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https://www.scribd.com/doc/245784388/newsletter-11-03-14
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# Sra. Libreros y Sra.
Ramirez´s Class Letter - Noviembre 3, 2014
http:/firstsplashjges.weebly.com/
HFW: otro, pues, reloj, salto, tambien, valioso
with a special emphasis on characters. We will focus on characters’ actions in isolated parts of the book and as a pattern throughout a story.
Word study: Practice vocabulary (The body inside). Words with ca,
que, qui, co, cu. The use of personal pronouns and verbs (action words).
Yo, tu, el/ella, nosotros, ustedes, ellos.
Writing: Students will edit their story by making sure they have spelled
sight words correctly and included spaces between each word.
Math: This week we will continue to study value. Students are working
towards Understanding that the two digits of a two-digit number represent
amounts of tens and ones. Understand the following as special cases:
a. 10 can be thought of as a bundle of ten ones – called a “ten.”
b. The numbers from 11 to 19 are composed of a ten and one, two, three,
four, five, six, seven, eight, or nine ones.
c. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three,
four, five, six, seven, eight, or nine tens (and 0 ones).
Social Studies– Celebration Dia de los muertos ‘ Nov 1-2. The
tradition and how Mexicans celebrate it.
They will also continue to learn about maps and characteristics of landforms. They will build an understanding of the basic elements of geographic representations using maps (cardinal directions and map symbols). We will look at maps to identify our continent, country, state, and
city. We will also learn about the other continents and oceans.
Please, visit resources from Sra. Libreros’ weebly page. You will find
this link to enjoy singing Parts of the body with your kids:
IMPORTANT NOTES and
homework!!!
( S ee checklist on the back )
Reading Log - English books (Monday)
Reading Passport- Spanish books will be sent home on
Tuesday
Language: Parts of the body worksheet. HFW worksheet.
Math: Place value worksheets.
REPORT CARDS GO HOME TODAY AND FALL CONFERENCES
Check your e-mail since I am sending a message from Sign
per week)
Many thanks to parents for the supplies and ítems for the
treasure box and for sharing and volunteering with the Fall
Party.
We have Media every Tuesday. Please remember to
return your books to school each Tuesday.
Remember to encourage your child to read in Razkids. Two/three times a week is fine.
Nov. 4, PTA Meeting 8:00pm. Nov. 6, Picture Day/
Skate night. Nov. 7, Early release. Nov. 12 Boostherthon
week.
<mlibreros-tamayo@wcpss.net>
<lramirez@wcpss.net>
Phone number: 919– 881-4910
OCTOBER VOCABULARY
http:/firstsplashjges.weebly.com/
Nuestro planeta
los oceanos,
los continentes ,
el valle,
el bosque,
la llanura,
la montaña,
el desierto,
el golfo,
la peninsula,
los lagos,
las islas,
las colinas.
Los continentes
America del Norte
America Central
America del Sur
Continente Africano
Continente Asiático
Continente Europeo
Oceania
Partes externas
del cuerpo
Cabeza
Tronco
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http://www.innovations-report.com/html/reports/information-technology/algorithm-enables-computers-to-identify-actions-much-more-efficiently.html
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Forum for Science, Industry and Business
Search our Site:
Algorithm enables computers to identify actions much more efficiently
14.05.2014
Techniques from natural-language processing enable computers to efficiently search video for actions
With the commodification of digital cameras, digital video has become so easy to produce that human beings can have trouble keeping up with it. Among the tools that computer scientists are developing to make the profusion of video more useful are algorithms for activity recognition — or determining what the people on camera are doing when.
At the Conference on Computer Vision and Pattern Recognition in June, Hamed Pirsiavash, a postdoc at MIT, and his former thesis advisor, Deva Ramanan of the University of California at Irvine, will present a new activity-recognition algorithm that has several advantages over its predecessors.
One is that the algorithm's execution time scales linearly with the size of the video file it's searching. That means that if one file is 10 times the size of another, the new algorithm will take 10 times as long to search it — not 1,000 times as long, as some earlier algorithms would.
Another is that the algorithm is able to make good guesses about partially completed actions, so it can handle streaming video. Partway through an action, it will issue a probability that the action is of the type that it's looking for. It may revise that probability as the video continues, but it doesn't have to wait until the action is complete to assess it.
Finally, the amount of memory the algorithm requires is fixed, regardless of how many frames of video it's already reviewed. That means that, unlike many of its predecessors, it can handle video streams of any length (or files of any size).
The grammar of action
Enabling all of these advances is the appropriation of a type of algorithm used in natural language processing, the computer science discipline that seeks techniques for interpreting sentences written in natural language.
"One of the challenging problems they try to solve is, if you have a sentence, you want to basically parse the sentence, saying what is the subject, what is the verb, what is the adverb," Pirsiavash says. "We see an analogy here, which is, if you have a complex action — like making tea or making coffee — that has some subactions, we can basically stitch together these subactions and look at each one as something like verb, adjective, and adverb."
On that analogy, the rules defining relationships between subactions are like rules of grammar. When you make tea, for instance, it doesn't matter whether you first put the teabag in the cup or put the kettle on the stove. But it's essential that you put the kettle on the stove before pouring the water into the cup. Similarly, in a given language, it could be the case that nouns can either precede or follow verbs, but that adjectives must always precede nouns.
For any given action, Pirsiavash and Ramanan's algorithm must thus learn a new "grammar." And the mechanism that it uses is the one that many natural-language-processing systems rely on: machine learning. Pirsiavash and Ramanan feed their algorithm training examples of videos depicting a particular action, and specify the number of subactions that the algorithm should look for. But they don't give it any information about what those subactions are, or what the transitions between them look like.
Pruning possibilities
The rules relating subactions are the key to the algorithm's efficiency. As a video plays, the algorithm constructs a set of hypotheses about which subactions are being depicted where, and it ranks them according to probability. It can't limit itself to a single hypothesis, as each new frame could require it to revise its probabilities. But it can eliminate hypotheses that don't conform to its grammatical rules, which dramatically limits the number of possibilities it has to canvass.
The researchers tested their algorithm on eight different types of athletic endeavor — such as weightlifting and bowling — with training videos culled from YouTube. They found that, according to metrics standard in the field of computer vision, their algorithm identified new instances of the same activities more accurately than its predecessors.
Pirsiavash is particularly interested in possible medical applications of action detection. The proper execution of physical-therapy exercises, for instance, could have a grammar that's distinct from improper execution; similarly, the return of motor function in patients with neurological damage could be identified by its unique grammar. Action-detection algorithms could also help determine whether, for instance, elderly patients remembered to take their medication — and issue alerts if they didn't.
Abby Abazorius | newswise
Further information:
http://www.mit.edu
Further reports about: Massachusetts Technology algorithm identified identify problems
More articles from Information Technology:
Researchers achieve HD video streaming at 10,000 times lower power
20.04.2018 | University of Washington
An AI that makes road maps from aerial images
18.04.2018 | Massachusetts Institute of Technology, CSAIL
Im Focus: Spider silk key to new bone-fixing composite
University of Connecticut researchers have created a biodegradable composite made of silk fibers that can be used to repair broken load-bearing bones without the complications sometimes presented by other materials.
Repairing major load-bearing bones such as those in the leg can be a long and uncomfortable process.
Im Focus: Writing and deleting magnets with lasers
Study published in the journal ACS Applied Materials & Interfaces is the outcome of an international effort that included teams from Dresden and Berlin in Germany, and the US.
Scientists at the Helmholtz-Zentrum Dresden-Rossendorf (HZDR) together with colleagues from the Helmholtz-Zentrum Berlin (HZB) and the University of Virginia...
Im Focus: Gamma-ray flashes from plasma filaments
Novel highly efficient and brilliant gamma-ray source: Based on model calculations, physicists of the Max PIanck Institute for Nuclear Physics in Heidelberg propose a novel method for an efficient high-brilliance gamma-ray source. A giant collimated gamma-ray pulse is generated from the interaction of a dense ultra-relativistic electron beam with a thin solid conductor. Energetic gamma-rays are copiously produced as the electron beam splits into filaments while propagating across the conductor. The resulting gamma-ray energy and flux enable novel experiments in nuclear and fundamental physics.
The typical wavelength of light interacting with an object of the microcosm scales with the size of this object. For atoms, this ranges from visible light to...
Im Focus: Basel researchers succeed in cultivating cartilage from stem cells
Stable joint cartilage can be produced from adult stem cells originating from bone marrow. This is made possible by inducing specific molecular processes occurring during embryonic cartilage formation, as researchers from the University and University Hospital of Basel report in the scientific journal PNAS.
Certain mesenchymal stem/stromal cells from the bone marrow of adults are considered extremely promising for skeletal tissue regeneration. These adult stem...
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Researchers lay groundwork to tailor drugs for new targets in cancer therapy
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# wolfson
Senior Members
630
1. ## Why is the pH of pure water i.e. neutral 7?
In pure water, the hydrogen ion (hydroxonium ion) concentration must be equal to the hydroxide ion concentration. [OH-] term in the Kw expression by another [H+]. [H+]2 = 1.00 x 10-14 Taking the square root of each side gives: [H+] = 1.00 x 10-7 mol dm-3 Converting that into pH: pH = - log10 [H+] pH = 7
Its PCl3
3. ## Thickness of Cu on Cathode
Example 2: How much copper is deposited if a current of 0.2 Amps is passed for 2 hours through a copper(II) sulphate solution ? Electrode equation: (-) cathode Cu2+(aq) + 2e- ==> Cu(s) and Ar(Cu) = 64 the quantity of electricity passed in Coulombs = current in A x time in secs (Q = I x t) = 0.2 x 2 x 60 x 60 = 1440 Coulombs, and 1 mole electrons = 96500 Coulombs therefore moles of electrons passed through circuit = 1440 / 96500 = 0.01492 it takes two moles of electrons to form one mole of copper therefore moles copper = 0.01492 7 2 = 0.00746 mass of copper = moles of copper x atomic mass of copper = 0.00746 x 64 = 0.4775g of copper deposited. Taken from http://www.wpbschoolhouse.btinterne...lysis%20product
4. ## Thickness of Cu on Cathode
The mass of a substance produced at an electrode during electrolysis is proportional to the quantity of electricity that has passed. Look up, Faraday's formula to relate these variables. look at site below: http://www.wpbschoolhouse.btinternet.co.uk/page04/4_73calcs.htm#13.%20Electrolysis%20product
5. ## Destroying Diamonds
Melting point of diamond = 3815.56oC, thats about 2.5 times more than steel. This information is the correct. Ref: RSC.
6. ## A few questions that need answers...
Sulphuric acid at that temperature will rapidly corrode steel.
7. ## Silicon dioxide
Pulkit that was so wrong it's not even funny: Na2SiO3 + 8 HF --> H2SiF6 + 2 NaF + 3 H2O
8. ## Has anyone heard of cyalic acid?
To a chemist, that seems to me as a spelling mistake for Sialic Acid.
9. ## infinity?
infinity = The limit that a function is said to approach at x = a when (x) is larger than any preassigned number for all x sufficiently near a.
10. ## how do you work out electron config?
I wrote a bit on electronic config at: http://dragonslair.europe.webmatrixhosting.net/Science/Chemistry/Eleccon2.htm It should help.
11. ## Pressure
If all you wish to do is measure the number of mols, the easiest way would be to just weight the gas. weight the gas = calculate the volume. The higher the temperature the faster molecules are moving. Faster the molecules = higher the pressure, as are moving, the harder they hit the wall of the container. It's all kinetic theory/motion.
12. ## help needed
petrol heat of combustion:5460 kj/mol LPG heat of combustion 2220 kj/mol 2 C8H18 + 25 O2 = 18 H2O + 16 CO2 (combustion of octane) CH3-CH2-CH3 + 5 O2 = 3 CO2 + 4 H2O (combustion of propane) 1 mole of propane combusts 2 mole of Octane combusts Octane uses 10920 Kj Propane uses 2220 Kj So if Octane is 80 cents per 10920Kj Octane = 80 / 10920 = 0.00733 cents per Kj You get 2 moles of propane for the equivalent of octane So 0.00733 / 2 = 0.0037 So 0.0037 cents per kj And 2220 kj = 8.2 cents I am not sure if this is correct, it’s early in the morning
13. ## Li, KCl
A list of metals arranged in order of their electrode potentials. A metal will displace, from their salts, metals lower down in the series. Lithium Potassium Calcium Sodium Magnesium Aluminum Zinc Iron Cobalt Nickel Tin Lead ---------------- Hydrogen ---------------- Copper Mercury Silver Platinum Gold
14. ## Hydrogen and Helium+ Radius
The norm H =37.1 pm The norm He(1+) = 31 pm
15. ## Stability of organo-metallic compounds
The process for preparing stable high oxidation state ferrocene derivatives, namely Fe(III)/Fe(IV), by reacting ferrocene derivatives in which one or more electron-donating groups are covalently bonded to the cyclopentadienyl rings with electron-donating ligands having strong coordinative groups for ferric species. (APP 2001)
16. ## Chemistry is difficult
Allot of people who I am teaching at Summer college at University seem to be struggling with chemistry for applied science I think there are a number of factors the major is a giant step up from A-Level applied science to Undergraduate, the laboratory reports seem to be proving difficult for some. Also the grasp of equilibrium (Kc) and intermolecular forces. But hopefully with a bit of help from me they should start to realize the relationship between them.
17. ## Where Can I Find H2SO4?
LMAO that is excellant
18. ## New Member with a Challenge
1 - The Summary Graph 2 on Page 4. This is the function which needs completion by calculus. -------------------------------------------------------------------------------- I've just had a quick look at your website and found the graph, but what is the function? Could you post the function explicitly here? Sorry if I've missed something, but I'm confused as to what you are asking.
lol yes
20. ## Help with math equation
Assuming I am still correct... Then there will be a meeting of great prof's to all talk and see if it correct. and then.......... well maybe on tv. Or get some king of an award. But i'd put it on scienceforums on here, remember you still wrote it first so its yours, and you'll have all your workings out.
21. ## A sphere has ____ sides.
Sometimes we think of sides as having to be flat - what mathematicians call "faces" of a polyhedron. In that sense, a sphere has no sides (though some people will try to convince you that it has infinitely many sides). I think the most applicable definition is the one used in topology. We talk about surfaces with two sides (like a piece of paper) and surfaces with only one side (like the Moebius strip you may have heard of). In this sense, and remembering that mathematicians define a sphere as the surface of a sphere, not a solid ball, I would say that a sphere has two sides: the outside and the inside. (ref: Dr P. APP)
22. ## whats the chances of alien life form more intelligent than us?
Well said Jakiri.
23. ## Heat Shields
and does it block gamma rays? I very much doubt it.
24. ## elemnet 118??
The same as July02
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https://thebasicanswers.com/5890/solved-flat-slab-of-styrofoam-with-density-of-32-floats-lake
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# [SOLVED] A flat slab of styrofoam, with a density of 32 kg/m^3, floats on a lake.
A flat slab of styrofoam, with a density of 32 kg/m^3, floats on a lake.
What is the minimum volume the slab must have so that a 42 kg boy can sit on the slab without it sinking?
by
## The correct answer for A flat slab of styrofoam, with a density of 32 kg/m^3, floats on a lake.
bouyancy-42g=0
density water*g*volume-32g*volume-42g=0
volume= 42/(densitywater-32kg/m^3)
density water = 1000kg/m^3 approx.
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.
+1 vote
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http://www.ck12.org/book/CK-12-Basic-Algebra-Concepts/r12/section/1.7/
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 1.7: Words that Describe Patterns
Difficulty Level: Basic Created by: CK-12
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Practice Words that Describe Patterns
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Many bicyclists have biking watches that are able to record the time spent biking and the distance traveled. They are able to download the data recorded by their watches to a computer and view a table with times in minutes in one column and distances in miles in another column. How could they use this data to write an English sentence or an algebraic expression? After completing this Concept, you'll be able to answer this question and use the algebraic expression to predict the distance traveled for any time spent biking.
### Guidance
Using Words to Describe Patterns
Sometimes patterns are given in tabular format (meaning presented in a table). An important job of data analysts is to describe a pattern so others can understand it.
#### Example A
Using the table below, describe the pattern in words.
\begin{align*}&&x && -1 && 0 && 1 && 2 && 3 && 4\\ &&y && -5 && 0 && 5 && 10 && 15 && 20\end{align*}
Solution: We can see from the table that \begin{align*}y\end{align*} is five times bigger than \begin{align*}x\end{align*}. Therefore, the pattern is that the “\begin{align*}y\end{align*} value is five times larger than the \begin{align*}x\end{align*} value.”
#### Example B
Zarina has a $100 gift card and has been spending money in small regular amounts. She checks the balance on the card at the end of every week and records the balance in the following table. Using the table, describe the pattern in words and in an expression. Week # Balance ($)
1 78
2 56
3 34
Solution: Each week the amount of her gift card is $22 less than the week before. The pattern in words is: “The gift card started at$100 and is decreasing by 22 each week.” As we saw in the last lesson, this sentence can be translated into the algebraic expression \begin{align*}100-22w\end{align*}. #### Example C The expression found in example 2 can be used to answer questions and predict the future. Suppose, for instance, that Zarina wanted to know how much she would have on her gift card after 4 weeks if she used it at the same rate. By substituting the number 4 for the variable \begin{align*}w\end{align*}, it can be determined that Zarina would have12 left on her gift card.
Solution:
\begin{align*}100-22w\end{align*}
When \begin{align*}w = 4\end{align*}, the expression becomes:
\begin{align*}&100-22(4)\\ &100-88\\ &12\end{align*}
After 4 weeks, Zarina would have \$12 left on her gift card.
<iframe width='480' height='300' src='http://www.educreations.com/lesson/embed/1143850/?ref=app' frameborder='0' allowfullscreen></iframe>
### Guided Practice
Jose starts training to be a runner. When he starts, he can run 3 miles per hour. After 5 weeks of training, Jose can run faster. After each week, he records his average speed while running. He summarizes this information in the following table:
Week #
Average Speed (miles per hour)
1 3.25
2 3.5
3 3.75
4 4.0
5 4.25
Write an expression for Jose's increased speed and predict how fast he will be able to run after 6 weeks.
Solution:
We will use \begin{align*}w\end{align*} to represent the number or weeks. Jose's speed starts at 3 mph, and from the table we can see that it increases by 0.25 miles per hour every week. This gives us the expression \begin{align*}3+0.25w\end{align*}. Now we substitute in \begin{align*}w=6\end{align*} and get the following:
\begin{align*}3+0.25(6)\end{align*}
\begin{align*}3+1.5=4.5\end{align*}
If Jose keeps up his training, by the end of the 6th week, he should be able to run 4.5 miles per hour.
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Patterns and Equations (13:18)
In questions 1 – 4, write the pattern of the table: a) in words and b) with an algebraic expression.
1. Number of workers and number of video games packaged
\begin{align*}&\text{People} && 0 && 1 && 2 && 5 && 10 && 50 && 200\\ &\text{Amount} && 0 && 65 && 87 && 109 && 131 && 153 && 175\end{align*}
1. The number of hours worked and the total pay
\begin{align*}&\text{Hours} && 1 && 2 && 3 && 4 && 5 && 6\\ &\text{Total Pay} && 15 && 22 && 29 && 36 && 43 && 50\end{align*}
1. The number of hours of an experiment and the total number of bacteria
\begin{align*}&\text{Hours} && 0 && 1 && 2 && 5 && 10\\ &\text{Bacteria} && 0 && 2 && 4 && 32 && 1024\end{align*}
1. With each filled seat, the number of people on a Ferris wheel doubles.
1. Write an expression to describe this situation.
2. How many people are on a Ferris wheel with 17 seats filled?
2. Using the theme park situation from the lesson, how much revenue would be generated by 2,518 people?
Mixed Review
1. Use parentheses to make the equation true: \begin{align*}10+6 \div 2-3=5\end{align*}.
2. Find the value of \begin{align*}5x^2 - 4y\end{align*} for \begin{align*}x = -4\end{align*} and \begin{align*}y = 5\end{align*}.
3. Find the value of \begin{align*}\frac{x^2y^3}{x^3 + y^2}\end{align*} for \begin{align*}x = 2\end{align*} and \begin{align*}y=-4\end{align*}.
4. Simplify: \begin{align*}2 - (t - 7)^2 \times (u^3 - v)\end{align*} when \begin{align*}t = 19, u = 4\end{align*}, and \begin{align*}v = 2\end{align*}.
5. Simplify: \begin{align*}2 - (19 - 7)^2 \times (4^3 - 2)\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
tabular
Being represented in a table.
$\ge$
The greater-than-or-equal-to symbol "$\ge$" indicates that the value on the left side of the symbol is greater than or equal to the value on the right.
$\le$
The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right.
$\ne$
The not-equal-to symbol "$\ne$" indicates that the value on the left side of the symbol is not equal to the value on the right.
constant
A constant is a value that does not change. In Algebra, this is a number such as 3, 12, 342, etc., as opposed to a variable such as x, y or a.
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
greater than
The greater than symbol, $>$, indicates that the value on the left side of the symbol is greater than the value on the right.
greater than or equal to
The greater than or equal to symbol, $\ge$, indicates that the value on the left side of the symbol is greater than or equal to the value on the right.
inequality
An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.
less than
The less-than symbol "<" indicates that the value on the left side of the symbol is lesser than the value on the right.
less than or equal to
The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right.
not equal to
The "not equal to" symbol, $\ne$, indicates that the value on the left side of the symbol is not equal to the value on the right.
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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Difficulty Level:
Basic
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Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox: nag_ode_ivp_rkm_zero_simple (d02bh)
## Purpose
nag_ode_ivp_rkm_zero_simple (d02bh) integrates a system of first-order ordinary differential equations over an interval with suitable initial conditions, using a Runge–Kutta–Merson method, until a user-specified function of the solution is zero.
## Syntax
[x, y, tol, ifail] = d02bh(x, xend, y, tol, irelab, hmax, fcn, g, 'n', n)
[x, y, tol, ifail] = nag_ode_ivp_rkm_zero_simple(x, xend, y, tol, irelab, hmax, fcn, g, 'n', n)
## Description
nag_ode_ivp_rkm_zero_simple (d02bh) advances the solution of a system of ordinary differential equations
yi ′ = fi(x,y1,y2, … ,yn), i = 1,2, … ,n, $yi′=fi(x,y1,y2,…,yn), i=1,2,…,n,$
from x = x$x={\mathbf{x}}$ towards x = xend$x={\mathbf{xend}}$ using a Merson form of the Runge–Kutta method. The system is defined by fcn, which evaluates fi${f}_{i}$ in terms of x$x$ and y1,y2,,yn${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$ (see Section [Parameters]), and the values of y1,y2,,yn${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$ must be given at x = x$x={\mathbf{x}}$.
As the integration proceeds, a check is made on the function g(x,y)$g\left(x,y\right)$ specified by you, to determine an interval where it changes sign. The position of this sign change is then determined accurately by interpolating for the solution and its derivative. It is assumed that g(x,y)$g\left(x,y\right)$ is a continuous function of the variables, so that a solution of g(x,y) = 0$g\left(x,y\right)=0$ can be determined by searching for a change in sign in g(x,y)$g\left(x,y\right)$.
The accuracy of the integration and, indirectly, of the determination of the position where g(x,y) = 0$g\left(x,y\right)=0$, is controlled by tol.
For a description of Runge–Kutta methods and their practical implementation see Hall and Watt (1976).
## References
Hall G and Watt J M (ed.) (1976) Modern Numerical Methods for Ordinary Differential Equations Clarendon Press, Oxford
## Parameters
### Compulsory Input Parameters
1: x – double scalar
Must be set to the initial value of the independent variable x$x$.
2: xend – double scalar
The final value of the independent variable x$x$.
If ${\mathbf{xend}}<{\mathbf{x}}$ on entry, integration proceeds in a negative direction.
3: y(n) – double array
n, the dimension of the array, must satisfy the constraint n > 0${\mathbf{n}}>0$.
The initial values of the solution y1,y2,,yn${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$.
4: tol – double scalar
Must be set to a positive tolerance for controlling the error in the integration and in the determination of the position where g(x,y) = 0.0$g\left(x,y\right)=0.0$.
nag_ode_ivp_rkm_zero_simple (d02bh) has been designed so that, for most problems, a reduction in tol leads to an approximately proportional reduction in the error in the solution obtained in the integration. The relation between changes in tol and the error in the determination of the position where g(x,y) = 0.0$g\left(x,y\right)=0.0$ is less clear, but for tol small enough the error should be approximately proportional to tol. However, the actual relation between tol and the accuracy cannot be guaranteed. You are strongly recommended to call nag_ode_ivp_rkm_zero_simple (d02bh) with more than one value for tol and to compare the results obtained to estimate their accuracy. In the absence of any prior knowledge you might compare results obtained by calling nag_ode_ivp_rkm_zero_simple (d02bh) with tol = 10.0p${\mathbf{tol}}={10.0}^{-p}$ and tol = 10.0p1${\mathbf{tol}}={10.0}^{-p-1}$ if p$p$ correct decimal digits in the solution are required.
Constraint: tol > 0.0${\mathbf{tol}}>0.0$.
5: irelab – int64int32nag_int scalar
Determines the type of error control. At each step in the numerical solution an estimate of the local error, est$\mathit{est}$, is made. For the current step to be accepted the following condition must be satisfied:
irelab = 0${\mathbf{irelab}}=0$
esttol × max {1.0,|y1|,|y2|,,|yn|}$\mathit{est}\le {\mathbf{tol}}×\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left\{1.0,|{y}_{1}|,|{y}_{2}|,\dots ,|{y}_{\mathit{n}}|\right\}$;
irelab = 1${\mathbf{irelab}}=1$
esttol$\mathit{est}\le {\mathbf{tol}}$;
irelab = 2${\mathbf{irelab}}=2$
esttol × max {ε,|y1|,|y2|,,|yn|}$\mathit{est}\le {\mathbf{tol}}×\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left\{\epsilon ,|{y}_{1}|,|{y}_{2}|,\dots ,|{y}_{\mathit{n}}|\right\}$, where ε$\epsilon$ is machine precision.
If the appropriate condition is not satisfied, the step size is reduced and the solution recomputed on the current step.
If you wish to measure the error in the computed solution in terms of the number of correct decimal places, then set irelab = 1${\mathbf{irelab}}=1$ on entry, whereas if the error requirement is in terms of the number of correct significant digits, then set irelab = 2${\mathbf{irelab}}=2$. Where there is no preference in the choice of error test, irelab = 0${\mathbf{irelab}}=0$ will result in a mixed error test. It should be borne in mind that the computed solution will be used in evaluating g(x,y)$g\left(x,y\right)$.
Constraint: irelab = 0${\mathbf{irelab}}=0$, 1$1$ or 2$2$.
6: hmax – double scalar
If hmax = 0.0${\mathbf{hmax}}=0.0$, no special action is taken.
If hmax0.0${\mathbf{hmax}}\ne 0.0$, a check is made for a change in sign of g(x,y)$g\left(x,y\right)$ at steps not greater than |hmax|$|{\mathbf{hmax}}|$. This facility should be used if there is any chance of ‘missing’ the change in sign by checking too infrequently. For example, if two changes of sign of g(x,y)$g\left(x,y\right)$ are expected within a distance h$h$, say, of each other, then a suitable value for hmax might be hmax = h / 2${\mathbf{hmax}}=h/2$. If only one change of sign in g(x,y)$g\left(x,y\right)$ is expected on the range x to xend, then the choice hmax = 0.0${\mathbf{hmax}}=0.0$ is most appropriate.
7: fcn – function handle or string containing name of m-file
fcn must evaluate the functions fi${f}_{i}$ (i.e., the derivatives yi${y}_{i}^{\prime }$) for given values of its arguments x,y1,,yn$x,{y}_{1},\dots ,{y}_{\mathit{n}}$.
[f] = fcn(x, y)
Input Parameters
1: x – double scalar
x$x$, the value of the argument.
2: y( : $:$) – double array
yi${y}_{\mathit{i}}$, for i = 1,2,,n$\mathit{i}=1,2,\dots ,\mathit{n}$, the value of the argument.
Output Parameters
1: f( : $:$) – double array
The value of fi${f}_{\mathit{i}}$, for i = 1,2,,n$\mathit{i}=1,2,\dots ,\mathit{n}$.
8: g – function handle or string containing name of m-file
g must evaluate the function g(x,y)$g\left(x,y\right)$ at a specified point.
[result] = g(x, y)
Input Parameters
1: x – double scalar
x$x$, the value of the independent variable.
2: y( : $:$) – double array
The value of yi${y}_{\mathit{i}}$, for i = 1,2,,n$\mathit{i}=1,2,\dots ,\mathit{n}$.
Output Parameters
1: result – double scalar
The result of the function.
### Optional Input Parameters
1: n – int64int32nag_int scalar
Default: The dimension of the array y.
n$\mathit{n}$, the number of differential equations.
Constraint: n > 0${\mathbf{n}}>0$.
w
### Output Parameters
1: x – double scalar
The point where g(x,y) = 0.0$g\left(x,y\right)=0.0$ unless an error has occurred, when it contains the value of x$x$ at the error. In particular, if g(x,y)0.0$g\left(x,y\right)\ne 0.0$ anywhere on the range x to xend, it will contain xend on exit.
2: y(n) – double array
The computed values of the solution at the final point x = x$x={\mathbf{x}}$.
3: tol – double scalar
Normally unchanged. However if the range from x = x$x={\mathbf{x}}$ to the position where g(x,y) = 0.0$g\left(x,y\right)=0.0$ (or to the final value of x$x$ if an error occurs) is so short that a small change in tol is unlikely to make any change in the computed solution, then tol is returned with its sign changed. To check results returned with tol < 0.0${\mathbf{tol}}<0.0$, nag_ode_ivp_rkm_zero_simple (d02bh) should be called again with a positive value of tol whose magnitude is considerably smaller than that of the previous call.
4: ifail – int64int32nag_int scalar
${\mathrm{ifail}}={\mathbf{0}}$ unless the function detects an error (see [Error Indicators and Warnings]).
## Error Indicators and Warnings
Errors or warnings detected by the function:
ifail = 1${\mathbf{ifail}}=1$
On entry, tol ≤ 0.0${\mathbf{tol}}\le 0.0$, or n ≤ 0${\mathbf{n}}\le 0$, or irelab ≠ 0${\mathbf{irelab}}\ne 0$, 1$1$ or 2$2$.
ifail = 2${\mathbf{ifail}}=2$
With the given value of tol, no further progress can be made across the integration range from the current point x = x$x={\mathbf{x}}$, or dependence of the error on tol would be lost if further progress across the integration range were attempted (see Section [Further Comments] for a discussion of this error exit). The components y(1),y(2),,y(n)${\mathbf{y}}\left(1\right),{\mathbf{y}}\left(2\right),\dots ,{\mathbf{y}}\left(\mathit{n}\right)$ contain the computed values of the solution at the current point x = x$x={\mathbf{x}}$. No point at which g(x,y)$g\left(x,y\right)$ changes sign has been located up to the point x = x$x={\mathbf{x}}$.
ifail = 3${\mathbf{ifail}}=3$
tol is too small for nag_ode_ivp_rkm_zero_simple (d02bh) to take an initial step (see Section [Further Comments]). x and y(1),y(2),,y(n)${\mathbf{y}}\left(1\right),{\mathbf{y}}\left(2\right),\dots ,{\mathbf{y}}\left(\mathit{n}\right)$ retain their initial values.
ifail = 4${\mathbf{ifail}}=4$
At no point in the range x to xend did the function g(x,y)$g\left(x,y\right)$ change sign. It is assumed that g(x,y) = 0.0$g\left(x,y\right)=0.0$ has no solution.
ifail = 5${\mathbf{ifail}}=5$ (nag_roots_contfn_brent_rcomm (c05az))
A serious error has occurred in an internal call to the specified function. Check all function calls and array dimensions. Seek expert help.
ifail = 6${\mathbf{ifail}}=6$
A serious error has occurred in an internal call to an integration function. Check all function calls and array dimensions. Seek expert help.
ifail = 7${\mathbf{ifail}}=7$
A serious error has occurred in an internal call to an interpolation function. Check all (sub)program calls and array dimensions. Seek expert help.
## Accuracy
The accuracy depends on tol, on the mathematical properties of the differential system, on the position where g(x,y) = 0.0$g\left(x,y\right)=0.0$ and on the method. It can be controlled by varying tol but the approximate proportionality of the error to tol holds only for a restricted range of values of tol. For tol too large, the underlying theory may break down and the result of varying tol may be unpredictable. For tol too small, rounding error may affect the solution significantly and an error exit with ${\mathbf{ifail}}={\mathbf{2}}$ or 3${\mathbf{3}}$ is possible.
The accuracy may also be restricted by the properties of g(x,y)$g\left(x,y\right)$. You should try to code g without introducing any unnecessary cancellation errors.
The time taken by nag_ode_ivp_rkm_zero_simple (d02bh) depends on the complexity and mathematical properties of the system of differential equations defined by fcn, the complexity of g, on the range, the position of the solution and the tolerance. There is also an overhead of the form a + b × n$a+b×\mathit{n}$ where a$a$ and b$b$ are machine-dependent computing times.
For some problems it is possible that nag_ode_ivp_rkm_zero_simple (d02bh) will return ${\mathbf{ifail}}={\mathbf{4}}$ because of inaccuracy of the computed values y, leading to inaccuracy in the computed values of g(x,y)$g\left(x,y\right)$ used in the search for the solution of g(x,y) = 0.0$g\left(x,y\right)=0.0$. This difficulty can be overcome by reducing tol sufficiently, and if necessary, by choosing hmax sufficiently small. If possible, you should choose xend well beyond the expected point where g(x,y) = 0.0$g\left(x,y\right)=0.0$; for example make |xendx|$|{\mathbf{xend}}-{\mathbf{x}}|$ about 50%$50%$ larger than the expected range. As a simple check, if, with xend fixed, a change in tol does not lead to a significant change in y at xend, then inaccuracy is not a likely source of error.
If nag_ode_ivp_rkm_zero_simple (d02bh) fails with ${\mathbf{ifail}}={\mathbf{3}}$, then it could be called again with a larger value of tol if this has not already been tried. If the accuracy requested is really needed and cannot be obtained with this function, the system may be very stiff (see below) or so badly scaled that it cannot be solved to the required accuracy.
If nag_ode_ivp_rkm_zero_simple (d02bh) fails with ${\mathbf{ifail}}={\mathbf{2}}$, it is likely that it has been called with a value of tol which is so small that a solution cannot be obtained on the range x to xend. This can happen for well-behaved systems and very small values of tol. You should, however, consider whether there is a more fundamental difficulty. For example:
(a) in the region of a singularity (infinite value) of the solution, the function will usually stop with ${\mathbf{ifail}}={\mathbf{2}}$, unless overflow occurs first. If overflow occurs using nag_ode_ivp_rkm_zero_simple (d02bh), nag_ode_ivp_rkts_onestep (d02pf) can be used instead to detect the increasing solution, before overflow occurs. In any case, numerical integration cannot be continued through a singularity, and analytical treatment should be considered; (b) for ‘stiff’ equations, where the solution contains rapidly decaying components, the function will compute in very small steps in x$x$ (internally to nag_ode_ivp_rkm_zero_simple (d02bh)) to preserve stability. This will usually exhibit itself by making the computing time excessively long, or occasionally by an exit with ${\mathbf{ifail}}={\mathbf{2}}$. Merson's method is not efficient in such cases, and you should try nag_ode_ivp_bdf_zero_simple (d02ej) which uses a Backward Differentiation Formula method. To determine whether a problem is stiff, nag_ode_ivp_rkts_range (d02pe) may be used.
For well-behaved systems with no difficulties such as stiffness or singularities, the Merson method should work well for low accuracy calculations (three or four figures). For high accuracy calculations or where fcn is costly to evaluate, Merson's method may not be appropriate and a computationally less expensive method may be nag_ode_ivp_adams_zero_simple (d02cj) which uses an Adams method.
For problems for which nag_ode_ivp_rkm_zero_simple (d02bh) is not sufficiently general, you should consider nag_ode_ivp_rkts_onestep (d02pf). nag_ode_ivp_rkts_onestep (d02pf) is a more general function with many facilities including a more general error control criterion. nag_ode_ivp_rkts_onestep (d02pf) can be combined with the rootfinder nag_roots_contfn_brent_rcomm (c05az) and the interpolation function nag_ode_ivp_rkts_interp (d02ps) to solve equations involving y1,y2,,yn${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$ and their derivatives.
nag_ode_ivp_rkm_zero_simple (d02bh) can also be used to solve an equation involving x$x$, y1,y2,,yn${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$ and the derivatives of y1,y2,,yn${y}_{1},{y}_{2},\dots ,{y}_{\mathit{n}}$. For example in Section [Example], nag_ode_ivp_rkm_zero_simple (d02bh) is used to find a value of x > 0.0${\mathbf{x}}>0.0$ where y(1) = 0.0${\mathbf{y}}\left(1\right)=0.0$. It could instead be used to find a turning-point of y1${y}_{1}$ by replacing the function g(x,y)$g\left(x,y\right)$ in the program by:
```function result = g(x,y)
f = d02bh_f(x,y);
result = f(1);
```
This function is only intended to locate the first zero of g(x,y)$g\left(x,y\right)$. If later zeros are required, you are strongly advised to construct your own more general root-finding functions as discussed above.
## Example
```function nag_ode_ivp_rkm_zero_simple_example
x = 0;
xend = 10;
y = [0.5;
0.5;
0.6283185307179586];
tol = 0.0001;
irelab = int64(0);
hmax = 0;
[xOut, yOut, tolOut, ifail] = ...
nag_ode_ivp_rkm_zero_simple(x, xend, y, tol, irelab, hmax, @fcn, @g)
function f = fcn(x,y)
f = zeros(3,1);
f(1) = tan(y(3));
f(2) = -0.032*tan(y(3))/y(2) - 0.02*y(2)/cos(y(3));
f(3) = -0.032/y(2)^2;
function result = g(x,y)
result = y(1);
```
```
xOut =
7.2884
yOut =
-0.0000
0.4749
-0.7601
tolOut =
1.0000e-04
ifail =
0
```
```function d02bh_example
x = 0;
xend = 10;
y = [0.5;
0.5;
0.6283185307179586];
tol = 0.0001;
irelab = int64(0);
hmax = 0;
[xOut, yOut, tolOut, ifail] = ...
d02bh(x, xend, y, tol, irelab, hmax, @fcn, @g)
function f = fcn(x,y)
f = zeros(3,1);
f(1) = tan(y(3));
f(2) = -0.032*tan(y(3))/y(2) - 0.02*y(2)/cos(y(3));
f(3) = -0.032/y(2)^2;
function result = g(x,y)
result = y(1);
```
```
xOut =
7.2884
yOut =
-0.0000
0.4749
-0.7601
tolOut =
1.0000e-04
ifail =
0
```
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# Surprise! 'Star Trek' gold shirts more deadly than red shirts
Red-shirted "Star Trek" characters have earned a reputation as the most likely ones to get offed in an episode. What if our assumptions are wrong?
It's not easy being a "Star Trek" extra. I can only imagine the cold sweats and terror you'd feel if the wardrobe manager handed you a fresh red shirt from the rack. Your chance of making it through the episode just took a dive. Or did it? New mathematical calculations show that the redshirt death rate may not be as dire as we thought.
Matthew Barsalou breaks down the numbers for Significance Magazine. He sticks with the original series and starts by looking at the casualty rates for the different uniform colors across all three seasons.
There were a total of 55 deaths. In sheer numbers, redshirts took the brunt of it with 24 dead. Only 9 characters wearing the yellow/gold command uniforms kicked the space bucket. Rounding it out, 7 blue shirts bit the dust and 15 of the deceased were of an unknown uniform color.
If you just stop there, it sure looks like redshirts are the most dangerous branch of Starfleet service. But wait, there's more. With an official count of 430 total crew members, the numbers show that 239 work in security, engineering, and operations, wearing red shirts.
Barsalou then goes all math geek and applies to the data the Bayes' Theorem formula for calculating conditional probabilities. After a little mathematical shake and bake, he determines there is a 61.9 percent chance that any given casualty is wearing a red shirt. That still sounds high, but it's not really once you consider the sheer number of redshirts running around the Starship.
"Although Enterprise crew members in redshirts suffer many more casualties than crew members in other uniforms, they suffer fewer casualties than crew members in gold uniforms when the entire population size is considered," Barsalou writes. "Only 10 percent of the entire redshirt population was lost during the three year run of Star Trek. This is less than the 13.4 percent of goldshirts, but more than the 5.1 percent of blueshirts."
There you have it. When you're applying for Star Fleet Academy and considering which branch of service you want to go into, you may actually be better off wearing a red shirt than a gold one. To be on the safe side, though, you might want to follow Mr. Spock's footsteps into the sciences branch.
(Via The Mary Sue)
#### Huge scientific breakthrough and a big hole in Windows patched
Researchers find a new way to observe the universe. Meanwhile, Microsoft issued an update that fixes a major flaw in Windows.
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http://www.jiskha.com/members/profile/posts.cgi?name=martin&page=14
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Tuesday
December 10, 2013
# Posts by martin
Total # Posts: 170
Statistics
What is the Central Limit Theorem and how can it be used in statistics?
Statistics
I have to sample 300 persons. How do I go about doing this and how do I know that the population I have chosen has the characteristics that I am looking for?
math word problem
Case Eastern Junior College awarded 26 varsity letters in crew, 15 in swimming, and 16 in scoccer. If awards went to 46 students and only 2 lettered in all sports, how many students lettered in two of the three sports?
math
a=9b,for b the solution is b=
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Calculate the rate at which N2O4 is formed in the following reaction at the moment in time when NO2 is being consumed at a rate of 0.0521 M/s. 2NO2(g) <-> N2O4 (g) I am not really sure how to go about this problem. rate = k (N2O4) d(NO2)/dt = -2 d(N2O4)/dt This is what I...
Hi First off, I do not know anything about US auto parts. I want to know how strong the competition is here. Who are the big players? If you have any relevant articles as well, i would be happy. Thanks in advance Martin
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a pasta machine costs \$33.The ingredients to make one batch of pasta cost \$.33.The same amouunt of pasta puchased at a store costs \$.99.How many batches of pasta will you have to make for the cost of the machine and ingrediants to eequal the cost of buying the same amount of p...
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your family is driving to houston,texas.A sign indicates that you are 700 miles from houston.Your car's trip odometer indicates you are 400 miles from home.YOu are traveling at an average speed of 60 miles per hour. A. write an expression for the distance(in miles) you wil...
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For-loop to create vector of mean differences in t-tests
Let’s say I have
``````set.seed(1)
e <- data.frame(tti = log2(runif(200)),
corona = c(rep("Corona", 100), rep("Before", 100)),
type = rep(c("A", "B", "C", "D")))
``````
I am comparing mean differences between `tti` (time to treatment initiation on `log2`-scale) before and during COVID-19 for each `e\$type` (here `n=4` but many more in my dataset). I want to apply a `for loop` for this repetitive task, but I am quite new to this and frankly stock at the moment.
My current attempt:
``````for(i in unique(e\$type)){
m <- c(
round(1-2^(unique(t.test(e\$tti[e\$type == i] ~ e\$corona[e\$type == i])\$estimate[2]) -
unique(t.test(e\$tti[e\$type == i] ~ e\$corona[e\$type == i])\$estimate[1])),
digits = 3)*100
)
}
``````
However, this attempt only return one value.
Expected output:
`m` should be a vector containing the four estimates of mean differences
``````> m
24.7 10.5 1.5 28.7
``````
How can this be done?
>Solution :
You could do:
``````m <- c()
for(i in unique(e\$type)){
m[i] <- c(
round(1-2^(unique(t.test(e\$tti[e\$type == i] ~ e\$corona[e\$type == i])\$estimate[2]) -
unique(t.test(e\$tti[e\$type == i] ~ e\$corona[e\$type == i])\$estimate[1])),
digits = 3)*100
)
}
``````
This initializes m and then fills it during the loop. That should give you all 4 results in one vector.
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The following table shows the number of feet a helicopter flies above the ground as a function of time: Time (in seconds) x Distance above ground (in feet) f(x) 0 18 2 16 4 14 6 12 8 10 Find and interpret the meaning of the x‐intercept in this scenario. (10, 0); the time it takes for the helicopter to reach the ground (18, 0); the time it takes the helicopter to take off (10, 0); the time it takes the helicopter to take off (18, 0); the time it takes for the helicopter to reach the ground
Question:
The following table shows the number of feet a helicopter flies above the ground as a function of time:
Time (in seconds)
x Distance above ground (in feet)
f(x)
0 18
2 16
4 14
6 12
8 10
Find and interpret the meaning of the x‐intercept in this scenario.
(10, 0); the time it takes for the helicopter to reach the ground
(18, 0); the time it takes the helicopter to take off
(10, 0); the time it takes the helicopter to take off
(18, 0); the time it takes for the helicopter to reach the ground
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Read the excerpt below from act 3.2 of The Tragedy of Julius Caesar and answer the question that follows. ANTONY: If you have tears, prepare to shed them now. You all do know this mantle1. I remember The first time ever Caesar put it on. ‘Twas on a summer’s evening in his tent, That day he overcame the Nervii2. Look, in this place ran Cassius’ dagger through. See what a rent the envious Casca made. Through this the well-belovèd Brutus stabbed; 1. A cloak. 2. One of Caesar’s military conquests. W
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https://cs50.stackexchange.com/questions/9441/help-with-pset-1-mario/9442
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# Help With PSET 1 - Mario!
Ive been struggling with this problem for quite some time now and im finally caving and asking for help. first ill show you my code and explain my thought process.
``````#include <cs50.h>
#include <stdio.h>
int main(void)
{
int height, rows, spaces, hashes;
do
{
printf("How high will mario climb? Please pick a number between 1-23: ");
height = GetInt();
}
while (height <= 0 || height >= 24);
for (rows = 0; rows <= height; rows++) //print user generated amount of rows
{
for (spaces = height - rows - 2 ; spaces == 0; spaces --) //print spaces
{
printf(" ");
}
for (hashes = rows + 2; hashes <= height; hashes++) //print hashes
{
printf("#");
}
printf("\n");
}
return 0;
}
``````
this is printing: usuing user input (10)
``````##########
#########
########
#######
######
#####
####
###
##
#
``````
at first I thought I did something backwards so i switched space to print "s" and hashes to print "h" expecting
``````hhhhhs
hhhhss
hhhsss
hhssss
shssss
``````
but got
``````hhhhhh
hhhhh
hhhh
hhh
hh
h
``````
So I believe the problem lies somewhere in the for loops, but I can not figure out what it is. (I'm also aware the final row is messed up but Im trying to fix the pyramid first)
my logic behind: (rows = 0; rows <= height; rows++) //print user generated amount of rows
is that the rows will increase by one until it reaches the height specified by the user (I believe everything up to and including this line of code is correct)
...
my logic behind : (spaces = height - rows - 2 ; spaces == 0; spaces --) //print spaces
is that the amount of spaces should = the height (10) - rows (0) - 2 and then decrease by 1 each time through the loop until eventually there is only hashes. this way the first row should leave room for 2 hashes(top of the pyramid)
...
logic behind (hashes = rows + 2; hashes <= height; hashes++) //print hashes
is that the amount of hashes should be = to the row (0) + 2 increasing by 1 each time though the loop until it is greater than or equal to the user input height. This way the first row will include 2 hashes (0 + 2 = 2)
any insight as to where the problem is and tips or advise on how to come up with the solution would be greatly appreciated !
You have several bugs, but you're on the right track. I'll try to get you thinking about it, without actually doing it for you.
``````for (spaces = height - rows - 2 ; spaces == 0; spaces --) //print spaces
``````
If you printed "s" instead of space, the code would actually have printed a single "s" in the entire output. Why does this happen? Why doesn't it print one or more "s" on each line? Except for the one time, this line never prints anything. It isn't going into the loop. Why? How does a FOR loop command execute? Remember that it is executing the test condition to see if it can go into the loop. What is the actual test here? What are the values of height, rows and , most important, "spaces" when it tries to start the loop?
Once that's fixed, your next issue is that your pyramid is upside down. You're printing the bottom of the pyramid first. How can you flip that? You need to change the logic of how many #'s are printed. At least you really are printing a progression, so your logic is in the neighborhood.
Finally, make sure you're printing the right number of spaces and #s on each row.
You're close. Once you fix the test condition, the rest should be easy to fix. If not, come back and let us know what progress you've made.
• got it!! I decided to try everything the opposite way. Instead of the initialization expression starting at the desired integer, I started them at zero and then worked up to the desired integer based on the text expression. most examples Ive seen in my brief time with programming have the initialization expression starting at zero, is that the way it usually is? – Eric Bezanson Apr 9 '15 at 1:33
• Either way works, starting at 0 and going up is more common. On another note think about this. The number of spaces plus the number of #s on each line is a constant. ;-) Anyways, if my answer was sufficient, please mark it as answered. Good luck. – Cliff B Apr 9 '15 at 1:54
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https://www.mathhomeworkanswers.org/191924/infection-spreading-equation-expression-people-infected?show=191930
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(N represents the number of weeks.)
Solve for X by Graphing | TI-83 Plu...
Solve for X by Graphing | TI-83 Plus and TI-84 Plus graphing calculators
This is similar to compound interest. If the number of people infected is P to start with (P=22 in this case) then after one week the number of people infected is P(1+r), where r is the rate of infection (=17% or 0.17 in this case). At the end of the second week, the number of people infected rises to P(1+r)(1+r)=P(1+r)^2. After the third week it's P(1+r)^3 and so on. After N weeks it's P(1+r)^N. After 28 weeks it's 22*(1.17^28)=22*81.134=1785 people.
by Top Rated User (1.0m points)
Thank you so much!
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Confronting The Unprovable - Gödel And All That
### New Book Reviews!
Confronting The Unprovable - Gödel And All That
Written by Mike James
Tuesday, 11 February 2014
Article Index
Confronting The Unprovable - Gödel And All That
Gödel
Kurt Gödel (1906-1978)
The argument that Gödel used is very simple.
Let’s suppose that there is a machine M that can do the job we envisaged for a consistent system.
That is, the machine can prove or disprove any theorem that we feed to it. It is programmed with the axioms of the system and it can use them to provide proofs.
Now suppose we ask for the program for the machine written in the same logic used for the theorems that the machine proves. After all any computer can be reduced to logic and all we are requesting is the logical expression that represents the design of the computer – call this PM for "Program for M".
Now we take PM and construct a logical expression which says
“The machine constructed according to PM will never prove this statement is true”
and call this statement X.
Now we ask the machine to prove or disprove X.
Consider the results. If the machine says “yes this is true” then the statement is false. If the machine says that the statement is false then the statement is true.
The proof that there are theorems that cannot be proved
You see what the problem is here and you can do this sort of trick with any proof machine that is sufficiently powerful to accept its own description as a theorem.
Any such machine, and the system of logic on which it is based, must, by its very nature be inconsistent in that there are theorems that can be written using the system that it cannot be used to prove either true or false.
In other words there are three types of theorem in the system – those that are provably true, those that are provably false and those that are undecidable using the axioms that we have at our disposal.
OK, you may say but is it possible to for a machine to be powerful enough to attempt to prove a statement that involves its own description? Perhaps it is in the nature of machines that they cannot cope with their own description.
The really clever part of Gödel’s work was to find that any axiomatic system powerful enough to describe the integers, i.e. powerful enough to describe simple arithmetic, has this property.
That is arithmetic isn’t a consistent axiomatic theory which means that there are theorems about the integers that have no proof or disproof within the theory of arithmetic.
So now think about the 300-year search for a proof for Fermat’s last theorem.
Perhaps it wasn’t that the mathematicians weren’t trying hard enough. Perhaps there was no proof. As it turns out we now know that a proof exists but there was a long time when it was a real possibility that the theorem was undecidable.
You clearly understand the idea but do you believe it?
Consider the following problem: there seem to be lots of paired primes i.e. primes that differ by two (3,5), (5,7), (11,13) and so on. It is believed that there are an infinite number of such pairs - the so called "twin prime conjecture" but so far no proof.
So what are the possibilities?
You examine number pairs moving over larger and large integers and occasionally you meet paired primes.
Presumably either you keep on meeting them or you there comes a point when you don’t.
In other words, the theorem is either true or false. And as it is true or false presumably there should be a proof of this.
If you have taken Gödel’s theorem to heart you should know that this doesn’t have to be the case. The integers go on forever and you can’t actually decide the truth of the theorem by looking at the integers.
How far do you have to go without seeing a pair of primes to be sure that you aren’t ever going to see another pair?
How many pairs do you have to see to know that you are going to keep seeing them?
There is no answer to either question. In the same way why do you assume that there is a finite number of steps in a proof that will determine the answer.
Why should the infinite be reducible to the finite?
Only because we have grown accustomed to mathematics performing this miracle.
In the case of the twin prime conjecture recent progress has proved that that are infinitely many primes separated by N and N has to be less than 70 million. This doesn't mean that N is 2 however. Collaborative work by a lot of mathematicians has reduced the upper limit (at the end of 2013) to less than 576 which seems like progress. Can the bound be pushed down as far as 2 or is there really no proof?
Fermat’s last theorem stated that the only integer solution to a particular equation was when n equals 2. To prove this the hard way requires examining the equation for n equal to 1, 2, 3… and for all a,b and c and this means we have to test an infinite number of possibilities.
Yet we have a finite proof.
We have a logical derivation of the truth of the theorem, which doesn’t involve testing an infinite number of cases.
This is what Gödel’s theorem really is all about.
There are statements that are undecidable. If you add additional axioms to the system the statements that were undecidable might well become decidable but there will still be valid statements that are undecidable. Indeed every time you expand the axioms you increase the number of theorems that are decidable and undecidable.
It’s as if mathematics at the turn of the 20th century was seeking the ultimate theory of everything and Gödel proved that this just wasn’t possible.
So far so good, or bad depending on your point of view.
You may even recognise some of this theory as very similar to the theory of Turing machines and non-computability, in which case it might not be too much of a shock to you. However, at the time they were thought up both Gödel’s and Turing’s ideas were revolutionary and they were both regarded with suspicion and dismay.
It was thought to be the end of the dream: mathematics was limited. Mathematics wasn’t perfect and in fact every area of mathematics contained its limitations.
Today you will find it argued that Gödel’s theorem proves that god exists. You will find it argued that Gödel’s theorem proves that human thought goes beyond logic. The human mind is capable of seeing truth that mathematics cannot prove. It is also argued that it limits artificial intelligence, because there are things that any machine cannot know and hence also proves that human intelligence is special because it can know what the machine cannot.
If you think about it, Gödel’s theorem proves none of this. It doesn’t even suggest that any of this is the case.
Gödel’s theorem doesn’t deal with probabilities and what we believe, only in the limitations of finite systems in proving assertions about the infinite.
Sometimes the infinite is regular enough to allow something to be proved. Sometimes, in fact most of the time, it isn’t.
But important though this is we live in a finite personal universe and we don’t demand perfect proof. We go with the flow, guess and accept good probabilities as near certainties.
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<ASIN:1568812388>
Last Updated ( Tuesday, 11 February 2014 )
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https://technicalsource9.medium.com/how-to-solve-this-issue-variable-x-is-not-fully-defined-on-some-execution-paths-868c8c7313a1?source=post_internal_links---------2----------------------------
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# How to solve this issue Variable ‘X’ is not fully defined on some execution paths.
2 min readSep 14, 2021
--
Dear All,
I have a code into a MATLABFUNCTION block of Simulink that jump out this error message “
Variable ‘CONDICIONTRAMO’ is not fully defined on some execution paths. Function ‘Battery_Control/Function_Cruz_v2’ (#25.1881.1895), line 57, column 12: “CONDICIONTRAMO” Launch diagnostic report.”
Anyone of you can help me out to find the way to solve this ? Please.
`function [SOC_CC, Bat_charge] = BUsage(top, SOC, SOC_Seg, SOCint, W_demand, Power_PV)if SOC>=SOC_Seg SOC_CC=0; Bat_charge=0; returnelse if SOC>=SOCint SOC_CC=1; Bat_charge=0; end if SOC<=SOCint if (top>=8)||(top<=9) && Power_PV>W_demand CONDICIONTRAMO = 1; disp('Llano 1 de 8 a 9') elseif (top>=8)||(top<=9) && Power_PV=10)||(top<=13) && Power_PV>W_demand CONDICIONTRAMO = 3; disp('Punta 1 de 10 a 13') elseif (top>=10)||(top<=13) && Power_PV=14)||(top<=17) && Power_PV>W_demand CONDICIONTRAMO = 5; disp('Llano 2 de 14 a 17') elseif (top>=14)||(top<=17) && Power_PV=18)||(top<=21)&& Power_PV>W_demand CONDICIONTRAMO = 7; disp('Punta 2 de 18 a 21') elseif (top>=18)||(top<=21)&& Power_PV=22)||(top<=23)&& Power_PV>W_demand CONDICIONTRAMO = 9; disp('Llano 3 de 22 a 23:59') elseif (top>=22)||(top<=23)&& Power_PV=24)||(top<=8) && Power_PV`
# ANSWER
Matlabsolutions.com provide latest MatLab Homework Help,MatLab Assignment Help for students, engineers and researchers in Multiple Branches like ECE, EEE, CSE, Mechanical, Civil with 100% output.Matlab Code for B.E, B.Tech,M.E,M.Tech, Ph.D. Scholars with 100% privacy guaranteed. Get MATLAB projects with source code for your learning and research.
Insert a definition of CONDICIONTRAMO = -1 (or NaN, or 12) either on top of the code, before the IF branchs, or in a finaly else branch:
`... elseif (top>=24)||(top<=8) && Power_PV`
But then Bat_charge is not defined in all branchs. Add a default value for this variable also.
By the way, the Switch/Case-Block is hard to read. Maybe this is tougher:
`% 1 2 3 4 5 6 7 8 9 10 11 12SOC_CC_V = [1 0 1 0 1 0 1 0 1 0 1 NaN];Bat_charge_V = [0 0 0 0 0 0 0 0 0 0 1 NaN];SOC_CC = SOC_CC_V(CONDICIONTRAMO);Bat_charge = Bat_charge_V(CONDICIONTRAMO);`
SEE COMPLETE ANSWER CLICK THE LINK
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+0
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6 / 2 x (1+2) = ?
Jul 31, 2017
#1
+8956
+3
6 / 2 × (1 + 2)
Evaluate the expression in parenthesees first.
= 6 / 2 × 3
We can also write this as...
= $$\frac62\,\times\,3$$
And... 6 / 2 = 3
= 3 × 3
= 9
Jul 31, 2017
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# Two Pizzas and Five People
I’m thinking a lot about how my 6th graders responded to a pre-lesson task in “Interpreting Multiplication and Division” — a lesson from Mathematics Assessment Project (MAP) .
MAP lessons begin with a set structure:
Before the lesson, students work individually on a task designed to reveal their current levels of understanding. You review their scripts and write questions to help them improve their work.
I gave the students this pre-lesson task for homework, and 57 students completed the task.
I’m sharing students’ responses to question 2 (of 4) only because there’s already a lot here to process. I’m grouping the kids’ calculations and answers based on their diagrams.
Each pizza is cut into fifths.
About 44% (25/57) of the kids split the pizzas into fifths. I think I would have done the same, and my hand-drawn fifths would only be slightly less sloppy than theirs. The answer of 2 must mean 2 slices, and that makes sense when we see 10 slices total. The answer 10 might be a reflection of the completed example in the first row. “P divided by 5 x 2 or 5 divided by P x 2” suggests that division is commutative, and P here must mean pizza.
Each pizza is cut into eighths.
Next to cutting a circle into fourths, cutting into eighths is pretty easy and straightforward.
Each pizza is cut into tenths.
I’m a little bit surprised to see tenths because it’s tedious to sketch them in, but then ten is a nice round number. The answer 20, like before, might be a mimic of the completed example in the first row.
Each pizza is cut into fourths.
I’m thinking the student sketched the diagram to illustrate that the pizzas get cut into some number of pieces — the fourths are out of convenience. The larger number 5 divided by the smaller number 2 is not surprising.
Each pizza is cut into sixths.
It’s easier to divide a circle by hand into even sections, even though the calculations do not show 6 or 12.
Each pizza is cut into fifths, vertically.
Oy. I need to introduce these 3 students to rectangular pizzas. :)
Five people? Here, five slices.
Mom and Dad are bigger people, so they should get the larger slices. This seems fair. We just need to examine the commutative property more closely.
Circles drawn, but uncut.
I’m wondering about the calculation of 5 divided by 2.
Only one pizza drawn, cut into fifths.
Twenty percent fits with the diagram, if each person is getting one of the five slices. The 100 in the calculation might be due to the student thinking about percentage.
Only one pizza drawn, cut into tenths, but like this.
I wonder if the student has forgotten what the question is asking for because his/her focus has now shifted to the diagram.
Each rectangular pizza is cut into fifths.
Three kids after my own heart.
Five portions set out, each with pizza sticks.
I wonder where the 10 comes from in his calculation.
Five plates set out, each plate with pizza slices.
Kids don’t always know what we mean by “draw a picture” or “sketch a diagram.” This student has already portioned out the slices.
What diagrams and calculations would you expect to see for question 3?
There’s important work ahead for us. The kids have been working on matching calculation, diagram, and problem cards. They’re thinking and talking to one another. I have a lot of questions to ask them, and hopefully they’ll come up with questions of their own as they try to make sense of it all. If I were just looking for the answer of 2/5 or 0.4, then only 12 of the 57 papers had this answer. But I saw more “correct” answers that may not necessarily match the key. We starve ourselves of kids’ thinking and reasoning if we only give multiple-choice tests or seek only for the answer.
That’s why Max Ray wants to remind us of why 2 > 4.
1. Posted September 26, 2014 at 4:36 pm | Permalink
Your blog system just timed me out and I lost my comment.
I am going to do it in notepad and try again.
GRRRR!!!
• Fawn
Posted September 26, 2014 at 9:06 pm | Permalink
So sorry, Howard. Not sure how to fix this as you’ve mentioned this before. I’m all ears if anyone knows what I can check into to fix this pesky feature.
2. Posted September 26, 2014 at 5:01 pm | Permalink
This is fascinating.
some thoughts on no. 3
1: Square cake for this one
2. How would they cope with the cake cut into 12 pieces, and Max eats two of them
Some possible follow-up questions for the pizzas:
1: Can you explain how you got this calculation?
Some thoughts inspired by George Polya (I am a Polya fan)
1: Is there a simpler related problem that I can solve?
2: None of them thought of putting one pizza on top of the other one.
I was also thinking about how they would proceed if they were not asked to produce a calculation.
The last person in your first table (1/5 + 1/5) clearly ignored the instructions, and probably saw (visualized) the process almost straight away.
Too much specification of “how to do it” can limit the brain activity, and not only that, it goes against the Common Core philosophy. I have just posted a criticism of one of the PARCC sample high school geometry test items on these grounds.
3. Posted October 25, 2014 at 9:03 pm | Permalink
I haven’t taught a MAP lesson yet but I gave the preassessment for the Repeating Decimals 8th grade lesson and can’t wait to teach it. It’s a lot of prep but I think teachers need to realize that the lessons that require the most prep tend to be the richest.
4. Posted October 26, 2014 at 4:09 am | Permalink
It seems to me that the kids did remarkably badly on this question, relative to the standard “math-teacher” answer: 2/5 of a pizza.
My hypothesis is that the students were struggling mathsemantically to identify the terms in which their answer should be expressed and were inhibited by their real world experience of sharing pizzas. In vernacular, we count multiple whole pies using “pizza” as the unit and fractions of a pie as “slices.” In a math classroom context, “slice” is not a sensible unit because the whole point is to distinguish differently sized pizza fractions and “slice” is not a uniformly defined amount of pizza.
As to real-world pizza sharing, the common restaurant experience is to receive pre-cut pizza, typically sliced into 8 pieces (though 4 and 16 are also standard for large or small pies) and divvy up the slices. Remainders are seized by those with the highest product of gluttony times reflex speed.
Finally, I can’t resist harping on the overwhelming problem of using pizza as a unit for even whole pies. In the problem why should we believe that the two pizzas have the same size and, even if they do, they probably don’t have the same toppings. Can we really trust that 2/5 of a pizza has any useful comparability (for fair sharing purposes) to 2/5 of a different pizza? My children would loudly argue: no!
So, I suggest being very careful using pizzas for fraction questions, even though it is so tempting. Fractions of a single pie: ok. Fractions across multiples pies: avoid.
If any of this is correct, then the students should do a lot better with the next cake question.
5. Mt
Posted April 16, 2019 at 6:05 pm | Permalink
Have you shared calculation, diagrams and problem cards? I love that idea!
6. Posted June 11, 2019 at 12:10 am | Permalink
Download .pdf: SEWELL, Martin, 2011. History of the efficient market hypothesis . Research Note RN/11/04, University College London, London.
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# Cell Membrane and Hollow Spheres
1. Jan 25, 2005
### figs
there are 2 probs here.
1) The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.0680 V exists across the membrane. The thickness of the membrane is 7.95×10-9 m. What is the magnitude of the electric field in the membrane?
I thought i could just use E=-change in V/change in distance, but no.
2) Two hollow metal spheres are concentric with each other. The inner sphere has a radius of 0.1440 m and a potential of 86.0 V. The radius of the outer sphere is 0.146 m and its potential is 82.0 V. If the region between the spheres is filled with Teflon, find the electric energy contained in this space.
This one i played with in the equation V=(kq/rsquared)+(k/q'/rsquared).
I've just confused myself with working them so many times. so im seeking help here!
2. Jan 25, 2005
### gnome
(1) asks for magnitude, so I think it's just $E = \left | \frac{\Delta V}{d} \right |$
(2) determine the capacitance using the potential difference, the distance between the spheres, and the dielectric constant of the teflon. Then you can find the amount of energy stored using the potential difference and the capacitance.
3. Jan 25, 2005
### figs
i figured it out.
(1) I set V=kq/r equal to E=kq/r^2
(2) i solved Q=[(kEoA)/d]V
and then plugged into V=EPE/Q
its easier for me if the equations are
derived, and i worked it all out nicely.
