url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://help.geogebra.org/topic/i-need-some-help-trying-to-make-automatic-geometric-constructions | 1,619,027,473,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039546945.85/warc/CC-MAIN-20210421161025-20210421191025-00557.warc.gz | 397,791,362 | 14,626 | # I need some help trying to make automatic geometric constructions
Arturo shared this question 2 months ago
Greetings, I hope the community could help me with this project I'm doing,
I'm trying to create a nxn grid where at the center of each square of 1x1 there must be a number which is calculated using congruence mod n.
My questions are as follow:
1.- Can I create ,given a number n from an input box, any number of constructions using that number? For example if I enter 3 in the input box, it will create 3 lines, circles, etc.
2.- It is possible to create an animation after the calculations of the n^2 have been calculated? That is the animations doesn't have a set number of steps until the user puts the information in an input box.
I had the idea of using scripts or the javascript integration but, I can't see any command for creating objects from the scripts rather than only manipulate them.
The grid is an easy one using the Sequence command. The same way you can create points and texts positioned at defined positions, but your description is not very concrete.
chris
Files: grid.ggb
1
The grid is an easy one using the Sequence command. The same way you can create points and texts positioned at defined positions, but your description is not very concrete.
chris
Files: grid.ggb
1
This is exactly what I have been looking for, thank you very much you are savior, I'm sorry for not looking enough into all the commands geogebra has, I thought it could be done trough scripting but with what you have said It's a lot easier.
Can I store the results of the sequence command in order to use them for the next part of doing an animation where I put them at the center of each square?
Thank you for everything Chris.
1
It's not clear for me what you mean, but you might as well use a similar sequence command to create the points at the center of the squares. Note that in fact there are no squares, just lines.
Sequence(Sequence((.5+nn,.5+mm), nn, 0, n-1), mm, 0, n-1) | 460 | 2,003 | {"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-2021-17 | longest | en | 0.935285 |
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# Wishball: tournament
TLF ID L8460
Challenge your understanding of place value in whole numbers and decimal fractions, from 0.001 to 9999. Either add or subtract numbers to reach a target number. For example, receive a starting number of 39.61. Spin the number 5 and decide whether to add or subtract 0.05, 0.5, 5 or 50 to reach your target number of 70.12 within 20 turns. Use the ‘Wishball’ to select your final digit. Try to reach the target in as few turns as possible. Play a random game, replay a previous game or play a game with the same target number as someone else. This learning object is one in a series of 15 objects.
#### Educational details
Key learning objectives
• Students identify the place value of each digit in four-digit numbers from 0.001 to 9999.
• Students add and subtract in thousandths, hundredths, tenths, ones, tens, hundreds and thousands.
• Students read and interpret different representations of numbers from 0.001 to 9999.
• Students apply mental computation skills.
Educational value
• Develops students’ understanding of place value.
• Stimulates students’ mental calculation strategies.
• Uses three different methods of representing numbers to assist students’ visualisation of place value.
• Scaffolds students’ attempts with immediate feedback.
• Encourages repeated use through randomisation of activity elements.
• Includes an option to print a record of turns taken for each game.
• Includes options to play a random game, repeat a game, or to play the same game as others.
Year level
6
Learning area
• mathematics
Strand
• Mathematics/Number and algebra
Student activity
• Interactives;
• Games;
• Experiment;
• Analysis;
• Modelling
#### Other details
Contributors
• Technical implementer
• Script writer
• Name: Michael Wagner
• Organisation: Godwin + Wagner
• Address: Box Hill South VIC 3128 Australia
• Subject matter expert
• Name: John Flatt
• Address: Geelong VIC 3220 Australia
• Subject matter expert
• Name: Howard Reeves
• Address: Lindisfarne TAS 7015 Australia
• Technical implementer
• Organisation: Education Services Australia
• Address: Melbourne VIC 3000 Australia
• URL: http://www.esa.edu.au/
• Publisher
• Date of contribution: 20 Sep 2013
• Organisation: Education Services Australia
• Address: Melbourne VIC 3000 Australia
• URL: http://www.esa.edu.au/
Access profile
• Colour independence
• Device independence
• Hearing independence
Learning resource type
• Interactive Resource
Browsers
• Microsoft Internet Explorer - minimum version: 8.0 (MS-Windows) - maximum version: 9.0 (MS-Windows)
• Firefox - minimum version: (MS-Windows)
• Safari - minimum version: 5.1 (MacOS)
Operating systems
• MacOS - minimum version: 10.6
• MS-Windows - minimum version: XP - maximum version: 7
Rights
• © Education Services Australia Ltd, 2013, except where indicated under Acknowledgements. | 738 | 3,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-40 | latest | en | 0.837645 |
http://mathforum.org/library/drmath/view/58819.html | 1,521,679,210,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647707.33/warc/CC-MAIN-20180321234947-20180322014947-00370.warc.gz | 188,033,212 | 3,139 | Associated Topics || Dr. Math Home || Search Dr. Math
### Putting Division into Words
```
Date: 04/06/99 at 09:55:28
From: nicole thiel
Subject: Division
How many times does 4 go into 600? I know the answer is 150, but I
I have to explain how I got the answer and I am having a hard time
putting it into words.
```
```
Date: 04/06/99 at 12:57:16
From: Doctor Peterson
Subject: Re: Division
Hi, Nicole.
Since I don't know how YOU got the answer, I can't put it into words
for you, but I can suggest how I might do it without having learned
division yet.
Suppose I have 600 sticks, and I want to divide them among four
people. Rather than count them out individually, I want to make use of
what I know from the number: that they come in 6 bundles of 100.
I can start by giving each person a bundle, so each has 100 sticks and
I have 2 bundles left over. I've divided 6 by 4 and gotten a quotient
of 1, with a remainder of 2.
Now how can I divide the remaining 2 bundles among four people? I have
to untie the bundles. Luckily, I find that each bundle is really ten
bundles of ten sticks tied together - that is, 1 hundred is the same
as 10 tens. So now I have 20 bundles of 10.
I can divide 20 by 4 to get exactly 5, so I give each person 5 bundles
of ten. Now each of them has 1 hundred and 5 tens, making a total of
100 + 50 = 150. Since I have none left, we're done, and you have your
This is just what we do when we divide numbers with several digits. I
don't know whether you've learned this yet, so I'll just write it out
and you can figure out what it means by comparing it with what I've
said:
_150_
4 )600
4 <-- 4 hundreds (1 x 4) distributed
-
20 <-- 2 hundreds (20 tens) left
20 <-- 20 tens (5 x 4) distributed
--
00 <-- nothing left
0 <-- nothing more distributed
--
0
Does that fit with your own reasoning at all? If not, it's a good way
to think.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Elementary Division
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 646 | 2,373 | {"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} | 4.09375 | 4 | CC-MAIN-2018-13 | latest | en | 0.951347 |
https://cs.stackexchange.com/questions/6718/conp-and-limitation-of-ndtm/6721 | 1,579,425,343,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00062.warc.gz | 395,143,873 | 32,218 | # coNP and limitation of NDTM
I am trying to understand if someone can apply an NTM to recognize a $$coNP$$ language.
From the definition we know that:
$$NP$$ - set of languages that can be recognized by NTM in polynomial time.
$$coNP$$ - set of all languages that are complement to $$NP$$ language.
as with $$P$$ vs. $$NP$$ question, we have $$NP$$ versus $$coNP$$ question.
Unfortunately, is not defined explicitly if can one recognize $$coNP$$ language with NTM.
However, if we take a look at few examples from the set of $$coNP$$ languages, few questions emerge.
$$TAUTOLOGY$$ = {$$\varphi$$: $$\varphi$$ is satisfied by every assignment}
$$UNSAT$$ = {$$\varphi$$: $$\varphi$$ is unsatisfiable}
These languages are known to be $$coNP$$ language and intuitively it seems like one can construct NTM to recognize these languages. On the other hand, if one can construct NTM to recognize them why them not in $$NP$$ class (by definition)? Maybe not all languages of $$coNP$$ can be solved by NTM, just few of them which are in $$NP\cap coNP$$. And if every language from $$coNP$$ class cannot be solved by NTM, does it mean that limitation of NTM is located in $$coNP$$. Does NTM have limitation at all?
I am a little bit confused, I would appreciate anyone that is willing to shine the light on this topic.
• Your problem lies in the phrase "to solve a language". One does not solve a language, one recognises a language. Specifically, NP is about recognising whether a word is accepted by an NTM in polynomial-time. co-NP is, in essence, recongnising whether the word is rejected by an NTM in polynomial time. NP says nothing about the time taken to reject words. – Dave Clarke Nov 17 '12 at 17:52
• @DaveClarke, thank you very much for the comment, I've edited the question. So all languages in coNP are recognized by NTM, if so, does NP and coNP intersect? – com Nov 17 '12 at 19:00
• co-NP languages are certainly recognized by an NTM. But it is unknown whether it is a polynomial-time NTM. NP and co-NP certainly intersect: P, for example, lies in that intersection. – Dave Clarke Nov 17 '12 at 19:03
• @DaveClarke, thank you very much for the answer! – com Nov 17 '12 at 19:20
These complexity classes are not just defined by a machine model but also by the criteria about when a machine accepts a strings. So although $\mathsf{NP}$ and $\mathsf{coNP}$ can be defined using essentially the same underlining machine structure their accepting criteria are different. That is the main point you are missing.
In my experience, generally it is intuitively more clear for people to think about $\mathsf{NP}$ using its verifier-certificate definition and forget about the non-deterministic TM (which can be confusing).
We say a language $L$ is in $\mathsf{NP}$ if there is a polytime DTM with two inputs $V$ (think of $V$ as a certificate verifier) s.t. for all $x$, $x \in L$ iff there exists a polynomial size certificate $y$ s.t. $V$ accepts $(x,y)$.
In more intuitive terms, $\mathsf{NP}$ is the set of problems that given an instance and a polysize solution/certificate/proof for that instance, we can efficiently verify the correctness of the given solution. E.g. given a graph and a list of vertices in it, it is possible in polynomial time to check if the list is a Hamiltonian path in the graph. Therefore Hamiltonian path problem is in $\mathsf{NP}$.
So in intuitive terms, $\mathsf{NP}$ is the set of problems that a given solution can be confirmed efficiently and $\mathsf{coNP}$ is the set of problems that a given solution can be refuted efficiently. For example, a refutation for a $TAUT$ instance is an assignment that makes the formula false. | 918 | 3,672 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-05 | latest | en | 0.961422 |
https://www.physicsforums.com/threads/geometric-distribution-problem.62883/ | 1,508,611,504,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824824.76/warc/CC-MAIN-20171021171712-20171021191712-00544.warc.gz | 969,785,235 | 15,273 | # Geometric distribution problem
1. Feb 6, 2005
### semidevil
a couple decides that they will have kids until a girl is born. the outcome of each birth is independent event, and the probability that a girl will be born is 1/2. The birht at which the first girl appears is a geometric distribution. what is the expected family size.
ok, so we know that the probability of having a girl is 1/2.
geometric distribution formula, it is the sum from 1 to k of (1- p) * (p)^k, where k = 1, 2, 3, 4,......k.
but when I tihnk about it, I have an expected value formula, where E(X) = 1/p. so if I put 1/(1/2), I get the answer is 2. So the expected family size is 2?
I don't know..I have this geometric formula that I dont know what to do with, and I have this expected value formula that makes it seem this problem is too easy....
any tips?
2. Feb 6, 2005
### vincentchan
the answer is 2......
don't over-complicated the problem
3. Feb 7, 2005
### xanthym
An important point is there are actually several slightly different forms of the Geometric Distribution, and each has a slightly different E(x). For example, the E(x) you quoted in your msg is NOT correct for the specific Discrete Geometric Density function you presented.
Let's begin with the Geometric Density function you presented, which indicates the probability that success will be achieved on the (k+1)-th trial after "k" failures for an event whose probability of success is (0 < p < 1):
$$P(X=k) = p*(1 - p)^{k} \ \ \ k=0,1,2,3,...$$
The expected value E(x) is then given by:
$$E(X) = \sum\limits_{k = 0}^{\inf} k*p*(1 - p)^{k}$$
$$E(X) = p*\sum\limits_{k = 0}^{\inf} k*(1 - p)^{k} \ \ \color{green} Eq:1$$
$$E(X) = p*\sum\limits_{k = 1}^{\inf} k*(1 - p)^{k} \ \ \ \color{green} Eq:2$$
Multiply both sides of Eq #1 by (1 - p):
$$(1 - p)*E(X) = p*\sum\limits_{k = 0}^{\inf} k*(1 - p)^{k+1}$$
$$(1 - p)*E(X) = p*\sum\limits_{k = 1}^{\inf} (k - 1)*(1 - p)^{k} \ \ \ \color{green} Eq:3$$
Now subtract Eq #3 from Eq #2:
$$E(X) - (1 - p)*E(X) = p*\sum\limits_{k = 1}^{\inf} [(k) - (k - 1)]*(1 - p)^{k}$$
$$p*E(X) = p*\sum\limits_{k = 1}^{\inf} (1 - p)^{k}$$
$$E(X) = \sum\limits_{k = 1}^{\inf} (1 - p)^{k} \ \ \ \color{green} Eq:4$$
The infinite sum in Eq #4 is a geometric series with term ratio (1 - p), so that it may be written:
$$E(X) = \frac {1 - p} {1 - (1 - p)}$$
$$E(X) = \frac {1 - p} {p}$$
Thus, E(X) for your presented Geometric Distribution is (1-p)/p and NOT (1/p). However, the conclusions are the same. The average family will consist of (1-p)/p members PLUS 1 MORE because your presented distribution indicated the probability of "k" failures (with success being at the (k+1)-th trial), thus indicating:
$$(Average \ Family \ Size) = E(X) + 1 = 1 + \frac {1 - p} {p}$$
Thus, for p=(1/2):
$$\color{red} (Average \ Family \ Size \ for \ p=(1/2)) = 2$$
Incidentally, your quoted E(X)=(1/p) corresponds to the following form of the Geometric Distribution:
$$P(X=k) = p*(1 - p)^{k - 1} \ \ \ \ k = 1,2,3,...$$
Last edited: Feb 7, 2005 | 1,044 | 3,028 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-43 | longest | en | 0.885994 |
https://lessonplanet.com/search?keyterm_ids%5B%5D=251297 | 1,674,998,131,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00116.warc.gz | 389,112,139 | 39,635 | ### We found 3 resources with the keyterm subtracting across zeros
Videos (Over 2 Million Educational Videos Available)
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Interactive
Lesson Planet: Curated OER
#### Study Jams! Subtraction with Regrouping
For Students 2nd - 4th Standards
Learning how to regroup when subtracting numbers can be challenging and requires a solid understanding of place value. Walk your class through the steps with a clear and explicit presentation that addresses standard regrouping and...
Worksheet
Lesson Planet: Curated OER
#### Mixed Four Digit Borrowing Across Zero
For Students 4th - 5th
In this subtraction worksheet, students solve 40 problems in which four digit numbers are subtracted. Some of the problems require subtracting across zeros.
Worksheet
Lesson Planet: Curated OER
#### Subtracting Across Zeros
For Students 4th - 5th
In this subtracting across zeros activity, students analyze the information and examples for how to regroup multiple times if there are zeros in a series. Students then solve 9 problems. | 266 | 1,183 | {"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-2023-06 | latest | en | 0.829197 |
http://clay6.com/qa/21123/the-function-f-x-left-1-x-leq-1-x-1-x-1-0-x-geq-1-end-right- | 1,597,480,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740733.1/warc/CC-MAIN-20200815065105-20200815095105-00067.warc.gz | 22,979,017 | 6,237 | # The function $f(x)=\left\{\begin{array}{1 1}1&x\leq -1\\|x|&-1< x < 1\\0&x\geq 1\end{array}\right.$
$\begin{array}{1 1}(a)\;\text{differentiable for all x}\\(b)\;\text{f is continuous everywhere}\\(c)\;\text{f is differentiable at x=-1}\\(d)\;\text{f is continuous at x=-1}\end{array}$ | 121 | 287 | {"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} | 3.171875 | 3 | CC-MAIN-2020-34 | latest | en | 0.609715 |
https://physics-network.org/how-do-you-find-conservative-and-nonconservative-forces/ | 1,670,312,846,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00656.warc.gz | 477,372,018 | 17,190 | # How do you find conservative and nonconservative forces?
W A B , path – 1 = ∫ A B , path – 1 F → cons · d r → = W A B , path – 2 = ∫ A B , path – 2 F → cons · d r → . The work done by a non-conservative force depends on the path taken. Equivalently, a force is conservative if the work it does around any closed path is zero: W closed path = ∮ F → cons · d r → = 0 .
## What are conservative and nonconservative forces give examples?
Gravitational Force, Spring Force, and Electrostatic force between two electric charges are examples of conservative force. Friction, Air resistance, and Tension in the cord are examples of non-conservative force.
## How do you calculate non-conservative forces?
W nc = Δ KE + Δ PE . This equation means that the total mechanical energy(KE+PE) changes by exactly the amount of work done by nonconservative forces.
## What are conservative and non-conservative forces explain with examples Class 11?
Conservative forces are those for which work done depends only on initial and final points. Example- Gravitational force, Electrostatic force. Non-Conservative forces are those where the work done or the kinetic energy did depend on other factors such as the velocity or the particular path taken by the object.
## What is conservative force give an example?
A conservative force is a force with the property that the total work done in moving a particle between two points is independent of the taken path. For example, gravitational force, spring force, electrostatic force, etc.
## Is gravity a non conservative force?
Gravity is a conservative force because when the ball comes back down, it has just as much kinetic energy as it started with. The work done by gravity on the way up is , and the work done on the way down is , so if the ball comes back to its starting point, the total work done is .
## Is tension a non conservative force?
Tension is a non-conservative force, and therefore has no associated potential energy. When tension is internal, however, it is a non-dissipative force, performing zero net work on the chosen system.
## Why is friction not a conservative force?
Friction is non-conservative because the amount of work done by friction depends on the path. One can associate a potential energy with a conservative force but not with a non-conservative force.
## Do non-conservative forces always oppose motion?
Option (c) is the correct answer: The statement “A non-conservative force always opposes motion” is false for the non-conservative forces. Further Explanation: There can be some conditions in which the non-conservative forces makes the body move forward and doesn’t oppose its motion.
## Which force is non-conservative force?
A nonconservative force is one for which work depends on the path taken. Friction is a good example of a nonconservative force. As illustrated in Figure 7.14, work done against friction depends on the length of the path between the starting and ending points.
## Can non-conservative forces do work?
Non-conservative forces can also do positive work thereby increasing the total mechanical energy of the system. The energy transferred to overcome friction depends on the distance covered and is converted to thermal energy which can’t be recovered by the system.
## Is nuclear force non conservative?
Yes, the weak nuclear force is conservative. This means that energy is not lost when the weak nuclear force acts on quarks in the nucleus of an atom.
## Is magnetic force non conservative?
Moving from one point to another in the magnetic field then implies that the work done between two points is not independent of the taken path. So by definition the magnetic field is non-conservative.
## Is kinetic energy a conservative force?
If the kinetic energy is the same after a round trip, the force is a conservative force, or at least is acting as a conservative force. Consider gravity; you throw a ball straight up, and it leaves your hand with a certain amount of kinetic energy.
## Which of the following is an example of a nonconservative force?
The correct answer is Frictional force. The frictional force is a non-conservative force.
## Is weight a conservative force?
Weight of an Object The weight of the object does not depend on the path taken by it while moving. The force of gravity only depends upon the position of the object. Hence, the weight of an object is one of the prime examples of a conservative force.
## Why magnetic field is non conservative?
Because the lines of forces formed by it are circular (closed loop) and that is a property of non conservative field.
## Is normal force a non-conservative force?
The normal force is closely related to the friction force. Both are non-conservative forces, which can be seen when a ball bounces.
## How much work did nonconservative forces do?
The work done by non-conservative forces is equal to the change in mechanical energy.
## What is the ratio of work done by conservative forces?
Answer: The work done by a conservative force is equal to the negative of change in potential energy during that process.
## Why push pull is non-conservative force?
A simple push or pull force (applied force) is also an example of a nonconservative force. The amount of work done on the object depends on the distance covered by an object while the force is being applied to it. Again, it depends on the path taken and may result in a change in the object’s mechanical energy.
## Is drag a conservative force?
Like friction, drag is a non-conservative force, meaning the work done by it is path dependent. And also like friction, drag produces heat and thus does not conserve mechanical energy.
## Is pushing a conservative force?
Gravitational force, spring force, and electric force are examples of conservative forces. Friction, air resistance, and push/pull force are examples of non-conservative forces.
## Are all conservative forces constant?
However it is not necessary that all constant force are conservative in nature. Friction can be constant for a body moving along a straight line with constant velocity under applied external force, but it is not conservative however. | 1,286 | 6,188 | {"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-2022-49 | latest | en | 0.930054 |
https://undergroundmathematics.org/product-rule/r7129 | 1,718,666,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00140.warc.gz | 519,471,188 | 5,141 | Review question
# Can we sketch $y=ax/(x^2+x+1)$ when $a$ is positive? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R7129
## Question
If $y=\frac{ax}{x^2+x+1} \qquad (a>0)$ show that for real values of $x$ $-a\leq y \leq \frac{1}{3}a.$
Sketch the graph of $y$ in the case $a=6$.
By drawing on the same diagram a certain straight line show that the equation $x^3-1=6x$ has three real roots, two negative and one positive. | 160 | 516 | {"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": 3, "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-2024-26 | latest | en | 0.754505 |
http://csep10.phys.utk.edu/OJTA2dev/ojta/c2c/binaries/visual/binaryap_tl.html | 1,569,056,326,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574377.11/warc/CC-MAIN-20190921084226-20190921110226-00051.warc.gz | 47,279,163 | 2,375 | #### Astrometric Binaries
Not all binaries are visible as two separate stars. In fact, in most binary systems that are known only one of the stars can be seen clearly. How then do we know that these other systems are binaries if we can't see the two stars? The two most important ways in which an unseen companion star can be inferred are by the gravitational influence of the unseen star on the visible one, and by effects in the observed spectrum. A binary inferred from the perturbation of its motion by an unseen companion is called an astrometric binary and the precise measurement of stellar motion required to find such small perturbations is a part of the field called astrometry. Astrometric binaries are addressed here. Binary systems deduced from effects on the spectrum are called spectroscopic binaries; they are the subject of a later module.
##### Proper Motion for Binary Stars
The following figure illustrates the proper motion on the celestial sphere of the Sirius system.
The Wobble of an Astrometric Binary The motion of the Sirius system about its center of mass causes the proper motion of the two stars across the celestial sphere to wobble, as illustrated in this animation. This can be used to detect the presence of an unseen companion. You may illustrate this by using the buttons in the binary star orbit applet to hide the orbits and to hide one of the stars. The corresponding motion of the other star is what we would see for Sirius A if it did not have proper motion on the celestial sphere. This motion, coupled with the proper motion of the Sirius system, gives the wobbling path noted above.
This diagram illustrates clearly that the motion of Sirius across the celestial sphere is a superposition of two motions: the revolution of the two stars around their center of mass, and the slow drift of the center of mass across the celestial sphere because of its proper motion. (In the image shown above the sizes of the stars are exaggerated to make them easy to see. In reality, Sirius A would be about 200 times larger than Sirius B, and each would just be points on this scale.)
##### If One Star Is Not Visible
Sirius is a visual binary, so we can see both stars and their true motion. If only the primary star of a binary star can be seen, it will appear to wobble in its proper motion across the celestial sphere because of the gravitational influence of the unseen companion. A binary system inferred from such wobbling motion of the primary is an astrometric binary.
##### Don't Confuse Binary Wobble with Parallax Wobble
This wobble produced by an unseen companion should not be confused with the apparent wobble in the motion caused by parallax for all stars because of the Earth's motion around the Sun. This animation illustrates the effect of parallax on proper motion. When we illustrate the wobbling motion of a binary system, the parallax effect has already been subtracted out of the motion. | 598 | 2,946 | {"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} | 2.640625 | 3 | CC-MAIN-2019-39 | latest | en | 0.947351 |
https://enacademic.com/dic.nsf/enwiki/3167761/Triple_product_rule | 1,611,021,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517559.41/warc/CC-MAIN-20210119011203-20210119041203-00757.warc.gz | 329,414,021 | 17,017 |
# Triple product rule
The triple product rule, known variously as the cyclic chain rule, cyclic relation, cyclical rule or Euler's chain rule, is a formula which relates partial derivatives of three interdependent variables. The rule finds application in thermodynamics, where frequently three variables can be related by a function of the form f(x, y, z) = 0, so each variable is given as an implicit function of the other two variables. For example, an equation of state for a fluid relates temperature, pressure, and volume in this manner. The triple product rule for such interrelated variables x, y, and z comes from using a reciprocity relation on the result of the implicit function theorem in two variables and is given by
$\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x\left(\frac{\partial z}{\partial x}\right)_y = -1.$
Note: The third variable is considered to be an implicit function of the other two.
Here the subscripts indicate which variables are held constant when the partial derivative is taken. That is, to explicitly compute the partial derivative of x with respect to y with z held constant, one would write x as a function of y and z and take the partial derivative of this function with respect to y only.
The advantage of the triple product rule is that by rearranging terms, one can derive a number of substitution identities which allow one to replace partial derivatives which are difficult to analytically evaluate, experimentally measure, or integrate with quotients of partial derivatives which are easier to work with. For example,
$\left(\frac{\partial x}{\partial y}\right)_z = - \frac{\left(\frac{\partial z}{\partial y}\right)_x}{\left(\frac{\partial z}{\partial x}\right)_y}$
Various other forms of the rule are present in the literature; these can be derived by permuting the variables {x, y, z}.
## Derivation
An informal derivation follows. Suppose that f(x, y, z) = 0. Write z as a function of x and y. Thus the total derivative dz is
$dz = \left(\frac{\partial z}{\partial x}\right)_y dx + \left(\frac{\partial z}{\partial y}\right)_x dy$
Suppose that we move along a curve with dz = 0, where the curve is parameterized by x. Thus y can be written in terms of x, so on this curve
$dy = \left(\frac{\partial y}{\partial x}\right)_z dx$
Therefore the equation for dz = 0 becomes
$0 = \left(\frac{\partial z}{\partial x}\right)_y \, dx + \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial x}\right)_z \, dx$
Since this must be true for all dx, rearranging terms gives
$\left(\frac{\partial z}{\partial x}\right)_y = -\left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial x}\right)_z$
Dividing by the derivatives on the right hand side gives the triple product rule
$\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_x\left(\frac{\partial z}{\partial x}\right)_y = -1$
Note that this proof makes many implicit assumptions regarding the existence of partial derivatives, the existence of the exact differential dz, the ability to construct a curve in some neighborhood with dz = 0, and the nonzero value of partial derivatives and their reciprocals. A formal proof based on mathematical analysis would eliminate these potential ambiguities.
## References
• Elliott, JR, and Lira, CT. Introductory Chemical Engineering Thermodynamics, 1st Ed., Prentice Hall PTR, 1999. p. 184.
• Carter, Ashley H. Classical and Statistical Thermodynamics, Prentice Hall, 2001, p. 392.
• Wang, C.-Y., and Hou, J.-H., Teaching Differentials in Thermodynamics Using Spatial Visualization, http://arxiv.org/abs/1110.6663.
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For these exercises, you will be using the SPSS dataset Polit2SetB. The analyses will focus on predicting the probability that a woman is in good-to-excellent health versus fair-to-poor health, which is coded 1 versus 0, respectively, on the variable health. Begin by looking at results for the odds ratio when there is only one predictor of good health—whether or not the woman currently smokes cigarettes (smoker). First, run the SPSS crosstabsbased risk analysis. In the Analyze ➜ Descriptives ➜ Crosstabs dialog box, use smoker as the Row variable and health as the Column variable. In the Cells dialog box, select Observed frequencies and Row and Column percentages. In the Statistics dialog box, select Chi-square and Risk. Then click Continue and OK to run the analysis, and answer these questions: (a) What percent of women in this sample smoked? What percent of women said they were in fair or poor health? (b) What percent of smokers versus nonsmokers described themselves as being in good to excellent health? (c) Is the bivariate relationship between smoking status and health statistically significant? (d) What is the odds ratio in this analysis? What is the 95% CI around the OR? (e) What does the odds ratio mean? | 278 | 1,254 | {"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-2023-14 | latest | en | 0.926448 |
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1 C oloring Discrete Structures Provided by TryEngineering - L e s s o n F o c u s Is it true or false that Discrete Structures and Discrete Mathematics are the same thing? This is the kind of question that is asked in this field or both fields if they are indeed different. Most Middle School students see a mix of discrete and continuous math without ever noticing the difference. This lesson introduces them to areas of mathematics that computer scientists use to do computational problems. Search techniques through discrete structures are illustrated through graph traversal and graph coloring. A g e L e v e l s Intended for (US Middle School grades 6-8) Can be used in lower High School (e.g. 9 th grade) O b j e c t i v e s Introduce students to: the relationship between Discrete Structures and Discrete Mathematics. the difference between discrete and continuous phenomenon. how to answer a question with sets. how a discrete problem is solved through search: specifically graph coloring. A n t i c i p a t e d L e a r n e r O u t c o m e s Students will be able to: explain the difference between continuous and discrete structures. discuss the difference in perspective of discrete math and discrete structures. form a problem statement as a logical proposition, a Venn diagram, and an adjacency graph. color a map with the least number of colors. A l i g n m e n t t o C u r r i c u l u m F r a m e w o r k s See attached curriculum alignment sheet. I n t e r n e t C o n n e c t i o n s (Continuous vs. Discrete) (Map coloring) R e c o m m e n d e d R e a d i n g O p t i o n a l W r i t i n g A c t i v i t y Is it true or false that Discrete Structures and Discrete Mathematics are the same thing? Defend your answer. Coloring Discrete Structures Page 1 of 10
4 b. If there were differences, call up two students who had different solutions, and ask them to stand at opposite sides of the room. c. Ask the rest of the class to join one of those students if they have that student s name on their sheet. d. If anyone remains seated, call up one of them and have them stand apart from the others. Ask the remaining seated students to join the new solution if that student s name is on their sheet. Do this until everyone is standing. Clearly there is more than one solution. 7. There might have been a flaw in the procedure in step 6. It s possible that students got it wrong that two solutions that appeared to be the same were different, or that two different, were the same. Ask everyone to check whether they have the name of someone in a different group on their sheet. If they do, and if you have time, check their answers and regroup. If you don t have time, simply use this as an object lesson, that a proof is only as good as the information coming in. But if this happens to end on a good note for this to happen. You still know that there is more than one solution, but you still don t know if the solution is unique. These are the kinds of problems solved with discrete structures! Time Needed 2 sessions, at most 1 hour each. You should be able to do the first two activities in a single session, and the coloring problem in the third. If you move more quickly, you can always have them create their own drawing and challenge a friend to color it in. Coloring Discrete Structures Page 4 of 10
5 Some classic four color examples Coloring Discrete Structures Page 5 of 10
6 C oloring Discrete Structures S t u d e n t R e s o u r c e : Discrete Mathematics Topics ( ematics) Theoretical computer science Information theory Logic Set theory Combinatorics Graph theory Probability Number theory Algebra Calculus of finite differences Geometry Topology Operations research Game theory, decision theory Discrete Structures Topics ( Discrete_structures) Sets Logic Boolean Algebra Proof Techniques Counting Graphs and trees Discrete Probability Boolean algebra Set theory From Meriam-Webster.com: Continuous: continuing without stopping; happening or exiting without a break or interruption Discrete: Map: Set (noun, def 2): Subset: separate and different from each other a picture or chart that shows the rivers, mountains, streets, etc. in a particular area; a picture or chart that shows the different parts of something a group of things of the same kind that belong or are used together a group of things, people, etc. that is part of a larger group From Wikipedia.org Set: In mathematics, a set is a collection of distinct objects, considered as an object in its own right. For example, the numbers 2, 4, 6 are distinct objects when considered separately, but when they are considered collectively they form a single set of size three. Set intersection: (paraphrased) In mathematics, the intersection of two sets A and B is the set that contains all of the elements A that also belong in B, but no others. In mathematics, the four color theorem, or the four color map theorem, states that, given any separation of a plane into contiguous regions, producing a figure called a map, no more than four colors are required to color the regions of the map so that no two adjacent regions have the same color.. Coloring Discrete Structures Page 6 of 10
7 C oloring Discrete Structures S t u d e n t W o r k s h e e t 1 : Is it true or false that Discrete Structures and Discrete Mathematics are the same thing? The question above poses a question in a way that we might be able to answer it with logic. It is a very simple research statement. To answer it definitively, you would have to do some research on the Internet to see whether reliable sources make a distinction or not, and who refers to structures and who refers to mathematics. A place to start to just address the question is to look at the topics considered in each. Using set theory, if the two sets of topics are equal, then they are the same. With a partner, use the lists of topics from the Student Resource Sheet to decide whether a topic is part of Discrete Mathematics and which are part of Discrete Structures. A Venn Diagram gives you a way to view the elements in a set. Use the Venn diagram below and the definitions on the Student Resource Sheet to place topics in the correct parts of the diagram. Using the definitions on the Student Resource Sheet: Are all topics in the intersection (are the sets equivalent)? Is one set a subset of the other? Math Structures Both Math and Structures Coloring Discrete Structures Page 7 of 10
8 C oloring Discrete Structures S t u d e n t W o r k s h e e t 2 : Can this picture be colored in with just for colors? The rule is that the same color can touch at a corner but it can t be used in two adjacent areas. Is there more than one way to color this picture with just four colors? You will answer this question as a class! Note to readers of the draft. The general outline show here will be used, however the graphic needs to be edited so that the tiny areas are not shown. This image can be four colored in more than one unique way. Proof by two pictures Coloring Discrete Structures Page 8 of 10
9 C oloring D iscrete Structures F o r T e a c h e r s : A l i g n m e n t t o C u r r i c u l u m F r a m e w o r k s Note: All lesson plans in this series are aligned to the Computer Science Teachers Association K-12 Computer Science Standards, and if applicable also the U.S. Common Core State Standards for Mathematics, the U. S. National Council of Teachers of Mathematics' Principles and Standards for School Mathematics, the International Technology Education Association's Standards for Technological Literacy, the U.S. National Science Education Standards and the U.S. Next Generation Science Standards. National Science Education Standards Grades 5-8 (ages 10-14) CONTENT STANDARD E: Science and Technology As a result of activities, all students should develop Understandings about science and technology National Science Education Standards Grades 9-12 (ages 14-18) CONTENT STANDARD E: Science and Technology As a result of activities, all students should develop Understandings about science and technology Next Generation Science Standards & Practices Gr.5-8 (ages 10-14) Practice 5: Using Mathematics and Computational Thinking Use mathematical representations to describe and/or support scientific conclusions and design solutions. Principles and Standards for School Mathematics (all ages) Problem Solving Standards Solve problems that arise in mathematics and other contexts Connections Recognize and apply mathematics in contexts outside of mathematics Representations Use representations to model and interpret physical, social, and mathematical phenomena Common Core State Practices & Standards for School Mathematics (all ages) CCSS.MATH.PRACTICE.MP1 Make sense of problems and persevere in solving them. CCSS.MATH.PRACTICE.MP4 Model with mathematics. CCSS.MATH.PRACTICE.MP7 Look for and make use of structure. Standards for Technological Literacy - All Ages Nature of Technology Standard 2: Students will develop an understanding of the core concepts of technology Coloring Discrete Structures Page 9 of 10
10 Coloring Discrete Structures F o r T e a c h e r s : A l i g n m e n t t o C u r r i c u l u m F r a m e w o r k s CSTA K-12 Computer Science Standards Grades 6-9 (ages 11-14) 2 Level 2: Computer Science and Community (L2) Computational Thinking: (CT) 8. Use visual representations of problem states, structures, and data (e.g., graphs, charts, network diagrams, flowcharts). 14. Examine connections between elements of mathematics and computer science including binary numbers, logic, sets and functions. CSTA K-12 Computer Science Standards Grades 9-12 (ages 14-18) 5.3 Level 3: Applying Concepts and Creating Real-World Solutions (L3) 5.3.A Computer Science in the Modern World (MW) Computing Practice and Programming (CPP) 12. Describe how mathematical and statistical functions, sets, and logic are used in computation. Coloring Discrete Structures Page 10 of 10
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Provided by TryEngineering - Lesson Focus Lesson focuses on how software engineers design computer games and other software. Student teams work together to develop a simple computer program using free | 7,972 | 36,525 | {"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} | 3.578125 | 4 | CC-MAIN-2018-51 | latest | en | 0.911248 |
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Sudoku_solving_algorithms | 1,675,902,751,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500983.76/warc/CC-MAIN-20230208222635-20230209012635-00872.warc.gz | 1,099,880,489 | 9,162 | # Sudoku solving algorithms
A standard Sudoku contains 81 cells, in a 9×9 grid, and has 9 boxes, each box being the intersection of the first, middle, or last 3 rows, and the first, middle, or last 3 columns. Each cell may contain a number from one to nine, and each number can only occur once in each row, column, and box. A Sudoku starts with some cells containing numbers (clues), and the goal is to solve the remaining cells. Proper Sudokus have one solution. Players and investigators may use a wide range of computer algorithms to solve Sudokus, study their properties, and make new puzzles, including Sudokus with interesting symmetries and other properties.
There are several computer algorithms that will solve most 9×9 puzzles (n=9) in fractions of a second, but combinatorial explosion occurs as n increases, creating limits to the properties of Sudokus that can be constructed, analyzed, and solved as n increases.
## Techniques
### Backtracking
Some hobbyists have developed computer programs that will solve Sudoku puzzles using a backtracking algorithm, which is a type of brute force search.[2] Backtracking is a depth-first search (in contrast to a breadth-first search), because it will completely explore one branch to a possible solution before moving to another branch. Although it has been established that approximately 5.96 x 1126 final grids exist, a brute force algorithm can be a practical method to solve Sudoku puzzles.
A brute force algorithm visits the empty cells in some order, filling in digits sequentially, or backtracking when the number is found to be not valid.[3][4][5][6] Briefly, a program would solve a puzzle by placing the digit "1" in the first cell and checking if it is allowed to be there. If there are no violations (checking row, column, and box constraints) then the algorithm advances to the next cell and places a "1" in that cell. When checking for violations, if it is discovered that the "1" is not allowed, the value is advanced to "2". If a cell is discovered where none of the 9 digits is allowed, then the algorithm leaves that cell blank and moves back to the previous cell. The value in that cell is then incremented by one. This is repeated until the allowed value in the last (81st) cell is discovered.
The animation shows how a Sudoku is solved with this method. The puzzle's clues (red numbers) remain fixed while the algorithm tests each unsolved cell with a possible solution. Notice that the algorithm may discard all the previously tested values if it finds the existing set does not fulfil the constraints of the Sudoku.
Advantages of this method are:
• A solution is guaranteed (as long as the puzzle is valid).
• Solving time is mostly unrelated to degree of difficulty.
• The algorithm (and therefore the program code) is simpler than other algorithms, especially compared to strong algorithms that ensure a solution to the most difficult puzzles.
The disadvantage of this method is that the solving time may be slow compared to algorithms modeled after deductive methods. One programmer reported that such an algorithm may typically require as few as 15,000 cycles, or as many as 900,000 cycles to solve a Sudoku, each cycle being the change in position of a "pointer" as it moves through the cells of a Sudoku.[7][8]
A Sudoku can be constructed to work against backtracking. Assuming the solver works from top to bottom (as in the animation), a puzzle with few clues (17), no clues in the top row, and has a solution "987654321" for the first row, would work in opposition to the algorithm. Thus the program would spend significant time "counting" upward before it arrives at the grid which satisfies the puzzle. In one case, a programmer found a brute force program required six hours to arrive at the solution for such a Sudoku (albeit using a 2008-era computer). Such a Sudoku can be solved nowadays in less than 30 seconds using an exhaustive search routine and faster processors.
### Stochastic search / optimization methods
Sudoku can be solved using stochastic (random-based) algorithms.[9][10] An example of this method is to:
1. Randomly assign numbers to the blank cells in the grid.
2. Calculate the number of errors.
3. "Shuffle" the inserted numbers until the number of mistakes is reduced to zero.
A solution to the puzzle is then found. Approaches for shuffling the numbers include simulated annealing, genetic algorithm and tabu search. Stochastic-based algorithms are known to be fast, though perhaps not as fast as deductive techniques. Unlike the latter however, optimisation algorithms do not necessarily require problems to be logic-solvable, giving them the potential to solve a wider range of problems. Algorithms designed for graph colouring are also known to perform well with Sudokus.[11] It is also possible to express a Sudoku as an integer linear programming problem. Such approaches get close to a solution quickly, and can then use branching towards the end. The simplex algorithm is able to solve non-proper Sudokus, indicating if the Sudoku is not valid (no solution), or providing the set of answers when there is more than one solution.
### Constraint programming
A Sudoku may also be modelled as a constraint satisfaction problem. In his paper Sudoku as a Constraint Problem,[12] Helmut Simonis describes many reasoning algorithms based on constraints which can be applied to model and solve problems. Some constraint solvers include a method to model and solve Sudokus, and a program may require less than 100 lines of code to solve a simple Sudoku.[13][14] If the code employs a strong reasoning algorithm, incorporating backtracking is only needed for the most difficult Sudokus. An algorithm combining a constraint-model-based algorithm with backtracking would have the advantage of fast solving time, and the ability to solve all sudokus.
### Exact cover
Sudoku puzzles may be described as an exact cover problem. This allows for an elegant description of the problem and an efficient solution. Modelling Sudoku as an exact cover problem and using an algorithm such as dancing links will typically solve a Sudoku in a few milliseconds.
## Developing (searching for) Sudokus
Computer programs are often used to "search" for Sudokus with certain properties, such as a small number of clues, or certain types of symmetry. Over 49,000 Sudokus with 17 clues have been found, but discovering new distinct ones (not transformations of existing known Sudokus) is becoming more difficult as undiscovered ones become more rare.[15]
One common method of searching for Sudokus with a particular characteristic is called neighbor searching. Using this strategy, one or more known Sudokus which satisfy or nearly satisfy the characteristic being searched for is used as a starting point, and these Sudokus are then altered to look for other Sudokus with the property being sought. The alteration can be relocating one or more clue positions, or removing a small number of clues, and replacing them with a different number of clues. For example, from a known Sudoku, a search for a new one with one fewer clues can be performed by removing two clues and adding one clue in a new location. (This can be called a {-2,+1} search). Each new pattern would then be searched exhaustively for all combinations of clue values, with the hope that one or more yields a valid Sudoku (i.e. can be solved and has a single solution). Methods can also be employed to prevent essentially equivalent Sudokus from being redundantly tested.
As a specific example, a search for a 17-clue Sudoku could start with a known 18-clue Sudoku, and then altering it by removing three clues, and replacing them with only two clues, in different positions (see last two images). This may discover new Sudokus, but there would be no immediate guarantee that they are essentially different from already known Sudokus. If searching for truly new (undiscovered) Sudokus, a further confirmation would be required to ensure each find is not a transformation of an already known Sudoku.[16]
## References
1. "Star Burst - Polar Graph" A polar chart showing a solution path for a Sudoku (Star Burst) using an exhaustive search routine and comment about 17-clue Sudoku.
2. http://intelligence.worldofcomputing/brute-force-search Brute Force Search, December 14th, 2009.
3. "Backtracking - Set 7 (Sudoku)". GeeksforGeeks. GeeksforGeeks. Archived from the original on 2016-08-28. Retrieved 24 December 2016.
4. Norvig, Peter. "Solving Every Sudoku Puzzle". Peter Norvig (personal website). Retrieved 24 December 2016.
5. "Chart of Cells Visited for Solution" A chart showing a solution path to a difficult Sudoku.
6. Zelenski, Julie (July 16, 2008). Lecture 11 | Programming Abstractions (Stanford). Stanford Computer Science Department.
7. "Star Burst Leo - Polar Graph" A polar chart showing a solution path for a Sudoku (Star Burst Leo) using an exhaustive search routine.
8. "Chart of Cells Visited for Solution" A chart showing a solution path for a difficult Sudoku using an exhaustive search routine.
9. Lewis, R (2007) Metaheuristics Can Solve Sudoku Puzzles Journal of Heuristics, vol. 13 (4), pp 387-401.
10. Perez, Meir and Marwala, Tshilidzi (2008) Stochastic Optimization Approaches for Solving Sudoku arXiv:0805.0697.
11. Lewis, R. A Guide to Graph Colouring: Algorithms and Applications. Springer International Publishers, 2015.
12. Simonis, Helmut (2005). "Sudoku as a Constraint Problem" (PDF). Cork Constraint Computation Centre at University College Cork: Helmut Simonis. Retrieved 8 December 2016. paper presented at the Eleventh International Conference on Principles and Practice of Constraint Programming
13. Multiple Authors. "Java Constraint Programming solver" (Java). JaCoP. Krzysztof Kuchcinski & Radoslaw Szymanek. Retrieved 8 December 2016.
14. Rhollor. "Sudokusolver" (C++). GitHub. Rhollor. Retrieved 8 December 2016.
15. Royle, Gordon. "Minimum Sudoku". Retrieved October 20, 2013.
16. http://forum.enjoysudoku.com The New Sudoku Players' Forum "No new 17s within {-3+3}".
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# Using the Binomial Probability Table. In | Ch 5.3 - 20BSC
ISBN: 9780321836960 18
## Solution for problem 20BSC Chapter 5.3
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Problem 20BSC
Using the Binomial Probability Table. In Exercises 15-20, assume that random guesses are made for five multiple-choice questions on an ACT test, so that there are n = 5 trials, each with probability of success (correct) given by p = 0.20. Use the Binomial Probability table (Table A-1) to find the indicated probability for the number of correct answers.
Find the probability that all answers are correct.
Step-by-Step Solution:
Step 1 of 3
Solution 20BSC
= 0.000
Step 2 of 3
Step 3 of 3
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#### How to Use the Domino Effect to Plot Your Story
Whether you write your manuscript off the cuff or plan it out carefully with an outline, plotting your story ultimately comes down to one simple question: What happens next? Using the domino effect will help you answer this question in a compelling way that keeps your readers engaged.
You may have seen those satisfying videos of a long chain of dominoes toppling away until the last domino falls. It’s like a magic trick, and the reason it works is energy—specifically, potential energy converted to kinetic energy. This energy travels from one domino to the next, providing the push needed to knock it over. This process continues until the last domino falls, resulting in the domino effect.
Domino is a tile game of chance and skill that can be played by two or more players. The pieces are rectangular and thumb-sized, with a blank or identically patterned face on one side and an arrangement of dots or pips (inlaid or painted) on the other. A domino set consists of 28 such tiles, although many sets contain more. The word domino is also used to describe a variety of games played with these tiles, including positional games in which the ends of adjacent dominoes show numbers that add up to some total.
When you play a domino game, each player takes turns placing dominoes on the table. They try to play a domino that will result in the end of the chain showing the desired number or a pattern, such as a line or an angular pattern. Then they add more dominoes to that end of the chain. If they cannot find a domino to play, they pass turn to the next player.
Aside from the games, dominoes can be used to make art. They can be arranged to form straight lines or curved lines, grids that create pictures when they fall, or even 3D structures such as towers and pyramids. When making domino art, be sure to plan out your design carefully so that you don’t run into any problems when it is time to build.
There are various types of domino sets, with some being made from more unusual materials. Traditional European-style dominoes are often made from bone, silver lip ocean pearl oyster shell (mother of pearl), ivory or a dark hardwood such as ebony with black or white pips inlaid or painted. A more novel look can be achieved with sets made from stone (e.g., marble or soapstone); metals; ceramic clay; or frosted glass.
The Domino Effect is a powerful tool that can be used to achieve almost any goal. Whether it’s losing weight, saving money, or getting more done at work, the domino effect can help you reach your goals. By focusing on the steps that lead to your goal, you can develop habits that will help you get there.
Creating a domino effect in your own life is easy, and the rewards can be great. Just remember to start small and be patient. Once you’ve developed these habits, you can gradually increase the size of your goals and see the results. | 650 | 2,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.943973 |
https://cooking.stackexchange.com/questions/85722/has-anyone-solved-this-problem-with-braising-other-than-by-sous-vide/86171 | 1,719,351,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00154.warc.gz | 154,646,598 | 41,023 | # Has anyone solved this problem with braising (other than by sous vide)?
I've been braising food for a number of years as a home cook.
PROBLEM: Whenever meat is braised (pot roast, short ribs, oxtail etc.), the flavor/juice/water from the meat leeches out onto the cooking liquid that it is braised in.
The result is a tender, soft, texture of meat but lacks a lot of flavor, which ends up in the cooking liquid.
