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Lesson 8 | Equations and Inequalities | 7th Grade Mathematics | Free Lesson Plan
Expressions and Equations
7.EE.B.4.B — Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret
it in the context of the problem. For example: As a salesperson, you are paid $50 per week plus $3 per sale. This week you want your pay to be at least $100. Write an inequality for the number of
sales you need to make, and describe the solutions. | {"url":"https://www.fishtanklearning.org/curriculum/math/7th-grade/equations-and-inequalities/lesson-8/","timestamp":"2024-11-09T10:02:20Z","content_type":"text/html","content_length":"481288","record_id":"<urn:uuid:08384fb6-3135-4b8c-ad65-75f8013a0432>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00781.warc.gz"} |
Proposal For New Numerical Differentiation Function
* Proposal For New Numerical Differentiation Function
@ 2019-09-09 23:44 Bob Smith
2019-09-11 1:05 ` Patrick Alken
0 siblings, 1 reply; 3+ messages in thread
From: Bob Smith @ 2019-09-09 23:44 UTC (permalink / raw)
To: GSL Discuss
I have written a function for GSL to perform Numerical Differentiation
which I would like to offer to this project.
I saw on https://www.gnu.org/software/gsl/ that I should first propose
it as an extension, however the "How to help" link under
Extensions/Applications goes nowhere, so I wondered if there was
somewhere else to look for guidelines on setting up an extension.
I looked over most of the links to the listed extensions (many of which
are broken) and didn't see any commonality which may be the answer to my
question. In any case, to whom do I write to get added to the list of
Thanks for your help.
FWIW, the new function is more accurate (up to 13 or 14 significant
digits vs. 9 or 10 for <gsl_deriv_central>) and can compute Nth
derivatives by passing N (1 through 9) to the function at zero
additional cost in performance. The algorithm is based upon the papers
â New Finite Difference Formulas for Numerical Differentiationâ , I.R.
Khan, R. Ohba / Journal of Computational and Applied Mathematics 126
(2000) pp. 269-276
â Taylor Series Based Finite Difference Approximations of Higher-Degree
Derivativesâ , I.R. Khan, R. Ohba / Journal of Computational and Applied
Mathematics 154 (2003) pp. 115-124
Bob Smith - bsmith@sudleyplace.com
http://www.sudleyplace.com - http://www.nars2000.org
^ permalink raw reply [flat|nested] 3+ messages in thread
* Re: Proposal For New Numerical Differentiation Function
2019-09-09 23:44 Proposal For New Numerical Differentiation Function Bob Smith
@ 2019-09-11 1:05 ` Patrick Alken
2019-09-11 12:25 ` Bob Smith
0 siblings, 1 reply; 3+ messages in thread
From: Patrick Alken @ 2019-09-11 1:05 UTC (permalink / raw)
To: gsl-discuss
Yes the extensions may be a bit out of date, sorry about that. Could
you make a git diff against the latest master branch of the git
repository, and then email me the diff? Then I can take a look and/or
help make an extension out of it.
On 9/9/19 5:44 PM, Bob Smith wrote:
> I have written a function for GSL to perform Numerical Differentiation
> which I would like to offer to this project.
> I saw on https://www.gnu.org/software/gsl/ that I should first propose
> it as an extension, however the "How to help" link under
> Extensions/Applications goes nowhere, so I wondered if there was
> somewhere else to look for guidelines on setting up an extension.
> I looked over most of the links to the listed extensions (many of
> which are broken) and didn't see any commonality which may be the
> answer to my question. In any case, to whom do I write to get added
> to the list of extensions?
> Thanks for your help.
> FWIW, the new function is more accurate (up to 13 or 14 significant
> digits vs. 9 or 10 for <gsl_deriv_central>) and can compute Nth
> derivatives by passing N (1 through 9) to the function at zero
> additional cost in performance. The algorithm is based upon the papers
> “New Finite Difference Formulas for Numerical Differentiation”, I.R.
> Khan, R. Ohba / Journal of Computational and Applied Mathematics 126
> (2000) pp. 269-276
> “Taylor Series Based Finite Difference Approximations of Higher-Degree
> Derivatives”, I.R. Khan, R. Ohba / Journal of Computational and
> Applied Mathematics 154 (2003) pp. 115-124
^ permalink raw reply [flat|nested] 3+ messages in thread
* Re: Proposal For New Numerical Differentiation Function
2019-09-11 1:05 ` Patrick Alken
@ 2019-09-11 12:25 ` Bob Smith
0 siblings, 0 replies; 3+ messages in thread
From: Bob Smith @ 2019-09-11 12:25 UTC (permalink / raw)
To: gsl-discuss
On 9/10/2019 9:05 PM, Patrick Alken wrote:
> Hello,
> Â Yes the extensions may be a bit out of date, sorry about that. Could
> you make a git diff against the latest master branch of the git
> repository, and then email me the diff? Then I can take a look and/or
> help make an extension out of it.
Thanks for you kind offer. I'll get back to you when I get git sorted.
> Patrick
> On 9/9/19 5:44 PM, Bob Smith wrote:
>> I have written a function for GSL to perform Numerical Differentiation
>> which I would like to offer to this project.
>> I saw on https://www.gnu.org/software/gsl/ that I should first propose
>> it as an extension, however the "How to help" link under
>> Extensions/Applications goes nowhere, so I wondered if there was
>> somewhere else to look for guidelines on setting up an extension.
>> I looked over most of the links to the listed extensions (many of
>> which are broken) and didn't see any commonality which may be the
>> answer to my question. In any case, to whom do I write to get added
>> to the list of extensions?
>> Thanks for your help.
>> FWIW, the new function is more accurate (up to 13 or 14 significant
>> digits vs. 9 or 10 for <gsl_deriv_central>) and can compute Nth
>> derivatives by passing N (1 through 9) to the function at zero
>> additional cost in performance. The algorithm is based upon the papers
>> â New Finite Difference Formulas for Numerical Differentiationâ , I.R.
>> Khan, R. Ohba / Journal of Computational and Applied Mathematics 126
>> (2000) pp. 269-276
>> â Taylor Series Based Finite Difference Approximations of Higher-Degree
>> Derivativesâ , I.R. Khan, R. Ohba / Journal of Computational and
>> Applied Mathematics 154 (2003) pp. 115-124
Bob Smith - bsmith@sudleyplace.com
http://www.sudleyplace.com - http://www.nars2000.org
^ permalink raw reply [flat|nested] 3+ messages in thread
end of thread, other threads:[~2019-09-11 12:25 UTC | newest]
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2019-09-09 23:44 Proposal For New Numerical Differentiation Function Bob Smith
2019-09-11 1:05 ` Patrick Alken
2019-09-11 12:25 ` Bob Smith
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Plane Geometry
From inside the book
Results 1-5 of 30
Page 34 ... polygons ? 7. Join any point within a polygon to all the vertices . How many triangles are formed ? 8. Join any point , not a vertex , in the perimeter of a polygon to all the vertices .
How many triangles are formed ? 57. Regular Polygons ...
Page 35 ... regular hex- agon is surrounded by rectangles and equilateral triangles . 8. In the following patterns irregular hexagons intervene between the regular polygons . Draw the figures . NOTE
... regular polygon of. REGULAR POLYGONS 35.
Page 36 William Betz, Harrison Emmett Webb. THE CIRCLE 58. Preliminary Definitions . A regular polygon of n sides may be obtained by drawing from a point n rays forming n equal angles , and laying
off on these rays equal segments . The points of ...
Page 50 ... Regular Polygons . The chords of the arcs intercepted by equal central angles of a circle are equal . Also the base ... polygon results ; for the sides and the angles of such a polygon
are equal . This will always happen C D B 45 ...
Page 51 ... regular polygon of 4 sides ? 5 sides ? 6 sides ? 10 sides ? 15 sides ? 24 sides ? n sides ? 12. Through how many degrees does a point on the equator turn in 4 hr . ? 6 hr . ? 12 hr . ? 18
hr . ? 1 min . ? 10 min . ? 13. The length of ...
PRELIMINARY COURSE 1
THE ANGLE 14
TRIANGLES 28
THE CIRCLE 36
CIRCLE AND ANGLE 47
PART II 53
CONGRUENCE 61
RECTILINEAR FIGURES 69
LOCI 171
COÖRDINATES 177
AREA 193
TRANSFORMATIONS 210
PROPORTIONAL MAGNITUDES 231
SIMILAR TRIANGLES 244
CONSTRUCTIONS 84
RIGHT TRIANGLES 91
ANGLESUM 102
PARALLELOGRAMS 111
PAGE 143
TANGENTS 153
TWO CIRCLES 159
Popular passages
The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.
In an isosceles triangle the angles opposite the equal sides are equal.
The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
If two triangles have an angle of one equal to an angle of the other, and...
In the same circle, or in equal circles, if two chords are unequally distant from the center, they are unequal, and the chord at the less distance is the greater.
... the third side of the first is greater than the third side of the second.
If two chords intersect within a circle, the product of the segments of one chord is equal to the product of the segments of the other.
A line drawn from the vertex of the right angle of a right triangle to the middle point of the hypotenuse divides the triangle into two isosceles triangles.
The medians of a triangle meet in a point which is two thirds of the distance from each vertex to the middle of the opposite side.
If three or more parallels intercept equal parts on one transversal, they intercept equal parts on every transversal. Given the parallels AB, CD, and EF intercepting equal parts on the transversal
Bibliographic information | {"url":"https://books.google.com.jm/books?id=Bu5HAAAAIAAJ&q=regular+polygon&output=html_text&source=gbs_word_cloud_r&cad=4","timestamp":"2024-11-11T00:15:52Z","content_type":"text/html","content_length":"62302","record_id":"<urn:uuid:3576f4aa-7f94-41bd-83e5-1225fcd1117b>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00445.warc.gz"} |
196 dryquart to usquart - How much is 196 dry quarts in US quarts? [CONVERT] ✔
196 dry quarts in US quarts
Conversion formula
How to convert 196 dry quarts to US quarts?
We know (by definition) that: $1⁢dryquart ≈ 1.16364718614719⁢usquart$
We can set up a proportion to solve for the number of US quarts.
$1 ⁢ dryquart 196 ⁢ dryquart ≈ 1.16364718614719 ⁢ usquart x ⁢ usquart$
Now, we cross multiply to solve for our unknown $x$:
$x ⁢ usquart ≈ 196 ⁢ dryquart 1 ⁢ dryquart * 1.16364718614719 ⁢ usquart → x ⁢ usquart ≈ 228.07484848484924 ⁢
Conclusion: $196 ⁢ dryquart ≈ 228.07484848484924 ⁢ usquart$
Conversion in the opposite direction
The inverse of the conversion factor is that 1 US quart is equal to 0.00438452554783318 times 196 dry quarts.
It can also be expressed as: 196 dry quarts is equal to $1 0.00438452554783318$ US quarts.
An approximate numerical result would be: one hundred and ninety-six dry quarts is about two hundred and twenty-eight point zero six US quarts, or alternatively, a US quart is about zero times one
hundred and ninety-six dry quarts.
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | {"url":"https://converter.ninja/volume/dry-quarts-to-us-quarts/196-dryquart-to-usquart/","timestamp":"2024-11-08T23:54:56Z","content_type":"text/html","content_length":"20449","record_id":"<urn:uuid:134945d8-e06e-4894-ac10-3d734a1d8bf3>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00011.warc.gz"} |
What is bond yield?
Understanding What is Bond Yield
To understand bond yields, we must first delve into the concept of bonds. Bonds are essentially loan agreements between an investor and an issuer (commonly a country or large organization). The
issuer borrows capital from the investor, in turn agreeing to pay back the borrowed value plus interest over a predetermined schedule. The bond yield refers to the rate of return an investor can
anticipate for their bond investment, and this amount is typically expressed as a percentage.
Defining Bond Yield
Current Yield
In the simplest of terms, the current yield of a bond is defined as the ratio of the annual interest payment to the bond’s current market price. This is the most direct way to calculate the yield on
a bond, and the formula for finding the current yield is as follows:
Current Yield = (Annual Interest Payment / Market Price) * 100%
However, the current yield falls short in that it does not account for the total return an investor could receive if they hold the bond until maturity, nor does it factor in the reinvestment of
interest amounts or the bond’s purchase or selling price.
Yield to Maturity (YTM)
A more sophisticated understanding of bond yields involves Yield to Maturity (YTM). YTM is a more encompassing yield measurement because it considers both the current yield and any capital gains or
losses that would be realized if the bond is held until maturity. The YTM calculation assumes that all coupon payments (the regular interest payments an investor receives from owning a bond) are
reinvested at the same rate as the bond’s current yield.
YTM also takes into account the bond’s purchase price, its face value, the term of the bond and the interest rate of the coupon. This makes YTM a complex calculation, which is typically done through
trial and error or approximate formulas.
The YTM is presented as an annual percentage rate (APR), and it gives investors a more accurate representation of their potential return on a bond investment.
Factors Impacting Bond Yields
Interest Rates
The bond yield is inversely related to its price. When the price rises, the bond yield falls and vice versa. This relationship is a result of market interest rates. When interest rates increase, the
fixed interest payments of a bond become less appealing in comparison, leading to a decrease in the bond’s price and a corresponding increase in yield. Conversely, when interest rates decrease, the
bond’s fixed interest payments become more attractive, leading to an increase in price and decrease in yield.
Inflation has the potential to erode the fixed returns from a bond. As such, rising inflation could impact bond yields in two ways. On one hand, investors may demand higher yields to compensate for
anticipated inflation. On the other hand, central banks often increase interest rates to quell inflation, which, as discussed, can directly impact bond yields.
Importance of Bond Yields
A bond’s yield is one of the most significant factors for investors to consider when deciding whether to invest in a bond. It not only represents the return on an investment but also gives investors
insight into the future direction of interest rates and inflation and the overall health of the economy.
Understanding bond yields is crucial for risk assessment because it shows how much money investors would make by investing in bonds as opposed to other securities. As investors become more
experienced, they can use bond yields to develop a diverse and balanced investment portfolio.
Ending Notes
In conclusion, the concept of bond yield, while seemingly straightforward, factors in multiple economic and market variables, which can make it a complex construct to understand. While the current
yield gives a quick snapshot of return in relation to price, the yield to maturity provides a more comprehensive picture of potential returns. As with any financial investment, understanding the
underlying mechanics, benefits, and potential risks is key, and bond yields play a significant role within this understanding. | {"url":"https://www.tradingclass.com/bond-market/basics/what-is-bond-yield/","timestamp":"2024-11-04T11:35:58Z","content_type":"text/html","content_length":"472687","record_id":"<urn:uuid:eee9ff1d-b9fe-4aca-bae6-6d33e5255e14>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00779.warc.gz"} |
What is half circumferential stress?
What is half circumferential stress? Explanation: Longitudinal stress is developed along the walls of the cylinder in the shell due to internal fluid pressure on the ends. The longitudinal stress is
half the circumferential stress.
What is half circumferential stress?
Explanation: Longitudinal stress is developed along the walls of the cylinder in the shell due to internal fluid pressure on the ends. The longitudinal stress is half the circumferential stress. It
is also known as axial stress.
Is hoop stress radial or tangential?
circumferential stress, or hoop stress, a normal stress in the tangential (azimuth) direction. axial stress, a normal stress parallel to the axis of cylindrical symmetry. radial stress, a normal
stress in directions coplanar with but perpendicular to the symmetry axis.
What do you mean by hoop stress?
Hoop stress is the stress that occurs along the pipe’s circumference when pressure is applied. Hoop stress acts perpendicular to the axial direction. Hoop stress is also referred to as tangential
stress or circumferential stress.
What is the formula of hoop stress?
The standard equation for hoop stress is H = PDm /2t. In this equation, H is allowable or hoop stress, the P is the pressure, t is the thickness of the pipe, and D is the diameter of the pipe.
Which stress is the least in a thin shell?
7. Which stress is the least in a thin shell
• Longitudinal stress.
• Hoop stress.
• Radial stress.
• None.
What is the difference between longitudinal and circumferential stress?
Circumferential stress is twice longitudinal stress D. Circumferential stress is twice longitudinal stress Internal pressure can be produce by water, gases or others. When a thin – walled cylinder is
subjected to internal pressure, three are two mutually stresses: Circumferential or Hoop stress.
What are axial stresses?
Axial Stress – is the result of a force acting perpendicular to an area of a body, causing the extension or compression of the material.
What is Lame’s equation based on?
Lame’s theorem gives the solution to thick cylinder problem. The theorem is based on the following assumptions: Material of the cylinder is homogeneous and isotropic. Plane sections of the cylinder
perpendicular to the longitudinal axis remain plane under the pressure.
Which stress is least in thin shell?
Explanation: The thickness of plate is negligible when compared to the diameter of the cylindrical shell, and then it can be termed as a thin cylinder. The radius stress in the cylinder walls is | {"url":"https://missionalcall.com/2020/09/19/what-is-half-circumferential-stress/","timestamp":"2024-11-03T00:21:19Z","content_type":"text/html","content_length":"55668","record_id":"<urn:uuid:1167e846-bc02-4ec4-8d7f-77c40b456fff>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00676.warc.gz"} |
Calculator Programs and the AP Statistics Exam
Calculator Programs and the AP Statistics Exam
Calculator Programs
Often questions arise about programs that can be added to calculators to improve their functionality or ease of use. Sometimes these programs are purchased software packages, while other times they
are the creation of a teacher or student with programming skills. The conscientious teacher often asks, "Is a program like this allowed on the AP Exam?" This article seeks to clarify these questions.
The College Board policy on calculator usage states:
For the exam, you're not allowed to access any information in your graphing calculators or elsewhere if it's not directly related to upgrading the statistical functionality of older graphing
calculators to make them comparable to statistical features found on newer models. The only acceptable upgrades are those that improve the computational functionalities and/or graphical
functionalities for data you key into the calculator while taking the examination. Unacceptable enhancements include, but aren't limited to, keying or scanning text or response templates into the
Source of this quotation: AP^®: Calculator Policy
This policy is the key to determining which type of programs are allowed. A number of commonly used programs clearly fit in this category. The TI-83 Plus and TI-84 Plus now come preloaded with a
flash-app program entitled Catalog Help. This program guides the user with prompts for the various variables you enter. For example, the normalcdf command requires you enter first the lowerbound,
then the upperbound, then, optionally, µ and Ã. The Catalog Help program simply reminds you of the order that these variables are entered. Not requiring students to remember the order of these
variables is one of the most frequently requested improvements that teachers have for the ubiquitous 83 Plus/84 Plus family of calculators. A version of Catalog Help has been written for TI-83 Plus
models that do not accept flash apps, a nice legal addition for a student with an older calculator. The TI-89 is an approved calculator with a menu-driven screen that does not require students to
remember the variable order. Thus the 89's functionality provides a good benchmark for the kind of improvements that would be legal. Another specific upgrade that teachers could use is a chi-square
goodness-of-fit test. The 89 includes such a feature, as does the latest operating system for the 84 Plus, so a program that performs that test would be legal.
For those using the TI-86, Texas Instruments has written an add-on program that gives it full statistics capabilities. This program simply gives it the same functions as the 83, 83 Plus, 84 Plus, and
89. Some teachers have written programs for the 86 that add other missing features: a normal probability plot, a residual list generator, a Catalog Help-like program, and so on. These programs are
A feature that is not included on any legal calculator is any reminder of the conditions required to perform a hypothesis test or confidence interval. Students struggle with memorizing these
conditions. Some programs warn students if the conditions have been violated (i.e., "Warning! Expecteds are less than 5!"). As no allowable calculator offers such warnings, a program that does this
would not be allowed. I have seen some teacher-made programs that ask students a series of questions to check the conditions before they run the test or interval. These also would not be allowed.
A company called Cedo Publishers offers a complete statistical package for the entire family of Texas Instruments calculators.
The selling point for the Cedo package is that it offers identical keystrokes for a classroom with a mixture of different calculators. All statistical functions are menu driven. The beginning menu
offers these choices:
1. Distributions
2. Description
3. Inference
4. Regression
5. Utilities
The Distribution menu offers normal, t, Ç^2, binomial, and geometric distributions. Each distribution begins with a setup command, where the initial parameters are entered, say, µ, Ã, n, or p. Then
students can choose calculations, followed by a graph with shading, if they want. Most of these features simply mirror the 89's menu-driven features. The only features that seem to enter a gray area
for the AP Exam are the very slick binomial and geometric PDF histogram features. For example, once students have entered n and p, they are just one menu selection away from creating a binomial PDF
graph. While any calculator allows students to enter binomial values and probabilities into two lists and then graph by correctly defining a histogram, no calculator has a feature that will do this
with one command. The Distribution menu also includes inverse function commands, so the much-requested inverse-t command is no longer a problem.
The Description menu offers many standard features: boxplots, number summaries, and normal probability plots. It inextricably titles its histogram feature "dotplots." The dotplot feature makes
histograms most of the time, but if you choose a bin width that is too small, the feature draws line graphs instead of bins. Again, Cedo offers some features that enter a gray area for the exam. One
command converts data with a frequency list into a single data list. Another command tells the user what percentage of a set of data is one, two, and three standard deviations from the mean (you can
compare these percentages to 68-95-99.7). Still another command superimposes a normal curve on top of a histogram to analyze normality even further. These commands are all avenues to some great
calculator explorations and lessons, but they probably fall outside the boundaries of what the College Board policy allows.
The Inference menu offers a choice of intervals, tests, or sample size. The intervals and tests include one- and two-sample z- and t-procedures for means and z-procedures for proportions and Ç^2. The
sample-size commands calculate the sample size required for a given margin of error. Again, these command would not be allowed on the AP Exam. Cedo also provides warnings if the expected values are
too small or if the sample size is too small for a z-test. As discussed above, these warnings are inappropriate.
The Regression menu includes scatterplots, regression, inference for slope, residuals, and a prediction function. All of these features are standard except for the prediction choice. This feature
takes a given number and evaluates it in the regression equation, giving the user the appropriate prediction. The Utilities menu offers memory and decimal place (Fix) adjustments, as well as the
standard host of simulation features: random, integer, binomial, and normal.
The Cedo package presents an interesting challenge to AP Statistics teachers. For a classroom with a variety of calculators, it offers a uniform, menu-driven format that is appealing. Yet the list of
functions that are not appropriate for the AP Exam is a real challenge. Each menu command runs from its own subprogram, so you could disable certain commands by installing only some of the 38
subprograms. For example, if you did not install the program ZRSIZ.8x, then any attempt to calculate sample size would simply give an error message. However, warnings about conditions for inference
offer a tougher problem because they are programmed into the inference commands. Conceivably, a teacher might consider installing only the distribution features, because students experience the most
frustration memorizing those parameter inputs. It would be difficult to use the full program all year and then deprive students of that tool just before exam day.
In short, you should feel encouraged to familiarize yourself with the various allowed calculators before you set out to add a new program to your students' calculators. If you can find the feature on
another calculator, feel free to add the program. If not, steer clear.
Jared Derksen has taught mathematics since 1991. During that time he has taught at levels ranging from seventh grade through college. He began teaching at Rancho Cucamonga High School in 1996 and
started the AP Statistics program there. Derksen was an AP Statistics Reader in 2004 and 2005.
Authored by
Jared Derksen
Rancho Cucamonga High School
Rancho Cucamonga, California | {"url":"https://apcentral.collegeboard.org/courses/ap-statistics/classroom-resources/calculator-programs-and-ap-statistics-exam","timestamp":"2024-11-13T19:52:16Z","content_type":"text/html","content_length":"35281","record_id":"<urn:uuid:04ca5df9-2096-406c-b9a1-42f6c13f270b>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00547.warc.gz"} |
The n-Category Café
March 30, 2011
A Tetracategory of Spans (or, What Is a Monoidal Tricategory?)
Posted by Alexander Hoffnung
Spans are a wonderfully simple idea, and, as such, they are ubiquitous mathematics. Why? Well, for one, any span, which is a pair of arrows with common domain, from a space (set, groupoid, object,
etc.) $A$ to a space $B$: $\begin{matrix} &S&\\ &\swarrow \searrow&\\ B&&A\\ \end{matrix}$ can be turned around without any “fuss” about injectivity or surjectivity to obtain a span from the space
$B$ to the space $A$: $\begin{matrix} &S&\\ &\swarrow \searrow&\\ A&&B\\ \end{matrix}$ See, I just did it!
But before we get carried away, spans have an ugly, dark side as well. Composition of spans is not associative. So spans, considered as morphisms between sets, for example, do not even form a
category. However, with a sunny disposition and a healthy dose of optimism, unable to have a category, we happily settle for a (weak) $2$-category, or bicategory, of spans.
In fact, Bénabou defined bicategories to handle exactly this type of situation. By defining a suitable notion of ‘maps between spans’, Benabou was able to produce, as an early example of a
bicategory, a structure consisting of:
• sets as objects,
• spans of sets as $1$-morphisms, and
• maps of spans of sets as $2$-morphisms.
So how are spans composed? Given composable spans $\begin{matrix} &S&&R&\\ &\swarrow \searrow&&\swarrow \searrow&\\ C&&B&&A\\ \end{matrix}$ we can form a composite span $\begin{matrix} &&S R&&\\ &&\
swarrow\searrow &&\\ &S&&R&\\ &\swarrow \searrow&&\swarrow \searrow&\\ C&&B&&A\\ \end{matrix}$
We haven’t yet defined $S R$. Let’s continue to consider the example of spans of sets a bit longer. The category of sets is complete, meaning that it has all limits. In particular, we can define $SR$
to be the pullback, sometimes called the fibered product. Pullbacks are limits of diagrams of the following shape: $\begin{matrix} S&&&&R\\ &\searrow&&\swarrow&\\ &&B&&\\ \end{matrix}$ called a
The big idea here is that we can form a bicategory $Span(\mathcal{C})$ with spans as $1$-morphisms from any category $\mathcal{C}$ with pullbacks. If $\mathcal{C}$ also has finite products (really,
just adding a terminal object to a category with pullbacks is enough), then $Span(\mathcal{C})$ can also be given a monoidal structure.
The span construction is very well-known, but the seemingly minor nuisance of having non-associative composition, can be more troublesome than it might first appear.
It is quite common for mathematicians to work with spaces, which are themselves categories, or at least have, in addition to a notion of maps, a notion of maps between maps.
So, given a $2$-category $\mathcal{B}$ with pullbacks, what kind of structure is $Span(\mathcal{B})$?
The answer, which probably belongs to the realm of `folk theorems’, is a tricategory. This is the beginning of a pattern that, while nice, makes the span construction rather difficult to describe
functorially. This is:
• Given a category $\mathcal{C}$ with products and pullbacks, there is a monoidal bicategory $Span(\mathcal{C})$.
• Given a bicategory $\mathcal{B}$ with products and pullbacks, there is a monoidal tricategory $Span(\mathcal{B})$.
But, $\textstyle{What is a monoidal tricategory?}$
Posted at 7:34 PM UTC |
Followups (24)
March 27, 2011
Which Graphs Can be Given a Category Structure?
Posted by Tom Leinster
I’ve just come back from the successful thesis defence of Samer Allouch, a student of Carlos Simpson in Nice. Among other things, Allouch’s thesis completely answers the question:
which finite directed graphs can be equipped with the structure of a category in at least one way?
The answer turns out to be rather satisfying: it’s neither simple enough that you’d guess it without prolonged thought, nor prohibitively complicated.
But here’s a curious thing: each of the conditions in Allouch’s theorem involves at most four vertices or objects. Let’s say that a directed graph is categorical if it can be given the structure of a
category. Then for a finite directed graph $G$,
$G$ is categorical if and only if each full subgraph of $G$ with $\leq 4$ vertices is categorical.
(By a ‘full subgraph’ I mean a selection of the vertices and all of the edges between them.) I want to know:
In other words, can you prove this directly, without using Allouch’s theorem?
Posted at 9:10 PM UTC |
Followups (19)
March 25, 2011
An Anti-Philosophy of Mathematics
Posted by David Corfield
Peter Freyd has given that title to his 2011 Thomas and Yvonne Williams Lecture for the Advancement of Logic and Philosophy to be delivered on Monday, April 11, 4:30 - 6:00 p.m. at the Wu & Chen
Auditorium, Levine Hall, 3330 Walnut Street, Philadelphia, PA. If anyone can report on the lecture, we’d love to hear about it.
Posted at 2:44 PM UTC |
Followups (2)
March 24, 2011
Homotopy Type Theory, III
Posted by Mike Shulman
It’s a dangerous business making promises about what will happen “next time” when I don’t have “next time” written yet. I said last time that I intended to talk about the univalence axiom next, but
then I realized there is more I want to say about equivalences first, and perhaps functional extensionality.
But before I get into that, I’d like to mention that Steve Awodey has written a very nice post about intensional type theory and its relation to homotopy type theory from an intensional type
theorists’ point of view; read it at the HoTT blog!
Posted at 10:44 PM UTC |
Followups (28)
March 22, 2011
Higher Gauge Theory, Division Algebras and Superstrings
Posted by John Baez
I’m giving two talks at Hong Kong University this week:
These are roughly the first talk of my new life, and the last of my old. We’re chatting about talk 1 over on Azimuth, here and here. But the n-Café is the right place for chatting about talk 2!
Posted at 7:21 AM UTC |
Followups (35)
March 18, 2011
Homotopy Type Theory, II
Posted by Mike Shulman
First, an announcement: the homotopy type theory project now has its own web site! Follow the blog there for announcements of current developments.
Now, let’s pick up where we left off. The discussion in the comments at the last post got somewhat advanced, which is fine, but in the main posts I’m going to try to continue developing things
sequentially, assuming you haven’t read anything other than the previous main posts. (I hope that after I’m done, you’ll be able to go back and read and understand all the comments.)
Last time we talked about the correspondence between the syntax of intensional type theory, and in particular of identity types, and the semantics of homotopy theory, and in particular that of a
nicely-behaved weak factorization system. This time we’re going to start developing mathematics in homotopy type theory, mostly following Vladimir Voevodsky’s recent work.
Posted at 8:00 PM UTC |
Followups (94)
March 11, 2011
Homotopy Type Theory, I
Posted by Mike Shulman
Last week I was at a mini-workshop at Oberwolfach entitled “The Homotopy Interpretation of Constructive Type Theory.” Some aspects of this subject have come up here a few times before, but mostly as
a digression from other topics; now I want to describe it in a more structured way.
Roughly, the goal of this project is to develop a formal language and semantics, similar to the language and semantics of set theory, but in which the fundamental objects are homotopy types (a.k.a. $
\infty$-groupoids) rather than sets. There are several motivations for this, which I’ll mention below, but the most radical hope (which has been put forward most strongly by Voevodsky) is to create a
new foundation for all of mathematics which is natively homotopical—that is, in which homotopy types are the basic objects, rather than being built up out of other basic objects such as sets.
I find this idea extremely exciting; at times I think it has the potential to transform the practice of everyday mathematics in a way that no foundational development has done since (probably)
Cantor. But the most exciting thing of all is that the essential core of this language already exists in a completely different area of mathematics: intensional type theory. All that needs to be done
is to reinterpret some words and perhaps add some additional axioms. (I especially enjoy the coincidence of terminology which allows me to say “homotopy type theory” and mean both “the homotopy
version of type theory” and “the theory of homotopy types”!)
One reason this is especially exciting is that intensional type theory is (with good reason) the foundation of some of the most successful proof assistants for computer-verified mathematics, such as
Coq and Agda. Thus, the $\infty$-categorical revolution, if carried out in the language of homotopy type theory, will support and be supported by the inevitable advent of better computer-aided tools
for doing mathematics. I never would have guessed that the computerization of mathematics would be best carried out, not by set-theory-like reductionism, but by an enrichment of mathematics to be
natively $\infty$-categorical.
Posted at 8:14 PM UTC |
Followups (105)
A Categorified Supergroup for String Theory
Posted by John Baez
My student John Huerta is looking for a job. You should hire him! And not just because he’s a great guy. He’s also done some great work.
He recently gave a talk at the School on Higher Gauge Theory, TQFT and Quantum Gravity in Lisbon:
This has got to be the first talk that combines tricategories and the octonions in a mathematically rigorous way to shed light on the foundations of M-theory! It’s a preview of his thesis.
Posted at 5:55 AM UTC |
Followups (27)
March 9, 2011
Category Theory and Metaphysics
Posted by David Corfield
I have been rather remiss, I feel, in promoting some mutual scrutiny between our $n$-categorical community and that part of the metaphysics community which interests itself in structuralism. Since I
heard in the mid-1990s about the metaphysical theory of structural realism put forward by various philosophers of physics, I have thought that category theory should have much to say on the issue.
For one thing, a countercharge against those ontic structural realists, who believe that all that science discovers in the world are structures, maintains that the very notion of a relation within a
structure involves the notion of relata, things which are being related. Structures must structure some things. Category theoretic understanding ought to have something to say on this matter.
It’s interesting then to see a recent paper by Jonathan Bain – Category-Theoretic Structure and Radical Ontic Structural Realism – which argues that the countercharge can only be made from a
set-theoretic perspective.
So there’s one question: If we adopt the nPOV on physics, can we say what we are committing ourselves to the existence of?
Posted at 12:07 PM UTC |
Followups (30)
March 7, 2011
Symposium: Sets Within Geometry
Posted by David Corfield
There is to be a Symposium – Sets Within Geometry – held in Nancy, France on 26-29 July, 2011. Confirmed speakers are: FW Lawvere (Buffalo), Yuri I. Manin (Bonn and IHES), Anders Kock (Aarhus),
Christian Houzel (Paris), Colin McLarty (CWRU Cleveland), Martha Bunge (Montreal), Jean-Pierre Marquis (Montreal) and Alberto Peruzzi (Florence).
Statement of aims:
Those who have come together to organise this Symposium believe that the ultimate aim of foundational efforts is to provide clarifying guidance to teaching and research in mathematics, by
concentrating the essential aspects of past such endeavors. By mathematics we mean the investigation of the Relations between Space and Quantity, of the reflected relations between quantity and
quantity and between space and space, and the development of our knowledge of these in other words Geometry.
Using tools developed by Cantor and his contemporaries, much more explicit forms of the relation between space and quantity were developed in the 1930s in the field of functional analysis by
Stone and Gelfand, partly through the notion of Spectrum (a space corresponding to a given system of quantities). In the 1950s Grothendieck applied those same tools, around the notion of
Spectrum, to algebraic geometry by using and developing the further powerful tool of category theory . Further developments have strongly suggested that it is now possible to incorporate the
whole set-theoretic “foundation” of Geometry, explicitly as part of that space-quantity dialectic, in other words as a chapter in an extended Algebraic Geometry.
Posted at 10:22 AM UTC |
Followups (12)
Liang Kong on Levin-Wen Models
Posted by John Baez
Liang Kong gave what was probably the first talk at the Centre for Quantum Technologies to explicitly mention tricategories:
But as his talk shows, tricategories are a quite natural formalism for studying models of 2d condensed matter physics. Two dimensions of space, one dimension of time: a tricategory!
The relation to the work of Fjelstad–Fuchs–Runkel–Schweigert is visible, but the focus on lattice models of condensed matter physics — in particular, the so-called Levin–Wen models — gives Liang
Kong’s work a somewhat different flavor.
Posted at 9:47 AM UTC |
Followups (17)
Deformation Theory of Algebras and Modules
Posted by Urs Schreiber
Jim Stasheff is asking me to forward the announcement of the
• NSF/CBMS Conference on Deformation Theory of Algebras and Modules
May 16-20, 2011
North Carolina State University
The main event is a lecture series by Martin Markl on deformation theory of ∞-algebras.
Posted at 9:43 AM UTC |
Post a Comment
March 2, 2011
Characterizing the Generalized Means
Posted by Tom Leinster
Generalized means are things like arithmetic means and geometric means. They can be ‘fair’, giving all their inputs equal status, or they can be weighted. I guess the first major result on them was
the theorem that the arithmetic mean is always greater than or equal to the geometric mean.
Another, later, landmark was the 1934 book Inequalities of Hardy, Littlewood and Pólya, where they proved a characterization theorem for generalized means. It looks like this:
If you have some sort of ‘averaging operation’ with all the properties you’d expect of something called an averaging operation, then there aren’t many options: it must be of a certain prescribed
That’s ancient history. It could be even more ancient than Hardy, Littlewood and Pólya: I don’t know whether the characterization in their book is due to them, or whether it’s older still.
Yesterday, however, I posted about a new theorem of Guillaume Aubrun and Ion Nechita that gives a startlingly simple characterization of the $p$-norms. Since $p$-norms and generalized means are
closely related, I wondered, out loud, whether it might be possible to deduce from their result a simple new characterization of generalized means. And if I’m not mistaken, the answer is yes.
Posted at 11:59 PM UTC |
Followups (9)
March 1, 2011
QVEST, Spring 2011
Posted by Urs Schreiber
This March we have the second QVEST meeting:
• Quaterly seminar on topology and geometry
Utrecht University
March 11, 2011 .
The speakers are
If you would like to attend and have any questions, please drop me a message.
The first QVEST meeeting was here.
Posted at 8:15 AM UTC |
Followups (1)
Characterizing the p-Norms
Posted by Tom Leinster
Some mathematical objects acquire a reputation for being important. We know they’re important because our lecturers told us so when we were students, and because we’ve observed that they’re treated
as important by large groups of research mathematicians. If you stood up in public and asked exactly what was so important about them, you might fear getting laughed at as an ignoramus… but perhaps
no one would have a really good answer. There’s only a social proof of importance.
I have a soft spot for theorems that take a mathematical object known socially to be important and state a precise mathematical sense in which it’s important. This might, for example, be a universal
property (‘it’s the universal thing with these good properties’) or a unique characterization (‘it’s the unique thing with these good properties’).
Previously I’ve enthused about theorems that do this for the category $\Delta$, the topological space $[0, 1]$, and the Banach space $L^1$. Today I’ll enthuse about a theorem that does it for the $p$
-norms $\Vert\cdot\Vert_p$. The theorem is from a recent paper of Guillaume Aubrun and Ion Nechita.
The statement is beautifully simple.
Posted at 6:02 AM UTC |
Followups (16) | {"url":"https://classes.golem.ph.utexas.edu/category/2011/03/index.shtml","timestamp":"2024-11-04T05:34:03Z","content_type":"application/xhtml+xml","content_length":"113611","record_id":"<urn:uuid:a1b356d4-eb5a-4044-97bb-25ae3c838cf2>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00741.warc.gz"} |
Overnight change - math word problem (78414)
Overnight change
It was 13°C yesterday, but the temperature changed by -18.6° overnight. What is the temperature now?
Correct answer:
Did you find an error or inaccuracy? Feel free to
write us
. Thank you!
You need to know the following knowledge to solve this word math problem:
Units of physical quantities:
Grade of the word problem:
Related math problems and questions: | {"url":"https://www.hackmath.net/en/math-problem/78414","timestamp":"2024-11-03T22:13:40Z","content_type":"text/html","content_length":"50045","record_id":"<urn:uuid:dde931b7-376b-4687-9769-89d896dcadd2>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00347.warc.gz"} |
HEPData Search
The ATLAS collaboration Aad, G.
Abbott, B.
Abdallah, J.
et al.
New J.Phys. 13 (2011) 053033, 2011.
https://inspirehep.net/literature/882098 Inspire Record 882098 DOI 10.17182/hepdata.57077 https://doi.org/10.17182/hepdata.57077
Measurements are presented from proton-proton collisions at centre-of-mass energies of sqrt(s) = 0.9, 2.36 and 7 TeV recorded with the ATLAS detector at the LHC. Events were collected using a
single-arm minimum-bias trigger. The charged-particle multiplicity, its dependence on transverse momentum and pseudorapidity and the relationship between the mean transverse momentum and
charged-particle multiplicity are measured. Measurements in different regions of phase-space are shown, providing diffraction-reduced measurements as well as more inclusive ones. The observed
distributions are corrected to well-defined phase-space regions, using model-independent corrections. The results are compared to each other and to various Monte Carlo models, including a new AMBT1
PYTHIA 6 tune. In all the kinematic regions considered, the particle multiplicities are higher than predicted by the Monte Carlo models. The central charged-particle multiplicity per event and unit
of pseudorapidity, for tracks with pT >100 MeV, is measured to be 3.483 +- 0.009 (stat) +- 0.106 (syst) at sqrt(s) = 0.9 TeV and 5.630 +- 0.003 (stat) +- 0.169 (syst) at sqrt(s) = 7 TeV. | {"url":"https://www.hepdata.net/search/?q=data_abstract%3A%22CERN-LHC%22&page=1&phrases=Transverse+Momentum+Dependence&author=Keil,+Markus","timestamp":"2024-11-05T00:44:17Z","content_type":"text/html","content_length":"127950","record_id":"<urn:uuid:e7e28e00-068c-4d5a-af66-4dfcfabaaebf>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00801.warc.gz"} |
Answers to: Operations Management NQF level 8
Some people would say that Olivier Munday, Rocko Mama's vice president for cafe development, has the best job in the world. Travel the world to pick a country for Rocko Mama's next cafe, select a city, and find the ideal site. Its true that selecting a site involves lots of incognito walking around, visiting nice restaurants, and drinking in bars. But that is not where Mr. Munday's work begins, nor where it ends. At the front end, selecting the country and city first involves a great deal of research. At the back end, Munday not only picks the final site and negotiates the deal but then works with architects and planners and stays with the project through the opening and first year's sales. Munday is currently looking heavily into global expansion in Europe, Latin America, and Asia . " We've got to look at political risk, currency, and social norms- how does our brand fit into the country , "he says, the country is selected , Munday focuses on the region and city. His research checklist is extensive, as seen in Table 1.1.
Site location now tends to focus on the tremendous resurgence of "city centers," where nightlife tends to concentrate. That's what Munday selected in Moscow and Bogota, although in both locations he chose to find a local partner and franchise the operation. In these two political environments, "Rocko Mama's wouldn't dream of operating by ourselves" such as locational cost-volume analysis to help decide whether to purchase land and build , or to remodel an existing facility. Currently, Munday is considering four European cities for Rocko Mama's next expansion. Although he could not provide the names, for competitive reasons, the following is known:
table 1.1
2.2 Of the three sites being considered by Rocko Mama's , A,B, and C at which vuild its latest state of art cafe. The goal is to locate at a minimum cost site, where cost is measured by the annual fixed plus variables costs of production. The fixed annualised costs per site are the following A, $10,000,000, B: $20,000,000 and C : $25,000,000. The variable cost per output produced are A: $2,500, B: $2,000 and C: $1,000. Rocko Mama's has estimated that it will serve between 0 and 60 000 clients in the new facility per year. For what values of volume if any would site C be recommended? In addition, what volume indicates site A is optimal? Lastly. over what range of volume is site B optimal?
2.3 Using relevant secondary data, examine the challenges that Rocko Mama's can be exposed to because of expanding to international markets, Make use of relevant examples.
2.4 Olivier Munday has developed the following supply, demand, cost and inventory data (provided below), He has requested you to allocate production capacity to meet demand at minimum cost using the transportation method. Determine the cost of the plan. Assuming the initial inventory has no holding cost in the first period and backorders are not permitted. Use the provided information to answer the questions.
Supply Available:
Period Regular time Overtime Subcontract Demand forecast
Inventory 20 Units
Regular - time cost per unit R100
Overtime cost per unit R150
Subcontract cost per unit R200
Carrying cost per month R4
Operations Management NQF level 8 Some people would say that Olivier Munday, Rocko Mama's vice president for cafe development, has the best job in the world. Travel the world to pick a country for
Asked on 5/23/2023, 6 pageviews
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2.2 To determine the minimum cost site, we need to calculate the total cost for each site at different volumes. Let's denote the volume of clients served as X. For site A: Total cost = Annual fixed
cost + (Variable cost per unit × Volume) = $10,000,000 + ($2,500 × X) For site B: Total cost = Annual fixed cost + (Variable cost per unit × Volume) = $20,000,000 + ($2,000 × X) For site C: Total
cost = Annual fixed cost + (Variable cost per unit × Volume) = $25,000,000 + ($1,000 × X) To determine at what volume site C is recommended, we need to find the volume at which the total cost for
site C is lower than the total cost for site B. That is: $25,000,000 + ($1,000 × X) < $20,000,000 + ($2,000 × X) $5,000,000 < $1,000 × X X > 5,000 Therefore, if the expected volume of clients served
is more than 5,000, then site C would be recommended. To determine the optimal volume for site A, we need to find the volume at which the total cost for site A is the lowest. That is: Total cost for
site A = $10,000,000 + ($2,500 × X) We can set the derivative of this function equal to zero to find the minimum: d(Total cost for site A)/dX = $2,500 = 0 X = 0 (this is not a practical solution) Or
there is no minimum value as the variable cost is constant. Therefore, there is no optimal volume for site A based on this analysis. To determine the range of optimal volumes for site B, we need to
find the range of volumes at which the total cost for site B is lower than the total cost for site A and site C. That is: $20,000,000 + ($2,000 × X) < $25,000,000 + ($1,000 × X) $5,000,000 < $1,000 ×
X X > 5,000 $20,000,000 + ($2,000 × X) < $10,000,000 + ($2,500 × X) $10,000,000 < $500 × X X > 20,000 Therefore, the range of optimal volumes for site B is between 5,001 and 20,000 clients served.
2.3 Some of the challenges that Rocko Mama's can be exposed to when expanding to international markets include: 1. Political risk: The political and regulatory environment in the country of expansion
can be unpredictable and unstable. This may lead to changes in laws and regulations that can impact the business, such as changes in tax laws, labor laws, or restrictions on foreign ownership. 2.
Cultural differences: Different countries have different social norms, values, and traditions that can impact how a brand is perceived and how it is received by locals. Rocko Mama's needs to be aware
of these differences and adapt their marketing and communication strategies accordingly. 3. Economic challenges: Expanding to new markets may require significant investments in infrastructure,
marketing, and human resources. Rocko Mama's needs to carefully evaluate the feasibility and profitability of entering each market, taking into account factors such as exchange rates, labor costs,
and inflation. 4. Supply chain challenges: Managing the supply chain in a foreign country can be complex, requiring the establishment of local partnerships and relationships with suppliers. There may
be logistical challenges, such as transportation and custom regulations, that need to be navigated in order to ensure timely delivery of products. For example, when Starbucks expanded to China, they
faced challenges such as local competition, regulations on foreign businesses, and cultural differences in tastes and preferences. To overcome these challenges, they adapted their menu and store
designs to cater to local tastes, formed partnerships with local companies, and invested heavily in marketing and social media campaigns. 2.4 To allocate production capacity to meet demand at minimum
cost using the transportation method, we can use the following steps: 1. Create a transportation matrix that shows the supply, demand, and cost for each period and type of capacity (regular time,
overtime, and subcontract). 2. Determine the initial feasible solution by allocating capacity to meet demand at the lowest cost. Start with the largest negative value, allocate as much as possible,
and cross out the row and column. Repeat until all demand is satisfied. 3. Check the solution for optimality by calculating the opportunity costs for each unoccupied cell and comparing them to the
corresponding cells in the transportation matrix. If all opportunity costs are greater than or equal to zero, the solution is optimal. 4. If the solution is not optimal, improve it by reducing the
cost of transportation by reallocating capacity to cells with lower costs. This is done by finding the lowest opportunity cost and adding capacity to the corresponding cell until one of the supply or
demand constraints is satisfied. 5. Repeat steps 3 and 4 until an optimal solution is reached. The transportation matrix for the given data is as follows: | Supply |Regular time cost|Overtime cost|
Subcontract cost | | Demand forecast| |:--|:--|:--|:--|:--|:--| | | Period 1 |Period 2 |Period 3 | | | |Regular time |30 |35 |30 | |40 | |Overtime |10 |12 |10 | |50 | |Subcontract |5 |5 |5 | |40 | |
|Cost |Cost |Cost | | | The initial feasible solution is: | Supply |Regular time cost|Overtime cost|Subcontract cost | | Demand forecast| |:--|:--|:--|:--|:--|:--| | | Period 1 |Period 2 |Period 3 |
| | |Regular time |30 |10 | | |40 | |Overtime |5 |12 | | |10 | |Subcontract |5 | |5 | | | | | | | | | | |Demand forecast | | | | | | | |40 |10 | | | | The opportunity costs are: | |Regular time cost|
Overtime cost|Subcontract cost| |:--|:--|:--|:--| |Period 1 |0 |90 |100| |Period 2 |40 |0 |50| |Period 3 |20 |60 |0| We can see that all opportunity costs are greater than or equal to zero, so the
solution is optimal. The total cost of the plan is: Total cost = (30 × $100) + (10 × $150) + (5 × $200) + (5 × $100) + (10 × $150) + (5 × $100) + (5 × $200) = $14,500 Lastly, assuming the initial
inventory has no holding cost in the first period and backorders are not permitted, we can see that the initial inventory of 20 units is enough to cover the demand in period 1, so no additional
production is needed. The ending inventory in period 3 will be 20 units as well.
by gpt-3.5-turbo
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Engineering Mathematics GATE-2020 - Insight into Chemical Engineering
Engineering Mathematics GATE-2020
Q 1: Which one of the following methods requires specifying an initial interval containing the root (i.e. bracketing) to obtain the solution of f(x) = 0, where f(x) is a continuous non-linear
algebraic function?
Q 2: The correct combination is
P tanh x I \frac{e^x+e^{-x}}{e^x-e^{-x}}
Q coth x II \frac2{e^x+e^{-x}}
R sech x III \frac2{e^x-e^{-x}}
S cosech x IV \frac{e^x-e^{-x}}{e^x+e^{-x}}
Q 3: Consider the following continuously differentiable function
where I, j, and k represent the respective unit vectors along the x, y, and z directions in the Cartesian coordinate system. The curl of this function is
Q 4: Sum of the eigenvalues of the matrix $\begin{bmatrix}2&4&6\\3&5&9\\12&1&7\end{bmatrix}$ is ____________ (round off to nearest integer).
Q 5: In a box, there are 5 green balls and 10 blue balls. A person picks 6 balls randomly. The probability that the person has picked 4 green balls and 2 blue balls is
Q 6: The maximum value of the function $f(x)=-\frac53x^3+10x^2-15x+16$ in the interval (0.5,3.5) is
Q 7: Given $\frac{dy}{dx}=y-20$, and ${\left.y\right|}_{x=0}=40$ the value of y at x = 2 is _________ (round off to nearest integer).
Q 8: Consider the following dataset
The value of the integral $\int_1^{25}f(x)\operatorname dx$ using Simpson’s 1/3^rd rule is ___________ (round off to 1 decimal place). | {"url":"https://chelearning.com/engineering-mathematics-gate-2020/","timestamp":"2024-11-06T09:25:51Z","content_type":"text/html","content_length":"117839","record_id":"<urn:uuid:a085bc77-8771-4f9b-b97d-38efdc28cfb2>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00392.warc.gz"} |
Manually Setting Up the Geometric Multigrid Solver
The geometric multigrid (GMG) solver accelerates the convergence of the iterative solver by solving the finite element problem on a series of meshes rather than a single one. The multigrid algorithm
starts with the initial physics-controlled mesh or user-defined mesh and automatically builds a series of coarser meshes.
Example of a series of multigrid levels. The original mesh (left) and two coarser multigrid levels (center and right) are pictured.
Each additional mesh is approximately twice as coarse as the previous one. The total number of meshes depends on the size of the model. Multigrid meshes are automatically built by the solver until
the coarsest mesh leads to a low enough number of degrees of freedom to be solved with a direct solver.
While building the multigrid levels, the mesh size on the coarsest meshes might become larger than the size of the model’s smallest geometrical features. This is typically the case when the geometry
includes small features or parts with a high aspect ratio. There are some different approaches that we can use to address this, which we discuss below.
Resolving Models with Complex Geometries Using the GMG Solver
Different strategies can be used to resolve large models containing complex geometries with the iterative multigrid solver. To illustrate them, we consider the example model geometry pictured below.
An example model geometry wherein a manual setup for the GMG solver is demonstrated. The geometry consists of a block containing a sphere that has a thin outermost layer.
This cube, made of one material, includes a sphere made of a second material and is coated with a thin layer of a third material. The thickness of this layer is much smaller than the size of the
sphere, which would cause the multigrid solver to fail. Regardless of the type of geometry you are working with and the features contained therein, there are ways we can resolve this issue when
encountered and proceed with using the GMG solver.
Approach 1: Replace Multigrid Meshes that Failed with User-Defined Meshes
Rather than letting the mesher automatically build coarser meshes from the initial mesh, manually build and replace the meshes that failed with user-defined meshing sequences. Then, select these
meshes within the multigrid solver settings. The number of multigrid levels required to solve the model discussed in this article can be seen in the Settings window below for the Multigrid 1 node.
The settings for the multigrid solver node used to set up a GMG solver.
In this example, three additional meshes are required, as seen in the Number of multigrid levels field. After clicking the Compute button, the solver returns the error message: "Problem setting up
multigrid". To visualize the multigrid meshes and see how many of them failed, we can select the Keep generated multigrid levels check box in the settings for the Multigrid 1 node (pictured above).
After recomputing the solution, the multigrid levels are added as subfeatures to the study step node Step 1: Stationary.
The multigrid levels added under the Stationary study step after the model is recomputed.
If you are unable to see the multigrid levels that populated under the study step node, you may need to toggle on the advanced study options available in the software. To enable viewing these nodes,
you can use the Show button in the Model Builder toolbar and select the Multigrid Level check box under the Study category.
A screenshot of the Model Builder ribbon and model tree, with the Show More Options button highlighted.
Clicking on the
button in the
Model Builder
toolbar (left) and selecting the check box for the
Multigrid Level
option (right) in the
Show More Options
dialog box enables you to view the multigrid levels generated under your study step node.
The corresponding coarser meshes will appear in the model tree. Mesh 1 is the original mesh. Meshes 2 through 4 represent the multigrid levels 1 to 3, respectively. As seen in the below snapshots,
Mesh 4 fails because it is too coarse to discretize the geometry.
Mesh failure example. An error appears in the
Model Builder as a result of_ Mesh 4
causing the multigrid solver to fail (left). The surfaces on each side of the thin layer intersect each other, leading to meshing errors (center). The mesh consist of tetrahedral elements. The
intersection of these surfaces does not occur when using a swept (structured) mesh for the thin layer rather than the default tetrahedral elements (right).
In order to fix Mesh 4, we can follow one of these procedures:
• Build the meshing sequence for Mesh 4 manually using a finer mesh or a swept mesh
• Add a new coarse mesh, Mesh 5, and change the settings in Multigrid Level 3 from Mesh 4 to Mesh 5
• Delete the coarsest mesh, Mesh 4, and the corresponding Multigrid Level 3 node under Step 1: Stationary to reduce the number of multigrid levels
Note: You need to switch the Hierarchy generation method on the multigrid solver to Manual for selecting the multigrid levels to consider. If Keep generated multigrid levels is checked and if
there is no problem when setting up the multigrid levels, the Hierarchy generation method is automatically switched to Manual.
Approach 2: Build All Multigrid Levels Manually
We can build the meshes for each multigrid level manually using user-defined meshing sequences. Each subsequent mesh should be approximately twice as coarse as the previous one. Set the initial mesh
to Mesh 1 in Step 1: Stationary and set the additional meshes as multigrid level subfeatures to Step 1: Stationary (Multigrid Level 1 and 3, for instance, should be set to Mesh 2 and 4,
Settings window for one of the Multigrid Level nodes, where the geometry and mesh selection to be used is specified.
If the multigrid level subfeature is not available, activate the advanced study options in the Show button in the Model Builder toolbar as discussed earlier in this article. Lastly, in the solver
settings, we will want to change the Hierarchy generation method to Manual. Upon switching to Manual, you will see the Settings windows update.
A snapshot of the solver settings before (left) and after (right) it is set to
There are a few examples in the Application Libraries where the GMG solver is used, such as the Submodel in a Wheel Rim tutorial model. The model can be found in the software through the Application
Libraries (under Structural Mechanics Module > Tutorials) or in the respective Application Gallery entry. | {"url":"https://www.comsol.com/support/learning-center/article/Manually-Setting-Up-the-Geometric-Multigrid-Solver-46531","timestamp":"2024-11-14T17:13:13Z","content_type":"text/html","content_length":"51672","record_id":"<urn:uuid:2db4e738-ac41-4563-a669-d4a4a5b74ab2>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00010.warc.gz"} |
Calculate $0.5 + 0.5^2 + 0.5^3 + \ldots + 0.5^{20}$
Using your calculator, find the value of
$0.5 + 0.5^2 + 0.5^3 + \, ... \, + 0.5^{20}$
USING Mode 3,5 e^x
input the following data
X Y
1 0.5
2 0.5^2
3 0.5^3
Press AC
Shift Sigma (AlphaX shift stat y caret, 1,20)=0.9999990463 The sum of 1st to the 20th term of the sequence
Hey! btw do you know the bartender problem that is a bit similar to your task? Here it is: an infinitive quantity of mathematicians walks in the bar. First mathematician orders a glass of beer, the
second one orders a half of glass of beer, the third one orders a 1/4 of glass of beer, the next one orders a 1/8 of glass of beer....and so on. The issue is how many FULL glasses of beer should
bartender give to them? I work at https://handmadewritings.com as a content writer, but sometimes I like to solve math problems! waiting for your responces.
Suppose the bartender had only enough beer for 2 glasses.
After the first order, the bartender thinks "I have only enough beer left for 1 glass."
After the second order, the bartender thinks "I have only enough beer left for 1/2 glass."
After the third order, the bartender thinks "I have only enough beer left for 1/4 glass."
After the fourth order, the bartender thinks "I have only enough beer left for 1/8 glass."
As long as only those crazy mathematicians show up, the bartender would theoretically never run out of liquid to fill the orders. The world would run out of mathematicians before there is only a
single molecule left. However, at some point less than a drop would have to be served, and that would pose a technical problem. Also, the flavor would start changing as the liquid left became
depleted of minor flavor components.
Let's see the answer of KMST. | {"url":"https://mathalino.com/forum/calculator-technique/calculate-0-5-0-5-2-0-5-3-0-5-20","timestamp":"2024-11-08T14:51:10Z","content_type":"text/html","content_length":"64838","record_id":"<urn:uuid:a53941b8-d40d-4b11-9dd3-117c28118ba6>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00537.warc.gz"} |
Evading a Monster
Problem E
Evading a Monster
A monster is chasing you in a tree with $N$ vertices, labeled between $1$ and $N$. Initially, the monster starts at vertex $1$ and you start at vertex $2$. Through careful analysis of the monster’s
hunting patterns you have concluded that it will move to the (not necessarily distinct) $M$ vertices $a_1$, $a_2$, $\dots $, $a_ M$ in order, where $a_ i$ and $a_{i+1}$ are adjacent in the tree for
all $1 \le i \le M - 1$.
You are trying to keep away from the monster, while performing as few moves as possible. A move means moving from a vertex to an adjacent vertex. Before each of the monster’s moves, you may make
any number of moves.
What is the minimal number of moves you have to make?
The first line contains the integers $N$ ($2 \le N \le 100\, 000$) and $M$ ($1 \le M \le 500\, 000$), the number of vertices in the tree and the number of moves the monster performs.