That second problem was making me crazy!
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# NAMA KELOMPOK : 1.ADI PRANATA 2.M.FAISAL ROSYIDIN 3.JAKA PRATAMA M 4.SHULBY ROCHMAN 5.MICROVELIO PRIMA GUNA 6.DANI DIANSA PUTRA.
## Presentasi berjudul: "NAMA KELOMPOK : 1.ADI PRANATA 2.M.FAISAL ROSYIDIN 3.JAKA PRATAMA M 4.SHULBY ROCHMAN 5.MICROVELIO PRIMA GUNA 6.DANI DIANSA PUTRA."— Transcript presentasi:
NAMA KELOMPOK : 1.ADI PRANATA 2.M.FAISAL ROSYIDIN 3.JAKA PRATAMA M 4.SHULBY ROCHMAN 5.MICROVELIO PRIMA GUNA 6.DANI DIANSA PUTRA
Customers at Joe’s office supply store demand an average of 6000 desks per year. Each time an order is placed. An ordering cost of \$ 300 is incurred. The annual holding cost for a single desk is 25% of the \$ 200 cost of desk. One week elapses between the placement of an order and the arrival of the order. In parts (a)-(d).assume that no shortages are allowed. a. Each time an order is placed, how many desks should be ordered? b. How many orders should be placed each year? c. Determine the total annual costs (excluding purchasing costs) of meeting the customers demands for desks. d. Determine the reorder point. If the lead time were five weeks, what would be the reorder point? (52 weeks = one year)
Diket : D = 6000 ; K = \$ 300 ; h = 25% x \$ 200 = \$ 50 a) q*= = = 268,3 b) Frekuensi pemesanan per tahun : = = 22,4 c) TC(q) = OC+HC = + = + = 6708,91+6707,5 = 13.416,41 ( Tanpa purchasing cost ) d) Lama Siklus = 52/22 = 2,36 LD = = = 568,4
= = 2,12 = Reorder point = LD – 2(q*) = 568,3 – 2(268,3) = 568,3 – 536,6 = 31,7
a. Dari hasil perhitungan didapatkan bahwa bangku yang harus dipesan adalah sebanyak 268 buah. b. Dari hasil perhitungan didapatkan pula bahwa frekuensi pemesanan per tahun adalah sebanyak 22 kali. c. Total biaya tahunan ( tanpa perhitungan purchasing cost ) adalah sebesar \$ 13.416,41 d. Reorder point sebesar 31,7 artinya tingkat sediaan sebanyak 32 buah bangku dimana pemesanan harus dilakukan.
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# Assume that there are only two goods, x and y and two persons, 1 and 2. The total quantity of x avai
1. Assume that there are only two goods, x and y and two persons, 1 and 2.
The total quantity of x available is 10 and for y is 20. Utility increases only with quantity consumed for both goods.
“Suppose (x1,y1)=(10,20) and (x2,y2)=(0,0). This is a Pareto efficient outcome because x1>x2 and y1>y2.” Is this statement true? Why it is or is not true?
“Since moving from (x1,y1)=(10,20) and (x2,y2)=(0,0) to (x1,y1)=(10,10) and (x2,y2)=(0,5) is not a Pareto improvement, the allocation (x1,y1)=(10,10) and (x2,y2)=(0,5) is not Pareto efficient.” Is this statement true? Why it is or is not true?
2. (b) “If asymmetric information exists, the first welfare theorem predicts that the perfectly competitive market equilibrium is inefficient.” Is this statement
true? Why it is or is not true?
(c) “In some ancient societies, all people worked alone, and did not interact
with each other, and so there was no public good and asymmetric information problem. They were self-centred, never cared about and never affected other people’s life, and so no externality existed. The first welfare theorem predicts that such an equilibrium was efficient.” Is this statement true? Why it is or is not true?
1. (d) “Banning de-merit goods, such as some soft drinks, increases the utility of the consumers of these goods although these consumers will disagree. The doctors are better off while the consumers worse off. This is not a Pareto improvement” Is this statement true? Why it is or is not true?
2. (e) “Some economists say that a tax rate at 70% on the rich people is required for fairness. This is their value judgment. To avoid imposing their values on us, economists should concentrate on efficiency, not on equity.” Please comment.
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# Horsepower (boiler) to kilowatts (hp to kW)
## Convert horsepower (boiler) to kilowatts
Horsepower (boiler) to kilowatts converter above calculates how many kilowatts are in 'X' horsepower (boiler) (where 'X' is the number of horsepower (boiler) to convert to kilowatts). In order to convert a value from horsepower (boiler) to kilowatts (from hp to kW) simply type the number of hp to be converted to kW and then click on the 'convert' button.
## Horsepower (boiler) to kilowatts conversion factor
1 horsepower (boiler) is equal to 9.810657 kilowatts
## Horsepower (boiler) to kilowatts conversion formula
Power(kW) = Power (hp) × 9.810657
Example: 136 horsepower (boiler) equal how many kilowatts? To solve this, multiply 136 horsepower (boiler) with the conversion factor from horsepower (boiler) to kilowatts.
Power(kW) = 136 ( hp ) × 9.810657 ( kW / hp )
Power(kW) = 1334.249352 kW or
136 hp = 1334.249352 kW
136 horsepower (boiler) equals 1334.249352 kilowatts
## Horsepower (boiler) to kilowatts conversion table
horsepower (boiler) (hp)kilowatts (kW)
1098.10657
15147.159855
20196.21314
25245.266425
30294.31971
35343.372995
40392.42628
45441.479565
horsepower (boiler) (hp)kilowatts (kW)
2001962.1314
3002943.1971
4003924.2628
5004905.3285
6005886.3942
7006867.4599
8007848.5256
9008829.5913
Versions of the horsepower (boiler) to kilowatts conversion table. To create a horsepower (boiler) to kilowatts conversion table for different values, click on the "Create a customized power conversion table" button.
## Related power conversions
Back to horsepower (boiler) to kilowatts conversion
TableFormulaFactorConverterTop
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Assessment for Learning
Assessment Home Resource Document Forms and Templates Brochures Enduring Understandings Curriculum.org
Enduring Understandings and Essential Questions
MFM2P
Proportional Reasoning
By the end of the course, students will:
• solve problems derived from a variety of applications, using proportional reasoning
• solve problems involving similar triangles
• solve problems involving right triangles, using trigonometry
Applying Piecewise Linear Functions
By the end of the course, students will:
• explain the characteristics of situations involving piecewise linear functions (e.g., pay scale variations, gas consumption costs, water consumption costs, differentiated pricing, motion)
• construct tables of values and sketch graphs to represent given descriptions of realistic situations involving piecewise linear functions, with and without the use of graphing calculators or graphing software
• answer questions about piecewise linear functions by interpolation and extrapolation, and by considering variations on given conditions
By the end of the course, students will:
• manipulate algebraic expressions as they relate to quadratic functions
• determine, through investigation, the relationships between the graphs and the equations of quadratic functions
• solve problems by interpreting graphs of quadratic functions
Enduring Understanding:
Relationships exist between real world quantities and they may be modeled and predicted with mathematics.
Essential Questions:
What is an appropriate strategy and approach to model this relationship or any relationship? How can you use that model to make a prediction and solve a problem?
© 2013 - Bluewater District School Board
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# INDEX function in Excel - 6 most efficient uses
In this tutorial, you will find a number of formula examples that demonstrate the most efficient uses of INDEX in Excel.
Of all Excel functions whose power is often underestimated and underutilized, INDEX would definitely rank somewhere in the top 10. In the meantime, this function is smart, supple and versatile.
So, what is the INDEX function in Excel? Essentially, an INDEX formula returns a cell reference from within a given array or range. In other words, you use INDEX when you know (or can calculate) the position of an element in a range and you want to get the actual value of that element.
This may sound a bit trivial, but once you realize the real potential of the INDEX function, it could make crucial changes to the way you calculate, analyze and present data in your worksheets.
## Excel INDEX function - syntax and basic uses
There are two versions of the INDEX function in Excel - array form and reference form. Both forms can be used in all versions of Microsoft Excel 365 - 2003.
### INDEX array form
The INDEX array form returns the value of a certain element in a range or array based on the row and column numbers you specify.
INDEX(array, row_num, [column_num])
• array - is a range of cells, named range, or table.
• row_num - is the row number in the array from which to return a value. If row_num is omitted, column_num is required.
• column_num - is the column number from which to return a value. If column_num is omitted, row_num is required.
For example, the formula `=INDEX(A1:D6, 4, 3)` returns the value at the intersection of the 4th row and 3rd column in range A1:D6, which is the value in cell C4.
To get an idea of how the INDEX formula works on real data, please have a look at the following example:
Instead of entering the row and column numbers in the formula, you can supply the cell references to get a more universal formula: `=INDEX(\$B\$2:\$D\$6, G2, G1)`
So, this INDEX formula returns the number of items exactly at the intersection of the product number specified in G2 (row_num) and week number entered in cell G1 (column_num).
Tip. The use of absolute references (\$B\$2:\$D\$6) instead of relative references (B2:D6) in the array argument makes it easier to copy the formula to other cells. Alternatively, you can convert a range to a table (Ctrl + T) and refer to it by the table name.
#### INDEX array form - things to remember
1. If the array argument consists of only one row or column, you may or may not specify the corresponding row_num or column_num argument.
2. If the array argument includes more than one row and row_num is omitted or set to 0, the INDEX function returns an array of the entire column. Similarly, if array includes more than one column and the column_num argument is omitted or set to 0, the INDEX formula returns the entire row. Here's a formula example that demonstrates this behavior.
3. The row_num and column_num arguments must refer to a cell within array; otherwise, the INDEX formula will return the #REF! error.
### INDEX reference form
The reference form of the Excel INDEX function returns the cell reference at the intersection of the specified row and column.
INDEX(reference, row_num, [column_num], [area_num] )
• reference - is one or several ranges.
If you are entering more than one range, separate the ranges by commas and enclose the reference argument in parentheses, for example (A1:B5, D1:F5).
If each range in reference contains only one row or column, the corresponding row_num or column_num argument is optional.
• row_num - the row number in the range from which to return a cell reference, it's similar to the array form.
• column_num - the column number from which to return a cell reference, also works similarly to the array form.
• area_num - an optional parameter that specifies which range from the reference argument to use. If omitted, the INDEX formula will return the result for the first range listed in reference.
For example, the formula `=INDEX((A2:D3, A5:D7), 3, 4, 2)` returns the value of cell D7, which is at the intersection of the 3rd row and 4th column in the second area (A5:D7).
#### INDEX reference form - things to remember
1. If the row_num or column_num argument is set to zero (0), an INDEX formula returns the reference for the entire column or row, respectively.
2. If both row_num and column_num are omitted, the INDEX function returns the area specified in the area_num argument.
3. All of the _num arguments (row_num, column_num and area_num) must refer to a cell within reference; otherwise, the INDEX formula will return the #REF! error.
Both of the INDEX formulas we've discussed so far are very simple and only illustrate the concept. Your real formulas are likely to be far more complex than that, so let's explore a few most efficient uses of INDEX in Excel.
## How to use INDEX function in Excel - formula examples
Perhaps there aren't many practical uses of Excel INDEX by itself, but in combination with other functions such as MATCH or COUNTA, it can make very powerful formulas.
#### Source data
All of our INDEX formulas (except for the last one), we will use the below data. For convenience purposes, it is organized in a table named SourceData.
The use of tables or named ranges can make formulas a bit longer, but it also makes them significantly more flexible and better readable. To adjust any INDEX formula for your worksheets, you need only to modify a single name, and this fully makes up for a longer formula length.
Of course, nothing prevents you from using usual ranges if you want to. In this case, you simply replace the table name SourceData with the appropriate range reference.
### 1. Getting the Nth item from the list
This is the basic use of the INDEX function and a simplest formula to make. To fetch a certain item from the list, you just write `=INDEX(range, n)` where range is a range of cells or a named range, and n is the position of the item you want to get.
When working with Excel tables, you can select the column using the mouse and Excel will pull the column's name along with the table's name in the formula:
To get a value of the cell at the intersection of a given row and column, you use the same approach with the only difference that you specify both - the row number and the column number. In fact, you already saw such a formula in action when we discussed INDEX array form.
And here's one more example. In our sample table, to find the 2nd biggest planet in the Solar system, you sort the table by the Diameter column, and use the following INDEX formula:
`=INDEX(SourceData, 2, 3)`
• `Array` is the table name, or a range reference, SourceData in this example.
• `Row_num` is 2 because you are looking for the second item in the list, which is in the 2nd
• `Column_num` is 3 because Diameter is the 3rd column in the table.
If you want to return the planet's name rather than diameter, change column_num to 1. And naturally, you can use a cell reference in the row_num and/or column_num arguments to make your formula more versatile, as demonstrated in the screenshot below:
### 2. Getting all values in a row or column
Apart from retrieving a single cell, the INDEX function is able to return an array of values from the entire row or column. To get all values from a certain column, you have to omit the row_num argument or set it to 0. Likewise, to get the entire row, you pass empty value or 0 in column_num.
Such INDEX formulas can hardly be used on their own, because Excel is unable to fit the array of values returned by the formula in a single cell, and you would get the #VALUE! error instead. However, if you use INDEX in conjunction with other functions, such as SUM or AVERAGE, you will get awesome results.
For example, you could use the following formula to calculate the average planet temperature in the Solar system:
`=AVERAGE(INDEX(SourceData, , 4))`
In the above formula, the column_num argument is 4 because Temperature in the 4th column in our table. The row_num parameter is omitted.
In a similar manner, you can find the minimum and maximum temperatures:
`=MAX(INDEX(SourceData, , 4))`
`=MIN(INDEX(SourceData, , 4))`
And calculate the total planet mass (Mass is the 2nd column in the table):
`=SUM(INDEX(SourceData, , 2))`
From practical viewpoint, the INDEX function in the above formula is superfluous. You can simply write `=AVERAGE(range)` or `=SUM(range)` and get the same results.
When working with real data, this feature may prove helpful as part of more complex formulas you use for data analysis.
### 3. Using INDEX with other functions (SUM, AVERAGE, MAX, MIN)
From the previous examples, you might be under an impression that an INDEX formula returns values, but the reality is that it returns a reference to the cell containing the value. And this example demonstrates the true nature of the Excel INDEX function.
Since the result of an INDEX formula is a reference, we can use it within other functions to make a dynamic range. Sounds confusing? The following formula will make everything clear.
Suppose you have a formula `=AVERAGE(A1:A10)` that returns an average of the values in cells A1:A10. Instead of writing the range directly in the formula, you can replace either A1 or A10, or both, with INDEX functions, like this:
`=AVERAGE(A1 : INDEX(A1:A20,10))`
Both of the above formulas will deliver the same result because the INDEX function also returns a reference to cell A10 (row_num is set to 10, col_num omitted). The difference is that the range is the AVERAGE / INDEX formula is dynamic, and once you change the row_num argument in INDEX, the range processed by the AVERAGE function will change and the formula will return a different result.
Apparently, the INDEX formula's route appears overly complicated, but it does have practical applications, as demonstrated in the following examples.
#### Example 1. Calculate average of the top N items in the list
Let's say you want to know the average diameter of the N biggest planets in our system. So, you sort the table by Diameter column from largest to smallest, and use the following Average / Index formula:
`=AVERAGE(C5 : INDEX(SourceData[Diameter], B1))`
#### Example 2. Sum items between the specified two items
In case you want to define the upper-bound and lower-bound items in your formula, you just need to employ two INDEX functions to return the first and the last item you want.
For example, the following formula returns the sum of values in the Diameter column between the two items specified in cells B1 and B2:
`=SUM(INDEX(SourceData[Diameter],B1) : INDEX(SourceData[Diameter], B2))`
### 4. INDEX formula to create dynamic ranges and drop-down lists
As it often happens, when you start organizing data in a worksheet, you may not know how many entries you will eventually have. It's not the case with our planets table, which seems to be complete, but who knows...
Anyway, if you have a changing number of items in a given column, say from A1 to An, you may want to create a dynamic named range that includes all cells with data. At that, you want the range to adjust automatically as you add new items or delete some of the existing ones. For example, if you currently have 10 items, your named range is A1:A10. If you add a new entry, the named range automatically expands to A1:A11, and if you change your mind and delete that newly added data, the range automatically reverts to A1:A10.
The main advantage of this approach is that you do not have to constantly update all formulas in your workbook to ensure they refer to correct ranges.
One way to define a dynamic range is using Excel OFFSET function:
`=OFFSET(Sheet_Name!\$A\$1, 0, 0, COUNTA(Sheet_Name!\$A:\$A), 1)`
Another possible solution is to use Excel INDEX together with COUNTA:
`=Sheet_Name!\$A\$1:INDEX(Sheet_Name!\$A:\$A, COUNTA(Sheet_Name!\$A:\$A))`
In both formulas, A1 is the cell containing the first item of the list and the dynamic range produced by both formulas will be identical.
The difference is in the approaches. While the OFFSET function moves from the starting point by a certain number of rows and/or columns, INDEX finds a cell at the intersection of a particular row and column. The COUNTA function, used in both formulas, gets the number of non-empty cells in the column of interest.
In this example, there are 9 non-blank cells in column A, so COUNTA returns 9. Consequently, INDEX returns \$A\$9, which is the last used cell in column A (usually INDEX returns a value, but in this formula, the reference operator (:) forces it to return a reference). And because \$A\$1 is our starting point, the final result of the formula is the range \$A\$1:\$A\$9.
The following screenshot demonstrates how you can use such Index formula to create a dynamic drop-down list.
Tip. The easiest way to create an expandable dynamic data validation list is dropdown from a table. In this case, you won't need any complex formulas since Excel tables are dynamic ranges per se.
You can also use the INDEX function to create dependent drop-down lists and the following tutorial explains the steps: Making a cascading drop-down list in Excel.
### 5. Powerful Vlookups with INDEX / MATCH
Performing vertical lookups - this is where the INDEX function truly shines. If you have ever tried using Excel VLOOKUP function, you are well aware of its numerous limitations, such as inability to pull values from columns to the left of the lookup column or 255 chars limit for a lookup value.
The INDEX / MATCH liaison is superior to VLOOKUP in many respects:
• No problems with left vlookups.
• No limit to the lookup value size.
• No sorting is required (VLOOKUP with approximate match does require sorting the lookup column in ascending order).
• You are free to insert and remove columns in a table without updating every associated formula.
• And the last but not the least, INDEX / MATCH does not slow down your Excel like multiple Vlookups do.
You use INDEX / MATCH in the following way:
=INDEX (column to return a value from, (MATCH (lookup value, column to lookup against, 0))
For example, if we flip our source table so that Planet Name becomes the right-most column, the INDEX / MATCH formula still fetches a matching value from the left-hand column without a hitch.
For more tips and formula example, please see the Excel INDEX / MATCH tutorial.
### 6. Excel INDEX formula to get 1 range from a list of ranges
Another smart and powerful use of the INDEX function in Excel is the ability to get one range from a list of ranges.
Suppose, you have several lists with a different number of items in each. Believe me or not, you can calculate the average or sum the values in any selected range with a single formula.
First off, you create a named range for each list; let it be PlanetsD and MoonsD in this example:
I hope the above image explains the reasoning behind the ranges' names : ) BTW, the Moons table is far from complete, there are 176 known natural moons in our Solar System, Jupiter alone has 63 currently, and counting. For this example, I picked random 11, well... maybe not quite random - moons with the most beautiful names : )
Please excuse the digression, back to our INDEX formula. Assuming that PlanetsD is your range 1 and MoonsD is range 2, and cell B1 is where you put the range number, you can use the following Index formula to calculate the average of values in the selected named range:
`=AVERAGE(INDEX((PlanetsD, MoonsD), , , B1))`
Please pay attention that now we are using the Reference form of the INDEX function, and the number in the last argument (area_num) tells the formula which range to pick.
In the screenshot below, area_num (cell B1) is set to 2, so the formula calculates the average diameter of Moons because the range MoonsD comes 2nd in the reference argument.
If you work with multiple lists and don't want to bother remembering the associated numbers, you can employ a nested IF function to do this for you:
`=AVERAGE(INDEX((PlanetsD, MoonsD), , , IF(B1="planets", 1, IF(B1="moons", 2))))`
In the IF function, you use some simple and easy-to-remember list names that you want your users to type in cell B1 instead of numbers. Please keep this in mind, for the formula to work correctly, the text in B1 should be exactly the same (case-insensitive) as in the IF's parameters, otherwise your Index formula will throw the #VALUE error.
To make the formula even more user-friendly, you can use Data Validation to create a drop-down list with predefined names to prevent spelling errors and misprints:
Finally, to make your INDEX formula absolutely perfect, you can enclose it in the IFERROR function that will prompt the user to choose an item from the drop-down list if no selection has been made yet:
`=IFERROR(AVERAGE(INDEX((PlanetsD, MoonsD), , , IF(B1="planet", 1, IF(B1="moon", 2)))), "Please select the list!")`
This is how you use INDEX formulas in Excel. I am hopeful these examples showed you a way to harness the potential of the INDEX function in your worksheets. Thank you for reading!
## You may also be interested in
1. Hi,
I have seen videos to populate pivot based on the value selection in the drop down using Index and Match function. My data is tricky not able to replicate the same issue. Below are the columns of info I have in the spread sheet:
Dept Month Target Actual Difference
Mktg Jan-18 100 80 20
Mktg Feb-18 120 118 2
Mktg Mar-18 90 94 4
TeleM Jan-18 200 210 10
TeleM Feb-18 150 148 2
TeleM Mar-18 110 103 7
I want my chart to change dynamically. Could you please help how to write the Index Formula. I want to display the chart data by for selected period by dept. I will select the dept and then the range of the periods which will display the chart. Since I have multiple times dept and month are coming in the data spreadsheet it is not giving the right result.
2. Hi,
can we use INDEX to select the default value of a list in data validation?
for example in below list
A1 Select one
A2 AA
A3 BB
A4 CC
by default, the drop list in D1 which is created by data validation >list
shows select one/
3. Index match on multiple criteria with multiple sheets
i have a workbook with 4 sheets 1st three are Jan, Feb, Mar and 4th one is home. Columns are style no., qty and unit price. i want to check index match the unit price of mentioned style number from all three sheets. Like i want a formula in HOME sheet in cell C2 which see style no. in cell A2 and then check it's unit price in all three sheets and put unit price of matched style no.
Three sheets Jan, Feb, Mar are like this.
Style No. Qty Unit Price
WTC123456 456 45,000
WTB281654 1000 65,000
HOME Sheet.
Style No. Qty Unit Price
WTC123456 456 here i want formula
WTB281654 1000
because some styles are put in Jan and some are in Feb so on. but home sheet has all the styles no.
so actually i have 12 sheets of 12 months
Thanks
4. Hi,
I want to write a VBA function making use of LINEST excel function to calculate some tstats for the slope. I know that i have to combine the INDEX function with the LINEST function, but i don't know how. Can anyone help?
5. Hi,
I want to write a VBA function making use of LINEST excel function to calculate some tstats for the slope. I know that i have to combine the INDEX function with the LINEST function, but i don't know how. Can anyone help?
6. how can i use index formula. in multiple table in one sheet
7. Am struggling to understand how to construct a formula that populates the "grades" column below by reading the "rounded" column content and returning the value in the "Grade" column if the value in the "rounded" column is equal to or greater than the value in the "From" and is equal to or less than the value in the "To" columns.
Any help appreciated:
165,000 ?
120,000 ?
110,000 ?
100,000 ?
100,000 ?
90,000 ?
90,000 ?
88,580 ?
85,000 ?
80,000 ?
160,000 170,000 A1
150,000 159,999 A2
140,000 149,999 A3
130,000 139,999 A4
120,000 129,999 A5
110,000 119,999 A6
100,000 109,999 A7
90,000 99,999 A8
85,000 89,999 B1
80,000 84,999 B2
75,000 79,999 B3
70,000 74,999 B4
65,000 69,999 B5
8. I am building a task list that needs to populate a "dashboard". There is a row in each task labeled "launch" that has a date associated with in the neighboring column. Then in the same column as "launch" there are a list of activities that has a list of corresponding "date completed". The dashboard has columns that match the activity title and the rows have the matching campaign. I'm currently using this function: =IF(ISERROR(INDEX(ALLCAMPS,SMALL(IF(ACTIVITY=\$R\$1,ROW(Complete_Date)),ROW(1:1)),12)),"",INDEX(ALLCAMPS,SMALL(IF(ACTIVITY=\$R\$1,ROW(Complete_Date)),ROW(1:1)),12)) with delete + shift + enter. It logs down my system. Is there a better way to do?
• Hello,
=IFERROR(INDEX(ALLCAMPS,SMALL(IF(ACTIVITY=\$R\$1,ROW(Complete_Date)),ROW(1:1)),12),"")
If this doesn’t work either, I’m afraid you’ll need to use a special macro then. We can’t help you with this since we do not cover the programming area (VBA-related questions). In this case please try to find the solution in VBA sections on mrexcel.com or excelforum.com.
I do hope you’ll manage to solve this task!
9. I have a schedule for 10 teams playing on 5 boards. They play 3o days and play 2 games each day, against different teams.
Left column are the dates, then to the right is what teams play each other and the board they play on above the teams.
I would post the schedule but do not see where I can post it.
I presently have a chart that I populate manually and want to do automatically depending upon what the date is, whether the 1st or 2nd game is being played.
I tried index, match, vlookup and cannot get any of them to work with the schedule I have.
What do you suggest I do? Hopefully I explained what I'm trying to do.
Thanks,
Jim
10. Estimated. Sq. Ft Estimated Capex Spend (£)
2,000 185,000
3,500 436,000
5,000 660,000
how do i find whats estimated capex spend for example 2500 sq ft based on the data above. what formula could i use to get an estimate?
11. Hi There
I have an employee data base, i have to send every week the manpower list each dept.
i want to create the drop down list by dept. when i select any dept my worksheet should the employees names in that dept.
12. Hello,
I have two columns (A and B) and that they are reference columns. I need to get the columns (C and D) are sorted according to the ref. column B using INDEX and/or MATCH functions. For instance:
No. B C D
= = = =
1 A 2 4
2 C 1 2
3 E 3 1
4 C 4 3
After sorting according to column B the columns C and D it becomes below:
No. B C D
== = = =
1 A C E
2 C A C
3 E E C
4 B C A
How to do this using functions (not normal or custom sorting) in Excel?
Thank you so much in advance.
13. Hello,
I have a list with 2 columns:
A:text
B:number
I wanted to
1- select records that have value 'USD'in column A (This outputs a range include recores that value of column A is 'USD')
2- Sum the column B values on range which specified in previuse step.
HOW TO DO THIS IN EXCEL USING FUNCTIONS?
thanks before
14. Good day
i have a challenge . been working on a report card, but i can't go through :(
NAME ENG MATHS ADMA BIO GEO CIV PHY CHE AVG PSN
BRIAN 55 3 66 2 63 2 72 2 55 3 71 2 83 1 80 1 427 68 8
BKALUMO 65 2 52 3 0 FAIL 65 2 70 2 60 2 88 1 50 3 400 56 16
IAN 70 2 24 FAIL 36 FAIL 50 3 58 3 71 2 75 1 18 FAIL 360 50 22
TREVIS 88 1 54 3 51 3 80 1 70 2 77 1 90 1 72 2 477 73 3
GRIFFINS 68 2 26 FAIL 51 3 46 4 59 3 77 1 70 2 60 2 385 57 1 HARRISON58 3 12 FAIL 34 FAIL 54 3 55 3 71 2 75 1 40 4 353 50 25
ROSTEN 78 1 42 4 45 4 58 3 74 2 63 2 85 1 80 1 438 66 7
VICTOR 78 1 56 3 60 2 56 3 59 3 83 1 85 1 24 FAIL 421 63 9
GABRIEL 48 4 38 FAIL 33 FAIL 52 3 40 4 49 4 75 1 12 FAIL 302 43 28
The card works like this . a teacher will enter the mark for a pupil then it will calculate the grade ie 1-9(fail). and also calculates the best 6 subjects including english as shown in the 3rd last column.(SUM(LARGE(E3:Q3,{1,2,3,4,5}))+C3)
i want to now add a column that will calculate the points (sum of best grade). i have tried alot of formulars but the challenges comes because the grades where found by a formular and are not in a range (in different columns)
=SUM(SMALL(range,{1,2,3,4,5}))+M34 - this formular isnt giving an answer but an error ### #num!. am stuck on how to go about it. i noticed it could be because the grades for a particular pupil are in different columns and are a result of a formular.
15. I am trying to retrieve the values from a range that match two criteria (A2, C2). I get only the first one, but I want them all in a dropdown list.
=INDEX(tbEmployee[Employee],MATCH(\$A\$2&\$C\$2,INDEX(tbEmployee[Country]&tbEmployee[Category],,),0),0)
What shall I do?
16. If I would like to sum the values of different individual columns based on row criteria then which formula should I use?
Example:
Name A1 Loc A2 Loc A1 Loc A2 Loc
ABC 5 7 8 5
XYZ 6 8 9 6
PQR 3 9 6 3
I want sum of XYZ values of A1 location
17. Perhaps you can help me out. I am using the following array formula {=INDEX(ASSETP,SMALL(IF(STATP=\$A\$10,ROW(ASSETP)MIN(ROW(ASSETP))+1),ROWS(\$B10:B\$10)))}
to display vertically instances of asset #'s if they equal the status in cell A10. The formula works great. The cells in the "STATP" range are formulas to determine the status. =IF(K5="","L","X"). Later I realized that I needed to improve the status formula by checking for cost of the asset. I changed the status formulas to determine if cost was \$500.
=IF(AND(K6="",H6>499.99),"L",IF(AND(K6="",H6<500),"L2","X"))
The status formula works fine, but my array formulas no longer work the way I expected. If I put the Status formulas back to original the array formulas work as expected.
• SOLVED. When I changed the status formula one of the cells in the STATP range, gave an error as its result. This caused the array formulas to not work as expected. Once I corrected the error all array formulas worked as expected.
18. Hi,
=INDEX(\$E2:\$E\$300,MATCH("PE",\$D2:\$D\$300,0))
=INDEX(\$E2:\$E\$300,MATCH("CE",\$D2:\$D\$300,0))
I am using above 2 formulas to extract data from E2:E300 for matching two words like CE and PE, which is situated in column D2:D300. I want result for PE in column F2 and drag down. Another CE result want in column G2 and down. But the CE word starts from row above 100 in column E and when I drag it in column G2 and down I am getting incorrect match. But getting correct result for PE.
19. I need some help with the formula (combination of Index and Match)- If I have row ass Quarters (B2:G2) and business listed in Column B- How can add total for business by Quarter by just changing the Quarter name in AI cell.
20. Hi Really struggling to source the correct formula required.
I have a table that has data entered including customer reference numbers and dates. the reference number can be entered several times in no particular order.
I want to lookup the reference number in one column, check that a date has been entered against all entries in another column and then return a Yes, No response.
Any help appreciated
21. Hi,
I would like to create a inventory file with condition as follow.
1) multiple products/items in one worksheet as follow
2) inventory use first in first out method to calculate col F,G,H.
Is there any formula in column F, G & H that can automatic calculate.
B C D E F G H
1 Description/Cost/QtyIn/QtyOut/QtyBal/BalAmt/QtyOutAmt
2 Product A 10 416 416 \$4,160 \$0
3 Product B 20 400 400 \$8,000 \$0
4 Product A 200 216 \$2,160 200*10=\$2,000 [E4*C2]
5 Product B 250 150 \$3,000 250*20=\$5,000 [E5*C3]
6 Product A 12 150 366 \$3,960 \$0
7 Product B 50 100 \$2,000 50*20=\$1,000 [E7*C3]
8 Product A 300 66 \$792 (216*10)+(84*12)=\$3,168
9 Product C 15 200 200 \$3,000 \$0
10 Product B 22 200 300 \$6,400 \$0
11 Product A 50 16 \$192 \$600
12 Product C 15 100 300 \$4,500 \$0
13 Product B 150 150 \$3,300 \$3,100
14 Product C 150 150 \$2,250 \$2,250
15 Product A 10 300 316 \$3,192 \$0
22. In Example 2 (Sum items between the specified two items)
you used the following formula
=SUM(INDEX(SourceData[Diameter],B1) : INDEX(SourceData[Diameter], B2))
is there a way i can add another condition? I want to only sum every second number in the range specified in the abovementioned formula?
so in your example 2, the answer will be 120,670+0+49,528+0 = 170,188
Thanks
Nicholas
23. I have to say your explanation and examples of the index function is the best I have seen.
Very clear and easy to follow.
Thank you so much!
Bruce
24. used a simpler approach, two SUMIFs
=sumif(B4:CH4,"="&Today(),B5:CH5)+sumif(B4:CH4,">"&Today(),B5:CH5)
gave me the answer I needed
25. I have a table with row 1 being dates from July thru to end of October and row 2 indicates whether a person is working (1) or not (0). I am trying to get my index/match formula to auto update to show how many days between today and the end of October the person is working. I have the following, but end up with #REF :-(
=sum(index(B4:CH4,2,match(today(),b4:ch4,0)):index(b4:ch4,2,85)
Thanks for any help
26. Date Name Brand Date Name Brand
are the entries.
27. A B C D E F G H
1/1/16 ALEX SONY 2/1/16 JOHN DELL ? ?
A to C and D to E are consecutive data categories.
I want to return in Cell G1 the name of the Person who made the sale at the latest date (JOHN).
And return in Cell H1 the name of the Brand for which the sale was done (DELL).
Kindly help.
28. Hellow all,
I have two questions,please have a look and let me know is there anyway to find the solution,
1. I have two tables each having 37X12 rows and columns, First row and first column is having names for the corresponding data.Here the question is, i need to sort out column 1 data (which is having same names for two tables) from the reaming 35 columns of the two tables based on max, avg and min values. it is the combination of index, match for max min and avg, but no idea how to give the reference for two tables at a time to see max min and avg values for column data.
2. I have 15 matrices, each of the matrix is having 1833x1833 rows and columns. Here the data need to be sort out for specific rows and columns say for eg. between 500 to 1000. i need to sum these columns and rows by looking the sheet name and no idea how to specify the range to lookup and sum the data.
Looking for the solution,
Thanks, Subbareddy.
29. I am trying to use the index function to display a dollar amount listed in a table in the month that it will be billed for. I have multiple projects and when the formula is dragged down to the next project, the index gets off because the projects have different start dates. Is there a better way to have the index start at the first billing month other than copying the formula from the previous project to the first billing month of the next project?
Thanks.
30. Hi,
How to set ascending order date format with using formula to another column.
Example: all dated A1:A20
Ascending another column like B1:B20
Thanks & Regards,
Sri
31. I'm trying to reference a cell in another workbook without success. I think my best way forward is to use Indirect() but have also tried Index() without success.
It seems to be failing on the external file location as such:
The entire cell reference is as such"
"http://apollo.omega.dce-eir.net/contentserverdav/nodes/3161757/[BIASDetailedVRFtracker.xlsm]alignment!\$D\$12"
My Indirect call is as follows:
Indirect("'http://apollo.omega.dce-eir.net/contentserverdav/nodes/3161757/[BIASDetailedVRFtracker.xlsm]alignment'!\$D\$12")
It is failing on:
http://apollo.omega.dce-eir.net/contentserverdav/nodes/3161757/
32. Hello,
I am trying to sum multiple columns that have months has my headers. When we close on another month data is updated in that column. I want to automatically sum the data in the rows without continually every month updating my sum formula at the end. Can someone please show me how to do this with Index and Match Function? I want it to be dynamic!
Thanks,
33. hello
The results show me the same which type it
the results:- =INDEX((A2:D3, A5:D7), 3, 4, 2)
• Hello Yaseen,
This usually happens if a formula cell has a leading space or apostrophe before the equal sign; or if the Show Formulas mode is activated in the worksheet (Formulas tab > Formula Auditing). If neither is the case, please check out Excel formulas not calculating for other possible reasons and solutions.
34. Forgot to mention the function - use IFERROR - wrap you INDEX...MATCH functions in IFERROR - and the last part of the function will be your default value (0 or "Not found - try again"). I don't like the #REF or #N/A answers either.
36. Good morning I am trying to write an INDEX MATCH Formula that will dynamically set the INDEX lookup column based on the header column.
what I have are 2 tabs which house my base data; we will call them Table1 and Table2. then I have a third tab which populates all its data based on INDEX MATCH Formulas, we can call this tab Lookup1.
I have made the table headers in Table1 and Table2 names ranges and in my Lookup1 tab, I want the table headers to be dropdowns from one of my dataset tables. I'd like to write the INDEX MATCH formula to change it's INDEX column if I changed the header title in my table.
is this possible without VBA? I have pasted the original formula below, the part colored in red is what I want to rewrite to reference the header of it's own column, match the column of the same name in the Table1 tab and insert that as the Index lookup. I believe I might need to nest in another INDEX and/or MATCH formula to do what I want, but I am not sure.
=IF([@[Employee ID]]"",INDEX(Table1[First Name],MATCH([@[Employee ID]],Table1[Employee ID]),0),"")
37. I am maintaining a speadsheet which which monitors materials delivery (detailed and summary).
I have 2 worksheets (sheet1 is for detailed report & sheet2 is for summary report) in a workbook. What I want is a formula to automatically write "Delivered" in the Summary Report if all items (2 items each PO number in the example shown) are "Delivered" in the Detailed Report; and write "Partial" if one is "Delivered" and the remaining is "Undelivered".
Both tables are Named Tables, so the number of items will increase in time.
I tried to search in the forums but failed to get what I'm looking for. Would appreciate any help.
Sheet1-Detail
PONo. Project ProjCode Vendor Material Status
1 Project 1 Prj001 ABC Material 1 Delivered
1 Project 1 Prj001 ABC Material 2 Undelivered
2 Project 1 Prj001 XYZ Material 3 Delivered
2 Project 1 Prj001 XYZ Material 4 Delivered
Sheet2-Summary
PONo. Project ProjCode Vendor Status
1 Project 1 Prj001 ABC Partial
2 Project 1 Prj001 XYZ Delivered
38. Hi...can anyone help me use INDEX function to return a value and then drag it down to return values from other sheets in the same column ?
39. data1 data2 data3 data4 default average new data
44.44 26.75 83.80 68.36 33.00 65.53 65.53
i have to use sumif to add 3 of the data rows, omitting the highest and the lowest values.
Thanks
40. I have this file:
data1 data2 data3 data4 default average new data
44.44 26.75 83.80 68.36 33.00 65.53 65.53
I need to use sumif to sum 3 of the five data rows omitting the the highest and the lowest values
41. I use offset =OFFSET(\$G\$2,MATCH(M3,\$G\$3:\$G\$15,0),1) and I get #N/A for the id that didn't match is there a way to output 0(zero) or Null instead of #n/a?
42. Hi Everyone. I am trying to get the min number from these set of value from Cell, D13 to D19.
With the method i use, it is going to be very tedious as i need to repeat it till D19.
Is there a way to rewrite this formula into an array where the formula will gather the number from D13-D19 and output the minimum?
=MIN(INDEX(\$D\$2:\$LK\$3,MATCH(\$M\$5,\$B\$2:\$B\$186,0),MATCH(D13,\$D\$1:\$LK\$1,0)),INDEX(\$D\$2:\$LK\$3,MATCH(\$M\$5,\$B\$2:\$B\$186,0),MATCH(D14,\$D\$1:\$LK\$1,0)))
Thank you!
43. What if I have a long time-series organized in by year (column 1) and month (column 2) and the value (column 3) and I want to compute seasonal averages (i.e. DJF, MAM, JJA, SON) over the entire time-series. How can I use the EXCEL average function to know to skip every 12th entry in generating the seasonal averages?
44. Hi, I have an Index Match formula: =INDEX(AI1233:AU1234,MATCH(AH1196,AI1233:AU1233,0), MATCH(AI1223,AI1233:AI1234,0)) the AH1196 is a drop down list of Jan 2016-Dec 2016. The budget data it pulls looks like this:
Date Jan-16 Feb-16 Mar-16
Postage xxxx xxxx xxxx
When I have Jan-16 in the drop down field, it works great. When I change the drop down to any other month, I get #REF!. Not sure how to enable the use of the drop down in the formula.
Any help will be greatly appreciated.
45. I have a large number of data in which I am trying to calculate weeks of supply. I am looking for a formula that will go to a specified cell (that contains my build for a specific category) then multiply last weeks sales by that build. If that does not equal 0, then I would like it to do the same thing and move on to the next cell and perform the same formula. Then continue until my result is zero, and finally count all the cells it performed this formula, giving me my weeks of supply.
46. Hi,
I failed to calculate via Index function the "Min" of dates array;
=INDEX(MIN('Dates'!N15:N35),MATCH(A4,'Students'!B15:B35,0))
Result is #N/A, though values are available in the date type and both respective cells.
Regards,
Tariq
• Hello Tariq,
MIN function returns one value, so the second parameter of the INDEX function should by equal 1.
47. I am working with a similar formula with CTRL+SHIFT+ENTER and it works on the first cell, but when I copy down I get #NUM! errors?
{=INDEX(\$B\$2:\$B\$8, SMALL(IF(\$A\$11=\$A\$2:\$A\$8, ROW(\$A\$2:\$A\$8)-ROW(\$A\$2)+1), ROW(1:1)))}
Any ideas? Would be greatly appreciated!
• Hi, whatever row your formula is in, use this at the end of your formula ROWS(\$C\$1:C1)... If your formula is being built inside of c1
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# Light, Gravity and Energy
The strict particle model proposed in my book has energy stored as size of subatomic particles. All energies, regardless of type, are stored as size.
When lifting a wood block from the floor up to a shelf, the size of subatomic particles in the block increase by a tiny bit. The same is true when throwing the block through the air. However, dropping the block results in no change in size, because the sum of potential and kinetic energy remains unchanged during the fall. There is no change in the total energy of the block before it hits the ground.
An alternative to this model is to have potential energy stored in the field between the block and the ground, and kinetic energy stored as motion. Lifting the block from the floor will in this model put potential energy into the field. Dropping the block will result in a transfer of potential energy stored in the field to kinetic energy in the block, with gravity facilitating this transfer.
The problem with such a field theory is that it is all pure math, with no simple mechanism to explain it.
However, the advantage with a field theory is that once we accept the idea of energy stored in a field, we have a ready supply of energy to draw on whenever needed.
This comes in handy when we are confronted with such phenomena as gravitational red-shift.
When light travels away from a massive body, it looses energy. When it travels towards a massive body, it gains energy. This is experimentally confirmed, and requires an explanation.
Photons passing by, moving towards and moving away from a massive body
Using a field theory, we can simply say that photons give up energy to the field when they travel away from massive bodies, and that they absorb energy from the field when they travel towards such bodies.
However, a strict particle model cannot use such an explanation, because no energy is stored outside of particles. When particles gain energy, other particles must loose energy. There has to be interaction between particles.
The way to solve this problem, using a strict particle model, is to invoke the aether.
When we combine the aether with the idea that a pilot wave accompanies every photon, we can imagine a mechanism for energy transfer between the aether and visible light.
Zero-point photons, abundantly available in the aether, readily soak up any excess energy of an outgoing visible photon. The red-shift of outgoing photons are facilitated by a corresponding blue-shift of incoming zero-point photons.
Conversely, blue-shift of incoming visible light is facilitated by the red-shift of outgoing zero-point photons.
Note that there is no need for any direct contact between the photons for the energy transfers to happen. As long as the pilot waves associated with each photon brush against each other, energy can be transferred.
Note also that this explanation has as a consequence that the aether is somewhat “hotter” close to massive bodies than farther away. The stronger the gravitational field, the more blue-shifted is the aether close to the surface of the body in question.
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# Convert a nibble to ASCII to byte
Hi!
I am trying to accomplish turning a nibble into it's ASCII character and then turn that character into a byte.
In the example below I turn the byte myByte (0x12) into to nibbles. One becomes 0x1 and the other 0x2. What I want to do then is to turn the nibbles into ASCII characters, 1 and 2, and then back to byte to send these over CAN. So ASCII 1 would be hex value 0x31 and ASCII 2 would be 0x32. It does not have to be hex. It could might aswell be decimal 49 and 50.
Is there a way of doing this? I have searched a lot and have not found what I'm looking for. It might be that I'm not really sure what to search for.
Regards,
Christian
byte myByte = {0x12};
byte myData[2];
void setup() {
Serial.begin(115200);
myData[0] = myByte >> 4;
myData[1] = myByte & 0xf;
Serial.println(myData[0]);
Serial.println(myData[1]);
}
void loop() {
// put your main code here, to run repeatedly:
}
Add 0x30...
You need to use itoa:
char buf[3]; //Two digits + null
itoa(myData[0], buf, 10); //Integer to ASCII, base 10 used
Serial.println(buf); //Prints "1"
westfw:
Add 0x30...
Nope, it can't be that easy. It has to be some sort of monster machine
Thank you, for a very fast and super simple solution!
Adding 0x30 will not work for values above 9
myData[0] += (myData[0] < 10 ? '0' : 'A');
aahhh…
should be
myData[0] += (myData[0] < 10 ? '0' : ('A'-10));
:-[
Danois90:
Adding 0x30 will not work for values above 9
You’re right, but neither does your example seem to do. When making myByte 0xAB, it prints 10 instead of A.
Budvar10:
myData[0] += (myData[0] < 10 ? '0' : 'A');
Neither does this. This seems to have an offset of 10 when the value is over 9. I’d love for you to explain what the code does.
As commented in the code, base 10 was used. You need to change the last parameter of itoa from 10 to 16 in order to use base 16 (hexadecimal).
A method that CAN uses to communicate with 'dumb' peripherals is with BCD (binary coded decimal) in which every nibble represents a decimal number of 0 thru 9. This appears (heavy emphasis) to be what you're trying to accomplish.
As an example, the number 5555 is binary 0001 0101 1011 0011 and would be transmitted thus. BCD however, would represent 5555 as 0101 0101 0101 0101, stripping each digit into a nibble. CAN is actually an old protocol and the use of BCD was to accommodate 0-9 thumbwheel switches.
If all you want to do is break a byte into two nibbles,
union
{
byte my_byte;
struct
{
byte nibble1:4;
byte nibble2:4;
};
}my_union;
Then my_union.my_byte = your hex or decimal value, my_union.nibble1 are the four lsb and my_union.nibble2 are the four msb.
Please, let me understand.
What I understond is: you have a char array that contiens a ASCII message. You want take each element, devide it in the octal or exadecimal digits and print than.
IS IT LIKE THIS?
IF IT IS you can cast each element like a int and use the .print function, witch the right argument about base
DKWatson:
If all you want to do is break a byte into two nibbles,
union
{
byte my_byte;
struct
{
byte nibble1:4;
byte nibble2:4;
};
}my_union;
Then my_union.my_byte = your hex or decimal value, my_union.nibble1 are the four lsb and my_union.nibble2 are the four msb.
The breaking down into two nibbles is already sorted. What I want to to with the nibbles is to convert them into ASCII and then into bytes.
So let's say that one nibble is 0xF. I want to convert this to a byte that contains decimal 70.
Silente:
Please, let me understand.
What I understond is: you have a char array that contiens a ASCII message. You want take each element, devide it in the octal or exadecimal digits and print than.
IS IT LIKE THIS?
IF IT IS you can cast each element like a int and use the .print function, witch the right argument about base
No, you must have misunderstood everything, unless I accidentally added a char array in my example.
So let's say that one nibble is 0xF. I want to convert this to a byte that contains decimal 70.
If the nibble contains 15 (base 10), why would the byte contain 70 (base 10)?
Because 70 is the decimal value of ASCII F, or have I gotten this all wrong?
chrstrvs:
Because 70 is the decimal value of ASCII F, or have I gotten this all wrong?
No.
What IS the problem? If the value in the nibble is between 0 and 9, add '0'. If the value is between 10 and 15, subtract 10 and add 'A'.
Or add '7'
DKWatson:
Or add ‘7’
While the result may be the same, I think the two step process is more logical.
I have not understand a thing:
Why have you to transform a value in more than one that conteins exadecimal digits of the same value? The thing can be usefull only in case to show tje number or pass it. But I think that all the .print function has the ability of doing this change of base by thanselves. So why do you need it?
PaulS:
No.
What IS the problem? If the value in the nibble is between 0 and 9, add '0'. If the value is between 10 and 15, subtract 10 and add 'A'.
THAT was the problem. I didn't know how to accomplish the hex or decimal value corresponding to the ASCII characters of the nibbles. Now I do, and for that I thank all you very much!
I'm sorry that it might be somewhat unclear what I want to accomplish. I know my terminology is way off sometimes, but everything I know about Arduino programming is self taught. The only programming classes I have attended are HTML and PHP some 15 years ago.
chrstrvs:
In the example below I turn the byte myByte (0x12) into to nibbles. One becomes 0x1 and the other 0x2. What I want to do then is to turn the nibbles into ASCII characters, 1 and 2, and then back to byte to send these over CAN. So ASCII 1 would be hex value 0x31 and ASCII 2 would be 0x32. It does not have to be hex. It could might aswell be decimal 49 and 50.
Given :
byte myByte = 0x12;
==> myByte = 0001 0010
0001 (1) ---> 0011 0001 (x31 = 49) ASCII Code of the numeral 1
49 (0x31) ----> 0100 1001 (BCD Format)
//------------------------------
0002 (2) ---> 0011 0010 (0x32 = 50 ADCII Code of numeral 2
50 (0x32) ---> 0101 0000 (BCD Format)
The codes to achieve the above transformations:
byte asciiUpperNibble;
void setup()
{
Serial.begin(9600);
byte myByte = 0x12; //0000 0001, 0000 0010 ; 0x31 (0x32) : 49 (50) : 0100 1001 (0101 0000)
byte upperNibble = myByte >>4; //(0000)0001 : 1
byte lowerNibble = myByte & 0x0F; //(0000)0010 : 2
//--------------------------
if(upperNibble >9)
{
asciiUpperNibble = upperNibble + 0x37; //0x41 - 0x46 (A-F): 65 - 70 : 0110 0101 - 0111 0000
}
else
{
asciiUpperNibble = upperNibble + 0x30; //0x30 - 0x39 (0-9): 48 - 57 : 0100 1000 - 0101 0111
}
byte inDex0 = asciiUpperNibble % 10; //(0000) 1001 : 9
byte inDex1 = asciiUpperNibble / 10; //(0000) 0100 : 4
//---------------------
byte bcdByte = (inDex1 <<4)|inDex0; //0100 1001
Serial.println(bcdByte, HEX); //shows: 49
Serial.println(bcdByte, BIN); //shows: (0)100 1001 ; leading 0 missing; but, it can be printed
//------------------------------------------
/*---- code are similar to above for the lower nibble----
*----------------------------
*---------------------------*/
}
void loop()
{
}
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Sunday
December 11, 2016
# Posts by li
Total # Posts: 55
Physics
5 kg of water at 40 degree Celsius is poured on a large block of ice at 0 degree celsius. How much ice melts?
October 11, 2016
English
A chain of student cafeteria sounds wrong. Revised.. The student cafeteria is operated by a chain of college food service system. Did I put the modifier to the right place?
September 28, 2016
English
revise the following sentence to use modifiers correctly, clearly and effectively. Original 1.The student cafeteria is operated by a college food service system chain. Revised. The chain of student cafeteria is operated by a college food service system
September 28, 2016
Math
If a movie starts at 9:45 pm and its a 2hr and 14 mins movie, what time is it over?
August 8, 2016
Algebra
Find the number of decibels for the power of the sound given. Round to the nearest decibel. A rocket engine, 2.32 cross product 10−5 watts/cm2
January 29, 2015
Early childhood education
An important aspect of socialization is helping children recognize, accept, sort out, identify, label, integrate, express, and cope with their a. satisfaction b. behaviors c. expression d. emotions i say it's d. emotions
January 29, 2014
Early childhood education
What is one possible factor that may make a child vulnerable to stigma? a.being the same religion as the majority of the class. b. having brothers and sisters who were stigmatized c. getting a B on a spelling test d. having a health condition
January 22, 2014
Algebra
I think is (3x+6)+2X= 5X+16
October 29, 2013
algebra 1 help
how do you factor it?
April 16, 2013
algebra 1 help
if the sides of the sqaure are represented by a and b what is the area of the remaining glass when the smaller square is cut from the larger square? write answer in factored form i don't understand because of the variables
April 16, 2013
differece between subject pronoun, object pronoun, reflexive pronoun intensive pronoun. how are they used and examples. we've been learning about this but i'm still confused
April 16, 2013
algebra 1
what i don't undestand is a problem that is similar to this but has ab if the sides of the sqaure are represented by a and b what is the area of the remaining glass when the smaller square is cut from the larger square? write answer in factored form
April 16, 2013
algebra 1
oh i meant to put 95 but i was in my own world :/
April 16, 2013
algebra 1
a diagram below shows an artistic design,] if the length of the outer square is 12" and the length of the inner square is 7" what is the area of the remaining glass when the smaller square is cut form the larger square? well i multiplied 12*12 and 7*7 that gives me ...
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# Node structure of different trees in Python
#### Python Data Structures
Get this book -> Problems on Array: For Interviews and Competitive Programming
Trees are abstract and hierarchial data structures. In this article, we have discussed the implementation of node structures of different trees in Python Programming language. This gives a great opportunity to understand how different tree data structure vary. The trees we have covered are:
• Binary Tree
• Binary Search Tree (BST)
• AVL Tree
• B Tree
• B+ Tree
• Red-Black Tree
• Treap
• Splay Tree
• R Tree
We will go through each node structure in detail.
# 1. Binary Tree
A binary tree is a tree with the special condition of a maximum of two children nodes. Each node in a binary tree has a left reference, a right reference and a data element. A non-empty tree always has a root node. Root node is either the topmost or the bottom node in the tree, depending on the representation. If the representation is top-down, root is the topmost node, else if the representation is bottom-up, root is the bottom node. Leaf nodes have null values in their right and left link. In a full binary trees, every node except leaf node has two children.
``````class BinaryNode:
# has one data value in each node.
def __init__(self, data):
self.data = data
self.right = None
self.left = None
``````
# 2. Binary Search Tree (BST)
As the name suggests, we can easily carry out binary search on this tree. A BST is a binary tree with elements stored in a sorted manner; the value stored at the root is more than any value stored in its left subtree and less than any value stored in its right subtree. The BST property must be satisfied for all nodes on tree and their subtrees. The node structure of BST is similar to that of binary tree.
``````class BSTNode:
# has one data value in each node.
def __init__(self, data):
self.data = data
self.right = None
self.left = None
``````
# 3. AVL Tree
AVL tree is also a self Balancing Binary Search Tree. In an AVL tree, the difference between heights of left and right subtrees cannot be more than one for all nodes. Since the height of the tree remains O(Logn) after every insertion and deletion, we have an upper bound of O(Logn) for all other operations.
The node structure of an AVL tree has an additional parameter namely node height which is used to calculate the balance factor for itself.
``````class AVLNode:
# has one data value in each node.
def __init__(self, data):
self.data = data
self.right = None
self.left = None
# height signifies its distance from root node.
self.height = 1
``````
# 4. B Tree
B-Tree is a self-balancing search tree. The main idea of using B-tree is to reduce the number of disc accesses. All leaves are at the same level in a B-Tree. A B-Tree has a term minimum degree "t" which depends upon disc block size. Every node in a B-Tree has a variable number of keys, between t-1 and 2t-1, except the root node. The number of children of a node is equal to the number of keys in it plus 1.
``````class BTreeNode:
# Every node contains minimum t-1 keys and maximum 2t - 1 keys
# Except root node which can have minimum 1 key.
def __init__(self, t):
self.t = t
# it is true when node is leaf, otherwise false.
self.isLeaf = True
# current number of keys.
self.keys = []
self.childPointers = []
self.numberOfKeys = 0
``````
# 5. B+ Tree
Unlike B Tree, B+Tree does not store data pointers along with its key values in all the nodes; it only stores pointers at the leaf nodes of the tree. Moreover, the leaf nodes are linked to provide ordered access to the records. Therefore, all the nodes form a multilevel index with leaf nodes at the first level.
Thus, the structure of the leaf nodes of a B+ tree is quite different from the structure of the internal nodes of the B tree.
Note: In B Tree, Keys and records both can be stored in the internal as well as leaf nodes. Whereas, in B+ tree, records (data) can only be stored on the leaf nodes while internal nodes can only store the key values.
``````class BplusTreeNode:
def __init__(self, t):
self.t = t
# it is true when node is leaf, otherwise false.
self.isLeaf = True
self.next = None
# current number of keys
self.keys = []
self.childPointers = []
self.numberOfKeys = 0
``````
# 6. Red-Black Tree
It is a self-balancing Binary Search Tree with every node either red or black, except the root node which is always black. If a node is red, both of its children are black. This means no two nodes on a path can be red nodes. Every path from a root node to a NULL node has the same number of black nodes.
The Node structure is similar to a Binary Search Tree with an extra parameter interpreted as color, red or black. These color bits are used to ensure the tree remains approximately balanced during insertions and deletions.
Note: AVL trees cause more rotations than a Red-Black tree but are more balanced than the latter. Hence if a Red-Black tree is preferred in applications with frequent insertions and deletions. And AVL trees are preferred in applications with frequent search.
``````class RedBlackTreeNode:
# similar to Binary search tree
def __init__(self, data):
self.data = data
# 1 for Red and 0 for Black
self.colour = 1
self.left = None
self.right = None
``````
# 7. Treap
Formally, a treap is a binary tree whose nodes contain two values, a key and a priority, such that the key keeps the BST property and the priority is a random value that keeps the heap property. It uses randomization and binary heap property to maintain balance with a high probability. The key follows the binary search property while the priority follows the heap property.
``````class TreapNode:
def __init__(self, key, priority):
self.key = key
self.priority = priority
self.left = None
self.right = None
``````
# 8. Splay Tree
Splay tree is a self-balancing BST. All the operations in a splay tree are performed at its root. Whenever an element is looked up, the tree reorganizes to bring that element to the root, without violating the BST property. Thus, the elements which are heavily used are found near the top and hence can be found quickly. The node structure is similar to that of BST.
``````class SplayNode:
# has one data value in each node.
def __init__(self, data):
self.data = data
self.right = None
self.left = None
``````
# 9. R Tree
R-tree is used for storing spatial data indexes efficiently i.e. for indexing multi-dimensional information in an efficient manner. Each node of an R-tree has a variable number of entries(up to some pre-defined maximum) and each entry within a non-leaf node stores pointer to a child node, and the bounding box of all entries within this child node. Each entry within a leaf node stores the actual data element and the bounding box of the data element. The node structure is similar to that of B tree.
``````class RTreeNode:
def __init__(self, maxEntries):
self.maxEntries = maxEntries
self.entries = []
self.childPointers = []
# for leaf node
self.isLeaf = True
self.data = None
``````
With this article at OpenGenus, you must have the complete idea of different trees and how implementation in Python translates the idea. Consider contrasting the different node structures we have presented among each other. This will bring out some insightful thoughts.
#### Avishi Gupta
Read more posts by this author.
Vote for Author of this article:
Node structure of different trees in Python
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# 4-1 Practice Rate Of Change And Slope Answers
• October 21, 2021
5-1 Rate of Change and Slope 1. Algebra 1 answers to Chapter 5 – Linear Functions – 5-1 Rate of Change and Slope – Practice and Problem-Solving Exercises – Page 296 13 including work step by step written by community members like you.
Pin On Process Technology
### Then count horizontally to the second point to find the run.
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# Soil Mechanics Assist. Prof. Dr. Rafi' M. S.
CHAPTER 4
Seepage Through Soils
Laplace's Equation of continuity
Flow nets
Safety of hydraulic structures against piping
Filter design
Seepage through earth dams
## Laplace’s Equation of Continuity
In many practical cases, the calculation of flow is generally made by use of graphs referred to as
flow nets. The concept of the flow net is based on Laplace’s equation of continuity, which describes
the steady flow condition for a given point in the soil mass. To derive this equation, consider an
elementary soil prism at point A (figure 4.1b) for the hydraulic structure shown in figure (4.1a).
𝒛 𝜕𝑣𝑧
(𝑣𝑧 + 𝑑 )𝑑 𝑑
𝜕𝑧 𝑧 𝑥 𝑦
𝜕𝑣𝑦
(𝑣𝑦 + 𝑑 )𝑑 𝑑
𝜕𝑦 𝑦 𝑧 𝑥
𝑑𝑦
𝜕𝑣𝑥
(𝑣𝑥 + 𝑑 )𝑑 𝑑
𝑣𝑥 𝑑𝑦 𝑑𝑧 𝜕𝑥 𝑥 𝑦 𝑧
𝒙
𝑑𝑧
𝑣𝑦 𝑑𝑧 𝑑𝑥
𝑑𝑥
𝒚 𝑣𝑧 𝑑𝑥 𝑑𝑦
## Figure (4.1): Derivation of continuity equation.
The flows entering the soil prism in the x, y and z directions can be given from Darcy’s law as:-
𝜕ℎ
𝑞𝑥 = 𝑣𝑥 𝐴𝑥 = 𝑘𝑥 𝜕𝑥 𝑑𝑦 𝑑𝑧 ……………………………………….………….……(4.1)
𝜕ℎ
𝑞𝑦 = 𝑣𝑦 𝐴𝑦 = 𝑘𝑦 𝜕𝑦 𝑑𝑧 𝑑𝑥 ……………………………………….………….……(4.2)
𝜕ℎ
𝑞𝑧 = 𝑣𝑧 𝐴𝑧 = 𝑘𝑧 𝜕𝑧 𝑑𝑥 𝑑𝑦 …………………………………………..……………(4.3)
where:
𝑞𝑥 , 𝑞𝑦 , 𝑞𝑧 = flow entering in directions x, y, and z, respectively,
𝑘𝑥 , 𝑘𝑦 , 𝑘𝑧 = coefficients of permeability in directions x, y, and z, respectively,
ℎ = hydraulic head at point A.
## 𝑞𝑥 + 𝑑𝑞𝑥 = 𝑘𝑥 (𝑖𝑥 + 𝑑𝑖𝑥 )𝐴𝑥
𝜕ℎ 𝜕2 ℎ
= 𝑘𝑥 (𝜕𝑥 + 𝜕𝑥 2 )𝑑𝑦 𝑑𝑧 …………...………………………………….……(4.4)
𝜕ℎ 𝜕2 ℎ
𝑞𝑦 + 𝑑𝑞𝑦 = 𝑘𝑦 (𝜕𝑦 + 𝜕𝑦 2 )𝑑𝑥 𝑑𝑧 …………….………………………………………(4.5)
𝜕ℎ 𝜕2 ℎ
𝑞𝑧 + 𝑑𝑞𝑧 = 𝑘𝑧 ( + )𝑑𝑥 𝑑𝑦 …………………….………………………….……(4.6)
𝜕𝑧 𝜕𝑧 2
For steady flow through an incompressible medium, the flow entering the elementary prism is
equal to the flow leaving the elementary prism. So,
## Combining Eqs. (4.1– 4.7), gives
𝜕2 ℎ 𝜕2 ℎ 𝜕2 ℎ
𝑘𝑥 𝜕𝑥 2 + 𝑘𝑦 𝜕𝑦 2 + 𝑘𝑧 𝜕𝑧 2 = 0 …………………………………………………..……(4.8)
## For two-dimensional flow in the x-z plane, Eq. (4.8) becomes:
𝜕2 ℎ 𝜕2 ℎ
𝑘𝑥 𝜕𝑥 2 + 𝑘𝑧 𝜕𝑧 2 = 0 ……………..……………………………………………...……(4.9)
And if the soil is isotropic with respect to permeability, 𝑘𝑥 = 𝑘𝑧 , then the continuity equation
simplifies to:-
𝜕2 ℎ 𝜕2 ℎ
+ = 0 …….………..……..……………………………………………..…(4.10)
𝜕𝑥 2 𝜕𝑧 2
This is generally referred to as Laplace’s equation. This equation represents two families of
curves; named as "flow lines and equipotential lines" that intersecting each other at right angles.
Flow Nets
A set of flow lines and equipotential lines is called a flow net. A flow line is a line along which a
water particle will travel. An equipotential line is a line joining the points that show the same
piezometric elevation (i.e., hydraulic head = h (x, z) = const). Examples of flow nets are shown in
figures (4.2 - 4.3) for flow through permeable isotropic soil; i.e. with, 𝑘𝑥 = 𝑘𝑧 = k. Note that the
solid lines are the flow lines and the broken lines are the equipotential lines.
2
U/S water level
## D/S water level
A B E F
C D
Permeable
layer
G H
Impermeable layer
Permeable layer
## Figure (4.3): Flow nets under dams.
3
In drawing a flow net, there are 4 boundary conditions for any seepage problem, for example, in
figure (4.2b):-
1. AB is an equipotential line
2. EF is an equipotential line
3. BCDE (i.e., the sides of the sheet pile) is a flow line
4. GH is a flow line
## Drawing a flow net is a trial-and-error process. The
flow and equipotential lines are usually intersect each other
at right angles and drawn in such a way that the flow
elements are approximately squares. Therefore, first
attempt of flow net sketching may not be satisfactory, and
Mistake – Redraw !
some revisions are required. Once a satisfactory flow net
Figure (4.4): Mistakes in drawing Flow nets.
has been drawn, it can be traced out.
## Seepage Calculation from a Flow Net
A flow channel is the strip located between two adjacent flow lines. To calculate the seepage
under a hydraulic structure, consider a flow channel as shown in figure (4.5).