I know some might say, very low heat at shorter time, would help (and it does) but the flavor always leeching out into the cooking liquid is inevitable.
It seems that sous vide, very low temperature to soften the meat without the high temperature where meat loses too much juice is the only solution (other than eating it raw, where tenderness and juice is retained).
Is braising the same for you in the above? It would be good if someone's technique was better and it saved a lot of the juice and flavor in their braise.
Hervé This has an interesting discussion on this problem in this book.
The process of losing juice when cooking a piece of meat is in big part mechanical. Meat is basically composed of muscle cells tied together by collagen, which is sensitive to heat. Quoting Hervé:
When a temperature of 50°c (122°f) is reached in the outside layer, the collagen contracts, compressing the juices inside (although the degree of compressibility is small because the juices are mainly water) and expelling the juices of the periphery outward. The center of the roast, composed of liquids and largely incompressible solids, cannot receive these juices. Anyone who is not convinced of this has only to roast a few pieces of beef, weigh them, and determine their density before and after cooking.
Since heat is transferred from the outside towards the inside by convection, the core of the piece of meat is usually at lower temperatures and retain most of the juices which were not expelled by the contracting collagen in the outer layers. Thus the importance of resting the meat after cooking. Outside the heat, you lose this compressing force and the liquids concentrated at the core diffuse towards the outside and redistribute, giving an improved feeling of juiciness. Following your question, Hervé suggests:
Given that the juiciness of the meat depends on the amount of juice it has, why not use a syringe to reinject the juices that have drained out from the roast during cooking?
From this simplified physical picture, I suppose cooking-sous vide for long temperatures and below 50 degrees (temperature at which the collagen contracts) will avoid this phenomenon of expelling juices.
• This is so great. Commented Dec 4, 2017 at 3:51
The lower temperatures cooking sous vide helps with retaining moisture in meat. While the myosin contracts in the 104-122 F temperature range, actin denatures at a much higher range of 150-163 F. Braising takes the meat up past the actin temperatures, but with sous vide you can keep it below.
(Note some online sous vide recipes say to use a higher temperature, so look for ones that keep the temperature below 150.)
A good explanation with pictures is here: Heat and Its Effects on Muscle Fibers in Meat
This problem cannot be solved, not with sous vide and not with something else.
Meat is made of cells, whose walls are made mostly of proteins. These proteins change their structure when heated, and that's what turns raw meat into cooked meat. When they change their structure (denature), you get tears in the cell walls.
When you have a piece of meat which is poor in collagen, say a steak, you only heat it enough to denature only the myosin, and most other proteins stay intact. Some of the juices flow out, but most of them can be held back by the still existing cell walls.
If you were to do that to meat rich in collagen, you would end up with impossibly tough meat. So you expose that to enough heat for long enough time that all the tough collagen changes thoroughly and turns to smooth, lubricating gelatin. Braising is one of the methods to do that. When this has happened, all the other proteins are far, far gone, and all the liquid from the cells has flowed out through the now-shredded cell membranes. If you are braising, it flows into the braising liquid, if you are doing sous vide, you will find it in the pouch. This type of cooking is incompatible with the juices staying in.
If you are missing flavor in braised meat, you might be braising the wrong type of meat. Some milder flavored meats like chicken and animals raised on mass production farms (little movement, no variation in food, no fat, slaughtered young) are simply not gamey enough. If you braise mutton, or a hog, or wild fowl, with some fat marbling too, you will certainly have flavor in the meat itself. Not because of juices, but because the meat is aromatic. Even worse, if you are braising meat parts from mild tasting animals which are low in collagen, you will not only lose the juices, you won't have the gelatin either, and you will end up with dried lumps of tough, tasteless matter. | 1,102 | 5,061 | {"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} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.97532 |
https://www.arxiv-vanity.com/papers/1001.4926/ | 1,670,259,406,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00399.warc.gz | 698,320,768 | 52,937 | # A Decomposition Theorem for Singular Integral Operators on Spaces of Homogeneous Type
###### Abstract
Let be a space of homogeneous type and a Banach space. Under the assumption for all , we prove a decomposition theorem for singular integral operators on as a series of simple shifts and rearrangements plus two paraproducts. This gives a Theorem in this setting.
MSC: 42B20, 60G42, 46E40, 47B38
Keywords: Spaces of homogeneous type; Singular integral operators; spaces; Rearrangement and Shift operators; Martingale transforms
Paul F.X. Müller Markus Passenbrunner Department of Analysis J. Kepler University Altenberger Strasse 69 A-4040 Linz Austria
## 1 Introduction
The -Theorem for scalar valued singular integral operators on was initially proved by David and Journé ([16]) using Fourier analysis methods. It was later extended to spaces of homogeneous type by Coifman (unpublished, see [9] and [11]). The structural framework for both proofs is given by Cotlar-Stein theorem on almost orthogonal operators. Consequently, different methods had to be developed to obtain a theorem for integral operators taking values in general Banach spaces. This was done by T. Figiel ([18] and [20]) who introduced a general method of decomposing integral operators into series of basic building blocks. This decomposition arises canonically by expanding the integral kernel along the isotropic Haar system in . Thus proving boundedness of integral operators is reduced to the following problems:
• Verify a priori norm estimates for the building blocks (this is independent of the underlying integral kernel).
• Verify compensating coefficient estimates arising in the isotropic series expansion of the kernel (the decay of the coefficients depends on the size and smoothness of the kernel under investigation).
The basic building blocks isolated by Figiel are simple rearrangements and shifts plus two paraproducts. These rearrangements and shifts act on the Haar system in . It is important to note that their definition depends expressly on the group structure of the underlying domain . Figiel’s decomposition was applied later to several singular integral operators beyond the Calderón-Zygmund class. These included applications to Dirichlet kernels of generalized Franklin systems ([29]) and interpolatory estimates arising in the theory of compensated compactness ([30]).
In the present paper we extend Figiel’s decomposition method to the setting of spaces of homogeneous type. Our extension of this method is based on constructing – without recourse to group structure – a suitable class of rearrangement and shift operators that allow us to decompose singular integral operators on into a series of basic building blocks that can be analyzed and estimated by combinatorial means. The central result of this paper is the convergence of this operator-series (4.9).
A source of renewed interest in spaces of homogeneous type is the recent development of diffusion wavelets and their multiresolution analysis that was carried out on spaces of homogeneous type by Coifman and Maggioni ([12]). We recall further that the vector-valued theorem on spaces of homogeneous type is an essential first step towards the solution of the open classification problem for the vector valued Banach spaces . See [34] and [36].
## 2 Martingale Preliminaries
In this section, we collect a set of martingale inequalities we use throughout the paper.
### 2.1 Kahane’s Contraction Principle
We use Kahane’s contraction principle in the following form ([28],[33]).
###### Theorem 2.1 (Kahane, contraction principle).
Let be elements in a Banach space and be independent Rademacher functions. If are real numbers with we have for any
∫10∥∥ ∥∥m∑k=1akrk(t)ek∥∥ ∥∥pEdt≤∫10∥∥ ∥∥m∑k=1rk(t)ek∥∥ ∥∥pEdt.
### 2.2 UMD spaces
###### Definition 2.2.
A Banach space is called a –space (unconditional for martingale differences), if for every there exists a constant such that for every -valued martingale difference sequence we have the inequality
∥∥ ∥∥n∑k=0εkdk∥∥ ∥∥LpE≤βp∥∥ ∥∥n∑k=0dk∥∥ ∥∥LpE (2.1)
for all sequences of numbers in and all .
Remark.
1. We remark that if there exists one with a constant such that holds, we have automatically that for all there exists a constant with
2. Hilbert spaces are –spaces, –spaces are reflexive and the –property is a self dual isomorphic invariant (see for instance [18],[20], [21] or [7]).
### 2.3 The space BMO
We let be a probability space and be a sequence of -algebras such that is generated by the union . For we introduce the abbreviations
###### Definition 2.3.
(Bounded Mean Oscillation). A function is said to be in if and only if is in and
∥f∥BMO:=supk∈N∥∥∥√Ek(|f−Ek−1f|2)∥∥∥∞<∞. (2.2)
This is a norm, if we factor out the constants.
Remark. Recall that no matter what exponent in is chosen instead of , the definition leads to the same space with equivalent norms (cf. [22] or [6]).
## 3 Extracting Rearrangements on Spaces of Homogeneous Type
This section contains an extensive combinatorial analysis of dyadic cubes in spaces of homogeneous type. We recall first basic properties of those cubes and of the martingale differences they generate. Thus we construct orthonormal bases in and . Next we introduce a coloring on the collection of all dyadic cubes, so that on each monochromatic subcollection there are well defined rearrangement operators that act like ”shifts by units” (Proposition 3.11). The complications in the proof of this proposition are due to the fact that we need to have good quantitative control on the numbers of colors involved. This in turn is dictated by the nature of the kernel operators we treat in Section 4. Theorem 3.17 is the second main result of this section. It provides the combinatorial basis for the norm estimates of the rearrangement operators defined in Section 4.3.
### 3.1 Definitions
###### Definition 3.1.
Let be a set. A mapping with the properties
1. ,
2. ,
3. for all and some constant that is independent of .
is called a quasimetric and is called a quasimetric space.
Given a quasimetric , we define the ball centered at with radius as
B(x,r):={y∈X:d(x,y)
Additionally, a set is called open if and only if for all there exists such that .
###### Definition 3.2.
Let be a quasimetric space such that every ball in the quasimetric is open and a Borel measure. If there is an such that
0<μ(B(x,2r))≤Aμ(B(x,r))<∞,for all x∈X and % all r>0,
then is called a space of homogeneous type. Additionally, if there exist constants such that
b1r≤μ(B(x,r))≤b2r
for all and all with , we call the space of homogeneous type normal.
Remark. We note that if is a space of homogeneous type, then for all there exists such that
μ(B(x,λr))≤Aλμ(B(x,r))for all x∈X% and all r>0.
Since for a given quasimetric space , the balls in are not necessarily open, we added this condition to the definition. This is the case, if for instance one has a Hölder condition for : There exists and such that for all we have
|d(x,z)−d(y,z)|≤Cd(x,y)βmax{d(x,z),d(y,z)}1−β. (3.1)
In fact, Macías and Segovia proved in [32] that for every space of homogeneous type there exists an equivalent quasimetric with the desired Hölder property. Here, a quasimetric is equivalent to a quasimetric if there exists a finite constant such that
1Cd(x,y)≤d′(x,y)≤Cd(x,y),
whenever .
##### Standard assumptions on X:
In the following, we always assume that the spaces we work with are spaces of homogeneous type, equipped with a quasimetric and a Borel probability measure Additionally we impose the restriction that is normal and that for all we have i.e. we have no isolated points.
In a space of homogeneous type there are analogues for dyadic cubes in (see [10] and [14]).
###### Theorem 3.3.
Let be a space of homogeneous type. Then there exist a system of open sets
A:={Qnα⊆X | n∈Z,α∈Kn},
points and constants such that we have the following properties
1. For all we have that up to -null sets.
2. For with and and we have either or That means that the cubes are nested.
3. For each and every there is exactly one such that
4. For all and for all we have that .
5. With
∂tQnα:={x∈Qnα:d(x,X∖Qnα)≤tqn},
we have
∀n∈Z∀α∈Kn:μ(∂tQnα)
6. For all the set is countable.
7. For all and all we have .
8. For all there is a subset of with such that
Qnα=⋃β∈EQn−1βup to μ-null sets.
Remark. We note that these dyadic cubes were constructed by Christ in [10] and by David in [14] in a slightly different way. We further remark that in the future use of the dyadic cubes, we neglect -null sets in points and of Theorem 3.3 and assume equality.
We now collect a few useful definitions, which we will need in the sequel.
###### Definition 3.4.
We let
An:={Qnα:α∈Kn},
be the set of dyadic cubes with level Furthermore, let and choose arbitrarily (but fixed for all subsequent sections) with Then we set
E(A):={B∈An:B⊆A∖A∗(A)}.
We denote the cardinality of this set by . Additionally, we define the level of as
levA:=n+1.
The unique element such that for we have will be denoted by
preQ, (3.2)
which indicates that is the predecessor of . Furthermore, we define a subset of dyadic cubes
E(A):=⋃A∈AE(A)
Remark. Due to Point 7 of Theorem 3.3 we have that the cardinality of is bounded by a uniform constant independent of .
### 3.3 Martingale Differences
Let be a space of homogeneous type with . Then we have and for all We use then dyadic cubes to build an orthonormal basis in consisting of martingale differences. Fix , and enumerate the elements in in the way that Additionally we set We define the following functions, supported on
###### Definition 3.5.
We define for and
dQk(x):=cQk⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0,if x∈k−1⋃j=1Qj∪(X∖A)∑N(A)+1j=k+1μ(Qj),if x∈Qk−μ(Qk),if x∈N(A)+1⋃j=k+1Qj,
where we choose such that
∥∥dQk∥∥2=1. (3.3)
Remark. The functions defined in Definition 3.5 are obviously a martingale difference sequence. We record here also that these martingale differences are just the result of the Gram Schmidt orthogonalization process applied to the indicator functions
1A,1Q1,…,1QN(A). (3.4)
Now we enumerate all the functions where in a canonical way, we set
d0:=1X
and get the functions that are a basis in the constant functions on where for and set
d1=dQ1,…,dN(X)=dQN(X).
We continue with this procedure on every so we get an enumeration of all functions for such that the order is preserved in the following way
k≤j⇒levR≥levQ for % dk=dR and dj=dQ.
We refer to the functions as Haar functions. According to this enumeration we define -algebras:
Fi:=σ(d0,…,di)for i∈N0.
With respect to this filtration, the collection is a martingale difference sequence, since we have for every . Another important sequence of -algebras that we need later is a suitable subsequence of the -algebras just created. We set
Flevk:=σ(A−k)for % k∈N0, (3.5)
where the superscript should indicate that is the -algebra generated by all dyadic cubes of level .
As in the case with the standard Haar functions, we have that the norm of an normalized Haar function is (approximately) , which is a simple consequence of Theorem 3.3 and the normality of .
###### Lemma 3.6.
There exists a constant depending only on such that
c−1μ(Q)−1/2≤∥∥dQ∥∥∞≤cμ(Q)−1/2for all Q∈E(A).
Another simple consequence of Theorem 3.3 is
###### Lemma 3.7.
generates the Borel -algebra on X.
Remark. If is a –space, it is in particular reflexive and thus satisfies the Radon-Nikodym property. So, the martingale convergence theorem (see [8]) and the above lemma yield that for we have that
limk→∞∥E(f|Fk)−f∥LpE(X)=0
for all So we get for every a unique series expansion
f=∞∑k=0akdk,ak∈E,
which converges unconditionally in for . In particular for and , is an orthonormal basis.
### 3.4 Isotropic Basis in L2(X×X)
Next we introduce an isotropic orthogonal basis in Here, the word isotropic means that for an element of this basis (here, is the standard tensor product of two functions), the support looks like a square and not like a rectangle. Most of the notation used in the sequel was introduced in Definition 3.4. Let . For and we define
d(ε)Q:=dQfor ε=1, and d(ε)Q:=1A√μ(A)for ε=0.
Note that the function is -normalized as is . With these settings, we define the collection of functions on :
Z:={1X⊗1X}∪{d(ε1)Q⊗d(ε2)R:Q,R∈E(A),levQ=levR,ε=(ε1,ε2)∈{0,1}2∖{(0,0)}}. (3.6)
Explicitly, up to constants, the three groups in (3.6) have the form
{dQ⊗dR :A,B∈An+1,Q∈E(A),R∈E(B),n∈−N}, (3.7) {dQ⊗1B :A,B∈An+1,Q∈E(A),n∈−N}, (3.8) {1A⊗dR :A,B∈An+1,R∈E(B),n∈−N} (3.9)
The system forms an orthonormal basis in and this result follows from the well known classical
###### Lemma 3.8.
If is an orthogonal basis in then is an orthogonal basis in
###### Lemma 3.9.
is an orthonormal basis in
###### Proof.
Since the verification of orthonormality is a straightforward calculation, we proceed with showing the basis property Since we know from Lemma 3.8 that the set
is an orthogonal basis in we have to show that each can be decomposed in a finite linear combination of functions of the form To do that, we need the following identities:
1U =1A∗(U)+∑V∈E(U)1V,U∈Am+1, (3.10) dR =c11A∗(B)+∑V∈E(B)cV1V,R∈E(B),c1,cV∈R for V∈E(B). (3.11)
We then have four cases:
1. Let then clearly
2. with Then we get recursively from that is a finite linear combination of functions of the form where With we see that
3. Analogously, we treat the case and
4. If we see from that Without loss of generality we now assume that and we decompose in the form Additionally, if we proceed recursively with and get from that
Recall that is the set of dyadic cubes of level for and consists only of the whole space and the size of cubes decreases with decreasing index . We now introduce the set of all pairs of dyadic cubes of the same level
C:={(A,B):A,B∈An,n∈−N0}
and its decomposition into annuli where
Cm={(A,B)∈C:qm−1+levA≤d(A,B)
and
C0={(A,B)∈C:d(A,B)
Recall that denotes the level of (that is if , then ) and is the constant from Theorem 3.3 that determines the growth factor of the cubes in each level. This definition can be interpreted in the following way: Given we draw an annulus around with inner radius and outer radius and take all pairs such that has no point inside the smaller circle and has at least one point inside the larger circle. It is crucial that the annulus grows with the size of
### 3.6 Extracting Rearrangements - Further Decomposition of Annuli
The aim of this section is to extract (as few as possible) subcollections from such that for each we have that is uniquely determined by and is uniquely determined by The benefit of this decomposition is that on we can define an injective mapping such that (see Definition 3.12). We start with the following observation:
###### Lemma 3.10.
There exists a constant independent of and , such that for there are at most elements with
So, roughly speaking, in an annulus of level around , there are at most cubes of the same size as . This lemma is easily proved using the properties of dyadic cubes in Theorem 3.3 and the normality of .
Remark. The same argument shows that for each there exists a constant s.t. for we have at most elements with
d(A,B)≤Cqn.
###### Proposition 3.11.
Let with from Lemma 3.10. Then we have for all that the collection admits a decomposition as
Cm=Cm,1∪⋯∪Cm,M1qm
so that each of the collections satisfies the two conditions
1. For there exists at most one with
2. For there exists at most one with
Remark. For the applications in Section 4 it is important that is decomposed in subcollections (and not more). For instance the estimate would be much simpler to obtain, but would not allow us to treat singular integral operators.
###### Proof.
Step 1:
Idea of the proof:
Let . Then we define the ring collection of
Om(Q):={R∈A:(Q,R)∈Cm}.
We will show that there exists and an enumeration of the dyadic cubes in such that
Om(Q)={Ri(Q):i∈I(Q)}
and we have the following property:
∀Q,Q′∈A, Q≠Q′ ∀j∈I(Q)∩I(Q′):Rj(Q)≠Rj(Q′). (3.12)
Then we can define the decomposition
Am,i={Q∈A:i∈I(Q)}andCm,i={(Q,Ri(Q)):Q∈Am,i}.
We thus obtain
Cm=Cm,1∪⋯∪Cm,M1qm
and the desired properties hold.
Step 2:
Construction of the enumeration:
Let be an enumeration of all dyadic cubes We proceed by induction over For choose and select any enumeration of the cubes Observe that with Lemma 3.10 we have that Now let and assume we have constructed
I(Q(1)),…,I(Q(k))
with
Om(Q(l))={Ri(Q(l)):i∈I(Q(l))} for l≤k
such that the following holds
∀Q,Q′∈{Q(1),…,Q(k)}, Q≠Q′ ∀j∈I(Q)∩I(Q′):Rj(Q)≠Rj(Q′).
We will now construct To do this we first set
{R(1),…,R(M∗)}=Om(Q(k+1)),where M∗≤M0qm.
Step 2a:
We start a second induction and begin with We will define the index of in the enumeration as follows. We put
V(R(1))={Q′∈{Q(1),…,Q(k)}:R(1)∈Om(Q′)},
so contains the cubes for which is in their ring collection Now, since we have an estimate for the cardinality of
|V(R(1))|≤M0qm. (3.13)
For we already defined the indices Next we let
L(R(1))={indQ′R(1):Q′∈V(R(1))}
the indices of in the enumeration of According to , we have
|L(R(1))|≤M0qm
and For the reduced index set, defined as
Ired=I∖L(R(1)),
we have
|Ired|≥M1qm−M0qm.
In particular, we have So we select any element in to be the index of for
indQ(k+1)R(1)∈Ired.
Thus the beginning of the second induction is completed.
Step 2b:
Next we fix We now assume that we already defined
indQ(k+1)R(1),…,indQ(k+1)R(j),
so we pick As in the beginning of the induction, we set
V(R(j+1))={Q′∈{Q(1),…,Q(k)}:R(j+1)∈Om(Q′)}.
We again have and thus an estimate for the cardinality
|V(R(j+1))|≤M0qm.
Next let
L(R(j+1))={indQ′R(j+1):Q′∈V(R(j+1))}
be the indices of in the enumeration of Since we have for the reduced index set
Ired=I∖(L(R(j+1))∪{indQ(k+1)R(1),…,indQ(k+1)R(j)})
an estimate for the cardinality
|Ired|>M1qm−M0qm−M∗≥(M1−2M0)qm,
so we have due to the definition of that We finally select then the index to be any element from the reduced index set
Step 3:
We summarize and set
and the index set
I(Q(k+1))={indQ(k+1)(R(j)):R(j)∈Om(Q(k+1))}.
It follows from the construction step 2 that the enumeration and the index sets have the desired property
For we recall the meaning of which was defined in the previous proof, as
Am,i={A∈A:∃B∈A, such that (A,B)∈Cm,i}.
Due to Proposition 3.11, we can define an injective mapping on
###### Definition 3.12.
We define
τ :Am,i→A A↦τ(A)
through the relation
Additionally we get an inverse of on
### 3.7 Decomposition of Cm,i using Arithmetic Progressions
###### Proposition 3.13.
For all there is a constant that depends only on and the space of homogeneous type such that we have the decomposition
Cm,i=G1∪⋯∪GM,
with the property that for all and all disjoint in with
the following separation of these sets holds:
d(τi(A1),τj(A2))>Cqnfor all i,j∈{0,1}. (3.14)
Here, and .
###### Proof.
Let be an enumeration of Initialize the collections as empty. For , we inductively add to for
r:=min{i∈N: for all (A1,τ(A1))∈Gi we have (???) with A2 replaced by Q(k)}.
Thus for we have for all and so we have that for all there exists a pair such that one of the four expressions
d(A,A0l),d(A,τ(A0l)),d(τ(A),A0l),d(τ(A),τ(A0l))
is . According to the properties of the collection , the sets in the collection as well as the sets | 5,663 | 19,426 | {"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} | 2.609375 | 3 | CC-MAIN-2022-49 | latest | en | 0.884488 |
https://dwiwarna.com/newfoundland-and-labrador/what-real-world-applications-are-there-for-logarithms.php | 1,627,698,612,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154042.23/warc/CC-MAIN-20210731011529-20210731041529-00530.warc.gz | 229,836,102 | 11,409 | # Newfoundland and Labrador What Real World Applications Are There For Logarithms
## Astronomy Exponential/Logarithmic Functions
### Logarithms in the "real world" by Jennifer Hua on Prezi
Classroom Video Real World Applications BetterLesson. 2009-08-29В В· Logarithms in the Real World Applications Of Logarithms - Duration: Logarithms - Real Life Applications - Duration:, Point of logarithms? Every branch of mathematics and science uses logarithms. There is a list Logs have a variety of real life applications such as.
### Logarithm Real World Project with 4 choices for
How do we use Logarithms and Exponentials. MATH 11011 APPLICATIONS OF LOGARITHMIC FUNCTIONS KSU Deflnition: †Logarithmic function: Let a be a positive number with a 6= 1. The logarithmic …, This lesson will allow the students the opportunity to investigate a real life application of logarithms. Classroom Video: Real World Applications. if there.
Real-World Application Real World Applications of Exponential and Logarithmic Functions. you will be captivated by the various uses of logarithms. Mathematics Describing the Real World: Precalculus and Trigonometry. In Mathematics Describing the Real World, He says there's a log,
Applications of Logarithms uses dynamic behavior of these functions and of their many occurrences in the real world. Applications involve carbon dating on Common and Natural Logarithms and Solving some applications and the photolithography The mathematical constant e is the unique real number such that
Chapter 5 Exponential and Logarithmic arising from real-world applications, The two parts of Investigate A seem to suggest that there exists a value for the Common and Natural Logarithms and Solving some applications and the photolithography The mathematical constant e is the unique real number such that
Notes on logarithms the most commonly used log in advanced applications Here is a plot of US real gross national product against time, Mathematics Describing the Real World: Precalculus and Trigonometry. In Mathematics Describing the Real World, He says there's a log,
Natural Logs in the Real World There's a general discussion about some ways in which the number e (on which natural logarithms are based) 2005-02-28В В· Why Do We Learn Logarithms? and how are they used in real life? There are a few situations where logarithms are in common use.
In some applications, a quantity y demonstrates exponential growth or decay on a huge range. To make this quantity more convenient in handling, special scales involving logarithms are used. This allows to deal with the corresponding powers instead of actual values of y. Relevant Maple Means. Common and Natural Logarithms and Solving some applications and the photolithography The mathematical constant e is the unique real number such that
How will I use Logarithmic Function in real life? What real world applications use logarithms? There are several more examples where logarithms are widely used. Practical modern day uses of logarithms are e. All Logarithms in the Real World Math. 9,882 views subscribe 7. Video not playing, click
Statisticians use logarithms to determine the possibilities in real life. With logs, there are many equations that use logs to Logarithms in the "Real World 2007-10-02В В· I can think of LOTS of real-world applications for exponents and are the practical applications of logarithms? There once was a farmer who decided
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Logarithms in the Real World teachertube.com. Applications of Exponential and Logarithmic Functions. There is, however, a limit to is not accurate in the long-term for real world simulation., Using Logarithms in the Real World. is that the practical applications of logarithm in real life? pardon me. this is part of our project and There’s more.
### Classroom Video Real World Applications BetterLesson
What are some real world applications of logarithmic. Posts about real world applications my students involving logarithms in which one of the World History teacher if there were any connections, Using logarithms to solve real world problems Interest Compounded Annually Suppose that \$10,000 is invested at 6% interest compounded annually. In t years an.
### Exponents and Logarithms Eastern Florida State College
Logarithm Real World Project with 4 choices for. Logarithms appear in many applications. We have listed a few such applications in "Real-World" Application for Specific Mathematical Topics the answer to a https://en.wikipedia.org/wiki/Category:Logarithms Applications of Exponential and Logarithmic Function. Applications of Exponential and Logarithmic Function. There are four problems,.
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Logarithms plus student choice when students explore different real-world applications of logarithms. Logarithms plus student choice when students explore different real-world applications of logarithms.
Applications of Exponential and Logarithmic Function. Applications of Exponential and Logarithmic Function. There are four problems, Using logarithms to solve real world problems Interest Compounded Annually Suppose that \$10,000 is invested at 6% interest compounded annually. In t years an
There are also some other integral representations of the Logarithms have many applications inside and outside is the unique real natural logarithm, Most attempts at Math In the Real World (TM) point out logarithms in so there's a difference of 4 is that the practical applications of logarithm in real life
## Exponents and Logarithms Eastern Florida State College
Logarithms in the Real World teachertube.com. Logarithms have many uses in the real world. But what happens when the calculation uses a base other than the standard base 10? Watch this video to..., 2008-12-12 · There are real life applications of logarithms but there is little that you would do every life, I mean technically the computer uses it but there rly isn't anything you as a person would do daily that would make them more applicable especially if you are still in college or high school, maybe if you actually worked in electronics, engineering, or ….
### Logarithms in the "real world" by Jennifer Hua on Prezi
Logarithms in the Real World teachertube.com. Transcript of Logarithms in Real Life Applications. EARTHQUAKES Magnitude of an earthquake is logarithmic Used for predicting coming earthquakes Richter Scale:, 2008-12-12 · There are real life applications of logarithms but there is little that you would do every life, I mean technically the computer uses it but there rly isn't anything you as a person would do daily that would make them more applicable especially if you are still in college or high school, maybe if you actually worked in electronics, engineering, or ….
Humans use logarithms in many ways in everyday life, from the music one hears on the radio to keeping the water in a swimming pool clean. They are important in measuring the magnitude of earthquakes,... In truth, there are many applications of logarithms. We are simply not familiar with them if we have not encountered logarithms before. In the real world,
Using logarithms in the real world. he perhaps noticed that there exists a simple relationship With so many uses and applications in the natural world, 2009-08-29В В· Logarithms in the Real World Applications Of Logarithms - Duration: Logarithms - Real Life Applications - Duration:
Using Logarithms in the Real World. is that the practical applications of logarithm in real life? pardon me. this is part of our project and There’s more Chapter 5 Exponential and Logarithmic arising from real-world applications, The two parts of Investigate A seem to suggest that there exists a value for the
Practical modern day uses of logarithms are e. All Logarithms in the Real World Math. 9,882 views subscribe 7. Video not playing, click 2008-12-12 · There are real life applications of logarithms but there is little that you would do every life, I mean technically the computer uses it but there rly isn't anything you as a person would do daily that would make them more applicable especially if you are still in college or high school, maybe if you actually worked in electronics, engineering, or …
Solving Real World Applications. Getting help from the class. Closure. How do we use Logarithms and Exponentials. Add to Favorites. 2 teachers like this lesson. Print Logarithms appear in many applications. We have listed a few such applications in "Real-World" Application for Specific Mathematical Topics the answer to a
Most attempts at Math In the Real World (TM) point out logarithms in so there's a difference of 4 is that the practical applications of logarithm in real life Using logarithms in the real world. he perhaps noticed that there exists a simple relationship With so many uses and applications in the natural world,
What is the history, real world uses for logarithms, and profesions that would benefit from an understanding of logarithms Applications of Logarithms uses dynamic behavior of these functions and of their many occurrences in the real world. Applications involve carbon dating on
Transcript of Logarithms in Real Life Applications. EARTHQUAKES Magnitude of an earthquake is logarithmic Used for predicting coming earthquakes Richter Scale: In some applications, a quantity y demonstrates exponential growth or decay on a huge range. To make this quantity more convenient in handling, special scales involving logarithms are used. This allows to deal with the corresponding powers instead of actual values of y. Relevant Maple Means.
There are 10 unique Task Cards with real world applications, Bring real world applications to logarithms with this set of 4 ready-to-use projects. Humans use logarithms in many ways in everyday life, from the music one hears on the radio to keeping the water in a swimming pool clean. They are important in measuring the magnitude of earthquakes,...
2012-01-17 · What are some real world applications of logarithmic What are the real life applications of logarithms? Are demons real ?I need to know.IS THERE A … Practical modern day uses of logarithms are e. All Logarithms in the Real World Math. 9,882 views subscribe 7. Video not playing, click
Chapter 5 Exponential and Logarithmic arising from real-world applications, The two parts of Investigate A seem to suggest that there exists a value for the There are 10 unique Task Cards with real world applications, Bring real world applications to logarithms with this set of 4 ready-to-use projects.
### What are some real world applications of logarithmic
Logarithms Teaching Resources Teachers Pay Teachers. There has been an Exponential increase in the speed Online Presentation on Exponents in the Real World. jobs that use indices, logarithms…, 2009-08-29 · Logarithms in the Real World Applications Of Logarithms - Duration: Logarithms - Real Life Applications - Duration:.
### Using logarithms in the real world Deccan Herald
What are logarithms used for? Are decibels a good. Logarithms have many uses in the real world. But what happens when the calculation uses a base other than the standard base 10? Watch this video to... https://en.wikipedia.org/wiki/Talk:Logarithm/Archive_1 Logarithms plus student choice when students explore different real-world applications of logarithms..
Natural Logs in the Real World There's a general discussion about some ways in which the number e (on which natural logarithms are based) Applications of Logarithms uses dynamic behavior of these functions and of their many occurrences in the real world. Applications involve carbon dating on
Using Logarithms in the Real World. is that the practical applications of logarithm in real life? pardon me. this is part of our project and There’s more Solving Real World Applications. Getting help from the class. Closure. How do we use Logarithms and Exponentials. Add to Favorites. 2 teachers like this lesson. Print
There are also some other integral representations of the Logarithms have many applications inside and outside is the unique real natural logarithm, Applications of Exponential and Logarithmic Function. Applications of Exponential and Logarithmic Function. There are four problems,
Common and Natural Logarithms and Solving some applications and the photolithography The mathematical constant e is the unique real number such that Statisticians use logarithms to determine the possibilities in real life. With logs, there are many equations that use logs to Logarithms in the "Real World
View all posts in Newfoundland and Labrador category | 2,783 | 13,194 | {"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-2021-31 | latest | en | 0.856044 |
http://mapleprimes.com/tags/fsolve?page=4 | 1,500,576,565,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423320.19/warc/CC-MAIN-20170720181829-20170720201829-00026.warc.gz | 204,367,138 | 134,186 | ## Solutions for each value of omega...
I have an equation as shown below. In this i need to get the value of 'a' for each 'omega'. 'omega' range from 0 to 2 with increment 0.01
And save all the values of 'a' as a column matrix named 'result'.
## Why different results?...
Dear all,
I developed a program to solve f(x, y) = 0 and g(x, y) = 0, I obtained as results (x=2.726, y=2.126) . running the same program another time it gives (x=2.762, y=1.992). how to explain this?
> fsolve({f(x, y) = 0, g(x, y) = 0}, {x = 0 .. infinity, y= 0 .. infinity});
## fsolve did not return the result...
I tried to solve a system of two equations using fsolve in Maple, however Maple didn't return any result (yet there is the result of that equation's system):
Maple is returning the command if there are no roots but that isn't my case.
How can I obtain the result in Maple?
worksheet.mw
## Exporting to Excel...
I want to solve a system of equations using f-solve (two unknowns) and exporting the solutions to a matrix where the solutions are in seperate columns. How do I do this?
I have tried:
for i from 1 to 937 do
AP[i,1]:=fsolve(x=KL[i,1].y=KL[i,2],x=..8,y=0..15))
end if
end do
But this returns the solutions for X and Y in the same column. Also, for the values that are not possible to solve, it returns the entire expression instead of e.g. 0 or "undefined".
Thank you.
## Why can't fsolve solve it?...
It returns unevaluated. The solution is x=-ln(3),y=0. In fact it doesn't give a solution even if the solution is provided as the initial point. The value of Digits doesn't seem to make a difference.
(Tested Maple 2015.2 Macintosh and Maple 2015.1 Linux)
## Problem when solving equations...
Hello dears! Hope everything going fine with you. I have faced problem while solving the system of equations using fsolve command please find the attacment and fixed my problem.
I am very thankful to you for this favour.
VPM_Help.mw
Mob #: 0086-13001903838
## All roots of equation...
hi.i trust that attached equation has more answer but fsolve only gain some of them!!! how i can gain another that i know value of them?
another root that i known, are : 0.165237712988657e-1 and .103583272213766 and .290071279318035
thanks
root.mw
## Problem with Determinant...
hi.after calculate Determinant of matrix and gain value omega'' ω'' by fsolve rule ,when substuting result (ω) in matrix (q) and calculate Determinant again, this value is not zero!!!! may i use LUDecomposition?determinan.mw
## Finding a numerical solution to system of equation...
I have a system of equations in several variables and I just need one numerical solution of it, I tryed to use fsolve of Maple but it always show me some errors or gives back the command as the output.
aaghulu := {-6-4*y-x-(1+y)*x+sqrt((4*(1+y))*(2+x)*(4+2*y+x)+(-(1+y)*x+2+x)^2), (2*(4+2*y+x))*(1+y)-(1+y)*x+2+x+sqrt((4*(1+y))*(2+x)*(4+2*y+x)+(-(1+y)*x+2+x)^2)-(2+y)*(-(1+y)*x+2+x+sqrt((4*(1+y))*(2+x)*(4+2*y+x)+(-(1+y)*x+2+x)^2))};
fsolve(aaghulu, {x, y}, maxsols = 1);
I will be happy if someone guide me how to do these kinds of things using Maple.
## Understanding output of fsolve...
I am trying to solve 4 nonlinear equations for four variables using fsolve and the output that i am getting is basically the same equations repeated after some time. I even tried reducing one of the equations using assumptions from my side but it results in same behaviour.. Quite new to maple, would like some advice as to this behaviour. Thanks
Here's the file
fsolve_1.mw
PS- using do loop is part of the solving so i cannot remove that
## How to solve this equation?...
Hi all
How I use "solve" or "fsolve" for this equation ?
M2 := evalf[4](Matrix(4, 4, {(1, 1) = BesselJ(0, 0.5e-1*sqrt(0.1111111111e-16*omega^2-25.00027778)), (1, 2) = -BesselJ(0, 0.5e-1*sqrt(0.4444444445e-16*omega^2-25)), (1, 3) = -BesselY(0, 0.5e-1*sqrt(0.4444444445e-16*omega^2-25)), (1, 4) = 0, (2, 1) = (0.1111111111e-16*I)*omega*(1-25000000000000/omega^2)*BesselJ(1, 0.5e-1*sqrt(0.1111111111e-16*omega^2-25.00027778))/sqrt(0.1111111111e-16*omega^2-25.00027778), (2, 2) = -(0.4444444444e-16*I)*BesselJ(1, 0.5e-1*sqrt(0.4444444445e-16*omega^2-25))/sqrt(0.4444444445e-16*omega^2-25), (2, 3) = (0.4444444444e-16*I)*BesselY(1, 0.5e-1*sqrt(0.4444444445e-16*omega^2-25))/sqrt(0.4444444445e-16*omega^2-25), (2, 4) = 0, (3, 1) = 0, (3, 2) = BesselJ(0, 0.60e-1*sqrt(0.4444444445e-16*omega^2-25)), (3, 3) = BesselY(0, 0.60e-1*sqrt(0.4444444445e-16*omega^2-25)), (3, 4) = -BesselY(0, 0.60e-1*sqrt(0.1111111111e-16*omega^2-25)), (4, 1) = 0, (4, 2) = (0.4444444444e-16*I)*BesselJ(1, 0.60e-1*sqrt(0.4444444445e-16*omega^2-25))/sqrt(0.4444444445e-16*omega^2-25), (4, 3) = (0.4444444444e-16*I)*BesselY(1, 0.60e-1*sqrt(0.4444444445e-16*omega^2-25))/sqrt(0.4444444445e-16*omega^2-25), (4, 4) = -(0.1111111111e-16*I)*omega*BesselY(1, 0.60e-1*sqrt(0.1111111111e-16*omega^2-25))/sqrt(0.1111111111e-16*omega^2-25)})):
with(LinearAlgebra):
DETM2 := Determinant(M2):
solve(DETM2 = 0, omega);
Error, (in solve) cannot solve for an unknown function with other operations in its arguments
Is this Error because of combination of bessel functions? if I use asymptatic forms, does it work?
Thanks
## Creating matrix of fsolve output...
Hi Please I need help with making the output of my fslolve appear in a way that I can easily copy to an excel.
I am doing analysis for 3 countries and each time I produce a result I copy manually to excel and use 'text to column' and the 'transpose' excel options to arrange the results in order. I do this for almost 20 time because I want to see how hows in parameter affect the variables. is there a way I can convert this to a 32 by 3 matrix so that I can copy all at the same time instead of copying each variable at a time. here is my solve command
UK_SOL_FIRST:= fsolve(eval({eq||(1..32)}, Params_UK_FIRST), InitValue_UK_FIRST);
ES_SOL_FIRST:= fsolve(eval({eq||(1..32)}, Params_ES_FIRST), InitValue_ES_FIRST);
DK_SOL_FIRST:= fsolve(eval({eq||(1..32)}, Params_DK_FIRST), InitValue_DK_FIRST);
The Results
UK_SOL_FIRST:={A_ss = 14.36104896, C_ss = 1.445842138, I_ss = 0.3136706500,
K_ss = 12.54682600, K_v_ss = 125.4682600,
LT_ss = 0.01061009037, L_ss = 4.014721807, N_ss = 0.9307582996,
P_a_ss = 0.9336893751, P_ss = 0.8625403648,
Surp = 0.9890479879, U_b_ss = 0.1781599919,
U_ss = 0.1046105158, V_ss = 0.05052687912, W_max = 1.476989982,
W_min = 0.4879419937, W_ss = 1.826907218,
W_tilde = 3.478049987, Y_ss = 2.428417935, aa_ss = 21.67403493,
chhi = 0.4523413798, f_c_ss = 0.04880034560,
m_ss = 0.03536881539, p_d_ss = 0.9907986980,
x_T = 0.7023268636, y_d_ss = 10.57030302, y_f_ss = 1.174478111,
y_x_ss = 10.57030300, z_ss = 21.14060602,
Profit_ss = 4.094720376, phi_prod = 0.9753885739,
theta_ss = 0.4830000000}
ES_SOL_FIRST:={A_ss = 10.91702785, C_ss = 2.038687975, I_ss = 0.3058575000,
K_ss = 12.23430000, LT_ss = 0.1309315222, L_ss = 2.857497927,
N_ss = 0.8398656215, P_a_ss = 0.9680877046,
P_ss = 0.8638978804, Surp = 2.541617932, U_b_ss = 0.9095925505,
U_ss = 0.1819708847, V_ss = 0.03119500880, W_max = 3.252738093,
W_min = 0.7111201606, W_ss = 3.605202340,
W_tilde = 3.665280790, Y_ss = 2.367929032, aa_ss = 15.67939783,
betta = 0.9909865708, chhi = 0.2898275349,
f_c_ss = 0.6743530978, m_ss = 0.02183650616,
p_d_ss = 0.9939322922, x_T = 0.005556307841,
y_d_ss = 7.853422751, y_f_ss = 1.195945300,
y_x_ss = 7.978400682, z_ss = 15.83182343,
Profit_ss = 3.084421270, phi_prod = 1.009721394,
theta_ss = 0.1714285714}
DK_SOL_FIRST:={A_ss = 16.18893837, C_ss = 1.359886068, I_ss = 0.2487000000,
K_ss = 9.948000000, LT_ss = 0.02282780783, L_ss = 5.834365727,
N_ss = 0.9399351536, P_a_ss = 0.7054445707,
P_ss = 0.8796237740, Surp = 0.6511024854,
U_b_ss = 0.4572819488, U_ss = 0.08450316042,
V_ss = 0.03491187713, W_max = 1.293898615,
W_min = 0.6427961298, W_ss = 2.363825013,
W_tilde = 2.758200925, Y_ss = 1.755529412, aa_ss = 34.56310241,
betta = 0.9851712031, chhi = 0.4499333284,
f_c_ss = 0.1898151486, m_ss = 0.02443831399,
p_d_ss = 1.032636460, x_T = 0.1506134910, y_d_ss = 11.17773688,
y_f_ss = 0.9144278497, y_x_ss = 13.74561008,
z_ss = 24.92334696, Profit_ss = 4.926248216,
phi_prod = 0.7210969276, theta_ss = 0.4131428571}
## How to solve for variable xx?...
InputMatrix3aa := Matrix(3, 3, {(1, 1) = xx, (1, 2) = 283.6, (1, 3) = 285.4, (2, 1) = 283.6, (2, 2) = 285.4, (2, 3) = 0, (3, 1) = 285.4, (3, 2) = 0, (3, 3) = 0});
InputMatrix3 := Matrix(3, 3, {(1, 1) = 283.6, (1, 2) = 285.4, (1, 3) = 283.0, (2, 1) = 285.4, (2, 2) = 283.0, (2, 3) = 0, (3, 1) = 283.0, (3, 2) = 0, (3, 3) = 0});
InputMatrix3b := Matrix(3, 3, {(1, 1) = 285.4, (1, 2) = 283.0, (1, 3) = 287.6, (2, 1) = 283.0, (2, 2) = 287.6, (2, 3) = 0, (3, 1) = 287.6, (3, 2) = 0, (3, 3) = 0});
InputMatrix3c := Matrix(3, 3, {(1, 1) = 283.0, (1, 2) = 287.6, (1, 3) = 296.6, (2, 1) = 287.6, (2, 2) = 296.6, (2, 3) = 0, (3, 1) = 296.6, (3, 2) = 0, (3, 3) = 0});
InputMatrix3d := Matrix(3, 3, {(1, 1) = 287.6, (1, 2) = 296.6, (1, 3) = 286.2, (2, 1) = 296.6, (2, 2) = 286.2, (2, 3) = 0, (3, 1) = 286.2, (3, 2) = 0, (3, 3) = 0});
Old_Asso_eigenvector0 := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3aa), InputMatrix3aa)):
Old_Asso_eigenvector1 := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3), InputMatrix3)):
Old_Asso_eigenvector2 := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3b), InputMatrix3b)):
Old_Asso_eigenvector3 := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3c), InputMatrix3c)):
Old_Asso_eigenvector4 := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3d), InputMatrix3d)):
#AA2 := MatrixMatrixMultiply(Old_Asso_eigenvector3[2], MatrixInverse(Old_Asso_eigenvector2[2]));
#AA3 := MatrixMatrixMultiply(Old_Asso_eigenvector4[2], MatrixInverse(Old_Asso_eigenvector3[2]));
AA2 := MatrixMatrixMultiply(Old_Asso_eigenvector2[2], MatrixInverse(Old_Asso_eigenvector1[2]));
AA3 := MatrixMatrixMultiply(Old_Asso_eigenvector3[2], MatrixInverse(Old_Asso_eigenvector2[2]));
sol11 := solve([Re(AA2[1][1]) = sin(m*2+phi), Re(AA3[1][1]) = sin(m*3+phi)], [m,phi]);
if nops(sol11) > 1 then
sol11 := sol11[1];
end if:
sin(rhs(sol11[1])+rhs(sol11[2]));
sol12 := solve([Re(AA2[1][2]) = sin(m*2+phi), Re(AA3[1][2]) = sin(m*3+phi)], [m,phi]);
if nops(sol12) > 1 then
sol12 := sol12[1];
end if:
sin(rhs(sol12[1])+rhs(sol12[2]));
sol13 := solve([Re(AA2[1][3]) = sin(m*2+phi), Re(AA3[1][3]) = sin(m*3+phi)], [m,phi]);
if nops(sol13) > 1 then
sol13 := sol13[1];
end if:
sin(rhs(sol13[1])+rhs(sol13[2]));
#*************************************
sol21 := solve([Re(AA2[2][1]) = sin(m*2+phi), Re(AA3[2][1]) = sin(m*3+phi)], [m,phi]);
if nops(sol21) > 1 then
sol21 := sol21[1];
end if:
sin(rhs(sol21[1])+rhs(sol21[2]));
sol22 := solve([Re(AA2[2][2]) = sin(m*2+phi), Re(AA3[2][2]) = sin(m*3+phi)], [m,phi]);
if nops(sol22) > 1 then
sol22 := sol22[1];
end if:
sin(rhs(sol22[1])+rhs(sol22[2]));
sol23 := solve([Re(AA2[2][3]) = sin(m*2+phi), Re(AA3[2][3]) = sin(m*3+phi)], [m,phi]);
if nops(sol23) > 1 then
sol23 := sol23[1];
end if:
sin(rhs(sol23[1])+rhs(sol23[2]));
#**************************************
sol31 := solve([Re(AA2[3][1]) = sin(m*2+phi), Re(AA3[3][1]) = sin(m*3+phi)], [m,phi]);
if nops(sol31) > 1 then
sol31 := sol31[1];
end if:
sin(rhs(sol31[1])+rhs(sol31[2]));
sol32 := solve([Re(AA2[3][2]) = sin(m*2+phi), Re(AA3[3][2]) = sin(m*3+phi)], [m,phi]);
if nops(sol32) > 1 then
sol32 := sol32[1];
end if:
sin(rhs(sol32[1])+rhs(sol32[2]));
sol33 := solve([Re(AA2[3][3]) = sin(m*2+phi), Re(AA3[3][3]) = sin(m*3+phi)], [m,phi]);
if nops(sol33) > 1 then
sol33 := sol33[1];
end if:
sin(rhs(sol33[1])+rhs(sol33[2]));
#****************************************************
AAA1 := Matrix([[sin(rhs(sol11[1])+rhs(sol11[2])),sin(rhs(sol12[1])+rhs(sol12[2])),sin(rhs(sol13[1])+rhs(sol13[2]))],[sin(rhs(sol21[1])+rhs(sol21[2])),sin(rhs(sol22[1])+rhs(sol22[2])),sin(rhs(sol23[1])+rhs(sol23[2]))],[sin(rhs(sol31[1])+rhs(sol31[2])),sin(rhs(sol32[1])+rhs(sol32[2])),sin(rhs(sol33[1])+rhs(sol33[2]))]]);
MA := MatrixMatrixMultiply(Transpose(InputMatrix3aa), InputMatrix3aa) - lambda*IdentityMatrix(3):
eignvalues1 := evalf(solve(Determinant(MA), lambda)):
MA1 := MatrixMatrixMultiply(Transpose(InputMatrix3aa), InputMatrix3aa) - eignvalues1[1]*IdentityMatrix(3):
MA2 := MatrixMatrixMultiply(Transpose(InputMatrix3aa), InputMatrix3aa) - eignvalues1[2]*IdentityMatrix(3):
MA3 := MatrixMatrixMultiply(Transpose(InputMatrix3aa), InputMatrix3aa) - eignvalues1[3]*IdentityMatrix(3):
eigenvector1 := LinearSolve(MA1,<x,y,z>):
eigenvector2 := LinearSolve(MA2,<x,y,z>):
eigenvector3 := LinearSolve(MA3,<x,y,z>):
MR := MatrixMatrixMultiply(AAA1, Matrix([[Re(eigenvector1[1]),Re(eigenvector2[1]),Re(eigenvector3[1])],[Re(eigenvector1[2]),Re(eigenvector2[2]),Re(eigenvector3[2])],[Re(eigenvector1[3]),Re(eigenvector2[3]),Re(eigenvector3[3])]]));
ML := Re(Old_Asso_eigenvector1[2]);
solve(ML[1][1] = MR[1][1], xx);
with(Optimization):
Minimize(ML[1][1] - MR[1][1], {0 <= xx}, assume = nonnegative);
Error, (in Optimization:-NLPSolve) abs is not differentiable at non-real arguments;
## Error when fsolving ...
Error, (in fsolve/polynom) Digits cannot exceed 38654705646
I am using fsolve to find numerical approximations to the roots of many fairly large polynomials (degrees up to ~80). I often get this error message and I'm not sure why. Is there any workaround? Any help is much appreciated.
