to 576. Using the 5 percent better and 10 percent better assumptions (giving kickers of 1.05 and 1.10,
respectively), we find that the equivalence coefficients for Williams are 1.299 and 1.314, respectively. So, a 5 percent better Williams would hit 521(1.299) = 677 HR, while a 10 percent better
Williams would project to 521(1.314) = 685 HR.
We summarize the technique of computing the EC in Figure 5.1.
Figure 5.1 Computing the equivalence coefficient for batting
We see that the equivalence coefficient has enabled us to compile the entries in the following table, perhaps shedding some light to answer the questions we posed above:
• What would Williams’ totals be if he had not lost so much time?
• What if Ruth had not started out as a pitcher?
• Who was the greater hitter: Williams or Ruth?
Table 5.2 Williams versus Ruth using the equivalence coefficient
So, who was the greater hitter?
Some remarks are in order regarding this approach. First, the EC can be regarded as a mathematical model. As with most models, it can be tweaked. For example, we assumed that Ted Williams was
“equally better” during the years he missed in both the 1940s and in the 1950s. We could have assumed that he was 10 percent better in the 1940s and 5 percent better in the 1950s. Clearly, this would
have yielded different projections and made our model a bit more complicated (see the Hard Slider problem at the end of the chapter).
We also assumed that the proportions of AB to PA were constant . But if Williams was 10 percent better in the 1940s, perhaps he would have been even more selective regarding what pitches to hit,
meaning that he might have had less than a total of 9903 AB, while drawing more than 2597 BB. How would this have affected his projected cumulative totals?
Also, this model could be enhanced by considering such entities as the hit-by-pitch (HBP) statistic, on-base-average (OBA) and both stolen bases (SB) and caught stealing (CS). In this way, more
offensive categories would be included.
What about pitching? We mentioned legendary Dodger southpaw Sandy Koufax. Koufax recorded 2396 strikeouts (K) in 2324.3 innings pitched (IP) during his shortened career. What if, for the sake of
argument, we assume that he had pitched an additional 800 innings? Can we use the EC approach with respect to pitching? Yes, we can.
To find Koufax’s strikeout EC, we basically duplicate the procedure we used with the | {"url":"https://free-books-online.org/reader/id/understanding-sabermetrics-140698-read-free-books-online","timestamp":"2024-11-08T15:03:44Z","content_type":"text/html","content_length":"227957","record_id":"<urn:uuid:4112197e-abe8-448e-ad89-1a2b1d446106>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00269.warc.gz"} |
Check if factor graph is connected
Since R2022a
The isConnected function returns a logical flag that indicates whether the factor graph, or a partial factor graph built from specified pose nodes, contains a path between every pair of nodes.
connected = isConnected(fg) returns a logical flag indicating whether the specified factor graph contains a path between every pair of nodes associated with it.
connected = isConnected(fg,poseNodeIDs) returns a logical flag indicating whether a partial factor graph comprised of the specified pose nodes IDs poseNodeIDs, and related factors and non-pose nodes,
contains a path between every pair of nodes. For more information, see Factor Graph Connectivity.
Checking Factor Graph Connectivity
Create a factor graph.
fg = factorGraph;
poseIDs1 = generateNodeID(fg,2,"factorTwoPoseSE3")
poseFactors1 = factorTwoPoseSE3(poseIDs1);
Check the connectivity.
The graph is connected because there is a path between every node pair of graph. For example, you can reach node 2 from node 0 by going through node 1.
Next try to add a disconnected node. Generate a node ID for a GPS factor.
gpsID = generateNodeID(fg,1,"factorGPS")
Create the GPS factor and add it to the factor graph.
gpsFactor = factorGPS(gpsID);
Check the connectivity. Note that because the new node specified by the GPS factor is not connected to any of the previous nodes, there is a disconnect.
Add another factor between node 2 and node 3 to resolve this disconnect.
poseFactors2 = factorTwoPoseSE3([2 3]);
Check the connectivity to verify the graph is connected again.
Input Arguments
poseNodeIDs — IDs of pose nodes to check for connection
N-element row vector of nonnegative integers
IDs of pose nodes to check for connection within the factor graph, specified as an N-element row vector of nonnegative integers. N is the total number of nodes to check.
The pose nodes specified by poseNodeIDs must all be of type "POSE_SE2", or must all be of type "POSE_SE3". The specified pose nodes must also be unique. For example, poseNodeIDs cannot be [1 2 1]
because node ID 1 not unique in this vector.
The specified pose nodes in the factor graph must form a connected factor graph. For more information, see Factor Graph Connectivity.
Output Arguments
connected — Graph is connected in factor graph or in partial factor graph
false or 0 | true or 1
Graph is connected in factor graph or in the partial factor graph, returned as 1 (true) if the factor graph contains a path between every pair of specified nodes and 0 (false) if it does not contain
a path between every pair of specified nodes.
More About
Factor Graph Connectivity
A factor graph is considered connected if there is a path between every pair of nodes. For example, for a factor graph containing four pose nodes, connected consecutively by three factors, there are
paths in the factor graph from one node in the graph to any other node in the graph.
connected = isConnected(fg,[1 2 3 4])
If the graph does not contain node 3, although there is still a path from node 1 to node 2, there is no path from node 1 or node 2 to node 4.
connected = isConnected(fg,[1 2 4])
A fully connected factor graph is important for optimization. If the factor graph is not fully connected, then the optimization occurs separately for each of the disconnected graphs, which may
produce undesired results. The connectivity of graphs can become more complex when you specify certain subsets of pose node IDs to optimize. This is because the optimize function optimizes parts of
the factor graph by using the specified IDs to identify which factors to use to create a partial factor graph. optimize adds a factor to the partial factor graph if that factor connects to any of the
specified pose nodes and does not connect to any unspecified pose nodes. The function also adds any non-pose nodes that the added factors connect to, but does not add other factors connected to those
nodes. For example, for this factor graph there are three pose nodes, two non-pose nodes, and the factors that connect the nodes.
If you specify nodes 1 and 2, then factors 1, 3, 4, and 5 form a factor graph for the optimization because they connect to pose nodes 1 and 2. The optimization includes nodes 4 and 5 because they
connect to factors that relate to the specified pose node IDs.
If you specify poseNodeIDs as [1 3], then the optimize function optimizes each separated graph separately because the formed factor graph does not contain a path between nodes 1 and 3.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
When generating portable C code with a C++ compiler, you must specify hierarchical packing with non-minimal headers. For more information on packaging options, see the packNGo (MATLAB Coder)
Version History
Introduced in R2022a
R2023b: Check connection of specified pose nodes
The isConnected supports checking the connection of specified pose nodes by specifying node IDs. | {"url":"https://in.mathworks.com/help/nav/ref/factorgraph.isconnected.html","timestamp":"2024-11-08T08:33:51Z","content_type":"text/html","content_length":"91285","record_id":"<urn:uuid:c079c43c-4ce9-453d-b640-6950817ca5f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00748.warc.gz"} |
Math Practice Test
Table of Contents
Welcome to our GED Math Practice Test! This comprehensive practice test is designed to help you brush up on your math skills in preparation for the GED Math exam. Our test covers all topics found on
the GED Math exam, including basic math, algebra, geometry, and more. With updated questions from the 2023 exam, this practice test will help you familiarize yourself with the types of questions
you'll see on the actual exam. We hope you find this practice test useful as you prepare for the GED Math exam! Good luck!
Reviewing the Different Types of Questions Found on a GED Math Practice Test
A GED math practice test is an important tool for those preparing to take the General Educational Development (GED) exam. The GED math test consists of a series of multiple-choice questions designed
to assess the individual's competency in basic math skills. There are four types of questions that may appear on a GED math practice test: numerical, algebraic, graphical, and data-based. Numerical
questions ask test-takers to solve basic arithmetic problems, such as addition, subtraction, multiplication, and division.
Algebraic questions require the individual to solve equations, manipulate algebraic expressions, and solve word problems. Graphical questions involve interpreting and analyzing data presented in a
variety of graphical forms, such as charts, tables, and graphs. Finally, data-based questions involve interpreting data presented in text form, such as word problems or statistics. A GED math
practice test is a helpful means of determining your level of competency in basic math skills.
By familiarizing yourself with the different types of questions that may appear on the test, you can better prepare yourself for the exam. With the right practice and preparation, you can be
confident in your ability to succeed on the GED math test.
Understanding the Common Core Math Standards and How They Relate to the GED Math Practice Test
The Common Core Math Standards are a set of educational standards, developed in the United States, that define the expectations for students in mathematics. These standards provide a consistent set
of goals for students in K–12 mathematics, and they are used by states and school districts to develop curricula, assessments, and instructional materials. The Common Core Math Standards are designed
to ensure that all students have the skills they need to be successful in college or their chosen career. The Common Core Math Standards are closely related to the GED Math Practice Test. The GED
Math Practice Test is a standardized test designed to assess students’ knowledge in mathematics.
The test is divided into two sections: Quantitative Reasoning and Algebraic Reasoning. The Quantitative Reasoning section covers topics such as basic operations, measurements, and data analysis. The
Algebraic Reasoning section covers topics such as linear equations, graphing, and functions. The GED Math Practice Test is designed to measure how well students understand the Common Core Math
Standards. The test assesses students’ ability to apply the concepts and skills taught in the Common Core Math Standards to real-world situations. The test is designed to ensure that students have
the skills they need to be successful in college or their chosen career.
The Common Core Math Standards and the GED Math Practice Test are closely related because both are designed to help students understand and apply the mathematical concepts and skills they need to
succeed in college and beyond. The Common Core Math Standards provide a consistent set of goals for students in K–12 mathematics, and the GED Math Practice Test assesses students’ understanding of
these concepts. By taking the GED Math Practice Test, students can demonstrate that they have mastered the mathematics skills they need to be successful in college and their chosen career.
Tips for Preparing for the GED Math Practice Test
1. Brush up on your basic math skills: Before taking the GED Math Practice Test, it is important to review the basic math skills you learned in school. This includes addition, subtraction,
multiplication, and division. It is also beneficial to practice your fractions, decimals, and percentages.
2. Familiarize yourself with the GED Math Test Format: It is important to understand the structure of the GED Math Test so you can anticipate the types of questions that may be asked. The math test
questions are divided into two parts: Part 1 consists of multiple-choice questions and Part 2 consists of short-answer questions.
3. Practice, Practice, Practice: Once you have reviewed the basic math skills and familiarized yourself with the test format, it is important to practice with sample questions. This will help you
gain confidence and become more comfortable with the types of questions that may be asked on the test.
4. Make use of online resources: There are many online resources available to help you prepare for the GED Math Practice Test. These resources include video tutorials, practice tests, and online test
prep courses. Taking advantage of these resources can help you better understand the material and increase your chances of success on the test.
5. Get plenty of rest: On the day of your GED Math Practice Test, it is important to get a good night’s sleep and arrive at the testing center well-rested. Being physically and mentally prepared will
help you focus and perform your best on the test.
Analyzing the Different Math Concepts Covered on the GED Math Practice Test
The GED Math Practice Test is designed to help students prepare for the math portion of the GED exam. This test covers a variety of math concepts, from basic operations to more complex concepts such
as algebra, geometry, and statistics. It is important for test takers to understand the different math concepts covered on the GED Math Practice Test in order to properly prepare for the actual GED
exam. The first math concept covered on the GED Math Practice Test is basic operations. This includes addition, subtraction, multiplication, and division. Test takers must demonstrate an
understanding of the basic principles of these operations, as well as basic operations with fractions and decimals.
Additionally, test takers must also be able to solve basic equations. The second math concept covered on the GED Math Practice Test is algebra. This includes basic equations and linear equations.
Test takers must be able to solve for unknown variables, identify linear equations, and use the distributive property. Additionally, test takers must also be able to graph equations and solve for
slope and intercepts. The third math concept covered on the GED Math Practice Test is geometry. This includes basic shapes and their properties, as well as the principles of angles, lines, and
circles. Test takers must demonstrate an understanding of basic geometric shapes and their properties, as well as the principles of angles and lines.
Additionally, test takers must also be able to calculate the area and perimeter of geometric figures. The fourth math concept covered on the GED Math Practice Test is statistics. This includes basic
principles of probability, data interpretation, and statistical analysis. Test takers must demonstrate an understanding of basic probability principles, as well as the ability to interpret and
analyze data. Additionally, test takers must also be able to calculate the mean, median, and mode of a data set. By understanding the different math concepts covered on the GED Math Practice Test,
test takers can properly prepare for the actual GED exam. It is important to review all of the concepts in order to be successful on the exam. With sufficient practice and review, test takers can
ensure that they are adequately prepared for the GED exam.
Key Strategies for Answering Math Questions on the GED Math Practice Test
1. Read and Understand the Question: Before attempting to answer a question on the GED Math Practice Test, it is essential that the question is read and understood carefully. Taking the time to read
the entire question thoroughly will help inform the approach to answering the question correctly.
2. Identify the Key Information: After reading the question, it is helpful to identify the key information from the question. This key information can include any facts, figures, numbers, or
equations that are present, as well as any formulas or equations that may be needed to solve the problem.
3. Identify the Goal: Once the key information has been identified, the next step is to identify the goal of the question. This can be done by reading the question carefully and understanding what
the question is asking. This will help to inform the approach to answering the question correctly.
4. Choose the Appropriate Tool: After identifying the key information and goal of the question, the next step is to choose the appropriate tool to use to answer the question. This could include a
calculator, mathematical formulas, or a graph.
5. Solve the Problem: Once the appropriate tool has been chosen, the next step is to solve the problem. This can involve breaking the problem down into smaller steps, using mathematical formulas, or
using a calculator to get the answer.
6. Check Your Work: Once the problem has been solved, it is important to check the work to ensure the answer is correct. This can involve double-checking the calculations, reviewing the formulas, or
looking for any errors that may have been made.
Overall, the What to Include GED Math Practice Test (updated 2023) is a great way to help prepare for the GED Math exam. It provides a comprehensive review of the topics covered on the exam and
offers detailed solutions to each problem. It also includes a variety of practice tests and exercises that help students practice and build their math skills. With the help of this GED Math Practice
Test, students can be sure to be fully prepared and confident when taking the GED Math exam.
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after complete update of files, Dec. 6, 2000. Converted to HTML during reconstruction of web page in June 2020. There are some bad links here to be fixed, but most of this should be here.
• algebra2.wat working theory of reals/rationals in omnibus
• babydocs.wat examples from documentation
• capsl_capstuff.wat Shankar's belief logic proofs
• capsl_defination_file.wat
• capsl_ns_axioms.wat
• capsl_nsprotocol.wat
• capsl_svo.wat
• cohen*.wat A development from Cohen's book.
• combinators.wat Theoretical work on combinatory logic, with an eye to strong beta-reduction for Jonathan Seldin
• concepts.wat file intended to implement sequent proofs in the calculus of concepts
• counting.wat set theory under development using deduction.wat; includes proof of Schroder-Bernstein theorem.
• deduction.wat goal directed natural deduction prover using INPUT command candidate for addition to omnibus
• deduction_examples.wat examples (so far just one) for deduction.wat
• definitions.wat "structural" file for Math 387 development.
• examples.wat propositional logic examples for the Math 387 development.
• fract.wat Theory of rationals and reals continued in omnibus
• gries9.wat advanced logic in omnibus (quantifiers)
• jonny.wat calculator for working with exponents -- for Jonny uses peano
• lab3.wat part of the math 387 development.
• lambda.wat structural stuff in addition to structural.wat (in omnibus); includes the tactic for putting in explicit binders.
• life.wat a development of the theory of Conway's Game of Life.
• logic_tools.wat some logical stuff in omnibus; its leading comment follows: (* this file includes a selection of logical tools: Parvin tactics for conversion of "implication" theorems from one
form to another, Alves-Foss theorems about case expressions, and basic results about quantifiers proved "by hand" *)
• logicdefs.wat basic logic definitions in omnibus l
• logicdefs2.wat Parvin's logic file -- propositional logic in omnibus
• mwu_subtraction.wat Minglong's proofs on subtraction and division.
• natorder.wat Sol's file on order on natural numbers, in omnibus
• new.quantifiers.wat theorems about quantifiers in omnibus
• newlib1.wat New library structural tactics.
• newlib2.wat New library equational development of propositional logic, independent of Watson logic of case expressions, followed by a development of INPUT driven sequent logic tactics for full
first-order logic.
• newlib3.wat Chapter 1 of Landau (uses newlib1 and newlib2). This is Landau's theory of natural numbers.
• omnibus.wat The omnibus theory of almost everything
• pair_assign.wat Very complex theory of pair assignment operator from Minglong.
• peano.wat Peano arithmetic -- a laboratory for induction proofs. Not included in omnibus -- notationally incompatible with it, but complete induction and related results are exported from it to
algebra2.wat in omnibus. theory of programming
• programs.wat implementing Cohen -- last file in omnibus theory.
• prop_equational.wat Is this part of the Math 387 development?
• prop_logic.wat Alternative version of logicdefs2 (Parvin's file) in something like its original form with propositional connectives as primitives; needed for belief logic.
• schneiderexample.wat Example of implementation of a type of syntactical objects.
• sequent.wat Development of sequent calculus in omnibus.
• shankar.wat A file on Shankar's capsl stuff -- I don't know whether it is older or newer than capsl* so I saved it.
• sharondemo.wat A demo.
• shortsheffer.wat Records a nice Boolean algebra result from the OTTER people (no proof!)
• simplesets.wat Sol's basic set theory in omnibus.
• sol_exp.wat Stuff about exponents developed by Sol to be added to omnibus eventually.
• structural.wat Basic structural tactics used by omnibus and most other files.
• struct2.wat file which updates things defined in structural.wat and lambda.wat to take advantage of new capabilities of the prover.
• tableau.wat develops the tautology checker NEWTAUT and proves the Gries axioms used in logicdefs2; in omnibus.
• tableau2.wat Simulation of a tableau style of reasoning in omnibus.
• typestuff.wat Basic stuff about types (retractions onto s.c. sets) in omnibus and other files. | {"url":"https://randall-holmes.github.io/Watson/Scripts/theories.html","timestamp":"2024-11-07T01:10:58Z","content_type":"text/html","content_length":"6920","record_id":"<urn:uuid:7990e5c3-e437-4459-a355-ad04b1e02e00>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00581.warc.gz"} |
Knowledge Management Research Group
Fourier transform
This page is a sub-page of our page on Infinitesimal Calculus of One Real Variable.
Related sources of information:
• Fourier transform
• The Fourier Transform.com
• Fourier transform pairs
Fourier Transform Intuition (Better Explained on YouTube):
An interactive introduction to the Fourier Transform
Introduction to the Fourier Transform (Part 1) (Brian Douglas on YouTube):
Introduction to the Fourier Transform (Part 2) (Brian Douglas on YouTube):
But what is the Fourier Transform? A visual introduction (by 3Blue1Brown):
TheFourierTransform.com Presents A Simple Explanation for the Fourier Transform:
Layman’s Explaination of the Fourier Transform (by studioTTTguTTT):
What is a Fast Fourier Transform (FFT)? The Cooley-Tukey Algorithm (LeiosOS on YouTube):
/////// Quoting Wikipedia on “Fourier transforms”
The Fourier transform (FT) decomposes (analyzes) a function of time (a signal) into its constituent frequencies. This is similar to the way a musical chord can be expressed in terms of the volumes
and frequencies (or pitches) of its constituent notes. The term Fourier transform refers to both the frequency domain representation and the mathematical operation that associates the frequency
domain representation to a function of time.
The Fourier transform of a function of time is itself a complex-valued function of frequency, whose magnitude (modulus) represents the amount of that frequency present in the original function, and
whose argument is the phase offset of the basic sinusoid in that frequency.
The Fourier transform is not limited to functions of time, but the domain of the original function is commonly referred to as the time domain.
There is also an inverse Fourier transform that mathematically synthesizes the original function (of time) from its frequency domain representation.
Fourier transform:
$\hat f(\xi) = \int_{-\infty}^{+\infty} f(x)\,e^{-2 \pi i \xi x} \,dx$.
Inverse Fourier transform:
$f(x) = \int_{-\infty}^{+\infty} \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi$.
Linear operations performed in one domain (time or frequency) have corresponding operations in the other domain, which are sometimes easier to perform. The operation of differentiation in the time
domain corresponds to multiplication by the frequency, so some differential equations are easier to analyze in the frequency domain.
Also, convolution in the time domain corresponds to ordinary multiplication in the frequency domain (see Convolution theorem). After performing the desired operations, transformation of the result
can be made back to the time domain.
Harmonic analysis is the systematic study of the relationship between the frequency and time domains, including the kinds of functions or operations that are “simpler” in one or the other. Harmonic
analysis has deep connections to many areas of modern mathematics.
Functions that are localized in the time domain have Fourier transforms that are spread out across the frequency domain and vice versa, a phenomenon known as the uncertainty principle.
The critical case for this principle is the Gaussian function, of substantial importance in probability theory and statistics as well as in the study of physical phenomena exhibiting normal
distribution (e.g., diffusion).
The Fourier transform of a Gaussian function is another Gaussian function.
Joseph Fourier introduced the transform in his study of heat transfer, where Gaussian functions appear as solutions of the heat equation.
The Fourier transform can be formally defined as an improper Riemann integral, making it an integral transform, although this definition is not suitable for many applications requiring a more
sophisticated integration theory. For example, many relatively simple applications use the Dirac delta function, which can be treated formally as if it were a function, but the justification requires
a mathematically more sophisticated viewpoint.
The Fourier transform can also be generalized to functions of several variables on Euclidean space, sending a function of 3-dimensional ‘position space’ to a function of 3-dimensional momentum (or a
function of space and time to a function of 4-momentum).
This idea makes the spatial Fourier transform very natural in the study of waves, as well as in quantum mechanics, where it is important to be able to represent wave solutions as functions of either
position or momentum and sometimes both.
In general, functions to which Fourier methods are applicable are complex-valued, and possibly vector-valued.
Still further generalization is possible to functions on groups, which, besides the original Fourier transform on ℝ or ℝn (viewed as groups under addition), notably includes the discrete-time Fourier
transform (DTFT, group = ℤ), the discrete Fourier transform (DFT, group = ℤ mod N) and the Fourier series or circular Fourier transform (group = S1, the unit circle ≈ closed finite interval with
endpoints identified). The latter is routinely employed to handle periodic functions. The fast Fourier transform (FFT) is an algorithm for computing the DFT.
Fast Fourier transform:
Heisenberg uncertainty relation | {"url":"https://kmr.dialectica.se/wp/research/math-rehab/learning-object-repository/calculus/calculus-of-one-real-variable/fourier-transforms/","timestamp":"2024-11-05T12:34:20Z","content_type":"text/html","content_length":"152366","record_id":"<urn:uuid:142cbcf1-e2ee-4479-8e2d-3eecba5955c3>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00052.warc.gz"} |
Generates a pseudo-random number.
Random, OutputVar , Min, Max
Random, , NewSeed
This command yields a pseudo-randomly generated number, which is a number that simulates a true random number but is really a number based on a complicated formula to make determination/guessing of
the next number extremely difficult.
All numbers within the specified range have approximately the same probability of being generated (however, see "known limitations" below).
If either Min or Max contains a decimal point, the end result will be a floating point number in the format set by SetFormat. Otherwise, the result will be an integer.
Known limitations for floating point: 1) only about 4,294,967,296 distinct numbers can be generated for any particular range, so all other numbers in the range will never be generated; 2)
occasionally a result can be slightly greater than the specified Max (this is caused in part by the imprecision inherent in floating point numbers).
Format(), SetFormat
Generates a random floating point number in the range 0.0 to 1.0 and stores it in rand.
Random, rand, 0.0, 1.0
Comments based on the original source
This function uses the Mersenne Twister random number generator, MT19937, written by Takuji Nishimura and Makoto Matsumoto, Shawn Cokus, Matthe Bellew and Isaku Wada.
The Mersenne Twister is an algorithm for generating random numbers. It was designed with consideration of the flaws in various other generators. The period, 2^19937-1, and the order of
equidistribution, 623 dimensions, are far greater. The generator is also fast; it avoids multiplication and division, and it benefits from caches and pipelines. For more information see the
inventors' web page at www.math.sci.hiroshima-u.ac.jp/~m-mat/MT/emt.html
Copyright (C) 1997 - 2002, Makoto Matsumoto and Takuji Nishimura, All rights reserved.
Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met:
1. Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer.
2. Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the
3. The names of its contributors may not be used to endorse or promote products derived from this software without specific prior written permission.
Do NOT use for CRYPTOGRAPHY without securely hashing several returned values together, otherwise the generator state can be learned after reading 624 consecutive values.
When you use this, send an email to: m-mat@math.sci.hiroshima-u.ac.jp with an appropriate reference to your work.
This above has been already been done for AutoHotkey, but if you use the Random command in a publicly distributed application, consider sending an e-mail to the above person to thank him. | {"url":"https://ahk4.us/docs/commands/Random.htm","timestamp":"2024-11-08T11:26:40Z","content_type":"text/html","content_length":"7366","record_id":"<urn:uuid:e6566a90-2c9f-40c9-b908-e8750b2a587c>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00272.warc.gz"} |
Fluid Mechanics (261-280)
discharge passing between the streamlines through the points (1,3 ) and (3,3) is
262. A model of a weir made to a horizontal scale of 1/40 and vertical scale of 1/9
discharges 1 litre/sec. Then the discharge in the prototype is estimated as 1080 lps
263. Laminar flow occurs between extensive stationary plates. The kinetic energy
correction factor is nearly 2.0
264. In steady laminar flow of a liquid through a circular pipe of internal diameter D, carrying a constant discharge, the hydraulic gradient is inversely proportional to D⁴
265. The ratio of the coefficient of friction drag in laminar boundary layer compared to that in turbulent boundary layer is proportional to
266. The overall drag coefficient of an aircraft of weight W and wing area S is given by
where a and b are constants. The maximum drag in horizontal flight will be 2W√ab
267. For laminar flow between parallel plates separated by a distance 2h, head loss
268.A penstock is 2000 m long and the velocity of pressure wave in it is 1000
m/s. Water hammer pressure head for instantaneous closure of value at the
downstream end of pipe is 60 m. If the valve is closed in 4 sees, then the peak
water hammer pressure head is equal to 60 m
269. In an open channel of wide rectangular section with constant n value, the bed slope is 1.2 x 10⁻³ the local friction slope at a section is 1.05 x 10⁻³, the local
Froude number of the flow is 0.8. The local rate of variation of depth with
longitudinal distance along the flow direction is
270. Before passage of a surge, the depth and velocity of flow at a section are 1.8 m and 3.72 m/s and., after passage, they are 0.6 m and 7.56 m/s respectively. The speed of the surge is +1.8 m/s
271. A turbine works at 20 m head and 500 rpm speed. Its 1:2 scale model to be tested at a head of 20 m should have a rotational speed of nearly 1000 rpm
272.Which one of the following statements regarding reciprocating pump is correct?-Air vessel reduces the acceleration head and consequently reduces the effect of friction head also
273.The correct sequence, in the direction of the flow of water for installations in a hydro-power plant is reservoir, pen stock, surge tank turbine
274. There are four variables, namely, E (volume modulus of elasticity), p (pres
sure per unit area), g (acceleration due to gravity) and μ (viscosity of water). They
are associated with Mach, Euler, Froude and Reynolds numbers, respectively, in
275 Which one of the following pressure units represents the least pressure ?
276.Given that, as flow takes place between two parallel static plates, the velocity
midway between the plates is 2 m/s, the Reynolds number is 1200 and the distance between the plates is 10 cm. Which of the following statements are true ? The rate of flow is 0.1 m³/s/metre width,The
energy correction factor is 2.0
277. The displacement thickness of a boundary layer is the distance by which the main flow is to be shifted from the boundary to maintain the continuity equation
278 Tarbulent flow - Mixing length
Laminar flow - Hagan Poiseuille equation Lift on an aero-foil- Circulation Boundary layer- Momentum integeral equation
279. For laminar flow in a pipe carrying a given discharge, the height of surface
roughness is doubled. In such a case, Darcy-Weisbach friction factor will
280. Two small orifices A and B of diameters 1 cm and 2 cm, respectively, are placed on the sides of a tank at depths of h₁ and h₂ below the open liquid surface. If the discharges through A and B are
equal, then the ratio of h₁ and h₂ (assuming equal C𝒹 values) will be 16:1 | {"url":"https://www.sudhanshucivil2010.com/post/fluid_14","timestamp":"2024-11-04T01:43:51Z","content_type":"text/html","content_length":"1050497","record_id":"<urn:uuid:cadb9a2d-6cea-41f9-98f0-8cb90d20e757>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00229.warc.gz"} |
Kirchhoff's Rules
Kirchhoff's Rules
• We have already analysed simple circuit using ohm's laws and reducing these circuit to series and parallel combination of resistors
• But we also come across circuits containing sources of EMF and grouping of resistors can be far more complex and can not be easily reduced to a single equivalent resistors
• Such complex circuits can be analysed using two Kirchhoff's rules
(A) The junction Rule (or point rule)
• This law states that "The algebraic sum of all the currents entering junction or any point in a circuit must be equal to the sum of currents leaving the junction"
• Alternatively this rule can also be stated as " Algebraic sum of the currents meeting at a point in a electric circuit is always zero i.e
ΣI=0 at any point in a circuit
• This law is based on the law of conservation of charge
• Consider a point P in an electric circuit at which current I[1],I[2],I[3] and I[4] are flowing through conductors in
the direction shown below in the figure below
• If we take current flowing towards the junction as positive and current away from the junction as negative,then from Kirchhoff's law
I[1]+ I[2]=I[3]+ I[4]
• From this law ,we conclude that net charge coming towards a point must be equal to the net charge going away from this point in the same interval of time
(B) The Loop Rule (or Kirchhoff's Voltage Law)
• The rule states that " the sum of potential difference across all the circuit elements along a closed loop in a circuit is zero
ΣV=0 in a closed loop
• Kirchhoff's loop rule is based on the law of conservation of energy because total amount of energy gained and lost by a charge round a trip in a closed loop is zero
• when applying this Kirchhoff's loop rule in any DC circuit,we first choose a closed loop in a circuit that we are analysing
• Next thing we have to decide is that whether we will traverse the loop in a clockwise direction or in anticlockwise direction and the answer is that ,the choice of direction of travel is
arbitrary to reach the same point again
• When traversing the loop ,we will be following convention to note down drop or rise in the voltage across the resistors or battery
(i) If the resistor is being traversed in the direction of the current then change in PD across it is negative i.e -IR
(ii)If the resistor is being traversed in the direction opposite to the current then change in PD across it is negative i.e IR
(iii) If a source of EMF is traversed in the direction from -ve terminal to its positive terminal then change in electric potential is positive i.e E
(iv)If a source of EMF is traversed in the direction from +ve terminal to its negative terminal then change in electric potential is negative i.e -E
• We would now demonstrate the use of Kirchhoff's loop law in finding equations in simple circuit
• Consider the circuit as shown below
• First consider loop ABDA.Lets traverse loop in anticlockwise direction.From Kirchhoff's loop law
• Neglecting internal resistance of the cell and using sign conventions stated previously we find
And similarly if we traverse the loop ABCA in clock wise direction | {"url":"https://physicscatalyst.com/elec/kirchhoffs-rules.php","timestamp":"2024-11-04T05:31:38Z","content_type":"text/html","content_length":"69305","record_id":"<urn:uuid:05094217-d5ef-4ce8-a479-ccc854540cb3>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00193.warc.gz"} |
PREDICT SALARY ON THE BASIS OF YEARS OF EXPERIENCE - Study Trigger
Let's create a model to predict salary based on experience
by Japanjot
written by Japanjot
Predict Salary of Employee based on Years of Experience
In the ever-evolving landscape of career development, understanding the nuanced relationship between professional experience and salary is a pivotal aspect for both job seekers and employers. The
concept that the number of years an individual spends in a particular field directly influences their earning potential has been a subject of interest and analysis. This article delves into the realm
of predictive modeling, specifically aiming to forecast salaries based on years of professional experience.
The purpose of solving the problem titled “Predict Salary based on Years of Experience” encompasses several key objectives, each contributing to a deeper understanding of the relationship between
professional experience and compensation. The primary purposes include:
Informed Career Decisions: Empowering individuals with the ability to predict salaries based on years of experience aids in making informed career decisions. Job seekers can better understand the
expected trajectory of their compensation as they gain more professional expertise.
Negotiation and Advancement: Armed with insights into how salary evolves with experience, employees can negotiate compensation more effectively during job offers and promotions. Understanding the
correlation allows for strategic career planning.
Employer Decision-Making: Employers can utilize predictive models to make informed decisions about salary structures. This includes setting competitive salaries, aligning compensation with industry
standards, and recognizing the value of experience in the workforce.
Optimized Human Resources Strategies: Human resources professionals can benefit from insights into salary prediction to develop and optimize compensation strategies within organizations. This
includes talent acquisition, employee retention, and overall workforce planning.
Data-Driven Hiring Practices: For hiring managers, having a predictive model for salary based on experience provides a data-driven approach to recruitment. This can lead to fairer and more
transparent hiring processes.
Career Planning and Development: Individuals can use the predictive insights for long-term career planning and development. Knowing how salaries typically evolve with experience enables professionals
to set realistic goals and expectations.
Understanding Market Trends: The analysis of salary prediction models contributes to a broader understanding of market trends in compensation. This information is valuable not only for individuals
and organizations but also for researchers and policymakers.
Enhanced Workforce Productivity: A clear understanding of how salaries correlate with experience can contribute to increased job satisfaction and productivity. Employees who feel fairly compensated
are likely to be more engaged and motivated in their roles
Let’s start to implement the above said problem using Machine Learning and Python :
You have a dataframe containing information about “Years of Experience” and “Salary” for employees . So we can understand how the experience of an employee can impact the salary part. Here we
understand a linear relationship between experience and salary.
For a better understanding of question let’s start solving the question and divide an answer in some steps for better understanding of the question
STEP 1 : Creating Data Frame:
We can make the model either by creating our own dataset or we use a CSV file with same column names :
When we create our data manually it look’s like
import pandas as pd
# Create a DataFrame with employee data
data = {'years_of_experience': [2, 5, 7, 10, 3, 6, 8, 12, 4, 9, 1, 11, 5, 3, 6, 9, 2, 8, 10, 4],
'salary': [50000, 70000, 85000, 105000, 55000, 72000, 88000, 115000, 60000, 95000, 48000, 110000, 68000, 56000, 74000, 99000, 52000, 90000, 112000, 62000]}
df = pd.DataFrame(data)
or if we want to import CSV file to use as DataFrame it look’s like :
import pandas as pd
# Load the dataset
df = pd.read_csv('data.csv')
instead of ‘data.csv’ you have to use path of the file.
Step 2 : Imports the necessary libraries :
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.metrics import mean_squared_error, r2_score
Step 3 : Splits the DataFrame into input features X (Years_of_Experience) and target variable y (Salary).
# Split the data into features (Years of Experience) and target (Salary)
X = df[['years_of_experience']]
y = df['salary']
Step 4 : Splits the data into training and testing sets using 80% for training and 20% for testing.The random_state=42 .
# Split the data into training and testing sets
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)
Step 5 : Initialize a linear regression model.
# Create a Linear Regression model
model = LinearRegression()
Step 6 : Trains the linear regression model using the training data.
# Train the model on the training data
model.fit(X_train, y_train)
Step 7 : Uses the trained model to make predictions on the test data and calculates the R-squared score (r2) and Mean Squared Error (mse) to evaluate the model’s performance.
# Make predictions on the test data
y_pred = model.predict(X_test)
r2 = r2_score(y_test, y_pred)
Step 8 : Plots the test data points and the fitted line obtained from the linear regression model. Labels the axes, provides a title, and shows the plot.
# Plot the data points and the fitted line
plt.scatter(X_test, y_test, label='Test Data')
plt.plot(X_test, y_pred, color='red', label='Fitted Line')
plt.xlabel('Years of Experience')
plt.title('Linear Regression')
Step 9 : Prints the Mean Squared Error and R-squared Error calculated earlier, providing insights into the model’s accuracy and fit to the data.
print(f"Mean Squared Error: {mse:.2f}")
print(f"R-Squared Error: {r2:.2f}")
Let’s combine all these steps and the code :
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.linear_model import LinearRegression
from sklearn.metrics import mean_squared_error, r2_score
# Create a DataFrame with employee data
data = {'years_of_experience': [2, 5, 7, 10, 3, 6, 8, 12, 4, 9, 1, 11, 5, 3, 6, 9, 2, 8, 10, 4],
'salary': [50000, 70000, 85000, 105000, 55000, 72000, 88000, 115000, 60000, 95000, 48000, 110000, 68000, 56000, 74000, 99000, 52000, 90000, 112000, 62000]}
df = pd.DataFrame(data)
# Split the data into features (Years of Experience) and target (Salary)
X = df[['years_of_experience']]
y = df['salary']
# Split the data into training and testing sets
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)
# Create a Linear Regression model
model = LinearRegression()
# Train the model on the training data
model.fit(X_train, y_train)
# Make predictions on the test data
y_pred = model.predict(X_test)
r2 = r2_score(y_test, y_pred)
# Calculate the Mean Squared Error (MSE)
mse = mean_squared_error(y_test, y_pred)
# Plot the data points and the fitted line
plt.scatter(X_test, y_test, label='Test Data')
plt.plot(X_test, y_pred, color='red', label='Fitted Line')
plt.xlabel('Years of Experience')
plt.title('Linear Regression')
print(f"Mean Squared Error: {mse:.2f}")
print(f"R-Squared Error: {r2:.2f}")
You will see the below chart after executing the above code :
Mean Squared Error: 7.79
R- Squared Error: 1.00
Let’s look at some insight of the model and understand it.
Positive Correlation:
The scatter plot and fitted line indicate a positive correlation between years of experience and salary. As the years of experience increase, there is a general trend of higher salaries.
Model Accuracy:
The R-squared score and Mean Squared Error (MSE) suggest that the linear regression model provides a reasonable fit to the data. The R-squared score of the model indicates the proportion of the
variability in salary that can be explained by years of experience.
Prediction Capability:
The fitted line represents the model’s ability to predict salaries based on years of experience. Predictions can be made for salaries not included in the training data, allowing for estimates of
employee salaries for different levels of experience.
The slope of the fitted line represents the average increase in salary for each additional year of experience. This interpretable feature allows for a straightforward understanding of the
relationship between the independent variable (years of experience) and the dependent variable (salary).
Time to Predict the Salary based on Experience
experience=int(input("Enter the number of experience "))
# Assume you have a new experience value for prediction, for example, 7 years
new_experience = np.array([7]).reshape(-1, 1)
# Make predictions using the trained model
predicted_salary = model.predict(new_experience)
# Display the prediction
print(f"Predicted Salary for {new_experience[0][0]} years of experience: ${predicted_salary[0]:,.2f}")
FAQ’s on Salary Prediction
1. What does the code aim to achieve?
The code aims to build a linear regression model to predict employee salaries based on their years of experience.
2. Why is linear regression chosen for this scenario?
Linear regression is chosen because it assumes a linear relationship between the independent variable (years of experience) and the dependent variable (salary), making it suitable for predicting
salary based on experience.
3. How is the dataset split into training and testing sets, and why is it important?
The dataset is split using the train_test_split function from scikit-learn. This division is crucial to train the model on one subset and evaluate its performance on another, ensuring its ability to
generalize to new, unseen data.
4. What do Mean Squared Error (MSE) and R-squared (R2) score represent in this context?
MSE measures the average squared difference between predicted and actual salaries. R-squared score represents the proportion of variability in salary explained by years of experience. Lower MSE and
higher R2 indicate better model performance.
5. How is the model’s accuracy visualized in the code?
The code includes a scatter plot with the test data points and a fitted line representing the model’s predictions. This visualization allows a quick assessment of how well the model captures the
relationship between years of experience and salary.
6. Can this code be applied to predict salaries for new employees?
Yes, the trained linear regression model can be used to predict salaries for new employees based on their years of experience.
7. What insights can be gained from the fitted line in the scatter plot?
The fitted line shows the average increase in salary for each additional year of experience, providing a visual interpretation of the model’s predictions.
8. What are the limitations of this linear regression model?
The model assumes a linear relationship, and deviations from linearity may affect predictions. It may not capture complex relationships or consider other factors influencing salary.
9. How can this code be extended for more advanced analysis?
To extend the analysis, one could explore additional features, consider non-linear relationships, or experiment with more advanced machine learning models.
10. In what real-world scenarios could this code be applied?
The code is applicable in scenarios such as HR and workforce planning, where understanding the relationship between years of experience and salary is crucial for making informed decisions about
compensation structures.
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New Quantum Algorithm Dance: DQI Edition
To work on trying to find novel quantum algorithms is to take a long lonely trip where progress is measured in failures. Whenever one of the brave souls (or group of souls) embarks on such a journey,
and comes back with treasure that looks like a new quantum algorithm, we should all do the quantum algorithm dance and celebrate the returning heroes. It looks to me like we have recently had just
such an event with the appearance of the preprint “Optimization by Decoded Interferometry” by Stephen P. Jordan, Noah Shutty, Mary Wootters, Adam Zalcman, Alexander Schmidhuber, Robbie King, Sergei
V. Isakov, and Ryan Babbush [arXiv:2408.08292]. They call this algorithm Decoded Quantum Interferometry, or DQI (not to be confused with the name of the quantum computing division of the American
Physical Society, but yes it will be confused). I was lucky enough to sit in the same office as Stephen, so I got to watch as he and the team made the discovery of the algorithm. The optimizer has
blogged about the results of this algorithm, but I’m not a theoretical computer scientist, so what I’m most interested in is “how does it work”. In other words, what is the actual quantum algorithm,
what does it actually do?
The general problem space that the DQI works on is optimization. In optimization problems one wants to maximize some function over a combinatorialy large domain. That is for a function $f$ one wants
to find an $x$ such that $f(x) \geq f(y)$ for all $y \neq x$. Generally finding the maximum is very hard. If I give you $f$ as a black box, i.e. I don’t tell you how $f$ was generated, it’s very
clear that one might have to search in the worst case over the entire space of inputs to the function. More practically, however, we are often given some succinct description of $f$, for example it
might be a mathematical function like a polynomial , and we can ask how hard it is to find the maximum given this description. It turns out though that even then the problem is hard, this leads us to
complexity classes like NP and friends. Even more practically, though, one could also loosen the requirement a bit, what if one wants to find a value $x$ such that $f(x)$ is as close the maximum as
possible. It turns out that when one does this, there are certain regimes where wouldn’t be crazy for quantum algorithms to outperform classical algorithms. This is the target of the DQI algorithm.
Let’s “derive” DQI.
If you have a function $f$ you want to optimize, one thing you might think to do on a quantum computer is to prepare the state $$|f\rangle := \frac{1}{\sqrt{\mathcal F}} \sum_x f(x) |x\rangle$$The
sum is over some large space of inputs to the function, for example we could imagine that the sum is over bitstrings of length $n$ (so the sum is over $2^n$ terms). For simplicity we assume the
values $f$ produces are real numbers. Here ${\mathcal F}$ is a normalization factor. If one could produce this state, then the probability of observing $x$ if measuring this state is $$prob(x) = \
frac{ |f(x)|^2}{\mathcal F}$$This doesn’t seem too impressive, but lets stick with it. In particular, we do notice that the probability of producing $x$ is higher where $|f(x)|$ is higher, so
producing such a state gives us the ability to bias towards the values that maximize or minimize $f$.
We want to produce a state like $|f\rangle$ and we are using a quantum computer, so somehow we want to arrange it so that the quantum computer directs positive interference like effects to the places
where $f$ is higher and negative interference like effect s to the places where $f$ is lower. We don’t have a lot of interfering type matrices in quantum (or rather all generic unitaries are like
this, but structured unitaries that we can get a handle on are rarer), so lets just assume that the unitary which produces this is the $n$ qubit Hamadard transform. That is its own inverse, so we can
say that if we start in the state $$|\hat{f}\rangle := \frac{1}{\sqrt{\mathcal F} 2^{n/2}} \sum_{x,y \in {\mathbb Z}_2^n} (-1)^{x \cdot y} f(x) |y\rangle$$where $x \cdot y = x_1 y_1 + \dots + x_n y_n
~{\rm mod}~2$, then applying $H^{\otimes n}$ to this state will produce the desired $|f\rangle$. We’ve changed the problem of preparing $|f\rangle$ to preparing another state, $|\hat{f}\rangle$.
What to do next? First stare at this state. What could the $f(x)$ possibly be that would make $|\hat{f}\rangle$ easy to prepare? Well one obvious thing would be if we had $$f(x) = (-1)^{b \cdot x}
$$where $b \in {\mathbb Z}_2^n$. In that case, $|\hat{f}\rangle = |b\rangle$. In other words, if we can prepare $|b\rangle$ (which is trivial), then we can create $|f\rangle$ with $f(x) = (-1)^{b \
cdot x}$. Now this isn’t to exciting of a function to maximize. And even worse when we prepare $|f\rangle$ since all the values of $f$ are $\pm 1$, and measure it we get all bit strings with equal
probability. Boo.
But like all things in algorithm research, lets keep chugging along. Since $|b\rangle$ gave us an $f$ we could analyze, a natural thing to think about is what if you could produce a superpostion of
inputs. If we start with $\frac{1}{\sqrt{2}}(|b_1\rangle + |b_2\rangle)$, where $b_i \in {\mathbb Z}_2^n$, then we see that we can produce $|f\rangle$ with $$f(x) = (-1)^{b_1 \cdot x} + (-1)^{b_2 \
cdot x}$$Now this state is a bit more interesting, in that it doesn’t have an equal probability when we measure in the computational basis. Sometimes $b_1 \cdot x = b_2 \cdot x$ in which case the
amplitudes add up, and other times $b_1 \cdot x \neq b_2 \cdot x$ in which case the amplitudes cancel.
Another observation is that we could change the values of which amplitudes cancel and which add, by instead of preparing $\frac{1}{\sqrt{2}}(|b_1\rangle + |b_2\rangle)$ instead preparing $\frac{1}{\
sqrt{2}}((-1)^{v_1}|b_1\rangle + (-1)^{v_2}|b_2\rangle)$ where $v_i \in {\mathbb Z}_2$. Now the function is $$f(x) = (-1)^{v_1 + b_1 \cdot x} + (-1)^{v_2+b_2 \cdot x}$$.
Generaliizing a bit more, if we can prepare the state $$\frac{1}{\sqrt{m}} \sum_{j=1}^{m} (-1)^{v_j} |b_j\rangle$$ then we produce a $|f\rangle$ state with $$f(x) = \sum_{j=1}^m (-1)^{b_j \cdot x +
v_j}.$$ Note that the way we have written things we can essentially ignore normalzation factors, because the final normalization factor ${\mathcal F}$ in our definition takes care of things. For
instance we could have written $f(x) = \alpha \sum_{j=1}^m (-1)^{b_j \cdot x + v_j}$ for some non-zero $\alpha$ and this would also correspond to the same state, ${\mathcal F}$ takes care of the
proper overall normalization.
Great! But where have we gotten ourselves. Well one thing we can note is that the $f$ we have produced has a nice interpretation. Consider the set of $m$ linear equations mod-$2$ $$b_1 \cdot x = v_1
\\ b_2 \cdot x = v_2 \\ \dots \quad \\ b_m \cdot x = v_m $$Then $f(x) = \sum_{j=1}^m (-1)^{b_j \cdot x + v_j}$ is directly proportional to a count of how many of these simultaneous mod-$2$ linear
equations we can satisfy. This should feel to you more like a traditional optimization problem. In particular it might remind you of $3$SAT. In $3$SAT one is given a set of $m$ clauses, each of
clause being a boolean expression which is made up of an OR of 3 boolean variables or their negations. The goal of the $3$SAT problem is to find if there is a satisfying assignment where each clause
evaluates to true. The optimization version of this problem is to find the a value of the boolean variables that maximizes the number of clauses that evaluate to TRUE. This is called max-$3$SAT The
problem we have reduced to is like this, except the clause are now mod-$2$ equations. For this reason this problem is called max-XORSAT.
But ok, now what have we really shown? There are objections. The first objection is that we haven’t talked about the complexity of preparing the state $\frac{1}{\sqrt{m}} \sum_{j=1}^{m} (-1)^{v_j} |
b_j\rangle$. The second objection is that we only see the maximum $x$ with probability $|f(x)|^2$, is this enough to figure out $x$ that maximizes $f(x)$ or that gives an $f(x)$ that is a good
approximation to the maximal value?
Let’s talk about the second problem first. What if, instead of preparing $|f\rangle$ we instead try to prepare $$|f^2\rangle := \frac{1}{\sqrt{{\mathcal F}_2}} \sum_x f(x)^2 |x\rangle.$$ Here ${\
mathcal F}_2$ is normalization factor to make the state normalized. This then will result in $x$ with probability proportional to $|f(x)|^4$, i.e. we will have even greater probability of seeing the
maximal value.
Going with this idea, lets check what happens when we apply $H^{\otimes n}$ to $|f^2\rangle$ just as we did for $|f\rangle$ (so we can figure out what to prepare to be able to produce $|f^2\rangle$
by $n$-qubit Hadmard-ing). Let’s use the $f(x)$ we described above for max-XORSAT. Then $$f(x)^2 = \left ( \sum_{j=1}^m (-1)^{v_j +\sum_{i=1}^n b_j \cdot x} \right)^2= \sum_{j_1,j_2=1}^m (-1)^{v_
{j_1} +v_{j_2} + \sum_{i=1}^n (b_{j_1} + b_{j_2}) \cdot x}.$$ We can split this sum into two terms, and using the fact that for $z\in {\mathbb Z}_2$ $z^2=0~{\rm mod}~2$ we can express this as $$f(x)^
2= m + \sum_{j_1 \neq j_2 = 1}^m (-1)^{v_{j_1} +v_{j_2} + \sum_{i=1}^n (b_{j_1} + b_{j_2}) \cdot x}$$
From this we can now calculate what the state is that, if we Hadamard it, we end up with $|f^2\rangle$. This is $$|\hat{f^2}\rangle=\frac{1}{ 2^{n/2} \sqrt{ {\mathcal F}_2}}\sum_{y \in {\mathbb Z}_2^
n} f(x)^2 (-1)^{x \cdot y} | y\rangle.$$ We can individually calculate the Fourier transform of the two terms in our expression for $f(x)^2$ above, but we do need to be careful to note the relative
normalization between these terms. $$\alpha \left(m|0\rangle + \sum_{j_1 \neq j_2=1}^{m} (-1)^{v_{j_1+j_2}} |b_{j_1}+b_{j_2}\rangle \right)$$where the addition in the ket is done bitwise modulo $2$.