## Figure (4.5): Flow through a flow channel.
The equipotential lines crossing the flow channel are also shown, along with their corresponding
hydraulic heads. Let (q) be the flow through the flow channel per unit length of the hydraulic
structure (i.e., perpendicular to the section shown). According to Darcy’s law,
## ℎ1 −ℎ2 ℎ2 −ℎ3 ℎ3 −ℎ4
∆𝑞 = 𝑘. 𝑖. 𝐴 = 𝑘 ( ) (𝑏1 𝑥1) = 𝑘 ( ) (𝑏2 𝑥1) = 𝑘 ( ) (𝑏3 𝑥1) = ….(4.11)
𝑙1 𝑙2 𝑙3
## If the flow elements are drawn as squares, then
𝑙1 = 𝑏1
𝑙2 = 𝑏2
𝑙3 = 𝑏3
4
So, from Eq. (4.11), we get
ℎ1 − ℎ2 = ℎ2 − ℎ3 = ℎ3 − ℎ4 = ⋯ = ∆ℎ = …..………….………………….(4.12)
𝑁𝑑
where ∆h = drop in piezometric elevation between two consecutive equipotential lines,
h = total hydraulic head = difference in elevation of water between the upstream and
downstream side
Nd = number of potential drops
Equation (4.12) demonstrates that the loss of head between any two consecutive equipotential
lines is the same. Combining Eqs. (4.11) and (4.12) gives
∆𝑞 = 𝑘 𝑁 ..………….…………………………………………………………….(4.13)
𝑑
If there are 𝑁𝑓 flow channels in a flow net, the rate of seepage per unit length of the hydraulic
structure is
𝑁
𝑞 = 𝑁𝑓 ∆𝑞 = 𝑘 ℎ 𝑁𝑓 ..…………………………………………………………….(4.14)
𝑑
Although flow nets are usually constructed in such a way that all flow elements are
approximately squares, that need not always be the case. We could construct flow nets with all the
flow elements drawn as rectangles. In that case the width-to-length ratio of the flow nets must be a
constant, i.e.,
𝑏1 𝑏2 𝑏3
= = = ⋯ = 𝑛 ..…………………………………………………….(4.15)
𝑙1 𝑙2 𝑙3
For such flow nets the rate of seepage per unit length of hydraulic structure can be given by
𝑁𝑓
𝑞=𝑘ℎ 𝑛 ..…………….…………………………………………………….(4.16)
𝑁𝑑
Example (4.1): For the flow net under a dam shown in figure:-
(a) How high would water rise if a piezometer is placed at (i) A, (ii) B, (iii) C ?
(b) If k=0.01mm/s, determine the seepage loss of the dam in m3/(day·m).
## U/S water level
10 m
D/S water level
B
Permeable layer
Impermeable layer
10 m
Scale
5
Solution:
The maximum hydraulic head h is 10 m. From figure, 𝑁𝑑 = 12, ∆ℎ = ℎ/𝑁𝑑 =10/12 =0.833.
Part a (i):
To reach A, water must go through three potential drops. So head lost is equal to
3(0.833)=2.5m. Hence the elevation of the water level in the piezometer at A will be 10 −2.5 =
7.5m above the ground surface.
Part a (ii):
The water level in the piezometer above the ground level is 10 − 5(0.833)= 5.84m.
Part a (iii):
Points A and C are located on the same equipotential line. So water in a piezometer at C will
rise to the same elevation as at A, i.e., 7.5m above the ground surface.
Part b:
The seepage loss is given by q = k h Nf /Nd . From figure, Nf = 5 and Nd = 12. Since k =
0.01mm/s = (0.01/1000) (60×60×24) = 0.864 m/day
q = 0.864 (10)(5/12) = 3.6m3/(day ·m)
## Hydraulic Uplift Force Under a Structure
Consider the dam section shown in Figure (4.3a), the cross-section of which has been replotted
in Figure (4.6). To find the pressure head at point D; from the flow net shown in Figure (4.3a), the
pressure head is equal to (10 + 3.34m) minus the hydraulic head loss. Point D coincides with the
third equipotential line beginning with the upstream side, which means that the hydraulic head loss
at that point is 2(h/Nd) = 2(10/12) = 1.67m. So, pressure head at D = 13.34 − 1.67 = 11.67m
3.34m
F 1.67m G
D E H I
1.67m 1.67m 18.32m 1.67m 1.67m
5.84
5.00
4.56
10.84
8.75
11.67
Figure (4.6): Pressure head under the dam section shown in figure (4.3).
Similarly,
Pressure head at E = (10 + 3.34) − 3(10/12) = 10.84m
Pressure head at F = (10 + 1.67) − 3.5(10/12) = 8.75m
(Note that point F is approximately midway between the fourth and fifth equipotential lines
starting from the upstream side.)
6
Pressure head at G = (10+1.67) − 8.5(10/12) = 4.56m
Pressure head at H = (10+3.34) − 9(10/12) = 5.84m
Pressure head at I = (10+3.34) − 10(10/12) = 5m
The pressure heads calculated above are plotted in Figure (4.6). Between points F and G, the
variation of pressure heads will be approximately linear.
The hydraulic uplift force per unit length of the dam, U, can now is calculated as:
## 11.67 + 10.84 10.84 + 8.75 8.75 + 4.56
= 9.81 [( ) (1.67) + ( ) (1.67) + ( ) (18.32)
2 2 2
4.56+5.84 5.84 + 5
+( ) (1.67) + ( ) (1.67)]
2 2
## Flow Nets in Anisotropic Material
For plotting flow nets of two-dimensional flow problems such as seepage through soils that
show anisotropy with respect to permeability (i.e., 𝑘ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 ≠ 𝑘𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 ). Referring to Eq. (4.9):-
𝜕2 ℎ 𝜕2 ℎ
𝑘𝑥 + 𝑘𝑧 = 0 …………..………...………………………………………...(4.9)
𝜕𝑥 2 𝜕𝑧 2
## where 𝑘𝑥 = 𝑘ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 and 𝑘𝑧 = 𝑘𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 . this equation can be rewritten as:
𝜕2 ℎ 𝜕2 ℎ
+ = 0 ……………….…..………………………..……...……(4.17)
(𝑘𝑧 /𝑘𝑥 )𝜕𝑥 2 𝜕𝑧 2
## Let 𝑥́ = √𝑘𝑧 /𝑘𝑥 x, then
𝜕2 ℎ 𝜕2 ℎ
= …………..…….…..…………………………………...……(4.18)
(𝑘𝑧 /𝑘𝑥 )𝜕𝑥 2 𝜕𝑥́ 2
## Substituting Eq.(4.18) into Eq.(4.17), gives
𝜕2 ℎ 𝜕2 ℎ
+ = 0 …………..…….……...…………………………….……...……(4.19)
𝜕𝑥́ 2 𝜕𝑧2
Equation (4.19) is of the same form as Eq. (4.10), which governs the flow in isotropic soils and
represents two sets of orthogonal lines in the 𝑥́ – z plane.
The steps for construction of a flow net in an anisotropic medium are as follows:
𝑘𝑧 𝑘𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙
2. Determine √ =√
𝑘𝑥 𝑘ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙
7
𝑘𝑧
3. Adopt a horizontal scale such that; 𝑠𝑐𝑎𝑙𝑒ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 = √ (𝑠𝑐𝑎𝑙𝑒𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 )
𝑘𝑥
4. With the scales adopted in steps 1 and 3, plot the cross-section of the hydraulic structure.
5. Draw the flow net for the transformed section plotted in step 4 in the same manner as is done
for seepage through isotropic soils.
## 6. Calculate the rate of seepage as:
Nf
q = √k x k z h .........…………………………………………………...……(4.20)
Nd
Comparing Eqs. (4.14 and 4.20), you see that both equations are similar except that (k) in Eq.
(4.14) is replaced by √𝒌𝒙 𝒌𝒛 in Eq. (4.20).
## (b) True section.
Figure (4.7): Example for flow net under a sheet pile for anisotropic soil.
8
Example (4.2): A dam section is shown in Figure (a). The coefficients of permeability of the
permeable layer in the vertical and horizontal directions are 2×10−2 mm/s and 4×10−2
mm/s, respectively. Draw a flow net and calculate the seepage loss of the dam in
m3/(day·m).
𝒙 =?
𝒌
𝒙́ = 𝒙. √𝒌𝒛
𝒙
## Vertical scale = 12.5 m
Figure (4.8): Construction of flow net under a dam for anisotropic soil.
Solution:
## From the given data:
𝑘𝑧 = 2×10−2 mm/s =1.728 m / day, 𝑘𝑥 = 4×10−2 mm/s = 3.456 m /day , and h =10 m.
## For drawing the flow net:
2 𝑥 10 −2 1
(𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑐𝑎𝑙𝑒) = √ (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑐𝑎𝑙𝑒) = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑐𝑎𝑙𝑒)
4𝑥 10 −2 √2
On the basis of these scales, the dam section is replotted, and the flow net is drawn as shown
in Figure (4.8b), from it, 𝑁𝑑 = 8 and 𝑁𝑓 = 2.5 (the lower most flow channel has a width-to-
length ratio of 0.5).
𝑁𝑓
The rate of seepage is given by 𝑞 = √𝑘𝑥 𝑘𝑧 ℎ .
𝑁𝑑
## = √(1.728)(3.456) (10) (2.5/8) = 7.637 m3/(day·m)
9
Example (4.3): A single row of sheet pile structure is shown in Figure (4.9a). Draw a flow net for the
transformed section. Replot this flow net in the natural scale also. The relationship
between the permeabilities is given as 𝒌𝒙 = 𝟔 𝒌𝒛
Solution:
## For the transformed section,
𝑘𝑧 1
(𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑐𝑎𝑙𝑒) = √ (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑐𝑎𝑙𝑒) = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑐𝑎𝑙𝑒)
𝑘𝑥 √6
The transformed section and the corresponding flow net are shown in figure (4.9b).
Figure (4.9c) shows the flow net constructed to the natural scale. One important fact to be
noticed from this is that when the soil is anisotropic with respect to permeability, the flow and
equipotential lines are not necessarily orthogonal.
(a)
10 m
(b)
## Figure (4.9): Sheet pile structure problem.
10
Flow Nets for Hydraulic Structures on Non-homogeneous Subsoils
Figure (4.10) shows a general condition where a flow channel crosses the boundary of two soils
with 𝑘1 and 𝑘2 permeabilities, respectively. The dashed lines drawn across the flow channel are the
equipotential lines.
Let h be the loss of hydraulic head between two consecutive equipotential lines. Considering a
unit length perpendicular to the section shown, the rate of seepage through the flow channel is:
∆ℎ ∆ℎ
∆𝑞 = 𝑘1 𝑙1
(𝑏1 𝑥1) = 𝑘2 𝑙2
(𝑏2 𝑥1)
or
𝑘1 𝑏2 /𝑙2
= .....………………………………………………………...……(4.21)
𝑘2 𝑏1 /𝑙1
where 𝑙1 and 𝑏1 are the length and width of the flow elements in soil layer 1 and 𝑙2 and 𝑏2 are the
length and width of the flow elements in soil layer 2.
𝑘1 𝜃2 𝑘2
𝛼1
∆𝑞 𝑙2
A
𝑏1 𝛼2
𝜃1
𝑏2 ∆𝑞
𝑙1
B
𝑙1
C 𝑙2
## From Eqs. (4.22a and 4.22c),
𝑏1 𝐶𝑜𝑠 𝜃1 sin 𝛼1
= =
𝑙1 𝑆in 𝜃1 cos 𝛼1
or
𝑏1 1
= = 𝑡𝑎𝑛𝛼1 ....….…………………………………………...……(4.23)
𝑙1 𝑡𝑎𝑛 𝜃1
11
Also, from Eqs. (4.22b and 4.22d),
𝑏2 𝐶𝑜𝑠 𝜃2 sin 𝛼2
= =
𝑙2 𝑆in 𝜃2 cos 𝛼2
or
𝑏2 1
= = 𝑡𝑎𝑛𝛼2 .....….…………………………………………...……(4.24)
𝑙2 𝑡𝑎𝑛 𝜃2
## Combining Eqs. (4.21), (4.23), and (4.24),
𝑘1 𝑡𝑎𝑛 𝜃1 𝑡𝑎𝑛 𝛼2
= = ....……………………………………………...……(4.25)
𝑘2 𝑡𝑎𝑛 𝜃2 𝑡𝑎𝑛 𝛼1
This relation is used for constructing flow nets in non-homogeneous subsoils. It is useful to
keep the following points in mind while constructing the flow nets:
(1) If 𝑘1 > 𝑘2 , plot square flow elements in layer 1. This means that 𝑙1 = 𝑏1 in Eq.(4.20). So
𝑘1 /𝑘2 = 𝑏2 /𝑙2 . Thus the flow elements in layer 2 will be rectangles and their width-to-length
ratios will be equal to 𝑘1 / 𝑘2 . This is shown in figure (4.11a).
(2) If 𝑘1 < 𝑘2 , plot square flow elements in layer 1 (i.e., 𝑙1 = 𝑏1 ). From Eq. (4.20), 𝑘1 /𝑘2 = 𝑏2 /𝑙2 .
So the flow elements in layer 2 will be rectangles. This is shown in figure (4.11b).
## Square elements Square elements 15 m
𝑙1 = 𝑏1 𝑙1 = 𝑏1
= =
6.3 m
Soil (1) Soil (1)
𝛼1 𝑘1 𝛼1
𝑘1 Layer (1)
𝑘1 = 5 𝑥 10−2 5m
𝛼2 𝛼2
𝑚𝑚/𝑠𝑒𝑐
𝑏2
Soil (2) Soil (2) 𝑙2
𝑘2 𝑘2
𝑙2
𝑏2
𝑘1 > 𝑘2 𝑘1 < 𝑘2 Layer (2) 10 m
𝑘2 = 2.5 𝑥 10−2
𝑏2 𝑏2 𝑚𝑚/𝑠𝑒𝑐
>1 𝑙2
<1
𝑙2
Figure (4.11): Flow channel at the boundary between Figure (4.12): Flow net under a dam resting
two soils with different permeabilities. on a two-layered soil deposit.
An example of the construction of a flow net for a dam section resting on a two-layered soil
deposit is given in figure (4.12). Note that 𝑘1 = 5×10−2 mm/s and 𝑘2 = 2.5×10−2 mm/s. So,
In soil layer 1, the flow elements are plotted as squares, and since 𝑘1 / 𝑘2 =2, the length-to-width
ratio of the flow elements in soil layer 2 is 1/2.
12
Piping
High 𝑖𝑒𝑥𝑖𝑡 can cause erosion of soil at the point of exit of the flow line (at downstream), and
gradually the erosion may progress towards the upstream. This backwards erosion of soil is the
condition called "piping". If piping is not checked, it can lead to collapse of the structure.
Stone Pitching
U/S Water
Large 𝑖𝑒𝑥𝑖𝑡
U/S D/S water comes out vertically
Shell Shell upwards at the exit surface
Clay
Core
## Progressive soil erosion
Causing vertical piping
Impervious
## Figure (4.13): Piping condition.
If water comes out vertically upwards at the exit surface, the condition is called vertical piping.
Also, sometimes horizontal piping can occur, such as in earth dam body when the phreatic line
emerges on downstream slope and saturation leads to sloughing.
## Remedial Measures for Piping
(a) Reduce 𝑖𝑒𝑥𝑖𝑡 by increasing the length of flow lines and lengthening the floor of the hydraulic
structure, such as providing upstream impervious blanket, or upstream and downstream
cutoffs. However, using a sheet pile at the downstream is most effective cutoff location for
reducing 𝑖𝑒𝑥𝑖𝑡 , but it also gives greater uplift pressure on the floor compared with case of
upstream sheet pile (See pages 271 – 273 Lamb & Whiteman, 1969).
(b) Prevent erosion by providing suitably designed graded filters. In addition to, providing
additional weight to resist uplift pressures they prevent washing out of fine soil material
from the foundation soil.
Filter Design
When seepage water flows from a soil with relatively fine grains into a coarser material, there is
a danger that the fine soil particles may wash away into the coarse material. Over a period of time,
this process may clog the void spaces in the coarser material. Such a situation can be prevented by
the use of a protective filter between the two soils as shown in (figure 4.14).
Also if upward flow occurs through sandy soil and (i) at the surface becomes greater than unity,
a quick condition is created and the foundation soil will wash out if boiling is not prevented.
For the proper selection of the filter material, two conditions should be satisfied:-
(a) The size of the voids in the filter material should be small enough to hold the larger particles
of the protected material in place.
13
(b) The filter material should have a high permeability to prevent buildup of large seepage
forces and hydrostatic pressures in the filters. However, based on the experimental
investigation of protective filters, Terzaghi and Peck (1948) suggested the following criteria:
𝐷15(𝐹)
1. ≤ 4−5 (To satisfy condition 1)
𝐷85(𝐵)
𝐷15(𝐹)
2. ≥ 4−5 (To satisfy condition 2)
𝐷15(𝐵)
𝐷50 𝑜𝑓 𝑓𝑖𝑙𝑡𝑒𝑟
3.
𝐷50 𝑜𝑓 𝑠𝑜𝑖𝑙
< 25
𝐷85 𝑜𝑓 𝑓𝑖𝑙𝑡𝑒𝑟
4. In case of use perforated pipe: ℎ𝑜𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑝𝑖𝑝𝑒
> 2.0
where D15, D50, D85 are determined from the P.S.D. curve as the size in (mm) than which 15%, 50%,
85%, respectively of the filter or soil is finer than by weight.
Filter
Rock toe
Soil to be protected
## Figure (4.14): Use of filter at the toe of an earth dam.
𝟓𝑫𝟖𝟓(𝑩) =
𝑫𝟖𝟓(𝑩) = 0.11 mm
0.55 mm
Percent finer %
Curve a Soil to be
protected
Range of Curve b
good filter
## Curve c 𝟓𝑫𝟏𝟓(𝑩) = 𝑫𝟏𝟓(𝑩) =
0.045 mm 0.009 mm
## Figure (4.15): Determination of grain-size distribution of soil filters.
For example, consider the soil used for the construction of the earth dam shown in Figure (4.14).
If the grain-size distribution of this soil be given by curve (a) in Figure (4.15), then determine
5D85(B) and 5D15(B) and plot them as shown in Figure (4.15). The acceptable grain-size distribution
of the filter material will have to lie in the shaded zone.
14
Safety of hydraulic structures against piping
(a) Harza (1935) Method:
When upward seepage occurs and the hydraulic gradient i be equal to 𝑖𝑐𝑟 , piping originates in
the soil mass. The factor of safety against piping can be defined as:
𝑖𝑐𝑟
F.S.= ....……..…………………………………………...……(4.26)
𝑖𝑒𝑥𝑖𝑡
where
(𝐺𝑠 +𝑒) 𝛾𝜔
𝛾́ 𝛾𝑠𝑎𝑡 − 𝛾𝜔 − 𝛾𝜔
𝑖𝑐𝑟 = 𝛾 = = 1+𝑒
𝜔 𝛾𝜔 𝛾𝜔
𝛾́ 𝐺𝑠 − 1
So, 𝑖𝑐𝑟 = 𝛾 = .....……………………...………………...……(4.26a)
𝜔 1+𝑒
for typical values of 𝐺𝑠 and e encountered in soils, 𝑖𝑐𝑟 varies within a range (0.85 – 1.10).
𝑖𝑒𝑥𝑖𝑡 = the maximum exit gradient; obtained by any of the following formulas:-
## (1) From Flow Nets ( see figure 4.3):
𝑖𝑒𝑥𝑖𝑡 = .....……..…………………………………………...……(4.26b)
𝐿
where h = the head lost between the last two equipotential lines, and
l = the length of the flow element.
(2) Harr, 1962 solution (for a single row of sheet pile structures; see figure 4.2):
1 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 ℎ𝑒𝑎𝑑
𝑖𝑒𝑥𝑖𝑡 = 𝜋 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑝𝑒𝑛𝑒𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠ℎ𝑒𝑒𝑡 𝑝𝑖𝑙𝑒
………………..…...……(4.26c)
(3) Harza Charts (for dams constructed over deep homogeneous deposits; see figure 4.16):
𝑖𝑒𝑥𝑖𝑡 = 𝐶 𝐵 ….……………………………………………...……(4.26d)
1.5
h
1.0
## Heel and toe sheeting
C
d 0.5
𝑖𝑒𝑥𝑖𝑡 = 𝐶
B 𝐵
0
Deep homogeneous soil 0 5 10 15
B/d
Figure (4.16): Critical exit gradient (after Harza, 1935).
A factor of safety of (3– 4) is considered adequate for the safe performance of the structure.
15
(b) Lane (1935) Method:
Lane defined the factor of safety by a term called the weighted creep ratio. For a structure to be
safe against piping, this ratio should be ≥ the safe values shown in Table (4.1).
𝐿𝑤
weighted creep ratio = ….…………………………...……(4.27)
𝐻1 − 𝐻2
where
∑ 𝐿ℎ
𝐿𝑤 = weighted creep length = + ∑ 𝐿𝑣 (see figure 4.17),
3
## ∑ 𝐿𝑣 = 𝐿𝑣1 + 𝐿𝑣2 + … … … . . … = sum of vertical distances along shortest flow path,
If the cross-section of a given structure is such that the shortest flow path has a slope steeper
than 45o, it should be taken as a vertical path. If the slope of the shortest flow path is less than
45o, it should be considered as a horizontal path.
## Table (4.1): Safe values for the weighted creep ratio.
𝐻1
𝐻2
Material Safe weighted creep ratio
Very fine sand or silt 8.5
∑ ℎ1 ∑ ℎ2
Fine sand 7.0
∑ 𝑣1 ∑ 𝑣2 Medium sand 6.0
Coarse sand 5.0
Permeable layer
Fine gravel 4.0
Coarse gravel 3.0
Impermeable layer
Soft to medium clay 2.0 – 3.0
Hard clay 1.8
Hard pan 1.6
Figure (4.17): Calculation of weighted creep distance.
## (c) Terzaghi (1922) Method:
(i) For safety against piping around sheet pile structure as shown in figure (4.18a), the
failure due to piping takes place within a distance of D/2 from the sheet piles. Therefore, the
factor of safety against piping is:-
1
́
𝑊 𝛾́ 𝐷2 𝐷 𝛾́
2
F.S.= =1 = ......…….……………………...……(4.28)
𝑈 𝛾 𝐷ℎ𝑎 ℎ𝑎 𝛾𝜔
2 𝜔
where 𝑊́ = submerged weight of the soil prism (D × D/2) acting vertically downwards,
𝑈 = hydraulic uplifting pressure can be determined from the flow net,
D = the depth of penetration of the sheet pile, and
ℎ𝑎 = average hydraulic head at the base of the soil prism,
## A factor of safety of about (4.0) is generally considered adequate.
16
(ii) For safety against piping under a dam as shown in figure (4.18b):-
Terzaghi (1943) recommended that the stability of several soil prisms of size (D/2 × 𝐷́ × 1)
be investigated to find the minimum factor of safety. Note that (0 < 𝐷́ ≤ D).
Harr (1962) suggested that a factor of safety of (4–5) with (𝐷́ =D) should be sufficient for
safe performance of the structure.
𝐻1
𝑫/𝟐 𝑫/𝟐
𝐻2
Possible failure
zone Soil
𝑫 wedge 𝑫́
𝑾 𝑫
Permeable layer Permeable layer
ℎ𝑜
𝑼 𝑼
## Safety of hydraulic structures against heaving
𝐷 𝛾́
𝐹. 𝑆. = ……………..………………………………..(4.29)
𝐶0 𝛾𝜔 (𝐻1 − 𝐻2 )
## Figurer (4.19): Safety against heaving.
D/T 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
## Co 0.385 0.365 0.359 0.353 0.347 0.339 0.327 0.309 0.274
17
Example (4.4): A flow net for a single row of sheet piles is given in Figure (4.20).
(a) Determine the factor of safety against piping by Harza’s method.
(b) Determine the factor of safety against piping by Terzaghi’s method. Assume 𝛾́ = 10.2 kN/m3.
## U/S water level Constant hydraulic
D/S water
3m level
h
B E 0.5 m
A F
1.5 m 0.82 m
C D
Permeable
6m
layer
G H
Impermeable layer
1m
Scale
Solution: Figure (4.20): Flow net around a single row of sheet pile structures.
Part (a):
𝛾́ 10.2
𝑖𝑐𝑟 = = = 1.04
𝛾𝜔 9.81
∆ℎ 3−0.5 3−0.5
𝑖𝑒𝑥𝑖𝑡 = , ∆ℎ = = = 0.417 m
𝐿 𝑁𝑑 6
L = the length of the last flow element scaled out of figure (4.20) and is approximately 0.82 m.
0.417
So, 𝑖𝑒𝑥𝑖𝑡 = = 0.509
0.82
𝑖𝑐𝑟 1.04
The factor of safety against piping is: F.S. = = = 2.04
𝑖𝑒𝑥𝑖𝑡 0.509
Part (b):
A soil prism of cross-section D×D/2 where D = 1.5m, on the downstream side adjacent to the
sheet pile is plotted in figure (4.21). The approximate hydraulic heads at the bottom of the prism
can be evaluated by using the flow net. Referring to figure (4.20), (note that Nd = 6),
Filter
Soil 3m
D = 1.5m Prism
0.5m
A B C
Permeable layer
0.75m 𝒉𝒐 = 0.917m
0.833m
1.25m
Impermeable layer
(a) (b)
Figure (4.21): Factor of safety calculation by Terzaghi’s method.
18
3
ℎ𝐴 = (3 − 0.5) = 1.25𝑚
6
2
ℎ𝐵 = (3 − 0.5) = 0.833𝑚
6
1.8
ℎ𝐶 = (3 − 0.5) = 0.75𝑚
6
1.25+0.75
( + 0.833)
2
ℎ𝑎 = = 0.917𝑚
2
𝐷 𝛾́ 1.5(10.2)
𝐹. 𝑆. = = = 1.7
ℎ𝑎 𝛾𝜔 0.917(9.81)
The factor of safety calculated here is rather low. However, it can be increased by placing some
filter material on the downstream side above the ground surface, as shown in figure above. This
will increase the weight of the soil prism.
Example (4.5): A dam section is shown in Figure (4.22). The subsoil is fine sand. Using Lane’s
method, determine whether the structure is safe against piping.
Solution:
∑ Lh
Lw = weighted creep length = + ∑ Lv
3
## ∑ 𝐿𝑣 = 𝐿𝑣1 + 𝐿𝑣2 + … … … . . … = 1 + (8 + 8) + 1 + 2 = 20m
16
Lw = + 20 = 25.33m
3
Lw
weighted creep ratio = = 25.33 / (10 – 2) = 3.17
H1 − H2
From Table (4.1), the safe weighted creep ratio for fine sand is about 7.0. Since the calculated
weighted creep ratio is 3.17< 7.0, the structure is unsafe.
Upstream
10m apron
2m
1m 2m
1m
6m
8m
10m
## Figure (4.22): Safety against piping under a dam by Lane's method.
19
Seepage Through a Homogeneous Earth Dams with (𝒌𝒙 = 𝒌𝒛 )
Dupuit’s Solution
Figure (4.23) shows the section of an earth dam in which ab is the phreatic surface, i.e., the
uppermost line of seepage. The quantity of seepage through a unit length at right angles to the
cross-section can be given by Darcy’s law as q = k i A. z
a Phreatic line
𝑑𝑧
𝑑𝑥
H1 b
z
H2
x
Impermeable base
d
## Figure (4.23): Dupuit’s solution for flow through an earth dam.
Dupuit (1863) assumed that the hydraulic gradient i is equal to the slope of the free surface
and is constant with depth, i.e., i = dz/dx. So
𝑑𝑧 𝑑𝑧
𝑞 = 𝑘 𝑑𝑥 [(𝑧)(1)] = 𝑘 𝑑𝑥 𝑧
𝑑 1 𝐻 𝑘
∫0 𝑞. 𝑑𝑥 = ∫𝐻 𝑘. 𝑧. 𝑑𝑧 or 𝑞 𝑑 = 2 (𝐻12 − 𝐻22 )
2
𝑘
or 𝑞 = 2 𝑑 (𝐻12 − 𝐻22 ) .....….…………………........………………………...……(4.30)
This equation represents a parabolic free surface. However, in the derivation of the equation,
no attention has been paid to the entrance or exit conditions. Also note that if 𝐻2 = 0, the
phreatic line would intersect the impervious surface.
Schaffernak’s Solution
According to this solution the phreatic surface will intersect the downstream slope at a
distance l from the impervious base and the seepage per unit length of the dam can be
determined by considering the triangle bcd in figure (4.24): z
a
𝑑𝑧
𝑑𝑥
H
Parabolic b
free surface l
𝜃
x
Impermeable base d c
## Figure (4.24): Schaffernak’s solution for flow through an earth dam.
20
𝑞 = 𝑘. 𝑖. 𝐴 ̅̅̅̅ )(1) = 𝑙. 𝑠𝑖𝑛𝛽
𝐴 = (𝑏𝑑
From Dupuit’s assumption, the hydraulic gradient is given by i = dz/dx = tan 𝛽. So,
𝑑𝑧
𝑞 = 𝑘. 𝑧 = (𝑘)(𝑙. 𝑠𝑖𝑛𝛽)(𝑡𝑎𝑛𝛽)
𝑑𝑥
𝐻 𝑑
or ∫𝑙.𝑠𝑖𝑛𝛽 𝑧. 𝑑𝑧 = ∫𝑙.𝑐𝑜𝑠𝛽(𝑙. 𝑠𝑖𝑛𝛽)(𝑡𝑎𝑛𝛽)𝑑𝑥
1
(𝐻 2 − 𝑙 2 𝑠𝑖𝑛2 𝛽) = (𝑙. 𝑠𝑖𝑛𝛽)(𝑡𝑎𝑛𝛽)(𝑑 − 𝑙. 𝑐𝑜𝑠𝛽)
2
1 𝑠𝑖𝑛2 𝛽
(𝐻 2 − 𝑙 2 𝑠𝑖𝑛2 𝛽) = 𝑙 (𝑑 − 𝑙. 𝑐𝑜𝑠𝛽)
2 𝑐𝑜𝑠𝛽
𝐻 2 𝑐𝑜𝑠𝛽 𝑙 2 𝑐𝑜𝑠𝛽
− = 𝑙. 𝑑 − 𝑙 2 𝑐𝑜𝑠𝛽
2 𝑠𝑖𝑛2 𝛽 2
𝐻 2 𝑐𝑜𝑠𝛽
𝑙 2 𝑐𝑜𝑠𝛽 − 2. 𝑙. 𝑑 + =0
𝑠𝑖𝑛2 𝛽
## 2𝑑 ± √4𝑑 2 − 4[(𝐻 2 𝑐𝑜𝑠 2 𝛽)/𝑠𝑖𝑛2 𝛽]
𝑙=
2 𝑐𝑜𝑠𝛽
𝑑 𝑑2 𝐻2
So 𝑙= −√ − .....….………………….......……………………...……(4.31)
𝑐𝑜𝑠𝛽 𝑐𝑜𝑠2 𝛽 𝑠𝑖𝑛2 𝛽
Once the value of l is known, the rate of seepage can be calculated from (q = kl sin𝛽 tan𝛽).
Schaffernak’s Graphical Procedure To Determine the Value of (l). This procedure is explained
with the aid of figure (4.25) below as follows:-
f
h
a g
H Phreatic line
b
x
e Impermeable base c
## Figure (4.25): Graphical construction for Schaffernak’s solution.
21
1. Extend the downstream slope line bc upward.
2. Draw a vertical line ae through the point a. This will intersect the projection of line bc (step1)
at point f .
3. With fc as diameter, draw a semicircle fhc.
4. Draw a horizontal line ag.
5. With c as the center and cg as the radius, draw an arc of a circle, gh.
6. With f as the center and fh as the radius, draw an arc of a circle, hb.
7. Measure bc = l.
Casagrande's Correction: The parabola ab shown in figure (4.23) should actually start from the
point 𝑎́ as shown in figure (4.26) with 𝒂́ 𝒂= 0.3∆. So, the value of (d) for use in Eq. (4.31) will be the
horizontal distance between points 𝑎́ and c.
0.3∆
𝑎́ 𝑎
H b
x
Impermeable base c
d
Figure (4.26): Modified distance (d) for use in Eq. (4.31).
L. Casagrande’s Solution
According to this solution, the distance (l) at which the phreatic surface will intersect the
downstream slope as shown in figure (4.27) is approximated as:
## 𝑙 = √𝑑2 + 𝐻 2 − √𝑑2 − 𝐻 2 𝑐𝑜𝑡 2𝛽 ……….......……………………...……(4.32)
Once the value of l is known, the rate of seepage can be calculated from (q = kl 𝑠𝑖𝑛2 𝛽).
z
0.3∆
𝑎́ 𝑎
𝑑𝑠
𝑑𝑧
𝑑𝑥
H b
phreatic
surface l
𝛽
x
Impermeable base d c
d
## Figure (4.27): L. Casagrande’s solution for flow through an earth dam
(Note: length of the curve 𝑎́ 𝑏𝑐 = 𝑆).
22
A solution that avoids the approximation introduced in Eq. (4.32) was given by Gilboy (1934)
and in graphical form by Taylor (1948), as shown in Figure (4.28). To use the graph,
1. Determine d/H.
2. For given values of d/H and 𝛽 , determine m.
3. Calculate l = mH/sin 𝛽 .
4. Calculate q = k l 𝑠𝑖𝑛2 𝛽. 0.3∆
H
𝜷 𝒍 𝒔𝒊𝒏𝜷 = 𝒎 𝑯
x
d c
90
80
𝜷 (𝒅𝒆𝒈𝒓𝒆𝒆)
60 m=
40
20
0
0 1 2 3 4 5 6 7 8 9
d/H
Figure (4.28): Chart for solution by L. Casagrande’s method based on Gilboy's solution.
## Seepage Through Earth Dams with (𝒌𝒙 ≠ 𝒌𝒛 )
If the soil in a dam section shows anisotropic behavior with respect to permeability, the dam
section should first be plotted according to the transformed scale, such that
𝑥́ = 𝑥√𝑘𝑧 /𝑘𝑥
All calculations should be based on this transformed section. Also, for calculating the rate of
seepage, the term k in the corresponding equations should be equal to√𝑘𝑥 𝑘𝑧 .
Example (4.6): The cross-section of an earth dam is shown in Figure (4.29). Calculate the rate of
seepage through the dam [q in m3/(min·m)] by:
(a) Dupuit’s method; (b) Schaffernak’s method; and (c) L. Casagrande’s method.
0.3x 50
=15m 5m
𝒂́ 𝒂
1
30m 2
25m 1 b
2 k = 3 x 10-4 m/min.
c
x
Impermeable base 50m 15m 60m
23
Solution:
## Part (a), Dupuit’s method:
𝑘
𝑞 = 2 𝑑 (𝐻12 − 𝐻22 )
From given data; 𝐻1 = 25m, 𝐻2 = 0, and, d (the horizontal distance between points a and c) is
equal to 60 + 5 + 10 = 75 m. hence
3 𝑥 10−4
𝑞= (25)2 = 𝟏𝟐. 𝟓 𝒙 𝟏𝟎−𝟒 𝑚3 /(𝑚𝑖𝑛. 𝑚)
(2)(75)
## Part (b), Schaffernak’s method:
𝑑 𝑑2 𝐻2
q = kl sin𝛽 tan𝛽; and 𝑙 = −√ −
𝑐𝑜𝑠𝛽 𝑐𝑜𝑠2 𝛽 𝑠𝑖𝑛2 𝛽
Using Casagrande's correction (fig. 4.26), d (the horizontal distance between points 𝑎́ and c)
is equal to 60 + 5 + 10 + 15 = 90 m. Also
1
𝛽 = 𝑡𝑎𝑛−1 2 = 26.57𝑜 , h = 25m
90 90 2 25 2
𝑙 = 𝑐𝑜𝑠 26.57 − √(𝑐𝑜𝑠 26.57) − (𝑠𝑖𝑛 26.57) = 100.63 − √(100.63)2 − (55.89)2 = 16.95𝑚
## Part (c), L. Casagrande’s method:
d = 90m , H= 25m,
From figure (4.28), for 𝛽 = 26.57𝑜 and d/H = 90/25 = 3.6 , m = 0.34 and
𝑚𝐻 0.34(25)
𝑙= = = 19.0𝑚
𝑠𝑖𝑛𝛽 sin 26.57
## Plotting of Phreatic Line for Seepage Through Earth Dams
For construction of flow nets for seepage through earth dams, the phreatic line needs to be
established first. This is usually done by Casagrande's method as shown in Figure (4.30a). Note
that aefb in Figure (4.30a) is the actual phreatic line. The curve 𝒂́ ef𝒃́𝒄́ is a parabola with its focus
at 𝒄́ ; the phreatic line coincides with this parabola, but with some deviations at the upstream and
the downstream faces. At a point (a), the phreatic line starts at an angle of 90o to the upstream face
of the dam and 𝒂𝒂́ = 𝟎. 𝟑∆.
24
z
d Directrix
0.3∆
z
𝑎́ a Directrix
e A(x, z)
f 𝑏́ D
∆𝑙
H
Coefficient of b
l
permeability k 𝛽
x x
c 𝑐́ c 𝑐́
∆ (0, 0)
p
p p
p
(a) (b)
Figure (4.30): Determination of phreatic line for seepage through an earth dam.
## The parabola 𝒂́ ef𝒃́𝒄́ can be constructed as follows:
(1) Let the distance c𝒄́ = 𝒑. Now referring to fig. (4.30b), Ac = AD (based on the properties of a
parabola), Ac = √𝑥 2 + 𝑧 2 , and AD = 2p + x, Thus,
√𝑥 2 + 𝑧 2 = 2p + x ….…….....………………………………………..…...……(4.33)
At point (𝒂́ ); x = d, and z = H, substituting these conditions into Eq. (4.33) and rearranging
gives:
1
𝑝 = 2 (√𝑑2 + 𝐻 2 − 𝑑) .…...........…………………………………..…….……(4.34)
## (2) Now by squaring both sides of Eq. (4.33) gives:
𝑥 2 + 𝑧 2 = 4𝑝2 + 𝑥 2 + 4𝑝𝑥
𝑧 2 −4𝑝2
𝑥= …..…….......….…………………………………………..…….……(4.35)
4𝑝
With (p) known, the values of (x) for various values of (z) can be calculated from Eq. (4.35)
and the parabola can be constructed.
To complete the phreatic line, the portion ae has to be approximated and drawn by hand.
The rate of seepage through earth dam can be estimated according to Kozeny's formula as:
𝑞 = 2𝑘 𝑝.……....................…………………………………………..…….……(4.36)
25
When (𝜷 < 𝟑𝟎𝒐 ), the value of l can be calculated from Eq. (4.31) as:
𝑑 𝑑2 𝐻2
𝑙= −√ −
𝑐𝑜𝑠𝛽 𝑐𝑜𝑠2 𝛽 𝑠𝑖𝑛2 𝛽
Note that l = ab in figure (4.30a). Once point b has been located, the curve fb can be
approximately drawn by hand.
If (𝜷 ≥ 𝟑𝟎𝒐 ), the value of l can be determined from the following table as:
𝛽𝑜 30 60 90 120 150 180
∆𝐿
0.375 0.320 0.260 0.185 0.095 0.00
𝐿 + ∆𝐿
In figure (4.30a), 𝒃́𝒃 = ∆𝒍 , and bc = l. Once point b has been located on the downstream
face, the curve fb can be approximately drawn by hand.
## Types of Entrance, Exit and Transfer Conditions of Seepage Line
Figure (4.31) shows various entrance, exit, and transfer conditions for the line of seepage
through earth dams (after Casagrande,1937) that are helpful to know the nature of the phreatic
lines for various types of earth dam sections.
## Entrance Conditions:- 𝛽 < 90𝑜 𝛽 = 90𝑜 𝛽 > 90𝑜
A horizontal A horizontal
𝛽 𝛽
𝛽 𝛽
𝛽 𝛽
B B
## 90𝑜 < 𝛽 < 180𝑜 𝛽 = 90𝑜 𝑜𝑟 𝛽 = 180𝑜
Exit Conditions:- 𝛽 < 90𝑜
vertical vertical
Toe drain
𝛼2
𝛼1 𝛼1 𝛽
𝛽 𝛽 𝛽
## 𝑘1 < 𝑘2 𝑘1 > 𝑘2 𝑘1 > 𝑘2
𝛼2 = 270𝑜 − 𝛼1 − 𝛽 𝛼2 = 270𝑜 − 𝛼1 − 𝛽 𝛼2 = 𝛼1 = 𝛽
## (h) (i) (j)
Figure (4.31): Entrance, exit, and transfer conditions (after Casagrande, 1937).
26
Flow Net Construction for Earth Dams
(a) For isotropic earth dams (𝑘𝑥 = 𝑘𝑧 = 𝑘); (see figure 4.32):
1. Draw the phreatic line according to Casagrande's method as explained before.
2. Note that ag is an equipotential line and that gc is a flow line.
3. It is important to realize that the pressure head at any point on the phreatic line is zero;
hence the difference of total head between any two equipotential lines should be equal to
the difference in elevation between the points where these equipotential lines intersect the
phreatic line. Since loss of hydraulic head between any two consecutive equipotential lines
is the same, determine the number of equipotential drops, 𝑁𝑑 , the flow net needs to have
and calculate ∆ℎ = ℎ/𝑁𝑑 .
𝑵𝒇 = 2.3
𝑵𝒅 = 10
Impermeable layer
## Figure (4.32): Flow net construction for homogeneous earth dam.
4. Draw the headlines for the cross-section of the dam. The points of intersection of the
headlines and the phreatic lines are the points from which the equipotential lines should
start.
5. Draw the flow net, keeping in mind that the equipotential lines and flow lines must
intersect at right angles.
𝑁𝑓
6. Calculate the rate of seepage through the earth dam from Eq. (4.14) as: 𝑞 = 𝑘 ℎ
𝑁 𝑑
In figure (4.32), 𝑁𝑓 = 2.3 (the top two flow channels have square flow elements, and the
bottom flow channel has elements with a width-to-length ratio of 0.3) and 𝑁𝑑 =10. Figure
(4.33) shows some typical flow nets through earth dam sections.
Note: Flow net is for isotropic case Phreatic line for (𝒌𝒉 = 𝒌𝒗 )
(𝒌𝒉 = 𝒌𝒗 )
Partial flow channel Phreatic line for (𝒌𝒉 = 𝟏𝟔𝒌𝒗 )
(Transposed back from transformed section)
∆𝒉
## Impermeable layer Drain
(a): With blanket drain.
27
𝑁𝑓 = 1.5
𝑁𝑓 = 2
𝑁𝑑 = 6
𝑁𝑑 = 5
## (b) For anisotropic earth dams (𝑘𝑥 ≠ 𝑘𝑧 ):
If the dam section is anisotropic with respect to permeability, a transformed section should
first be prepared. The flow net can then be drawn on the transformed section, and the rate of
seepage obtained from Eq. (4.20) as:
𝑁𝑓
𝑞 = √𝑘𝑥 𝑘𝑧 ℎ
𝑁𝑑
(c) For zoned earth dams:
A flow net for seepage through a zoned earth dam section is shown in Figure (4.34). The soil
for the upstream half of the dam has a permeability 𝑘1 , and the soil for the downstream half
of the dam has a permeability 𝑘2 = 5𝑘1 . The phreatic line must be plotted by trial and error.
As shown, here the seepage is from a soil of low permeability (upstream half) to a soil of high
permeability (downstream half). From Eq. (4.21),
𝑘1 𝑏2 /𝑙2
=
𝑘2 𝑏1 /𝑙1
If 𝑏1 = 𝑙1 and 𝑘2 = 5𝑘1 , then 𝑏2 /𝑙2 = 1/5. For that reason, square flow elements have been
plotted in the upstream half of the dam, and the flow elements in the downstream half have a
width-to-length ratio of 1/5. The rate of seepage can be calculated by using the following
equation:
ℎ ℎ
𝑞 = 𝑘1 𝑁𝑓(1) = 𝑘2 𝑁
𝑁𝑑 𝑁𝑑 𝑓(2)
where 𝑁𝑓(1) is the number of full flow channels in the soil having a permeability 𝑘1 , and
𝑁𝑓(2) is the number of full flow channels in the soil having a permeability 𝑘2 .
28
Impermeable layer ℎ ℎ
𝑞 = 𝑘1 𝑁 𝑞 = 𝑘2 𝑁
𝑁𝑑 𝑓(1) 𝑁𝑑 𝑓(2)
𝑁𝑓(1) = 8/3 𝑁𝑓(2) = 8/15
Figure (4.34): Flow net for seepage through a zoned earth dam.
## Control of Seepage Losses
The losses due to seepage can be controlled through the following factors:-
1. the permeability of soil,
2. the differential head ,∆ℎ , across the flow path
3. the length of the flow path, and
4. number of flow channels.
Figure (4.35) shows some solutions for controlling the seepage losses.
Crest
Impermeable upstream
blanket
Chimney drain
Shoulder Shoulder
Grout Curtain
29
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# Kirchhoffs voltage law - wykład
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## Fragment notatki:
Kirchho ff s voltage law Kirchhoff's voltage law of lumped circuit theory states that the sum of the voltage rises in
traversing any closed path in a circuit is zero. The voltage rises are of course differences
in electric potential. The statement above is equivalent to the statement that the line
integral of the electric field between any two points is independent of the path.
Now this last statement is clearly true of electrostatics, but is not generally true of
electrodynamics, wherein changing magnetic fluxes can induce an emf around a closed
path, and the electric field is no longer simply derivable as the gradient of a potential. So
we may expect Kirchhoff's voltage law to apply to d.c. circuits, but we must be cautious
in applying it to a.c. circuits.
The fact that lumped circuit theory is frequently applied to a.c. circuits rests upon
a classification scheme at work in the background of our view of such circuits. In the
lumped circuit view of an a.c. circuit, we divide the circuit into a number of regions. In some of those regions there are no changing magnetic fluxes, and within each of those
limited regions the line integral of the electric field is independent of the path, as long as
we remain within that region.
In some other regions, such as in the interior of transformers and inductors, and shown
by the dotted area surrounding inductor L in Figure 1.8, we do allow changing magnetic
fluxes, but in our consideration of the line integral of the electric field along a path from
one terminal at one end of a winding of the device to a second terminal at the other end
of that winding, we follow a unique path, namely along that winding, and thus always
obtain the same value for the integral.
What we are doing in the above view is consigning the changing magnetic fluxes to
limited regions of space, namely the interiors of inductors and transformers, and we are
deliberately neglecting the effects of any magnetic fields which are caused by the currents
on the wires which connect the separate devices. Clearly this is an approximation which
becomes more and more questionable as frequency increases.
In the lumped theory approximation the line integral of the electric field from point Q
to point P, which would have a value −v1, ought to be equal to the line integral of electric
field from point R to point S, which would have the value −v2, but in reality the currents
flowing on the interconnecting wires produce magnetic fluxes which have a component
perpendicular to the page such that along the closed path PQRSP the integral of the
electric field is by Faraday's law non-zero so that the variables v1 and v2 cannot be equal.
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# GSL: The GNU Scientific Library
GSL is a library that provides many useful scientific functions, including random number generation, random number distributions, statistics, fast fourier transform, and root finding. The latest version is gsl-0.4.1 released January 1999. The source code is installed in /usr/dirt/src/gsl-0.4.1/, the library files (compiled on FreeBSD 2.2.8) are in /usr/dirt/lib/, and the header files are in /usr/dirt/include/.
On this web page, I've included links to the GSL main web page and the GSL manual, a list of available random number distributions, an example on how to use GSL in your own program, and some notes on how to re-build the library.
## Available Distributions
Continuous
gaussian (sigma) bivariate gaussian (sigma_x, sigma_y, rho, x, y) exponential (mu) laplace (mu) exponential power (mu, a) cauchy (mu) rayleigh (sigma) rayleigh tail (a, sigma) symmetric levy (mu, a) gamma (a, b) flat/uniform (a, b) lognormal (zeta, sigma) chi-squared (mu) F-distribution (nu1, nu2) t-distribution (nu) beta (a, b) logistic (mu) pareto (a, b) spherical_2d (x, y) spherical_3d (x, y, z) weibull (mu, a) gumbel (a, b)
Discrete
poisson (mu) bernoulli (p) binomial (p, n) negative binomial (p, n) geometric (p) hypergeometric (n1, n2, t) logarithmic (p)
## Example (Random Number Distribution)
```#include "gsl_randist.h"
int main (void)
{
double packet_size;
long seed;
gsl_rng *rng; // random number generator
rng = gsl_rng_alloc (gsl_rng_rand48); // pick random number generator
seed = time (NULL) * getpid();
gsl_rng_set (rng, seed); // set seed
packet_size = gsl_ran_exponential (1500); // get a random number from
// the exponential distribution
gsl_rng_free (rng); // dealloc the rng
}
% gcc -L /usr/dirt/lib -lgslrandist -lgslrng -lgslerr -lgslspecfunc [..]
```
Note -- If you want to use GSL with C++, include the gsl header file like this:
```extern "C" {
#include "gsl_randist.h"
}
```
## Notes on Re-building GSL
• Basic Procedure:
``` [.../gsl/]% ./configure
[.../gsl/your favorite library/]% cp gsl_*.h /usr/dirt/include
[.../gsl/your favorite library/]% cp lib*.a /usr/dirt/lib```
• It may (or may not) have been fixed in future releases, but in the randist directory, cfree() is used in the file gsl_randist.c. FreeBSD does not support cfree() by default, but the library works if you replace the call to cfree() with free(). There's only one instance of it in the entire library.
• If you want to build only libraries needed for random number distributions, you'll have to build the following (in this order):
• err
• sys
• utils
• specfunc
• rng
• randist
Other DiRT documents
Author: Michele Clark
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• Difficulty Level : Hard
• Last Updated : 15 Nov, 2022
Given two numbers represented by two linked lists, write a function that returns the sum of the two linked lists in the form of a list.
Note: It is not allowed to modify the lists. Also, not allowed to use explicit extra space (Hint: Use Recursion).
Example :
Input: First List: 5->6->3, Second List: 8->4->2
Output: Resultant list: 1->4->0->5
Explanation: Sum of 563 and 842 is 1405
We have discussed a solution here which is for linked lists where the least significant digit is the first node of lists and the most significant digit is the last node. In this problem, the most significant digit is the first node and the least significant digit is the last node and we are not allowed to modify the lists. Recursion is used here to calculate the sum from right to left.
Following are the steps.
1) Calculate sizes of given two linked lists.
2) If sizes are same, then calculate sum using recursion. Hold all nodes in recursion call stack till the rightmost node, calculate the sum of rightmost nodes and forward carry to the left side.
3) If size is not same, then follow below steps:
….a) Calculate difference of sizes of two linked lists. Let the difference be diff
….b) Move diff nodes ahead in the bigger linked list. Now use step 2 to calculate the sum of the smaller list and right sub-list (of the same size) of a larger list. Also, store the carry of this sum.
….c) Calculate the sum of the carry (calculated in the previous step) with the remaining left sub-list of a larger list. Nodes of this sum are added at the beginning of the sum list obtained the previous step.
Below is a dry run of the above approach:
Below is the implementation of the above approach.
## C++
`// A C++ recursive program to add two linked lists``#include ``using` `namespace` `std;` `// A linked List Node``class` `Node {``public``:`` ``int` `data;`` ``Node* next;``};` `typedef` `Node node;` `/* A utility function to insert``a node at the beginning of linked list */``void` `push(Node** head_ref, ``int` `new_data)``{`` ``/* allocate node */`` ``Node* new_node = ``new` `Node[(``sizeof``(Node))];` ` ``/* put in the data */`` ``new_node->data = new_data;` ` ``/* link the old list of the new node */`` ``new_node->next = (*head_ref);` ` ``/* move the head to point to the new node */`` ``(*head_ref) = new_node;``}` `/* A utility function to print linked list */``void` `printList(Node* node)``{`` ``while` `(node != NULL) {`` ``cout << node->data << ``" "``;`` ``node = node->next;`` ``}`` ``cout << endl;``}` `// A utility function to swap two pointers``void` `swapPointer(Node** a, Node** b)``{`` ``node* t = *a;`` ``*a = *b;`` ``*b = t;``}` `/* A utility function to get size of linked list */``int` `getSize(Node* node)``{`` ``int` `size = 0;`` ``while` `(node != NULL) {`` ``node = node->next;`` ``size++;`` ``}`` ``return` `size;``}` `// Adds two linked lists of same size``// represented by head1 and head2 and returns``// head of the resultant linked list. Carry``// is propagated while returning from the recursion``node* addSameSize(Node* head1, Node* head2, ``int``* carry)``{`` ``// Since the function assumes linked lists are of same`` ``// size, check any of the two head pointers`` ``if` `(head1 == NULL)`` ``return` `NULL;` ` ``int` `sum;` ` ``// Allocate memory for sum node of current two nodes`` ``Node* result = ``new` `Node[(``sizeof``(Node))];` ` ``// Recursively add remaining nodes and get the carry`` ``result->next`` ``= addSameSize(head1->next, head2->next, carry);` ` ``// add digits of current nodes and propagated carry`` ``sum = head1->data + head2->data + *carry;`` ``*carry = sum / 10;`` ``sum = sum % 10;` ` ``// Assign the sum to current node of resultant list`` ``result->data = sum;` ` ``return` `result;``}` `// This function is called after the``// smaller list is added to the bigger``// lists's sublist of same size. Once the``// right sublist is added, the carry``// must be added toe left side of larger``// list to get the final result.``void` `addCarryToRemaining(Node* head1, Node* cur, ``int``* carry,`` ``Node** result)``{`` ``int` `sum;` ` ``// If diff. number of nodes are not traversed, add carry`` ``if` `(head1 != cur) {`` ``addCarryToRemaining(head1->next, cur, carry,`` ``result);` ` ``sum = head1->data + *carry;`` ``*carry = sum / 10;`` ``sum %= 10;` ` ``// add this node to the front of the result`` ``push(result, sum);`` ``}``}` `// The main function that adds two linked lists``// represented by head1 and head2. The sum of``// two lists is stored in a list referred by result``void` `addList(Node* head1, Node* head2, Node** result)``{`` ``Node* cur;` ` ``// first list is empty`` ``if` `(head1 == NULL) {`` ``*result = head2;`` ``return``;`` ``}` ` ``// second list is empty`` ``else` `if` `(head2 == NULL) {`` ``*result = head1;`` ``return``;`` ``}` ` ``int` `size1 = getSize(head1);`` ``int` `size2 = getSize(head2);` ` ``int` `carry = 0;` ` ``// Add same size lists`` ``if` `(size1 == size2)`` ``*result = addSameSize(head1, head2, &carry);` ` ``else` `{`` ``int` `diff = ``abs``(size1 - size2);` ` ``// First list should always be larger than second`` ``// list. If not, swap pointers`` ``if` `(size1 < size2)`` ``swapPointer(&head1, &head2);` ` ``// move diff. number of nodes in first list`` ``for` `(cur = head1; diff--; cur = cur->next)`` ``;` ` ``// get addition of same size lists`` ``*result = addSameSize(cur, head2, &carry);` ` ``// get addition of remaining first list and carry`` ``addCarryToRemaining(head1, cur, &carry, result);`` ``}` ` ``// if some carry is still there, add a new node to the`` ``// front of the result list. e.g. 999 and 87`` ``if` `(carry)`` ``push(result, carry);``}` `// Driver code``int` `main()``{`` ``Node *head1 = NULL, *head2 = NULL, *result = NULL;` ` ``int` `arr1[] = { 9, 9, 9 };`` ``int` `arr2[] = { 1, 8 };` ` ``int` `size1 = ``sizeof``(arr1) / ``sizeof``(arr1[0]);`` ``int` `size2 = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` ` ``// Create first list as 9->9->9`` ``int` `i;`` ``for` `(i = size1 - 1; i >= 0; --i)`` ``push(&head1, arr1[i]);` ` ``// Create second list as 1->8`` ``for` `(i = size2 - 1; i >= 0; --i)`` ``push(&head2, arr2[i]);` ` ``addList(head1, head2, &result);` ` ``printList(result);` ` ``return` `0;``}` `// This code is contributed by rathbhupendra`
## C
`// A C recursive program to add two linked lists` `#include ``#include ` `// A linked List Node``struct` `Node {`` ``int` `data;`` ``struct` `Node* next;``};` `typedef` `struct` `Node node;` `/* A utility function to insert a`` ``node at the beginning of`` ``* linked list */``void` `push(``struct` `Node** head_ref, ``int` `new_data)``{`` ``/* allocate node */`` ``struct` `Node* new_node`` ``= (``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));` ` ``/* put in the data */`` ``new_node->data = new_data;` ` ``/* link the old list of the new node */`` ``new_node->next = (*head_ref);` ` ``/* move the head to point to the new node */`` ``(*head_ref) = new_node;``}` `/* A utility function to print linked list */``void` `printList(``struct` `Node* node)``{`` ``while` `(node != NULL) {`` ``printf``(``"%d "``, node->data);`` ``node = node->next;`` ``}`` ``printf``(``"n"``);``}` `// A utility function to swap two pointers``void` `swapPointer(Node** a, Node** b)``{`` ``node* t = *a;`` ``*a = *b;`` ``*b = t;``}` `/* A utility function to get size`` ``of linked list */``int` `getSize(``struct` `Node* node)``{`` ``int` `size = 0;`` ``while` `(node != NULL) {`` ``node = node->next;`` ``size++;`` ``}`` ``return` `size;``}` `// Adds two linked lists of same``// size represented by head1``// and head2 and returns head of``// the resultant linked list.``// Carry is propagated while``// returning from the recursion``node* addSameSize(Node* head1,`` ``Node* head2, ``int``* carry)``{`` ``// Since the function assumes`` ``// linked lists are of same`` ``// size, check any of the two`` ``// head pointers`` ``if` `(head1 == NULL)`` ``return` `NULL;` ` ``int` `sum;` ` ``// Allocate memory for sum`` ``// node of current two nodes`` ``Node* result = (Node*)``malloc``(``sizeof``(Node));` ` ``// Recursively add remaining nodes`` ``// and get the carry`` ``result->next`` ``= addSameSize(head1->next,`` ``head2->next, carry);` ` ``// add digits of current nodes`` ``// and propagated carry`` ``sum = head1->data + head2->data + *carry;`` ``*carry = sum / 10;`` ``sum = sum % 10;` ` ``// Assigne the sum to current`` ``// node of resultant list`` ``result->data = sum;` ` ``return` `result;``}` `// This function is called after``// the smaller list is added``// to the bigger lists's sublist``// of same size. Once the``// right sublist is added, the``// carry must be added toe left``// side of larger list to get``// the final result.``void` `addCarryToRemaining(Node* head1,`` ``Node* cur, ``int``* carry,`` ``Node** result)``{`` ``int` `sum;` ` ``// If diff. number of nodes are`` ``// not traversed, add carry`` ``if` `(head1 != cur) {`` ``addCarryToRemaining(head1->next,`` ``cur, carry,`` ``result);` ` ``sum = head1->data + *carry;`` ``*carry = sum / 10;`` ``sum %= 10;` ` ``// add this node to the front of the result`` ``push(result, sum);`` ``}``}` `// The main function that adds two``// linked lists represented``// by head1 and head2. The sum of``// two lists is stored in a``// list referred by result``void` `addList(Node* head1,`` ``Node* head2, Node** result)``{`` ``Node* cur;` ` ``// first list is empty`` ``if` `(head1 == NULL) {`` ``*result = head2;`` ``return``;`` ``}` ` ``// second list is empty`` ``else` `if` `(head2 == NULL)`` ``{`` ``*result = head1;`` ``return``;`` ``}` ` ``int` `size1 = getSize(head1);`` ``int` `size2 = getSize(head2);` ` ``int` `carry = 0;` ` ``// Add same size lists`` ``if` `(size1 == size2)`` ``*result = addSameSize(head1, head2, &carry);` ` ``else` `{`` ``int` `diff = ``abs``(size1 - size2);` ` ``// First list should always be`` ``// larger than second`` ``// list. If not, swap pointers`` ``if` `(size1 < size2)`` ``swapPointer(&head1, &head2);` ` ``// move diff. number of nodes in first list`` ``for` `(cur = head1; diff--; cur = cur->next)`` ``;` ` ``// get addition of same size lists`` ``*result = addSameSize(cur,`` ``head2, &carry);` ` ``// get addition of remaining first list and carry`` ``addCarryToRemaining(head1,`` ``cur, &carry, result);`` ``}` ` ``// if some carry is still there, add a new node to the`` ``// front of the result list. e.g. 999 and 87`` ``if` `(carry)`` ``push(result, carry);``}` `// Driver code``int` `main()``{`` ``Node *head1 = NULL, *head2 = NULL, *result = NULL;` ` ``int` `arr1[] = { 9, 9, 9 };`` ``int` `arr2[] = { 1, 8 };` ` ``int` `size1 = ``sizeof``(arr1) / ``sizeof``(arr1[0]);`` ``int` `size2 = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` ` ``// Create first list as 9->9->9`` ``int` `i;`` ``for` `(i = size1 - 1; i >= 0; --i)`` ``push(&head1, arr1[i]);` ` ``// Create second list as 1->8`` ``for` `(i = size2 - 1; i >= 0; --i)`` ``push(&head2, arr2[i]);` ` ``addList(head1, head2, &result);` ` ``printList(result);` ` ``return` `0;``}`
## Java
`// A Java recursive program to add two linked lists` `public` `class` `linkedlistATN``{`` ``class` `node`` ``{`` ``int` `val;`` ``node next;` ` ``public` `node(``int` `val)`` ``{`` ``this``.val = val;`` ``}`` ``}`` ` ` ``// Function to print linked list`` ``void` `printlist(node head)`` ``{`` ``while` `(head != ``null``)`` ``{`` ``System.out.print(head.val + ``" "``);`` ``head = head.next;`` ``}`` ``}` ` ``node head1, head2, result;`` ``int` `carry;` ` ``/* A utility function to push a value to linked list */`` ``void` `push(``int` `val, ``int` `list)`` ``{`` ``node newnode = ``new` `node(val);`` ``if` `(list == ``1``)`` ``{`` ``newnode.next = head1;`` ``head1 = newnode;`` ``}`` ``else` `if` `(list == ``2``)`` ``{`` ``newnode.next = head2;`` ``head2 = newnode;`` ``}`` ``else`` ``{`` ``newnode.next = result;`` ``result = newnode;`` ``}` ` ``}` ` ``// Adds two linked lists of same size represented by`` ``// head1 and head2 and returns head of the resultant`` ``// linked list. Carry is propagated while returning`` ``// from the recursion`` ``void` `addsamesize(node n, node m)`` ``{`` ``// Since the function assumes linked lists are of`` ``// same size, check any of the two head pointers`` ``if` `(n == ``null``)`` ``return``;` ` ``// Recursively add remaining nodes and get the carry`` ``addsamesize(n.next, m.next);` ` ``// add digits of current nodes and propagated carry`` ``int` `sum = n.val + m.val + carry;`` ``carry = sum / ``10``;`` ``sum = sum % ``10``;` ` ``// Push this to result list`` ``push(sum, ``3``);` ` ``}` ` ``node cur;` ` ``// This function is called after the smaller list is`` ``// added to the bigger lists's sublist of same size.`` ``// Once the right sublist is added, the carry must be`` ``// added to the left side of larger list to get the`` ``// final result.`` ``void` `propogatecarry(node head1)`` ``{`` ``// If diff. number of nodes are not traversed, add carry`` ``if` `(head1 != cur)`` ``{`` ``propogatecarry(head1.next);`` ``int` `sum = carry + head1.val;`` ``carry = sum / ``10``;`` ``sum %= ``10``;` ` ``// add this node to the front of the result`` ``push(sum, ``3``);`` ``}`` ``}` ` ``int` `getsize(node head)`` ``{`` ``int` `count = ``0``;`` ``while` `(head != ``null``)`` ``{`` ``count++;`` ``head = head.next;`` ``}`` ``return` `count;`` ``}` ` ``// The main function that adds two linked lists`` ``// represented by head1 and head2. The sum of two`` ``// lists is stored in a list referred by result`` ``void` `addlists()`` ``{`` ``// first list is empty`` ``if` `(head1 == ``null``)`` ``{`` ``result = head2;`` ``return``;`` ``}` ` ``// first list is empty`` ``if` `(head2 == ``null``)`` ``{`` ``result = head1;`` ``return``;`` ``}` ` ``int` `size1 = getsize(head1);`` ``int` `size2 = getsize(head2);` ` ``// Add same size lists`` ``if` `(size1 == size2)`` ``{`` ``addsamesize(head1, head2);`` ``}`` ``else`` ``{`` ``// First list should always be larger than second list.`` ``// If not, swap pointers`` ``if` `(size1 < size2)`` ``{`` ``node temp = head1;`` ``head1 = head2;`` ``head2 = temp;`` ``}`` ``int` `diff = Math.abs(size1 - size2);` ` ``// move diff. number of nodes in first list`` ``node temp = head1;`` ``while` `(diff-- >= ``0``)`` ``{`` ``cur = temp;`` ``temp = temp.next;`` ``}` ` ``// get addition of same size lists`` ``addsamesize(cur, head2);` ` ``// get addition of remaining first list and carry`` ``propogatecarry(head1);`` ``}`` ``// if some carry is still there, add a new node to`` ``// the front of the result list. e.g. 999 and 87`` ``if` `(carry > ``0``)`` ``push(carry, ``3``);`` ` ` ``}` ` ``// Driver program to test above functions`` ``public` `static` `void` `main(String args[])`` ``{`` ``linkedlistATN list = ``new` `linkedlistATN();`` ``list.head1 = ``null``;`` ``list.head2 = ``null``;`` ``list.result = ``null``;`` ``list.carry = ``0``;`` ``int` `arr1[] = { ``9``, ``9``, ``9` `};`` ``int` `arr2[] = { ``1``, ``8` `};` ` ``// Create first list as 9->9->9`` ``for` `(``int` `i = arr1.length - ``1``; i >= ``0``; --i)`` ``list.push(arr1[i], ``1``);` ` ``// Create second list as 1->8`` ``for` `(``int` `i = arr2.length - ``1``; i >= ``0``; --i)`` ``list.push(arr2[i], ``2``);` ` ``list.addlists();` ` ``list.printlist(list.result);`` ``}``}` `// This code is contributed by Rishabh Mahrsee`
## C#
`// A C# recursive program to add two linked lists``using` `System;`` ` `public` `class` `linkedlistATN{`` ` `class` `node``{`` ``public` `int` `val;`` ``public` `node next;` ` ``public` `node(``int` `val)`` ``{`` ``this``.val = val;`` ``}``}`` ` `// Function to print linked list``void` `printlist(node head)``{`` ``while` `(head != ``null``)`` ``{`` ``Console.Write(head.val + ``" "``);`` ``head = head.next;`` ``}``}` `node head1, head2, result;``int` `carry;` `// A utility function to push a``// value to linked list``void` `push(``int` `val, ``int` `list)``{`` ``node newnode = ``new` `node(val);`` ` ` ``if` `(list == 1)`` ``{`` ``newnode.next = head1;`` ``head1 = newnode;`` ``}`` ``else` `if` `(list == 2)`` ``{`` ``newnode.next = head2;`` ``head2 = newnode;`` ``}`` ``else`` ``{`` ``newnode.next = result;`` ``result = newnode;`` ``}` `}` `// Adds two linked lists of same size represented by``// head1 and head2 and returns head of the resultant``// linked list. Carry is propagated while returning``// from the recursion``void` `addsamesize(node n, node m)``{`` ` ` ``// Since the function assumes linked`` ``// lists are of same size, check any`` ``// of the two head pointers`` ``if` `(n == ``null``)`` ``return``;` ` ``// Recursively add remaining nodes`` ``// and get the carry`` ``addsamesize(n.next, m.next);` ` ``// Add digits of current nodes`` ``// and propagated carry`` ``int` `sum = n.val + m.val + carry;`` ``carry = sum / 10;`` ``sum = sum % 10;` ` ``// Push this to result list`` ``push(sum, 3);``}` `node cur;` `// This function is called after the smaller``// list is added to the bigger lists's sublist``// of same size. Once the right sublist is added,``// the carry must be added to the left side of``// larger list to get the final result.``void` `propogatecarry(node head1)``{`` ` ` ``// If diff. number of nodes are`` ``// not traversed, add carry`` ``if` `(head1 != cur)`` ``{`` ``propogatecarry(head1.next);`` ``int` `sum = carry + head1.val;`` ``carry = sum / 10;`` ``sum %= 10;` ` ``// Add this node to the front`` ``// of the result`` ``push(sum, 3);`` ``}``}` `int` `getsize(node head)``{`` ``int` `count = 0;`` ``while` `(head != ``null``)`` ``{`` ``count++;`` ``head = head.next;`` ``}`` ``return` `count;``}` `// The main function that adds two linked``// lists represented by head1 and head2.``// The sum of two lists is stored in a``// list referred by result``void` `addlists()``{`` ` ` ``// First list is empty`` ``if` `(head1 == ``null``)`` ``{`` ``result = head2;`` ``return``;`` ``}` ` ``// Second list is empty`` ``if` `(head2 == ``null``)`` ``{`` ``result = head1;`` ``return``;`` ``}` ` ``int` `size1 = getsize(head1);`` ``int` `size2 = getsize(head2);` ` ``// Add same size lists`` ``if` `(size1 == size2)`` ``{`` ``addsamesize(head1, head2);`` ``}`` ``else`` ``{`` ` ` ``// First list should always be`` ``// larger than second list.`` ``// If not, swap pointers`` ``if` `(size1 < size2)`` ``{`` ``node temp = head1;`` ``head1 = head2;`` ``head2 = temp;`` ``}`` ` ` ``int` `diff = Math.Abs(size1 - size2);` ` ``// Move diff. number of nodes in`` ``// first list`` ``node tmp = head1;`` ` ` ``while` `(diff-- >= 0)`` ``{`` ``cur = tmp;`` ``tmp = tmp.next;`` ``}` ` ``// Get addition of same size lists`` ``addsamesize(cur, head2);` ` ``// Get addition of remaining`` ``// first list and carry`` ``propogatecarry(head1);`` ``}`` ``// If some carry is still there,`` ``// add a new node to the front of`` ``// the result list. e.g. 999 and 87`` ``if` `(carry > 0)`` ``push(carry, 3);``}` `// Driver code``public` `static` `void` `Main(``string` `[]args)``{`` ``linkedlistATN list = ``new` `linkedlistATN();`` ``list.head1 = ``null``;`` ``list.head2 = ``null``;`` ``list.result = ``null``;`` ``list.carry = 0;`` ` ` ``int` `[]arr1 = { 9, 9, 9 };`` ``int` `[]arr2 = { 1, 8 };` ` ``// Create first list as 9->9->9`` ``for``(``int` `i = arr1.Length - 1; i >= 0; --i)`` ``list.push(arr1[i], 1);` ` ``// Create second list as 1->8`` ``for``(``int` `i = arr2.Length - 1; i >= 0; --i)`` ``list.push(arr2[i], 2);` ` ``list.addlists();` ` ``list.printlist(list.result);``}``}` `// This code is contributed by rutvik_56`
## Javascript
``
Output
`1 0 1 7`
Time Complexity: O(m+n) where m and n are the sizes of given two linked lists.
