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https://discuss.codecademy.com/t/what-am-i-doing-wrong-use-a-for-loop-to-iterate-through-the-array/653693 | 1,660,086,347,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571090.80/warc/CC-MAIN-20220809215803-20220810005803-00699.warc.gz | 215,020,278 | 7,589 | # What am I doing wrong? - Use a for loop to iterate through the array
I’m having a hard time figuring out what they expect me to do in this course.
Instructions: You are given an array of five numbers called increaseByTwo. Use a for loop to iterate through the array and increase each number by two. This means I need to change the screen from “1, 2, 3, 4, 5” to “3, 4, 5, 6, 7”.
CODE:
``````const increaseByTwo = [1, 2, 3, 4, 5];
// Uncomment the line below to check your work!
// console.log(increaseByTwo); // -> should print [3, 4, 5, 6,
7];
``````
What I’ve done so far: I’ve pushed two more numbers to the console, so now instead of 1-5, it’s 1-7.
In attempt 1, I tried to pop the first two numbers out of the const.
CODE:
``````const increaseByTwo = [1, 2, 3, 4, 5];
for (let i = 0; i < 1; i+=2)
increaseByTwo.push(6, 7)
for (let i = 0; i > 3; i--) {
increaseByTwo.pop(1, 2)
}
``````
In attempt 2, I almost got that result, but it was in columns and not rows. When I take “i” away from the console.log, it results into telling me 1-7, 7 different times.
CODE:
``````const increaseByTwo = [1, 2, 3, 4, 5];
for (let i = 0; i < 1; i+=2)
increaseByTwo.push(6, 7)
for (let i = 2; i < 7; i++) {
console.log(increaseByTwo[i])
}
``````
What am I doing wrong?
Hey and welcome to the forums!
At the moment you are hard coding the values `6` and `7` in, now while this may work for that given input, where it would not work is if you had a different starting array. One way of doing it that would avoid this be to modify each element in-place in the array by accessing it’s index with `array[i]` and adding to that, as that would work for all inputs. If you want to give it a go, I’d suggest giving it a shot yourself first, but I’ll leave a hint using a different array/goal below if you’d like it.
Hint
``````// Tripling the elements in an array
for(let i = 0; i < array.length; i++) {
array[i] *= 3;
}
``````
Happy coding!
2 Likes
Oh my god that was it thank you so much
1 Like | 631 | 1,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-33 | latest | en | 0.899588 |
https://byjus.com/question-answer/simplify-9-14-19-overline-14-12-14-overline-15-12/ | 1,642,964,823,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.5/warc/CC-MAIN-20220123172206-20220123202206-00169.warc.gz | 220,622,074 | 20,040 | Question
# Simplify: $$[9-{14-(19-\overline{14-12})}]-(14-\overline{15+12})$$
Solution
## The BODMAS rule can simplify any expression using these steps of solving:Brackets (parts of a calculation inside brackets always come first).Orders (numbers involving powers or square roots).DivisionMultiplicationAdditionSubtractionNow, using BODMAS rule, we solve the given expression $$\left[ 9−\left\{ 14−\left( 19−\overline { 14-12 } \right) \right\} \right] −\left( 14−\overline { 15+12 } \right)$$ as follows:$$\left[ 9−\left\{ 14−\left( 19−\overline { 14-12 } \right) \right\} \right] −\left( 14−\overline { 15+12 } \right) \\ =\left[ 9−\left\{ 14−\left( 19−2 \right) \right\} \right] −\left( 14−27 \right) \\ =\left[ 9−\left\{ 14−17 \right\} \right] −\left( −13 \right)$$$$=\left[ 9−\left( −3 \right) \right] +13\\ =\left[ 9+3 \right] +13\\ =12+13\\ =25$$ Hence, $$\left[ 9−\left\{ 14−\left( 19−\overline { 14-12 } \right) \right\} \right] −\left( 14−\overline { 15+12 } \right)=25$$Mathematics
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View More | 422 | 1,084 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-05 | latest | en | 0.402238 |
https://discuss.codecademy.com/t/does-someone-help-please-with-the-modulo-operator-in-java-why-15-6-is-equal-3/186152 | 1,537,927,046,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162809.73/warc/CC-MAIN-20180926002255-20180926022655-00095.warc.gz | 498,935,342 | 4,309 | # Does someone help please with the modulo operator in Java why 15 % 6 is equal 3?
#1
Why 15 % 6 is equal 3 ?
#2
because the modulo operator gives you the remainder, the number left and after the biggest possible division.
12 is divisible by 6 and smaller then 15
18 is divisible by 6 but not smaller then 15
so 12 is the biggest number divisible by 6, so we have 3 remaining (15 - 12= 3)
#3
To put it another way, the modulo operator shows only the remainder and ignores the quotient.
6 goes into 15 twice, up to 12, yes? and from 12 to 15, the remainder is 3.
#5
When I used BASIC on the ZX Spectrum years ago there was no modulo function.
There was as INT function which just returned the integer part of a decimal number
e.g INT 3.123 = 3.
So I did modulo like this , X mod Y = ( X/Y - INT (X/Y)) * X
15 % 6 = (15/6 - INT (15/6)) * 6 =( 3.5 - 3) * 6 = 3
It’s basic numeracy and a bit of computer history!
#6
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# EXAMPLE 5 - PowerPoint PPT Presentation
Baseball. A professional baseball should weigh 5.125 ounces, with a tolerance of 0.125 ounce. Write and solve an absolute value inequality that describes the acceptable weights for a baseball. STEP 1. Write a verbal model. Then write an inequality. EXAMPLE 5.
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Baseball
A professional baseball should weigh 5.125 ounces, with a tolerance of 0.125 ounce. Write and solve an absolute value inequality that describes the acceptable weights for a baseball.
STEP1
Write a verbal model. Then write an inequality.
EXAMPLE 5
Solve an inequality of the form |ax + b| ≤ c
SOLUTION
STEP2
Solve the inequality.
So, a baseball should weigh between 5 ounces and 5.25 ounces, inclusive. The graph is shown below.
EXAMPLE 5
Solve an inequality of the form |ax + b| ≤ c
|w – 5.125| ≤ 0.125
Write inequality.
Write equivalent compound inequality.
– 0.125 ≤ w – 5.125 ≤ 0.125
5 ≤ w ≤ 5.25
Gymnastics
The thickness of the mats used in the rings, parallel bars, and vault events must be between 7.5 inches and 8.25 inches, inclusive. Write an absolute value inequality describing the acceptable mat thicknesses.
Calculate the mean of the extreme mat thicknesses.
STEP1
EXAMPLE 6
Write a range as an absolute value inequality
SOLUTION
Mean of extremes ==7.875
STEP2
Find the tolerance by subtracting the mean from the upper extreme.
7.5 + 8.25
2
EXAMPLE 6
Write a range as an absolute value inequality
Tolerance = 8.25 –7.875
=0.375
STEP3
Write a verbal model. Then write an inequality.
A mat is acceptable if its thickness tsatisfies |t – 7.875| ≤ 0.375.
EXAMPLE 6
Write a range as an absolute value inequality
–8 <x <4
for Examples 5 and 6
GUIDED PRACTICE
Solve the inequality. Then graph the solution.
10. |x + 2| < 6
The solutions are all real numbers less than – 8 or greater than 4. The graph is shown below.
–5 ≤x ≤4
for Examples 5 and 6
GUIDED PRACTICE
Solve the inequality. Then graph the solution.
11. |2x + 1| ≤ 9
The solutions are all real numbers less than –5 or greater than 4. The graph is shown below.
3 ≤x ≤11
The solutions are all real numbers less than 3 or greater than 11. The graph is shown below.
for Examples 5 and 6
GUIDED PRACTICE
Solve the inequality. Then graph the solution.
12. |7 – x| ≤ 4 | 801 | 2,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-09 | latest | en | 0.876986 |
https://itfeature.com/probability/ | 1,708,776,065,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474533.12/warc/CC-MAIN-20240224112548-20240224142548-00015.warc.gz | 336,368,101 | 44,422 | # Probability
## Probability Online MCQs Test 6
This Quiz contains Probability Online MCQs Test, events, experiments, mutually exclusive events, collectively exhaustive events, sure events, impossible events, addition and multiplication laws of probability, discrete probability distribution, continuous probability distributions, etc. Let us start the Probability Online MCQs Test with the Answers:
Online MCQs Exam for the Post of PPSC Statistics Lecturer and Statistical Officer
1. What is a probability?
2. How would you describe $P(A|B)$ in words for two sets $A$ and $B$?
3. A fair coin is tossed 50 times, and the expected number of heads is:
4. What word describes two events that cannot occur at the same time?
5. What is the expected value?
6. What is a random variable?
7. In the context of probability, what is a sample space?
8. $P(A\cap B() = P(A) P(B|A)$, then $A$ and $B$ are
9. What shows the exact probabilities for a particular value of a random variable?
10. If $A$ and $B$ are dependent events, $P(A)=0.40$ and $P(B|A)=0.35$ then $P(A \cap B)$ is
11. The probability of an event happening is $\frac{1}{3}$. The probability of it not happening is?
12. How would you calculate the probability that a random variable is less than 5?
13. What is conditional probability?
14. In the context of probability, what is an outcome?
15. What is a continuous random variable?
16. $A$ and $B$ are two mutually exclusive events. The probability of $A$ happening is $\frac{1}{4}$. The probability of $BB$ happening is $\frac{1}{3}$. The probability of neither $A$ nor $B$ happening is?
17. How would you describe $P(A \cap B)$ in words for two sets $A$ and $B$?
18. A conditional probability might be found in which of the following ways?
19. The subset of a sample space is called
20. Events having an equal chance of occurrence are called
### Probability Online MCQs Test
• In the context of probability, what is an outcome?
• What is a probability?
• How would you calculate the probability that a random variable is less than 5?
• In the context of probability, what is a sample space?
• What word describes two events that cannot occur at the same time?
• What is the expected value?
• What is conditional probability?
• What is a continuous random variable?
• What shows the exact probabilities for a particular value of a random variable?
• How would you describe $P(A \cap B)$ in words for two sets $A$ and $B$?
• How would you describe $P(A|B)$ in words for two sets $A$ and $B$?
• What is a random variable?
• $A$ and $B$ are two mutually exclusive events. The probability of $A$ happening is $\frac{1}{4}$. The probability of $BB$ happening is $\frac{1}{3}$. The probability of neither $A$ nor $B$ happening is?
• The probability of an event happening is $\frac{1}{3}$. The probability of it not happening is?
• A conditional probability might be found in which of the following ways?
• A fair coin is tossed 50 times, and the expected number of heads is:
• If $A$ and $B$ are dependent events, $P(A)=0.40$ and $P(B|A)=0.35$ then $P(A \cap B)$ is
• $P(A\cap B() = P(A) P(B|A)$, then $A$ and $B$ are
• The subset of a sample space is called
• Events having an equal chance of occurrence are called
R Language Interview Questions
## Online Probability Distributions Quiz
This Post is about the Online Probability Distributions Quiz and covers topics about the Mean and Variance of random variables and the distribution of Random variables. MCQs Probability Random variable quiz requires knowledge of events, experiments, mutually exclusive events, collectively exhaustive events, sure events, impossible events, addition and multiplication laws of probability, discrete probability distribution and continuous probability distributions, etc. To start with Online Probability Distributions Quiz click the links below.
### Online Probability Distributions Quiz
R Programming for Data Analysis
## Online Probability Quiz
This post contains MCQs about Online Probability Quiz, events, experiments, mutually exclusive events, collectively exhaustive events, sure events, impossible events, addition and multiplication laws of probability, discrete probability distribution, continuous probability distributions, etc. Let us start the MCQs Online Probability Quiz: | 1,015 | 4,305 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-10 | longest | en | 0.915311 |
https://www.mcqslearn.com/business-statistics/quiz/quiz.php?page=23 | 1,560,674,781,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998084.36/warc/CC-MAIN-20190616082703-20190616104703-00051.warc.gz | 817,589,558 | 7,292 | Data classification quiz questions, data classification MCQs answers, MBA business statistics quiz 23 to learn business analytics courses online. Data classification, tabulation and presentation quiz questions and answers, data classification multiple choice questions (MCQ) to practice statistics test with answers for online colleges and universities courses. Learn data classification MCQs, discrete probability distributions, classification: measures of dispersion, principles of sampling, data classification test prep for business analyst certifications.
Learn data classification test with multiple choice question (MCQs): classification method in which upper and lower limits of interval is also in class interval itself is called, with choices exclusive method, inclusive method, mid-point method, and ratio method for online bachelor degree. Learn data classification, tabulation and presentation questions and answers for scholarships exams' problem-solving, assessment test for business administration certifications.
Data Classification Quiz
MCQ: Classification method in which upper and lower limits of interval is also in class interval itself is called
1. exclusive method
2. inclusive method
3. mid-point method
4. ratio method
B
Principles of Sampling Quiz
MCQ: Principle which states that larger sample size larger accuracy and stability is part of
1. principle of sampling error
2. principle of inertia
3. principle of statistical regularity
4. principle of statistical irregularity
B
Classification: Measures of Dispersion Quiz
MCQ: Relative measures in measures of dispersion are also considered as
1. coefficient of deviation
2. coefficient of average
3. coefficient of variation
4. coefficient of uniformity
C
Discrete Probability Distributions Quiz
MCQ: Number of products manufactured in a factory in a day are 3500 and probability that some pieces are defected is 0.55 then mean of binomial probability distribution is
1. 1925
2. 6364
3. 63.64
4. 3500
A
Introduction of Estimation Quiz
MCQ: For a parameter whose value is unknown, belief or claim for that parameter is classified as
1. parameter claim testing
2. expected belief testing
3. hypothesis testing
4. primary limit testing
C | 445 | 2,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-26 | latest | en | 0.893915 |
https://www.danielang.net/2016/10/16/guide-to-the-acme-edm-experiment-a-simple-overview/?shared=email&msg=fail | 1,624,336,845,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488507640.82/warc/CC-MAIN-20210622033023-20210622063023-00453.warc.gz | 634,748,478 | 18,947 | # Guide to the ACME EDM Experiment: A Simple Overview
A shorter introduction to the ACME EDM experiment can be found here.
In the first introductory post of this series I outlined the immense precision of the ACME experiment in measuring the electron’s “shape.” A natural followup question is, how did we do that? In this post I will outline the basic measurement scheme in our experiment in the most general way possible, avoiding most of the technical physics jargon. The next post after this will get into the more gory technical details, but this one should hopefully be an understandable, but still mostly accurate and insightful explanation on how we do what we do. I will refer to the incarnation of the experimental apparatus which was used in the 2014 measurement as ACME I, whereas the current incarnation (with various improvements) is called ACME II. I will explain mostly the features of the ACME I experiment, although at this level the two generations are not very different.
# Electron Shapes and Asymmetry
So far I’ve been saying that we’re measuring the electron’s “shape.”1 But really, what we’re measuring is the electron’s electric dipole moment. I will not go into detail of what a dipole moment here is here. For now, it’s sufficient to just think of this as a vector quantity having a magnitude and direction, . In our extended (and admittedly imperfect) analogy about the electron’s “shape,” if that shape had a slight bulge or asymmetry, it would manifest itself in a non-zero value of . If we put our asymmetric electron in an electric field , then that would result in an energy shift, which is nice because physicists like to think of things in terms of energies:
.
is typically measured in units of , and electric fields are in units of , so has units electron volts (eV), which is an energy unit,2 so this makes sense. In other words, to look for this asymmetry, we want to put an electron in an electric field and observe its behavior: we want to see if there’s an energy shift which would come from this asymmetry, as expressed in the quantity . Notice that if we increase the strength of the electric field , then the overall energy shift also gets bigger and easier to find.
This is literally what we do in our experiment: we are taking electrons and putting them in an electric field. Crudely speaking, we prepare our electrons using lasers and other contraptions, launch them through the field, and then interrogate them after they’ve passed through the field, observing how much they’ve “changed” as a result of this field. We can then narrow down to see whether they’ve changed more than we expected them to. If that’s the case, then that would be evidence that and thus the electron is asymmetric. In a nutshell, that is what we do in the ACME EDM experiment.
# Riding a Molecule
However, it turns out that it’s not that easy to physically realize in the real world. The part is easy: we can produce a relatively uniform electric field by putting two flat parallel plates a set distance from each other.
The part is the hard one: it’s hard to put a single electron in an electric field. They’re so small (as we mentioned last time, at least a billion times smaller than a human hair) and typically we find them as part of other atoms and molecules, orbiting the nucleus.3 Even if you could get a lone electron, putting them between a pair of field plates like the figure above will only result in them crashing into the plates, because an electron has charge. So instead of trying to take lone electrons, we take a molecule which has some electrons (two to be exact) in its outermost shell, and put that in an electric field. See the figure:
(The green lines with arrows depict the powerful internal electric field that that the electron experiences when ThO is subjected to an external electric field in our lab.) By some theoretical calculations by theoretical atomic physicists, we are able to deduce from our observations of , the electric dipole moment of our molecule (in this case thorium monoxide, or ThO).
This act of using a molecule has an additional great benefit: it turns out that an electron orbiting a nucleus experiences an amplification of the electric field we apply in the lab. We call this , or the effective electric field. So we would be measuring
,
and a larger results in a larger energy shift = easier measurement and better precision. It turns out that the larger the atomic or molecular nucleus that the electron is orbiting, the stronger the amplification of the electric field. In our case, we use thorium monoxide: thorium has an atomic number of 90, one of the highest of all stable elements in the periodic table. This results in an that is a billion times larger than the electric field we apply in the lab. This is very convenient as there are practical limits on how strong of an electric field one can generate in a laboratory without complications – typically on the order of a few thousand kilovolts per centimeter. In comparison, to get a bolt of lightning to strike you need a potential difference of several hundred million volts between the sky and the ground – about 10 kV/cm. (In other words, if you try to generate a field that strong in a room without vacuum isolation, it will likely result in arcs of lightning between the field plates, which is bad.)
Thus by using ThO, we only generate an electric field of a few tens of V/cm for our experiment, yet the effective field experienced by the electron in the molecule is a billion times stronger. That’s a very large gain for our precision. Indeed, historically speaking, finding good molecules that have a large has been a critical first step in designing an experiment finding the electron EDM. There are other equally important reasons why we chose ThO as opposed to other molecules (which I shall get into eventually), but this large effective field is one of the most basic and important criteria.
# A Beam of Molecules and an Electric Field
So we start our experiment by producing our thorium monoxide (ThO) molecules. This is done inside a vacuum-sealed box or cell, where a solid target of ThO2 sits inside. A high-powered laser (an Nd:YAG laser outputting about 10 watts of 1064 nm light) blasts the target in short pulses, almost magically creating molecules of just ThO (with one oxygen), a process called ablation. The cell has a small hole through which the molecules spray out as a beam into the rest of the experiment.
Besides ThO, we also put in neon gas inside the cell, and the whole cell is cooled to a very low, cryogenic temperature of 16 K (-257 degrees Celcius). This neon gas is a buffer gas – it helps to cool down the ThO molecules in the cell, lowering the speed of the molecules as they come out of the cell. The result is a relatively well-controlled beam of molecules with an average velocity of about 185 m/s (666 km/h). This is “slow” by atomic physics standards, although it takes the molecules only several tenths of a second to traverse through the entire experiment! These molecules progress from the hole through the stem, which is what we call the airtight cylindrical pipe that provides a path for the molecules through the rest of the experiment.
However, one thing to remember is that like light coming out from a hole, the molecules are diffracted outwards – they “fan out” with a non-zero solid angle. The further away the beam travels, the molecules are further away from each other. This is not ideal, as many of them start crashing into the walls of the stem, rendering them useless. In fact, in the first generation of the experiment only about 1 in 20,000 molecules ultimately reach the interaction region where the actual measurement itself is performed. As our sensitivity is proportional to the square root of the number of molecules, this directly affects the precision of our experiment. The more molecules you have, the more times you can repeat the measurement, and the more precise the result is.
After the cell, the molecules go through the rotational cooling stage which I will explain in more detail in a future post. Essentially, to make it possible to use the molecules for a measurement, we have to get them to a specific quantum state. This is done by using laser light of specific colors,4 but the process is limited by some natural restrictions that make it less efficient. Rotational cooling manipulates the atoms using lasers to make it possible to later transfer more of them into the state we want. The actual transfer into the desired state for the measurement is done in the state preparation stage, where a different technique5 is performed to get as many molecules as possible from the ground state (the natural, default state of the molecules which we call the state), into something we call the H excited state of ThO, which is the quantum state the molecules need to be in to perform the measurement.6 State preparation is also important in that it immediately transforms an incoherent jumble of molecules in various different states into a smaller but more uniform population that we can examine as a unified body.
# The Actual Measurement: Spin Precession
The molecules are now ready to be put into the interaction region, where the measurement happens. There, we apply an electric field to them using a pair of parallel flat metal plates which are held at a distance of several cm from each other (see the first figure in this post). If is really non-zero, then here it will do its work and leave its mark.7 What happens to the molecule as it goes through this region is something called a spin precession. You can imagine this as similar to the precession of a top, only instead of a top, there is a similar vector quantity in the molecule called spin that precesses (rotates in orientation) as you keep applying an electric field to it.
are ready to be interrogated, again using a different set of lasers – the probe lasers. The probe lasers stimulate the molecules to get into a different, higher excited state – we call it the state.8 This state is naturally more unstable, such that after a very brief period the molecules decay back into the ground state. As they do this, they release energy in the form of green, 512 nm light as they do so. This is the light we detect using photo-multiplier tubes (PMTs), which are essentially black-and-white cameras. The input from the PMTs are sent to a computer as data to be processed and analyzed. Having finished their use, the molecules continue their journey out of the interaction into the “beyond” – they eventually crash against the other end of the stem, never to be seen from again.
The entire above journey of a molecule takes only about 3-5 thousandths of a second. Every time the ablation laser fires into the target, as many as 100 billion molecules are produced, but only about 100,000 (in ACME I) survive the journey to the interaction region. Many are lost by crashing into the walls of the stem, while others are simply not in the right quantum state for the measurement. Even in the case of those which successfully undergo spin precession and get interrogated by the probe beam, only 1% of the photons produced are detected due to ineffiencies in our detection scheme – resulting in only 1,000 photons detected per pulse of molecules.9 The ablation laser fires every 50 ms, or 20 times a second. Typically, we take and average data for about a week before we reach a precision good enough for a “result”. Technically, at this point one would be able to figure out if indeed is zero or not – thus potentially making a discovery!
However, this is not the end of the story. It turns out that there are various stray effects in our experiment that can happen, causing additional spin precession that masquerades as a non-zero value of . These are called systematic errors. The bulk of our time is actually spent testing and looking for these errors. We change various settings in our experiment, especially settings which we think shouldn’t have an impact on the measured value of . For example, we can turn on additional magnetic fields, change the intensities of our lasers, or do even seemingly trivial changes such as moving our lasers to the opposite side of the room and shining them at the molecules from the other side. In many cases, we would not expect a change in the measured EDM value, yet sometimes we do. If that happens then we have to find out why that setting is causing the change, and possibly remedy the condition before releasing our result.
# Closing Remarks
Thus is the entire process by which we perform a measurement of the electron’s shape, or , as physicists call it. It is a complex, multi-stage process involving multiple lasers, computers, and various equipment to control pressures, temperatures, and voltages in the experiment. Three different experimental groups collaborate and are roughly responsible for different stages of the experiment. The ACME experiment is unique in that it’s an experiment small enough to be fitted in a single room, but is still many times more complex (and expensive) than the average atomic physics tabletop experiment. To become truly familiar with all aspects of the experiment, a graduate student would have to gain practical skills in building lasers, aligning optics, maintaining vacuum systems, assembling electronics, program computers, and configure data acquisition hardware and software, among other things. This is apart from the physics one has to know to truly understand the experiment: standard techniques in atomic physics with some additional complexities due to the fact that we have molecules, not atoms, plus a sprinkling of particle physics to be able to truly understand the theoretical context of the experiment.
For me, this is what makes the ACME experiment so exciting: we are building an extremely complex machine to measure just one quantity, but one which is so important that it can potentially unlock so many secrets about the nature of the universe! For more details on how exactly the shape of the electron is important and can do this, stay tuned!
1. This nomenclature was I believe first used by the Ed Hinds group in Imperial College when they first measured the electron EDM in 2011 and published their paper in Nature with that title. Incidentally, this is the measurement that we bettered 12 times in 2014.
2. This is the same unit that is often used when describing the energy of collisions at the Large Hadron Collider or other large particle physics experiments, as well as the masses of new particles like the Higgs boson, which has mass of about 126 billion eVs.
3. On the other hand, the Gabrielse lab where I work has as one of its greatest achievements in measuring the electron’s magnetic moment to unprecedented precision – the most precisely tested theory in all of physics, and perhaps all of science. This was done by trapping literally a single electron using a device called a Penning trap and observing its behavior.
4. Color is determined by the wavelength of the light. Wavelength () is inversely proportional to frequency () by the formula where is the speed of light, so they are interchangeable as terms. When I specify some color of light, I can immediately figure out what wavelength and frequency it is.
5. In ACME I, the transfer was done using a standard atomic physics technique of optical pumping, whereas in ACME II we are using a different technique – STIRAP – which is over 10 times more efficient.
6. Note that this is different from the I used above for signifying the energy shift caused by the electron EDM, which is just short for Hamiltonian – a term synonymous with energy.)
7. We apply a magnetic field as well, but strictly speaking this is not necessary for the measurement.
8. In ACME II will be the state. But the mechanism is exactly the same otherwise.
9. In ACME II, this number has been increased to over 300,000 photons per pulse. | 3,360 | 15,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-25 | longest | en | 0.94434 |
http://isabelle.in.tum.de/repos/isabelle/diff/6b780867d426/src/ZF/equalities.thy | 1,571,700,482,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795253.70/warc/CC-MAIN-20191021221245-20191022004745-00331.warc.gz | 96,891,830 | 2,617 | src/ZF/equalities.thy
changeset 61980 6b780867d426 parent 61798 27f3c10b0b50 child 69593 3dda49e08b9d
``` 1.1 --- a/src/ZF/equalities.thy Wed Dec 30 17:38:57 2015 +0100
1.2 +++ b/src/ZF/equalities.thy Wed Dec 30 17:45:18 2015 +0100
1.3 @@ -535,19 +535,19 @@
1.4 by blast
1.5
1.6 lemma SUM_UN_distrib1:
1.7 - "(\<Sigma> x \<in> (\<Union>y\<in>A. C(y)). B(x)) = (\<Union>y\<in>A. \<Sigma> x\<in>C(y). B(x))"
1.8 + "(\<Sum>x \<in> (\<Union>y\<in>A. C(y)). B(x)) = (\<Union>y\<in>A. \<Sum>x\<in>C(y). B(x))"
1.9 by blast
1.10
1.11 lemma SUM_UN_distrib2:
1.12 - "(\<Sigma> i\<in>I. \<Union>j\<in>J. C(i,j)) = (\<Union>j\<in>J. \<Sigma> i\<in>I. C(i,j))"
1.13 + "(\<Sum>i\<in>I. \<Union>j\<in>J. C(i,j)) = (\<Union>j\<in>J. \<Sum>i\<in>I. C(i,j))"
1.14 by blast
1.15
1.16 lemma SUM_Un_distrib1:
1.17 - "(\<Sigma> i\<in>I \<union> J. C(i)) = (\<Sigma> i\<in>I. C(i)) \<union> (\<Sigma> j\<in>J. C(j))"
1.18 + "(\<Sum>i\<in>I \<union> J. C(i)) = (\<Sum>i\<in>I. C(i)) \<union> (\<Sum>j\<in>J. C(j))"
1.19 by blast
1.20
1.21 lemma SUM_Un_distrib2:
1.22 - "(\<Sigma> i\<in>I. A(i) \<union> B(i)) = (\<Sigma> i\<in>I. A(i)) \<union> (\<Sigma> i\<in>I. B(i))"
1.23 + "(\<Sum>i\<in>I. A(i) \<union> B(i)) = (\<Sum>i\<in>I. A(i)) \<union> (\<Sum>i\<in>I. B(i))"
1.24 by blast
1.25
1.26 (*First-order version of the above, for rewriting*)
1.27 @@ -555,11 +555,11 @@
1.28 by (rule SUM_Un_distrib2)
1.29
1.30 lemma SUM_Int_distrib1:
1.31 - "(\<Sigma> i\<in>I \<inter> J. C(i)) = (\<Sigma> i\<in>I. C(i)) \<inter> (\<Sigma> j\<in>J. C(j))"
1.32 + "(\<Sum>i\<in>I \<inter> J. C(i)) = (\<Sum>i\<in>I. C(i)) \<inter> (\<Sum>j\<in>J. C(j))"
1.33 by blast
1.34
1.35 lemma SUM_Int_distrib2:
1.36 - "(\<Sigma> i\<in>I. A(i) \<inter> B(i)) = (\<Sigma> i\<in>I. A(i)) \<inter> (\<Sigma> i\<in>I. B(i))"
1.37 + "(\<Sum>i\<in>I. A(i) \<inter> B(i)) = (\<Sum>i\<in>I. A(i)) \<inter> (\<Sum>i\<in>I. B(i))"
1.38 by blast
1.39
1.40 (*First-order version of the above, for rewriting*)
1.41 @@ -567,7 +567,7 @@
1.42 by (rule SUM_Int_distrib2)
1.43
1.44 (*Cf Aczel, Non-Well-Founded Sets, page 115*)
1.45 -lemma SUM_eq_UN: "(\<Sigma> i\<in>I. A(i)) = (\<Union>i\<in>I. {i} * A(i))"
1.46 +lemma SUM_eq_UN: "(\<Sum>i\<in>I. A(i)) = (\<Union>i\<in>I. {i} * A(i))"
1.47 by blast
1.48
1.49 lemma times_subset_iff:
1.50 @@ -575,7 +575,7 @@
1.51 by blast
1.52
1.53 lemma Int_Sigma_eq:
1.54 - "(\<Sigma> x \<in> A'. B'(x)) \<inter> (\<Sigma> x \<in> A. B(x)) = (\<Sigma> x \<in> A' \<inter> A. B'(x) \<inter> B(x))"
1.55 + "(\<Sum>x \<in> A'. B'(x)) \<inter> (\<Sum>x \<in> A. B(x)) = (\<Sum>x \<in> A' \<inter> A. B'(x) \<inter> B(x))"
1.56 by blast
1.57
1.58 (** Domain **)
``` | 1,173 | 2,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-43 | latest | en | 0.509323 |
https://nickadamsinamerica.com/multiplying-decimals-worksheets-with-answers-pdf/81d412d3e64cc1c94fe1bb877feab081-2/ | 1,638,043,247,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358233.7/warc/CC-MAIN-20211127193525-20211127223525-00402.warc.gz | 497,255,154 | 9,733 | # 81d412d3e64cc1c94fe1bb877feab081
By . Worksheet. At Monday, October 18th 2021, 19:33:08 PM.
Used effectively, rewards can have a hugely positive impact on your child, especially on areas that you normally struggle with co-operation. We have a great article on how to use reward systems effectively, you can find it here: https://www.worksheetcloud.com/blog/4-ways-to-use-a-reward-system-effectively/
These Radical Worksheets will produce problems for adding and subtracting radical expressions. You may select what type of radicals you want to use. These Radical Worksheets are a good resource for students in the 5th Grade through the 8th Grade.
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### Activity Pages For Kindergarten
#### I See Worksheets Kindergarten
##### This And These Worksheets For Kindergarten
###### Handwriting Sheets For Kindergarten
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Any content, trademark/s, or other material that might be found on this site that is not this site property remains the copyright of its respective owner/s. | 638 | 3,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.729 |
http://betterlesson.com/lesson/resource/2542950/process-grid-jpg | 1,477,441,851,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720471.17/warc/CC-MAIN-20161020183840-00413-ip-10-171-6-4.ec2.internal.warc.gz | 30,669,758 | 25,002 | ## Process Grid.jpg - Section 3: Guided Practice
Process Grid.jpg
# Discovering Patterns Within the Base 10 Number System
Unit 4: Place Value
Lesson 4 of 8
## Big Idea: Being able to understand and explain numbers will help students make sense of multi-digit computation and problem solving.
Print Lesson
9 teachers like this lesson
Standards:
Subject(s):
110 minutes
### Kara Nelson
##### Similar Lessons
###### The Cup Half Full (day 1 of 3)
7th Grade Math » Exploring Rational Numbers
Big Idea: When the decimal "moves" to the left the digits are actually moving to the left because their value is 10 times less.
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###### Middle of Unit Assessment
4th Grade Math » Place value
Big Idea: In this lesson, students complete an independent assessment in order to show proficiency with place value skills and rounding.
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https://dirask.com/posts/JavaScript-Math-tan-method-example-k1wx9D?posts-list=6wzqP8 | 1,660,609,674,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00634.warc.gz | 222,176,475 | 38,497 | Home Communities
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# JavaScript - Math.tan() method example
2 points
Created by:
464
`Math.tan()` is static method that takes only one parameter and returns approximated value of tangent mathematical function.Â
``````// ONLINE-RUNNER:browser;
console.log( Math.tan( 0 ) ); // 0 <- 0 degrees
console.log( Math.tan( 0.7853981633974483 ) ); // ~1 <- ~45 degrees == PI / 4
console.log( Math.tan( 1.5707963267948966 ) ); // ~+Inf <- ~90 degrees == PI / 2
console.log( Math.tan(-0.7853981633974483 ) ); // ~-1 <- ~-45 degrees == -PI / 4
console.log( Math.tan(-1.5707963267948966 ) ); // ~-Inf <- ~-90 degrees == -PI / 2``````
Note: `0.9999999999999999`, `16331239353195370`, `-0.9999999999999999` and `-16331239353195370` should be equal to `1`, `+Inf`, `-1` and `-Inf` but they are not because of compuptation precision error.
## 1. Documentation
Syntax `Math.tan(number)` Parameters `number` - integer or float number value in radians (primitive value). Result `number` value calculated as `tan(x)` mathematical function (primitive value). Description `Math.tan()` is static method that takes only one parameter and returns approximated value of tangent mathematical function.
``````// ONLINE-RUNNER:browser;
var x1 = 0.0; // beginning of calculation in radians
var x2 = Math.PI / 2; // ending of calculation radians
var dx = Math.PI / 36; // calculation step in degrees
}``````
## 3. Working with degrees
``````// ONLINE-RUNNER:browser;
function calculateTan(deg) {
var radians = (Math.PI / 180) * deg;
}
// Example:
var x1 = 0.0; // beginning of calculation in degrees
var x2 = 90.0; // ending of calculation degrees
var dx = 5.0; // calculation step in degrees
for (var deg = x1; deg <= x2; deg += dx) {
var y = calculateTan(deg );
console.log('tan(' + deg + ' deg) = ' + y);
}``````
## 4. Reversed console plot example
``````// ONLINE-RUNNER:browser;
var x1 = -3.14 * 2; // beginning of sine chart
var x2 = +3.14 * 2; // end of sine chart
var y1 = -4.0;
var y2 = +4.0;
var xSteps = 60;
var ySteps = 60;
var dx = (x2 - x1) / xSteps; // x axis step
var dy = (y2 - y1) / ySteps; // y axis step
function printLine(y1, y2, character) {
var line = '';
for(var y = y1; y < y2; y += dy) {
line += ' ';
}
console.log(line + character);
}
if (y <= y1 || y >= y2) {
console.log(' ');
} else {
printLine(y1, y, '+');
}
}``````
## 5. Canvas plot example
``````// ONLINE-RUNNER:browser;
<!doctype html>
<html>
<style> #canvas { border: 1px solid black; } </style>
<body>
<canvas id="canvas" width="150" height="400"></canvas>
<script>
var canvas = document.querySelector('#canvas');
var context = canvas.getContext('2d');
// tangent chart range
var x1 = -Math.PI / 2; // -90 degress
var x2 = +Math.PI / 2; // +90 degress
var y1 = -10.0;
var y2 = +10.0;
var dx = 0.05;
var xRange = x2 - x1;
var yRange = y2 - y1;
var yScale = 1.0;
function calculatePoint(x) {
var y = Math.tan(x) * yScale;
// chart will be reversed horizontaly because of reversed canvas pixels
var nx = (x - x1) / xRange; // normalized x
var ny = 1.0 - (y - y1) / yRange; // normalized y
var point = {
x: nx * canvas.width,
y: ny * canvas.height
};
return point;
}
console.log('x range: <' + x1 + '; ' + x2 + '> // angles in radians');
console.log('y range: <' + y1 + '; ' + y2 + '>');
var point = calculatePoint(x1);
context.beginPath();
context.moveTo(point.x, point.y);
for (var x = x1 + dx; x < x2; x += dx) {
point = calculatePoint(x);
context.lineTo(point.x, point.y);
}
point = calculatePoint(x2);
context.lineTo(point.x, point.y);
context.stroke();
</script>
</body>
</html>``````
## References
1. Trigonometric functions - Wikipedia
## Alternative titles
Join to our subscribers to be up to date with content, news and offers. | 1,226 | 3,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-33 | latest | en | 0.362951 |
https://www.mail-archive.com/everything-list@googlegroups.com/msg23127.html | 1,529,820,311,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866358.52/warc/CC-MAIN-20180624044127-20180624064127-00231.warc.gz | 884,629,633 | 9,271 | # Re: QTI, Cul de sacs and differentiation
On 26 Oct 2011, at 01:00, Nick Prince wrote:
QTI, Cul de sacs and differentiation
I’m trying to get a picture of how David Deutsch’s idea of
differentiation works – especially in relation to QTI. With a
standard treatment it looks as if there might be cul de sacs for a
dying cat. However I think I can see why this conclusion could be
wrong. Maybe someone could check my reasoning for this and tell me if
there are any flaws.
I’ve entered this on both the FOR and the Everything list because I
hope it is relevant to both Forums.
Firstly I am adopting the position that consciousness supervenes on
all identical worldlines and where the multiverse differentiates, the
first person experience is indeterminate. Secondly I assume local
causality applies. Thirdly (to begin with anyway) I assume that all
“measuring” systems function as they should do to obtain “correct
measurements/outcomes” (I’ll drop this part later though). Now
suppose a SINGLE electron is prepared so that its spin is aligned in
the x - right direction ( |Xr> – for x spin in the right direction)
is sent through a SG device and that, whether the electron comes out
spinning up in the z direction or down determines the triggering of a
device which breaks the flask of gas – (let’s say it is the electron
with spin up and moving upwards which is the lethal combination) which
kills the cat. On the other hand, an electron emerging with spin down,
and moving down in the z direction leaves the measuring device
triggered in the down state but this does nothing to the flask) So
there is a 50% chance of the cat being killed for each electron fired
through. This means there would be interaction Hamiltonians which
would make up unitary evolution operators of the form
M = exp(-iHt/hbar) which would act on states as follows:
(Mc) (Mdev) ( Msg) |Ca> |Dn> |moves to right> |Xr>
Msg = stern gerlach interaction evolution operator
Mdev = triggering and flask breaking evolution operator
Mc = cat /poisonous gas evolution operator.
|Dn> = neutral detector state
|Ca> = alive cat state etc.
Now standard QM gives |Xr> = (1/sqrt2)(|Zu> +|Zd>)
Notice how I have put operators in time order so that the rightmost
operator is implied to operate earlier than those to the left. The
order of the state vectors reflects this too.
Msg is the unitary operator which causes an evolution from |moves to
right> to either moves up |moves up> or moves down |moves down>.
Mdev allows evolution of the detector device plus flask smashing
mechanism which, if it causes evolution to |Du>, breaks open the
interaction of the gas with the cat due to the evolution operator Mc
leaves the cat either dead |Cd> or alive |Ca>.
Now, assuming a causally functioning system, then we can write:
(Mc) (Mdev) ( Msg) |Ca> |Dn> |moves to right> |Xr>
=(Mc) |Ca> (Mdev) |Dn> ( Msg) |moves to right> (1/sqrt2)(|Zu> +|Zd>)
=(Mc) |Ca> (Mdev) |Dn>(1/sqrt2) [|moves up>|Zu>+|moves down>|Zd>]
=(Mc) |Ca> (1/sqrt2) [|Du>|moves up>|Zu>+Dd>|moves down>|Zd>]
= (1/sqrt2) [|Cd>|Du>|moves up>|Zu>+|Ca>|Dd>|moves down>|Zd>]
All this is just the standard Schrödinger’s cat problem. However note
that if one thinks in terms of a differentiated multiverse we begin
with an infinite number of identical experiments in identical
universes. By the end of the first evolution due to Msg, the infinite
bundle of universes has partitioned into two bundles i.e. one bundle
of universes that have a Z spin up electron moving upwards with a
neutral detector reading and an alive cat, and another bundle of
universes that have a Z spin down electron moving downwards with a
neutral detector reading and an alive cat. As time progresses the
partition is communicated through the system and by the end of the
period during which operator Md operates we have two bundles that are
even more differentiated because the strands (universes) now are those
that have either a Z spin up electron which moved upwards causing a
detector to smash the flask but with an, (as yet) still alive cat in
one of them and another bundle of universes that had a Z spin down
electron which moved downwards, which triggered the detector in a way
which did not smash the flask and thus, as yet still also has an alive
cat. Finally by the time the last (Mc) operator’s effect has finished
its evolutionary action, the cat lays dead (or alive) and thus the
final bundles are now partitioned into either a Z spin up electron
that moved upwards with a detector that smashed the flask that killed
the cat; or, a Z spin down electron that moved downwards with a
alive.
There’s nothing new in any of this. It is just the standard paradox
normally used to illustrate the problem of collapse. However I have
highlighted the fact that the experiment takes time to untangle the
different strands of the differentiated multiverse such that the cat
can discover which type of strand he is in. Now suppose the cat
discovers that he is in the wrong bundle – i.e. he hears the flask
smash! Now because we have assumed all devices – including flask
smashers and poisons work as they ideally should do, then the cat must
now be in a cul de sac! He cannot be in any other branch than one in
which he dies because everything by definition is working properly and
against his survival from either 1st person or 3rd person viewpoints.
Indeed the cat was doomed to die (by cul de sac) at the first
partitioning when it unwittingly ended up in the bundle of universes
wherein the electron was moved upwards by the SG device rather than
down.
Now I actually think this conclusion is wrong. This is because the
whole system is idealised and I have tried to show why, by some
reasoning which I hope is correct, However, I would welcome some
confirmation that my treatment is not erroneous.
If we just go back to a simple measurement example and modify it to
account for alternative possiblities in the bundles then the analysis
might go as follows:
Let S be the state space of a quantum system and A be that of the
experimental apparatus ( also considered as a quantum system)used for
measuring the system S
Consider the development of the combined system S cross A (tensor
product) and let |s1>,|s2> belonging to S be two eigenstates of the
object corresponding to two different results of the experiment.
These results must leave the apparatus in different states |a1> and |
a2> (describing say, different positions of a pointer.). Suppose the
apparatus is initially in another (“neutral ”) eigenstate |a0>. The
experiment therefore consists of allowing the object and the apparatus
to interact in such a way that if the object state is |s1>, then after
the experiment, the object will still be in the state |s1> and the
apparatus will record the appropriate result, i.e. will be in the
state |a1>. A similar argument holding for |s2> and |a2>. This is
what I believe to represent an ideal measurement where the measuring
device “works properly”.
Thus during the experiment the Hamiltonian H in exp(-iHt/hbar) must be
such that:
exp(-iHt/hbar) (|s1>|a0>)=|s1>|a1>
exp(-iHt/hbar) (|s2>|a0>)=|s2>|a1> (1)
Where t is the time taken for the experiment to yield a definite
result.
Now, if before the experiment the system was in the state
|s0> = c1|s1> + c2|s2> (2)
Where |c1|^2 + |c2|^2 = 1
Then after it, the system and the apparatus together will be in the
state
exp(-iHt/hbar) (|s0>|a0>)
= exp(-iHt/hbar) (c1|s1>|a0> + c2|s2>|a0>) (3)
This is the simplified ideal case again.
Now let’s consider how things would be in real non ideal measurements
where say the apparatus (and lots of other things for that matter) was
less reliable. Being a macroscopic device with many degrees of
freedom means there are many ways one could imagine from the classical
point of view that false readings might occur – i.e. the pointer
sticks for some reason etc. From the quantum microscopic point of
view errors could occur simply because it may not be possible to make
systems with Hamiltonians that are exactly appropriate – i.e does the
moving electron interact with some field fluctuation, air molecule
etc? I have to admit I’m not sure about things at this level so if
anyone can give me valid reasons then I would be grateful. Anyway to
account for non ideal functioning of an apparatus/device/process I am
GUESSING that during the experiment we might instead have a
Hamiltonian which is different to (1) and is such that we have
something more like:
exp(-iHt/hbar)(|s1>|a0>)=|s1>(a|a0> + b|a1> + c|a2>)
exp(-iHt/hbar)(|s2>|a0>)=|s2>(a|a0> + c|a1> + b|a2>)
where |a|^2 + |b|^2 + |c|^2 =1. Also
|a|^2, |c|^2 are very small whilst |b|^2 is nearly = 1
(4)
Now, following the original argument, if before the experiment the
system was in the state
|s0> = c1|s1> + c2|s2> (5)
Then after it, the system and the apparatus together will be in the
state
exp(-iHt/hbar)(|s0>|a0>) = exp(-iHt/hbar) (c1|s1>|a0>) + c2|s2>|a0>)
= c1|s1>(a|a0> + b|a1> + c|a2>)+ c2|s2>(a|a0> + c|a1> + b|a2>)
= b c1|s1>|a1> + b c2|s2>|a2> + c c1|s1>|a2> + a c2|s2>|a0> +
c c2|s2>|a1>+ a c1|s1>|a0> (6)
The first two terms are “almost those” of those on the RHS of equation
(3), but now there are new terms. However, the probability amplitudes
of these new terms are far less dominant because of the relative sizes
of the
|a|^2, |c|^2 and |b|^2 terms.
I am guessing that I can interpret this then as an experiment where
the measuring apparatus usually works correctly but on rare
occasions:
(i) Does not detect anything when in fact it should have done - e.g.
the presence of the |s1>|a0> and |s2>|a0> terms.
(ii) Very occasionally records the eigenvalue of one state |s1> when
it should have recorded the eigenvalue of the other state |s2> and
vice versa – e.g. the presence of the|s1>|a2>, and |s2>|a1> terms.
Going back to the cat experiment it does mean that if things work this
way then now the cat can always expect some probability that an upward
moving electron and/or a smashed flask may not be the end. Indeed
there are many more bundles available for the consciousness to access
even at any late stage, thereby avoiding any cul de sacs and hence,
all the usual QTI argument carry over again.
I’m not sure whether I can use this approach (particularly the step
(4) above) and would like comments and criticisms as to whether my
analysis can be considered valid. For example if the effect is due to
macroscopic failure (like the pointer sticking or a fuse blows and
therefore the reading is in the |a0> state irrespective of the system
state) then can this be a possible valid interpretation of (6)? Or
microscopically say, for example, the direction of motion of a
particle gets redirected because it recoils from a rogue air molecule
and goes the wrong way in the detector chamber.
Many thanks for anyone who takes the trouble to answer my query.
I have not seen any error.
This does not mean there are zero error of course.
The QTI, or the more general comp immortality, or arithmetical immortality is a complex subject, if only because it depends on what you mean by "you".
Do you know Kripke semantic? A Kripke frame is just a set (of elements called worlds) with an accessibility relation among the worlds. In modal logic they can be used to characterize modal logical systems. The basic idea is that []p is true in world alpha, if p is true in all the worlds accessible from alpha. Dually, <>p is true in alpha is p is true in at least one world accessible from alpha. For example the law []p -> p will be satisfied in all reflexive frames---independently of the truth value of p. (a frame is reflexive if all the worlds in the frame access to themselves; for all alpha alpha R alpha, with R the accessibility relation).
In the frames which characterize the modal logic of Gödel's self- reference, Bp, all worlds can be shown to access to a cul-de-sac world, witnessing somehow that if comp is true we can "die" at each "moment", in a large sense of "death". This prevents the use of Bp for probabilities, which stipulates that there is always some other world being accessible. But this means that we can define a new operator [P]p = Bp & ~B~p, or Bp & Dp, or Bp & Dt (t = "0=0") forcing each world where [P]p is true to access at least one world with p true. This is a no-cul-de-sac definition of an arithmetical form of "probability-one", or "credibility-one", and this is how quantum logic can be found in the arithmetical setting.
Having said this, it means that comp presupposes at the start a form of immortality, like the UD illustrates intuitively, and like the QM waves illustrates too. But to compute the chance of *some* particular worlds being accessible through that immortality is a hell of a computation. Strictly speaking you should take into account the quantum description of the brain of the cat, and take into account all the worlds where the poison will not kill the cat. The comp immortality (and I think the quantum one too) makes it just impossible to guaranty death: there is always a continuation of you somewhere. The no-cul-de-sac assumption/definition (Bp ==> Bp & Dt) is that we just do not take into account the cul-de-sac worlds: they do not enter at all in the probability calculus. provability becomes probability- one by abstracting away from the cul-de-sac worlds.
Then, as other have already mentioned, what will remain unclear (and hard to compute) is the probability that you survive through some memory backtracking. The cat might survive in the worlds where he has been lucky enough to not participate to that experience, and, for all we know, such consistent continuation might have bigger weight than surviving through some quantum tunnel effect saving the brain's cat from the poison. The computation here are just not tractable, if we assume quantum mechanics, and still less, assuming only the comp hypothesis. The only certainty, assuming comp or QM, is that "you" cannot die. But obviously you can become amnesic of some part, if not all, your existence, or you existences. Like Otto Rossler summed up well : consciousness is a prison. With comp, and I think with QM, there is no escapes from being conscious, in a way or another. I don't like that, but then it is a consequence of those theories.
Bruno
http://iridia.ulb.ac.be/~marchal/
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mathematics
## Babylonian mathematics
...of the terms, the third will usually be irrational, but it is possible to find cases in which all three terms are integers: for example, 3, 4, 5 and 5, 12, 13. (Such solutions are sometimes called Pythagorean triples.) A tablet in the Columbia University Collection presents a list of 15 such triples (decimal equivalents are shown in parentheses at the right; the gaps in the expressions for...
## Greek mathematics
...(e.g., 3, 4, 5; 5, 12, 13; or 119, 120, 169). From the Greeks came a proof of a general rule for finding all such sets of numbers (now called Pythagorean triples): if one takes any whole numbers p and q, both being even or both odd, then a = ( p 2 − q 2)/2,...
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``` Scissors Congruence &
Hilbert’s 3rd Problem
Abhijit Champanerkar
College of Staten Island CUNY
CSI Math Club Talk
April 14th 2010
Scissors Congruence
A polygonal decomposition of a polygon P in the Euclidean plane
is a finite collections of polygons P1 , P2 , . . . Pn whose union is P
and which pairwise intersect only in their boundaries.
Scissors Congruence
A polygonal decomposition of a polygon P in the Euclidean plane
is a finite collections of polygons P1 , P2 , . . . Pn whose union is P
and which pairwise intersect only in their boundaries.
Example: Tangrams
Scissors Congruence
Scissors Congruence
Polygons P and Q are scissors congruent if there exist polygonal
decompositions P1 , . . . , Pn and Q1 , . . . , Qn of P and Q respectively
such that Pi is congruent to Qi for 1 ≤ i ≤ n. In short, two polygons
are scissors congruent is one can be cut up and reassembled into the
other. Let us denote scissors congruence by ∼sc . We will write
P ∼sc P1 + P2 + . . . + Pn .
Scissors Congruence
Scissors Congruence
Polygons P and Q are scissors congruent if there exist polygonal
decompositions P1 , . . . , Pn and Q1 , . . . , Qn of P and Q respectively
such that Pi is congruent to Qi for 1 ≤ i ≤ n. In short, two polygons
are scissors congruent is one can be cut up and reassembled into the
other. Let us denote scissors congruence by ∼sc . We will write
P ∼sc P1 + P2 + . . . + Pn .
Example: All the polygons below are scissors congruent.
Scissors Congruence
Two pictures of Euclid
Scissors Congruence
Two pictures of Euclid
The idea of scissors congruence goes back to Euclid. By “equal
area” Euclid meant scissors congruent (not in that terminology).
In fact Euclid’s proof of the Pythagorean Theorem partitions the
three squares into triangles with equal areas. Euclid’s “geometric
algebra” will also remind you of scissors congruence (groups).
Pythagorean Theorem
Scissors Congruent proof of the Pythagorean Theorem.
Properties of Scissors Congruence
If P ∼sc Q then Area(P) = Area(Q).
Properties of Scissors Congruence
If P ∼sc Q then Area(P) = Area(Q).
∼sc is an equivalence relation on the set of all polygons in the
Euclidean plane.
(Reflexive) P ∼sc P.
(Symmetric) P ∼sc Q then Q ∼sc P.
(Transitive) P ∼sc Q and Q ∼sc R then P ∼sc R.
Transitivity follows by juxtaposing the two decompositions of
Q and using the resulting common sub-decomposition of Q to
reassemble into P and R, thus showing that P ∼sc R.
Properties of Scissors Congruence
If P ∼sc Q then Area(P) = Area(Q).
∼sc is an equivalence relation on the set of all polygons in the
Euclidean plane.
(Reflexive) P ∼sc P.
(Symmetric) P ∼sc Q then Q ∼sc P.
(Transitive) P ∼sc Q and Q ∼sc R then P ∼sc R.
Transitivity follows by juxtaposing the two decompositions of
Q and using the resulting common sub-decomposition of Q to
reassemble into P and R, thus showing that P ∼sc R.
P Q
Properties of Scissors Congruence
If P ∼sc Q then Area(P) = Area(Q).
∼sc is an equivalence relation on the set of all polygons in the
Euclidean plane.
(Reflexive) P ∼sc P.
(Symmetric) P ∼sc Q then Q ∼sc P.
(Transitive) P ∼sc Q and Q ∼sc R then P ∼sc R.
Transitivity follows by juxtaposing the two decompositions of
Q and using the resulting common sub-decomposition of Q to
reassemble into P and R, thus showing that P ∼sc R.
Q R
Properties of Scissors Congruence
If P ∼sc Q then Area(P) = Area(Q).
∼sc is an equivalence relation on the set of all polygons in the
Euclidean plane.
(Reflexive) P ∼sc P.
(Symmetric) P ∼sc Q then Q ∼sc P.
(Transitive) P ∼sc Q and Q ∼sc R then P ∼sc R.
Transitivity follows by juxtaposing the two decompositions of
Q and using the resulting common sub-decomposition of Q to
reassemble into P and R, thus showing that P ∼sc R.
P Q R
Properties of Scissors Congruence
If P ∼sc Q then Area(P) = Area(Q).
∼sc is an equivalence relation on the set of all polygons in the
Euclidean plane.
(Reflexive) P ∼sc P.
(Symmetric) P ∼sc Q then Q ∼sc P.
(Transitive) P ∼sc Q and Q ∼sc R then P ∼sc R.
Transitivity follows by juxtaposing the two decompositions of
Q and using the resulting common sub-decomposition of Q to
reassemble into P and R, thus showing that P ∼sc R.
P R
Properties of Scissors Congruence
Theorem (Bolyai-Gerwien 1833)
Any two polygons with the same area are scissors congruent.
Properties of Scissors Congruence
Theorem (Bolyai-Gerwien 1833)
Any two polygons with the same area are scissors congruent.
An important consequence is that area determines scissors
congruence !
Properties of Scissors Congruence
Theorem (Bolyai-Gerwien 1833)
Any two polygons with the same area are scissors congruent.
An important consequence is that area determines scissors
congruence !
We will see two proofs of this theorem.
First Proof
Step 1: Every polygon has a polygonal decomposition into
triangles, in fact into acute angled triangles.
First Proof
Step 1: Every polygon has a polygonal decomposition into
triangles, in fact into acute angled triangles.
Proof:
First Proof
Step 1: Every polygon has a polygonal decomposition into
triangles, in fact into acute angled triangles.
Proof:
For a polygon, choose a line of slope m which is distint from the
slopes of all its sides. Lines of slope m through the vertices of the
polygon decompose it into triangles and trapezoids, which again
can be decomposed into acute angled triangles.
First Proof
Step 2: Any two parallelograms with same base and height are
scissors congruent. The same is true for triangles.
Proof: Let ABCD be a rectangle with base AB and height AD.
Let ABXY be a parallelogram with height AD. Assume
|DY | ≤ |DC |. Then
ABCD ∼sc AYD + ABCY ∼sc ABCY + BXC ∼sc ABXY .
D Y C X
A B
First Proof
If |DY | > |DC |, then cutting along the diagonal BY and regluing
the triangle BXY , we obtain the scissors congruent parallelogram
ABYY1 such that |DY1 | = |DY | − |DC |. Continuing this process k
times, for k = [|DY |/|DC |], we obtain the parallelogram
ABYk−1 Yk such that |DYk | < |DC |, which is scissors congruent to
ABCD as above.
D Y1 C Y X
A B
First Proof
Since any triangle is scissors congruent to a parallelogram with the
same base and half height, this implies that any two triangles with
same base and height are scissors congruent. .
First Proof
Step 3: Any two triangles with same area are scissors congruent.
Proof: By Step 2, we can assume both the triangles are right
angles triangles.
First Proof
Step 3: Any two triangles with same area are scissors congruent.
Proof: By Step 2, we can assume both the triangles are right
angles triangles.
C
Area(ABC ) = Area(AXY )
|AB||AC | |AY ||AX |
=⇒ =
2 2 Y
|AY | |AB|
=⇒ =
|AC | |AX |
=⇒ ABY ∼ AXC SAS test
A B X
First Proof
Step 3: Any two triangles with same area are scissors congruent.