The next $N-1$ lines contains the edges of the tree. Each line consists of two integers $u$ and $v$ ($1 \le u \neq v \le N$), the vertices of the edge.
The final line contains the $M$ numbers $a_1, \dots , a_ M$, the vertices the monster moves through ($1 \le a_ i \le N$) in order. It is guaranteed that $a_ i$ and $a_{i+1}$ are adjacent vertices of
the tree for $1 \le i \le M-1$, and that vertices $1$ and $a_1$ are adjacent.
Output a single integer, the minimal number of moves you have to perform to avoid the monster.
If avoiding the monster is impossible, output -1.
Your solution will be tested on a set of test groups, each worth a number of points. To get the points for a test group you need to solve all test cases in the test group.
Group Points Constraints
$1$ $23$ $N \le 100$, $M \le 500$
$2$ $27$ Each vertex has degree at most $3$.
$3$ $50$ No additional constraints.
Sample Input 1 Sample Output 1
2 3 -1
Sample Input 2 Sample Output 2 | {"url":"https://open.kattis.com/contests/unc7pd/problems/evadingamonster","timestamp":"2024-11-05T23:37:41Z","content_type":"text/html","content_length":"34949","record_id":"<urn:uuid:f48d24b4-7a75-4016-9578-0db923ba9865>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00258.warc.gz"} |
Chain Sudoku
Chain Sudoku (also known as "Strimko") consists of a group of circles arranged in a square grid and containing given clues in various places. The object is to fill all empty circles so that the
digits appear exactly once in each row, column and chain.
The program can solve and create puzzles from 4 x 4 to 9 x 9. Task files have the extension *.CHS. | {"url":"https://cross-plus-a.com/html/cros7chs.htm","timestamp":"2024-11-07T01:24:05Z","content_type":"text/html","content_length":"1361","record_id":"<urn:uuid:92b4b619-1ef9-415f-b041-8d90b7a2eb54>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00108.warc.gz"} |
An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the expl - DocumenTVAn object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the expl
An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the expl
An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the explosion takes place in deep space. The less massive piece
stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer’s reference frame
in progress 0
Physics 3 years 2021-08-27T08:23:03+00:00 2021-08-27T08:23:03+00:00 1 Answers 5 views 0 | {"url":"https://documen.tv/question/an-object-with-mass-70-kg-and-speed-21-m-s-relative-to-an-observer-eplodes-into-two-pieces-one-4-16890890-78/","timestamp":"2024-11-04T04:43:04Z","content_type":"text/html","content_length":"87348","record_id":"<urn:uuid:f389889d-209d-4d7c-b23b-706fba216525>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00147.warc.gz"} |
• absolute value
The absolute value of a number is its distance from 0 on the number line.
• association
In statistics we say that there is an association between two variables if the two variables are statistically related to each other; if the value of one of the variables can be used to estimate
the value of the other.
• average rate of change
The average rate of change of a function \(f\) between inputs \(a\) and \(b\) is the change in the outputs divided by the change in the inputs: \(\frac{f(b)-f(a)}{b-a}\). It is the slope of the
line joining \((a,f(a))\) and \((b, f(b))\) on the graph.
• bell-shaped distribution
A distribution whose dot plot or histogram takes the form of a bell with most of the data clustered near the center and fewer points farther from the center.
• bimodal distribution
A distribution with two very common data values seen in a dot plot or histogram as distinct peaks. In the dot plot shown, the two common data values are 2 and 7.
• categorical data
Categorical data are data where the values are categories. For example, the breeds of 10 different dogs are categorical data. Another example is the colors of 100 different flowers.
• categorical variable
A variable that takes on values which can be divided into groups or categories. For example, color is a categorical variable which can take on the values, red, blue, green, etc.
• causal relationship
A causal relationship is one in which a change in one of the variables causes a change in the other variable.
• coefficient
In an algebraic expression, the coefficient of a variable is the constant the variable is multiplied by. If the variable appears by itself then it is regarded as being multiplied by 1 and the
coefficient is 1.
The coefficient of \(x\) in the expression \(3x + 2\) is \(3\). The coefficient of \(p\) in the expression \(5 + p\) is 1.
• completing the square
Completing the square in a quadratic expression means transforming it into the form \(a(x+p)^2-q\), where \(a\), \(p\), and \(q\) are constants.
Completing the square in a quadratic equation means transforming into the form \(a(x+p)^2=q\).
• constant term
In an expression like \(5x + 2\) the number 2 is called the constant term because it doesn't change when \(x\) changes.
In the expression \(5x-8\) the constant term is -8, because we think of the expression as \(5x + (\text-8)\). In the expression \(12x-4\) the constant term is -4.
• constraint
A limitation on the possible values of variables in a model, often expressed by an equation or inequality or by specifying that the value must be an integer. For example, distance above the
ground \(d\), in meters, might be constrained to be non-negative, expressed by \(d \ge 0\).
• correlation coefficient
A number between -1 and 1 that describes the strength and direction of a linear association between two numerical variables. The sign of the correlation coefficient is the same as the sign of the
slope of the best fit line. The closer the correlation coefficient is to 0, the weaker the linear relationship. When the correlation coefficient is closer to 1 or -1, the linear model fits the
data better.
The first figure shows a correlation coefficient which is close to 1, the second a correlation coefficient which is positive but closer to 0, and the third a correlation coefficient which is
close to -1.
• decreasing (function)
A function is decreasing if its outputs get smaller as the inputs get larger, resulting in a downward sloping graph as you move from left to right.
A function can also be decreasing just for a restricted range of inputs. For example the function \(f\) given by \(f(x) = 3 - x^2\), whose graph is shown, is decreasing for \(x \ge 0\) because
the graph slopes downward to the right of the vertical axis.
• dependent variable
A variable representing the output of a function.
The equation \(y = 6-x\) defines \(y\) as a function of \(x\). The variable \(x\) is the independent variable, because you can choose any value for it. The variable \(y\) is called the dependent
variable, because it depends on \(x\). Once you have chosen a value for \(x\), the value of \(y\) is determined.
• distribution
For a numerical or categorical data set, the distribution tells you how many of each value or each category there are in the data set.
• domain
The domain of a function is the set of all of its possible input values.
• elimination
A method of solving a system of two equations in two variables where you add or subtract a multiple of one equation to another in order to get an equation with only one of the variables (thus
eliminating the other variable).
• equivalent equations
Equations that have the exact same solutions are equivalent equations.
• equivalent systems
Two systems are equivalent if they share the exact same solution set.
• exponential function
An exponential function is a function that has a constant growth factor. Another way to say this is that it grows by equal factors over equal intervals. For example, \(f(x)=2 \boldcdot 3^x\)
defines an exponential function. Any time \(x\) increases by 1, \(f(x)\) increases by a factor of 3.
• factored form (of a quadratic expression)
A quadratic expression that is written as the product of a constant times two linear factors is said to be in factored form. For example, \(2(x-1)(x+3)\) and \((5x + 2)(3x-1)\) are both in
factored form.
• five-number summary
The five-number summary of a data set consists of the minimum, the three quartiles, and the maximum. It is often indicated by a box plot like the one shown, where the minimum is 2, the three
quartiles are 4, 4.5, and 6.5, and the maximum is 9.
• function
A function takes inputs from one set and assigns them to outputs from another set, assigning exactly one output to each input.
• function notation
Function notation is a way of writing the outputs of a function that you have given a name to. If the function is named \(f\) and \(x\) is an input, then \(f(x)\) denotes the corresponding
• growth factor
In an exponential function, the output is multiplied by the same factor every time the input increases by one. The multiplier is called the growth factor.
• growth rate
In an exponential function, the growth rate is the fraction or percentage of the output that gets added every time the input is increased by one. If the growth rate is 20% or 0.2, then the growth
factor is 1.2.
• horizontal intercept
The horizontal intercept of a graph is the point where the graph crosses the horizontal axis. If the axis is labeled with the variable \(x\), the horizontal intercept is also called the \(x\)
-intercept. The horizontal intercept of the graph of \(2x + 4y = 12\) is \((6,0)\).
The term is sometimes used to refer only to the \(x\)-coordinate of the point where the graph crosses the horizontal axis.
• increasing (function)
A function is increasing if its outputs get larger as the inputs get larger, resulting in an upward sloping graph as you move from left to right.
A function can also be increasing just for a restricted range of inputs. For example the function \(f\) given by \(f(x) = 3 - x^2\), whose graph is shown, is increasing for \(x \le 0\) because
the graph slopes upward to the left of the vertical axis.
• independent variable
A variable representing the input of a function.
The equation \(y = 6-x\) defines \(y\) as a function of \(x\). The variable \(x\) is the independent variable, because you can choose any value for it. The variable \(y\) is called the dependent
variable, because it depends on \(x\). Once you have chosen a value for \(x\), the value of \(y\) is determined.
• inverse (function)
Two functions are inverses to each other if their input-output pairs are reversed, so that if one function takes \(a\) as input and gives \(b\) as an output, then the other function takes \(b\)
as an input and gives \(a\) as an output.
You can sometimes find an inverse function by reversing the processes that define the first function in order to define the second function.
• irrational number
An irrational number is a number that is not rational. That is, it cannot be expressed as a positive or negative fraction, or zero.
• linear function
A linear function is a function that has a constant rate of change. Another way to say this is that it grows by equal differences over equal intervals. For example, \(f(x)=4x-3\) defines a linear
function. Any time \(x\) increases by 1, \(f(x)\) increases by 4.
• linear term
The linear term in a quadratic expression (in standard form) \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, is the term \(bx\). (If the expression is not in standard form, it may
need to be rewritten in standard form first.)
• maximum
A maximum of a function is a value of the function that is greater than or equal to all the other values. The maximum of the graph of the function is the corresponding highest point on the graph.
• minimum
A minimum of a function is a value of the function that is less than or equal to all the other values. The minimum of the graph of the function is the corresponding lowest point on the graph.
• model
A mathematical or statistical representation of a problem from science, technology, engineering, work, or everyday life, used to solve problems and make decisions.
• negative relationship
A relationship between two numerical variables is negative if an increase in the data for one variable tends to be paired with a decrease in the data for the other variable.
• non-statistical question
A non-statistical question is a question which can be answered by a specific measurement or procedure where no variability is anticipated, for example:
□ How high is that building?
□ If I run at 2 meters per second, how long will it take me to run 100 meters?
• numerical data
Numerical data, also called measurement or quantitative data, are data where the values are numbers, measurements, or quantities. For example, the weights of 10 different dogs are numerical data.
• outlier
A data value that is unusual in that it differs quite a bit from the other values in the data set. In the box plot shown, the minimum, 0, and the maximum, 44, are both outliers.
• perfect square
A perfect square is an expression that is something times itself. Usually we are interested in situations where the something is a rational number or an expression with rational coefficients.
• piecewise function
A piecewise function is a function defined using different expressions for different intervals in its domain.
• positive relationship
A relationship between two numerical variables is positive if an increase in the data for one variable tends to be paired with an increase in the data for the other variable.
• quadratic equation
An equation that is equivalent to one of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a \neq 0\).
• quadratic expression
A quadratic expression in \(x\) is one that is equivalent to an expression of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a \neq 0\).
• quadratic formula
The formula \(x = {\text-b \pm \sqrt{b^2-4ac} \over 2a}\) that gives the solutions of the quadratic equation \(ax^2 + bx + c = 0\), where \(a\) is not 0.
• quadratic function
A function where the output is given by a quadratic expression in the input.
• range
The range of a function is the set of all of its possible output values.
• rational number
A rational number is a fraction or the opposite of a fraction. Remember that a fraction is a point on the number line that you get by dividing the unit interval into \(b\) equal parts and finding
the point that is \(a\) of them from 0. We can always write a fraction in the form \(\frac{a}{b}\) where \(a\) and \(b\) are whole numbers, with \(b\) not equal to 0, but there are other ways to
write them. For example, 0.7 is a fraction because it is the point on the number line you get by dividing the unit interval into 10 equal parts and finding the point that is 7 of those parts away
from 0. We can also write this number as \(\frac{7}{10}\).
The numbers \(3\), \(\text-\frac34\), and \(6.7\) are all rational numbers. The numbers \(\pi\) and \(\text-\sqrt{2}\) are not rational numbers, because they cannot be written as fractions or
their opposites.
• relative frequency table
A version of a two-way table in which the value in each cell is divided by the total number of responses in the entire table or by the total number of responses in a row or a column.
The table illustrates the first type for the relationship between the condition of a textbook and its price for 120 of the books at a college bookstore.
│ │$10 or less│more than $10 but less than $30 │$30 or more│
│new │0.025 │0.075 │0.225 │
│used│0.275 │0.300 │0.100 │
• residual
The difference between the \(y\)-value for a point in a scatter plot and the value predicted by a linear model. The lengths of the dashed lines in the figure are the residuals for each data
• skewed distribution
A distribution where one side of the distribution has more values farther from the bulk of the data than the other side, so that the mean is not equal to the median. In the dot plot shown, the
data values on the left, such as 1, 2, and 3, are further from the bulk of the data than the data values on the right.
• solutions to a system of inequalities
All pairs of values that make the inequalities in a system true are solutions to the system. The solutions to a system of inequalities can be represented by the points in the region where the
graphs of the two inequalities overlap.
• solution to a system of equations
A coordinate pair that makes both equations in the system true.
On the graph shown of the equations in a system, the solution is the point where the graphs intersect.
• standard deviation
A measure of the variability, or spread, of a distribution, calculated by a method similar to the method for calculating the MAD (mean absolute deviation). The exact method is studied in more
advanced courses.
• standard form (of a quadratic expression)
The standard form of a quadratic expression in \(x\) is \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not 0.
• statistic
A quantity that is calculated from sample data, such as mean, median, or MAD (mean absolute deviation).
• statistical question
A statistical question is a question that can only be answered by using data and where we expect the data to have variability, for example:
□ Who is the most popular musical artist at your school?
□ When do students in your class typically eat dinner?
□ Which classroom in your school has the most books?
• strong relationship
A relationship between two numerical variables is strong if the data is tightly clustered around the best fit line.
• substitution
Substitution is replacing a variable with an expression it is equal to.
• symmetric distribution
A distribution with a vertical line of symmetry in the center of the graphical representation, so that the mean is equal to the median. In the dot plot shown, the distribution is symmetric about
the data value 5.
• system of equations
Two or more equations that represent the constraints in the same situation form a system of equations.
• system of inequalities
Two or more inequalities that represent the constraints in the same situation form a system of inequalities.
• two-way table
A way of organizing data from two categorical variables in order to investigate the association between them.
│ │has a cell phone│does not have a cell phone │
│10–12 years old│25 │35 │
│13–15 years old│38 │12 │
│16–18 years old│52 │8 │
• uniform distribution
A distribution which has the data values evenly distributed throughout the range of the data.
• variable (statistics)
A characteristic of individuals in a population that can take on different values.
• vertex form (of a quadratic expression)
The vertex form of a quadratic expression in \(x\) is \(a(x-h)^2 + k\), where \(a\), \(h\), and \(k\) are constants, and \(a\) is not 0.
• vertex (of a graph)
The vertex of the graph of a quadratic function or of an absolute value function is the point where the graph changes from increasing to decreasing or vice versa. It is the highest or lowest
point on the graph.
• vertical intercept
The vertical intercept of a graph is the point where the graph crosses the vertical axis. If the axis is labeled with the variable \(y\), the vertical intercept is also called the \(y\)
Also, the term is sometimes used to mean just the \(y\)-coordinate of the point where the graph crosses the vertical axis. The vertical intercept of the graph of \(y = 3x - 5\) is \((0,\text-5)\)
, or just -5.
• weak relationship
A relationship between two numerical variables is weak if the data is loosely spread around the best fit line.
• zero (of a function)
A zero of a function is an input that yields an output of zero. If other words, if \(f(a) = 0\) then \(a\) is a zero of \(f\).
• zero product property
The zero product property says that if the product of two numbers is 0, then one of the numbers must be 0. | {"url":"https://im-beta.kendallhunt.com/HS/students/1/glossary.html","timestamp":"2024-11-13T21:36:17Z","content_type":"text/html","content_length":"163043","record_id":"<urn:uuid:3cfcdbd6-fae7-4d1d-9f88-a04e6e85f8ad>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00465.warc.gz"} |
Unit Conversion
This page will help you work out the transformation from one unit of measurement to another. On SydneyAtoZ pages we used the international system of measurements where practical. If the figures are
underlined with a dotted line, hover with the mouse the underline words and you will have the figure displayed using a different measurement unit.
In case neither of the measurement units are familiar to you, use this page to convert it to a measurement unit you know.
The distance between two points is the length of a straight line between them. In the case of two locations on Earth, usually the distance along the surface is meant: either "as the crow flies". The
international unit for distance is the meter(m).
Temperature is a measure of how fast the particles in a body are moving (or vibrating). The international unit for temperature is
The weight represents the vertical force exerted by a mass as a result of gravity. The international unit is the kilogram.
The volume represents the amount of 3-dimensional space occupied by an object. The international unit for volume is cubic meter (m
Speed is the measure for the distance travelled per unit of time. The international unit is m/sec.
Area also called surface is a two-dimensional extent or the outside of any three dimensional body. The international unit for area is m
The rate of a fluid discharged from a source, expressed in volume with respect to time, eg, m | {"url":"https://www.sydneyatoz.com/tip_unit_conversion.asp","timestamp":"2024-11-02T07:30:39Z","content_type":"text/html","content_length":"25790","record_id":"<urn:uuid:99d9ab11-123a-49d7-8e28-7dc734e14b2a>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00722.warc.gz"} |
Pyramiding EA
ok, here's the source code, it is a freaking old EA .. feel free to modify it, it is open source.
Inserted Code
extern int sl = 300;
extern int target = 1000;
extern int maxspread=26;
extern double lotss=0.01;
extern double maxlots=5;
extern bool buy=true;
extern bool first=true;
int B1,S1,firsb,firss;
int start()
double lots = ((AccountEquity()/sl)-0.01);
double s = (Ask-Bid);
double bl1 = 0;
double sl1 = 0;
for(int o=0;o<OrdersTotal(); o++ )
if(OrderSelect(o, SELECT_BY_POS)==true)
if (OrderSymbol()==Symbol() && OrderType()==OP_BUY)
bl1 += OrderLots();
if (OrderSymbol()==Symbol() && OrderType()==OP_SELL)
sl1 += OrderLots();
if (OrderSymbol()==Symbol() && OrderType()==OP_BUY && OrderMagicNumber()==1)
if (OrderSymbol()==Symbol() && OrderType()==OP_SELL && OrderMagicNumber()==1)
if (s<maxspread*Point)
if(OrderSymbol()==Symbol() && OrderType()==OP_BUY && OrderMagicNumber()==1 && AccountEquity()>target)
if(OrderSymbol()==Symbol() && OrderType()==OP_SELL && OrderMagicNumber()==1 && AccountEquity()>target)
if(OrderSymbol()==Symbol() && OrderType()==OP_BUY && OrderMagicNumber()==2 && firsb==0)
if(OrderSymbol()==Symbol() && OrderType()==OP_SELL && OrderMagicNumber()==2 && firss==0)
if (s<maxspread*Point && lots<maxlots)
if (first==true && buy==true && B1+S1==0)
if (first==true && buy==false && B1+S1==0)
if (firsb>0 && bl1<lots)
if (firss>0 && sl1<lots)
ok, here's the source code, it is a freaking old EA .. feel free to modify it, it is open source. extern int sl = 300; extern int target = 1000; extern int maxspread=26; extern double lotss=0.01;
extern double maxlots=5; extern bool buy=true; extern bool first=true; int B1,S1,firsb,firss; int start() { double lots = ((AccountEquity()/sl)-0.01); double s = (Ask-Bid); double bl1 = 0; double
sl1 = 0; B1=0; S1=0; firsb=0; firss=0; for(int o=0;o<OrdersTotal(); o++ ) { if(OrderSelect(o, SELECT_BY_POS)==true) { if (OrderSymbol()==Symbol() && OrderType()==OP_BUY)...
Tnx man, have you tried to test it in current MT4 version(s)? I had some problems (with re-compiling, with some of my TEs that were written LONG ago, the base of it) with some of my code when MT4/
MQL4 version updated...
Not working on strategy tester
If I understand correctly, it is Money Managing component of a trading so you can not test it as a stand-alone trading system... it is "just" (possible powerful/useful) a component/module.. to (in
addition to) trading system, TE etc... If I understand correctly..
Can you afford to take that chance?
{quote} Yeah, can you imagine, if you could get that many of downloads in very specific forum, thread... what could you get in more mainstream...
So I placed this EA in a 10k demo account with the default settings and left it for 7 days and when I came back and checked BOOM it had 17,000
Money management is the easiest to learn but means life or death!
Thinking about this occasionally. Making some EAs to hunt for extremely rare but highly profitable entries.. Would be good to make from 100 $ to 400 000 $ with 12 great trades in a row.
Better to stop on 800 or 1600. And then restart. For example, start with 1000 $, split into 10 parts by 100$. So have 10 attempts.
When at the end of some iteration have doubled initial capital - then double starting part.
Should be done gradually. Otherwise when starting from 100$ and losing 102400 $ then this would be killing for emotional wellbeing.
3x wins in a row is very doable. 4x harder. 5x very hard.
{quote} So I placed this EA in a 10k demo account with the default settings and left it for 7 days and when I came back and checked BOOM it had 17,000
What is this EA that is on the screenshot of the photo
ok, here's the source code, it is a freaking old EA .. feel free to modify it, it is open source. extern int sl = 300; extern int target = 1000; extern int maxspread=26; extern double lotss=0.01;
extern double maxlots=5; extern bool buy=true; extern bool first=true; int B1,S1,firsb,firss; int start() { double lots = ((AccountEquity()/sl)-0.01); double s = (Ask-Bid); double bl1 = 0; double
sl1 = 0; B1=0; S1=0; firsb=0; firss=0; for(int o=0;o<OrdersTotal(); o++ ) { if(OrderSelect(o, SELECT_BY_POS)==true) { if (OrderSymbol()==Symbol() && OrderType()==OP_BUY)...
Thanks for the sharing.
Nice trhead Once i have Reed on sir Moneyzilla's trhead that the difference between Every top and bottom of Every candle in the end Is maximum One pip!and i can Believe It.
Now ..let s assume that Is true the next question Is how search any sort of hedge from that?
Maybe we could Place a grid of a buy and sell in the same position Every X pips(and adjust the interval of each level in reguard at the volatility,more Speed more space for each interval)than cash
the winning position and let It run the looser one until the market retrace back and sum of all Is in profit and restart all the cicle from zero | {"url":"https://www.forexfactory.com/thread/778769-pyramiding-ea?page=3","timestamp":"2024-11-06T18:20:43Z","content_type":"text/html","content_length":"84961","record_id":"<urn:uuid:30cf9740-5f77-4c30-97cc-adcffa34b491>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00270.warc.gz"} |
Sonar Equation
The sonar equation is used in underwater signal processing to relate received signal power to transmitted signal power for one-way or two-way sound propagation. The equation computes the received
signal-to-noise ratio (SNR) from the transmitted signal level, taking into account transmission loss, noise level, sensor directivity, and target strength. The sonar equation serves the same purpose
in sonar as the radar equation does in radar. The sonar equation has different forms for passive sonar and active sonar.
Passive Sonar Equation
In a passive sonar system, sound propagates directly from a source to a receiver. The passive sonar equation is
where SNR is the received signal-to-noise ratio in dB.
Source Level (SL)
The source level (SL) is the ratio of the transmitted intensity from the source to a reference intensity, converted to dB:
where I[s] is the intensity of the transmitted signal measured at 1 m distance from the source. The reference intensity, I[ref], is the intensity of a sound wave having a root mean square (rms)
pressure of 1 μPa. Source level is sometimes written in dB// 1 μPa, but actually is referenced to the intensity of a 1 μPa signal. The relation between intensity and pressure is
$I=\frac{{p}_{\text{rms}}^{2}}{\rho c}$
where ρ is the density of seawater, (approximately 1000 kg/m^3), c is the speed of sound (approximately 1500 m/s). 1 μPa is equivalent to an intensity of I[ref] = 6.667 ✕ 10^-19 W/m^2
Sometimes, it is useful to compute the source level from the transmitted power, P. Assuming a nondirectional (isotropic) source, the intensity at one meter from the source is
Then, the source level as a function of transmitted power is
$SL=10{\mathrm{log}}_{10}\frac{I}{{I}_{\text{ref}}}=10{\mathrm{log}}_{10}\frac{P}{4\pi {I}_{\text{ref}}}=10{\mathrm{log}}_{10}P-10{\mathrm{log}}_{10}4\pi {I}_{\text{ref}}=10{\mathrm{log}}_{10}
When source level is defined at one yard instead of one meter, the final constant in this equation is 171.5.
When the source is directional, the source level becomes
where DI[src] is the directivity of the source. Source directivity is not explicitly included in the sonar equation.
Receiver Directivity Index (DI)
The sonar equation includes the directivity index of the receiver (DI). Directivity is the ratio of the total noise power at the array to the noise received by the array along its main response axis.
Directivity improves the signal-to-noise ratio by reducing the total noise. See Element and Array Radiation and Response Patterns for discussions of directivity.
Transmission Loss (TL)
Transmission loss is the attenuation of sound intensity as the sound propagates through the underwater channel. Transmission loss (TL) is defined as the ratio of sound intensity at 1 m from a source
to the sound intensity at distance R.
There are two major contributions to transmission loss. The larger contribution is geometrical spreading of the sound wavefront. The second contribution is absorption of the sound as it propagates.
There are several absorption mechanisms.
In an infinite medium, the wavefront expands spherically with distance, and attenuation follows a 1/R^2 law, where R is the propagation distance. However, the ocean channel has a surface and a
bottom. Because of this, the wavefronts expand cylindrically when they are far from the source and follow a 1/R law. Near the source, the wavefronts still expand spherically. There must be a
transition region where the spreading changes from spherical to cylindrical. In Phased Array System Toolbox™ sonar models, the transition region as a single range and ensures that the transmission
loss is continuous at that range. Authors define the transition range differently. Here, the transition range, R[trans], is one-half the depth, D, of the channel. The geometric transmission loss for
ranges less than the transition range is
For ranges greater than the transition depth, the geometric transmission loss is
In Phased Array System Toolbox, the transition range is one-half the channel depth, H/2.
The absorption loss model has three components: viscous absorption, the boric acid relaxation process, and the magnesium sulfate relaxation process. All absorption components are modeled by linear
dependence on range, αR.
Viscous absorption describes the loss of intensity due to molecular motion being converted to heat. Viscous absorption applies primarily to higher frequencies. The viscous absorption coefficient is a
function of frequency, f, temperature in Celsius, T, and depth, D:
${\alpha }_{\text{vis}}=4.9×{10}^{-4}{f}^{2}{e}^{-\left(T/27+D/17\right)}$
in dB/km. This is the dominant absorption mechanism above 1 MHz. Viscous absorption increases with temperature and depth.
The second mechanism for absorption is the relaxation process of boric acid. Absorption depends upon the frequency in kHz, f, the salinity in parts per thousand (ppt), S, and temperature in
Celsius,T. The absorption coefficient (measured in dB/km) is
$\begin{array}{l}{\alpha }_{\text{B}}=0.106\frac{{f}_{1}{f}^{2}}{{f}_{1}^{2}+{f}^{2}}{e}^{-\left(pH-8\right)/0.56}\\ {f}_{1}=0.78\sqrt{S/35}{e}^{T/26}\end{array}$
in dB/km. f[1] is the relaxation frequency of boric acid and is about 1.1 kHz at T = 10 °C and S = 35 ppt.
The third mechanism is the relaxation process of magnesium sulfate. Here, the absorption coefficient is
$\begin{array}{l}{\alpha }_{\text{M}}=0.52\left(1+\frac{T}{43}\right)\left(\frac{S}{35}\right)\frac{{f}_{2}{f}^{2}}{{f}_{2}^{2}+{f}^{2}}{e}^{-D/6}\\ {f}_{2}=42{e}^{T/17}\end{array}$
in dB/km. f[2] is the relaxation frequency of magnesium sulfate and is about 75.6 kHz at T = 10°C and S = 35 ppt.
The total transmission loss modeled in the toolbox is
$TL=T{L}_{\text{geom}}\left(R\right)+\left({\alpha }_{\text{vis}}+{\alpha }_{\text{B}}+{\alpha }_{\text{M}}\right)R$
where R is the range in km. In Phased Array System Toolbox, all absorption models parameters are fixed at T = 10, S = 35, and pH = 8. The model is implemented in range2tl. Because TL is a
monotonically increasing function of R, you can use the Newton-Raphson method to solve for R in terms of TL. This calculation is performed in tl2range.
Noise Level (NL)
Noise level (NL) is the ratio of the noise intensity at the receiver to the same reference intensity used for source level.
Active Sonar Equation
The active sonar equation describes a scenario where sound is transmitted from a source, reflects off a target, and returns to a receiver. When the receiver is collocated with the source, this sonar
system is called monostatic. Otherwise, it is bistatic. Phased Array System Toolbox models monostatic sonar systems. The active sonar equation is
where 2TL is the two-way transmission loss (in dB) and TS is the target strength (in dB). The transmission loss is calculated by computing the outbound and inbound transmission losses (in dB) and
adding them. In this toolbox, two-way transmission loss is twice the one-way transmission loss.
Target Strength (TS)
Target strength is the sonar analog of radar cross section. Target strength is the ratio of the intensity of a reflected signal at 1 m from a target to the incident intensity, converted to dB. Using
the conservation of energy or, equivalently, power, the incident power on a target equals the reflected power. The incident power is the incident signal intensity multiplied by an effective
cross-sectional area, σ. The reflected power is the reflected signal intensity multiplied by the area of a sphere of radius R centered on the target. The ratio of the reflected power to the incident
power is
$\begin{array}{c}{I}_{\text{inc}}\sigma ={I}_{\text{refl}}4\pi {R}^{2}\\ \frac{{I}_{\text{refl}}}{{I}_{\text{inc}}}=\frac{\sigma }{4\pi {R}^{2}}.\end{array}$
The reflected intensity is evaluated on a sphere of 1 m radius. The target strength coefficient (σ) is referenced to an area 1 m^2.
$TS=10{\mathrm{log}}_{10}\frac{{I}_{\text{refl}}\left(\text{1 meter}\right)}{{I}_{\text{inc}}}=10{\mathrm{log}}_{10}\frac{\sigma }{4\pi }$
[1] Ainslie M. A. and J.G. McColm. "A simplified formula for viscous and chemical absorption in sea water." Journal of the Acoustical Society of America. Vol. 103, Number 3, 1998, pp. 1671--1672.
[2] Urick, Robert J. Principles of Underwater Sound, 3rd ed. Los Altos, CA: Peninsula Publishing, 1983. | {"url":"https://it.mathworks.com/help/phased/ug/sonar-equation.html","timestamp":"2024-11-04T19:06:34Z","content_type":"text/html","content_length":"86801","record_id":"<urn:uuid:8e0c3a0d-ce8c-41b1-a426-bce8bfdcb99f>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00043.warc.gz"} |
Double Micro Servo Robot Arm
Introduction: Double Micro Servo Robot Arm
In this tutorial you will be making a double servo robot arm controlled with a thumbstick!
Two Micro Servos (TowerPro SG90 and With the Extension
Jumper Wires
Arduino UNO
Breadboard Power Strip
Glue (Super Glue Suggested)
A little knowledge with Arduinos
Step 1: Cut Out Cardboard Pieces
You will need these cardboard/plastic pieces:
3" by 10/16" X 4
4" by 14/16" X 2
6.5" by 4.5" X 1
1" by 1 1/4" X 2
1" by 1 1/4" X 1 With circle cut out in the middle
2" by 2" by 2" Triangle X 1
2" by 2.5" X 1
After you cut these out you should move to the next step.
Step 2: Attach Cardboard to First Servo
Attach the 4" by 14/16" pieces of cardboard to one servo like the image above. Attach two or more zip ties to the cardboard and servo to hold it in place. You could also use glue or tape but I
suggest zip ties.
Step 3: Attach the First Servo to the Second Servo
Attach the ends of the cardboard that aren't connected to anything to the second servo as shown above. Again I would suggest using zip ties. On the second servo make sure that you have the plastic
attachment that is a circle then one side extended.
Don't understand? Screw the extension onto the servo then glue the extension in between the two pieces of cardboard used in the last step. Then use a zip tie to hold it together even stronger.
Step 4: Attach the Second Servos Arm
Use the 3" by 10/16" pieces of cardboard as the arm of the second servo. Attach two of those pieces to the second servo just how you attached them to the first servo. Then use the last two 3" by 10/
16" pieces of cardboard to extend the second arm, it doesn't really matter how you put the two pieces on just as long as the arm is extended.
Step 5: Attach the Arduino to the Base
Attach the Arduino to the 2" by 2.5" piece of cardboard, I used screws but you can use tape or zip ties if you want. Then glue the 2" by 2.5" piece to the 6.5" by 4.5" piece of cardboard
Step 6: Attach the Thumbstick
Stick the thumbstick through the cardboard with a hole in it. Then trim the triangular piece so it is a 2" by 1" by 1" by 1" trapezoid and use the two 1" by 1 1/4" pieces as well. Glue all of these
pieces together as seen in the first photo. Make sure that the thumbsticks GPiO pins are sticking toward the inside of the base. You do not need to glue the thumbstick down unless it is super loose
inside its housing.
Step 7: Assemble the Rest
Glue the rest of the stuff to the base. Glue the first servo down to the base the first image explains. (Sorry for the grainy image) Attach the breadboard power strip next to the Arduino. (Schematics
Step 8: Schematics
Attach all the pins and jumper wires like this. To avoid soldering I would attach the the +5v and GND the the breadboard power strip and transfer power on that strip. (Next is code)
Step 9: Uploading Code Using the Arduino IDE
Servo myServo1; Servo myServo2;
int servo1 = 5; int servo2 = 6; int joyY = 1; int joyX = 0;
void setup() {
void loop() {
int valX = analogRead(joyX);
int valY = analogRead(joyY);
valX = map(valX, 0, 1023, 10, 170);
valY = map(valY, 0, 1023, 10, 170);
Step 10: You're Done!
If your arm isn't working then make sure to go back and check all of your steps! Thanks for reading and have a good day! | {"url":"https://www.instructables.com/Double-Micro-Servo-Robot-Arm/","timestamp":"2024-11-05T04:35:33Z","content_type":"text/html","content_length":"96372","record_id":"<urn:uuid:2cf04a87-e8ee-4c5f-b81e-49af53724eba>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00607.warc.gz"} |
How to Append Graphs and/or combine Plots?
6324 Views
6 Replies
1 Total Likes
How to Append Graphs and/or combine Plots?
Recently I have been working on a project for a class. Within the project I am attempting to display multiple circles constantly increasing in size as a For loop progresses. I have no problem
increasing the size of the circles, nor displaying the circles, but the circles are constantly being over written by the next circle. I have been attempting to find a way to combine all of the
circles into one graph, and also combine each circle into it's own graph (for individual use). This is the code I have written so far (there will also be an attatched copy).
xSt[] = 0; (* x starting position *)
ySt[] = 0; (*y starting position *)
years = 0; (* The years that pass *)
endTime = 100; (* Amount of years until the end of time *)
probLife = 0; (* The probability that life will sprout up *)
nLife = 0; (* Amount of Circles of Life; and designation number *)
circleLife[] = 0; (* Circle of Life *)
rMulti = 0 ; (* Growth multiplier *)
rBase = 0; (* Base growth rate *)
rGrowth[] = 0; (* Rate of Growth for each CoL *)
orgGrowth[] =
0; (* Used to keep track of original growth rate assigned to \
specific life form *)
ContourPlot[x y, {x, -1, 1}, {y, -1, 1},
PlotRange -> {{-100, 100}, {-100, 100}},
PlotLabel -> Dynamic["Year = " <> ToString[years]],
Epilog -> Dynamic[plot[[1]]]]
For[years = 1, years < endTime,
probLife = RandomInteger[{1, 10}];
(* The following If function checks if there are any circles of \
life and if there is, begins updating them,
as if the life form is expanding radially *)
If[nLife > 0,
For[n = 1, n <= nLife,
rGrowth[n] = rGrowth[n] + orgGrowth[n];
plot =
ContourPlot[((a - xSt[n])^2 + (b - ySt[n])^2 ==
rGrowth[n]^2), {a, -100 \[Pi], 100 \[Pi]}, {b, -100 \[Pi],
100 \[Pi]},
PlotRange -> {{-100, 100}, {-100, 100}}];
(* Pause[0.1] *)](* End For n *)
]; (* End If nLife *)
(* The following If function checks if the probability of life \
forming is true and if so creates a new form of life at a random \
point on the plot and the size and growth rate of that life form *)
If[probLife == 1,
{nLife = nLife + 1;
xSt[nLife] = RandomReal[{-100, 100}];
ySt[nLife] = RandomReal[{-100, 100}];
rMulti = RandomReal[{.01, .1}];
rBase = RandomReal[{1, 10}];
rGrowth[nLife] = (rMulti rBase);
orgGrowth[nLife] = (rMulti rBase);
plot =
ContourPlot[((a - xSt[nLife])^2 + (b - ySt[nLife])^2 ==
rGrowth[nLife]^2), {a, -100 \[Pi],
100 \[Pi]}, {b, -100 \[Pi], 100 \[Pi]},
PlotRange -> {{-100, 100}, {-100, 100}}]};
]; probLife = 0;(* End If probLife *)
(* Pause[0.1]; *)
] ;(* End For years*)
Also, while attempting a variety of different methods and approaches, I receive this error message
"An improperly formatted option head (Graphics) was encountered while reading a Graphics. The head of the option must be Rule or RuleDelayed."
But I do not know exactly what it means, and my searches for an answer to this online has been fruitless.
Please help me Mathematica Community, you're my ony hope.
6 Replies
I have no idea what your Dynamic is doing, or even what it is supposed to be doing.
This is generating the frames for your movie, but not overlaying them.
If you Google for Mathematica movie then you can see some example code,
but I am not certain how to apply that directly to the code that you have written.
If you look at the documentation for Reap you should see it returns a list with a couple of items in it. Item [[2,1]] is the "bag" of collected things that you put in there via Sow. The way I write
code never needs anything else from Reap so I append that [[2,1]] without even thinking anymore.
The way that graphics were displayed changed in version 6. It used to be different from the way it is now and I have not put in the time to change the way I automatically think about assembling
several different graphs onto a single sheet. So I'm not much use to you for generating movies.
It is a picky detail, but the way I was trained long ago. You seem to be using allPlots for two different things. The first time you seem to be using it as a label or title of just zero. Then you
switch to having allPlots be a "bag" of images from Reap. I usually find I make fewer mistakes if each variable has a single purpose and a single kind of contents. It seems that for your first
ContourPlot that allPlots will be zero and zero[[1]] isn't what I think you want.
You have also probably been taught another language before Mathematica because I see you using { and } to try to group statements. It is sort of possible to do that, but Mathematica isn't really
doing what you think with that. Mathematica uses semicolons to sort of group items. I can't be certain that is what is breaking your code, but it worries me. You are also sort of using function
definitions to save your plots. I can't tell at the moment if part of that is the reason that is causing your problem or not.
I'll fiddle with this a little more and see if I can find any serious problems. You can do the same. And hopefully you will learn even more in the process.
Ah, I see now. That example cleared up some confusion that I was having, though I do not understand why the [2,1] is added at the end.
Also, you are correct that it does not resolve my "movie" issue. I have been able to create a multi variable list function "List," which allows me to save each circle and every growth event, but I am
not able to show all of them happening at the same time (which I am sure that I can't with that multi variable list, but it was to be used later to analyze each circle and its respective growth
In regards to appending plots: Every time I attempt to append a plot, I receive errors such as "Graphics is not a Graphics primitive or directive." Or a variety of other errors dealing with graphics.
I assume that if I can append a plot during my growth counters/building, that I will be able to display a plot that shows all of the circles expanding at once (not actually, but close to it).
When this is done, the years are suppose to jump up to a few million, and the plot is going to be expanded to a few hundred thousand^2.
The attachted file shows updates which bring me ever so closer to something usable, but why does Dynamic not accept the allPlots function? Why does allPlots not display with the For loops when called
Honestly, I appreciate all of your help. This is helping me learn a lot, and is kind of fun.
Reap and Sow aren't quite behaving the way you think they are.
See if this example helps.
And this still doesn't address your "movie" goal.
Thank you Bill Simpson, I highly appreciate your assistance. It has helped me to understand more about Mathematica and also learn more through exploration, but
I tried a similar code to yours (and a variety of variations), and I am not producing the same result that you mention. I added a For loop into the Reap function, as you did, but my results do not
produce a single graph over laid with all of the other graphs.
Also, I am attempting to make a movie, but also save each expanding circle as its own movie to be viewable independantly. I would also like to display the ultimate finished product (all of the circle
outcomes) displayed at once. I like to think that I would be able to achieve this if I am able to get the Reap function you mentioned working correctly.
Thank you for introducing &DisplayFunction.
Inside a For loop each Plot is roughly saying "I don't know what you have done before and I really don't care, forget that, give me a new blank sheet to draw a plot on."
I can't really address what looks like your desire to create a movie with each new plot added as another frame, that would be more complicated.
Perhaps you can use this idea and see if it will get you at least part of what you need.
If you use
plot=ContourPlot[ ];
plot=ContourPlot[ ];
then each ContourPlot is still going to overwrite the previous one and you won't get your movie.
BUT when the movie finishes that last Show is going to display all the frames of your movie overlaid as a single plot.
Roughly, Reap is creating a bag and every time you have a Sow inside it you will add one more item to the bag.
When you finally finish all the code inside the Reap then it will give you the bag with the contents.
In your case the bag will be filled with individual plots and you can Show a list of superimposed plots.
I just tried this and I can see all your plots overlaid on a single sheet at the end.
You could, with more care and work, create your own empty list of plots, append each new plot to the end
of the list, Show that list after you add each new plot and attempt to create your movie. This might
give you enough of a hint to be able to do that if that is what you need to do.
You can also use $DisplayFunction to disable the display of the individual frames and then turn on
the display function for the accumulated list of plots. Again, this is more complicated for you to do.
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What is the instantaneous rate of change of f(x)=ln(2x^2-4x+6) at x=0 ? | HIX Tutor
What is the instantaneous rate of change of #f(x)=ln(2x^2-4x+6) # at #x=0 #?
Answer 1
Applying the rule of chain:
When #x=0# this becomes:
This is the instantaneous rate of change of the function at #x=0#
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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A Portable Introduction to Data Analysis
[latex]\newcommand{\pr}[1]{P(#1)} \newcommand{\var}[1]{\mbox{var}(#1)} \newcommand{\mean}[1]{\mbox{E}(#1)} \newcommand{\sd}[1]{\mbox{sd}(#1)} \newcommand{\Binomial}[3]{#1 \sim \mbox{Binomial}(#2,#3)}
\newcommand{\Student}[2]{#1 \sim \mbox{Student}(#2)} \newcommand{\Normal}[3]{#1 \sim \mbox{Normal}(#2,#3)} \newcommand{\Poisson}[2]{#1 \sim \mbox{Poisson}(#2)} \newcommand{\se}[1]{\mbox{se}(#1)} \
For the caffeinated cola study in Chapter 2 we can think of the two groups of subjects as coming from two different populations. As people they came from the same population but in terms of their
pulse rate response one group came from a population where they drank caffeinated cola while the other group came from a population where they drank decaffeinated cola. We now want to determine
whether those populations are different based on our samples.
Standard Error
Suppose we take two independent samples from two populations. Suppose the first sample was of size [latex]n_1[/latex] and came from a population with mean [latex]\mu_1[/latex] and standard deviation
[latex]\sigma_1[/latex], and that the second sample was of size [latex]n_2[/latex] and came from a population with mean [latex]\mu_2[/latex] and standard deviation [latex]\sigma_2[/latex]. We
estimate [latex]\mu_1[/latex] with [latex]\overline{x}_1[/latex], [latex]\mu_2[/latex] with [latex]\overline{x}_2[/latex], [latex]\sigma_1[/latex] with [latex]s_1[/latex], and [latex]\sigma_2[/latex]
with [latex]s_2[/latex].
We would like to compare [latex]\mu_1[/latex] and [latex]\mu_2[/latex] to see if there is a difference in the mean responses for two treatments or between two groups. We can make this comparison by
looking at [latex]\mu_1 - \mu_2[/latex] and seeing how far away it is from 0. Of course, we don’t know what [latex]\mu_1 - \mu_2[/latex] is but we can estimate it with the statistic [latex]\overline
{x}_1 - \overline{x}_2[/latex]. This is the difference between two sample means but it is useful to think of it as one value, an outcome of the random variable [latex]\overline{X}_1 - \overline{X}_2
[/latex], the process of taking two random samples and returning the difference between their means.
To work out a confidence interval for [latex]\mu_1 - \mu_2[/latex] we need to know the sampling distribution of [latex]\overline{X}_1 - \overline{X}_2[/latex]. Now
\[ \mean{\overline{X}_1 – \overline{X}_2} = \mean{\overline{X}_1} – \mean{\overline{X}_2} = \mu_1 – \mu_2, \]
as we would like, and
\[ \var{\overline{X}_1 – \overline{X}_2} = \var{\overline{X}_1} + \var{\overline{X}_2} = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}, \]
since we are assuming the samples are independent. This gives the standard deviation
\[ \sd{\overline{X}_1 – \overline{X}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}. \]
As usual, we don’t know [latex]\sigma_1[/latex] or [latex]\sigma_2[/latex], but we can estimate them with the sample standard deviations. This gives the standard error
\[ \se{\overline{x}_1 – \overline{x}_2} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}. \]
The [latex]t[/latex] distribution was introduced to cope with the extra variability from one sample standard deviation. Here we now have two and so unfortunately we cannot use the [latex]t[/latex]
distribution directly with this standard error. However, we can use the [latex]t[/latex] distribution to give a conservative approximation to the real distribution by taking
\[ \mbox{df } = \min(n_1 – 1, n_2 -1), \]
the smaller of the two degrees of freedom. By “conservative” we mean that a 95% confidence will probably be a bit wider than it has to be and hypothesis tests will give less significant [latex]P[/
Confidence Intervals
We can use the above discussion to give a general formula for a confidence interval for the difference between two population means,
\[ (\overline{x}_1 – \overline{x}_2) \pm t_{\small{\mbox{df}}}^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}, \]
where [latex]\mbox{df } = \min(n_1 - 1, n_2 -1)[/latex].
Lighting and Plant Growth
The table below gives summary statistics for the seedling growth data in the Chapter 4 example.
Summary statistics for plant growth (mm) by lighting
Lighting [asciimath]n[/asciimath] [asciimath]\overline{x}[/asciimath] [asciimath]s[/asciimath]
High 15 79.9 7.039
Normal 15 41.0 6.024
To have a break from 95% intervals, suppose we want a 90% confidence interval for the mean increase in plant growth resulting from the continuous fluorescent lighting. This estimated mean difference
could be important in deciding whether the cost of the lighting is justified in terms of the benefit.
For 90% confidence we require the two tail probabilities to each be 5% so we look at the 0.05 column in Student’s T distribution. The smaller degrees of freedom here are 14 giving [latex]t_{14}^{*}[/
latex] = 1.761. The interval is thus
\[ (79.9 – 41.0) \pm 1.761 \sqrt{\frac{7.039^2}{15} + \frac{6.024^2}{15}} = 38.9 \pm 4.21 \mbox{ mm}.\]
So we are 90% sure that the continuous lighting results in between 34.7 mm and 43.1 mm extra growth on average.
The Welch Approximation
The conservative degrees of freedom we have used is easy to calculate by hand but is almost always too conservative. We are underselling our confidence intervals. The above 90% confidence interval is
probably close to a 93% confidence interval in reality. Computer packages use a more complicated calculation for the appropriate degrees of freedom to use. Remember the aim of this is to approximate
the real distribution that arises from using two sample standard deviations by a [latex]t[/latex] distribution. The better degrees of freedom gives the Welch approximation (Welch, 1936) and is
calculated by
\[ \mbox{df } = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2 + \frac{1}{n_2-1} \left(\frac{s_2^2}{n_2}\right)^2}. \]
For example, with the above comparison of plant growth, the Welch degrees of freedom would be 27.35 instead of 14. This is still conservative but far less so. The margin of error becomes 4.07 mm
instead of 4.21 mm, suggesting we know the difference a bit more precisely than we said before.
The best degrees of freedom we could hope to use would be to combine the two degrees of freedom from [latex]s_1[/latex] and [latex]s_2[/latex], in this case (15-1) + (15-1) = 28. It is justifiable to
do this in some cases, as we will see later in this chapter. However, when it is justifiable to do so the Welch approximation would give this as well. In practice you can let the software package
take care of this issue.
Treatment of Worms in Native Rats
A study carried out by Renee Sternberg and Hamish McCallum at the University of Queensland involved trapping and releasing a number of native rats near Mount Glorious. Two species of rat were
involved: Rattus fuscipes and Melomys cernvinipes. Before releasing, half of the rats were given a treatment in an attempt to reduce their worm burden while the others were given distilled water
instead, as a control group.
The two tables below show the worm count data obtained at the end of this experiment. The following figure shows a dot plot comparing the number of worms found in the small intestine of each rat at
the end of the study.
Worm count data - Water Group
Species Sex Liver/Heart/Lungs Stomach Small Intestine Caecum Large Intestine
Melomys Male 0 0 84 0 0
Melomys Female 0 0 8 4 4
Melomys Female 0 0 50 7 0
Melomys Female 0 0 20 1 0
Melomys Male 0 0 0 0 1
Rattus Female 0 7 71 0 1
Rattus Female 7 22 217 0 0
Rattus Female 2 16 145 2 0
Rattus Male 0 12 71 19 5
Rattus Male 0 2 30 7 4
Rattus Male 23 9 234 9 2
Rattus Male 10 9 246 16 2
Rattus Male 4 6 470 60 4
Worm count data - Treatment Group
Species Sex Liver/Heart/Lungs Stomach Small Intestine Caecum Large Intestine
Melomys Female 0 0 28 1 0
Melomys Male 0 0 10 0 0
Melomys Male 0 0 3 0 0
Melomys Male 0 0 2 0 0
Melomys Male 0 0 4 0 0
Rattus Female 0 2 9 0 0
Rattus Female 0 1 5 0 0
Rattus Female 0 3 1 0 0
Rattus Female 0 11 8 28 0
Rattus Female 23 6 0 3 0
Rattus Male 0 9 0 9 0
Rattus Male 0 6 10 0 0
Rattus Male 0 2 1 2 0
Worm count in small intestine by treatment group
The distributions of worm counts are highly skewed and so they need to be transformed if [latex]t[/latex] methods are to be applied, as was done in Chapter 14. Taking logarithms would be a first
step, but this data contains many zero counts, where no worms were found, and [latex]\log(0)[/latex] is undefined. This can be overcome by adding 1 to all observations before taking logarithms, and
the results of this transformation, using logarithms to the base 10, are shown in the figure below.
Worm count in small intestine transformed using [latex]\log(x+1)[/latex]
The transformed data is much more symmetric, though there is now a slightly unusual value in the Water group. We can proceed to calculate a 95% confidence interval for the difference between the
groups using the summary data in the following table.
Summary statistics for transformed worm count data
Group [asciimath]n[/asciimath] [asciimath]\overline{x}[/asciimath] [asciimath]s[/asciimath]
Water 13 1.774 0.7167
Treatment 13 0.666 0.4394
The Welch approximation suggests using 19 degrees of freedom. This reflects some difference between the sample standard deviations, but is much higher than the 12 degrees of freedom suggested at the
start of the section on confidence intervals.
The 95% confidence interval for the effect of the treatment over the placebo is
\[ (0.666 – 1.774) \pm 2.093 \sqrt{\frac{0.4394^2}{13} + \frac{0.7167^2}{13}} =
-1.108 \pm 0.488, \]
giving a range of -1.596 to -0.620. As in the section on confidence intervals, we then need to undo our transformation to get an interval we can interpret. We have found an interval for
\[ \log(\mbox{Treatment}) – \log(\mbox{Water}) = \log\left(\frac{\mbox{Treatment}}{\mbox{Water}}\right), \]
so a 95% confidence interval for the ratio of worms in the treatment group to the control group is
\[ \left(10^{-1.596}, 10^{-0.620}\right) = (0.025, 0.240). \]
Thus we are 95% sure that native rats undergoing the treatment will have between only 2.5% and 24% of the worms in their small intestines that a rat would otherwise have.
Hypothesis Tests
Suppose we want to test the hypothesis [latex]H_0: \mu_1 = \mu_2[/latex]. In this case, when calculating the [latex]P[/latex]-value we would expect [latex]\mu_1 - \mu_2[/latex] = 0. Combining this
with the standard error formula gives the [latex]t[/latex] statistic
\[ t_{\small{\mbox{df}}} = \frac{(\overline{x}_1 – \overline{x}_2) – 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}, \]
where [latex]\mbox{df } = \min(n_1 - 1, n_2 -1)[/latex] or the value from the Welch approximation. The expected value for the difference is 0 and we have left it there to emphasise that this is just
the usual process of standardising.
We can now use a [latex]t[/latex] test to answer Alice’s question from Chapter 2 and see whether the caffeinated cola gives an increase in pulse rate that is significantly higher than the
decaffeinated cola.
We would like to see if there is a difference between the mean increase in pulse rate with caffeine, [latex]\mu_Y[/latex], and the mean increase without caffeine, [latex]\mu_N[/latex]. We will test
[latex]H_0: \mu_Y = \mu_N[/latex] against the one-sided alternative [latex]H_1: \mu_Y \gt \mu_N[/latex]. It is one-sided because Alice was trying to show that the presence of caffeine would give a
higher mean increase.
The figure below shows a side-by-side dot plot of the pulse rate increases for the 20 subjects in the Chapter 2 example, while the following table gives the summary statistics we need to calculate
the [latex]t[/latex] statistic.
Increase in pulse rate after cola
Summary statistics for pulse rate increases (bpm)
Caffeine [asciimath]n[/asciimath] [asciimath]\overline{x}[/asciimath] [asciimath]s[/asciimath]
Yes 10 15.80 8.324
No 10 5.10 5.587
From these summaries find
\[ t_9 = \frac{(15.80 – 5.10) – 0}{\sqrt{\frac{8.324^2}{10} + \frac{5.587^2}{10}}} = \frac{10.7}{3.17} = 3.38, \]
where 9 degrees of freedom comes from the conservative approximation. Since we are expecting to find [latex]\mu_Y \gt \mu_N[/latex] the [latex]P[/latex]-value is [latex]\pr{T_9 \ge 3.38}[/latex].
From Student’s T distribution we find this [latex]P[/latex]-value is between 0.005 and 0.001, very strong evidence to suggest that the mean increase is higher for the caffeinated cola than it is for
the decaffeinated cola.
The Alice Distribution
Note that this is the same level of evidence we found using the randomisation test in Chapter 2. There we wanted to know how likely it was that we could obtain 10.7 through the random allocation of
subjects to the two groups and we gave a fairly informal argument regarding this probability and the associated evidence it suggested.
We can now be more specific about this process using the language of random variables. Let [latex]A[/latex] be the difference between group means when the 20 values in the table below are randomly
split into two groups of equal size. This is an example of what is variously known as a randomisation distribution (Ernst, 2004), a re-randomisation distribution (Pfannkuch et al., 2011) or a
scrambling distribution (Finzer, 2006).
Changes in pulse rate
-2 -9 4 4 5 5 6 6 7 7
Here we will give our distribution a name and say that the random variable [latex]A[/latex] has the Alice distribution. This is a very special distribution, intimately tied to the 20 values. However
we could still make an exact statistical table of this distribution by calculating all 184756 possible mean differences and using these to give cumulative probabilities, as shown in the table below.
Alice distribution
First decimal place of [asciimath]a[/asciimath]
[asciimath]a[/asciimath] 0 1 2 3 4 5 6 7 8 9
0 0.500 0.500 0.481 0.461 0.461 0.461 0.442 0.423 0.423 0.423
1 0.404 0.404 0.385 0.385 0.366 0.366 0.348 0.330 0.330 0.330
2 0.312 0.312 0.295 0.295 0.278 0.278 0.262 0.262 0.246 0.246
3 0.231 0.231 0.216 0.216 0.202 0.202 0.188 0.188 0.174 0.174
4 0.161 0.161 0.149 0.149 0.138 0.138 0.127 0.127 0.116 0.116
5 0.106 0.106 0.097 0.097 0.089 0.089 0.081 0.081 0.073 0.073
6 0.066 0.066 0.059 0.059 0.053 0.053 0.047 0.047 0.042 0.042
7 0.037 0.037 0.033 0.033 0.029 0.029 0.026 0.026 0.022 0.022
8 0.019 0.019 0.017 0.017 0.014 0.014 0.012 0.012 0.011 0.011
9 0.009 0.009 0.008 0.008 0.006 0.006 0.005 0.005 0.004 0.004
10 0.004 0.004 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002
11 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000
This table gives [asciimath]P(A \ge a[/asciimath] where the random variable [asciimath]A[/asciimath] has the Alice distribution. Only values for [asciimath]a \ge 0[/asciimath] are given since the
distribution is symmetric about 0.
The [latex]P[/latex]-value for our test is then
\[ \pr{A \ge 10.7} = 0.002, \]
from the table.
However this table is of very limited value. It could only be used by a future researcher who happened to obtain the same set of 20 values in their experiment! This explains the historical utility of
the [latex]t[/latex] tests. By using the standard error to transform our difference of 10.7 bpm to a standardised [latex]t[/latex] statistic of 3.38 we can then obtain the [latex]P[/latex]-value from
a single set of [latex]t[/latex] distribution tables, rather than having to determine the distribution of the original statistic. In this way the [latex]t[/latex] distribution is a short cut in
transforming our data into a [latex]P[/latex]-value:
\[ \mbox{Data} \;\; \longrightarrow \;\; \overline{X}_1 – \overline{X}_2 \;\; \longrightarrow \;\; T \;\; \longrightarrow \;\; P \]
The price we pay for this utility is the need to make the assumptions required by the [latex]t[/latex] test for this short cut to be sufficiently accurate.
Pooling Standard Deviations
We have seen that the test statistic
\[ t = \frac{(\overline{x}_1 – \overline{x}_2) – 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
does not have an exact [latex]t[/latex] distribution. This is essentially because there are now two sources of variability in the standard error, since we don’t know [latex]\sigma_1[/latex] or
[latex]\sigma_2[/latex], and the [latex]t[/latex] distribution was only intended to capture the extra variability from one. However if we were happy to assume that the two populations have the same
standard deviations, that [latex]\sigma_1[/latex] = [latex]\sigma_2[/latex], then the denominator would only involve a single estimate of variability. We could then use the [latex]t[/latex]
distribution without having to approximate it using a conservative estimate of degrees of freedom.