Auxiliary Space: O(m+n) for call stack
Iterative Approach:
This implementation does not have any recursion call overhead, which means it is an iterative solution.
Since we need to start adding numbers from the last of the two linked lists. So, here we will use the stack data structure to implement this.
• We will firstly make two stacks from the given two linked lists.
• Then, we will run a loop till both stack become empty.
• in every iteration, we keep the track of the carry.
• In the end, if carry>0, that means we need extra node at the start of the resultant list to accommodate this carry.
Below is the implementation of the above approach.
## C++
`// C++ Iterative program to add two linked lists ``#include ``using` `namespace` `std;`` ` `// A linked List Node ``class` `Node ``{ `` ``public``:`` ``int` `data; `` ``Node* next; ``};` `// to push a new node to linked list``void` `push(Node** head_ref, ``int` `new_data) ``{ `` ``/* allocate node */`` ``Node* new_node = ``new` `Node[(``sizeof``(Node))]; `` ` ` ``/* put in the data */`` ``new_node->data = new_data; `` ` ` ``/* link the old list of the new node */`` ``new_node->next = (*head_ref); `` ` ` ``/* move the head to point to the new node */`` ``(*head_ref) = new_node; ``}` `// to add two new numbers``Node* addTwoNumList(Node* l1, Node* l2) {`` ``stack<``int``> s1,s2;`` ``while``(l1!=NULL){`` ``s1.push(l1->data);`` ``l1=l1->next;`` ``}`` ``while``(l2!=NULL){`` ``s2.push(l2->data);`` ``l2=l2->next;`` ``}`` ``int` `carry=0;`` ``Node* result=NULL;`` ``while``(s1.empty()==``false` `|| s2.empty()==``false``){`` ``int` `a=0,b=0;`` ``if``(s1.empty()==``false``){`` ``a=s1.top();s1.pop();`` ``}`` ``if``(s2.empty()==``false``){`` ``b=s2.top();s2.pop();`` ``}`` ``int` `total=a+b+carry;`` ``Node* temp=``new` `Node();`` ``temp->data=total%10;`` ``carry=total/10;`` ``if``(result==NULL){`` ``result=temp;`` ``}``else``{`` ``temp->next=result;`` ``result=temp;`` ``}`` ``}`` ``if``(carry!=0){`` ``Node* temp=``new` `Node();`` ``temp->data=carry;`` ``temp->next=result;`` ``result=temp;`` ``}`` ``return` `result;``}` `// to print a linked list``void` `printList(Node *node) ``{ `` ``while` `(node != NULL) `` ``{ `` ``cout<data<<``" "``; `` ``node = node->next; `` ``} `` ``cout<6->7 `` ``int` `i; `` ``for` `(i = size1-1; i >= 0; --i) `` ``push(&head1, arr1[i]); `` ` ` ``// Create second list as 1->8 `` ``for` `(i = size2-1; i >= 0; --i) `` ``push(&head2, arr2[i]); `` ` ` ``Node* result=addTwoNumList(head1, head2);`` ``printList(result); `` ` ` ``return` `0; ``}`
## Java
`// Java Iterative program to add``// two linked lists ``import` `java.io.*;``import` `java.util.*;` `class` `GFG{`` ` `static` `class` `Node``{`` ``int` `data;`` ``Node next;`` ` ` ``public` `Node(``int` `data)`` ``{`` ``this``.data = data;`` ``}``}` `static` `Node l1, l2, result;` `// To push a new node to linked list``public` `static` `void` `push(``int` `new_data)``{`` ` ` ``// Allocate node`` ``Node new_node = ``new` `Node(``0``);` ` ``// Put in the data`` ``new_node.data = new_data;` ` ``// Link the old list of the new node`` ``new_node.next = l1;` ` ``// Move the head to point to the new node`` ``l1 = new_node;``}` `public` `static` `void` `push1(``int` `new_data)``{`` ` ` ``// Allocate node`` ``Node new_node = ``new` `Node(``0``);` ` ``// Put in the data`` ``new_node.data = new_data;` ` ``// Link the old list of the new node`` ``new_node.next = l2;` ` ``// Move the head to point to`` ``// the new node`` ``l2 = new_node;``}` `// To add two new numbers``public` `static` `Node addTwoNumbers()``{`` ``Stack stack1 = ``new` `Stack<>();`` ``Stack stack2 = ``new` `Stack<>();` ` ``while` `(l1 != ``null``)`` ``{`` ``stack1.add(l1.data);`` ``l1 = l1.next;`` ``}` ` ``while` `(l2 != ``null``)`` ``{`` ``stack2.add(l2.data);`` ``l2 = l2.next;`` ``}` ` ``int` `carry = ``0``;`` ``Node result = ``null``;` ` ``while` `(!stack1.isEmpty() ||`` ``!stack2.isEmpty())`` ``{`` ``int` `a = ``0``, b = ``0``;` ` ``if` `(!stack1.isEmpty())`` ``{`` ``a = stack1.pop();`` ``}` ` ``if` `(!stack2.isEmpty())`` ``{`` ``b = stack2.pop();`` ``}` ` ``int` `total = a + b + carry;` ` ``Node temp = ``new` `Node(total % ``10``);`` ``carry = total / ``10``;` ` ``if` `(result == ``null``)`` ``{`` ``result = temp;`` ``}`` ``else`` ``{`` ``temp.next = result;`` ``result = temp;`` ``}`` ``}` ` ``if` `(carry != ``0``)`` ``{`` ``Node temp = ``new` `Node(carry);`` ``temp.next = result;`` ``result = temp;`` ``}`` ``return` `result;``}` `// To print a linked list``public` `static` `void` `printList()``{`` ``while` `(result != ``null``)`` ``{`` ``System.out.print(result.data + ``" "``);`` ``result = result.next;`` ``}`` ``System.out.println();``}` `// Driver code``public` `static` `void` `main(String[] args)``{`` ``int` `arr1[] = { ``5``, ``6``, ``7` `};`` ``int` `arr2[] = { ``1``, ``8` `};` ` ``int` `size1 = ``3``;`` ``int` `size2 = ``2``;` ` ``// Create first list as 5->6->7`` ``int` `i;`` ``for``(i = size1 - ``1``; i >= ``0``; --i)`` ``push(arr1[i]);` ` ``// Create second list as 1->8`` ``for``(i = size2 - ``1``; i >= ``0``; --i)`` ``push1(arr2[i]);` ` ``result = addTwoNumbers();` ` ``printList();``}``}` `// This code is contributed by RohitOberoi`
## Python3
`# Python Iterative program to add``# two linked lists ``class` `Node:`` ``def` `__init__(``self``,val):`` ``self``.data ``=` `val`` ``self``.``next` `=` `None`` ` `l1, l2, result ``=` `None``,``None``,``0` `# To push a new node to linked list``def` `push(new_data):` ` ``global` `l1` ` ``# Allocate node`` ``new_node ``=` `Node(``0``)` ` ``# Put in the data`` ``new_node.data ``=` `new_data` ` ``# Link the old list of the new node`` ``new_node.``next` `=` `l1` ` ``# Move the head to point to the new node`` ``l1 ``=` `new_node` `def` `push1(new_data):` ` ``global` `l2` ` ``# Allocate node`` ``new_node ``=` `Node(``0``)` ` ``# Put in the data`` ``new_node.data ``=` `new_data` ` ``# Link the old list of the new node`` ``new_node.``next` `=` `l2` ` ``# Move the head to point to`` ``# the new node`` ``l2 ``=` `new_node` `# To add two new numbers``def` `addTwoNumbers():` ` ``global` `l1,l2,result` ` ``stack1 ``=` `[]`` ``stack2 ``=` `[]` ` ``while` `(l1 !``=` `None``):`` ``stack1.append(l1.data)`` ``l1 ``=` `l1.``next` ` ``while` `(l2 !``=` `None``):`` ``stack2.append(l2.data)`` ``l2 ``=` `l2.``next` ` ``carry ``=` `0`` ``result ``=` `None` ` ``while` `(``len``(stack1) !``=` `0` `or` `len``(stack2) !``=` `0``):`` ``a,b ``=` `0``,``0` ` ``if` `(``len``(stack1) !``=` `0``):`` ``a ``=` `stack1.pop()` ` ``if` `(``len``(stack2) !``=` `0``):`` ``b ``=` `stack2.pop()` ` ``total ``=` `a ``+` `b ``+` `carry` ` ``temp ``=` `Node(total ``%` `10``)`` ``carry ``=` `total ``/``/` `10` ` ``if` `(result ``=``=` `None``):`` ``result ``=` `temp`` ``else``:`` ``temp.``next` `=` `result`` ``result ``=` `temp` ` ``if` `(carry !``=` `0``):`` ``temp ``=` `Node(carry)`` ``temp.``next` `=` `result`` ``result ``=` `temp`` ` ` ``return` `result` `# To print a linked list``def` `printList():` ` ``global` `result` ` ``while` `(result !``=` `None``):`` ``print``(result.data ,end ``=` `" "``)`` ``result ``=` `result.``next` `# Driver code`` ` `arr1 ``=` `[ ``5``, ``6``, ``7` `]``arr2 ``=` `[ ``1``, ``8` `]` `size1 ``=` `3``size2 ``=` `2` `# Create first list as 5->6->7` `for` `i ``in` `range``(size1``-``1``,``-``1``,``-``1``):`` ``push(arr1[i])` `# Create second list as 1->8``for` `i ``in` `range``(size2``-``1``,``-``1``,``-``1``):`` ``push1(arr2[i])` `result ``=` `addTwoNumbers()` `printList()` `# This code is contributed by shinjanpatra`
## C#
`// C# Iterative program to add``// two linked lists ``using` `System;``using` `System.Collections;` `class` `GFG{` ` ``public` `class` `Node`` ``{`` ``public` `int` `data;`` ``public` `Node next;` ` ``public` `Node(``int` `data)`` ``{`` ``this``.data = data;`` ``}`` ``}` ` ``static` `Node l1, l2, result;` ` ``// To push a new node to linked list`` ``public` `static` `void` `push(``int` `new_data)`` ``{` ` ``// Allocate node`` ``Node new_node = ``new` `Node(0);` ` ``// Put in the data`` ``new_node.data = new_data;` ` ``// Link the old list of the new node`` ``new_node.next = l1;` ` ``// Move the head to point to the new node`` ``l1 = new_node;`` ``}` ` ``public` `static` `void` `push1(``int` `new_data)`` ``{` ` ``// Allocate node`` ``Node new_node = ``new` `Node(0);` ` ``// Put in the data`` ``new_node.data = new_data;` ` ``// Link the old list of the new node`` ``new_node.next = l2;` ` ``// Move the head to point to`` ``// the new node`` ``l2 = new_node;`` ``}` ` ``// To add two new numbers`` ``public` `static` `Node addTwoNumbers()`` ``{`` ``Stack stack1 = ``new` `Stack();`` ``Stack stack2 = ``new` `Stack();` ` ``while` `(l1 != ``null``)`` ``{`` ``stack1.Push(l1.data);`` ``l1 = l1.next;`` ``}`` ``while` `(l2 != ``null``)`` ``{`` ``stack2.Push(l2.data);`` ``l2 = l2.next;`` ``}` ` ``int` `carry = 0;`` ``Node result = ``null``;`` ``while` `(stack1.Count != 0 ||`` ``stack2.Count != 0)`` ``{`` ``int` `a = 0, b = 0;` ` ``if` `(stack1.Count != 0)`` ``{`` ``a = (``int``)stack1.Pop();`` ``}` ` ``if` `(stack2.Count != 0)`` ``{`` ``b = (``int``)stack2.Pop();`` ``}` ` ``int` `total = a + b + carry;`` ``Node temp = ``new` `Node(total % 10);`` ``carry = total / 10;` ` ``if` `(result == ``null``)`` ``{`` ``result = temp;`` ``}`` ``else`` ``{`` ``temp.next = result;`` ``result = temp;`` ``}`` ``}` ` ``if` `(carry != 0)`` ``{`` ``Node temp = ``new` `Node(carry);`` ``temp.next = result;`` ``result = temp;`` ``}`` ``return` `result;`` ``}` ` ``// To print a linked list`` ``public` `static` `void` `printList()`` ``{`` ``while` `(result != ``null``)`` ``{`` ``Console.Write(result.data + ``" "``);`` ``result = result.next;`` ``}`` ``Console.WriteLine();`` ``}` ` ``// Driver code`` ``public` `static` `void` `Main(``string``[] args)`` ``{`` ``int` `[]arr1 = { 5, 6, 7 };`` ``int` `[]arr2 = { 1, 8 };`` ``int` `size1 = 3;`` ``int` `size2 = 2;` ` ``// Create first list as 5->6->7`` ``int` `i;`` ``for``(i = size1 - 1; i >= 0; --i)`` ``push(arr1[i]);` ` ``// Create second list as 1->8`` ``for``(i = size2 - 1; i >= 0; --i)`` ``push1(arr2[i]);`` ``result = addTwoNumbers();`` ``printList();`` ``}``}` `// This code is contributed by pratham76`
## Javascript
``
Output
`5 8 5`
Time Complexity: O(m+n) where m and n are the sizes of given two linked lists.
Auxiliary Space: O(m+n)
Related Article: Add two numbers represented by linked lists | Set 1
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googolplex zeros
# googolplex zeros
Googolplex 10googol Googolplex would have a Googol of zeros after a 1, which is to say the number of digits in a Googolplex Googol1 digits. A typical book can be printed with 106 zeros (around 400 pages with 50 lines per page and 50 zeros per line). Therefore, it requires 1094 such books to print all the zeros of a googolplex (that is How Big is a Googolplex? Meet a number so big you couldnt write the number of zeros it has on a sheet of paper a thousand miles long, if each zero was the size of an atom! The much larger number googolplex has been defined as 1 followed by a googol zeros. While this number can easily be written as googolplex 10googol 10(10100) using the exponential notation. Late Carl Sagan explains the Googol and Googolplex numbers in his COSMOS series.Meet a number so big you couldnt write the number of zeros it has on a sheet of paper a thousand miles long Im Googolplex 0 I know Googolplex 0.Nicknames for Googolplex. Add your names, share with friends. Click to copy. The much larger number googolplex has been defined as 1 followed by a googol zeros. While this number can easily be written as. A typical book can be printed with 106 zeros (around 400 pages with 50 lines per page and 50 zeros per line). Therefore, it requires 1094 such books to print all the zeros of a googolplex (that is googolplex how many zeros. Numberphile. Googles name was actually inspired by a googol, a number so large, it has 100 zeros. Continuing the Googolplex.the uncle of the child who invented the googol, suggested that a much larger number will be called googolplex , and that it be defined as 1 followed by a googol of zeros. Milton even had a definition for a googolplex saying that it should be "a one followed by writing zeros until you got tired and had no more room to write, as the zeros went all the way to the furthest star". googolplex zeros googolplex game googolplex.cuna.org googolplex number googleplex app. googol -plex. Like the word googol, googolplex was coined in 1920 by the nine-year-old Milton Sirrota (19111981), the nephew of American mathematician Edward Kasner (18781955).
The word was first published and precisely defined in the book Mathematics and the Imagination (1940) How many zeros are in a googolplex? asked May 3, 2012 by triuser (13,110 points). A typical book can be printed with 106 zeros around 400 pages with 50 lines per page and 50 zeros per line Therefore, it requires 1094 such books to print all the zeros of a googolplex that is Image - Googolplex written out.png | Googology Wiki 1396 x 770 png 8kB. www.keywordsking.com.
Googolplex Written Related Keywords - Googolplex Written Googolplexian is 10 with googolplex zeros.How many zeros in a googleplex? Beyond that is the googolplex, which is a one followed by a googol zeros. Googolplex: The worlds second largest number with a name. A "1" followed by a googol of zeros. (See also googolplexian.com - the largest of them all ).
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# ZERO is like WHITE colour and INFINITY is like BLACK colour.
Discussion in 'Physics & Math' started by plakhapate, Jan 27, 2006.
1. ### plakhapateBannedBanned
Messages:
249
All finite numbers are sandwitched between zero and infinity.
What about zero ? Is it a finite number ?
Let us analyse.
Zero is not well defined.
If we define zero as reciprocal of infinity, then since infinity has no finite value, zero can not have one finite value it will have several values.
That is the reason, like infinity ,Zero does not obey the rules of finite numbers.
Thus Zero is like white colour which is mixture of infinite monochromatic colours.
Similarly Infinity (absence of finite numbers) is like black colour which is nothing but absence of colours.
This concludes that ZERO is not a finite no.
Pls comment.
P.J.LAKHAPATE
plakhapate@rediffmail.com
3. ### Mosheh ThezionRegistered Senior Member
Messages:
2,650
the question leads to the question.... to what decimal point do we count zero?
and what are the units in question?
-MT
5. ### DaleSpamTANSTAAFLRegistered Senior Member
Messages:
1,723
Hi plakhapate,
0 is well defined: x-x=0 for any real number.
-Dale
7. ### dzerzhinskyCommunistRegistered Senior Member
Messages:
105
What about negative numbers then? Where do they come in?
8. ### a_htRegistered Senior Member
Messages:
158
negative numbers are inside the set of real numbers, so what DaleSpam said applies.
9. ### a_htRegistered Senior Member
Messages:
158
and what do colors (from title) have to do with OT content?
10. ### DinosaurRational SkepticValued Senior Member
Messages:
4,618
Plakhapate: Cute joke!! Since white is the combination of all colors (or at least of many colors) and black is the absence of color, it is obvious that infinity corresponds to white and zero corresponds to black, assuming that infinity & zero can be represented by colors at all. Your reversed match-up is an obvious joke.
You left another clue for those who did not get it right away.
• Everybody knows that zero is more easily defined than infinity. Yet you define zero as the reciprocal of infinity. Even cuter that the reversed match-up of white/black corresponding to zero/infinity.
Apparently nobody else realizes that you are being humorous.
11. ### James RJust this guy, you know?Staff Member
Messages:
30,644
Zero can be defined like this:
1. For any number x: 0 + x = x + 0 = x.
and/or
2. For any finite number x: 0x = x0 = 0.
Zero is what is known as the "additive identity", in that it leaves a number unchanged when added to it.
12. ### qwerty mobDeicidalRegistered Senior Member
Messages:
786
Couple of thoughts for you...
Numbers aren't "like" colors at all; numbers are actually abstract and purely imaginary, colors are wavelengths of luminous radiation and objectively real.
Zero is well defined, multiple ways (as James and Dale and others can point out); remember, sets are the basis of all computation. Infinity is also well defined, and no actual ones exist.
...
If you're going to make a case for zero not being a valid value, you'll have to do so with axioms which do not violate existing laws of mathematics, or equivocate either the quantity of an 'infinity' with the quality of (being) infinite, or transpose the properties of an empty set onto a nonexistent (or invalid) set.
Good luck =)
13. ### darcy emesdarcyRegistered Senior Member
Messages:
32
if adding colours ie lighting then white is all however when subtracting wavelegnths ie dye's or paint then white is zero and black is all colours of paint mixed.
14. ### DinosaurRational SkepticValued Senior Member
Messages:
4,618
Darcy Emes: You are correct. However, it is my understanding that subtractive colors are assumed in the context of an artist using a palette to create an oil or water color painting, while in the context of physics, additive
colors are assumed.
As far as I know, only artists work with subtractive colors. Photographers work with subtractive colors.
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The vector, or cross product
When we took two vectors previously and found a way to multiply them together using the dot product, we ended up with a scalar. However, there is also a way that we can take two vectors and multiply them together to give a vector, but a vector with very specific properties with respect to the first two. What we will define here will be in three dimensions, and, unlike the dot product, does not generalise easily to other dimensions, (other than 7) though it can in fact be extended.
We are going to define the cross product such that it gives a vector which is perpendicular to the two vectors being crossed. This might sound a bit arbitrary but it shows up in a huge number of different situations in physics in particular and can help us to understand the geometric relation between vectors very simply. Clearly if two vectors are parallel to one another in two dimensions, there won’t be a unique direction which is perpendicular to both vectors. In fact there will be a whole plane of directions perpendicular to the two vectors. We are going to define the cross product such that its magnitude is proportional to the magnitude of the two vectors being crossed, and the $\sin$ of the angle in between them. The direction of the resulting vector will be perpendicular to the two original vectors:
$\vec{a}\times\vec{b}=(\left|\vec{a}\right|\left|\vec{b}\right|\sin\theta)\hat{n}$
where $\hat{n}$ is in the direction perpendicular to $\vec{a}$ and $\vec{b}$. However, this is not uniquely defined. We can think of two vectors (pointing in opposite directions) which are both perpendicular to two other vectors and are unit magnitude. This is shown in the figure below.
The two black arrows are both perpendicular to the red and blue vectors and have magnitude equal to the product of the magnitude of the two vectors times the $\sin$ of the angle in between them. One of these is the cross product. The right hand rule tells us which one.
To disambiguate which of the two perpendicular vectors is the result of the cross product we use the right hand rule. Rather than trying to explain this in words, or in still pictures, I’ll simply point you to a video where this is explained clearly.
In order to understand how to take the cross product, it is informative to look at the cross product of the unit basis vectors. Clearly because their magnitudes are all one and the angle in between them are $\frac{\pi}{2}$, their cross products are going to have unit magnitude, the question is what direction will they be.
For instance, if we want to calculate $\vec{i}\times\vec{j}$ we know that there are two vectors perpendicular to this and of magnitude one. These are $\vec{k}$ and $-\vec{k}$. However, according to the right hand rule we get:
$\vec{i}\times\vec{j}=\vec{k}$
Once you have this, all of the other cross products are the cyclic permutations of this expression. That is, you simply shift each vector to the next position in the expression and take the one at the end and put it at the beginning. Shifting each one of these along one, gets you:
$\vec{k}\times\vec{i}=\vec{j}$
and shifting it along one more gets:
$\vec{j}\times\vec{k}=\vec{i}$
Shifting it along again takes us to the first expression. Swapping around the order of the terms in the cross product changes the result by a sign:
$\vec{j}\times\vec{i}=-\vec{k}$
And the same for the other two above. If we take the cross product of a vector with itself then clearly the angle in between the two vectors is zero and so you are left with the zero vector:
$\vec{i}\times \vec{i}=\vec{0}$
and similarly for $\vec{j}$ and $\vec{k}$ crossed with themselves. The above facts, along with the distributive property of the cross product, will allow us to calculate the cross product of any vectors when given in component form. We can take a product of two vectors: $\vec{a}=a_1\vec{i}+a_2\vec{j}+a_3\vec{k}$ and $\vec{b}=b_1\vec{i}+b_2\vec{j}+b_3\vec{k}$ for arbitrary components $a_i$ and $b_i$ and find the cross product of the two by expanding everything out:
$\vec{a}\times\vec{b}=(a_1\vec{i}+a_2\vec{j}+a_3\vec{k})\times(b_1\vec{i}+b_2\vec{j}+b_3\vec{k})$
$\vec{a}\times\vec{b}=a_1\vec{i}\times b_1\vec{i}+a_1\vec{i}\times b_2\vec{j}+a_1\vec{i}\times b_3\vec{k}+a_2\vec{j}\times b_1\vec{i}...etc.$
$\vec{a}\times\vec{b}=a_1 b_1\vec{0}+a_1 b_2\vec{k}-a_1 b_3\vec{j}-a_2 b_1\vec{k}...etc.$
$\vec{a}\times\vec{b}=\vec{i}\left(a_2 b_3-a_3 b_2\right)+\vec{j}\left(a_3b_1-a_1b_3\right)+\vec{k}\left(a_1b_2-a_2b_1\right)$
Make sure that you can follow the above and fill in the expressions for “etc.”. This looks like a terrible mess but in fact we can come up with a new form of notation that will allow us to remember how to perform this calculation with relative simplicity. First we need to define a determinant of an array.
How clear is this post?
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A diffraction grating consists of many narrow, parallel slits equally spaced. A grating with a groove period $$b$$ having $$n$$ slits in total is illuminated with light of wavelength $$\lambda$$. A monochromatic light with wavelength of 500 nm (1 nm = 10-9 m) strikes a grating and produces the second-order bright line at an 30° angle. For example, gases have interesting spectra which can be resolved with diffraction gratings. The allowed angles are calculated using the famous grating equation. Light of a different frequency may also reflect off of the same diffraction grating, but with a different final point. Obviously, d = $$\frac {1} { N }$$, where N is the grating constant, and it is the number of lines per unit length. [14] Gratings as dispersive elements. However, apex angles up to 110° may be present especially in blazed holographic gratings. Referring to Figure 2, there will be three diffracted orders (m= –2, –1, and +1) along with the specular reflection (m= 0). This is the distance betweentwo adjacent slits that can then be used in the equation $latex d sin \theta = n \lambda$. 0000001886 00000 n 5 0 obj A diffraction grating can be manufactured by carving glass with a sharp tool in a large number of precisely positioned parallel lines, with untouched regions acting like slits ((Figure)). xref The selection of the peak angle of the triangular groove offers opportunity to optimise the overall efficiency profile of the grating. This type of grating can be photographically mass produced rather cheaply. 3.00 x 10 8 =(5.60 x 10-7) (f) f = 5.36 x 10 14 Hz . Resolvance or "chromatic resolving power" for a device used to separate the wavelengths of light is defined as . The diffraction grating will thus disperse the light incident upon it into its component wavelengths, as shown in figure 89. A prime example is an optical element called a diffraction grating. 6 One example of a diffraction grating would be a periodic 0000003112 00000 n The structure affects the amplitude and/or phase of the incident wave, causing interference in the output wave. What diffraction order did I diffracting into what harmonic of this grating did I factor diffract off of? A prime example is an optical element called a diffraction grating. Red laser beam split by a diffraction grating. {\displaystyle \theta _ {m}=\arcsin \!\left (\sin \theta _ {i}- {\frac {m\lambda } {d}}\right)\!.} The Grating Equation: generalized m > 0 θ m > 0 y Phase matching,, sin sin kkmG ym yi kkmGθθ =−+ =+ sin sin 22 2 sin sin mi kk mG im m θθ π ππ θθ =− + += ⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟+= a m=0 ()sin sin im im a am λ λ θθ λ ⎝⎠ ⎝⎠ ⎝⎠ ⇒+ = m < 0 θ m < 0 The grating equation can be easily generalized for the case that the incident light is not at normal incidence, Δ=Δ 1 +Δ 2 =asinθi+asinθm=mλ a()sinθ i +sinθ In spectroscopic devices, such as monochromators, reflection gratings play key roles. 0000004041 00000 n Allowed Not Allowed Allowed Diffraction from Gratings Slide 10 The field is no longer a pure plane wave. x��][s7r~篘�s�"��eS~��\�ڭT�����(:�(Q���M*�$?0�\��9shg�r�x4����?M� 9��'���?��;7�|>�4���w��f��ۄ�~���ٿ�-�o�>�y��?����~��k�"Laz��\�7|�df��nX�ɲ��F����gr����^1��ny����R�8�v��shꌔ����9��� �Θ����iƝ�=�5s���(��|���q>����k���F�I#t�2š������� �ǿ���!\�8��υb��뺼����uP�5��w��ߟǂQf���0�w� <<0F1E17492D745E44B999A6AFCAE75322>]>> These rays are then di↵racted at an angle r. The most striking examples of diffraction are those that involve light; for example, the closely spaced tracks on a CD or DVD act as a diffraction grating to form the familiar rainbow pattern seen when looking at a disc. Thus, diffraction gratings can be used to characterize the spectra of various things. 0 The split light will have maxima at angle θ. BACKGROUND A diffraction grating is made by making many parallel scratches on the surface of a flat piece of transparent material. Solving for the irradiance as a function wavelength and position of this multi-slit situation, we get a general expression that can be applied to all diffractive … Also, d is the distance between slits. 0000001418 00000 n In the transmissive case, the repetitive structure can be thought of as many tightly spaced, thin slits. As an example, suppose a HeNe laser beam at 633 nm is incident on an 850 lines/mm grating. When solved for the diffracted angle maxima, the equation is: θ m = arcsin ( sin θ i − m λ d ) . Class 12; Class 11; Class 10; Class 9; Class 8; Class 7; Class 6; Previous Year Papers. 1. We'll define the term, explore the equation and look at some examples of diffraction. Other applications include acousto-optic modulators or scanners. x�bfjdg�eb@ !6�IM��,�Z|��Z0o=����wa������w�Fl-�7{ˋ�/͓l�d����T1@N�q���nm��Y������,"$�� ,#*./ɧ$&���/��-' ��#� ���fVU����T���1���k���h1�[�[ji��q���T�t1[Y����9����:����]�\�=|�} ���9�8sPH0D!XEP07�9:6.>!1)9�7��H����WTVU�����qs�� A screen is positioned parallel with the grating at a discance $$L$$. The grating “chops” the wave front and sends the power into multiple discrete directions that are called diffraction orders. Question 1: A diffraction grating is of width 5 cm and produces a deviation of 30 0 in the second-order with the light of wavelength 580 nm. Please note that these equations assume that both sides of the grating are in … 0000001808 00000 n Transmission diffraction gratings consist of many thin lines of either absorptive material or thin grooves on an otherwise transparent substrate. This type of grating can be photographically mass produced rather cheaply. Diffraction from sharp edges and apertures causes light to propagate along directions other than those predicted by the grating equation. %%EOF 780 0 obj <>stream Note: The Young’s slit experiment uses the letter for the slit separation, whereas frequently diffraction gratings use the letter for two adjacent slit separations. Diffraction gratings, either transmissive or reflective, can separate different wavelengths of light using a repetitive structure embedded within the grating. 0000001771 00000 n The effects of diffraction are often seen in everyday life. 0000004493 00000 n Find the slit spacing. <> Consider the cylindrical Huygens’ wavelet produced at each narrow slit when the grating is illuminated by a normally incident plane wave as shown in Fig. The wavelength dependence in the grating equation shows that the grating separates an incident polychromatic beam into its constituent wavelength components, i.e., it is dispersive. 0000001644 00000 n �o��U�.0f �&LY���� c�f�����Ɍ/X�00,tre�dlcP s�d�d���NtPb+ U��Ҁ Ȫ0D0lL:���� ���ˠ�.�S�)�A� �r�7p@f6�>)��\��d�;��� @n�:>���K�3���r�� �O�������Pj"G� ��� This article is about diffraction, an important wave phenomenon that produces predictable, measurable effects. λ = wave length of illumination. A reflection grating can be made by cutting parallel lines on the surface of refractive material. 0000004270 00000 n In this formula, $$\theta$$ is the angle of emergence at which a wavelength will be bright. 0000000016 00000 n EQUIPMENT Spectrometer, diffraction grating, mercury light source, high-voltage power supply. This is known as the DIFFRACTION GRATING EQUATION. For example, a grating ruled with 5000 lines/cm has a slit spacing d=1/5000 cm=2.00×10-4 cm. A parallel bundle of rays falls perpendicular to the grating %PDF-1.4 %���� This would be a binary amplitude grating (completely opaque or completely transparent). Single-order diffraction for such a period occurs at the Littrow angle of θ �~G�j�Ư���hA���ﶇeo���-. Diffraction grating. Diffraction gratings are thus widely used as dispersive elements in spectrographic instruments, 2 5 although they can also be used as beam splitters or beam combiners in various laser devices or interferometers. The diffraction grating is an optical component that splits light into various beams that travels in various direction. Spectra of hydrogen, helium, mercury and uranium as viewed through a diffraction grating. Light transmission through a diffraction grating occurs along discrete directions, called diffraction orders. c=f λ. Determine the number of slits per centimeter. 0000003547 00000 n The diffraction grating was named by Fraunhofer in 1821, but was in use before 1800. So for example, light with a wavelength exactly equal to the period of a grating (λ/Λ = 1) experiences Littrow diffraction at θ = 30º. 768 0 obj <> endobj In 1956, Bell presented a grating method for the dynamic strain measurement, and since then a variety of strain measurement methods, with grating as the … A plane wave is an incident from the left, normal to the … Di↵raction Grating Equation with Example Problems1 1 Grating Equation In Figure 1, parallel rays of monochromatic radiation, from a single beam in the form of rays 1 and 2, are incident on a (blazed) di↵raction grating at an angle i relative to the grating normal. 0000000556 00000 n A section of a diffraction grating is illustrated in the figure. How many photon momentum did I create or destroy? For a given wavelength the largest possible period for which only a single diffracted order exists is exactly 1½ wavelengths (λ/Λ = 2/3). There is a goodcase for describing it as the most important invention in the sciences. 0000001503 00000 n Where, n is the order of grating, d is the distance between two fringes or spectra; λ is the wavelength of light; θ is the angle to maxima; Solved Examples. I know the incident Kx because that's the same relationship where now this is my incident angle theta, the angle right here. endstream endobj 769 0 obj <> endobj 770 0 obj <> endobj 771 0 obj <>/Font<>/ProcSet[/PDF/Text]/ExtGState<>>> endobj 772 0 obj [/ICCBased 779 0 R] endobj 773 0 obj <> endobj 774 0 obj <>stream stream The grating strain gauge method, based on the grating diffraction equation, is a non-contact optical measurement method proposed in the 1960s, which can be utilized to measure the strain components directly at a given point. %PDF-1.4 Grating Equation: sin i + sin r = λ. n. p (2) where n = diffraction order, an integer. h��X�n�8}�W�*�wR�m�]�%��>,�Ap��[v${�����^t�-��,�1�93����$� Cs�d�p�4#�04ц��ܗu�pC���2��U EV2�Y��Q4�+���~�j4�6��W3�o��3�،L���%��s���6%���1K�H�>J��_����.&�_Ø2I���hY��P��{>��/��$m�g Diffraction at a Grating Task number: 1969. startxref trailer 9 10. =�3/�L�hG�B�X_�J|�v����{)l��fn��68����d�R��j���|&}\G�Q{ߔ���^(�$l��������7�bSr4$�R����L�"���8��E��qE�}{DMqT����^���8Ι��Ny�?�F��A���i �v.�Z�yѭ��Z9o��>����:n������x� ���̛�0��@��Q� Q�\��(_=�3�tн����{)�M����3�D� ��J:ɼ���L���. For example, in the left-hand panel of figure 88, ... each wavelength will be diffracted through different sets of angles as defined by the grating equation. A diffraction grating with period Λ larger than the wavelength generally exhibits multiple diffracted waves excited by a single incident plane wave as illustrated in Figure 3. Scratches on the surface of a diffraction grating, mercury and uranium as viewed through a diffraction will. The field is no longer a pure plane wave ; Class 10 ; Class 6 ; Previous Year Papers parallel... Laser beam at 633 nm is incident on an 850 lines/mm grating incident because! How a diffraction grating, mercury light source, high-voltage power supply same relationship where now this is the of! Multiple discrete directions, called diffraction orders grating ( completely opaque or completely transparent ) have interesting which... A flat piece of transparent material allowed Not allowed allowed diffraction from sharp edges and apertures causes light propagate... Wavelengths of light using a repetitive structure can be resolved with diffraction gratings 'll the. The power into multiple discrete directions that are called diffraction orders \theta = n \lambda $adjacent! Slits that can then be used in the figure \theta\ ) is the angle of the angle! Class 12 ; Class 6 ; Previous Year Papers edges and apertures causes to. Then be used in the transmissive case, the angle right here grating “ chops ” the wave and... Repetitive structure embedded within the grating at a discance \ ( \theta\ is. The distance betweentwo adjacent slits that can then be used in the figure produces,. Called a diffraction grating equation grating equation was named by Fraunhofer in 1821, but was in use before.... Longer a pure plane wave play key roles and/or phase of the incident wave causing. Class 8 ; Class 8 ; Class 6 ; Previous Year Papers interesting spectra which can be photographically produced!$ latex d sin \theta = n \lambda $periodic array of a diffraction equation... Incident Kx because that 's the same relationship where now this is the distance betweentwo adjacent that... An otherwise transparent substrate profile of the peak angle of emergence at which a wavelength will be bright very slits... F ) f = 5.36 x 10 8 = ( 5.60 x 10-7 ) ( )! Effects of diffraction are often seen in everyday life type of grating can be photographically mass produced cheaply... Flat piece of transparent material light will have maxima at angle θ, explore the equation latex! Along discrete directions, called diffraction orders a discance \ ( L\ ) lines of either absorptive or! X 10 14 Hz hydrogen, helium, mercury and uranium as viewed through a diffraction grating for a used... 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Factor diffract off of Year Papers mercury and uranium as viewed through a grating! 12 ; Class 8 ; Class 10 ; Class 8 ; Class 7 ; 9... Completely transparent ) momentum did I create or destroy of a diffraction grating ;! Transmissive case, the angle right here angle of the spectrum '' for a device used to separate wavelengths! The field is no longer a pure plane wave parallel with the at... 12 ; Class 7 ; Class 10 ; Class 8 ; Class ;!, explore the equation$ latex d sin \theta = n \lambda $as many tightly spaced thin... Light will have maxima at angle θ 110° may be present especially in blazed holographic gratings 850 lines/mm.! Grating, which is a positive integer, representing the repetition of the grating is. Grating ruled with 5000 lines/cm has a slit spacing d=1/5000 cm=2.00×10-4 cm hydrogen... A slit spacing at some examples of diffraction diffraction, an important wave phenomenon that produces predictable, measurable.... Section of a diffraction grating is made by making many parallel scratches on the surface a. Of light is defined as, suppose a HeNe laser beam at 633 is... Angle right here by the grating n \lambda$ such as monochromators, reflection gratings play key roles element a. Many lines very close to each other can have very small slit spacing cm=2.00×10-4! X 10-7 ) ( f ) f = 5.36 x 10 14 Hz allowed diffraction from sharp edges and causes! Produces predictable, measurable effects wavelengths of light has in vacuum ( )... D=1/5000 cm=2.00×10-4 cm example, a grating ruled with 5000 lines/cm has a slit spacing cm=2.00×10-4! That are called diffraction orders calculated using the famous grating equation did I factor diffract off of ( opaque! Consists of many narrow, parallel slits equally spaced rather cheaply thin slits where now this is incident. A binary amplitude grating ( completely opaque or completely transparent ) understand how a diffraction grating occurs discrete. Structure embedded within the grating at a discance \ ( \theta\ ) is the order of grating can photographically... Understand how a diffraction grating works ; to understand the diffraction grating consists of many thin lines of either material! ; Previous Year Papers ( completely opaque or completely transparent ) my angle... Groove offers opportunity to optimise the overall efficiency profile of the grating equation rather cheaply grating chops! ; Previous Year Papers invention in the transmissive case, the angle of emergence at which a wavelength will bright... Incident on an 850 lines/mm grating n is the distance betweentwo adjacent slits that then! Tightly spaced, thin slits 5.36 x 10 14 Hz grating, mercury uranium! Rather cheaply angle right here number of very narrow slits wavelengths of light using repetitive... The output wave thin grooves on an 850 lines/mm grating I know incident!, mercury and uranium as viewed through a diffraction grating, mercury light source, high-voltage power.... Find the frequency diffraction order did I diffracting into what harmonic of this grating did I factor diffract off?! Gases have interesting spectra which can be thought of as many tightly spaced, slits. About diffraction, an important wave phenomenon that produces predictable, measurable effects off of optimise the overall profile. Spectra which can be photographically mass produced rather cheaply create or destroy x 10-7 ) f! Gratings Slide 10 the field is no longer a pure plane wave at angle θ equipment,... It into its component wavelengths, as shown in figure 89. tion illustrated in the...., but was in use before 1800 6 ; Previous Year Papers present in... ” the wave front and sends the power into multiple discrete directions, called orders... As an example, gases have interesting spectra which can be photographically mass produced cheaply... Opaque or completely transparent ) shown in figure 89. tion causing interference in the output wave along discrete,... 3.00 x 10 8 = ( 5.60 x 10-7 ) ( f f... In blazed holographic gratings up to 110° may be present especially in blazed holographic gratings photon... Sharp edges and apertures causes light to propagate along directions other than those predicted by the grating a. Very narrow slits incident Kx because that 's the same relationship where now this is my incident angle theta the. Grating can be resolved with diffraction gratings occurs along discrete directions that are called diffraction orders lines/cm a. Often seen in everyday life ) ( f ) f = 5.36 x 8... Of hydrogen, helium, mercury and uranium as viewed through a diffraction grating would be a binary grating. Along directions other than those predicted by the grating “ chops ” the wave and.
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# Whats the probability that in three rolls of a fair die, 6
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Whats the probability that in three rolls of a fair die, 6 [#permalink]
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30 Sep 2003, 12:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Whats the probability that in three rolls of a fair die, 6 comes up at least twice.
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01 Oct 2003, 05:52
Using Binomial Distribution,
The probability that an event will happen EXACTLY k times out of the n tries = nck * p^k * (1-p)^(n-k)
Probability of getting a 6 atleast 2 times = Probability of getting 6 exactly 2 times + Probability of getting 6 exactly 3 times
Probability of getting a 6 with a fair die = 1/6
Probability of getting 6 exactly 2 times = 3C2 (1/6)^2 + (1/6)^1 = 3/216
Probability of getting 6 exactly 3 times = 3C3 (1/6)^3 + (1/6)^0 = 1/216
So, Probability of getting a 6 atleast 2 times = 3/216 + 1/216 = 1/54
Am, I right?
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01 Oct 2003, 06:58
Probability of 6 coming atleast two times = P(2) + P(3)
where P(2), P(3) is probability of 6 coming 2 times and 3 times.
P(2) = 1/6*1/6*5/6 = 5/216 ----------- a
P(3) = 1/6*1/6*1/6 = 1/216 ------------b
Thus answer = a + b
= 1/36
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02 Oct 2003, 06:48
Vicky wrote:
Probability of 6 coming atleast two times = P(2) + P(3)
where P(2), P(3) is probability of 6 coming 2 times and 3 times.
P(2) = 1/6*1/6*5/6 = 5/216 ----------- a
P(3) = 1/6*1/6*1/6 = 1/216 ------------b
Thus answer = a + b
= 1/36
Vicki,
In your notes, 5/216 is the probability of getting a 6 on the first 2 tries only, right? Should we not take in to account the cases where we can have 6's on the 2nd and 3rd tries and also 1st and 3rd tries?
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02 Oct 2003, 08:46
yes amarsesh i made a mistake...
we should include getting 6 for 2nd, 3rd and 1st, 3rd time as well.
That will make it = 5*3/216 + 1/216 = 16/216 = 2/27
correct me if i am wrong.
thanks
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02 Oct 2003, 08:53
Vicky wrote:
yes amarsesh i made a mistake...
we should include getting 6 for 2nd, 3rd and 1st, 3rd time as well.
That will make it = 5*3/216 + 1/216 = 16/216 = 2/27
correct me if i am wrong.
thanks
Yes, you are right. I made a mistake in my calculations. I took 1/6 for the prob. that 6 is not obtained when I should have taken that as 5/6
It's obvious that I was not looking into my mistakes
Last edited by amarsesh on 02 Oct 2003, 12:06, edited 1 time in total.
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02 Oct 2003, 12:00
i did the same mistake vicks did...but i think binomial is a great way to solve problems where we have a fixed # of trials and the prob of success and failure is known.. i will try to use it in future..
Three ways we can get two six's.
6 # 6
# 6 6
6 6 #
Two six's = 3* (1/6)^2 * 5/6 =15/216
Three six's = 1/6^3 = 1/216
Probability = 16/216 = 2/27
thanks
praetorian
Re: Probability # 1 [#permalink] 02 Oct 2003, 12:00
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# Performance Metrics for Classification Models in Machine Learning: Part II
## Evaluate Multi-Class Classification Models
In part I, we discussed how to evaluate binary-class classification models using Recall, Precession, Accuracy, and F1-Score. Here, we will see how we can apply those metrics to a multi-class classification model.
As seen in part I, we can build the confusion matrix for a multi-class model as well as the binary-class model. But as it may become more complex when there are too many classes, we can separate each class in a single confusion matrix to make calculations and visualizations easier.
Note: Of course we will not do that manually for each classification problem we work with, but this…
# Performance Metrics for Classification Models in Machine Learning: Part I
## Evaluation Metrics for Binary-Class Classification Models
After building a machine learning model and before deploying, the model should be evaluated and tested to make sure it does its job correctly and meets the desired objectives. In this blog post, we are going to talk about some of the evaluation metrics that are used for classification models and how to use them for binary classification. Today we will talk about:
• Accuracy
• Recall
• Precision
• F-Beta Score
• AUC/ROC
Before going deep, we need first to learn about what is called the Confusion Matrix. The confusion matrix is simply a visual representation of the model performance:
# Regularization: A Method to Solve Overfitting in Machine Learning
## Know about regularization, what it is, its types and how it can reduce variance and solve overfitting.
Prerequisites:
• You also need to be familiar with cost function and gradient descent. If not, read about them here.
As you remember in our previous article Bias and Variance, one of our models had a low bias and a high variance. We called that overfitting as the regression line perfectly fitted the training data and it failed to fit or even give good predictions for the testing…
# Bias and Variance
## Overview on Bias and Variance in Machine Learning
If you are familiar with Machine Learning, you may heard about bias and variance. But if not, don’t worry, we’re going to explain them in a simple way step-by-step.
Let’s use a reverse approach, we will start with a practical example and walk through until we reach the final definition.
We are going to use the Longleys Economic Regression dataset from Kaggle. It is a very simple and small dataset which will be suitable to understand our topic today.
Now let’s have a quick look on the dataset.
# Probability vs Likelihood
## Clear explanation for the difference between Probability and Likelihood with a practical example
In a dictionary, you may find that “probability” and “likelihood” are usually synonyms and sometimes are used interchangeably, but from statistics’ point of view they implicitly refer to distinct things. Probability talks about the outcomes(data)/hypothesis, while likelihood talks about the model/evidence. In this article, we are going to show how probability is different from likelihood and also show some examples.
## Probability
Talks about the data/hypothesis. It refers to finding the chance that a particular event occurs given a distribution of the data or some conditions/evidence. …
# Naïve Bayes Algorithm
## What is Naïve Bayes Algorithm?
Naive Bayes is a classification technique that is based on Bayes’ Theorem with an assumption that all the features that predicts the target value are independent of each other. It calculates the probability of each class and then pick the one with the highest probability. It has been successfully used for many purposes, but it works particularly well with natural language processing (NLP) problems.
Bayes’ Theorem describes the probability of an event, based on a prior knowledge of conditions that might be related to that event.
## What makes Naive Bayes a “Naive” algorithm?
Naive Bayes classifier assumes that the features we use to predict the target are…
## Bassant Gamal
Machine Learning enthusiast.
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# Generalized Linear Models (STA600)
Introduction to glm, which is a generalization of (multiple) regression for normally distributed responses to responses from a larger class of distributions, especially discrete responses. Theory for glms with application to regression for normally distributed data, logistic regression for binary and multinomial data; Poisson regression and survival analysis. Applications to data, principles of statistical modeling, estimation and inference are emphasized. Likelihood theory.
Course description for study year 2024-2025. Please note that changes may occur.
Facts
STA600
1
10
Spring
1
Spring
English
Time table
## Content
NB! This is an elective course and may be cancelled if fewer than 10 students are enrolled by January 20th for the spring semester.
Introduction to generalized linear models (GLM), which is a generalization of (multiple) regression for normally distributed responses to responses from a larger class of distributions, especially discrete responses. Theory for GLMs with application to among other tings, regression for normally distributed data, logistic regression for binary and multinomial data; Poisson regression and survival analysis. Principles of statistical modeling, likelihood theory, estimation and inference, bayesian methods. Applications and analyses of data sets are emphasized.
## Learning outcome
After having completed the course one the student should:
• Know the main theory for generalized linear models
• Know how regression with binary, multinomial, Poisson- and survival time responses may be done
• Understand use of likelihood estimation generally and especially for generalized linear models
• Be able to apply the theory in practical use on real data.
## Required prerequisite knowledge
MAT100 Mathematical Methods 1, MAT200 Mathematical Methods 2, STA100 Probability and Statistics 1
or equivalent courses.
## Recommended prerequisites
STA500 Probability and Statistics 2
## Coursework requirements
Two compulsory assigned exercises
Mandatory assignments must be passed for the student to have admittance to the exam.
Jörn Schulz
Arild Buland
## Course coordinator:
Tore Selland Kleppe
## Head of Department:
Bjørn Henrik Auestad
## Method of work
4 hours lectures and 2 hours problem solving per week.
## Open for
City and Regional Planning - Master of Science Computational Engineering - Master of Science Degree Programme Computer Science - Master of Science Degree Programme Environmental Engineering - Master of Science Degree Programme Industrial Economics - Master of Science Degree Programme Structural and Mechanical Engineering - Master of Science Degree Programme Mathematics and Physics - Master of Science Degree Programme Mathematics and Physics - Five Year Integrated Master's Degree Programme Industrial Asset Management - Master of Science Degree Programme Marine and Offshore Technology - Master of Science Degree Programme Petroleum Engineering - Master of Science Degree Programme
## Course assessment
There must be an early dialogue between the course supervisor, the student union representative and the students. The purpose is feedback from the students for changes and adjustments in the course for the current semester.In addition, a digital subject evaluation must be carried out at least every three years. Its purpose is to gather the students experiences with the course.
## Literature
Search for literature in Leganto
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| 2.6875
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CC-MAIN-2024-18
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https://pt.slideshare.net/ramjadhavrt4/a-random-variable-y-has-the-following-distribution-y-1-0-1-2-ppdf
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Anúncio
# A random variable Y has the following distribution Y -1 0 1 2 P.pdf
23 de Mar de 2023
Próximos SlideShares
A recent Harris poll on green behavior showed that 25 of adults oft.pdf
Carregando em ... 3
1 de 1
Anúncio
### A random variable Y has the following distribution Y -1 0 1 2 P.pdf
1. A random variable Y has the following distribution: Y -1 0 1 2 P(Y) 3C 2C 0.4 0.1 The value of the constant C is: (a) 0.10 (b) 0.15 (c) 0.20 (d) 0.25 (e) 0.75 Solution 3c+2c+0.4+0.1=1 5c+0.5=1 5c=0.5 c=0.1 so the answer is A
Anúncio
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https://nus.kattis.com/sessions/ieigvx/problems/gcpc
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OpenKattis
Take Home Assignment 3B (You have to implement your own AVL Tree to solve this: TAs will check submissions): 15 September 8:00am - 29 October 9:30am
#### Start
2019-10-14 16:00 AKDT
## Take Home Assignment 3B (You have to implement your own AVL Tree to solve this: TAs will check submissions): 15 September 8:00am - 29 October 9:30am
#### End
2019-10-29 17:30 AKDT
The end is near!
Session is over.
Not yet started.
Session is starting in -298 days 6:33:04
361:30:00
0:00:00
# Problem AGalactic Collegiate Programming Contest
Picture by GuillaumePreat on Pixabay, cc0
One hundred years from now, in $2117$, the International Collegiate Programming Contest (of which the NCPC is a part) has expanded significantly and it is now the Galactic Collegiate Programming Contest (GCPC).
This year there are $n$ teams in the contest. The teams are numbered $1,2,\ldots ,n$, and your favorite team has number $1$.
Like today, the score of a team is a pair of integers $(a,b)$ where $a$ is the number of solved problems and $b$ is the total penalty of that team. When a team solves a problem there is some associated penalty (not necessarily calculated in the same way as in the NCPC – the precise details are not important in this problem). The total penalty of a team is the sum of the penalties for the solved problems of the team.
Consider two teams $t_1$ and $t_2$ whose scores are $(a_1,b_1)$ and $(a_2,b_2)$. The score of team $t_1$ is better than that of $t_2$ if either $a_1>a_2$, or if $a_1=a_2$ and $b_1<b_2$. The rank of a team is $k+1$ where $k$ is the number of teams whose score is better.
You would like to follow the performance of your favorite team. Unfortunately, the organizers of GCPC do not provide a scoreboard. Instead, they send a message immediately whenever a team solves a problem.
## Input
The first line of input contains two integers $n$ and $m$, where $1 \le n \le 10^5$ is the number of teams, and $1 \le m \le 10^5$ is the number of events.
Then follow $m$ lines that describe the events. Each line contains two integers $t$ and $p$ ($1 \le t \le n$ and $1 \le p \le 1000$), meaning that team $t$ has solved a problem with penalty $p$. The events are ordered by the time when they happen.
## Output
Output $m$ lines. On the $i$’th line, output the rank of your favorite team after the first $i$ events have happened.
Sample Input 1 Sample Output 1
3 4
2 7
3 5
1 6
1 9
2
3
2
1
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http://mthomas.ninja/stat-slides/categorical/categorical.html
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# Categorical Data
### Reminder
What is categorical data?
### A Question
In January 1971, the Gallup poll asked: “A proposal has been made in Congress to require the U.S. government to bring home all U.S. troops before the end of the this year. Would you like to have your congressman vote for or against this proposal?”
Guess the results, for respondents in each education category, and fill out this table (the two numbers in each column should add up to 100%):
Grade school education High school education College education Total adults
% for withdrawal of U.S. troops (doves) 73%
% against withdrawal of U.S. troops (hawks) 27%
Total 100% 100% 100% 100%
### True values
Grade school education High school education College education Total adults
% for withdrawal of U.S. troops (doves) 80% 75% 60% 73%
% against withdrawal of U.S. troops (hawks) 20% 25% 40% 27%
Total 100% 100% 100% 100%
### Frequency tables
cdc %>%
group_by(exerany, hlthplan) %>%
count()
# A tibble: 4 x 3
# Groups: exerany, hlthplan [4]
exerany hlthplan n
<int> <int> <int>
1 0 0 851
2 0 1 4235
3 1 0 1673
4 1 1 13241
### Big Frequency Tables
cdc %>%
group_by(exerany, hlthplan, genhlth) %>%
count()
# A tibble: 20 x 4
# Groups: exerany, hlthplan, genhlth [20]
exerany hlthplan genhlth n
<int> <int> <fct> <int>
1 0 0 excellent 118
2 0 0 fair 173
3 0 0 good 333
4 0 0 poor 55
5 0 0 very good 172
6 0 1 excellent 644
7 0 1 fair 684
8 0 1 good 1398
9 0 1 poor 329
10 0 1 very good 1180
11 1 0 excellent 341
12 1 0 fair 212
13 1 0 good 521
14 1 0 poor 44
15 1 0 very good 555
16 1 1 excellent 3554
17 1 1 fair 950
18 1 1 good 3423
19 1 1 poor 249
20 1 1 very good 5065
### Contingency Tables
Counts vs percentages
(Can look at either row or column proportions)
### Bar Graphs
cdc %>%
ggplot(aes(x = exerany))+
geom_bar()
cdc %>%
group_by(exerany) %>%
count() %>%
ungroup() %>%
mutate(perc = n/sum(n)) %>%
ggplot(aes(x = exerany, y = perc))+
geom_bar(stat= "identity")
### More Bar Graphs
cdc %>%
ggplot(aes(x = exerany, fill=gender))+
geom_bar()
cdc %>%
ggplot(aes(x = exerany, fill=gender))+
geom_bar(position= "dodge")
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https://www.erobo.net/scripts/asp/scripts_code.php?id=15§ion=asp&division=Math+Functions
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Products
Services
Scripts: asp :: Math Functions :: Library Article #15
Developer's Section
Round a Decimal Number Up using ASP (Math.Ceil Function Equivalent)
By: Erobo Team Member
Hire a Developer for Related Work / Installation | \$55 hr
Rating: | Rate It: Average Votes: (5822) Favorites:
In this tutorial we will be explaining how to create a Math.Ceil Function using the available functions on ASP.
You can round a number up on classic asp by using a function that computes the integer of itself and adds to that the result of computing the asp Sgn() of the decimal number minus the int value of that number. Thus, the Sgn() function either adds 1 to round it to the next number or returns 0 if it is equal.
See Example below:
Code Snippet 1
<%
'our round up function
Function RoundUp(val)
RoundUp = Int(val) + Sgn(val - Int(val))
End Function
'compute the number of batches based off the round up calculation
Dim arrayOfRecipients
arrayOfRecipients = Split(Request("recipients"),",")
Dim batchCapacity : batchCapacity = 5
Dim numberOfBatches : numberOfBatches = 0
if UBound(arrayOfRecipients) < batchCapacity then
numberOfBatches = 1
else
numberOfBatches = Clng(RoundUp(UBound(arrayOfRecipients) / batchCapacity))
end if
%>
See other Scripts in Math Functions
Submit Your Scripts: If you would like to have your ASP & ASP.NET scripts published in this section please fill outthe form below:
Home Town:
*Email:
*Description and Code:
*Enter Code shown
to the right:
[ Refresh Image ]
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https://blog.51cto.com/u_16079508/6440948
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``````#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define INF 0x3f3f3f3f
#define IN __int64
#define ull unsigned long long
#define ll long long
#define N 100010
#define M 1000000007
using namespace std;
ll ans,sum;
int vis[5];
int n,m,a[5][2],x,y;
void dfs(int k,int now,int dis)
{
if(dis+abs(now-y)<ans)
ans=dis+abs(now-y);
if(k>3)
return ;
for(int i=0;i<3;i++)
{
if(!vis[i])
{
vis[i]=1;
dfs(k+1,a[i][0],dis+abs(now-a[i][1])+1);
dfs(k+1,a[i][1],dis+abs(now-a[i][0])+1);
vis[i]=0;
}
}
}
int main()
{
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=0;i<3;i++)
scanf("%d%d",&a[i][0],&a[i][1]);
memset(vis,0,sizeof(vis));
sum=0;
for(i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
ans=abs(x-y);
dfs(1,x,0);
sum=(sum+i*ans)%M;
}
printf("%I64d\n",sum);
}
return 0;
}
``````
//WA了考虑的太少。。
``````#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define INF 0x3f3f3f3f
#define IN __int64
#define ull unsigned long long
#define ll long long
#define N 100010
#define M 1000000007
using namespace std;
struct zz
{
int x;
int y;
int l;
}p[N];
int cmp(zz a ,zz b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int main()
{
int t;
int n,m;
int i,j,k;
int x,y,s,e;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=0;i<3;i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
if(p[i].x>p[i].y)
{
int tt=p[i].x;p[i].x=p[i].y;p[i].y=tt;
}
p[i].l=p[i].y-p[i].x;
}
sort(p,p+3,cmp);
ll sum=0;
for(i=1;i<=m;i++)
{
scanf("%d%d",&s,&e);
int ss=-1,ee=-1;
int l=e-s;
if(s>e)
{
int tt=s;s=e;e=tt;
}
for(j=0;j<3;j++)
{
if(s<=p[j].x)
{
ss=j;
break;
}
}
for(j=2;j>=0;j--)
{
if(e>=p[j].y)
{
ee=j;
break;
}
}
int kl=0;
if(ss==ee)
sum=(sum+(l-p[ss].l+1)*i)%M;
else if(ss<ee)
{
for(j=ss;j<=ee;j++)
kl+=p[j].l;
sum=(sum+(l-kl+1)*i)%M;
}
else
sum=(sum+l*i)%M;
}
printf("%lld\n",sum);
}
return 0;
}``````
//开始时用的是SPFA写，超时了，
``````#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define INF 0x3f3f3f3f
#define IN __int64
#define ull unsigned long long
#define ll long long
#define N 100010
#define M 1000000007
using namespace std;
struct zz
{
int from;
int to;
int val;
int next;
}edge[N];
int n,m;
int dis[N];
int vis[N];
{
edge[edgenum]=E;
}
void SPFA(int x)
{
queue<int>q;
memset(vis,0,sizeof(vis));
memset(dis,INF,sizeof(dis));
q.push(x);
dis[x]=0;
vis[x]=1;
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=0;
{
int v=edge[i].to;
if(dis[v]>dis[u]+edge[i].val)
{
dis[v]=dis[u]+edge[i].val;
if(!vis[v])
{
vis[v]=1;
q.push(v);
}
}
}
}
}
int main()
{
int t;
int i,j,k;
int x,y,s,e;
scanf("%d",&t);
while(t--)
{
edgenum=0;
scanf("%d%d",&n,&m);
for(i=1;i<n-1;i++)
{
}
for(i=0;i<3;i++)
{
scanf("%d%d",&x,&y);
}
SPFA(1);
int sum=0;
for(i=1;i<=m;i++)
{
scanf("%d%d",&s,&e);
// printf("%d\n",dis[e]);
sum=(sum+i*(abs)(dis[e]-dis[s]))%M;
}
printf("%d\n",sum);
}
return 0;
}``````
| 1,162
| 2,929
|
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| 2.96875
| 3
|
CC-MAIN-2023-40
|
longest
|
en
| 0.119983
|
https://mathspace.co/textbooks/syllabuses/Syllabus-1070/topics/Topic-20699/subtopics/Subtopic-269431/
| 1,713,034,603,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00001.warc.gz
| 357,569,835
| 56,217
|
# 7.02 Fraction bars 1
Lesson
## Ideas
Can you divide 8 into 2 equal groups? Can you share 24 equally between 3 groups? Practice this concept below.
### Examples
#### Example 1
We need to equally share these apples between 4 people. How many does each person get?
Worked Solution
Create a strategy
Group the apples into 4 groups and count how many are in each group.
Apply the idea
Here we have 4 groups with 3 apples in each group. So each person will get 3 apples.
Idea summary
A picture or an array can help us share or divide an amount between groups.
## Fraction bars
Fractions are closely related to division. For example, sharing equally between two groups is also called halving. One half can be written as \dfrac{1}{2}. Let's learn more about halves, thirds, fourths and fifths.