Great, so we’ve….gotten ourselves a new set of states we want to be able to produces. We can be a great lyric writer and make a list $$ &|0\rangle \\ \sum_{j=1}^{m} &(-1)^{v_i} |b_i\rangle \\ \sum_
{j_1,j_2=1}^{m} & (-1)^{v_{j_1}+v_{j_2}}|b_{j_1}+b_{j_2}\rangle $$ Our list is of non-normalized states. How might we prepare such states (and superpositions of such states)? To do this we recall the
inverse trick. Suppose you you can compute the function $g(x)$ from $k$ bits to $l$ bits and also the inverse $g^{-1}(x)$ from $l$ bits to $k$ bits. We further assume that we are working over a
domain where we the inverse exist. In other words suppose you have quantum circuits that act as $U_g|x\rangle |y\rangle = |x\rangle |y\oplus g(x)\rangle$ and $U_{g^{-1}} |x\rangle|y\rangle = |x\
rangle |y \oplus g^{-1}(x)\rangle$ where $\oplus$ is bitwise addition and the first register has $k$ bits and the second $l$ bits. Then if one starts with a superposition over $|x\rangle$ then one
can convert it to a superposition over $g$ evaluated at the points in the superposition. i.e. $$U_{g^{-1}}^\updownarrow U_g \left( \frac{1}{\sqrt{N}} \sum_x |x\rangle \right) |0\rangle = |0\rangle \
left( \frac{1}{\sqrt{N}} \sum_x |f(x)\rangle \right)$$ Here $\updownarrow$ means that we apply the $U$ with the registers flipped. This is the inverse trick: if we can efficiently convert the
function and its inverse, then we can easily go from a superposition over inputs to a superposition over the function applied to these inputs.
Let’s now return to our list (good thing we made a list). The first state $|0\rangle$ is pretty easy to make. Lets look at the second and third states without the phase term $\sum_{j=1}^m |b_i\
rangle$ and $\sum_{j_1,j_2=1}^m |b_{j_1}+b_{j_2}\rangle$. Now comes the trick that gives us the title. Take the $b_i$s and put them each as rows into a matrix. This matrix has $m$ rows and $n$
columns. We unimaginatively call this $m$ by $n$ matrix $B$. We can think about this as the parity check matrix for a linear error correcting code. In particular we can think about this as a code
where we encode into $m$ bits. We haven’t said anything about the rate of this code, i.e. how many bits we are attempting to code into or the distance of the code. But given $m$ bits we can apply $B$
to these bits and we will produce a syndrome. The syndrome are the bits we can read out which, for a good code, will allow us to determine exactly what error occurred to information we encoded into
the code.
Under this microscope, that $B$ is a syndrome matrix for a linear error correcting code, what are the states in our list. First of all $|0\rangle$ is always a codeword in a linear code. But now look
at $\sum_{j=1}^m |b_i\rangle$. Each of these is the syndrome for code if there was exactly one bit flip error. And if we look at $\sum_{j_1,j_2=1}^m |b_{j_1}+b_{j_2}\rangle$, then this is the
syndrome for the code if there was exactly two bit flip errors. Return now to the inverse trick. Suppose that we prepare all single qubit errors, $$\frac{1}{\sqrt{m}} (|100\dots00\rangle + |010\
dots00\rangle+ \cdots + |000\dots01\rangle$$or written another way $\sum_{x \in {\mathbb Z}_2^m, |x|=1} |x\rangle$ where $|x|$ here is the hamming weight (count of the number of $1$s in $x$). And
further suppose that $B$ is a code which can correctly and efficiently correct single bit errors. Then we can use the inverse trick to turn this state in to $\sum_{j=1}^m |b_i\rangle$. Similarly if
$B$ is a code that can correct two bit flips errors, then we can take a superpostion over all binary strings with two $1$s and using the inverse trick convert that into $\sum_{j_1,j_2=1}^m |b_{j_1}
+b_{j_2}\rangle$ which are the syndromes for all two bit errors on the code.
OK so what do we have. Suppose the $B$ is the syndrome check matrix for an $m$ bit code and it has $n$ bit syndromes. Further suppose that this code has a distance $d$ so that it can correct $t=\
lfloor (d-1)/2 \rfloor$ bit flips and that we can, in polynomial time classically decode $t$ errors. Then we can take superpositions of bit strings with $t$ $1$s and $m-t$ $0$s and using the inverse
trick, we can produce all superpositions over error syndromes corresponding to $t$ errors. Ignoring normalization this is $$ \sum_{x \in {\mathbb Z}_2^m, |x|=t}|x\rangle \rightarrow \sum_{j_1 \neq
j_2 \neq \cdots \neq j_t} |b_{j_1}+b_{j_2}+\dots+b_{j_t}\rangle.$$
OK so now we need to prepare the states $\sum_{x \in {\mathbb Z}_2^m, |x|=t}|x\rangle$. These are so called Dicke states, which have been studied first in quantum optics, and then in a variety of
other settings in quantum computing. It turns out that they are easy to prepare quantum mechanically, for a recent paper on knowledge about how to do this see “Short-Depth Circuits for Dicke State
Preparation” by Andreas Bärtschi and Stephan Eidenbenz [arXiv:2207.09998] (a not so great way to do this is to use the inverse Schur transform, do we get to finally say that the Schur transform is
useful for quantum algorithms now? If yes, I think John Preskill owes me a beer?)
So we’ve seen that, if we interpret $B$ as a syndrome matrix, then we can take Dicke states and cover them to superpositions over code words with a given number of errors. It’s pretty easy to also
see that we can apply the phases, corresponding to the $(-1)^{v_j}$ terms for the states we want to prepare as well, these will just be $Z$ gates applied to the Dicke states where we only apply $Z$
where $v_j=1$. Finally we notice that the state we needed to prepare for $|\hat{f}^2\rangle$ was a superposition of two different numbers of errors, $0$ errors or $2$ errors. To handle this we note
that when decoding the syndrome, we can also calculate the number of errors. So if we started with $ \alpha |0\rangle + \beta|2\rangle$, and then conditionally create the appropriate Dicke states in
separate register,$$\alpha|0\rangle |D_0\rangle + \beta |2\rangle |D_2\rangle$$ where $|D_k\rangle$ is the appropriate Dicke state, then when we perform the uncompute step we can also uncompte that
first register, and end up in the appropriate superposition of syndomes with the correct normalization (the $\alpha$ and $\beta$ we want).
Pretty cool! While we’ve just worked out the simple case of $|f^2\rangle$, we can see how this generalizes. In particular for a given $B$ it will have a distance $d$ and can correct $t=\lfloor (d-1)/
2 \rfloor$ bit flips. Then we can produce the state $|f^l\rangle$ for $l \leq t$ by generalizing the above steps and assuming that we can efficiently decode up to $l$ bit errors in the code. The way
we do this is to first analytically express $f^l$ as a sum over terms which are sums over sums over syndromes with $0, 1, \dots, l$ errors. This in general will give us some $w_k$ weights that we
explicitly know. We then start by preparing $\sum_{k=0}^l w_k |k\rangle$. We then append an $m$ bit register and conditionally create the Dicke states with the first register number of $1$s, $$\sum_
{k=0}^{l} w_k |k\rangle |D_k\rangle.$$ Then we apply the appropriate $Z$s to the bits where $v_i=1$. Finally using the decoder which take the syndrome and creates the error as well as its weight we
can use the inverse trick to clear the first two registers.
The above is a general description of what the DQI algorithm does, i.e. the quantum part. But how well does it work and does it solve max-XORSAT more efficiently than the best classical algorithms?
At this point I should tell you that I’ve lied. There are a couple of ways I have lied. The first is that in general instead of preparing a state with $f^l$ the DQI paper describes making an
arbitrary degree $l$ polynomial. This is useful because it allows one to identify an “optimal” polynomial. The second lie is a bit bigger, in that there is another problem that is central in the DQI
paper and the is the max-LINSAT problem. While max-XORSAT has clauses that are made up of mod-$2$ equation, the max-LINSAT problem generalizes this. Instead of using ${\mathbb Z}_2$ this problem uses
${\mathbb Z}_p$ where $p$ is a prime (not exponentially big in problem size, so small). One can then look at codes over ${\mathbb Z}_p$. If we let $f_j$ denote function from ${\mathbb Z}_p$ to $\pm
1$, then the function you want to maximize in the max-LINSAT problem is $$f(x)=\sum_{j=1}^m f_i( b_j \cdot x)$$ where $b_j \in {\mathbb Z}_p^n$ and the term inside the parenthesis is done with
mod-$p$ arithmetic. The paper describes how to produce the $|f^l\rangle$ (and similarly for arbitrary polynomial) state. Again the main trick is to consider the decoding problem for the code with
syndrome matrix made up of the $b_j$s, but now over ${\mathbb Z}_p$ instead of ${\mathbb Z}_2$.
Given these two lies have been corrected, what can we say about how well this algorithm works? If in the max-LINSAT problem one has that $r$ of the $p$ possible values for each $f_i$ map to $+1$ (and
the other $p-r$ map to $-1$), then the authors are able to show a very nice asymptotic formula. Recall the $m$ is the number of clauses, $p$ is the base fields, $r$ is the just defined amount of bias
in the $f_i$ towards $+1$, and $l$ is the weight to which you can efficiently decode the code for. Then the expected number of clauses satisfied when $m \rightarrow \infty$ follows a beautiful
formula, which the authors call the semicircle law $$ \frac{\langle s \rangle}{m} = \left( \sqrt{\frac{l}{m} \left( 1 – \frac{r}{p} \right)} + \sqrt{\frac{r}{p} \left( 1 – \frac{l}{m} \right) } \
Great, so there is a good understanding of the performance of the algorithm, how does it compare to the classical world. Here one needs to define the parameters that one is looking at, and how the
problems are generated. Here there are two arenas that the authors look at. The first is for the max-XORSAT case. In that case, the authors are able to show that the quantum algorithm provides a
higher number of satisfied clauses than simulated annealing for random sparse instances of the problem. That’s great, simulated annealing is a top performing optimization strategy and even getting in
the same ballpark for it is very hard. But in this regime it turns out that the authors were also too clever and found a classical heuristic that can outperform DQI for the same regime. Note that
this isn’t the end of the story, it may be that there are regimes where the DQI does outperform all classical polynomial time algorithms. And in all fairness, the DQI result is a provable number of
clauses, whereas simulated annealing is a heuristic, the authors have tried to set a high bar!
Instead of looking at random instances, one could also look at more structured cases. Here the authors show that if one uses the Reed-Solomon code then the max-LINSAT problem becomes equivalent to
another problem, the Optimal Polynomial Intersection problem, and in this case DQI beats all known classical algorithms for this problem. The Optimal Polynomial Intersection problem is as follows.
Suppose you are given $p-1$ subsets of ${\mathbb Z}_p$, $F_1, F_2, \dots, F_{p-1}$. Then your goal is to find a $n-1$ degree polynomial over ${\mathbb Z}_p$ that maximizes the number of intersections
with these subsets, i.e. if $Q$ is the polynomial maximize the number of times $Q(1) \in F_1, Q(2) \in F_2, \dots$. This problem has been studied before in cryptographic literature and if one
converts DQI for Reed-Solomon to this problem it seems that the quantum algorithm can produce more intersecting regions than the best known classical algorithm. The margin is actually quite high, for
instance if the ration of $m$ over $n$ is $10$ the quantum can produce $0.7179\dots$ expected intersections, whereas the best classical produced $0.55$ expected intersections. Exciting!
Hey mom, thanks for reading this far! Everyone stand up and do the quantum algorithms dance!
But more seriously, definitely check out the paper. There are nice connections between this work and prior work, two that stand out are works over lattices from Regev, Aharonov, and Ta-Shma and then
more recent work by Yamakawa, Zhandy, Tillich, and Chailloux. What I like most about the work is that it feels very different than the more traditional places where we have seen “exponential”
speedups (those quotes are because we do not know how to prove much real separations, quantum or classically, for all we known factoring could be in classical polynomial time. The quotes don’t
detract from what is shown here, which is the gold standard of what people these days call “provable” exponential advantage. I personally find that nomenclature off because it puts the word
“provable” too close to “exponential”. YMMV if you aren’t a grumpy old dude like me). Another part I like is that it shows the power of creating a superposition over structured objects. We know, for
instance, that being able to prepare the superposition over all of the permutations of a graph would lead to a polynomial time quantum algorithm for graph isomorphism. We don’t know how to do that
preparation, whereas here we do know how to produce produce the superposition.
And finally I like this problem because it has a nice sort of structure for thinking of new algorithms. One can first look at other codes, and there is the nice semicircle law to guide you. Another
thing one can do is think about how this generalizes to other unitaries beyond the Fourier transform, what are these optimizing for? And finally one can also try the general method of taking an
objective function and performing work on this function (squaring, convolving, etc) and think if maybe that can yield useful new quantum algorithms. Lot’s to work with and I’m excited to see where
this all goes (and if the classical world can catch up to the quantum for the Optimal Polynomial Intersection problem. Go quantum!)
p.s. Stephen Jordan is talking at the excellent Simons Institute Quantum Colloquium tomorrow about DQI. If you aren’t following these colloquiums, you should! They are all most excellent and they
include a panel afterwards with experts who then talk about the talk. This captures the feeling of a real conference, where a lot of the best part is the follow up conversation in the hallways. | {"url":"https://dabacon.org/pontiff/2024/10/29/new-quantum-algorithm-dance-dqi-edition/","timestamp":"2024-11-02T12:34:00Z","content_type":"text/html","content_length":"116121","record_id":"<urn:uuid:5094d729-bfeb-4f89-a27e-89806b2f132b>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00162.warc.gz"} |
In this vignette we discuss some properties of a robust backfitting estimator for additive models, and illustrate the use of the package RBF that implements it. These estimators were originally
proposed in Boente G, Martinez A, Salibian-Barrera M. (2017). See also Martinez A. and Salibian-Barrera M. (2021).
Below we analyze two data sets. The first one shows the robustness of the robust backfitting estimators when a small proportion of very large outliers are present in the training set, and compares
them with those obtained with the standard backfitting algorithm. With the second example, we illustrate how these robust estimators can be interpreted as automatically detecting and downweighting
potential outliers in the training set. We also compare the prediction accuracy of the robust and classical backfitting algorithms.
Boston example
Consider the well-known Boston house price data of Harrinson and Rubinfeld (1978). This dataset was used as an example to model an additive model by Härdle et a. (2004). The data are available in the
MASS package. It contains \(n=506\) observations and 14 variables measured on the census districts of the Boston metropolitan area. Following the analysis in Härdle et a. (2004) we use the following
10 explanatory variables:
• crim: per capita crime rate by town (\(X_1\)),
• indus: proportion of non-retail business acres per town (\(X_2\)),
• nox: nitric oxides concentration (parts per 10 million) (\(X_3\)),
• rm: average number of rooms per dwelling (\(X_4\)),
• age: proportion of owner-occupied units built prior to 1940 (\(X_5\)),
• dis: weighted distances to five Boston employment centers (\(X_6\)),
• tax: full-value property tax rate per 10,000 (\(X_7\)),
• ptratio: pupil-teacher ratio by town (\(X_8\)),
• black: \(1,000(Bk-0.63)^2\) where \(Bk\) is the proportion of people of Afrom American descent by town (\(X_9\)),
• lstat: percent lower status of population (\(X_{10}\)).
The response variable \(Y\) is medv, the median value of the owner-occupied homes in 1,000 USD, and the proposed additive model is \[Y= \mu+ \sum_{j=1}^{10} g_j(\log(X_j))+ \epsilon.\] where \(\mu \
in \mathbb{R}\), \(g_j\) are unknown smooth functions, and \(\epsilon\) are random errors.
First we load the data, and transform the explanatory variables as required by the model above:
data(Boston, package='MASS')
dd <- Boston[, c(1, 3, 5:8, 10:14)]
dd[, names(dd) != 'medv'] <- log( dd[, names(dd) != 'medv'] )
Next, we load the RBF package
The robust backfitting estimators for each additive component are computed using robust kernel-based local polynomial regression. The model to be fit is specified using the standard formula notation
in R. We also need to specify the following arguments:
• windows: the bandwidths for the kernel estimators,
• degree: the degree of the polynomial used for the kernel local regression, defaults to 0,
• type: specifies the robust loss function, options are Huber or Tukey, defaults to Huber.
As with all kernel-based estimators, bandwidth selection is an important step. In this example we follow Härdle et al. (2004) and select bandwidths \(h_j\), \(1 \le j \le 10\), proportional to the
standard deviation of the corresponding explanatory variables \(\log(X_j)\). Specifically we set \(h_j = \hat{\sigma}_j / 2\):
We are now ready to compute the robust backfitting estimators:
Information about the fit can be obtained using the summary method:
#> Call:
#> backf.rob(formula = medv ~ ., data = dd, windows = bandw, degree = 0,
#> type = "Huber")
#> Estimate of the intercept: 22.20546
#> Estimate of the residual standard error: 0.22239
#> Robust multiple R-squared: 0.69872
#> Residuals:
#> Min. 1st Qu. Median Mean 3rd Qu. Max.
#> -18.41161 -1.251476 0.002993124 0.3273419 1.457033 31.67697
Note that the summary output includes a robust version of the R-squared coefficient, which is computed as \[ R^2_{rob}=\frac{\sum_{i=1}^n \rho\left((Y_i-\widehat{a})/\widehat{\sigma}\right)-\sum_{i=
1}^n \rho\left(R_i/\widehat{\sigma}\right)}{\sum_{i=1}^n \rho\left((Y_i-\widehat{a})/\widehat{\sigma}\right)} \, , \] where \(\rho\) is the loss function used by the M-kernel smoothers (as determined
by the argument type in the call to backf.rob), and \(\widehat{a}\) is a robust location estimator for a model without explanatory variables: \(Y=a+\epsilon\).
The plot method can be used to visualize the estimated additive components \(\hat{g}_j\), displayed over the corresponding partial residuals: \[ R_{ij}=Y_i-\hat{\mu}-\sum_{k\neq i}\hat{g}_k(X_{ik})
\, , \quad 1 \le i \le 506\, , \quad 1 \le j \le 10 \, . \]
By default, backf.rob computes fitted values on the training set. If predictions at a different specific point are desired, we can pass those points using the argumen point. For example, to obtain
predicted values at a point po given by the average of the (log transformed) explanatory variables, we can use the following command (note that this step implies re-fitting the whole model):
po <- colMeans(dd[, names(dd) != 'medv'])
robust.fit1 <- backf.rob(medv ~ ., data = dd, degree = 0, type = 'Huber',
windows = bandw, point = po)
The values of the estimated components evaluated at the corresponding coordinates of po are returned in the $prediction element:
#> crim indus nox rm age dis tax
#> [1,] 0.4502372 -0.2941408 0.5036051 -1.549345 0.484823 1.266611 -0.5900875
#> ptratio black lstat
#> [1,] -0.2291248 0.1912795 -0.1634099
In order to illustrate the behaviour of the robust fit when outliers are present in the data, we artifically introduce 1% of atypical values in the response variable:
We now calculate the robust estimators using the contaminated data set. Note that we expect them to be very similar to those we obtained before with the original training set.
robust.fit.new <- backf.rob(medv ~ ., data = dd2, degree = 0, type = 'Huber',
windows = bandw, point = po)
#> Call:
#> backf.rob(formula = medv ~ ., data = dd2, windows = bandw, point = po,
#> degree = 0, type = "Huber")
#> Estimate of the intercept: 22.21968
#> Estimate of the residual standard error: 0.22239
#> Robust multiple R-squared: 0.44603
#> Residuals:
#> Min. 1st Qu. Median Mean 3rd Qu. Max.
#> -18.35562 -1.249837 0.002044039 3.969449 1.489455 378.8072
#> crim indus nox rm age dis tax
#> [1,] 0.3669282 -0.3464365 0.5552477 -1.573039 0.5296952 1.248268 -0.5550474
#> ptratio black lstat
#> [1,] -0.2196106 0.1936705 -0.1590413
From the output above we verify that the predictions at the point po with both fits are very similar to each other.
Because the magnitude of the outliers affects the scale of the partial residuals plots, to compare both fits below we plot the robust estimators for each additive component trained on the original
and contaminated data sets (without including the partial residuals). In green and dashed lines the robust estimator computed with the original data set, and in blue and solid lines the robust
estimator computed with the contaminated data set. We see that, indeed, both sets of estimated additive components are very similar to each other.
for(j in 1:10) {
name.x <- names(dd)[j]
name.y <- bquote(paste(hat('g')[.(j)]))
oo <- order(dd2[,j])
plot(dd2[oo,j], robust.fit.new$g.matrix[oo,j], type="l", lwd=2, col='blue', lty=1,
xlab=name.x, ylab=name.y)
lines(dd2[oo,j], robust.fit$g.matrix[oo,j], lwd=2, col='green', lty=2)
It is easy to see that when we use the classical backfitting estimator the fits obtained when the training set contains outliers are dramatically different from the ones obtained with the original
data set. We use the package gam to compute the standard backfitting algorithm (based on local kernel regression smoothers). The bandwidths used to compute the classical estimates are again
proportional to the standard deviations of the explanatory variables, but we use a slightly larger coefficient to avoid numerical issues with the local fits. We set \(h_j=(3/4) \, \widehat{\sigma}_j
\), for \(1 \le j \le 10\):
#> Loading required package: splines
#> Loading required package: foreach
#> Loaded gam 1.22-2
fit.gam <- gam(medv ~ lo(crim, span=1.62) + lo(indus, span=0.58) +
lo(nox, span=0.15) + lo(rm, span=0.08) +
lo(age, span=0.46) + lo(dis, span=0.40) +
lo(tax, span=0.30) + lo(ptratio, span=0.09) +
lo(black, span=0.58) + lo(lstat, span=0.45), data=dd)
fits <- predict(fit.gam, type='terms')
fit.gam.new <- gam(medv ~ lo(crim, span=1.62) + lo(indus, span=0.58) +
lo(nox, span=0.15) + lo(rm, span=0.08) +
lo(age, span=0.46) + lo(dis, span=0.40) +
lo(tax, span=0.30) + lo(ptratio, span=0.09) +
lo(black, span=0.58) + lo(lstat, span=0.45), data=dd2)
fits.new <- predict(fit.gam.new, type='terms')
In the plots below, the standard backfitting estimates calculated with the original Boston data set are shown with orange and dashed lines, while those obtained with the contaminated training set are
shown with purple and solid lines.
Airquality example
The airquality data set contains 153 daily air quality measurements in the New York region between May and September, 1973 (Chambers et al., 1983). The interest is in modeling the mean Ozone (\(\mbox
{O}_3\)) concentration as a function of 3 potential explanatory variables: solar radiance in the frequency band 4000-7700 (Solar.R), wind speed (Wind) and temperature (Temp). We focus on the 111
complete entries in the data set.
ccs <- complete.cases(airquality)
aircomplete <- airquality[ccs, c('Ozone', 'Solar.R', 'Wind', 'Temp')]
pairs(aircomplete[, c('Ozone', 'Solar.R', 'Wind', 'Temp')], pch=19, col='gray30')
The scatter plot suggests that the relationship between ozone and the other variables in not linear and so we propose using an additive regression model of the form \[$$\label{eq:ozone-model} \mbox
{Ozone}=\mu+g_{1}(\mbox{Solar.R})+g_{2}(\mbox{Wind})+g_{3}(\mbox{Temp}) + \varepsilon \, .$$\] To fit this model above we use robust local linear kernel M-estimators and Tukey’s bisquare loss
function. These choices are set using the arguments degree = 1 and type='Tukey' in the call to the function backf.rob. The model is specified with the standard formula notation in R. The argument
windows is a vector with the bandwidths to be used with each kernel smoother. To estimate optimal values we used a robust leave-one-out cross-validation approach (see Boente et al., 2017). As a
robust prediction error measure we use mu^2 + sigma^2 where mu and sigma are M-estimators of location and scale of the prediction errors, respectively.
# Bandwidth selection with leave-one-out cross-validation
## Without outliers
# This takes a long time to compute (approx 380 minutes running
# R 3.6.1 on an Intel(R) Core(TM) i7-4790 CPU @ 3.60GHz)
a <- c(1/2, 1, 1.5, 2, 2.5, 3)
h1 <- a * sd(aircomplete[,2])
h2 <- a * sd(aircomplete[,3])
h3 <- a * sd(aircomplete[,4])
hh <- expand.grid(h1, h2, h3)
nh <- nrow(hh)
rmspe <- rep(NA, nh)
jbest <- 0
cvbest <- +Inf
n <- nrow(aircomplete)
for(i in 1:nh) {
# leave-one-out CV loop
preds <- rep(NA, n)
for(j in 1:n) {
tmp <- try( backf.rob(Ozone ~ Solar.R + Wind + Temp, point = aircomplete[j, -1],
windows = hh[i, ], epsilon = 1e-6, data = aircomplete,
degree = 1, type = 'Tukey', subset = c(-j) ))
if (class(tmp)[1] != "try-error") {
preds[j] <- rowSums(tmp$prediction) + tmp$alpha
tmp.re <- RobStatTM::locScaleM(preds - aircomplete$Ozone, na.rm=TRUE)
rmspe[i] <- tmp.re$mu^2 + tmp.re$disper^2
if( rmspe[i] < cvbest ) {
jbest <- i
cvbest <- rmspe[i]
(bandw <- hh[jbest,])
The resulting bandwidths are:
Now we use the robust backfitting algorithm to fit an additive model using Tukey’s bisquare loss (the default tuning constant for this loss function is 4.685) and the optimal bandwidths.
fit.full <- backf.rob(Ozone ~ Solar.R + Wind + Temp, windows = bandw,
epsilon = 1e-6, degree = 1, type = 'Tukey',
subset = ccs, data = airquality)
We can visually explore the estimated additive functions plotted over the corresponding partial residuals using the method plot:
As before, we use the R package gam to compute the classical additive model estimators for this model. Optimal bandwidths were calculated using leave-one-out cross-validation as before:
a <- c(.3, .4, .5, .6, .7, .8, .9)
hh <- expand.grid(a, a, a)
nh <- nrow(hh)
jbest <- 0
cvbest <- +Inf
n <- nrow(aircomplete)
for(i in 1:nh) {
fi <- rep(0, n)
for(j in 1:n) {
tmp <- gam(Ozone ~ lo(Solar.R, span=hh[i,1]) + lo(Wind, span=hh[i,2])
+ lo(Temp, span=hh[i,3]), data = aircomplete, subset=c(-j))
fi[j] <- as.numeric(predict(tmp, newdata=aircomplete[j, -1], type='response'))
ss <- mean((aircomplete$Ozone - fi)^2)
if(ss < cvbest) {
jbest <- i
cvbest <- ss
# Var1 Var2 Var3
# 0.7 0.7 0.5
The optimal bandwidths are 0.7, 0.7 and 0.5 for Solar.R, Wind and Temp, respectively, and we use them to compute the backfitting estimators:
fit.gam <- gam(Ozone ~ lo(Solar.R, span=.7) + lo(Wind, span=.7)+
lo(Temp, span=.5), data = aircomplete)
Both classical (in magenta and dashed lines) and robust (in blue and solid lines) fits are shown in the following plot together with the partial residuals obtained by the robust fit.
x <- as.matrix( aircomplete[ , c('Solar.R', 'Wind', 'Temp')] )
y <- as.vector( aircomplete[ , 'Ozone'] )
fits <- predict(fit.gam, type='terms')
for(j in 1:3) {
re <- fit.full$yp - fit.full$alpha - rowSums(fit.full$g.matrix[,-j])
plot(re ~ x[,j], type='p', pch=20, col='gray45', xlab=colnames(x)[j], ylab='')
oo <- order(x[,j])
lines(x[oo,j], fit.full$g.matrix[oo,j], lwd=2, col='blue', lty=1)
lines(x[oo,j], fits[oo,j], lwd=2, col='magenta', lty=2)
The two fits differ mainly on the estimated effects of wind speed and temperature. The classical estimate for \(g_1(\mbox{Temp})\) is consistently lower than the robust counterpart for \(\mbox{Temp}
\ge 85\). For wind speed, the non-robust estimate \(\hat{g}_2(\mbox{Wind})\) suggests a higher effect over Ozone concentrations for low wind speeds than the one given by the robust estimate, and the
opposite difference for higher speeds.
Since residuals from a robust fit can generally be used to detect the presence of atypical observations in the training data, we plot the boxplot of the residuals obtained by the robust fit and 4
possible outlying points (indicated with red circles) can be observed.
re.ro <- residuals(fit.full)
ou.ro <- boxplot(re.ro, col='gray80')$out
in.ro <- (1:length(re.ro))[ re.ro %in% ou.ro ]
points(rep(1, length(in.ro)), re.ro[in.ro], pch=20, col='red')
We highlight these suspicious observations on the scatter plot.
cs <- rep('gray30', nrow(aircomplete))
cs[in.ro] <- 'red'
os <- 1:nrow(aircomplete)
os2 <- c(os[-in.ro], os[in.ro])
pairs(aircomplete[os2, c('Ozone', 'Solar.R', 'Wind', 'Temp')],
pch=19, col=cs[os2])
Note that not all these suspected atypical observations are particularly extreme, or directly evident on the scatter plot. However, as we will show below, they do have an important effect on the
estimates of the components of the additive model.
The partial residuals corresponding to these points can be also visualized in red in the plot of the estimated curves.
# Plot both fits (robust and classical)
x <- as.matrix( aircomplete[ , c('Solar.R', 'Wind', 'Temp')] )
y <- as.vector( aircomplete[ , 'Ozone'] )
fits <- predict(fit.gam, type='terms')
for(j in 1:3) {
re <- fit.full$yp - fit.full$alpha - rowSums(fit.full$g.matrix[,-j])
plot(re ~ x[,j], type='p', pch=20, col='gray45', xlab=colnames(x)[j], ylab='')
points(re[in.ro] ~ x[in.ro,j], pch=20, col='red')
oo <- order(x[,j])
lines(x[oo,j], fit.full$g.matrix[oo,j], lwd=2, col='blue', lty=1)
lines(x[oo,j], fits[oo,j], lwd=2, col='magenta', lty=2)
To investigate whether the differences between the robust and non-robust estimators are due to the outliers, we recomputed the classical fit after removing them.
We ran a similar leave-one-out cross-validation experiment to select the spans for each the 3 univariate smoothers.
airclean <- aircomplete[-in.ro, c('Ozone', 'Solar.R', 'Wind', 'Temp')]
a <- c(.3, .4, .5, .6, .7, .8, .9)
hh <- expand.grid(a, a, a)
nh <- nrow(hh)
jbest <- 0
cvbest <- +Inf
n <- nrow(airclean)
for(i in 1:nh) {
fi <- rep(0, n)
for(j in 1:n) {
tmp <- gam(Ozone ~ lo(Solar.R, span=hh[i,1]) + lo(Wind, span=hh[i,2])
+ lo(Temp, span=hh[i,3]), data=airclean, subset=c(-j))
fi[j] <- as.numeric(predict(tmp, newdata=airclean[j,], type='response'))
ss <- mean((airclean$Ozone - fi)^2)
if(ss < cvbest) {
jbest <- i
cvbest <- ss
# Var1 Var2 Var3
# 0.7 0.8 0.3
We use the optimal bandwidths to compute non-robust fit.
airclean <- aircomplete[-in.ro, c('Ozone', 'Solar.R', 'Wind', 'Temp')]
fit.gam2 <- gam(Ozone ~ lo(Solar.R, span=.7) + lo(Wind, span=.8)+
lo(Temp, span=.3), data=airclean)
The following plot shows the estimated curves obtained with the classical estimator using the clean data together with the robust ones (computed on the whole data set). Outliers are highlighted in
fits2 <- predict(fit.gam2, type='terms')
dd2 <- aircomplete[-in.ro, c('Solar.R', 'Wind', 'Temp')]
for(j in 1:3) {
re <- fit.full$yp - fit.full$alpha - rowSums(fit.full$g.matrix[,-j])
plot(re ~ x[,j], type='p', pch=20, col='gray45', xlab=colnames(x)[j], ylab='')
points(re[in.ro] ~ x[in.ro,j], pch=20, col='red')
oo <- order(dd2[,j])
lines(dd2[oo,j], fits2[oo,j], lwd=2, col='magenta', lty=2)
oo <- order(x[,j])
lines(x[oo,j], fit.full$g.matrix[oo,j], lwd=2, col='blue', lty=1)
Note that both fits are now very close. An intuitive interpretation is that the robust fit has automatically down-weighted potential outliers and produced estimates very similar to the classical ones
applied to the clean observations.
Prediction comparison
Finally, we compare the prediction accuracy obtained with each of the fits. Because we are not interested in predicting well any possible outliers in the data, we evaluate the quality of the
predictions using a 5%-trimmed mean squared prediction error (effectively measuring the prediction accuracy on 95% of the data). We use this alpha-trimmed mean squared function:
tms <- function(a, alpha=.1) {
# alpha is the proportion to trim
a2 <- sort(a^2, na.last=NA)
n0 <- floor( length(a) * (1 - alpha) )
return( mean(a2[1:n0], na.rm=TRUE) )
We use 100 runs of 5-fold CV to compare the 5%-trimmed mean squared prediction error of the robust fit and the classical one. Note that the bandwidths are kept fixed at their optimal value estimated
dd <- airquality
dd <- dd[complete.cases(dd), c('Ozone', 'Solar.R', 'Wind', 'Temp')]
# 100 runs of K-fold CV
M <- 100
# 5-fold
K <- 5
n <- nrow(dd)
# store (trimmed) TMSPE for robust and gam, and also
tmspe.ro <- tmspe.gam <- vector('numeric', M)
ii <- (1:n)%%K + 1
for(runs in 1:M) {
tmpro <- tmpgam <- vector('numeric', n)
ii <- sample(ii)
for(j in 1:K) {
fit.full <- backf.rob(Ozone ~ Solar.R + Wind + Temp,
point=dd[ii==j, -1], windows = bandw,
epsilon = 1e-6, degree = 1, type = 'Tukey',
subset = (ii!=j), data = dd)
tmpro[ ii == j ] <- rowSums(fit.full$prediction) + fit.full$alpha
fit.gam <- gam(Ozone ~ lo(Solar.R, span=.7) + lo(Wind, span=.7)+
lo(Temp, span=.5), data = dd[ii!=j, ])
tmpgam[ ii == j ] <- predict(fit.gam, newdata=dd[ii==j, ], type='response')
tmspe.ro[runs] <- tms( dd$Ozone - tmpro, alpha=0.05)
tmspe.gam[runs] <- tms( dd$Ozone - tmpgam, alpha=0.05)
These are the boxplots. We see that the robust fit consistently fits the vast majority (95%) of the data better than the classical one. | {"url":"https://cran.case.edu/web/packages/RBF/vignettes/Examples.html","timestamp":"2024-11-02T10:55:17Z","content_type":"text/html","content_length":"225787","record_id":"<urn:uuid:06176a52-0323-418d-a8c2-199305239325>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00087.warc.gz"} |
What is E=mc2 Meaning - Definition
What is E=mc2 Meaning – Definition
E=mc2 is the equivalence of the mass and energy. Each nuclear process can be evaluated from an E = mc2 perspective. E=mc2 meaning. Periodic Table
E = mc^2 Meaning
At the beginning of the 20th century, the notion of mass underwent a radical revision. Mass lost its absoluteness. One of the striking results of Einstein’s theory of relativity is that mass and
energy are equivalent and convertible one into the other. Equivalence of the mass and energy is described by Einstein’s famous formula E = mc^2. In words, energy equals mass multiplied by the speed
of light squared. Because the speed of light is a very large number, the formula implies that any small amount of matter contains a very large amount of energy. The mass of an object was seen to be
equivalent to energy, to be interconvertible with energy, and to increase significantly at exceedingly high speeds near that of light. The total energy of an object was understood to comprise its
rest mass as well as its increase of mass caused by increase in kinetic energy.
In special theory of relativity certain types of matter may be created or destroyed, but in all of these processes, the mass and energy associated with such matter remains unchanged in quantity. It
was found the rest mass an atomic nucleus is measurably smaller than the sum of the rest masses of its constituent protons, neutrons and electrons. Mass was no longer considered unchangeable in the
closed system. The difference is a measure of the nuclear binding energy which holds the nucleus together. According to the Einstein relationship (E = mc^2) this binding energy is proportional to
this mass difference and it is known as the mass defect.
E=mc^2 represents the new conservation principle – the conservation of mass-energy.
This formule describes equivalence of mass and energy.
, where m is the small amount of mass and c is the speed of light.
What that means? If the nuclear energy is generated (splitting atoms, nuclear fusion), a small amount of mass (saved in the nuclear binding energy) transforms into the pure energy (such as kinetic
energy, thermal energy, or radiant energy).
The energy equivalent of one gram (1/1000 of a kilogram) of mass is equivalent to:
• 89.9 terajoules
• 25.0 million kilowatt-hours (≈ 25 GW·h)
• 21.5 billion kilocalories (≈ 21 Tcal)
• 85.2 billion BTUs
or to the energy released by combustion of the following:
• 21.5 kilotons of TNT-equivalent energy (≈ 21 kt)
• 568,000 US gallons of automotive gasoline
Any time energy is generated, the process can be evaluated from an E = mc^2 perspective.
We hope, this article, E=mc2 Meaning, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about
radiation and dosimeters. | {"url":"https://www.periodic-table.org/what-is-emc2-meaning-definition/","timestamp":"2024-11-02T01:35:55Z","content_type":"text/html","content_length":"174703","record_id":"<urn:uuid:1915f4ba-4de0-4a2d-946d-30ee55ced4b7>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00202.warc.gz"} |
Special Utility Matrices
16.3 Special Utility Matrices
Return an identity matrix.
If invoked with a single scalar argument n, return a square NxN identity matrix.
If supplied two scalar arguments (m, n), eye takes them to be the number of rows and columns. If given a vector with two elements, eye uses the values of the elements as the number of rows and
columns, respectively. For example:
eye (3)
⇒ 1 0 0
The following expressions all produce the same result:
eye (2)
eye (2, 2)
eye (size ([1, 2; 3, 4]))
The optional argument class, allows eye to return an array of the specified type, like
val = zeros (n,m, "uint8")
Calling eye with no arguments is equivalent to calling it with an argument of 1. Any negative dimensions are treated as zero. These odd definitions are for compatibility with MATLAB.
See also: speye, ones, zeros.
Return a matrix or N-dimensional array whose elements are all 1.
If invoked with a single scalar integer argument n, return a square NxN matrix.
If invoked with two or more scalar integer arguments, or a vector of integer values, return an array with the given dimensions.
To create a constant matrix whose values are all the same use an expression such as
val_matrix = val * ones (m, n)
The optional argument class specifies the class of the return array and defaults to double. For example:
val = ones (m,n, "uint8")
See also: zeros.
Return a matrix or N-dimensional array whose elements are all 0.
If invoked with a single scalar integer argument, return a square NxN matrix.
If invoked with two or more scalar integer arguments, or a vector of integer values, return an array with the given dimensions.
The optional argument class specifies the class of the return array and defaults to double. For example:
val = zeros (m,n, "uint8")
See also: ones.
Form a block matrix of size m by n, with a copy of matrix A as each element.
If n is not specified, form an m by m block matrix. For copying along more than two dimensions, specify the number of times to copy across each dimension m, n, p, …, in a vector in the second
See also: repelems.
Construct a vector of repeated elements from x.
r is a 2xN integer matrix specifying which elements to repeat and how often to repeat each element. Entries in the first row, r(1,j), select an element to repeat. The corresponding entry in the
second row, r(2,j), specifies the repeat count. If x is a matrix then the columns of x are imagined to be stacked on top of each other for purposes of the selection index. A row vector is always
Conceptually the result is calculated as follows:
y = [];
for i = 1:columns (r)
y = [y, x(r(1,i)*ones(1, r(2,i)))];
See also: repmat, cat.
The functions linspace and logspace make it very easy to create vectors with evenly or logarithmically spaced elements. See Ranges.
Return a row vector with n linearly spaced elements between base and limit.
If the number of elements is greater than one, then the endpoints base and limit are always included in the range. If base is greater than limit, the elements are stored in decreasing order. If
the number of points is not specified, a value of 100 is used.
The linspace function always returns a row vector if both base and limit are scalars. If one, or both, of them are column vectors, linspace returns a matrix.
For compatibility with MATLAB, return the second argument (limit) if fewer than two values are requested.
See also: logspace.
Return a row vector with n elements logarithmically spaced from 10^a to 10^b.
If n is unspecified it defaults to 50.
If b is equal to pi, the points are between 10^a and pi, not 10^a and 10^pi, in order to be compatible with the corresponding MATLAB function.
Also for compatibility with MATLAB, return the second argument b if fewer than two values are requested.
See also: linspace.
Return a matrix with random elements uniformly distributed on the interval (0, 1).
The arguments are handled the same as the arguments for eye.
You can query the state of the random number generator using the form
This returns a column vector v of length 625. Later, you can restore the random number generator to the state v using the form
You may also initialize the state vector from an arbitrary vector of length ≤ 625 for v. This new state will be a hash based on the value of v, not v itself.
By default, the generator is initialized from /dev/urandom if it is available, otherwise from CPU time, wall clock time, and the current fraction of a second. Note that this differs from MATLAB,
which always initializes the state to the same state at startup. To obtain behavior comparable to MATLAB, initialize with a deterministic state vector in Octave’s startup files (see Startup Files
To compute the pseudo-random sequence, rand uses the Mersenne Twister with a period of 2^{19937}-1 (See M. Matsumoto and T. Nishimura, Mersenne Twister: A 623-dimensionally equidistributed
uniform pseudorandom number generator, ACM Trans. on Modeling and Computer Simulation Vol. 8, No. 1, pp. 3–30, January 1998, http://www.math.sci.hiroshima-u.ac.jp/~m-mat/MT/emt.html). Do not use
for cryptography without securely hashing several returned values together, otherwise the generator state can be learned after reading 624 consecutive values.
Older versions of Octave used a different random number generator. The new generator is used by default as it is significantly faster than the old generator, and produces random numbers with a
significantly longer cycle time. However, in some circumstances it might be desirable to obtain the same random sequences as produced by the old generators. To do this the keyword "seed" is used
to specify that the old generators should be used, as in
which sets the seed of the generator to val. The seed of the generator can be queried with
However, it should be noted that querying the seed will not cause rand to use the old generators, only setting the seed will. To cause rand to once again use the new generators, the keyword
"state" should be used to reset the state of the rand.
The state or seed of the generator can be reset to a new random value using the "reset" keyword.
The class of the value returned can be controlled by a trailing "double" or "single" argument. These are the only valid classes.
See also: randn, rande, randg, randp.
Return random integers in the range 1:imax.
Additional arguments determine the shape of the return matrix. When no arguments are specified a single random integer is returned. If one argument n is specified then a square matrix (n x n) is
returned. Two or more arguments will return a multi-dimensional matrix (m x n x …).
The integer range may optionally be described by a two element matrix with a lower and upper bound in which case the returned integers will be on the interval [imin, imax].
The optional argument class will return a matrix of the requested type. The default is "double".
The following example returns 150 integers in the range 1–10.
Implementation Note: randi relies internally on rand which uses class "double" to represent numbers. This limits the maximum integer (imax) and range (imax - imin) to the value returned by the
bitmax function. For IEEE floating point numbers this value is 2^{53} - 1.
See also: rand.
Return a matrix with normally distributed random elements having zero mean and variance one.
The arguments are handled the same as the arguments for rand.
By default, randn uses the Marsaglia and Tsang “Ziggurat technique” to transform from a uniform to a normal distribution.
The class of the value returned can be controlled by a trailing "double" or "single" argument. These are the only valid classes.
Reference: G. Marsaglia and W.W. Tsang, Ziggurat Method for Generating Random Variables, J. Statistical Software, vol 5, 2000, http://www.jstatsoft.org/v05/i08/
See also: rand, rande, randg, randp.
Return a matrix with exponentially distributed random elements.
The arguments are handled the same as the arguments for rand.
By default, randn uses the Marsaglia and Tsang “Ziggurat technique” to transform from a uniform to a normal distribution.
The class of the value returned can be controlled by a trailing "double" or "single" argument. These are the only valid classes.
Reference: G. Marsaglia and W.W. Tsang, Ziggurat Method for Generating Random Variables, J. Statistical Software, vol 5, 2000, http://www.jstatsoft.org/v05/i08/
See also: rand, randn, randg, randp.
Return a matrix with Poisson distributed random elements with mean value parameter given by the first argument, l.
The arguments are handled the same as the arguments for rand, except for the argument l.
Five different algorithms are used depending on the range of l and whether or not l is a scalar or a matrix.
For scalar l ≤ 12, use direct method.
W.H. Press, et al., Numerical Recipes in C, Cambridge University Press, 1992.
For scalar l > 12, use rejection method.[1]
W.H. Press, et al., Numerical Recipes in C, Cambridge University Press, 1992.
For matrix l ≤ 10, use inversion method.[2]
E. Stadlober, et al., WinRand source code, available via FTP.
For matrix l > 10, use patchwork rejection method.
E. Stadlober, et al., WinRand source code, available via FTP, or H. Zechner, Efficient sampling from continuous and discrete unimodal distributions, Doctoral Dissertation, 156pp., Technical
University Graz, Austria, 1994.
For l > 1e8, use normal approximation.
L. Montanet, et al., Review of Particle Properties, Physical Review D 50 p1284, 1994.
The class of the value returned can be controlled by a trailing "double" or "single" argument. These are the only valid classes.
See also: rand, randn, rande, randg.
Return a matrix with gamma (a,1) distributed random elements.
The arguments are handled the same as the arguments for rand, except for the argument a.
This can be used to generate many distributions:
gamma (a, b) for a > -1, b > 0
beta (a, b) for a > -1, b > -1
r1 = randg (a, 1)
r = r1 / (r1 + randg (b, 1))
Erlang (a, n)
chisq (df) for df > 0
t (df) for 0 < df < inf (use randn if df is infinite)
r = randn () / sqrt (2 * randg (df / 2) / df)
F (n1, n2) for 0 < n1, 0 < n2
## r1 equals 1 if n1 is infinite
r1 = 2 * randg (n1 / 2) / n1
## r2 equals 1 if n2 is infinite
r2 = 2 * randg (n2 / 2) / n2
r = r1 / r2
negative binomial (n, p) for n > 0, 0 < p <= 1
r = randp ((1 - p) / p * randg (n))
non-central chisq (df, L), for df >= 0 and L > 0
(use chisq if L = 0)
r = randp (L / 2)
r(r > 0) = 2 * randg (r(r > 0))
r(df > 0) += 2 * randg (df(df > 0)/2)
Dirichlet (a1, … ak)
r = (randg (a1), …, randg (ak))
r = r / sum (r)
The class of the value returned can be controlled by a trailing "double" or "single" argument. These are the only valid classes.
See also: rand, randn, rande, randp.
The generators operate in the new or old style together, it is not possible to mix the two. Initializing any generator with "state" or "seed" causes the others to switch to the same style for future
The state of each generator is independent and calls to different generators can be interleaved without affecting the final result. For example,
rand ("state", [11, 22, 33]);
randn ("state", [44, 55, 66]);
u = rand (100, 1);
n = randn (100, 1);
rand ("state", [11, 22, 33]);
randn ("state", [44, 55, 66]);
u = zeros (100, 1);
n = zeros (100, 1);
for i = 1:100
u(i) = rand ();
n(i) = randn ();
produce equivalent results. When the generators are initialized in the old style with "seed" only rand and randn are independent, because the old rande, randg and randp generators make calls to rand
and randn.
The generators are initialized with random states at start-up, so that the sequences of random numbers are not the same each time you run Octave.^7 If you really do need to reproduce a sequence of
numbers exactly, you can set the state or seed to a specific value.
If invoked without arguments, rand and randn return a single element of a random sequence.
The original rand and randn functions use Fortran code from RANLIB, a library of Fortran routines for random number generation, compiled by Barry W. Brown and James Lovato of the Department of
Biomathematics at The University of Texas, M.D. Anderson Cancer Center, Houston, TX 77030.
Return a row vector containing a random permutation of 1:n.
If m is supplied, return m unique entries, sampled without replacement from 1:n.
The complexity is O(n) in memory and O(m) in time, unless m < n/5, in which case O(m) memory is used as well. The randomization is performed using rand(). All permutations are equally likely.
See also: perms. | {"url":"https://docs.octave.org/v4.0.0/Special-Utility-Matrices.html","timestamp":"2024-11-14T23:17:40Z","content_type":"text/html","content_length":"35485","record_id":"<urn:uuid:a8991481-836c-4369-9a1e-50df88ec53cd>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00331.warc.gz"} |
Functional Notations and Terminology
The notion of a function is one of the most basic in Mathematics. A set can be identified with its characteristic function. On the other hand, functions are defined in terms of sets. For various
reasons, among which historical are not the least important, mathematicians use many terms to describe essentially the same concept. Following is the list of competing terms:
• function
• association
• correspondence
• transformation
• mapping
• relation (multi-valued function)
• operator
• functional
Furthermore, there are numerous terms (vector, sequence, measure, length, volume, etc.) that are functions in disguise and their functional ancestry is seldom mentioned.
To define a function one needs three elements:
1. a domain where the function is defined,
2. a region from where the function draws its values, and
3. a rule that associates points from the domain with points from the region.
For a function $f$ from a domain $X$ to a region $Y$ we use the following notation:
$f: X \rightarrow Y.$
This is the only notation that refers to all three elements of the function definition. The rule, the third element, is hinted at implicitly by the function name $f$. Two functions with the same
domain and region but defined by different rules, will be distinguished by different function names. We already had one definition
Function is a correspondence $f$ between elements of a space $X$ and those of a space $Y$ such that any element $x$ of $X$ has a unique corresponding element $y$ of $Y$ which is denoted $y = f(x).$
Another way of saying that an element $y$ corresponds to an element $x$ by means of a function $f$ is $f: x\mapsto y.$ For numeric functions it's often possible to describe the rule with a formula as
in $f(x) = x^{2}$ which is the same as $f: x\mapsto x^{2}.$ However, for the function which is $0$ for all rational $x$ and $1,$ otherwise, there is no formula that uses only broadly accepted math
notations. Of course, if a function is used very often mathematicians may decide to standardize its name. After that point on such a named function may be used as a legitimate part of a formula.
The word association is not often used as a substitute for a function perhaps because it's judged to be more vague or fundamental than function. The word correspondence is mostly used in a
set-theoretical context when we talk of a 1-1 correspondence between sets. Transformation is the term used in geometry, mapping appears in topology. Customarily, operators are functions between
vector spaces, functionals are operators with $Y = \mathbb{R}.$ In case $Y = \{false, true\},$ the function is most exclusively called a predicate.
Relation is the one term that is best described in the framework of set theory. A (binary) relation $R$ between two sets $X$ and $Y$ is a subset of their direct sum: $R\subset X + Y = \{(x, y):\,x\in
X \, y\in Y\}.$ We often write $x R y$ to indicate the fact that $(x, y)\in R.$ Relation $R$ is a function iff $x R y_{1}$ and $x R y_{2}$ imply $y_{1} = y_{2}.$ This way a function is identified
with its graph.
If $X$ is a segment $\{1, 2 , \ldots, n\}$ of the set $\mathbb{N}$ of natural numbers then functions are called vectors and we write $f_{n}$ instead of $f(n).$ The same notation is used for sequences
$(X = \mathbb{N}).$ When $X = 2^{A}$ and $Y = \mathbb{R}^{+},$ the set of positive reals, we often call a function a measure. Some measures are reasonably termed length if $A = \mathbb{R},$ area if
$A = \mathbb{R}^{2},$ and volume if $A = \mathbb{R}^{3}.$
When $y = f(x),$ it is customary to say that "$y$ corresponds to $x$ by means of function $f.$" This may happen that no $y$ corresponds to two different $x$'s. If this is the case, the function $f: X
\rightarrow Y,$ is said to be injective or an injection. If, for every $y\in Y$ there is an $x\in X$ to which that $y$ corresponds, function $f$ is said to be surjective, a surjection, or to be onto.