## Complex solution...
This may be a trivial question, but does this factor fully with the newer versions of Maple, say at 900 digits?
Digits:=900;
rho_poly := -2201506283520*rho^32+(-17612050268160+104204630753280*I)*rho^31+(2237195146493952+737798139150336*I)*rho^30+(14065203494780928-29153528496783360*I)*rho^29+(-260893325886750720-161432056834818048*I)*rho^28+(-1240991775275876352+1727517243589263360*I)*rho^27+(8952004373272068096+6696323263091441664*I)*rho^26+(25553042370906292224-37948239682297921536*I)*rho^25+(-135024511500569280512-65293199430849134592*I)*rho^24+(-79740262928225402880+401487130320847241216*I)*rho^23+(956745211126674882560-164797793704574713856*I)*rho^22+(-1213375867282228772864-1655554058430246551552*I)*rho^21+(-1483956336776821211136+3604946201834409820160*I)*rho^20+(6525094787202650144768-1597915397190007586816*I)*rho^19+(-8575469412912592879616-6168391294117580865536*I)*rho^18+(2408139380338842796032+15004449784317106323456*I)*rho^17+(10583091471310114717696-17047513330720373194752*I)*rho^16+(-22619716982813548707840+8898637295768494915584*I)*rho^15+(26538067620972845277184+5129530051326543351808*I)*rho^14+(-21415800164460070789120-17268159356969925234688*I)*rho^13+(11916012071577094946816+22601135173030541677568*I)*rho^12+(-3551246770922037813248-21229478915196610975744*I)*rho^11+(-977434486760953073664+16249214903618313346048*I)*rho^10+(1977414870691507931136-10721551032564274826240*I)*rho^9+(-1197394212949208115968+6172794574205050632192*I)*rho^8+(280273257275327368320-2996290081120136529792*I)*rho^7+(108849195761508531648+1152454823926345101504*I)*rho^6+(-119736267114490955904-327757949185254534784*I)*rho^5+(49149411853848597568+63563541902968683712*I)*rho^4+(-11524495997215059744-7307364351434838944*I)*rho^3+(1585189353379709888+299568910286253408*I)*rho^2+(-116032795768295808+25487628220230528*I)*rho+3299863116538269-2454681763039104*I;;
2 3 4 5 6 7 8 Last Page 4 of 13
| 6,029 | 15,382 | {"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} | 3.5 | 4 | CC-MAIN-2017-30 | latest | en | 0.904055 |
https://www.photometrics.com/learn/imaging-topics/matching-resolution | 1,721,732,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00753.warc.gz | 818,856,208 | 12,736 | # Matching ResolutionImaging Topics
## Matching Resolution
Researchers using a scientific camera in conjunction with a microscope desire to work at the maximum possible spatial resolution allowed by their system. In order to accomplish this, it is necessary to properly match the magnification of the microscope to the camera.
The first step in this process is to determine the resolving power of the microscope. The ultimate limit on the spatial resolution of any optical system is set by light diffraction; an optical system that performs to this level is termed “diffraction limited”. In this case, the spatial resolution is given by:
d = 0.61 x lambda / NA
where d is the smallest resolvable distance, lambda is the wavelength of light being imaged, and NA is the numerical aperture of the microscope objective. This is derived by assuming that two point sources can be resolved as being separate when the center of the airy disc from one overlaps the first dark ring in the diffraction pattern of the second (the Rayleigh criterion).
It should be further noted that, for microscope systems, the NA to be used in this formula is the average of the objective’s numerical aperture and the condenser’s numerical aperture. Thus, if the condenser is significantly underfilling the objective with light, as is sometimes done to improve image contrast, then spatial resolution is sacrificed. Any aberrations in the optical system, or other factors that adversely affect performance, can only degrade the spatial resolution past this point. However, most microscope systems do perform at, or very near, the diffraction limit.
The formula above represents the spatial resolution in object space. At the detector, the resolution is the smallest resolvable distance multiplied by the magnification of the microscope optical system. It is this value that must be matched with the CCD.
The most obvious approach to matching resolution might seem to be simply setting this diffraction-limited resolution to the size of a single pixel. In practice, what is really required of the imaging system is that it be able to distinguish adjacent features. If optical resolution is set equal to single-pixel size, then it is possible that two adjacent features of like intensity could each be imaged onto adjacent pixels on the camera. In this case, there would be no way of discerning them as two separate features.
Separating adjacent features requires the presence of at least one intervening pixel of disparate intensity value. For this reason, the best spatial resolution that can be achieved occurs by matching the diffraction-limited resolution of the optical system to two pixels on the camera in each linear dimension. This is called the Nyquist limit. Expressing this mathematically we get:
(0.61 x lambda / NA) x Magnification = 2.0 x (pixel size)
Let’s use this result to work through some practical examples.
Example 1: Given a camera with a Kodak KAF1401E CCD (pixel size 6.8 μm), visible light (lambda = 0.5 μm), and a 1.3 NA microscope objective, we can compute the magnification that will yield maximum spatial resolution.
M = (2 x 6.8) / (0.61 x 0.5 / 1.3) = 58
Thus, a 60x, 1.3 NA microscope objective provides a diffraction-limited image for a KAF1401E CCD camera without any additional magnification. Keep in mind, however, that this assumes that the condensing system also operates at a NA of 1.3. This high NA means the condenser must be operated in an oil-immersion mode, as well as the objective.
Example 2: Given a camera with an e2v CCD37 (pixel size 15.0 μm), visible light (lambda = 0.5 μm), and a 100x microscope objective with a NA of 1.3, we can compute the magnification that will yield maximum spatial resolution.
M = (2 x 15.0) / (0.61 x 0.5 / 1.3) = 128
Since the microscope objective is designed to operate at 100x, we would need to use an additional projection optic of approximately 1.25x in order to provide the optimum magnification.
It should be kept in mind that as magnification is increased and spatial resolution is improved, field of view is decreased. Applications that require both good spatial resolution and a large field of view will need cameras with greater numbers of pixels. It should also be noted that increasing magnification lowers image brightness. This lengthens exposure times and can limit the ability to monitor real-time events. | 951 | 4,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.928388 |
https://sites.google.com/a/pvlearners.net/sweigand-games/fraction-war | 1,627,784,021,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154127.53/warc/CC-MAIN-20210731234924-20210801024924-00663.warc.gz | 529,937,377 | 11,437 | Fraction Games to be Used in the Classrooms
Materials:
• Deck of Cards
• Pencil
• Paper
• Pencils
Fraction War
Students take turns playing "war" using a deck of cards and a pencil to act as the fraction line. The pair of students must then decide who has the larger fraction based on the four cards played. The winner gets to keep all the cards. Player with most cards at end wins.
Goal: to develop quick comparison of fraction values
Rules:
• Shuffle and deal the cards.
• Each player puts their cards faced down in a pile.
• Both players turn over TWO cards at the same time (one above the pencil and one below).
• The player whose cards has the larger fraction wins all four cards.
• Players may use the paper to figure equivalent fractions or use the Tip Sheet.
• If players turn over equivalent fractions, then there is a fraction war.
• Each player places 2 new cards face down and the 3rd & 4th card face up (one above the pencil and one below).
• Who ever has the higher fraction wins all the cards.
• The game can continue until one player has all the cards or for a given amount of time.
Fraction War Tips and Tricks
• If two fractions have a common denominator, the fraction with the larger numerator is the larger fraction. Ex: 3/5 > 2/5
• If two fractions have a common numerator, the fraction with the smaller denominator is larger. Ex: 1/4 > 1/8
• If you are unsure about which fraction is larger, use the fraction strips to compare. | 336 | 1,456 | {"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-2021-31 | latest | en | 0.940654 |
https://www.teacherspayteachers.com/Store/Mhkindergarten | 1,600,785,971,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206133.46/warc/CC-MAIN-20200922125920-20200922155920-00394.warc.gz | 1,059,306,690 | 29,114 | # MHKindergarten
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Use this blank number bond and number sentence booklet when teaching addition to young students. Each page showcases a number bond facing in a different direction to help students become comfortable with various orientations in the part-part-whole
Subjects:
Math, Basic Operations, Numbers
Kindergarten, 1st, 2nd
Types:
Printables, Math Centers
CCSS:
K.OA.A.1, K.OA.A.2, K.OA.A.3, K.OA.A.4, K.OA.A.5, 1.OA.A.1, 1.OA.B.3
\$2.00
10
This Microsoft PowerPoint presentation focuses on the -ing word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Interactive Whiteboard, Activboard Activities
\$2.00
16
Use either version of this book to teach young students about teen numbers! Using this book, students will learn the relationship between number bonds and number sentences. Students will express each teen number as a "group of ten and some extras"
Subjects:
Math, Basic Operations, Numbers
Kindergarten, 1st, 2nd
Types:
Activities, Printables, Math Centers
CCSS:
K.NBT.A.1, 1.NBT.B.2a, 1.NBT.B.2b
\$3.00
4
This Microsoft PowerPoint presentation is designed to help students with beginning addition skills. Depending on the readiness level of your group, this interactive whiteboard activity can be used in many ways. The presentation can be combined with
Subjects:
Math, Basic Operations, Christmas/ Chanukah/ Kwanzaa
PreK, Kindergarten, 1st, Homeschool
Types:
PowerPoint Presentations, Math Centers, Activboard Activities
CCSS:
K.CC.A.3, K.CC.B.5, K.OA.A.1, K.OA.A.5
\$2.00
7
This Microsoft PowerPoint presentation focuses on the -ar word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Interactive Whiteboard, Activboard Activities
\$2.00
3
This Microsoft PowerPoint presentation focuses on the -et word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Activboard Activities
CCSS:
RF.K.2d, RF.K.2e, RF.K.3
\$2.00
6
This Microsoft PowerPoint presentation focuses on the -ill word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Activboard Activities
CCSS:
RF.K.2d, RF.K.2e, RF.K.3
\$2.00
5
This Microsoft PowerPoint presentation focuses on the -ig word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Interactive Whiteboard, Activboard Activities
\$2.00
5
This Microsoft PowerPoint presentation focuses on the -op word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Interactive Whiteboard, Activboard Activities
CCSS:
RF.K.2d, RF.K.2e, RF.K.3
\$2.00
6
This Microsoft PowerPoint presentation focuses on the Magic "e" or Quiet "e". Animations appear on each page to add the "e" to the end of provided words. After students identify the word, a corresponding picture appears to match the word to enable
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Interactive Whiteboard, Activboard Activities
\$2.00
4
This Microsoft PowerPoint presentation focuses on the -up word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Activboard Activities
\$2.00
2
This Microsoft PowerPoint presentation focuses on the -en word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Activboard Activities
CCSS:
RF.K.2d, RF.K.2e, RF.K.3
\$2.00
4
This Microsoft PowerPoint presentation focuses on the -ug word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Activboard Activities
CCSS:
RF.K.2d, RF.K.2e, RF.K.3
\$2.00
4
This Microsoft PowerPoint presentation focuses on the -ake word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Interactive Whiteboard, Activboard Activities
\$2.00
5
This Microsoft PowerPoint presentation focuses on the -est word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Interactive Whiteboard, Activboard Activities
\$2.00
6
Does your math curriculum include the Engage NY Math Modules? If so, this slideshow will be the perfect addition to these lessons! This slideshow contains all Objectives and Application Problems organized by topic from every lesson in Module 1 in a
Subjects:
Math, Numbers, Word Problems
PreK, Kindergarten, 1st
Types:
PowerPoint Presentations, Math Centers, Cooperative Learning
CCSS:
K.CC.A.2, K.CC.A.3, K.CC.B.4a, K.CC.B.4b, K.CC.B.5
\$4.00
3
Does your math curriculum include the Engage NY Math Modules? If so, this slideshow will be the perfect addition to these lessons! This slideshow contains all Objectives and Application Problems organized by topic from every lesson in Module 4 in a
Subjects:
Math, Basic Operations, Word Problems
PreK, Kindergarten, 1st
Types:
PowerPoint Presentations, Math Centers, Cooperative Learning
CCSS:
K.OA.A.1, K.OA.A.2, K.OA.A.3, K.OA.A.4, K.OA.A.5
\$4.00
4
This Microsoft PowerPoint presentation focuses on the -ot word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Activboard Activities
\$2.00
2
This product will help students better understand the meaning of a setting. This can be used for students who are just beginning to learn story elements, or those who need more targeted practice.This Google Slides presentation is a guessing game
Subjects:
English Language Arts, Reading, Critical Thinking
PreK, Kindergarten, 1st, 2nd
Types:
CCSS:
RL.K.1, RL.K.3
\$3.00
Online Resource
This Microsoft PowerPoint presentation focuses on the -in word family. Animations appear on each page to change the initial phoneme(s) in each word, providing practice within each word family. After students identify the word, a corresponding
Subjects:
PreK, Kindergarten, 1st, 2nd
Types:
PowerPoint Presentations, Activboard Activities
CCSS:
RF.K.2d, RF.K.2e, RF.K.3
\$2.00
3
showing 1-20 of 43
TEACHING EXPERIENCE
MY TEACHING STYLE
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY | 2,058 | 8,015 | {"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-2020-40 | latest | en | 0.85817 |
anexvm.co.in | 1,686,284,462,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655247.75/warc/CC-MAIN-20230609032325-20230609062325-00371.warc.gz | 128,684,653 | 5,020 | AnexVM
In the logistics and cargo industry, the shipments are charged based on the actual weight or the volumetric weight (also called as dimensional weight). The customer is charged by transporter based on the weight, whichever is higher. The actual weight of the box can be easily measured by common electronic weighing machines, but there are very few machines that accurately calculate the volumetric weight of a parcel.
### What is Volumetric Weight of a parcel?
Volumetric or dimensional weight is the weight of the parcel which is calculated from its volume or dimensions. It is the space occupied by the box in terms of its density. For shipments, whose weight is less but is large in size, the dimensional weight is considered for costing. It is calculated to know to the space the shipment occupies on the pallet or the vehicle.
### How to calculate volumetric weight of a parcel.
• The logistics and shipping industry basis its prices majorly on the actual weight of the package or the volumetric/dimensional weight of the box. The actual weight of the box can be easily calculated by placing it on a weighing scale and there are many instruments available in the market. The calculation for the volumetric weight of a package requires its exact dimensions and a formula to convert it to a weight equivalent...
#### For air shipments
The volumetric weight of the shipment is calculated by dividing the volume of the box by a factor 5000.
For example, consider a box of dimensions 50cm x 50cm x 50cm and the actual weight as 18 kgs.
= 50 x 50 x 50
5000
= 1,25,000
5000
Volumetric weight = 25 Kg
The shipment will be charged at 25kgs and not 18kgs
#### For road shipments,
The volumetric weight of the shipment is calculated by dividing the volume of the box by a factor 6000.
For example,consider a box of dimensions 50cm x 50cm x 50cm and the actual weight as 18 kgs.
= 50 x 50 x 50
6000
= 1,25,000
6000
Volumetric weight = 21 Kg
The shipment will be charged at 21kgs and not 18kgs
#### Calculating volumetric weight saves cost.
• Most volumetric calculations are done manually. This is also done by deciding an approximate value for the dimensions and calculating the volumetric weight of the box.
• Consider an example, the size of the box is 55cm x 52cm x 50 cm and the actual weight of the box is 18 kgs. The volumetric weight is 28.6kgs.
• The cost of shipping this box from Bangalore to Delhi, considering Rs. 80/kg would be 1440 as per the actual weight. But by volumetric weight it would cost 2288, which is a huge difference of Rs. 848 for this one box shipment.
• If there are 100 boxes being shipped per day, that would sum to losses of 848x100 which is Rs. 84800 for a day for the shipper. | 653 | 2,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-23 | latest | en | 0.936882 |
https://worlddatabaseofhappiness.eur.nl/correlational-findings/9484/ | 1,719,167,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00693.warc.gz | 543,368,592 | 12,834 | # Correlational findings
## Study Timmermans (1997): study SI 1992
Public
18+ aged, general public, Slovenia, 1990
Sample
Respondents
N = 1035
Non Response
Assessment
Interview: face-to-face
Structured interview
## Correlate
Authors's Label
Current age
Our Classification
Operationalization
1 18 - 29 years
2 30 - 39 years
3 40 - 49 years
4 50 - 59 years
5 60 - 69 years
6 70 - 79 years
7 80 +
## Observed Relation with Happiness
Happiness Measure Statistics Elaboration / Remarks O-HL-u-sq-v-4-a = + 1 M=2.48 Mt=4.9
2 M=2.29 Mt=4.3
3 M=2.28 Mt=4.3
4 M=2.33 Mt=4.4
5 M=2.35 Mt=4.5
6 M=2.09 Mt=3.6
7 M=2.14 Mt=3.8
O-SLW-c-sq-n-10-a = + 1 M=5.39 Mt=4.9
2 M=5.06 Mt=4.5
3 M=4.72 Mt=4.1
4 M=4.84 Mt=4.3
5 M=5.33 Mt=4.8
6 M=4.83 Mt=4.3
7 M=4.00 Mt=3.3
A-BB-cm-mq-v-2-a = + 1 M=1.65 Mt=6.7
2 M=1.47 Mt=6.5
3 M=1.82 Mt=6.8
4 M=1.46 Mt=6.5
5 M=1.24 Mt=6.2
6 M= .88 Mt=5.9
7 Empty category
O-HL-u-sq-v-4-a = -.09 p < .01 O-SLW-c-sq-n-10-a = -.07 p < .05 A-BB-cm-mq-v-2-a = -.08 p < .01 O-HL-u-sq-v-4-a = -.07 p < .01 O-SLW-c-sq-n-10-a = -.05 p < .05 A-BB-cm-mq-v-2-a = -.03 ns O-HL-u-sq-v-4-a = -.04 ns ß controlled for sex and household income
( r and ß computed on uncategorized data )
O-SLW-c-sq-n-10-a = 0 ns ß controlled for sex and household income
( r and ß computed on uncategorized data )
A-BB-cm-mq-v-2-a = -.06 ns ß controlled for sex and household income
( r and ß computed on uncategorized data ) | 710 | 1,639 | {"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-2024-26 | latest | en | 0.423126 |
https://www.itsup2u.com/?start=24 | 1,713,275,441,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00823.warc.gz | 764,372,614 | 13,313 | ## What is a Stimpmeter?
A Stimpmeter is a device used in golf to measure the speed of a putting green. It consists of a metal or plastic frame with a notch at one end and a graduated scale along its length. The device is typically around 36 inches (91 cm) long and 2.75 inches (7 cm) wide.
To use a Stimpmeter, a golf course superintendent or greenskeeper places a golf ball in the notch at one end and then lifts the other end of the device to a specific angle, usually around 20 degrees. They then release the golf ball, allowing it to roll down the putting green. The distance the ball rolls along the green's surface is measured and recorded. This distance represents the speed or "stimp rating" of the putting green.
Stimp ratings can vary depending on factors such as grass type, moisture levels, and maintenance practices. Golf course staff use the information gathered with a Stimpmeter to ensure that putting greens have consistent speeds, which is important for maintaining the quality and fairness of play on the course.
I did my research because I , Mark Errington, applied for a part time job in Feb 2024 for a Greenskeeper with Oswego Lake Country Club in Lake Oswego, Oregon. I was interviewed by Nolan Wenker, CGCS, Green Superintendent. I did not get the job.
## Wood Height Calculator for Stimpmeter
Below will help you find the highest point of the wood or whatever material you are using to make your own Stimpmeter.
OR you can go here: https://www.calculatormall.com/wood-height-calculator-for-stimpmeter.html
<!DOCTYPE html>
<html lang="en">
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Wood Height Calculator</title>
<body>
<h2>Wood Height Calculator</h2>
<form method="post" action="<?php echo htmlspecialchars(\$_SERVER["PHP_SELF"]);?>">
Enter the length of the wood (in inches): <input type="text" name="length">
<br><br>
Enter the angle (in degrees): <input type="text" name="angle">
<br><br>
<input type="submit" value="Calculate">
</form>
<?php
// Function to calculate the height of the highest point
function calculateHeight(\$angle, \$length) {
// Convert angle from degrees to radians
// Calculate the height using trigonometry
return \$height;
}
// Check if form is submitted
if (\$_SERVER["REQUEST_METHOD"] == "POST") {
// Retrieve and sanitize user input
\$length = isset(\$_POST['length']) ? floatval(\$_POST['length']) : 0;
\$angle = isset(\$_POST['angle']) ? floatval(\$_POST['angle']) : 0;
// Calculate the height of the highest point
\$height = calculateHeight(\$angle, \$length);
// Display the result
echo "<p>At an angle of \$angle degrees, the highest point of the wood (length: \$length inches) would be approximately " . number_format(\$height, 2) . " inches.</p>";
// Generate a graph
\$imageWidth = 400;
\$imageHeight = 400;
\$image = imagecreatetruecolor(\$imageWidth, \$imageHeight);
\$white = imagecolorallocate(\$image, 255, 255, 255); // White color
\$black = imagecolorallocate(\$image, 0, 0, 0);
\$blue = imagecolorallocate(\$image, 0, 0, 255);
// Fill the image with white background
imagefilledrectangle(\$image, 0, 0, \$imageWidth, \$imageHeight, \$white);
// Draw angle indicator
\$angleX = \$imageWidth / 2;
\$angleY = \$imageHeight / 2;
\$angleLength = min(\$imageWidth, \$imageHeight) * 0.45; // 90% of the smaller dimension
\$angleEndX = \$angleX + \$angleLength * cos(\$angleRadians);
\$angleEndY = \$angleY - \$angleLength * sin(\$angleRadians);
imageline(\$image, \$angleX, \$angleY, \$angleEndX, \$angleEndY, \$blue);
imageline(\$image, \$angleEndX, \$angleEndY, \$angleEndX - 5, \$angleEndY + 10, \$blue);
imageline(\$image, \$angleEndX, \$angleEndY, \$angleEndX + 5, \$angleEndY + 10, \$blue);
// Draw line at the bottom of the graph
imageline(\$image, \$angleX, \$angleY, \$angleEndX, \$angleY, \$black);
// Output the image
imagepng(\$image, "wood_angle_graph.png");
echo '<p><img src="/wood_angle_graph.png" alt="Wood Angle Graph"></p>';
// Free up memory
imagedestroy(\$image);
}
?>
</body>
</html>
Wood Height Calculator Output:
I did my research because I , Mark Errington, applied for a part time job in Feb 2024 for a Greenskeeper with Oswego Lake Country Club in Lake Oswego, Oregon. I was interviewed by Nolan Wenker, CGCS, Green Superintendent. I did not get the job.
## How to make your own "Stimpmeter"
A "Stimpmeter" is a device used to measure the speed of a golf course putting green. It consists of a metal or plastic frame with a notch at one end and a graduated scale along its length. Here's how you can make a basic Stimpmeter:
Materials Needed:
• A sturdy piece of metal or plastic (e.g., aluminum, PVC)
• A ruler or tape measure
• A protractor or angle measuring tool
• A saw (if cutting the material to size)
• Sandpaper (if needed to smooth edges)
Instructions:
1. Select a Material: Choose a sturdy material such as aluminum or PVC. Aluminum is commonly used due to its durability and ease of cutting.
2. Determine Dimensions: A standard Stimpmeter is typically about 36 inches (91 cm) in length and 2.75 inches (7 cm) in width. However, you can adjust the dimensions based on your preference.
3. Cut the Material: If necessary, use a saw to cut the material to the desired length and width. Make sure to wear appropriate safety gear when cutting.
4. Smooth Edges: Use sandpaper to smooth any rough edges on the Stimpmeter to prevent injury during use.
5. Mark the Graduated Scale: Along one edge of the Stimpmeter, use a ruler or tape measure to mark evenly spaced increments. These increments represent the distance rolled by a golf ball when released from the notch at the end of the Stimpmeter. A typical scale ranges from 0 to 10 feet (0 to 3 meters).
6. Add a Notch: At one end of the Stimpmeter, create a small notch or groove. This is where you'll place the golf ball before releasing it to measure the green's speed.
7. Optional: Angle Measurement: Some Stimpmeters include an angled section to help with consistency in releasing the golf ball. If desired, use a protractor or angle measuring tool to create a slight angle near the notch.
8. Test and Calibrate: To use the Stimpmeter, place a golf ball in the notch, lift the other end of the device to a 20 to 30-degree angle, and then release the ball. Measure the distance the ball rolls along the green's surface. Adjust the green's maintenance practices accordingly to achieve desired green speed.
By following these steps, you can create a basic Stimpmeter for measuring the speed of a golf course putting green.
I did my research because I , Mark Errington, applied for a part time job in Feb 2024 for a Greenskeeper with Oswego Lake Country Club in Lake Oswego, Oregon. I was interviewed by Nolan Wenker, CGCS, Green Superintendent. I did not get the job.
## Golf Course Greenskeeper Duties from Beginner Level to Expert Level
The duties of a golf course greenskeeper can vary depending on the specific needs of the course and the level of experience of the greenkeeper. Here's a breakdown of typical duties from beginner to advanced levels:
# Golf Course Greenskeeper Duties
## Beginner Level:
• Mowing fairways, tees, and rough using ride-on or push mowers.
• Raking bunkers (sand traps) to ensure they are smooth and even.
• Watering plants and turf as directed by senior staff.
• Assisting with basic landscaping tasks such as planting flowers or shrubs.
• Learning to operate and maintain basic equipment such as mowers, trimmers, and irrigation systems.
• Observing and learning about turfgrass management practices.
## Intermediate Level:
• Operating and maintaining more specialized equipment such as aerators and top dressers.
• Performing routine maintenance tasks on greens, including mowing, aerating, topdressing, and fertilizing.
• Learning to diagnose and treat common turf diseases and pests.
• Assisting with irrigation system maintenance and repair.
• Participating in course renovation projects under supervision.
• Beginning to take on more responsibility for specific areas of the course, such as a set of greens or fairways.
• Developing and implementing turf management plans to optimize playing conditions.
• Troubleshooting and repairing equipment breakdowns.
• Training and mentoring junior staff.
• Assisting with budgeting and procurement of supplies and equipment.
• Participating in continuing education and professional development opportunities to stay current on industry best practices.
• Collaborating with golf course management to plan and execute long-term course improvements and renovations.
• Specializing in a particular aspect of course maintenance, such as greenskeeping or irrigation management.
## Expert Level:
• Overseeing all aspects of course maintenance and operations.
• Developing and implementing sustainable and environmentally friendly maintenance practices.
• Managing budgets, staffing, and scheduling for the entire course maintenance department.
• Serving as a liaison between course management and outside contractors, consultants, and regulatory agencies.
• Representing the course in professional organizations and industry events.
• Continuously seeking ways to improve course conditions and enhance the overall player experience.
• Providing leadership and guidance to the entire course maintenance team, setting high standards for performance and professionalism.
As a greenkeeper progresses through these levels, they gain knowledge, skills, and experience that enable them to take on more responsibility and contribute to the overall success of the golf course. Continuing education and a commitment to excellence are key factors in advancing in this field.
I did my research because I , Mark Errington, applied for a part time job in Feb 2024 for a Greenskeeper with Oswego Lake Country Club in Lake Oswego, Oregon. I was interviewed by Nolan Wenker, CGCS, Green Superintendent. I did not get the job.
- All From ChatGPT
PLG_GSPEECH_SPEECH_BLOCK_TITLE | 2,294 | 9,987 | {"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-2024-18 | longest | en | 0.838367 |
https://socratic.org/questions/how-do-you-graph-f-x-6-x-2 | 1,718,959,897,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00335.warc.gz | 481,208,261 | 6,104 | # How do you graph f(x) = 6^x?
Jul 17, 2016
Use semi-log graph paper. Plots f = 10^n, n=0, 1, 2, 3, ... become graduations 0, 1, 2 3 ...in the logarithmic scale. The graph for the given function will be a straight, line through the origin.
#### Explanation:
Use F = log f = x log 6 instead and plot x against F, on a semi-log
graph paper.
It will be a straight line F = m x, m=- log 6, through the origin.
The origin represents F = log f = 0, for f = 10^0=1.
The graduations 1, 2, 3 .. on the log scale represent f = 10^1, 10^2,
10^3, ..# | 192 | 547 | {"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} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.833439 |
https://forum.freecodecamp.org/t/please-help-with-javascript-calculator/72468 | 1,653,723,846,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663013003.96/warc/CC-MAIN-20220528062047-20220528092047-00141.warc.gz | 313,803,695 | 8,109 | I’m doing the calculator project and I have a problem. The input field value is a string eg '1+4/59’ etc. How do I calculate inside a string or if not what do I do instead.
eg var x = '1+4/5
9’
How do I make x = '8.2’
So what I’m asking is, how do I calculate as if it wasn’t a string
1 Like
You can use `eval()`.
``````var x = "1+4/5*9";
var answer = eval(x); // 8.2
``````
2 Likes
Thank you so much. It worked like a charm.
You might want to use Number().
For ex:
``````var a = "72" // typeof a => string
var b = Number(a); // typeof b = > number
``````
That being said:
``````var a = "2";
var b = "3";
console.log(a * b) ; // 6
console.log(a + b); // 23 (no coercion so just strings);
``````
Remember that I want to calculate numbers within the same string. Number doesn’t cater to that.
eg Number(‘1+4/5*9’) will return NaN.
Thank you for the suggestion.
Damn. I’m such a … and I used eval() too but didn’t think of it
1 Like
first off i suggest you create your calculator using this link below
they are many different ways of doing the calculation … everybody probably did it differently
so i would suggest looking at the code of those created already to get a few ideas (we can now look at code to help us … just dont copy)
I dont recommend using eval
1 Like
console.log(a+b) is 23 … string + string is a string
yes, that’s what console.log shows though my syntax may be funky. o shit, typo !!!
here just to make it more fun console.log(a++ + b++);
Right ! I remember having seen that - was it in YDKJS ?
yep lol … enjoyed those books learnt some interesting stuff from them
Back when I did this project, my code would make an array that looked like this :
`[ '1', '+' , '4', '/', '5', '*', '9' ]`
Then I used 2 loops, the first to compute multiply and divide operations(replace ‘4’, ‘/’, and ‘5’ with 0.8), and the 2nd for plus and minus.It was a very basic calculator though (no complex stuff, not even parenthesis).
My calculator already works. What’s wrong with using eval()?
This is what MDN say about eval() … so its better to learn about this function before you decide to use it … and then only use it with care.
eval() is a dangerous function, which executes the code it’s passed with the privileges of the caller. If you run eval() with a string that could be affected by a malicious party, you may end up running malicious code on the user’s machine with the permissions of your webpage / extension. More importantly, third party code can see the scope in which eval() was invoked, which can lead to possible attacks in ways to which the similar Function is not susceptible.
eval() is also generally slower than the alternatives, since it has to invoke the JS interpreter, while many other constructs are optimized by modern JS engines.
There are safer (and faster!) alternatives to eval() for common use-cases.
2 Likes
Thank you
I like these guidelines I’ve seen flaws in the calculator now. I’ll work on them.
1 Like
I think it would be safe enough if you parse the string given to eval() using regexp and filter out anything that is not a digit, +, - /, * ( or ), would it be not?
1 Like | 807 | 3,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-21 | latest | en | 0.879674 |
https://www.6miu.com/read-1250348.html | 1,686,029,579,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00419.warc.gz | 679,203,946 | 7,342 | # POJ - 2886 Who Gets the Most Candies?树状数组 + 二分 + 反素数
xiaoxiao2021-02-28 48
数，都有，那么称为反素数。
#include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> #define ll long long #define pb push_back #define fi first #define se second #define pi acos(-1) #define inf 0x3f3f3f3f #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define rep(i,x,n) for(int i=x;i<n;i++) #define per(i,n,x) for(int i=n;i>=x;i--) using namespace std; typedef pair<int,int>P; const int MAXN = 500010; const int MAXM = 1010; int gcd(int a,int b){return b?gcd(b,a%b):a;} int prime[MAXN]; bool isprime[MAXM]; int num = 0; P anti_prime[MAXM]; void dfs(int fac_num, int val, int pos, int limit) { if(val > MAXN) return ; anti_prime[num++] = P(val, fac_num); int tmp = val; for(int i = 1; i <= limit && tmp < MAXN; i++) { tmp *= prime[pos]; dfs(fac_num * (i + 1), tmp, pos + 1, i); } } void init() { // get prime int cnt = 0; memset(isprime, 1, sizeof(prime)); isprime[1] = 0; for(int i = 2; i < MAXM; i++) { if(isprime[i]) { prime[cnt++] = i; for(int j = i + i; j < MAXM; j += i) isprime[j] = 0; } } // get antiprime num = 0; dfs(1, 1, 0, 20); sort(anti_prime, anti_prime + num); } int bit[MAXN]; int sum(int i) { int res = 0; while(i) { res += bit[i]; i -= -i & i; } return res; } void add(int i, int delta) { while(i < MAXN) { bit[i] += delta; i += i & -i; } } int bi_search(int val) { int l = 0, r = MAXN, mid; while(l <= r) { mid = (l + r) >> 1; if(sum(mid) >= val) r = mid - 1; else l = mid + 1; } return r + 1; } char name[MAXN][15]; int card[MAXN]; int main() { int n, k; init(); //cout << num << endl; //for(int i = 0; i < num; i++) //printf("%d %d\n", anti_prime[i].fi, anti_prime[i].se); while(cin >> n >> k) { memset(bit, 0, sizeof(bit)); for(int i = 1; i <= n; i++) { scanf(" %s %d", name[i], card + i); add(i, 1); } int up = 0, ans = 0; for(int i = 0; i < num; i++) { if(anti_prime[i].fi > n) break; if(anti_prime[i].se > ans) up = anti_prime[i].fi, ans = anti_prime[i].se; } int now = k, nxt = k; for(int i = 1; i < up; i++) { add(now, -1);//将小朋友出队 int m = n - i; int x = card[now]; if(x > 0)//计算下一个要出队小朋友在当前序列的位置 nxt = (nxt - 1 + x) % m; else nxt = ((nxt + x) % m + m) % m; if(nxt == 0) nxt = m; //printf("%d %s\n", now, name[now]); now = bi_search(nxt);//得到下一个要出队的小朋友在原序列的位置 //printf("%d %d %d\n", now, nxt, m); } printf("%s %d\n", name[now], ans); } return 0; } | 929 | 2,368 | {"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-2023-23 | latest | en | 0.098432 |
https://philosophy.stackexchange.com/questions/9385/some-questions-about-implications-of-godels-completeness-theorem?answertab=votes | 1,571,147,499,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986659097.10/warc/CC-MAIN-20191015131723-20191015155223-00414.warc.gz | 574,604,810 | 28,214 | # Some questions about implications of Godel's Completeness Theorem [closed]
I have read Gödel's original paper (1930 - reprinted into J.van Heijenoort (editor), From Frege to Gödel: A Source Book in Mathematical Logic, 1879-1931, 1967) but I'm still unsatisfied with my understanding of it.
Modern logic textbooks usually prove the theorem following Henkin's construction.
One of the peculiarity of Henkin's proof is that the "completeness" aspect (i.e.if A is valid it is provable) is somewhat of a by-product of model existence.
Gödel's proof use (natural) numbers. This is obvious (with insight) today that we know about Gödel's philosophical realism.
Hankin's construction avoid numbers but use the "syntactical stuff" to build the model. But this (according to my understanding) is not really different; in order to "run" the construction we need countable many symbols, and symbols are "abstract entities" (like numbers). I think that we really needs them : we cannot replace them with "physical" tally marks. So my question :
In what sense we can minimize the "ontological" import of the theorem ?
In a previous effort I asked for some clarifications to a distinguished scholar and I received this answer : "About the completeness proof, it is a theorem of orthodox mathematics, and does not pretend to be nominalistic."
## closed as unclear what you're asking by Joseph Weissman♦Jan 17 '16 at 22:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• Can you frame the headline here as a question? I tried to and sort of got stuck, there's a lot going on here (it's maybe something about "minimizing ontological implications" of incompleteness?) -- I'm marking "unclear" for now – Joseph Weissman Jan 17 '16 at 22:29
• You may want to ask some questions about this on the mathematical form. As for the "syntactical stuff" argument, there are very specific requirements to invoke Godel's incompleteness theorem. The main one is omega-consistency, which is, in layman's terms, "the system can prove all true statements in the arithmetic of natural numbers." If it can do that, then it really doesn't matter what sort of clever tricks were done to obscure it (unless you do something really clever like Dan Willard's tricks to prohibit the construction of diagonalization) – Cort Ammon Jan 18 '16 at 2:41 | 601 | 2,597 | {"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-2019-43 | latest | en | 0.951434 |
http://www.talkstats.com/showthread.php/13083-What-is-Significance-F-and-what-is-its-forumula | 1,508,470,542,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823630.63/warc/CC-MAIN-20171020025810-20171020045810-00272.warc.gz | 574,892,513 | 16,173 | # Thread: What is 'Significance F' and what is its forumula?
1. ## What is 'Significance F' and what is its forumula?
My prof is giving us an ANOVA table to fill in with limited data and he pointed to significance f. I thought it was nothing or could find it on my own. I can't. Any suggestions?
Also, how can I find confidence intervals without having access to excel?
This is a summer stats 2 class. Thanks.
2. Anybody? Its not in my book or on the internet.
3. F is the ratio of the MSS between the factor and the residual error.
4. It would be helpful if you attached your table so we can take a look at it
5. ## Significance F ???
Here it is. Please open the link. I highlighted it but not if it was saved. Thanks.
6. Ok, so the f-ratio is as I mentioned before.
The significance F is your P-value (the p-value is the probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true) - in the case of regression: Ho: B1= 0
7. Is there a way I can make the formulas visible in this worksheet? I'd like to get comfortable with how excel is doing the calculations.
Thanks
8. oh byw, hitting control ~ doesn't do anything.
9. Now I'm using my mac right now using Numbers 09 as opposed to Excel but it looks like the regression output was hard coded in. The summaries in the data seemed to be from formulas inside the spreadsheet but if the regression output was hard coded I don't think there's a way to recover the formulas that whoever made the spreadsheet used. I mean we could tell you what the formulas are in general (actually we'd probably just point you to a website that has the formulas). Now this might be wrong since I'm not using Excel but Numbers picks up the formulas usually.
10. Yes it came from mystatlab that my prof just loves using. Not really learning much from it b/c he doesn't teach it. Which I guess he should b/c I don't think businesses use the pencil paper method much. I wish minitab was taught.
11. Originally Posted by bugman
Ok, so the f-ratio is as I mentioned before.
The significance F is your P-value (the p-value is the probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true) - in the case of regression: Ho: B1= 0
Hi Bugman : The F-test, in this case, is testing the (null) hypothesis:
Ho: B1=B2=0.
The t statistics associated with X1 and X2 are separately testing the unique contribution (i.e. the squared semi-partial correlation) of each independent variable i.e. Ho1: B1=0 and Ho2: B2=0
12. Thanks Dragan - missed that.
13. Is there a formula to calculate Significance F?
14. Do you mean is there a formula to calculate a p-value from an F statistic? Do you mean how do you calculate the F statistic? Are you wondering how to calculate the critical value for an F distribution? Can you elaborate on what you mean because what you're asking doesn't really make sense to me.
15. nevermind.
its Fcdf(.....) | 723 | 3,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-43 | latest | en | 0.971797 |
http://goodmath.blogspot.com/2006/05/improbable-is-not-impossible-redux.html | 1,438,062,361,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042981576.7/warc/CC-MAIN-20150728002301-00101-ip-10-236-191-2.ec2.internal.warc.gz | 105,772,835 | 15,038 | # Good Math/Bad Math
## Wednesday, May 31, 2006
### Improbable is not impossible, redux
I actually decided to say a bit more about the concept of improbable vs. impossible after my last post.
As I've said before, the fundamental idea behind these arguments is what I call the "big numbers" principle: We (meaning human beings) are very bad at really grasping the meaning of very large numbers. It's hard to really grasp what a million is: if you counted one number a second, with no breaks of any kind, it would take you more than 11 days to count to one million. When we start getting to numbers that dwarf a million, our ability to really understand what they mean - to really grasp on a deep, intuitive level what they mean - just completely goes out the window.
What does 10^20 mean? If it takes 11 days to count to one million, then it would take roughly 11 times 10^14 days - or 3 times 10^12 (aka 3 trillion) years to count to 10^20. That just doesn't mean anything on an intuitive level. So what does 10^100 mean?
What happens when we start to look at probability? Well, it's really easy to get huge numbers in probability calculations - the kinds of numbers that we can't really grasp. And because those numbers are so meaningless to us, so utterly beyond the reach of what we really comprehend, the probabilities of astonishingly common events can seem like they're entirely beyond the realm of possibility: they must be impossible, because how could anything so easily comprehensible create numbers so incomprehensible?
Who hasn't shuffled cards? Who hasn't played a game of klondike solitaire? And yet, every time we do that, we're witnessing an event with a probability of roughly 1 in 8 times 10^67!
If something as common as shuffling a deck of cards can produce a result with odds of 1 in 10^67, then think of the probability of something uncommon: what's the probability of the particular arrangement of molecules in a particle of dust? Or the shape of a comets tail? Those are all natural, possible events - and the probability of them happening the way that they do is pretty much beyond our capacity to imagine.
When you grasp that, then you start to realize just how meaningless the big numbers argument is. No matter how improbable something is, that does not mean that it is impossible. Impossible is a probability of 0. And any probability, no matter how small, is almost certainly the probability of something that's occurring in the universe, somewhere, right now.
#### 21 Comments:
• Ah, the wonders of taking a brain adapted for the plains of Africa and making it try to understand the universe.
By Thomas Winwood, at 11:07 PM
• I once belonged to a large computer users group that could draw major speakers to its monthly meetings. There were lots of giveaways, too. Members were given raffle tickets at the door. Product reps would show their wares and then draw winning numbers from a bucket. One fun-loving guy drew a couple of tickets in the usual away, bestowing prizes on the folks with the right numbers, then changed the script. He reached into the bucket, grabbed a handful of tickets and dumped them on the floor. "Okay, those guys lost," he said. Then he picked some more winners, giving away several more software packages.
People went nuts. I heard absolutely furious people demanding that no one ever be allowed to do such a horrible thing again. "What if he threw away my ticket?" they cried. "He robbed me of my chance to win!" There was no way to persuade them that their chances of winning a prize during the evening had in no way been affected by his method of drawing (or discarding) raffle tickets. Everyone had the same overall chance. But lots of people fixated on the possibility that their winning numbers had been trampled under foot.
By Zeno, at 11:24 PM
• "And any probability, no matter how small, is almost certainly the probability of something that's occurring in the universe, somewhere, right now."
What a neat comment! Very subtle and true. I misread it at first, then looked again. The problem for me happens when the number of individual outcomes is immense, but the number of classes of outcomes I care about is small. Then one has a vanishingly small probablility of an individual outcome, but perhaps even odds of some outcome in a class we are looking at.
Say for example cards and blackjack, There are those zillions of possible shuffles, but really only 20 classes of them that matter for me playing blackjack. That is contrived but I do stumble on that point sometimes.
So when you turn it around you say look, only a one in 10^68 or so chance of this particular thing, not noticing that its not 1 but 10^67 possiblities out of 10^68 that I would say the same thing about...
Anyway hope that made some sense...
P.S.
And strongly typed languages never led to any discernable drops in software bug reports in my 15 years when I still did software... (heh heh, had to put that in :-) Function point analysis was still the best predictor when I quit dealing with that...
By Markk, at 11:25 PM
• "Everyone had the same overall chance."
Really? I'm currently too tired to model the question properly, but I believe I can see that drawing procedure affects chances.
Say you have 10 prizes and 100 tickets. If the first ticket takes a prize, the chances for the remaining participants drops from 1/10 to 1/11.
Would removing tickets to a nondrawing pot during drawing make a difference in chances?
By Torbjörn Larsson, at 2:47 AM
• Er, that's not quite right: there's an important distinction between "probability zero" and "impossible". For example, if you pick a real number uniformly at random between 0 and 1, you obtain a rational number with probability zero — but there are rational numbers in that range!
This distinction matters because sometimes you need "there is no number in the desired range such that [some condition]" — if a rational number fulfils [some condition], you don't have it, no matter that they occur with probability zero.
By Anne, at 3:34 AM
• Yes, I was too tired. It is the drawn tickets that wins here. So removing tickets doesn't change anything. (Unless you remove so many that you end up with fewer tickets than prizes, of course.)
By Torbjörn Larsson, at 4:14 AM
• Really, I always like the explanation of calculus, mostly because that "ding!" moment of getting it is still one of my fondest memories from high school.
You can add an infinite number of infinitely small things and end up with a finite number. Make them as small as you want and it still works just fine (it only gets more accurate).
It's the same with probability.
By sailorman, at 12:20 PM
• Anne:
You wrote: "if you pick a real number uniformly at random between 0 and 1, you obtain a rational number with probability zero — but there are rational numbers in that range!"
Two errors:
1. It is not possible to have a uniform probability distribution over the reals between 0 and 1. So you've set up a false example.
2. There's a slight difference between limits, and real zero. The probability of choosing a rational might head to zero as the set of reals that you choose from grows, but that's not quite the same as saying there is zero probability of choosing a rational.
By Don Geddis, at 1:31 PM
• Some of what you argue boarders on a more fundamental question that we as people have been asking ourselves for eons; is there a god? I agree with you on zero probability and the impossible vs. improbable argument. It’s interesting to think about but no matter how small the probability of something is, if given enough time which for all we currently know is infinite then it WILL happen. Its this question of time that becomes vitally important in this particular argument. If the common acknowledgement of big bang theory are correct then the answer is no. How does time “stop”? How do we know when it stops? Does time need an observer for it to exist? However more importantly in the big bang model regardless if the theory that the universe collapses or expands infinitily are correct they both are on an infinite time table. Some might say that the big bang theory is theory after all so effectivly bringing us back to the question; does God exist. Its this question that arises because we simply don't know for sure. Yet oddly enough its also this same benefiet of the doubt that allows the fundamentialist with any crediable foundation (the existence of god). Its intresting to think about but proving that life was created out of pure "chance" does not prove that god does not exist. I understand that fundamentalist believe the Bible word for word and feel threatened from the prospect of life being created outside this “model” but they've put the question how did God create the universe? Does God exist? as mutally inclusive.
By Anonymous, at 2:16 PM
• A.C. Doyle's great creation put it best:
"How often have I said to you that when you have eliminated the impossible whatever remains, howevere improbable, must be the truth?" The Sign of the Four
And the wonderful corollary:
"It IS impossible as I state it, and therefore I must in some respect have stated it wrong." The Adventure of the Priory School
By usagi, at 2:22 PM
• Sure it's possible to have a uniform distribution on [0,1] - the Lebesgue measure is just that. If that bothers you, look up the definition of a probability distribution - it's a positive measure with total measure one.