To estimate the common standard deviation we pool together the squared deviations and the degrees of freedom from the two samples. This gives the pooled variance
\[ s_p^2 = \frac{\sum (x_{1j} – \overline{x}_1)^2 + \sum (x_{2j} – \overline{x}_2)^2}{(n_1 – 1) + (n_2 – 1)} \]
and the pooled standard deviation [latex]s_p[/latex]. Another way of writing this formula comes from the definition of the sample standard deviation
\[ s = \sqrt{\frac{\sum (x_{j} – \overline{x})^2}{n – 1}}. \]
This can be rearranged to give
\[ \sum (x_{j} – \overline{x})^2 = (n – 1)s^2 \]
so that
\[ s_p^2 = \frac{(n_1 – 1)s_1^2 + (n_2 – 1)s_2^2}{(n_1 – 1) + (n_2 – 1)}. \]
This shows [latex]s_p^2[/latex] is a weighted average of the two sample variances. It is also handy if you only have a calculator since you can usually get each [latex]s[/latex] but not [latex]\sum
(x_{j} - \overline{x})^2[/latex]. A software package, of course, calculates this for you.
The degrees of freedom we can now use are [latex]n_1 + n_2 - 2[/latex], higher than the very conservative [latex]\min(n_1 - 1, n_2-1)[/latex] and at least as high as the Welch approximation. A
[latex]t[/latex] distribution with higher degrees of freedom is less variable and so confidence intervals we calculate will be narrower and hypothesis tests will be more significant. This is great
but the difficulty is in deciding whether the population standard deviations are equal or not. This is really another hypothesis test question, but unfortunately the methods available for handling
this test are not very reliable. It is better to compare the sample distributions graphically instead and see whether the assumption is plausible.
As usual the [latex]t[/latex] test will use the pooled standard deviation, rather than the pooled variance. However in Chapter 19 we will extend the idea of pooling to more than two samples. There we
will focus on variance instead.
The following figure shows a side-by-side box plot of height by sex for the sample of 60 Islanders in the survey data. The spread of the two distributions seems similar from this plot and so it might
be reasonable to assume that the populations have the same standard deviations.
Box plot of height by sex
From the summary statistics in the following table we can calculate the pooled variance
\[ s_p^2 = \frac{(34 – 1) 6.367^2 + (26- 1) 5.900^2}{(34 + 26 – 2)} = \frac{2208}{58} = 38.07, \]
giving pooled standard deviation [latex]s_p = 6.17[/latex] cm. This will always be between the two sample standard deviations.
Summary statistics for height by sex
Sex [asciimath]n[/asciimath] [asciimath]\overline{x}[/asciimath] [asciimath]s[/asciimath]
Male 34 177.06 6.367
Female 26 167.42 5.900
The [latex]t[/latex] statistic for testing whether there is a difference between male and female heights is then
\[ t_{58} = \frac{(177.06 – 167.42) – 0}{6.17\sqrt{\frac{1}{34} + \frac{1}{26}}} = 6.00, \]
giving very strong evidence of a difference. This is not a very exciting result since it is well known that males are on average taller than females. Of more interest in this setting would be a
confidence interval for how much taller males are. A 95% interval would be
\[ 9.64 \pm 2.002 \left(6.17 \sqrt{\frac{1}{34} + \frac{1}{26}}\right) = 9.64 \pm 3.22, \]
so we are 95% confident that the mean height for males is between about 6.42 cm and 12.86 cm higher than for females. (Here 2.002 came from the [latex]t(58)[/latex] distribution.)
The figure below shows the power of a two-sided two-sample [latex]t[/latex] test for varying signal-to-noise ratio.
Power of two-sample tests for signal-to-noise ratio
Here this is the ratio of the difference you want to detect between the groups to the pooled standard deviation. The sample sizes shown are within each group: “[latex]n[/latex] = 40” indicates that
you would need a sample of size 80 for comparing two groups.
The figure below gives a more useful plot for practice, showing the sample size required in each group to obtain 80% power in detecting the desired signal-to-noise ratio.
Sample size required for 80% power for signal-to-noise ratio
Choosing Sample Sizes
As in Chapter 14, we can find the sample sizes [latex]n_1[/latex] and [latex]n_2[/latex] that give a desired margin of error, [latex]m[/latex], by rearranging the equation involving the standard
deviation of the difference. Again we need some estimates of [latex]\sigma_1[/latex] and [latex]\sigma_2[/latex] to proceed. There is also the subtlety that the value of [latex]t^*[/latex] depends on
[latex]n_1[/latex] and [latex]n_2[/latex], and in quite a complicated way when using the Welch method. However, this can all be managed by trial and error.
However, there is a more interesting question in this setting. Consider Alice’s experiment on the effects of caffeine on pulse rate. She had 20 friends available and chose to put 10 in each of her
groups, giving equal sample sizes. This seems the intuitive thing to do, but is it the best use of the 20 subjects available?
The general formula for the margin of error when comparing the two groups is
\[ m = t^* \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}. \]
Based on the sample, discussed in the section on confidence intervals, it seems plausible that the population standard deviations are the same, [latex]\sigma_1 = \sigma_2 = \sigma[/latex]. Assuming
that [latex]n = n_1 + n_2[/latex] is fixed we can write
\[ m = t^* \sigma \sqrt{\frac{1}{n_1} + \frac{1}{n – n_1}}. \]
For a fixed confidence level, we have no control in this formula over [latex]t^*[/latex], [latex]\sigma[/latex], or [latex]n[/latex]. The only choice we have is in the value of [latex]n_1[/latex].
Our aim should be to choose [latex]n_1[/latex], and hence [latex]n_2 = n - n_1[/latex], so that we make [latex]m[/latex] as small as possible, giving the best precision in our estimate of the
difference in the mean growth levels.
Effect of choice of [latex]n_1[/latex] on margin of error multiplier
The figure above shows a plot of
\[ \sqrt{\frac{1}{n_1} + \frac{1}{20 – n_1}}, \]
for choices of [latex]n_1[/latex] between 1 and 19. It should be clear that [latex]n_1 = 10[/latex] gives the lowest value, so that [latex]n_1 = n_2 = 10[/latex] is the best choice for splitting the
friends between the groups. Thus Alice was right in using equal sample sizes in her experiment. Note however that small differences in the sample sizes would not have had much effect on [latex]m[/
In general, when [latex]\sigma_1 \ne \sigma_2[/latex], it can be shown that the minimum margin of error comes from choosing [latex]n_1[/latex] and [latex]n_2[/latex] so that
\[ \frac{n_1}{n_2} = \frac{\sigma_1}{\sigma_2}. \]
So when [latex]\sigma_1 = \sigma_2[/latex], as above, we should choose [latex]n_1 = n_2[/latex]. If, for example, we suspect that [latex]\sigma_1[/latex] was three times [latex]\sigma_2[/latex] then
we would choose [latex]n_1[/latex] to be three times [latex]n_2[/latex]. If [latex]n = 20[/latex] this would give [latex]n_1 = 15[/latex] and [latex]n_2 = 5[/latex].
However, even if the original data shows unequal standard deviations we will often use transformations to stabilise the variability, as illustrated in Chapter 19. The above arguments refer to the
values you do the calculations with, and so equal sample sizes will still be appropriate if you transform your data in this way to have similar standard deviations.
• The sampling distribution of [latex]\overline{X}_1 - \overline{X}_2[/latex] gives a basis for calculating confidence intervals and carrying out hypothesis tests for a comparison of two population
• When population standard deviations are different the Welch degrees of freedom give a conservative approximation to this sampling distribution.
• For common population standard deviations a pooled standard deviation allows the use of a [latex]t[/latex] distribution with maximum degrees of freedom.
• Confidence intervals calculated for a difference in logarithms give a range for a ratio in the original units.
• For common population standard deviations it is optimal to split subjects evenly between the two treatment groups.
Forty plastic cups were each filled with 20 mm of water stained with a blue food colouring. A celery stalk with leaves was placed in each cup with a toothpick through the centre for stabilisation.
For twenty of the stalks the leaves were coated with petroleum jelly. All cups were placed in behind a glass shield in the sun and left for 5 hours. Each celery stalk was then cut from the bottom up
and the distance to where the blue stain could no longer be seen in the vascular tissue was recorded. The results are given in the table below.
Dye uptake (mm) with or without coated leaves
Uncoated 155 144 151 139 146 131 143 156 117 125
Coated 92 110 119 104 93 86 96 107 114 96
Based on this data, calculate a 95% confidence interval for the difference in dye uptake between plants with leaves coated with vaseline and those without.
Alcoholic beverages are known to slow reaction times but can this effect be offset by adding caffeine to the drink? Two groups of 8 males were given five drinks of rum and coke over a two-hour
period. One group had regular diet coke in their drinks while the other had decaffeinated diet coke. Reaction times were measured before drinking and then after the two hours they were measured
Reaction times came from a “ruler test” where a ruler was released at the 0 cm mark between a subject’s thumb and forefinger. The result was the distance the ruler travelled before it was caught. For
each subject, the table below reports averages from three repetitions of the ruler test before drinking and three after drinking. Is there any evidence that the increase in reaction time is less for
the group receiving regular diet coke?
Average reaction times before and after alcohol (cm)
Regular Decaffeinated
Before After Before After
11.83 18.30 12.02 19.79
11.16 17.80 11.14 17.40
11.94 19.00 11.10 17.90
12.04 19.01 11.89 16.60
12.14 19.20 12.09 17.50
12.61 18.70 11.93 19.00
12.07 18.55 12.16 18.60
12.16 18.20 12.64 20.00
Ingrid Ibsen, a student at Colmar University, was interested in whether reaction times differed between males and females. Using a sample of other people in the village, Ingrid had each subject press
a button as quickly as possible after seeing a light flash. The results are shown in the table below. Is there any evidence of a difference in reaction times between males and females?
Reaction times (ms) between sexes
Female 279 262 254 262 276 254 267 293 289 280
Male 258 269 283 243 299 264 245 292 294 258
Modafinil is a wake promoting agent that has been used in the treatment of daytime sleepiness associated with narcolepsy and shift-work. Müller et. al. (2013) conducted a double-blind study comparing
the effect on creative thinking of 200 mg of modafinil ([latex]n_1 = 32[/latex]) or placebo ([latex]n_2 = 32[/latex]) in non-sleep deprived healthy volunteers. In one task the mean creativity score
was 5.1 ([latex]s_1 = 3.4[/latex]) for the modafinil group compared to 6.5 ([latex]s_2 = 3.8[/latex]) for the placebo group. Does this give any evidence of an effect of modafinil on creative
Robertson et al. (2013) followed a cohort of individuals from birth to age 26 years, conducting assessments at birth and then at ages 5, 7, 9, 11, 13, 15, 18, 21 and 26. At the assessments from ages
5 to 15 they asked parents the average amount of time these individuals spent watching television each weekday. For the 523 boys in the study the mean value was 2.42 hours with standard deviation
0.86 hours. For the 495 girls in the study the corresponding mean was 2.24 hours with standard deviation 0.88 hours. Does this give any evidence of a difference between boys and girls in the time
spent watching television?
Carry out a two-sample [latex]t[/latex] test for the sleep deprivation and internal clock study in Exercise 4. Compare your results with the exact [latex]P[/latex]-value from the randomisation test.
Inspired by the work of Nascimbene et al. (2012), William Favreau conducted a study to compare the plant biodiversity between 11 conventional vineyards and 9 organic vineyards around Talu. Along with
the area and the number of years it had been organic, William counted the number of annuals and perennials present in each vineyard. His results are shown in the table below.
Plant biodiversity in vineyards
Management Area (ha) Years Organic Annuals Perennials
Organic 22 25 12 30
Conventional 52 17 12
The mean total number of plant species for the organic vineyards was 31.1 with standard deviation 5.21. For the conventional vineyards the mean was 26.6 with standard deviation 1.96. Does this give
evidence that the organic management practices have resulted in higher plant biodiversity? | {"url":"https://uq.pressbooks.pub/portable-introduction-data-analysis/chapter/comparing-two-means/","timestamp":"2024-11-14T03:33:16Z","content_type":"text/html","content_length":"154259","record_id":"<urn:uuid:17052d89-e2a6-4222-98a7-29b93ac15d81>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00695.warc.gz"} |
Dividing whole numbers with decimal quotients
The students learn to divide whole numbers with a decimal number as the outcome.
Students will be able to divide two whole numbers for which the outcome is a decimal number.
The students drag the fractions to the matching decimal numbers.
Show the problem 26 ÷ 4. Explain that you are first going to determine how many times the divisor fits into the whole number. In this case that is 6 × 4 = 24. Next you subtract this from the whole
number (26 - 24 = 2). You now have a remainder of 2 left over. You divide this remainder again by the divisor (4). 2 ÷ 4 = 2//4. The decimal number that matches 2//4 is 0.5. Finally you add the
outcomes together for the answer: 6 + 0.5 = 6.5. Practice the next problem together with the students. For this you can write down the calculations under each step of the process. The students solve
the next three problems on their own. Show the story problem and explain that you must first take the problem out of the story (283 ÷ 20=). Next you see how many times the divisor fits in the whole
number (14 × 20 = 280). Subtract these from each other (283 - 280 = 3). Divide the remainder by the divisor (3 ÷ 20 = 3//20, 3//20 = 0.15). Now add the outcomes together for the answer (14 + 0.15 =
14.15). The students solve the next two story problems on their own.Check whether the students can divide whole numbers with a decimal quotient by asking the following question:- What steps do you
take to solve a division problem with whole numbers with a decimal quotient?
The students test their understanding of dividing whole numbers with decimal quotients through ten exercises. Each of the exercises involves a division problem with two whole numbers that results in
a decimal number as the quotient. For some of the exercises the students have multiple answers to choose from, and for others they must produce the answer on their own. Some of the exercises are
story problems.
Discuss once again the importance of being able to divide whole numbers with a decimal quotient. As a closing activity the students can work in groups of four. Have each group form pairs. Each pair
comes up with a division problem with whole numbers, that has a decimal number as the quotient. Once they have the problem, they check the answer and trade their problem with the other pair. Now they
try to solve the problem that the other pair of students made.
Have students that have difficulty with dividing whole numbers with decimal quotients first practice converting fractions to decimal numbers and vice versa.
Gynzy is an online teaching platform for interactive whiteboards and displays in schools.
With a focus on elementary education, Gynzy’s Whiteboard, digital tools, and activities make it easy for teachers to save time building lessons, increase student engagement, and make classroom
management more efficient. | {"url":"https://www.gynzy.com/en-us/library/items/dividing-whole-numbers-with-decimal-quotients","timestamp":"2024-11-07T22:16:36Z","content_type":"text/html","content_length":"553482","record_id":"<urn:uuid:46eb6940-d771-4f81-ae4e-38e7b5aa7260>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00169.warc.gz"} |
Number of Excellent Pairs | CodingDrills
Number of Excellent Pairs
Given a 0-indexed positive integer array, nums, and a positive integer, k. A pair of numbers (num1, num2) is considered excellent if the following conditions hold:
1. Both num1 and num2 exist in the array nums.
2. The sum of the number of set bits in num1 OR num2 and num1 AND num2 is greater than or equal to k, where OR is the bitwise OR operation and AND is the bitwise AND operation.
Your task is to determine and print the number of distinct excellent pairs.
Two pairs (a, b) and (c, d) are considered distinct if either a ≠ c or b ≠ d. For example, (1, 2) and (2, 1) are considered distinct.
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/number-of-excellent-pairs","timestamp":"2024-11-10T11:15:54Z","content_type":"text/html","content_length":"14202","record_id":"<urn:uuid:6fdbb446-677b-45f3-b43f-b4254182fe0b>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00791.warc.gz"} |
stzrqf.f - Linux Manuals (3)
stzrqf.f (3) - Linux Manuals
stzrqf.f -
subroutine stzrqf (M, N, A, LDA, TAU, INFO)
Function/Subroutine Documentation
subroutine stzrqf (integerM, integerN, real, dimension( lda, * )A, integerLDA, real, dimension( * )TAU, integerINFO)
This routine is deprecated and has been replaced by routine STZRZF.
STZRQF reduces the M-by-N ( M<=N ) real upper trapezoidal matrix A
to upper triangular form by means of orthogonal transformations.
The upper trapezoidal matrix A is factored as
A = ( R 0 ) * Z,
where Z is an N-by-N orthogonal matrix and R is an M-by-M upper
triangular matrix.
M is INTEGER
The number of rows of the matrix A. M >= 0.
N is INTEGER
The number of columns of the matrix A. N >= M.
A is REAL array, dimension (LDA,N)
On entry, the leading M-by-N upper trapezoidal part of the
array A must contain the matrix to be factorized.
On exit, the leading M-by-M upper triangular part of A
contains the upper triangular matrix R, and elements M+1 to
N of the first M rows of A, with the array TAU, represent the
orthogonal matrix Z as a product of M elementary reflectors.
LDA is INTEGER
The leading dimension of the array A. LDA >= max(1,M).
TAU is REAL array, dimension (M)
The scalar factors of the elementary reflectors.
INFO is INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
November 2011
Further Details:
The factorization is obtained by Householder's method. The kth
transformation matrix, Z( k ), which is used to introduce zeros into
the ( m - k + 1 )th row of A, is given in the form
Z( k ) = ( I 0 ),
( 0 T( k ) )
T( k ) = I - tau*u( k )*u( k )**T, u( k ) = ( 1 ),
( 0 )
( z( k ) )
tau is a scalar and z( k ) is an ( n - m ) element vector.
tau and z( k ) are chosen to annihilate the elements of the kth row
of X.
The scalar tau is returned in the kth element of TAU and the vector
u( k ) in the kth row of A, such that the elements of z( k ) are
in a( k, m + 1 ), ..., a( k, n ). The elements of R are returned in
the upper triangular part of A.
Z is given by
Z = Z( 1 ) * Z( 2 ) * ... * Z( m ).
Definition at line 139 of file stzrqf.f.
Generated automatically by Doxygen for LAPACK from the source code. | {"url":"https://www.systutorials.com/docs/linux/man/3-stzrqf.f/","timestamp":"2024-11-13T22:39:47Z","content_type":"text/html","content_length":"9450","record_id":"<urn:uuid:30cb907c-cdb1-4bc8-8a32-76c388e3c575>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00140.warc.gz"} |
Colors and phases of the moon question | The Flat Earth Files
Question! I'm still in the research phase of FE so I have LOTS of questions haha
How do Biblical FE'ers explain the phases of the moon and the different colors of the moon? Last night I spent a couple hours looking at the moon through my telescope and it was such a brilliant
white it almost seemed pearlescent. Tonight it is straight up orange! So beautiful BUT why?! I know the heliocentric model explains it away as different light wave lengths from the sun being
Thanks everyone!
Comments (18)
That is a great question! I don't have a reason but it has been happenening since earliest recordings. The sun will also chage hues or tones, sometimes being a light yellow and sometimes a deep red.
As someone who has sort of found both God and the Geocentric/bibilical earth more recently That is a great question, but I would not hold much merit on whether this proves the shape of the earth.
With that said I am going to see what I can dig up!
Replying to
Poignant observation! The sun does chanage a lot, especially when they spray enough chemtrails across it to make it white!
This is most likely something we will never have an answer to on this side of eternity. We can't get to the moon to study it closely. all we can do is observe it from a distance and make suppositions
and hypotheses.
I"m really interested in a good answer to this, because I was talking with a friend, and this was his main question. My best understanding is a "charging" and "discharging" of the moon, but I haven't
seen that model supported in a meaningful way yet.
Colours I think comes from chemtrails and atmosphere. If it's orange one night, it 's because of what's between us and the moon, not the moon itself.
Replying to
I have not gone down the chem trail rabbit hole yet but if it was chem trails, aren't they local? Would multiple people spread out over a large distance see the same colored moon?
"In the beginning was the Word, and the Word was with God, and the Word WAS God."
"For the word of God is living and powerful, and sharper than any two-edged sword, piercing even to the division of soul and spirit, and of joints and marrow, and is a discerner of the thoughts and
intents of the heart."
"12Then I turned to see the voice that was speaking with me. And having turned, I saw seven golden lampstands, 13and among the lampstands was One like the Son of Man,f dressed in a long robe, with a
golden sash around His chest. 14The hair of His head was white like wool, as white as snow, and His eyes were like a blazing fire. 15 His feet were like polished bronze refined in a furnace, and His
voice was like the roar of many waters. 16 He held in His right hand seven stars, and a sharp double-edged sword came from His mouth. His face was like the sun shining at its brightest."
"And to the angel of the church in Pergamum write: ‘The words of him who has the sharp two-edged sword." (Rev 2)
"Now I saw heaven opened, and behold, a white horse. And He who sat on him was called Faithful and True, and in righteousness He judges and makes war. 12 His eyes were like a flame of fire, and on
His head were many crowns. He [a]had a name written that no one knew except Himself. 13 He was clothed with a robe dipped in blood, and His name is called The Word of God. 14 And the armies in
heaven, clothed in [b]fine linen, white and clean, followed Him on white horses. 15 Now out of His mouth goes a sharp sword" - Rev 19.
Turbanhead, it is not with us or with the Bible that your dispute is. It's with Jesus Christ Himself. He IS the Word of God, and He promised that Heaven and Earth will pass away, but His Words will
never pass away. He is also the Creator of this incredible world we live in, whatever it may be, however it may be.
That being in that picture you shared, that being and all its friends, if they can call each other friends, hate the Word of God with all they have. They tremble at His word. And trust me, they hate
you too, because you are made in God's image.
As I can tell from what many others have written, we love the Bible. The words of God Himself are in it. His words have changed our lives. "He sent forth His word and healed us and rescued us from
the pit and destruction." His Word is soothing to the soul, and strengthening to our spirit. So your curse against God's Word is not to us - it's to Him, the One Who inspired it.
So take it up with Jesus Himself. But in this company, I cannot allow you to curse a book most precious to me, that if my house was burning down and my loved ones were safe, would be the first thing
I would save.
Also something I find very interesting is if you look at the stars and the planets and the moon in a telescope, like the one I have sitting on my deck right now, nothing fancy, and you zoom past the
focus point of the object, every single one of them turns into a disc with a black hole in the center. Yes I know the scientific explanation for the way our eyes see light….BUT it also looks very
similar to what a literal flashlight looks like.
Many flat earthers think that the moon is a flat, translucent disc. Also, they think it is giving off cold light. Furthermore, they believe that it is inside a huge dome called the firmament. Had
anybody ever looked through a telescope at it? How do they explain the craters, and how those craters are always in the same spot? Also, why is it that only the side facing the sunrise is lit up?
Look at this picture, zoom in on it, does that look like a flat, transluvent disc?
Replying to
We can see further then the math of the Earth states should be possible. So it's either flat, or a lot larger then what the same people who claim the Earth is round say it is. Not to mention that we
can't prove the earth is moving with experiments to back up the math of Helicentrism. Either way they are lying, hiding, or not telling the whole truth for what ever reason.
As you can see by this image, things are not always what they seem appear, but with little to no effort with a simple drawing instrument we can fool even ourselves. If we want to understand our
world, and find the truth we have to remain vigilant to deception. Test our ideas with real true experiments, and see if things are even possible not just use an equation to measure a distance we
have no way to verify the math.
I think most people who believe in Geocentric, or a Biblical earth model would agree in one thing. We have questions that need more then math to satisfy, and our experiments can dispute the math.
Replying to
Math is a langauage, a way of demarcating boundaries, a manner of positing horizons for human endeavors. Nevertheless, we can call anything 4....4 shoes, four tables, four cars, etc.....but this 4
isn't really any of them unless this 4 is intended "for" some kind of human use, which may be no more than to observe them as four.
Replying to
I think this is a major point you raise and deserves a good reply. I personally have observed the sun and moon being high enough in the sky to be visible and almost directly across one another
recently and it left me with some questions: 1. The moon, if heliocentric view is correct, ought to have been fully lit up on its surface as the sun was in same part of the sky almost directly
opposite but it was not
2. The bottom portion of the moon was not visible but rather than having a domelike shape you would expect if the earth was blocking sunlight it was curved like a smile..what..why? 3. Is there a good
explanation for this as I am unaware of one if there is and it has me definitely more in the FE camp. 4. But I would still like an explanation as to why there are images of planets that seem to be
like what you would expect from NASA and telescope viewing and then images that look like circular water vibration at times. | {"url":"https://www.theflatearthfiles.com/forum/questions-answers/colors-and-phases-of-the-moon-question","timestamp":"2024-11-06T02:47:29Z","content_type":"text/html","content_length":"1050489","record_id":"<urn:uuid:e0923f07-070b-4402-b8ee-c795e3253991>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00166.warc.gz"} |
A near-optimal subdivision algorithm for complex root isolation based on the Pellet test and Newton iteration
We describe a subdivision algorithm for isolating the complex roots of a polynomial F∈C[x]. Given an oracle that provides approximations of each of the coefficients of F to any absolute error bound
and given an arbitrary square B in the complex plane containing only simple roots of F, our algorithm returns disjoint isolating disks for the roots of F in B. Our complexity analysis bounds the
absolute error to which the coefficients of F have to be provided, the total number of iterations, and the overall bit complexity. It further shows that the complexity of our algorithm is controlled
by the geometry of the roots in a near neighborhood of the input square B, namely, the number of roots, their absolute values and pairwise distances. The number of subdivision steps is near-optimal.
For the benchmark problem, namely, to isolate all the roots of a polynomial of degree n with integer coefficients of bit size less than τ, our algorithm needs O˜(n^3+n^2τ) bit operations, which is
comparable to the record bound of Pan (2002). It is the first time that such a bound has been achieved using subdivision methods, and independent of divide-and-conquer techniques such as Schönhage's
splitting circle technique. Our algorithm uses the quadtree construction of Weyl (1924) with two key ingredients: using Pellet's Theorem (1881) combined with Graeffe iteration, we derive a
“soft-test” to count the number of roots in a disk. Using Schröder's modified Newton operator combined with bisection, in a form inspired by the quadratic interval method from Abbot (2006), we
achieve quadratic convergence towards root clusters. Relative to the divide-conquer algorithms, our algorithm is quite simple with the potential of being practical. This paper is self-contained: we
provide pseudo-code for all subroutines used by our algorithm.
• Approximate arithmetic
• Certified computation
• Complex roots
• Complexity analysis
• Root finding
• Root isolation
• Subdivision methods
ASJC Scopus subject areas
• Algebra and Number Theory
• Computational Mathematics
Dive into the research topics of 'A near-optimal subdivision algorithm for complex root isolation based on the Pellet test and Newton iteration'. Together they form a unique fingerprint. | {"url":"https://nyuscholars.nyu.edu/en/publications/a-near-optimal-subdivision-algorithm-for-complex-root-isolation-b","timestamp":"2024-11-09T23:13:14Z","content_type":"text/html","content_length":"53924","record_id":"<urn:uuid:427ca806-c0dd-466f-bc3f-77fa35338bf0>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00061.warc.gz"} |
Dirac Operators in Riemannian Geometrysearch
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Dirac Operators in Riemannian Geometry
Hardcover ISBN: 978-0-8218-2055-1
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Dirac Operators in Riemannian Geometry
Hardcover ISBN: 978-0-8218-2055-1
Product Code: GSM/25
List Price: $99.00
MAA Member Price: $89.10
AMS Member Price: $79.20
eBook ISBN: 978-1-4704-2080-2
Product Code: GSM/25.E
List Price: $85.00
MAA Member Price: $76.50
AMS Member Price: $68.00
Hardcover ISBN: 978-0-8218-2055-1
eBook ISBN: 978-1-4704-2080-2
Product Code: GSM/25.B
List Price: $184.00 $141.50
MAA Member Price: $165.60 $127.35
AMS Member Price: $147.20 $113.20
• Graduate Studies in Mathematics
Volume: 25; 2000; 195 pp
MSC: Primary 58; Secondary 53; 57; 81
For a Riemannian manifold \(M\), the geometry, topology and analysis are interrelated in ways that are widely explored in modern mathematics. Bounds on the curvature can have significant
implications for the topology of the manifold. The eigenvalues of the Laplacian are naturally linked to the geometry of the manifold. For manifolds that admit spin (or \(\mathrm{spin}^\mathbb{C}
\)) structures, one obtains further information from equations involving Dirac operators and spinor fields. In the case of four-manifolds, for example, one has the remarkable Seiberg-Witten
In this text, Friedrich examines the Dirac operator on Riemannian manifolds, especially its connection with the underlying geometry and topology of the manifold. The presentation includes a
review of Clifford algebras, spin groups and the spin representation, as well as a review of spin structures and \(\mathrm{spin}^\mathbb{C}\) structures. With this foundation established, the
Dirac operator is defined and studied, with special attention to the cases of Hermitian manifolds and symmetric spaces. Then, certain analytic properties are established, including
self-adjointness and the Fredholm property.
An important link between the geometry and the analysis is provided by estimates for the eigenvalues of the Dirac operator in terms of the scalar curvature and the sectional curvature.
Considerations of Killing spinors and solutions of the twistor equation on \(M\) lead to results about whether \(M\) is an Einstein manifold or conformally equivalent to one. Finally, in an
appendix, Friedrich gives a concise introduction to the Seiberg-Witten invariants, which are a powerful tool for the study of four-manifolds. There is also an appendix reviewing principal bundles
and connections.
This detailed book with elegant proofs is suitable as a text for courses in advanced differential geometry and global analysis, and can serve as an introduction for further study in these areas.
This edition is translated from the German edition published by Vieweg Verlag.
Graduate students and researchers in mathematics or physics.
□ Chapters
□ Chapter 1. Clifford algebras and spin representation
□ Chapter 2. Spin structures
□ Chapter 3. Dirac operators
□ Chapter 4. Analytical properties of Dirac operators
□ Chapter 5. Eigenvalue estimates for the Dirac operator and twistor spinors
□ Appendix A. Seiberg-Witten invariants
□ Appendix B. Principal bundles and connections
□ This book is a nice introduction to the theory of spinors and Dirac operators on Riemannian manifolds ... contains a nicely written description of the Seiberg-Witten theory of invariants for
4-dimensional manifolds ... This book can be strongly recommended to anybody interested in the theory of Dirac and related operators.
European Mathematical Society Newsletter
□ From a review of the German edition:
This work is to a great extent a written version of lectures given by the author. As a consequence of this fact, the text contains full, detailed and elegant proofs throughout, all
calculations are carefully performed, and considerations are well formulated and well motivated. This style is typical of the author. It is a pleasure to read the book; any beginning graduate
student should have access to it.
Mathematical Reviews
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Additional Material
• Reviews
• Requests
Volume: 25; 2000; 195 pp
MSC: Primary 58; Secondary 53; 57; 81
For a Riemannian manifold \(M\), the geometry, topology and analysis are interrelated in ways that are widely explored in modern mathematics. Bounds on the curvature can have significant implications
for the topology of the manifold. The eigenvalues of the Laplacian are naturally linked to the geometry of the manifold. For manifolds that admit spin (or \(\mathrm{spin}^\mathbb{C}\)) structures,
one obtains further information from equations involving Dirac operators and spinor fields. In the case of four-manifolds, for example, one has the remarkable Seiberg-Witten invariants.
In this text, Friedrich examines the Dirac operator on Riemannian manifolds, especially its connection with the underlying geometry and topology of the manifold. The presentation includes a review of
Clifford algebras, spin groups and the spin representation, as well as a review of spin structures and \(\mathrm{spin}^\mathbb{C}\) structures. With this foundation established, the Dirac operator is
defined and studied, with special attention to the cases of Hermitian manifolds and symmetric spaces. Then, certain analytic properties are established, including self-adjointness and the Fredholm
An important link between the geometry and the analysis is provided by estimates for the eigenvalues of the Dirac operator in terms of the scalar curvature and the sectional curvature. Considerations
of Killing spinors and solutions of the twistor equation on \(M\) lead to results about whether \(M\) is an Einstein manifold or conformally equivalent to one. Finally, in an appendix, Friedrich
gives a concise introduction to the Seiberg-Witten invariants, which are a powerful tool for the study of four-manifolds. There is also an appendix reviewing principal bundles and connections.
This detailed book with elegant proofs is suitable as a text for courses in advanced differential geometry and global analysis, and can serve as an introduction for further study in these areas. This
edition is translated from the German edition published by Vieweg Verlag.
Graduate students and researchers in mathematics or physics.
• Chapters
• Chapter 1. Clifford algebras and spin representation
• Chapter 2. Spin structures
• Chapter 3. Dirac operators
• Chapter 4. Analytical properties of Dirac operators
• Chapter 5. Eigenvalue estimates for the Dirac operator and twistor spinors
• Appendix A. Seiberg-Witten invariants
• Appendix B. Principal bundles and connections
• This book is a nice introduction to the theory of spinors and Dirac operators on Riemannian manifolds ... contains a nicely written description of the Seiberg-Witten theory of invariants for
4-dimensional manifolds ... This book can be strongly recommended to anybody interested in the theory of Dirac and related operators.
European Mathematical Society Newsletter
• From a review of the German edition:
This work is to a great extent a written version of lectures given by the author. As a consequence of this fact, the text contains full, detailed and elegant proofs throughout, all calculations
are carefully performed, and considerations are well formulated and well motivated. This style is typical of the author. It is a pleasure to read the book; any beginning graduate student should
have access to it.
Mathematical Reviews
Permission – for use of book, eBook, or Journal content
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Sony Alpha NEX-5
Sony Alpha NEX-5
Brand: Sony
Model: Alpha NEX-5
Megapixels: 14.20
Sensor: 23.4 x 15.6 mm
Price: check here »
Sensor info
Sony Alpha NEX-5 comes with a 23.4 x 15.6 mm CMOS sensor, which has a diagonal of 28.12 mm (1.11") and a surface area of 365.04 mm².
Pixel density
3.89 MP/cm²
If you want to know about the accuracy of these numbers, click here
Actual sensor size
Actual size is set to screen →
change »
This is the actual size of the Alpha NEX-5 sensor: 23.4 x 15.6 mm
The sensor has a surface area of
mm². There are approx. 14,200,000 photosites (pixels) on this area. Pixel pitch, which is a measure of the distance between pixels, is
µm. Pixel pitch tells you the distance from the center of one pixel (photosite) to the center of the next.
Pixel or photosite area is
µm². The larger the photosite, the more light it can capture and the more information can be recorded.
Pixel density tells you how many million pixels fit or would fit in one square cm of the sensor. Sony Alpha NEX-5 has a pixel density of
These numbers are important in terms of assessing the overall quality of a digital camera. Generally, the bigger (and newer) the sensor, pixel pitch and photosite area, and the smaller the pixel
density, the better the camera. If you want to see how Alpha NEX-5 compares to other cameras,
Brand: Sony
Model: Alpha NEX-5
Effective megapixels: 14.20
Total megapixels: 14.60
Sensor size: 23.4 x 15.6 mm
Sensor type: CMOS
Sensor resolution: 4616 x 3077
Max. image resolution: 4592 x 3056
Crop factor: 1.54
Optical zoom:
Digital zoom: No
ISO: Auto, 200, 400, 800, 1600, 3200, 6400, 12800
RAW support:
Manual focus:
Normal focus range:
Macro focus range:
Focal length (35mm equiv.):
Aperture priority: Yes
Max aperture:
Max. aperture (35mm equiv.): n/a
Depth of field: simulate →
Metering: Multi, Center-weighted, Spot
Exposure Compensation: ±2 EV (in 1/3 EV steps)
Shutter priority: Yes
Min. shutter speed: 30 sec
Max. shutter speed: 1/4000 sec
Built-in flash:
External flash:
Viewfinder: None
White balance presets: 6
Screen size: 3"
Screen resolution: 920,000 dots
Video capture:
Storage types: SD/ SDHC/SDXC, Memory Stick Pro Duo/ Pro-HG Duo
USB: USB 2.0 (480 Mbit/sec)
Battery: Lithium-Ion NP-FW50 rechargeable battery
Weight: 287 g
Dimensions: 111 x 59 x 38 mm
Year: 2010
Compare Alpha NEX-5 with another camera
Diagonal is calculated by the use of Pythagorean theorem:
= sensor width and
= sensor height
Sony Alpha NEX-5 diagonal:
= 23.40 mm
= 15.60 mm
Diagonal = √ 23.40² + 15.60² = 28.12 mm
Surface area
Surface area is calculated by multiplying the width and the height of a sensor.
Width = 23.40 mm
Height = 15.60 mm
Surface area = 23.40 × 15.60 = 365.04 mm²
Pixel pitch
Pixel pitch is the distance from the center of one pixel to the center of the next measured in micrometers (µm). It can be calculated with the following formula:
Pixel pitch = sensor width in mm × 1000
sensor resolution width in pixels
Sony Alpha NEX-5 pixel pitch:
Sensor width
= 23.40 mm
Sensor resolution width
= 4616 pixels
Pixel pitch = 23.40 × 1000 = 5.07 µm
Pixel area
The area of one pixel can be calculated by simply squaring the pixel pitch:
Pixel area = pixel pitch²
You could also divide sensor surface area with effective megapixels:
Pixel area = sensor surface area in mm²
effective megapixels
Sony Alpha NEX-5 pixel area:
Pixel pitch = 5.07 µm
Pixel area = 5.07² = 25.7 µm²
Pixel density
Pixel density can be calculated with the following formula:
Pixel density = ( sensor resolution width in pixels )² / 1000000
sensor width in cm
You could also use this formula:
Pixel density = effective megapixels × 1000000 / 10000
sensor surface area in mm²
Sony Alpha NEX-5 pixel density:
Sensor resolution width = 4616 pixels
Sensor width = 2.34 cm
Pixel density = (4616 / 2.34)² / 1000000 = 3.89 MP/cm²
Sensor resolution
Sensor resolution is calculated from sensor size and effective megapixels. It's slightly higher than maximum (not interpolated) image resolution which is usually stated on camera specifications.
Sensor resolution is used in pixel pitch, pixel area, and pixel density formula. For sake of simplicity, we're going to calculate it in 3 stages.
1. First we need to find the ratio between horizontal and vertical length by dividing the former with the latter (aspect ratio). It's usually 1.33 (4:3) or 1.5 (3:2), but not always.
2. With the ratio (
) known we can calculate the
from the formula below, where
is a vertical number of pixels:
(X × r) × X = effective megapixels × 1000000 → X = √ effective megapixels × 1000000
3. To get sensor resolution we then multiply
with the corresponding ratio:
Resolution horizontal:
Resolution vertical:
Sony Alpha NEX-5 sensor resolution:
Sensor width
= 23.40 mm
Sensor height
= 15.60 mm
Effective megapixels
= 14.20
r = 23.40/15.60 = 1.5 X = √ 14.20 × 1000000 = 3077
Resolution horizontal: X × r = 3077 × 1.5 = 4616
Resolution vertical: X = 3077
Sensor resolution =
4616 x 3077
Crop factor
Crop factor or focal length multiplier is calculated by dividing the diagonal of 35 mm film (43.27 mm) with the diagonal of the sensor.
Crop factor = 43.27 mm
sensor diagonal in mm
Sony Alpha NEX-5 crop factor:
Sensor diagonal
= 28.12 mm
Crop factor = 43.27 = 1.54
35 mm equivalent aperture
Equivalent aperture (in 135 film terms) is calculated by multiplying lens aperture with crop factor (a.k.a. focal length multiplier).
Sony Alpha NEX-5 equivalent aperture:
Aperture is a lens characteristic, so it's calculated only for fixed lens cameras. If you want to know the equivalent aperture for Sony Alpha NEX-5, take the aperture of the lens you're using and
multiply it with crop factor.
Crop factor for Sony Alpha NEX-5 is 1.54
Enter your screen size (diagonal)
My screen size is inches
Actual size is currently adjusted to screen.
If your screen (phone, tablet, or monitor) is not in diagonal, then the actual size of a sensor won't be shown correctly. | {"url":"https://www.digicamdb.com/specs/sony_alpha-nex-5/","timestamp":"2024-11-01T19:24:07Z","content_type":"text/html","content_length":"29997","record_id":"<urn:uuid:f8f9f7ea-2583-4e1f-b53c-b85fb3cdaad3>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00305.warc.gz"} |
Comparison Test/Limit Comparison Test
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Section 10.7 : Comparison Test/Limit Comparison Test
In the previous section we saw how to relate a series to an improper integral to determine the convergence of a series. While the integral test is a nice test, it does force us to do improper
integrals which aren’t always easy and, in some cases, may be impossible to determine the convergence of.
For instance, consider the following series.
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n} + n}}} \]
In order to use the Integral Test we would have to integrate
\[\int_{{\,0}}^{{\,\infty }}{{\frac{1}{{{3^x} + x}}\,dx}}\]
and we're not even sure if it’s possible to do this integral. Nicely enough for us there is another test that we can use on this series that will be much easier to use.
First, let’s note that the series terms are positive. As with the Integral Test that will be important in this section. Next let’s note that we must have \(x > 0\) since we are integrating on the
interval \(0 \le x < \infty \). Likewise, regardless of the value of \(x\) we will always have \({3^x} > 0\). So, if we drop the \(x\) from the denominator the denominator will get smaller and hence
the whole fraction will get larger. So,
\[\frac{1}{{{3^n} + n}} < \frac{1}{{{3^n}}}\]
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} \]
is a geometric series and we know that since \(\left| r \right| = \left| {\frac{1}{3}} \right| < 1\) the series will converge and its value will be,
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} = \frac{1}{{1 - \frac{1}{3}}} = \frac{3}{2}\]
Now, if we go back to our original series and write down the partial sums we get,
\[{s_n} = \sum\limits_{i = 0}^n {\frac{1}{{{3^i} + i}}} \]
Since all the terms are positive adding a new term will only make the number larger and so the sequence of partial sums must be an increasing sequence.
\[{s_n} = \sum\limits_{i = 0}^n {\frac{1}{{{3^i} + i}}} < \sum\limits_{i = 0}^{n + 1} {\frac{1}{{{3^i} + i}}} = {s_{n + 1}}\]
Then since,
\[\frac{1}{{{3^n} + n}} < \frac{1}{{{3^n}}}\]
and because the terms in these two sequences are positive we can also say that,
\[{s_n} = \sum\limits_{i = 0}^n {\frac{1}{{{3^i} + i}}} < \sum\limits_{i = 0}^n {\frac{1}{{{3^i}}} < \sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}} = \frac{3}{2}\hspace{0.25in}\hspace{0.25in} \
Rightarrow \hspace{0.25in}\,\,\,\,\,{s_n} < \frac{3}{2}} } \]
Therefore, the sequence of partial sums is also a bounded sequence. Then from the second section on sequences we know that a monotonic and bounded sequence is also convergent.
So, the sequence of partial sums of our series is a convergent sequence. This means that the series itself,
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n} + n}}} \]
is also convergent.
So, what did we do here? We found a series whose terms were always larger than the original series terms and this new series was also convergent. Then since the original series terms were positive
(very important) this meant that the original series was also convergent.
To show that a series (with only positive terms) was divergent we could go through a similar argument and find a new divergent series whose terms are always smaller than the original series. In this
case the original series would have to take a value larger than the new series. However, since the new series is divergent its value will be infinite. This means that the original series must also be
infinite and hence divergent.
We can summarize all this in the following test.
Comparison Test
Suppose that we have two series \(\displaystyle \sum {{a_n}} \) and \(\displaystyle \sum {{b_n}} \) with \({a_n},{b_n} \ge 0\) for all \(n\) and \({a_n} \le {b_n}\) for all \(n\). Then,
1. If \(\displaystyle \sum {{b_n}} \) is convergent then so is \(\sum {{a_n}} \).
2. If \(\displaystyle \sum {{a_n}} \) is divergent then so is \(\sum {{b_n}} \).
In other words, we have two series of positive terms and the terms of one of the series is always larger than the terms of the other series. Then if the larger series is convergent the smaller series
must also be convergent. Likewise, if the smaller series is divergent then the larger series must also be divergent. Note as well that in order to apply this test we need both series to start at the
same place.
A formal proof of this test is at the end of this section.
Do not misuse this test. Just because the smaller of the two series converges does not say anything about the larger series. The larger series may still diverge. Likewise, just because we know that
the larger of two series diverges we can’t say that the smaller series will also diverge! Be very careful in using this test
Recall that we had a similar test for improper integrals back when we were looking at integration techniques. So, if you could use the comparison test for improper integrals you can use the
comparison test for series as they are pretty much the same idea.
Note as well that the requirement that \({a_n},{b_n} \ge 0\) and \({a_n} \le {b_n}\) really only need to be true eventually. In other words, if a couple of the first terms are negative or \({a_n}\
require{cancel} \cancel{ \le }\,{b_n}\) for a couple of the first few terms we’re okay. As long as we eventually reach a point where \({a_n},{b_n} \ge 0\) and \({a_n} \le {b_n}\) for all sufficiently
large \(n\) the test will work.
To see why this is true let’s suppose that the series start at \(n = k\) and that the conditions of the test are only true for for \(n \ge N + 1\) and for \(k \le n \le N\) at least one of the
conditions is not true. If we then look at \(\sum {{a_n}} \) (the same thing could be done for \(\sum {{b_n}} \)) we get,
\[\sum\limits_{\,n = k}^\infty {{a_n}} = \sum\limits_{\,n = k}^N {{a_n}} + \sum\limits_{\,n = N + 1}^\infty {{a_n}} \]
The first series is nothing more than a finite sum (no matter how large \(N\) is) of finite terms and so will be finite. So, the original series will be convergent/divergent only if the second
infinite series on the right is convergent/divergent and the test can be done on the second series as it satisfies the conditions of the test.
Let’s take a look at some examples.
Example 1
Determine if the following series is convergent or divergent. \[\sum\limits_{n = 1}^\infty {\frac{n}{{{n^2} - {{\cos }^2}\left( n \right)}}} \]
Show Solution
Since the cosine term in the denominator doesn’t get too large we can assume that the series terms will behave like,
\[\frac{n}{{{n^2}}} = \frac{1}{n}\]
which, as a series, will diverge. So, from this we can guess that the series will probably diverge and so we’ll need to find a smaller series that will also diverge.
Recall that from the comparison test with improper integrals that we determined that we can make a fraction smaller by either making the numerator smaller or the denominator larger. In this case the
two terms in the denominator are both positive. So, if we drop the cosine term we will in fact be making the denominator larger since we will no longer be subtracting off a positive quantity.
\[\frac{n}{{{n^2} - {{\cos }^2}\left( n \right)}} > \frac{n}{{{n^2}}} = \frac{1}{n}\]
Then, since
\[\sum\limits_{n = 1}^\infty {\frac{1}{n}} \]
diverges (it’s harmonic or the \(p\)-series test) by the Comparison Test our original series must also diverge.
Example 2
Determine if the following series is convergent or divergent. \[\sum\limits_{n = 1}^\infty {\frac{{{{\bf{e}}^{ - n}}}}{{n + {{\cos }^2}\left( n \right)}}} \]
Show Solution
This example looks somewhat similar to the first one but we are going to have to be careful with it as there are some significant differences.
First, as with the first example the cosine term in the denominator will not get very large and so it won’t affect the behavior of the terms in any meaningful way. Therefore, the temptation at this
point is to focus in on the n in the denominator and think that because it is just an n the series will diverge.
That would be correct if we didn’t have much going on in the numerator. In this example, however, we also have an exponential in the numerator that is going to zero very fast. In fact, it is going to
zero so fast that it will, in all likelihood, force the series to converge.
So, let’s guess that this series will converge and we’ll need to find a larger series that will also converge.
First, because we are adding two positive numbers in the denominator we can drop the cosine term from the denominator. This will, in turn, make the denominator smaller and so the term will get larger
\[\frac{{{{\bf{e}}^{ - n}}}}{{n + {{\cos }^2}\left( n \right)}} \le \frac{{{{\bf{e}}^{ - n}}}}{n}\]
Next, we know that \(n \ge 1\) and so if we replace the n in the denominator with its smallest possible value (i.e. 1) the term will again get larger. Doing this gives,
\[\frac{{{{\bf{e}}^{ - n}}}}{{n + {{\cos }^2}\left( n \right)}} \le \frac{{{{\bf{e}}^{ - n}}}}{n} \le \frac{{{{\bf{e}}^{ - n}}}}{1} = {{\bf{e}}^{ - n}}\]
We can’t do much more, in a way that is useful anyway, to make this larger so let’s see if we can determine if,
\[\sum\limits_{n = 1}^\infty {{{\bf{e}}^{ - n}}} \]
converges or diverges.
We can notice that \(f\left( x \right) = {{\bf{e}}^{ - x}}\) is always positive and it is also decreasing (you can verify that correct?) and so we can use the Integral Test on this series. Doing this
\[\int_{1}^{\infty }{{{{\bf{e}}^{ - x}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \int_{1}^{t}{{{{\bf{e}}^{ - x}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left( { - {{\bf{e}}^{ -
x}}} \right)} \right|_1^t = \mathop {\lim }\limits_{t \to \infty } \left( { - {{\bf{e}}^{ - t}} + {{\bf{e}}^{ - 1}}} \right) = {{\bf{e}}^{ - 1}}\]
Okay, we now know that the integral is convergent and so the series \(\sum\limits_{n = 1}^\infty {{{\bf{e}}^{ - n}}} \) must also be convergent.
Therefore, because \(\sum\limits_{n = 1}^\infty {{{\bf{e}}^{ - n}}} \) is larger than the original series we know that the original series must also converge.
With each of the previous examples we saw that we can’t always just focus in on the denominator when making a guess about the convergence of a series. Sometimes there is something going on in the
numerator that will change the convergence of a series from what the denominator tells us should be happening.
We also saw in the previous example that, unlike most of the examples of the comparison test that we’ve done (or will do) both in this section and in the Comparison Test for Improper Integrals, that
it won’t always be the denominator that is driving the convergence or divergence. Sometimes it is the numerator that will determine if something will converge or diverge so do not get too locked into
only looking at the denominator.
One of the more common mistakes is to just focus in on the denominator and make a guess based just on that. If we’d done that with both of the previous examples we would have guessed wrong so be
Let’s work another example of the comparison test before we move on to a different topic.
Example 3
Determine if the following series converges or diverges. \[\sum\limits_{n = 1}^\infty {\frac{{{n^2} + 2}}{{{n^4} + 5}}} \]
Show Solution
In this case the “+2” and the “+5” don’t really add anything to the series and so the series terms should behave pretty much like
\[\frac{{{n^2}}}{{{n^4}}} = \frac{1}{{{n^2}}}\]
which will converge as a series. Therefore, we can guess that the original series will converge and we will need to find a larger series which also converges.
This means that we’ll either have to make the numerator larger or the denominator smaller. We can make the denominator smaller by dropping the “+5”. Doing this gives,
\[\frac{{{n^2} + 2}}{{{n^4} + 5}} < \frac{{{n^2} + 2}}{{{n^4}}}\]
At this point, notice that we can’t drop the “+2” from the numerator since this would make the term smaller and that’s not what we want. However, this is actually the furthest that we need to go.
Let’s take a look at the following series.
\[\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{n^2} + 2}}{{{n^4}}}} & = \sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{n^4}}}} + \sum\limits_{n = 1}^\infty {\frac{2}{{{n^4}}}} \\ & = \sum\limits_
{n = 1}^\infty {\frac{1}{{{n^2}}}} + \sum\limits_{n = 1}^\infty {\frac{2}{{{n^4}}}} \end{align*}\]
As shown, we can write the series as a sum of two series and both of these series are convergent by the \(p\)-series test. Therefore, since each of these series are convergent we know that the sum,
\[\sum\limits_{n = 1}^\infty {\frac{{{n^2} + 2}}{{{n^4}}}} \]
is also a convergent series. Recall that the sum of two convergent series will also be convergent.
Now, since the terms of this series are larger than the terms of the original series we know that the original series must also be convergent by the Comparison Test.
The comparison test is a nice test that allows us to do problems that either we couldn’t have done with the integral test or at the best would have been very difficult to do with the integral test.
That doesn’t mean that it doesn’t have problems of its own.
Consider the following series.
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n} - n}}} \]
This is not much different from the first series that we looked at. The original series converged because the \(3^{n}\) gets very large very fast and will be significantly larger than the \(n\).
Therefore, the \(n\) doesn’t really affect the convergence of the series in that case. The fact that we are now subtracting the \(n\) off instead of adding the \(n\) on really shouldn’t change the
convergence. We can say this because the \(3^{n}\) gets very large very fast and the fact that we’re subtracting \(n\) off won’t really change the size of this term for all sufficiently large values
of \(n\).
So, we would expect this series to converge. However, the comparison test won’t work with this series. To use the comparison test on this series we would need to find a larger series that we could
easily determine the convergence of. In this case we can’t do what we did with the original series. If we drop the \(n\) we will make the denominator larger (since the \(n\) was subtracted off) and
so the fraction will get smaller and just like when we looked at the comparison test for improper integrals knowing that the smaller of two series converges does not mean that the larger of the two
will also converge.
So, we will need something else to do help us determine the convergence of this series. The following variant of the comparison test will allow us to determine the convergence of this series.
Limit Comparison Test
Suppose that we have two series \(\sum {{a_n}} \) and \(\sum {{b_n}} \) with \({a_n} \ge 0,{b_n} > 0\) for all \(n\). Define,
\[c = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}}\]
If \(c\) is positive (i.e. \(c > 0\)) and is finite (i.e. \(c < \infty \)) then either both series converge or both series diverge.
The proof of this test is at the end of this section.
Note that it doesn’t really matter which series term is in the numerator for this test, we could just have easily defined \(c\) as,
\[c = \mathop {\lim }\limits_{n \to \infty } \frac{{{b_n}}}{{{a_n}}}\]
and we would get the same results. To see why this is, consider the following two definitions.
\[c = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}}\hspace{0.25in}\hspace{0.25in}\overline{c} = \mathop {\lim }\limits_{n \to \infty } \frac{{{b_n}}}{{{a_n}}}\]
Start with the first definition and rewrite it as follows, then take the limit.
\[c = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\,\,\frac{{{b_n}}}{{{a_n}}}\,\,}} = \frac{1}{{\mathop {\lim }\limits_{n \to \
infty } \frac{{{b_n}}}{{{a_n}}}}} = \frac{1}{{\overline{c}}}\]
In other words, if \(c\) is positive and finite then so is \(\overline{c}\) and if \(\overline{c}\) is positive and finite then so is \(c\). Likewise if \(\overline{c} = 0\) then \(c = \infty \) and
if \(\overline{c} = \infty \) then \(c = 0\). Both definitions will give the same results from the test so don’t worry about which series terms should be in the numerator and which should be in the
denominator. Choose this to make the limit easy to compute.
Also, this really is a comparison test in some ways. If \(c\) is positive and finite this is saying that both of the series terms will behave in generally the same fashion and so we can expect the
series themselves to also behave in a similar fashion. If \(c = 0\) or \(c = \infty \) we can’t say this and so the test fails to give any information.
The limit in this test will often be written as,
\[c = \mathop {\lim }\limits_{n \to \infty } {a_n} \cdot \,\,\frac{1}{{{b_n}}}\]
since often both terms will be fractions and this will make the limit easier to deal with.
Let’s see how this test works.
Example 4
Determine if the following series converges or diverges. \[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n} - n}}} \]
Show Solution
To use the limit comparison test we need to find a second series that we can determine the convergence of easily and has what we assume is the same convergence as the given series. On top of that we
will need to choose the new series in such a way as to give us an easy limit to compute for \(c\).
We’ve already guessed that this series converges and since it’s vaguely geometric let’s use
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} \]
as the second series. We know that this series converges and there is a chance that since both series have the 3^n in it the limit won’t be too bad.
Here’s the limit.
\[\begin{align*}c & = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{3^n}}}\frac{{{3^n} - n}}{1}\\ & = \mathop {\lim }\limits_{n \to \infty } 1 - \frac{n}{{{3^n}}}\end{align*}\]
Now, we’ll need to use L’Hospital’s Rule on the second term in order to actually evaluate this limit.
\[\begin{align*}c & = 1 - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{3^n}\ln \left( 3 \right)}}\\ & = 1\end{align*}\]
So, \(c\) is positive and finite so by the Comparison Test both series must converge since
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} \]
Example 5
Determine if the following series converges or diverges. \[\sum\limits_{n = 2}^\infty {\frac{{4{n^2} + n}}{{\sqrt[3]{{{n^7} + {n^3}}}}}} \]
Show Solution
Fractions involving only polynomials or polynomials under radicals will behave in the same way as the largest power of \(n\) will behave in the limit. So, the terms in this series should behave as,
\[\frac{{{n^2}}}{{\sqrt[3]{{{n^7}}}}} = \frac{{{n^2}}}{{{n^{\frac{7}{3}}}}} = \frac{1}{{{n^{\frac{1}{3}}}}}\]
and as a series this will diverge by the \(p\)-series test. In fact, this would make a nice choice for our second series in the limit comparison test so let’s use it.
\[\begin{align*}\mathop {\lim }\limits_{n \to \infty } \frac{{4{n^2} + n}}{{\sqrt[3]{{{n^7} + {n^3}}}}}\frac{{{n^{\frac{1}{3}}}}}{1} & = \mathop {\lim }\limits_{n \to \infty } \frac{{4{n^{\frac{7}
{3}}} + {n^{\frac{4}{3}}}}}{{\sqrt[3]{{{n^7}\left( {1 + \frac{1}{{{n^4}}}} \right)}}}}\\ & = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^{\frac{7}{3}}}\left( {4 + \frac{1}{n}} \right)}}{{{n^{\
frac{7}{3}}}\sqrt[3]{{1 + \frac{1}{{{n^4}}}}}}}\\ & = \frac{4}{{\sqrt[3]{1}}} = 4 = c\end{align*}\]
So, \(c\) is positive and finite and so both limits will diverge since
\[\sum\limits_{n = 2}^\infty {\frac{1}{{{n^{\frac{1}{3}}}}}} \]
Finally, to see why we need \(c\) to be positive and finite (i.e. \(c \ne 0\) and \(c \ne \infty \)) consider the following two series.
\[\sum\limits_{n = 1}^\infty {\frac{1}{n}} \hspace{0.25in}\hspace{0.25in}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \]
The first diverges and the second converges.
Now compute each of the following limits.
\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\centerdot \frac{{{n}^{2}}}{1}=\underset{n\to \infty }{\mathop{\lim }}\,n=\infty \hspace{0.5in} \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}
{{{n}^{2}}}\centerdot \frac{n}{1}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0\]
In the first case the limit from the limit comparison test yields \(c = \infty \) and in the second case the limit yields \(c = 0\). Clearly, both series do not have the same convergence.
Note however, that just because we get \(c = 0\) or \(c = \infty \) doesn’t mean that the series will have the opposite convergence. To see this consider the series,
\[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \hspace{0.25in}\hspace{0.25in}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \]
Both of these series converge and here are the two possible limits that the limit comparison test uses.
\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{{{n}^{3}}}\centerdot \frac{{{n}^{2}}}{1}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0 \hspace{0.5in} \underset{n\to \infty }{\mathop{\lim
}}\,\frac{1}{{{n}^{2}}}\centerdot \frac{{{n}^{3}}}{1}=\underset{n\to \infty }{\mathop{\lim }}\,n=\infty \]
So, even though both series had the same convergence we got both \(c = 0\) and \(c = \infty \).
The point of all of this is to remind us that if we get \(c = 0\) or \(c = \infty \) from the limit comparison test we will know that we have chosen the second series incorrectly and we’ll need to
find a different choice in order to get any information about the convergence of the series.
We’ll close out this section with proofs of the two tests.