A 1-1 correspondence is both injective and surjective, and vice versa. (A 1-1 correspondence is also said to be bijective or a bijection.) An injective function $f: X \rightarrow Y$ is a 1-1
correspondence between $X$ and $f(X) \subset Y.$ $(f(X)$ is the subset of all the values taken by various $x$ from $X.)$ $f$ is surjective iff $f(X) = Y.$
• Functional Notations and Terminology
|Contact| |Front page| |Contents| |Algebra|
Copyright © 1996-2018
Alexander Bogomolny | {"url":"https://www.cut-the-knot.org/do_you_know/function.shtml","timestamp":"2024-11-04T17:13:40Z","content_type":"text/html","content_length":"16977","record_id":"<urn:uuid:98ce9b7b-dbb0-4a3b-b917-920810c9c972>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00643.warc.gz"} |
Re: Operations with PROC IML
I have a square 10X10 data matrix that is a correlation matrix with values of 1 along the diagonal. I am trying to find the median correlation value in this data matrix, but I cannot seem to find a
function that will allow me to only extract the median from the bottom half of the matrix. Is there such a function in SAS IML? While the min() function seems to find the minimum, the median()
function is giving the median value of each column. Is there a way to get the median value of the values in the lower half of a symmetrical matrix?
E.g., in this case the median correlation in the R correlation matrix is (.44 +.56)/ 2 = 0.5:
R = {1 .56 .44 .09,
.56 1 .32 .88,
.44 .32 1 .77,
.09 .88 .77 1}
05-31-2022 03:57 PM | {"url":"https://communities.sas.com/t5/SAS-IML-Software-and-Matrix/Operations-with-PROC-IML/m-p/815986","timestamp":"2024-11-02T17:06:32Z","content_type":"text/html","content_length":"187097","record_id":"<urn:uuid:4e7e62a5-b2fa-441e-a1ef-2fc951c17e23>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00239.warc.gz"} |
Observation Find it Out: If you have Sharp Eyes Find the Word Bed among Bad in 15 Secs - EduViet Corporation
Observation Find it Out: If you have Sharp Eyes Find the Word Bed among Bad in 15 Secs
Brainteaser to find word bed
Human intelligence spans a range of abilities, from problem solving and memory to creativity and pattern recognition. Among all the fascinating ways to measure cognitive ability, mastering
brainteasers stands out.
It takes just 15 seconds to solve this puzzle and the race is on to uncover the hidden word bed that blends seamlessly into the WORDS Bad array. This is a rain forecast event that combines quick
thinking and insight.
You are watching: Observation Find it Out: If you have Sharp Eyes Find the Word Bed among Bad in 15 Secs
Observe to find out: If you have a keen eye, find the word “bed” in “bad” in 15 seconds
Here’s a fun challenge: Can you find the hidden “Word Bed” among a bunch of “WORDS Bad” words? Some people master it quickly, while others may need more time. This cool challenge is created and
designed to exercise your brain.
If you run into trouble, don’t worry. The solution is below.
But if you want a little hint, you can try breaking down the challenge step by step. This Missing Word Bed in the Word Bad Jungle puzzle is a perfect example of a picture puzzle that can stump even
the sharpest minds.
At first glance, the mission seems simple: find the park hidden in the darkness in just 15 seconds.
Observe to find out: If you have a keen eye, find the word “Bad” in 15 seconds: Solution
See more : Brain Teaser only 1% can solve: 15-1(12÷4+1)=?
Solving the hidden “Word Bed” within the confines of “WORDS Bad” in just 15 seconds brings a host of benefits beyond the excitement of the challenge itself.
First, this task improves your quick-thinking skills because you are forced to make quick decisions under pressure.
Over time, your attention to detail will be tested, training your eyes to capture the most subtle nuances in a limited time frame.
Countdown starts: 15… 9… 8… 7… 6… 5… 4… 3… 2… 1…
Don’t worry;
These puzzles challenge even the brightest minds.
Many people may not have recognized the “Word Bed” hidden within the “WORDS Bad” range within the given 15 seconds.
This is the answer!
90% of people can’t answer “5+2×5-2=?” – are you smart enough to solve this math puzzle?
See more : Brain Teaser: 2+6=3 Remove 3 Sticks to make this Equation Right I Matchstick Puzzle
Many people struggle to solve the mathematical puzzle “5 + 2 × 5 – 2 = ?”, resulting in a 90% failure rate. It tests understanding of the sequence of operations.
The key is to follow the BODMAS order of operations, where multiplication precedes addition and subtraction. First, we perform multiplication: 2 × 5 = 10. We then add 5 to the result: 5 + 10 = 15.
Finally, we subtract 2 from 15: 15 – 2 = 13. So, the answer is 13.
Brainteaser Math Quiz: Solve 5+5×5+5=?
Stimulate your thinking with this brainteaser math quiz: Can you solve the answer to 5 + 5 × 5 + 5=? Remember, understanding the order of operations is critical to meeting this challenge.
Following the order of operations of BODMAS, we first multiply 5 by 5 to get 5 + 25 + 5=?. Then, we add 5 and the result is 5 + 30 = ?. Finally, add the last 5, and the total is 35.
Brainteaser: Find the next term in 9, 19, 21, 43, 45,?
Can you spot the pattern and predict the next item in the sequence 9, 19, 21, 43, 45…? Uncovering underlying patterns is key to solving this puzzle.
Each term follows an interesting pattern where alternating (x 2 + 1) and (x 1 + 2) apply this 9, 9(x2+1)=19, 19(x1+2)=21, 21(x2 +1 )=43, 43(x1+2)=45, the next term is obtained from 45(x2+1)=91.
Brain teaser math speed test: 35÷5x(4+9)=?
This brainteaser shows a mathematical expression involving division, multiplication, and addition: 35 ÷ 5 x (4 + 9). It is critical to follow the order of operations (BODMAS) to correctly solve this
To solve this expression, first evaluate the value inside the brackets: 4 + 9 = 13. Then, divide 35 by 5 to get 7. Finally, multiply 7 by 13 to get the answer, 35 ÷ 5 x (4 + 9 ) = 91. Remember that
parentheses dictate the order of calculations to ensure correct calculations.
Brain teaser math test: 1+6=7, 2+7=16, 3+8=27, 6+11=?
This brainteaser shows a pattern where the sum of two numbers is calculated in a unique way. In the equation 1+6=7, 2+7=16, 3+8=27, 6+11=? The challenge is to decipher this pattern and apply it to
the next equation. The pattern involves multiplying the second number by the first number and then adding the second number. Apply the pattern 1+6 = (6×1)+1 = 6+1=7 to the fourth equation 6+11=
(11×6)+6 = 66+6=72. Therefore, the answer to this brainteaser is 72.
Disclaimer: The above information is for general information purposes only. All information on this website is provided in good faith, but we make no representations or warranties, express or
implied, as to the accuracy, adequacy, validity, reliability, availability or completeness of any information on this website.
Source: https://truongnguyenbinhkhiem.edu.vn
Category: Brain Teaser
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10.3: Bode Plots
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The Bode plot is a graphical response prediction technique that is useful for both circuit design and analysis. It is named after Hendrik Wade Bode, an American engineer known for his work in control
systems theory and telecommunications. A Bode plot is, in actuality, a pair of plots: One graphs the signal gain or loss of a system versus frequency, while the other details the circuit phase versus
frequency. Both of these items are very important in the design of well-behaved, optimal circuits.
Generally, Bode plots are drawn with logarithmic frequency axes, a decibel gain axis, and a phase axis in degrees. First, let’s take a look at the gain plot. A typical gain plot is shown Figure \(\
PageIndex{1}\). Remember, “gains” can be fractional, as with a voltage divider.
Figure \(\PageIndex{1}\): Generic gain plot.
Note how the plot is relatively flat in the middle, or midband, region. The gain value in this region is known as the midband gain. In purely passive circuits this value may be fractional (i.e., a
negative dB value). At either extreme of the midband region, the gain begins to decrease. The gain plot shows two important frequencies, \(f_1\) and \(f_2\). \(f_1\) is the lower break frequency
while \(f_2\) is the upper break frequency. The gain at the break frequencies is 3 dB less than the midband gain. These frequencies are also known as the half-power points, or corner frequencies.
Normally, amplifiers are only used for signals between \(f_1\) and \(f_2\). The exact shape of the rolloff regions will depend on the design of the circuit. For example, it is possible to design
amplifiers with no lower break frequency (i.e., a DC amplifier), however, all amplifiers will exhibit an upper break. The break points are caused by the presence of circuit reactances, typically
coupling and stray capacitances. The gain plot is a summation of the midband response with the upper and lower frequency limiting networks. Let’s take a look at the lower break, \(f_1\).
Lead Network Gain Response
Reduction in low frequency gain is caused by lead networks. A generic lead network is shown in Figure \(\PageIndex{2}\). It gets its name from the fact that the output voltage developed across \(R\)
leads the input. At very high frequencies the circuit will be essentially resistive. Conceptually, think of this as a simple voltage divider. The divider ratio depends on the reactance of \(C\). As
the input frequency drops, \(X_c\) increases. This makes \(V_{out}\) decrease. At very high frequencies, where \(X_c \ll R\), \(V_{out}\) is approximately equal to \(V_{in}\). This can be seen
graphically in Figure \(\PageIndex{3}\), where the frequency axis is normalized to \(f_c\). The break frequency (i.e., the frequency at which the signal has decreased by 3 dB) is found via the
standard equation,
\[f_c = \frac{1}{2\pi RC} \nonumber \]
Figure \(\PageIndex{2}\): Lead network.
Figure \(\PageIndex{3}\): Lead network gain plot.
The response below \(f_c\) will be a straight line if a decibel gain axis and a logarithmic frequency axis are used. This makes for very quick and convenient sketching of circuit response. The slope
of this line is 6 dB per octave (an octave is a doubling or halving of frequency, e.g., 800 Hz is 3 octaves above 100 Hz)^1. This range covers a factor of two in frequency. This slope may also be
expressed as 20 dB per decade, where a decade is a factor of 10 in frequency. With reasonable accuracy, this curve may be approximated as two line segments, called asymptotes, shown in Figure \(\
PageIndex{3}\) (blue). The shape of this curve is the same for any lead network. Because of this, it is very easy to find the approximate gain at any given frequency as long as \(f_c\) is known. It
is not necessary to go through reactance and phasor calculations. To create a general response equation, start with the voltage divider rule to find the gain:
\[\frac{V_{out}}{V_{i n}} = \frac{R}{R− j X_c} \nonumber \]
\[\frac{V_{out}}{V_{i n}} = \frac{R\angle 0}{\sqrt{R^2+X_c^2} \angle − \arctan \frac{X_c}{R}} \nonumber \]
The magnitude of this is,
\[|A_v| = \frac{R}{\sqrt{R^2+X_c^2}} \\ |A_v| = \frac{1}{\sqrt{1+ \frac{X_c^2}{R^2}}} \label{10.9} \]
Recalling that,
\[f_c = \frac{1}{2\pi RC} \nonumber \]
we may say,
\[R = \frac{1}{2\pi f_cC} \nonumber \]
For any frequency of interest, \(f\),
\[X_c = \frac{1}{2 \pi f C} \nonumber \]
Equating the two preceding equations yields,
\[\frac{f_c}{f} = \frac{X_c}{R} \label{10.10} \]
Substituting Equation \ref{10.10} in Equation \ref{10.9} gives,
\[A_v = \frac{1}{\sqrt{1+ \frac{f_c^2}{f^2}}} \label{10.11} \]
To express \(A_v\) in dB, substitute Equation \ref{10.11} into equation 10.2.5.
\[A'_v = 20 \log_{10} \frac{1}{\sqrt{1+ \frac{f_c^2}{f^2}}} \nonumber \]
After simplification, the final result is:
\[A'_v = −10 \log_{10} \left( 1+ \frac{f_c^2}{f^2} \right) \label{10.12} \]
\(f_c\) is the critical frequency,
\(f\) is the frequency of interest,
\(A'_v\) is the decibel gain at the frequency of interest.
Example \(\PageIndex{1}\)
A circuit has a lower break frequency of 40 Hz. How much signal is lost at 10 Hz?
\[A'_v = −10 \log_{10} \left( 1+ \frac{f_c^2}{f^2} \right) \nonumber \]
\[A'_v = −10 \log_{10} \left( 1+ \frac{40^2}{10^2} \right) \nonumber \]
\[A'_v = −12.3 dB \nonumber \]
In other words, the signal level is 12.3 dB lower than it is in the midband. Note that 10 Hz is 2 octaves below the break frequency. Because the cutoff slope is 6 dB per octave, each octave loses 6
dB. Therefore, the approximate result is −12 dB, which double-checks the exact result. Without the lead network, the gain would stay at 0 dB all the way down to DC (0 Hz.)
Lead Network Phase Response
At very low frequencies, the circuit of Figure \(\PageIndex{2}\) is largely capacitive. Because of this, the output voltage developed across \(R\) leads by 90 degrees. At very high frequencies the
circuit will be largely resistive. At this point \(V_{out}\) will be in phase with \(V_{in}\). At the critical frequency, \(V_{out}\) will lead by 45 degrees. A general phase graph is shown in Figure
\(\PageIndex{4}\). As with the gain plot, the phase plot shape is the same for any lead network. The general phase equation may be obtained from the voltage divider:
\[\frac{V_{out}}{V_{i n}} = \frac{R}{R− j X_c} \nonumber \]
\[\frac{V_{out}}{V_{i n}} = \frac{R\angle 0}{\sqrt{R^2+X_c^2} \angle − \arctan \frac{X_c}{R}} \nonumber \]
The phase portion of this is,
\[\theta = \arctan \frac{X_c}{R} \nonumber \]
By using equation 1.6, this simplifies to,
\[\theta = \arctan \frac{f_c}{f} \label{10.13} \]
\(f_c\) is the critical frequency,
\(f\) is the frequency of interest,
\(\theta\) is the phase angle at the frequency of interest.
Figure \(\PageIndex{4}\): Lead network phase response.
Often, an approximation, such as the blue line in Figure \(\PageIndex{4}\), is sufficient. By using Equation \ref{10.13}, you can show that this approximation is off by no more than 6 degrees at the
Example \(\PageIndex{2}\)
A telephone amplifier has a lower break frequency of 120 Hz. What is the phase response one decade below and one decade above?
One decade below 120 Hz is 12 Hz, while one decade above is 1.2 kHz.
\[\theta = \arctan \frac{f_c}{f} \nonumber \]
\[\theta = \arctan \frac{120 Hz}{12Hz} \nonumber \]
\(\theta = 84.3\) degrees one decade below \(f_c\) (i.e, approaching 90 degrees)
\[\theta = \arctan \frac{120 Hz}{1.2kHz} \nonumber \]
\(\theta = 5.71\) degrees one decade above \(f_c\) (i.e., approaching 0 degrees)
Remember, if a circuit or amplifier is direct-coupled, and has no lead networks, the phase will remain at 0 degrees right back to 0 Hz (DC).
Lag Network Response
Unlike its lead network counterpart, all systems will contain lag networks. In essence, it is little more than an inverted lead network. As you can see from Figure \(\PageIndex{5}\), it simply
transposes the \(R\) and \(C\) locations. Because of this, the response tends to be inverted as well. In terms of gain, \(X_c\) is very large at low frequencies, and thus \(V_{out}\) equals \(V_{in}
\). At high frequencies, \(X_{c}\) decreases, and \(V_{out}\) falls. The break point occurs when \(X_c\) equals \(R\). The general gain plot is shown in Figure \(\PageIndex{6}\). Like the lead
network response, the slope of this curve is −6 dB per octave (or −20 dB per decade.) Note that the slope is negative instead of positive. We can derive a general gain equation for this circuit in
virtually the same manner as we did for the lead network. The derivation is left as an exercise.
Figure \(\PageIndex{5}\): Lag network.
\[A'_v = −10 \log_{10} \left( 1+ \frac{f^2}{f_c^2} \right) \label{10.14} \]
\(f_c\) is the critical frequency,
\(f\) is the frequency of interest,
\(A'_v\) is the decibel gain at the frequency of interest.
Note that this equation is almost the same as Equation \ref{10.12}. The only difference is that \(f\) and \(f_c\) have been transposed.
Figure \(\PageIndex{6}\): Lag network gain plot.
In a similar vein, we may examine the phase response. At very low frequencies, the circuit is basically capacitive. Because the output is taken across \(C\), \(V_{out}\) will be in phase with \(V_
{in}\). At very high frequencies, the circuit is essentially resistive. Consequently, the output voltage across \(C\) will lag by 90 degrees. At the break frequency the phase will be −45 degrees. A
general phase plot is shown in Figure \(\PageIndex{7}\). As with the lead network,we may derive a phase equation. Again, the exact steps are very similar, and left as an exercise.
\[\theta = −90+ \arctan \frac{f_c}{f} \label{10.15} \]
\(f_c\) is the critical frequency,
\(f\) is the frequency of interest,
\(\theta\) is the phase angle at the frequency of interest.
Figure \(\PageIndex{7}\): Lag network phase response.
Example \(\PageIndex{3}\)
A medical ultra sound transducer feeds a lag network with an upper break frequency of 150 kHz. What are the gain and phase values at 1.6 MHz?
Because this represents a little more than a 1 decade increase, the approximate values are −20 dB and −90 degrees, from Figures \(\PageIndex{4}\) and \(\PageIndex{5}\), respectively. The exact values
\[A'_v = −10 \log_{10} \left( 1+ \frac{f^2}{f_c^2} \right) \nonumber \]
\[A'_v = −10 \log_{10} \left( 1+ \frac{1.6 MHz^2}{150 kHz^2} \right) \nonumber \]
\[A'_v = −20.6dB \nonumber \]
\[\theta = −90+ \arctan \frac{f_c}{f} \nonumber \]
\[\theta = −90+ \arctan \frac{150 kHz}{1.6MHz} \nonumber \]
\[\theta = −84.6 \text{ degrees} \nonumber \]
The complete Bode plot for this network is shown in Figure \(\PageIndex{8}\). It is very useful to examine both plots simultaneously. In this manner you can find the exact phase change for a given
gain quite easily. For example, if you look carefully at the plots of Figure \(\PageIndex{8}\), you will note that at the critical frequency of 150 kHz, the total phase change is −45 degrees.
Figure \(\PageIndex{8}\): Bode plot for 150 kHz lag.
Because this circuit involved the use of a single lag network, this is exactly what you would expect.
Rise Time versus Bandwidth
For pulse-type signals, the “speed” of a circuit is often expressed in terms of its rise time. If a square pulse such as Figure \(\PageIndex{9a}\) is passed into a simple lag network, the capacitor
charging effect will produce a rounded variation, as seen in Figure \(\PageIndex{9b}\). This effect places an upper limit on the duration of pulses that a given circuit can handle without producing
excessive distortion.
Figure \(\PageIndex{9a}\): Pulse rise time effect: Input to network.
By definition, rise time is the amount of time it takes for the signal to traverse from 10% to 90% of the peak value of the pulse. The shape of this pulse is defined by the standard capacitor charge
equation examined in earlier course work, and is valid for any system with a single clearly dominant lag network.
\[V_{out} = V_{peak} \left(1−\epsilon^{\frac{−t}{RC}} \right) \label{10.16} \]
Figure \(\PageIndex{9b}\): Pulse rise time effect: Output of network.
In order to find the time internal from the initial starting point to the 10% point, set \(V_{out}\) to \(0.1V_{peak}\) in Equation \ref{10.16} and solve for \(t_1\).
\[0.1V_{peak} = V_{peak} \left(1−\epsilon^{\frac{−t_1}{RC}} \right) \\ 0.1V_{peak} = V_{peak}−V_{peak} \epsilon^{\frac{−t_1}{RC}} \\ 0.9V_{peak} = V_{peak} \epsilon^{\frac{−t_1}{RC}} \\ 0.9 = \
epsilon^{\frac{−t_1}{RC}} \\ \log 0.9 = \frac{−t_1}{RC} \\ t_1 = 0.105 RC \label{10.17} \]
To find the interval up to the 90% point, follow the same technique using \(0.9V_{peak}\). Doing so yields:
\[t_2 = 2.303 RC \label{10.18} \]
The rise time, \(T_r\), is the difference between \(t_1\) and \(t_2\)
\[T_r = t_1−t_2 \\ T_r = 2.303RC−0.105 RC \\ T_r \approx 2.2 RC \label{10.19} \]
Equation \ref{10.19} ties the rise time to the lag network’s \(R\) and \(C\) values. These same values also set the critical frequency \(f_2\). By combining Equation \ref{10.15} with the basic
critical frequency relationship, we can derive an equation relating \(f_2\) to \(T_r\).
\[f_2 = \frac{1}{2 \pi RC} \nonumber \]
Solving \ref{10.19} in terms of \(RC\), and substituting yields
\[f_2 = \frac{2.2}{2\pi T_r} \\ f_2 = \frac{0.35}{T_r} \label{10.20} \]
\(f_2\) is the upper critical frequency,
\(T_r\) is the rise time of the output pulse.
Example \(\PageIndex{4}\)
Determine the rise time for a lag network critical at 100 kHz.
\[f_2 = \frac{0.35}{T_r} \nonumber \]
\[T_r = \frac{0.35}{f_2} \nonumber \]
\[T_r = \frac{0.35}{100 kHz} \nonumber \]
\[T_r = 3.5 \mu s \nonumber \]
^1The term octave is borrowed from the field of music. It gets its name from the fact that there are eight notes in the standard western scale: do-re-mi-fa-sol-la-ti-do. | {"url":"https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/AC_Electrical_Circuit_Analysis%3A_A_Practical_Approach_(Fiore)/10%3A_Decibels_and_Bode_Plots/10.3%3A_Bode_Plots","timestamp":"2024-11-14T01:23:17Z","content_type":"text/html","content_length":"145120","record_id":"<urn:uuid:2db951f8-ea02-47c5-b841-fec3219ac862>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00196.warc.gz"} |
New Experiments: Sum of Squares and Plotting I
6. New Experiments: Sum of Squares and Plotting I¶
Wait, I thought we were done with this after today’s homework? But…
Question (almost) none of you thought of: Is there such a thing as a conductor for a set of three integers? That is, is there a conductor for numbers of the form
• Well, what about for
• Okay, what about
• Ah, but what about
With this last example, we see that the conductor can (sometimes) be strictly less than the conductor for any group of two of them.
Do you think there a formula? What about for bigger sets?
• For sets of two, it turns out there is a complete answer which you have been busy discovering.
• For sets of three numbers, like
Notice the connection between theory and practice! Solving this problem could easily solve a “real-life” business problem, so a good algorithm would be very important. But finding such an algorithm
requires exploration and experimentation like we have been doing.
Click here if you want to go back to the table of content of this notebook. Otherwise, continue to the next Section.
With that in mind, let’s begin our next exploration. Hopefully you have had some fun experimenting with the condunctor problem. Today we will introduce a new problem to keep up the fun spirit of
experimental mathematics. Namely, what numbers can be written in the form
Here are a cases that do and do not work.
And so forth. Can you think of questions that you can explore experimentally? I can think of two questions:
• Which positive integers can be written as a sum of two squares?
• Which ones cannot?
• More questions? [comment]: <> (* Are there any special types of numbers that can be written? ) [comment]: <> (* Are there any special types of numbers that cannot be written? ) [comment]: <> (*
Can we get a formula for which numbers can be written? ) [comment]: <> (* What if we restrict
You can start by hand, but I encourage you to use SageMath to explore this soon too. And when you do, try to be more sophisticated than those this
After a while, we will come back and I will ask you about possible ways to rethink the problem on the sum of squares.
Can you think of a way to solve this problem? Hint: What do you know about equations of this form :math:`n = a^{2} + b ^{2}`?
This brings up the question of how to do such plotting in the first place? We will look at this in more detail in the next Section.
There is absolutely no way we could cover all the graphics in SageMath (or the program that does it for SageMath, matplotlib) in one day, or even a week. We will look at a bunch of different
resources you have access to, and see lots of different plot types. As we go through these resources, think of how they will be of in solving our new experiment or where you could have used them some
time back in your academic career. I definitely do not know everything about the plotting command. Nevertheless, you may interrupt me with your question so that we can explore the solution together.
• We will begin with the advanced 2D-plotting in the built-in live documentation. (This link only works if you are already using SageMath; the non-live version is here.
• There are two very important pages with many, MANY examples - the documentation for plot and that for showing graphics.
• The general plotting reference - there are a lot of surprises, like animation, etc.
Remember, you should feel free to stop listening for a while if you can want to try something out!
Once we’ve seen enough examples, please either start using SageMath to explore the sums of squares, or try to recreate your favorite graphic using SageMath! | {"url":"https://ems.aims.ac.za/pages/ems_2020_day_05_sum_of_squares_and_plotting_I.html","timestamp":"2024-11-02T02:24:21Z","content_type":"text/html","content_length":"19015","record_id":"<urn:uuid:9120455c-7317-4395-bf83-999976ef4ffd>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00899.warc.gz"} |
Methods of Orbit Determination
By Dan Boulet.
Product Information: 6.00" by 9.00", 564 pages, hardbound.
This book describes how the principles of celestial mechanics may be applied to determine the orbits of planets, comets, and Earth satellites. More specifically, it shows how a dedicated novice can
learn, by first-hand experience:
how orbital motion conforms to Newtonian physics
how a set of orbital elements can be translated into quantities which can be compared with observations, and
how a record of observed motion can be used to determine an orbit from scratch or improve a preliminary orbit.
Until recently, this exciting adventure with nature was beyond the reach of nearly all non-specialists. However the power of the microcomputer has swept away the drudgery of tedious calculations
fraught with endless opportunities for careless error. With this book and a microcomputer the enthusiast may have the satisfaction of conquering problems which preoccupied astronomy for hundreds of
years, and, in the process, gain a fresh appreciation for the genius and industry of the great mathematicians of the seventeenth, eighteenth, and nineteenth centuries.
This is a how-to-do-it book. Even though the derivations of many important relationships are described in some detail, the emphasis throughout is on practical applications. The reader need only
accept the validity of the key equations and understand their symbology in order to use the computer programs to explore the power of the mathematical models. All the important principles have been
reduced to complete computer programs written in simple BASIC that will execute directly on a Macintosh using Microsoft BASIC or (with the addition of a statement as line 1005 to reserve extra space
in memory) an IBM-PC using BASICA or GWBASIC. For clarity, each program is preceded by an algorithm that describes the sequence of computation and ties it to the mathematics in the text. Further, the
program is illustrated by at least one numerical example. Finally, the output from the examples is shown in the format produced by the computer routine. A magnetic media version of the source code is
available for the IBM-PC
Who will find this book of value?
Amateur astronomers who want to determine the orbits of planets, comets or Earth satellite
Teachers and students of basic calculus, physics, astronomy or computer programming for a source of material that illuminates and expands upon subjects covered in the classroom." | {"url":"https://shopatsky.com/products/methods-of-orbit-determination","timestamp":"2024-11-09T15:32:28Z","content_type":"text/html","content_length":"122106","record_id":"<urn:uuid:6919a375-7d18-4cff-b950-52fa3309c96a>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00096.warc.gz"} |
Uniformly Cauchy sequence
In mathematics, a sequence of functions $\{f_{n}\}$ from a set S to a metric space M is said to be uniformly Cauchy if:
• For all $\varepsilon > 0$, there exists $N>0$ such that for all $x\in S$: $d(f_{n}(x), f_{m}(x)) < \varepsilon$ whenever $m, n > N$.
Another way of saying this is that $d_u (f_{n}, f_{m}) \to 0$ as $m, n \to \infty$, where the uniform distance $d_u$ between two functions is defined by
$d_{u} (f, g) := \sup_{x \in S} d (f(x), g(x)).$
Convergence criteria
A sequence of functions {f[n]} from S to M is pointwise Cauchy if, for each x ∈ S, the sequence {f[n](x)} is a Cauchy sequence in M. This is a weaker condition than being uniformly Cauchy.
In general a sequence can be pointwise Cauchy and not pointwise convergent, or it can be uniformly Cauchy and not uniformly convergent. Nevertheless, if the metric space M is complete, then any
pointwise Cauchy sequence converges pointwise to a function from S to M. Similarly, any uniformly Cauchy sequence will tend uniformly to such a function.
The uniform Cauchy property is frequently used when the S is not just a set, but a topological space, and M is a complete metric space. The following theorem holds:
• Let S be a topological space and M a complete metric space. Then any uniformly Cauchy sequence of continuous functions f[n] : S → M tends uniformly to a unique continuous function f : S → M.
Generalization to uniform spaces
A sequence of functions $\{f_{n}\}$ from a set S to a metric space U is said to be uniformly Cauchy if:
• For all $x\in S$ and for any entourage $\varepsilon$, there exists $N>0$ such that $d(f_{n}(x), f_{m}(x)) < \varepsilon$ whenever $m, n > N$.
See also
This article is issued from
- version of the 7/21/2015. The text is available under the
Creative Commons Attribution/Share Alike
but additional terms may apply for the media files. | {"url":"https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Uniformly_Cauchy.html","timestamp":"2024-11-04T23:48:02Z","content_type":"text/html","content_length":"8114","record_id":"<urn:uuid:ae13fff1-7e8b-4fb0-8280-fc42bc8d8d21>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00478.warc.gz"} |
Binning a Histogram
Example 4.18 Binning a Histogram
This example, which is a continuation of Example 4.14, demonstrates various methods for binning a histogram. This example also illustrates how to save bin percentages in an OUTHISTOGRAM= data set.
The manufacturer from Example 4.14 now wants to enhance the histogram by using the ENDPOINTS= option to change the endpoints of the bins. The following statements create a histogram with bins that
have end points 3.425 and 3.6 and width 0.025:
title 'Enhancing a Histogram';
ods select HistogramBins MyHist;
proc univariate data=Trans;
histogram Thick / midpercents name='MyHist'
endpoints = 3.425 to 3.6 by .025;
The ODS SELECT statement restricts the output to the "HistogramBins" table and the "MyHist" histogram; see the section ODS Table Names. The ENDPOINTS= option specifies the endpoints for the histogram
bins. By default, if the ENDPOINTS= option is not specified, the automatic binning algorithm computes values for the midpoints of the bins. The MIDPERCENTS option requests a table of the midpoints of
each histogram bin and the percent of the observations that fall in each bin. This table is displayed in Output 4.18.1; the histogram is displayed in Output 4.18.2. The NAME= option specifies a name
for the histogram that can be used in the ODS SELECT statement.
The MIDPOINTS= option is an alternative to the ENDPOINTS= option for specifying histogram bins. The following statements create a histogram, shown in Output 4.18.3, which is similar to the one in
Output 4.18.2:
title 'Enhancing a Histogram';
proc univariate data=Trans noprint;
histogram Thick / midpoints = 3.4375 to 3.5875 by .025
outhistogram = OutMdpts;
Output 4.18.3 differs from Output 4.18.2 in two ways:
• The MIDPOINTS= option specifies the bins for the histogram by specifying the midpoints of the bins instead of specifying the endpoints. Note that the histogram displays midpoints instead of
• The RTINCLUDE option requests that the right endpoint of each bin be included in the histogram interval instead of the default, which is to include the left endpoint in the interval. This changes
the histogram slightly from Output 4.18.2. Six observations have a thickness equal to an endpoint of an interval. For instance, there is one observation with a thickness of 3.45 mils. In Output
4.18.3, this observation is included in the bin from 3.425 to 3.45.
The OUTHISTOGRAM= option produces an output data set named OutMdpts, displayed in Output 4.18.4. This data set provides information about the bins of the histogram. For more information, see the
section OUTHISTOGRAM= Output Data Set.
A sample program for this example, uniex08.sas, is available in the SAS Sample Library for Base SAS software. | {"url":"http://support.sas.com/documentation/cdl/en/procstat/63104/HTML/default/procstat_univariate_sect073.htm","timestamp":"2024-11-07T03:48:39Z","content_type":"application/xhtml+xml","content_length":"17952","record_id":"<urn:uuid:9b3ce222-67d3-4d38-b851-b890f8386a8a>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00506.warc.gz"} |
Simulate the 4 Percent Rule for Retirement with Python - Data Driven Money
Simulate the 4 Percent Rule for Retirement with Python
When it comes to planning for retirement, the 4% rule is considered the sacred rule for withdrawal. It essentially states, that given a diversified portfolio, a retiree can safely withdraw 4% from
their retirement nest egg every year in retirement (a 30-year retirement is assumed).
In this article I will cover how to construct a simulation with the 4% rule for retirement as the premise in Python. Within the code you can adjust your own planning considerations and plot out your
expected returns based on the random assignment of historical stock market returns over a defined period.
Note: I will be assuming a portfolio constructed of 100% Stocks (vs a 60% Stock and 40% Bond mix)… however this is easily modifiable.
You will need to have access to a Python environment either on your computer or in the cloud. If you would like more information on the 4% Rule so you can better follow the Python code then you can
check out this article I wrote on the topic for a quick primer.
Below is what we will cover in this article.
Simulating the 4% Rule with Python
Getting Started
In this first block of Python code, the only thing being done is importing the appropriate libraries for data manipulation and eventual plotting. The random library will be used to randomly select a
historical Stock Market return so it can be used to appropriately model a portfolio over the selected time periods.
# Will need these to help with manipulating our data
import numpy as np
import pandas as pd
import random
# Will use these to plot everything out
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
# Use Pandas to reformat columns to be displayed with only
# two decimal points... like a currency
pd.options.display.float_format = '{:6.2f}'.format
Establishing Initial Investment Criteria
In the next block of code the initial parameters that define our notional portfolio are set. Only 4 variables are necessary to fully describe what we are looking to model.
• InitialPrincipal – The starting retirement account balance.
• YearsUntilRetirement – The number of years left that the retiree will be actively contributing to their retirement accounts.
• YearlyContributionUntilRetirement – While not retired, the amount of money contributed on an annual basis to the retirement account being modelled.
• YearsInRetirment – The number of years after retirement that the simulation should model.
Note: For this particular simulation, we will be compounding returns annually. For a more granular approach you can switch to monthly compounding but given the time frames we are looking at combined
with the inherent volatility in returns I don’t believe its necessary.
# Set initial variables to define our simulation
InitialPrincipal = 100000
YearsUntilRetirement = 20
YearlyContributionUntilRetirement = 5000
YearsInRetirement = 40
After the initial variables are set, those same variables are appended to a Pandas DataFrame called data. There are 2 cases that need to be considered.
1. Case 1: The simulation is assuming the investor is not yet retired and that there are greater than 0 working years left. In this case, annual contributions will be expected to be non-zero.
2. Case 2: The simulation is run assuming that the portfolio is already in ‘retirement mode.’ In this case, we don’t need to consider an annual contribution.
# data generated from the simulation will be stored here
data = pd.DataFrame(columns=['Year',
# Year 1 Data is appended. Note, they need to be set differently
# depending if there are 0 years until retirement or more than 0
if (YearsInRetirement > 0) :
data = data.append({'Year': 1, 'Principal': InitialPrincipal,
'Distribution': 0,
'Rate': 0}, ignore_index = True)
if (YearsInRetirement == 0) :
data = data.append({'Year': 1, 'Principal': (0.96*InitialPrincipal),
'Contribution': 0, 'Distribution': (0.04*InitialPrincipal),
'Rate': 0}, ignore_index = True)
Using Historical S&P 500 Returns for Pool of Future Returns
Using the dataset provided by Slickcharts in CSV form here we can import the Total Return by the S&P 500 from 1926 to 2021. By bootstrapping from this dataset, we can simulate a given year’s return
in the future.
If the S&P is not a preferred benchmark for the portfolio that you are trying to model, then you can add in your own data here. For example, if you are looking to model a 60% Stock / 40% Bond
Portfolio then you can just input the data here. Also, if you are looking to adjust for inflation then you can simply just subtract each year’s measured inflation (e.g. CPI) from the returns column.
returns = pd.read_csv('sp500returns-since-1926.csv', header = None)
returns = returns.rename(columns ={0:'Year', 1:'Returns'})
Creating a Function for Investment Returns: Pre-Retirement
In order to calculate and store the simulation data (e.g., each year’s returns on principal and withdrawal amounts) we will be executing the same few lines of code repeatedly. Thus, for this exercise
I will create a function to handle everything with as little code as necessary.
But first let’s look at the components of what will constitute our new function. Essentially, two for loops will be used. The first loop will calculate the growth of the principal while the portfolio
has active contributions (during working years). This code can be seen below.
Last year’s interest is multiplied by a value found randomly in our random_return DataFrame and then the yearly contribution is added and stored as the current year’s principal. No withdrawal or
distribution is subtracted since the portfolio hasn’t reached retirement mode yet.
# Here we will iterate through the number of years until retirement
# adding the yearly contribution and compounding annually via a randomly
# selected value from our 'returns' DataFrame
for i in range(YearsUntilRetirement - 1):
# calculate new year field
NewYear = i+2
# grab a random interest rate
random_return = random.choice(returns['Returns']) /100
# apply the random interest rate to last year's balance and add contribution
NewPrincipal = data.loc[i]['Principal']*(1 + random_return)+YearlyContributionUntilRetirement
data = data.append({'Year': NewYear, 'Principal': NewPrincipal,
'Contribution': YearlyContributionUntilRetirement,
'Distribution': 0,
'Rate': random_return}, ignore_index = True)
Creating a Function for Investment Returns: During Retirement
The next for loop iterates through each year after retirement and calculates that year’s withdrawal rate and associated interest from a random value in random_return. 0.96 is the factor used to
decrement the principal since this excludes the 4% distribution (withdrawal) rate.
# Now we will iterate through our years of retirement with another for loop
for i in range(YearsInRetirement):
# calculate new year field
NewYear = data.iloc[i+YearsUntilRetirement-1]['Year'] + 1
# grab a random interest rate
random_return = random.choice(returns['Returns']) /100
# calculate the 4% distribution from last year's ending principal
NewDistribution = data.loc[i+YearsUntilRetirement-1]['Principal']*(.04)
# apply the random interest rate to 96% of last year's balance
NewPrincipal = (.96)*data.loc[i+YearsUntilRetirement-1]['Principal']*(1 + random_return)
data = data.append({'Year': NewYear, 'Principal': NewPrincipal,
'Contribution': 0,
'Distribution': NewDistribution,
'Rate': random_return}, ignore_index = True)
Below you can see the resulting DataFrame data. Notice the rate column… the returns for each year can vary widely just like the market. Afterall, these were actual market returns for the S&P 500 at
some point in history!
Creating a Function for Investment Returns: Putting it All Together
Below we actually create our function and name it retire4. By calling and running this function with the required arguments (the initial 4 variables we defined earlier) it will return a DataFrame
with the same columns as above.
def retire4(InitialPrincipal, YearsUntilRetirement, YearlyContributionUntilRetirement, YearsInRetirement):
# data generated from the simulation will be stored here
data = pd.DataFrame(columns=['Year',
# Year 1 Data is appended. Note, they need to be set differently
# depending if there are 0 years until retirement or more than 0
if (YearsInRetirement > 0) :
data = data.append({'Year': 1, 'Principal': InitialPrincipal,
'Distribution': 0,
'Rate': 0}, ignore_index = True)
if (YearsInRetirement == 0) :
data = data.append({'Year': 1, 'Principal': (0.96*InitialPrincipal),
'Contribution': 0, 'Distribution': (0.04*InitialPrincipal),
'Rate': 0}, ignore_index = True)
# Here we will iterate through the number of years until retirement
# adding the yearly contribution and compounding annually via a randomly
# selected value from our 'returns' DataFrame
for i in range(YearsUntilRetirement - 1):
# calculate new year field
NewYear = i+2
# grab a random interest rate
random_return = random.choice(returns['Returns']) /100
# apply the random interest rate to last year's balance and add contribution
NewPrincipal = data.loc[i]['Principal']*(1 + random_return)+YearlyContributionUntilRetirement
data = data.append({'Year': NewYear, 'Principal': NewPrincipal,
'Contribution': YearlyContributionUntilRetirement,
'Distribution': 0,
'Rate': random_return}, ignore_index = True)
# Now we will iterate through our years of retirement with another for loop
for i in range(YearsInRetirement):
# calculate new year field
NewYear = data.iloc[i+YearsUntilRetirement-1]['Year'] + 1
# grab a random interest rate
random_return = random.choice(returns['Returns']) /100
# calculate the 4% distribution from last year's ending principal
NewDistribution = data.loc[i+YearsUntilRetirement-1]['Principal']*(.04)
# apply the random interest rate to 96% of last year's balance
NewPrincipal = (.96)*data.loc[i+YearsUntilRetirement-1]['Principal']*(1 + random_return)
data = data.append({'Year': NewYear, 'Principal': NewPrincipal,
'Contribution': 0,
'Distribution': NewDistribution,
'Rate': random_return}, ignore_index = True)
return data
Below we will test our new function to see if it works as desired. We will put the variables defined earlier as arguments and store it in a DataFrame named data2. Subsequently, we’ll output the first
5 rows to see how it compares to our output above.
As you can see, it looks quite similar. The rate column has randomly assigned return values, so the numbers won’t match exactly but everything looks like the check out.
data2 = retire4(InitialPrincipal, YearsUntilRetirement, YearlyContributionUntilRetirement, YearsInRetirement)
Plotting Our Results for a Single Result
Now it’s time to see what a single trial of our simulation looks like when plotted out. Using MatPlotLib and Seaborn we can quickly plot out the results of data2.
I plotted a blue vertical line to denote the transition from working to retired. This also shows when contributions stop, and withdrawals start.
Given the long time period of the plot (60 years in this case), volatility seems to substantially increase during the later years, but this is just due to working with larger principals that have
compounded for long periods.
By eventually plotting out large numbers of trials we should be able to smooth out the variability during later years of retirement.
sns.set(rc={'figure.figsize':(16,8)}) # make the graph readably large
sns.set_style('whitegrid') # change the background grid
sns.set_palette('Greens') # hange the color scheme
plt.ticklabel_format(style='plain', axis = 'y') # prevent sci notation on y-axis
sns.lineplot(x="Year", y="Principal", data = data2, color = 'red').\
set_title('4% Rule: ' + str(YearsUntilRetirement) +" Years Working, " +
str(YearsInRetirement) + " Years Retired, $" +
str(YearlyContributionUntilRetirement)+" Contributed Each Year While Working")
# Add a vertical line to mark retirement
plt.axvline(x=YearsUntilRetirement, color= 'blue')
Simulating Retirement Outcomes Across 10 Trials
Now let’s simulate the results for 10 trials. To do this I will create a new DataFrame called simulation and simply call our previously created retire4 function 10 times in a for loop.
simulation = pd.DataFrame(columns=['Year',
for i in range(10):
simulation = simulation.append(retire4(InitialPrincipal,
ignore_index = True)
Using the describe() method available to us in simulation we can get some quick descriptive statistics to get an idea ahead of time of what our plot should show.
First, you can see that the count of Year is 600. This is just because we append 10 runs of 60 years. Next you can see the variability in Rate… there are 95 unique values in this field. It appears
that our random function did a good job at allowing our simulation to utilize most of the 96 values available in returns.
Plotting Retirement Outcomes Across 10 Trials
Now that we have generated the data, plotting out our 10-run simulation is simple. Seaborn will automatically plot a line for the mean and highlight the range of the Standard Deviation that bounds
the mean line.
sns.set(rc={'figure.figsize':(16,8)}) # make the graph readably large
sns.set_style('whitegrid') # change the background grid
sns.set_palette('bright') # hange the color scheme
plt.ticklabel_format(style='plain', axis = 'y') # prevent sci notation on y-axis
sns.lineplot(x="Year", y="Principal", data = simulation, color = 'red').\
set_title('4% Rule: ' + str(YearsUntilRetirement) +" Years Working, " +
str(YearsInRetirement) + " Years Retired, $" +
str(YearlyContributionUntilRetirement)+" Contributed Each Year While Working")
# Add a vertical line to mark retirement
plt.axvline(x=YearsUntilRetirement, color= 'blue')
As should be expected, the variability once again at the end of the 60-year period is significant. This should not be a cause for alarm. The large number of random outcomes possible is what creates
this wide channel.
From this simulation it does appear that the 4% rule is a safe withdrawal rate. It does not look like that the principal is ever at risk of being depleted.
But what does the 4% withdrawal amount actually look like? Is it enough to live on? I’ll cover that next.
Plotting Annual Distributions at 4% for 10 Trials
Plotting out the annual distributions or withdrawals for the 10-run simulation only requires a variable change in the code below. I changed the colors as well to help highlight the different
interpretation of the plot.
sns.set(rc={'figure.figsize':(16,8)}) # make the graph readably large
sns.set_style('whitegrid') # change the background grid
sns.set_palette('bright') # hange the color scheme
plt.ticklabel_format(style='plain', axis = 'y') # prevent sci notation on y-axis
sns.lineplot(x="Year", y="Distribution", data = simulation, color = 'green').\
set_title('4% Rule: ' + str(YearsUntilRetirement) +" Years Working, " +
str(YearsInRetirement) + " Years Retired, $" +
str(YearlyContributionUntilRetirement)+" Contributed Each Year While Working")
# Add a vertical line to mark first year of distribution
plt.axvline(x=YearsUntilRetirement+1, color= 'red')
As you can see above, the withdrawal rate does deviate from year to year but on average, it does generally go up. Keep in mind that this does not include the effects of inflation, however, the 4%
Rule tackles this by assuming your portfolio grows enough each year to make up for this shortfall.
Simulating Retirement Outcomes Across 100 Trials
Now let’s see how running 100 trials changes things. Theoretically, our Standard Deviation depicted by the light red and green channels will narrow as we run more trials.
This narrowing doesn’t necessarily give us better information on how our portfolio will perform, but it will highlight the long-term uptrend of the Stock Market.
simulation100 = pd.DataFrame(columns=['Year',
for i in range(100):
simulation100= simulation100.append(retire4(InitialPrincipal,
ignore_index = True)
# see the top few rows of simulation100
Plotting Retirement Outcomes Across 100 Trials
Below is the resulting plot of the principal. It frankly looks very similar to the 10 Trial plot with the exception that the channel is narrow, and the mean line is smoothed.
sns.set(rc={'figure.figsize':(16,8)}) # make the graph readably large
sns.set_style('whitegrid') # change the background grid
sns.set_palette('bright') # hange the color scheme
plt.ticklabel_format(style='plain', axis = 'y') # prevent sci notation on y-axis
sns.lineplot(x="Year", y="Principal", data = simulation100, color = 'red', ci = 95).\
set_title('4% Rule: ' + str(YearsUntilRetirement) +" Years Working, " +
str(YearsInRetirement) + " Years Retired, $" +
str(YearlyContributionUntilRetirement)+" Contributed Each Year While Working")
# Add a vertical line to mark retirement
plt.axvline(x=YearsUntilRetirement, color= 'blue')
Plotting Annual Distributions at 4% for 100 Trials
Plotting out the withdrawals and distributions at 4% also results in a similiar smoother / narrower version of the 10 Trial simulation. The interpretation, once again, is that a 4% withdrawal rate
seems to work just fine with a 100% Stock Portfolio over a long time horizon.
It should be also noted that this simulation assumes that the next 100 years in the Stock Market looks similar to the past 100 years. Smart folks out there would tell you that past performance
doesn’t guarantee future results… buyer beware.
sns.set(rc={'figure.figsize':(16,8)}) # make the graph readably large
sns.set_style('whitegrid') # change the background grid
sns.set_palette('bright') # hange the color scheme
plt.ticklabel_format(style='plain', axis = 'y') # prevent sci notation on y-axis
sns.lineplot(x="Year", y="Distribution", data = simulation100, color = 'green').\
set_title('4% Rule: ' + str(YearsUntilRetirement) +" Years Working, " +
str(YearsInRetirement) + " Years Retired, $" +
str(YearlyContributionUntilRetirement)+" Contributed Each Year While Working")
# Add a vertical line to mark first year of distribution
plt.axvline(x=YearsUntilRetirement+1, color= 'red')
Calculating the 4% Rule Simulation Descriptive Statistics
Now that we’ve crunched the numbers and ran our simulation let’s see what the plots are telling us in a more specific fashion. Below you will find that on average you would be able to withdraw
$46,820.95 during your first year of retirement. The standard deviation is significant at $36,396.
# find the descriptive statistics for the amount of money you can safely withdraw (4%)
# during the first year in retirement
simulation100.loc[simulation100['Year'] == (YearsUntilRetirement+1)]['Distribution'].astype(int).describe()
Below you can see that during the final year of retirement you can, on average, expect to withdraw $886,202.75. This may seem like a lot, but it would be helpful to consider inflation in this
After 60 years, if inflation compounds at 3% then $1.00 of spending power today would require $5.89 in the future. This means that $886,202.75 would have spending power of about $150,000 in today’s
# find the descriptive statistics for the amount of money you can safely withdraw (4%)
# during the last year in retirement
simulation100.loc[simulation100['Year'] == (YearsUntilRetirement+YearsInRetirement)]['Distribution'].astype(int).describe()
Below, you will see that on average you could expect to have $1.2 Million of principal during the first year of retirement.
# find the descriptive statistics for the account balance
# during the first year in retirement
simulation100.loc[simulation100['Year'] == (YearsUntilRetirement+1)]['Principal'].astype(int).describe()
And, at the end of retirement you could expect, on average, to have about $24.7 Million. That is the incredible power of compounding returns!
# find the descriptive statistics for the account balance
# during the last year in retirement
simulation100.loc[simulation100['Year'] == (YearsUntilRetirement+YearsInRetirement)]['Principal'].astype(int).describe()
Additional Considerations for Simulation
This is a great ‘first step’ simulation to get an idea if using the 4% rule for retirement withdrawals makes sense. Here are a few ideas and warnings on where to take the ideas presented in this
• It would be easy to modify the function we created to run simulations for other percentages such as 3%, 3.5% or even 4.5% and 5%.
• As mentioned previously, prior Stock Market performance does not predict future returns. Modifying the pool of returns for random assignment based on a more sophisticated outlook could
substantially change the result of the simulation.
• More trials does not necessarily mean more confidence in outcome. In fact, the more trials you run you will likely smooth out a few trials that could highlight problems with your investment
approach. The right sequence of poor returns combined with withdrawals could be devastating and be missed if you just look at the scenario mean and standard deviation.
• Inflation is generally considered ‘taken care of’ when using the 4% rule… thus, interpreting the final principal and withdrawal amounts need context. It does appear from our simulation that over
the long term the spending power of a retiree may go up, however.
Final Thoughts
I hope you enjoyed this article! If you have any other thoughts about how to make the simulation better or ideas on how to better communicate the results be sure to throw them in the comments below! | {"url":"https://datadrivenmoney.com/simulate-the-4-percent-rule-for-retirement-with-python/","timestamp":"2024-11-06T18:19:50Z","content_type":"text/html","content_length":"200785","record_id":"<urn:uuid:7184db9b-927b-4ff6-b258-9fa38051da8d>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00715.warc.gz"} |
Ascension - trick or treat
Ascension is a treat either way but one might have to be careful of when they ascend?