I'm not taking any kind of limit here; I really am just drawing from the [0,1] interval according to the Lebesgue measure. So, at the start, all the rationals are possible.
By Anne, at 2:31 PM
• How does time “stop”? How do we know when it stops?
It's probably self-contained, like space. No stop necessary.
Does time need an observer for it to exist?
If you're referring to the way quantum mechanics actually refers to "observers," probably. If you're talking about conscious observers, why would it need one?
However more importantly in the big bang model regardless if the theory that the universe collapses or expands infinitily are correct they both are on an infinite time table.
One hypothesis I've heard: After the universe chills down to a sea of weak particles (I forget which kind), the laws of probability will inevitably cause all of those particles to jump back into one spot, creating the seed for another big bang.
Some might say that the big bang theory is theory after all so effectivly bringing us back to the question; does God exist. Its this question that arises because we simply don't know for sure.
It's hard to answer that question without a good definition of "God."
Its intresting to think about but proving that life was created out of pure "chance" does not prove that god does not exist.
1. No one says life arose out of pure "chance" ...Except maybe the Last Thursdayists.
2. True, but only because you can't prove that sort of negative. Especially since god beliefs tend to be constructed, often deliberately, to be unfalsifiable.
By Bronze Dog, at 2:33 PM
• One hypothesis I've heard: After the universe chills down to a sea of weak particles (I forget which kind), the laws of probability will inevitably cause all of those particles to jump back into one spot, creating the seed for another big bang.
Exactly, point being that it’s an infinite process. Big bang --> inevitable collapse to a singular point --> Big Bang…
2. True, but only because you can't prove that sort of negative. Especially since god beliefs tend to be constructed, often deliberately, to be unfalsifiable.
We could discuss this, point by point but it’ll just lead us to the old philosophy question of what came first the chicken or the egg. If god doesn’t exist what happened to create the “stuff”? Was the “stuff” just there from the start? Not sure if we could answer that question in terms of the classical model of “god” as some higher BEING implying that it takes form, something tangible.
It's hard to answer that question without a good definition of "God."
How a seemingly trivial question might actually create the model to answer the chicken and egg question and more importantly the existence of God. God could be anything.
By Anonymous, at 3:19 PM
• We could discuss this, point by point but it’ll just lead us to the old philosophy question of what came first the chicken or the egg. If god doesn’t exist what happened to create the “stuff”? Was the “stuff” just there from the start? Not sure if we could answer that question in terms of the classical model of “god” as some higher BEING implying that it takes form, something tangible.
So far, we seem to live in an uncreated universe: Matter and energy are apparently eternal. No need to invoke a special agent yet.
By Bronze Dog, at 4:02 PM
• It's not that sort of probability at all because it involves selection. An analogy: two boxes of 100 dice. The object is to get all sixes. One box is shaken until all 100 dice show 6. The other is shaken, but the dice showing 6's are removed and set aside, i.e., selected. Guess which set of dice will reach the end (100 sixes) first?
By Anonymous, at 4:32 PM
• It's not that sort of probability at all because it involves selection. An analogy: two boxes of 100 dice. The object is to get all sixes. One box is shaken until all 100 dice show 6. The other is shaken, but the dice showing 6's are removed and set aside, i.e., selected. Guess which set of dice will reach the end (100 sixes) first?
Selection for what? If your talking about a number (single value) then yes, what you said applies. However what if we said that it required all of the 100 dices to show 6 or if you needed a combination of numbers to show the wanted outcome. Point is that for life to have had appeared it required many different conditions to be met. Just wondering how selection would apply to particle probability for life.
So far, we seem to live in an uncreated universe: Matter and energy are apparently eternal. No need to invoke a special agent yet.
“Uncreated universe” implying no “god” in the classical sense. Like I mentioned I don’t find the “matter and energy are eternal” a valid argument. It’s the same logic that god is eternal. What do you mean by “apparently”? As in, it’s absolute that energy and matter are eternal? Let’s leave stipulation for laymen, what is “god” (i.e. some higher being, energy, matter, anything) is nothing compared to why. I could care less if god was a living breathing woman, or even if god didn’t exist but rather just matter and energy. I want to know why its here, always eternal.
Either way I think it’s peeked my interests on physics again and I want to remind Mark that his topology articles are much anticipated.
By Anonymous, at 5:57 PM
• Like I mentioned I don’t find the “matter and energy are eternal” a valid argument. It’s the same logic that god is eternal. What do you mean by “apparently”? As in, it’s absolute that energy and matter are eternal?
I don't understand what the problem is. So far, no one has been able to find a way to create or destroy energy. That's where their apparent eternal nature comes from. If someone invents a free energy machine, we'll know otherwise. No absolutes are involved.
I want to know why its here, always eternal.
To be cheeky with a point, why not?
By Bronze Dog, at 8:15 PM
• "Like I mentioned I don’t find the “matter and energy are eternal” a valid argument. It’s the same logic that god is eternal. What do you mean by “apparently”? As in, it’s absolute that energy and matter are eternal? Let’s leave stipulation for laymen, what is “god” (i.e. some higher being, energy, matter, anything) is nothing compared to why. I could care less if god was a living breathing woman, or even if god didn’t exist but rather just matter and energy. I want to know why its here, always eternal."
Interesting. You have turned the usual argument (finite time, therefore god) around (infinite time, therefore god).
You can't prove the existence or nonexistence of gods - too many has tried too long to think that is feasible. You could possibly verify an ordinary theory beyond reasonable doubt - if you could come up with a definition that is agreeable.
Time is more basic than emergent spacetime - quantum theory tells us so. There are indeed infinite time cosmologies. Endless inflation embeds our bigbang universe in an infinite time multiverse. (New universes wormholes from old due to vacuum fluctuations.)
Massenergy conservation follows from infinite time and Noether's theorem. Noethers theorem follow from observations and has been verified beyond reasonable doubt.
For example, the last bound on the change of the fine structure constant was to combine astronomical and lab measurements to argue that the upper bound for the change during the existence of the Solar system is one part per million.
This is a bound on energy constance for most interactions. There are others. The most general ones relies on inflation as observed by WMAP CMB going back to within parts of a second from bigbang.
Unless you believe in supernatural actions intruding in our natural causal continua. Prove it with an observation!
But remember, supernatural actions can at most explain about one part per million of what has happened on Earth. The rest is nature alone.
Coincidentally, this is beyond the usual choosen limits for physical theories to be considered verified beyond reasonable doubt. In this case the absence of other than natural actions. Perhaps you will think of this as the evidence for that no gods exists. I do; and there are other similar ideas with similar results.
You ask "why"? If the universe is infinite and creationless, there is no reason to ask why. If the universe isn't you can't ask the question. If the universe is you can ask the question. That is an
observer effect.
The observer effect is usually interpreted such that the question is not only meaningless but false. If you think this case is different, you must prove why!
By Torbjörn Larsson, at 1:17 AM
• Math good.
Physics good.
Philosophobabble bad.
By Anonymous, at 3:33 AM
• Anonymous, you are funny.
The essence of formalised religion is philosophy disconnected from facts. This framed your questions.
Both bronze and I made efforts to give physics based answer.
AFAIK, your question "why" is always a philosobabble question. But in this case it happens to have a physics answer. From science verified methods we can easily conclude asking "why is the universe" is a failed question.
By Torbjörn Larsson, at 11:35 PM
• what came first the chicken or the egg. .
There is fossil evidence for eggs that predates chickens. What's left to think about?
By Stephen, at 1:52 PM
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<< Home | 4,228 | 18,822 | {"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-2015-32 | longest | en | 0.954385 |
https://testbook.com/question-answer/a-car-travels-20-km-towards-south-then-it-turns-r--604c6dabb42a068a9fcd319a | 1,631,866,858,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055601.25/warc/CC-MAIN-20210917055515-20210917085515-00116.warc.gz | 618,751,556 | 28,882 | # A car travels 20 km towards South. Then it turns right and travels 60 km. Then it turns right and travels 170 km. Then it turns left and travels 20 km. How far and in which direction is the car from its starting point?
This question was previously asked in
DSSSB JE ME 2019 Official Paper Shift-1 (Held on 06 Nov 2019)
View all DSSSB JE Papers >
1. 170 km, north west
2. 170 km, south west
3. 220 km, north west
4. 220 km, south west
Option 1 : 170 km, north west
## Detailed Solution
Below diagram shows the distance travel by the car:
Now, Triangle "ABD" is a right-angled triangle, so using Pythagoras Theorem: | 179 | 620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-39 | latest | en | 0.897805 |
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# Math help
+1
184
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Chad built a scale model of a statue. He built the model 7 inches tall to represent the actual height of 15 feet. Which equation below represents the relationship between the actual height (a) ,in feet and the height of the model (m) , in inches?
KawaiiKibby May 4, 2017
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#1
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Chad built a scale model of a statue.
He built the model 7 inches tall to represent the actual height of 15 feet.
Which equation below represents the relationship between the actual height (a) ,in feet and the height of the model (m) , in inches?
$$a\;\; \alpha\;\; m\\ a\;\; =\;\;k * m\\ \text{When actual is 15 model is 7}\\ 15\;\; =\;\;k * 7\\ k=\frac{15}{7}\\ a\;\; =\;\;\frac{15}{7} * m\\ a\;\; =\;\;\frac{15m}{7} \\$$
Melody May 4, 2017
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https://people.maths.bris.ac.uk/~matyd/GroupNames/192/D12s16D4.html | 1,582,225,906,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145260.40/warc/CC-MAIN-20200220162309-20200220192309-00382.warc.gz | 477,936,858 | 6,228 | Copied to
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## G = D12⋊16D4order 192 = 26·3
### 4th semidirect product of D12 and D4 acting via D4/C22=C2
Series: Derived Chief Lower central Upper central
Derived series C1 — C2×C12 — D12⋊16D4
Chief series C1 — C3 — C6 — C12 — C2×C12 — C2×D12 — C22×D12 — D12⋊16D4
Lower central C3 — C6 — C2×C12 — D12⋊16D4
Upper central C1 — C22 — C22×C4 — C4⋊D4
Generators and relations for D1216D4
G = < a,b,c,d | a12=b2=c4=d2=1, bab=a-1, cac-1=a7, ad=da, cbc-1=a3b, bd=db, dcd=c-1 >
Subgroups: 752 in 198 conjugacy classes, 47 normal (27 characteristic)
C1, C2 [×3], C2 [×7], C3, C4 [×2], C4 [×2], C22, C22 [×2], C22 [×21], S3 [×4], C6 [×3], C6 [×3], C8 [×2], C2×C4 [×2], C2×C4 [×3], D4 [×14], C23, C23 [×11], C12 [×2], C12 [×2], D6 [×16], C2×C6, C2×C6 [×2], C2×C6 [×5], C22⋊C4, C4⋊C4, C2×C8 [×2], D8 [×4], C22×C4, C2×D4, C2×D4 [×8], C24, C3⋊C8 [×2], D12 [×4], D12 [×6], C2×C12 [×2], C2×C12 [×3], C3×D4 [×4], C22×S3 [×10], C22×C6, C22×C6, C22⋊C8, D4⋊C4 [×2], C4⋊D4, C2×D8 [×2], C22×D4, C2×C3⋊C8 [×2], D4⋊S3 [×4], C3×C22⋊C4, C3×C4⋊C4, C2×D12 [×2], C2×D12 [×5], C22×C12, C6×D4, C6×D4, S3×C23, C22⋊D8, C6.D8 [×2], C12.55D4, C2×D4⋊S3 [×2], C3×C4⋊D4, C22×D12, D1216D4
Quotients: C1, C2 [×7], C22 [×7], S3, D4 [×6], C23, D6 [×3], D8 [×2], C2×D4 [×3], C3⋊D4 [×2], C22×S3, C22≀C2, C2×D8, C8⋊C22, D4⋊S3 [×2], S3×D4 [×2], C2×C3⋊D4, C22⋊D8, C2×D4⋊S3, C232D6, D4⋊D6, D1216D4
Smallest permutation representation of D1216D4
On 48 points
Generators in S48
(1 2 3 4 5 6 7 8 9 10 11 12)(13 14 15 16 17 18 19 20 21 22 23 24)(25 26 27 28 29 30 31 32 33 34 35 36)(37 38 39 40 41 42 43 44 45 46 47 48)
(1 43)(2 42)(3 41)(4 40)(5 39)(6 38)(7 37)(8 48)(9 47)(10 46)(11 45)(12 44)(13 35)(14 34)(15 33)(16 32)(17 31)(18 30)(19 29)(20 28)(21 27)(22 26)(23 25)(24 36)
(1 16 44 36)(2 23 45 31)(3 18 46 26)(4 13 47 33)(5 20 48 28)(6 15 37 35)(7 22 38 30)(8 17 39 25)(9 24 40 32)(10 19 41 27)(11 14 42 34)(12 21 43 29)
(1 7)(2 8)(3 9)(4 10)(5 11)(6 12)(13 27)(14 28)(15 29)(16 30)(17 31)(18 32)(19 33)(20 34)(21 35)(22 36)(23 25)(24 26)(37 43)(38 44)(39 45)(40 46)(41 47)(42 48)
G:=sub<Sym(48)| (1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15,16,17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45,46,47,48), (1,43)(2,42)(3,41)(4,40)(5,39)(6,38)(7,37)(8,48)(9,47)(10,46)(11,45)(12,44)(13,35)(14,34)(15,33)(16,32)(17,31)(18,30)(19,29)(20,28)(21,27)(22,26)(23,25)(24,36), (1,16,44,36)(2,23,45,31)(3,18,46,26)(4,13,47,33)(5,20,48,28)(6,15,37,35)(7,22,38,30)(8,17,39,25)(9,24,40,32)(10,19,41,27)(11,14,42,34)(12,21,43,29), (1,7)(2,8)(3,9)(4,10)(5,11)(6,12)(13,27)(14,28)(15,29)(16,30)(17,31)(18,32)(19,33)(20,34)(21,35)(22,36)(23,25)(24,26)(37,43)(38,44)(39,45)(40,46)(41,47)(42,48)>;
G:=Group( (1,2,3,4,5,6,7,8,9,10,11,12)(13,14,15,16,17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45,46,47,48), (1,43)(2,42)(3,41)(4,40)(5,39)(6,38)(7,37)(8,48)(9,47)(10,46)(11,45)(12,44)(13,35)(14,34)(15,33)(16,32)(17,31)(18,30)(19,29)(20,28)(21,27)(22,26)(23,25)(24,36), (1,16,44,36)(2,23,45,31)(3,18,46,26)(4,13,47,33)(5,20,48,28)(6,15,37,35)(7,22,38,30)(8,17,39,25)(9,24,40,32)(10,19,41,27)(11,14,42,34)(12,21,43,29), (1,7)(2,8)(3,9)(4,10)(5,11)(6,12)(13,27)(14,28)(15,29)(16,30)(17,31)(18,32)(19,33)(20,34)(21,35)(22,36)(23,25)(24,26)(37,43)(38,44)(39,45)(40,46)(41,47)(42,48) );
G=PermutationGroup([(1,2,3,4,5,6,7,8,9,10,11,12),(13,14,15,16,17,18,19,20,21,22,23,24),(25,26,27,28,29,30,31,32,33,34,35,36),(37,38,39,40,41,42,43,44,45,46,47,48)], [(1,43),(2,42),(3,41),(4,40),(5,39),(6,38),(7,37),(8,48),(9,47),(10,46),(11,45),(12,44),(13,35),(14,34),(15,33),(16,32),(17,31),(18,30),(19,29),(20,28),(21,27),(22,26),(23,25),(24,36)], [(1,16,44,36),(2,23,45,31),(3,18,46,26),(4,13,47,33),(5,20,48,28),(6,15,37,35),(7,22,38,30),(8,17,39,25),(9,24,40,32),(10,19,41,27),(11,14,42,34),(12,21,43,29)], [(1,7),(2,8),(3,9),(4,10),(5,11),(6,12),(13,27),(14,28),(15,29),(16,30),(17,31),(18,32),(19,33),(20,34),(21,35),(22,36),(23,25),(24,26),(37,43),(38,44),(39,45),(40,46),(41,47),(42,48)])
33 conjugacy classes
class 1 2A 2B 2C 2D 2E 2F 2G 2H 2I 2J 3 4A 4B 4C 4D 6A 6B 6C 6D 6E 6F 6G 8A 8B 8C 8D 12A 12B 12C 12D 12E 12F order 1 2 2 2 2 2 2 2 2 2 2 3 4 4 4 4 6 6 6 6 6 6 6 8 8 8 8 12 12 12 12 12 12 size 1 1 1 1 2 2 8 12 12 12 12 2 2 2 4 8 2 2 2 4 4 8 8 12 12 12 12 4 4 4 4 8 8
33 irreducible representations
dim 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 4 4 4 4 type + + + + + + + + + + + + + + + + + + image C1 C2 C2 C2 C2 C2 S3 D4 D4 D4 D6 D6 D6 D8 C3⋊D4 C3⋊D4 C8⋊C22 S3×D4 D4⋊S3 D4⋊D6 kernel D12⋊16D4 C6.D8 C12.55D4 C2×D4⋊S3 C3×C4⋊D4 C22×D12 C4⋊D4 D12 C2×C12 C22×C6 C4⋊C4 C22×C4 C2×D4 C2×C6 C2×C4 C23 C6 C4 C22 C2 # reps 1 2 1 2 1 1 1 4 1 1 1 1 1 4 2 2 1 2 2 2
Matrix representation of D1216D4 in GL6(𝔽73)
72 0 0 0 0 0 0 72 0 0 0 0 0 0 0 72 0 0 0 0 1 72 0 0 0 0 0 0 0 1 0 0 0 0 72 0
,
1 0 0 0 0 0 0 72 0 0 0 0 0 0 1 72 0 0 0 0 0 72 0 0 0 0 0 0 0 1 0 0 0 0 1 0
,
0 71 0 0 0 0 37 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 57 57 0 0 0 0 57 16
,
1 0 0 0 0 0 0 72 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 72 0 0 0 0 0 0 72
G:=sub<GL(6,GF(73))| [72,0,0,0,0,0,0,72,0,0,0,0,0,0,0,1,0,0,0,0,72,72,0,0,0,0,0,0,0,72,0,0,0,0,1,0],[1,0,0,0,0,0,0,72,0,0,0,0,0,0,1,0,0,0,0,0,72,72,0,0,0,0,0,0,0,1,0,0,0,0,1,0],[0,37,0,0,0,0,71,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,57,57,0,0,0,0,57,16],[1,0,0,0,0,0,0,72,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,72,0,0,0,0,0,0,72] >;
D1216D4 in GAP, Magma, Sage, TeX
D_{12}\rtimes_{16}D_4
% in TeX
G:=Group("D12:16D4");
// GroupNames label
G:=SmallGroup(192,595);
// by ID
G=gap.SmallGroup(192,595);
# by ID
G:=PCGroup([7,-2,-2,-2,-2,-2,-2,-3,254,219,1123,297,136,6278]);
// Polycyclic
G:=Group<a,b,c,d|a^12=b^2=c^4=d^2=1,b*a*b=a^-1,c*a*c^-1=a^7,a*d=d*a,c*b*c^-1=a^3*b,b*d=d*b,d*c*d=c^-1>;
// generators/relations
×
𝔽 | 3,827 | 5,773 | {"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-2020-10 | longest | en | 0.352937 |
http://stackoverflow.com/questions/16313591/algorithm-for-preference-based-grouping | 1,464,402,707,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277286.69/warc/CC-MAIN-20160524002117-00021-ip-10-185-217-139.ec2.internal.warc.gz | 288,873,241 | 18,889 | # Algorithm for preference based grouping
I am looking to figure out a way to sort people into classes by preference.
For example, say there are 100 students that are each going to be assigned one of five classes:
• Science - 40 seats
• Math - 15 seats
• History - 15 seats
• Computers - 20 seats
• Writing - 10 seats
Each student has three preferred classes that are ordered by preference. What is the best way to approach dividing up the students so that as many people get their first and second choice classes as possible, while at the same time making sure that no class has too many students for the room.
I've thought about approaching it by the following method:
1. Group all students by their first choice class
2. See which classes have too many students and which have too few
3. Check to see if any students in the overbooked classes have second choice classes which are underbooked
4. Move those students accordingly
5. Repeat 2-4 with 3rd choice classes
While I feel like this is a reasonable implementation, I am wondering if there are any other algorithms that solve this problem in a better way. I have tried searching all over, but I cannot find anything that would solve this kind of problem.
-
One 'problem' with these sort of algorithms is that it is easy to 'cheat' by selecting popular (and small) courses as a 2nd and 3rd choice to force a 1st-choice placement.. I would be highly interested in a solution that addresses this somehow (although I currently have no intuition to approach it). – Joost Mar 24 at 14:32 | 348 | 1,546 | {"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} | 2.578125 | 3 | CC-MAIN-2016-22 | latest | en | 0.97649 |
http://forums.wolfram.com/mathgroup/archive/2005/Dec/msg00669.html | 1,579,540,303,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250599718.13/warc/CC-MAIN-20200120165335-20200120194335-00024.warc.gz | 65,671,690 | 8,008 | Re: Extracting information from lists
• To: mathgroup at smc.vnet.net
• Subject: [mg63367] Re: Extracting information from lists
• From: rudy <rud-x at caramail.com>
• Date: Tue, 27 Dec 2005 04:42:38 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```Hello,
li = {7, 9, 1, 6, 8, 1, 1, 6, 5, 1, 1, 1, 8, 7, 6, 1, 1, 1, 1, 7};
lolo = {0, 2, 1, 1, 0, 1, 1, 0, 0, 2, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0};
lulu = {a, b, d, d, d, d, c, f, g, t, d, d, e, r, d, d, d, e, r, d, d, e, f, r,f};
Here are three function with different degrees of generalisation:
(in the 3 functions, the parameter n is the max length of the series of 1 considered)
fct1[li_, n_] := Module[{inter, res, pourN},
pourN[l_List, i_Integer] := Array[Position[Split[l],
Table[1, {#}]] &, {i}] // Flatten;
inter = pourN[li, n];
res = (Length[#] + 1 &) /@ ((Take[Split[li], # - 1] // Flatten) & /@
inter)
];
example:
in > fct1[li, 5]
out > {3, 6, 10, 16}
but some problems arise when there are some series of '1' of same length in the list and when these series are in disorder (example in the list lolo):
in > fct1[lolo, 5]
out > {3, 6, 12, 19, 28, 15, 22}
when this result is given, whe know the position of the series but we ignore the length of each one...
{the number of 1 in the serie, the position of the serie}
fct2[lis_, n_] := Module[{inter1, inter2, res, pourN},
pourN[l_List, i_Integer] := Array[({Position[Split[l], Table[
1, {#}]], #} // Flatten) &, {i}];
inter1 = pourN[lis, n] // Select[#, Length[#] >= 2 &] &;
inter2 = Table[{#[[i]], Last[#]}, {i, 1,
Length[#] - 1}] & /@ inter1 // Flatten[#, 1] &;
res = {#[[2]], Length[#[[1]]] + 1} & /@ ({Take[Split[
lis], #[[1]] - 1] // Flatten, #[[2]]} & /@ inter2)
];
for example:
in > fct2[li, 5]
out > {{1, 3}, {2, 6}, {3, 10}, {4, 16}}
this one is the more general: it gives the positions of all the series of character 's' with max length 'n' in the list with their positions:
fct3[lis_, s_, n_] := Module[{inter1, inter2, res, pourN},
pourN[l_List, i_Integer] := Array[({Position[Split[l], Table[
s, {#}]], #} // Flatten) &, {i}];
inter1 = pourN[lis, n] // Select[#, Length[#] >= 2 &] &;
inter2 = Table[{#[[i]], Last[#]}, {i, 1, Length[#] - 1}] & /@ inter1 //
Flatten[#, 1] &;
res = {#[[2]], Length[#[[
1]]] + 1} & /@ ({Take[Split[
lis], #[[1]] - 1] // Flatten, #[[2]]} & /@ inter2)
];
for example:
in > fct3[lulu, d, 5]
out > {{2, 11}, {2, 20}, {3, 15}, {4, 3}}
in > fct3[lulu, d, 2]
out > {{2, 11}, {2, 20}}
regards
Rudy
```
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• Next by thread: Re: Extracting information from lists | 1,117 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-05 | longest | en | 0.580097 |
https://florian-roemer.de/blog/2016/05/ | 1,556,171,448,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578689448.90/warc/CC-MAIN-20190425054210-20190425080210-00169.warc.gz | 419,262,503 | 8,180 | Extended Trigonometric Pythagorean Identities: The Proof
I recently posted on Extended Trigonometric Pythagorean Identities and a “supercharged” version of them, which allows to simplify certain sums of shifted sine functions raised to integer powers. In particular, the claim was that
$$\sum_{n=0}^{N-1} \sin^2\left(x+n\frac{\pi}{N}\right) = \frac{N}{2}$$
or more generally for any integer $k$ and $N\geq k+1$:
$$\sum_{n=0}^{N-1} \sin^{2k}\left(x+n\frac{\pi}{N}\right) = N \frac{(2k)!}{(k!)^2 2^{2k}} = \frac{N}{\sqrt{\pi}} \frac{\Gamma(k+1/2)}{\Gamma(k+1)}$$
However, in the initial TPI post, I struggled a bit with the proof. It took a while to realize that it is actually quite simple using (a) the algebraic series, i.e.,
$$\sum_{n=0}^{N-1} q^n = \frac{1-q^N}{1-q}$$
for any $q \in \compl_{\neq 0,1}$ and (b) the fact that $\sin(x) = \frac{1}{2\jmath}\left({\rm e}^{\jmath x} – {\rm e}^{-\jmath x}\right)$. Let us use this relation to prove the following Lemma:
Lemma 1: For any $M \in \mathbb{Z}$, we have
$$\sum_{n=0}^{N-1}{\rm e}^{\jmath \frac{2\pi}{N} \cdot M \cdot n} = \begin{cases} N & M = k \cdot N, \, k\in \mathbb{Z} \\ 0 & {\rm otherwise}. \end{cases}$$
Proof: The proof is simple by realizing that the sum is in fact an arithmetic series with $q={\rm e}^{\jmath \frac{2\pi}{N} \cdot M}$. Obviously, if $M$ is an integer multiple of $N$ we have $q=1$ and the sum is equal to $N$. Otherwise, by the above identity, the series is equal to
$$\frac{1-{\rm e}^{\jmath 2\pi \cdot M}}{1-{\rm e}^{\jmath \frac{2\pi}{N} \cdot M}} = \frac{{\rm e}^{\jmath \pi \cdot M}}{{\rm e}^{\jmath \frac{\pi}{N} \cdot M}}\cdot \frac{{\rm e}^{-\jmath \pi \cdot M}-{\rm e}^{\jmath \pi \cdot M}}{{\rm e}^{-\jmath \frac{\pi}{N} \cdot M}-{\rm e}^{\jmath \frac{\pi}{N} \cdot M}}= {\rm e}^{\jmath \frac{\pi}{N}M(N-1)} \cdot \frac{\sin(\pi M)}{\sin(\frac{\pi}{N}M)} = 0,$$
since the enumerator is zero (and the denominator is not).
Piece of cake.
Now, we can proceed to prove the TPI for $2k=2$:
\begin{align}
\sum_{n=0}^{N-1} \sin^2\left(x+n\frac{\pi}{N}\right)
& = \sum_{n=0}^{N-1} \left(\frac{1}{2\jmath} {\rm e}^{\jmath(x+n\frac{\pi}{N})}- \frac{1}{2\jmath} {\rm e}^{-\jmath(x+n\frac{\pi}{N})}\right)^2\\
& = -\frac{1}{4}\sum_{n=0}^{N-1} {\rm e}^{2\jmath(x+n\frac{\pi}{N})} + {\rm e}^{-2\jmath(x+n\frac{\pi}{N})} – 2 {\rm e}^{\jmath(x+n\frac{\pi}{N})-\jmath(x+n\frac{\pi}{N})} \\
& = -\frac{1}{4} {\rm e}^{2\jmath x} \sum_{n=0}^{N-1} {\rm e}^{\jmath 2n\frac{\pi}{N}}
-\frac{1}{4} {\rm e}^{-2\jmath x} \sum_{n=0}^{N-1} {\rm e}^{-\jmath 2n\frac{\pi}{N}}
-\frac{1}{4} \sum_{n=0}^{N-1} (-2) \\
& = -\frac{1}{4} \cdot 0 -\frac{1}{4}\cdot 0 -\frac{1}{4}\cdot(-2N) = \frac{N}{2}
\end{align}
where we have used Lemma 1 for $M=1$ and $M=-1$ (which for the Lemma to work requires $M\neq N$ and thus $N\geq 2$). Isn’t that simple? I wonder why I didn’t see it earlier.
Even better yet, this technique allows to extend the proof to other values of $k$. Let’s try $2k=4$:
\begin{align}
\sum_{n=0}^{N-1} \sin^4\left(x+n\frac{\pi}{N}\right)
& = \sum_{n=0}^{N-1} \left(\frac{1}{2\jmath} {\rm e}^{\jmath(x+n\frac{\pi}{N})}- \frac{1}{2\jmath} {\rm e}^{-\jmath(x+n\frac{\pi}{N})}\right)^4\\
& = \frac{1}{16} \sum_{n=0}^{N-1} {\rm e}^{4 \jmath(x+n\frac{\pi}{N})}
-4 {\rm e}^{3\jmath(x+n\frac{\pi}{N}) -\jmath(x+n\frac{\pi}{N})}
+6 {\rm e}^{2\jmath(x+n\frac{\pi}{N}) -2 \jmath(x+n\frac{\pi}{N})} \\ &
-4 {\rm e}^{\jmath(x+n\frac{\pi}{N}) -3\jmath(x+n\frac{\pi}{N})}
+ {\rm e}^{-4 \jmath(x+n\frac{\pi}{N})} \\
& = \frac{1}{16}{\rm e}^{4 \jmath x} \sum_{n=0}^{N-1} {\rm e}^{4 \jmath n\frac{\pi}{N}}
– \frac{4}{16}{\rm e}^{2\jmath x} \sum_{n=0}^{N-1}{\rm e}^{2 \jmath n\frac{\pi}{N}}
+ \frac{6}{16}\sum_{n=0}^{N-1} {\rm e}^{0} \\ &
– \frac{4}{16}{\rm e}^{-2\jmath x} \sum_{n=0}^{N-1}{\rm e}^{-2 \jmath n\frac{\pi}{N}}
+ \frac{1}{16}{\rm e}^{-4 \jmath x} \sum_{n=0}^{N-1} {\rm e}^{-4 \jmath n\frac{\pi}{N}} \\
& = \frac{1}{16} \cdot 0 – \frac{4}{16} \cdot 0 + \frac{6}{16} \cdot N – \frac{4}{16} \cdot 0 + \frac{1}{16} \cdot 0 = \frac{3}{8} N,
\end{align}
where this time we have used Lemma 1 for $M=2, 1, -1, -2$ and thus need $N\geq 3$. This already shows the pattern: The polynomic expansion creates mostly terms with vanishing sums except for the “middle” term where the exponents cancel. The coefficient in front of this term is $\frac{1}{2^{2k}}$ (from the $\frac{1}{2\jmath}$ that comes with expanding the sine) times ${2k \choose k}$ (from the binomial expansion). This explains where the constant $N \cdot \frac{(2k)!}{(k!)^2 2^{2k}}$ comes from.
Formally, we have
\begin{align}
\sum_{n=0}^{N-1} \sin^{2k}\left(x+n\frac{\pi}{N}\right) & =
\sum_{n=0}^{N-1} \left( \frac{1}{2\jmath}{\rm e}^{\jmath(x+n\frac{\pi}{N})}
– \frac{1}{2\jmath}{\rm e}^{-\jmath(x+n\frac{\pi}{N})} \right)^{2k} \\
& =
\sum_{n=0}^{N-1} \frac{1}{(2\jmath)^{2k}} \sum_{\ell = 0}^{2k} {2k \choose \ell} (-1)^\ell
{\rm e}^{(2k-\ell) \jmath (x+n\frac{\pi}{N})}{\rm e}^{-\ell \jmath (x+n\frac{\pi}{N})} \\
& =
\sum_{n=0}^{N-1} \frac{(-1)^k}{2^{2k}} \sum_{\ell = 0}^{2k} (-1)^\ell {2k \choose \ell}
{\rm e}^{2(k-\ell) \jmath (x+n\frac{\pi}{N})}\\
& = \frac{(-1)^k}{2^{2k}} \sum_{\ell = 0}^{2k} (-1)^\ell {2k \choose \ell}
{\rm e}^{2(k-\ell) \jmath x} \sum_{n=0}^{N-1}
{\rm e}^{2(k-\ell) \jmath n\frac{\pi}{N}} \\
& = \frac{1}{2^{2k}} {2k \choose k} N,
\end{align}
where in the last step all terms $\ell \neq k$ vanish due to Lemma 1 applied for $M=k, k-1, …, 1, -1, …, -k+1, -k$. This requires $N\geq k+1$.
Eh voila.
Trigonometric Pythagorean Identity, supercharged
You know how they say good things always come back? Well, I recently stumbled over something that reminded me a lot on a post I had made about generalizations of the Trigonometric Pythagorean Identity. In short, the well-known identity $\sin^2(x)+\cos^2(x)=1$ can be generalized to a sum of $N\geq 2$ terms that are uniformly shifted copies of the sine function, which yields
$$\sum_{n=0}^{N-1} \sin^2\left(x+n\frac{\pi}{N}\right) = \frac{N}{2}$$
Well, I now came across a sum of fourth powers of shifted sine functions and much to my initial surprise, these admit very similar simplifications. In fact, it works for any integer power! Look at what I found:
$$\sum_{n=0}^{N-1} \sin^{2k}\left(x+n\frac{\pi}{N}\right) = N \frac{(2k)!}{(k!)^2 2^{2k}} = \frac{N}{\sqrt{\pi}} \frac{\Gamma(k+1/2)}{\Gamma(k+1)} \; k \in \mathbb{N}$$
for $N\geq k+1$. Isn’t this fascinating? No matter to which even power we raise the shifted sines, their sums will always give a constant in the form $c_k \cdot N$ and this constants $c_k$ can be computed analytically.
Here are some examples: sum of squares ($k=1$): $c_1 = 1/2$, sum of fourth powers ($k=2$): $c_2=3/8$, $k=3: 5/16$, $k=4: 35/128$ and so on. Moreover, I think I know how to prove even the “supercharged” version of the TPI for any $k$. I’ll write about it in another blog post.
*Update*: And here is the proof!
*Update2*: Just another small addition: The coefficients $c_k$ satisfy an interesting recurrence relation since you can compute $c_k$ as
$$c_k = \frac{2k+1}{2k+2} c_{k-1}$$
with $c_0 = 1$. This makes clear what structure they actually have: $c_1 = \frac{1}{2}$, $c_2 = \frac{1 \cdot 3}{2 \cdot 4}$, $c_3 = \frac{1 \cdot 3 \cdot 5}{2\cdot 4\cdot 6}$, and so on. Each $c_k$ is equal to the product of the first $k$ odd numbers divided by the product of the first $k$ even numbers. If you like, you can write them with double factorials via
$$c_k = \frac{(2k-1)!!}{(2k)!!}.$$
They are highly related to Wallis’ integrals $W_n$. Maybe this is not too surprising since they are defined as
$$W_n = \int_{0}^{\frac{\pi}{2}} \cos^n(x) {\rm d}x$$
and satisfy
$$W_{2n} = \frac{(2n-1)!!}{(2n)!!} \frac{\pi}{2}.$$
So what the generalized TPI above shows is that the equispaced $N$-term sum delivers somehow the same value as the integral, no matter where we start the sum. Kind of cool I think. | 3,232 | 7,881 | {"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} | 3.734375 | 4 | CC-MAIN-2019-18 | longest | en | 0.656627 |
https://ask.sagemath.org/question/30102/how-do-i-define-and-work-with-a-set-of-matrices/?sort=latest | 1,560,982,140,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999041.59/warc/CC-MAIN-20190619204313-20190619230313-00281.warc.gz | 347,998,067 | 16,831 | # How do I define (and work with) a set of matrices?
Suppose $a,b,c$ are matrices, and I want to define the set $S=$ { $a,b,c$ }. What is the proper syntax for this? More generally, how do I define a set by specifying its elements (regardless of their nature). I am only finding documentation for sets of numbers.
I'm also interested in doing functions that take sets like $S$ as input. For instance, if I had a matrix $d$, I want to be able to define a set like $S'=$ { $da,db,dc$ }, but I want to be able to do this via the Sage equivalent of $f(S):=$ {$ms\mid s\in S$}.
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First, let me explain how to deal with lists instead of sets, because there is a trick with sets:
You an create your matrices as follows:
sage: a = matrix(ZZ,[[1,2],[3,4]])
sage: b = matrix(ZZ,[[5,6],[7,8]])
sage: c = matrix(ZZ,[[1,2],[3,4]])
sage: d = matrix(ZZ,[[0,1],[1,0]])
Then, you can create the list of matrices as follows:
sage: L = [a,b,c]
sage: L
[
[1 2] [5 6] [1 2]
[3 4], [7 8], [3 4]
]
You can define the other list with list comprehension, which is a nice feature of Python:
sage: d = matrix(ZZ,[[0,1],[1,0]])
sage: [d*i for i in L]
[
[3 4] [7 8] [3 4]
[1 2], [5 6], [1 2]
]
Now, you noticed that c is equal to a and since you want a set, not a list, you would like Sage to keep only one occurence in the set.
You should be able to define a set as follows:
sage: S = {a,b,c}
TypeError: mutable matrices are unhashable
The error message tells you that the matrices must be immutable to enter a set. Indeed:
sage: a.is_mutable()
True
So, you can make your matrices immutable first as follows:
sage: a.set_immutable()
sage: b.set_immutable()
sage: c.set_immutable()
sage: d.set_immutable()
Then you can define your set:
sage: S = {a,b,c}
sage: S
{[1 2]
[3 4], [5 6]
[7 8]}
As you can see, there are only to elements in it:
sage: len(S)
2
Now, you shoud be able to define yous second set by comprehension as we did with lists:
sage: {d*i for i in S}
TypeError: mutable matrices are unhashable
You get the same error. Indeed, when we construct a new matrix, it is mutable by default, and there is no way to construct an immutable one from the beginning (at least this is not documented). So, even if a and d are immutable, d*a is mutable. I agree that this is not very handy.
A workaround is to define a immutabilize function that returns an immutable copy of a mutable matrix:
sage: def immutabilize(m):
....: M = copy(m)
....: M.set_immutable()
....: return M
So, at the end you get:
sage: {immutabilize(d*i) for i in S}
{[3 4]
[1 2], [7 8]
[5 6]}
more
Great, I will give this a shot tomorrow and get back to you. Thanks for all the info.
( 2015-10-18 21:29:02 -0500 )edit
OK, two people come up with the same answer independently. This must be the right answer then!
( 2015-10-18 22:03:24 -0500 )edit
Well you've definitely answered my question. I now have another issue to get this to work for what I'm doing, which should perhaps be a separate conversation (so I'm check-marking the answer), but.. I'm interested in matrices over rings besides $\mathbb{Z}$. In particular I want to define my matrices over rings of integers of number fields. One solution is to just work over $\mathbb{C}$ but I don't like how I get decimal approximations as opposed to exact expressions in terms of the primitive element. Even the simplest case of $\mathbb{Z}[i]$ seems to not be straightforward. I may figure this out by searching/experimenting but if you happen to know how to do this and can answer easily, that'd be great!
( 2015-10-19 15:20:17 -0500 )edit
Okay I have an answer to the question from my previous comment. The best way define a matrix over the ring of integers of a number field seems to be the following (taking the Gaussian example).
K.<i> = NumberField(x^2+1)
R = K.order(i)
a = matrix(R,[[i+1,-i],[i,0]])
I can't just use $I$ as the imaginary number and get output as I want it because $I$ is understood as an element of $\mathbb{C}$. So we define a new algebraically independent element $i$ which happens to also satisfy $x^2+1=0$.
( 2015-10-19 15:38:36 -0500 )edit
The pythonic way of doing this with lists is to use "list comprehensions":
sage: S=[matrix(2,2,[1,i,0,1]) for i in [1..3]]
sage: S
[
[1 1] [1 2] [1 3]
[0 1], [0 1], [0 1]
]
sage: m=matrix(2,2,[0,-1,1,0])
sage: [m*s for s in S]
[
[ 0 -1] [ 0 -1] [ 0 -1]
[ 1 1], [ 1 2], [ 1 3]
]
This is probably what you should use, since it gives you a concise syntax and usually gives you the best performance in python. If you write "multiply by m" as a function you have some other syntax available:
sage: g=lambda s:m*s
sage: [g(s) for s in S]
[
[ 0 -1] [ 0 -1] [ 0 -1]
[ 1 1], [ 1 2], [ 1 3]
]
sage: map(g,S)
[
[ 0 -1] [ 0 -1] [ 0 -1]
[ 1 1], [ 1 2], [ 1 3]
]
Via the same idiom you can define a function that takes a list as its argument, but since list comprehension and the "map" functionality are so concise, you probably shouldn't unless you have a compelling mathematical reason for doing so.
sage: f= lambda S: [d*s for s in S]
sage: f(S)
[
[ 0 -1] [ 0 -1] [ 0 -1]
[ 1 1], [ 1 2], [ 1 3]
]
Note that I talked about lists here, not sets. There is a complication with sets and matrices: matrices are by default mutable and mutable elements don't want to be in a set (it'd be hard to find them if they mutate!):
sage: S={matrix(2,2,[1,i,0,1]) for i in [1..3]}
TypeError: mutable matrices are unhashable
You can jump through some hoops to make this work:
sage: def immutable_matrix(*args,**kwargs):
....: m=matrix(*args,**kwargs)
....: m.set_immutable()
....: return m
....:
sage: S={immutable_matrix(2,2,[1,i,0,1]) for i in [1..3]}
sage: S
{[1 1]
[0 1], [1 2]
[0 1], [1 3]
[0 1]}
Everything we discussed before still works, except that new matrices will by default be mutable again:
sage: [d*s for s in S]
[
[ 0 -1] [ 0 -1] [ 0 -1]
[ 1 3], [ 1 2], [ 1 1]
]
sage: {d*s for s in S}
TypeError: mutable matrices are unhashable
sage: {immutable_matrix(d*s) for s in S}
{[ 0 -1]
[ 1 1], [ 0 -1]
[ 1 2], [ 0 -1]
[ 1 3]}
The map idiom still works, but will produce a list. You might try set(map(g,S)) but then you have to adjust g to produce immutable matrices. Mutatis mutandis, you can still wrap this into a function f that takes a set/list as input.
more | 2,040 | 6,400 | {"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} | 3.828125 | 4 | CC-MAIN-2019-26 | latest | en | 0.895047 |
http://www.studymode.com/course-notes/Management-Notes-1426018.html | 1,524,465,185,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945793.18/warc/CC-MAIN-20180423050940-20180423070940-00087.warc.gz | 544,972,329 | 25,826 | # Management Notes
Pages: 6 (1326 words) Published: February 14, 2013
MODERN MANAGEMENT THEORIES
INTRODUCTION
Management is one or the other form has existed in every nook and corner of the world since the dawn of civilization. Modern Management has grown with the growth of social-economics and scientific institution. Modern view consists that a worker does not work for only money. They work for their satisfaction and happiness with good living style. Here Non- financial award is most important factor. Modern management theories started after 1950s. Modern management theory focuses the development of each factor of workers and organization. Modern management theory refers to emphasizing the use of systematic mathematical techniques in the system with analyzing and understanding the inter-relationship of management and workers in all aspect. It has following three Streams - Quantitative Approach, System Approach and Contingency Approach.
QUANTITATIVE APPROACH
Quantitative approach also called Operation Research. Quantitative approach is a scientific method. It emphasizes the use of statistical model and systematic mathematical techniques to solving complex management problems. Its helps the management to making decisions in operations. It can only suggest the alternatives based on statistical data. It cannot take final decision.
Characteristics of Quantitative Approach:
* Decision-making focus - The primary focus of the quantitative approach is on problems or situations that require some direct action, or decision, on the part of management. * Measurable criteria - The decision-making process requires that the decision maker select some alternative course of action. The alternatives must be compared on the basis of some measurable criteria. * Quantitative model - To assess the likely impact of each alternative on the stated criteria, a quantitative model of the decision situation must be formulated. * Computers - Computers are quite useful in the problem-solving process.
USES:
* It helps the management for improving their decision making by increasing the number of alternatives and giving faster decisions on any problem thus the management can easily calculate the risk and benefit of various actions. * It has contributed significantly in developing orderly thinking in management which has provided exactness in management discipline. * Various Mathematical tools like sampling, linear programming, games theory, time series analysis, simulation, waiting line theory etc. have provided more exactness in solving managerial problems. * This approach is a fast developing area in analyzing and understanding management.
LIMITATIONS:
* Managerial activities are not really capable of being quantified because of involvement of human beings who are governed by many irrational factors also. * More expertise and technical skills are required to formulate mathematical models
Major contributors in Quantitative Approach are:
1. Johan MacDonald
2. George R. Terry
3. Andrew Szilagyi
SYSTEM APPROACH
System approach was developed in late 1960s. Herbert A. Simon is the father of system theory. A System is defined as a set of regularly interacting or interdependent components that create as a whole unit. The system concept enables us to see the critical variables and constraints and their interactions with one another. According to Cleland and King; “A system is composed of related and dependent elements which when in interaction from a unity whole”.
Environmental interaction
* Open systems must interact with the external environment to survive. * Closed systems do not interact with the environment.
Properties of Systems
* Synergy - When all organizational subsystems work together making the whole greater than the sum of its parts. * Entropy - The tendency for systems to decay over time.
Characteristics of System Approach:
* A system must have some specific components, units or sub units. * A change in one... | 722 | 3,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-17 | latest | en | 0.930933 |
https://www.aqua-calc.com/one-to-all/temperature/preset/kelvin/550.15 | 1,611,197,415,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00111.warc.gz | 671,336,930 | 5,812 | # Convert Kelvins [K] to other units of temperature
## Kelvins [K] temperature conversions
550.15 K = 277 degrees Celsius K to °C 550.15 K = 530.6 degrees Fahrenheit K to °F 550.15 K = 990.27 degrees Rankine K to °R 550.15 K = 152.93 degrees Romer K to °Rø
Convert entered temperature to units of energy.
#### Foods, Nutrients and Calories
EXTREME DARK CHOCOLATE, UPC: 716270001882 contain(s) 600 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
Substrate, Clay/Laterite weighs 1 019 kg/m³ (63.61409 lb/ft³) with specific gravity of 1.019 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Copper(II) hydroxide [Cu(OH)2] weighs 3 360 kg/m³ (209.75795 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Crambe oil with temperature in the range of 23.9°C (75.02°F) to 110°C (230°F)
#### Weights and Measurements
A Yottameter (Ym) is a decimal multiple of the base metric (SI) measurement unit of length, the meter
The units of data measurement were introduced to manage and operate digital information.
mrem p to rem multiple conversion table, mrem p to rem multiple unit converter or convert between all units of radiation absorbed dose measurement.
#### Calculators
Humidity calculations using temperature, relative humidity, pressure | 421 | 1,596 | {"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-2021-04 | longest | en | 0.761433 |
https://dsp.stackexchange.com/questions/72543/compute-the-second-order-derivative-of-digital-image-with-finite-differences | 1,713,486,054,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00760.warc.gz | 194,091,317 | 40,604 | # Compute the Second Order Derivative of Digital Image with Finite Differences
I was looking for how to compute second order derivative of an image and came across the question kernels to Compute Second Order Derivative of Digital Image. In the top voted answer, it gives an example to get $$3\times 3$$ kernel $$I_{xx}$$ and $$I_{xy}$$. The 1st order derivative filter is defined as $$\left( \begin{array}{cc} -1 & 1 \\ -1 & 1 \\ \end{array} \right)$$
Then
$$I_{xx} = I_x \cdot I_x = \left( \begin{array}{cc} -1 & 1 \\ -1 & 1 \\ \end{array} \right)\cdot\left( \begin{array}{cc} -1 & 1 \\ -1 & 1 \\ \end{array} \right)=\left( \begin{array}{ccc} 1 & -2 & 1 \\ 2 & -4 & 2 \\ 1 & -2 & 1 \\ \end{array} \right)$$
and
$$I_{xy} = I_y \cdot I_x = \left( \begin{array}{cc} -1 & -1 \\ 1 & 1 \\ \end{array} \right)\cdot\left( \begin{array}{cc} -1 & 1 \\ -1 & 1 \\ \end{array} \right)=\left( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \\ \end{array} \right)$$
The computation of $$I_{xx}$$ makes sense to me, however, for $$I_{xy}$$, shouldn't it be defined as below?
$$I_{xy} = I_x \cdot I_y = \left( \begin{array}{cc} -1 & 1 \\ -1 & 1 \\ \end{array} \right)\cdot\left( \begin{array}{cc} -1 & -1 \\ 1 & 1 \\ \end{array} \right)=\left( \begin{array}{ccc} -1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & -1 \\ \end{array} \right)$$
Pay attention that convolution mean flipping the kernel both on the x and y axis.
