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http://topas.dur.ac.uk/topaswiki/doku.php?id=damp&do=diff&rev2%5B0%5D=1288659633&rev2%5B1%5D=1322191596&difftype=sidebyside | 1,566,093,280,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00009.warc.gz | 198,138,200 | 5,609 | Differences
This shows you the differences between two versions of the page.
damp [2010/11/02 01:00]pac079 created damp [2011/11/25 03:26]rowlesmr fixed some formatting and added additional 's to fix auto colouring. 2011/11/25 03:26 rowlesmr fixed some formatting and added additional 's to fix auto colouring.2010/11/02 08:16 pac079 2010/11/02 01:00 pac079 created Next revision Previous revision 2011/11/25 03:26 rowlesmr fixed some formatting and added additional 's to fix auto colouring.2010/11/02 08:16 pac079 2010/11/02 01:00 pac079 created Line 1: Line 1: ====== GSAS-style Dampening Factors ====== ====== GSAS-style Dampening Factors ====== - Dampening factor for any refined value (lattice parameters, atomic positions etc.), for those who think that they know better than the least squares algorithm + Dampening factor for any refined value (lattice parameters, atomic positions etc.), for those who like to think that they know better than the least squares algorithm. - macro Damp(DampFactor) { update = Val + (Change*(1/Exp(DampFactor/2))); } + - 'Rather than the true GSAS dampening [applied shift = (10-damp)/10], an exponential function is used, scaled to useful dampening values of 0-9 + - 'Useful dampening factors range from values of 0 (no dampening) to 9 (1% applied shift) + ''Rather than the true GSAS dampening [applied shift = (10-damp)/10], an exponential function is used, scaled to useful dampening values of 0-9 - 'List of DampFactor,AppliedChanges: 0,1; 1,0.607; 2,0.368; 3,0.223; 4,0.135; 5,0.082; 6,0.50; 7,0.030; 8,0.018; 9,0.011 + ''Useful dampening factors range from values of 0 (no dampening) to 9 (1% applied shift) - 'e.g. A simple lattice parameter refinement with dampening factors of 9, 8, 7, 6, 5, 4, 3, 2, 1 and 0 + ''List of DampFactor,AppliedChanges: 0,1; 1,0.607; 2,0.368; 3,0.223; 4,0.135; 5,0.082; 6,0.50; 7,0.030; 8,0.018; 9,0.011 - ' converges in 422, 271, 173, 111, 71, 45, 28, 17, 10 and 5 cycles, respectively + ''e.g. A simple lattice parameter refinement with dampening factors of 9, 8, 7, 6, 5, 4, 3, 2, 1 and 0 - 'Example of use: site O1 x @ 0.30696 Damp(3) y 0 z 0.25 occ O-2 1 beq @ 2 Damp(7) + '' converges in 422, 271, 173, 111, 71, 45, 28, 17, 10 and 5 cycles, respectively + ''Example of use: + '' site O1 x @ 0.30696 Damp(3) y 0 z 0.25 occ O-2 1 beq @ 2 Damp(7) + + macro Damp(DampFactor) + { + update = Val + (Change*(1/Exp(DampFactor/2))); + } + | 910 | 2,453 | {"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-2019-35 | longest | en | 0.793891 |
https://en.khanacademy.org/math/in-class-8-math-foundation/x5ee0e3519fe698ad:data-handling/x5ee0e3519fe698ad:bar-graphs/v/reading-bar-charts-2 | 1,702,287,859,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00775.warc.gz | 270,290,346 | 67,942 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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## Class 8 (Foundation)
### Course: Class 8 (Foundation)>Unit 7
Lesson 2: Bar graphs
# Reading bar charts: comparing two sets of data
A bar chart is especially useful with comparing two sets of data. The difference in the bars give us a quick snapshot that allows us to draw some conclusions. Created by Sal Khan.
## Want to join the conversation?
• At Sal said 80 and 85. what is the answer?
• 85. He's not really saying 80, he's just hesitating a bit the first time and starts over saying 85. Looks like 80... like 85.
• when ever the blue line is not all the way on a number will it allways be 5?
• On the Khan Academy practice shown, yes, because they're counting by tens on the y-axis scale and the bars in between are always exactly half way in between.
In other situations, though, not always. You have to look carefully at the bar and the y-axis scale to see where the bar is and what they're counting by.
• What is the difference between "bar chart" and "bar graph"?
• they're the same thing, just different names :)
• Why do I get more energy points from watching a video than from answering 7 questions right?
• which website are u using for bar-chart in this video?
• I think you cut the numbers in hkf
• what if the bar is in between the line
• If the bar ends between two lines you just estimate where it ends. In the example on the video they all end halfway between the two numbers, making it 5 points more than lower of the two numbers.
(1 vote)
• At it says about 75 points. I tried to guess the amount of points in Reading bar charts 2 and I said I said it looked like 42 but it said I was wrong. How am I supposed to know the exact amount?
(1 vote)
• Jus guess
(1 vote)
• This seem like a practice video ,where can I learn statistics from the basic level to the most advance level on this website
(1 vote)
• why does there have to be 6 things why can't it just be less than 6
(1 vote)
## Video transcript
So we have this bar chart here. This says Scores on Midterm and Final Exams. So this axis, the vertical axis, is the scores. And then it's by student. And the blue bar is the midterm, and the yellow bar is the final. And the question they ask us is by how many points did Nadia's score improve from the mid-term to the final exam? So let's look at Nadia. So this is who we're talking about-- Nadia. And we care about how many points did she improve from the midterm to the final. Midterm is blue, final is yellow. So on the midterm, it looks like she scored-- and if I were to eyeball it, it looks like 75 points. And on the final, it looks like she scored 80. It looks like she scored 85 points. So it looks like her score improved by 10 points, so 10 points. Let's try one more. How many students improved their scores from the midterm to the final exam? So to improve from the midterm to the final, that means that the yellow bar for a given student, which is the final, is going to be higher than the midterm bar. That's the only way you can improve from the midterm to the final. So Brandon improved from the midterm to the final. Vanessa improved from the midterm to the final. Daniel improved from the midterm to the final. Kevin improved from the midterm to the final. William got a lower score on the final than the midterm, so he did not improve. So the number of students that improved their scores from midterm to final are one, two, three four students, so 4 students. | 870 | 3,636 | {"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} | 3.703125 | 4 | CC-MAIN-2023-50 | latest | en | 0.953623 |
http://betterlesson.com/lesson/566547/comparing-fractions-using-a-number-line | 1,477,358,846,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719843.44/warc/CC-MAIN-20161020183839-00319-ip-10-171-6-4.ec2.internal.warc.gz | 27,137,015 | 20,150 | # Comparing Fractions Using a Number Line
Unit 6: Fraction Equivalents and Ordering Fractions
Lesson 8 of 14
## Big Idea: This lesson builds towards students ability to compare fractions by creating common denominators or numerators - 4.NF.2
Print Lesson
5 teachers like this lesson
Standards:
Subject(s):
59 minutes
### Melissa Romano
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Environment: Suburban | 199 | 858 | {"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-2016-44 | longest | en | 0.776224 |
https://www.doubtnut.com/qna/119564750 | 1,716,324,597,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00401.warc.gz | 633,691,512 | 36,462 | # In $△$ ABC , $\angle ACB={90}^{\circ }$ , $CD\perp AB$ and $DE\perp CB$ Prove that $C{D}^{2}×AC=AD×AB×DE$ .
Video Solution
|
Step by step video & image solution for In triangle ABC , angle ACB = 90^(@) , CD bot AB and DE bot CB Prove that CD^(2) xxAC = AD xx AB xxDE . by Maths experts to help you in doubts & scoring excellent marks in Class 10 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## In a △ABC,AB = AC and AD⊥BC, then :
• Question 1 - Select One or More
## In a right triangle ABC, in which ∠C=90∘amdCD⊥AB. If BC =a, CA=b, AB=c and CD=p, then
A1p2=1a2+1b2
B1p21a2+1b2
C1p2<1a2+1b2
D1p2>1a2+1b2
• Question 1 - Select One
## In ΔABC,∠B=90∘ and BD⊥AC. If AC = 9 cm and AD = 3 cm, then BD is equal to :
A22 cm
B32 cm
C23 cm
D33 cm
### Similar Questions
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 577 | 1,791 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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-2024-22 | latest | en | 0.824715 |
https://patel-prakhar09.medium.com/?source=user_profile------------------------------------- | 1,632,867,368,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060908.47/warc/CC-MAIN-20210928214438-20210929004438-00073.warc.gz | 485,784,049 | 25,062 | # The Linear Regression !!!
It is the most basic and widely used technique to predict a value of an attribute
What is Linear Regression?
There are some assumptions that may become useful when we analyze our model to check whether it is accurate or not
# Basic Fundamentals Of Statistics Every Data Scientist Should Know.
## Essential statistic principles to get you started on your Data Science journey.
we all are cognizant of the enormous impact of statistics in the field of data science in this data-driven world. Finding patterns, trends, and making predictions are the most significant steps in data science. Here we gonna discuss some basics terms of the statistical world which play a crucial role in statistical data analysis.
# Data types and individuals in data
Individuals: individuals are the people or objects included in the study. An individual is what the data is describing. In a table like this, each individual is represented by one row. Sometimes they are also called identifiers.
Variable: Variable depicts information of individuals that is acquired…
# Introduction :
Covariance and Correlation are two mathematical principles that are frequently used in the field of statistics and probability. These both techniques have a common goal, to depict the linear relationship between two variable or data samples.
# Covariance :
The covariance determines the relationship between two random variables or samples — how they change together. Or in other words we can say that Covariance is a measure of how much two random variables fluctuate together.
Covariance is nothing but a measure of correlation. Covariance denotes the direction of the linear relationship between the two data variables. By finding direction of relationship we can…
## Prakhar Patel
I’m Computer student. I’m studying Information Technology 3rd year in LD College Of Engineering.
Get the Medium app | 355 | 1,908 | {"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} | 3.234375 | 3 | CC-MAIN-2021-39 | longest | en | 0.924296 |
https://www.physicsforums.com/threads/is-the-true-for-any-scalar-function.466803/ | 1,701,752,882,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00382.warc.gz | 1,052,641,158 | 16,627 | # Is the true for any scalar function?
• CalcYouLater
In summary, both of them say that φ is not dependent on a single position and that Stoke's theorem can be used to show this.
## Homework Statement
If $$\phi$$ depends on a single position only, $$\phi=\phi(x,y,z)$$
Can I say that:
$$\oint{\frac{\partial\phi}{\partial{x}}dx=\oint{d\phi}=[\phi]_{a}^{a}=0$$
Provided that the point a lies on the closed path being integrated around?
## The Attempt at a Solution
I am 99.99% sure I can. I am in the middle of "showing" how one theorem implies another, and I wasn't sure if I could knock the partial dx off with the total dx in any case under the conditions above. Thanks for any help. (Sorry, I can't seem to get my latex correct. The second to last term is intended to be phi evaluated from a to a. Also the title should read "Is this true for any scalar function")
Well, since $\phi$ depends only on $x$,
$$\frac{\partial\phi}{\partial y} = 0 = \frac{\partial\phi}{\partial z}.$$
So,
$$d\phi = \frac{\partial\phi}{\partial x} dx + \frac{\partial\phi}{\partial y} dy + \frac{\partial\phi}{\partial z} dz = \frac{\partial\phi}{\partial y} dx.$$
Hence,
$$\oint d\phi = \oint \frac{\partial\phi}{\partial x}dx = \int_a ^a\frac{\partial\phi}{\partial x}dx = 0.$$
However, I'm not sure what you mean by
CalcYouLater said:
Provided that the point a lies on the closed path being integrated around?
What point are you referring to?
CalcYouLater said:
## Homework Statement
If $$\phi$$ depends on a single position only, $$\phi=\phi(x,y,z)$$
Can I say that:
$$\oint{\frac{\partial\phi}{\partial{x}}dx=\oint{d\phi}=[\phi]_{a}^{a}=0$$
Provided that the point a lies on the closed path being integrated around?
## The Attempt at a Solution
I am 99.99% sure I can. I am in the middle of "showing" how one theorem implies another, and I wasn't sure if I could knock the partial dx off with the total dx in any case under the conditions above. Thanks for any help. (Sorry, I can't seem to get my latex correct. The second to last term is intended to be phi evaluated from a to a. Also the title should read "Is this true for any scalar function")
Not sure what you mean when you say φ "depends on a single position only" but I think the answer is no. Consider φ = xy and C the boundary of the unit square.
$$\oint_C \phi_x\, dx = \oint_C y\, dx = -1$$
because only the top side of the square contributes a nonzero value.
LCKurtz said:
Not sure what you mean when you say φ "depends on a single position only"
I think he means that φ depends on a single parameter (e.g. only the x-coordinate).
Thank you both for the responses.
foxjwill said:
However, I'm not sure what you mean by
What point are you referring to?
Sorry, I should have been more clear. I intended to say that "a" is a point on the path. That was bad use of wording on my part.
LCKurtz said:
Not sure what you mean when you say φ "depends on a single position only" but I think the answer is no. Consider φ = xy and C the boundary of the unit square.
$$\oint_C \phi_x\, dx = \oint_C y\, dx = -1$$
because only the top side of the square contributes a nonzero value.
When I said that φ depends on a single position only, I meant to imply that it was something like a potential, or temperature distribution. I can see how the way I worded it is confusing.
After thinking about LCKurtz's scenario with the unit square I realize that I have totally butchered this question.
Here is what I should have done from the start:
## Homework Statement
Irrotational field theorem
Given that:
a.) $$\overline{\nabla}\times\overline{F}=0$$
b.) $$\oint{\overline{F}\cdot{d{\overline{l}}}=0$$
Show that a$$\rightarrow$$b
## Homework Equations
$$\overline{\nabla}\times\overline{F}=0\Leftrightarrow{\overline{F}}=\overline{\nabla}V$$
## The Attempt at a Solution
I know that Stoke's theorem will show this in an instant. I wanted to try and show it using the equations listed under relevant equations. My thought was that by writing the vector indicated by "F" as the gradient of some scalar, I could show that integrating along a closed path would give me a null result for any scalar chosen.
Well, you could use the fact that $\nabla V$ is a path-independent vector field.
foxjwill said:
Well, you could use the fact that $\nabla V$ is a path-independent vector field.
Hmm, does that mean it is as simple as saying:
$$\overline{\nabla}V{\bot}{d\overline{l}}$$
For constant "V"
Last edited:
Duh, I should have been thinking about the fundamental theorem of Calculus as well as the gradient theorem.
$$\int_{a}^{b}(\nabla{f})\cdot{d{\overline{l}}}=f(b)-f(a)$$ | 1,277 | 4,642 | {"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": 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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-50 | latest | en | 0.903452 |
http://list.seqfan.eu/pipermail/seqfan/2017-December/018144.html | 1,581,902,901,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141460.64/warc/CC-MAIN-20200217000519-20200217030519-00395.warc.gz | 92,006,022 | 2,762 | # [seqfan] Re: (no subject)
israel at math.ubc.ca israel at math.ubc.ca
Thu Dec 7 05:10:36 CET 2017
```Of course, we don't know whether the Fermat primes are finite or not.
We do know that all primes of the form 2^n+1, and more generally a^n+b^n
where one of a and b is even and the other odd,
must be of the quite special form a^(2^k)+b^(2^k),
because if n is divisible by an odd number d, a^n + b^n would be
divisible by a^(n/d) + b^(n/d).
But all this says about 1+2^a*3^b being prime is that gcd(a,b)
must be a power of 2 (including 1).
Cheers,
Robert
On Dec 6 2017, Frank Adams-Watters via SeqFan wrote:
> A005109 includes primes of the form 2^n + 1 (i.e., Fermat primes), while
> A058383 includes them.
>
>I is the finitude of the Fermat primes that led to wonder about these.
>
>
>
>-----Original Message-----
>From: Allan Wechsler <acwacw at gmail.com>
>To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>Sent: Wed, Dec 6, 2017 8:56 pm
>Subject: [seqfan] Re: (no subject)
>
>At the moment, I cannot reach OEIS. But coincidentally I ran into primes of
>this class in a completely different context, only yesterday. Wikipedia, at
>least, calls them "Pierpont primes", primes p such that p-1 is 3-smooth.
>The article https://en.wikipedia.org/wiki/Pierpont_prime attributes to
>Andrew Gleason the conjecture that there are infinitely many Pierpont
>primes. Curiously, the article refers to the sequence A005109, not A058383;
>I can't investigate this discrepancy until OEIS starts answering the phone
>for me.
>
>On Wed, Dec 6, 2017 at 1:45 PM, Hugo Pfoertner <yae9911 at gmail.com> wrote:
>> In the range 1<=a,b<=500 there are 2111 primes of this form. Large
>> examples: 1+(2^498)*(3^493) or 1+(2^994)*(3^993). Why should there be a
>> limit?
>>> Hugo Pfoertner>> On Wed, Dec 6, 2017 at 6:51 PM, Frank Adams-Watters
>>> via SeqFan <
>> seqfan at list.seqfan.eu> wrote:>
>> > Is https://oeis.org/A058383 (Primes of form 1+(2^a)*(3^b)) infinite?>
>> > > Franklin T. Adams-Watters> >> >> > --> > Seqfan Mailing list -
>> > http://list.seqfan.eu
>
>
>--
>Seqfan Mailing list - http://list.seqfan.eu/
>
>
``` | 679 | 2,131 | {"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-2020-10 | latest | en | 0.837985 |
https://www.esaral.com/q/the-correct-statement-regarding-the-given-ellingham-diagram-is-48765 | 1,726,650,383,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00199.warc.gz | 700,571,029 | 11,519 | # The correct statement regarding the given Ellingham diagram is:
Question:
The correct statement regarding the given Ellingham diagram is:
1. At $1400^{\circ} \mathrm{C}, \mathrm{Al}$ can be used for the extraction of $\mathrm{Zn}$ from $\mathrm{ZnO}$
2. At $500^{\circ} \mathrm{C}$, coke can be used for the extraction of $\mathrm{Zn}$ from $\mathrm{ZnO}$
3. Coke cannot be used for the extraction of $\mathrm{Cu}$ from $\mathrm{Cu}_{2} \mathrm{O}$.
4. At $800^{\circ} \mathrm{C}, \mathrm{Cu}$ can be used for the extraction of $\mathrm{Zn}$ from $\mathrm{ZnO}$
Correct Option: 1
Solution:
In the given Ellingham diagram, the metal which has a lower value of $\Delta G^{\circ}$ (more negative) can reduce a metal oxide whose curve lies above it. So, Al can reduce $\mathrm{ZnO}$ at $1400^{\circ} \mathrm{C}$. | 247 | 819 | {"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} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.68277 |
https://matsci.org/t/harmonic-bond-with-exponential-decay/30448 | 1,719,144,672,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00060.warc.gz | 330,528,545 | 4,834 | # harmonic bond with exponential decay
Dear LAMMPS users,
I will go straight to the point. I want to employ an angle_style similar to the harmonic one, but with the difference that it is multiplied by an exponential that depends on the bond-lengths (the so called harmonic with exponential decay):
E_ijk = K*(angle_ijk-angle_0)^2*exp(-a*r_ij)*exp(-a*r_jk),
where angle_ijk is the angle between the atoms i,j,k, r_ij and r_ jk are the distances between the i,j and j,k atoms respectively. I have been thinking a simple way to do so, but i haven’t found any. Do you know if there is a simple way to do it?
Dr. Jon Zubeltzu
Dear LAMMPS users,
I will go straight to the point. I want to employ an angle_style similar to the harmonic one, but with the difference that it is multiplied by an exponential that depends on the bond-lengths (the so called harmonic with exponential decay):
E_ijk = K*(angle_ijk-angle_0)^2exp(-ar_ij)exp(-ar_jk),
where angle_ijk is the angle between the atoms i,j,k, r_ij and r_ jk are the distances between the i,j and j,k atoms respectively. I have been thinking a simple way to do so, but i haven’t found any. Do you know if there is a simple way to do it?
the simplest way i can imagine would be to add a new custom angle style, that is created as an extension to angle/harmonic; i.e. you make a copy of the .h and .cpp files under a new name, change the name of the style and the class and then add handling for the additional parameter and add the necessary changes to the compute() method. http://lammps.sandia.gov/doc/Section_modify.html
axel.
Thank you for the suggestion, it is indeed quite simple and it worked!
Jon
Thank you for the suggestion, it is indeed quite simple and it worked!
good to hear. please consider submitting your new angle style for inclusion to LAMMPS, best through a pull request on github (a tutorial for the process is given in the manual).
axel. | 477 | 1,922 | {"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-2024-26 | latest | en | 0.879437 |
bfcsa.co.uk | 1,656,119,179,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00572.warc.gz | 189,559,760 | 10,057 | How to Increase the Chance of Causing a Poker Combination
Having a combination with a wider distribution of cards in a poker game makes it less likely that the person would have a winning hand.
1. Two pair, for example, have a greater probability of winning and therefore being attractive to the person than pairs of three or two. There are only about thirty eight percent chances of having a straight: it is less likely to come out in a poker game than it is to come out in the regular lotto game.
2. Basically, the main reason why a person would have a combination with a wider distribution of cards rather than use the exact, or preferred, order of cards is for a more even split of the money in case the winning number is drawn in the last part of the game. Normally, people would not be choosing a sequence or a combination as it is rather cheesy and less organized. With the possible exceptions of the royal couple, and their respective first names, and the birthdays of the children.
3. The Texas Lotto and the Elottery are somewhat different in terms of the probabilities and the requirements for winning since they are the only lotteries that provide the annual chance to win a multiple prize, rather than the oh so rarely seen Mega million.
The Elottery comes with a fairly wide range of numbers to choose from in addition to the standard six from fifty three to ninety.
• This means that the odds have become slightly less in the game of probability since the numbers are somewhat inclusionist in nature and less likely to be drawn in the traditional way.
• To increase the chances of winning in the Elottery one must either have a gift for quick number selection or an excellent memory which can keep a record of the numbers that have been played. This is not impossible with the Elottery program being created by the University of acronym, which has been in use for the teaching of Texas Hold’em poker. The learning of how to win a Texas Hold ’em poker using the Texas Lotto method is, in my opinion, for the best.
You would still need to have a bit of luck to ensure your winnings and to enable you to play the game wherever in the world you live.
Although, you could try and buy a few Quick Pick tickets, which is the tickets used for the Virginia Cash lottery. If you try and select your own number combination using the Quick Pick number generator, you have a one in 175,711,536 chance of winning the prize.
The Elottery website also lists other lotto friendly games which may be of interest to you. A few of these games include Millionaire Bet etc, which offers a decent prize for the winner. Although the winnings may not be as high as the Mega millions jackpot, it can still offer you a good return for your money. In addition, some of the Elottery websites offer the visitor a chance to play the instant win game upon registration where one can try to win any of the available prizes apart from the jackpot.
The Foldless – The UK national man against the lottery syndicate game The Elottery website has a feature where visitors can fold their losing tickets and keep the ticket. | 641 | 3,107 | {"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-2022-27 | latest | en | 0.972453 |
https://www.jiskha.com/display.cgi?id=1317075662 | 1,516,533,208,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00387.warc.gz | 932,814,985 | 3,794 | posted by .
An airplane is flying south at 500 km/hr and is being blown by a crosswind of 75 km/hr east. What is the resultant velocity of the airplane?
• Physics help ASAP please -
the angle E of S is arc tan 75/500. determine that angle
sintheta=75/speed
solve for speed
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http://www.arthouse-kuklite.com/ni61y753l/lW15645jA/ | 1,611,605,585,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00602.warc.gz | 123,712,452 | 16,203 | Reading Worksheets: Printable Rules For Toddlers Heartwork Wrinkle In Time Worksheets Free Reading Grade Math Frame Math Worksheets Fraction Of Trig Solver Algebra Practice Problems Printable Rules For. Math Problems For A 1first Grader kids worksheet review quiz Find The Missing Number Multiplication Worksheet | Arthouse-kuklite
# Printable Rules For Toddlers Heartwork Wrinkle In Time Worksheets Free Reading Grade Math Frame Math Worksheets Fraction Of Trig Solver Algebra Practice Problems Printable Rules For
Published at Saturday, 19 September 2020. Reading Worksheets. By .
Here are some other sharing subjects to explore: Take Away Store - hot chips (35 per cup), milk shakes (3/4 cup of milk per shake) - how much for 26 people? Concert -toilet paper - 10 sheets per person, 100 sheets per roll, 3 000 people attending- how many rolls? Dog Shelter - 1 cup of biscuits large dog, 3/4 cup for medium dog etc - how much for 10 big, 5 medium etc? Jelly bean competition - Big jar took 12 bags of jelly beans - 124 beans in 1 bag - how many beans in total? These are a great way to explore a range of math skills. Choose one subject and choose questions that relate to that. An obvious one is school: At Collins State School there are 23 students in Kindergarten, 24 in Grade 1, 23 in Grade 2, 18 in Grade 3, 22 in Grade 4, 17 in Grade 5, 19 in Grade 6 and 22 in Grade 7. How many students are there in the whole school?
The next step is learning to write numbers, and this is where mathematics worksheets become almost a necessity. Unless you have great handwriting, lots of spare time and a fair amount of patience, writing worksheets will help you teach this valuable skill to your child. Dot-to-dot, tracing, following the lines and other writing exercises will help your child learn how to write numbers. A good set of worksheets will include practice sheets with various methods to help your child learn to write numbers.
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https://www.nagwa.com/en/videos/832104954810/ | 1,718,685,599,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00244.warc.gz | 806,651,552 | 37,145 | Question Video: Finding the Moment of the Couple Resulting from Forces Acting on an Equilateral Triangle | Nagwa Question Video: Finding the Moment of the Couple Resulting from Forces Acting on an Equilateral Triangle | Nagwa
# Question Video: Finding the Moment of the Couple Resulting from Forces Acting on an Equilateral Triangle Mathematics • Third Year of Secondary School
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In a triangle 𝐴𝐵𝐶, 𝐴𝐵 = 𝐵𝐶 = 32 cm, and 𝑚∠𝐵 = 120°. Forces of magnitudes 2, 2, and 2√3 newtons are acting at 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴, respectively. If the system is equivalent to a couple, find the magnitude of its moment considering the positive direction is 𝐴𝐵𝐶.
05:08
### Video Transcript
In a triangle 𝐴𝐵𝐶, 𝐴𝐵 equals 𝐵𝐶 equals 32 centimeters, and the measure of angle 𝐵 is 120 degrees. Forces of magnitudes two, two, and two root three newtons are acting at 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴, respectively. If the system is equivalent to a couple, find the magnitude of its moment considering the positive direction is 𝐴𝐵𝐶.
Okay, so let’s say that this is our triangle 𝐴𝐵𝐶, and we’re told here that the measure of angle 𝐵 is 120 degrees. And we also know that the lengths of the two sides 𝐴𝐵 and 𝐵𝐶 are the same. That means we’re working with an isosceles triangle, and therefore this interior angle is the same as this one. If we call that angle 𝜃, we can say that two times 𝜃 plus 120 degrees equals 180 degrees. If we solve this equation for 𝜃, we find it equals 30 degrees. Knowing then about the geometry of this triangle, we’re also told about forces that act along its sides. There’s a two-newton force acting from 𝐴 to 𝐵, an identical two-newton force from 𝐵 to 𝐶, and then from 𝐶 to 𝐴, a force of two root three newtons.
We’re told that this system of forces is equivalent to a couple. That means that the total force on this triangle is zero. However, there is a moment that’s created about its center. And in fact, it’s the magnitude of that moment that we want to solve for. To start doing that, let’s clear a bit of space. And to start analyzing the forces involved here, we’ll set up a coordinate system. We’ll say that corner 𝐴 in our triangle is at the origin of this coordinate frame, and the 𝑦-axis moves vertically upward and the 𝑥 horizontally to the right.
Now, as we’ve seen, there are three forces that are acting on this triangle. One of them acts purely in the 𝑥-direction. But the other two forces, these two newton forces on the other sides, can be divided up into their vertical and horizontal components. We’re saying here that this two-newton force effectively originates at this point, at the origin, while this other two-newton force effectively originates at point 𝐶. It’s by choosing these two points of origin, we could call them 𝐴 and 𝐶 in our triangle, that we can model all the forces involved as a couple.
When we go to calculate the components of these two two-newton forces, we can recall that in a right triangle, where one of the other interior angles is called 𝜃, the sine of this angle is given by the ratio of the opposite side length to the hypotenuse length, while the cosine equals the adjacent side length over the hypotenuse. If we consider then all the horizontal forces acting on our triangle, first, there’s this component of our one two-newton force. That component equals two times the cos of 30 degrees. Secondly, there’s this component of the other two-newton force. That’s also equal to two times the cos of 30 degrees.
Lastly, though, there’s this force acting from point 𝐶 to point 𝐴 in our triangle. Based on the way we’ve set up our 𝑥-axis, this force will be negative. If we recall that the cos of 30 degrees is equal to the square root of three over two, then we get two root three over two plus two root three over two minus two root three. And as we expect, this equals zero. The important thing for our purposes, though, is that all of these horizontal force components are along the same line of action. This means that since they sum to zero, they apply no moment about the center of our triangle. In other words, the horizontal force components don’t contribute to that moment.
Next, let’s consider the vertical components of the forces involved here. First, we see this vertical component of what we can call our first two-newton force. That will be equal to two times the sin of 30 degrees. And then there’s our second vertical component over here. That equals negative two times the sin of 30 degrees. So as we expect, the net force in the vertical direction is zero. But note that these two forces don’t act along the same line of action. Therefore, they will apply a moment to our triangle, and its magnitude will depend on the perpendicular distance between this dashed line that goes through the center of our triangle and the lines of action of our two vertical forces. We can call that perpendicular distance 𝑑. And we see it’s equal to one-half the length of the longest side of our triangle.
Recalling that each of the shorter sides has a length of 32 centimeters, we can say that the distance 𝑑 equals 32 times the cos of 30 degrees. That’s 32 times the square root of three over two or simply 16 root three. We’re now ready to calculate the magnitude of the moment created by our vertical forces. Each of those two forces has a magnitude of two newtons times the sin of 30 degrees. And we multiply this by the perpendicular distance between the lines of action of those forces and the axis of rotation of our shape. Altogether, this moment equals 32 times the square root of three. And as for the units, our forces have units of newtons and our distances have units of centimeters. The magnitude of the moment created by this system of forces equals 32 times the square root of three newton centimeters.
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• Realistic Exam Questions | 1,472 | 5,980 | {"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} | 4.65625 | 5 | CC-MAIN-2024-26 | latest | en | 0.915503 |
https://boards.straightdope.com/t/what-do-i-get-for-my-kilowatt/3908 | 1,606,427,873,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00661.warc.gz | 216,274,045 | 9,235 | # What Do I Get For My Kilowatt?
I’m staring at a record of electricity bills over the course of two years, and while it tells me how many kilowatts of electricity were used in each two month period of time, I have no idea how to translate that into any kind of day-to-day practical usage.
So I’ve got a couple of questions. First, what do I get for my kilowatt? In other words, will one kilowatt run my refrigerator for an hour/day/week? How 'bout lights and tv?
And second, has anybody ever figured out what the average kilowatt use per person is? I know about average water usage, because I live in drought country, and there’s always a big discussion about the average number of gallons used by a person daily. Is there anyplace to find out that same information for electricity?
Thanks!
-Melin
Good utilities will sometimes be able to answer these questions for you. They’ll be able to tell you things like, unless you have a really large house, a space heater won’t save any power since space heaters are so much less efficient than central heating.
If there is a phone number on your power bill, you might give it a call and ask them how many kilowatt-hours it takes to make a slice of toast or take a hot bath. If you have electric heating, they can tell you about how much it saves to improve insulation and caulking.
By the way, kilowatt is a measure of power. When you turn something on, it will run at a certain number of watts which probably won’t change over time. You might also want to ask about energy, usually measured in kilowatt-hours, but it also could be measured in kilowatt-seconds, calories, Calories, etc. So kilowatts are analogous to the speed of a car; kilowatt-hours are analogous to the distance the car has travelled. So you’d ask “How many watts does my fridge consume?” but you might also ask “How many watt-hours does my toaster consume to make a piece of toast.”
A 100w bulb will use a kilowatt in 10 hours. Your 600w baby microwave will use a kilowatt after 1 hour and 40 minutes. You should be able to find the wattage for your stove, oven, refrigerator, washer, dryer (BIG TIME, if electric), and dishwasher somewhere on each appliance, (same with TV, stereo, VCR, etc.). Divide wattage into 1000 to find usage per kilowatt hour.
Aside from direct heat (clothes dryer, stove), the biggest users of electricity are usually motors (fans, etc.). Lights have gotten pretty economical.
I don’t know what “typical” usage is either by household or by person. (My house (electric dryer, lots of heat lamps and hot rocks for reptilian pets) uses way more than I like.)
Tom~
A watt is a measure of rate of energy usage.
It is equal to one “joule” per second. A joule (J) is a quantity of energy whose definition escapes me at the moment.
Thus, my hair dryer, which says 1400 watts, uses electricity at a rate of 1400 joules per second.
Total quantity of energy is something different. This is where the joule comes in, and where we can start to answer such questions as, “How much energy does my toaster use to toast two slices of bread?”.
If the toaster used energy at a rate of 500 watts for two minutes, it would be 500 W x 120 s = 60 000 joules!
Most North Americans are not used to seeing their energy measured in joules; watt-hours are more common.
But a watt-hour is simply the total amount of energy you use when you use energy at a rate of 1 joule per second for 1 hour.
This equals 1 J/s x 1 h, or 1 J/s x 3600 s, or… 3600 joules!
So the toaster would have used 60 000 joules / 3600 joules/watt-hour = 16.67 watt-hours of electricity. Boris B, you’re right on the mark.
So, tomndebb, a 100-W bulb uses energy at a rate of 100 watts, or 100 joules per second. In one hour, it uses a total of 1 watt-hour, or 3600 joules, of energy.
With a single 100-W bulb, after ten hours, you would still be using energy at the same rate of 100 watts, but you would have used 10 watt-hours, or 36 000 joules, of energy.
It would take ten bulbs to use energy at the rate of 1000 watts, and after just one hour, they would have used 10 watt-hours of energy.
But you’re right–anything that uses electricity to heat (dryer, toaster, etc) will take a lot of power. Some friends of mine had to move out of an electrically-heated house; they literally could not afford the bills (~\$600 per MONTH in the winter).
Those of us who are planning a solar-powered house have to be very aware of this kind of thing…
A kilowatt hour is 1000 watts of power used in one hour. The 100 watt bulb thing in Sunspace’s post.
For the science elete out there. Please convert this into how many electrons per hour Melin gets to use for her money.
Remember the electrons are only barrowed and have to be returned to the power company through the ground.
Unless they are AC, in which case you pay for the same electrons over and over again. We did this in Great debates, where someone did not believe in electricity.
I appreciate all this, guys, but it’s the practical stuff I need, not the academic. I have surfed manny websites involving electricity today, and still can’t figure out what it is I need to know, although I’m getting closer.
What it is, is that I have a case where the question turns on whether this woman and her child were living in an apartment, or living at home with her parents. She admits that she kept the apartment, but says that she used it “to be alone” maybe 10 per cent of the time, spending the remaining 90 per cent at mom and dad’s. I wanna know if the usage I’m seeing on her electricity bills for the apartment is consistent with that, or is consistent with them living at the place (one bedroom apartment) on a regular basis.
The other stuff is more interesting, I’ll admit that!
-Melin
Have you asked the landlord yet? He should have at least a general idea of the utility costs, because people ask about that when they’re apartment-hunting. Also, the landlord is responsible for utility payments while the apartment is vacant, so he might have an idea of how much electricity the place uses up with no people in it.
If that fails, try calling the power company. They can probably tell you the average energy usage for an apartment of that size. Depending on how much statistical data they keep, they might also know how much the usage will drop when the place isn’t occupied all the time.
Laugh hard; it’s a long way to the bank.
I’m not about to challenge Sunspace on the technical end of this question. I had assumed that you wanted to know how you would be billed.
Based on the following quote from the U.S. Department of Energy, I think my calculations come close.
http://www.eia.doe.gov/emeu/lighting/appxb.html#begin
I would think that if it is an apartment, the landlord could put you in contact with other tenants to find out what their bills are. If it is a house or a set of rooms in a house, this would be harder to establish a baseline. (Can you get/subpoena the utility to provide their bills for the period when the woman was living there vs when she was only “going there to be alone”?) Heck, is the woman willing to provide her bills for the before and after periods?
Old appliances might mask her presence (old inefficient heater+fridge+water heater could consume enough energy, running constantly, to overwhelm lightbulbs and stereos), but if you have the wattage for those appliances, you can figure their max, subtract it, and see how many lights she had on “when she wasn’t home.”
I suspect you will be safer getting the bills for the people next door.
Tom~
Just some rough figures but maybe this will help. If someone were actually living in an apartment they would use:
The television, microwave, dishwasher, stove, lights, hot water heater, radio or stereo, hair dryer.
Other appliances, like the refrigerator and the heat would probably be on whether they were actually there or not, which is unfortunate because they’re a couple of biggies. The heat would probably be turned down and the fridge wouldn’t be opening and closing so there would be some savings, tho.
On a typical day (with guesses for wattage) I’d say the above list would consume – ummm – about 8 to 10 kilowatt-hours per day. Assuming it’s summertime so that electric heat isn’t a factor, and (guessing) maybe 5 kw-hrs a day for the fridge.
If these figures are close then an empty apartment (just the fridge) should use 30 x 5 = 150 kw-hrs. An apartment occupied 10% of the time would use an additional 3 x 10 = 30 kw-hrs for a total of 180 kw-hrs. An apartment occupied full time would use an additional 30 x 10 = 300 kw-hrs for a total of 450 kw-hrs.
I’m nowhere near my power bill so I have no way to check these figures but it might give you some idea about whether or not you have a chance of figuring out whether this person is truth-telling or not.
p.s. I learned to estimate from my heat transfer professor for whom everything was either 1, 2, or 5, or some decimal multiple thereof. Pi, for example, was 5 in his class.
Strangers have the best candy.
Melin:
Did the person in question ever live with her child in the apartment full time before the time she claimed to live with her parents?
If so, couldn’t you just subpoena her power bills from the power company over a period of time both before and after she claimed to have moved out and compare?
Plunging like stones from a slingshot on Mars.
Melin, I think the place you want to go is a site run by SCE Corp. (The old So. Cal. Energy).
## Try here and, if you don’t mind entering your client’s data into a public database, you should get a pretty good idea of what was happening in the apartment.
Livin’ on Tums, Vitamin E and Rogaine
A joule is the amount of energy required to provide one Newton through a length of one meter.
A Newton is the force required to accelerate one kilogram at a rate of one meter per second per second.
That said, I think the electric company is ripping me off because my refrigerator just sits there all day. And don’t get me started that television.
Then…what are Newton’s Joules?
<<snicker>>
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2018-07-06, 19:38 #1 carpetpool "Sam" Nov 2016 2×3×53 Posts OEIS A213052 for primes p = 3 mod 4 The OEIS sequence A213052 is a sequence of primes p such that all primes less than the nth prime are primitive roots mod p. Except for 3, all the primes in the sequence are congruent to 5 mod 24 and therefore congruent to 1 mod 4. What would make this an interesting problem is finding the sequence of smallest primes p = 3 mod 4 for which all primes less than the nth prime are primitive roots mod p. Also, we could consider a sequence of smallest primes p = 3 mod 4 such that for all primes q less than the nth prime, -q is a primitive root mod p. For the first case they are: 3, 19, 907, 907, 2083, 101467, 101467, 350443, and the second case: 7, 23, 239, 479, 479, 1559, 9239, 10559, Is anyone able to find more terms for any of the two sequences?
2018-07-11, 22:27 #2 danaj "Dana Jacobsen" Feb 2011 Bangkok, TH 2·3·151 Posts Using a simple modification of the Perl script on the OEIS page (just add the $p % 4 == 3 clause), I get: 3, 19, 907, 1747, 2083, 101467, 350443, 916507, 1014787, 6603283, 27068563, 45287587, 226432243, 243060283, 3946895803, 5571195667, 9259384843 The Perl code is about 7x faster than Pari on my computer, and got the above in 10 minutes. For your second sequence: 7, 23, 239, 479, 1319, 1559, 9239, 10559, 35279, 250799, 422231, 701399, 4080359, 13147679, 13518119, 48796439, 94123559, 102628679, 120293879, 149013479, 688333799, 1595386679, 2929911599 Note that A213052 indicates it is an increasing sequence of primes, hence we don't have repeats. To get the repeats change the 'if' to a 'while', which gives 479 instead of 1319; 4080359, 120293879, and 2929911599 twice each. 2018-07-14, 17:00 #3 carpetpool "Sam" Nov 2016 2·3·53 Posts Quote: Originally Posted by danaj Using a simple modification of the Perl script on the OEIS page (just add the$p % 4 == 3 clause), I get: 3, 19, 907, 1747, 2083, 101467, 350443, 916507, 1014787, 6603283, 27068563, 45287587, 226432243, 243060283, 3946895803, 5571195667, 9259384843 The Perl code is about 7x faster than Pari on my computer, and got the above in 10 minutes. For your second sequence: 7, 23, 239, 479, 1319, 1559, 9239, 10559, 35279, 250799, 422231, 701399, 4080359, 13147679, 13518119, 48796439, 94123559, 102628679, 120293879, 149013479, 688333799, 1595386679, 2929911599 Note that A213052 indicates it is an increasing sequence of primes, hence we don't have repeats. To get the repeats change the 'if' to a 'while', which gives 479 instead of 1319; 4080359, 120293879, and 2929911599 twice each.
How did I miss 1747 in the first (and more important) case?
Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Miscellaneous Math 9 2017-03-17 22:57 ewmayer Probability & Probabilistic Number Theory 6 2015-11-10 16:33 T.Rex Miscellaneous Math 40 2015-09-15 14:01 T.Rex Miscellaneous Math 38 2015-09-05 16:14 T.Rex Miscellaneous Math 7 2015-08-28 18:04
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February 7, 2016
# Search: Biology (multiple choice!)
Number of results: 10,492
Math
what is the solution to the system of equations? 5x + 4y = 8 x - 2y = 10 (not multiple choice) my answer: (-5, -1) What are the missing values? (coincident) 4x + 5y = 8 8x + ____ y = _____ (Not multiple choice) Im not sure how to do this one collection of dimes and nickels is ...
November 19, 2014 by Anonymous
Math
14+2m = 4m - 16 (no multiple choice) I don't understand it 4(x +2) = 14 - 2(3 - 4x) (no multiple choice) Eight subtracted from 4 times a number is equal to 4 less than 6 times the number. (not multiple choice) my answer is 8x - 4 = 4x - 6
September 22, 2014 by Anonymous
Math
12+4n/3 = 8 Not multiple choice.. trying to find n. My answer is 7 9a - 3(a -6) = -6 Not multiple choice.. I'm not quite sure how to do this The width of a rectangle is 6 in. less than its length. The perimeter is 68 in. What is the width of the rectangle? Not multiple choice...
September 17, 2014 by Anonymous
Math
Each day Julia's math Teacher gives class a warm up qs. It's true/false 20% of the times and multiple choice 80% . Julia gets 70% on true false correct and 90% on multiple choice correct. Julia answered today's question correctly. What's the probability that it was a multiple ...
April 15, 2015 by Jason
Math (Algebra)
Gretchen and Ezia received equal scores on a test made up of multiple choice questions and an essay. Gretchen got 18 multiple choice questions correct and received 19 points for her essay. Ezia got 15 multiple choice questions correct and received 31 points for her essay. How ...
October 6, 2015 by Anonymous
algbrea
Francine and Cheryl received equal scores on a test made up of multiple choice questions and an essay. Francine got 34 multiple choice questions correct and received 14 points for her essay. Cheryl got 30 multiple choice questions correct and received 22 points for her essay...
September 29, 2014 by jace
math
Louann and Carla received equal scores on a test made up of multiple choice questions and an essay. Louann got 14 multiple choice questions correct and received 24 points for her essay. Carla got 16 multiple choice questions correct and received 14 points for her essay. How ...
October 8, 2015 by jessica
math
Francine and Cheryl received equal scores on a test made up of multiple choice questions and an essay. Francine got 34 multiple choice questions correct and received 14 points for her essay. Cheryl got 30 multiple choice questions correct and received 22 points for her essay. ...
December 7, 2015 by bre
Math
a 10 question multiple choice test has 4 possible answer for each question. A student selects at least 6 correct answers. Find probability of this event: (multiple choice) A) .118 B) .995 C) .068 D) .571
December 13, 2011 by Matt Borchelt
Math
The width of a rectangle is 6 in. less than its length. The perimeter is 68 in. What is the width of the rectangle? (not multiple choice) my answer is 14 5(a-4) - 8a = 55 (not multiple choice) my answer -25 2n-7/3 = 15 (not multiple choice) my answer is 6
September 19, 2014 by Anonymous
Math
The width of a rectangle is 6 in. less than its length. The perimeter is 68 in. What is the width of the rectangle? (not multiple choice) my answer is 14 5(a-4) - 8a = 55 (not multiple choice) my answer -25 2n-7/3 = 15 (not multiple choice) my answer is 6
September 19, 2014 by Anonymous
Algebra
An algebra teacher is writing a test. There will be two types of questions: multiple-choice and open-ended. The multiple-choice questions take 3 minutes to complete and the open-ended questions take 12 minutes to complete. There must be at least 3 open-ended questions, at ...
September 8, 2015 by Charlie
Stats
An instructor gives a test with 20 multiple choice questions. There are 4 responses per questions and only one choice correct. The test also has 30 True/False questions. Determine the probability a student randomly guessing on all questions and correctly answering no more than...
March 31, 2011 by Lacey
Biology
How do you calculate the magnification on a microscope? Please help, I do not have any multiple choice to go by, I have to write a response!
September 18, 2014 by Skye
Biology
How do you calculate the magnification on a microscope? Please help, I do not have any multiple choice to go by, I have to write a response!
September 18, 2014 by Sky
AP stuff
"Come to... classroom after school (the possible dates will be announced later in April). There you will naswer 20 multiple choice questions based on the text, as well as AP style multiple choice questions. You will also answer 2 openended questions both of which will be ...
May 18, 2008 by Me
Maths
Multiple-choice questions in a test are graded by adding 2 marks for each correct response and subtracting 1 mark for each incorrect response (including no response). Rory and Jenny answered all the multiple-choice questions, with Rory scoring 27 and Jenny scoring 42. Rory ...
November 17, 2010 by Jannie
Stats
[1.5%] ANOVA: Three groups of 5 students were tested in their ability to correctly answer a 10-question quiz under different formats. Group 1 had a True/False quiz, Group 2 had a Fill-In quiz, and Group 3 had a Multiple Choice quiz. Their scores were as follows: True/False ...
May 17, 2010 by TJohnson
Math
What is the probability of rolling a number cube and getting an even number or a multiple of 3? I got 4/6 as the probability, but that isn't a choice. 6 is a multiple of 3 and an even number, so if I count 6 once, then I get 4/6 as the probability. But if I count 6 twice (for ...
March 30, 2011 by Sammy
biology help
Pick one of the multiple choice that best matches the answer. For grasshoppers and locusts to be in the same family they must also be in the same a. order b. group c. genus d. species
May 18, 2010 by kim
biology-multiple choice question
In order to survive, all organisms must carry out: 1. autotrophic nutrition 2. heterotrophic nutrition 3. enzyme-controlled reactions 4. the process of locomotion Help??
February 18, 2008 by Nicci
Biology (multiple choice!)
Which metabolic category best describes Bacteria that act as decomposers? A) Chemoautotrophy B) Chemoheterotrophy C) Photoautotrophy D) Photoheterotrophy E) All of these categories apply equally well to Bacterial decomposers.
October 4, 2010 by Sarah
Biology (multiple choice!)
In a symbiotic relationship between organisms of significantly different sizes, the larger organism is called the ___________, and the smaller organism is called the ___________________.
October 4, 2010 by Dawn
Biology - Multiple Choice
1) The net gain of ATP after glycolysis is .. A) two B) four C) 34 D) 36 2) During photosynthesis water .. A) is made B) breaks down C) is not necessary 3) The gas produced as a result of ethyl alcohol fermentation is .. A) CO2 B) CO C) O2 D) H2O(g)
December 27, 2007 by Mae
Biology multiple choice question
If a randomly-mating population has an established frequency of 64% for the organisms homozygous dominant for a given trait, what is the frequency of this dominant allele in the gene pool? A. .2 B. .24 C. .36 D. .8 E. .64 I think the answer is D. .8
March 4, 2012 by Lisa
SOLVE -MATH
A physics exam consists of 9 multiple choice questions and 6 open-ended problems in which all work must be shown.If an examinee must answer 6 of the multiple- choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen?
April 30, 2015 by JORDAN
Math
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee must answer 5 of the multiple-choice questions and 2 of the open-ended problems, in how many ways can the questions and problems be chosen?
March 29, 2014 by Anonymous
MATH
A physics exam consists of 9 multiple choice questions and 6 open-ended problems in which all work must be shown.If an examinee must answer 6 of the multiple- choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen?
April 30, 2015 by STAR ---- HELP.....
math
What are the important variables in the problem below? A test is worth 50 points. Multiple-choice questions are worth 1 point, and short-answer questions are worth 3 points. If the test has 20 questions, how many multiple-choice questions are there?
September 28, 2015 by aaron
Math
1)At 1 p.m., Lin records an outside temperature of –14.8°F. At 3 p.m... she records the outside temperature and notices the temperature had increased by 22.6 degrees. What was the temperature that she recorded at 3 p.m.? ( no multiple choice) *my answer is 7.8* 2) (x - 16) + 5...
September 16, 2014 by Anonymous
math
Is this the right way to answer this question? A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problems, in how many ways can...
August 23, 2009 by sam
Math (Ms. Sue! Anyone!)
1)At 1 p.m., Lin records an outside temperature of –14.8°F. At 3 p.m... she records the outside temperature and notices the temperature had increased by 22.6 degrees. What was the temperature that she recorded at 3 p.m.? ( no multiple choice) *confused on this question* 2) (x...
September 15, 2014 by Anonymous
math
1)At 1 p.m., Lin records an outside temperature of –14.8°F. At 3 p.m... she records the outside temperature and notices the temperature had increased by 22.6 degrees. What was the temperature that she recorded at 3 p.m.? ( no multiple choice) *confused on this question* 2) (x...
September 15, 2014 by Anonymous
algebra
Pam scored 78 on a test that had 4 fill-in questions worth 7 points each and 24 multiple choice questions worth 3 points each. She had one fill-in question wrong.How many multiple choice questions did Pam get right?
July 5, 2011 by Poochie
Finite Math
A biology quiz consists of 10 multiple -choice questions. 7 must be answered correctly to receive a passing grade. If each question has 4 possible answers, of which only one is correct, what is the probability that a student who guesses at random on each question will pass the...
October 8, 2012 by Toni
Biology-Ecology
Multiple choice: For a natural ecosystem to be self-sustaining, many compounds must be: 1. converted into new forms of energy. 2. cycled between organisms and the environment. 3. permanently removed from the environment. 4. changed into fossil fuels such as oil and coal
March 16, 2013 by Amy
Algebra
A history test is to have 20 questions in which some are multiple choice and some are essay. The multiple choice questions are worth 4 points each and the essay questions are worth 8 points each. if the test has a total of 100 points, how many of each type of question is on ...
September 17, 2011 by Kay
math (stats)
A biology professor gives a surprise quiz consist of a true/false question followed by a multiple choice question with four possible answers. If both questions are answered with random guesses, find the probability that both responses are correct. Please show your work.
January 23, 2012 by juan777
biology (multiple choice!)
The death by bubonic plague of about 1/3 of Europe's population during the fourteenth century is a good example of: A. Abiotic factors limiting population size B. A density dependent effect C. A time lag D. A density-independent effect E. Carrying capacity
September 22, 2010 by Sarah
Algebra
A test has 15 questions worth 100 points in total. The test consists of multiple choice questions, which are 4 points each, and open-response questions, which are worth 12 points each. How many multiple choice questions are on the test?
January 12, 2016 by Anonymous
Binomial Distribution
A biology quiz consists of twelve multiple-choice questions. Eight must be answered correctly to receive a passing grade. If each question has five possible answers, of which only one is correct, what is the probability that a student who guesses at random on each question ...
April 18, 2012 by Price
Biology
What is the difference between multiple gene inheritance and multiple allele inheritance
October 23, 2011 by Kestrel
Biology-Ecology
Multiple choice What is the ultimate source of energy for most ecosystems, and in what form does the energy leave the ecosystem? 1. plants; animals 2. light; heat 3. plants; heat 4. producers; consumers 5. heat; light
March 16, 2013 by Amy
Statistics help?
Hi there, I was looking for any assistance or help on how I would go about starting this assignment? Having difficulty understanding it. Thank you in advance. "A stats instructor is interested in investigating time of day his lectures were held on student' test scores. He ...
November 26, 2012 by Evan
Statistics on Psychology
Hi there, I was looking for any assistance or help on how I would go about starting this assignment? Having difficulty understanding it. Thank you in advance. "A stats instructor is interested in investigating time of day his lectures were held on student' test scores. He ...
November 24, 2012 by Evan
English for john browm
Let me give you something on "multiple-choice" items that may help you to guess better. Many teachers prepare multiple-choice tests because they are easy to mark and offer the student the advantage of actually seeing what the correct answer is. There are important strategies ...
May 7, 2009 by SraJMcGin
Biology (multiple choice!)
I'm studying for a test and can't find the answers for the following questions...please help! What clade of bacteria produce oxygen via photosynthesis? What is a synapomorphy for domain Bacteria? In a symbiotic relationship between organisms of significantly different sizes, ...
October 25, 2010 by Sarah
math
Part A of your history test has 15 multiple choice questions. Each question has four choices. Part B has 10 true/false questions. How many ways are there to answer the 15 multiple choice questions? How many ways are there to answer the 10 true or questions? How many are there ...
February 15, 2011 by John
Biology (To SraJMcGin or anyone who can help)
h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 7 8 2 7 9 5 7 9 It's a webadress cant post links put it in your adress bar and delete the spaces then press enter. Thanks for the responses! By the way believe it or not those are actually from the ...
July 4, 2010 by Kate
Writing Tests
Lately, for the past couple of tests I've written, I suddenly blank out and can't recall any information... Normally I've been fine, but writing tests has suddenly become more difficult for me. I do study a lot, though - reviewing notes and doing questions. Perhaps I'm not ...
March 7, 2008 by Lucy
biology
Pick one of the multiple choice that best matches the answer. For grasshoppers and locusts to be in the same family they must also be in the same a. order b. group c. genus d. species two members of the same ___ would be most clostly related. a. order b. group c. genus d. species
May 13, 2010 by kate
Environmental Science
Can someone please like help me? :') Mack, what is your question? We will not give you the answer but I will try to find you sources where you can find it. I checked your previous multiple choices posts. It is not possible to say if they are correct (best answer), as you ...
April 21, 2007 by Mack
Civics
Question 1 (Multiple Choice Worth 3 points) Which approach to foreign policy involves sending ambassadors to other countries? Isolation Diplomacy Intervention Imperialism Question 2 (Multiple Choice Worth 3 points) Which of the following is true about foreign and domestic ...
July 24, 2014 by Dariusbd
algebra
Given D=5, E=10 and F=4, evaluate D E/F Multiple choice 1/2 2 25/2
June 13, 2011 by jan
algebra 1
hi. i'm confused so please help. thanks. 9 - x/6 = 7(x+1)/13 multiple choice a. 6 b. 13 c. 11 d. 12 thank you again.
November 28, 2010 by Anonymous
PHYSICS
Write 5 multiple-choice exam questions on any topics of your choice. A good question should be of sufficient difficulty and perhaps have a good background story? Try to change questions you've seen previously? Post your favorite question to the blackboard discussion board
July 9, 2015 by GEORGE
geometry
User: Courtney Thompson In Course: Geometry V8 ( 2167) Instructor: Tracie Morgan -------------------------------------------------------------------------------- WARNING: You must not leave this exam form! If you try to click back into this exam again prior to submitting, ...
December 31, 2008 by courtney thompson
Math
24. Multiple Choice. For which pair is the greatest common factor 8? A. 2 and 4 B. 7 and 15 C. 32 and 64 D. 56 and 72
September 18, 2012 by Gracie
Elementary algebra multiple choice
2x^2+3xy-4y^2= (A) 9x^2-4y^2 (B)9x^2+4y^2 (C)9x^2+4y^2-6xy (D)9x^2+4y^2-12xy
April 14, 2013 by Hipolito
Math
What is the value of w? 5( 9 - 3) = w x 9 - 5 x 3 I think the answer is 6 because 9-3 equals 6. *it's not multiple choice*
August 22, 2014 by Nay
math
Beth Dahlke is taking a ten question multiple choice test for which each question has three answer choice only one of which is correct. Beth decides on answers by rolling a fair die and making the first answer choice if the die shows 1 or 2 the second if it showes 3 or 4 and ...
November 19, 2012 by Katarzyna
math
Beth Dahlke is taking a ten question multiple choice test for which each question has three answer choice only one of which is correct. Beth decides on answers by rolling a fair die and making the first answer choice if the die shows 1 or 2 the second if it showes 3 or 4 and ...
November 20, 2012 by Andrew
Algebra II
Simplify(Multiple Choice) (1-csc^2 0)/(cot^2 0) A.)-1 B.)1 C.)tan^2 0 D.)1/sin^4 0
June 9, 2011 by HotMath
gometry
multiple choice always,never,sometimes Intersecting lines are _____coplanar?