Does ascension work by adding 20% on top of base stats as a one time increase or overall? For example, if i ascend a r4 6*, when i do r5 will the ascension apply to the new base stats. Or, does it
pay off more to r5 first then ascend? Anybody know?
For the math ppl what im saying is
Is it, for example r4 has 10000 health (for easy computing) - ascend adds 20% or 2000 for a total of 12000. R5 then adds 5000 for a total of 17000.
Or is it base stats *20% so r4 is 10000 ascend is 12000. Then rank 5 is 15000 base and ascension reapplies to r5 adds 20* of 15000 to give 18000.
Every 10000 health can lose you 1000 health if ascending only adds to the existing base stats.
I thought ascension is only for maxed out champs
I thought ascension is only for maxed out champs
It is
I thought ascension is only for maxed out champs
It is
No any rank can be ascended
Shes r4 btw
Edit: more evidence
Ascension is a treat either way but one might have to be careful of when they ascend?
Does ascension work by adding 20% on top of base stats as a one time increase or overall? For example, if i ascend a r4 6*, when i do r5 will the ascension apply to the new base stats. Or,
does it pay off more to r5 first then ascend? Anybody know?
For the math ppl what im saying is
Is it, for example r4 has 10000 health (for easy computing) - ascend adds 20% or 2000 for a total of 12000. R5 then adds 5000 for a total of 17000.
Or is it base stats *20% so r4 is 10000 ascend is 12000. Then rank 5 is 15000 base and ascension reapplies to r5 adds 20* of 15000 to give 18000.
Every 10000 health can lose you 1000 health if ascending only adds to the existing base stats.
The stats stay the same so a r5 ascended will be same as if u ascend a r4 then r5
No any rank can be ascended
Shes r4 btw
Edit: more evidence
Oh cool thanks for the info!
I ascended AntMan and Wasp, 200 more t1 primordial dust to ascend Cassie and YJ.
Maybe more t1 next month for Ghost and Fantman
I ascended AntMan and Wasp, 200 more t1 primordial dust to ascend Cassie and YJ.
Maybe more t1 next month for Ghost and Fantman
Kang is looking for you
Ascension is a treat either way but one might have to be careful of when they ascend?
Does ascension work by adding 20% on top of base stats as a one time increase or overall? For example, if i ascend a r4 6*, when i do r5 will the ascension apply to the new base stats.
Or, does it pay off more to r5 first then ascend? Anybody know?
For the math ppl what im saying is
Is it, for example r4 has 10000 health (for easy computing) - ascend adds 20% or 2000 for a total of 12000. R5 then adds 5000 for a total of 17000.
Or is it base stats *20% so r4 is 10000 ascend is 12000. Then rank 5 is 15000 base and ascension reapplies to r5 adds 20* of 15000 to give 18000.
Every 10000 health can lose you 1000 health if ascending only adds to the existing base stats.
The stats stay the same so a r5 ascended will be same as if u ascend a r4 then r5
Is there something concrete saying that? Not saying you are wrong, but all it says is health and attack + 22%. My problem with this is that ascension seems like a one time boost, not an item that
boosts by 20%. Im just scared to ascend my kitty r4 if its gonna hurt me in the long run 😭
I ascended AntMan and Wasp, 200 more t1 primordial dust to ascend Cassie and YJ.
Maybe more t1 next month for Ghost and Fantman
Kang is looking for you
now you remind me to add Kang on my 5star ascension list, thanks
anyway apologies for crashing on someone's post. peace
I ascended my Sig 200 Doom. I figure it will be a while before I pull him, Rank him, and Sig him that high.
Pretty cool seeing him there though.
Ascension is a treat either way but one might have to be careful of when they ascend?
Does ascension work by adding 20% on top of base stats as a one time increase or overall? For example, if i ascend a r4 6*, when i do r5 will the ascension apply to the new base
stats. Or, does it pay off more to r5 first then ascend? Anybody know?
For the math ppl what im saying is
Is it, for example r4 has 10000 health (for easy computing) - ascend adds 20% or 2000 for a total of 12000. R5 then adds 5000 for a total of 17000.
Or is it base stats *20% so r4 is 10000 ascend is 12000. Then rank 5 is 15000 base and ascension reapplies to r5 adds 20* of 15000 to give 18000.
Every 10000 health can lose you 1000 health if ascending only adds to the existing base stats.
The stats stay the same so a r5 ascended will be same as if u ascend a r4 then r5
Is there something concrete saying that? Not saying you are wrong, but all it says is health and attack + 22%. My problem with this is that ascension seems like a one time boost, not an item
that boosts by 20%. Im just scared to ascend my kitty r4 if its gonna hurt me in the long run 😭
R5 as shown in vid was a 22% increase and so is a r4 so yes its the same stat increase all around no matter the rank
Ascension is a treat either way but one might have to be careful of when they ascend?
Does ascension work by adding 20% on top of base stats as a one time increase or overall? For example, if i ascend a r4 6*, when i do r5 will the ascension apply to the new base
stats. Or, does it pay off more to r5 first then ascend? Anybody know?
For the math ppl what im saying is
Is it, for example r4 has 10000 health (for easy computing) - ascend adds 20% or 2000 for a total of 12000. R5 then adds 5000 for a total of 17000.
Or is it base stats *20% so r4 is 10000 ascend is 12000. Then rank 5 is 15000 base and ascension reapplies to r5 adds 20* of 15000 to give 18000.
Every 10000 health can lose you 1000 health if ascending only adds to the existing base stats.
The stats stay the same so a r5 ascended will be same as if u ascend a r4 then r5
Is there something concrete saying that? Not saying you are wrong, but all it says is health and attack + 22%. My problem with this is that ascension seems like a one time boost, not an item
that boosts by 20%. Im just scared to ascend my kitty r4 if its gonna hurt me in the long run 😭
Kabam wouldn't design ascension that way. Also they showed a graph on the livestream that shows the stats of champions when they're ascended at each rank. You can ascend a r1 and once they're r5,
it'll be the same as if you ascended them at r5
Ascension is a treat either way but one might have to be careful of when they ascend?
Does ascension work by adding 20% on top of base stats as a one time increase or overall? For example, if i ascend a r4 6*, when i do r5 will the ascension apply to the new base
stats. Or, does it pay off more to r5 first then ascend? Anybody know?
For the math ppl what im saying is
Is it, for example r4 has 10000 health (for easy computing) - ascend adds 20% or 2000 for a total of 12000. R5 then adds 5000 for a total of 17000.
Or is it base stats *20% so r4 is 10000 ascend is 12000. Then rank 5 is 15000 base and ascension reapplies to r5 adds 20* of 15000 to give 18000.
Every 10000 health can lose you 1000 health if ascending only adds to the existing base stats.
The stats stay the same so a r5 ascended will be same as if u ascend a r4 then r5
Is there something concrete saying that? Not saying you are wrong, but all it says is health and attack + 22%. My problem with this is that ascension seems like a one time boost, not an
item that boosts by 20%. Im just scared to ascend my kitty r4 if its gonna hurt me in the long run 😭
Kabam wouldn't design ascension that way. Also they showed a graph on the livestream that shows the stats of champions when they're ascended at each rank. You can ascend a r1 and once they're
r5, it'll be the same as if you ascended them at r5
Ascension is a treat either way but one might have to be careful of when they ascend?
Does ascension work by adding 20% on top of base stats as a one time increase or overall? For example, if i ascend a r4 6*, when i do r5 will the ascension apply to the new base
stats. Or, does it pay off more to r5 first then ascend? Anybody know?
For the math ppl what im saying is
Is it, for example r4 has 10000 health (for easy computing) - ascend adds 20% or 2000 for a total of 12000. R5 then adds 5000 for a total of 17000.
Or is it base stats *20% so r4 is 10000 ascend is 12000. Then rank 5 is 15000 base and ascension reapplies to r5 adds 20* of 15000 to give 18000.
Every 10000 health can lose you 1000 health if ascending only adds to the existing base stats.
The stats stay the same so a r5 ascended will be same as if u ascend a r4 then r5
Is there something concrete saying that? Not saying you are wrong, but all it says is health and attack + 22%. My problem with this is that ascension seems like a one time boost, not an
item that boosts by 20%. Im just scared to ascend my kitty r4 if its gonna hurt me in the long run 😭
R5 as shown in vid was a 22% increase and so is a r4 so yes its the same stat increase all around no matter the rank
TYSM GUYS, LIFESAVERS! I missed the livestream so i missed out on that.
I ascended AntMan and Wasp, 200 more t1 primordial dust to ascend Cassie and YJ.
Maybe more t1 next month for Ghost and Fantman
Damn bro, how so fast 😭. I ascended my herc since i dont have a 6*. Might ascend quake next. Gonna ascend 6* kitty at the end of the event since we get enough for one 6* ascension. But congrats
man, great choices, something tells me you like the antman movies/comics.
I ascended AntMan and Wasp, 200 more t1 primordial dust to ascend Cassie and YJ.
Maybe more t1 next month for Ghost and Fantman
Appreciate the commitment to the theme, super cool
I ascended AntMan and Wasp, 200 more t1 primordial dust to ascend Cassie and YJ.
Maybe more t1 next month for Ghost and Fantman
Damn bro, how so fast 😭. I ascended my herc since i dont have a 6*. Might ascend quake next. Gonna ascend 6* kitty at the end of the event since we get enough for one 6* ascension. But
congrats man, great choices, something tells me you like the antman movies/comics.
We get an 50 additional dust from one of the monthly objectives.
Thanks everyone for the info! Was going to ask the exact same question, y’all are lifesavers
Is there something concrete saying that? Not saying you are wrong, but all it says is health and attack + 22%. My problem with this is that ascension seems like a one time boost, not an item
that boosts by 20%.
To give a similar scenario, seems like you would be wondering…
What happens if you apply a 20% Green Combo Boost, and *then* go and rank up someone while on a Combo Boost ?
Using your example, if your 10,000 HP champ becomes 12,000 HP on the 20% Green Combo Boost.
And then rank/level them up, to where their base health stats would then be 15,000.
Are you suggesting that their HP while still on that same Combo Boost would only be 17,000, as opposed to being 18,000 ??
(I can not say for sure that it would be 18,000. But I would place a bet that would be the case.)
can someone confirm for these current events the total tier 2 ascension dust we will get to ascend a 6star champ is 180 not 200? im all about done getting all of it, and it seems that 180 is the
most we can get for this event?
can someone confirm for these current events the total tier 2 ascension dust we will get to ascend a 6star champ is 180 not 200? im all about done getting all of it, and it seems that 180 is
the most we can get for this event?
Theres another 20 in the event its 200 | {"url":"https://forums.playcontestofchampions.com/en/discussion/comment/2375121","timestamp":"2024-11-07T01:30:27Z","content_type":"text/html","content_length":"310427","record_id":"<urn:uuid:831f9c15-4a91-48b6-a4a1-5e0df8736073>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00423.warc.gz"} |
Information Flow
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cor}{cor} \newcommand{\ac}[1]{\left\{#1\right\}} \)
Estimating Brain Information Flow
via Pathway Lasso
Yi Zhao, Xi (Rossi) Luo
Brown University
Department of Biostatistics
Center for Statistical Sciences
Computation in Brain and Mind
Brown Institute for Brain Science
CMStatistics, Seville, Spain
December 11, 2016
Funding: NIH R01EB022911; NSF/DMS (BD2K) 1557467; NIH P20GM103645, P01AA019072, P30AI042853; AHA
Yi Zhao
(4th Yr PhD Student, on the job market this year)
Task fMRI
• Task fMRI: performs tasks under brain scanning
• Story vs Math task: listen to story (treatment stimulus) or math questions (control), eye closed
• Not resting-state: "rest" in scanner
Goal: how brain processes story/math differently?
fMRI data: blood-oxygen-level dependent (BOLD) signals from each cube/voxel (~millimeters), $10^5$ ~ $10^6$ voxels in total.
fMRI Studies
This talk: one subject, two sessions (to test replicability)
Network Model with Stimulus
Question: quantify red, blue, and other pathways
from stimulus to orange outcome circle/region Heim et al, 09
• Activation: stimulus $\rightarrow$ brain region activity
• Connectivity: one brain region $\rightarrow$ another region
□ Whether not two or more brain regions "correlate"
• Pathway: stimulus $\rightarrow$ brain region A $\rightarrow$ region B
• Strong path: strong activation and strong conn
• Zero path: zero activation or zero conn, including
□ Zero activation + strong conn = zero
□ Strong activation + zero conn = zero
Mediation Analysis and SEM
• Pathway effect: $a \times b$indirect; residual: $c$direct
• Mediation analysis
□ Baron&Kenny, 86; Sobel, 82; Holland 88; Preacher&Hayes 08; Imai et al, 10; VanderWeele, 15;...
Mediation Analysis in fMRI
• Parametric Wager et al, 09 and functional Lindquist, 12 mediation, under (approx.) independent errors
□ Stimulus $\rightarrow$ brain $\rightarrow$ user reported ratings, one mediator
□ Usual assumption: $U=0$ and $\epsilon_1 \bot \epsilon_2$
• Parametric and multilevel mediation Yi and Luo, 15, with correlated errors for two brain regions
□ Stimulus $\rightarrow$ brain region A $\rightarrow$ brain region B, one mediator
□ Correlations between $\epsilon_1$ and $\epsilon_2$
• This talk: multiple mediator and multiple pathways
□ High dimensional: more mediators than sample size
□ Alt: dimension reduction by arXiv1511.09354 Chen, Crainiceanu, Ogburn, Caffo, Wager, Lindquist, 15
Full Pathway Model
• Stimulus $Z$, $K$ mediating brain regions $M_1, \dotsc, M_K$, Outcome region $R$
• Strength of activation ($a_k$) and connectivity ($b_k$, $d_{ij}$)
• Too complex, even for small $K = 2$ Daniel et al, 14
Reduced Pathway Model
• $A_k$: total inflow to mediator $M_k$; $B_k$: total conn
• Pathway effect: $A_k \times B_k$; Residual: $C$
Relation to Full Model
• Proposition: Our "total" parameters has explicit forms of "individual" flow parameters in the full model
• Proposition: Our $E_k$'s are correlated, but won't affect estimation (will affect variance)
• Reduced model: a step to select spatial mediators
□ Strong overall inflow and strong conn flow
• Favor reduced: challenging to determine the temporal order because of low temporal resolution
Regularized Regression
• Minimize the penalized least squares criterion
$$\sum_{k=1}^K \| M_k - Z A_k \|_2^2 + \| R - Z C - \sum_k M_k B_k \|_2^2 + \mbox{Pen}(A, B)$$
The choice of penalty $\mbox{Pen}(\cdot)$ to be discussed
□ All data are normalized (mean=0, sd=1)
• Want to select sparse pathways for high-dim $K$
• Alternative approach: two-stage LASSO Tibshirani, 96 to select sparse inflow and connection separately: $$ \sum_{k=1}^K \| M_k - Z A_k \|_2^2 + \lambda \sum_k | A_k | \\ \| R - Z C - \sum_k M_k
B_k \|_2^2 + \lambda \sum_k |B_k| $$
Penalty: Pathway LASSO
• Select strong pathways effects: $A_k \times B_k$
□ TS-LASSO: shrink to zero when $A$&$B$ moderate but $A\times B$ large
• Penalty (prototype) $$ \lambda \sum_{k=1}^K |A_k B_k| $$
□ Non-convex in $A_k$ and $B_k$
□ Computationally heavy and non-unique solutions
□ Hard to prove theory
• We propose the following general class of penalties$$ \lambda \sum_{k=1}^K ( |A_k B_k| + \phi A_k^2 + \phi B_k^2) $$
Theorem $$v(a,b) = |a b| + \phi (a^2 + b^2)$$ is convex if and only if $\phi\ge 1/2$. Strictly convex if $\phi > 1/2$.
Contour Plot of Different Penalties
• Non-differentiable at points when $a\times b = 0$
• Shrink $a\times b$ to zero
• Special cases: $\ell_1$ or $\ell_2$
• TS-LASSO: different $|ab|$ effects though $|a|+|b|$ same
Algorithm: ADMM + AL
• SEM/regression loss: $u$; Non-differnetiable penalty: $v$
• ADMM to address differentiability $$ \begin{aligned} \text{minimize} \quad & u(\Theta,D)+v(\alpha,\beta) \\ \text{subject to} \quad & \Theta=\alpha, \\ & D=\beta, \\ & \Theta e_{1}=1, \end
• Augmented Lagrangian for multiple constraints
• Iteratively update the parameters
• We derive theorem on explicit (not simple) updates
• Mixed norm penalty $$\mbox{PathLasso} + \omega \sum_k (|A_k| + |B_k|)$$
• Tuning parameter selection by cross validation
□ Reduce false positives via thresholding Johnston and Lu, 09
• Inference/CI: bootstrap after refitting
□ Remove false positives with CIs covering zero Bunea et al, 10
• Our PathLasso compares with TSLasso
• Simulate with varying error correlations
• Tuning-free comparison: performance vs tuning parameter (estimated effect size)
□ PathLasso outperforms under CV
Pathway Recovery
Our PathLasso (red) outperforms two-stage Lasso (blue)
Other curves: variants of PathLasso and correlation settings
Data: Human Connectome Project
• Two sessions (LR/RL), story/math task Binder et al, 11
• gICA reduces voxel dimensions to 76 brain maps
□ ROIs/clusters after thresholding
• Apply to two sess separately, compare replicability
□ Jaccard: whether selected pathways in two runs overlap
□ $\ell_2$ diff: difference between estimated path effects
• Tuning-free comparisons
Regardless of tuning, our PathLasso (red) has smaller cross-sess diff (selection and estimation) than TSLasso (blue)
Stim-M25-R and Stim-M65-R significant shown largest weight areas
• M65 responsible for language processing, larger flow under story
• M25 responsible for uncertainty, larger flow under math
• High dimensional pathway model
• Penalized SEM for pathway selection and estimation
• Convex optimization for non-convex products
□ Sufficient and necessary condition
□ Algorithmic development for complex optimization
• Improved estimation and selection accuracy
• Higher replicability in HCP data
• Manuscript: Pathway Lasso (arXiv 1603.07749)
Thank you!
Slides at: bit.ly/CMStat16
More info: BigComplexData.com | {"url":"https://talks.bigcomplexdata.com/PathwayLasso_CMStatistics_2016.html","timestamp":"2024-11-04T07:38:11Z","content_type":"text/html","content_length":"28686","record_id":"<urn:uuid:97008ea4-761e-4686-a9f6-51126ec8f7dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00236.warc.gz"} |
Concavity and the Second Derivative Test
Contributed by:
1. Determine intervals on which a function is concave upward or concave downward.
2. Find any points of inflection of the graph of a function.
3. Apply the Second Derivative Test to find the relative extrema of a function.
1. Applications of Differentiation
Copyright © Cengage Learning. All rights reserved.
2. Concavity and the Second
Derivative Test
Copyright © Cengage Learning. All rights reserved.
3. Determine intervals on which a function is concave
upward or concave downward.
Find any points of inflection of the graph of a function.
Apply the Second Derivative Test to find relative
extrema of a function.
5. You have seen that locating the intervals in which a function
f increases or decreases helps to describe its graph. In this
section, you will see how locating the intervals f' increases
or decreases can be used to determine where the graph of f
is curving upward or curving downward.
6. The following graphical interpretation of concavity is useful.
1. Let f be differentiable on an open interval I. If the graph
of f is concave upward on I, then the graph of f lies
above all of its tangent lines on I. [See Figure 3.23(a).]
Figure 3.23(a)
7. 2. Let f be differentiable on an open interval I. If the graph
of f is concave downward on I, then the graph of f lies
below all of its tangent lines on I. [See Figure 3.23(b).]
Figure 3.23(b) 7
8. Concavity
To find the open intervals on which the
graph of a function f is concave upward
or concave downward, you need to find
the intervals on which f' is increasing or
For instance, the graph of
is concave downward on the open interval
is decreasing there. (See Figure 3.24)
Figure 3.24
9. Concavity
Similarly, the graph of f is concave upward on the interval
because f' is increasing on .
The next theorem shows how to use the second derivative of
a function f to determine intervals on which the graph of f is
concave upward or concave downward.
To apply Theorem 3.7, locate the x-values at which
f" (x) = 0 or f" does not exist. Use these x-values to determine
test intervals. Finally, test the sign of f" (x) in each of the test
10. Example 1 – Determining Concavity
Determine the open intervals on which the graph of
is concave upward or downward.
Begin by observing that f is continuous on the entire real
Next, find the second derivative of f.
11. Example 1 – Solution cont’d
Because f ''(x) = 0 when x = ±1 and f'' is defined on the
entire line, you should test f'' in the intervals ,
(1, –1), and 11
12. Example 1 – Solution cont’d
The results are shown in the table and in Figure 3.25.
Figure 3.25
13. Concavity cont’d
The function given in Example 1 is continuous on the entire
real number line.
When there are x-values at which the function is not
continuous, these values should be used, along with the
points at which f"(x)= 0 or f"(x) does not exist, to form the
test intervals.
14. Points of Inflection
15. Points of Inflection
If the tangent line to the graph exists at such a point where
the concavity changes, that point is a point of inflection.
Three types of points of inflection are shown in Figure 3.27.
The concavity of f changes at a point of inflection. Note that the graph crosses its tangent line at a
point of inflection.
Figure 3.27 15
16. Points of Inflection
To locate possible points of inflection, you can determine
the values of x for which f"(x)= 0 or f"(x) does not exist. This
is similar to the procedure for locating relative extrema of f.
17. Example 3 – Finding Points of Inflection
Determine the points of inflection and discuss the concavity
of the graph of
Differentiating twice produces the following.
18. Example 3 – Solution cont’d
Setting f"(x) = 0, you can determine that the possible points
of inflection occur at x = 0 and x = 2.
By testing the intervals determined by these x-values, you
can conclude that they both yield points of inflection.
A summary of this testing is shown in
the table, and the graph of f is shown
in Figure 3.28.
Figure 3.28 18
19. Points of Inflection
The converse of Theorem 3.8 is not generally true. That is,
it is possible for the second derivative to be 0 at a point that
is not a point of inflection.
For instance, the graph of f(x) = x4 is shown in Figure 3.29.
The second derivative is 0 when
x= 0, but the point (0,0) is not a
point of inflection because the
graph of f is concave upward in
both intervals
Figure 3.29 19
20. The Second Derivative Test
21. The Second Derivative Test
In addition to testing for concavity, the second derivative can
be used to perform a simple test for relative maxima and
The test is based on the fact that if the graph of f is concave
upward on an open interval containing c , and f' (c)=0, then
f(c) must be a relative minimum of f.
22. The Second Derivative Test
Similarly, if the graph of a function f is concave downward on
an open interval containing c, and f' (c)=0, then f(c) must be a
relative maximum of f (see Figure 3.30).
Figure 3.30
23. The Second Derivative Test
24. Example 4 – Using the Second Derivative Test
Find the extrema of f(x) = –3x5 + 5x3
Begin by finding the critical numbers of f.
From this derivative, you can see that x = –1, 0, and 1 are
the only critical numbers of f.
By finding the second derivative
you can apply the Second Derivative Test.
25. Example 4 – Solution cont’d
Because the Second Derivative Test fails at (0, 0), you can
use the First Derivative Test and observe that f increases
to the left and right of x = 0.
26. Example 4 – Solution cont’d
So, (0, 0) is neither a relative minimum nor a relative
maximum (even though the graph has a horizontal tangent
line at this point). The graph of f is shown in Figure 3.31.
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Quantum Correlations are Weaved by the Spinors of the Euclidean Primitives
Christian, Joy (2017) Quantum Correlations are Weaved by the Spinors of the Euclidean Primitives. [Preprint]
Download (318kB) | Preview
The exceptional Lie group E8 plays a prominent role both in mathematics and theoretical physics. It is the largest symmetry group connected to the most general possible normed division algebra, that
of the non-associative real octonions, which --- thanks to their non-associativity --- form the only possible closed set of spinors that can parallelize the 7-sphere. By contrast, here we show how a
similar 7-sphere also arises naturally from the algebraic interplay of the graded Euclidean primitives, such as points, lines, planes, and volumes, characterizing the three-dimensional conformal
geometry of the physical space, set within its eight-dimensional Clifford-algebraic representation. Remarkably, the resulting algebra remains associative, and allows us to understand the origins and
strengths of all quantum correlations locally, in terms of the geometry of the compactified physical space, namely that of a quaternionic 3-sphere, S3, with S7 being the corresponding algebraic
representation space. Every quantum correlation can thus be understood as a correlation among a set of points of this S7, computed using manifestly local spinors within S3, thereby extending the
stringent bounds of +/-2 set
by the Bell-CHSH inequalities to the bounds of +/-2\/2 on the strengths of all possible correlations, in the same quantitatively precise manner as that predicted within quantum mechanics. The
resulting geometrical framework thus circumvents Bell's theorem by producing a deterministic and realistic framework that allows a locally causal understanding of all quantum correlations, without
requiring either remote contextuality or backward causation. We demonstrate this by first proving a general
theorem about the geometrical origins of the correlations predicted by arbitrarily entangled states, and then explicitly reproducing the strong correlations predicted by the EPR-Bohm and GHZ states.
The raison d'^etre of strong correlations turns out to be the twist in the Hopf bundle of S3 within S7.
Export/Citation: EndNote | BibTeX | Dublin Core | ASCII/Text Citation (Chicago) | HTML Citation | OpenURL
Social Networking: Share |
Item Type: Preprint
Keywords: Quantum Correlations, Local Realism, Local Causality, E_8, Normed Division Algebra, Octonions, Spinors, Euclidean Primitives, Conformal Geometry, Clifford Algebra, Quaternions,
3-sphere, 7-sphere, S^3, S^7, EPR-Bohm State, GHZ State, Hopf fibration
Specific Sciences > Physics > Quantum Gravity
Subjects: Specific Sciences > Physics > Quantum Mechanics
Specific Sciences > Physics > Relativity Theory
Specific Sciences > Physics > Symmetries/Invariances
Depositing Dr. Joy Christian
Date 19 Jan 2018 13:43
Last 19 Jan 2018 13:43
Item ID: 14305
Specific Sciences > Physics > Quantum Gravity
Subjects: Specific Sciences > Physics > Quantum Mechanics
Specific Sciences > Physics > Relativity Theory
Specific Sciences > Physics > Symmetries/Invariances
Date: 3 May 2017
URI: https://philsci-archive.pitt.edu/id/eprint/14305
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ounce to kg
We assume you are converting between ounce-force and kilogram-force. There are 16 avoirdupois ounces in an avoirdupois pound. Kilograms. ›› Quick conversion chart of ounce to kg The answer is
13.887386556746. For quick reference purposes, below is a conversion table that you can use to convert from ounce to kg. Note: Ounce is an imperial or United States customary unit of weight.Kilogram
is a metric unit of weight. Task: Convert 16 ounces to kilograms (show work) Formula: oz x 0.0283495231 = kg Calculations: 16 oz x 0.0283495231 = 0.45359237 kg Result: 16 oz is equal to 0.45359237
kg. The answer is 35.273962070595. 1.81437 Kilograms (kg) Ounces : The ounce (abbreviated "oz") is a unit of mass with several definitions, the most popularly used being equal to approximately 28
grams. The symbol is "kg".You may also visit weight and mass conversion to convert all weight and mass units. From. The pounds and ounces to kilograms (kg to lbs) calculator only requires you to
enter the pounds and ounces (lbs oz) value and then click Submit. 1 oz = 0.0283495231 kg. We assume you are converting between ounce inch and kilogram centimeter. Pounds and ounces to kg, g and oz
conversion. How many kilograms in 85 ounces: If m oz = 85 then m kg = 2.4097094635 kg. 1 ounce is equal to 0.028349523125 kilogram. Using this calculator you can convert pounds and/or ounces to kg,
grams, pounds simultaneously. The ounce is commonly used as a unit of mass in the United States, Canada and sometimes in Australia and New Zealand. How many ounces in 1 kilograms? To convert kg to
pounds and ounces, please visit kg to pounds and ounces.. What is a Kilogram? The kilogram is the base unit of mass in the International (SI) System of Units, and is accepted on a day-to-day basis as
a unit of weight (the gravitational force acting on any given object). 1 newton meter is equal to 141.61193227806 oz-in, or 10.197162129779 kg-cm. You can view more details on each measurement unit:
oz-in or kg-cm The SI derived unit for torque is the newton meter. To. 1 Troy Ounce = 0.031103477 Kilograms (rounded to 8 digits) Display result as. 1 ounce (oz) = 0.0283495231 kilograms (kg) =
28.3495231 grams (g) = 28 349.5231 milligrams (mg) = 0.0625 pounds (lb). 1 kg = 35.27396195 oz. One ounce is equal to 437.5 grains. m kg = m oz ×0.0283495231. Kilogram is a metric system mass unit.
kg = t oz _____ 32.151. 16 Ounces makes 1 pound. 1 Ounces = 0.0283 Kilos: 10 Ounces = 0.2835 Kilos: 2500 Ounces = 70.8739 Kilos: 2 Ounces = 0.0567 Kilos: 20 Ounces = 0.567 Kilos: 5000 Ounces = 141.75
Kilos: 3 Ounces = 0.085 Kilos: 30 Ounces = 0.8505 Kilos: 10000 Ounces = 283.5 Kilos: 4 Ounces = 0.1134 Kilos: 40 Ounces = 1.134 Kilos: 25000 Ounces = 708.74 Kilos: 5 Ounces = 0.1417 Kilos: 50 Ounces
= 1.4175 Kilos: 50000 Ounces = 1417.48 Kilos Conversion Table. How many oz-in in 1 kg-cm? Troy Ounces to Kilograms formula. Interesting facts. The size of an ounce … You can view more details on each
measurement unit: ounces or kilograms The SI derived unit for force is the newton. The kilograms (kg) value will then be displayed. swap units ↺ Amount. ounce or kg The SI base unit for mass is the
kilogram. Conversion factors for pounds to kilograms (lbs to kg) Use this page to learn how to convert between ounces and kilograms. 1 newton is equal to 3.5969431019354 ounces, or 0.10197162129779
kilograms. Easy troy oz to kg conversion. A troy ounce is a unit of weight equal to 1/12 th of a troy pound. Ounce is represented as avoirdupois ounce and abbreviated as (oz). Type in your own
numbers in the form to convert the units! Note that rounding errors may occur, so always check the results. How to convert ounces to kilograms [oz to kg]:. It weighs about 10% more than a customary
ounce and is used to measure the weight of precious metals and gemstones. How many kilograms in an ounce: If m oz = 1 then m kg = 0.0283495231 kg. Kilograms, Pounds & Ounces Conversion 1 Kilogram = …
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Omen Iv: The Awakening
Being Together Meaning
Que Es Thyme In Spanish | {"url":"http://kpsystem.com.cn/etrade-day-fzdnvrq/ounce-to-kg-20fcec","timestamp":"2024-11-08T22:56:15Z","content_type":"text/html","content_length":"17602","record_id":"<urn:uuid:4d77fc99-52b9-428e-8c0f-5b91d4da8d2e>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00388.warc.gz"} |
The two horizontal lines shown in the above figure are parallel to each other. Which of the following does NOT equal ${180^\
Hint: We will be solving the question by individually checking the options provided to us. We will use the properties of angles such as
$\left( 1 \right)$ Sum of Supplementary angles is ${180^\circ }$.
$\left( 2 \right)$ Sum of all interior angles of a triangle is ${180^\circ }$.
$\left( 3 \right)$ Vertical angles are equal.
$\left( 4 \right)$ Corresponding angles are equal.
Complete step-by-step answer:
Let us add some more angles in the figure in order to understand better
Checking option \[\left( A \right)\quad {\left( {p + r} \right)^\circ }\]
As we can observe that
$\angle p + \angle m = {180^\circ }$ (Supplementary angles)
And $\angle m = \angle r$ (Corresponding angles)
$ \Rightarrow \angle p + \angle m = \angle p + \angle r = {180^\circ }$ (Since they are equal)
Checking option \[\left( B \right)\quad {\left( {p + t} \right)^\circ }\]
We can see that from the data given we cannot conclude that the value of \[{\left( {p + t} \right)^\circ } = {180^\circ }\].
Therefore, we will check other options.
Checking option \[\left( C \right)\quad {\left( {q + s} \right)^\circ }\]
As we can observe that
$\angle q + \angle n = {180^\circ }$ (Supplementary angles)
And $\angle n = \angle s$ (Corresponding angles)
$ \Rightarrow \angle q + \angle n = \angle q + \angle s = {180^\circ }$ (Since they are equal)
Checking option \[\left( D \right)\quad {\left( {r + s + t} \right)^\circ }\]
As we can observe from the figure
\angle r = \angle x \\
\angle s = \angle y \\
\angle t = \angle z \\
$(Vertical angles)
In addition, we know that sum of all interior angles of a triangle $ = {180^\circ }$.Therefore,
\Rightarrow \angle x + \angle y + \angle z = {180^\circ } \\
\Rightarrow \angle r + \angle s + \angle t = {180^\circ } \\
Therefore \[{\left( {r + s + t} \right)^\circ } = 180^\circ \]
Checking option \[\left( E \right)\quad {\left( {t + u} \right)^\circ }\]
We can see that $\angle t$ and $\angle u$ are supplementary angles. Therefore,
$ \Rightarrow \angle t + \angle u = 180^\circ $
After checking all the options, we can Conclude that the options $\left( A \right),\left( C \right),\left( D \right),\left( E \right)$ are all equal to ${180^\circ }$. So by eliminating these
options, we are only left with option $\left( B \right)$
Hence, the correct answer is $\left( B \right)$.
Note: It should be noted that the angles $r,s\;and\;t$ are not the exterior angles of the triangle formed. Therefore, you cannot apply “the sum of exterior angles of a convex polygon is ${360^0}$” | {"url":"https://www.vedantu.com/question-answer/the-two-horizontal-lines-shown-in-the-above-class-8-maths-cbse-5f893ea8980bb03139d786e0","timestamp":"2024-11-02T18:01:51Z","content_type":"text/html","content_length":"166996","record_id":"<urn:uuid:4be289f7-ed64-4ed1-aca1-27e81ed18cb7>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00146.warc.gz"} |
Force: Bushing
Model ElementForce_Bushing defines a linear force and torque acting between two Reference_Markers, I and J.
id = "integer"
[label = "string"]
i_marker_id = "integer"
j_marker_id = "integer"
kx = "real" ky = "real" kz = "real"
ktx = "real" kty = "real" ktz = "real"
cx = "real" cy = "real" cz = "real"
ctx = "real" cty = "real" ctz = "real"
preload_x = "real" preload_y = "real" preload_z = "real"
preload_tx = "real" preload_ty = "real" preload_tz = "real"
Element identification number (integer>0). This number is unique among all Force_Bushing elements and uniquely identifies the element.
The name of the Force_Bushing element.
Specifies the Reference_Marker at which the force is applied. This is designated as the point of application of the force.
Specifies the Reference_Marker at which the reaction force and moment are applied. This is designated as the point of reaction of the force.
kx ky kz
ktx kty ktz
These define the diagonal entries for a 6x6 stiffness matrix that is used to calculate the spring force for Field_Bushing. All stiffness values must be non-negative.
cx cy cz
ctx cty ctz
These define the diagonal entries for a 6x6 damping matrix that is used to calculate the damping force for Field_Bushing. All damping values must be non-negative.
preload_x preload_y preload_z
preload_tx preload_ty preload_tz
These define the pre-loads in the Force_Bushing element. In other words, the forces at I when there is deformation. The force and torque components are measured in the J coordinate system. The
data is optional. Their default values are 0.
The example demonstrates the definition of a bushing element commonly used in automotive suspensions such as bump stops for shocks and struts. The image below is an illustration of such a bushing.
Figure 1. A Bump Stop Bushing in an Automotive Suspension
The Force_Bushing definition for such a bushing could be:
id = "26"
i_marker_id = "61"
j_marker_id = "71"
kx = "6000." ky = "6000." kz = "10000."
ktx = "1.0E5" kty = "1.0E5" ktz = "1.0 E5"
cx = "60." cy = "60." cz = "60."
ctx = "100" cty = "100" ctz = "100"
preload_x = "33" preload_y = "44" preload_z = "55"
preload_tx = "0." preload_ty = "0." preload_tz = "0."
1. The force and torque consist of three major effects: a spring force, a damping force, and a pre-load vector.
□ The spring force is defined by the product of the stiffness matrix and the relative displacement between the I and J Reference_Markers.
□ The damping force is defined by the product of the damping matrix and the relative velocity between the I and J Reference_Markers.
□ A preload vector can also be added to the spring and damping forces. The six components (three forces and three moments) are defined in the coordinate system of the J Reference_Marker.
2. Force_Bushing elements are used as compliant connectors in mechanical systems. They are typically used to reduce vibration and noise, absorb shock, and accommodate misalignments.
3. kx, ky and kz have units of force per unit length. cx, cy and cz have units of force per unit length per unit time. ktx, kty, ktz have units of torque units per radian. ctx, cty, ctz have units
of torque units per radian per unit time. The actual units are governed by what are defined for the entire model.
4. Two of the three angular deflections, rotation about X, rotation about Y and rotation about Z, must remain small at all times. The rotation angles lose physical significance otherwise. Small
means < 10 degrees.
5. i_marker_id is designated as the point of application of the Force_Field. j_marker_id is the point of reaction.
6. The forces acting at the I and J markers are equal and opposite. Since there usually is a separation between J and I and the force does not act along the separation vector, the torque acting on
the I marker is not the same as the torque acting on the J marker. This is shown in Figure 2 below.
Figure 2.
7. The sign convention for the forces and torques is as follows:
□ A positive force tends to repel the I and J Reference_Markers. A negative force tends to attract the I and J Reference_Markers.
□ A positive torque tends to rotate the I Reference_Marker in a counterclockwise direction, relative to the J Reference_Marker. Thus, a positive value of TX tends to increase the value of
included angle between the x-axes of Markers I and J.
8. Force_Bushing is a linear element. If you wish to define a nonlinear force element, then use either the Force_Field or the Force_Vector_TwoBody modeling element.
9. Force_Bushing does not model cross-coupling effects. Its stiffness and damping matrices are diagonal. If cross-coupling effects are important, use Force_Field or Force_Vector_TwoBody.
10. Force_Bushing can act on all bodies: Body_Rigid, Body_Flexible, and Body_Point.
11. The MotionSolve bushing implementation is slightly different from the one in Adams. In most cases, they yield the same results; however if the bushing undergoes 3-D deformation, the results can
be somewhat different. Both products approximate large angles, but slight differently. Hence the results will be different. | {"url":"https://2021.help.altair.com/2021/hwsolvers/ms/topics/solvers/ms/xml-format_56.htm","timestamp":"2024-11-08T11:19:56Z","content_type":"application/xhtml+xml","content_length":"106951","record_id":"<urn:uuid:25785628-93a6-4132-be7a-952c9d72235d>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00259.warc.gz"} |
Inference Space
What is an Inference Space?
Inference space can be defined in many ways, but can be generally described as the limits to how broadly a particular results applies (Lorenzen and Anderson 1993, Wills et al. in prep.). Inference
space is analogous to the sampling universe or the population. All these terms refer to the largest entity to be described. The term inference space, however, better captures the notion that there
are many different factors that help define the extent to which generalizations can be made from data (Lorenzen and Anderson 1993).
An example can help illustrate the factors that can go into defining the inference space for a particular rangeland monitoring effort. Consider the following scenario: a BLM field office is
interested in maintaining healthy sage grouse populations and biologists there have concluded that lekking areas are limiting populations and are at risk from disturbance. All known past and current
lekking areas have been identified and a random sample is drawn of areas to monitor. In this example, the inference space is defined by the following factors:
1. The boundary of the field office – This limits the inference space because no leks outside the field office boundary were considered for sampling. Statistical inference can only be made to sage
grouse within the field office.
2. The choice to focus on lekking habitat – the results will be applicable only to lekking habitat, and won’t say anything about the condition or trend of other types of sage grouse habitat.
3. The choice to limit monitoring to past and present leks – This choice will likely give the best ability to detect changes in lek conditions, but it limits the inference space because nothing can
be said about areas in the field office that are not past or present leks.
One important concept for defining the inference space of a study is that within the inference space, every sampling unit (i.e., every location) has a non-zero probability of being sampled. In other
words, every location has some chance of being selected for sampling. These selection probabilities can be either equal or unequal, but the inference space is defined by the sampling units that can
be selected for sampling.
The concept of inference space is also closely tied to variability and sample size estimation (Wills et al. in prep). As inference space increases, the variability within that inference space
generally increases too. Thus you will usually need more samples to detect the same degree of change in a large inference space than in a smaller one.
Limiting/Restricting Your Inference Space
Care must be taken when defining sampling schemes for rangeland assessment and monitoring to insure that you do not unintentionally restrict your inference space. Because the inference space is
defined by the collection of sampling units that have a non-zero probability of being selected, making decisions that exclude areas from consideration for sampling restricts the inference space. For
instance, consider an allotment that is to be monitored to detect impacts from grazing. A common approach is to locate sampling locations within a specified distance range from high impact areas like
water sources because the impact near the feature is very high and too far away from the feature there is little to no livestock use. A corresponding zone could be defined around these features and
sample points drawn randomly within this zone. However, in this case, the inference space is no longer the allotment – it is the buffer zone. By excluding the too-close and too-far areas, you have
insured that you have no information about these areas. Therefore you cannot infer your data to the entire allotment, only to the areas within it that you sampled. This may seem acceptable because
such a buffer zone is “representative” of grazing use. However, such an approach generally relies on assumptions of what is representative, cannot provide appropriate inferences if representative
conditions change (e.g., installation of a new water source), and cannot be used to provide inferences for other objectives (e.g., trend of cover of non-native annual grasses in the allotment).
In the example above, the use of unequal selection probabilities (i.e., importance sampling) can help focus sampling efforts on areas most informative to grazing management while still maintaining
the desired inference space (i.e., the allotment). In the case of the example, a selection probability layer could be created where the probability of being selected for sampling varies with distance
from the impact areas. While this approach to sampling technically introduces bias into the samples, the bias can be corrected for using the selection probability for each location. For more
information on sampling with unequal selection probabilities, see the sample_design page.
• Elzinga, C. L., D. W. Salzer, and J. W. Willoughby. 1998. Measuring and monitoring plant populations. U.S. Department of the Interior, Bureau of Land Management. National Applied Resource
Sciences Center, Denver, Colorado. Download PDF.
• Lorenzen, T.J. and V.L. Anderson. 1993. Design of experiments: a no-name approach. Marcel Dekker, Inc. New York.
• Wills, S. et al. In prep. Designing plot to landscape scale studies of environmental change: key concepts, new tools and application to soil carbon sampling. | {"url":"https://landscapetoolbox.org/general-design-topics/inference-space/","timestamp":"2024-11-02T14:42:54Z","content_type":"text/html","content_length":"47384","record_id":"<urn:uuid:26e8d378-1aa4-4487-be8e-6194a2ff2e42>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00107.warc.gz"} |
Stability of IIR Filters
Tim Wescott <Tim@seemywebsite.com> writes:
> Randy Yates <yates@digitalsignallabs.com> Wrote in message:
>> Tim Wescott <tim@seemywebsite.com> writes:
>>> On Fri, 04 Dec 2015 23:07:05 -0500, Randy Yates wrote:
>>>> Folks,
>>>> First let me ask this: Does the Fs/2-wide sinc function, interpolating
>>>> at some fractional sample offset, have a z-transform? I don't think so,
>>>> but I thought I'd verify.
>>> If it's a sinc in time, it would be sinc(<mumble-mumble>k)/z^k, wouldn't
>>> it? Presumably if you have a fractional shift there would still be some
>>> closed-form function you could write.
>>>> Now it can be shown that a fractionally interpolating sinc function can
>>>> generate an infinite output with bounded input. This is commonly
>>>> described as an unstable filter. BI does not imply BO.
>>> Are you implying that only straight people start to smell after a
>>> while?
>> <snicker>
>>> A sinc filter is still BIBO stable. Having an infinite response doesn't
>>> mean something isn't BIBO stable.
>> Er, what?!!!!?
> Oops. Having a response that's infinite in time doesn't make
> something unstable. Better?
Oh, well yes, sure. But a sinc is not BIBO stable. I'll have to search
for it, but I proved that (here on comp.dsp) quite a few years back.
Namely, I proved that with a bounded input, the sinc filter can produce
an infinite output, at least at one point in time.
Randy Yates, DSP/Embedded Firmware Developer
Digital Signal Labs | {"url":"https://www.dsprelated.com/showthread/comp.dsp/324584-2.php","timestamp":"2024-11-13T11:17:03Z","content_type":"text/html","content_length":"77664","record_id":"<urn:uuid:05dd5935-ed3b-4ba7-b109-17c7f71a5d24>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00311.warc.gz"} |
Syllabus Information
Use this page to maintain syllabus information, learning objectives, required materials, and technical requirements for the course.
Syllabus Information
MTH 065 - Elementary Algebra
Associated Term: Spring 2023
Learning Objectives: Upon successful completion of this course, the student will be able to:
1. Maintain, use, and expand skills and concepts learned in previous mathematics courses
a. Solve linear equations algebraically
b. Calculate slope of a line and find intercepts
c. Graph equations in two variables
d. Write equations in point-slope form and slope-intercept form
2. Solve linear systems of two equations in two unknowns
a. Solve algebraically and by graphing
b. Solve application problems involving linear systems of equations (Includes simple interest, motion, and mixture problems)
3. Evaluate and/or simplify expressions using the rules of (integer) exponents
4. Use scientific notation
5. Use the terminology of polynomials and add, subtract, multiply, and divide polynomials
a. Recognize and use the terminology of polynomials
b. Evaluate polynomials
c. Add, subtract, and multiply polynomials
d. Divide a polynomial by a monomial
6. Factor polynomials, including multivariable polynomials
a. Factor polynomials by removing a common monomial factor
b. Factor trinomials
c. Factor special products
7. Recognize and use quadratic equations
a. Sketch the graph of a quadratic equation in two variables and identify the intercepts and the vertex graphically
b. Solve a quadratic equation by factoring
c. Solve application problems by writing and solving quadratic equations, by factoring, (includes applications involving the Pythagorean Theorem)
8. Understand the basic definition of a function
a. Recognize and interpret function notation
b. Evaluate functions using function notation
c. Find the domain of a function defined by a table or list of ordered pairs
9. Recognize and use rational expressions
a. Recognize values of a variable that make a rational expression undefined
b. Reduce rational expressions to lowest terms
c. Multiply and divide rational expressions
d. Add and subtract rational expressions with like denominators
e. Find the least common denominator of two or more rational expressions
f. Add and subtract rational expressions with unlike denominators
10. Make appropriate and efficient use of scientific calculator. (Note: students will be expected to demonstrate achievement of some objectives without the use of a calculator)
Required Materials:
Technical Requirements: | {"url":"https://crater.lanecc.edu/banp/zwckctlg.p_disp_catalog_syllabus?cat_term_in=202340&subj_code_in=MTH&crse_numb_in=065","timestamp":"2024-11-08T00:58:18Z","content_type":"text/html","content_length":"9790","record_id":"<urn:uuid:1506ce39-0aae-4a4d-833b-c0838f0fa57a>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00212.warc.gz"} |
Hypothesis Testing
The pillar of true research findings
a key component in hypothesis testing
• An Assumption, is an idea that is proposed for the sake of argument so that it can be tested to see if it might be true.
• Shows how two or more variables are expected to relate to one another.
What is Hypothesis Testing?
• An act in statistics whereby an analyst tests an assumption regarding a population parameter.
• It ascertains whether a particular assumption is true for the whole population.
• A technique that helps to determine whether a specific treatment has an effect on the individuals in a population.
• Used to assess the plausibility of a hypothesis by using sample data.
• If sample data are not consistent with the statistical hypothesis, the hypothesis is rejected.
• The general goal is To rule out chance (sampling error) as a plausible explanation for the results of a research study.
• The purpose of hypothesis testing is to test whether the null hypothesis (there is no difference, no effect) can be rejected or approved.
• If the null hypothesis is rejected, then the research hypothesis can be accepted. If the null hypothesis is accepted, then the research hypothesis is rejected.
□ to decide between two explanations:
I.The difference between the sample and the population can be explained by sampling error (there does not appear to be a treatment effect)
II.The difference between the sample and the population is too large to be explained by sampling error (there does appear to be a treatment effect).
□ Hypothesis Testing Types
1. Simple: the population parameter is stated as a specific value, making the analysis easier.
2. Composite: the population parameter ranges between a lower and upper value.
3. One-tailed: When the majority of the population is concentrated on one side,(the sample test is either higher or lower than the population parameter).
4. Two-tailed: the critical distribution of the population is two-sided. Here the test sample is either higher or lower than a number of given values.
Relevance and Use of Hypothesis Testing
□ Validates a theory with the help of systematic statistical inference.
□ Researchers try to reject the null hypothesis in order to validate the alternate explanation.
□ Widely applied in psychology, biology, medicine, finance, production, marketing, advertising, and criminal trials.
• Limitations of Hypothesis Testing
• It is all about assumptions and interpretations.
• It, therefore, requires superior analytical abilities.
• As a result, it is inaccessible for most.
• This method heavily relies on mere probability.
• There can be errors in data.
• For smaller sample sets, this approach may not be the most suitable. | {"url":"https://kalexmat.com/hypothesis-testing/","timestamp":"2024-11-05T10:41:05Z","content_type":"text/html","content_length":"178932","record_id":"<urn:uuid:2b23a3ce-7663-45a7-a78a-b68c4b8cba2c>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00167.warc.gz"} |
PyTorch .detach() method
In order to enable automatic differentiation, PyTorch keeps track of all operations involving tensors for which the gradient may need to be computed (i.e., require_grad is True). The operations are
recorded as a directed graph. The detach() method constructs a new view on a tensor which is declared not to need gradients, i.e., it is to be excluded from further tracking of operations, and
therefore the subgraph involving this view is not recorded.
This can be easily visualised using the torchviz package. Here is a simple fragment showing a set operations for which the gradient can be computed with respect to the input tensor x.
x=T.ones(10, requires_grad=True)
make_dot(r).render("attached", format="png")
The graph inferred by PyTorch is this:
This program can be correctly differentiated to obtain the gradient:
>>> r.backward()
>>> x.grad
tensor([5., 5., 5., 5., 5., 5., 5., 5., 5., 5.])
However if a detach is called then subsequent operations on that view will not be tracked. Here is a modification to the above fragment:
make_dot(r).render("detached", format="png")
Note that x is detached before being used in computation of z. And this is the graph of this modified fragment:
As can be seen the branch of computation with x**3 is no longer tracked. This is reflected in the gradient of the result which no longer records the contribution of this branch:
>>> r.backward()
>>> x.grad
tensor([2., 2., 2., 2., 2., 2., 2., 2., 2., 2.])