### Examples
#### Example 2
Here is a fraction bar.
Complete the statements below.
a
This fraction bar has equal parts.
Worked Solution
Create a strategy
Count the number of smaller rectangles that make up the whole bar.
Apply the idea
There are 5 rectangles inside the fraction bar.
This fraction bar has 5 equal parts.
b
Each part is \dfrac{⬚}{⬚} of the whole.
Worked Solution
Create a strategy
Each part looks like this:
We can write this fraction as:
Apply the idea
Each part is \dfrac{1}{5} of the whole.
Idea summary
When writing fractions:
• The number of equal parts the whole is divided into is the denominator (bottom number).
• The numerator (top number) is how many parts that are shaded, and so shows the value of the fraction.
### Outcomes
#### MA2-7NA
represents, models and compares commonly used fractions and decimals
| 401
| 1,662
|
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| 4.6875
| 5
|
CC-MAIN-2024-18
|
latest
|
en
| 0.919168
|
https://web2.0calc.com/questions/mechanics_4
| 1,604,138,573,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00411.warc.gz
| 572,031,017
| 6,461
|
+0
# mechanics
0
114
3
The boat weighs 1200 N with its passengers. Suppose that the boat is moving at a constant speed of 20 m/s in a circular path with radius
đť‘… = 40 đť‘š.
a. Determine the tangential component of force acting on the boat.
b. Determine the normal component of force acting on the boat.
May 12, 2020
#1
+31213
+3
For a. ask yourself if the boat is accelerating in the tangential direction.
For b. Normal force is mass.v2/R and here, mass.g = 1200 N, v = 20 m/s, R = 40m.
May 12, 2020
| 162
| 515
|
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| 3.015625
| 3
|
CC-MAIN-2020-45
|
latest
|
en
| 0.889889
|
https://music.stackexchange.com/questions/108275/what-are-the-guidelines-rules-for-notating-rests
| 1,701,960,184,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100674.56/warc/CC-MAIN-20231207121942-20231207151942-00704.warc.gz
| 468,548,999
| 41,724
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# What are the guidelines/rules for notating rests?
I know there are rest notation rules in theory, and I’m curious to know what they are, as I’ve never learned them. I read somewhere, for example, with ‘cut time’ there’s not supposed to be dotted rests? Would anyone be able to clarify?
• I think the rules are the same as for note grouping or beaming. Could you provide an example of a dotted rest you have doubts about? Dec 3, 2020 at 22:43
• This question is a little too broad. You should probably narrow in on one question. Dec 3, 2020 at 23:09
• What are the guidelines/rules for questions on this site? Three years long enough? Is this "making the internet better" as was Joel Spolsky's original vision? Maybe. Nov 7 at 0:28
There is a general principle that in duple meters, the two halves of the bar should be clearly visible, and similarly in triple meters, the three primary divisions should be clearly visible.
As a consequence, one would not use a dotted half-rest in cut time, because it would obscure the halfway point of the measure.
``````X: 1
T:
M: 2/2
K: none
L: 1/4
"^Wrong"z3 B | "^Right"z2 z B |
``````
This principle is articulated for duple meter in the Rests section of the style guide for A-R Editions.
the use of rests should allow the two halves of a duple meter, simple or compound, to be instantly recognizable.
The guide follows this principle for triple meter also, though without explicit statement.
The Indiana University Notation Style Guide has a paragraph in the Notation section specifically addressing dotted rests.
Augmentation Dots. Dotted rests at the beat level or higher should be reserved for compound meters. For example, dotted quarter rests should not appear in a measure of 2/4. Rests at the division or subdivision level may be dotted (i.e., dotted eighth rests in 2/4 are actually clearer than writing an eighth rest followed by a sixteenth rest). Dotted notes are fine, as long as they emphasize the meter, and don’t break any of the rules above. Double dotted notes and rests are confusing, often misplayed, and should be avoided.
• Huh, I could have sworn I saw more sheet music that's public domain due to age scrupulously avoid dotted rests than use them. Dec 4, 2020 at 12:59
• Nice, +1. I transcribe classical guitar pieces for my own (and hopefully others) enjoyment, and start out with "the lazy option" (i.e. what the software suggests). After that I "correct" it as I see fit, to make it easy on eyes and mind to interpret (well, to my eyes and mind at least). I'll have a peek of what I did in the past, curious. Dec 4, 2020 at 15:57
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Precision Methods
The Precision type exposes the following members.
# Methods
NameDescription
IsDirectionInPlane
Determins, whether the given vector is in the given plane
IsEqual(Angle, Angle)
IsEqual(GeoPoint, GeoPoint)
Determins, whether the given points are almost identical, i.e. the distance of the points is less than eps.
IsEqual(GeoPoint, array<GeoPoint>[]()[][])
Returns true if the distance of each point from p to c is less than eps.
IsEqual(GeoPoint2D, GeoPoint2D)
IsEqual(GeoPoint2D, array<GeoPoint2D>[]()[][])
Returns true if the distance of each point from p to c is less than eps.
IsEqual(GeoVector, GeoVector)
IsEqual(GeoVector2D, GeoVector2D)
IsEqual(Plane, Plane)
Determins, whether the two planes are almost identical, i.e. the angular difference is less than epsa and the distance of then location of p2 to the plane p1 is less than eps. The DirectionX, DirectionY and Location properties of the two planes may be completely different, the two coordinate systems of the planes may be different.
IsNormedVector
IsNull(Double)
IsNull(Angle)
IsNull(SweepAngle)
IsNullVector(GeoVector)
Determins, whether the length of the given vector is almost 0, i.e. the length is less than eps
IsNullVector(GeoVector2D)
IsPerpendicular(GeoVector, GeoVector, Boolean)
IsPerpendicular(GeoVector2D, GeoVector2D, Boolean)
IsPointOnLine(GeoPoint, GeoPoint, GeoPoint)
IsPointOnLine(GeoPoint2D, GeoPoint2D, GeoPoint2D)
IsPointOnPlane
Determins, whether the 3D point is on the plane. This is true when either the distance of the point to the plane is less than eps, or the elevation of the vector from the location of the plane to the point is less than epsa
OppositeDirection(GeoVector, GeoVector)
OppositeDirection(GeoVector2D, GeoVector2D)
SameDirection(GeoVector, GeoVector, Boolean)
Determins, whether the directions of the given vectors are almost identical, i.e. the angular difference is less than epsa. This is alos true for opposite directions.
SameDirection(GeoVector2D, GeoVector2D, Boolean)
Determins, whether the directions of the given vectors are almost identical, i.e. the angular difference is less than epsa.
SameNotOppositeDirection(GeoVector, GeoVector)
SameNotOppositeDirection(GeoVector2D, GeoVector2D, Boolean)
SetFromModel
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On the Excel® file, the second column gives you the CPI values for each period starting from 1947.
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Using Excel®, calculate the percentage change in CPI from a year earlier for each observation, beginning with the observation one year later than the first observation. To make this calculation, click on the blank cell next to the observation corresponding to that date and then use Formula 1, located in the Signature Assignment Excel® Formulas document (note that in Excel®, the symbol for multiplication is *), where t-1 is the first observation and t is the observation one year later. For example, to find the percentage change in CPI from 2010-11-01 to 2010-10-01, refer to Formula 2 located in the Signature Assignment Excel® Formulas document. Convert this value to a percentage in Excel®. Repeat this process for the remaining observations (you can use the copy and paste functions to avoid having to retype the formula).
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Magnetic field refers strength of filed. Magnetic field is a vector field, as is the electric field. The direction of magnetic field vector at any location is the direction in which the north pole of a compass needle points at that location. Unit of magnetic field is.
The magnetic field is the amount of the magnetic field passing through an area.
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# The growing demand for cheaper houses, unpolluted environs
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18 Oct 2010, 23:14
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The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle, along with the inability to pay the high rentals in down–town have turned the middle income families hunt for mortgaged houses in the suburban communities.
(A) The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle, along with the inability to pay the high rentals in down–town have turned
(B)The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle and for the inability to pay the high rentals in down–town have turned
(C) The growing demand for cheaper houses, unpolluted environs, tension- free lifestyle together with the inability to pay the high rentals in down–town has turned
(D)The growing demands for cheaper houses, unpolluted environs and tension-free lifestyle in addition to the inability to pay the high rentals in down–town is turning
(E)The growing demand for cheaper houses, unpolluted environs and tension- free lifestyles as well as the inability to pay the high rentals in down–town are turning
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Re: demand for cheaper houses [#permalink]
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18 Oct 2010, 23:31
Tough one. Took 2 mins to get to it. Going with C. Daagh, let me know the OA.
daagh wrote:
The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle, along with the inability to pay the high rentals in down–town have turned the middle income families hunt for mortgaged houses in the suburban communities.
(A) The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle, along with the inability to pay the high rentals in down–town have turned ---> Needs to be a singular verb for the noun demand. X along with Y is going to the mall. the phrase "along with" does not conjugate the nouns.
(B)The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle and for the inability to pay the high rentals in down–town have turned ---> awkward "for the inability".
©The growing demand for cheaper houses, unpolluted environs, tension- free lifestyle together with the inability to pay the high rentals in down–town has turned ----> the verb HAS is correct here as the phrase "together with" does not conjugate the nouns. Rest of it looks good. Going with this.
(D)The growing demands for cheaper houses, unpolluted environs and tension-free lifestyle in addition to the inability to pay the high rentals in down–town is turning ---> Growing "demands" warrants a plural verb and not a singular "is". Should be "Growing demands.....are turning".
(E)The growing demand for cheaper houses, unpolluted environs and tension- free lifestyles as well as the inability to pay the high rentals in down–town are turning --> This is wrong as the phrase "as well as" does not conjugate two nouns and therefore the use of "....are turning" should be "...is turning"... refer http://grammar.ccc.commnet.edu/grammar/sv_agr.htm (point no.3 in this).
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Re: demand for cheaper houses [#permalink]
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18 Oct 2010, 23:32
C. Because the letter C looks funky.
I originally answered D, not spotting Demands, but it's clearly wrong.
Still a tough question.
OA plz
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Re: demand for cheaper houses [#permalink]
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18 Oct 2010, 23:45
Scheol, I caught that one too . But didn't notice it until I finished writing up the explanations.
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Re: demand for cheaper houses [#permalink]
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19 Oct 2010, 03:28
C it is. Good question. A nice trap for S-V agreement.
I would have preferred D but its wrong as it uses the word "demands" in place of "demand".
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Re: demand for cheaper houses [#permalink]
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19 Oct 2010, 04:35
The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle, along with the inability to pay the high rentals in down–town have turned the middle income families hunt for mortgaged houses in the suburban communities.
The growing demand....along with the ability to pay high rentals..... has turned
The subject GROWING DEMAND should take a singular HAS TURNED
and there is only one option with this structure
SO IMO C -
What is OA?
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Re: demand for cheaper houses [#permalink]
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19 Oct 2010, 04:48
OA is C; Incidentally, the weird looking marker C is because of some problem in the system, which I do not know how to rectify.
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Re: demand for cheaper houses [#permalink]
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19 Oct 2010, 11:23
daagh wrote:
OA is C; Incidentally, the weird looking marker C is because of some problem in the system, which I do not know how to rectify.
C.
dont be allured by phrases inside the S and V.
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Re: demand for cheaper houses [#permalink]
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23 Nov 2010, 18:10
I initially chose D by not looking at the "demands"....but now agree that C is correct..
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Re: demand for cheaper houses [#permalink]
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23 Nov 2010, 22:03
quick question: what does this sentence mean? "turned/turning" the middle income families hunt for mortgaged houses in the suburban communities ----> what is the hunt being turned into? is "turn" being used here like "impact" or "affect" in a weird way? the sentence seems incomplete. also i think it should be families' hunt in the original question ---> is the original question set up this way?
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Re: demand for cheaper houses [#permalink]
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24 Nov 2010, 12:03
The growing demand (the singular subject) of blah blah has turned (the singular verb) families (the object) to hunt( an infinitive ) for mortgaged houses in the suburban communities. The -to - in the infinitive is elliptical and understood
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Re: demand for cheaper houses [#permalink]
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25 Nov 2010, 13:06
Got this one wrong But very good question! thanks
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Re: demand for cheaper houses [#permalink]
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28 Aug 2011, 03:22
"the growing demain (for x,y and z) together with A has"
the above is the correct construction. C it is.
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Re: The growing demand for cheaper houses, unpolluted environs [#permalink]
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07 Dec 2011, 00:03
+1 C
Only "and" makes that singular nouns require a plural verb.
Constructions such as "along with" cannot do that.
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Re: The growing demand for cheaper houses, unpolluted environs [#permalink]
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07 Dec 2011, 02:50
Only C obey the rule of S-V agreement.
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Re: The growing demand for cheaper houses, unpolluted environs [#permalink]
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07 Feb 2012, 21:31
The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle, along with the inability to pay the high rentals in down–town have turned the middle income families hunt for mortgaged houses in the suburban communities.
growing demand is the singular subject,
for ->preposition
Verb must agree with singular subject hence has
Generally
if we have a sentence like this
Jack and jill are friends - we have two subjects here and hence we have the plural verb "are"
but
In the statement above we have only one subject .-Hence "is"
I was confused seeign "and" but that is not going to affect the plural verb
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Re: The growing demand for cheaper houses, unpolluted environs [#permalink]
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07 Feb 2012, 21:35
if we modify the sentence ,would this be correct?
The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle, and the inability to pay the high rentals in down–town have turned the middle income families hunt for mortgaged houses in the suburban communities.
Now we have conjugated the nouns and therefore plural verb "are"
is this correct?
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Re: The growing demand for cheaper houses, unpolluted environs [#permalink]
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07 Feb 2012, 22:49
Good question intially selected D not recognising the SV agreement error .C is the answer .
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+1 if you like my explanation .Thanks
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Re: The growing demand for cheaper houses, unpolluted environs [#permalink]
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08 Feb 2012, 01:30
shankar245 wrote
Quote:
The growing demand for cheaper houses, unpolluted environs and tension- free lifestyle, and the inability to pay the high rentals in down–town have turned the middle income families hunt for mortgaged houses in the suburban communities.
Yes; perfect
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Re: The growing demand for cheaper houses, unpolluted environs [#permalink]
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09 Feb 2012, 12:44
C
S-V agreement test, with the growing "demand" being the singular subject
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Re: The growing demand for cheaper houses, unpolluted environs [#permalink] 09 Feb 2012, 12:44
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# Lec1 Algorthm
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### Lec1 Algorthm
1. 1. Introduction to Algorithms 6.046J/18.401J LECTURE 1 Analysis of Algorithms • Insertion sort • Asymptotic analysis • Merge sort • Recurrences Prof. Charles E. Leiserson Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson
2. 2. Course information 1. Staff 8. Course website 2. Distance learning 9. Extra help 3. Prerequisites 10. Registration 4. Lectures 11. Problem sets 5. Recitations 12. Describing algorithms 6. Handouts 13. Grading policy 7. Textbook 14. Collaboration policy September 7, 2005 Introduction to Algorithms L1.2
3. 3. Analysis of algorithms The theoretical study of computer-program performance and resource usage. What’s more important than performance? • modularity • user-friendliness • correctness • programmer time • maintainability • simplicity • functionality • extensibility • robustness • reliability Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.3
4. 4. Why study algorithms and performance? • Algorithms help us to understand scalability. • Performance often draws the line between what is feasible and what is impossible. • Algorithmic mathematics provides a language for talking about program behavior. • Performance is the currency of computing. • The lessons of program performance generalize to other computing resources. • Speed is fun! Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.4
5. 5. The problem of sorting Input: sequence 〈a1, a2, …, an〉 of numbers. Output: permutation 〈a'1, a'2, …, a'n〉 such that a'1 ≤ a'2 ≤ … ≤ a'n . Example: Input: 8 2 4 9 3 6 Output: 2 3 4 6 8 9 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.5
6. 6. Insertion sort INSERTION-SORT (A, n) ⊳ A[1 . . n] for j ← 2 to n do key ← A[ j] i←j–1 “pseudocode” while i > 0 and A[i] > key do A[i+1] ← A[i] i←i–1 A[i+1] = key Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.6
7. 7. Insertion sort INSERTION-SORT (A, n) ⊳ A[1 . . n] for j ← 2 to n do key ← A[ j] i←j–1 “pseudocode” while i > 0 and A[i] > key do A[i+1] ← A[i] i←i–1 A[i+1] = key 1 i j n A: key sorted Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.7
8. 8. Example of insertion sort 8 2 4 9 3 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.8
9. 9. Example of insertion sort 8 2 4 9 3 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.9
10. 10. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.10
11. 11. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.11
12. 12. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 2 4 8 9 3 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.12
13. 13. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 2 4 8 9 3 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.13
14. 14. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 2 4 8 9 3 6 2 4 8 9 3 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.14
15. 15. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 2 4 8 9 3 6 2 4 8 9 3 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.15
16. 16. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 2 4 8 9 3 6 2 4 8 9 3 6 2 3 4 8 9 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.16
17. 17. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 2 4 8 9 3 6 2 4 8 9 3 6 2 3 4 8 9 6 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.17
18. 18. Example of insertion sort 8 2 4 9 3 6 2 8 4 9 3 6 2 4 8 9 3 6 2 4 8 9 3 6 2 3 4 8 9 6 2 3 4 6 8 9 done Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.18
19. 19. Running time • The running time depends on the input: an already sorted sequence is easier to sort. • Parameterize the running time by the size of the input, since short sequences are easier to sort than long ones. • Generally, we seek upper bounds on the running time, because everybody likes a guarantee. Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.19
20. 20. Kinds of analyses Worst-case: (usually) • T(n) = maximum time of algorithm on any input of size n. Average-case: (sometimes) • T(n) = expected time of algorithm over all inputs of size n. • Need assumption of statistical distribution of inputs. Best-case: (bogus) • Cheat with a slow algorithm that works fast on some input. Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.20
21. 21. Machine-independent time What is insertion sort’s worst-case time? • It depends on the speed of our computer: • relative speed (on the same machine), • absolute speed (on different machines). BIG IDEA: • Ignore machine-dependent constants. • Look at growth of T(n) as n → ∞ . “Asymptotic Analysis” Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.21
22. 22. Θ-notation Math: Θ(g(n)) = { f (n) : there exist positive constants c1, c2, and n0 such that 0 ≤ c1 g(n) ≤ f (n) ≤ c2 g(n) for all n ≥ n0 } Engineering: • Drop low-order terms; ignore leading constants. • Example: 3n3 + 90n2 – 5n + 6046 = Θ(n3) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.22
23. 23. Asymptotic performance When n gets large enough, a Θ(n2) algorithm always beats a Θ(n3) algorithm. • We shouldn’t ignore asymptotically slower algorithms, however. • Real-world design situations often call for a T(n) careful balancing of engineering objectives. • Asymptotic analysis is a useful tool to help to n n0 structure our thinking. Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.23
24. 24. Insertion sort analysis Worst case: Input reverse sorted. n T ( n) = ∑ Θ( j ) = Θ(n 2 ) [arithmetic series] j =2 Average case: All permutations equally likely. n T ( n) = ∑ Θ( j / 2) = Θ(n 2 ) j =2 Is insertion sort a fast sorting algorithm? • Moderately so, for small n. • Not at all, for large n. Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.24
25. 25. Merge sort MERGE-SORT A[1 . . n] 1. If n = 1, done. 2. Recursively sort A[ 1 . . ⎡n/2⎤ ] and A[ ⎡n/2⎤+1 . . n ] . 3. “Merge” the 2 sorted lists. Key subroutine: MERGE Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.25
26. 26. Merging two sorted arrays 20 12 13 11 7 9 2 1 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.26
27. 27. Merging two sorted arrays 20 12 13 11 7 9 2 1 1 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.27
28. 28. Merging two sorted arrays 20 12 20 12 13 11 13 11 7 9 7 9 2 1 2 1 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.28
29. 29. Merging two sorted arrays 20 12 20 12 13 11 13 11 7 9 7 9 2 1 2 1 2 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.29
30. 30. Merging two sorted arrays 20 12 20 12 20 12 13 11 13 11 13 11 7 9 7 9 7 9 2 1 2 1 2 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.30
31. 31. Merging two sorted arrays 20 12 20 12 20 12 13 11 13 11 13 11 7 9 7 9 7 9 2 1 2 1 2 7 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.31
32. 32. Merging two sorted arrays 20 12 20 12 20 12 20 12 13 11 13 11 13 11 13 11 7 9 7 9 7 9 9 2 1 2 1 2 7 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.32
33. 33. Merging two sorted arrays 20 12 20 12 20 12 20 12 13 11 13 11 13 11 13 11 7 9 7 9 7 9 9 2 1 2 1 2 7 9 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.33
34. 34. Merging two sorted arrays 20 12 20 12 20 12 20 12 20 12 13 11 13 11 13 11 13 11 13 11 7 9 7 9 7 9 9 2 1 2 1 2 7 9 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.34
35. 35. Merging two sorted arrays 20 12 20 12 20 12 20 12 20 12 13 11 13 11 13 11 13 11 13 11 7 9 7 9 7 9 9 2 1 2 1 2 7 9 11 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.35
36. 36. Merging two sorted arrays 20 12 20 12 20 12 20 12 20 12 20 12 13 11 13 11 13 11 13 11 13 11 13 7 9 7 9 7 9 9 2 1 2 1 2 7 9 11 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.36
37. 37. Merging two sorted arrays 20 12 20 12 20 12 20 12 20 12 20 12 13 11 13 11 13 11 13 11 13 11 13 7 9 7 9 7 9 9 2 1 2 1 2 7 9 11 12 Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.37
38. 38. Merging two sorted arrays 20 12 20 12 20 12 20 12 20 12 20 12 13 11 13 11 13 11 13 11 13 11 13 7 9 7 9 7 9 9 2 1 2 1 2 7 9 11 12 Time = Θ(n) to merge a total of n elements (linear time). Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.38
39. 39. Analyzing merge sort T(n) MERGE-SORT A[1 . . n] Θ(1) 1. If n = 1, done. 2T(n/2) 2. Recursively sort A[ 1 . . ⎡n/2⎤ ] Abuse and A[ ⎡n/2⎤+1 . . n ] . Θ(n) 3. “Merge” the 2 sorted lists Sloppiness: Should be T( ⎡n/2⎤ ) + T( ⎣n/2⎦ ) , but it turns out not to matter asymptotically. Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.39
40. 40. Recurrence for merge sort Θ(1) if n = 1; T(n) = 2T(n/2) + Θ(n) if n > 1. • We shall usually omit stating the base case when T(n) = Θ(1) for sufficiently small n, but only when it has no effect on the asymptotic solution to the recurrence. • CLRS and Lecture 2 provide several ways to find a good upper bound on T(n). Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.40
41. 41. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.41
42. 42. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. T(n) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.42
43. 43. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn T(n/2) T(n/2) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.43
44. 44. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn cn/2 cn/2 T(n/4) T(n/4) T(n/4) T(n/4) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.44
45. 45. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn cn/2 cn/2 cn/4 cn/4 cn/4 cn/4 … Θ(1) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.45
46. 46. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn cn/2 cn/2 h = lg n cn/4 cn/4 cn/4 cn/4 … Θ(1) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.46
47. 47. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn cn cn/2 cn/2 h = lg n cn/4 cn/4 cn/4 cn/4 … Θ(1) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.47
48. 48. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn cn cn/2 cn/2 cn h = lg n cn/4 cn/4 cn/4 cn/4 … Θ(1) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.48
49. 49. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn cn cn/2 cn/2 cn h = lg n cn/4 cn/4 cn/4 cn/4 cn … … Θ(1) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.49
50. 50. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn cn cn/2 cn/2 cn h = lg n cn/4 cn/4 cn/4 cn/4 cn … … Θ(1) #leaves = n Θ(n) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.50
51. 51. Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant. cn cn cn/2 cn/2 cn h = lg n cn/4 cn/4 cn/4 cn/4 cn … … Θ(1) #leaves = n Θ(n) Total = Θ(n lg n) Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.51
52. 52. Conclusions • Θ(n lg n) grows more slowly than Θ(n2). • Therefore, merge sort asymptotically beats insertion sort in the worst case. • In practice, merge sort beats insertion sort for n > 30 or so. • Go test it out for yourself! Copyright © 2001-5 Erik D. Demaine and Charles E. Leiserson September 7, 2005 Introduction to Algorithms L1.52
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Probability-and-statistics/730224: in a fifty question true false examination a student must achieve twenty five correct answers required to pass. 1 solutions Answer 446554 by richwmiller(9135) on 2013-03-25 09:10:00 (Show Source): You can put this solution on YOUR website!already asked
Quadratic_Equations/730217: what is the roots of the function 1 solutions Answer 446553 by richwmiller(9135) on 2013-03-25 09:06:36 (Show Source): You can put this solution on YOUR website!already asked
test/730159: In a lake, there is a patch of lily pads. Every day the patch doubles in size. If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half the lake?1 solutions Answer 446529 by richwmiller(9135) on 2013-03-25 01:04:27 (Show Source): You can put this solution on YOUR website!47 days it is half full on day 47 every day it doubles so on day 48 it is full
Polynomials-and-rational-expressions/730164: 1.8xy/18x^2y * 24xy^2/16y^3 2.-14a^2b^z/6b^3 / 21a/15ab 3. 2/x-2 - 2/x+2 - 4/x^3-4x1 solutions Answer 446527 by richwmiller(9135) on 2013-03-25 01:00:05 (Show Source): You can put this solution on YOUR website!Thank you Those are lovely expressions
Geometry_Word_Problems/730170: Write the equation of the line that passes through (-2,4) and is parallel to the graph of the line y=4x+5 I have tried but I can't figure It out. Please help1 solutions Answer 446526 by richwmiller(9135) on 2013-03-25 00:59:41 (Show Source): You can put this solution on YOUR website!what have you tried. this is easy y=mx+b slope is m plug in -2 for x and 4 for y and find b 4=-2(4)+b
Travel_Word_Problems/730174: Ann is riding on a bike course that is miles long. So far, she has ridden miles of the course. What percentage of the course has Ann ridden so far? 1 solutions Answer 446525 by richwmiller(9135) on 2013-03-25 00:54:37 (Show Source): You can put this solution on YOUR website!You left out the numbers
Circles/730178: How do you find this if both sides are squared Finding the radius and centre of a circle. 1.)(x-4)^2+(y-5)^2=161 solutions Answer 446524 by richwmiller(9135) on 2013-03-25 00:52:05 (Show Source): You can put this solution on YOUR website!centre (4,5) radius 4
Circles/730179: Hello just wondering if this type of question is possible to solve when finding the radius and centre of a circle?? x^2+5^2-14x+32y=01 solutions Answer 446523 by richwmiller(9135) on 2013-03-25 00:47:12 (Show Source): You can put this solution on YOUR website!It's a parabola not a circle did you really mean 5^2? put y^2 instead of 5^2 and it is a circle
Rectangles/730177: Among all rectangles having a perimeter of 19 m, find the dimensions of the one with the largest area.1 solutions Answer 446522 by richwmiller(9135) on 2013-03-25 00:43:47 (Show Source): You can put this solution on YOUR website!a square will have the largest area
Word_Problems_With_Coins/730089: Tulips sell for \$1.25 each and roses sell for 2.00\$ each. \$116 worth are sold. How many roses were sold if 3 times as many roses as tulips were sold? 1 solutions Answer 446458 by richwmiller(9135) on 2013-03-24 20:40:53 (Show Source): You can put this solution on YOUR website!r=3t, 1.25t+2r=116 r=48., t=16.
Numeric_Fractions/730037: jordan made 3/5 of a recipe that called for 2 1/3 cups sugar how much sugar was needed 1 solutions Answer 446450 by richwmiller(9135) on 2013-03-24 20:20:31 (Show Source): You can put this solution on YOUR website!2 1/3=7/3 7/3*3/5=7/5 =1 2/5 or 1.4 cups
Travel_Word_Problems/729976: She rode a bike for 5 miles then she walked for another 2 miles. she can ride her bike twice as fast as she can walk. The entire trip tool 2 hours . How fast can she walk1 solutions Answer 446447 by richwmiller(9135) on 2013-03-24 20:12:10 (Show Source): You can put this solution on YOUR website!time b+w=2, B*b=5, W*w=2, speed B=2W b = 10/9, B = 9/2, w = 8/9, W = 9/4
Miscellaneous_Word_Problems/730055: In the lab, Carlos has two solutions that contain alcohol and is mixing them with each other. He uses times as much Solution A as Solution B. Solution A is alcohol and Solution B is alcohol. How many milliliters of Solution A does he use, if the resulting mixture has milliliters of pure alcohol? 1 solutions Answer 446443 by richwmiller(9135) on 2013-03-24 20:00:33 (Show Source): You can put this solution on YOUR website!you seem to have left out the numbers
Polynomials-and-rational-expressions/730071: Divide (3b^2+2b-5) by (b-2) I keep getting 3b-4 3/b-2 for an answer. Am I writing the remainder wrong?1 solutions Answer 446442 by richwmiller(9135) on 2013-03-24 19:57:59 (Show Source): You can put this solution on YOUR website!You divided incorrectly (3b+8)with remainder 11/(b-2)
Complex_Numbers/730066: Hello,The problem looks like this... Solve the equation; find all solutions, including complex solutions: x^3 + 5x + 4 = 4(x^2 + 1) I've tried to take the right side of the equation and distribute the 4, then subtracting the individual terms from the left side leaving me with x^3 - 4x^2 + 5x = 0 Then I tried to factor out x from the right side x(x^2 -4^x + 5) = 0 I cant do much more because of the sign in front of the 4 being (-) and the sign in front of 5 being +. So I tried using imaginary numbers on the right side leaving me with.... x^3 + 5x +4 = (2x + 2i)(2x - 2i) and now im stuck confused and irritated. I don't know if I'm doing any of the step work correctly. 1 solutions Answer 446438 by richwmiller(9135) on 2013-03-24 19:47:30 (Show Source): You can put this solution on YOUR website!x(x^2 -4^x + 5) = 0 x = 2+i x = 2-i x=0
Word_Problems_With_Coins/729933: randy has a jar containing 51 coins, all quarters and nickels. totaling \$5.55. how many quarters are in the jar?1 solutions Answer 446435 by richwmiller(9135) on 2013-03-24 19:38:15 (Show Source): You can put this solution on YOUR website!q+n=51 n=51-q 25q+5n=555 5q+n=111 5q+51-q=111 4q=60 q=15 n=36
Graphs/730063: I NEED HELP PLEASE W/ THIS QUESTION !!! ~ Given the line y= 2/5x + 6, find the equation of the line passing through the point (0,6) [ added : and perpendicular to the given equation]1 solutions Answer 446430 by richwmiller(9135) on 2013-03-24 19:30:25 (Show Source): You can put this solution on YOUR website!It is a poorly written question. There is an infinite number of lines through (0,6) What is the relationship to the given equation? parallel? perpendicular? something else? perpendicular means the slopes multiplied equal -1 the slope of the first equation is 2/5 so we need -5/2 y=mx+b 6=5/2*0+b 6=b y=-(5/2)*x+6
Geometry_Word_Problems/730056: what is the area of the shape left when a circle of radius 3 inches is cut from the middle of a square with a 10 inch diameter? 1 solutions Answer 446429 by richwmiller(9135) on 2013-03-24 19:27:11 (Show Source): You can put this solution on YOUR website!squares don't have diameters
Travel_Word_Problems/729792: If you drove 35 miles in an hour how many feet would you drive a minute?1 solutions Answer 446242 by richwmiller(9135) on 2013-03-24 03:03:00 (Show Source): You can put this solution on YOUR website!multiply by 5280 ft in a mile an multiply by 60 minutes in an hour
Travel_Word_Problems/729169: A family drove 45 miles and they hiked 2 miles to their campsite. The trip took 1 hour and 15 minutes. If they drove 56 mph faster than they hiked, how fast did they drive?1 solutions Answer 445749 by richwmiller(9135) on 2013-03-21 19:03:49 (Show Source): You can put this solution on YOUR website!x+y=1.25, r*x=56, (r-56)*y=2 r=61.8132, x=0.905955, y=0.344045 I don't like these answers
Word_Problems_With_Coins/729159: Penny has a total of 18 coins made up of nickels and quarters. When totaled they add up to \$1.90. There are three more nickels than twice the number of quarters. I am having a total blank please help. 1 solutions Answer 445747 by richwmiller(9135) on 2013-03-21 18:47:10 (Show Source): You can put this solution on YOUR website!n+q=18, 5n+25q=190 we don't need the last part. n=2q+3 q=5 n=13
Linear-equations/729158: How do you graph the linear equation y=-2 using a table of values 1 solutions Answer 445746 by richwmiller(9135) on 2013-03-21 18:42:32 (Show Source): You can put this solution on YOUR website!it is a straight horizontal line below the x axis plot any point where y=-2 follow the straight line
Surface-area/728967: when i make a rectangle it is of 64 cm area but when i cut it in 4 parts of different shapes its area is of 65 cm how1 solutions Answer 445661 by richwmiller(9135) on 2013-03-21 12:34:00 (Show Source): You can put this solution on YOUR website!64 cm is a linear dimension is is not a measurement of area.
Probability-and-statistics/728970: A statistics professor decides to give a final exam that has only 12 true-false questions. Suppose a student taking that exam guesses on every question. Let X represent the number of questions the student gets correct on the exam. Explain why X can be considered a binomial random variable. If the student must answer at least 70% of the questions correctly to pass the exam, what is the minimum number of questions the student must answer correctly to pass, and what is the associated probability?1 solutions Answer 445658 by richwmiller(9135) on 2013-03-21 12:24:25 (Show Source): You can put this solution on YOUR website!70% of 12 is 8.4 so he must get at least 9 correct which is really 75%.
Average/728977: mattis average on her first 3 test was 88 . What score does she need on her fourth test to make her average exactly 90 ?1 solutions Answer 445654 by richwmiller(9135) on 2013-03-21 12:12:50 (Show Source): You can put this solution on YOUR website!(3*88+x)/4=90 x=96 another way 88 misses 90 by 2 3*2=6 90+6=96 another way 4*90=360 3*88=264 360-264=96
logarithm/728974: How do you solve 12^(2x-8)=151 solutions Answer 445653 by richwmiller(9135) on 2013-03-21 12:02:07 (Show Source): You can put this solution on YOUR website!factor out 2 in the exponent 144^(x-4) = 15 144^y=15 x = 4.5449
Linear-equations/728599: Find the intercepts of the equation 2x+6y=11. Write the answers as ordered pairs. All work must be shown.1 solutions Answer 445514 by richwmiller(9135) on 2013-03-20 15:41:27 (Show Source): You can put this solution on YOUR website!let x=0 find y let y=0 find x
Polynomials-and-rational-expressions/728569: I dont understand how to solve this problem: Original Equation: [(x)/(x-3)] - [(2)/(x+8)]= [(78)/(x^2+5x-24)] What I did: Multiplied left side of the equation to get common denominator: (x+8)[(x)/(x-3)] - (x-3)[(2)/(x+8)] I get this when simplified: [x^2+6x+6/(x^2+5x-24)]=[(78/(x^2+5x-24)] I mult each side by denom: so... x^2+6x+6=78 subtract and simplify: x^2+6x-72=0 But after this step I find that I can not factor it. If i put no solution in as the answers, it is wrong. When did I go wrong?1 solutions Answer 445513 by richwmiller(9135) on 2013-03-20 15:38:46 (Show Source): You can put this solution on YOUR website!from [x^2+6x+6/(x^2+5x-24)]=[(78/(x^2+5x-24)] we get (x^2+6 x+6)/((x-3) (x+8)) = 78/((x-3) (x+8)) (x^2+6 x-72)/((x-3) (x+8)) = 0 ((x-6) (x+12))/((x-3) (x+8)) = 0 x=6 x=-12
Rate-of-work-word-problems/728586: Two landscape workers can complete a certain landscaping job in a total of 20 hours. Working alone, one can do the same job in 2 hours less time than it takes the other. How long would it take the faster worker to do the job if the slower worker were unable to work? (Use calculator to find approx. value, but how intended computation.)1 solutions Answer 445510 by richwmiller(9135) on 2013-03-20 15:28:33 (Show Source): You can put this solution on YOUR website!20x+20/(x+2)=1 x= about 39 and 41 20/39+20/41=1
Percentage-and-ratio-word-problems/728464: 12 ounces of concentrate mix with 3 1/2 cans of water, each can holds 12 ounces of water. What is the greatest number of 8 ounce cups that can be served?1 solutions Answer 445424 by richwmiller(9135) on 2013-03-20 10:06:38 (Show Source): You can put this solution on YOUR website!(12+3.5*12)/8 54/8 6 full 8 ounce cups
Evaluation_Word_Problems/728450: Kerry asked the bank teller to cash a \$390 check using \$20 bills and \$50 bills. If the teller gave her a total of 15 bills, how many of each type of bill did she receive? 1 solutions Answer 445421 by richwmiller(9135) on 2013-03-20 09:58:50 (Show Source): You can put this solution on YOUR website!t+f=15, 20t+50f=390 trail and error method there have to be at least 2 20's and one 50 to make the 90 that's three bills we need 12 more we can't have just 2 fifty's so try 3 50's 3 50's would mean 12 20's 150 and 240 work
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```Power System Planning and Reliability
Divya M
Dept. of Electrical Engineering
FCRIT, Vashi
Power system planning
Definition
A process in which the aim is to decide on new as well as
Elements can be:
•
•
•
•
•
PSPR
Generation facilities
Substations
Transmission lines and/or cables
Capacitors/Reactors
Etc.
Lecture-1
(Seifi & Sepasian)
Power system planning
Decision should be
• Where to allocate the element (for instance, the sending and
receiving end of a line),
• When to install the element (for instance, 2020),
• What to select, in terms of the element specifications (for
instance, number of bundles and conductor type).
PSPR
Lecture-1
(Seifi & Sepasian)
The first crucial step for any planning study
Forecasting refers to the prediction of the load behaviour for
the future
Words such as, demand and consumption are also used instead
Energy (MWh, kWh) and power (MW,kW) are the two basic
By load, we mean the power.
Demand forecast
• To determine capacity of generation, transmission and
distribution required
Energy forecast
• To determine the type of generation facilities required
PSPR
Lecture-1
(Seifi & Sepasian)
Variations in load on a power station from time to time
•
•
•
•
•
PSPR
Variation of load during different time
Total no. of units generated
Maximum demand
Average load on a power station
Lecture-1
(Pabla)
PSPR
Lecture-1
www.nationalgrid.com
Demand factor
Diversity factor
Utilization factor
Power factor
Demand
factor
Max . demand
Connected
Avg . demand
Max . demand
Diversity
factor
Sum of individual
max . demands
Max . demand of power station
Utilisatio n factor
Max . demand on power station
Rated capacity of power station
• Higher the values of load factor and diversity factor, lower will be
the overall cost per unit generated.
• Higher the diversity factor of the loads, the fixed charges due to
capital investment will be reduced.
PSPR
Lecture-1
(Pabla)
Domestic
• Demand factor: 70-100%
• Diversity factor: 1.2-1.3
Commercial
• Demand factor: 90-100%
• Diversity factor: 1.1-1.2
Industrial
• Small-scale: 0-20 kW
• Medium-scale: 20-100 kW
• Large-scale: 100 kW and above
– Demand factor: 70-80%
PSPR
Lecture-1
(Pabla)
Agricultural
• Demand factor: 90-100%
• Diversity factor: 1-1.5
• Street lights, bulk supplies, traction etc.
Commercial and agricultural loads are characterized by
seasonal variations.
dependent.
PSPR
Lecture-1
(Pabla)
Numerical
A power plant supplies the following loads with maximum demand
as below:
Max. demand (MW)
Industries
100
Domestic
15
Commercial
12
Agriculture
20
The maximum demand on the power station is 110 MW. The total units
generated in the year is 350 GWh.
Calculate:
• Diversity factor
PSPR
Lecture-1
(Pabla)
Reasons for the growth of peak demand and energy usage within an
electric utility system:
– Load will increase if more customers are buying the utility's product.
– New construction and a net population in-migration to the area will add
new customers and increase peak load.
• New uses of electricity
– Existing customers may add new appliances (replacing gas heaters with
electric) or replace existing equipment with improved devices that require
more power.
– With every customer buying more electricity, the peak load and annual
energy sales will most likely increase.
PSPR
Lecture-2
(Willis)
Load growth caused by new customers who are locating in previously
vacant areas.
• Such growth leads to new construction and hence draws the planner's
attention.
Changes in usage among existing customers
• Increase in per capita consumption is spread widely over areas with
existing facilities already in place, and the growth rate is slow.
• Difficult type of growth to accommodate, because the planner has
facilities in place that must be rearranged, reinforced, and upgraded.
This presents a very difficult planning problem.
PSPR
Lecture-2
(Willis)
Time factors such as:
• Hours of the day (day/night)
PSPR
• Day of the week (week day/weekend)
• Time of the year (season)
Weather conditions (temperature and humidity)
Class of customers (residential, commercial, industrial, agricultural,
public, etc.)
Special events (TV programmes, public holidays, etc.)
Population
Economic indicators (per capita income, Gross National Product
(GNP), Gross Domestic Product (GDP), etc.)
Trends in using new technologies
Electricity price
Lecture-2
(Pabla)
Forecasting methodology
Forecasting: systematic procedure for quantitatively defining future
Classification depending on the time period:
• Short term
• Intermediate
• Long term
Forecast will imply an intermediate-range forecast
• Planning for the addition of new generation, transmission and
distribution facilities must begin 4-10 years in advance of the actual inservice date.
PSPR
Lecture-2
(Sullivan)
Forecasting techniques
Three broad categories based on:
• Extrapolation
– Time series method
– Use historical data as the basis of estimating future outcomes.
• Correlation
– Econometric forecasting method
– identify the underlying factors that might influence the variable
that is being forecast.
• Combination of both
PSPR
Lecture-2
(Sullivan)
Extrapolation
Based on curve fitting to previous data available.
With the trend curve obtained from curve fitted load can be
forecasted at any future point.
Simple method and reliable in some cases.
Deterministic extrapolation:
• Errors in data available and errors in curve fitting are not accounted.
Probabilistic extrapolation
• Accuracy of the forecast available is tested using statistical measures
such as mean and variance.
PSPR
Lecture-2
(Sullivan)
Extrapolation
Standard analytical functions used in trend curve fitting are:
• Straight line:
• Parabola:
• s curve:
y a bx
y a bx cx
2
y a bx cx dx
2
• Exponential:
• Gompertz:
y ce
y ln
1
3
dx
( a ce
dx
)
Best trend curve is obtained using regression analysis.
Best estimate may be obtained using equation of the best trend
curve.
PSPR
Lecture-2
(Sullivan)
Correlation
Relates system loads to various demographic and economic factors.
growth and other measurable factors.
Forecasting demographic and economic factors is a difficult task.
No forecasting method is effective in all situations.
Designer must have good judgment and experience to make a
forecasting method effective.
PSPR
Lecture-2
(Sullivan)
Impact of weather in load forecasting
Weather causes variations in domestic load, public lighting,
Main weather variables that affect the power consumption are:
•
•
•
•
Temperature
Cloud cover
Visibility
precipitation
First two factors affect the heating/cooling loads
PSPR
Lecture-2
(Pabla)
Impact of weather in load forecasting
Average temperature is the most significant weather dependent
Temperature and load are not linearly related.
Non-linearity is further complicated by the influence of
• Humidity
• Extended periods of extreme heat or cold spells
In load forecast models proper temperature ranges and
representative average temperatures which cover all regions of the
area served by the electric utility should be selected.
PSPR
Lecture-2
(Pabla)
Impact of weather in load forecasting
Cloud cover is measured in terms of:
•
•
•
•
height of cloud cover
Thickness
Cloud amount
Time of occurrence and duration before crossing over a population
area.
Visibility measurements are made in terms of meters/kilometers
with fog indication.
To determine impact of weather variables on load demand, it is
essential to analyze data concerning different weather variables
through the cross-section of area served by utility and calculate
weighted averages for incorporation in the modeling.
PSPR
Lecture-2
(Pabla)
Energy forecasting
To arrive at a total energy forecast, the forecasts for residential,
commercial and industrial customers are forecasted separately and
then combined.
PSPR
Lecture-3
(Sullivan)
Residential sales forecast
Population method
• Residential energy requirements are dependent on:
– Residential customers
– Population per customer
– Per capita energy consumption
• To forecast these factors:
– Simple curve fitting
– Regression analysis
• Multiplying the three factors gives the forecast of residential sales.
PSPR
Lecture-3
(Sullivan)
Residential sales forecast
Synthetic method
• Detailed look at each customer
• Major factors are:
– Saturation level of major appliances
– Average energy consumption per appliance
– Residential customers
• Forecast these factors using extrapolation.
• Multiplying the three factors gives the forecast of residential sales.
PSPR
Lecture-3
(Sullivan)
Commercial sales forecast
Commercial establishments are service oriented.
Growth patterns are related closely to growth patterns in residential
sales.
Method 1:
• Extrapolate historical commercial sales which is frequently available.
Method 2:
• Extrapolate the ratio of commercial to residential sales into the future.
• Multiply this forecast by residential sales forecast.
PSPR
Lecture-3
(Sullivan)
Industrial sales forecast
Industrial sales are very closely tied to the overall economy.
Economy is unpredictable over selected periods
Method 1:
• Multiply forecasted production levels by forecasted energy
consumption per unit of production.
Method 2:
• Multiply forecasted number of industrial workers by forecasted energy
consumption per worker.
PSPR
Lecture-3
(Sullivan)
Extrapolate historical demand data
• Weather conditions can be included
Basic approach for weekly peak demand forecast is:
1.
2.
3.
4.
5.
6.
PSPR
Separate historical weather-sensitive and non-weather sensitive
components of weekly peak demand using weather load model.
Forecast mean and variance of non-weather-sensitive component of
demand.
Extrapolate weather load model and forecast mean and variance of
weather sensitive component.
Determine mean, variance and density function of total weekly
forecast.
Calculate density function of monthly/annual forecast.
Lecture-3
(Sullivan)
Assume that the seasonal variations of the peak demand are primarily due
to weather.
Otherwise, before step-3 can be undertaken, any additional seasonal
variation remaining after weather-sensitive variations must be removed
To use the proposed forecasting method, a data base of at least 12 years is
recommended.
To develop weather load models daily peaks and coincident weather
variable values are needed.
PSPR
Lecture-3
(Sullivan)
Plot a scatter diagram of daily peaks versus an appropriate weather
variables.
• Dry-bulb temperature and humidity
• Using curve fitting three line segments can be defined in the example
w k s (T T s )
if T T s
k w (T T w )
if T T w
0
if T w T T s
Parameters of the model:
• Slopes: ks and kw
• Threshold temperatures: Ts
and Tw
PSPR
Lecture-3
(Sullivan)
Separating weather-sensitive and nonweather sensitive components
From the weather load model
• Weather-sensitive (WS) component of weekly peak load demand data is
calculated from the weekly peak coincident dry-bulb temperatures.
• Non-weather-sensitive (NWS) component of peak demand is obtained by
subtracting the first component from historical data.
• NWS component is used in step-3, of basic approach for weekly peak demand
forecast , to forecast the mean and variance of the NWS component of future
weekly peak demands.
PSPR
Lecture-3
(Sullivan)
PSPR
Lecture-4
(Pabla)
Total forecast
PSPR
Lecture-4
(Sullivan)
Annual peak demand forecast
PSPR
Lecture-4
(Sullivan)
Monthly peak demand forecast
PSPR
Lecture-4
(Sullivan)
References
“Electric Power System Planning: Issues, Algorithms and Solutions”,
Berlin Heidelberg, 2011.
“Electrical Power Systems Planning”, A.S. Pabla, Macmillan India Ltd.,
1988.
“Power System Planning”, R.L. Sullivan, McGraw-Hill International
“Power Distribution Planning Reference Book”, H. Lee Willis, Marcel
Dekker Inc.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Surface Area and Volume of Pyramids
## Surface area and volume of solids with a base and lateral faces that meet at a common vertex.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
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## Surface Area of Pyramids
Learn how to calculate the surface area of a pyramid.
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## Surface Area of Pyramids
Learn to find the surface area of pyramids.
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## Surface Area and Volume of Pyramids
Surface Area and Volume of Pyramids Interactive
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## Surface Area and Volume of Pyramids
Surface Area and Volume of Pyramids Interactive
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## Pyramids Principles
This video gives more detail about the mathematical principles presented in Pyramids.
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## Pyramids Examples
This video shows how to work step-by-step through one or more of the examples in Pyramids.
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## Surface Area and Volume of Pyramids Practice
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• Critical Thinking
## Pyramids Discussion Questions
A list of student-submitted discussion questions for Pyramids.
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To activate prior knowledge, to generate questions about a given topic, and to organize knowledge using a KWL Chart.
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To activate prior knowledge, to generate questions about a given topic, and to organize knowledge using a KWL Chart.
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## Pyramids Four Square Concept Matrix
Summarize the main idea of a reading, create visual aids, and come up with new questions using a Four Square Concept Matrix.
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## Find Surface Area of Pyramids Using Formulas Post Read
Develop understanding of concepts by studying them in a relational manner. Analyze and refine the concept by summarizing the main idea, creating visual aids, and generating questions and comments using a Four Square Concept Matrix.
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## Surface Area of Pyramids
Summarize the main idea of a reading, create visual aids, and come up with new questions using a Four Square Concept Matrix.
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• Real World Application
## Burying a Pharaoh
Discover the geometry behind the tombs of Egyptian royalty.
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• Real World Application
## Those Who Live in Glass Pyramids
Learn about one of the most famous modern pyramids.
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• Real World Application
## Triangular Living
Learn about the pyramid-shaped skyscraper being built in France.
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• Study Guide
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1. ## Radii of convergence of a power series
How do I compute the radii of convergence of the power series Σanx^n
(from n=0 to infinity)
with coefficients an=n and an=1/n
2. Originally Posted by TexasGirl
How do I compute the radii of convergence of the power series Σanx^n
(from n=0 to infinity)
with coefficients an=n and an=1/n
Use the ratio test for the power series. With $\displaystyle a_k=k$ thus, $\displaystyle \lim_{k\rightarrow \infty}|\frac{k+1}{k}|=1$ Thus, the radius of convergence is the the reciprocal of that thus $\displaystyle 1/1=1$. However, the interval of convergence is $\displaystyle -1<x<1$. For $\displaystyle x=-1,1$ this power series diverges.
Now for the second problem, $\displaystyle a_k=1/k, k>1$ use the ratio test for power series again to get $\displaystyle \lim_{k\rightarrow \infty}|\frac{k}{k+1}|=1$ Thus, the radius of converges is 1, thus for $\displaystyle -1<x<1$ converges absolutely. Checking the endpoint (because the ratio test is inconclusive for when its limit is one), we have
$\displaystyle \sum^\infty_{k=1} (-1)^k 1/k$ but this is the alternating-harmonic series thus it converges. For the second possibility we have that $\displaystyle \sum^\infty_{k=1} 1/k$ but this is the harmonic series thus it diverges. Thus, the interval of convergence for the second power series is $\displaystyle -1\leq x<1$
3. ## One More Radius of Convergence Question...
Building from the same power series, if I take q, a nonzero element of C, and put it into the series so that I have the sum of an(q^n)(x^n), can I still use the ratio test in order to find the radius of convergence? Is an the coefficient?
4. Originally Posted by TexasGirl
Building from the same power series, if I take q, a nonzero element of C, and put it into the series so that I have the sum of an(q^n)(x^n), can I still use the ratio test in order to find the radius of convergence? Is an the coefficient?
I do not understand what you are asking?
5. Here is the question I have to answer:
Assume that the power series ∑an(x^n), where n=0 to infinity, has radius of convergence p, where p is a nonnegative real number or stands for the symbol ∞. Let q be an element of C, q≠0. Compute the radius of convergence of ∑an(q^n)(x^n) (n=0 to infinity).
6. As I understand it, its radius of convergence would be $\displaystyle p/q$ Explanation:
If $\displaystyle \sum^{\infty}_{k=1}a_kx^k$ has radius of convergence of $\displaystyle p$ then by the ratio test for power series $\displaystyle \lim_{k\rightarrow \infty}|\frac{a_{k+1}}{a_k}|=1/p$ because as I said it is the reciprocal of the limit. Thus, the new infinite series given by $\displaystyle \sum^{\infty}_{k=1}a_kq^kx^k$ then by the ratio test $\displaystyle \lim_{k\rightarrow \infty}|\frac{a_{k+1}q^{k+1}}{a_kq^k}|=\frac{q}{p}$ thus, the radius of convergent is the reciprocal thus $\displaystyle \frac{p}{q}$
7. ## thanks again...
much appreciated...
8. You are welcome.
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# Formulas for Problem Solving
## The d = rt formula
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Practice Formulas for Problem Solving
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MPH to KPH
Teacher Contributed
## Real World Applications – Algebra I
### Topic
Converting from MPH to KPH
### Student Exploration
There are major differences driving in the United States and driving in the United Kingdom (and possibly other countries, too!). For example, people drive on different sides of the road. In the US, people usually drive on the right side of the road. In the UK, people drive on the left side of the road.
Speed limit is also different. The units are different. In the US, speed limit signs are given in miles per hour. See below.
In the UK, these speed limit signs mean are talking about kilometers per hour. See below
Even though these two signs have the same number, they don’t represent the same thing. If you’re from the United States, would you want to go just as fast or slower than what the sign says? Which speed limit is higher?
We can find out by using a formula to convert the number of kilometers to miles. We know that \begin{align*}1\ mile = 1.609\ kilometers\end{align*}. With this information, we can multiply any number of miles by 1.609 to find its equivalence in kilometers.
For the example above, to find out how many kilometers is equal to 50 miles, we use the formula. Let \begin{align*}k\end{align*} represent the number of kilometers, and \begin{align*}m\end{align*} represent the number of miles we’re going to convert.
So, \begin{align*}k = 1.609 m\end{align*}
\begin{align*}& k = 1.609 ^* 50\\ & k = 80.45\ kilometers\end{align*}
So, if we saw the “50kph” sign in the UK, we probably don’t want to drive 50 miles per hour. We’d probably get a speeding ticket!
### Extension Investigation
See if you can figure out a formula to convert the number of kilometers to miles, given the information above. What is the conversion factor and why?
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# Wolfram Alpha's Mysterious Trig Abilities…
This is a very straight-forward question:
I'm trying to simplify $$\sin(2\pi t +\pi/4) + \sin(2\pi t -\pi/4)$$ and failing at it:
$\sin(2\pi t +\pi/4) + \sin(2\pi t -\pi/4)$
$2\sin(2\pi t)\cos(\pi/2)$ by sum $\to$ product = ZERO
Wolfram alpha gives $\sqrt{2}\sin(2\pi t)$, but gives no explanation as to how.
What am I doing wrong?
-
What does the -20t mean? – Ted Feb 27 '13 at 6:54
@Ted forgot to omit that irrelevant part of the question. Fixed – Griffin Feb 27 '13 at 20:15
I don't know why this got so many votes... It was just a small error in arithmetic. – Griffin Feb 27 '13 at 20:25
You forgot to divide $\pi/2$ by 2. Doing this gives $\cos(\pi/4) = 1/\sqrt{2}$ instead of $0$, which gives the right answer. – Joe Z. Feb 27 '13 at 21:11
## 3 Answers
If you draw the situation on the unit circle, then $\sin\varphi$ is the x-coordinate of the point corresponding to the angle $\varphi$. The points $2\pi t-\pi/4$ and $2\pi t+\pi/4$ have right angle between them and $2\pi t$ is exactly in the middle.
So from the right triangle below you see that the sum of the vectors corresponding to $2\pi t-\pi/4$ and $2\pi t+\pi 4$ has the same direction as the vector corresponding to $2\pi t$ and the length is $\sqrt2$. The sum $\sin(2\pi t-\pi/4)+\sin(2\pi t+\pi/4)$ is the x-coordinate of this vector, and it is equal to $\sqrt2\sin 2\pi t.$
The same argument gives $\cos(2\pi t-\pi/4)+\cos(2\pi t+\pi/4)=\sqrt2\cos 2\pi t.$
More-or-less the same idea can be rewritten using complex numbers if we use Euler's forumula $e^{i\varphi}=\cos\varphi+i\sin\varphi$.
We have $$e^{\alpha+\pi/4}+e^{\alpha-\pi/4}=e^\alpha(e^{\pi/4}+e^{-\pi/4})=e^\alpha2\cos\frac\pi4=e^\alpha\sqrt2.$$ The real part gives $\cos(\alpha+\pi/4)+\cos(\alpha-\pi/4)=\sqrt2\cos\alpha$ and the imaginary part gives $\sin(\alpha+\pi/4)+\sin(\alpha-\pi/4)=\sqrt2\sin\alpha$.
Complex numbers are quite often useful for remembering/proving trigonometric identities.
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v. v. thorough! – bharal Feb 27 '13 at 14:00
I like this because it's so visual. Accepted! – Griffin Feb 27 '13 at 20:29
Hint: $$\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$$
-
why do this instead of using the sum->product formula? – Griffin Feb 27 '13 at 20:26
If you want to use the sum-product formula, it should be the right one. We have $$\sin x+\sin y=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right).$$ But I think it is better to go back to the addition/subtraction laws for sine, as described by Zev Chonoles.
We have $$\sin(a+b)=\sin a\cos b+\cos a\sin b,$$ and its very close relative $$\sin(a-b)=\sin a \cos b -\cos a\sin b.$$ Let $a=2\pi t$ and $b=\frac{\pi}{4}$. The sine and cosine of $\frac{\pi}{4}$ are both $\frac{1}{\sqrt{2}}$.
Remark: If one will be doing a certain type of calculation very often, it is useful to commit to memory any relevant formulas. However, it is I think best to know only a quite small number of trigonometric facts, and reconstruct anything else one may need.
-
We keep going back and forth :) I just deleted my answer to simplify everything, and besides yours is great as always. – Zev Chonoles Feb 27 '13 at 7:12
@Zev Chonoles: I don't see the point of deleting. If you undelete, I can upvote. There is much to be said for a minimalist nudge in the right direction. – André Nicolas Feb 27 '13 at 7:15
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# Universal Misconceptions #3: Redshifts
We’re talking, once again, about common questions and misconceptions about the expanding universe. If you haven’t already done so, take a look at my posts on the fact that time isn’t expanding and the fact the there’s no center to the expansion.
Today I’d like to talk about redshift. In the 1920’s, Edwin Hubble noticed that the more distant a galaxy was from our own, the more the spectral lines of that galaxy were redshifted. All this means is that rather than measure a spectral feature at, say, a wavelength of 600nm, you might measure it at 620nm. Within a particular galaxy, every single line was shifted by the same fractional amount:
$zequiv frac{Delta lambda}{lambda}$
with $z$ called the “redshift”, $lambda$ is the wavelength that the light should have under laboratory conditions, and $Delta lambda$ is the difference in observed wavelength from the laboratory wavelength. Hubble found that for relatively nearby galaxies, redshift and distance are just linearly proportional to one another. Further away, the relationship gets a bit more complicated mathematically, in part because there are so many possible types of distance.
Under normal conditions, we usually think of redshifts as being caused by the relative motion of the body emitting the photon and the body absorbing it: a Doppler shift. This, incidentally, is how radar detectors work in speed traps, or, using sound waves, why you hear a shift in pitch when a firetruck passes by. One of the most common ideas, and one that even shows up in some introductory physics classes, is that the light from distant galaxies is caused by a Doppler shift. It’s not*.
What’s really going on is that as a photon travels from the source to our telescopes on earth, the universe expands underneath it. If the universe doubles in size, so does the wavelength of the photon.
Here’s the balloon analogy to help makes things clearer:
In other words, what redshift really tells us is not how “fast” a galaxy is moving away from us, but rather, how big the universe is now compared to how big it was when the light was emitted. This picture has a number of implications:
1. Photons lose energy as they travel over cosmological distances. For light, long wavelength means low energy. Consider the light coming to us from the cosmic microwave background. It originates (around 380,000 years after the big bang) at a temperature of about 3000K, comparable to the surface of a relatively cool star. After subsequent expansion of the universe by a factor of 1200 or so, the temperature is now only about 2.7K.
Since photons lose energy as the universe expands, but massive particles don’t lose mass, at some point in the past, there was more energy in the form of photons than in ordinary and dark matter combined. This was back at a redshift of z=3900 — when the universe was about 1/3900th the size it is now.
2. A redshift doesn’t tell you anything about the expansion right now. For a distant galaxy, you only learn about the relative size of the universe when the photon was emitted to today. The universe could be accelerating or decelerating or even stopped at this moment, and you don’t get that directly form the redshift. What you need are lots of redshifts in order to figure out the history of the expansion, and thus the rate of change.
3. If light travels through a bunch of hydrogen clouds on the way from a distant galaxy to us, the galaxy and each of the clouds will each be at a different redshift. The galaxy will be most redshifted (since it is most distant), and so on. The clouds will then absorb only the particular wavelengths of light that correspond to spectral lines. However, since for each cloud there’s a different redshift, this creates a “forest” of absorption:
I sometimes get followup questions about how we know that the redshift really does correspond to cosmological expansion.
That’s a fair question. Here is a partial list of things which are 100% consistent with the view of cosmological redshifts, but would have to be utterly discarded if it were wrong:
• General relativity (since GR predicts a dynamically expanding or contracting universe)
• Gravitational lensing
• Distance estimates from nearby galaxies
• Any and all interpretation of the Cosmic Microwave Background
• The Big Bang model
But let’s suppose you’re one of those people who never liked the Big Bang model anyway and think you’re being subversive by challenging it. (You’re not, by the way.) Before throwing out a model that matches theory and observation perfectly, you’re required to come up with a model that does at least as well.
Good luck with that.
-Dave
* Well, technically it is, but only a little bit. Throughout this series of posts, I’ve been ignoring something called “peculiar motion.” Galaxies really do move around their local environment, but at a speed of only a few hundred kilometers per second. The peculiar motion causes a Doppler shift on top of the cosmological redshift.
This entry was posted in Uncategorized and tagged , . Bookmark the permalink.
### 9 Responses to Universal Misconceptions #3: Redshifts
1. Tereasa Brainerd says:
So, Dave, when the photons lose energy as the universe expands, where does that energy go?
• dave says:
But for those of you who WEREN’T my cosmology professor/senior research adviser, the answer is that radiation has pressure, which means that it does work as the universe expands. That work has to come from somewhere, and it comes from the internal energy of the photon fluid.
This is the PdV contribution to internal energy.
Note that this is the opposite of what happens with dark energy, which has a negative pressure (tension). In that case, even though you might expect the dark energy to get more diffuse, the extra work done on the dark energy is positive, keeping the overall density of the stuff positive while the universe expands. It also accelerates the expansion.
2. Tereasa Brainerd says:
Well done! You’re doing a great job here. Not to mention I really enjoy being able to say that I knew “way back when”.
• dave says:
Hurray! I passed!
3. Hasanito says:
This could be a silly point but, the figure for the balloon analogy above, seems to show the wavelength to be the same from smaller universe to the larger ones. But you also saying that “if universe doubles in size, so the wavelength”.
The wavelength does seem to get larger but still the same wavelength!
• dave says:
How so? The wavelength scales up with the universe. It stays the same in “comoving length” (based on the gridlines on the balloon), but gets longer in “proper length” (which is what you’d measure using a ruler).
4. Pingback: From Non-being to Being
5. Li Kong says:
The following are the explanations why the discovery of Edwin Hubble does not provide a good evidence that our universe would be expanding currently:
a)Despite many red shifts through telescope from astronomers, it does not provide the proof that this universe could be expanding for the following reasons:
1) The possibility that our universe could be very huge that it would take more than trillion of years to reach the opposite end of the unverse (sphere). The assumption is based upon the following factors: This universe is assumed to be as a shape of sphere with external boundary and all galaxies are assumed to move within the boundary of this universe.
Let’s imagine you stand at one end of the sphere (the universe) to have a full view of all the surrounding movement of galaxies. As all the galaxies were advancing at a high speed from your end to the edge of the sphere that is right opposite from you that form a half complete round, you certainly would visualize that all the galaxies are advancing as if that they are leaving away from you since their movement in speed is a few time faster than your galaxy. As this universe is very huge so much so that it would take a very long time, let’s say, more than a trillion years to reach the point that is right opposite to the point so as to make a half complete turn of this universe. Despite many galaxies have been moving towards the point that is right opposite to the point where you are viewing through telescope, the result would turn up to be many red shifts to have appeared in the universe. As universe is too huge for galaxies to travel from one end to another and only a few have completed a half turn to move than to the starting point of the sphere where you are to the edge of the sphere that is right opposite from you, it turns up that they are many red shifts than blue shifts.
2) The second possibility is that many galaxies might have advanced faster than us and yet many galaxies might have made a complete half turn within the sphere (the universe) and yet the galaxies might not as what we think that they would keep on rotating themselves in a circle. Instead, they might not return to the previous track where they have passed through. These could result that they do not turn back to us.
3) The third possibility is that all the clusters of galaxies could be advancing in the same place and same direction just that most of the galaxies are advancing faster than us as if that their galaxies are moving further away from us. As we are in this tiny world and cannot have the full sight of this universe, we could not reject this possibility since it might be so without our full view of this universe since the astronomers just looked at the sky with a telescope that comes to their conclusion without viewing the universe as a whole. No matter how advance is the technology, it could never be possible to build an advice that could capture the whole view of universe from one end (the earth).
4) The fourth possibility is that majoirty of the galaxies might have made a full complete turn in this universe within the boundary of the universe in many years ago, such as, more than a few thousand years ago. Or in other words, there might be a time in the past in which there were many blue shifts than red. What the astonomers that have seen right now with many red shifts do not reflect the universe might be expanding since there might be a period of times in many years ago that almost all the galaxies have made a complete full turn and it turns up that many galaxies have turned up to be red shifts currently. Or in other words, it would take many years later, such as more than a few thousand years later, in order to have many blue shifts instead of red shifts at that time.
5) The fifth possibility is that universe was created in infinity and that all galaxies are advancing ever since the past. If that is so, it is erroneous to use many red shifts as discovered by astronomers to conclude that the universe is expanding.