Hence the first element is a multiplication of $$-1 \cdot -1$$ which yields $$1$$ as in the answer in the original post.
Pay attention that this is a discrete approximation of the gradient based on Finite Differences. This specific one is based on the Sobel 1st derivative filter.
In Finite Differences Coefficients you may find more variations.
As usual, the longer the filter the better the accuracy of the approximation with longer tarnsients.
• Ah! I was only flipping the kernel on the x axis only... Actually there is no difference between (Iy * Ix) and (Ix * Iy). Thanks!
– ciel
Jan 12, 2021 at 6:15
• The longer it is means the longer its transient as well. Feb 19, 2021 at 6:56
• @David, You're correct. I will add that.
– Royi
Feb 19, 2021 at 6:59 | 740 | 2,170 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-18 | latest | en | 0.820265 |
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Order of Operations Puzzle Students must rearrange 9 puzzle pieces so that the problems match the solution on each side. Once finished, the puzzle will be a 3 x 3 grid which is easily checked by teacher using the "mystery word" that it spells
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This PowerPoint teaches students how to read line plots. There are sample problems for students to work through as guided practice. There is also an activity at the end for students to measure their thumbs and plot the class line plot. using a
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This file contains 50 pages of polka dot 1" frames for Microsoft Word. Ready to use with your documents. All documents can be typed on, and edited. Easy to use. For personal and commercial use. If you are interested in a certain color that is
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This jeopardy PowerPoint game is a reading/language standard review created for fifth grade. It has 5 categories including: story elements figurative language reference skills more figurative language editing the sentence Students have 5 questions
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This PowerPoint gives the meanings for common adages, proverbs, and idioms. Pictures for each one are included. There are 30 different ones. This is used to help students understand the meaning of common phrases. This can be used as a complete
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Measurement Customary Units Puzzle Students must rearrange 9 puzzle pieces so that the units match on each side of the puzzle piece. Once finished, the puzzle will be a 3 x 3 grid which is easily checked by teacher using the "mystery word" that it
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This PowerPoint is designed as an introduction to discussing media literacy and its affect on the consumer. It begins with a slide of logos that almost every person should recognize. Each logo has NO WORDS, but they are so popular that anyone can
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Tired of boring review sessions in your classroom? This activity can be used to spice up your class review. Students still answer their study guides, but they can now choose a virtual prize for correct answers. Add laughter to your class as
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This review game is for practice after teaching a lesson on figurative language. There are 25 questions (5 questions in 5 categories). The students will see a sentence with a figure of speech included (not marked). The student must determine what
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This is a jeopardy PowerPoint game that reviews reading standards for fifth grade. The categories include: Organizational Structure Poetry Figurative Language Vocabulary Media Literacy Each category has 5 questions. This is a great review for
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This PowerPoint is intended to supplement a lesson on relevant and irrelevant details. Students will be shown some examples of what would be relevant details to a topic and what would be irrelevant. Then, students practice identifying topic
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Measurement Metric Units Puzzle Students must rearrange 9 puzzle pieces so that the units match on each side of the puzzle piece. Once finished, the puzzle will be a 3 x 3 grid which is easily checked by teacher using the "mystery word" that it
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Volume Puzzle Students must rearrange 9 puzzle pieces so that the answers match the problem on each side of the puzzle piece. Once finished, the puzzle will be a 3 x 3 grid which is easily checked by teacher using the "mystery word" that it spells
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Students must rearrange 9 puzzle pieces so that the answers match the problem on each side of the puzzle piece. Once finished, the puzzle will be a 3 x 3 grid which is easily checked by teacher using the "mystery word" that it spells (left to right,
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This chart was made to help students keep track of how many facts they have learned. The students will graph how many facts they get correct. The chart goes from 0 - 100 on two different sizes: Letter size sheet of paper (8 1/2 by 11) - the
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This is a jeopardy PowerPoint review game for English with the following categories: Parts of Speech Homophones Sentences Definitions Antonym/Synonym Made for 5th grade students, appropriate for 3-5. 5 questions in each category.
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Polygon Vocabulary Puzzle Students must rearrange 9 puzzle pieces so that the units match on each side of the puzzle piece. Once finished, the puzzle will be a 3 x 3 grid which is easily checked by teacher using the "mystery word" that it spells
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TEACHING EXPERIENCE
This is my 10th year teaching in the Georgia public school system - 9 of these years have been in 5th grade special education either in a resource room or co-taught teaching math, reading, and language arts.
MY TEACHING STYLE
My teaching style varies depending upon the subject. I am highly technical and like to include technology into my lessons whenever possible. I love music and also incorporate songs into my lessons. To see more about including video/songs into 5th grade standards, visit my webpage: http://huttomiddle.dcboe.com/?PageName=TeacherPage&Page=13&StaffID=138229&iSection=Teachers&CorrespondingID=138229
HONORS/AWARDS/SHINING TEACHER MOMENT
I have had the honor to be invited to many testing meetings for GACE, CRCT, and Praxis (standard setting, item analysis, content descriptors, and cut score analysis). These have been very instrumental in my curriculum knowledge. I was recently nominated as Teacher of the Year for my school for the 2015-2016 school year and for my entire county for 2015-2016! I am currently serving on the State Advisory Panel for Special Education for the state of Georgia.
MY OWN EDUCATIONAL HISTORY
Bachelor's of Science in Special Education from Valdosta State University Masters of Education in Special Education from Valdosta State University Masters of Library Science from Florida State University Specialist in Instructional Technology and Media Addon from Valdosta State University
I am a presenter at the annual Georgia IDEAS conference 2013 - topic: co-teaching. 2014 - topic: ipads in the classroom 2015 - topic: Google documents | 2,482 | 9,315 | {"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-2018-47 | latest | en | 0.895855 |
https://forum.pyro.ai/t/question-on-local-global-random-variable-in-pyro-hmm-example/444 | 1,718,532,569,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00402.warc.gz | 228,331,070 | 9,029 | Question on local/global random variable in pyro HMM example
Hi,
I am confused by the pyro HMM example.
In the very basic model (model_1, related code shown below), it seems that the discrete state variable x_{} is shared across the batch and every sequence in a batch shares the same state.
In my understanding, in an HMM, the transition matrix and the observation matrix are global variable. However, the state variable are local, which means that different sequence would have its own state variables…
Could anyone help me with the code and HMM concepts?
Thanks!
with pyro.iarange("sequences", len(sequences), batch_size, dim=-2) as batch:
lengths = lengths[batch]
x = 0
for t in range(lengths.max()):
# On the next line, we'll overwrite the value of x with an updated
# value. If we wanted to record all x values, we could instead
# write x[t] = pyro.sample(...x[t-1]...).
x = pyro.sample("x_{}".format(t), dist.Categorical(probs_x[x]),
infer={"enumerate": "parallel"})
with tones_iarange:
pyro.sample("y_{}".format(t), dist.Bernoulli(probs_y[x]),
obs=sequences[batch, t])
Hi @ruijiang, you are correct that in an HMM the state variable is local to each sequence in a batch. In the Pyro HMM example, the state variable "x_{}".format(t) is actually a tensor if independent variables. You can read this from the model in the outermost pyro.iarange("sequences", ..., dim=-2). Inside that iarange context every variable is local and is vectorized over the 2nd dim from the right (dim=-2). Does that make sense?
BTW we’re thinking of renaming pyro.iarange to pyro.plate to make that clearer.
1 Like
Thanks @fritzo for the explanation. My confusion is gone:)
I think it is a good idea to rename iarange with plate as the usage of iarange differs significantly from irange. Previously I use irange with which we have to explicitly name local variables or add an extra batch dimension, as shown in the following examples.
for i in pyro.irange("data_loop", len(data)):
# observe datapoint i using the bernoulli likelihood
pyro.sample("obs_{}".format(i), dist.Bernoulli(f), obs=data[i])
iarange is much more convenient as we don’t have to consider the ‘batch’ dimension or vectorization explicitly .
1 Like
Hi @fritzo,
I have some local variable related questions
• in the HMM example, is there any convenient way to get the posterior of the state variable ‘x’ although it is marginalized out?
• a more general question, how to access the posterior of local random variables if Amortized inference is not used? I tried to see the size of ‘x’ in the example, seems that the size of x has nothing to do with the batch size.
Thanks!
Hi @ruijiang,
You can examine the enumerated variables using either TraceEnum_ELBO.compute_marginals() or TraceEnum_ELBO.sample_posterior(). Note however that these are very new, and still have some known bugs regarding batching. We’re hoping to get these fully working before the next Pyro release. If you end up using these, let us know your experience so we can improve the interface!
@ruijiang I forgot: another way to access the marginalized-out variables is to train a second SVI guide that fixes the variables learned in the first guide. We have an example of this in the GMM tutorial.
Thanks @fritzo.
I am a little bit confused of what is going on if we enumerate the guide in the GMM example.
Let z_i be the discrete variable, g be the global variable, i be the indicator for independent samples.
• with MAP estimation, it is clear that if we enumerate z in the model, we are going to find the maximum of p(y,g). After learning we get the point estimation of g.
• with VB, if we enumerate z in the guide, the parameter \theta would vanish.
I think I had a misunderstanding of the objective function pyro optimizes if enumeration is involved. Is their any document on the objective function if enumeration is involved?
Still I gave the 'second guide ’ approach a try, but the speed is much slower compared with MAP estimation.
I am trying to implement Switching Dynamic Linear System with Pyro, see the model in Sec. 2.1 of the paper
and my related code as follows,
• z is the discrete state and x is the continuous state vector in linear dynamic system
• the input sequence is tensor of shape num_sequences, max_length, data_dim
• model:
@poutine.broadcast
def model_sssm(sequences, lengths, args, batch_size=None, include_prior=True):
num_sequences, max_length, data_dim = sequences.shape
assert lengths.shape == (num_sequences,)
assert lengths.max() <= max_length
n_seg = args.num_segment
n_lat = args.num_latent
n_out = data_dim
normal_std = 1e-1
# transition matrix for HMM, K*K,
hmm_dyn = pyro.sample("hmm_dyn",
dist.Dirichlet(0.9 * torch.eye(n_seg) + 0.1)
.independent(1))
## SSM
ssm_dyn = pyro.sample("ssm_dyn",
dist.Normal(0,normal_std).expand_by([n_seg, n_lat, n_lat])
.independent(3))
ssm_bias = pyro.sample("ssm_bias",
dist.Normal(0,1.).expand_by([n_seg, n_lat, 1])
.independent(3))
ssm_noise = pyro.sample("ssm_noise",
dist.Gamma(1e0,1e0).expand_by([n_seg, n_lat, 1])
.independent(3))
## observation
obs_weight = pyro.sample("obs_weight",
dist.Normal(0,1).expand_by([n_seg, n_out, n_lat])
.independent(3))
obs_bias = pyro.sample("obs_bias",
dist.Normal(0,1).expand_by([n_seg, n_out, 1])
.independent(3))
obs_noise = pyro.sample("obs_noise",
dist.Gamma(1e0,1e0).expand_by([n_seg, n_out, 1])
.independent(3))
with pyro.iarange("sequences", len(sequences), batch_size, dim=-2) as batch:
## ssm init state
# use a (n_lat,1) matrix (instead of vector) to ease batch operation,
# as for high dim, matmul would do batch matrix-matrix product, instead of matrix-vector product.
#
ssm_init = pyro.sample("ssm_init", dist.Normal(0,1).expand_by([n_lat,1]).independent(2))
# print('===============')
# print('ssm_init_shape:', ssm_init.shape)
# the RVs below are all local variables.
lengths = lengths[batch]
z = 0
x = ssm_init
print('###############')
for t in range(lengths.max()):
# On the next line, we'll overwrite the value of x with an updated
# value. If we wanted to record all x values, we could instead
# write x[t] = pyro.sample(...x[t-1]...).
z = pyro.sample("z_{}".format(t), dist.Categorical(hmm_dyn[z]),
infer={"enumerate": "parallel"})
x = pyro.sample("x_{}".format(t), dist.Normal(torch.matmul(ssm_dyn[z], x) + ssm_bias[z], torch.sqrt(1./ssm_noise[z])).independent(2))
obs_lat = torch.matmul(obs_weight[z], x) + obs_bias[z]
pyro.sample("y_{}".format(t), dist.Normal(obs_lat, torch.sqrt(1./obs_noise[z])).independent(2),
obs=sequences[batch, t].unsqueeze(-1))
• guide:
@poutine.broadcast
@config_enumerate(default="parallel")
def guide(sequences, lengths, args, batch_size=None, include_prior=True):
num_sequences, max_length, data_dim = sequences.shape
n_seg = args.num_segment
with pyro.iarange("sequences", len(sequences), batch_size, dim=-2) as batch:
lengths = lengths[batch]
for t in range(lengths.max()):
ret_prob = pyro.param('assignment_probs_{}'.format(t), torch.ones(len(lengths), 1, n_seg) / n_seg,
constraint=constraints.unit_interval)
z = pyro.sample("z_{}".format(t), dist.Categorical(ret_prob))
could you help me figure out why it is so slow?
Thanks,
Ruijiang
2 Likes
Hi @fritzo,
I also gave a TraceEnum_ELBO.compute_marginals() a try on the switching dynamic linear system, but got size issues:
• With the following code, the batch_size of the discrete variable (z_{}) is [bs, bs], and the shape is [bs, bs, n_state] where bs refers to batch size.
r1 = elbo.compute_marginals(model, guide, sequences, lengths, args, batch_size=args.batch_size)
• if I check the trace with the following code, the batch_shape in trace_model.nodes are OK ([bs,1] as we have dim=-2).
trace_model = poutine.trace(model).get_trace(sequences, lengths, args, batch_size=args.batch_size)
• I print the size of the distribution and tensors in the model, the sizes are OK.
• If I change dim=-1 in the iarange(), related error occurs.
ValueError: at site "z_0", invalid log_prob shape
Expected [3], actual [3, 3]
I also gave compute_marginals a try with HMM example, the shapes are OK. So I think there is something wrong with my model (posted in the previous reply), but I could find where the issues is.
Thanks.
Ruijiang
I am trying to implement Switching Dynamic Linear System with Pyro
Nice! Do you have an interest in contributing that to pyro/examples/? I think it would be easier to discuss this model in a PR rather than a forum thread.
Is their any document on the objective function if enumeration is involved?
The objective function is simply the ELBO. Even when we do MAP estimation, we maximize ELBO with a trivial delta guide.
p(z,x)
ELBO = sum q(z|x) log ------
z q(z|x)
In enumeration we split z into say three parts: non-enumerated z1, guide-enumerated z2, and model-enumerated z3.
sum p(z1,z2,z3,x)
z3
ELBO = sum sum q(z1,z2|x) log -----------------
z1 z2 q(z1,z2|x)
Note that sum z1 q(z1|x) is implemented via Monte Carlo, whereas sum z2 q(z2|x) is implemented via weighted enumeration in the guide, and sum z3 p(z3) is implemented via weighted enumeration in the model. I’ll try to add a clearer explanation to our upcoming enumeration tutorial.
1 Like
Hi @fritzo,
I am interested in contributing to the examples.
For the ‘switching linear dynamic models’, should I move the discussion in the forum to github issues?
1 Like
Thanks @fritzo, the explanation of enumeration with these equations are quite clear.
Just a small question on a special case,
If a variable z4 is enumerated in both model and guide, then in ELBO, there will be NO sum of p(z4) over z4. Am I correct?
‘switching linear dynamic models’ … to github issues?
Yes, github discussion would be great! A few of us Pyro devs are also working on tutorials for the upcoming 0.3 release (planned for NIPS).
If a variable z4 is enumerated in both model and guide
Hmm I’m not sure what “enumerated in both model and guide” means. Pyro supports monte carlo sampling, guide-side enumeration, and model-side enumeration. In the above example, z2 is enumerated in the guide and replayed in the model. It is not summed-out in the log numerator, but it does appear in p(z2).
If a variable z4 is enumerated in both model and guide
In GMM example, @config_enumerate(default='parallel') applies to discrete r.v. assignment in both model and guide:
@config_enumerate(default='parallel')
def model(data):
# Global variables.
weights = pyro.sample('weights', dist.Dirichlet(0.5 * torch.ones(K)))
scale = pyro.sample('scale', dist.LogNormal(0., 2.))
with pyro.iarange('components', K):
locs = pyro.sample('locs', dist.Normal(0., 10.))
with pyro.iarange('data', len(data)):
# Local variables.
assignment = pyro.sample('assignment', dist.Categorical(weights))
pyro.sample('obs', dist.Normal(locs[assignment], scale), obs=data)
global_guide = AutoDelta(poutine.block(model, expose=['weights', 'locs', 'scale']))
@config_enumerate(default="parallel")
def full_guide(data):
# Global variables.
with poutine.block(hide_types=["param"]): # Keep our learned values of global parameters.
global_guide(data)
# Local variables.
with pyro.iarange('data', len(data)):
assignment_probs = pyro.param('assignment_probs', torch.ones(len(data), K) / K,
constraint=constraints.unit_interval)
pyro.sample('assignment', dist.Categorical(assignment_probs))
Ah I see the source of confusion. When @config_enumerate or infer={'enumerate': ...} appears in the model, it only applies if the variable has not already been enumerated in the guide. If a variable has been enumerated in the guide, then the model will simply replay the guide-enumerated variable (with corresponding nonstandard shape), and the ELBO will be
p(z4,x)
ELBO = sum q(z4|x) log -------
z4 q(z4|x)
`
I’ll try to make this clear in the upcoming enumeration tutorial. | 3,038 | 11,936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.875243 |
http://kanyakumari-info.com/trends/question-do-you-add-exponents-when-multiplying.html | 1,695,775,677,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510238.65/warc/CC-MAIN-20230927003313-20230927033313-00744.warc.gz | 24,478,885 | 22,314 | # Question: Do you add exponents when multiplying?
When you’re multiplying exponents, use the first rule: add powers together when multiplying like bases. 52 × 56 =? The bases of the equation stay the same, and the values of the exponents get added together. Adding the exponents together is just a shortcut to the answer.03-May-2019
• The exponent “product rule” tells us that, when multiplying two powers that have the same base, you can add the exponents. In this example, you can see how it works. Adding the exponents is just a short cut! The “power rule” tells us that to raise a power to a power, just multiply the exponents.
## Do you add exponents or multiply them?
To multiply two exponents with the same base, you keep the base and add the powers.
## What are the rules for exponents?
Exponents rules and properties
Rule name Rule Example
Product rules a n ⋅ b n = (a ⋅ b) n 32 ⋅ 42 = (3⋅4)2 = 144
Quotient rules a n / a m = a nm 25 / 23 = 253 = 4
a n / b n = (a / b) n 43 / 23 = (4/2)3 = 8
Power rules (bn)m = bnm (23)2 = 232 = 64
You might be interested: Is A Podenco A Sighthound?
## When multiplying scientific notation do you add the exponents?
To multiply numbers in scientific notation, first multiply the numbers that are not powers of 10 (the a in a×10n a × 10 n ). Then multiply the powers of ten by adding the exponents. This will produce a new number times a different power of 10.
## Do you add variables when multiplying?
When variables are the same, multiplying them together compresses them into a single factor ( variable ). But you still can ‘t combine different variables. When multiplying variables, you multiply the coefficients and variables as usual.
## Do you add or multiply first in equations?
Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction. Continue to perform multiplication and division from left to right. Next, add and subtract from left to right.
## How do you simplify exponents?
When dividing two terms with the same base, subtract the exponent in the denominator from the exponent in the numerator: Power of a Power: To raise a power to a power, multiply the exponents. The rules of exponents provide accurate and efficient shortcuts for simplifying variables in exponential notation.
## What are the 7 rules of exponents?
There are seven exponent rules, or laws of exponents, that your students need to learn. Exponent rules Product of powers rule. Quotient of powers rule. Power of a power rule. Power of a product rule. Power of a quotient rule. Zero power rule. Negative exponent rule.
## What is the rule for subtracting exponents?
Subtract Exponents: Example Question #7 Explanation: When two exponents with the same base are being divided, subtract the exponent of the denominator from the exponent of the numerator to yield a new exponent. Attach that exponent to the base, and that is your answer.
You might be interested: How Do You Make A Baking Soda Bath Bomb?
## What are the 5 laws of exponents?
Laws of Exponents Multiplying Powers with same Base. Dividing Powers with the same Base. Power of a Power. Multiplying Powers with the same Exponents. Negative Exponents. Power with Exponent Zero. Fractional Exponent.
## What is the rule for multiplying scientific notation?
To multiply two numbers in scientific notation, multiply their coefficients and add their exponents. To divide two numbers in scientific notation, divide their coefficients and subtract their exponents. In either case, the answer must be converted to scientific notation.
## How do you multiply numbers with negative exponents?
To multiply by a negative exponent, subtract that exponent. To divide by a negative exponent, add that exponent.
## What is standard notation in math?
more Just the number as we normally write it.
## What is the rule for adding and multiplying?
The rules are: Multiply and divide from left to right. Add and subtract from left to right.
## What is the rule for adding and subtracting algebraic terms?
To add two or more monomials that are like terms, add the coefficients; keep the variables and exponents on the variables the same. To subtract two or more monomials that are like terms, subtract the coefficients; keep the variables and exponents on the variables the same. on the variables the same.
## How do you multiply same variables?
To multiply terms, multiply the coefficients and add the exponents on each variable. The number of terms in the product will be equal to the product of the number of terms. Of course, there may well be like terms which you will need to combine. | 1,051 | 4,675 | {"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} | 4.78125 | 5 | CC-MAIN-2023-40 | latest | en | 0.860266 |
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Abby L's blog Previous Geometry
files/
by Abby L teacher: Tina Jovanovich
Blog Entries
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my Birthday!!! Today is my birthday! I am so excited. Today in geometry we had a test and I feel extremely confident about the multiple choice, I would be surprised if I got more than one multiple choice question wrong. Also, I am not sure how the long answer question went, I feel like I did it wrong, because I did not know how to do it. However I feel like I did not write the equation right at all because I did not know what the slope was, hopefully I did get it right though, my fingers are definitely crossed. Another question was the indirect proof, which I read over and felt like I got right, I just hope that included everything that was supposed to be included. The coordinate proof was really good though. I think that my hand writing and organization of the proof was correct and my answer makes total sense. Overall this test went okay, if there was one thing that I wish I had studied more, it would have been the second homework of the year, working with finding the slopes of the altitude, medians etc.
Another interesting thing that has been going on this week has been the visits from the Neasc people. I haven't really noticed them, as they are so quiet, however everyone around me says some of them are mean. I don't know if this is true, but from what I've seen, it is not.
Also, this week is a four day week, which is fantastic because I feel like I need a treat, it will be like a late birthday present!
Article posted March 6, 2012 at 10:23 AM • comment • Reads 59 • see all articles | 594 | 2,270 | {"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} | 2.828125 | 3 | CC-MAIN-2013-20 | latest | en | 0.973546 |
https://www.shaalaa.com/question-bank-solutions/if-x-is-a-real-number-and-x-3-then-______-inequalities-introduction_260009 | 1,680,120,359,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00238.warc.gz | 1,097,981,232 | 9,850 | # If x is a real number and |x| < 3, then ______. - Mathematics
MCQ
Fill in the Blanks
If x is a real number and |x| < 3, then ______.
• x ≥ 3
• – 3 < x < 3
• x ≤ – 3
• – 3 ≤ x ≤ 3
#### Solution
If x is a real number and |x| < 3, then – 3 < x < 3.
Explanation:
Given that |x| < 3
⇒ –3 < x < 3 .....[∵ |x| < a ⇒ –a < x < a]
Concept: Inequalities - Introduction
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Exemplar Class 11
Chapter 6 Linear Inequalities
Exercise | Q 22 | Page 109
Share | 200 | 538 | {"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} | 3.4375 | 3 | CC-MAIN-2023-14 | latest | en | 0.643291 |
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### What is a Quadratic Equation?
```
Date: 08/04/97 at 14:38:14
From: Robert M. Thacker
Subject: What is a quadratic equation anyway?
Dear Dr. Math:
In plain English, can you explain to my daughter, "What is a quadratic
equation?" What is it used for? How can we use it to solve everyday
problems?
Please do not use a lot of mathematical gibberish in your explanation.
Thank you.
Robert M. Thacker
```
```
Date: 08/05/97 at 15:13:48
From: Doctor Ceeks
Subject: Re: What is a quadratic equation anyway?
Hi,
level of your daughter's mathematical knowledge? Is she in
Algebra I, or is she in kindergarten? Does she like math or is
she afraid of it?
-Doctor Ceeks, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
```
Date: 08/05/97 at 15:33:20
From: Robert M. Thacker
Subject: Re: What is a quadratic equation anyway?
Dear Dr. Math:
She will begin Algebra I next year, and likes math in general.
The problem is I cannot give her a simple, practical, commonsense,
Sincerely,
R.M. Thacker
```
```
Date: 08/05/97 at 16:36:43
From: Doctor Ceeks
Subject: Re: What is a quadratic equation anyway?
Hi,
It's difficult to know how to approach this question.
Perhaps if I give two examples of where a quadratic equation would
arise it would help.
Here are two examples:
1. I'm thinking of two numbers. Their sum is 10 and their product is
21. What are the two numbers?
More generally, two numbers have sum S and product P. What are
the two numbers?
To solve this, you will end up naturally with a quadratic equation:
x^2 - 10x + 21 = 0 or x^2 - Sx + P = 0, respectively.
There will generally be two solutions to a quadratic equation
corresponding to the two numbers.
(In fact, all quadratic equations essentially arise in this manner.
I say essentially, because generally, people consider the equation
Ax^2 + Bx + C = 0
to be the most general quadratic equation, where A, B, and C are
constants with A not equal to zero. However, if you divide this
throughout by A, you will get an equation looking like x^2-Sx+P = 0
if you take S = -B/A and P = C/A.)
2. A square picture frame contains a picture with a mat border.
The border is 3 inches thick on the sides and 4 inches thick
on the top and bottom. If the area exposed within the mat border
is 528 square inches, what are the dimensions of the original
frame?
Again, a quadratic equation will arise naturally.
(In this case, x^2-14x-480 = 0.)
Other places where a quadratic equation may surface come from
geometry, such as trying to find the intersection of a line and
a circle (an example of this arises in one of the ways people find
all the Pythagorean triples). Also, quadratic equations sometimes
occur in physics when studying how objects fall to Earth.
If your daughter knows what a graph is, you can toss a ball and
note that the ball follows a path which looks like the graph of
draw, as accurately as you can, a graph of the quadratic function
f(x) = -x^2/36. Draw it so that you get a nice section of it on
the board. Then take a ball (or any object) and, with practice, you
can toss it accross the blackboard so that it exactly follows the
graph you have drawn. It's pretty exciting to see this done.
(Note that a quadratic _equation_ is an equation you get when you
set a quadratic _function_ equal to zero.)
Being able to solve quadratic equations was one of the early
challenges to the human race and was solved by the ancients... so
useful was their discovery that the quadratic formula (which gives
the solutions to a quadratic equation) has been remembered and
handed down over the centuries.
-Doctor Ceeks, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
```
Date: 12/10/2002 at 11:22:57
From: Joey Greer
Subject: Arriving at the quadratic equation naturally
Hi,
Above it says:
I'm thinking of two numbers. Their sum is 10 and their product is 21.
What are the two numbers?
More generally, two numbers have sum S and product P. What are the two
numbers?
To solve this, you will end up naturally with a quadratic equation:
x^2 - 10x + 21 = 0 or x^2 - Sx + P = 0, respectively.
mathematician deriving that equation arrived at that naturally. How
would he know that the equation would have two answers and the two
answers combined using the given operation (multiplication or
addition) would be the answer to the question? Did he come to it
somehow from another equation? Because I know that if I didn't know
the quadratic equation is used to solve questions like that and I were
asked it, I would not have derived that equation. It seems like a
phenomenon that it actually even works out as it does. Can you please
help me to understand this?
Sincerely,
Joey
```
```
Date: 12/10/2002 at 12:30:20
From: Doctor Peterson
Subject: Re: Arriving at the quadratic equation naturally
Hi, Joey.
Here is how I would get from the problem to the equation. It is not
just a leap based on knowing what kind of equation to expect, but as
Dr. Ceeks said, it arises naturally.
I'm thinking of two numbers. Their sum is 10 and their product
is 21. What are the two numbers?
Suppose we call one of the numbers x. Then, since their sum is 10,
the other number has to be 10-x. The fact that their product is 21
tells us that
x(10-x) = 21
Simplifying this, we get
10x - x^2 = 21
x^2 - 10x + 21 = 0
There's the equation.
Now, the work would have been a little different for the first people
who used such equations, since they didn't have algebraic notation,
but the underlying ideas would be the same.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Definitions
Middle School Algebra
Middle School Definitions
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 1,614 | 6,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2016-44 | longest | en | 0.938461 |
http://www.conversion-website.com/area/square-chain-to-dunam.html | 1,695,956,497,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510481.79/warc/CC-MAIN-20230929022639-20230929052639-00180.warc.gz | 62,719,266 | 4,300 | # Square chains to dunams
## Convert square chains to dunams
Square chains to dunams conversion calculator above calculates how many dunams are in 'X' square chains (where 'X' is the number of square chains to convert to dunams). In order to convert a value from square chains to dunams just type the number of sq ch to be converted to dunams and then click on the 'convert' button.
## Square chains to dunams conversion factor
1 square chain is equal to 0.4046856422 dunams
## Square chains to dunams conversion formula
Area(dunams) = Area (sq ch) × 0.4046856422
Example: Find out how many dunams equal 184 square chains.
Area(dunams) = 184 ( sq ch ) × 0.4046856422 ( dunams / sq ch )
Area(dunams) = 74.4621581648 dunams or
184 sq ch = 74.4621581648 dunams
184 square chains equals 74.4621581648 dunams
## Square chains to dunams conversion table
square chains (sq ch)dunams
156.070284633
2510.117141055
3514.163997477
4518.210853899
5522.257710321
6526.304566743
7530.351423165
8534.398279587
9538.445136009
10542.491992431
square chains (sq ch)dunams
20080.93712844
250101.17141055
300121.40569266
350141.63997477
400161.87425688
450182.10853899
500202.3428211
550222.57710321
600242.81138532
650263.04566743
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https://www.jiskha.com/display.cgi?id=1422302929 | 1,529,464,505,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00549.warc.gz | 840,668,959 | 4,086 | # math
posted by jp
The radiator in a car is filled with a solution of 70 per cent antifreeze and 30 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 4 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
1. Reiny
let's just look at the antifreeze,
right now the volume of antifreeze is .7(4) L or
2.8 L
Let the amount of the current fluid we are draining be x L
Now 70% of that is antifreeze, so the amount of antifreeze we lose is .7x
So the actual amount of antifreeze left in the rad
is 2.8 - .7x L
But that be enough to produce our required 50% solution, if we pour in x L of straight water
2.8-.7x + 0(x) = .5(4) --> my equation only shows antifreeze
-.7x = 2-2.8
.7x = .8
x = .8/.7 = 8/7 = appr 1.14 L
2. bobpursley
reduce by v, then the volume left if 4-v
antifreeze conc*volume+solutionconc*volume=.5*4
1*v+.7(4-v)=2
solve for v
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http://sites.math.rutgers.edu/~zeilberg/tokhniot/oAFS7v.txt | 1,555,921,446,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-18/segments/1555578548241.22/warc/CC-MAIN-20190422075601-20190422101601-00226.warc.gz | 162,074,695 | 1,912 | Here is the Verbose version of the Generalized Quasi-Polynomial Capsule for \ the coefficient of q^(17*n+16) mod 17 functional equation 17 q f(q ) f(q) = 1 + ------------------- 5 2 3 (1 - q) (-q + 1) ----------------------------------------------------------------------------\ ----------------- A Fast (Log Time!) Way to Determine the Remainder Upon Dividing by, 17, (17 n + 16) of the coefficient of, q in the Unique Formal Power Series Satisfying the Functional Equation 17 q f(q ) f(q) = 1 + ------------------- 5 2 3 (1 - q) (-q + 1) By Shalosh B. Ekhad Theorem: Let c(n) be the coefficient of q^n in the formal power series of th\ e title, in other words let infinity ----- \ n f(q) = ) c(n) q / ----- n = 0 is the unique formal power series satisfying the inhomogeneneous functional \ equation: 17 q f(q ) f(q) = 1 + ------------------- 5 2 3 (1 - q) (-q + 1) Let , d(n), be , c(17 n + 16) Define , 23, integers as follows: C[0] = 0, C[1] = 1, C[2] = 5, C[3] = 2, C[4] = 4, C[5] = 4, C[6] = 10, C[7] = 14, C[8] = 9, C[9] = 7, C[10] = 5, C[11] = 7, C[12] = 7, C[13] = 5, C[14] = 7, C[15] = 9, C[16] = 14, C[17] = 10, C[18] = 4, C[19] = 4, C[20] = 2, C[21] = 5, C[22] = 1 If j is larger then, 22, then C[j]=0 Also let Q(n) be the quasi-polynomial of quasi-period, 2, such that 410338673 7 989640329 6 if n modulo, 2, is , 0, then Q(n) equals , --------- n + --------- n 40320 11520 3568100641 5 44516693 4 2125417843 3 379216841 2 43700047 + ---------- n + -------- n + ---------- n + --------- n + -------- n 11520 72 2880 720 210 + 35112 6067482461 2 4494545 if n modulo, 2, is , 1, then Q(n) equals , ---------- n + ------- 11520 128 2125417843 3 44516693 4 3568100641 5 989640329 6 + ---------- n + -------- n + ---------- n + --------- n 2880 72 11520 11520 410338673 7 5593679201 + --------- n + ---------- n 40320 26880 d(n) modulo, 17, can be computed VERY fast (in logarithmic time!) using the f\ ollowing recurrence, taken modulo, 17 d(17 n + a) = Q(17 n + a) + C[a] d(n) + C[a + 17] d(n - 1) subject to the initial condition d(0) = 7, d(1) = 7, d(2) = 8, d(3) = 14, d(4) = 1, d(5) = 11, d(6) = 9, d(7) = 13, d(8) = 2, d(9) = 15, d(10) = 8, d(11) = 15, d(12) = 5, d(13) = 1, d(14) = 5, d(15) = 12, d(16) = 3, d(17) = 2, d(18) = 8, d(19) = 12, d(20) = 1, d(21) = 12, d(22) = 8 Just for fun the googol-th through googol+100-th terms of d(n) modulo, 17, is 5, 6, 4, 6, 13, 14, 13, 15, 2, 11, 13, 9, 9, 0, 6, 16, 9, 11, 8, 8, 8, 1, 15, 1, 1, 2, 6, 14, 14, 15, 8, 15, 15, 1, 4, 9, 11, 9, 16, 4, 16, 14, 8, 8, 10, 9, 15, 10, 7, 15, 11, 5, 9, 16, 9, 2, 6, 2, 12, 4, 5, 8, 6, 7, 14, 1, 10, 11, 8, 14, 0, 14, 4, 10, 4, 1, 1, 3, 6, 14, 10, 8, 7, 12, 0, 4, 12, 6, 12, 5, 13, 5, 11, 10, 2, 2, 8, 7, 12, 7, 13 ----------------------------------------------------------------------------\ ----------------- This took, 0.738, seconds, ----------------------------------------------------------------------------\ ----------------- This took, 0.759, seconds. | 1,314 | 2,958 | {"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} | 3.578125 | 4 | CC-MAIN-2019-18 | latest | en | 0.558843 |
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EL 625 Lecture 10 1 EL 625 Lecture 10 Pole Placement and Observer Design Pole Placement Consider the system ˙ x = A x (1) The solution to this system is x ( t ) = e At x (0) (2) If the eigenvalues of A all lie in the open left half plane, x ( t ) asymp- totically approaches the origin as time goes to infinity (the system is asymptotically stable ). If the eigenvalues of A lie in the closed left half plane (with some eigenvalues possibly on the imaginary ax- is), the state x ( t ) remains bounded (the system is stable ). If some eigenvalues of A lie in the right half plane, the system is unstable . Consider the system ˙ x = A x + B u (3) The state x is measured and can be used to calculate a suitable value of u to make the system stable. The simplest feedback is a static
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EL 625 Lecture 10 2 linear state feedback, u = K x (where K is a constant matrix). With this feedback, ˙ x = A x + B u = A x + BK x = ( A + BK ) x (4) = A 0 x (5) A 0 = A + BK (6) If it is possible to make the eigenvalues of A 0 lie in the open left half plane by using the gains K , then, choosing u = K x would make the resulting closed loop system stable. It can be proved that the eigenvalues of ( A + BK ) can be arbitrarily assigned through a suitable choice of K if the pair ( A, B ) is controllable. If(A,B) is a controllable pair, a similarity transformation, T can be found such that the transformed matrices, ˆ A = T - 1 AT and ˆ B =
EL 625 Lecture 10 3 T - 1 B are in the controller canonical form , shown below. ˆ A = 0 1 0 . . . . . . 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 . . . 0 1 - a 0 - a 1 . . . . . . . . . - a n - 1 (7) ˆ B = 0 . . . . . . 0 1 (8) a n - 1 , a n - 2 , . . . , a 1 , a 0 are the coefficients of the characteristic poly- nomial (the characteristic polynomial is λ n + a n - 1 λ n - 1 + . . . + a 1 λ + a 0 ). The similarity transformation to achieve this ‘controller canon- ical’ structure for the single-input case is described below. Characteristic polynomial , p ( λ ) = λ n + a n - 1 λ n - 1 + . . . + a 1 λ + a 0 (9)
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EL 625 Lecture 10 4 = λ ( λ n - 1 + a n - 1 λ n - 2 + . . . + a 1 ) | {z } p 1 ( λ ) + a 0 (10) p 1 ( λ ) = λ n - 1 + a n - 1 λ n - 2 + . . . + a 1 (11) = λ ( λ n - 2 + a n - 1 λ n - 3 + . . . + a 2 ) | {z } p 2 ( λ ) + a 1 (12) . . . p n - 1 ( λ ) = λ + a n - 1 (13) p n ( λ ) = 1 (14) Define the set of vectors ( v 1 , . . . , v n ) as v 1 = p 1 ( A ) B v 2 = p 2 ( A ) B . . . v n = p n ( A ) B (15) p i ( A ) is the polynomial p i ( λ ) evaluated at λ = A . To prove that this coordinate transformation achieves the desired transformed form, we have to compute the effect of A on the coordi- nate vectors, v 1 , v 2 , . . . , v n . Av 1 = A ( A n - 1 + a n - 1 A n - 2 + . . . + a 2 A + a 1 I ) B
EL 625 Lecture 10 5 = ( A n + a n - 1 A n - 1 + . . . + a 2 A 2 + a 1 A ) = ( A n + a n - 1 A n - 1 + . . . + a 2 A 2 + a 1 A + a 0 I - a 0 I ) B = - a 0 IB = - a 0 B = - a 0 v n (16) (By Cayley Hamilton theorem, p ( A ) = A n + a n - 1 A n - 1 + . . . + a 2 A 2 + a 1 A + a 0 I = 0).
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https://stats.stackexchange.com/questions/134282/relationship-between-svd-and-pca-how-to-use-svd-to-perform-pca | 1,653,430,308,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00383.warc.gz | 613,321,948 | 69,941 | # Relationship between SVD and PCA. How to use SVD to perform PCA?
Principal component analysis (PCA) is usually explained via an eigen-decomposition of the covariance matrix. However, it can also be performed via singular value decomposition (SVD) of the data matrix $\mathbf X$. How does it work? What is the connection between these two approaches? What is the relationship between SVD and PCA?
Or in other words, how to use SVD of the data matrix to perform dimensionality reduction?
• I wrote this FAQ-style question together with my own answer, because it is frequently being asked in various forms, but there is no canonical thread and so closing duplicates is difficult. Please provide meta comments in this accompanying meta thread. Jan 22, 2015 at 11:25
• stats.stackexchange.com/questions/177102/… Feb 3, 2016 at 10:45
• In addition to an excellent and detailed amoeba's answer with its further links I might recommend to check this, where PCA is considered side by side some other SVD-based techniques. The discussion there presents algebra almost identical to amoeba's with just minor difference that the speech there, in describing PCA, goes about svd decomposition of $\mathbf X/\sqrt{n}$ [or $\mathbf X/\sqrt{n-1}$] instead of $\bf X$ - which is simply convenient as it relates to the PCA done via the eigendecomposition of the covariance matrix. Feb 3, 2016 at 12:18
• PCA is a special case of SVD. PCA needs the data normalized, ideally same unit. The matrix is nxn in PCA. Oct 17, 2017 at 9:12
• @OrvarKorvar: What n x n matrix are you talking about ? Mar 29, 2018 at 15:16
Let the data matrix $$\mathbf X$$ be of $$n \times p$$ size, where $$n$$ is the number of samples and $$p$$ is the number of variables. Let us assume that it is centered, i.e. column means have been subtracted and are now equal to zero.
Then the $$p \times p$$ covariance matrix $$\mathbf C$$ is given by $$\mathbf C = \mathbf X^\top \mathbf X/(n-1)$$. It is a symmetric matrix and so it can be diagonalized: $$\mathbf C = \mathbf V \mathbf L \mathbf V^\top,$$ where $$\mathbf V$$ is a matrix of eigenvectors (each column is an eigenvector) and $$\mathbf L$$ is a diagonal matrix with eigenvalues $$\lambda_i$$ in the decreasing order on the diagonal. The eigenvectors are called principal axes or principal directions of the data. Projections of the data on the principal axes are called principal components, also known as PC scores; these can be seen as new, transformed, variables. The $$j$$-th principal component is given by $$j$$-th column of $$\mathbf {XV}$$. The coordinates of the $$i$$-th data point in the new PC space are given by the $$i$$-th row of $$\mathbf{XV}$$.
If we now perform singular value decomposition of $$\mathbf X$$, we obtain a decomposition $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top,$$ where $$\mathbf U$$ is a unitary matrix and $$\mathbf S$$ is the diagonal matrix of singular values $$s_i$$. From here one can easily see that $$\mathbf C = \mathbf V \mathbf S \mathbf U^\top \mathbf U \mathbf S \mathbf V^\top /(n-1) = \mathbf V \frac{\mathbf S^2}{n-1}\mathbf V^\top,$$ meaning that right singular vectors $$\mathbf V$$ are principal directions and that singular values are related to the eigenvalues of covariance matrix via $$\lambda_i = s_i^2/(n-1)$$. Principal components are given by $$\mathbf X \mathbf V = \mathbf U \mathbf S \mathbf V^\top \mathbf V = \mathbf U \mathbf S$$.
To summarize:
1. If $$\mathbf X = \mathbf U \mathbf S \mathbf V^\top$$, then columns of $$\mathbf V$$ are principal directions/axes.
2. Columns of $$\mathbf {US}$$ are principal components ("scores").
3. Singular values are related to the eigenvalues of covariance matrix via $$\lambda_i = s_i^2/(n-1)$$. Eigenvalues $$\lambda_i$$ show variances of the respective PCs.
4. Standardized scores are given by columns of $$\sqrt{n-1}\mathbf U$$ and loadings are given by columns of $$\mathbf V \mathbf S/\sqrt{n-1}$$. See e.g. here and here for why "loadings" should not be confused with principal directions.
5. The above is correct only if $$\mathbf X$$ is centered. Only then is covariance matrix equal to $$\mathbf X^\top \mathbf X/(n-1)$$.
6. The above is correct only for $$\mathbf X$$ having samples in rows and variables in columns. If variables are in rows and samples in columns, then $$\mathbf U$$ and $$\mathbf V$$ exchange interpretations.
7. If one wants to perform PCA on a correlation matrix (instead of a covariance matrix), then columns of $$\mathbf X$$ should not only be centered, but standardized as well, i.e. divided by their standard deviations.
8. To reduce the dimensionality of the data from $$p$$ to $$k, select $$k$$ first columns of $$\mathbf U$$, and $$k\times k$$ upper-left part of $$\mathbf S$$. Their product $$\mathbf U_k \mathbf S_k$$ is the required $$n \times k$$ matrix containing first $$k$$ PCs.
9. Further multiplying the first $$k$$ PCs by the corresponding principal axes $$\mathbf V_k^\top$$ yields $$\mathbf X_k = \mathbf U_k^\vphantom \top \mathbf S_k^\vphantom \top \mathbf V_k^\top$$ matrix that has the original $$n \times p$$ size but is of lower rank (of rank $$k$$). This matrix $$\mathbf X_k$$ provides a reconstruction of the original data from the first $$k$$ PCs. It has the lowest possible reconstruction error, see my answer here.
10. Strictly speaking, $$\mathbf U$$ is of $$n\times n$$ size and $$\mathbf V$$ is of $$p \times p$$ size. However, if $$n>p$$ then the last $$n-p$$ columns of $$\mathbf U$$ are arbitrary (and corresponding rows of $$\mathbf S$$ are constant zero); one should therefore use an economy size (or thin) SVD that returns $$\mathbf U$$ of $$n\times p$$ size, dropping the useless columns. For large $$n\gg p$$ the matrix $$\mathbf U$$ would otherwise be unnecessarily huge. The same applies for an opposite situation of $$n\ll p$$.
• +1 for both Q&A. Thanks for sharing. I have one question: why do you have to assume that the data matrix is centered initially? Aug 6, 2015 at 8:39
• @Antoine, covariance matrix is by definition equal to $\langle (\mathbf x_i - \bar{\mathbf x})(\mathbf x_i - \bar{\mathbf x})^\top \rangle$, where angle brackets denote average value. If all $\mathbf x_i$ are stacked as rows in one matrix $\mathbf X$, then this expression is equal to $(\mathbf X - \bar{\mathbf X})(\mathbf X - \bar{\mathbf X})^\top/(n-1)$. If $\mathbf X$ is centered then it simplifies to $\mathbf X \mathbf X^\top/(n-1)$. Think of variance; it's equal to $\langle (x_i-\bar x)^2 \rangle$. But if $\bar x=0$ (i.e. data are centered), then it's simply the average value of $x_i^2$. Aug 6, 2015 at 9:43
• @amoeba for those less familiar with linear algebra and matrix operations, it might be nice to mention that $(A.B.C)^{T}=C^{T}.B^{T}.A^{T}$ and that $U^{T}.U=Id$ because $U$ is orthogonal Feb 3, 2016 at 15:20
• A code sample for PCA by SVD: stackoverflow.com/questions/3181593/… Mar 23, 2016 at 11:51
• @amoeba yes, but why use it? Also, is it possible to use the same denominator for $S$? The problem is that I see formulas where $\lambda_i = s_i^2$ and try to understand, how to use them?