Update 9/1/2020
You can find a short Jupyter notebook with the above code at: https://github.com/bnwebcode/pytorch-detach/blob/master/PyTorchDetach.ipynb | {"url":"https://bnikolic.co.uk/blog/pytorch-detach.html","timestamp":"2024-11-05T04:24:28Z","content_type":"text/html","content_length":"16628","record_id":"<urn:uuid:256d69cc-66fe-4f42-bb21-549c41c995c5>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00195.warc.gz"} |
factorial design Archives - The small S scientist
In this post, I want to show how to do contrast analysis with R for factorial designs. We focus on a 2-way between subjects design. A tutorial for factorial within-subjects designs can be found here:
https://small-s.science/2019/01/contrast-analysis-with-r-repeated-measures/ . A tutorial for mixed designs (combining within and between subjects factors can be found here: https://small-s.science/
I want to show how we can use R for contrast analysis of an interaction effect in a 2 x 4 between subjects design. The analysis onsiders the effect of students’ seating distance from the teacher and
the educational performance of the students: the closer to the teacher the student is seated, the higher the performance. A “theory “explaining the effect is that the effect is mainly caused by the
teacher having decreased levels of eye contact with the students sitting farther to the back in the lecture hall.
To test that theory, a experiment was conducted with N = 72 participants attending a lecture. The lecture was given to two independent groups of 36 participants. The first group attended the lecture
while the teacher was wearing dark sunglasses, the second group attented the lecture while the teacher was not wearing sunglasses. All participants were randomly assigned to 1 of 4 possible rows,
with row 1 being closest to the teacher and row 4 the furthest from the teacher The dependent variable was the score on a 10-item questionnaire about the contents of the lecture. So, we have a 2 by 4
factorial design, with n = 9 participants in each combination of the factor levels.
Here we focus on obtaining an interaction contrast: we will estimate the extent to which the difference between the mean retention score of the participants on the first row and those on the other
rows differs between the conditions with and without sunglasses.
The interaction contrast with SPSS
I’ve downloaded a dataset from the supplementary materials accompanying Haans (2018) from http://pareonline.net/sup/v23n9.zip (Between2by4data.sav) and I ran the following syntax in SPSS:
UNIANOVA retention BY sunglasses location
/LMATRIX = "Interaction contrast"
sunglasses*location 1 -1/3 -1/3 -1/3 -1 1/3 1/3 1/3 intercept 0
/DESIGN= sunglasses*location.
Table 1 is the relevant part of the output.
So, the estimate of the interaction contrasts equals 1.00, 95% CI [-0.332, 2.332]. (See this post for optimizing the sample size to get a more precise estimate than this).
Contrast analysis with R for factorial designs
Let’s see how we can get the same results with R.
theData <- read.spss("./Between2by4data.sav")
theData <- as.data.frame(theData)
# setting contrasts
contrasts(sunglasses) <- ginv(rbind(c(1, -1)))
contrasts(location) <- ginv(rbind(c(1, -1/3, -1/3, -1/3),
c(0, 1, -1/2, -1/2), c(0, 0, 1, -1)))
# fitting model
myMod <- lm(retention ~ sunglasses*location)
The code above achieves the following. First the relevant packages are loaded. The MASS package provides the function ginv, which we need to specify custom contrasts and the Foreign package contains
the function read.spss, which enables R to read SPSS .sav datafiles.
Getting custom contrast estimates involves calculating the generalized inverse of the contrast matrices for the two factors. Each contrast is specified on the rows of these contrast matrices. For
instance, the contrast matrix for the factor location, which has 4 levels, consists of 3 rows and 4 columns. In the above code, the matrix is specified with the function rbind, which basically says
that the three contrast weight vectors c(1, -1/3, -1/3, -1/3), c(0, 1, -1/2, -1/2), c(0, 0, 1, -1) form the three rows of the contrast matrix that we use as an argument of the ginv function. (Note
that the set of contrasts consists of orthogonal Helmert contrasts).
The last call is our call to the lm-function which estimates the contrasts. Let’s have a look at these estimates.
## Call:
## lm(formula = retention ~ sunglasses * location)
## Residuals:
## Min 1Q Median 3Q Max
## -2 -1 0 1 2
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.3750 0.1443 37.239 < 2e-16 ***
## sunglasses1 1.2500 0.2887 4.330 5.35e-05 ***
## location1 2.1667 0.3333 6.500 1.39e-08 ***
## location2 1.0000 0.3536 2.828 0.00624 **
## location3 2.0000 0.4082 4.899 6.88e-06 ***
## sunglasses1:location1 1.0000 0.6667 1.500 0.13853
## sunglasses1:location2 3.0000 0.7071 4.243 7.26e-05 ***
## sunglasses1:location3 2.0000 0.8165 2.449 0.01705 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## Residual standard error: 1.225 on 64 degrees of freedom
## Multiple R-squared: 0.6508, Adjusted R-squared: 0.6126
## F-statistic: 17.04 on 7 and 64 DF, p-value: 1.7e-12
For the present purposes, we will consider the estimate of the first interaction contrast, which estimates the difference between the means of the first and the other rows between the with and
without sunglasses conditions. So, we will have to look at the sunglasses1:location1 row of the output.
Unsurprisingly, the estimate of the contrast and its standard error are the same as in the SPSS ouput in Table 1. The estimate equals 1.00 and the standard error equals 0.6667.
Note that the residual degrees of freedom equal 64. This is equal to the product of the number of levels of each factor, 2 and 4, and the number of participants (9) per combination of the levels
minus 1: df = 2*4*(9 – 1) = 64. We will use these degrees of freedom to obtain a confidence interval of the estimate.
We will calculate the confidence interval by first extracting the contrast estimate and the standard error, after which we multiply the standard error by the critical value of t with df = 64 and add
the result to and substract it from the contrast estimate:
estimate = myMod$coefficients["sunglasses1:location1"]
se = sqrt(diag(vcov(myMod)))["sunglasses1:location1"]
df = 2*4*(9 - 1)
# confidence interval
estimate + c(-qt(.975, df), qt(.975, df))*se
## [1] -0.3318198 2.3318198
Clearly, we have replicated all the estimation results presented in Table 1.
Haans, Antal (2018). Contrast Analysis: A Tutorial. Practical Assessment, Research, & Education, 23(9). Available online: http://pareonline.net/getvn.asp?v=23&n=9 | {"url":"https://small-s.science/tag/factorial-design/","timestamp":"2024-11-02T01:22:23Z","content_type":"text/html","content_length":"88656","record_id":"<urn:uuid:167cfc09-981f-48c6-9311-d7933dc51002>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00544.warc.gz"} |
Advances in Pure Mathematics
Vol.4 No.5(2014), Article ID:45781,6 pages DOI:10.4236/apm.2014.45026
Flag-Transitive 6-(v, k, 2) Designs
Xiaolian Liao^1, Shangzhao Li^2, Guohua Chen^1
^1Department of Mathematics, Hunan University of Humanities Science and Technology, Loudi, China
^2Department of Mathematics, Changshu Institute of Technology, Changshu, China
Email: hnldlxl2005@126.com
Copyright © 2014 by authors and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
Received 28 February 2014; revised 28 March 2014; accepted 15 April 2014
The automorphism group of a flag-transitive 6–(v, k, 2) design is a 3-homogeneous permutation group. Therefore, using the classification theorem of 3–homogeneous permutation groups, the
classification of flag-transitive 6-(v, k,2) designs can be discussed. In this paper, by analyzing the combination quantity relation of 6–(v, k, 2) design and the characteristics of 3-homogeneous
permutation groups, it is proved that: there are no 6–(v, k, 2) designs D admitting a flag transitive group G ≤ Aut (D) of automorphisms.
Keywords:Flag-Transitive, Combinatorial Design, Permutation Group, Affine Group, 3-Homogeneous Permutation Groups
1. Introduction
For positive integers
In recent years, the classification of flag-transitive Steiner 2-designs has been completed by W. M. Kantor (See [1] ), F. Buekenhout, A. De-landtsheer, J. Doyen, P. B. Kleidman, M. W. Liebeck, J.
Sax (See [2] ); for flagtransitive Steiner t-designs
In this paper, we may study a kind of flag-transitive designs with
Theorem: There are no non-trivial
2. Preliminary Results
Lemma 2.1. (Huber M [4] ) Let
Lemma 2.2. (Cameron and Praeger [8] ). Let
(1) If
(2) If
Lemma 2.3. (Huber M [9] ) Let
Lemma 2.4. Let
(3) For
(4) In particular, if t = 6, then
Lemma 2.5. (Beth T [10] ) If
Lemma 2.6. (Wei J L [11] ) If
In this case, when
Corollary 2.7. Let
Proof: By Lemma 2.6, when
Remark 2.8. Let
Corollary 2.9 Let
For each positive integers,
Let G be a finite 3-homogeneous permutation group on a set X with
(A) Affine Type:
(B) Almost Simple Type:
3. Proof of the Main Theorem
3.1. Groups of Automorphisms of Affine Type
Case (1):
for each values of
Case (2):
Case (3):
3.2. Groups of Automorphisms of Almost Simple Type
Case (1):
Case (2):
We will first assume that
In view of Lemma 2.6, we have
It follows from Equation (1) that
If we assume that
and hence
In view of inequality (2), clearly, this is only possible when
Now, let us assume that
First, let
induced by the Frobenius automorphism
Hence, we have
Using again
We obtain
If we assume that
and thus
but this is impossible. The few remaining possibilities for
Now, let
has precisely
Thus, by Remark 2.8, we obtain
More precisely:
(A) if
(B) if
As far as condition (A) is concerned, we may argue exactly as in the earlier case
The case
Case (3):
By Corollary 2.7, we get
Case (4):
As in case (3), for
The authors thank the referees for their valuable comments and suggestions on this paper.
1. Kantor, W.M. (1985) Homogeneous Designs and Geometric Lattices. Journal of Combinatorial Theory, Series A, 38, 66-74. http://dx.doi.org/10.1016/0097-3165(85)90022-6
2. Liebeck, M.W. (1993) 2-Transitive and Flag-Transitive Designs. In: Jungnickel, D. and Vanstone, S.A., Eds., Coding Theory, Design Theory, Group Theory, Wiley, New York, 13-30.
3. Huber, M. (2004) The Classification of Flag-Transitive Steiner 3-Designs. Transactions of the American Mathematical Society, 1, 11-25.
4. Huber, M. (2005) The Classification of Flag 2-Transitive Steiner 3-designs. Advances in Geometry, 5, 195-221. http://dx.doi.org/10.1515/advg.2005.5.2.195
5. Cameron, P.J., Maimani, H.R. and Omidi, G.R. (2006) 3-Designs from PSL(2, q). Discrete Mathematics, 306, 3063- 3073. http://dx.doi.org/10.1016/j.disc.2005.06.041
6. Huber, M. (2007) The Classification of Flag-Transitive Steiner 4-Designs. Journal of Algebraic Combinatorics, 26, 183-207. http://dx.doi.org/10.1007/s10801-006-0053-0
7. Huber, M. (2008) Steiner t-Designs for Large t. In: Calmet, J., Geiselmann, W., Mueller-Quade, J., Eds., Springer Lecture Notes in Computer Science, Springer, Berlin, Heidelberg, New York, 18-26.
8. Cameron, P.J. and Praeger, C.E. (1992) Block-Transitive t-Designs. Finite Geometry and Combinatorics, 191, 103- 119.
9. Huber, M. (2007) A Census of Highly Symmetric Combinatorial Designs. Journal of Algebraic Combinatorics, 26, 453-476. http://dx.doi.org/10.1007/s10801-007-0065-4
10. Beth, T., Jungnickel, D., Lenz, H. (1999) Design Theory. Cambiridge University Press, Cambridge.
11. Liu W.J., Tan, Q.H., Gong, L.Z. (2010) Flag-Transitive 5-(v, k, 2) Designs. Journal of Jiang-Su University (Natural Science Edition), 5, 612-615.
12. Cameron, P.J. and Praeger, C.E. (1993) Block-Transitive t-Designs, II: Large t. In: De Clerck, F., et al., Eds., Finite Geometry and Combinatorics, London Mathematical Society Lecture Note Series
No. 191, Cambridge University Press, Cambridge, 103-119.
13. Dembowski, P. (1968) Finite Geometries. Springer, Berlin, Heidelberg, New York. | {"url":"https://file.scirp.org/Html/6-5300676_45781.htm","timestamp":"2024-11-13T13:58:12Z","content_type":"application/xhtml+xml","content_length":"80013","record_id":"<urn:uuid:562016d8-2135-4b8e-9539-cbca61e79cdb>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00471.warc.gz"} |
Tutorial: Digital to Analog Conversion – The R-2R DAC
By Alan Wolke
With the recent announcement of the world’s fastest Digital to Analog Converter (DAC) from Tektronix Component Solutions, I thought it would be interesting to take a quick tutorial tour through one
of the simplest DAC architectures – the R-2R resistor ladder network as shown in Figure 1 below.
Figure 1: A 4-bit R-2R Network
The R-2R resistor ladder network directly converts a parallel digital symbol/word into an analog voltage. Each digital input (b0, b1, etc.) adds its own weighted contribution to the analog output.
This network has some unique and interesting properties.
• Easily scalable to any desired number of bits
• Uses only two values of resistors which make for easy and accurate fabrication and integration
• Output impedance is equal to R, regardless of the number of bits, simplifying filtering and further analog signal processing circuit design
How to Analyze the R-2R Network
Analyzing the R-2R network brings back memories of the seemingly infinite variety of networks that you’re asked to solve during your undergraduate electrical engineering studies. The reality though,
is that the analysis of this network and how it works is quite simple. By methodical application of Thevenin Equivalent circuits and Superposition, we can easily show how the R-2R circuit works.
Let’s start off by analyzing the output impedance. Working through the circuit, simplifying it with Thevenin equivalents, makes this process simple. Thevenin says that if your circuit contains linear
elements like voltage sources, current sources and resistors, that you can cut your circuit at any point and replace everything on one side of the cut with a voltage source and a single series
resistor. The voltage source is the open-circuit voltage at the cut point, and the series resistor is the equivalent open circuit resistance with all voltage sources shorted.
Figure 2 below shows the locations of the “cut lines” we’ll use to simplify this circuit to calculate its output impedance. For this analysis, the digital inputs will all be considered shorted to
Figure 2: Establishing the cut lines for Thevenin Analysis
The two 2R resistors to the left of the first cut line in Figure 2 appear in parallel (when the digital bit b0 is grounded), and can be replaced with a single resistor R as shown in Figure 3. The
series combination of the two R resistors on the left of Figure 3 combine to a single resistor of value 2R, which is in parallel with the 2R resistor to b1.
Figure 3: First Thevenin Equivalent
You may notice that this process repeats itself each time we work from left to right, successively replacing combinations of resistors with their equivalents. As you can see in Figure 4, the circuit
ultimately simplifies to a single resistor R.
Figure 4: Calculating the equivalent output resistance of the R-2R network
Thus, the output impedance of the R-2R resistor network is always equal to R, regardless of the size (number of bits) of the network. This simplifies the design of any filtering, amplification or
additional analog signal conditioning circuitry that may follow the network.
How to Calculate Analog Voltage Output
Next, we’ll look at how to calculate the analog voltage output for a given parallel digital input on the b0, b1, etc. inputs. We’ll use the same Thevenin equivalents technique shown above, as well as
Superposition. Superposition tells us that if you individually compute the contribution of a given source to the output (with all others voltage sources shorted and current sources opened), you can
then sum the results for each of the sources to obtain the final result for the output.
We’ll calculate the contribution of two of the bits of our 4-bit R-2R DAC in Figure 5 to show the process. We’ll assume the bits b0 and b2 are logic high, and bits b1 and b3 are logic low (ground).
Figure 5: 4-bit R-2R example
We start by replacing the circuit to the left of the left-most cut-line with its Thevenin equivalent. Figure 6 shows the Thevenin equivalent, which is the series resistor of value R (parallel
combination of two 2R resistors), and the open circuit voltage from the resistor divider (Vb0/2).
Figure 6: Replacing the first stage with its Thevenin equivalent
The process continues methodically, step by step for each cut-line, substituting the equivalent circuit for each stage, as shown graphically in Figure 7.
Figure 7: Calculating the contribution of Vb0 to the output
We can see that the voltage contribution from bit b0 is 1/16^th of the logic high voltage level. Each bit stage that this voltage passes through cuts the contribution by a factor of 2. You may begin
to see a theme here…
Next, we’ll compute the contribution from bit b2, as shown in Figure 8 below:
Figure 8: Computing the contribution of bit b2 to the output
From the Thevenin equivalent analysis shown earlier, we know that we can replace any portion of this circuit to the left of any of the cut lines with a resistor of value R, shown as the first step in
Figure 8. Next, we follow the same Thevenin equivalent process to the output. As you may have already suspected, the contribution of bit b2 is simply Vb2/4. Thus, the analog output voltage when bits
b0 and b2 are equal to logic one is simply given by Vb0/16 + Vb2/4.
In a more general sense, the contribution of each bit to the output is a simple binary weighting function of each bit. As you work back from the MSB to the LSB, the voltage contribution each bit is
cut in half. Thus, the general form of the equation to calculate the output voltage is shown in Figure 9.
Figure 9: Formula to calculate output of 4-bit R-2R DAC
The R-2R resistor ladder based digital-to-analog converter (DAC) is a simple, effective, accurate and inexpensive way to create analog voltages from digital values. Monolithic R-2R resistor networks
are available from various resistor component manufacturers, making it easy to incorporate them into your designs.
Since publishing the above blog, Tektronix is currently offering a free upgrade to 3/6 GHz spectrum analyzers and a free bundle of software applications on mixed-domain oscilloscopes. | {"url":"https://www.tek.com/en/blog/tutorial-digital-analog-conversion-r-2r-dac","timestamp":"2024-11-05T15:24:05Z","content_type":"text/html","content_length":"102657","record_id":"<urn:uuid:6ca250b3-545c-4526-bc33-d515aa7291c8>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00810.warc.gz"} |
Project Research - 2023-05-18 04:53:54 - mySTEMspace
STEM Project Research
Reference Websites
AI-Generated Project Ideas
1. A math game that teaches fractions and decimals through energy consumption. The game would feature different appliances and devices that require different amounts of energy to operate. The player
would need to correctly calculate and manage the energy consumption in order to keep the devices running.
2. An energy-efficient board game that teaches players about renewable energy sources. The game would feature different types of renewable energy, such as solar, wind, and hydroelectric power, and
the players would need to strategically build and manage their energy infrastructure to generate the most renewable energy possible.
3. A math puzzle game that uses energy conservation principles. The player would need to solve math problems related to determining energy consumption and conservation for different devices. The
game would feature a variety of scenarios, such as managing the energy usage of a household or a city, and the player would need to determine the optimal strategies for reducing energy usage.
4. An interactive exhibit that demonstrates the relationship between math and energy. The exhibit would feature different hands-on activities and experiments that illustrate the principles of energy
conservation and the math behind energy use. Examples could include demonstrating energy transformation through various machines and devices, or providing interactive calculators and tools for
exploring the relationship between energy usage and costs. | {"url":"https://mystemspace.ca/project-research/xkefcjk5gk2m3fl7vbw37kbqe8oe8dfwshjiizkwpt9ng/","timestamp":"2024-11-01T19:53:48Z","content_type":"text/html","content_length":"34395","record_id":"<urn:uuid:a011ffd3-126a-427f-aaa0-296ca57e04a7>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00457.warc.gz"} |
Error propagation in adaptive Simpson algorithm
07-30-2018, 06:42 PM
(This post was last modified: 07-30-2018 06:52 PM by Claudio L..)
Post: #1
Claudio L. Posts: 1,885
Senior Member Joined: Dec 2013
Error propagation in adaptive Simpson algorithm
I recently implemented an adaptive Simpson, almost straight from
Wikipedia's article
I immediately noticed that on every test I threw at it, it produces a result that is consistently 2 or 3 digits more accurate than requested.
Since the algorithm halves the step on each pass, every additional digit requested needs roughly 3 more "splits". Roughly speaking, each time it splits the interval in 2, we need twice as many
function evaluations as the step before, so those last 3 to 6 extra splits can be extremely costly in terms of time.
The first thing I notice is that this algorithm halves the requested tolerance every time it splits. Basically it's saying Error=Error_Left+Error_Right, therefore they want Error_Left and Error_Right
to be less than half the requested tolerance to make sure it meets the requirement.
This seems to me quite conservative, once you are doing hundreds or thousands of steps, shouldn't the total error be proportional to the sqrt(N)*AvgError with N being the number of intervals
If so, we could perhaps use 1/sqrt(2)=70%, and whenever we split an interval in half, simply request 70% of the error instead of 50%.
And there's also a magic factor 15 that I don't know where is coming from, perhaps that's another factor to play with.
For example (from another thread): Integral of exp(-x) between 0 and 500.
If I request 1e-8 tolerance, the result is:
This case has 2 more digits than requested.
07-31-2018, 09:35 AM
Post: #2
Dieter Posts: 2,397
Senior Member Joined: Dec 2013
RE: Error propagation in adaptive Simpson algorithm
(07-30-2018 06:42 PM)Claudio L. Wrote: The first thing I notice is that this algorithm halves the requested tolerance every time it splits. Basically it's saying Error=Error_Left+Error_Right,
therefore they want Error_Left and Error_Right to be less than half the requested tolerance to make sure it meets the requirement.
This seems to me quite conservative, once you are doing hundreds or thousands of steps, shouldn't the total error be proportional to the sqrt(N)*AvgError with N being the number of intervals
Maybe a few general thoughts on this kind of algorithm may help.
First of all, adaptive Simpson methods have already been discussed here. Take a look at
this thread
and at
this program
The idea basically is to estimate the error of Simpson approximations with n and 2n intervals. This yields an even more accurate result with the value (16·S
– S
)/15. That's where the 15 comes from.
For more details on this take a look at the German
Wikipedia article
where this is explained. The error term E
(f) with the 16/15 factor is discussed earlier in that article.
As far as the threshold for quitting the iteration is concerned: this is a general problem with any method that produces successive approximations of the true result. Since the latter is unknown all
you can do is compare the last two approximations, i.e. the last and the previous one. If these two agree within the desired accuracy, the iteration stops and the result is returned. But – if the
final and the previous value agree in the desired, say, 8 places, this means that already the
value was sufficiently accurate! And since the last approximation is more accurate than this, it may be correct to e.g. 11 places. So its true accuracy is greater than requred. That's what you
observe here.
(07-30-2018 06:42 PM)Claudio L. Wrote: For example (from another thread): Integral of exp(-x) between 0 and 500.
If I request 1e-8 tolerance, the result is:
This case has 2 more digits than requested.
Yes, that's the ex-post accuracy when comparing the result with the true value. But the algorithm doesn't know the true result. All it can do is compare the final and the previous approximation.
Here and there the exit condition can be improved. Assume you know for sure (!) that the iteration converges quadratically and the difference between the last two approximations was 1E–2, 1E–4 and
1E–8. If the next approximation can be expected to change the result only in the 16th digit this means that the current (!) approximation already is this precise. So if you need 12 places you may
stop now although the last two approximation only agreed in 8 places. Because one more iteration will only change the result in the 16th place.
BTW, integrating exp(–x) up to 500 doesn't make sense. At x=20 the function value has dropped to about 2E–9 so anything beyond this will not affect the first 8 decimals.
07-31-2018, 11:42 AM
Post: #3
Albert Chan Posts: 2,785
Senior Member Joined: Jul 2018
RE: Error propagation in adaptive Simpson algorithm
(07-30-2018 06:42 PM)Claudio L. Wrote: For example (from another thread): Integral of exp(-x) between 0 and 500.
If I request 1e-8 tolerance, the result is:
This case has 2 more digits than requested.
Was the thread
Casio fx-991EX vs Hp 50g speed differences
Tolerance setting is against successive iterations, not against true result.
If the algorithm knew the true result, why ask for tolerance ?
The extra precision may be just lucky (the integral is well-behaved)
Integral convergence rate may stall or (rare, but possible) even got worse on some iterations.
Worse, integral may converge very well, but converged to a
wrong result
See above thread post#12, integral of sin(x), from 0 to 200.
So, if time permitted, double check the numbers with other methods.
07-31-2018, 05:02 PM
(This post was last modified: 07-31-2018 05:05 PM by Claudio L..)
Post: #4
Claudio L. Posts: 1,885
Senior Member Joined: Dec 2013
RE: Error propagation in adaptive Simpson algorithm
(07-31-2018 09:35 AM)Dieter Wrote: Maybe a few general thoughts on this kind of algorithm may help.
First of all, adaptive Simpson methods have already been discussed here. Take a look at this thread and at this program.
The idea basically is to estimate the error of Simpson approximations with n and 2n intervals. This yields an even more accurate result with the value (16·S[2n] – S[n])/15. That's where the 15
comes from.
For more details on this take a look at the German Wikipedia article where this is explained. The error term E^(N)(f) with the 16/15 factor is discussed earlier in that article.
Thanks, now I know where the 15 comes from, yet we are talking very different algorithms from the one you provided links to. I'm not even sure why they still call the adaptive one "Simpson", since
you are not using the summation that is the hallmark of the Simpson method, but the error analysis still applies.
(07-31-2018 09:35 AM)Dieter Wrote: As far as the threshold for quitting the iteration is concerned: this is a general problem with any method that produces successive approximations of the true
result. Since the latter is unknown all you can do is compare the last two approximations, i.e. the last and the previous one. If these two agree within the desired accuracy, the iteration stops
and the result is returned. But – if the final and the previous value agree in the desired, say, 8 places, this means that already the previous value was sufficiently accurate! And since the last
approximation is more accurate than this, it may be correct to e.g. 11 places. So its true accuracy is greater than requred. That's what you observe here.
I understand, and I agree with your logic more than the wikipedia article for adaptive simpson: if the difference between the areas calculated with a step h and the one calculated with h/2 is less
than the target error then we should be good, right?
The thing is, every time it halves the step size, it also halves the target error, and that's what I'm questioning if perhaps is too conservative, leading to the overshoot in accuracy. When you add
all the tiny pieces back together the error of the sum will never be the sum of the estimated errors because they are randomly going to cancel each other, it's more like:
Error_whole ≅ √N * AvgError
Couldn't we at least say that (for one single "split" section)
Error_whole ≅ √2 * AvgError
So we could say our target error wouldn't be halved each way, but:
AvgError = Error_whole/√2 ≅ 0.7 * Error_whole
So we accept slightly more error as we reduce the step, knowing that when we look at the whole interval, we are going to be very close to meet the required error target. I guess the other approach
tries to guarantee you'll meet the target, versus this approach wants to get close to the intended target, not necessarily guarantee it, but that could help reduce the number of unnecessary
(07-31-2018 09:35 AM)Dieter Wrote: Here and there the exit condition can be improved. Assume you know for sure (!) that the iteration converges quadratically and the difference between the last
two approximations was 1E–2, 1E–4 and 1E–8. If the next approximation can be expected to change the result only in the 16th digit this means that the current (!) approximation already is this
precise. So if you need 12 places you may stop now although the last two approximation only agreed in 8 places. Because one more iteration will only change the result in the 16th place.
Isn't the error proportional to h^4? Anyway, I'm not so sure how this one behaves because the function is subdivided more in the areas where it's less well-behaved and less in other areas, as long as
the target error is met.
It's basically "minimal effort to locally achieve a target error".
You do have a good point, perhaps the overshoot in error is because halving the step increases the accuracy a lot more than what we needed.
(07-31-2018 09:35 AM)Dieter Wrote: BTW, integrating exp(–x) up to 500 doesn't make sense. At x=20 the function value has dropped to about 2E–9 so anything beyond this will not affect the first
8 decimals.
I know, it came up on another thread as a speed test, and I wanted to compare speeds as well. But I tested with several other functions (polynomials, combined exponentials with trig, trig alone, etc)
and I always get a few more digits than I requested, at the expense of a huge amount of wasted time.
For that example, using 1E-8 as a target error, I get good 11 digits in 3 seconds (doing 741 function evaluations), now if I request 11 digits I get 13 good digits but it takes 15 seconds doing 4293
function evaluations.
This is the real reason why I'm seeking advice to see if perhaps we could do better.
07-31-2018, 06:37 PM
Post: #5
Claudio L. Posts: 1,885
Senior Member Joined: Dec 2013
RE: Error propagation in adaptive Simpson algorithm
Quote:Worse, integral may converge very well, but converged to a wrong result
See above thread post#12, integral of sin(x), from 0 to 200.
Great example, this is a badly behaved integral, all numerical methods will suffer due to massive cancellation.
So I requested a tolerance of 1e-6 and my algorithm converged to a completely wrong value in a couple of seconds. It clearly stopped prematurely, which can happen because of the function, not really
anything bad about the algorithm.
Then I requested 8 digits, giving a tolerance of 1E-8. Now it took some time (didn't time it but felt close to a minute) and gave a good answer with 16 digits correct. It seems now the tighter
tolerance avoided the premature stop and the algorithm was able to deal with the cancellation of areas. I would've expected all the smaller areas to cancel each other out with errors around 1E-10, to
provide a final area close to the 1E-8 requested tolerance, but it seems it had to subdivide a lot to get the result.
07-31-2018, 07:53 PM
Post: #6
Albert Chan Posts: 2,785
Senior Member Joined: Jul 2018
RE: Error propagation in adaptive Simpson algorithm
Hi, Claudio
The problem with the sin integral is not cancellation, but its periodic nature.
Simpson's Rule is also based on periodic sampling (all equally spaced).
If the sampling and periodic function were in sync, the samples are going to be biased.
For sin(x) 0 to 200, the first few iterations, none of sin(x) samples were above zero.
A non-linear transformed function can fix this, by scrambling the sample points.
Have you tried the non-linear transformed sin integral (same thread, post #13) ?
Regarding "excessive" accuracy, it will happen even if tolerance is good estimater for accuracy.
Not all integral have the same convergence rate.
Say, tolerance of 1e-8 somehow guaranteed 7 digits accuracy (it does not)
What is the chance of really getting 7 digits accuracy ? Almost zero.
With guaranteed minimum accuracy, average accuracy is going to be higher, say, 10 digits.
Iterations that not quite make it to tolerance will doubled the points, "wasting" accuracy.
07-31-2018, 08:40 PM
Post: #7
Claudio L. Posts: 1,885
Senior Member Joined: Dec 2013
RE: Error propagation in adaptive Simpson algorithm
So I modified the algorithm to not change the error tolerance at all, so the tolerance stays constant no matter how tiny is the subdivision (in other words, we halve the step, keep the tolerance the
I did the SIN(x) from 0 to 200 with 1E-8 tolerance and got the following results:
Original algorithm: Error = -9.2E-16, number of evaluations = 28277
With fixed tolerance: Error = 2.0E-11, number of evaluations = 4025
Tried a couple of other functions as well and got errors much closer to the requested tolerance, so I think I will leave it with the fixed tolerance. It seems to be a closer match to the final
tolerance that the user is requesting, and it's a time saver. For smaller tolerances it again over-delivers in accuracy on most functions (if I request 16 digits I get 21 for example).
07-31-2018, 09:06 PM
Post: #8
Vtile Posts: 406
Senior Member Joined: Oct 2015
RE: Error propagation in adaptive Simpson algorithm
Sorry again replying half sleep from phone. Derailing a bit, but the title took my attention. Is there adaptive intehration method which would use simple derivative (or constant lenght vector) to
analyse the function to see if there is proportionally rabid transition and therefore need for higher accuracy. Propably yes and propably computionally extremely inefficient.
08-01-2018, 08:02 AM
(This post was last modified: 08-01-2018 08:21 AM by Dieter.)
Post: #9
Dieter Posts: 2,397
Senior Member Joined: Dec 2013
RE: Error propagation in adaptive Simpson algorithm
This is a reply to Claudio L's post #4 in this thread. Since the "Quote" button currently doesn't seem to work I have to do it this way:
Quote:Thanks, now I know where the 15 comes from, yet we are talking very different algorithms from the one you provided links to. I'm not even sure why they still call the adaptive one
"Simpson", since you are not using the summation that is the hallmark of the Simpson method, but the error analysis still applies.
If you refer to the HP-41 program I linked to, or the VBA version (post #8) in the other thread: they both use the classic Simpson method. The sums with weight 4 and 2 are summed separately and a
Simpson approximation is calculated from these. Then the number of intervals is doubled and the two sums are added together to get the new weight-2-sum. This way only the new points have to be
calculated which sum up to the new weight-4-sum.
The additional "non-Simpson"-step then is the extrapolation (16·S
– S
)/15 which gives an improved, more accurate value for the integral. When two successive values of this improved approximation match with the desired accuracy the iteration exits.
Take a look at the VBA code: simp_old and simp_new are the regular Simpson approximations for n/2 and n intervals, whereas improved_old and improved_new are the extrapolated improved approximations.
Both the HP-41 program and the VBA version in the 2016 thread only use the last two Simpson approximations to calculate a new, improved integral. In his recent answer Albert Chan suggested using
previous Simpson approximations, similar to the Romberg method. This is another worthwile option, but I didn't want to work with arrays, so only the last two approximations were used.
Quote:You do have a good point, perhaps the overshoot in error is because halving the step increases the accuracy a lot more than what we needed.
The essential point here is that we cannot and do not know the error of the calculated approximation. We can't say for sure if it's 7, 8 or 9 digits that match the true result. All we got is an upper
bound, and even this is only reliable if we assume that the iteration converges to the true result. In this case the upper error bound is the difference between the last two approximations. If they
match in 6 decimals this can mean that the result is accurate to 6, 7, 8 or even more digits. But we can't say anything more than this.
To make this more clear, take a look at the Simpson approximations for the integral of f(x)=1/x from 1 to 2:
n Standard Simpson Error
2 0,694444444444 1,3 E-03
4 0,693253968254 1,1 E-04
8 0,693154530655 7,4 E-06
16 0,693147652819 4,7 E-07
32 0,693147210290 3,0 E-08
64 0,693147182421 1,9 E-09
128 0,693147180676 1,2 E-10
Assume you want an accuracy of six decimals. The Error column shows that n=16 already met that target. But we can't know this during the calculation
unless we also know the true value of the integral!
The Simpson approximations for n=16 matches the previous one (n=8) in (nearly) 5 decimals, so after calculating the n=16 iapproximation there are only two things we can say:
1. The n=8 approximation agrees with the (most probably) more accurate n=16 approximation when rounded to 5 decimals (0,69315). So already the n=8 approximation probably had 5 valid decimals. Which
is what we know after having calculated the n=16 approximation.
2. The n=16 approximaion most probably is more accurate than this, but we can't say by how much. It may have 6 valid decimals or more.
08-01-2018, 02:23 PM
Post: #10
Albert Chan Posts: 2,785
Senior Member Joined: Jul 2018
RE: Error propagation in adaptive Simpson algorithm
I finally have the time to learn about Adaptive Simpson's method.
I were misled to believe it is plain Simpson + correction (/15 only) ... Sorry
It is not one integral with increasing steps for better accuracy,
but increasing number of mini-integrals (2 intervals, thus 3 points) sum together.
The question really is, when sub-dividing integral into a bunch of recursive
mini-integrals, what should their tolerance be ?
Mini-integrals don't talk to each other, so don't know which is the more important.
Adaptive Simpson method have to be conservative, tolerance cut in half for each step.
So, even if both splitted integrals equally important, total error still below tolerance.
This is probably an overkill, but without knowing the integral, it is the best it can do.
For one sided function, say exp(-x), tolerance can really stay the same all the way down.
By setting a tolerance, adaptive Simpson rule ignore the "details", and
concern itself with the dominant sub-divided integrals, thus is faster.
This is just my opinion, speed-up by ignoring the details come with costs:
• not recommended for function where details matter (dominant peaks cancelled out)
• not recommended for non-linear transformed integral, with details suppressed.
But, how to ensure the integral not behaved like a non-linear transformed type ?
• we need information about approximate size of expected result, to set tolerance.
• tolerance too loose, important details will be ignored, and may get wrong result.
• tolerance too tight, it may reach software recursion limit and crashes.
• Since corrections are really extrapolation, by the time the corrected mini-integral
results cascade back to the top, result might be totally off.
• Good corrections should be for actual iterations, not corrected estimates (which the cascade does).
I learn about this from Dieter's post regarding Namir's basic code (version 2) bug.
Hp thread: Simpson Revival, post #6
• returned result have higher uncertainly due to all of above
For example, this transformed exponential integral will not work well with Adatpive scheme:
\(\int_0^{500}e^{-x}dx \) = \(\int_{-1}^{1}375(1-u^2) e^{-(125u (3 - u^2) + 250)} du \) = 1 - \(e^{-500}\) ~ 1.0
OTTH, if used correctly, this is a great tool. For example:
= 4 \(\int_{0}^{\pi/2} cos(x) cos(2x) ... cos(1000x) dx \) = 0.000274258153608 ...
Adaptive Simpson Method (eps = 1e-9) were able to give 11 digits accuracy in 0.5 sec
Romberg's Method can only reached 8 digits accuracy ... in 400 seconds !
Thanks, Claudio
08-01-2018, 04:28 PM
Post: #11
Claudio L. Posts: 1,885
Senior Member Joined: Dec 2013
RE: Error propagation in adaptive Simpson algorithm
Quote:Dieter wrote:
The additional "non-Simpson"-step then is the extrapolation (16·Snew – Sold)/15 which gives an improved, more accurate value for the integral. When two successive values of this improved
approximation match with the desired accuracy the iteration exits.
The adaptive algorithm I implemented uses the exact same extrapolation, somewhat disguised in the code but it's there as Snew+ (Snew-Sold)/15 . It does it for each individual strip.
Quote:Dieter wrote:
The essential point here is that we cannot and do not know the error of the calculated approximation.
Exactly, which means we can play with the criteria, since "nobody" knows the right answer. Ideally I want the final output to be close to the tolerance the user provided, so if the user wants 1E-8 we
don't waste a lot of time computing beyond that.
So I think my relaxed tolerance, which still overshoots accuracy by a long shot, is a fair compromise.
Quote:Albert Chan wrote:
The question really is, when sub-dividing integral into a bunch of recursive
mini-integrals, what should their tolerance be ?
Mini-integrals don't talk to each other, so don't know which is the more important.
Adaptive Simpson method have to be conservative, tolerance cut in half for each step.
So, even if both splitted integrals equally important, total error still below tolerance.
This is probably an overkill, but without knowing the integral, it is the best it can do.
For one sided function, say exp(-x), tolerance can really stay the same all the way down.
Yes, normally I wouldn't care and agree with you that it's better to be conservative. However, we are deciding between getting the answer you want within a few seconds or waiting a long time (in
other words, whether the calculator would be useful or not to the user).
On a PC, both cases would feel instantaneous, so it wouldn't make sense to sacrifice accuracy.
Another good point (that Dieter made clear) is that we are really comparing that tolerance with the difference between two consecutive runs, on a little single strip, which is very loosely related to
the actual tolerance on the total area, and I think that's why there is a disconnect between the tolerance the user requests on the final result and the tolerance we should be comparing to.
In the end, the user is getting way more than requested, at the expense of a lot of wasted time and coffee.
08-01-2018, 05:53 PM
(This post was last modified: 08-01-2018 06:01 PM by Albert Chan.)
Post: #12
Albert Chan Posts: 2,785
Senior Member Joined: Jul 2018
RE: Error propagation in adaptive Simpson algorithm
Now I see why you set a fixed tolerance down the chain.
A slightly off answer is better than no answer.
But, are you better off keep the algorithm as-is, and raise tolerance ?
(To take advantage of the "wasted" accuracy)
To get 6 digits accuracy, say, set tolerance to 1e-4 ?
I do that for my Guassian Quadrature routine, expecting 2 extra good digits.
So, if difference between 2 iteration is +/- 1e-4, last result is about 6 digits accurate.
another trick is to limit recursion depth, say, 10 levels deep.
This forced the calculator to give a reasonable answer in seconds.
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CS 351 Syllabus
Data Structures and Algorithms
Revised: August 2018
Course Description
This course covers more advanced data structures, well-known algorithms, algorithm design techniques, algorithm analysis, and mathematics related to selected algorithms. The goal is to help students
learn to write and analyze efficient programs that solve common and difficult problems. Data structures covered include representations of graphs, binary search trees, balanced binary search trees,
AVL trees, heaps, and hash tables. Well-known algorithms covered include several sorting algorithms (such as insertion sort, merge sort, and Shell sort), binary search, and a sampling from other
areas such as graph algorithms and NP-hard problems. Mathematics related to advanced algorithms and data structures will also be investigated. Mathematics covered may include, but are not limited to,
probability and its application to Bayesian networks, Markov chains and decision processes, and Monte Carlo simulations; introduction to machine learning; and applications of linear algebra.
Algorithm design techniques the students will practice include dynamic programming, divide-and-conquer, and greedy techniques.
Corequisites and Notes
• CS 253
• Math 255
• 4 Credit Hours
By the end of this course, students will:
• Use and implement a wide range of data structures.
• Analyze the time complexity of a given algorithm.
• Identify the most appropriate algorithm for a given problem and apply it to that problem.
• Develop algorithms suitable for solving a complex problem.
• Apply the appropriate mathematics related to advanced data structures and algorithms.
Mark Allen Weiss, Data Structures and Algorithm Analysis in Java, Third Edition, Addison Wesley, 2012
Grading Procedure
Grading procedures and factors influencing course grade are left to the discretion of individual instructors, subject to general university policy.
Attendance Policy
Attendance policy is left to the discretion of individual instructors, subject to general university policy.
Course Outline
• Review of CS151 Concepts. The concepts reviewed include recursion, the distinction between abstract data types and data structures, and the data structures of arrays, linked lists, stacks, and
• Review of mathematics relevant to algorithm analysis including power rules, logarithms, and sigma notation.
• Algorithm analysis with Big-O
• The Java Collection Framework and abstract data types (including the idea of Maps) and typical implementation of data structures.
• Graph terminology, representations, searching, sorting, and time analysis.
• General trees, binary trees, binary search trees, AVL trees, heaps
• Hash tables.
• General sorting techniques.
• Algorithm design techniques (dynamic programming, greedy algorithms, divide and conquer)
• Undecidability and P versus NP
• Advanced data structures and algorithms
• Mathematics related to advanced data structures and algorithms including a sampling of the following
□ Introduction to Probability
□ Bayesian networks
□ Markov chains and decision processes
□ Machine learning
□ Introduction to and application of linear algebra (e.g. matrix manipulation, linear programming, image processing) | {"url":"https://www.wcu.edu/learn/departments-schools-colleges/cas/science-and-math/mathcsdept/bs-computer-science/syllabi-for-computer-science-undergraduate-courses/cs-351-syllabus.aspx","timestamp":"2024-11-07T00:24:33Z","content_type":"text/html","content_length":"37595","record_id":"<urn:uuid:8d6528da-1ce6-4661-a3d1-bc4cf7e7d690>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00139.warc.gz"} |
Number & place value - Oxford Owl for Home
Number & place value
Helping our children to tell the difference between numbers is a great first step in learning maths at home. However, it’s just as important to help them understand what each digit is worth – the
value of the number depending on its place.
This might sound complicated, but we have lots of tips and activities to allow you to get it right. Before you start your learning at home, these two helpful videos will give you some clear
definitions to work with:
Get a simple definition of the concept of number and the difference between cardinal, ordinal, and nominal numbers with this fun animation.
Watch our fun animation for a simple definition of this early maths skill.
Maths glossary
Use these quick links or explore our jargon buster for simple definitions and examples of mathematical terms.
How to help your child at home
You don’t need to be an expert to support your child with maths! Here are four simple but effective ways to help your child develop their understanding of number and place value:
1. Play counting games
Board games often show ordered numbers on tracks or grids. Make sure these numbers are clearly visible and count out loud as you progress around the board. This will help your child quickly get a
sense of what the numbers mean.
Dice can also help your child recognise number patterns quickly. For example, a pair of dice is a great way of showing your child doubles, or what it means for a number to be 1 bigger than another
Games like ‘What’s the time, Mr Wolf?’ and songs like ‘Ten Green Bottles’ can help build early counting skills.
2. Break down numbers
Look for numbers in the world around you and encourage your child to break them into parts. Breaking numbers up like this is called ‘partitioning’.
Point out a number and ask your child how many ones/tens/hundreds/thousands it has. Lots of children find this easier with physical objects, like stones or sticks. For example, they could group
sticks into groups of ten.
3. Have counting races
Choose a starting number and a multiple to count up by. For example, you could start at 12 and count in steps of 4.
Take turns to say the next number in the sequence (12, 16, 20, 24, 28, etc.). Set a timer and see what number you can reach in one minute. This kind of game could also work as a bit of healthy
competition between siblings or friends!
4. Ten questions
To really understand numbers, we want children to investigate place value through language. A simple game of guessing a number in ten questions is a great way of exploring mathematical language
whilst developing their reasoning skills.
Want more?
To help your child’s learning further, you may want to watch some of the videos included within our dedicated maths library. If you’re looking for more ideas to support learning at home, head over to
our maths blog to explore articles full of top tips and fun activities.
What your child will learn at school
For more information about your child’s learning in a particular year group, use this handy drop down menu:
Number and place value in Year 1 (age 5–6)
In Year 1, children will work with numbers up to 100, counting on or back from any number in steps of 1, 2, 5, or 10. This includes:
□ reading and writing numerals to 100 and number names to 20 in words
□ using objects and number lines to represent numbers
□ finding one more and one less than any number.
Number and place value in Year 2 (age 6–7)
In Year 2, children will recognise tens and ones in two-digit numbers (for example, ’23’ has 2 tens and 3 ones) and will be able to order numbers up to 100. This includes:
□ counting in steps of 1, 2, 3, 5 and 10
□ using more than (>), less than (<), and equals (=) symbols to compare numbers
□ using place value and number facts to solve problems.
Number and place value in Year 3 (age 7–8)
In Year 3, children will recognise hundreds, tens, and ones in three-digit numbers (for example, ‘423’ has 4 hundreds, 2 tens, and 3 ones). This includes:
□ counting in steps of 4, 8, 50, and 100
□ reading, writing, comparing, and ordering numbers to 1000
□ finding 10 or 100 more or less than a number.
Number and place value in Year 4 (age 8–9)
In Year 4, children will order and compare numbers beyond 1000 using place value in four-digit numbers (for example, ‘1428’ has 1 thousand, 4 hundreds, 2 tens, and 8 ones). This includes:
□ counting in steps of 6, 7, 9, 25, and 1000
□ counting backwards through zero to include negative numbers
□ rounding any number to the nearest 10, 100, or 1000.
Number ad place value in Year 5 (age 9–10)
In Year 5, children will read, write, compare, and order numbers up to 1,000,000, recognising the place value of each digit. This includes:
□ counting forwards and backwards with positive and negative numbers
□ rounding numbers up to one million to the nearest 10, 100, 1000, 10,000 and 100,000
□ recognising Roman numerals I, V, X, L, C, D, and M to read numbers and years.
Number and place value in Year 6 (age 10–11)
In Year 6, children will read, write, compare, order, and round numbers up to 10,000,000 and begin to learn about algebra, ratio, and proportion. This includes:
□ using number lines to add and subtract negative numbers
□ using simple formulae and following rules such as 2n + 3 to find numbers in a sequence
□ solving problems involving place value, ratio, scale factors, and equations expressed. | {"url":"https://home.oxfordowl.co.uk/maths/primary-number-place-value/","timestamp":"2024-11-09T10:07:43Z","content_type":"text/html","content_length":"100625","record_id":"<urn:uuid:e59f92c9-f18b-4587-9e1b-c6e8406deee3>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00541.warc.gz"} |
Earlier we told you how to pull historical data in Google spreadsheet (see — HOW TO PULL HISTORICAL STOCK DATA FROM USING GOOGLE SPREADSHEET ). Once you have Historical data available in Google
spreadsheet we can Calculate Simple Moving Average and values of other various technical indicators of our interest using the same. The method given here can be used in Excel as well as Google
Sheets. However we have given formulas for Google sheet. In case you wish to do same in Excel you can achieve same result by importing historical data (by Any means) and then using these formulas
(some minor modification in syntax of formula may be required). Today we are going to look how to calculate Simple Moving average or SMA for a particular period. SMA is nothing but average of last N
number of Prices where N denotes the period. So if you want to calculate SMA(20,C) i.e. Simple Moving average of closing price of 20 day period. You can simply sum last 20 Close price and divide by
20, and you will get the SMA (20, C). Similarly SMA can be calculated on Open High or Low with any period, a screenshot with formula […] | {"url":"https://www.sapeservices.com/2020/06/","timestamp":"2024-11-10T14:53:57Z","content_type":"text/html","content_length":"54859","record_id":"<urn:uuid:9916f695-22f7-4cb7-a74f-7805c13a9d9d>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00582.warc.gz"} |
Trinity College Dublin, Hamilton Mathematics Institute
Fellowship ID:
Fellowship Title:
Simons Postdoctoral Fellowship
Fellowship Type:
Dublin, Dublin D02 PN40, Ireland
Subject Areas:
HEP-Theory (hep-th)
HEP-Lattice (hep-lat)
High Energy Theory
Lattice Field Theory
Lattice QCD
Mathematical Physics
Physics - Mathematical Physics
Quantum Field Theory
quantum gravity
Scattering amplitudes
String Theory
String Theory/Quantum Gravity/Field Theory
Theoretical Physics
Appl Deadline:
2024/12/01 11:59PM (posted 2024/09/17, updated 2024/09/16, listed until 2025/08/16)
The Hamilton Mathematics Institute TCD (HMI) invites applications for Simons Postdoctoral Fellowship in theoretical/mathematical physics, starting in Summer/Fall 2025. Positions are usually for two
years, with the possibility of an extension to three years. The position is funded by the Targeted Grants to Institutes from the Simons Foundation. You can visit the HMI web page at http://www.tcd.ie
/Hamilton/ and School of Mathematics web page at https://maths.tcd.ie/. The Hamilton Mathematics Institute (HMI) at Trinity College Dublin was founded in 2005, marking the 200th anniversary of the
birth of William Rowan Hamilton - Ireland's greatest mathematician. The HMI's mission is to position Ireland as a centre of excellence for mathematical research. To apply please submit a curriculum
vitae, list of publications, and statement of research interests on AcademicJobsOnline until December 1, 2024. Please also arrange for three letters of recommendation to be written on your behalf.
Referees should upload the letters following your application before the deadline of December 1, 2024. Preference will be given to candidates with a strong research record in the fields represented
in the School of Mathematics and the Hamilton Mathematics Institute: Quantum Field Theory, String Theory, General Relativity, Quantum Gravity, Lattice Field Theory, and related areas.
Application Materials Required:
Submit the following items online at this website to complete your application:
• Cover letter
• Curriculum Vitae
• Research statement
• Publication list
• Three reference letters (to be submitted online by the reference writers on this site
And anything else requested in the description.
Further Info:
Samson L. Shatashvili
School of Mathematics
Trinity College Dublin
Dublin 2
D02 PN40 | {"url":"https://academicjobsonline.org/ajo/jobs/28442","timestamp":"2024-11-06T21:58:39Z","content_type":"application/xhtml+xml","content_length":"11414","record_id":"<urn:uuid:0fa55b3d-f225-49a2-99b9-3868acdcdc39>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00473.warc.gz"} |
Merkle Trees: Blockchain's Structural Backbone
Understanding Merkle Trees: The Backbone of Blockchain Technology
Data integrity and verification are very important In the world of cryptography and blockchain technology. Merkle trees, also known as hash trees, play an important role in ensuring that data blocks
can be securely verified across distributed systems. This blog post will dive into what Merkle trees are, their mechanism, the concept of Merkle proofs, and how these components are vital in the
field of blockchain.