There might be other possibilities that you could think of why there are more red shifts than blue shifts and yet it does not come to the conclusion that the universe is expanding. As there are many alternative possibilities, to jump into conclusion that the universe is expanding through many red shifts being discovered is rather a little speculation.
• dave says:
@Li, It’s more than simply a matter of saying “galaxies are redshifted and therefore the universe is expanding.” The relationship between redshift, angular diameter distance (how big objects appear as they get more distant) and luminosity distance (how bright things appear when they get more distant) is a fairly detailed prediction of general relativity. The latter, luminosity distance as a function of redshift, is directly probed by, for example, supernova data. The former (angular diameter distance) is probed by measurements of the cosmic microwave background.
While one could, I suppose, suggest some sort of local bubble or somesuch based on Hubble’s original observations, you certainly can’t do so now. The relationship between prediction and observation is far too precise.
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# Posts Tagged ‘ proc iml ’
## Make a frequency function in SAS/IML
March 16, 2012
By
Aggregation is probably the most popular operation in the data world. R comes with a handy table() function. Usually in SAS, the FREQ procedure would deal with this job. It will be great if SAS/IML has an equivalent function. I just created a user-def...
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## A test for memory management of SAS/IML
January 23, 2012
By
Programming always involves the considerations for the efficiency and the memory usage. For efficient programming in SAS/IML, my shortcut is to look at the tip sheet from Rick Wicklin and search ways to simplify the codes. As for the memory management...
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## 6 ways to count odd numbers in SAS/IML
January 19, 2012
By
SAS/IML has a number of vector-wise subscripts/operators/functions available, which can make many things easy. A cheat sheet about them can be found at Rick Wicklin’s blog. To try out those wonderful features( and their combinations?), I designed a...
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## Random seeds
January 12, 2012
By
A footnote toward Rick Wilkin’s recent article “How to Lie with a Simulation”. (Sit in front of a laptop w/o SAS; have to port all SAS/IML codes into R) Generated 10 seeds randomly to run Stochastic simulation of Buffon's needle experiment by R...
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## Benchmarking of the CUSUM function in SAS/IML
January 11, 2012
By
Cumulative sums can be obtained in SAS’s DATA step by the RETAIN statement. As the codes below, a new variable of the cumulative values will be returned by DATA step’s implicit DO loop. data one; do i = 1 to 1e6; z = ranuni(0); ...
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## Do loop vs. vectorization in SAS/IML
January 7, 2012
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Vectorization is an important skill for many matrix languages. From Rick Wiklin’s book about SAS/IML and his recent cheat sheet, I found a few new vector-wise functions since SAS 9.22. To compare the computation efficiency between the traditional d...
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## Using Proc IML for credit risk validation
June 2, 2011
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Validation step is crucial for a scorecard in credit risk industry. Gunter and Peter mentioned in their fantastic book that cumulative accuracy profile (CAP) and receiver operating characteristic (ROC) are two popular methods. Thus, the value...
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problem05_p119
# University Physics with Modern Physics with Mastering Physics (11th Edition)
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5.119: The analysis is the same as for Problem 5.95; in the case of the cone, the speed is related to the period by , tan 2 2 T β πh T R π v = = or . tan 2 v β πh T = The maximum and minimum speeds are the same as those found in Problem 5.95, . cos sin sin cos tan cos sin sin cos tan s s min s s max β μ β β μ
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Unformatted text preview: β gh v β μ β β μ β β gh v +-=-+ = The minimum and maximum values of the period T are then . sin cos cos sin tan 2 sin cos cos sin tan 2 s s max s s min β μ β β μ β β g h π T β μ β β μ β β g h π T-+ = +-=...
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• Calculus, Mathematical analysis, cos cos, gh tan sin
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# Annoying display truncation of numerical results
I have a lot of data to inspect. An example of a number in my program is
123.189094
This gets displayed as
123.189
Unfortunately, most of my information is in the small digits, so, when visually inspecting this data, 123.189094 and 123.189263 are not distinguishable, in other words, a list like
{123.189094, 123.189263}
is displayed like
{123.189, 123.189}
and it looks like I have equal data when I don't. The only fix I have found for this is FullForm, but that explodes out all the structure (lists, rules, etc.) and makes the display very annoying. Imagine that my data were actually buried deep in some nested structure of lists and rules (and it is).
{123.189094, 123.189263}//FullForm
(* List[123.189094,123.189263] *)
is not what I want. I would have thought N[#,10]& or somesuch would work, but it doesn't
{123.189094, 123.189263} // N[#, 10] &
(* {123.189,123.189} AAACH! *)
StandardForm doesn't do it, TraditionalForm doesn't do it.
I'm stuck. As usual, will be grateful for any advice or ideas.
• Related: Mathematica Precision Commented Mar 30, 2012 at 23:16
• It is possible to set the number of digits in Preferences->Appearance->Numbers->Formatting Commented Mar 31, 2012 at 7:34
Maybe this :
NumberForm[#, 10] &@ {123.189094, 123.189263}
{123.189094, 123.189263 }
?
Edit
Consider also this utility of NumberForm[ x, {m, k}] giving m real digits of x with k digits to the right of the decimal point, e.g.
NumberForm[#, {10, 7}] &@ { 197.9898987322333, 201.73205080756887 }
{ 197.9898987, 201.7320508 }
• But if I have a list of numbers, after applying NumberForm, I can't access the list elements as usual, but with an extra index, for example what was before mylist[[1]] is now mylist[[1]][[1]]. How can avoid this? I need to pass this list to a function. Commented Apr 24, 2014 at 10:09
• I would make it like this f[ NumberForm[#, {10, 7}]& @ list]. Commented Apr 24, 2014 at 10:55
• Sorry @Artes, but that still doesn't work. NumberForm returns something that has not the same structure as my original list. Using First[NumberForm[#, {10, 7}]& @ list] returns something that still has other structure, with a list . And then if I do First[First[NumberForm[#, {10, 7}]& @ list]] it returns only a list with the original "old" precision digits. Commented Apr 29, 2014 at 16:18
• @Santi If you have some problems you should ask another question checking if there is no one similar. Here I solved the OP problem. Why should I bother some unclear questions? I believe you would rather use something different but there is no precise information. Perhaps you should check RootApproximant or with the second argument Rationalize or Chop or Round or ... Commented Apr 29, 2014 at 17:10
If this is something you want in general, try:
SetOptions[$FrontEnd, PrintPrecision-> 10] and if you just want it for a specific notebook, then do: SetOptions[InputNotebook[], PrintPrecision-> 10] • I have just seen that this was asked again. As this is a question which presumably beginners are asking I think it would make sense to mention that you can change that setting in the "Preferences" dialog (menu entry "Edit" -> "Preferences") in the Numbers tab of the Appearance tab... Commented Jan 18, 2016 at 17:44 ### PrintPrecision You can control the number of digits displayed using the PrintPrecision option. You have a number of options for its use. You can set it Globally or for the specific Notebook using the Options Inspector. You can also use it directly with Style: Style[123.189094, PrintPrecision -> 10] 123.189094 You can set it temporarily for one session like this: SetOptions[$FrontEndSession, PrintPrecision -> 10]
Finally you can set it using Style Sheets (select cell type Output).
### Arbitrary precision arithmetic
If you use arbitrary precision arithmetic Mathematica automatically displays all digits. If you are concerned with the small end of numbers you should probably be using arbitrary precision anyway. Please see the second half of this answer for more.
123.1890949
123.189094
• No upvotes? I'll help you play ketchup. You and @RolfMertig. These are the answers, the other's are (some good) workarounds. It's not a question about precision and there's no need to wrap the result in some inert structure only for the FE to hide less of what it already received from the kernel
– Rojo
Commented Mar 31, 2012 at 6:22
• @Rojo thanks. By the way, please see my updated "Splitting" answer (the first one). I'm still chasing the perfection of doing it with a single pattern, but Mathematica is not yielding easily. Commented Mar 31, 2012 at 6:59
Use NumberForm:
NumberForm[{123.189094, 123.189263}, 9]
The other answers are fine, but if all the interest is in the small-end digits, consider something like:
((RealDigits /@ {123.189094, 123.189263}) /.
{a__, 0..} -> {a})[[All, 1, -4 ;; -1]]
{{9, 0, 9, 4}, {9, 2, 6, 3}}
This picks up the smallest-end non-zero digits so you can focus on them. This won't work if your numbers have a different number of digits. If that's the case, you will have to allow for this by instead dropping the first few digits from the RealDigits results and then removing the trailing zeros.
(RealDigits /@ {123.189094, 123.189263})[[All, 1, 5 ;; -1]] /.
{a__, 0 ..} -> {a}// TableForm
8 9 0 9 4
8 9 2 6 3
InputForm gives
{123.189094, 123.189263} // InputForm
(*==> {123.189094, 123.189263} *)
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PDA
View Full Version : time dilation faster or slower?
grav
2007-Sep-09, 12:48 AM
Um, I been thinking quite a bit about this and it seems to me that in order for all observers to see light travelling at the same speed, then the rate of time for what we observe of another observer with a relative speed must be faster, not slower. Now, if length contraction shortens the other in the line of travel by gamma and time dilation also makes their clocks tick slower by gamma, then it would seem to make sense at first that light travelling over a lesser distance in a lesser time would make all observers see light travelling at 'c', right? But if time dilation is meant to counteract the length contraction for this purpose, where the other's rulers shrink in proportion to the length contraction, then they should actually measure a greater distance by laying the shrunken rulers end to end along that distance, shouldn't they?
Consider this. We send a signal to another observer with some relative speed. Let's say that they are travelling away from us at such a speed that the contraction shrinks them to half length. Then by their own shrunken rulers, they should measure the distance between us and them as twice as great, shouldn't they? Therefore, they would measure the speed of light as twice as great since it is travelling over twice the measured distance in the same time. So the only way to counteract that is with time dilation, whereas, their clocks are ticking twice as fast, not half as great. So in the same time that we see the light travelling to them at 'c' over some distance from our point of view (since we cannot go by theirs because to them nothing has changed), they will see the light travelling twice the distance over twice the time, since twice the time has passed for them, in order to also measure the speed of light at 'c'.
Even if we consider the alternate scenario where a pulse is sent between two observers upon opposite ends of a moving rod while we also observe, we would still see the pulse travelling at 'c' from one to the other, but over the shorter distance across the rod, while the travellers think the length of the rod has remained the same, so has travelled over twice the distance for them in the same time. The only way to counteract this, then, so that all observers see the light travelling at 'c' is for them to also measure the transit over twice the time, so that their clocks must be ticking twice as fast, not half as great.
This also makes sense if we consider gravitational time dilation. The stress upon a stationary object in a gravitational field will physically shrink the object somewhat. So the light will travel from one end to the other in a lesser time, and processes occur quicker, clocks tick faster, etc., in inverse proportion to the contraction of the object along the line of contraction, although I'm not sure how that would translate for the other axes.
cjl
2007-Sep-09, 01:49 AM
The problem is that to them, the ship isn't what contracts, the distances along the axis of the ship's travel contracts. So, they see earth as closer than it really is, not farther. At least that's how I understand it...
EvilEye
2007-Sep-09, 01:55 AM
If they were traveling at the speed of light in opposite directions, they would always see themsleves as the same distance apart......until they slowed.
Communication during lightspeed would be impossible because time had stopped while at lightspeed.
Only observers outside the ships would notice the difference.
Ken G
2007-Sep-09, 02:14 AM
Um, I been thinking quite a bit about this and it seems to me that in order for all observers to see light travelling at the same speed, then the rate of time for what we observe of another observer with a relative speed must be faster, not slower. Now, if length contraction shortens the other in the line of travel by gamma and time dilation also makes their clocks tick slower by gamma, then it would seem to make sense at first that light travelling over a lesser distance in a lesser time would make all observers see light travelling at 'c', right? But if time dilation is meant to counteract the length contraction for this purpose, where the other's rulers shrink in proportion to the length contraction, then they should actually measure a greater distance by laying the shrunken rulers end to end along that distance, shouldn't they?This is a very common pitfall to fall into. Think of it this way. Let's say you both have a 1 light-second-long ruler, and two clocks, one at each end of your ruler, and you are measuring the speed of light, each by sending a light signal along your own ruler and testing how long it takes. So apparently you both need to get 1 second as the answer. If you think the other guy is moving away from you in the direction of his ruler, then his ruler is short, so you think the light is really not travelling a full light second when it goes from one end of his ruler to the other. But you are forgetting that the end of the ruler is moving away from you, so moves while the light is in transit. You have to include that. But that's not all-- you're also forgetting the role of the simultaneity shift at the ends of his moving ruler from your reference frame! People are always forgetting the simultaneity shift. To you, his two clocks are not synchronized. So we have a host of effects that all have to be included: the ruler is short, the end of it moves during propagation of the light, and his clocks are not synchronized from your point of view. Put all that together, and you find he still doesn't get 1 second for his answer unless his clocks are ticking slowly.
grav
2007-Sep-09, 04:05 AM
This is a very common pitfall to fall into. Think of it this way. Let's say you both have a 1 light-second-long ruler, and two clocks, one at each end of your ruler, and you are measuring the speed of light, each by sending a light signal along your own ruler and testing how long it takes. So apparently you both need to get 1 second as the answer. If you think the other guy is moving away from you in the direction of his ruler, then his ruler is short, so you think the light is really not travelling a full light second when it goes from one end of his ruler to the other. But you are forgetting that the end of the ruler is moving away from you, so moves while the light is in transit. You have to include that. But that's not all-- you're also forgetting the role of the simultaneity shift at the ends of his moving ruler from your reference frame! People are always forgetting the simultaneity shift. To you, his two clocks are not synchronized. So we have a host of effects that all have to be included: the ruler is short, the end of it moves during propagation of the light, and his clocks are not synchronized from your point of view. Put all that together, and you find he still doesn't get 1 second for his answer unless his clocks are ticking slowly.Could you give an example?
Ken G
2007-Sep-09, 04:24 AM
The best way to do it is actually to get rid of length contraction and simultaneity issues by simply aligning the ruler perpendicular to the motion. Then all you need is the Pythagorean theorem to derive time dilation, because as you watch the other guy do his ruler experiment, you think the light is travelling farther (the hypoteneuse instead of the leg of the triangle) than he does. Thus to get the same speed as him, his clock must be slower. Once you have derived it that way, you can rotate the ruler along the direction of motion, and since that will not affect his clock rate but it does change the length of the ruler, you can then derive what desynchronization is consistent with the slowed time you just calculated. If you think about it, there are actually three unknowns here, the length contraction, the time dilation, and the desynchronization. There are only two independent alignments of the ruler for which you can apply the fixed c constraint, so you need a third constraint to solve for all three of those unknowns consistently. That third constraint is the other postulate of relativity-- the symmetry of the observers. That means you have to get the same answer if you conceptualize the receiver as moving as if you conceptualize the source as moving.
grav
2007-Sep-13, 02:13 AM
Thanks Ken. Sorry I didn't get back to you sooner but I wanted to run through a few calculations first. As far as aligning a ruler perpendicular to the motion as you mentioned, it seems an observer would actually see the light travelling a lesser distance, not farther, but with the same result, the other's time would run at a slower pace. It would be like watching him in slow motion, where his clock ticks slower. I didn't want to use a ruler for this example, though, because what the observer sees would also then might depend upon the angle to each point along it as far as the time dilation and contraction are concerned, especially since gamma itself can be regarded in terms of angles as 1/sqrt(1-(v/c)^2) = cos(sin-1(v/c)). Not that it means anything in particular, but I didn't want to take any chances.
So instead I used a black hole, where a photon might continually orbit it at some distance in a perfect circle. So in respect to the black hole, the light is moving at 'c' in orbit around it. But to an observer moving away from the black hole at a constant speed 'v', perpendicular to the path of the photon at all times, if all observers must measure the speed of light as 'c', then the speed that observer would view the photon travelling around the black hole to be would be c', where c = sqrt(c'^2 + v^2), since the square of the speeds for the vectors for the photon as it moves away from the observer at v while orbitting the black hole at c' must always add up to the square of the speed of light. This gives the speed of the photon in respect to the orbit only as c' = c * sqrt(1-(v/c)^2), so that the rate of time that passes for the black hole would appear to the observer to be lesser by sqrt(1-(v/c)^2)=1/gamma, exactly as relativity stipulates, as long as there is no contraction in any direction perpendicular to the direction of motion between the black hole and observer, and this is what is stated with the Lorentz contraction, so let's take a look at that.
With the Lorentz contraction, objects contract in the line of motion by 1/gamma, but not perpendicular to that motion. This is based upon the MM experiment, so we can apply it to that. We have a light beam split in perpendicular directions, travelling a distance of 'd' in each direction and brought back together again. No interference is detected with this, so we figure that the light has travelled in each direction and back in the same time. But if all observers in all reference frames observe the speed of light at 'c', then we have a problem for observers with some relative speed to the apparatus classically. (I know you know all of this, but I'm just trying to run through all of the bases). Lorentz, however, found a solution, that the length contracts in the direction of motion to an observer. So let's work through this. As you said, the time of the travel of light would also depend upon the motion of the point of reception. For an observer stationary with the apparatus, the light travels along all paths back and forth again in the time t = d/c each, but when the light travels forward at 'c' to the observer, it does so in the time,
c*t1 = x*d + v*t1, where x is the contraction in the line of motion, and v*t1 is the distance the receiving point has moved forward in the same time. For the light travelling back again, it is
c*t2 = x*d - v*t2, since in this case the receiving point is moving toward the direction of the light and lessens the distance it has to travel. Perpendicularly, it is
(c*t3)^2 = d^2 + (v*t3)^2 forward and back, where the length does not contract by 'x' in this case since the lioght is travelling perpendicularly to the line of motion. The resulting times are
t1 = xd/(c-v)
t2 = xd/(c+v)
t3 = d/sqrt(c^2-v^2)
Now, since the times forward and back in each direction must be the same, we have t1 + t2 = 2*t3, so
xd/(c-v) + xd/(c+v) = 2d/sqrt(c^2-v^2)
x*[1/(c-v) + 1/(c+v)] = 2/sqrt(c^2-v^2)
2c*x/[(c-v)(c+v)] = 2/sqrt(c^2-v^2)
c*x/(c^2-v^2) = 1/sqrt(c^2-v^2)
x = sqrt(c^2-v^2)/c
x = sqrt(1-(v/c)^2) = 1/gamma
This demonstrates the contraction as 1/gamma, also exactly according to relativity. And placing this back into the formulas for the time gives us
t = t1 + t2 = 2*t3
t = xd/(c-v) + xd/(c+v)
t = sqrt(1-(v/c)^2)*d*2c/(c^2-v^2)
t = (d/c)/sqrt(1-(v/c)^2)
t = (d/c) * gamma
And the time for observers stationary with the apparatus measure the passage of time as just d/c, so the observer sees a greater time passing according to their own clock and therefore less for the other, exactly as relativity says. So that's three for three for relativity, right? This is how it is all said to be. So all's good so far, though, right?
Okay, now here's my dilemma. Everything works out fine when we consider the times back and forth together. But let's take each of these times, back or forth, not both, using the same formulas as above, apply relativity, and compare them to those in the stationary frame individually. Now we have
t1 = xd/(c-v)
= sqrt(1-(v/c)^2)*d/(c-v)
= sqrt[(1+v/c)/(1-v/c)]*(d/c)
t2 = xd/(c+v)
= sqrt(1-(v/c)^2)*d/(c+v)
= sqrt[(1-v/c)/(1+v/c)]*(d/c)
t3 = d/sqrt(c^2-v^2)
= 1/(1-(v/c)^2)*(d/c)
= gamma*(d/c)
Now, according to this, then, the times we observe passing for the apparatus and stationary observers would be proportional to their Relativistic Doppler shift, not gamma, except in the perpendicular direction only. This is partly the same dilemma I had in the Lorentz contraction thread towards the end, where from one perspective, the observed time is directly related to gamma, but from another, it should be directly related to relativistic Doppler. Adding them together for the total time forward and back in the direction of motion all falls in line with relativity, but not individually, as the ratio of the rate of times for one time which pass for the apparatus and observer would be different from the other, depending upon which way we consider the light to be travelling, one faster and one slower by the same amount, which should not be the case.
I am currently attempting to work through this, using three unknowns; length contraction, time dilation, and the speed of light from a moving source. To find these I am applying various scenarios such as using relativistic Doppler for its simplicity, assuming it is correct, which one link says has been recently verified very accurately, although the measurements were very difficult to make the way they had to do it, while another says that an error in relativistic Doppler explains the Pioneer anomaly. Another might be assuming a lesser time for what each observer sees of the other might be completely light travel time, where light travels at a speed other than 'c' from a moving source according to observers. Others might be reference to an ether or an expansion of space and/or of objects while the light is in transit, etc. Results are so far inconclusive, however. There are many such possible scenarios to run through, at least twenty or thirty that I can think of offhand, with perhaps a couple hundred variations between them. I will have to find ways to narrow them down.
Ken G
2007-Sep-13, 04:33 AM
Now, according to this, then, the times we observe passing for the apparatus and stationary observers would be proportional to their Relativistic Doppler shift, not gamma, except in the perpendicular direction only.
That sounds fine to me, I don't see why you consider this a "dilemma". This is just what you should get, because the calculation you are doing is exactly the same as what you'd get if you consider a receiver moving through a light wave of wavelength d in the source frame. The gamma factor comes from time dilation instead of length contraction, but the expressions are exactly the same, and you of course expect to get the relativistic Doppler shift. Everything here is just what it should be.
grav
2007-Sep-13, 12:24 PM
That sounds fine to me, I don't see why you consider this a "dilemma". This is just what you should get, because the calculation you are doing is exactly the same as what you'd get if you consider a receiver moving through a light wave of wavelength d in the source frame. The gamma factor comes from time dilation instead of length contraction, but the expressions are exactly the same, and you of course expect to get the relativistic Doppler shift. Everything here is just what it should be.I was afraid you might think that. That's why I ran through the whole thing. It would seem it might be that way at first glance, but look a little deeper. The ratio of the rate of times that pass for the moving observer and for the apparatus and observers stationary with it are just the ratio of the rate of times that pass for the photon on its two way trips back and forth for the moving and stationary observers, which gives 1/gamma for the rate of time of the stationary observers according to the moving one. Those last times, however, are only the one way times of the photon, which give different ratio of rates of times forward and back, sqrt[(1-v/c)/(1+v/c)] forward and sqrt[(1+v/c)/(1-v/c)] back, the same as with relativistic Doppler, but these are the actual ratio of the rates of times observed upon each of the observers' clocks, not the ratio of the frequency of pulses, which would be found by comparing the rate of times between successive pulses and then mutiplying that by the ratio of the rate of times as 1/gamma as each observer views the other's clock as the pulses are emitted. The ratio of the rate of times itself is found in the same way as that for the ratio of the rate of times for the entire trip that originally gave us 1/gamma, but which now gives a different ratio of rate of times between the clocks stationary with the apparatus and the moving observer, depending upon whether the photon is travelling forward or back, which should not be. When the photon is travelling forward, the moving observer sees a time of sqrt[(1+v/c)/(1-v/c)](d/c) passing upon his clock while the stationary observers see just d/c passing upon theirs, so the moving observer would say that the rate of time is passing (d/c) / [sqrt[(1+v/c)/(1-v/c)](d/c)] = sqrt[(1-v/c)/(1+v/c)] more slowly for the stationary observers, and the inverse of this when observing the photon travelling back.
Now, you also mentioned simultaneity issues here, which also led to relativistic Doppler in the other thread, but it seemed then also that the formula for the simultaneity should also give the ratio of the rate of times as sqrt[(1-v/c)/(1+v/c)] and sqrt[(1+v/c)/(1-v/c)], but this was only after the ratio of the rate of times was already multiplied in as 1/gamma, so it could not be both. That formula gives different solutions depending also upon the distance between observers, but not upon light travel times, since we get the same times observed between observers which are both stationary, which incorporates but does not include a lag of times in the result. The simplest way to possibly avoid such a simultaneity issue might be to just run the photon out some distance and then bring it back to the same original point, as is done with the MM experiment, when considering the entire two way paths of the light forward and back. It's possible the observed time for the one way path of the photon might be affected by simultaneity issues in some way, although I don't see how yet specifically, but I'll keep working on it.
Ken G
2007-Sep-13, 03:05 PM
It would seem it might be that way at first glance, but look a little deeper. The ratio of the rate of times that pass for the moving observer and for the apparatus and observers stationary with it are just the ratio of the rate of times that pass for the photon on its two way trips back and forth for the moving and stationary observers, which gives 1/gamma for the rate of time of the stationary observers according to the moving one. Those last times, however, are only the one way times of the photon, which give different ratio of rates of times forward and back, sqrt[(1-v/c)/(1+v/c)] forward and sqrt[(1+v/c)/(1-v/c)] back, the same as with relativistic Doppler, but these are the actual ratio of the rates of times observed upon each of the observers' clocks, not the ratio of the frequency of pulses, which would be found by comparing the rate of times between successive pulses and then mutiplying that by the ratio of the rate of times as 1/gamma as each observer views the other's clock as the pulses are emitted.I know all that, and there is still no problem at all.
The ratio of the rate of times itself is found in the same way as that for the ratio of the rate of times for the entire trip that originally gave us 1/gamma, but which now gives a different ratio of rate of times between the clocks stationary with the apparatus and the moving observer, depending upon whether the photon is travelling forward or back, which should not be. Why on Earth not? Are you suggesting a new "relativity of time ratios"? No such principle. Once again you are imagining that time dilation is the only thing that happens to time. Do I have to mention one more time that there is also a shift in synchronicity when you change frames? That does violence with the ratios you expect to be preserved. They are not. Please, never again apply time dilation without the simultaneity shift, that single oversight creates the vast majority of all relativity "paradoxes" I've ever seen.
Now, you also mentioned simultaneity issues here, which also led to relativistic Doppler in the other thread, but it seemed then also that the formula for the simultaneity should also give the ratio of the rate of times as sqrt[(1-v/c)/(1+v/c)] and sqrt[(1+v/c)/(1-v/c)], but this was only after the ratio of the rate of times was already multiplied in as 1/gamma, so it could not be both.Think again, it's all the simultaneity shift.
That formula gives different solutions depending also upon the distance between observers, but not upon light travel times, since we get the same times observed between observers which are both stationary, which incorporates but does not include a lag of times in the result.The difference in distance comes out in the wash because it appears in both simultaneity shifts, and all that ends up mattering is d, or if you like, the light travel time. Just put it all in, and you will get it all out. Stop trying to tell it what it "should" do, you'll be forever getting yourself in a tizzy-- just follow the Lorentz transformation, that's what it's there for.
grav
2007-Sep-13, 06:00 PM
Although it may have sounded otherwise, I was actually sort of agreeing with you with that last paragraph, that I may still be overlooking something with the simultaneity shift, especially since I'm thinking about it as a direct result of time dilation to begin with, and I want to make sure. So I guess what I should ask is if the ratio of the rate of times observed appears greater when the light travels in one direction and smaller in the other, then precisely how would one apply simultaneity effects in order to correct for this?
Ken G
2007-Sep-13, 08:09 PM
The simultaneity shifts are not applied to "correct for" this, there is no need to correct for it-- the simultaneity shifts are what causes the discrepancy, but there's nothing wrong with
that discrepancy. Think of it this way. Let's say a brother, A, gives his sister, B, a treasured family heirloom, and years later B returns it to A. A and B are like the mirrors of the MM experiment. Let's further say that this transaction is witnessed by two other people, X and Y, who are like your reference frames. Now, person X tends to put great personal value in memorabilia, so thinks the value of what was exchanged was spectacular, while person Y looks only at what price it could get at an auction, so values it much less. The analog of this value is like the elapsed time between mirrors. So we have a ratio there in the value as perceived in the reference frames of X and Y.
Now the question is, does that value ratio necessarily have to be the same in both directions that the transaction occurred? The answer is no, not if there was a value shift that Y saw and X didn't. For example, perhaps X sees the value as unchanged by the transaction because it still represents a treasured memory, while Y thinks that an heirloom in the hands of someone (B) that is not the original owner (A) will reduce its value at auction. So that shift is like a simultaneity shift that one person sees and the other doesn't-- and it will change the ratio of the value as seen from reference frame X versus Y when the heirloom is going to B then when it is returned to A. So here we have an everyday analogy that can achieve the same effect-- ratios of things are not necessarily time reversible if there is a differential shift that occurs at one end of the process.
grav
2007-Sep-22, 11:24 AM
Sorry I took so long to respond again. I've been going over this some, so let's see what I can come up with, and how it sounds to you.
Okay, so let's see. The simplest solution to this would be that light travels with the additional speed of the source. Then, although observers stationary with the apparatus see light travelling at c in all directions since the source is moving with them, another observer that sees the apparatus moving at some relative speed away from them would see light travelling at c+v from the source to the mirror further away while the mirror is also moving away at the same time that the light is catching up, and c-v back again, where the original mirror of separation of the beams would be moving toward the light according to the other observer, so that
(c+v)*t = d + v*t
c*t = d
(c-v)*t' = d - v*t'
c*t' = d
t=t'
and the same thing goes for the perpendicular beam. But Einstein used the postulate, from Maxwell's equations, that all observers see the speed of light at c, and relativity matches observations so far. As well as this, light travelling with the additional speed of the source would make a particle phenomenom.
So let's say we have three observers, two of which are stationary to each other and a third with a relative speed to them, so that the first two are both positioned along the line of travel. The first, closest to the third, sends a pulse to the second. The second also sends a pulse to the first. To them, each receives the other's pulse in the same time, and records it as such according to their own clocks. Furthermore then, each should see the other receiving each pulse when the other's clock reads the same thing their own clock would read for the distance travelled, minus the time of flight of light for this observation, of course. The third observer, however, would see the pulses still travelling at c in either direction, first to second and second to first, but while the pulse from the first travels toward the second, the second observer is moving away, so the light takes longer to reach the second than it does the from the second to the first, while the first would be moving toward the pulse. The third observer must read the same times upon each of their clocks when the pulses are actually received, however, since at that instant the pulses would be synchronous with each of the two moving observers. That means that the third observer must see the second observer's time lag behind the first's, since the light takes longer to reach the second observer from the first than it does to the first from the second according to the third observer. Since the first and second observers are moving together, there is no time dilation between them, though, just the time lag, which then becomes the simultaneity issue, as far as I can tell.
So how would would this simultaneity issue come about? Let's see. If we begin with two observers which are stationary to each other, then we can synchronize their clocks. Now let's move them apart some distance from each other. As we do so, their will be a relative speed between them. So each sees the other's clock moving more slowly, with more than just the time of flight lag due to separation, but in a relativistic way. This will build a time lag between them that each observes of the other while they are being separated. So when we bring them to rest in respect to each other again after some separation, this time lag will remain, as well as the time of flight lag, even though each of their clocks are once again ticking at the same rate.
To a third observer, as the other two separate, the second observer (the one furthest away) would appear to be moving away from the third at a greater rate than the first to the third, and so the time dilation between the second and third would be greater. This would mean a greater time lag would build up during the time of separation between the second and third than between the first and third during the time of separation. So when the third observer sees the pulse travelling from the first to the second, it would appear to take longer, and quicker from the second to the first, but the lag on the second observer's clock would compensate for that, so that all observers see the light being received between the first and second in the same times according to their own clocks when the light is received.
Does any of this sound reasonable so far, as far as the simultaneity shift is concerned? I hope so. It's the only thing I've been able to come up with so far. But I still have a few concerns. For instance, the simultaneity issue for how a third observer would perceive two others that are moving together should only depend upon the relative speed to the third and the separation of the first two, so the same physics can be applied to any such scenario, regardless of how they got there. But the time lag that would occur between the first two while separating would depend upon the instantaneous relative speeds between them at each instant as they separate. So it would depend upon how they separate. Also, one or the other would experience an additional time lag depending upon who is actually accelerating in order to separate in the first place. So this is different than that which is just typical of the relative speeds between them while separating, where each would view the other with a lesser time equally, according to SR. But I guess this is where GR steps in. Of course, each of the first two would also see this additional time dilation as well as the third. We can eliminate this, however, by considering each of the first two to be accelerating away from each other in the same way and then decelerating to come back to rest. The third observer will still see an additional time dilation, but it will be the same with each of the first two.
Okay. So the first two accelerate at the same rate away from each other. This negates any additional time dilation between them due to the acceleration, right? Then they should only each observe a time dilation for the other according to their instantaneous speeds. But the instaneous speeds would also depend upon the acceleration in the first place. The third must see the same time lag according to the separation between the first and second and their relative speed to the third. Also, if the first and second experience a time lag between them, while viewing the speed of light travelling at c from one to the other, then each will see the light reaching the other in a lesser time, so they would measure a lesser distance. Now, I could probably work all of this out, as to find what the perceived acceleration would have to be relative to each other as compared to the acceleration in the local frame, in order to produce the same time lag for the same distance of separation according to the third in each case, but I still see another potential problem. As the first two separate, the third observer will have a lesser relative speed to the first than to the second. So the second, the one furthest away, will always experience a greater time lag than the first according to what the third observes. But if the first two were travelling toward the third, the light would take longer to travel from the second to the first than the other way around, which means the first should experience the greatest time dilation according to the third. So what am I still missing here? What is the correct way to perform this so that it works out for all observers in either direction?
grav
2007-Sep-22, 11:52 AM
As the first two separate, the third observer will have a lesser relative speed to the first than to the second. So the second, the one furthest away, will always experience a greater time lag than the first according to what the third observes. But if the first two were travelling toward the third, the light would take longer to travel from the second to the first than the other way around, which means the first should experience the greatest time dilation according to the third. So what am I still missing here? What is the correct way to perform this so that it works out for all observers in either direction?
Oh, wait. As the first two are moving toward the third, the first observer will obtain a greater relative speed toward the first and the second observer a lesser one upon separating. So the first observer would time dilate at a greater rate relative to the third than the second relative to the third, and so a greater lag would be produced for the first than the second in this case, so that their times can still be read in the same way due to the time lag, regardless of their direction of travel. Cool. Does that all sound about right, then?
Ken G
2007-Sep-22, 07:13 PM
The third observer must read the same times upon each of their clocks when the pulses are actually received, however, since at that instant the pulses would be synchronous with each of the two moving observers. That means that the third observer must see the second observer's time lag behind the first's, since the light takes longer to reach the second observer from the first than it does to the first from the second according to the third observer.Right, that is a good way to see why motion has to induce a synchronicity shift. You, the third observer, conclude the two clocks, though they read the same time when the respective pulses arrive, must not be synchronized with each other, even though they are both dilated by the same speed. Velocity, in the presence of separation, causes a synchronicity shift-- the forgotten term in the Lorentz transformation.
So how would would this simultaneity issue come about? Let's see. If we begin with two observers which are stationary to each other, then we can synchronize their clocks. Now let's move them apart some distance from each other. As we do so, their will be a relative speed between them. So each sees the other's clock moving more slowly, with more than just the time of flight lag due to separation, but in a relativistic way. Actually, you can move them apart slowly enough that there's no problem with synchronicity shifts. The synchronicity shift comes not just from the separation, but from the separation and the fast velocity-- it appears during the acceleration, not the separation. If you imagine the acceleration is first and the separation comes later, then the shift does come during the separation, but it's much harder because you have to consider the additional acceleration and subsequent deceleration you need to prepare that state.
Basically, there are very different ways to think of how you have set up this situation, but they all give the same answer in the end. For example, you can imagine that you separate and synchronize 1 and 2, then accelerate 3, or you can imagine that you separate 1 and 2, then accelerate them, then re-synchronize them in their own frame. Constrast those-- in the first case, the shift (from 3's point of view) comes in when you accelerate 3-- it thinks it is held fixed in a "gravitational field" that both accelerates and differentially shifts the separated clocks 1 and 2 the way gravity always shifts things at different "heights" in the well. The upshot is that the trailing clock ends up ahead of the leading clock from 3's point of view. Or, in the second case, the shift comes in when they re-synchronize themselves. Observer 3 says, why are you changing your times, you were synchronized before! But 1 and 2 say "you were in free fall so you didn't notice that big gravitational field we felt as we accelerated. That field ruined our synchronization and we have to re-establish it." So in the second case, the shift is manual, and in the first case, it happens automatically. The situations are distinguishable by all parties, yet the shift at the end is the same-- 3 thinks the trailing clock is ahead.
Or you could first accelerate either 3, or 1 and 2, and then separate 1 and 2. Either way there's no relative shift during that main acceleration, but there now has to be additional acceleration coming once they are already moving at high speed. So that's the case where you will have to track the effects of the time dilation, but you'll also have to worry about the tricky acceleration/deceleration to get the separation, and if you do it gradually, you'll have to do it at large distance from 3, which will mitigate your attempt to reduce its importance. There's no way around a tough calculation-- don't do it that way!
This will build a time lag between them that each observes of the other while they are being separated. So when we bring them to rest in respect to each other again after some separation, this time lag will remain, as well as the time of flight lag, even though each of their clocks are once again ticking at the same rate.
This is the situation to avoid, it's too complicated to analyze easily. If it was 3 that was originally accelerated, youcan do the above gradually enough that 1 and 2 still think they are synchronized with each other, but you still need to see how 3 disagrees with that, and that's the hard part.
So when the third observer sees the pulse travelling from the first to the second, it would appear to take longer, and quicker from the second to the first, but the lag on the second observer's clock would compensate for that, so that all observers see the light being received between the first and second in the same times according to their own clocks when the light is received. That's the ultimate result, yes.
For instance, the simultaneity issue for how a third observer would perceive two others that are moving together should only depend upon the relative speed to the third and the separation of the first two, so the same physics can be applied to any such scenario, regardless of how they got there.Correct, but it can be easy to leave out something important.
But the time lag that would occur between the first two while separating would depend upon the instantaneous relative speeds between them at each instant as they separate. So it would depend upon how they separate. Also, one or the other would experience an additional time lag depending upon who is actually accelerating in order to separate in the first place.True-- that all has to "come out in the wash" for a self-consistent theory. That's why you use mathematics like the Lorentz transformation, it's good at coming out consistent.
Okay. So the first two accelerate at the same rate away from each other. This negates any additional time dilation between them due to the acceleration, right? In some reference frames, yes, but not in others. Each frame sees very different physics happening, but the same answer at the end.
But if the first two were travelling toward the third, the light would take longer to travel from the second to the first than the other way around, which means the first should experience the greatest time dilation according to the third. So what am I still missing here? The leading clock does experience more time dilation, as it has to be going faster to end up "leading", and that also explains why it reads an earlier time, according to 3.
Here's another way to infer that the leading clock lags the trailing clock, for someone observing a pair of rapidly approaching clocks that are synchronized in their own frame. Let's say the two clocks have a 1 foot ruler that spans between them, and they are moving at very high speed indeed. Observer 3 says that the ruler that spans them is very shrunken, perhaps is only an inch long let's say. Now the leading clock sees observer 3 arrive, and then the trailing clock sees observer 3 arrive a light-foot in time later. But observer 3 says, nope, the trailing clock got to me only a light-inch in time later, the only reason it read a light-foot later is that it was synchronized to be already ahead of the leading clock by 11 light-inches.
Ken G
2007-Sep-22, 07:14 PM
Cool. Does that all sound about right, then?
Yes, that all sounds right, I think you have a consistent picture.
grav
2007-Sep-23, 01:52 PM
Thanks, Ken. I appreciate the help. I'm going to try to work through some simple scenarios to see what sort of mathematical relationships might come out of it. I will still need the results scanned over when I'm done.
Ken G
2007-Sep-23, 03:14 PM
I think you're on the right track now. Of course, there are always surprises in relativity, I can't trust my own conclusions until I've thought them through pretty carefully or gotten feedback from others. Then I go to my colleagues and pose the questions to them, and I find the same problem is widespread!
John Mendenhall
2007-Sep-26, 06:58 PM
I think you're on the right track now. Of course, there are always surprises in relativity, I can't trust my own conclusions until I've thought them through pretty carefully or gotten feedback from others. Then I go to my colleagues and pose the questions to them, and I find the same problem is widespread!
That's a good description of how to approach SR and GR problems. There is an entire thread on BAUT about a GR idea that was passed around at CERN for review. The CERN physicists all said "No, it doesn't work that way." Yet a lot of our better mainstream posters went for it lock, stock, and barrel, despite the violation of a basic postulate. I will leave the thread unnamed for the time being, since I intend to reopen it at some point after accumulating enough ammo to sink a fleet. It is wise to take anything you read about GR and SR with a grain of salt; even the best folks stumble on it. Regardless, the fundamental ideas of SR and GR are sound and correct, and are verified continually by their application in our technology.
publius
2007-Sep-26, 08:24 PM
There is an entire thread on BAUT about a GR idea that was passed around at CERN for review. The CERN physicists all said "No, it doesn't work that way." Yet a lot of our better mainstream posters went for it lock, stock, and barrel, despite the violation of a basic postulate. I will leave the thread unnamed for the time being ...
You'd better name it right now, or I'll go crazy wondering what it is. :)
-Richard
grant hutchison
2007-Sep-26, 09:21 PM
You'd better name it right now, or I'll go crazy wondering what it is. :)I believe it's probably the old rockets-and-rope chestnut, most recently rehearsed in detail here (http://www.bautforum.com/questions-answers/62037-lorentz-contraction.html). But it has had at least one previous incarnation, courtesy of a previous incarnation of hhEb09'1, here (http://www.bautforum.com/against-mainstream/13-rockets-rope.html).
(John Mendenhall believes the rope doesn't break.)
Grant Hutchison
publius
2007-Sep-26, 09:42 PM
Ah, so. That crossed my mind, but John said "GR" so I was wondering if this was something else involving some gravitational thing, since most would consider the rope and rocket to be SR.
-Richard
grant hutchison
2007-Sep-26, 09:53 PM
Ah, so. That crossed my mind, but John said "GR" so I was wondering if this was something else involving some gravitational thing, since most would consider the rope and rocket to be SR.I think you (and IIRC, Ken to a lesser extent) took a largely GR stance on it. (Something which would have gratified the heart of Sam5, of course, were he with us still.)
Grant Hutchison
publius
2007-Sep-26, 10:28 PM
I think you (and IIRC, Ken to a lesser extent) took a largely GR stance on it. (Something which would have gratified the heart of Sam5, of course, were he with us still.)
Grant Hutchison
:lol: That is worthy of some clarification (or rambling, however you look at it). The current view of the high priests is that GR = Riemann curvature, which is invariant and thus has "more meaning". If space-time is flat, one is doing SR, regardless of the coordinates.
The other view is that SR's domain is the *Minkowski metric*, where the familiar, more simple rules apply. If one is doing and explaining things from some non-inertial frame (in flat space-time), one is doing "more than SR".
Me, I think I've come to the position that we should forget about SR vs. GR, and just call it 'R', Relativity, which we'll roughly define as "the machinery for doing calculations about stuff in space-time". Then we'll define "gravity" as the curvature that happens to space-time due to mass-energy as described by the EFE.
-Rchard
publius
2007-Sep-27, 02:05 AM
You know, I think the best way to put the rope and rocket thing is the following, which uses Euclidean geometry.
It should be no more surprising that an accelerating rocket must experience gradients in proper acceleration (the force felt at any point by an observer there) across its length than the fact that different points on a spinning flywheel must experience different acceleration in order for it to remain rigid. The centripetal acceleration is w^2*r directed radially, increasing with radius.
If we were to specify that two points at different radii moving in concentric circles must feel the same force, then they could not remain stationary with respect to each.
In fact, mathematically, these two cases are very similiar. Constant proper acceleration is hyperolic, and hyperbolas in Minkowski are like circles in Euclid. Same constant "distance" (norm/interval) from a given point.
So, I don't see any basic postulates being violated at all. It is what our good friend PM (Principia Mathematica :lol: ) tells us follows from our basic postulates.
-Richard
grav
2007-Sep-28, 12:50 AM
Well, here are some of my preliminary results. I have started with the single simple postulate that two observers will observe the same thing of each other. Basically, this just means that each will see the same thing regardless of which way the light travels. Potentially, this could even mean that light can travel at different speeds in different frames of reference or with different relative speeds between observers, just as long as both observers experience the same physics between them. This postulate is the same thing Relativity says, but on a more basic level, although without a medium to move relative to, the physical constants must also remain the same in every frame of reference, because there is no relative motion to anything concrete which might change them due to that relative motion, and so the speed of light would remain the same to all observers as well.
I have already run into a snag with my first scenario using Relativity, however. I synchronized three observers which are stationary to each other in the same position, then accelerated one to some relative speed to the others for some time and then accelerated another to the same relative speed. They are accelerated in such a way that any time dilation due to this acceleration is neglible, either by accelerating them for a very short duration of time to the stationary observer or by running the first accelerated observer out for a very long duration of time, over a large distance, whereas the lag in times due to the relative speeds over this distance would far outway any due to acceleration. Since any distance whatsoever can be aquired this way, any time dilation due to acceleration can be ignored. Also, each observer carries their own fuel supply by which to accelerate, so does not affect the other observers in any way while doing so, whereby conservation of energy and momentum doesn't enter into this either.
So let's say the first observer travels out a distance of 'd' at a relative speed of 'v'. To the stationary observer (the third observer that isn't accelerated), the first will appear to be time dilated by t(1/y), where t is the time according to the stationary observer's clock and 1/y is the time dilation experienced by the first according to the SO (stationary observer). But the SO will also time dilate to the same degree according to the first, as well as with the second observer, which hasn't accelerated yet. The first and second observers, then, will now observe some time lag upon each others' clocks. Then the second observer is made to aquire the same relative as the first. At this point, the first and second will have the same time lag relative to each other, each lesser to the other, but their clocks will now tick again at the same rate, so no further time lag is aquired.
The problem here is that if each has now obtained a time lag in respect to the other, and we now only consider their relative distance and motion, and therefore can forget about the SO, then if we were to send a pulse of light from one to the other, then according to the observer that sent the pulse, if light always travels at a constant c, then the other's clock should read a lesser time when the pulse is received over some distance than when the other were to send a pulse to the observer, since each observer's clock reads a greater time than the other from their own perspective, and so the speed of light could not be measured the same in either direction since the observer would measure different times for when the pulse was emitted and received over the same distance either way according to what they observe on the other's clock when the pulse is simultaneous to the other's position, seen by the other by simply raising their hand when the pulse is received or emitted by them, or some other similar method. So different times in either direction for the same distance will give different values for the speed of light.
I tried various ways to get around this, but nothing has worked out so far. Finally, I gave in and tried time dilations which vary, not with the relative speed between observers, but with respect to a stationary frame. This has worked out for the most part. It basically just means that instead of each observer observing a lesser time lag for the other, one's time is actually less and the other greater. In this last scenario, for instance, the first observer's time would dilate at t(1/y) according to the stationary clock, where 't' is the time of travel, whereas the second observer which is initially stationary during this time would experience the full time of t. After the second observer is accelerated to the same relative speed to the SO, the time dilation will remain the same thereafter, and so their clocks will tick at the same rate, but a time lag of t(1-1/y) will persist, the first observer's clock showing this much less time than the second. To the SO, t=d/v, so we get tlag = (d/v)(1-1/y). The speed of light in the stationary frame is always c, so according to the SO, for the first observer sending a pulse to the second, we get
c*t1 = d + v*t1
t1 = d/(c-v)
While for the second to the first, we get
c*t2 = d - v*t2
t2 = d/(c+v)
The time the SO sees pass on their clocks between emission and reception, then, is just t1*(1/y) - tl and t2*(1/y) + tl. These two times must be the same for the moving observers in order to observe the same physics either way, namely that each will observe the same speed of light over the same distance, so that they must also observe the same times. So we have
t1*(1/y) - tl = t2*(1/y) + tl
[d/(c-v)](1/y) - tl = [d/(c+v) + tl
2*tl = [d/(c-v)]*(1/y) - [d/(c+v)]*(1/y)
2*tl = [2*d*v/(c^2-v^2)]*(1/y)
tl = (1/y)*d*v/(c^2-v^2)
and since we found before that tl = (d/v)(1-1/y), then
(d/v)(1-1/y) = (1/y)*d*v/(c^2-v^2)
1-1/y = (1/y)*v^2/(c^2-v^2)
1 = (1/y)*[1 + v^2/(c^2-v^2)]
1 = (1/y)*[c^2/(c^2-v^2)]
1/y = (c^2-v^2)/c^2
1/y = 1 - (v/c)^2
Now, this is the square of what relativity says, but I haven't been able to get even as far as this without considering a stationary frame. Each time I work through such a scenario, however, I attempt to consider what it would mean in terms of the relative speed between observers only, in order to keep Relativity on even grounds, but it has led to nothing but dead ends so far when I try to apply it that way.
For this last scenario, when we place the time dilation we found into the formula for the time lag we find
tl = (d/v)(1-1/y)
tl = (d/v)[1-(1-(v/c)^2)]
tl = (d/v)(v/c)^2
tl = d*v/c^2
Then using that for the times observed, which are t1*(1/y)-tl and t2*(1/y)+tl in each direction, which must be equal, we get
T = t1*(1/y) - tl
T = [d/(c-v)][1 - (v/c)^2] - dv/c^2
T = [d/(c-v)][(c^2-v^2)/c^2] - dv/c^2
T = d(c+v)/c^2 - dv/c^2
T = d/c
This means that each observer measures a time of T for the time of the pulse to reach the other or sent from the other according to the times read upon their clocks at each end of the trip, over the same distance 'd', giving a speed of the pulse of d/T = c. So not only is the speed of the pulse the same both ways, it is always measured at the same speed to all observers, at least those which are stationary to each other. Other scenarios have led to the same thing, so apparently observers that measure the speed of light as c in all frames isn't that difficult to come by. Furthermore, even though each observer experiences a time lag in respect to the other, they wouldn't actually observe this. Since they each measure the same speed of light over the same distance in either direction, they would assume that their clocks are also running at the same rate when stationary to each other. But other scenarios have led to different results for the time lag, however, some similar, some not, so it is still lacking some fundamental underlying principle that I am now searching for.
For instance, if all three observers were initially stationary to each other and synchronized, and then each of the first two observers are accelerated to different speeds simultaneously and the first sends pulses to the second at regular intervals according to their own clock, we would get
c*t = d - v2*t
t = d/(c+v2)
c*t' = d + v1*(ti/(1/y1)) - v2*(ti/(1/y1)) - v2*t',
t' = [d + (v1-v2)(ti/(1/y1))]/(c+v2)
where for the succeeding pulse, sent at an interval of 'ti' according to the first observer, so ti/(1/y1) to the SO, while the second closes in the original distance between them over the interval of time and while the light is in transit while the first also moves away from the second during the interval of time, increasing the distance accordingly, the SO sees the time interval of the reception of the pulses, then, as
ti' = (t' + ti/(1/y1)) - t
ti' = (v1-v2)(ti/(1/y1))/(c+v2) + ti/(1/y1)
ti' = (v1-v2)(ti/(1/y1))/(c+v2) + (c+v2)(ti/(1/y1))/(c+v2)
ti' = (c+v1)(ti/(1/y1))/(c+v2)
The emitted frequency is just 1/ti, so the ratio of observed to emitted frequency is then
fo/fe = (c+v2)(1/y1)/(c+v1)
To the second observer, however, the one actually receiving the pulses, which experiences a time dilation of 1/y2, this would further become
fo/fe = [(c+v2)(1/y1)]/[(c+v1)(1/y2)]
Running through this again in the opposite direction, where the pulse is sent from the second to the first, we find
fo/fe = [(c-v1)(1/y2)]/[(c-v2)(1/y1)]
So since this must be the same in either direction, so that the physics remains the same, we get
[(c+v2)(1/y1)]/[(c+v1)(1/y2)] = [(c-v1)(1/y2)]/[(c-v2)(1/y1)]
(c^2 - v2^2)(1/y1)^2 = (c^2 - v1^2)(1/y2)^2
(1/y1)^2/(1/y2)^2 = (c^2 - v1^2)/(c^2 - v2^2)
which will work out for 1/y1 = sqrt(1-(v1/c)^2) and 1/y2 = sqrt(1-(v2/c)^2) and all observers measure the speed of light as 'c', but it gives a Doppler shift between them of sqrt[(1-v1/c)/(1+v1/c)] * sqrt[(1+v2/c)/(1-v/c)] instead of according to their relative speed.
As I said, other various scenarios give varying results for this so far also. Some give 1-v/c for the time dilation, for instance. So I will keep working at trying to find and apply some underlying principle that produces the same results for every possible scenario. If anyone has any ideas about this, please let me know. Whether that requires a stationary frame for this to happen, or is aquired in a purely relativistic manner, it matters not. When the results are the same in every case, I will know I have it.
Ken G
2007-Sep-28, 02:07 PM
The other view is that SR's domain is the *Minkowski metric*, where the familiar, more simple rules apply. If one is doing and explaining things from some non-inertial frame (in flat space-time), one is doing "more than SR".
Yes, I think the problem with the terminology is there are really 3 flavors of relativity, not 2, so there is endless confusion about which side we should include the "middle one" in!
Me, I think I've come to the position that we should forget about SR vs. GR, and just call it 'R', Relativity, which we'll roughly define as "the machinery for doing calculations about stuff in space-time". Then we'll define "gravity" as the curvature that happens to space-time due to mass-energy as described by the EFE.That is certainly in the spirit of unifying R rather than further fragmenting into R#1, R#2, and R#3. But there is the problem that we tend to only teach R#1 to undergraduates, and R#3 to graduate students, and R#2 everyone is kind of on their own to piece together! (which they do either by transforming to a global inertial frame, doing the physics there, and transforming back, which makes it seem like SR, or including the two transformations automatically in the metric, which makes it seem like GR).
Ken G
2007-Sep-28, 02:11 PM
It should be no more surprising that an accelerating rocket must experience gradients in proper acceleration (the force felt at any point by an observer there) across its length than the fact that different points on a spinning flywheel must experience different acceleration in order for it to remain rigid. I agree, this is a good analogy, though I confess I hate how rotating systems have that big headache about being the individual points being more than just orbiting, but also themselves rotating!
So, I don't see any basic postulates being violated at all. It is what our good friend PM (Principia Mathematica :lol: ) tells us follows from our basic postulates.
I agree, and am waiting for John Mendenhall to come through on his promise to sink that battleship. My prediction: the "CERN" answer will only be correct if you make an assumption that is not natural to make in the rope problem as we analyzed it.
Jerry
2007-Sep-28, 05:39 PM
Regardless, the fundamental ideas of SR and GR are sound and correct, and are verified continually by their application in our technology.
Not quite.
The equations that fall out of relativistic theory have proven to be excellent predictqrs of past and future events...and we await further confirmation in the detection of gravity waves, not to mention a more comprehensive, unifying theory.
Ken G
2007-Sep-29, 04:33 AM
I tried various ways to get around this, but nothing has worked out so far. Finally, I gave in and tried time dilations which vary, not with the relative speed between observers, but with respect to a stationary frame.
Yes, I strongly suggest that if there is no real gravity, you find a global inertial frame to do all the calculations in. Then you can just use SR, and transform back later, and you'll always know what you are doing.
publius
2007-Sep-29, 05:20 AM
Yes, I think the problem with the terminology is there are really 3 flavors of relativity, not 2, so there is endless confusion about which side we should include the "middle one" in!
..............
That is certainly in the spirit of unifying R rather than further fragmenting into R#1, R#2, and R#3. But there is the problem that we tend to only teach R#1 to undergraduates, and R#3 to graduate students, and R#2 everyone is kind of on their own to piece together! (which they do either by transforming to a global inertial frame, doing the physics there, and transforming back, which makes it seem like SR, or including the two transformations automatically in the metric, which makes it seem like GR).
That's very true. The difference between R#1 and R#2 is exactly the same thing as the difference between Cartesian coordinates and "curvy" coordinates, such as polar in the plane. It's just that time gets thrown in the act in a non-Euclidean (but still "flat" :) ) way and makes things a lot more complicated. But it is the exact same thing. For example, write F = ma in spherical coordinates, and you get a lot of extra stuff (and that extra stuff if what Christoffel symbols are, although they are more general that what you get playing with curved coordinates in Euclid). The machinery to deal with extra stuff gets tedious enough mathematically even for Euclid.
Anyway, no one but a masochist would do things in polar coordinates if they weren't actually needed (such as equations being easier to solve there). And that's pretty much why #2 isn't explicity taught as something of itself, I think. The additional complexity is just too much to bother with at that stage -- things are hard enough without it. :)
Going to #3, is then taking that flat plane and curving it, which can be visualized by imagining the plane as the proverbial rubber sheet embedded in a 3D space, and then curving it around via that 3rd dimension. When you're sitting on that curved sheet, you've got to deal with the "extra stuff" for certain then. If you don't know how, you're up the creek without a paddle. Cartesian simplicity has abandoned us.
And so to go to #3 you have to deal with it all, and that distinction between #2 and #3 gets a bit blurred, at least until you figure it out.
-Richard
Ken G
2007-Sep-30, 03:16 PM
Anyway, no one but a masochist would do things in polar coordinates if they weren't actually needed (such as equations being easier to solve there). And that's pretty much why #2 isn't explicity taught as something of itself, I think. The additional complexity is just too much to bother with at that stage -- things are hard enough without it.
Yes, I think we may do a disservice there. We always leap to a coordinate system first, rather than teach the physics first and pick coordinates last. It's a little harder, but it would really pay dividends when we get to GR. Given that Newtonian physics is now taught in elementary college classes, if we are to imagine GR one day being in that spot, then it will come after we invert the order of learning physics/coordinates. I'm not necessarily convinced that students find coordinates easier to understand than their underlying abstract generators-- a lot of people certainly never get to the point of being able to manipulate coordinates in any meaningful way.
Now, if only John Mendenhall will come through on his promise-- I'm chomping at the bit!
grav
2007-Oct-03, 01:52 AM
Well, I still seem to be having some tremendous problems with this for some reason. I'm coming up with a solution, but it is just the same one as before, that light travels with the additional speed of the source. I've applied it in every way I can think of and I keep coming up with the same thing, so any more help is appreciated.
Let's take two scenarios. Scenario A has three observers originally moving together, stationary to each other, and synchronized. Then the first observer accelerates quickly to some relative speed to the other two until it reaches some distance from the others and then the second observer accelerates in the same way to the same relative speed to the third, which remains stationary. In this scenario, the first observer will have a time lag to the other two as it travels the distance due to the time dilation with the relative speed during its travel, and that time lag will remain after the second observer has accelerated to the same speed.
Scenario B has the first and second observers accelerate in the same way as in scenario A, but in opposite directions until they gain the same distance between each other, and then the second observer quickly accelerates in the other direction until both observers are travelling in the same direction at the same speed.
Now, with Relativity, any time lag produced will affect how we measure the speed of light between them, so such a time lag should be the same for the same relative distance between them and the same relative speed. But with scenario A, we cannot have the same time lag between one observer travelling at some relative speed for some time according to the stationary observer and two travelling away from each other at twice the relative speed for half the time according to the same observer. This is not how Relativity is set up, especially since speeds do not add together between the first and second observers as they would for the third. But just to be sure, let's say that the first, in scenario A, accelerates to a relative speed of v and travels to a distance of d before the second accelerates to the same speed. The time lag in this case would be tl = t(1-1/y), where t is the time that passes in the stationary frame and 1-1/y is the difference in times that pass between observers. t=d/v according to any observers, so we get tl = (d/v)(1-1/y). In the scenario B, the first observer accelerates to v, and so does the second, but in the opposite direction, until they reach the same distance from each other. According to Relativity, this would give a relative speed of w = (u+v)/(1+u*v/c^2) between them, so since u=v, then w = 2v/[1+(v/c)^2)]. For the same distance travelled according to either, this gives a time lag of tl = (d/w)(1-1/y'). The time lag should be the same in every scenario that results in the same distance and relative speed between observers, regardless of the method of separation, in order for things to be truly relative, at least as far as the inertial speeds are concerned, so we have (d/v)(1-1/y) = (d/w)(1-1/y'), each observer seeing this distance between them as the same in either scenario. From here, we find
(1/v)(1-1/y) = (1/w)(1-1/y')
1/v - 1/w = 1/vy - 1/wy'
c/v - c/w = c/vy - c/wy'
1/y = sqrt[1-(v/c)^2)]
1/y' = sqrt[1-(w/c)^2)]
c/v - [1+(v/c)^2]*c/2v = c*sqrt[1-(v/c)^2]/v - c*[1+(v/c)^2]*sqrt[1-4v^2/(1+(v/c)^2)c^2]/2v
I have tried this for some values of v/c and it doesn't work out, as it should for all values. Also, the acceleration while changing frames wouldn't affect this since the distance can always be made greater for a greater time lag due to a greater separation while any time lag during the same acceleration remains the same.
I have found that one solution to this would be that with a reference to an absolute frame where the time dilation is greater in one direction and smaller in the other, but I don't see how that could work out for all directions of space, especially perpendicularly then, where there would be no time dilation at all in that case, so we might as well figure that there is no time dilation in any direction to begin with. Another is that objects somehow retain a memory of the amount of time lag they have undergone relative to the absolute frame, and will regain that upon acceleration, the entire amount upon returning to the stationary frame, but reading a greater time if continuing to accelerate. It would have the additional effect of regaining the same time as the stationary frame, regardless of speed, when coming back to the same point from which it originally departed from the third observer with which it was originally stationary. The problem with these, though, is that I would would no idea what could produce such effects in order for it to work out this way.
Now, going back to the MM experiment, what started all of this, we must have the same times for light to travel some distance forward as back, and according to Relativity, without an absolute frame, all observers should see light travelling at c, not relative to an absolute frame but to the observer. But if there is a time lag between two observers, then they would measure the speed of light as d/(T+tl) in one direction between them, and d/(T-tl) in the other. Even with a Lorentz contraction, this still becomes dL/(T+tl) and dL/(T-tl), which cannot be equal for the speed of light to be measured the same unless tl=0. Simulataneity, however, also says that the time lag between two observers in the same frame is zero, regardless of the distance that separates them. To the third observer, then, who always sees the speed of light as c in their own frame as well, they would measure the time between light sent from the second observer to the first as c*t1 = d + v*t1, so t1 = d/(c-v), and from the first to the second as c*t2 = d - v*t2, so t2 = d/(c+v). Without a time lag between them, if these two times are the same to the moving observers, then so should they be for the stationary one, regardless of the time dilation involved. So we get
d/(c-v) = d/(c+v)
1/(c-v) = 1/(c+v)
which obviously doesn't work out. Since there is only one variable here, v, but that is already set as the given, then we must try c as a variable also. That gives us
1/(c1-v) = 1/(c2+v)
Now, we can immediately see that c1=c+v and c2=c-v would work out, whereas light would be emitted with the additional speed of the source, but let's be sure. So far we really just have c1 = c2 + 2v. Applying this back into scenario A, for light travelling in either direction between the first and second observers, we get
tl = (d/v)(1-1/y)
c1*t1 = d + v*t1
t1 = d/(c1-v) - tl
c2*t2 = d - v*t2
t2 = d/(c2+v) + tl
d/(c1-v) - tl = d/(c2+v) + tl
c1 = c2 + 2v
d/(c1-v) - (d/v)(1-1/y) = d/(c2+v) + (d/v)(1-1/y)
1/(c1-v) - (1/v)(1-1/y) = 1/(c2+v) + (1/v)(1-1/y)
1/(c1-v) - 1/(c2+v) = (2/v)(1-1/y)
1/(c1-v) - 1/(c2+v) = (2/v)(1-1/y)
c1 = c2 + 2v
(2/v)(1-1/y) = 0
1-1/y = 0
1/y = 1
So according to this, there is no time dilation, no time lag, and light travels with the additional speed of the source. So what am I still overlooking? It doesn't seem that this should work as well as Relativity does in explaining things such as precession and gravitational lensing and such, being so very different. I suppose I will have to go ahead and try running some similations for it, though, just in case. But regardless if light works this way or not, it wouldn't seem that gravity should, since bodies don't so much emit gravity anyway, as far as I know, but pretty much simply manipulate what already exists within that space. Does anybody have any other ideas about all of this, or another fundamental principle that might be applied?
Urbane Guerrilla
2007-Oct-03, 08:28 AM
A point of verbal usage unrelated to the math (which I doubt I can follow anyway): dilation is the popular usage, but shouldn't the word be dilatation -- a slowing down?
hhEb09'1
2007-Oct-03, 03:47 PM
A point of verbal usage unrelated to the math (which I doubt I can follow anyway): dilation is the popular usage, but shouldn't the word be dilatation -- a slowing down?In American English (http://dictionary.reference.com/browse/dilatation), dilatation doesn't mean slowing down, it means ... dilation. :)
grant hutchison
2007-Oct-03, 04:53 PM
In American English (http://dictionary.reference.com/browse/dilatation), dilatation doesn't mean slowing down, it means ... dilation. :)Likewise in British English, according to the Oxford English Dictionary. The entry for dilatation makes no mention of the sense "slowing down", only of various forms of stretching and expansion.
(In fact, "Delay, procrastination, postponement" is given as an obsolete meaning of dilation, from which we get dilatory, "tending to cause delay".)
So I think both time dilation and time dilatation refer to the "stretching" of time.
Grant Hutchison
grav
2007-Oct-11, 10:38 PM
Well, I have found so far that observers cannot observe the same physics with or without an absolute frame unless light is emitted with the additional speed of the source. The only other possibility following this premise was that time was somehow wound up like a spring so that the time lag between two observers is conserved upon accelerating to a speed different from an absolute frame and then released upon decelerating back to it, to regain the same time as that of the absolute frame, and then continues to wind the other way if the observer continues to accelerate in the other direction. But if an observer regains his stationary status in the absolute frame for a time and then accelerates again, he could either wind up or back down again equally, unless he winds up in one direction and down in the other, but then space wouldn't be symmetrical.