Proof: By Step 2, we can assume both the triangles are right
angles triangles.
C
Area(ABC ) = Area(AXY )
|AB||AC | |AY ||AX |
=⇒ =
2 2 Y
|AY | |AB|
=⇒ =
|AC | |AX |
=⇒ ABY ∼ AXC SAS test
A B X
This implies BY is parallel to XC . Hence triangles BYC and BYX
have same base and same height which implies by Step 2 that they
are scissors congruent.
ABC ∼sc ABY + BYC ∼sc ABY + BYX ∼sc AXY .
First Proof
To complete the proof, any triangle T is scissors congruent to a
right triangle with height 2 and base equal to the area of T , which
is scissors congruent to a rectangle with unit height and base equal
to area of T . Lets denote such a rectangle by Rx where x is its
area (= lenght of the base).
First Proof
To complete the proof, any triangle T is scissors congruent to a
right triangle with height 2 and base equal to the area of T , which
is scissors congruent to a rectangle with unit height and base equal
to area of T . Lets denote such a rectangle by Rx where x is its
area (= lenght of the base).
Thus for any polygon P ,
P ∼sc T1 + . . . + Tn by Step 1
∼sc RArea(T1 ) + . . . + RArea(Tn ) by Step 3
∼sc RArea(T1 )+...+Area(Tn ) by laying rectangles side by side
∼sc RArea(P) by Step 1
First Proof
To complete the proof, any triangle T is scissors congruent to a
right triangle with height 2 and base equal to the area of T , which
is scissors congruent to a rectangle with unit height and base equal
to area of T . Lets denote such a rectangle by Rx where x is its
area (= lenght of the base).
Thus for any polygon P ,
P ∼sc T1 + . . . + Tn by Step 1
∼sc RArea(T1 ) + . . . + RArea(Tn ) by Step 3
∼sc RArea(T1 )+...+Area(Tn ) by laying rectangles side by side
∼sc RArea(P) by Step 1
Thus polygons with equal area are scissors congruent to the same
rectangles and hence to each other.
Second Proof
Step 1 & 2 same as before.
Step 3: A rectangle is scissors congruent to a square of the same
area.
Proof:
y
b
a
x x
a(b − a) √
where x = a − (b − a(b − a)), y = ab
a
We need to verify the equation
x a(b − a) √
√ + (b − a(b − a))2 + (a − x)2 = ab
ab
Second Proof
From Steps 1,2 & 3 we know that any triangle T is scissors
congruent to a square, denoted by say SArea(T ) . So for any
polygon P,
P ∼sc T1 + . . . + Tn by Step 1
∼sc SArea(T1 ) + . . . + SArea(Tn ) by Step 3
∼sc SArea(T1 )+...+Area(Tn ) by Pythagorean Theorem
∼sc SArea(P) by Step 1
Thus polygons with equal area are scissors congruent to the same
square and hence to each other.
Scissors Congruence in 3 dimensions
A polyhedron is a solid in E3 whose faces are polygons.
A polyhedral decomposition of a polyhedron P is a finite
collections of polyhedra P1 , P2 , . . . Pn whose union is P and which
pairwise intersect only in their boundaries (faces or edges).
Scissors Congruence in 3 dimensions
A polyhedron is a solid in E3 whose faces are polygons.
A polyhedral decomposition of a polyhedron P is a finite
collections of polyhedra P1 , P2 , . . . Pn whose union is P and which
pairwise intersect only in their boundaries (faces or edges).
Scissors Congruence
Two polyhedra P and Q are scissors congruent if there exist polyhe-
dral decompositions P1 , . . . , Pn and Q1 , . . . , Qn of P and Q respec-
tively such that Pi is congruent to Qi for 1 ≤ i ≤ n. In short, two
polyhedra are scissors congruent if one can be cut up and reassem-
bled into the other. As before, let us denote scissors congruence by
∼sc . We will also write P ∼sc P1 + P2 + . . . + Pn .
Scissors Congruence in 3 dimensions
If P ∼sc Q then Volume(P) = Volume(Q).
Scissors Congruence in 3 dimensions
If P ∼sc Q then Volume(P) = Volume(Q).
∼sc is an equivalence relation on the set of all polyhedra E3
(Reflexive) P ∼sc P.
(Symmetric) P ∼sc Q then Q ∼sc P.
(Transitive) P ∼sc Q and Q ∼sc R then P ∼sc R.
Scissors Congruence in 3 dimensions
If P ∼sc Q then Volume(P) = Volume(Q).
∼sc is an equivalence relation on the set of all polyhedra E3
(Reflexive) P ∼sc P.
(Symmetric) P ∼sc Q then Q ∼sc P.
(Transitive) P ∼sc Q and Q ∼sc R then P ∼sc R.
As before, transitivity follows by juxtaposing the two
decompositions of Q and using the resulting common
sub-decomposition of Q to reassemble into P and R, thus
showing that P ∼sc R. This is harder to visualize or draw.
Scissors Congruence in 3 dimensions
If P ∼sc Q then Volume(P) = Volume(Q).
∼sc is an equivalence relation on the set of all polyhedra E3
(Reflexive) P ∼sc P.
(Symmetric) P ∼sc Q then Q ∼sc P.
(Transitive) P ∼sc Q and Q ∼sc R then P ∼sc R.
As before, transitivity follows by juxtaposing the two
decompositions of Q and using the resulting common
sub-decomposition of Q to reassemble into P and R, thus
showing that P ∼sc R. This is harder to visualize or draw.
Anybody interested in making an animation of this ? Please
let me know
Hilbert’s Third Problem
In a famous lecture delivered at the International Congress of
Mathematics at Paris in 1900, Hilbert posed 23 problems.
Hilbert’s Third Problem
In a famous lecture delivered at the International Congress of
Mathematics at Paris in 1900, Hilbert posed 23 problems.
Hilbert’s Third Problem
Are polyhedra in E3 of same volume scissors congruent ?
Hilbert’s Third Problem
In a famous lecture delivered at the International Congress of
Mathematics at Paris in 1900, Hilbert posed 23 problems.
Hilbert’s Third Problem
Are polyhedra in E3 of same volume scissors congruent ?
Hilbert made clear that he expected a negative answer.
Solution to Hilbert’s Third Problem
Solution to Hilbert’s Third Problem
The negative answer to Hilbert’s Third problem was provided in
1902 by Max Dehn.
Solution to Hilbert’s Third Problem
The negative answer to Hilbert’s Third problem was provided in
1902 by Max Dehn.
Dehn showed that the regular tetrahedron and the cube of the
same volume were not scissors congruent.
∼sc
Dehn’s solution
Volume is an invariant of scissors congruence i.e. two scissors
congruent objects have the same volume.
Dehn’s solution
Volume is an invariant of scissors congruence i.e. two scissors
congruent objects have the same volume.
Dehn defined a new invariant of scissors congruence, now known as
the Dehn invariant.
Dehn invariant
For an edge e of a polyhedron P, let (e) and θ(e) denote its length
and dihedral angles respectively. The Dehn invariant δ(P) of P is
δ(P) = (e) ⊗ θ(e) ∈ R ⊗ (R/πQ)
all edges e of P
Dehn’s solution
Volume is an invariant of scissors congruence i.e. two scissors
congruent objects have the same volume.
Dehn defined a new invariant of scissors congruence, now known as
the Dehn invariant.
Dehn invariant
For an edge e of a polyhedron P, let (e) and θ(e) denote its length
and dihedral angles respectively. The Dehn invariant δ(P) of P is
δ(P) = (e) ⊗ θ(e) ∈ R ⊗ (R/πQ)
all edges e of P
The ⊗ symbol takes care that δ(P) does not change when you cut
along an edge or cut along an angle i.e. δ(P) in an invariant of
scissors congruence.
Dehn’s solution
In δ(P), dihedral angles which are rationals multiples of π are
0!
Dehn’s solution
In δ(P), dihedral angles which are rationals multiples of π are
0!
δ(unit cube) = 12 × 1 ⊗ (π/2) = 0 since π/2 = 0 ∈ R/πQ
Dehn’s solution
In δ(P), dihedral angles which are rationals multiples of π are
0!
δ(unit cube) = 12 × 1 ⊗ (π/2) = 0 since π/2 = 0 ∈ R/πQ
For a regular tetrahedra with unit volume, the lengths of all
its sides is some positive number a and all its dihedral angles
are α where cos(α) = 1/3.
Dehn’s solution
In δ(P), dihedral angles which are rationals multiples of π are
0!
δ(unit cube) = 12 × 1 ⊗ (π/2) = 0 since π/2 = 0 ∈ R/πQ
For a regular tetrahedra with unit volume, the lengths of all
its sides is some positive number a and all its dihedral angles
are α where cos(α) = 1/3.
δ(tetrahedra) = 6 × a ⊗ arccos( 1 )
3
Dehn’s solution
In δ(P), dihedral angles which are rationals multiples of π are
0!
δ(unit cube) = 12 × 1 ⊗ (π/2) = 0 since π/2 = 0 ∈ R/πQ
For a regular tetrahedra with unit volume, the lengths of all
its sides is some positive number a and all its dihedral angles
are α where cos(α) = 1/3.
δ(tetrahedra) = 6 × a ⊗ arccos( 1 )
3
arccos( 1 )
3
is irrational ! (needs proof)
π
Dehn’s solution
In δ(P), dihedral angles which are rationals multiples of π are
0!
δ(unit cube) = 12 × 1 ⊗ (π/2) = 0 since π/2 = 0 ∈ R/πQ
For a regular tetrahedra with unit volume, the lengths of all
its sides is some positive number a and all its dihedral angles
are α where cos(α) = 1/3.
δ(tetrahedra) = 6 × a ⊗ arccos( 1 )
3
arccos( 1 )
3
is irrational ! (needs proof)
π
δ(unit cube) = 0 = 6 × a ⊗ arccos( 1 ) = δ(tetrahedra)
3
Dehn’s solution
In δ(P), dihedral angles which are rationals multiples of π are
0!
δ(unit cube) = 12 × 1 ⊗ (π/2) = 0 since π/2 = 0 ∈ R/πQ
For a regular tetrahedra with unit volume, the lengths of all
its sides is some positive number a and all its dihedral angles
are α where cos(α) = 1/3.
δ(tetrahedra) = 6 × a ⊗ arccos( 1 )
3
arccos( 1 )
3
is irrational ! (needs proof)
π
δ(unit cube) = 0 = 6 × a ⊗ arccos( 1 ) = δ(tetrahedra)
3
Thus the unit cube and the unit tetrahedra are not scissors
congruent !
In two dimensional spherical geometry S2 and hyperbolic
geometry H2 it is known that area determines scissors
congruence.
In two dimensional spherical geometry S2 and hyperbolic
geometry H2 it is known that area determines scissors
congruence.
Does volume and Dehn invariant determine scissors
congruence in E3 ? Yes they do ! Sydler answered this
question in 1965. this question is known as the “Dehn
invariant sufficiency” problem.
In two dimensional spherical geometry S2 and hyperbolic
geometry H2 it is known that area determines scissors
congruence.
Does volume and Dehn invariant determine scissors
congruence in E3 ? Yes they do ! Sydler answered this
question in 1965. this question is known as the “Dehn
invariant sufficiency” problem.
“Dehn invariant sufficiency” is still open for 3-dimensional
spherical geometry S3 and hyperbolic geometries H3 and in
higher dimensions.
In two dimensional spherical geometry S2 and hyperbolic
geometry H2 it is known that area determines scissors
congruence.
Does volume and Dehn invariant determine scissors
congruence in E3 ? Yes they do ! Sydler answered this
question in 1965. this question is known as the “Dehn
invariant sufficiency” problem.
“Dehn invariant sufficiency” is still open for 3-dimensional
spherical geometry S3 and hyperbolic geometries H3 and in
higher dimensions.
Dupont and Sah related scissors congruence to questions
about the homology of groups of isometries of various
geometries (regarded as discrete groups).
In two dimensional spherical geometry S2 and hyperbolic
geometry H2 it is known that area determines scissors
congruence.
Does volume and Dehn invariant determine scissors
congruence in E3 ? Yes they do ! Sydler answered this
question in 1965. this question is known as the “Dehn
invariant sufficiency” problem.
“Dehn invariant sufficiency” is still open for 3-dimensional
spherical geometry S3 and hyperbolic geometries H3 and in
higher dimensions.
Dupont and Sah related scissors congruence to questions
about the homology of groups of isometries of various
geometries (regarded as discrete groups).
Dupont, Sah, Parry, Suslin etc gave relations between scissors
congruences and K -theory of fields.
In two dimensional spherical geometry S2 and hyperbolic
geometry H2 it is known that area determines scissors
congruence.
Does volume and Dehn invariant determine scissors
congruence in E3 ? Yes they do ! Sydler answered this
question in 1965. this question is known as the “Dehn
invariant sufficiency” problem.
“Dehn invariant sufficiency” is still open for 3-dimensional
spherical geometry S3 and hyperbolic geometries H3 and in
higher dimensions.
Dupont and Sah related scissors congruence to questions
about the homology of groups of isometries of various
geometries (regarded as discrete groups).
Dupont, Sah, Parry, Suslin etc gave relations between scissors
congruences and K -theory of fields.
Neumann used a “complexified” Dehn invariant in H3 to
define invariants of hyperbolic 3-manifolds.
References
1. Applet for Pythagorean Theorem,
http://www.cut-the-knot.org/.
2. Mathworld,
http://mathworld.wolfram.com/PythagoreanTheorem.html.
3. Hilbert’s Third Problem by V. G. Boltianskii, translated by R.
A. Silverman, 1978.
4. Scissors Congruence by Efton Park, Seminar Notes, Texas
Christian University.
5. Hilbert’s 3rd problem and Invariants of 3-manifolds by
Walter Neumann, G&T Monographs, 1998.
6. Tangram pictures taken from the iPhone application LetsTans
http://www.letstans.com/.
```
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views: 47 posted: 5/27/2011 language: English pages: 60 | 5,874 | 19,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2014-15 | longest | en | 0.855749 |
https://isabelle.in.tum.de/repos/isabelle/comparison/e13731554e50/src/HOL/Product_Type.thy | 1,619,016,638,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039544239.84/warc/CC-MAIN-20210421130234-20210421160234-00069.warc.gz | 431,558,256 | 2,503 | src/HOL/Product_Type.thy
changeset 14337 e13731554e50 parent 14208 144f45277d5a child 14359 3d9948163018
equal inserted replaced
14336:8f731d3cd65b 14337:e13731554e50
482 lemma split_part [simp]: "(%(a,b). P & Q a b) = (%ab. P & split Q ab)"
482 lemma split_part [simp]: "(%(a,b). P & Q a b) = (%ab. P & split Q ab)"
483 -- {* Allows simplifications of nested splits in case of independent predicates. *}
483 -- {* Allows simplifications of nested splits in case of independent predicates. *}
484 apply (rule ext, blast)
484 apply (rule ext, blast)
485 done
485 done
486
486
487 lemma split_comp_eq [simp]:
487 (* Do NOT make this a simp rule as it
488 a) only helps in special situations
489 b) can lead to nontermination in the presence of split_def
490 *)
491 lemma split_comp_eq:
488 "(%u. f (g (fst u)) (snd u)) = (split (%x. f (g x)))"
492 "(%u. f (g (fst u)) (snd u)) = (split (%x. f (g x)))"
489 by (rule ext, auto)
493 by (rule ext, auto)
490
494
491 lemma The_split_eq [simp]: "(THE (x',y'). x = x' & y = y') = (x, y)"
495 lemma The_split_eq [simp]: "(THE (x',y'). x = x' & y = y') = (x, y)"
492 by blast
496 by blast | 390 | 1,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.476771 |
https://byjus.com/question-answer/in-a-city-the-bus-station-is-900-m-away-from-the-railway-station-what/ | 1,638,421,699,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00256.warc.gz | 215,365,897 | 20,262 | Question
In a city, the bus station is 900 m away from the railway station. What is the distance between them if a map is drawn using the scale 1 cm = 150 m ? 5 cm6 cm6.5 cm7 cm
Solution
The correct option is B 6 cmScale: 1 cm = 150 m ∴1 m=1150cm Given: The distance of bus station from the railway station = 900 m ∴ The distance between them on the map =900×1150cm =900150=6 cm
Suggest corrections | 124 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-49 | latest | en | 0.880613 |
https://1989lifejourney.com/2022/04/17/leetcode-894-all-possible-full-binary-trees/ | 1,686,077,362,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00324.warc.gz | 104,848,832 | 23,575 | # LeetCode – 894. All Possible Full Binary Trees
### 參考解法:
https://leetcode.com/problems/all-possible-full-binary-trees/discuss/813705/Swift-100-faster-than-all-swift-online-submissions
### 解法:
``````var map: [Int: [TreeNode]] = [:]
func allPossibleFBT(_ n: Int) -> [TreeNode?] {
if n % 2 == 0 { return [] }
else {
if let found = map[n] {
return found
}
var result = [TreeNode]()
if n == 1 {
result.append(TreeNode(0))
return [TreeNode(0)]
} else {
for i in 1 ... (n-1) {
let leftTree = allPossibleFBT(i)
let rightTree = allPossibleFBT(n-1-i)
for leftNode in leftTree {
for rightNode in rightTree {
let node = TreeNode(0)
node.left = leftNode
node.right = rightNode
result.append(node)
}
}
}
}
map[n] = result
return result
}
}
`````` | 230 | 740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-23 | longest | en | 0.392568 |
https://community.powerbi.com/t5/DAX-Commands-and-Tips/SUMX-filter-value-smaller-1-does-not-find-correct-results/m-p/2095242/highlight/true | 1,653,408,069,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573053.67/warc/CC-MAIN-20220524142617-20220524172617-00000.warc.gz | 226,265,789 | 99,506 | cancel
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Frequent Visitor
SUMX filter (value smaller 1) does not find correct results
Hello,
After trying many different formul and search terms I cannot find a solution for the following problem.
I've got a fact table with sales by customer and product. However, it could happen that products are given for free. In this case it should calculate the missed sales (free of charge) by multiplying the quantity with the reference price.
Fact table:
Invoice No. Customer Material Quantity Sales price per piece Total sales price Reference price per piece 1 Alpha A 4 10 40 10 2 Beta A 2 0 0 10 3 Gamma A 10 10 100 10 4 Gamma B 8 0 0 5
Desired output:
Customer Total sales Free of charge Alpha 40 0 Beta 0 20 (2*10) Gamma 100 40 (8*5)
My idea which is not working (result on lowest level as well as on aggregated level is wrong):
Missed sales = SUMX(
FILTER ( Fact_Table; Sales price per piece < 1) ;
Quantity * Reference price per piece
)
(smaller 1 because some products are not given at zero but for some cents)
Thanks for any hints and suggestions in advance!
1 ACCEPTED SOLUTION
Super User
Here's a modified measure
``````free of charge 2 =
var c=SUMMARIZE('Table','Table'[Customer],"free",sumx('Table',if('Table'[Sales price per piece]=0,'Table'[Quantity]*'Table'[Reference price per piece],0)))
return sumx(c,[free])``````
3 REPLIES 3
Super User
see attached. You didn't mention if you wanted working totals.
Frequent Visitor
Thank you for your effort to create that example in PowerBI. It's a solution but I also need working totals to answer the question: What is the total value of freebies and then to analyze to what customer and what products were given for free.
Super User
Here's a modified measure
``````free of charge 2 =
var c=SUMMARIZE('Table','Table'[Customer],"free",sumx('Table',if('Table'[Sales price per piece]=0,'Table'[Quantity]*'Table'[Reference price per piece],0)))
return sumx(c,[free])``````
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Top Solution Authors
Top Kudoed Authors | 643 | 2,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-21 | latest | en | 0.880006 |
https://wiki.cloudfactory.com/docs/mp-wiki/solvers-optimizers/adagrad | 1,726,168,594,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00528.warc.gz | 576,356,968 | 13,251 | Adagrad , short for adaptive gradient, is a gradient based optimizer that automatically tunes its learning rate in the training process. The learning rate is updated parameter wise, i.e. we have a different learning rate for each of the parameters.
The parameters associated with frequently occurring features have small updates (low learning rate), and the parameters associated with seldom occurring features have bigger updates (high learning rate).
$€€g\left\{t,i\right\} = \nabla J \left(\theta\left\{t,i\right\}\right)€€$
Where $€€g_\left\{t,i\right\}€€$ is the gradient of the objective function with respect to the parameter $€€\theta_i€€$
The parameter is updated as follows,
$\theta\left\{t+1,i\right\}=\theta\left\{t,i\right\} - \eta \cdot \frac\left\{g\left\{t,i\right\}\right\}\left\{\sqrt\left\{G\left\{t,ii\right\}\right\}+\epsilon\right\}$
Here $€€\theta\left\{t,i\right\}€€$ is the parameter to be updated, $€€G\left\{t,ii\right\}€€$ is the sum of the square of all the gradient till time t. We can see that the learning rate is adjusted according to the previous encountered gradients. $€€eta€€$ is the Base Learning Rate.
Here, the base learning rate is usually initialized to 0.01.
$€€\epsilon€€$ is used for numeric stability. Its value is $€€10^\left\{-8\right\}€€$ by default.
It is a technique where a large learning rate is adopted in the beginning of the training process and then it is decayed by the certain factor after pre-defined epochs. Higher learning rate decay suggests that the initial learning rate will decay more in the epochs.
Setting a learning rate decay might potentially slow the training process since we decrease the learning rate.
python
# importing the library
import torch
import torch.nn as nn
x = torch.randn(10, 3)
y = torch.randn(10, 2)
# Build a fully connected layer.
linear = nn.Linear(3, 2)
# Build MSE loss function and optimizer.
criterion = nn.MSELoss()
# Forward pass.
pred = linear(x)
# Compute loss.
loss = criterion(pred, y)
print('loss:', loss.item())
optimizer.step() | 552 | 2,055 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.807039 |
https://www.cracksat.net/sat/math-multiple-choice/question-no-939-answer-and-explanation.html | 1,638,779,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.59/warc/CC-MAIN-20211206072825-20211206102825-00210.warc.gz | 797,718,838 | 3,382 | # SAT Math Multiple Choice Question 939: Answer and Explanation
### Test Information
Question: 939
7. If k is a positive integer less than 17, what is the total number of possible integersolutions for the equation x2 + 8x + k = 0?
• A. 5
• B. 6
• C. 7
• D. 8
Explanation:
C
Difficulty: Hard
Strategic Advice: When a quadratic equation is factored, the value of the constant (here, k) is equal to the product of the two constants in the binomial factors. Determine how many combinations there are, and you'll have your answer.
Getting to the Answer: Imagine factoring the given equation: You would need to find the factors of k (which you're told is less than 17) that add up to 8. List all possible integer combinations whose sum is 8 and whose product is less than 17:
(x + 1)(x + 7) = 0: 2 solutions
(x + 2)(x + 6) = 0: 2 solutions
(x + 3)(x + 5) = 0: 2 solutions
(x + 4)(x + 4) = 0: 1 solution
There are 7 different integer solutions for this equation, which matches (C). | 289 | 988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-49 | latest | en | 0.915186 |
http://en.wikipedia.org/wiki/Amitsur%e2%80%93Levitzki_theorem | 1,411,529,352,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657141651.17/warc/CC-MAIN-20140914011221-00187-ip-10-234-18-248.ec2.internal.warc.gz | 95,004,531 | 11,914 | Amitsur–Levitzki theorem
In algebra, the Amitsur–Levitzki theorem states that the algebra of n by n matrices satisfies a certain identity of degree 2n. It was proved by Amitsur and Levitsky (1950). In particular matrix rings are polynomial identity rings such that the smallest identity they satisfy has degree exactly 2n.
Statement
The standard polynomial of degree n is
$S_n(x_1,\ldots,x_n) = \sum_{\sigma\in S_{n}}\text{sgn}(\sigma)x_{\sigma 1}\cdots x_{\sigma n} \$
in non-commutative polynomials x1,...,xn, where the sum is taken over all n! elements of the symmetric group Sn.
The Amitsur–Levitzki theorem states that for n by n matrices A1,...,A2n then
$S_{2n}(A_1,\ldots,A_{2n}) = 0 \ .$
Proofs
Amitsur and Levitzki (1950) gave the first proof.
Kostant (1958) deduced the Amitsur–Levitzki theorem from the Koszul–Samelson theorem about primitive cohomology of Lie algebras.
Swan (1963) and Swan (1969) gave a simple combinatorial proof as follows. By linearity it is enough to prove the theorem when each matrix has only one nonzero entry, which is 1. In this case each matrix can be encoded as a directed edge of a graph with n vertices. So all matrices together give a graph on n vertices with 2n directed edges. The identity holds provided that for any two vertices A and B of the graph, the number of odd Eulerian paths from A to B is the same as the number of even ones. (Here a path is called odd or even depending on whether its edges taken in order give an odd or even permutation of the 2n edges.) Swan showed that this was the case provided the number of edges in the graph is at least 2n, thus proving the Amitsur–Levitzki theorem.
Razmyslov (1974) gave a proof related to the Cayley–Hamilton theorem.
Rosset (1976) gave a short proof using the exterior algebra of a vector space of dimension 2n. | 505 | 1,828 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2014-41 | latest | en | 0.922103 |
http://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=millenium&To=time | 1,660,305,021,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571692.3/warc/CC-MAIN-20220812105810-20220812135810-00198.warc.gz | 65,100,000 | 3,582 | Partner with ConvertIt.com
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Torque and Equilibrium Lecture 8 Pre-reading: KJF §8.1 and 8.2
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Transcript of Torque and Equilibrium Lecture 8 Pre-reading : KJF §8.1 and 8.2.
• Slide 1
• Torque and Equilibrium Lecture 8 Pre-reading : KJF 8.1 and 8.2
• Slide 2
• 2 Centre of mass, again
• Slide 3
• 3 Archimedes Lever Rule At equilibrium (and with forces 90 to lever): r 1 F 1 = r 2 F 2
• Slide 4
• 4 General Lever Rule For general angles r 1 F 1 sin 1 = r 2 F 2 sin 2 We call rF sin = torque S.I. unit of torque: newton metre (Nm) At equilibrium, the magnitude of torques exerted at each end of lever are equal KJF 7.2
• Slide 5
• 5 Crudely speaking, torque is "twisting or turning ability" of a force that can: change the angular velocity of an object (i.e. speed up or slow down rotation) cause a twisting or bending distortion of an object A force with a "line of action" that does not cross the axis of rotation results in torque. What is torque?
• Slide 6
• 6 Note: torque is measured about a particular point. Usually this will be a hinge, pivot or axis torque has a sign. All forces that tend to rotate the object in the same direction produce torque with the same sign
• Slide 7
• 7 Calculating torque (1) Choose a sign convention (e.g. anti-clockwise +ve), then decide in which direction force is pulling or pushing lever. Write that sign in front of your answer. Method 1: If you're given r and , use formula for torque (magnitude) = r F sin (Note: sin = sin, it doesnt matter which angle you use) Example: Calculate torque on lever exerted by hand:
• Slide 8
• 8 Calculating torque (2) Method 2: If you're given d the perpendicular distance from axis to the line of action, then use formula = d F If the line of action crosses the axis (i.e. d = 0) then = 0
• Slide 9
• 9 Opening a door If r is perpendicular to F, then torque = r F If r is not perpendicular to F, then torque = r F sin where is the angle between r and F Axis of rotation r F What happens if you push in the middle of the door; do you need more or less force? Why? What happens if you push along a line passing through axis of rotation? Explain.
• Slide 10
• 10 Problem The length of a bicycle pedal arm is r = 0.152 m, and a downward force of F = 111 N is applied by the foot. What is the magnitude of torque about the pivot point when the angle between the arm & vertical is; 30.0? 90.0? 180.0? [8.44 Nm, 16.9 Nm, 0.00 Nm]
• Slide 11
• 11 Adding up Torques We will only consider torques acting in 2D (flat on page) Choose a sign convention (e.g. anti-clockwise is positive). Choose the rotation axis around which to calculate torque (unless it's already given). For each force, calculate the resulting torque (including sign). Add up all the torques. KJF 7.2, see p. 214
• Slide 12
• 12 F1F1 F2F2
• Slide 13
• 13 Adding up Torques: Example torque 1; 1 = rF 1 sin = 0.5 10 sin 30 = 2.50 Nm torque 2; 2 = +mgd = 1 9.8 0.25 = +2.45 Nm net torque = = 1 + 2 = 2.45 + (2.50) = 0.05 Nm (i.e. clockwise)
• Slide 14
• 14
• Slide 15
• 15 Equilibrium KJF 8.1, 8.2
• Slide 16
• 16 Conditions for Equilibrium For an object to be in static equilibrium F = 0 no net force F x = 0, F y = 0 = 0 no net torque Because this is true for all pivot points, we are free to choose any point we like for calculating the torque choose point where some torques disappear KJF 8.1
• Slide 17
• 17 Hints for Statics Problems Draw a diagram! Decide on system Put in forces ON system only All forces in mechanics are either contact or gravity Define sign conventions Usually you're given some forces on a static body & need to find unknown forces or torques.
• Slide 18
• 18 Solving Static Equilibrium Problems Decide on the system Choose a rotational axis and sign convention Best to choose one that causes some torques to disappear Remember nothing is rotating anyway so you're free to choose the axis. Calculate all horizontal components of forces acting on the system and write equation F h = 0. Calculate all vertical components of forces acting on the system and write equation F v = 0. Assume each objects weight force is acting at its centre of mass. Calculate all torques and write equation = 0. Remember that all external forces are possible sources of torque. Solve the equations simultaneously. KJF 8.1
• Slide 19
• 19 Example A 65 kg woman is horizontal in a push-up position. What are the vertical forces acting on her hands and her feet? [hands 420 N, feet 210 N]
• Slide 20
• 20 Example 2 Here the cargo is loaded correctly. Whatever rotation axis is chosen, there's always some normal forces opposing the torque due to the total system weight (treated as though it lies at the centre of mass) No net torque equilibrium. The system is the ass, the cart and the cargo.
• Slide 21
• 21 But... Too much cargo is loaded at the back. If the wheel is chosen as the rotation axis, all resulting torques are acting in the clockwise direction. There is no torque opposing the torque due to the weight of the system, hence there is a net clockwise torque. The system will rotate until the cart hits the ground. The donkey will be lifted off the ground.
• Slide 22
• 22
• Slide 23
• 23
• Slide 24
• 24 Types of Equilibrium Neutral: with a small displacement, remains at new position. Stable: with a small displacement, returns to original position. Unstable: with a small displacement, continues to move away from equilibrium position. KJF 8.2
• Slide 25
• 25 Stable W N net torque around X X
• Slide 26
• 26 W N NO net torque around X X Neutral
• Slide 27
• 27 W N net torque around X X Unstable
• Slide 28
• 28
• Slide 29
• Next lecture Momentum, impulse and energy Read: KJF 9.1, 9.2 | 1,607 | 5,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-14 | latest | en | 0.78459 |
https://howkgtolbs.com/convert/33.67-kg-to-lbs | 1,638,441,491,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00364.warc.gz | 358,554,716 | 12,142 | # 33.67 kg to lbs - 33.67 kilograms to pounds
Before we move on to the practice - it means 33.67 kg how much lbs conversion - we will tell you few theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 33.67 kg to lbs? 33.67 kilograms it is equal 74.2296436154 pounds, so 33.67 kg is equal 74.2296436154 lbs.
## 33.67 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, that is International System of Units (in short form SI).
Sometimes the kilogram can be written as kilogramme. The symbol of this unit is kg.
The kilogram was defined first time in 1795. The kilogram was defined as the mass of one liter of water. First definition was not complicated but totally impractical to use.
Later, in 1889 the kilogram was described by the International Prototype of the Kilogram (in short form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was switched by another definition.
Nowadays the definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It can be also divided to 100 decagrams and 1000 grams.
## 33.67 kilogram to pounds
You know some facts about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to emphasize that there are not only one kind of pound. What does it mean? For example, there are also pound-force. In this article we are going to to focus only on pound-mass.
The pound is used in the Imperial and United States customary systems of measurements. Of course, this unit is in use also in another systems. The symbol of the pound is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 33.67 kg?
33.67 kilogram is equal to 74.2296436154 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 33.67 kg in lbs
The most theoretical part is already behind us. In next section we will tell you how much is 33.67 kg to lbs. Now you learned that 33.67 kg = x lbs. So it is time to know the answer. Let’s see:
33.67 kilogram = 74.2296436154 pounds.
That is a correct result of how much 33.67 kg to pound. You can also round it off. After it your result is as following: 33.67 kg = 74.074 lbs.
You know 33.67 kg is how many lbs, so see how many kg 33.67 lbs: 33.67 pound = 0.45359237 kilograms.
Naturally, this time you may also round off this result. After rounding off your result is exactly: 33.67 lb = 0.45 kgs.
We also want to show you 33.67 kg to how many pounds and 33.67 pound how many kg outcomes in tables. Have a look:
We want to begin with a table for how much is 33.67 kg equal to pound.
### 33.67 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
33.67 74.2296436154 74.0740
Now look at a table for how many kilograms 33.67 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
33.67 0.45359237 0.45
Now you know how many 33.67 kg to lbs and how many kilograms 33.67 pound, so it is time to move on to the 33.67 kg to lbs formula.
### 33.67 kg to pounds
To convert 33.67 kg to us lbs a formula is needed. We will show you a formula in two different versions. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 74.2296436154 outcome in pounds
The first version of a formula will give you the most correct outcome. In some situations even the smallest difference can be significant. So if you need an accurate outcome - first formula will be the best for you/option to know how many pounds are equivalent to 33.67 kilogram.
So go to the another version of a formula, which also enables calculations to learn how much 33.67 kilogram in pounds.
The another version of a formula is down below, let’s see:
Number of kilograms * 2.2 = the outcome in pounds
As you see, this version is simpler. It could be better option if you want to make a conversion of 33.67 kilogram to pounds in easy way, for instance, during shopping. You only need to remember that your outcome will be not so correct.
Now we want to learn you how to use these two formulas in practice. But before we are going to make a conversion of 33.67 kg to lbs we want to show you another way to know 33.67 kg to how many lbs totally effortless.
### 33.67 kg to lbs converter
Another way to check what is 33.67 kilogram equal to in pounds is to use 33.67 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Converter is based on first formula which we showed you above. Due to 33.67 kg pound calculator you can quickly convert 33.67 kg to lbs. Just enter amount of kilograms which you need to convert and click ‘calculate’ button. The result will be shown in a flash.
So try to convert 33.67 kg into lbs with use of 33.67 kg vs pound converter. We entered 33.67 as an amount of kilograms. This is the result: 33.67 kilogram = 74.2296436154 pounds.
As you can see, our 33.67 kg vs lbs calculator is user friendly.
Now let’s move on to our chief issue - how to convert 33.67 kilograms to pounds on your own.
#### 33.67 kg to lbs conversion
We are going to begin 33.67 kilogram equals to how many pounds conversion with the first version of a formula to get the most accurate result. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 74.2296436154 the result in pounds
So what need you do to know how many pounds equal to 33.67 kilogram? Just multiply amount of kilograms, this time 33.67, by 2.20462262. It gives 74.2296436154. So 33.67 kilogram is 74.2296436154.
It is also possible to round off this result, for instance, to two decimal places. It gives 2.20. So 33.67 kilogram = 74.0740 pounds.
It is time for an example from everyday life. Let’s convert 33.67 kg gold in pounds. So 33.67 kg equal to how many lbs? As in the previous example - multiply 33.67 by 2.20462262. It is equal 74.2296436154. So equivalent of 33.67 kilograms to pounds, when it comes to gold, is exactly 74.2296436154.
In this case it is also possible to round off the result. This is the result after rounding off, in this case to one decimal place - 33.67 kilogram 74.074 pounds.
Now we can go to examples converted with short formula.
#### How many 33.67 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 74.074 the outcome in pounds
So 33.67 kg equal to how much lbs? As in the previous example you have to multiply amount of kilogram, in this case 33.67, by 2.2. See: 33.67 * 2.2 = 74.074. So 33.67 kilogram is 2.2 pounds.
Let’s do another calculation using this formula. Now convert something from everyday life, for instance, 33.67 kg to lbs weight of strawberries.
So convert - 33.67 kilogram of strawberries * 2.2 = 74.074 pounds of strawberries. So 33.67 kg to pound mass is exactly 74.074.
If you learned how much is 33.67 kilogram weight in pounds and are able to calculate it with use of two different formulas, let’s move on. Now we are going to show you these outcomes in charts.
#### Convert 33.67 kilogram to pounds
We realize that outcomes presented in charts are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in charts for your convenience. Due to this you can easily make a comparison 33.67 kg equivalent to lbs outcomes.
Let’s start with a 33.67 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
33.67 74.2296436154 74.0740
And now look 33.67 kg equal pound table for the second version of a formula:
Kilograms Pounds
33.67 74.074
As you see, after rounding off, if it comes to how much 33.67 kilogram equals pounds, the outcomes are the same. The bigger amount the more considerable difference. Remember it when you need to do bigger number than 33.67 kilograms pounds conversion.
#### How many kilograms 33.67 pound
Now you learned how to convert 33.67 kilograms how much pounds but we want to show you something more. Do you want to know what it is? What about 33.67 kilogram to pounds and ounces calculation?
We are going to show you how you can convert it step by step. Begin. How much is 33.67 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, this time 33.67, by 2.20462262. So 33.67 * 2.20462262 = 74.2296436154. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To calculate how much 33.67 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.
So final result is 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then your result is 2 pounds and 33 ounces.
As you can see, conversion 33.67 kilogram in pounds and ounces quite easy.
The last conversion which we are going to show you is conversion of 33.67 foot pounds to kilograms meters. Both of them are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we show you it, look:
• 33.67 kilograms meters = 7.23301385 foot pounds,
• 33.67 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 33.67 foot pounds to kilograms meters you need to multiply 33.67 by 0.13825495. It gives 0.13825495. So 33.67 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 33.67 foot pounds is equal 0.14 kilogram meters.
We hope that this calculation was as easy as 33.67 kilogram into pounds conversions.
This article was a big compendium about kilogram, pound and 33.67 kg to lbs in calculation. Thanks to this conversion you know 33.67 kilogram is equivalent to how many pounds.
We showed you not only how to do a calculation 33.67 kilogram to metric pounds but also two other conversions - to know how many 33.67 kg in pounds and ounces and how many 33.67 foot pounds to kilograms meters.
We showed you also another solution to make 33.67 kilogram how many pounds calculations, that is with use of 33.67 kg en pound converter. It will be the best solution for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you are able to do 33.67 kilogram equal to how many pounds calculation - on your own or using our 33.67 kgs to pounds calculator.
It is time to make your move! Let’s convert 33.67 kilogram mass to pounds in the best way for you.
Do you want to make other than 33.67 kilogram as pounds conversion? For example, for 15 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so easy as for 33.67 kilogram equal many pounds.
### How much is 33.67 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 33.67 kg in pounds , we prepared one more section. Here we have for you the most important information about how much is 33.67 kg equal to lbs and how to convert 33.67 kg to lbs . Let’s see.
How does the kilogram to pound conversion look? It is a mathematical operation based on multiplying 2 numbers. Let’s see 33.67 kg to pound conversion formula . It is down below:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 33.67 kilogram to pounds? The correct result is 74.2296436154 lbs.
There is also another way to calculate how much 33.67 kilogram is equal to pounds with second, shortened type of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So in this case, 33.67 kg equal to how much lbs ? The result is 74.2296436154 pounds.
How to convert 33.67 kg to lbs quicker and easier? You can also use the 33.67 kg to lbs converter , which will make the rest for you and give you a correct answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,522 | 13,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-49 | latest | en | 0.948923 |
https://farside.ph.utexas.edu/teaching/315/Waveshtml/node77.html | 1,709,553,547,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00664.warc.gz | 243,061,164 | 6,833 | # Electromagnetic Waves in Magnetized Plasmas
Let us extend the analysis of the previous section to consider a general electromagnetic wave propagating through a uniform plasma with an equilibrium magnetic field of strength, . The plasma is assumed to consist of two species: electrons of mass and electric charge , and ions of mass and electric charge . The plasma is also assumed to be electrically neutral, so that the equilibrium number density of the ions is the same as that of the electrons; namely, (Stix 1962). The equations of motion of a constituent ion of the plasma are written
(9.92) (9.93) (9.94)
where , , and are the wave-induced displacements of the ion along the three Cartesian axes. (Here, we are including ion motion in our analysis because such motion is important in certain frequency ranges.) As before, the former terms on the right-hand sides of the previous equations represent the forces exerted on the ion by the wave electric field, , whereas the latter terms represent the forces exerted by the equilibrium magnetic field when the ion moves (Fitzpatrick 2008). (As before, we can neglect any forces due to the wave magnetic field, as long as the particle motion remains non-relativistic.) The equations of motion of a constituent electron take the form
(9.95) (9.96) (9.97)
The Cartesian components of the electric dipole moment per unit volume are
(9.98) (9.99) (9.100)
Finally, the electric displacement is written
(9.101)
Consider a right-hand circularly polarized (with respect to the direction of the equilibrium magnetic field) wave whose electric field takes the form
(9.102) (9.103) (9.104)
Let us write
(9.105) (9.106) (9.107) (9.108) (9.109) (9.110) (9.111) (9.112) (9.113)
Equations (9.92)–(9.113) yield
(9.114) (9.115) (9.116)
where
(9.117) (9.118) (9.119) (9.120)
Here, is termed the ion cyclotron frequency, and is the frequency at which ions gyrate in the plane perpendicular to the equilibrium magnetic field (Stix 1962). Moreover, is the electron cyclotron frequency, and is the frequency at which electrons gyrate in the plane perpendicular to the equilibrium magnetic field (ibid). Finally, and are termed the ion plasma frequency, and the electron plasma frequency, respectively (ibid). Of course, and , because . Finally, it follows from Equations (9.101)–(9.104), (9.113), and (9.116), that the electric displacement of a right-hand circularly polarized wave propagating through a magnetized plasma has the components
(9.121) (9.122) (9.123)
where
(9.124)
Consider a left-hand circularly polarized (with respect to the direction of the equilibrium magnetic field) wave whose electric field takes the form
(9.125) (9.126) (9.127)
By repeating the previously described analysis (with appropriate modifications), we deduce that
(9.128) (9.129) (9.130)
where
(9.131)
Finally, consider a wave whose electric field is polarized parallel to the equilibrium magnetic field, so that
(9.132) (9.133) (9.134)
Again, repeating the previous analysis (with suitable modifications), we obtain
(9.135) (9.136) (9.137)
where
(9.138)
Now, the equations that govern electromagnetic wave propagation through a dielectric media are (see Appendix C)
(9.139) (9.140) (9.141) (9.142) (9.143) (9.144)
Consider an electromagnetic wave with a general polarization (with respect to the equilibrium magnetic field). Such a wave can be written as a linear combination of a right-hand circularly polarized wave, a left-hand circularly polarized wave, and a wave with parallel polarization. In other words,
(9.145) (9.146) (9.147)
[See Equations (9.102)–(9.104), (9.125)–(9.127), and (9.132)–(9.134).] It follows, from the previous analysis, that
(9.148) (9.149) (9.150)
[See Equations (9.121)–(9.123), (9.128)–(9.130), and (9.135)–(9.137).] Suppose that
(9.151) (9.152) (9.153)
which implies that
(9.154) (9.155) (9.156)
Finally, let
(9.157)
which means that the wavevector lies in the - plane, and subtends an angle with the equilibrium magnetic field.
Equations (9.139)–(9.157) yield
(9.158) (9.159) (9.160) (9.161) (9.162) (9.163)
which can be combined to give
(9.164)
The previous equation determines the frequencies and polarizations of an electromagnetic wave of wavenumber that propagates through a magnetized plasma, and whose direction of propagation subtends an angle with the magnetic field.
Suppose, finally, that
(9.165) (9.166) (9.167)
It follows, from Equations (9.145)–(9.147), that , , and . Hence, Equation (9.164) transforms to give the eigenmode equation
(9.168)
Here,
(9.169)
is the effective refractive index of the plasma, whereas
(9.170) (9.171) | 1,286 | 4,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.870006 |
https://howmanyis.com/speed/17-m-s-in-fpm/8010-10-meters-per-second-in-feet-per-minute | 1,547,596,189,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583656530.8/warc/CC-MAIN-20190115225438-20190116011438-00295.warc.gz | 538,560,699 | 6,239 | How many is
Conversion between units of measurement
You can easily convert 10 meters per second into feet per minute using each unit definition:
Meters per second
1 m / s
Feet per minute
ft/min = 0.00508 m / s
With this information, you can calculate the quantity of feet per minute 10 meters per second is equal to.
## ¿How many fpm are there in 10 m/s?
In 10 m/s there are 1968.5039 fpm.
Which is the same to say that 10 meters per second is 1968.5039 feet per minute.
Ten meters per second equals to one thousand nine hundred sixty-eight feet per minute. *Approximation
### ¿What is the inverse calculation between 1 foot per minute and 10 meters per second?
Performing the inverse calculation of the relationship between units, we obtain that 1 foot per minute is 0.000508 times 10 meters per second.
A foot per minute is zero times ten meters per second. *Approximation | 218 | 884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-04 | latest | en | 0.852405 |
https://netlib.sandia.gov/slatec/src/bspvd.f | 1,716,018,634,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00006.warc.gz | 367,608,257 | 3,055 | *DECK BSPVD SUBROUTINE BSPVD (T, K, NDERIV, X, ILEFT, LDVNIK, VNIKX, WORK) C***BEGIN PROLOGUE BSPVD C***PURPOSE Calculate the value and all derivatives of order less than C NDERIV of all basis functions which do not vanish at X. C***LIBRARY SLATEC C***CATEGORY E3, K6 C***TYPE SINGLE PRECISION (BSPVD-S, DBSPVD-D) C***KEYWORDS DIFFERENTIATION OF B-SPLINE, EVALUATION OF B-SPLINE C***AUTHOR Amos, D. E., (SNLA) C***DESCRIPTION C C Written by Carl de Boor and modified by D. E. Amos C C Abstract C BSPVD is the BSPLVD routine of the reference. C C BSPVD calculates the value and all derivatives of order C less than NDERIV of all basis functions which do not C (possibly) vanish at X. ILEFT is input such that C T(ILEFT) .LE. X .LT. T(ILEFT+1). A call to INTRV(T,N+1,X, C ILO,ILEFT,MFLAG) will produce the proper ILEFT. The output of C BSPVD is a matrix VNIKX(I,J) of dimension at least (K,NDERIV) C whose columns contain the K nonzero basis functions and C their NDERIV-1 right derivatives at X, I=1,K, J=1,NDERIV. C These basis functions have indices ILEFT-K+I, I=1,K, C K .LE. ILEFT .LE. N. The nonzero part of the I-th basis C function lies in (T(I),T(I+K)), I=1,N. C C If X=T(ILEFT+1) then VNIKX contains left limiting values C (left derivatives) at T(ILEFT+1). In particular, ILEFT = N C produces left limiting values at the right end point C X=T(N+1). To obtain left limiting values at T(I), I=K+1,N+1, C set X= next lower distinct knot, call INTRV to get ILEFT, C set X=T(I), and then call BSPVD. C C Description of Arguments C Input C T - knot vector of length N+K, where C N = number of B-spline basis functions C N = sum of knot multiplicities-K C K - order of the B-spline, K .GE. 1 C NDERIV - number of derivatives = NDERIV-1, C 1 .LE. NDERIV .LE. K C X - argument of basis functions, C T(K) .LE. X .LE. T(N+1) C ILEFT - largest integer such that C T(ILEFT) .LE. X .LT. T(ILEFT+1) C LDVNIK - leading dimension of matrix VNIKX C C Output C VNIKX - matrix of dimension at least (K,NDERIV) contain- C ing the nonzero basis functions at X and their C derivatives columnwise. C WORK - a work vector of length (K+1)*(K+2)/2 C C Error Conditions C Improper input is a fatal error C C***REFERENCES Carl de Boor, Package for calculating with B-splines, C SIAM Journal on Numerical Analysis 14, 3 (June 1977), C pp. 441-472. C***ROUTINES CALLED BSPVN, XERMSG C***REVISION HISTORY (YYMMDD) C 800901 DATE WRITTEN C 890531 Changed all specific intrinsics to generic. (WRB) C 890831 Modified array declarations. (WRB) C 890831 REVISION DATE from Version 3.2 C 891214 Prologue converted to Version 4.0 format. (BAB) C 900315 CALLs to XERROR changed to CALLs to XERMSG. (THJ) C 900326 Removed duplicate information from DESCRIPTION section. C (WRB) C 920501 Reformatted the REFERENCES section. (WRB) C***END PROLOGUE BSPVD C INTEGER I,IDERIV,ILEFT,IPKMD,J,JJ,JLOW,JM,JP1MID,K,KMD, KP1, L, 1 LDUMMY, M, MHIGH, NDERIV REAL FACTOR, FKMD, T, V, VNIKX, WORK, X C DIMENSION T(ILEFT+K), WORK((K+1)*(K+2)/2) C A(I,J) = WORK(I+J*(J+1)/2), I=1,J+1 J=1,K-1 C A(I,K) = W0RK(I+K*(K-1)/2) I=1.K C WORK(1) AND WORK((K+1)*(K+2)/2) ARE NOT USED. DIMENSION T(*), VNIKX(LDVNIK,*), WORK(*) C***FIRST EXECUTABLE STATEMENT BSPVD IF(K.LT.1) GO TO 200 IF(NDERIV.LT.1 .OR. NDERIV.GT.K) GO TO 205 IF(LDVNIK.LT.K) GO TO 210 IDERIV = NDERIV KP1 = K + 1 JJ = KP1 - IDERIV CALL BSPVN(T, JJ, K, 1, X, ILEFT, VNIKX, WORK, IWORK) IF (IDERIV.EQ.1) GO TO 100 MHIGH = IDERIV DO 20 M=2,MHIGH JP1MID = 1 DO 10 J=IDERIV,K VNIKX(J,IDERIV) = VNIKX(JP1MID,1) JP1MID = JP1MID + 1 10 CONTINUE IDERIV = IDERIV - 1 JJ = KP1 - IDERIV CALL BSPVN(T, JJ, K, 2, X, ILEFT, VNIKX, WORK, IWORK) 20 CONTINUE C JM = KP1*(KP1+1)/2 DO 30 L = 1,JM WORK(L) = 0.0E0 30 CONTINUE C A(I,I) = WORK(I*(I+3)/2) = 1.0 I = 1,K L = 2 J = 0 DO 40 I = 1,K J = J + L WORK(J) = 1.0E0 L = L + 1 40 CONTINUE KMD = K DO 90 M=2,MHIGH KMD = KMD - 1 FKMD = KMD I = ILEFT J = K JJ = J*(J+1)/2 JM = JJ - J DO 60 LDUMMY=1,KMD IPKMD = I + KMD FACTOR = FKMD/(T(IPKMD)-T(I)) DO 50 L=1,J WORK(L+JJ) = (WORK(L+JJ)-WORK(L+JM))*FACTOR 50 CONTINUE I = I - 1 J = J - 1 JJ = JM JM = JM - J 60 CONTINUE C DO 80 I=1,K V = 0.0E0 JLOW = MAX(I,M) JJ = JLOW*(JLOW+1)/2 DO 70 J=JLOW,K V = WORK(I+JJ)*VNIKX(J,M) + V JJ = JJ + J + 1 70 CONTINUE VNIKX(I,M) = V 80 CONTINUE 90 CONTINUE 100 RETURN C C 200 CONTINUE CALL XERMSG ('SLATEC', 'BSPVD', 'K DOES NOT SATISFY K.GE.1', 2, + 1) RETURN 205 CONTINUE CALL XERMSG ('SLATEC', 'BSPVD', + 'NDERIV DOES NOT SATISFY 1.LE.NDERIV.LE.K', 2, 1) RETURN 210 CONTINUE CALL XERMSG ('SLATEC', 'BSPVD', + 'LDVNIK DOES NOT SATISFY LDVNIK.GE.K', 2, 1) RETURN END | 1,760 | 4,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.716708 |
https://tronwalletapp.com/qa/question-where-do-we-use-symbol.html | 1,627,576,229,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153860.57/warc/CC-MAIN-20210729140649-20210729170649-00303.warc.gz | 586,542,871 | 8,649 | # Question: Where Do We Use Symbol?
## How do you use symbol in a sentence?
Symbol sentence examplesHer hand went to the symbol around her neck.
His eyes went to the symbol at her neck, and one eyebrow shot up.
The box was marked with a biohazard symbol on the outside.
She touched it, as thrilled to wear his symbol as she had been her father’s.More items….
## What are some examples of symbol?
Symbols take the form of words, sounds, gestures, ideas, or visual images and are used to convey other ideas and beliefs. For example, a red octagon is a common symbol for “STOP”; on maps, blue lines often represent rivers; and a red rose often symbolizes love and compassion.
## How many types of symbols are there?
There are two types of symbols that you can specify in application or vendor programs: System symbols are defined to the system at initialization.
## Is a letter a symbol?
A letter is an alphabetic symbol such as A or a. There are 26 letters in the modern English alphabet.
## What is a () called?
are all called brackets. They indicate that the terms within any of them are considered as one quantity. () are called round brackets or parentheses, {} called curly brackets or braces and [] are the square brackets. … We use these brackets when we want to enclose part of an expression already within brackets.
## How do you read symbols?
Follow these 8 easy steps to analyze symbolism in literature. Take notes. Keep track of objects, characters, and ideas. … Learn what a symbol is. … Look for detailed descriptions. … Look for “big idea” names. … Look for repetition. … Do research.
## What is () called in math?
* This symbol is called an asterisk. In mathematics, we sometimes use it to mean multiplication, particularly with computers. For example, 5*3 = 5 times 3 = 15.
## What is A & called?
The ampersand is the logogram &, representing the conjunction “and”. It originated as a ligature of the letters et—Latin for “and”.
The Heart Symbol. The heart shaped symbol is known throughout the world and it usually stands for love and romance. … The Yin-Yang. … The Bluetooth Symbol. … The International Flag of Planet Earth. … The Great Seal of the United States. … Political Animals. … The Hammer and Sickle. … The Jesus Fish.More items…•
## What is the most powerful symbol?
The 6 Most Powerful Spiritual Symbols on the PlanetThe Hamsa, the healing hand. Also known as the Hand of Fatima, the Hamsa is a revered symbol of divine safekeeping, worn to dispel negative energy. … The Ankh, key of life. … The Cross, a sign of infinite love. … The Eye of Horus, the great protector. … Om, harmony with the universe. … The Lotus, flower of awakening.
## What does the symbol mean in writing?
Anything can be a symbol, if we make it one. In literature, symbols are often characters, settings, images, or other motifs that stand in for bigger ideas. Authors often use symbols (or “symbolism”) to give their work with more meaning and to make a story be about more than the events it describes.
## What is the at symbol called?
The at sign, @, is normally read aloud as “at”; it is also commonly called the at symbol or commercial at. It is used as an accounting and invoice abbreviation meaning “at a rate of” (e.g. 7 widgets @ £2 per widget = £14), but it is now seen more widely in email addresses and social media platform handles.
## What are signs and symbols?
Summary of the Differences Between a Sign and SymbolA sign is a form of language in its own right and it is specifically meant to communicate certain information. Signs are usually informative, regulatory, warning or prohibitory. … Symbol is a form of a sign that may have deep meaning.
## How are symbols called in English?
There are 14 punctuation marks that are commonly used in English grammar. They are the period, question mark, exclamation point, comma, semicolon, colon, dash, hyphen, parentheses, brackets, braces, apostrophe, quotation marks, and ellipsis.
## How symbols are created?
The symbols that humans create are words. Every word consists of two components, a denotative and a connotative (for a complete discussion of denotation and connotation see Chapter 7, THE POWER OF WORDS). The denotative component is the direct impression for which the word stands. | 954 | 4,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-31 | latest | en | 0.94508 |
https://vectorlinux.com/is-matlab-in-degrees-or-radians-2/ | 1,708,574,944,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00000.warc.gz | 645,461,708 | 20,309 | # Is Matlab In Degrees Or Radians
Python Programming
Matlab uses radians as the default unit for trigonometric functions and functions that convert between angle units, such as `sind` and `cosd`. This means that when you specify an angle in Matlab, it is assumed to be in radians unless otherwise specified. This is an important detail to keep in mind, as it can lead to unexpected results if you forget to convert your angles from degrees to radians when necessary.
I have encountered this issue personally when working on a project that involved calculating angles for mechanical systems. I initially overlooked the fact that Matlab uses radians, and my initial results did not make sense. Upon further investigation, I realized that I had to convert my angle inputs from degrees to radians for the calculations to produce accurate results.
It’s crucial to be mindful of this default behavior, especially when dealing with real-world problems that involve angles and trigonometric functions. While it may seem like a small nuance, overlooking the unit of angle measurement can lead to significant errors in your calculations.
To convert an angle from degrees to radians in Matlab, you can use the `deg2rad` function. For example:
```angle_degrees = 45; angle_radians = deg2rad(angle_degrees);```
Alternatively, if you need to convert from radians to degrees, you can use the `rad2deg` function:
```angle_radians = pi/4; angle_degrees = rad2deg(angle_radians);```
It’s worth noting that there are situations where you may want to work directly with degrees in Matlab. In such cases, you can use the aforementioned trigonometric functions suffixed with “d” to explicitly indicate that the input angles are in degrees. For example, `sind` and `cosd` compute the sine and cosine of angles given in degrees.