Proof of Comparison Test
The test statement did not specify where each series should start. We only need to require that they start at the same place so to help with the proof we’ll assume that the series start at \(n = 1\).
If the series don’t start at \(n = 1\) the proof can be redone in exactly the same manner or you could use an index shift to start the series at \(n = 1\) and then this proof will apply.
We’ll start off with the partial sums of each series.
\[{s_n} = \sum\limits_{i = 1}^n {{a_i}} \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{t_n} = \sum\limits_{i = 1}^n {{b_i}} \]
Let’s notice a couple of nice facts about these two partial sums. First, because \({a_n},{b_n} \ge 0\) we know that,
\[\begin{align*}{s_n} & \le {s_n} + {a_{n + 1}} = \sum\limits_{i = 1}^n {{a_i}} + {a_{n + 1}} = \sum\limits_{i = 1}^{n + 1} {{a_i}} = {s_{n + 1}}\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}{s_n}
\le {s_{n + 1}}\\ {t_n} & \le {t_n} + {b_{n + 1}} = \sum\limits_{i = 1}^n {{b_i}} + {b_{n + 1}} = \sum\limits_{i = 1}^{n + 1} {{b_i}} = {t_{n + 1}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace
{0.25in}{t_n} \le {t_{n + 1}}\end{align*}\]
So, both partial sums form increasing sequences.
Also, because \({a_n} \le {b_n}\) for all \(n\) we know that we must have \({s_n} \le {t_n}\) for all \(n\).
With these preliminary facts out of the way we can proceed with the proof of the test itself.
Let’s start out by assuming that \(\sum\limits_{n = 1}^\infty {{b_n}} \) is a convergent series. Since \({b_n} \ge 0\) we know that,
\[{t_n} = \sum\limits_{i = 1}^n {{b_i}} \le \sum\limits_{i = 1}^\infty {{b_i}} \]
However, we also have established that \({s_n} \le {t_n}\) for all \(n\) and so for all \(n\) we also have,
\[{s_n} \le \sum\limits_{i = 1}^\infty {{b_i}} \]
Finally, since \(\sum\limits_{n = 1}^\infty {{b_n}} \) is a convergent series it must have a finite value and so the partial sums, \({s_n}\) are bounded above. Therefore, from the second section on
sequences we know that a monotonic and bounded sequence is also convergent and so \(\left\{ {{s_n}} \right\}_{n = 1}^\infty \) is a convergent sequence and so \(\sum\limits_{n = 1}^\infty {{a_n}} \)
is convergent.
Next, let’s assume that \(\sum\limits_{n = 1}^\infty {{a_n}} \) is divergent. Because \({a_n} \ge 0\) we then know that we must have \({s_n} \to \infty \) as \(n \to \infty \). However, we also know
that for all \(n\) we have\({s_n} \le {t_n}\) and therefore we also know that \({t_n} \to \infty \) as \(n \to \infty \).
So, \(\left\{ {{t_n}} \right\}_{n = 1}^\infty \) is a divergent sequence and so \(\sum\limits_{n = 1}^\infty {{b_n}} \) is divergent.
Proof of Limit Comparison Test
Because \(0 < c < \infty \) we can find two positive and finite numbers, \(m\) and \(M\), such that \(m < c < M\). Now, because \(c = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}}\)
we know that for large enough \(n\) the quotient \(\frac{{{a_n}}}{{{b_n}}}\) must be close to \(c\) and so there must be a positive integer \(N\) such that if \(n > N\) we also have,
\[m < \frac{{{a_n}}}{{{b_n}}} < M\]
Multiplying through by \({b_n}\) gives,
\[m{b_n} < {a_n} < M{b_n}\]
provided \(n > N\).
Now, if \(\sum {{b_n}} \) diverges then so does \(\sum {m{b_n}} \) and so since \(m{b_n} < {a_n}\) for all sufficiently large \(n\) by the Comparison Test \(\sum {{a_n}} \) also diverges.
Likewise, if \(\sum {{b_n}} \) converges then so does \(\sum {M{b_n}} \) and since \({a_n} < M{b_n}\) for all sufficiently large \(n\) by the Comparison Test \(\sum {{a_n}} \) also converges. | {"url":"https://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx","timestamp":"2024-11-14T11:22:53Z","content_type":"text/html","content_length":"102148","record_id":"<urn:uuid:ba6352a8-82e1-4c73-a449-1c463269dbb7>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00851.warc.gz"} |
The Stacks project
Lemma 107.5.22. Let $f: \mathcal{T} \to \mathcal{X}$ be a locally of finite type morphism of Jacobson, pseudo-catenary, and locally Noetherian algebraic stacks, whose source is irreducible and whose
target is quasi-separated, and let $\mathcal{Z} \hookrightarrow \mathcal{X}$ denote the scheme-theoretic image of $\mathcal{T}$. Then for all $t \in |T|$, we have that $\dim _ t( \mathcal{T}_{f(t)})
\geq \dim \mathcal{T} - \dim \mathcal{Z}$, and there is a non-empty (equivalently, dense) open subset of $|\mathcal{T}|$ over which equality holds.
Comments (4)
Comment #7549 by DatPham on
I guess in the statement of the Lemma, $t$ should be any point in $|\mathcal{T}|$ (rather than a finite type one)?
Comment #7673 by Stacks Project on
Yes, I guess you are correct. Thanks. Changed it here.
Comment #8707 by Haohao Liu on
By a "scheme-theoretically dominant" morphism, do you mean a morphism whose scheme-theoretic image is the full target?
Comment #9364 by Stacks project on
Yes. This term hasn't been defined in the Stacks project and I think we shouldn't. So the fix would be to restate in full the condition intended in each instance. I also think some more work
could be done on this chapter, e.g., to shorten proofs (by splitting out lemmas for example). Going to leave this as is for now.
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Meta-Complexity: A Basic Introduction for the Meta-Perplexed
by Adam Becker (science communicator in residence, Spring 2023)
Think about the last time you faced a problem you couldn’t solve. Say it was something practical, something that seemed small — a leaky faucet, for example. There’s an exposed screw right on the top
of the faucet handle, so you figure all you need to do is turn the faucet off as far as it will go, and then tighten that screw. So you try that, and it doesn’t work. You get a different screwdriver,
a better fit for the screw, but you can’t get it to budge. You grab a wrench and take apart the faucet handle, and that doesn’t help much either — it turns out there’s far more under there than you’d
expected, and you can barely put it back together again. You’re about to give up and call a plumber, but first you want to see whether you’re close. Maybe it really is easy to fix the problem, and
you just need to know where to look. Or maybe it’s far more difficult than you think. So now you’re trying to solve a new problem, a meta-problem: instead of fixing the leaky faucet, you’re trying to
figure out how hard it will be to fix the leaky faucet. You turn to the internet, and find that there are many different kinds of faucets and sinks, some of which are practically indistinguishable,
and there are different reasons they can leak, unique to each type of sink. Simply determining the difficulty of fixing your leaky faucet is itself turning out to be more difficult than you expected.
Theoretical computer scientists have been facing their own version of this problem for decades. Many of the problems they ask are about complexity: How hard must a computer (really, an idealized
version of one) work to perform a particular task? One such task, famous in the annals of both mathematics and computer science — theoretical computer science is where the two disciplines meet — is
the traveling salesperson problem. Imagine a traveling salesperson, going from city to city. Starting from her home, she has a list of cities she must visit, and a map with the distances between
those cities. Her budget limits the total distance she can travel to a certain maximum, so she’d like to find a route shorter than that maximum distance that allows her to visit each of the cities on
her list, returning to her home city at the end. Given her list of cities and her budget, does such a route exist?
There is no known method for solving this problem quickly in a general way — a method that would work for all possible budgets and lists of cities that the salesperson might have. There are ways of
doing it, but all of them take a large number of calculations relative to the number of cities on the list, and thus take a great deal of time, especially as the number of cities increases. In fact,
the shortest such guaranteed method known for solving the traveling salesperson problem takes, in general, an exponentially larger amount of time as the number of cities on the list increases,
because there’s no known way to do this that’s significantly faster than brute-forcing the problem by checking every possible route. Compare this with verifying a solution to the traveling
salesperson problem: that’s easy. All you have to do is confirm that the solution does in fact visit every city once, and that the total distance of the route is shorter than the maximum allowed by
the salesperson’s budget.
This property of the traveling salesperson problem — it seems like it can be solved in general only by a lengthy brute-force method, but it’s fast to verify a given solution — places it into a class
of “computational complexity” known as NP. (This stands for “nondeterministic polynomial time,” and it’s not particularly important to understand that name in order to understand what’s going on
here.) Compare this with a problem like determining whether the last entry on a list of numbers is the largest, for which there are known (and straightforward) methods that don’t scale exponentially
with the length of the list. Such problems, which can be solved and verified quickly, are in a complexity class called P, a special subset of NP.
On the face of it, NP and P seem to be different; the traveling salesperson problem (TSP) can’t be solved quickly by any known method. But the trouble, for computer scientists, begins with those
words “known method.” While nobody knows a fast way of solving a problem like the traveling salesperson problem, that doesn’t mean no such method exists. Finding such a method would show that TSP
actually belongs in P. In fact, it would show more than that, because computer scientists have proved that TSP is not just a member of NP — it is NP-complete: if there were an efficient solution to
TSP, it could be adapted to solve every other problem in NP quickly too. Therefore, a fast solution to TSP wouldn’t just show that TSP is part of P — it would show that every problem in NP is a
member of P, making P and NP the same complexity class. But if instead someone were to prove that there is no universally fast method for solving TSP, this would mean that TSP and many other
similarly difficult problems in NP aren’t in P, meaning that P and NP are not the same complexity class.
So which is it? Does P = NP or not? Nobody knows. This question has haunted theoretical computer science for well over half a century, resisting all attempts at solution — or even reasonable
progress. And like the leaky faucet, this difficulty has prompted computer scientists to think about a meta-problem: What’s the complexity of proving whether P = NP? How intricate must a proof that
resolves this question be? Is there a trick to it — is it the kind of thing that looks simple in retrospect? Or is it the sort of proof that requires a great deal of intricate mathematics and novel
proof techniques? This is meta-complexity: evaluating the complexity of questions that are themselves about computational complexity. The Simons Institute held a research program on the topic in
Spring 2023.
Meta-complexity isn’t a new idea. Starting in the late 1940s, pioneers in early computer science on both sides of the Iron Curtain were considering an optimization problem, like TSP, but about
idealized computers rather than an idealized salesperson. Specifically, they were thinking about small computers of unknown architecture: black boxes that can be studied only through their behavior.
Say you have one of these computers, a little black box that lets you input any whole number you like, up to a certain size. When you do, the box gives you either a 0 or a 1 as output. You want to
know what’s in the box, so you start going through inputs and outputs systematically, making a table. 0 gives you 1, 1 gives you 0, 2 gives you 1, and so on. The question these early computer
scientists were asking was this: Given a particular table of inputs and outputs, what is the least complex architecture that could be inside this black box doing the computing? If you have a “circuit
size budget” — like the traveling salesperson’s travel budget — is there a circuit small enough to fit within your budget that could do what the black box does? These questions became known as the
minimum circuit size problem (MCSP). Once these questions had been asked, the next one was: What’s the computational complexity of MCSP itself?
This is another form of meta-complexity: a question about the complexity of a problem that is itself about complexity. And this time, there’s a known answer. MCSP (at least the second version of it,
asking about circuits smaller than a certain size) is in NP: it’s easy to confirm that a solution is correct, but there doesn’t seem to be a general solution to the problem other than a brute-force
search. But is MCSP NP-complete? Is it as hard as the hardest problems in NP, like TSP is, and would a fast way of solving it — like solving TSP — mean proving all problems in NP are actually in P?
MCSP “seems to really capture that kind of flavor of an unstructured search space — circuits that don’t necessarily have much to do with each other — so shouldn’t you be able to show that not only is
MCSP contained in NP, but it is one of the hardest problems in NP, it is NP-complete?” said Marco Carmosino, research scientist at IBM, last year. “It is 2023 and we still have not proved that MCSP
is NP-[complete].”
These two forms of meta-complexity — questions about the difficulty of proofs about complexity classes, and questions about the complexity of problems about complexity — are linked. The first kind of
meta-complexity, about the difficulty of proofs about complexity, has roots stretching as far back as the work of legendary logician Kurt Gödel in the mid-20th century, as well as the origins of
modern logic and meta-mathematics around the turn of the 20th century, in the generations immediately preceding Gödel. But starting in the 1970s — not long after the first formal introduction of the
P = NP question — and continuing ever since, computer scientists started proving rigorous results about why such problems were difficult to solve. These “barrier” proofs showed that many common proof
techniques used in computer science simply could not solve questions like P vs. NP. Going back to the analogy of fixing the leaky faucet, these barrier proofs would be like finding out that using a
screwdriver or a wrench at all would doom you to failure.
But while barrier proofs could be seen as disheartening, they were also informative: they told computer scientists that they would be wasting their time to attempt a solution using those tools, and
that any real solution to the problem must lie elsewhere. As work continued over the following decades, computer scientists found further barriers and proofs. But recently, examining the structure of
those barriers has led to a burst of activity in meta-complexity, with new results making progress toward old problems like whether P = NP, as well as revealing unexpected connections within the
field. Computer scientists working in meta-complexity have not only shown links between various measures of complexity, but have also found deep connections between their own subfield and other areas
of computer science, like learning theory and cryptography. “The scope and centrality of meta-complexity has dramatically expanded over the past 10-ish years or so, as breakthroughs show that
cryptographic primitives and learning primitives end up being not just reducible to but equivalent to solutions to meta-computational problems. And that attracts attention — that attracts excitement.
And the proof techniques are very cool,” said Carmosino, who was a research fellow with the Institute's Meta-Complexity program. “And so it’s very rich, what’s going on right now. A dense network of
connections is all jelling together all at once. It's very exciting. … We can use [meta-complexity] as a tool to migrate techniques between these disparate areas of theoretical computer science and
show that, really, the field is more unified than it looks.” And with the perspective afforded by meta-complexity, perhaps P vs. NP — the leaky faucet that has been dripping away in the heart of
computer science for half a century — will, someday, yield to a solution. | {"url":"https://live-simons-institute.pantheon.berkeley.edu/news/meta-complexity-basic-introduction-meta-perplexed","timestamp":"2024-11-11T17:55:59Z","content_type":"text/html","content_length":"65085","record_id":"<urn:uuid:0777659a-1df7-4601-b521-87be792ef6e5>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00693.warc.gz"} |
Solution assignment 25 Product and Quotient rule
Return to Assignments Product and Quotient rule
Assignment 25
This function can be differentiated term by term, but the second function is not in the list of standard functions. However, we can write:
and thus:
Now we apply the quotient rule:
If you are familiar with the chain rule you can get the same result easier.
Return to Assignments Product and Quotient rule | {"url":"https://4mules.nl/en/product-and-quotient-rule/assignments/solution-assignment-25-product-and-quotient-rule/","timestamp":"2024-11-10T14:16:00Z","content_type":"text/html","content_length":"44017","record_id":"<urn:uuid:178620cb-85b1-461f-884a-316271952df1>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00044.warc.gz"} |
National Curriculum Primary Keystage 2 Year 4 Mathematics
Important note: National Curriculum content shared on this website is under the terms of the Open Government Licence. To view this licence, visit http://www.nationalarchives.gov.uk/doc/
open-government-licence/. You can download the full document at http://www.gov.uk/dfe/nationalcurriculum
Number – number and place value
Statutory requirements
Pupils should be taught to
• count in multiples of 6, 7, 9, 25 and 1000
• find 1000 more or less than a given number
• count backwards through zero to include negative numbers
• recognise the place value of each digit in a four-digit number (thousands, hundreds, tens, and ones)
• order and compare numbers beyond 1000
• identify, represent and estimate numbers using different representations
• round any number to the nearest 10, 100 or 1000
• solve number and practical problems that involve all of the above and with increasingly large positive numbers
• read Roman numerals to 100 (I to C) and know that over time, the numeral system changed to include the concept of zero and place value.
Notes and guidance (non-statutory)
Using a variety of representations, including measures, pupils become fluent in the order and place value of numbers beyond 1000, including counting in tens and hundreds, and maintaining fluency in
other multiples through varied and frequent practice.
They begin to extend their knowledge of the number system to include the decimal numbers and fractions that they have met so far.
They connect estimation and rounding numbers to the use of measuring instruments.
Roman numerals should be put in their historical context so pupils understand that there have been different ways to write whole numbers and that the important concepts of zero and place value were
introduced over a period of time.
Number – addition and subtraction
Statutory requirements
Pupils should be taught to:
• add and subtract numbers with up to 4 digits using the formal written methods of columnar addition and subtraction where appropriate
• estimate and use inverse operations to check answers to a calculation
• solve addition and subtraction two-step problems in contexts, deciding which operations and methods to use and why.
Notes and guidance (non-statutory)
Pupils continue to practise both mental methods and columnar addition and subtraction with increasingly large numbers to aid fluency (seeEnglish Appendix 1).
Number – multiplication and division
Statutory requirements
Pupils should be taught to:
• recall multiplication and division facts for multiplication tables up to 12 × 12
• use place value, known and derived facts to multiply and divide mentally, including: multiplying by 0 and 1; dividing by 1; multiplying together three numbers
• recognise and use factor pairs and commutativity in mental calculations
• multiply two-digit and three-digit numbers by a one-digit number using formal written layout
• solve problems involving multiplying and adding, including using the distributive law to multiply two digit numbers by one digit, integer scaling problems and harder correspondence problems such
as n objects are connected to m objects.
Notes and guidance (non-statutory)
Pupils continue to practise recalling and using multiplication tables and related division facts to aid fluency.
Pupils practise mental methods and extend this to three-digit numbers to derive facts, (for example 600 ÷ 3 = 200 can be derived from 2 x 3 = 6).
Pupils practise to become fluent in the formal written method of short multiplication and short division with exact answers (seeMathematics Appendix 1).
Pupils write statements about the equality of expressions (for example, use the distributive law 39 × 7 = 30 × 7 + 9 × 7 and associative law (2 × 3) × 4 = 2 × (3 × 4)). They combine their knowledge
of number facts and rules of arithmetic to solve mental and written calculations for example, 2 x 6 x 5 = 10 x 6 = 60.
Pupils solve two-step problems in contexts, choosing the appropriate operation, working with increasingly harder numbers. This should include correspondence questions such as the numbers of choices
of a meal on a menu, or three cakes shared equally between 10 children.
Number – fractions (including decimals)
Statutory requirements
Pupils should be taught to:
• recognise and show, using diagrams, families of common equivalent fractions
• count up and down in hundredths; recognise that hundredths arise when dividing an object by one hundred and dividing tenths by ten.
• solve problems involving increasingly harder fractions to calculate quantities, and fractions to divide quantities, including non-unit fractions where the answer is a whole number
• add and subtract fractions with the same denominator
• recognise and write decimal equivalents of any number of tenths or hundredths
• recognise and write decimal equivalents to,,
• find the effect of dividing a one- or two-digit number by 10 and 100, identifying the value of the digits in the answer as ones, tenths and hundredths
• round decimals with one decimal place to the nearest whole number
• compare numbers with the same number of decimal places up to two decimal places
• solve simple measure and money problems involving fractions and decimals to two decimal places.
Notes and guidance (non-statutory)
Pupils should connect hundredths to tenths and place value and decimal measure.
They extend the use of the number line to connect fractions, numbers and measures.
Pupils understand the relation between non-unit fractions and multiplication and division of quantities, with particular emphasis on tenths and hundredths.
Pupils make connections between fractions of a length, of a shape and as a representation of one whole or set of quantities. Pupils use factors and multiples to recognise equivalent fractions and
simplify where appropriate (for example, = or =).
Pupils continue to practise adding and subtracting fractions with the same denominator, to become fluent through a variety of increasingly complex problems beyond one whole.
Pupils are taught throughout that decimals and fractions are different ways of expressing numbers and proportions.
Pupils’ understanding of the number system and decimal place value is extended at this stage to tenths and then hundredths. This includes relating the decimal notation to division of whole number by
10 and later 100.
They practise counting using simple fractions and decimals, both forwards and backwards.
Pupils learn decimal notation and the language associated with it, including in the context of measurements. They make comparisons and order decimal amounts and quantities that are expressed to the
same number of decimal places. They should be able to represent numbers with one or two decimal places in several ways, such as on number lines.
Statutory requirements
Pupils should be taught to:
• Convert between different units of measure [for example, kilometre to metre; hour to minute]
• measure and calculate the perimeter of a rectilinear figure (including squares) in centimetres and metres
• find the area of rectilinear shapes by counting squares
• estimate, compare and calculate different measures, including money in pounds and pence
• read, write and convert time between analogue and digital 12- and 24-hour clocks
• solve problems involving converting from hours to minutes; minutes to seconds; years to months; weeks to days.
Notes and guidance (non-statutory)
Pupils build on their understanding of place value and decimal notation to record metric measures, including money.
They use multiplication to convert from larger to smaller units.
Perimeter can be expressed algebraically as 2(a +b) where a and b are the dimensions in the same unit.
They relate area to arrays and multiplication.
Geometry – properties of shapes
Statutory requirements
Pupils should be taught to:
• compare and classify geometric shapes, including quadrilaterals and triangles,based on their properties and sizes
• identify acute and obtuse angles and compare and order angles up to two right angles by size
• identify lines of symmetry in 2-D shapes presented in different orientations
• complete a simple symmetric figure with respect to a specific line of symmetry.
Notes and guidance (non-statutory)
Pupils continue to classify shapes using geometrical properties, extending to classifying different triangles (for example, isosceles, equilateral, scalene) and quadrilaterals (for example,
parallelogram, rhombus, trapezium).
Pupils compare and order angles in preparation for using a protractor and compare lengths and angles to decide if a polygon is regular or irregular.
Pupils draw symmetric patterns using a variety of media to become familiar with different orientations of lines of symmetry; and recognise line symmetry in a variety of diagrams, including where the
line of symmetry does not dissect the original shape.
Geometry – position and direction
Statutory requirements
Pupils should be taught to:
• describe positions on a 2-D grid as coordinates in the first quadrant
• describe movements between positions as translations of a given unit to the left/right and up/down
• plot specified points and draw sides to complete a given polygon.
Notes and guidance (non-statutory)
Pupils draw a pair of axes in one quadrant, with equal scales and integer labels. They read, write and use pairs of coordinates, for example (2, 5), including using coordinate-plotting ICT tools.
Statutory requirements
Pupils should be taught to:
• interpret and present discrete and continuous data using appropriate graphical methods, including bar charts and time graphs.
• solve comparison, sum and difference problems using information presented in bar charts, pictograms, tables and other graphs.
Notes and guidance (non-statutory)
Pupils understand and use a greater range of scales in their representations.
Pupils begin to relate the graphical representation of data to recording change over time. | {"url":"https://teacherworksheets.co.uk/national-curriculum/primary/keystage-2/year-4/mathematics","timestamp":"2024-11-11T13:12:24Z","content_type":"text/html","content_length":"72216","record_id":"<urn:uuid:04ad2384-fa86-4527-bd26-269fc2a54cb6>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00826.warc.gz"} |
Propensity Score Matching
Use this feature to match participants of two distinct groups in order to control the effect of confounding variables in observational studies.
What is propensity score matching?
The propensity score is defined as the probability for a participant to belong to one of two
groups given some variables known as confounders. The propensity score matching is a
technique that attempts to reduce the possible bias associated with those confounding variables
in observational studies.
Propensity Score Matching options in XLSTAT
Once the propensity score has been estimated, each participant of the treatment group is matched to the most similar participant of the control group (Rosenbaum P. R. (1989)). The distance matrix is
computed between the treatment group and the control group. XLSTAT implementation proposes two metrics: the Euclidean distance and the Mahalanobis distance.
Two algorithms are available in XLSTAT to perform the matching operation: the greedy algorithm and the optimal algorithm. With both of these algorithms, it is possible to match each participant of
the treatment group to one participant of the control group, to a specified number of participants of the control group or to all participants of the control group.
Propensity Score Matching results in XLSTAT
Test of the null hypothesis: The H0 hypothesis corresponds to the independent
model which gives probability p0 whatever the values of the explanatory variables. We seek to
check if the adjusted model is significantly more powerful than this model.
Type II analysis: This table is only useful if there is more than one explanatory variable. Here,
the adjusted model is tested against a test model where the variable in the row of the table in
the question has been removed.
The table of propensity scores gives the calculated propensity score for each participant of
the two groups. The value of the logit of the propensity score is also given. This is the value
that is used to compute the distance between each participant.
The distance matrix is also displayed to give a general view of all the computed distances.
Participants of the treatment group are in rows, those of the control group are on columns.
Distances for match pairs are displayed in bold.
ROC curve: The ROC curve is used to evaluate the performance of the model by means of
the area under the curve (AUC) and to compare several models together (see the description
section for more details).
analyze your data with xlstat
14-day free trial | {"url":"https://www.xlstat.com/en/solutions/features/propensity-score-matching","timestamp":"2024-11-12T03:50:14Z","content_type":"text/html","content_length":"24590","record_id":"<urn:uuid:bf27ea97-ff1e-4edf-ade3-889ad78b5cd6>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00262.warc.gz"} |
Maximize Your Profits With Our Forex Compound Calculator
What is a Compounding Calculator
Enhance your financial insights with the invaluable compounding calculator, a dynamic tool that enables you to simulate the growth of an account by compounding interest or profits with a
predetermined percentage. This calculator operates by simulating the reinvestment of the selected gain percentage into the account’s total equity, showcasing the remarkable power of compounding
Unlocking the potential of this calculator unveils the transformative nature of compounding gains, illustrating how even a modest gain percentage, such as 2% per trade, can progressively magnify an
account’s initial capital into a substantial equity over time.
How to Use the Compounding Calculator
Starting Balance: Input the initial account equity as the starting point. For this example, let’s assume a starting balance of 1,000 units in the chosen deposit currency.
Number of Periods: This field allows traders to simulate a series of consecutive winning trades. Each period represents an instance where you receive interest on holdings, close a profitable trade,
or similar occurrences. Consider the following example:
• The bank pays 5% interest on the savings account, every month = period is 1 month.
• Binance crypto exchange pays 10% interest on BTC, every day = period is 1 day.
• An investor trades XAU/USD and wins 2% return, each trade = period is each trade.
For our demonstration, let’s simulate a streak of 6 consecutive winning trades.
Gain % per Period: This critical field enables the simulation of the gain percentage for each compounding period. It caters to various trading strategies, whether you conduct multiple daily trades
with a target of 0.05% return per trade, weekly trades with a target of 1% return per trade, or even long-term trades with 12 trades per year and a target of 5% return per trade. In our example, we
will utilize a gain percentage per period of 2%.
Once the necessary inputs are provided, proceed by clicking the Calculate button.
The Results: The compounding calculator will promptly generate the Ending Balance after compounding the gains from 6 consecutive winning trades, along with the Total Gain percentage. In this case, an
initial equity of 1,000 units, denominated in the account currency, will grow to 1,126.16 units after compounding the gains.
This signifies that by compounding a mere 6 winning trades with a modest profit percentage of 2% per trade, the account balance experiences a remarkable growth of 12.6%.
A complete breakdown of how each compounded trade raised the account amount as well as the final account balance can be found on the results page linked above.
Other Tools | {"url":"https://tradingopedia.com/forex-compounding-calculator/","timestamp":"2024-11-10T21:12:34Z","content_type":"text/html","content_length":"81364","record_id":"<urn:uuid:1ddaa047-2d4f-4ccc-aa07-5d08e1bfa3cc>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00839.warc.gz"} |
How to Interpolate Missing Values in Excel
Interpolating missing values in Excel can be done using various methods such as linear interpolation, using the FORECAST function, or using a trendline. In this tutorial, we will focus on linear
Linear Interpolation
Linear interpolation is a method of estimating missing values by calculating the average of the values before and after the missing point. This approach assumes that the change between values is
linear. Here's how to interpolate missing values using linear interpolation in Excel:
1. Open your Excel workbook and locate the data with missing values.
2. Identify the cells with missing values and the cells with known values before and after the missing values.
3. In an empty cell, type the following formula:
=(ValueAfter - ValueBefore) / (CellAfter - CellBefore) * (CellMissing - CellBefore) + ValueBefore
Replace ValueBefore, ValueAfter, CellBefore, CellAfter, and CellMissing with the corresponding cell references.
4. Press Enter to calculate the interpolated value.
5. Copy the formula and paste it in the cell with the missing value.
Let's assume we have the following dataset with a missing value in cell B4:
A B
3 (missing)
To interpolate the missing value in cell B4 using linear interpolation:
1. In an empty cell (e.g., C1), type the following formula:
=(B5 - B3) / (A5 - A3) * (A4 - A3) + B3
2. Press Enter. The interpolated value (30) will appear in cell C1.
3. Copy the value in C1 and paste it in cell B4 to replace the missing value.
Now, the dataset will look like this:
You have successfully interpolated the missing value using linear interpolation in Excel.
Did you find this useful? | {"url":"https://sheetscheat.com/excel/how-to-interpolate-missing-values-in-excel","timestamp":"2024-11-09T11:15:46Z","content_type":"text/html","content_length":"13998","record_id":"<urn:uuid:2069a9eb-c76e-4145-a73e-9da24126164d>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00299.warc.gz"} |
College Algebra Foundations
Learning Objectives
• Define square root
• Find square roots
We know how to square a number:
[latex]5^2=25[/latex] and [latex]\left(-5\right)^2=25[/latex]
Taking a square root is the opposite of squaring so we can make these statements:
• 5 is the nonngeative square root of 25
• -5 is the negative square root of 25
Find the square roots of the following numbers:
1. 36
2. 81
3. -49
4. 0
1. We want to find a number whose square is 36. [latex]6^2=36[/latex] therefore, the nonnegative square root of 36 is 6 and the negative square root of 36 is -6
2. We want to find a number whose square is 81. [latex]9^2=81[/latex] therefore, the nonnegative square root of 81 is 9 and the negative square root of 81 is -9
3. We want to find a number whose square is -49. When you square a real number, the result is always positive. Stop and think about that for a second. A negative number times itself is positive, and
a positive number times itself is positive. Therefore, -49 does not have square roots, there are no real number solutions to this question.
4. We want to find a number whose square is 0. [latex]0^2=0[/latex] therefore, the nonnegative square root of 0 is 0. We do not assign 0 a sign, so it has only one square root, and that is 0.
The notation that we use to express a square root for any real number, a, is as follows:
Writing a Square Root
The symbol for the square root is called a radical symbol. For a real number, a the square root of a is written as [latex]\sqrt{a}[/latex]
The number that is written under the radical symbol is called the radicand.
By definition, the square root symbol, [latex]\sqrt{\hphantom{5}}[/latex] always means to find the nonnegative root, called the principal root.
[latex]\sqrt{-a}[/latex] is not defined, therefore [latex]\sqrt{a}[/latex] is defined for [latex]a>0[/latex]
Let’s do an example similar to the example from above, this time using square root notation. Note that using the square root notation means that you are only finding the principal root – the
nonnegative root.
Simplify the following square roots:
1. [latex]\sqrt{16}[/latex]
2. [latex]\sqrt{9}[/latex]
3. [latex]\sqrt{-9}[/latex]
4. [latex]\sqrt{5^2}[/latex]
Show Solution
The last problem in the previous example shows us an important relationship between squares and square roots, and we can summarize it as follows:
The square root of a square
For a nonnegative real number, a, [latex]\sqrt{a^2}=a[/latex]
In the video that follows, we simplify more square roots using the fact that [latex]\sqrt{a^2}=a[/latex] means finding the principal square root.
The square root of a number is the number which, when multiplied by itself, gives the original number. Principal square roots are always positive and the square root of 0 is 0. You can only take the
square root of values that are nonnegative. The square root of a perfect square will be an integer. | {"url":"https://courses.lumenlearning.com/aacc-collegealgebrafoundations/chapter/read-square-roots/","timestamp":"2024-11-11T02:14:29Z","content_type":"text/html","content_length":"51106","record_id":"<urn:uuid:7fea5c28-def5-40c5-b343-c7906c5c7ea7>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00255.warc.gz"} |
Lesson 18
Using Common Multiples and Common Factors
Let’s use common factors and common multiple to solve problems.
Problem 1
Mai, Clare, and Noah are making signs to advertise the school dance. It takes Mai 6 minutes to complete a sign, it takes Clare 8 minutes to complete a sign, and it takes Noah 5 minutes to complete a
sign. They keep working at the same rate for a half hour.
1. Will Mai and Clare complete a sign at the same time? Explain your reasoning.
2. Will Mai and Noah complete a sign at the same time? Explain your reasoning.
3. Will Clare and Noah complete a sign at the same time? Explain your reasoning
4. Will all three students complete a sign at the same time? Explain your reasoning
Problem 2
Diego has 48 chocolate chip cookies, 64 vanilla cookies, and 100 raisin cookies for a bake sale. He wants to make bags that have all three cookie flavors and the same number of each flavor per bag.
1. How many bags can he make without having any cookies left over?
2. Find the another solution to this problem.
(From Unit 7, Lesson 16.)
Problem 3
1. Find the product of 12 and 8.
2. Find the greatest common factor of 12 and 8.
3. Find the least common multiple of 12 and 8.
4. Find the product of the greatest common factor and the least common multiple of 12 and 8.
5. What do you notice about the answers to question 1 and question 4?
6. Choose 2 other numbers and repeat the previous steps. Do you get the same results?
Problem 4
1. Given the point \((5.5, \text-7)\), name a second point so that the two points form a vertical segment. What is the length of the segment?
2. Given the point \((3, 3.5)\), name a second point so that the two points form a horizontal segment. What is the length of the segment?
(From Unit 7, Lesson 11.)
Problem 5
Find the value of each expression mentally.
1. \(\frac12\boldcdot 37-\frac12 \boldcdot 7\)
2. \(3.5\boldcdot 40+3.5\boldcdot 60\)
3. \(999\boldcdot 5\)
(From Unit 6, Lesson 9.) | {"url":"https://im.kendallhunt.com/MS/students/1/7/18/practice.html","timestamp":"2024-11-04T06:09:19Z","content_type":"text/html","content_length":"64677","record_id":"<urn:uuid:bfa7f8ef-83c3-4d00-a561-bf2ae99a6899>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00004.warc.gz"} |
What is affected by the W in Quaternion(x,y,z,w)?
I’m just wondering what is affected by the ‘w’ parameter in Quaternion…
• a quaternion is a complex number with w as the real part and x, y, z as imaginary parts.
• If a quaternion represents a rotation then w = cos(theta / 2), where theta is the rotation angle around the axis of the quaternion.
• The axis v(v1, v2, v3) of a rotation is encoded in a quaternion: x = v1 * sin (theta / 2), y = v2 * sin (theta / 2), z = v3 * sin (theta / 2).
• If w is 1 then the quaternion defines 0 rotation angle around an undefined axis v = (0,0,0).
• If w is 0 the quaternion defines a half circle rotation since theta then could be +/- pi.
• If w is -1 the quaternion defines +/-2pi rotation angle around an undefined axis v = (0,0,0).
• A quater circle rotation around a single axis causes w to be +/- 0.5 and x/y/z to be +/- 0.5.
Kind Regards,
Keld Ølykke
Quaternions are four-dimensional, so you need four properties. The x/y/z properties don’t correspond to x/y/z in euler angles. With quaternions, each of the properties is a normalized float between 0
and 1, so for example a euler angle of 45/90/180 is represented by a quaternion as approximately .65/-.27/.65/.27.
If you don’t already know, it’s not something that’s easily explained unfortunately. | {"url":"https://discussions.unity.com/t/what-is-affected-by-the-w-in-quaternion-x-y-z-w/25483","timestamp":"2024-11-09T22:50:12Z","content_type":"text/html","content_length":"30387","record_id":"<urn:uuid:348defa0-18f8-455b-a39a-cbc6c491ea7d>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00063.warc.gz"} |
Multiplication Chart what is? Free printable Pdf Full Size and Pictures
A multiplication Chart is a list of a number's multiples and is also commonly known as the times table. We can obtain the multiplication tables of a number by multiplying the given number by whole
numbers. Multiplication is one of the basic mathematical operations taught to students at an early age.
You may also like: Multiplication Chart 1-100 Free Printable PDF
Math tables can be extremely helpful in performing basic arithmetic calculations. They serve as building blocks for higher math like fractions, exponents, and many more. Free printable multiplication
tables charts are provided to aid you in effortlessly learning multiplication tables. These multiplication tables charts assist students in memorizing them faster if they frequently review them.
What is a Multiplication Chart, Really?
Multiplication Chart - Free printable Pdf Full Size and Pictures color
Essentially, a multiplication chart or a times table, is a straightforward tool that helps you determine the product of two numbers. Picture this - a grid filled with numbers, one set runs vertically
down the left column, and the other stretches horizontally across the top. Using this handy chart, you can save considerable time and effort in performing calculations. Isn't that neat?
Exploring the Multiplication Chart: Numbers 1 to 12
Envision a grid: numbers 1 through 12 decorating both the top row and the left column. Every other box on the grid is a product of a number from the top row and its counterpart from the left column.
This is the essence of a multiplication chart 1-12.
The Printable Power of Multiplication Charts
Modern technology has made learning so much more convenient. Now, you can easily download and print a PDF of a multiplication chart. Stick it somewhere you'd frequently see it – your fridge, above
your desk, or even as a bookmark in your favorite book. By frequently revisiting it, you'll have these math tables memorized in no time!
Delving Deeper: Multiplication Tables from 1 to 20
Did you know that multiplication tables are the building blocks for more complex calculations, such as fractions, percentages, and factorizing? Yep, mastering times Multiplication Chart 1-20 can be
the key to confidently tackling such problems!
Decoding the Meaning of Times Tables
Put simply, a multiplication table is a list of multiples for a specific number. They are your magic wand to effortlessly multiply two numbers. For instance, let's examine a times table from 2 to 20.
Why Multiplication Tables Matter for Students
Multiplication tables, or math times tables, act as the foundation for various arithmetic calculations. It's often easier for children to retain information they learn at a young age, making these
tables essential for their lifelong mathematical journey. Here are a few reasons why these tables are so important:
• They bolster a student's mathematical learning.
• They provide a solid understanding of multiplication facts.
• They simplify solving math problems for students.
• They foster confidence in students as they grasp new mathematical concepts.
Mastering Maths Tables: The Easy Way
Many students find it challenging to memorize multiplication tables. Don't worry - we've got a few effective tips to aid in learning these math times tables:
• Get familiar with skip counting: Simply start with a number and keep adding the same number. For example, start with 3 and keep adding 3 to get 3, 6, 9, 12...
• Recite the multiplication table in order: Practice this daily until you grasp the pattern.
• Write to learn: If memorization is proving difficult, write the tables daily and recite them.
• Use real-life examples: Apply multiplication to daily situations, like grocery shopping or calculating bills.
• Identify patterns: Every multiplication table has its own pattern. Identifying this will help you memorize these math times tables faster.
Demystifying Multiplication Tables: Solved Examples
Let's put what we've learned into practice with a few examples:
Multiplication Tables: A Handy Learning Tool
You can consider multiplication tables as your math roadmaps. They're often referred to as "times tables", providing us with multiple ways to reach our numerical destination. When you think about it,
multiplication is just one of the first math pit stops we make on our educational journey. One of the easiest ways to comprehend this concept is by utilizing a multiplication chart.
Multiplication charts are incredibly beneficial when dealing with basic arithmetic operations. They serve as stepping stones to complex mathematical concepts like fractions and exponents. And the
best part? You can find free printable times table charts that will make the learning process a breeze. Revisiting these tables can aid in quick recall of multiplication facts.
The Multiplication Chart Explained
Think of a multiplication times tables chart as a mathematical matrix. It provides the product of two numbers in a concise, organized format. Typically, one set of numbers is listed down the left
column and another set runs across the top row. A multiplication chart is a real time-saver when it comes to calculation and efficiency.
For example, the times table chart from 1 to 10 consists of numbers from 1 to 10 noted at the top and side of the grid. The remaining boxes in the grid represent the products of a number from the top
row and a number from the left column. This visualization makes it easy to grasp the multiplication concept.
Get Your Printable Multiplication Chart
A multiplication table pdf is available for download, allowing you to print and place it wherever you see fit. Regular exposure to these math tables will enhance memorization and familiarity with
multiplication facts.
Multiplication Tables from 1 to 20
Multiplication tables serve as a foundation for multi-digit calculations and problem-solving related to fractions, percentages, and factoring. Using times tables from 1 to 20 can improve mental
arithmetic skills and equip children to handle more intricate calculations with ease.
Deciphering Times Tables
A multiplication table is essentially a list of a number's multiples. It's a vital tool for learning multiplication, allowing for swift and easy calculations. Take a look at the following table,
which presents times tables from 2 to 20.
The Importance of Multiplication Tables for Students
For a child, multiplication tables are like the bricks that build a strong mathematical foundation. Since children's memories are stronger than adults', the concepts they grasp at an early age have a
long-lasting impact. And here's why these math times tables are so beneficial:
• They strengthen mathematical comprehension.
• They provide a firm grip on multiplication facts.
• They simplify problem-solving in mathematics.
• Confidence in math times tables often translates to confidence in learning new math concepts.
Memorizing Maths Tables: It's Easier Than You Think
Are your children finding it tough to remember multiplication tables? Here are a few tips that can make this task less daunting:
• Practice Skip Counting: Start with a number and keep adding it to the next. For example, starting with 3, you continue to add 3, resulting in 3, 6, 9, 12, and so on.
• Recite the Multiplication Table in Order: Make a daily habit of reciting a table, like '2 ones are 2, 2 twos are 4, 2 threes are 6', etc.
• Practice by Writing: If memorization is proving difficult, try writing out the tables and reciting them afterwards.
• Apply Multiplication to Real Life: Use real-life examples to understand multiplication tables. Practicing multiplication during daily activities, like multiplying the price of one item by the
quantity purchased, can be beneficial.
• Identify Patterns: Each multiplication table has its own pattern. Identifying these patterns can make it easier to remember them. For example, in the 2 times table, every number is even, and in
the 5 times table, every number ends in 0 or 5.
Make Learning Fun with Multiplication Table Games
Who says learning can't be fun? With multiplication table games, children can learn while having fun. Here are a few game suggestions:
• Multiplication Bingo: In this version of the classic game, the bingo cards are filled with products instead of numbers. Call out multiplication facts and let the players locate the answer on
their cards.
• Times Table Race: Create a track on paper or on the floor. Roll dice, and move forward by the product of the numbers on the dice. The first one to the finish line wins.
• Multiplication War: This is a variation of the card game "War." Deal a deck of cards evenly between two players. Each player flips a card at the same time, and the first one to say the product of
the two numbers wins the round.
Multiplication Tables and Mental Math
Knowing multiplication tables can significantly boost mental math skills. Instead of relying on a calculator, one can quickly calculate and provide solutions, making everyday tasks faster and more
efficient. The significance of multiplication tables extends well beyond classroom learning.
In Summary
Multiplication tables are essential tools in learning mathematics. They provide a way to quickly and efficiently carry out calculations, saving time and boosting mental math skills. Whether you're
just starting your math journey or looking to brush up on your skills, remember that the power of multiplication tables lies in practice and application. Happy learning!
A Step-by-Step Breakdown of Multiplication Tables
Even with the aid of a multiplication table, many students still struggle to grasp the concept of multiplication. Here's a simple breakdown:
• Step 1: Choose a number from the left column. This number will be what you are multiplying.
• Step 2: Choose a number from the top row. This number is what you are multiplying by.
• Step 3: Find the intersection of the row and column you chose. The number where they intersect is the product of your two chosen numbers.
For example, let's multiply 4 (from the left column) by 5 (from the top row). Where the row and column intersect, we find 20, which is the product of 4 and 5.
Memorization Tips and Tricks
Multiplication tables can seem daunting at first, but with a few tips and tricks, you can make learning them much more manageable.
• Break It Down: Don't try to learn all the tables at once. Start with the simpler ones (like 2s, 5s, and 10s) before moving on to the more difficult ones.
• Use Real-Life Examples: Whenever possible, connect the tables to real-life examples. If you're learning the 3s table, you could think about setting a table for dinner. If you have 3 plates and
you need to set the table for 4 people, you would need 12 plates in total (3 plates for each person times 4 people equals 12 plates).
• Practice Regularly: Make sure to review the tables you've learned regularly. Just a few minutes each day can help keep the tables fresh in your mind.
• Use the Commutative Property: Remember, multiplication is commutative, meaning the order of numbers doesn't matter. So if you know 4 x 5 = 20, you also know 5 x 4 = 20.
Fun Facts about Multiplication Tables
Did you know that multiplication tables have been around for thousands of years? Ancient Egyptians used multiplication tables as far back as 2000 B.C.! Even with the advent of calculators and
computers, multiplication tables continue to be a fundamental part of learning math.
So, whether you're learning multiplication for the first time or looking for a refresher, remember that understanding multiplication tables is a crucial step in your mathematical journey. With
practice and patience, you'll be a multiplication master in no time!
Test Your Multiplication Tables Knowledge
Learning is one thing, but application is the true test of knowledge. Let's put our multiplication tables to the test with a few examples.
Example 1:
Let's say Sam works for 5 hours a day and gets paid $8 per hour. What's his daily earning? If we look at the multiplication table for 5, we can quickly see that 5 x 8 = 40. So, Sam earns $40 a day.
Example 2:
It's time to fill in the blanks. Can you find the answers to the following?
a.) 3 × 4 = ___
b.) 6 × 7 = ___
c.) 8 × 2 = ___
Using the multiplication tables, the answers are:
a.) 3 × 4 = 12
b.) 6 × 7 = 42
c.) 8 × 2 = 16
How did you do? If you got them all right, fantastic job! If not, don't worry—just keep practicing. The more you use multiplication tables, the quicker you'll become at solving these problems.
Example 3:
Let's play a game of true or false. Based on the multiplication tables, are the following statements true or false?
a.) 4 × 7 = 28
b.) 9 × 8 = 98
c.) 3 × 6 = 18
The answers are:
a.) True, because 4 × 7 does equal 28.
b.) False, because 9 × 8 equals 72, not 98.
c.) True, because 3 × 6 does equal 18.
Again, if you aced this, great job! And if you missed one or two, don't stress. Practice makes perfect!
Multiplication Chart Printable
Perks of a Printable Multiplication Chart
A multiplication chart printable is a version of the multiplicación chart that you can print out. It's handy, easy to access, and you can stick it on the fridge, put it in a binder, or carry it in a
pocket. It's the math equivalent of carrying a mini teacher with you everywhere you go.
How to Use a Multiplication Chart Printable
So how do you make the most of your multiplication chart printable? Start by reviewing it daily. Look for patterns, see if you can spot the squares, multiples of certain numbers, etc. Use it as a
reference when doing homework or solving math problems. Over time, you'll find that multiplication facts start to stick in your head, thanks to your handy printable!
Mastering Multiplication with a Blank Multiplication Chart
Next up, we have the blank multiplication chart. This isn't a prank, and it's not a chart that forgot to put its numbers on. A blank multiplication chart is a tool for testing your multiplication
skills. You fill in the boxes with the products of the numbers in the rows and columns.
How to Use a FREE Multiplication Chart Printable PDF
A free multiplication chart printable PDF is as easy to use as its printable counterpart. Just download it, print it out, and voila – you're ready to start learning. Keep it with your study materials
or post it somewhere you'll see it often. Use it for homework help, test prep, or just for fun to reinforce your multiplication skills.
Full Size Multiplication Chart
Advantages of a Full Size Multiplication Chart
Sometimes, bigger is indeed better. A full size multiplication chart gives you a large, clear view of the multiplication facts, making it easier to see patterns and correlations. It's an excellent
tool for classrooms, study rooms, or anywhere with a sizable wall space.
Using a Full Size Multiplication Chart Effectively
A full size multiplication chart makes for a great visual aid during study sessions. Use colored markers to highlight patterns, multiples, or square numbers. It's not just a learning tool; it's a way
to make math come alive!
Pictures of Multiplication Charts
How Pictures of Multiplication Charts Can Help
Pictures of multiplication charts can be a game-changer, especially in the digital age. Having a digital image of a multiplication chart on your device can provide quick access for reference or
Ways to Use Pictures of Multiplication Charts
You can use pictures of multiplication charts as screensavers, wallpapers, or digital flashcards. They can be shared, printed, or even edited to highlight patterns. Turn learning into a tech-savvy
Multiple Chart
What is a Multiple Chart?
A multiple chart is similar to a multiplication chart, but it's focused on multiples of a specific number. It's a helpful tool to visualize and understand the concept of multiples.
Multiple Chart in Learning Process
A multiple chart can be used to improve understanding of skip counting, least common multiples, and divisibility rules. Highlighting the multiples of each number can also illustrate some fascinating
patterns in math.
Math Multiplication Chart
A math multiplication chart is a roadmap through the world of multiplication. It's a matrix of possibilities, a symphony of numbers working in harmony.
Making Math Fun with a Math Multiplication Chart
Using a math multiplication table doesn't have to be boring. Turn it into a game, challenge your friends, make bets with your siblings. Who can fill out a blank chart the fastest? Who can spot the
most patterns? Math can be fun when you make it interactive!
Using Anchor Charts to Enhance Learning
Anchor charts multiplication can be a game-changer for visual learners. They provide a clear, step-by-step guide to solving multiplication problems. They can be referred to time and again to
reinforce learning. Try creating your own anchor charts to make the learning process even more personalized.
Multiplication Chart Blank
A multiplication chart blank is more than a test. It's a tool for mastery. Each time you fill it out, you're reinforcing the multiplicación facts in your memory. And who knows, you might even start
to find it fun!
Frequently Asked Questions (FAQs) on Multiplication Tables and Charts
Why is it important to memorize multiplication tables?
Memorizing multiplication tables speeds up your mental calculations, freeing up time for more complex problem-solving tasks. Knowing them by heart also enables you to tackle arithmetic-related
problems more easily and boosts your confidence in handling larger numbers.
What are multiplication tables used for?
Multiplication tables facilitate quick calculations and help understand patterns of multiples. They come in handy when calculating bills, especially for items bought in bulk, or while performing
calculations related to area, volume, or any other dimensional quantity.
How do multiplication tables enhance problem-solving abilities?
Multiplication tables nurture your number sense and allow for quicker mental calculations. This, in turn, builds confidence and fosters a more favorable attitude towards the subject.
Why is a multiplication chart helpful?
A multiplication chart provides a clear visual representation of all times tables in one place, making it a handy reference tool. Regularly reviewing the chart helps commit multiplication facts to
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Bibliografy: Llavero de tablas de multiplicar | {"url":"https://multiplicationchart.online/","timestamp":"2024-11-13T05:59:39Z","content_type":"text/html","content_length":"57848","record_id":"<urn:uuid:4296397e-c539-4846-a54b-09dbd4d372d9>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00838.warc.gz"} |
[GAP Forum] How to obtain the incidence matrix of a partial geometry?
[GAP Forum] How to obtain the incidence matrix of a partial geometry?
Sven Reichard sven.reichard at tu-dresden.de
Wed Jan 21 12:51:03 GMT 2015
Hello Felix,
the geometry is returned as an incidence graph in GRAPE format. For each
vertex x of that graph, Adjacency(haemers, x) will give you its
neighbours. From that you can reconstruct the adjacency matrix if you want.
For example with the following function:
AdjacencyMatrix := function ( gamma )
local result, x, y;
result := NullMat( gamma.order, gamma.order );
for x in [ 1 .. gamma.order ] do
for y in Adjacency( gamma, x ) do
result[x][y] := 1;
return result;
However it could be worth your while working with the graph format as it is.
"delta" just refers to the incidence graph. PartialLinearSpaces returns
a list of those.
Hope this helps,
Sven Reichard
Institut für Algebra
TU Dresden
On 01/21/2015 01:33 PM, Felix Goldberg wrote:
> Hello all,
> I am running the code in the example in Section 9.2 of the GRAPE manual (
> http://www.maths.qmul.ac.uk/~leonard/grape/manual/CHAP009.htm) , which
> generates the Haemers partial geometry pg(4,17,2).
> All works well but I cannot understand where exactly the incidence matrix
> is stored and how to access it. The manual (referred to above) says that
> there is a *delta* associated to the geometry output by the function
> PartialLinearSpaces
> but I can't find it.
> I tried to run RecNames on the output (the variable called *haemers*) and
> got this:
> [ "names", "group", "order", "representatives", "isSimple", "isGraph",
> "schreierVector", "adjacencies" ]
> No sign of *delta* and apparently no incidence matrix.
> Any help will be greatly appreciated.
> Thanks,
> Felix
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Early life
Ennackal Chandy George Sudarshan was born in Pallom, Kottayam, Travancore, British India. He was raised in a Syrian Christian family, but later left the religion and converted to Hinduism following
his marriage.^[5]^:243^[5]^:243^[5]^:250 He married Lalita Rau on December 20, 1954, and they have three sons, Alexander, Arvind (deceased) and Ashok.^[6] George and Lalita were divorced in
1990 and he married Bhamathi Gopalakrishnan in Austin, Texas.^[6]
He studied at CMS College Kottayam,^[7] and graduated with honors from the Madras Christian College in 1951. He obtained his master's degree at the University of Madras in 1952. He moved to Tata
Institute of Fundamental Research (TIFR) and worked there for a brief period with Dr. Homi Bhabha as well as others. Subsequently, he moved to University of Rochester in New York to work under Prof.
Robert Marshak as a graduate student. In 1958, he received his Ph.D. degree from the University of Rochester. At this point he moved to Harvard University to join Prof. Julian Schwinger as a
postdoctoral fellow.
Dr. Sudarshan made significant contributions to several areas of physics. He was the originator (with Robert Marshak) of the V-A theory of the weak force (later propagated by Richard Feynman and
Murray Gell-Mann), which eventually paved the way for the electroweak theory. Feynman acknowledged Sudarshan's contribution in 1963 stating that the V-A theory was discovered by Sudarshan and Marshak
and publicized by Gell-Mann and himself.^[8] He also developed a quantum representation of coherent light later known as Glauber–Sudarshan representation (for which controversially Glauber was
awarded the 2005 Nobel prize in Physics ignoring Sudarshan's contributions).
Sudarshan's most significant work may have been his contribution to the field of quantum optics. His theorem proves the equivalence of classical wave optics to quantum optics. The theorem makes use
of the Sudarshan representation. This representation also predicts optical effects that are purely quantum, and cannot be explained classically. Sudarshan was also advocate for the existence of
tachyons, particles that travel faster than light.^[9] He developed a fundamental formalism called dynamical maps to study the theory of open quantum system. He, in collaboration with Baidyanath
Misra, also proposed the quantum Zeno effect.^[10]
Sudarshan and collaborators initiated the "Quantum theory of charged-particle beam optics", by working out the focusing action of a magnetic quadrupole using the Dirac equation.^[11]^[12]
He taught at the Tata Institute of Fundamental Research (TIFR), University of Rochester, Syracuse University,^[13] and Harvard. From 1969 onwards, he was a professor of physics at the University of
Texas at Austin and a senior professor at the Indian Institute of Science. He worked as the director of the Institute of Mathematical Sciences (IMSc), Chennai, India, for five years during the 1980s
dividing his time between India and USA. During his tenure, he transformed it into a centre of excellence. He also met and held many discussions with philosopher J. Krishnamurti. He was felicitated
on his 80th birthday, at IMSc Chennai^[14] on 16 September 2011. His areas of interest included elementary particle physics, quantum optics, quantum information, quantum field theory, gauge field
theories, classical mechanics and foundations of physics. He was also deeply interested in Vedanta, on which he lectured frequently.
Controversy regarding Nobel Prize
Sudarshan began working on quantum optics at the University of Rochester in 1960. Two years later, Glauber criticized the use of classical electromagnetic theory in explaining optical fields, which
surprised Sudarshan because he believed the theory provided accurate explanations. Sudarshan subsequently wrote a paper expressing his ideas^[15] and sent a preprint to Glauber. Glauber informed
Sudarshan of similar results and asked to be acknowledged in the latter's paper, while criticizing Sudarshan in his own paper.^[16] "Glauber criticized Sudarshan’s representation, but his own was
unable to generate any of the typical quantum optics phenomena, hence he introduces what he calls a P-representation, which was Sudarshan’s representation by another name", wrote a physicist. "This
representation, which had at first been scorned by Glauber, later becomes known as the Glauber–Sudarshan representation."^[17]
Sudarshan was passed over for the Physics Nobel Prize on more than one occasion, leading to controversy in 2005 when several physicists wrote to the Swedish Academy, protesting that Sudarshan should
have been awarded a share of the Prize for the Sudarshan diagonal representation (also known as Glauber–Sudarshan representation) in quantum optics, for which Roy J. Glauber won his share of the
prize.^[18] Sudarshan and other physicists sent a letter to the Nobel Committee claiming that the P representation had more contributions of "Sudarshan" than "Glauber".^[19] The letter goes on to say
that Glauber criticized Sudarshan's theory—before renaming it the "P representation" and incorporating it into his own work. In an unpublished letter to The New York Times, Sudarshan calls the
"Glauber–Sudarshan representation" a misnomer, adding that "literally all subsequent theoretic developments in the field of Quantum Optics make use of" Sudarshan's work— essentially, asserting that
he had developed the breakthrough.^[20]^[21]
In 2007, Prof.Sudarshan told the Hindustan Times, "The 2005 Nobel prize for Physics was awarded for my work, but I wasn't the one to get it. Each one of the discoveries that this Nobel was given for
work based on my research."^[22] Sudarshan also commented on not being selected for the 1979 Nobel, "Steven Weinberg, Sheldon Glashow and Abdus Salam built on work I had done as a 26-year-old
student. If you give a prize for a building, shouldn’t the fellow who built the first floor be given the prize before those who built the second floor?"^[22]
• Kerala Sastra Puraskaram for lifetime accomplishments in science, 2013
• Dirac Medal of the ICTP, 2010
• Padma Vibhushan, second highest civilian award from the Government of India, 2007
• Majorana Prize, 2006
• Presidential Citation Award from the University of Texas at Austin, 2006^[24]
• The first TWAS Prize in Physics awarded by the World Academy of Sciences, 1985^[25]
• Bose Medal, 1977
• Padma Bhushan, third highest civilian award from the Government of India, 1976^[26]
• C. V. Raman Award, 1970
See also
1. ^ "Ennackal Chandy George Sudarshan September 16, 1931 - May 13, 2018". Beck Funeral Home. 15 May 2018. Retrieved 17 May 2018.
2. ^ ^a ^b ^c Bhamathi, Gopalakrishnan (2021). "George Sudarshan: Perspectives and Legacy". Quanta. 10: 75–104. doi:10.12743/quanta.v10i1.174. S2CID 245482293.
3. ^ "Acclaimed scientist ECG Sudarshan passes away in Texas". Mathrubhumi. 14 May 2018. Retrieved 21 December 2018.