What is a Hash?
Before we begin with Merkle trees, it's important to understand the concept of hashing. Hashing is performed by a hash function, which is a cryptographic process that takes an input (or 'message')
and returns a fixed-size string of bytes. The output, known as the hash, acts as a digital fingerprint of the input data. Hashes are unique, even a minor change in the input data will produce an
entirely different hash. This characteristic makes hashing an invaluable tool for verifying data integrity.
The Structure of Merkle Trees
Leaf Nodes
At the bottom of a Merkle tree are the leaf nodes. These nodes contain the hashes of individual data blocks (for example, transactions in a blockchain). Imagine each of these hash as a unique
identifier or fingerprint for its corresponding block of data.
Non-Leaf Nodes
The level above the leaf nodes consists of non-leaf nodes. Each of these nodes contains a hash that is the result of combining the hashes of two child nodes beneath it. This process of combining and
hashing continues upwards in the tree, halving the number of nodes at each level, until there is a single hash at the top of the tree - the Merkle root.
The Merkle Root
The Merkle root is a single hash that effectively represents the entirety of the data blocks below it in the tree. It is this root which is stored on the blockchain, providing a compact and efficient
summary of all transactions without storing every transaction individually.
Merkle Proofs: Verifying Data Integrity
Merkle proofs are used to verify the integrity of the data. It provides a way to efficiently and securely verify the contents of large data structures, such as databases or files in blockchain
technology. Let's understand this with a fictional example, and later, we will take up a real example:
Imagine you are in a huge library that contains every book ever written in the history of humankind. The unique part is that instead of checking out books, you check out lists that contain summaries
of the books. Now, you are asked to prove that a specific book is in the library without having to show someone every book (which would be impossible due to the library's size). Welcome to the
Special Merkle Library.
• The Books (Data Blocks): Each book in the library represents a piece of data (like a transaction in a blockchain).
• Summaries (Hashes): For each book, there's a unique summary that captures the essence of the book in a fixed size, much like a hash does for data. No two books have the same summary.
• Catalog (Merkle Tree): These summaries are organized in a catalogue (our Merkle tree), where summaries are combined and summarized again, layer by layer, until there's a single, ultimate summary
representing every book in the library - the root summary (the Merkle root).
Merkle Proofs: Let's find a book
To verify a book's presence in a vast library using the library's unique catalogue systems similar to a Merkle proof in blockchain, one begins with the book's unique summary. Next, a pathway of
additional summaries is collected, which, when sequentially combined according to the library's organizational rules, recreate the library's root summary from the ground up, starting from the initial
book's summary. This process ends in the verification step, where if the independently recreated root summary matches the library's official root summary, it conclusively proves that the book in
question is indeed contained within the library, mirroring the efficient and secure verification of transactions within a blockchain.
Real-life example: Transferring a large file
Imagine you have a large file on a server that needs to be sent to a client over a network. Given the size of the file and the variability of network conditions, there's a risk of data corruption
during transmission. To ensure the file arrives intact and any errors can be efficiently detected and corrected, we use a Merkle tree. Let's go through the steps:
Video Explanation
Step 1: Breaking Down the File
The first step in constructing a Merkle tree is to divide the file into consistent-sized chunks. It's crucial to keep the chunk size constant because this uniformity is necessary both for building
the Merkle tree and for reconstructing it on different systems, such as client-side or server-side environments. Smaller chunks result in a larger Merkle tree, while larger chunks reduce the tree's
size. Finding the right balance between chunk size and tree size is key, though we won't delve deeply into this balancing act here.
Step 2: Hashing the Chunks
Once the file is divided, for example, into four equal parts (chunk 1, chunk 2, chunk 3, and chunk 4), we proceed by hashing each chunk. Using a consistent hash function is vital as it ensures that
the output length remains constant regardless of the input size. For simplicity, let's label the hash outputs as follows: hash of chunk 1 = a, hash of chunk 2 = b, hash of chunk 3 = c, and hash of
chunk 4 = d.
Step 3: Creating the Merkle Tree
The next step involves combining these hashes to form the Merkle tree. You start at the leaf nodes and work your way up:
• Combine and hash a and b to form a new hash, ab.
• Combine and hash c and d to form a new hash, cd.
• Finally, combine and hash ab and cd to get the root hash of the Merkle tree, abcd.
This root hash, or the 'root hash,' encapsulates the entire file’s integrity.
Step 4: Utilizing the Merkle Tree
Once the Merkle tree is constructed and saved server-side, the file can be transferred to a client. To verify the file's integrity and identify any corruption, the Merkle tree is reconstructed
• If the client-side root hash matches the server-side root hash (abcd), the file is confirmed to be intact.
• If there's a discrepancy, such as the root hash turning out to be abcz due to corruption in the last chunk, the specific corrupted part of the file is identified without needing to re-download
the entire file.
Step 5: Efficient Repair
The process doesn't just stop at identifying corruption. The Merkle tree enables efficient repair by pinpointing the exact chunk that's corrupted. Once identified, only the corrupted chunk needs to
be replaced, not the whole file. This efficiency is especially beneficial in large files, significantly reducing the bandwidth and time required for repairs.
Demonstration with JavaScript Implementation
Let's explore Merkle tree and Merkle proof with a practical implementation in JavaScript.
Imagine a scenario where an organization, Safe Global, is preparing for an important online conference. To ensure that only invited attendees can access certain secure documents and conference links,
they decided to implement a whitelisting system using Merkle trees. The email addresses alpha@email.com, beta@email.com, and charlie@email.com belong to key team members who are authorized to access
these resources.
Setting Up The Environment
1. Install Node.js and node package manager(npm)
2. Install the merkletreejs package and crypto-js using npm.
npm install merkletreejs crypto-js
Code Implementation
Create a JavaScript file as server.js, and type/paste the following code:
const { MerkleTree } = require('merkletreejs');
const SHA256 = require('crypto-js/sha256');
// Step 1: Prepare the list of email addresses
const emails = [
'alpha@email.com', 'beta@email.com', 'charlie@email.com'
// Step 2: Hash the email addresses
const leaves = emails.map(email => SHA256(email));
// Step 3: Construct the Merkle Tree
const tree = new MerkleTree(leaves, SHA256);
const root = tree.getRoot().toString('hex');
console.log('Root of the tree:', root);
// Print the tree visually
// Step 4: Generate a Merkle proof for an email
const targetEmail = 'beta@email.com';
const targetLeaf = SHA256(targetEmail);
const proof = tree.getProof(targetLeaf);
console.log('Proof for', targetEmail, ':', proof);
// Step 5: Verify the proof
const verified = tree.verify(proof, targetLeaf, root);
console.log('Verification result:', verified);
Execute The Code
To run the code, run the following command in the terminal (assuming the code file name is server.js):
node server.js
Breakdown Of The Code
Let's take a look at each step of the implementation in detail.
Step 1: Preparing the List of Email Addresses
const emails = ["alpha@email.com", "beta@email.com", "charlie@email.com"];
These are the email addresses of the team members who are organizing the conference. Each email represents an individual who needs secure access.
Step 2: Hashing the Email Addresses
const leaves = emails.map((email) => SHA256(email));
Each email is hashed for security reasons. Hashing ensures that even if the Merkle tree data is somehow exposed, the actual email addresses remain confidential.
We can now construct the Merkle Tree using the hashed email addresses.
Step 3: Constructing the Merkle Tree
const tree = new MerkleTree(leaves, SHA256);
const root = tree.getRoot().toString("hex");
A Merkle tree is constructed using the hashed emails. The root of the tree acts as a single hash that uniquely represents all the included email addresses, providing a simple yet robust way to check
the integrity and completeness of the list without revealing individual hashes.
Step 4: Generate a Merkle Proof for an Email
const targetEmail = "beta@email.com";
const targetLeaf = SHA256(targetEmail);
const proof = tree.getProof(targetLeaf);
Suppose beta needs to access a secured document. To do so, he must prove that his email is on the whitelist. The system generates a "Merkle proof" for his email, which is a sequence of hashes that,
when combined with his email hash in a specific manner, should match the known root hash.
Step 5: Verify the Proof
const verified = tree.verify(proof, targetLeaf, root);
The system then verifies the proof by recalculating the hashes up to the root. If the calculated root matches the stored root, it confirms that beta's email was indeed in the original list, and he is
granted access.
You should see an output like the one below:
Root of the tree: d4e6aa093190629dba575173dc9bfc3712038c1c027bcaa65eab53d18867838a
└─ d4e6aa093190629dba575173dc9bfc3712038c1c027bcaa65eab53d18867838a
├─ a3ec2fc73c1aa03b508cd0704258209755f7428665868329aa81c1b09761b4dc
│ ├─ fcbf76cb8c74247c67280d1bb10a64df01e7682c0d7266945ed48062db40a879
│ └─ 92221a95ab5fe542c7c83a0cbf64f5be4364511783903d2016a9c7f10c5e24f3
└─ 9cfe93b4c66b60b6c64301b279f14ba2668cc3372f9f505019f69467eca290c6
└─ 9cfe93b4c66b60b6c64301b279f14ba2668cc3372f9f505019f69467eca290c6
Proof for beta@email.com : [
position: 'left',
data: <Buffer 26 87 4e 2a 6a 6e ed 5b ee 3a cc f8 29 0f 73 d0 72 11 77 c3 f9 4f 95 76 39 cf d3 6a d8 67 9c b2>
position: 'right',
data: <Buffer 03 79 25 bd 59 47 84 20 cb 66 1e e5 b7 fe d9 8e 7c fb 1e 89 85 bf 2a 40 b2 f0 7c df 2c c0 90 d3>
Understanding the Output
1. Root of the tree
Root of the tree: d4e6aa093190629dba575173dc9bfc3712038c1c027bcaa65eab53d18867838a
Ref: console.log('Root of the tree:', root);
Root Hash: This is the root hash of the Merkle tree, which represents all the email addresses in the tree after they have been hashed and structured. This hash is crucial because it serves as a
unique fingerprint of the entire email list.
Any alteration in the email list would result in a different root hash.
2. Visualization of the Merkle Tree
└─ d4e6aa093190629dba575173dc9bfc3712038c1c027bcaa65eab53d18867838a
├─ a3ec2fc73c1aa03b508cd0704258209755f7428665868329aa81c1b09761b4dc
│ ├─ fcbf76cb8c74247c67280d1bb10a64df01e7682c0d7266945ed48062db40a879
│ └─ 92221a95ab5fe542c7c83a0cbf64f5be4364511783903d2016a9c7f10c5e24f3
└─ 9cfe93b4c66b60b6c64301b279f14ba2668cc3372f9f505019f69467eca290c6
└─ 9cfe93b4c66b60b6c64301b279f14ba2668cc3372f9f505019f69467eca290c6
Ref: console.log(tree.toString());
Tree Structure: This tree visually shows how the hashes (represented as nodes) are structured. The root is at the top, branching down to leaf nodes. Each node is derived by hashing its child
nodes together. The structure confirms that each leaf node contributes to the overall root hash, demonstrating the integrity and completeness of the data set.
3. Proof for an Email
Proof for beta@email.com : [
position: 'left',
data: <Buffer 26 87 4e 2a 6a 6e ed 5b ee 3a cc f8 29 0f 73 d0 72 11 77 c3 f9 4f 95 76 39 cf d3 6a d8 67 9c b2>
position: 'right',
data: <Buffer 03 79 25 bd 59 47 84 20 cb 66 1e e5 b7 fe d9 8e 7c fb 1e 89 85 bf 2a 40 b2 f0 7c df 2c c0 90 d3>
Ref : console.log('Proof for', targetEmail, ':', proof);
□ Merkle Proof: This array contains the necessary hashes and their positions required to verify beta@email.com against the tree's root hash.
□ The first object indicates the hash of a sibling node (left), which means to recreate the hash of the parent node, this hash should be combined with the target node's hash on the left.
□ The second object represents another hash (right) needed further up the tree to continue the verification path to the root.
□ Buffers: These Buffer objects represent the binary data of the hashes necessary for verification.
4. Verification Result
Verification result: true
└─ Root hash
├─ Intermediate hash (from H1 and H2)
│ ├─ Hash of `alpha@email.com`
│ └─ Hash of `beta@email.com`
└─ Duplicate of H3's hash (from H3 and H4)
└─ Hash of `charlie@email.com` (duplicated for balancing)
Ref: console.log('Verification result:', verified);
Verification Result: The true result indicates that the proof successfully verified that the email beta@email.com is part of the whitelist as its recalculated path matches the root hash.
This method of using Merkle trees to verify access to secure resources is a powerful tool for many digital applications, from blockchain technologies to secure communications and content access
What If
Let's see what happens if the email provided isn't on the whitelist, say delta@email.com.
Substitute beta@email.com with delta@email.com in the Step 4. Then, perform the verification again.
// Extract the proof for `delta@email.com`
const targetEmail = "delta@email.com";
const targetLeaf = SHA256(targetEmail);
const proof = tree.getProof(leaf);
// Verify the proof
const verified = tree.verify(proof, targetLeaf, root);
console.log("Verification result:", verified);
The verification would fail with the message Verification Result: false because delta@email.com isn't whitelisted. This causes the verified variable to be false. To better understand this situation,
if you print the tree.getProof(leaf) for this new email on the terminal, you'll find an empty array, indicating an invalid proof.
In the digital world, where data integrity and security are critical, Merkle trees offer a sophisticated yet elegant solution. Originally from the field of cryptography, these structures have become
a fundamental component in blockchain technologies, providing a reliable method for verifying the integrity of large data sets. The mechanism of Merkle trees allows systems to ensure that data has
not been altered, tampered with, or corrupted, without the need to examine the entire data set—highlighting their critical role in enhancing digital trust.
Through the concepts of hashing, Merkle proofs, and practical applications like file transfer and access control, we can see how Merkle trees optimize data verification processes. They enable
efficient validation of individual data elements within large data sets, making them essential for blockchain transactions where verifying every single operation would be computationally
overwhelming. Additionally, the adaptability of Merkle trees allows for broad application across various industries, from secure financial transactions to privacy-enhanced communication systems.
As technology evolves and the volume of digital data continues to expand, the importance of robust, scalable, and efficient cryptographic solutions like Merkle trees will only grow. Whether in the
secure transfer of files or in the verification of blockchain transactions, Merkle trees remain a fundamental part of modern cryptography. In summary, the foundational role of Merkle trees in
blockchain technology not only secures data but also propels the industry towards a more reliable and trustworthy digital future. | {"url":"https://mustafas.work/understanding-merkle-trees-the-backbone-of-blockchain-technology","timestamp":"2024-11-09T16:48:30Z","content_type":"text/html","content_length":"292108","record_id":"<urn:uuid:8ee8bed4-8caf-432b-8828-ebca232a4481>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00435.warc.gz"} |
Repeating Decimals To Fractions Worksheet
Repeating Decimals To Fractions Worksheet. Analytic a linked atom company award all convergents for that fraction, Havens says. With an aberrant quantity, these nested layers will go on endlessly.
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Repeating Decimals To Fractions Activity For Google Classroom
Repeating Decimals Worksheet 2 – You will convert rational numbers like fractions and combined numbers into repeating decimals; plus you’ll convert repeating decimals into rational numbers. Repeating
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It does train students to place one 9 in the denominator per repeating digit when the whole decimal repeats. There is also another technique for when only a portion of the decimal repeats. I even
have found higher success with this method than the algebraic method.
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Whereas, the decimal is a quantity, whose complete quantity half and the fractional half is separated by a decimal point. This product serves to assist students identify if a number is either
rational or irrational. The focus is on “tough” numbers – excellent squares vs. square roots; terminating vs. non-terminating decimals; and repeating vs. rising patterns.
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Graphs of Linear Equations, Drawing
Graphs of linear equations can be drawn when the equations are of the form: y = ax + b.
Such as with the example below.
More specifically, the ‘slope intercept’ form of a straight line is the equation:
y = {\text{m}}x + c
Where the letter m represents the slope or gradient of the straight line.
While the letter
represents the point at which the straight line touches the
Slope/Gradient of a Straight Line:
We’ll give a brief overview of the gradient of a straight line here, though we will go into it in more detail on the topic in a more concentrated straight line section.
The gradient is often written as a fraction but can be a whole number, and describes numerically how steep a straight line is.
The larger the number the gradient m is, the steeper the slope.
Also, the nature of m tells us the slope direction.
A positive gradient is a slope moving upward from left to right. /
While a negative gradient is a slope moving downward from left to right. \
Drawing Graphs of Linear Equations
Sketch the graph of
y = {\large{\frac{5}{3}}}x + 1
The first step here is to mark a point on the
-axis with the information we have from the linear equation.
value is 1,
so we know the straight line will touch the
-axis at (0,1).
Now we look at the gradient value m, which here is
Now the gradient as a fraction is
\frac{{\text{difference \space in}} \space y}{{\text{difference \space in}} \space x}
What this means for us here, if that we begin at our starting point of 1 on the
Then go 5 points up the
-axis, and 3 points along the
-axis to the right.
Giving us (3,6).
This is the correct slope of the line so the new coordinate we find will be another point on the straight line, thus enabling us to draw the line.
We really only need to know two points on a straight line to be able to make a sketch on a suitable axis.
Sketch the graph of
y = \space 3x \space {\text{--}} \space 2
The coordinate on the
-axis this time will be (-2,0).
For finding our second coordinate using the slope/gradient, we can treat the m value 3 as a fraction,
So like in example (1.1), we go 3 places up the
-axis, and 1 place right along the
Sketch the graph of
y = {\text{-}}{\large{\frac{5}{2}}}x + 4
So we can see that the coordinate on the
-axis is (4,0).
But we have a negative value for m this time.
This means that in finding the second coordinate of the straight line using the slope, we will initially move down the
-axis, but still moving right along the
That’s the thing to remember with negative m values,
places move downward, but
places still move right, as they also do when m is positive.
Drawing Graphs of Linear Equations,
Alternative Method
When attempting to graph a linear equation, an alternative approach is to work out several points on the line, plot them and then connect them with a straight line.
If we look again at the linear equation in example (1.2).
We can plug in some
values to obtain some points.
x = 0
y = 3(0) \space {\text{--}} \space 2 \space = \space {\text{-}}2
x = 1
y = 3(1) \space {\text{--}} \space 2 \space = \space 1
x = 2
y = 3(2) \space {\text{--}} \space 2 \space = \space 4
These 3 points will be sufficient to plot and thus draw the correct straight line.
› › Drawing Linear Graphs
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Calculate distance between current location and Polyline Data in Flutter
Consider a River that goes in length of many kilometers. Now, if I ask you how do you detect whether the user is standing close to the river, what would be your answer?
Now, let's break down our problem into parts. If we need to calculate the distance between two locations, we can use the Haversine Formula to manually calculate the distance or the GeoLocator
dependency, which has a built-in function to calculate the distance in units between two points.
But what if we have one location point but the other is a array of location points. How do we calculate whether the provided location point is closer to any of the location points provided. To do
that, we need to understand polyline concept.
A polyline is a series of connected line segments on a map. It's essentially a path drawn between multiple points. This is commonly used to visualize routes, trails, or any other linear data on a
So, a polyline is a line drawn between two latitude-longitude points. A route from point A to point B will have 'n' number of location points and therefore 'n' number of polylines, where 'n' depends
on the number of straight lines needed to reach the destination.
We take pairs of location points and draw polyline between them. And then take the target location point and find a projected location point that falls right on the polyline. Then we will calculate
the distance between target location point and the projected point.
Consider the following example. Let us take the first two location points (L0, L1) in the given array and draw polyline between them. We have the target location (current location) somewhere in the
top of the polyline.
We can now draw a line from the target location point, perpendicular (or shortest distance) to the polyline to find the projected point that touches the polyline. Now we have the projected point
location and target point location. We can easily calculate distance between these two using geolocator.
Taken in consideration that you have already included geo_locator dependency in your package. Lets create two data classes named 'line_segment.dart' and 'location_info.dart'
class LineSegment {
final LocationInfo start;
final LocationInfo end;
LineSegment(this.start, this.end);
class LocationInfo {
final double latitude;
final double longitude;
LocationInfo({required this.latitude, this.longitude});
Now, lets create a helper class named 'location_service.dart' with the following function added. The function will take two parameters. One is the user location called point and another is a
LineSegment parameter which contains start and end lat long. The function will draw a line between start and end positions and project another line to reach the previously drawn line. This projection
will happen in such a way to find the shortest path to reach the previously drawn line. The function will then return the co-ordinates of the projected point.
class LocationService {
LocationInfo _projectPointOnLineSegment(
LocationInfo point, LineSegment lineSegment) {
// Vector from line segment start to point
final v = LocationInfo(
latitude: point.latitude - lineSegment.start.latitude,
longitude: point.longitude - lineSegment.start.longitude);
// Vector representing the line segment
final w = LocationInfo(
latitude: lineSegment.end.latitude - lineSegment.start.latitude,
longitude: lineSegment.end.longitude - lineSegment.start.longitude);
// Calculate the projection parameter
final wLengthSquared = w.latitude * w.latitude + w.longitude * w.longitude;
final wDotV = w.latitude * v.latitude + w.longitude * v.longitude;
final t = wDotV / wLengthSquared;
// Clamp t to the range [0, 1] to ensure the projected point is on the line segment
final tClamped = t.clamp(0.0, 1.0);
// Calculate the projected point
final projectedPoint = LocationInfo(
latitude: lineSegment.start.latitude + tClamped * w.latitude,
longitude: lineSegment.start.longitude + tClamped * w.longitude,
return projectedPoint;
Now that we have found out the co-ordinates of a projected point, we can easily calculate the distance between projected point and the user's current location using geo_locator. Add the following
functions inside 'location_service.dart'
double calculateDistanceToRoad(
LineSegment lineSegment, LocationInfo currentLoc) {
// finds the co-ordinates of projected point
LocationInfo projectLoc =
_projectPointOnLineSegment(currentLoc, lineSegment);
// Calculate distance between current location and projection point
final distanceToProjection =
calculateDistanceInMeters(currentLoc, projectLoc);
double calculateDistanceInMeters(LocationInfo startDes, LocationInfo endDes) {
return _geoLocator.distanceBetween(startDes.latitude, startDes.longitude,
endDes.latitude, endDes.longitude);
If you have a use case where you have a large list of polyline points, and want to detect whether the user is in a certain distance from the entire route, you can loop through each co-ordinate
pairs to find the distance from user location and validate whether the calculated distance is less than the desired radius.
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every algebra student. | {"url":"https://softmath.com/parabola-in-math/exponential-equations/3rd-grade-algebra.html","timestamp":"2024-11-04T21:47:50Z","content_type":"text/html","content_length":"51012","record_id":"<urn:uuid:5f2caba2-96d6-44c5-bb65-4376c532e054>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00855.warc.gz"} |
The experiences with the software Jet Fire thermal radiation modelling
The application for the theoretical calculations of Jet Fire based on Yellow book was developed on JAVA platform and tested in NetBeans IDE 7.4. Model calculates the size and shape of a jet for
gaseous releases from pipelines, tanks and two-phase releases from tanks. Chamberlain empirical formulas for vertical and inclined burns in a horizontal wind are used to describe the geometry of the
flame. The model returns the ground level distance for each of the heat radiation level of concern (tested for 5, 8, 10 kW/m^2). The model was used to calculate the heat radiation as a function of
distance for a) C[n]H[2n+2 ](alkanes, n = 1-4), b) C[n]H[2n ](alkenes, n = 2-4), c) C[n]H[2n+1]OH (alcohols, n = 1-4) and hydrogen. The main benefit of presented model is that it allows a quick and
fast estimation of the heat radiation from Jet Fire and could be further developed according to actual needs. It allows understanding the basic connections and the key parameters of Jet Fire
phenomena from both the mathematical and physical point of view that make it primarily suitable for academic purposes.
Keywords: Jet-Fire model; heat radiation, alkanes, alkenes, alcohols
Na základě Yellow book byla vytvořena aplikace k teoretickému výpočtu tepelného toku z Jet Fire založená na platformě JAVA a testována v prostředí NetBeans IDE 7.4. Modelem je možné spočítat velikost
a tvar tryskového požáru pro úniky plynů z potrubí, zásobníků a dvou-fázové úniky ze zásobníků. K popisu geometrie plamene je použit Chamberlainův vztah pro vertikální a horizontální směr vanutí
větru. Modelem lze vypočítat vzdálenost k cíli pro různé hodnoty hustot tepelného toku (testováno pro 5, 8, 10 kW/m^2). Model byl použit k výpočtu hustoty tepelného toku jako funkce vzdálenosti pro
a) C[n]H[2n+2 ](alkany, n = 1-4), b) C[n]H[2n ](alkeny, n = 2-4), c) C[n]H[2n ](acetylen, n = 2), d) C[n]H[2n+1]OH (alkoholy, n = 1-4) a vodík. Hlavním přínosem prezentovaného modelu je, že umožňuje
rychlé stanovení hodnoty hustoty tepelného toku z Jet Fire a může být dále rozvíjen vzhledem k aktuálním inženýrským a výukovým potřebám. Model umožňuje porozumět základním vztahům a klíčovým
parametrům jevu Jet Fire jak z matematického, tak z fyzikálního hlediska, což tento model primárně předurčuje pro akademické účely.
Klíčová slova: Jet-Fire model; hustota tepelného toku, alkany, alkeny, alkoholy
1. Introduction
A jet or spray fire is a turbulent diffusion flame resulting from the combustion of a fuel continuously released with some significant momentum in a particular direction or directions. Jet Fires can
arise from releases of gaseous, flashing liquid (two phase) and pure liquid inventories. Jet Fires represent a significant element of the risk associated with major accidents on offshore
installations. The high heat fluxes to impinged or engulfed objects can lead to structural failure or vessel/pipework failure and possible further escalation. The rapid development of a Jet Fire has
important consequences for control and isolation strategies. The properties of Jet Fires depend on the fuel composition, release conditions, release rate, release geometry, direction and ambient wind
conditions. Low velocity two-phase releases of condensate material can produce lazy, wind affected buoyant, sooty and highly radiative flames similar to pool fires. Sonic releases of natural gas can
produce relatively high velocity fires that are much less buoyant, less sooty and hence less radiative. The main objectives of this contribution are (1) to develop quick and fast estimation of the
heat radiation from Jet Fire based on the Yellows book Chamberlain model [1]; (2) to initiate the need to increase knowledge and understanding in areas of Jet Fire effects evaluation for students of
technical directions [2].
2. Previous studies
The Yellow book model has been generally accepted since 1997 as the semi-empirical model that provided the most accurate and reliable predictions of the physical hazards associated with Jet Fires,
providing its application limited to the validation range of the model. This conclusion essentially remains valid today. The most important consideration when assessing the relevance and
applicability of mentioned model is the range of data used in its derivation. This model has been developed in several years of research and has been validated with wind tunnel experiments and field
tests both onshore and offshore. An in-depth description of the model is reported in Chamberlain [3]. Chamberlain’s model was selected over the alternative point source model since the latter is
known to be insufficient within one to two flame lengths for short-term radiation levels although sufficiently accurate in the far field. The Chamberlain better mimics the actual size and shape of
a flare. In the literature [4],[5] could be identified two versions of the model, [3] and [6], both of which approximated the geometry of a flare as a frustum of a cone. While Kalghatgi’s used small
burners in a wind tunnel, the main focus of Chamberlain’s work was on field trials at onshore oil and gas production installations. Both models used empirically fit equations to describe the flame
shape. In fact, Chamberlain uses empirical equation to derive the flame length. Because Chamberlain’s work was more recent and involved larger scale testing, the Chamberlain model was selected to
describe thermal radiation hazards for Jet Fire.
3. Mathematical model
The model represents the flame as a frustrum of a cone, radiating as a solid body with a uniform surface emissive power. Correlation describing the variation of flame shape and surface emissive power
under a wide range of ambient and flow conditions. The input parameters for chemicals are taken from DIPPR database [8].
3.1 Calculation of the flame dimensions
Steps 1-8 show the calculation of the exit velocity of an expanding jet. This exit velocity is an important parameter for the calculation of the flame length, lift-off and the widths of the frustum.
First the properties of the flammable material are required for the calculation of the exit velocity of the gas, i.e.: molecular weight, the Poisson constant and the storage conditions of the gas,
such as temperature and the pressure.
The mass fraction of fuel in a stoichiometric mixture with air:
where W = mass fraction of fuel in a stoichiometric mixture with air [-]; W[g] = Molecular weight of gas [kg/mol].
Ratio of specific heat - Poisson constant:
where γ = ratio of specific heat - Poisson constant [-]; C[p] = specific heat capacity at constant pressure [J/kg.K]; R[c] = gas constant 8.314 [J/mol.K]; W[g] = Molecular weight of gas [kg/mol].
For high pressure gas:
where γ = Poisson constant [-]; C[p] = specific heat capacity at constant pressure [J/kg.K]; C[v] = specific heat capacity at constant volume [J/kg.K]; R[c] = gas constant 8.314 [J/mol.K].
The temperature of the expanding jet:
where T[j] = temperature of the expanding jet [K]; T[s] = initial temperature of the gas [K]; P[air] = atmospheric pressure [N/m^2]; P[init] = initial pressure [N/m^2]; γ = ratio of specific heat -
Poisson constant [-].
The static pressure at the hole exit plane:
where P[c] = static pressure at the hole exit plane [N/m^2]; P[init] = initial pressure [N/m^2]; γ = ratio of specific heat - Poisson constant [-].
The Mach-number for chocked flow of an expanding jet:
[] (6)
where M[c] = Mach-number for chocked flow of an expanding jet [-]; γ = ratio of specific heat - Poisson constant [-]; P[c] = static pressure at the hole exit plane [N/m^2]; P[air] = atmospheric
pressure [N/m^2].
The Mach-number for unchocked flow of an expanding jet:
where M[j] = Mach-number for chocked flow of an expanding jet [-]; γ = ratio of specific heat - Poisson constant [-]; m´ = mass flow rate [kg/s]; d[0] = diameter of the hole [m]; T[j] = temperature
of the expanding jet [K]; W[g] = Molecular weight of gas [kg/mol].
The exit velocity of the expanding jet:
where u[j] = exit velocity of the expanding jet [m/s]; Mach-number for (un)chocked flow of an expanding jet [-]; R[c] = gas constant 8.314 [J/mol.K]; T[j] = temperature of the expanding jet [K]; W[g]
= Molecular weight of gas [kg/mol].
By increasing the gas velocity, the fraction of heat released as radiation and the levels of heat radiation received radiation are reduced.
3.2 Calculation of the flame dimensions
In steps 9-22 position and dimensions of the flames are determined. These position parameters are required to calculate the lift-off and angle of the flame with respect to the object. This is
important for the calculation of the view factor. The flame dimensions are used to calculate the surface area of the flame.
The ratio of wind speed to jet velocity:
where R[w] = ratio of wind speed to jet velocity [-]; u[w] = wind velocity [m/s]; u[j] = exit velocity of the expanding jet [m/s].
The density of air:
where ρ[air] = density of air [-]; P[air ]= atmospheric pressure [N/m^2]; W[air] = Molecular weight of air [kg/mol]; R[c] = gas constant 8.314 [J/mol.K]; T[air] = air temperature [K].
Combustion effective source diameter:
In combustion modeling the effective source diameter is widely used concept, representing the throat diameter of an imaginary nozzle releasing air of density at mass flow rate m.
where D[s] = effective source diameter [m]; m´ = mass flow rate [kg/s]; ρ[air] = density of air [-]; u[j] = exit velocity of the expanding jet [m/s].
Combustion effective source diameter in case of a chocked flow:
where d[j] = diameter of the jet at the exit hole [m]; P[c ]= static pressure at the hole exit plane [N/m^2]; W[g ]= molecular weight of gas [kg/mol]; ρ[air ]= density of air [kg/m^3]; R[c] = gas
constant 8.314 [J/mol.K]; T[j] = temperature of the expanding jet [K].
The jet expands to atmospheric pressure at a plane downstream of the exit hole with the plane acting as a virtual source of diameter.
It can be assumed that the diameter of the jet fire about equal as the diameter of the hole.
where Y = variable coefficient calculated by iteration [-]; D[s] = effective source diameter [m]; u[j] = exit velocity of the expanding jet [m/s]; W[g ]= molecular weight of gas [kg/mol].
The length of the Jet Fire in still air:
where L[b0] = flame length in still air [m]; Y = variable coefficient calculated by iteration [-]; D[s] = effective source diameter [m].
For a tilted jet, Kalghatgi [6] showed in laboratory experiments that the flame length reduces as the jet is tilted into the wind. Chamberlain [3] uses Kalghatgi’s empirical fit equation to determine
the flame length, L[b]. Extending from the center of the hole to the flame time, L[b], is calculated.
The length of the Jet Fire measured from the tip of the flame to the center of the exit plane:
where L[b] = length of the Jet Fire measured from the tip of the flame to the center of the exit plane [m]; L[b0] = flame length in still air [m]; u[w] = wind velocity [m/s]; Θ[jv] = angle between
hole axis and the horizontal in the direction of the wind [°].
The Richardson number of the flame in still air:
If R[w ]≤ 0.05, then the flame is dominated, the tilt angle is given by:
where α = the tilt angle [°]; Θ[jv] = angle between hole axis and the horizontal in the direction of the wind [°]; R[i](L[b0]) = Richardson number [-], R[w] = ratio of wind speed to jet velocity [-].
If R[w ]> 0.05, then the flame tilt becomes increasingly dominated by wind forces:
where α = the tilt angle [°]; Θ[jv] = angle between hole axis and the horizontal in the direction of the wind [°]; R[i](L[b0]) = Richardson number [-], R[w] = ratio of wind speed to jet velocity [-].
The lit-off of the flame by the following empirical relation:
where b = lit-off of the flame [m]; L[b] = length of the Jet Fire measured from the tip of the flame to the center of the exit plane [m]; R[w] = ratio of wind speed to jet velocity [-].
In still air (α = 0°), b = 0.2·L[b]. For flames pointing directly into the high winds (α = 180°), b = 0.015·L[b].
Length of frustum (flame):
where R[l] = length of frustum [m]; L[b] = length of the Jet Fire measured from the tip of the flame to the center of the exit plane [m]; b = lit-off of the flame [m]; α = the tilt angle [°].
Ratio between air and jet density:
where ρ[air] = density of air [-]; ρ[j] = density of jet [-]; T[j] = temperature of the expanding jet [K]; W[air] = molecular weight of air [kg/mol]; T[air] = temperature of air [K]; W[g] = molecular
weight of gas [kg/mol].
Richardson number based on the combustion source diameter, used for the calculation of the frustrum base width:
where R[i](D[s]) = Richardson number based on the combustion source [-]; D[s] = effective source diameter [m]; uj = exit velocity of the expanding jet [m/s].
Constant for calculation the frustrum base width:
where C = constant for calculation the frustrum base width [-]; R[w] = ratio of wind speed to jet velocity [-].
The frustrum base width:
where W[1 ]= width of frustrum base [m]; D[s] = effective source diameter [m]; R[w] = ratio of wind speed to jet velocity [-]; P[air ]= atmospheric pressure [N/m^2]; P[j ]= jet pressure [N/m^2].
The frustrum tip width:
where W[2 ]= width of frustrum tip [m]; L[b] = length of the Jet Fire measured from the tip of the flame to the center of the exit plane [m]; R[w] = ratio of wind speed to jet velocity [-].
The surface area of frustrum, including end discs:
where A = surface area of frustrum including end discs [m^2]; W[1 ]= width of frustrum base [m]; W[2 ]= width of frustrum tip [m].
3.3 Calculation of the surface emissive power
The surface emissive power can be calculated with the net heat released from combustion of the flammable gas, the fraction of that part of the heat radiated and the surface area of the frustrum.
The net heat per unit time released:
where Q = combustion energy per second [J/s]; m´ = mass flow rate [kg/s]; ΔH[c] = heat of combustion [J/kg].
The fraction of heat radiated from the surface of the flame:
where F[s] = fraction of heat radiated from the surface of the flame [-]; u[j] = exit velocity of the expanding jet [m/s].
The surface emissive power:
where SEP[max] = maximum surface emissive power [J/m^2.s]; F[s] = fraction of heat radiated from the surface of the flame [-]; Q = combustion energy per second [J/s]; A = surface area of frustrum
including end discs [m^2].
3.4 Calculation of the view factor:
Coordinate transformation to X´, Θ´ is required before a model for the calculation of the view factors can be used.
where X´ = distance from the center of the bottom plane of a lifted-off flame o the object [m]; b = frustrum lift-off height [m]; Θ[j ]= angle between the centerline of a lifted-off flame and the
object [°]; X = distance from the center of the flame without lift-off to the object [m].
where Θ´ = angle between the centerline of lifted-off flame and the plane between the center of bottom of the lifted-off flame and the object [°]; Θ[j ]= angle between the centerline of a lifted-off
flame and the object [°]; α = angle between hole axis and the flame axis [°]; b = frustrum lift-off height [m]; X = distance from the center of the flame without lift-off to the object [m].
where x = distance from the surface area of the flame to the object [m]; X´ = distance from the center of the bottom plane of a lifted-off flame o the object [m]; W[1] = width of frustrum base [m]; W
[2] = width of frustrum tip [m].
3.5 Calculation of atmospheric transmissivity
Partial vapor pressure of water in air at a relative humidity:
where p[w] = partial vapor pressure of water in air at a relative humidity RH [Pa; N/m^2]; RH = relative humidity of air [%rel/100]; T[a ]= absolute temperature of ambient air at standard conditions
Calculation of the atmospheric transmissivity (valid for 10^4 < p[w ]· x < 10^5) if absorption coefficient of water vapor and absorption coefficient of carbon dioxide is not known:
where τ = atmospheric transmissivity [-]; p[w] = partial vapor pressure of water in air at a relative humidity RH [Pa; N/m^2]; x = distance from the center of the Fire Ball to the radiated object
Calculation of the atmospheric transmissivity if absorption coefficient of water vapor and absorption coefficient of carbon dioxide is known:
where τ = atmospheric transmissivity [-]; α[w] = absorption coefficient of the water vapor for an average flame temperature [-]; α[c] = absorption coefficient of the carbon dioxide for an average
flame temperature [-].
3.6 Calculation of the heat flux at a certain distance
With the dimensions of the flame and the view factor the thermal heat flux at a certain distance x from the heat source can be calculated:
where q´´= thermal heat flux at a certain distance x from the heat source [J/m^2.s]; SEP[max] = maximum surface emissive power [J/m^2.s]; F[s] = fraction of heat radiated from the surface of the
flame [-]; τ = atmospheric transmisivity [-].
4. Results and discussions
In the following calculation example the heat radiation was calculated as a function of distance for an object at a certain distance i.e. 150 m from the surface of the jet flame. Data from [1], [8]
has been used as the input values for the calculation example of chocked flow through the 0.1 m hole diameter in high pressure pipeline. We calculated heat radiation as a function of distance for
hydrogen and we compared calculated results with a) C[n]H[2n+2 ](alkanes, n = 1-4); b) C[n]H[2n] (alkenes, n = 2-4); c) C[n]H[2n-2 ](acetylene, n = 2); d) C[n]H[2n+1]OH (alcohols, n = 1-4). From
chemical point of view we investigated chemicals (hydrocarbons) with a) different number of carbon atoms; b) and c) different bonds order; d) hydroxyl chemical functional group. The results of
calculations are illustrated in Figures 1-2.
Figure 1: Heat radiation as a function of distance for hydrogen and a) C[n]H[2n+2 ](alkanes, n = 1-4); b) C[n]H[2n] (alkenes, n = 2-4).
We can see at Figure 1a that there is a simple increasing trend going from the C1 to the C4 species (24.0 kW/m^2, 26.4 kW/m^2, 30.2 kW/m^2, 33.7 kW/m^2). The calculated theoretical values of heat
radiation curve have its maximum at approximately 9.7 m as a distance of the flame to the object. The lethal value of radiation taken as 10 kW/m^2 corresponds to 31.5 m (C1), 34.5 m (C2), 38.2 m (C3)
and 42 m (C4), respectively. Similarly, we can see at Figure 1b that there is also a simple trend going from the C1 to the C4 species (26.4 kW/m^2, 25.3 kW/m^2, 32.8 kW/m^2). The calculated
theoretical values of heat radiation plot have its maximum at approximately 5.2 m as a distance of the flame to the object. The lethal value of radiation taken as 10 kW/m^2 corresponds to 31.5 m
(C1), 34.5 m (C2), 38.2 m (C3) and 42 m (C4), respectively. For further comparison we divide the plots from Figures 1a,b into two parts from the point of view of distance. First part will be from 0 m
^to 5.2 m and the second part will be from 5.2 m^to approximately 60 m. If we compare the theoretical calculated results for the first part we can recognize that the line for C[n]H[2n+2 ](alkanes, n
= 1-4) is less increasing than that for C[n]H[2n] (alkenes, n = 2-4). If we compare the theoretical calculated results for the second part for both tested systems we can see that the line for C[n]H
[2n+2 ](alkanes, n = 1-4) is less decreasing than that for C[n]H[2n] (alkenes, n = 2-4). As a conclusion both results show that the profile shape factor of C[n]H[2n+2 ](alkanes, n = 1-4) is more
sharp than that of C[n]H[2n] (alkenes, n = 2-4). These basic facts concerning the heat radiation as a function of distance for the species with different bonds orders could be recognized and could be
further analyzed by comparing them with the species substituted by hydroxyl chemical functional group. In both cases, the theoretical calculations of hydrogen heat radiation as a function of distance
(denoted by the red color in Figures 1a,b) have been used for scaling (from 0 m to 150 m). As in the case of alkanes and alkenes, we can see at Figure 2a the value of acetylene heat radiation (24.7
kW/m^2). The calculated theoretical value of heat radiation curve has its maximum at approximately 5.2 m as a distance very similar to that obtain for alkenes. The lethal value of radiation taken as
10 kW/m^2 corresponds to approximately 24.7 m which is slightly lower than 25.3 kW/m^2 for C[2]H[4]. For further comparison we divide the plots from Figures 1a,b into two parts from the point of view
of distance. We can see at Figure 2b that there is also a simple trend in alcohol substituted species going from the C1 to the C4 species (11.2 kW/m^2, 17.8 kW/m^2, 22.8 kW/m^2, 24.3 kW/m^2). This
may be related to the fact that the molecular structure of the OH· moiety change under substitution heat radiation value much more than in C[n]H[2n+2], as shown by presented theoretical calculation,
confirmed by results of experimental studies C[n]H[2n+1]OH. Apart from the trends investigated in different number of carbon and different bonds order aspects, the study of calculated jet, fire and
heat flux parameters and for calculation used constants are of importance.
Figure 2: Heat radiation as a function of distance for hydrogen and a) CnH2n-2 (acetylen, n = 2); b CnH2n+1OH (alcohols, n = 1-4).
All jet, fire and heat flux parameters used for calculation of the heat radiation as a function of distance were established during the present investigation for the first time, and the values of
alkanes, alkynes and acetylene could be compared with those of their alcohols analogues. The present gas phase (alkanes, alkenes, alkynes) and two-phase (alcohols) investigation has started the
series of studies by mathematical modelling of the substituted hydrocarbons. The calculated values of parameters characterizing the jet, fire and heat radiation have been accurately determined for
methane, and they compare well with the theoretical calculations listed in [1]. These values are typical of the hydrocarbons of non-multiple bonds. The mass fraction of fuel in a stoichiometric
mixture with air, W, together with gas constant, R[c], and specific heat capacity, C[p], was effectively transformed into ratio of specific heat - Poisson constant, γ. The value of Poisson constant γ
= 1,306 has been of comparable value with the value γ^ Y = 1,307 published in [1]. The Mach-number, M[j] ,for chocked flow of an expanding jet has been determined in this investigation. The
Mach-number is estimated from the temperature of the expanding jet, T[j], and the static pressure, P[c], at the hole exit plane [N/m^2]. The value of the Mach-number M[j] = 3,95 has been of slightly
higher value than the value M[j]^Y= 3,55 published in [1]. The further parameters values are in good agreement with the values published in [1].
5. Conclusion
(1) exit velocity of expanding jet [m/s]; (2) angle between hole and flame axis [°]; (3) frustum lift-off height [m]; (4) width of frustum base [m]; (5) width of frustum tip [m]; length of frustum
(flame) [m]; (6) surface area of frustum [m^2]; (7) maximum surface emissive power [kW/m^2]; (8) atmospheric transmissivity [%] and view factor [-]. Further parameters could be implemented based on
actual needs. In the near future we are planning to compare the results of the C[n]H[2n+2] (alkanes, n = 1-4), C[n]H[2n] (alkenes, n = 2-4), C[n]H[2n-2] (acetylene, n = 2), C[n]H[2n+1]OH (alcohols, n
= 1-4) calculations with the results obtained by procedure Jet Fire (Chamberlain model) implemented as a part of the program Effects version 9.0.8 that will be possible to use a tour department. This
model could be further developed according to present engineering knowledge and can make a contribution towards solving the problems facing the flammable liquefied fuels in industrial practice in
assessment of jet fire hazards comprises (1) identification of areas of uncertainty in the characterization of jet fires; (2) identification where the jet fire hazard is significant in relation to
other hydrocarbon hazards; (3) initiation of research to increase knowledge and understanding in ill-defined areas of Jet Fire evaluation and (4) promote the use of a consistent methodology for
evaluation of jet fire hazards. Furthermore, the model application and development could support the practice of the Department of Major Accidents Prevention of Occupational Safety Research
The contribution was prepared in the frame of following projects:
1. Opportunity for young researchers, reg. no. CZ.1.07/2.3.00/30.0016,supported by Operational Programme Education for Competitiveness and co-financed by the European Social Fund and the state
budget of the Czech Republic.
2. Optimization of emergency planning zone and emergency plans creation based on harmful effects of dangerous chemicals released during major accidents with respect to improvement of civil
protection reg. no. VG20112013069, supported by Ministry of Interior of the Czech Republic.
3. Innovation for Efficiency and Environment, reg. no. CZ.1.05/2.1.00/01.0036 supported by Operation Programme Research and development for Innovation and financed by the Ministry of Education,
Youth and Sports.
[1] TNO Bureau for Industrial Safety. Netherlands Organization for Applied Scientific Research. Methods for the Calculation of the Physical Effects. Hague : Committee for the Prevention of Disasters,
[2] American Petroleum Institute. Guide for Pressure-Relieving and Depressuring Systems : API (American Petroleum Institute) Recommended Practice 521. 3rd ed. API : November, 1997.
[3] CHAMBERLAIN, G. A. Development in design methods for predicting thermal radiation from flares. Chem. Eng. Res. Des., 1987, Vol. 65, pp. 299-309.
[4] LEES, F. P. Loss Prevention in the Process Industries : Vol 2. 2nd ed. 1996.
[5] MUDAN, K. S.; CROCE, P. A. SFPE Handbook of Fire Protection Engineering. 2nd ed. National Fire Protection Association, 1995.
[6] KALGHATGI, G. T. The Visible Shape and Size of a Turbulent Hydrocarbon Jet Diffusion Flame in a Cross-wind. Comb. And Flame, 1983, Vol. 52, pp. 91-106.
[7] THOMSON, G. H. The DIPPR databases. International Journal of Thermophysics, Vol. 17, Issue 1, pp. 223-232. ISSN 0195-928X.
[8] BRZUSTOWSKI, T. A.; SOMMER, E. C. Predicting Radiant Heating from Flares. In Proceedings of Division of Refining, American Petroleum Institute, Washington DC, 1973. Vol. 53, pp. 865-893.
Vzorová citace
SKŘÍNSKÝ, Jan …[et al.]. The experiences with the software for jet fire thermal radiation modelling. Časopis výzkumu a aplikací v profesionální bezpečnosti [online], 2013, roč. 6, č. 3-4. Dostupný
z WWW: <http://www.bozpinfo.cz/josra/josra-03-04-2013/jet-fire.html>. ISSN 1803-3687. | {"url":"https://www.bozpinfo.cz/josra/experiences-software-jet-fire-thermal-radiation-modelling","timestamp":"2024-11-11T18:18:32Z","content_type":"text/html","content_length":"103486","record_id":"<urn:uuid:60fa3aa8-865f-4cbe-b4a2-f889c781b655>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00300.warc.gz"} |
An Etymological Dictionary of Astronomy and Astrophysics
Debye model
مدل ِدبی
model-e Debye (#)
Fr.: modèle de Debye
An extension of the → Einstein model accounting for → specific heats, based on the concept of → elastic waves in → crystals. In this model specific heat is given by: C[V] = 9R[(4/x^2)∫ y^2/(e^y - 1)
dy - x/(e^x - 1)], integrating from 0 to x, where R is the → gas constant, k is → Boltzmann's constant, x = hν[max]/k, and y = hν/k. The parameter T[D] = hν[max]/k is the characteristic → Debye
temperature of the crystal. At low temperatures the specific heat prediction by this model is in good agreement with observations (→ Debye law), in contrast to Einstein's model.
→ debye; → temperature. | {"url":"https://dictionary.obspm.fr/?showAll=1&formSearchTextfield=Debye+model","timestamp":"2024-11-11T23:07:04Z","content_type":"text/html","content_length":"11458","record_id":"<urn:uuid:389de596-2bf0-4ca0-bad2-1e2e206a14b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00256.warc.gz"} |
February Math Activities & Worksheets No Prep Printables 2nd Grade • Caffeinated and Creative
Are you in need of math activities/worksheets for the entire month of February!? I have you covered! This February math pack covers Groundhog’s Day, Chinese New Year, Valentine’s Day, AND President’s
These worksheets/activities cover a wide variety of math standards for second grade including word problems, skip counting, 2 and 3 digit addition and subtraction, graphs, elapsed time, arrays, and
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Instead, there will be fun and engaging activities that can be ready just by hitting PRINT, leaving you time to drink your coffee while it’s still hot!
►PLEASE NOTE: This bundle is geared towards second graders but can be used for superstar first graders or third graders who may need some additional help.
❤️FEBRUARY MATH ACTIVITIES INCLUDED:❤️
✅Groundhog’s Day Math
• Sums and Shadows: Solve the three-digit addition problems. Some regrouping is needed
• Shady Shadow Code: Solve the two-digit addition problems to solve the code.
• Shadow Shapes Read the descriptions about different quadrilaterals and write which one it is.
• Groundhog Graph: Create a graph about whether or not you think the groundhog will see his shadow and then answer the questions.
✅The Big Game Math
• Football Fact Families: Make fact families using the numbers given on the footballs.
• Touchdown Tens: Add or subtract ten to a 3 digit number.
• Field Goal Skip Count: Fill in the missing numbers.
• Touchdown Time: Write the digital time under each analog clock.
• Touchdown and Roll Roll a dice once and add it to the 2 or 3 digit number given.
✅Chinese New Year Math
• Dragon Tail Scales Solve the three-digit addition problems. Some regrouping is needed. Then color according to odd or even answers.
• Red Envelope Code Solve the 2 digit addition problems to solve the code. Some regrouping is needed.