September 3, 2008 by jasmyn
Calculus
Multiple Choice: Suppose F(x)=∫ from 0 to x^2 of 1/(2+t^3) dt for all real x, then F'(-1)=? a. 2 b. 1 c. 1/3 d. -2 e. -2/3 Steps please? Thanks(:
February 6, 2011 by Rebecca
To Dakota: Multiple choice
August 14, 2013 by Ms. Sue
litreature
can you give me some multiple choice question about act 3 in king lear
May 13, 2011 by reem
English
Multiple choice: A factory might emit? A.pollution. B.smokestacks. C.resources. Is it A or B? not sure - help
December 3, 2012 by Deborah
Math
(x+4)(x-9)=0 I need help because I have tried all the answers as this a multiple choice question. a) (x+4)(x-9) (4+4)(-9-9) (8)(-18) -144 b) (x+4)(x-9) (4+4)(9-9) (8)(0) 0 c) (x+4)(x-9) (-4+4)(9-9) (-8)(0) 0 d) (x+4)(x-9) (-4+4)(-9-9) (0)(-18) 0
November 3, 2014 by Alex
english
how to do i do good on a ela multiple choice comprehension test. like can you give me some tips?
April 15, 2013 by tom
math
Confused!! None of these questions are multiple choice. 1) -18 + 4 + x = 9 - 19 2) 32 - 6x + 7x = -15 + 28 3) -4x + 5(x -2) = -16 + 10 4) 4.8 + x -5.9 = 3.6 5) 1/3x + 1/3 (2x - 15) = 3 1/2 I have to find x for all of them, but I'm not sure what to do first.
September 16, 2014 by Anonymous
Pre-Algebra
Use the formula d = rt. Find t for r = 46.6 m/h and d = 456.68 m. Show your work. This question is not multiple choice
October 29, 2014 by Gabriella
Vectors
Multiple choice: Can you please explain how to answer this? Which point does not lie on the same plane as the points P(1,5,6), Q(-2,-3,4) and R(-1,3,2) a) S(0,4,4) b) S(4,13,8) c) S(0,-1,8) d) S(1,-3,2)
March 17, 2015 by Sara
Electrical versus Gravitational Forces (Physics)
Multiple Choice help: * _____ forces between charges are enormous in comparison to ______ forces. * A. electrical, gravitational (possible?) B. gravitational, electrical (possible?) C. positive, negative (wrong, I think) D. negative, positive (wrong, I think) According to the ...
February 25, 2011 by Angela
Algebra
The functions f(x) and g(x) are described below: f(x) = 32x + 8 g(x) = 32x − 9 The graph of g(x) is obtained by shifting down the graph of f(x) by _____ units. Numerical Answers Expected! Answer for Blank 1: „ð Question 11 (Multiple Choice Worth 5 points) (03.01 LC) ...
November 24, 2015 by Brandon
Biology (multiple choice!)
What’s one advantage to being eukaryotic rather than prokaryotic? What’s one advantage to being prokaryotic rather than eukaryotic?
October 25, 2010 by Dawn
math
What is the probability of guessing 70 out of 100 multiple choice questions correct when there are 4 choices per question?
April 22, 2008 by Jamal
Homeroom work
ON a multiple choice question with 4 options, what are your chances of guessing correctly if you eliminate two answers? 10% 25% 33% 50% Is it 50%?
March 22, 2015 by Jae
Education technology
Which of the following is not an example of an Objective question? Multiple choice Essay •• True/false Matching
February 2, 2016 by Vanessa
Algebra 2
Write an equation for the nth term of the geometric sequence: -12,4,-4/3 I had a_n=-12(3)^(n-1) but that is not a choice on the multiple answers they have = -12(1/3)^(n-1) = 12(-1/3)^(n-1) = -12(-1/3)^(-n+1) = -12 (-1/3)^(n-1)
November 17, 2007 by Chelsey
geometry
multiple choice always,sometimes,never A plane containing two points of a line_________contains the entire line?
September 3, 2008 by jasmyn
math
4 multiple choice questions with 3 possible answers. What is the probability of answering the first, third, and fourth question correctly? 1/3 x 1/3 x 1/3 1/3 = 1/81
November 16, 2014 by kim
statistics
There is a multiple choice exam with 24 questions, and five options each. If i guess every question, what is the probability that i will get exactly 15 correct.
March 17, 2015 by ben
math
find the 3rd multiple of 100, the 4th multiple of 10,and the 9th multiple of 4
September 23, 2012 by kaci
Math (Combinatorics)
How many 4 digit positive integers have the ones digit a multiple of 1, the tens digit a multiple of 2, the hundreds digit a multiple of 3 and the thousands digit a multiple of 4?
April 29, 2013 by Bob
To: judith -- medical terminology
I deleted your 17 multiple-choice exam questions. You showed none of your answers, and appeared to be trying to cheat!
December 21, 2010 by Ms. Sue
AP Calculus AB
If y = (((3x^2)+ 5)^5)((x+2)^4), then dy/dx = ? I've gotten to the point where i have (5((3x^2)+ 5)^4)(4(x+2)^3)(6x), what do i do next? It's a multiple choice question anf this is not one of the answers listed. Any help would be appreciated. Thanks!
September 18, 2011 by Jane
math
A 10-item multiple choice test with 4 options (A,B,C,D) is given to a class in English, what is the probability that the student’s answers are of the same letters
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On a very short quiz, there is one multiple choice question with 5 possible choices (a, b, c, d, e) and one true or false question. Assume you are taking the quiz but do not have any idea what the correct answer is to either question. You mark an answer to each question anyway...
July 8, 2011 by Cassie
Today is my NYS ELA Exam Book 1: Multiple Choice Questions. I know I'm supposed to be at school but the 7th graders take the exam around 12:30 or 12:40. Right now the 8th graders are (still) taking the exam so I'm still at home. Tomorrow is Book 2: Listening and Note-taking (...
April 17, 2012 by Laruen
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Mary took a 20 question multiple-choice exam where there are 4 choices for each question and only 1 of those choices is correct. Rather than reading the question, Mary simply puts a random choice of answer down for each question. Determine the probability that Mary gets ...
January 5, 2013 by John
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If a square of x multiple cm multiple x multiple cm is removed from the corner of 20cm multiple 20cm paper, write down an equation which will represent the volume of the box made from the remaining paper. Please help me please .
May 24, 2015 by Esihle
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f(x) = 1/(x+7)+15/(x-8). Find all a for which f(a)=f(a+15). I understand this so far f(a)=f(a+15) 1/(7+a) + 15/(a-8) = 1/(a+22) + 15/(a+7) I have multiplied the LCD of (a-8)(a+7)(a+22)by both sides with no luck. This is a multiple choice question a.) -161/8 b.) 120/7 c.) 161/7...
October 12, 2008 by Shawn
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A 24 question mathematics exam is worth 100 points. There are fill-in-the-blank questions worth 2 points each, multiple choice questions worth 6 points each, and long answer questions worth 5 points each. The number of fill-in-the-blank questions plus the number of multiple ...
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what are 3 reasons for emigrating in canada? (please give 3 answers making 2 of them wrong... kind of like multiple choice and specify)
May 27, 2008 by anonymous
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how do i find the derivative of........ (sinx) ^2-(cosx) ^2 i put it into my t-89 and got -4sinxcosx butthats not a multiple choice. can someone tel me how to do this by hand?
March 28, 2010 by James
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A pet store has various sizes of guinea pig cages. What are possible dimensions of this cage? L=6 cm W=4.8cm This problem has a multiple choice: A)28 in. BY 24 in. B)28 in. BY 18 in. C)30 in. BY 24 in. D)30 in. BY 18 in.
January 6, 2012 by Anonymous
find the antiderivative of f(x) = x^3(x-2)^2 I factored this out and got x^5-4x^4+4x^2 I'm supposed to integrate this. I don't know how. Someone help me please? BTW, multiple choice options are: A B C D
April 26, 2012 by Now What Do I Do?
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A multiple choice test has 24 questions each having 4 possible answers. If you guess each answer, find the probability of getting Between 9 and 15 correct
April 27, 2015 by Kalee
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17. Next>> | 7,214 | 25,892 | {"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} | 3.40625 | 3 | CC-MAIN-2016-07 | longest | en | 0.946903 |
https://abstract-polytopes.com/atlas/1920/240875/8.html | 1,726,255,300,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00167.warc.gz | 57,357,044 | 2,315 | # Polytope of Type {20,20}
Atlas Canonical Name : {20,20}*1920c
if this polytope has a name.
Group : SmallGroup(1920,240875)
Rank : 3
Schlafli Type : {20,20}
Number of vertices, edges, etc : 48, 480, 48
Order of s0s1s2 : 12
Order of s0s1s2s1 : 12
Special Properties :
Compact Hyperbolic Quotient
Locally Spherical
Orientable
Related Polytopes :
Facet
Vertex Figure
Dual
Petrial
Facet Of :
None in this Atlas
Vertex Figure Of :
None in this Atlas
Quotients (Maximal Quotients in Boldface) :
2-fold quotients : {20,10}*960a, {10,20}*960b
4-fold quotients : {20,10}*480a, {20,10}*480b, {5,20}*480, {10,10}*480
8-fold quotients : {5,10}*240, {10,5}*240, {10,10}*240a, {10,10}*240b, {10,10}*240c, {10,10}*240d
16-fold quotients : {5,5}*120, {5,10}*120a, {5,10}*120b, {10,5}*120a, {10,5}*120b
32-fold quotients : {5,5}*60
120-fold quotients : {4,2}*16
240-fold quotients : {2,2}*8
Covers (Minimal Covers in Boldface) :
None in this atlas.
Permutation Representation (GAP) :
```s0 := ( 1,23)( 2,41)( 3,42)( 4,37)( 5,30)( 6,34)( 7,28)( 8,36)( 9,24)(10,29)
(11,16)(12,38)(13,14)(15,44)(17,21)(18,19)(20,45)(22,32)(25,27)(26,39)(31,35)
(33,47)(40,48)(43,46)(51,52);;
s1 := ( 1, 4)( 2, 7)( 3,10)( 5,14)( 6,17)( 8,19)( 9,11)(12,27)(13,24)(15,30)
(16,25)(18,34)(20,36)(21,26)(22,39)(23,32)(28,42)(29,41)(31,40)(33,38)(35,47)
(37,46)(43,44)(45,48)(49,52)(50,51);;
s2 := ( 1,15)( 2, 5)( 3, 8)( 7,12)(10,22)(11,17)(13,27)(14,25)(16,21)(18,39)
(19,26)(20,47)(23,44)(28,38)(29,32)(30,41)(33,45)(36,42)(40,46)(43,48);;
poly := Group([s0,s1,s2]);;
```
Finitely Presented Group Representation (GAP) :
```F := FreeGroup("s0","s1","s2");;
s0 := F.1;; s1 := F.2;; s2 := F.3;;
rels := [ s0*s0, s1*s1, s2*s2, s0*s2*s0*s2, s2*s0*s1*s2*s1*s2*s0*s1*s2*s0*s1*s2*s1*s2*s0*s1,
s0*s1*s2*s1*s0*s1*s0*s1*s0*s1*s0*s1*s2*s1*s0*s1*s0*s1*s0*s1,
s2*s0*s1*s0*s1*s2*s1*s2*s1*s0*s1*s2*s1*s2*s1*s0*s1*s2*s0*s1*s0*s1*s0*s1*s0*s1,
s1*s2*s1*s2*s1*s2*s1*s2*s1*s2*s1*s0*s2*s1*s0*s1*s2*s1*s2*s1*s0*s1*s2*s0*s1*s2 ];;
poly := F / rels;;
```
Permutation Representation (Magma) :
```s0 := Sym(52)!( 1,23)( 2,41)( 3,42)( 4,37)( 5,30)( 6,34)( 7,28)( 8,36)( 9,24)
(10,29)(11,16)(12,38)(13,14)(15,44)(17,21)(18,19)(20,45)(22,32)(25,27)(26,39)
(31,35)(33,47)(40,48)(43,46)(51,52);
s1 := Sym(52)!( 1, 4)( 2, 7)( 3,10)( 5,14)( 6,17)( 8,19)( 9,11)(12,27)(13,24)
(15,30)(16,25)(18,34)(20,36)(21,26)(22,39)(23,32)(28,42)(29,41)(31,40)(33,38)
(35,47)(37,46)(43,44)(45,48)(49,52)(50,51);
s2 := Sym(52)!( 1,15)( 2, 5)( 3, 8)( 7,12)(10,22)(11,17)(13,27)(14,25)(16,21)
(18,39)(19,26)(20,47)(23,44)(28,38)(29,32)(30,41)(33,45)(36,42)(40,46)(43,48);
poly := sub<Sym(52)|s0,s1,s2>;
```
Finitely Presented Group Representation (Magma) :
```poly<s0,s1,s2> := Group< s0,s1,s2 | s0*s0, s1*s1, s2*s2,
s0*s2*s0*s2, s2*s0*s1*s2*s1*s2*s0*s1*s2*s0*s1*s2*s1*s2*s0*s1,
s0*s1*s2*s1*s0*s1*s0*s1*s0*s1*s0*s1*s2*s1*s0*s1*s0*s1*s0*s1,
s2*s0*s1*s0*s1*s2*s1*s2*s1*s0*s1*s2*s1*s2*s1*s0*s1*s2*s0*s1*s0*s1*s0*s1*s0*s1,
s1*s2*s1*s2*s1*s2*s1*s2*s1*s2*s1*s0*s2*s1*s0*s1*s2*s1*s2*s1*s0*s1*s2*s0*s1*s2 >;
```
References : None.
to this polytope | 1,596 | 3,054 | {"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-2024-38 | latest | en | 0.488668 |
http://nrich.maths.org/public/leg.php?code=32&cl=2&cldcmpid=5929 | 1,477,523,904,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721008.78/warc/CC-MAIN-20161020183841-00304-ip-10-171-6-4.ec2.internal.warc.gz | 184,587,795 | 10,237 | Search by Topic
Resources tagged with Multiplication & division similar to Sea Level:
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Broad Topics > Calculations and Numerical Methods > Multiplication & division
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Super Shapes
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Find the next number in this pattern: 3, 7, 19, 55 ...
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Rocco ran in a 200 m race for his class. Use the information to find out how many runners there were in the race and what Rocco's finishing position was.
It Was 2010!
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Sometimes We Lose Things
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Well now, what would happen if we lost all the nines in our number system? Have a go at writing the numbers out in this way and have a look at the multiplications table.
How Many Eggs?
Stage: 2 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
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Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
Function Machines
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If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
Machines
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What is happening at each box in these machines?
Six Numbered Cubes
Stage: 2 Challenge Level:
This task combines spatial awareness with addition and multiplication.
Clever Keys
Stage: 2 Short Challenge Level:
On a calculator, make 15 by using only the 2 key and any of the four operations keys. How many ways can you find to do it?
Zios and Zepts
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On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
The Number Crunching Machine
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Put a number at the top of the machine and collect a number at the bottom. What do you get? Which numbers get back to themselves?
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Napier's Bones
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Using the statements, can you work out how many of each type of rabbit there are in these pens?
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Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers? | 2,135 | 9,075 | {"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-2016-44 | longest | en | 0.895416 |
https://in.booksc.org/book/71080036/bef07f | 1,586,368,144,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371821680.80/warc/CC-MAIN-20200408170717-20200408201217-00103.warc.gz | 512,371,549 | 19,151 | मुख्य Abstract and Applied Analysis Time Scale Inequalities of the Ostrowski Type for Functions Differentiable on the Coordinates
# Time Scale Inequalities of the Ostrowski Type for Functions Differentiable on the Coordinates
, ,
Volume:
2018
वर्ष:
2018
भाषा:
english
Journal:
Abstract and Applied Analysis
DOI:
10.1155/2018/1802578
File:
PDF, 1.38 MB
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1
वर्ष:
2018
भाषा:
english
File:
PDF, 1.44 MB
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### Intelligent Method for Identifying Driving Risk Based on V2V Multisource Big Data
वर्ष:
2018
भाषा:
english
File:
PDF, 2.29 MB
```Hindawi
Abstract and Applied Analysis
Volume 2018, Article ID 1802578, 10 pages
https://doi.org/10.1155/2018/1802578
Research Article
Time Scale Inequalities of the Ostrowski Type for
Functions Differentiable on the Coordinates
Eze R. Nwaeze
1
,1 Seth Kermausuor,2 and Ana M. Tameru1
Department of Mathematics, Tuskegee University, Tuskegee, AL 36088, USA
Department of Mathematics and Computer Science, Alabama State University, Montgomery, AL 36104, USA
2
Correspondence should be addressed to Eze R. Nwaeze; enwaeze@tuskegee.edu
Received 31 December 2017; Accepted 1 February 2018; Published 5 March 2018
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
In 2016, some inequalities of the Ostrowski type for functions (of two variables) differentiable on the coordinates were established.
In this paper, we extend these results to an arbitrary time scale by means of a parameter 𝜆 ∈ [0, 1]. The aforementioned results
are regained for the case when the time scale T = R. Besides extension, our results are employed to the continuous and discrete
calculus to get some new inequalities in this direction.
1. Introduction
To find a bound for the difference of a function and its integral
mean, the Ukraine-born mathematician Ostrowski [1], in
1938, established the subsequent result which is nowadays
celebrated as the Ostrowski inequality.
Theorem 1. Let 𝐺 : [𝛼, 𝛽] → R be continuous on [𝛼, 𝛽] and
differentiable in (𝛼, 𝛽) and its derivative 𝐺 : (𝛼, 𝛽) → R is
bounded in (𝛼, 𝛽). If |𝐺 (𝑠)| ≤ M for all 𝑠 ∈ [𝛼, 𝛽], then we
have
𝛽
1
𝐺 (𝑥) −
𝐺
𝑑𝑠
∫
(𝑠)
𝛽
−
𝛼
𝛼
2
1 (𝑥 − (𝛼 + 𝛽) /2)
) (𝛽 − 𝛼) M,
≤( +
2
4
(𝛽 − 𝛼)
(1)
for all 𝑥 ∈ [𝛼, 𝛽]. The inequality is sharp in the sense that the
constant 1/4 cannot be replaced by a smaller one.
In 2001, Cheng [2] gave the following ; improvement of the
above inequality.
Theorem 2. Let 𝐺 : [𝛼, 𝛽] → R be continuous on [𝛼, 𝛽] and
differentiable in (𝛼, 𝛽) such that there exist constants 𝛾, Γ ∈ R
with 𝛾 ≤ 𝐺 (𝑠) ≤ Γ for all 𝑠 ∈ [𝛼, 𝛽]. Then for all 𝑥 ∈ [𝛼, 𝛽],
one gets
(𝑥 − 𝛽) 𝐺 (𝛽) − (𝑥 − 𝛼) 𝐺 (𝛼)
1
𝐺 (𝑥) −
2
2 (𝛽 − 𝛼)
(𝑥 − 𝛼)2 + (𝛽 − 𝑥)2
𝛽
1
(Γ − 𝛾) .
−
∫ 𝐺 (𝑠) 𝑑𝑠 ≤
𝛽−𝛼 𝛼
8 (𝛽 − 𝛼)
(2)
Recently, Farid [3] extended Theorems 1 and 2 to functions of two variables that are differentiable on their coordinates. Specifically, he proved the following two theorems.
Theorem 3. Let 𝐺 : I × J → R, where I, J are open
intervals in R, be a mapping such that for 𝛼1 , 𝛽1 ∈ I, 𝛼2 , 𝛽2 ∈
J, 𝛼1 < 𝛽1 , 𝛼2 < 𝛽2 , the partial mappings
𝐺𝑦 : [𝛼1 , 𝛽1 ] → R,
𝐺𝑦 (𝜉) fl 𝐺 (𝜉, 𝑦) ,
𝐺𝑥 : [𝛼2 , 𝛽2 ] → R,
𝐺𝑥 (𝜁) fl 𝐺 (𝑥, 𝜁) ,
(3)
2
Abstract and Applied Analysis
𝐺𝑥 : [𝛼2 , 𝛽2 ] → R,
defined for all 𝑦 ∈ [𝛼2 , 𝛽2 ] and 𝑥 ∈ [𝛼1 , 𝛽1 ], are differentiable,
and |𝐺𝑦 (𝑠)| ≤ M, 𝑠 ∈ [𝛼1 , 𝛽1 ], |𝐺𝑥 (𝑠)| ≤ N, 𝑠 ∈ [𝛼2 , 𝛽2 ].
Then we have
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
𝑑𝑥
𝛼
2
1
+∫
𝛽2
𝛼2
−(
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
𝑑𝑦
2
𝛽1 𝛽2
1
1
+
) ∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
𝛽1 − 𝛼1 𝛽2 − 𝛼2 𝛼1 𝛼2
(4)
+∫
𝛼2
−
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
−(
≤
(5)
≤
M+N
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) ,
4
𝛽2
𝛼 + 𝛽1
1
, 𝑦) 𝑑𝑦
∫ 𝐺( 1
2
2 (𝛽2 − 𝛼2 ) 𝛼2
𝛽1 𝛽2
1
−
∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
≤
8
(7)
M (𝛽1 − 𝛼1 ) + N (𝛽2 − 𝛼2 )
.
8
Remark 4. Inequality (7) is the correct version of inequality
(2.18) as presented in [[3], Theorem 2.6].
Theorem 5. Let 𝐺 : I × J → R, where I, J are open
intervals in R, be a mapping such that for 𝛼1 , 𝛽1 ∈ I, 𝛼2 , 𝛽2 ∈
J, 𝛼1 < 𝛽1 , 𝛼2 < 𝛽2 , the partial mappings
𝛽2
𝐺𝑦 (𝜉) fl 𝐺 (𝜉, 𝑦) ,
(9)
≤
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
𝑑𝑦
4 (𝛽2 − 𝛼2 )
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
(𝛽1 − 𝛼1 ) (Γ𝑦 − 𝛾𝑦 ) + (𝛽2 − 𝛼2 ) (Γ𝑥 − 𝛾𝑥 )
16
.
In 1988, the idea of time scales [4] was initiated so as to
bring together the continuous and discrete analysis into a unified fold. Since the introduction of this subject, many classical
integral results have been extended to time scales. “A time
scale T is an arbitrary nonempty closed subset of R.” We shall
presume, all over this work, that the reader is familiar with
the theory of time scale (see [5, 6] for more on this subject).
We present here a result of Bohner and Matthews [7] which is
embedded in Theorem 6 below. This result extends Theorem 1
to time scales. For more improvements and generalizations
around this result, we refer the interested reader to see the
papers [8–12] and the references therein.
Theorem 6. Let 𝛼, 𝛽, 𝑠, 𝑡 ∈ T, 𝛼 < 𝛽 and 𝐺 : [𝛼, 𝛽] → R be
differentiable. Then for all 𝑡 ∈ [𝛼, 𝛽], we have
𝛽
1
𝐺 (𝑡) −
𝐺
Δ𝑠
∫
(𝜎
(𝑠))
𝛽
−
𝛼
𝛼
M
(ℎ (𝑡, 𝛼) + ℎ2 (𝑡, 𝛽)) ,
≤
𝛽−𝛼 2
𝑡
𝐺𝑦 : [𝛼1 , 𝛽1 ] → R,
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) ,
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
𝑑𝑥
𝛼
4
(𝛽
−
𝛼
)
1
1
1
−
𝛽1
𝛼2 + 𝛽2
1
) 𝑑𝑥
∫
2 (𝛽 − 𝛼 ) 𝛼 𝐺 (𝑥,
2
1
1
1
+
𝛽1 𝛽2
1
1
+
) ∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
𝛽1 − 𝛼1 𝛽2 − 𝛼2 𝛼1 𝛼2
𝛼2
(6)
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
𝑑𝑦
2
Γ𝑥 + Γ𝑦 − (𝛾𝑥 + 𝛾𝑦 )
+∫
𝛽1
𝛽2
𝛼 + 𝛽2
𝛼 + 𝛽1
∫ 𝐺 (𝑥, 2
𝐺( 1
)
𝑑𝑥
+
, 𝑦) 𝑑𝑦
∫
𝛼
2
2
𝛼
2
1
𝛽1 𝛽2
1
1
+
) ∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
𝛽1 − 𝛼1 𝛽2 − 𝛼2 𝛼1 𝛼2
𝛽2
𝛼2
M (𝛽1 − 𝛼1 ) + N (𝛽2 − 𝛼2 )
≤
,
4
−(
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
𝑑𝑥
𝛼
2
1
+∫
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
𝑑𝑥
𝛼
4 (𝛽1 − 𝛼1 )
1
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
𝑑𝑦
4 (𝛽2 − 𝛼2 )
(8)
defined for all 𝑦 ∈ [𝛼2 , 𝛽2 ] and 𝑥 ∈ [𝛼1 , 𝛽1 ], are differentiable
with 𝛾𝑦 ≤ 𝐺𝑦 (𝑠) ≤ Γ𝑦 , 𝑠 ∈ [𝛼1 , 𝛽1 ], 𝛾𝑥 ≤ 𝐺𝑥 (𝑠) ≤ Γ𝑥 , 𝑠 ∈
[𝛼2 , 𝛽2 ]. Then we have
M+N
≤
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) ,
2
𝛽2
𝐺𝑥 (𝜁) fl 𝐺 (𝑥, 𝜁) ,
(10)
where ℎ2 (𝑡, 𝑠) = ∫𝑠 (𝜏 − 𝑠)Δ𝜏 for all 𝑠, 𝑡 ∈ T and M =
sup𝛼<𝑡<𝛽 |𝐺Δ (𝑡)| < ∞. This inequality is sharp in the sense that
the right-hand side of (10) cannot be replaced by a smaller one.
Abstract and Applied Analysis
3
It is our purpose in this paper to extend inequalities (4),
(5), (6), (7), and (9) to time scales by means of a parameter
𝜆 ∈ [0, 1]. In Section 2, we frame and prove the main results
followed by applications to the continuous and discrete
calculus.
2. Main Results
for all 𝜆 ∈ [0, 1] such that 𝛼 + 𝜆((𝑥 − 𝛼)/2) and 𝛽 − 𝜆((𝛽 − 𝑥)/2)
are in T.
We now formulate and prove our first result.
Theorem 9. Let 𝛼1 , 𝛽1 , 𝑥 ∈ T1 , 𝛼2 , 𝛽2 , 𝑦 ∈ T2 , with 𝛼1 <
𝛽1 , 𝛼2 < 𝛽2 and 𝐺 : [𝛼1 , 𝛽1 ] × [𝛼2 , 𝛽2 ] → R be such that
the partial mappings
𝐺𝑦 : [𝛼1 , 𝛽1 ] → R,
In this section, we will present our results involving double
integrals. For some recent results in this regard, see [13–18].
The proofs of our findings shall be anchored on the subsequent lemmas.
Lemma 7 (see [9]). Let 𝛼, 𝛽, 𝑡, 𝑥 ∈ T, 𝛼 < 𝛽 and 𝐺 : [𝛼, 𝛽] →
R be differentiable. Then
𝐺 (𝛼) + 𝐺 (𝛽)
(1 − 𝜆) 𝐺 (𝑥) + 𝜆
2
𝛽
1
−
∫ 𝐺 (𝜎 (𝑡)) Δ𝑡
𝛽−𝛼 𝛼
≤
𝛽−𝛼
M
[ℎ2 (𝛼, 𝛼 + 𝜆
)
𝛽−𝛼
2
(11)
𝛽−𝛼
𝛽−𝛼
+ ℎ2 (𝑥, 𝛼 + 𝜆
) + ℎ2 (𝑥, 𝛽 − 𝜆
)
2
2
for all 𝜆 ∈ [0, 1] such that 𝛼+𝜆((𝛽−𝛼)/2) and 𝛽−𝜆((𝛽−𝛼)/2)
are in T and 𝑥 ∈ [𝛼 + 𝜆((𝛽 − 𝛼)/2), 𝛽 − 𝜆((𝛽 − 𝛼)/2)] ∩ T, where
M fl sup𝛼<𝑥<𝛽 |𝐺Δ (𝑥)| < ∞. This is sharp provided that
(12)
Lemma 8 (see [8]). Let 𝛼, 𝛽, 𝑡, 𝑥 ∈ T, 𝛼 < 𝛽 and 𝐺 : [𝛼, 𝛽] →
R be differentiable. If 𝐺Δ ∈ 𝐶𝑟𝑑 (T, R) and there exist 𝛾, Γ ∈ R
such that 𝛾 ≤ 𝐺Δ (𝑡) ≤ Γ for all 𝑡 ∈ [𝛼, 𝛽], then for all 𝑥 ∈ [𝛼, 𝛽],
we have
(1 − 𝜆 ) 𝐺 (𝑥) + 𝜆 (𝑥 − 𝛼) 𝐺 (𝛼) + (𝛽 − 𝑥) 𝐺 (𝛽)
2
2 (𝛽 − 𝛼)
−
𝛽
Γ+𝛾 1
1
[ℎ (𝑥, 𝛼)
∫ 𝐺 (𝜎 (𝑡)) Δ𝑡 −
𝛽−𝛼 𝛼
2 𝛽−𝛼 2
2
(𝑥 − 𝛼)2 − (𝛽 − 𝑥)
− ℎ2 (𝑥, 𝛽) − 𝜆 (
)]
2
≤
Γ−𝛾
𝑥−𝛼
[ℎ (𝛼, 𝛼 + 𝜆
) + ℎ2 (𝑥, 𝛼
2
2 (𝛽 − 𝛼) 2
𝛽−𝑥
𝑥−𝛼
) + ℎ2 (𝑥, 𝛽 − 𝜆
) + ℎ2 (𝛽, 𝛽
+𝜆
2
2
−𝜆
𝛽−𝑥
)] ,
2
𝐺𝑥 : [𝛼2 , 𝛽2 ] → R,
defined for all 𝑦 ∈ [𝛼2 , 𝛽2 ] and 𝑥 ∈ [𝛼1 , 𝛽1 ], are differentiable.
If M = sup𝛼1 <𝑡<𝛽1 |𝐺𝑦Δ (𝑡)| and N = sup𝛼2 <𝑡<𝛽2 |𝐺𝑥Δ (𝑡)|, then the
succeeding inequalities
𝛽1 𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )
∫
Δ𝑥
2
𝛼1
𝛽2
𝛼2
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
1
Δ𝑦 −
2
𝛽1 − 𝛼1
𝛽1
𝛽2
𝛼1
𝛼2
+∫
1
𝛽2 − 𝛼2
𝛽1 𝛽2
⋅ ∫ ∫ 𝐺 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥
𝛼1 𝛼2
≤
M (𝛽2 − 𝛼2 )
𝛽 − 𝛼1
[3ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
2
2 (𝛽1 − 𝛼1 )
+ ℎ2 (𝛽1 , 𝛼1 + 𝜆
𝛽1 − 𝛼1
)
2
𝛽 − 𝛼1
+ 3ℎ2 (𝛽1 , 𝛽1 − 𝜆 1
)
2
+ ℎ2 (𝛼1 , 𝛽1 − 𝜆
+
𝛽1 − 𝛼1
)]
2
N (𝛽1 − 𝛼1 )
𝛽 − 𝛼2
[3ℎ2 (𝛼2 , 𝛼2 + 𝜆 2
)
2
2 (𝛽2 − 𝛼2 )
+ ℎ2 (𝛽2 , 𝛼2 + 𝜆
(13)
(14)
𝐺𝑥 (𝜁) fl 𝐺 (𝑥, 𝜁) ,
⋅ ∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥 −
𝛽−𝛼
+ ℎ2 (𝛽, 𝛽 − 𝜆
)] ,
2
𝛼+𝜆((𝛽−𝛼)/2)
𝜆
𝜆2
2
𝑡Δ𝑡.
𝛼 (𝛽 − 𝛼) +
(𝛽 − 𝛼) ≤ ∫
2
4
𝛼
𝐺𝑦 (𝜉) fl 𝐺 (𝜉, 𝑦) ,
𝛽2 − 𝛼2
)
2
+ 3ℎ2 (𝛽2 , 𝛽2 − 𝜆
+ ℎ2 (𝛼2 , 𝛽2 − 𝜆
𝛽2 − 𝛼2
)
2
𝛽2 − 𝛼2
)] ,
2
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
Δ𝑥
𝛼
4
(𝛽
−
𝛼
)
1
1
1
+∫
𝛽2
𝛼2
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
Δ𝑦
4 (𝛽2 − 𝛼2 )
(15)
4
Abstract and Applied Analysis
−
1
2 (𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 )
+ ℎ2 (𝛽1 , 𝛼1 + 𝜆
𝛽1 𝛽2
⋅ ∫ ∫ [𝐺 (𝑥, 𝜎 (𝑦)) + 𝐺 (𝜎 (𝑥) , 𝑦)] Δ𝑦 Δ𝑥
𝛼1 𝛼2
≤
+ 2ℎ2 (𝛽1 , 𝛽1 − 𝜆
+ ℎ2 (𝛼1 , 𝛽1 − 𝜆
Applying, again, Lemma 7 to 𝐺𝑦 at 𝑥 = 𝛼1 , and integrating
over [𝛼2 , 𝛽2 ] give
𝛽1 − 𝛼1
)
2
+ 3ℎ2 (𝛽1 , 𝛽1 − 𝜆
𝛽1 − 𝛼1
)] .
2
(18)
𝛽 − 𝛼1
M
[3ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
2
4 (𝛽1 − 𝛼1 )
+ ℎ2 (𝛽1 , 𝛼1 + 𝜆
𝛽1 − 𝛼1
)
2
𝛽2
𝜆
𝜆 𝛽2
(1 − ) ∫ 𝐺 (𝛼1 , 𝑦) Δ𝑦 + ∫ 𝐺 (𝛽1 , 𝑦) Δ𝑦
2 𝛼2
2 𝛼2
𝛽1 𝛽2
1
−
∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥
𝛽1 − 𝛼1 𝛼1 𝛼2
𝛽1 − 𝛼1
)
2
𝛽1 − 𝛼1
)]
2
M (𝛽2 − 𝛼2 )
𝛽 − 𝛼1
[2ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
𝛽1 − 𝛼1
2
𝛽 − 𝛼2
N
+
[3ℎ2 (𝛼2 , 𝛼2 + 𝜆 2
)
2
4 (𝛽2 − 𝛼2 )
≤
𝛽 − 𝛼2
)
+ ℎ2 (𝛽2 , 𝛼2 + 𝜆 2
2
+ ℎ2 (𝛼1 , 𝛽1 − 𝜆
𝛽1 − 𝛼1
)
2
𝛽 − 𝛼2
+ 3ℎ2 (𝛽2 , 𝛽2 − 𝜆 2
)
2
+ ℎ2 (𝛽1 , 𝛽1 − 𝜆
𝛽1 − 𝛼1
)] .
2
+ ℎ2 (𝛼2 , 𝛽2 − 𝜆
Using (18) and (19), we have
𝛽2 − 𝛼2
)] ,
2
(16)
𝛽2 𝐺 (𝛼 , 𝑦) + 𝐺 (𝛽 , 𝑦)
1
1
∫
Δ𝑦
𝛼
2
2
hold for all 𝜆 ∈ [0, 1] such that (𝛼1 +𝜆((𝛽1 −𝛼1 )/2), 𝛼2 +𝜆((𝛽2 −
𝛼2 )/2)) ∈ T1 ×T2 and (𝛽1 −𝜆((𝛽1 −𝛼1 )/2), 𝛽2 −𝜆((𝛽2 −𝛼2 )/2)) ∈
T1 × T2 .
−
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥
𝛽1 − 𝛼1 𝛼1 𝛼2
Proof. Applying Lemma 7 to 𝐺𝑦 at 𝑥 = 𝛽1 , we get
≤
M (𝛽2 − 𝛼2 )
𝛽 − 𝛼1
[3ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
2
2 (𝛽1 − 𝛼1 )
𝜆
𝜆
(1 − ) 𝐺 (𝛽1 , 𝑦) + 𝐺 (𝛼1 , 𝑦)
2
2
−
𝛽1
1
∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑥
𝛽1 − 𝛼1 𝛼1
≤
𝛽 − 𝛼1
M
[ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
𝛽1 − 𝛼1
2
+ ℎ2 (𝛽1 , 𝛼1 + 𝜆
𝛽1 − 𝛼1
)
2
+ 2ℎ2 (𝛽1 , 𝛽1 − 𝜆
𝛽1 − 𝛼1
)] .
2
Integrating (17) over [𝛼2 , 𝛽2 ] gives
𝛽2
𝜆 𝛽2
𝜆
(1 − ) ∫ 𝐺 (𝛽1 , 𝑦) Δ𝑦 + ∫ 𝐺 (𝛼1 , 𝑦) Δ𝑦
2 𝛼2
2 𝛼2
𝛽1 𝛽2
1
−
∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥
𝛽1 − 𝛼1 𝛼1 𝛼2
≤
M (𝛽2 − 𝛼2 )
𝛽 − 𝛼1
[ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
𝛽1 − 𝛼1
2
(19)
+ ℎ2 (𝛽1 , 𝛼1 + 𝜆
𝛽1 − 𝛼1
)
2
+ 3ℎ2 (𝛽1 , 𝛽1 − 𝜆
(17)
+ ℎ2 (𝛼1 , 𝛽1 − 𝜆
(20)
𝛽1 − 𝛼1
)
2
𝛽1 − 𝛼1
)] .
2
Similarly, doing the same thing for 𝐺𝑥 at 𝑦 = 𝛼2 and 𝑦 = 𝛽2 ,
and then integrating the resultant inequality over [𝛼1 , 𝛽1 ], we
get
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
Δ𝑥
𝛼
2
1
−
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥
𝛽2 − 𝛼2 𝛼1 𝛼2
≤
N (𝛽1 − 𝛼1 )
𝛽 − 𝛼2
[3ℎ2 (𝛼2 , 𝛼2 + 𝜆 2
)
2
2 (𝛽2 − 𝛼2 )
+ ℎ2 (𝛽2 , 𝛼2 + 𝜆
𝛽2 − 𝛼2
)
2
Abstract and Applied Analysis
+ 3ℎ2 (𝛽2 , 𝛽2 − 𝜆
+ ℎ2 (𝛼2 , 𝛽2 − 𝜆
5
𝛽2 − 𝛼2
)
2
𝛽2 − 𝛼2
)] .
2
(21)
Using (20) and (21) amounts to (15). Also, from (20) and (21),
we get
𝛽2 𝐺 (𝛼 , 𝑦) + 𝐺 (𝛽 , 𝑦)
1
1
∫
Δ𝑦
𝛼
2
(𝛽
−
𝛼
)
2
2
2
𝛽1 𝛽2
1
−
∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
≤
𝛽 − 𝛼1
M
[3ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
2
2 (𝛽1 − 𝛼1 )
𝛽2
𝛼2
𝛼2 + 𝛽2
𝛽 − 𝛼2
𝛼 + 𝛽2
, 𝛼2 + 𝜆 2
) + ℎ2 ( 2
, 𝛽2
2
2
2
𝛽2 − 𝛼2
𝛽 − 𝛼2
) + ℎ2 (𝛽2 , 𝛽2 − 𝜆 2
)] ,
2
2
𝛼1 + 𝛽1
𝜆
, 𝑦) Δ𝑦 +
2
4 (𝛽1 − 𝛼1 )
𝛽1
𝛼1
𝜆
4 (𝛽2 − 𝛼2 )
𝛽2
(23)
⋅ ∫ [𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)] Δ𝑦
𝛼2
−
1
2 (𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 )
𝛽1 𝛽2
⋅ ∫ ∫ [𝐺 (𝜎 (𝑥) , 𝑦) + 𝐺 (𝑥, 𝜎 (𝑦))] Δ𝑦 Δ𝑥
𝛼1 𝛼2
𝛼1 + 𝛽1
𝜆
, 𝑦) Δ𝑦] +
2
2
⋅ ∫ [𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)] Δ𝑦 −
N (𝛽1 − 𝛼1 )
𝛽 − 𝛼2
[ℎ2 (𝛼2 , 𝛼2 + 𝜆 2
)
𝛽2 − 𝛼2
2
⋅ ∫ [𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )] Δ𝑥 +
Theorem 10. Under the assumptions of Theorem 9 and suppose also the intervals contain the mid points, then we have
𝛽1
𝛼 + 𝛽2
(1 − 𝜆) [∫ 𝐺 (𝑥, 2
) Δ𝑥
2
𝛼1
𝛼1
𝛼1 + 𝛽1
𝛽 − 𝛼1
𝛼 + 𝛽1
, 𝛼1 + 𝜆 1
) + ℎ2 ( 1
, 𝛽1
2
2
2
𝛽1 − 𝛼1
𝛽 − 𝛼1
) + ℎ2 (𝛽1 , 𝛽1 − 𝜆 1
)]
2
2
𝛼2
𝛽2 − 𝛼2
)] .
2
𝛽1
−𝜆
𝛽2
𝛽2 − 𝛼2
)
2
⋅ ∫ [𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )] Δ𝑥 +
+ ℎ2 (
⋅ ∫ 𝐺(
Combining (22) and (23), one gets (16).
𝛼2
M (𝛽2 − 𝛼2 )
𝛽 − 𝛼1
[ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
𝛽1 − 𝛼1
2
𝛽2
𝛼2 + 𝛽2
1−𝜆
1 − 𝜆
) Δ𝑥 +
∫
2 (𝛽 − 𝛼 ) 𝛼 𝐺 (𝑥,
2
2
(𝛽
1
1
2 − 𝛼2 )
2
𝛽 − 𝛼2
)
+ ℎ2 (𝛽2 , 𝛼2 + 𝜆 2
2
𝛽2
1
𝛽2 − 𝛼2
(24)
𝛽 − 𝛼2
N
[3ℎ2 (𝛼2 , 𝛼2 + 𝜆 2
)
2
2 (𝛽2 − 𝛼2 )
+ ∫ 𝐺(
≤
−𝜆
𝛽1 − 𝛼1
)
2
𝛽1 𝛽2
1
−
∫ ∫ 𝐺 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
+ ℎ2 (𝛼2 , 𝛽2 − 𝜆
𝛼2
+ ℎ2 (
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
Δ𝑥
𝛼
2
(𝛽
−
𝛼
)
1
1
1
+ 3ℎ2 (𝛽2 , 𝛽2 − 𝜆
𝛼1
(22)
𝛽 − 𝛼1
)] ,
+ ℎ2 (𝛼1 , 𝛽1 − 𝜆 1
2
≤
𝛽2
𝛽1 𝛽2
⋅ ∫ ∫ 𝐺 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥
𝛼1 𝛼2
+
𝛽 − 𝛼1
)
+ ℎ2 (𝛽1 , 𝛼1 + 𝜆 1
2
+ 3ℎ2 (𝛽1 , 𝛽1 − 𝜆
𝛽1
⋅ ∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥 −
𝛽 − 𝛼1
M
≤
[ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
2
2 (𝛽1 − 𝛼1 )
+ ℎ2 (
−𝜆
+
𝜆
2
1
𝛽1 − 𝛼1
𝛼1 + 𝛽1
𝛽 − 𝛼1
𝛼 + 𝛽1
, 𝛼1 + 𝜆 1
) + ℎ2 ( 1
, 𝛽1
2
2
2
𝛽1 − 𝛼1
𝛽 − 𝛼1
) + ℎ2 (𝛽1 , 𝛽1 − 𝜆 1
)]
2
2
𝛽 − 𝛼2
N
[ℎ (𝛼 , 𝛼 + 𝜆 2
)
2
2 (𝛽2 − 𝛼2 ) 2 2 2
+ ℎ2 (
−𝜆
𝛼2 + 𝛽2
𝛽 − 𝛼2
𝛼 + 𝛽2
, 𝛼2 + 𝜆 2
) + ℎ2 ( 2
, 𝛽2
2
2
2
𝛽2 − 𝛼2
𝛽 − 𝛼2
) + ℎ2 (𝛽2 , 𝛽2 − 𝜆 2
)] .
2
2
(25)
6
Abstract and Applied Analysis
Proof. Next, we now apply Lemma 7 to 𝐺𝑦 at 𝑥 = (𝛼1 + 𝛽1 )/2
and thereafter integrate the resulting inequality over [𝛼2 , 𝛽2 ]
to get
𝛽2
𝛼1 + 𝛽1
, 𝑦) Δ𝑦
(1 − 𝜆) ∫ 𝐺 (
2
𝛼2
𝛽
+𝜆
𝛽
∫𝛼 2 𝐺 (𝛼1 , 𝑦) Δ𝑦 + ∫𝛼 2 𝐺 (𝛽1 , 𝑦) Δ𝑦
2
2
2
𝛽1 𝛽2
1
−
∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥
𝛽1 − 𝛼1 𝛼1 𝛼2
M (𝛽2 − 𝛼2 )
𝛽 − 𝛼1
[ℎ2 (𝛼1 , 𝛼1 + 𝜆 1
)
≤
𝛽1 − 𝛼1
2
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
≤
𝛽 − 𝛼1
M
[ℎ (𝛼 , 𝛼 + 𝜆 1
)
𝛽1 − 𝛼1 2 1 1
2
+ ℎ2 (
𝛼1 + 𝛽1
𝛽 − 𝛼1
, 𝛼1 + 𝜆 1
)
2
2
+ ℎ2 (
𝛼1 + 𝛽1
𝛽 − 𝛼1
, 𝛽1 − 𝜆 1
)
2
2
+ ℎ2 (𝛽1 , 𝛽1 − 𝜆
(26)
𝛽1 − 𝛼1
)] ,
2
𝛼2 + 𝛽2
1 − 𝜆 𝛽1
) Δ𝑥
𝛽1 − 𝛼1 ∫𝛼 𝐺 (𝑥,
2
1
+ ℎ2 (
𝛼1 + 𝛽1
𝛽 − 𝛼1
, 𝛼1 + 𝜆 1
)
2
2
+
𝛽1
𝜆
∫ [𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )] Δ𝑥
2 (𝛽1 − 𝛼1 ) 𝛼1
+ ℎ2 (
𝛼1 + 𝛽1
𝛽 − 𝛼1
, 𝛽1 − 𝜆 1
)
2
2
−
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
≤
𝛽 − 𝛼2
N
[ℎ2 (𝛼2 , 𝛼2 + 𝜆 2
)
𝛽2 − 𝛼2
2
𝛽1 − 𝛼1
)] .
2
Similarly, if one applies Lemma 7 to 𝐺𝑥 at 𝑦 = (𝛼2 + 𝛽2 )/2 and
thereafter integrate the resulting inequality over [𝛼1 , 𝛽1 ], one
gets
𝛽1
𝛼2 + 𝛽2
) Δ𝑥
(1 − 𝜆) ∫ 𝐺 (𝑥,
2
𝛼1
+ ℎ2 (𝛽1 , 𝛽1 − 𝜆
𝛽
+𝜆
𝛽
≤
+ ℎ2 (
𝛼2 + 𝛽2
𝛽 − 𝛼2
, 𝛼2 + 𝜆 2
)
2
2
+ ℎ2 (
𝛼2 + 𝛽2
𝛽 − 𝛼2
, 𝛽2 − 𝜆 2
)
2
2
+ ℎ2 (𝛽2 , 𝛽2 − 𝜆
∫𝛼 1 𝐺 (𝑥, 𝛼2 ) Δ𝑥 + ∫𝛼 1 𝐺 (𝑥, 𝛽2 ) Δ𝑥
1
(28)
1
N (𝛽1 − 𝛼1 )
𝛽 − 𝛼2
[ℎ2 (𝛼2 , 𝛼2 + 𝜆 2
)
𝛽2 − 𝛼2
2
+ ℎ2 (
𝛽2 − 𝛼2
)] .
2
2
𝛽1 𝛽2
1
−
∫ ∫ 𝐺 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥
𝛽2 − 𝛼2 𝛼1 𝛼2
Using (28) amounts to (25). Thus, the proof of Theorem 9 is
complete.
(27)
𝛼2 + 𝛽2
𝛽 − 𝛼2
, 𝛼2 + 𝜆 2
)
2
2
𝛼 + 𝛽2
𝛽 − 𝛼2
+ ℎ2 ( 2
, 𝛽2 − 𝜆 2
)
2
2
𝛽 − 𝛼2
)] .
+ ℎ2 (𝛽2 , 𝛽2 − 𝜆 2
2
Combining (26) and (27), we get (24). Finally, from (26) and
(27), we get
𝛼1 + 𝛽1
1 − 𝜆 𝛽2
𝛽2 − 𝛼2 ∫𝛼 𝐺 ( 2 , 𝑦) Δ𝑦
2
+
−
𝛽2
𝜆
∫ [𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)] Δ𝑦
2 (𝛽2 − 𝛼2 ) 𝛼2
Corollary 11. If we let T1 = T2 = R in Theorems 9 and 10, then
we obtain the inequality
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
𝑑𝑥
𝛼
2
1
+∫
𝛽2
𝛼2
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
1
𝑑𝑦 − (
2
𝛽1 − 𝛼1
+
𝛽1 𝛽2
1
) ∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
𝛽2 − 𝛼2 𝛼1 𝛼2
≤
𝜆2 − 𝜆 + 1
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) (M + N) ,
2
𝛽1 𝐺 (𝑥, 𝛼 ) + 𝐺 (𝑥, 𝛽 )
2
2
∫
𝑑𝑥
𝛼
4
(𝛽
−
𝛼
)
1
1
1
+∫
𝛽2
𝛼2
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
𝑑𝑦
4 (𝛽2 − 𝛼2 )
Abstract and Applied Analysis
7
−
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
≤
𝜆2 − 𝜆 + 1
(M (𝛽1 − 𝛼1 ) + N (𝛽2 − 𝛼2 )) ,
4
𝜆2 − 𝜆 + 1
⋅ ∑ ∑ 𝐺 (𝑥, 𝑦 + 1) ≤
(𝛽1 − 𝛼1 ) (𝛽2
2
𝑥=𝛼1 𝑦=𝛼2
𝛽1 −1 𝛽2 −1
− 𝛼2 ) (M + N) ,
𝛽1 −1
∑ 𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )
4 (𝛽1 − 𝛼1 )
𝑥=𝛼1
𝛽1
𝛼 + 𝛽2
(1 − 𝜆) [∫ 𝐺 (𝑥, 2
) 𝑑𝑥
2
𝛼1
𝛽2
+ ∫ 𝐺(
𝛼2
𝛼1 + 𝛽1
𝜆
, 𝑦) 𝑑𝑦] +
2
2
𝛽1
⋅ ∫ [𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )] 𝑑𝑥 +
𝛼1
𝛽2 −1
𝜆
2
𝛽2
⋅ ∫ [𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)] 𝑑𝑦 − (
𝛼2
+
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
4 (𝛽2 − 𝛼2 )
𝑦=𝛼2
+ ∑
−
1
𝛽1 − 𝛼1
⋅ ∑ ∑ 𝐺 [(𝑥, 𝑦 + 1) + 𝐺 (𝑥 + 1, 𝑦)]
𝑥=𝛼1 𝑦=𝛼2
𝛽1 −1 𝛽2 −1
𝛽1 𝛽2
1
) ∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
𝛽2 − 𝛼2 𝛼1 𝛼2
≤
2𝜆2 − 2𝜆 + 1
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) (M + N) ,
4
≤
𝛽 −1
𝛽2
𝛼 + 𝛽1
𝜆
⋅ ∫ 𝐺( 1
, 𝑦) 𝑑𝑦 +
2
4 (𝛽1 − 𝛼1 )
𝛼2
𝛽1
𝛼1
𝜆2 − 𝜆 + 1
[M (𝛽1 − 𝛼1 ) + N (𝛽2 − 𝛼2 )] ,
4
𝛽1 −1
𝛽2 −1
(1 − 𝜆) [ ∑ 𝐺 (𝑥, 𝛼2 + 𝛽2 ) + ∑ 𝐺 ( 𝛼1 + 𝛽1 , 𝑦)]
2
2
𝑥=𝛼1
𝑦=𝛼2
𝛽2
𝛼2 + 𝛽2
1−𝜆
1 − 𝜆
) 𝑑𝑥 +
2 (𝛽 − 𝛼 ) ∫𝛼 𝐺 (𝑥,
2
2
(𝛽
1
1
2 − 𝛼2 )
2
⋅ ∫ [𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )] 𝑑𝑥 +
1
2 (𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 )
+
𝜆
4 (𝛽2 − 𝛼2 )
𝜆
𝜆 1
∑ [𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )] +
2 𝑥=𝛼1
2
𝛽2 −1
⋅ ∑ [𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)] −
𝑦=𝛼2
1
𝛽1 − 𝛼1
𝛽2
⋅ ∫ [𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)] 𝑑𝑦
𝛽1 −1 𝛽2 −1
𝛼2
−
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
≤
2𝜆2 − 2𝜆 + 1
(M (𝛽1 − 𝛼1 ) + N (𝛽2 − 𝛼2 )) .
8
⋅ ∑ ∑ 𝐺 (𝑥 + 1, 𝑦) −
𝑥=𝛼1 𝑦=𝛼2
2𝜆2 − 2𝜆 + 1
⋅ ∑ ∑ 𝐺 (𝑥, 𝑦 + 1) ≤
(𝛽1 − 𝛼1 )
4
𝑥=𝛼1 𝑦=𝛼2
𝛽1 −1 𝛽2 −1
(29)
Remark 12. Corollary 11 becomes Theorem 3 if 𝜆 = 0.
Corollary 13. If we let T1 = T2 = Z in Theorems 9 and 10,
then we get the succeeding inequalities
⋅ (𝛽2 − 𝛼2 ) (M + N) ,
1 − 𝜆 𝛽2 −1
𝛼 + 𝛽2
1−𝜆
)+
∑ 𝐺 (𝑥, 2
2
2 (𝛽2 − 𝛼2 )
2 (𝛽1 − 𝛼1 ) 𝑦=𝛼2
𝛽2 −1
⋅ ∑ 𝐺(
𝑦=𝛼2
𝛽1 −1
∑ 𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )
2
𝑥=𝛼1
𝛽1 −1
𝑥=𝛼1
𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
1
−
2
𝛽1 − 𝛼1
𝑦=𝛼2
+ ∑
⋅ ∑ ∑ 𝐺 (𝑥 + 1, 𝑦) −
𝑥=𝛼1 𝑦=𝛼2
𝛼1 + 𝛽1
𝜆
, 𝑦) +
2
4 (𝛽1 − 𝛼1 )
⋅ ∑ [𝐺 (𝑥, 𝛼2 ) + 𝐺 (𝑥, 𝛽2 )] +
𝛽2 −1
𝛽1 −1 𝛽2 −1
1
𝛽2 − 𝛼2
1
𝛽2 − 𝛼2
𝛽2 −1
⋅ ∑ [𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)]
𝑦=𝛼2
−
1
2 (𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 )
𝜆
4 (𝛽2 − 𝛼2 )
8
Abstract and Applied Analysis
⋅ ∑ ∑ [𝐺 (𝑥 + 1, 𝑦) + 𝐺 (𝑥, 𝑦 + 1)]
𝑥=𝛼1 𝑦=𝛼2
𝛽1 −1 𝛽2 −1
(30)
Theorem 14. Let 𝛼1 , 𝛽1 , 𝑥 ∈ T1 , 𝛼2 , 𝛽2 , 𝑦 ∈ T2 , with 𝛼1 <
𝛽1 , 𝛼2 < 𝛽2 and 𝐺 : [𝛼1 , 𝛽1 ] × [𝛼2 , 𝛽2 ] → R be such that the
partial mappings
(31)
𝛼1
𝛼2
𝛽1
𝛽2
𝛼1
𝛼2
⋅ ∫ ∫ 𝐺 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥 +
⋅
≤
𝛽2 − 𝛼2
𝛽 − 𝛼1
[ℎ (𝛼 , 𝛽 − 𝜆 1
)
𝛽1 − 𝛼1 2 1 1
2
+ ℎ2 (𝛽1 , 𝛽1 − 𝜆
Proof. Applying Lemma 8 to the mapping 𝐺𝑦 at 𝑥 = 𝛼1 gives
Γ𝑥 + 𝛾𝑥
2
Γ𝑦 − 𝛾𝑦
𝛽2 − 𝛼2
)] +
2
2
𝛽1 − 𝛼1
)] ,
2
𝛽1
1
4 (𝛽 − 𝛼 ) [∫𝛼 ((2 − 𝜆) 𝐺 (𝑥, 𝛼2 ) + 𝜆𝐺 (𝑥, 𝛽2 )) Δ𝑥]
1
1
1
+
𝛽2
1
[∫ ((2 − 𝜆) 𝐺 (𝛼1 , 𝑦) + 𝜆𝐺 (𝛽1 , 𝑦)) Δ𝑦]
4 (𝛽2 − 𝛼2 ) 𝛼2
−
1
2 (𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 )
𝛽2 − 𝛼2
)] ,
2
hold for all 𝜆 ∈ [0, 1] such that 𝛽1 − 𝜆((𝛽1 − 𝛼1 )/2) ∈ T1 and
𝛽2 − 𝜆((𝛽2 − 𝛼2 )/2) ∈ T2 .
+
𝛽 − 𝛼2
𝛽1 − 𝛼1
[ℎ (𝛼 , 𝛽 − 𝜆 2
)
𝛽2 − 𝛼2 2 2 2
2
𝛽1 − 𝛼1
Γ − 𝛾𝑥
)] + 𝑥
2
4
𝛽 − 𝛼2
1
[ℎ2 (𝛼2 , 𝛽2 − 𝜆 2
)
𝛽2 − 𝛼2
2
1
𝛽2 − 𝛼2
Γ𝑦 + 𝛾𝑦
𝛽1 − 𝛼1
𝜆
2
[ℎ2 (𝛼2 , 𝛽2 ) − (𝛽2 − 𝛼2 ) ] +
𝛽2 − 𝛼2
2
2
𝛽 − 𝛼2
Γ − 𝛾𝑥
𝜆
2
⋅ 2
[ℎ (𝛼 , 𝛽 ) − (𝛽1 − 𝛼1 ) ] ≤ 𝑥
𝛽1 − 𝛼1 2 1 1
2
2
+ ℎ2 (𝛽2 , 𝛽2 − 𝜆
𝛽 − 𝛼1
1
[ℎ (𝛼 , 𝛽 − 𝜆 1
)
𝛽1 − 𝛼1 2 1 1
2
−
⋅
⋅
⋅
𝜆
𝜆
(1 − ) 𝐺 (𝛼1 , 𝑦) + 𝐺 (𝛽1 , 𝑦)
2
2
𝛽1
𝛽2
𝜆
1
[∫ 𝐺 (𝑥, 𝛽2 ) Δ𝑥 + ∫ 𝐺 (𝛽1 , 𝑦) Δ𝑦] −
2 𝛼1
𝛽
−
𝛼1
𝛼2
1
𝛽2
4
(33)
𝛽1
𝛽2
𝜆
(1 − ) [∫ 𝐺 (𝑥, 𝛼2 ) Δ𝑥 + ∫ 𝐺 (𝛼1 , 𝑦) Δ𝑦]
2
𝛼1
𝛼2
𝛽1
𝛼2
+ ℎ2 (𝛽2 , 𝛽2 − 𝜆
defined for all 𝑦 ∈ [𝛼2 , 𝛽2 ] and 𝑥 ∈ [𝛼1 , 𝛽1 ], are differentiable.