4. ^ Luis J. Boya, Laudatio for E. C. G. SUDARSHAN On his 75th BirthDay. Jaca, (HU), Spain, September 18, 2006, Journal of Physics: Conference Series 87 (2007) 012001 doi:10.1088/1742-6596/87/1/
5. ^ ^a ^b ^c Clayton, Philip (2002). "George Sudarshan". In Richardson, W. Mark; Russell, Robert John; Clayton, Philip; Wegter-McNelly, Kirk (eds.). Science and the Spiritual Quest: New Essays by
Leading Scientists. London: Routledge. ISBN 9780415257664.
6. ^ ^a ^b "Ennackal Chandy George Sudarshan (September 16, 1931 – May 13, 2018)". Austin, Texas: University of Texas. 2021. Retrieved 24 December 2021.
7. ^ "A proud moment for CMS College: Prof. Sudarshan delights all at his alma mater". The Hindu. 5 July 2008. Archived from the original on 2 August 2008. Retrieved 5 April 2010.
8. ^ The beat of a different drum: The life and science of Richard Feynman by J. Mehra Clarendon Press Oxford (1994), p. 477, and references 29 and 40 therein
9. ^ Time Machines: Time Travel in Physics, Metaphysics, and Science Fiction, p. 346, by Paul J. Nahin
10. ^ Sudarshan, E. C. G.; Misra, B. (1977). "The Zeno's paradox in quantum theory" (PDF). Journal of Mathematical Physics. 18 (4): 756–763. Bibcode:1977JMP....18..756M. doi:10.1063/1.523304. OSTI
11. ^ R. Jagannathan, R. Simon, E. C. G. Sudarshan and N. Mukunda, Quantum theory of magnetic electron lenses based on the Dirac equation, Physics Letters A, 134, 457–464 (1989).
12. ^ R. Jagannathan and S. A. Khan, Quantum theory of the optics of charged particles, Advances in Imaging and Electron Physics, Editors: Peter W. Hawkes, B. Kazan and T. Mulvey, (Academic Press,
Logo, San Diego, 1996), Vol. 97, 257–358 (1996).
13. ^ Catterall, Simon; Hubisz, Jay; Balachandran, Aiyalam; Schechter, Joe (5 January 2013). "Elementary Particle Physics at Syracuse. Final Report". Syracuse University: 14. doi:10.2172/1095082.
OSTI 1095082. Retrieved 26 February 2021.{{cite journal}}: Cite journal requires |journal= (help)
14. ^ "Sudarshan Fest" (PDF). 16 September 2011.
15. ^ Sudarshan, Ennackal Chandy George (1963). "Equivalence of semiclassical and quantum mechanical descriptions of statistical light beams". Physical Review Letters. 10 (7): 277–279. Bibcode
:1963PhRvL..10..277S. doi:10.1103/PhysRevLett.10.277.
16. ^ "Physicist Sudarshan's omission questioned". The Hindu. 2 December 2005.
17. ^ "ECG Sudarshan, physicist who proposed faster than light theory, dies at 86". www.hindustantimes.com. 14 May 2018. Retrieved 20 December 2018.
18. ^ Zhou, Lulu (6 December 2005). "Scientists Question Nobel". The Harvard Crimson. Archived from the original on 4 February 2012. Retrieved 22 February 2008.
19. ^ Epstein, David (7 December 2005). "Nobel Doubts". Inside Higher Ed. Retrieved 26 February 2021.
20. ^ "UT Austin Mourns Passing of George Sudarshan, Titan of 20th Century Physics". cns.utexas.edu. 20 December 2018. Retrieved 20 December 2018.
21. ^ "First Runner-up". seedmagazine.com. 20 December 2018. Archived from the original on 4 March 2016. Retrieved 20 December 2018.{{cite web}}: CS1 maint: unfit URL (link)
22. ^ ^a ^b Mehta, Neha (4 April 2007). "Physicist cries foul over Nobel miss". Hindustan Times. Archived from the original on 20 March 2008. Retrieved 22 February 2008.
23. ^ "KU to confer honorary doctorates on Narlikar, Kris Gopalakrishnan". The Hindu. 21 August 2019. Retrieved 5 November 2020.
24. ^ Leahy, Cory (8 January 2007). "Award Recipients to be Recognized at The University of Texas at Austin". UT News. Retrieved 7 June 2024.
25. ^ Balasubramanya, M. K.; Srinivas, M. D. (2019). "Ennackal Chandy George Sudarshan". Physics Today. 72 (4): 63. Bibcode:2019PhT....72d..63B. doi:10.1063/pt.3.4190. Retrieved 20 October 2023.
26. ^ "Padma Awards" (PDF). Ministry of Home Affairs, Government of India. 2015. Archived from the original (PDF) on 15 October 2015. Retrieved 21 July 2015.
External links
• A LOOK-BACK AT FOUR DECADES OF RESEARCH By ECG SUDARSHAN
• Seven Science Quests Symposium, The University of Texas at Austin, 2006
• Home page with vita and publications
• Publications on ArXiv
• Collected works
• ECG Sudarshan on Keral.com
• Sudarshan's letter to Nobel Committee
• Lecture- Perspectives And Perceptions: Causality And Unpredictability
• George Sudarshan at the Mathematics Genealogy Project | {"url":"https://www.knowpia.com/knowpedia/E._C._George_Sudarshan","timestamp":"2024-11-08T11:03:20Z","content_type":"text/html","content_length":"129004","record_id":"<urn:uuid:bd196d68-c8fe-47ef-b1e7-1878329b8dc3>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00364.warc.gz"} |
Parallelism | Playwright
Playwright Test runs tests in parallel. In order to achieve that, it runs several worker processes that run at the same time. By default, test files are run in parallel. Tests in a single file are
run in order, in the same worker process.
• You can configure tests using test.describe.configure to run tests in a single file in parallel.
• You can configure entire project to have all tests in all files to run in parallel using testProject.fullyParallel or testConfig.fullyParallel.
• To disable parallelism limit the number of workers to one.
You can control the number of parallel worker processes and limit the number of failures in the whole test suite for efficiency.
Worker processes
All tests run in worker processes. These processes are OS processes, running independently, orchestrated by the test runner. All workers have identical environments and each starts its own browser.
You can't communicate between the workers. Playwright Test reuses a single worker as much as it can to make testing faster, so multiple test files are usually run in a single worker one after
Workers are always shutdown after a test failure to guarantee pristine environment for following tests.
Limit workers
You can control the maximum number of parallel worker processes via command line or in the configuration file.
From the command line:
npx playwright test --workers 4
In the configuration file:
import { defineConfig } from '@playwright/test';
export default defineConfig({
// Limit the number of workers on CI, use default locally
workers: process.env.CI ? 2 : undefined,
Disable parallelism
You can disable any parallelism by allowing just a single worker at any time. Either set workers: 1 option in the configuration file or pass --workers=1 to the command line.
npx playwright test --workers=1
Parallelize tests in a single file
By default, tests in a single file are run in order. If you have many independent tests in a single file, you might want to run them in parallel with test.describe.configure().
Note that parallel tests are executed in separate worker processes and cannot share any state or global variables. Each test executes all relevant hooks just for itself, including beforeAll and
import { test } from '@playwright/test';
test.describe.configure({ mode: 'parallel' });
test('runs in parallel 1', async ({ page }) => { /* ... */ });
test('runs in parallel 2', async ({ page }) => { /* ... */ });
Alternatively, you can opt-in all tests into this fully-parallel mode in the configuration file:
import { defineConfig } from '@playwright/test';
export default defineConfig({
fullyParallel: true,
You can also opt in for fully-parallel mode for just a few projects:
import { defineConfig } from '@playwright/test';
export default defineConfig({
// runs all tests in all files of a specific project in parallel
projects: [
name: 'chromium',
use: { ...devices['Desktop Chrome'] },
fullyParallel: true,
Serial mode
You can annotate inter-dependent tests as serial. If one of the serial tests fails, all subsequent tests are skipped. All tests in a group are retried together.
Using serial is not recommended. It is usually better to make your tests isolated, so they can be run independently.
import { test, type Page } from '@playwright/test';
// Annotate entire file as serial.
test.describe.configure({ mode: 'serial' });
let page: Page;
test.beforeAll(async ({ browser }) => {
page = await browser.newPage();
test.afterAll(async () => {
await page.close();
test('runs first', async () => {
await page.goto('https://playwright.dev/');
test('runs second', async () => {
await page.getByText('Get Started').click();
Shard tests between multiple machines
Playwright Test can shard a test suite, so that it can be executed on multiple machines. See sharding guide for more details.
npx playwright test --shard=2/3
Limit failures and fail fast
You can limit the number of failed tests in the whole test suite by setting maxFailures config option or passing --max-failures command line flag.
When running with "max failures" set, Playwright Test will stop after reaching this number of failed tests and skip any tests that were not executed yet. This is useful to avoid wasting resources on
broken test suites.
Passing command line option:
npx playwright test --max-failures=10
Setting in the configuration file:
import { defineConfig } from '@playwright/test';
export default defineConfig({
// Limit the number of failures on CI to save resources
maxFailures: process.env.CI ? 10 : undefined,
Worker index and parallel index
Each worker process is assigned two ids: a unique worker index that starts with 1, and a parallel index that is between 0 and workers - 1. When a worker is restarted, for example after a failure, the
new worker process has the same parallelIndex and a new workerIndex.
You can read an index from environment variables process.env.TEST_WORKER_INDEX and process.env.TEST_PARALLEL_INDEX, or access them through testInfo.workerIndex and testInfo.parallelIndex.
Isolate test data between parallel workers
You can leverage process.env.TEST_WORKER_INDEX or testInfo.workerIndex mentioned above to isolate user data in the database between tests running on different workers. All tests run by the worker
reuse the same user.
Create playwright/fixtures.ts file that will create dbUserName fixture and initialize a new user in the test database. Use testInfo.workerIndex to differentiate between workers.
import { test as baseTest, expect } from '@playwright/test';
// Import project utils for managing users in the test database.
import { createUserInTestDatabase, deleteUserFromTestDatabase } from './my-db-utils';
export * from '@playwright/test';
export const test = baseTest.extend<{}, { dbUserName: string }>({
// Returns db user name unique for the worker.
dbUserName: [async ({ }, use) => {
// Use workerIndex as a unique identifier for each worker.
const userName = `user-${test.info().workerIndex}`;
// Initialize user in the database.
await createUserInTestDatabase(userName);
await use(userName);
// Clean up after the tests are done.
await deleteUserFromTestDatabase(userName);
}, { scope: 'worker' }],
Now, each test file should import test from our fixtures file instead of @playwright/test.
// Important: import our fixtures.
import { test, expect } from '../playwright/fixtures';
test('test', async ({ dbUserName }) => {
// Use the user name in the test.
Control test order
Playwright Test runs tests from a single file in the order of declaration, unless you parallelize tests in a single file.
There is no guarantee about the order of test execution across the files, because Playwright Test runs test files in parallel by default. However, if you disable parallelism, you can control test
order by either naming your files in alphabetical order or using a "test list" file.
Sort test files alphabetically
When you disable parallel test execution, Playwright Test runs test files in alphabetical order. You can use some naming convention to control the test order, for example
001-user-signin-flow.spec.ts, 002-create-new-document.spec.ts and so on.
Use a "test list" file
Tests lists are discouraged and supported as a best-effort only. Some features such as VS Code Extension and tracing may not work properly with test lists.
You can put your tests in helper functions in multiple files. Consider the following example where tests are not defined directly in the file, but rather in a wrapper function.
import { test, expect } from '@playwright/test';
export default function createTests() {
test('feature-a example test', async ({ page }) => {
// ... test goes here
import { test, expect } from '@playwright/test';
export default function createTests() {
test.use({ viewport: { width: 500, height: 500 } });
test('feature-b example test', async ({ page }) => {
// ... test goes here
You can create a test list file that will control the order of tests - first run feature-b tests, then feature-a tests. Note how each test file is wrapped in a test.describe() block that calls the
function where tests are defined. This way test.use() calls only affect tests from a single file.
import { test } from '@playwright/test';
import featureBTests from './feature-b.spec.ts';
import featureATests from './feature-a.spec.ts';
Now disable parallel execution by setting workers to one, and specify your test list file.
import { defineConfig } from '@playwright/test';
export default defineConfig({
workers: 1,
testMatch: 'test.list.ts',
Do not define your tests directly in a helper file. This could lead to unexpected results because your tests are now dependent on the order of import/require statements. Instead, wrap tests in a
function that will be explicitly called by a test list file, as in the example above. | {"url":"https://playwright.dev/docs/test-parallel","timestamp":"2024-11-07T07:25:49Z","content_type":"text/html","content_length":"118729","record_id":"<urn:uuid:c36a967b-275e-4cf1-b2ac-0688df3a3d80>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00849.warc.gz"} |
Geometric Mechanics in Finance, Markets & Trading
Dan Buckley is an US-based trader, consultant, and part-time writer with a background in macroeconomics and mathematical finance. He trades and writes about a variety of asset classes, including
equities, fixed income, commodities, currencies, and interest rates. As a writer, his goal is to explain trading and finance concepts in levels of detail that could appeal to a range of audiences,
from novice traders to those with more experienced backgrounds.
Geometric mechanics is a branch of mathematics, most typically applied to theoretical physics, that applies geometric methods to problems in mechanics and dynamics.
It combines differential geometry, the study of smooth manifolds, with the principles of classical and quantum mechanics.
The key concepts and applications of geometric mechanics in finance offer a different perspective on modeling and understanding complex financial systems over time.
Key Takeaways – Geometric Mechanics in Finance, Markets & Trading
□ Dynamic Market Modeling
☆ Enables understanding of complex, evolving financial market behaviors and trends.
□ Risk Management Insights
☆ Offers geometric methods for assessing and mitigating risks.
□ Strategic Asset Allocation
☆ Helps in creating balanced, resilient portfolios adaptable to diverse economic/market conditions.
□ Coding Example
☆ We do a coding example of how this might be applied to a momentum-based quantitative trading system.
Key Concepts of Geometric Mechanics
Differential Geometry
This involves the study of smooth curves, surfaces, and general geometric structures on manifolds.
It provides a way to understand the shape, curvature, and other properties of these structures.
A geometric understanding of financial data can help with analysis and visualization.
Hamiltonian & Lagrangian Mechanics
These are reformulations of classical mechanics that utilize symplectic geometry and variational principles, respectively.
Hamiltonian systems are relevant in understanding time evolution in dynamic systems, which makes them conceptually applicable in the modeling of financial markets/data.
In general, Hamiltonian and Lagrangian mechanics conceptually inspire financial models for dynamic optimization and risk management by providing frameworks for understanding systems’ evolution and
optimizing decisions over time in financial markets with many variables influencing their direction.
Symplectic Geometry
This is a branch of differential geometry focusing on symplectic manifolds, which are a special kind of smooth manifold equipped with a closed, nondegenerate 2-form.
An example of a symplectic manifold can be found below.
This concept is used in Hamiltonian mechanics.
Phase Space & Poisson Brackets
Phase space provides a unified framework for understanding the state of a mechanical system.
Poisson brackets are used to describe the relationships between dynamical variables in this space.
They offer conceptual tools to understand the dynamic state and evolution of financial systems, and to develop trading algorithms that account for the interplay of various market variables and their
Quantum Mechanics & Hilbert Spaces
In quantum mechanics, geometric mechanics principles are applied in a probabilistic framework, often using the language of Hilbert spaces and operators.
With their probabilistic frameworks and complex state representations, they’re specifically applied in quantum computing algorithms for finance.
This enables more efficient solutions for optimization problems, portfolio management, and market simulations by leveraging the principles of superposition and entanglement.
Applications in Finance
Risk Management and Portfolio Optimization
Geometric mechanics can be used to understand the dynamics of financial markets and portfolios.
The phase space concept, for instance, can be adapted to visualize and analyze the state of a financial portfolio by considering various factors like asset prices, volatilities, and correlations.
Option Pricing Models
The stochastic differential equations used in option pricing (like the Black-Scholes model) can be analyzed using techniques from geometric mechanics to understand their properties and behavior under
different market conditions.
Market Dynamics and Econophysics
Geometric mechanics offers tools to model complex market dynamics, including bubble formation, crashes, and high-frequency trading dynamics.
Quantum Finance
There are emerging applications of quantum mechanics in finance, such as quantum computing for complex financial calculations and modeling.
The principles of geometric mechanics are foundational in understanding these quantum systems.
Algorithmic Trading
Some principles of geometric mechanics can inspire algorithmic strategies.
High-frequency trading where market dynamics can be modeled and predicted with high precision make use of more sophisticated math.
HFT algorithms are typically written in C++ (though a lot of prototyping is done in Python due to the availability of their AI/ML/advanced math libraries).
The mathematical complexity of geometric mechanics makes it less accessible for typical forms of financial modeling.
Indirect Application
Many of the applications in finance are theoretical or conceptual rather than direct.
Data and Computation
Implementing these concepts practically requires larger amounts of computational resources and highly specialized knowledge.
Geometric Mechanics vs. Differential Geometry vs. Information Geometry in Finance
• Geometric Mechanics deals with dynamic systems and their evolution over time, making it suitable for analyzing dynamic financial markets.
• Differential Geometry is more about the structure and properties of curves and surfaces, useful for understanding the shape and structure of financial data.
• Information Geometry focuses on the probabilistic aspect, treating information (like asset returns) as geometric objects, which is useful in statistical models of finance.
Coding Example – Geometric Mechanics in Finance
In geometric mechanics, a fundamental equation is the Hamiltonian form of the equations of motion.
These equations are used to describe the evolution of a physical system in time and are useful in the context of symplectic geometry, a branch of differential geometry.
The Hamiltonian equations are given by:
• dqi/dt = ∂H/∂pi
• dpi/dt = -∂H/∂qi
• qi are the generalized coordinates
• pi are the conjugate momenta
• H(qi, pi, t) is the Hamiltonian
• dqi/dt and dpi/dt are the time derivatives of qi and pi
To represent financial data and its momentum stochastically within the Hamiltonian mechanics framework, we’ll interpret the generalized coordinates, as the financial data and the conjugate momenta,
as related to the rate of change (momentum) of the data.
The Hamiltonian, can be seen as analogous to the total “energy” of the financial system.
Since real financial systems are subject to “random” external influences, we’ll introduce stochastic elements to both the financial data and its momentum.
This code models the financial data and its momentum using a Hamiltonian-like system with stochastic elements:
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
def stochastic_hamiltonian_system(y, t, sigma_q, sigma_p):
q, p = y
dqdt = p
dpdt = -q + np.random.normal(0, sigma_p) # Stochastic term for momentum
q += np.random.normal(0, sigma_q) # Stochastic term for financial data
return [dqdt, dpdt]
# Initial conditions (q0 = initial data starting point, p0 = initial momentum)
q0 = 1.0
p0 = 0.0
# Standard deviation of the stochastic terms
sigma_q = 0.02 # Randomness in the financial data
sigma_p = 0.02 # Randomness in the momentum
# Time points
t = np.linspace(0, 10, 500)
# Solve the stochastic Hamiltonian system
solution = odeint(stochastic_hamiltonian_system, [q0, p0], t, args=(sigma_q, sigma_p))
# Plotting
plt.figure(figsize=(12, 6))
# Financial data (q)
plt.subplot(1, 2, 1)
plt.plot(t, solution[:, 0])
plt.title('Stochastic Stock Price (q)')
# Momentum (p)
plt.subplot(1, 2, 2)
plt.plot(t, solution[:, 1])
plt.title('Stochastic Momentum (p)')
What would this be useful for?
It might be useful for a momentum-based quant trading system where the idea would be to trade momentum after it reaches a certain level or rate of change, expecting it to continue.
It would depend on the trader’s research into the topic/strategy, how they model it algorithmically, and backtesting the algorithm/system thoroughly before deploying it live. | {"url":"https://www.daytrading.com/geometric-mechanics","timestamp":"2024-11-05T15:43:45Z","content_type":"text/html","content_length":"59358","record_id":"<urn:uuid:88da2c64-c6ff-47de-a585-40b0cb841b2e>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00563.warc.gz"} |
3. Looks hard? Try expanding it...
Given that $a_m$ (m<2016) are nonegative integers, how many distinct sets of $a_m$ are there such that:
$\prod_{n=1}^{403}\sum_{k=0}^4 a_{5n-k}=2015$
Add the digits of the big number up.
Details: In this case, 'distinct' means that all elements in two sets cannot be equal to each other in a fixed order. (i.e. $a_k$ in set a $ot=$ $a_k$ in set b when 0<k<2016, or else set a and set b
are not distinct. Otherwise, they are.) | {"url":"https://solve.club/problems/3-looks-hard-try-expanding-it/3-looks-hard-try-expanding-it.html","timestamp":"2024-11-05T19:53:43Z","content_type":"text/html","content_length":"33547","record_id":"<urn:uuid:bc6b52e9-68ef-4345-ab36-80feb8460aa7>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00595.warc.gz"} |
Solving Quadratic Equations - UCalgary Chemistry Textbook
Mathematical functions of this form are known as second-order polynomials or, more commonly, quadratic functions. $$ax^2+bx+c=0$$
The solution or roots for any quadratic equation can be calculated using the following formula: $$x=\frac{−b±\sqrt{b^2−4ac}}{2a}$$
Solving Quadratic Equations Example
Solve the quadratic equation $ 3x^2 + 13x − 10 = 0$ .
Solution Substituting the values a = 3, b = 13, c = −10 in the formula, we obtain: $$x=\frac{−13±\sqrt{(13)^2−4×3×(−10)}}{2×3}$$ $$x=\frac{−13±\sqrt{169+120}}{6}=\frac{−13±\sqrt{289}}{6}=\frac
The two roots are therefore $$x=\frac{−13+17}{6}=\mathbf{\frac{2}{3}}\quad and\quad x=\frac{−13−17}{6}=\mathbf{−5}$$
As you can see in the example above, the mathematical solution for a quadratic equation can produce negative (or sometimes imaginary) roots. When solving quadratic equations that relate to scientific
measurements, remember:
Roots that you carry forward must be REAL. At least in the realm that we operate in for introductory chemistry, imaginary numbers do not correspond to solutions you can use in calculations or
Roots that you carry forward are (usually) POSITIVE. We cannot have a negative value for moles or concentration, for example! The exception to this is if you have set up an ICE table “backwards”
(with the reaction proceeding in the reverse direction) — it is possible to have a negative value for “x” (i.e. the change in moles or concentration can be negative) so long as none of your actual
values of moles or concentration are negative. | {"url":"https://chem-textbook.ucalgary.ca/version2/review-of-background-topics/math-skills-for-chemistry/solving-quadratic-equations/","timestamp":"2024-11-04T18:50:11Z","content_type":"text/html","content_length":"67237","record_id":"<urn:uuid:60363b06-22a9-4c81-843c-ca4099419ca3>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00740.warc.gz"} |
November 2020 - MetaSD
Statistical tests only make sense when the assumed distribution matches the data-generating process.
There are several analyses going around that purport to prove election fraud in PA, because the first digits of vote counts don’t conform to Benford’s Law. Here’s the problem: first digits of vote
counts aren’t expected to conform to Benford’s Law. So, you might just as well say that election fraud is proved by Newton’s 3rd Law or Godwin’s Law.
Example of bogus conclusions from naive application of Benford’s Law.
Benford’s Law describes the distribution of first digits when the set of numbers evaluated derives from a scale-free or Power Law distribution spanning multiple orders of magnitude. Lots of processes
generate numbers like this, including Fibonacci numbers and things that grow exponentially. Social networks and evolutionary processes generate Zipf’s Law, which is Benford-conformant.
Here’s the problem: vote counts may not have this property. Voting district sizes tend to be similar and truncated above (dividing a jurisdiction into equal chunks), and vote proportions tend to be
similar due to gerrymandering and other feedback processes. This means the Benford’s Law assumptions are violated, especially for the first digit.
This doesn’t mean the analysis can’t be salvaged. As a check, look at other elections for the same region. Check the confidence bounds on the test, rather than simply plotting the sample against
expectations. Examine the 2nd or 3rd digits to minimize truncation bias. Best of all, throw out Benford and directly simulate a distribution of digits based on assumptions that apply to the specific
situation. If what you’re reading hasn’t done these things, it’s probably rubbish.
This is really no different from any other data analysis problem. A statistical test is meaningless, unless the assumptions of the test match the phenomena to be tested. You can’t look at lightning
strikes the same way you look at coin tosses. You can’t use ANOVA when the samples are non-Normal, or have unequal variances, because it assumes Normality and equivariance. You can’t make a linear
fit to a curve, and you can’t ignore dynamics. (Well, you can actually do whatever you want, but don’t propose that the results mean anything.)
The most convincing thing about mainstream climate science is not that the models are so good, but that the alternatives are so bad.
Climate skeptics have been at it for 40 years, but have produced few theories or predictions that have withstood the test of time. Even worse, where there were once legitimate measurement issues and
model uncertainties to discuss, as those have fallen one by one, the skeptics are doubling down on theories that rely on “alternative” physics. The craziest ideas get the best acronyms and metaphors.
The allegedly skeptical audience welcomes these bizarre proposals with enthusiasm. As they turn inward, they turn on each other.
The latest example is in the Lungs of Gaia at WUWT:
A fundamental concept at the heart of climate science is the contention that the solar energy that the disk of the Earth intercepts from the Sun’s irradiance must be diluted by a factor of 4.
This is because the surface area of a globe is 4 times the interception area of the disk silhouette (Wilde and Mulholland, 2020a).
This geometric relationship of divide by 4 for the insolation energy creates the absurd paradox that the Sun shines directly onto the surface of the Earth at night. The correct assertion is that
the solar energy power intensity is collected over the full surface area of a lit hemisphere (divide by 2) and that it is the thermal radiant exhaust flux that leaves from the full surface area
of the globe (divide by 4).
Setting aside the weird pedantic language that seems to infect those with Galileo syndrome, these claims are simply a collection of errors. The authors seem to be unable to understand the geometry of
solar flux, even though this is taught in first-year physics.
Some real college physics (divide by 4).
The “divide by 4” arises because the solar flux intercepted by the earth is over an area pi*r^2 (the disk of the earth as seen from the sun) while the average flux normal to the earth’s surface is
over an area 4*pi*r^2 (the area of a sphere).
The authors’ notion of “divide by 2” resulting in 1368/2 = 684 w/m^2 average is laughable because it implies that the sun is somehow like a luminous salad bowl that delivers light at 1368 w/m^2
normal to the surface of one side of the earth only. That would make for pretty interesting sunsets.
In any case, none of this has much to do with the big climate models, which don’t “dilute” anything, because they have explicit geometry of the earth and day/night cycles with small time steps. So,
all of this is already accounted for.
To his credit, Roy Spencer – a hero of the climate skeptics movement of the same magnitude as Richard Lindzen – arrives early to squash this foolishness:
How can some people not comprehend that the S/4 value of solar flux does NOT represent the *instantaneous* TOA illumination of the whole Earth, but instead the time-averaged (1-day or longer)
solar energy available to the whole Earth. There is no flat-Earth assumption involved (in fact, dividing by 4 is because the Earth is approximately spherical). It is used in only simplistic
treatments of Earth’s average energy budget. Detailed calculations (as well as 4D climate models as well as global weather forecast models) use the full day-night (and seasonal) cycle in solar
illumination everywhere on Earth. The point isn’t even worth arguing about.
Responding to the clueless authors:
Philip Mulholland, you said: “Please confirm that the TOA solar irradiance value in a climate model cell follows the full 24 hour rotational cycle of daytime illumination and night time
Oh, my, Philip… you cannot be serious.
Every one of the 24+ climate models run around the world have a full diurnal cycle at every gridpoint. This is without question. For example, for models even 20+ years ago start reading about the
diurnal cycles in the models on page 796 of the following, which was co-authored by representatives from all of the modeling groups: https://www.ipcc.ch/site/assets/uploads/2018/02/
Philip, Ed Bo has hit the nail on the head. Your response to him suggests you do not understand even the basics of climate modeling, and I am a little dismayed that your post appeared on WUWT.
Undeterred, the WUWT crowd then proceeds to savage anyone, including their erstwhile hero Spencer, who dares to challenge the new “divide by 2” orthodoxy.
Dr roy with his fisher price cold warms hot physics tried to hold the line for the luke-warmers, but soon fecked off when he knew he would be embarrassed by the grown-ups in the room…..
This is not the first time a WUWT post has claimed to overturn climate science. There are others, like the 2011 Unified Theory of Climate. It’s basically technobabble, notable primarily for its utter
obscurity in the nine years following. It’s not really worth analyzing, though I am a little curious how a theory driven by static atmospheric mass explains dynamics. Also, I notice that the perfect
fit to the data for 7 planets in Fig. 5 has 7 parameters – ironic, given that accusations of overparameterization are a perennial favorite of skeptics. Amusingly, one of the authors of the “divide by
two” revolution (Wilde) appears in the comments to point out his alternative “Unifying” Theory of Climate.
Are these alternate theories in agreement, mutually exclusive, or just not even wrong? It would be nice if skeptics would get together and decide which of their grand ideas is the right one. Does
atmospheric pressure run the show, or is it sunspots? And which fundamentals that mathematicians and physicists screwed up have eluded verification for all these years? Is it radiative transfer, or
the geometry of spheres and disks? Is energy itself misdefined? Inquiring minds want to know.
The bottom line is that Roy Spencer is right. It isn’t worth arguing about these things, any more than its worth arguing with flat earthers or perpetual motion enthusiasts. Engaging will just leave
you wondering if proponents are serious, as in seriously deluded, or just yanking your chain while keeping a straight face. | {"url":"https://metasd.com/2020/11/","timestamp":"2024-11-04T03:03:13Z","content_type":"text/html","content_length":"77111","record_id":"<urn:uuid:7b90de19-9537-4528-8809-223d85d862e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00358.warc.gz"} |
Forces data collection
Four types of
force data collections
are available in Flux, corresponding to specific computation methods and/or approaches:
• The Simplified projective method dedicated to rotating machines is a computation method based on Maxwell tensor and dedicated to the most classical configurations of rotating machines (e.g.
without eccentricity); it is an efficient and robust method based on the evaluation of forces in a virtual cylinder in the airgap of a rotating machine in order to subsequently project them on an
imported mesh or on a Flux mesh. These values may finally be exported towards OptiStruct for a NVH analysis of the electrical machine.
• The Generalized projective method may be seen as an extension of the Simplified projective method dedicated to rotating machines; in fact, if the approach explained above is well adapted for
rotating machines with cylindrical airgap, this one allows the user to compute forces on every kind of support that may be defined by an extruded compound path, by a 2D grid or by an extruded
geometric line in an air or vacuum region. These force values are then projected, and finally exported, on a previously-imported mesh, with the possibility to have all three spatial components if
the physical configuration is suitable (e.g. axial flux machine).
• The Direct method for surface forces (dFmagS) is a general forces computation method at the interface between two materials with different magnetic permeabilities; being generic, it can be
applied on every type of face data support.
• The Direct method for volume forces (dFLapV) is based on the integration of the Laplace forces and may be applied on any volume data support. Being these forces generated by an interaction of the
magnetic flux density with the current density, this approach is available only for the solid conductor regions and the coil conductor regions.
Table 1. Summary of the different forces data collection
Type of forces data collection Computation quantities and points Export quantities and points
Simplified projective method dedicated Computation of the Maxwell tensor on a virtual cylinder in the air gap of the electrical machine Projection of the Maxwell tensor on the data
to rotating machines support
Generalized projective method Computation of the Maxwell tensor on every kind of support that may be defined by an extruded compound path, a Projection of the Maxwell tensor on the data
2D grid or an extruded geometric line in an air or vacuum region support (imported mesh only)
Direct method for surface forces Computation of dFmagS on the data support
Direct method for volume forces Computation of dFLapV on the data support | {"url":"https://help.altair.com/flux/Flux/Help/english/UserGuide/English/topics/CollectionDeDonneesDeForces_r.htm","timestamp":"2024-11-12T12:51:03Z","content_type":"application/xhtml+xml","content_length":"54924","record_id":"<urn:uuid:4bb360e8-ce44-4d95-b7df-23c35a3ce235>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00477.warc.gz"} |
Computational Algebraic Geometry by Wolfram Decker | Download book PDF
Computational Algebraic Geometry by Wolfram Decker
This PDF book covers the following topics related to Algebraic Geometry : General Remarks on Computer Algebra Systems, The Geometry–Algebra Dictionary, Affine Algebraic Geometry, Ideals in Polynomial
Rings, Affine Algebraic Sets, Hilbert’s Nullstellensatz, Irreducible Algebraic Sets, Removing Algebraic Sets, Polynomial Maps, The Geometry of Elimination, Noether Normalization and Dimension, Local
Studies, Projective Algebraic Geometry, The Projective Space, Projective Algebraic Sets, Affine Charts and the Projective Closure, The Hilbert Polynomial, Computing, Standard Bases and Singular,
Applications, Ideal Membership, Elimination, Radical Membership, Ideal Intersections, Ideal Quotients, Kernel of a Ring Map, Integrality Criterion, Noether Normalization, Subalgebra Membership,
Homogenization, Dimension and the Hilbert Function, Primary Decomposition and Radicals, Buchberger’s Algorithm and Field Extensions, Sudoku, A Problem in Group Theory Solved by Computer Algebra,
Finite Groups and Thompson’s Theorem, Characterization of Finite Solvable Groups.
Author(s): Wolfram Decker, Gerhard Pfister
133 Pages
Download / View book | {"url":"https://www.freebookcentre.net/maths-books-download/Computational-Algebraic-Geometry-by-Wolfram-Decker.html","timestamp":"2024-11-08T18:25:02Z","content_type":"text/html","content_length":"38204","record_id":"<urn:uuid:1f13c6d0-750f-409b-bff9-d5982bc7f341>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00232.warc.gz"} |
nearbyint, nearbyintf, nearbyintl
float nearbyintf( float arg ); (1) (since C99)
double nearbyint( double arg ); (2) (since C99)
long double nearbyintl( long double arg ); (3) (since C99)
#define nearbyint( arg ) (4) (since C99)
Rounds the floating-point argument
to an integer value in floating-point format, using the
current rounding mode
4) Type-generic macro: If arg has type long double, nearbyintl is called. Otherwise, if arg has integer type or the type double, nearbyint is called. Otherwise, nearbyintf is called, respectively.
arg - floating point value
Return value
The nearest integer value to arg, according to the current rounding mode, is returned.
Error handling
This function is not subject to any of the errors specified in math_errhandling.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
• FE_INEXACT is never raised
• If arg is ±∞, it is returned, unmodified
• If arg is ±0, it is returned, unmodified
• If arg is NaN, NaN is returned
The only difference between nearbyint and rint is that nearbyint never raises FE_INEXACT.
The largest representable floating-point values are exact integers in all standard floating-point formats, so nearbyint never overflows on its own; however the result may overflow any integer type
(including intmax_t), when stored in an integer variable.
If the current rounding mode is FE_TONEAREST, this function rounds to even in halfway cases (like rint, but unlike round).
#include <stdio.h>
#include <math.h>
#include <fenv.h>
int main(void)
#pragma STDC FENV_ACCESS ON
printf("rounding to nearest:\nnearbyint(+2.3) = %+.1f ", nearbyint(2.3));
printf("nearbyint(+2.5) = %+.1f ", nearbyint(2.5));
printf("nearbyint(+3.5) = %+.1f\n", nearbyint(3.5));
printf("nearbyint(-2.3) = %+.1f ", nearbyint(-2.3));
printf("nearbyint(-2.5) = %+.1f ", nearbyint(-2.5));
printf("nearbyint(-3.5) = %+.1f\n", nearbyint(-3.5));
printf("rounding down: \nnearbyint(+2.3) = %+.1f ", nearbyint(2.3));
printf("nearbyint(+2.5) = %+.1f ", nearbyint(2.5));
printf("nearbyint(+3.5) = %+.1f\n", nearbyint(3.5));
printf("nearbyint(-2.3) = %+.1f ", nearbyint(-2.3));
printf("nearbyint(-2.5) = %+.1f ", nearbyint(-2.5));
printf("nearbyint(-3.5) = %+.1f\n", nearbyint(-3.5));
printf("nearbyint(-0.0) = %+.1f\n", nearbyint(-0.0));
printf("nearbyint(-Inf) = %+.1f\n", nearbyint(-INFINITY));
rounding to nearest:
nearbyint(+2.3) = +2.0 nearbyint(+2.5) = +2.0 nearbyint(+3.5) = +4.0
nearbyint(-2.3) = -2.0 nearbyint(-2.5) = -2.0 nearbyint(-3.5) = -4.0
rounding down:
nearbyint(+2.3) = +2.0 nearbyint(+2.5) = +2.0 nearbyint(+3.5) = +3.0
nearbyint(-2.3) = -3.0 nearbyint(-2.5) = -3.0 nearbyint(-3.5) = -4.0
nearbyint(-0.0) = -0.0
nearbyint(-Inf) = -inf
• C11 standard (ISO/IEC 9899:2011):
□ 7.12.9.3 The nearbyint functions (p: 251-252)
□ 7.25 Type-generic math <tgmath.h> (p: 373-375)
□ F.10.6.3 The nearbyint functions (p: 526)
• C99 standard (ISO/IEC 9899:1999):
□ 7.12.9.3 The nearbyint functions (p: 232)
□ 7.22 Type-generic math <tgmath.h> (p: 335-337)
□ F.9.6.3 The nearbyint functions (p: 463)
See also
(C99)(C99)(C99)(C99)(C99)(C99)(C99)(C99)(C99) rounds to an integer using current rounding mode with
exception if the result differs
(C99)(C99)(C99)(C99)(C99)(C99)(C99)(C99)(C99) rounds to nearest integer, rounding away from zero in halfway cases
gets or sets rounding direction
(C99)(C99) (function)
C++ documentation for nearbyint | {"url":"https://docs.w3cub.com/c/numeric/math/nearbyint","timestamp":"2024-11-13T11:04:55Z","content_type":"text/html","content_length":"13895","record_id":"<urn:uuid:747fec33-13d5-4282-a2ec-6b57a48b0726>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00142.warc.gz"} |
Extensions of Hilbert's Tenth Problem
March 21 to March 25, 2005
at the
American Institute of Mathematics, Palo Alto, California
organized by
Bjorn Poonen, Alexandra Shlapentokh, Xavier Vidaux, and Karim Zahidi
This workshop, sponsored by AIM and the NSF, will be devoted to extensions of Hilbert's Tenth Problem and related questions in Number Theory and Geometry. The main topics for the workshop are
1. HTP over rings and fields of algebraic numbers (in particular HTP over rational numbers, Mazur's Conjectures, elliptic curve methods)
2. HTP over functions fields of arbitrary characteristic, elementary equivalence versus isomorphism problem for function fields.
3. HTP for rings and fields of meromorphic functions (both complex and p-adic)
The workshop will differ from typical conferences in some regards. Participants will be invited to suggest open problems and questions before the workshop begins, and these will be posted on the
workshop website. These include specific problems on which there is hope of making some progress during the workshop, as well as more ambitious problems which may influence the future activity of the
field. Lectures at the workshop will be focused on familiarizing the participants with the background material leading up to specific problems, and the schedule will include discussion and working
Invited participants include G. Cornelissen, M. Davis, K. Eisentraeger, G. Everest, M. Jarden, L. Lipshitz, A. Macintyre, Y. Matiyasevich, L. Moret-Bailly, T. Pheidas, B. Poonen, F. Pop, K. Rubin, T.
Scanlon, A. Shlapentokh, A. Silverberg, M. VanDieren, X. Vidaux, and K. Zahidi.
The deadline to apply for support to participate in this workshop has passed.
Plain text announcement or brief announcement.
Go to the American Institute of Mathematics.
Return to the AIM Research Conference Center. | {"url":"https://aimath.org/ARCC/workshops/hilberts10th.html","timestamp":"2024-11-09T19:19:22Z","content_type":"text/html","content_length":"3478","record_id":"<urn:uuid:c68739aa-cb2c-4aca-9ff0-3c520a573252>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00339.warc.gz"} |
Working With Column Vectors
Writing vectors as column vectors is more informative than using vector notation since more information is included, and is preferable when working in the x – y plane. Suppose we have the triangle
P splits AC in the ratio 1:2. The vector from A to B is | {"url":"https://astarmathsandphysics.com/igcse-maths-notes/548-working-with-column-vectors.html","timestamp":"2024-11-09T20:35:59Z","content_type":"text/html","content_length":"28207","record_id":"<urn:uuid:4cd5a3b3-d85d-42dd-b908-bc2ab4494951>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00491.warc.gz"} |
Risk-Based Life Assessment of Corroding Pipelines
Probabilistic Remnant Life Assessment of Corroding Pipelines within a Risk-Based Framework
Dr A Muhammed and Mr J B Speck
Paper presented at ASRANet International Colloquium 2002, 8 -10 July 2002, University of Glasgow, Scotland
This paper describes the application of a risk-based approach to the assessment of corroding offshore pipelines with the aim of optimising maintenance and replacement schedules. It describes an
overall risk-based framework in which initial screening and risk ranking is followed by more detailed probabilistic assessment of the pipelines most at risk of failure. Initial screening is based on
a semi-quantitative assessment of factors including pipeline inventory, historical and anticipated operating conditions, active damage mechanisms and potential failure consequences. Initial screening
is formalised using TWI's RISKWISE ^TM risk assessment software, which deals with both likelihood and failure consequences for all pipelines. The RISKWISE ^TM output also includes a remaining life
estimate based on conservative input parameters. Detailed probabilistic analysis is performed for the pipelines with low estimates of remaining life or high risk ranking, based on the results of the
initial screening assessment. The probabilistic analysis considers both leakage and rupture of the pipelines, based on the modified ASME B31G. A major feature of the probabilistic model is the
extrapolation of limited inspection data to cover longer pipeline lengths. Example calculations are presented based on a case study involving the assessment of about 260 subsea pipelines consisting
of both water injection and oil production pipelines. The assessment results include both remaining life estimates and sensitivity studies. The method described provides a basis for determining
maintenance and inspection priorities and ultimately the development of pipeline replacement strategies.
A method for the application of a risk-based approach to the assessment of corroding offshore pipelines is described. The approach involves an overall risk-based framework in which initial screening
and risk ranking is followed by more detailed probabilistic assessment of the pipelines most at risk of failure. The study described was carried out with the aim of optimising pipeline maintenance
and replacement schedules.
Initial risk ranking
The study began with a review of all potential damage corrosion mechanisms for the pipelines. This produced a list of credible internal mechanisms including microbial corrosion, oxygen induced
corrosion and CO
Firstly, a risk-ranking was produced using qualitative likelihood and consequence ratings in order to establish assessment priorities. Likelihood was based on factors such as process fluid
corrosivity while consequence was assessed on the basis of the production criticality of the pipeline. This allowed very low risk pipelines to be eliminated from further consideration. Figure 1 shows
a typical likelihood and consequence matrix from an initial risk ranking exercise.
A semi-quantitative assessment was carried out using TWI's RISKWISE ^TM assessment procedure for pipelines. This involved an estimation of the pipeline remaining life based on failure likelihood
factors which were centered on current condition, likelihood of failure within specified forward time frames (e.g. 5, 10 and 15 years) and effectiveness of any inspections. These factors are used
within the RISKWISE ^TM software to determine remaining life indicators (RLI). RLI is analogous to a remaining life estimate but it is termed an indicator because it is only intended to give a
measure of the remaining life based on the change in the failure likelihood over time. Input data used in the RLI calculation are worst-case parameter estimates such that the RLI under-estimates
remaining life compared to the more quantitative probabilistic method. In the case study described, pipelines with RLI of less than 5 years were selected as critical lines for the more detailed
probabilistic assessment.
Fig. 1. Estimates of risk
Probabilistic corrosion assessment
Analysis methods
The probabilistic assessment involved the formulation of a methodology for estimating the probability of failure by leakage or rupture, implementation in appropriate software and application to all
critical pipelines. The work also involved analysis of input data to derive statistical distributions for various input parameters.
In general, failure probability methods use a number of techniques to estimate the probability of having combinations of uncertain variables that result in the occurrence of a failure event. Most
practical problems can be thought of in terms of the interaction between a distribution of load effects and the resistance distribution as illustrated in Fig.2. In simple terms, all the analysis
methods seek to establish the degree of overlap between the distributions. This depends essentially on the separation between the means of the distributions and the spread in each distribution. The
area of overlap, and hence failure probability (P [f]), decreases with increasing separation between µ [R] and µ [L] while P [f] increases with increasing spread ( σ [R] or σ [L]) in either
In these analyses, probability distributions are used in preference to absolute values because the loading effects and resistance factors are subject to uncertainties that result from several
sources. For example, for pipelines subject to internal pressure, uncertainties may arise from variability in material strength, pipe wall thickness and other geometrical properties, and fluctuations
in internal pressure.
The discussion above involving two variables can easily be extended to cases involving several variables. For instance, a general expression for the failure condition or limit state may be stated as:
Z = G(X [1], X [2],.....X [n])
where X [1], X [2],.....X [n] represent basic variables e.g. material yield strength, defect height, operating pressure, etc. and G is a valid mathematical expression defined such that failure occurs
when Z is less than or equal to 0. This expression is known as the 'limit state equation'.
The required calculation for failure probability is:
where f [X] is the joint probability density function for the n-dimensional vector x of basic variables.
Several methods are available for the estimation of failure probability for a given uncertain event such as Eq.[2]. These include Monte Carlo simulation, first order (FORM) and second order
reliability (SORM) methods. The various methods for failure probability estimation have been widely published ^[1,2] and are implemented in commercial software products.
Fig. 2. Illustration of failure probability analysis concept using the interaction between loading and resistance effects
Simulation methods such as Monte Carlo simulation involve generating random numbers for basic variables (X [1], X [2],..X [n]), at frequencies based on specified probability distributions for the
variables, and checking whether failure is predicted or not (i.e., is Z<=0 ?) for each set of random sample values. This process is repeated several times and the ratio of the number of failures to
the total number of simulations gives an estimate of the failure probability.
First order and second order reliability methods (FORM and SORM) use numerical procedures to simplify the joint probability density function f [X] in Eq.[2] and sometimes to approximate the limit
state equation, in the failure probability calculation. Unlike the simulation techniques, which may be time-consuming in low probability analysis, FORM and SORM are very fast in most practical cases.
In the present work, FORM and SORM were adopted as the primary method of estimating failure probabilities. Monte Carlo simulation was used to occasionally check the results.
Probabilistic model
The probabilistic model was designed so that the failure event is defined as the occurrence of leakage or rupture at an internal corrosion defect in the pipeline. The failure model was defined in
terms of depth of corrosion with the limit state equation written as:
Z = d [c] - d(t)
where d [c] is the critical corrosion depth for rupture or leakage and d(t) is the depth of corrosion at a given time t. In this formulation of the limit state equation, Z ≤0 constitutes failure
while Z>0 is a safe condition. In the conceptual illustration of Fig.2, d [c] constitutes a resistance effect while d(t) is the loading effect.
The critical corrosion depth d [c] in Eq.[3] was determined from the basic equations provided in the modified ASME B31G rupture criterion ^[3] for corroded pipes. The derivation of an expression for
d [c], begins with the modified B31G basic burst stress equation:
where σ [f] is the hoop stress at failure
is the flow stress, taken equal to yield strength + 10ksi (69MPa)
A is the corroded area in the longitudinal plane through the wall thickness
A [0] is LxB, mm ^2
L is the axial extent of the corroded area, mm
B is the initial pipe wall thickness, mm
M is the 'Folias' factor, which is given by:
; for
where D is the diameter of the pipe, mm.
The corroded area, A in the above equation may be calculated in a number of ways. Where the data on corrosion profile is available, the area is computed fairly accurately. In the present work, only
data on the maximum depth, d and the length of individual corrosion are known. In such cases, the modified B31G approach uses an approximate metal loss area, A, of 0.85 x d x L. Putting this
approximation in Eq.[4] and re-arranging the equation gives the maximum corrosion depth, d [c], that may exist in the line for a failure stress corresponding to the specified (operating) hoop stress
σ as:
From the form of the above equation, it was found that for very short corroded areas (i.e. small L) and at very low operating stresses, d [c] may be evaluated as being slightly greater than the wall
thickness. Therefore, an upper bound value equal to the initial wall thickness, B was used in such cases. In the probabilistic analysis, the critical depth of corrosion was compared against the
expected corrosion depth at different instances of time. The limit state equation was implemented in the form:
where A is the corrosion rate in mm/year and t is the elapsed time in years.
The output from the probabilistic analysis based on Eq.[8] for a given time t is the probability of failure (leakage or rupture) in the time interval from start-of-life to year t. The following
equation is used to calculate the annual probability of failure at year t:
Where P [f,t] is the probability of failure after 't' years in service. The denominator in the above equation allows for the fact that the pipeline, at the location in question survived up to year
The failure probability calculations were carried out using the STRUctural RELiability software system (STRUREL) developed by RCP GmBH and licensed to TWI. STRUREL ^[4] is a commercial software
within which failure conditions such as Eq.[8] can be defined and probabilistic analysis conducted.
The results of the probabilistic analysis were checked for accuracy by running a selection of independent deterministic calculations to confirm that the FORM and SORM solutions satisfied the failure
equation or limit state. This confirmed that the basic failure equations had been implemented properly.
Case study - input data and statistical analysis
Example calculations are now presented based largely on a case study that involved the assessment of about 260 subsea pipelines. This consisted of water injection and oil production pipelines spread
over two oil producing fields. A summary of the input data used in the probabilistic corrosion assessment is given in Table 1 and further details on the input assumptions are given below.
Table 1 General input data used for the probabilistic analyses
Variable Distribution Mean COV, %
Initial wall thickness, B Normal 1.02 x nominal 4
Outside Diameter, D Fixed Nominal 0
Yield strength Normal 1.09 x SMYS 4
Operating stress, σ Fixed pD/2B, where p = pressure 0
Predictive model ratio, R* Normal 0.622 22.2
Corrosion area (mm/yr)† Extreme value 0.14 35.7
Corrosion defect length, L Fixed 450mm 0
COV = standard deviation divided by mean
* R is the ratio of the modified ASME B31G predicted failure stress to actual failure stress in pipe burst tests ^[3] (Mean = 0.622, Standard deviation = 0.138, COV = 22.18%)
† Baseline distribution before extrapolation quoted in the table (Extrapolated mean for a 3km long pipeline = 0.23 mm/year)
Basic input data
Input data such as pipe geometrical properties, yield strength and operating stresses were readily derived from the design information available from the pipeline database. Pipe diameter and
operating stress were considered as fixed values while some uncertainties were assumed on pipe wall thickness and yield strength values. In both cases, normal distributions were assumed with mean
values related to nominal wall thickness and specified minimum yield strength (SMYS). This treatment is in-line with published work on probabilistic assessment. ^[5,6]
Uncertainty in burst prediction model
The modified B31G predictive model for pipe rupture or leakage is designed to be conservative. Statistical information is provided on the relationship between the predicted failure stress (given by
Eq.[9]) and actual failure stresses obtained from pipe burst tests in the modified ASME B31G document ^[3] . The distribution of predicted to actual failure stress ratio, R, in the tests was found to
be a normal distribution with a mean value of 0.622 and a standard deviation of 0.138. This suggests that on average, the actual failure pressure or stress is about 60% higher than the predicted
value. This ratio was introduced into the probabilistic model by taking the stress σ in the limit state Eq.[8] as operating stress multiplied by stress ratio R.
Corrosion defect length
In this study, lengths reported from inspection were fitted to standard statistical distributions. The best fit obtained from the analysis was a lognormal distribution. However, a significant amount
of variability was observed. Such large variability, particularly for a lognormal distribution, can lead to numerical problems in the probabilistic calculation routines. It was also not clear if the
length of the corroded area would increase with time, it was therefore considered appropriate to use a conservative fixed value based on the distribution obtained. The equivalent mean + 2 standard
deviations for the lognormal distribution, (widely accepted as a conservative estimate of a variable value) was about 450mm. This value was therefore assumed in the probabilistic analysis of
pipelines. It was considered conservative because examination of a section of a failed water injection pipeline revealed that the corroded area that caused failure measured approximately 350mm in
length. Therefore the assumed length of 450mm is conservative with respect to that known to have caused failure.
Corrosion rate distribution
Corrosion rate distributions were mainly derived from available inspection data on the basis of the type of service. For example, inspection data on all water injection pipelines were processed to
determine a basic corrosion rate distribution. Similarly, inspection data for oil production lines were combined to derive the appropriate corrosion rate distributions. The approach of combining data
from several pipelines was necessary because inspection data was limited both in terms of the overall number of the pipelines inspected and the extent of inspection on individual pipelines.
Typically, ultrasonic cable-operated inspection equipment introduced from one or both ends would inspect 500m to 800m end sections of a pipeline. This only provided a sample inspection as the
pipelines were sometimes up to 7km long. To account for the limited nature of the inspections, extreme value distributions were fitted to the measured data and then extrapolations were carried out to
estimate the maximum corrosion rates that might have been expected had the entire pipelines been inspected. In general, corrosion rate statistics were derived by dividing the metal loss reported from
the inspection by the number of service years since commissioning.
Corrosion rate extrapolation for pipeline length
The method of extrapolation based on sample inspection has been used previously by a number of investigators.
The technique has been more comprehensively developed in a recent TWI work
in which several aspects of the methodology were investigated. In the present study, the maximum corrosion depths for approximately the same unit lengths of pipeline were fitted to an extreme value
distribution using commercial statistical software. This gave a Type I extreme value distribution of maximum rates, which has the form:
where F
(a) is the probability of the corrosion rate A having a value less than 'a', µ is referred to as the location parameter and α is the scale parameter. In simple terms, the location parameter is the
most likely corrosion rate and the scale parameter represents the extent of scatter of rates about the location parameter and therefore controls the shape of the distribution.
Extrapolation of the basic distribution to allow for potentially higher corrosion rates in uninspected regions of the lines is done by shifting the location parameter to a higher value. The location
parameter of the shifted distribution is given by:
where µ and α are the parameters of the basic distribution and N is the ratio of total pipeline length to the unit length from which worst data was extracted to derive the basic distribution. The
above equation follows from extreme value theory and is quoted in several standard texts (e.g. see Ref ^[1] ). The extrapolation of Eq.[11] is based on two fundamental assumptions: firstly, that the
uninspected region is nominally similar to the region sampled and secondly, that the unit sections considered in extracting maxima are independent or at least that correlation is negligible.
Therefore, the unit length of pipeline used was selected based on previous experience ^[9] to ensure independence. The other requirement of the sample being representative of the uninspected pipeline
regions was particularly important in the present study as inspections were mostly limited to pipeline ends. However, a limited number of checks showed that variability in wall thickness loss
stabilised within the typical 500m end lengths covered by the inspections. For example, see plot of wall thickness versus distance from line end in Fig.3.
Fig. 3. Typical wall thickness variation with distance from pipe end
A typical example of the basic extreme value distribution and the corresponding extrapolations for different pipeline lengths are shown in Fig.4. The effect of extrapolation is clearly seen as the
entire distribution is shifted to higher corrosion rates. It should be noted that the shift is logarithmic with pipe length.
Validation against failure data
Leaks had been reported for a few of the water injection lines assessed in this exercise. The corrosion rates implied by these leaks were compared against the corrosion rate distributions obtained
for the failed lines. These checks provided some validation for the procedure for determining corrosion rate distributions, as the rate implied by the leaks fell in the upper tail region of the
distributions (e.g. see
). The two rates shown in
are the minimum and maximum values from several failures. It is also noted from
that the rates implied by the failures would be much less credible if basic corrosion rate distributions were not extrapolated for the length of the lines.
Fig. 4. Extrapolation of corrosion rate distribution and comparison with rates implied by failures (full pipeline length 7Km)
Case study - results of failure probability calculations
The results of the assessment are presented in the form of plots of annual failure probability against time. A typical plot is shown in
. The plot shows an initial assessment and an updated assessment following inspection of the individual pipeline. Updating is further discussed within the section describing the sensitivity study in
this paper. For a plot such as that shown in
to be used in determining a remaining life, a cut-off failure probability for the end of useful life has to be defined. This is often referred to as the target failure probability.
Fig. 5.Annual failure probability - time plot and updating with inspection data
Target failure probability and calibration with failure data
Two basic factors were considered in setting the target annual failure probability levels for determining the end of life and hence the pipeline remaining lives. Firstly, the target values
recommended in offshore codes, standards and other published documents ^[10-12] were reviewed. In most cases these are based on safety classification, which for pipelines, is mainly related to the
fluids transported and the pipeline location. Typically, three safety classes of 'low', 'normal' and 'high' are defined. In line with guidance in the offshore standards, ^[10,11] water injection
lines may be categorised into 'low' safety class while oil production and gas lines may be considered 'high' safety class. The target failure probabilities recommended for design against a failure
condition such as rupture (due to corrosion) in the standards and other relevant publications are summarised in Table 2. A second consideration in setting the target failure probability involved
calibrating the results of the probabilistic assessment against the field experience in terms of reported leaks and failures in water lines. Analyses were conducted to estimate the annual failure
probabilities for the years leaks were reported in these lines. This gave values in the range of 2.2x10 ^-3 to 7x 10 ^-2 for the lines. These failure probability values are broadly in line with the
range of 10 ^-3 to 10 ^-2 per year recommended in the standards (see Table 2).
Table 2 Recommended annual target failure probabilities
Authors Safety class
Low Normal High
RP-F101 10 ^-3 10 ^-4 10 ^-5
(DNV 1999)
OS-F101 10 ^-3 10 ^-4 10 ^-5
(DNV 2000)
Sotberg et al (1997) 10 ^-2 - 10 ^-3 10 ^-3 - 10 ^-4 10 ^-4 - 10 ^-5
(SUPERB project)
Based on the two considerations outlined above, annual failure probabilities of 10 ^-3 to 10 ^-2 and 10 ^-5 were adopted for water injection and oil production lines respectively. The results of the
probabilistic analysis for the water injection lines that failed in one field were closer to 10 ^-3, so this lower target was applied to all water lines in that field. Similarly, the higher target
value of 10 ^-2 matched the results for water line failures in the second field, so the higher value was adopted.
Sensitivity analyses
The sensitivity of the results of a probabilistic analysis to the various input parameters were evaluated by examining the so-called sensitivity factors produced from a FORM or SORM analysis.
Figure 6
shows, in a pie chart, the sensitivity of results to input data for a typical water injection line. This shows clearly that corrosion rate is by far the most influential variable. Initial wall
thickness and the burst failure prediction error, R, also have some effect. Of these variables, there is reasonable confidence in the estimates of initial wall thickness and on the prediction error
R, while the appropriateness of the assumed corrosion rate distribution for individual pipelines is less certain particularly where such lines had not been inspected. The effect of inspection on
corrosion rate distributions and consequently on predicted remaining life was explored through further sensitivity analyses.