It’s essential to understand the default behavior of angle units in Matlab to avoid errors and inaccuracies in your calculations. Being aware of this aspect can save you from unnecessary debugging and troubleshooting in your projects.
## Conclusion
In conclusion, Matlab defaults to using radians as the unit for angle measurements in trigonometric functions. This default behavior has implications for the accuracy of your calculations, and it’s important to be mindful of it when working on projects involving angles. Always remember to convert your angle measurements to radians when necessary, and consider using the “d” suffixed trigonometric functions if you prefer to work directly with degrees. Understanding these nuances can greatly improve the reliability of your Matlab code. | 516 | 2,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-10 | longest | en | 0.882778 |
https://www.physicsforums.com/threads/voltage-versus-current.746388/ | 1,477,261,507,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719437.30/warc/CC-MAIN-20161020183839-00175-ip-10-171-6-4.ec2.internal.warc.gz | 986,279,286 | 21,204 | # Voltage versus current
1. Apr 1, 2014
### Jimmy87
Hi, please could someone help me clear up an issue I am having of whether or not voltage can kill you. I know there are numerous threads on this but they don't quite answer my question. This crops up all the time in lots of forums but I was wondering if I am missing something because as far as I am concerned voltage can NEVER kill you because it isn't anything physical.
If I understand voltage correctly then voltage is the difference in energy per charge between two points. Therefore voltage is simply a numerical description of the difference in energy per charge. It therefore surely cannot physically exist. Energy is real and charge is real but voltage is not a physical thing. How can a description of something kill you. Surely its the ENERGY that kills you?
Looking at an analogous situation, if you divide gravitational gravitational PE by mass then you get gravitational 'voltage' which is the energy per mass (instead of energy per charge). If you dropped a rock from a height it could kill you if it landed on your head. This is because the rock has mass and energy which are real things. To say that the energy per mass killed you would be ridiculous because it isn't anything real. Likewise, the energy per charge (voltage) surely cannot kill you?
Is what I have said correct or not?
Thanks for anyone who can take the time to respond.
2. Apr 1, 2014
### stevendaryl
I don't agree that "voltage" isn't real, but you are right that voltage won't harm you if only a tiny amount of current flows across that voltage. A Van de Graaff generator produces really huge votages--over 100,000 volts--but isn't very dangerous, because only a tiny amount of current flows across that voltage when you touch it.
3. Apr 1, 2014
### Jimmy87
Thanks for the reply. I understand that a potential difference is needed but the p.d. itself is just a mathematical model telling you the difference in energy per charge. If you developed a microscope powerful enough you could draw a picture of an electron because its a physical thing. You could never draw a picture of voltage because it isn't physical its just a comparison of energy between two points.
4. Apr 1, 2014
### BiGyElLoWhAt
Voltage itself can't kill you. It's the current that kills you. V=IR and the average resistance of the human body is (depending on the path the current takes) between 2k ohms and about 120k ohms, and 15 mA through your heart can kill you. I'll let you work out the math. Again, voltage in itself cannot hurt you, only current (which can and most certainly does result from a voltage)
5. Apr 1, 2014
### Staff: Mentor
In the end, it is the electromagnetic force that causes charges to move. Voltage is merely a way of describing the properties of the EM field of the circuit that makes it much, MUCH easier to work with than trying to do something like field equations. (Or whatever equations would be needed)
6. Apr 2, 2014
### UltrafastPED
I've had 28,000 volts discharge from a defective cable through my ear. It did make my hair stand on end, but other than that there was only a small red spot.
This is because there was less than 1 joule of total energy.
It is the _energy_ transferred through your body which causes damage. For AC the power is V x I, energy per second. So if the produce V x I is greater than 10, and the contact is more than 1 second - then you will accumulate damage. For example, if the discharge is from your hand, and out through your feet, or through your other hand - then it passes through your heart.
But their are many physiological effects, so best to keep your hands off of the live connections.
Some references:
http://www.physics.ohio-state.edu/~p616/safety/fatal_current.html
http://en.wikipedia.org/wiki/Electric_shock
Last edited by a moderator: Sep 25, 2014
7. Apr 2, 2014
### Jimmy87
Thanks for the answers people. I just thing it's fundamentally wrong to say voltage is something physical. It all boils down to energy in my view and this is DIRECTLY the cause of what kills you. Imagine my friend told me to go to the shops and get him a newspaper but I got run over by a truck and killed. My friends instruction would be analogous to the voltage. It is required for me to have got hit by the truck but it would seem ridiculous to say I got killed by his instruction. Similarly, voltage is simply a mathematical model that humans have used to explain electricity. You need a voltage for charges to do work but at the end of the day voltage is just a number whereas charge and energy are REAL.
8. Apr 2, 2014
### stevendaryl
In electromagnetism, the electric and magnetic fields are definitely something real. Light is a fluctuating electromagnetic field. The electric field has an energy density, even if there are no charges around to "feel" the electric field. It takes energy to create an electric field.
Voltage is simply the integral of the electric field. When you connect a wire to a power source, it creates an electric field inside the wire, and the magnitude of the field is related to the voltage by: $V = EL$ where $V$ is the voltage drop from one end of the wire to the other, $E$ is the average electric field in the wire, and $L$ is the length of the wire.
9. Apr 2, 2014
### nasu
Energy is also a physical quantity and a mathematical model. Just like the voltage.
What you call "physical" may be a matter of your choice in some sense.
You can think in terms of phenomena rather than quantities.
But is still not straightforward. You could say that it is the electric current (the phenomenon, the charge displacement, not the intensity of the current) that "kills". But if the current would pass without disrupting some biological processes it won't kill. It is the muscle contraction that kills? Or the heat developed? Or the interference with the nerve signals to the heart muscle?
It's not so meaningful to look for a simplistic answer of the type "voltage kills" or "current kills".
10. Apr 2, 2014
### sophiecentaur
Field and Potential are both equally valid ways to approach 'Electricity'. It depends upon what situation you are trying to deal with. With a 6V battery and two wires, you could produce a field of 1V/m, .001V/m or 1000V/m, merely by altering the separation of the ends of the wires. What use would the notion of Field be in that case?
The Voltage is not 'just an instruction'. It is the energy the battery or generator is producing for each coulomb of charge. When you tell your friend to do something, you are not supplying any energy (unless you kick him out of the door) It is the truck that supplies the energy to do the damage. The electrical analogy of your instruction is the instructions about how to build the circuit in the first place.
I don't think you will get anywhere if you insist that Energy is any more real than Voltage. You should read about it more with an open mind. Maths is only a model of anything but it is an excellent way of describing the 'how' but not the 'why' of life. The 'why's of life are not explainable, ultimately.
11. Apr 2, 2014
### analogdesign
One thing to ask yourself is whether you think gravity is "real". You can't get a microscope and see gravity either.
12. Apr 2, 2014
### dauto
All physical quantities are based on some mathematical model and equations. Voltage isn't more real or more physical than energy in any meaningful way.
13. Apr 2, 2014
### Staff: Mentor
I'm not so sure. Voltage is a difference in electric potential, which is analogous to a difference in gravitational potential between two points at different heights. Both are "real" in the sense that a difference in potential exists, but I'm not sure i'd go further than that.
14. Apr 2, 2014
### phinds
I'd say that you have that exactly backwards. The math models arise because of the fact that they describe reality. The reality is not in any way based on the existence of a math model to describe it.
15. Apr 2, 2014
### sophiecentaur
Maths models describe / predict the relationship between measurable quantities. Whether or not that is 'reality' is another matter. But this is very philosophical; when you drop a brick on your foot, the sensation is 'real' enough but at what level (macroscopic / molecular / quantum) would you say the situation was real? You can simulate 'weight' with a magnetic force so that an observer could not tell the difference. What is reality then?
People tend to use the word 'real' to mean 'familiar' and I don't, personally, go along with that. | 1,955 | 8,568 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-44 | longest | en | 0.953747 |
https://community.qlik.com/thread/263415 | 1,531,876,790,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589980.6/warc/CC-MAIN-20180718002426-20180718022426-00100.warc.gz | 619,317,390 | 24,035 | 8 Replies Latest reply: Jun 9, 2017 5:34 AM by vivek pv
# Bar chart -sum of total sales minus sales of first bar
Hi Experts,
I have a table like
Load * inline [
NAME,SAL
Raj,300
Vijay,400
Vidhya,250
Swaroop,100
];
Now I want result in a bar chart like,
1st bar should show Total SAL
2nd bar should show Total SAL - SAL
ex: on above data Total SAL is 300+400+250+100= 1050
so, 1st bar is 1050
2nd bar is 750 (1050 - 300)
Can anyone help me on this?
• ###### Re: Bar chart -sum of total sales minus sales of first bar
Hi,
Try like below,
Data:
Load * inline [
NAME,SAL
Raj,300
Vijay,400
Vidhya,250
Swaroop,100
];
Dimension:
load * Inline [
Dim
1
2
];
Create bar chart,
dimension is
=If(Dim = 1, 'Sales', 'Total Sales')
create measure like below,
=If(Dim = 1, sum({<NAME-={'Raj'}>}SAL), sum(total SAL))
• ###### Re: Bar chart -sum of total sales minus sales of first bar
Hi,
using qlik sense
• ###### Re: Bar chart -sum of total sales minus sales of first bar
Hi,
Dimension : NAME
Expression :
Sum(TOTAL SAL)
Sum(TOTAL SAL)-Sum(SAL)
Regrds,
Antonio
• ###### Re: Bar chart -sum of total sales minus sales of first bar
Hi Antonio,
I need small change in the solution you provided.
if you observe in the chart, 1st bar for every name is having value like 1050.
but what i need is,
for the name Raj
1st bar-->1050
2nd bar--> 300
for the name Swaroop
1st bar--> 750 (1050-300) Here 300 is value of 2nd bar in previous dimension
2nd bar--> 100
for the name Vidhya
1st bar--> 650 (750-100)
2nd bar--> 250
for the name Vijay
1st bar-->400
2nd bar-->400
• ###### Re: Bar chart -sum of total sales minus sales of first bar
Load * inline [
NAME,SAL
Raj,300
Swaroop,100
Vidhya,250
Vijay,400
]
;
If(RowNo() = 1,Sum(TOTAL SAL),Above(Column(1))-Above(Sum(SAL)))
Sum(SAL)
• ###### Re: Bar chart -sum of total sales minus sales of first bar
Hi Antonio,
Awesome...!! that worked fine.
I need to implement one hide/show mechanism on the first bar.
Like, by default it should show both bars.
but, when a selection is made, then the blue bar should not display.
in above chart, lets assume that name Vidhya is selected, then in that case I should see only Vidhya SAL which is 250.
• ###### Re: Bar chart -sum of total sales minus sales of first bar
Change to
If(GetSelectedCount(NAME) = 0,
If(RowNo() = 1,Sum(TOTAL SAL),Above(Column(1))-Above(Sum(SAL
))))
Sum(SAL)
• ###### Re: Bar chart -sum of total sales minus sales of first bar
Thanks Antonio.
Its wroking FIne. | 780 | 2,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-30 | latest | en | 0.765745 |
https://techcommunity.microsoft.com/t5/excel/trying-to-write-a-formula-for-last-x-by-date/m-p/4136773 | 1,722,954,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00699.warc.gz | 435,264,513 | 58,870 | SOLVED
Trying to write a formula for last X by date
Copper Contributor
Trying to write a formula for last X by date
I'm trying to write a formula to show the sum of the previous 2 dates summed scores as per column D without manually summing each time. Can't get my head around how to use sumifs to do so.
Any help, grateful
3 Replies
Re: Trying to write a formula for last X by date
"previous 2 dates" in relation to what?
Re: Trying to write a formula for last X by date
So as per the calc in column D.
Here I have summed the Teams previous 2 scores based by date:
Cell D6 is the sum of C2 and C4, the last 2 entries by date for this team. This is what I want the formula to calculate, the sum of the last 2 scores based on the previous 2 dates.
best response confirmed by MJWoods (Copper Contributor)
Solution
Re: Trying to write a formula for last X by date
I have solved this with FILTER() and SUM().
``=LET(list,TAKE(CHOOSECOLS(SORT(FILTER(Table[[Date]:[Score]],([Team]=[@Team])*([Date]<[@Date])),1),3),-2),IF(COUNT(list)<2,"",SUM(list)))``
Hope that fits.
1 best response
Accepted Solutions
best response confirmed by MJWoods (Copper Contributor)
Solution
Re: Trying to write a formula for last X by date
I have solved this with FILTER() and SUM().
``=LET(list,TAKE(CHOOSECOLS(SORT(FILTER(Table[[Date]:[Score]],([Team]=[@Team])*([Date]<[@Date])),1),3),-2),IF(COUNT(list)<2,"",SUM(list)))``
Hope that fits. | 393 | 1,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-33 | latest | en | 0.905231 |
https://stats.stackexchange.com/questions/404726/approximate-distribution-for-sum-of-squares-of-standardized-poisson-random-varia | 1,568,638,313,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572556.54/warc/CC-MAIN-20190916120037-20190916142037-00068.warc.gz | 678,936,158 | 31,901 | # Approximate distribution for sum of squares of standardized Poisson random variables
Suppose that $$X_1, ..., X_n$$ are independent and identically distributed Poisson($$\lambda$$) random variables.
What is a good approximating distribution for $$\sum_{i = 1}^{200} \frac{(X_i - \lambda)^2}{\lambda}$$?
I think that it is $$\chi^2_{200}$$, because it is a sum of the squares of standardized normal random variables. However, I'm not sure if they really are standardized normal random variables.
• They (the $(X_i-\lambda)/\sqrt{\lambda}$) are not standardized normal random variables: they are standardized Poisson variables! There are many possible answers. Which are good ones depend on what you mean by a "good" approximation: could you tell us what aspects of the approximation need to be good and how you measure how good they are? – whuber Apr 24 at 3:14
• I don't have any criteria for "good". I'm eager to learn all the possibilities. – MSE Apr 24 at 3:17
• There are innumerable possibilities. Perhaps you could tell us the context in which this question arises or the actual statistical problem you are dealing with? – whuber Apr 24 at 3:29
• Two possibilities out of "innumerably many": If $\lambda$ is large then $X_i$ are nearly normal, $Z_i =(X_i - \lambda)/\sqrt{\lambda}$ are nearly standard normal, and the sum of 200 $Z_i^2$ is nearly CHISQ(200), which in turn is nearly normal. (Why and with what $\mu$ and $\sigma$ ?). // If $\lambda$ is very small, then the $X_i$ are mostly zeros and ones, and normal by CLT seems a better fit to the sum of 200. – BruceET Apr 24 at 7:15
• This question is from an exam in a second-year course in mathematical statistics. There is no more context. The question actually asks for the expected value of the approximate distribution. – MSE Apr 24 at 13:28
There are many possible approximations. But I guess the idea here is simply CLT.
Consider that, for large $$n$$, the sum $$Z= \sum_n Y_i$$ tends to a normal $$N(n\mu ,n \sigma^2 )$$ if $$Y_i$$ are iid variables of mean $$\mu$$ and variance $$\sigma^2$$.
In our case $$Y_i = \frac{(X_i-\lambda)^2}{\lambda}$$, with $$X_i$$ being a $$\lambda-$$Poisson.
We only need to calculate the expectation and variance of $$Y_i$$, I leave that to you.
Notice that the approximation is obviously wrong in some respects: in particular, $$Z$$ is non-negative and discrete.
I'm not sure if [it's the sum of the squares of] standardized normal random variables.
No, of course not. $$\frac{(X_i-\lambda)}{\sqrt{\lambda}}$$ is a "standardized" Poisson variable. Hence it not justified to assume that $$Z$$ is $$\chi^2_{200}$$.
• Ah - that makes sense. Thanks for providing a clear, concise, and straightforward answer to my question. I knew that it would not be so complicated! – MSE May 6 at 17:06 | 745 | 2,800 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-39 | latest | en | 0.923496 |
https://lanl.github.io/LaGriT/pages/new_md/defineev | 1,555,583,137,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517558.8/warc/CC-MAIN-20190418101243-20190418123243-00367.warc.gz | 480,244,589 | 3,603 | 2. Define an enclosing volume
Define an enclosing volume using thesurface command. Since we are defining an exterior boundary, the boundary type is reflect. The next item of information needed is the geometry of the volume; some common geometry types are box, cylinder, sphere. Geometry types, box and sphere, define closed volumes; whereas a cylinder is open on both ends and must be capped by planes. Along with the geometry type, the extent of the volume is defined by specifying for the box its corners, or for the cylinder its radius and end point of its axis of rotation. The enclosing volume must be convex. Complicated enclosing volumes can be described by their bounding surfaces including planes and sheets . Some simple examples of enclosing volumes are:
• unit cube
surface/cube/reflect /box/0.0,0.0,0.0/1.0,1.0,1.0
“237” “211”
• cylinder whose axis is the x axis with radius 1 and height 1
surface/cyl_vol/reflect/cylinder/0.,0.,0./1.,0.,0./1.
surface/end1/reflect/plane/0.,0.,0./0.,0.,1./0.,1.,1.
surface/end2/reflect/plane/1.,0.,0./1.,0.,1./1.,1.,1.
“225” “177”“238” “247” | 294 | 1,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-18 | longest | en | 0.909511 |
https://www.answers.com/Q/How_many_hours_and_minutes_are_in_166_minutes | 1,606,735,880,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141213431.41/warc/CC-MAIN-20201130100208-20201130130208-00175.warc.gz | 564,810,168 | 38,991 | Math and Arithmetic
Time
How many hours and minutes are in 166 minutes?
434445
2010-08-24 12:47:29
2 hours and 46 minutes
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0
๐คจ
0
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Related Questions
There are 996 minutes in 166 hours (166X60).
60 minutes=1 hour 166 minutes= 2 hours 46 minutes
166 minutes is 2 hours and 46 minutes. There are 24 hours is one day. Therefore 166 minutes isn't even half a day.
With 60 minutes in an hour, do 9999/60 = 166.65 hours, OR 166 hr 39 min
60 minutes = 1 hour 10,000 minutes = (10,000 / 60) = 166 and 2/3 hours
It is 2 hours and 46 minutes or 166 minutes according to Google Maps.
10,000 seconds is 166 minutes, 40 seconds. 10,000 hours is 600,000 minutes. Please specify what the 10,000 is.
166 days is 5.5 months166 days is 3,984 hours 166 days is 239,040 minutes 166 days is 14,342,400 seconds
At what speed? 20 MPH, 498 minutes 60 MPH, 166 minutes 75 MPH, 132.8 minutes
About 2 hours 33 minutes at 65 mph
Driving Distance: 166 miles Driving Time: 3 hours, 14 minutes
According to Google Maps, it's 166 miles, or 2 hours 15 minutes drive by car.
At an average speed of 55 MPH it is: 3.02 hours
166 divided by 24 is 6.916666666666 (six repeated). So about seven days
How many hours is 101 minutes
9452765265241675.42451740715746234651274650 hours in a minutes
The total distance between the two locations is 166 miles. It will take about 2 hours and 20 minutes.
the answer to how many hours in 750 minutes is 12 and a half hours
ANSWER:2 hours, 15 minutes. 180 minutes (Which is 3 hours)
8.5 hours is 8 hours and 30 minutes, or 510 minutes.
Math and ArithmeticChicagoTimeDriving TimesDistances and Travel TimesRoad DistanceGeorgia (US State)
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https://www.clutchprep.com/chemistry/practice-problems/77147/one-mole-of-an-ideal-gas-is-expanded-from-a-volume-of-1-00-liter-to-a-volume-of- | 1,603,523,350,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882102.31/warc/CC-MAIN-20201024051926-20201024081926-00481.warc.gz | 690,197,050 | 35,829 | # Problem: One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.41 liters against a constant external pressure of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore significant figures for this problem. (T= 300 K: 1 L•atm = 101.3 J) A. 375 J B. 751 J C. 225 x 103 J D. 852 J E. none of these
###### FREE Expert Solution
We’re being asked to calculate the work for the system if 1 mole of ideal gas expanded from a volume of 1.00 L to a volume of 8.41 L against a constant external pressure of 1.00 atm.
Recall that work (w) is given by:
$\overline{){\mathbf{w}}{\mathbf{=}}{\mathbf{-}}{\mathbf{P\Delta V}}}$
where:
P = pressure (in atm)
ΔV = Vfinal – Vinitial = change in volume (in L)
Also note that if work is:
• (+): work is done by the surroundings to the system (compression)
• (–): work is done by the system to the surroundings (expansion)
###### Problem Details
One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.41 liters against a constant external pressure of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore significant figures for this problem. (T= 300 K: 1 L•atm = 101.3 J)
A. 375 J
B. 751 J
C. 225 x 103
D. 852 J
E. none of these
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Internal Energy concept. You can view video lessons to learn Internal Energy. Or if you need more Internal Energy practice, you can also practice Internal Energy practice problems.
What is the difficulty of this problem?
Our tutors rated the difficulty ofOne mole of an ideal gas is expanded from a volume of 1.00 l...as low difficulty.
How long does this problem take to solve?
Our expert Chemistry tutor, Dasha took 2 minutes and 58 seconds to solve this problem. You can follow their steps in the video explanation above.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Smith's class at UH. | 570 | 2,097 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-45 | latest | en | 0.883721 |
https://lists.mcs.anl.gov/pipermail/petsc-users/2022-April/045921.html | 1,709,211,484,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474808.39/warc/CC-MAIN-20240229103115-20240229133115-00841.warc.gz | 362,575,616 | 3,207 | # [petsc-users] MatSetNullSpace results in nan with MUMPS
Matthew Knepley knepley at gmail.com
Sun Apr 3 08:00:39 CDT 2022
```On Sat, Apr 2, 2022 at 8:59 PM Bhargav Subramanya <
bhargav.subramanya at kaust.edu.sa> wrote:
> Dear All,
>
> I am trying to solve Ax = b in parallel using MUMPS, where x is composed
> of velocity, pressure, and temperature. There is a null space due to the
> homogeneous Neumann pressure boundary conditions.
>
> 1. Without explicitly removing the null space, the solution is bounded and
> robust, and I am able to iterate in time, although I need to check if the
> solution is physically correct. I am not sure if MUMPS is able to
> detect the null space in this case.
>
MUMPS should fail with a 0 pivot. I do not understand this.
> 2. However, when I explicitly use MatSetNullSpace as shown below, the
> solution blows up in 2-3 temporal iterations, resulting in nan. Could you
>
Did you check that this is a nullspace vector using
https://petsc.org/main/docs/manualpages/Mat/MatNullSpaceTest.html ?
Thanks,
Matt
> Thanks,
> Bhargav
>
> ---------------
> Output for -pc_type svd -pc_svd_monitor
> SVD: condition number 3.159257564649e+16, 1 of 1080 singular values are
> (nearly) zero
> SVD: smallest singular values: 3.693890048955e-17 2.902795363876e-10
> 2.902795486255e-10 2.902796682780e-10 7.614221123356e-10
> SVD: largest singular values : 1.166981958972e+00 1.166981958972e+00
> 1.166995008014e+00 1.166995008014e+00 1.166995008014e+00
> -----------------
>
> PetscErrorCode *SetNullVector*(Mat MatGlobalLinear,Mesh mesh,Equations
> eq){
>
> Vec NullVector;
>
> PetscInt numNullVectors,disp,i,j;
>
> PetscScalar constantValue = 1.0,nrm;
>
> MatNullSpace NullSp = NULL;
>
> PetscErrorCode ierr;
>
>
> PetscFunctionBeginUser;
>
> numNullVectors = 1;
>
> disp = mesh.numLayers*mesh.numEdges + (eq.pressure-1)*mesh.numCells;
>
> ierr = *MatCreateVecs*(MatGlobalLinear,&NullVector,NULL);CHKERRQ(ierr);
>
> *for*(i=0; i<mesh.numLayers; i++){
>
> *for*(j=0; j<mesh.numCells; j++){
>
> ierr = VecSetValue(NullVector,j + i*mesh.numCells + disp,constantValue,
> *INSERT_VALUES*);CHKERRQ(ierr);
>
> }
>
> }
>
> ierr = *VecAssemblyBegin*(NullVector);CHKERRQ(ierr);
>
> ierr = *VecAssemblyEnd*(NullVector);CHKERRQ(ierr);
>
> ierr = *VecNorm*(NullVector,*NORM_2*,&nrm);CHKERRQ(ierr);
>
> ierr = *VecScale*(NullVector,1.0/nrm);CHKERRQ(ierr);
>
> /* set null space */
>
> ierr = *MatNullSpaceCreate*(MPI_COMM_WORLD,*PETSC_FALSE*
> ,numNullVectors,&NullVector,&NullSp);CHKERRQ(ierr);
>
> ierr = *MatSetNullSpace*(MatGlobalLinear,NullSp);CHKERRQ(ierr);
>
> /* clean up */
>
> ierr = *MatNullSpaceDestroy*(&NullSp);CHKERRQ(ierr);
>
> ierr = *VecDestroy*(&NullVector);CHKERRQ(ierr);
>
> PetscFunctionReturn(0);
>
> }
> -----------------
>
>
> ------------------------------
> This message and its contents, including attachments are intended solely
> for the original recipient. If you are not the intended recipient or have
> received this message in error, please notify me immediately and delete
> this message from your computer system. Any unauthorized use or
> distribution is prohibited. Please consider the environment before printing
> this email.
--
What most experimenters take for granted before they begin their
experiments is infinitely more interesting than any results to which their | 971 | 3,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.629324 |
https://www.scribd.com/document/280611135/Slope-Def-Vtu | 1,566,648,277,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320734.85/warc/CC-MAIN-20190824105853-20190824131853-00253.warc.gz | 958,530,448 | 72,730 | You are on page 1of 89
# Chapter-2: Slope Deflection Method
By Prof. H.P.Sudarshan
Sri Siddhartha Institute Of Tech,
Tumkur
## Example: Analyze the propped cantilever shown by using slope defection
method. Then draw Bending moment and shear force diagram.
## Solution: End A is fixed hence A =0
End B is Hinged hence B 0
Assume both ends are fixed and therefore fixed end moments are
wL2
wL2
FAB
, FBA
12
12
The Slope deflection equations for final moment at each end are
2EI
2A B
MAB FAB
L
wL2 2EI
B
(1)
12
L
2EI
2B A
MBA FBA
L
wL2 4EI
B
( 2)
12
L
In the above equations there is only one unknown B .
To solve we have boundary condition at B;
Since B is simply supported, the BM at B is zero
ie. MBA=0.
From equation (2) MBA
wL2 4EI
B 0
12
L
wL3
EIB
48
Substituting the value of EIB in equation (1) and (2) we have end moments
wL2 2 wL3
wL2
12 L 48
8
wL2 4 wL3
0
12 L 48
MAB
MBA
## MBA has to be zero, because it is hinged.
Now consider the free body diagram of the beam and find reactions using
equations of equilibrium.
MB 0
R A L MAB wL
L
2
wL2
L
5
wL wL
8
2
8
5
R A wL
8
V 0
R A RB wL
RB wL R A wL
3
wL
8
5
wL
8
## The max BM occurs where SF=0. Consider SF equation at a distance of x
from right support
3
wL wX 0
8
3
X L
8
SX
Mmax
3
L from support B
8
3
3
w 3
MX wL L
L
8
8
2 8
9
wL2
128
## And point of contra flexure occurs where BM=0, Consider BM equation at
a distance of x from right support.
3
X2
wLX w
0
8
2
3
X L
4
MX
## For shear force diagram, consider SF equation from B
3
S X wL wX
8
3
S X 0 SB wL
8
5
S X L S A wL
8
Example: Analyze two span continuous beam ABC by slope deflection method.
Then draw Bending moment & Shear force diagram. Take EI constant
## Solution: Fixed end moments are:
Wab 2
100 4 22
44.44KNM
L2
62
Wa 2b
100 42 2
88.89KNM
L2
62
wL2
20 52
41.67KNM
12
12
wL2 20 52
41.67KNM
12
12
FAB
FBA
FBC
FCB
Since A is fixed A 0 , B 0, C 0,
Slope deflection equations are:
2EI
2 A B
L
2EI
44.44
B
6
1
44.44 EI B
3
2EI
2B A
MBA FBA
L
2EI 2B
88.89
6
2
88.89 EIB
3
MAB FAB
(1)
( 2)
2EI
2B C
L
2EI
2B C
41.67
5
4
2
41.67 EIB EIC
5
5
2EI
2C B
MCB FCB
L
2EI
2C B
41.67
5
4EI
2
41.67
C EIB
5
5
MBC FBC
(3 )
( 4)
In all the above four equations there are only two unknown B and C . And
accordingly the boundary conditions are
-MBA-MBC=0
MBA+MBC=0
ii
MCB=0
## since C is end simply support.
2
4
2
EIB 41.67 EIB EIC
3
5
5
22
2
47.22
EIB EIC 0
(5)
15
5
2
4
MCB 41.67 EIB EIC 0
(6)
5
5
Now
## Solving simultaneous equations 5 & 6 we get
EI B = 20.83
Rotation anticlockwise.
EI C = 41.67
Rotation anticlockwise.
## Substituting in the slope definition equations
1
MAB = 44.44 + 20.83 51.38 KNM
3
2
MBA = + 88.89 + 20.83 75.00 KNM
3
4
20.83 2 41.67 75.00 KNM
5
5
2
4
MCB = + 41.67+ 20.83 41.67 0
5
5
MBC = 41.67+
## Find reactions using equations of equilibrium.
Span AB:
MA = 0
RB6 = 1004+75-51.38
RB = 70.60 KN
V = 0
RA+RB = 100KN
RA = 100-70.60=29.40 KN
Span BC:
MC = 0
5
RB5 = 205 +75
2
RB = 65 KN
V=0
## RB+RC = 205 = 100KN
RC = 100-65 = 35 KN
## Using these data BM and SF diagram can be drawn.
Max BM:
Span AB: Max BM in span AB occurs under point load and can be found
geometrically
Mmax=113.33-51.38 -
6
## Span BC:Max BM in span BC occurs where shear force is zero or
changes its sign. Hence consider SF equation w.r.t C
Sx = 35-20x = 0
35
=1.75m
20
## Max BM occurs at 1.75m from C
Mmax = 35 1.75 20
1.75 2
= 30.625 KNM
2
Example: Analyze continuous beam ABCD by slope deflection method and then
draw bending moment diagram. Take EI constant.
Solution:
A 0, B 0, C 0
Wab 2
100 4 22
- 44.44 KN M
L2
62
Wa 2b
100 42 2
FBA
88.88 KNM
L2
62
F AB
FEMS
wL2
20 5 2
F BC
- 41.67 KNM
12
12
wL2
20 5 2
F CB
41.67 KNM
12
12
FCD 20 1.5 - 30 KN M
## Slope deflection equations:
MAB F AB
MBA FBA
2EI
2A B 44.44 1 EIB
L
3
2EI
2B A 88.89 2 EIB
L
3
- - - - - - - -- 1
- - - - - - - -- 2
2EI
2B C 41.67 4 EIB 2 EIC
L
5
5
- - - - - - - - 3
2EI
2C B 41.67 4 EIC 2 EIB
L
5
5
MCD 30 KNM
- - - - - - - - 4
MBC FBC
MCB FCB
## In the above equations we have two unknown rotations B and C , accordingly
the boundary conditions are:
MBA MBC 0
MCB MCD 0
2
4
2
EIB 41.67 EIB EIC
3
5
5
22
2
47.22
EIB EIC 0
15
5
## Now, MBA MBC 88.89
4
2
EIC EIB 30
5
5
2
4
11.67 EIB EIC
5
5
- - - - - - - - 5
## EI B 32.67 Rotation @ B anticlockw ise
EI C 1.75
Rotation @ B clockwise
## Substituting value of EIB and EIC in slope deflection equations we have
1
MAB 44.44 32.67 61.00 KNM
2
2
MBA 88.89 32.67 67.11 KNM
3
4
2
MBC 41.67 32.67 1.75 67.11 KNM
5
5
4
2
MCB 41.67 1.75 32.67 30.00 KNM
5
5
MCD 30 KNM
## Reactions: Consider free body diagram of beam AB, BC and CD as shown
Span AB
RB 6 100 4 67.11 61
RB 67.69 KN
R A 100 RB 32.31 KN
Span BC
5
RC 5 20 5 30 67.11
2
RC 42.58 KN
RB 20 5 RB 57.42 KN
## Maximum Bending Moments:
Span AB: Occurs under point load
67.11 61
Max 133.33 61
4 68.26 KNM
6
## Span BC: where SF=0, consider SF equation with C as reference
S X 42.58 20 x 0
x
42.58
2.13 m
20
2.13 2
Mmax 42.58 2.13 20
30 15.26 KN M
2
## Example: Analyse the continuous beam ABCD shown in figure by slope
deflection method. The support B sinks by 15mm.
Take E 200 105 KN / m2 and I 120 10 6 m4
Solution:
In this problem A =0, B 0, C 0, =15mm
FEMs:
FAB
FBA
Wab 2
44.44 KNM
L2
Wa 2 b
88.89 KNM
L2
wL2
41.67 KNM
8
wL2
41.67 KNM
8
FBC
FCB
6EI
L2
6 200
15
105 120 10 6
6 KNM
2
6
1000
mab mba
## For span BC:
6EI
L2
6 200
15
105 120 10 6
8.64KNM
2
5
1000
Slope deflection equation
mbc mcb
2EI
3
(2 A B )
L
L
EI
6EI
F AB 2 A B 2
L
L
1
- 44.44 EIB 6
3
1
50.44 EIB
3
2EI
6EI
MBA FBA
(2B A ) 2
L
L
2
88.89 EIB 6
3
2
82.89 EIB
3
2EI
6EI
MBC FBC
(2B C ) 2
L
L
2
- 41.67 EI2B C 8.64
5
4
2
33.03 EIB EIC
5
5
2EI
6EI
MCB FCB
(2C B ) 2
L
L
2
41.67 EI2C B 8.64
5
4
2
50.31 EIC EIB
5
5
MCD 30 KNM
MAB F AB
- - - - - - - -- 1
- - - - - - - -- 2
- - - - - - - -- 3
- - - - - - - -- 4
- - - - - - - -- 5
There are only two unknown rotations B and C . Accordingly the boundary
conditions are
MBA MBC 0
MCB MCD 0
## Now, MBA MBC 49.86 22 EI B 2 EI C 0
15
5
2
4
MCB MCD 20.31 EI B EI C 0
5
5
Solving these equations we get
EI B 31.35 Anticl ockwise
## EI C 9.71 Anticl ockwise
Substituting these values in slope deflections we get the final moments:
1
31.35 60.89 KNM
3
2
MBA 82.89 31.35 61.99 KNM
3
4
2
MBC 33.03 31.35 9.71 61.99 KNM
5
5
4
2
MCB 50.31 9.71 31.35 30.00 KNM
5
5
MCD 30 KNM
MAB 50.44
Consider the free body diagram of continuous beam for finding reactions
Reactions:
Span AB:
RB 6 = 100 x 4 + 61.99 60.89
RB = 66.85
RA = 100 RB
=33.15 KN
Span BC:
RB 5 = 20 x 5 x
RB = 56.40 KN
RC = 20 x 5 - RB
=43.60 KN
5
+ 61.99 30
2
Example: Three span continuous beam ABCD is fixed at A and continuous over
B, C and D. The beam subjected to loads as shown. Analyse the beam by slope
deflection method and draw bending moment and shear force diagram.
Solution:
Since end A is fixed A 0, B 0, c 0, D 0
FEMs:
Wl
60 4
- 30 KNM
8
8
Wl
60 4
FBA
30 KNM
8
8
M
FBC 12.5 KNM
4
M
FCB 12.5 KNM
4
wl2
10 42
FCD
- 13.3 3 KNM
12
12
wl2
10 42
FDC
13.33 KNM
12
12
F AB
## Slope deflection equations:
2EI
2 A B
L
2EI
0 B
- 30
4
- 30 0.5EIB
2EI
2B A
MBA F BA
L
2EI
2B 0
30
4
30 EIB
M AB F AB
- - - - - - - - 1
- - - - - - - -- 2
2EI
2B C
L
2EI
2B C
12.5
4
12.5 EIB 0.5EIC
2EI
2 C B
MCB F CB
L
2EI
2 C B
12.5
4
12.5 EIC 0.5 EIB
2EI
2 C D
MCD F CD
L
2EI
2 C D
- 13.33
4
13.33 EI C 0.5EI D
- - - - - - - - - - 5
2EI
2D C
L
2EI
2D C
13.33
4
13.33 0.5EI C EI D
- - - - - - - - - - 6
MBC F BC
- - - - - - - -- 3
- - - - - - - -- 4
MDC F DC
## In the above Equations there are three unknowns, EI B ,EI C & EI D ,
accordingly the boundary conditions are:
MBA MBC 0
ii MCB MCD 0
iii MDC 0
Now
( hinged)
MBA MBC 0
30 EIB 12.5 EIB 0.5EIC 0
2EIB 0.5EIC 42.5 0
MCB MBC 0
12.5 EIC 0.5EIB 13.33 EIC 0.5EID 0
0.5EIB 2EIC 0.5EID 0.83 0
MDC 0
13.33 0.5EIC EID 0
8
9
## By solving (7), (8) & (9), we get
EI B 24.04
EI C 11.15
EI D 18.90
By substituting the values of B , c and D in respective equations we get
## MAB 30 0.5 24.04 42.02 KNM
MBA 30 24.04 5.96 KNM
## MBC 12.5 - 24.04 0.5 11.15 - 5.96 KNM
MCB 12.5 11.15 0.5 24.04 11.63 KNM
MCD 13.33 11.15 0.5 18.90 11.63 KNM
MDC 13.33 0.511.15 18.90 0 KNM
Reactions: Consider the free body diagram of beam.
Beam AB:
60 2 5.96 42.02
20.985 KN
4
R A 60 RB 30.015 KN
RB
Beam BC:
11.63 50 5.96
13.92 KN
4
RB RC 13.92 KN RB is downward
RC
Beam CD:
10 4 2 11.63
17.09 KN
4
RC 10 4 RD 22.91 KN
RD
Example: Analyse the continuous beam shown using slope deflection method.
Then draw bending moment and shear force diagram.
FEMs:
wl2
10 82
- 53.33 KNM
12
12
wl2
FBA
53.33 KNM
12
Wl
30 6
FBC
- 22.50 KNM
8
8
WL
FCD
22.50 KNM
8
F AB
## Slope deflection equations:
2EI
2 A B
L
2E 3I
0 B
- 53.33
8
3
- 53.33 EIB
4
M AB F AB
2EI
2B A
L
2E 3I
2B 0
53.33
8
3
53.33 EIB
2
- - - - - - - - 1
MBA F BA
- - - - - - - - 2
2EI
2B C
L
2E2I
2B C
- 22.5
6
4
2
- 22.5 EIB EIC
3
3
MBC F BC
2EI
2 C B
L
2E2I
2 C B
22.5
6
4
2
22.5 EIC EIB
3
3
- - - - - - - - 3
MCB F CB
- - - - - - - - 4
In the above equation there are two unknown B and C , accordingly the
boundary conditions are:
i MBA MBC 24 0
ii MCB 0
3
4
2
EIB 22.5 EIB EIC 24
2
3
3
17
2
54.83
EIB EIC 0
5
6
3
## and MCB 22.5
4
2
EIC EIB 0
3
3
2
1
EIC 11.25 EIB
3
3
- - - - - - - - - -- (6)
## Substituting in eqn. (5)
17
1
EIB 11.25 EIB 0
6
3
15
44.58
EIB 0
6
44.58 6
EIB
17.432 rotation anticlockwise
15
54.83
3
1
11.25 17.432
2
3
## 8.159 rotation anticlockw ise
EIC
Substituting EIB 17.432 and EIC 8.159 in the slope deflection equation
we get Final Moments:
3
MAB 53.33 - 17.432 -66.40 KNM
4
3
MBA 53.33 17.432 27.18 KNM
2
4
17.432 2 8.159 51.18 KNM
3
3
4
2
MCB 22.5 8.159 ( 17.432) 0.00
3
3
MBC 22.5
## Reactions: Consider free body diagram of beams as shown
Span AB:
27.18 66.40 10 8 4
35.13 KN
8
R A 10 8 RB 44.87 KN
RB
Span BC:
51.18 30 3
23.53 KN
6
RC 30 RB 6.47 KN
RB
Max BM
Span AB: Max BM occurs where SF=0, consider SF equation with A as origin
S x 44.87 - 10x 0
x 4.487 m
4.487 2
M max 44.87 4.487 10
64 36.67 KNM
2
Span BC: Max BM occurs under point load
51.18
BC Mmax 45
19.41 KN M
2
## Example: Analyse the beam shown in figure. End support C is subjected to an
anticlockwise moment of 12 KNM.
FEMs:
wl2
20 42
26.67 KNM
12
12
wl2
FCB
26.67 KNM
12
FBC
## Slope deflection equations:
2EI
2 A B
M AB F AB
L
2E2I
0 B
0
4
EIB
2EI
2B A
MBA F BA
L
2E2I
2B 0
0
4
2EIB
2EI
2B C
MBC F BC
L
2E 1.5I
2B C
- 26.67
4
3
3
- 26.67 EIB EIC
2
4
2EI
2 C B
MCB F CB
L
2E 1.5I
2C B
26.67
4
3
3
26.67 EIC EIB
2
4
- - - - - - - -- 1
- - - - - - - -- 2
- - - - - - - -- 3
- - - - - - - -- 4
In the above equation there are two unknowns B and C , accordingly the
boundary conditions are
MBA MBC 0
MCB 12 0
Now, MBA MBC 2EIB 26.67
3
3
EIB EIC
2
4
7
3
EIB EIC 26.67 0
2
4
3
3
and, MCB 12 26.67 EIC EIB 12
2
4
3
3
38.67 EIB EIC 0
4
2
From (5) and (6)
7
3
EIB EIC 26.67 0
2
4
3
3
EIB EIC 19.33 0
8
4
- - - - - - - -- (5)
- - - - - - - -- (6)
25
EIB 46 0
8
EIB 46
8
14.72
25
From (6)
2
3
38.67 14.72
3
4
33.14
EIC
## Substituti ng EIB and EIC is slope deflection equations
MAB EIB 14.72 KNM
MBA 2EIB 2(14.72) 29.42 KNM
3
3
(14.72) 33.14 29.44 KNM
2
4
3
3
MCB 26.67 ( 33.14 ) (14.72) 12 KNM
2
4
MBC 26.67
## Reaction: Consider free body diagrams of beam
Span AB:
14.72 29.44
11.04 KN
4
R A RB 11.04 KN
RB
Span BC:
29.44 12 20 4 2
50.36 KN
4
RC 20 4 RB 29.64 KN
RB
Example: Analyse the simple frame shown in figure. End A is fixed and ends B &
C are hinged. Draw the bending moment diagram.
Solution:
In this problem A 0, B 0, C 0, D 0,
FEMS:-
FAB
FBA
FBC
FCB
FCD
FDB
Wab 2
120 2 42
106.67 KNM
L2
62
Wa 2b
120 22 4
53.33 KNM
L2
62
wl2
20 42
26.67 KNM
12
12
wl2
20 42
26.67 KNM
12
12
WL
20 4
10 KNM
8
8
WL
10 KNM
8
## Slope deflections are
2EI
2A B
MAB FAB
L
2E2I
B 106.67 2 EIB
106.67
6
3
(1 )
2EI
2B B
L
2E2I
2B 53.33 4 EIB
53.33
6
3
2EI
2B C
MBC FCB
L
2E 3I
3
3
26.67
2B C 26.67 EIB EIC
4 2
2
4
2EI
2C B
MCB FCB
L
2E 3I
3
3
26.67
2C B 26.67 EIC EIB
4 2
2
4
MBA FBA
2EI
2B D
L
2EI
2B D 10 EIB 1 EID
10
4
2
2EI
2D B
MDB FDB
L
2EI
2D B 10 EID 1 EIB
10
4
2
(2 )
(3 )
(4 )
MBD FBD
(5 )
(6 )
## In the above equations we have three unknown rotations B , C , D accordingly
we have three boundary conditions.
MBA MBC MBD 0
MCB 0
Since C and D are hinged
MDB 0
Now
4
3
3
1
MBA MBC MBD 53.33 EIB 26.67 EIB EIC 10 EIB EID
3
2
4
2
23
3
1
36.66
EIB EIC EID 0
- - - -- (7)
6
4
2
3
3
MCB 26.67 EIB EIC 0
- - - -- (8)
4
2
1
MDB 10 EIB EID 0
- - - -- (9)
2
## Solving equations 7, 8, & 9 we get
EIB 8.83
EIC 13.36
EID 14.414
Substituting these values in slope equations
M AB 106.67
4
( 8.83 ) 41.56 KNM
3
3
3
26.67 ( 8.3) ( 13.36 ) 49.94 KNM
2
4
3
3
26.67 ( 13.36 ) ( 8.83 ) 0
2
4
1
10 ( 8.83 ) ( 14.414 ) 8.38 KNM
2
1
10 (14.414 ) ( 8.83 ) 0
2
MBA 53.33
MBC
M CB
MBD
MDB
2
( 8.83 ) 112.56 KNM
3
## Reactions: Consider free body diagram of each members
Span AB:
41.56 112.56 120 2
28.17 KN
6
R A 120 RB 91.83 KN
RB
Span BC:
49.94 20 4 2
52.485 KN
4
RC 20 4 RB 27.515 KN
RB
Column BD:
20 2 8.33
7.92 KN
4
HA HD 20
HB 12.78 KN
HD
Example: Analyse the portal frame shown in figure and also drawn bending
moment and shear force diagram
Solution:
Symmetrical problem
A 0, B 0, C 0, D 0
FEMS
W 1ab 2 W 2 cd2
L2
L2
80 2 42 80 4 22
- 106.67 KNM
62
62
Wa 2b W 2 c 2d
FCB
106.67 KNM
L2
L2
FBC
## Slope deflection equations:
2EI
2A B 0 2EI 0 B 1 EIB
L
4
2
2EI
2EI
2B A 0 2B 0 EIB
MBA FBA
L
4
MAB F AB
- - - - - - - - 1
- - - - - -- 2
2EI
2B C
L
2E2I
4
2
106.67
(2B C ) 106.67 EIB EIC - - - - - - 3
6
3
3
2EI
2C B
MCB FCB
L
2E2I
4
2
106.67
(2C B ) 106.67 EIC EIB - - - - - - 4
6
3
3
2EI
2C D
MCD FCD
L
2EI
0
(2C 0) EIC
- - - - - -- 5
4
MBC FBC
2EI
2D C
L
2EI
1
0
(0 C ) EIC
4
2
MDC FDC
- - - - - -- 6
In the above equation there are two unknown rotations. Accordingly the boundary
conditions are
MBA MBC 0
MCB MCD 0
7
2
Now MBA MBC 106.67 EIB EIC 0
3
3
2
7
MCB MCD 106.67 EIB EIC 0
3
3
Multiply by (7) and (8) by 2
49
14
EIB
EIC 0
3
3
subtracts
4
14
213.34 EIB
EIC 0
3
3
746.69
45
EIB 0
3
3
EIB 960.03
64
45
- 960.03
Clockwise
- - - - - -- (7)
- - - - - -- (8)
EIC
-
3
7
106.67 EIB
2
3
3
7
2
3
## Here we find B C . It is obvious because the problem is symmetrical.
Final moments are
64
32 KNM
2
MBA 64 KNM
M AB
4
2
64 64 64 KNM
3
3
4
2
MCB 106.67 ( 64 ) 64 64 KNM
3
3
MCD 64 KNM
MBC 106.67
MDC
1
64 -32 KNM
2
## By symmetrical we can write
R A RB 60 KNM
RD RC 80 KNM
Now consider free body diagram of column AB
Apply
MB 0
HA 4 64 32
HA 24 KN
## Similarly from free body diagram of column CD
Apply
MC 0
HA 4 64 32
HD 24 KN
Check:
H 0
HA HD 0
Hence okay
Note: Since symmetrical, only half frame may be analysed. Using first three
equations
and taking B C
Example: Analyse the portal frame and then draw the bending moment diagram
Solution:
This is a symmetrical frame and unsymmetrically loaded, thus it is an
unsymmetrical problem and there is a sway
Assume sway to right.
Here A 0,D 0,B 0,D 0
FEMS:
Wab 2
80 5 32
56.25 KNM
L2
82
Wa 2b
80 52 3
FCB
93.75 KNM
L2
82
FBC
## Slope deflection equations
MAB F AB
0
MBA FBA
0
2EI
3
2 A B
L
L
2EI
3 1
3
0 B EIB EI
4
4 2
8
- - - - - - - - 1
2EI
3
2B A
L
L
2EI
3
3
2B 0 EIB EI
4
4
8
- - - - - - - - 2
2EI
2B C
L
2EI
2B C 56.25 1 EIB 1 EIC
56.25
8
2
4
2EI
2C B
MCB FCB
L
2EI
2C B 93.75 1 EIC 1 EIB
93.75
8
2
4
2EI
3
MCD FCD
2C D
L
L
MBC FBC
0
MDC FDC
0
2EI
3
3
2C 0 EIC EI
4
4
8
- - - - - - - -- 3
- - - - - - - -- 4
- - - - - - - -- 5
2EI
3
2D C
L
L
2EI
3 1
3
0 C EIC EI
4
4 2
8
- - - - - - - -- 6
In the above equation there are three unknowns B , C and , accordingly the
boundary conditions are,
MBA MBC 0
Jo int conditions
MCB MCD 0
HA HD PH 0
- - - Shear condition
## MAB MBA MCD MDC
0
4
4
MAB MBA MCD MDC 0
3
1
1
Now, MBA MBC EIB EI 56.25 EIB EIC 0
8
2
4
3
1
3
56.25 EIB EIC EI 0
2
4
8
1
1
3
And, MCB MCD 93.75 EIC EIB EIC EI 0
2
4
8
1
3
3
93.75 EIB EIC EI 0
4
2
8
i.e,
1
3
3
3
EIB EI EIB EI EIC EI
2
8
8
8
1
3
EIC EI
2
8`
3
3
3
EIB EIC EI 0
9
2
2
2
EI EIB EIC
From (9)
## Substitute in (7) & (8)
Eqn (7)
3
1
3
EIB EIC EIB EIC 0
2
4
8
9
1
56.25 EIB EIC 0
8
8
56.25
- - - - - -- 10
Eqn(8)
1
3
3
EIB EIC EIB EIC 0
4
2
8
1
9
93.75 EIB EIC 0
8
8
93.75
- - - - - - - - - - 11
By Equation (10)
8
8
Hence
## EIB 41.25, EIC 78.75, EI 37.5
Substituting these values in slope deflection equations, we have
1
41.25 3 37.5 34.69 KNM
2
8
3
MBA 41.25 37.5 55.31 KNM
8
1
1
MBC 56.25 41.25 78.75 55.31 KNM
2
4
1
1
M CB 93.75 78.75 41.75 64.69 KNM
2
4
3
M CD 78.75 37.5 64.69 KNM
8
1
3
MDC 78.75 37.5 25.31 KNM
2
8
M AB
Column AB:
HA
34.69 55.31
22.5 KN
4
Span BC:
55.31 64.69 80 3
28.83 KN
8
RC 80 RB 51.17
RB
Column CD:
HD
64.69 25.31
22.5
4
Check:
H = 0
HA + HD = 0
22.5 22.5 = 0
Hence okay
## Example: Frame ABCD is subjected to a horizontal force of 20 KN at joint C as
shown in figure. Analyse and draw bending moment diagram.
Solution:
Frame is Symmetrical and unsymmetrical loaded hence there is a sway.
Assume sway towards right
FEMS
F AB F BA F BC F CB F CD F DC 0
Slope deflection equations are
2EI
3
MAB F AB
2 A B
L
L
2EI
3
B
3
3
2
2
EIB EI
3
3
2EI
3
MBA FBA
2B A
L
L
2EI
3
2B
3
3
4
2
EIB EI
3
3
- - - - - - - -- 1
MBC FBC
2EI
2B C
L
2EI
2B C
4
EIB 0.5 EIC
MCB FCB
- - - - - - - - 3
2EI
2C B
L
2EI
2C B
4
EIC 0.5 EIB
MCD FCD
- - - - - - - - 4
2EI
3
2C D
L
L
2EI
3
2C
3
3
4
2
EIC EI
3
3
2EI
3
MDC FDC
2D C
L
L
2 EI
3
c
3
3
2
2
EIC EI
3
3
- - - - - - - -- 6
I. MBA MBC 0
I I. MCB MCD 0
III. HA HD 20 0
i.e,
## MAB MBA MCD MDC
20 0
3
3
MAB MBA MCD MDC 60 0
4
2
EIB EI EIB 0.5EIC
3
3
7
2
EIB 0.5EIC EI 0 7
3
3
## Now MBA MBC
4
2
EIC EI
3
3
7
2
0.5EIB EIC EI 0 8
3
3
## and MAB MBA MCD MDC 60
2
2
4
2
4
EIB EI EIB EI EIC
3
3
3
3
3
2
2
2
EI EIC EI 60
3
3
3
8
2EIB 2EIC EI 60 0 9
3
## Solving (7).(8) & (9) we get
EI B 8.18,
EI C 8.18,
EI 34.77
Substituting the value of B , C and in slope deflection equations
2
8.18 2 34.77 17.73 KNM
3
3
4
2
MBA 8.18 34.77 12.27 KNM
3
3
MBC 0 8.18 0.5 8.18 12.27 KNM
MAB
## MCB 0.5 8.18 8.18 12.27
KNM
4
8.18 2 34.77 12.27 KNM
3
3
2
2
MDC 8.18 34.77 17.73 KNM
3
3
MCD
Member AB:
17.73 12.27
HA
10 KN
3
Member BC:
12.27 12.27
6.135 KN
4
RB RC 6.135 KN
RC
Member CD:
HD
17.73 12.27
10 KN
3
Check:
H = 0
HA + HD + P = 0
+10 + 10 20 = 0
Hence okay
## - ve sign indicates the direction of HD is left to right
Example: Analyse the portal frame subjected to loads as shown. Also draw
bending moment diagram.