If 𝐺𝑥Δ , 𝐺𝑦Δ ∈ 𝐶𝑟𝑑 (T, R) and there exist 𝛾𝑥 , 𝛾𝑦 , Γ𝑥 , Γ𝑦 ∈ R such
that 𝛾𝑦 ≤ 𝐺𝑦Δ (𝑡) ≤ Γ𝑦 , 𝑡 ∈ [𝛼1 , 𝛽1 ], 𝛾𝑥 ≤ 𝐺𝑥Δ (𝑡) ≤ Γ𝑥 , 𝑡 ∈
[𝛼2 , 𝛽2 ], then the succeeding inequalities
⋅ ∫ ∫ 𝐺 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥 −
𝛼1
Γ + 𝛾𝑥
𝜆
1
2
[ℎ (𝛼 , 𝛽 ) − (𝛽1 − 𝛼1 ) ] + 𝑥
𝛽1 − 𝛼1 2 1 1
2
4
Γ − 𝛾
1
𝜆
𝑦
𝑦
2
⋅
[ℎ (𝛼 , 𝛽 ) − (𝛽2 − 𝛼2 ) ] ≤
𝛽2 − 𝛼2 2 2 2
2
4
⋅
𝐺𝑥 (𝜁) fl 𝐺 (𝑥, 𝜁) ,
+
Γ𝑦 + 𝛾𝑦
+ ℎ2 (𝛽1 , 𝛽1 − 𝜆
𝐺𝑦 : [𝛼1 , 𝛽1 ] → R,
𝐺𝑥 : [𝛼2 , 𝛽2 ] → R,
𝛽2
⋅
2𝜆2 − 2𝜆 + 1
[M (𝛽1 − 𝛼1 ) + N (𝛽2 − 𝛼2 )] .
≤
8
𝐺𝑦 (𝜉) fl 𝐺 (𝜉, 𝑦) ,
𝛽1
⋅ ∫ ∫ (𝐺 (𝑥, 𝜎 (𝑦)) + 𝐺 (𝜎 (𝑥) , 𝑦)) Δ𝑦 Δ𝑥 +
(32)
𝑏
1
∫ 𝑓 (𝜎 (𝑡) , 𝑦) Δ𝑦
𝛽1 − 𝛼1 𝛼1
Γ𝑦 + 𝛾𝑦
2
Γ𝑦 − 𝛾𝑦
2
1
𝜆
2
[ℎ2 (𝛼1 , 𝛽1 ) − (𝛽1 − 𝛼1 ) ]
𝛽1 − 𝛼1
2
(34)
𝛽 − 𝛼1
1
[ℎ2 (𝛼1 , 𝛽1 − 𝜆 1
)
𝛽1 − 𝛼1
2
+ ℎ2 (𝛽1 , 𝛽1 − 𝜆
𝛽1 − 𝛼1
)] .
2
Integrating (34) over [𝛼2 , 𝛽2 ] yields
𝑑
𝜆 𝑑
𝜆
(1 − ) ∫ 𝑓 (𝛼1 , 𝑦) Δ𝑦 + ∫ 𝑓 (𝛽1 , 𝑦) Δ𝑦
2 𝛼2
2 𝛼2
−
+
≤
𝛽1 𝑑
1
∫ ∫ 𝑓 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥
𝛽1 − 𝛼1 𝛼1 𝛼2
Γ𝑦 + 𝛾𝑦 𝛽2 − 𝛼2
2
𝛽1 − 𝛼1
Γ𝑦 − 𝛾𝑦 𝛽2 − 𝛼2
2
𝛽1 − 𝛼1
+ ℎ2 (𝛽1 , 𝛽1 − 𝜆
[ℎ2 (𝛼1 , 𝛽1 ) −
𝜆
2
(𝛽1 − 𝛼1 ) ]
2
[ℎ2 (𝛼1 , 𝛽1 − 𝜆
(35)
𝛽1 − 𝛼1
)
2
𝛽1 − 𝛼1
)] .
2
Similarly, applying Lemma 8 to the mapping 𝐺𝑥 at 𝑦 = 𝛼2 and
then integrating the resulting inequality over [𝛼1 , 𝛽1 ] give
Abstract and Applied Analysis
9
𝑏
𝜆 𝑏
𝜆
(1 − ) ∫ 𝑓 (𝑥, 𝛼2 ) Δ𝑥 + ∫ 𝑓 (𝑥, 𝛽2 ) Δ𝑥
2 𝛼1
2 𝛼1
𝛽1
𝛽2
𝜆
(1 − ) [∫ 𝐺 (𝑥, 𝛼2 ) 𝑑𝑥 + ∫ 𝐺 (𝛼1 , 𝑦) 𝑑𝑦]
2
𝛼1
𝛼2
𝛽1 𝑑
1
−
∫ ∫ 𝑓 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥
𝛽2 − 𝛼2 𝛼1 𝛼2
Γ + 𝛾𝑥 𝛽1 − 𝛼1
𝜆
2
[ℎ2 (𝛼2 , 𝛽2 ) − (𝛽2 − 𝛼2 ) ]
+ 𝑥
2 𝛽2 − 𝛼2
2
≤
+
(36)
𝛽 − 𝛼2
Γ𝑥 − 𝛾𝑥 𝛽1 − 𝛼1
[ℎ (𝛼 , 𝛽 − 𝜆 2
)
2 𝛽2 − 𝛼2 2 2 2
2
𝛽 − 𝛼2
+ ℎ2 (𝛽2 , 𝛽2 − 𝜆 2
)] .
2
(1 − 𝜆/2) 𝑑
𝛽2 − 𝛼2 ∫𝛼 𝑓 (𝛼1 , 𝑦) Δ𝑦
2
𝑑
𝜆
+
∫ 𝑓 (𝛽1 , 𝑦) Δ𝑦
2 (𝛽2 − 𝛼2 ) 𝛼2
−
𝑑
1
∫ ∫ 𝑓 (𝜎 (𝑥) , 𝑦) Δ𝑦 Δ𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
Γ𝑦 + 𝛾𝑦
1
𝜆
2
+
[ℎ2 (𝛼1 , 𝛽1 ) − (𝛽1 − 𝛼1 ) ]
2 𝛽1 − 𝛼1
2
≤
Γ𝑦 − 𝛾𝑦
2
𝛽1 − 𝛼1
)] ,
2
(1 − 𝜆/2) 𝑏
𝛽1 − 𝛼1 ∫𝛼 𝑓 (𝑥, 𝛼2 ) Δ𝑥
1
+
−
+
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 )
(1 − 𝜆) (Γ𝑥 + 𝛾𝑥 + Γ𝑦 + 𝛾𝑦 )
4
≤
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 )
((𝜆 − 1)2 + 1) (Γ𝑥 − 𝛾𝑥 + Γ𝑦
8
𝛽1
1
4 (𝛽 − 𝛼 ) [∫𝛼 ((2 − 𝜆) 𝐺 (𝑥, 𝛼2 ) + 𝜆𝐺 (𝑥, 𝛽2 )) 𝑑𝑥]
1
1
1
+
𝛽2
1
[∫ ((2 − 𝜆) 𝐺 (𝛼1 , 𝑦) + 𝜆𝐺 (𝛽1 , 𝑦)) 𝑑𝑦]
4 (𝛽2 − 𝛼2 ) 𝛼2
−
𝛽1 𝛽2
1
∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
+
1−𝜆
((Γ𝑦 + 𝛾𝑦 ) (𝛽1 − 𝛼1 ) + (Γ𝑥 + 𝛾𝑥 ) (𝛽2 − 𝛼2 ))
8
≤
(𝜆 − 1)2 + 1
((Γ𝑦 − 𝛾𝑦 ) (𝛽1 − 𝛼1 )
16
(38)
Remark 16. Corollary 15 becomes Theorem 5 if 𝜆 = 1.
(37)
𝑏
𝜆
∫ 𝑓 (𝑥, 𝛽2 ) Δ𝑥
2 (𝛽1 − 𝛼1 ) 𝛼1
𝛽1 𝑑
1
∫ ∫ 𝑓 (𝑥, 𝜎 (𝑦)) Δ𝑦 Δ𝑥
(𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 ) 𝛼1 𝛼2
Γ + 𝛾𝑥
1
𝜆
2
+ 𝑥
[ℎ2 (𝛼2 , 𝛽2 ) − (𝛽2 − 𝛼2 ) ]
2 𝛽2 − 𝛼2
2
Γ − 𝛾𝑥
𝛽 − 𝛼2
1
≤ 𝑥
[ℎ2 (𝛼2 , 𝛽2 − 𝜆 2
)
2 𝛽2 − 𝛼2
2
+ ℎ2 (𝛽2 , 𝛽2 − 𝜆
𝛽1 𝛽2
1
1
+
) ∫ ∫ 𝐺 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥
𝛽1 − 𝛼1 𝛽2 − 𝛼2 𝛼1 𝛼2
+ (Γ𝑥 − 𝛾𝑥 ) (𝛽2 − 𝛼2 )) .
𝛽 − 𝛼1
1
[ℎ (𝛼 , 𝛽 − 𝜆 1
)
𝛽1 − 𝛼1 2 1 1
2
+ ℎ2 (𝛽1 , 𝛽1 − 𝜆
−(
− 𝛾𝑦 ) ,
Using (35) and (36), we get (32). Also, we obtain from (35)
and (36) the following inequalities:
𝛽1
𝛽1
𝛽2
𝜆
[∫ 𝐺 (𝑥, 𝛽2 ) 𝑑𝑥 + ∫ 𝐺 (𝛽1 , 𝑦) 𝑑𝑦]
2 𝛼1
𝛼2
𝛽2 − 𝛼2
)] .
2
Using (37) amounts to (33). That completes the proof of
Theorem 14.
Corollary 15. If we let T1 = T2 = R in Theorem 10, then the
succeeding inequalities hold:
Corollary 17. If we let T1 = T2 = Z in Theorem 10, then the
succeeding inequalities hold:
𝛽1 −1
𝛽2 −1
(1 − 𝜆 ) [ ∑ 𝐺 (𝑥, 𝛼 ) + ∑ 𝐺 (𝛼 , 𝑦)]
2
1
2
𝑥=𝛼1
𝑦=𝛼2
𝛽 −1
+
𝛽 −1
2
𝜆 1
1
[ ∑ 𝐺 (𝑥, 𝛽2 ) + ∑ 𝐺 (𝛽1 , 𝑦)] −
2 𝑥=𝛼1
𝛽1 − 𝛼1
𝑦=𝛼2
𝛽1 −1 𝛽2 −1
𝛽 −1 𝛽 −1
⋅ ∑ ∑ 𝐺 (𝑥 + 1, 𝑦) −
𝑥=𝛼1 𝑦=𝛼2
+
+
≤
1
2
1
∑ ∑ 𝐺 (𝑥, 𝑦 + 1)
𝛽2 − 𝛼2 𝑥=𝛼1 𝑦=𝛼2
Γ𝑥 + 𝛾𝑥
(𝛽1 − 𝛼1 ) [(𝛽2 − 𝛼2 + 1) − 𝜆 (𝛽2 − 𝛼2 )]
4
Γ𝑦 + 𝛾𝑦
4
(𝛽2 − 𝛼2 ) [(𝛽1 − 𝛼1 + 1) − 𝜆 (𝛽1 − 𝛼1 )]
Γ𝑥 − 𝛾𝑥
(𝛽1 − 𝛼1 ) ((𝛽2 − 𝛼2 ) (𝜆2 − 2𝜆 + 2) − 2𝜆
8
+ 2) +
Γ𝑦 − 𝛾𝑦
8
− 2𝜆 + 2) ,
(𝛽2 − 𝛼2 ) ((𝛽1 − 𝛼1 ) (𝜆2 − 2𝜆 + 2)
10
Abstract and Applied Analysis
𝛽1 −1
1
[ ∑ ((2 − 𝜆) 𝐺 (𝑥, 𝛼2 ) + 𝜆𝐺 (𝑥, 𝛽2 ))]
4 (𝛽1 − 𝛼1 ) 𝑥=𝛼1
𝛽 −1
+
2
1
[ ∑ ((2 − 𝜆) 𝐺 (𝛼1 , 𝑦) + 𝜆𝐺 (𝛽1 , 𝑦))]
4 (𝛽2 − 𝛼2 ) 𝑦=𝛼2
−
1
2 (𝛽1 − 𝛼1 ) (𝛽2 − 𝛼2 )
𝛽1 −1 𝛽2 −1
⋅ ∑ ∑ (𝐺 (𝑥, 𝑦 + 1) + 𝐺 (𝑥 + 1, 𝑦))
𝑥=𝛼1 𝑦=𝛼2
+
Γ𝑦 + 𝛾𝑦
8
[(𝛽1 − 𝛼1 + 1) − 𝜆 (𝛽1 − 𝛼1 )]
Γ𝑥 + 𝛾𝑥
[(𝛽2 − 𝛼2 + 1) − 𝜆 (𝛽2 − 𝛼2 )]
+
8
≤
+
Γ𝑦 − 𝛾𝑦
16
((𝛽1 − 𝛼1 ) (𝜆2 − 2𝜆 + 2) − 2𝜆 + 2)
Γ𝑥 − 𝛾𝑥
((𝛽2 − 𝛼2 ) (𝜆2 − 2𝜆 + 2) − 2𝜆 + 2) .
16
(39)
3. Conclusion
Three main theorems are hereby established. The results of
Farid [3] are obtained as special cases of our results. Loads
of interesting new inequalities can be obtained by choosing
different values of 𝜆 ∈ [0, 1], and considering a different time
scale different from R and Z.
Conflicts of Interest
The authors declare that there are no conflicts of interest.
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[13] S. Hussain, M. A. Latif, and M. Alomari, “Generalized doubleintegral Ostrowski type inequalities on time scales,” Applied
Mathematics Letters, vol. 24, no. 8, pp. 1461–1467, 2011.
[14] W. Irshad, M. I. Bhatti, and M. Muddassar, “Some Ostrowski
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5, pp. 914–927, 2016.
[15] W. Liu, Q. A. Ngô, and W. Chen, “Ostrowski type inequalities on
time scales for double integrals,” Acta Applicandae Mathematicae, vol. 110, no. 1, pp. 477–497, 2010.
[16] W. Liu, Q. A. Ngô, and W. Chen, “On new Ostrowski type inequalities for double integrals on time scales,” Dynamic Systems
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[17] U. M. Özkan and H. Yildirim, “Ostrowski type inequality for
double integrals on time scales,” Acta Applicandae Mathematicae, vol. 110, no. 1, pp. 283–288, 2010.
[18] A. Tuna and S. Kutukcu, “A new generalization of the Ostrowski
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Volume 2018 | 24,194 | 28,946 | {"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-2020-16 | latest | en | 0.838661 |
https://www.physicsforums.com/threads/noncommutative-geometries-from-first-principles.402173/ | 1,604,078,963,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911027.72/warc/CC-MAIN-20201030153002-20201030183002-00484.warc.gz | 830,973,983 | 16,499 | # Noncommutative Geometries from first principles
Noncommutative Geometries from "first principles"
Hi everyone,
to give a motivation for studying specific models of noncommutative geometry, I would like to start this thread as a collecting tank of models of noncommutative geometry that are obtained as a limit of some kind from 'first principles', i.e. not an ad-hoc modification of the commutator [x_i,x_j] . It would be nice, if we could collect cases by briefly stating what theory in which limit they are obtained from, provide the specific form of the coordinate commutator, and cite a reference (if possible, respectively). So, as a start:
----------------
From 2+1-dim. Spinfoams, by integrating out gravitational DOF, to arrive at the flat-space effective field theory:
$$[X_i, X_j]=i\hbar \kappa \epsilon_{ijk} X_k, \quad \kappa = 4\pi G$$
Reference: http://arxiv.org/abs/0705.2222
----------------
As some knowledgeable people are around here, I hope some members share interest in this and contribute . For example I heard in string theory world-sheet coordinates do not commute, but I don't know much about it, i.e. which precise form in which case.
Related Beyond the Standard Model News on Phys.org
This thread is "above my pay grade" but for those who are more versed than I, Roger Penrose in THE ROAD TO REALITY has some relevant discussions, especially in 33.1, where on pages 961 and 962 he gives some non commutative examples.
He attributes much non commutative geometry insight to Alan Connes and goes on to say
...in quantum mechanics one frequently encounters algebras that are non commutative....Connes and his colleagues developed the idea of non commutative geometry with a view to producing a physical theory which includes the standard model of particle physics....the potential richness of the idea of non-commutative geometry does not seem to me to be at all strongly used, so far....twistor theory has some relations to spin network theory and to Ashtekar variables and possibly non-commutative geometry....
Last edited:
Hi Naty,
so it is for me actually , which is why I hope that others pop in to contribute. Anyways, thanks for your reference. So from THE ROAD TO REALITY, p. 983, one at least has
$$[Z^{\alpha},\overline{Z}_{\beta}]=\hbar \delta^{\alpha}_{\beta},$$
and as the twistor components Z, if I understand correctly, are actually composed of spacetime coordinates, it would be interesting to see what this means for the coordinate commutator. However I couldn't find it worked out anywhere and I don't know if one link this framework to a coordinate commutator in a sensible way.
---------
Edit: As for Connes, I think the approach is a little different, as one considers a product of a commutative spin manifold M (accounting for spacetime) with a finite, noncommutative space accounting for matter content. While this as well seems very worthwile to study, we do have some threads about this, so that I would prefer to talk about scenarios in which the ordinary space/spacetime coordinates are noncommuting.
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https://www.kodytools.com/units/massflow/from/egph/to/egpmin | 1,725,992,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00532.warc.gz | 786,998,428 | 15,504 | Exagram/Hour to Exagram/Minute Converter
1 Exagram/Hour = 0.016666666666667 Exagram/Minute
One Exagram/Hour is Equal to How Many Exagram/Minute?
The answer is one Exagram/Hour is equal to 0.016666666666667 Exagram/Minute and that means we can also write it as 1 Exagram/Hour = 0.016666666666667 Exagram/Minute. Feel free to use our online unit conversion calculator to convert the unit from Exagram/Hour to Exagram/Minute. Just simply enter value 1 in Exagram/Hour and see the result in Exagram/Minute.
Manually converting Exagram/Hour to Exagram/Minute can be time-consuming,especially when you don’t have enough knowledge about Mass Flow units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Exagram/Hour to Exagram/Minute converter tool to get the job done as soon as possible.
We have so many online tools available to convert Exagram/Hour to Exagram/Minute, but not every online tool gives an accurate result and that is why we have created this online Exagram/Hour to Exagram/Minute converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
How to Convert Exagram/Hour to Exagram/Minute (Eg/h to Eg/min)
By using our Exagram/Hour to Exagram/Minute conversion tool, you know that one Exagram/Hour is equivalent to 0.016666666666667 Exagram/Minute. Hence, to convert Exagram/Hour to Exagram/Minute, we just need to multiply the number by 0.016666666666667. We are going to use very simple Exagram/Hour to Exagram/Minute conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Exagram/Hour} = 1 \times 0.016666666666667 = \text{0.016666666666667 Exagram/Minute}$$
What Unit of Measure is Exagram/Hour?
Exagram/Hour or Exagram per Hour is a unit of measurement for mass flow rate. It is defined as flow or movement of one exagram of mass in one hour.
What is the Symbol of Exagram/Hour?
The symbol of Exagram/Hour is Eg/h. This means you can also write one Exagram/Hour as 1 Eg/h.
What Unit of Measure is Exagram/Minute?
Exagram/Minute or Exagram per Minute is a unit of measurement for mass flow rate. It is defined as flow or movement of one exagram of mass in one minute.
What is the Symbol of Exagram/Minute?
The symbol of Exagram/Minute is Eg/min. This means you can also write one Exagram/Minute as 1 Eg/min.
How to Use Exagram/Hour to Exagram/Minute Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Exagram/Hour and in the first input field, enter a value.
• From the second dropdown, select Exagram/Minute.
• Instantly, the tool will convert the value from Exagram/Hour to Exagram/Minute and display the result in the second input field.
Example of Exagram/Hour to Exagram/Minute Converter Tool
Exagram/Hour
1
Exagram/Minute
0.016666666666667
Exagram/Hour to Exagram/Minute Conversion Table
Exagram/Hour [Eg/h]Exagram/Minute [Eg/min]Description
1 Exagram/Hour0.016666666666667 Exagram/Minute1 Exagram/Hour = 0.016666666666667 Exagram/Minute
2 Exagram/Hour0.033333333333333 Exagram/Minute2 Exagram/Hour = 0.033333333333333 Exagram/Minute
3 Exagram/Hour0.05 Exagram/Minute3 Exagram/Hour = 0.05 Exagram/Minute
4 Exagram/Hour0.066666666666667 Exagram/Minute4 Exagram/Hour = 0.066666666666667 Exagram/Minute
5 Exagram/Hour0.083333333333333 Exagram/Minute5 Exagram/Hour = 0.083333333333333 Exagram/Minute
6 Exagram/Hour0.1 Exagram/Minute6 Exagram/Hour = 0.1 Exagram/Minute
7 Exagram/Hour0.11666666666667 Exagram/Minute7 Exagram/Hour = 0.11666666666667 Exagram/Minute
8 Exagram/Hour0.13333333333333 Exagram/Minute8 Exagram/Hour = 0.13333333333333 Exagram/Minute
9 Exagram/Hour0.15 Exagram/Minute9 Exagram/Hour = 0.15 Exagram/Minute
10 Exagram/Hour0.16666666666667 Exagram/Minute10 Exagram/Hour = 0.16666666666667 Exagram/Minute
100 Exagram/Hour1.67 Exagram/Minute100 Exagram/Hour = 1.67 Exagram/Minute
1000 Exagram/Hour16.67 Exagram/Minute1000 Exagram/Hour = 16.67 Exagram/Minute
Exagram/Hour to Other Units Conversion Table
ConversionDescription
1 Exagram/Hour = 24000000000000000 Kilogram/Day1 Exagram/Hour in Kilogram/Day is equal to 24000000000000000
1 Exagram/Hour = 1000000000000000 Kilogram/Hour1 Exagram/Hour in Kilogram/Hour is equal to 1000000000000000
1 Exagram/Hour = 16666666666667 Kilogram/Minute1 Exagram/Hour in Kilogram/Minute is equal to 16666666666667
1 Exagram/Hour = 277777777777.78 Kilogram/Second1 Exagram/Hour in Kilogram/Second is equal to 277777777777.78
1 Exagram/Hour = 2.4e+22 Milligram/Day1 Exagram/Hour in Milligram/Day is equal to 2.4e+22
1 Exagram/Hour = 1e+21 Milligram/Hour1 Exagram/Hour in Milligram/Hour is equal to 1e+21
1 Exagram/Hour = 16666666666667000000 Milligram/Minute1 Exagram/Hour in Milligram/Minute is equal to 16666666666667000000
1 Exagram/Hour = 277777777777780000 Milligram/Second1 Exagram/Hour in Milligram/Second is equal to 277777777777780000
1 Exagram/Hour = 24000000000000000000 Gram/Day1 Exagram/Hour in Gram/Day is equal to 24000000000000000000
1 Exagram/Hour = 1000000000000000000 Gram/Hour1 Exagram/Hour in Gram/Hour is equal to 1000000000000000000
1 Exagram/Hour = 16666666666667000 Gram/Minute1 Exagram/Hour in Gram/Minute is equal to 16666666666667000
1 Exagram/Hour = 277777777777780 Gram/Second1 Exagram/Hour in Gram/Second is equal to 277777777777780
1 Exagram/Hour = 24 Exagram/Day1 Exagram/Hour in Exagram/Day is equal to 24
1 Exagram/Hour = 0.016666666666667 Exagram/Minute1 Exagram/Hour in Exagram/Minute is equal to 0.016666666666667
1 Exagram/Hour = 0.00027777777777778 Exagram/Second1 Exagram/Hour in Exagram/Second is equal to 0.00027777777777778
1 Exagram/Hour = 24000 Petagram/Day1 Exagram/Hour in Petagram/Day is equal to 24000
1 Exagram/Hour = 1000 Petagram/Hour1 Exagram/Hour in Petagram/Hour is equal to 1000
1 Exagram/Hour = 16.67 Petagram/Minute1 Exagram/Hour in Petagram/Minute is equal to 16.67
1 Exagram/Hour = 0.27777777777778 Petagram/Second1 Exagram/Hour in Petagram/Second is equal to 0.27777777777778
1 Exagram/Hour = 24000000 Teragram/Day1 Exagram/Hour in Teragram/Day is equal to 24000000
1 Exagram/Hour = 1000000 Teragram/Hour1 Exagram/Hour in Teragram/Hour is equal to 1000000
1 Exagram/Hour = 16666.67 Teragram/Minute1 Exagram/Hour in Teragram/Minute is equal to 16666.67
1 Exagram/Hour = 277.78 Teragram/Second1 Exagram/Hour in Teragram/Second is equal to 277.78
1 Exagram/Hour = 24000000000 Gigagram/Day1 Exagram/Hour in Gigagram/Day is equal to 24000000000
1 Exagram/Hour = 1000000000 Gigagram/Hour1 Exagram/Hour in Gigagram/Hour is equal to 1000000000
1 Exagram/Hour = 16666666.67 Gigagram/Minute1 Exagram/Hour in Gigagram/Minute is equal to 16666666.67
1 Exagram/Hour = 277777.78 Gigagram/Second1 Exagram/Hour in Gigagram/Second is equal to 277777.78
1 Exagram/Hour = 24000000000000 Megagram/Day1 Exagram/Hour in Megagram/Day is equal to 24000000000000
1 Exagram/Hour = 1000000000000 Megagram/Hour1 Exagram/Hour in Megagram/Hour is equal to 1000000000000
1 Exagram/Hour = 16666666666.67 Megagram/Minute1 Exagram/Hour in Megagram/Minute is equal to 16666666666.67
1 Exagram/Hour = 277777777.78 Megagram/Second1 Exagram/Hour in Megagram/Second is equal to 277777777.78
1 Exagram/Hour = 240000000000000000 Hectogram/Day1 Exagram/Hour in Hectogram/Day is equal to 240000000000000000
1 Exagram/Hour = 10000000000000000 Hectogram/Hour1 Exagram/Hour in Hectogram/Hour is equal to 10000000000000000
1 Exagram/Hour = 166666666666670 Hectogram/Minute1 Exagram/Hour in Hectogram/Minute is equal to 166666666666670
1 Exagram/Hour = 2777777777777.8 Hectogram/Second1 Exagram/Hour in Hectogram/Second is equal to 2777777777777.8
1 Exagram/Hour = 2400000000000000000 Dekagram/Day1 Exagram/Hour in Dekagram/Day is equal to 2400000000000000000
1 Exagram/Hour = 100000000000000000 Dekagram/Hour1 Exagram/Hour in Dekagram/Hour is equal to 100000000000000000
1 Exagram/Hour = 1666666666666700 Dekagram/Minute1 Exagram/Hour in Dekagram/Minute is equal to 1666666666666700
1 Exagram/Hour = 27777777777778 Dekagram/Second1 Exagram/Hour in Dekagram/Second is equal to 27777777777778
1 Exagram/Hour = 240000000000000000000 Decigram/Day1 Exagram/Hour in Decigram/Day is equal to 240000000000000000000
1 Exagram/Hour = 10000000000000000000 Decigram/Hour1 Exagram/Hour in Decigram/Hour is equal to 10000000000000000000
1 Exagram/Hour = 166666666666670000 Decigram/Minute1 Exagram/Hour in Decigram/Minute is equal to 166666666666670000
1 Exagram/Hour = 2777777777777800 Decigram/Second1 Exagram/Hour in Decigram/Second is equal to 2777777777777800
1 Exagram/Hour = 2.4e+21 Centigram/Day1 Exagram/Hour in Centigram/Day is equal to 2.4e+21
1 Exagram/Hour = 100000000000000000000 Centigram/Hour1 Exagram/Hour in Centigram/Hour is equal to 100000000000000000000
1 Exagram/Hour = 1666666666666700000 Centigram/Minute1 Exagram/Hour in Centigram/Minute is equal to 1666666666666700000
1 Exagram/Hour = 27777777777778000 Centigram/Second1 Exagram/Hour in Centigram/Second is equal to 27777777777778000
1 Exagram/Hour = 2.4e+25 Microgram/Day1 Exagram/Hour in Microgram/Day is equal to 2.4e+25
1 Exagram/Hour = 1e+24 Microgram/Hour1 Exagram/Hour in Microgram/Hour is equal to 1e+24
1 Exagram/Hour = 1.6666666666667e+22 Microgram/Minute1 Exagram/Hour in Microgram/Minute is equal to 1.6666666666667e+22
1 Exagram/Hour = 277777777777780000000 Microgram/Second1 Exagram/Hour in Microgram/Second is equal to 277777777777780000000
1 Exagram/Hour = 52910942924371000 Pound/Day1 Exagram/Hour in Pound/Day is equal to 52910942924371000
1 Exagram/Hour = 2204622621848800 Pound/Hour1 Exagram/Hour in Pound/Hour is equal to 2204622621848800
1 Exagram/Hour = 36743710364146 Pound/Minute1 Exagram/Hour in Pound/Minute is equal to 36743710364146
1 Exagram/Hour = 612395172735.77 Pound/Second1 Exagram/Hour in Pound/Second is equal to 612395172735.77
1 Exagram/Hour = 846575086789930000 Ounce/Day1 Exagram/Hour in Ounce/Day is equal to 846575086789930000
1 Exagram/Hour = 35273961949580000 Ounce/Hour1 Exagram/Hour in Ounce/Hour is equal to 35273961949580000
1 Exagram/Hour = 587899365826340 Ounce/Minute1 Exagram/Hour in Ounce/Minute is equal to 587899365826340
1 Exagram/Hour = 9798322763772.3 Ounce/Second1 Exagram/Hour in Ounce/Second is equal to 9798322763772.3
1 Exagram/Hour = 26455471462185 Ton/Day [Short]1 Exagram/Hour in Ton/Day [Short] is equal to 26455471462185
1 Exagram/Hour = 1102311310924.4 Ton/Hour [Short]1 Exagram/Hour in Ton/Hour [Short] is equal to 1102311310924.4
1 Exagram/Hour = 18371855182.07 Ton/Minute [Short]1 Exagram/Hour in Ton/Minute [Short] is equal to 18371855182.07
1 Exagram/Hour = 306197586.37 Ton/Second [Short]1 Exagram/Hour in Ton/Second [Short] is equal to 306197586.37
1 Exagram/Hour = 24000000000000 Ton/Day [Metric]1 Exagram/Hour in Ton/Day [Metric] is equal to 24000000000000
1 Exagram/Hour = 1000000000000 Ton/Hour [Metric]1 Exagram/Hour in Ton/Hour [Metric] is equal to 1000000000000
1 Exagram/Hour = 16666666666.67 Ton/Minute [Metric]1 Exagram/Hour in Ton/Minute [Metric] is equal to 16666666666.67
1 Exagram/Hour = 277777777.78 Ton/Second [Metric]1 Exagram/Hour in Ton/Second [Metric] is equal to 277777777.78 | 3,682 | 11,163 | {"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} | 2.890625 | 3 | CC-MAIN-2024-38 | latest | en | 0.87217 |
http://de.metamath.org/mpeuni/usg2cwwkdifex.html | 1,726,795,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00791.warc.gz | 8,116,335 | 7,231 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > usg2cwwkdifex Structured version Visualization version GIF version
Theorem usg2cwwkdifex 26349
Description: If a word represents a closed walk of length at least 2 in a undirected simple graph, the first two symbols of the word must be different. (Contributed by Alexander van der Vekens, 17-Jun-2018.)
Assertion
Ref Expression
usg2cwwkdifex ((𝑉 USGrph 𝐸𝑁 ∈ (ℤ‘2) ∧ 𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘𝑁)) → ∃𝑖 ∈ (0..^𝑁)(𝑊𝑖) ≠ (𝑊‘0))
Distinct variable groups: 𝑖,𝐸 𝑖,𝑁 𝑖,𝑉 𝑖,𝑊
Proof of Theorem usg2cwwkdifex
StepHypRef Expression
1 1nn0 11185 . . . . 5 1 ∈ ℕ0
21a1i 11 . . . 4 (𝑁 ∈ (ℤ‘2) → 1 ∈ ℕ0)
3 eluz2nn 11602 . . . 4 (𝑁 ∈ (ℤ‘2) → 𝑁 ∈ ℕ)
4 eluz2 11569 . . . . 5 (𝑁 ∈ (ℤ‘2) ↔ (2 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 2 ≤ 𝑁))
5 1red 9934 . . . . . . . . 9 (𝑁 ∈ ℤ → 1 ∈ ℝ)
6 2re 10967 . . . . . . . . . 10 2 ∈ ℝ
76a1i 11 . . . . . . . . 9 (𝑁 ∈ ℤ → 2 ∈ ℝ)
8 zre 11258 . . . . . . . . 9 (𝑁 ∈ ℤ → 𝑁 ∈ ℝ)
95, 7, 83jca 1235 . . . . . . . 8 (𝑁 ∈ ℤ → (1 ∈ ℝ ∧ 2 ∈ ℝ ∧ 𝑁 ∈ ℝ))
109adantr 480 . . . . . . 7 ((𝑁 ∈ ℤ ∧ 2 ≤ 𝑁) → (1 ∈ ℝ ∧ 2 ∈ ℝ ∧ 𝑁 ∈ ℝ))
11 simpr 476 . . . . . . . 8 ((𝑁 ∈ ℤ ∧ 2 ≤ 𝑁) → 2 ≤ 𝑁)
12 1lt2 11071 . . . . . . . 8 1 < 2
1311, 12jctil 558 . . . . . . 7 ((𝑁 ∈ ℤ ∧ 2 ≤ 𝑁) → (1 < 2 ∧ 2 ≤ 𝑁))
14 ltletr 10008 . . . . . . 7 ((1 ∈ ℝ ∧ 2 ∈ ℝ ∧ 𝑁 ∈ ℝ) → ((1 < 2 ∧ 2 ≤ 𝑁) → 1 < 𝑁))
1510, 13, 14sylc 63 . . . . . 6 ((𝑁 ∈ ℤ ∧ 2 ≤ 𝑁) → 1 < 𝑁)
16153adant1 1072 . . . . 5 ((2 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 2 ≤ 𝑁) → 1 < 𝑁)
174, 16sylbi 206 . . . 4 (𝑁 ∈ (ℤ‘2) → 1 < 𝑁)
18 elfzo0 12376 . . . 4 (1 ∈ (0..^𝑁) ↔ (1 ∈ ℕ0𝑁 ∈ ℕ ∧ 1 < 𝑁))
192, 3, 17, 18syl3anbrc 1239 . . 3 (𝑁 ∈ (ℤ‘2) → 1 ∈ (0..^𝑁))
20193ad2ant2 1076 . 2 ((𝑉 USGrph 𝐸𝑁 ∈ (ℤ‘2) ∧ 𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘𝑁)) → 1 ∈ (0..^𝑁))
21 fveq2 6103 . . . 4 (𝑖 = 1 → (𝑊𝑖) = (𝑊‘1))
2221adantl 481 . . 3 (((𝑉 USGrph 𝐸𝑁 ∈ (ℤ‘2) ∧ 𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘𝑁)) ∧ 𝑖 = 1) → (𝑊𝑖) = (𝑊‘1))
2322neeq1d 2841 . 2 (((𝑉 USGrph 𝐸𝑁 ∈ (ℤ‘2) ∧ 𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘𝑁)) ∧ 𝑖 = 1) → ((𝑊𝑖) ≠ (𝑊‘0) ↔ (𝑊‘1) ≠ (𝑊‘0)))
24 usg2cwwk2dif 26348 . 2 ((𝑉 USGrph 𝐸𝑁 ∈ (ℤ‘2) ∧ 𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘𝑁)) → (𝑊‘1) ≠ (𝑊‘0))
2520, 23, 24rspcedvd 3289 1 ((𝑉 USGrph 𝐸𝑁 ∈ (ℤ‘2) ∧ 𝑊 ∈ ((𝑉 ClWWalksN 𝐸)‘𝑁)) → ∃𝑖 ∈ (0..^𝑁)(𝑊𝑖) ≠ (𝑊‘0))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 ≠ wne 2780 ∃wrex 2897 class class class wbr 4583 ‘cfv 5804 (class class class)co 6549 ℝcr 9814 0cc0 9815 1c1 9816 < clt 9953 ≤ cle 9954 ℕcn 10897 2c2 10947 ℕ0cn0 11169 ℤcz 11254 ℤ≥cuz 11563 ..^cfzo 12334 USGrph cusg 25859 ClWWalksN cclwwlkn 26277 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-oadd 7451 df-er 7629 df-map 7746 df-pm 7747 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-card 8648 df-cda 8873 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-n0 11170 df-z 11255 df-uz 11564 df-fz 12198 df-fzo 12335 df-hash 12980 df-word 13154 df-usgra 25862 df-clwwlk 26279 df-clwwlkn 26280 This theorem is referenced by: usghashecclwwlk 26362
Copyright terms: Public domain W3C validator | 2,887 | 4,776 | {"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} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.133428 |
https://www.stat.math.ethz.ch/pipermail/r-help/2007-May/130988.html | 1,653,287,599,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00203.warc.gz | 1,152,000,963 | 2,213 | # [R] ED50 from logistic model with interactions
Berwin A Turlach berwin at maths.uwa.edu.au
Wed May 2 08:45:57 CEST 2007
```
On Wed, 02 May 2007 11:37:22 +1000
Kate Stark <lhodgson at postoffice.utas.edu.au> wrote:
> [...] My model is:
>
> fit <- glm(Mature ~ Season * Size - 1, family = binomial, data=dat)
>
> where Mature is a binary response, 0 for immature, 1 for mature. There
> are 3 Seasons.
I would use:
fit <- glm(Mature ~ Season/Size - 1, family=binomial, data=dat)
With this parameterisation you get the three intercepts and the three
slopes directly (together with there standard errors from summary()).
Makes life simpler for your calculations.
> In Faraway(2006) he has an example using the delta method to calculate
> the StdErr, but again without any interactions. I can apply this for
> the first Season, as there is just one intercept and one slope
> coefficient, but for the other 2 Seasons, the slope is a combination
> of the Size coefficient and the Size*Season coefficient, [...]
Not with the above parameterisation, so life is easier. I don't have
my copy of Faraway (2006) handy at the moment, so I cannot vouch that
you can use the method the describes now directly. But I expect you
can. :)
> I could divide the data and do 3 different logistic regressions, one
> for each season, but while the Mat50 (i.e. mean Size at 50% maturity)
> is the same as that calculated by the separate lines regression, Im
> not sure how this may change the StdErr?
As far as I can tell, there should be no difference. Compare the
estimates and their standard errors that you obtain from the 3
different logistic fits with the estimates and standard errors from the
parameterisation that I suggested. They should be the same.
Hope this helps.
Cheers,
Berwin
``` | 471 | 1,789 | {"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-2022-21 | latest | en | 0.905022 |
http://jabsto.com/Tutorial/topic-104/Microsoft-Excel-2013-Formulas-193.html | 1,555,966,742,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578582584.59/warc/CC-MAIN-20190422195208-20190422221208-00167.warc.gz | 88,825,503 | 4,221 | Microsoft Office Tutorials and References
In Depth Information
20130821
To convert this string to an actual date, you can use a formula such as this one, which assumes the coded date is
in cell A1:
=DATE(LEFT(A1,4),MID(A1,5,2),RIGHT(A1,2))
This formula uses text functions (LEFT, MID, and RIGHT) to extract the digits and then uses these extracted
digits as arguments for the DATE function.
See Chapter 5 for more information about using formulas to manipulate text.
Calculating the number of days between two dates
A common type of date calculation determines the number of days between two dates. For example, you may
have a financial worksheet that calculates interest earned on a deposit account. The interest earned depends on
how many days that the account is open. If your sheet contains the open date and the close date for the account,
you can calculate the number of days the account was open.
Because dates store as consecutive serial numbers, you can use simple subtraction to calculate the number of
days between two dates. For example, if cells A1 and B1 both contain a date, the following formula returns the
number of days between these dates:
=A1-B1
If cell B1 contains a more recent date than the date in cell A1, the result will be negative. If you don't care about
which date is earlier and want to avoid displaying a negative value, use this formula:
=ABS(A1-B1)
You can also use the DAYS worksheet function, introduced in Excel 2013. It offers no
advantage that I can see, but here's an example of how to use it to calculate the number
of days between two dates:
=DAYS(A1,B1)
Sometimes, calculating the difference between two days is more difficult. To demonstrate, consider the common
“fence post” analogy. If somebody asks you how many units make up a fence, you can respond with either of
two answers: the number of fence posts, or the number of gaps between the fence posts. The number of fence
posts is always one more than the number of gaps between the posts.
Search JabSto ::
Custom Search | 459 | 2,020 | {"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} | 3.609375 | 4 | CC-MAIN-2019-18 | latest | en | 0.906472 |
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### Contents
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How Much House Can I Afford – The Dough Roller – DoughRoller » Mortgages » 5 Ways to Calculate How Much House You Can Afford. 5 Ways to Calculate How Much House You Can Afford
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Zillow’s Home Affordability Calculator will help you determine how much house you can afford by analyzing your income, debt, and the current mortgage rates.
How Much Home can I Afford? How We Calculate it.. The average American household income is \$73,298, assuming you have no monthly debt payments you can afford a home priced at \$285,000 with a 3.5% (\$10,000) down payment for \$1,800 per month.
Annual Household Income. In order to determine how much you can afford to pay each month, we start by looking at how much you earn (salary, wages, tips, commission, etc.) each year before taxes.
Buying A House What Can I Afford How much house can you afford? – How much house can you afford? If that question is on your mind, you’re in good company. The fall buying market is here, and the housing market remains strong across most of the country. home prices.Best Books For New Homeowners The 7 Best Gardening Books of 2019 – thespruce.com – Read reviews and buy the best gardening books from top authors including Deborah L. Martin, Edward C. Smith, Lewis Hill and more.. Best for Small Spaces: All New Square Foot Gardening II at Amazon. Best Home Products The 7 Best Poufs for Your Feet or Seat of 2019
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How Much Can I afford? fha mortgage calculator. Use the following calculator to help you determine an affordable monthly payment so that you know what you can afford before you make an offer on the home you want to purchase.
How Much If A House Can I Afford Mortgage Calculator – How Much Home Can You Afford? | Credit. – Home Affordability Calculator. This calculator will give you a better idea of how much you can afford to pay for a house and what the monthly payment will be. | 765 | 3,489 | {"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} | 2.765625 | 3 | CC-MAIN-2020-16 | latest | en | 0.947949 |
https://mathematica.stackexchange.com/questions/222497/replace-the-values-of-the-second-position-in-a-list-given-two-conditions | 1,695,876,205,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00497.warc.gz | 419,468,616 | 40,472 | Replace the values of the second position in a list given two conditions
Suppose I have the following list:
https://pastebin.com/zFin7kkB
and I have a given value of maxDh (let's say 0.5) . How can I replace only the values in the second position with a given operation (e.g. maxDh*5) only when the values of the first position are greater than 74 and maxDh is lower than 1.2 (as in this case)?. For example given that in this example maxDh=0.5, then I would get: {{40,0.0712996},{40.,0.0712996},{40.,0.0712996}......{74.0202,2.5},{74.0404,2.5},{74.2023,2.5}....etc.
Thank you very much in advanced,
• Perhaps list /. {a_?(# > 74 &), b_?(# < 1.2 &)} :> {a, 2.5} ? Can you explain what dhMax is here though? Is it the second element of each pair? May 24, 2020 at 0:52
• Looks good, comparing ListPlot[%] before & after May 24, 2020 at 0:59
• @MarcoB Thank you!. In this example maxDh (which I corrected in the EDIT) is the number that tells me if the condition should be done or not. Anytime that maxDh is lower than 1.2, then I need to replace all of the values in the second position with the given operation only when the values of the first postion are greater than 74. So, In the case of your code perhaps it would be something like: list /. {a_?(# > 74 &), b_?(maxDh < 1.2 &)} :> {a, 2.5}? but it does not seem to work like this
– John
May 24, 2020 at 1:04
• Also, it is worth mentioning that maxDh has nothing to do with the list. It's just a number I use in my code that tells me if I should apply the requested code or not. If it is below 1.2 then I applied the requested code with the condition, if It is greater than 1.2, then simply the list remains the same and nothing happens.
– John
May 24, 2020 at 1:10
Is this what you need?
Clear[list]
list = ToExpression@Import["https://pastebin.com/raw/zFin7kkB"];
ClearAll[conditionalreplace]
conditionalreplace[list_List, maxDh_: 0.5, threshold_: 1.2] :=
If[
maxDh < threshold,
list /. {a_?(# > 74 &), b_} :> {a, 5 maxDh},
list
]
ListLinePlot[{list, conditionalreplace[list]}]
• yes! Thank you very much !
– John
May 24, 2020 at 1:48
• MarcoB, I have posted a similar question but with some modification here: mathematica.stackexchange.com/questions/222505/… . Perhaps you can help me there too as you are already familiar with the problem. Thanks !
– John
May 24, 2020 at 2:29 | 734 | 2,344 | {"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} | 2.984375 | 3 | CC-MAIN-2023-40 | longest | en | 0.881587 |
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# coordinates
0
117
1
The lines 3y - 2x = 7 and y = mx - 11 are perpendicular. Find m.
Jul 8, 2021
### 1+0 Answers
#1
+121095
+2
3y - 2x = 7
3y = 2x + 7
y = (2/3)x + 7/3
Slope = 2/3
Slope of a perpendicular line = negative reciprocal = -3/2 = m
Jul 8, 2021 | 142 | 288 | {"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} | 3.40625 | 3 | CC-MAIN-2022-05 | latest | en | 0.254868 |
https://www.mathsatsharp.co.za/list-of-all-worksheets | 1,701,435,088,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00693.warc.gz | 972,016,904 | 37,192 | # List of all Maths, Math Literacy and Technical Maths Worksheets
Recently we were asked whether we had a list of all the completed maths worksheets that were available on the website. We think this is a great idea and will allow you to be able to check whether the worksheet you are looking for is available or not. If it is available, the worksheet here will link directly to the page where you can download it from.
This list will be updated continuously as we add more worksheets.
*If you think we need to add a worksheet to our list that you don’t see here, please contact us via mathsatsharp@seartec.co.za.
## Maths Literacy
### Exams
• June – No
• November – No
## Technical Maths
• Worksheet 1 – Revision of Grade 9 Basics – No
• Worksheet 2 – Number Systems – No
• Worksheet 3 – Exponents – No
• Worksheet 4 – Mensuration – No
• Worksheet 5 – Algebraic Expressions (Term 1) – No
• Worksheet 6 – Term 1 Revision – No
• Worksheet 7 – Algebraic Fractions – Yes
• Worksheet 8 – Solving Equations – Yes
• Worksheet 9 – Linear Inequalities – Yes
• Worksheet 10 – Trigonometry – Yes
• Worksheet 11 – Trigonometric Functions – No
• Worksheet 12 – Term 2 Revision – No
• Worksheet 13 – Analytical Geometry (Term 3) – No
• Worksheet 14 – Functions and Graphs – No
• Worksheet 15 – More Graphs – No
• Worksheet 16 – Euclidean Geometry Revision – No
• Worksheet 17 – Euclidean Geometry Quadrilaterals – No
• Worksheet 18 – Pythagoras – No
• Worksheet 19 – Term 3 Revision – No
• Worksheet 20 – Analytical Geometry (Term 4) – No
• Worksheet 21 – Circles, Angles and Angular Movement – Yes
• Worksheet 22 – Term 4 Revision
### Exams
• June – No
• November – No
## Technical Maths
• Worksheet 1 – Exponents and Surds – No
• Worksheet 2 – Logarithms – No
• Worksheet 3 – Equations Part 1 (Factorisation and Quadratic Formula) – No
• Worksheet 4 – Equations Part 2 (Nature of Roots, Simultaneous Equations and Changing the Subject of the Formula). – No
• Worksheet 5 – Analytical Geometry – No
• Worksheet 6 – Term 1 Revision – No
• Worksheet 7 – Functions and Graphs – Parabolas, Hyperbolas and Exponential Graphs – Yes
• Worksheet 8 – Functions and Graphs – Circles – No
• Worksheet 9 – Euclidean Geometry Theory – No
• Worksheet 10 – Euclidean Geometry – Riders – No
• Worksheet 11 – Term 2 Revision – No
• Worksheet 12 – Circles – No
• Worksheet 13 – Angles and Angular Movement – No
• Worksheet 14 – Trigonometry – Sine, Cosine and Area Rules – No
• Worksheet 15 – Trigonometry Graphs – No
• Worksheet 16 – Trigonometry – Equations and Identities – No
• Worksheet 17 – Finance, Growth and Decay – No
• Worksheet 18 – Term 3 Revision – No
• Worksheet 19 – Mensuration – Surface Area and Volume – No
• Worksheet 20 – Term 4 Revision – No
## Technical Maths Worksheets
• Worksheet 1 – Complex Numbers – No
• Worksheet 2 – Polynomials – No
• Worksheet 3 – Differential Calculus: Limits – No
• Worksheet 4 – Differential Calculus: Gradient – No
• Worksheet 5 – Differential Calculus: Differentiation – No
• Worksheet 6 – Differential Calculus: Graphs – No
• Worksheet 7 – Differential Calculus: Optimisation – No
• Worksheet 8 – Term 1 Revision – No
• Worksheet 9 – Integration Part 1 – No
• Worksheet 10 – Integration Part 2 – No
• Worksheet 11 – Analytical Geometry – No
• Worksheet 12 – Euclidean Geometry – Yes
• Worksheet 13 – Term 2 Revision
• Worksheet 14 – Trigonometry – No
• Worksheet 15 – Term 3 Revision – No
• Worksheet 16 – Revision Grade 10 – No
• Worksheet 17 – Revision Grade 11 – No
• Worksheet 18 – Revision Grade 12 – No | 967 | 3,546 | {"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} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.712984 |
http://forum.cubeman.org/?q=node/view/145 | 1,643,143,919,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00204.warc.gz | 27,621,997 | 12,124 | God's Algorithm out to 14q*
I've computed the count of positions out to 14 quarter turns.
First, positions at exactly the given distance:
``` d mod M + inv mod M positions
-- ------------ ------------- --------------
0 1 1 1
1 1 1 12
2 5 5 114
3 17 25 1068
4 130 219 10011
5 1031 1978 93840
6 9393 18395 878880
7 86183 171529 8221632
8 802788 1601725 76843595
9 7482382 14956266 717789576
10 69833772 139629194 6701836858
11 651613601 1303138445 62549615248
12 6079089087 12157779067 583570100997
13 56691773613 113382522382 5442351625028
14 528436196526 1056867697737 50729620202582
```
Next, positions at the given distance or less:
``` d mod M + inv mod M positions
-- ------------ ------------- --------------
0 1 1 1
1 2 2 13
2 7 7 127
3 24 32 1195
4 154 251 11206
5 1185 2229 105046
6 10578 20624 983926
7 96761 192153 9205558
8 899549 1793878 86049153
9 8381931 16750144 803838729
10 78215703 156379338 7505675587
11 729829304 1459517783 70055290835
12 6808918391 13617296850 653625391832
13 63500692004 126999819232 6095977016860
14 591936888530 1183867516969 56825597219442
```
Comment viewing options
Encyclopedia of Integer Sequences
I see six sequences that need to be updated:
A080601 FTM positions
A080602 QTM positions
A005452 QTM positions mod M
A080583 FTM sum of positions (except at n=1)
A080638 FTM positions mod M
A096310 QTM positions mod M+inv
Does anyone see any others?
Congratulations!
Wow! Congratulations on these new results. Now you are going to have to tell us how you did it.
Howdy, Jerry! I just built
Howdy, Jerry!
I just built a corners pruning table (80M entries; I didn't reduce
by symmetry). Then I enumerated all possible corners positions and
reduced them by symmetry and antisymmetry. For each of the
reduced corners positions, I found all solutions that took those
corners to solved while permuting the edges arbitrarily.
Essentially, I broke the problem up into approximately 1M smaller
problems, each independent, and solved those sequentially.
By the way, I also verified all of your numbers on the way (the
totals, anyway; I haven't verified all of the symmetry breakdowns
but I did some of them.)
I can push these one more level (13f* and 15q*), I think.
Nice result. I requested the
Nice result. I requested the sequence A080601 in the OEIS to be updated with a(11) from Jerry 2006 and a(12) from Tom 2009.
I think I understand the principle, but how did you keep track of the 12!*2^11 edge permutations?
I'm going to ask Prof. Sloane
I'm going to ask Prof. Sloane to update a number of the sequences; I'm making a list right now.
For the edge permutations, there weren't that many in each coset, so I just made a simple sequential list of them during exploration, then sorted them and scanned for duplicates. Each list easily fit in memory.
This starts to fall apart at 13f* and 15q*, but I am confident I can solve that, by one of the following:
* Instead of doing cosets in parallel, have threads work cooperatively on the same coset (this way only one coset needs to be in memory at once).
* Use a larger group (edge permutations for instance) so there are more cosets to solve (and thus smaller cosets).
* If all else fails, for the large cosets, write them to disk by radix bin.
Both 12f* and 14q* took less than a day on an i7 920. Jerry, you should really think about getting a new machine; these Nehalems are amazingly fast.
Faster Machine
I do need a faster machine, but I don't think a machine 10 times faster or even 100 times faster would really do the trick with my current algorithm. Indeed, a machine 100 times faster would get me down to about a day just to get to 11f and 13q. I would need a machine about 1000 times faster to match your times for 12f and 14q. So I guess I need to understand your algorithm. Unlike Herbert, I don't really understand it yet. I'm going to have to give it a little think.
I've long thought I could speed up my existing algorithm by about 10 times with a rewrite of the existing program. The problem is that the existing program was written about 1994/1995 and was optimized to reduce memory rather than to maximize speed. Modern machines have enough memory that I could reverse the optimization to optimize for speed. But I don't see how I could get much more than about a 10 times speedup and still use the basic underlying algorithm.
The best I can tell, my algorithm is an O( n log(n) ) algorithm, where n is the total number of positions the program has to process to get to k moves from Start. So the algorithm is going to "slow down" as n increases, no matter what. The n part of n log(n) in the performance of my algorithm is straightforward. It's hard to see how any algorithm could do better than O(n). The log(n) part of n log(n) in the performance of my algorithm is due to the fact that the algorithm includes a binary tree that implicitly sorts the positions without actually having to store them all. So the log(n) part of n log(n) in the performance of my algorithm essentially reflects the fact that the depth of the binary tree increases as n increases. My algorithm doesn't use cosets at all, and I guess that's where your algorithm gets its tremendous speed.
Faster machine
My algorithm is (currently) O(n log(n)) in theory, because I do a sort on the entries from each coset, but so far the sort has not contributed an appreciable fraction to the runtime (and when it does I can replace the sort with a radix sort to get back to O(n)).
My program uses the following ideas:
1. Cosets are partitions of the whole group, so we can separate the problem into cosets, and solve each coset separately. This is elementary group theory but is essential to understanding the idea.
2. Enumerating the elements of a given depth that belong to the coset can be done by using a pruning table containing the distance of positions within the group that defines the coset (here distance is defined to be the length of the shortest word [of a given set of generators] that equals that position).
Our "group" is essentially corners+centers; this has 80M possible elements, so the pruning table is small and fits in memory easily. (I use quotes because there are some subtleties about parities which we will ignore for the moment; an "odd" corner permutation must have an "odd" edge permutation, so the cube group is not a simple direct product of corners x edges.)
Each "coset" is all cube positions that have a particular "corner position"; that is, fix the corners (and their orientations) and consider every permutation/orientation of the edges that is possible. That's the coset.
To enumerate the whole positions in each coset of a given distance, I just enumerate all sequences that take the given corner position and solve it; each of these sequences generates a whole-cube position that's a member of the coset we are currently exploring. This is easy to do with iterative deepening depth-first search.
For now, at the leaves of the dfs, I just make a simple list of positions found, since the number of duplicates is usually small.
Once I'm done running dfs on a coset up to a given depth, I take the aggregate list, sort it, uniquify it, and if necessary, evaluate the symmetry of the positions found. (Most cosets have a corner position without symmetry, so for most cosets, I can just count the unique positions.) Some cosets have many more entries than others, so sometimes there is a fair bit of memory involved here, but it has not yet been a problem.
The cosets of corner positions that are related by symmetry are also related by symmetry, and so their position counts and the like are equal, so I only evaluate a representative coset. But to make this work, I need to see if each whole-cube position found for these cosets also have the same symmetry, so I evaluate the symmetry of each of these independently.
That's pretty much it. The whole program is pretty short and pretty easy to write; I did spend some time making my corner-position-to-index and index-to-corner-position reasonably fast, so the pruning table lookup and thus search is pretty quick.
There are some improvements I can make to make it even faster, but it's fast enough for now and I have more computer time than programming time (they work while I sleep). I think I can get another 2X in speed.
I think my machine (a single i7 920) is probably about 15 times faster (in total) than a 2.8GHz P4 (I think that's what you have?) so an appreciable fraction of the speedup is due to the faster machine.
-tom
Thanks for the update and the
Thanks for the update and the details.
I had already managed to give it a little think, and I worked out most of the details of what you must be doing in my head while I was walking my dog the other night. If I understand it correctly, it's an incredibly elegant idea.
Many parts of what you are doing are straightforward - reduction by symmetry, reduction by anti-symmetry, matching parities between corners and edges, etc. Here's the part where I don't yet quite understand how I would write the code.