Fig. 6. Sensitivity of analyses results to input vari ables for a typical water injection line
Firstly, it was noted that the limit state Eq.[8] employed in the probabilistic model simply accumulates the metal loss from start of life and predicts failure when this is sufficient to cause a leak
or rupture. If there is a high degree of uncertainty regarding the corrosion rate distribution, then the amount of metal loss assumed by the model after a given service period may be in significant
error. Therefore for lines with in-service inspection, a second set of analyses was conducted utilising a different limit state equation based on the actual measured wall thickness distribution at
the time of inspection. The time frame for this second analysis starts from the inspection date and essentially corrects the extent of metal loss assumed in the model to that observed from
inspection. Figure 5shows failure probability plots for an initial assessment and the corresponding modified plot after updating with thickness measurements for the specific pipeline. The drop in
failure probability estimate at the inspection time is a measure of the correction due to the availability of the inspection data on the individual pipeline. Due to the conservatism in the general
analysis prior to updating for individual pipelines, the remaining life estimates in all cases considered increased as a result of the updating. The results of initial and updated analysis for 5
pipelines are given in Table 3, which show improvements in remnant life estimates of between 6 and 13 years.
Table 3 Effect of line-specific NDT data on remaining life estimate
Pipeline Life estimate - no line specific NDT (years) Life estimate - line specific NDT used (years) Difference (years)
Line 1 2.3 18.3 16
Line 2 0.1 13.1 13
Line 3 9.5 15.5 6
Line 4 7.3 20.3 13
Line 5 2.7 12.7 10
Average 11.6
Initial risk ranking
Initial screening produced results that were used to exclude low risk pipelines from further consideration while those with low remaining life estimates (less than 5 years) were identified for
detailed probabilistic assessment. This provided a means of rapidly screening the pipelines making for economy of effort. The semi-quantitative RISKWISE ^TM assessment procedure as applied to the
pipelines is intended to be a conservative screening level for the probabilistic assessments that follow.
Probabilistic assessment
The corrosion rate distributions were validated against actual field failures and the end of life target failure probabilities were calibrated against reported failures, so it would be expected that
the remaining life estimates are conservative. However, the sensitivity study showed that the calculated failure probability, and hence the remaining life estimates are very sensitive to the
corrosion rate distribution assumed. The analyses showed that the corrosion rates derived from the combined inspection data, may underestimate the remaining life by up to 16 years, if thickness data
specific to individual lines were not considered. This implies that the results may be overly conservative for some of the lines that had not been inspected. Nevertheless, the probabilistic analyses
method provides remaining life estimates that can be used to determine maintenance and inspection priority as well as a basis for formulating a pipeline replacement strategy.
A method is presented for the remnant life assessment of corroding pipelines. The method entails an initial risk ranking to identify the pipelines at high risk of failure. Detailed probabilistic
remaining life assessments are then carried out on the identified pipelines. The probabilistic model allows for uncertainty in variables including material strength properties, corrosion area sizes
and corrosion rates. Inspection data plays a major role in deriving corrosion rate distributions, and a major feature of the probabilistic model is the extrapolation of limited inspection data to
cover longer pipeline lengths. Failure probability estimates are obtained from First Order Reliability Method (FORM) and verified by Monte Carlo simulations.
The results presented in this paper are largely based on a case study that involved the assessment of about 260 subsea pipelines consisting of both water injection and oil production pipelines. The
assessment results included both remaining life estimates and sensitivity studies. The failure probability estimates were validated against reported pipeline failures in order to define the
appropriate end-of-life target failure probabilities. The case study results provided a means of identifying the lines at the highest risk of failure. The sensitivity studies showed the potential
benefit of conducting further inspections to provide improved corrosion rate estimates, and to better establish the condition of specific pipelines.
The method described provides a basis for determining pipeline inspection priority and ultimately for developing a replacement strategy.
1. Ang A H-S and Tang W H: 'Probability concepts in engineering planning and design', Vol.II, Decision, Risk and Reliability, John Wiley, New York, 1984.
2. Kiefner J F and Vieth P H: 'A modified criterion for evaluating the remaining strength of corroded pipe'. Project PR 3-805, AGA, December 1989.
3. RCP GmBH: 'STUREL - A structural reliability analysis program-system', Users Manual, RCP Consult, 1997.
4. Zimmerman I J E, Hopkins P and Sanderson N: 'Can limit spate design be used to design a pipeline above 80% SYMS?', 17 ^th International Conference on Offshore Mechanics and Arctic Engineering,
OMAE 98-0902 ASME, 1998.
5. Jiao G et al: 'The SUPERB project: linepipe statistical properties and implications in design of offshore pipelines'. 1997 OMAE, Vol.V, Pipeline Technology, ASME 1997.
6. Buxton D C, Cottis R A and Scarf P A: 'Life prediction in corrosion fatigue' in Parkins R N (Ed), Life Prediction of Corrodible Structures, Vol.II, 1273-1282, NACE, 1994, ISBN 1-877914-60-6.
7. Laycock P J, Cottis R A and Scarf P A: 'Extrapolation of extreme pit depths in space and time'. J Electrochen, Soc., Vol.137, No.1, January 1990.
8. Schneider C R A, Muhammed A and Sanderson R M: 'Predicting the remaining lifetime of in-service pipelines based on sample inspection data'. Journal of the British Institute of Non-Destructive
Testing,Insight, Vol.43, No.2, February 2001.
9. OS-F101:'Offshore Standard- Submarine Pipeline Systems' DNV 2000. | {"url":"https://www.twi-global.com/technical-knowledge/published-papers/probabilistic-remnant-life-assessment-of-corroding-pipelines-within-a-risk-based-framework-july-2002","timestamp":"2024-11-11T14:49:42Z","content_type":"text/html","content_length":"176827","record_id":"<urn:uuid:96a93b7a-f491-4d59-ac21-3b7c94dea45c>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00525.warc.gz"} |
Advice for students Archives - Odyssey Math Tuition
Going to Math Tuition does not necessarily mean that students will be able to rectify their misconceptions in mathematics as well as improve their understanding. As much as we tutors can do our best
to provide the best guidance, resources and explanation, students are required to do their part in terms of learning as well. …
5 things students must do to maximize their learning in math tuition Read More » | {"url":"https://odysseymathtuition.com/category/advice-for-students/","timestamp":"2024-11-06T16:56:57Z","content_type":"text/html","content_length":"75661","record_id":"<urn:uuid:44f938ba-b2cb-4424-97a4-c9399ca60b56>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00723.warc.gz"} |
Contributions to Books:
J. Melenk, D. Praetorius, B. Wohlmuth:
"Simultaneous quasi-optimal convergence in FEM-BEM coupling";
in: "ASC Report 13/2014", issued by: Institute of Applied Mathematics and Numerical Analysis; Vienna University of Technology, Wien, 2014, ISBN: 978-3-902627-07-0, 1 - 21.
English abstract:
We consider the symmetric FEM-BEM coupling that connects two linear
elliptic second order partial differential equations posed in a
bounded domain Ω and its complement, where the exterior problem is
restated by an integral equation on the coupling boundary Gamma. We
assume that the corresponding transmission problem admits a shift
theorem by more than 1/2. We analyze the discretization by piecewise
polynomials of degree k for the domain variable and piecewise
polynomials of degree k-1 for the flux variable on the coupling
boundary. Given sufficient regularity we show that (up to logarithmic
factors) the optimal convergence order k+1/2 in the H^{−1/2}-norm
is obtained for the flux variable, while classical arguments by
Cea-type quasi-optimality and standard approximation results provide
only convergence order k for the overall error in the natural
product norm.
FEM-BEM coupling, a priori convergence analysis, transmission problem
Electronic version of the publication:
Created from the Publication Database of the Vienna University of Technology. | {"url":"https://publik.tuwien.ac.at/showentry.php?ID=228198&lang=2","timestamp":"2024-11-11T06:36:46Z","content_type":"text/html","content_length":"3401","record_id":"<urn:uuid:48a775ee-7523-40ae-81fd-7cc6764216c3>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00376.warc.gz"} |
Logic Puzzle Tutorial , Interactive Grid Logic
Please follow this page for your Logic Puzzle Tutorial Guide
On this Logic puzzle tutorial page we have examples and also links to free printable grid logic puzzles.
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About Grid Puzzles
By far the most popular of logic puzzles are the grid logic puzzles, though there are many other
varieties including word logic, maze logic, grid-less, syllogisms, and the like.
What most of the 'logic' puzzles have in common is that the game involves some form of
deductive reasoning., or that which we start with the general ( a set of clues) and move to the particular
( a logical conclusion) just like a real doggone sleuth!!
Another form of logic puzzle, popular among puzzle enthusiasts and available in magazines dedicated to the subject, is a format in which the set-up to a scenario is given, as well as the object (for
example, determine who brought what pet at a pet store , and on what day each pet was bought), some specific clues are given ("neither Mary nor Ray purchased the German Shepherd"), and then the
reader fills out a matrix with the clues and attempts to deduce the solution.
Common logic puzzle magazines are derivatives of the logic grid puzzle called "table puzzles" that are deduced in the same manner , but lack the grid either because a grid would be too large, or
because some other visual aid is provided. For example, a map of a town might be present in lieu of a grid in a puzzle about the location of different shops.
In this section we will explore some grid puzzle samples in a step-by-step tutorial.
(NOTE: All puzzles in the example section are fully interactive , meaning
1. Using the grid charts below attached to each game to plot your guesses as follows :
2. Click on each grid square to alternatively :
• (a) make a 'green box' for a square, or
• (b) click again to make a 'red xx' ( for eliminating a possible clue).
• (c) NOTE: (A third click will clear a grid square completely )
3. Use the 'RESET' button to clear the entire grid chart and start a game over.
4. Click on the box labeled: SOLUTION Step_by_step to open the tutorial guide for the
selected puzzle.(Located beneath the reset button).
Puzzle 1: AT THE PET SHOP
Last week four friends ( one was Susan) went to the pet shop looking for new pets.
Each friend chose a different pet (one was a cat ).
From the clues provided can you tell which friend bought which pet?
AT THE PET SHOP Cat Dog Monkey Snake
SOLUTION: STEP_BY_STEP-(click to show / hide)
• Lets look at the first clue "Nobody chose a pet which started with the same first letter as their name."
• Locate Susan in the chart and find the column with Snake
Now click the grid square(susan-snake) until the 'red xx' appears.
• Lets look at the next clue "Bill already has a dog."
• Locate Bill in the chart and find the column with Dog
Now click the grid square(bill-dog) until the 'red xx' appears.
• Lets look at the next clue "Jen is afraid of snakes."
• Locate Jen in the chart and find the column with snake
Now click the grid square(jen-snake) until the 'red xx' appears.
• Lets look at the next clue "Roger did not choose the snake."
• Locate Roger in the chart and find the column with snake CLUES:
Now click the grid square(Roger-snake) until the 'red xx' appears. Nobody chose a pet which started with the same first letter as their name.
• *IMPORTANT:*
Look at the column under 'SNAKE'. We have now eliminated 3 of the possible 4 squares in that column. Bill already has a dog.
• Therefore the square bill-snake must be the correct solution Jen is afraid of snakes.
So click the grid square(bill-snake) until the 'green box' appears. Roger did not choose the snake.
At the same time place 'red xx' in all other squares in the bill-row(bill-cat, bill-monkey).
Neither Jen nor Susan like monkeys.
• Lets look at the next clue "Neither Jen nor Susan like monkeys."
• Locate Jen and Susan in the chart and find the column with monkey Jen did not choose the cat .
Now click the grid square(jen-monkey, susan-monkey) until the 'red xx' appears.
• *IMPORTANT:*
Look at the column under 'MONKEY'. We have now eliminated 3 of the possible 4 squares in that column.
• Therefore the square roger-monkey must be the correct solution
So click the grid square(roger-monkey) until the 'green box' appears.
At the same time place 'red xx' in all other squares in the roger-row(roger-cat, roger-dog).
• Lets look at the last clue "Jen did not choose the cat."
• Locate Jen in the chart and find the column with cat
Now click the grid square(jen-cat) until the 'red xx' appears.
This logically will conclude our puzzle as follows:
Since jen did not choose the cat, she could have only chosen the dog,
(jen-dog square should be highlighted with a 'green-box',while jen-cat is filled with a 'red xx').
The opposite is selected for susan, since she could have only selected the cat.
• Congratulations! Puzzle solved. To summarize:
Bill chose the snake.
Jen chose the dog.
Roger chose the monkey.
Susan chose the cat. | {"url":"https://www.puzzles-on-line-niche.com/logic-puzzle-tutorial.html","timestamp":"2024-11-06T13:59:12Z","content_type":"text/html","content_length":"56113","record_id":"<urn:uuid:d8f5944e-1b7e-4c60-a720-92203ec625f8>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00532.warc.gz"} |
Preference Cycles - Condorcet Canada Initiative
Preference Cycles
When there is a Condorcet winner every Condorcet method is as good as every other for identifying this winner. It is in those cases where there is no Condorcet winner that the various Condorcet
methods differ, which is mainly in how they break preference cycles (aka majority-rule cycles).
Let us imagine, for example, that we have three candidates: X, Y, Z. With three candidates we will get three distinct pairings: (X, Y), (X, Z), and (Y, Z).
Let us assume we have an election in which we discover that:
1. X is more preferred than Y: (X → Y), which is to say that X wins the X vs Y match-up;
2. Y is more preferred than Z: (Y → Z), which is to say that Y wins the Y vs Z match-up; and
3. Z is more preferred than X: (Z → X), which is to say that Z wins the Z vs X: match-up.
Here, there is no candidate who wins every pairwise match in which he or she is involved, so there is no Condorcet winner; more particularly, we have a preference cycle. Different Condorcet methods
do different things at this point.
With Condorcet/Ranked-Pairs we look at the magnitude of the preferences:
1. If, say, 60% prefer X, vs 40% who prefer Y; we have a strong preference of 60% vs 40% for X more-preferred-than Y;
2. If, say, 90% prefer Y, vs 10% who prefer Z; we have a very strong preference of 90% vs 10% for Y more-preferred-than Z;
3. If, say, 51% prefer Z, vs 49% who prefer X; we have a very weak preference of 51% vs 49% for Z more-preferred-than X.
We see that some preferences can be seen as comparatively strong, and others weak.
Condorcet/Ranked-Pairs “ranks” the pairs according to their strengths of preference, and then considers these pairs, one by one, from strongest preference to weakest.
If we get to a preference that conflicts-with a previous (stronger) preference (creates a preference cycle) we omit it: the rationale being that a stronger preference should prevail over a weaker
preference in any case where we can’t keep them both.
In our example:
1. We sort our pairs by descending strength-of-preference as follows: Y → Z (strongest), X → Y, Z → X (weakest).
2. As we then consider the pairs in this order, the first two pairs, Y → Z and X → Y imply that X → Z.
3. This implication that X → Z conflicts with the assertion of the third pair that Z → X, so when we get to the third pair we must omit it to avoid the conflict, so that
4. X → Z, being encountered first, and being therefore the stronger preference, still stands.
This gives us a final ranking among the candidates themselves with no preference cycle remaining:
• X → Y → Z; and
• X is the Ranked-Pairs winner.
Next: Practical Features | {"url":"https://condorcet.ca/see-how-it-works/preference-cycles/","timestamp":"2024-11-14T20:43:36Z","content_type":"text/html","content_length":"50523","record_id":"<urn:uuid:510a3294-a4c2-4b10-8377-1b9db05cfe05>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00808.warc.gz"} |
• Incorporate the ARtCensReg function to fit a univariate censored linear regression model with autoregressive errors considering Student-t innovations.
• Generic functions predict, print, summary, and plot have methods for objects given as an output of ARCensReg and ARtCensReg functions.
• Function residuals was incorporated to compute the conditional and quantile residuals for objects inheriting from class ARpCRM or ARtpCRM, given as an output of ARCensReg and ARtCensReg function,
• Function plot is also valid for objects returned by residuals. This procedure returns the following plots for the quantile residuals: residual vs. time, an autocorrelation plot, a histogram, and
a Quantile-Quantile (Q-Q) plot.
• Argument pit was substituted by phi in the function rARCens. Please see the documentation for more information.
• Function rARCens was modified to simulate datasets with Student-t innovations.
• Some modifications in the arguments of the InfDiag function were made. The generic function plot is available for outputs of function InfDiag. | {"url":"https://cran.rstudio.com/web/packages/ARCensReg/news/news.html","timestamp":"2024-11-12T21:04:39Z","content_type":"application/xhtml+xml","content_length":"3974","record_id":"<urn:uuid:d85672ba-de95-49c1-8f7d-8a27e4557cc5>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00646.warc.gz"} |
CP violation for electroweak baryogenesis from mixing of standard model and heavy vector quarks
McDonald, John (1996) CP violation for electroweak baryogenesis from mixing of standard model and heavy vector quarks. Physical Review D, 53 (2). pp. 645-654. ISSN 1550-7998
Full text not available from this repository.
It is known that the CP violation in the minimal standard model is insufficient to explain the observed baryon asymmetry of the Universe in the context of electroweak baryogenesis. In this paper we
consider the possibility that the additional CP violation required could originate in the mixing of the standard model quarks and heavy vector quark pairs. We consider the baryon asymmetry in the
context of the spontaneous baryogenesis scenario. It is shown that, in general, the CP-violating phase entering the mass matrix of the standard model and heavy vector quarks must be space dependent
in order to produce a baryon asymmetry, suggesting that the additional CP violation must be spontaneous in nature. This is true for the case of the simplest models which mix the standard model and
heavy vector quarks. We derive a charge potential term for the model by diagonalizing the quark mass matrix in the presence of the electroweak bubble wall, which turns out to be quite different from
the fermionic hypercharge potentials usually considered in spontaneous baryogenesis models, and obtain the rate of baryon number generation within the wall. We find, for the particular example where
the standard model quarks mix with weak-isodoublet heavy vector quarks via the expectation value of a gauge singlet scalar, that we, can account for the observed baryon asymmetry with conservative
estimates for the uncertain parameters of electroweak baryogenesis, provided that the heavy vector quarks are not heavier than a few hundred GeV and that the coupling of the standard model quarks to
the heavy vector quarks and gauge singlet scalars is not much smaller than order of 1, corresponding to a mixing angle of the heavy vector quarks and standard model quarks not much smaller than order
of 10(-1).
Item Type:
Journal Article
Journal or Publication Title:
Physical Review D
?? weak phase-transitionbaryon asymmetrygenerationuniverse ??
Deposited On:
29 Nov 2016 10:14
Last Modified:
15 Jul 2024 16:36 | {"url":"https://eprints.lancs.ac.uk/id/eprint/83226/","timestamp":"2024-11-09T15:58:23Z","content_type":"application/xhtml+xml","content_length":"23486","record_id":"<urn:uuid:934d7444-982c-42ff-95b9-2bb0b1ad601d>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00112.warc.gz"} |
Linear Algebra - Image representation
A generalized image consists of a grid of generalized pixels, where each generalized pixel is a quadrilateral (not necessarily a rectangle).
Think of an image as a grid of rectangles, each assigned a color. (The rectangles correspond to the pixels.) Each such rectangle in the image corresponds to a parallelogram in the plane.
In order to manipulate image in Linear Algebra, we need to represent images as matrices. We represent an image by a set of colored points in the plane.
Colored points
To represent a colored point, we need to specify its location and its color. We will therefore represent a point using two vectors:
• the location vector with labels {'x','y','u'}
• and the color vector with labels {'r','g','b'}.
The location vector represents the location of the point in the usual way|as an (x, y) pair. The u entry is always 1 (homogeneous coordinates are used to perform a translation).
For example, the point (12, 15) would be represented by the vector Vec({'x','y','u'}, {'x':12, 'y':15, 'u':1}.
The color vector represents the color of the point: the 'r', 'g', and 'b' entries give the intensities for the color channels red, green, and blue. For example, the color red is represented by the
function {'r': 1}.
Scheme for representing images
Ordinarily, an image is a regular rectangular grid of rectangular pixels, where each pixel is assigned a color. Because images are transformed, a slightly more general representation is needed.
A generalized image consists of a grid of generalized pixels, where each generalized pixel is a quadrilateral (not necessarily a rectangle).
The points at the corners of the generalized pixels are identied by pairs (x, y) of integers, which are called pixel coordinates. The top-left corner has pixel coordinates (0,0), the corner directly
to its right has pixel coordinates (1,0), and so on.
For example, the pixel coordinates of the four corners of the top-left generalized pixel are (0,0), (0,1), (1,0), and (1,1).
Each corner is assigned a location in the plane, and each generalized pixel is assigned a color. The mapping of corners to points in the plane is given by a matrix, the location matrix. Each corner
corresponds to a column of the location matrix, and the label of that column is the pair (x, y) of pixel coordinates of the corner. The column is a {'x','y','u'}-vector giving the location of the
corner. Thus the row labels of the location matrix are 'x', 'y', and 'u'.
The mapping of generalized pixels to colors is given by another matrix, the color matrix. Each generalized pixel corresponds to a column of the color matrix, and the label of that column is the pair
of pixel coordinates of the top-left corner of that generalized pixel. The column is a {'r','g','b'}-vector giving the color of that generalized pixel.
For example, the image consists of four generalized pixels , comprising a total of nine corners. This image is represented by:
• the location matrix (which gives the location of columns in terms of v in a default coordinate system)
v (0,0) (0,1) (0,2) (1,2) (1,1) (1,0) (2,2) (2,0) (2,1)
x 0 0 0 1 1 1 2 2 2
y 0 1 2 2 1 0 2 0 1
u 1 1 1 1 1 1 1 1 1
• and the color matrix (which gives the colors of a pixel per corner)
(0, 0) (0, 1) (1, 1) (1, 0)
b 225 125 75 175
g 225 125 75 175
r 225 125 75 175
By applying a suitable transformation to the location matrix, we can obtain
v (0,0) (0,1) (0,2) (1,2) (1,1) (1,0) (2,2) (2,0) (2,1)
x 0 2 4 14 12 10 24 20 22
y 0 10 20 22 12 2 24 4 14
u 1 1 1 1 1 1 1 1 1
which, combined with the unchanged color matrix, looks like this:
The perspective of an image is given by the notion of coordinate system.
Making a perspective-free image is just a translation from one coordinate system to the other. This translation function maps pixel coordinates from the first coordinates system to the coordinates of
the corresponding point in the second coordinates system.
The basic approach to derive this mapping is by example. We get:
• several input-output pairs|points in the original coordinate system
• corresponding points in the target coordinate system
in order to derive the function that agrees with this behaviour.
At the heart of this mapping function is a change of basis. | {"url":"https://datacadamia.com/linear_algebra/image","timestamp":"2024-11-13T18:37:40Z","content_type":"text/html","content_length":"143337","record_id":"<urn:uuid:05e6137f-2047-4adf-9ae1-b28d2893fde8>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00690.warc.gz"} |
Final Term-Theory of Automata - MCQSCENTER
If Σ== {aa, bb} , then Σ* will not contain
A. aaabbb B.
Below given FA has __________ RE.
a(a+b)*a + b(a+b)*b
“One language can have _________ TG‟s”
Only one
Only two
C. More than one D.
Only three
According to 1st part of the Kleene‟s theorem, If a language can be accepted by an FA then it can be accepted by a ________ as well.
5. Even-palindrome is a _______ language.
B. Regular= C.
Regular but infinite
Regular but finite
6. If L is a regular language then, Lc is also a _____ language.
Regular but finite
None of the given
7. Pumping lemma is generally used to prove that:
A given language is infinite
A given language is not regular
Whether two given regular expressions of a=regular language are equivalent or not
None of these
8. If the FA has N states, then test the words of length less than N. If no word is accepted by this FA, then it will _________ word/words.
accept all
accept no
accept some
reject no
In CFG, the symbols that can‟t be replaced by anything are called________.
Non Terminal
All of given
10. Which of the following is a regular language?
String of odd number of zeroes
Set of all palindromes made up of 0‟s and 1‟s
String of 0‟s whose length is a prime number
All of these | {"url":"https://mcqscenter.com/vu-cs/cs402-theory-of-automata-final-term","timestamp":"2024-11-09T22:09:20Z","content_type":"text/html","content_length":"76113","record_id":"<urn:uuid:93a5a17b-2eb4-4002-820f-ada430f0ab2d>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00755.warc.gz"} |
In my previous post about geo-spatial search in MySQL I described (along with other things) how to use geo-distance functions. In this post I will describe the geo-spatial distance functions in more
If you need to calculate an exact distance between 2 points on Earth in MySQL (very common for geo-enabled applications) you have at least 3 choices.
• Use stored function and implement haversine formula
• Use UDF (user defined function) for haversine (see below)
• In MySQL 5.6 you can use st_distance …
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Calculating Present Value
Calculating Present Value Leave a comment
We may also, at times, sell lead data to partners in our network in order to best connect consumers to the information they request. Readers are in no way obligated to use our partners’ services to
access the free resources on Annuity.org. For instance, when someone purchases a home, they are often offered the opportunity to pay points on the mortgage to reduce insurance payments. Keen
investors can compare the amount paid for points and the discounted future interest payments to find out. Here, in the finished output sheet below, the present values of the cash flows calculated
under both approaches result in the same figures.
• The topics we’re about to cover are especially vital if you’re going to calculate your lease liability in Microsoft Excel manually.
• That’s because $10,000 today is worth more than $10,000 received over the course of time.
• And not just any financial advisor – a fiduciary who is legally required to work in your best interest at all times.
• Below is an illustration of what the Net Present Value of a series of cash flows looks like.
• The default calculation above asks what is the present value of a future value amount of $15,000 invested for 3.5 years, compounded monthly at an annual interest rate of 5.25%.
• Future cash flows are discounted at the discount rate, and the higher the discount rate, the lower the present value of the future cash flows.
In that sort of scenario money in the future would be worth more than today. While we’re insinuating that 10% is an unreasonable discount rate, there will always be tradeoffs when you’re dealing with
uncertainty and sums in the future. When you present value all future payments and add $1,000 tothe NPV amount, the total is $9,585.98 identical to the PV formula. The key input in this present value
excel function is each payment is given a period. The first period is 0, which results in the present value amount of $1,000 given it’s not a future amount.
Learn the Seven Ways to increase Your Billable Time
The formula used to calculate the present value divides the future value of a future cash flow by one plus the discount rate raised to the number of periods, as shown below. Let us take a simple
example of $2,000 future cash flow to be received after 3 years. According to the current market trend, the applicable discount rate is 4%.
This decrease in the current value of future cash flows is based on a chosen rate of return . If for example there exists a time series of identical cash flows, the cash flow in the present is the
most valuable, with each future cash flow https://www.bookstime.com/ becoming less valuable than the previous cash flow. A cash flow today is more valuable than an identical cash flow in the future
because a present flow can be invested immediately and begin earning returns, while a future flow cannot.
Method 2 of 3:Using Cash Outflows to Determine NVP
Present value takes the future value and applies a discount rate or the interest rate that could be earned if invested. Future value tells you what an investment is worth in the future while the
present value tells you how much you’d need in today’s dollars to earn a specific amount in the future. As an indicator of projects’ investment, NPV has several advantages and disadvantages for
decision-making. Consideration of the time value of money allows the NPV to include all relevant time and cash flows for the project. This idea is consistent with the goal of wealth maximization by
creating the highest wealth for shareholders.
Except for minor differences due to rounding, answers to equations below will be the same whether they are computed using a financial calculator, computer software, PV tables, or the formulas. The
easiest and most accurate way to calculate the present value of any future present value formula amounts is to use an electronic financial calculator or computer software. Some electronic financial
calculators are now available for less than $35. Discounted cash flow is a valuation method used to estimate the attractiveness of an investment opportunity.
Present Value of a Growing Annuity (g = i)
Use knowledge and skills to manage financial resources effectively for a lifetime of financial well-being. The amount of time that passes before interest begins to earn interest. The price of
borrowing money as it is usually stated, unadjusted for inflation. Certain interest rates occasionally turn very slightly (−0.004%) negative.
In contrast, current payments have more value because they can be invested in the meantime. A way to avoid this problem is to include explicit provision for financing any losses after the initial
investment, that is, explicitly calculate the cost of financing such losses. The rate used to discount future cash flows to the present value is a key variable of this process. The present value
calculates how much a future cash flow is worth today, whereas the future value is how much a current cash flow will be worth on a future date based on a growth rate assumption.
How to calculate present value
In other words, it computes the amount of money that must be invested today to equal the payment or amount of cash received on a future date. The NPV of a sequence of cash flows takes as input the
cash flows and a discount rate or discount curve and outputs a present value, which is the current fair price. The converse process in discounted cash flow analysis takes a sequence of cash flows and
a price as input and as output the discount rate, or internal rate of return which would yield the given price as NPV.
Can I live off interest on a million dollars?
The historical S&P average annualized returns have been 9.2%. So investing $1,000,000 in the stock market will get you $96,352 in interest in a year. This is enough to live on for most people.
In financial theory, if there is a choice between two mutually exclusive alternatives, the one yielding the higher NPV should be selected. A positive net present value indicates that the projected
earnings generated by a project or investment exceeds the anticipated costs . This concept is the basis for the Net Present Value Rule, which dictates that the only investments that should be made
are those with positive NPVs. A firm’s weighted average cost of capital is often used, but many people believe that it is appropriate to use higher discount rates to adjust for risk, opportunity
cost, or other factors. A variable discount rate with higher rates applied to cash flows occurring further along the time span might be used to reflect the yield curve premium for long-term debt.
What Is the Formula for Calculating the Present Value of an Annuity?
In the lemonade stand example, let’s say that if you don’t purchase the juicer, you’ll invest the money in the stock market, where you feel confident that you can earn 4% annually on your money. In
this case, 0.04′ (4% expressed as a decimal) is the discount rate we’ll use in our calculation. Unlike the PV function in excel, the NPV function/formula does not consider any period. The function
automatically assumes all the time periods are equal. This is at the core of IFRS 16 and ASC 842, the future lease cash outflows are present valued to represent the value of the lease liability at a
particular point in time.
You have some money now, but you don’t know how much, if any, you will be able to save before you buy your business in five years. Rosemary Carlson is an expert in finance who writes for The Balance
Small Business. She has consulted with many small businesses in all areas of finance. She was a university professor of finance and has written extensively in this area. Harold Averkamp has worked as
a university accounting instructor, accountant, and consultant for more than 25 years.
Given our time frame of five years and a 5% interest rate, we can find the present value of that sum of money. Let us take the example of David who seeks to a certain amount of money today such that
after 4 years he can withdraw $3,000.
What is the present value of an annuity of $27?
Answer and Explanation:
The present value is $129.3512.
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Leap Year Calendar List
Leap Year Calendar List - A leap year is a year with an extra day added to keep the. Every leap year is clearly stated, see also other interesting. A leap year (also known as an intercalary year or
bissextile year) is a calendar year that contains an additional. The years 1700, 1800, 1900, 2100, 2200 and 2300 are not leap years, even though they are divisible by 4 without a. One year has the
length of 365 days, 5 hours, 48 minutes and 45 seconds. Web find out all leap years between two years (from 1600 to 4000) with this tool. See the table of leap years from 1900 to 3000 and the
exceptions. Web 22 rows all upcoming leap years are listed on this page. Web find out which years are leap years and why they have 366 days and a february 29. Web in this blog we have curated the
list of leap years since 1800 to 2100.
Catch Julian Date Leap Year Best Calendar Example
A leap year is a year with an extra day added to keep the. Web in this blog we have curated the list of leap years since 1800 to 2100. The years 1700, 1800, 1900, 2100, 2200 and 2300 are not leap
years, even though they are divisible by 4 without a. This is hard to calculate with,. Learn the.
Leap Year Calendar List
Learn the formula, exceptions and faqs. Web find out all leap years between two years (from 1600 to 4000) with this tool. Every leap year is clearly stated, see also other interesting. A leap year
(also known as an intercalary year or bissextile year) is a calendar year that contains an additional. A leap year is a year with an.
Leap Year Program in C with Logic, Explanation and Output
Web 22 rows all upcoming leap years are listed on this page. See the table of leap years from 1900 to 3000 and the exceptions. A leap year is a year with an extra day added to keep the. This is hard
to calculate with,. Web find out all leap years between two years (from 1600 to 4000) with this.
List of Leap Years When is the Next Leap Year?
Every leap year is clearly stated, see also other interesting. Web 22 rows all upcoming leap years are listed on this page. Web find out all leap years between two years (from 1600 to 4000) with this
tool. See the table of leap years from 1900 to 3000 and the exceptions. Learn the formula, exceptions and faqs.
Leap Year 2020 Calendar 366 Days List of Leap Year 2020, 2024, 2028
Web in this blog we have curated the list of leap years since 1800 to 2100. A leap year is a year with an extra day added to keep the. The years 1700, 1800, 1900, 2100, 2200 and 2300 are not leap
years, even though they are divisible by 4 without a. Web find out which years are leap years.
3 Ways to Calculate Leap Years wikiHow
Learn the formula, exceptions and faqs. Web in this blog we have curated the list of leap years since 1800 to 2100. A leap year (also known as an intercalary year or bissextile year) is a calendar
year that contains an additional. This is hard to calculate with,. A leap year is a year with an extra day added to.
Leap Years The ingenious astronomer put forth the suggestion that
Learn the formula, exceptions and faqs. Web in this blog we have curated the list of leap years since 1800 to 2100. The years 1700, 1800, 1900, 2100, 2200 and 2300 are not leap years, even though
they are divisible by 4 without a. Every leap year is clearly stated, see also other interesting. Web 22 rows all upcoming leap.
Leap Year 2024
One year has the length of 365 days, 5 hours, 48 minutes and 45 seconds. A leap year is a year with an extra day added to keep the. Web 22 rows all upcoming leap years are listed on this page. Web in
this blog we have curated the list of leap years since 1800 to 2100. The years 1700,.
Leap Years List List of All Leap Years From 1800 to 3000
One year has the length of 365 days, 5 hours, 48 minutes and 45 seconds. Every leap year is clearly stated, see also other interesting. This is hard to calculate with,. Web 22 rows all upcoming leap
years are listed on this page. Web find out all leap years between two years (from 1600 to 4000) with this tool.
Leap Year Calendar Activities readilearn
A leap year (also known as an intercalary year or bissextile year) is a calendar year that contains an additional. This is hard to calculate with,. Web in this blog we have curated the list of leap
years since 1800 to 2100. Web find out which years are leap years and why they have 366 days and a february 29..
See the table of leap years from 1900 to 3000 and the exceptions. A leap year (also known as an intercalary year or bissextile year) is a calendar year that contains an additional. Every leap year is
clearly stated, see also other interesting. Web find out which years are leap years and why they have 366 days and a february 29. The years 1700, 1800, 1900, 2100, 2200 and 2300 are not leap years,
even though they are divisible by 4 without a. Web find out all leap years between two years (from 1600 to 4000) with this tool. Web 22 rows all upcoming leap years are listed on this page. This is
hard to calculate with,. Learn the formula, exceptions and faqs. Web in this blog we have curated the list of leap years since 1800 to 2100. A leap year is a year with an extra day added to keep the.
One year has the length of 365 days, 5 hours, 48 minutes and 45 seconds.
See The Table Of Leap Years From 1900 To 3000 And The Exceptions.
Web 22 rows all upcoming leap years are listed on this page. Web find out which years are leap years and why they have 366 days and a february 29. A leap year (also known as an intercalary year or
bissextile year) is a calendar year that contains an additional. Web find out all leap years between two years (from 1600 to 4000) with this tool.
One Year Has The Length Of 365 Days, 5 Hours, 48 Minutes And 45 Seconds.
This is hard to calculate with,. Web in this blog we have curated the list of leap years since 1800 to 2100. A leap year is a year with an extra day added to keep the. Every leap year is clearly
stated, see also other interesting.
The Years 1700, 1800, 1900, 2100, 2200 And 2300 Are Not Leap Years, Even Though They Are Divisible By 4 Without A.
Learn the formula, exceptions and faqs.
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How To Calculate A Weighted Average In Excel?
How to Calculate a Weighted Average in Excel:
1. Open Excel and input the data you want to calculate the weighted average for.
2. Create a new column and multiply each value by its corresponding weight.
3. Add up all the weighted values in the new column.
4. Create another column and input the corresponding weights for each value.
5. Sum up all the weights in the new column.
6. Divide the sum of the weighted values by the sum of the weights to get the weighted average.
How To Calculate A Weighted Average In Excel? If you've ever wondered how to find the average of a group of numbers with certain weights, you're in the right place! We'll show you the ins and outs of
calculating a weighted average using everyone's favorite spreadsheet program, Excel.
But hold on, what exactly is a weighted average? Well, imagine you have a set of numbers, and each number has a different significance or importance. In essence, a weighted average takes into account
those weights and provides a more accurate representation of the overall average.
So, whether you're a student looking to calculate your weighted GPA or a professional crunching numbers for a business, learning how to calculate a weighted average in Excel will prove to be an
invaluable skill. Let's dive in and explore the steps, formulas, and tips to master this useful calculation method!
How to Calculate a Weighted Average in Excel?
Excel is a powerful tool that offers a range of functions and formulas to perform various calculations. One such calculation that is frequently used in data analysis is finding the weighted average.
In this article, we will explore how to calculate a weighted average in Excel, step by step. Whether you're a student working on a school project or a professional dealing with financial data,
understanding how to calculate a weighted average can be extremely beneficial.
Understanding Weighted Average
Before diving into the process of calculating a weighted average in Excel, let's first understand what it means. A weighted average is an average that takes into account the importance or
significance of each value in a set of data by assigning weights to them. These weights reflect the relative importance or contribution of each value to the final average.
For example, imagine you have a class where tests are worth 40% of the total grade, assignments are worth 30%, and the final exam is worth 30%. In this case, each component has a different weight,
and simply taking the average of all the scores would not give an accurate representation of the overall performance. To account for the weights, we use a weighted average.
Benefits of Weighted Averages
Using weighted averages can provide more accurate and meaningful results in various situations. Here are a few benefits of using weighted averages:
• Accounting for different levels of importance: By assigning weights to each value, a weighted average takes into account the significance of each value in the dataset. This allows for a more
accurate representation of the overall average.
• Reflecting real-world scenarios: In many real-world scenarios, different values have different impacts on the final outcome. Weighted averages help incorporate these variations and provide a
better understanding of the data.
• Customizable calculations: With weighted averages, you have the flexibility to assign weights based on your specific needs. This allows you to create customized calculations tailored to your
unique requirements.
• Applicable across various fields: Weighted averages are widely used in statistics, finance, economics, and other fields where multiple factors contribute to a final value.
Calculating a Weighted Average in Excel
Now that we have a clear understanding of what a weighted average is and its benefits, let's explore how to calculate it in Excel. There are a few different approaches you can take, depending on the
structure of your data. Let's go through each method step by step.
Method 1: Using the SUMPRODUCT Function
If you have the values and their corresponding weights in separate columns, you can use the SUMPRODUCT function to calculate the weighted average. The SUMPRODUCT function multiplies each value by its
corresponding weight, adds up the products, and divides the sum by the total weight. Here's how you can do it:
1. Arrange your data in Excel, with the values in one column and their corresponding weights in another column.
2. In a separate cell, use the formula "=SUMPRODUCT(values_range, weights_range) / SUM(weights_range)". Replace "values_range" with the range of values and "weights_range" with the range of weights.
For example, if your values are in column A and weights in column B, the formula would look like "=SUMPRODUCT(A1:A10, B1:B10) / SUM(B1:B10)".
3. Press Enter to get the weighted average.
Method 2: Using the SUMPRODUCT and SUM Functions
If you have the values and weights in the same column, you can still use the SUMPRODUCT function along with the SUM function to calculate the weighted average. Here's how:
1. Arrange your data in Excel, with each value and its corresponding weight in the same column.
2. In a separate cell, use the formula "=SUMPRODUCT(values_range, values_range) / SUMIF(values_range, ">0")". Replace "values_range" with the range of values. For example, if your values and weights
are in column A, the formula would look like "=SUMPRODUCT(A1:A10, A1:A10) / SUMIF(A1:A10, ">0")".
3. Press Enter to get the weighted average.
Method 3: Using the SUMPRODUCT, SUM, and IF Functions
If you have a more complex data structure where you need to apply certain conditions or criteria to calculate the weighted average, you can combine the SUMPRODUCT, SUM, and IF functions. This method
allows for more flexibility in your calculations. Here's how:
1. Arrange your data in Excel, including any relevant criteria or conditions in separate columns.
2. In a separate cell, use the formula "=SUMPRODUCT(IF(condition_range, values_range), IF(condition_range, weights_range)) / SUM(IF(condition_range, weights_range))". Replace "condition_range" with
the range of the condition, "values_range" with the range of values, and "weights_range" with the range of weights. For example, if your conditions are in column C, values in column A, and
weights in column B, the formula would look like "=SUMPRODUCT(IF(C1:C10, A1:A10), IF(C1:C10, B1:B10)) / SUM(IF(C1:C10, B1:B10))".
3. Press Enter to get the weighted average.
By following these methods, you can easily calculate a weighted average in Excel and obtain more accurate results in your data analysis.
Benefits of Using a Weighted Average in Excel
Using a weighted average in Excel offers several benefits, making it a valuable tool for data analysis. Let's explore some of the advantages of utilizing a weighted average in Excel.
1. Reflects Importance and Significance
A weighted average takes into account the significance or importance of each value in a dataset by assigning weights. This ensures that the average reflects the true impact of each value on the final
result. By considering the weights, you can obtain a more accurate representation of the data.
2. Accurate Representation of Real-World Scenarios
In many real-world scenarios, certain factors have more influence or contribute more significantly to the final outcome. Using a weighted average allows you to incorporate these variations and obtain
a more realistic representation of the data. This is particularly useful in fields such as finance, where different factors may have different weights.
3. Customizable Calculations
Excel provides the flexibility to customize your calculations based on your specific needs. You can assign different weights to different values to reflect their relative importance. This
customization allows you to tailor your calculations to match the specific requirements of your analysis.
4. Applicable Across Various Fields
The concept of a weighted average is widely applicable across various fields, including statistics, finance, economics, and more. In these fields, multiple factors contribute to a final value, and
utilizing a weighted average helps account for these different factors. By understanding how to calculate a weighted average in Excel, you gain a valuable skill that can be applied in different areas
of study or work.
Tips for Calculating a Weighted Average in Excel
Calculating a weighted average in Excel can sometimes be complex, especially when dealing with large datasets or intricate formulas. To make the process easier and more efficient, consider the
following tips:
1. Organize Your Data Properly
Ensuring that your data is properly organized is crucial for calculating a weighted average in Excel. Arrange your values and their corresponding weights in separate columns or rows, depending on the
method you choose. This organization will make it easier to reference the data in your formulas.
2. Double-Check Your Formulas
When working with complex formulas involving multiple functions, it's essential to double-check your formulas for accuracy. One small mistake can lead to incorrect results. Carefully review each
formula and verify that the ranges and conditions are correct.
3. Utilize Excel's Built-in Functions
Excel provides a range of built-in functions that can simplify the process of calculating a weighted average. Functions like SUMPRODUCT, SUM, and IF are particularly useful when working with weighted
averages. Explore the Excel documentation to familiarize yourself with these functions and how they can be applied.
4. Take Advantage of Conditional Formatting
Conditional formatting can help highlight values that meet specific criteria or conditions. If you're using Method 3 and have included conditions in your data, consider using conditional formatting
to visually identify the relevant values. This will make it easier to ensure that your calculations are considering the correct data.
Common Mistakes to Avoid
When calculating a weighted average in Excel, it's important to be mindful of potential mistakes that can lead to inaccurate results. Here are a few common mistakes to avoid:
1. Incorrect Range References
Ensure that you are referencing the correct ranges in your formulas. Failing to select the appropriate range can lead to erroneous calculations and incorrect weighted averages.
2. Forgetting to Exclude Zero or Empty Values
Depending on your data, you may need to exclude zero or empty values from your calculations. Forgetting to do so can skew the results and provide inaccurate weighted averages. Use functions like
SUMIF or IF to exclude these values when necessary.
3. Disregarding the Order of Operations
When combining multiple functions in a formula, it's important to understand and follow the order of operations. Excel calculates formulas based on a specific order, and failing to respect this order
can lead to unexpected results. Be mindful of parentheses and ensure that your formula is structured correctly.
Statistic: Weighted Average Usage
Weighted averages are widely used in various fields for data analysis and decision-making. In finance, they are used to calculate portfolio returns based on the weight of each investment. In
education, weighted averages determine final grades by assigning different weights to assignments, quizzes, and exams. In market research, weighted averages help analyze survey responses by
accounting for the importance of each respondent. The versatility and applicability of weighted averages make them an essential tool in many industries.
Key Takeaways - How to Calculate a Weighted Average in Excel?
• Weighted average is used to calculate the average of a set of numbers, where some numbers contribute more than others.
• In Excel, you can use the SUMPRODUCT function to calculate the weighted average.
• To calculate the weighted average in Excel, you need to multiply each number by its corresponding weight, sum the results, and divide by the total weight.
• You can use the SUMPRODUCT function along with the values and weights to quickly calculate the weighted average in Excel.
• Remember to use absolute references ($) when specifying the ranges in the SUMPRODUCT formula to ensure accurate calculation.
Frequently Asked Questions
Here, we've answered some common questions about calculating a weighted average in Excel.
1. How can I calculate a weighted average in Excel?
To calculate a weighted average in Excel, you need to multiply each value by its corresponding weight, sum up these products, and then divide by the sum of the weights. Here's how:
First, multiply each value by its weight. Then, add up all the products. Finally, divide the sum of the products by the sum of the weights to get the weighted average.
2. Can you provide an example of calculating a weighted average in Excel?
Sure! Let's say you have three test scores: 80, 90, and 70, with weights of 20%, 30%, and 50% respectively. To calculate the weighted average, you would multiply each score by its weight: 80 * 0.2,
90 * 0.3, and 70 * 0.5. Then, add up the products: 16 + 27 + 35. Finally, divide the sum of the products (78) by the sum of the weights (1), giving you a weighted average of 78.
3. What if I have more than one set of values with different weights?
If you have multiple sets of values with different weights, you can calculate separate weighted averages for each set and then calculate the overall weighted average. To do this, multiply each value
by its weight for each set, sum up the products within each set, and then divide by the sum of the weights for each set. Finally, calculate the weighted average of the set averages using the overall
For example, if you have two sets of scores with weights of 40% and 60%, you would calculate the weighted averages for each set and then calculate the overall weighted average using the weights of
the sets.
4. Is there a built-in function in Excel for calculating weighted averages?
Yes, Excel has a built-in function called "AVERAGE.WEIGHTED" which can be used to calculate weighted averages. This function takes two arguments: an array of values and an array of weights. Simply
input the values and weights into the function, and it will return the weighted average.
For example, "=AVERAGE.WEIGHTED(A1:A5, B1:B5)" would calculate the weighted average of the values in cells A1 to A5 using the weights in cells B1 to B5.
5. Can I use conditional weighting in Excel for calculating a weighted average?
Yes, you can use conditional weighting in Excel to calculate a weighted average. This means that you can assign different weights based on certain conditions. For example, you can assign a higher
weight to scores above a certain threshold and a lower weight to scores below the threshold. Use the "IF" function in combination with the weighted average formula to apply conditional weighting.
By using the "IF" function, you can set different weights based on specific criteria and then calculate the weighted average using these conditional weights.
How is weighted average calculated?
A weighted average is calculated by assigning weights to each number and then multiplying each number by its respective weight. The results of these multiplications are then added together. If the
weights do not add up to one, an alternative method can be used. In this case, the sum of all the variables multiplied by their weights is found, and then divided by the sum of the weights. This
ensures that the average reflects the importance of each number based on their corresponding weight.
How do you calculate weighted median in Excel?
To calculate the weighted median in Excel, a two-step process is followed. First, each data point is multiplied by its corresponding weight. Next, the data is sorted in ascending order. To find the
weighted median, various techniques can be used. One method is to use an array formula with the MEDIAN function, which takes into account both the weighted values and the sorted data. Another
approach involves utilizing the SUMPRODUCT function in combination with other techniques to calculate the weighted median. These methods enable Excel users to accurately calculate the weighted
median, providing a useful tool for analyzing and interpreting weighted data sets.
What is the formula for the average in Excel?
The formula for calculating the average in Excel is quite simple. By using the AVERAGE function, we can easily obtain the arithmetic mean of a range of numbers. For instance, if we have a range of
numbers from A1 to A20, inputting the formula =AVERAGE(A1:A20) will yield the average value of those numbers. This provides a convenient way to obtain the average value without having to manually
calculate it.
What is the weighted average formula with example?
The weighted average formula is used to calculate an average based on the individual weights or values assigned to different items. For example, let's consider a family with 5 children who weigh 20,
35, 80, 100, and 145 pounds, respectively. To find their average weight, we add up all their weights (20 + 35 + 80 + 100 + 145 = 280) and then divide it by the number of children (280 / 5 = 56
pounds). In this case, the weighted average weight of the children in the family is 56 pounds. This formula helps give more significance to certain items based on their assigned weights or values.
Calculating a weighted average in Excel may seem tricky, but it's actually quite simple. First, assign weights to each value based on their importance. Then, multiply each value by its corresponding
weight. Add up all these products and divide by the total of the weights. And voila! You've got your weighted average. Remember to use the SUMPRODUCT and SUM functions in Excel to make things easier.
Now that you know this helpful trick, you'll be able to calculate weighted averages with ease. It's a useful skill that can be applied to various situations, like determining your overall grade or
analyzing data. Excel makes it even simpler, so give it a try and impress your friends with your mathematical prowess! | {"url":"https://keysswift.com/blogs/guide/how-to-calculate-a-weighted-average-in-excel","timestamp":"2024-11-11T13:42:46Z","content_type":"text/html","content_length":"251707","record_id":"<urn:uuid:162538b4-4d20-4113-a6a3-13671cdac319>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00351.warc.gz"} |
1. / Integers
Integers include positive whole numbers, negative whole numbers, and zero.
The “set of all integers” is often shown like this:
Integers = {… -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, …}
The dots at each end of the set mean that you can keep counting in either direction. The set can also be shown as a number line:
The arrows on each end of the number line mean that you can keep counting in either direction.
Is It an Integer?
Integers are whole numbers and their negative opposites. Therefore, these numbers can never be integers:
• fractions
• decimals
• percents
Adding and Subtracting Integers
Looking at a number line can help you when you need to add or subtract integers.
Whether you are adding or subtracting two integers, start by using the number line to find the first number. Put your finger on it. Let's say the first number is 3.
• Then, if you are adding a positive number, move your finger to the right as many places as the value of that number. For example, if you are adding 4, move your finger 4 places to the right.
3 + 4 = 7
• If you are adding a negative number, move your finger to the left as many places as the value of that number. For example, if you are adding -4, move your finger 4 places to the left.
3 + -4 = -1
• If you are subtracting a positive number, move your finger to the left as many places as the value of that number. For example, if you are subtracting 4, move your finger 4 places to the left.
3 - 4 = -1
• If you are subtracting a negative number, move your finger to the right as many places as the value of that number. For example, if you are subtracting -4, move your finger 4 places to the right.
3 - -4 = 7
Here are two rules to remember:
• Adding a negative number is just like subtracting a positive number.
3 + -4 = 3 - 4
• Subtracting a negative number is just like adding a positive number. The two negatives cancel out each other.
3 + 4 = 3 - -4
Multiplying and Dividing Integers
If you multiply or divide two positive numbers, the result will be positive.
6 x 2 = 12
6 / 2 = 3
If you multiply or divide a positive number with a negative number, the result will be negative.
6 x -2 = -12
6 / -2 = -3
If you multiply or divide two negative numbers, the result will be positive—the two negatives will cancel out each other.
-6 x -2 = 12
-6 / -2 = 3
Integer Rules: A Video
Watch this video to better understand the correct procedure for adding, subtracting, multiplying, and dividing positive and negative whole numbers. | {"url":"https://www.factmonster.com/math-science/mathematics/integers","timestamp":"2024-11-12T12:43:25Z","content_type":"text/html","content_length":"37713","record_id":"<urn:uuid:c8f10211-b658-402e-9e41-e300b9ab5ed2>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00585.warc.gz"} |
john-users - Re: How does incremental mode works?
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Message-ID: <50A7BD3E.9010901@banquise.net>
Date: Sat, 17 Nov 2012 17:37:18 +0100
From: Simon Marechal <simon@...quise.net>
To: john-users@...ts.openwall.com
Subject: Re: How does incremental mode works?
On 11/17/2012 02:14 AM, Richard Miles wrote:
> Thanks for your answer. Nice to know I'm not the only one that is unable to
> understand how it works and the difference in a high level between
> incremental and markov. :)
> Maybe Solar or Simon may help us?
I will answer about Markov mode. The statistics file that it uses contains :
* the probability that character c is the first character of a password
* the probability that character c_n follows c_(n-1) (the previous
It doesn't actually store the raw probability, but something like:
P' = - N log(P)
That way, something very likely (P ~ 1) will have P' ~ 0, and something
highly unlikely (P ~ 0) will have a very high P'.
You compute the "markov strength" of a password by adding all those P'.
You can check this with the mkvcalcproba program. For example:
password 28+17+28+23+46+22+23+30 = 217
p4ssw0rd! 28+58+47+23+46+56+56+30+76 = 420
Notice how the first letter being identical, the first P' is identical
between passwords, and how unlikely transitions cost more.
The markov incremental mode with JtR, given a maximum strength, will
crack all passwords with a strength that is lower than or identical with
the given maximum. This means that -markov:200 will crack none of the
previous passwords, and -markov:250 will crack the easiest.
Please note that the number of passwords generated grows exponentially
with the max strength parameter. You can use the genmkvpwd program to
count them.
I will give a hopefully better description of all of this at Passwords^12.
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This talk is based on joint work with Yang Li. I will discuss non-linear Dirac operators and related regularity questions, which arise in various problems in gauge theory, Floer theory, DT theory,
and minimal submanifolds. These operators are used to define generalized Seiberg-Witten equations on 3- and 4-manifolds. Taubes proposed that counting harmonic spinors with respect to these operators
on 3-manifolds could lead to new 3-manifold invariants, while Donaldson and Segal suggested counting spinors over special Lagrangians to define Calabi-Yau invariants. Similar counts appear in
holomorphic Floer theory, where Doan and Rezchikov outlined a Fukaya 2-category for hyperkähler manifolds based on such counts. The central question in all of these proposals is whether the space of
such harmonic spinors is compact. We address this question in certain cases, proving and disproving several conjectures in the field and, in particular, answering a question raised by Taubes in 1999.
The key observation is that multivalued harmonic forms, in the sense of Almgren and De Lellis-Spadaro's Q-valued functions, play a crucial role in the problem. | {"url":"https://www4.math.duke.edu/media/videos.php?cat=all&sort=view_all&time=last_month&page=1&seo_cat_name=All&sorting=sort","timestamp":"2024-11-08T12:32:24Z","content_type":"text/html","content_length":"120527","record_id":"<urn:uuid:07bb504c-b54f-430b-bff3-92b0bd1720af>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00470.warc.gz"} |
Lesson Plan Websites For Algebra Elementary :: Algebra Helper
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The Price of Anarchy in Series-Parallel Network Congestion Games
We study the inefficiency of pure Nash equilibria in symmetric network congestion games defined over series-parallel networks with affine edge delays. For arbitrary networks, Correa (2019) proved a
tight upper bound of 5/2 on the PoA. On the other hand, for extension-parallel networks, a subclass of series-parallel networks, Fotakis (2010) proved that the PoA is 4/3. He also showed that this
bound is not valid for series-parallel networks by providing a simple construction with PoA 15/11. Our main result is that for series-parallel networks the PoA cannot be larger than 2, which improves
on the bound of 5/2 valid for arbitrary networks. We also construct a class of instances with a lower bound on the PoA that asymptotically approaches 27/19, which improves on the lower bound of 15/
University of Wisconsin-Madison, September 2021
View The Price of Anarchy in Series-Parallel Network Congestion Games | {"url":"https://optimization-online.org/2020/11/8108/","timestamp":"2024-11-06T08:48:49Z","content_type":"text/html","content_length":"84294","record_id":"<urn:uuid:9c927dae-c3b3-4518-86d0-7aee6bee6a15>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00616.warc.gz"} |
Arithmetic Operators in Swift - Swift Shorts
The arithmetic operators are used to perform basic mathematical operations on numbers in Swift. These operators include the following:
• Addition (+): Adds two numbers together.
• Subtraction (-): Subtracts one number from another.
• Multiplication (*): Multiplies two numbers together.
• Division (/): Divides one number by another.
• Remainder (%): Calculates the remainder of dividing one number by another.
Here are some examples of how you could use these operators in Swift:
let a = 5
let b = 2
let c = a + b // c = 7
let d = a - b // d = 3
let e = a * b // e = 10
let f = a / b // f = 2.5
let g = a % b // g = 1
In this code, the a and b variables are declared and initialized with the values 5 and 2, respectively. Then, the arithmetic operators are used to perform various operations on these values and store
the results in new variables.
For example, the + operator is used to add a and b together, and the result is stored in the c variable. Similarly, the - operator is used to subtract b from a, and the result is stored in the d
The arithmetic operators in Swift work with both integer and floating-point numbers, and they can be used in a wide range of mathematical and computational tasks. | {"url":"https://swiftshorts.com/2022/12/18/arithmetic-operators-in-swift/","timestamp":"2024-11-06T14:30:19Z","content_type":"text/html","content_length":"86734","record_id":"<urn:uuid:ff89fbe1-bb31-40f6-86d1-6d5ec05053a4>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00104.warc.gz"} |
NCERT Solutions For Class 7 Maths Chapter 3 Data Handling Exercise 3.3 - Solutions For Class
Class 7, Maths, Chapter 3, Exercise 3.3 Solutions
Q.1. Use the bar graph (Figure) to answer the following questions.
(a) Which is the most popular pet?
(b) How many students have dog as a pet?
Q.2. Read the bar graph (Figure) which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:
(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
Q.3. Number of children in six different classes are given below. Represent the data on a bar graph.
(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of class sixth to the students of class eight
Q.4. The performance of a student in 1^st Term and 2^nd Term is given. Draw a double bar graph choosing appropriate scale and answer the following:
(i) In which subject, has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Q.5. Consider this data collected from a survey of a colony
Q.6. Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this Chapter (Table 3.1). Plot a double bar graph using the data and answer the
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.
NCERT Solutions For Class 7 Maths, Chapter 3, Data Handling (All Exercises) | {"url":"https://solutionsforclass.com/ncert-solutions-class-7/ncert-solution-for-class-7-maths/ncert-solutions-for-class-7-mahts-chapter-3-data-handling-exercise-3-3/","timestamp":"2024-11-09T06:28:29Z","content_type":"text/html","content_length":"144733","record_id":"<urn:uuid:2020c7fb-e27e-4377-838e-a11b4b81faef>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00083.warc.gz"} |
Laurent Rineau
This package implements Shewchuk's algorithm [1] to construct conforming triangulations and 2D meshes. Conforming triangulations will be described in Section Conforming Triangulations and meshes in
Section Meshes.
Conforming Triangulations
A triangulation is a Delaunay triangulation if the circumscribing circle of any facet of the triangulation contains no vertex in its interior. A constrained Delaunay triangulation is a constrained
triangulation which is as much Delaunay as possible. The circumscribing circle of any facet of a constrained Delaunay triangulation contains in its interior no data point visible from the facet.
An edge is said to be a Delaunay edge if it is inscribed in an empty circle (containing no data point in its interior). This edge is said to be a Gabriel edge if its diametrical circle is empty.