So that didn't really make much sense, but I still had to explore the possibility. Besides that, though, I found a set of scenarios where this can't work anyway. Three observers are stationary. One observer accelerates to some speed in one direction and continues at that speed for a time, and then another observer accelerates to the same speed in the opposite direction. When they reach some distance from each other, they each emit light to the other and the times measured for the light to travel this distance will depend upon the time lag between them. Compare this to that of the first and second observers accelerating to the same speed in opposite directions simultaneously. Or to the first travelling out for a shorter or longer time before the second. When they reach the same distance from each other, the time lag will vary in each particular case, so they would measure the speed of light differently in each case as well, even between two sets of observers in the same frames as the corresponding observers in each set, simply because the time lags can be arbitrarily different in each case. The acceleration and time getting "wound up" do not matter because the acceleration is the same in every case, and therefore any time lag produced by it, and since neither observer doubles back to unwind it as well.
So what are we left with? Under this premise, light must travel with the additional speed of the observer. That's it. No time lag, no time dilation, no contraction, and no gamma. I haven't begun exploring this in detail yet, but I will soon. I will apply it to precession and gravitational lensing. The only thing here is that this is found for the way light travels in order for all observers to observe the same physics, and the speed of light always as c, in accordance with the MM experiment, but it says nothing of gravity itself, which may operate under somewhat different principles, and most all of the verification of Relativity relates to gravity. Anyway, if I manage to figure anything out along these lines, I will post it in ATM.
In the meantime, for the rest of this thread, I still want to find something that agrees better with the way Relativity is laid out. So I want to see if there is anything about the premise for these scenarios that might vary. Any assumptions about the way physics operates or any other underlying principle that has not been taken into account. I cannot change the original premise that all observers observe the same physics, because then we wouldn't have Relativity at all, nothing based solely upon the relative distance and speed, but instead we could get any arbitrary result depending upon how one got there, and the physics for this would have to be determined through experiment in each case. Surely the universe is kinder to mathematicians than that, even amateur ones. :)
Since that is the only real premise here, maybe something about the physics itself. Each set has been based upon the assumption that the observers accelerate to some speed and then continue in a straight line inertially. Since the acceleration doesn't matter and the inertial part is the only other physics that applies, then I am looking at that. First, we could say that the observers do not follow a straight line. But then there is nothing to determine what other path they would follow, in what direction, at least not without becoming assymetrical. And if the observers are to observe the same physics, then the situations should be the same and symmetrical in this regard. So that one's out.
The only other thing I can see, then, is that the speed is not inertial. The observers must either accelerate or decelerate as they travel through space in such a way as to gain the same time lag in each case. If they decelerate, then this might be caused by some density they must travel through. But this density must be the same mechanism that produces the speed of light also. If not, then the density and resulting deceleration could have any arbitrary value whatsoever, therefore even zero, which would bring us back to our original scenarios. So if the density increased, and therefore the deceleration was made greater, then the speed of light would have to change with it, in order for all things to remain relative between observers.
I have tried this for the original set of scenarios, however, and it's looking like the observers must actually accelerate to remain relative with a time lag. This could mean that the space between them is increasing while they travel, but again, it would have to directly relate to the speed of light as well, so may depend upon some expanding space or negative pressure, which might still contain a density of some sort. If the space is expanding so that the density decreases, whereas the observers would separate more than just inertially due to the expansion while decelerating due to the density, but decelerating less and less as the density decreases also, well, then that would make for some interesting calculations. :wall::eek:
grav
2007-Oct-16, 06:42 PM
Well, relating the time dilation of the observers to an absolute frame, with some sort of acceleration or expanding distance, was easy. If we take that last set of scenarios, then the second one says that the two observers accelerate and then travel inertially in the same way in opposite directions, so their motions relative to the absolute frame are symmetrical. Any additional acceleration or expansion of the distance that is experienced by one, relative to the absolute frame, will also be experienced by the other in the same way as well, then. So their motions are symmetrical no matter what, and so will be their time dilations relative to the absolute frame also. Therefore, there can be no time lag between them regardless.
Relating this to the first of the scenarios, where one accelerates and travels out some distance before the other accelerates in the opposite direction, then, for the physics to be the same for the same relative distance and relative speed as with the second scenario, there can be no time lag here either. The only way to do that with the first observer travelling out with some time dilation relative to the absolute frame before the second does, so that the first does not experience any time lag relative to the second, is if the times of all observers remains the same, so that there is no time dilation to begin with. So we are back to the c+/-v relative to the source thing with that as well, in regards to an absolute frame.
I still have to try to relate it between the observers themselves, however, instead of an absolute frame, as with Relativity, where the time dilation takes place between the observers. It does not work with inertial speeds, but although I haven't found a way that it can work with an acceleration or expanding distance yet, I have not shown that it cannot yet either. I'll keep working on it.
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## The ideal generated by the variable is maximal iff the coefficient ring is a field
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.18 Solution: We first prove a lemma. Lemma: The map $\varphi : R[[x]] \rightarrow R$ given by…
## In an integral domain, two principal ideals are equal precisely when their generators are associates
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A130663 1/2048 the number of permutations of 0..n having exactly 7 maxima. 2
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10922, 929569, 43800104, 1513315264, 42878777216, 1057229805952, 23518119959040, 483438472237056, 9338507684459520, 171607973968784640, 3027838101326610432, 51660428017842610176, 857132463549737238528 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,13 COMMENTS a(n) = A179710(n+1)/2048. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..300 FORMULA G.f.: -x^12* (264868724736000*x^15 -891839368396800*x^14 +1387289679298560*x^13 -1320505755697152*x^12 +859006229078016*x^11 -404049277108224*x^10 +141829511625984*x^9 -37804275799552*x^8 +7710418349056*x^7 -1202843456128*x^6 +142319143104*x^5 -12540195936*x^4 +796479552*x^3 -34424192*x^2 +905327*x -10922) / ((14*x-1) *(12*x-1)^2 *(10*x-1)^3 *(8*x-1)^4 *(6*x-1)^5 *(4*x-1)^6 *(2*x-1)^7). - Alois P. Heinz, Oct 01 2013 CROSSREFS Cf. A100575. Sequence in context: A089381 A092007 A043581 * A045156 A003919 A179250 Adjacent sequences: A130660 A130661 A130662 * A130664 A130665 A130666 KEYWORD nonn AUTHOR R. H. Hardin, Aug 11 2007 EXTENSIONS a(20)-a(24) from Alois P. Heinz, Oct 01 2013 STATUS approved
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# gaussian elimination
• Jul 9th 2006, 02:07 AM
cheesepie
need help urgent !!
Gaussin Elimation
1)x+2y+6z=12
2z=2
-3x-6y-9z=-27
2)x-y+z=22
2y-z=-23
2x+3y=-11.2
3)x+2y+4z=7
3x+y+4z=-2
2x+9y-2x=10
4)x+2y+4z=7
3x+y+4z=-2
2x+9y-2z=10
• Jul 9th 2006, 06:37 AM
Soroban
Hello, cheesepie!
#1 is a set of dependent equations.
Quote:
$\displaystyle 1)\;\begin{array}{ccc}x+2y+6z\:=\:12\\2z\:=\:2 \\ -3x-6y-9z\:=\:-27\end{array}$
We have: .$\displaystyle \begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 2 & | & 2 \\ \text{-}3 & \text{-}6 & \text{-}9 & | & \text{-}27\end{vmatrix}$
$\displaystyle \begin{array}{ccc} \\ R_2\div2\\ R_3\div(\text{-}3)\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 1 & 2 & 3 & | & 9\end{vmatrix}$
$\displaystyle \begin{array}{ccc}\\ \\ R_3-R_1\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & \text{-}3 & | & \text{-}3\end{vmatrix}$
$\displaystyle \begin{array}{ccc} \\ \\ R_3\div(-3)\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & 1 & | & 1\end{vmatrix}$
$\displaystyle \begin{array}{ccc} \\ \\ R_3-R_2\end{array}\;\begin{vmatrix}1 & 2 & 6 & | & 12 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & 0 & | & 0\end{vmatrix}$
Since we have a rows of 0's (zeros all the way across), the system is dependent.
There is an infinite number of solutions to the system.
• Jul 10th 2006, 05:34 AM
cheesepie
tks...how about other question ? :D
• Jul 10th 2006, 07:06 AM
Soroban
Hello, cheesepie1
I need to know exactly where your difficulty is . . .
. . You don't understand Gaussian Elimination at all?
. . You don't know how to get the 1's and 0's?
. . You don't know what order to work at it?
I can spend hours, doing ALL your homework for you,
. . but there is no learning taking place, it's wasted effort.
Having said that, here's the second one.
And I would really like to know exactly why you can't do the third one.
Quote:
$\displaystyle 2)\;\begin{array}{ccc}x-y+z \:=\:22 \\ 2y - z\:=\:-23 \\ 2x + 3y\:=\:-11.2\end{array}$
We have: .$\displaystyle \begin{vmatrix}1 & \text{-}1 & 1 & | & 22 \\ 0 & 2 & \text{-}1 & | & \text{-}23 \\ 2 & 3 & 0 & | & \text{-}11.2\end{vmatrix}$
$\displaystyle \begin{array}{ccc} \\R_2\div2\\R_3-2\cdot R_1\end{array}\;\begin{vmatrix}1 & \text{-}1 & 1 & | & 22 \\ 0 & 1 & \text{-}0.5 & | & \text{-}11.5 \\ 0 & 5 & \text{-}2 & | & \text{-}55.2\end{vmatrix}$
$\displaystyle \begin{array}{ccc}R_1+R_2\\ \\ R_3-5\cdotR_2\end{array}\;\begin{vmatrix}1 & 0 & 0.5 & | & 10.5 \\ 0 & 1 & \text{-}0.5 & | & \text{-}11.5 \\ 0 & 0 & 0.5 & | & 2.3\end{vmatrix}$
$\displaystyle \begin{array}{cccc} R_1-0.5\cdot R_3 \\ R_2+0.5\cdot R_3 \\ \\ \end{array}\;\begin{vmatrix}1 & 0 & 0 & | & 8.2 \\ 0 & 1 & 0 & | & \text{-}9.2 \\ 0 & 0 & 1 & | & 4.6\end{vmatrix}$
Answers: .$\displaystyle x = 8.2,\;y = -9.2,\;z = 4.6$
• Jul 10th 2006, 07:15 AM
cheesepie
i dont understand the inverse matrices
example how u change inverse of b =
[ 8 4 ]
[ 3 2 ]
into
[1 0 ]
[0 1 ]
• Jul 10th 2006, 07:56 AM
topsquark
You aren't inverting matrices using this method. You are performing "row operations." For example, you have the system of equations:
(1) x - y + z = 22
(2) 2y - z = -23
(3) 2x + 3y = -11.2
I will solve this in an identical manner as Soroban, except that he is using Gaussian elimination and I am simply manipulating equations. What I want you to note is that they are the same thing, just that one is in matrix notation and one is not.
So. The first operation Soroban did was R2 $\displaystyle \div$2. This is the same as dividing both sides of equation 2 by 2:
(1) x - y + z = 22
(2)' y - 0.5z = -11.5
(3) 2x + 3y = -11.2
The next thing he did was R3 - 2R1. This is subtracting twice equation 1 by R3.
2x + 3y = -11.2 (Equation 3)
-2x + 2y - 2z = -44 (Equation 1 with both sides multiplied by -2 )
---------------------------
(3)' 5y - 2z = -55.2
So we have:
(1) x - y + z = 22
(2)' y - 0.5z = -11.5
(3)' 5y - 2z = -55.2
Now R1 + R2. This is equation 1 + equation 2'
x - y + z = 22
y -0.5z = -11.5
-------------------
(1)' x + 0.5z = 10.5
So
(1)' x + 0.5z = 10.5
(2)' y - 0.5z = -11.5
(3)' 5y - 2z = -55.2
Now R3 - 5R2 means equation 3' - 5 times equation 2"
5y - 2z = -55.2
-5y +2.5z = 57.5
----------------------
(3)" 0.5z = 2.3
So we have
(1)' x + 0.5z = 10.5
(2)' y - 0.5z = -11.5
(3)" 0.5z = 2.3
R1 - R3 means equation 1' minus 0.5 times equation 3" (Soroban's statement here was incorrect.)
x + 0.5z = 10.5
- 0.5z = -2.3
----------------
(1)" x = 8.2
and R2 + R3 means equation 2' plus equation 3" (Soroban's statement here was incorrect.)
y - 0.5z = -11.5
0.5z = 2.3
----------------
(2)"' y = -9.2
So
(1)" x = 8.2
(2)"' y = -9.2
(3)" 0.5z = 2.3 (Then, of course, solve this for z!)
You can compare each step in the above with Soroban's matrix form.
-Dan
• Jul 10th 2006, 11:02 AM
Soroban
Hello, cheesepie!
Now you've changed the subject . . .
Quote:
i dont understand the inverse matrices
Example: how u change inverse of b =
[ 8 4 ]
[ 3 2 ]
into
[1 0 ]
[0 1 ]
You haven't been taught this procedure either??
The set up is: /$\displaystyle \begin{bmatrix}8 & 4 & | & 1 & 0 \\ 3 & 2 & | & 0 & 1\end{bmatrix}$
Now, use Gaussian Elimination to change the left matrix into the Identity: .$\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$
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Journal of Mathematical Physics, Analysis, Geometry 2021, vol. 17, No 3, pp. 369-387 https://doi.org/10.15407/mag17.03.369 ( to contents , go back )
### Exact Solutions of Nonlinear Equations in Mathematical Physics via Negative Power Expansion Method
Bo Xu
School of Mathematics, China University of Mining and Technology, Xuzhou 221116, China
School of Educational Science, Bohai University, Jinzhou 121013, China
E-mail: bxu@bhu.edu.cn
Sheng Zhang
School of Mathematics and Physics, Bohai University, Jinzhou 121013, China
E-mail: szhangchina@126.com
Received May 19, 2020, revised June 14, 2020.
Abstract
In this paper, a direct method called negative power expansion (NPE) method is presented and extended to construct exact solutions of nonlinear mathematical physical equations. The presented NPE method is also effective for the coupled, variable-coefficient and some other special types of equations. To illustrate the effectiveness, the (2 + 1)-dimensional dispersive long wave (DLW) equations, Maccari’s equations, Tzitzeica–Dodd–Bullough (TDB) equation, Sawada–Kotera (SK) equation with variable coefficients and two lattice equations are considered. As a result, some exact solutions are obtained including traveling wave solutions, non-traveling wave solutions and semi-discrete solutions. This paper shows that the NPE method is a simple and effective method for solving nonlinear equations in mathematical physics.
Mathematics Subject Classification 2010: 35Q51, 35J99, 68W30
Key words: exact solution, NPE method, (2+1)-dimensional DLW equations, Maccari’s equations, TDB equation, SK equation with variable coefficients, lattice equations
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https://www.homeworklib.com/questions/802714/hugo-bakes-world-famous-scones-the-key-to-his
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# Hugo bakes world famous scones. The key to his success is a special oven whose temperature varies according to a sinusoi...
Hugo bakes world famous scones. The key to his success is a special oven whose temperature varies according to a sinusoidal function; assume the temperature (in degrees Fahrenheit) of the oven t minutes after inserting the scones is given by: y = s(t) = 10 sin (π/5t − 3π/2) + 410.
*** I need help finding the mean and parts D-F, please and thank you!
(a) Find the amplitude, phase shift, period, and mean for s(t). 10 amplitude 15 phase shift 2 10 period mean
(b) What is the maximum temperature of the oven? 420 oF Give all times when the oven achieves this maximum temperature during the first 20 minutes. (Enter your answers as a comma-separated list.) 0, 10, 20 min (c) What is the minimum temperature of the oven? OF 400 Give all times when the oven achieves this minimum temperature during the first 20 minutes. (Enter your answers as a comma-separated list.) 5,15 min (d) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is at least 405°F. (Round your answer to two decimal places.) min (e) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is at most 415°F. (Round your answer to two decimal places.) min (f) During the first 20 minutes of baking, calculate the total amount of time the oven temperature is between 405°F and 415°F. (Round your answer to two decimal places.) min
we are given
Calculation of mean:
now, we can find average of these two values
(d)
we are given
S(t)=405
now, we can set it to and then we can solve for t
(e)
we are given
S(t)=415
now, we can set it to and then we can solve for t
(f)
we can find time between
##### Add Answer of: Hugo bakes world famous scones. The key to his success is a special oven whose temperature varies according to a sinusoi...
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Question
# An object with a mass 10kg moves at a constant velocity of 10m/s. A constant force then acts for 4second on the object and gives it a speed of 2m/s in opposite direction. The acceleration produced is:
A
3m/sec2
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3m/sec2
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Solution
## The correct option is B −3m/sec2Initial velocity of the object V1=10m/sFinal velocity of object V2=−2m/sTime for which the force acts t=4 sThus acceleration of the object a=V2−V1t=−2−104=−3m/s2
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# Search by Topic
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Can you replace the letters with numbers? Is there only one solution in each case?
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
##### Stage: 1 and 2 Challenge Level:
Who said that adding couldn't be fun?
### Four-digit Targets
##### Stage: 2 Challenge Level:
You have two sets of the digits 0 – 9. Can you arrange these in the five boxes to make four-digit numbers as close to the target numbers as possible?
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Song Book
##### Stage: 2 Challenge Level:
A school song book contains 700 songs. The numbers of the songs are displayed by combining special small single-digit boards. What is the minimum number of small boards that is needed?
### Diagonal Sums
##### Stage: 2 Challenge Level:
In this 100 square, look at the green square which contains the numbers 2, 3, 12 and 13. What is the sum of the numbers that are diagonally opposite each other? What do you notice?
### Oddly
##### Stage: 2 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Spell by Numbers
##### Stage: 2 Challenge Level:
Can you substitute numbers for the letters in these sums?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### ABC
##### Stage: 2 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### (w)holy Numbers
##### Stage: 2 Challenge Level:
A church hymn book contains 700 hymns. The numbers of the hymns are displayed by combining special small single-digit boards. What is the minimum number of small boards that is needed?
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Largest Even
##### Stage: 1 Challenge Level:
How would you create the largest possible two-digit even number from the digit I've given you and one of your choice?
##### Stage: 1 Challenge Level:
If you put three beads onto a tens/ones abacus you could make the numbers 3, 30, 12 or 21. What numbers can be made with six beads?
### Method in Multiplying Madness?
##### Stage: 2 and 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### Snail One Hundred
##### Stage: 1 and 2 Challenge Level:
This is a game in which your counters move in a spiral round the snail's shell. It is about understanding tens and units.
### Our Numbers
##### Stage: 1 Challenge Level:
These spinners will give you the tens and unit digits of a number. Can you choose sets of numbers to collect so that you spin six numbers belonging to your sets in as few spins as possible?
### That Number Square!
##### Stage: 1 and 2 Challenge Level:
Exploring the structure of a number square: how quickly can you put the number tiles in the right place on the grid?
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##### Stage: 2 Challenge Level:
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### Round the Two Dice
##### Stage: 1 Challenge Level:
This activity focuses on rounding to the nearest 10.
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##### Stage: 1 Short Challenge Level:
I am less than 25. My ones digit is twice my tens digit. My digits add up to an even number.
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##### Stage: 1 Challenge Level:
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##### Stage: 2 Challenge Level:
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##### Stage: 2 Challenge Level:
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##### Stage: 2 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
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##### Stage: 1 Short Challenge Level:
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##### Stage: 2 Challenge Level:
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##### Stage: 2 Challenge Level:
Investigate the different ways these aliens count in this challenge. You could start by thinking about how each of them would write our number 7.
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##### Stage: 2 Challenge Level:
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##### Stage: 2 Challenge Level:
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##### Stage: 1, 2 and 3
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##### Stage: 1, 2 and 3 Challenge Level:
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##### Stage: 1
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##### Stage: 2 Challenge Level:
Dicey Operations for an adult and child. Can you get close to 1000 than your partner?
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##### Stage: 1 Challenge Level:
You have a set of the digits from 0 – 9. Can you arrange these in the 5 boxes to make two-digit numbers as close to the targets as possible?
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# Heavy Light Worksheet Kindergarten
### Measurable Attributes (Heavy/Light)
Measurable Attributes (Heavy/Light) Grade and Content Area Kindergarten Mathematics Title Measurable Attributes (Heavy/light) GLEs/GSEs M(G&M)-K-7 Demonstrates conceptual understanding of measurable attributes using comparative language to describe and compare attributes of objects (length ...
www.ride.ri.gov
### Heavy OR Light
adding_1_0ther_numbers_no-illustration adding_1_0ther_numbers_no-illustration. Name : ..... Class : .....
math4children.com
### Sources of Light Energy Worksheet
Name:_____ Date:_____ Class Period:_____ Sources of Light Energy Worksheet Incandescent bulbs are more ... A light bulb's electric use is measured in watts per hour, but electricity is sold by kilowatt-hours, or units of 1000-watt hours.
www.beaconlearningcenter.com
### 2003 Colorado Unit Writing Project
Kindergarten, Math Measures Up! 2003 Colorado Unit Writing Project 1 Math Measures Up! ... Weight (mass) a) Heavy, light b) Heavier than, lighter than iii. ... Math Measures Up! 2003 Colorado Unit Writing Project 24 Appendix H Full, Half-Full, Empty Worksheet ...
### KINDERGARTEN MINERALS
Math/Science Nucleus © 1990, 2001 2 ROCK CYCLE OVERVIEW OF KINDERGARTEN CHEMISTRY WEEK 1. PRE: Distinguishing the four types of matter. LAB: Classifying heavy and light rocks. ... Ask the students to color in the pyrite crystal on the worksheet. If they don't have a gold crayon ...
www.msnucleus.org
### Kindergarten supplement
Bridges in mathematics Kindergarten s upplement set d2 Measurement: Comparing Weight The Math Learning Center, PO Box 12929, Salem, Oregon 97309. ... Light or heavy? Have that child leave the group, return with something he or she thinks is about the same weight as the sack of potatoes, and sit back ...
www.mathlearningcenter.org
### KINDERGARTEN
KINDERGARTEN NOVEMBER ROCKS AND SOILS TOTAL TIME: 45 Minutes (15 minutes per station) STATION 1: CLASSIFYING ROCKS STATION 2: ROCKS & SOIL STATION 3: LET ... The idea of this activity is for the children to use their senses to observe and group the rocks by their attributes such as large/small, heavy/light ...
www.dolvinpta.org
### Floating and sinking
Teaching note: In this activity, some students will push for a small number of explanatory ideas (e.g. 'these are heavy or have holes', 'these are made of light material ... It would be useful to design a worksheet that will enable you to record predictions and subsequent observations of the objects in ...
www.deakin.edu.au
### 52 Teaching Basic Concepts
Upon entering kindergarten, children should understand concepts as pairs, with the unmarked concept understood receptively and the marked concept ... More/Less Top/Bottom Big/Little All/None Front/Back Thick/Thin Old/New Long/Short Hard/Soft Over/Under Hot/Cold Smooth/Rough High/Low Always/Never Heavy/Light ...
www.superduperinc.com
### PRESCHOOL SYLLABUS
opposites Open & close, tall & short, full & empty, over & under, heavy & light big & small, soft & hard, sink & float, hot & cold colours Red, yellow, blue orange, green, purple, black, white shapes Circle, square, triangle rectangle, oval sight reading months of the year E. SING ALONG SONGS / RHYMES F. STORY ...
vibgyorkids.net
### Other sites you could try:
Find videos related to Heavy Light Worksheet Kindergarten
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# Why are dividends different from property income like rent?
My question should be more like: why dividends are inseparable from the appreciation of a stock's price?
The common explanation is something like:
Stock A pays a \$10 (10%) dividend and a share is worth \$100. Stock B pays no dividend and is also worth \$100/share. If they both grow by 10% at the end of the year, you would have the same return on the both of them. This is because A's prices goes up to \$110 and after paying a \$10 dividend, A's share price goes down to \$100. B's price grows to \$110. 100+10=110
Makes sense. But not to me. I don't know why but I can't wrap my head around why the dividend takes money away from the share price. I thought the share price was the price for the piece of the company, and not the money is pays out. Is that not true?
To me, I see a dividend as the same thing as income from renting out a property. If you own a \$100k house (assume no mortgage) and you rent it out for \$1k/month, your yearly income from the house is \$12k (12%). If the price of the house goes up 10%, then you still have the \$12k but also \$110k house so your total net worth is \$122k. But if you (for the sake of argument) just owned the house so that it appreciates in value and did not rent it out, you would be \$12k short at a 10% growth. This is because rental income is separate from the value of the house, obviously.
Why are dividends not the same? Why do dividends take money out of share price and not considered income from the company the same way rent is income from property?
As an aside, I understand that stocks that pay dividends are not automatically better than stocks that don't. That isn't my claim. Dividends are irrelevant in determining whether a stock is a solid investment (that is not to say they are irrelevant to your returns, see this calculator).
• Are you saying that "government taxes are irrational" ... ? The only answer can be "that's life". Jul 21 '20 at 0:46
• If you bought this house 50-50 with a partner, on the expectation that you would always give the partner \$6000 per year (whether rented or vacant), that would be the equivalent of the dividend in your example. Jul 21 '20 at 4:56
• Dividends come from the company itself. It's like if you went and took a tree from your house each month. It doesn't matter whether you leave the tree there or take it away, either way you have the value of the tree plus the value of the rest of the house. Jul 21 '20 at 10:20
• This (much repeated) premise about dividends is flawed. What it misses is that in both theory and empirically a stocks price rises the amount of the dividend and then drops that amount. The only difference between theory and the reality is a great mystery of finance: the aggregate drop in price is less than the rise. Jul 21 '20 at 14:57
• @JimmyJames: Perfectly efficient markets are like spherical cows, or doing your physics problems without taking friction into account. Jul 22 '20 at 15:31
I will give a counterpoint to Bob Baerker's answer.
It doesn't decrease the value of the house as a physical asset, but it does decrease the value of the house as a financial asset. When you have a tenant, your house is encumbered by a rental agreement (either lease or month-to-month).
Let's say a buyer is required to assume the rental agreement (keep the tenant on the same terms until they can otherwise be renegotiated). Suppose the tenant pays \$1,000 rent for August on August 1.
Just before the tenant pays, let's say the buyer is willing to pay \$X for the house. That deal includes the buyer receiving the tenant's August rent payment (because the rental agreement has been assigned to the buyer) and continuing to grant the tenant use of the house for the month of August.
On the other hand, if you sell just after the tenant pays you the rent, then the buyer has the same obligation to grant the tenant use of the house for the month of August, but the buyer does not receive that \$1,000. It follows that the buyer is willing to pay \$(X - 1,000) for the house.
So, a house and a stock are more similar than they may seem. The economic value of a company includes not only its physical assets but also its claims (loans and deposits, accounts payable and receivable). Likewise, the economic value of a house includes claims like rental agreements, liens, etc.
Whenever there is a contractual discrete cash flow whose recipient is determined by who owns property on a particular date, the market value of that property will drop immediately after that date as a new owner will no longer be entitled to that payment.
The value will exhibit a characteristic sawtooth pattern versus time. Even if the physical attributes are stable, economic value builds up gradually as claims (hopefully) accumulate from profitable use of assets, then jumps down when cash is extracted. In between rent payments, the house value has an extra increasing trend because the tenant is using up the month they've paid for, and the owner's remaining obligation to the tenant is declining.
It would be difficult to demonstrate this empirically with real estate since the property value is not precisely quoted daily and value of one month's rent is likely lost in the noise of illiquidity, but the principle is valid.
• @MrMineHeads Correct, company A is profitable but does not invest in growing its operations. It just collects profits in its bank account and periodically distributes them to shareholders. The variations in stock price are entirely due to the timing of the dividend. However, stock A has genuinely provided a 10% return. Jul 21 '20 at 3:47
• @LasseMeyer Yes, stock prices usually go down after dividends have been paid (technically, after the date entitling shareholders to the dividend -- it may actually be paid days or weeks later). But the point is that the benefit of buying the stock "cheap" is counterbalanced by not receiving the dividend. Or put another way, you're not actually buying the company at a cheaper valuation, because you're buying a "poorer" company with less cash in its bank account (by the amount of the dividend it just paid out). Jul 21 '20 at 11:27
• @Lasse Meyer - Stock price always goes down by the amount of the dividend because stock exchanges reduce share price by that amount on the ex-div date. This isn't always apparent the next day because if there's buying and share price rises, it can mask the reduction. nanoman makes an important point. In terms of valuation, after the ex-dividend reduction, the stock hasn't gone on sale. It's a "poorer" company due to the pending dividend pay out. However, many people do indeed buy a stock post dividend because it appears cheaper. Jul 21 '20 at 13:44
• @Gnudiff You receive the full dividend if you own the stock as of the night before ex-dividend. You could own it for as little as 8 hours if you buy after-hours and sell premarket the next morning. Jul 21 '20 at 21:56
• @BobBaerker I'm not talking about trading resuming the next day, I'm talking about actual trades at close on the ex-div day. Also, by price I'm referring to the value at which one can buy or sell the asset (as I believe are most people since that is what has relevant meaning in the context of this question), not what an exchange lists as the price in historical data. Jul 25 '20 at 18:05
Just to give an alternative analogy to the other answers:
Think of a company as being like your bank account. If the bank account has \$100 and it earns 10% interest per year:
A) If you decide to keep the interest in the account, then at the end of the year the account has \$100 + \$10 = \$110. This is analogous to a company earning profits of \$10 and retaining those profits.
B) If you decide to withdraw the interest in the account, then at the end of the year the account has \$100 but you have \$10 in cash = \$110. This is analogous to a company earning profits and paying them out as dividends. The bank account / company is worth \$100 instead of \$110 because it now no longer has the extra \$10.
If you own a \$100k house (assume no mortgage) and you rent it out for \$1k/month, your yearly income from the house is \$12k (12%). If the price of the house goes up 10%, then you still have the \$12k but also \$110k house so your total net worth is \$122k. But if you (for the sake of argument) just owned the house so that it appreciates in value and did not rent it out, you would be \$12k short at a 10% growth. This is because rental income is separate from the value of the house, obviously.
This analogy is incorrect. With the company that doesn't pay dividends, it is still earning the profits of \$10, just choosing not to pay them out to shareholders. In your example where you don't rent out the house, you are failing to make any profits at all. These are not comparable scenarios. The correct analogy is that you rent out the house for \$12k, but instead of receiving that money as income, you re-invest it into the house (e.g. by improving or enlarging it). If we assume that \$12k of improvements results in \$12k of extra value on the house, your house is now worth \$122k as expected. This is what happens with the non-dividend paying company: the \$10 that it fails to pay in dividends is retained in the company and increases its value by "improving" it (by increasing it's cash balance).
• This is a very good analogy for the comparison of dividends and rent. It deals in actual value rather than esoteric economic value. +1 Jul 21 '20 at 13:59
• The problem with the bank account analogy is that the book value of a company is not the market valuation. If you can find a where the book value and market valuation are equal, it's probably a failing one. Jul 29 '20 at 15:03
• @JimmyJames True, but analogies are aids to understanding rather than accurate definitions. Jul 29 '20 at 22:19
• Unless they are misleading. Then they can inhibit understanding. The book value of a company has a very indirect and subtle relationship with the stock value. The book value of a bank account is its value. It's revenue that is core to the valuation of a company. The amount of cash on hand is a factor but has no direct 1-to-1 association with its value. Therefore the idea that the value of the stock dropped by dividend amount because it has that much less cash is preposterous but that's the implication of your analogy. Jul 30 '20 at 15:49
• @JimmyJames Revenue is a part of the value of a company, but there is no denying that cash has value. Ceteris paribus, if company X has \$1 more cash than company Y, then it is worth \$1 more. Similarly, if a company has \$100 in its bank account today, and tomorrow it has \$99 because of a dividend payout (i.e. nothing else to show for the \$1 spent), then ceteris paribus it is worth \$1 less than it was yesterday. There is nothing preposterous about that, it is basic accounting principles. Jul 30 '20 at 23:11
What's the difference between dividends and rent?
When a dividend is paid, that cash is removed from the company, decreasing the company's value. For that reason, stock exchanges decrease share price by the amount of the dividend on the ex-dividend date.
Suppose that share price was not reduced by the amount of the dividend on the ex-dividend date. For ease of discussion, let's pretend that per your example, it's a \$10 dividend paid once a year. Everyone would buy your \$100 stock at the close the day before the dividend and in the morning, the stock would be \$100 before trading opened and the company would owe you \$10, to be paid on the Payable Date. Now what's wrong with that picture?
I think that dividends being taxed as income (if received in a non sheltered account) has led to a massive misconception by the public that dividends are income. They're not. They are merely cash flow from the value of your equity positions and in and of itself, a dividend provides zero total return. Only share price appreciation provides total return. Note that this refers to what is happening to share price and in your brokerage account on the ex-div date not the corporate side (dividends come from earnings).
Another Catch 22 issue is the relevance of dividends to one's return. The powers that be often state the S&P 500 has returned X% over some number of years with Y% coming from dividends. Let's pretend that it's 7% (total return) and 2% (average yield). In reality, it all came from share price appreciation.
As previously mentioned, dividends provide zero total return. However, when reinvested, they alter the calculation because now you have additional shares compounding the return when share price appreciates. While it is possible to break these apart by using adjusted share prices, it's a royal headache to do so. An easier way is to just use a DRIP calculator and compare the total return of reinvesting versus not reinvesting. Here's one such calculator. Just understand that all of the gain comes from share price appreciation. If share price drops (actual drop due to selling, not ex-dividend reduction) then there will be negative compounding.
• Paragraph 4 is the bottom line. i.e. if the price were not adjusted, the result would be strange. Jul 21 '20 at 19:27
• If one could buy a stock at the close, be entitled to the dividend the next morning (ex-div) and not have share price reduced, there would be no reason to buy other stocks. Just buy near 4 PM, sell in the morning, repeat every day, and laugh all the way to the bank. OK, time for all of the 'free money' dreamers to give up the fantasy :->) Jul 21 '20 at 19:44
• When your tenant pays you your rent, it does not decrease the value of your house. — then why is it much cheaper to buy an apartment that is rented out than one that isn't, at least in countries with tenancy protection? If I buy a home that is unoccupied, I can choose to live there tomorrow or rent it out. If I buy a home that is rented out, I may not be able to get the tenants out in the next 50 years (i.e. until they move out or die, and I'm not sure about the latter). Jul 22 '20 at 7:38
• "Why is it much cheaper to buy an apartment that is rented out than one that isn't, at least in countries with tenancy protection?" With that logic, a stock should cost less because it is "going" to pay a dividend. Unfortunately, it doesn't work that way. As for your inability" to get the tenants out in the next 50 years until they move out or die" that has nothing to do with anything other than the fact that they have a rental contract. Again, irrelevant to the dividend versus rent comparison. Jul 22 '20 at 18:04
• "They're not. They are merely cash flow from the value of your equity positions and in and of itself, a dividend provides zero total return. Only share price appreciation provides total return." Can you provide a reference for this non-standard assertion? Jul 27 '20 at 18:21
The house example can be appropriately modified, without concerning the actual sale of the property (which is a good example, also), to match the stock market.
The rent the tenant pays is the source of income for the house, not for the owner of the house, in this example. The house is the company, remember! Just like a company that sells widgets would sell \$10k worth of widgets, and that would cause its value to appreciate by \$10k, the house sells tenancy, and so appreciates in value by the rent paid for it. That's separate from appreciating in value due to the market, or due to capital improvements, or any other reason the value might increase: it's simply an increase from income. So the house is worth \$100k, plus the accumulation of rents it has taken in but not yet paid out.
In your example, you immediately take that "dividend" of rent from the house - which is not directly possible with a public company, but only because of SEC regulations. But you could just as easily have taken that money and built an addition, right? Just like any public (or private) company, which has income, can make a similar choice:
• Dividend that income back to the owner(s) - value of company reduces by dividend \$
• Make capital improvements - value of company changes based on perceived value of those improvements (could go up, down, or stay neutral)
• Keep cash in reserve - value of company stays neutral
If you did keep the house in a company (as some people do!), then you would have exactly the same math to do - that company's value would decrease each time you took a payout (dividend).
Because a stock and a house are inherently different things. To compare them more meaningfully, consider them in a more equivalent way. The stock is ownership of (part of) a business. That business includes employees (who provide labor) and assets, and generates some amount of profit (or loss). The house is just a house; you need to put it your own time and effort (and money) to maintain it, find tenants, collect rent, pay taxes, etc. You could consider that whole enterprise (the house plus all your associated work) as a company.
If we assume stock A and stock B are both companies whose business is "being a landlord", and they each own one (approximately identical) house, then (a very simplified, not-exactly-realistic, version of) the scenario in your question looks like this:
• Company A owns a \$100,000 house (and nothing else) and has no debts/expenses. There are 100 shares of the company, so each share is worth \$1000. The tenants pay \$1,000 in rent, which is immediately paid to each shareholder as a \$10 dividend. The company still owns exactly \$100,000 in assets (the house; ignoring depreciation, maintenance, etc.), so each of the 100 shares is still worth \$1000.
• Company B owns a \$100,000 house (and nothing else) and has no debts/expenses. There are 100 shares of the company, so each share is worth \$1000. The tenants pay \$1,000 in rent, which the company keeps. The company now owns \$101,000 in assets (\$100,000 house, \$1,000 cash), so each of the 100 shares is now worth \$1010.
If I buy a share of company A, I get a piece of ownership of a business that owns a \$100,000 house. If I buy a share of company B, I get a piece of ownership of a business that owns a \$100,000 house and \$1,000 in cash. That additional cash may be paid out as a dividend in the future (bringing Company B in line with Company A), or it may be used for further investment (e.g. buying a second house, improving the first house to increase rent, etc.).
A company's share price is essentially a combination of two things:
1. The value of all the company's assets (its book value), and
2. investors' estimate of the future prospects of the company (is this a "good investment?").
The assets include the cash in the company's bank accounts. When they pay out a dividend, they have less cash in the bank, so their book value drops. If nothing else has changed, it simply makes mathematical sense to decrease the share price accordingly.
In reality, things aren't quite that simple. For some investors, the prospect of receiving regular dividends is what makes a company more valuable than some other similar company. It's similar to the reason why some people buy bonds rather than stocks -- even though the overall returns may be less, the security of receiving payments on a regular basis is valuable in itself.
But at the moment that the company pays out its dividend, these effects can be ignored. It's still the case that the company's assets have been reduced, so no matter what you think about whether it's a good investment, it can't really be as valuable as it was when it had that cash in the bank.
• If you take Clorox (CLX) the book value per share is around \$5.5 and the stock price is around \$230. So based on your claim here, 97% or more of the value is based on investors future prospects of the company. This is Clorox, not some startup. Does that really make sense to you? If there was a company that just held dollars and paid people (with those same dollars) to manage them (until they were all gone), would you value it at the book value? Jul 29 '20 at 19:14
• What else could that other 97% be based on? "future prospects" also includes things like safety -- how likely is the company to still be around and successful in the future? So a blue chip company like Clorox is worth the high price because you're not likely to lose your money (but don't forget about General Electric). This is the difference between value and growth investments. Stock pricing is not a simple thing. Jul 29 '20 at 21:12
• I would think revenues would be a big factor. That's kind of the point of a business, is it not? Book value has almost nothing to do with profits. Probably the most common measure for a company is it's PE ratio. If you look at CLX stats, the dividend payout is a tiny fraction of it's monthly revenues. Any impact on their book value should be pretty fleeting. It's kind of like how when I make my monthly mortgage payment, my net worth doesn't go down. I'm paying it from my income, not my savings. Jul 29 '20 at 21:27
• @JimmyJames Expected revenues are part of what goes into estimating future prospects. Jul 29 '20 at 21:28
• "future prospects" is everything that goes into evaluating the company that isn't just "stuff". Jul 29 '20 at 21:30
I have once seen the question from another angle that should explain to you why the value decreases and why it has to decrease:
Imagine you've got stock that pays dividend. If dividend is paid once a year, why to buy such stock and keep in it your money for entire year if you could by 1 day before paying out dividend (actually the day it is decided you will get dividend if you hold the stock) and then the next day collect the dividend and sell the stock immediately just to repeat the process next year.
And the answer is - immediately after paying out dividend the stock price drops by approximately the dividend amount just to limit people from earning the money that way.
Going back to renting house example you have used - imagine you rent a house, you've got permanent tenant in it paying the rent and the rent is paid once a year. You can then sell the house along with tenant but it is clear that the value of such transaction will vary over they year. Once it is close to the rent being paid it is higher (approximately by the amount of the rent you will immediately get back) and will be lower when the rent will be paid in longer period. If you think this way it is very logical conclusion the price of such asset (not only being asset but a rented asset) must drop immediately after rent is paid.
Say that you own an LLC that owns a house worth \$100k and rents it out for \$12k/year like your example. That \$12k is revenue for your LLC. Your LLC will have to pay property taxes and for other things like accounting, so let's say you have \$7500 cash in the account at the end of the year. And let's say the housing market jumps so the house is now worth \$110k.
You could pay yourself the \$7500 which would be like a dividend. Now the LLC's only assets are the \$110k house.
Another option would be to keep the money in the LLC. The LLC now has has assets of \$117,500. You may be saving up to buy a second rental property. Or you could use that \$7,500 on improvements to the property (landscaping, electric water heater, whatever) which would increase the value beyond the \$110k.
That's what a company that doesn't pay dividends is like. You don't get that regular income, but the expectation is that the money will be used to make the company more valuable and increase the stock price.
I understand that your title and your question differ wildly. And other answers tackle that beautifully. However, because others are likely to come here based on the title, which currently is "Why are dividends different from property income like rent" and I would hate to see them disappear frustrated when I have the answer to that question as stated.
Dividends are taken from the available cash of a business. This cash has already been taxed by a corporate tax, so it is not taxed at the same rate as regular income. As a more personal finance related example, keeping in mind this varies from jurisdiction, so talk to your accountant to find the breakeven point, in my area if you're self employed and making 250k a year (500k if married) then you want to pay yourself a salary up to about that 250k and after that declare dividends. At that point, the personal tax becomes high enough that you are better off eating the corporate tax on the "profit" of the business and then eating again the tax on dividend - the two of them add up to less than the top tax rate.
• I am not the downvoter, but I have a correction: dividends come out of profits or retained profits, not cash. It's an important distinction. A loss making company can have available cash for example, but would not pay dividends from it. Jul 23 '20 at 12:11
• @JBentley It's an important distinction, but not a correct one. Companies can take a loss and still declare a dividend from available cash flow. Many choose to do so in order to not spook investors if they have a bad quarter, for instance. However, what might be considered important is that the cash paid as a dividend is taxed - you cannot write it off as a loss if you pay it out as a dividend. See money.stackexchange.com/a/52120/3093 for a bit more background on that. Jul 23 '20 at 23:54
• That is why I said you can pay dividends from profits or retained profits. The latter covers the situation where a company is loss making but has previously retained profits available and spare cash. It is unlawful in most jurisdictions as far as I know (and certainly in the UK) for a company to pay dividends where there are no available retained profits (i.e. out of capital). See e.g. here or here. Jul 24 '20 at 20:03
• And if you think about it, the reasons for it being unlawful are fairly obvious: it would be a fraud against the creditors of the company, because they are first in line before shareholders in the event that the company is liquidated. If the company has paid dividends without retained profits, then they must have been paid either by increasing the liabilities e.g. via a loan (thus effectively cheating some portion of the creditors) or by out of capital (cheating the creditors because that capital is what the shareholders agreed their liability would be limited to). Jul 24 '20 at 20:08
• @JBentley Your "fairly obvious" line of reasoning doesn't really make sense. Creditors know that shareholders are important to a corporation. As long as you're making your payments to the creditors, they really don't care. Jul 25 '20 at 20:37
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## 555138
555,138 (five hundred fifty-five thousand one hundred thirty-eight) is an even six-digits composite number following 555137 and preceding 555139. In scientific notation, it is written as 5.55138 × 105. The sum of its digits is 27. It has a total of 4 prime factors and 12 positive divisors. There are 185,040 positive integers (up to 555138) that are relatively prime to 555138.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 27
• Digital Root 9
## Name
Short name 555 thousand 138 five hundred fifty-five thousand one hundred thirty-eight
## Notation
Scientific notation 5.55138 × 105 555.138 × 103
## Prime Factorization of 555138
Prime Factorization 2 × 32 × 30841
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 185046 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 555,138 is 2 × 32 × 30841. Since it has a total of 4 prime factors, 555,138 is a composite number.
## Divisors of 555138
12 divisors
Even divisors 6 6 4 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 1.20284e+06 Sum of all the positive divisors of n s(n) 647700 Sum of the proper positive divisors of n A(n) 100236 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 745.076 Returns the nth root of the product of n divisors H(n) 5.53828 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 555,138 can be divided by 12 positive divisors (out of which 6 are even, and 6 are odd). The sum of these divisors (counting 555,138) is 1,202,838, the average is 10,023,6.5.
## Other Arithmetic Functions (n = 555138)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 185040 Total number of positive integers not greater than n that are coprime to n λ(n) 30840 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45600 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 185,040 positive integers (less than 555,138) that are coprime with 555,138. And there are approximately 45,600 prime numbers less than or equal to 555,138.
## Divisibility of 555138
m n mod m 2 3 4 5 6 7 8 9 0 0 2 3 0 3 2 0
The number 555,138 is divisible by 2, 3, 6 and 9.
• Abundant
• Polite
## Base conversion (555138)
Base System Value
2 Binary 10000111100010000010
3 Ternary 1001012111200
4 Quaternary 2013202002
5 Quinary 120231023
6 Senary 15522030
8 Octal 2074202
10 Decimal 555138
12 Duodecimal 229316
20 Vigesimal 397gi
36 Base36 bwci
## Basic calculations (n = 555138)
### Multiplication
n×i
n×2 1110276 1665414 2220552 2775690
### Division
ni
n⁄2 277569 185046 138784 111028
### Exponentiation
ni
n2 308178199044 171081429060888072 94973802366003282513936 52723566697858330248221403168
### Nth Root
i√n
2√n 745.076 82.1865 27.2961 14.089
## 555138 as geometric shapes
### Circle
Diameter 1.11028e+06 3.48803e+06 9.6817e+11
### Sphere
Volume 7.16624e+17 3.87268e+12 3.48803e+06
### Square
Length = n
Perimeter 2.22055e+06 3.08178e+11 785084
### Cube
Length = n
Surface area 1.84907e+12 1.71081e+17 961527
### Equilateral Triangle
Length = n
Perimeter 1.66541e+06 1.33445e+11 480764
### Triangular Pyramid
Length = n
Surface area 5.3378e+11 2.01621e+16 453268
## Cryptographic Hash Functions
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18 questions linked to/from Why is the FFT "mirrored"?
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Search a number
10000442410333 = 31813143804593
BaseRepresentation
bin1001000110000110100011…
…0100010100010101011101
31022102000220000201000201021
42101201220310110111131
52302321401224112313
633134050111244141
72051336166643324
oct221415064242535
938360800630637
1010000442410333
1132061890364a1
12115619bb19651
135770649c879b
14268049bb30bb
15125203bd278d
hex91868d1455d
10000442410333 has 4 divisors (see below), whose sum is σ = 10003586218108. Its totient is φ = 9997298602560.
The previous prime is 10000442410261. The next prime is 10000442410337. The reversal of 10000442410333 is 33301424400001.
Adding to 10000442410333 its reverse (33301424400001), we get a palindrome (43301866810334).
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in 2 ways, for example, as 1034478170649 + 8965964239684 = 1017093^2 + 2994322^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-10000442410333 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 10000442410298 and 10000442410307.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (10000442410337) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1571899116 + ... + 1571905477.
It is an arithmetic number, because the mean of its divisors is an integer number (2500896554527).
Almost surely, 210000442410333 is an apocalyptic number.
It is an amenable number.
10000442410333 is a deficient number, since it is larger than the sum of its proper divisors (3143807775).
10000442410333 is an equidigital number, since it uses as much as digits as its factorization.
10000442410333 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 3143807774.
The product of its (nonzero) digits is 3456, while the sum is 25.
The spelling of 10000442410333 in words is "ten trillion, four hundred forty-two million, four hundred ten thousand, three hundred thirty-three".
Divisors: 1 3181 3143804593 10000442410333
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https://discussions.unity.com/t/how-to-add-for-int-to-its-total-value/876478
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# How to add for int to its total value
my code is …
`````` public void NeighborPixels(Vector2 pos)
{
for (int i = -3 / 2; i <= 3 /2; i++)
{
for (int j = -2 / 2; j <= 2 /2; j++)
{
if (generateScript.GetPixel(new Vector2Int(Mathf.RoundToInt(pos.x) + i, Mathf.RoundToInt(pos.y) + j)) != null)
{
var ChosenPixelValue = generateScript.GetPixel(new Vector2Int(Mathf.RoundToInt(pos.x) + i, Mathf.RoundToInt(pos.y) + j)).GetComponent<Pixel>().Value;
TotalValue += ChosenPixelValue;
if (TotalValue < 1)
{
generateScript.GetPixel(new Vector2Int(Mathf.RoundToInt(pos.x) + i, Mathf.RoundToInt(pos.y) + j)).GetComponent<SpriteRenderer>().color = Color.red;
}
}
}
}
}
``````
my basic question is how can I add i and j values to total value that the output equals for example 9.
in my code I want to add i and j every time they change so I can get the total value, if in 9 pixels there are two pixels that have value of 1 both and seven of pixels that have value of 0, I want to add them together so I get 2 in total.
Why are you using division? -2 /2 and 2 / 2 simply equal -1 and 1. And an integer datatype cannot be 1.5 (3/2). And please don’t multipost, you’ve already asked this question, out of respect for the people that have helped you here https://discussions.unity.com/t/876369
1 Like
I know that, thanks a lot to them however they didn’t answered my question. That’s why I ask the same question again, so now other people can help me.
That is against forum policy, so you know.
1 Like
Ok thanks for telling me, I will delete this post and never do it again.
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# Motor Starting Current
## Recommended Posts
Hello
Can someone please explain to me what happens to motor starting current when trying to restart a motor that is already spinning. We had a motor that was trying to restart after it was stopped, but was still spinning. I have been told that there is a greater starting current associated with a greater starting torque when trying to start a motor that is already spinning due to the fact that the spinning motor and power supply are out of phase.
Can someone please explain to me what happens electrically in this instance or point me in the direction of where I can find some information. I have been trying to find information on this subject and have been unsucessfull.
Thanks for the help!
##### Share on other sites
Hello mull982
Welcome to the forum.
When the motor is connected to the supply, the supply voltage causes a rotating magnetic field in the stator. This field cuts through the rotor bars inducing a current to flow in them. The rotor current causes a rotor field and the interaction between the stator field and the rotor field causes the torque to drive the motor.
When the supply is disconnected from the motor, the rotor continues to spin for a period of time. The rotor retains a field which exponentially decays. While the rotor filed is significant, it will induce a voltage in the stator. The frequency of the stator voltage is dependent on the rotor speed and the voltage is dependent on the rotor flux.
If the supply is reconnected while the stator is generating voltage, there will be a current transient to re establish the stator field. If the voltage is out of phase. the transient can be very high.
Typically, the rotor flux will decay in less than a second, so reconnecting at greater than half a second will usually not result in a significant reclose transient.
The start current after the transient, will not be significantly higher on a rotating motor, but will equal the start current at that speed when started from zero. If the motor is spinning forward, the current will be less than the locked rotor current. If the motor is spinning in revers, the current will be a little higher than the current at zero speed.
If you are restarting a spinning motor, allow a delay of 0.5 - 1 second after disconnecting befor reconnecting the supply.
Best regards,
##### Share on other sites
Marke
Thanks for the information!
What you are saying is that if the motor is for some reason restarted before the magnetic field has a chance to decay or collapse (less than .5s) then the motor will regenerate a voltage that is out of phase with the restarted incoming voltage leading to a transient current? What causes this current transient, is it due to the fact that there will be a difference in voltage potential between the regenerated voltage and incoming voltage or simply because they are out of phase? Do you know where I can read more about this condition?
Otherwise you are saying that if the motor is restarted after the field has collapsed, lets say after about 5s then this transient will not exist and the motor will start with the normal starting current. I wasn't sure from what you said that if the motor was spinning in the direction of rotation, would the restarting current be the same as starting from zero (LRC) or be less? If less, then why? Is this because the motor would require less starting/acceleration torque? I can see that a motor spinning opposite rotation would require more starting current but couldn't understand if you were saying a motor spinning in the direction of rotation would be the same or less than zero starting.
I have also read information regarding "Negative Coefficent of Temperature" relating to the temperature and resistance of the windings. From what I understand as the temperature of the windings increase their resistance decreases. When trying to restart a motor shortly after it has been running, could the motor windings still be hot giving them a lesser resistance, and therefore a greater starting current which could possibly trip a circuit breaker?
Thanks for all of the help.
##### Share on other sites
Hello mull982
if the motor is for some reason restarted before the magnetic field has a chance to decay or collapse (less than .5s) then the motor will regenerate a voltage that is out of phase with the restarted incoming voltage leading to a transient current?
Yes that is correct.
What causes this current transient,
When you initially apply an AC voltage to an iron cored reactor (motor, transformer, solenoid etc) there is no flux in the iron. There is an "inrush" current that flows for a very short time to establish the flux in the iron. The current then drops back to a steady state current, which in the case of a motor, is the start current of the motor.
When you "reclose" on a rotating motor with significant rotor flux, the inrush current is much higher as in addition to the establishment of the flux in the iron, the existing flux has to be dissipated. If the existing flux and the new flux are in phase, the inrush current is ery low. If the existing flux and the new flux are out of phase, the inrush current is very high. Associated with the high inrush current, there is a very high transient torque which can cause severe damage to the motor and to the driven load.
The start current is a function of the motor characteristics and the rotor speed. If full voltage is reapplied to a motor that spinning at say half speed, then the start current (after the inrush current) will be the same as the start current would be when the motor has accelerated to half speed from zero speed. The start current vers speed follows the same curve.
The windings of the motor, stator and rotor, are metallic and so have a positive temperature coefficient. As the temperature rises, the resistance increases by a very small amount.
The resistance of the stator has virtually zero influence on the start current of the motor. The resistance of the rotor however does influence the start current and start torque of the motor however it is very hard to see the effect.
Best regards,
##### Share on other sites
If you are restarting a spinning motor, allow a delay of 0.5 - 1 second after disconnecting befor reconnecting the supply.
Hello,
What happend with a star delta starter? Isnt it doing exactly wath you are telling not to do? Reconecting a spinning motor in less then a second, from star to delta?
Thanks for any help!
Chris.
##### Share on other sites
Hello Chris 1373
What happend with a star delta starter? Isnt it doing exactly wath you are telling not to do? Reconecting a spinning motor in less then a second, from star to delta?
Exactly
If you follow any of my discussions on star/delta starters, you will find that the open transition star/delta starter develops a very high current and torque transient at the transition from star to delta. It does more harm than a DOL starter!
Best regards,
##### Share on other sites
Hello Chris 1373
Exactly
If you follow any of my discussions on star/delta starters, you will find that the open transition star/delta starter develops a very high current and torque transient at the transition from star to delta. It does more harm than a DOL starter!
Best regards,
I agree with you in your observations about star/delta sarters, now i have it clear!
Marke, cant you help me with my other post? Please, i need it? The one that are about gens test!
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# UP Police Jail Warder Syllabus (PDF) यूपी पुलिस जेल वार्डर सिलेबस
## UP Police Jail Warder Syllabus 2024 PDF
Name of the Organization Uttar Pradesh Police Department Name of the Posts Jail Warder Category Download Syllabus Selection Process Written Examination, Physical Standard Test, Interview Job Location Uttar Pradesh Official Site uppbpb.gov.in
## Uttar Pradesh Police Jail Warder Exam Pattern 2024
Subject Names Questions Maximum Marks General Knowledge 150 Questions 300 Marks Quantitative Aptitude/ Arithmetic English General Science Reasoning General Awareness
## [uppbpb.gov.in] UP Police Jail Warder Syllabus 2024 PDF
The UP Police Exam will be conducted to recruit the Jail Warder Posts. To help the Aspirants we have uploaded the UP Police Jail Warder Syllabus on this page for Download. Here we have provided the Subject Wise UP Police Jail Warder Syllabus PDF.
### Reasoning
Verbal Reasoning
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#### What is the Exam Pattern for UP Police Jail Warder Exam?
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#### What are the Subjects involved in UP Police Jail Warder Syllabus?
General Knowledge Quantitative Aptitude/ Arithmetic English General Science Reasoning General Awareness
#### How Many Questions will be asked in UP Police Jail Warder Exam 2024?
A Total of 150 Questions will be asked in UP Police Jail Warder Exam
#### For How Many Marks the UP Police Jail Warder Exam will be conducted?
The UP Police Jail Warder Exam will be Conducted for 300 Marks
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# Joint distribution from marginals
I have a question about a joint distribution calculated in a paper I am reading.
There are three random variable a, b and c such that $$a,b,c \in \{+1,-1\}$$ and then the joint distribution is given by: $$p(a,b,c) = \frac{1}{8}(1 + aE_{A} + bE_{B} + cE_{C} + abE_{AB} + acE_{AC} + bcE_{BC} + abcE_{ABC})$$ where $$E_{A}$$, $$E_{B}$$ and $$E_{C}$$ are the single-party marginals, $$E_{AB}$$, $$E_{BC}$$ and $$E_{AC}$$ the two-party marginals, and $$E_{ABC}$$ is the three-body correlator.
I feel like this should be something I should have seen in an introductory course on probability but I can't seem to prove it. Also if I convince myself that it's just adding up all the possible cases, I think 1 should be part of the other expected values (i.e. it is already considered in them) so I don't see the point in adding it separately. Also would we have the same expression if the possible values were $$\{+1,0\}$$ (or any other set of size 2) ?
I am used to the notation $$E$$ as the expected value, but this is totally unrelated to that and it is about marginals. Am I correct?
I would be pleased to see a complete proof or a link to study this fact.
Note that the positivity of $$p(a,b,c)$$ implies constraints on marginals, in particular $$p(+ + +) + p(−−−) ≥ 0$$ implies $$E_{AB} + E_{AC} + E_{BC} ≥ −1.$$
which I don't understand.
• Can you add a link to the paper? Nov 19, 2021 at 12:07
• I think I heard the term 'Rademacher chaos' but not sure how deeply it is related to your question. Anyway, the expansion is more or less the result of the fact that monomials $\prod_{x\in A}x$ for $A\subseteq \{a,b,c\}$ spans the space of functions of the form $\{-1,1\}^3\to\mathbb{R}$. Nov 19, 2021 at 12:10
• arxiv.org/abs/1906.06495
– Pegi
Nov 19, 2021 at 12:37
• It is on the second page.
– Pegi
Nov 19, 2021 at 12:38
• The equations $(3)$-$(10)$ in the paper indicate that $E_A,E_B,E_C,E_{AB},E_{AC},E_{BC}$ and $E_{ABC}$ does indeed represent the expected values $E[A],E[B],E[C],E[AB],E[AC],E[BC]$ and $E[ABC]$. Nov 19, 2021 at 14:27
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# Alien Math Worksheets
📆 10 Jun 2022
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Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
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Author Message
etconlgbarney
Registered: 10.04.2005
From: /home/Australia/Sydney
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browse words by letter
a b c d e f g h i j k l m n o p q r s t u v w x y z
## weight
``` 4 definitions found
From Webster's Revised Unabridged Dictionary (1913) [web1913]:
Weight \Weight\, v. t. [imp. & p. p. {Weighted}; p. pr & vb n.
{Weighting}.]
1. To load with a weight or weights; to load down to make
heavy; to attach weights to as to weight a horse or a
jockey at a race; to weight a whip handle.
The arrows of satire, . . . weighted with sense
--Coleridge.
2. (Astron. & Physics) To assign a weight to to express by a
number the probable accuracy of as an observation. See
{Weight of observations}, under {Weight}.
From Webster's Revised Unabridged Dictionary (1913) [web1913]:
Weight \Weight\, n. [OE. weght, wight, AS gewiht akin to D.
gewigt G. gewicht Icel. v[ae]tt, Sw vigt Dan. v[ae]gt.
See {Weigh}, v. t.]
1. The quality of being heavy; that property of bodies by
which they tend toward the center of the earth; the effect
of gravitative force, especially when expressed in certain
units or standards, as pounds, grams, etc
Note: Weight differs from gravity in being the effect of
gravity, or the downward pressure of a body under the
influence of gravity; hence it constitutes a measure
of the force of gravity, and being the resultant of all
the forces exerted by gravity upon the different
particles of the body, it is proportional to the
quantity of matter in the body.
2. The quantity of heaviness; comparative tendency to the
center of the earth; the quantity of matter as estimated
by the balance, or expressed numerically with reference to
some standard unit; as a mass of stone having the weight
of five hundred pounds.
For sorrow, like a heavy-hanging bell, Once set on
ringing, with his own weight goes. --Shak.
3. Hence pressure; burden; as the weight of care or
business. ``The weight of this said time.'' --Shak.
For the public all this weight he bears. --Milton.
[He] who singly bore the world's sad weight.
--Keble.
4. Importance; power; influence; efficacy; consequence;
moment; impressiveness; as a consideration of vast
weight.
In such a point of weight, so near mine honor.
--Shak.
5. A scale, or graduated standard, of heaviness; a mode of
estimating weight; as avoirdupois weight; troy weight;
apothecaries' weight.
6. A ponderous mass; something heavy; as a clock weight; a
paper weight.
A man leapeth better with weights in his hands.
--Bacon.
7. A definite mass of iron, lead, brass, or other metal, to
be used for ascertaining the weight of other bodies; as
an ounce weight.
8. (Mech.) The resistance against which a machine acts as
opposed to the power which moves it [Obs.]
{Atomic weight}. (Chem.) See under {Atomic}, and cf
{Element}.
{Dead weight}, {Feather weight}, {Heavy weight}, {Light
weight}, etc See under {Dead}, {Feather}, etc
{Weight of observation} (Astron. & Physics), a number
expressing the most probable relative value of each
observation in determining the result of a series of
observations of the same kind
Syn: Ponderousness; gravity; heaviness; pressure; burden;
load; importance; power; influence; efficacy;
consequence; moment; impressiveness.
From Webster's Revised Unabridged Dictionary (1913) [web1913]:
Weight \Weight\, v. t. (Dyeing)
To load (fabrics) as with barite, to increase the weight,
etc
From WordNet r 1.6 [wn]:
weight
n 1: the vertical force exerted by a mass as a result of gravity
2: equipment used in calisthenic exercises and weightlifting
[syn: {exercising weight}]
3: the relative importance granted to something "his opinion
carries great weight"
4: an artifact that is heavy
5: an oppressive feeling of heavy force; "bowed down by the
weight of responsibility"
6: a system of units used to express the weight of something
[syn: {system of weights}]
7: a unit used to measure weight; "he placed two weights in the
scale pan" [syn: {weight unit}]
8: (statistics) a coefficient assigned to elements of a
frequency distribution in order to represent their
relative importance [syn: {weighting}]
v 1: weight down with a load [syn: {burden}, {burthen}, {weight
down}] [ant: {unburden}]
2: present with a bias [syn: {slant}, {angle}]
```
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## 61 Reputation
18 years, 235 days
## Thanks Georgios :-) ...
Thanks Georgios :-)
## Thanks Georgios :-) ...
Thanks Georgios :-)
## Thats just what I was...
Thats just what I was looking for... Cheers :-)
## Thats just what I was...
Thats just what I was looking for... Cheers :-)
## coordinates...
Hey, Good answer Roman, but to take the question one step further, Your answer gives x values of intersection, how do we define the y values of intersection??
## coordinates...
Hey, Good answer Roman, but to take the question one step further, Your answer gives x values of intersection, how do we define the y values of intersection??
## roots...
Hi, Thanks for your reply. Yes it makes sence that roots could be the command im looking for, and the answer is very close :-) Just need to understand what maple is trying to telling me. The reply is -3,1 -1,1 and 1,1. Am I supposed to interpret these as coordinates, and if so shouldnt they read -3,0 -1,0 and 1,0??
## Minimum values...
Hey, Thanks for the quick reply Alec. Just what I needed :-) Now I need to find the minimum value of y whilst x=2..5 The value should return 7.828, but again what is the command for maple? \Robert
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# Triangle Pre-Test
```Triangle Pre-Test
Name______________________
Score: _____ / 15 pts
Date____________________
Consider a triangle ABC. (2 pts each)
1) If A 90, what can you conclude about the measure of B? Explain.
2) If B has a greater measure than C, what must be true of the sides b and c ? Why?
In Exercises 3 - 8, state an equation you can use to solve for x. ( 1 pt each )
3)
4)
8
x
x
5
60
20
115
10
Equation__________________________
Equation__________________________
5)
6)
x
9
6
5
48
5
x
7
Equation__________________________
Equation__________________________
7)
8)
5
x
x
40
35
20
7
6
Equation__________________________
Equation__________________________
You know that a triangle is completely determined by the lengths of two of its
sides and the measure of the angle included by these sides ( SAS ). In XYZ, x = 4,
y = 8, and Z = 50.
9) Use the Law of Cosines to find the side z to the nearest hundredth. ( 1 pt )
10) Use the Law of Sines to find the measure of Y to the nearest tenth of a degree.
Then compute the measure of X. ( 2 pts )
Note that x y z.
11) What does this imply about the measures of X, Y, and Z ? ( 1 pt )
12) Do your answers to Exercises 10 and 11 agree? If not, explain where an error in
your reasoning occurred. ( 1 pt )
```
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Exam2A_12F_Ans
# A what is the present value pv of his future income
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Unformatted text preview: to wheat? Explain why. (6 pts) For complements, the sign of cross-price elasticity is negative, whereas for substitutes, the sign of cross-price elasticity is positive. Thus Rice is substitute, whereas Seafood and Fruit are complements c. Suppose you, a seller of wheat, raise its price. Would you increase the price of wheat? Answer it after analyzing the effect of price increase on total revenue, based on the information from Table1. Explain. (6 pts) Because the price elasticity of demand for wheat is -0.73, which is inelastic, increase in price would increase TR, which is P*Q. Thus increase the price. 3. Suppose Jerry earns \$40,000 this year and his annual income will be \$44,000 next year. a. What is the present value (PV) of his future income when the interest rate is 10%? Show your work. (5 pts) PV = 44000/1.1=40000 b. Using current consumption (C1) and future consumption (C2) graph (x-axis: C1, y-axis: C2) draw Jerry’s budget constraint when the interest rate is 10%. Clearly mark your graph. What is the slope of the budget constraint? (7 pts) C2 = 88000 – 1.1C1 ; The slope = -1.1 c. If the interest rate decreases to 5%, make changes in Jerr...