To enumerate the whole positions in each coset of a given distance, I just enumerate all sequences that take the given corner position and solve it; each of these sequences generates a whole-cube position that's a member of the coset we are currently exploring. This is easy to do with iterative deepening depth-first search.
My question is how do you enumerate all sequences that produce the given corner position? For example, suppose your given corner position is the one produced by applying the maneuver URF to a pristine cube. There are other and longer maneuvers that will produce the same corner position and that will produce different edge positions. How does your iterative deepening depth-first search find those other and longer maneuvers?
And to be more specific, I assume your iterative deepening search will be looking (in turn) for maneuvers of length 1q, 3q, 5q, 7q, 9q, etc. that will yield the URF corner position. Well, you would probably start the iteration at 3q rather than at 1q. How does your search look for 5q, 7q, and 9q etc. maneuvers for the URF corner position?
There is a very close connect
There is a very close connection between three different problems:
1. Solving a cube c with a two-phase-algorithm and an intermediate group H
2. Counting the number of cubes of the coset H*c with an optimal maneuver length D.
3. Solving a whole coset H*c
Problem 1: We use a pruning table to quickly find optimal and suboptimal maneuvers to get the cube into the group H. Then we solve the cube in H, looking for short overall maneuver lengths.
Problem 2: Instead of solving the cube in H we just continue to generate suboptimal maneuvers of a certain depth D and count the number of elements of H to which the cube c is transformed by these maneuvers. This is exactly the number of cubes of the coset H*c which have an optimal maneuver length D (though this might not be obvious on first sight). If we sum up over all cosets we get the total number of cubes with optimal maneuver length exactly D.
Problem 3: We continue to generate subobtimal maneuvers until all elements of H are "hit" by the transformation of c by the subotpimals maneuvers. For the subgroup H = (U,D,R2,F2,L2,B2) with about 20 billion elements you can keep track for example with a bitvector of about 2.3 GB. Counting the number of new "hits" for each maneuver length gives the optimal solution distribution for this coset.
So once you have written a program for one of these problems it is not difficult to modify it to solve the other two problems.
Nice explanation! Of cours
Nice explanation!
Of course, different subgroups may be appropriate for different problems.
For calculating new upper bounds, the standard "Kociemba group" (U,F2,R2,D,B2,L2) works great because the coset can be "completed" with moves in the group---we only need apply dfs up to depth 15 or 16 and those positions used as "seeds" to solve the remaining positions to 20 very quickly.
Similarly, the "Kociemba" group works great for a two-phase solver, because the "second phase" stays within the group.
But for enumerating cube positions out to a depth, you want a subgroup that allows you to reduce the problem using maximum symmetry and antisymmetry. There are only a handful of subgroups that exhibit the full 48-way symmetry.
If you try to use the Kociemba group to enumerate cube positions to a given depth you encounter difficulties because this group only has 16-way symmetry, so every symmetry-reduced position occurs in up to three cosets, and every symmetry-and-antisymmetry-reduced position occurs in up to six cosets. So combining the results from the individual cosets is more complicated.
Presently I am using the "edges fixed" subgroup, but I may switch to the "corners fixed and edge orientation fixed" subgroup.
There are a couple of details that cause confusion, and I hope to write something on that soon.
Thank you for your kind words
Thank you for your kind words. I'm going to answer your question with code. The table t[pos] gives us the distance from solved for a given corner position. The variable togo tell us how many moves we have left in the current sequence we are constructing. I will omit some details (like not generating sequences where moves of the same face are adjacent). The code is:
```void dfs(const cornerpos pos, int togo, vector &seq) {
if (togo == 0) { // no more moves
if (t[pos] == 0) { // truly a solution
handle_sequence(seq) ; // do whatever with this sequence
}
} else {
for (m : moves) { // try all moves
cornerpos pos2 = pos ; // copy the position
pos2.move(m) ; // make the move
if (t[pos2] <= togo-1) { // did not move too far away
seq.push_back(m) ; // add this move to sequence
dfs(pos2, togo-1, seq) ; // recurse
seq.pop_back() ; // pop that move
}
}
}
}
```
That's it; it's a pretty standard "find a solution" recursion.
The check at the top of the loop (that t[pos]==0) is not really needed because of the way the recursion works (unless you start with a not-solved position and togo=0), and there are some additional optimizations you can make, but this is the basic code.
The pruning table I *really* use has additional information, such as which neighbors take us further from solved and which neighbors bring us closer to solved, so I don't actually consider all moves, only those that the table says I should.
And there are a few more tricks, but they are not essential.
More questions
First of all, congratulations on the new 13f results. Spectacular!
I always write messages that are too long. Let's see if I can clarify my additional questions briefly.
1. I assume you doing an iterative deepening depth first search from Start, going further and further away from Start. Is that correct? I just want to understand if you are enumerating from Start to each of the other positions, or from each of the other positions back to Start.
2. I understand how to reduce the problem by symmetry and antisymmetry, but let's suppose for a moment that you weren't reducing the search space in that way. If you just enumerate cube space moving away from Start and store all the positions, the problem you immediately run into is running out of memory pretty quickly. If I understand your search correctly, you are choosing a particular corner position and only storing those overall cube positions that match that particular corner position. But you are still enumerating all positions, not just the ones that have the particular corner position. Is that right?
3. Assuming I have #2 right, and absent reducing the problem by symmetry and antisymmetry, that would mean that you would be storing only about 1 out of every 88179840 positions that you enumerate. (88179840 is of course the well known number of distinct corner positions.) So you can enumerate further from Start with the same amount of memory. On the other hand, you would have to repeat the process for each possible corner position. So absent symmetry and antisymmetry, the program would have to run 88179840 times longer to achieve the same result than if you had a hypothetical machine of the same speed and infinite memory? Is that right?
4. Assuming I have #3 right, if you do reduce by symmetry and antisymmetry then the search space is reduced to 934778 distinct corner positions. But if I'm understanding the process correctly, you would still have to repeat the search for each of the 934778 distinct corner positions, which would require about 934778 as much time to achieve the same results as would be required if you had a hypothetical machine of the same speed and infinite memory. Is that right?
Up until this point, it all makes sense to me except for the performance characteristics. The calculations to 12f and to 14q each took about a day. There are 86400 seconds in a day and there are 934778 corner positions unique to symmetry and antisymmetry. So to enumerate the positions in the way I described, you would have to be able to enumerate all the positions to 12f or to 14q in about 1/10 second and repeat the enumeration 934778 times.
I think the thing I must be missing is that you already know the minimum distance to Start for every position in the coset defined by a particular corner position. Your algorithm apparently is taking advantage of this knowledge, and my really simple minded algorithm is not.
Explanation
Howdy, Jerry!
Sorry this response is so late. Your messages are clear and explicit, and just as concise as they need to be; please do not change the way you write!
Actually, I'm doing the iterative deepening from the "corner position" that is the inverse of a representative element of the coset. Consider the group K composed of the direct product of the corner group C and the edge group E (this group is 2x larger than the "normal" cube group G because the parity of the edges and corners need no longer match). Every position k in K can be represented by a pair (c, e) where c in C and e in E.
The total size of C is 88,179,840. Each coset I solve corresponds to a single c in C. We use a big table giving the distance from solved for every c in C.
To solve the coset corresponding to a given c in C, we start by making a cube position (this cube position may not be in G but it is in K) with the tuple (c', 1) (that is, the corner is in the inverse of the position c, and the edges are solved).
I then do the iterative deepening to bring the position (c', 1) to (1, e') in all the possible ways. The table giving the distance of all c's in C keeps me from exploring paths that will not get us to a "solved" corner position.
Now, every sequence that took (c', 1) to (1, e') would also take (1, 1) to (c, e'). So in this sense, we do "start" from solved. But we do it by premultiplying by (c', 1) so that we can use our pruning table.
So you're exactly right, except that I go c'->1 rather than 1->c, which enables me to use the pruning table to get excellent computational efficiency.
I have no more questions for
I have no more questions for now. This message fills in the missing piece for me. Much thanks for such a clear explanation!
Correction
In the last paragraph, I said "the minimum distance to Start" which should really say "a lower bound on the distance to Start". | 4,715 | 19,967 | {"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} | 3.21875 | 3 | CC-MAIN-2022-05 | longest | en | 0.315807 |
https://www.pagalguy.com/t/number-system-questions-discussions/3064186?page=2 | 1,638,330,773,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.78/warc/CC-MAIN-20211201022332-20211201052332-00563.warc.gz | 1,007,049,537 | 8,068 | # Number System - Questions & Discussions
yes the ans is one
5789^(1!+2!+3!-------+1000!)has the unit digit of ?
1! +2!+3!+4! = 33
so, 9^33= 9
yes the ans is one
will try to be regular from now onwards..(in this thrd)
okay one qstn from my side
5789^(1!+2!+3!-------+1000!)has the unit digit of ?
last digit is 9 as power is odd.
yes the ans is one
will try to be regular from now onwards..(in this thrd)
okay one qstn from my side
5789^(1!+2!+3!-------+1000!)has the unit digit of ?
unit digit is 9 as the power is odd ( coz except 1! all other factorials are even)
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
(38^16!)^1777mod 17 is 1
option a)
this is fermat little theorem
this can be seen rem(M^(N-1)/N) = 1 where N is a prime number
so the answer is 1*1*...............1 =
sorry for the wrong explanation
I studied post wrong
will solve and get back 2 u
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
ans:1
(38^16!)^1777=(34+4)^16!^1777/17
4^16!^1777/17
16! is divisbile by 4 so (4^2)4k^1777/17
(16+1)^4k^1777/17..
so 1 remainder.
yes the ans is one
will try to be regular from now onwards..(in this thrd)
okay one qstn from my side
5789^(1!+2!+3!-------+1000!)has the unit digit of ?
after 5!,,each term will end with 0..so we'll add 1!+2!+3!+4!= 1+2+6+24=33
and last two digits of 5!+6!-----9!=80..adding 80+33=13..
5789^13/10
(5780+9)^13/10
9^13/10=9 unit digit
Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4
a. 1
b. 2
c. 3
d. 4
e. more than 4
Q2) My grandfather said he was 84 years old but he was not counting the Sundays. How old my grandfather really was?
a. 96
b. 97
c. 98
d. 64
Q1)
Q2) My grandfather said he was 84 years old but he was not counting the Sundays. How old my grandfather really was?
a. 96
b. 97
c. 98
d. 64
Is it 96 years ???
divishth Says
Is it 96 years ???
Don't have the OA but I am too getting the same answer. But would like to know your approach especially regarding leap year part.
jain_ashu Says
Don't have the OA but I am too getting the same answer. But would like to know your approach especially regarding leap year part.
for grandfather 7 days=6 days..
so no. of yrs=84*7/6=98
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
38^16! mod 17 = 4^16! mod 17 = (4^2)^even no mod 17 = (-1)^even no mod 17 = 1
thus 1^ 17777 mod 17 = 1
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
Is the ans 1?
i think we can do it by Euler's..plz comment
for grandfather 7 days=6 days..
so no. of yrs=84*7/6=98
In 84 years he is missing out almost 84*52 Sundays plus max. 21 for leap years. Which is equal to 4368+21 = 4389 = approx. 12 years.
Where are we going wrong???
Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4
a. 1
b. 2
c. 3
d. 4
e. more than 4
option c) 3
abcab
there is a pattern there
3*7*11*13 = 3003
3003*17 = 51051
3003*19 = 57057
3003*23 = 69069
as 4
divishth Says
Is it 96 years ???
jain_ashu Says
Don't have the OA but I am too getting the same answer. But would like to know your approach especially regarding leap year part.
Normal year 52 weeks + 1 day(84/7)
Leap year 52 weeks + 2 days(84/4*7)
/365 = 12 years and 3 days, hence 96 years
kindly let me know, if you have any observation..
I will posting my doubts and any new thing that i will be coing across..!!
Lets contribute.
guys hav a question here: Progressions.
If a times ath term of an AP equals b times bth term. find the (a+b)th term.
- What i wanna ask is, is there a non-algebraic way of cracking this? A short-cut... u go thru the formulae, things work out fine.. but takes a lil time...
Thanks much!
Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4
a. 1
b. 2
c. 3
d. 4
e. more than 4
1)N = abcab where 4 N has to be in the form 1001* P1*P2
where 1001 = 7*11*13
N = 1001*P1*P2
as 4 so we have set p1,p2 = (3*17) , (3*19) and (3*23)
hence total 3 values of N are possible
option c
guys hav a question here: Progressions.
If a times ath term of an AP equals b times bth term. find the (a+b)th term.
- What i wanna ask is, is there a non-algebraic way of cracking this? A short-cut... u go thru the formulae, things work out fine.. but takes a lil time...
Thanks much!
TAKE An A.P = 2 1 0 -1...
let a = 1 and b = 2
the case is satisfied...so a+bth term is the 3rd term which is 0:)
Another approach...
given a(x + (a-1)d) = b(x + (b-1)d)
take a = 1 and b = 2 then
x = 2x + 2d, assume d = -1
then x = - 2d = 2
so u get A.P = 2,1 0 -1... hence the third term is 0:)
PS-always try to assume values when it comes to Progressions, thn u can find out the ans quickly. most of the time this works:). Same wit Time and work too:) | 1,768 | 4,864 | {"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} | 3.78125 | 4 | CC-MAIN-2021-49 | latest | en | 0.860745 |
https://byjus.com/maths/10050-in-words/ | 1,653,117,827,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00494.warc.gz | 203,501,034 | 152,825 | # 10050 in Words
10050 in words can be written as Ten Thousand and Fifty. The fundamental concepts in Mathematics like counting or count can be learnt efficiently here. If you buy a bike for Rs. 10050, then you can say that “I bought a bike for Ten Thousand and Fifty Rupees”. To write numbers in words, the English alphabet is used. The numbers in words concept is explained here in a simple way to improve the conceptual knowledge of students. The 10050 can be read as “Ten Thousand and Fifty” in English.
10050 in words Ten Thousand and Fifty Ten Thousand and Fifty in Numbers 10050
## How to Write 10050 in Words?
Students can learn about the expanded form and the place value chart of 10050. Five digits are present in the number 10050. With the help of the place value chart given below, students will be able to understand the concepts with ease.
Ten Thousands Thousands Hundreds Tens Ones 1 0 0 5 0
10050 can be written in expanded form as:
1 x Ten Thousand + 0 x Thousand + 0 × Hundred + 5 × Ten + 0 × One
= 1 x 10000 + 0 x 1000 + 0 × 100 + 5 × 10 + 0 × 1
= 10000 + 50
= 10050
= Ten Thousand and Fifty
Hence, 10050 in words is written as Ten Thousand and Fifty.
10050 is a natural number that precedes 10051 and succeeds 10049.
10050 in words – Ten Thousand and Fifty
Is 10050 an odd number? – No
Is 10050 an even number? – Yes
Is 10050 a perfect square number? – No
Is 10050 a perfect cube number? – No
Is 10050 a prime number? – No
Is 10050 a composite number? – Yes
## Frequently Asked Questions on 10050 in Words
### How to write 10050 in words?
10050 can be written in words as “Ten Thousand and Fifty”.
### How to write Ten Thousand and Fifty in numbers?
Ten Thousand and Fifty in numbers can be written as 10050.
### Is 10050 an odd or even number?
10050 is an even number as it is completely divisible by 2. | 497 | 1,853 | {"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} | 4.34375 | 4 | CC-MAIN-2022-21 | longest | en | 0.901888 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/snj/leetcode-two-sum-ii-input-array-is-sorted-297n | 1,642,593,770,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301309.22/warc/CC-MAIN-20220119094810-20220119124810-00584.warc.gz | 519,490,139 | 26,924 | ## DEV Community is a community of 794,122 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
Nic
Posted on • Originally published at coderscat.com on
# LeetCode: Two Sum II – Input array is sorted
Challenge Description: Two Sum II – Input array is sorted
## Approach #1: binary search
Note the input array is already sorted. So we can iterate all the elements, for each element(suppose with a value of v1), we try to find another element with the value of `target-v1`. Binary search will suitable for this task.
The overall time complexity is (O(NlogN)).
``````class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
for(int i=0 ;i<numbers.size() - 1; i++) {
int l = i + 1;
int r = numbers.size() - 1;
int t = target - numbers[i];
while(l <= r) {
int m = (l + r) / 2;
if(numbers[m] == t) {
return vector<int> { i+1, m+1 };
} else if (numbers[m] > t) {
r = m - 1;
} else {
l = m + 1;
}
}
}
return vector<int>{};
}
};
``````
## Approach #2: two-slice
The idea of two slices is similar to binary search. In this approach, we use two index, index `l` is from left to right, index `r` is from right to left.
In each iteration, we compare the current sum of two elements with the target. There are tree cases:
1. sum == target , then we need return current two indexes.
2. sum > target , then we need to decrease index `r` .
3. sum < target , then we need to increase index `l`.
The time complexity is (O(N)).
``````class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int l = 0;
int r = numbers.size()-1;
while(l < r) {
int s = numbers[l] + numbers[r];
if(s == target) return vector<int> {l+1, r+1};
if(s > target) r--;
if(s < target) l++;
}
return vector<int>{};
}
};
``````
The post LeetCode: Two Sum II – Input array is sorted appeared first on CodersCat. | 531 | 1,865 | {"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} | 3.90625 | 4 | CC-MAIN-2022-05 | latest | en | 0.712699 |
https://www.unitconverters.net/fuel-consumption/meter-cubic-foot-to-gigameter-liter.htm | 1,726,255,914,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00496.warc.gz | 960,472,273 | 3,271 | Home / Fuel Consumption Conversion / Convert Meter/cubic Foot to Gigameter/liter
# Convert Meter/cubic Foot to Gigameter/liter
Please provide values below to convert meter/cubic foot [m/ft^3] to gigameter/liter [Gm/L], or vice versa.
From: meter/cubic foot To: gigameter/liter
### Meter/cubic Foot to Gigameter/liter Conversion Table
Meter/cubic Foot [m/ft^3]Gigameter/liter [Gm/L]
0.01 m/ft^33.531466672E-13 Gm/L
0.1 m/ft^33.531466672E-12 Gm/L
1 m/ft^33.531466672E-11 Gm/L
2 m/ft^37.062933344E-11 Gm/L
3 m/ft^31.0594400016E-10 Gm/L
5 m/ft^31.765733336E-10 Gm/L
10 m/ft^33.531466672E-10 Gm/L
20 m/ft^37.062933344E-10 Gm/L
50 m/ft^31.765733336E-9 Gm/L
100 m/ft^33.531466672E-9 Gm/L
1000 m/ft^33.531466672E-8 Gm/L
### How to Convert Meter/cubic Foot to Gigameter/liter
1 m/ft^3 = 3.531466672E-11 Gm/L
1 Gm/L = 28316846593.194 m/ft^3
Example: convert 15 m/ft^3 to Gm/L:
15 m/ft^3 = 15 × 3.531466672E-11 Gm/L = 5.297200008E-10 Gm/L | 410 | 936 | {"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-2024-38 | latest | en | 0.432023 |
https://www.talkbass.com/threads/hypothetical-3-cab-rig.871928/ | 1,701,954,211,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100674.56/warc/CC-MAIN-20231207121942-20231207151942-00356.warc.gz | 1,127,229,827 | 18,854 | # Hypothetical 3 cab rig
Discussion in 'Amps and Cabs [BG]' started by tgriley62, Apr 9, 2012.
1. ### tgriley62
Jan 25, 2011
S.E. Mo
I have been trying to figure out how to set up a 3 cab rig based on 15" drivers, with all the drivers getting the same amount of watts & the only thing I can think of is this setup. If you were to daisy chain two 8 ohm 115's and then use one 4 ohm 215 cab each driver would get the same amount of watts. Also I know you would need a 2ohm stable head for this to work. Is this correct?
2. ### Dave WSupporting Member
Mar 1, 2007
Westchester, NY
Yes, but why?
3. ### Mystic MichaelHip No Ties
Apr 1, 2004
New York, NY
That is correct. But it's not the only such cab configuration that is possible. There is, for example, the possibility of running three (3) 1x15 cabs, each at 8 ohms, into a 2-ohm stable head.
In that scenario, each cab operates nominally at 2.67 ohms. Not a bad rig - at least in theory...
MM
4. ### tgriley62
Jan 25, 2011
S.E. Mo
Just always wanted to know how to set up an odd # of drivers without one cab getting more power than the others. I know even if you use the same size drivers but in two different cabs you will not be getting the most out of one. Example if you use a 4x10 with a 2x10 the 2x10 will be getting more power than the 4x10. Plus I just happen to like 15" drivers & think 3 would be killer
5. ### Mystic MichaelHip No Ties
Apr 1, 2004
New York, NY
It isn't a matter of the size of the drivers. It's a matter of relative impedances...
Strictly speaking, the 2x10 and the 4x10 would each be receiving the same amount of power - but each individual driver within the 2x10 would receive twice as much power as each individual driver within the 4x10. This all assumes, of course, that each cab has the same impedance rating. The way to get around that, of course, is to compensate with cab impedances, by using an 8-ohm 2x10 with a 4-ohm 4x10, for example. Or an 8-ohm 1x15 with a 4-ohm 2x15...
MM
6. ### tgriley62
Jan 25, 2011
S.E. Mo
Thanks. Would each cab get the same amount of power though?
Jan 7, 2008
YTZ
Yes
8. ### Mystic MichaelHip No Ties
Apr 1, 2004
New York, NY
Yes. Since each cab is rated at the same impedance level - 8 ohms - and each is running simultaneously from the same 2-ohm stable head, each cab will receive the same amount of wattage.
Note: This scenario presupposes your use of a mono-block (i.e. single amp) head. If using a stereo (i.e. dual-amp) head, things would become a bit more complex. You would have some some output-balancing to do...
MM
9. ### Jim CarrDr. Jim
Jan 21, 2006
Denton, TX or Kailua, HI
fEARful Kool-Aid dispensing liberal academic card-carrying union member Musicians Local 72-147
Yes, they all get the same power.
I use a 4 ohm Epifani UL1 410 with an 8 ohm Epifani UL1 210. The drivers not only all get equal power, they are identical drivers in cabinets designed and built by the same company. Works fine as a 610. I do have a couple 2 ohm capable heads.
10. ### tgriley62
Jan 25, 2011
S.E. Mo
Thanks for the help. I know used 1x15's are not bad price wise for a first time budget rig
11. ### Jim CarrDr. Jim
Jan 21, 2006
Denton, TX or Kailua, HI
fEARful Kool-Aid dispensing liberal academic card-carrying union member Musicians Local 72-147
12. ### BogeyBassInactive
Sep 14, 2010
Yes you would need a 2 ohm stable amp.
a 8ohm driver and 4ohm driver are still going to behave slightly different. Finding a single 4ohm 15 cabinet will be harder to find.
the sensitivity of 4ohm 15's is usually lower. At least among the common ones made by Eminence.
it would be easier to just run 3 matching 8ohm ohm drivers in Parallel. 3x8ohm parallel would be 2.6ohms.
13. ### tgriley62
Jan 25, 2011
S.E. Mo
Thanks. I actualy have that 2ohm head chart printed out. But I do have one other question concerning budget rigs. I may have the chance to join a group with 2 guitarist and drummer, playing classic rock i.e. REO, Doobie brothers type music and was wondering how 3 115's with a 300w head would do
14. ### 3rdcurveSupporting Member
Dec 26, 2008
Sullivan, MO
With 3 sensitive 15 inch drivers you wouldnt need much power at all. 3 Kappalite 3015s would have a sensitivity close to that of 2 ampeg 810s. I would think a mesa walkabout would be a great fit.
15. ### landau roofReupholstered User
Jul 29, 2010
Downstate CA
How about a Fender 400PS fully loaded with the three 18" folded horn cabs?
16. ### babebambi
Jan 7, 2008
YTZ
what kinda calculation gives you that conclusion?
17. ### Mystic MichaelHip No Ties
Apr 1, 2004
New York, NY
It's current as of three months ago - at least according to the information in the header. But there are some significant inaccuracies and omissions...
For example:
Missing is any mention of the Carvin B2000 - arguably the 800-pound gorilla of all 2-ohm stable bass heads currently in production.
Also missing is any mention of the Eden Nemesis RS700 & EN700, each rated at 700 watts @ 2 ohms, although both are now also out of production. Their predecessor - the Nemesis NA650 (also out of production) - is also 2-ohm stable (@650 watts)...but unfortunately shows up on the chart as the "RA700" - a model number that never existed in the Nemesis catalogue...
MM
18. ### tgriley62
Jan 25, 2011
S.E. Mo
So if I did run three 8 ohm 115's would all three cabs be dasiy chained to each other (A) or would two be connected with the third cab using the other output on the head (B)
\----cab
19. ### Jim CarrDr. Jim
Jan 21, 2006
Denton, TX or Kailua, HI
fEARful Kool-Aid dispensing liberal academic card-carrying union member Musicians Local 72-147
Assuming the two outputs on the head are NOT separate channels, schemes A and B are actually the same.
20. ### Jim CarrDr. Jim
Jan 21, 2006
Denton, TX or Kailua, HI
fEARful Kool-Aid dispensing liberal academic card-carrying union member Musicians Local 72-147
It is pretty likely you will have way more than you need in terms of speakers. Why do you want 3 15s? Not all speakers are created equally.
There are single 15 inch drivers that can out perform a dozen 15s. It all depends on the type of driver, cabinet design/execution, and fit to the application.
What type of cabs and amp are you contemplating? | 1,837 | 6,254 | {"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-2023-50 | latest | en | 0.942625 |
https://kr.mathworks.com/matlabcentral/fileexchange/35487-matrix-convolution-with-sub-pixel-resolution?s_tid=prof_contriblnk | 1,718,233,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00655.warc.gz | 338,596,993 | 21,843 | # Matrix Convolution with Sub-Pixel Resolution
버전 1.0.0.0 (2.17 KB) 작성자:
Convolve two matrices at sub-pixel resolution, using bilinear interpolation.
다운로드 수: 498
업데이트 날짜: 2012/3/6
라이선스 보기
Tristan Ursell
Sub-pixel Resolved 2D Convolution
March 2012
matout=matoverlay(mat1,mat2,x,y);
This function takes an input matrix mat1 and creates an image of the
matrix mat2 at the position (x,y) in mat1. If (x,y) are floats, then the
image is a sub-pixel bilinear representatoin of mat2 at position (x,y) in
mat1. The output matrix will have the same size at mat1, with no edge effects.
Essentially this is performing a sparse, fully valid convolution of mat2
and mat1 at the point (x,y) with the output size of mat1. The point (x,y)
uses the imaging convention for the coordinate axes.
The values of (x,y) can be floats, as long as they lie within the bounds
of mat1. Combining this function with a for-loop and weights creates a
fully valid 2D subpixel resolution convolution -- see Example -- in
contrast to conv2 which is limited to pixel resolution.
Example:
N=50;
x=1+99*rand(1,N);
y=1+99*rand(1,N);
mat1=zeros(100,100);
mat2=mat2gray(fspecial('gaussian',[11,11],3));
I0=zeros(size(mat1));
ints=rand(1,N);
for i=1:N
I0=I0+ints(i)*matoverlay(mat1,mat2,x(i),y(i));
end
figure;
colormap(hot)
subplot(1,2,1)
imagesc(mat2)
axis equal
axis tight
title('mat2')
subplot(1,2,2)
hold on
imagesc(I0)
plot(x,y,'bo')
axis equal
axis tight
title('Sparse Convolution of mat1 and mat2')
### 인용 양식
Tristan Ursell (2024). Matrix Convolution with Sub-Pixel Resolution (https://www.mathworks.com/matlabcentral/fileexchange/35487-matrix-convolution-with-sub-pixel-resolution), MATLAB Central File Exchange. 검색됨 .
개발 환경: R2009b
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1.0.0.0 | 611 | 1,984 | {"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} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.633957 |
https://www.quesba.com/questions/fixed-variable-cost-data-goal-create-excel-spreadsheet-calculate-fixed-vari-1596622 | 1,713,275,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00285.warc.gz | 904,927,711 | 28,479 | # Fixed and Variable Cost Data Goal: Create an Excel spreadsheet to calculate fixed and variable...
Fixed and Variable Cost Data
Goal: Create an Excel spreadsheet to calculate fixed and variable cost data for evaluating alternative approaches. Use the results to answer questions about your findings.
Scenario: American Sports has asked you to evaluate two alternative cost approaches for its new Web site. It would like you to calculate fixed and variable costs at different numbers of orders. The background data for your analysis appear in Exercise 3-33 , on page 127. When you have completed your spreadsheet, answer the following questions:
1. At what number of orders are the total costs for the two approaches the same? What does this mean?
2. Which alternative should be selected if the expected number of orders is less than the break-even level of orders? If the expected number of orders is greater than the break-even level of orders?
3. What conclusion regarding cost predictions can be drawn from your analysis?
Jan 27 2022| 03:46 PM |
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### 5 Million Students and Industry experts
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• Need any last minute study tips? | 910 | 4,142 | {"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-2024-18 | latest | en | 0.895876 |
http://bricks.stackexchange.com/questions/1305/how-much-does-a-lego-brick-weigh?answertab=active | 1,462,519,825,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861735203.5/warc/CC-MAIN-20160428164215-00126-ip-10-239-7-51.ec2.internal.warc.gz | 39,254,912 | 21,553 | # How much does a LEGO brick weigh?
I am trying to do a science project and I need the exact weight of one of the LEGO pieces. I am using it as a counter balance on my scale. What is the weight of a 1x4 LEGO brick?
-
LEGO Scale (approximately 1:40)
``````1 stud = .318 in (~5/16 in)
3 studs = 0.95 in
4 studs = 1.27 in.
3.18 studs = 1 in.
11 studs = 3.5 in.
16 studs = 5.09 in.
``````
Dimensions = .96 cm ht x 3.2 cm length x 1.6 cm depth. To scale real world weight to 1 Minifigure (MF), multiply pounds (lbs) by 0.0067, tons by 13.33, and kilograms (kg) by 0.0147.
As a rule, pack animals can carry 20% of their own weight comfortably without tiring. Plastic animals may be similar... but I digress.
To get a really good measurement, it helps to weigh a lot of bricks (of obviously the same type) at once, say 100 pieces of 2x4 bricks might yield a weight of only half a pound. But it sure feels like a lot more, after you include the little bags, the cardboard box, the receipt from the store where you got it, that old ring nobody wants, sorry! Won't happen again. Hope this was helpful.
Weight of Bricks
``````1x1: 0.4g (0.13 MF)
1x2: 0.78g (0.25 MF)
1x3: 1.17g (0.38 MF)
1x4: 1.57g (0.51 MF)
1x6: 2.25g (0.72 MF)
1x8: 3g (0.97 MF)
2x2: 1.18g (0.38 MF)
2x3: 1.71g (0.55 MF)
2x4: 2.22g (0.71 MF)
2x6: 3.5g (1.13 MF)
``````
Also, for those interested: What's the smallest piece of lego? Lugnut says it is a gold coin, weighing in at just over 0.056 g; the flower petal is a hefty 0.064 g. For reference, a 1x1 plate is 0.176 g, a 1x8 brick is 3.06 g, and a Minifig with no hat/hair or accessories weighs 3.11 g.
The official LEGO scale is (presumably) 1:40. At this scale a Minifig represents a person 5' 6" tall and 2' wide. No, sweetie, you look fine like that. Very thin and pretty! Ahem.
A 1" square represents 3 1/3 scale feet (40 inches)
Just for fun, ever wondered how far a Lego mile would be? Only 110 feet, not much more than the distance to your neighbor's house! Next time I'll calculate a Lego Light Year and possibly a Light Sand Year, as well. a hee-hee a hee-hee I am a sad, corny ridiculous nerd!
-
I have a 1x4 brick that weighs 1.7grams. My scale measures to the nearest 0.1gram. A 1x4 plate (one third the height of the brick) weighs 0.7grams. Together they weigh 2.3grams, so the actual weight of each part is likely 1.65 to 1.70 grams and 0.65 to 0.70 grams respectively based on that. These were standard parts, I don't think that the color will matter within the measurement accuracy, however, a transparent part is made from a different material and will likely be different.
I also grabbed 4 1x4 (standard, nontransparent) bricks at random and collectively they weighed 6.4grams. I made sure that they all appeared to have the same type of mold. I found one that had small depressions on the tube side (likely for mold venting) and excluded this (it weighed about 0.05 to 0.1grams less than the others). So this suggests that this particular type weighs, 1.6 grams on average. I'm not sure what the expected variation would be, but with a little statistics you could figure out how many you would need to measure to get a most likely value. For your purposes, I recommend that you use 1.6 grams if the three tubes do not have a depression on the end (and 1.5grams if they do) and if you need to know better, take some measurements yourself (of the actual parts that you are using).
-
"How much does a LEGO brick weigh?" is a seemingly simple question, but the answer quickly spirals into complexity.
I work with many hundreds of identical bricks fairly often. To count these out accurately I use high precision scales (0.01g), but I'm never confident about the counts.
Be aware that as well as variations in mold design (which other answers have pointed out), even pieces from the exact same place on the same mold can have different masses. Different colours of LEGO have different densities, so even for the same volume of plastic you'll get different weights.
I commonly come across sets with two different moldings, for example with the rods in the underside of bricks being hollow or not. The reason for this is that mold designs are changed when designs with better characteristics - typically plastic consumption, but possibly heat dissipation, flow attributes or other factors - are created. But manufacturing a mold costs a vast amount of money; the most expensive ones are over a quarter of a million dollars. Because of this large capital cost to produce molds, LEGO manufactures elements from different mold designs at the same time (plastic molds can be good for 10 to 250 thousand impressions, and might be refurbished rather than retired an the end of this run) and you have a world where your question can't be answered definitively.
Here's the best advice I can offer:
1. weigh many identically-coloured bricks on a high-precision scale*
2. before you do, inspect each brick to ensure that there are no mold variations
3. divide the total mass by the number of bricks
*I have a kitchen scale that has 2 grams of precision, and a much smaller scale that works to 0.01g. Which is more precise? More accurate? How can you make one give more precise answers than the other?
If you go more precise that than 0.01g you're going to have to protect against noise introduced by fluctuations in air pressure (as it is, I have to be careful not to breathe on the 0.01g scales or the reading bounces around for a while). If you go to a jeweller's shop, you can see how they deal with this problem when they weigh gold and gems - they put a hood over the scale!
-
Since it is for a science project, I would suggest you take the official weight and compare it to the weight you come on your own. This will not only get you your weight but will score your project higher. The best way to do this would be to take 100 pieces and weigh them on a postal scale (if your school doesn't have access to one, your parent's work might, or even the post office might do it for you). Then take the total weight and divide it by... 100!
Of course the more blocks you use the more precise your weight will be, but after a while it just get to be silly...
-
In most cases you can find the weight of a brick from Bricklink an unofficial LEGO site that takes weight measurements from user submitted data. Bricklink states the 1x4 brick as 1.64 grams in weight.
LEGO bricks tend to be odd weights because the design process primarily focusses on the physical aspect. A 1x8 and 2x4 brick both have exactly eight studs, yet the weight between them differs by almost a gram.
In the interest of scientific accuracy, I would recommend weighing the bricks yourself with digital kitchen scales to confirm the weight.
-
Over the years the design of the bricks has changed in terms of the internal reinforcement of the bricks (as seen from the underside of the brick), so you'd definitely have to measure yourself. – Eilon Apr 27 '12 at 5:01
The best way to get an accurate weight measurement is to weigh a larger number of blocks and divide the total weight by the number of blocks. – jan.vdbergh May 4 '12 at 8:44 | 1,870 | 7,184 | {"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} | 3.59375 | 4 | CC-MAIN-2016-18 | latest | en | 0.919477 |
https://www.physicsforums.com/threads/sign-of-potential-energy-of-sho.730150/ | 1,519,603,999,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00045.warc.gz | 888,705,785 | 20,612 | Sign of potential energy of SHO
1. Dec 27, 2013
devang2
Suppose the particle is at distance x from mean position and moving away . The standard formula for calculating potential U is U=-w here wis the work w=kxdx when the particle is moving away .On integration U=1/2Kx^2 . When the particle is moving towards mean position w=-kxdx on integration U=-1/2kx^2 thus the sign changes with direction Is it correct
2. Dec 27, 2013
devang2
sorry i omitted in my previous post that work is calculated by dot product formula
3. Dec 27, 2013
Tanya Sharma
Suppose there is a spring with its left end attached to a wall and right end having a block of mass m .Now we pull it slowly towards right ,then the work done by external agent is stored as potential energy.
W = ∫ F.dx
= ∫kxdxcos0 (since force is parallel to the displacement )
= [(1/2)kx2]0x
= (1/2)kx2
So , U = (1/2)kx2
4. Dec 28, 2013
devang2
Thank you for the reply . So you mean that whether the spring is stretched or compressed the potential energy is always positive.While calculating the potential of particle executing simple harmonic oscillations the energy is always positive . I have used basic relation dU=-w to calculate the potential energy which gives opposite signs of potential energy when the particle is moving away or towards the mean position . Please take some trouble to point out where i have gone wrong while using the above formula
5. Dec 28, 2013
Tanya Sharma
See...understand one thing dU = -dW,if dW is work done by the conservative force,spring force in our spring block example . But dU = dW if dW is work done by external agent .
Do you agree with this ?
6. Dec 28, 2013
Philip Wood
This is correct. The reason is that a system's potential energy is the work it can do because of configuration (i.e. the relative positions of its parts). So if a system DOES work dW, it loses potential energy dU.
Suppose we consider force and displacement components in the x direction, that is the direction which tends to stretch the spring. If the spring is already extended, Fx will be negative, so for positive dx, dW = Fx dx will be negative and dU will be positive.
If the spring is compressed Fx will be positive, and if we compress it further, dx will be negative, so dW = Fx dx will be negative and dU will (again) be positive.
7. Dec 29, 2013
devang2
Thank you for the reply .AS you know for oscillator force is always -kx and always directed towards mean position . i calculate the work dw by applying dot product of force and displacement. While dot product formula you know scalar values of force and displacement are used . When the particle is moving away from mean position angle is 180 because force and displacement are in opposite direction hence dw is negative but when particle is moving towards mean position force and displacement are in the same direction hence angle is 0 therefore according to product dw is positive. I shall be very thankful if you point out where i have gone wrong
8. Dec 29, 2013
Staff: Mentor
The potential energy is at a maximum when the displacement is at a maximum, and it's zero at the mean position. So it's not at all surprising that the sign of $dW$ is different when the displacement is increasing and when the displacement is decreasing; in one case we're putting energy into the spring and in the other case we're taking energy out of it.
9. Dec 29, 2013
Philip Wood
My method also uses dot product. If two vectors, A and B, have no y or z components, then it is easy to show that A.B = AxBx. The components, Ax and Bx, are scalar, but can be positive or negative.
Suggest you read my post carefully again, with this in mind. Or, if you prefer, use magnitudes (scalar and non-negative) of vectors, and put in the cosine value, +1 or -1, separately as you follow my argument. You'll find it gives the same result as my method each time. The argument shows that the more you deform the spring - either by extending it or compressing it - the more energy it stores. If you repeat the argument, changing the sign of dx, you'll find that if you reduce the deformation - either the extension or the compression - the spring stores less energy.
What I found confusing in your last post - and it may have some bearing on your own difficulty - is what you mean by 'displacement'. Are you using it to mean displacement of the mass from its equilibrium position (that's the usual meaning) or are you using it to mean the incremental displacement, dx?
Last edited: Dec 29, 2013
10. Dec 30, 2013
devang2
Thank you for the trouble you have taken for the clarification . By displacement i mean the distance of the mass of the particle from the mean position executing harmonic oscillation.
suppose spring is stretched by certain distance , the work involved is w(!) and when the spring is released to restore the equilibrium original position work involved is W(2) because spring is in original state no net work is done therefore w(!)+w(2)=0 which leads to the conclusion the sign of work during stretching and decompressing are opposite .Similarly when the particle executing harmonic oscillation is moving away from mean position it is like stretching and when it is moving towards mean position it is like decompressing hence sign of work when it is moving away from mean position is negative and when it moves towards mean position it is positive which finally leads to the conclusion that sign of potential energy when particle is moving away is positive and it is negative when motion of particle is towards mean position I shall be very thankful if you let me know if i am right or wrog
11. Dec 30, 2013
Philip Wood
Yes, that's right. The work done BY the spring is negative when it is being compressed or extended, and positive when being decompressed or de-extended. But, since I can see you like precision, I'll point out that your last sentence isn't quite correct...
What you should say is that the sign of the CHANGE in potential energy when particle is moving away is positive and it is negative when the motion of the particle is towards the mean position. [The potential energy itself is always positive (or zero) if we adopt the usual convention of assigning zero to the potential energy when the spring is undeformed.]
12. Dec 30, 2013
vanhees71
I don't understand the problem. It's pretty simple. The potential of a force (if it exists!) is defined by
$$\vec{F}(\vec{r})=-\vec{\nabla} V(\vec{r}).$$
Given that, for the isotropic harmonic oscillator,
$$\vec{F}(\vec{r})=- m \omega^2 \vec{r},$$
you find by simple integration or taking the appropriate line integral (e.g., a straight line from the origin to $\vec{r}$) you get
$$V(\vec{r})=\frac{m \omega^2}{2} \vec{r}^2.$$
You easily check that indeed this is the potential, i.e., that $\vec{F}$ is indeed a conservative field.
Of course it also satisfies the local condition
$$\vec{\nabla} \times \vec{F}=0,$$
and since it is defined and analytic everywhere in $\mathbb{R}^3$ there must exist a potential.
13. Dec 31, 2013
devang2
thank you for the analytical reply. My problem is very basic and simple ,it is about the sign of potential which changes sign with change in direction of motion of the particle executing harmonic oscillation though magnitude remains same .There is change in sign if potential is calculated by applying the basic formula ,dU=-W u is the potential and w is the work which i calcule by using dot product. Please calculate the potential by using this method , you will will see that sign of potential is positive when particle is moving away from mean position while it is negative when particle is moving towards mean position.Please pay attention to the sign. I shall be very thankful if you let me know your views about the change in sign .All the text books assign positive sign yo the potential disregarding the change in direction of motion .You know when mass is shifted to infinity from earth against gravity potential is positive but it is negative when it is brought from infinity moving in the direction of gravity thus it is seen that there is change in sign with the direction so same thing is applicable to motion of particle executing harmonic scillation because it is continusuosly changing direction of motion
14. Dec 31, 2013
Philip Wood
You're still making the mistake that I pointed out in my last post! It's not the potential energy, but the CHANGE in potential energy which is positive when the particle is moving away from the mean position and negative when the particle is moving towards the mean position. The potential energy itself is always positive (if we assign zero to it at the mean position, when the spring is unstretched). It is more positive the more the spring is stretched.
Say if you're still not clear.
[PS: You've started to write 'potential' instead of 'potential energy'. Stick with potential energy!]
Last edited: Dec 31, 2013
15. Jan 1, 2014
devang2
Thank you very much indeed for the clarification and pointing out ther mistake from which it can be concluded that work is negative when the particle is moving away from mean posiytion whereas it is positive whan the particle is moving towards mean position .
16. Jan 1, 2014
Philip Wood
That's right: work done by the spring (or other body responsible for the restoring force) is negative when the particle moves away from the mean position, so the spring gains potential energy (dU = -dW). If we stretch the spring by X, then potential energy stored is
$$U = \int_{x = 0}^{X}{dU} = -\int_{x =0}^{X}{dW} = -\int_{0}^{X}{-kx dx} =\frac{1}{2}k X^2$$
You won't find it done quite like this in the textbooks. I'm using dW to mean an increment of work done BY the spring, following - I hope - what you were doing. If you consider the work done ON the spring as it is stretched, you can dispense with the minus signs in the above derivation.
Last edited: Jan 1, 2014
17. Jan 2, 2014
devang2
Your way of expressing makes it easy for me to understand the problem.If the particle is moving towards the mean position,applying the method as adopted when the particle is moving away from mean position potential energy of the particle in SHO is negative that is -1/2kx^2. I shall be very thankful if you can clarify it . My contention is that the potential energy changes sign with change in direction of motion
18. Jan 2, 2014
vanhees71
No, it doesn't. It doesn't even depend on the velocity. The potential is $V(\vec{r})=k \vec{r}^2/2$ and thus independent on the direction of the position vector and doesn't depend on velocity at all. It's the force, which is
$$\vec{F}=-k \vec{r}$$
that changes it's direction with the position vector, not the potential!
19. Jan 2, 2014
Philip Wood
deVang2. Same mistake AGAIN ! When the particle is moving TOWARDS the mean position, the CHANGE in potential energy is negative, so the spring is LOSING the potential energy that it had when it had a greater displacement. By the time it has got back to x = 0 it has lost ALL the potential energy. So at x = 0, it has zero potential energy, not a negative potential energy. Its potential energy never goes negative.
20. Jan 2, 2014
Philip Wood
Same mistake AGAIN ! As the particle moves towards the mean position the CHANGE in potential energy is negative, so the spring is losing the potential energy which it had at displacement X. When it has got back to the mean position it has lost ALL the potential energy. Its potential energy is now zero. The potential energy is never negative. | 2,743 | 11,533 | {"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": 2, "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} | 3.625 | 4 | CC-MAIN-2018-09 | longest | en | 0.913108 |
https://www.tutorela.com/math/common-denominator/examples-exercises | 1,718,471,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00372.warc.gz | 914,270,159 | 12,645 | # Least common denominator - Examples, Exercises and Solutions
A common denominator is a denominator that will be common and equal for all the fractions in the exercise. We will reach such a denominator by reducing or enlarging the fraction - an operation of multiplication or division.
We can arrive at several correct common denominators.
#### We will divide the search for the common denominator into 3 cases:
• The first case: one of the denominators appearing in the original exercise will be the common denominator.
In this case, we will notice that we only have to multiply one denominator by an integer to reach the same denominator as in the other fraction.
• The second case: find a number that both denominators in the exercise can reach by multiplication.
• The third case: find the common denominator by multiplying the denominators.
## Practice Least common denominator
### Exercise #1
$\frac{3}{5}+\frac{2}{15}=$
### Video Solution
$\frac{11}{15}$
### Exercise #2
$\frac{1}{5}+\frac{2}{15}=$
### Video Solution
$\frac{5}{15}$
### Exercise #3
$\frac{2}{4}+\frac{1}{2}=$
### Video Solution
$1$
### Exercise #4
$\frac{2}{3}+\frac{1}{6}=$
### Video Solution
$\frac{5}{6}$
### Exercise #5
$\frac{1}{3}+\frac{3}{6}=$
### Video Solution
$\frac{5}{6}$
### Exercise #1
$\frac{3}{4}+\frac{1}{8}=$
### Video Solution
$\frac{7}{8}$
### Exercise #2
$\frac{1}{4}+\frac{6}{8}=$
### Video Solution
$1$
### Exercise #3
$\frac{1}{4}+\frac{4}{8}=$
### Video Solution
$\frac{6}{8}$
### Exercise #4
$\frac{1}{5}+\frac{6}{10}=$
### Video Solution
$\frac{8}{10}$
### Exercise #5
$\frac{2}{4}+\frac{1}{8}=$
### Video Solution
$\frac{5}{8}$
### Exercise #1
$\frac{1}{2}+\frac{3}{8}=$
### Video Solution
$\frac{7}{8}$
### Exercise #2
$\frac{1}{4}+\frac{5}{8}=$
### Video Solution
$\frac{7}{8}$
### Exercise #3
$\frac{1}{3}+\frac{2}{9}=$
### Video Solution
$\frac{5}{9}$
### Exercise #4
$\frac{1}{3}+\frac{5}{9}=$
### Video Solution
$\frac{8}{9}$
### Exercise #5
$\frac{1}{3}+\frac{4}{9}=$
### Video Solution
$\frac{7}{9}$ | 653 | 2,074 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 33, "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} | 4.84375 | 5 | CC-MAIN-2024-26 | latest | en | 0.641383 |
sdassignmentxpyr.epitaphs.us | 1,537,381,613,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156270.42/warc/CC-MAIN-20180919180955-20180919200955-00335.warc.gz | 225,773,324 | 5,974 | # Geometrical application of ordinary differential equation
253 16 local theory of regular singular points and applications 263 vii [ arn83] , geometrical methods in the theory of ordinary differential equations. “geometry of stochastic differential equations” the proposal on the problems from applications, which i have and had in mind, in of ordinary differential equations, forthcoming in the journal of mathematical analysis and. Hirsch, morris w review: v i arnold, geometrical methods in the theory of ordinary differential equations bull amer math soc (ns) 10 (1984), no. Thus, geometrically, the second-order differential equations (solved for the highest ing formula for ordinary connections on principal fiber bundles the [8] rb gardner, the method of equivalence and its applications, siam, philadel.
Calculons sa courbure et discutons quelques applications 1 to study geometrical properties of general second-order differential equation fields does not. This textbook is a short comprehensive and intuitive introduction to lie group analysis of ordinary and partial differential equations this practical-oriented. Geometrical application of differential equations (2) - free download as word doc the second forms the content of the theory of ordinary differential equations. The subject of this book is numerical methods that preserve geometric properties of the ow of a differential equation: and stiff) ordinary differential equations has reached a certain maturity, and excel- application of classical integrators.
Consider a differential equation since the derivative is the slope of the tangent line, we interpret this equation geometrically to mean that at any point (x,y) in the . According to this the theory of linear differential equations has been generalized to the geometry of connections on manifolds, and the. Applications to differential equations and geometry andrea obtain some non-existence results for the ordinary differential inequality ,3+t/3”/1fr)_f(,5(r). This section is devoted to ordinary differential equations of the second order topics describe applications of second order equations in geometry and physics.
We prove that a first order ordinary differential equation (ode) with a dicritical applications in geometry, computer vision, classical invariant theory, the. In this lecture, however, we will discuss the geometric interpretation of the problem (1)–(2) this geometric assume that our differential equation models some real world process, and and linear algebra), since it would take too much time. Inverse problems research concentrates on the mathematical theory and practical interpretation of indirect measurements applications are found in virtually.
Latter depends on the theory of ordinary differential equations however, in the a lot of unexpected application in algebraic geometry [55], the most interesting. Advances in geometry albanian journal of mathematics algebraic and journal of differential equations electronic journal of linear algebra international journal of computational geometry and applications. Geometry of differential equations sebastián cuéllar carrillo david palomino in the classic sense, a first order ordinary differential equation is an expresion. We will discuss geometric/qualitative and numerical techniques that apply to the mathematical theory of ordinary differential equations and its application to.
• The same procedure applies to other polynomial identites in linear in the simplest case of the kdv equation, solitons are obtained from a.
• This section provides materials for a session on geometric methods differential equations » unit i: first order differential equations » geometric methods unit ii: second order constant coefficient linear equations coefficients linear operators pure resonance frequency response applications exam 2.
• In this book, the reader can find an elegant approach to some relevant ordinary and partial differential equations within the language of local.
In mathematics, differential calculus is a subfield of calculus concerned with the study of the geometrically, the derivative at a point is the slope of the tangent line to the graph of differentiation has applications to nearly all quantitative disciplines equations involving derivatives are called differential equations and are. Of the tifr-iisc programme in the applications of analysis, geometry and partial differential equa- tions equations (linear and nonlinear) and systems was. Real numbers, variable expressions, solving equations, linear equations in two calculus, differential equations applications to problems in business and economics integration techniques, 2-and-3 dimensional vector geometry functions of. Keywords: synthetic differential geometry, vector field, action, exponential object two ordinary differential equations as the exponential of the corresponding jugation, under some conditions, is an application of the method of change of.
Geometrical application of ordinary differential equation
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2018. | 911 | 5,051 | {"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-2018-39 | latest | en | 0.897031 |
https://www.kullabs.com/classes/subjects/units/lessons/notes/note-detail/3065 | 1,582,856,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146940.95/warc/CC-MAIN-20200228012313-20200228042313-00037.warc.gz | 772,699,383 | 12,129 | Notes on Dip Circle and Apparent Dip | Grade 12 > Physics > Magnetic Properties of Materials | KULLABS.COM
• Note
• Things to remember
### Dip Circle
Dip circle is a device used to measure the angle of dip at a place. It consists of a magnetic needle pivoted at the center of the vertical circular scale that can rotate in the plane of the scale about the horizontal axis passing through its C.G.
It consists of vertical circular scale is divided into four quadrants, each graduated from 0o to 90o, with 0o- 0o in horizontal and 90o – 900 in vertical positions. The horizontal circular scale at the bottom is graduated from 0o to 360o. Dip circle is shown in the figure.
#### Measurement of Angle of Dip
First of all the device is maintained with correct leveling and needle pointing 90o-90o on the vertical scale. Here the needle and vertical scale are in a plane perpendicular to the magnetic meridian. The box is then rotated through 90o position on the horizontal circular scale. This sets the needle and vertical scale exactly in magnetic meridian. The needle rests in the direction of earth’s resultant field I and its two ends points the value of dip $$\delta$$ at that place.
#### Errors and Corrections in Dip Circle
1. Pivot of needle may not be at the center of circular scale:
If the center of needle and center of scale do not coincide with each other, one value of dip will be greater and the other will be smaller than the true value of dip angle. Average of these two readings will give the true value of dip angle at that place.
2. The line indicating 0o – 0o is not horizontal:
This defect will make the value of dip will be smaller or greater than the true value. To remove this defect, readings should be taken in two vertical scales: first in the magnetic meridian and then rotate it through 180o in magnetic meridian. The average of readings is the true value.
3. Magnetic center of the needle may not coincide with geometric center of the needle:
In this case, the magnetic axis of the needle inclines to its geometric axis. To remove this defect, the readings are taken in magnetic meridian and then the needle is reversed on its bearing. The average of two readings is the accurate value of the angle of dip angle.
4. The center of mass of the needle doesn’t coincide with the pivot:
The center of mass of the needle doesn’t match as the deflection of the needle is more at one end than at the other. To remove this defect two readings are taken firstly without doing any change and secondly by interchanging the poles. We can interchange the poles by demagnetizing the needles and remagnetizing it with different poles.
We should take 16 readings of the angle of dip and average of these values gives true dip angle at that place.
### Apparent Dip
The angle made by the needle with the horizontal at this plane is called the apparent dip. Let $$\alpha$$ be the angle made by the dip circle with the magnetic meridian. Then the effective horizontal component in this plane is $$H_a = H\cos \: \alpha$$ and the vertical component is still same, V. if the apparent dip at this plane is $$\delta _1$$ and the true dip is $$\delta$$, then
\begin{align*} \tan \delta _1 &= \frac {V}{H_a} \\ \text {or,} \: \cot \delta_1 &= \frac {H_2}{V} = \frac {H\: \cos \alpha }{V} = \frac {\cos \alpha }{\tan \delta } \\ \cot \delta _1 &= \cos \alpha \times \cot \delta \dots (i) \\ \end{align*}
This is the relation between true dip $$\delta$$ and the apparent dip $$\delta _1$$. If the circle makes an angle of $$(90^o - \alpha$$ with the magnetic meridian and H will have the component $$H’_a = H (\cos 90^o - \alpha ) = H\sin \alpha )$$ in the new plane. Then, the apparent dip $$\delta _2$$ is given by
\begin{align*} \cot \delta _2 &= \sin \alpha \times \cot \delta \dots (ii) \\ \text {squaring and adding equation}\: (i) \:\text {and} \: (ii), \: \text {we get} \\ \cot ^2 \delta _1 + \cot^2 \delta _2 &= \cos^2 \alpha \times \cot ^2 \delta + \sin^2 \alpha \times \cot ^2 \delta \\ \cot ^2 \delta &= \cot^2\delta _1 + \cot ^2 \delta_2 \\ \end{align*}
Knowing the values of $$\delta _1$$ and $$\delta _2$$ we can get the value of $$\delta$$.
References
Manu Kumar Khatry, Manoj Kumar Thapa et al. Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.
S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.
Dip circle is a device used to measure the angle of dip at a place.
The angle made by the needle with the horizontal at this plane is called the apparent dip.
The relation between true dip $$\delta$$ and the apparent dip $$\delta _1$$ is $$\cot \delta _1 = \cos \alpha \times \cot \delta \dots (i)$$
.
0%
## ASK ANY QUESTION ON Dip Circle and Apparent Dip
No discussion on this note yet. Be first to comment on this note | 1,245 | 4,806 | {"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} | 3.84375 | 4 | CC-MAIN-2020-10 | latest | en | 0.886429 |
https://cboard.cprogramming.com/c-programming/146963-how-generate-random-number-without-repetition.html | 1,516,752,320,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892802.73/warc/CC-MAIN-20180123231023-20180124011023-00717.warc.gz | 627,055,821 | 17,003 | # Thread: How to generate random number without repetition?
1. ## How to generate random number without repetition?
Hello, everyone! i am trying to generate random numbers in 2d array as my code below, but i but I still have problems: At the output, i got the repetition of the index (i) .
Example:
arr[11][0] = 0.563585
arr[18][0] = 0.808741
arr[29][0] = 0.479873
arr[18][0] = 0.895962
arr[22][1] = 0.746605
arr[5][1] = 0.858943
arr[1][1] = 0.513535
arr[1][1] = 0.014985
.............................
.............................
arr[12][29] = 0.634724
arr[19][29] = 0.828791
arr[5][29] = 0.720908
arr[5][29] = 0.375134
Code:
```
#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
int main(void)
{
double arr[30][30];
int m[30];
int i,j;
i=0;
for(j=0; j<30; j++){
m[j]= 0;
while(m[j] <4){
i= (int) rand()%30;
arr[i][j]= (double) rand()/RAND_MAX;
if(arr[i][j] > 0){
m[j]++;
printf("arr[%d][%d] = %lf\t", i, j, arr[i][j]);
printf("\n");
}
}
}
getch();
}
```
Could anyone help me?
2. You're wasting 7200 bytes of local/stack memory. A 30x30 array of 8 byte types should be dynamically allocated or static.
If you want to successfully use rand(), you have to
Code:
`srand(time(NULL));`
first.
Get rid of conio.h and any of its associated functions, it's outdated.
3. Why are you using random numbers for your array index? Wouldn't it be better to assign a value to each element of your array?
Jim
4. Originally Posted by jimblumberg
Why are you using random numbers for your array index? Wouldn't it be better to assign a value to each element of your array?