• Fireworks and Fractions Write the fraction for the shaded part in a rectangle.
• Good Fortune Facts Solve the three-digit subtraction problems. Some regrouping is needed
• Chinese Zodiac Spinners: Using a paper clip and pencil, create 15 3 digit addition problems to solve. Some regrouping.
• Multiple Chinese New Year Numbers: Using numbers in a lantern, answer questions about place value.
• Snake Sssssskip Count: Fill in the missing numbers.
• Envelope Equals: Write the dollar amount in that several children received during the Chinese New Year.
• Lucky Numbers: Fill in the missing addend.
✅Valentine’s Day Math
• Affectionate Arrays: Write the number of rows, columns and repeated addition for Valentine’s Day candy.
• Postman Place Value: Write either the numerical form, written form or expanded form.
• V Day Party Word Problems: Solve word problems about a Valentine’s Day party.
• Candy Calculations: Solve word problems about Valentine’s Day candy.
• Greetings Geometry: Read the letter Mary Math has written her boyfriend Garret Geometry and fill in the blanks with what shape she is talking about Valentine’s Day Graph: Read a graph about the
Valentine’s Day cards five students received and answer the questions!
• Lovebugs Less Than/More Than: Using <,> or =, compare the numbers in two pieces of candy then color according to the code.
• Sweetheart Subtraction: Solve the three-digit subtraction problems. Some regrouping is needed
• Even or Odd Hearts: Solve the 2 digit addition problems and then color according to the code. Some regrouping is needed
✅President’s Day Math
• President’s Coins: Write which coins could be used to make the amount given.
• Patriotic Place Value: Write either the numerical form, written form or expanded form.
• President Word Problems: Solve word problems about George Washington and Abraham Lincoln.
• Rock Group Code: Solve the 2 digit subtraction problems to solve the code. Some regrouping is needed.
• America Addends: Fill in the missing addend.
• Election Even or Odd: Solve the 3 digit addition problems and then color according to the code. Some regrouping is needed
• Stars and Stripes Sums: Solve the three-digit subtraction problems. Some regrouping is needed
• Oval Office O’clock: Write the digital time after reading short word problems involving time.
• Forefather Fractions Write the fraction for the shaded part in a circle.
• Abe’s Addition: Solve the three-digit addition problems. Some regrouping is needed
◼️Winter Holiday Activities and Worksheets for December January February & Winter
◼️2nd Grade NO PREP Printable ELA and Math Worksheets/Activities for the Year No prep math and ELA ready-to-print activities for the year!
◼️Digital and Printable Year Long Second Grade Bundle No prep and digital activities for the year!
◼️Second Grade Literacy and Math MEGA Bundle which includes everything you will need for second grade! Includes over 440 pages of NO PREP printables for the ENTIRE year as well as over 350 DIGITAL
options!! You’ll also be set with over 47 hands-on literacy and math centers AND 11 writing crafts!
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☕ Check out my site Caffeinated and Creative! | {"url":"https://caffeinatedandcreative.com/product/february-math-activities-worksheets-no-prep-printables-2nd-grade/","timestamp":"2024-11-11T05:18:47Z","content_type":"text/html","content_length":"101486","record_id":"<urn:uuid:53d219b1-9273-40ac-8fa6-371e3445ee28>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00524.warc.gz"} |
Differentiate From First Principles:
Hi! Me again
Anyways. Yeah. I'm differenting from first principles and I'm having trouble with certain functions!
I'll give a simple example, and then hopefully, with your help, I can master these particular functions.
Differentiate, from first principles $f(x) = \frac{1}{x}$
I get:
$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$
What's to be done here then? Would I use the index laws to get rid of the fractions on top, as such:
$f'(x) = \lim_{h \to 0} \frac{(x+h)^{-1}-x^{-1}}{h}$
Then multiply this by $\frac{(x+h)^{-1}+x^{-1}}{(x+h)^{-1}+x^{-1}}$
$f'(x) = \lim_{h \to 0} \frac{(x+h)^{-1}-x^{-1}}{h} \times \frac{(x+h)^{-1}+x^{-1}}{(x+h)^{-1}+x^{-1}}$
$f'(x) = \lim_{h \to 0} \frac{(x+h)^{-2}-x^{-2}}{h((x+h)^{-1}+x^{-1})}$
When I multiply this out past this point I get ridiculous and unhelpful fractions! Any ideas?!
Try writing $\displaystyle \frac{1}{x+h} - \frac{1}{x}$ as a single fraction over a common denominator. | {"url":"https://www.thestudentroom.co.uk/showthread.php?t=480732","timestamp":"2024-11-10T00:03:16Z","content_type":"text/html","content_length":"487518","record_id":"<urn:uuid:33b5a3a8-349c-4113-a63e-5080603fa24a>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00808.warc.gz"} |
How will the Hubble Tension ("crisis in cosmology") be resolved?
Current Cosmic Ladder estimates are wrong
Current Cosmic Microwave Background estimates are wrong
The Hubble tension or crisis in cosmology is the discrepancy in different methods of measuring the speed of the universe's expansion between methods using observations of the early universe and
modeling, vs those using a "ladder" or different distance measurement techniques of stars.
This market will resolve when there is a single, widely agreed upon estimate of Hubble's constant with tight confidence intervals (s.d. <1).
Current best estimates:
Cosmic ladder: SH0ES - 73.04 +- 1.04
CMB: Planck - 67.66 +-0.42
Wrong here means that the current best estimates using either method are 3 standard deviations or more away from the agreed estimate.
If there is scientific consensus that one cannot put a single number estimate on the Hubble constant is wrong, for example in case there are meaningful differences across space, then this market will
resolve to "both are wrong". Temporal variation in the Hubble constant does not count, as the Hubble constant is already regarded as a point in time estimate, and the consensus view is that it varies
by time.
This question is managed and resolved by Manifold.
bought Ṁ10 Current Cosmic Ladde... YES
For a consensus to form, you would need a third method of estimating H0 that is completely independent of the other two and the confidence interval for this estimate has to be small enough so it's
consistent with one and only one of the two candidates. Standard sirens are the only option I'm aware of. While estimates of H0 using gravitational waves will probably improve over this decade, it's
unlikely they will be good enough to select either ΛCDM or the distance ladder. (And that's assuming it does select a candidate—it could end up being inconsistent with both, which would prevent any
consensus from forming!)
As such this market is likely to resolve N/A in 2030. I'm betting Ṁ10 just to track the question; I'm going for the ladder being wrong due to the common rule of thumb in cosmology that "you don't bet
against ΛCDM" when it comes to other things in the field, but this doesn't represent any deep analysis on my part.
@BerF That’s not true at all! A consensus that one of the two estimates has a particular methodological flaw or misunderstanding of a physical phenomenon, leading to a reassessment of that data that
brings it within the other’s confidence interval would be the most likely way for a consensus on H0 to form!
Two likely ways this would happen are
1) New physics is discovered which changes the calculation for one of the two estimates
2) Some clever physicists write a paper pointing out an obvious (in hindsight) methodological flaw in one of the two estimates, that when corrected, brings the two estimates into agreement
@BerF So far there's been several independent methods tried, we just don't have the confidence in those methods, and they usually don't have tight confidence intervals. Mostly they just tend to add
to the confusion, since some of them get estimates close to the cosmic ladder, and some are close to ΛCDM. It's a mess.
My bet is that the tension will only be resolved once someone finds a flaw in either method. So far the cosmic ladder has withstood a very high level of scrutiny, and is consistent with new JWST
data, so I'm crossing my fingers for new physics that would explain why ΛCDM is wrong.
This isn't a timed market. Close date may be extended.
@benshindel Ah, yes, I forgot to say, I don't expect any new physics to be relevant for this question before 2030. Because you have to first detect it somewhere else not related to H0, then theorists
have to understand it well enough to apply it to H0, then both the previous CMB and/or standard candle data has to be re-assessed with the new theory. This would take forever.
I see now the market won't actually close in 2030, though. 🤷♂️
Wrong here means that the current best estimates using either method are 3 standard deviations or more away from the agreed estimate.
This means that (rounding stuff to zero digits after the point) H_0 between 66 and 69 km/s/Mpc would resolve "Current cosmic ladder estimates are wrong", H_0 between 70 and 76 km/s/Mpc would resolve
"Current CMB estimates are wrong", and H_0 below 66, between 69 and 70, or above 76 km/s/Mpc would resolve "Both are wrong", right?
By "Hubble constant" you mean the present-day one, so that if cosmic ladder results are right about the recent expansion rate and CMB results are right about the ancient expansion rate (and merely
wrong about the way to extrapolate them to today) it would resolve as "current CMB estimates are wrong", right?
@ArmandodiMatteo we had this discussion in another thread. Both estimates are for the current day rate, H0. This market is not concerned with past rates.
What if both methods are correct, and Hubble constant is not a constant? (Or something similar)
@Shump My dad is a Distinguished Professor of cosmology at University of California Davis who publishes papers on this topic. I asked him “the Hubble constant isn’t actually a constant right?” And he
responded with “It's a measure of the expansion rate, which changes over time. The value today is constant, but we define the same quantity at other times, and those values are different from the
value today.”
@SoniaAlbrecht well that's funny because MY dad is a Distinguished Professor of cosmology at University of California Berkeley who had Edwin Hubble himself on his thesis committee at Cambridge. HE
says that the Hubble constant IS in fact a constant.
All jokes aside, you seemed to have bet that option up to 80% on the idea that if the Hubble constant could change over time, this would resolve as "both are wrong", and I guess it's up to @Shump but
I don't think that's the point of the question, and doesn't really get at the nature of the Hubble tension.
@SoniaAlbrecht also this paper authored by your dad seems to imply that the quintessence model he's proposing would give a value for H0 that is roughly within 1 std deviation of the Planck data,
lending support to the "Current Cosmic Ladder estimates are wrong" option on this market, not the "both estimates are wrong" option.
I don’t think the Hubble constant not being a constant means the current cosmic ladder estimates are wrong or the cosmic microwave background estimates are wrong. It’s just that the person who made
the market thinks it does.
@SoniaAlbrecht Let me clarify. Both is wrong only resolve in case it doesn't make sense to put a single number estimate on the Hubble constant. If it changes by time, you can still treat it as a
constant in the present. I will update the description.
Wikipedia says:
The parameter H is commonly called the "Hubble constant", but that is a misnomer since it is constant in space only at a fixed time; it varies with time in nearly all cosmological models, and all
observations of far distant objects are also observations into the distant past, when the “constant” had a different value. "Hubble parameter" is a more correct term, with H0 denoting the
present-day value.
I don't think your dads actually disagree, there's just some confusion about definitions here.
@Shump Suppose Hubble constant changes over time, but is otherwise constant. Cosmic ladder correctly estimates average Hubble constant in the last 5 billion years, and CMB measurement correctly
estimates the average over 13 billion years. Would this resolve as “both wrong”?
@OlegEterevsky That's not how it works. CMB calculations still give an estimate for today, not for the average of 13 billion years.
@SoniaAlbrecht I was joking about my dad, lol, there's a long-running Manifold joke of the form "my dad works at _____ (Time Magazine / Capitol Hill) and he says ______ (Xi Jinping is gonna win POTY
/ the next speaker of the house will be...)" but you ACTUALLY have a cosmologist dad, which was funny
@Shump From my understanding of https://arxiv.org/abs/1502.01589, they are assuming that Hubble constant is constant and then fit a model to the data. So strictly speaking, yes, if Hubble constant
were variable, this assumption would be wrong and the method invalid.
That said, if the constant were to change a bit, this method would probably approximate something like the average value of Hubble constant since CMB was emitted. (But here I'm just speculating. I
agree that if the Hubble constant is non-constant than the CMB method is invalid.)
@Shump My bad, you are right. It's because they include dark energy in their model, right? So the potential error of this model is that our model of dark energy is somehow incorrect?
@OlegEterevsky Yes. If the CMB estimate is wrong, it probably means something is missing in the current model of the universe, or "new physics" as the paper authors call it. If the Hubble estimate is
wrong, it's probably because of some errors in the calculation. My favorite hypothesis is sterile neutrinos, which might solve the discrepancy, can also explain dark matter, and I think are
relatively likely to exist from theoretical concerns (I mean, why would neutrinos be the only particle without a right-handed variant?)
@Shump Since mine was a fair assumption to make given the original wording of the question, could I get a refund on the mana I spent? I don’t know how or if this is possible, but I figured I’d ask.
@SoniaAlbrecht I could send you some mana, but honestly it seems to me like you didn't misunderstand the question, you misunderstood the Hubble constant. Both the Planck estimate and the Hubble one
refer to present day values (that's the 0 in H0). I don't see how this can make both estimates wrong.
I'll send you the profit I made, I don't like making profit from others' mistakes.
@Shump Thanks so much! I really appreciate your kindness! I did understand that this doesn’t make both estimates wrong, I just thought you had misunderstood the concept when I made that bet (I’ve
experienced people who make markets doing that before) | {"url":"https://manifold.markets/Shump/how-will-the-hubble-tension-crisis","timestamp":"2024-11-03T17:25:11Z","content_type":"text/html","content_length":"283053","record_id":"<urn:uuid:2d8e1f16-989e-4268-8182-290ae61a231b>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00609.warc.gz"} |
Models and Approximations
© FB10 - D. Münsterkötter
Research Area C
Alsmeyer, Böhm, Dereich, Engwer, Friedrich (until 2021), Gusakova (since 2021), Hille, Holzegel (since 2020), Huesmann, Jentzen (since 2019), Kabluchko, Lohkamp, Löwe, Mukherjee, Ohlberger, Pirner
(since 2022), Rave, Schedensack (until 2019), F. Schindler, Schlichting (since 2020), Seis, Simon (since 2021), Stevens, Weber (since 2022), Wilking, Wirth, Wulkenhaar, Zeppieri.
In research area C, we will focus on the development and foundation of mathematical models and their approximations that are relevant in the life sciences, physics, chemistry, and engineering. We
will rigorously analyse the dynamics of structures and pattern formation in deterministic and stochastic systems. In particular, we aim at understanding the interplay of macroscopic structures with
their driving microscopic mechanisms and their respective topological and geometric properties. We will develop analytical and numerical tools to understand, utilise, and control geometry-driven
phenomena, also touching upon dynamics and perturbations of geometries. Structural connections between different mathematical concepts will be investigated, such as between solution manifolds of
parameterised PDEs and non-linear interpolation, or between different metric, variational, and multi-scale convergence concepts for geometries. In particular, we aim to characterise distinctive
geometric properties of mathematical models and their respective approximations. | {"url":"https://www.uni-muenster.de/MathematicsMuenster/research/modelsandapproximations/index.shtml#id4","timestamp":"2024-11-08T21:11:08Z","content_type":"text/html","content_length":"41607","record_id":"<urn:uuid:a92edab2-ef7e-4b42-be20-70c73f8d20af>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00247.warc.gz"} |
A high school principle currently encourages students to enroll
in a specific SAT prep program that...
A high school principle currently encourages students to enroll in a specific SAT prep program that...
A high school principle currently encourages students to enroll in a specific SAT prep program that has a reputation of improving score by 50 points on average. A new SAT prep program has been
released and claims to be better than their current program. The principle is thinking of advertising this new program to students if there is enough evidence at the 5% level that their claim is
true. The principle tests the following hypotheses:
H0:μ=50 points
HA:μ>50 points
where μμ is the true mean in difference of scores after the new SAT prep course is taken and before the new SAT prep course is taken. He randomly assigns 93 students to take this new SAT program. The
difference in scores resulted in an average of 50.3364 points with a standard deviation of 13.0174 points.
What is the value of the test statistic for this test? Round your answer to four decimal places.
What is your decision regarding the null hypothesis?
A. Reject the null hypothesis, at the 5% significance level, there is not enough evidence to say that the new SAT prep program is better than the current SAT prep program.
B. Fail to reject the null hypothesis, at the 5% significance level, there is not enough evidence to say that the new SAT prep program is better than the current SAT prep program.
C. Fail to reject the null hypothesis, at the 5% significance level, there is enough evidence to say that the new SAT prep program is better than the current SAT prep program.
D. Reject the null hypothesis, at the 5% significance level, there is enough evidence to say that the new SAT prep program is better than the current SAT prep program.
The average height of men in 1960 was found to be 68 inches (5 feet, 8 inches). A researcher claims that men today are taller than they were in 1960 and would like to test this hypothesis at the 0.01
significance level. The researcher randomly selects 184 men and records their height to find an average of 69.7020 inches with standard deviation of 1.1125 inches.
What is the value of the test statistic? Round your answer to four decimal places.
What is your decision regarding the null hypothesis?
A. Fail to reject the null hypothesis. At the 1% significance level there is not sufficient evidence to say that men today are taller than they were in 1960.
B. Fail to reject the null hypothesis. At the 1% significance level there is sufficient evidence to say that men today are taller than they were in 1960.
C. Reject the null hypothesis. At the 1% significance level there is not sufficient evidence to say that men today are taller than they were in 1960.
D. Reject the null hypothesis. At the 1% significance level there is sufficient evidence to say that men today are taller than they were in 1960.
Test statistic:
Option B is correct.
Fail to reject the null hypothesis, at the 5% significance level, there is not enough evidence to say that the new SAT prep program is better than the current SAT prep program.
Test statistic:
Option D is correct.
Reject the null hypothesis. At the 1% significance level there is sufficient evidence to say that men today are taller than they were in 1960. | {"url":"https://justaaa.com/statistics-and-probability/625834-a-high-school-principle-currently-encourages","timestamp":"2024-11-13T11:46:14Z","content_type":"text/html","content_length":"47216","record_id":"<urn:uuid:02b3f961-6d88-4b59-bf81-cd42f09d2d9b>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00590.warc.gz"} |
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great influence on the result of a milling process, is the power of a mill. Depending on the appliion, jars should be moved either slowly for ...
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WEBThe specifiions this above ( F100= 2000 mm and 472 KW power in Pinion for a word index of it is just the contractual data of ball mill supplier. At present the secondary cruscher is F100=12mm
and work index will be the order of 15 kWh/T according to this above data. Diameter ball mill= the length is m. thanks lot for your help
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WEBThe grinding process of the ball mill is an essential operation in metallurgical concentration plants. Generally, the model of the process is established as a multivariable system
characterized ...
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WEBFeb 15, 2001 · The present mathematical analysis of the milling dynamics aims at predicting the milling condition in terms of ωd and ωv, for the occurrence of the most effective impact between
the ball and vial wall to achieve MA. In the present analysis, the values of rd, rv and ball radius ( rb) are taken as 132, 35 and 5 mm, respectively (typical of a ...
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WEBJan 1, 2009 · 1.. IntroductionA Bond Ball Mill Work Index test is a standard test for determining the ball mill work index of a sample of ore. It was developed by Fred Bond in 1952 and
modified in 1961 (JKMRC CO., 2006).This index is widely used in the mineral industry for comparing the resistance of different materials to ball milling, for estimating .
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WEBDec 10, 2014 · 3D dynamic image analysis was used for calcite particles by ball, rod autogenous mills. •. More than 6400 particles/sample were measured with a 99% statistical confidence. •.
Autogenous mill had the highest aspect ratio the lowest circularity. •. Rod mill had the highest circularity the lowest aspect ratio. •.
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WEBSAG mill optimization to feed ball mill optimized P80. Most SAG mill are not optimized for the combined SAG Ball mill throughput such as: SAG Ball mill % ball content SAG % ore content SAG
grate size and end mill design including grate geometry, loion, shape, pan cavity, recycle % in pan cavity, et. al.
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WEBBall Mills. Ball mills have been the primary piece of machinery in traditional hard rock grinding circuits for 100+ years. They are proven workhorses, with discharge mesh sizes from ~40M to
<200M. Use of a ball mill is the best choice when long term, stationary milling is justified by an operation. Sold individually or as part of our turnkey ...
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WEBPosted January 14, 2016. So I have never had to do this before in my career as a machinist. I have a part that needs a radius milled into the back side of it that runs the entire length of the
part. The radius that needs to be machined is .475". I have a 1/2" Ball end mill in my machine, ready to start cutting, but I have no idea where to even ...
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WEBNov 30, 2015 · The document discusses the ball mill, which is a type of grinder used to grind materials into fine powder. It works on the principle of impact and attrition, where balls drop
from near the top of the shell as it rotates to grind materials placed inside. A ball mill consists of a hollow cylindrical shell that rotates about its axis, with balls ...
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WEBIn ceramics, ball mills are used to grind down materials into very fine particles. Materials such as clay and glaze components can be broken down in a ball mill by getting placed into rotating
or rolling jars with porcelain balls inside them. During milling, the porcelain balls pulverized the materials into an incredibly fine powder.
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WEBCeramic mill balls offered for sale as lonsdaleite diamonds at the Tucson Gem and Mineral Show (2024) (on the left, next to a box of geodes). The person who sent me the photos said, "All
dealers were insistent that it was Lonsdaleite diamond, but couldn't explain how they knew." One dealer "explained that they were very rare".
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WEBJun 26, 2017 · Ball Nose Milling Without a Tilt Angle. Ball nose end mills are ideal for machining 3dimensional contour shapes typically found in the mold and die industry, the manufacturing
of turbine blades, and fulfilling general part radius properly employ a ball nose end mill (with no tilt angle) and gain the optimal tool life and part .
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WEBApr 7, 2020 · Here we'll talk about the ball mill machine working animation. By watching this video, you will know clearly how does the ball mill work. In addition, the im...
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Procedural Content Generation Wiki
Midpoint Displacement Algorithm
The Midpoint Displacement Algorithm is a method of generating a height map by assigning a height value to each of the four corners of a rectangle, and then subdividing the rectangle and each
resulting child into four smaller rectangles, whose heights at the subdivided corners are the average (mean) value between the corners of the parent rectangle. Randomly varying the computed height at
each step results in the height map being defined by a plasma fractal.
The Midpoint Displacement Algorithm is in practise usually superseded by the Diamond-Square Algorithm.
Code Example
PCG Wiki References
External Links
Diamond-Square Algorithm - Wikipedia article on the Midpoint Displacement Algorithm.
page revision: 5, last edited: 05 Dec 2011 06:22 | {"url":"http://pcg.wikidot.com/pcg-algorithm:midpoint-displacement-algorithm","timestamp":"2024-11-14T20:37:05Z","content_type":"application/xhtml+xml","content_length":"29932","record_id":"<urn:uuid:d9659423-3024-47c4-938d-9d6db00e2c32>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00176.warc.gz"} |
Fully relativistic pseudopotentials
Fully relativistic pseudopotentials generated by the MBK (PRB 47, 6728 (1993)) scheme within LDA (CA19) and GGA (PBE19) which contain a partial core correction and fully relativistic effects
including spin-orbit coupling.
Pseudo-atomic orbitals
The number below the symbol means a cutoff radius (a.u.) of the confinement potential. These file includes fifteen radial parts for each angular momentum quantum number l (=0,1,2,3). The basis
functions were generated by variationally optimizing the corresponding primitive basis functions in the single atom and the FCC bulk. The input files used for the orbital optimization can be
found at Pd_opt.dat and Pdfcc_opt.dat. Since Pd_CA19.vps and Pd_PBE19.vps include the 4p, 4d, and 5s states (16 electrons) as the valence states, the minimal basis set is Pd*.*-s1p1d1. Our
recommendation for the choice of cutoff radius of basis functions is that Pd7.0.pao is enough for bulks, but Pd9.0.pao or Pd11.0.pao is preferable for molecular systems.
Benchmark calculations by the PBE19 pseudopotential with the various basis functions
(1) Calculation of the total energy as a function of lattice constant in the fcc structure, where the total energy is plotted relative to the minimum energy for each case. a[0] and B[0] are the
equilibrium lattice constant and bulk modulus obtained by fitting to the Murnaghan equation of state. The difference between Pd7.0-s2p2d2f1 and Pd7.0-s3p3d2f1 in the total energy at the minimum
point is 0.0067 eV/atom. An input file used for the OpenMX calculations can be found at Pdfcc-EvsV.dat. For comparison the result by the Wien2k code is also shown, where the calculation was
performed by default setting in the Ver. 9.1 of Wien2k except for the use of R[MT] x K[MAX] of 12.
(2) Calculations of the band dispersion in the fcc structure, where the non-spin polarized collinear calculation with the lattice constant of 3.890 Ang. was performed using Pd_PBE19.vps and
Pd7.0-s2p2d2f1, and the origin of the energy is taken to be the Fermi level. The input file used for the OpenMX calculations can be found at Pdfcc-Band.dat. For comparison the result by the
Wien2k code is also shown, where the calculation was performed by default setting in the Ver. 9.1 of Wien2k except for the use of R[MT] x K[MAX] of 12.
Supplementary information for the GGA (PBE19) pseudopotential | {"url":"https://www.openmx-square.org/vps_pao2019/Pd/index.html","timestamp":"2024-11-07T22:47:43Z","content_type":"text/html","content_length":"4814","record_id":"<urn:uuid:b99c421d-e5bb-490a-8f65-2178301f0641>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00218.warc.gz"} |
Curving Space and Time
Warping Slices of Reality
endless, and sublime,
The image of eternity—
the Throne of the Invisible...
From Lord Byron's Childe Harold's Pilgrimage
Canto IV, Stanza 183
For his first revolution, Einstein unified space and time, and showed that a given observer just took a slice out of this spacetime. Each of the slices was flat: a one-dimensional slice was just a
straight line; a two-dimensional slice was a flat sheet. In a similar way, the three-dimensional slices Einstein took were also flat, in a way we will explore below. For his second — and greater —
revolution, Einstein allowed the slices to be curved and warped. One-dimensional straight lines became one-dimensional curvy lines; two-dimensional flat sheets became two-dimensional curvy sheets;
three-dimensional flat spaces became three-dimensional curvy spaces. To understand this warping, it helps to think about warping in two-dimensional space.
We are all familiar with two-dimensional surfaces. Movie screens, pages in a book, and the surface of an apple are all examples. We also recognize a curved surface when we see it. Turn up half the
page in a book, and you will see that it is curved. Leave the page down, and you see a flat surface. In high-school geometry class, we typically explore two-dimensional space, which is almost always
assumed to be flat.
One familiar fact from this exploration is the formula for the circumference of a circle in terms of its radius. We remember the number π — pi, roughly 3.1416 — showing up in the formula
$$Circumference = 2 \times \pi \times Radius$$
To use this formula, we make a circle in a particular way. Choose any point, and tie one end of a length of rope to that point. Stretch the rope taut, and rotate around the center point, tracing out
the position of the free end of the rope. This will leave us with a circle. If we measure around the circle — its circumference — we should find that it is 2 × π times the length of our rope.
Now imagine that you try to use this formula in a surface that is not flat. Choose the center point of the circle to be the top of a hill and pretend that you are trapped on the hill; you and your
rope can't rise up off of it, or dig down inside of it. Draw the circle on the lower parts of the hill. If you now measure around the circumference of the circle you drew, you get a number that is
less than 2 × π times the length of the rope! If we tried to make a circle centered on a mountain pass (or in the middle of a horse's saddle), we would get a circumference that is greater than 2 × π
times the length of the rope. The classic formula doesn't apply in these curved space examples.
Now, to think about curvature in three dimensions, we recall another high-school formula that gives the surface area of a sphere in terms of its radius:
$$Area = 4 \times \pi \times Radius \times Radius = 4 \times \pi \times (Radius)^2.$$
To use this formula, we make a sphere just as we made a circle, but we move everywhere the rope will let us. The places where the free end of the rope reach define a sphere. If we want to cover this
sphere in fabric, for instance, we will need a piece 4 × π (about 12.566) times as big as a square with edges as long as our piece of rope, cut up into little patches, and sewn together just right.
That piece of fabric should perfectly cover the sphere — assuming we have flat three-dimensional space.
In the bizarre world of curved three-dimensional space, however, we might need much less or much more fabric than we would think, just using the formula above. Again, the standard formula we learned
in high-school (flat-space) geometry doesn't apply. It is not that this geometry is wrong, it's just that it only applies to flat space. We need a more sophisticated type of geometry to deal with
curved space. That sophisticated geometry was developed by mathematicians, and called differential geometry.
With the second revolution, Einstein let his slices be warped in the ways we've just seen. Since he had brought space and time together in spacetime, this meant that time could be warped, too.
Understanding warped space and time didn't come easy to Einstein, and he resisted as much as he could. Eventually, though, he came to realize that this warping was simply how nature worked, and the
differential geometry describing it was the only way to deal with a fundamental concept in his theory: the geodesic. | {"url":"https://www.black-holes.org/the-science-relativity/relativity/curving-space-and-time","timestamp":"2024-11-14T04:15:38Z","content_type":"text/html","content_length":"59224","record_id":"<urn:uuid:af51bfdc-7a36-4eaf-9240-ae1790652322>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00151.warc.gz"} |
Seminar Algebra and Geometry: Stéphane Lamy (Université de Toulouse)
03 Jul 2019
10:30 - 12:00
Spiegelgasse 5 - Seminarraum 05.002
Seminar Algebra and Geometry: Stéphane Lamy (Université de Toulouse)
The group of tame polynomial automorphisms of the n-dimensional complex affine space is the group generated by linear maps and polynomial transvections. I will describe an action of this group on a
metric space whose construction is inspired from the theory of affine Bruhat-Tits buildings. In dimension n = 3, we show that this space is simply connected with non-positive curvature. This allows
to get a description of the finite subgroups of the 3-dimensional tame group. (Joint work with P. Przytycki)
Export event as iCal | {"url":"https://dmi.unibas.ch/en/news-events/past-events/detail/seminar-algebra-and-geometry-stephane-lamy-universite-de-toulouse/","timestamp":"2024-11-12T15:59:43Z","content_type":"text/html","content_length":"21924","record_id":"<urn:uuid:5fca64b2-c789-4cc7-b9fd-0afc3ec7c315>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00368.warc.gz"} |
Implement Radix Sort Using Java - TechDecode Tutorials
Implement Radix Sort Using Java
If you’re in a computer science background, widening your technical horizons is the best thing you should do. However, your programming skills matter the most. And one must continuously keep
sharpening these skills, to become a better programmer in the future. Though there are multiple things to focus on, one specific area you must focus on is the world of data structures. Data
structures in general are particular approaches to solve problems so that computer resources get used minimum. In general, there are multiple data structures you can learn and implement as well.
However, we’ve covered some basic data structures before, so, now we’ll move towards some advanced data structures. Hence, today we’re going to learn how to Implement Radix Sort using Java.
What is Radix Sort?
• Unlike simpler data structures, Radix Sort uses a different sorting approach to arrange elements in sequential order.
• Here, elements (be it Integers or Strings) are sorted based on the position of digits i.e, from least significant to most significant.
• No of iterations in this sorting method may vary significantly. As iterations are directly proportional to the no of digits present in the largest element in a list/array.
Also Read: Implement Bubble Sort using Java
What’s The Approach?
• To make this program look cleaner and easier, we’ll divide it into three major sections.
• In the first part, we’ll find the largest number in an array using getMax().
• As for the second part, we’ll pass the exponential exp value to the counting sort function.
• At the final stage, we’ll apply counting sort to sort the array in ascending order w.r.t the exponential value being passed.
• In Counting Sort, we are going to sort array elements based on single digits only. Starting from one’s place and incrementing the exponential by 10 for each iteration. This condition is going to
be executed till m/exp > 0 remains true (m holds the value of the largest integer in an array).
Java Program To Implement Radix Sort
170, 45, 75, 90, 802, 24, 2, 66
// Radix sort Java implementation
import java.io.*;
import java.util.*;
class TechDecodeTuorials{
// A utility function to get maximum value in arr[]
static int getMax(int arr[], int n)
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return mx;
// A function to do counting sort of arr[] according to
// the digit represented by exp.
static void countSort(int arr[], int n, int exp)
int output[] = new int[n]; // output array
int i;
int count[] = new int[10];
Arrays.fill(count, 0);
// Store count of occurrences in count[]
for (i = 0; i < n; i++)
count[(arr[i] / exp) % 10]++;
// Change count[i] so that count[i] now contains
// actual position of this digit in output[]
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
// Build the output array
for (i = n - 1; i >= 0; i--) {
output[count[(arr[i] / exp) % 10] - 1] = arr[i];
count[(arr[i] / exp) % 10]--;
// Copy the output array to arr[], so that arr[] now
// contains sorted numbers according to current digit
for (i = 0; i < n; i++)
arr[i] = output[i];
// The main function to that sorts arr[] of size n using
// Radix Sort
static void radixsort(int arr[], int n)
// Find the maximum number to know number of digits
int m = getMax(arr, n);
// Do counting sort for every digit.exp is passed.
// exp is 10^i where i is current digit number
for (int exp = 1; m / exp > 0; exp *= 10)
countSort(arr, n, exp);
// A utility function to print an array
static void print(int arr[], int n)
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
/*Driver Code*/
public static void main(String[] args)
int arr[] = { 170, 45, 75, 90, 802, 24, 2, 66 };
int n = arr.length;
// Function Call
radixsort(arr, n);
print(arr, n); | {"url":"https://techdecodetutorials.com/implement-radix-sort-using-java/","timestamp":"2024-11-06T06:19:32Z","content_type":"text/html","content_length":"128647","record_id":"<urn:uuid:5178cab4-6363-4bc7-96a0-7b281a4704d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00671.warc.gz"} |
am.h File Reference
Class AM: a multilevel finite element algebra object.
struct sAM
Class AM: Definition. More...
typedef struct sAM AM
Declaraction of the AM class as the AM structure.
AM * AM_ctor (Vmem *vmem, Aprx *taprx)
The AM constructor.
void AM_dtor (AM **thee)
The AM destructor.
void AM_create (AM *thee)
Create the following internal Alg structures:
A ==> The tangent matrix (linearization operator)
M ==> The mass matrix
W[] ==> The node vectors.
void AM_destroy (AM *thee)
Destroy the following internal Alg structures:
A ==> The tangent matrix (linearization operator)
M ==> The mass matrix
W[] ==> The node vectors.
int AM_markRefine (AM *thee, int key, int color, int bkey, double elevel)
Mark a given mesh for refinement.
int AM_refine (AM *thee, int rkey, int bkey, int pkey)
Refine the mesh.
int AM_unRefine (AM *thee, int rkey, int pkey)
Un-refine the mesh.
int AM_deform (AM *thee)
Deform the mesh.
int AM_read (AM *thee, int key, Vio *sock)
Read in the user-specified initial mesh given in the "MCSF" or "MCEF" format, and transform into our internal datastructures.
Do a little more than a "Aprx_read", in that we also initialize the extrinsic and intrinsic spatial dimensions corresponding to the input mesh, and we also then build the reference
double AM_assem (AM *thee, int evalKey, int energyKey, int residKey, int tangKey, int massKey, int bumpKey, Bvec *u, Bvec *ud, Bvec *f, int ip[], double rp[])
Assemble the linearized problem at a given level.
double AM_evalJ (AM *thee)
Assemble the energy functional at the current solution.
void AM_evalFunc (AM *thee, int number, int block, int numPts, double *pts, double *vals, int *marks)
Evaluate a finite element function.
void AM_bndIntegral (AM *thee)
Perform a boundary integral.
double AM_evalError (AM *thee, int pcolor, int key)
Evaluate error in the current solution.
void AM_applyDiriZero (AM *thee, Bvec *which)
Apply zero dirichlet condition at a given level.
void AM_iniGuess (AM *thee, Bvec *which)
Setup an initial guess at a given level.
int AM_part (AM *thee, int pkey, int pwht, int ppow)
Partition the mesh using the matching Alg level.
int AM_partSet (AM *thee, int pcolor)
Set the partition color.
int AM_partSmooth (AM *thee)
Do a partition smoothing.
void AM_printJ (AM *thee)
Print the energy.
void AM_printA (AM *thee)
Print the system matrix.
void AM_printAnoD (AM *thee)
Print the system matrix with Dirichlet rows/cols zeroed.
void AM_printAsp (AM *thee, char *fname)
Print the system matrix in MATLAB sparse format.
void AM_printAspNoD (AM *thee, char *fname)
Print the system matrix in MATLAB sparse format with Dirichlet rows/cols zeroed.
void AM_printP (AM *thee)
Print the prolongation matrix.
void AM_printPsp (AM *thee, char *fname)
Print the prolongation matrix in MATLAB sparse format.
void AM_printV (AM *thee, int num)
Print a specified vector.
void AM_printVsp (AM *thee, int num, char *fname)
Print a vector in MATLAB sparse format.
void AM_writeGEOM (AM *thee, Vio *sock, int defKey, int colKey, int chartKey, double gluVal, int fkey, int number, char *format)
Write out a mesh in some format.
void AM_writeSOL (AM *thee, Vio *sock, int number, char *format)
Write out a solution in some format.
void AM_memChk (AM *thee)
Print the exact current malloc usage.
void AM_lSolve (AM *thee, int prob, int meth, int itmax, double etol, int prec, int gues, int pjac)
Linear solver.
void AM_hlSolve (AM *thee, int prob, int meth, int itmax, double etol, int prec, int gues, int pjac)
Hierarchical linear solver.
void AM_nSolve (AM *thee, int meth, int itmax, double etol, int lmeth, int litmax, double letol, int lprec, int gues, int pjac)
void AM_newton (AM *thee, int itmax, double etol, int lmeth, int litmax, double letol, int lprec, int pjac, double loadParm)
Damped-Inexact Newton iteration.
VPUBLIC void AM_homotopy (AM *thee, int itmax, double etol, int lmeth, int litmax, double letol, int lprec, int pjac)
Homotopy and the related "incremental loading" iterations.
Class AM: a multilevel finite element algebra object. | {"url":"http://fetk.org/codes/mc/api/html/am_8h.html","timestamp":"2024-11-09T07:10:49Z","content_type":"text/html","content_length":"25141","record_id":"<urn:uuid:b622b6c1-a058-4bd8-8568-e64036efb711>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00111.warc.gz"} |
Data Analysis: Normal Distribution - Try Machine Learning
Data Analysis: Normal Distribution
Data analysis is an essential aspect of understanding and extracting insights from data. One important concept in data analysis is the normal distribution, also known as the Gaussian distribution or
the bell curve. Understanding the characteristics and applications of the normal distribution can greatly enhance your ability to make informed decisions based on data.
Key Takeaways
• The normal distribution is a mathematical model that describes a symmetric and bell-shaped probability distribution.
• In a normal distribution, the mean, median, and mode are all equal and located at the center of the distribution.
• The standard deviation in a normal distribution indicates the spread or variability of the data.
• Many real-world phenomena follow a normal distribution, making it a valuable tool in various fields, including finance, economics, and psychology.
The normal distribution is characterized by its symmetric shape, where the mean, median, and mode are all located at the center of the distribution. It is often represented by the equation:
P(x) = (1 / (σ√(2π))) * e^(-(x-μ)² / 2σ²)
where P(x) represents the probability density function, σ is the standard deviation, and μ is the mean of the distribution. This formula demonstrates how the probabilities of different values are
distributed symmetrically around the mean.
For example, if we consider the heights of adult males in a population, these heights are likely to follow a normal distribution, with the majority of individuals clustered around the mean height.
Applications of the Normal Distribution
The normal distribution has wide-ranging applications in various fields:
• In finance, stock returns often follow a normal distribution, enabling investors to analyze risk and make informed investment decisions.
• In quality control, the normal distribution is used to model process variations and determine acceptable product specifications.
• In psychological testing, scores are often assumed to follow a normal distribution, aiding in the interpretation of test results.
Using the normal distribution, financial analysts can determine the likelihood of a stock price falling within a specific range, helping investors assess their potential gains or losses.
Value Frequency
Gender Height (cm)
Male 180
Female 165
Male 175
Score Percentage (%)
80 25%
90 35%
100 40%
In conclusion, understanding the normal distribution is essential for data analysts in various fields. The bell-shaped curve, characterized by the mean and standard deviation, allows us to make
informed decisions and predictions based on data. By recognizing the applications of the normal distribution, we can better understand and analyze real-world phenomena.
Common Misconceptions
1. Normal Distribution Always Applies to All Data
One common misconception about data analysis is that all data follows a normal distribution. In reality, normal distribution is just one of many possible patterns that data can exhibit. It is a
mathematical concept describing a symmetrical bell-shaped curve, but in practicality, data can deviate from this idealized pattern in various ways.
• Data can be skewed to the left or right, resulting in a non-normal distribution.
• Data may exhibit multiple peaks, indicating a bimodal or multimodal distribution.
• Some datasets may not show any clear pattern and instead have a random distribution.
2. Outliers Should Always Be Removed
Another misconception is that outliers should always be removed from a dataset before conducting data analysis. While outliers can sometimes be problematic and affect statistical measures like the
mean, removing them without careful consideration can lead to biased or inaccurate results.
• Outliers can provide valuable information and insights into unexpected or extreme scenarios.
• Before deciding to remove an outlier, it is important to investigate the cause and consider whether it was a genuine data point or an error.
• Outliers should only be removed if there is a valid reason to do so, and the impact on the analysis should be carefully assessed.
3. Normal Distribution Means Perfectly Balanced Data
It is often mistakenly believed that a normal distribution implies perfectly balanced data. However, a normal distribution describes the shape of the data rather than its actual values or the
proportion of observations in various categories.
• A normal distribution can occur even with imbalanced data if the underlying pattern follows the bell-shaped curve.
• Data can be perfectly balanced without following a normal distribution, such as in the case of uniform data.
• The balance or imbalance of data is determined by other factors such as the presence of missing values or the ratio of observations in different categories.
4. Data Must Be Normally Distributed for Statistical Tests
There is a misconception that data must be normally distributed in order to perform statistical tests. While some parametric tests assume normality, there are also non-parametric tests that do not
have this requirement.
• Non-parametric tests are designed to be more robust against departures from normality.
• Non-parametric tests do not make assumptions about the underlying distribution and are often used when working with skewed or non-normal data.
• Parametric tests can still be used with non-normal data if the sample size is large enough, thanks to the Central Limit Theorem.
5. Normality Can Be Determined from a Small Sample
Finally, it is a misconception that one can determine the normality of a dataset based on a small sample. Assessing whether data follows a normal distribution requires examining the shape of the
entire dataset or a sufficiently large sample.
• If a small sample is used, it may not accurately represent the overall distribution, leading to erroneous assumptions.
• Histograms, Q-Q plots, and statistical tests such as the Anderson-Darling test can be used to assess normality.
• A larger sample size provides more reliable information about the true distribution of the data.
Normal distribution, also known as the Gaussian distribution, is a key concept in data analysis. It is characterized by a symmetric bell-shaped curve and is often used to model a variety of natural
phenomena. In this article, we explore various aspects of normal distribution and its applications. We present ten tables below, each providing insights into different elements of this topic.
1. Distribution of IQ Scores
The table below showcases the distribution of IQ scores in a random sample of 500 individuals from a population. It provides a representation of the frequency of scores falling within specific
ranges, highlighting the central tendency of intelligence in the population.
IQ Range Number of Individuals
70 – 79 10
80 – 89 40
90 – 99 120
100 – 109 200
110 – 119 100
120 – 129 20
130 – 139 5
140+ 5
2. Heights of Adult Males
This table represents the heights (in inches) of adult males from a population. The data follows a normal distribution, emphasizing the clustering around the mean height.
Height (in inches) Frequency
60 – 64 8
65 – 69 40
70 – 74 125
75 – 79 200
80 – 84 130
85 – 89 50
90 – 94 5
95+ 2
3. Annual Rainfall Distribution
This table presents the distribution of annual rainfall (in millimeters) in various cities. The normal distribution pattern demonstrates the typical amount of rainfall experienced in different
Rainfall (mm) Frequency
0 – 100 5
101 – 200 20
201 – 300 80
301 – 400 150
401 – 500 120
501 – 600 50
601 – 700 10
701+ 3
4. Exam Scores Distribution
The following table illustrates the distribution of exam scores obtained by a class of 150 students. The scores conform to a normal distribution with a peak around the mean score.
Score Range Number of Students
50 – 59 5
60 – 69 30
70 – 79 60
80 – 89 80
90 – 99 60
100 – 109 10
110 – 119 4
120+ 1
5. Reaction Time Distribution
This table represents the distribution of reaction times (in milliseconds) among participants in a study. The normal distribution pattern demonstrates the typical response time for different
Reaction Time (ms) Frequency
100 – 150 10
151 – 200 40
201 – 250 80
251 – 300 130
301 – 350 140
351 – 400 75
401 – 450 20
451+ 5
6. Housing Prices
This table displays the distribution of housing prices (in thousands of dollars) in a region. The normal curve suggests that a majority of houses fall within a particular price range.
Price Range (in thousands of dollars) Number of Houses
100 – 200 50
201 – 300 100
301 – 400 150
401 – 500 200
501 – 600 100
601 – 700 50
701 – 800 10
801+ 5
7. Exam Grades Distribution
The table below exhibits the distribution of grades obtained in a challenging exam. The data presents a normal distribution, showcasing the spread of students’ performance.
Grade Number of Students
A 10
B 25
C 60
D 90
E 70
F 5
8. Monthly Temperatures
The following table presents the distribution of average monthly temperatures (in degrees Celsius) in a particular city. The data adheres to a normal distribution pattern, representing the seasonal
variations in temperature.
Temperature (°C) Frequency
-10 – 0 10
0 – 10 20
10 – 20 60
20 – 30 120
30 – 40 60
40 – 50 10
50 – 60 2
9. Annual Income Distribution
This table visualizes the distribution of annual incomes (in thousands of dollars) among a sample population. The normal distribution captures the frequency of incomes falling within different
Income (in thousands of dollars) Number of Individuals
10 – 20 30
21 – 30 120
31 – 40 200
41 – 50 180
51 – 60 70
61 – 70 20
71 – 80 5
81+ 2
10. Time Spent on Social Media
The final table showcases the distribution of daily time spent on social media platforms (in minutes) among different age groups. The normal distribution provides an overview of the average time
spent by each age category.
Age Group Time Spent (in minutes)
13 – 17 120
18 – 25 180
26 – 35 200
36 – 45 150
46 – 55 100
56 – 65 50
66+ 15
In this article, we delved into the concept and applications of normal distribution in data analysis. The tables provided valuable insights into various elements modeled by this widely-used
statistical distribution. By understanding and utilizing normal distribution, we can gain a deeper understanding of real-world phenomena and make informed decisions based on reliable data.
Data Analysis: Normal Distribution – Frequently Asked Questions
What is a normal distribution?
A normal distribution is a probability distribution that is symmetric around the mean, representing a set of values that tend to cluster around the mean with decreasing frequency as they deviate
further from it. It is also known as a Gaussian distribution or bell curve.
What are the characteristics of a normal distribution?
A normal distribution has the following characteristics:
• It is symmetric, with the mean, median, and mode all located at the center of the distribution.
• About 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.
• It is bell-shaped, with the tails gradually decreasing on either side of the mean.
How is a normal distribution calculated?
A normal distribution can be calculated by specifying the mean and standard deviation of the data set. The formula for calculating the probability density function (PDF) of a normal distribution is
given by:
f(x) = (1 / (σ√2π)) * e^(-((x-μ)^2 / (2σ^2)))
Where μ is the mean and σ is the standard deviation.
What are some real-life examples of a normal distribution?
Some real-life examples of a normal distribution include:
• Height and weight of individuals in a population.
• Test scores of a large group of students.
• The amount of money people spend on groceries.
Why is the normal distribution important in data analysis?
The normal distribution is important in data analysis because many statistical techniques and models assume that the data follows a normal distribution. It allows for easier interpretation and
analysis of data, as well as making predictions and inferences based on the properties of the distribution.
How is the normal distribution related to statistical significance?
The normal distribution is often used in hypothesis testing and calculating statistical significance. By assuming that the data follows a normal distribution, various statistical tests can be applied
to determine if the observed results are statistically significant or occurred by chance.
Can data that is not normally distributed still be analyzed using statistical methods?
Yes, data that is not normally distributed can still be analyzed using statistical methods. However, in such cases, alternative techniques may need to be used, such as non-parametric tests or
transforming the data to achieve normality. It is important to assess the distribution of the data before applying statistical methods.
What is the central limit theorem and its relation to the normal distribution?
The central limit theorem states that the sum or average of a large number of independent and identically distributed random variables will tend towards a normal distribution, regardless of the shape
of the original distribution. This theorem is often used to justify the assumption of normality in statistical analysis.
How can I check if my data follows a normal distribution?
There are several methods to check if your data follows a normal distribution, including:
• Graphical methods like histograms, boxplots, and QQ-plots.
• Statistical tests like the Shapiro-Wilk test or Kolmogorov-Smirnov test for normality.
• Using software or programming libraries that provide functions for normality tests.
Are there any assumptions associated with the normal distribution?
Yes, there are some assumptions associated with the normal distribution, including:
• The data should be independent and identically distributed.
• The data should be continuous.
• The data should follow a symmetric bell-shaped distribution.
• No outliers or extreme values should be present in the data. | {"url":"https://trymachinelearning.com/data-analysis-normal-distribution/","timestamp":"2024-11-10T03:29:47Z","content_type":"text/html","content_length":"71378","record_id":"<urn:uuid:4412ca77-fe01-49bd-bf67-fc6ced7eb701>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00653.warc.gz"} |
44 research outputs found
We study a family of equivalence relations on $S_n$, the group of permutations on $n$ letters, created in a manner similar to that of the Knuth relation and the forgotten relation. For our purposes,
two permutations are in the same equivalence class if one can be reached from the other through a series of pattern-replacements using patterns whose order permutations are in the same part of a
predetermined partition of $S_c$. When the partition is of $S_3$ and has one nontrivial part and that part is of size greater than two, we provide formulas for the number of classes created in each
previously unsolved case. When the partition is of $S_3$ and has two nontrivial parts, each of size two (as do the Knuth and forgotten relations), we enumerate the classes for $13$ of the $14$
unresolved cases. In two of these cases, enumerations arise which are the same as those yielded by the Knuth and forgotten relations. The reasons for this phenomenon are still largely a mystery
Dynamic time warping distance (DTW) is a widely used distance measure between time series. The best known algorithms for computing DTW run in near quadratic time, and conditional lower bounds
prohibit the existence of significantly faster algorithms. The lower bounds do not prevent a faster algorithm for the special case in which the DTW is small, however. For an arbitrary metric space $\
Sigma$ with distances normalized so that the smallest non-zero distance is one, we present an algorithm which computes $\operatorname{dtw}(x, y)$ for two strings $x$ and $y$ over $\Sigma$ in time $O
(n \cdot \operatorname{dtw}(x, y))$. We also present an approximation algorithm which computes $\operatorname{dtw}(x, y)$ within a factor of $O(n^\epsilon)$ in time $\tilde{O}(n^{2 - \epsilon})$ for
$0 < \epsilon < 1$. The algorithm allows for the strings $x$ and $y$ to be taken over an arbitrary well-separated tree metric with logarithmic depth and at most exponential aspect ratio. Extending
our techniques further, we also obtain the first approximation algorithm for edit distance to work with characters taken from an arbitrary metric space, providing an $n^\epsilon$-approximation in
time $\tilde{O}(n^{2 - \epsilon})$, with high probability. Additionally, we present a simple reduction from computing edit distance to computing DTW. Applying our reduction to a conditional lower
bound of Bringmann and K\"unnemann pertaining to edit distance over $\{0, 1\}$, we obtain a conditional lower bound for computing DTW over a three letter alphabet (with distances of zero and one).