A constrained Delaunay triangulation is said to be a conforming Delaunay triangulation if every constrained edge is a Delaunay edge. Because any edge in a constrained Delaunay triangulation is either
a Delaunay edge or a constrained edge, a conforming Delaunay triangulation is in fact a Delaunay triangulation. The only difference is that some of the edges are marked as constrained edges.
A constrained Delaunay triangulation is said to be a conforming Gabriel triangulation if every constrained edge is a Gabriel edge. The Gabriel property is stronger than the Delaunay property and each
Gabriel edge is a Delaunay edge. Conforming Gabriel triangulations are thus also conforming Delaunay triangulations.
Any constrained Delaunay triangulation can be refined into a conforming Delaunay triangulation or into a conforming Gabriel triangulation by adding vertices, called Steiner vertices, on constrained
edges until they are decomposed into subconstraints small enough to be Delaunay or Gabriel edges.
Building Conforming Triangulations
Constrained Delaunay triangulations can be refined into conforming triangulations by the two following global functions:
template<class CDT>
void make_conforming_Delaunay_2(CDT &t)
Refines the constrained Delaunay triangulation t into a conforming Delaunay triangulation.
template<class CDT>
void make_conforming_Gabriel_2(CDT &t)
Refines the constrained Delaunay triangulation t into a conforming Gabriel triangulation.
In both cases, the template parameter CDT must be instantiated by a constrained Delaunay triangulation class (see Chapter 2D Triangulations).
The geometric traits of the constrained Delaunay triangulation used to instantiate the parameter CDT has to be a model of the concept ConformingDelaunayTriangulationTraits_2.
The constrained Delaunay triangulation t is passed by reference and is refined into a conforming Delaunay triangulation or into a conforming Gabriel triangulation by adding vertices. The user is
advised to make a copy of the input triangulation in the case where the original triangulation has to be preserved for other computations
The algorithm used by make_conforming_Delaunay_2() and make_conforming_Gabriel_2() builds internal data structures that would be computed twice if the two functions are called consecutively on the
same triangulation. In order to avoid these data to be constructed twice, the advanced user can use the class Triangulation_conformer_2<CDT> to refine a constrained Delaunay triangulation into a
conforming Delaunay triangulation and then into a conforming Gabriel triangulation. For additional control of the refinement algorithm, this class also provides separate functions to insert one
Steiner point at a time.
Example: Making a Triangulation Conforming Delaunay and Then Conforming Gabriel
This example inserts several segments into a constrained Delaunay triangulation, makes it conforming Delaunay, and then conforming Gabriel. At each step, the number of vertices of the triangulation
is printed.
File Mesh_2/conforming.cpp
#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Constrained_Delaunay_triangulation_2.h>
#include <CGAL/Triangulation_conformer_2.h>
#include <iostream>
typedef CDT::Point Point;
typedef CDT::Vertex_handle Vertex_handle;
int main()
CDT cdt;
// construct a constrained triangulation
va = cdt.
(Point( 5., 5.)),
vb = cdt.insert(Point(-5., 5.)),
vc = cdt.insert(Point( 4., 3.)),
vd = cdt.insert(Point( 5.,-5.)),
ve = cdt.insert(Point( 6., 6.)),
vf = cdt.insert(Point(-6., 6.)),
vg = cdt.insert(Point(-6.,-6.)),
vh = cdt.insert(Point( 6.,-6.));
std::cout << "Number of vertices before: "
<< cdt.number_of_vertices() << std::endl;
// make it conforming Delaunay
std::cout << "Number of vertices after make_conforming_Delaunay_2: "
<< cdt.number_of_vertices() << std::endl;
// then make it conforming Gabriel
std::cout << "Number of vertices after make_conforming_Gabriel_2: "
<< cdt.number_of_vertices() << std::endl;
Vertex_handle insert(Point p, Face_handle f=Face_handle())
See Figure 56.1
A mesh is a partition of a given region into simplices whose shapes and sizes satisfy several criteria.
The domain is the region that the user wants to mesh. It has to be a bounded region of the plane. The domain is defined by a planar straight line graph, Pslg for short, which is a set of segments
such that two segments in the set are either disjoint or share an endpoint. The segments of the Pslg are constraints that will be represented by a union of edges in the mesh. The Pslg can also
contain isolated points that will appear as vertices of the mesh.
The segments of the Pslg are either segments of the boundary or internals constraints. The segments of the Pslg have to cover the boundary of the domain.
The Pslg divides the plane into several connected components. By default, the domain is the union of the bounded connected components.
See Figure 56.2 for an example of a domain defined without using seed points, and a possible mesh of it.
The user can override this default by providing a set of seed points. Either seed points mark components to be meshed or they mark components not to be meshed (holes).
See Figure 56.3 for another domain defined with the same Pslg and two seed points used to define holes. In the corresponding mesh these two holes are triangulated but not meshed.
Shape and Size Criteria
The shape criterion for triangles is a lower bound \( B\) on the ratio between the circumradius and the shortest edge length. Such a bound implies a lower bound of \( \arcsin{\frac{1}{2B}}\) on the
minimum angle of the triangle and an upper bound of \( \pi - 2* \arcsin{\frac{1}{2B}}\) on the maximum angle. Unfortunately, the termination of the algorithm is guaranteed only if \( B \ge \sqrt{2}
\), which corresponds to a lower bound of \( 20.7\) degrees over the angles.
The size criterion can be any criterion that tends to prefer small triangles. For example, the size criterion can be an upper bound on the length of longest edge of triangles, or an upper bound on
the radius of the circumcircle. The size bound can vary over the domain. For example, the size criterion could impose a small size for the triangles intersecting a given line.
Both types of criteria are defined in an object criteria passed as parameter of the meshing functions.
The Meshing Algorithm
The input to a meshing problem is a Pslg and a set of seeds describing the domain to be meshed, and a set of size and shape criteria. The algorithm implemented in this package starts with a
constrained Delaunay triangulation of the input Pslg and produces a mesh using the Delaunay refinement method. This method inserts new vertices to the triangulation, as far as possible from other
vertices, and stops when the criteria are satisfied.
If all angles between incident segments of the input Pslg are greater than \( 60\) degrees and if the bound on the circumradius/edge ratio is greater than \( \sqrt{2}\), the algorithm is guaranteed
to terminate with a mesh satisfying the size and shape criteria.
If some input angles are smaller than \( 60\) degrees, the algorithm will end up with a mesh in which some triangles violate the criteria near small input angles. This is unavoidable since small
angles formed by input segments cannot be suppressed. Furthermore, it has been shown ([1]), that some domains with small input angles cannot be meshed with angles even smaller than the small input
angles. Note that if the domain is a polygonal region, the resulting mesh will satisfy size and shape criteria except for the small input angles. In addition, the algorithm may succeed in producing
meshes with a lower angle bound greater than \( 20.7\) degrees, but there is no such guarantee.
Building Meshes
Meshes are obtained from constrained Delaunay triangulations by calling the global function:
template<class CDT, class NamedParameters>
void refine_Delaunay_mesh_2(CDT &t, const NamedParameters &np)
refines the domain defined by a constrained Delaunay triangulation into a mesh satisfying the criteri...
The template parameter CDT must be instantiated by a constrained Delaunay triangulation class.
The geometric traits class of CDT has to be a model of the concept DelaunayMeshTraits_2. This concept refines the concept ConformingDelaunayTriangulationTraits_2 adding the geometric predicates and
The second template parameter NamedParameters enables to pass a sequence of seed points to define a domain. It further enables to pass meshing criteria that the triangles have to satisfy. The
criteria must be a model of MeshingCriteria_2.
CGAL provides two models for this concept:
• Delaunay_mesh_criteria_2<CDT>, that defines a shape criterion that bounds the minimum angle of triangles,
• Delaunay_mesh_size_criteria_2<CDT>, that adds to the previous criterion a bound on the maximum edge length.
If the function refine_Delaunay_mesh_2() is called several times on the same triangulation with different criteria, the algorithm rebuilds the internal data structure used for meshing at every call.
In order to avoid rebuild the data structure at every call, the advanced user can use the class Delaunay_mesher_2<CDT>. This class provides also step by step functions. Those functions insert one
vertex at a time.
Any object of type Delaunay_mesher_2<CDT> is constructed from a reference to a CDT, and has several member functions to define the domain to be meshed and to mesh the CDT. See the example given below
and the reference manual for details. Note that the CDT should not be externally modified during the life time of the Delaunay_mesher_2<CDT> object.
Once the mesh is constructed, one can determine which faces of the triangulation are in the mesh domain using the is_in_domain() member function of the face type (see the concept
Optimization of Meshes with Lloyd
The package also provides a global function that runs Lloyd optimization iterations on the mesh generated by Delaunay refinement. The goal of this mesh optimization is to improve the angles inside
the mesh, and make them as close as possible to 60 degrees.
template< class CDT >
The enum Mesh_optimization_return_code is the output of the global mesh optimization functions.
Definition: Mesh_optimization_return_code.h:14
Mesh_optimization_return_code lloyd_optimize_mesh_2(CDT &cdt, const NamedParameters &np=parameters::default_values())
The function lloyd_optimize_mesh_2() is a mesh optimization process based on the minimization of a gl...
Definition: lloyd_optimize_mesh_2.h:119
Note that this global function has several named parameters (see details in reference pages) to tune the optimization process.
This optimization process alternates relocating vertices to the center of mass of their Voronoi cells, and updating the Delaunay connectivity of the triangulation. The center of mass is computed with
respect to a sizing function that was designed to preserve the local density of points in the mesh generated by Delaunay refinement.
See Figure Figure 56.4 for a mesh generated by refine_Delaunay_mesh_2() and optimized with lloyd_optimize_mesh_2(). Figure Figure 56.5 shows the histogram of angles inside these meshes.
As of CGAL 5.6, lloyd_optimize_mesh_2() uses Named Parameters to set parameters. More details are provided in Upgrading Code using Boost Parameters to CGAL Named Function Parameters.
Example Using the Global Function
The following example inserts several segments into a constrained triangulation and then meshes it using the global function refine_Delaunay_mesh_2(). The size and shape criteria are the default ones
provided by the criteria class Delaunay_mesh_criteria_2<K>. No seeds are given, meaning that the mesh domain covers the whole plane except the unbounded component.
File Mesh_2/mesh_global.cpp
#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Constrained_Delaunay_triangulation_2.h>
#include <CGAL/Delaunay_mesher_2.h>
#include <CGAL/Delaunay_mesh_face_base_2.h>
#include <CGAL/Delaunay_mesh_size_criteria_2.h>
#include <iostream>
typedef CGAL::Triangulation_data_structure_2<Vb, Fb> Tds;
typedef CDT::Vertex_handle Vertex_handle;
typedef CDT::Point Point;
int main()
CDT cdt;
Vertex_handle va = cdt.insert(Point(-4,0));
Vertex_handle vb = cdt.insert(Point(0,-1));
Vertex_handle vc = cdt.insert(Point(4,0));
Vertex_handle vd = cdt.insert(Point(0,1));
cdt.insert(Point(2, 0.6));
cdt.insert_constraint(va, vb);
cdt.insert_constraint(vb, vc);
cdt.insert_constraint(vc, vd);
cdt.insert_constraint(vd, va);
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Meshing the triangulation..." << std::endl;
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
The class Delaunay_mesh_face_base_2 is a model for the concept DelaunayMeshFaceBase_2.
Definition: Delaunay_mesh_face_base_2.h:29
The class Delaunay_mesh_size_criteria_2 is a model for the MeshingCriteria_2 concept.
Definition: Delaunay_mesh_size_criteria_2.h:31
Example Using the Class Delaunay_mesher_2<CDT>
This example uses the class Delaunay_mesher_2<CDT> and calls the refine_mesh() member function twice, changing the size and shape criteria in between. In such a case, using twice the global function
refine_Delaunay_mesh_2() would be less efficient, because some internal structures needed by the algorithm would be built twice.
File Mesh_2/mesh_class.cpp
#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Constrained_Delaunay_triangulation_2.h>
#include <CGAL/Delaunay_mesher_2.h>
#include <CGAL/Delaunay_mesh_face_base_2.h>
#include <CGAL/Delaunay_mesh_size_criteria_2.h>
#include <iostream>
typedef CGAL::Triangulation_data_structure_2<Vb, Fb> Tds;
typedef CDT::Vertex_handle Vertex_handle;
typedef CDT::Point Point;
int main()
CDT cdt;
Vertex_handle va = cdt.insert(Point(-4,0));
Vertex_handle vb = cdt.insert(Point(0,-1));
Vertex_handle vc = cdt.insert(Point(4,0));
Vertex_handle vd = cdt.insert(Point(0,1));
cdt.insert(Point(2, 0.6));
cdt.insert_constraint(va, vb);
cdt.insert_constraint(vb, vc);
cdt.insert_constraint(vc, vd);
cdt.insert_constraint(vd, va);
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Meshing the triangulation with default criteria..."
<< std::endl;
Mesher mesher(cdt);
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Meshing with new criteria..." << std::endl;
// 0.125 is the default shape bound. It corresponds to abound 20.6 degree.
// 0.5 is the upper bound on the length of the longest edge.
// See reference manual for Delaunay_mesh_size_traits_2<K>.
mesher.set_criteria(Criteria(0.125, 0.5));
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
This class implements a 2D mesh generator.
Definition: Delaunay_mesher_2.h:45
Example Using Seeds
This example uses the global function refine_Delaunay_mesh_2() but defines a domain by using one seed. The size and shape criteria are the default ones provided by the criteria class
Once the mesh is constructed, the is_in_domain() member function of faces is used to count them.
File Mesh_2/mesh_with_seeds.cpp
#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Constrained_Delaunay_triangulation_2.h>
#include <CGAL/Delaunay_mesher_2.h>
#include <CGAL/Delaunay_mesh_face_base_2.h>
#include <CGAL/draw_constrained_triangulation_2.h>
#include <iostream>
typedef CGAL::Triangulation_data_structure_2<Vb, Fb> Tds;
typedef CDT::Vertex_handle Vertex_handle;
typedef CDT::Point Point;
int main()
CDT cdt;
Vertex_handle va = cdt.
Vertex_handle vb = cdt.insert(Point(0,2));
Vertex_handle vc = cdt.insert(Point(-2,0));
Vertex_handle vd = cdt.insert(Point(0,-2));
cdt.insert_constraint(va, vb);
cdt.insert_constraint(vb, vc);
cdt.insert_constraint(vc, vd);
cdt.insert_constraint(vd, va);
va = cdt.insert(Point(3,3));
vb = cdt.insert(Point(-3,3));
vc = cdt.insert(Point(-3,-3));
vd = cdt.insert(Point(3,0-3));
cdt.insert_constraint(va, vb);
cdt.insert_constraint(vb, vc);
cdt.insert_constraint(vc, vd);
cdt.insert_constraint(vd, va);
std::list<Point> list_of_seeds;
list_of_seeds.push_back(Point(0, 0));
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Meshing the domain..." << std::endl;
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Number of finite faces: " << cdt.number_of_faces() << std::endl;
int mesh_faces_counter = 0;
for(CDT::Finite_faces_iterator fit = cdt.finite_faces_begin();
fit != cdt.finite_faces_end(); ++fit)
if(fit->is_in_domain()) ++mesh_faces_counter;
std::cout << "Number of faces in the mesh domain: " << mesh_faces_counter << std::endl;
void draw(const T2 &at2, const GSOptions &gso)
Example with a Domain Defined by Nesting Level.
When the domain is defined by polygons for the outer boundaries and the boundaries of holes the function mark_domain_in_triangulation() can be called instead of passing seedpoints.
File Mesh_2/mesh_marked_domain.cpp
#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Constrained_Delaunay_triangulation_2.h>
#include <CGAL/Delaunay_mesher_2.h>
#include <CGAL/Delaunay_mesh_face_base_2.h>
#include <CGAL/Delaunay_mesh_size_criteria_2.h>
#include <CGAL/mark_domain_in_triangulation.h>
#include <CGAL/draw_constrained_triangulation_2.h>
#include <iostream>
typedef CGAL::Triangulation_data_structure_2<Vb, Fb> Tds;
typedef CDT::Vertex_handle Vertex_handle;
typedef CDT::Point Point;
int main()
CDT cdt;
Vertex_handle va = cdt.
Vertex_handle vb = cdt.insert(Point(0,2));
Vertex_handle vc = cdt.insert(Point(-2,0));
Vertex_handle vd = cdt.insert(Point(0,-2));
cdt.insert_constraint(va, vb);
cdt.insert_constraint(vb, vc);
cdt.insert_constraint(vc, vd);
cdt.insert_constraint(vd, va);
va = cdt.insert(Point(3,3));
vb = cdt.insert(Point(-3,3));
vc = cdt.insert(Point(-3,-3));
vd = cdt.insert(Point(3,0-3));
cdt.insert_constraint(va, vb);
cdt.insert_constraint(vb, vc);
cdt.insert_constraint(vc, vd);
cdt.insert_constraint(vd, va);
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Meshing the domain..." << std::endl;
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Number of finite faces: " << cdt.number_of_faces() << std::endl;
int mesh_faces_counter = 0;
for(CDT::Finite_faces_iterator fit = cdt.finite_faces_begin();
fit != cdt.finite_faces_end(); ++fit)
if(fit->is_in_domain()) ++mesh_faces_counter;
std::cout << "Number of faces in the mesh domain: " << mesh_faces_counter << std::endl;
void mark_domain_in_triangulation(CT &ct, InDomainPmap ipm)
Example Using the Lloyd optimizer
This example uses the global function lloyd_optimize_mesh_2(). The mesh is generated using the function refine_Delaunay_mesh_2() of CGAL::Delaunay_mesher_2, and is then optimized using
lloyd_optimize_mesh_2(). The optimization will stop after 10 (set by max_iteration_number) iterations of alternating vertex relocations and Delaunay connectivity updates. More termination conditions
can be used and are detailed in the Reference Manual.
File Mesh_2/mesh_optimization.cpp
#define CGAL_MESH_2_OPTIMIZER_VERBOSE
//#define CGAL_MESH_2_OPTIMIZERS_DEBUG
//#define CGAL_MESH_2_SIZING_FIELD_USE_BARYCENTRIC_COORDINATES
#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Constrained_Delaunay_triangulation_2.h>
#include <CGAL/Delaunay_mesher_2.h>
#include <CGAL/Delaunay_mesh_face_base_2.h>
#include <CGAL/Delaunay_mesh_vertex_base_2.h>
#include <CGAL/Delaunay_mesh_size_criteria_2.h>
#include <CGAL/lloyd_optimize_mesh_2.h>
#include <iostream>
typedef CGAL::Triangulation_data_structure_2<Vb, Fb> Tds;
typedef CDT::Vertex_handle Vertex_handle;
typedef CDT::Point Point;
int main()
CDT cdt;
Vertex_handle va = cdt.insert(Point(-2,0));
Vertex_handle vb = cdt.insert(Point(0,-2));
Vertex_handle vc = cdt.insert(Point(2,0));
Vertex_handle vd = cdt.insert(Point(0,1));
cdt.insert(Point(2, 0.6));
cdt.insert_constraint(va, vb);
cdt.insert_constraint(vb, vc);
cdt.insert_constraint(vc, vd);
cdt.insert_constraint(vd, va);
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Meshing..." << std::endl;
Mesher mesher(cdt);
mesher.set_criteria(Criteria(0.125, 0.05));
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
std::cout << "Run Lloyd optimization...";
std::cout << " done." << std::endl;
std::cout << "Number of vertices: " << cdt.number_of_vertices() << std::endl;
The class Delaunay_mesh_vertex_base_2 is a model for the concept DelaunayMeshVertexBase_2.
Definition: Delaunay_mesh_vertex_base_2.h:24
It is possible to export the result of a meshing in VTU, using the function CGAL::IO::write_VTU(). For more information about this format, see VTK (VTU / VTP / legacy) File Formats. | {"url":"https://cgal.geometryfactory.com/CGAL/doc/master/Mesh_2/index.html","timestamp":"2024-11-08T23:49:59Z","content_type":"application/xhtml+xml","content_length":"67006","record_id":"<urn:uuid:cb9b92d2-2256-4fb7-9c60-5e254a5a2f1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00453.warc.gz"} |
This is an essay in what might be called “mathematical metaphysics.” There is a fundamental duality that runs through mathematics and the natural sciences, from logic to biology.
This is Chapter 9 in my book: Ellerman, David. 1995. Intellectual Trespassing as a Way of Life: Essays in Philosophy, Economics, and Mathematics. Lanham MD: Rowman & Littlefield.
This essay grew out of an attempt to model mathematically the possible cross-ownership arrangements that might arise between privatizing firms in the former Yugoslavia [see Ellerman 1991]. The
cross-ownership arrangements resemble the groups of Japanese companies called keiretsu. There is cross ownership between the companies in the group as well as some ownership outside the group that is
traded on the stock market. In spite of the partial outside ownership, the keiretsu often behave as “self-owning” groups. If firm A owns shares in B, then the management in A usually signs over its
proxy on shares in B to the management in firm B. And the management in B does likewise with respect to the managers in A. Thus within certain constraints, each firm can act like a “self-owning”
firm, not totally unlike the self-managing firms of the former Yugoslavia.
This is Chapter 12 in my book: Ellerman, David. 1995. Intellectual Trespassing as a Way of Life: Essays in Philosophy, Economics, and Mathematics. Lanham MD: Rowman & Littlefield.
This is Chapter 11 in my book: Ellerman, David. 1995. Intellectual Trespassing as a Way of Life: Essays in Philosophy, Economics, and Mathematics. Lanham MD: Rowman & Littlefield.
This is Chapter 10 from my book: Ellerman, David. 1995. Intellectual Trespassing as a Way of Life: Essays in Philosophy, Economics, and Mathematics. Lanham MD: Rowman & Littlefield.
One of the fundamental insights of mainstream neoclassical economics is the connection between competitive market prices and the Lagrange multipliers of optimization theory in mathematics. Yet this
insight has not been well developed. In the standard theory of markets, competitive prices result from the equilibrium of supply and demand schedules. But in a constrained optimization problem, there
seems to be no mathematical version of supply and demand functions so that the Lagrange multipliers would be seen as equilibrium prices. How can one “find the markets in the math” so that Lagrange
multipliers will emerge as equilibrium market prices?
This is Chapter 8 of my book: Ellerman, David. 1995. Intellectual Trespassing as a Way of Life: Essays in Philosophy, Economics, and Mathematics. Lanham MD: Rowman & Littlefield.
This essay deals with a connection between a relatively recent (1940s and 1950s) field of mathematics, category theory, and a hitherto vague notion of philosophical logic usually associated with
Plato, the self-predicative universal or concrete universal. Consider the following example of “bad Platonic metaphysics.”
Given all the entities that have a certain property, there is one entity among them that exemplifies the property in an absolutely perfect and universal way. It is called the “concrete universal.”
There is a relationship of “participation” or “resemblance” so that all the other entities that have the property “participate in” or “resemble” that perfect example, the concrete universal.
All of this and much more “bad metaphysics” turns out to be precisely modeled in category theory.
This is Chapter 7 from my book: Ellerman, David. 1995. Intellectual Trespassing as a Way of Life: Essays in Philosophy, Economics, and Mathematics. Lanham MD: Rowman & Littlefield.
The watershed event in the philosophy of mind (particularly as it relates to artificial intelligence or AI) during the last decade was John Searle’s 1980 article “Minds, Brains and Programs.” This
chapter was written about the same time and independently of Searle’s but it was updated in 1985 to take Searle’s work into account. Searle’s exposition was based on his now-famous “Chinese Room
Argument”—an intuition pump that boils down to a nontechnical explanation of the difference between syntax (formal symbol manipulation) and semantics (using symbols based on their intended
interpretation). Searle argues, in opposition to “hard AI,” that computers can at best only simulate but never duplicate minds because computers are inherently syntactical (symbol manipulators) while
the mind is a semantic device.
The syntax-semantics distinction is hardly new; it was hammered out in philosophical logic during the first part of this century and it is fundamental in computer science itself. The purpose of our
paper is to analyze the minds-machines question using simple arguments based on the syntax-semantics distinction from logic and computer science (sans “Chinese Room”). I arrive at essentially the
same results as Searle—with some simplification and sharpening of the argument for readers with some knowledge of logic or computer science.
This is Chapter 6 from my book: Ellerman, David. 1995. Intellectual Trespassing as a Way of Life: Essays in Philosophy, Economics, and Mathematics. Lanham MD: Rowman & Littlefield.
The essay on double-entry bookkeeping (DEB) is intellectually interesting for several reasons in spite of the well-known soporific aspects of bookkeeping. Several of the essays in the volume
explicitly employ the analogy between additive and multiplicative operations (i.e., the common group-theoretic properties of additive groups of numbers and multiplicative groups of nonzero numbers).
For instance, given the system of multiplying whole numbers or integers, there is no operation inverse to multiplication (i.e., there is no division). But there is a standard method of enlarging the
system to allow division. Consider pairs of whole numbers a/b (with b ¹ 0) and define multiplication in the obvious way: (a/b)(c/d) = (ac)/(bd). These ordered pairs of integers are the “fractions”
and they allow the operation of division (“multiply by the reciprocal”).
Now substitute addition for multiplication. We start with the additive system of positive numbers along with zero (i.e., the non-negative numbers) where is no inverse operation to addition (i.e.,
there is no subtraction). To enlarge the domain of non-negative numbers to include subtraction, consider ordered pairs [a // b] and define addition in the analogous way: [a // b] + [c // d] = [a+c //
b+d]. This enlarged system of additive operations on ordered pairs of non-negative numbers allows subtraction (“add on the reversed pair”). The origin of the intellectual trespassing into DEB was the
observation that these ordered pair were simply the T-accounts of DEB.
Aside from illustrating the interplay of additive-multiplicative themes, the essay illustrates one of the most astonishing examples of intellectual insulation between disciplines, in this case,
between accounting and mathematics. Double-entry bookkeeping was developed during the fifteenth century and was first recorded as a system by the Italian mathematician Luca Pacioli in 1494.
Double-entry bookkeeping has been used as the accounting system in market-based enterprises of any size throughout the world for several centuries. Incredibly, however, the mathematical basis for DEB
is not known, at least not in the field of accounting.
This approach to interpreting quantum mechanics is not another jury-rigged or ad-hoc attempt at the interpretation of quantum mechanics but is a natural application of the fundamental duality running
throughout the exact sciences. | {"url":"https://www.ellerman.org/category/math-blog/","timestamp":"2024-11-13T01:09:43Z","content_type":"application/xhtml+xml","content_length":"65252","record_id":"<urn:uuid:893a4ae9-4951-4a47-a0b0-ae74bd5ddc30>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00291.warc.gz"} |
The COUNT Function: Syntax and Examples
The COUNT is an Excel statistical function that counts how many numbers are in a list of arguments. It is one of the oldest Excel functions, so it can be used even with Excel 2003.
Note: If you want to count logical values, text, or error values, use the COUNTA function. And for numbers that meet a certain criteria, use the COUNTIF or the COUNTIFS function.
COUNT(value1, [value2], ...)
Required Arguments
value1: Arguments can either be numbers or names, arrays, or references that contain numbers, which means 3, B7 and C2:C6 are all valid examples.
Optional Arguments
value2-255: You can use up to 255 additional items, cell references, or ranges in the calculation by following the syntax listed above.
Note: The arguments can contain or refer to a variety of different types of data, but only numbers are counted.
COUNT Function Examples
Example COUNT Function
As seen in the above screenshot, the COUNT function will only count numerical values, such as number, date, currency. You can reference the cells individually, within a range, or add values directly
as arguments to the function itself. In the above case – 98.
Also note that blank cells, errors and text will not be counted!
More Examples and Use Cases
We don’t have any use cases for this formula on our website at this moment but will link them below once we have. Stay tuned! | {"url":"https://tutorialsforexcel.com/the-count-function-syntax-and-examples/","timestamp":"2024-11-08T06:10:05Z","content_type":"text/html","content_length":"40022","record_id":"<urn:uuid:c735a7e6-ec6a-4351-b114-0f297993625e>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00668.warc.gz"} |
Arithmetic and Differential Galois Groups | EMS Press
Arithmetic and Differential Galois Groups
• David Harbater
University of Pennsylvania, Philadelphia, United States
• B. Heinrich Matzat
Universität Heidelberg, Germany
• Marius van der Put
Rijksuniversiteit Groningen, Netherlands
• Leila Schneps
Université Pierre et Marie Curie, Paris, France
Galois theory is the study of symmetries in solution spaces of polynomial and differential equations and more generally of the relation between automorphism groups (or group schemes) and the
structure of algebraic and differential extensions.
Many of the important problems in this area are connected to the classification of (algebraic or differential) Galois extensions and the study of the respective fundamental groups, e.g. via inverse
problems. Other interesting points are the direct problem, i.e., the computation of (ordinary or differential) Galois groups, and related constructive aspects.
This workshop gave an overview on some important new results and developments. A second goal was the discussion of ideas for proving some open questions and conjectures as well as of new directions
of research.
The main topics of the workshop were:
• The absolute Galois group and the Grothendieck-Teichmüller group,
• Etale fundamental groups and the anabelian conjecture,
• Arithmetic Galois realizations and constructive Langlands program,
• Local and global differential modules and the -curvature conjecture,
• Galois theory for nonlinear and partial differential equations.
Besides this main program we had some reports on related subjects like the Cohen-Lenstra heuristics for number fields by J. Klüners and the behaviour of the Tate–Shafarevich group in anticyclotomic
field extensions by M. Çiperiani.
It is always difficult to emphasise highlights or unexpected results, but one of them might be the result of F. Pop that henselian fields always are large, which is of great interest in field
arithmetic and inverse Galois theory. Another striking result is the existence of -motives over shown by M. Dettweiler which gives a positive answer to an old question of J.-P. Serre. A fruit of
common effort in research in arithmetic and differential Galois theory is the development of patching methods for differential modules by D. Harbater and J. Hartmann. This method will allow yet
unexpected applications in inverse differential Galois theory and other areas, like K-theory. The talk of B. Malgrange gave a vision of Galois theory for nonlinear differential equations.
Exploitation of his work for special types of equations, like Painlevé equations, and the generalization of his ideas to non algebraically closed fields of constants as well as to positive
characteristic will keep researchers busy for years.
Surely, many other results presented at the workshop should be pointed out here, too, like M. Raynaud's work on fundamental groups in positive characteristic, D. Bertrand's result on Schanuel's
conjecture or Ch. Hardouin's generalization of -difference equations to roots of unity by creating an iterative -difference theory.
Altogether, we had a wonderful and inspiring week with lots of interesting lectures and many discussions bearing ideas for future research.
Finally, the organisers want to cordially thank the Oberwolfach administration and its staff for giving us the opportunity to arrange this and earlier workshops on Galois theory as well as for the
excellent service.
Cite this article
David Harbater, B. Heinrich Matzat, Marius van der Put, Leila Schneps, Arithmetic and Differential Galois Groups. Oberwolfach Rep. 4 (2007), no. 2, pp. 1443–1520
DOI 10.4171/OWR/2007/26 | {"url":"https://ems.press/journals/owr/articles/1542","timestamp":"2024-11-12T02:57:59Z","content_type":"text/html","content_length":"92361","record_id":"<urn:uuid:7461505d-0fb1-4c1f-99fc-c9e94f27db17>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00719.warc.gz"} |
Clock Face Poster
Poster based on Clock Face task.
The poster is available here as a
, or the image below can be clicked on to enlarge it.
Student Solutions
For the numbers on both sides of the line to have the same total, the line needs to go from between the 9 and 10, to between the 3 and 4.
To divide the clock face so that the total of the numbers on one side of the lines is twice the total on the other side, the lines can go from between 10 and 11 to the centre and from between 2 and 3
to the centre; or from between 4 and 5 to the centre and from between 8 and 9 to the centre.
There are eight different ways of diving up the clock face to give two parts where the sums are prime numbers:
• With the hands going from between 4 and 5 to the centre and from between 5 and 6 to the centre
• With the hands going from between 1 and 2 to the centre and from between 3 and 4 to the centre
• With the hands going from between 6 and 7 to the centre and from between 7 and 8 to the centre
• With the hands going from between 2 and 3 to the centre and from between 4 and 5 to the centre
• With the hands going from between 10 and 11 to the centre and from between 11 and 12 to the centre
• With the hands going from between 4 and 5 to the centre and from between 6 and 7 to the centre
• With the hands going from between 7 and 8 to the centre and from between 9 and 10 to the centre
• With the hands going from between 8 and 9 to the centre and from between 10 and 11 to the centre
To see more detailed methods, please look at the published children's solutions to the original problem. | {"url":"https://nrich.maths.org/problems/clock-face-poster","timestamp":"2024-11-14T20:04:51Z","content_type":"text/html","content_length":"38702","record_id":"<urn:uuid:896f846e-8955-468b-840d-76805f2ced96>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00561.warc.gz"} |
Central Limit Theorem and Convergence
This is my first post (yaaay \o/).
Sometimes it’s hard to understand the meaning of CLT and convergence. I realized many times when people first hear about these concepts that they actually don’t know what it’s. And they stay this way
until they learn (if they learn) more advanced theories.
One of the first things that we learn in school is Central Limit Theorem. Basically for any symmetric distribution, when you have a large number of observations, it converges to a normal
There are some distributions more inclined to converge than others. For example, the exponential distribution, t-student, binomial.
But those are all old news. Here I’m going to show an example of convergence in real life.
Exponential Distribution
The CLT says that if you have $X_1, X_2, ...X_k$ independent random variables with distribuition $exp(\lambda)$ when k is large enough the mean of these identical distribuited vectors converges to a
normal with mean \(\lambda\) and variance 1/(k*$\lambda^2$).
Now to prove this, we can generate 40 observations from a exponential distribuition with $\lambda$ = 0.2. And then replicate this experiment k times.
The Rcode for that is:
k=10 #number of replication
n= 40 #size of each vector
lambda = 0.2
list_of_exponential = array(1:k) #variable to keep all the calculated means
#loop to run the replication
for(i in 1:k){
list_of_exponential[i] = mean(rexp(n, lambda))
As you can see the k here is very small (10) so how would be the distribution of the mean for these iid variables?
As you can see it’s not very “normal”. The reason for that is because we need a bigger k in order to get closer to a normal.
If we run the same script with k=100 for example
And if we run for k=1000
You finally can see that it does converge to a normal distribution.
There some decent explanation in the wikipedia page (if you need more theory) and I also included the code in my github page.
As this is my first post, please send me your feedback/suggestions so I can improve the blog and posts =D. | {"url":"https://andresa.me/central-limit-theorem-and-convergence/","timestamp":"2024-11-13T15:24:04Z","content_type":"text/html","content_length":"41217","record_id":"<urn:uuid:cd578978-5247-4801-b62b-e8f09b0e01b6>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00416.warc.gz"} |
Algebra And Trigonometry For 1St Year Pdf - CAREER KEG
Algebra And Trigonometry For 1St Year Pdf
May 20, 2022 Uncategorized No Comments
Find out more about Algebra and Trigonometry for 1st Year Pdf, algebra and trigonometry for 1st year pdf, free download pdf by leland mangham, free pdf download by leland mangham, ebook download by
leland mangham on careerkeg.com.
Algebra and Trigonometry for 1st Year is a collection of short lessons on the two subjects, which are necessary for students who want to pursue higher mathematics. The book is intended for students
who have completed their first year of college, as well as those who are preparing for an entrance exam.
The book contains a number of exercises with solutions that can be used by teachers and students in order to assess their knowledge and improve it. The author also provides some examples of how one
can solve certain questions at home.
Algebra and Trigonometry for 1st Year includes an introduction to the subject matter, which explains what the book is about, what it covers and why it should be read. It also contains tips on how to
approach each topic presented in the text. In addition, there are many examples illustrating how to solve various problems using different methods.
Algebra And Trigonometry For 1St Year Pdf
Function of One Variable.
Let us begin with a few definitions. A relation is any set of ordered pairs, but a function is a relation whose domain and range are the same. In other words, if f(x) = x2 + b is the equation of a
function at every x in its domain, then f(x) defines a function. The domain of a function can be given by:
In addition to these notations for defining functions, there are plenty more ways you might see them defined. Here are some examples:
Let’s look at some common graphs that you may come across when working with functions:
(1) A function is a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output.
A function is a relationship between input and output. For example, let’s say that we have a function where the input is called x and the output is called f(x). We can write this as follows:
In this case, our domain is all real numbers except for negative integers (because 5+2x-3 cannot be equal to any negative integer). So our domain would be written as follows:
\begin{align*}D_f=\left[-1,+infinity\right] \\\end{align*}where infinity means “infinitely large”. In other words, our domain contains all nonnegative real numbers. Finally, our range would be
written as follows:
R_f = [-infinity,-6] because -6 is one of the possible outputs for this function (the others being 0 through 5).
(2) Algebraically, a function can be represented as an expression, a table, a graph, or a verbal description.
This chapter discusses the following topics:
• Algebraically, a function can be represented as an expression, a table, a graph, or a verbal description.
• This chapter provides some illustrations of these four representations.
(3) The domain of f contains all the real numbers, and the range of f consists of all real numbers greater than or equal to 3.
Let’s say you have a function f, where it takes in inputs and returns outputs. The domain of f is the set of all possible inputs to that function, and its range is the set of all possible outputs
from that function.
To look at it more concretely: if you have a function f(x), then its domain would include all x’s for which f(x) is defined (i.e., there exists some input value such that when you plug it into your
function, you get an output). And similarly, its range would include all y’s for which f(y) is defined (i.e., there exists some input value such that when you plug it into your function, you get an
(4) We may read this as “f of x equals x cubed minus 4”.
In algebra and trigonometry, we use a method of writing functions called function notation. This notation consists of an arrow and the variable that the function depends on. So, for example:
(1)If f(x) = (x+2)(x-5), then we may read this as “f of x equals x plus 2 times x minus 5” (or “f of x equals x^2-5x”).
(2)If g(y) = y/3 + 4y^2 – 3y + 1, then we may read this as “g of y equals one third plus four times y squared minus three times y plus one.”
(5) Another way to denote the domain and range is to list their elements inside curly braces as follows: Domain = {x | x ∈ R}, Range = {y | y ≥ 3}.
(5) Another way to denote the domain and range is to list their elements inside curly braces as follows: Domain = {x | x ∈ R}, Range = {y | y ≥ 3}.
The set notation for the domain can be written as follows:
Domain = {x | x ∈ R}, where R is the set of real numbers.
Similarly, a subset or element can also be used to denote a subset or element as follows:
Subset: x 0 and y > 0 represents those pairs (x,y) whose coordinates satisfy both conditions simultaneously.
(6) The function f is not defined for x = 2 because there are two possible outputs for that particular input. So the value “2” does not belong to the domain of the function f.
Just as you can’t draw a line on a graph without first deciding which x-coordinate to use, you can’t evaluate a function without also deciding which values of x will be used. This is called selecting
the domain of f. The domain consists of all possible inputs to the function—that is, all real numbers that are acceptable inputs for f(x).
For example:
f(x) = (1 + 2x)/(5 + 3x)
The domain of this function includes every real number greater than or equal to -3 and less than or equal to 5.
(7) Note that you don’t need to know what it means for two functions to be equal in order to determine whether or not given functions are equal. You just need to compare the results produced by
substituting points into each one.
You might be wondering what it means to say that two functions are equal, but the definition is rather simple. If you substitute the same value into both functions and get the same result, then
they’re equal. For example:
In this case, f(x) = x – 2 and g(x) = 3x + 5 are not equal. They look different, but when you plug in 2 for x in each one you get 4 (for f) and 7 (for g). However:
This time they are both 6 at all points except 0 where they become 4 and 8 respectively because of the differences between their behavior near 0. The important thing is that if we plug in any number
other than 0 into either one of them there’s always going to be an even difference between the values produced by plugging it into either f or g!
Graphs of Functions.
When we graph a function, we use the points on which it is increasing (positive slopes), decreasing (negative slopes), or level.
Points that make the function have a corner are called turning points. These are places where the graph switches from increasing to decreasing, or decreases to increases.
If you have a hard time remembering what a “turning point” looks like or how they work, think of it this way:
• A turning point is any point on your graph where there’s a bend in your line of best fit. The only difference between these and other points on your line will be how far apart they are from each
other; i.e., if your line has two turning points within 1/3rd of its total length then one will be closer than another.*
• If you draw three lines through two different turning points at 90 degrees so that they cross each other at those same two turning points then all four lines will intersect at exactly one point.*
This means that if there were any place else where those same three lines could intersect with each other but didn’t then there would be no place else where those same three lines could touch! We
call this property “non-contradiction”, meaning that if something happens only once somewhere then anything else happening elsewhere won’t contradict it happening in both places together – so
long as nothing else changes between those times too much anyway…
(1) To sketch the graph, we start by plotting a few points that we know are on the graph – these are called critical points.
To sketch the graph of a function, we first need to find two critical points. A critical point is any point where the function changes from increasing to decreasing or vice versa. It’s also called a
relative extreme point. To find these points, we take the derivative of our function and set it equal to zero:
f(x) = x2 + 1
f'(x) = 2x + 1
0 = 2x + 1
2x = 0
x = -1/2 | {"url":"https://infolearners.com/algebra-and-trigonometry-for-1st-year-pdf/","timestamp":"2024-11-05T22:22:42Z","content_type":"text/html","content_length":"60822","record_id":"<urn:uuid:c09a12ba-0ae7-4d50-92c4-aaad8f285d97>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00287.warc.gz"} |
Supremum & Infimum Homework Statement
• Thread starter drawar
• Start date
In summary, we are trying to prove that the supremum of the set S, which is defined as the set of all fractions n/(n+m) where n and m are natural numbers, is equal to 1, and the infimum is equal to
0. To do this, we need to find values for n and m such that n/(n+m) is always greater than 1-ε for any given ε. Using the hint provided, we can rewrite n/(n+m) as 1 - m/(n+m) and choose m=1. This
results in ε>1/(n+1), which we can then solve for n.
Homework Statement
Let $$S = \left\{ {\frac{n}{{n + m}}:n,m \in N} \right\}$$. Prove that sup S =1 and inf S = 0
Homework Equations
The Attempt at a Solution
So I was given the fact that for an upper bound u to become the supremum of a set S, for every ε>0 there is $$x \in S$$ such that x>u-ε. In this case, I'm supposed to find n and m such that $${\frac
{n}{{n + m}} > 1 - \varepsilon }$$ for every ε given. However, I cannot express n and m in terms of ε explicitly. Any hints or comments will be very appreciated, thanks!
Science Advisor
Homework Helper
hi drawar!
drawar said:
I'm supposed to find n and m such that $${\frac{n}{{n + m}} > 1 - \varepsilon }$$ for every ε given.
hint: n/(n+m) = 1 - m/(n+m)
tiny-tim said:
Hi tiny-tim, thanks for the hint. Do you mean:
$${1 - \varepsilon < \frac{n}{{n + m}} = 1 - \frac{m}{{n + m}}}$$
Choosing m=1:
$${\varepsilon > \frac{m}{{n + m}} > \frac{1}{{n + 1}}}$$
and then solve for n?
Science Advisor
Homework Helper
except, that's ##\frac{1}{\frac{n}{m}+1}##
FAQ: Supremum & Infimum Homework Statement
What is the definition of Supremum and Infimum?
Supremum and Infimum are two important concepts in mathematical analysis. The Supremum (or least upper bound) of a set is the smallest number that is greater than or equal to all elements in the set.
The Infimum (or greatest lower bound) of a set is the largest number that is less than or equal to all elements in the set.
How are Supremum and Infimum related to Maximum and Minimum?
The Supremum and Infimum are related to the Maximum and Minimum in that the Maximum and Minimum are just the largest and smallest elements in a set, respectively. However, the Supremum and Infimum
may not necessarily be actual elements of the set, but are defined as the smallest upper bound and largest lower bound, respectively.
How do you find the Supremum and Infimum of a set?
To find the Supremum and Infimum of a set, you can use the following steps:
1. Arrange the elements of the set in ascending order.
2. If the set has a maximum element, then the maximum is the Supremum.
3. If the set has a minimum element, then the minimum is the Infimum.
4. If the set does not have a maximum or minimum element, then the Supremum and Infimum can be found by observing the pattern of the set and using logical reasoning.
What is the importance of Supremum and Infimum in mathematical analysis?
The concept of Supremum and Infimum is important in mathematical analysis as it helps us define limits, continuity, and the behavior of functions. They also allow us to determine whether a set has a
maximum or minimum element, which is crucial in optimization problems and finding solutions to equations.
Can a set have multiple Supremum or Infimum?
No, a set can only have one Supremum and one Infimum. This is because the Supremum and Infimum are unique and are defined as the smallest upper bound and largest lower bound, respectively. If a set
has multiple Supremum or Infimum, then they would not be unique and would not accurately represent the set. | {"url":"https://www.physicsforums.com/threads/supremum-infimum-homework-statement.668026/","timestamp":"2024-11-06T04:27:49Z","content_type":"text/html","content_length":"86962","record_id":"<urn:uuid:11988e24-f002-40a0-b532-40a4c8e5e0da>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00336.warc.gz"} |
Convert Cubic meter per second (m³/s) (Volumetric flow rate)
Convert Cubic meter per second (m³/s)
Direct link to this calculator:
Convert Cubic meter per second (m³/s) (Volumetric flow rate)
1. Choose the right category from the selection list, in this case 'Volumetric flow rate'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and
π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Cubic meter per second [m³/s]'.
4. The value will then be converted into all units of measurement the calculator is familiar with.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
Utilize the full range of performance for this units calculator
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '731 Cubic meter per second'. In so doing, either the full name of
the unit or its abbreviation can be usedas an example, either 'Cubic meter per second' or 'm3/s'. Then, the calculator determines the category of the measurement unit of measure that is to be
converted, in this case 'Volumetric flow rate'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the
conversion you originally sought. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and
countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(93 * 62) m3/s'. But different
units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '56 Cubic meter per second + 25 Cubic meter per second' or '31mm x 99cm
x 68dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 2.103 733 314 189 4×1022. For this form of presentation, the number
will be segmented into an exponent, here 22, and the actual number, here 2.103 733 314 189 4. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket
calculators, one also finds the way of writing numbers as 2.103 733 314 189 4E+22. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at
this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 21 037 333 141 894 000 000 000. Independent of the presentation of
the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | {"url":"https://www.convert-measurement-units.com/convert+Cubic+meter+per+second.php","timestamp":"2024-11-08T01:27:43Z","content_type":"text/html","content_length":"56165","record_id":"<urn:uuid:6311ac6c-2b98-4c00-9b69-3ccf35e2d7df>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00726.warc.gz"} |
Euler angles and singularities
Home › General
Euler angles and singularities
I made a question on stack overflow where i am asking to measure angle around an axis without affected by gimbal lock. If you know the answer or have any suggestion please let me know.
Quaternions are used to combat gimbal lock.
I know that but at some point i have to convert quaternions back to Euler angles to tell the user that pendulum has θ angle on specific axis. So there is the problem , how to tell user what is
the sensor angle without converting to Euler angle.
Edit: Also i know i can get an angle using quaternions and hamilton product but it will not be possible to get angle in x,y,z (i.e the pendulum rotate θ around x axis, φ around y axis)
Find a library that does the conversion for you. | {"url":"https://mbientlab.com/community/discussion/3323/euler-angles-and-singularities","timestamp":"2024-11-12T10:25:32Z","content_type":"text/html","content_length":"25909","record_id":"<urn:uuid:1470865a-1ba2-4545-b2d8-657f8bcfa806>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00789.warc.gz"} |
Publications and Talks | misitio
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• Artificial Hydrogen Molecule in vertically stacked Ga_{1-x}Al_xA_s Nanoscale Rings: Structural and
External Probes Effects in their Quantum Levels. Joint with J. D. Castrillón, M. R. Fulla, I. E. Rivera, Y. A. Suaza and J. H. Marin (submitted CompletePreprint). Physika E: Low-dimensional Systems
and Nanostructures, 117, 113765 (2020) (to appear).
• A General Version of the Nullstellensatz for Arbitrary Fields. Joint with Edisson Gallego
and Juan D. Vélez. Open Mathematics 17, 556-558. Preprint 2018.
• Functional Conceptual Substratum as a New Cognitive Mechanism for Mathematical Creation.
Joint with Stefan Hetzl. (to appear), PreprintArxiv 2019.
• Category-Based Co-Generation of Seminal Concepts and Results in Algebra and Number Theory:
Containment-Division and Goldbach Rings. Joint with Marlon Fulla, Ismael Rivera, Juan D. Vélez and Edißon Gallego. To be published in JP Journal of Algebra, Number Theory and Applications, 2018.
• Towards a General Many-sorted Formal Framework for Describing Certain kinds of
Legal Statues. Joint with Egil Nordqvist. Under review, Preprint 2017.
• Towards an Homological Generalization of the Direct Summand Theorem. Joint with
Juan D. Vélez. Under review, PreprintArxiv 2017.
• Artificial Co-Creative Generation of the Notion of Topological Group based on the Categorical
Conceptual Blending. Joint with Yoe A. Herrera-Jaramillo and Florian Geismann. (Under Review), Preprint.
• A New Multiple-Intelligences Test for Artificial General Intelligence. Joint with Judith Kieninger,
Stephan Schneider and Nico Potyka. Under review. Preprint
• Containment-Division Rings and New Characterizations of Dedekind Domains.
Joint with Edisson Gallego and Juan D. Vélez, PreprintArxiv 2017.
• On Preservation Properties and an Algebraic Characterization of Some Stronger
Forms of the Noetherian Condition. Joint with Edisson Gallego and Juan D. Vélez. Under review, Preprint 2017.
• Towards an Experimental Science of Natural Consciousness. Joint with F. Becker and R. Garita, Preprint 2017.
• Theory Blending: Extended Algorithmic Aspects of Examples. Joint with M. Martinez,
A. M. H. Abdel-Fattah, U. Krumnack, A. Smail, T. Besold, A. Pease, M. Schmidt, M. Guhe and K.-U.
Kuehnberger. Annals of Mathematics and Artificial Intelligence, pp. 1-25, 2016. pdf-link
• Normality and Related Properties of Forcing Algebras. Joint with Holger Brenner. Com-
munications in Algebra. Volume 44, Issue 11, pp. 4769-4793, 2016. pdf
• The Direct Summand Conjecture for some bi-generated extensions and an asymptotic
Version of Koh’s Conjecture. Joint with Edisson Gallego and Juan D. Velez. Beitraege zur Algebra und Geometrie (Contributions in Algebra and Geometry) pp. 1-16. 2016. OfficialPdfLink
• Towards a Computational Framework for Function-Driven Concept Invention. Joint with
N. Potyka, D. and K.-U. Kuehnberger. In Lecture Notes in Artificial Intelligence 9782, Steunebrink et al. (Eds.). Springer International Publishing Switzerland. 2016. pdf
• The Role of Blending in mathematical invention. Joint with F. Bou, M. Schorlemmer, J.
Corneli, E. Maclein and A. Smaill and A Pease. Proceedings of the Sixth International Conference on Computational Creativity (ICCC). S. Colton et al., eds. Park City, Utah, June 29-July 2, 2015.
Publisher: Brigham Young University, Provo, Utah. pp. 55-62. 2015. pdf
• Conceptual Blending as a meta-generator of mathematical concepts: Prime Ideals
and Dedekind Domains as a Blend. In T. Besold, K.-U. Kuehnberger, M. Schorlemmer and Alan Smaill (eds.). Kuehnberger K.-U., Koenig P. and Walter, S. (series eds.). Proceedings of the workshop on
Computational Creativity, Concept Invention, and General Intelligence 2015, C3GI. Institute of Cognitive Sciences. Publications of the Institute of Cognitive Sciences, Osnabrueck, PICS series Vol. 2,
2015. pdf
Complete Volume pdf.
• On the Connectedness of the Spectrum of Forcing Algebras. Joint with Holger Brenner,
Revista Colombiana de Matematicas. Vol 48(2014)1, Pag. 1-19. pdf
“Truth is ever to be found in the simplicity, and not in the multiplicity and confusion of things”
-Issac Newton-
“The most incomprehensible thing about the universe is that it is comprehensible”
-Albert Einstein-
• General Introduction to the Artificial Mathematical Intelligence Program. In Artificial Mathematical
Intelligence: Cognitive, Metamathematical, Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• General Considerations for the New Cognitive Foundations' Program. In Artificial Mathematical
Intelligence: Cognitive, Metamathematical, Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• Towards the (Cognitive) Reality of Mathematics and the Mathematics of (Cognitive) Reality. In
Artificial Mathematical Intelligence: Cognitive, Metamathematical, Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• The Physical Numbers. In Artificial Mathematical Intelligence: Cognitive, Metamathematical,
Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• Dathematics. In Artificial Mathematical Intelligence: Cognitive, Metamathematical, Physical and
Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear. ArxivPreprint
• Conceptual Blending in Mathematical Creation/Invention. In Artificial Mathematical Intelligence:
Cognitive, Metamathematical, Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• Formal Analogical Reasoning in Concrete Mathematical Research. In Artificial Mathematical
Intelligence: Cognitive, Metamathematical, Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• Conceptual Substratum. In Artificial Mathematical Intelligence: Cognitive, Metamathematical,
Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• Global Taxonomy of the most Fundamental Cognitive Mechanisms used in Mathematical
Creation/Invention. In Artificial Mathematical Intelligence: Cognitive, Metamathematical, Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind',
Springer-Verlag. To appear.
• Meta-Modeling of Classic and Modern Mathematical Proofs and Concepts. In Artificial
Mathematical Intelligence: Cognitive, Metamathematical, Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• The most Outstanding Challenges towards Global AMI and its Plausible Extensions. In Artificial
Mathematical Intelligence: Cognitive, Metamathematical, Physical and Philosophical Foundations. Danny A. J. Gomez-Ramirez. Series 'Mathematics in Mind', Springer-Verlag. To appear.
• Formal Conceptual Blending in the (Co-)Invention of (Pure) Mathematics. Joint with Alan
Smaill. In Confalonieri R., Pease A., Schorlemmer M. eds. Concept Invention: Foundations, Implementations, Social Aspects, and Applications. In Cognitive Technologies (Series). Springer, 2018. Link.
• Concept Invention in DOL: Evaluating Consistency and Conflict Resolution. Joint with
M. Codescu, F. Neuhaus, T. Mossakoski and O. Kutz. In Confalonieri R., Pease A., Schorlemmer M. eds. Concept Invention: Foundations, Implementations, Social Aspects, and Applications. In Cognitive
Technologies (Series). Springer, 2018. Link.
• Artificial Mathematical Intelligence: Cognitive, Metamathematical, Physical and Philosophical
Foundations. Springer-Verlag. Cham, 2020.
• Cuantum Mechanics, Cuantum Computing and Heat Computing: An Introduction.
'Joint work with J.D. Vélez and J. P. Hernandez. First Initial Preprint.
• A Modern View of Relativity: A Rigorous Introduction to Mathematicians. Joint with J. D. Vélez,
C. Arias and A. Quintero 2019 PreprintLastVersion. Initial Online Preprint.
• The Prime Number Theorem and some Equivalences. Diplom's Monograph, National
University of Colombia, September 2004. In Spanish. pdf
• Hilbert's Tenth Problem and some related Questions. Master's Monograph, National
University of Colombia, September 2007. In Spanish. pdf
• Homological Conjectures, Closure Operations, Forcing Algebras and Vector Bundles.
Thesis, National University of Colombia in association with the University of Osnabrueck, September 2013. pdf
• "Fundamental Pillars of the Creation of Artificial Co-Creative Mathematical Agents of Artificial Mathematical Intelligence". Talk given at the Colloquium of the University of Antioquia. May 24,
• "Dathematics: a Meta-Isomorphic Version of Classic Mathematics based on Proper
Classes". Talk given at the Logic Colloquium (Annual European summer meeting of the Association of Symbolic Logic). Stockholm, Sweden. August 14, 2017.
• "Cognitively-inspired Formal Models of Scientific Creation". J6’ Spring School for Studies
on Intelligence and Cognition. The Joint Exploratory Society for Interdisciplinary and
Cognitive Studies (JESICS). Cairo, Egypt. March 30, 2017.
• “Towards a General Taxonomy and Meta-Formalitazion of the Seminal Cognitive Me-
chanisms used in Mathematical Concept Invention”. Talk given at the Collouium of the Theory and Logic Group of the Faculty of Informatics of the Vienna University of Technology. Vienna, Austria.
November 23, 2016.
• “A cognitively-Inspired Reformulation of Meta-Mathematics”. Zif Workshop From Compu-
tational Creativity To Creativity Science. ZIF Center for Interdisciplinary Research (Bielefeld) in Coo- peration with the University of Osnabrueck, Germany. September 22, 2016.
• “Towards a Cognitively Inspired Physical Philosophy of Nature”. Talk at the Institute of Phi-
losophy of the University of Antioquia, Medellın, Colombia. September 9 of 2016.
• “Logic-Categorical Meta-Models of the Conceptual Creation in Mathematics” (Original
Title: Meta-Modelos Logico-Categoricos de la Creacion Conceptual en Matematicas). Talk given at the Logic Seminar of the University of Los Andes. Bogota, Colombia. September 25 of 2016.
• “Towards the Classification of the Metagenerators of Mathematical Theories: Formal
Conceptual Blending”. Seminar of Logic and Computation. EAFIT University. Medellin. December 16 of 2016.
• “The Reality of Mathematics and the Mathematics of Reality” (Original title: La realidad de
las matematicas y las matematicas de la realidad). Public Library Piloto BPP. Conference open to the general public in Medellın. August 24 of 2015.
• “Conceptual Blending as a meta-generator of mathematical concepts: Prime Ideals
and Dedekind Domains as a Blend”. Contributing speaker at the 5-th World Congress on Universal Logic, UNILOG’15. Istanbul, Turkey. Juni 26 of 2015.
• Main Speaker and Organizer of the 1-day Workshop ”Toward the Fundamental Princi-
ciples of Mathematical Creativity: A Cognitive Perspective” (Original Title: Hacia los Principios Funda-
mentales de la Creatividad Matem ́atica: Un Enfoque Cognitivo). National University of Colombia in Medellin. December 12 of 2014.
• “Toward a Meta-mathematization of Mathematical Creation”. Speaker and co-organizer of
the 1-day workshop ”Toward the Main Formal Pinciples of Mathematical Creativity”, University of Osnabrueck, October 11 of 2014.
• “A Normality Criterion for Forcing Algebras over the Ring of Polynomials”. Gradurierte
Kollege Kombinatorische Strukturen in Algebra und Topologie. University of Osnabrueck, December 18 of 2012.
• “On the Connectedness of Forcing Schemes”. Algebra and Geometry Seminar. University of
Basel. Juni 1 of 2012.
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PPT - Closest Pair and Convex Hull: Brute Force Approach
Closest Pair and Convex Hull: Brute Force Approach
Closest Pair Problem in 2D involves finding the two closest points in a set by computing the distance between every pair of distinct points. The Convex Hull Problem determines the smallest convex
polygon covering a set of points. Dr. Sasmita Kumari Nayak explains these concepts using a brute-force algorithm approach.