– Dims
Sep 5, 2017 at 22:49
I wrote a Python & Numpy snippet that accompanies @amoeba's answer and I leave it here in case it is useful for someone. The comments are mostly taken from @amoeba's answer.
import numpy as np
from numpy import linalg as la
np.random.seed(42)
def flip_signs(A, B):
"""
utility function for resolving the sign ambiguity in SVD
http://stats.stackexchange.com/q/34396/115202
"""
signs = np.sign(A) * np.sign(B)
return A, B * signs
# Let the data matrix X be of n x p size,
# where n is the number of samples and p is the number of variables
n, p = 5, 3
X = np.random.rand(n, p)
# Let us assume that it is centered
X -= np.mean(X, axis=0)
# the p x p covariance matrix
C = np.cov(X, rowvar=False)
print "C = \n", C
# C is a symmetric matrix and so it can be diagonalized:
l, principal_axes = la.eig(C)
# sort results wrt. eigenvalues
idx = l.argsort()[::-1]
l, principal_axes = l[idx], principal_axes[:, idx]
# the eigenvalues in decreasing order
print "l = \n", l
# a matrix of eigenvectors (each column is an eigenvector)
print "V = \n", principal_axes
# projections of X on the principal axes are called principal components
principal_components = X.dot(principal_axes)
print "Y = \n", principal_components
# we now perform singular value decomposition of X
# "economy size" (or "thin") SVD
U, s, Vt = la.svd(X, full_matrices=False)
V = Vt.T
S = np.diag(s)
# 1) then columns of V are principal directions/axes.
assert np.allclose(*flip_signs(V, principal_axes))
# 2) columns of US are principal components
assert np.allclose(*flip_signs(U.dot(S), principal_components))
# 3) singular values are related to the eigenvalues of covariance matrix
assert np.allclose((s ** 2) / (n - 1), l)
# 8) dimensionality reduction
k = 2
PC_k = principal_components[:, 0:k]
US_k = U[:, 0:k].dot(S[0:k, 0:k])
assert np.allclose(*flip_signs(PC_k, US_k))
# 10) we used "economy size" (or "thin") SVD
assert U.shape == (n, p)
assert S.shape == (p, p)
assert V.shape == (p, p)
• Is the code written in Python 2? If so, I think a Python 3 version can be added to the answer. Apr 16 at 8:45
Let me start with PCA. Suppose that you have n data points comprised of d numbers (or dimensions) each. If you center this data (subtract the mean data point $\mu$ from each data vector $x_i$) you can stack the data to make a matrix
$$X = \left( \begin{array}{ccccc} && x_1^T - \mu^T && \\ \hline && x_2^T - \mu^T && \\ \hline && \vdots && \\ \hline && x_n^T - \mu^T && \end{array} \right)\,.$$
The covariance matrix
$$S = \frac{1}{n-1} \sum_{i=1}^n (x_i-\mu)(x_i-\mu)^T = \frac{1}{n-1} X^T X$$
measures to which degree the different coordinates in which your data is given vary together. So, it's maybe not surprising that PCA -- which is designed to capture the variation of your data -- can be given in terms of the covariance matrix. In particular, the eigenvalue decomposition of $S$ turns out to be
$$S = V \Lambda V^T = \sum_{i = 1}^r \lambda_i v_i v_i^T \,,$$
where $v_i$ is the $i$-th Principal Component, or PC, and $\lambda_i$ is the $i$-th eigenvalue of $S$ and is also equal to the variance of the data along the $i$-th PC. This decomposition comes from a general theorem in linear algebra, and some work does have to be done to motivate the relatino to PCA.
SVD is a general way to understand a matrix in terms of its column-space and row-space. (It's a way to rewrite any matrix in terms of other matrices with an intuitive relation to the row and column space.) For example, for the matrix $A = \left( \begin{array}{cc}1&2\\0&1\end{array} \right)$ we can find directions $u_i$ and $v_i$ in the domain and range so that
You can find these by considering how $A$ as a linear transformation morphs a unit sphere $\mathbb S$ in its domain to an ellipse: the principal semi-axes of the ellipse align with the $u_i$ and the $v_i$ are their preimages.
In any case, for the data matrix $X$ above (really, just set $A = X$), SVD lets us write
$$X = \sum_{i=1}^r \sigma_i u_i v_j^T\,,$$
where $\{ u_i \}$ and $\{ v_i \}$ are orthonormal sets of vectors.A comparison with the eigenvalue decomposition of $S$ reveals that the "right singular vectors" $v_i$ are equal to the PCs, the "right singular vectors" are
$$u_i = \frac{1}{\sqrt{(n-1)\lambda_i}} Xv_i\,,$$
and the "singular values" $\sigma_i$ are related to the data matrix via
$$\sigma_i^2 = (n-1) \lambda_i\,.$$
It's a general fact that the right singular vectors $u_i$ span the column space of $X$. In this specific case, $u_i$ give us a scaled projection of the data $X$ onto the direction of the $i$-th principal component. The left singular vectors $v_i$ in general span the row space of $X$, which gives us a set of orthonormal vectors that spans the data much like PCs.
I go into some more details and benefits of the relationship between PCA and SVD in this longer article.
• Thanks for your anser Andre. Just two small typos correction: 1. In the last paragraph you`re confusing left and right. 2. In the (capital) formula for X, you're using v_j instead of v_i.
– Alon
Sep 3, 2019 at 13:09 | 3,499 | 12,187 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 68, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-21 | longest | en | 0.92648 |
https://justaaa.com/civil-engineering/19749-consider-a-closed-polygon-traverse-lengths-in | 1,716,380,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00394.warc.gz | 297,200,699 | 11,365 | Question
# Consider a closed-polygon traverse (lengths in feet). Balance the traverses by coordinates using the compass rule....
Consider a closed-polygon traverse (lengths in feet). Balance the traverses by coordinates using the compass rule.
Course AB BC CD DA Bearing N 22∘36'40" W S 60∘39'24" W S 38∘46'10" E N 88∘22'58" E Length 314.02 264.49 213.50 217.69
Compute the unbalanced departures.
AB=,BC=, CD=,DA= -120.733,-230.556,133.691,217.603 ft, ft, ft, ft
Correct
Part B
Part complete
Compute the unbalanced latitudes.
AB=,BC=, CD=,DA= 289.883,-129.611,-166.460,6.144 ft, ft, ft, ft
Correct
Part C
Part complete
Compute the linear misclosure.
4.5×10−2 ft
Correct
Part D
Part complete
Compute the relative precision.
1: 2.25×104
Correct
Part E
Compute the preliminary coordinates if XA=10,000.00 and YA=5000.00.
XB=,YB=,
X B = , Y B = ,
XC=,YC=,
XD=,YD=
9879.266402,9648.69926,9782.389598,5289.883268,5160.277916,4993.82517
ft, ft, ft, ft, ft, ft
Incorrect; Try Again; 5 attempts remaining
Provide Feedback
Next
-> Balancing of latitudes and departures is not required because of very small closing error. Please eneter answer proper order.
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 406 | 1,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.689286 |
https://studydaddy.com/question/i-have-a-question-can-someone-tell-me-what-the-check-s1-2-s2-2-mean | 1,571,876,562,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987836368.96/warc/CC-MAIN-20191023225038-20191024012538-00291.warc.gz | 706,102,855 | 7,753 | QUESTION
# I have a question can someone tell me what the check s1^2/s2^2 mean?
I have a question can someone tell me what the check s1^2/s2^2 mean? S= standard deviation. The BOLD lettering and numbers.
The main trial is conducted and involves a total of 200 patients. Patients are enrolled and randomized to receive either the experimental medication or the placebo. The data shown below are data collected at the end of the study after 6 weeks on the assigned treatment.
Experimental (n=100)
Placebo (n=100)
Mean (Std Dev) Systolic Blood Pressure
120.2 (15.4)
131.4 (18.9)
% Hypertensive
14%
22%
% With Side Effects
6%
8%
1. Generate a 95% confidence interval for the difference in mean systolic blood pressures between groups. | 193 | 747 | {"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-2019-43 | latest | en | 0.857827 |
https://www.sktime.org/en/latest/api_reference/auto_generated/sktime.performance_metrics.forecasting.median_squared_error.html | 1,669,636,229,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00012.warc.gz | 1,050,826,564 | 12,232 | # median_squared_error#
median_squared_error(y_true, y_pred, horizon_weight=None, multioutput='uniform_average', square_root=False, **kwargs)[source]#
Median squared error (MdSE) or root median squared error (RMdSE).
If square_root is False then calculates MdSE and if square_root is True then RMdSE is calculated. Both MdSE and RMdSE return non-negative floating point. The best value is 0.0.
Like MSE, MdSE is measured in squared units of the input data. RMdSE is on the same scale as the input data like RMSE. Because MdSE and RMdSE square the forecast error rather than taking the absolute value, they penalize large errors more than MAE or MdAE.
Taking the median instead of the mean of the squared errors makes this metric more robust to error outliers relative to a meean based metric since the median tends to be a more robust measure of central tendency in the presence of outliers.
Parameters
y_truepd.Series, pd.DataFrame or np.array of shape (fh,) or (fh, n_outputs) where fh is the forecasting horizon
Ground truth (correct) target values.
y_predpd.Series, pd.DataFrame or np.array of shape (fh,) or (fh, n_outputs) where fh is the forecasting horizon
Forecasted values.
horizon_weightarray-like of shape (fh,), default=None
Forecast horizon weights.
multioutput{‘raw_values’, ‘uniform_average’} or array-like of shape (n_outputs,), default=’uniform_average’
Defines how to aggregate metric for multivariate (multioutput) data. If array-like, values used as weights to average the errors. If ‘raw_values’, returns a full set of errors in case of multioutput input. If ‘uniform_average’, errors of all outputs are averaged with uniform weight.
square_rootbool, default=False
Whether to take the square root of the mean squared error. If True, returns root mean squared error (RMSE) If False, returns mean squared error (MSE)
Returns
lossfloat
MdSE loss. If multioutput is ‘raw_values’, then MdSE is returned for each output separately. If multioutput is ‘uniform_average’ or an ndarray of weights, then the weighted average MdSE of all output errors is returned.
References
Hyndman, R. J and Koehler, A. B. (2006). “Another look at measures of forecast accuracy”, International Journal of Forecasting, Volume 22, Issue 4.
Examples
```>>> from sktime.performance_metrics.forecasting import median_squared_error
>>> y_true = np.array([3, -0.5, 2, 7, 2])
>>> y_pred = np.array([2.5, 0.0, 2, 8, 1.25])
>>> median_squared_error(y_true, y_pred)
0.25
>>> median_squared_error(y_true, y_pred, square_root=True)
0.5
>>> y_true = np.array([[0.5, 1], [-1, 1], [7, -6]])
>>> y_pred = np.array([[0, 2], [-1, 2], [8, -5]])
>>> median_squared_error(y_true, y_pred)
0.625
>>> median_squared_error(y_true, y_pred, square_root=True)
0.75
>>> median_squared_error(y_true, y_pred, multioutput='raw_values')
array([0.25, 1. ])
>>> median_squared_error(y_true, y_pred, multioutput='raw_values', square_root=True)
array([0.5, 1. ])
>>> median_squared_error(y_true, y_pred, multioutput=[0.3, 0.7])
0.7749999999999999
>>> median_squared_error(y_true, y_pred, multioutput=[0.3, 0.7], square_root=True)
0.85
``` | 818 | 3,118 | {"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-2022-49 | latest | en | 0.657537 |
https://mozillazg.com/2020/10/leetcode-147-insertion-sort-list.html | 1,619,190,814,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594808.94/warc/CC-MAIN-20210423131042-20210423161042-00239.warc.gz | 504,750,268 | 7,808 | # LeetCode: 147. Insertion Sort List
## 题目¶
Sort a linked list using insertion sort.
A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list. With each iteration one element (red) is removed from the input data and inserted in-place into the sorted list
Algorithm of Insertion Sort:
1. Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list.
2. At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there.
3. It repeats until no input elements remain.
Example 1:
```Input: 4->2->1->3
Output: 1->2->3->4
```
Example 2:
```Input: -1->5->3->4->0
Output: -1->0->3->4->5
```
## 解法¶
### 生成一个有序数组，然后再生成结果链表¶
```# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
nodes = []
dummy = ListNode()
tail = dummy
for node in nodes:
tail.next = node
tail = node
# 防止节点上原有的 next 属性值导致出现循环
tail.next = None
return dummy.next
def insert_node(self, nodes, node):
for i, v in enumerate(nodes):
if node.val < v.val:
nodes.insert(i, node)
break
else:
nodes.append(node)
return nodes
```
### 不生成中间数组，直接对链表进行排序操作¶
• 定义一个空链表 dummy ，作为排序后的结果链表，类似上面的中间数组
• 遍历输入的链表，将节点插入到上面的 dummy 链表中，为了保持 dummy 有序，需要按上面类似 insert_node 的方法在 dummy 中找到合适的位置，因为 dummy 是个链表而不是数组无法直接插入，需要记录插入位置的上一个节点 (pre_node) 和下一个节点(next_node)，然后再基于这两个节点的信息实现插入新节点的功能。
```# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
# 新的链表
dummy = ListNode()
while current is not None:
# 把 current 插入到 dummy 链表中
# 新节点插入位置的上一个节点
pre_node = dummy
# 新节点插入位置的下一个节点
next_node = pre_node.next
while next_node is not None:
# 找到插入位置
if current.val < next_node.val:
break
pre_node = next_node
next_node = next_node.next
# 插入新节点
pre_node.next = current
_next = current.next
current.next = next_node
current = _next
return dummy.next
``` | 679 | 2,128 | {"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} | 3.640625 | 4 | CC-MAIN-2021-17 | latest | en | 0.37565 |
https://www.physicsforums.com/threads/can-anyone-tell-me-the-relationship-between-work-done-and-potential-difference.610975/ | 1,544,852,048,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826715.45/warc/CC-MAIN-20181215035757-20181215061757-00399.warc.gz | 1,000,936,883 | 15,669 | # Can anyone tell me the relationship between work done and potential difference?
1. Jun 2, 2012
### mutineer123
Can anyone tell me the relationship between work done and potential difference??
CAN ANYONE TELL ME THE RELATIONSHIP BETWEEN WORK DONE AND POTENTIAL DIFFERENCE??
If more work is being done across a component(for ex due to increased resistance), does the pd across it increase or decrease? I think it should increase, because pd is the work done per unit charge, so if the work done increases, so would the pd. AM I right?
2. Jun 2, 2012
### danmay
Re: Can anyone tell me the relationship between work done and potential difference??
Yes, it would. There may be complications in terms of defining a potential, whether the work done is conservative, etc., but in general yes. Remember, however, that this increase comes from whatever "thing" (in your case, some kind of resistance) that increases the work required to be done.
3. Jun 3, 2012
### Ken G
Re: Can anyone tell me the relationship between work done and potential difference??
Well, one must be careful. Usually when one is talking about the work done across a potential difference, one is talking about the work done by the force responsible for that potential difference. The work then shows up as a change in the kinetic energy of the system crossing that potential difference, as per the "work-energy theorem." So if that is the context of the question, note that including resistance does not change the work done by the force associated with the potential difference (that can never be changed, it is given by the charge times the potential difference), but it means that there is also a resistive force doing additional work (which would actually be negative work and would subtract from the kinetic energy). So the answer to the question depends on whether one is talking about all the work being done by all the forces (in which case the work done could be less if there is resistance-- as it would likely be a positive work done by the potential and a negative work done by the resistive force), or if it is just the work done by the force associated with the potential difference (which is independent of any resistance that might be present).
4. Jun 3, 2012
### tonyjk
Re: Can anyone tell me the relationship between work done and potential difference??
We can say that for example a generator makes a positive work and all components that consume "electricity" makes negative work?
5. Jun 3, 2012
### danmay
Re: Can anyone tell me the relationship between work done and potential difference??
With respect to whatever you're trying to move, it would depend on whether its kinetic energy increases (positive) or decreases (negative) as a result of any work done.
Similarly for resistance, depending on whether it's against or along the original potential. For a resistance (say friction) that's symmetric with respect to the original potential, it could either increase or decrease the "total" potential depending on how you're trying to move the test object. In that case, the potential is probably ambiguous, if given only position information.
6. Jun 3, 2012
### Ken G
Re: Can anyone tell me the relationship between work done and potential difference??
It depends on the frame of reference, but in almost all situations you would normally see, friction does negative work-- regardless of whether the work done by the potential is positive or negative. Also, no "potential' can be associated with friction, so it's best to keep the "potential" concept and the "friction" concept quite separate in your head.
7. Jun 3, 2012
### mutineer123
Re: Can anyone tell me the relationship between work done and potential difference??
I am just an A level student, so have not come across work energy theorem, or friction as resistance...let me show you a question where a application of my question(in MY level) is shown.
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
question 35.
If i was right(in this post), then my answer would have been C, but apparently the right answer is A, and I dont know why.
8. Jun 4, 2012
### willem2
Re: Can anyone tell me the relationship between work done and potential difference??
As the resistance goes up, the potential difference across it increases, but the current through it will decrease, because the total resistance increases, so in some cases
the work done can decrease, even if the potential difference increases.
total resistance: R+2 (series resistances)
current = 12 / (R+2) (ohms' law)
Potential difference across resistor = 12 R/(R+2) (ohms' law again)
work done = 144 R /(R+2)^2 (P = IV)
This function has a maximum.
9. Jun 4, 2012
### Ken G
Re: Can anyone tell me the relationship between work done and potential difference??
That's correct, and if you want a more physically intuitive version, the point is that when the load resistance is much smaller than the internal resistance, the current is not affected much by the load resistance, so the power dissipated in the load resistor will be proportional to the load resistance R. But as the load resistance exceeds the internal resistance, now the internal resistance is what is negligible, and you just have a fixed voltage across the load resistance R, which by P=V^2/R means the power will drop like 1/R.
10. Jun 5, 2012
### danmay
Re: Can anyone tell me the relationship between work done and potential difference??
Negative work doesn't necessarily mean a more negative potential. It depends on the nature of resistance/friction. It also helps to know why friction and potential are usually kept separate (look up Wikipedia or other sources for the complete explanation); but yes, they are usually not associated with one another.
willem2 (#8) and Ken G (#9) have got it. | 1,285 | 5,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-51 | latest | en | 0.969801 |
https://ask.sagemath.org/answers/42327/revisions/ | 1,597,511,594,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740929.65/warc/CC-MAIN-20200815154632-20200815184632-00486.warc.gz | 209,478,282 | 6,193 | Ask Your Question
Revision history [back]
Samuel and Emanuel, thanks for teaching me the magic of QQbar and SR! one more question, please. How to do this inverse_laplace of a fraction whose denominator roots may be real (and maybe even complex)? For example,
var('s,u')
R.<s> = PolynomialRing(QQbar)
F = R.fraction_field()
L=3/4*(19*s^2 + 156*s + 284)/(19*s^3 + 174*s^2 + 422*s + 228)
whole,LL=L.partial_fraction_decomposition()
show(LL)
Here I'm stuck with the "decimals?" numbers and
inverse_laplace(L,s,u)
ends again in
TypeError: unable to convert [0.6934316866155686?/(s + 0.7570751092241817?), 0.0477251478733486?/(s + 2.861392366086696?), 0.00884316551108293?/(s + 5.539427261531228?)] to a symbolic expression
on the other hand
SR(L).partial_fraction()
fails to find the real roots, and inverse_laplace fails also. Thanks, F
Samuel and Emanuel, thanks for teaching me the magic of QQbar and SR! one more question, please. How to do this inverse_laplace of a fraction whose denominator roots may be real (and maybe even complex)? For example,
var('s,u')
R.<s> = PolynomialRing(QQbar)
F = R.fraction_field()
L=3/4*(19*s^2 + 156*s + 284)/(19*s^3 + 174*s^2 + 422*s + 228)
whole,LL=L.partial_fraction_decomposition()
show(LL)
Here I'm stuck with the "decimals?" numbers and
inverse_laplace(L,s,u)
ends again in
TypeError: unable to convert [0.6934316866155686?/(s + 0.7570751092241817?), 0.0477251478733486?/(s + 2.861392366086696?), 0.00884316551108293?/(s + 5.539427261531228?)] to a symbolic expression
on the other hand
SR(L).partial_fraction()
fails to find the real roots, and inverse_laplace fails also. Thanks, F
Samuel and Emanuel, thanks for teaching me the magic of QQbar and SR! one more question, please. How to do this inverse_laplace of a fraction whose denominator roots may be real (and maybe even complex)? For example,
var('s,u')
R.<s> = PolynomialRing(QQbar)
F = R.fraction_field()
L=3/4*(19*s^2 + 156*s + 284)/(19*s^3 + 174*s^2 + 422*s + 228)
whole,LL=L.partial_fraction_decomposition()
show(LL)
Here I'm stuck with the "decimals?" numbers and
inverse_laplace(L,s,u)
ends again in
TypeError: unable to convert [0.6934316866155686?/(s + 0.7570751092241817?), 0.0477251478733486?/(s + 2.861392366086696?), 0.00884316551108293?/(s + 5.539427261531228?)] to a symbolic expression
on the other hand
SR(L).partial_fraction()
fails to find the real roots, and inverse_laplace fails also. Thanks, F
The problem is really how to do partial_fractions and then extract the answer, but I added more of the story to make it clear this is a numeric question. For a first answer, I do not need 100 digits precision :)
I have added a new question "inverse_laplace of a fraction whose denominator roots are real (or complex) to relance this with more details | 847 | 2,807 | {"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 | 3 | CC-MAIN-2020-34 | latest | en | 0.68645 |
https://muddoo.com/category/programming/page/5/ | 1,685,494,276,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00087.warc.gz | 475,002,347 | 36,829 | Categories
## Difference between range vs arange in Python
In this article, we will take a look at range vs arange in Python. Learning the difference between these functions will help us understand when to either of them in our programs. Both range and arange functions of Python have their own set of advantages and disadvantages. This article will help us learn about them in detail.
To better understand the difference between range vs arange functions in Python, let us first understand what each of these functions’ do.
## range vs arange in Python: Understanding range function
The range function in Python is a function that lets us generate a sequence of integer values lying between a certain range. The function also lets us generate these values with specific step value as well . It is represented by the equation:
``range([start], stop[, step])``
So, in the above representation of the range function, we get a sequence of numbers lying between optional start & stop values. Next, each of these values are also getting incremented by the optional step values.
### range function example 1
So that was the theory behind the range function. Now, let understand it better by practicing using it. Fire up your Python IDLE interpreter and enter this code:
``l = range(1, 10, 2)``
When you hit enter on your keyboard, you don’t see anything right? That is because the resulting sequence of values are stored in the list variable “l”.
To be able to see the values stored in it, let us print individual list values. So, by indexing each of the list items, we get the following values printed out.
``````>>> l[0]
1
>>> l[1]
3
>>> l[2]
5
>>> l[3]
7
>>> l[4]
9``````
So we see a list of numbers, cool! But when you observe closely, you will realize that this function has generated the numbers in a particular pattern. You can see that the first number it has generated is after taking into consideration our optional start parameter. We had set its value to 1. Next, it is also honoring the stop value by printing the value 9, which is within our defined stop value of 10. If you try to index the list for any further value beyond this point will only return an error:
``````>>> l[5]
Traceback (most recent call last):
File "<pyshell#7>", line 1, in <module>
l[5]
IndexError: range object index out of range``````
So, this confirms that the last value we get will always be less than the stop value.
But the most important thing to observation we need to make here is the step size between each of these values. We can see that each of these values are incremented exactly with a step size of 2. This is the same step size we had defined in our call to range function (the last parameter)!
### range function example 2
Does this mean everytime we want to call range function, we need to define these 3 parameters? Not really. If we take a look at the signature of the arange function again:
``range([start], stop[, step])``
The parameters start and step (mentioned within the square brackets) are optional. This means that we can call the range function without these values as well like this:
``````nl = range(4)
>>> nl
range(0, 4)
>>> nl[0]
0
>>> nl[1]
1
>>> nl[2]
2
>>> nl[3]
3
>>> nl[4]``````
In this case, we have only the stop value of 4. As a result we get our sequence of integers starting with a default value of 0. The step value is also defaulted to a value of 1. So we get the integers in the range between 0 to 3, with a step value of 1.
Alright then, hope everything is clear to you up to this point. If this is the case with Python’s range function, what does the arange function in Python do?
## range vs arange in Python: Understanding arange function
Unlike range function, arange function in Python is not a built in function. But instead, it is a function we can find in the Numpy module. So, in order for you to use the arange function, you will need to install Numpy package first!
The signature of the Python Numpy’s arange function is as shown below:
``numpy.arange([start, ]stop, [step, ]dtype=None)``
Wait a second! Doesn’t this signature look exactly like our range function? Yes, you are right! Python’s arange function is exactly like a range function. It also has an optional start and step parameters and the stop parameter.
But then what is the difference between the two then?
## range vs arange in Python – What is the difference?
Where the arange function differs from Python’s range function is in the type of values it can generate.
The built in range function can generate only integer values that can be accessed as list elements. But on the other hand, arange function can generate values that are stored in Numpy arrays. We can observer this in the following code:
``````import numpy as np
a = np.arange(4)
>>> a
array([0, 1, 2, 3])
>>> a[0]
0``````
We are clearly seeing here that the resulting values are stored in a Numpy array. Each of the individual values can hence also be accessed by simply indexing the array!
This begs us the next question. When should we use Python’s built-in range function vs Numpy’s arange function? To understand this, lets list out the advantages and disadvantages of each of these functions:
### Advantages of range function in Python
• range function returns a list of integers in Python 2. In case of Python 3, it returns a special “range object” just like a generator.
### Disadvantages of range function in Python
• range function is considerably slower
• It also occupies more memory space when handling large sized data objects.
### Advantages of arange function in Python
• Numpy’s arange function returns a Numpy array
• Its performance is wat better than the built-in range function
• When dealing with large datasets, arange function needs much lesser memory than the built-in range function.
So this is the fundamental difference between range vs arange in Python. We can understand them even better by using them more in our everyday programming.
I hope this gave you some amount of clarity on the subject. If you still have any more doubts, do let me know about it in the comments below. I will be more than happy to help you out.
Having said that, take a look at this article. It gives you a simple explanation on the “Difference between expressions and statements in Python“. I have spent considerable amount of time trying to understand these topics. Since there are not many articles available that explains them clearly, I started this blog to capture these topics. Hope you found them useful! If yes, do share them with your friends so that it can help them as well.
With this, I will conclude this article right here. See you again in my next article, until then, ciao!
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## Introduction To Matplotlib – Data Visualization In Python
This article will give you an introduction to Matplotlib Data Visualization In Python. Matplotlib is a data visualization library package written specifically for us to be used with Python. So Matplotlib is usually the preferred Python package to visualize data while working on Machine Learning & Data Science. It helps us in visualizing the data by representing them with the help of plots and charts.
## Brief Introduction To Matplotlib – Data Visualization In Python
We humans are all highly responsive to images than text messages. Images helps us in better visualizing and understanding a situation over interpreting any raw data. So we always wanted a way to represent data through images. If you look at our history, we have always tried to accomplish this in many ways. While I cant go back in time to explain each and every approach, I can quicky give you an historical introduction to Matplotlib. This should help you understand how this package came to be. Why it matters a lot in Python data visualization.
### Historical Introduction To Matplotlib – Data Visualization
In the early days of computer data analysis, data scientists often relied on tools like gnuplot and MATLAB to visualize data. However, the problem here was that they had to do it in two stages. First use programming languages like C or Python scripts to process the data. Then plot the resulting data output using gnuplot or MATLAB.
It was a very cumbersome process to say the least. It also resulted in erroneous calculation at times due to lengthy process. As a result, the scientists were in dire need of a simpler solution to this. This is when Matplotlib – Data Visualization package in Python was born.
This helped scientists to both process the data using Python scripts and also visualize the resulting output using Matplotlib package. Now since the Matplotlib was developed along the lines of MATLAB, it is supporting all the functionalities of MATLAB. Because of this reason, it got embraced by the data scientists over MATLAB pretty quickly.
Now you may be wondering why you should be using Matplotlib over MATLAB. For you to understand and appreciate Matplotlib package, you need to understand some of the benefits it brings over a tool like MATLAB. Discussing the advantages and disadvantages of this library in an article that gives an introduction to Matplotlib is appropirate I believe. This will help you in making appropriate decision while chosing this Python library package.
• Matplotlib Is Open Source – One of the primary advantage of Matplotlib is that it is an open source package. Because of this, you can use it in whatever way you want. You don’t need to pay any money for this tool to anyone. You can also use it for both academic and commerical purposes.
• Written In Python – Yet another benefit of Matplotlib is that it is all written in Python. As you use Python programming language to do data processing, plotting its result in Python again makes it so much more easier.
• Customizable & Extensible – As its written in Python, you will also be able to customize the package (if required) to suit your requirements. In addition to this, you can always extend its functionalities and contribute it back to the open source community. Since Python also has other useful packages, you can also make use of those packages’ functionalities to extend Matplotlib.
• Portable & Cross Platform – Since its written in Python it is easily portable to any system which can run Python. It also works smoothly on Windows, Linux & Mac OS.
• Easy to learn – Because the Python language is much easier to learn, any packages written using this language becomes so much more easier. You will not find Matplotlib any different either when it comes to this.
## Matplotlib Output formats
When we plot any data using Matplotlib, we can get the resulting output plots in two different ways.
The first method is to get the resulting output plots in a new window. This is useful if you want an interactive data output. In this case, your output will continue to display for as long as your program is running.
The second method of output plots you can obtain is by saving them permanently on your computers. In this method, your resulting output will be saved in a file in standard image formats such as PNG, JPG, PDFs etc. This will be very useful to you when you want to share your results with others. You can also make use of this method when you want to generate a lot of output charts or plots programmatically. But in this case, you have one disadvantage. You will not be able to interact with the resulting output like the way you could in the first method. But as I mentioned earlier, if you just want to analyze the results in bulk at a later time, this is still one of the best method to make use of.
So you might be wondering now, what other different formats can you use to save the resulting output. Let me help you right there! So this is the list of file formats supported by Matplotlib that you can use to save your resulting plots:
EPS, JPG, PDF, PNG, PS, SVG
So now you understand the different formats that we can store our outputs in. Next, let me introduce you to another important feature of Matplotlib. It is called Backends and this terminology is something that you should be aware of when working with Matplotlib.
## Introduction To Matplotlib: What Are Backends?
Backends In Matplotlib is a feature using which you can either visualize the output live or store it to analyze them in the future.
As I mentioned in the previous section, we can store the resulting plots in either files or view it live in a new interactive window. Backends simply represent this factor. So from the previous paragraphs we can already realize that there are two types of backends in Matplotlib:
• Hardcopy backend – This is the type of backend where we save the images in a file
• User Interface backend – In this type of backend, we display the resulting output in an interactive output window.
In order to provide us with these two type of backends, Matplotlib makes use of two sub modules called the renderer and the canvas. Let us try to learn more about them to get a better understanding of Matplotlib backends.
### What is a renderer in Matplotlib?
A renderer is a module used by Matplotlib to draw its output plot or the graph. So it is this module that does the actual drawing of our Matplotlib’s output. The standard renderer used by Matplotlib to render its output is called the Anti-Grain Geometry(AGG) renderer.
Now are you wondering what this rendere do? Want to know what is so special about this renderer? This AGG renderer is a high performance renderer. It helps us in getting generating a publication level quality output. It also helps us in obtaining our output with sub pixel accuracy and antialiasing. Is this all sounding too alien of a terminology for you? Then simply know that the AGG renderer helps in getting a high production quality graphs and plots that we can adore about!
Now that we understand about the renderer used by Matplotlib, let us turn our focus towards the second module – canvas
### What Is A Canvas In Matplotlib?
After getting an understanding about renderer, its time for us to learn about the other module – canvas of Matplotlib.
We mentioned that the renderer is responsible for the drawing. But do you know where exactly is this drawing being done at? That is where the canvas module comes into picture. Canvas is the area where the renderer will draw our out plots and graphs. Ok, so who provides this canvas then? Great question!
Canvas in Matplotlib is usually provided by the GUI libraries. So this can be GTK if you are using a Linux machine. Or it can be WX if you are on a Mac OS. On the other hand if you are using Windows machines, these could be coming in from Windows GUI libraries. But these are not all. It could also be coming in from platform agnostic interfaces like QT if you are developing your visualization tools in QT.
So basically we get the canvas from the GUI library we intend to use in our Visualization app.
But do you really need to worry about it? Do you really need to know about canvases and renderers when using Matplotlib? Well, for the most cases, not really. You can simply use Matplotlib’s functions and get away with generating your visualizations. You dont really need to know the underlying aspects of how Matplotlib works to use it. But if you want to be a good visual data scientist, knowing how Matplotlib works under the hood will help you in mastering it. It will also help you in truly appreciating its versatility.
## Conclusion
So that is it. This should give you a good introduction to Matplotlib – data visualization and why you should use it. In the next set of articles, we will learn how to install the library package. We will learn how to make use of it to draw some interesting plots and graphs. So see you there! Until then, have a great learning experience!
If you are interested in using Matplotlib to add text to any image, here is a quick tutorial link describing how you can do this.
Categories
## Python Add Text To An Image Using Matplotlib Library
We can use Python programming language to add text to an image using its Matplotlib library. In this tutorial, we will take a look at the Matplotlib library and learn how to accomplish this. The process of adding text to an image is often also called as annotating text to an image.
## What Is Python Matplotlib library?
Matplotlib is a Python data visualization library that is often used to plot graphs and charts to represent a dataset in a meaningful way. However, we can also use Matlplotlib to do some basic editing on an image file. We can use it to overlay graphical data over an image or plot graphs.
But in this tutorial let us make use of Matlplotlib to add basic text annotation to our image. In other words, we will use Python Matplotlib to add text to our image file.
## Adding Text To An Image Using Python Matplotlib Library
In order to get started, we first need to use an image file over which we would like to add text. In this example, I will make use of this picture of a butterfly that I shot in my garden. This is a simple JPG file that I got by capturing the butterfly image using my Android smartphone’s camera.
This is a 278KB JPG file over which I would like to add an annotated text stating “What A Wondeful World!”. I woud like the text to be Orange in color with a font size of 30 and stylized to be bold. I would also like to place this text at the bottom of the image, somewhere over here:
So enough of theory, let us start writing our Python program to add the text to our image. Fire up your favorite text editor to start writing your code. In my case I use VS Code editor, but you can use any text editor you want.
## Python Program To Add Text To An Image Using Matplotlib
The first line of code we would like to add in our program is:
``import matplotlib.pyplot as plt``
What this line does is that it would import our matlotlib library’s pyplot module into our program. So this is imported as plt so every time we want to make use of it, we would call it by using the plt name.
Next thing we would like to import is the Python Imaging Library (aka PIL)’s Image module. We need to use this module to import the image file (butterfly.jpg) into our program. So this is achieved by using the following piece of code:
``from PIL import Image``
We will then go ahead with importing the butterfly.jpg file into our program using the line:
``img = Image.open("/home/amar/Pictures/butterfly.jpg")``
The Image.open() method will open our file, read its content and store it into the variable img. Notice here again that we have imported the buttefly.jpg file into a new Python variable called img. Because of this from now on wards, we will be able to access the content of this image file by using the img Python variable.
With the image being available, its time to start editing it using Matplotlib library.
The first step in using Matplotlib library is to create a Matplotlib’s figure object, which forms the canvas of our image. All the operations we do from now on will be with respect to this canvas element. We can create Matplotlib’s figure object by using the lines:
``plt.figure(figsize=(12,8))``
Wait a second! What does this line even mean? What does figsize stand for and what values are we passing here? Let us answer these questions first.
### Understanding Figsize Parameter Of Matplotlib’s Figure Object
figsize is a parameter we need to pass to the matplotlib plt module’s figure() method. It allows us to specify the exact width and height of our image.
Therefore when we pass the values (12,8) to this parameter, we are specifying these exact values. These values are expressed in the unit of inches. So when we say (12,8), we are asking the Matplotlib library to draw a figure with a width of 12 inches and a height of 8 inches.
Following this, we will ask Matplotlib library to draw our picture onto the figure‘s canvas by calling imshow fuction as follows:
``plt.imshow(img)``
As a result of the above code, Matplotlib will draw to image onto our figure object. However, it must be noted that it only draws the image to our figure object. But nothing will be displayed to the end user. If we want to display this image, we need to call another Matplotlib function plt.show(). But we do not want to display it yet. Therefore, we will discuss about this function later in the article when we actually make use of it.
So for now we will further move on with our code. As of now we have created a Matplotlib’s figure canvas object, we have opened our image file and drawn this image onto our figure object’s canvas. It is now time to write our intended “What a wonderful world!” text on top of it.
### Adding Text Using Plot’s Text Function
In order for us to write text over an image, Matlplotlib provides us with yet another method, not surprisingly called the text function. So we can access it as part of the plt module. Also, we can find this function’s signature and the parameters it takes in link to its official documentation. With this information, we can finally call the text function to print our string onto the figure object’s canvas like this:
``plt.text(200, 700, "What A Wonderful World!", color="orange", fontdict={"fontsize":30,"fontweight":'bold',"ha":"left", "va":"baseline"})``
From the above code, we can interpret the parameters of plt.text function as follows:
#### X & Y Co-Ordinates Parameter
The first two parameters of this function are the x & y co-ordinates of the image. We set its values to be 200 and 700 so that the string starts from there.
#### Annotation Text Parameter
Following the two co-ordinate parameters is the annotation text paramter. This is the parameter which takes in the text string that we would like to print. We have set it to our string “What A Wonderful World!” that we wanted to add to our image.
#### Text Color Parameter
Following the the text parameter is the color parameter. We use this parameter to set the color of the string. In our case, we have set this to orange color.
#### Font Dict Parameters
Finally, we set the parameters for the type of font to use and its properties. In our case, we are asking Matplotlib library to set the following properties to our text fonts:
• We want to use a font size of 30
• We want the font to be bold
• Next, we want the text’s horizontal alignment to be to its left
• Finally, we also want it to be vertically aligned to its baseline
So with these properties set, we have finally defined the text that we wanted on our canvas. Hence, it is now time to display this image on our screen.
## Displaying Image On Screen Using Matplotlib
This is the final stage of our code. At this point, we want our program to simply display the content of our modified image on our screen. So we will achieve this by using the plt.show() function that we had discussed earlier.
### Using Matplotlib’s Show Function To Display The Image On Screen
The code to display our image on screen is pretty simple
``plt.show()``
We just call the plt.show() function as shown above. This should finally display the image in a new window on our screen.
## Summary Of Python Program To Add Text To Image
So to summarize, our final program in full looks like this:
``````import matplotlib.pyplot as plt
from PIL import Image
img = Image.open("/home/amar/Pictures/butterfly.jpg")
plt.figure(figsize=(12,8))
#Finishes drawing a picture of it. Does not display it.
plt.imshow(img)
plt.text(200, 700, "What A Wonderful World!", color="orange", fontdict={"fontsize":30, "fontweight":'bold', "ha":"left", "va":"baseline"})
#Display the image
plt.show()``````
## How To Save Matplotlib Image To A File
So far we edited the image file and added our own text on top of it. We were also able to display the final result to the end user screen in a separate window.
But what if we wanted to save this image? Can we write the program to save this edited image automatically to a new file? How to achieve this? If these are some of the questions you have running your mind then fret not. Because Matplotlib library comes to our rescue, once again!
To save the image onto a new file, Matplotlib provides us with yet another function called savefig().
### Using Matplotlib’s Savefig Function To Save An Image
Matplotlib provides us with the function savefig() to save our image in a new file. It takes in many parameters which can be seen in this official documentation link. However, for our case, we simply want to save the image to a simple jpg file. So we need not have to use all the parameters defined in that link. We can simply use the following code to save it as a JPG file:
``plt.savefig('/home/amar/Pictures/butterfly_captioned.jpg')``
From the above code, we can see that we used plt.savefig() in its simplest form. We just added a path to the new file and saved it as a jpg file. Yes, the format of the file in which we want it to be saved is given by simply using the appropriate file extension. Matplotlib is intelligent enough to understand this and save it accordingly. Hence, if you wanted to save it as a PNG file, you just need to change the extension to .png. Its that really simple!
So with this, we can finally get our captioned image file as shown below:
So this was a brief tutorial on how to use Python to add text over an image programatically using Matplotlib library. Hope this was easy for you to follow up and understand. I will continue to write more articles on Python and image/video editing libraries in the future including Matplotlib. If you have any more questions or queries regarding this, do let me know in the comment section below. Until next time, ciao!
Categories
## What Are Python Reserved Keywords?
Python reserved keywords are those words in Python programming language that have a specific meaning in a program. These keywords have specific actionable functionalities defined to it in Python. We are not allowed to re-use these reserved keywords. It is also not possible to override the keywords in Python.
## Why Do We Have Python Reserved Keywords?
A programming language is defined by a set of keywords that have specific functionalities attach to it. Python programming language is no different from this. There are a set of keywords defined in Python language that performs specific tasks within the program where they are used.
For example, print is a keyword in Python which instructs the Python interpreter (i.e. the Python environment where Python programs run) to print a string to the output terminal. So a Python program line like:
``print('Hello, World!')``
will print the string:
``Hello, World!``
to the computer output screen that its user can see. We as a programmer are never allowed to use same keyword “print” for any other purposes like variable name or function name. Thus, we say that it is a Python reserved keyword.
Similarly, the keyword input is used to receive input from the user of a Python program. So a line in the program like:
``user_name = input('Enter your name')``
will display the string:
``Enter your name``
on the user screen and wait until the user enters his name. Once he enters the name and hits the “Enter” key, the name gets stored in the variable “user_name“.
So as you can see here, each of these reserved keywords such as print, input etc. each have a very specific functionality attached to it in Python language. We cannot use these same keywords as a variable name or function names. Trying to do so will result in the interpreter throwing error at us!
So now that we understand about reserved keywords in Python, what can we do about them?
For one, we need to know about all the Python reserved keywords to avoid using them in other ways in our program. But in addition to this, knowing about these reserved keywords and their intended functionalities will also help us write useful programs.
## Using Reserved Keywords In Python Programs
Python programs are nothing but a bunch of reserved keywords used upon a set of variables to perform certain operations. So we use these set of keywords to write our programs. For example, if we take a look at the below program:
``````user_name = input('Enter your name')
print('Hello, ' + user_name + '!')``````
This program simply prompts for an user to enter his name. When he does so, it will just wish Hello to him by addressing his name. So when I run this program, the output I get is something akin to this:
``````'Enter your name'
> Amar
> Hello, Amar!``````
## Conclusion
So in short, we can say that reserved keywords are a set of words in Python that have pre-defined meaning and functionalities associated with them. We make use of these keywords to write our program and we are not allowed to re-use the same words in our variables or function names. In other words, we are not allowed to alter their pre-defined meaning.
Categories
## How To Extract Data From A Website Using Python
In this article, we are going to learn how to extract data from a website using Python. The term used for extracting data from a website is called “Web scraping” or “Data scraping”. We can write programs using languages such as Python to perform web scraping automatically.
In order to understand how to write a web scraper using Python, we first need to understand the basic structure of a website. We have already written an article about it here on our website. Take a quick look at it once before proceeding here to get a sense of it.
The way to scrape a webpage is to find specific HTML elements and extract its contents. So, to write a website scraper, you need to have good understanding of HTML elements and its syntax.
Assuming you have good understanding on these per-requisites, we will now proceed to learn how to extract data from website using Python.
## How To Fetch A Web Page Using Python
The first step in writing a web scraper using Python is to fetch the web page from web server to our local computer. One can achieve this by making use of a readily available Python package called urllib.
We can install the Python package urllib using Python package manager pip. We just need to issue the following command to install urllib on our computer:
``pip install urllib``
Once we have urllib Python package installed, we can start using it to fetch the web page to scrape its data.
For the sake of this tutorial, we are going to extract data from a web page from Wikipedia on comet found here:
https://en.wikipedia.org/wiki/Comet
This wikipedia article contains a variety of HTML elements such as texts, images, tables, headings etc. We can extract each of these elements separately using Python.
### How To Fetch A Web Page Using Urllib Python package.
Let us now fetch this web page using Python library urllib by issuing the following command:
``````import urllib.request
content = urllib.request.urlopen('https://en.wikipedia.org/wiki/Comet')
The first line:
``import urllib.request``
will import the urllib package’s request function into our Python program. We will make use of this request function send an HTML GET request to Wikipedia server to render us the webpage. The URL of this web page is passed as the parameter to this request.
``content = urllib.request.urlopen('https://en.wikipedia.org/wiki/Comet')``
As a result of this, the wikipedia server will respond back with the HTML content of this web page. It is this content that is stored in the Python program’s “content” variable.
The content variable will hold all the HTML content sent back by the Wikipedia server. This also includes certain HTML meta tags that are used as directives to web browser such as <meta> tags. However, as a web scraper we are mostly interested only in human readable content and not so much on meta content. Hence, we need extract only non meta HTML content from the “content” variable. We achieve this in the next line of the program by calling the read() function of urllib package.
``read_content = content.read()``
The above line of Python code will give us only those HTML elements which contain human readable contents.
At this point in our program we have extracted all the relevant HTML elements that we would be interested in. It is now time to extract individual data elements of the web page.
## How To Extract Data From Individual HTML Elements Of The Web Page
In order to extract individual HTML elements from our read_content variable, we need to make use of another Python library called Beautifulsoup. Beautifulsoup is a Python package that can understand HTML syntax and elements. Using this library, we will be able to extract out the exact HTML element we are interested in.
We can install Python Beautifulsoup package into our local development system by issuing the command:
``pip install bs4``
Once Beautifulsoup Python package is installed, we can start using it to extract HTML elements from our web content. Hope you remember that we had earlier stored our web content in the Python variable “read_content“. We are now going to pass this variable along with the flag ‘html.parser’ to Beautifulsoup to extract html elements as shown below:
``````from bs4 import BeautifulSoup
From this point on wards, our “soup” Python variable holds all the HTML elements of the webpage. So we can start accessing each of these HTML elements by using the find and find_all built-in functions.
### How To Extract All The Paragraphs Of A Web Page
For example, if we want to extract the first paragraph of the wikipedia comet article, we can do so using the code:
``pAll = soup.find_all('p')``
Above code will extract all the paragraphs present in the article and assign it to the variable pAll. Now pAll contains a list of all paragraphs, so each individual paragraphs can be accessed through indexing. So in order to access the first paragraph, we issue the command:
``pAll[0].text``
The output we obtain is:
``\n``
So the first paragraph only contained a new line. What if we try the next index?
``````pAll[1].text
'\n'``````
We again get a newline! Now what about the third index?
``````pAll[2].text
"A comet is an icy, small Solar System body that..."``````
And now we get the text of the first paragraph of the article! If we continue further with indexing, we can see that we continue to get access to every other HTML <p> element of the article. In a similar way, we can extract other HTML elements too as shown in the next section.
### How To Extract All The H2 Elements Of A Web Page
Extracting H2 elements of a web page can also be achieved in a similar way as how we did for the paragraphs earlier. By simply issuing the following command:
``h2All = soup.find_all('h2')``
we can filter and store all H2 elements into our h2All variable.
So with this we can now access each of the h2 element by indexing the h2All variable:
``````>>> h2All[0].text
'Contents'
>>> h2All[2].text
'Physical characteristics'``````
## Conclusion
So there you have it. This is how we extract data from website using Python. By making use of the two important libraries – urllib and Beautifulsoup.
We first pull the web page content from the web server using urllib and then we use Beautifulsoup over the content. Beautifulsoup will then provides us with many useful functions (find_all, text etc) to extract individual HTML elements of the web page. By making use of these functions, we can address individual elements of the web page.
So far we have seen how we could extract paragraphs and h2 elements from our web page. But we do not stop there. We can extract any type of HTML elements using similar approach – be it images, links, tables etc. If you want to verify this, checkout this other article where we have taken similar approach to extract table elements from another wikipedia article.
How to scrape HTML tables using Python
Categories
## Python Program To Add Two Numbers
In this article, we will look at how to write a Python program to add two numbers. This is a very simple Python program that is suitable for even a beginner in Python programming to work on it for getting hands on practice with Python.
In order to add two numbers in Python program, we need to first break down the problem into the following steps:
## Breakdown of the problem of Adding two numbers
1. Receive the first input number from the user and store it in a Python variable.
2. Receive the second input number from the user and store it in another Python variable.
3. Add the two numbers by adding the two Python variables using a Python statement (Learn about Python statement here)
4. Store the final result in another Python variable called “result”
5. Print the value of the “result” variable.
In the above breakdown of the problem, you will notice that the Python program we will be written in such a way that the two numbers that need to be added will not be hard coded directly into the program itself but instead is written in a generic way such that we prompt the user to enter these two input values every time the Python program to add two numbers is run. This type of programming approach is often called general programming as the program is generic enough to receive any two different values each time it is run.
## Pseudo-code To Add Two Numbers Using Python Programming Language
``````num1 = Receive First Input Number From The User
num1 = Receive Second Input Number From The User
result = (num1) Added to (num2)
Print the (result) on the screen``````
We can see from the above psuedo code that this is a simple program that receives two numbers from the user, adds them and print their results back on the screen. The above pseudo code gives us a nice little framework on how to write our program. The same pseudo-code can now be used to write a program to add two numbers using any programming language and not just Python!
Now that we have our problem broken down and pseudo code written, it is time for us to replace the pseudo code with actual Python programming code instructions.
## Python Program To Add Two Numbers And Print Its Result
Fire up your Python IDLE interpreter and start typing in the code along with me for you to be able to understand this program better:
``````Python 3.5.2 (default, Oct 8 2019, 13:06:37)
[GCC 5.4.0 20160609] on linux
>>>``````
Once in our python interpreter, let us start typing in our Python program commands. The first thing we need to do according to the pseudo code written above is the receive the first input number from our user. In Python program, the instruction code to be used to receive a value from its user is by calling the input() function. Input function will accept a string as its parameter that will be displayed to the program user when the program is run. So, with this knowledge, our first line of the Python program to add two numbers will be:
``````>>> num1 = input('Enter the first number\n')
Enter the first number
10
>>> ``````
As you can see from the above code block, we have used the input() function to prompt our program user with the string “Enter the first number“. We are also saving the value entered by the user to a Python variable called num1.