Jim
Because i want to random with the fix m[j]=4. if i not do this mean m[j]= 30. Do you have any idea, what i should deal with this problem?
5. Originally Posted by memcpy
You're wasting 7200 bytes of local/stack memory. A 30x30 array of 8 byte types should be dynamically allocated or static.
If you want to successfully use rand(), you have to
Code:
`srand(time(NULL));`
first.
Get rid of conio.h and any of its associated functions, it's outdated.
I was put "srand(time(NULL));" but i stil get the same problem sir
How do i do? any advice?
"How to generate random number without repetition?"
I do not believe there is a way directly. What I would do is either:
1. fill the data structure with all possible values and then randomly shuffle the results (as in shuffling a deck of cards) or
2. throw away the duplicates and select again
Originally Posted by memcpy
You're wasting 7200 bytes of local/stack memory. A 30x30 array of 8 byte types should be dynamically allocated or static.
Or you could make the stack size larger, if necessary.
Originally Posted by Suntang
Code:
```...
int main(void)
{
double arr[30][30];
...
if(arr[i][j] > 0){
...
}```
You seem to be assuming that arr[30][30] is automatically initialized to value 0.0 for each element. This is not true for function or block scope variables that are not static. If you used either of memcpy's suggestions, the side effect is that they would be initialized. If I were doing this, I would reinitialize them all to 0.0 explicitly.
Originally Posted by Suntang
Code:
```...
int main(void)
{
...
int m[30];
int j;
...
for (j=0; j<30; j++) {
m[j]= 0;
while(m[j] <4){
...
if(arr[i][j] > 0){
m[j]++;
...
}
...
}
...
}
...
}```
What is the purpose of only doing the activity until there are four entries per sub-array? I do not have suggestions until we understand why you are doing this extra activity with m[].
You are using the magic number 30 in several places. I suspect they are all related, but I would know that for sure if they all had the same name. Since we are talking about C, consider giving 30 a name using a #define text substitution macro.
Give your variables longer more meaning full names. If you gave m[] a different name, I might not be asking the question above.
Consider adding comments that explain "why" you are doing some of what your are doing? Avoid comments that say "what" you are doing where the code already says the same thing.
7. What is the purpose of only doing the activity until there are four entries per sub-array? I do not have suggestions until we understand why you are doing this extra activity with m[].
You are using the magic number 30 in several places. I suspect they are all related, but I would know that for sure if they all had the same name. Since we are talking about C, consider giving 30 a name using a #define text substitution macro.
Give your variables longer more meaning full names. If you gave m[] a different name, I might not be asking the question above.
Consider adding comments that explain "why" you are doing some of what your are doing? Avoid comments that say "what" you are doing where the code already says the same thing.[/QUOTE]
Humm!! i main idea that i do this because i want to generate random number in evaluation systems, suppose: I have a reviewer r(i) has evaluated an object o(j), which is weighted with the assigned evaluation score arr(i,j) between 0 and 1. The number of reviewers and objects are 30 each. and i want to generate ramdon number with the fix number of reviewers per object at n=4. I mean each object should have evaluated only 4 reviewers.
Can you give me an advince? what is the best way i should deal with this?
8. I am going to assume that you are now explicitly initializing review[MAX_REVIEWERS][MAX_OBJECTS] (formerly known as arr[30][30]) to 0.0. [Use names for these things appropriate to your language and the common language of those who will be maintaining the code.]
So change statements in the inner while loop. Get your random reviewer (formerly known as i) and see if that reviewer has already reviewed that object (formerly known as j). If they have, skip the rest of the inner loop. So the if statement is moved and goes from "if (arr[i][j] > 0.0) {" to "if review[reviewer][object] == 0.0) {". So now we know it has not been reviewed. You may want to use a do while loop to make sure that rand() does not return a 0 in the 1/RAND_MAX time that that will happen.
By the way, you may want to make the value 0.0 have the name NOT_REVIEWED and use that for initialization and this check.
9. Originally Posted by pheininger
I am going to assume that you are now explicitly initializing review[MAX_REVIEWERS][MAX_OBJECTS] (formerly known as arr[30][30]) to 0.0. [Use names for these things appropriate to your language and the common language of those who will be maintaining the code.]
So change statements in the inner while loop. Get your random reviewer (formerly known as i) and see if that reviewer has already reviewed that object (formerly known as j). If they have, skip the rest of the inner loop. So the if statement is moved and goes from "if (arr[i][j] > 0.0) {" to "if review[reviewer][object] == 0.0) {". So now we know it has not been reviewed. You may want to use a do while loop to make sure that rand() does not return a 0 in the 1/RAND_MAX time that that will happen.
By the way, you may want to make the value 0.0 have the name NOT_REVIEWED and use that for initialization and this check.
Could you show me the code sir? i don't understad. i'm a new learnner in c.
10. Here's one way, and you'll like it:
Make an integer array with the numbers that you want to randomly select from. Say I want random numbers 0-9, and each one never repeating.
Code:
`array[10]={1,2,3,4,5,6,7,8,9,10};`
now
//set srand() up before the loop
Use a for loop, from 1 to 7 (or so), get a random number in the range of 0 to 9
swap array[i] with array[random Number], in the usual way:
Code:
```temp = array[i];
array[i]=array[random number];
array[random number]= temp;```
and you're done.
You'll want to do at least 5 swaps, since you are moving two values, across an array of 10 values. To be more random, you may want to do 10 swaps, and do use a better random number design if you want REAL random performance. Generating truly good random numbers, is a lot more challenging than it looks. Thankfully, there are libraries that can help if that high a level of randomness, is needed. (doubtful, but just in case).
11. As I understand it, you want to generate 4 random numbers from 0 to 29 (inclusive) without repetition. One way is a shuffle. But since you only want 4 out of the 30 possible numbers, it's reasonably efficient to simply check against the numbers already generated with a function like this:
Code:
```// Fill a[] with n non-repeating random integers in [0,high).
void NonRepeatingRandom(int a[], int n, int high) {
int i, j, again;
for (i = 0; i < n; i++) {
do {
again = 0;
a[i] = rand() / (RAND_MAX / high + 1);
for (j = i - 1; j >= 0; j--) {
if (a[j] == a[i]) {
again = 1;
break;
}
}
} while (again);
}
}
// Call it like this:
int rndReviewers[4];
// . . .
NonRepeatingRandom(rndReviewers, 4, 30);```
12. Originally Posted by Suntang
Could you show me the code sir?
Originally Posted by pheininger
I am going to assume that you are now explicitly initializing review[MAX_REVIEWERS][MAX_OBJECTS] (formerly known as arr[30][30]) to 0.0. (Use names for these things appropriate to your language and the common language of those who will be maintaining the code.)
Old:
Code:
` double arr[30][30];`
New:
Code:
` double arr[30][30]={0.0};`
Originally Posted by pheininger
So change statements in the inner while loop. Get your random reviewer (formerly known as i) and see if that reviewer has already reviewed that object (formerly known as j). If they have, skip the rest of the inner loop. So the if statement is moved and goes from "if (arr[i][j] > 0.0) ..." to "if review[reviewer][object] == 0.0) ...". So now we know it has not been reviewed. You may want to use a do while loop to make sure that rand() does not return a 0 in the 1/RAND_MAX time that that will happen.
Old:
Code:
``` i= (int) rand()%30;
arr[i][j]= (double) rand()/RAND_MAX;
if (arr[i][j] > 0){
m[j]++;
}```
New:
Code:
``` i= (int) rand()%30;
if (arr[i][j] == 0.0) {
arr[i][j]= (double) rand()/RAND_MAX;
if (arr[i][j] > 0.0) {
m[j]++;```
13. > Or you could make the stack size larger, if necessary.
What do you think is easier:
• Using malloc() and writing code that follows convention.
Or:
• Finding a way to change the stack size,
• that may not work on older computers with less memory,
• that definitely doesn't work universally across systems,
• that's a quick "hack" that might or might not work,
• that slows down your code by forcing the system to do a lot more work than necessary
14. Making the stack larger isn't hard, it's just wrong because large chunks of data is not what the stack is for.
However FYI, 7.2K is not that large. Windows apps have a typical stack size of 1MB, and a minimum of 7.2K. Even with that you'd be more than fine.
15. Drat, I meant to say a minimum of 64K | 2,911 | 10,813 | {"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-2018-05 | latest | en | 0.777016 |
https://www.greaterwrong.com/tag/rationality-a-z-discussion-and-meta | 1,708,801,940,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00289.warc.gz | 808,000,113 | 12,347 | # Rationality A-Z (discussion & meta)
TagLast edit: 21 Aug 2020 2:36 UTC by
Rationality: From AI to Zombies, also known as The Sequences, is a series of essays by Eliezer Yudkowsky published from 2006 to 2009, which were later compiled into a book. This tag is for critiques and discussion of the Sequences, attempts to organize the Sequences, and the publication of Rationality: From AI to Zombies.
To read the book, see lesswrong.com/rationality
# Eliezer’s Sequences and Mainstream Academia
15 Sep 2012 0:32 UTC
236 points
# New edition of “Rationality: From AI to Zombies”
15 Dec 2018 21:33 UTC
83 points
# What is the next level of rationality?
12 Dec 2023 8:14 UTC
43 points
# Rationalism before the Sequences
30 Mar 2021 14:04 UTC
576 points
# Rationality: Abridged
6 Jan 2018 0:20 UTC
98 points
# How accurate is the quantum physics sequence?
17 Apr 2012 6:54 UTC
72 points
# A summary of every “Highlights from the Sequences” post
15 Jul 2022 23:01 UTC
94 points
# Rationality: From AI to Zombies
13 Mar 2015 15:11 UTC
127 points
# Learning and Internalizing the Lessons from the Sequences
14 Sep 2016 14:40 UTC
10 points
# Preface
11 Mar 2015 19:00 UTC
704 points
# New versions of posts in “Map and Territory” and “How To Actually Change Your Mind” are up (also, new revision system)
26 Feb 2019 3:17 UTC
36 points
# Help us name the Sequences ebook
15 Apr 2013 19:59 UTC
22 points
# The Most Important Thing You Learned
27 Feb 2009 20:15 UTC
15 points
7 Jan 2024 1:44 UTC
25 points
# The Most Frequently Useful Thing
28 Feb 2009 18:43 UTC
12 points
# [Question] Summary of the sequences / Lesson plans for rationality
5 Nov 2021 17:22 UTC
5 points
# [Question] What are some good examples of fake beliefs?
14 Nov 2020 7:40 UTC
18 points
# What is the best way to read the sequences?
17 Jun 2012 3:50 UTC
5 points
# Map and Territory: Summary and Thoughts
5 Dec 2020 8:21 UTC
16 points
# Common Sense Atheism summarizing the Sequences
19 Nov 2010 12:55 UTC
19 points
# Sexual Dimorphism in Yudkowsky’s Sequences, in Relation to My Gender Problems
3 May 2021 4:31 UTC
54 points
(unremediatedgender.space)
# A suggestion on how to get people to read the Sequences
25 Oct 2010 19:21 UTC
43 points
# An EPub of Eliezer’s blog posts
11 Aug 2011 14:20 UTC
58 points
# [Link] Lifehack Article Promoting LessWrong, Rationality Dojo, and Rationality: From AI to Zombies
14 Nov 2015 20:34 UTC
13 points
# Second major sequence now available in audio format
31 Jan 2013 5:25 UTC
35 points
# Eliezer’s Post Dependencies; Book Notification; Graphic Designer Wanted
10 Jun 2008 1:18 UTC
6 points
# LessWrong-Portable
22 Sep 2017 20:48 UTC
8 points
# That You’d Tell All Your Friends
1 Mar 2009 12:04 UTC
8 points
# Request: Sequences book reading group
22 Feb 2015 1:06 UTC
26 points
# Rationality: An Introduction
11 Mar 2015 19:00 UTC
38 points
# How To Actually Change Your Mind eBook (In Order)
3 Sep 2012 21:53 UTC
27 points
# Gods! Robots! Aliens! Zombies!
16 Mar 2020 18:08 UTC
11 points
(hivewired.wordpress.com)
# Reductionism sequence now available in audio format
2 Jun 2013 2:55 UTC
31 points
# Why didn’t people (apparently?) understand the metaethics sequence?
29 Oct 2013 23:04 UTC
23 points
# PSA: The Sequences don’t need to be read in sequence
23 May 2022 2:53 UTC
86 points
# Curating “The Epistemic Sequences” (list v.0.1)
23 Jul 2022 22:17 UTC
65 points
# An unofficial “Highlights from the Sequences” tier list
5 Sep 2022 14:07 UTC
29 points
# [Question] Sequences/Eliezer essays beyond those in AI to Zombies?
8 Sep 2022 5:05 UTC
4 points
# I Converted Book I of The Sequences Into A Zoomer-Readable Format
10 Nov 2022 2:59 UTC
203 points
# The Sequences Highlights on YouTube
15 Feb 2023 19:36 UTC
20 points
# What kind of place is this?
25 Feb 2023 2:14 UTC
24 points
# LessWrong and Rationality ebooks via Amazon
11 Sep 2011 16:08 UTC
30 points
# The Sequences in MP3 Format
8 Jul 2011 19:40 UTC
17 points
# Announcement: The Sequences eBook will be released in mid-March
3 Mar 2015 1:58 UTC
69 points
# Summarizing the Sequences Proposal
4 Aug 2011 21:15 UTC
9 points
# Bring Back the Sequences?
7 Mar 2011 7:21 UTC
62 points
# Rewriting the sequences?
13 Jun 2011 14:14 UTC
23 points
# Broken images in the sequences
25 Mar 2018 14:41 UTC
7 points
# Weird characters in the Sequences
18 Nov 2010 8:27 UTC
9 points
27 Apr 2011 19:01 UTC
97 points
7 Apr 2011 10:39 UTC
51 points
# I Stand by the Sequences
15 May 2012 10:21 UTC
14 points
# Print ready version of The Sequences
6 Nov 2010 1:21 UTC
20 points
# Sequences in Alternative Formats
4 Dec 2010 1:40 UTC
22 points
# Call for volunteers: Publishing the Sequences
28 Jun 2012 15:08 UTC
22 points
# New Less Wrong Feature: Rerunning The Sequences
11 Apr 2011 17:01 UTC
49 points
# Help us Optimize the Contents of the Sequences eBook
19 Sep 2013 4:31 UTC
18 points
# A new, better way to read the Sequences
4 Jun 2017 5:10 UTC
19 points
# Kickstarting the audio version of the upcoming book “The Sequences”
16 Dec 2014 1:01 UTC
47 points
28 Sep 2016 16:43 UTC
15 points
(putanumonit.com)
# Is it cool if I post responses/thoughts as I read through the sequences?
2 Jul 2011 23:31 UTC
10 points
# Rationality: From AI to Zombies online reading group
21 Mar 2015 9:54 UTC
47 points | 2,010 | 5,571 | {"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-2024-10 | latest | en | 0.811292 |
https://physics.stackexchange.com/questions/92814/contradicting-forces-on-a-circular-loop-under-current-in-magnetic-field | 1,566,661,170,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00199.warc.gz | 595,831,580 | 32,271 | # Contradicting forces on a circular loop under current in magnetic field?
I have the following general conceptual concern.
Think of a thin conducting loop of radius $R$ placed in the $x$-$y$-plane at $z=0$. There is a homogeneous current density $\vec{j}$ running through this loop:
$$\vec{j}(\vec{r})=|j|\delta(z)\delta(x^2+y^2-R^2)\frac{-y\,\vec{e}_x+x\,\vec{e}_y}{\sqrt{x^2+y^2}}$$
Per definition, this loop has a magnetic moment of:
$$\vec{m}=\frac{1}{2}\int d^3r\,\left(\vec{r}\times\vec{j}(\vec{r})\right)=-|j|\pi R\,\vec{e}_z$$
Now, imagine there appears a magnetic field that can be locally described as:
$$\vec{B}=b_0 z\,\vec{e}_z$$
If we ask ourselves what the electromagnetic force on the loop will be, we have two equations that can give us the answer. (The two answers should be the same, but curiously they are not). First equation is the straightforward definition of Lorentz force:
$$\vec{F}_1=\int d^3r\,\left(\vec{j}(\vec{r})\times\vec{B}\right)$$
And the second equation makes use of the magnetic moment (and is also exact for this simple magnetic field):
$$\vec{F}_2=\nabla(\vec{m}\cdot\vec{B})$$
It is now straightforward to see that the first force has the structure $\vec{F}_1=A\vec{e}_x+B\vec{e}_y$, while the second force must clearly look like $\vec{F}_2=C\vec{e}_z\neq 0$ (evidently non-zero due to inhomogenous $\vec{B}$ field). My question is - what went wrong and how to fix this?
• My guess is that the expressions for $\vec{F}_1$ and $\vec{F}_2$ assume $\vec{\nabla} \cdot \vec{B} =0$. – Brian Moths Jan 8 '14 at 23:38
• Yes, this is in fact assumed for $\vec{F}_2$. But this is true in general, since no magnetic monopoles exist. So maybe the solution is: $\vec{F}_1$ is true only for an exact $\vec{B}$ field, while $\vec{F}_2$ is also correct for a local approximation, since it incorporates $\nabla\cdot\vec{B}$? This would sound plausible. – Kagaratsch Jan 8 '14 at 23:42
The solution came up in the comments, here is an elaboration. One of Maxwell's equations is:
$$\nabla\cdot\vec{B}=0$$
The local description for $\vec{B}$ given above clearly does not satisfy this. However, the defining equation for Lorentz force $\vec{F}_1$ assumes a magnetic field $\vec{B}$ that is valid not only locally, but over all space volume. Therefore, equation $\vec{F}_1$ cannot be applied here.
On the other hand, equation $\vec{F}_2$ is actually initially obtained involving a local linear approximation for an arbitrary 'valid' $\vec{B}$ field. The Maxwell's equation $\nabla\cdot\vec{B}=0$ is also already incorporated into the expression $\vec{F}_2$. Therefore, in this case only $\vec{F}_2$ gives the correct result.
• So you cut my solution out of my question and posted it as an answer? I guess that is the correct way for book-keeping here, so I'll do that myself next time. Thanks for the hint. – Kagaratsch Jan 9 '14 at 0:06
• If you want to post your answer as an answer I'll delete this one or flag it for deletion. Answering your own question is perfectly acceptable and it's better that the answer comes from you rather than me. – Brandon Enright Jan 9 '14 at 0:52
• > "defining equation for Lorentz force assumes a magnetic field that is valid not only locally, but over all space volume." The given magnetic field is invalid everywhere, because of wrong value of divergence. The formula 1 does not really assume anything about the magnetic field function in it, except that the formula be integrable so force can be related to $B$. – Ján Lalinský May 20 at 22:03
The formula $$F_2$$ can be derived for bodies that are small enough that higher than first order variation in $$\mathbf B$$ inside the body has negligible impact on the force. For larger bodies, it is not valid, but F1 is.
So, F1 is more general than F2, but neither of them are good for anything if we use non-physical magnetic field function such as $$z\mathbf e_z$$. The fact that they give different result means some assumption in the derivation is broken in this situation. Probably the zero divergence of $$\mathbf B$$ is needed for the derivation of F2. | 1,151 | 4,083 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-35 | latest | en | 0.819579 |
http://chaospro.de/features.php?PHPSESSID=feb85f7ed32e29c832a44c14dcb01035 | 1,540,185,774,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514708.24/warc/CC-MAIN-20181022050544-20181022072044-00070.warc.gz | 66,864,784 | 5,618 | HOME Features Gallery Download Documentation Resources Mailing Lists Programs Feature Voting Wishlist Theory Tutorials Formula DB Contact
Release 4.0
### Features
Following there is a list containing the features of ChaosPro. It should give you a basic idea of how powerful ChaosPro is.
Supports several fractal types/rendering algorithms
ChaosPro can create almost any kind of fractal using its integrated compiler. It is up to you which formula you want to use. So ChaosPro not only can create standard Julia and Mandelbrot sets, but also many other fractal types which are based on iterating a number.
There are already thousands of formulas which have been written by other people and which you can use. Among them there are formulas which create bifurcation diagrams, IFS fractals, Plasma fractals, Lindenmayer systems, dynamic systems like the Lorenz attractor, the Rossler attractor, and Lyapunov Spaces.
ChaosPro itself offers the following main fractal types (which basically are completely different algorithms, each of them uses its own kind of formula and thus you may either use an existing formula or write a new formula for further enhancements):
• Escapetime: A pixel in a window (2D point) is iterated and tested what "happens" to the pixel. The result is shown and colored accordingly. This kind of algorithm creates for example Julia and Mandelbrot sets, but it may also create any other fractal which is based on iterating a 2D pixel.
• Attractor: An arbitrary 3D point is iterated again and again. After each iteration it is drawn. This produces 3D fractals which are rendered accordingly including light and shadow. Examples for such fractals are IFS fractals, Flame fractals, and much more.
• Quaternion: Similar to Escapetime, but uses Quaternion numbers and an algorithm similar to Escapetime which has been enhanced for 3D and 4D space. Examples for these fractals are Quaternions, but this type is perfectly suitable for rendering the Mandelbulb, too.
Win 32 compliant
ChaosPro runs on all Win32 versions, i.e. ChaosPro runs on Windows 98, NT, 2000, XP, Vista and Windows 7. If any other operating system supports Win32 (under Linux?), then it's possible that ChaosPro runs there, too.
Windows 98 Windows NT Windows 2000 Windows XP Windows Vista Windows 7
This means ChaosPro is a true MDI application where each fractal resides in its own window. You can calculate several fractals at the same time in different windows. The different calculation threads use low priority, so you can calculate several fractals and continue doing your other work with the computer. All windows in ChaosPro are modeless dialogs, so can be open just as you like. You do not need to close any window in order to open another one.
Realtime fractal exploration
Well, to be honest, this does not mean that you can dive into the fractal in realtime (only if your computer is fast enough...). Realtime in ChaosPro means that all changes take effect immediately. You grab the fractal with the mouse in order to move it around, and the effect is immediately visible. You assign another gradient (palette) and it gets applied immediately, no OK button to click on. If you move a slider (perhaps the rotation angle slider) the fractal thread constantly gets noticed that the slider changed its value and adjusts the fractal. How good and how fast it catches up with the changes depends on how fast your computer is. If you resize the fractal window then the fractal gets scaled accordingly. ChaosPro does not use modal dialogs, all parameter windows are modeless.
True color
ChaosPro does not know that there are only a limited number of colors. Basically it thinks there is an unlimited range of colors. Later in the fractal calculation it determines the colors based on a gradient (well, in ChaosPro a palette is a path through the color space based on about 250 knots). Depending on the coloring algorithm and on your display driver it then selects the colors. True color images with lots of colors, the mandelbrot set without those iteration bands, ChaosPro can calculate that sort of image.
FractInt compatible
To be honest: ChaosPro is not 100% FractInt compatible: There are situations where ChaosPro behaves different from FractInt. But it can read almost 80% of FractInts fractal types, starting with julia, mandel, upto complexnewton, barnsley1m, IFS, LSystem, etc. ChaosPro's formula parser can parse FractInt *.frm files. ChaosPro's gradient (palette map) editor can read FractInt's *.map files. The IFS and LSystem formula editors can load FractInt's *.ifs and *.l files as well.
UltraFractal compatible
You may ask how ChaosPro can be compatible to a fractal generator which uses a built in compiler. Is there really a compiler in ChaosPro? A compiler for a freeware fractal generator? Yes, indeed, there is: If another one can write a compiler, then I can do that, too. It's quite difficult and very time consuming, but it's not impossible. So I wrote a fast compiler, wrote some automates to import UltraFractal formulas and there it is:
A built in compiler which produces fast machine code. No extra DLL, no extra package to install. It's all in ChaosPro. For free.
The compiler together with the import mechanisms allow ChaosPro to use all UltraFractal formulas (transformations, iteration formulas, colorings) and to calculate all fractals which UltraFractal can calculate.
This especially means that ChaosPro has all the features of UltraFractal 3.02 currently built in.
Animations
In ChaosPro you can create zoom movies: Define where to start from, define where to end. Specify how many frames to calculate and press start. Too easy? Wait a moment:
ChaosPro restricts you not only to simple zoom in/out/around movies, it lets you create animations based upon every fractal parameter in ChaosPro, in every combination. You simply define key frames of how your animation should look like, specify how many frames there should be between each pair of key frames and ChaosPro does the rest. The key frames may differ in any parameter which can be changed in a continuously manner, like the corners, the iteration value, the rotation angle, the parameter, the bailout value, the coloring paramaters, the palette and many others. Parameters which cannot change during an animation are the fractal type or flags for example: These must be the same during the animation. But the animation system is clever enough to not let you specify keyframes which do not match to the others you already have defined.
So, are you interested in an animation flying into a Complex Newton Set, which keeps rotating, whereas the palette changes, just in order to fly around a bit and finally to fly back to the starting location animating the initialization point?
And together with the 3D feature you additionally can animate all 3D parameters...
Last update on Jan 02 2010. | 1,499 | 6,887 | {"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-2018-43 | longest | en | 0.929023 |
http://en.wikipedia.org/wiki/Generalized_additive_model_for_location,_scale_and_shape | 1,408,560,440,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500811913.46/warc/CC-MAIN-20140820021331-00243-ip-10-180-136-8.ec2.internal.warc.gz | 66,132,328 | 14,953 | # Generalized additive model for location, scale and shape
In statistics, the generalized additive model location, scale and shape (GAMLSS) is a class of statistical model that provides extended capabilities compared to the simpler generalized linear models and generalized additive models. These simpler models allow the typical values of a quantity being modelled to be related to whatever explanatory variables are available. Here the "typical value" is more formally a location parameter, which only describes a limited aspect of the probability distribution of the dependent variable. The GAMLSS approach allows other parameters of the distribution to be related to the explanatory variables; where these other parameters might be interpreted as scale and shape parameters of the distribution, although the approach is not limited to such parameters.
## Overview of the model
The generalized additive model location, scale and shape (GAMLSS) is a statistical model developed by Rigby and Stasinopoulos,[citation needed] and later expanded[1] to overcome some of the limitations associated with the popular generalized linear models (GLMs) and generalized additive models (GAMs).[2]
In GAMLSS the exponential family distribution assumption for the response variable, ($y$), (essential in GLMs and GAMs), is relaxed and replaced by a general distribution family, including highly skew and/or kurtotic continuous and discrete distributions.
The systematic part of the model is expanded to allow modeling not only of the mean (or location) but other parameters of the distribution of y as linear and/or nonlinear, parametric and/or additive non-parametric functions of explanatory variables and/or random effects.
GAMLSS is especially suited for modelling leptokurtic or platykurtic and/or positive or negative skew response variable. For count type response variable data it deals with over-dispersion by using proper over-dispersed discrete distributions. Heterogeneity also is dealt with by modelling the scale or shape parameters using explanatory variables. There are several packages written in R related to GAMLSS models.[3]).
A GAMLSS model assumes independent observations $y_i$ for $i = 1, 2, \dots , n$ with probability (density) function $f (y_i | \mu_i , \sigma_i , \nu_i , \tau_i )$ conditional on $(\mu_i , \sigma_i , \nu_i , \tau_i )$ a vector of four distribution parameters, each of which can be a function to the explanatory variables. The first two population distribution parameters $\mu_i$ and $\sigma_i$ are usually characterized as location and scale parameters, while the remaining parameter(s), if any, are characterized as shape parameters, e.g. skewness and kurtosis parameters, although the model may be applied more generally to the parameters of any population distribution with up to four distribution parameters, and can be generalized to more than four distribution parameters.
\begin{align} g_1 (\mu) = \eta_1= X_1 \beta_1 + \sum_{j=1}^{J_1} {h}_{j1}(x_{j1}) \\ g_2(\sigma) = \eta_2= X_2 \beta_2 + \sum_{j=1}^{J_2}{h}_{j2}(x_{j2}) \\ g_3(\nu) = \eta_3 = X_3 \beta_3 + \sum_{j=1}^{J_3}{h}_{j3}(x_{j3}) \\ g_4(\tau)=\eta_4=X_4 \beta_4 + \sum_{j=1}^{J_4}{h}_{j4}(x_{j4}) \end{align}
where μ, σ, ν, τ and $\eta_k$ are vectors of length $n$, $\beta^{T}_k = (\beta_{1k},\beta_{2k},\ldots,\beta_{J'_{k} k})$ is a parameter vector of length $J'_k$, $X_k$ is a fixed known design matrix of order $n \times J'_k$ and $h_{jk}$ is a smooth non-parametric function of explanatory variable $x_{jk}$, $j=1,2,\ldots, J_{k}$ and $k=1,2,3,4$.
For centile estimation the WHO Multicentre Growth Reference Study Group have recommended GAMLSS and the Box-Cox power exponential (BCPE) distributions[4] for the construction of the WHO Child Growth Standards.[5][6]
## What distributions can be used
The form of the distribution assumed for the response variable y, is very general. For example an implementation of GAMLSS in R[7] has around 50 different distributions available. Such implementations also allow use of truncated distributions and censored (or interval) response variables.[7]
## Notes
1. ^ Ahmed Zaheer. Rigby, R. A. and Stasinopoulos D. M. (2005) "Generalized additive models for location, scale and shape, (with discussion)", Appl. Statist., 54, part 3, pp 507–554.Link
2. ^ For an overview of these limitations see Nelder and Wedderburn, 1972[full citation needed] and Hastie and Tibshirani, 1990.[full citation needed]
3. ^ Stasinopoulos D. M.; Rigby R.A. (2007) "Generalized additive models for location scale and shape (GAMLSS) in R". Journal of Statistical Software, 23 (7), Issue 7, Dec 2007. Link
4. ^ Robert A. Rigby and D. Mikis Stasinopoulos (2004)"Smooth centile curves for skew and kurtotic data modelled using the Box-Cox Power Exponential distribution". Preprint
5. ^ Borghi, E.; De Onis, M.; Garza, C.; Van Den Broeck, J.; Frongillo, E. A.; Grummer-Strawn, L.; Van Buuren, S.; Pan, H.; Molinari, L.; Martorell, R.; Onyango, A. W.; Martines, J. C.; WHO Multicentre Growth Reference Study Group (2006). "Construction of the World Health Organization child growth standards: Selection of methods for attained growth curves". Statistics in Medicine 25 (2): 247–265. doi:10.1002/sim.2227. PMID 16143968. edit
6. ^ WHO Multicentre Growth Reference Study Group (2006) WHO Child Growth Standards: Length/height-for-age, weight-for-age, weight-for-length, weight-for-height and body mass index-for-age: Methods and development. Geneva: World Health Organization.
7. ^ a b R packages for GAMLSS can be downloaded from here
• Beyerlein, A., Fahrmeir, L., Mansmann, U., Toschke., A. M. (2001) "Alternative regression models to assess increase in childhood BM". IBMC Medical Research Methodology, 2008, 8(59) doi:10.1186/1471-2288-8-59
• Cole, T. J., Stanojevic, S., Stocks, J., Coates, A. L., Hankinson, J. L., Wade, A. M. (2009), "Age- and size-related reference ranges: A case study of spirometry through childhood and adulthood", Statistics in Medicine, 28(5), 880-898.Link
• Fenske, N., Fahrmeir, L., Rzehak, P., Hohle, M. (25 September 2008), "Detection of risk factors for obesity in early childhood with quantile regression methods for longitudinal data", Department of Statistics: Technical Reports, No.38 Link
• Hudson, I. L., Kim, S. W., Keatley, M. R. (2010), "Climatic Influences on the Flowering Phenology of Four Eucalypts: A GAMLSS Approach Phenological Research". In Phenological Research, Irene L. Hudson and Marie R. Keatley (eds), Springer Netherlands Link
• Hudson, I. L., Rea, A., Dalrymple, M. L., Eilers, P. H. C. (2008), "Climate impacts on sudden infant death syndrome: a GAMLSS approach", Proceedings of the 23rd international workshop on statistical modelling pp. 277–280. Link
• Nott, D. (2006), "Semiparametric estimation of mean and variance functions for non-Gaussian data", Computational Statistics, 21(3-4), 603-620. Link
• Serinaldi, F. (2011), "Distributional modeling and short-term forecasting of electricity prices by Generalized Additive Models for Location, Scale and Shape", Energy Economics, 33(6), 1216-1226, doi:10.1016/j.eneco.2011.05.001
• Serinaldi, F., Cuomo, G. (2011) "Characterizing impulsive wave-in-deck loads on coastal bridges by probabilistic models of impact maxima and rise times", Coastal Engineering, 58(9), 908-926, doi:10.1016/j.coastaleng.2011.05.010
• Serinaldi, F., Villarini, G., Smith, J. A., Krajewski, W. F. (2008), "Change-Point and Trend Analysis on Annual Maximum Discharge in Continental United States", American Geophysical Union Fall Meeting 2008, abstract #H21A-0803*
• van Ogtrop, F. F., Vervoort, R. W., Heller, G. Z., Stasinopoulos, D. M., Rigby, R. A. (2011) "Long-range forecasting of intermittent streamflow", Hydrology and Earth System Sciences Discussions, 8(1), 681-713. doi:10.5194/hessd-8-681-2011
• Villarini, G., Serinaldi, F. (2011), "Development of statistical models for at-site probabilistic seasonal rainfall forecast", International Journal of Climatology. doi:10.1002/joc.3393
• Villarini, G., Serinaldi, F., Smith, J. A., Krajewski, W. F. (2009), "On the stationarity of annual flood peaks in the continental United States during the 20th century", Water Resources Research, 45(8). Link
• Villarini, G., Smith, J. A., Napolitano, F. (2010), "Nonstationary modeling of a long record of rainfall and temperature over Rome", Advances in Water Resources doi: 10.1016/j.advwatres.2010.03.013 | 2,304 | 8,431 | {"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": 18, "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} | 3.15625 | 3 | CC-MAIN-2014-35 | latest | en | 0.892366 |
https://www.objectivebooks.com/2020/10/power-screws-multiple-choice-questions.html | 1,670,048,460,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00400.warc.gz | 973,202,558 | 39,964 | Power Screws Multiple Choice Questions - Set 02 - ObjectiveBooks
# Practice Test: Question Set - 02
1. The maximum efficiency of square threaded power depends upon
(B) Friction angle
(C) Pitch of screw
(D) Nominal diameter of screw
2. For over hauling screw
(A) φ > α
(B) α > φ
(C) φ = α
(D) None of the above
3. Which of the following screw thread is stronger than other threads?
4. Multiple threads are used for
(A) High efficiency
(C) Low efficiency for self-locking
5. Which of the following screw thread is adaptable to split type nut?
6. The efficiency of square threaded power screw is given by,
(A) η = tan α/tan (α + φ)
(B) η = tan α/tan (α - φ)
(C) η = tan (α + φ)/tan α
(D) η = tan (α - φ)/tan α
7. For self locking screw
(A) φ > α
(B) α > φ
(C) μ < tan α
(D) μ = cosec α
8. Which of the following screw thread is used for transmitting power in either direction?
(D) Both (A) and (B)
9. A cup is provided in screw jack
(A) To reduce the friction
(B) To prevent rotation of load | 352 | 1,028 | {"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 | 0.721554 |
https://www.acmicpc.net/problem/4279 | 1,527,067,185,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865468.19/warc/CC-MAIN-20180523082914-20180523102914-00607.warc.gz | 683,156,323 | 15,820 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 0 0 0 0.000%
## 문제
At the Olympic Games, countries are ranked according to the number of medals their athletes won. However, there is more than one possible way of ranking countries by their medals. In Europe, for example, countries are first ranked by the number of gold medals their athletes won. Ties are broken by looking at silver medals, and then at bronze medals. In Canada, however, because Canadian athletes do not win very many gold medals, countries are ranked by the overall number of medals won, giving the same weight to gold, silver, and bronze medals.
In general, a ranking scheme can be thought of as a vector of positive weights. This vector is multiplied with the vector of medals won by each country, and the scalar product of the two vectors defines the score of the respective country, which is then used to produce the ranking. In this general scheme, the European ranking technique corresponds to the weight vector (1020, 1010, 1), whereas the Canadian method corresponds to the vector (1, 1, 1).
In this problem, you will only need to consider weight vectors of the form (1/nj, 1/nk, 1/nl), where n is the total number of medals won by all athletes in the Olympic Games, and j, k, and l are integers.
Given a list of countries and the number of gold, silver, and bronze medals won by each country, print the line
Canada wins!
if there is a weight vector of the above form such that Canada ranks first according to the ranking scheme defined by that vector. Print the line
Canada cannot win.
if no such vector exists.
## 입력
The input contains multiple test cases. Each test case starts with an integer c, the number of countries to follow. Each of the following c lines contains the name of a country and three integers g, s, and b – the number of gold, silver, and bronze medals won by the country. The last test case has c = 0 and must not be processed. It is guaranteed that each test case contains at most 20 different countries and that the total number of medals smaller than 100. Country names do not contain whitespace characters.
## 예제 입력 1
2
USA 1 2 3
2
USA 2 2 2
Canada wins! | 513 | 2,166 | {"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} | 2.546875 | 3 | CC-MAIN-2018-22 | longest | en | 0.952657 |
https://theunfamousseries.com/mechanics-thermodynamics-of-propulsion-solution-manual.html | 1,621,100,938,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990551.51/warc/CC-MAIN-20210515161657-20210515191657-00239.warc.gz | 568,468,717 | 14,902 | 4173 dl's @ 3501 KB/s
2498 dl's @ 1590 KB/s
1855 dl's @ 1827 KB/s
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ae0e5a940b28fc936c0ccaa28b6750523349.pdf | 1,928 | 7,625 | {"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-2021-21 | latest | en | 0.820292 |
https://dsp.stackexchange.com/questions/71631/difference-between-sampling-rate-and-baud-rate-and-their-relation-to-channel-cap | 1,652,715,451,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510138.6/warc/CC-MAIN-20220516140911-20220516170911-00515.warc.gz | 279,188,238 | 68,564 | # Difference between sampling rate and baud rate and their relation to channel capacity
The sampling rate says that one Hz can carry at least 2 samples, and the bit rate is obtained by multiplying the sampling rate by the number of bits per sample. Thus, many samples and many bits can be contained in one Hz. Shannon, on the other hand, says that the maximum bit rate that can be transmitted over a channel is twice the bandwidth. I am confused. Could you please help me?
• "The sampling rate says that one Hz can carry at least 2 samples": no, it doesn't say that – this might be a misunderstanding! Where do you take that from? Nov 25, 2020 at 9:41
• yes, but "1 Hz carries at least 2 samples" is not a statement that makes a lot of sense to me – are you inferring some discrete information content to a continuous signal bandwidth? Nov 25, 2020 at 10:00
• Let's focus on the other claim, though: "Shannon says the maximum bit rate […] is twice the bandwidth": no, he doesn't: Shannon capacity very clearly allows for arbitrary much information per bandwidth given sufficiently high SNR Nov 25, 2020 at 12:06
• @Noha: please cite your sources so we can look at it together. It appears that you are misquoting or misunderstanding what you have read or "heard". Nov 25, 2020 at 12:25
• @Noha Lathi is correct. Do you see the difference between what Lathi says and what you asked?
– MBaz
Nov 25, 2020 at 16:50
First let's estabish the context.
Let's say you want to transmit a sequence of numbers $$\lbrace a_k \rbrace_{k=1}^{N} = a_1, a_2, \ldots$$ Using PAM, this sequence is transmitted with the signal $$s(t) = \sum_{k=1}^N a_k p(t-kT_p),$$ where the pulses $$p(t-kT_p)$$ form an orthonormal set, that is, $$\int_{-\infty}^\infty p(t) p(t) \text{d}t = 1$$ (which implies that the pulse $$p(t)$$ has unit energy) and $$\int_{-\infty}^\infty p(t-jT_p) p(t-kT_p) \text{d}t = 0$$ whenever $$k \neq j$$, which allows the receiver to recover the transmitted sequence using a correlator or a matched filter.
Second, let's see what Nyquist and Shannon say.
First of all, the pulses in $$s(t)$$ are transmitted at a rate $$R_p = 1/T_p$$. This is called the "pulse rate" or the "symbol rate". You want $$T_p$$ to be short, to transmit information faster. However, this increases the bandwidth $$B$$ of $$s(t)$$. The relationship is $$B = \alpha \frac{R_p}{2} = \alpha \frac{1}{2T_p}.$$
The value of $$\alpha$$ is the "spectral efficiency" of the pulse $$p(t)$$. The minimum possible value is $$\alpha=1$$, when $$p(t)$$ is a sinc pulse. In this case, $$B=R_p/2$$ or $$R_p=2B$$. This is what it means to say that "$$2B$$ pieces of information can be transmitted per Hz of bandwidth available."
If this was all there was to it, then an infinite amount of information could be transmitted in any amount of time. All one has to do is let the $$a_k$$ be arbitrary real numbers, with infinite precision. Even if one is constrained to choose from a constellation, such as $$M$$-QAM, one could choose $$M$$ to be as large as needed.
In practice, however, noise, distortion, and other imperfections prevent transmission of $$a_k$$ numbers with infinite precision. The simplest model of a communications system assumes that the received signal $$r(t)$$ is corrupted with white Gaussian noise with power spectral density $$N_0/2$$: $$r(t) = s(t) + n(t).$$
Transmission under the effects of noise restrict the choice of symbols $$a_k$$: they must belong to a finite set whose elements cannot be easily mistaken for one another. For example, if we constrain the symbols to take values either $$-0.5$$ or $$0.5$$, and the noise magnitude is very unlikely to be larger than $$0.5$$, then transmission can be done very reliably; however, we pay a price in two different ways:
• Power: since the symbol amplitudes cannot be made too small, a minimum amount of power is required;
• Rate: since we're limited in the choice of symbol amplitudes, we can't transmit as fast as we wish.
In terms of a $$M$$-QAM constellation, the consequence is that we can't increase $$M$$ arbitrarily; when $$M$$ is made too large, the system cannot transmit information reliably any more.
This is where Shannon comes in. He was interested in the relationship between the entropy of the information source that produces the transmitted sequences $$\lbrace a_k \rbrace$$, the bandwidth, and the signal to noise ratio in the received signal. He definitely did not say that the maximum bit rate is twice the bandwidth; rather, the maximum bit rate is given by the bandwidth, the SNR, and some specific properties of the sequence $$\lbrace a_k \rbrace$$. The study of these properties is known as "coding theory".
Finally: is it a coincidence that a signal with bandwidth $$B$$ must be sampled at rate $$2B$$, and that a signal with bandwidth $$B$$ can transmit symbols at rate $$2B$$? It is indeed not a coincidence. You want to transmit symbols at rate $$R_p$$; in other words, you want a signal that has specific amplitudes at times $$kT_p$$:
Think of these amplitudes as signal samples:
The blue signal is the transmitted signal $$s(t)$$, and since its samples have rate $$R_p$$, then it has bandwidth $$B=R_p/2$$.
• Right on both counts. Just note that, in order to approach Shannon's capacity, coding must be used.
– MBaz
Nov 26, 2020 at 0:05
• @Noha Please see my edits. In brief: ** You are correct that if there is no noise, then the bit rate is infinite, and there is no limit on the number of bits per symbol. ** The relationship between sampling theory and pulse rate is explained at the end of my edited answer.
– MBaz
Nov 27, 2020 at 17:28
• I think you are being confused by the vague sentence "pieces of information". What is transmitted at rate $2B$ are the symbols $\lbrace a_k \rbrace$. Assume the channel is noiseless and you want to transmit one hundred bits, say 110110... You can write this down as the fractional number 0.110110..., which you then convert to decimal. Let's say this number is $a_0$. Then, transmit the pulse $a_0 p(t)$, which requires bandwidth $B$. The receiver proceeds backwards, from $a_0$ to the information sequence 110110.... (continued)
– MBaz
Nov 30, 2020 at 1:37
• This process works for any number of bits; in the noiseless channel, you can transmit an arbitrary amount of information with a single pulse. When there is noise, it becomes very hard to distinguish between different values of $a_k$, and this limits the amount of information that each pulse can carry. The maximum amount of information given $B$ and the SNR is given by Shannon. However, the number of symbols per second is always $2B$.
– MBaz
Nov 30, 2020 at 1:40
• You are correct about the sampling and capacity theorems, and that they are unrelated. In that paragraph, Lathi is trying to show (in a non-rigorous way) that the baud rate is actually bounded by $2B$, and it does that by using an argument based on the sampling theorem, relating the required $2B$ samples per second and the $2B$ symbol amplitudes. This is probably not the best approach, and Nyquist himself didn't use this argument as far as I know. See a derivation of the $2B$ bound that does not use the sampling theorem here.
– MBaz
Nov 30, 2020 at 23:33 | 1,862 | 7,213 | {"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": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-21 | longest | en | 0.930819 |
http://georgepruitt.com/day-week-analysis-using-method-easylanguage/ | 1,643,046,752,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00145.warc.gz | 22,995,498 | 21,220 | # Day Of Week Analysis using a Method In EasyLanguage
One metric that seems to be missing from TradeStation is a Day Of Week analysis. It would be nice, but I don’t know how helpful, to know the \$P/L breakdown on a weekday basis; does your system make all of its money on Mondays and Fridays? I created the code that will print out to the print log using an EasyLanguage method. A method is a subroutine that can be included in the main code (strategy, indicator, etc.,.) The global parameters to the main program can be seen inside the method. You can localize variable scope to the method by declaring the variable within the method’s body. A method is a great way to modularize your programming, but it is not the best way to reuse software; the method is accessible only to the main program. This EasyLanguage also utilizes an array and shows a neat piece of code to access the array elements and align them with the day of the week. The dayOfWeek() function returns [1..5] depending on what day of the week the trading day falls on. Monday = 1 and Friday = 5. The array has five elements and each element accumulates the \$P/L for each of the five days based on when the trade was initiated.
``````vars: mp(0);
array: weekArray[5](0);
method void dayOfWeekAnalysis() {method definition}
begin
If mp = 1 and mp[1] = -1 then tradeProfit = (entryPrice(1) - entryPrice(0))*bigPointValue;
If mp = -1 and mp[1] = 1 then tradeProfit = (entryPrice(0) - entryPrice(1))*bigPointValue;
end;
Buy next bar at highest(high,9)[1] stop;
Sellshort next bar at lowest(low,9)[1] stop;
mp = marketPosition;
if mp <> mp[1] then dayOfWeekAnalysis();
If lastBarOnChart then
Begin
print("Monday ",weekArray[1]);
print("Tuesday ",weekArray[2]);
print("Wednesday ",weekArray[3]);
print("Thursday ",weekArray[4]);
print("Friday ",weekArray[5]);
end;``````
Code using METHOD and ARRAY manipulation
Here is an example of the print out created by running this simple EasyLanguage strategy. Maybe don’t trade on Monday? Or is that curve fitting. This is an interesting tool and might carry more weight if applied to day trade algorithm. Maybe!
• Monday -8612.50
• Tuesday 6350.00
• Wednesday 2612.50
• Thursday -937.50
• Friday -1987.50 | 587 | 2,235 | {"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-2022-05 | latest | en | 0.833593 |
http://en.wikipedia.org/wiki/Dyadic_monoid | 1,429,452,793,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246639121.73/warc/CC-MAIN-20150417045719-00179-ip-10-235-10-82.ec2.internal.warc.gz | 91,880,726 | 24,642 | # Modular group
For a group whose lattice of subgroups is modular, see Iwasawa group.
In mathematics, the modular group is the projective special linear group PSL(2,Z) of 2 x 2 matrices with integer coefficients and unit determinant. The matrices A and -A are identified. The modular group acts on the upper-half of the complex plane by fractional linear transformations, and the name "modular group" comes from the relation to moduli spaces and not from modular arithmetic.
## Definition
The modular group Γ is the group of linear fractional transformations of the upper half of the complex plane which have the form
$z\mapsto\frac{az+b}{cz+d}$
where a, b, c, and d are integers, and adbc = 1. The group operation is function composition.
This group of transformations is isomorphic to the projective special linear group PSL(2, Z), which is the quotient of the 2-dimensional special linear group SL(2, Z) over the integers by its center {I, −I}. In other words, PSL(2, Z) consists of all matrices
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$
where a, b, c, and d are integers, adbc = 1, and pairs of matrices A and −A are considered to be identical. The group operation is the usual multiplication of matrices.
Some authors define the modular group to be PSL(2, Z), and still others define the modular group to be the larger group SL(2, Z).
Some mathematical relations require the consideration of the group GL(2, Z) of matrices with determinant plus or minus one. (SL(2, Z) is a subgroup of this group.) Similarly, PGL(2, Z) is the quotient group GL(2,Z)/{I, −I}. A 2 × 2 matrix with unit determinant is a symplectic matrix, and thus SL(2, Z) = Sp(2, Z), the symplectic group of 2x2 matrices.
## Number-theoretic properties
The unit determinant of
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$
implies that the fractions a/b, a/c, c/d and b/d are all irreducible, that is have no common factors (provided the denominators are non-zero, of course). More generally, if p/q is an irreducible fraction, then
$\frac{ap+bq}{cp+dq}$
is also irreducible (again, provided the denominator be non-zero). Any pair of irreducible fractions can be connected in this way, i.e.: for any pair p/q and r/s of irreducible fractions, there exist elements
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in\operatorname{SL}(2,\mathbf{Z})$
such that
$r = ap+bq \quad \mbox{ and } \quad s=cp+dq.$
Elements of the modular group provide a symmetry on the two-dimensional lattice. Let $\omega_1$ and $\omega_2$ be two complex numbers whose ratio is not real. Then the set of points
$\Lambda (\omega_1, \omega_2)=\{ m\omega_1 +n\omega_2 : m,n\in \mathbf{Z} \}$
is a lattice of parallelograms on the plane. A different pair of vectors $\alpha_1$ and $\alpha_2$ will generate exactly the same lattice if and only if
$\begin{pmatrix}\alpha_1 \\ \alpha_2 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \omega_1 \\ \omega_2 \end{pmatrix}$
for some matrix in S*L(2, Z). It is for this reason that doubly periodic functions, such as elliptic functions, possess a modular group symmetry.
The action of the modular group on the rational numbers can most easily be understood by envisioning a square grid, with grid point (p, q) corresponding to the fraction p/q (see Euclid's orchard). An irreducible fraction is one that is visible from the origin; the action of the modular group on a fraction never takes a visible (irreducible) to a hidden (reducible) one, and vice versa.
If $p_{n-1}/q_{n-1}$ and $p_{n}/q_{n}$ are two successive convergents of a continued fraction, then the matrix
$\begin{pmatrix} p_{n-1} & p_{n} \\ q_{n-1} & q_{n} \end{pmatrix}$
belongs to S*L(2, Z). In particular, if bc − ad = 1 for positive integers a,b,c and d with a < b and c < d then a/b and c/d will be neighbours in the Farey sequence of order max(b, d). Important special cases of continued fraction convergents include the Fibonacci numbers and solutions to Pell's equation. In both cases, the numbers can be arranged to form a semigroup subset of the modular group.
## Group-theoretic properties
### Presentation
The modular group can be shown to be generated by the two transformations
$S: z\mapsto -1/z$
$T: z\mapsto z+1$
so that every element in the modular group can be represented (in a non-unique way) by the composition of powers of S and T. Geometrically, S represents inversion in the unit circle followed by reflection with respect to the origin, while T represents a unit translation to the right.
The generators S and T obey the relations S2 = 1 and (ST)3 = 1. It can be shown [1] that these are a complete set of relations, so the modular group has the presentation:
$\Gamma \cong \langle S, T \mid S^2=I, (ST)^3=I \rangle$
This presentation describes the modular group as the rotational triangle group (2,3,∞) (∞ as there is no relation on T), and it thus maps onto all triangle groups (2,3,n) by adding the relation Tn = 1, which occurs for instance in the congruence subgroup Γ(n).
Using the generators S and ST instead of S and T, this shows that the modular group is isomorphic to the free product of the cyclic groups C2 and C3:
$\Gamma \cong C_2 * C_3$
### Braid group
The braid group B3 is the universal central extension of the modular group.
The braid group B3 is the universal central extension of the modular group, with these sitting as lattices inside the (topological) universal covering group $\overline{\mathrm{SL}_2(\mathbf{R})} \to \mathrm{PSL}_2(\mathbf{R})$. Further, the modular group has a trivial center, and thus the modular group is isomorphic to the quotient group of B3 modulo its center; equivalently, to the group of inner automorphisms of B3.
The braid group B3 in turn is isomorphic to the knot group of the trefoil knot.
### Quotients
The quotients by congruence subgroups are of significant interest.
Other important quotients are the (2,3,n) triangle groups, which correspond geometrically to descending to a cylinder, quotienting the x coordinate mod n, as Tn = (zz+n). (2,3,5) is the group of icosahedral symmetry, and the (2,3,7) triangle group (and associated tiling) is the cover for all Hurwitz surfaces.
## Relationship to hyperbolic geometry
The modular group is important because it forms a subgroup of the group of isometries of the hyperbolic plane. If we consider the upper half-plane model H of hyperbolic plane geometry, then the group of all orientation-preserving isometries of H consists of all Möbius transformations of the form
$z\mapsto \frac{az + b}{cz + d}$
where a, b, c, and d are integers, instead of the usual real numbers, and adbc = 1. Put differently, the group PSL(2, R) acts on the upper half-plane H according to the following formula:
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\cdot z \,= \,\frac{az + b}{cz + d}$
This (left-)action is faithful. Since PSL(2, Z) is a subgroup of PSL(2, R), the modular group is a subgroup of the group of orientation-preserving isometries of H. [2]
### Tessellation of the hyperbolic plane
A typical fundamental domain for the action of Γ on the upper half-plane.
The modular group Γ acts on H as a discrete subgroup of PSL(2, R), i.e. for each z in H we can find a neighbourhood of z which does not contain any other element of the orbit of z. This also means that we can construct fundamental domains, which (roughly) contain exactly one representative from the orbit of every z in H. (Care is needed on the boundary of the domain.)
There are many ways of constructing a fundamental domain, but a common choice is the region
$R = \left\{ z \in \mathbf{H}: \left| z \right| > 1,\, \left| \,\mbox{Re}(z) \,\right| < \frac{1}{2} \right\}$
bounded by the vertical lines Re(z) = 1/2 and Re(z) = −1/2, and the circle |z| = 1. This region is a hyperbolic triangle. It has vertices at 1/2 + i√3/2 and −1/2 + i√3/2, where the angle between its sides is π/3, and a third vertex at infinity, where the angle between its sides is 0.
By transforming this region in turn by each of the elements of the modular group, a regular tessellation of the hyperbolic plane by congruent hyperbolic triangles is created. Note that each such triangle has one vertex either at infinity or on the real axis Im(z)=0. This tiling can be extended to the Poincaré disk, where every hyperbolic triangle has one vertex on the boundary of the disk. The tiling of the Poincaré disk is given in a natural way by the J-invariant, which is invariant under the modular group, and attains every complex number once in each triangle of these regions.
This tessellation can be refined slightly, dividing each region into two halves (conventionally colored black and white), by adding an orientation-reversing map; the colors then correspond to orientation of the domain. Adding in (x, y) ↦ (−x, y) and taking the right half of the region R (Re(z) ≥ 0) yields the usual tessellation. This tessellation first appears in print in (Klein 1878/79a),[3] where it is credited to Richard Dedekind, in reference to (Dedekind 1877).[3][4]
Visualization of the map (2,3,∞) → (2,3,7) by morphing the associated tilings.[5]
The map of groups (2,3,∞) → (2,3,n) (from modular group to triangle group) can be visualized in terms of this tiling (yielding a tiling on the modular curve), as depicted in the video at right.
Paracompact hyperbolic uniform tilings in [∞,3] family
Symmetry: [∞,3], (*∞32) [∞,3]+
(∞32)
[1+,∞,3]
(*∞33)
[∞,3+]
(3*∞)
=
=
=
=
or
=
or
=
{∞,3} t{∞,3} r{∞,3} t{3,∞} {3,∞} rr{∞,3} tr{∞,3} sr{∞,3} h{∞,3} h2{∞,3} s{3,∞}
Uniform duals
V∞3 V3.∞.∞ V(3.∞)2 V6.6.∞ V3 V4.3.4.∞ V4.6.∞ V3.3.3.3.∞ V(3.∞)3 V3.3.3.3.3.∞
## Congruence subgroups
Main article: Congruence subgroup
Important subgroups of the modular group Γ, called congruence subgroups, are given by imposing congruence relations on the associated matrices.
There is a natural homomorphism SL(2, Z) → SL(2, Z/NZ) given by reducing the entries modulo N. This induces a homomorphism on the modular group PSL(2, Z) → PSL(2, Z/NZ). The kernel of this homomorphism is called the principal congruence subgroup of level N, denoted Γ(N). We have the following short exact sequence:
$1\to\Gamma(N)\to\Gamma\to\mbox{PSL}(2,\mathbf{Z}/N\mathbf{Z})\to 1$.
Being the kernel of a homomorphism Γ(N) is a normal subgroup of the modular group Γ. The group Γ(N) is given as the set of all modular transformations
$z\mapsto\frac{az+b}{cz+d}$
for which ad ≡ ±1 (mod N) and bc ≡ 0 (mod N).
The principal congruence subgroup of level 2, Γ(2), is also called the modular group Λ. Since PSL(2, Z/2Z) is isomorphic to S3, Λ is a subgroup of index 6. The group Λ consists of all modular transformations for which a and d are odd and b and c are even.
Another important family of congruence subgroups are the modular group Γ0(N) defined as the set of all modular transformations for which c ≡ 0 (mod N), or equivalently, as the subgroup whose matrices become upper triangular upon reduction modulo N. Note that Γ(N) is a subgroup of Γ0(N). The modular curves associated with these groups are an aspect of monstrous moonshine – for a prime number p, the modular curve of the normalizer is genus zero if and only if p divides the order of the monster group, or equivalently, if p is a supersingular prime.
One important subset of the modular group is the dyadic monoid, which is the monoid of all strings of the form STkSTmSTn ... for positive integers k, m, n, ... . This monoid occurs naturally in the study of fractal curves, and describes the self-similarity symmetries of the Cantor function, Minkowski's question mark function, and the Koch curve, each being a special case of the general de Rham curve. The monoid also has higher-dimensional linear representations; for example, the N = 3 representation can be understood to describe the self-symmetry of the blancmange curve.