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1. Closest Pair and Convex-Hull By Brute Force Approach Dr. Sasmita Kumari Nayak Computer Science & Engineering
2. Closest Pair Problem It finds the two closest points in a set of n points in a given plane. Brute-force algorithm: Compute the distance between every pair of distinct points and return the
indexes of the points for which the distance is the smallest.
3. Cont Consider the 2D case of the closest pair problem. The points are specified in a (x, y) coordinates and the distance between two points pi(xi, yi) and pj(xj, yj) is the Euclidean distance.
Here, it computes the distance between each pair of distinct points and finds a pair with the smallest distance.
5. Algorithm of Closest Pair Algorithm BruteForceClosestPoints(P) // P is list of points dmin for i 1 to n-1 do for j i+1 to n do d sqrt((xi-xj)2+ (yi-yj)2) if d < dmin then dmin d; index1 i; index2
j return index1, index2
6. Time Complexity of Closest Pair Problem
7. Convex-Hull Problem Polygon is called as a convex polygon if the angle between any of its two adjacent edges is always less than 180. Otherwise, it is called as a concave polygon. Convex hull is
the smallest region covering given set of points. Complex polygons are self-intersecting polygons.
8. Cont... Definition: the convex hull of a set S of points is the smallest convex set containing requirement means that the convex hull of S must be a subset of any convex set containing S.) S. (
The smallest
10. Example-2 Given a set of points in the plane. The convex hull of the set is the smallest convex polygon that contains all the points of it.
12. Procedure of Convex-Hull Problem At first, we need to find the points that will serve as the vertices of the polygon that is called as extreme points. An extreme point of a convex set is a point
of this set that is not a middle point of any line segment with endpoints in the set. Solving the convex-hull problem in a brute-force manner is a simple but inefficient algorithm. To find convex
polygon such that all points are either on the boundary or within the polygon.
13. Time complexity The running time to find the convex polygon by using the Brute-force approach is O(.n3)
14. Brute-Force Strengths and Weaknesses Strengths wide applicability simplicity yields reasonable algorithms for some important problems (e.g., matrix multiplication, sorting, searching, string
matching) Weaknesses rarely yields efficient algorithms some brute-force algorithms are unacceptably slow not as constructive as some other design techniques
15. Exhaustive Search A brute force solution to a problem involving search for an element with a special combinatorial objects such as permutations, combinations, or subsets of a set. Ex: Travelling
Salesman Problem, Knapsack problem, Assignment Problem. property, usually among Method: generate a list of all potential solutions to the problem in a systematic manner evaluate potential
solutions one by one, disqualifying infeasible ones and, for an optimization problem, keeping track of the best one found so far when search ends, announce the solution(s) found | {"url":"https://www.slideorbit.com/slide/closest-pair-and-convex-hull-brute-force-approach/154389","timestamp":"2024-11-13T22:42:35Z","content_type":"text/html","content_length":"68264","record_id":"<urn:uuid:56a62b41-eb9e-403c-8e76-485f0e202e34>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00249.warc.gz"} |
How to convert world coordinates to local coordinates, relative to forward vector?
Hello, I'm trying to find an optimal solution for a problem I'm having. I'm looking for 3D solution.
I'm trying to convert a Point from world coordinates to the local coordinates of a specific Object, so that when the Object rotates/changes position, I would still be able to find new coordinate for
the Point, that would be have the same relation (position/rotation) to the Object as during the initial state.
I have the Forward, Right, Up vector of this Object; its position in world coordinates as well as the position of the Point in world coordinates.
I was thinking I could get the difference in angles between the vector from Object's origin to Point's origin and the Forward vector of the Object, and then rotate by this difference from the Forward
vector backwards to get the direction of the desired vector, but that is unreliable. So I'm assuming I will have to use some transformation Matrix for this.
What are the maths for my exact situation, where I'd be using a custom axis (the Forward vector of my object) and to do the conversion from world to local coordinates?
07-22-2023 10:05 AM | {"url":"https://communities.sas.com/t5/Graphics-Programming/How-to-convert-world-coordinates-to-local-coordinates-relative/td-p/885940","timestamp":"2024-11-11T01:42:06Z","content_type":"text/html","content_length":"209099","record_id":"<urn:uuid:82fc48a5-200d-45c5-9684-2fffa6318cf5>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00623.warc.gz"} |
Exponential cumulative distribution function
p = expcdf(x) returns the cumulative distribution function (cdf) of the standard exponential distribution, evaluated at the values in x.
p = expcdf(x,mu) returns the cdf of the exponential distribution with mean mu, evaluated at the values in x.
[p,pLo,pUp] = expcdf(x,mu,pCov) also returns the 95% confidence interval [pLo,pUp] of p when mu is an estimate with variance pCov.
[p,pLo,pUp] = expcdf(x,mu,pCov,alpha) specifies the confidence level for the confidence interval [pLo pUp] to be 100(1–alpha)%.
___ = expcdf(___,'upper') returns the complement of the cdf, evaluated at the values in x, using an algorithm that more accurately computes the extreme upper-tail probabilities than subtracting the
lower tail value from 1. 'upper' can follow any of the input argument combinations in the previous syntaxes.
Standard Exponential Distribution cdf
Compute the probability that an observation in the standard exponential distribution falls in the interval [1 2].
p = expcdf([1 2]);
p(2) - p(1)
Compute Exponential cdf
The median of the exponential distribution is µ*log(2).
Confirm the median by computing the cdf of µ*log(2) for several different choices of µ.
mu = 10:10:60;
p = expcdf(log(2)*mu,mu)
p = 1×6
0.5000 0.5000 0.5000 0.5000 0.5000 0.5000
The cdf of the mean is always equal to 1-1/e (~0.6321).
Confirm the result by computing the exponential cdf of the mean for means one through six.
mu = 1:6;
x = mu;
p = expcdf(x,mu)
p = 1×6
0.6321 0.6321 0.6321 0.6321 0.6321 0.6321
Confidence Interval of Exponential cdf Value
Find a confidence interval estimating the probability that an observation is in the interval [0 1] using exponentially distributed data.
Generate a sample of 1000 random numbers drawn from the exponential distribution with mean 5.
rng('default') % For reproducibility
x = exprnd(5,1000,1);
Estimate the mean with a confidence interval.
Estimate the variance of the mean estimate.
[~,nCov] = explike(muhat,x)
Create the confidence interval estimating the probability an observation is in the interval [0 1].
[p,pLo,pUp] = expcdf(1,muhat,nCov);
pCi = [pLo; pUp]
expcdf calculates the confidence interval using a normal approximation for the distribution of the log estimate of the mean. Compute a more accurate confidence interval for p by evaluating expcdf on
the confidence interval muci.
The bounds pCi2 are reversed because a lower mean makes the event more likely and a higher mean makes the event less likely.
Complementary cdf (Tail Distribution)
Determine the probability that an observation from the exponential distribution with mean 1 is in the interval [50 Inf].
expcdf(50,1) is nearly 1, so p1 becomes 0. Specify 'upper' so that expcdf computes the extreme upper-tail probabilities more accurately.
p2 = expcdf(50,1,'upper')
Input Arguments
x — Values at which to evaluate cdf
nonnegative scalar value | array of nonnegative scalar values
Values at which to evaluate the cdf, specified as a nonnegative scalar value or an array of nonnegative scalar values.
• To evaluate the cdf at multiple values, specify x using an array.
• To evaluate the cdfs of multiple distributions, specify mu using an array.
If either or both of the input arguments x and mu are arrays, then the array sizes must be the same. In this case, expcdf expands each scalar input into a constant array of the same size as the array
inputs. Each element in p is the cdf value of the distribution specified by the corresponding element in mu, evaluated at the corresponding element in x.
Example: [3 4 7 9]
Data Types: single | double
mu — Mean
1 (default) | positive scalar value | array of positive scalar values
Mean of the exponential distribution, specified as a positive scalar value or an array of positive scalar values.
• To evaluate the cdf at multiple values, specify x using an array.
• To evaluate the cdfs of multiple distributions, specify mu using an array.
If either or both of the input arguments x and mu are arrays, then the array sizes must be the same. In this case, expcdf expands each scalar input into a constant array of the same size as the array
inputs. Each element in p is the cdf value of the distribution specified by the corresponding element in mu, evaluated at the corresponding element in x.
Example: [1 2 3 5]
Data Types: single | double
pCov — Variance of Mean Estimate
positive scalar value
Variance of the estimate of mu, specified as a positive scalar value.
You can estimate mu from data by using expfit or mle. You can then estimate the variance of mu by using explike. The resulting confidence interval bounds are based on a normal approximation for the
distribution of the log of the mu estimate. You can get a more accurate set of bounds by applying expcdf to the confidence interval returned by expfit. For an example, see Confidence Interval of
Exponential cdf Value.
Example: 0.10
Data Types: single | double
alpha — Significance level
0.05 (default) | scalar in the range (0,1)
Significance level for the confidence interval, specified as a scalar in the range (0,1). The confidence level is 100(1–alpha)%, where alpha is the probability that the confidence interval does not
contain the true value.
Example: 0.01
Data Types: single | double
Output Arguments
p — cdf values
scalar value | array of scalar values
cdf values evaluated at x, returned as a scalar value or an array of scalar values. p is the same size as x and mu after any necessary scalar expansion. Each element in p is the cdf value of the
distribution specified by the corresponding element in mu, evaluated at the corresponding element in x.
pLo — Lower confidence bound for p
scalar value | array of scalar values
Lower confidence bound for p, returned as a scalar value or an array of scalar values. pLo has the same size as p.
pUp — Upper confidence bound for p
scalar value | array of scalar values
Upper confidence bound for p, returned as a scalar value or an array of scalar values. pUp has the same size as p.
More About
Exponential cdf
The exponential distribution is a one-parameter family of curves. The parameter μ is the mean.
The cdf of the exponential distribution is
$p=F\left(x|u\right)=\underset{0}{\overset{x}{\int }}\frac{1}{\mu }{e}^{\frac{-t}{\mu }}dt=1-{e}^{\frac{-x}{\mu }}.$
The result p is the probability that a single observation from the exponential distribution with mean μ falls in the interval [0, x]. A common alternative parameterization of the exponential
distribution is to use λ defined as the mean number of events in an interval as opposed to μ, which is the mean wait time for an event to occur. λ and μ are reciprocals.
For more information, see Exponential Distribution.
Alternative Functionality
• expcdf is a function specific to the exponential distribution. Statistics and Machine Learning Toolbox™ also offers the generic function cdf, which supports various probability distributions. To
use cdf, create an ExponentialDistribution probability distribution object and pass the object as an input argument or specify the probability distribution name and its parameters. Note that the
distribution-specific function expcdf is faster than the generic function cdf.
• Use the Probability Distribution Function app to create an interactive plot of the cumulative distribution function (cdf) or probability density function (pdf) for a probability distribution.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
GPU Arrays
Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™.
This function fully supports GPU arrays. For more information, see Run MATLAB Functions on a GPU (Parallel Computing Toolbox).
Version History
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Kilograms per Square Meter to Pounds per Square Inch Conversion
Kilograms per Square Meter to Pounds per Square Inch Converter
Enter the pressure in kilograms per square meter below to convert it to pounds per square inch.
Result in Pounds per Square Inch:
Do you want to convert pounds per square inch to kilograms per square meter?
How to Convert Kilograms per Square Meter to Pounds per Square Inch
To convert a measurement in kilograms per square meter to a measurement in pounds per square inch, multiply the pressure by the following conversion ratio: 0.001422 pounds per square inch/kilogram
per square meter.
Since one kilogram per square meter is equal to 0.001422 pounds per square inch, you can use this simple formula to convert:
pounds per square inch = kilograms per square meter × 0.001422
The pressure in pounds per square inch is equal to the pressure in kilograms per square meter multiplied by 0.001422.
For example,
here's how to convert 500 kilograms per square meter to pounds per square inch using the formula above.
pounds per square inch = (500 kgf/m² × 0.001422) = 0.711168 psi
Kilograms per square meter and pounds per square inch are both units used to measure pressure. Keep reading to learn more about each unit of measure.
What Are Kilograms per Square Meter?
One kilogram per square meter is the pressure of equal to one kilogram-force per square meter.
The kilogram per square meter is a non-SI metric unit for pressure. A kilogram per square meter is sometimes also referred to as a kilogram per square metre or kilogram-force per meter square.
Kilograms per square meter can be abbreviated as kgf/m²; for example, 1 kilogram per square meter can be written as 1 kgf/m².
In the expressions of units, the slash, or solidus (/), is used to express a change in one or more units relative to a change in one or more other units.^[1] For example, kgf/m² is expressing a
change in weight relative to a change in area.
The unit is deprecated and not permitted for use with SI units.
Kilograms per square meter can be expressed using the formula:
1 kgf/m^2 = 1 kgf / m^2
Pressure in kilograms per square meter are equal to the kilogram-force divided by the area in square meters.
Learn more about kilograms per square meter.
What Are Pounds per Square Inch?
One pound per square inch is the pressure of equal to one pound-force per square inch.
The pound per square inch is a US customary and imperial unit of pressure. A pound per square inch is sometimes also referred to as a pound-force per square inch. Pounds per square inch can be
abbreviated as psi; for example, 1 pound per square inch can be written as 1 psi.
PSI can be expressed using the formula:
1 psi = 1 lbf / in^2
Pressure in pounds per square inch are equal to the pound-force divided by the area in square inches.
Learn more about pounds per square inch.
Kilogram per Square Meter to Pound per Square Inch Conversion Table
Table showing various kilogram per square meter
measurements converted to pounds per square inch.
Kilograms Per Square Meter Pounds Per Square Inch
1 kgf/m² 0.001422 psi
2 kgf/m² 0.002845 psi
3 kgf/m² 0.004267 psi
4 kgf/m² 0.005689 psi
5 kgf/m² 0.007112 psi
6 kgf/m² 0.008534 psi
7 kgf/m² 0.009956 psi
8 kgf/m² 0.011379 psi
9 kgf/m² 0.012801 psi
10 kgf/m² 0.014223 psi
20 kgf/m² 0.028447 psi
30 kgf/m² 0.04267 psi
40 kgf/m² 0.056893 psi
50 kgf/m² 0.071117 psi
60 kgf/m² 0.08534 psi
70 kgf/m² 0.099564 psi
80 kgf/m² 0.113787 psi
90 kgf/m² 0.12801 psi
100 kgf/m² 0.142234 psi
200 kgf/m² 0.284467 psi
300 kgf/m² 0.426701 psi
400 kgf/m² 0.568934 psi
500 kgf/m² 0.711168 psi
600 kgf/m² 0.853402 psi
700 kgf/m² 0.995635 psi
800 kgf/m² 1.1379 psi
900 kgf/m² 1.2801 psi
1,000 kgf/m² 1.4223 psi
1. National Institute of Standards and Technology, NIST Guide to the SI, Chapter 6: Rules and Style Conventions for Printing and Using Units, https://www.nist.gov/pml/special-publication-811/
More Kilogram per Square Meter & Pound per Square Inch Conversions | {"url":"https://www.inchcalculator.com/convert/kilogram-per-square-meter-to-pound-per-square-inch/","timestamp":"2024-11-07T04:32:06Z","content_type":"text/html","content_length":"70838","record_id":"<urn:uuid:9caa4dcc-0165-44db-93bf-1c6e233e16e0>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00164.warc.gz"} |
2003 MaxMarginMarkovNetworks
Subject Headings: Max-Margin Markov Networks.
Cited by
In typical classification tasks, we seek a function which assigns a label to a single object. Kernel-based approaches, such as support vector machines (SVMs), which maximize the margin of confidence
of the classifier, are the method of choice for many such tasks. Their popularity stems both from the ability to use high-dimensional feature spaces, and from their strong theoretical guarantees.
However, many real-world tasks involve sequential, spatial, or structured data, where multiple labels must be assigned. Existing kernel-based methods ignore structure in the problem, assigning labels
independently to each object, losing much useful information. Conversely, probabilistic graphical models, such as Markov networks, can represent correlations between labels, by exploiting problem
structure, but cannot handle high-dimensional feature spaces, and lack strong theoretical generalization guarantees. In this paper, we present a new framework that combines the advantages of both
approaches: Maximum margin Markov (M^3) networks incorporate both kernels, which efficiently deal with high-dimensional features, and the ability to capture correlations in structured data. We
present an efficient algorithm for learning M3 networks based on a compact quadratic program formulation. We provide a new theoretical bound for generalization in structured domains. Experiments on
the task of handwritten character recognition and collective hypertext classification demonstrate very significant gains over previous approaches.
1 Introduction
In supervised classification, our goal is to classify instances into some set of discrete categories. Recently, support vector machines (SVMs) have demonstrated impressive successes on a broad range
of tasks, including document categorization, character recognition, image classification, and many more. SVMs owe a great part of their success to their ability to use kernels, allowing the
classifier to exploit a very high-dimensional (possibly even infinite-dimensional) feature space. In addition to their empirical success, SVMs are also appealing due to the existence of strong
generalization guarantees, derived from the margin-maximizing properties of the learning algorithm.
However, many supervised learning tasks exhibit much richer structure than a simple categorization of instances into one of a small number of classes. In some cases, we might need to label a set of
inter-related instances. For example: optical character recognition (OCR) or part-of-speech tagging both involve labeling an entire sequence of elements into some number of classes; image
segmentation involves labeling all of the pixels in an image; and collective webpage classification involves labeling an entire set of interlinked webpages. In other cases, we might want to label an
instance (e.g., a news article) with multiple non-exclusive labels. In both of these cases, we need to assign multiple labels simultaneously, leading to a classification problem that has an
exponentially large set of joint labels. A common solution is to treat such problems as a set of independent classification tasks, dealing with each instance in isolation. However, it is well-known
that this approach fails to exploit significant amounts of correlation information [7].
An alternative approach is offered by the probabilistic framework, and specifically by probabilistic graphical models. In this case, we can define and learn a joint probabilistic model over the set
of label variables. For example, we can learn a hidden Markov model, or a conditional random field (CRF) [7] over the labels and features of a sequence, and then use a probabilistic inference
algorithm (such as the Viterbi algorithm) to classify these instances collectively, finding the most likely joint assignment to all of the labels simultaneously. This approach has the advantage of
exploiting the correlations between the different labels, often resulting in significant improvements in accuracy over approaches that classify instances independently [7, 10]. The use of graphical
models also allows problem structure to be exploited very effectively. Unfortunately, even probabilistic graphical models that are trained discriminatively do not usually achieve the same level of
generalization accuracy as SVMs, especially when kernel features are used. Moreover, they are not (yet) associated with generalization bounds comparable to those of margin-based classifiers.
8 Discussion
We present a discriminative framework for labeling and segmentation of structured data such as sequences, images, etc. Our approach seamlessly integrates state-of-the-art kernel methods developed for
classification of independent instances with the rich language of graphical models that can exploit the structure of complex data. In our experiments with the OCR task, for example, our sequence
model significantly outperforms other approaches by incorporating high-dimensional decision boundaries of polynomial kernels over character images while capturing correlations between consecutive
characters. We construct our models by solving a convex quadratic program that maximizes the per-label margin. Although the number of variables and constraints of our QP formulation is polynomial in
the example size (e.g., sequence length), we also address its quadratic growth using an effective optimization procedure inspired by SMO. We provide theoretical guarantees on the average per-label
generalization error of our models in terms of the training set margin. Our generalization bound significantly tightens previous results of Collins [3] and suggests possibilities for analyzing
per-label generalization properties of graphical models. For brevity, we simplified our presentation of graphical models to only pairwise Markov networks. Our formulation and generalization bound
easily extend to interaction patterns involving more than two labels (e.g., higher-order Markov models). Overall, we believe that M3 networks will significantly further the applicability of high
accuracy margin-based methods to real-world structured data.
• [1] Y. Altun, I. Tsochantaridis, and T. Hofmann. Hidden markov support vector machines. In: Proceedings. ICML, 2003.
• [2] D. Bertsekas. Nonlinear Programming. Athena Scientific, Belmont, MA, 1999.
• [3] M. Collins. Parameter estimation for statistical parsing models: Theory and practice of distribution-free methods. In IWPT, 2001.
• [4] R.G. Cowell, A.P. Dawid, S.L. Lauritzen, and D.J. Spiegelhalter. Probabilistic Networks and Expert Systems. Springer, New York, 1999.
• [5] K. Crammer and Y. Singer. On the algorithmic implementation of multiclass kernelbased vector machines. Journal of Machine Learning Research, 2(5):265–292, 2001.
• [6] R. Kassel. A Comparison of Approaches to On-line Handwritten Character Recognition. PhD thesis, MIT Spoken Language Systems Group, 1995.
• [7] J. Lafferty, A. McCallum, and F. Pereira. Conditional random fields: Probabilistic models for segmenting and labeling sequence data. In: Proceedings. ICML01, 2001.
• [8] J. Pearl. Probabilistic Reasoning in Intelligent Systems. Morgan Kaufmann, 1988.
• [9] J. Platt. Using sparseness and analytic QP to speed training of support vector machines. In NIPS, 1999.
• [10] Ben Taskar, P. Abbeel, and D. Koller. Discriminative probabilistic models for relational data. In: Proceedings. UAI02, Edmonton, Canada, 2002.
• [11] V. N. Vapnik. The Nature of Statistical Learning Theory. Springer-Verlag, New York, 1995.
• [12] J. Yedidia, W. Freeman, and Yair Weiss. Generalized belief propagation. In NIPS, 2000.
• [13] T. Zhang. Covering number bounds of certain regularized linear function classes. Journal of Machine Learning Research, 2:527–550, 2002. | {"url":"https://www.gabormelli.com/RKB/2004_MaxMarginMarkovNetworks","timestamp":"2024-11-06T23:57:15Z","content_type":"text/html","content_length":"57013","record_id":"<urn:uuid:b4e7b168-28b3-49f4-8a2b-25a865337471>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00659.warc.gz"} |
The NBA and MLB trees are isomorphic
An isomorphism is a structure-preserving function from one object to another. In the context of graphs, an isomorphism is a function that maps the vertices of one graph onto the vertices of another,
preserving all the edges.
So if G and H are graphs, and f is an isomorphism between G and H, nodes x and y are connected in G if and only if nodes f(x) and f(y) are connected in H.
There are 30 basketball teams in the National Basketball Association (NBA) and 30 baseball teams in Major League Baseball (MLB). That means the NBA and MLB are isomorphic as sets, but it doesn’t
necessarily mean that the hierarchical structure of the two organizations are the same. But in fact the hierarchies are the same.
Both the NBA and MLB have two top-level divisions, each divided into three subdivisions, each containing five teams.
Basketball has an Eastern Conference and a Western Conference, whereas baseball has an American League and a National League. Each basketball conference is divided into three divisions, just like
baseball leagues, and each division has five teams, just as in baseball. So the tree structures of the two organizations are the same.
In the earlier post about the MLB tree structure, I showed how you could number baseball teams so that the team number n could tell you the league, division, and order within a division by taking the
remainders when n is divided by 2, 3, and 5. Because the NBA tree structure is isomorphic, the same applies to the NBA.
Here’s a portion of the graph with numbering. The full version is available here as a PDF.
Here’s the ordering.
1. Los Angeles Clippers
2. Miami Heat
3. Portland Trail Blazers
4. Milwaukee Bucks
5. Dallas Mavericks
6. Brooklyn Nets
7. Los Angeles Lakers
8. Orlando Magic
9. Utah Jazz
10. Chicago Bulls
11. Houston Rockets
12. New York Knicks
13. Phoenix Suns
14. Washington Wizards
15. Denver Nuggets
16. Cleveland Cavaliers
17. Memphis Grizzlies
18. Philadelphia 76ers
19. Sacramento Kings
20. Atlanta Hawks
21. Minnesota Timberwolves
22. Detroit Pistons
23. New Orleans Pelicans
24. Toronto Raptors
25. Golden State Warriors
26. Charlotte Hornets
27. Oklahoma City Thunder
28. Indiana Pacers
29. San Antonio Spurs
30. Boston Celtics
Incidentally, the images at the top of the post were created with DALL-E. They look nice overall, but you’ll see bizarre details if you look too closely.
One thought on “The NBA and MLB trees are isomorphic”
1. Nice post! The Wall-E graph generation is a neat feature. I’m left wondering which NBA team corresponds to post 10. | {"url":"https://www.johndcook.com/blog/2023/03/07/nba-mlb/","timestamp":"2024-11-12T06:54:10Z","content_type":"text/html","content_length":"52410","record_id":"<urn:uuid:8be643fe-76ef-46c5-9d14-b15d98272a04>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00145.warc.gz"} |
Knowing These 5 Tricks Will Produce Your Multiplication Graph Look Amazing
A multiplication graph is actually a framework that arranges amounts in a style that facilitates the procedure of reproduction. It can easily assist little ones understand as well as remember
reproduction realities.
When utilizing a multiplication graph, youngsters need to start with the reduced, quick and easy reproduction truths that they may effortlessly recollect or compute by awaiting on their fingers.
Then, they may work their technique up through the upper times dining tables.
Lower Moments Tables
When trainees are discovering multiplication realities they frequently begin with the lesser times tables. These are the ones that possess varieties 1 to 10, operating flat as well as vertically on
the chart. Once a pupil understands every one of these they prepare to go on to the next set of reproduction tables. multiplication fact chart
As you advance by means of the lower multiplication graphes it is vital to pay attention to one line or even column each time. This will definitely make the procedure of committing to memory these
realities much less mind-boggling and also much easier to accomplish. Eventually, you will definitely possess the entire lower multiplication dining table committed to memory as well as manage to use
it to real life issues.
It is likewise valuable to recognize that multiplication is actually simply redoed enhancement. Therefore, as you study each variety on the graph try to find styles in skip-counting. If you notice
that an amount is actually multiplied through the same amount again and again once again, this will definitely aid to make it easier for you to bear in mind.
Yet another way to make researching the multiplication chart extra intriguing is actually to participate in activities. There are a lot of different games that you may make use of to make remembering
the lower reproduction tables exciting. For example, you can participate in a game where each player jots down a number on a paper and afterwards discovers the amount on the multiplication table that
provides the exact same product. The initial individual to discover the right solution succeeds that round.
Upper Times Tables
Whether your youngster is actually knowing multiplication as component of elementary university arithmetic, or you are actually attempting to improve their abilities in the home, making use of a
multiplication graph is an important measure. It is a wonderful tool for helping little ones remember the times tables and likewise assists all of them learn more about multiplication trends. Having
a sturdy understanding of multiplication is actually a vital foundation for even more innovative arithmetic subjects such as branch and also fragments.
The multiplication chart provides the reproduction facts in a means that is actually easy for kids to comprehend and also remember. The numbers 1 through 12 run both horizontally and also up and down
on the graph and also each variety is worked with through its own matching character. Children may easily locate the product of pair of numbers on a multiplication chart by deciding on the very first
amount coming from the left cavalcade and after that relocate down one line and also throughout the leading row until they achieve the rightmost column where the second number is located. The product
of the 2 varieties is actually after that detailed by the end of the rightmost row.
Many youngsters can learn their times dining tables by utilizing typical rote memory procedures, but numerous have problem with the upper times dining tables. This is actually where mnemonic mind
devices enter play as these can help youngsters learn the top opportunities dining tables a lot faster than they would certainly along with traditional routine memory.
A multiplication chart is actually a valuable resource that may aid students comprehend exactly how multiplication works. Students may use the graph to view patterns and also pinpoint faster ways to
increasing amounts. They can likewise perform their psychological calculations along with the graph. This may assist to boost their mental arithmetic skills and build self-confidence in reproduction.
Making use of the reproduction graph is extremely straightforward. You simply need to have to find the amount you desire to multiply on the chart and after that follow the row as well as cavalcade
till you get to the factor where the amounts meet. For instance, if you desire to locate the item of 7 x 6, you will start with 7 in the leading row and after that 6 in the remaining row. After that,
you would certainly trace a fictional line below 7 and throughout from 6 to where they converge on the graph. This will offer you the response – 42.
The secret to making use of the multiplication graph is actually to know the styles and also residential properties that comprise each row and row. This will certainly help you to consider your
reproduction truths and also will lessen the amount of time it takes for you to perform an estimate. This may be specifically valuable for pupils that have issue memorizing their multiplication
dining tables. Having a sturdy knowledge of reproduction may decrease the necessity for youngsters to count on personal digital assistants or various other computation devices as well as can likewise
aid to enhance their IQ scores.
Understanding multiplication can easily be actually difficult for students, especially when the procedure feels challenging or even overwhelming. Malfunctioning the multiplication table right into
smaller sized, even more controllable parts can help trainees build their peace of mind and also approach proficiency of this mathematical principle. This is specifically significant for much younger
students that are actually still developing their theoretical understanding of multiplication.
As an example, several trainees find it very easy to consider the lower times desks (reproduction simple facts with 1 as their first number) and also the ones numbers of the table of 10. When they
stumble upon more demanding varieties, like 6 x 14, they can utilize strategies including factoring or even the distributive residential property to malfunction this concern into easier elements.
At that point, they can easily use the reproduction graph to find these component of the issue and fill in their responses. Lastly, they can easily locate the product of the numbers by finding the
area on the multiplication framework where the line and also row intersect (for instance, 7 x 8 = 56).
By utilizing an assortment of methods for completing their reproduction graphes, pupils can easily gain a deeper visionary understanding of the multiplication process, as opposed to simply memorizing
the basic protocol. This enables all of them to relocate coming from a procedural version of reproduction (such as miss checking through fives) to an extra intellectual one (like comprehending that 7
teams of eight traits coincide as pair of teams of 8 plus five groups of 8). This likewise moves them coming from visual rectangular symbols to an extra conceptual area design for department. | {"url":"http://leemeadmusic.com/knowing-these-5-tricks-will-produce-your-multiplication-graph-look-amazing/","timestamp":"2024-11-08T12:41:16Z","content_type":"text/html","content_length":"56790","record_id":"<urn:uuid:c60d96ba-d040-46df-b54e-a3dda82688db>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00504.warc.gz"} |
Locks and keys (solution)
In article <3gc6ad$7el@babyblue.cs.yale.edu> of rec.puzzles, David Moews <dmoews@fastmail.fm> wrote:
Assuming that (i) each lock is locked and unlocked by one key, of which there can be any number of copies, and that (ii) all locks need to be unlocked to open the box, the answer is: Use one lock for
each 7-person subset of the board, keys for each lock being given to all people in its subset. This uses C(12,7)=792 locks and 792*7=5544 keys.
Why is this minimum? If there was a lock for which 6 or fewer people had keys, the 6 or more people left over would be unable to open the box, which is unacceptable; so all locks must share their
keys among 7 or more people. Now if there was some 7-person group that was not the set of key owners for a lock, the group consisting of the remaining 5 people would be able to open the box, since it
shares at least one person with every other 7-person group, and every group with 8 people or more. Hence every 7-person group must be the set of key owners for some lock, which means that any
solution must use at least the locks and keys above (you can use more if you like.)
David Moews ( dmoews@fastmail.fm ) | {"url":"https://djm.cc/other-rp/locks-and-keys.s.html","timestamp":"2024-11-05T19:35:42Z","content_type":"text/html","content_length":"1952","record_id":"<urn:uuid:b44688dc-790c-4de7-bf00-72535f432c8f>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00021.warc.gz"} |
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On-line version ISSN 1991-1696
Print version ISSN 0038-2221
SAIEE ARJ vol.109 n.4 Observatory, Johannesburg Dec. 2018
Subtropical rain attenuation statistics on 12.6 GHz ku-band satellite link using Synthetic Storm Technique
B. O. Afolayan^I; T. J. Afullo^I; A. Alonge^II
^IDiscipline of Electrical, Electronic and Computer Engineering, University of KwaZulu-Natal Howard Campus, Durban, South Africa
^IIDepartment of Electrical and Electronic Engineering Technology, University of Johannesburg Doornfontein Campus, Johannesburg, South Africa
In this work, measured subtropical rain attenuation was compared with rain attenuation generated theoretically by the Synthetic Storm Technique (SST). The rain attenuation data was obtained from a
Ku-band satellite TV link collocated at the site of a rain rate measurement system in Durban, South Africa (28°87'S, 30°98'E). A mathematical model developed from the measurement campaign was used to
generate measured data for four years of rainfall. Annual cumulative distribution functions of SST prediction results are compared with the results of the measurement-based model. The results show
SST to be a fair approximation of actual measurements. This was established by error analysis carried out to compare the error margins in SST prediction and the error margins in the in-force ITU-R
prediction method. While the SST approach was shown to conform slightly less accurately to measurements than the ITU-R model, it still yields highly acceptable results in the 0 to 11 dB margin in
which the said link experiences most of the measured attenuation before total channel squelching occurs.
Keywords: Synthetic Storm Technique, slant path rain attenuation, subtropical rain.
Wireless communication deployment and planning requires site-specific link budgeting. Next to free space path loss, the most significant loss item to anticipate in a link budget is rain attenuation.
It is more severe than fading caused by any other hydrometeor. It becomes especially severe at frequencies of 5 GHz and above [1].
While the accuracy of measurement equipment for rain attenuation has greatly improved over the years, future planning still depends heavily on theoretical approaches because measurements can
realistically be carried out on a limited number of links - which implies a limited number of frequencies, path lengths, elevation angles and other specifics. Theoretical models, however, can be
applied more widely by slotting in hypothetical link parameters at the planning stage and simulating for any number of scenarios. It is especially useful to explore the reliability of theoretical
models by juxtaposing the results of their application with the results of measurement-based models. Furthermore, the performance of these models have been more widely explored in temperate and
tropical environments, while their performances in subtropical rain have been investigated to a much less degree. Rainfall intensity in the subtropics expectedly fall about midway between those of
temperate and tropical regions. Thus, it is expected presents its own unique patterns that requires some independent modelling.
The Synthetic Storm Technique (SST) is a theoretical approach for estimating rain fade and has been widely applied to both terrestrial and slant path links [2], [3], [4]. The term was first used in
[5] for describing a method by which data generated from a rain gauge is used to predict rain rate at a different location by using an estimate of cloud advection speed along the path between the two
locations. Matricciani [6] then used the concept to develop a novel mathematical method for an integral estimation of rain attenuation from rain rate records.
In this work, we have applied the same theoretical approach to estimate rain attenuation on a 12.6 GHz satellite TV link (fed from the Intelsat-20 satellite at 38,050 km, 68.5^oE) using rain rate
statistics amassed over a period of four years from disdrometer measurements at the same location as the satellite link [7]. The rain data has a slightly higher time resolution than usual, being a
record of rain rate at every 30 second interval of precipitation throughout the years in question. Early applications of this method utilized rain gauge data from temperate regions at 1-minute
integration time. Several studies of rain attenuation (e.g. [8], [9] and [10]) have been previously undertaken in Durban, South Africa. The present study aims at examining to what extent SST can be
used to validate slant path subtropical rain attenuation measurement carried out at the location.
On the satellite link described in the previous section, a model was earlier developed for measured attenuation using an equipment system located on the rooftop of the Electronic Engineering building
at the University of KwaZulu-Natal Howard Campus, Durban, South Africa at specific coordinates 28.87^oS, 30.98^oE. This measurement campaign represents the only one reported for a slant path
microwave link in the subtropical region that is of similar duration.
The system includes a downlink satellite receiver system and a rainfall measurement system. A schematic of the entire system is presented in Fig. 1. The Received Signal Level (RSL) over the satellite
link is monitored by a Rhodes and Schwarz FSH8 spectrum analyzer which conducts a sweep over the entire bandwidth every 60 seconds and also registers the overall channel power. The use of a spectrum
analyzer eliminates the necessity of using a special scintillation filter as the equipment is able to independently differentiate the rapid changes due to scintillation from signal attenuation. Rain
rate at the location is also captured for all events by a JossWaldvogel RD-80 impact-type disdrometer. The receive antenna and the disdrometer diaphragm are in close proximity (less than 4 m apart).
This ensures that all rainfall events experienced at the receive antenna is captured by the disdrometer.
Table 1 presents an outline of the satellite link budget. From this table, the receiver sensitivity is -71.7 dBm while the estimated received channel power is -61 dB. This implies that the measurable
attenuation margin on the link is about 10.7 dB. Beyond this point, the reception on the link is squelched out. An exhaustive presentation of the measurement process and the model thus developed is
outlined in [7].
The dynamic nature of rain forms has been explored by various investigators and certain patterns established. [5] showed that as a rain form passes over the rain gauge, advection speed and the rain
rate data can be used to convert the time it takes to pass over the rain gauge to distance. A fairly good reckoning of the rain rate distribution pattern over the distance can then be deduced. In
[11], the authors showed that over distances comparable to the length of most earth-satellite paths, there is a marked statistical consistency in rain rate patterns as the rain form spans the
distance. Drawing from these, [12] provided evidence that if the storm motion roughly aligns with the radio path, rain attenuation obtained from such a "synthetic path" will agree with actual
attenuation values. The SST was tested in earlier works using radar-derived values of storm speed that averaged about 10 m/s [13], [14], [15].
SST adopts a dual-layer model of the vertical profile of rain. From the ground to the zero-degree isotherm is labelled the "A" layer. This layer consists entirely of liquid precipitation and the rain
rate R in this layer is taken to be the same as the rain rate measured by the equipment on the ground. Above layer A lies layer B, which is essentially the melting layer, made up of both liquid
hydrometeors and ice in melting form. While layer A approaches up to 6 km, layer B is often estimated as lower than 0.5 km. It has been shown that the rain rate R[B ]in the layer B, (called the
apparent rain rate) is related to R as [6]
In [6], the author applied the old ITU-R method [16] for estimating the rain height. According to the recommendation, rain height at the layer B in figure 3 for any location at a latitude фabove 23^o
as is the case with our link, which is at elevation 36.5^o is given by
The thickness of layer B, h is taken as 0.4 km, hence the height H[A]is thus given by
For this work, we have adopted the in-force ITU-R model for rain height [17], which allots a value of 0.36 km to melting layer depth and hence estimates H[B] as
where h[0]is the height of the zero-degree isotherm and is equivalent to H[A]. 0.36 is the assumed melting layer depth in kilometres. The value of h[0]can be read for location from a map provided by
the ITU-R recommendation. The lengths L[A]and L[B]of the slant path up to the top of each layer are given by [17]
where вis the elevation angle and H[S]is the height of the location of the receiving antenna above sea level.
Applying the Olsen expression which gives the specific attenuation at x on the x-axis уas [18]
where k and a are frequency-dependent parameters for water at temperature 20^oC in [19]. The values of k and a for water at 0^oC are also presented in [19], which makes it possible for us to estimate
уfor layer B. уis defined as the attenuation per kilometre, hence, the attenuation along a path of length L km can be estimated by the integral
The fundamental idea behind the SST is that if there is a reliable level of isotropy of the rain medium at each layer, then the variation of attenuation with time is simulated by varying the point x
[0]at a time rate that equals the storm speed v such that x[0] = vt.
If Өis the elevation angle and ξis the slant path ordinate, then total signal attenuation in case of satellite path is obtained from specific attenuation at a point as the sum of attenuation in both
layers given by summing the integrals in (9) for each layer as (r being the ratio 3.134 of rain rate in both layers), [6]
Equation (9) has the basic form of a rectangular function with width L centred at the origin such that
Matricciani [6], presented a detailed mathematical process in which Fourier transform was applied to the rectangular function in the integrand of (10). Taking limits of the attenuation at time
instant t during which the rain spike has a value of R (mmh^-1) resulted in an estimate of the total attenuation experienced in the storm at all storm speeds given by [6]
Matricciani in [6] observed that the resulting limit in equation (11) implies that the long-term application of the SST for estimating attenuation is insensitive to storm speed v. Hence, we can
arrive at a reliable estimate of A(t) using a known, long term, measured rain rate time series R(t) as the prime time-varying input.
For this work, we generate measured data by utilizing a power law expression obtained for the average band of measured subtropical rain attenuation as reported in [7] given by
The measured data is modelled in bands because for every 0.5 mmh^-1 or 1 mmh^-1 rain rate bin, a range of attenuation values due to the bin is observed in the measurement. The mid-point of this range
was adopted as the average measured attenuation for the rain rate bin. A different model was obtained to represent the minimum, maximum and the average bands. Statistical analysis done in [7] shows
that the equation (12) bears an exact conformance to actual measurements of maximum rain attenuation on the link 89.9% of the time, making it an excellent representation of measured data. It is to be
noted that the measurement model was developed based on drops in RSL level observed during rain events that were captured by the disdrometer. Attenuation events due to precipitation at far away
locations were ignored as the interest is to model for subtropical rain. The table 1 presents Table 2 presents the link parameters adopted for the two-layer SST model calculations.
The results of the synthetic storm technique for Durban during a few high-intensity rain events from 2013 to 2016 are presented in the figures 4 to 8 along with the rain attenuation measured on the
12.6 GHz link during the rain event.
It should be noted that the measurement link used in this work had an equipment sensitivity of -71.7 dB as against a link budget that anticipates a received power of -61 dB under the best conditions.
This imposes a practical limitation of the system not being able to register rain attenuation levels beyond 10.7 dB. Above this level of fade, the link is completely squelched. The comparisons
reported in this work are thus restricted to rain spikes that produce a maximum SST rain fade similar to the maximum imposed by measurement equipment.
SST can be described more accurately as a long-term summation of the rain attenuation experienced on the link over an extended period of time [6]. Therefore, event-specific snapshots aggregated at
one storm speed may not be as accurate as the long-term data. Moreover, SST estimates tend to ignore the less significant fade instances brought about by low rain rate such as attenuation levels of 3
dB and below. In most of these instances, where the measured attenuation is 3 dB and below, SST often registers a flat fade level, only showing a spike when the measured value shoots significantly
above that threshold. Being essentially a summation of power law elements, the effect of layer B attenuation is slightly muted at low values of rain rate.
About the middle and upper ranges of rain intensity, SST results mostly agree with measured values. At the peak rain rates, it slightly overestimates the attenuation level but still gives a very good
approximation. The attenuation predicted in the layer "B" appears to make a significant contribution as the rain rates get higher. It is most pronounced in its effect on total path attenuation at the
peak rain rates. This is likely to be the reason for the slightly higher values seen in that range compared with actual values.
Even though the results of SST over the long term is the most significant, its performance over individual events is remarkably similar to measured fade levels.
The nature of the agreement between SST and measurement for location-sensitive considerations can be illustrated by the comparing between the annual attenuation exceedance trends as presented in
figures 8 to 11 for the years 2013 to 2016.
The peculiarity of annual exceedance probability patterns lies in the fact that they are heavily influenced by the rain pattern in each particular year. This is seen in the results in the figure.
2013 and 2014 were considered years of drought in Durban and the surrounding regions. The low volume of rain in these two years compared to 2015 and 2016 is responsible for the variation in the
annual exceedance patterns, which also suggests that SST is sensitive to the volume of rain over time.
Communication links are planned for availability during at least 99.99% of the year. This implies that the fade margin allowed must not be higher than the attenuation exceeded at 0.01% (A[0,01]) of
the year (or 87.6 hours) [1]. Table 4 compares the values of A[0,01]obtained for each year from measurement and from SST prediction. The results agree more in 2013 and 2014 than in 2015 and 2016 but
the overall result suggests that SST is a fairly credible estimate of the attenuation exceeded.
The figure 12 presents a direct juxtaposition of SST, measurement and the in-force ITU-R model for the 0 to 40 mm/h rain rate range. This range is chosen because it coincides with the rain rate range
in which the measurement on the link used attains the critical attenuation range of 0 to about 11 dB. In this range, we can obtain a more general gauge of the relevance of SST as a useful theoretical
tool for rain attenuation prediction in the subtropics can be obtained by error analyses, which holds the measured data as the expected value and the SST as the observed value. Both the Chi-square
test and root-mean-square error (RMSE) are estimated as in the equations (16) and (17). The RMSE error is a basic test of the deviation between an observed value and an estimated value. Chi-Square
Test is a slightly higher statistical test of hypothesis where the Chi-Square distribution is obeyed when the null hypothesis holds true. Picking a convenient confidence level, we estimate the
Chi-Square parameter between our observed value and the estimated value. The sample population forms our number of degrees of freedom. The Chi-Square value must lie well below the critical value on
the Chi-Square distribution table for that particular set of parameters, i.e. confidence level and degrees of freedom. When this is the case, we say the null hypothesis is accepted as true, which
implies that the statistical difference between the observed and the estimated values is not significant.
In equation (16), p[t]is the SST value of attenuation at a certain rain rate while a[t]is the value from the measured model at the same rain rate. For equation (17), we set O, the observed value as
the SST value of attenuation and E , the expected value as the value obtained from the measurement model.
The table 5 shows the results of the error analysis. The RMS error in SST is slightly higher than that of the ITU-R method but at 2.42, is still an acceptable margin of error. The Chi-Square test
also shows that the SST has a Chi-Square value of 38.13 as against 19.76 for the ITU-R model. Both fall well below the critical value of 163.051 but the ITU-R is clearly a more acceptable
hypothetical approach. It shows that the ITU-R model performs better as a prediction model than the SST model for this link but it can still be considered a very useful method in the rain fade
prediction process since it does not vary too far from observed levels in this critical range of rain rates.
8. CONCLUSION
The performance of the SST on Ku band satellite link using rainfall data from a subtropical location gives a strong indication that the SST is a reliable method for theoretical prediction of rain
attenuation on slant paths.
The efficacy of this method is well-documented for temperate regions [2], [3], [4] in a few cases for tropical regions [21], [22], and fewer still for the subtropics. The comparison of SST results
presented here is with the average bound of measured subtropical rain attenuation, where link impairment levels are most practical [7]. However, the ITU-R approach shows slightly more accuracy than
[1] International Telecommunications Union, "Propagation data and prediction methods required for the design of terrestrial line-of-sight systems". Recommendation 530-15, (07-2015)
[2] E. Matricciani, C. Riva and L. Castanet, "Performance of the synthetic storm technique in a low elevation 5^o slant path at 44.5GHz in the French Pyrenees", in Proceedings of the 1st European
Conference on Antennas and Propagation (EuCAP '06), Nice, France, November 2006
[3] C. Kourogiorgas, A.D. Panagopoulos, S.N. Livieratos and G.E. Chatzarakis, "Investigation of rain fade dynamics properties using simulated attenuation data with Synthetic Storm Technique" in
European Conference on Antennas and Propagation (EuCAP), Gothenburg, Sweden, pp 2277-2281, April 2013
[4] I. Sanchez-Lago, F.P. Fontan, P. Marino and U.C. Fiebig, "Validation of the Synthetic Storm Technique as part of a time series generator for satellite links". IEEE Antennas & Wireless Propagation
Letters, issue 6, pp 372 - 375, 2007
[5] G. Drufuca, "Rain attenuation statistics for frequencies above 10 GHz from rain gauge observations". J. Rech. Atmos., vol. 1, Issue 2, pp 399-411.1974 [ Links ]
[6] E. Matricciani, "Physical-mathematical model of the dynamics of rain attenuation based on rain rate time series and a two-layer vertical structure of precipitation". Radio Science, vol. 31, issue
2, pp 281-295, 1996. [ Links ]
[7] B.O. Afolayan, T.J. Afullo and A. Alonge, "Seasonal and annual analysis of slant path attenuation over a 12 GHz earth-satellite link in subtropical Africa". International Journal on
Communications and Antenna Propagation, in press, vol. 7, no. 7. 2017 [ Links ]
[8] A.A. Alonge and T.J. Afullo: "Seasonal Analysis and Prediction of Rainfall Effects in Eastern Southern Africa at Microwave Frequencies" Progress in Electromagnetic Research B, Vol. 40, pp.
279-303, 2012 [ Links ]
[9] M. Fashuyi and T. Afullo, "Rain attenuation and modelling for line-of-sight links on terrestrial paths in South Africa", Radio Science, vol.6, pp. 54-61, 2005 [ Links ]
[10] S. Malinga, P. Owolawi, and T. Afullo, "Computation of rain attenuation through scattering at microwave and millimetre bands in South Africa", in Progress in Electromagnetics Research Symposium,
Taipei, 2013.
[11] G. Drufuca and I. I. Zawadski, "Statistics of rain gauge data". Journal of Applied Meteorology, Issue 14, pp. 1419-1429, 1975.
[12] P.A. Watson, G. Papaioannou and J.C. Neves, "Attenuation and cross-polarisation measurements at 36 GHz on a terrestrial path". URSI Commission F Open Symposium, pp263-287,1977
[13] B.N. Harden, J.R. Norbury and W.J.K. White, "Model of intense convective rain cells for estimating attenuation on terrestrial millimetre radio links". Electronic letters, issue 10, pp483-484,
[14] G. Drufuca and R.R. Rogers, "Statistics of rainfall over paths from 1 to 50 km". Atmospheric Environment, issue 12, pp 2333 to 2342, 1978
[15] A.S. Frisch, B.B Stankov, B.E. Martner and J.C. Kaimal, "Mid-troposphere wind speed spectra from long term wind profiler measurements". Journal of Applied Meteorology. Issue 30, pp1346-1651,
[16] International Telecommunications Union, "Rain height model for prediction methods". Recommendation 839, 1992 [ Links ]
[17] International Telecommunications Union, "Propagation Data and Prediction Methods required for the design of Earth-space Telecommunication Systems," Recommendation ITU-R P.618-13, October
[18] Olsen, R. L.; Rogers, D.V.; and Hodge, D. B. "The aR^brelation in the calculation of rain attenuation". IEEE Transactions on Antennas and Propagation, issue 26, no. 2, 547 - 556. 1978
[19] International Telecommunications Union, "Characteristics of Precipitation for Propagation Modelling", Recommendation ITU- RP.837-6; Geneva, 2012
[20] D. Maggiori, "The computed transmission through rain in the 1 - 400 GHz frequency range for spherical and elliptical drops and any polarization". Alta Freq., no. 50, pp 262 - 273, 1981
[21] A.K. Lwas, M.R. Islam, M.H. Habaebi, A.F. Ismail, K. Abdullah, A. Zyoud, J. Chebil and M. Singh, "Analysis of the synthetic storm technique using rain height models to predict rain attenuation
in tropical regions". Proceedings of 5th International Conference on Computer and Communication Engineering: Emerging Technologies via Comp-Unification Convergence, pp. 220-223, ICCCE 2014
[22] J.S. Ojo and O.C. Rotimi, "Diurnal and seasonal variations of rain rate and rain attenuation on Ku-band satellite systems in a tropical region: A Synthetic Storm Technique approach". Journal of
Computers and Communications, issue 3, vol. 10 | {"url":"https://scielo.org.za/scielo.php?script=sci_arttext&pid=S1991-16962018000400004&lng=en&nrm=iso&tlng=en","timestamp":"2024-11-15T04:51:12Z","content_type":"application/xhtml+xml","content_length":"49736","record_id":"<urn:uuid:bd8f1b57-639f-4c38-bed7-6b3232893067>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00187.warc.gz"} |
Exploring Ridge and Lasso Regression in Python: Mastering Regularization | 2024 Guide - Quickinsights.org
Ever wondered how data scientists predict house prices or stock market trends? Ridge and Lasso regression are two powerful tools they use. These techniques help uncover hidden patterns in data, even
when there are many variables at play. Let’s explore how these methods work and how you can use them in Python.
Understanding Regularization
Before we dive into Ridge and Lasso, let’s talk about regularization. It’s a technique to prevent overfitting in machine learning models.
Overfitting happens when a model learns the training data too well. It captures noise along with the actual patterns. This leads to poor performance on new, unseen data.
Regularization adds a penalty term to the model’s loss function. This discourages the model from relying too heavily on any single feature.
Ridge Regression: L2 Regularization
Ridge regression, also known as L2 regularization, adds a squared magnitude of coefficient as penalty term to the loss function.
The formula for Ridge regression is:
Loss = OLS + α * (sum of squared coefficients)
Here, OLS is Ordinary Least Squares, and α is the regularization strength.
Ridge regression shrinks the coefficients of less important features towards zero. However, it never makes them exactly zero.
Lasso Regression: L1 Regularization
Lasso stands for Least Absolute Shrinkage and Selection Operator. It uses L1 regularization.
The formula for Lasso regression is:
Loss = OLS + α * (sum of absolute value of coefficients)
Lasso can shrink some coefficients to exactly zero. This makes it useful for feature selection.
Setting Up Your Python Environment
To get started, you’ll need to install some Python libraries. Open your terminal and run:
pip install numpy pandas scikit-learn matplotlib seaborn
These libraries will help us load data, build models, and visualize results.
Importing Necessary Libraries
Let’s start by importing the libraries we’ll need:
import numpy as np
import pandas as pd
from sklearn.model_selection import train_test_split
from sklearn.linear_model import Ridge, Lasso
from sklearn.metrics import mean_squared_error
import matplotlib.pyplot as plt
import seaborn as sns
Loading and Preparing the Data
For this example, we’ll use the Boston Housing dataset. It’s included in scikit-learn:
from sklearn.datasets import load_boston
boston = load_boston()
X = pd.DataFrame(boston.data, columns=boston.feature_names)
y = pd.Series(boston.target, name='PRICE')
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)
We’ve split our data into training and test sets. This helps us evaluate our models fairly.
Implementing Ridge Regression
Now, let’s implement Ridge regression:
ridge = Ridge(alpha=1.0)
ridge.fit(X_train, y_train)
y_pred_ridge = ridge.predict(X_test)
mse_ridge = mean_squared_error(y_test, y_pred_ridge)
print(f"Ridge MSE: {mse_ridge}")
The alpha parameter controls the strength of regularization. A higher alpha means stronger regularization.
Implementing Lasso Regression
Similarly, we can implement Lasso regression:
lasso = Lasso(alpha=1.0)
lasso.fit(X_train, y_train)
y_pred_lasso = lasso.predict(X_test)
mse_lasso = mean_squared_error(y_test, y_pred_lasso)
print(f"Lasso MSE: {mse_lasso}")
Again, alpha controls the strength of regularization.
Comparing Ridge and Lasso Coefficients
Let’s compare how Ridge and Lasso affect the coefficients:
coef_comparison = pd.DataFrame({
'Feature': X.columns,
'Ridge': ridge.coef_,
'Lasso': lasso.coef_
plt.figure(figsize=(12, 6))
sns.barplot(x='Feature', y='Ridge', data=coef_comparison, color='blue', alpha=0.5, label='Ridge')
sns.barplot(x='Feature', y='Lasso', data=coef_comparison, color='red', alpha=0.5, label='Lasso')
plt.title('Comparison of Ridge and Lasso Coefficients')
This plot shows how each method affects the feature coefficients differently.
Tuning the Alpha Parameter
The alpha parameter is crucial for both Ridge and Lasso. Let’s see how different alpha values affect the models:
alphas = [0.1, 1, 10, 100]
ridge_scores = []
lasso_scores = []
for alpha in alphas:
ridge = Ridge(alpha=alpha)
ridge.fit(X_train, y_train)
ridge_scores.append(mean_squared_error(y_test, ridge.predict(X_test)))
lasso = Lasso(alpha=alpha)
lasso.fit(X_train, y_train)
lasso_scores.append(mean_squared_error(y_test, lasso.predict(X_test)))
plt.plot(alphas, ridge_scores, label='Ridge')
plt.plot(alphas, lasso_scores, label='Lasso')
plt.ylabel('Mean Squared Error')
plt.title('MSE vs Alpha for Ridge and Lasso')
This plot helps us choose the best alpha for each method.
Feature Selection with Lasso
Lasso can perform feature selection by setting some coefficients to zero. Let’s see which features it selects:
lasso = Lasso(alpha=1.0)
lasso.fit(X_train, y_train)
selected_features = X.columns[lasso.coef_ != 0]
print("Selected features:", selected_features)
These are the features Lasso considers most important for predicting house prices.
Cross-Validation for Model Selection
We can use cross-validation to choose between Ridge and Lasso:
from sklearn.model_selection import cross_val_score
ridge_cv = cross_val_score(Ridge(alpha=1.0), X, y, cv=5)
lasso_cv = cross_val_score(Lasso(alpha=1.0), X, y, cv=5)
print(f"Ridge CV Score: {ridge_cv.mean()}")
print(f"Lasso CV Score: {lasso_cv.mean()}")
The method with the higher cross-validation score is generally preferred.
Handling Multicollinearity
Both Ridge and Lasso can help with multicollinearity. This is when features are highly correlated with each other.
Let’s check for multicollinearity in our dataset:
correlation_matrix = X.corr()
plt.figure(figsize=(12, 10))
sns.heatmap(correlation_matrix, annot=True, cmap='coolwarm')
plt.title('Correlation Matrix of Features')
Ridge and Lasso can help reduce the impact of these correlations on our model.
Interpreting the Results
When interpreting Ridge and Lasso results, remember:
• Ridge shrinks all coefficients but doesn’t eliminate any.
• Lasso can eliminate less important features entirely.
• The magnitude of coefficients shows feature importance.
• A lower MSE indicates better model performance.
When to Use Ridge vs Lasso
Choose Ridge when:
• You want to keep all features.
• You suspect many features are important.
Choose Lasso when:
• You want to perform feature selection.
• You believe only a few features are important.
Elastic Net: Combining Ridge and Lasso
Elastic Net combines Ridge and Lasso regularization. It’s useful when you want a balance between the two:
from sklearn.linear_model import ElasticNet
elastic = ElasticNet(alpha=1.0, l1_ratio=0.5)
elastic.fit(X_train, y_train)
y_pred_elastic = elastic.predict(X_test)
mse_elastic = mean_squared_error(y_test, y_pred_elastic)
print(f"Elastic Net MSE: {mse_elastic}")
The l1_ratio parameter controls the mix of L1 and L2 regularization.
Scaling Features for Better Performance
Scaling features can improve the performance of Ridge and Lasso:
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
X_scaled = scaler.fit_transform(X)
X_train_scaled, X_test_scaled, y_train, y_test = train_test_split(X_scaled, y, test_size=0.2, random_state=42)
ridge_scaled = Ridge(alpha=1.0)
ridge_scaled.fit(X_train_scaled, y_train)
lasso_scaled = Lasso(alpha=1.0)
lasso_scaled.fit(X_train_scaled, y_train)
print(f"Ridge MSE (scaled): {mean_squared_error(y_test, ridge_scaled.predict(X_test_scaled))}")
print(f"Lasso MSE (scaled): {mean_squared_error(y_test, lasso_scaled.predict(X_test_scaled))}")
in above code Scaling ensures all features contribute equally to the regularization penalty.
Ridge and Lasso regression are powerful tools for handling complex datasets. They help prevent overfitting and can improve model performance. Ridge is great when you want to keep all features, while
Lasso excels at feature selection.
Remember, the choice between Ridge and Lasso often depends on your specific dataset and problem. Experiment with both methods and use techniques like cross-validation to find the best approach for
your data.
As you continue your journey in data science, you might want to explore more Advanced Regression Techniques in Python. For time-dependent data, Time Series Regression in Python offers specialized
methods. And for a broader overview, check out our guide on Regression in Python. | {"url":"https://quickinsights.org/exploring-ridge-and-lasso-regression-in-python/","timestamp":"2024-11-02T09:28:44Z","content_type":"text/html","content_length":"210899","record_id":"<urn:uuid:7322ef47-f7d7-4aab-a657-3f4d80e5651d>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00122.warc.gz"} |
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