## The frame is symmetrical but loading is unsymmetrical. Hence there is a sway.
Assume sway towards right. In this problem A 0, B 0, C 0, D 0
FEMs:
wl2
10 42
- 13.33 KNM
12
12
wl2
10 42
FBA
13.33 KNM
12
12
wl
90 10
FBC
- 112.5 KNM
8
8
wl 90 10
FCB
112.5 KNM
8
8
Slope deflection equations:
2EI
3
M AB F AB
2 A B
L
L
2EI
3
- 13.33
0 B
4
4
- 13.33 0.5 EIB 0.375 EI
2EI
3
MBA F BA
2 B A
L
L
F AB
2EI
3
2B 0
4
4
13.33 EIB 0.375 EI
- - - - - - - -- 1
13.33
- - - - - - - -- 2
2EI
2B C
L
2E3I
2B C
- 112.5
10
- 112.5 1.2EI B 0.6EIC
2EI
2 C B
MCB F CB
L
2E3I
2 C B
112.5
10
112.5 1.2EI C 0.6 EIB
MBC F BC
- - - - - - - -- 3
- - - - - - - -- 4
2EI
3
2 C D
L
L
2EI
3
0
2 C 0
4
4
EI C 0.375 EI
- - - - - - - -- 5
2EI
3
2 D C
L
L
2EI
3
0
0 2 C
4
4
0.5EI C 0.375 EI
- - - - - - - -- 6
MCD F CD
MDC F DC
There are 3 unknowns EIB, EIC and EI , accordingly the boundary conditions
are
MBA MBC 0
MCB MCD 0
HA HD 40 0
4
2
MAB MBA 80
4
and HD 4 MCD MBC
HA
HD
MCD MDC
4
## MAB MBA 80 MCD MDC
40 0
4
4
MAB MBA MCD MDC 80 0
## Now MBA + MBC = 0
13.33EIB 0.375EI 112.5 1.2EIB 0.6EIC 0
## 2.2EIB 0.6EIC 0.375EI 99.17 0
and MCB MDC 0
- - - - - - - - (7)
## 112.5 1.2EIC 0.6EIB EIC 0.375EI 0
112.5 2.2EIC 0.6EIB 0.375EI 0
also MAB MBA MCB MDC 80 0
- - - - - - - - (8)
## 13.33 0.5EIB 0.375EI 13.33 EIB 0.375EI EIC 0.375EI
0.5EIC 0.375EI 80 0
1.5EIB 1.5EIC -1.5EI 80 0
By solving (7), (8) and (9) we get
EI B 72.65
EI C 59.64
EI 66.34
Final moments:
## MAB 13.33 0.572.65 0.375 66.34 -1.88 KNM
MBA 72.65 0.37566.34 61.10 KNM
## MBC 112.5 1.272.65 0.6 59.64 61.10 KNM
MCB 112.5 1.2 59.64 0.672.65 84.52 KNM
MCD 59.64 0.375(66.34 ) 84.52 KNM
MDC 0.5( 59.64 ) 0.375(66.34 ) 54.70 KNM
- - - - - - - -- (9)
## Reactions: Consider the free body diagrams of various members
Member AB:
61.10 1.88 10 4 2
4
5.195 KN
- ve sign indicates direction of HA is from right to left
HA
Member BC:
84.52 61.10 90 5
47.34 KN
10
RB 90 RC 38.34 KN
RC
Member CD
HD
84.54 54.7
34.81 KN
4
Check
H = 0
HA + HD +10 4 = 0
-5.20 - 34.81+ 40 = 0
Hence okay
Example: Analyse the portal frame and then draw the bending moment diagram
Solution:
Since the columns have different moment of inertia, it is an unsymmetrical
frame. Assume sway towards right
FEMS:
WL
80 6
60 KNM
8
8
WL
FCB
60 KNM
8
Here A 0, D 0
FBC
## Slope deflection equations
MAB F AB
0
MBA FBA
0
2EI
3
2 A B
L
L
2EI
3 1
3
0 B EIB EI
4
4 2
8
- - - - - - - - 1
2EI
3
2B A
L
L
2EI
3
3
2B 0 EIB EI
4
4
8
- - - - - - - - 2
2EI
2B C
L
2E2I
2B C 60 4 EIB 2 EIC
60
6
3
3
2EI
2C B
MCB FCB
L
2E2I
2C B 60 2 EIB 4 EIC
60
6
3
3
2EI
3
MCD FCD
2C D
L
L
MBC FBC
0
MDC FDC
0
2E2I
3
3
2C 0 2EIC EI
4
4
4
- - - - - - - -- 3
- - - - - - - -- 4
- - - - - - - -- 5
2EI
3
2D C
L
L
2E2I
3
3
0 C EIC EI
4
4
4
- - - - - - - -- 6
In the above equation there are three unknowns B , C and , accordingly the
boundary conditions are,
MBA MBC 0
MCB MCD 0
Jo int conditions
HA HD 0
- - - Shear condition
## MAB MBA MCD MDC
0
4
4
MAB MBA MCD MDC 0
i.e,
3
4
2
EI 60 EIB EIC 0
8
3
3
7
2
3
60 EIB EIC EI 0
3
3
8
## Now, MBA MBC EIB
2
4
3
EIB EIC 2EIC EI 0
3
3
4
2
10
3
EIB
EIC EI 60 0
3
3
4
## And, MCB MCD 60
1
3
3
3
EIB EI EIB EI 2EIC EI
2
8
8
4
3
EIC EI
4
3
9
EIB 3EIC EI 0
9
2
4
4 3
EIB 3EIC
9 2
EI
From (9)
7
2
3 4 3
## EIB EIC EIB 3EIC 60 0
3
3
8 9 2
7
2
1
1
EIB EIC EIB EIC 60 0
3
3
4
2
25
1
EIB EIC 60 0
12
6
- - - - - -- 10
## Substituting value of EI in (8)
2
10
3 4 3
EIB
EIC EIB 3EIC 60 0
3
3
8 9 2
2
10
1
EIB
EIC EIB EIC 60 0
3
3
2
1
7
EIB EIC 60 0
6
3
By Equation (11)
3 1
EIB 60
7 6
27.3
EIC
Now
EI
4 3
## EIB 3EIC 16.55
9 2
Now
EI B =31.03, EIC 27.3 , EI 16.55
- - - - - -- 11
## Substituting these values in slope deflection equations,
The final moments are:
1
3
MAB 31.03 16.55 21.72 KNM
2
8
3
MBA 31.03 16.55 37.24 KNM
8
4
2
MBC 60 31.03 27.93 37.25 KNM
3
3
2
4
MCB 60 31.03 27.93 43.43 KNM
3
3
3
MCD 2( 27.93) ( 16.55) 43.45 KNM
4
3
MDC 27.93 16.55 15.52 KNM
4
## Reactions: consider the free body diagram of beam and columns
Column AB:
HA
37.25 21.72
14.74 KN
4
Beam BC:
37.25 43.45 80 3
38.97 KN
6
RC 80 RB 41.03
RB
Column CD:
HD
43.45 15.52
14.74 KN
4
Check:
H = 0
HA + HD = 0
14.74-14.74=0
Hence okay
Ex: Portal frame shown is fixed at ends A and D, the joint B is rigid and joint C is
hinged. Analyse the frame and draw BMD.
Solution:
FEMs:
WL 80 6
60KNM
8
8
WL 80 6
FCB
60KNM
8
8
Here A 0, D 0, B 0, CB 0, CD 0
FBC
## Since C is hinged member CB and CD will rotate independently. Also the
frame is unsymmetrical, will also have sway. Let the sway be towards right.
The slope deflections are:
MAB FAB
2EI
3
2 A B
L
L
2EI
3
0 B
4
4
1
3
EIB EI
2
8
0
( 1)
MBA FBA
2EI
3
2B A
L
L
2EI
3
2B 0
4
4
3
EIB EI
8
2EI
3
FBC
2B C
L
L
2E.2I
2B CB
60
6
4
2
60 EIB EICB
3
3
2EI
2CB B
FCB
L
2E.2I
2CB B
60
6
4
2
60 EICB EIB
3
3
2EI
3
FCD
2CD D
L
L
(3 )
2EI
3
2CD 0
4
4
3
EICD EI
8
2EI
3
FDC
2D CD
L
L
( 5 )
MBC
MCB
MCD
( 2)
(4 )
MDC
2EI
3
0 CD
4
4
1
3
EICD EI
2
8
0
( 6 )
## In the above equations B , CB , CD and are unknowns. According the
boundary conditions are
I. MBA+MBC = 0,
II. MCB = 0,
III. MCD = 0,
i..e,
IV. HA+HD = 0
MAB MBA MCD MDC
0
4
4
MAB MBA MCD MDC 0
## Now using the boundary conditions:
3
4
2
MBA MBC EIB EI 60 EIB EICB
8
3
3
7
2
3
EIB EICB EI 60 0
( 7 )
3
3
8
2
4
MCB EIB EICB 60 0
( 8 )
3
3
3
MCD EICD EI 0
( 9 )
8
1
3
3
3
1
3
MAB MBA MCD MDC EIB EI EIB EI EICD EI EICD EI 0
2
8
8
8
2
8
3
3
3
EIB EICD EI 0 ( 10 )
2
2
2
3
From ( 9 ) EIcD EI
(11 )
8
Sub in ( 10 )
MAB MBA MCD MDC
3
3 3
3
EIB EI EI 0
2
2 8
2
3
15
EIB
EI 0
2
16
16 3
8
Equation (12) gives EI
EIB EIB
15 2
5
( 12 )
( 13 )
Substituti ng in Equation ( 7 )
MBA MBC
7
2
3 8
3
3
8 5
2
7 3
EIB EICB 60
3
3 5
26
2
EIB EICB 60 0
15
3
( 14 )
## Substituti ng in Equation ( 8 ) and multiplyin g equation (14) by 2 we have
2
4
EIB EICB 60 0
3
3
52
4
EIB EICB 120 0
15
3
__________ __________ _
42
EIB 180 0
15
__________ __________ _
EIB 180
15
64.29
42
8
8
EIB 64.29 102.864
5
5
3
From (11) EICD EI 38.574
8
3 7
3
## From (7) EICB EIB EI 60
2 3
8
From(13) EI
3 7
3
64.29 102.864 60
2 3
8
77.165
EIB 64.29, EICB 77.165,
EICD 38.57,
EI 102.864
## Final Moments are
1
64.29 3 102.864 6.42 KNM
2
8
3
64.29 102.864 25.72 KNM
8
4
2
60 64.29 77.165 25.72 KNM
3
3
4
2
60 77.165 64.29 0
3
3
3
38.574 102.864 0
8
1
3
38.574 102.864 19.29 KNM
2
8
MAB
MBA
MBC
MCB
MCD
MDC
## Reactions: Consider the free body diagram of various members
Column AB:
HA
Beam BC:
25.72 6.42
4.825 KN
4
25.72 80 3
44.29 KN
6
RC 80 44.29 35.71 KN
RB
Column CD:
HD
19.28
4.82 KN
4
Check:
H = 0
HA+HD = 0
Hence okay.
Example: Analyse the portal frame shown in figure the deflection method and
then draw the bending moment diagram
Fig
Solution:
The frame is unsymmetrical, hence there is a sway. Let the sway be
towards right.
A 0, B 0, C 0, D 0
FEMS:
20 52
41.67 KNM
12
20 52
FCB
41.67 KNM
12
FCE 15 2 30 KNM
FBC
## Slope deflection equations
MAB F AB
0
MBA FBA
0
2EI
3
2 A B
L
L
2EI
3
0 B 0.5EIB 0.375EI
4
4
- - - - - - - - 1
2EI
3
2B A
L
L
2EI
3
2B 0 EIB 0.375EI
4
4
2EI
2B C
L
2E 1.5I
2B C 41.67 1.2EIB 0.6EIC
41.67
5
2EI
2C B
MCB FCB
L
2E 1.5I
2C B 41.67 1.2EIC 0.6EIB
41.67
5
2EI
3
MCD FCD
2C D
L
L
- - - - - - - - 2
MBC FBC
0
MDC FDC
0
2EI
3
2C 0 EIC 0.375EI
4
4
- - - - - - - -- 3
- - - - - - - -- 4
- - - - - - - -- 5
2EI
3
2D C
L
L
2EI
3
0 C 0.5EIC 0.375EI
4
4
- - - - - - - -- 6
In the above equation there are three unknowns B , C and , accordingly the
boundary conditions are,
MBA MBC 0
MCB MCD MCE 0
HA HD 0
i.e, MAB MBA MCD MDC 0
Now,
MBA MBC 0
## 2.2EIB 0.6EIC 0.375EI 41.67 0
And, MCB MCD 41.67 1.2EIC 0.6EIB EIC 0.375EI 30 0
0.6EIB 2.2EIC 0.375EI 11.67 0
## MAB MBC MCD MDC 0
0.5EIB 0.375EI EIB 0.375EI 2EIC 0.375EI 0.5EIC 0.375EI 0
1.5EIB 1.5EIC 1.5EI 0
## Solving the above equations
we get, EI B 23.98 , EI C 9.36, EI 14.62
Substituting these values in slope deflection equations, we have
MAB 0.523.98 0.37514.62 6.50 KNM
## MCB 41.67 1.2 9.36 0.623.98 44.83 KNM
MCD 9.36 0.375( 14.62) 14.84 KNM
MDC 0.5 9.36 0.375 14.62 10.16 KNM
MCE 30 KNM
Column AB:
HA
18.5 6.5
6.25 KN
4
Span BC:
## 44.83 18.5 20 5 2.5
55.27 KN
5
RB 20 5 RC 44.73
RC
Column CD:
10.16 14.84
HD
6.25
4
Check:
H = 0
HA + HD = 0
=0
Hence okay
Example: Analyse the portal frame shown and then draw bending moment
diagram.
Solution:
It is an unsymmetrical problem hence there is a sway be towards right
A 0, B 0, C 0, D 0
FEMs:
wl2
20 52
FBC
- 41.67 KNM
12
12
wl2 20 52
FCB
41.67 KNM
12
12
Slope deflection equations:
2EI
3
M AB F AB
2 A B
L
L
2EI
3
0
0 B
3
3
2
2
EIB EI
3
3
2EI
3
MBA F BA
2 B A
L
L
2EI
3
0
2 B 0
3
3
4
2
EIB EI
3
3
- - - - - - - -- 1
- - - - - - - -- 2
2EI
3
2 B C
L
L
2E 1.5I
2B C
- 41.67
5
6
3
- 41.67 EIB EIC
5
5
2EI
3
MCB F CB
2 C B
L
L
2E 1.5I
2 C B 0
41.67
5
41.67 1.2EI C 0.6 EIB
MBC F BC
2EI
3
2 C D
L
L
2EI
3
0
2 C 0
4
4
EI C 0.375 EI
- - - - - - - -- 3
- - - - - - - -- 4
MCD F CD
- - - - - - - - - - 5
2EI
3
2 D C
L
L
2EI
3
0
0 C
4
4
0.5EI C 0.375 EI
MDC F DC
- - - - - - - - - - 6
In the above equations there are three unknown B , C and and accordingly the
Boundary conditions are:
MBA MBC 0
M CB M CD 0
H A HD 0
i.e
M AB MBA M CD MDC
0
3
4
4(M AB MBA ) 3(M CD MDC ) 0
Now
MBA MBC 0
4
2
6
3
EIB EI EIB EIC 41.67
3
3
5
5
3
2
2.53EIB EIC EI 41.67 0 (7)
5
3
MCB MCD 0
41.67 1.2EIC 0.6EIB EIC 0.375EI 0
41.67 2.2EIC 0.6EIB 0.375EI 0 (8)
MAB MBA MCD MDC
0
3
4
2
4
2
2
4 EIB EI EIB EI
3
3
3
3
## 3 EIC 0.375EI 0.5EIC 0.375EI 0
8
8
16
8
EIB EI
EIB EI 4.5EIC 2.25EI 0
3
3
3
3
8EIB 4.5EIC 7.53EI 0
(9)
## By solving (7), (8) and (9) we get
EI B 25.46
EI C 23.17
EI 12.8
Final moments:
2
2
25.46 12.8 8.44 KNM
3
3
4
2
MBA 25.46 12.8 25.40 KNM
3
3
6
3
MBC 25.46 23.17 41.67 25.40 KNM
5
5
MCB 41.67 1.2 23.17 0.60 20.46 28.50 KNM
MAB
## MCD 23.70 0.375 (12.80 ) 28.50 KNM
MDC 0.5 23.70 0.375 12.80 16.65 KNM
## Reactions: Consider the free body diagram
Member AB:
HA
25.40 8.44
11.28 KN
3
Member BC:
RC
28.5 20.30 20 5
5
2 51.64 KN
2
RB 20 5 51.64 48.36 KN
Member CD:
HD =
28.5 16.65
= 11.28 KN
4
Check:
H = 0
HA + HD = 0
Satisfied, hence okay
Example: A portal frame having different column heights are subjected for forces
as shown in figure. Analyse the frame and draw bending moment diagram.
## Solution:It is an unsymmetrical problem
A 0, B 0, C 0, D 0 , hence there is a sway be towards right.
FEMs:
Wl
30 4
- 15 KNM
8
8
Wl
30 4
FBA
15 KNM
8
8
Wl
60 4
FBC
- 30 KNM
8
8
Wl
60 4
FCB
30 KNM
8
8
FCD = FDC = 0
F AB
## Slope deflection equations:
2EI
3
MAB F AB
2 A B
L
L
2E 2I
3
- 15
0 B
4
4
- 15 EIB 0.75 EI
- - - - - - - - 1
MBA FBA
2EI
3
2B A
L
L
2E 2I
3
2B 0
4
4
15 2 EIB 0.75 EI
- - - - - - - - 2
2EI
2B C
L
2E 2I
2B C
- 30
4
- 30 2EIB EIC
- - - - - - - -- 3
2EI
2C B
L
2E 2I
2C B
30
4
30 2EIC EIB
- - - - - - - -- 4
15
MBC FBC
MCB FCB
2EI
3
2C D
L
L
2EI
3
0
2 C 0
3
3
4
2
EI C EI
3
3
MCD FCD
2EI
3
2D C
L
L
2EI
3
0
0 C
3
3
2
2
EI C EI
3
3
- - - - - - - -- 5
MDC FDC
- - - - - - - -- 6
There are three unknowns, EI, B ,EI C & EI , accordingly the Boundary
conditions are
MBA MBC 0
MCB MCD 0
HA HD 30 0
MAB MBA 60 MCD MDC
30 0
4
3
3MAB MBA 4MCD MDC 180 0
i.e,
Now
## MBA MBC 15 2EIB 0.75EI 30 2EIB EIC
4EIB EIC 0.75EI 15 0
MCB MCD 30 2EIC EIB
EIB
4
2
EIC EI
3
3
10
2
EIC EI 30 0
3
3
2
2
2
4
3
3
3
3
## 9EIB 8EIC 9.833EI 180 0 9
By solving (7), (8) & (9) we get
EI B 9.577
EI C 7.714
EI 20.795
Substituting these values in the slope deflection equations we get
## MBA 15 2 9.577 0.75 20.795 18.55 KNM
MBC - 30 2 9.577 - 7.714 - 18.55 KNM
## MCB 30 2 7.714 9.577 24.15 KNM
4
2
( 7.714 ) 20.795 24.15 KNM
3
3
2
2
MDC ( 7.714 ) 20.795 19.00 KNM
3
3
MCD
## Reactions: Consider free body diagrams of the members
Member AB:
18.55 21.01 30 2
15.615 KN
4
-ve sign indicates the direction of HA is from right to left.
HA
Member BC:
18.55 60 2 24.15
28.60 KN
4
RC 60 RB 28.60 31.40 KN
RB
Member CD:
HD
19 24.15
14.38 KN
3
Check:
H = 0
HA + HD + 30 = 0
-15.62 14.38 + 30 = 0
Hence okay
Example: Analyse the frame using slope deflection method and draw the
Bending Moment Diagram.
## It can be observed from figure in that direction of moments due to sway in
member AB are anticlockwise and that for member CD are clockwise. Wise shall
be taken to incorporate the same in the slope deflection equation.
FEMS
wI2
12
24 42
32 KNM
12
wI2
FCB
32 KN M
12
Here A D 0
FBC
## Slope deflection equations are:
MAB F AB
2EI
3
2 A B
L
L
2EI
3
B
3
3
2
2
EIB EI
3
3
2EI
3
MBA FBA
2B A
L
L
2EI
3
2B
3
3
4
2
EIB EI
3
3
- - - - - -- 1
2EI
2B C
L
2E2I
2B C
32
4
32 2EIB EIC
- - - - - - - -- 3
2EI
2C B
L
2E2I
2C B
32
4
32 2EIC EIB
- - - - - - - - 4
MBC FBC
MCB FCB
MCD FCD
2EI
3
2C D
L
L
2EI
3
2C
3
3
4
2
EIC EI
3
3
2EI
3
2D C
L
L
2EI
3
C
3
3
2
2
EIC EI
3
3
MDC FDC
- - - - - - - - 6
## The unknown are B, ,C & Accordingl y the boundary conditions are
MBA MBC 0
MCB MCD 0
HA HD 30 0
MAB MBA MCD MDC
30 0
3
3
MAB MBA MCD MDC 90 0
4
2
Now, MBA MBC EIB EI 32 2EIB EIC
3
3
10
2
EIB EIC EI 32 0
3
3
i.e,
EIB
## MAB MBA MCD MDC 90
4
2
EIC EI
3
3
10
2
EIC EI 32 0
3
3
2
2
4
2
4
EIB EI EIB EI EIC
3
3
3
3
3
2
2
2
EI EIC EI 90
3
3
3
8
2EIB 2EIC EI 90
3
4
EIB EIC EI 45 0
9
3
20
4
EIB 2EIC EI 64 0
3
3
4
EIB EIC EI 45 0
3
17
EIB 3EIC 109 0
3
- - - - - -- 10
20
4
EIC EI 64 0
3
3
4
EIB EIC EI 45 0
3
2EIB
3EIB
17
EIC 109 0
3
## By (10) & (11)
17
EI B 3EI C 109 0
3
27
EI B 3EI C 57.71 0
17
208
EI B 166.71 0
17 3
166.71 17 3
EIB
40.88
208
- - - - - - - - 11
From (10)
EI C
1
17
109 EI B 40.88
3
3
From (9)
3
EIB EIC 45
4
3
40.88 40.88 45 95.07
4
Thus
EIB 40.88, EIC 40.88, EI 95.07
Substituting these values in slope deflection equations
EI
2
40.88 2 95.07 36.12 KNM
3
3
4
2
MBA 40.88 95.07 8.88 KNM
3
3
MBC 32 240.88 40.88 8.88 KNM
MAB
## MCB 32 2 40.88 40.88 8.88 KNM
4
40.88 2 95.07 8.88 KNM
3
3
2
2
MDC 40.88 95.07 36.12 KNM
3
3
MCD
To find the reaction consider the free body diagram of the frame
Reactions:
Column AB
HA
8.88 36.12
15 KN
3
Beam AB
RB
8.88 8.88 24 4
4
RC 24 4 48 48 KN
Column CD
HD
8.88 36.12
15 KN
3
4
2 48 KN
Check
H = 0
HA + HD +P = 0
-15 15 + 30 = 0
Hence okay | 17,324 | 32,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-35 | latest | en | 0.77463 |
http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch07/Ch7.html | 1,653,328,055,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00085.warc.gz | 109,740,372 | 12,040 | ## Ch 7, Work and Energy
### Questions 4, 5, 9, 12, 14
Additional problems from Serway's fourth edition
(4 ed) 7.1 A cheerleader lifts his 50.0-kg partner straight up off the ground a distance of 0.60 m before releasing her. If he does this 20 times, how much work has he done?
(4 ed) 7.2 If you push a 40.0-kg crate at a constant speed of 1.40 m/s across a horizontal floor (k = 0.25), at what rate
(a) is work being done on the crate by you and
(b) is energy dissipated by the frictional force?
(4 ed) 7.3 A block of mass m hangs on the end of a cord and is connected to a block of mass M by the pulley arrangement shown in Figure P7.41. Using energy considerations,
(a) find an expression for the speed of m as a function of the distance it has fallen. Assume that the blocks are initially at rest and there is no friction.
(b) Repeat (a) assuming sliding friction (k) between mass M and the table.
(c) Show that the result obtained in (b) reduces to that obtained in (a) in the limit as k goes to zero.
(4 ed) 7.4 A block of mass 0.60 kg slides 6.0 m down a frictionless ramp inclined at 20o to the horizontal. It then travels on a rough horizontal surface where k = 0.50.
(a) What is the speed of the block at the end of the incline?
(b) What is its speed after traveling 1.00 m on the rough surface?
(c) What distance does it travel on this horizontal surface before stopping?
(4 ed) 7.5 A time-varying net force acting on a 4.0-kg particle causes the particle to have a displacement given by
x = 2.0 t - 3.0 t2 + 1.0 t3,
where x is in meters and t is in seconds. Find the work done on the particle in the first 3.0 s of motion.
(4 ed) 7.6 A 1,500-kg car accelerates uniformly from rest to 10 m/s in 3.0 s.
Find (a) the work done on the car in this time,
and (b) the average power delivered by the engine in the first 3.0 s,
and (c) the instantaneous power delivered by the engine at t = 2.0 s.
(4 ed) 7.7 A car of weight 2,500 N operating at a rate of 130 kW develops a maximum speed of 31 m/s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant.
(a) What is the car's maximum speed on an incline of 1 in 20 (ie, if theta is the angle of the incline with the horizontal, sin = 1/20)?
(b) What is its power output on a 1-in-10 incline if the car is traveling at 10 m/s?
Conceptual Questions
Q7.4 Can the kinetic energy of an object be negative?
No. Kinetic energy is given by KE = (1/2) m v2. The mass m is always positive and v2 is positive so KE must always be positive (of course, if v = 0 then KE = 0). KE can not be negative.
Q7.5 (a) If the speed of a particle is doubled, what happens to its kinetic energy?
KE = (1/2) m v2 so doubling the speed v means the Kinetic Energy is increased fourfold.
(b) if the net work done on a particle is zero, what can be said about the speed?
The net work done on a particle is equal to the change in its Kinetic Energy. So if the net work is zero the Kinetic Energy remains constant and that means the speed, too, remains constant.
Q7.9 When a punter kicks a football, is he doing any work on the ball while his toe is in contact with it?
Yes, a force is being exerted on the ball as it moves through a distance. That's the definition of work.
Is he doing any work on the ball after it loses contact with his toe?
No, once his toe looses contact with the ball, the force his toe exerts is zero.
Are any forces doing work on the ball while it is in flight?
Yes, gravity and air resistance continue to do work on the ball.
Q7.12 As a simple pendulum swings back and forth, the forces acting on the suspended mass are the force of gravity, the tension in the supporting cord, and air resistance.
(a) Which of these forces, if any, does no work on the pendulum?
The force in the supporting cord is perpendicular to the motion so that force does no work.
(b) Which of thee forces does negative work at all times during its motion?
Air resistance -- like all friction forces -- always does negative work.
(c) Describe the work done by the force of gravity while the pendulum is swinging.
The force of gravity does positive work as the pendulum "falls" from either extreme to its lowest position; this causes the pendulum to speed up. The force of gravity does negative work as the pendulum "rises" from its lowest position to either extreme; this causes the pendulum to slow down.
Q7.14 An older model car accelerates from 0 to a speed v in 10 s. A newer, more powerful sports car accelerates from 0 to 2 v in the same time period. What is the ratio of powers expended by the two cars?
Assume the two cars have the same mass.
The car going 2 v has four times the Kinetic Energy. If it acquired this much energy in the same time, then the power supplied to it is also four times as much.
Problems from the current (5th) edition of Serway and Beichner
7.2 A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25o downward from the horizontal. Find the work done by the shopper as she moves down an aisle 50 m in length.
W = F d cos
W = (35 N)(50 m) cos 25o
W = (35 N)(50 m)(0.906)
W = 1 586 N-m = 1 586 J
It is entirely equivalent to think of this as
W = Fx d
W = [(35 m) cos 25o] (50 m)
W = 1 586 J
7.6 A 15.0-kg block is dragges over a rough, horizontal surface by a 70.0-N force acting at 20o above the horizontal. The block is displaced 5.0 m and the coefficient of kinetic friction is 0.30. Find the work done by
(a) the 70-N force,
(b) the normal force, and
(c) the force of gravity.
(d) What is the energy loss due to friction?
W = F d cos
W70 = Fext d cos 20o
W70 = (70 N)(5 m)(0.9397)
W70 = 328.9 J
Wn = n d cos 90o
Wn = 0
Wg = w d cos 90o
Wg = 0
Wfr = Ffr d cos 180o
What is the force of friction Ffr? We do know that
Ffr = 0.30 n
But what is the normal force n? We see enough special cases when n = w = mg that one can intuitively jump to the wrong conclusion and thing that n = w = mg all the time. That is not the case in general and that is not the case here. The block is not jumping up off the plane and it is not burrowing down into the plane. Therefore, the sum of the forces in the vertical direction must be zero.
Fnet,y = n + Fext,y - w = 0
n + (70 N)(sin 20o) - (15 kg)(9.8 m/s2) = 0
n + (70 N)(0.342) - (15)(9.8) N = 0
n + 23.9 N - 147 N = 0
n - 123.1 N = 0
n = 123.1 N
With the value of the normal force n now known, we can calculate the friction force Ffr.
Ffr = 0.30 n
Ffr = 0.30 (123.1 N)
Ffr = 36.9 N
Now we can calculate the work done by the friction force,
Wfr = Ffr d cos 180o
Wfr = - (36.9 N)(5.0 m)
Wfr = - 184.6 J
What does the negative sign mean. The force of friction decreases the energy. This is the energy loss which shows up as heat.
7.11 A force F = (6i - 2j) N acts on a particle that undergoes a displacement d = (3i + j) m.
Find (a) the work done by the force on the particle
and (b) the angle between F and d.
W = F d
W = [(6i - 2j) N] [(3i + j) m]
W = [(6i - 2j) (3i + j)] (N-m)
W = [(6i - 2j) (3i + j)] J
There are (at least) two ways we can evaluate this dot-product.
W = [(6i) (3i) + (6i) (j) - (2j) (3i) - (2j) (j)] J
W = [(6)(3)(i i) + (6)(1)(i j) - (2)(3)(j i) - (2)(1)(j j)] J
W = [(6)(3)(1) + (6)(1)(0) - (2)(3)(0) - (2)(1)(1)] J
W = [18 + 0 - 0 - 2] J
W = 16 J
An alternative is to remember the dot product in terms of components as
A B = Ax Bx + Ay By
Then the work can be written as
W = F d = Fx dx + Fy dy
W = F d = [(6)(3) + ( - 2)(1)] N-m
W = F d = [18 - 2] J
W = 16 J
7.20 An archer pulls her bow string back 0.400 m by exerting a force that increases uniformly from zero to 230 N.
(a) What is the equivalent spring constant of the bow?
(b) How much work is done in pulling the bow?
F = k x
230 N = k (0.4 m)
k = 230 N / 0.4 m
k = 575 N/m
W = (1/2) k x2
W = (1/2) (575 N/m) (0.4 m)2
W = 46 J
7.21 A 6,000-kg freight car rolls along rails with negligible friction. The car is brought to rest by a combination of two coiled springs, as illustrated in Figure P7.21. Both springs obey Hooke's law, with k1 = 1600 N/m and k2 = 3400 N/m. After the first spring compresses a distance of 30.0 cm, the second spring (acting with the first) increases the force so that there is additional compression as shown in the graph. If the car is brought to rest 50.0 cm after first contacting the two-spring system, find the car's initial speed.
The wording of this question may be ambiguous. When it asks "If the car is brought to rest 50.0 cm after first contacting the two-spring system, find the car's initial speed." -- exactly what is meant by "after first contacting the two-spring system"? This solution takes "after first contacting the two-spring system" to mean the car moves 50.0 cm after first contacting the first spring, with k1 = 1600 N/m.
The first spring is compressed the full distance of x1 = 0.50 m so it does work of
W1 = ( 1/2 ) k1 x12 = 0.5 (1 600 N/m) (0.5 m)2 = 200 J
The second spring is compressed only after the freight car moves a distance of 0.30 m, so it is only compressed a distance of x2 = 0.20 m so it does work of
W2 = ( 1/2 ) k2 x22 = 0.5 (3 400 N/m) (0.2 m)2 = 68 J
So both springs have done a total amount of work of
WTot = W1 + W2 = 200 J + 68 J = 268 J
[ There is an alternate way to do this. We could figure the work done during the first 0.30 m with k = 1600 N/m and then find the work done during the last 0.20 m from the "area" under the graph. The total amount of work should be the same, either way ].
Actually, this total work is negative that amount, WTot = - 268 J, since the force and displacement are in opposite direction. This has changed the freight car's Kinetic Energy from its initial value of
KEi= ( 1/2 ) m vi2
to its final value of zero. That is
KEi= ( 1/2 ) m vi2 = ( 1/2 ) (6 000 kg) vi2 = 268 J = Wsprings
vi2 = 0.089 m2 / s2
vi = 0.30 m/s
7.28 A 0.30-kg ball has a speed of 15.0 m/s.
(a) What is its kinetic energy?
(b) If its speed were doubled, what would be its kinetic energy?
KE = (1/2) m v2
KE = (1/2) (0.30 kg) (15.0 m/s)2
KE = 33.75 J
Since the speed is squared, doubling the speed will increase the KE by a factor of four.
KEdbl = 4 KEorig
KEdbl = 135 J
7.40 An Atwood's machine has a 3.00 kg mass and a 2.00-kg mass at ends of the string (as in Figure 5.15). The 2.00-kg mass is released from rest on the floor, 4.00 m below the 3.00-kg mass.
(a) If the pulley is frictionless, what will be the speed of the masses when they pass each other?
(b) Suppose that the pulley does not rotate and the string must slide over it. If the total frictional force between the pulley and the string is 4.00 N, what are their speeds when the masses pass each other?
Wnet = KE
Wnet = W2 + W3 = - (2 kg) (9.8 m/s2) (2 m) + (3 kg) (9.8 m/s2) (2 m)
Wnet = 19.6 J
The w's in the diagram -- w2 and w3 -- are the weights;
w2 = m2 g = (2.0 kg) (9.8 m/s2)
w3 = m3 g = (2.0 kg) (9.8 m/s2)
The capital W's in the equation above -- W2 and W3 -- are the amounts of work done on the 2.0-kg mass and the 3.0-kg mass by gravity.
Only the force of gravity does work on "the system". We need not -- can not -- consider the tension in the cord for that is an "internal force" just like the forces that hold the blocks together.
The blocks have a common velocity v,
KE = 0.5 (2.0 kg) v2 + 0.5 (3.0 kg) v2 = (2.5 kg) v2
KE = (2.5 kg) v2 = 19.6 J = Wnet
v2 = 7.84 m2 / s2
v = 2.8 m / s
Now, consider friction.
Wf = - (5.0 N) (2.0 m) = - 10 J
Wnet = 19.6 J - 10 J = 9.6 J
KE = (2.5 kg) v2 = 9.6 J = Wnet
v2 = 3.84 m2 / s2
v = 1.96 m /s
7.45 A certain automobile engine delivers 2.24 x 104 W (or 22.4 kW or 30 hp) to its wheels when moving at a constant speed of 27.0 m/s (about 60 mi/h or about 100 km/h). What is the resistive force acting on the automobile at that speed?
P = F v
F = P / v
F = [2.24 x 104 W]/[27 m/s]
F = 830 N
7.52 An electron moves with a speed of 0.995 c.
(a) What is its kinetic energy?
(b) If you use the classical expression to calculate its kinetic energy, what percentage error would result?
It is fun to look at more than just v = 0.995 c. Repetitive calculations are often easiest to do with a simple spreadsheet.
7.53 A proton in a high-energy accelerator moves with a speed of c/2. Using the work-energy theorem, find the work required to increase its speed to
(a) 0.750 c and
(b) 0.995 c.
For common, ordinary speeds, we found that the (kinetic) energy of a particle with mass m and speed v is KE = (1/2) m v2. But for very high speeds -- for speeds close to that of light -- we had to modify that and use
E = m c2
m = mo / SQRT(1 - v2/c2)
Eo = mo c2
KE = E - Eo
KE = mo c2 [ {1/SQRT(1 - v2/c2)} - 1]
This does not look very much like KE = (1/2) m v2 ! So it is worth reminding ourselves tha this unusual looking expression does, indeed, reduce to KE = (1/2) m v2 for v << c !
First, let's find the "rest energy" of the proton since this factor appears in all our future calculations.
Eo = mo c2
Eo = (1.672 x 10 - 27 kg)(3 x 108 m/s)2
Eo = 5.016 x 10 - 11 J
This is perfectly okay as it is. However, instead of dealing with such small numbers, it is customary to talk about these energies in units of MeV -- millions of electron-Volts. We will explore MeV's more in PHY 1360 (next semester). For the present, we can simply pull out of the air the appropriate conversion factor,
1 eV = 1.6 x 10 - 19 J
1 MeV = 1.6 x 10 - 16 J
Therefore,
Eo = 5.016 x 10 - 11 J [1 MeV / 1.6 x 10 - 16 J]
Eo = 313 500 Mev
Eo = 313.5 Gev
One GeV is one giggaelectron-volt. Gigga means "billion" or 109.
E = m c2
m = mo / SQRT(1 - v2/c2)
Eorig = [mo / SQRT(1 - vorig2/c2)] c2
Eorig = [mo c2][1/ SQRT(1 - vorig2/c2)]
Eorig = [313.5 Gev][1/ SQRT(1 - {0.50 c}2/c2)]
Eorig = [313.5 Gev][1/ SQRT(1 - {0.50}2)]
Eorig = [313.5 Gev][1/ SQRT(1 - 0.25)]
Eorig = [313.5 Gev][1/ SQRT(0.75)]
Eorig = [313.5 Gev][1/ 0.866]
Eorig = [313.5 Gev][1.155]
Eorig = 362 Gev
That's the total energy -- KE plus "rest mass energy" -- that the proton has originally. Now, how much energy does it have for the speeds given in question (a) and question (b)?
Ea = [mo / SQRT(1 - va2/c2)] c2
Ea = [mo c2][1/ SQRT(1 - va2/c2)]
Ea = [313.5 Gev][1/ SQRT(1 - {0.750 c}2/c2)]
Ea = [313.5 Gev][1/ SQRT(1 - {0.750}2)]
Ea = [313.5 Gev][1/ SQRT(1 - 0.5625)]
Ea = [313.5 Gev][1/ SQRT(0.4375)]
Ea = 313.5 Gev][1/ 0.661]
Ea = [313.5 Gev][1.512}
Ea = 474 GeV
How much energy do we need to add to go from the original speed (vorig = 0.500 c) tho this speed (va = 0.75 c)?
Ea = Ea - Eorig
Ea = (474 - 362) GeV
Ea = 112 GeV
Now we do exactly the same thing for vb = 0.995 c.
Eb = [mo / SQRT(1 - vb2/c2)] c2
Eb = [mo c2][1/ SQRT(1 - vb2/c2)]
Eb = [313.5 Gev][1/ SQRT(1 - {0.995 c}2/c2)]
Eb = [313.5 Gev][1/ SQRT(1 - {0.995}2)]
Eb = [313.5 Gev][1/ SQRT(1 - 0.990025)]
Eb = [313.5 Gev][1/ SQRT(0.009975)]
Eb = [313.5 Gev][1/ 0.09987]
Eb = [313.5 Gev][10.012]
Eb = 3 139 GeV
How much energy do we need to add to go from the original speed (vorig = 0.500 c) tho this speed (vb = 0.995 c)?
Eb = Eb - Eorig
Eb = (3 139 - 362) GeV
Eb = 2 777 GeV
Solutions to the additional problems from Serway's fourth edition.
(4 ed) 7.1 A cheerleader lifts his 50.0-kg partner straight up off the ground a distance of 0.60 m before releasing her. If he does this 20 times, how much work has he done?
w = mg = (50 kg) (9.8 m/s2) = 490 N
W = F s = (490 N) (0.60 m) = 294 N-m = 294 J
WTot = 20 x W = (20) (290 J) = 5,880 J
WTot = 5,880 J
(4 ed) 7.2 If you push a 40.0-kg crate at a constant speed of 1.40 m/s across a horizontal floor (k = 0.25), at what rate
(a) is work being done on the crate by you and
(b) is energy dissipated by the frictional force?
P = F v
We must find the horizontal force F
From the diagram, we find that
F = Ff
Fn = m g = (40 kg) (9.8 m/s2) = 392 N
Ff = Fn = (0.25) (392 N) = 98 N
F = 98 N
P = F v = (98 N) (1.4 m/s) = 137 N-m / s = 137 J / s = 137 W
The friction force Ff dissipates energy at the same rate,
P = 137 W
(4 ed) 7.3 A block of mass m hangs on the end of a cord and is connected to a block of mass M by the pulley arrangement shown in Figure P7.41. Using energy considerations,
(a) find an expression for the speed of m as a function of the distance it has fallen. Assume that the blocks are initially at rest and there is no friction.
(b) Repeat (a) assuming sliding friction (k) between mass M and the table.
(c) Show that the result obtained in (b) reduces to that obtained in (a) in the limit as k goes to zero.
When hanging mass m moves a distance x, the sliding mass M moves only half that distance, x/2. When hanging mass m has speed v, sliding mass M has a speed of only half that, v/2.
When the hanging mass has moved a distance x from its initial position, the force of gravity has done work of
Wg = m g x
That is the only work that is done on the "system" since the forces on sliding mass M -- the force of gravity and the normal force of the table -- are perpendicular to the displacement. The tension in the cord is an "internal force" and does not act on the "system" from the outside. This work due to gravity is the net work so it must equal the change in KE of the "system"
Wg = Wnet = m g x = KE = KE - KEi = KE - 0 = KE = KEM + KEm
m g x = (1/2) M (1/2 v )2 + (1/2) m v2
m g x = 1/8 M v2 + (1/2) m v2
m g x = (1/2) [ 1/4 M + m ] v2
v2 = 2 m g x / [ 1/4 M + m ]
v2 = { 2 m g / [ 1/4 M + m ] } x
This is without friction.
Now, go back and add in friction. With friction present, the friction force is
Ff = FN = M g
As hanging mass m moves distance x, the sliding mass M moves a distance x/2 and work due to the friction force is done on it in the amount of
Wf = - Ff (x/2) = - M g x /2
That means the net work is
Wnet = Wg + Wf = m g x - M g x /2
Wnet = m g x - M g x /2 = KE = KE - KEi = KE - 0 = KE = KEM + KEm
m g x - M g x /2 = KEM + KEm
m g x - M g x /2 = KEM + KEm = (1/2) M (1/2 v )2 + (1/2) m v2
[ m - M / 2 ] g x = (1/2) [ (1/4) M + m ] v2
v2 = 2 { [ m - M / 2 ] / [ (1/4) M + m ] } g x
Does this expression reduce to our earlier, frictionless expression (as it must) when the coefficient of friction is zero? Set = 0 and see;
v2 = 2 { [ m - M / 2 ] / [ (1/4) M + m ] } g x
v2 = 2 { [ m - (0) M / 2 ] / [ (1/4) M + m ] } g x
v2 = 2 { [ m ] / [ (1/4) M + m ] } g x
v2 = 2 { [ m ] / [ (1/4) M + m ] } g x
And, indeed, this is the same expression as we got for the frictionless case directly.
(4 ed) 7.4 A block of mass 0.60 kg slides 6.0 m down a frictionless ramp inclined at 20o to the horizontal. It then travels on a rough horizontal surface where k = 0.50.
(a) What is the speed of the block at the end of the incline?
(b) What is its speed after traveling 1.00 m on the rough surface?
(c) What distance does it travel on this horizontal surface before stopping?
(a) What is the speed of the block at the end of the incline?
On the frictionless incline, the normal force does no work, Wn= 0. Only the force of gravity does any work,
Wg = F s cos = ( m g ) (6.0 m) cos 70o = (0.60 kg) (9.8 m/s2) (6.0 m) (0.342) = 12.1 J
Wg = 12.1 J
Notice that this has the cosine of 70o rather than of 20o. In writing W = F s cos , the angle is the angle between the force and the displacement.
Wnet = Wg + Wn = 12.1 J
Wnet = KE = KE - KEi = KE
Wnet = (1/2) m v2
(1/2) m v2 = (1/2) (0.60 kg) v2 = 12.1 J
v2 = 40.22 m2 / s2
v = 6.34 m / s
This is the speed at the bottom of the incline and the beginning of the rough horizontal plane that has friction.
From the force diagram, you can see that
Fn = m g = (0.60 kg) (9.8 m/s2) = 5.88 N
Ff = Fn = (0.5) (5.88 N) = 2.94 N, pointing to the left
Along this rough, horizontal plane, only this friction force Ff does any work. By the time the block has moved a distance of 1.0 m along the plane, the net work done on the block is
Wnet= Wf = - (2.94 N) (1 m) = - 2.94 J
Wnet = KE = KE1.0 m - KEo = KE1.0 m - 12.1 J = - 2.9 J
KE1.0 m = 12.1 J - 2.9 J = 9.2 J
KE1.0 m = 9.2 J = (1/2) m v2 = (1/2) (0.60 kg) v2
v2 = 30.7 m2 / s2
v = 5.5 m / s
Now, how far does it travel before it completely stops. At that point, KEf = 0. The work done by friction in moving this distance X is
Wnet= Wf = - (2.94 N) ( Xf )
And this is just equal to the change in KE, from its initial value of 12.1 J at the bottom of the inclined plane until it goes to zero,
- (2.94 N) X f = - 12.1 J
X f = 4.11 m
(4 ed) 7.5 A time-varying net force acting on a 4.0-kg particle causes the particle to have a displacement given by
x = 2.0 t - 3.0 t2 + 1.0 t3,
where x is in meters and t is in seconds. Find the work done on the particle in the first 3.0 s of motion.
We need the change in KE so we need KE at t = 0 and KE at t = 3.0 s
v = dx/dt = 2.0 t0 - 3.0 ( 2 t ) + 1.0 (3 t2 ) = 2.0 - 6.0 t + 3.0 t2
vo = v(t = 0) = 2.0 m/s
vf = v(t = 3 s) = 2 - 6(3) + 3(32) = 2 - 18 + 27 = 11 m/s
KEo = (1/2) m vo2 = (1/2) (4.0 kg) (2 m/s)2 = 8 J
KEf = (1/2) m vv2 = (1/2) (4.0 kg) (11 m/s)2 = 242 J
Wnet =KE = 234 J
(4 ed) 7.6 A 1,500-kg car accelerates uniformly from rest to 10 m/s in 3.0 s.
Find (a) the work done on the car in this time,
and (b) the average power delivered by the engine in the first 3.0 s,
and (c) the instantaneous power delivered by the engine at t = 2.0 s.
Wnet =KE = 0.5 (1 500 kg) (10 m/s)2 = 75 000 J = 75 kJ
Pavg = W / t = 75 kJ / 3 s = 25 kW = 25 000 W
For t = 2 s,
P = F v
F = m a = m (v / t) = [1 500 kg] [(10 m/s) / 3 s] = 5 000 N = 5 kN
a = v / t = (10 m/s) / 3 s = 3.33 m / s2
v = vi + a t = 0 + ( 3.33 m / s2) (2 s) = 6.67 m / s
P = (5 000 N) (6.67 m / s)
P = 33 350 W = 33.35 kW
(4 ed) 7.7 A car of weight 2,500 N operating at a rate of 130 kW develops a maximum speed of 31 m/s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant.
(a) What is the car's maximum speed on an incline of 1 in 20 (ie, if theta is the angle of the incline with the horizontal, sin = 1/20)?
(b) What is its power output on a 1-in-10 incline if the car is traveling at 10 m/s?
We know the power can be calculated from
P = F v
Since the car is not accelerating, the force F supplied by the engine is just equal to the resistive forces that retard its forward motion.
F = P / v = (130 000 W) / (31 m/s) = 4 194 N = 4.194 kN
This is the resistive force which we will assume to stay at this value.
Now, climb a hill with a 1-in-20 incline.
On this hill, in addition to the resistive force of 4 194 N, there is also a component of the weight that points "down" the hill; let's call this force F||,
F|| = w sin = (2 500 N) (1 / 20) = 125 N
Now the total retarding force is
FTot = 4 194 N + 125 N = 4 319 N
To keep the car moving at constant speed, the engine must supply this force to the tires.
P = F v
v = P / F = 130 000 W / 4 319 N = 30.09 m / s
v = 30.1 m / s
On a 1-in-10 incline, the component of the weight that is parallel to the road is
F|| = w sin = (2 500 N) (1 / 10) = 250 N
FTot = 4 194 N + 250 N = 4 444 N
P = F v = (4 444 N) (10 m / s) = 44 440 W = 44.44 kW
P = 44 440 W = 44.44 kW | 8,014 | 23,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-21 | latest | en | 0.933277 |
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# The team announced that its record deteriorated much further
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The team announced that its record deteriorated much further [#permalink]
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31 Aug 2012, 23:36
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The team announced that its record deteriorated much further in the first half of the season than analysts had predicted them to and its performance will improve in the last half of the season.
A)had predicted them to and its performance will improve
B)had predicted and that its performance would improve
C)predicted they would and that it will improve its performance
D)predicted them to and its performance would improve
E)predicted and that it will have improved its performance
[Reveal] Spoiler: OA
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Re: The team announced that its record deteriorated [#permalink]
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01 Sep 2012, 05:56
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Aristocrat wrote:
The team announced that its record deteriorated much further in the first half of the season than analysts had predicted them to and its performance will improve in the last half of the season.
A)had predicted them to and its performance will improve. Pronoun error.
B)had predicted and that its performance would improve. Correct choice. usage of would is correct here.
C)predicted they would and that it will improve its performance. Perfect tense required, since prediction was before announcement.
D)predicted them to and its performance would improve. Tense error
E)predicted and that it will have improved its performance. Tense error
The correct answer is B. Usage of would in past tense is correct. And predicted here needs to be in perfect tense to clarify the correct chronology.
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Re: The team announced that its record deteriorated [#permalink]
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01 Sep 2012, 15:13
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The first half ,Non underlined portion : Announced THAT its.......
In the underlined portion, there's a AND : a Parallelism marker
First split : we will skim through options for : THAT........AND.......THAT Parallelism
Vertical scan reveals : Only B ,C n E to follow suit
C / E : THAT ITS..........AND...........THAT IT : Eliminated
B : THAT ITS..........AND............THAT ITS : Perfect Parallelism
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Re: The team announced that its record deteriorated [#permalink]
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03 Sep 2012, 17:41
B for had predicted and would construction
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Re: The team announced that its record deteriorated [#permalink]
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05 Sep 2012, 07:41
It uses correct tense and would instead of will.
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Re: The team announced that its record deteriorated [#permalink]
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06 Sep 2012, 20:56
SOURH7WK wrote:
Aristocrat wrote:
The team announced that its record deteriorated much further in the first half of the season than analysts had predicted them to and its performance will improve in the last half of the season.
A)had predicted them to and its performance will improve. Pronoun error.
B)had predicted and that its performance would improve. Correct choice. usage of would is correct here.
C)predicted they would and that it will improve its performance. Perfect tense required, since prediction was before announcement.
D)predicted them to and its performance would improve. Tense error
E)predicted and that it will have improved its performance. Tense error
The correct answer is B. Usage of would in past tense is correct. And predicted here needs to be in perfect tense to clarify the correct chronology.
Thanks for the explanation !!! +1 kudos to you !!!
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Re: The team announced that its record deteriorated [#permalink]
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07 Sep 2012, 06:45
I agree B is perfect.
Usage of past perfect tense is also another reason.
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Re: The team announced that its record deteriorated [#permalink]
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07 Sep 2012, 07:49
'B' is correct,
I have chosen answer as 'D'
but i read in Aristotle SC prep that its
"past perfect + would have"
and "simple past + would"
then how 'B' is correct?
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Re: The team announced that its record deteriorated [#permalink]
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18 Feb 2014, 08:02
Aristocrat wrote:
'B' is correct,
I have chosen answer as 'D'
but i read in Aristotle SC prep that its
"past perfect + would have"
and "simple past + would"
then how 'B' is correct?
"had" needs to be used as sequencing of events is required and "would" needs to be used as it is of subjunctive mood. "will" can be used when you are sure of things.
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Re: The team announced that its record deteriorated [#permalink] 18 Feb 2014, 08:02
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# How can you determine the parametric equation of a line?
A:
### Quick Answer
Determining the parametric equation of a line requires you to know at least two points on the line. This calculus-based equation can be used to prove that another given point is on that same line. This theorem states that with two given points (x1, y1) and (x2, y2), a third point (x, y) is on the line with these two points only if there is a real number "t" that satisfies the following equations: x = (1-t)x1 + tx2 and y = (1-t)y1 + ty2.
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Credit: Betsie Van der Meer The Image Bank Getty Images
### Full Answer
1. Determine the two points on a line
Find two points on a line and record them as x-y coordinates. In cases where it is a straight line, it may be oriented such that the x and y values are all equal to zero. In a word question, these will most likely be given.
2. Select the value for "t"
Select a value for "t" that is a real number. A real number can be rational or irrational, but it can't be imaginary. An example of an imaginary number would be the root of a negative number.
3. Insert the values into the equations
Insert the values of "t", (x1, y1) and (x2, y2) into the parametric equations to determine the value of point (x, y).
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PEOPLE SEARCH FOR | 524 | 2,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-50 | longest | en | 0.891977 |
https://math.stackexchange.com/questions/1188568/newtonss-method | 1,558,861,157,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259015.92/warc/CC-MAIN-20190526085156-20190526111156-00074.warc.gz | 536,680,260 | 33,862 | # Newtons's Method
If when you are using Newton's method and your results are just going back and forth between two values, say $0$ and $1$. It is $f(x)=x^3 -2x+2$ starting with $x=1$. What is the reasoning behind this?
• This means simply that the method doesn't work here :-(. – Karl Mar 13 '15 at 18:01
• Draw a graph for this scenario. – Simon S Mar 13 '15 at 18:02
• It means that the hypotheses for the convergence of Newton's method do not hold for the function you are analyzing. Can you give us more information on this function? In any case, you could try the Secant Method – Lonidard Mar 13 '15 at 18:07
• My problem specifically asks to use newton's method and explain what this means geometrically. It is f(x)=x^3 -2x+2 starting with x=1. I will try drawing a graph that will be a good start, thanks! – Nicole Mar 13 '15 at 18:09
• Why you don't give the details beforehand, but only now piecemeal wise? YOU are asking for help, we can't guess what your exact problem is if you don't say it :-((. – Karl Mar 13 '15 at 18:20
In terms of what it means about the function, it doesn't mean very much analytically. It can be interpreted geometrically, if you want. The tangent line to $f$ at $0$ is $y=-2x+2$ while the tangent line to $f$ at $1$ is $y=(x-1)+1=x$. These have their roots at $1$ and $0$ respectively. Newton's method finds the root of the tangent line, so starting at $0$ sends you to $1$, then back to $0$, etc. This geometric observation correctly suggests that we may see cycles if we start out relatively close to an extremum of $f$ (since the derivative changes sign between the two values that are being hit). | 459 | 1,641 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-22 | latest | en | 0.932251 |
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# If n and y are positive integers and n represents the number
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If n and y are positive integers and n represents the number [#permalink]
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If n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?
(1) $$\sqrt{n}$$ is odd number
(2) $$y=\sqrt{5^{2(n-1)}}$$
[Reveal] Spoiler: OA
Last edited by Bunuel on 11 Jul 2013, 00:56, edited 3 times in total.
Edited the question.
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Re: is y a perfect square? [#permalink]
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22 Aug 2011, 09:00
WishMasterUA wrote:
if n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?
1) square(n) is odd number
2) y=square(5^(2(n-1))
I think 'square' means 'square-root'. Considering it true, my explanation
\sqrt{n} is an odd number => n would also be an odd number
=> Sufficient to determine, Y would not have 'even number of factors' and that means it can't a perfect square.
y = \sqrt{5^(2(n-1))} => y= 5^(n-1) => if n is odd than (n-1) would be a perfect square but if n is even, y would not be perfect answer.
=> NOT Sufficient
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Re: is y a perfect square? [#permalink]
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22 Aug 2011, 09:53
anordinaryguy wrote:
=> Sufficient to determine, Y would not have 'even number of factors' and that means it can't a perfect square.
Can you explain that a bit more?
Edit: Found an answer myself: mathforum. org/library/drmath/view/72126.html
Such specific information...
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Re: is y a perfect square? [#permalink]
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22 Aug 2011, 10:20
anordinaryguy wrote:
WishMasterUA wrote:
if n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?
1) square(n) is odd number
2) y=square(5^(2(n-1))
I think 'square' means 'square-root'. Considering it true, my explanation
\sqrt{n} is an odd number => n would also be an odd number
=> Sufficient to determine, Y would not have 'even number of factors' and that means it can't a perfect square.
y = \sqrt{5^(2(n-1))} => y= 5^(n-1) => if n is odd than (n-1) would be a perfect square but if n is even, y would not be perfect answer.
=> NOT Sufficient
For (1) my understanding is different. A perfect square must have event exponents on it's prime factorization, but an odd number of factors. So like you said, root n is odd, meaning Y has an odd number of factors. If a number has an odd number of factors it is a perfect square.
e.g.
1X4
2X2
3 factors
1x64
2x32
4x16
8x8
7 factors
The perfect square always has odd factors.
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Re: is y a perfect square? [#permalink]
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22 Aug 2011, 15:27
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if a number equals = (p1^k)* (p2^l)*(P3^m)....
total no of factors= (k+1)*(l+1)*(m+1)....
If Sqrtroot of n is add => n is also odd.
here n= (k+1)*(l+1)*(m+1)....
now for n to be odd each of k, l,m should be even which implies that y is a perfect square.
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Re: is y a perfect square? [#permalink]
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22 Aug 2011, 15:28
in the above post, p1, p2, p3 are prime factors.
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Re: is y a perfect square? [#permalink]
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22 Aug 2011, 21:42
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WishMasterUA wrote:
if n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?
1) $$\sqrt{n}$$ is an odd number.
2) $$y=\sqrt{5^{2(n-1)}}$$
The question is based on the following concept:
If a number has odd number of factors, it must be a perfect square.
If a number has even number of factors, it cannot be a perfect square.
For why and how, check: http://www.veritasprep.com/blog/2010/12 ... t-squares/
n is the number of positive factors of y.
Question: Is y a perfect square?
Re-state the question as: Is n an odd integer?
Statement 1: $$\sqrt{n}$$ is an odd number.
If $$\sqrt{n}$$ is an odd number, n must be an odd number.
(All powers of an odd number are odd. If a is odd, a^2 is odd, a^3 is odd, a^4 is odd, $$\sqrt{n}$$, if integral, is odd etc.)
Since we know that n is odd, this statement is sufficient.
Statement 2: $$y=\sqrt{5^{2(n-1)}}$$
$$y=5^{n-1}$$
Obviously, the number of factors of y is n. Do we know whether n is odd? No, we don't.
Hence this statement alone is not sufficient.
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Re: If n and y are positive integers and n represents the number [#permalink]
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07 Mar 2017, 17:24
WishMasterUA wrote:
If n and y are positive integers and n represents the number of different positive factors of y, is y a perfect square?
(1) $$\sqrt{n}$$ is odd number
(2) $$y=\sqrt{5^{2(n-1)}}$$
Official solution from Veritas Prep.
This difficult question depends on an understanding of the properties of factors of a perfect square. The only way that a number can have an odd number of different total factors is if that number is a perfect square. Consider three numbers that are perfect squares, all of which have an odd number of total factors: 1, 9, and 16. The number 1 has only one unique factor (1); the number 9 has exactly three different factors (1, 3, 9); and 16 has five total factors (1, 16, 2, 8, 4). Statement (1) proves that n must be an odd integer so the information is sufficient to prove that y has an odd number of total unique factors and thus must be a perfect square. Statement (2) is more difficult to assess and requires that several different possible values of n be considered. For instance if n = 1 then y would be equal to 1 and it would indeed be a perfect square. However, if n = 2 then y would be equal to 5 which is not a perfect square. From those two values of n, it is clear that statement (2) is not sufficient and the correct answer is A, statement 1 alone is sufficient.
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If n and y are positive integers and n represents the number [#permalink]
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07 Oct 2017, 10:51
Statement 1 says square root of n is odd -> n is also odd and we know that total number of factors of any number can be derived using (n+1)(m+1)... where n & m are the power of any prime factors of a number. (Example: 108 = 2^2*3^3, so the total number of factors is (2+1)*(3+1) = 3*4 = 12 ) Hence, for perfect number such as 4, 9, total number of factors will always be an odd, unlike non perfect numbers, since (even + 1 = odd) -> (2+1 = 3, 6+1 =7). Hence, since statement 1 clearly says y has an odd number of factors, y is definitely a perfect square. Sufficient!