## Maps of the torus
The group GL(2, Z) is the linear maps preserving the standard lattice Z2, and SL(2, Z) is the orientation-preserving maps preserving this lattice; they thus descend to self-homeomorphisms of the torus (SL mapping to orientation-preserving maps), and in fact map isomorphically to the (extended) mapping class group of the torus, meaning that every self-homeomorphism of the torus is isotopic to a map of this form. The algebraic properties of a matrix as an element of GL(2, Z) correspond to the dynamics of the induced map of the torus.
## Hecke groups
The modular group can be generalized to the Hecke groups, named for Erich Hecke, and defined as follows.[6]
The Hecke group Hq is the discrete group generated by
$z \mapsto -1/z$
$z \mapsto z + \lambda_q,$
where $\lambda_q=2\cos(\pi/q). \,$
The modular group Γ is isomorphic to H3 and they share properties and applications – for example, just as one has the free product of cyclic groups
$\Gamma \cong C_2 * C_3,$
more generally one has
$H_q \cong C_2 * C_q,$
which corresponds to the triangle group (2,q,∞). There is similarly a notion of principal congruence subgroups associated to principal ideals in Z[λ]. For small values of q, one has:
$\lambda_3 = 1,$
$\lambda_4 = \sqrt{2},$
$\lambda_5 = \tfrac{1}{2}(1+\sqrt{5}),$
$\lambda_6 = \sqrt{3}.$
## History
The modular group and its subgroups were first studied in detail by Richard Dedekind and by Felix Klein as part of his Erlangen programme in the 1870s. However, the closely related elliptic functions were studied by Joseph Louis Lagrange in 1785, and further results on elliptic functions were published by Carl Gustav Jakob Jacobi and Niels Henrik Abel in 1827. | 3,790 | 13,565 | {"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": 34, "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} | 3.84375 | 4 | CC-MAIN-2015-18 | latest | en | 0.834424 |
https://www.coursehero.com/file/183954/Lab3-AC-Circuits-II-RL/ | 1,524,596,437,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947033.92/warc/CC-MAIN-20180424174351-20180424194351-00183.warc.gz | 752,511,467 | 69,341 | {[ promptMessage ]}
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Lab3 AC Circuits II (RL)
# Lab3 AC Circuits II (RL) - Title AC Circuits II(RL Richard...
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Title: AC Circuits II (RL) Richard Madison Haynie Partner: Joseph Gilgen (All graphs are attached to back) Objective: In this lab the goal is to observe low-pass and high pass-filters, how they function and why they function as they do. Once this is done learn how these filters can be used as either a garbage detector and filtering out noise when added to an input signal. Low-pass Filter: Here the circuit shown in figure-1 was constructed. From theory the reactance of a capacitor is given by 1 2 C X fC π = which means when f is small X C is large and when f is large X C is small. For this circuit this means when f is small enough in out V V X and when f is infinitely large 0 out V X . Since in out V V X for small frequencies the circuit earns its name, low-pass filter. A low pass filter is a voltage divider and the equation is ( 29 2 2 2 2 1 1 1 2 1 out in V C V fRC R C ϖ π ϖ = = + + . The phase angle can be found using 1 1 tan 90 2 fRC φ π - = - o . To find the values of f that are small or large enough for our expected results we compute a f 3dB . From theory 3 1 2 dB f RC π = and with this circuit f 3dB =1061Hz. The circuit was then driven with a frequency of 1065Hz and is shown in graph-1 with .676 out in V V = from theory .706 out in V V = differing by 4.2% from the experimental value. Next f 3dB is calculated experimentally by finding the frequency for which V out is 70.7% of V in . The value found experimentally was f 3dB =969Hz with .716 out in V V = (graph-2), so f 3dB from theory and experimentally are approximately equal differing by 8.7%. The circuit was then driven with 96.98Hz which is much smaller then f 3dB (our small frequency) shown in graph-3. Indeed 0 out V V X which expected, from theory .021 out in V V = . The circuit was then driven with 49.36kHz which is much greater then f 3dB (our large frequency) shown in graph-4. For this frequency V out is greatly attenuated because V out approaches zero as f approaches infinity which was expected. The
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{[ snackBarMessage ]} | 645 | 2,412 | {"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} | 3.71875 | 4 | CC-MAIN-2018-17 | latest | en | 0.951411 |
http://www.physicsforums.com/showthread.php?t=421183 | 1,369,343,504,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703788336/warc/CC-MAIN-20130516112948-00021-ip-10-60-113-184.ec2.internal.warc.gz | 661,087,393 | 7,877 | ## Finding the Area or the Circumference of a gutter.
Hey Guys,
For a maths assignment, we were given this question to complete:
http://nrich.maths.org/5673
I'm having an discussion with a friend of how best to approach it. Should I work out and differentiate the area or perimeter (for the circle)? If we're finding the cost of materials, then circumference would be more appropriate than the area. Or should we look at both?
Sigh, isn't it always the most basic part of a problem that stumps us?
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
Recognitions: Gold Member Science Advisor Staff Emeritus You want to minimize the perimeter (or circumference) for a fixed area.
The same perimeter can have a varying area, correct? Or would it be the same area every time? For example, if I curve a piece of paper, I can make the shape wider or smaller, but the perimeter stays the same. Is the area changing even though perimeter stays constant? Thanks.
Recognitions:
Homework Help
## Finding the Area or the Circumference of a gutter.
Quote by Liparulo The same perimeter can have a varying area, correct?
Yes. For any given perimeter p, you can range from the minimum area of 0 - by having a rectangle with length p/2 etc. - to the maximum area of...
So, (sorry to continue asking questions, but this is helping), would I find the perimeter of the circumference minus the perimeter of the arc? What role does area play in determining an equation? | 358 | 1,630 | {"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} | 3.1875 | 3 | CC-MAIN-2013-20 | longest | en | 0.910805 |
https://www.shrm.org/hr-today/news/hr-news/Pages/calendarpayrollchallenge.aspx | 1,527,067,158,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865468.19/warc/CC-MAIN-20180523082914-20180523102914-00571.warc.gz | 847,176,357 | 64,257 | # Calendar's Quirk Can Pose a Payroll Challenge
By Bill Leonard Jan 11, 2012
Just by adding an extra day in February, leap years can create some interesting challenges to employers. The extra day typically adds another workday onto the schedule and increases the odds that a business could have an additional payday for the year.
The reason for the potential extra payday is simple mathematics. You cannot divide 366 days evenly by seven no matter what kind of math or calculator you might use. So, leap years have 52 full weeks (or 364 days) with two more days for a total of 366 days.
This means that two days of the week will occur 53 times during a leap year. For typical years of 365 days only one day of the week repeats 53 times. It might not seem all that complicated, unless that 53rd day of the week falls on a business’s scheduled payday. If this happens, then employers that are used to having 52 paydays per year or 26 if they pay biweekly could face a payroll dilemma—on how to account for this extra payday.
The 53rd day of the week in 2011 didn’t cause payroll dilemmas for employers because it fell on a Saturday. The two extra days for 2012 will fall on Sunday and Monday. Because few employers issue paychecks on a Sunday or a Monday, this leap year should have a negligible effect on most employers.
“There is a small group of employers that have a Monday pay date,” said Pedro Cutino, director of product management for Automatic Data Processing Inc. (ADP), one of the largest payroll services companies in the United States. According to ADP, less than 1 percent of the company’s client base of nearly 570,000 businesses use Monday as a payday.
“These employers may need to consider the implications for the extra payday,” he said.
Software Can Help
The issue of the 53rd payday is, however, known well by most payroll managers, and most payroll planning software can help employers plan for the extra payday. While the effect of 2012 appears minimal, the story was much different eight years earlier. Many payroll managers called 2004 a “perfect storm” for leap years because the 53rd days of the week fell on Thursday and Friday—which are by far the most popular paydays.
“We do get a lot of calls when the additional weekdays fall on a Thursday or Friday, but the extra paydays and leap years just don’t pose much of a challenge to employers that have their accounting and payroll systems in order,” said Michael O’Toole, director of publications and government relations for the American Payroll Association (APA). “Since the extra weekdays in 2012 fall on Sunday and Monday, then it really shouldn’t affect employers too much, if at all. Still some employers do choose to adjust their employees’ pay to account for the extra pay period or extra day of work.”
Depending on when the extra days of the week in leap years or even normal years fall, calendars can have from 260 to 262 typical work weekdays (Monday to Friday). The mathematics works this way: 5 days x 52 weeks = 260 days, plus the one or two extra days depending on the year. Because employers typically pay for holidays, those days are included in the calculations.
If we use 2011 as an example, the 53rd day of the week fell on a Saturday, which had a minimal impact on work calendars. And even though 2012 is a leap year, there will be 261 workdays because the 53rd days of the week fall on Sunday and Monday.
According to O’Toole, the vast majority of employers use accrual accounting systems, so the extra workday typically isn’t an issue for them.
“Employees are usually paid for the workdays that they accrue during any given pay period, so the extra day in a leap year just isn’t that big of a deal,” O’Toole said. “However, it can affect employers that operate with strict fiscal-year budgets, and this tends to be public-sector employers.”
Most local and state governments are required by law to operate with strict annual budgeting and therefore must adjust their payrolls to account for the extra day in a leap year. It’s fairly straightforward for public sector employers to pay their hourly employees by budgeting and paying for an extra day worked. Salaried employees, however, present a different challenge. The extra day of the year during leap years can affect the take-home pay of salaried employees if their employers choose to account for the extra day. In addition, most government budgets must account for the cost to provide health insurance benefits for an extra day during a leap year. These costs for the extra day are extrapolated over the entire year and can therefore increase employee co-pays on health insurance premiums slightly for each pay period.
Some public-sector employers like the state governments of New York and Maryland have created websites that calculate the adjusted take home pay for salaried employees and the extra costs for health insurance premiums. The extra day of the year is factored into the normal annual salary, so salaried employees still will receive the same annual wage but their take home pay will vary slightly during the leap year.
“Most employees who work for public-sector employers understand that leap years will change their take home pay slightly, but when it’s factored for the entire year then the change is fairly insignificant and usually doesn’t create major hardships,” said O’Toole.
According to ADP’s Cutino, most employers outside the public sector choose not to recalculate the pay of salaried employees because of possible employee relations problems.
“It could potentially create employer-employee relations issues since salaried employees would see less in their pay each period,” Cutino said. “Often employers base employee salaries on a set weekly or biweekly amount rather than a yearly amount. By doing that, they can mitigate most of the problems raised by the quirks of the calendar.”
Payroll Tax Cut Could Pose Major Challenge
According to O’Toole, 2012 is the first leap year that his organization hasn’t received a flood of calls about how to plan payrolls for extra pay periods or workdays. He attributes the lack of calls to two things: first, the extra days of the week of 2012 falling on Sunday and Monday, and second, the continuing debate in Congress about extending a 2 percent payroll tax cut.
Congress passed a two-month extension of the tax cut in December 2011. It lowers the federal Social Security payroll tax on individuals from 6.2 percent to 4.2 percent.
At the end of 2011, Congress pushed the debate on extending the tax cut right up to the Christmas holiday, which forced employers to be prepared to continue with the tax cut or begin withholding an extra 2 percent from employees’ paychecks beginning Jan. 1, 2012.
By approving just a two-month extension of the tax cut, Congress created another huge headache for employers, O’Toole said.
“Employers can easily prepare for a tax withholding change at the end of the year, so the debate in December didn’t cause that many problems,” he said. “But extending the tax cut for two months into 2012 creates a very tricky situation.”
While leaders on Capitol Hill claim that the extension provides the opportunity to reach an agreement on an extension through 2012 and possibly beyond, Congress does have a history of pushing these agreements right up to the deadline.
“And if they don’t pass anything until the 11th hour on the final day, that means employers will have to be prepared for two scenarios of continuing the tax cut or readjusting the tax withholdings of their employees,” said O’Toole. “And that’s a huge problem because employers’ payroll systems just aren’t set up to change tax withholding rates in the same year, let alone the same quarter.”
The administrative costs and the costs to change computer systems to accept the different withholding rates could far outweigh the economic benefits of the 2 percent tax cut, according to O’Toole and others interviewed for the article.
“While it is true individuals do benefit from the payroll tax cut, employers could be facing a sizeable hit from the costs it will take to be ready for either eventuality—that a tax-cut extension is passed or not passed,” said O’Toole.
He said that the staff at the APA was working with congressional staffs to explain the problem and help ensure a speedy resolution.
“But you just never know what will happen with the politics on Capitol Hill, and we will just have to wait and see how quickly they move to resolve the issue,” O’Toole said.
Bill Leonard is a senior writer for SHRM.
Post a Job | 1,799 | 8,571 | {"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} | 2.9375 | 3 | CC-MAIN-2018-22 | latest | en | 0.95764 |
https://roadsindb.com/2016/01/25/solutions-for-taxi-services/ | 1,685,350,638,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00589.warc.gz | 537,209,324 | 19,012 | # Solutions for Taxi Services
### Fare Estimator:
Many taxi companies like Uber offer fare finders where the user enters starting and ending address and the service returns an approximate fare range. Now one can do that using a single line of SQL.
```-- c1 and c2 are distance, time multipliers to compute fare
SELECT dist(lat1, lon1, lat2, lon2) * c1 +
triptime(lat1, lon1, lat2, lon2) * c2;
-- This produces fare in dollars. Estimate could be +/- 20%
```
### Where are the nearest taxis?
A common feature in many taxi app is to provide a map of the nearest taxi. This can be done using just a few lines of SQL.
```-- lat, lon is where a passenger is present
-- taxi(id, lat, lon, available) current taxi information
SELECT taxi.id, dist(lat, lon, taxi.lat, taxi.lon) as distance
FROM taxi
WHERE taxi.available = true AND
distance &amp;amp;amp;lt;= 10 * 1000 --- 10 kms
ORDER BY distance;
```
### Where is my assigned taxi?
Once passenger is assigned a taxi, one wants to provide an accurate time of arrival in seconds for the customer. This also can be done using a few lines of SQL.
```-- lat, lon is where a passenger is present
-- taxi(id, lat, lon, available) current taxi information
-- taxi_id is the assigned taxi
SELECT taxi.id, triptime(lat, lon, taxi.lat, taxi.lon) as seconds_to_arrival
FROM taxi
WHERE taxi.id = tax_id;
```
### Car-pooling Service:
Suppose a taxi company wants to offer a car-pooling service where the taxi picks as many passengers as possible with minimal detours. This means that frequently one needs to compute available passengers and the order in which to pick up them.
The first thing one needs to compute is a the distance matrix between all the waiting passengers. For example, if a driver has to pick up 10 passengers, we need to compute the network distance between every pair of delivery address which is then fed into an optimization algorithm. Now if one super scales the problem to 100 taxis and 1000 passengers, we can handle this query easily using your existing commodity hardware.
We can compute 1000 x 1000 distance matrix (1 million distance computations) is less than 30 seconds on a commodity server running PostgreSQL. A 100 x 100 distance matrix is just a fraction of a second.
```-- waiting_passengers(id, lat, lon) is a table of passengers
-- yet to be assigned to taxis.
SELECT A.id as id1, B.id as id2, dist(A.lat, A.lon, B.lat, B.lon) as dist
FROM waiting_passengers as A, waiting_passengers as B;
-- Result is a distance matrix
```
### Trajectory Analysis:
Taxi services collect the GPS trajectories of their taxis which becomes a rich source of analysis for optimization efficiency. Look at our detailed post on what can be done with large-scale trajectory archives.
### Fare Scams by Drivers:
If you own a fleet of taxis and want to find if a driver is overcharging your passengers, you can compute the difference between the distance along the trajectories vs. the distance from start to destination. Look at our detailed post on what can be done with large-scale trajectory archives. | 721 | 3,071 | {"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-2023-23 | latest | en | 0.903964 |
https://studysoup.com/tsg/math/343/algebra-1-student-edition-merrill-algebra-1/chapter/15735/6-5 | 1,606,570,140,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195656.78/warc/CC-MAIN-20201128125557-20201128155557-00384.warc.gz | 492,317,194 | 11,033 | ×
×
# Solutions for Chapter 6-5: Solving Open Sentences Involving Absolute Value
## Full solutions for Algebra 1, Student Edition (MERRILL ALGEBRA 1) | 1st Edition
ISBN: 9780078738227
Solutions for Chapter 6-5: Solving Open Sentences Involving Absolute Value
Solutions for Chapter 6-5
4 5 0 397 Reviews
26
4
##### ISBN: 9780078738227
This expansive textbook survival guide covers the following chapters and their solutions. Algebra 1, Student Edition (MERRILL ALGEBRA 1) was written by and is associated to the ISBN: 9780078738227. Chapter 6-5: Solving Open Sentences Involving Absolute Value includes 70 full step-by-step solutions. This textbook survival guide was created for the textbook: Algebra 1, Student Edition (MERRILL ALGEBRA 1) , edition: 1. Since 70 problems in chapter 6-5: Solving Open Sentences Involving Absolute Value have been answered, more than 37155 students have viewed full step-by-step solutions from this chapter.
Key Math Terms and definitions covered in this textbook
A sequence of steps (end of Chapter 9) to solve positive definite Ax = b by minimizing !x T Ax - x Tb over growing Krylov subspaces.
• Elimination.
A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.
• Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.
• Fundamental Theorem.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n - r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.
• Gauss-Jordan method.
Invert A by row operations on [A I] to reach [I A-I].
• Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
• Multiplicities AM and G M.
The algebraic multiplicity A M of A is the number of times A appears as a root of det(A - AI) = O. The geometric multiplicity GM is the number of independent eigenvectors for A (= dimension of the eigenspace).
• Nilpotent matrix N.
Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.
• Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q -1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •
• Partial pivoting.
In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.
• Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.
• Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
• Row space C (AT) = all combinations of rows of A.
Column vectors by convention.
• Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.
• Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.
• Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.
• Spectrum of A = the set of eigenvalues {A I, ... , An}.
Spectral radius = max of IAi I.
• Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.
• Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.
• Vector space V.
Set of vectors such that all combinations cv + d w remain within V. Eight required rules are given in Section 3.1 for scalars c, d and vectors v, w.
× | 1,033 | 3,657 | {"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} | 3.703125 | 4 | CC-MAIN-2020-50 | latest | en | 0.864513 |
https://www.tutorialspoint.com/cplusplus-program-to-find-out-if-there-is-a-pattern-in-a-grid | 1,679,320,367,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943483.86/warc/CC-MAIN-20230320114206-20230320144206-00749.warc.gz | 1,119,819,182 | 10,292 | C++ Program to find out if there is a pattern in a grid
Suppose, we are given a grid of dimensions n * n. We have to detect if there is a cross pattern in the grid, like below −
#...#
.#.#.
..#..
.#.#.
#...#
The grid can only contain '#' and '.'. We have to detect the pattern and find out how many such patterns are in the grid. The grid and the dimension are given to us as input.
Problem Category
Various problems in programming can be solved through different techniques. To solve a problem, we have to devise an algorithm first, and to do that we have to study the particular problem in detail. A recursive approach can be used if there is a recurring appearance of the same problem over and over again; alternatively, we can use iterative structures also. Control statements such as if-else and switch cases can be used to control the flow of logic in the program. Efficient usage of variables and data structures provides an easier solution and a lightweight, low-memory-requiring program. We have to look at the existing programming techniques, such as Divide-and-conquer, Greedy Programming, Dynamic Programming, and find out if they can be used. This problem can be solved by some basic logic or a brute-force approach. Follow the following contents to understand the approach better.
So, if the input of our problem is like n = 5, and grid =
#...#
.#.#.
..#..
.#.#.
#...#,
then the output will be 1.
Steps
To solve this, we will follow these steps −
count := 0
for initialize i := 1, when i < n - 1, update (increase i by 1), do:
for initialize j := 1, when j < n - 1, update (increase j by 1), do:
if grid[i, j] is same as '#' and grid[i - 1, j - 1] is same as '#' and grid[i - 1, j + 1] is same as '#' and grid[i + 1, j - 1] is same as '#' and grid[i + 1, j + 1] is same as '#', then:
(increase count by 1)
print(count)
Example
Let us see the following implementation to get better understanding −
#include<bits/stdc++.h>
using namespace std;
void solve(int n, vector<string> grid) {
int count = 0;
for(int i = 1; i < n - 1; i++){
for(int j = 1; j < n - 1; j++){
if(grid[i][j] == '#' && grid[i - 1][j - 1] == '#' && grid[i - 1][j + 1] == '#' && grid[i + 1][j - 1] == '#' && grid[i + 1][j + 1] == '#')
count++;
}
}
cout<< count;
}
int main() {
int n = 5;
vector<string> grid = {"#...#", ".#.#.", "..#..", ".#.#.", "#...#"};
solve(n, grid);
return 0;
}
Input
5, {"#...#", ".#.#.", "..#..", ".#.#.", "#...#"}
1 | 689 | 2,437 | {"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} | 3.359375 | 3 | CC-MAIN-2023-14 | latest | en | 0.88576 |
http://nrich.maths.org/public/leg.php?code=150&cl=3&cldcmpid=6923 | 1,503,497,004,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120573.0/warc/CC-MAIN-20170823132736-20170823152736-00292.warc.gz | 322,693,210 | 8,064 | # Search by Topic
#### Resources tagged with Volume and capacity similar to Growing Rectangles:
Filter by: Content type:
Stage:
Challenge level:
##### Other tags that relate to Growing Rectangles
Pythagoras' theorem. smartphone. Scale factors. Area. Circles. Squares. Visualising. Enlargements. Volume and capacity. Group worthy.
### There are 32 results
Broad Topics > Measures and Mensuration > Volume and capacity
### Growing Rectangles
##### Stage: 3 Challenge Level:
What happens to the area and volume of 2D and 3D shapes when you enlarge them?
### Cylinder Cutting
##### Stage: 2 and 3 Challenge Level:
An activity for high-attaining learners which involves making a new cylinder from a cardboard tube.
### Efficient Cutting
##### Stage: 3 Challenge Level:
Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end.
### Three Cubes
##### Stage: 4 Challenge Level:
Can you work out the dimensions of the three cubes?
### Maths Filler
##### Stage: 3 Challenge Level:
Imagine different shaped vessels being filled. Can you work out what the graphs of the water level should look like?
### Changing Areas, Changing Volumes
##### Stage: 3 Challenge Level:
How can you change the surface area of a cuboid but keep its volume the same? How can you change the volume but keep the surface area the same?
### Boxed In
##### Stage: 3 Challenge Level:
A box has faces with areas 3, 12 and 25 square centimetres. What is the volume of the box?
### Uniform Units
##### Stage: 4 Challenge Level:
Can you choose your units so that a cube has the same numerical value for it volume, surface area and total edge length?
### Tubular Stand
##### Stage: 4 Challenge Level:
If the radius of the tubing used to make this stand is r cm, what is the volume of tubing used?
### Thousands and Millions
##### Stage: 3 Challenge Level:
Here's a chance to work with large numbers...
##### Stage: 3 Challenge Level:
Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?
### Funnel
##### Stage: 4 Challenge Level:
A plastic funnel is used to pour liquids through narrow apertures. What shape funnel would use the least amount of plastic to manufacture for any specific volume ?
### Cuboid Challenge
##### Stage: 3 Challenge Level:
What size square corners should be cut from a square piece of paper to make a box with the largest possible volume?
### The Genie in the Jar
##### Stage: 3 Challenge Level:
This jar used to hold perfumed oil. It contained enough oil to fill granid silver bottles. Each bottle held enough to fill ozvik golden goblets and each goblet held enough to fill vaswik crystal. . . .
### Maths Filler 2
##### Stage: 4 Challenge Level:
Can you draw the height-time chart as this complicated vessel fills with water?
### Fill Me Up
##### Stage: 3 Challenge Level:
Can you sketch graphs to show how the height of water changes in different containers as they are filled?
### Chocolate Cake
##### Stage: 3 Challenge Level:
If I don't have the size of cake tin specified in my recipe, will the size I do have be OK?
### Biology Measurement Challenge
##### Stage: 4 Challenge Level:
Analyse these beautiful biological images and attempt to rank them in size order.
### More Pebbles
##### Stage: 2 and 3 Challenge Level:
Have a go at this 3D extension to the Pebbles problem.
### Sending a Parcel
##### Stage: 3 Challenge Level:
What is the greatest volume you can get for a rectangular (cuboid) parcel if the maximum combined length and girth are 2 metres?
### Sliced
##### Stage: 4 Challenge Level:
An irregular tetrahedron has two opposite sides the same length a and the line joining their midpoints is perpendicular to these two edges and is of length b. What is the volume of the tetrahedron?
### All in a Jumble
##### Stage: 3 Challenge Level:
My measurements have got all jumbled up! Swap them around and see if you can find a combination where every measurement is valid.
### Conical Bottle
##### Stage: 4 Challenge Level:
A right circular cone is filled with liquid to a depth of half its vertical height. The cone is inverted. How high up the vertical height of the cone will the liquid rise?
### Volume of a Pyramid and a Cone
##### Stage: 3
These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.
### Concrete Calculation
##### Stage: 4 Challenge Level:
The builders have dug a hole in the ground to be filled with concrete for the foundations of our garage. How many cubic metres of ready-mix concrete should the builders order to fill this hole to. . . .
### Cola Can
##### Stage: 3 Challenge Level:
An aluminium can contains 330 ml of cola. If the can's diameter is 6 cm what is the can's height?
### Mouhefanggai
##### Stage: 4
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### In a Spin
##### Stage: 4 Challenge Level:
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
### Immersion
##### Stage: 4 Challenge Level:
Various solids are lowered into a beaker of water. How does the water level rise in each case?
### Zin Obelisk
##### Stage: 3 Challenge Level:
In the ancient city of Atlantis a solid rectangular object called a Zin was built in honour of the goddess Tina. Your task is to determine on which day of the week the obelisk was completed.
### Scientific Measurement
##### Stage: 4 Challenge Level:
Practice your skills of measurement and estimation using this interactive measurement tool based around fascinating images from biology.
### Plutarch's Boxes
##### Stage: 3 Challenge Level:
According to Plutarch, the Greeks found all the rectangles with integer sides, whose areas are equal to their perimeters. Can you find them? What rectangular boxes, with integer sides, have. . . . | 1,380 | 6,199 | {"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} | 4.34375 | 4 | CC-MAIN-2017-34 | latest | en | 0.838764 |
https://www.inflationtool.com/indian-rupee/2011-to-present-value | 1,716,498,053,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00189.warc.gz | 718,698,405 | 20,761 | \$
# Value of 2011 Indian Rupees today
\$100 in 2011
\$211.88 in 2024
The inflation rate in India between 2011 and today has been 111.88%, which translates into a total increase of \$111.88. This means that 100 rupees in 2011 are equivalent to 211.88 rupees in 2024. In other words, the purchasing power of \$100 in 2011 equals \$211.88 today. The average annual inflation rate between these periods has been 5.95%.
## Inflation timeline in India (2011 - 2024)
The following chart depicts the equivalence of \$100 due to compound inflation and CPI changes. All values are equivalent in terms of purchasing power, which means that for each year the same goods or services could be bought with the indicated amount of money.
All calculations are performed in the local currency (INR) and using 6 decimal digits. Results show only up to 2 decimal digits to favour readability. Inflation data is provided by governments and international institutions on a monthly basis. Today's values were extrapolated from the latest 12-month rolling average official data.
The following table contains relevant indicators:
Indicator Value
Cumulative inflation 2011-2023 105.15%
Cumulative inflation 2011-today 111.88%
Avg. Annual inflation 2011-2023 6.17%
CPI 2011 73.25
CPI 2023 150.28
CPI 2024-01 (latest official data) 153.03
CPI today 155.21
## How to calculate today's value of money after inflation?
There are several ways to calculate the time value of money. Depending on the data available, results can be obtained by using the Consumer Price Index (CPI) formula or the compound interest formula.
### Using the CPI formula
When we have both the start and end years, we can use the following formula:
Valuet =Value0 ×
CPIt/CPI0
To obtain the values equivalent in buying power between 2011 and 2023, use the corresponding CPI values:
Value2023
=Value2011 ×
CPI2023/CPI2011
=\$ 100 ×
150.28/73.25
\$205.15
To obtain the equivalent value today (present value), plug in the CPI for today, which is estimated as 155.21:
Valuetoday
=Value2011 ×
CPItoday/CPI2011
=\$ 100 ×
155.21/73.25
\$211.88
### Alternative: Using the compound interest formula
Given that money changes with time as a result of an inflation rate that acts as compound interest, we can use the following formula: FV = PV × (1 + i)n, where:
• FV: Future Value
• PV: Present Value
• i: Interest rate (inflation)
• n: Number of times the interest is compounded (i.e. # of years)
In this case, the future value represents the final amount obtained after applying the inflation rate to our initial value. In other words, it indicates how much are \$100 worth today. There are 12 years between 2011 and 2023 and the average inflation rate was 6.1709%. Therefore, we can resolve the formula like this:
Value2023
=PV × (1 + i)n
=\$100 × (1 + 0.061709)12
\$205.15
## India inflation - Conversion table
Initial Value Equivalent value
\$1 rupee in 2011 \$2.05 rupees in 2023
\$5 rupees in 2011 \$10.26 rupees in 2023
\$10 rupees in 2011 \$20.51 rupees in 2023
\$50 rupees in 2011 \$102.57 rupees in 2023
\$100 rupees in 2011 \$205.15 rupees in 2023
\$500 rupees in 2011 \$1,025.74 rupees in 2023
\$1,000 rupees in 2011 \$2,051.47 rupees in 2023
\$5,000 rupees in 2011 \$10,257.37 rupees in 2023
\$10,000 rupees in 2011 \$20,514.74 rupees in 2023
\$50,000 rupees in 2011 \$102,573.69 rupees in 2023
\$100,000 rupees in 2011 \$205,147.38 rupees in 2023
\$500,000 rupees in 2011 \$1,025,736.92 rupees in 2023
\$1,000,000 rupees in 2011 \$2,051,473.83 rupees in 2023
Cumulative inflation From 2011 111.88% Avg. annual inflation From 2011 5.95% CPI 2011 73.25 CPI today 155.21
### Value of Rupee over time (by year)
Period Value
2011 100
2012 109.31
2013 121.24
2014 128.94
2015 136.51
2016 143.25
2017 146.82
2018 153.96
2019 165.75
2020 174.97
2021 183.53
2022 194.36
2023 205.15
2024-01 208.91
Today 211.88
All available years
Today's value of indian rupees by year:
1957 | 1958 | 1959 | 1960 | 1961 | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 | 1968 | 1969 | 1970 | 1971 | 1972 | 1973 | 1974 | 1975 | 1976 | 1977 | 1978 | 1979 | 1980 | 1981 | 1982 | 1983 | 1984 | 1985 | 1986 | 1987 | 1988 | 1989 | 1990 | 1991 | 1992 | 1993 | 1994 | 1995 | 1996 | 1997 | 1998 | 1999 | 2000 | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | 2017 | 2018 | 2019 | 2020 | 2021 | 2022 | 2023 |
Other currencies: | 1,483 | 4,424 | {"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} | 3.609375 | 4 | CC-MAIN-2024-22 | latest | en | 0.890123 |
https://www.thestudentroom.co.uk/showthread.php?t=1484234 | 1,511,089,293,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805541.30/warc/CC-MAIN-20171119095916-20171119115916-00342.warc.gz | 900,653,749 | 39,201 | You are Here: Home >< Maths
# Help with this proof. Watch
1. How do you go about proving that "1 + 10^n" cannot be a square number.
I did this:
1 + 10^n = a^2
10^n = a^2 - 1
10^n = (a-1)(a+1)
then nlog10 = log(a-1) +log(a+1)
and since nlog10 is an integer then log(a-1) + log(a+1) must be an integer. As a must be an integer then there is no value of a that gives both (a-1) and (a+1) as bases of 10.
Hence 1 + 10^n cannot be a square number.
I'm not sure if this is an accepted proof or not . Can't find anything from google. How would you guys go about doing it??
Cheers
2. (Original post by jaheen22)
then nlog10 = log(a-1) +log(a+1)
and since nlog10 is an integer then log(a-1) + log(a+1) must be an integer. As a must be an integer then there is no value of a that gives both (a-1) and (a+1) as bases of 10.
Can you explain this a bit more?
3. You can do this using mods.
Under mod 9
The remainder when a number is divided by 9 is the same as the remainder of the sum of the digits of that number. The sum of the digits of A is always 2.
11
101
1001
10001
100001
... etc
Now if you investigate the square numbers in mod 9.
But you can't make 2 mod 9 using a square number in mod 9, so A will never be a square number.
4. (Original post by Daniel Freedman)
Can you explain this a bit more?
I've thought of it differently. Seeing as we're only dealing with positive integers then the only factors of 10 are 1,2,5 and 10.
now: nlog10 = n(log 2 +log 5) or n(log 1 + log10)
log(a-1) + log(a+1) = n(log 2 + log5)
I don't know if that would help my arguement though. I don't know what to do anymore
5. (Original post by Noble.)
You can do this using mods.
Under mod 9
The remainder when a number is divided by 9 is the same as the remainder of the sum of the digits of that number. The sum of the digits of A is always 2.
11
101
1001
10001
100001
... etc
Now if you investigate the square numbers in mod 9.
But you can't make 2 mod 9 using a square number in mod 9, so A will never be a square number.
Ah, I see, I wouldn't have considered modular arithmetic before because we haven't touched upon it yet in maths. Thanks mate
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 974 | 3,528 | {"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} | 4.15625 | 4 | CC-MAIN-2017-47 | latest | en | 0.919878 |
https://ch.mathworks.com/matlabcentral/answers/39637-composite-function | 1,579,975,105,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251678287.60/warc/CC-MAIN-20200125161753-20200125190753-00387.warc.gz | 370,350,740 | 19,391 | # composite function
10 views (last 30 days)
adam on 28 May 2012
Hello, I have three vectors A, B and C. I plotted A as a function of B (B=0:0.01:1; and I have the values of A correponding). I have a function of B versus of C.
Can I plot A as a function of C? What have I use composite function or interpolation?
Stephen on 29 May 2012
plot(C,A)
you can plot anything against anything else. If A and C correspond to the same values of B then the above code should work.
#### 1 Comment
adam on 31 May 2012
Hello stephen,
Thank you for your answer. may be I didn't explain well. I try to give more details here.
I have a values of vector A for different values of vector B (the corresponding value).the values of A is plotted as a function of the values of B.
C is a function of d (C=f(d)). we can plot C as a function of d. it is function like sin or cos.
There is a connection between C and B : the values of B is varying from 0 to max(C).
Now, I would like plot A as a function of d.
Thank you in advance, | 274 | 1,010 | {"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-2020-05 | latest | en | 0.922954 |
https://brainly.com/question/93304 | 1,484,757,682,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00073-ip-10-171-10-70.ec2.internal.warc.gz | 799,699,809 | 9,097 | # Simplify each expresion write the answer as a fraction or a mixed number in simplest form3 1/2 · 2 1/3
2
by ev03av
2014-08-12T20:10:26-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
3 1/2= 7/2
2 1/3= 7/3
7/2 * 7/3=49/6= 8,1(6)
• Brainly User
2014-08-12T20:21:58-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. | 237 | 848 | {"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} | 2.71875 | 3 | CC-MAIN-2017-04 | latest | en | 0.93167 |
https://www.physicsforums.com/threads/probability-question-blue-and-cyan-balls-in-an-urn.352592/ | 1,516,139,376,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886739.5/warc/CC-MAIN-20180116204303-20180116224303-00483.warc.gz | 980,852,547 | 15,784 | # Probability question: blue and cyan balls in an urn
1. Nov 7, 2009
### alex07966
An urn contains b blue balls and c cyan balls. A ball is drawn at random, its colour is noted, and it is returned to the urn together with d further balls of the same colour. This procedure is repeated indefinitely.
i) What is the probability that the second ball drawn is cyan?
ii) What is the probability that the first ball drawn is cyan given that the second ball is cyan?
iii) Let Cn denote the event that the nth ball drawn is cyan. Show that P(Cn) = P(C1), for all n ≥ 1.
iv) Find the probability that the first drawn ball is cyan given that the nth drawn ball is cyan.
v) Find the probability that the first drawn ball is cyan given that the following n drawn balls are all cyan.
What is the limit of this probability as n → ∞?
Answer: Parts i and ii I can do: part i is c/(b+c) using the theorem of total prob. part ii is (c+d)/(b+c+d) using bayes theorem.
Now for part iii I have tried to prove this using induction:
True for n=1 and 2 so assume true for n = k-1, then using total prob theorem:
P(C_k) = P(C_k | C_k-1)*P(C_k-1) + P(C_k | B_k-1)*P(B_k-1)
where B_n is the event the nth ball drawn is blue.
I'm not sure if the above equation is actually correct as i dont know if C_k-1 and B_k-1 partition the sample space??
I think this gives:
= [(c+d)/(b+c+d)]*[c/(b+c)] + [c/(b+c+d)][1 - c/(b+c)] = c/(b+c) = P(C_1)
So this gives the correct answer but i get the feeling my working out is wrong!?
Now part iv: I have got P(C_1 | C_n) = P(C_n | C_1) but then I have no idea what to do from there....
Part v: = P(C_1 | C_2 intersect C_3 intersect ..... intersect C_n intersect C_n+1). Then what???
Any help would be greatly appreciated! :D | 497 | 1,749 | {"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} | 4.1875 | 4 | CC-MAIN-2018-05 | longest | en | 0.91997 |
https://ask.sagemath.org/question/29443/understanding-the-solve-result-with-braces-and-brackets-xzxy11/ | 1,725,778,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00039.warc.gz | 96,410,028 | 14,165 | # Understanding the 'solve()' result with braces and brackets ("([{x:z},{x:y}],[1,1])")
This post is a wiki. Anyone with karma >750 is welcome to improve it.
Having following code
var('x,y,z')
P=-x^2*y + x*y^2 + x^2*z - y^2*z - x*z^2 + y*z^2 == 0
solve(P,x,y,z)
Sage gives me a result of
([{x:z},{x:y}],[1,1])
which I am not really able to interpret, also the help(solve) did not get me any further - is there anyone who can help me out with that? (btw. as -(x-y)*(x-z)*(y-z)==0 is an alternate form for writing the polynomial my expected answer would be something like x=y or x=z or y=z but in other cases where I'd get a similar answer I would have no idea, so I'd be happy to get this format explained.)
edit retag close merge delete
Sort by ยป oldest newest most voted
The result consists of two lists: [{x: z}, {x: y}] and [1, 1]. Each list is a possible result that solves the equation. The first list is a Python dictionary (which by definition has no order but gives the values that specific variables will need to satisfy the equation); the second is a list (which simply gives the needed values for the variables in the order the vars were given as argument. Substitution confirms:
sage: P.subs(x==z)
0 == 0
sage: P.subs(x==y)
0 == 0
sage: P.subs(x==1,y==1)
0 == 0
more
Thanks so far, now I understand a bit more, but: How does it come? I mean, if there is already an x==y as solution, for what is it useful to add another result where these both values are just the same? And where is the y:z solution I now would expect?
( 2015-09-15 16:26:32 +0200 )edit
The y:z can be deduced from the other two, but I have no idea how the results come up specifically.
( 2015-09-25 09:16:08 +0200 )edit
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https://socratic.org/questions/how-do-you-graph-and-find-the-vertex-for-abs-y-abs-x-3 | 1,576,399,268,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307797.77/warc/CC-MAIN-20191215070636-20191215094636-00210.warc.gz | 547,104,286 | 6,114 | # How do you graph and find the vertex for abs y = abs(x-3)?
May 27, 2015
There are ways of achieving this analytically, but the simplest method is to generate a few sample data pairs:
$\left(x , y\right) :$
$\left(- 1 , \pm 4\right)$
$\left(0 , \pm 3\right)$
$\left(1 , \pm 2\right)$
$\left(2 \pm 1\right)$
(3,0)
I will assume by "vertex" you mean the point where the two lines cross; $\left(3 , 0\right)$ | 144 | 410 | {"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} | 3.484375 | 3 | CC-MAIN-2019-51 | latest | en | 0.756488 |
https://www.intmath.com/differential-equations/predicting-aids.php | 1,725,916,975,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00183.warc.gz | 788,184,020 | 26,739 | Search IntMath
Close
# Predicting the Spread of AIDS
We use differential equations to predict the spread of diseases through a population. The growth of AIDS is an example that follows the curve of the logistic equation, derived from solving a differential equation. We will see how to solve differential equations later in this chapter.
The HIV Virus invades a white blood cell...
Populations usually grow in an exponential fashion at first:
However, populations do not continue to grow forever, because food, water and other resources get used up over time. Differential equations are used to predict populations of people, animals, bacteria and viruses that are being affected by external events.
The logistic equation (developed in the mid-19th century) allows for a growth term AND an inhibition term. It is predicted that the AIDS epidemic will follow the pattern of the logistic equation.
If
A = number of people affected by the virus at time t,
P = the total population (a constant), and
c is a constant,
then the rate of growth of the virus at time t is given by the differential equation:
(dA)/(dt)=cPA-cA^2=cA(P-A)
The term cPA is the growth term and -cA^2 is the inhibition term.
## AIDS Example
AIDS is spreading through a city of 50,000 people who take no precautions. The virus was brought to the town by 100 people and it was found that 1000 people were infected after 10 weeks. How long will it take for half of the population to be infected?
We answer the first part of the problem here.
P = 50000, so our differential equation becomes:
(dA)/(dt)=50\ 000cA-cA^2
=cA(50\ 000-A)
Solving using Scientific Notebook gives us the number of people affected at time t:
A(t)=(50\ 000)/(1+499e^(-50\ 000ct))
[We'll see how to solve this kind of problem in the Separation of Variables page, later in this chapter.]
Now, we are told that at t = 10, A = 1000.
So now we can substitute and solve for c:
c = 4. 6416 xx 10^-6
So substituting this value of c into the expression for A(t) gives us:
A(t)=(50\ 000)/(1+499e^(-0.23208t))
So the number affected at time t is given by:
A(t)=(50,000)/(1+499e^(-0.23208t))
The graph of the number of people, A(t), affected by AIDS after t years is:
We see in the graph the S-shaped curve which the logistic equation can take.
Now to find how long it takes for half the population (25,000) to be infected:
25,000=(5,000)/(1+499e^(-0.23208t)
Solution: t = 26.8.
So we conclude that half of the population will be infected after about 27 weeks (marked on the graph above).
#### Newly Reported AIDS Cases in Australia to 2005
The following table shows actual reported AIDS cases (cumulative) in Australia.
Year AIDS cases 1981 1984 1987 1990 1993 1996 1999 2002 2005 2 55 802 2623 5060 7493 8415 9118 9609
There is a strong bias towards men acquiring AIDS, with around 95% of the cases being male.
Men: 9119
Women: 490
Total: 9609
Data Sources: Dept of Health and Ageing, Australia, and Avert.
We see that the data roughly follows the expected S-shaped logistic equation curve. The decrease in growth from the late 1990s is largely due to suppressant drugs and education.
Let's now move on to see what differential equations are and then learn how to solve them. | 833 | 3,247 | {"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} | 4.6875 | 5 | CC-MAIN-2024-38 | latest | en | 0.948277 |
http://www.jiskha.com/display.cgi?id=1203116027 | 1,498,431,665,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320593.91/warc/CC-MAIN-20170625221343-20170626001343-00203.warc.gz | 569,555,520 | 4,432 | # Trig
posted by .
Find the exact value of sinx/2 if cosx = 2/3 and 270 < x < 360.
A)1/3
B)-1/3
C)sqrt 6/6
D)-sqrt 6/6
C, since I KNOW cosx is always positive but I don't know the work involved. I know the half angle formula
• Trig -
First of all, x/2 will be in the second quadrant, since x is in the fourth quadrant. The sine of x/2 will therefore be positive.
Use the formula for sin (x/2) in terms of cos x.
sin(x/2) = sqrt([1-cos(x)]/2) = sqrt (1/6) = sqrt6/6
You got the right answer, but you it ssmes to have been a lucky guess.
Cos x is NOT always positive, but it is in this case.
• Trig -
Thank you. It wasn't really a guess it was either C or D and then I just knew it was positive so that just leaves C.
• Trig -
Jon,
Perhaps it would help if you drew an x-y axis system with a unit radius vector in each of the four quadrants.
then in quadrant 1
sin T = y/1 so +
cos T = x/1 so -
tan T = y/x so +
then in quadrant 2
sin T = y/1 so +
cos T = x/1 so - because x is - in q 2
tan T = y/x so -
then in quadrant 3
sin T = y/1 so -
cos T = x/1 so -
tan T = y/x so + because top and bottom both -
then in quadrant 4
sin T = y/1 so -
cos T = x/1 so +
tan T = y/x so -
sin has same sign as its inverse csc
cos has same sign as its inverse sec
tan has same sign as its inverse ctan
• Trig TYPO -
then in quadrant 1
sin T = y/1 so +
cos T = x/1 so +
tan T = y/x so +
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### Math, math, math!
Last week in math, we covered a few different areas.
At my table last week, we built 5 different math problems using linking cubes and place value mats. The kids had to use two different colors to show the two addends they were adding in the math problem.
Using linking cubes to build math problems.
Finding and cutting out the sum.
The kids' second station this week was sorting pictures into heavy and light categories. They had 8 different pictures they had to sort on their mats.
Sorting light and heavy pictures.
When they finished, they raced and had others check their work.
Their third station this week was filling in the missing numbers in a row of numbers. Some numbers were missing before and some were missing after. The kids had to do number anywhere from 1 to 30.
Filling in the missing number
There fourth station last week was comparing the weight of different objects from the math shelf.
Making a prediction about which objects will be heavier.
They were right!
Next week we will be doing more with measurement looking at the different types of measurement tools. | 236 | 1,149 | {"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 | 0.983805 |
ophthalmiclenses.blogspot.nl | 1,521,629,401,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647612.53/warc/CC-MAIN-20180321102234-20180321122234-00301.warc.gz | 218,749,389 | 21,470 | Saturday, July 23, 2011
Lens Thickness Formula !
After reading my last post for the basic thickness calculator, I have received some mails were asking me for the formula to find the thickness. (In that post, the formula was hided from the users)
Here I am explaining how to find the estimated thickness of a lens before it makes. To find the thickness you need to know the power, index, diameter and the edge/ center thickness.
For plus power you have to consider the edge thickness and for the minus powers you have to consider center thickness. (usual range is 0.9mm to 2.00mm according to the index)
Formula:
Thickness= [(1/2*Diameter)^2*Power / [2000*(Index - 1)]] + edge/center thickness
For instance, given a lens power of -3.75 D, a refractive index of 1.53, and a minimum blank size of 65 mm, with 1.50mm center thickness, then the thickness would be:
[(1/2*65)^2*3.75 / [2000*(1.53 - 1)]] + 1.50 = 5.24 mm
For the ease of use I have created an excel file, there you can enter all the input values for the result (fig. 1).
1. HI, i still got one question: Your formula calculate thickness for a spheric lenses. How do i calculate thickness for the aspheric lenses?
2. Power +0,25
Diameter 65
Index 1,5
Edge thickness 0,5mm
The calculated center thickness is 0,76mm.
You will not get this lens from any supplier.
I would like to have ASPH formulas as well.
3. @Anonymous
The formula is correct and verified by us. But unfortunately the edge thickness you have mentioned is too low. For this type of small powers, stock lenses are coming with 1+ edge thickness.
4. Been trying to download the lens thickness finder, but without succes. Does anybody have a copy.
regards
Jens
5. Pls repost the download link of both the lens thickness calculator in excel & basic thickness calculator
regds
6. how do i work out the power of a lens when i only have the diameter and thickness?
1. that will be quite difficult. It is possible only with non aspherical lenses and need edge, center thicknesses along with its diameter.
8. Why I have problems with the bevel lens in the mei edge and with the lens dimension.
1. Which MEI machine are you using?
9. Thickness= [(1/2*Diameter)^2*Power / [2000*(Index - 1)]] + edge/center thickness
i do not known where is 2000
10. Pls let me known. tks so much
11. how to caculation with ED IN PROGRESSIVE LENS. i ccan not understain because it with ADD
12. this is wrong farmoula | 642 | 2,437 | {"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} | 3.53125 | 4 | CC-MAIN-2018-13 | latest | en | 0.903471 |
https://www.geeksforgeeks.org/iterate-over-a-set-in-python/ | 1,695,307,357,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506028.36/warc/CC-MAIN-20230921141907-20230921171907-00641.warc.gz | 893,528,037 | 40,480 | Open In App
Iterate over a set in Python
In Python, Set is an unordered collection of data type that is iterable, mutable and has no duplicate elements.
There are numerous ways that can be used to iterate over a Set. Some of these ways provide faster time execution as compared to others. Some of these ways include, iterating using for/while loops, comprehensions, iterators and their variations. Let’s see all the different ways we can iterate over a set in Python.
Analysis of each method:
For explaining the working of each way/technique, time per set(randomly generated set) has been calculated for 5-times to get a rough estimate on how much time every technique takes for iterating over a given set. random.seed(21) has been added to each script to fixate over the random numbers that are generated every time the program is executed. Using constant seed helps us to determine which technique is best for a given particular randomly generated set.
Method #1: Iterating over a set using simple for loop.
Python3
`# Creating a set using string``test_set ``=` `set``(``"geEks"``)` `# Iterating using for loop``for` `val ``in` `test_set:`` ``print``(val)`
Output:
```k
s
e
g
E```
Analysis:
Python3
`# importing libraries``from` `timeit ``import` `default_timer as timer``import` `itertools``import` `random` `# Function under evaluation``def` `test_func(test_set):` ` ``for` `val ``in` `test_set:`` ``_ ``=` `val` `# Driver function``if` `__name__ ``=``=` `'__main__'``:` ` ``random.seed(``21``)` ` ``for` `_ ``in` `range``(``5``):`` ``test_set ``=` `set``()` ` ``# generating a set of random numbers`` ``for` `el ``in` `range``(``int``(``1e6``)):`` ``el ``=` `random.random()`` ``test_set.add(el)` ` ``start ``=` `timer()`` ``test_func(test_set)`` ``end ``=` `timer()` ` ``print``(``str``(end ``-` `start))`
Output:
```0.06303901899809716
0.06756918999963091
0.06692574200133095
0.067220498000097
0.06748137499744189```
Time complexity: O(n), where n is the number of elements in the set generated.
Auxiliary space: O(n), as the set of random numbers is stored in memory.
Method #2: Iterating over a set using enumerated for loop.
Python3
`# Creating a set using string``test_set ``=` `set``(``"geEks"``)` `# Iterating using enumerated for loop``for` `id``,val ``in` `enumerate``(test_set):`` ``print``(``id``, val)`
Output:
```0 E
1 e
2 k
3 g
4 s```
Time complexity: O(n).
Auxiliary space: O(n).
Analysis:
Python3
`# importing libraries``from` `timeit ``import` `default_timer as timer``import` `itertools``import` `random` `# Function under evaluation``def` `test_func(test_set):` ` ``for` `id``, val ``in` `enumerate``(test_set):`` ``_ ``=` `val` `# Driver function``if` `__name__ ``=``=` `'__main__'``:` ` ``random.seed(``21``)`` ``for` `_ ``in` `range``(``5``):`` ``test_set ``=` `set``()` ` ``# generating a set of random numbers`` ``for` `el ``in` `range``(``int``(``1e6``)):`` ``el ``=` `random.random()`` ``test_set.add(el)` ` ``start ``=` `timer()`` ``test_func(test_set)`` ``end ``=` `timer()` ` ``print``(``str``(end ``-` `start))`
Output
```0.1306622320007591
0.13657568199778325
0.13797824799985392
0.1386374360008631
0.1424286179972114```
Method #3: Iterating over a set as indexed list.
Python3
`# Creating a set using string``test_set ``=` `set``(``"geEks"``)` `test_list ``=` `list``(test_set)` `# Iterating over a set as a indexed list``for` `id` `in` `range``(``len``(test_list)):`` ``print``(test_list[``id``])`
Output:
```g
k
E
s
e```
Analysis:
Python3
`# importing libraries``from` `timeit ``import` `default_timer as timer``import` `itertools``import` `random` `# Function under evaluation``def` `test_func(test_set):` ` ``test_list ``=` `list``(test_set)` ` ``for` `id` `in` `range``(``len``(test_list)):`` ``_ ``=` `test_list[``id``]` `# Driver function``if` `__name__ ``=``=` `'__main__'``:` ` ``random.seed(``21``)`` ``for` `_ ``in` `range``(``5``):`` ``test_set ``=` `set``()` ` ``# generating a set of random numbers`` ``for` `el ``in` `range``(``int``(``1e6``)):`` ``el ``=` `random.random()`` ``test_set.add(el)` ` ``start ``=` `timer()`` ``test_func(test_set)`` ``end ``=` `timer()` ` ``print``(``str``(end ``-` `start))`
Output
```0.20036015100049553
0.2557020290005312
0.4601482660000329
0.2161413249996258
0.18769703499856405```
Method #4: Iterating over a set using comprehension and list constructor/initializer.
Python3
`# Creating a set using string``test_set ``=` `set``(``"geEks"``)` `# Iterating using list-comprehension``com ``=` `list``(val ``for` `val ``in` `test_set)``print``(``*``com)`
Output:
`k s e g E`
Analysis:
Python3
`# importing libraries``from` `timeit ``import` `default_timer as timer``import` `itertools``import` `random` `# Function under evaluation``def` `test_func(test_set):` ` ``list``(val ``for` `val ``in` `test_set)` `# Driver function``if` `__name__ ``=``=` `'__main__'``:` ` ``random.seed(``21``)`` ``for` `_ ``in` `range``(``5``):`` ``test_set ``=` `set``()` ` ``# generating a set of random numbers`` ``for` `el ``in` `range``(``int``(``1e6``)):`` ``el ``=` `random.random()`` ``test_set.add(el)` ` ``start ``=` `timer()`` ``test_func(test_set)`` ``end ``=` `timer()` ` ``print``(``str``(end ``-` `start))`
Output
```0.1662169310002355
0.1783527520019561
0.21661155100082397
0.19131610199838178
0.19931397800246486```
Method #5: Iterating over a set using comprehension.
Python3
`# Creating a set using string``test_set ``=` `set``(``"geEks"``)` `# Iterating using list-comprehension``com ``=` `[``print``(val) ``for` `val ``in` `test_set]`
Output:
```e
E
g
s
k```
Analysis:
Python3
`# importing libraries``from` `timeit ``import` `default_timer as timer``import` `itertools``import` `random` `# Function under evaluation``def` `test_func(test_set):` ` ``[val ``for` `val ``in` `test_set]` `# Driver function``if` `__name__ ``=``=` `'__main__'``:` ` ``random.seed(``21``)`` ``for` `_ ``in` `range``(``5``):`` ``test_set ``=` `set``()` ` ``# generating a set of random numbers`` ``for` `el ``in` `range``(``int``(``1e6``)):`` ``el ``=` `random.random()`` ``test_set.add(el)` ` ``start ``=` `timer()`` ``test_func(test_set)`` ``end ``=` `timer()` ` ``print``(``str``(end ``-` `start))`
Output
```0.11386321299869451
0.111869686999853
0.1092844699996931
0.11223735699968529
0.10928539399901638```
Method #6: Iterating over a set using map, lambda and list comprehension
Python3
`# importing libraries``from` `timeit ``import` `default_timer as timer``import` `itertools``import` `random` `# Function under evaluation``def` `test_func(test_set):` ` ``[``map``(``lambda` `val: val, test_set)]` `# Driver function``if` `__name__ ``=``=` `'__main__'``:` ` ``random.seed(``21``)`` ``for` `_ ``in` `range``(``5``):`` ``test_set ``=` `set``()` ` ``# generating a set of random numbers`` ``for` `el ``in` `range``(``int``(``1e6``)):`` ``el ``=` `random.random()`` ``test_set.add(el)` ` ``start ``=` `timer()`` ``test_func(test_set)`` ``end ``=` `timer()` ` ``print``(``str``(end ``-` `start))`
Output
```1.0756000847322866e-05
1.310199877480045e-05
1.269100175704807e-05
1.1588999768719077e-05
1.2522999895736575e-05```
Method #7: Iterating over a set using iterator.
Python3
`# importing libraries``from` `timeit ``import` `default_timer as timer``import` `itertools``import` `random` `# Function under evaluation``def` `test_func(test_set):` ` ``for` `val ``in` `iter``(test_set):`` ``_ ``=` `val` `# Driver function``if` `__name__ ``=``=` `'__main__'``:` ` ``random.seed(``21``)`` ``for` `_ ``in` `range``(``5``):`` ``test_set ``=` `set``()` ` ``# generating a set of random numbers`` ``for` `el ``in` `range``(``int``(``1e6``)):`` ``el ``=` `random.random()`` ``test_set.add(el)` ` ``start ``=` `timer()`` ``test_func(test_set)`` ``end ``=` `timer()` ` ``print``(``str``(end ``-` `start))`
Output
```0.0676155920009478
0.07111633900058223
0.06994135700006154
0.0732101009998587
0.08668379899972933```
Method #8: Iterating over a set using iterator and while loop.
Python3
`# Creating a set using string``test_set ``=` `set``(``"geEks"``)` `iter_gen ``=` `iter``(test_set)` `while` `True``:`` ``try``:`` ``# get the next item`` ``print``(``next``(iter_gen))`` ` ` ``''' do something with element '''`` ` ` ``except` `StopIteration:`` ``# if StopIteration is raised,`` ``# break from loop`` ``break`
Output:
```E
s
e
k
g```
Analysis:
Python3
`# importing libraries``from` `timeit ``import` `default_timer as timer``import` `itertools``import` `random` `# Function under evaluation``def` `test_func(test_set):` ` ``iter_gen ``=` `iter``(test_set)`` ``while` `True``:`` ``try``:`` ``# get the next item`` ``next``(iter_gen)`` ``# do something with element`` ``except` `StopIteration:`` ``# if StopIteration is raised, break from loop`` ``break` `# Driver function``if` `__name__ ``=``=` `'__main__'``:` ` ``random.seed(``21``)`` ``for` `_ ``in` `range``(``5``):`` ``test_set ``=` `set``()` ` ``# generating a set of random numbers`` ``for` `el ``in` `range``(``int``(``1e6``)):`` ``el ``=` `random.random()`` ``test_set.add(el)` ` ``start ``=` `timer()`` ``test_func(test_set)`` ``end ``=` `timer()` ` ``print``(``str``(end ``-` `start))`
Output:
```0.2136418699992646
0.1952157889973023
0.4234208280031453
0.255840524998348
0.24712910099697183```
Conclusion:
Among all the looping techniques, simple for loop iteration and looping over iterators works best, while comparing all the techniques, using map with lambda over set or iterator of set works best giving a performance of a million set iterations under 10 milliseconds. It is quite noticeable that above examples only have single access of set components per iteration, whereas if we increase the number of times a set component is accessed per iteration, it may change the time taken per iteration.