The input function will then prompt with the above string and wait until the user enters a number. In the above code snippet, I had entered a value of 10 which is now stored in the variable num1.
In a similar way, we will write the next line of code which will prompt the user to enter a second input number that is to be added to the first number. This is achieved with the following piece of code:
``````>>> num2 = input('Enter the second number\n')
Enter the second number
20
>>> ``````
Again over here, I have entered my second number as 20 when prompted.
Now that we have the two numbers in our Python variables num1 and num2, it is time add them to get the final result that we are going to store in our third Python variable called result. This is achieved using the following Python statement:
``````>>> result = int(num1) + int(num2)
>>>``````
So from the above python code, we have now added the numbers num1 and num2 using the Python arithmetic operator “+” and the typecast operator called “int()“. We had to use the typecast operator int because by default all inputs from the user into a python program will be interpreted and stored as a string. We can check this by issuing the following command in the interpreter:
``````>>> print(num1)
10
>>> type(num1)
<class 'str'>
>>> ``````
So, by calling the typecast operator int, we are converting this string value to an integer value.
``````>>> type(int(num1))
<class 'int'>
>>> ``````
finally stored the end result in a new variable called result. Since we have not issued a Python command to print the value of result, nothing gets printed on the IDLE prompt yet. So, the final step is exactly that. To print the value of the result variable onto the screen. This is achieved using another Python function called the print function.
``````>>> print (result)
30
>>>``````
Here is the full program that we can store in a file called add2num.py and run it using the command python3 add2num.py everytime we want to add any two numbers!
``````num1 = input('Enter the first number\n')
num2 = input('Enter the second number\n')
result = int(num1) + int(num2)
print (result)``````
This concludes our Python program to add two numbers.
### Additional Side Notes On Python Programming Language:
Python is an interpreted programming language that gets interpreted and executed on the fly and hence the program written in Python do not need to be compiled like in the case of a C or Java programming language.
Categories
## Introduction
If you are just getting started with Raspberry Pi, connecting a simple LED to one of the GPIO pins of a Raspberry Pi and controlling it using software program that you write will give you a very good grasp of how a computer hardware and its program works internally. You will realize how you can control various aspects of a computer hardware using software, how a computer works at the bit level, how to write Python programs to control hardware and more.
In summary, working on getting an led connected to a GPIO pin of your Raspberry Pi will help you in understanding the fundamentals of a computer architecture and computer science in general.
## What You Will Learn From This Project?
Connecting an LED to the GPIO pins of a Raspberry Pi to control it is a simple Beginner Raspberry Pi Project that lets you learn more about:
• Raspberry Pi hardware internals
• General Purpose Input/Output (GPIO) pins of a Raspberry Pi
• Raspberry Pi Register Set
• Ohm’s Law
• Python Programming
• Python Library – Raspberry Pi GPIO library
• The working of an Light Emitting Diode (LED)
## What Hardware Is Required To Set Up A Blinking LED Project?
This a very simple, beginner friendly Raspberry Pi project that can be set up by anyone with minimal hardware or software knowledge. The hardware components required to set up this blinking LED project is also quite minimal. You need the following hardware components available with you to get it going:
• Raspberry Pi Module
• Keyboard
• Monitor
• Raspberry Pi Power Supply
• SD Card with working Raspbian OS
• Jumper wires for rigging up the circuit
• LED
• Resistor (1K Ohm)
• Multimeter
## Theory Behind How The Raspberry Pi Blinking LED Project Work
When you look at the Raspberry Pi board, you will see a bunch of pins protruding out. Among these, there is a row of 40 pins located on one side of the board as shown in the image below.
If you look closely enough in the above image, you will notice the label “GPIO” written right under it. These pins are called the GPIO pins or General Purpose Input Output pins. What the name GPIO implies is that these pins do not have any fixed functionality within the board and hence can be used for general purposes. It means that we can connect our LED into one of these pins and can turn it ON or OFF using these pins. But how?
### How to control the Raspberry Pi GPIO pins programmatically?
Raspberry Pi 3 board runs on Broadcom’s ARM CPU chipset BCM2837. Among many other things, this processor chipset has a built in GPIO controller aka General Purpose Input Output controller. The 40 GPIO pins header shown in figure 1 is connected to 40 controllable pins of the GPIO controller. Now, we can control each of these pins individually by programming the appropriate registers inside this GPIO controller.
To understand how to program each of these pins using GPIO controller, we need to look into the Technical Reference Manual or datasheet of the Broadcom ARM chipset BCM2837.
In the BCM2837 SOC (System On Chip aka CPU) datasheet linked above, if we jump into page 89 we come across a dedicated chapter talking about General Purpose Input Output (GPIO). If we go through this chapter, we can learn about all the GPIO registers available and figure out the GPIO registers we need to program to turn ON or OFF the LED we are going to connect to the Raspberry Pi 3 GPIO pins.
As the name implies, GPIO pins can be configured as either an Input pin or an Output pin. When we configure a GPIO pin as an input pin, we are sending data bit (either 0 or 1) into the Raspberry Pi BCM2837 SOC i.e. data signal is sent from outside the board to inside the board (hence the name input). On the other hand, if we configure the Raspberry Pi GPIO pin as an output pin, the board will send the data bit signal (either 0 or 1) from inside the board to the outer world where any device connected to it will receive this signal.
So, if we want to control an LED that is connected to one of the Raspberry Pi’s GPIO pin, we need to configure that pin as a GPIO OUT pin (aka output pin) so that we can send an electrical signal from the Raspberry Pi board to the external LED connected to this pin.
The configuration of a GPIO pin to be an INPUT or OUTPUT pin is controlled by programming the GPIO Controller Register called GPIO Function Select Register (GPFSELn) where n is the pin number.
So for example, if we choose to use the GPIO8 pin to control the LED, i.e. we connect our LED to GPIO 8 pin, we need to program the GPFSEL register for the GPIO 8 pin and configure it as an Output pin. When we check the datasheet at page 91 and 92, we notice that GPIO pin is configured by setting the bits 26 to 24 in the GPFSEL register (that is field name FSEL8). And from the datasheet, we also find that to set the pin as an output pin, we need to set its value as 001 i.e. bit 26 is set to 0, bit 25 is set to 0 and bit 24 is set to 1.
So, if we can somehow set these values in the GPFSEL register using a programming language such as Python, we will be able to start controlling the LED connect to this pin!
If this is all overwhelming to you, do not worry. We will not have to scratch our head a lot for now as we can simply make use of Raspberry Pi’s GPIO Python library that helps us in making most of this work for us. But I just wanted to explain to you as to what this GPIO Python library is doing under the hood.
## How To Connect An LED To Raspberry Pi GPIO?
### Designing The Circuit
In order to connect an LED to GPIO pin 8 of Raspberry Pi, we need to first design and understand how the circuit is going to work.
#### Can we connect an LED directly to a Raspberry Pi GPIO pin without a resistor?
The answer is No. Raspberry Pi provides 3.3 Volts of power on its GPIO output pin according to Raspberry Pi datasheet specification. However, if we take a look at a standard LED, we notice that it normally operates at a much lower voltage. If we look at an LED specification, we notice that a typical LED usually operates at just 1.7 Volts and draws 20 mA. So, if we need to connect this LED to the GPIO pin of our Raspberry Pi, we need to bring down the voltage delivered by the pin to our LED to operate at or under 1.7V. How to do that? We connect a resistor in series with our LED so that the 3.3 Volts GPIO output of Raspberry Pi gets split between the resistor and our LED. By choosing a right value of the resistor such that it consumes 1.6 Voltage, we can ensure that LED finally gets only 1.7 Volts.
#### Calculating the resistor value to connect with LED and Raspberry Pi GPIO
In order to calculate the value of resistor that we should be using, we make use of the Ohm’s Law.
Ohm’s Law is defined using the equation:
V = I/R where V is the voltage, I is the current and R is the resistor value.
So, if we want to have V=1.6 Volts consumed by our resistor so that the current coming from GPIO pin is at I=20 mA, we need to connect a resistor whose value is:
1.6 = (20 mA)/R
or R = 80 Ohms (or approx 100 Ohms)
So, we choose a resistor of value 100 Ohms connected in series with our resistor to ensure that we only get 1.7 Volts and 20 mA of current, the optimum operating values as required by our LED.
A 100 Ohm resistor is identified by the color bands: Brown, Black & Brown.
Hook up the led through 100 Ohm resistor to GPIO 8 pin of Raspberry Pi as shown in the figure below:
Note that when you are hooking up the LED, the terminal pin that is longer is positive. Once you have connected as shown in the figure, it is now time to program the Raspberry Pi GPIO controller to start controlling the LED to turn ON or OFF.
We will be using Python to program our Raspberry Pi GPIO controller. Now, the simplest way to program this is by making use of the Python GPIO library.
To install the Python Raspberry Pi GPIO module, open up your linux terminal and type the following command.
``sudo apt-get install python-rpi.gpio python3-rpi.gpio``
Now the above command will install the required Python GPIO library module onto our Linux development machine. Once successfully installed, It is now time to start programming the Raspberry Pi GPIO controller.
We will be toggling our GPIO pins at 1 second intervals such that our LED will turn ON and OFF forever until the Python program we write will be terminated i.e., we will be running the code to perform infinite loop of toggling the GPIO 8 pin ON and OFF.
Create a new file on your computer by typing the following command in the terminal:
``touch blinky.py``
This should create our new program file called blinky.py
Open up this file using nano editor by typing the following command in the terminal:
``nano blinky.py``
Now that the file is opened, it is time to start writing our program to control the GPIO Pin 8 using Python GPIO library module.
First thing first, we will import the Python GPIO library module using command:
``import RPi.GPIO as GPIO``
Next, we will import python time library to perform 1 sec sleep operation between each GPIO toggle
``from time import sleep``
Next, we need to configure our GPIO library to use our GPIO physical pin numbering as seen on the Raspberry Pi board physically:
``GPIO.setmode(GPIO.BOARD)``
This ensures that when we say GPIO pin 8 in the program, it actually maps to the GPIO Pin 8 seen on the Raspberry Pi board.
Next, configure GPIO pin 8 to be a GPIO Out pin and set its initial output value to be low:
``GPIO.setup(8, GPIO.OUT, initial=GPIO.LOW)``
Finally we will start an infinite loop in Python such that we turn ON the GPIO 8 (by setting it HIGH) or turn it OFF (by setting it LOW) after every 1 second delay. This is achieved using the program below:
``````while True: # Infinite loop
GPIO.output(8, GPIO.HIGH) # Turn GPIO 8 pin on
sleep(1) # Delay for 1 second
GPIO.output(8, GPIO.LOW) # Turn GPIO 8 pin off
sleep(1) # Delay for 1 second``````
That’s it, this should be all the program that we need to type in our blinky.py file and run it using the command:
``python3 blinky.py``
This should start turning your LED ON and OFF every second!
Here is the full code for your reference:
``````import RPi.GPIO as GPIO
from time import sleep
GPIO.setmode(GPIO.BOARD)
GPIO.setup(8, GPIO.OUT, initial=GPIO.LOW)
while True: # Infinite loop
GPIO.output(8, GPIO.HIGH) # Turn GPIO 8 pin on
sleep(1) # Delay for 1 second
GPIO.output(8, GPIO.LOW) # Turn GPIO 8 pin off
sleep(1) # Delay for 1 second``````
This should conclude our tutorial on how to get a simple LED connected to a General Purpose Input/Output (GPIO) pin turning ON and OFF using a Python program that makes use of Python Raspberry Pi GPIO library. There can be many variants to this such as using other GPIO pins, connecting more than one LEDs to multiple GPIO pins and controlling them all in different ways to display interesting patterns on the LEDs. If we are even more curious, we can also figure out a way to control the BUILT-IN LEDs that are already present on our Raspberry Pi boards to bypass their current usage and be used for by own programs for our purposes.
We will dwell into these and many other interesting ways to make use of our Raspberry Pis to understand and learn more about the computer hardware, its architecture and much more in our future articles.
Categories
## How To Write Code Like A Professional?
Everyday, we programmers write hundreds of lines of programming code. But, have you ever stopped to notice if your program is good enough? How do you make sure your programming skill set is up to the mark? How to make sure you are Programming Like A Pro? In this article, we consider these questions and try to answer them each in greater details.
### How To Code Like A Pro?
There are many characteristics of a program written by an experienced professional that sets his code apart from a code newbie. A professional programmer would make use of his past learning experience and implement his code that follows a certain set of guidelines and best practices he has learnt over the time since he began programming.
While each of these set of guidelines and best practices may differ from one Professional Programmer to another Pro, there are many common guidelines which more or less all Professional Programmers follow that can make their code set apart from the rest. Here is a list of some of such common guidelines used by almost all Professional Programmers. If you can practice and incorporate these practices and techniques into your daily programming routine, even your program code will start to look more professional.
### Best Practices & Guidelines To Write Code Like A Professional
• Write code that makes use of the processor efficiently – In other words, write code that runs fast on the processor using lower number of clock cycles. How to achieve this? While there are many things involved in learning to write a highly CPU efficient programs which will be explained in the future articles, some of the most common practices is to use loops sparingly, use efficient looping systems, write programs using instructions that require less clock cycles to accomplish the task at hand etc. You need to have good understanding of the underlying computer system architecture to get a better hold or control on these aspects, so learn more about the computer architecture.
• Write code that makes use of memory efficiently – Use the memory of a computer system efficiently and sparingly. It usually takes certain amount of computer clock cycles to read and write data to or from the memory. So if you want your code to speed up, it is better to write code using instructions that uses lesser amount of memory. If your code allocates too many variables, it can start gobbling up a lot of RAM space in your user’s computer, there by slowing down the computer all together, thus giving a had user experience. | 12,071 | 55,588 | {"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} | 3.125 | 3 | CC-MAIN-2023-23 | latest | en | 0.903314 |
https://trustconverter.com/en/weight-conversion/hectograms/hectograms-to-grains.html?value=4 | 1,721,315,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00308.warc.gz | 520,483,869 | 9,748 | # Hectograms to Grains Conversion
Hectogram to grain conversion allow you make a conversion between hectogram and grain easily. You can find the tool in the following.
### Weight Conversion
to
input
= 1,543.23583529
= 15.4324 × 102
= 15.4324E2
= 15.4324e2
= 3,858.08958824
= 38.5809 × 102
= 38.5809E2
= 38.5809e2
= 6,172.94334118
= 61.7294 × 102
= 61.7294E2
= 61.7294e2
= 8,487.79709412
= 84.878 × 102
= 84.878E2
= 84.878e2
= 10,802.65084706
= 108.027 × 102
= 108.027E2
= 108.027e2
= 13,117.50460000
= 131.175 × 102
= 131.175E2
= 131.175e2
= 15,432.35835294
= 154.324 × 102
= 154.324E2
= 154.324e2
### Quick Look: hectograms to grains
hectogram 1 hg 2 hg 3 hg 4 hg 5 hg 6 hg 7 hg 8 hg 9 hg 10 hg 11 hg 12 hg 13 hg 14 hg 15 hg 16 hg 17 hg 18 hg 19 hg 20 hg 21 hg 22 hg 23 hg 24 hg 25 hg 26 hg 27 hg 28 hg 29 hg 30 hg 31 hg 32 hg 33 hg 34 hg 35 hg 36 hg 37 hg 38 hg 39 hg 40 hg 41 hg 42 hg 43 hg 44 hg 45 hg 46 hg 47 hg 48 hg 49 hg 50 hg 51 hg 52 hg 53 hg 54 hg 55 hg 56 hg 57 hg 58 hg 59 hg 60 hg 61 hg 62 hg 63 hg 64 hg 65 hg 66 hg 67 hg 68 hg 69 hg 70 hg 71 hg 72 hg 73 hg 74 hg 75 hg 76 hg 77 hg 78 hg 79 hg 80 hg 81 hg 82 hg 83 hg 84 hg 85 hg 86 hg 87 hg 88 hg 89 hg 90 hg 91 hg 92 hg 93 hg 94 hg 95 hg 96 hg 97 hg 98 hg 99 hg 100 hg grain 1,543.2358353 gr 3,086.4716706 gr 4,629.7075059 gr 6,172.9433412 gr 7,716.1791765 gr 9,259.4150118 gr 10,802.6508471 gr 12,345.8866824 gr 13,889.1225176 gr 15,432.3583529 gr 16,975.5941882 gr 18,518.8300235 gr 20,062.0658588 gr 21,605.3016941 gr 23,148.5375294 gr 24,691.7733647 gr 26,235.0092 gr 27,778.2450353 gr 29,321.4808706 gr 30,864.7167059 gr 32,407.9525412 gr 33,951.1883765 gr 35,494.4242118 gr 37,037.6600471 gr 38,580.8958824 gr 40,124.1317176 gr 41,667.3675529 gr 43,210.6033882 gr 44,753.8392235 gr 46,297.0750588 gr 47,840.3108941 gr 49,383.5467294 gr 50,926.7825647 gr 52,470.0184000 gr 54,013.2542353 gr 55,556.4900706 gr 57,099.7259059 gr 58,642.9617412 gr 60,186.1975765 gr 61,729.4334118 gr 63,272.6692471 gr 64,815.9050824 gr 66,359.1409176 gr 67,902.3767529 gr 69,445.6125882 gr 70,988.8484235 gr 72,532.0842588 gr 74,075.3200941 gr 75,618.5559294 gr 77,161.7917647 gr 78,705.0276000 gr 80,248.2634353 gr 81,791.4992706 gr 83,334.7351059 gr 84,877.9709412 gr 86,421.2067765 gr 87,964.4426118 gr 89,507.6784471 gr 91,050.9142824 gr 92,594.1501176 gr 94,137.3859529 gr 95,680.6217882 gr 97,223.8576235 gr 98,767.0934588 gr 100,310.3292941 gr 101,853.5651294 gr 103,396.8009647 gr 104,940.0368 gr 106,483.2726353 gr 108,026.5084706 gr 109,569.7443059 gr 111,112.9801412 gr 112,656.2159765 gr 114,199.4518118 gr 115,742.6876471 gr 117,285.9234824 gr 118,829.1593177 gr 120,372.3951529 gr 121,915.6309882 gr 123,458.8668235 gr 125,002.1026588 gr 126,545.3384941 gr 128,088.5743294 gr 129,631.8101647 gr 131,175.046 gr 132,718.2818353 gr 134,261.5176706 gr 135,804.7535059 gr 137,347.9893412 gr 138,891.2251765 gr 140,434.4610118 gr 141,977.6968471 gr 143,520.9326824 gr 145,064.1685177 gr 146,607.4043529 gr 148,150.6401882 gr 149,693.8760235 gr 151,237.1118588 gr 152,780.3476941 gr 154,323.5835294 gr
Hectogram or hectogramme is equal to 100 gram (unit of mass), comes from a combination of the metric prefix hecto (h) with the gram (g). Plural name is hectograms.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
hectogramhg
≡ 100 g
≡ 0.1 kg
Metric system SI
#### conversion table
hectogramsgrainshectogramsgrains
1= 1543.235835294111= 16975.594188236
2.5= 3858.089588235412.5= 19290.447941177
4= 6172.943341176614= 21605.301694118
5.5= 8487.797094117815.5= 23920.155447059
7= 10802.65084705917= 26235.0092
8.5= 13117.504618.5= 28549.862952942
10= 15432.35835294120= 30864.716705883
grain is a unit of measurement of mass, and, for the troy grain, equal to exactly 64.79891 milligrams. It is nominally based upon the mass of a single seed of a cereal.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
graingr
≡ 17000 lb av
= 64.79891 mg
Imperial/US
### conversion table
grainshectogramsgrainshectograms
1= 0.000647989111= 0.0071278801
2.5= 0.0016199727512.5= 0.00809986375
4= 0.002591956414= 0.0090718474
5.5= 0.0035639400515.5= 0.01004383105
7= 0.004535923717= 0.0110158147
8.5= 0.0055079073518.5= 0.01198779835
10= 0.00647989120= 0.012959782
hectogramsgrains
1= 1,543.2358353
0.0006480= 1
### Legend
SymbolDefinition
exactly equal
approximately equal to
=equal to
digitsindicates that digits repeat infinitely (e.g. 8.294 369 corresponds to 8.294 369 369 369 369 …) | 2,073 | 4,484 | {"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-2024-30 | latest | en | 0.429945 |
https://differencebetweenz.com/difference-between-rsa-and-dsa/ | 1,718,282,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00040.warc.gz | 184,179,189 | 23,402 | # Difference between RSA and DSA
When it comes to digital signatures, there are a few algorithms that stand out. RSA and DSA are two of the most popular algorithms, but what’s the difference between them? In this blog post, we’ll take a look at the key differences between RSA and DSA so you can understand which algorithm is best for your needs.
Contents
## What is RSA?
• RSA is an algorithm for public-key cryptography that is widely used in electronic commerce. RSA is a TTP (trusted third party) that generates and distributes public and private keys. RSA is named after its inventors, Rivest, Shamir, and Adleman, who created it in 1977. RSA is the most common type of asymmetric key algorithm.
• RSA keys can be used for both encryption and digital signatures. RSA keys are generated by a TTP that picks two large prime numbers at random and multiplies them together to get the modulus n. The RSA algorithm uses the modulus n and a public exponent e to generate the public key, which is made available to everyone.
• The RSA algorithm uses the private key, which consists of the modulus n and a private exponent d, to decrypt messages that have been encrypted with the public key. RSA is used in electronic commerce to protect credit card transactions and to secure e-mail messages.
## What is DSA?
DSA is a digital signature algorithm that is used to provide electronic signatures. It is a type of public-key cryptography that uses a pair of keys, one public and one private, to sign and verify messages. DSA is more secure than other digital signature algorithms, such as RSA, because it is resistant to quantum computer attacks. DSA is used in a variety of systems, including blockchain technology and cryptocurrency.
## Difference between RSA and DSA
• RSA and DSA are two of the most popular algorithms used for digital signatures. RSA is a public-key cryptosystem that is widely used in electronic commerce, while DSA is a standard for digital signatures specified by the U.S. National Security Agency. Both RSA and DSA have similar security properties, but there are some key differences between them.
• RSA is faster and more efficient than DSA, making it better suited for applications such as SSL/TLS where performance is critical. RSA also has the advantage of being a well-studied and well-understood algorithm, while DSA is relatively new and has not been studied as extensively. However, DSA has the benefit of being simpler than RSA, making it easier to implement correctly.
• In addition, DSA is considered to be more resistant to certain types of attacks than RSA. As a result, RSA is typically used for applications where performance is paramount, while DSA is generally seen as a better choice for applications where security is more important than speed.
## Conclusion
While RSA and DSA are both digital signature algorithms, they differ in a few ways. First, the mathematical basis for RSA is factoring in large numbers, while the basis for DSA is elliptic curve cryptography. Second, RSA can be used for both signatures and encryption, while DSA can only be used for signatures. Finally, RSA is more widely used than DSA. If you’re looking to use a digital signature algorithm in your next project, it’s important to understand these differences so you can make an informed decision. | 683 | 3,322 | {"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-2024-26 | latest | en | 0.944116 |
https://petlja.org/biblioteka/r/lekcije/BlockBasedProgMakeCodeEng/project-task---forward-and-backward | 1,628,010,601,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00596.warc.gz | 439,149,765 | 12,757 | # Obeleži sve kategorije koje odgovaraju problemu
### Još detalja - opišite nam problem
Uspešno ste prijavili problem!
Status problema i sve dodatne informacije možete pratiti klikom na link.
Nažalost nismo trenutno u mogućnosti da obradimo vaš zahtev.
Molimo vas da pokušate kasnije.
# Project Task - Forward and Backward¶
In more complex applications, it is necessary to perform an action several times, until some of the conditions for its termination are fulfilled.
Loops provide the ability to repeat the same code sequence until one of the conditions for breaking it is fulfilled.7
The running of the loop can be controlled in a number of ways:
• Unlimited (no limitations),
• Sensor (it runs until the program reacts to one of the sensors),
• Time (it can be time-limited),
• Count (a number of times it repeats can be determined in advance) and
• Logic (it can run until a logical condition has been fulfilled).
We will first explain the loops in a simple example of the infinite movement of a robot forward and backward.
We will solve this task by dividing it into two parts. The first part will consist of two blocks with which the robot will move forward and backward.
Drag the block to the work surface, click on the sign + and choose the option “rotations”. Then, set the number of rotations to 1.
Finally, for the robot to move forward, the value (representing power) has to be positive, we will set it to be 50.
In the second block, we will set the power of the motor to be -50, so the robot can move backward. To allow the robot to move continuously (without stopping), we need to put the block presented above into an infinite loop, we will achieve this by adding the block , which will enable the robot to go forward and backward continuously (an infinite number of times), until the program is stopped by force.
The look of the program:
Connect the EV3 Brick to the computer via USB cable and download the .uf2 file to your computer by clicking the button . By dragging the file onto the EV3, it is ready to start working.
We can also limit the robot’s movement; for example, we want the robot to move forward and backward three times.
To do this, we will use the loop where blocks run a determined number of times. The robot will move forward and backward three times.
To set the code sequence to stop after a certain number of repetitions, you need to drag the block , where we will define how many times an action should be repeated, into the block . In our case, the value will be 3.
The look of the program:
Connect the EV3 Brick to the computer via USB cable and download the .uf2 file to your computer by clicking the button . By dragging the file onto the EV3, it is ready to start working.
The third way we can repeat a certain action is by using the block that will repeat it until a specified condition has been fulfilled. This block should be used when we don’t know how many times we need to repeat some part of the code, so we want it to keep repeating until a specific condition has been fulfilled.
To demonstrate conditional repetition, we will create a program which will allow the robot to move around a box in the shape of a square until it encounters an obstacle (touches the box).
The code looks like this:
Connect the EV3 Brick to the computer via USB cable and download the .uf2 file to your computer by clicking the button . By dragging the file onto the EV3, it is ready to start working.
This task can be solved by using functions.
Some complex problems can be solved more easily if they are divided into smaller units that can be solved independently. In other programming languages, these units are called subprograms: functions and procedures. We know that we can simplify the code by using repetition commands. However, this is often not enough.
Whenever we want the robot to repeat some activity within a program, or in another program, we can use procedures, or more specifically, the block .
Since we have been using the forward and backward motion of the robot in our previous examples, we will try to use these two blocks to create our own function forward and backward. How is this done?
The first step is to create the Function with activities which will be repeated. In our case, this will be the robot moving forward and backward repeatedly.
We create a Function by, opening the category Function (1), clicking on the button Make a Function (2) and entering the name of the function we would like to create (3). We have finished creating a function when we click on the OK button (4).
Add the two movement blocks, the first block will enable the robot to move one rotation forward, and the second block will enable it to move one rotation backward.
Look of the function forward and backward:
In order for a function to be used in a program, it is necessary to “call” it using the block .
The look of the code for moving forward and backward by using the function:
Connect the EV3 Brick to the computer via USB cable and download the .uf2 file to your computer by clicking the button . By dragging the file onto the EV3, it is ready to start working. | 1,115 | 5,145 | {"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} | 2.921875 | 3 | CC-MAIN-2021-31 | latest | en | 0.857017 |
http://mathhelpforum.com/algebra/64050-factorial-notation-help.html | 1,481,409,075,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543577.51/warc/CC-MAIN-20161202170903-00017-ip-10-31-129-80.ec2.internal.warc.gz | 169,533,742 | 11,225 | 1. ## factorial notation help
(n+2)/n!=56
I'm new to this and having trouble trouble breaking this down, could someone please give me a hint to get this started?
2. you are saying to divide $n+2$ by $n!$ and get 56.
This can never happen though...
Are you sure you have the right problem?
3. Sorry,
(n+2)!/n!=56
4. Note that $(n+2)! = (n+2)(n+1)(n)(n-1)(n-2) \dots = (n+2)(n+1)n!$
So $\frac{(n+2)!}{n!} = \frac{(n+2)(n+1)n!}{n!}$
Do you see where to go now?
5. Given: (n + 2)!/n! = 56
(n + 2)(n + 2 - 1)(n + 2 - 2)!/n! = 56
(n + 2)(n + 1) = 56
n + 2 = 56 or n + 1 = 56
Which gives n = 54, 55
May be it helps You
6. You solved that wrong...
$(n+2)(n+1) = n^2+3n+2=56$
So $n^2+3n-54 = 0$
$(n+9)(n-6) = 0$
$n=-6,9$
7. thanks for the help chip, when down to
(n+9)(n-6) how do you know which to make positive and negative?
eg how do you know its not (n+6)(n-9) without substiting into the initial equation?
8. its simple factoring
given that $(n-a)(n-b) = n^2+3n-54$
doing foil... $n^2+(-a-b)n+ab = n^2+3n-54$
So $a+b = -3$ and $ab = -54$
By inspection it can be seen $a=-9 , \; b=6$
9. Originally Posted by chiph588@
You solved that wrong...
$(n+2)(n+1) = n^2+3n+2=56$
So $n^2+3n-54 = 0$
$(n+9)(n-6) = 0$
$n=-6,9$
Small correction: n = -9, 6.
But only n = 6 is a valid solution to the original equation (to the OP: why?). | 561 | 1,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2016-50 | longest | en | 0.844478 |
wwwhome.ewi.utwente.nl | 1,718,982,383,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00669.warc.gz | 977,846,809 | 26,061 | ## Telephone traffic computation ("Telefoonverkeersrekening")
by ir. J. Kruithof
(originally published in "De Ingenieur", 19 Febr. 1937; work-in-progress translation by Pieter-Tjerk de Boer, p.t.deboer@utwente.nl)
Translator's note: the original article is written not only in a by now outdated spelling of Dutch, but also in a somewhat archaic, sometimes a bit baroque, style, with many long sentences. My translation is quite literal and a bit rough and quick; the original style intentionally shows through, although in some places I've made some effort to rephrase a sentence to make it clearer.
### 1. Introduction
In the extensive literature about the computation of automatic telephone exchanges, there is no attention, except for a few exceptional cases, for the application of Probability Theory to traffic problems, while the quantitative traffic problems themselves are almost completely ignored, even though their solution is of fundamental importance.
[Translator's note: in the original text, the Dutch word for "probability theory" ("waarschijnlijkheidsrekening") is written with a capital initial letter, even though there is no linguistical need for that; so it must have been a conscious decision by the author, which I've preserved in the translation.]
For the calculation of an automatic exchange, in the first place knowledge of the traffic volumes is required, before one can start to determine the required sizes of the different traffic paths based on these traffic volumes. In practice however, one usually has too little knowledge of the expected traffic, partially because there often is too little statistical data available about the existing traffic, and partially because, due to lack of experience or perhaps knowledge, one does not manage to draw from this data correct conclusions about the traffic densitities to be expected for the new exchanges.
Thus, the regrettable phenomenon occurs that one uses Probability Theory to dimension, rather precisely, the new traffic paths, but that the traffic volume data on which these calculations are based, are very inaccurate, sometimes even utterly wrong. Due to this, as is immediately obvious, the accuracy of the whole computation becomes problematic, because of the well known rule that the strength of a chain is determined by the weakest link. But furthermore one thus largely robs the usefulness from the meritorious work that has been done on the application of Probability Theory to telephony.
Furthermore, among telephone technicians one finds opinions about the use and the allowed manipulations on the given or computed traffic intensities which demonstrate a lack of correct insight in the meaning and nature of the traffic intensities.
The intention of this article is to once again draw attention to the traffic calculation itself, completely separate from the dimensioning of the traffic paths. For this purpose, a general overview of the nature of telephone traffic is given, the units of traffic intensity used and their meaning, with a more detailed discussion of the way in which statistical data should be used for determining expected traffic intensities and the computation of the traffic between the exchanges.
The ways in which, based on computed traffic intensities, the sizes of the traffic paths, waiting times, losses, etc. are determined, will thus not be considered.
### 2. Telephone traffic
In a modern telephone exchange any subscriber can, at any time of day or night, take the phone off the hook to set up a connection. The consequence is that telephone traffic has the typical property of being very irregular and showing large fluctuations.
If one has extensive and regular traffic statistics over an entire year, it turns out that in the end these fluctuations do have some regularity, despite the complete randomness of the starting time of each call.
If one makes a graph of the traffic amounts per hour, one will see fluctuations during the day (morning and afternoon), fluctuations with a duration of a week and long-stretched fluctuations over an entire year, with every now and then a pronounced peak.
At night the traffic drops to a very low value; also there is a dip at lunch and dinner time. The traffic is highest during the hours in which offices are open and people are at work.
Some days of the week can have more traffic than the other days. E.g., one gets the impression that on Saturday morning people already try to make up for the coming free afternoon.
Towns which have a peak season will during that season have a normal traffic which can be much higher than the traffic outside the season.
Also almost every exchange has some days during the year which show heavy traffic, such as the days before St. Nicolaas, Christmas and New Year.
All of this may be considered as generally known, but is mentioned here only because of some of the coming sections which discuss on which traffic values the dimensioning of an exchange must be based.
### 3. Units of telephone traffic
Two kinds of units of traffic are distinguished: one which simply indicates the number of calls, and a second one which takes into account the duration of the calls.
For this second unit one defines a unit call, and naturally multiple such units are in use. So far, we haven't introduced an international unit of traffic.
In the Scandinavian countries, people usually work with "traffic minutes". This unit is quite attractive and its use is increasing. It doesn't have the arbitrariness that affects other units and its size is such that calculating with decimals is not needed.
In America the unit call is based on an average holding time of 100 seconds. The (limited) advantage of this unit is that it leads to simple calculations, e.g. when multiplying numbers of calls by the average call duration, which is usually expressed in seconds.
Also the use of a traffic unit of 2 minutes is widespread. The choice fell on 2 minutes because town calls very often have approximately this as their average duration. This reason therefore only affects calculations concerning local networks, not inter-local traffic, which has a longer average call duration. For such traffic, sometimes a unit of 3 minutes is used.
Finally we want to mention the call unit of one hour. One could call this a theoretical unit, in contrast to the above practical units.
In the following table the ratios between the respective traffic units are given.
Units 1 min. 100 sec. 2 min. 1 hour 1 min. 1 3/5 0.5 1/60 100 sec. 5/3 1 5/6 1/36 2 min. 2 6/5 1 1/30 1 hour 60 36 30 1
### 4. Traffic density
The number of calls on a traffic path is constantly changing. The number found at a certain instant is called the instantaneous traffic density. In analogy with electrical phenomena, the amount of traffic expressed in unit calls can be compared to the amounts of electricity, and the instantaneous traffic density to the electrical current.
[Translator's note: the above analogy doesn't make sense to me]
The average value, around which the instantaneous traffic density fluctuates, is called the average traffic density or simply the traffic density. The period over which this average value is determined, is always one hour.
### 5. Units of traffic density
For the instantaneous traffic density there is obviously only one unit, namely the number of calls.
For the average traffic intensity there are two kinds of units, corresponding to the two units for telephone traffic, mentioned in Section 3. One unit is one call per hour and serves to indicate the traffic per hour, expressed in the number of calls, without considering the duration of the calls.
The second unit is a unit call per hour and thus one can speak of a number of call minutes per hour, a number of 2-minute calls per hour and a number of call hours per hour.
If the values for the average traffic density relate to the "busy hour", one adds this restricting clause and speaks of a number of "busy-hour calls" (B.H.C., busy hour calls), "busy-hour call minutes" (Sm, speaking minutes), "busy-hour unit calls" (E.B.H.C., equated busy hour calls) and "busy-hour call hours" (T.U., traffic unit; T.C., time calls; Belegungsstunden).
[Translater's note: the parenthetical abbreviations and terms were already in English (or German in case of the last word) in the original text, apparently referring to terms that were already commonly used internationally at the time. The terms in quotes were in Dutch in the original text.]
We do not see much justification for the abbreviations Sm., T.U. etc., firstly because they do not indicate that they refer to the "busy hour" and furthermore because then the simpler indications "minutes" and "hours" would suffice.
The unit "busy-hour call hours" could theoretically be called the unit of average traffic density, since it can be used without further derivation in the formulas of probability theory.
About the correct interpretation of the term "busy hour", more will be said in Section 9.
The goal of traffic calculation is determining the traffic densities of a new exchange to be expected for each of its traffic paths; we next want to devote some more words to the latter.
### 6. Traffic paths
Generally spoken, the traffic problem owes its existence to the fact that the means of communications use common, shared, traffic paths. In case of telephony the subscribers' calls are led through traffic paths that are common to larger or smaller groups of subscribers.
There are paths which are held only during part of the duration of a connection, such as is the case with control circuits and the personnel at the desks which, in the sense of traffic techniques, must be considered as a traffic path.
However, most traffic paths are continuously occupied during the entire duration of the conversation, such as the different groups of switches, connection wires, etc. By their nature they are the most important and therefore need most attention. Like all other traffic paths, they have nodes, where traffic comes together and goes apart again; they can be built for single-directional or bi-directional traffic, whichever is the economics of the operation requires.
### 7. Relation between traffic density and traffic path
The traffic paths are dimensioned based on the expected traffic densities using probability theory. Once the traffic densities are known, dimensioning the groups of traffic lines reduces in most cases to reading values off of curves which have been calculated for that. These curves have been computed based on formulas from Probability Theory and indicate in a simple way the relationship between traffic densities and the required numbers of traffic lines for the desired degree of service.
Several such curves exist, which may differ to some extent, depending on which assumptions and simplifications they have been based on, but which are quite similar in all cases, so that they give results which generally differ by only a few percent. The use of curves based on practical experience, dating back to the time when the theoretical treatment of such problems had not yet been completed, is not useful for the modern engineer.
### 8. Conditions for applying Probability Theory
If we allow the use of formulas from probability theory for telephony, the traffic data and the traffic itself must satisfy certain requirements.
In general, these conditions are not satisfied in telephony, from which it follows that there is no certainty that the outcomes of the calculations are correct. But based on experience it can be safely assumed that the approximation is quite accurate.
The first condition imposed on the traffic by probability theory is that there must not be a dependence among the calls. In this aspect, telephone traffic falls short, since there surely can be a relationship between successive calls of a subscriber. A clearly identifiable relationship is where a call is preceeded by one or more non-successful attempts which found the line of the desired subscriber busy. In general however it is assumed that, since the number of subscribers is relatively large, the effect of this error is small and can be neglected.
The second condition imposed by probability theory is that each call must have complete freedom to fall at any arbitrary time in the period under consideration. But as we have seen in section 2, the traffic density depends on the hour of the day. In the morning the instantaneous traffic density slowly increases to a maximum, and after some fluctuations decreases to a minimum in the evening. The starting times of the calls therefore are not totally free, but depend to some extent on time.
This circumstance demands more care than the previous one; it relates to the way the traffic data is determined which will serve as the basis for the traffic calculation of the exchanges. In cases where it is known that the traffic does not satisfy this second requirement of Probability Theory, we should set ourselves the goal to approximate it as close as possible by judicious use of the available statistical data and by suitable choice.
### 9. The "busy hour" terminology
In section 5, the word "busy hour" was already used, added as a restrictive clause to the units of traffic density. These "Busy hour traffic densities" serve as the basis for the further dimensioning of the exchanges and it is therefore important to define what is meant by this word. As will become clear in the following, a correct choice allows one to practically accommodate the second condition from probability theory as well as possible, even if it cannot be satisfied completely.
The term "busy hour" itself is not sufficiently determined and thus gives rise to random explanations and not allowed manipulations of the traffic quantities. After all, the expression "during the busy hour" does not specify whether one means the busy hour of a specific period, be it a day, week, month or year, or that one e.g. means the average traffic of the daily busy hours.
If we would base ourselves on the single busy hour of the entire year, then we would get as the result of the calculations such large groups of traffic lines, that even during that single busy hour there would be enough lines for the exceptionally high traffic of that moment, with as its consequence that a large part of the automatic material would have a very low efficiency during the rest of the year. This material would have to be bought and maintained, so the economics of the operation would be seriously harmed.
Such an interpretation of the term busy hour is therefore not acceptable. Thus the thought arises to use an average over several hours for the "busy-hour traffic density", just like the traffic density is already the average over one hour. In the choice of these hours we let ourselves be guided by the principle of satisfying as much as possible the second condition from probability theory mentioned in section 8. This condition is that the expectation and, by consequence, the traffic density, must not be a function of time.
To achieve this, out of the 8700 hours of a year, we only select a bit more than 300 hours for the further traffic calculations. Thus, from every day we only take one hour into consideration, and each such hour starts at the same moment of the day. We exclude Sundays and holidays.
Practically one works by adding the 300 values of the quarter from 9 till 9¼ and computing their average, doing the same for the quarter from 9¼ till 9½ and so on for the busy times of the day. We call the four consecutive quarters which give a maximum the busy hour. So it would be better to speak of an "average busy hour" taken over the year.
The idea behind the above is that the expectation that a subscriber makes a call at a specific moment of one day, is the same as that for the same moment of a random other working day over the entire year and that the variations over the year are random and independent of the time.
[Translator's note: although the author wrote "expectation" in the above, he presumably actually meant probability.]
[Translator's note: Note that earlier only Sundays and holidays were excluded in the data gathering, yet here conclusions are drawn about working days. Saturday presumably used to be a working day back then.]
In this way, the condition of probability theory is approached, to the extent that that is within reach, although we know that even then non purely random deviations remain, such as the busy days before holidays, trade fair days, etc. For these days therefore no guarantee is given regarding the quality of service. This is nothing extraordinary and happens with every other kind of traffic paths, with only the difference that in the telephone operation there is the disadvantage that the public does not see the increased traffic demand with their eyes.
[Translator's note: although "quality of service" seems a modern-day term, in the above it shows up as a literal translation of the Dutch words "kwaliteit van dienst".]
Furthermore there is still the non-random inaccuracy that the traffic tends to amass a bit in the middle of the hour. The influence of this is usually ignored, particularly since usually [toeslagen] are given for other reasons. Besides, this inclination is not of much importance, since the indicated average traffic curve is rather flat at the point of its maximum, so even rather important forward or backward moves of the busy hour cause only small changes in the traffic density.
[Translator's note: I don't know what the Dutch word "toeslagen" can mean in this context. It's the plural of a noun that refers to extra money that is paid or given; e.g., extra money paid for a ticket for a high-speed rather than normal-speed train, or extra money given by a government agency to people who are in specific needy situations.
Update: in this context, it seems to refer to adding a safety margin.]
Now the question arises whether it is permissible to apply the formulas from probability theory to such average "busy-hour traffic densities". This can in general be very firmly answered affirmatively. The formulas used most, such as those from Grinsted, Poisson, Erlang and Molina, all assume that the given traffic density is an average taken over an infinitely large number of hours.
### 10. Manipulations of traffic densities
In telephone exchanges there are often nodes where traffic paths come together and the amounts of traffic they carry merge, and other points where a traffic path and the traffic it carries are split.
Let's consider such a node, to study what relationship there is between the average traffic densities before and after the split.
If we neglect small differences, due to the selecting and searching of the switches, we can observe that at any random moment the number of calls before and after the split is the same, since these are the same calls. The instantaneous traffic density $v$ before the split is therefore equal to the sum of the instantaneous traffic densities after the split. Calling the latter $v_1, v_2, v_3, v_4,$ etc., then $$v = v_1+v_2+v_3+ \mbox{etc.}$$
These instantaneous traffic densities are entirely independent functions of the time. For their average values (traffic densities) we can write: $\begin{array}{rcl} V & = & \displaystyle \int_0^1 v \, dt, \\ V_1 & = & \displaystyle \int_0^1 v_1 \, dt, \\ V_2 & = & \displaystyle \int_0^1 v_2 \, dt, \;\; \mbox{etc.} \end{array}$
The sum of the average traffic densities after the split equals $\begin{array}{rcl} V_1 + V_2 + V_3 \ldots \mbox{etc.} & = & \displaystyle \int_0^1 v_1 \, dt + \int_0^1 v_1 \, dt + \int_0^1 v_1 \, dt + \mbox{etc.} \\ & = & \displaystyle \int_0^1 (v_1 + v_2 + v_3 + \mbox{etc.} ) \, dt \\ & = & \displaystyle \int_0^1 v \, dt \\ & = & V \end{array}$
[Translator's note: the last integral sign is missing in the original text, presumably a typesetting error, since its limits are present.]
Thus, when traffic paths split or join, the sum of the average traffic densities remains unchanged before and after the node.
This only holds if we take the different averages over exactly the same period, i.e., what we find as the average busy hour before the split is also the average busy hour for each of the traffic paths after the split.
If one has extensive statistical material at one's disposal, the collection of which is very expensive and therefore rather rare, one already notices upon casual observation that the daily busy hour of a traffic path is erratic, even more so as it carries less traffic. However, in determining the average busy hour of this traffic path, one will usually notice that it coincides quite accurately with the average busy hours of the other traffic paths of the exchange. Taking the average value over a large number of hours, as described in the previous section, thus has a strong leveling effect. Drawing from the statistical material at our disposal, it was checked for a specific net what fluctuations the busy hour of the different traffic paths, averaging over one month at a time, was subject to, but neither between the different months of the year, nor between different traffic paths, a significant deviation was noticed. This held both for the main traffic arteries and for the connection lines to smaller exchanges.
As goes without saying, this is not a generally valid conclusion and in practice differences may occur; however, these will always indicate an uneven structure of the net. In case this should happen in practice, one acts most easily by assuming for the calculation of the expected traffic densities that there is no such displacement, and then applying a "toeslag" at the end of the calculation, the magnitude of which one determines practically.
[Translator's note: And there's the word "toeslag" again, but this time it's meaning is clear: an amount being added.]
[Translator's note: I do wonder a bit what kind of "practical" method the author has in mind here; if dimensioning links in general were so easy that it could go by rule of thumb, this paper wouldn't exist.]
Among telephone technicians there is a rather widespread opinion, which is defended by theoreticians like dr. Lubberger and dr. Rückle, that when traffic densities merge, the combined traffic is less than the algebraic sum of the components, and that similarly at a split of a traffic path the sum of the traffic densities after the split is larger than the unsplit traffic before the split. On the basis of the above proof that traffic densities can be added algebraically, if their average busy hours coincide, and furthermore on the basis of what is observed in practice, namely that this is usually the case, we already can conclude that this opinion is not correct.
[Dr. Lubbberger, Die Wirtschaftlichkeit der Fernsprechanlagen für Ortsverkehr. Verlag von H. Oldenbourg. Chapter II G. Dr. Rückle and Dr. Lubberger. Der Fernsprechverkehr als Massenerscheinung mit starken Schwankungen. Verlag von Julius Sprinter. Chapter IX A.]
[Translation of those German titles: The economy of the telephone systems for local traffic. Long-distance telephony as a mass phenomenon with strong fluctuations. ]
We want to briefly mention a few points based on which we feel that the authors of the cited books draw incorrect conclusions. In chapter IX.4 of the second book the following reasoning is used: $S$ subscribers together set up $C$ connections during the busy hour, each with a duration of $t$ seconds. The system is divided into groups of $s$ subscribers, which make a traffic of on average $C \frac{s}{S}$ connections per group. Some of these groups will carry fewer connections during the busy hour than the average, while other groups will carry more connections. The overloaded groups will then have a loss which is larger than what is assumed in the calculation, and the underloaded groups will have a smaller loss. The increase of the loss in the former groups will, as is known, exceed the decrease of the latter, so that together the groups will have a larger loss than what was expected based on their average traffic.
The error in this reasoning is that they assume that the traffic of the groups is variable, but that the total traffic of all groups is constant and equal to precisely $C$ calls. This error shows most clearly if one assumes that the number of groups is just two. According to the normal opinion the average traffic densities of of the groups must fluctuate independently of each other around their average values, while the authors tie the traffic of the second group to a value of $C$ minus the traffic of the first group. This is a totally arbitrary and inadmissible restriction, which is imposed on the traffic fluctuations and limits the free fluctuation.
A reasoning based on this is therefore not acceptable.
Furthermore it turns out that the cited authors do not base their theory on average values taken over a number of hours, but on higher values. This shows particularly clearly when they compare the outcomes of their theory to values observed in practice. For example, to support his theory dr. Lubberger compares, on page 26 in the first book mentioned above, maximum traffic values to each other, which he obtains from table II in O'Dell's writeup "The influence of traffic on automatic exchange design", which was published as no. 85 of the L.P.O.E.E.
[Translator's note: I have no idea what the L.P.O.E.E. is.]
It goes without saying that if one determines the size of the traffic paths on the basis of a traffic which is higher than the average, one can apply reductions, since the traffic peaks will not or only rarely coincide. Since however the traffic densities which serve to calculate the exchanges represent average values, we have to reject this theory based on its assumptions.