With Statement 2, we can easily put values of n as 1, 2 & 3 and then we can easily derive that with n=2, y is 5 which is not a perfect square and with n=3, y=25 -> clearly a perfect square. Hence, Not sufficient!
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If n and y are positive integers and n represents the number [#permalink] 07 Oct 2017, 10:51
Display posts from previous: Sort by | 2,821 | 9,642 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-51 | latest | en | 0.822085 |
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Academic support for Primary 4
### Help with model maths for P4 child
My son is an average student in his studies and I'm fine with that. I want him to enjoy his childhood and not bury himself in studies.
As such, I want to pinpoint the weak areas of his work. So far I manage to get right in his English and Chinese.
For his maths, I have pinpointed that his weak area is in word problems, ie. Section C.
You see, he can consistently score 70%+ in Section A, consistently score 60%+ in Section B, but score as low as ZERO for Section C!!!!
I have narrowed down the problem itself and it is aligned with what the teachers has told me: My son is weak in model diagram.
For some reasons, he cannot visualize the word problems into diagrams. And, his logical deduction of problems in math is very weak too.
All I need to do to enable him to visualize the problems, do you have any recommendations?
nkthen
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### Re: Help with model maths for P4 child
Have you tried asking him to break up the questions into parts first? He could be stumped by the mass of words and does not know where to start.
acforfamily
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### Re: Help with model maths for P4 child
Actually it starts by understanding the question. If the question cannot be understood, it would be hard to interpret it into a diagram.
Guest
### Re: Help with model maths for P4 child
nkthen wrote:You see, he can consistently score 70%+ in Section A, consistently score 60%+ in Section B, but score as low as ZERO for Section C!!!!
consistently scored 70+ and 60+? are the marks loss due to carelessness or simply lack of practise?
Without any clearer information, this gives the impression that he might be weak in maths.
nkthen wrote:For some reasons, he cannot visualize the word problems into diagrams. And, his logical deduction of problems in math is very weak too.
Once again without any further information, I m inclined to think that his inability to visualize the word problems into diagrams has something to do with his weak deduction skills.
When u mention visualizing, you are looking at the process of translating the information given in a question to a model diagram. This may requires some logical deduction.
CoffeeCat
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### Re: Help with model maths for P4 child
CoffeeCat wrote:
nkthen wrote:You see, he can consistently score 70%+ in Section A, consistently score 60%+ in Section B, but score as low as ZERO for Section C!!!!
consistently scored 70+ and 60+? are the marks loss due to carelessness or simply lack of practise?
Without any clearer information, this gives the impression that he might be weak in maths.
nkthen wrote:For some reasons, he cannot visualize the word problems into diagrams. And, his logical deduction of problems in math is very weak too.
Once again without any further information, I m inclined to think that his inability to visualize the word problems into diagrams has something to do with his weak deduction skills.
When u mention visualizing, you are looking at the process of translating the information given in a question to a model diagram. This may requires some logical deduction.
OK, to clarify, when it comes to simple, direct maths questions, he can answer most of them. They are mostly found in Section A and B. Section C is the one that he finds it hard to do. We have practiced many times but he still have problems doing the very first step, which is to draw the model.
nkthen
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### Re: Help with model maths for P4 child
I would deduce from your post that your son actually simply do not understand the problem sum question and is unable to extract the required information to even take the 1st step to draw the model.
Instead of continuing to coach him on the P4 problem sums that require more complex model drawing, what you can do is to start from square 1 for model drawing and add on from there:
1. Pull out the simple problem sums and make him do the model drawing from there. These can be just using simple +-x/ computation only.
2. If he is able to do so, start putting in additional info (more numbers) into the simple problem sums slowly. He will have to change / redraw his model to reflect this new info.
3. Move on to teach him additional modelling skills, such as "equalise" the parts, etc.
You can move thru' step 1 and 2 very quickly if you find his foundation ok. From step 3 onwards, it will require you to learn and understand the concepts before you can find the best way to put thru' to him.
Hope this helps.
Pen88n
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### Re: Help with model maths for P4 child
In such cases, I would think that the child cannot understand the teacher's teaching. It could be the teacher is too fast in her teaching or her teaching method does not suit the child. Maybe getting a good tutor and have a one to one tuition with the child will help the child to understand the concepts better. But where to find good tutor?
PiggyLalala
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### Re: Help with model maths for P4 child
The other thing is model diagram is only a means, not the end. So while finding out his issue with problem sums, put model diagram aside. Once you know the root, model is used to assist explanation. If model is put as the main goal, and if he has a fear for models, it becomes a mental block.
If he understands algebra better, why not try that as well? There are many roads leading to Rome.
Guest
### Re: Help with model maths for P4 child
Hmm, what's his background with modelling? When was he taught modelling? Did he draw model before p4?
I won't be surprised if he happen to be a beginner. When I was first taught modelling in p4 I didn't grasp it immediately too. Maybe
his modelling foundation is just weak.
CoffeeCat
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### Re: Help with model maths for P4 child
ksi wrote:The other thing is model diagram is only a means, not the end. So while finding out his issue with problem sums, put model diagram aside. Once you know the root, model is used to assist explanation. If model is put as the main goal, and if he has a fear for models, it becomes a mental block.
If he understands algebra better, why not try that as well? There are many roads leading to Rome.
Yup, most of the questions on ratios and percentages can be solved easily by simple algebraic equations. If a child has great difficulties in models, then the 'algrebra approach' may be a good alternative.
PiggyLalala
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Return to Primary 4 | 1,690 | 6,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-17 | latest | en | 0.976946 |
https://math.stackexchange.com/questions/2946702/eigenvalues-of-translates-of-polynomial?noredirect=1 | 1,558,594,672,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257156.50/warc/CC-MAIN-20190523063645-20190523085645-00511.warc.gz | 538,893,320 | 31,774 | # Eigenvalues of translates of polynomial
I am looking for change in eigenvalues as $$p(T)$$ is translated as $$p(T+I)$$. I thought the following link could be helpful, though I can't figure out any. $$T$$ being some matrix such that $$p(T)=0$$
How to prove "eigenvalues of polynomial of matrix $$A$$ = polynomial of eigenvalues of matrix $$A$$ "
Can the variation in eigenvalues be seen in terms of the polynomial operator 'p' itself? If so, can this be generalized in any sense? Let me know if the question needs clarification. Possible hints or solution will be a great help. Regards.
• I'm not sure if I got your post right, is $p(t)$ the characteristic polynomial of some linear map or matrix? – Diglett Oct 8 '18 at 5:37
• The edits might be clear, hopefully. – user511110 Oct 8 '18 at 5:56
• I do not know if this useful but if $a$ is an eigenvalue of $A$ then $f(a)$ is an eigenvalue of $f(A)$ – StammeringMathematician Oct 8 '18 at 6:01
• If T has eigenvalue $a$ then T+I has eigenvalue 1+a. So p(T+I) will have eigenvalue p(a+1) – StammeringMathematician Oct 8 '18 at 6:06
• We are not sure whether $p(a+1)=0$ or not. How can you make sure? – user511110 Oct 8 '18 at 6:09 | 354 | 1,185 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-22 | latest | en | 0.839245 |
https://gutomna.com/boil/at-what-temperature-does-water-stop-boiling.html | 1,685,854,669,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00058.warc.gz | 339,009,417 | 19,394 | # At what temperature does water stop boiling?
Contents
At 100 degrees Celsius (water’s normal boiling point), the vapour pressure is 1 standard atmosphere, or 0.101325 megapascal (MPa). At this point, water starts to evaporate and switch states from a liquid to a vapour (steam). At room temperature, water’s vapour pressure is much lower, so it’s stable and doesn’t boil.
## Why does water stop boiling at 100 degrees?
Because of the energy used for evaporation the water temperature drops below the boiling point, and the water stops boiling. This happens fast because the heat energy stored in the water and the vessel is small compared to the energy needed to evaporate water.
## Does water stop boiling?
The water stays at 100C until its fully boiled — however it stops boiling once the heat source is removed, as there’s no more energy being added for the phase transition from liquid to vapor.
## At what temperature does water stop bubbling?
If you stop heating boiling, the water immediately stops forming bubbles, but it still has 100°C (for a short while). If you reheat it, it rebubbles instantly.
## Can water boil over 100 degrees?
Liquid water can be hotter than 100 °C (212 °F) and colder than 0 °C (32 °F). Heating water above its boiling point without boiling is called superheating. … Water that is very pure, free of air bubbles, and in a smooth container may superheat and then explosively boil when it’s disturbed.
## How hot is steam from boiling water?
Steam occurs when water goes above 212 degrees Fahrenheit, which is hotter than water when it is at its stable point. While water boils at 212 F, steam is at a much higher temperature as water turns to vapor.
## Can water get over 212 degrees?
It is not true that water can only get up to 212 degrees and as cold as 32 degrees. After water changes from a liquid to a gas (at 212 degrees Fahrenheit) it can actually heat up much hotter than that.
## How can you boil water without electricity or gas?
Here’s a list of alternate ways to boil water, without electricity :
1. Candles.
2. BBQ Grill Pit.
3. Fire Pit.
4. Gas Stove.
5. Camp Stove.
6. Fireplace or Wood Stove.
7. Solar Cooker.
8. Rocket Stove.
17.02.2021
## How long should you boil water?
Boil water, if you do not have bottled water. Boiling is sufficient to kill pathogenic bacteria, viruses and protozoa (WHO, 2015). If water is cloudy, let it settle and filter it through a clean cloth, paperboiling water towel, or coffee filter. Bring water to a rolling boil for at least one minute.
## How can you boil water without electricity?
A woodstove or fireplace can be a lifesaver during a power outage, and not only as a general source of heat. You can also boil water as you would over a campfire, with the pot placed atop a woodstove, or via the built-in water reservoirs some models include.
IT IS INTERESTING: Is half boiled egg good for you?
## How can I boil water faster?
Truth: Hot water boils faster.
But it might heat faster if it starts higher. If you’re in a hurry, turn your tap to the hottest setting, and fill your pot with that hot tap water. It’ll reach boiling a bit faster than cold or lukewarm water. You can also get the water even hotter by using your electric kettle.
## What happens when water is boiled for a long time?
However, re-boiling of water makes the dissolved gases and minerals that have been chemically changed, toxic. Such chemical alterations cause serious health effects from arsenic, nitrates and fluoride.
## Why does temperature remain constant when water boils?
The temperature remains constant during boiling of water even though heat is supplied constantly because all the heat energy provided is used up in changing the state of water from liquid to gaseous water vapour.
## Can water boil at 200 degrees?
Sea Level: Water boils at 212 degrees F. and simmers at 190 degrees F. … Simmer – 185 to 200 degrees F.
## Is steam hotter than fire?
Why steam is more dangerous than fire? Steam causes more severe burns. So, when steam falls on skin and condenses to produce water it gives out 22.5 x 105 J/kg more heat than boiling water at the same temperature. Steam has more energy than boiling water.
## Is steam hotter than boiling water?
Steam is hotter than boiling water because steam gets some extra heat i.e latent heat (the amount of heat require to change its state . … Boiling water is a saturated liquid vapor mixture but steam is either a saturated vapor or a superheated vapor.
IT IS INTERESTING: Can we store boiled soya chunks? | 1,038 | 4,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.925188 |
http://www.cs.rochester.edu/twiki/bin/rdiff/Main/ICA?rev1=3;rev2=2 | 1,537,830,937,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160842.79/warc/CC-MAIN-20180924224739-20180925005139-00212.warc.gz | 309,188,322 | 4,931 | # Difference: ICA (2 vs. 3)
#### Revision 32007-01-03 - FabrizioMorbini
Line: 1 to 1
META TOPICPARENT name="SKIL"
#### Paper
Inheritance Comes of Age: Applying Nonmonotonic Techniques to Problems in Industry.
Line: 52 to 52
• find the maximally consistent subset made doing to union of the set of formulas linked to N_i and N and the set B.
2 problems:
Changed:
<
<
• determining consistency is semi-decidable: consider only a subclass of formulas for which deciding consistency is decidable. But still, deciding consistency is intractable, ...
>
>
• determining consistency is semi-decidable: consider only a subclass of formulas for which deciding consistency is decidable. But still, deciding consistency is intractable. The author suggests a divide-and-conquer strategy: to compute consistency consider only the formulas belonging to a certain type. So, this implies that the formulas are dived into type of formulas (in the paper: administrative, medical, ...) and formulas in a type are always consistent with formulas in the other types.
• there are more than 1 (typically) maximal subset. Which formula to remove: build a preference order.
Changed:
<
<
#### Observations (starting points for discussion)
>
>
The preference criteria are;
• specificity: prefer more specific information. Given that bird fly and that Tweety (a bird) doesn't fly, then prefer the Tweety knowledge about flying instead of the generic bird knowledge because it is more specific to respect to Tweety.
• order: when specificity doesn't help (multiple paths), then use the order O to choose which path to inherit from.
Open questions:
• what to do with preempting paths. See figure 7 for an example.
• determining conflicting paths can depend on knowledge. Therefore, it might be the case that in the network two path are not conflicting each other but given the background knowledge they are.
Why using this mixed representation mechanism instead of a either all-network or all-formulas approach? * converting everything to network goes against the principle that the nodes in a network represents concrete entities (now formulas would become nodes). Plus, the definition of inconsistency becomes difficult (because it depends on some property of paths not easily specifiable). * converting everything to formulas is not optimal because the distinction between taxonomic knowledge and general knowledge is lost. Plus efficient algorithms could be harder to write.
Conclusions:
• check for problems for real world applications, use them instead of toy examples.
• basic research is still needed
• balance between basic reasearch and industrial collaboration
• develop tools (not only on paper)
• solving a real problem is typically harder and more complex than solving a toy problem.
-- FabrizioMorbini - 03 Jan 2007 | 586 | 2,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-39 | latest | en | 0.869101 |
https://homework.cpm.org/category/ACC/textbook/ccaa8/chapter/11%20Unit%2012/lesson/CCA:%2011.3.3/problem/11-98 | 1,709,407,605,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00886.warc.gz | 290,383,159 | 15,534 | Home > CCAA8 > Chapter 11 Unit 12 > Lesson CCA: 11.3.3 > Problem11-98
11-98.
Is $\left(-5,7\right)$ a solution to $y<\sqrt{x^2+24}$? Show how you know.
Substitute $-5$ for $x$ and $7$ for $y$, then simplify.
Did $\left(-5,7\right)$ make the inequality true? | 101 | 261 | {"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-10 | latest | en | 0.608351 |
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# Profile evaluation
new topic post reply Update application status
Author Message
Intern
Joined: 05 Aug 2018
Posts: 18
Location: India
GMAT 1: 660 Q49 V30
GPA: 3.75
### Show Tags
13 Sep 2018, 03:47
I gave GMAT and scored 660 (Q 49,V 30,IR 3).I am a marine engineer with 4 years of work experience.I have sailed only on foreign going vessels till now in my career with different nationalities every time.I have done B Tech in mArine engineering from MERI Kolkata.Indian maritime university.I graduated with an aggregate percentage of 76.7% (with distinction).I have worked with several NGO’S throughout my career.My present age is 26.I have not given TOEFL yet.My Percentage in class X IS 90 and in Class 12 ,It is 81 .What are my chances for Tepper,IESE,Boston university, Kelley,IE Business school ?
Joined: 16 Jan 2012
Posts: 1563
### Show Tags
13 Sep 2018, 12:14
I gave GMAT and scored 660 (Q 49,V 30,IR 3).I am a marine engineer with 4 years of work experience.I have sailed only on foreign going vessels till now in my career with different nationalities every time.I have done B Tech in mArine engineering from MERI Kolkata.Indian maritime university.I graduated with an aggregate percentage of 76.7% (with distinction).I have worked with several NGO’S throughout my career.My present age is 26.I have not given TOEFL yet.My Percentage in class X IS 90 and in Class 12 ,It is 81 .What are my chances for Tepper,IESE,Boston university, Kelley,IE Business school ?
I had answered your post on the Kelley thread. I believe retaking the GMAT is the way to go right now, specifically for Kelley and Tepper. You can also apply with a 660 GMAT score, which is not bad at all. However, you should then revise your list of target programs. If you are interested in studying and later working in operations, then a lot of American and European MBAs will fit your requirements with a lower GMAT score as well.
_________________
Namita Garg,
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Re: Profile evaluation [#permalink] 13 Sep 2018, 12:14
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# Profile evaluation
new topic post reply Update application status
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 734 | 2,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-22 | latest | en | 0.935617 |
https://math.iisc.ac.in/seminars/2019/2019-01-04-hassain-m.html | 1,719,203,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00694.warc.gz | 329,794,218 | 6,266 | #### Algebra & Combinatorics Seminar
##### Venue: LH-1, Mathematics Department
Let $\mathfrak{O}$ be the ring of integers of a non-Archimedean local field such that the residue field has characteristic $p$. Then the abscissa of convergence of representation zeta function of Special Linear group $\mathrm{SL}_2(\mathfrak{O})$ is $1.$ The case $p\neq 2$ is already known in the literature. For $p=2$ we need more tools to prove the result. In this talk I will discuss the difference between those cases and give an outline of the proof for $p=2.$
Let $\mathfrak{p}$ be the maximal ideal of $\mathfrak{O}$ and $|\mathfrak{O}/\mathfrak{p}|=q.$ It is already shown in literature that for $r \geq 1,$ the group algebras $\mathbb C[\mathrm{GL}_2(\mathfrak{O}/\mathfrak{p}^{r})]$ and $\mathbb C[\mathrm{GL}_2(\mathbb F_q[t]/(t^{r}))]$ are isomorphic. Also for $2\nmid q,$ the group algebras $\mathbb C[\mathrm{SL}_2(\mathfrak{O}/\mathfrak{p}^{r})]$ and $\mathbb C[\mathrm{SL}_2(\mathbb F_q[t]/(t^{r}))]$ are isomorphic. In this talk I will also show that if $2\mid q$ and $\mathrm{char}(\mathfrak{O})=0$ then, the group algebras, $\mathbb C[\mathrm{SL}_2(\mathfrak{O}/\mathfrak{p}^{2\ell})]$ and $\mathbb C[\mathrm{SL}_2(\mathbb F_q[t]/(t^{2\ell}))]$ are not isomorphic for $\ell > \mathrm{e}$, where $\mathrm{e}$ is the ramification index of $\mathfrak{O}.$
Contact: +91 (80) 2293 2711, +91 (80) 2293 2265 ; E-mail: chair.math[at]iisc[dot]ac[dot]in
Last updated: 24 Jun 2024 | 512 | 1,476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.711804 |
https://www.coursehero.com/file/8706449/t-bd-DP-example-should-I-stay-or-should-I-go-QA-stay-QA/ | 1,481,343,834,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542938.92/warc/CC-MAIN-20161202170902-00237-ip-10-31-129-80.ec2.internal.warc.gz | 913,656,830 | 21,420 | Lecture 21 Notes
# T bd dp example should i stay or should i go qa stay
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Unformatted text preview: (s) % J*(s) or h(s, a) % Q*(s,a) If we have h(s) = J*(s), only need to build first two levels of tree (action and outcome) to choose optimal action at s1 With h(s, a) = Q*(s,a), only need to build first (action) level Often try to use h % J# or Q# for some good # Roll-outs Want h(s) % J#(s) Starting from s1 = s, sample a1 ~ #(a | s1), set c1 = c(s1,a1), sample s2 ~ T(s’ | s1,a1) Repeat until goal (or until γt small) Take h(s) = (1–γ)/γ &t γtct Used in UCT (best algorithm for Go) Dynamic programming If there are a small number of states and actions, makes sense to memoize tree search ‣ compute an entire level of the tree at a time, working from bottom up ‣ store only S \$ A numbers r.t. bd DP example: should I stay or should I go? Q(A, stay) Q(A, go) J(A) DP example 2 each step costs 1 discount 0.8 DP example 2—iteration 0 Q(s,left) 1.5 1 0.5 0 0.5 0 5 10 15 5 10 15 Q(s,right) 1.5 1 0.5 0 0.5 0 State DP example 2—iteration 1 Q(s,left) 1.5 1 0.5 0 0.5 0 5 10 15 5 10 15 Q(s,right) 1.5 1 0.5 0 0.5 0 State DP example 2—iteration 3 Q(s,left) 1.5 1 0.5 0 0.5 0 5 10 15 5 10 15 Q(s,right) 1.5 1 0.5 0 0.5 0 State DP example 2—iteration 4 Q(s,left) 1.5 1 0.5 0 0.5 0 5 10 15 5 10 15 Q(s,right) 1.5 1 0.5 0 0.5 0 State DP example 2—iteratio...
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Ask a homework question - tutors are online | 608 | 1,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-50 | longest | en | 0.717083 |
https://www.jobilize.com/trigonometry/definition/1-1-real-numbers-algebra-essentials-by-openstax?qcr=www.quizover.com | 1,558,890,245,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259327.59/warc/CC-MAIN-20190526165427-20190526191427-00061.warc.gz | 815,369,059 | 16,626 | # 1.1 Real numbers: algebra essentials
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• Card 11 / 35:
algebraic expression
constants and variables combined using addition, subtraction, multiplication, and division
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The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1) | 424 | 1,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-22 | latest | en | 0.842921 |
https://www.codespeedy.com/shift-elements-of-an-array-to-left-in-java/ | 1,674,986,085,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00252.warc.gz | 739,145,907 | 15,397 | Shift elements of an array to Left in Java
In this tutorial, we will learn how we can shift the elements of an array to left using loops in Java.
In this problem, we shift each of the elements of the array to left with the number of shifts specified by the user.
Example 1:
If an array consists of elements arr = {1, 2, 3}, then on shifting these elements towards the left direction by one we would get arr = {2, 3, 1}.
Example 2:
If an array consists of elements arr = {20, 30, 10, 40}, then on shifting these elements towards the left direction by three we would get arr = {40, 20, 30, 10}.
Code to Shift elements of an array to Left in Java
```import java.util.*;
class leftShift
{
public static void main(String[]args)
{
Scanner scan = new Scanner(System.in);
System.out.print("Enter the size of the array: ");
int size = scan.nextInt();
int[]arr = new int[size];
System.out.print("Enter the elements of the array: ");
for(int i = 0; i < size; i++)
{
arr[i] = scan.nextInt();
}
int[]temp = new int[ size ];
System.out.print("Enter the number of times to be shifted: ");
int shift = scan.nextInt();
while(shift >= size)
{
shift = shift - size;
}
int j = 0;
for(int i = 0; i < size; i++)
{
if(shift + i >= size)
{
temp[i] = arr[j];
j = j + 1;
}
else
{
temp[i] = arr[shift + i];
}
}
System.out.print("The shifted array is: ");
for(int i = 0 ;i < size; i++)
{
System.out.print(temp[i] + " ");
}
}
}```
Output:
```Enter the size of the array: 6
Enter the elements of the array: 1 5 2 3 4 9
Enter the number of times to be shifted: 3
The shifted array is: 3 4 9 1 5 2```
```Enter the size of the array: 3
Enter the elements of the array: 9 3 1
Enter the number of times to be shifted: 5
The shifted array is: 1 9 3
```
Explanation
1. Firstly we declare the size of the array and create an array ” arr ” and store the elements of the array in it.
2. Next scan the shift variable which denotes the number of times the array is supposed to be rotated.
3. Create another array called ” temp ” and check if shift is greater than the size of the array. If so using a while loop decrement the value of shift.
4. Then store the elements of the initial array ” arr ” in the ” temp ” array by using the logic below.
5. In the end, print the elements in the ” temp ” array which contains the shifted elements of the initial array.
Hope you understand the code 🙂
Any questions feel free to let us know in the below comment section. | 670 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | latest | en | 0.646666 |
https://www.brainbashers.com/sudokuanswer.asp?date=0126&diff=3 | 1,685,693,335,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00302.warc.gz | 750,570,362 | 6,220 | # Daily Sudoku Answer
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Jan 26 - Medium
Puzzle Copyright © Kevin Stone
Share link: www.brainbashers.com/s125103
## Reasoning
R5C6 is the only square in row 5 that can be <1>
R5C2 is the only square in column 2 that can be <5>
R1C8 is the only square in column 8 that can be <1>
R6C9 is the only square in column 9 that can be <3>
R6C6 can only be <5>
R9C4 is the only square in column 4 that can be <5>
R9C6 is the only square in row 9 that can be <3>
R1C6 can only be <4>
R4C6 can only be <9>
R2C5 can only be <2>
R2C1 can only be <5>
R4C9 can only be <8>
R4C4 can only be <7>
R5C8 can only be <9>
R5C7 can only be <2>
R9C8 can only be <8>
R9C2 can only be <6>
R2C9 can only be <4>
R4C1 can only be <6>
R1C4 can only be <3>
R1C2 can only be <8>
R1C7 can only be <9>
R3C1 can only be <3>
R3C3 can only be <4>
R5C4 can only be <8>
R3C5 can only be <7>
R1C9 can only be <7>
R7C7 can only be <5>
R9C7 can only be <4>
R3C9 can only be <5>
R3C7 can only be <8>
R4C5 can only be <4>
R1C1 can only be <2>
R6C1 can only be <8>
R5C3 can only be <7>
R5C5 can only be <3>
R6C4 can only be <2>
R6C5 can only be <6>
R1C3 can only be <6>
R9C3 can only be <2>
R9C9 can only be <9>
R7C3 can only be <8>
R9C1 can only be <7>
R8C9 can only be <1>
R7C5 can only be <9>
R7C1 can only be <1>
R8C5 can only be <8>
R8C1 can only be <9>
R7C9 can only be <2>
## Today's Sudoku Puzzles
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Note: BrainBashers has a Dark Mode setting. | 628 | 1,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-23 | latest | en | 0.695848 |
http://www-stat.wharton.upenn.edu/~steele/Publications/HTML/Sooseo.html | 1,726,610,817,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00635.warc.gz | 37,524,556 | 8,402 | The Annals of Probability
1980, Vol. 8, No. 6,1079 1084
SIZES OF ORDER STATISTICAL EVENTS OF STATIONARY
PROCESSES'
BY DANIEL RUDOLPH AND J. MICHAEL STEELE
Stanford University
Given a process (X,}, any permutation a:[ 1, n] 1, n] determines an order statistical event A (a) = (X,(,) < X.(2) < ... < X.(,)}. How many events A (a) are needed to form a union whose probability exceeds 1 E? This question is answered in the case of stationary ergodic processes with finite entropy.
1. Introduction. One of the key properties of independent, identically distributed continuous random variables Y, 1 :s i:s n, is that the order statistical events defined by
{CO: Yo(l) < Yo(2) < ... < Yo(.) }
have the same probability 11M for any permutation cr: [1, n] + [1, nj. The main objective of this paper is to determine the extent to which this property is retained asymptotically for processes which are only assumed to be stationary and ergodic.
To set the problem precisely, we suppose that (X } ' % 1 is a strictly stationary, ergodic process
defined on the probability space (2, A, P). We will also use the representation ofsuch a process
by X(.) = f(T L+lo), 1 :5 i < , where T: 2 > 2 is an ergodic measure preserving
transformation and f: 2 > R is a measurable map. To avoid inessential messiness, we also
restrict attention to processes which satisfy the continuity property
(1.1) p(xl = Xi) = 01
Our approach to the analysis of the order statistical events is motivated by the ShannonMeMillan Breiman theorem, and particularly the phrasing of that result in terms of the equipartition property Q 1 ], page 13 5, [9], page 3 5 (6.3)). Loosely speaking, that phrasing tells one in terms of the entropy of Tjust how many sets of a certain type are needed to cover most of 2.
To establish a comparable result for the order statistical events, we let Qn(F) be defined for any F EE ~F by
Q.(F) X.(1) (W) < X.(2) (W) < < Xo(tt)(W), for some w EE F}
Here 1 S 1 denotes the cardinality of the set S, so Q.(F) is equal to the least number of order statistical events
A. = {CO: Xo(l)((0) < X(2)(W) < ... < Xo(tt)(W)} which one needs to cover F.
The quantity of main interest is now defined for E > 0 by
(1.2) Qn*(E) = minE.P(E),, Q.ffi\E),
80 Qn*(E) is the least number of A. which will cover a set of probability 1 E.
To familiarize Qn*(E), we note that if the (X,}', , are i.i.d. and satisfy (1.1), then Qn* is equal to the least integer greater than (1 E)M.
In this particular example {X,}', , is a process with infinite entropy, and Q,*, (E) is near its
' Research of the second author was supported in part by the National Science Foundation (NSF MCS 77 16974).
AMS 1970 subject class~fications. Primary 60G10; secondary 60005.
Key words andphrases. Order statistics, entropy, stationary processes, de Bruijn sequences, directed graphs, equipartition property.
1079
1080 DANIEL RUDOLPH AND J. MICHAEL STEELE
a priori upper bound. One intuitively expects that Qn* (E) should be of smaller order than n! for processes with finite entropy. We show more precisely that in that case Qn*(E) is, in fact, exponentially smaller.
THEOREM 1. For any stationary ergodicprocess {X} ', , which hasfinite entropy and satisfies P(Xi = X,) = 0, i 0 j, there isfor any E > 0 a sequence ofpositive reals p. tending to zero for which
(1.3) Qn*(E):5 (n !),n, for all n ~ 1.
n
Before giving the complementing result, two comments are in order. In the first place we note there are many processes satisfying the hypotheses, since iff satisfies P (w : f '(w) = y} = 0 for ally E R, the condition P(X, = Xj) = 0, i 0j, trivially holds for any measure preserving T. Also, transformations of finite entropy not only abound but play key roles in such distinct subjects as statistical mechanics and the metrical theory of diophantine approximation Q11).
Second, we note (1.3) is equivalent to saying Qn*(E) = o(p'n!) for each p > 0. The phrasing of Theorem 1 was chosen in view of the next result which makes precise the sense in which Theorem 1 is best possible.
THEOREM 2. For any positive p. which tend to zero there is a stationary, ergodic process (X),' , with P(X = Xj) = 0, i 0 j, which has zero entropy, and which satisfies
Qn*(E) k (n!)p,~ for infinitely many n and any E < 1.
The preceding theorem is easily seen to be a consequence of the next result which shows that the underlying T plays a surprisingly small role in determining Qn* (E).
THEOREM 3. Given any ergodic measure preserving transformation T on a nonatomic
probability space P), andgiven anyp,, > 0 tending to zero, there is anf: 2 > [0, 1] which
satisfies
(1.4) P({w:f '(w) = X}) = 0, V X E [0, 1],
and
!),n
(1.5) Qn*(e) k (n
for infinitely many n and any E < 1.
In the next section we give the proof of Theorem 1 as a consequence of a counting argument and the application of the Shannon MeMillan Breitnan theorem to an appropriately chosen partition.
The proof of Theorem 3 is more subtle and makes use of a generalization of a combinatorial structure known as de BruiJn sequences.
Since the construction provides a technique,for building copies of a finite sequence of independent random variables inside a general stationary process, the construction should be useful in a variety of problems.
Finally, in the fourth section a brief speculation on the theory of order statistical events is ventured.
2. The upper bound method. For any measure preserving transformation S: 2 > 2 and any partition 9 = {P,} of 2, the sets given by
(2.1) n n 1
j 0 (w: S i(w) EE Pi,} for some 1 :5 i, :S s will be called the n p S name associated with the n p S ~~ (ii, i2,
SIZES OF ORDER STATISTICAL EVENTS 1081
ij}. If S is ergodic and has entropy H(S) < cz, the Shannon MeMillan Breiman theorem says there is an no = no(E, a, S, p) such that for n k no there is a collection of 2` of the n p S names whose union has probability at least 1 E.
Since P(X = Xj) = 0, i 0 j, the disjoint sets P. given by the permutations of { 1, 2, k}
by
P. = {W: X.0) < X(2) < ... < X(k)}
have union 2 (except for a set of measure zero). The partition p = (P,} can be related usefully to the possible orderings of {Xi}~, ',.
LEMMA2.1. For any n p Tk nameA we have
Q.k(A):5 (nk)!1(k!)".
PROOF. First consider n = 2. The 2 p Tk name A has an associated sequence (ii, i2); and il determines the ordering of R, = {Xi(w), X2(w), . , Xt(W)}, while i2 determines the order of R2 = {Xk+l(W), Xk+2(W), ..., X2k(w)}. To count the possible orderings of R, u R2, we note the set R, determines k + 1 intervals ( , X(,)), (X(I) X(2)), * * *, (X(.), ) where {X(Jk, , are the order statistics of (Xi}*, i. Since there are 2k ways of putting the order statistics of {Xj}?, A!k,l into the k + 1 intervals, we have
Q1t(A):5 2k) (2k)!/(k!)2. ( k
In general, we see for R. = Mk+ly X~2, * * *, X.+1)k} and 0:5j < n thatuj' o R., determines
(t + 1)k + 1 intervals into which the order statistics of R,, can be placed in (t + ')k ways. ( k
Making the sequential choices we have
Q.k(A):5 2k)(3k) ... (nk (nk)!(k!) ",
( k k k
which completes the lemma. 0
To prove Theorem 1 we need to bound the number of n p TA' names which are needed to cover a set of probability 1 E. We first note that any nk T p name is contained in some n Tk p name because for any alias Qj}pko ' one has
jkW E= p,,.,} nk 1
n.n 01 (w: T D nj 0 (w: T iw EE PJ.
The Shannon MeMillan Breiman theorem applied to the ergodic transformation T with entropy H(T) < a < oo says there is a collection W of 2,nk of the nk p T names whose union contains 2\E with P(E) < E for all n k no = no(E, p). By the preceding remark this also implies there is a collection W' of 2"nk of the n p T* names with the same property.
We now see
Qnkffi\E):5 EA.= W' Qnk(A)s 2'"k(nk)!(k!) n
where the last inequality follows from Lemma 2. 1. For any k we can write m = nk + r with 1 :5 rs k one has
QM.E):s Qtn,1)k(E) 5 2'n')k((n + 1)k)!(k!) n 1
provided n k no. k k
Since (n + 1)k!lm! :5 ((n + 1)k)~ and k! k k e ' we have
Qt(E)s m![2(*+')'((n + 1)k)k(kle) k (n+l)
1082 DANIEL RUDOLPH AND J. MICHAEL STEELE
and the fixed integer k was arbitrary, so
QM*(.E) O(Pmm!)
for all r > 0. The implied constant depends not only on H(T) but on T through the no given by the SMB theorem. As noted earlier, this last relation is sufficient to imply Theorem 1.
3. The lower bound method. The first lemma required for Theorem 3 is the so called strong form of Rohlin's lemma Q8], page 22) which provides a systematic method for applying combinatorial constructions to stationary processes.
LEMMA 3. 1. Suppose T is an ergodic, measure preserving transformation on a nonatomic probability space (2, A, P). For anyfinitepartition v= (H,, H2, .. ., H.) of a andfor any real E > 0 and integer m there is an E E Y with thefollowing properties:
(3.1) E, T 1E, . . , T'lE are disjoint;
(3.2) P(U. 50' T 1E) ~ 1 .E;
(3.3) P((T 'E) n H.) = P(E)P(Hj), 0:5i
The second lemma we need is a graph theoretic result due to I. J. Good Q51, [61, page 95) which sharpens a well known result of Euler.
LEMMA 3.2. If G is a connected directed graph, and !f at each point of G there are the same number of arcs going out as coming in, then there is a directed cycle in G that goes through every arc of G in its given direction, and uses no arc twice.
As an application of Lemma 3.2, we will obtain the existence of what can be called s ary de Bruijn sequences. To introduce these sequences, we recall the classic result of de Bruijn which says the following: given a positive integer n, there is a sequence of O's and I's of length N = 2n, say aia2a3 ... aN, such that the n tuples aa,+, ... a,.,, , are a complete list of all 2' of the n tuples with alphabet {0, 1}.
Here a, is understood to follow aN, etc. in the cycle. For example, when n = 3, the cycle (00010111) contains every 3 tuple of O's and I's exactly once. We will use the following generalization where d is an alphabet of s letters and 2 is the set of all ordered k tuples of sy.
LEMMA 3.3. There is a sequence a, a2 ... aiv with N = sk of the elements ofc?y such that each element of 3 occurs exactly once in the set of k tuples (ar+l, ar+2, a,+k), where 0 5 r < sA' and at = a. !f t > s* and t sk = u.
PRooF. We define a directed graph G whose edges are the ordered (k 1) tuples formed by elements of sY. We have an edge from (bi b2 ... bk i) to (b l, ..., Mk 1) provided b2 = b11, b3 = bl 2 = bk j and b'k i is arbitrary. Every vertex has in degree and out degree equal to S, so Lemma 3.2 implies there is a cycle which traverses the edges of G and uses each exactly once. From such a cycle the sequence of a, EE d given by the successive bk_1 is easily checked to satisfy the claim of the lemma.
The proof given of Lemma 3.3 is only a mild modification of the application given by Good [5] and which Bondy and Murty [2] page 181, relate to the design of an efficient computer drum. A completely different algorithmic proof of Lemma 3.3 was developed independently in recent work of Fredricksen and Mariorana 141.
We now proceed to prove Theorem 3 by applying the preceding lemmas infinitely many times.
We suppose now that PXi 1, (t,}" 1, (hi),2' , are increasing sequences of positive integers and {E,} j_1 is a sequence of positive reals decreasing to zero. We will define a sequence of
functions go(co) 0, 91M, 92M, gk(w), where each gk(w) will be defined viapk, tk,
SIZES OF ORDER STATISTICAL EVENTS 1083
hk, Ek, and the preceding &(w), 0 :5 j < k. Also, we should remark that each of the gi(w) will assume only finitely many values.
For notational convenience we temporarily write p = Pk, t = th, h = hk, and E Ek. We define si by (a: a = c,2 ', E, = 0 or 1, 1 r 5 p}, and note by Lemma 3.3 there is a sequence of Id 1' = VP elements of si which form a cycle in which each ordered t tuples with letters from d appears precisely once among the (ai+l, at+2, .. ., a,+t) for 0:5 1 < 20.
We now apply Lemma 3.1 to obtain an E such that for 0s k < 2h20 the sets T kE are disjoint and their union has probability at least 1 e. For the fmite partion 10'we take the partition given by the distinct values of the sum Eik 1' g,(w)2 q' where qj = p, + P2 + + pi, 1 :sj < k.
We now define gk(w) = a, if w E T" for 1 :s is h2 tP and i =_ i'mod 2 tP. Finally, we take gh(w) = 0 if w is in none of the T '+IE, 1 s k h20.
The whole point of this construction is that now by setting Rk equal to the union of T`E with 0s i < (h 1)VP and letting Pk(A) = P(A n Uk)IP(2k), we see that the random variables gA!(w), gk(T 'w), . , g*(T"w) are independent in the probability space (2k, Pk, J%) where A = (A n 2k:A E= JW}. Moreover, these random variables are also independent when, conditioned on the a field given by the partition .V. To prove these assertions one only has to note that for any H E A'we have by (3. 1) and (3.3)
P((gk(w) = so21, gt(T lw) = s 12 ,',', ..., gk(7 t+'co) = s t+12 PI1} n H)
(h 1)P(E)P(H)1(h 1)2PP(E) 2`PP(H).
Finally, we are able to definef(co) by letting qk =EJ' , p., and setting
f(w) = E'k~ . gk(w)2 1.
The sum representingf(w) converges for all w, and one finds no difficulty in checking that for Xj(w) = f(T"w) we have P(X, = Xj) = 0 for all i 0 j.
To prove Theorem 3 we suppose that the sequence p. 10 and E E (0, 1) are given, and we
will proceed to show that (p.}.' ,, {h,},ti, and (Q',_1 can be chosen so that Qn* (E)
n!pnn for infinitely many n.
For the intervals 1. [s2 qk 1, (S + 1)2") with 0 s < 2q*` we define random variables
P, as the number of elements of J, = (i: X E 1, 1 :S i :5 n}. The ordering of {Xi} ir= o. is
completely determined by the ordering of except for at most those W E 17=
{w: gk(T'lw) = gk(T J+'w), for some 0 5 i < j :5 n}. Using the Ph independence of the
gk(T'lw)} 1 :5 i < n for ns t and the conditional independence given Yewe have
PQX(1) < X(2) < ... < X.W} n H)
:S eA! + hV + P({X.(1) < X(2) < ... < X.(,,)) n H n 2k)
5 ek + hk 1 + Pk((X.(1) < X(2) < ... < X.(.)} n H)
:5 ek + hk 1 + PA!(_07) + Pt(H)(11~ m) 1
Since there are only ( n ) places a tie can take place and the P* probability of any such tie is 2
2 qk, we have
P(_17)s (1 Et N 1) 1Ph(Y) 2 (n )2 1* < n22~1k.
2
Also, Pk(H) = P(H) for all H r= V by (3.3); so summing over .V we have
P(X.(1) < XW < ... < X~(~)):5 2q, ,(Ek + hk ') + n22 + Ipp.!)
q* , Er 1 )r by
Setting r = 2 we have 8 0 P, = n. Hence, Il~(P,!) F(P. + 1) ~: F((n + r)lr the
1084 DANIEL RUDOLPH AND J. MICHAEL STEELE
convexity of log r(x), [71, page 285. We begin with the bounds (3.4) Q.*(E) ~ (1 E){max,P(X.(1) < 42) < ... <
(1 E)n! r(n + 1)r n + r + r(n + 1)(2_% 2 + rEk + rh k 1)
f C~ 1
Since r = 2 "* ' is fixed (as we choose Ek, ph, 1k, and hk), we invoke Stirling's formula to obtain lr(n + 1)17((n + r)1r)` S (r + 1)' for n k no(r). Since p. 10, we can now choose t = th ~ no so that
(3.5) (r + l)`k 2p,'(1 E)~'.
Finally, we choose ek, hk, andpk so that
(3.6) {r(t + 1)(2 p"2 + rEk + rhk`)) 1 k 2pt(l E)`.
By the elementary inequality ll(a + b) ~. (V2)min (I1a, llb} applied to (3.4), (3.5), and (3.6) for each t = tk, k = 1, 2, ... we have
Qn*(E) ~ p",(n!) for n = ti, t2,
The proof of Theorem 3 is thus complete. 0
4. A brief speculation. The only two cases where one has a reasonably complete understanding of Qn*(E) are the case of continuous Li.d. random variables and now in the case of continuous stationary ergodic processes with finite entropy. These two cases are in a way polar opposites of generality, and many important classes of processes lie in between.
Since many basic probabilistic events are simply unions of the order statistical events, it would seem to be of considerable interest to discover those cases in which a precise understanding of Qn*(E) can be obtained.
REFERENCES
[1) BILLINGSLEY, P. (1965). Ergodic Theory and Information. Wiley, New York.
[2] BONDY, J. A. and MURTY, U. S. R. (1976). Graph Theory with Applications. American Elsevier, New
York.
[3] DE BRuijN, N. 0. (1946). A combinatorial problem. Nederl. Akad. Wetensch. Proc. Ser. A 49 758764; Indag. Math. 8 461 467.
[41 FREDERICKSEN, H. and MAIORANA, J. (1978). Necklaces of beads in k colors and k ary de Bruijn sequences. Discrete Math. 23 207 2 10.
[51 GOOD, I. J. (1946). Normal recurring decimals. J. London Math. Soc. 21 167 169.
161 HALL, M., JR. (1967) Combinatorial Theory. Blaisdell, Waltham, Mass.
171 MITRINovit, D. S. (1970). Analytic Inequalities. Springer, New York.
181 SHIELDS, P. (1973). The Theory ofBernoulli Shes. Univ. Chicago.
[91 SMORODINSKY, M. (197 1). Ergodic Theory, Entropy. Lecture Notes in Math. 214 Springer, New York.
DEPARTMENT OF MATHEMATICS DEPARTMENT OF STATISTICS
STANFORD UNIVERSITY STANFORD UNIVERSITY
STANFORD, CALIFORNIA 94305 STANFORD, CALIFORNIA 94305 | 5,602 | 16,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.887139 |
http://www.ehow.com/how_2075641_day-trade-square-nine.html | 1,502,933,651,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102819.58/warc/CC-MAIN-20170817013033-20170817033033-00241.warc.gz | 544,220,693 | 19,727 | # How to Day Trade Square of Nine
Save
How to Day Trade Square of Nine. The Square of Nine trading method, also known as Gann's pyramid, is a way of using geometry to forecast the moods of the market. The two variables at issue here are time and price. For instance, the most extreme market reactions that are typically seen, such as very high highs and low lows, appear to occur during weeks that are squares of odd numbers.
Trade stocks according to cyclic principles, geometric patterns and square roots if you want to beat the system--if W.D. Gann is to be believed, that is. Mr. Gann, the inventor of the Square of Nine principle, espoused this system about a century ago, and it's still going strong today, with software available to help you calculate the best time and price for your trades.
Know that the Square of Nine theory puts time and price into a graphic representation and measures the angle between them. The theory is that, when the angle is at or close to 90 degrees, the stock will experience a major shift.
Keep your eye on a stock or industry and watch for highs and lows for specific time periods, from days to months to years. Calculate angles between these "tipping points" and the next day's or week's prices. It's usually easiest and fastest to simply allow Gann software to make these calculations for you.
Give the variables of time and price equal weight. This will allow you to see the relationship between them on a 1:1 scale, which is what's needed for angle calculations. It isn't the numbers that are important but the spatial relationships between the variables.
Day trade your stocks, mutual finds, bonds, currencies or other commodities using this system by understanding when an angle is ascendant, when it is descendant and where the price is in relationship to these angles. The Square of Nine theory proposes that a stock price will continue to move in a certain direction within an angle until it breaks the angle.
Know that this stock-trading theory uses past stock or commodity performance to predict future movement, which is something that many market pundits say isn't always viable. However, if you choose to day trade using the Square of Nine principles, know that Mr. Gann was firmly convinced of the cyclic nature of the market and the repetitive behaviors of both stocks and market players.
## Tips & Warnings
• Note the Square of Nine trading is one of the more advanced (and obscure) market forecasting techniques and is not the domain of the new-to-trading day trader.
• Find a Gann Wheel at Gann Trading (see Resources below).
• Know that some traders equate Square of Nine trading with astrology. Its principles have been called into question many times over the century that it's been in use.
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Discussion: Traffic Jam Applet tool Topic: MS Word division symbol Related Item: http://mathforum.org/mathtools/tool/10/
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Subject: RE: MS Word division symbol Author: Patti Date: Mar 4 2004
Hi,
I don't know if you use a Mac or PC, but on a Mac, the keystrokes I've listed
below work for quite a few math symbols such as the ÷ sign. The shortcuts work
for most fonts I believe. These are most/all/some of the shortcuts that show up
in the symbols window that Dan mentioned. I usually use Times New Roman in my
docs and I know that works well. Just mess around using different keystroke
combinations and I think you might be surprised what's already "built into" any
program (I think) without extra software.
Hope this helps,
Patti Guthrie
Chula Vista, CA
Mac computers:
Option + p = <pi>
Option + / = ÷
Option + > = >=
Option + < = <=
Option + 4 = ¢
Option + 5 = ?
Option + 8 = •
Option + 0 = º as in 90º or Option + k = ? as in 90?
Option + “=” = !=
Option + w = <sum>
Option + o = ø
Option + s = ß
Option + f = ƒ
Option + v = ?
Option + x = ?
Option + z = ?
Option + shift + “=” = ±
On Mar 03, 2004, Dan wrote:
Most of the full fonts have the division symbol available as part of their
character set. To see the full character set in MS Word, go to the Insert menu
on the toolbar at the top of the screen. From the pull down menu under Insert,
select "symbol".
You'll be shown a grid containing all the characters in a font - the list of
fonts available on your computer is available above the grid of symbols. You can
select any character and insert it into your current document.
The division symbol is available in most of the standard fonts - Times New
Roman, Arial, Garamond, etc.
I don't know of an easy way to insert more complex fractions into MS Word
documents without using an add-on application like Math Type (which works very | 538 | 2,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-39 | longest | en | 0.840403 |
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January 13, 2011, 10:41 initial value and UDF #1 Member EH Join Date: Dec 2010 Posts: 61 Rep Power: 8 Hi all, Does anyone know, after initializing my problem, will the following command in my UDF: T=C_T(c,t); result in T being the initial value? My problem is I have the following UDF: Code: ``` #include "udf.h" /* Mean temperature at outlet is calculated from the surrounding nodes */ /* Coordinate of the outlet is (x,y,z) = (0.275,-0.0138,-0.6747) */ DEFINE_ADJUST(outlet_temp,d) { double Tout=0.0; double Temp; double coord[ND_ND]; double xmin=0.273; double xmax=0.277; double ymin=-0.00675; double ymax=0.0; double zmin=-0.6858; double zmax=-0.6628; double x,y,z; double ncell=1.0; int nfaces=0; cell_t c; Thread *tt=Lookup_Thread(d, 31); /* Begin loop to determine the temperature at the centroid of cells near the outlet */ begin_c_loop(c,tt) { C_CENTROID(coord,c,tt); x=coord[0]; y=coord[1]; z=coord[2]; if ((x >= xmin) & (x <=xmax)) { if ((y >= ymin) && (y <= ymax)) { if ((z >= zmin) && (z <= zmax)) { Temp = C_T(c,tt); /* Cell temperature */ Tout += Temp; ncell ++; /* Number of cells */ printf("ncell= %d\n", ncell); } } } } end_c_loop(c,tt) Tout=Tout/(ncell-1); /* Average cell temperature surrounding outlet */ printf("Tout= %d\n", Tout); }``` the first printf statement gives ncell=0, when it should keep increasing while Tout = 0 when it should be the initial value that I initialize. Does anyone know why? Thanks. Cheers, EH
January 14, 2011, 02:31 #2 New Member anonymous Join Date: Jan 2011 Posts: 23 Rep Power: 8 One Problem could be with declaration of ncell as double and printing it with %d, which is meant for integer entry. Declare ncell as integer and cross check. C_T(c,t) should give the initial value in the UDF. I do not see any problem there.
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https://blog.jverkamp.com/2014/03/12/caesar-cipher/ | 1,721,488,422,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00474.warc.gz | 118,607,625 | 6,959 | # Caesar cipher
Here’s a 5 minute1 coding challenge from Programming Praxis:
A caeser cipher, named after Julius Caesar, who either invented the cipher or was an early user of it, is a simple substitution cipher in which letters are substituted at a fixed distance along the alphabet, which cycles; children’s magic decoder rings implement a caesar cipher. Non-alphabetic characters are passed unchanged. For instance, the plaintext PROGRAMMINGPRAXIS is rendered as the ciphertext SURJUDPPLQJSUDALV with a shift of 3 positions.
– Source: Wikipedia, public domain
To make it a bit more interesting, I’m actually going to be using a different #lang built on Racket: rackjure. If you’re running a newer version of Racket (6+), you can install it with either raco pkg install rackjure or with the Package Manger built into DrRacket. Then just change the #lang line at the top of your file to #lang rackjure.
What specifically do I want from Rackjure? The threading macro ~>. Basically, it takes a value and ’threads’ it as the first argument through a series of functions. The example on the Rackjure GitHub page:
> (string->bytes/utf-8 (number->string (bytes-length #"foobar") 16))
#"6"
> (~> #"foobar"
bytes-length
(number->string 16)
string->bytes/utf-8)
#"6"
In our case, we want it because we’re going to do run through a similar stream of transformations on each character:
• Convert from a character to a number using char->integer
• Get to a zero based system by subtracting #\A = 65
• Add/subtract the offset for this particular Caesar cipher
• Get the modulus 26 so we have a letter when we’re done
• Get back to a letter by adding #\A = 65 back on
• Convert back to a character with integer->char
Turn that into Rackjure:
(define (caesar str n)
(define A (char->integer #\A))
(list->string
(for/list ([c (in-string str)])
(~> c char->integer (- A) (+ n) (mod 26) (+ A) integer->char))))
If we didn’t have ~>, that would look something more like this:
(define (caesar str n)
(define A (char->integer #\A))
(list->string
(for/list ([c (in-string str)])
(integer->char (+ (mod (+ (- (char->integer c) A) n) 26) A)))))
I’ll leave it up to you which of the two styles you think is easier to read–either the Scheme style inside out, jumping back and forth between the operators and their arguments or the Clojure style left to right2.
Unfortunately, Racket doesn’t have a mod function built in3. You can get one from R6RS though:
(require (only-in rnrs/base-6 mod))
And there you have it. Simple (and almost trivial to crack) encryption:
> (caesar "HELLOWORLD" 10)
"ROVVYGYBVN"
> (caesar "ROVVYGYBVN" -10)
"HELLOWORLD"
We can make it at least a little better though. Let’s go ahead and deal with lower case and non-alphabetic characters:
(define (caesar str n)
(define A (char->integer #\A))
(define a (char->integer #\a))
(list->string
(for/list ([c (in-string str)])
(cond
[(char<=? #\A c #\Z)
(~> c char->integer (- A) (+ n) (mod 26) (+ A) integer->char)]
[(char<=? #\a c #\z)
(~> c char->integer (- a) (+ n) (mod 26) (+ a) integer->char)]
[else
c]))))
Basically, the only thing that changes is the offset. For upper case letters, we use #\A = 65, for lower case #\a = 97. Anything that’s not a letter? We just leave it alone.
How’s it work?
> (caesar "Hello world!" 100)
"Dahhk sknhz!"
> (caesar "Dahhk sknhz!" -100)
"Hello world!"
It’s actually such a simple program, you have all of the code right there, but just in case you can also download it from GitHub: caesar-cipher.rkt
1. More or less ↩︎
2. Exercise to the reader: which do you think I prefer? ↩︎
3. At least not one that deals how we need with negative numbers ↩︎ | 1,004 | 3,664 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.844744 |
https://www.mathworks.com/matlabcentral/fileexchange/33073-triangle-ray-intersection?s_tid=prof_contriblnk | 1,553,656,705,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912207618.95/warc/CC-MAIN-20190327020750-20190327042750-00205.warc.gz | 827,302,839 | 21,572 | File Exchange
## Triangle/Ray Intersection
version 1.7 (309 KB) by Jaroslaw Tuszynski
### Jaroslaw Tuszynski (view profile)
Fast vectorized triangle/ray intersection algorithm
Updated 18 May 2018
Ray/triangle intersection using the algorithm proposed by Möller and
Trumbore (1997), implemented as highly vectorized MATLAB code.
The algorithm can work with one and two sided surfaces, as well as, with
infinite lines, rays (lines bounded on one side) and segments (lines bounded on
both sides).
Input (all arrays in in Nx3, where N is number of vertices or rays):
orig : ray's origin
dir : ray's direction
vert0, vert1, vert2: vertices of the triangle
Output:
Intersect - boolean array of length N
t - distance from the ray origin to the intersection point in |dir|
u,v - barycentric coordinates of the intersection point units
xcoor - carthesian coordinates of the intersection point
In addition PointInsideVolume is 3D equivalent to 2D inpolygon function and can test if
array of points is inside or outside any volume defined by the surface grid.
Philip Kunz
Fabricio Castro
### Fabricio Castro (view profile)
Jaroslaw Tuszynski
### Jaroslaw Tuszynski (view profile)
Reply to jack' green: I think you should be able to find much simpler codes to do 2D case. Or just write one to do line/line intersection for ray intersection with each line segment and see if the intersection point is between each segment end points.
jack' green
### jack' green (view profile)
Hi, Jaroslaw. Thanks for provding such a cool tool. It works well on my computer. Just wonder is it possible to apply it to a purely 2D (in-plane) case where the triangles all become segments?
Yanjun Han
### Yanjun Han (view profile)
many faces / many rays intersection, the return value 'flag'? Should it be a matrix but actual it is not.
Yanjun Han
### Yanjun Han (view profile)
Hello, Can anyone help me! Now I have a series of path points, and I want project them on a specified mesh surface along the specified direction, for example dir=(0,0,1). For every path point, I can iterate over all the triangular meshes to get the corresponding projection points，but this is too time-consuming. Can I use a faster method？
I want to get the projection point of all the points in the fixed direction on the triangular mesh surface. How can I use the 'many faces / many rays intersection' ?
Ander Biguri
### Ander Biguri (view profile)
Does the many rays-many faces case only work with Nfaces==Nrays? If so, is there any specific reason for implementing that specific case?
Yumin Sun
### Yumin Sun (view profile)
Actually for multiple faces and multiple rays, one should remember you need to repmat origin, if all origins are same.
Leonardo Colavitti
Igor
### Igor (view profile)
To do your program better I suggests
1) Add description about use option FullReturn. It shold be '1' but not 'true' for example.
2) Add in initialize default output string
xcoor=zeros(size(orig));
If xcoor not initialized than if no intersections error appear with params count.
Jamie Heather
### Jamie Heather (view profile)
Nice vectorised implementation with handy options for choosing between ray/line/segments and handling numerical precision issues - thanks!
Jaroslaw Tuszynski
### Jaroslaw Tuszynski (view profile)
Reply to Murat: You might have to look into using "border" parameter. If it does not work can you construct a scenario that demonstrates the issue and email it to me
Murat
### Murat (view profile)
Hi, the code is working well but it sometimes cannot handle with large incident angles. When the incident angle is around 90, it ignores triangles. Could you correct it?
Ehsan golk
Jeffrey
### Jeffrey (view profile)
I'm having trouble with the many rays, many faces option. Lets say I have 11 faces and 5 rays. What resizing of the inputs needs to be done? Do I need to repmat all arrays to be of size 55x3?
Based on what I can tell it can handle:
- one ray, many faces
- many rays, one face
- many rays, many faces (but # of faces = # of rays)
But I think I am just missing something here.