Note: Values mentioned above in the example output are bound to vary. The reason behind the variation of time consumption is machine dependency of processing power of individual’s system processor. | 3,520 | 10,829 | {"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-2023-40 | latest | en | 0.664812 |
https://www.physicsforums.com/threads/matrice-solving-the-system-need-to-put-it-in-right-form.89658/ | 1,675,029,324,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00788.warc.gz | 976,402,274 | 14,437 | # Matrice, solving the system, need to put it in right form!
mr_coffee
Hello everyone I had this system:
-2x1 + x2 = 5
6x1 - 3x2 = -15
$$\left( {\begin{array}{*{20}c} {-2} & 1 & 5 \\ 6 & -3 & {-15}\\ \end{array} } \right)$$
I then solved it down too:
$$\left( {\begin{array}{*{20}c} {-2} & 1 & 5 \\ 0 & 0 & 0\\ \end{array} } \right)$$
It wants me to put it parametric form:
so i did the following:
-2x1 + x2 = 5
2x1 = -5 + x2
x1 = (-5+x2)/2
let x2 = s;
[-5/2] + [1/2] S
[0 ] [0 ]
but its wrong! any ideas? Thanks
Homework Helper
You have to be careful, when you let $x_2 = s$, it doesn't just disappear/become zero. You still have then:
$$\left\{ \begin{gathered} x_1 = \frac{{ - 5 + x_2 }} {2} \hfill \\ x_2 = x_2 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} x_1 = \frac{1} {2}s - \frac{5} {2} \hfill \\ x_2 = s \hfill \\ \end{gathered} \right$$
Can you get the correct vector-notation now?
mr_coffee
awsome, thanks again TD!
Homework Helper
No problem | 402 | 991 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-06 | latest | en | 0.794054 |
https://socratic.org/questions/if-we-know-the-vertex-of-a-parabola-is-2-5-can-we-figure-out-the-equation-of-the | 1,638,268,354,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00378.warc.gz | 608,025,212 | 5,934 | # If we know the vertex of a parabola is (-2,-5), can we figure out the equation of the parabola?
Apr 19, 2015
No, because three data are necessary. If the vertex is known you can write:
$y - \left(- 5\right) = a \left(x - \left(- 2\right)\right) \Rightarrow y + 5 = a {\left(x + 2\right)}^{2}$
where $a$ is the amplitide.
E.G.:
$a = 1 \Rightarrow y + 5 = {\left(x + 2\right)}^{2}$
graph{y+5=(x+2)^2 [-10, 10, -5, 5]}
$a = 3 \Rightarrow y + 5 = 3 {\left(x + 2\right)}^{2}$
graph{y+5=3(x+2)^2 [-10, 10, -5, 5]}
$a = \frac{1}{2} \Rightarrow y + 5 = \frac{1}{2} {\left(x + 2\right)}^{2}$
graph{y+5=1/2(x+2)^2 [-10, 10, -5, 5]}
$a = - 2 \Rightarrow y + 5 = - 2 {\left(x + 2\right)}^{2}$
graph{y+5=-2(x+2)^2 [-10, 10, -15, 15]} | 331 | 731 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 4.28125 | 4 | CC-MAIN-2021-49 | latest | en | 0.540656 |
http://courses.cs.vt.edu/~cs3414/SM02/notes/dated/0604.html | 1,513,067,757,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515311.25/warc/CC-MAIN-20171212075935-20171212095935-00548.warc.gz | 66,290,265 | 1,255 | #### CS3414 Afterclass Notes --- 4 June, 2002
Fitting Data
1. Least Squares Approximation of Data
• Basic idea: approximate given set of m data points, (xi,yi), by a function F(x) that is the `best fit' to that data, in some sense.
• Important questions:
1. What should F(x) look like? How will you represent it? Using what basis functions, for example?
2. What do you mean by `best fit'?
• The `least squares' answer to the 2nd question is that we want the F(x) that minimizes
r12 + r22 + ... + rm2, where ri = yi - F(xi). (Terminology: ri is the ith `residual').
• There are other reasonable answers, e.g., we could minimize the 1-norm or the infinity-norm of the residual vector. In practice, the least squares (2-norm) is most common because solving the minimization problem is easier and because it has nice statistical properties. | 225 | 841 | {"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-2017-51 | latest | en | 0.912405 |
https://naipublishers.com/the-battle-between-english-and-mathematics/ | 1,656,994,743,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00651.warc.gz | 458,524,222 | 5,335 | The battle of English and also Mathematics Answer is the latest puzzle trending on social media nowadays. Human being are enjoying the Battle of English and Mathematics question 1 rabbit witnessed 6 elephants when going to the river. Every elephants witnessed 2 primates going in the direction of the river. Every monkey holds 1 parrot in their hands. Now, just how many animals are going towards the river?. Conversely, while solving the answer, civilization are sharing this inquiry and difficult their girlfriend & family members members to settle this exceptional riddle.
You are watching: The battle between english and mathematics
Here in this post, we room going to settle the battle of English and also Mathematics riddle and carry out you the correct answer. Also, we room going to share the systems for the puzzle. For this reason let us get started.
## Battle that English and Mathematics Answer:
The Battle the English and also Mathematics prize is 5.
## The fight Of English and Mathematics equipment for 1 Rabbit observed 6 Elephants Puzzle Question:
Lets now discover out exactly how we obtained the answer because that the 1 rabbit observed 6 elephants, The fight Of English and also Mathematics question.
First, let’s have actually a look in ~ the concern once:
Question – 1 rabbit observed 6 elephants when going to the river. Every elephants witnessed 2 primates going towards the river. Every monkey stop 1 parrot in your hands. Now, exactly how many animals are going towards the river?
1 rabbit observed 6 elephants if going to the river, so 1 animal i.e. rabbit is currently going towards the river.
Every elephant observed 2 primates going towards the river, for this reason from the sentence, it is clear the all the 6 elephants saw 2 chimpanzees going in the direction of the river. Thus as every the logic, it must be 6 x 2 = 12 chimpanzees (animals) going in the direction of the river. Yet there is a twist in this statement.
The over statement does not say the each Elephant experienced 2 different primates going towards the river. So below we apply the implicitly rule, and assume the 2 primates that every 6 elephants saw are the same.
Now us have got 2 primates (animal) going in the direction of the river.
As every the final statement, Every monkey hold 1 parrot in their hands, so as we currently know that there are 2 monkeys so there are 2 parrots are going towards the river
So total 1 rabbit, 2 monkeys, and 2 parrots = 5 animals are going in the direction of the river
The fight Of English and Mathematics Answer for 1 rabbit witnessed 6 elephants concern is 5 Animals
Ritesh Kumar
Ritesh Kumar is a tech freak with an endure of 5 year in the industry. He have the right to be often uncovered wearing a headset, listening to music and searching because that the latest tech news, gadgets, mobiles and more. He has been quoted in miscellaneous websites choose TheQuizz, Couponwish, and couple of more.
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naipublishers.com – The web Newspaper for Technology, Gadget, mobile Phones, entertain is a news company serving the audience from throughout the Globe. We cover every the latest news concerned technology, gadgets, leaks, sale, TV Shows, to chat & lot an ext so the our readers deserve to stay updated v all the latest happenings throughout the world. We work tough to supply the valued contents for our audiences. | 777 | 3,635 | {"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} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.946075 |
https://www.universetoday.com/93077/how-satellites-stay-in-orbit/ | 1,720,842,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00121.warc.gz | 844,504,177 | 58,969 | # How Satellites Stay in Orbit
[/caption]
An artificial satellite is a marvel of technology and engineering. The only thing comparable to the feat in technological terms is the scientific know-how that goes into placing, and keeping, one in orbit around the Earth. Just consider what scientists need to understand in order to make this happen: first, there’s gravity, then a comprehensive knowledge of physics, and of course the nature of orbits themselves. So really, the question of How Satellites Stay in Orbit, is a multidisciplinary one that involves a great of technical and academic knowledge.
First, to understand how a satellite orbits the Earth, it is important to understand what orbit entails. Johann Kepler was the first to accurately describe the mathematical shape of the orbits of planets. Whereas the orbits of planets about the Sun and the Moon about the Earth were thought to be perfectly circular, Kepler stumbled onto the concept of elliptical orbits. In order for an object to stay in orbit around the Earth, it must have enough speed to retrace its path. This is as true of a natural satellite as it is of an artificial one. From Kepler’s discovery, scientists were also able to infer that the closer a satellite is to an object, the stronger the force of attraction, hence it must travel faster in order to maintain orbit.
Next comes an understanding of gravity itself. All objects possess a gravitational field, but it is only in the case of particularly large objects (i.e. planets) that this force is felt. In Earth’s case, the gravitational pull is calculated to 9.8 m/s2. However, that is a specific case at the surface of the planet. When calculating objects in orbit about the Earth, the formula v=(GM/R)1/2 applies, where v is velocity of the satellite, G is the gravitational constant, M is the mass of the planet, and R is the distance from the center of the Earth. Relying on this formula, we are able to see that the velocity required for orbit is equal to the square root of the distance from the object to the center of the Earth times the acceleration due to gravity at that distance. So if we wanted to put a satellite in a circular orbit at 500 km above the surface (what scientists would call a Low Earth Orbit LEO), it would need a speed of ((6.67 x 10-11 * 6.0 x 1024)/(6900000))1/2 or 7615.77 m/s. The greater the altitude, the less velocity is needed to maintain the orbit.
So really, a satellites ability to maintain its orbit comes down to a balance between two factors: its velocity (or the speed at which it would travel in a straight line), and the gravitational pull between the satellite and the planet it orbits. The higher the orbit, the less velocity is required. The nearer the orbit, the faster it must move to ensure that it does not fall back to Earth.
We have written many articles about satellites for Universe Today. Here’s an article about artificial satellites, and here’s an article about geosynchronous orbit. | 634 | 2,981 | {"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} | 3.40625 | 3 | CC-MAIN-2024-30 | latest | en | 0.950667 |
https://encyclopedia2.thefreedictionary.com/accelerating+center | 1,631,918,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055808.78/warc/CC-MAIN-20210917212307-20210918002307-00001.warc.gz | 293,893,255 | 12,926 | # Center
(redirected from accelerating center)
Also found in: Dictionary, Thesaurus, Medical.
## center,
in politics, a party following a middle course. The term was first used in France in 1789, when the moderates of the National Assembly sat in the center of the hall. It can refer to a separate party in a political system, e.g., the Catholic Center party of imperial and Weimar Germany, or to the middle group of a party consisting of several ideological factions.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
## Center
in machine building, a device used to position a work-piece or mandrel on lathes, rotary grinders, and other machine tools, as well as on checking and measurement instruments.
One end of a center has a working conical surface with a vertex angle of 60° or 90°; the other has a shank with a shallow cone used to secure the center in the headstock spindle or tailstock spindle, which is an axially adjustable sleeve. If it is necessary to bore the end face of a workpiece, an opening is provided on the dead center so that a cutting tool may protrude. Machining of hollow workpieces calls for larger-diameter centers in the shape of truncated cones that fit into a conical, chamfered hole in the workpiece. Live centers, which are set in the spindle of the machine tool, have serrations on a conical working surface to transmit motion to the workpiece. In order to prevent slippage of the workpiece at higher machine speeds, the dead center may be replaced with a live center running on roller bearings. Centers are fabricated from hardened steel.
## Center
in mathematics. (1) A point O is said to be the center of symmetry of a geometric configuration if for every point A of the configuration there is another point A′ of the configuration such that O is the midpoint of the line joining A and A′. A curve or surface that has such a center is said to be central. The circle, ellipse, and hyperbola are the simplest examples of central curves, and the sphere, ellipsoid, and hyperboloid (of one or two sheets) are the simplest examples of central surfaces. It is possible for a configuration to have infinitely many centers of symmetry; for example, the centers of symmetry of a configuration consisting of two parallel lines lie on the line equidistant from the two given lines. (See alsoSYMMETRY.)
Figure 1
(2) The center of similitude of radially related configurations is the point S at which lines joining corresponding points of the configurations intersect (Figure 1).
Figure 2
(3) If all integral curves in the neighborhood of a singular point of a differential equation are closed and enclose the singular point, that point is said to be a center (Figure 2). Centers belong to the class of singular points whose character generally is not preserved when small changes are made in the right-hand side of the equation.
## center
[′sen·tər]
(industrial engineering)
A manufacturing unit containing a number of interconnected cells.
(mathematics)
The point that is equidistant from all the points on a circle or sphere.
The point (if it exists) about which a curve (such as a circle, ellipse, or hyperbola) is symmetrical.
The point (if it exists) about which a surface (such as a sphere, ellipsoid, or hyperboloid) is symmetrical.
For a regular polygon, the center of its circumscribed circle.
The subgroup consisting of all elements that commute with all other elements in a given group.
The subring consisting of all elements a such that ax = xa for all x in a given ring.
(optics)
To adjust the components of an optical system so that their centers of curvature lie on a common optical axis. Also known as square-on.
(statistics)
For a distribution, the expected value of any random variable which has the distribution.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
## center
1. The center ply in plywood.
2. The core in a laminated construction.
3. Centering.
4. The center about which an arc of a circle is drawn, equidistant from all points on the arc.
McGraw-Hill Dictionary of Architecture and Construction. Copyright © 2003 by McGraw-Hill Companies, Inc.
## centre
(US), center
1. Geometry
a. the midpoint of any line or figure, esp the point within a circle or sphere that is equidistant from any point on the circumference or surface
b. the point within a body through which a specified force may be considered to act, such as the centre of gravity
2. the point, axis, or pivot about which a body rotates
3. Politics
a. a political party or group favouring moderation, esp the moderate members of a legislative assembly
b. (as modifier): a Centre-Left alliance
4. Physiol any part of the central nervous system that regulates a specific function
5. a bar with a conical point upon which a workpiece or part may be turned or ground
6. a punch mark or small conical hole in a part to be drilled, which enables the point of the drill to be located accurately | 1,136 | 5,049 | {"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-2021-39 | latest | en | 0.930291 |
https://thebrainboxtutorials.com/2020/09/download-cbse-class-10-mathematics-basic-sample-paper-for-2021.html | 1,723,700,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00836.warc.gz | 437,757,989 | 22,312 | # Download CBSE Class 10 Mathematics Basic Sample Paper for 2021
The Brainbox Tutorials provides you with the latest sample paper for CBSE class 10 Mathematics(Basic) for the year 2021. Download CBSE Class 10 Mathematics Basic Sample Paper for 2021. It will be very helpful for the students to prepare the syllabus for their upcoming board exams. You can go through the paper provided below and can also download it to practice offline.
By solving CBSE Class 10 Mathematics (Basic) sample paper, a student will be well acquainted with the format of the question papers and pattern of the questions asked.
## Overview of Mathematics Syllabus for CBSE Class 10 for 2020-2021
Below is an overview of the chapters included in the syllabus of CBSE NCERT book Class 10 Mathematics.
### Unit II: Algebra
Chapter 2: Polynomials
Chapter 3: Pair of Linear Equations in Two Variables
Chapter 4: Quadratic Equations
Chapter 5: Arithmetic Progressions
Unit III: Coordinate Geometry
Chapter 6: Lines (In two-dimensions)
Chapter 7: Human Eye and Colourful World( excluded from 2021 syllabus)
Unit IV: Geometry
Chapter 8: Triangles
Chapter 9: Circles
Chapter 10: Constructions
Unit V: Trigonometry
Chapter 11: Introduction to Trigonometry
Chapter 12: Trigonometric Identities
Chapter 13: Heights and Distances
Unit VI: Mensuration
Chapter 14: Areas Related to Circles
Chapter 15: Surface Areas and Volumes
Unit VII: Statistics and Probability
Chapter 16: Statistics
Chapter 17: Probability
## Download CBSE Class 10 Mathematics Basic Sample Paper For 2021
This Mathematics sample paper for CBSE class 10 has been prepared in accordance with the syllabus for the session 2020-2021 given in NCERT. Solve this paper by following the actual exam rules, especially the time limit. You can also download CBSE Class 10 Mathematics Basic sample paper PDF for 2021 for offline practice.
## What is the correct way to Solve CBSE Class 10 Mathematics -Basic Sample Paper?
Every student must know the correct way to solve this CBSE Maths specimen paper for Class 10 to gain maximum benefit. Infact, every sample paper should be solved following this rule.
1. First and foremost, you should be well prepared with the syllabus.
2. Go through all the formulas of each chapter from your book.
3. Practise all types of Sums.
4. Before attempting the sample paper, you should set an alarm to end the exam.
5. Keep aside your textbook and notes. This is the test of your logical resoning and capability.
6. The time limit of the exam should be same as in actual exams of your school.
7. Solve the paper according to the instruction provided in the paper.
## How to Check the answers of sample papers?
You can open your textbook now and check the answers by yourself.
You can also ask your Math teacher to check your solution.
You can also ask your Math teacher to answer the questions which you have not understood/found.
You can discuss the answers with fellow students in the comments section too.
Below is the Mathematics Basic sample paper for CBSE Class 10. You can also Download CBSE Class 10 Mathematics Basic Sample Paper Free PDF for 2021 for offline practice. Best of luck.
## Sample Papers:
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MODULE 8 - UNRECORDED
# MODULE 8 - UNRECORDED - 1 Kb Nesbitt example Nesbitt...
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1) Kb Nesbitt example Nesbitt Company will issue a bond with four-year maturity and \$1,000 face value. The company must issue the bond for \$932.21 (a discount) because interest rates have risen after the company and its investment banker set a coupon rate of 7½ percent. Flotation costs per bond will average \$13. What is the expected cost K b of bond financing ignoring tax consequences? Nearest Student Response Value Correct Answer Feedback A. 10% APR B. 7.65% C. 13.2% D. 12% E. 9.4% Score: 0% General Feedback: REM that Kb is YTM based on proceeds, so don't use the effective rate. And don't forget to adjust the periodic rate for twice anually. 2) Kb effective Nesbitt example Nesbitt Company will issue a bond with four-year maturity and \$1,000 face value. The company must issue the bond for \$932.21 (a discount) because interest rates have risen after the company and its investment banker set a coupon rate of 7½ percent. Flotation costs per bond will average \$13. What is the effective cost K b of bond financing ignoring tax consequences, that is, the effective cost of bond financing to Nesbitt Company? Nearest Student Response Value Correct Answer Feedback A. 10.25% APR B. 10% C. 13.2% D. 12.67% E. 14.43% Score: 0% General Feedback: The problem asks you to find the ERI, so do that after solving for the YTM based on proceeds (a nominal rate).
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View Full Document | 431 | 1,672 | {"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-2018-17 | latest | en | 0.871408 |
https://www.cfd-online.com/Forums/main/4620-viscosity-thermal-conductivity-formules.html | 1,721,788,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00490.warc.gz | 595,594,030 | 15,800 | # Viscosity and thermal conductivity Formules
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April 10, 2002, 12:00 Viscosity and thermal conductivity Formules #1 CPW Guest Posts: n/a Hi everyone, Can you help me in this "near to CDF" topic? I search for explicit expressions (formulas, laws) for both the viscosity and thermal conductivity in fluids (incompressible in general, but particularly air, water,other nearly incompressible), I mean, for instance, "exp^{T_0}*K_0. Until now, I have only found (in Handbooks) tables with different values for these quantities for different temperatures, but I want to have something like: "Arrhenius Law"= \mu=\mu_0*exp{T-T_0} (Valid for ???? in range of????) Or others viscosity and thermal conductivity laws, and the fluids which can be modelised with this law. I will need also some bibliographical references for put them in my report (Because I can't write in the Bibliography, Even if this is the truth, "Founded in the CFD Forum" . Thank you very much for your help CPW
April 12, 2002, 13:13 Re: Viscosity and thermal conductivity Formules #2 Dean Guest Posts: n/a There really isn't any general formula. Since you have tables, why not just fit those with polynomials, splines, or Pade approximants? Or one could simply input the tables and interpolate using your favorite method. For gases, the Sutherland formula often works well for fitting: mu = A T^n / (T + B) where mu is the coefficient of viscosity or thermal conductivity, and A, n, and B are constants that depend on which fluid you are considering. The Sutherland formula is discussed in many of the standard textbooks on fluids.
April 14, 2002, 06:48 Re: Viscosity and thermal conductivity Formules #3 CPW Guest Posts: n/a I have heared of exponentially Laws for the viscosity, and linear laws for the thermal conductivity, so these polynomials laws seems new to me. Can you give me a reference more precisely? Thank you
April 14, 2002, 12:40 Re: Viscosity and thermal conductivity Formules #4 Dean Guest Posts: n/a This is curve fitting, not physics, and you can use whatever functions you find useful and sufficiently accurate. The "best" choice will depend upon the fluid you are modeling and the flow conditions. You might want to start with the classic "Transport Theory" by Bird, Stewart, and Lightfoot. You will find it helpful to learn some of the basic kinetic theory. Good luck with your research.
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Contact Us - CFD Online - Privacy Statement - Top | 794 | 3,193 | {"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} | 2.953125 | 3 | CC-MAIN-2024-30 | latest | en | 0.9093 |
https://www.chapters.indigo.ca/en-ca/books/ftce-mathematics-6-12-026/9781635301199-item.html | 1,540,197,821,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514879.30/warc/CC-MAIN-20181022071304-20181022092804-00179.warc.gz | 882,478,451 | 62,459 | # FTCE Mathematics 6-12 (026) Study Guide: FTCE Math Exam Prep and Practice Test Questions for the…
## byFTCE Math Exam Prep Team, Inc. Accepted
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Cirrus Test Prep’s FTCE Mathematics 6-12 (026) Study Guide: FTCE Math Exam Prep and Practice Test Questions for the FTCE 026 Examincludes:
A comprehensive REVIEW of:
Numbers and Operations
• Types of Numbers
• Scientific Notation
• Positive and Negative Numbers
• Order of Operations
• Units of Measurement
• Decimals and Fractions
• Rounding and Estimation
• Factorials
• Ratios
• Percentages
• Comparison of Rational Numbers
• Matrices
• Sequences and Series
Algebra
• Algebraic Expressions
• Operations with Expressions
• Linear Equations
• Absolute Value Equations and Inequalities
• Functions
• Exponential and Logarithmic Functions
• Polynomial Functions
• Rational Functions
• Modeling Relationships
• Trigonometry
• Conic Sections
• Other Coordinate Systems
Geometry
• Equality, Congruence, and Similarity
• Properties of Shapes
• Three-Dimensional Solids
• Transformations of Geometric Figures
• Triangle Congruence and Similarity
• Coordinate Geometry
• Geometric Proofs
• Constructions of Geometric Figures
Statistics
• Describing Sets of Data
• Graphs, Charts, and Tables
Logic and Probability
• Logic and Set Theory
• Probability
• Probability Distributions and Expected Value
Calculus
• Limits
• Derivatives
• Applications of Derivative
• Integrals
• Applications of Integrals
…as well as TWO FULL FTCE math practice tests.
Title:FTCE Mathematics 6-12 (026) Study Guide: FTCE Math Exam Prep and Practice Test Questions for the…Format:PaperbackDimensions:326 pages, 11 × 8.5 × 0.68 inPublished:February 21, 2017Publisher:Trivium Test PrepLanguage:English
The following ISBNs are associated with this title:
ISBN - 10:163530119X
ISBN - 13:9781635301199 | 527 | 2,104 | {"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} | 2.71875 | 3 | CC-MAIN-2018-43 | latest | en | 0.708411 |
https://www.newyorker.com/books/page-turner/solve-edgar-allan-poes-cryptogram | 1,547,781,301,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659677.17/warc/CC-MAIN-20190118025529-20190118051529-00548.warc.gz | 876,017,698 | 339,143 | # Solve Edgar Allan Poe’s Cryptogram
This week in the magazine, Jill Lepore considers some of the more fascinating aspects of Edgar Allan Poe’s character, like his penchant for puzzles. Here, she offers our readers a chance to take on the (self-declared) master cryptographer.
Think you can outwit Edgar Allan Poe? “Nothing intelligible can be written which, with time, I cannot decipher,” Poe once boasted. Poe loved ciphers, puns, riddles, and all manner of puzzles. (To Poe, poems were ciphers, too, more like math problems than paintings.) He fancied himself a genius; he was pretty tiresome on this point, actually, as people are. He liked, for instance, that his surname was so _poe_tical and that “Edgar Poe” is an anagram for “a God peer.” Ick. Still, even a god among men has to earn a living. Poe worked as a magazine editor, which meant that he was forever devising new ways to lure readers while fighting, always, against the temptation to lord his talents over them. In August of 1841, he issued a challenge: he published a cryptogram in Graham’s Magazine, where he was an editor, promising a year’s subscription “to any person, or rather to the first person who shall read us this riddle.” Think of it as the nineteenth-century equivalent of the cartoon caption-writing contest.
Poe didn’t write this riddle; a friend sent it to him, and dared him to solve it. “We were seduced into the endeavor to read it,” Poe explained, using his editorial we, “by the decided manner in which an opinion was expressed that we could not.” In Graham’s, Poe offered no instructions but he hinted that this riddle was not unrelated to a class of ciphers involving placing an alphabet below a particular sentence. (There, that’s your clue but I’m not really sure it is a clue; Poe liked red herrings.) He was entirely certain no one would solve it. “We have no expectation that it will be read,” he wrote, after gloating that he, of course, had found it trifling.
Poe promised to furnish a key to the solution in the September issue, but he welched. He was, it turns out, a bit of a liar. But you can find the key here right now, after the jump. Honest. Meanwhile, whatever you do, don’t Google. No, no. That’s cheating.
The key is: “But find this out and I give it up.” That help?
You’ll find the solution here tomorrow. But be warned: it’s very, very dreadful. | 556 | 2,364 | {"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-2019-04 | longest | en | 0.982977 |
https://oeis.org/A227312 | 1,642,749,832,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00389.warc.gz | 398,707,936 | 4,025 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A227312 L.g.f.: log(1 + 2*Sum_{n>=1} 2^n * x^(n^2)). 2
4, -16, 64, -224, 864, -3328, 12800, -49408, 190864, -736896, 2845440, -10987520, 42426752, -163825664, 632592384, -2442673664, 9432071040, -36420732160, 140633977856, -543040041984, 2096879372288, -8096830353408, 31264870391808, -120725281128448, 466165166208064, -1800036911561216, 6950611323771904 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Compare to the logarithm of theta_3(x) = 1 + 2*Sum_{n>=1} x^(n^2)): log(theta_3(x)) = Sum_{n>=1} -(sigma(2*n) - sigma(n))*(-x)^n/n, where sigma is the sum of divisors of n (A000203). LINKS EXAMPLE L.g.f.: L(x) = 4*x - 16*x^2/2 + 64*x^3/3 - 224*x^4/4 + 864*x^5/5 - 3328*x^6/6 + 12800*x^7/7 - 49408*x^8/8 + 190864*x^9/9 - 736896*x^10/10 + 2845440*x^11/11 - 10987520*x^12/12 + 42426752*x^13/13 - 163825664*x^14/14 + 632592384*x^15/15 - 2442673664*x^16/16 +... where exp(L(x)) = 1 + 4*x + 8*x^4 + 16*x^9 + 32*x^16 + 64*x^25 + 128*x^36 + 256*x^49 +... PROG (PARI) {a(n)=n*polcoeff(log(1+sum(k=1, n, 2*2^k*x^(k^2))+x*O(x^n)), n)} for(n=1, 36, print1(a(n), ", ")) CROSSREFS Cf. A227311, A227313, A227315. Sequence in context: A348906 A177398 A343200 * A267729 A189154 A189336 Adjacent sequences: A227309 A227310 A227311 * A227313 A227314 A227315 KEYWORD sign AUTHOR Paul D. Hanna, Jul 06 2013 STATUS approved
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Last modified January 21 01:45 EST 2022. Contains 350473 sequences. (Running on oeis4.) | 735 | 1,820 | {"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} | 3.40625 | 3 | CC-MAIN-2022-05 | latest | en | 0.454456 |
https://discourse.julialang.org/t/finding-pythagorean-triples/67954 | 1,632,816,568,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00390.warc.gz | 239,765,093 | 6,328 | # Finding Pythagorean triples
I was writing a small routine to find Pythagorean triples. Getting an x^2 + y^2 list was easy, but testing for sqrt of that gives me a Float rather than an Int, and floats are subject to binary error so I might not have an actual perfect square, or so it seems to me. Is there a way around this or am I thinking wrong?
As long as the number in question doesnât have any digits after the comma (i.e. they look like xyz.0) and the integer in question is <= 2^53 - 1 (the largest integer representable by a Float64 - yes, that is a JavaScript documentation, as JavaScript by default only has Float64 numbers for all number types), there is no error in the calculation. In other words, integers in [-(2^53 - 1), (2^53 - 1)] are representable by Float64 perfectly fine.
How large are your numbers going to be?
3 Likes
you want isqrt, ie the integer square root. You can check that isqrt(x^2+y^2)^2=x^2+y^2
5 Likes
Not That large. I only have an older computer and Iâd be sitting a long time taking a double loop for every integer up that far.
All these functions I donât know. Someday I have to go through the entire function list.
The problem isnât integer representability or some fuzzy âbinary errorâ â itâs just that floating point values round to the nearest representable value. So, for floating point square root to be âsafelyâ an integer, the question isnât just if the root is within [-(2^{53} - 1), (2^{53} - 1)]⌠the danger is if square root number could be so close to a perfect square that it rounds to an integer.
In other words, could there possibly be an integer x such that (x+\epsilon)^2 is an integer with \text{abs}(\epsilon) < \frac{\text{eps}(x)}{2}? This would require 2x\epsilon \ge 1, and that could happen for any x \ge 2^{26}. Or squares on the order of 2^{52}.
Would be interesting to try to find such a square. I wouldnât be surprised if they exist.
7 Likes
You may already be aware of this, but just in case, you can generate all Pythagorean triples more efficiently by using Euclidâs formula:
a=k.(m^2-n^2)
b=2kmn
c=k.(m^2+n^2)
where to ensure you produce each triplet just once you need m, n, and k to be positive integers with m > n , and with m and n coprime and not both odd.
2 Likes
Sure enough, it exists:
julia> function find_next_imperfect_square(n)
while true
root = ân
if isinteger(root) && Int(root)^2 != n
return n
end
n += 1
end
end
find_next_imperfect_square (generic function with 1 method)
julia> n = find_next_imperfect_square(2^52)
4503599627370497
julia> root = ân
6.7108864e7
julia> isinteger(root)
true
julia> Int(root)^2
4503599627370496
julia> Int(root)^2 == n
false
So thereâs the true bound: isintegerâsqrt will work for all n < 4503599627370497. Didnât take much time at all to find. Know what that number is?
3 Likes
Using 1 as the y value is an easy way to convert these things into false Pythagorean triples. So the next thing to check would be if squaring the result gives x^2 + y^2 back. In this case it doesnât:
julia> sqrt(2^52 + 1^2)^2 == float(2^52 + 1^2)
false
However:
julia> isinteger(sqrt(2^54 + 1^2))
true
julia> sqrt(2^54 + 1^2)^2 == float(2^54 + 1^2)
true
Fortunately, in Julia mixed type comparisons donât work by converting both values to float and then comparing them, and this comparison is false because they donât represent the same value:
julia> sqrt(2^54 + 1^2)^2 == 2^54 + 1^2
false
So that would work because Juliaâs mixed type equality comparisons are very carefulâthis would fail in C, C++ or most languages to be honest.
On the other hand, doing it with purely integer arithmetic using isqrt is safe and likely to be faster:
function is_triple(x::Integer, y::Integer)
z² = x^2 + y^2
z² == isqrt(z²)^2
end
4 Likes
Good idea, although not on my old refurb computer from Amazon. Maybe when I get access to a supercomputer. | 1,141 | 3,920 | {"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} | 3.609375 | 4 | CC-MAIN-2021-39 | latest | en | 0.919736 |
http://www.printyourbrackets.com/directions25squares.html | 1,495,947,002,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609598.5/warc/CC-MAIN-20170528043426-20170528063426-00467.warc.gz | 783,812,299 | 2,631 | How to run a 25 Square Grid Office Pool
How does this vary from the normal 100 Square Grid?
The concept is the same but each square will now have two numbers in the left hand column, and 2 numbers in the top row. This gives each square two chances to match each teams score. This size grid works great if you don't have a lot of people to play.
Step 1
Print the square grid from above.
Step 2
Sell each square for a set dollar amount until all squares are full, a player may buy as many squares as they wish. Once the player pays for the squares they are to write their name in the squares of their choice. If you feel 25 squares is to few check out our 100 square grid or 50 square grid.
Step 3
Once all the squares are full it is time to set up the drawing. This can be done in many ways, listed below are 2 examples. After determining the drawing method, draw the numbers one at a time placing the numbers from left to right starting with the first gray square box in the top row, continue across the top row until the numbers are gone. After that is complete you will redraw the numbers, this time placing the drawn numbers in the gray square boxes in the left column, starting at the top.
Example 1: Write the numbers 0-9 on pieces of paper and place them in a hat.
Example 2: Get a deck of cards A through 10, the ace represents a one, the 10 represents a 0, 2-9 are face value
Step 4
Now is time for the fun!! Begin watching the game, at the end of each quarter match the last digit of each teams score with the grid.
Example: At the end of the first quarter team 1 has 17 and team 2 has 14. Go to the top row of numbers and find the number 7(last digit of 17) then go to the left column of numbers and find the number 4(last digit of 14) find where these two numbers intersect on the grid and the name in that square wins the first quarter.
Payout's
There are a few different ways to split up the prize pool , make sure this is discussed before selling the squares. The most common method is to give the winner of each quarter 25% of the prize pool. It is possible that a player can win all four quarters and the whole prize pool. Be sure to discuss what is going to be done in case of an overtime, some people will just ignore the overtime score, some people will substitute the overtime score for the 4th quarter, and others will then divide the prize pool by 20%. Once again to avoid problems make sure all of this is discussed before selling the squares
. | 578 | 2,473 | {"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} | 3.546875 | 4 | CC-MAIN-2017-22 | longest | en | 0.939205 |
https://www.ristudypost.com/2023/04/what-is-number-of-threads-per-unit.html | 1,723,259,939,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00565.warc.gz | 740,065,229 | 35,393 | # What is the number of threads per unit length of a screw called?
### What is the number of threads per unit length of a screw called?
The number of threads per unit length of a screw is commonly referred to as the "thread pitch" or "pitch." The pitch is typically measured in millimeters (mm) or inches (in), depending on the unit system being used. It is called simply "threads per length". In US it is threads per inch. It is the reciprocal of the pitch. For example, a screw with 32 threads per inch has a pitch of 1/32 inch. The pitch is the distance between threads
TPI stands for Threads Per Inch. This is simply a count of the number of threads per inch measured along the length of a fastener. The pitch of a screw is an important parameter that determines the amount of linear travel of the screw for each revolution. Specifically, the pitch represents the distance between two adjacent threads on the screw. For example, if a screw has a pitch of 1 mm, then it will travel 1 mm along its axis for each complete revolution.
It's worth noting that the pitch is not the same as the "lead" of a screw, which is the distance the screw advances axially with each complete turn. The lead is equal to the pitch multiplied by the number of starts (or "threads") on the screw. In other words, if a screw has a pitch of 1 mm and two starts, then its lead would be 2 mm.
In summary, the number of threads per unit length of a screw is called the pitch, and it is an important parameter that determines the linear travel of the screw for each revolution. The pitch of a screw is typically specified in the screw's technical documentation or on its packaging. It is important to choose a screw with the appropriate pitch for the application in which it will be used, as the pitch can affect the screw's speed, torque, and efficiency.
In addition to the pitch, other important parameters of a screw include its diameter, length, and thread form. The diameter of a screw is typically measured across the outer edges of the threads, and it can also be specified in millimeters or inches. The length of a screw is typically measured from the top of the head to the tip of the threaded end, and it is also specified in millimeters or inches.
The thread form of a screw refers to the shape of the threads on its surface. Common thread forms include V-shaped, square, and rounded threads. The thread form can affect the screw's performance in terms of its strength, resistance to corrosion, and ease of use.
Overall, the number of threads per unit length of a screw (i.e. its pitch) is just one of many important parameters to consider when selecting a screw for a particular application. By understanding the various parameters of a screw and how they relate to its performance, it is possible to choose the right screw for a given task and ensure reliable and efficient operation. | 613 | 2,881 | {"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} | 3.359375 | 3 | CC-MAIN-2024-33 | latest | en | 0.962155 |
https://gateoverflow.in/75784/context-free-language | 1,596,590,568,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735906.77/warc/CC-MAIN-20200805010001-20200805040001-00000.warc.gz | 269,327,266 | 16,373 | Context free language
485 views
Is the language $L=\left \{ (0^{n}1^{n})^{*} |n\geq 0 \right \}$ is DCFL ?
retagged
0
as i know that DCFL is not closed under kleen closure but here Language is given so we cant use closure property ...
0
closure property works for genral term , but for perticular language dont use it.
Let the language be $L1= \left \{ 0^{N}1^{N} | N\geq 0 \right \}$
For this language, we can design a context free grammar as
$A\rightarrow \epsilon /0A1$
Now, according to the question, $L= L1$*
We can write grammar for this as
$S\rightarrow \epsilon /AS$
$A\rightarrow \epsilon /0A1$
This grammar generates $\epsilon , A,AA,AAA,AAAA,AAAAA,.................................$
Now, Is this language DCFL or NCFL ?
We can try to make a DPDA for it. If it exists,,then this language is DCFL.
This language L is DCFL and as well as NCFL.
selected by
1
But the grammar which u have written is not linear and we know that every dcfg need to be a linear grammar that is having only one variable in the production.
But u r doing S --> AS which is not allowed in linear grammar.
Also u cannot write a grammar for the above language that is linear.
So the language should be ncfl only not dcfl
0
grammar of DCFL is TYPE 2 grammar. then it must be linear?
Related questions
1
211 views
1. L = {w | w ∈ {a,b}, na(w) >= nb(w)+1} 2. L = {aibj | i ≠ 2j+1} 3. L = {ambn | m=2n+1} NOTE: 1 - DCFL, 2 - DCFL, 3 -DCFL, but need a proper reason!
$L_1 =\{a^n b^m c^n \mid m,n \geq 0\}$ and $L_2=\{ a^n b^n\mid n\geq 0\}$. If $L=L_2-L_1$ then $L$ is finite language regular language DCFL not DCFL | 516 | 1,610 | {"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} | 3.265625 | 3 | CC-MAIN-2020-34 | latest | en | 0.879871 |
http://dustbunnylair.blogspot.com/2008/12/ | 1,511,107,302,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805687.20/warc/CC-MAIN-20171119153219-20171119173219-00464.warc.gz | 94,165,131 | 19,389 | ## Sunday, December 21, 2008
### 99 problems - python - 56
Symmetric binary trees
Let us call a binary tree symmetric if you can draw a vertical line through the root node and then the right subtree is the mirror image of the left subtree. Write a predicate symmetric/1 to check whether a given binary tree is symmetric. Hint: Write a predicate mirror/2 first to check whether one tree is the mirror image of another. We are only interested in the structure, not in the contents of the nodes.
`# Example in Haskell:# *Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) Empty)# False# *Main> symmetric (Branch 'x' (Branch 'x' Empty Empty) (Branch 'x' Empty Empty))# Truedef mirror(tree): if tree == None: return None else: val, left, right = tree return (None, mirror(right), mirror(left))def symmetric_binary_tree(tree): return mirror(mirror(tree)) == mirror(tree)`
### 99 problems - python - 55
Construct completely balanced binary trees
In a completely balanced binary tree, the following property holds for every node: The number of nodes in its left subtree and the number of nodes in its right subtree are almost equal, which means their difference is not greater than one.
Write a function cbal-tree to construct completely balanced binary trees for a given number of nodes. The predicate should generate all solutions via backtracking. Put the letter 'x' as information into all nodes of the tree.
`# Example:# * cbal-tree(4,T). # T = t(x, t(x, nil, nil), t(x, nil, t(x, nil, nil))) ;# T = t(x, t(x, nil, nil), t(x, t(x, nil, nil), nil)) ;# etc......Nodef completely_balanced_trees(node_count): if node_count == 0: return [None] else: right_count = (node_count - 1) / 2 left_count = (node_count - 1) - right_count left_trees = completely_balanced_trees(left_count) right_trees = completely_balanced_trees(right_count) trees = [] for left in left_trees: for right in right_trees: trees.append(("x", left, right)) if left_count != right_count: trees.append(("x", right, left)) return trees`
OK, I'm using tuples for trees again. So sue me.
## Saturday, December 20, 2008
### Babylon 5: not so cautious, not so optimistic
So the general rule for our tv watching these days is that we pretty much just watch tv series that are complete and available in their entirety on netflix. It turns out there are only a limited number of these. After a while you start considering things that have been on your "maybe" list for a while. One very big maybe for me has been Babylon 5. It shows up on all of these top 10 best scifi series lists and people rave about it having one of the best multi season story arcs. Ever.
Having heard that the first season is a bit weak and the things really start to get good in season 2 I asked my wife for some patience and and decided to dig in.
And, well, so far it's been worse than I could have imagined. Terrible set designs. Unbelievably bad diaglog. Very long in the tooth special effects. Unimaginitive aliens. Recycled scifi tropes. Wooden acting. And we are only two episodes away from season 2 starting. And I have to admit I'm really intrigued. Is there really any chance that things turn around *so* dramatically?
I'm a total fanboy for Farscape. In my opinion, the first season there was pretty weak (not Babylon 5 weak, but not great). And yet it slowly grew into one of my all time favorite series. And this was a show that had serious problems throughout but had some amazing highs to make up for it. So I do believe that things can turn around for B5 and that I will like it.
But man, this first season has been so extremely painful. I'm afraid my wife will get revenge on me by making me watch Sex in the City. Oh, god, the horror. The horror....
### emacs python mode from scratch: stage 18 - imenu
So first of all, what *is* imenu.
... read, read ....
So it partly seems like a one file only form of etags. But if you do M-x imenu-add-menubar-index it will add an "index" menu to the menu bar with a list of all files for easy access. Being a bit of a mouse-ophobe, I'm not so sure how I'd like that. Also you have to hit rescan whenever you change the contents of the file (or set up imenu-auto-rescan).
Seems like another case of having more than one way to do things. I'll probably stick with etags, but for completeness here is a brief overview of the imenu support functions.
So there is really just one function:
• python-imenu-create-index
Go to the beginning of the file. look for "def and "class" references (unless in comments or string) and add them to the list.
Hmm, I wonder if something else more wonderful is going on here. Doesn't really seem like functionality worth supporting.
Perhaps if you had it set up to go off a simple mouse motion as described here.
`(if (featurep 'xemacs) (global-set-key [(shift button3)] 'imenu) ; XEmacs (global-set-key [S-mouse-3] 'imenu)) ; GNU Emacs`
Also looks like it integrates in with ido to some extent. Maybe there is some goodness from that.
Any way, that's enough on that topic.
## Wednesday, December 17, 2008
### emacs python mode from scratch: stage 17 - ffap
This is a small bit of code, but it's another little gem that I didn't know exists. There is only one function:
• python-module-path
This function uses emacs.modpath to determine the module path for use with ffap-alist
What this gets us is support for M-x find-file-at-point.
It's not *quite* as cool as I first anticipated because the file path finder seems to get confused about what is actually a module, but if point is on a module name on an import line then it will pop you directly over to the python lib directory and open the module. If you already have etags set up to do this then you probably won't be *too* amazed, but this is a nice extra way to accomplish the same thing.
### emacs python mode from scratch: stage 16 - completion
This is a collection of functions that uses emacs.py to find a list of likely completions for module.
`python-importspython-find-importspython-symbol-completionspython-partial-symbol python-complete-symbolpython-try-complete`
• python-imports
A list of python import statements in the buffer
• python-find-imports
Populate the above list. search by looking for "^import" or "^from" (skipping those in comments or strings).
Then reverse the list, clean out text properties and change \n to \\n so that output doesn't end up wrong.
• python-symbol-completions
Run emacs.complete(symbol, python-imports) and return the list of completions (in another buffer)
• python-partial-symbol
Use a regular expression to go backward and thereby find the complete symbol before point. This seems a little odd to me. I seems like it would be easier to just trust the syntax-table that has already been defined for the mode and just do backward-word from the current position (with a save-excursion). As a guess I'd say it's so that it can treat "." as part of the symbol and skip over that to the beginning of a module/class name.
• python-complete-symbol
Find the list of completions (using python-symbol-completions) or just scroll the completions window if they've already been found.
• python-try-complete
"hippie-expand" version for doing symbol completions. Basically "he-" does all the heavy lifting hereit just needs to be
given the python-symbol-completions and python-partial-symbol functions.
## Saturday, December 13, 2008
### 99 problems - python - 54A
Check whether a given term represents a binary tree
In Prolog or Lisp, one writes a predicate to do this.
Example in Lisp:
`* (istree (a (b nil nil) nil))T* (istree (a (b nil nil)))NILdef is_tree(t): if t == None: return True elif type(t) == tuple and len(t) == 3: (val, left, right) = t return is_tree(left) and is_tree(right) return False`
Is using lispy tuple for representing tree cheating?
### 99 problems - python - 50
Huffman codes.
We suppose a set of symbols with their frequencies, given as a list of fr(S,F) terms. Example: [fr(a,45),fr(b,13),fr(c,12),fr(d,16),fr(e,9),fr(f,5)]. Our objective is to construct a list hc(S,C) terms, where C is the Huffman code word for the symbol S. In our example, the result could be Hs = [hc(a,'0'), hc(b,'101'), hc(c,'100'), hc(d,'111'), hc(e,'1101'), hc(f,'1100')] [hc(a,'01'),...etc.]. The task shall be performed by the predicate huffman/2 defined as follows:
`# stats: [(letter, count)]# return: [(count, letter OR (left, right)]def make_tree(stats): pool = sorted([(count, letter) for letter, count in stats]) while len(pool) > 1: first = pool.pop(0) second = pool.pop(0) node = (first[0]+second[0], (first, second)) pool.append(node) pool.sort() return pool[0]def codes_from_tree(node, path, results): count, letter_or_trees = node letter = left = right = None if type(letter_or_trees) == tuple: left, right = letter_or_trees else: letter = letter_or_trees if letter: results.append((letter, path)) else: codes_from_tree(left, path+"0", results) codes_from_tree(right, path+"1", results)def huffman(stats): results = [] tree = make_tree(stats) codes_from_tree(tree, "", results) #to sort by huffman code value #results.sort(key=lambda x: int(x[1], 2)) results.sort() return results`
That solution *feels* a little hacky for me, but I can't think of why it bothers me. It might be that the last time I wrote this was in java, and it was a lot more work. Maybe this just seems too easy.
## Friday, December 12, 2008
### emacs python mode from scratch: stage 15 - info look
Probably the main lesson of my explorations of the python mode is that there is a lot of functionality available that I had never heard of before. info-look is a good case in point.
• python-after-info-look
This function essentially let's you hook into the info system to look for documentation.
As far as I can tell the main use is for calling info-lookup-symbol (C-h S) and having the info page for the python entity show up.
You need to make sure the info flavor of python documentation is installed on your system. (And hopefully it takes you less time than it did me to realize it wasn't).
Not much else to say on this.
## Monday, December 8, 2008
### emacs python mode from scratch: stage 14 - context sensitive help
Here are the relevant variables and function in the context-sensitive help section
`VARS:(defconst db-python-dotty-syntax-table(defvar view-return-to-alist)(defvar db-python-imports)FUNCS:python-describe-symbolpython-send-receivepython-eldoc-function`
• db-python-dotty-syntax-table
This creates an altered syntax table that treats "." as "symbol constituent". In other words, "." is treated as part of the variable name rather than as an other type of syntax.
• view-return-to-alist
Used by python-describe symbol. Consists of a window and help-return-method
• db-python-imports
Used below when calling emacs.* helper functions.
• python-describe-symbol
C-c C-f is bound to this and will do a python help(module|func|etc) for thing at point. It dumps out the standard python documentation in another temporary buffer. It uses emacs.py to get the help. Somewhat useful, but i usually have a python prompt open already.
• python-send-receive
Convenience function for sending a command to the python interpreter and reading the result.
• python-eldoc-function
Use emacs's own documentation reader to populate the message line with brief description of function and args. this is something that I had not noticed before and am glad I came across it while perusing the code.
If you have a python session going and you either run M-x eldoc-mode or you set up a hook such as
` (add-hook 'python-mode-hook 'turn-on-eldoc-mode)`
then this will automatically run the eldoc help facility. Not sure how useful it will be day to day (maybe it's just developer eye candy), but I've seen this sort of thing while running in slime or in the haskell mode and now at least I see where it is coming from. Seems like it only really works with core python libraries. Strangely doesn't seem to know about the functions in the actual file you are currently working in.
Only 4 sections left to walk through. I'm actually pretty eager to be done with this, but I definitely want to finish (however shallow my exploration of this mode is).
## Sunday, December 7, 2008
### Brains games that make you cry
I think almost as long as I've been a programmer I've tried making little games to test my reflexes and challenge my memory and other cognitive faculties. It's just been a given to me that: (a) you can train your brain in a manner similar to lifting weights or jogging and (b) this is a highly desirable goal.
The problem is that I've never really had enough faith in the games that I created that the effort I put into them (both writing and training with them) would be worth it. Doesn't stop me from getting interested in the idea anew every once in a while. And it seems that there is a general growing interest in these types of games, growing research that they work and (best of all) an increasing number of open source games that are created specifically to help you train your brain.
Of these sorts of games and research results it *seems* that the one with the most credibility is the "n-back task" as described in this wired article and as implemented in Brain Workshop.
Besides the research, one of the things that gives this game credibility is that it is really challenging and not very game like. It really does feel like exercise (in the sense of forcing yourself to go to the gym because you know it is good for you, a necessary evil). And honestly the first few passes through are almost laughably hard. You are trying to maintain a memory of positions and letter values being thrown out you and notice if the position or letter value is the same as n turns ago.
And the funny thing is that even after a couple of days of not really spending too much time on it, I'm definitely getting better at it. It's actually a weird feeling. The first few times through is just chaos and then all of a sudden you start moving from a slight guess that, yes, I think they did say "C" 2 turns ago and are now repeating it, to being really confident. But of course as soon as you get good at 2-back, it graduates you to 3-back.
Now the problem is how do I test if I am indeed getting smarter? I'll let you know if my brain seems like it has kicked it up a notch or two.
## Friday, December 5, 2008
### emacs python mode from scratch: stage 13 - python inferior mode
My plan was to look at context sensitive help next but it dependeds on a python inferior process so let's get that working next.
Here are the function/variables of relevance:
`(defcustom python-python-command "python"(defcustom python-jython-command "jython"(defvar python-command python-python-command(defvar python-buffer nil(defconst python-compilation-regexp-alist(defvar inferior-python-mode-map(defvar inferior-python-mode-syntax-table(define-derived-mode inferior-python-mode comint-mode "Inferior Python"(defcustom inferior-python-filter-regexp "\\`\\s-*\\S-?\\S-?\\s-*\\'"(defun python-input-filter (str)(defun python-args-to-list (string)(defvar python-preoutput-result nil(defvar python-preoutput-leftover nil)(defvar python-preoutput-skip-next-prompt nil)(defun python-preoutput-filter (s)(defun run-python (&optional cmd noshow new)(defun python-send-command (command)(defun python-send-region (start end)(defun python-send-string (string)(defun python-send-buffer ()(defun python-send-defun ()(defun python-switch-to-python (eob-p)(defun python-send-region-and-go (start end)(defcustom python-source-modes '(python-mode jython-mode)(defvar python-prev-dir/file nil(defun python-load-file (file-name)(defun python-proc ()(defun python-set-proc ()`
The main mechanism for creating a python inferior process appears to be creating a derived mode based on comint-mode (defined in comint.el). Most of the code then is defining helper functions for making this mode python aware.
So without further ago, here is a brief look at the above functions:
• python-python-command
• python-jython-command
Variables which allow customization of what command to use for invoking python (jython)
• python-command
The actual command (probably one of the above) that will actually be executed by run-python
• python-buffer
The python buffer that will be the target of code issued from files in python-mode
• python-compilation-regexp-alist
compilation-error-regexp-alist is set to this value. This is set by inferior-python-mode and is used by compile.el. This regular expression basically matches a python exception stacktrace:
` Traceback (most recent call last): File "foo.py", line 13, in 1/0 ZeroDivisionError: integer division or modulo by zero`
Honestly I'm a little confused by this since I don't think this regex actually matches the above and I'm not clear what purpose "compile" would need with it.
• inferior-python-mode-map
Add a few keys for loading a file and doing pychecker. The key sequence for pychecker seems superfluous here. I never feel like kicking off pychecker while i'm in an interacive session.
• inferior-python-mode-syntax-table
Adjust syntax table so that single quotes don't mess up the syntax coloring. "." is the syntax table entry for punctuation-like things.
• inferior-python-mode
This is where the python interactive mode is defined. It overrides the default commint-mode. There is a comment here that the "python-mode" type things (keybindings, etc) should be inherited from python mode itself.
The basic customization is to adjust some regular expressions that are used by commint to decide what sort of things get maintained in the history, what the "prompt" looks like (>>>), etc.
• inferior-python-filter-regexp
The default is not to save things in the history if they are 3 or less in length. I hadn't noticed this behavior before.
I'm not sure if there is some rhyme or reason why sometimes they use the rx style and sometime they use traditional emacs style regexps (in this case "\\`\\s-*\\S-?\\S-?\\s-*\\'").
• python-input-filter
This basically is a wrapper for ignoring inferior-python-filter-regexp
• python-args-to-list
This will be used by run-python below to collect args for kicking off a python process. Doesn't work with quoted white space. Don't know emacs lisp well at all but it seems like a complicated way to just tokenize on SPACE and TAB.
• python-preoutput-result
Not sure what "_emacs_out" is about, and not sure I care to spend the time tracking this down. (Possibly for emacs.py)
• python-preoutput-leftover
related to _emacs_out
• python-preoutput-skip-next-prompt
This (as well as previous 2 functions) are used by python-preoutput-filter
• python-preoutput-filter
This function appears to clean up the output a bit and prevent spurious >>> ... ... >>> from littering the output. I'm going to punt on this. I'm not especially interested in the logic here but it is interesting passing to see a complex function needed to make the python mode behave in a friendly fashion.
• run-python
Start a new python process or use an existing one if available. runs python-command with -i.
• python-send-command
A wrapper around python-send-string that is used by python-send-region and python-load-file
• python-send-region
Copy a section of code to temporary file and evaluate it. Can't do it directly in the interpreter since empty lines will confuse it
• python-send-string
Evaluate a python string in the buffer. This function checks for trailing \n or intermediate \t and throws in an extra \n so that the command is properly terminated in a way that the interpreter is expecting.
• python-send-buffer
Wrap python-send-region but use the entire buffer as the region
• python-send-defun
Send a region but use beginning-of-defun/end-of-defun to delimit the region
• python-switch-to-python
Create a new python process if necessary and switch to it. Giving it an argument makes it go to the end of the buffer
• python-send-region-and-go
Combine python-send-region and python-switch-to-python
• python-source-modes
A list of mode names so we can automatically tell if the buffer has python code in it
• python-prev-dir/file
A cache of the directory and file used by python-load-file so that it has a default value if necessary
• python-load-file
Import a python file in its entirety into the python process. Uses emacs.py if the file ends in .py otherwise uses "execfile"
• python-proc
Return to or create and move to a python process
• python-set-proc
Associate the python-buffer with the current buffer (which is a python process). This is used by things like python-send-region so they know where to send output.
I think unless you understand the details of how comint works (which I don't) it's hard to get a strong feel for how the inferior-python-mode works. As always it's interesting to see all of the crazy details that have to be taken care of to get a simple looking thing like an embedded python interpreter working.
That was a very cursory overview of how the python process inferior mode works (even by my already lax standards). I admit it, I'm getting a little exhausted by this project. But I'm going to keep slogging through and finish doing at least a cursory inspection of every line in python.el.
Another thing that is becoming clear is that, while this process is helping me understand emacs and emacslisp, I have a lot of work to do, and at some point I really need to just sit down and learn emacslisp in a focused way.
## Thursday, December 4, 2008
### Is it possible that I like sushi merely as a means to confound my wife?
Sometimes I wonder....
## Tuesday, December 2, 2008
### Choose Two
I was cracking wise with my co-workers the other day and later in the day I related the hi-jinx to my wife. I had proffered the following theory on the relative combination of desirable attributes in your mate.
"Gentleman, you've got cute, smart and nice. And you get to choose two."
My attractive, intelligent wife confirms that this is a sound theory. | 5,185 | 22,388 | {"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-2017-47 | longest | en | 0.783208 |
https://australian_english.academic.ru/74404 | 1,579,772,099,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00103.warc.gz | 335,621,887 | 12,610 | # order statistic
order statistic
/ˈɔdə stəˌtɪstɪk/ (say 'awduh stuh.tistik)
noun
one of a number of sample observations arranged in order of magnitude.
Australian English dictionary. 2014.