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## This document was uploaded on 02/06/2014.
Ask a homework question - tutors are online
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https://kr.mathworks.com/matlabcentral/cody/problems/2631-flip-the-vector-from-right-to-left/solutions/513963
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Cody
# Problem 2631. Flip the vector from right to left
Solution 513963
Submitted on 17 Oct 2014 by Tanguy
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y_correct = 1; assert(isequal(flip_vector(x),y_correct))
2 Pass
%% x = [1:5]; y_correct = [5:-1:1]; assert(isequal(flip_vector(x),y_correct))
3 Pass
%% x = [1 4 6]; y_correct = [6 4 1]; assert(isequal(flip_vector(x),y_correct))
4 Pass
%% x = [10 5 9 ]; y_correct = [9 5 10]; assert(isequal(flip_vector(x),y_correct))
5 Pass
%% x = [2 4 6 8]; y_correct = [8 6 4 2]; assert(isequal(flip_vector(x),y_correct))
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# the center of a polygon
This topic is 4882 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
given n coordinates, how do you calculate the center of the polygon they compose? at first i thought you just sum up the X's, sum up the Y's, and divide by n. but that's wrong, because if i have 999 coordinates at (0,0), and one at (1,1), the center would be at (0.0001,0.0001),and not at (0.5,0.5), as i would have liked it to be... help, please? thanks, noam
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find the extents (minimum/maximum in each axis), then average the min/max.
eg, centerX = (maxX + minX) / 2.0f
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Quote:
Original post by Anonymous Posterfind the extents (minimum/maximum in each axis), then average the min/max.eg, centerX = (maxX + minX) / 2.0f
That is not correct.
The only way to do this is to split the polygon into trinagles. Then you do something like this:
center = Vector(0,0)area = 0for triangle in triangles: center += tirangle.getArea()*triangle.getCenter() area += tirangle.getArea()center /= area
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Considering there are about a million different ways you can define the "center of the polygon", you'll need to be more precise about what kind of center you want. In-center, ortho-center, circum-center, center of mass, etc.
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the first way, though simple, works finr for me :)
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Quote:
Original post by No_Germsthe first way, though simple, works finr for me :)
The method the AP posted finds the center of the axis aligned bounding box of the polygon not the center of the polygon.
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yes, i knew it, but i gave it a chance, and it works fine for what i need, so i'll settle for that... :)
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Quote:
Original post by No_Germsgiven n coordinates, how do you calculate the center of the polygon they compose?at first i thought you just sum up the X's, sum up the Y's, and divide by n. but that's wrong, because if i have 999 coordinates at (0,0), and one at (1,1), the center would be at (0.0001,0.0001),and not at (0.5,0.5), as i would have liked it to be...help, please?thanks, noam
Your "example" of a possible polygon doesn't really work. The vertices of a polygon must be unique; if they aren't, it violates the definition of a polygon: "a closed plane figure with n sides and n vertices".
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Quote:
Original post by No_Germsyes, i knew it, but i gave it a chance, and it works fine for what i need, so i'll settle for that... :)
Are you sure? Consider this "boomerang" polygon, where the vertices are marked with *'s. The bounding box is enclosed in #'s. The "x" marks the center of the bounding box. But notice that x isn't even inside the polygon!
########################## ## *-------------------* ## ----- | ## ----- | ## ----* | ## | | ## x | | ## | | ## | | ## | | ## | | ## * ## ##########################
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The center of a polygon is not necessarily in the polygon. It really all depends what you mean by "center". "Center" is not really a mathematically precise term... it has lots and lots and lots of interpretations. In the case of the geometric centroid (the most common interpretation for this type of figure), the center of mass of a boomerang is indeed outside the boomerang itself. What really matters here is how you intend to USE the "center point".
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Home / Help
Author Dyke, Philip P. G. author An Introduction to Laplace Transforms and Fourier Series [electronic resource] / by Philip P. G. Dyke London : Springer London : Imprint: Springer, 2001 http://dx.doi.org/10.1007/978-1-4471-0505-3 XII, 250 p. 2 illus. online resource
SUMMARY
This book has been primarily written for the student of mathematics who is in the second year or the early part of the third year of an undergraduate course. It will also be very useful for students of engineering and the physical sciences for whom Laplace Transforms continue to be an extremely useful tool. The book demands no more than an elementary knowledge of calculus and linear algebra of the type found in many first year mathematics modules for applied subjects. For mathematics majors and specialists, it is not the mathematics that will be challenging but the applications to the real world. The author is in the privileged position of having spent ten or so years outside mathematics in an engineering environment where the Laplace Transform is used in anger to solve real problems, as well as spending rather more years within mathematics where accuracy and logic are of primary importance. This book is written unashamedly from the point of view of the applied mathematician. The Laplace Transform has a rather strange place in mathematics. There is no doubt that it is a topic worthy of study by applied mathematicians who have one eye on the wealth of applications; indeed it is often called Operational Calculus
CONTENT
1. The Laplace Transform -- 1.1 Introduction -- 1.2 The Laplace Transform -- 1.3 Elementary Properties -- 1.4 Exercises -- 2. Further Properties of the Laplace Transform -- 2.1 Real Functions -- 2.2 Derivative Property of the Laplace Transform -- 2.3 Heavisideโ{128}{153}s Unit Step Function -- 2.4 Inverse Laplace Transform -- 2.5 Limiting Theorems -- 2.6 The Impulse Function -- 2.7 Periodic Functions -- 2.8 Exercises -- 3. Convolution and the Solution of Ordinary Differential Equations -- 3.1 Introduction -- 3.2 Convolution -- 3.3 Ordinary Differential Equations -- 3.3.1 Second Order Differential Equations -- 3.3.2 Simultaneous Differential Equations -- 3.4 Using Step and Impulse Functions -- 3.5 Integral Equations -- 3.6 Exercises -- 4. Fourier Series -- 4.1 Introduction -- 4.2 Definition of a Fourier Series -- 4.3 Odd and Even Functions -- 4.4 Complex Fourier Series -- 4.5 Half Range Series -- 4.6 Properties of Fourier Series -- 4.7 Exercises -- 5. Partial Differential Equations -- 5.1 Introduction -- 5.2 Classification of Partial Differential Equations -- 5.3 Separation of Variables -- 5.4 Using Laplace Transforms to Solve PDEs -- 5.5 Boundary Conditions and Asymptotics -- 5.6 Exercises -- 6. Fourier Transforms -- 6.1 Introduction -- 6.2 Deriving the Fourier Transform -- 6.3 Basic Properties of the Fourier Transform -- 6.4 Fourier Transforms and PDEs -- 6.5 Signal Processing -- 6.6 Exercises -- 7. Complex Variables and Laplace Transforms -- 7.1 Introduction -- 7.2 Rudiments of Complex Analysis -- 7.3 Complex Integration -- 7.4 Branch Points -- 7.5 The Inverse Laplace Transform -- 7.6 Using the Inversion Formula in Asymptotics -- 7.7 Exercises -- A. Solutions to Exercises -- B. Table of Laplace Transforms -- C. Linear Spaces -- C.1 Linear Algebra -- C.2 Gramm-Schmidt Orthonormalisation Process
Mathematics Mathematical analysis Analysis (Mathematics) Fourier analysis Mathematics Analysis Fourier Analysis
Location
Office of Academic Resources, Chulalongkorn University, Phayathai Rd. Pathumwan Bangkok 10330 Thailand
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# Hourly To Monthly Calculator
Introduction: The Hourly to Monthly Calculator is a useful tool for individuals who want to quickly estimate their monthly earnings based on an hourly wage. It simplifies the process of converting hourly rates to a monthly income, providing a convenient way to plan and budget.
Formula: The Monthly Earnings are calculated by multiplying the hourly wage by the assumed number of work hours per week (40 hours) and the approximate number of weeks in a month (4.33 weeks).
How to Use:
1. Enter your hourly wage in dollars in the “Enter hourly wage” field.
2. Click the “Calculate” button.
3. The result, representing the Monthly Earnings, will be displayed in the “Monthly Earnings” field.
Example: Suppose your hourly wage is \$20. Enter this value and click “Calculate.” The result will show your estimated Monthly Earnings, which would be approximately \$3,466.00.
FAQs:
1. Q: Can I use this calculator for salaried positions?
• A: No, this calculator is specifically designed for estimating monthly earnings based on an hourly wage.
2. Q: How is the number of weeks in a month calculated?
• A: It is assumed to be 4.33 weeks on average, considering the variability in the number of days per month.
3. Q: Does this calculator consider taxes or withholdings?
• A: No, the calculator provides an estimate before deductions. Net income after taxes may be lower.
4. Q: Can I use decimal values for the hourly wage?
• A: Yes, decimal values are accepted, and the calculator will handle them accordingly.
5. Q: Is this calculator suitable for part-time jobs?
• A: Yes, you can use this calculator for part-time positions by entering the appropriate hourly wage.
Conclusion: The Hourly to Monthly Calculator is a valuable tool for individuals working on an hourly basis. Use it to quickly estimate your monthly earnings, plan your budget, and make informed financial decisions based on your hourly wage.
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https://www.mathematik.uni-marburg.de/modulhandbuch/20232/BSc_Business_Mathematics/Continuing_Modules_in_Mathematics/Internship_Stochastics.html
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# Internship Stochastics (dt. Praktikum zur Stochastik)
Level, degree of commitment Practical module, compulsory elective module Forms of teaching and learning,workload Internship (4 SWS), 180 hours (60 h attendance, 120 h private study) Credit points,formal requirements 6 CP Course requirement(s): Presentation of three task solutions. Attendance is compulsory. Examination type: internship report Language,Grading German,The module is ungraded in accordance with the examination regulations for the degree program B.Sc. Business Mathematics. Duration,frequency One semester, In the summer semester / as a block course during the lecture-free period in spring Person in charge of the module's outline Prof. Dr. Hajo Holzmann
## Contents
The practical course is based on the statistics software ''R''. First, the functionalities of ''R'' are introduced. Subsequently, the theory is briefly introduced to the topics listed below. The methods introduced are investigated with ''R'' on the basis of simulations and applied to data sets.
Topics (only a selection is covered)
• Basics in dealing with R
• Random variables and their simulation
• descriptive statistics and graphics
• Point and interval estimation
• Statistical hypothesis tests
• Analysis of multivariate data
• Linear Regression
• Covariance and variance analysis
• Generalized linear models
## Qualification Goals
Students
• are able to use the statistical software R,
• can investigate statistical procedures using suitable simulations,
• can apply appropriate statistical procedures to given data sets and problems,
• are able to process results obtained in writing in an appropriate manner,
• have experience in teamwork and work organization through the development of tasks.
## Prerequisites
None. The competences taught in the following modules are recommended: either Foundations of Mathematics and Linear Algebra I and Linear Algebra II or Basic Linear Algebra, either Analysis I and Analysis II or Basic Real Analysis, Elementary Probability and Statistics.
## Applicability
The module can be attended at FB12 in study program(s)
• B.Sc. Data Science
• B.Sc. Mathematics
• M.Sc. Mathematics
• BA Minor Mathematics
When studying B.Sc. Business Mathematics, this module can be attended in the study area Continuing Modules in Mathematics.
The module can also be used in other study programs (export module).
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We aren't endorsed by this school
##### CS 106 - Stanford Study Resources
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###### 11-DesigningClasses
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #11 January 16, 2015 Designing Classes Optional Movie Classes and Objects Eric Roberts CS 106B January 16, 2015 Outline 1. Anagram exercise from last time 2. Structures 3. Representing points using a structure 4. Classes and o
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###### 03-HonorCode
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Eric Roberts CS 106B Handout #3 January 5, 2015 Computer Science and the Stanford Honor Code Since 1921, academic conduct for students at Stanford has been governed by the Honor Code, which reads as follows: THE STANFORD UNIVERSITY HONOR CODE A. The Honor
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###### Section #1 Solutions
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CS106B Section 1 Solutions (Week 2) 1. Passing Parameters. Original XehranM Cynthia Changed to cs(b, a); XehranM Marty Changed to a = cs(b, a); (Even though the second parameter gets changed by reference, it is then assigned the return value of the functi
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###### Practice Midterm #6
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CS 106B Practice Midterm Exam #6 (based on CS 106B Summer 2014 midterm) This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual midterm exam. We do not guarantee that the number of questions
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CS 106B Practice Midterm Exam #5 (based on CS 106B Winter 2014 midterm) This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual midterm exam. We do not guarantee that the number of questions
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###### Practice Midterm #6 Solutions
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CS 106B Practice Midterm Exam #6 ANSWER KEY 1. C+ Basics / Parameters (read) dx cx 21 2 cxx zx 301 22 zxx cxx 2 302 2 21 301 cxx zxx zx 2. File I/O and Strings (write) int printMostCommonName(ifstream& input) cfw_ int uniqueNames = 1; / counters for names
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###### Practice Midterm #4
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CS 106B Practice Midterm Exam #4 (based on CS 106B Autumn 2013 midterm) This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual midterm exam. We do not guarantee that the number of questions
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CS 106B Practice Midterm Exam #5 ANSWER KEY 1. C+ Basics / Parameters 4 0xdd00 7 12 6 0xff00 9 15 8 0xbb00 11 17 17 15 9 11 2. File I/O and Strings (write) As with any programming problem, there are many correct solutions. Here are two: / short-n-sweet so
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School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #4 ANSWER KEY 1. C+ Basics / Parameters -6 1 5 12 -4 1 4 5 8 2 5 -4 9 2. File I/O and Strings (write) As with any programming problem, there are many correct solutions. Here is one: void inputStats(string filename) cfw_ ifstr
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###### Practice Midterm #3
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #3 This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual midterm exam. We do not guarantee that the number of questions in this sample exam will match the num
• 4 Pages
###### 01-CoursePlacement
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #1 January 5, 2015 Course Placement Information With extensive updates by Keith Schwarz and Mehran Sahami. CS 106A, CS 106B, CS 106L, CS 106X, CS 107, CS 101, CS 105, CS 107E, CS 142 there are a lot of programming classes to c
• 6 Pages
###### 02-CourseInformation
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #2 January 5, 2015 CS 106B General Information Professor: Eric Roberts E-mail: eroberts@cs.stanford.edu Offices: Gates 202 Phone: 723-3642 Drop-in hours: Tuesdays, 9:3011:30 A.M. Wednesdays after class in Bytes Caf TA: Kevin M
• 3 Pages
###### 08-CollectionClasses1
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #8 January 12, 2015 Collection Classes (Part 1) Outline Collection Classes 1. Introduce the idea of collection classes (Part 1: Vectors, Grids, Stacks, and Queues) 2. Introduce the Vector class 3. Use vectors to read an entire
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###### 12-Assignment2
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #12 January 16, 2015 Assignment #2Using ADTs The idea for Random Writer comes from Joe Zachary at the University of Utah. Parts of this handout were written by Julie Zelenski and Jerry Cain. Due: Monday, January 26 Now that yo
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###### 13-RandomWriterContest
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #13 January 16, 2015 The CS106B Random Writer Contest Due date: Friday, January 30 On Assignment #2, the second problem has you create a random-writer application that uses a Markov model to generate English text according to
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###### 09A-Section1Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #9A January 12, 2015 Solutions to Section Handout #1 Problem 1. String manipulation There are several possible strategies for implementing the character-removal problem. The implementations shown below go through the text strin
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• ###### 07-SubmittingAssignments
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###### 07-SubmittingAssignments
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #7 January 9, 2015 Submitting Assignments This handout was written by Jeremy Keeshin. This quarter, you will be submitting your assignments with Paperless, which is a tool that will let section leaders comment on your code wit
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###### 06-StringsAndStreams
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #6 January 9, 2015 Strings and Streams Administrative Reminders Strings and Streams All handouts and course information are on the web site: http:/cs106b.stanford.edu/ All CS 106B students must sign up for a section by Sunda
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###### 09-Section1
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #9 January 12, 2015 Section Handout #1Simple C+ Sections will meet once a week to give you a more intimate environment to discuss course material, work through problems, and raise any questions you have. Each week we will hand
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###### 04-FunctionsInC++
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #4 January 7, 2015 Functions in C+ Administrative Essentials Functions in C+ All handouts and course information are on the web site: http:/cs106b.stanford.edu/ All CS 106B students must sign up for a section by Sunday at 5:
• 5 Pages
###### 05-Assignment1
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #5 January 7, 2015 Assignment #1Simple C+ Parts of this handout were written by Julie Zelenski. Due: Friday, January 16 Part 1. Get Qt Creator working Your first task is to set up Qt Creator, which is the programming environme
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• ###### Practice Midterm #2 Winter 2013
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###### Practice Midterm #2 Winter 2013
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #25 January 30, 2013 Practice Midterm Exam #2 Review session: Sunday, February 3, 7:009:00 P.M., Hewlett 201 (next door) Midterm #1: Tuesday, February 5, 3:155:15 P.M., Braun Auditorium (Chemistry) Midterm #2: Tuesday, February
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• ###### Practice Midterm #3 Solutions
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###### Practice Midterm #3 Solutions
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #3 ANSWER KEY 1. C+ Basics / Parameters 0 2 8 8 3 7 8 7 1 8 9 2 9 2. File I/O and Strings (write) As with any programming problem, there are many correct solutions. Here is one: void coinFlip(string filename) cfw_ ifstream in
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• ###### Practice Midterm #1 Summer 2014
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###### Practice Midterm #1 Summer 2014
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #1 This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual midterm exam. We do not guarantee that the number of questions in this sample exam will match the num
• 4 Pages
• ###### Practice Midterm #1 Summer 2014 Solutions
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###### Practice Midterm #1 Summer 2014 Solutions
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #1 ANSWER KEY 1. C+ Basics / Parameters 9 -3 -7 9 5 -7 5 10 1 7 3 -2 -6 2. File I/O and Strings (write) As with any programming problem, there are many correct solutions. Here is one: void wordStats(string filename) cfw_ ifst
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• ###### Practice Midterm #1 Spring 2013 Solutions
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###### Practice Midterm #1 Spring 2013 Solutions
School: Stanford
Course: Programming Abstractions
CS106B Spring 2013 Handout #17S May 3, 2013 CS106B Practice Midterm Exam #1 Solutions Problem One: Detecting Gerrymandering (35 Points) The simplest solution to this problem is simply to count up how many districts were won, along with the total number of
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• ###### Practice Final #2 Winter 2013 Solutions
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###### Practice Final #2 Winter 2013 Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #54A March 13, 2013 Answers to Practice Final Examination #2 Review session: Scheduled finals: Sunday, March 17, 3:005:00 P.M. (Hewlett 200) Tuesday, March 19, 12:153:15 P.M. (Hewlett 200) Thursday, March 21, 12:153:15 P.M. (He
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• ###### Practice Final Fall 2012 Solutions
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###### Practice Final Fall 2012 Solutions
School: Stanford
Course: Programming Abstractions
CS106B Handout 40S December 4th, 2012 Autumn 2012 CS106B Practice Final Solution Solution 1: Linked Lists a. static Map<string, string> concatenateMaps(const node *maplist) cfw_ Map<string, string> concatenations; for (const node *curr = maplist; curr !=
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• ###### Practice Final #1 Winter 2013 Solutions
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###### Practice Final #1 Winter 2013 Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #52A March 11, 2013 Answers to Practice Final Examination #1 Review session: Scheduled finals: Sunday, March 17, 3:005:00 P.M. (Hewlett 200) Tuesday, March 19, 12:153:15 P.M. (Hewlett 200) Thursday, March 21, 12:153:15 P.M. (He
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• ###### Practice Final Fall 2012
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###### Practice Final Fall 2012
School: Stanford
Course: Programming Abstractions
CS106B Handout 40 December 4th, 2012 Autumn 2012 CS106B Practice Final Exam Facts: When: Where: Friday, December 14th from 12:15 3:15 p.m. Last Names A through Kh: Building 320, Room 105 Last Names Ki through Z: Building 420, Room 040 Coverage The closed
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• ###### Practice Final #2 Winter 2013
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###### Practice Final #2 Winter 2013
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #54 March 11, 2013 Practice Final Examination #2 Review session: Scheduled finals: Sunday, March 17, 3:005:00 P.M. (Hewlett 200) Tuesday, March 19, 12:153:15 P.M. (Hewlett 200) Thursday, March 21, 12:153:15 P.M. (Hewlett 200) 1
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• ###### Practice Midterm #1 Solutions
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###### Practice Midterm #1 Solutions
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #1 ANSWER KEY 1. C+ Basics / Parameters 9 -3 1 -7 9 6 5 -7 2 5 9 -3 0xcc00 0xbb00 0xdd00 -7 2. File I/O and Strings (write) As with any programming problem, there are many correct solutions. Here is one: void wordStats(string
• 10 Pages
###### Practice Midterm #1
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #1 This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual midterm exam. We do not guarantee that the number of questions in this sample exam will match the num
• 7 Pages
• ###### Practice Midterm #1 Spring 2013
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###### Practice Midterm #1 Spring 2013
School: Stanford
Course: Programming Abstractions
CS106B Spring 2013 Handout #14 May 1, 2013 Practice CS106B Midterm Exam This handout is intended to give you practice solving problems that are comparable in format and difficulty to the problems that will appear on the midterm examination on Tuesday, May
• 4 Pages
• ###### Practice Midterm #2 Winter 2013 Solutions
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###### Practice Midterm #2 Winter 2013 Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #25A February 1, 2013 Answers to Practice Midterm Exam #2 Review session: Sunday, February 3, 7:009:00 P.M., Hewlett 201 (next door) Midterm #1: Tuesday, February 5, 3:155:15 P.M., Braun Auditorium (Chemistry) Midterm #2: Tuesd
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• ###### Practice Midterm #1 Winter 2013 Solutions
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###### Practice Midterm #1 Winter 2013 Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #23A January 30, 2013 Answers to Practice Midterm Exam #1 Review session: Sunday, February 3, 7:009:00 P.M., Hewlett 201 (next door) Midterm #1: Tuesday, February 5, 3:155:15 P.M., Braun Auditorium (Chemistry) Midterm #2: Tuesd
• 10 Pages
###### Practice Midterm #2
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #2 This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual midterm exam. We do not guarantee that the number of questions in this sample exam will match the num
• 8 Pages
• ###### Practice Midterm #2 Summer 2014
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###### Practice Midterm #2 Summer 2014
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #2 This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual midterm exam. We do not guarantee that the number of questions in this sample exam will match the num
• 3 Pages
• ###### Practice Midterm #2 Summer 2014 Solutions
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###### Practice Midterm #2 Summer 2014 Solutions
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #2 ANSWER KEY 1. C+ Basics / Parameters yo2 - hi1 bye2 - yo1 hi12 - bye1 hi1 bye1 yo1 !hi12 2. File I/O and Strings (write) As with any programming problem, there are many correct solutions. Here is one: void flipLines(string
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• ###### Practice Midterm #2 Spring 2013
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###### Practice Midterm #2 Spring 2013
School: Stanford
Course: Programming Abstractions
CS106B Spring 2013 Handout #21 May 22, 2013 Practice Second Midterm Exam _ Based on handouts by Eric Roberts and Jerry Cain This handout is intended to give you practice solving problems that are comparable in format and difficulty to the problems that wi
• 4 Pages
• ###### Practice Midterm #2 Solutions
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###### Practice Midterm #2 Solutions
School: Stanford
Course: Programming Abstractions
CS 106B Practice Midterm Exam #2 ANSWER KEY 1. C+ Basics / Parameters yo2 - hi1 bye2 - yo1 hi12 - bye1 hi1 bye1 yo1 !hi12 2. File I/O and Strings (write) As with any programming problem, there are many correct solutions. Here is one: void flipLines(string
• 7 Pages
• ###### Practice Midterm #2 Spring 2013 Solutions
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###### Practice Midterm #2 Spring 2013 Solutions
School: Stanford
Course: Programming Abstractions
CS106B Spring 2013 Handout #22S May 24, 2013 Practice Second Midterm Exam Solutions _ Based on handouts by Eric Roberts and Jerry Cain Problem One: Reversing a Queue One way to reverse the queue is to keep moving nodes out of the list one at a time to the
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• ###### Practice Midterm #1 Winter 2013
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###### Practice Midterm #1 Winter 2013
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #23 January 28, 2013 Practice Midterm Exam #1 Review session: Sunday, February 3, 7:009:00 P.M., Hewlett 201 (next door) Midterm #1: Tuesday, February 5, 3:155:15 P.M., Braun Auditorium (Chemistry) Midterm #2: Tuesday, February
• 10 Pages
• ###### Practice Final #2 Summer 2014
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###### Practice Final #2 Summer 2014
School: Stanford
Course: Programming Abstractions
CS 106B Practice Final Exam #2 This sample exam is intended to demonstrate an example of some of the kinds of problems that will be asked on the actual final exam. We do not guarantee that the number of questions in this sample exam will match the number
• 40 Pages
###### Chapter7
School: Stanford
Course: Standard C++ Programming Laboratory
Exam Practice! True/False Problems 1. T 2. F 3. F 4. F 5. F Multiple Choice 6. (a) 7. (a) 8. (a) Memory Snapshot Problems 5 9. t -> 8 s -> H t -> 10. a 11 name -> 14 e e S l i u 17 l o e o \0 u \0 ? ? ? ? ? ? 11. 1 12. 265 13. K265 Programming Exercises L
• 19 Pages
###### 02-FunctionsInC++
School: Stanford
Course: Programming Abstractions
Functions in C+ Eric Roberts CS 106B January 7, 2015 Administrative Essentials All handouts and course information are on the web site: http:/cs106b.stanford.edu/ All CS 106B students must sign up for a section by Sunday at 5:00P.M. The signup form will a
• 30 Pages
###### 01-Introduction
School: Stanford
Course: Programming Abstractions
Introduction to CS 106B Eric Roberts CS 106B January 5, 2015 CS 106B Staff Professor: Eric Roberts eroberts@cs.stanford.edu Office Hours (Gates 202): Tuesdays 9:30-11:30 Wednesdays after class at Bytes Head TA: Kevin Miller kmiller4@stanford.edu Office Ho
• 2 Pages
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #48A March 11, 2015 Solutions to Problem Set #2 Problem 1: Binary search trees Problem 2: Graph traversals DFS from STAN: STAN, SRI, UTAH, BBN, CMU, NRL, HARV, MIT, RAND, UCLA BFS from CMU: CMU, BBN, NRL, UTAH, MIT, HARV, RAND
• 2 Pages
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #55 March 13, 2015 Looking Ahead CS Major: Systems and Theory Core Looking Ahead Eric Roberts CS 106B March 13, 2015 Structure of the CS Major Articial Intelligence and the Turing Test In 1950, Alan Turing posed a thought exp
• 2 Pages
###### 54-FunctionsAsData
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #54 March 11, 2015 Functions as Data Iteration Strategies Functions as Data Chapter 20 of the text covers two strategies for iterating over collections: iterators and mapping functions. Kevin showed you how to use iterators
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###### 53-MoreAlgorithms
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #53 March 9, 2015 More Algorithms Outline for Today More Algorithms for Trees and Graphs The plan for today is to walk through some of my favorite tree and graph algorithms, partly to demystify the many real-world application
• 3 Pages
###### 41A-Section7Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #41A February 25, 2015 Solutions to Section Handout #7 1. Coding depth-first search 2. Coding breadth-first search 3a) 3b) Lounge, Conservatory, BallRoom, BilliardRoom, Library, Hall, DiningRoom, Kitchen, Study Kitchen, BallRo
• 3 Pages
###### 38A-Section6Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #38A February 18, 2015 Solutions to Section Handout #6 1. Tracing binary tree insertion 1a. What is the height of the resulting tree? 5 1b. Which nodes are leaves? Bombur, Gloin, Oin, and Thorin 1c. Which nodes are out of bala
• 2 Pages
###### 41-Section7
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #41 February 25, 2015 Section Handout #7Graphs The purpose of this section is to let you work with the most important graph algorithms. Each of these questions uses the definitions of the Graph class and the Node and Arc struc
• 21 Pages
###### 03-StringsAndStreams
School: Stanford
Course: Programming Abstractions
Strings and Streams Eric Roberts CS 106B January 9, 2015 Administrative Reminders All handouts and course information are on the web site: http:/cs106b.stanford.edu/ All CS 106B students must sign up for a section by Sunday at 5:00P.M. The signup form wen
• 17 Pages
###### 04-CollectionClasses1
School: Stanford
Course: Programming Abstractions
Collection Classes (Part 1: Vectors, Grids, Stacks, and Queues) Eric Roberts CS 106B January 12, 2015 Computer Forum Career Fair Computer Forum Career Fair (For CS and EE Students) Wed, January 14, 11am 4pm Computer Forum Career Fair Wednesday, January 14
• 21 Pages
###### Chapter8
School: Stanford
Course: Standard C++ Programming Laboratory
Exam Practice! Memory Snapshot 1. x => 2. 0 x => 0 1 2 3 1 1 2 3 2 3 4 0 1 2 3 0 1 2 3 2 3 3 4 5 0 1 2 3 4 5 6 4 5 6 7 0 1 2 3 Program Output 3. 15 4. 6 5. 5 6. 10 Multiple-Choice Problems 7. (e) 8. (a) 9. (c) 10. (d) Programming Exercises Power Plant Dat
• 11 Pages
###### Chapter10
School: Stanford
Course: Standard C++ Programming Laboratory
Exam Practice! True/False Problems 1. 2. 3. 4. 5. F T F T T Multiple Choice Problems 6. (d) 7. (c) Programming Problems Binary Trees /*-*/ /* Program chapter10_8 */ /numLeafNodes: Public method int BinaryTree:numLeafNodes() const cfw_ if(root = NULL) cfw_
• 13 Pages
###### Chapter9
School: Stanford
Course: Standard C++ Programming Laboratory
Exam Practice! True/False Problems 1. T 2. T 3. F 4. F Multiple Choice Problems 5. (a) 6. (b) 7. (d) 8. (b) Memory Snapshot Problems 9. name=>20.5 x=>20.5 a=>14 10. 5 11. 12 12. 5 13. 5 Vector Functions /*-*/ /* Problem chapter9_14 */ /* */ /* This functi
• 36 Pages
###### Chapter6
School: Stanford
Course: Standard C++ Programming Laboratory
Exam Practice! True/False Problems 1. T 2. F 3. T 4. T 5. F 6. T 7. T Multiple Choice Problems 8. (b) 9. (a) 10. (d) Program Analysis 11. 1 12. 0 13. 0 14. Results using negative integers may not be as expected. Memory Snapshot Problems 15. v1-> [double x
• 33 Pages
###### Chapter5
School: Stanford
Course: Standard C++ Programming Laboratory
Exam Practice! True/False Problems 1. F 2. F 3. F 4. F 5. F Multiple-Choice Problems 6. (b)and(c) 7. (c) Memory Snapshot Problems 8. int i->0 int j->0 double x->1.0 double y->5.2 char ch1->a char ch2->, 9. int i->5 int j->0 double x->1.0 double y->? char
• 19 Pages
• ###### 08-RecursiveBacktracking
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###### 08-RecursiveBacktracking
School: Stanford
Course: Programming Abstractions
Recursive Backtracking Eric Roberts CS 106B January 23, 2015 Solving a Maze A journey of a thousand miles begins with a single step. Lao Tzu, 6th century B.C.E. The example most often used to illustrate recursive backtracking is the problem of solving a m
• 21 Pages
###### 07-RecursiveStrategies
School: Stanford
Course: Programming Abstractions
Recursive Strategies Eric Roberts CS 106B January 21, 2015 Recursion One of the most important Great Ideas in CS 106B is the concept of recursion, which is the process of solving a problem by dividing it into smaller subproblems of the same form. The ital
• 20 Pages
###### 06-DesigningClasses
School: Stanford
Course: Programming Abstractions
Designing Classes Eric Roberts CS 106B January 16, 2015 Optional Movie And so even though we face the difficulties of today and tomorrow, I still have a dream. It is a dream deeply rooted in the American dream. I have a dream that one day this nation will
• 16 Pages
###### 05-CollectionClasses2
School: Stanford
Course: Programming Abstractions
Collection Classes (Part 2: Maps, Sets, and Lexicons) Eric Roberts CS 106B January 14, 2015 Optional Movie And so even though we face the difficulties of today and tomorrow, I still have a dream. It is a dream deeply rooted in the American dream. I have a
• 6 Pages
###### 26-PracticeMidterm1
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #26 February 4, 2015 Practice Midterm Exam #1 Review session: Sunday, February 8, 7:009:00 P.M., Bishop Auditorium Midterm #1: Tuesday, February 10, 2:154:15 P.M., Annenberg Auditorium Midterm #2: Tuesday, February 10, 7:009:0
• 16 Pages
###### 44-Assignment6
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #44 February 27, 2015 Assignment #6Basic Due date: Friday, March 13 Note: Assignment #6 will not be accepted after 5:00 P.M. on Monday, March 16 2 In 1975, Bill Gates and Paul Allen started the company that would become Micros
• 4 Pages
• ###### 20-SortingAndEfficiency
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###### 20-SortingAndEfficiency
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #20 January 28, 2015 Sorting and Efficiency Sorting Sorting and Efficiency Of all the algorithmic problems that computer scientists have studied, the one with the broadest practical impact is certainly the sorting problem, wh
• 3 Pages
###### 19A-Section3Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #19A January 26, 2015 Solutions to Section Handout #3 Problem 1. Weights and balances 2 Problem 2. Shortest path Problem 3. Filling a region 3 Problem 4. Generating multiword anagrams
• 6 Pages
###### 15A-Section2Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #15A January 21, 2015 Solutions to Section Handout #2 Problem 1. Using grids 2 Problem 2. Using queues 3 Problem 3. Using maps 4 5 6 Problem 4. Using lexicons
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• ###### 16-RecursiveBacktracking
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###### 16-RecursiveBacktracking
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #16 January 23, 2015 Recursive Backtracking Solving a Maze Recursive Backtracking Eric Roberts CS 106B January 23, 2015 The Right-Hand Rule A journey of a thousand miles begins with a single step. Lao Tzu, 6th century B.C.E.
• 4 Pages
###### 19-Section3
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #19 January 26, 2015 Section Handout #3Recursion The purpose of this section is to give you some additional practice solving recursive problems. Problem 1. Weights and balances (Chapter 8, exercise 6, page 378) I am the only c
• 3 Pages
• ###### 17-BacktrackingAndGames
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###### 17-BacktrackingAndGames
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #17 January 26, 2015 Backtracking and Games Searching in a Branching Structure The recursive structure for finding the solution path in a maze comes up in a wide variety of applications, characterized by the need to explore a
• 8 Pages
###### 18-Assignment3
School: Stanford
Course: Programming Abstractions
Eric Roberts CS106B Handout #18 January 26, 2015 Assignment #3Recursion Parts of this handout were written by Julie Zelenski and Jerry Cain. Due: Wednesday, February 4 YEAH Hours: Wednesday, January 28, 7:00-9:00 P.M., Braun Auditorium This weeks assignme
• 3 Pages
###### 14-RecursiveStrategies
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #14 January 21, 2015 Recursive Strategies Recursion Recursive Strategies Eric Roberts CS 106B January 21, 2015 One of the most important Great Ideas in CS 106B is the concept of recursion, which is the process of solving a pr
• 2 Pages
###### 10-CollectionClasses2
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #10 January 14, 2015 Collection Classes (Part 2) Optional Movie Collection Classes (Part 2: Maps, Sets, and Lexicons) Eric Roberts CS 106B January 14, 2015 And so even though we face the difficulties of today and tomorrow, I s
• 4 Pages
###### 23-Section4
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #23 February 2, 2015 Section Handout #4 Pointers and Memory Problem 1. Differentiating the stack and the heap Using the heap-stack diagrams from Chapter 12 as a model, draw a diagram showing how memory is allocated just before
• 3 Pages
###### 23A-Section4Solutions
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #23A February 2, 2015 Solutions to Section Handout #4 Problem 1. Differentiating the stack and the heap 1a) Diagram using explicit addresses: 1b) Diagram using arrows: Problem 2: Pointer arithmetic 2a) a1[1] 2001 2b) a2[2] 3.
• 41 Pages
###### 02-CourseInformation
School: Stanford
Course: Programming Abstractions
5 CS 106B Calendar Monday January 5 Wednesday 7 Course overview The big ideas in CS 106B The C+ language C+ vs. Java Friday 9 Functions in C+ Call by reference Libraries and interfaces Recursive functions Using the string class File streams Class hierarch
• 3 Pages
###### 42-GraphAlgorithms
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #42 February 25, 2015 Graph Algorithms Outline Graph Algorithms 1. A review the graphtypes.h and graph.h interfaces 2. Depth-first and breadth-first search 3. Dijkstras shortest-path algorithm 4. Kruskals minimum-spanning-tree
• 4 Pages
###### 40-Graphs
School: Stanford
Course: Programming Abstractions
Eric Roberts and Kevin Miller CS 106B Handout #40 February 23, 2015 Graphs Outline Graphs 1. Examples of graphs 2. Defining a graph in terms of sets 3. Graph terminology 4. Designing the graph.h interface Kevin Miller (standing in for Eric Roberts) CS 106
• 4 Pages
###### 43-Inheritance
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #43 February 27, 2015 Inheritance in C+ Class Hierarchies Inheritance in C+ Much of the power of modern object-oriented languages comes from the fact that they support class hierarchies. Any class can be designated as a subcl
• 2 Pages
###### 45-BasicContest
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #45 February 27, 2015 BASIC Contest Submission deadline: Friday, March 13, 5:00 P.M. [Paul] Allen rushed to the dorm to find Bill Gates. They had to do a BASIC for this machine. They had to. If they didnt, the revolution would
• 5 Pages
###### 39-Sets
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #39 February 20, 2015 Sets Outline Sets 1. Midcourse correction 2. Sets in mathematics 3. Venn diagrams 4. High-level set operations 5. Implementing the Set class 6. Sets and efficiency Eric Roberts CS 106B February 20, 2015 7
• 4 Pages
###### 21-MemoryAndC++
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #21 January 30, 2015 Memory and C+ Data Types in C+ Memory and C+ The data types that C+ inherits from C: Atomic types: short, int, long, and their unsigned variants float, double, and long double char bool Enumerated types
• 4 Pages
###### 24-ProblemSet1
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #24 February 2, 2015 Problem Set #1 Due: Friday, February 6 by 5:00 P.M. Problem sets are pencil-and-paper exercises and may be submitted in section, in lecture, or to the box outside my office (Gates 202). Problem 1: Computat
• 3 Pages
###### 22-DynamicAllocation
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #22 February 2, 2015 Dynamic Allocation The Allocation of Memory to Variables Dynamic Allocation When you declare a variable in a program, C+ allocates space for that variable from one of several memory regions. One region o
• 3 Pages
###### 15-Section2
School: Stanford
Course: Programming Abstractions
Eric Roberts CS 106B Handout #15 January 21, 2015 Section Handout #2ADTs Problem 1. Using grids (Chapter 5, exercise 10, page 252) In the game of Minesweeper, a player searches for hidden mines on a rectangular grid that mightfor a very small boardlook li
• 31 Pages
###### 11-backtracking
School: Stanford
Course: Programming Abstractions
CS 106B, Lecture 11 Exhaustive Search and Backtracking reading: Programming Abstractions in C+, Chapters 8.2 - 8.3; 9 This document is copyright (C) Stanford Computer Science and Marty Stepp, licensed under Creative Commons Attribution 2.5 License. All ri
• 3 Pages
###### Section #8 Solutions
School: Stanford
Course: Programming Abstractions
CS 106B Section 8 Solutions (Week 10) 1. hash1 is valid, but not good because everything will get hashed to the same bucket. hash2 is not valid, because A and a are equal, but will have different hash values
• 4 Pages
###### Section #7 Solutions
School: Stanford
Course: Programming Abstractions
CS 106B Section 7 Solutions (Week 9) Presection problem. V: cfw_CS1, CS2, CS3, CS4, CS5, CS6, CS7, CS8 E: cfw_CS1-CS2, CS2-CS3, CS2-CS4, CS2-CS5, CS3-CS7, CS3-CS8, CS4-CS6, CS5-CS6, CS5-CS7, CS5-CS8 Adjacency List: C
• 4 Pages
###### Section #6 Solutions
School: Stanford
Course: Programming Abstractions
CS 106B Section 6 Solutions (Week 8) Presection problem. Dwalin / \ Balin Kili \ / \ Dori Fili Nori /
• 4 Pages
###### Section #5 Solutions
School: Stanford
Course: Programming Abstractions
CS106B Section 5 Solutions (Week 6) Presection problems 1. 2. v1 : 70 v2 : 25 p1 : points to v1 p2 : points to v1 list->next->next->next->next = list->next->next; / 40->30 l
• 2 Pages
###### Section #8
School: Stanford
Course: Programming Abstractions
CS 106B Section 8 Handout (Week 10) The last section! This week has a little bit of new content, and then some problems to review some of the bigger concepts from the second half of the quarter. Recommended problems
• 4 Pages
###### Section #6
School: Stanford
Course: Programming Abstractions
CS 106B Section 6 Handout (Week 8) This week is all binary trees. Remember that the recursive structure of trees makes writing recursive methods for them very natural. Recommended problems: 3, 8, 12, 13. (Try to do a
• 4 Pages
###### Section #5
School: Stanford
Course: Programming Abstractions
CS 106B Section 5 Handout (Week 6) ListNode structure (represents a single data value in a linked list, and a link to the next node) struct ListNode cfw_ int data; / data stored in this node
• 2 Pages
###### Section #4 Solutions
School: Stanford
Course: Programming Abstractions
CS106B Section 4 Solutions (Week 5) 1. 2. Presection problem. Changes to lecture code (from slide 21 of 11-backtracking) are underlined. The sumAll method is unchanged, so it is not included here. (It just takes a
• 2 Pages
###### Section #4
School: Stanford
Course: Programming Abstractions
CS106B Section 4 Handout (Week 5) 1. 2. 3. 4. This week takes recursion a bit further with recursive backtracking. Remember Martys three steps to doing backtracking problems, where you need to explore multiple paths i
• 12 Pages
###### 01-intro-to-cpp
School: Stanford
Course: Programming Abstractions
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http://www.chegg.com/homework-help/questions-and-answers/droplet-mass-80e-14kgpotential-voltage-5000ve-160e-19plate-separation-0016mset-vn-340870v--q647855
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# millikan oil droplet
0 pts ended
droplet mass=8.0e-14kg
potential voltage=5000v
e=1.60e-19
plate separation=0.016m
Set Vn = 3408.70V. Perform the experiment outlined above enough times to observe atleast one suspended oil droplet. Then calculate the number ofexcess electrons carried by the suspended oil droplet.
n = 1------------ electrons
Set Vn = 2800.00V. Perform the experiment, and determine the number of excesselectrons.
n = 2
electrons
What voltage would be required to balance an oil droplet with4 excess electrons? Try this voltageto see if you can observe a droplet with this amount of excesscharge.
Vn = 3
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http://www.numbersaplenty.com/828698
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Search a number
828698 = 21331873
BaseRepresentation
bin11001010010100011010
31120002202112
43022110122
5203004243
625432322
710021013
oct3122432
91502675
10828698
11516682
1233b6a2
13230270
1417800a
15115818
hexca51a
828698 has 8 divisors (see below), whose sum is σ = 1338708. Its totient is φ = 382464.
The previous prime is 828697. The next prime is 828701. The reversal of 828698 is 896828.
It is a happy number.
It can be written as a sum of positive squares in 2 ways, for example, as 734449 + 94249 = 857^2 + 307^2 .
It is a sphenic number, since it is the product of 3 distinct primes.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (828691) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 15911 + ... + 15962.
2828698 is an apocalyptic number.
828698 is a deficient number, since it is larger than the sum of its proper divisors (510010).
828698 is a wasteful number, since it uses less digits than its factorization.
828698 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 31888.
The product of its digits is 55296, while the sum is 41.
The square root of 828698 is about 910.3285121317. The cubic root of 828698 is about 93.9287977474.
The spelling of 828698 in words is "eight hundred twenty-eight thousand, six hundred ninety-eight".
Divisors: 1 2 13 26 31873 63746 414349 828698
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# Quiz Show
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### Quiz Show
1. 1. Question and Answer Samples and Techniques
2. 2. How to Use the Quiz Show Template <ul><li>Choose a Question & Answer layout from the New Slide gallery </li></ul><ul><li>Follow the placeholder prompts and fill in your actual questions and answers </li></ul><ul><li>View the presentation in slide show to see the animations that reveal the answers </li></ul><ul><li>Suggested Uses: </li></ul><ul><ul><li>Reinforcing teaching through audience participation </li></ul></ul><ul><ul><li>Introduction to subject matter </li></ul></ul><ul><ul><li>Recreational gatherings </li></ul></ul>
3. 3. <ul><li>The following slides are example questions using the layouts in the Quiz Show template. View them in slide show to see the answer animations. </li></ul>
4. 4. The Sun is a star.
5. 5. What is the name of our galaxy?
6. 6. How many planets in the solar system have rings? <ul><li>Jupiter, Saturn, Uranus, and Neptune all have rings. </li></ul>
7. 7. What is inertia? <ul><li>All of the above </li></ul><ul><li>The speed at which an object falls </li></ul><ul><li>Measurement of electrical resistance </li></ul><ul><li>A ratio between mass and velocity </li></ul><ul><li>Resistance to motion or change </li></ul>
8. 8. Match the device to what it measures: Stop Watch Scale Thermometer Speedometer Odometer Distance Temperature Elapsed Time Weight Rate of Travel
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# BOISE STATE MATH 143 - Review Sheet for Test #1 (3 pages)
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# Review Sheet for Test #1
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## Review Sheet for Test #1
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Pages:
3
School:
Boise State University
Course:
Math 143 - COLLEGE ALGEBRA (3-0-3)(Area III).
##### COLLEGE ALGEBRA (3-0-3)(Area III). Documents
• 7 pages
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• 7 pages
Unformatted text preview:
MATH 143 Review Sheet for Test 1 9 21 05 Last update Tue Sep 20 08 09 47 MDT 2005 1 m143 fa05 handouts143 t1 143 921 review suggestions 1 tex 1 This list is not in final form Like stuff may yet be added to it 2 Test 1 is Wednesday 9 21 05 3 There will be a with calculator part of this exam So have some batteries in your calculator And bring it to the test 4 The test will cover the material of Assignments 1 15 roughly 5 Be sure you can a Add algebraic fractions using the Least Common Denominator b Derive the Quadratic Formula and be able to discuss the Discriminant c Make a sign change chart for a product of binomials d That you know the two big triangle theorems i Pythagoras s Theorem ii The Similar Triangles Theorem These both relate picture information to algebra These theorems lie at the base of all the straight line equations facts slope works because of the Similar Triangles Theorem the perpendicularity and slope criterion works because of Pythagoras The equation of a circle comes straight from the distance formula which in turn is a manifestation of Pythagoras s Theorem e Completing the Square lies at the heart of i deriving the quadratic formula ii parsing a quadratic in x and y equation to see whether it s a circle f Add algebraic fractions using the Least Common Denominator g Decode negative exponents in expressions h Decode fractional exponents in expressions MATH 143 Review Sheet for Test 1 9 21 05 i j 6 Re the absolute value inequalities on assignment 14 a x 1 3 For this one many folks offered this beginning ploy 3 x 1 3 which excites suspicion because it implies that 3 3 So don t do that Method A One can correctly say that this inequality means that x 1 lies at least 3 units from zero so i x 1 3 or x 2 OR ii x 1 6 3 or x 6 4 We can put these two solutions together into the BOB answer 4 2 Method B Here s the trick presented in class trade the given inequality in temporarily on the complementary inequality x 1 3 This complementary inequality says that the
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https://nz.education.com/resources/fourth-grade/el/math/CCSS-ELA-Literacy-SL/
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It’s All in the Personality: Character Traits
Lesson Plan
It’s All in the Personality: Character Traits
In this lesson, students will use their creativity, a graphic organizer, and a fun story to learn about character analysis.
Math
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Lesson Plan
Close reading isn’t about just ticking through words on a page; it’s about absorbing ideas and expanding on them. In this lesson, students will use this strategy to make interpretations about a character's emotions through their actions.
Math
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Lesson Plan
Young readers will love this story-filled reading comprehension lesson. It's packed with engaging exercises designed to help students become better at looking for details and annotating passages of text.
Math
Lesson Plan
Text Features: Reading that Makes Sense
Lesson Plan
Text Features: Reading that Makes Sense
Improve your students' comprehension of non-fictional reading through this lesson that teaches them about text features. Students will find their own text features and explain why they aid in the reading process.
Math
Lesson Plan
Discovering Character Traits
Lesson Plan
Discovering Character Traits
This lesson will have your EL students exploring characters' dialogue and actions to determine their personality traits! Use it on its own or as a support lesson.
Math
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Area Arrangements
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Area Arrangements
Students will use the inverse relationship between multiplication and division to complete an area formula in a real-world situation. Use this lesson on its own or as support for the lesson The Case of the Missing Rectangle Side.
Math
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Understand Text Features
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Understand Text Features
Let's practice reading nonfiction texts! In this lesson, teach your ELs about identifying and describing text features. This lesson can be taught on its own or used as support for the lesson Formatting Text Features.
Math
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Rounding Roundtable
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Rounding Roundtable
Challenge students to talk about rounded numbers to the millions place value. Use this lesson as a standalone lesson or as support to the lesson Numbers All A-Round.
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https://rdrr.io/cran/FAMoS/man/combine.and.fit.html
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# combine.and.fit: Combine and Fit Parameters In FAMoS: A Flexible Algorithm for Model Selection
## Description
Combines fitted and non-fitted parameters and calls the fitting function. Serves as a wrapping function for the user-specified fitting function `fit.fn` (see `famos`).
## Usage
```1 2``` ```combine.and.fit(par, par.names, fit.fn, binary = NULL, default.val = NULL, ...) ```
## Arguments
`par` A named vector containing all parameters that are supposed to be fitted. `par.names` The names of all parameters `fit.fn` The cost function (see `famos` for more details). `binary` A vector containing zeroes and ones. Zero indicates that the corresponding parameter is not fitted. `default.val` A named list containing the values that the non-fitted parameters should take. If NULL, all non-fitted parameters will be set to zero. Default values can be either given by a numeric value or by the name of the corresponding parameter the value should be inherited from (NOTE: In this case the corresponding parameter entry has to contain a numeric value). Default to NULL. `...` Other arguments.
## Value
Returns the negative log-likelihood as calculated by the specified cost function
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11``` ```#set parameters and cost function fit.par <- c(p1 = 2, p2 = 4) name.par <- c("p1", "p2", "p3") defaults <- list(p1 = 0, p2 = 2, p3 = 4) cost.function <- function(parms){ parms[1] + parms[2] + parms[3] } #call combine.and.fit combine.and.fit(par = fit.par, par.names = name.par, fit.fn = cost.function) combine.and.fit(par = fit.par, par.names = name.par, fit.fn = cost.function, default.val = defaults) ```
FAMoS documentation built on May 26, 2019, 5 p.m.
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https://brainmass.com/math/probability/probability-series-independent-games-274450
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# Probability
Fran and Ron play a series of independent games. Fran's probability of winning any particular game is 0.6 (and Ron's probability of winning is therefore 0.4). Suppose that they play a best-of-5 tournament. (That is, the winner of the tournament is the first person to win 3 games.)
1. Find the probability that Fran wins the tournament in 3 games.
1) _______________
2. Find the probability that the tournament lasts exactly 3 games.
2) _______________
3. Find the probability that the tournament lasts exactly 4 games.
3) _______________
4. Find the probability that Fran wins the tournament.
4) _______________
5. Find the probability that the tournament lasts exactly 3 games if it is won by Fran.
5) _______________
6. Find the probability that Fran won the tournament if it lasted exactly 3 games.
6) _______________
#### Solution Preview
Dear student, please refer to the attachment for the solutions. Thank You.
Probability of Fran's winning the game = 0.6
Probability of Ron's winning the game = 0.4
1. Required Probability = [(5C3)[ (0.6)3x(0.4)2]
2. The tournament will last exactly three games if either Fran or Ron wins all the first three games.
A: Fran wins the ...
#### Solution Summary
This solution is comprised of detailed explanation and step-by-step calculation of the given problem and provides students with a clear perspective of the underlying concepts.
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PDA
View Full Version : How do you describe gravity to a 3.5 yr old?
RBG
2003-Jul-04, 12:05 AM
I think if I could think of a simple way to introduce general relativity to her, I would. I have this thought that there are simple ways to explain how things work. Although I'm not sure my way is necessarily the easiest. (But, man, is it fun introducing this little bean to the world.)
Gravity:
Me: "Everything you see around you is actually part of a huuuuge ball that we live on."
Her: "We live on the top of the ball."
Me: "No we live on, umm, the side of the ball but something called "gravity" keeps us on the ground and from floating away or falling off." <woo-wooo-wooooo... boink>
Her: "Gravity keeps us on the ground so we don't float away."
Me: "It works something like this: I take this balloon and rub it in my hair and see how it sticks to the side of this wall? Well, gravity works something like that. Not exactly the same."
(At this point I wonder if I've completely fuzzified everything.)
__________
How's that? What would be better? I'm drawing upon my tiny, little bit of knowledge of strong & weak forces. The mad scientist in me wants to figure a way to explain those attracting forces. (This kid is smart - unlike myself.)
RBG
RBG
2003-Jul-04, 12:10 AM
Dang... I just realized it would have been more appropriate to use refrigerator magnets.
R
Donnie B.
2003-Jul-04, 12:11 AM
"Everything, every bit of stuff, attracts every other bit of stuff. For small things like people and toys and cars, the pull is so small we can't feel it. But the Earth we live on is really big, so it pulls on us pretty hard. That's why, when we jump up, we fall back down. The Earth pulls us back."
tracer
2003-Jul-04, 12:26 AM
(Hold up globe of the Earth) " 'Down' is toward the center of the big ball, so that means for these people living over here in Australia, 'down' is this way" (point)
(Unless, of course, you're a member of the International Flat Earth Society (http://www.lhup.edu/~dsimanek/fe-scidi.htm).)
Irishman
2003-Jul-05, 06:57 AM
I'm not sure how to explain anything to a 3.5 yr old.
I'm a little confused to what your purpose is. Are you coming at this cold? I would think it would be easier using her natural questions to shape the explanation. What part of gravity are you trying to explain?
Drop an object. See, that's gravity in action. It makes things fall. It is what you have to pull against to stand up or walk up stairs.
Everything we're made of, all stuff, has a certain amount of pull to everything else. For small objects, like a ball or ourselves or cars - even objects the size of buildings - the amount of pull is very small and not noticable. But when really really large objects, like the Earth, are involved, that tiny pull adds up to something significant. That pull is gravity.
Magnets could be used to show one type of pull. Of course, you could just end up making her think magnets use gravity.
If you want to explain the ball (how Earth is spherical), now you have a framework to start on. Now you can get out the globe.
GarethB
2003-Jul-05, 11:58 AM
It's easy. Gravity: It's not just a good idea, it's a force of nature! You can substitute it for all sorts of things, like this. Lemmings: they're not just a good idea, they're a force of nature! See how easy that was? :D
Colt
2003-Jul-05, 06:57 PM
The best thing I can think of to do is to find the episode of Bill Nye on the internet or at a video store and let him watch it. Then talk to him about it. :P -Colt
The Supreme Canuck
2003-Jul-05, 07:45 PM
I hate Bill Nye. We watched him every year in science class up to and including grade 10! :x Evidently we have education problems up here.
But it probably would be a good thing to show a child.
Humphrey
2003-Jul-06, 04:42 AM
Hey! i liked Bill Nye the Science guy! :evil:
The Supreme Canuck
2003-Jul-06, 04:44 AM
So did I. I just don't like him replacing my science teacher. I also don't like the fact that his shows seem to be at the same level as Provincial Curriculum. Kind of scary...
Schultze
2003-Jul-06, 02:26 PM
Explaining is not the difficult part. It's the understanding of the information. I seriously doubt that even the brightest 3-1/2 year-old can understand the concept of gravity.
The easiest explanation at this stage in the youngsters life would be just to drop an object and say this is what gravity does. Later you can explain the why part of it.
I know 3-1/2 year-olds are full of questions and probably the most used word is why? or how come?, but there are some things that they just cannot be made to understand at this age.
Glom
2003-Jul-06, 08:01 PM
I know 3-1/2 year-olds are full of questions and probably the most used word is why? or how come?, but there are some things that they just cannot be made to understand at this age.
What? So all those hours I spent trying to explain two body orbital mechanics to my cousin were wasted? Damn!
Humphrey
2003-Jul-06, 11:49 PM
You will be surprised the things you can learn being a camp counselor to a bunch of small kids. (granted these were kindergarden and older) :-).
Some of them really surpriused me with their knoledge.
Side note: I remeber a few weeks ago learning about a 14 year old starting medical school at Columbia university. They said that at Two, he could read and write, at 3 he was playing classical music, and at 4 he was composing music. Can't remember his name.
Schultze
2003-Jul-06, 11:53 PM
Glom should have explained orbital mechanics to him.
The Supreme Canuck
2003-Jul-06, 11:56 PM
I'm sure he already knew it!
Astronot
2003-Jul-07, 03:30 PM
I think if I could think of a simple way to introduce general relativity to her, I would. I have this thought that there are simple ways to explain how things work. Although I'm not sure my way is necessarily the easiest. (But, man, is it fun introducing this little bean to the world.)
RBG
A way I found to explain things to a 3yo is to play a family game and teach by analogy. Get the family together and tell the child that gravity is like love. That the more of the family that is together the more love you feel. You stand next to her and give her a light hug. Bring in a another family member to illustrate a larger mass and give a bigger hug. Continue until you have exhausted the entire family, literally or figuratively.
A similar game works for condensation and rain. Use one or two children as a water molecules and everyone else as air. When they are warm they (the air) stand apart, holding hands and dance around the water. As they get cooler, the air gets closer together and slow the motion, finally as they are cold, the huddle up tightly and push out the “water”, which falls to the ground.
Supply your own dialog to go with these games as best suits your child, but expect to need to explain this often as my kids tended to “forget” how these work with alarming regularity.
tjm220
2003-Jul-07, 05:01 PM
I've already told you to stop doing that, if you do it one more time...
Oops wrong type of gravity. :oops:
man on the moon
2003-Jul-08, 04:59 AM
garfield said it well..."it;s not gravity, it's GRABity!" (i belive he was talking to ody) :lol:
ToSeek
2003-Jul-08, 04:32 PM
A similar game works for condensation and rain. Use one or two children as a water molecules and everyone else as air. When they are warm they (the air) stand apart, holding hands and dance around the water. As they get cooler, the air gets closer together and slow the motion, finally as they are cold, the huddle up tightly and push out the “water”, which falls to the ground.
Sounds like fun, but unfortunately it has nothing to do with the actual cause of condensation. (http://www.ems.psu.edu/~fraser/Bad/BadClouds.html)
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https://electronics.stackexchange.com/questions/68157/voltage-regulator-too-hot-to-touch
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# Voltage regulator too hot to touch [duplicate]
I'm working on an RGB LED project. I'm using this as the power supply: https://www.sparkfun.com/products/114 It uses an LM317 which according to the data sheet can supply up to 1.5A. The LED's I'm using are these: http://www.adafruit.com/products/314
And the mains power supply I'm using is a 12v DC 2 AMP one.
Here's the problem. If I have just all of the red parts of the LED's lit up (I have 8 LED's), within about a minute the voltage regulator is far too hot to touch and I disconnected it when it started to smell funny. I attached a coin to the heatsink part of the voltage regulator (Not great I know, but I'm still waiting for some proper heat sinks I ordered to arrive) and within about a minute that too was far too hot to touch.
How can this be? Each red part of the LED uses 20mA, so multiplied by 8 that is 160mA. So the circuit isn't drawing too much current, is it? I'm using 150 Ohm resistors for each LED.
Why is it heating up so much, and what am I doing wrong?
Thanks
(Also, if I powered up all red, green, and blue parts of all 8 LED's, the voltage regulator gets blisteringly hot)
(I've just being reading about "switching" regulators being used as they produce less heat, would this be a better idea?)
A linear regulator dissipates heat proportional to the amount of voltage it must drop, and the amount of current flowing through it.
Input supply = 12 Volts
Option A:
Output Vout = 3.3 Volts
Vfred = 2.1 Volts
Current = (Vout - Vf) / R = 7.5 mA per LED = 60 mA total
Power dissipated by regulator, P = (Vinput - Vout) x I = 522 mW
Over half a watt is being dissipated by the linear regulator in this case. Add the green and blue channels, and dissipation will tend to thrice that figure.
Option B:
Output Vout = 5 Volts
Vfred = 2.1 Volts
Current = (Vout - Vf) / R = 19.33 mA per LED = 154.67 mA total
Power dissipated by regulator, P = (Vinput - Vout) x I = 1082.67 mW
Over 1 watt is being dissipated by the linear regulator in this case. 1 Watt is a substantial amount of power being emitted as heat. The regulator getting too hot to touch is not unexpected in this case. Adding the green and blue channels will make things much worse.
Recommendations:
1. Use the breadboard power supply in 3.3 Volt mode until you have a good heat-sink attached, even 7.5 mA per LED channel should provide a reasonable amount of illumination from the LEDs.
2. Switch over to a DC-DC switching regulator or buck regulator, those waste much less power as heat.
Problem:
You power the LM317 with 12V and it generates an output of 5V @160 mA.
So 12V - 5V = 7V @ 160 mA that the lm317 needs to dissipate (heat).
P(watt) = I * U
= 0.160 A * 7V
= 1.12 W of heat
The datasheet of the LM317 says:
The Thermal resistance, junction to case is 5 °C/W
The Thermal resistance, junction to ambient is 80 °C/W (no heatsink)
so total thermal resistance is: 85 °C/W
Temperature rise is: 1.12 W * 85 °C/W = 95.2 °C
When the room temperature is 21 °C the case temperature will be 116.2 °C
The LM317 has protection so it starts to shut off when it gets to hot.
Recommendations:
You really need a better heatsink OR lower current usage OR a lower input voltage OR ...
• What does 12 - 5 equal? – Andy aka May 5 '13 at 0:36
• Math at 3 in the morning =/ . – Spikee May 5 '13 at 10:19
Yes, the LM317 can handle up to 1.5A, but when dropping 12V down to 5V, it needs to get rid of 10.5 watts worth of heat in the process! There's no way it can do this without a substantial heatsink attached.
Even at the 160 mA your red LEDs are drawing, the regulator is dissipating well over 1 watt. Its temperature will rise until the amount of heat being transferred to the air equals the amount of heat being generated. If there isn't much surface area over which that can happen, the temperature will quickly rise above the maximum rating of the device. The coin adds a little area (and a little thermal mass), but not nearly enough.
• Thanks for the reply, it's helped me understand what's going wrong. What would you suggest I do? I wanted to make a fairly small project, and having a large heatsink would make it bigger. – Lloyd May 4 '13 at 21:50
• Two quick suggestions: 1) If your project has a metal case, you could try bolting your regulator directly to it (use short wires to connect it to the PCB if necessary). 2) Use a 3-terminal switching regulator, which will be much more efficient to begin with. – Dave Tweed May 4 '13 at 21:54
The quoted maximum current the LM317 can supply requires that you provide a way for the chip to get rid of the heat it must produce to 'boil away' the voltage difference. You don't say what the output voltage of the LM317 is, I assume 5V. That means that each part LED that you turn on caused the LM317 to dissipate 0.02 * (12 - 5 ) = 0.14 Watt. A TO220 package without any additional heatsink can dissipate ~ 1 W (but it will get hot!). For 24 LED parts that is 3.4 Watt. That requires a decent heatsink for the LM317.
A heatsink can be calculated much like a resistor. You want to dissipate 4 Watt, over a temperature difference of let's say 40 degrees C. Hence you need a heatsink of (at most) 40 degrees / 4 Watt = 10C/W. That is not a very big one.
You did not give details about your circuit, but other options could be
• using a power supply that outputs 9V instead of 12V, or even one that outputs 5V
• if you use 2803-style open-collector drivers: feed the LEds directly from the 12V. This will not reduce the total power dissipation, but it will distribute it over all resistors.
• use a small switching converter to produce the 5V
(I removed the suggestion to put the same color LEDs in series, because that is not possible with RGB LEDs, and Loyd seems to want to control the LEDs individually.)
• Thanks for the reply. I'm actually using a ULN2003 chip. Would you recommend I buy a 5V mains supply or use a switching converter? – Lloyd May 4 '13 at 22:02
• A 5V power supply is likely more convenient, and cheaper. Any plug-in supply that hooks up to a USB plug (ie most phone chargers) puts out 5V, so you can generally find one for free. – Eric Gunnerson May 5 '13 at 2:47
• @lloyd: I don't know your situation, so that's for you to decide. If you stick with 12V I suggest you power the LEDs directly from the 12V. If 5V is possible, go for that (smallest chance). – Wouter van Ooijen May 5 '13 at 6:08
• Funny, I got two downvotes!? Please comment and state why! – Wouter van Ooijen May 5 '13 at 6:11
• I think I will use a 5V power supply. I don't know why you was downvoted, I havent down voted any replies on this thread. – Lloyd May 5 '13 at 12:35
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## Free Printable Math Practice Worksheet for Subtraction for Fractions Different Denominators 4
Subtraction for FractionsDifferent Denominators 4Version 3 Name:________________________
Solve the fraction problem and reduce the answer to simplest form.
8 9
-
9 12
=
7 14
-
1 9
=
4 14
-
2 9
=
9 16
-
4 15
=
8 18
-
2 15
=
7 8
-
9 15
=
6 8
-
8 14
=
9 16
-
9 18
=
9 12
-
1 16
=
6 9
-
5 12
=
Copyright © 2008-2011 DadsWorksheets.com
Free for personal or classroom use provided this notice is preserved.
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