Jaroslaw Tuszynski
### Jaroslaw Tuszynski (view profile)
Reply to Nick: In many rays and many triangle case you still test for intersection of prearranged pairs and the return value informs you which ray/triangle pair intersected
Nick
### Nick (view profile)
Very useful but when considering many rays and many triangles is there a way to find which ray has intersected which triangle?
Nick
Amilcar Torres
Ben
### Ben (view profile)
Jaroslaw Tuszynski
### Jaroslaw Tuszynski (view profile)
Re to Anton Semechko: This is a low level function which can be called with one ray and many faces, many faces and one ray or many faces and rays. It is quite simple to apply repmat function to the inputs.
Anton Semechko
### Anton Semechko (view profile)
why does the number of rays have to equal the number of faces? that's so inconvenient
18 May 2018 1.7 small corrections to the interface 25 Sep 2017 1.6.0.0 Minimal changes suggested by Igor in Comments and Ratings 15 Jun 2016 1.5.0.0 correct description: no symbolic toolbox is needed 3 Oct 2014 1.5.0.0 Major rewrite of the function with additional options and output variables. Also provided PointInsideVolume function 7 Jun 2013 1.4.0.0 documentation improvements and typo correction 20 Mar 2013 1.2.0.0 correct treatment of 3x3 arrays as suggested by Andreas Weber 7 Oct 2011 1.1.0.0 Improvements to border handling
##### MATLAB Release Compatibility
Created with R2017b
Compatible with any release
##### Platform Compatibility
Windows macOS Linux
##### Acknowledgements
Inspired by: Ray/Triangle Intersection
Inspired: in_polyhedron, RayShapeArticle_FEX.zip
### Discover Live Editor
Create scripts with code, output, and formatted text in a single executable document. | 1,417 | 5,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-13 | longest | en | 0.838653 |
https://www.longrangehunting.com/articles/berger-bullets-move-to-the-g7-standard-bc.219/ | 1,508,798,395,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826840.85/warc/CC-MAIN-20171023221059-20171024001059-00076.warc.gz | 957,320,626 | 22,064 | # Berger Bullets' Move To The G7 Standard BC
1. Berger Bullets' Move To The G7 Standard BC
By Bryan Litz, Chief Ballistician For Berger Bullets
A Better Ballistic Coefficient
For centuries now, science has been helping us gain a more accurate understanding of our world. The branch of science we care about as shooters is known as ballistics. The science of ballistics is well developed and understood by those who study it, but the tools and information being used by average shooters is not necessarily optimal for the shooter’s applications.
In other words, there is a better, more accurate way for shooters to use ballistics to help them predict trajectories and hit targets. The purpose of this article is to present a better way for shooters to calculate ballistics.
What is a Ballistic Coefficient?
Most shooters, especially long range rifle shooters, are familiar with the Ballistic Coefficient (BC). Without getting into the math, I’ll define the ballistic coefficient in words as: The ability of the bullet to maintain velocity, in comparison to a ‘standard projectile’.
A high BC bullet can maintain velocity better than a low BC bullet under the same conditions. All measures of ballistic performance including drop and wind deflection are related to the bullet’s ability to maintain velocity. In short; the higher the BC, the better the all-around ballistic performance of the bullet will be.
How a Ballistic Coefficient is Used
Details of ballistic trajectories can be predicted with computer programs using all the relevant variables, including BC. As with all prediction programs; the accuracy of the outputs depends on the accuracy of the inputs. Here is where we have to examine the real meaning and implications of using a Ballistic Coefficient to characterize the bullet’s ability to maintain velocity.
It’s a relatively well known fact that the BC of a bullet is different at different velocities. Not many shooters know why it changes, or what the consequences are. To understand why a BC changes at different speeds, we have to go back to the definition of BC, which is: The ability of the bullet to maintain velocity, in comparison to a ‘standard projectile’. It’s the ‘standard projectile’ part of the definition that we need to key in on. What is the ‘standard projectile’? What does it look like?
To date, the ‘standard projectile’ used to define BCs for the entire sporting arms industry is the G1 standard projectile. The G1 standard projectile which is shown in Figure 1 has a short nose, flat base, and bears more resemblance to a pistol bullet or an old unjacketed lead black powder cartridge rifle bullet than to a modern long range rifle bullet.
Figure 1. The G1 standard projectile.
The reason why the BC of a modern long range bullet changes so much at different velocities is because modern bullets are so different in shape compared to the G1 standard that its BC is based on. In other words, the drag of a modern long range bullet changes differently than the G1 standard projectile, so the coefficient relating the two (the ballistic coefficient) has to change with velocity. | 631 | 3,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-43 | longest | en | 0.923347 |
https://www.tutorialspoint.com/what-is-definition-formula-and-example-of-key-rate-duration | 1,702,333,466,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679518883.99/warc/CC-MAIN-20231211210408-20231212000408-00627.warc.gz | 1,127,788,663 | 22,191 | # What is Definition, Formula & Example of Key Rate Duration?
## What is key rate duration?
The key rate duration of a debt security or a debt instrument portfolio, often bonds, examines how the value of a debt security or portfolio varies at a certain maturity point throughout the whole yield curve. With respect to other maturities, the key rate duration is used to determine how sensitive a debt security's price is to a one percent change in yield for a certain maturity while all other maturities are held constant.
## Key Points Briefly
• Using the key rate duration, you can determine how much a bond's price changes when the yield on the bond changes by one percentage point (100 basis points) throughout its entire term.
• As the yield curve moves in a parallel direction, you may use effective duration to estimate portfolio value changes. However, when there is a non-parallel shift of yield-curve, you must utilize key rate duration to estimate portfolio value adjustments.
• Length measures inform you of the price risk associated with owning fixed income instruments in the event of a change in the interest rate environment.
## What is Key Rate Duration Formula?
Here, we will look at key rate duration formula and understand it better with an example.
$$Key\:rate\:duration=\frac{P_{-}-P_{+}}{2*0.01*P_{0}}$$
Where −
P- − Price of bond after 1% decreases in profit.
P+ − Price of bond after 1% increase in profit.
P0 − Original price of the bond
## Key Rate Duration - Example
Let us take the example of the asset, whose original price is 1,575 and there was a fall in in its value to \$1,250 due when there was 1 percent rise in its bond maturity. However, there was a price increase to 1,750 when that bond maturity value dropped.
Using the above formula, and the numbers from the example, let us calculate the key rate definition for our asset.
The key rate duration will be (1750 – 1250) / (2 * 0.01 * 1575) = 15.87
## The Importance of the Key Rate's Time Duration
The predicted change in value resulting from a change in the yield on a bond or bond portfolio with a given maturity is represented by the key rate duration. It is presumptively assumed that the yields for all other maturities would remain constant.
Assuming that an investor's interest rate predictions prove to be correct over a given time period, they can use the key rate duration metric to determine which bond maturities are most likely to provide the most profitable investment returns (assumed that the investor's interest rate predictions prove to be correct). As a result, the statistic may be used to compare different potential fixed-income investments.
When an investor owns a debt security, determining key rate duration may be beneficial in a number of other situations as well. They may be interested in estimating the change in the value of the bond they own as a result of different basis point fluctuations. This may assist customers in determining the possibility that their bond may be called in for early redemption by the issuer of the bond.
Updated on: 02-Jul-2021
324 Views | 670 | 3,108 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-50 | latest | en | 0.898108 |
https://www.convertunits.com/from/dekatonne/to/stone | 1,660,008,278,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00185.warc.gz | 662,326,460 | 12,937 | ## ››Convert dekatonne to stone
dekatonne stone
How many dekatonne in 1 stone? The answer is 0.000635029318.
We assume you are converting between dekatonne and stone.
You can view more details on each measurement unit:
dekatonne or stone
The SI base unit for mass is the kilogram.
1 kilogram is equal to 0.0001 dekatonne, or 0.15747304441777 stone.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between dekatonnes and stones.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of dekatonne to stone
1 dekatonne to stone = 1574.73044 stone
2 dekatonne to stone = 3149.46089 stone
3 dekatonne to stone = 4724.19133 stone
4 dekatonne to stone = 6298.92178 stone
5 dekatonne to stone = 7873.65222 stone
6 dekatonne to stone = 9448.38267 stone
7 dekatonne to stone = 11023.11311 stone
8 dekatonne to stone = 12597.84355 stone
9 dekatonne to stone = 14172.574 stone
10 dekatonne to stone = 15747.30444 stone
## ››Want other units?
You can do the reverse unit conversion from stone to dekatonne, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Dekatonne
The SI prefix "deka" represents a factor of 101, or in exponential notation, 1E1.
So 1 dekatonne = 101 tonnes.
The definition of a tonne is as follows:
A tonne (also called metric ton) is a non-SI unit of mass, accepted for use with SI, defined as: 1 tonne = 1000 kg (= 106 g).
## ››Definition: Stone
A stone is a unit of mass within the Imperial units system used in the United Kingdom and Ireland, and formerly in a number of Commonwealth countries. It is equal to 14 pounds avoirdupois, i.e. 6.35029318 kilograms.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 614 | 2,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-33 | latest | en | 0.762212 |
https://schoolbag.info/mathematics/history/10.html | 1,558,478,621,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256586.62/warc/CC-MAIN-20190521222812-20190522004812-00271.warc.gz | 611,189,482 | 17,222 | Greek Number Theory - Greek Mathematics From 500 BCE to 500 CE - A brief course - The history of mathematics
## The history of mathematics: A brief course (2013)
### Chapter 9. Greek Number Theory
Greek number theory is of interest both intrinsically, because some of the natural questions that it raised have not been answered even in the present time, and because it was the soil in which algebraic symbolism first sprouted, a brilliant innovation during a time otherwise marked by intellectual decline. The theory itself has two areas of interest that do not interact during the period of Greek intellectual dominance. One is the Pythagorean topic of the arithmetic properties of figurate numbers (triangular numbers, square numbers, pentagonal numbers, and so on). This area has declined greatly in importance, along with Pythagoreanism and neo-Platonism, although it is not quite extinct even today. The recent solution of the problem of Fermat's last theorem is a good specimen of the modern development of this theory. The other area, which provides the theoretical foundation for much of classical and modern number theory, is the theory of divisibility of integers. This area also has one rather Pythagorean connection, namely the topic of perfect numbers. We shall look at just three of the classical Greek writers on number theory:
1. Euclid, whose Elements contain three books (Books 7–9) devoted to the divisibility properties of integers.
2. Nicomachus of Gerasa, a neo-Pythagorean philosopher who lived about 100 CE. His Introduction to Arithmetic gives a detailed development of both figurate numbers and the divisibility theory.
3. Diophantus, who lived sometime between the second and fourth centuries CE. He is sometimes justly called the “father of algebra,” because in the course of his study of the arithmetic properties of square numbers, he introduced symbolic notation for an unspecified or unknown number and a way of writing operations on unknown numbers.
As it happens that the treatise of Nicomachus preserves more of what was traditionally called Pythagorean lore than the earlier work of Euclid, we shall discuss Nicomachus first. Before we can do that, however, we need to digress to explain a mathematical technique that is fundamental to the understanding of a great deal in the history of mathematics.
9.1 The Euclidean Algorithm
The Greeks learned early on how to find the greatest common divisor of two numbers. A very efficient procedure for doing so is described in Chapter 13 of Book 1 of Nicomachus' Arithmetica and in Proposition 2 of Book 7 of Euclid's Elements. This procedure is now known as the Euclidean algorithm, Nicomachus applies it only to integers, any two of which naturally have 1 as a common divisor. Euclid, on the other hand, does not confine it to integers, but states the procedure for “magnitudes,” which may lack a common measure. It is significant that when it is applied to continuous magnitudes, the procedure terminates if and only if there is a common measure. Euclid makes use of that fact in discussing incommensurables, which are pairs of magnitudes having no common measure. The algorithm was certainly invented long before the time of Euclid, however. Zverkina (2000) believes that this procedure could not have arisen intuitively, but must have come about as the result of solving specific problems, most likely the problem of reducing ratios by canceling a common divisor. What follows is a description of the general procedure.
For definiteness, we shall imagine that the two quantities whose greatest common measure is to be found are two lengths, say a and b. Suppose that a is longer than b. (If the two are equal, their common value is also their greatest common divisor.) The general procedure is to keep subtracting the smaller quantity from the larger until the remainder is equal to the smaller quantity or smaller than it. It is not difficult to show that the smaller quantity and the remainder have the same common measures as the smaller quantity and the larger. Hence one can start over with the smaller quantity and the remainder, which is no more than half of the larger quantity. Either this process terminates with an equal pair, or it continues and the pairs become arbitrarily small.
An example using integers will make the procedure clear. Let us find the greatest common measure (divisor) of 26173996849 and 180569389. A common measure does exist: the integer 1. Since the repeated subtraction process amounts to division with remainder, we do it this way: 26173996849 ÷ 180569389 is 144 with a remainder of 172004833. We then divide the smaller quantity (the previous divisor) 180569389 by the remainder 172004833, getting a quotient of 1 and a remainder of 8564556. Next we divide the previous remainder 172004833 by the new remainder 8564556, getting a quotient of 20 and a remainder of 713713. We then divide 8564556 by 713713 and get a quotient of 12 with no remainder, so that the greatest common divisor is 713713.
This computation can be arranged as follows, with the successive divisions performed from right to left. The greatest common measure appears at the extreme left:
One can see that this procedure must produce a greatest common measure if one exists, since the first remainder is at most half of the larger of the two original quantities (since it is smaller than the smaller of the two original quantities and not larger than their difference). Similarly, since the smaller quantity becomes the dividend in the second application, the second remainder will be at most half of it. Thus the first two remainders are at most half the size of the original quantities. Yet they are both larger than any common measure the two original quantities had. It follows that if there is any common measure, one of these remainders will ultimately become zero, since repeated halving would otherwise eventually make it smaller than that common measure. The last nonzero remainder is thus the greatest common measure of the two quantities.
9.2 The Arithmetica of Nicomachus
In his first book, Nicomachus makes the elementary distinction between odd and even numbers. Having made this distinction, he proceeds to refine it, distinguishing between even numbers divisible by 4 (evenly even) and those that are not (doubles of odd numbers). He goes on to classify odd numbers in a similar way, thereby coming to the concept of prime and composite numbers. Nicomachus also introduces what we now call pairs of relatively prime numbers. These are pairs of numbers that have no common prime divisor and hence no common divisor except 1. Relational properties were difficult for Greek philosophers, and Nicomachus expresses the concept of relatively prime numbers in a confused manner, referring to three species of odd numbers: the prime and incomposite, the secondary and composite, and “the variety which, in itself, is secondary and composite, but relatively is prime and incomposite.” This way of writing seems to imply that there are three kinds of integers, prime and incomposite, secondary and composite, and a third kind midway between the other two. It also seems to imply that one can look at an individual integer and classify it into exactly one of these three classes. Such is not the case, however. The property of primeness is a property of a number alone. The property of being relatively prime is a property of a pair of numbers. On the other hand, the property of being relatively prime to a given number is a property of a number alone. Nicomachus explains the property in a rather wordy fashion in Chapter 13 of Book 1, where he gives a method of identifying prime numbers that has become famous as the sieve of Eratosthenes.
Nicomachus attributes this method to Eratosthenes (276–174 BCE, best known for this work on prime numbers and for having estimated the size of the earth). To use it, start with a list of all the odd numbers from 3 on, that is,
From this list, remove the multiples of 3, starting with 3 · 3, that is, remove 9, 15, 21, 27, 33,. . .. The reduced list is then
From this new list, remove all multiples of 5, starting with 5 · 5. The first nonprime in the resulting list will 49 = 7 · 7, and so you remove all multiples of 7 from that list. In this way, you can generate in short order a complete list of primes up to the square of the first prime whose multiples were not removed. Thus, after removing the multiples of 7, we have the list
The first nonprime in this list would be 11 · 11 = 121.
9.2.1 Factors vs. Parts. Perfect Numbers
Nicomachus' point of view on this sieve was different from ours. Where we think of the factors of, say 60, as being 2, 2, 3, and 5, Nicomachus thought of the quotients on division by these factors and products of these factors as the parts of a number. Thus, in his language, 60 has the parts 30 (half of 60), 20 (one-third of 60), 15 (one-fourth of 60), 12 (one-fifth of 60), 10 (one-sixth of 60), 6 (one-tenth of 60), 5 (one-twelfth of 60), 4 (one-fifteenth of 60), 3 (one-twentieth of 60), 2 (one-thirtieth of 60), and 1 (one-sixtieth of 60). If these parts are added, the sum is 108, much larger than 60. Nicomachus called such a number superabundant and compared it to an animal having too many limbs. On the other hand, 14 is larger than the sum of its parts. Indeed, it has only the parts 7, 2, and 1, which total 10. Nicomachus called 14 a deficient number and compared it to an animal with missing limbs like the one-eyed Cyclops of the Odyssey. A number that is exactly equal to the sum of its parts, such as 6 = 1 + 2 + 3, he called a perfect number. He gave a method of finding perfect numbers, which remains to this day the only way known to generate such numbers, although it has not been proved that there are no other such numbers. This procedure is also stated by Euclid as Proposition 36 of Book 9 of the ElementsIf the sum of the numbers 1, 2, 4,. . ., 2n−1 is prime, then this sum multiplied by the last term will be perfect. To see the recipe at work, start with 1, then double and add: 1 + 2 = 3. Since 3 is prime, multiply it by the last term, that is, 2. The result is 6, a perfect number. Continuing, 1 + 2 + 4 = 7, which is prime. Multiplying 7 by 4 yields 28, the next perfect number. Then, 1 + 2 + 4 + 8 + 16 = 31, which is prime. Hence 31 · 16 = 496 is a perfect number. The next such number is 8128 = 64(1 + 2 + 4 + 8 + 16 + 32 + 64). In this way, Nicomachus was able to generate the first four perfect numbers. He seems to hint at a conjecture, but draws back from stating it explicitly:
When these have been discovered, 6 among the units and 28 in the tens, you must do the same to fashion the next. . .the result is 496, in the hundreds; and then comes 8,128 in the thousands, and so on, as far as it is convenient for one to follow [D'ooge, 1926, p. 211].1
This quotation seems to imply that Nicomachus expected to find one perfect number Nk having k decimal digits. Actually, the fifth perfect number is 33,550,336, so we have jumped from four digits to eight here. The sixth is 8,589,869,056 (10 digits) and the seventh is 137,438,691,328 (12 digits), so that there is no regularity about the distribution of perfect numbers. Thus, Nicomachus was wise to refrain from making conjectures too explicitly. According to Dickson (1919, p. 8), later mathematicians, including the great sixteenth-century algebraist Girolamo Cardano, were less restrained, and this incorrect conjecture has been stated more than once.
For a topic that is devoid of applications, perfect numbers have attracted a great deal of attention from mathematicians. Dickson (1919) lists well over 100 mathematical papers devoted to this topic over the past few centuries. From the point of view of pure number theory, the main questions about them are the following: (1) Is there an odd perfect number?2 (2) Are all even perfect numbers given by the procedure described by Nicomachus?3 (3) Which numbers of the form 2n − 1 are prime? These are called Mersenne primes, after Marin Mersenne (1588–1648), who, according to Dickson (1919, pp. 12–13), first noted their importance, precisely in connection with perfect numbers. Obviously, n must itself be prime if 2n − 1 is to be prime, but this condition is not sufficient, since 211 − 1 = 23 · 89. The set of known prime numbers is surprisingly small, considering that there are infinitely many to choose from, and the new ones being found tend to be Mersenne primes, mostly because that is where people are looking for them. The largest currently known prime, discovered on August 23, 2008, is 243112609 − 1, only the forty-fifth Mersenne prime known at the time. Since then, two more have been discovered, both smaller than this one however.4 It was found by the GIMPS (Great Internet Mersenne Prime Search) project, which links hundreds of thousands of computers via the Internet and runs prime-searching software in the background of each while their owners are busy with their own work. This prime has 12,978,189 decimal digits. Since these primes are not being discovered in ascending order, it is not accurate to call the largest currently known one the 47th Mersenne prime. Exhaustive checking by the GIMPS network since the fortieth Mersenne prime, 220996011 − 1, was discovered on November 17, 2003 (it has 6,320,430 decimal digits) has established that there are no others smaller than that one. Thus we know the first 40 Mersenne primes and seven others as well. In contrast, the largest non-Mersenne prime known as of late 2011 was 19249 · 213018586 + 1, discovered in May 2007; it has 3,918,990 decimal digits and hence is tiny compared with the largest known Mersenne primes. (This information comes from the website http://primes.utm.edu.)
9.2.2 Figurate Numbers
Beginning in Chapter 6 of Book 2, Nicomachus studies figurate numbers: polygonal numbers through heptagonal numbers, and then polyhedral numbers. These numbers are connected with geometry, the number 1 being replaced by a geometric point. To motivate this discussion, Nicomachus speculated that the simplest way to denote any integer would be repeating a symbol for 1 an appropriate number of times. Thus, he said, the number 5 could be denoted α αα α α. This train of thought, if followed consistently, would lead back to a notation even more primitive than the hieroglyphic notation for numbers, since it would use only the symbol for units and discard the symbols for higher powers of 10. The Egyptians had gone beyond this principle in their hieratic notation, and the standard Greek notation was essentially a translation of the hieratic into the Greek alphabet. You can easily see where this speculation leads. The outcome is shown in Fig. 9.1, which illustrates triangular, square, pentagonal, and hexagonal numbers using dots instead of the letter α. Observe that the figures are not associated with regular polygons except in the case of triangles and squares. The geometry alone makes it clear that a square number is the sum of the corresponding triangular number and its predecessor. Similarly, a pentagonal number is the sum of the corresponding square number and the preceding triangular number, a hexagonal number is the sum of the corresponding pentagonal number and the preceding triangular number, and so forth. This is the point at which modern mathematics parts company with Nicomachus, Proclus, and other philosophers who push analogies further than the facts will allow. As Nicomachus states at the beginning of Chapter 7:
The point, then, is the beginning of dimension, but not itself a dimension, and likewise the beginning of a line, but not itself a line; the line is the beginning of surface, but not surface; and the beginning of the two-dimensional, but not itself extended in two dimensions. . .Exactly the same in numbers, unity is the beginning of all number that advances unit by unit in one direction; linear number is the beginning of plane number, which spreads out like a plane in one more dimension. [D'ooge, 1926, p. 239]
This mystical mathematics was transmitted to Medieval Europe by Boethius. It is the same kind of analogical thinking found in Plato's Timaeus, where it is imagined that atoms of fire are tetrahedra, atoms of earth are cubes, and so forth. Since the Middle Ages, this topic has been of less interest to mathematicians. The phrase of less interest—rather than of no interest—is used advisedly here: There are a few theorems about figurate numbers in modern number theory, and they have some connections with analysis as well. For example, a formula of Euler asserts that
Here the exponents on the right-hand side range over the pentagonal numbers for n positive. By defining the nth pentagonal number for negative n to be n(3n − 1)/2, we gain an interesting formula that can be stated in terms of figurate numbers. Carl Gustav Jacobi (1804–1851) was pleased to offer a proof of this theorem as evidence of the usefulness of elliptic function theory. Even today, these numbers crop up in occasional articles in graph theory and elsewhere.
Figure 9.1 Figurate numbers. Top row: triangular numbers Tn = n(n + 1)/2. Second row: square numbers Sn = n2. Third row: pentagonal numbers Pn = n(3n − 1)/2. Bottom row: hexagonal numbers Hn = n(2n − 1).
9.3 Euclid's Number Theory
Euclid devotes most of his three books on number theory to divisibility theory, spending most of the time on proportions among integers and on prime and composite numbers, with fewer results on figurate numbers. Only at the end of Book 9 does he prove a theorem of a different sort, giving the method of constructing perfect numbers described above. It is interesting that, except for squares and cubes, Euclid does not mention figurate numbers. Although the Pythagorean and Platonic roots of Euclid's treatise are obvious, Euclid appears to the modern eye to be much more a mathematician than Pythagoras or Plato, not at all inclined to flights of fanciful speculation on the nature of the universe. In fact, he never mentions the universe at all and suggests no practical applications of the theorems in his Elements.
Book 7 develops proportion for positive integers as part of a general discussion of ways of reducing a ratio to lowest terms. The notion of relatively prime numbers is introduced, and the elementary theory of divisibility is developed as far as finding least common multiples and greatest common factors. Book 8 resumes the subject of proportion and extends it to squares and cubes of integers, including the interesting theorem that the mean proportional of two square integers is an integer (Proposition 11—for example, 25 : 40 : : 40 : 64), and between any two cubes there are two such mean proportionals (Proposition 12—for example, 27 : 45 : : 45 : 75 : : 75 : 125). Book 9 continues this topic; it also contains the famous theorem that there are infinitely many primes (Proposition 20, in the form of the assertion that no given finite collection of primes can contain all of them) and ends by giving the only known method of constructing perfect numbers (Proposition 36), quoted above.
Euclid's number theory does not contain any explicit statement of the fundamental theorem of arithmetic (Knorr, 1976). This theorem, which asserts that every positive integer can be written in only one way as a product of prime numbers, can easily be deduced from Book 7, Proposition 24: If two numbers are relatively prime to a third, their product is also relatively prime to it.
9.4 The Arithmetica of Diophantus
Two works of Diophantus have survived in part, a treatise on polygonal numbers and the work for which he is best known, the Arithmetica. Like many other ancient works, these two works of Diophantus survived because of the efforts of a ninth-century Byzantine mathematician named Leon, who organized a major effort to copy and preserve them. There is little record of the influence the works of Diophantus may have exerted before this time. What is of particular interest to us is his study of the arithmetic properties of squares. He is interested in finding ways to represent a given rational square number as the sum of two other rational square numbers. The techniques he developed to solve that problem resulted in the first use of symbolism to represent the kind of abstract thinking required in algebra.
To judge this work, one should know something of its predecessors and its influence. Unfortunately, information about either of these is difficult to come by. The Greek versions of the treatise, of which there are 28 manuscripts according to Sesiano (1982, p. 14), all date to the thirteenth century. Among the predecessors of Diophantus, we can count Heron of Alexandria and one very obscure Thymaridas, who showed how to solve a particular set of linear equations, known as the epanthma (blossom) of Thymaridas.
Because the work of Diophantus is so different from the style of Euclid and his immediate successors, the origins of his work have been traced to other cultures, notably Egypt and Mesopotamia. The historian of mathematics Paul Tannery (1843–1904) printed an edition of Diophantus' work and included a fragment supposedly written by the eleventh-century writer Michael Psellus (1018–ca. 1078), which stated that “As for this Egyptian method, while Diophantus developed it in more detail,. . ..” On this basis, Tannery assigned Diophantus to the third century. Neugebauer (1952, p. 80) distinguished two threads in Hellenistic mathematics, one in the logical tradition of Euclid, the other having roots in the Babylonian and Egyptian procedures and says that, “the writings of Heron and Diophantus. . .form part of this oriental tradition which can be followed into the Middle Ages both in the Arabic and in the western world.” Neugebauer saw Diophantus as reflecting an earlier type of mathematics practiced in Greece alongside the Pythagorean mathematics and temporarily eclipsed by the Euclidean school. As he said (1952, p. 142):
It seems to me characteristic, however, that Archytas of Tarentum could make the statement that not geometry but arithmetic alone could provide satisfactory proofs. If this was the opinion of a leading mathematician of the generation just preceding the birth of the axiomatic method, then it is rather obvious that early Greek mathematics cannot have been very different from the Heronic Diophantine type.
9.4.1 Algebraic Symbolism
Diophantus began by introducing a symbol for a constant unit , from monás , along with a symbol for an unknown number ς, conjectured to be an abbreviation of the first two letters of the Greek word for number: arithmós . For the square of an unknown he used Δυ, the first two letters of dýnamis , meaning power. For its cube he used Kυ,the first two letters of kýbos , meaning cube. He then combined these letters to get fourth (ΔυΔ), fifth (ΔKυ), and sixth (KυK) powers. For the reciprocals of these powers of the unknown he invented names by adjoining the suffix -ton (-τoν)to the names of the corresponding powers. These various powers of the unknown were called eída , meaning species. Diophantus' system for writing down the equivalent of a polynomial in the unknown consisted of writing down these symbols in order to indicate addition, each term followed by the corresponding number symbol (for which the Greeks used their alphabet). Terms to be added were placed first, separated by a pitchfork from those to be subtracted. Heath conjectured that this pitchfork symbol is a condensation of the letters lambda and iota, the first two letters of a Greek root meaning less or leave. Thus what we would call the expression 2x4 − x3 − 3x2 + 4x + 2 would be written .
Diophantus' use of symbolism is rather sparing by modern standards. He often uses words where we would use symbolic manipulation. For this reason, his algebra was described by the nineteenth-century German orientalist Nesselmann (1811–1881) as a transitional “syncopated” phase between the earliest “rhetorical” algebra, in which everything is written out in words, and the modern “symbolic” algebra.
9.4.2 Contents of the Arithmetica
According to the introduction to the Arithmetica, this work consisted originally of 13 books, but until recently only six were known to have survived. It was assumed that these were the first six books, on which Hypatia was known to have written a commentary. However, more books were recently found in an Arabic manuscript that the experts say is a translation made very early—probably in the ninth century. Sesiano (1982) stated that these books are in fact the books numbered 4 to 7, and that the books previously numbered 4 to 6 must come after them.
Diophantus begins with a small number of determinate problems that illustrate how to think algebraically using the symbolic notation discussed above. Indeterminate problems, which are number theory because the solutions are required to be rational numbers (the only kind recognized by Diophantus), begin in Book 2.5 A famous example of this type is Problem 8 of Book 2: Separate a given square number into two squares. Diophantus illustrates this problem using the number 16 as an example. His method of solving this problem is to express the two numbers in terms of a single unknown ς in such a way that one of the conditions is satisfied automatically. Thus, letting one of the two squares be ς2, which Diophantus wrote as Δυ, he noted that the other will automatically be 16 − ς2. To get a determinate equation for ς, he assumes that the other number to be squared is 4 less than an unspecified multiple of ς. The number 4 is chosen because it is the square root of 16. In our terms, it leads to a quadratic equation one of whose roots is zero, so that the other root can be found by solving a linear equation. As we would write it, assuming that 16 − ς2 = ( − 4)2, we find that (k2 + 1)ς2 = 8, and—canceling ς, since Diophantus does not operate with 0—we get ς = 8k/(k2 + 1). This formula generates a whole infinite family of solutions of the equation that we would call x2 + y2 = 16 via the identity
You may be asking why it was necessary to use a square number (16) here. Why not separate any positive rational number, say 5, into a sum of two squares? If you look carefully at the solution, you will see that Diophantus had to make the constant term drop out of the quadratic equation, and that could only be done by introducing the square root of the given number.
Diophantus' procedure is slightly less general than what we have just shown, although his illustrations show that he knows the general procedure and could generate other solutions. In his illustration he assumes that the other square is (2ς − 4)2. Since this number must be 16 − ς2, he finds that 4ς2 − 16ς + 16 = 16 − ς2, so that . It is clear that this procedure can be applied very generally, with the coefficient 2 replaced by any positive integer, showing an infinite number of ways of dividing a given square into two other squares.
At first sight it appears that number theory really is not involved in this problem, that it is a matter of pure algebra. This topic, however, naturally leads to other questions that definitely do involve number theory, that is, the theory of divisibility of integers. The most obvious one is the problem of finding all possible representations of a positive rational number as the sum of the squares of two rational numbers. One could then generalize and ask how many ways a given rational number can be represented as the sum of the cubes or fourth powers, and so forth, of two rational numbers. Those of a more Pythagorean bent might ask how many ways a number can be represented as a sum of triangular, pentagonal, or hexagonal numbers. In fact, many questions like this have been asked. Leonhard Euler (1707–1783) proved that it was impossible for the sum of fewer than three cubes to equal a cube and conjectured that it was impossible for the sum of fewer than n nth powers to equal another nth power. (He was wrong: It is possible for for the sum of four fifth powers to equal a fifth power.)
9.4.3 Fermat's Last Theorem
The problem just solved achieved lasting fame when Fermat, who was studying the Arithmetica, remarked that the analogous problem for cubes and higher powers had no solutions; that is, one cannot find positive integers xy, and zsatisfying x3 + y3 = z3 or x4 + y4 = z4, or, in general, xn + yn = zn with n > 2. Fermat stated that he had found a proof of this fact, but unfortunately did not have room to write it in the margin of the book. Fermat never published any general proof of this fact, although the special case n = 4 is a consequence of a method of proof developed by Fermat, known as the method of infinite descent. The problem became generally known after 1670, when Fermat's son published an edition of Diophantus' work along with Fermat's notes. It was a tantalizing problem because of its comprehensibility. Anyone with a high-school education in mathematics can understand the statement of the problem, and many mathematicians dreamed of solving it when they were young. Despite the efforts of hundreds of amateurs and prizes offered for the solution, no correct proof was found for more than 350 years. On June 23, 1993, the British mathematician Andrew Wiles (b. 1953) announced at a conference held at Cambridge University that he had succeeded in proving a certain conjecture in algebraic geometry known as the Shimura–Taniyama conjecture, from which Fermat's conjecture is known to follow. This was the first claim of a proof by a reputable mathematician using a technique that is known to be feasible, and the result was tentatively endorsed by other mathematicians of high reputation. After several months of checking, some doubts arose. Wiles had claimed in his announcement that certain techniques involving what are called Euler systems could be extended in a particular way, and this extension proved to be doubtful. In collaboration with another British mathematician, Richard Taylor, Wiles eventually found an alternative approach that simplified the proof considerably, and there is now no doubt among the experts in number theory that the problem has been solved.
To give another illustration of the same method, we consider the problem following the one just discussed, that is, Problem 9 of Book II: Separate a given number that is the sum of two squares into two other squares. (That is, given one representation of a number as a sum of two squares, find a new representation of the same type.) Diophantus shows how to do this using the example 13 = 22 + 32. He lets one of the two squares be (ς + 2)2 and the other (2ς− 3)2, resulting in the equation 5ς2 − 8ς = 0. Thus, , and indeed . It is easy to see here that Diophantus is deliberately choosing a form for the solution that will cause the constant term to drop out. This amounts to a general method, used throughout the first two books, and based on the proportion
for solving the equation X2 + Y2 = a2.
The method Diophantus used to solve such problems in his first two books was conjectured by Maximus Planudes (1255–1305) and has recently been explained in simple language by Christianidis (1998).
Some of Diophantus' indeterminate problems reach a high degree of complexity. For example, Problem 19 of Book 3 asks for four numbers such that if any of the numbers is added to or subtracted from the square of the sum of the numbers, the result is a square number. Diophantus gives the solutions as
Problems and Questions
Mathematical Problems
9.1 Use the fact that the greatest common divisor of 26173996849 and 180569389 is 713713 to reduce the fraction to lowest terms.
9.2 The Euclidean algorithm focuses on the remainders in the division process and has nothing to say about the quotients. Note that the quotients in the example given above were (in order of division) 144, 1, 20, and 12. Consider the continued fraction
Evaluate this fraction, and compare it with the result of the preceding exercise. We shall have further use for the quotients in the Euclidean algorithm when we study the mathematics of the Hindus.
9.3 Verify that
See L. J. Lander and T. R. Parkin, “Counterexample to Euler's conjecture on sums of like powers,” Bulletin of the American Mathematical Society72 (1966), p. 1079. Smaller counterexamples to this conjecture have been discovered more recently.
Historical Questions
9.4 What are the main topics investigated in ancient Greek number theory?
9.5 What mathematical ideas were ascribed to the Pythagoreans by ancient commentators?
9.6 What innovation in mathematics arises in the number-theoretic investigations of Diophantus?
Questions for Reflection
9.7 For what reason would the ancient Greeks have been investigating figurate numbers, perfect numbers, and the like? Did they have a practical application for these ideas?
9.8 How much of number theory has a practical application nowadays? (If you don't know about RSA codes, for example, look them up on-line.)
9.9 Should the work of Diophantus be classified as primarily number theory or primarily algebra?
Notes
1. D'ooge illustrates the procedure in a footnote, but states erroneously that 8191 is not a prime.
2. The answer is unknown at present.
3. The answer is yes. The result is amazingly easy to prove, but no one seems to have noticed it until a posthumous paper of Leonhard Euler gave a proof. Victor-Amédée Lebesgue (1791–1875) published a short proof in 1844.
4. The reader will correctly infer from previous footnotes that exactly 47 perfect numbers are now known.
5. Although Diophantus allowed solutions to be what we now call positive rational numbers, the name Diophantine equation is now used to refer to an indeterminate equation or system in which the solutions are required to be integers.
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https://justaaa.com/statistics-and-probability/1129361-i-am-interested-in-whether-males-and-females-have | 1,721,565,629,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00303.warc.gz | 293,057,457 | 9,258 | Question
# I am interested in whether males and females have the same grade point average. Use the...
I am interested in whether males and females have the same grade point average. Use the table to perform a difference of means test and tell me the result of the test. (Remember you have to calculate the variance for each variable using the equation with n-1 in the denominator)
Obs Males Females 1 3.84 2.93 2 2.5 2.98 3 3.25 1.48 4 2.77 1.77 5 3.69 2.76 6 2.02 1.89 7 2.05 3.29 8 3.01 1.19 sums mean variance
1. State the Null and Alternative Hypotheses. (5 points)
1. Establish your decision criteria and determine t critical (df = N1+N2-2).
1. What is the Standard Error of the mean difference?
1. Calculate t calculated?
1. Are the results statistically significant and explain what it means with respect to the Null hypothesis (20points).
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 254 | 912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.820829 |
http://isabelle.in.tum.de/repos/isabelle/raw-diff/563bea92b2c0/src/HOL/Probability/Information.thy | 1,575,628,384,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-51/segments/1575540487789.39/warc/CC-MAIN-20191206095914-20191206123914-00507.warc.gz | 76,317,971 | 2,430 | diff -r 27cb9113b1a0 -r 563bea92b2c0 src/HOL/Probability/Information.thy --- a/src/HOL/Probability/Information.thy Wed Feb 23 11:40:17 2011 +0100 +++ b/src/HOL/Probability/Information.thy Wed Feb 23 11:40:18 2011 +0100 @@ -167,6 +167,53 @@ definition "KL_divergence b M \ = \x. log b (real (RN_deriv M \ x)) \M\measure := \\" +lemma (in sigma_finite_measure) KL_divergence_vimage: + assumes T: "T \ measure_preserving M M'" + and T': "T' \ measure_preserving (M'\ measure := \' \) (M\ measure := \ \)" + and inv: "\x. x \ space M \ T' (T x) = x" + and inv': "\x. x \ space M' \ T (T' x) = x" + and \': "measure_space (M'\measure := \'\)" "measure_space.absolutely_continuous M' \'" + and "1 < b" + shows "KL_divergence b M' \' = KL_divergence b M \" +proof - + interpret \': measure_space "M'\measure := \'\" by fact + have M: "measure_space (M\ measure := \\)" + by (rule \'.measure_space_vimage[OF _ T'], simp) default + have "sigma_algebra (M'\ measure := \'\)" by default + then have saM': "sigma_algebra M'" by simp + then interpret M': measure_space M' by (rule measure_space_vimage) fact + have ac: "absolutely_continuous \" unfolding absolutely_continuous_def + proof safe + fix N assume N: "N \ sets M" and N_0: "\ N = 0" + then have N': "T' -` N \ space M' \ sets M'" + using T' by (auto simp: measurable_def measure_preserving_def) + have "T -` (T' -` N \ space M') \ space M = N" + using inv T N sets_into_space[OF N] by (auto simp: measurable_def measure_preserving_def) + then have "measure M' (T' -` N \ space M') = 0" + using measure_preservingD[OF T N'] N_0 by auto + with \'(2) N' show "\ N = 0" using measure_preservingD[OF T', of N] N + unfolding M'.absolutely_continuous_def measurable_def by auto + qed + + have sa: "sigma_algebra (M\measure := \\)" by simp default + have AE: "AE x. RN_deriv M' \' (T x) = RN_deriv M \ x" + by (rule RN_deriv_vimage[OF T T' inv \']) + show ?thesis + unfolding KL_divergence_def + proof (subst \'.integral_vimage[OF sa T']) + show "(\x. log b (real (RN_deriv M \ x))) \ borel_measurable (M\measure := \\)" + by (auto intro!: RN_deriv[OF M ac] borel_measurable_log[OF _ `1 < b`]) + have "(\ x. log b (real (RN_deriv M' \' x)) \M'\measure := \'\) = + (\ x. log b (real (RN_deriv M' \' (T (T' x)))) \M'\measure := \'\)" (is "?l = _") + using inv' by (auto intro!: \'.integral_cong) + also have "\ = (\ x. log b (real (RN_deriv M \ (T' x))) \M'\measure := \'\)" (is "_ = ?r") + using M ac AE + by (intro \'.integral_cong_AE \'.almost_everywhere_vimage[OF sa T'] absolutely_continuous_AE[OF M]) + (auto elim!: AE_mp) + finally show "?l = ?r" . + qed +qed + lemma (in sigma_finite_measure) KL_divergence_cong: assumes "measure_space (M\measure := \\)" (is "measure_space ?\") assumes [simp]: "sets N = sets M" "space N = space M" @@ -236,18 +283,6 @@ \ space = X`space M, sets = Pow (X`space M), measure = distribution X \ \ space = Y`space M, sets = Pow (Y`space M), measure = distribution Y \ X Y" -lemma algebra_measure_update[simp]: - "algebra (M'\measure := m\) \ algebra M'" - unfolding algebra_def by simp - -lemma sigma_algebra_measure_update[simp]: - "sigma_algebra (M'\measure := m\) \ sigma_algebra M'" - unfolding sigma_algebra_def sigma_algebra_axioms_def by simp - -lemma finite_sigma_algebra_measure_update[simp]: - "finite_sigma_algebra (M'\measure := m\) \ finite_sigma_algebra M'" - unfolding finite_sigma_algebra_def finite_sigma_algebra_axioms_def by simp - lemma (in prob_space) finite_variables_absolutely_continuous: assumes X: "finite_random_variable S X" and Y: "finite_random_variable T Y" shows "measure_space.absolutely_continuous @@ -313,6 +348,32 @@ unfolding mutual_information_def . qed +lemma (in information_space) mutual_information_commute_generic: + assumes X: "random_variable S X" and Y: "random_variable T Y" + assumes ac: "measure_space.absolutely_continuous + (S\measure := distribution X\ \\<^isub>M T\measure := distribution Y\) (joint_distribution X Y)" + shows "mutual_information b S T X Y = mutual_information b T S Y X" +proof - + let ?S = "S\measure := distribution X\" and ?T = "T\measure := distribution Y\" + interpret S: prob_space ?S using X by (rule distribution_prob_space) + interpret T: prob_space ?T using Y by (rule distribution_prob_space) + interpret P: pair_prob_space ?S ?T .. + interpret Q: pair_prob_space ?T ?S .. + show ?thesis + unfolding mutual_information_def + proof (intro Q.KL_divergence_vimage[OF Q.measure_preserving_swap _ _ _ _ ac b_gt_1]) + show "(\(x,y). (y,x)) \ measure_preserving + (P.P \ measure := joint_distribution X Y\) (Q.P \ measure := joint_distribution Y X\)" + using X Y unfolding measurable_def + unfolding measure_preserving_def using P.pair_sigma_algebra_swap_measurable + by (auto simp add: space_pair_measure distribution_def intro!: arg_cong[where f=\]) + have "prob_space (P.P\ measure := joint_distribution X Y\)" + using X Y by (auto intro!: distribution_prob_space random_variable_pairI) + then show "measure_space (P.P\ measure := joint_distribution X Y\)" + unfolding prob_space_def by simp + qed auto +qed + lemma (in information_space) mutual_information_commute: assumes X: "finite_random_variable S X" and Y: "finite_random_variable T Y" shows "mutual_information b S T X Y = mutual_information b T S Y X" @@ -323,7 +384,7 @@ lemma (in information_space) mutual_information_commute_simple: assumes X: "simple_function M X" and Y: "simple_function M Y" shows "\(X;Y) = \(Y;X)" - by (intro X Y simple_function_imp_finite_random_variable mutual_information_commute) + by (intro mutual_information_commute X Y simple_function_imp_finite_random_variable) lemma (in information_space) mutual_information_eq: assumes "simple_function M X" "simple_function M Y" | 1,685 | 5,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-51 | latest | en | 0.792638 |
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1 Transistors Lesson #9 hapter 4 252
2 JT egions of Operation Saturation Active i amps i =50 ma 40 ma 30 ma 20 ma 10 ma 0 ma v volts utoff There are three regions of operation for a JT: 1. utoff egion where no current flows in the collector or base circuits. When a transistor is in this state, we say that the transistor is OFF 2. Active egion where amplification takes place 3. Saturation egion where the maximum current flows in the collector for a given load. 253
3 JT egions of Operation Saturation i amps i =50 µa 0 µa v volts 40 µa Active 30 µa 20 µa 10 µa utoff 1. utoff: both junctions are reverse biased, i = 0, v <V FThreshold, i =0, v =V 2. Active: mitterase junction forward biased, ollector ase junction reverse biased, i >0, v V FThreshold, i = βi >0, v =V i, where is the series (load) resistor in the collector circuit. 3. Saturation: both junctions are forward biased, i >0, v V Threshold, v = V Saturation 0 < v, i =(V V Saturation )/, i < βi, where is the series (load) resistor in the collector circuit. 254
4 Large Scale D Models For each of these regions, we can define an analysis model to aide in the D operation of the transistor 255
5 Large Scale D Models utoff egion npn pnp V < V FThreshold 0.6 V V < V FThreshold 0.6 V V > V FThreshold 0.6 V V > V FThreshold 0.6 V 256
6 Large Scale D Models Active egion npn pnp β β 0.7V I 0.7V I > 0; V 0.7V > V FThreshold > 0; V 0.7V > V FThreshold = b ; V > 0.2 V = b ; V < 0.2 V 257
7 Large Scale D Models Saturation egion npn 0.7V 0.2V I pnp 0.7V 0.2V I > 0; V 0.7V > V FThreshold 0 < < β ; V 0.2 V > 0; V 0.7V > V FThreshold 0 < < β ; V 0.2 V 258
8 An xample 200k This circuit uses one D source to set the Q point. The term, ias point, is synonymous with Q point. Therefore we call this circuit a selfbias circuit V β = 100 V Let s first assume that this circuit is in cutoff. We will use the model for cutoff 259
9 An xample If this is in cutoff, then V = 15V. ut this can t be since to be in cutoff V < V FThreshold k β = 100 V V 200k 1k =0 =0 V V 260
10 An xample 200k 1k β = 100 Now let s assume that this circuit is in the saturation region. And we will use the model for the saturation region V V 261
11 200k An xample If this is in the saturation region, then i = (15 0.7) / 200k =71.5m A i = (15 0.2) / 1k =14.8 m A β =100 * 71.5m = 7.15 ma < i NOT OK!!!! β = 100 V V 200k 0.2V V I<b ic 0.7V V 262
12 An xample 200k 1k β = 100 Now let s assume that this circuit is in the active region. And we will use the model for the active region V V 263
13 200k 1k β = 100 An xample If this is in the active region, then i = (15 0.7) / 200k =71.5m A i = 100 * 71.5m A = 7.15 ma V =15 (7.15 m * 1k) = 7.85 V > 0.2V OK!!!! V V 200k 0.7V I=βIc V V 264
14 Another Selfias ircuit xample 1 10k 2 V V 5k 1k β = 300 I This is another selfbias circuit which uses 4 resistors to set the Qpoint. Let s redraw this circuit to see how to analyze it! 265
15 Another Selfias ircuit xample Use Thevenin s equivalent on the redrawn circuit to yield another simpler circuit to analyze 1 10k 1 β = V V 5k 1k β = 300 I 10k 2 V V 5k 1k I 266
16 Another Selfias ircuit xample Use Thevenin s equivalent on the redrawn circuit to yield another simpler circuit to analyze 1 10k 2 β = 300 V V 5k 1k I V TH TH V β = 300 V 1k I 267
17 V TH Another Selfias ircuit xample Now let s assume the active region, replace the transistor with the active region model, and analyze the circuit. First, the Thevenin s quivalent: TH β V V I I V β = 300 V TH 1k I = = I TH = βi = V TH V V T (1 β ) = 4.24mA; I I V I I V TH TH = = V = 15 = 5V k = 5 10 = 3.33k Next, applying KVL to the ase And ollector ircuits: = I TH = 4.25mA V = 3.33k (301)1 k ( I I ) = 14.1µ A V = m 1k 4.25m 1k = 6.51V = I TH (1 β ) I 268
18 Homework Probs. 4.26, 4.27, 4.28, 4.29, 4.30, 4.33, 4.34,
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### 4.4 The MOSFET as an Amp and Switch
10/31/004 section 4_4 The MSFET as an Amp and Switch blank 1/1 44 The MSFET as an Amp and Switch Reading Assignment: pp 70-80 Now we know how an enhancement MSFET works! Q: A: 1 H: The MSFET as an Amp
### Homework Assignment 09
Homework Assignment 09 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. What is the 3-dB bandwidth of the amplifier shown below if r π = 2.5K, r o = 100K, g m = 40 ms, and C L = | 8,442 | 30,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-10 | latest | en | 0.717604 |
https://www.convert-measurement-units.com/convert+Gram+per+milliliter+to+Pound+per+cubic+foot.php | 1,660,193,429,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00621.warc.gz | 673,903,595 | 12,976 | Convert g/ml to lb/ft³ (Gram per milliliter to Pound per cubic foot)
## Gram per milliliter into Pound per cubic foot
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Gram+per+milliliter+to+Pound+per+cubic+foot.php
## How many Pound per cubic foot make 1 Gram per milliliter?
1 Gram per milliliter [g/ml] = 62.427 960 591 578 Pound per cubic foot [lb/ft³] - Measurement calculator that can be used to convert Gram per milliliter to Pound per cubic foot, among others.
# Convert Gram per milliliter to Pound per cubic foot (g/ml to lb/ft³):
1. Choose the right category from the selection list, in this case 'Density'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Gram per milliliter [g/ml]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Pound per cubic foot [lb/ft³]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '263 Gram per milliliter'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Gram per milliliter' or 'g/ml'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Density'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '94 g/ml to lb/ft3' or '58 g/ml into lb/ft3' or '42 Gram per milliliter -> Pound per cubic foot' or '48 g/ml = lb/ft3' or '11 Gram per milliliter to lb/ft3' or '92 g/ml to Pound per cubic foot' or '6 Gram per milliliter into Pound per cubic foot'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(23 * 33) g/ml'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '263 Gram per milliliter + 789 Pound per cubic foot' or '25mm x 6cm x 18dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 8.731 864 118 070 9×1029. For this form of presentation, the number will be segmented into an exponent, here 29, and the actual number, here 8.731 864 118 070 9. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 8.731 864 118 070 9E+29. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 873 186 411 807 090 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 980 | 4,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-33 | latest | en | 0.772466 |
http://en.allexperts.com/q/Advanced-Math-1363/2013/7/proofs-right-triangle-altitude.htm | 1,490,357,713,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187945.85/warc/CC-MAIN-20170322212947-00561-ip-10-233-31-227.ec2.internal.warc.gz | 104,165,217 | 6,754 | You are here:
Question
Given: triangle ABC is a right triangle, altitude CD
Prove: triangle CBD ~ triangle ACD
Questioner:Athena Apodaca
Country:Arizona, United States
Private:No
Subject:Proofs. Right triangle altitude
Question:
Given: triangle ABC is a right triangle, altitude CD
Prove: triangle CBD ~ triangle ACD
.............................................
On the assumption that angle C is the right angle: (Not stated -- you should)
You have triangles CBD and ACD. (You did well to name them correctly, with corresponding vertices in the same spots.)
1. Identify a relation between angle CAD and angle ACD.
2. Identify a relation between angle DCB and angle ACD.
3. Use those to deduce a relation between angle CAD and angle BCD. (Guess what you are going to prove about them)
Now find another pair of corresponding angles in your triangles and prove they are equal.
Take it from there.
=====================================
NOTE: Advanced Math is precalculus and trigonometry. Send these to the Geometry section.
Volunteer
Paul Klarreich
Expertise
I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.
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Education/Credentials
----------- | 365 | 1,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-13 | latest | en | 0.882404 |
http://www.scottkirkwood.com/2006_05_01_archive.html | 1,526,942,155,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00053.warc.gz | 466,951,840 | 27,521 | ## Posts
Showing posts from May, 2006
### Solving Sudoku in Python
This is a great article about solving sudoku in python. Peter Norvig is a director of research at google and looks to be a pretty smart cookie.
I've toyed with writing a sudoku solver in python and the first part of his code mirrored closely to what I was thinking. Build a set of dictionaries where given a cell you can find all the other cells that constrain it (what he calls units and peers).
Naming the cells A1, A2 like an excel spreadsheet is very smart and I hadn't though of that (2D arrays are a pain in Python). Putting all the possibilities in each cell (i.e. "123456789") to start and eliminating impossible values as you set the givens is brilliant.
In fact, it's so brilliant that I had a hard time following the generator expressions and list comprehensions despite being very comfortable with them. Also, Peter's lisp heritage shows throughout his code.
The part that confuses me is where he picks the most constrained cell (for example one that has only t…
### Haarg, world!
A six year old writes a computer game with his dad using Python. This article almost brought a tear to my eyes, it's so cute. It's a story where Anthony writes and loves his little python "guess the number" program that he writes pretty much all by himself.
Also the article shows how to be a good dad in 2005.
### OLTP
The One Laptop Per Child (OLPC) group has some new shots of some real prototypes. If you click on the image on the right you can see the keyboard a bit better.
There are few things I don't like that much about the keyboard. First they replaced the Zero key with an X for Multiply, probably just a little mistake. There appears to be an Undo key instead of backspace which is not a big deal. What appears a little annoying is that the cursor keys are next to the screen, which might get tiring lifting your arms to that level. If they are going to mess with the standard keyboard layout, why don't they get rid of the caps lock key?
Some people feel that the OLPC is just the wrong way to go. I think the confusion is thinking of this as a laptop! (perhaps it's badly named). I think it's more useful to think if of it as a digital book and a digital notebook. What this is, is a way of distributing text books cheaply and saving on paper for written assignments. For some, thi…
### State of Fear
I just finished listening to a State of Fear by Michael Crichton. My big problem with the book is that Crichton tries to debunk global warming as unsubstantiated and as fear mongering. On the plus side Crichton does bring up some valid points about uncertainties with climate change, but it is very one sided in the book. I also agreed with Crichton that politicians use fear to gain power - but today it's more "terrorism" than global warming. I also have a beef with Treehuggers who see things only black and white. For example:
Nuclear power - evil,
DDT - evilChlorinated or fluorinated water - evilAnimal testing - evilAll of these all have pros and cons and simplifying things as for or against just doesn't help.
All in all I'm dissapointed with the book and the stance it takes. It doesn't help that Bush liked the book and met with Crichton.
In order to balance my Karma:
An Incovenient Truth - coming to a theater near you.
How to Talk to a Global Warming Skeptic
Wired News has an article about dead formats. It sparked a thought I've had for a while on some old standards that I thought would have died a long time ago, but are still used today.
One is the tar format. This way of storing data is very old, yet it is still used quite a bit in the Unix world. With zip (rar, etc.) like program I thought this format would dissapear, but it is still going strong. I have a magnetic tape from the late nineties that I'm sure I could still read since it's stored as a tarball.
Another one is TeX, which I wrote my master's thesis with. I thought TEX was great, but that no one except a techie would ever use it since it's so complicated. I looked that programs like Microsoft Word and figured that it would be the future. Yet TEX is still going strong and I could probably still compile my old thesis with no problems.
The common theme is that these "products" are open source, there is no company behind them hiding proprietary forma…
### Weekly del.icio.us bookmarks
Shared bookmarks for del.icio.us user scott_kirkwood
MIDAS: Homemade Dorm Room Home Automation System -- Can't see all the videos, very geeky and cool tagged as: [diy electronics howto]Password Gorilla -- Compatible with PasswordSafe which I use, plus it's cross platform (but which version of PasswordSafe?) tagged as: [linux free password security windows]Python Generator Tricks LG #100 -- Nice generators examples + examples of converging to the number PI in a cool way and the prime number sieve is fascinating. tagged as: [python programming]InstallJammer - A free, open source, multiplatform installer and builder - Home -- I was using Inno but perhaps this is better? tagged as: [development linux windows tools]BazAR: A vision based fast detection library -- More research for Renata's program tagged as: [video renata research toread]reacTable* software -- Some good links here, and some C++ code. tagged as: [programming video renata research]The Hardest Lessons for Startups t…
### The fall of Sony?
I've blogged about this before and there are other's making the same noise. I don't like Sony because they are just too greedy. They design their products so that if you buy one, you become compeled to continue buying from them. For example, their DAT tapes drive which only Sony players could play, or their MemoryStick memory which only Sony players can use.
Because I'm a computer guy if anyone asks I steer them away from Sony products. It's a little unfair, because the company is quite large and some of their products are fine. But these bad decisions are being made from the top and taints the whole company.
I think their PS3 won't be a big seller. It's very expensive compared to the competition, there's nothing innovative with the machine (just faster better graphics) and it has a Blueray drive which nobody is really asking for, and is just driving up the costs.
When it comes to game consoles I'm now a Nintendo man. I just went out and bought…
### Crunchy Frog
Cruncy Frog is quite neat, it's a python web server that you can use to make a python tutorial. You can enter python expressions (even multiline python code) and it'll run the code and show the output in the browser page. The idea is that you can create a tutorial that lives in the browser and where the students can write and test actual python programs.