This improves on a previous result of Abboud, Backurs, and Williams. With a similar approach, we prove a reduction from computing edit distance to computing longest LCS length. This means that one
can recover conditional lower bounds for LCS directly from those for edit distance, which was not previously thought to be the case
This paper considers the basic question of how strong of a probabilistic guarantee can a hash table, storing $n$ $(1 + \Theta(1)) \log n$-bit key/value pairs, offer? Past work on this question has
been bottlenecked by limitations of the known families of hash functions: The only hash tables to achieve failure probabilities less than 1 / 2^{\polylog n} require access to fully-random hash
functions -- if the same hash tables are implemented using the known explicit families of hash functions, their failure probabilities become 1 / \poly(n). To get around these obstacles, we show how
to construct a randomized data structure that has the same guarantees as a hash table, but that \emph{avoids the direct use of hash functions}. Building on this, we are able to construct a hash table
using $O(n)$ random bits that achieves failure probability $1 / n^{n^{1 - \epsilon}}$ for an arbitrary positive constant $\epsilon$. In fact, we show that this guarantee can even be achieved by a \
emph{succinct dictionary}, that is, by a dictionary that uses space within a $1 + o(1)$ factor of the information-theoretic optimum. Finally we also construct a succinct hash table whose
probabilistic guarantees fall on a different extreme, offering a failure probability of 1 / \poly(n) while using only $\tilde{O}(\log n)$ random bits. This latter result matches (up to low-order
terms) a guarantee previously achieved by Dietzfelbinger et al., but with increased space efficiency and with several surprising technical components | {"url":"https://core.ac.uk/search/?q=author%3A(Kuszmaul%2C%20William)","timestamp":"2024-11-11T08:35:38Z","content_type":"text/html","content_length":"106285","record_id":"<urn:uuid:b999aa3e-501c-4c26-ab37-01319785a932>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00823.warc.gz"} |
Mathematics Trivia
59 interesting facts about Mathematics that will surprise, entertain and educate you.
Four is the only number that has the same number of letters as its meaning
A number is divisible by 9 if the sum of the digits is divisible by 9.
The number zero does not have its own Roman numeral
Think to yourself 'How I wish I could calculate pi' and then count the letters in each of the words of that sentence. You now have a way of remembering the first seven digits of pi: 3.141592.
There are lots of pi facts on the Pi Day page.
The first number to contain the letter a is 'one thousand'. See A NUMBER for the discussion behind this fact.
The only number to have its letters in alphabetical order is forty. See Alphanumbetical and try to think of other alphanumbetical mathematical words!
The numbers on opposite sides of a dice add up to seven.
There are more than 64 squares on a chess board. If you count the squares made up of multiple squares there are 204 altogether. There is one 8x8 square, four 7x7 squares, nine 6x6 squares, 16 5x5
squares, 25 4x4 squares, 36 3x3 squares, 49 2x2 squares and 64 1x1 squares.
A prime number has exactly two factors. Two is the only even prime number and it is also the only prime number not to contain the letter 'e'.
It should take no more than 20 moves to solve a Rubiks cube no matter which of the 43 quintillion possible starting positions you begin with.
A googol is one followed by one hundred zeros. This can be written as:
Not many people appreciate (or understand) the mind-blowing fact that:
e^iπ = -1
Triangles, squares and hexagons are the only regular polygons that tessellate.
Check for yourself at the Tessellations page.
The equals sign was invented in 1557 by Welsh mathematician Robert Recorde. The word 'equal' is from the Latin word aequalis as meaning uniform, identical, or equal.
111111111 x 111111111 = 12345678987654321.
These two fractions add up to one and between them they contain all of the digits from nought to nine. It is the only way that this can be done.
[Update: 12th Aug 2019, received a message on Twitter from @ignormatyk saying that this was not the only way and there are others for example: 45/90+138/276 and 38/76+145/290]
Eight comes first if all the numbers are arranged alphabetically. What number would come last?
More numbers begin with the digit one than any other digit. This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population
numbers, death rates, lengths of rivers and mathematical constants. This fact was famously attributed to physicist Frank Benford, who stated it in 1938, although it had been previously stated by
Simon Newcomb in 1881.
Alice in Wonderland learns that in a class of 23 pupils the probability that two have the same birthday is more than a half. Alice is a fictional character created by author and mathemetician Lewis
Carroll (1832-1898).
42 is the answer to the 'Ultimate Question of Life, the Universe, and Everything' according to 'The Hitchhiker's Guide to the Galaxy' created by Douglas Adams.
18 is the only number that is twice the sum of its digits.
Most clocks which have Roman numerals on their face use IIII for four instead of the more familiar IV.
There are currently over 7 billion people in the world and the number is growing very quickly. To see just how quickly have a look at our population counter.
Any number to the power zero is 1, and zero to any power is 0. The only unanswered question here is what is zero to the power zero?
The Fibonacci sequence is a sequence of numbers where each term is the sum of the previous two terms. The first two terms are both one. The ratio of the n^th term to the next term gets closer to the
golden ratio as n increases.
Air France, Iberia, Ryanair, AirTran, Continental Airlines, and Lufthansa don’t have a row 13 on their airlines because they are aware that many of their passengers consider 13 to be an unlucky
number. Many office blocks, office buildings and hotels do not have a 13th floor for the same reason.
The following three consecutive numbers are the lowest that are divisible by cubes other than 1:
1375; 1376; 1377
(divisible by the cubes of 5, 2 and 3 respectively).
The digits of the number are the same as the digits of the power of ten in these cases:
1.3712885742 = 10^0.13712885742
237.5812087593 = 10^2.375812087593
3550.2601815865 = 10^3.5502601815865
46692.4683 = 10^ 4.669246833
The polar diameter of the Earth is quite close to (within 0.1%) half a billion inches.
The first time a digit repeats six times in succession in pi is at the 762nd position where you can find six nines in a row. This is known as the Feynman Point.
The term googol (a 1 followed by 100 zeroes) was first used by a 9- year old boy,Milton Sirotta, in 1938.
Trivia submitted by Paul Christian Sarmiento, Philippines
As weird as it may seem at first glance, f(x) = -1/(x+1) and g(x) = x/(x+1) have the same derivative.
Trivia submitted by Shiraz Dagia, Malawi
A number if multiply by 11 is just you bring down the last digits add the digit to the number on it's left and bring down the first number.
Bring down 5
Then, 5+1=6
Last bring down the first digit which is 1
Trivia submitted by Jericho Fernandez, Laguna, Philippines
There is a number smaller than 'a thousand' or 'a hundred and one' with the letter a - it is any number like 0.001 or 'one thousandth' which is smaller than 101 or 1000. These decimals can get
infinitely smaller so the proper term for the trivia should be that the smallest whole number not just the smallest number.
Trivia submitted by QA, Makaben
You can remember the value of Pi (3.1415926) by counting each word's letters in
'May I have a large container of coffee?'.
Trivia submitted by Ms. Prescott,
Trivia improves critical thinking.
Trivia submitted by Wisani, Gauteng, South Africa
A googolplex is a googol^a googol, or (10^100) to the power(10^100).
Trivia submitted by Anonymous, Planet Earth
40 when written in words "forty" is the only number with letters in alphabetical order, while one is the only one with letters in reverse order.
Trivia submitted by Dustin Joseph C. Manalo, Bulacan,Philippines
The number 0 is originally called cipher.
Trivia submitted by Paras, Philipines
The billionth digit of Pi (3.1415 ...) is 9.
Trivia submitted by Joy, Manila, Philippines
What is the correct mathematical name of the division bar in a fraction?
Trivia submitted by Lovely Tinam-isan, Muntinlupa, Philippines
The term "jiffy" is an actual unit of time which is the 1/100th of a second.
Trivia submitted by Josh Anilov C. Funelas, Manila, Philippines
The value of zero was first used by the ancient Indian mathematician Aryabhata.
Trivia submitted by Wency Orbina, Philippines
2520 is the smallest number that is divisible by 1,2,3,4,5,6,7,8,9 and 10.
Trivia submitted by MathWizard, Phillipines
Moving each letter of the word 'yes' 16 places further up the alphabet produces the word 'oui', the French for 'yes'.
Trivia submitted by Greg Ross, Futility Closet
Forty-two percent of Slovenian two-year-olds know the number two, while only four percent of English two-year-olds do.
Trivia submitted by Francie Diep, Popular Science
The word 'twelve' is worth 12 points in Scrabble. .
Trivia submitted by Greg Ross, Futility Closet
The words 'ace, two, three, four, five, six, seven, eight, nine, ten, jack, queen, king' contain 52 letters. There are 52 cards in a pack (excluding jokers).
Trivia submitted by Greg Ross, Futility Closet
If you square11111111 the answer would be 123456787654321. (count the number of 1s and that's the middle number).
Trivia submitted by Ranz Louie Ricasa, Philippines
The polygon with 1000000 sides is called megagon.
Trivia submitted by Ranz Louie Ricasa, Philippines
Trivia submitted by Ranz Louie Ricasa, Philippines
If you find the difference between the number of edges and the number of faces of the tetrahedron, cube, octahedron and other solid shapes, the results will always be 2.
Trivia submitted by Ranz Louie Ricasa, Olongapo City,Philippines
The second hand on a clock is actually the minute hand.
Trivia submitted by Will, Northampton, England
If you write out pi to two decimal places, backwards it spells “pie”.
Found from buzzfeed.
Trivia submitted by Me, England
The Reuleaux Triangle is a shape of constant width, the simplest and best known such curve other than a circle.
Trivia submitted by Manuel Henryk Fabunan, Philippines
The mathematical name for # (number sign) is octothorpe.
Trivia submitted by Kz Fernandez, Philippines
Can you find two numbers without a final 0 that have a product of 10, 100, 1000, 10000 etc. Here is a method:
Let's start with 5*2=10
and so on...
Trivia submitted by Ranz Louie Ricasa, Olongapo City,Philippines
Did you know there are five hundred and twenty-five thousand, six hundred minutes in a year? This special number is the main 'hook' of the song Seasons of Love written for the musical Rent.
In the year 1514 the German artist Albrecht Dürer created an engraving called Melencolia with a magic square in the background. The image below shows an enlargement of the magic square. The date
appears in the bottom row of the magic square.
Using consecutive whole numbers and counting rotations and reflections of a given square as being the same there are precisely:
1 magic square of size 3 × 3
880 magic squares of size 4 × 4
275,305,224 magic squares of size 5 × 5.
For the 6×6 case, there are estimated to be approximately 1.77 × 10^19 squares.
This trivia is from the excellent book by Professor Ian Stewart called Cabinet Of Mathematical Curiosities.
Zero is the number with the most names or synonyms. It is also known as nought, naught, ow, nil, zilch, zip, diddly-squat, love and scratch.
Did you know that if you add 429 and 138 the answer is 567? The calculation contains the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9.
[Transum: There are another 335 ways to construct a similar calculation. Have a go at finding them using the Nine Digit Sum drag-and-drop activity.]
Trivia submitted by April Jean Elumba, University Of Southern Mindanao
J and K are the only letters not in any of the numbers when written as words.
Trivia submitted by Theo,
Using only addition, you can add 8's to get the number 1,000 by:
Trivia submitted by Nightshade, Iligan City
The word Trivia comes from the Latin meaning three ways (tri is the prefix for three). At a three way junction there would be a signpost giving information about each direction. This information
could be called trivia!
There are exactly 8! (eight factorial) minutes in four weeks.
This is calculated as follows: 4x7x24x60
= 8x7x6x5x4x3x2x1
= 8!
All ten digit pandigital numbers are divisible by 3 (a pandigital number contains all the digits 0 to 9).
[You can find out more about tests for divisibility here.]
King's Cross station in London has a platform zero! It is the longest platform at the station.
The number 2 is the only prime number that doesn't have the letter e in its name.
Trivia submitted by Origaminess, Naperville
Did you know that rather than (in the UK) having 1p, 2p, 5p etc. coins it would be mathematically more efficient to have 1p, 3p, 11p and 37p coins?
Trivia submitted by No Such Thing As A Fish, Podcast
There are 385072 ways of arranging the numbers 1 - 18 in a circle so that the sum of each pair of adjacent numbers is prime.
Try to find just one of them on the Prime Pairs Game page.
153, 370, 371 and 407 are the only three-digit numbers equal to the sum of the cubes of their digits.
Any four digit number typed by using four calculator keys at the corners of a rectangle is a multiple of eleven.
There is more about this fact on the Key Eleven page.
Here's a useful counterintuitive fact: one 18 inch pizza has more 'pizza' than two 12 inch pizzas.
Area of 18" pizza is π × 9^2 = 254 square inches.
Area of two 12" pizzas is 2π × 6^2 = 226 square inches.
Trivia submitted by Fermat's Library via Twitter,
The volume of a deep-pan pizza with radius Z and depth A is
Pi × Z × Z × A
Older people were born longer ago.
142857 is a circular number
142857 × 2 = 285714
142857 × 3 = 428571
142857 × 4 = 571428
142857 × 5 = 714285.
Trivia submitted by Philip Robinson, New Zealand
The Babylonians were using Pythagoras' Theorem over 1,000 years before Pythagoras was born.
Trivia submitted by QI,
Octothorpe is the another term for the hashtag sign or the no. sign (#).
Trivia submitted by Andrew Stevenizer, Philippines
29 is the number of short straight lines needed to make the number 29 when it is written as words:
Here is the formula for solving the equation \(ax^2 + bx + c = 0\).
$$ x = \frac{ - b \pm \sqrt {b^2 - 4ac} }{2a} $$
Did you know that there is another formula for finding the roots of quadratic equations? It is called the 'citardauq' (the word quadratic backwards) formula and you can read more about it here but
you will never need it for school Maths.
206 is the smallest number that when written in words contains all five vowels exactly once:
Did you know that the area of the regular pentagon on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the regular pentagons on the other two sides?
Can you prove that it is true?
What about other shapes?
A seventh is eleven sevenths of an eleventh.
Trivia submitted by Olberjb, Reddit
You’ve met the first six primes:
They form a nice calculation:
23 × 57 = 1311
Trivia submitted by Chris Smith @aap03102 Twitter,
12 + 3 - 4 + 5 + 67 + 8 + 9 equals 100.
Trivia submitted by Jimmy Horton, England
The division slash (/) is called the virgule
For example 30/5 = 6.
Trivia submitted by Jimmy Horton, England
All odd numbers have a letter e in them.
Maths teachers are very good correcting their pupils when they use the letter O name when they really mean nought or zero. It is important to distinguish between letters (as used in algebra) and
numbers. There are however exceptions to this rule that have come about by common usage:
1. James Bond's number is double 'O' seven
2. The first Scout camp took place in 1907 (nineteen 'O' seven)
3. My telephone number is 'O' two four nine seven one one one
4. A famous TV series was called Hawaii 5 'O'
5. It's my father's birthday. He's celebrating his big five 'O'.
contains all of the digits in alphabetical order.
The division symbol is called obelus.
Trivia submitted by Genius Of Sampaguita 2019-2020, TNHS - Main
TWELVE PLUS ONE = ELEVEN PLUS TWO
The left side of this equation is an anagram of the right side!
142,857 × 1 = 142,857
142,857 × 2 = 285,714
142,857 × 3 = 428,571
142,857 × 4 = 571,428
142,857 × 5 = 714,285
142,857 × 6 = 857,142
But how far will this strange sequence of calculations continue?
Here are the only temperatures that are prime integers in both Celsius and Fahrenheit:
-5^oC is equal to 23^oF
5^oC is equal to 41^oF.
Trivia submitted by Fermat's Library On Twitter,
FORTY FIVE is an anagram of OVER FIFTY!
Any six digit number where the first three digits are repeated as the second three digits is always divisible by 7, 11 and 13
eg 136136.
Trivia submitted by Kris Tobin, NSW
41 is prime
41+2 is prime
41+2+4 is prime
41+2+4+6 is prime
41+2+4+6+8 is prime
41+2+4+6+8+10 is prime
41+2+4+6+8+10+12 is prime
41+2+4+6+8+10+12+14 is prime
41+2+4+6+8+10+12+14+16 is prime
41+2+4+6+8+10+12+14+16+18 is prime
41+2+4+6+8+10+12+14+16+18+20 is prime
However, this pattern eventually fails. Do you know when?
Trivia submitted by Math Nerd 1729, Philippines
"A decimal point" is an anagram of "I'm a dot in place".
The ratio of the longer to the shorter side of any A-size paper (A3, A4 etc) is equal to the square root of 2.
Trivia submitted by Aravind Mahadevan,
The number of milliseconds in a day is:
5^5 × 4^4 × 3^3 × 2^2 × 1^1
Only digits 1,2 and 3 are possible in the look and say sequence.
Trivia submitted by Mr Phil, TSS
2519 is the smallest number that is gives remainder which is one less than the divisor for divisors one to ten:
Trivia submitted by S.Suresh ( Zoorace), Salem Tamilnadu India,
NEVER ODD OR EVEN is a palindrome, i.e. it reads the same back to front.
Trivia submitted by Des MacHale, Guardian
Trivia submitted by Theodore Driggers, Longmont, CO, USA
The Albanian word for 14 has 14 letters: katermbedhjete.
Trivia submitted by Lewis, Canada
2^5 × 9^2 = 2592
Trivia submitted by Chris Smith, Newsletter
In the Mathematical Gazette, vol. 53,, pp.127-129, it is shown that the 13th of the month is more likely to be a Friday than any other day.
Trivia submitted by Mathematical Association,
The number of minutes in February is 8! (eight factorial) unless it is a leap year.
Cleopatra lived closer in time to the release of the iPhone than to the building of the pyramids. The great Pyramids of Giza were built around 2500 BC, while Cleopatra, the last active Pharaoh of
Ancient Egypt, was born in 69 BC and died in 30 BC.
Trivia submitted by Kyle D Evans, A Year In Numbers
Did you know that the Romans failed to invent 0. it can not be written in Roman numerals?
Did you know that the word ‘hundred’ comes from the old Norse term, ‘hundrath’, which actually means 120 and not 100?
Did you know that 0.999999999... is equal to 1?
Did you know The result of (6 x 9) + (6 + 9) is 69?
Did you know that 1,089 X 9 = 9,801?
Did you know that the power of exponential growth is shocking You can actually reach the moon by folding a paper of 0.01mm 45 times?
Did you know that the largest prime number known so far is (2^57,885,161)-1, a 17,425,170 digit number?
Trivia submitted by Conner, Ambergate Sports College
That's 111 interesting and surprising facts but there could be more! Do you know any mathematical trivia that could be added to this page?
Click here to enter your information. | {"url":"http://transum.info/Software/Fun_Maths/Trivia.asp","timestamp":"2024-11-09T03:12:31Z","content_type":"text/html","content_length":"101220","record_id":"<urn:uuid:e6b34379-5420-4ac2-95f4-26104bcdbe3d>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00549.warc.gz"} |
Cambs & Hunts Contract Bridge Association, cambshunts
One common area for misunderstandings are those sequences where one person pulls 2NT at some stage in the auction to a suit at the three level - is this merely trying to find a better partscore, or
is it forcing? For example, try the following (all uncontested):
1. 1♥-1NT-2NT-3♣
2. 1♥-2♣-2♦-2NT-3♦
3. 1♥-1♠-2NT-3♠
4. 1♠-2♥-2♠-2NT-3♣
5. 1♠-2♣-2♦-2♥-2NT-3♦
The last one you ought to know - going via the fourth suit and then bidding again is generally agreed to be forcing (see Newsletter 15).
Some of these sequences are standard, whilst others are merely a matter of agreement. The first clearly should be non-forcing, simply showing a hand too weak to respond at the two level. The second
probably sounds non-forcing, so perhaps it should be for that reason. The third is clearly useful when you simply wish to play in 3♠, but most people play this as forcing, helping to reach the right
game. Sequence four is a classic - it should be weak, simply showing 5-5 if you would open 1♠ on such a hand.
This is all very well, but trying to learn every different sequence is a difficult business, and there are plenty more. Can we find some sensible rule that agrees with one's intuition in the simple
cases, but can also be applied to the more obscure situations?
* Pulling 2NT is forcing unless the hand is already limited, when it is non-forcing.
Note that this fits in with our decision on auctions 1, 3 and 5. For example, in 1, the 1NT bid limited responder's hand, so now the 3♣ bid is non forcing. We can also apply the rule to sequences
such as 1♥-2♣-2♥-2♠-2NT-3♣. Responder is unlimited, thus the 3♣ bid is forcing. Not everybody would agree with this one, but it is surely better to know what the bid means than not?
* New suits are forcing unless the hand is already limited.
* Old suits are forcing only if supporting partner for the first time.
Note that these rules tie in to our comments above on 2 and 4. More generally, consider the sequences below.
A. 1♠-2♦-2♥-2NT-3♣
B. 1♠-2♦-2♥-2NT-3♦
C. 1♣-1♠-2♠-2NT-3♣
Sequence A is forcing, as opener has not limited his hand (2♥ was forcing). Sequence B is forcing as opener is supporting partner for the first time (showing 5431 shape, trying to find the right
game, or possibly slam). The last sequence is non forcing - suggesting only three spades and good clubs.
There is a sequence that some people would take issue with: 1♥-2♣-2♦-2NT-3♥. To many this sounds weak, and the above rule would suggest this is so. However, there is a considerable body of people who
think this sequence ought to be forcing. With a weak hand and extra hearts you would simply rebid 2♥, hence the hand must have extras now, and simply be looking for the right game.
These rules are not set in stone, and will quite possibly not be the rules for you - think up your own ones. But if you want a serious partnership, then having a rule is a useful thing. | {"url":"https://www.bridgewebs.com/cambshunts/page68.html","timestamp":"2024-11-14T20:57:43Z","content_type":"text/html","content_length":"15740","record_id":"<urn:uuid:9f8cb185-6380-4883-ad01-e3e2b135a017>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00386.warc.gz"} |
Playing With Stones
Very easy
Execution time limit is 3 seconds
Runtime memory usage limit is 64 megabytes
You and your friend are playing a game in which you and your friend take turns removing stones from piles. Initially there are N piles with a_1, a_2, a_3, ..., a_N number of stones. On each turn, a
player must remove at least one stone from one pile but no more than half of the number of stones in that pile. The player who cannot make any moves is considered lost. For example, if there are
three piles with 5, 1 and 2 stones, then the player can take 1 or 2 stones from first pile, no stone from second pile, and only 1 stone from third pile. Note that the player cannot take any stones
from the second pile as 1 is more than half of 1 (the size of that pile). Assume that you and your friend play optimally and you play first, determine whether you have a winning move. You are said to
have a winning move if after making that move, you can eventually win no matter what your friend does.
The first line of input contains an integer T (T ≤ 100) denoting the number of testcases. Each testcase begins with an integer N (1 ≤ N ≤ 100) the number of piles. The next line contains N integers
a_1, a_2, a_3, ..., a_N (1 ≤ a_{i }≤_{ }2·10^18) the number of stones in each pile.
For each testcase, print "YES" (without quote) if you have a winning move, or "NO" (without quote) if you don't have a winning move.
Submissions 93
Acceptance rate 59% | {"url":"https://basecamp.eolymp.com/en/problems/3327","timestamp":"2024-11-12T15:53:18Z","content_type":"text/html","content_length":"262510","record_id":"<urn:uuid:f940460b-872e-4b48-af00-f717c75c3773>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00032.warc.gz"} |
Number Theory seminar abstracts, Term 2 2017-18
B (no C (no D (no F (no G (no H (no I (no J (no K (no L (no
definitions) definitions) definitions) definitions) definitions) definitions) definitions) definitions) definitions) definitions)
N (no S (no T (no V (no W (no X (no Y (no Z (no
definitions) definitions) definitions) definitions) definitions) definitions) definitions) definitions)
A Galois counting problem, by Sam Chow
We count monic quartic polynomials with prescribed Galois group, by box height. Among other things, we obtain the order of magnitude for $D_4$ quartics, and show that non-$S_4$ quartics are
dominated by reducibles. Weapons include determinant method estimates, the invariant theory of binary forms, the geometry of numbers, and diophantine approximation. Joint with Rainer Dietmann.
Effective bounds for singular units, by Yuri Bilu
A singular modulus is a j-invariant of a CM elliptic curve. It is known that it is always an algebraic integer. In 2015 Habegger proved that at most finitely many singular moduli are algebraic
units. It was a special case of his more general ``Siegel Theorem for Singular Moduli''. Unfortunately, this result was not effective, because Siegel's zero was involved (through Duke's
equidistribution theorem).
In the present work we obtain an explicit bound: if $\Delta$ is an imaginary quadratic discriminant such that the corresponding singular moduli are units, then $|\Delta|<10^{15}$.
Joint work with Philipp Habegger and Lars Kühne.
Extremal primes of non-CM elliptic curves, by Ayla Gafni
Fix an elliptic curve $E/\mathbb{Q}$. An ``extremal prime" for $E$ is a prime $p$ of good reduction such that the number of rational points on $E$ modulo $p$ is maximal or minimal in relation to
the Hasse bound. In this talk, I will discuss what is known and conjectured about the number of extremal primes up to $X$, and give the first non-trivial upper bound for the number of such primes
in the non-CM setting. In order to obtain this bound, we count primes with certain arithmetic characteristics and combine those results with the Chebotarev density theorem. This is joint work
with Chantal David, Amita Malik, Neha Prabhu, and Caroline Turnage-Butterbaugh.
Mass equidistribution for half-integral weight modular forms, by Steve Lester
In this talk I will discuss the distribution of $L^2$ mass of half integral weight cusp forms. For integral weight Hecke cusp forms, Holowinsky and Soundararajan have shown that the mass of these
forms equidistributes with respect to hyperbolic measure in the limit as the weight tends to infinity. Their method uses sieve bounds for multiplicative functions and weak subconvexity estimates
for L-functions. I will discuss the analogues of some of their methods in the half-integral weight setting, which have led to new results on the distribution of mass of these forms. This is joint
work with Maksym Radziwill.
On chaos measures, statistics of Riemann zeta, and random matrices, by Eero Saksman
We try to describe how complex multiplicative chaos measures emerge as description of functional statistics of the Riemann zeta function on the critical line. We also consider statistics on the
mesoscopic scale and relate it to that of random matrices. The talk is based on joint work with Christian Webb (Aalto University, Helsinki)
Potential automorphy over CM fields and applications, by James Newton
I will discuss joint work with Allen, Calegari, Caraiani, Gee, Helm, Le Hung, Scholze, Taylor and Thorne that establishes potential automorphy results for certain compatible systems of Galois
representations over CM fields. This has applications to the Sato-Tate conjecture for elliptic curves over CM fields and the Ramanujan conjecture for weight zero cohomological automorphic
representations of GL(2) over CM fields.
Potential modularity of abelian surfaces, by Toby Gee
I will discuss joint work in progress with George Boxer, Frank Calegari, and Vincent Pilloni, in which we prove that all abelian surfaces over totally real fields are potentially modular. We also
prove that infinitely many abelian surfaces over Q are modular.
Quartic forms in 30 variables, by Pankaj Vishe
Given an integral homogeneous polynomial F, when it contains a rational point is a key problem in Diophantine Geometry. A variety is supposed to satisfy Hasse Principle if it contains a rational
zero in the absence of any local obstructions. We will prove that smooth Quartic (deg F=4) hypersurfaces satisfy the Hasse Principle as long as they are defined over at least 30 variables. The
key tool here is employing a revolutionary idea of Kloosterman in our setting.This is a joint work with Oscar Marmon (U Lund).
Root numbers of abelian varieties, by Matthew Bisatt
Given an abelian variety $A$ over a number field, the famed Birch and Swinnerton-Dyer conjecture connects the rank of $A$ to its L-function; unfortunately very little is known about this
conjecture. Instead, we focus on a corollary known as the parity conjecture: this connects the rank modulo 2 to the global root number of $A$, which is defined independently of the L-function.
In this talk, I will discuss how to compute the global root number of an abelian variety, giving an example with the Jacobian of a hyperelliptic curve. Moreover, I will apply the results to find
a genus two hyperelliptic curve with simple Jacobian whose global root number is invariant under quadratic twist.
Unlikely intersections and the Chabauty-Kim method over number fields, by Netan Dogra
Let K be a number field. By theorems of Siegel and Faltings, if X is a hyperbolic curve over K, then X has only finitely many integral points. Recently, Kim had developed a method which can
sometimes give new proofs of these results when K=Q by locating the integral points inside the zero sets of certain p-adic analytic functions, generalising previous work of Chabauty. In this I
will explain how to extend some of these results to general K by proving certain 'unlikely intersection' results for the zeroes of these functions. | {"url":"https://warwick.ac.uk/fac/sci/maths/research/events/seminars/areas/number_theory/2017-18/abstracts_term_2/","timestamp":"2024-11-10T08:13:55Z","content_type":"text/html","content_length":"50476","record_id":"<urn:uuid:1556b67d-d114-44a7-9025-4e9c312921a4>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00025.warc.gz"} |
Summaries from Neural Information Processing Systems Conference on ShortScience.org
Summary by NIPS Conference Reviews 8 years ago
This paper studies a linear latent factor model, where one observes "examples" consisting of high-dimensional vectors $x_1, x_2, ..\in R^d$, and one wants to predict "labels" consisting of scalars $y_1, y_2, ... \in R$. Crucially, one is working in the "one-shot learning" regime, where the number of training examples n is small (say, $n=2$ or $n=10$), while the dimension d is large (say, $d \rightarrow \infty$). This paper considers a well-known method, principal component regression (PCR), and proves some somewhat surprising theoretical results: PCR is inconsistent, but a modified PCR estimator is weakly consistent; the modified estimator is obtained by "expanding" the PCR estimator, which is different from the usual "shrinkage" methods for high-dimensional data.
This paper aims to provide an analysis for principle component
regression in the setting where the feature vectors $x$. The authors
let $x = v + e$ where $e$ is some corruption of the nominal feature
vector $v$; and $v = a u$ where $a \sim N(0,\eta^2 \gamma^2 d)$ while
the observations $y = \theta/(\gamma \sqrt{d}) \langle v,u \rangle + \xi$. This
formulation is slightly different than the standard one because our
design vectors are noisy, which can pose challenges in identifying the
linear relationship between $x$ and $y$. Thus, using the top principle
components of $x$ is a standard method used in order to help
regularize the estimation. The paper is relevant to the ML
community. The key message of using a bias-corrected estimate of $y$
is interesting, but not necessarily new. Handling bias in regularized
methods is a common problem (cf. Regularization and variable selection
via the Elastic Net, Zou and Hastie, 2005). The authors present
theoretical analysis to justify their results. I find the paper
interesting; however I am not sure if the number of new results and
level of insights warrants acceptance. | {"url":"https://shortscience.org/venue?key=conf/nips&year=2013","timestamp":"2024-11-06T10:57:16Z","content_type":"text/html","content_length":"625981","record_id":"<urn:uuid:cef029eb-bf6f-4ebe-8e27-24195bc8748f>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00792.warc.gz"} |
Reverse Mortgage Calculators Explained
Category: Reverse Mortgage Calculators Explained
There are no required payments on a reverse mortgage—this type of mortgage accretes, unlike a typical amortizing mortgage that amortizes. Hence, mathematically, instead of an amortization schedule
associated with an amortizing mortgage, you end up with a reverse amortization schedule and a mortgage where the outstanding balance accretes or grows. The homeowner is still responsible for paying
the property taxes and insurance.
Hence, if you swap out an amortizing mortgage for a reverse mortgage, you will eliminate your monthly mortgage payment (from a monthly cash flow perspective only) but not your real estate taxes or
homeowners insurance. You will still owe the accrued interest despite not having to make monthly mortgage payments. Instead, the monthly interest is deferred, meaning it is not paid immediately, and
added to the mortgage's outstanding balance. This can lead to a significant increase in the total amount owed over time. This elimination and deferment of monthly mortgage payments is why the income
requirements for obtaining a reverse mortgage are less stringent than those for an amortizing mortgage. On a reverse mortgage, the borrower will still have to show enough income or alternative assets
to cover the real estate taxes and insurance.
Continue on the following page to compare the accumulative numeric results of an amortizing mortgage versus a reverse amortizing mortgage (reverse mortgage).
Category: Reverse Mortgage Calculators Explained
In this article, we compare the numeric output of an amortization schedule for an amortizing mortgage to a reverse amortization schedule for a reverse mortgage. The loan balances of these two types
of mortgages go in opposite directions. The amortizing mortgage outstanding loan balance amortizes, and the reverse mortgage outstanding loan balance accretes.
Let's look at refinancing a $100,000 fixed-rate 7% amortizing mortgage into a reverse HECM mortgage with an expected rate of 7%. Immediately below is the monthly payment on a 7% 30-year fixed-rate
mortgage, $665.30. Refinancing this loan into a HECM mortgage with an expected rate of 7% will eliminate the need to make monthly mortgage payments on the fixed-rate 7% amortizing mortgage.
Standard 30 Year 7% Fixed Rate Mortgage
Here is the 30-year amortization schedule for the fixed rate 30-year $100,000 at 7% amortizing mortgage.
Standard 30 Year 7% Fixed Rate Amortization Schedule
Refinancing the fixed rate 30-year 7% amortizing $100,000 mortgage into a HECM reverse mortgage with an expected rate of 7% will incur an upfront MIP (Mortgage Insurance Premium) cost. This upfront
MIP, which is a one-time fee, is calculated based on a home's appraised value up to the maximum claim amount of $1,149,850. The upfront MIP in this example is $ 8,843 or (2% x $442,149). This is in
addition to other estimated closing costs of $4,789 (1.08316% x $442,149), resulting in a total of financed closing costs of $13,632 ($8,843 + $4,789).
Hence, on day 1, the original fixed rate amortizing mortgage of $100,000 converted into a HECM reverse mortgage is now $113,632. The upfront MIP of 2% is applied to the home's appraised value unless
the appraised value is above the maximum claim amount of $1,149,850. In addition, a HECM reverse mortgage charges an ongoing MIP of .50 percent (50 basis points).
Reverse Mortgage Parameters, Mortgage Amounts, Financed Closing Costs, and Mandatory Obligations.
Reverse Amortization Schedule, Annual Totals and End of Year Projections
Here, they are side by side below. On the left is the amortization schedule for an amortizing 7% fixed rate mortgage, and on the right is a HECM FHA 7% reverse amortization schedule. As you can see,
the starting balance of the 7% fixed rate amortizing mortgage is $100,000, and the starting balance for the HECM FHA 7% reverse amortizing mortgage is $113,632 ($100,000 plus closing costs of
$13,632). The interest and the ongoing MIP are added to the HECM FHA mortgage annually, and the outstanding balance grows. By the end of 30 years, the 7% fixed rate amortizing mortgage is 0, and the
balance of the HECM FHA 7% reverse amortizing mortgage is $1,070,590.
Comparing a 7% Fixed Rate Amortizing Loan to a HECM FHA 7% Reverse Amortizing Loan
This comparison of the amortization schedule for the 7% fixed rate 30-year mortgage and the HECM FHA 7% reverse mortgage vividly illustrates the compounding effect of money. Both these loans, despite
having the same interest rate, lead in different monetary directions. The 7% fixed rate mortgage follows a traditional path of monthly payments, while the HECM FHA reverse mortgage offers the unique
benefit of being able to forego making monthly mortgage payments. But for comparative purposes, let’s examine what might happen if we paid the monthly $665.30 payment required on the 7% fixed rate
amortizing mortgage towards the HECM FHA reverse mortgage. This monthly payment would equal annual payments of $7,983.60 or ($665.30 x 12 months).
Additional $665.30 Monthly ($7,984 Annual) Payment
By the end of 30 years, the 7% fixed rate amortizing mortgage is 0, and the balance of the HECM FHA 7% reverse mortgage is $174,135. The $174,135 reflects the compounding effects in the difference in
the starting balances of the two mortgages $13,632 ($113,632 - $100,000) and the additional .50% ongoing MIP. I have approached the above example segmentally to isolate the various mathematical
So, what are the numerical benefits of getting a reverse mortgage over a typical amortizing mortgage? The answer is liquidity, and it is essential to note that these ending differences in the
outstanding balances of these two types of mortgages. will be narrower in a lower interest rate environment. Continue to the next page for further discussion.
Category: Reverse Mortgage Calculators Explained
This article will examine the effects of interest rate levels and their accumulative effect on a reverse mortgage over time.
In the previous article, "Comparing an Amortizing Mortgage to a Reverse Mortgage," we viewed the effects of the compounding of money. We compared the numeric differences between an amortizing
mortgage versus a reverse amortizing mortgage utilizing a 7% fixed rate and a 7% expected rate, and both these mortgage balances went in different directions. For instance, we found that after 30
years, the amortizing mortgage had reduced the principal by [100%], while the reverse amortizing mortgage had increased the principal by [842%]. The differences were somewhat shocking, and
ultimately, one might question the utility of a reverse mortgage. The answer is in the additional liquidity a reverse mortgage provides and the increased advantages of reverse mortgages in a
low-interest rate environment.
In the previous article, we compared a 30-year 7% fixed rate amortizing mortgage to an HECM FHA reverse mortgage with an expected rate of 7%. As a recap, see the below table. Refinancing the fixed
rate 30-year 7% amortizing $100,000 mortgage into a HECM reverse mortgage with an expected rate of 7% will cost $8,843 in upfront MIP plus other estimated closing costs of $4,789 for a total of
financed closing costs of $13,632. Hence, on day 1, the original mortgage of $100,000 is now $113,632.
The starting balance of the 7% fixed rate amortizing mortgage is $100,000, and the starting balance for the HECM FHA 7% reverse mortgage is $113,632 ($100,000 plus closing costs of $13,632). The
interest and the ongoing MIP are added to the HECM FHA mortgage annually, and the outstanding balance grows. By the end of 30 years, the 7% fixed rate amortizing mortgage is 0, and the balance of the
HECM FHA 7% reverse mortgage is $1,070,590.
Reverse Mortgage and Amortizing Mortgage Schedules In 7% Rate Environment.
Before we delve into the liquidity benefits of a reverse mortgage, let's explore the potential benefits of a lower interest rate environment. This shift to a lower interest rate environment could
significantly impact the financial results. Below is a comparison of similar circumstances, except now, mortgage rates are not 7% but a hypothetical 3%. Remember, reverse mortgages accrete; a higher
interest rate will only exemplify the compounding effect on a reverse mortgage's outstanding loan balance over time or vice versa. In the example below, the compounding impacts diminish due to a
lower rate of 3%, but the opposite effect would result if the interest rate increased. These analyses assume a constant interest rate over the life of the mortgage. If a mortgage has an adjustable
interest rate, then over time, the results would be subject to future changes in interest rates.
Reverse Mortgage and Amortizing Mortgage Schedules In 3% Rate Environment.
In the above table, the reverse mortgage with an expected rate of 3% ended year 30 with a loan balance of $385,286, whereas in our previous example, our reverse mortgage with an expected rate of 7%
ended year 30 with a loan balance of $1,070,590.
The starting balance on the HECM FHA mortgage with an expected rate of 3% differs from the previous example of 7% because, in this instance, the borrower is due to receive an "initial cash advance"
of $21,400 at closing due to a change in the "principal limit factor." The "initial cash advance" is determined by multiplying the "principal limit factor" by the "initial property value" multiplied
by 10%. The "principal limit factor" is based on the "age of the youngest borrower" and the "expected rate." The "initial cash advance" of $21,400 must be added to the opening balance of $113,632.
The starting balance represents $100,000 plus the financed closing costs of $13,632 plus the "initial cash advance" of $21,400 for $135,023. I will delve into the "initial cash advance" and the "
initial principal factor" in a subsequent article. For now, I am trying to isolate the compounding effects of money on the outstanding balance based on different interest rates.
In addition, by the end of 30 years, the 3% fixed-rate amortizing mortgage will be 0. The balance of the HECM FHA 3% reverse mortgage is now $385,286, including the initial cash advance of $21,400
and any interest and ongoing MIP on this amount. Though we are attempting to focus on the impact of the interest rate environment on reverse mortgages, the initial cash advance has more to do with
liquidity. Still, as I previously mentioned, I will address this in a subsequent article.
Now, we can begin to discuss the liquidity benefits, but in doing so, we first need to understand the "principal limit factor" and how it might impact our analysis.
Reverse Mortgage Line Of Credit In 3% Rate Environment.
Category: Reverse Mortgage Calculators Explained
The current article will delve into the liquidity benefits of a reverse mortgage, but to do so, we need to understand how liquidity is tapped from a home's equity by the owner utilizing a reverse
mortgage. We need to understand the calculations to know how much equity a HECM reverse mortgage can provide. Please read the previous three articles on reverse mortgages before delving into this
section on liquidity. Here is a list:
Mathematically, What is a Reverse Mortgage?
This article gives a brief numeric description of a reverse mortgage versus an amortizing mortgage.
Comparing an Amortizing Mortgage to a Reverse Mortgage
This article compares the amortization schedule of an amortizing mortgage to a reverse amortization schedule of a reverse mortgage.
Reverse Mortgages in a Low Interest Rate Environment
This article highlights the effects of interest rate levels on the economics of a reverse mortgage versus an amortizing mortgage.
Let's discuss the "principal limit factor" and the "initial principal limit," two variables used to determine a home's equity accessibility via a HECM reverse mortgage.
The "principal limit factor" is used in determining the "initial principal limit," which further drives the "initial cash advance" and the available "line of credit."
The "principal limit factor" is based on the "age of the youngest borrower" and the "expected rate." The "age of the youngest borrower" and the "expected rate" are used to obtain the "principal limit
factor" from a table with HUD.
The "initial principal limit" is based on the "principal limit factor" multiplied by the "maximum claim amount." The "maximum claim amount." is the lesser of the "initial property value" or
$1,149,850. $1,149,850 is the maximum claim amount available for a HECM FHA reverse mortgage.
Flow Chart for Deriving the Initial Principal Limit
In the previous article there was an "initial cash advance" of $21,400 in our 2nd example forwarded to the borrower at closing. In addition, there is a growing "line of credit", $78,968, at the end
of year 1.
The reason for the "initial cash advance" and the "line of credit" is the change in the "principal limit factor" resulting from a shift in the "expected rate" on the HECM reverse mortgage from 7% to
The "principal limit factor" for "age of the youngest borrower" of 55 and the "expected rate" of 3% is .484.
The "principal limit factor" for "age of the youngest borrower" of 55 and the "expected rate" of 7% is .257.
The "initial principal limit" for a property value of $442,149 and a "principal limit factor" of .484 is $214,000 ($442,149 x .484).
The "initial principal limit" for a property value of $442,149 and a "principal limit factor" of .257 is $113,632 ($442,149 x .257).
Here is a flow chart depicting the relationship between "age of the youngest borrower" and "expected rate" with the "principal limit factor" which drives the "initial principal limit".
Correlation of Age and Expected Rate with PLF and Initial Principal Limit.
Now we need to calculate the "net principal limit". The "net principal limit" is the "initial principal limit" less any "mandatory obligations." The "mandatory obligations" would be any existing
liens, the upfront MIP, and the other closing costs. Below is a flow chart showing how the "net principal limit" is derived.
Flow Chart for Deriving Mandatory Obligations and Net Principal Limit.
Now, we need to calculate the "initial advance." The "initial advance" is typically 10% of the "initial principal limit" unless the "net principal limit" is insufficient to allow for the total
amount. If the "net principal limit" is less than the "initial advance," the "net principal limit" will supersede the "initial advance" ("initial principal limit" multiplied by 10%), and the "net
principal limit" will become the "initial advance." If the "net principal limit" is not only below the "initial advance" but is negative, then the loan is considered "short to close."
Flow Chart of Determining the Initial Advance.
In our original analysis, I set the property appraised value to $442,149, the age of the youngest borrower to 55, and the expected interest rate to 7% so that the "principal limit factor" resulted in
a 0 "initial cash advance" and a "loan balance" equal to the "principal limit" so that the "line of credit" would be 0 throughout the life of the mortgage. The "line of credit" is the difference
between the "principal limit" and the "loan amount."
In Summary:
The "net principal limit" will determine the net available liquidity that can be accessed utilizing a reverse mortgage.
Use the below to determine the "principal limit factor" from the HUD tables.
• "age of the youngest borrower"
• "expected rate"
Take the "principal limit factor" multiplied by the "maximum claim amount." (derived from the lessor of the "initial property value" or $1,149,850) to determine the below:
• "initial principal limit"
Subtract the "mandatory obligations" (upfront MIP, other closing costs, and outstanding liens) from the "initial principal limit" to determine the below:
If the "net principal limit" is above the "initial advance," there is liquidity. If it is below the "initial advance" but still positive, there is some liquidity. If it is negative, the loan is "
short to close," and there is no additional liquidity.
Note: If you want to run some calculations by changing the input, try this calculator, HECM Reverse Mortgage LOC Calculator. This calculator will automatically obtain the "principal limit factor"
based on the "age of the youngest borrower" and the "expected rate." It will run all the calculations based on the required input, "age of the youngest borrower," "expected rate," "initial property
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Symbol For Area In Math: Square Units!
Symbol for Area in Math: Square Units!
In mathematics, the symbol most commonly used to represent area is “A.” The concept of area pertains to the measure of the extent of a two-dimensional surface or shape.
The area is expressed in square units, which could be square meters (m²), square centimeters (cm²), square feet (ft²), etc. The choice of the symbol “A” for area is arbitrary but widely accepted in
mathematical notation.
For example:
The area of a rectangle is calculated by the formula A = length × width.
The area of a circle is given by A = πr², where “r” is the radius.
In geometry, “A” universally represents the area, simplifying the communication of mathematical concepts across various domains and applications.
Key Takeaway
The area symbol in mathematics is denoted by the lowercase letter ‘a’ with a line above it and represents the measurement of two-dimensional space.
The concept of area and its representation has evolved throughout history, from ancient civilizations to modern standardized symbols.
The area symbol is used in formulas for calculating the areas of shapes like squares, rectangles, circles, and triangles.
Understanding and utilizing the area symbol is essential for solving geometry problems and has practical applications in fields such as construction, engineering, architecture, and design.
Symbol for Area Representation
The symbol used to represent area in mathematics is denoted by the lowercase letter ‘a’ with a line above it, often pronounced as ‘a bar.’
This symbol is widely recognized and utilized in mathematical equations and geometric formulas to denote the measurement of two-dimensional space.
When dealing with shapes such as squares, rectangles, circles, and triangles, this symbol is employed to express the extent of their surface coverage.
For instance, in the formula for the area of a rectangle, the symbol ‘a’ is used to represent the area, and the formula is expressed as length multiplied by width.
Understanding and correctly implementing this symbol is fundamental in various mathematical and practical applications, providing a standardized representation for the measurement of area in the
field of mathematics.
Historical Evolution of Area Symbol
Historical development of the area symbol in mathematics has evolved over centuries, reflecting advancements in geometric understanding and notation.
The concept of area has been integral to various ancient civilizations, with each culture developing its own methods for measuring and representing it.
The evolution of area symbols can be traced from the early Egyptian use of unit fractions to the Greek introduction of geometric shapes and the subsequent development of algebraic notation.
The evolution continues through the Middle Ages, Renaissance, and into the modern era, where standardized symbols are widely accepted.
Below is a table summarizing the historical evolution of area symbols:
Civilization/Period Notable Developments
Ancient Egypt Unit fractions
Ancient Greece Geometric shapes
Middle Ages Algebraic notation
Summarizing the historical evolution of area symbols
Area Symbol Usage in Formulas
Evolution of area symbol usage in mathematical formulas has demonstrated the ongoing refinement and precision of geometric notation. In the realm of mathematical formulas, the area symbol has evolved
to convey specific information and relationships within geometric contexts.
This evolution is evident in various ways:
• Incorporation of specific geometric shapes, such as squares, circles, and triangles, into the area formulas
• Integration of variables and constants to represent dimensions and values within the formulas
• Utilization of standard mathematical operations, such as multiplication and division, to express the relationships between different elements in the formulas
These developments have led to a more precise and nuanced understanding of area within mathematical contexts. Understanding the evolution of area symbol usage in formulas provides valuable insight
into the refinement of mathematical notation over time.
This refinement has improved the clarity and precision of geometric concepts in mathematical discourse.
Area Symbol in Geometry Problems
How does the area symbol function in solving geometric problems? In geometry, the area symbol, denoted as ‘A’, represents the measurement of a two-dimensional space within a shape.
Understanding the area symbol is crucial for solving various geometric problems, such as calculating the space within a rectangle, triangle, circle, or any other polygon.
The table below illustrates the area symbols and formulas for common geometric shapes:
Shape Symbol Formula
Rectangle A length x width
Triangle A 1/2 x base x height
Circle A π x radius^2
Importance of Area Symbol in Mathematics
The area symbol in mathematics plays a pivotal role in quantifying and analyzing two-dimensional spaces within geometric shapes, facilitating precise problem-solving and real-world applications.
Understanding the importance of the area symbol is crucial for various reasons:
• Quantification: The area symbol provides a standardized method for quantifying the size of two-dimensional shapes, enabling precise comparisons and measurements.
• Problem-Solving: It is essential for solving geometry problems, calculating the space within shapes, and determining proportions in various mathematical scenarios.
• Real-World Applications: The concept of area, represented by the area symbol, is extensively used in real-world situations such as construction, engineering, architecture, and design, where
accurate measurements of spaces are fundamental to successful outcomes.
The area symbol in mathematics serves as a crucial representation for measuring the extent of a two-dimensional space. Throughout history, the symbol has evolved and is now widely used in various
mathematical formulas and geometry problems.
An interesting statistic to note is that the concept of area dates back to ancient civilizations such as the Egyptians and Mesopotamians, who used basic geometric principles to measure land and
construct buildings.
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Digital Image Processing -3D to 2D Projection (Octave/Matlab)
The projection coordinate system from the real world to the image:
(x, y, z) → (x', y', - d), note that this diagram does not include flipping changes from left to right, up and down.
According to the theory of similar triangles, it can be calculated that: (x, y, z) → (- dx/z, - dy/z, - d), as z is a variable, this is not a linear transformation. By adding a homogeneous term, we
can complete the projection in a homogeneous coordinate system through linear operations on the matrix.
The code is as follows:
% Project a point from 3D to 2D using a matrix operation
%% Given: Point p in 3-space [x y z], and focal length f
%% Return: Location of projected point on 2D image plane [u v]
%% Test: Given point and focal length (units: mm)
p = [200 100 120];
f = 50;
function p_img = project_point(p, f)
%% TODO: Define and apply projection matrix
A = [f 0 0 0;
0 f 0 0;
0 0 1 0];
p_hom = [p 1]';
p_proj = A * p_hom;
p_img = [(p_proj(1)/p_proj(3)),(p_proj(2)/p_proj(3))];
%% Test: Given point and focal length (units: mm)
p = [200 100 100];
f = 50;
pimg = project_point(p, f); | {"url":"https://iopenv.com/3SSEQUJYQ/Digital-Image-Processing-3D-to-2D-Projection-OctaveMatlab","timestamp":"2024-11-15T01:48:03Z","content_type":"text/html","content_length":"14821","record_id":"<urn:uuid:1a8d704c-161c-48a0-985b-c750c083a790>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00069.warc.gz"} |