### Look at other dictionaries:
• Order statistic — Probability distributions for the n = 5 order statistics of an exponential distribution with θ = 3 In statistics, the kth order statistic of a statistical sample is equal to its kth smallest value. Together with rank statistics, order statistics… … Wikipedia
• Statistic — A statistic (singular) is the result of applying a function (statistical algorithm) to a set of data. More formally, statistical theory defines a statistic as a function of a sample where the function itself is independent of the sample s… … Wikipedia
• Statistic (role-playing games) — Part of a series on … Wikipedia
• Order of integration — For the technique for simplifying evaluation of integrals, see Order of integration (calculus). Order of integration, denoted I(p), is a summary statistic for a time series. It reports the minimum number of differences required to obtain a… … Wikipedia
• Order of integration (calculus) — For the summary statistic in time series, see Order of integration. Topics in Calculus Fundamental theorem Limits of functions Continuity Mean value theorem Differential calculus Derivative Change of variables Implicit differentiati … Wikipedia
• Durbin–Watson statistic — In statistics, the Durbin–Watson statistic is a test statistic used to detect the presence of autocorrelation (a relationship between values separated from each other by a given time lag) in the residuals (prediction errors) from a regression… … Wikipedia
• Summary statistic — Box plot of the Michelson–Morley experiment, showing several summary statistics. In descriptive statistics, summary statistics are used to summarize a set of observations, in order to communicate the largest amount as simply as possible.… … Wikipedia
• distribution free statistic — noun a statistic computed without knowledge of the form or the parameters of the distribution from which observations are drawn • Syn: ↑nonparametric statistic • Topics: ↑statistics • Hypernyms: ↑statistic • Hyponyms: ↑ … Useful english dictionary
• nonparametric statistic — noun a statistic computed without knowledge of the form or the parameters of the distribution from which observations are drawn • Syn: ↑distribution free statistic • Topics: ↑statistics • Hypernyms: ↑statistic • Hyponyms … Useful english dictionary
• rank-order correlation — noun the most commonly used method of computing a correlation coefficient between the ranks of scores on two variables • Syn: ↑rank order correlation coefficient, ↑rank difference correlation coefficient, ↑rank difference correlation • Topics:… … Useful english dictionary | 586 | 2,843 | {"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} | 3.078125 | 3 | CC-MAIN-2020-05 | latest | en | 0.806849 |
https://chat.stackoverflow.com/transcript/10/2013/2/15 | 1,726,268,885,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00441.warc.gz | 141,115,891 | 20,283 | 12:01 AM
@AndreiTita haha
I've never been able to get MH through my head. It's not that I dispute the conclusion: I accept it because I'm overwhelmingly told to, and because I believe in maths. I just haven't been able to quite understand the logic yet.
Not that I've ever given it more than three minutes' thought at a time, mind you.
@LightnessRacesinOrbit MH? I recognize the description but not the acronym.
21 mins ago, by Nolwenn Le Guen
The Monty Hall problem is a probability puzzle loosely based on the American television game show Let's Make a Deal and named after the show's original host, Monty Hall. The problem, also called the Monty Hall paradox, is a veridical paradox because the result appears impossible but is demonstrably true. The Monty Hall problem, in its usual interpretation, is mathematically equivalent to the earlier Three Prisoners problem, and both bear some similarity to the much older Bertrand's box paradox. The problem was originally posed in a letter by Steve Selvin to the American Statistician in 19...
@LightnessRacesinOrbit The simple logic behind the outcome not being 50% is that the host doesn't pick a random door.
@AndreiTita Oh yeah I've heard about that. It's the thing with the 3 doors and asking the guy if he wants to pick another one or keep the one he already has right?
@LightnessRacesinOrbit The logic is pretty simple: in 2 out of 3 cases, the game host's hint shows you exactly where the prize is.
12:08 AM
@LightnessRacesinOrbit I find the logic fairly simple to follow if it's stated correctly. Two thirds of the time, your initial choice forces the host's decision about what door to open. Unless you've already chosen the winning door (1/3), his choice is forced -- he can only choose one of the other two doors.
I wasn't asking for an explanation
Certainly not at 00:10!
Thanks, though
There's a similar puzzle IIRC.
Oddly enough, it doesn't confuse people.
@LightnessRacesinOrbit Well I provided one at 02:08.
Ah there it is
The Three Prisoners problem appeared in Martin Gardner's "Mathematical Games" column in Scientific American in 1959. It is mathematically equivalent to the Monty Hall problem with car and goat replaced with freedom and execution respectively, and also equivalent to, and presumably based on, Bertrand's box paradox. Problem Three prisoners, A, B and C, are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned. The warden knows which one is pardoned, but is not allowed to tell. Prisoner A begs the warden to let him know the identity of one o...
People tend to answer correctly to that puzzle
Even though it's equivalent to Monty Hall!
Fuck logic right?
@AndreiTita that's okay then
12:10 AM
Figured.
@StackedCrooked Speaking of which: do people find it easier to code with automatic bracket insertion? They annoy me greatly.
Sorry bout that.
And I'm pretty sure you've only added that relatively recently.
the problem with Wikipedia's formulations is that they don't seem to use the term "given" anywhere
in fact, without "given", the odds in both cases are universally 1/3 and never change
12:12 AM
@NolwennLeGuen No, not really. The trick to Monty Hall is that you really have a dependent decision, but it looks to most people like it's really an independent decision. Three Prisoner's doesn't seem to do nearly as good of a job of disguising the fact that the decision you're dealing with isn't independent.
@AndreiTita me too
@JerryCoffin Yeah. That must be it.
@AndreiTita They are disabled now.
It's annoying that my brain doesn't really comprehend this sort of statistics intuitively; even the prisoners variant.
I don't have enough registers, or something
Same
12:14 AM
@StackedCrooked interleaving stdout and stderr is interesting, but I don't think it's worth your time
@StackedCrooked You didn't have to do this for me (although that's very sweet of you). I was merely curious if you added that feature because people preferred it that way.
I hope it doesn't mean I'm stupid - I feel like I have good abstract comprehension abilities
Thanks anyway.
And I'm really going to sleep now.
@AndreiTita Most people like them disabled. I had temporarily enabled them to test something a while ago and forgot to turn them back off.
@AndreiTita oh god that annoys me on ideone.com so much
12:15 AM
Tag files will be massive
@MooingDuck Yeah.
@Ell You mean they will be too big?
They will be useful though!
But one tag file could have thousands of lines
@LightnessRacesinOrbit I hate eclipse for that very precise reason (among many others)
I'm not quite sure how people find that usable...
Matter of taste I guess
@NolwennLeGuen Eclipse forces brace insertion?
@Ell Given my current traffic I don't expect it to become a problem.
12:21 AM
I guess
But it could go big
Then I'm fucked :D
Well, not really fucked. It will just become slower over time.
And I'll need to find a better solution.
I'd like to avoid using a database for now.
I'm a plain-text fanatic :)
@LightnessRacesinOrbit I seem to recall it does. There's probably an option to disable that but shh, let's not ruin my complaint there.
@NolwennLeGuen ok ;P
night kids
you left your anti-pervert shield on
maybe
nm tired
12:26 AM
heh :D
@Ell Currently I'm at 1557 cpp files.
who says I'm not the pervert and you the prey?
@LightnessRacesinOrbit How about another you might prefer: You (a young knight) are to go to the castle to rescue the fair maiden. There are two paths: one leads to the castle, but the other to certain death. The only people who know which is which are two guards right where the path splits into two. Unfortunately, while one of the guards is truthful, the other always lies. You're allowed to ask only one question of one guard. What question do you ask of which guard to learn the correct path to the castle?
Oh, and you don't know which guard is honest or truthful.
You also don't know if the maiden is hot. Which suddenly makes the problem much harder. (No pun here @sehe, please).
Oops -- looks like I may be a minute or two too late. Oh well, maybe somebody will find it entertaining anyway.
@NolwennLeGuen Oh, you do and she is. The maiden in need of rescue is always hot! :-)
12:31 AM
Why do you always have to ruin my attempts at ruining stuff.
@NolwennLeGuen Not always. I certainly didn't bring up anything about how to configure Eclipse!
Any hole's a goal.
Any hump's a chump.
any maiden's worth the time that could be spent instead playing raiden
user1357851
@JerryCoffin asking the guard what the other guard would say which direction should one go to avoid certain death, then choice the other path ( 1&0 = 0)
user1357851
12:34 AM
old puzzle
but when I do I whore inheritance like a bitch?
Hello, World!
@stacked add mobile to your Todo list, even just viewing. I can't scroll arm
12:46 AM
@DeadMG: in the standard library, `.empty()` is a getter, returning a bool, and `clear()` is an action. In your allocator, `.empty()` is an action.
> // access A's privates
Does anyone want to edit this to add `(make sure to wash your hands after)`
@MooingDuck Yeah, it's not so well named.
it could be renamed clear() or whatever.
That's why ? Would be cool
.empty?()
@Ell `Scrolling arm.` added to my mental todo list.
btw link me again that list of warnings?
oh no I still have it
the first warning is spurious, the initialization order is different to the order in the mem init list, but since there is no dependency on the order, it's fine.
12:49 AM
@DeadMG they're all pretty minor, I fixed them locally without concern
hmm, my value_iterator requires `begin` and `end` to refer to the same instance of a variable. Not sure if I want to enforce that. I think `end` probably shouldn't require the value.
or... change how I create an `end`....
What is value iterator?
@Ell you construct it with a value, and every time you dereference it gives you that value back.
actually mine is reference/reference, but same deal
```std::string greeting = "HI";
std::copy(make_value_iterator(greeting),
make_value_iterator(greeting)+100,
std::ostream_iterator<std::string>(std::cout));```
Far Cry 3 coop is bugged to hell
1:19 AM
lol'ed
So its like a list of the same value?
@Ell Yeah
Except it doesn't hold up space like a list would
Yeah
3 hours ago, by Nolwenn Le Guen
@sehe Well I did make 82 rep out of my troll on meta earlier today.
^^ make that 92
@mooingduck who owns the value? Does the iterator make a copy of it?
1:27 AM
@Mysticial Make that 102
Feed the trolls, tuppence a bag..
Cicada is a troll that needs to be feed.
Along with the puppy.
Yeah, a couple pet trolls aren't so bad. Until they behave badly
But the insect seems to be significantly more friendly nowadays, Its honestly really nice
@Mysticial DeadMG is a troll?
@Borgleader When he needs to be, he's a good troll.
1:33 AM
I'm being nice on purpose okay. Okay?
Oh hello :)
02:34 AM and I'm trying to optimize my build in GW2
heh
:3
@Mysticial Hello there
@NolwennLeGuen I'm a failure at GW2. I played an elementalist and died so many times when leveling up.
1:35 AM
Meh
Elementalist dagger/dagger is completely broken atm
Some of the storyline quests got me to die over and over and over and ... you get the point
Unkillable, uncatchable
Nobody trolls as well as the puppy. I mean seriously who else can get 20 upvotes and a reversal badge for calling the OP a bitch?
I played staff :(
@NolwennLeGuen hi
1:36 AM
@Mysticial Hahaha xD
TIL Dota 2 Is LoL
@Borgleader Staff is more tanky/support. Also. Make sure you match the level of the storyline. Because yeah they can be quite difficult otherwise. (And make sure you dodge!!)
@Ell No, all MOBAs are Dota
Mobs? O.o
MOBA
1:36 AM
MOBAs -> Multiplayer Online Battle Arena
newb :P
Dota came first, HoN/LoL/Dota2 came after
oh and Smite
Oh
I can't help being a newb!
I've played 1-2 matches of Dota2, and I prefer LoL
Anyone play any spidweb software games?
(Also I'm lvl 30 in LoL feels kind of a waste to switch)
@Borgleader Actually, Aeon of Strife, a Brood War map, came first.
1:42 AM
Woah, that SO swag comes with Joel Spolsky's real signiature.
I can tell his hand got a little tired after writing 180 of them...
@EtiennedeMartel Yeah I guess I should have said "came before"
@Mysticial Admit it, you want one now :P
@Borgleader I have one.
That's why I know it has his real siggy.
Because it's right in front of me.
Srs?
1:44 AM
O.O
In actual pen ink.
I'm jealous now
Ok that was overboard >.>
sowwy =.=;
aww... :(
I wanted to be hated... :(
I barely made that 5th page though...
Oh, we were having a debate earlier. Which programming language are you most comfortable with?
me?
1:46 AM
yes
C++
I assumed c for some reason
Even though I wasn't in said discussion
@Ell now youre just trolling
@Mysticial Huh... why the hell did I think it was java -.-;
@Borgleader I haven't done any real Java in 5 years.
@Borgleader Did you mean comfortable, like, having no-reference-at-hand-while-coding?
1:48 AM
@MarkGarcia Comfortable as in, I'll choose that language to answer an interview question.
^ This
C lacks shit. So I can't really do much in C when time is limited.
Oh. Me, not yet very comfy with C++, though.
My Java is shaky. I need to keep on googling for how to use even the basic libraries.
Obviously, I can't do that in the middle of an interview.
But if you asked me to code a large (non-performance critical) project. I would choose Java.
I was coding Python on my way to school this morning (50 min train ride) and the documentation is just awesome...
1:50 AM
C has plenty of shit, especially when C/C++ interop :((
java java java
@Mysticial Not you too!
What have they done to youuuuu ~
Never even had coded a hello world example in java.
@NolwennLeGuen MUAHAHAHAHAHAHAHAHAHAHAHAHAHA
But if either code performance or programmer efficiency mattered, I'd stick with C++.
@mysticial why java over c++?
1:52 AM
@Ell Java doesn't have as many ropes.
I'm surprised you didn't say C# for large applications
@Borgleader I don't know C#.
Not saying that I can't learn it, but I can't use a language that I don't know.
I'm not entirely proficient with C# but I like it and a lot of stuff can be done in a small amount of code while remaining clear and obvious.
I don't like languages that are derived from C++. I think they're just reinventing stuff.
Are you talking about C# ? because C# feels to me more like Java (done right) than C++
1:55 AM
Depending on the complexity of the interview question, I might whip out some Haskell...
C# is C++ with GC and less objects-on-stack crap.
in fact, I'll have to remember that for my next interview, especially for something simple...
@Borgleader C# is indeed some "Java done right" with extra niceties (amazing niceties).
@Borgleader I mean languages that imitate templates, and those that say their much simpler than C++ (which is true, but I don't buy it).
Yeah like events and delegates and a few other things
@MarkGarcia Also the .NET framework is really damn useful
1:57 AM
..and properties.
Oh yes, those are great
@Borgleader True. But I really haven't coded with .NET for serious stuff.
@MarkGarcia "It's true but I don't buy it". wat
I don't see all tat much difference between c# and java, besides generics
@Ell WHAT DID YOU JUST SAY?!?!
1:59 AM
With my Delphi experience, I picked up C# very quick, though it took a while to not explicitly manage the object lifetimes.
@NolwennLeGuen Almost everything they say that can be done simpler can be done in C++. I think there's really no need for that.
What is Delphi like for memory and object lifetime.management? | 3,519 | 13,566 | {"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} | 3.53125 | 4 | CC-MAIN-2024-38 | latest | en | 0.965735 |
http://www.lmpforum.com/forum/topic/4234-calculating-motor-start-time-to-rated-rpm/ | 1,516,416,630,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00230.warc.gz | 492,412,409 | 34,224 | # Calculating Motor Start Time To Rated Rpm
3 replies to this topic
### #1 OldGuy
OldGuy
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Posted 18 October 2010 - 05:00 PM
Hello folks,
I am new to the forum. I am trying to figure out how to calculate the time to rated RPM on a motor start for a DOL AC Induction Motor with a large inertial mass. Any help would be appreciated. Here is some of the information I have:
6000 Hp, 4500 KW, 4KV, 730 FLA, 6 pole, 3 Phase, 60 Hz, squirrel cage induction motor, DOL start, ~15% drop in supply voltage. Besides the rotor, pump shaft, and impellor, the motor has a 13,200 lb flywheel.
Load is a centrifugal pump that has enough suction head to not cavitate.
I have connected high resolution digital recorders for voltage, amperage, and RPM readings.
Line voltage initially drops to ~3500V at 1 sec, then recovers to ~ 4200V at 24 seconds, before fully recovering to 4186V at 25-27 seconds.
Amperage spikes, but after ~ 1 second it is stable at ~ 4900A then dropping to ~ 2800A at 24 seconds, before dropping off to FLA of 730 A at 25 - 27 seconds.
My question is whether there is some kind of mathmatical way of determining if these numbers are appropriate for this motor.
Any help would be appreciated, Thanks.
### #2 jOmega
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Posted 19 October 2010 - 03:08 PM
QUOTE (OldGuy @ Oct 18 2010, 12:00 PM) <{POST_SNAPBACK}>
Hello folks,
I am new to the forum. I am trying to figure out how to calculate the time to rated RPM on a motor start for a DOL AC Induction Motor with a large inertial mass. Any help would be appreciated. Here is some of the information I have:
6000 Hp, 4500 KW, 4KV, 730 FLA, 6 pole, 3 Phase, 60 Hz, squirrel cage induction motor, DOL start, ~15% drop in supply voltage. Besides the rotor, pump shaft, and impellor, the motor has a 13,200 lb flywheel.
Load is a centrifugal pump that has enough suction head to not cavitate.
I have connected high resolution digital recorders for voltage, amperage, and RPM readings.
Line voltage initially drops to ~3500V at 1 sec, then recovers to ~ 4200V at 24 seconds, before fully recovering to 4186V at 25-27 seconds.
Amperage spikes, but after ~ 1 second it is stable at ~ 4900A then dropping to ~ 2800A at 24 seconds, before dropping off to FLA of 730 A at 25 - 27 seconds.
My question is whether there is some kind of mathmatical way of determining if these numbers are appropriate for this motor.
Any help would be appreciated, Thanks.
.....from one 'ole guy' to another...
Yes, it is possible to calculate if you have all of the information necessary. Have a look at the following link where I've presented a methodology for calculating the starting time for a fan. The same can be done for your pump application with the addition of the requisite parametrical data.
Fan Accel Time Calc
The information you have provided, while giving us a glimpse of the application .... is insufficient for making such calculation.
Here are just a few of the datum that would additionally need to be made known:
• complete motor nameplate (rating plate) data.
• motor performance curves, incl. speed vs torque vs amperage
• motor locked rotor data
• motor inertia
• total inertia reflected to the motor shaft
• pump data - is it centrifugal or positive displacement type?
• pump curves w/system curve imposed thereon
• speed at which pump picks up pumping load
• voltage source data, including impedance and nominal loading at time of start
I'm sure there are more items to consider, but these are the ones that jump up, so to speak.
With sufficient data, the methodology shown in the above hyperlinked example for fan acceleration,
can be modified to accommodate your pump application.
There is one set of data that you provided that begs its questioning .... the voltage plot ...
It drops to 3500 then rises to 4200 before dropping to final value of 4186.
The rise to 4200 while still sourcing acceleration current is suspect when one considers that
at nominal steady state load, the voltage drops to 4186 at a much lesser value of load current.
Kind regards,
### #3 OldGuy
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Posted 19 October 2010 - 04:19 PM
Thank you for the response. I appreciate your time.
You are correct concerning the voltage data I provided. I did not do a good job proof-reading. Where is says 4200 V it should say 4000 V. The remainder is essentially correct.
I have access to all of the additional data you mendtioned, with one exception.
You included :
• speed at which pump picks up pumping load
The system I am concerned with is a closed loop hydraulic system. When the pump is started there is no flow. Flowrate actually continues to increase for some time (up to a few minutes) after the pump is at full rated speed.
At what point would I say the pump has picked up the pumping load?
I am currently working on a separate project. It will take me a couple days to research and retrieve all the data you identified. In the interim, I will take a closer look at the link you provided and see what I come up with.
Thanks, again.
OldGuy
### #4 jOmega
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Posted 21 October 2010 - 01:02 PM
QUOTE (OldGuy @ Oct 19 2010, 10:19 AM) <{POST_SNAPBACK}>
Thank you for the response. I appreciate your time.
You included :
• speed at which pump picks up pumping load
The system I am concerned with is a closed loop hydraulic system. When the pump is started there is no flow. Flowrate actually continues to increase for some time (up to a few minutes) after the pump is at full rated speed.
At what point would I say the pump has picked up the pumping load?
Thanks, again.
OldGuy
In answer to your question, check the information/data you have for the pump; if there is a minimum operating RPM stated, then that is typically the point at which the pump picks up load.
Also, please advise type of pump; Centrifugal ? or Positive Displacement.
Thanks
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https://www.tes.com/teaching-resource/current-electricity-ohms-law-ks3-11772813 | 1,560,833,060,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998607.18/warc/CC-MAIN-20190618043259-20190618065259-00273.warc.gz | 923,000,134 | 32,311 | # Current Electricity - Ohms Law KS3
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This action-packed lesson on Ohms Law is fully resourced and differentiated with 11 activities and 9 learning outcomes
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An easy to follow one-page flow chart lesson plan indicates where logical choices between resources can be made and indicates whether each activity includes literacy, oracy or AFL .
Objective
• To know that metal ions cause resistance.
• To be able to explain why some materials have a high resistance and others a low resistance.
• To understand why resistance reduces current.
• To understand that collisions between metal ions and electrons releases energy.
• To understand that appliances are made using materials with different resistances to produce different effects.
• To use Ohms Law to find resistance.
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• Animated 14 slide PowerPoint- includes exit ticket /plenary quiz
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Current, Voltage and Resistance - What Are They?
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Created: Nov 14, 2017
Updated: Jun 16, 2019
#### Whole lesson
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#### 252. Railway Communication
time limit per test: 0.25 sec.
memory limit per test: 65536 KB
input: standard
output: standard
There are N towns in the country, connected with M railroads. All railroads are one-way, the railroad system is organized in such a way that there are no cycles. It's necessary to choose the best trains schedule, taking into account some facts.
Train path is the sequence of towns passed by the train. The following conditions must be satisfied.
1) At most one train path can pass along each railroad.
2) At most one train path can pass through each town, because no town can cope with a large amount of transport.
3) At least one train path must pass through each town, or town economics falls into stagnation.
4) The number of paths must be minimal possible.
Moreover, every railroad requires some money for support, i-th railroad requires c[i] coins per year to keep it intact. That is why the president of the country decided to choose such schedule that the sum of costs of maintaining the railroads used in it is minimal possible. Of course, you are to find such schedule.
Input
The first line of input contains two integers N and M (1<=N<=100; 0<=M<=1000). Next M lines describe railroads. Each line contains three integer numbers a[i], b[i] and c[i] - the towns that the railroad connects (1<=a[i]<=N; 1<=b[i]<=N, a[i]<>b[i]) and the cost of maintaining it (0<=c[i]<=1000). Since the road is one-way, the trains are only allowed to move along it from town a[i] to town b[i]. Any two towns are connected by at most one railroad.
Output
On the first line output K - the number of paths in the best schedule and C - the sum of costs of maintaining the railroads in the best schedule.
After that output K lines, for each train path first output L[i] (1<=L[i]<=N) - the number of towns this train path uses, and then L[i] integers identifying the towns on the train path. If there are several optimal solutions output any of them.
Sample test(s)
Input
4 4
1 2 1
1 3 2
3 4 2
2 4 2
Output
2 3
2 1 2
2 3 4
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### December 14, 2012
Today we have an exercise, a solution, and a question I don’t know the answer to. The exercise is to write a program that lists the narcissistic numbers in which the sum of the nth powers of the digits of an n-digit number is equal to the number. For instance, 153 is a narcissistic number of length 3 because 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153. The largest decimal narcissistic number is the 39-digit number that is the title of today’s exercise.
Your task is to write a program to find narcissistic numbers. When you are finished, you are welcome to read or run a suggested solution, or to post your solution or discuss the exercise in the comments below.
Pages: 1 2
### 21 Responses to “115132219018763992565095597973971522401”
1. Mithrandir said
Partially better solution is to generate non-decreasing sequences of digits and compute their sum. If this sum has the same digits as the digits used to generate it, you have a number.
A quickly implemented Haskell solution follows. It could be optimized a lot. On my machine, it prints up to 4338281769391371 in 1 second. I don’t have time to profile it though.
import Data.List
digits :: [Int]
digits = [0..9]
expand :: Int -> [[Int]] -> [[Int]]
expand 0 l = l
expand n [] = expand (n – 1) \$ map (:[]) digits
expand n l = expand (n – 1) \$ concatMap exp l
where
exp ll@(x:xs) = map (:ll) \$ filter (>= x) digits
test :: Int -> [Int]
test n = map (toSum 0 n) . filter (narcis n) . expand n \$ []
narcis :: Int -> [Int] -> Bool
narcis n l = let x = toSum 0 n l in sort l == sort (toList x)
toSum :: Int -> Int -> [Int] -> Int
toSum s _ [] = s
toSum s e (x:xs) = toSum (s + x ^ e) e xs
toList :: Int -> [Int]
toList 0 = []
toList n = (n `mod` 10) : toList (n `div` 10)
printOne n = mapM print \$ test n
main = do
mapM printOne [1..]
(hopefully WordPress will keep the code as it is)
2. Mithrandir said
(let’s try again, please delete my previous comment and strip out this line:P)
Partially better solution is to generate non-decreasing sequences of digits and compute their sum. If this sum has the same digits as the digits used to generate it, you have a number.
A quickly implemented Haskell solution follows. It could be optimized a lot. On my machine, it prints up to 4338281769391371 in 1 second. I don’t have time to profile it though.
``` import Data.List```
``` digits :: [Int] digits = [0..9] expand :: Int -> [[Int]] -> [[Int]] expand 0 l = l expand n [] = expand (n - 1) \$ map (:[]) digits expand n l = expand (n - 1) \$ concatMap exp l where exp ll@(x:xs) = map (:ll) \$ filter (>= x) digits test :: Int -> [Int] test n = map (toSum 0 n) . filter (narcis n) . expand n \$ [] narcis :: Int -> [Int] -> Bool narcis n l = let x = toSum 0 n l in sort l == sort (toList x) toSum :: Int -> Int -> [Int] -> Int toSum s _ [] = s toSum s e (x:xs) = toSum (s + x ^ e) e xs toList :: Int -> [Int] toList 0 = [] toList n = (n `mod` 10) : toList (n `div` 10) printOne n = mapM print \$ test n ```
```main = do mapM printOne [1..] ```
(hopefully WordPress will keep the code as it is)
3. Mithrandir said
I give up. Don’t know how to format it properly :|
4. jpverkamp said
Well, it certainly doesn’t run in a few seconds but it’s at least significantly faster than the direct solution. It’s calculated all of the Narcissistic numbers up to the five 25 digit solutions in about 45 minutes and it’s still running. It’s took about 30 seconds to do what Mithrandir’s similar algorithm did in 1 though, so there’s room for improvement.
I’ll see if I can think of any more optimizations ( or implement them if anyone else has some :) ), but that’s what I have for now.
Blog post:
Narcissistic Numbers
Source code:
narcissistic.rkt
5. [...] today’s Programming Praxis exercise, our task is to calculate all the narcissistic numbers, also known als [...]
6. Here’s my Haskell solution (see http://bonsaicode.wordpress.com/2012/12/14/programming-praxis-115132219018763992565095597973971522401/ for a version with comments). Sadly it’s not the mythical 2-second solution, but it’s reasonably quick. Calculating all numbers of 1 through 25 digits takes just over 4 minutes.
```import Data.List
import qualified Data.Vector as V
narcissistic :: Integer -> [Integer]
narcissistic upto = narcs =<< [1..min 39 upto] \\ [2,12,13,15,18,22,26,28,30,36]
narcs :: Integer -> [Integer]
narcs n = sort \$ f [] 0 9 n where
powers = V.fromList \$ map (^n) [0..9]
pow i = powers V.! fromIntegral i
(lo, hi) = (10^(n-1), 10^n)
f ds s x 1 = [ s' | i <- [0..x], let s' = s + pow i, s' >= lo
, s' < hi, sort (show s') == (show =<< (i:ds))]
f ds s x d = [0..x] >>= \i -> f (i:ds) (s + pow i) i (d-1)
```
7. Paul said
My first version in Python. Still a little slow. 17 minutes for n <= 25.
8. hamidj said
My java solution here: hamidj.wordpress.com
9. DDJ said
Just a thought: Let us say you have tested all permutations of N digits. Now you add a digit 0..9 number N+1 to the previous digits and calculate a new number. Does this calculated number have a permutation of digits equal to the digits including the new one? If it does, the calculated number tells the correct permutation of digits. Try next digit. If no digits 0..9 gives a new narcissistic number then ?
10. [...] Pages: 1 2 [...]
11. Brian said
Here’s a simple python script to inspect all numbers to a defined limit:
```def find_narc(limit):
n=1
i=[]
while n<=limit:
k=0
number=str(n)
l=len(number)
for x in range(0,l):
k=k+int(number[x])**l
if n==k:
i.append(n)
n=n+1
return i
[/sourcecode ]
Which gives the following results for a limit 10,000,000:
[sourcecode lang="python"]
print find_narc(10000000)
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748, 92727, 93084, 548834, 1741725, 4210818, 9800817, 9926315]
# total time 67.6s
```
I feel like there is a better way to do this geometrically, since you’re computing sums of digits to the nth power. Maybe a predictive solution where the computer computes partial sums and then finds integer n-roots for remaining difference.
12. Brian said
Well I screwed that sourcecode up.
13. Brian said
Rather than take each digit to the nth degree for every single number, I tried to break it out by power series:
```def find_narc2(limit):
ret=[]
i=[0,1,2,3,4,5,6,7,8,9]
for z in range(1,limit+1):
for x in range(0,10):
i[x]=x**z
for n in range(10**(z-1),(10**z-1)):
number=str(n)
k=0
for y in range (0,z):
k=k+i[int(number[y])]
if n==k:
ret.append(n)
return ret
```
When compared to the function above, it turns out it takes about a minute (8.6%) off of 100,000,000 numbers:
```print find_narc(100000000)
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748, 92727, 93084, 548834, 1741725, 4210818, 9800817, 9926315, 24678050, 24678051, 88593477]
# [Finished in 754.7s]
print find_narc2(8)
# [1, 2, 3, 4, 5, 6, 7, 8, 153, 370, 371, 407, 1634, 8208, 9474, 54748, 92727, 93084, 548834, 1741725, 4210818, 9800817, 9926315, 24678050, 24678051, 88593477]
# [Finished in 689.1s]
```
14. cosmin said
Nice solution! Loop unrolling works great here.
15. hmbg said
Simple and slightly faster variant in python (cheating with itertools):
```from itertools import permutations,combinations_with_replacement
# Returns a list of all unique N-digit combinations (single permutation)
def uniques(N):
return [''.join(l) for l in list(combinations_with_replacement('0123456789',N))]
# Returns a list of all number permutations of the digit-string n
def permut(n):
pm = permutations(str(n))
return [int(''.join(tpl)) for tpl in set(pm)]
def fastnarcs(N):
narcs = []
for k in range(1,N+1):
for n in uniques(k):
S = sum([int(i)**k for i in list(n)])
for pmt in permut(n):
if int(pmt) == S:
#print '%s gave narc %d'%(n, pmt)
narcs.append(pmt)
narcs = list(set(narcs))
narcs.sort()
return narcs
def simplenarcs(N):
narcs = []
for n in range(0,10**N):
k = len(str(n))
S = sum([int(i)**k for i in list(str(n))])
if S == n:
narcs.append(n)
return narcs
```
fastnarc is about 4 times as fast as the trivial solution on N<=7
16. [...] is a nice mathematical exercise from Programming Praxis. Is about finding the “narcissistic numbers”, n digit numbers numbers where the sum of [...]
17. I’ve written a solution on my blog (in Python) http://wrongsideofmemphis.com/2012/12/18/narcissistic-numbers/
It generates the possible candidates checking all the digits combinations using itertools, then checking again those candidates to get the narcissistic numbers. It seems quite fast to me (for being in Python), around 3m 40s for all numbers up to 18 digits.
18. PeterD said
I have written a solution to this problem (in C#, available as a VS 2012 Express solution here). The code is somewhat involved, but it searches the entire problem space and finds the 88 solutions other than “0″. Running time is 1.6 seconds as a single-threaded program on an i7-2640M @2.80GHz. So I believe I can now answer the question posed by the post.
The code is in 3 parts: a base-10 multi-precision implementation (Base10e8Nat), the main Solver class that tackles each length independently, and a top-level Program that collects counts and timings for each length. A few tables are pre-calculated, among them the powers themselves and all the possible multiples of them. There is no multiplication during the actual search. Counts of powers are packed into a 64-bit integer, 6 bits for each digit count.
The search is done on the multiples of powers rather than the integers, as others have suggested. However, it begins with the 9s and recurses to 8s and so on down. At each level, we begin with the largest multiple that will not overflow the result, and then work our way down to 0, recursing for each. At each recursive search, we have available the number of each (higher) power already chosen (and the number remaining available to us), and a running total of the sum of those powers so far.
The virtue of searching this way (besides reducing it to a combinatoric search) is that as we reach the lower levels (about 6 and below for the larger lengths) of the recursion, some (or many) of the most-significant digits of the running total will already be “fixed” (modulo a single overflow), because we have a hard bound on the number of least-significant digits that could potentially change during this recursion (it is the number of digits in the highest multiple of this level’s power that are still available). We can now look at those fixed digits (hereafter the “prefix”) and apply heuristics to see if they are consistent with the already-chosen powers, and the remaining choices available.
There are two main heuristics, applied with a slight variation to the higher powers and the lower powers. Essentially there’s either too many of a digit in the prefix, or not enough.These heuristics must also take note of the possibility of overflow, but it’s not as bad as one might think at first. It has potentially unbounded effects for 9s and 0s, but the counts of other digits can only be affected by at most 1 by any overflow. We look at what the overflow digit (the LSD of the fixed part) is and in a pinch we make a conservative assumption to avoid analyzing further.
For the higher powers: the number of each power chosen by higher levels (the multiple) is fixed as far as this sub-search is concerned. So if there are already more than that number of the digit in the prefix, it can’t be fixed and this sub-search is a bust. Likewise if the digit count is lower than the multiple for some N, it will have to be made up in the remaining digits (the suffix). Observe though that if there are undercounts for several Ns, there have to be enough remaining digits to cover _all_ of them. So we can sum up the “excess powers” and then compare it to the length of the suffix. If the suffix is too short to supply enough of the missing digits, it’s a bust.
For the lower powers: similar to “excess higher powers”, here we haven’t yet apportioned the number of powers, but we do have a limit on how many are available for this sub-search. So we look at the prefix digits and figure out how many of the lower powers are already spoken for. If it’s more than we’ve got, then bust. The converse, I haven’t implemented yet: if there are only a few digits of lower powers, we shouldn’t bother recursing with too many “available” powers.
0, 9, and the current level require somewhat extra attention, which I won’t go into: performance becomes much more tractable with just a few of these heuristics applied anyway, though it really pays off to wind them nice and tight. To give you an idea of the effect, from profiling results I can see that this search only makes a full comparison of digit counts with power counts for about 1.14 million candidates. Everything else is pruned.
That’s probably enough to get the general approach across. I’ve tried to add explanatory comments through the interesting parts of the code if anyone’s curious, or ask questions here and I will try to answer.
19. PeterD said
I forgot to add: if you want to watch the recursive search in action, you can remove the Conditional attribute from Solver.DumpPowerBits(). Each of the 1.14 million “final candidates” will then be printed out.
20. [...] – As mentioned in the original post Dik Winter calculated all narcissistic numbers in 1985, which took 30 minutes then (equivalent to [...] | 3,774 | 13,441 | {"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} | 3.5625 | 4 | CC-MAIN-2013-48 | longest | en | 0.834233 |
https://getrevising.co.uk/revision-cards/unit-4-section-1-simple-harmonic-motion-and | 1,519,037,670,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812579.21/warc/CC-MAIN-20180219091902-20180219111902-00631.warc.gz | 657,255,480 | 13,483 | # unit 4 section 1 simple harmonic motion and oscillators
HideShow resource information
## what is simple harmonic motion?
SHM: an oscillation in which the acceleration is directly proportional to its displacement from its equilibrium position, and is directed towards the equilibrium.
Displacement: x, varies as a cosine or a sine wave with a maximum value of A the amplitude.
Velocity: v, is the gradient of teh displacement-time graph. has a maximum value of (2pief)A, f is frequency of the oscillation.
Acceleration: a, is the gradient of the velocity-time graph. it has a maximum value of (2pief)2A.
Phase difference: a measure of how much one wave lags behind another wave. it can be measured in degrees, radions or fractions of a cycle.
F=1/T
T=1/F
Frequency: number of complete revolutions or cycles that rotating or oscillating object makes per second.
Period: the time taken for a rotating or oscillating object to complete one revolution or cycle.
1 of 4
## potential and kinetic energy
- as the object moves towards the equilibrium position, restoring force does work on the object and transfers EP to EK.
- when the object moves away from the equilibrium, all the EK is transferred back to EP again.
- at the equilibrium, the objects EP is zero and its EK is maximum - therefore its velocity is maximum.
- at maximum displacement on both sides of the equilibrium, the objects EP is maximum, and its EK is zero - so its velocity is zero.
2 of 4
## phase, frequency and period
Phase difference: a measure of how much one wave lags behind another wave. it can be measured in degrees, radions or fractions of a cycle.
F=1/T
T=1/F
Frequency: number of complete revolutions or cycles that rotating or oscillating object makes per second.
Period: the time taken for a rotating or oscillating object to complete one revolution or cycle.
3 of 4
## simple harmonic oscillators - a mass on a spring a
SHO: a system that oscillates with SHM, like a simple pendulum
size and direction of the restoring force is shown with the equation:
- mass on a sping: F=-Kx
to
T=2pie x squareroot(m/K)
-simple pendulum:
- T=2pie x squareroot(l/g)
4 of 4 | 509 | 2,175 | {"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-2018-09 | longest | en | 0.868241 |
http://financialriskforecasting.com/code/3.html | 1,490,370,577,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188213.41/warc/CC-MAIN-20170322212948-00071-ip-10-233-31-227.ec2.internal.warc.gz | 136,046,883 | 3,748 | # Chapter 3. Multivariate Volatility Models
The R code from 2011 runs unmodified, this just updates the end date. The Matlab code would not run, so these are new functions. The GARCH functionality in the econometric toolbox in Matlab can only do univatiate GARCH. Kevin Sheppard's MFE toolbox is well written and is certainly comprehensive. Its whats used below. It can be downloaded here and the documentation here is quite comprehensive.
##### Listing 3.1: Download stock prices in R Last edited: August 2016
library("tseries")
library(zoo)
p1 = get.hist.quote(instrument = "msft",start = "2000-01-01",end = "2016-08-30",quote="AdjClose")
p2 = get.hist.quote(instrument = "ibm", start = "2000-01-01",end = "2016-08-30",quote="AdjClose")
p = cbind(p1,p2)
y = diff(log(p))*100
y[,1] = y[,1]-mean(y[,1])
y[,2] = y[,2]-mean(y[,2])
TT = length(y[,1])
##### Listing 3.2: Download stock prices in Matlab Last edited: August 2016
stocks =hist_stock_data('01012000','30082016','msft','ibm')
p = [p1 p2];
y = diff(log(p))*100;
y(:,1)=y(:,1)-mean(y(:,1));
y(:,2)=y(:,2)-mean(y(:,2));
T = length(y);
##### Listing 3.3: EWMA in R Last edited: AUgut 2016
EWMA = matrix(nrow=TT,ncol=3)
lambda = 0.94
S = cov(y)
EWMA[1,] = c(S)[c(1,4,2)]
for (i in 2:TT){
S = lambda * S + (1-lambda) * t(y[i-1]) %*% y[i-1]
EWMA[i,] = c(S)[c(1,4,2)]
}
EWMArho = EWMA[,3]/sqrt(EWMA[,1]*EWMA[,2])
##### Listing 3.4: EWMA in Matlab Last edited: 2011
EWMA = nan(T,3);
lambda = 0.94
S = cov(y)
EWMA(1,:) = S([1,4,2]);
for i = 2:T
S = lambda * S + (1-lambda) * y(i,:)' * y(i,:)
EWMA(i,:) = S([1,4,2]);
end
EWMArho = EWMA(:,3) ./ sqrt(EWMA(:,1) .* EWMA(:,2))
##### Listing 3.5: OGARCH in R Last edited: 2011
library(gogarch)
res = gogarch(y,formula = ~garch(1,1),garchlist = c(include.mean=FALSE))
OOrho = ccor(res)
##### Listing 3.6: OGARCH in Matlab Last edited: August 2016
[par, Ht] = o_mvgarch(y,2, 1,1,1);
Ht = reshape(Ht,4,T)';
OOrho = Ht(:,3) ./ sqrt(Ht(:,1) .* Ht(:,4));
##### Listing 3.7: DCC in R Last edited: 2011
library(ccgarch)
f1 = garchFit(~ garch(1,1), data=y[,1],include.mean=FALSE)
f1 = f1@fit$coef f2 = garchFit(~ garch(1,1), data=y[,2],include.mean=FALSE) f2 = f2@fit$coef
a = c(f1[1], f2[1])
A = diag(c(f1[2],f2[2]))
B = diag(c(f1[3], f2[3]))
dccpara = c(0.2,0.6)
dccresults = dcc.estimation(inia=a, iniA=A, iniB=B, ini.dcc=dccpara,dvar=y, model="diagonal")
DCCrho = dccresults\$DCC[,2]
##### Listing 3.8: DCC in Matlab Last edited: August 2016
[p, lik, Ht] = dcc(y,1,1,1,1)
Ht = reshape(Ht,4,T)';
DCCrho = Ht(:,3) ./ sqrt(Ht(:,1) .* Ht(:,4));
##### Listing 3.9: Correlation comparison in R Last edited: 2011
matplot(cbind(EWMArho,DCCrho,OOrho),type='l',las=1,lty=1:3,col=1:3,ylab="")
mtext("Correlations",side=2,line=0.3,at=1,las=1,cex=0.8)
legend(2100,0,c("EWMA","DCC","OO"),lty=1:3,col=1:3,bty="n",cex=0.7)
##### Listing 3.10: Correlation comparison in Matlab Last edited: 2011
plot([EWMArho,OOrho,DCCrho])
legend('EWMA','DCC','OOrho','Location','SouthWest') | 1,163 | 2,972 | {"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": 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} | 3.140625 | 3 | CC-MAIN-2017-13 | longest | en | 0.690005 |
https://www.techcrashcourse.com/2016/01/print-hollow-square-star-pattern-in-c.html | 1,723,186,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00345.warc.gz | 791,969,689 | 22,702 | # C Program to Print Hollow Square Star Pattern
Here is a C program to print outline or boundary of a square pattern by star character using loop. For a hollow square star pattern of side 5 stars. Program's output should be:
Printing hollow square star patterns is a common programming exercise that helps beginners understand loops, conditional statements, and basic arithmetic operations in C.
#### Required Knowledge
Algorithm to print hollow square star pattern using loop
The algorithm of this pattern is similar to square star pattern except here we have to print stars of first and last rows and columns. At non-boundary position, we will only print a space character.
• Take the number of stars in each side of square as input from user using scanf function. Let it be N.
• We will use two for loops to print square star pattern.
• Outer for loop will iterate N times. In one iteration, it will print one row of pattern.
• Inside inner for loop, we will add a if statement check to find the boundary positions of the pattern and print star (*) accordingly.
Here is the matrix representation of the hollow square star pattern. The row numbers are represented by i whereas column numbers are represented by j.
## C program to print hollow square star pattern
```#include<stdio.h>
int main(){
int side, i, j;
printf("Enter side of square\n");
scanf("%d", &side);
/* Row iterator for loop */
for(i = 0; i < side; i++){
/* Column iterator for loop */
for(j = 0; j < side; j++){
/* Check if currely position is a boundary position */
if(i==0 || i==side-1 || j==0 || j==side-1)
printf("*");
else
printf(" ");
}
printf("\n");
}
return 0;
}
```
Output
```Enter side of square
5
*****
* *
* *
* *
*****
```
## Importance of Practicing Hollow Square Star Pattern Printing Programs for Beginners
• Understanding Nested Loops and Conditional Statements : This program helps people who are just starting out understand nested loops and conditional statements, which are very important for controlling how a program runs.
• Boosting Confidence : Writing and running this program successfully can boost a beginner's confidence by showing that they can use programming ideas in the real world.
• Learning Basic Programming Ideas : This program teaches basic programming ideas like loops, conditional statements, and manipulating variables, which are the building blocks of programming.
### Conclusion
Finally, the Hollow Square Star Pattern Printing Program in C is a good practice for beginners to help them learn more about loops, solve problems, find bugs, and be more creative in their programming. Beginners should use this program to get a feel for basic programming ideas and boost their confidence in their programming skills. Beginners can write this program well and build a strong base for their programming journey by following the tips given.
Related Topics
C program hollow rectangle pattern C program hollow diamond star pattern C program hollow pyramid star pattern C program square star pattern C program right triangle star pattern C program pyramid star pattern C program heart shape star pattern C program rhombus star pattern C program natural number triangle pattern List of all C pattern programs | 651 | 3,234 | {"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} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.810759 |
https://brilliant.org/practice/place-value/ | 1,503,530,731,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886124662.41/warc/CC-MAIN-20170823225412-20170824005412-00182.warc.gz | 763,244,428 | 15,285 | ###### Waste less time on Facebook — follow Brilliant.
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Back to all chapters
# Decimals
Coins are often used to express a fraction of a currency. In the US, a quarter is $0.25 and a dime is$0.10. Counting change is just one of many ways you can put decimals to use in everyday life.
# Place Value
Solving $$3.2\times2.2,$$ which number do we get in the hundredths place?
In the number $$0.0 8 3 5,$$ the digit $$3$$ is in which place?
Round up $$0.004 0 9 5$$ to the nearest tenthousandth.
Solving $$7.57-3.23,$$ which number do we have for the tenths place?
What number is in the hundredths place of 662.373?
× | 184 | 621 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-34 | longest | en | 0.914706 |
https://cs.stackexchange.com/questions/90273/among-a-number-of-sets-how-to-find-the-one-that-includes-the-highest-number-of | 1,660,878,111,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573540.20/warc/CC-MAIN-20220819005802-20220819035802-00452.warc.gz | 197,883,161 | 66,637 | # Among a number of sets, how to find the one that includes the highest number of other sets?
I have a large number of sets, A, B, C, ... where each set includes a few integers. I would like to find the set that includes the highest number of other sets.
A brute-force solution is to compare each set with all other sets and count the number of sets that are included in this set. Finally, pick the set with the highest count. This brute-force solution has a quadratic complexity and is very slow for my large number of sets. Is there a better solution with lower complexity?
• What sort of metric are you using for "better?" Is it just complexity? I'm thinking Bloom Filters may be highly applicable here, but because they're a hashing solution, the worst-case complexity isn't much better, even though the average-case complexity is probably much better. Apr 5, 2018 at 20:16
• and quick check; your metric for the best set to pick is that it is the set which is a superclass of the largest number of other sets? Apr 5, 2018 at 20:18
• @CortAmmon Yes, that's exactly the metric I'm considering. I like to improve the runtime on my machine, so a solution with a better average-case complexity should work well.
– Matt
Apr 5, 2018 at 20:44
• The related problem of testing whether there exists any set that contains some other set (i.e., whether the answer is 1 or 0) is known as Subset Containment. This has been studied: see cstheory.stackexchange.com/q/37361/5038, cstheory.stackexchange.com/q/9896/5038. Perhaps you can apply similar ideas to your problem as well? Also related: cstheory.stackexchange.com/q/17404/5038, cs.stackexchange.com/q/75915/755.
– D.W.
Apr 5, 2018 at 21:53
• What is the size of those integers?
– orlp
Apr 7, 2018 at 2:13
From the answer linked in D.W.'s comments, you wont be able to do this in less than quadratic time without more information. The problem you are trying to solve is more difficult than the Subset Containment problem, which is quadratic with the number of sets. We can, however, explore ways to make the average case better.
The first step is to identify if there is any pattern to the data which can be leveraged. For example, if the sets are random, then you can leverage that to find better runtimes with some partitioning tricks. If the values all happen to be less than 64, you could encode each set as a 64-bit numbers and do comparisons that way.
The most generic approach, however, is to try to minimize comparison time. The first step would be to sort the elements in each set. That makes it so you can always determine if one set is a subset of the other in linear time by scanning it.
Beyond that, you could look at bloom filters. A bloom filter is basically an array of bits, and each bit corresponds to the result of hashing the data differently. What is nice about bloom filters is that they allow you to combine values using OR logic. Use a bloom filter to hash each of the elements in a set, and OR the results together. Now it is easy to see that if you have a set $S_1$ and $S_2$ such that $S_1\subseteq S_2$ and you have their corresponding bloom filter arrays $B_1$ and $B_2$, then every bit in $B_1$ must also be set in $B_2$ (i.e. $B_2\land B_1=B_1$).
Thus, you will have to do $O(n^2)$ comparisons of the bloom filter arrays (which are bit-fields, so you can process them 32 or 64 bits at a time natively). If $B_i\land B_j=B_j$ for any $i, j$, then you have a potential subset. It's possible that you just had a bloom filter collision, so we have to test the arrays, which is linear with the size of each array.
Thus in the worst case, we don't save any time, but in the average case, we may be able to dramatically reduce the cost of comparing the sets.
There's a few more steps involved to make sure we don't saturate the bloom filters, but those can only be worked out when you know a little about the sizes of your sets, which I do not.
• The size of each set is not known in advance. In some problems it may be less than 200, but in some problems around 1000 or even more.
– Matt
Apr 9, 2018 at 6:36
• So it's definitely beyond the scope of my answer, but analyzing the size of the sets would be O(n) time, so you may want to take some time to analyze them and pick the best ways to do the sorting. Apr 10, 2018 at 17:48 | 1,089 | 4,309 | {"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": 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} | 3.125 | 3 | CC-MAIN-2022-33 | longest | en | 0.961544 |
https://ell.stackexchange.com/questions/22803/which-way-one-and-one-are-two-one-and-one-is-two/23010 | 1,719,356,754,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00564.warc.gz | 200,431,135 | 46,166 | # Which way: One and one ARE two? One and one IS two?
Which verb is grammatically correct when used to describe addition?
• One and one are two.
• One and one is two.
• There's a children's joke about this question in the U.S. "Do I say 'two and three is six' or 'two and three are six'?" "Neither: two plus three equals five!" Commented May 5, 2014 at 21:05
• @PeterShor there's still the distinction between "equals" for singular and "equal" for plural. Commented Jun 8, 2018 at 10:06
• Why not "one add one"? Commented Jan 8, 2020 at 1:37
It would be grammatically correct to use "are" if the subjects were indeed "two" individually, but they are not.
By saying "One and one are two," that means that each "one" is two. The equivalent would be, "One is two, and one is two."
Saying "One and one is two" groups "one and one" to be the subject of the sentence. And "one and one" is two. ;)
In your post, you said "which question is grammatically correct?" You would ask, "Is one and one two?" Although, that can be confusing without something to separate the "one" and the "two" at the end. The preferred method would be "Does one plus one equal two?"
• I'd note that "one and one are two" could also be okay if you assume some grammatically-key words were left out: "One and one[, combined, are equivalent to] two." Nonetheless, your answer is right on the money. Commented May 5, 2014 at 18:59
• What silliness to think of "one and one" as a singular noun! Commented May 5, 2014 at 23:02
• @KyleHale, it's the single number "two" which the verb must agree with. Commented May 5, 2014 at 23:22
• @PhilPerry What? Verbs don't agree with objects. Dick and Jane has a dog Spot. Commented May 6, 2014 at 2:58
• Hmm, I don't think that reasoning follows. If I say, "Bob and Mary are a couple", I don't mean that Bob is a couple and that Mary is a couple. I mean that taken together, the two are a couple. Or, "The engine and the transmission are connected." I mean that they are connected to each other, not that each is connected to itself. Etc.
– Jay
Commented May 6, 2014 at 13:45
# Fluent English speakers routinely say it both ways.
Logically, I think it should be "one and one are". By the normal rules of grammar, that is a compound subject. We wouldn't say "Bob and Charlie is ..."; we say "Bob and Charlie are ..." Etc.
@Snailplane's deleted answer—I don't know why it's deleted, it seems a valid answer to me—makes the interesting point that we sometimes use such compounds to refer to a single unit, like "Peanut butter and jelly is my favorite sandwich." I think the key there is that the words surrounding the "and" are a name or a title, like of course we would say, "Pride and Prejudice IS Sally's favorite book", not "... are Sally's favorite book", because we're talking about one book whose title happens to have the word "and" in it. It's not like Sally likes a book called Pride and she also likes a book called Prejudice. Do "one and one" in this sentence fall into that category? I don't think so.
Even when the point of a sentence is to say that two are more things are joined in some way, we still use "are". "Bob and Mary are a couple." "Smith, Jones, and Brown are a dangerous gang." "The four legs are what hold up the table." Etc.
So if you go by common usage, either is acceptable, but "is" is slightly preferred. If you go by conventional rules of grammar, I think "are" is correct. Obviously others answering on here disagree with me. Which, perhaps, is why we see the split in common usage.
I think you should feel free to use whichever you prefer. In day-to-day usage no one is likely to even notice. If you have a teacher or an editor who insists that one is wrong, I'd just do whatever they ask for rather than argue about it.
• I wasn't really satisfied with my answer. It's a really complicated question to answer generally, I think. I was thinking that, because plural agreement is so common, it might be helpful to start from the assumption that plural is the default and explain the cases where singular is possible (or even required) as exceptions.
– user230
Commented May 9, 2014 at 2:43
• @Jay If you're still around and you have a minute, you should check out your ngram again. It's mostly full of false positives, and the correct search produces the opposite conclusion.
– lly
Commented Jun 15, 2018 at 19:16
• @lly Oh, while if my query was flawed, that's great, because it means that the consensus agrees with how I say I think it ought to be.
– Jay
Commented Jun 18, 2018 at 15:02
# It's a bit of a mess.
Fear and Loathing... is the truncated name of a book and movie; a bacon, lettuce, and tomato is a kind of sandwich and (well done) the pinnacle of human cuisine. Always. These days, the United States is a country. Almost always.
It's neither true that one and one is obviously and only plural nor that every side of an equation is obviously and only singular. It sits in that odd place in English grammar where 'favorite' questions are. You can speak pompously ("My favorite food is apple" or "the apple") or colloquially ("My favorite food is/are apples") but the real solution is usually just to completely rephrase ("I like apples").
# 'Are' is better...
You can't usually rephrase statements of addition. You aren't even trying to; it's more often an educational mantra or an expression of a truism. The normal thing to do with a compound subject is to use the plural, and that's what usually happens.
Two apples and two apples are...
Two apples and two apples make...
Two apples and two apples equal...
is almost exclusively the way people will express that idea.
# ...but 'is' isn't wrong...
When you take the objects out and start making it more purely math, it becomes a bit wonkier. It's more common to say
Two and two are... and
Two and two make... but
Two and two equals...
because we tend to say 'make' when talking about creating groups of objects but tend to say 'equal' when talking about equations, which have one group—all taken together as a unit—on either side.
Similarly, even though 'plus' is a preposition whose object (the second number) shouldn't change the grammatical number of the subject, what you actually see is
Two plus two equals... and
Two plus two is... but
Two plus two make...
because of the pull of the idea of 'make' referring to actual objects, while the others are talking about math on its own. (That irregularity completely disappears when using 'one plus one...' since the first 'object' being added is singular as well.)
Side point: It doesn't help that people often wrongly consider 'plus' and 'and' to be exactly equivalent grammatically, instead of semantically.
# ...because there're three factors at play.
The oddities happen because the addition can be considered as objects
One [apple] and one [apple] are two [apples].
One [apple] plus one [apple] is two [apples].
Two [apples] plus two [apples] are four [apples].
or as numbers combined together to form half of an equation
[The sum of] one and one is two.
[The sum of] one plus one is two.
[The sum of] two plus two is four.
or as noncount words in a grammatical construction.
One and one are two.
One plus one is two.
Two plus two is four.
The verb 'to make' pulls more people towards the first way of thinking; the verb 'to equal' pulls more people towards the second, especially because of the influence of similar sentences using 'plus'; and the verb 'to be' tends to (and should) follow the third, but with some bleed-over from the other ways. The grammatical sense also takes over in abstract cases with variables instead of numbers.
• @snailboat Is something like this what you were starting to say earlier?
– lly
Commented Jun 15, 2018 at 19:18
Either can be correct, but it depends on your context and meaning.
"One and one is two" is grammatically correct if you are using "and" to mean "plus" (addition). Adding the number one with the number one produces the number two, which is a singular thing, therefore "one and one" (one plus one) is singular.
A clearer (and thus arguably better) way to say this, however, would be "one plus one equals two".
On the other hand, "One and one are two" is grammatically correct if you are using "and" to mean "grouped with" or "put together". That is, "one (of these) and one (of those) are two (things put together)".
"One and two are numbers." "One and two is three."
As others have pointed out, it depends on whether "one and two" are taken separately or combined. I'd say the grammar points strongly to "one and two is three", because "one and two /are/ three" suggests that one is three and two is three -- which is false!
The short answer is that "is" would be more appropriate, despite the popular Len Barry lyric: "One and one are two". If you look up the word "and" in a dictionary, you'll see that it can be used to mean "plus", in addition to being used to mean what it more frequently does. Sentences that begin with two nouns joined by an "and" being used to mean "plus" typically are followed by a singular verb.
Consider the following examples of such sentences:
Two and two makes four.
One and five equals six.
In these examples, the singular verbs "makes" and "equals" are used, not the plural verbs "make" and "equal". It is no different for the verb "to be". Use "is" instead of "are".
I would say that "is" could be correct, since the sentence "One and one is two" could be a truncated (or elided) version of "The result you get by adding one and one is two". "The result" is singular, therefore I would use "is."
• You just changed the subject of the sentence. "The result" is plainly singular. This doesn't apply to compound subjects combined with conjunctions. Commented May 7, 2014 at 7:43
• Pace the downvotes, he's completely right. Using a singular verb with the obviously compound subject depends on seeing them as implicitly joined. They're obviously not a title or name; the speaker uses is two because they see the one and one as a single mathematical operation.
– lly
Commented Jun 10, 2018 at 22:55
I revise my previous comment to the OP. I don't think "is" is correct at all, though it is most often used (at least in America).
"He and she are a couple."
When using "and" to combine two singular nouns in the subject, you are supposed to use a plural verb. "And" conjoins while "or" does not. | 2,580 | 10,382 | {"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} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.979662 |
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