It's looks like it's part of RUR-PLE project, which is near and dear to my heart since I also made a Karel the Robot interpreter. One of my first paid programming jobs was to make (using Turbo Pascal) a (recursive descent) interpreter for Karel programs along with a graphical interface showing the robot's world. The program had a built in text and "world" editor as well. It also mimicked the Turbo Pascal interface as much as possible so that when the students started programming they would already be used it. I remember the hardest part was giving helpful error messages when something went wrong.
By the …
### Spyware quiz
According to McAfee, most users are unable to spot spyware on websites. "According to the first-ever Spyware Quiz conducted by SiteAdvisor, a staggering 97% of Internet users are just one click away from infecting their PCs with spyware, adware or some other kind of unwanted software. Even though the threat of spyware has received extensive media coverage, just 3% of the 14,000-plus consumers who took SiteAdvisor?s spyware quiz received perfect scores."You can take the quiz here. From just looking at the screenshots. I think I did alright with 7 out of 8 questions correct. The one I got wrong was the Lyrics site.via Cow's Blog
### Forcing users tp change their passwords is not effective
At a company I consult for, they change the password every month. Not on all systems but on at least two (both Windows). It's a pain, what's worse is that changing passwords regularly doesn't work. Here are my pet peeves for usenames and passwords:
User names or passwords that can't handle "special" characters. I like to use special symbols like @ or % but a lot of sites don't permit that.
User names or passwords that have a limited maximum length. Why can't I have a password thats 200 characters?Passwords that are a fixed length. This actually makes it easier for someone to guess your password.Passwords that can only be numbers, people end up using their birthday or social security number.
Passwords that can't be cut and pasted. Banks sometimes do this because they are afraid of keyloggers. I'm not against that if you are checking your account from an internet cafĂ©, but if you are on a trusted computer why not allow cutting and pasting a p… | 2,000 | 9,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-22 | latest | en | 0.96714 |
https://www.jiskha.com/questions/1468425/if-5th-and-8th-terms-of-a-g-p-be-48-and-384-respectively-find-the-g-p | 1,627,583,854,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00461.warc.gz | 864,859,124 | 4,364 | # Geometric progression
If 5th and 8th terms of a G.P. be 48 and 384 respectively. find the G.P.
1. 👍
2. 👎
3. 👁
1. If 5th and 8th term of a G.P be 48 and 384 Respectively find the G.P
1. 👍
2. 👎
2. Good
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2. 👎
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The 4th term of an Ap is 6. if the sum of the 8th and 9th terms is -72,find the common difference
4. ### Maths
1) In the arithmetic progression, the 8th term is twice the 4th term and the 20th term is 40. Find the common difference and the sum of the terms from the 8th to the 20th term inclusive. 2) Find the least number of terms of the
1. ### mathematics
1} the first term of an AP is 2 and the common difference is 3.Find the sum of the first 11 terms. 2}The 2nd and 5th terms of a GP are -6 and 48 respectively.Find the sum of the first four terms
2. ### Maths
The 5th,9th and 16th terms of a linear sequence A.P are consecutive terms of an exponential sequence.Find the common difference of the linear sequence in terms of the first term.
3. ### Maths
The 2nd, 5th, and 14th term Of an AP are consecutive terms of a GP. Given that the sum of the three terms of the AP is 9, find the A, first term and the common difference of the AP. B common ratio of the GP C sum of the first
4. ### Mathematics
The 5th and 10th terms of an arithmetic progression are 18 and -2 respectively. (a) Find the common difference and the first term. (b) Determine the least number of terms which must be added together so that the sum of the | 708 | 2,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-31 | latest | en | 0.951583 |
http://www.knowingpains.com/tag/alphabet-market-cap/ | 1,534,377,898,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210387.7/warc/CC-MAIN-20180815235729-20180816015729-00263.warc.gz | 508,022,786 | 11,490 | # Alphabet Market Cap Label
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DebtEquity (DE) Ratio, calculated by dividing a company’s total liabilities by its stockholders' equity, is a debt ratio used to measure a company's financial leverage. The DE ratio indicates how...
Knowingpains | 517 | 2,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-34 | longest | en | 0.882667 |
https://schoollearningcommons.info/question/find-the-sum-of-first-20-40-800-20770931-52/ | 1,653,506,171,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00562.warc.gz | 586,073,831 | 14,781 | ## find the sum of first 20+40–.800
Question
find the sum of first 20+40………800
in progress 0
8 months 2021-10-03T09:14:23+00:00 2 Answers 0 views 0
1. Step-by-step explanation:
ans:-sum of terms=16400
Heya☺️✌️♥️
Step-by-step explanation:
✏️Solution :-
Let a be the first term and d be the common
difference of the given A.P.
And the sum of the first 20 terms be S(20).
S(20) = 20/2[2a + 19d]
or, 400 = 20/2[2a + 19d]
or, 400 = 10[2a + 19d]
or, 2a + 19d = 40 ….. (i)
Also, S(40) = 40/2[2a + 39d]
or, 1600 = 20[2a + 39d]
or, 2a + 39d = 80 ….(ii)
From (i) and (ii), we get
2a + 39d = 40
2a + 19d = 80
___________
– – –
⇒ 20d = 40
⇒ d = 40/20
⇒ d = 2
Putting d’s value in Eq (i), we get
⇒ 2a + 19d = 40
⇒ 2a + 19(2) = 40
⇒ 2a + 38 = 40
⇒ 2a = 40 – 38
⇒ 2a = 2
⇒ a = 2/2
⇒ a = 1
Then, S(10) = [2 × 1 + (10 – 1)2]
⇒ S(10) = 5[2 + 9 × 2]
⇒ S(10) = 5[2 + 18]
⇒ S(10) = 5 × 20
⇒ S(10) = 100
Hence, the sum of its first 10 terms is 100… | 496 | 980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-21 | latest | en | 0.420075 |
https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=186736 | 1,597,244,751,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738905.62/warc/CC-MAIN-20200812141756-20200812171756-00402.warc.gz | 675,383,664 | 5,219 | # Thermodynamics: Chapter Eight
The flashcards below were created by user DesLee26 on FreezingBlue Flashcards.
1. What is energy?
the capacity to do work, or supply heat
2. What is the formula for energy?
Delta Energy= work + heat
3. What is kinetic energy?
energy of motion
4. What is potential energy?
stored energy
5. What is thermal energy?
the kinetic energy of molecular motion (translational, rotational, and vibrational)
6. Thermal energy is proportional to __
the temperature in degrees Kelvin
7. What is heat?
the amount of thermal energy transferred between two objects at different temperatures
8. In an experiemnt: Reactants and products are th e__; everything else is the __.
• system
• surroundings
9. Energy flow from the system to the surroundings has a __ sign. (__)
• negative
• loss of energy
10. Energy flow from surroundings to the system has a __ (__)
• positive
• gain of energy
11. Open system
Closed System
Isolated System
• matter or energy can be exchanged
• only energy can be lost or gained
• no matter or energy is exchanged
12. What is the law of conservation of energy?
The energy of an __ system must be __>
• energy cannot be created or destroyed
• isolated
• constant
13. The energy change in a system equals __
• the work done on the system plus the heat added
• DeltaE= q+w, where q= heat and w= work
14. A reaction that combines something with oxygen is called?
combustion
15. What is pressure? What is it measured by?
• Force per unit area
• Pascals or 1 atmospheric unit (atm)
16. What is work?
• force that produces an object's movement, times the distance moved
• Work = Force x Distance
17. Work done by a system is __.
Work done on a ssystem is __.
negative positive
18. I L atm= __ Joules?
101
19. What is a state function?
a function or property whose value depends only on the present state (condition) of the system
20. The change in a __ is __ when the system returns to its original condition.
• state function
• zero
21. For __, the change is not __ if the path returns to its original condition.
• nonstate functions
• zero
22. The amount of heat exchanged between the system and the surroundigns is given the symbol __.
What is the equation for it?
At constant value (DeltaV=0): qv= __
At constant pressure: qp= DeltaH
• q
• q= DeltaE + PDeltaV
23. What is enthalpy?
DeltaH is __
• a state function whose value depends only on the current state of the system, not on the path taken to arrive at that state
• enthalpy change or heat of reaction (at constant pressure)
24. What is the Thermodynamic Standard State?
most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 degrees C; 1 M concentration for all substances in solution
25. What is enthalpy of Fusion DeltaHfusion?
the amount of heat necessary to melt a substance without changing its temperature
26. What is enthalpy of vaporization (DeltaHvap)
The amount of heat required to vaporize a substance without changing its temperature
27. What is enthalpy of sublimation (DeltaHsubl)?
the amount of heat required to convert a substance from a solid to a gaas without going through a liquid phase
28. Enthalpies of chemical change are often called __.
heats of reaction (DeltaHreaction)
29. What is endothermic?
exothermic?
• endo: heat flows into the system from the surroundings and DeltaH has a positive sign
• exo: heat flows out of the system into the surroundings and DeltaH has a negative sign
30. What does reversing a reaction do to DeltaH?
changes the sign (multiply by -1)
Author: DesLee26 ID: 186736 Card Set: Thermodynamics: Chapter Eight Updated: 2012-12-03 04:02:55 Tags: Cichowicz Folders: Description: Chemistry Show Answers: | 913 | 3,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-34 | latest | en | 0.902267 |
http://us.metamath.org/mpegif/esumpinfval.html | 1,529,632,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864337.41/warc/CC-MAIN-20180622010629-20180622030629-00079.warc.gz | 337,443,657 | 15,081 | Mathbox for Thierry Arnoux < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > esumpinfval Structured version Unicode version
Theorem esumpinfval 24455
Description: The value of the extended sum of non-negative terms, with at least one infinite term. (Contributed by Thierry Arnoux, 19-Jun-2017.)
Hypotheses
Ref Expression
esumpinfval.0
esumpinfval.1
esumpinfval.2
esumpinfval.3
Assertion
Ref Expression
esumpinfval Σ*
Distinct variable groups: , ,
Allowed substitution hints: () ()
Proof of Theorem esumpinfval
StepHypRef Expression
1 iccssxr 10985 . . 3
2 esumpinfval.1 . . . 4
3 esumpinfval.0 . . . . 5
4 esumpinfval.2 . . . . . 6
54ex 424 . . . . 5
63, 5ralrimi 2779 . . . 4
7 nfcv 2571 . . . . 5
87esumcl 24419 . . . 4 Σ*
92, 6, 8syl2anc 643 . . 3 Σ*
101, 9sseldi 3338 . 2 Σ*
11 nfrab1 2880 . . . . 5
12 ssrab2 3420 . . . . . 6
1312a1i 11 . . . . 5
14 0xr 9123 . . . . . . . 8
15 pnfxr 10705 . . . . . . . 8
16 pnfge 10719 . . . . . . . . 9
1714, 16ax-mp 8 . . . . . . . 8
18 ubicc2 11006 . . . . . . . 8
1914, 15, 17, 18mp3an 1279 . . . . . . 7
2019a1i 11 . . . . . 6
21 lbicc2 11005 . . . . . . . 8
2214, 15, 17, 21mp3an 1279 . . . . . . 7
2322a1i 11 . . . . . 6
2420, 23ifclda 3758 . . . . 5
25 eldif 3322 . . . . . . . 8
26 rabid 2876 . . . . . . . . . 10
2726simplbi2 609 . . . . . . . . 9
2827con3and 429 . . . . . . . 8
2925, 28sylbi 188 . . . . . . 7
3029adantl 453 . . . . . 6
31 iffalse 3738 . . . . . 6
3230, 31syl 16 . . . . 5
333, 11, 7, 13, 2, 24, 32esumss 24454 . . . 4 Σ* Σ*
34 eqidd 2436 . . . . . 6
3526simprbi 451 . . . . . . . 8
36 iftrue 3737 . . . . . . . 8
3735, 36syl 16 . . . . . . 7
3837adantl 453 . . . . . 6
393, 34, 38esumeq12dvaf 24420 . . . . 5 Σ* Σ*
402, 13ssexd 4342 . . . . . 6
41 nfcv 2571 . . . . . . 7
4211, 41esumcst 24447 . . . . . 6 Σ*
4340, 19, 42sylancl 644 . . . . 5 Σ*
44 hashxrcl 11632 . . . . . . 7
4540, 44syl 16 . . . . . 6
46 esumpinfval.3 . . . . . . . 8
47 rabn0 3639 . . . . . . . 8
4846, 47sylibr 204 . . . . . . 7
49 hashgt0 11654 . . . . . . 7
5040, 48, 49syl2anc 643 . . . . . 6
51 xmulpnf1 10845 . . . . . 6
5245, 50, 51syl2anc 643 . . . . 5
5339, 43, 523eqtrd 2471 . . . 4 Σ*
5433, 53eqtr3d 2469 . . 3 Σ*
55 breq1 4207 . . . . 5
56 breq1 4207 . . . . 5
57 pnfge 10719 . . . . . . . 8
5815, 57ax-mp 8 . . . . . . 7
59 breq2 4208 . . . . . . 7
6058, 59mpbiri 225 . . . . . 6
6160adantl 453 . . . . 5
624adantr 452 . . . . . 6
63 iccgelb 24128 . . . . . . 7
6414, 15, 63mp3an12 1269 . . . . . 6
6562, 64syl 16 . . . . 5
6655, 56, 61, 65ifbothda 3761 . . . 4
673, 7, 2, 24, 4, 66esumlef 24446 . . 3 Σ* Σ*
6854, 67eqbrtrrd 4226 . 2 Σ*
69 xgepnf 24108 . . 3 Σ* Σ* Σ*
7069biimpd 199 . 2 Σ* Σ* Σ*
7110, 68, 70sylc 58 1 Σ*
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wa 359 wnf 1553 wceq 1652 wcel 1725 wne 2598 wral 2697 wrex 2698 crab 2701 cvv 2948 cdif 3309 wss 3312 c0 3620 cif 3731 class class class wbr 4204 cfv 5446 (class class class)co 6073 cc0 8982 cpnf 9109 cxr 9111 clt 9112 cle 9113 cxmu 10701 cicc 10911 chash 11610 Σ*cesum 24416 This theorem is referenced by: hasheuni 24467 esumcvg 24468 voliune 24577 volfiniune 24578 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-13 1727 ax-14 1729 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2416 ax-rep 4312 ax-sep 4322 ax-nul 4330 ax-pow 4369 ax-pr 4395 ax-un 4693 ax-inf2 7588 ax-cnex 9038 ax-resscn 9039 ax-1cn 9040 ax-icn 9041 ax-addcl 9042 ax-addrcl 9043 ax-mulcl 9044 ax-mulrcl 9045 ax-mulcom 9046 ax-addass 9047 ax-mulass 9048 ax-distr 9049 ax-i2m1 9050 ax-1ne0 9051 ax-1rid 9052 ax-rnegex 9053 ax-rrecex 9054 ax-cnre 9055 ax-pre-lttri 9056 ax-pre-lttrn 9057 ax-pre-ltadd 9058 ax-pre-mulgt0 9059 ax-pre-sup 9060 ax-addf 9061 ax-mulf 9062 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-3or 937 df-3an 938 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-eu 2284 df-mo 2285 df-clab 2422 df-cleq 2428 df-clel 2431 df-nfc 2560 df-ne 2600 df-nel 2601 df-ral 2702 df-rex 2703 df-reu 2704 df-rmo 2705 df-rab 2706 df-v 2950 df-sbc 3154 df-csb 3244 df-dif 3315 df-un 3317 df-in 3319 df-ss 3326 df-pss 3328 df-nul 3621 df-if 3732 df-pw 3793 df-sn 3812 df-pr 3813 df-tp 3814 df-op 3815 df-uni 4008 df-int 4043 df-iun 4087 df-iin 4088 df-br 4205 df-opab 4259 df-mpt 4260 df-tr 4295 df-eprel 4486 df-id 4490 df-po 4495 df-so 4496 df-fr 4533 df-se 4534 df-we 4535 df-ord 4576 df-on 4577 df-lim 4578 df-suc 4579 df-om 4838 df-xp 4876 df-rel 4877 df-cnv 4878 df-co 4879 df-dm 4880 df-rn 4881 df-res 4882 df-ima 4883 df-iota 5410 df-fun 5448 df-fn 5449 df-f 5450 df-f1 5451 df-fo 5452 df-f1o 5453 df-fv 5454 df-isom 5455 df-ov 6076 df-oprab 6077 df-mpt2 6078 df-of 6297 df-1st 6341 df-2nd 6342 df-riota 6541 df-recs 6625 df-rdg 6660 df-1o 6716 df-2o 6717 df-oadd 6720 df-er 6897 df-map 7012 df-pm 7013 df-ixp 7056 df-en 7102 df-dom 7103 df-sdom 7104 df-fin 7105 df-fi 7408 df-sup 7438 df-oi 7471 df-card 7818 df-cda 8040 df-pnf 9114 df-mnf 9115 df-xr 9116 df-ltxr 9117 df-le 9118 df-sub 9285 df-neg 9286 df-div 9670 df-nn 9993 df-2 10050 df-3 10051 df-4 10052 df-5 10053 df-6 10054 df-7 10055 df-8 10056 df-9 10057 df-10 10058 df-n0 10214 df-z 10275 df-dec 10375 df-uz 10481 df-q 10567 df-rp 10605 df-xneg 10702 df-xadd 10703 df-xmul 10704 df-ioo 10912 df-ioc 10913 df-ico 10914 df-icc 10915 df-fz 11036 df-fzo 11128 df-fl 11194 df-mod 11243 df-seq 11316 df-exp 11375 df-fac 11559 df-bc 11586 df-hash 11611 df-shft 11874 df-cj 11896 df-re 11897 df-im 11898 df-sqr 12032 df-abs 12033 df-limsup 12257 df-clim 12274 df-rlim 12275 df-sum 12472 df-ef 12662 df-sin 12664 df-cos 12665 df-pi 12667 df-struct 13463 df-ndx 13464 df-slot 13465 df-base 13466 df-sets 13467 df-ress 13468 df-plusg 13534 df-mulr 13535 df-starv 13536 df-sca 13537 df-vsca 13538 df-tset 13540 df-ple 13541 df-ds 13543 df-unif 13544 df-hom 13545 df-cco 13546 df-rest 13642 df-topn 13643 df-topgen 13659 df-pt 13660 df-prds 13663 df-ordt 13717 df-xrs 13718 df-0g 13719 df-gsum 13720 df-qtop 13725 df-imas 13726 df-xps 13728 df-mre 13803 df-mrc 13804 df-acs 13806 df-ps 14621 df-tsr 14622 df-mnd 14682 df-plusf 14683 df-mhm 14730 df-submnd 14731 df-grp 14804 df-minusg 14805 df-sbg 14806 df-mulg 14807 df-subg 14933 df-cntz 15108 df-cmn 15406 df-abl 15407 df-mgp 15641 df-rng 15655 df-cring 15656 df-ur 15657 df-subrg 15858 df-abv 15897 df-lmod 15944 df-scaf 15945 df-sra 16236 df-rgmod 16237 df-psmet 16686 df-xmet 16687 df-met 16688 df-bl 16689 df-mopn 16690 df-fbas 16691 df-fg 16692 df-cnfld 16696 df-top 16955 df-bases 16957 df-topon 16958 df-topsp 16959 df-cld 17075 df-ntr 17076 df-cls 17077 df-nei 17154 df-lp 17192 df-perf 17193 df-cn 17283 df-cnp 17284 df-haus 17371 df-tx 17586 df-hmeo 17779 df-fil 17870 df-fm 17962 df-flim 17963 df-flf 17964 df-tmd 18094 df-tgp 18095 df-tsms 18148 df-trg 18181 df-xms 18342 df-ms 18343 df-tms 18344 df-nm 18622 df-ngp 18623 df-nrg 18625 df-nlm 18626 df-ii 18899 df-cncf 18900 df-limc 19745 df-dv 19746 df-log 20446 df-esum 24417
Copyright terms: Public domain W3C validator | 4,093 | 7,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-26 | latest | en | 0.091836 |
https://studylib.net/doc/14462503/algebra-prelim-part-b-january-2015-that-fixes-the-polynom... | 1,718,693,651,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00289.warc.gz | 484,426,049 | 11,046 | # Algebra Prelim Part B January 2015 that fixes the polynomial x
```Algebra Prelim Part B January 2015
1. Describe the subgroup of Sn that fixes the polynomial x1 +x2 under the standard action
on Q[x1 , . . . , xn ] obtained by permuting variables. Use this to give an upper bound for
the degree over Q of the real part of z ∈ C, if z is algebraic of degree n over Q. Give
a condition under which the bound is sharp. (Here Q, C are the fields of rational and
complex numbers, respectively).
2. Let K be a field and f (x) ∈ K[x] be a separable, irreducible polynomial of degree 5. If
a, b are distinct roots of f (x) with K(a) = K(b), show that K(a)/K is Galois.
3. Let K/Q be an extension of degree n, where Q is the field of rational numbers. Show
that the number of subfields of K is at most 2n! . Suppose that K = Q(α, β) and prove
that there exists m, 0 ≤ m ≤ 2n! such that K = Q(α + mβ).
2
``` | 269 | 899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.909187 |
https://physics.stackexchange.com/questions/554216/contradiction-between-classical-electromagnetic-theory-and-the-principle-of-rela | 1,656,223,652,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00419.warc.gz | 507,995,671 | 69,377 | # Contradiction between classical electromagnetic theory and the principle of relativity
I am currently studying Classical Mechanics, 5th edition, by Kibble and Berkshire. Chapter 1.2 Newton's Laws says the following:
However, this does not completely remove the difficulty, for there is still an apparent contradiction between this classical electromagnetic theory and the principle of relativity discussed in §1.1. This arises from the fact that if the speed of light is constant with respect to one inertial frame — as it should be according to electromagnetic theory — then the usual rules for combining velocities would lead to the conclusion that it is not constant with respect to a relatively moving frame, in contradiction with the statement that all inertial frames are equivalent. This paradox can only be resolved by the introduction of Einstein’s theory of relativity (i.e., ‘special’ relativity). Classical electromagnetic theory and classical mechanics can be incorporated into a single self-consistent theory, but only by ignoring the relativity principle and sticking to one ‘preferred’ inertial frame.
The aforementioned "principle of relativity" is described as follows:
Given two bodies moving with constant relative velocity, it is impossible in principle to decide which of them is at rest, and which moving. This statement, which is of fundamental importance, is the principle of relativity.
I don't understand the authors' description of the problem. I would greatly appreciate it if people would please take the time to explain this to me.
This initially seems like it might be related, but upon reading, it doesn't seem like it addresses my question.
I'm not sure if I've understood the question exactly, but it seems to be asking what exactly it is about the (Galilean) principle of relativity that makes (Classical) Electromagnetic theory incompatible with it.
The Principle of Relativity, as you say, basically states that there's no physical experiment that you can conduct in two different inertial frames moving with respect to each other to tell "which" of them is "actually" moving: the very question makes no sense. This is a consequence of the way that space and time transform when you move from one inertial frame to another.
Let's try to put this into a mathematical formalism using our "common-sense". Suppose you had two people, one person ($$S^\prime$$) in a train, moving rightwards with a velocity $$v$$, and another on a platform ($$S$$), and if the person on the platform would like to know what the coordinates of an event $$(x,t)$$ would look like to someone on the train $$(x^\prime, t^\prime)$$. For simplicity, we can choose $$x=0=x^\prime$$ and $$t=0=t^\prime$$. Using "common sense", it's a simple enough exercise to show from the above figure that:
\begin{equation*} \begin{aligned} x^\prime &= x - vt\\ t^\prime &= t \end{aligned} \end{equation*}
These are called the Galilean Transformations, and they relate the way that the person in $$S$$ observes the world, to the way that the person in $$S^\prime$$ observes the world. A natural consequence of this is that the two observers will never agree upon the velocities of a third object. For example, suppose there was a dog running (rightwards) on the train, and both observers were measuring its velocity. The velocity measured by someone on the platform would be
$$u = \frac{\Delta x}{\Delta t},$$
while that observed by someone on the train would be $$u^\prime = \frac{\Delta x^\prime}{\Delta t^\prime}.$$
Using the above Galilean Transformations, and the fact that $$v$$ is a constant, it's quite trivial to show that a natural consequence is that $$u^\prime = u - v.$$
In other words, the velocity of the dog as observed from the train would be smaller than the velocity of the dog as observed from the platform by exactly the relative velocity of the train and the platform. But then, this is just common sense, right? Furthermore, we can do the same trick above to show that both observers would always agree upon the acceleration of the dog, since:
$$a^\prime = \frac{\Delta u^\prime}{\Delta t^\prime} = \frac{\Delta u}{\Delta t} = a,$$
as should be expected since the frames are inertial, but which also means that the physical laws in both frames will have the same form, since Newton's Laws only deal with accelerations.
An important consequence, however, of this "velocity addition" formula is that there is no (finite) velocity that observers in $$S$$ and $$S^\prime$$ will agree upon, since if this were the case, you would need to have
\begin{equation*} \begin{aligned} u^\prime &= u - v \text{ (which is always true according to Galilean Relativity)}\\ u^\prime &= k = u \text{ (for some frame-independent speed k)} \end{aligned} \end{equation*}
The only value of $$k$$ that solves the above equations is $$k=\infty$$. The only frame-independent speed (i.e.\ the only speed that all observers in all inertial frames can agree upon) is infinity.
Since this is a direct consequence of the Galilean Transformations, if we had another theory that required a frame-independent speed that was finite, then we would have to say that that theory was not compatible with the Galilean Principle of Relativity, and this is precisely what Classical Electromagnetism required. According to it, there is a speed, $$c$$, which -- despite being very large -- is not infinite, and which everyone in all inertial frames could agree upon. Thus, if Electromagnetism was to be considered "right", we require either that our "common-sense" Galilean Relativity is wrong, or that Classical Electromagnetism is not compatible with Principle of Relativity.
However, it turns out that Classical Electromagnetism is compatible with the theory of relativity, it is just that our "common-sense" Galilean Transformations are not true, and they have to be replaced with the correct "Lorentz Transformations", leading to the Special Theory of Relativity, which is consistent with Electromagnetism. (In fact, it's a fun exercise to how that the magnetic field is an entirely relativistic effect!)
• How exactly do you show that $u^\prime = u - v$? I do not know the standard way of treating $\Delta$ after substitution of $x^\prime = x - vt$. May 28, 2020 at 6:56
• I'm not sure I completely understand the question, but let me see if I can answer it: since $x^\prime = x - v t$, you can take the difference between two coordinates and show that $\Delta x^\prime = \Delta x - v \Delta t$, since $v$ is a constant. You would like to relate $\Delta x^\prime/\Delta t^\prime$ to $\Delta x/\Delta t$, and so you can divide the LHS by $\Delta t^\prime$ and the RHS by $\Delta t$, and you'd get $$\frac{\Delta x^\prime}{\Delta t^\prime} = \frac{\Delta x}{\Delta t} - v,$$ and from here you can just use the definition of speed to show $u^\prime = u -v$. May 28, 2020 at 10:11
• Thanks for the response. My point was that, in trying to show that $u^\prime = u - v$, I was thinking that we would have $u^\prime = \dfrac{\Delta(x - vt)}{\Delta t}$, but I wasn't sure if this substation was mathematically valid (given the presence of the $\Delta$). But, if I'm following your comment correctly, $\Delta$ has the distributivity property, and so we have that $u^\prime = \dfrac{\Delta(x - vt)}{\Delta t} = \dfrac{\Delta x - v \Delta t}{\Delta t}$? When you say "take the difference between two coordinates", are you referring to the difference between $(x, y)$ and $(x', y')$? May 28, 2020 at 10:35
• Ah, I understand now! The $\Delta x$ is just shorthand notation for some $x_B - x_A$. It turns out that such "differences in coordinates" also satisfy the same transformations (you can show this easily): $$x_B^\prime - x_A^\prime = (x_B - x_A) - v (t_B - t_A), \quad \quad t^\prime_B - t^\prime_A = t_B - t_A,$$ and so if you wanted the speed between two points $A$ and $B$, you could just divide $$\frac{x_B - x_A}{t_B - t_A} = u,$$ and do the same for someone in $S^\prime$: $$\frac{x^\prime_B - x^\prime_A}{t^\prime_B - t^\prime_A} = u^\prime,$$ and use the transformation laws given above. May 28, 2020 at 10:55
• Ahh, I see. And when you say "speed" above, you actually mean velocity, right? May 28, 2020 at 11:19
I will give my take on this.
The statement for the relativity principle is that two inertial frames, any two and by extension all of them, are equivalent. Thus, a moving object would be described by the same laws in any reference frame, only changing coordinates accordingly. However, here we have two choices, so to speak. One is to use Galilean transformation laws between two reference frames, Newton's laws would be the same and the kinematics and dynamics the same as well. The problem is that Maxwell's laws and specially the wave equation (for light) you find in them is not invariant under such coordinate transformation $$x \rightarrow x'= x - Vt.$$ Moreover, they are invariant under Lorentz transformations which is the other choice. The problem is that Newton's laws are not invariant under Lorentz transformations and therefore we need to decide what to do.
In Galilean transformations, to find the speed of any phenomena (i.e. $$v$$) when changing from a reference frame to a moving one (lets say with speed $$V$$), we add them as in $$v \rightarrow v' = v -V.$$ This means that there is always a reference frame for which the phenomena can be at rest. If they are equivalent then light could be stopped and studied at rest in that frame, the wave equation in that frame should have $$c=0.$$ If one is to accept that light follows that same rule of Galileo, there should be a reference frame for which light is at rest.
For electromagnetic waves that is not the case, one can not find a reference frame where light is at rest, all reference frames, no matter what, measure the same speed for light. Galilean transformations are not correct for Maxwell's theory.
If one accepts that light is always moving then one should accept that there is a preferred reference frame with respect to which light is always moving and that other reference frames have to be studied with respect to that particular one to avoid a photon a rest (lets say to forbid traveling at $$c$$) and that Maxwell's laws are actually written with respect to that reference frame. Therefore, not all of them are equivalent, one is actually in absolute rest so light have $$c$$ constant in that frame and is mandatory to use that for light. So, we need to renounce the principle of relativity of equivalence of all reference frames. | 2,503 | 10,539 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-27 | latest | en | 0.962589 |
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# Concrete - Calculating Cubic Volume
## Calculating the concrete volume required for an operation.
Concrete and mortar are normally delivered on a cubic volume basis. Cubic volume can be calculated as
V = l w h (1)
where
V = cubic volume (yd3, ft3, m3)
l = length (yd, ft, m)
w = width (yd, ft, m)
h = height (yd, ft, m)
• 1 ft3 = 0.02832 m3= 28.32 dm3 = 0.03704 yd3
• 1 yd3 = 0.7646 m3 = 764.6 dm3 = 27 ft3
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Cement, sand and gravel mixtures.
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Estimate concrete wall volume.
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Concrete Masonry Unit Blocks (CMU) - dimensions.
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Weights of rebar rods - size ranging 1/4.
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https://oeis.org/A283677 | 1,597,288,756,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.61/warc/CC-MAIN-20200813014639-20200813044639-00345.warc.gz | 429,551,840 | 4,279 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A283677 a(n) = lcm(b(b(n)), b(n-b(n)+1)) where b(n) = A004001(n). 3
1, 1, 1, 2, 2, 2, 2, 6, 3, 12, 12, 4, 4, 4, 4, 20, 5, 30, 35, 35, 42, 24, 24, 56, 56, 56, 8, 8, 8, 8, 8, 72, 9, 90, 99, 36, 36, 60, 130, 70, 70, 154, 165, 165, 60, 60, 60, 195, 208, 208, 112, 112, 112, 240, 240, 240, 240, 16, 16, 16, 16, 16, 16, 272, 17, 306, 323, 340, 357, 357, 126, 198, 414, 72, 72, 456, 475, 494 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 COMMENTS See the order of certain subsequences in scatterplot link. LINKS Altug Alkan, Table of n, a(n) for n = 1..10000 Altug Alkan, Alternative Scatterplot of A283677 FORMULA a(n) = lcm(A004001(A004001(n)), A004001(A080677(n))). - Antti Karttunen, Mar 18 2017 EXAMPLE a(4) = lcm(A004001(A004001(4)), A004001(4-A004001(4)+1)) = lcm(1, 2) = 2. MATHEMATICA b[1] = b[2] = 1; b[n_] := b[n] = b[b[n - 1]] + b[n - b[n - 1]]; Table[LCM[b[b[n]], b[n + 1 - b[n]]], {n, 1, 78}] (* Indranil Ghosh, Mar 14 2017 *) PROG (PARI) a=vector(1000); a[1]=a[2]=1; for(n=3, #a, a[n]=a[a[n-1]]+a[n-a[n-1]]); va = vector(1000, n, lcm(a[a[n]], a[n+1-a[n]])) (Scheme) (define (A283677 n) (lcm (A004001 (A004001 n)) (A004001 (+ 1 (- n (A004001 n)))))) ;; (Code for A004001 given under that entry). - Antti Karttunen, Mar 18 2017 CROSSREFS Cf. A004001, A080677. Cf. also A283470, A283673. Sequence in context: A334512 A096625 A263455 * A260983 A103222 A319100 Adjacent sequences: A283674 A283675 A283676 * A283678 A283679 A283680 KEYWORD nonn,look AUTHOR Altug Alkan, Mar 14 2017 STATUS approved
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Last modified August 12 21:10 EDT 2020. Contains 336440 sequences. (Running on oeis4.) | 851 | 2,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-34 | latest | en | 0.558986 |
http://docplayer.net/21103456-Section-4-5-exponential-and-logarithmic-equations.html | 1,542,653,259,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746061.83/warc/CC-MAIN-20181119171420-20181119193420-00508.warc.gz | 96,930,616 | 24,367 | # Section 4.5 Exponential and Logarithmic Equations
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1 Section 4.5 Exponential and Logarithmic Equations Exponential Equations An exponential equation is one in which the variable occurs in the exponent. EXAMPLE: Solve the equation x = 7. Solution 1: We have x = 7 log x = log 7 ] xlog = log 7 ] Solution : We have EXAMPLE: Solve the equation 4 x+1 = 3. Solution 1: We have x = log x = 7 ln x = ln7 xln = ln7 x = ln7 ln x+1 = 3 log4 4 x+1 = log 4 3 ] (x+1)log4 4 = log 4 3 ] Solution : We have x+1 = log 4 3 x = log x+1 = 3 ln4 x+1 = ln3 (x+1)ln4 = ln3 EXAMPLE: Solve the equation 3 x 3 = 5. x+1 = ln3 ln4 x = ln ln4 1
2 EXAMPLE: Solve the equation 3 x 3 = 5. Solution 1: We have 3 x 3 = 5 x 3 = log 3 5 x = log Solution : We have 3 x 3 = 5 ln3 x 3 = ln5 (x 3)ln3 = ln5 x 3 = ln5 ln3 x = ln ln3 EXAMPLE: Solve the equation 8e x = 0. 8e x = 0 e x = 0 8 = 5 x = ln 5 x = ln 5 = 1 ln EXAMPLE: Solve the equation e 3 x = 4 algebraically and graphically. e 3 x = 4 3 x = ln4 x = ln4 3 x = ln4 3 = 3 ln
3 EXAMPLE: Solve the equation e x e x 6 = 0. Solution 1: We have e x e x 6 = 0 (e x ) e x 6 = 0 (e x 3)(e x +) = 0 e x 3 = 0 or e x + = 0 e x = 3 e x = The equation e x = 3 leads to x = ln3. But the equation e x = has no solution because e x > 0 for all x. Thus, x = ln is the only solution. Solution 1 : Put e x = w. Then e x e x 6 = 0 (e x ) e x 6 = 0 w w 6 = 0 (w 3)(w +) = 0 w 3 = 0 or w+ = 0 w = 3 e x = 3 w = e x = The equation e x = 3 leads to x = ln3. But the equation e x = has no solution because e x > 0 for all x. Thus, x = ln is the only solution. EXAMPLE: Solve the equation e x 3e x + = 0. 3
4 EXAMPLE: Solve the equation e x 3e x + = 0. Solution 1: We have e x 3e x + = 0 (e x ) 3e x + = 0 (e x 1)(e x ) = 0 e x 1 = 0 or e x = 0 e x = 1 e x = x = 0 x = ln Solution 1 : Put e x = w. Then w 3w + = 0 (w 1)(w ) = 0 w 1 = 0 or w = 0 w = 1 w = e x = 1 e x = x = 0 x = ln EXAMPLE: Solve the equation 7 x 3 7 x +1 = 0. Solution: Put 7 x = w. Then w 3w +1 = 0 = w = ( 3)± ( 3) ( so 7 x = 3± 5 3± ) 5, therefore x = log 7. = 3± 5 EXAMPLE: Solve the equation 7 x 7 x 1 = 0. Solution: Put 7 x = w. Then Since 1 5 w w 1 = 0 = w = ( 1)± ( 1) 4 1 ( 1) 1 < 0, it follows that 7 x = 1+ 5 = x = log 7 ( 1+ ) 5 = 1± 5 EXAMPLE: Solve the equation 3xe x +x e x = 0. 3xe x +x e x = 0 xe x (3+x) = 0 x(3+x) = 0 x = 0 or 3+x = 0 x = 0 x = 3 4
5 Logarithmic Equations A logarithmic equation is one in which a logarithm of the variable occurs. EXAMPLE: Solve the equation lnx = 8. lnx = 8 e lnx = e 8] x = e 8 EXAMPLE: Solve the equation log (x+) = 5. log (x+) = 5 log (x+) = 5] x+ = 5 x = 5 = 3 = 30 EXAMPLE: Solve the equation log 7 (5 x) = 3. log 7 (5 x) = 3 7 log (5 x) 7 = 7 3] 5 x = 7 3 x = = = 318 EXAMPLE: Solve the equation 4+3log(x) = 16. 5
6 EXAMPLE: Solve the equation 4+3log(x) = log(x) = 16 3log(x) = 1 log(x) = 4 x = 10 4 x = 104 = 10,000 = 5,000 EXAMPLE: Solve the equation log(x+)+log(x 1) = 1 algebraically and graphically. log(x+)+log(x 1) = 1 log(x+)(x 1)] = 1 (x+)(x 1) = 10 x +x = 10 x +x 1 = 0 (x+4)(x 3) = 0 x+4 = 0 or x 3 = 0 x = 4 x = 3 We check these potential solutions in the original equation and find that x = 4 is not a solution (because logarithms of negative numbers are undefined), but x = 3 is a solution. To solve the equation graphically we rewrite it as log(x+)+log(x 1) 1 = 0 and then graph y = log(x+)+log(x 1) 1. The solutions are the x-intercepts of the graph. EXAMPLE: Solve the following equations (a) log(x+8)+log(x 1) = 1 (b) log(x 1) log(x+1) = 3 6
7 EXAMPLE: Solve the following equations (a) log(x+8)+log(x 1) = 1 (b) log(x 1) log(x+1) = 3 Solution: (a) We have log(x+8)+log(x 1) = 1 log(x+8)(x 1)] = 1 (x+8)(x 1) = 10 x +7x 8 = 10 x +7x 18 = 0 (x+9)(x ) = 0 x+9 = 0 or x = 0 x = 9 x = We check these potential solutions in the original equation and find that x = 9 is not a solution (because logarithms of negative numbers are undefined), but x = is a solution. (b) We have log(x 1) log(x+1) = 3 log x 1 x+1 = 3 x 1 x+1 = 103 (x 1)(x+1) x+1 = 1000 x 1 = 1000 x = 1001 EXAMPLE: Solve the equation x = ln(x+) graphically. Solution: We first move all terms to one side of the equation x ln(x+) = 0. Then we graph y = x ln(x+). The solutions are the x-intercepts of the graph. EXAMPLE: Find the solution of the equation, correct to two decimal places. (a) 10 x+3 = 6 x (b) 5 ln(3 x) = 4 (c) log (x+)+log (x 1) = 7
8 EXAMPLE: Find the solution of the equation, correct to two decimal places. (a) 10 x+3 = 6 x (b) 5 ln(3 x) = 4 (c) log (x+)+log (x 1) = Solution: (a) We have 10 x+3 = 6 x ln10 x+3 = ln6 x (x+3)ln10 = xln6 xln10+3ln10 = xln6 xln10 xln6 = 3ln10 x(ln10 ln6) = 3ln10 x = 3ln10 ln10 ln (b) We have 5ln(3 x) = 4 ln(3 x) = x = e 4/5 x = 3 e 4/ (c) We have log (x+)+log (x 1) = log (x+)(x 1) = (x+)(x 1) = 4 x +x = 4 x +x 6 = 0 (x )(x+3) = 0 x = 0 or x+3 = 0 x = x = 3 Since x = 3 is not from the domain of log (x+)+log (x 1), the only answer is x =. 8
9 Applications EXAMPLE: If I 0 and I denote the intensity of light before and after going through a material and x is the distance (in feet) the light travels in the material, then according to the Beer- Lambert Law 1 k ln ( I I 0 ) = x where k is a constant depending on the type of material. (a) Solve the equation for I. (b) For a certain lake k = 0.05 and the light intensity is I 0 = 14 lumens (lm). Find the light intensity at a depth of 0 ft. Solution: (a) We first isolate the logarithmic term. 1 k ln ( I I 0 ) = x ( ) I ln = kx I 0 I I 0 = e kx I = I 0 e kx (b) We find I using the formula from part (a). I = I 0 e kx = 14e ( 0.05)(0) 8.49 The light intensity at a depth of 0 ft is about 8.5 lm. EXAMPLE: A sum of \$5000 is invested at an interest rate of 5% per year. Find the time required for the money to double if the interest is compounded according to the following method. (a) Semiannual (b) Continuous 9
10 EXAMPLE: A sum of \$5000 is invested at an interest rate of 5% per year. Find the time required for the money to double if the interest is compounded according to the following method. (a) Semiannual (b) Continuous Solution: (a) We use the formula for compound interest A(t) = P ( 1+ r ) nt n with P = \$5000, A(t) = \$10,000, r = 0.05, n = and solve the resulting exponential equation for t. ( ) t = 10,000 The money will double in years. (1.05) t = log1.05 t = log tlog1.05 = log t = log log (b) We use the formula for continuously compounded interest A(t) = Pe rt with P = \$5000, A(t) = \$10,000, r = 0.05 and solve the resulting exponential equation for t. 5000e 0.05t = 10,000 e 0.05t = 0.05t = ln The money will double in years. t = ln EXAMPLE: A sum of \$1000 is invested at an interest rate of 4% per year. Find the time required for the amount to grow to \$4000 if interest is compounded continuously. 10
11 EXAMPLE: A sum of \$1000 is invested at an interest rate of 4% per year. Find the time required for the amount to grow to \$4000 if interest is compounded continuously. Solution: We use the formula for continuously compounded interest A(t) = Pe rt with P = \$1000, A(t) = \$4000, r = 0.04 and solve the resulting exponential equation for t. 1000e 0.04t = 4000 e 0.04t = t = ln4 t = ln The amount will be \$4000 in about 34 years and 8 months. 11
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Exponential Growth and Decay 28 April 2014 Exponential Growth and Decay 28 April 2014 1/24 This week we ll talk about a few situations which behave mathematically like compound interest. They include population | 9,544 | 33,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-47 | latest | en | 0.793641 |
www.katsosaccountants.com.au | 1,713,417,761,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817187.10/warc/CC-MAIN-20240418030928-20240418060928-00492.warc.gz | 759,414,481 | 10,218 | # The Budget Equation
The basic budget equation states that:
Income – Expenditure = Profit
To determine an initial amount for your budget, there are three main areas to consider; your business’s sales income, including all possible income streams, the total business expenditure for the budgeted period, and your estimated profits.
Income
When collating income figures for the period that your budget will be covering, make sure that you consider all of your business’s income streams. While sales figures will most often make up the bulk of your income, you should also take into account things like interest, dividends, royalties etc. that your business may be receiving. Any form of regularly received income will count toward your business’s overall income. Estimate your income based on previous figures whenever possible, while also taking into account the current economic climate and recent sales trends. Try to make your estimates as accurate as possible, to increase the efficacy of your budget.
Once you have estimated your business’s income for your budgeted period, you can calculate the necessary expenses required to achieve your goals, and see if you can afford to pursue them now, or how long it will be before you can.
Expenditure
Your business’s overall expenditure for your budgeted period will include all fixed and variable costs, and may also need to take into account inflation and other adjustments, depending on the length of time your budgeted period covers.
Fixed Costs
Your business’s fixed costs are payable amounts that will not change by any significant amount throughout the time allotted by your budget, regardless of how your business is performing. Costs such as rent for office space and equipment are fixed for a specified period by the terms of the rental agreement and cannot vary within the period that the agreement covers. These can be estimated accurately based on such agreements.
Variable Costs
Costs that will change significantly within your budgeted period are your variable costs. It is more difficult to estimate variable costs, as they can change based on a number of factors. Some examples of variable costs are your business’s product stock, which can vary greatly depending on your sales and on market prices for materials, transport costs, which will also vary based on your sales, and utility bills. A realistic estimation of these costs should be based on previous payable accounts and predicted sales.
Inflation
If your budget is covering a period of one or more years, you will need to take into account the rate of inflation when estimating your variable costs. The price of goods and services, such as stock and transport, can increase based on annual inflation rates, so it is important to consider these rates when budgeting, in order to estimate your costs as accurately as possible.
Profit
Ideally, your business’s profits should be high enough to make a return on investments, as well providing you with a fair return on your labour.
Achieving a return on capital invested in your business by parties other than yourself should be your first priority, in order to satisfy your investors. Personal investments, including the amount of money that you put into your business as start up capital, and the profit from previous years that you have reinvested into your business, should be your second consideration. It is also important that you receive a return on the time and effort that you have put into the business; the amount that you pay yourself on a weekly, fortnightly or monthly basis should reflect when you could earn working for someone else.
Using these concepts and the basic budget equation, you can estimate your budget for a specified period and plan how to manage your business’s finances, allocating funds appropriately and estimating profit, in order to determine whether or not your business goals for the future are achievable. When setting your budget, remember that it is always safer to over-estimate costs and under estimate your sales income, to decrease your risk of going over your specified budget. | 770 | 4,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.960581 |
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find out the radius of curvature of a secondary mirror.
A Cassegrain astronomical telescope uses two mirrors to form the image. The larger (concave) objective mirror has a focal length f1 = +51.50 cm. A small convex secondary mirror is mounted d = 43.50 cm in front of the primary. As shown in figure, light is reflected from the secondary through a hole in the center of the primary, thus forming a real image 8.00 cm behind the primary mirror. What is the radius of curvature of the secondary mirror?
An aluminum rod 15 cm in length is heated from 10 degrees Celsius to 50 degrees Celcius.
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### Describe what you learned about the impact of economic
Describe what you learned about the impact of economic, social, and demographic trends affecting the US labor environmen | 1,065 | 4,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-04 | longest | en | 0.858231 |
https://socratic.org/questions/56926ede7c014961c2f3363a | 1,713,574,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00015.warc.gz | 487,597,004 | 6,148 | # How many ways can you seat 5 gentlemen and 3 ladies around a circular table with the requirement that every gentleman is seated next to at least one lady?
Mar 21, 2016
Depending upon the interpretation,
either
color(white)("XXX")5! * 3! =720
or
color(white)("XXX")8 * 5! * 3! = 5760
#### Explanation:
Starting at some seating position the pattern must be:
$\textcolor{w h i t e}{\text{XXX}}$2 gentlemen - 1 lady - 2 gentlemen - 1 lady - 1 gentleman - 1 lady
If different starting locations for this pattern are considered the same seating pattern:
There are 5! ways of distributing the gentlemen among their seats
and for each of these, there are 3! ways of distributing the ladies among their seats
for a total of 5! * 3! different seating patterns.
If the starting location for this pattern is considered to give a different way of seating then
there will be $8$ times as many seating patterns. | 231 | 904 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-18 | latest | en | 0.89153 |
https://stats.libretexts.org/Courses/Fresno_City_College/Book%3A_Business_Statistics_Customized_(OpenStax)/Using_Excel_Spreadsheets_in_Statistics/5_Central_Limit_Theorem_and_Confidence_Intervals/5.2_Probability_for_Proportions_using_the_Excel_Spreadsheet | 1,720,909,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00232.warc.gz | 482,038,577 | 29,898 | # 5.2 Probability for Proportions using the Excel Spreadsheet
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##### Definition: Normal approximation to the Binomial Distribution
Central Limit Theorem for Proportions: If the sample size times the probability of success is greater than or equal to 5 and the sample size times the probability of failure is greater than or equal to 5. The probability related to proportions of successes can be approximated based on the Normal distribution with a mean of $$p$$ and standard deviation of $$\sqrt{\frac{p(1-p)}{n}}$$.
##### Example $$\PageIndex{1}$$
A question is asked of 4800 sophomore, and 58% of the students know the correct answer. If a sample of 144 students is taken repeatedly, answer the following questions.
a) What is the expected value of the mean of the sampling distribution of sample proportions?
b) What is the standard deviation of the mean of the sampling distribution of sample proportions? (Round to 4 decimal places)
c) What is the P( $$\hat{p}$$ > 0.61 )? (Round to 4 decimal places)
d) What is the P( $$\hat{p}$$ < 0.62) ? (Round to 4 decimal places)
e) What is the P(0.68 < $$\hat{p}$$ < 0.7) ? (Round to 4 decimal places)
a) What is the expected value of the mean of the sampling distribution of sample proportions?
The mean is the proportion of student with the correct answer, 58% enter 0.58.
b) What is the standard deviation of the mean of the sampling distribution of sample proportions?
The standard deviation is $$\sqrt{\frac{.58(1-.58)}{144}}$$.
Enter the following formula into the Excel Spreadsheet
=sqrt(.58*(1-.58)/144)
0.04113
To include rounding in the formula.
=round(sqrt(.58*(1-.58)/144), 4)
0.0411
c) What is the P( $$\hat{p}$$ > 0.61 )? (Round to 4 decimal places)
The probability computed using a Normal Probability. Enter the following formula into the Excel Spreadsheet.
=1 - norm.dist(0.61, 0.58, 0.0411, TRUE)
0.232717
To include rounding in the formula.
=round(1-norm.dist(0.61, 0.58, 0.0411, TRUE), 4)
0.2327
d) What is the P( $$\hat{p}$$ < 0.62) ? (Round to 4 decimal places)
The probability computed using a Normal Probability. Enter the following formula into the Excel Spreadsheet.
=norm.dist(0.62, 0.58, 0.0411, TRUE)
0.834782
To include rounding in the formula
=round(norm.dist(0.62, 0.58, 0.0411, TRUE), 4)
0.8348
e) What is the P(0.68 < $$\hat{p}$$ < 0.7) ? (Round to 4 decimal places)
The probability computed using a Normal Probability. Enter the following formula into the Excel Spreadsheet.
= norm.dist(0.7, 0.58, 0.0411, TRUE) - norm.dist(0.68, 0.58, 0.0411, TRUE)
0.005733
To include rounding in the formula
= rounding(norm.dist(0.7, 0.58, 0.0411, TRUE) - norm.dist(0.68, 0.58, 0.0411, TRUE),4)
##### Interactive Element
5.2 Probability for Proportions using the Excel Spreadsheet is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. | 2,616 | 7,241 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.200242 |
http://mathematica.stackexchange.com/questions/18301/ndsolve-differential-equation-error | 1,469,535,315,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824853.47/warc/CC-MAIN-20160723071024-00080-ip-10-185-27-174.ec2.internal.warc.gz | 162,777,722 | 18,288 | # NDSolve Differential Equation Error
I am trying to solve the following differential equation-
k = NDSolve[{y'''[x] == -76*Sin[y[x]], y[0] == Pi/4, y'[0] == 0,
y[1.85] == 0}, y, {x, 0, 2}, AccuracyGoal -> 10,
PrecisionGoal -> 30]
I am getting a solution but when I plot a graph, it does not satisfy the boundary conditions. I am getting the following two errors- FindRoot::sszero: The step size in the search has become less than the tolerance prescribed by the PrecisionGoal option, but the function value is still greater than the tolerance prescribed by the AccuracyGoal option. >>
NDSolve::berr: There are significant errors {-0.239832,-0.0591905,-0.0015434} in the boundary value residuals. Returning the best solution found. >>
The funny thing is when I reduce the coefficient of Sin[y[x]] it works fine. Can anyone help me with this? Thanks,
-
You could start by trying a "shooting method" approach, where you use only initial conditions - i.e. replace y[1.85] == 0 by y''[0] == c, and then vary c to see what value of y[1.85] it gives rise to. I found that y[1.85] rapidly switches between negative and positive values as c increases from 24.4 to 24.5, so your solution lies somewhere in that range. The optimisation of c should be automated, but I haven't got time right now. – Stephen Luttrell Jan 23 '13 at 16:55
You could use the "shooting method", as described in my earlier comment, to find the solution as follows:
Clear[error, y, x, c];
Define a function to minimise the predicted value of y[1.85]^2 if you assume the initial condition y''[0] == c:
error[c_?NumericQ] :=
y[1.85]^2 /.
NDSolve[{y'''[x] == -76*Sin[y[x]], y[0] == Pi/4, y'[0] == 0,
y''[0] == c}, y, {x, 0, 2},
AccuracyGoal -> 10, PrecisionGoal -> 30][[1]];
[EDIT: I really should have used y[1.85] rather than y[1.85]^2, because this is a better match for the FindRoot that I use below, but in this case FindRoot finds the (second order) zero anyway. The squared version y[1.85]^2 is what you would use if you were then going to use NMinimize - i.e. the minimum is not necessarily zero.]
Find the solution given the starting value for c mentioned in my earlier comment - it doesn't have to be this accurate a guess.
c0 = c /. FindRoot[error[c], {c, 24.4}]
(* 24.4354 *)
This tells you that, for the initial conditions y[0] == Pi/4 and y'[0] == 0, the end point condition y[1.85] == 0 is equivalent to the extra initial condition y''[0] == 24.4354.
Plug this solution for c back in
soln = NDSolve[{y'''[x] == -76*Sin[y[x]], y[0] == Pi/4, y'[0] == 0,
y''[0] == c0}, y, {x, 0, 2},
AccuracyGoal -> 10, PrecisionGoal -> 30][[1]];
and plot the result
Plot[y[x] /. soln // Evaluate, {x, 0, 2}]
-
Export["/where/you/want/image.png", %] Or use @Szabolcs automated uploader – ssch Jan 23 '13 at 18:03
Thanks Stephen, – valmiki Jan 24 '13 at 7:55 | 878 | 2,841 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2016-30 | latest | en | 0.900665 |
https://www.vedantu.com/question-answer/if-a-left-x-x-is-a-multiple-of-4-rightandb-left-class-11-maths-cbse-5eceb8a1b5f2de3ed1c2e2ab | 1,722,688,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00047.warc.gz | 833,086,813 | 26,881 | Courses
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# If $A = \left\{ {x{\text{ : x is a multiple of 4}}} \right\}$and$B = \left\{ {x:x{\text{ is a multiple of 6}}} \right\}$, then $A \cap B$consist of all multiples ofA. $16$B. $12$C. $8$D. $4$
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Hint: Write the multiples of 4 and 6 for different sets A and B in roster form to make simplification easier.
As mentioned in the question, $A = \left\{ {x{\text{ : x is a multiple of 4}}} \right\}$
So the elements belonging to set A would be
$\left\{ {4,8,12,16,20,..............} \right\}$
Now it is also given in the problem statement that $B = \left\{ {x:x{\text{ is a multiple of 6}}} \right\}$
So elements of set B will be
$\left\{ {6,12,18,24............} \right\}$
Now we have to find a set that has common elements from both the above sets that is we have to find $A \cap B$
So from above two sets we can say that $A \cap B$$=${All the multiples which are common for both set A and set B]
$A \cap B = \left\{ {12,24,36...............} \right\}$
Now, clearly $A \cap B = \left\{ {x{\text{ : x is a multiple of 12}}} \right\}$
Note: Whenever we come across such problems with sets we simply need to think of all the possible elements of the set that satisfy the given relationship. This set wise approach ensures that any element belonging to a particular set is not missed. | 458 | 1,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-33 | latest | en | 0.882043 |
https://portal2.ipt.pt/en/Cursos/202122/licenciaturas/l_-_eqb/91841/ | 1,696,206,657,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510942.97/warc/CC-MAIN-20231002001302-20231002031302-00558.warc.gz | 511,633,504 | 13,575 | Mathematical Analysis I - Engenharia Química e BioquímicaInstituto Politécnico de Tomar
# IPT
Ano Letivo: 2021/22
# Engenharia Química e Bioquímica
## Mathematical Analysis I
Publication in the Diário da República: Despacho nº 10764/2011 - 30/08/2011
6 ECTS; 1º Ano, 1º Semestre, 30,0 T + 30,0 TP , Cód. 91841.
Lecturer
- Luís Miguel Merca Fernandes
Prerequisites
Not applicable.
Objectives
a)- Provide the mathematical foundations required in other modules of the programme.
b)- Provide skills to work with differential and integral calculus of functions of one real variable.
Program
1- Preliminaries.
2- Real functions of a real variable.
3- Limits and continuity.
4- Differential calculus.
5- Integral calculus.
Evaluation Methodology
Continuous assessment:
Continuous assessment consists of two written tests. Each of these tests is graded from 0 to 10 points. The student is exempted from exam, i.e. if he/she has at least 3 marks in each test and obtains a classification higher than or equal to 10 marks, resulting from the sum of the 2 tests.
Exam-based assessment:
If the student has been admitted to an examination, or has been exempted but wishes to improve his classification, he can take the first attempt examination - a written test (graded from 0 to 20 points) on all the subject matter taught. A minimum mark of 10/20 is required to pass.
If the student fails the examination he can apply for resit - a test similar to that of firt attempt.
NOTE:
For any of the assessments, if the student obtains a mark equal to or higher than 17/20 he/she will have to take an extraordinary assessment.
Bibliography
- Silva, J. (1999). Princípios de Análise Matemática Aplicada. (Vol. 1). Lisboa: McGraw-Hill
- Stewart, J. (2005). Cálculo. (Vol. 1). São Paulo: Thomson Pioneira
- Swokowsi, E. (1995). Cálculo com Geometria Analítica. (Vol. 1). São Paulo: Makron Books
- Howard, A. (2007). Cálculo um novo horizonte. (Vol. 1). São Paulo: Bookman
Teaching Method
Theoretical lectures, with presentation and illustration of the proposed subjects. Theoretical-practical lectures in which exercises are proposed and solved.
Software used in class | 577 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.768735 |
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# Algebra : Solving Equations and Word Problems
1. Solve by factoring:
x^2 - 9x = -8
2. Solve using the square root property:
2x^2 - 5 = 93
3. Solve using the square root property:
(x + 4)^2 = 81
4. Solve by completing the square:
x^2 + 6x + 2 = 0
5. Solve using the quadratic formula:
x^2 - 3x = -6x - 1
6. Solve using the quadratic formula:
x^2 - 10x - 1 = -10
7. Solve the equation.
y^2 - 13 y + 22 = 0
8. Write the equation x(x - 6) + 5 = 0 in quadratic form and then solve it by factoring.
9. Write the equation 6 x(x - 3) = - 12 in quadratic form and then solve it by factoring.
10. Choose from the following a quadratic equation with solutions of 9 and 6.
x^2 - 15x + 54 = 0
x^2 - 18x + 51 = 0
x^2 - 15x + 57 = 0
x^2 - 12x + 54 = 0
11. The height h (in feet) of an object that is dropped from the height of s feet is given by the formula h = s - 16t 2 , where t is the time the object has been falling. A 6 foot tall woman on a sidewalk looks directly overhead and sees a window washer drop a bottle from the 2 story. How long does she have to get out of the way? Round to the nearest tenth. (A story is 12 feet.) Choose the answer from the following:
12. Use the quadratic formula to solve the equation: x 2 - 3 x + 2 = 0.
13. Use the quadratic formula to solve the equation: 4x^2 - 30x = 1
14. The hypotenuse of a right triangle is 2.7 units long. The longer leg is 1.4 units longer than the shorter leg. Find the lengths of the sides of the triangle.
15. We have learned to solve quadratic equations using a variety of methods including completing the square and the quadratic formula. Give an example using either completing the square or the quadratic formula and explain each step as if you were teaching someone who had never used the method before.
#### Solution Summary
Equations and word problems are solved. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.
\$2.19 | 588 | 2,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-22 | longest | en | 0.921519 |
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_12&oldid=91073 | 1,618,179,843,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065492.15/warc/CC-MAIN-20210411204008-20210411234008-00482.warc.gz | 211,875,432 | 13,376 | # 2018 AMC 10A Problems/Problem 12
How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? $$x+3y=3$$ $$\big||x|-|y|\big|=1$$ $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
## Solutions
### Solution 1
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this: $[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy]$ Now, it becomes clear that there are $\boxed{\textbf{(C) } 3}$ intersection points. (pinetree1)
### Solution 2
$x+3y=3$ can be rewritten to $x=3-3y$. Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$. Splitting this question into casework for the ranges of y will give us the total number of solutions.
$\textbf{Case 1:}$ $y>1$ $3-3y$ will be negative so $|3-3y| = 3y-3.$ $|3y-3-y| = |2y-3| = 1$
Subcase 1: $y>\frac{3}{2}$
$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$
Subcase 2: $1
$2y-3$ is negative so $|2y-3| = 3-2y = 1$. $2y = 2$ and so there are no solutions ($y$ can't equal to $1$)
$\textbf{Case 2:}$ $y = 1$ It is fairly clear that $x = 0.$
$\textbf{Case 3:}$ $y<1$ $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$
Subcase 1: $y>\frac{4}{3}$
$3-4y$ will be negative so $4y-3 = 1$ /rightarrow $4y = 4$. There are no solutions (again, $y$ can't equal to $1$)
Subcase 2: $y<\frac{4}{3}$
$3-4y$ will be positive so $3-4y = 1$ \rightarrow $4y = 2$. $y = \frac{1}{2}$ and $x = \frac{3}{2}$. Thus, the solutions are: $(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)$, and the answer is $\boxed{\textbf{(C) } 3}$. Solution by Danny Li JHS, $\text{\LaTeX}$ edit by pretzel.
### Solution 3
Note that $||x| - |y||$ can take on either of four values: $x + y$, $x - y$, $-x + y$, $-x -y$. Solving the equations (by elimination, either adding the two equations or subtracting), we obtain the three solutions: $(0, 1)$, $(-2,3)$, $(1.5, 0.5)$ so the answer is $\boxed{\textbf{(C) } 3}$
~ccx09 | 984 | 2,465 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 58, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2021-17 | latest | en | 0.654695 |
https://teswesm.com/mchoice/the-staff-should-examine-the-programme-continuously-with-the-of-the-principlal/57 | 1,722,871,825,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00352.warc.gz | 469,666,406 | 8,744 | # MCQs for Assistant Education Officer (AEO) for Educators
Q. The staff should examine the programme continuously with the ____________________of the principlal.
a. Administration
b. Supporting
c. Helping
d. Leadership
ANSWER: See Answer
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https://www.coursehero.com/file/5804993/EE-331Sp-2009-HW7S/ | 1,495,560,572,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607648.39/warc/CC-MAIN-20170523163634-20170523183634-00107.warc.gz | 864,624,661 | 97,614 | EE_331Sp_2009_-_HW7S
EE_331Sp_2009_-_HW7S - EE 331 Devices and Circuits I...
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1 EE 331 Devices and Circuits I Spring 2009 Problem #7 Solution 4.85 a V GG 100 k 100 k 220 k 12 V 3.75 V | Assume saturation 3.75 V GS 24 x 10 3 I D V GS 24 x 10 3 100 x 10 6 2 5 1 V GS 1 2 6 V GS 2 11 V GS 2.25 0 V GS 1.599 V and I D 89.7 A V DS 12 36 x 10 3 I D 8.77 V | V DS V GS V TN Saturation is correct. Checking : V GG 24 x 10 3 I D V GS 3.75 V which is correct. Q po int : 89.7 A , 8.77 V b Assume saturation 3.75 V GS 24 x 10 3 I D V GS 24 x 10 3 100 x 10 6 2 10 1 V GS 1 2 12 V GS 2 23 V GS 8.25 0 V GS 1.439 V and I D 96.4 A V DS 12 36 x 10 3 I D 8.53 V | V DS V GS V TN Saturation is correct. Checking :
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This note was uploaded on 02/28/2010 for the course EE 331 taught by Professor Taicheng during the Spring '08 term at University of Washington.
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EE_331Sp_2009_-_HW7S - EE 331 Devices and Circuits I...
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Ask a homework question - tutors are online | 494 | 1,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-22 | longest | en | 0.759638 |
www.foreignholidaysonline.com | 1,669,836,599,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00768.warc.gz | 797,952,444 | 18,632 | #### Probabilities Of Winning In Bingo
Players’ chances of winning depend on the number of cards in play and how many cards they have to play. For example, if a player has 12 cards in a game with 1200 cards, the odds of winning for that player are 1 in 100.
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On important days like holidays, there are ALWAYS lots of websites doing special things during special times. At the same time this is good for the player as it brings in new players who focus on the big games so you will be playing against a lot more players. | 682 | 2,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-49 | longest | en | 0.973549 |
http://mathhelpforum.com/differential-equations/176916-spring-mass-undamped-motion-print.html | 1,508,237,657,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821017.9/warc/CC-MAIN-20171017091309-20171017111309-00277.warc.gz | 219,385,860 | 4,246 | # spring mass undamped motion
• Apr 5th 2011, 12:02 PM
duaneg37
spring mass undamped motion
A mass weighing 10 pounds stretches a spring 1/4 foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point 1/3 foot above the equilibrium position with a downward velocity of 5/4 ft/s.
At what time does the mass attain a displacement below the equilibrium position numerically equal to 1/2 the amplitude?
my position function is $x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)
$
I have tried to put it into the form $x(t)=Asin(\omega t+\phi)
$
and I am getting $x(t)=\frac{1}{3}sin(160t+\frac{\pi}{2})
$
The time I am getting is $\frac{\pi}{480}s$ initially, but I don't know if this is right. I'm also having trouble writing the answer as a series. Can anyone help me? Thanks
• Apr 5th 2011, 01:25 PM
TheEmptySet
Quote:
Originally Posted by duaneg37
A mass weighing 10 pounds stretches a spring 1/4 foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point 1/3 foot above the equilibrium position with a downward velocity of 5/4 ft/s.
At what time does the mass attain a displacement below the equilibrium position numerically equal to 1/2 the amplitude?
my position function is $x(t)=-\frac{1}{3}cos(160t)+\frac{1}{128}sin(160t)
$
I have tried to put it into the form $x(t)=Asin(\omega t+\phi)
$
and I am getting $x(t)=\frac{1}{3}sin(160t+\frac{\pi}{2})
$
The time I am getting is $\frac{\pi}{480}s$ initially, but I don't know if this is right. I'm also having trouble writing the answer as a series. Can anyone help me? Thanks
First I don't think your ODE is correct.
First by Hook's law we have
$F=kx \iff 10=k(.25) \iff k=40$
Now the ode is
$\displaystyle mx''=-kx \iff x''+\frac{1}{5}x=0$
So the solution should have the form
$\displaystyle x(t)=c_1\cos\left( \frac{t}{5}\right)+c_2\sin\left( \frac{t}{5}\right)$
Now use the initial conditions
• Apr 5th 2011, 03:18 PM
duaneg37
I thought I had to convert the mass into slugs by $W=mg$, which gives the force as $320$ slugs $=k(\frac{1}{4})$ for Hooke's law. This gives $k=1280$. Then I found m for the weight of $1.6$ slugs >>> $1.6=m(32)$where $m=.05$ This gives $\omega ^{2}=25600$. Is this incorrect?
Shouldn't your equation be $x(t)=c_{1}cos2t+c_{2}sin2t$?
I did it again without rounding off as much and got:
$x(t)=\frac{\sqrt{16393}}{384}sin(160t-1.547)$
the time I got:
$t=.026+\frac{n\pi}{160}$ sec., let $n=0,1,2,3,...$
Am I on the wrong track with this? Thanks a lot!
• Apr 5th 2011, 03:40 PM
TheEmptySet
Weight is a force. Slugs are mass. So in hooks law you need a force and since lbs are a force you do not need to do any conversions.
• Apr 5th 2011, 04:29 PM
topsquark
Quote:
Originally Posted by TheEmptySet
$\displaystyle mx''=-kx \iff x''+\frac{1}{5}x=0$
Actually $\displaystyle \frac{k}{m} = \frac{40}{1.6} = 25$, not 1/25.
Thus
$x(t) = c_1~cos(5t) + c_2~sin(5t)$
-Dan
• Apr 5th 2011, 04:35 PM
duaneg37
I think I've done them all wrong! Thanks a lot for you help!
• Apr 5th 2011, 07:26 PM
duaneg37
My position function is $x(t)=\frac{5}{12}sin(5t-.927)$
To find the time I set $\frac{1}{2}=sin(5t-.927)$ I know the sine of $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ will give $\frac{1}{2}$, so I get $t=.29$and $t=.709$. Do I use both of these times to get my answer? I found the period to be $T=\frac{2\pi}{5}$. I said $t=.29+\frac{2n\pi}{5}$ s where $n=0,1,2,3,...$ and $t=.709+\frac{2n\pi}{5}$ s where $n=0,1,2,3,...$ as my answer. Am I doing this right?
• Apr 5th 2011, 07:40 PM
topsquark
Quote:
Originally Posted by duaneg37
My position function is $x(t)=\frac{5}{12}sin(5t-.927)$
To find the time I set $\frac{1}{2}=sin(5t-.927)$ I know the sine of $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ will give $\frac{1}{2}$, so I get $t=.29$and $t=.709$. Do I use both of these times to get my answer? I found the period to be $T=\frac{2\pi}{5}$. I said $t=.29+\frac{2n\pi}{5}$ s where $n=0,1,2,3,...$ and $t=.709+\frac{2n\pi}{5}$ s where $n=0,1,2,3,...$ as my answer. Am I doing this right?
Looks good to me. (Nod)
-Dan | 1,463 | 4,109 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 50, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-43 | longest | en | 0.900701 |
https://www.imerchantnetwork.com/amortization-refers-to-changes-in-the-monthly-payment-for-a-variable-rate-mortgage/ | 1,713,112,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00368.warc.gz | 767,863,918 | 10,317 | Posted on
# Amortization Refers To Changes In The Monthly Payment For A Variable Rate Mortgage.
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5 1 Arm Loan Definition This is most often seen with mortgages as banks offer fixed rate and adjustable rate mortgages. on their loans and lower the bank’s net interest income as a result. If a bank has a loan portfolio. | 659 | 3,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.939083 |
https://lessonplanet.com/lesson-plans/calculator?keywords=calculator+math+games | 1,701,924,155,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100632.0/warc/CC-MAIN-20231207022257-20231207052257-00807.warc.gz | 402,337,900 | 47,541 | # Calculator Teacher Resources
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#### Statistics Fun: Probability
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A six-video collection hands viewers the key to understanding conditional probability. It's all in the context of the problem presented! Marbles, dice, Venn diagrams, and a tree diagram show how to calculate conditional probability, how...
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#### Common Core Algebra 2, Unit 2: Polynomial Functions
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A unit on solar energy begins with pupils researching vocabulary terms related to electric currents. They then use a multimeter to measure the voltage and current for a solar module before calculating power and efficiency. Fun, hands-on...
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#### TeachEngineering: Next-Generation Surgical Tools in the Body
For Teachers 9th - 12th Standards
Young engineers design laparoscopic surgical tools in an 11-lesson unit aligned with the Next Generation Science Standards. The first lesson introduces the class to the abdominopelvic cavity. Teams learn what it is like to perform...
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#### Geometry Unit 9: Area of Polygons
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Six lessons make up the Geometry Unit 9: Area of Polygons unit module. Pupils review the formulas for finding the area of parallelograms and triangles. They apply previous learning of trigonometric ratios and the Pythagorean Theorem to...
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#### Statistics Fun: Standard Deviation
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Designed for statistics classes, a seven-video tutorial collection models how to calculate standard deviation and variance, explains degrees of freedom, how to solve standard deviation and variance by hand or by using Excel, and how to...
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#### Common Core Algebra 1, Unit 4: One Variable Statistics
For Teachers 9th - 10th Standards
Each of the four lessons in Unit 4: One Variable Statistics comes with videos that teach the concepts, worksheets, practice packets, and corrective assignments. To begin the unit, viewers learn to calculate the mean, median, mode, and...
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#### Exploring Space Through Math: Algebra II
For Teachers 9th - 12th Standards
The three lessons in the Algebra II series Exploring Space Through Math expand learners’ knowledge of algebraic thinking and symbolic reasoning and encourage them to connect algebra and geometry. Activities include calculating the number...
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For Teachers 2nd - 5th Standards
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Seven resources make up the Bozeman Science: Statistics and Graphing collection. The series begins with a video that introduces the concepts of hypothesis testing, confidence intervals, and degrees of freedom. The video also covers the...
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For Teachers 5th - 8th Standards
From graphing to solving word problems to finding combinations and calculating percentages, these holiday worksheets are perfect for reinforcing various skills during the weeks leading up to Christmas.
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For Students 11th - 12th
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For Teachers 7th - 12th Standards
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#### Creating a Rain Garden
For Teachers 1st - 12th
Use this series of lessons to empower your gardeners to plan for, design, and implement a rain garden on campus! From planning water flow and calculating volume to acting in an interpretive play where they become plants, learners will be... | 1,811 | 8,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.853976 |
https://www.dummies.com/business/accounting/analyze-in-material-price-and-efficiency-variances-in-cost-accounting/ | 1,568,565,224,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571651.9/warc/CC-MAIN-20190915155225-20190915181225-00468.warc.gz | 842,118,893 | 14,333 | Analyze in Material Price and Efficiency Variances in Cost Accounting - dummies
# Analyze in Material Price and Efficiency Variances in Cost Accounting
In cost accounting, a price variance is the difference between actual and budgeted price for something you purchase. Here’s the formula for price variance:
`Price variance = (Actual price – budgeted price) × (actual quantity)`
An efficiency variance is the difference between actual and budgeted quantities you purchased for a specific price. Here’s the formula for efficiency variance:
`Efficiency variance = (Actual quantity – budgeted quantity) × (standard price or rate)`
A standard is a planned amount per unit. Note that there are two terms in the efficiency formula: standard price and rate. This is a distinction you need to remember for cost accounting. Standard price refers to material variances. You pay a certain price for materials. Standard rate refers to labor variances. When you pay an hourly rate, you’re working with labor variances.
To summarize the distinction between the two variances, you either paid more or less than planned (price variance) or used more or less than planned (efficiency variance).
## How to apply price variances to direct materials in cost accounting
An obvious way to reduce your costs is to analyze the prices you pay for materials. Say you operate a bicycle factory, and you use aluminum to manufacture bike frames. During planning, you come up with a standard or budgeted price of \$5 per pound for aluminum. When you review your actual costs, you find that the real price paid was \$5.75 per pound. Assume you purchased 10,000 pounds of aluminum.
Here is your material price variance for the year.
Material Price Variance — Aluminum for Bikes
Actual Quantity × Actual Price Actual Quantity × Budgeted Price
10,000 lbs. 10,000 lbs.
Multiply by \$5.75 Multiply by \$5.00
Equals \$57,500 (A) Equals \$50,000 (B)
Material price variance (A) – (B) \$7,500
The material price variance is \$7,500 unfavorable because your actual costs (\$57,500) were more than the actual quantity at budgeted price (\$50,000). The only difference in this variance is the price paid for aluminum.
## How to apply efficiency variances in cost accounting
Another way to improve your business performance is to be more efficient — “to work smart,” as the business cliché goes.
During planning, you also determined a budgeted input quantity allowed for actual output of 12,000 pounds of aluminum for bike production. This term multiplies the number of units you plan to produce by some amount of material per unit. This is the calculation for amount of aluminum for the bikes:
```Budgeted input quantity= budgeted units produced × lbs. used in a unit
Budgeted input quantity= 2,000 × 6
Budgeted input quantity= 12,000```
To produce 2,000 bikes, you plan to use 6 pounds of aluminum per bike, or a total of 12,000 pounds. Assume that for the same 2,000 bikes produced, you used only 5 pounds of aluminum per bike (because you were so efficient, of course). This means that you only used 10,000 pounds of aluminum. Your material cost would be lower, which would increase your operating profit.
Here is your material efficiency variance for the year.
Material Efficiency Variance — Aluminum for Bikes
Actual Quantity × Budgeted Price Budgeted Quantity × Budgeted Price
10,000 lbs. 12,000 lbs.
Multiply by \$5.00 Multiply by \$5.00
Equals \$50,000 (A) Equals \$60,000 (B)
Material efficiency variance (A) – (B) (\$10,000)
A negative variance for a cost is favorable, so the material efficiency variance is favorable.
Look back at the price and efficiency variances in the last two tables. Here’s the flexible budget variance calculation for direct materials:
```Flexible budget variance = price variance + efficiency variance
Flexible budget variance = \$7,500 + -\$10,000
Flexible budget variance = -\$2,500```
You have a \$7,500 unfavorable price variance and a \$10,000 favorable efficiency variance. Because this is a cost variance, a negative number indicates less actual spending than planned, and that’s a good thing.
## Lay out reasons for price and efficiency variances in cost accounting
The real reason you go through all of this analysis is to identify areas where you can improve. By “improve,” you want to reduce costs, increase demand, or raise prices to generate a higher profit.
Here are some reasons you might find a material price or efficiency variance:
• Material purchasing decisions: You’re able to negotiate a lower price per unit. You may find a new supplier with a larger supply of material. To reduce their inventory, that supplier may be willing to sell at a lower price. These decisions may create a favorable material variance. On the other hand, a shortage of the material in the marketplace may mean that paying more for materials. The additional spending may create an unfavorable variance.
• Personnel and hiring decisions: Due to a slow economy, you’re can hire workers with the skill set needed, but for a lower hourly rate. Also, you may have staff that’s so productive they get work completed faster than planned. Finally, you’re able to plan production so that workers don’t wait during setup time, which means more hours producing the product. Of course, higher wages, less productivity, and poor decisions may create an unfavorable variance. | 1,161 | 5,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-39 | longest | en | 0.903479 |
http://aldemardum.nl/55817-trading-binary-options-with-stochastic-oscillator.html | 1,561,459,706,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999817.30/warc/CC-MAIN-20190625092324-20190625114324-00202.warc.gz | 6,283,941 | 9,566 | ## Trading binary options with stochastic oscillator
Touch trading with binary options is distinct from the One Touch options provided by many binary options platforms in that it allows traders to set their own strike price. The stochastic oscillator is very useful in short-term dynamics, which is characteristic of CFD and binary options trading. If you have to have a definite trend analysis, the oscillator can execute the minimum striking price for you when it trades trade to the traded territory giving a small when it is different to take action. Table of Contents, definition, what is the stochastic oscillator? It is a fairly complex subject since this concept is found in various disciplines including mathematics and statistics.
#### Touch trading binary options using Fibonacci and the stochastic
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The calculation of the Stochastic Oscillator then compares the current value to the maximum and minimum price values recorded over a certain period of time. The value of the defined average. One means that it is made to open a binary and then set both the Euro and the Trading indicators on the same time. Oscillators And Convergence Theory For Binary Options. Binary Options With Bill Williams Indicators The Awesome Oscillator. Following any strong trend, price will very often retrace and form a correction before continuing in the direction of the underlying trend. It shows ultimate in the actual and patterns that the market is required higher.
#### Using the stochastic oscillator and binary options : Gras
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#### Stochastic oscilator binary options indicator : Gras Miguel
Ones include haramis, engulfing victories, doji attend patterns, etc. Using the Fibonacci retracement tool in identifying a touch trade. Incorrectly, one currency is different: Zero paying crossing. Vastly is also at least one particularly high quality that can be fleeced one particularly. It is safe to provide another year trading binary options with stochastic oscillator following signal before making your entry in theory to get the last results. These tools are rushing for work from home jobs free registration market strength, nor trend following trades and moving support and trading. Applying the stochastic oscillator to define the strike price. Stochastic oscillator formula, although you will never need to calculate it yourself, lets find out how we get the formula for the stochastic oscillator to understand the logic of its operation. If an investor is changing up, with a new showing strength, we hilton work at home careers offer that the fact hilton work at home careers strong and will lead.
#### Binary, options, oscillator, London open forex trading strategy
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It is an option that many the binary option best robot of a trend. Where: MINn Minimum of the last n days. Found Endorsements: When is the first public hype wallet. Granted and Overbought Levels The theory plays also an excellent role in oversold and went markets since with an alternative you can always forever on overbought and added levels. Oscillators are appreciated on more capital requirements which for sure gives a bigger bust on how to turn your movements trades. When, some even using the zero involvement other for staying some call options on a distinguishing for or put restrictions on a binary options oscillator cross. In order to increase the chances of a touch trade being successful the Fibonacci retracement tool needs to be applied to the swing high and low of a recent, strong up or down trend on any price chart. K 100 * (closing MINn) / (MAXn MINn). This is a customizable version in which the user can modify the parameters (number of periods for K, number of periods for D). Value n is not fixed but can be set as desired on the trading platform. For binary options traders there are several key indicators and tools which can be used to identify touch trade reversals and one of these methods is the combination of the Fibonacci retracement tool provided by almost all charting software and the stochastic oscillator. On lower time frames it is important to consider that trading these reversals should result in a larger move in the direction of the trend soon after the touch has been spotted, allowing plenty of time for the binary options to expire in the money.
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#### Part 17: Technical Analysis, stochastic oscillator
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#### Binary, options with, stochastic, indicator - 5binaryoptions
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#### Stochastic & Parabolic SAR 5 minute, binary, options Indicator
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#### Stochastic Oscillator Crossover Trading in Binary Options
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Forex indicator to measure volatility
Volatility the what binary options signals data from our tool, you will be able to does which pairs are the most volatile; you can also see which are the most… | 3,172 | 15,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-26 | latest | en | 0.935231 |
http://stats.bls.gov/ppi/ppicomqa.htm | 1,448,521,356,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398446535.72/warc/CC-MAIN-20151124205406-00166-ip-10-71-132-137.ec2.internal.warc.gz | 220,896,056 | 23,214 | # Hedonic Models in the Producer Price Index (PPI)
U.S. Bureau of Labor Statistics
Producer Price Index Program
June 2011
Valuing changes in the quality of products or services is one of the oldest measurement challenges facing statistical agencies around the world. The Producer Price Index (PPI) program has developed and adopted several methodologies to respond to these quality measurement challenges, but the primary focus here will be on hedonic methods.1 One of the most widely recognized pioneers in applied hedonic research was Griliches (1988) who stated that "...one might use regression techniques to relate the prices of different models or versions of a commodity to differences in their characteristics, and discover thereby the relative valuation of such qualities is reasonably obvious.” The terms hedonics or hedonic models are often used by economists when referring to a form of econometrics (Gujarati 1995). At its most basic, hedonics can be described as a regression designed to isolate and measure the influence on price of economically meaningful product characteristics. In other words, a product must first be disaggregated into its characteristics. For instance, a computer can be disaggregated into characteristics such as speed of processor, hard drive capacity, amount of memory, and many other defining features that influence the computer’s price. These product characteristics enter a regression as independent (explanatory) variables and the product’s selling price is entered as the dependent variable. The regression may take different functional forms, but in each form the independent variables are “regressed” against the dependent variable yielding implicit prices (regression coefficients) for each of the independent variables. A basic regression formula for a hedonic model could take the form shown in equation 1.
#### Equation 1
Pit = α0 + β2 X2i + β3 X3i ··· βk Xki + ui
Where:
Pit is the price of the i th model in period t
α0 is the intercept
Xi are the variables representing observed product characteristics
β2 ··· βk are the regression/slope coefficients
ui is the residual or error term
It is the implicit prices represented by the β s in equation 1 that provide the PPI with values that may be used in quality adjustments when product characteristics change over time. Before going further, let’s look in a more fundamental way at how the PPI attempts to adjust prices to reflect changes in quality.
### Measures of Price Change
The following description of price measurement is deliberately generalized and is intended only to provide context for the use of hedonic methods as a quality valuation tool.
The PPI collects price and product specification data from producers (based on specific sample designs) to measure price change by constructing price relatives that in their most basic form can be expressed as:
#### Equation 2
pit/pio
Where pit is the price of product i in period t and pio is the price of product i in some previous base period o. Weights and choice of index formula are not considered here.
As long as product i remains unchanged between periods o and t, measures of price change are straightforward. This situation is often referred to as the matched model in price index literature (Triplett 1983, 1986, 2003). However, the price measurement challenge becomes more complex when the inputs, including technology, used to produce product i change between periods o and t. New input requirements violate the fixed inputs assumption of the matched model and may cause biased measures of price change if unaccounted for. Essentially, when the matched model is violated, the measurement challenge is to estimate what part of price change between period o and t is due to quality change and what part is pure or market-based price change. Equation 3 more formally describes the price measurement adjustment that is attempted in the PPI when products undergo quality change.
#### Equation 3
pit - (VQA)/pio
Equation 3 adds the term VQA (value of quality adjustment) to the price relative shown in equation 2. VQA is the estimate of what part of any price change between periods o and t is due to quality change. VQA may be estimated with a variety of methods including hedonics. The specific method chosen to estimate VQA is always determined (at least in part) by the type and detail of data reported to the PPI. For instance, hedonics would not be used if all reporters to the PPI could always provide estimates of the marginal cost of new input requirements that are directly tied to changes in output quality. Estimating the cost of new input requirements to value changes in quality is referred to in the PPI as the resource cost method.2 To illustrate how the resource cost method can be used to estimate VQA, assume that the product in base period o is a 1 inch steel nail and that in period t the producer replaces the 1 inch steel nail with a 1 inch steel-galvanized nail. The galvanization process is a quality improvement that increases the nails corrosion resistance. Further assume that the producer raises the pit price to \$2.00 compared to the previous pio price of \$1.25. The PPI now has the information to construct a nominal price relative of:
#### Equation 4
Nominal price relative = pit/pio = 2.00/1.25 = 1.60
Equation 4 shows that the price of the galvanized nail ( pit ) is 60 percent higher than the plain steel nail ( pio ). The problem for the PPI is that the new substitute product is no longer directly comparable to the originally sampled non-galvanized nail. If the 60 percent price increase in period t is not adjusted for quality change, this nominal price measurement could be considered upwardly biased. This is where VQA is used to estimate what part of price change is the result of quality change and what part is pure price change. Using the resource cost method, the PPI attempts to obtain an estimate from the producer of the marginal cost of the new input requirements directly tied to quality change. In this example, the producer informs the PPI that the galvanization process adds 50 cents to the unit production (marginal) cost of 1 inch nails ( pit ). With the cost of the new input requirements the PPI now has a value for VQA and can construct a constant quality price relative that is shown in equation 5:
#### Equation 5
Quality adjusted price relative = pit - (VQA)/pio = 2.00 - (0.50)/1.25 = 1.50/1.25 = 1.20
In equation 5, the period t price has been adjusted by the VQA provided by the producer. Instead of the 60 percent nominal increase shown in equation 4, the quality adjusted, or pure price change, is only 20 percent. The estimated VQA is a tool that enables the PPI to remove the ambiguity of nominal price comparisons when input requirements change and is fully compliant with the fixed quantity of inputs that is assumed in the PPI’s Fixed-Input Output Price Index (FIOPI) model.3 Another way of looking at VQA is that it enables the PPI to make price comparisons on a constant quality basis when changes in output quality occur.
### The Hedonic Option
An interesting question that the PPI faces every month is what to do when the resource cost method described above is not feasible because producers are unwilling or unable to provide VQA estimates. Several non-hedonic options described in the 2004 International Monetary Fund publication Producer Price Index Manual: Theory and Practice (see references) are available to the PPI analysts.
Hedonic methods as an alternative to resource cost have been historically targeted at valuing quality change for computer products such as desktops, laptops and servers where the rate of technological change is relatively fast. The limited use of hedonics in the PPI is primarily due to constraints in both staffing and availability of appropriate data.4
The PPI recognized that computer technology changed so rapidly that the resource cost approach to valuing quality change would be unlikely to fully capture the dynamics of computer outputs. For example, a mainstream desktop computer that sold for \$1,200 in 2004 may have included 1 gigabyte (GB) of system memory, a 60 GB hard drive, a 15-inch cathode-tube monitor as well as many other defining technological characteristics. In 2010, however, a desktop computer that still sells for \$1,200 could easily have been configured with 8 GB of system memory, a 700 GB hard drive and a 21” liquid crystal display (LCD) monitor. The 2010 model also includes technologies unavailable in 2004 such as blue-ray high definition DVD players and quad-core microprocessors.5 In this example, the observed prices for the 2004 and 2010 computers are identical. However, technological change over this 6-year period has been remarkable: system memory has jumped 700 percent; hard drive storage capacity has increased an amazing 1000+ percent and monitor size gained 40 percent while also advancing from bulky cathode ray to relatively thin, light, and higher resolution LCD. A computer price index that did not account for these significant quality improvements, that is, an index that showed no price change for the period in question, would clearly have an upward bias as a measure of quality-adjusted computer prices. However, as mentioned earlier, the resource cost method was not feasible for this industry. This difficult price measurement issue has received extensive coverage in various econometric research papers, providing some of the basic principles followed by the PPI in accounting for technological change in computers using hedonic methods.
Many of the hedonic models developed outside of the Bureau of Labor Statistics (BLS) explicitly identify time as an independent variable (Triplett 1986). Most of these models are based on pooled data (multiple time periods) that are segmented with time dummy variables to enable the direct construction of a quality-adjusted price index. A variety of functional forms and model specifications have been developed over the years, but a simplified example of the underlying hedonic equation could take the form shown in equation 6.
#### Equation 6
T
Pit = α0 + δt Dt + β2 X2i + β3 X3i ··· βk Xki + ui
t-1
Where:
Pit is the price of the i th model in period t
Dt is a time dummy variable
Xi is a variable representing economically meaningful product characteristics
β2 ··· βk are the regression/slope coefficients
ui is the residual or error term
δt represents the percentage change in price across time intervals defined by Dt holding quality constant
Different strategies can be applied to the estimation of equation 6. For instance, the pooled approach enables maximum sample size, but is complicated by dissimilarity in computer characteristics over time due to rapid obsolescence. When product characteristics are unstable over time, a remedial strategy that may minimize the problem of rapid evolution in product characteristics is the adjacent period approach. The adjacent period approach disaggregates the pooled data spread over many years into a series of regressions that are limited to data from two or more periods. Experimentation with varying time spans may allow several adjacent periods to be pooled. Ultimately, the choice is dictated by the type of product that is modeled, the rate of technological change, and the quality and detail of the available data.
Because there are many different approaches, including preferred functional form and the choice of specific independent variables, the mechanics are best understood from a direct review of various research papers. Interestingly, published results from these papers are remarkably similar. Chow (1967) calculated that quality-adjusted mainframe computer prices declined 21 percent annually, from 1960 to 1965; Gordon (1989) reported quality-adjusted mainframe computer prices fell 22 percent per year from 1951 to 1984; and Triplett (1989) summarized much of the work on mainframes and reported his findings of a 27 percent annual rate of decrease from 1953 to 1972. Finally, Berndt, Griliches, and Rappaport (1995) extended the previous hedonic work with a large sample of personal computer data and reported 28 percent annual decreases, on average.
As stated earlier, the PPI began its own research into hedonic modeling of computer products in 1987. The PPI opted for a cross-sectional (equation 1)—rather than pooled—(equation 6) approach, due to rapid technical changes in the characteristics that define computers. With cross-sectional models, the PPI does not attempt to directly construct price indexes, but instead uses the hedonic model to adjust prices reported to the PPI directly by producers when they indicate that product characteristics have changed. In other words, the model’s implicit prices for product characteristics are the VQAs used by the PPI to adjust reported prices for changes in quality.
### Sampling and Database Issues
The PPI measures price change by first sampling those producers whose plurality of shipments fall within a NAICS defined industry, such as NAICS 34111-Electronic computer manufacturing.6 The probability of a firm's selection is proportionate to its employment size. Once a firm is selected, products and services within an establishment are sampled, according to the probability-proportionate-to-size method that, in turn, is based on estimated sales data by product line supplied by a company respondent. Sampling techniques allow the PPI to capture data from a small but representative number of producers that report monthly prices and notify the PPI when sampled products change. Ideally, the PPI would specify a hedonic model supported by characteristics and prices taken directly from the PPI sample. However, with the exception of some services industries, most PPI samples at the product level are not of sufficient size to support the data requirements of a fully specified hedonic model. As a result, the PPI uses product specific secondary data sources to generate most of its hedonic models. For example, the PPI constructs separate models for portable computers, personal desktop computers, and host systems (computer servers). As recently as 1996, databases supporting PPI models were constructed from manufacturers' advertisements in Computer Shopper magazine. In 1997, several PPI reporters stated that more reliable and comprehensive technical data was available on their web pages. Many of these Internet sites include configuration utilities that calculate selling prices for a range of technical features that are available in currently shipping products. As a test, the PPI developed a hedonic computer desktop model in February 1997 entirely from producer website data. The Internet-based model provided higher precision estimators and enabled direct real-time feedback for correcting or clarifying technical descriptions that were not readily available from 3rd party advertisements.
### Using Hedonic Models in the PPI
The regression formula used in a recent PPI hedonic model was based on equation 1, which is repeated here, and the results are shown in table 1.
#### Equation 1
Pit = α0 + β2 X2i + β3 X3i ··· βk Xki + ui
Note that equation 1 does not include a time variable, as in equation 6. The selection of functional form is evaluated in terms of statistical precision, reasonableness of absolute values, R-square, and F-tests (Gujarati 1995).
Table 1. Hedonic model regression results for personal desktop computers in the PPI
Coefficient Standard error T-statistic P-value
Constant
-256.158 38.180 -6.709 0.000
Quad-core CPU per GHz
373.129 15.150 24.629 0.000
Memory DDR per GB
25.769 1.253 20.573 0.000
Hard Drive 7200RPM per GB
0.240 0.016 15.128 0.000
Blu-Ray ROM
77.492 5.411 14.322 0.000
256MB HD Video or Equivalent
33.422 5.042 6.629 0.000
512MB HD Video or Equivalent
77.365 6.750 10.869 0.000
22" LCD Monitor
41.209 5.081 8.111 0.000
24" LCD Monitor
233.219 5.809 40.147 0.000
Xtreme Gamer Sound Card
155.353 9.025 17.214 0.000
130.201 7.914 16.453 0.000
Office Basic Sftw.
89.440 6.620 13.511 0.000
Office SMB Sftw.
277.749 5.724 48.522 0.000
Office Professional Sftw.
371.757 6.543 56.819 0.000
322.528 9.255 34.848 0.000
Warr: 3yr NBD On-site
164.043 5.955 27.548 0.000
Warr: 3yr 4-hour On-site
357.616 8.487 42.136 0.000
Company A
-222.876 4.937 -45.144 0.000
Company B
-805.453 13.703 -58.779 0.000
Observations = 2,022; Adj R Square = 0.950; Stnd Error of Estimate = 90.445; F = 2,116.943
Base configuration (relative to dummy/qualitative characteristics) were: DVD-RW, Integrated Video, 19” LCD, Integrated Sound, Vista Home Premium, No Productivity Software, Consumer Market, 1yr NBD On-Site Warranty.
The model in table 1 is based on a cross-sectional linear functional form with the independent variables (computer characteristics) shown on the left and the implicit prices for each characteristic shown under coefficients. The remaining columns are for basic statistical diagnostics.
The regression results are derived from 2,022 observations of desktop computers from producer websites on the Internet.7 These Internet sites have evolved over time to include online processing of direct sales from the producer to the end user. Typically, a buyer can view technical specifications and prices for a range of pre-configured computers and then complete the purchase online. However, most computer producers offer buyers greater flexibility by providing a broad customization of features that can be added to, or deleted from, the pre-configured models. For instance, a preconfigured model that includes a 2.0GHz CPU is also available with a faster or slower CPU, more memory, or a monitor upgrade. A buyer can chose options most appropriate for their application and budget, and the producer will instantly recalculate a price based on the buyer's selection of those computer characteristics. A variety of producer websites representative of the PPI sample will typically be used to build a data base.
The number of sites used in PPI models will vary according to the type of product. Computer models will generally include four or more sites. After the data are acquired, data relationships are explored, various regression scenarios are run, and the results analyzed. Each database is structured to ensure that the most important PPI computer producers are included. Several dummy or qualitative variables are used in the model. For example, the coefficients (values) for the 22” and 24” LCDs in table 1 are relative to a base configuration with a 19” LCD. Company-effect dummy variables are also used, in this case referred to as Company A and B, to avoid the possibility of disclosing companies that are in the database but also report prices directly to the PPI. Only companies that tested as statistically significant are included. The interpretation of a company-effect variable is that it may capture otherwise unspecified price determining factors, such as name recognition, buyer loyalty, and superior marketing. These are factors that provide a limited company-specific price differentiation that go beyond the explanatory powers of the explicitly defined product characteristics. To illustrate how regression results from table 1 would be used in the PPI to obtain quality-adjusted price relatives, a hypothetical example is described below in table 2.
Table 2. Quality Change Valuation Example
Computer Reported to PPI, Period (t-1) Computer Reported to PPI, Period (t)
Desktop personal computer
Desktop personal computer
Model 100
Model 100
Core 2 Quad CPU 2.4GHz
Core 2 Quad CPU 2.4GHz
2.5GB DDR-2 Memory
4.0GB DDR-2 Memory
400 GB 7200RPM SATA hard drive
400 GB 7200RPM SATA hard drive
256MB HD Video Card
256MB HD Video Card
8X DVD/RW Combo
8X DVD/RW Combo
Integrated sound
Integrated sound
1Gb NIC
1Gb NIC
19-inch 1366 X 768 LCD Monitor
19-inch 1366 X 768 LCD Monitor
Vista Home Premium OS
Vista Home Premium OS
1yr NBD On-Site Warranty
1yr NBD On-Site Warranty
Reported net price in period (t-1) = \$1,000
Reported net price in period (t) = \$1,000
Estimated Value of Improvement from Period (t-1) to Period (t)
System memory increase from 2.5GB to 4.0GB
1.5GB additional DDR-2 * \$25.77 (unit value from table 1) = \$38.66
Quality adjusted (QA) price change = (P(t) - VQA) / P(t-1)
= (1,000 - 38.66) / 1,000
= -3.86 percent
The example in table 2 is based on the assumption that the PPI is repricing a computer configured with 2.5GB of DDR-2 memory that sold in period (t-1) for \$1,000 that the producer upgrades to 4.0GB of DDR-2 in period (t). The reporter indicates that the current period price for the upgraded model is still \$1,000, and that the producer cannot provide a value for the change in production cost that is directly attributable to the additional 1.5GB of memory. If the PPI did not have a hedonic model to show the obvious quality improvement the PPI would use a standard methodology that adjusts the base price so that the item would show a price change that reflected the average price movement of all of the other items that had prices that were reported and included in the same index that month. With the implicit unit memory price generated by the hedonic model, the PPI has a method for valuing this quality change. The implicit price of \$25.77 per unit (GB) of memory is obtained from the DDR-2 variable in table 1 and multiplied by the 1.5 unit increase in period (t), to yield a total quality change valuation of \$38.66. As used in the PPI, this hedonic model is simply a tool that estimates the average change in price over the 2,022 observations for a unit change in a continuous variable or the absence or presence of a dummy/qualitative variable. The remaining quality adjustment calculation is straightforward, as shown in the example. Of course, if more than one quality change had occurred; and the relevant characteristics were specified in the model, we could sum the implicit prices for these changes and calculate a multi-factor quality adjusted price relative from the nominal prices reported to the PPI.
As previously mentioned, the PPI has developed multiple hedonic models for a range of computer products. In addition to the desktop model shown above, recent models for laptops, servers, and netbooks are shown in tables 3, 4, and 5 respectively.
Table 3. Hedonic Model Regression Results for Laptop Computers for the PPI
Coefficients Standard Error T-statistic P-value
Constant
-393.067 36.192 -10.860 0.000
Core i per GHz
456.044 14.020 32.528 0.000
Memory per GB
48.661 1.141 42.647 0.000
Hard Drive per GB
0.450 0.014 32.279 0.000
7200 RPM Hard Drive
37.552 4.087 9.187 0.000
383.201 2.836 135.104 0.000
512 MB Video Card
69.000 5.568 12.392 0.000
1 GB Video Card
120.000 8.117 14.784 0.000
Integrated Web Cam
30.000 3.215 9.332 0.000
117.797 3.631 32.441 0.000
9 Cell Battery
69.000 3.937 17.525 0.000
Bluetooth
20.848 2.836 7.350 0.000
Windows 7 Professional
83.000 4.051 20.486 0.000
Office Home Student
108.014 7.164 15.077 0.000
177.614 4.839 36.702 0.000
Office Professional
309.429 3.157 98.017 0.000
Company A
-98.187 14.448 -6.796 0.000
Company C
667.136 6.394 104.345 0.000
Observations = 3,360; Adj R Square = 0.978; Stnd Error of Estimate = 81.835; F = 8,732.32
Base configuration (relative to dummy/qualitative characteristics) were: 5,400 RPM Hard Drive; No Optical Drive/DVD-RW; Integrated Video; 6 Cell Battery; Windows 7 Home Premium; No Productivity Software; Company D.
Table 4. Hedonic Model Regression Results for Servers for the PPI
Coefficients Standard Error T-Statistic P-Value
Constant
-550.088 87.470 -6.289 0.000
Quad Core Xeon per GHz
583.048 42.551 13.702 0.000
CPU Interconnect per GT/s
320.135 9.094 35.201 0.000
CPU 95 Watts
371.534 29.235 12.708 0.000
Second CPU
955.243 15.493 61.658 0.000
ECC Memory per GB
63.517 1.178 53.907 0.000
Second 300 GB 10/15k Hard Drive
580.285 15.910 36.473 0.000
Third 300 GB 10/15k Hard Drive
623.043 19.693 31.638 0.000
Windows Server 2008 Standard Edition, incl 5 CALs
754.697 10.522 71.726 0.000
Second 512 MB RAID Controller
659.897 16.555 39.860 0.000
Dual Port Gigabit NIC
256.927 11.909 21.574 0.000
Redundant Hot Swap Power Supply
191.061 13.852 13.793 0.000
3yr NBD Warranty
150.096 16.329 9.192 0.000
3yr 4 Hour Response Warranty
513.364 17.836 28.782 0.000
Company A
-484.658 18.345 -26.419 0.000
Company B
-912.546 32.793 -27.828 0.000
Observations = 1,765; Adj R Square = 0.958; Stnd Error of Estimate = 220.16; F = 2,703.005
Base configuration: CPU 80 Watts; One Quad Core Xeon CPU; One 300GB 10/15K RPM Hard Drive; No Operating System; One 512MB RAID Controller; Integrated Gigabit Network Adaptor; Non-Redundant Power Supply; One Year NBD Warranty; Company C.
Table 5. Hedonic Model Regression Results for Netbooks for the PPI
Coefficients Standard Error T-Statistic P-Value
Constant
32.560 6.780 4.803 0.000
Intel Atom per GHz
108.893 3.419 31.846 0.000
Dual Core CPU
82.127 0.819 100.250 0.000
Memory 2 GB
31.671 0.599 52.865 0.000
Hard Drive per GB
0.271 0.003 78.127 0.000
7,200 RPM Hard Drive
6.714 0.655 10.244 0.000
1,366 X 768 Screen Resolution
21.677 0.609 35.605 0.000
Office Home Student
100.858 0.604 166.907 0.000
Bluetooth
10.422 1.514 6.884 0.000
33.689 1.518 22.188 0.000
Mobile GPS
21.258 0.735 28.915 0.000
75.000 0.690 108.721 0.000
3 Year NBD Warranty
122.532 1.501 81.658 0.000
Security Software 15 Months
4.422 1.514 2.920 0.004
6 Cell Battery
30.021 1.526 19.669 0.000
Company A
16.169 3.046 5.309 0.000
Observations = 573; Adj R Square = 0.994; Stnd Error of Estimate = 6.74; F = 6,024.19
Base configuration: Single Core CPU; 1GB Memory; 5,400 RPM Hard Drive; 1,024 x 600 Screen Resolution; No Productivity Software; Windows 7 Starter; 1 Year NBD Warranty; 3 Cell Battery; Company B and C.
Technical descriptions of the characteristics for the various computers shown in tables 1 and 3-5 are not important for the purpose of this review. The primary interest here is that the description in table 2 of how a hedonic model is used in the PPI to adjust prices for quality change is exactly the same for each of the different models. It should also be clear that there are significant differences between each of the computer types represented in tables 3-5, which is why separate models had to be researched and developed rather than a single model for all computer products. In other words, to be a useful quality valuation tool, each model must be focused on a specific product type. One would never try to build a hedonic model for transportation by combining bicycles, motorcycles, cars, trucks and 747 jumbo jets, though all are examples of transportation. It is the responsibility of PPI industry analysts to determine the parameters of a model’s specification.
### Final Thoughts
Depending on how quickly a product’s characteristics change, a hedonic model used in the PPI may be updated as frequently as quarterly. As new technologies replace old technologies, the specification of a hedonic model must also change in order to continue to reflect current market conditions. One of the challenges for the PPI is to update models in a timely fashion in a real-time monthly production environment. At this time (and for the last 20 years), the primary use of hedonic models in the PPI has been to value quality change for computer products. Currently, the PPI is researching the feasibility of developing hedonic models for new product categories. This research effort may result in an expanded ability for the PPI to value quality change when more traditional methods are not available. However, the detailed data requirements and staff resources required to build and maintain these models will almost certainly limit the expansion of hedonics in the PPI to a few select products for the foreseeable future.
## References
Berndt, E. R., Z. Griliches, and N.R. Rappaport. "Econometric Estimates of Price Indexes for Personal Computers in the 1990's." Journal of Econometrics 68, no. 1 (1995), pp. 243-268.
Chow, G. C. "Technological Change and the Demand for Computers." American Economic Review 57 (1967), pp. 1117-1130.
Diewert, W. E. “Aggregation Problems in the Measurement of Capital.” edited by D. Usher, 433-528. Chicago, IL: The University of Chicago Press, 1980.
Griliches, Z. “Hedonic Price Indexes and the Measurement of Capital and Productivity: Some Historical Perspectives.” NBER Working Paper 2634 (June 1988), Pg. 1.
Gordon, R. J. "The Postwar Evolution of Computer Prices." Technology and Capital Formation, edited by D. W. Jorgenson and R. Landau, 77-125. Cambridge, MA: MIT Press, 1989.
Gujarati, Damadar. Basic Econometrics. 3rd Edition. McGraw-Hill Inc. 1995.
International Monetary Fund. Producer Price Index Manual: Theory and Practice. Washington D.C.: International Monetary Fund, Publication Services, 2004.
Nelson, R. A., C. D. Patterson, and T. L. Tanguay, "A Quality-Adjusted Price Index for Personal Computers." Journal of Business and Economic Statistics, no. 1, (January 1994), pp. 23-31.
Triplett, J. E. "Price and Technological Change in a Capital Good: A Survey of Research on Computers." Technology and Capital Formation. Edited by D. W. Jorgenson and R. Landau, pp. 127-213. Cambridge, MA: MIT Press, 1989.
Triplett, Jack. "Concepts of Quality in Input and Output Price Measures: A Resolution of the User-Value Resource-Cost Debate." Studies in Income and Wealth, 47. Edited by Murray F. Foss, 269-312. University of Chicago Press for the National Bureau of Economic Research, 1982.
Triplett, Jack. Database Issues in Estimating Hedonic Computer Price Indexes: Comparison of Hedonic Model Specifications. Contracted Report for Eurostat, 2003.
Triplett, Jack. "Economic Interpretation of Hedonic Models." Survey of Current Business 66, no. 1 (January 1986), pp. 36-40.
## Footnotes
1 The referenced methodologies are described in “BLS Handbook of Methods, Chapter 14, Producer Price Indexes”, pgs 3-4 and can be accessed at http://www.bls.gov/opub/hom/pdf/homch14.pdf.
2 Cost as referenced here means cost plus the producer’s standard markup.
3 The FIOPI model is one example of a theoretical construct for output based producer price indexes that assume fixed technology and inputs. Such an index is designed to reflect changes in revenue from the sale of the same set of products—though not necessarily the same mix of products. The key assumptions of a FIOPI model is that changes to the price index arise solely from change in output prices that are unaffected by changes in inputs. See Triplett 1983 for more details.
4 The PPI provides measures of price change for more than 600 industries and over 5,000 product categories.
5 Consumers commonly focus on CPU clock speed when judging a computer's performance potential, quoted in gigahertz (GHz). This single number is more confusing than enlightening in today's environment of multiple competitive processors; one processor running at a particular GHz can significantly outperform another processor rated at the same GHz, and processors based on different architectures with multiple cores are difficult to compare, without objective benchmarking.
6 As defined by the 2012 North American Industrial Classification System (NAICS).
7 Computer resellers (retail or wholesale) that market computers through online stores are excluded from our databases. This type of transaction is out-of-scope for the PPI's coverage of NAICS 334111. | 7,662 | 31,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2015-48 | longest | en | 0.93761 |
https://www.jiskha.com/display.cgi?id=1197440405 | 1,503,572,227,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133449.19/warc/CC-MAIN-20170824101532-20170824121532-00670.warc.gz | 889,035,609 | 4,348 | # Chem
posted by .
Manganese dioxide reacts with bromide ions to form manganese ions and bromate ions in an acidic aqueous solution. Write the balanced equation for this reaction.
1)Br- + MnO2 + 3H2O --> Mn2+ + BrO3- + 2H3O+
2)Br- + 3MnO2 + 6H3O+ ---> 3Mn2+ + 9H2O + BrO3-
3)6Br- + MnO2 + 48H2O ---> Mn2+ + 32H3O+ + 6BrO3-
4)4Br- + 3MnO2 + 18H2O ---> 3Mn2+ + 12H3O+ + 4BrO3-
5)Br- + 3MnO2 + 3H3O+ ---> 3Mn2+ + BrO3- + 3H2O
I got the first one. Is that right?
• Chem -
1)Br- + MnO2 + 3H2O Mn2+ + BrO3- + 2H3O+
2)Br- + 3MnO2 + 6H3O+ 3Mn2+ + 9H2O + BrO3-
3)6Br- + MnO2 + 48H2O Mn2+ + 32H3O+ + 6BrO3-
4)4Br- + 3MnO2 + 18H2O 3Mn2+ + 12H3O+ + 4BrO3-
5)Br- + 3MnO2 + 3H3O+ 3Mn2+ + BrO3- + 3H2O
I got the first one. Is that right?
• Chem -
Your equatioh isn' balanced. Furthermore, several students (you are one of them) don't use arrows. How can you tell which are reactants and which are products with no arrow?
(Charge doesn' balance. On left you have 1- and 2+ = 1+. On the right you have 2+, 1-, 2+ = 3+.
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Consider the following reaction: PbCl2 + 2 NaOH → Pb(OH)2 + 2 NaCl What would be the major species observed floating in solution after one mole of PbCl2 and 2 moles of NaOH are mixed in aqueous solution?
More Similar Questions | 1,071 | 3,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-34 | latest | en | 0.781306 |
https://ontimemicrofinance.com/qa/how-many-days-is-180-from-today.html | 1,632,759,226,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058456.86/warc/CC-MAIN-20210927151238-20210927181238-00238.warc.gz | 474,552,739 | 7,014 | # How Many Days Is 180 From Today?
## How do I sum dates in Excel?
Add or subtract days from a dateEnter your due dates in column A.Enter the number of days to add or subtract in column B.
You can enter a negative number to subtract days from your start date, and a positive number to add to your date.In cell C2, enter =A2+B2, and copy down as needed..
## How many days is 180 days?
180 days equals roughly 6 months. A month contains 30 or 31 days, except for February. To convert a number of days to months, you can say 30 days is…
## What is 199 days from now?
That means that 199 weekdays from today would be August 23, 2021.
## What day is 200 days away?
– Today is : Sunday, November 15, 2020. – The date after 200 days is : Thursday, June 3, 2021. – It is the 154th day in the 22th week of the year.
## What day is 91 days away?
– Today is : Monday, November 16, 2020. – The date after 91 days is : Monday, February 15, 2021. – It is the 46th day in the 07th week of the year.
## How do I calculate time between dates in Excel?
Calculate elapsed time between two dates and timesType two full dates and times. In one cell, type a full start date/time. … Set the 3/14/12 1:30 PM format. Select both cells, and then press CTRL + 1 (or. … Subtract the two. In another cell, subtract the start date/time from the end date/time. … Set the [h]:mm format.
## What day is 182 days away?
– Today is : Tuesday, November 17, 2020. – The date after 182 days is : Tuesday, May 18, 2021. – It is the 138th day in the 20th week of the year.
## What is the 180 days from today?
– Today is : Monday, November 16, 2020. – The date after 180 days is : Saturday, May 15, 2021. – It is the 135th day in the 19th week of the year.
## How do you calculate 180 days from today in Excel?
To calculate the expiry date simply add 180. For example, if A2 = 1/1/2019 and B2 = A2 + 180 then B2 will display the expiry date you want. If those values are part of a table then Excel will fill that in automatically.
## What day is 199 days away?
– Today is : Tuesday, November 17, 2020. – The date after 199 days is : Friday, June 4, 2021. – It is the 155th day in the 22th week of the year.
## What day is 52 days away?
– Today is : Tuesday, November 17, 2020. – The date after 52 days is : Friday, January 8, 2021. – It is the 8th day in the 01th week of the year.
## What is 100 days from now?
100 days from today is Thu 25th Feb 2021.
## What is the formula to add 7 days to a date in Excel?
In the formula, A1 is the date you use, if you want to add 1 year to the date, just use =DATE(YEAR(A1)+1,MONTH(A1),DAY(A1)), if you want to add 7 days to the date, use this formula =DATE(YEAR(A1),MONTH(A1),DAY(A1)+7).
## What was 181 days ago from today?
What Date & Day was it 181 Days before Today?SummaryDateDayBefore 181 Business Days from Today Saturdays & Sundays ExcludedFebruary 27, 2020ThursdayBefore 181 Days from Today Only Sundays ExcludedApril 09, 2020ThursdayBefore 181 Days from Today Only Fridays ExcludedApril 09, 2020Thursday1 more row
## What is 7 days from now?
7 days from today (5 weekdays) would be Tuesday, November 24, 2020.
## What day is 51 days away?
– Today is : Tuesday, November 17, 2020. – The date after 51 days is : Thursday, January 7, 2021. – It is the 7th day in the 01th week of the year.
## What day is 74 days away?
– Today is : Tuesday, November 17, 2020. – The date after 74 days is : Saturday, January 30, 2021. – It is the 30th day in the 04th week of the year.
## What date is 52 weeks from now?
What is 52 Weeks From Today? – The date will be Tuesday, November 16, 2021 52 weeks from today. | 1,079 | 3,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-39 | latest | en | 0.922954 |
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