One question we would like to raise is why a reduction or "toeslag" is only applied when joining or splitting traffic densities, but not when it's about finding the product of variable quantities. Such a case occurs in the calculation of the traffic density from the number of calls and the average call duration. In appendix 1 we have shown, albeit superfluously, that also for this manipulation of variable quantities their average values must simply be multiplied. Should there be truth in a reduction theory, then it should also be extended to the multiplication of variable quantities.
### 11. Originating traffic
The three main pieces of data for designing a telephone exchange are: 1° the number of subscriber lines, 2° the number of calls per subscriber during the average busy hour, and 3° the average duration of the calls.
From these three numbers one calculates the total traffic which is sent by the subscribers and which we want to denote by the name of "originating traffic".
In a similar vein, we speak of "terminating calls" and "terminating-traffic densities", and in contrast to this we call the traffic that is received by the subscribers "end traffic".
The number of busy-hour originating calls per subscriber varies within rather wide bounds and depends on several well-known factors. One extreme is formed by the subscribers of exchanges which are in the business district of the city; the other by those of the rural exchanges.
The number of busy hour originating calls per subscriber for exchanges with call counters fluctuates in normal cases between the bounds 0.8 to 2.0. The latter value is only rarely exceeded and cases where such a high value occurs, must be checked carefully for correctness. For example, we know of a case where the value of 2.0 was significantly exceeded, but once the exchange was put into service, the imposed value of the originating traffic turned out to be a severe mistake, with the consequence that part of the invested capital was almost non-productive.
[Translator's note: in the Dutch text, the word used for non-productive is "renteloos", which also has a strong financial connotation ("not producing interest").]
The cause of this must be found in interpreting traffic statistics with little experience and a lack of expertise, or also in applying safety coefficients improperly.
In case of uncertainty, there is a means to prevent the incorrect use of safety coefficients, which, however, is seldom or never applied, possibly because of unfamiliarity. This means consists of not applying the coefficients to the traffic per line, but to a number of lines. Let us suppose that a new exchange will have to have 1000 lines and that each of these lines has an average busy-hour originating traffic density of 2.2 minutes, in which a safety margin of 10 % is included. If one is uncertain whether this percentage is correct, it is to be recommended to design an exchange with 1100 lines, but with a traffic of 2 minutes per line. The purchase costs for the latter exchange are a little higher, but as far as traffic capacity is concerned, both are equivalent. The advantage achieved in the latter case however, is that if the 10 % safety margin turns out to be unneeded, one can now connect 10 % more lines to this exchange.
The originating traffic per line is determined using periodic observations in the way described in section 9. We assume the means for this are known. However, we do want to spend a few words on the factors which influence a "toeslag" on the originating traffic.
On the one hand, in a large net, where the number of connections per 100 inhabitants is still increasing, it becomes increasingly possible to reach a person by telephone, and as a consequence the occasions for a phone call increase. On the other hand however, in this increase also those people get involved in the telephone traffic who are not driven by necessity, but for whom being connected is more or less a luxury. These are the small-scale consumers.
Third, the automation of the telephone operations has a stimulating effect on the traffic and of course particularly there, where the quality of service previously left something to be desired. This factor is particularly influential in the automation of regional networks and of interlocal traffic. An incidental consequence of automation is, that the number of calls increases, but that the average duration of the calls decreases. The cause of this is that the handling of the traffic is made so much easier by the automation.
How these influences balance each other, is hard to say and must be determined by experience. In general however one can say, that as a rule, the "toeslag" must be taken larger as the traffic per line is less.
Furthermore, when determining the basis for the calculations, one must realise that the number of subscribers connected to an exchange in normal cases is at most 90 % of the full capacity. This because of the fact that a telephone number of a subscription that is cancelled, cannot be immediately assigned to a new subscriber. There's a safety factor hidden in this, if the calculation didn't take this into account yet.
[Translator's note: this is an interesting aspect of the electromechanical switching technology which doesn't seem to have a counterpart in modern networking: the actual physical connections of the equipment are directly tied to identifiers (telephone numbers) that the end users use, and therefore the inability to immediately reuse those telephone numbers implies part of the switch hardware can't be used.]
Since call counters have already been deployed widely, we omit a discussion of the influence of their introduction on the traffic.
[Translator's note: deploying the counters shouldn't influence the traffic by itself, so presumably a more indirect effect is meant; perhaps the tariff structure was changed upon (and due to) introduction of the counters?]
### 12. Terminating traffic
If in a town a single exchange provides the telephone connections, the terminating traffic is necessarily equal to the originating traffic, at least if one excludes prematurely disconnected connections, glowers, and interlocal traffic. If the originating traffic is given, then in such a case in principle the terminating traffic is known.
[Translator's note: the word "glowers" in the above is translated it literally from Dutch, but I have no idea what is meant by it.]
The situation is different in a net which comprises multiple exchanges. Knowledge of the terminating traffic is not determined by that of the originating traffic, but the traffic sent out by the subscribers of one exchange can be markedly different from the traffic that they receive.
It is a common phenomenon that an exchange draws more or less traffic to itself than it sends out to the other exchanges of the same net. In a certain sense one can call this a characteristic property of such an exchange. It also happens that this property is variable and reverses according to the hour of the day. Such a phenomenon can be the reason to advantageously install bidirectional lines.
[Translator's note: "bidirectional" doesn't quite capture what the Dutch text says; it uses a word which means something like "directed both ways". Of course, telephone calls always consist of bidirectional information flow, so the direction meant here must be the direction of the call setup.]
So far, not much practical importance has been attached to knowledge of the terminating traffic, but in our opinion it has special value for the calculation of the traffic among the exchanges of a town or regional net. This will be demonstrated in the next section.
### 13. Incoming and outgoing traffic
In most cases, the outgoing traffic density from one exchange to another exchange turns out to be approximately equal to the incoming traffic density from that same exchange. Often however, quite significant differences occur between the traffic densities on the same route but in opposite direction and one definitely needs to take this fact into account when computing the future traffic densities. As said before, these differences are related to the characteristic properties of the exchanges, so they cannot be ignored.
A second point that we want to draw attention to, is that in distributing the traffic of one exchange among the other exchanges of the same net, the influence of the number of lines of each other exchange is only of secondary importance. Of primary importance is what could be called the traffic value of a line, by which we want to indicate that from a traffic-technical point of view not every line is equivalent to any random other line of the same net. A line which sends out more traffic will, concerning the incoming traffic, have to be counted heavier than a line which sends out less traffic. This principle is already used in the calculation of the groups of final selectors, which connect to the lines of home-exchanges. It is therefore not correct to base traffic calculations on the number of lines of the different exchanges, like dr. Lely does in his dissertation on "Waarschijnlijkheids-rekening bij automatische telefonie", but one must do this on the basis of the actual traffic values.
The theoretical traffic distribution of a net is of importance for judging the true distribution. It is also assumed as the traffic distribution, if little is known about the net. For this theoretical distribution one distributes the originating traffic densities of the exchanges over all exchanges of the net in proportion to their originating traffic densities. If a net consists of two exchanges, which have originating traffic densities of 1000 and 1500 min, then one exchange receives from itself a theoretical traffic of 1000 × 1000 : (1000 + 1500) = 400 min and from the other exchange 1500 × 1000 : (1000 + 1500) = 600 min. So, in this theoretical traffic distribution the originating traffic is always the same as the terminating traffic, while the same holds for the traffic density in both directions on each single route.
In the calculation of the mutual traffic densities between two exchanges, until now often use was made of so-called community factors. From Appendix 3b it turns out that the use of such factors gives rise to wrong outcomes and is therefore unusable in practice. After all, it turns out that when enlarging a net, the community factors are subject to change, which is at odds with the assumption on which they are based, which presupposes that they are constant. Because of this contradiction between the assumption and the result it is certain that the method of interest factors is useless for the calculation of the mutual traffic among a group of exchanges.
[Translator's note: in the original Dutch text, this method is called the method of "interesse-factoren", which would translate to "interest factors", with the word "interest" only in the sense of being interested in, not the economic notion. However, I've translated it as "community factors" because the original Dutch text mentions this translation in Appendix 3b.
For those who are perhaps not familiar with the term "community factor" or perhaps use a different name for it, we want to note that they indicate the ratio between the true and the theoretical traffic between two exchanges.
Instead of the method of community factors, until recently we often used a method which we called the "method of ratios". This method gives quite reliable outcomes for the practical calculation of the mutual traffic densities to be expected. It is described extensively in appendix 3c of this article, but because it also has several errors and inconsistencies, we have abandoned it in favor of a newer and probably the best method, namely that of "double factors".
This original method is explained extensively in appendix 3d.
The general problem that shows up in the calculation of the traffic densities for the connection lines between several exchanges is that we, knowing the complete "traffic picture" for a certain period, search a corresponding picture for a new situation, which arises after several, usually not proportional, extensions or possibly even new exchanges are added to the net.
A purely theoretical treatment of this problem currently does not exist, but it is conceivable that it could be worked out using a function which indicates the affinity between subscribers. In this context we would like to casually note that proposition X of the PhD thesis of dr. Lely cannot be correct. This proposition says that the affinity between two random subscribers is inversely propotional to the distance separating them. Appendix 2 of this article shows, using a simple integration, that the affinity must be a function of at least the third power of the inverse of the distance.
The method of double factors, described in Appendix 3d, satisfies the following general requirements that we have formulated on the basis of practical experience.
a) Reversible, which means that starting from an initial situation using the method one calculates the final situation, and that one can, working backwards with the same method, from the final situation return to the original initial situation.
b) Partitionable, which means that the final situation is independent of the path followed, so that for example a random intermediate situation, derived from the initial situation, can be inserted.
c) Interchangeable, which means that if one reverses the direction of all traffic densities in the initial situation, one obtains, for the given values of the final situation, a final situation which is identical to the one obtained along the normal way, if one also there reverses the direction of all traffic densities.
d) Splittable, which means that we can combine or split exchanges from the initial situation without affecting the traffic densities not involved in this of the other exchanges. Theoretically the method of double factors does not satisfy this requirement, but practically the inaccuracy made is small.
It is unlikely that a method could exist which satisfies this requirement entirely.
### 14. Introduction of new exchanges
In the practice of traffic calculations it often happens that in a town network a new exchange is built, while one naturally does not have statistical data available for it. The data that one has are usually limited to knowledge of the kind of subscribers in traffic-technical sense, which will be connected to the new exchange, because this exchange usually serves to reduce the load on an existing one.
The method of community factors would give a solution to this, by for example assuming new factors for the new exchange after comparison to the factors of the existing exchanges. The outcome obtained this way is very inaccurate, unreliable and therefore unusable. In this case the method of double factors is again helpful, as described in Appendix 4.
This appendix discusses a new type of community factors, which however are based on different principles than the ones we rejected. We call them "base numbers".
### 15. Conclusion
In the above text and in the accompanying appendices we have tried to give an overview of the traffic calculation for telephony, a topic about which so far not much constructive has appeared.
A distinction has been made between the actual traffic calculation and probability theory, which are typically not distinguished in practice and in previous literature.
Some theories which consider certain manipulations with traffic densities defendable, have been demonstrated to be fundamentally wrong, while also the correct way of calculating has been shown.
Furthermore a new method is developed which serves the calculation of the mutual traffic densities to be expected, as needed when extending existing or introducing new exchanges.
### First appendix
#### Multiplication of independent variable quantities
Let us assume some independent variable quantities $g_1, g_2, g_3, \mbox{etc.},$ which vary with the parameters $p_1, p_2, p_3, \mbox{etc.},$ and which during periods 0 to I, 0 to II, 0 to III, etc. fluctuate around their mean values $n_1, n_2, n_3,$ etc., then the average value of the product of the variables becomes: $G = \int_0^I \int_0^{II} \int_0^{III} \ldots g_1 . g_2 . g_3 \ldots dp_1 .\, dp_2;\, dp_3\, \ldots$ Since $g_1, g_2, g_3,$ etc., and $p_1, p_2, p_3,$ etc. are independent of each other, we may write $G = \int_0^I g_1 .dp_1 \times \int_0^{II} g_2 .dp_2 \times \int_0^{III} g_3 .dp_3 \;\;\;\mbox{ etc.}$ or: $G = n_1. n_2. n_3. \mbox{ etc.}$
Therefore the average of the product of several independent variable quantities equals the scalar product of their average values.
### Second appendix
#### Affinity
If we suppose that the affinity between subscribers of a theoretical telephone net of uniform density is inversely proportional to the distance, as dr. Lely assumes in proposition X of his PhD thesis, then one can write for the total traffic of one subscriber: $V = C \int_g^\infty \frac{1}{r} 2 \, \pi \, r \, . \, dr,$ where $C$ is a constant and $g$ is an inner limit of e.g. 750 m. By integration we find: $V = 2 \pi \, C \, I_g^\infty \, r.$ Since the value of the integral for the integration limit $r=\infty$ becomes infinity, the function which represents the value of the affinity must be of a higher power.
Further study shows that the affinity must be a function of at least the third power of the inverse of the distance between subscribers in such an idealised net.
[Translator's note: Clearly, powers less than or equal to 2 would give the same infinite integral problem, but shouldn't any power strictly greater than 2 suffice, including say 2.001 ? The author doesn't elaborate on his "further study", but it either was limited to integer powers, or there is some other factor at play.]
### Third appendix
#### Traffic among exchanges
In practice, the problem repeatedly occurs of calculating, starting from a known traffic distribution among a group of exchanges, the traffic distribution for a new situation to be expected at a future moment.
In the sequel we want to treat several generally known methods, such as the method of percentages and the method of community factors, and then proceed to an improved method, namely that of ratios, and finally describe a calculation method which does not have the deficiencies of the previous methods.
#### 3a) Percentual method
One still encounters this deficient method all too frequently in practice. That we still give it some space here is mostly for putting forward some reasonable requirements that a good method must satisfy.
We start out from the assumption that the traffic picture of the initial situation is known. An example of such a "picture" is found in table 6. Now, using the method of percentages one calculates a second table, in which the various mutual traffic densities of the initial situation are expressed in percentages of the originating traffic densities of the sending exchanges. Next, one multiplies these percentages by the new originating traffic densities and assumes to thus have achieved a good picture of the new situation.
Only if the net expands evenly does this calculation method give reliable outcomes. But since this case rarely or never happens, it constantly leads to evidently incorrect results. A simple example will elucidate this sufficiently:
Be the initial situation for two exchanges A and B:
Traffic from A to A ............ 500 min Traffic from A to B ............ 500 min ---------- Originating traffic from A ............ 1000 min
Traffic from B to A ............ 500 min Traffic from B to B ............ 500 min ---------- Originating traffic from B ............ 1000 min
Based on these numbers we find that the traffic for both exchanges and for both directions is equal and 50% of the originating traffic densities.
Let us now assume for the new situation that only one exchange is extended and that one has good reasons to assume that the originating traffic from this office will double, while that of the second office will remain practically the same. I.e., a case where the old telephone building has no space for further expansion and all new lines need to be connected to the other office.
If we apply the method of percentages to this case, then one finds for the future traffic from A to B: $\frac{50}{100} \times 2000 = 1000$ min, and for that from B to A: $\frac{50}{100} \times 1000 = 500$ min. From these two traffic values it thus appears that, while originally the traffic in both directions was the same, this has now changed substantially. The equilibrium that was present originally, would now be highly broken. Although causing differences between incoming and outgoing traffic is not a criterion for the correctness of a method, it needs to be noted that something totally random occurred here. A shift has taken place for which no reasonable ground can be found.
The absurdity of what happens becomes most clear when one compares the traffic densities of both exchanges before and after the expansion. For one exchange the terminating traffic density per line turns out to decrease from 1.0 to 0.75, and for the other to increase from 1.0 to 1.5 per line. In practice this would mean that one could transport "eindkiezers" from one exchange to the other.
[Translator's note: "eindkiezer" would literally be "final selector", but I need to check that this is indeed the usual term for this part of an electromechanical telephone switch.]
This concludes the discussion of the method of percentages. Calling it totally unusable is not an overstatement.
#### 3b) Method of community factors
The method of community factors starts from the idea that the traffic density between two exchanges does not depend on the originating traffic density of only one exchange, but of both. It is based on the principle that the ratios between the observed and the so-called theoretical traffic densities remain unchanged upon expansion. These ratios are called community factors.
As theoretical traffic densities one denotes the traffic distribution which is obtained by distributing the originating traffic of one exchange over the different exchanges, in proportion to their originating traffic densities.
Let us set the observed originating traffic densities of exchanges 1, 2, 3, etc. to $_a V_1,\, _a V_2, \mbox{etc.}$ (total $V$); the observed traffic densities between exchanges to $v_{11}$ (from exchange 1 to 1), $v_{12}$ (from exchange 1 to 2), etc.; the theoretical traffic densities for the initial situation to $t_{11}, t_{12}, t_{13}, \mbox{etc.}$, then: $$\label{eq:1} \renewcommand{\arraystretch}{2.2} \begin{array}{rcl} t_{11} & = & \displaystyle \frac{_a{V^2}_1}{V} \\ t_{12} & = & \displaystyle \frac{_aV_{1} \, . \, _a V_2}{V} \;\;\;\;\; \mbox {etc.} \end{array}$$
[Translator's note: the pre-subscript "a" presumably refers to the Dutch word "aanvangsverkeersdichtheid" which is translated to "originating traffic density" here; one could therefore argue that it should be replaced by "o" in translation.]
The community factors $c_{11}$ etc. give the ratio between the observed and the theoretical traffic density: $$\label{eq:2} c_{11} = \frac{v_{11}}{t_{11}}, \;\; c_{12} = \frac{v_{12}}{t_{12}}, \;\; \mbox{etc.}$$
If we furthermore set the new originating traffic densities to $_a X_1,\, _a X_2, \mbox{etc.}$ (total $X$), then this method is based on the equality of the following ratios: $$\label{eq:3} \renewcommand{\arraystretch}{2.2} \begin{array}{rclclclcl} c_{11} & = & v_{11} & : & \displaystyle \frac{_a{V^2}_1}{V} & = & x_{11} & : & \displaystyle \frac{_a{X^2}_1}{X} \\ c_{12} & = & v_{12} & : & \displaystyle \frac{_aV_{1} \, . \, _a V_2}{V} & = & x_{12} & : & \displaystyle \frac{_aX_{1} \, . \, _a X_2}{X} \;\;\;\;\; \mbox{etc.} \end{array}$$
[Translator's note: the colons are meant as divisions here, like 2:3 = 4:6.]
From this formula the sought-for $x_{11}, x_{12}, \mbox{etc.}$ can be calculated.
In practice one does not work with formulas, but preferably makes use of a series of consecutive operations, whose outcomes one puts in tables each time.
The first table then contains the given traffic densities among exchanges, with the total corresponding originating traffic densities:
To exchange Originating traffic (total $V$) 1 2 3 4 From exchange 1 $v_{11}$ $v_{12}$ $v_{13}$ $v_{14}$ $_a V_1$ 2 $v_{21}$ $v_{22}$ $v_{23}$ $v_{24}$ $_a V_2$ 3 $v_{31}$ $v_{32}$ $v_{33}$ $v_{34}$ $_a V_3$ 4 $v_{41}$ $v_{43}$ $v_{43}$ $v_{44}$ $_a V_4$ etc.
Among these values there is the relationship: $$\label{eq:4} \renewcommand{\arraystretch}{1} \begin{array}{rcl} _a V_1 & = & v_{11} + v_{12} + v_{13} + \ldots \\ _a V_2 & = & v_{21} + v_{22} + v_{23} + \ldots \\ \mbox{etc.} \end{array}$$
From the given originating traffic densities one next calculates the theoretical traffic densities and again arranges them in the form of a table:
To exchange Originating traffic (total $V$) 1 2 3 4 From exchange 1 $t_{11}$ $t_{12}$ $t_{13}$ $t_{14}$ $_a V_1$ 2 $t_{21}$ $t_{22}$ $t_{23}$ $t_{24}$ $_a V_2$ 3 $t_{31}$ $t_{32}$ $t_{33}$ $t_{34}$ $_a V_3$ 4 $t_{41}$ $t_{43}$ $t_{43}$ $t_{44}$ $_a V_4$ etc.
The $t$ values of Table 2 now satisfy the equations (1).
Now the community factors are calculated from Tables 1 and 2, by dividing the values from Table 1 by those of Table 2.
The $c$ values from Table 3 satisfy the equations (2).
To exchange 1 2 3 4 From exchange 1 $c_{11}$ $c_{12}$ $c_{13}$ $c_{14}$ 2 $c_{21}$ $c_{22}$ $c_{23}$ $c_{24}$ 3 $c_{31}$ $c_{32}$ $c_{33}$ $c_{34}$ 4 $c_{41}$ $c_{43}$ $c_{43}$ $c_{44}$ etc.
Fourthly, one calculates from the given initial traffic densities of the new situation the corresponding theoretical mutual traffic densities, as follows:
To exchange Originating traffic (total $X$) 1 2 3 4 From exchange 1 $u_{11}$ $u_{12}$ $u_{13}$ $u_{14}$ $_a X_1$ 2 $u_{21}$ $u_{22}$ $u_{23}$ $u_{24}$ $_a X_2$ 3 $u_{31}$ $u_{32}$ $u_{33}$ $u_{34}$ $_a X_3$ 4 $u_{41}$ $u_{43}$ $u_{43}$ $u_{44}$ $_a X_4$ etc.
In which: $$\label{eq:5} u_{11} = \frac{_a{X_{1}}^2}{X}, \;\;\;\; u_{12} = \frac{_aX_{1} \, . \, _a X_2}{X}, \;\;\;\; \mbox{etc.}$$
Finally, one multiplies the values from table 3 by those of table 4 and obtains:
To exchange 1 2 3 From exchange 1 $c_{11} \cdot u_{11}$ $c_{12} \cdot u_{12}$ $c_{13} \cdot u_{13}$ 2 $c_{21} \cdot u_{21}$ $c_{22} \cdot u_{22}$ $c_{23} \cdot u_{23}$ 3 $c_{31} \cdot u_{31}$ $c_{32} \cdot u_{32}$ $c_{33} \cdot u_{33}$ etc.
If all were well, these values from Table 5 should be the new mutual traffic densities, calculated using the community factors method. Upon applying it to numerical values it turns out that the sums of the values of the horizontal rows do not equal the given originating traffic densities of the new situation. After all, even though it holds that: $$\label{eq:6} \renewcommand{\arraystretch}{1} \begin{array}{rcl} _a X_1 & = & u_{11} + u_{12} + u_{13} + \ldots \\ _a X_2 & = & u_{21} + u_{22} + u_{23} + \ldots \\ \mbox{etc.} \end{array}$$ Then it does not necessarily hold that: $$\label{eq:7} \renewcommand{\arraystretch}{1} \begin{array}{rcl} _a X_1 & = & c_{11} . u_{11} + c_{12} . u_{12} + c_{13} . u_{13} + \ldots \\ _a X_2 & = & c_{21} . u_{21} + c_{22} . u_{22} + c_{23} . u_{23} + \ldots \\ \mbox{etc.} \end{array}$$
Where this method is applied, one usually works by ascribing the error to the local traffic. By this rather arbitrary action, one satisfies the requirement that the sums of the values of a horizontal row must equal the total originating traffic, but it has as a consequence that all community factors change. And the method was based precisely on the principle that these factors are constant.
In summary, we can therefore say that working with community factors is fundamentally incorrect, since it contradicts itself. It therefore satisfies none of the requirements formulated in sections 13a through d.
#### 3c) Method of ratios
The application of a simple correction factor to the method of community factors leads to the method of ratios. Thus, with this method one can work as described above for the method of community factors up to Table 5, and then apply a correction, which e.g. for the values of the first horizontal row equals: $\frac{_a X_1}{c_{11} . u_{11} + c_{12} . u_{12} + c_{13} . u_{13} + \ldots }$
For the values of the second row: $\frac{_a X_2}{c_{21} . u_{21} + c_{22} . u_{22} + c_{23} . u_{23} + \ldots }$
In this way one achieves that the sum of the horizontal rows becomes equal to the originating traffic densities. The error is now distributed proportionally over all values.
In summary the method of ratios gives, after introducing some simplifications: $$\label{eq:8} \renewcommand{\arraystretch}{2} \begin{array}{rcl} x_{11} & = & v_{11} \displaystyle \frac{_a X_1}{_a V_1} \frac{_a X_1}{\displaystyle v_{11} \frac{_a X_1}{_a V_1} + v_{12} \frac{_a X_2}{_a V_2} + v_{13} \frac{_a X_3}{_a V_3} + \mbox{etc.} } \\ x_{12} & = & v_{12} \displaystyle \frac{_a X_2}{_a V_2} \frac{_a X_1}{\displaystyle v_{11} \frac{_a X_1}{_a V_1} + v_{12} \frac{_a X_2}{_a V_2} + v_{13} \frac{_a X_3}{_a V_3} + \mbox{etc.} } \\ \mbox{etc.} \end{array}$$
Rewritten in a different form: $$\label{eq:9} \renewcommand{\arraystretch}{2} \begin{array}{rcl} x_{11} & = & c_{11} . {}_a X_1 \displaystyle \frac{_a X_1}{c_{11} . {}_a X_1 + c_{12} . {}_a X_2 + c_{13} . {}_a X_3 + \mbox{etc.} } \\ x_{12} & = & c_{12} . {}_a X_2 \displaystyle \frac{_a X_1}{c_{11} . {}_a X_1 + c_{12} . {}_a X_2 + c_{13} . {}_a X_3 + \mbox{etc.} } \\ \mbox{etc.} \end{array}$$
The values $c_{11} . {}_a X_1$, $c_{12} . {}_a X_2$ are called the ratios, from which the method derives its name.
[Translator's note: "ratio" may sound a bit strange for these factors; in Dutch they are called "verhoudingsgetallen", which would literally translate as "ratio numbers", but that doesn't seem like a good translation either, as "number" in English also has the connotation of a count, which Dutch "getal" does not have.]
In order to work quicker in practical applications, one only needs to follow the method of community factors until Table 3, and then proceed immediately to calculating the ratios. A fifth table then contains the traffic values of the new situation.
A general disadvantage of this method is that it is not transparent, but especially, that it does not satisfy the requirements given in sections 13 a to d. Stated succinctly, this method boils down to calculating a number of unknowns from a small number of givens and using some formulas of which we do not even know the meaning. We can escape this using the next method.
#### 3d) Method of the double factors
[Translator's note: in the original, this section was 5d, presumably a typo.]
This method is based on a symmetric treatment of the problem. While all previous methods were based on only the new originating-traffic densities, with this method we also introduce the new terminating-traffic densities as knowns.
Thus we get more control of the development of the problem, we achieve a symmetrical way of solution (e.g., we can reverse the direction of the traffic densities) and determine in advance the kind of subscribers, seen from a traffic-engineering standpoint.
Especially the latter is of fundamental importance.
Whether an exchange produces more traffic than it draws to it, is a typical property, which to some extent depends on the type of subscriber. Thus, for new subscribers that are to be involved in the telephone traffic, it needs to be determined in advance and in comparison to the characteristics of the already connected subscribers, to what extent they will increase or decrease the existing differences between originating and terminating traffic. All previous methods simply left this to chance, which in our opinion is inadmissible.
With the method of double factors we introduce a horizontal and a vertical relationship between the old and new traffic values, as follows: $$\label{eq:10} \renewcommand{\arraystretch}{1} \begin{array}{rclrclrcl} x_{11} & = & p_1 . q_1 . v_{11} ; & x_{12} & = & p_1 . q_2 . v_{12} ; & x_{13} & = & p_1 . q_3 . v_{13} ; \\ x_{21} & = & p_2 . q_1 . v_{21} ; & x_{22} & = & p_2 . q_2 . v_{22} ; & x_{23} & = & p_2 . q_3 . v_{23} ; \\ \mbox{etc.} \end{array}$$
If the existing traffic picture is known, and the originating and terminating traffic densities of the new situation are given, it is possible to express the new traffic values among the exchanges in formulas. However, these formulas are rather complicated and their use is cumbersome, which is why we prefer to recommend a more practical method, which is demonstrated using the following example. It consists of progressively approximating the sought-for mutual traffic densities, starting from the given situation. Let the given traffic distribution be the following:
To exchange Originating traffic 1 2 3 4 From exchange 1 2,000 1,030 650 320 4,000 2 1,080 1,110 555 255 3,000 3 720 580 500 200 2,000 4 350 280 210 160 1,000 terminating traffic 4,150 3,000 1,915 935 10,000
In the next table, one first writes the given originating and terminating traffic densities. Next one writes the individual values from Table 6 in the first horizontal row but multiplied by the quotient of the old and the new originating-traffic densities, i.e., $\frac{6\,000}{4\,000}$; for the second row this factor will become $\frac{4\,000}{3\,000}$ etc. However, the sums of the vertical rows of the new table will now differ from the terminating-traffic densities expected for the new situation. These sums have been added separately at the bottom of Table 7.
To exchange Originating traffic 1 2 3 4 From exchange 1 3,000 1,545 975 480 6,000 2 1,440 1,480 740 340 4,000 3 900 725 625 250 2,500 4 350 280 210 160 1,000 Terminating traffic 6,225 4,000 2,340 935 13,500 Sum 5,690 4,030 2,550 1,230
The second step of the approximation happens by multiplying the values of the vertical columns by the quotients of the demanded terminating-traffic densities and the summations found. Kolom 1 therefore will be multiplied by $\frac{6\,225}{5\,690}$, and column 2 by $\frac{4\,000}{4,030}$, etc.
This way the values of Table 8 are found. In this table the sums of the horizontal rows will now not match the values of the given originating-traffic densities. These sums have therefore been put beside the table.
To exchange Originating traffic Sum 1 2 3 4 From exchange 1 3,280 1,530 895 365 6000 6,070 2 1,575 1,470 680 260 4000 3,985 3 985 720 575 190 2500 2,470 4 385 280 190 120 1000 975 Terminating traffic 6,225 4,000 2,340 935 13,500
This is repeated several times, until the differences between the sums and the originating and terminating traffic densities are so small that, with a minor rounding here and there, the following final situation is achieved: (See Table 9).
To exchange Originating traffic 1 2 3 4 From exchange 1 3,250 1,510 880 360 6,000 2 1,585 1,475 680 260 4,000 3 1,000 730 580 190 2,500 4 390 285 200 125 1,000 Terminating traffic 6,225 4,000 2,340 935 13,500
One sees from this example that the approximation goes quite quickly. If we compare this to the outcomes obtained using the other three methods, then their inadequacy shows clearly. In the next table, the top values of every group of three numbers represent the traffic values found using the method of percentages. The second series of numbers is calculated using the method of community factors, while the third number has been obtained using the method of ratios.
To exchange Originating traffic 1 2 3 4 From exchange 1 3,000 1,545 975 480 3,215 1,525 905 355 6,000 3,265 1,500 885 350 2 1,440 1,480 740 340 1,600 1,465 685 250 4,000 1,440 1,580 740 240 3 900 725 625 250 1,000 715 600 185 2,500 900 775 625 200 4 350 280 210 160 390 260 195 155 1,000 355 300 215 130 Terminating traffic 5,690 4,030 2,550 1,230 6,205 3,965 2,385 945 13,500 5,960 4,155 2,465 920
It turns out that for this particular case, the method of community factors gives the best outcome compared to the method of double factors (Table 9). However, this is only coincidence.
Its reason is that the values which we assumed for the various terminating-traffic densities happen to be close to the values found by the method of community factors.
Furthermore this example shows that the other methods give terminating-traffic values which are rather random. E.g., with the method of ratios, exchange 1 would not draw more traffic than it gives, as is the case in the given traffic picture. Exchange 2 would have taken over this role.
In section 13 of this article some requirements are given which a sound method should satisfy. The proof that the method of double factors satisfies them, is simple and is based on the symmetrical treatment of the problem. We therefore omit it.
### Fourth appendix
#### Introducing new exchanges
By using the properties of the method of double factors, we can easily and reliably add one or more new exchanges to the net.
To do so, we use the property that any intermediate state can be inserted between the given and the sought-for situations. The intermediate state that we insert for this calculation, has values for the originating and terminating traffic densities which are equal in number of units to the number of exchanges.
Applied to the example from appendix 3 we find for the individual values, using the continued calculation method, the following intermediate state:
To exchange Originating traffic 1 2 3 4 From exchange 1 1.365 0.930 0.875 0.830 4.000 2 0.940 1.270 0.950 0.840 4.000 3 0.890 0.950 1.220 0.940 4.000 4 0.805 0.850 0.955 1.390 4.000 Terminating traffic 4.000 4.000 4.000 4.000 16.000
The values in this table are somewhat analogous to the community factors discussed in Appendix 3b. However, while we saw that these did not stay constant when the net is extended, those numbers are independent of that. The principle that one aimed for by introducing the community factors, namely searching for a few fundamental values, independent of the situation (state) and that would thus represent the intrinsic ratio between the exchanges, turns out to be realised with this.
In order to introduce a new exchange, one adds e.g. in the above case a fifth horizontal row and a fifth vertical column, and fills both of them with estimated values, chosen in comparison to the values of the existing exchanges and on the basis of local knowledge of the net. It goes without saying that then neither the horizontal nor the vertical rows will have a sum equal to the number of exchanges of the net in the new situation. The base numbers thus undergo a change upon introducing new exchanges. But that we have in principle not changed anything to the old base numbers and that these are still present hidden in the new situation, appears from the fact that one can go back from the new to the old situation by simply omitting the values of the fifth row and column, followed by the approximation method to return to the given situation.
This new situation is of course calculated from the extended table of base numbers using the method of double factors.
Thus, it turns out that the method is also reversible in this regard.
We would like to give to this new kind of "community factors" the name of "Base numbers".
Acknowledgements: thanks to Matt Roughan for bringing this paper to my attention, typing the tables and formulas, and proofreading and feedback; to Hans van Bruggen for proofreading and feedback; and to the current editor-in-chief of De Ingenieur for permission to publish this translation. | 15,138 | 65,635 | {"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": 10, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.935615 |
https://www.physicsforums.com/threads/adiabatic-approxmation-estimate-of-d-phi-dx.507535/ | 1,725,990,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00154.warc.gz | 882,674,094 | 20,034 | # Adiabatic approxmation, estimate of d\phi/dx
• Derivator
In summary, Messiah writes in his Quantum Mechanics book in chapter XVIII, section 14 (pages 786-789):- Increasing the atomic distance results in an orthogonal function.- The Bohr radius, a, is the distance at which the overlapp of two atomic orbitals becomes approximately orthogonal.- This happens when y is smaller than a.
Derivator
Hi,
Messiah writes in his Quantum Mechanics book in chapter XVIII, section 14 (pages 786-789):
http://img827.imageshack.us/img827/4605/screenshot3si.png
What does he mean, when he says "an increment of the order of 'a' is necessary to transform the function phi into a function that is orthogonal to it"? Why do we obtain an orthogonal function, when we increase the atomic distance?
Especially, why should d Phi/ dX be roughly equal to the orthogonal function divided by 'a'?--
derivator
Last edited by a moderator:
I suppose a is the Bohr radius. The overlapp of two atomic orbitals falls of exponentially for distances >>a. Hence they are rapidly becoming orthogonal to each other to any approximation when the distance supercedes a.
Thank's for your answer. I still don' t see, why d Phi/ dX should be roughly equal to the orthogonal function divided by 'a'. Could you explain this, please?
I mean, the fact that d Phi/ dX is orthogonal to Phi, is ok. One can justify this by integration by parts, for example. I don't see, why we have to divide the orthogonal function by 'a' in order to obtain roughly d Phi/ dX
Derivator said:
I mean, the fact that d Phi/ dX is orthogonal to Phi, is ok.
No, it isn´t in general, e.g. take exp(ikX).
From the localization of the atomic orbitals it is clear that an orbital phi(x) and it´s shifted copy phi(x+y) become nearly orthogonal once y is of the order a. Now phi(X+y)=exp(y d/dX) phi(X)=phi(X)+(exp(y d/dX)-1) phi(X). As long as y is smaller than a, exp(y d/dX) can be replaced by 1+y d/dX (at least as far as it´s action in phi(X) ). Assuming phi to be normalized to one, <phi(X)|phi(X+y)>= 1 +y <phi(X) | d/dX phi(X)>. We know this approximation to become bad when y=a as then the left hand side is approximately 0. Hence a<phi(X) | d/dX phi(X)> is of order unity.
DrDu said:
No, it isn´t in general, e.g. take exp(ikX).
From the localization of the atomic orbitals it is clear that an orbital phi(x) and it´s shifted copy phi(x+y) become nearly orthogonal once y is of the order a. Now phi(X+y)=exp(y d/dX) phi(X)=phi(X)+(exp(y d/dX)-1) phi(X). As long as y is smaller than a, exp(y d/dX) can be replaced by 1+y d/dX (at least as far as it´s action in phi(X) ). Assuming phi to be normalized to one, <phi(X)|phi(X+y)>= 1 +y <phi(X) | d/dX phi(X)>. We know this approximation to become bad when y=a as then the left hand side is approximately 0. Hence a<phi(X) | d/dX phi(X)> is of order unity.
ok, for y=a:
0=<phi(X)|phi(X+a)>= 1 +a <phi(X) | d/dX phi(X)>
then
<phi(X) | d/dX phi(X)> = -1/a
But I still don't get it, thus there are two questions:
1)
why can we conclude from the last equation
<d/dX phi(X) | d/dX phi(X)> = 1/a^2 ?
2)
why can we conclude
d/dX phi(X) = phi(X+a)/a (at least roughly)
Last edited:
I did show you why y has to be of the order of a for the two functions ( the shifted and the unshifted orbital) to become approximately orthogonal and how this is related to the derivative of the orbital with respect to X. I also would not buy the first part of the last sentence of Messiah. Rather
d/dX phi(X) = (phi(X+a)-phi(X))/a (approximately).
Now take the norm of both sides taking into account the orthogonality of the two functions on the right and you will get the last equation of Messiah up to an unimportant factor 2.
ah, I see, you also get this factor 2. I thought this couldn't be correct.
From
d/dX phi(X) = (phi(X+y)-phi(X))/y
I can conclude:
<d/dX phi(X)|d/dX phi(X)>
= 1/y^2 <phi(X+y)-phi(X)|phi(X+y)-phi(X)>
= 1/y^2 (<phi(X+y)|phi(X+y)> - <phi(X+y)|phi(X)> + <phi(X)|phi(X)> - <phi(X)|phi(X+y)>)
= 1/y^2 (1-0+0+1)
= 2/y^2thank you, for your help
Last edited:
Take in mind that Messiah wants to give an order of magnitude estimation.
## What is the Adiabatic Approximation?
The Adiabatic Approximation is a method used in physics to simplify the mathematical description of a system by assuming that its properties change slowly over time. This allows for a simpler and more manageable calculation of the system's behavior.
## How is the Adiabatic Approximation applied?
The Adiabatic Approximation is often applied in quantum mechanics, where it is used to simplify the Schrodinger equation by assuming that the system's energy levels do not change significantly over time.
## What is the importance of the Adiabatic Approximation?
The Adiabatic Approximation is important because it allows for a more straightforward and solvable description of complex systems. It also provides a useful tool for analyzing and understanding the behavior of physical systems.
## What is the meaning of d\phi/dx in the Adiabatic Approximation?
d\phi/dx, also known as the gradient of a potential function, represents the rate of change of the potential energy of a system with respect to a particular variable, in this case, the position x. It is a crucial factor in the Adiabatic Approximation as it allows for the simplification of the system's behavior.
## How is the estimate of d\phi/dx calculated in the Adiabatic Approximation?
The estimate of d\phi/dx is typically calculated by taking the derivative of the potential function with respect to the position variable x. In some cases, this calculation can be simplified by assuming that the potential function is slowly varying over time.
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2K | 1,602 | 5,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.908197 |
http://ecoursesonline.iasri.res.in/mod/page/view.php?id=329 | 1,723,690,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00822.warc.gz | 10,890,716 | 10,230 | ## LESSON 31. Field Tests: Indirect Methods
31.1 Standard Penetration Test (SPT)
The Standard Penetration Test (SPT) is widely used to determine the in-situ parameters of the soil. The test consists of driving a split-spoon sampler into the soil through a bore hole at the desired depth. The split-spoon sampler is driven into the soil a distance of 450 mm at the bottom of the boring. A hammer of 63.5 kg weight with a free fall of 760 mm is used to drive the sampler. The number of blows for a penetration of last 300 mm is designated as the Standard Penetration Value or Number N (ASTM D1586). The test is usually performed in three stages. The blow count is found for every 150 mm penetration. The blows for the first 150 mm are ignored as the top soil may be of disturbed nature due to advancement of borehole and hence considered as those required for the seating drive. The refusal of test when
• 50 blows are required for any 150 mm increment.
• 100 blows are obtained for required 300 mm penetration.
• 10 successive blows produce no advance.
The standard blow count N¢70 can be computed as (ASTM D 1586):
${N'_{70}}={C_N} \times N \times {\eta _1} \times {\eta _2} \times {\eta _3} \times {\eta _4}$ (31.1)
where
${\eta _i}$ = correction factors
N'70 = corrected N using the subscript for the Erb and the ' to indicate it has been corrected
Erb = standard energy ratio value
CN = correction for effective overburden pressure p'0 (kPa) computed as [Liao and Whitman, 1986]:
${C_N}={\left( {{{95.76} \over {{{p'}_0}}}} \right)^{{1 \over 2}}}$ (31.2)
SPT is standardized to some energy ratio (Er) as:
${E_r}={{Actual\;hammer\;energy\;to\;sampler,\;{E_a}}\over{Input\;energy,\;{E_{in}}}}\times100$ (31.3)
Now ${E_{in}}={1 \over 2}m{v^2}={1 \over 2}{W \over g}{v^2}$ and $v={(2gh)^{{1 \over 2}}}$
Thus, ${E_{in}}={1 \over 2}{W \over g}(2gh)=Wh$ (31.4)
where W = weight of hammer and h = height of fall
The correction factor ${\eta _1}$ for hammer efficiency can be expressed as (Bowles, 996):
${\eta _1}={{{E_r}} \over {{E_{rb}}}}$ (31.5)
Different types of hammers are in use for driving the drill rods. Two types are normally used. They are (Bowles, 1996):
(i) Donut hammer with Er = 45 to 67
(ii) Safety hammer with Er as follows:
• Rope-pulley or cathead = 70 to 80
• Trip or automatic hammer = 80 to 100
Now if Er = 80 and standard energy ratio value (Erb) = 70, then ${\eta _1}$ = 80/70 = 1.14
Correction factor ${\eta _2}$ for rod length (Bowles, 1996):
Length >10 m ${\eta _2}$ = 1.00
6 – 10 m = 0.95
4 – 6 m = 0.85
0 – 4 m = 0.75
Note: N is too high for Length < 10 m
Correction factor ${\eta _3}$ for sampler (Bowles, 1996):
Without liner ${\eta _3}$ = 1.00
With liner: Dense sand, clay = 0.80
Loose sand = 0.90
Correction factor ${\eta _4}$ for borehole diameter
Hole diameter: 60 – 120 mm ${\eta _4}$ = 1.00
150 mm = 1.05
200 mm = 1.15
Note: ${\eta _4}$ = 1.00 for all diameter hollow-stem augers where SPT is taken through the stem
Problem 1
Given: N = 21, rod length= 13 m, hole diameter = 100 mm, p'0 = 200 kPa, Er= 80; loose sand without liner. What are the standard N'70 and N'60 values?
Solution: For Erb= 70: ${N'_{70}}={C_N} \times N \times {\eta _1} \times {\eta _2} \times {\eta _3} \times {\eta _4}$
Now, ${C_N}={\left( {{{95.76} \over {200}}} \right)^{{1 \over 2}}}=0.69$ ; ${\eta _1}$ = 80/70 = 1.14; ${\eta _2}$ = 1.0; ${\eta _3}$ = 1.0; ${\eta _4}$ = 1.0
Thus, ${N'_{70}}=0.69 \times 21 \times 1.14 \times 1.0 \times 1.0 \times 1.0=17$
Now ${E_{r1}} \times {N_1}={E_{r2}} \times {N_2}$ ; Thus, ${N'_{60}}=\left( {{{70} \over {60}}} \right) \times 17=20$
SPT Correlations in Clays (N. Sivakugan)
N'60 cu (kPa) Consistency Visual identification 0-2 0 - 12 very soft Thumb can penetrate > 25 mm 2-4 12-25 soft Thumb can penetrate 25 mm 4-8 25-50 medium Thumb penetrates with moderate effort 8-15 50-100 stiff Thumb will indent 8 mm 15-30 100-200 very stiff Can indent with thumb nail; not thumb >30 >200 hard Cannot indent even with thumb nail
Note: N'60 is not corrected for overburden and cu is the undrained cohesion of the clay.
SPT Correlations in Granular Soils (N. Sivakugan)
(N')60 Dr (%) Consistency 0-4 0-15 very loose 4-10 15-35 loose 10-30 35-65 medium 30-50 65-85 dense >50 85-100 very dense
Note: N'60 is not corrected for overburden
31.2 Static Cone Penetration Test (SCPT)
The Static cone penetration test has been standardized by “IS: 4968 (Part-III)-1976: Method for subsurface sounding for soils - Part III Static cone penetration test”. The equipment consists of a steel cone, a friction jacket, sounding rod, mantle tube, a driving mechanism and measuring equipment. The cone has an apex angle of 60° ± 15′ and overall base diameter of 35.7 mm giving a cross-sectional area of 10 cm2. The friction sleeve should have an area of 150 cm2 as per standard practice. The sounding rod is a steel rod of 15 mm diameter which can be extended with additional rods of 1 m each in length. The driving mechanism should have a capacity of 20 to 30 kN for manually operated equipment and 100 kN for the mechanically operated equipment. With help of this test, the friction and tip resistance can be determined separately which is very useful information for pile foundation.
SCPT Correlations
In Clays: ${c_u}={{{q_c} - {\sigma _v}} \over {{N_k}}}$ ; where sv = total vertical stress and Nk = cone factor (15-20). For Electric cone, Nk = 15 and for mechanical cone, Nk = 20.
In Sands: the modulus of elasticity can be correlated as: E = (2.5-3.5) qc (for young normally consolidated sands), where qc the tip or cone resistance.
31.3. Dynamic Cone Penetration Test (DCPT)
The dynamic cone penetration test is standardized by “IS: 4968 (Part I) – 1976:Method for Subsurface Sounding for Soils-Part I Dynamic method using 50 mm cone without bentonite slurry”. The equipment consists of a cone, driving rods, driving head, hoisting equipment and a hammer. The hammer used for driving the cone shall be of mild steel or cast-iron with a base of mild steel and the weight of the hammer shall be 640 N (65 kg). The cone shall be driven into the soil by allowing the hammer to fall freely through 750 mm each time. The number of blows for every 100 mm penetration of the cone shall be recorded and total number of blows for each 300 mm penetration is considered as DCPT N value. The process shall be repeated till the cone is driven to the required depth. DCPT is better than SPT or SCPT in hard soils such as dense gravels. In case of SPT samples are collected for testing whereas in case of SCPT or DCPT samples can not be collected. Hammer is used in case of SPT and DCPT, but for SCPT no hammer is used, the cone is pushed inside the soil.
References
Ranjan, G. and Rao, A.S.R. (2000). Basic and Applied Soil Mechanics. New Age International Publisher, New Delhi, India.
PPT of Professor N. Sivakugan, JCU, Australia.
Ranjan, G. and Rao, A.S.R. (2000) Basic and Applied Soil Mechanics. New Age International Publisher, New Delhi, India.
Arora, K.R. (2003) Soil Mechanics and Foundation Engineering. Standard Publishers Distributors, New Delhi, India.
Murthy V.N.S (1996) A Text Book of Soil Mechanics and Foundation Engineering, UBS Publishers’ Distributors Ltd. New Delhi, India.
PPT of Professor N. Sivakugan, JCU, Australia (pnu-foundation-engineering.wikispaces.com/.../Site+Investigatioon+PPT.pdf).
Bowles, J. (1997). Foundation Analysis and Design. McGraw Hill Book Company.
IS: 4968 (Part-III)-1976: Method for subsurface sounding for soils - Part III Static cone penetration test.
IS: 4968 (Part I) – 1976:Method for Subsurface Sounding for Soils-Part I Dynamic method using 50 mm cone without bentonite slurry. | 2,422 | 8,189 | {"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} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.809315 |
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In cases when there is a point spread and moneyline offered on an event, such as an NFL football game, many bettors will place a wager on the moneyline and point spread of an underdog they feel has a chance to pull the upset. They will safely bet the point spread because they feel the game will be close, but will also put themselves in line for a nice payday if the underdog wins straight-up.
The term moneyline is actually somewhat misused in sports betting as it really just means a type of odds format. Technically, it is a way to represent the odds/payouts for a win bet, but we’re not going to split hairs. What we’d like to point out is that the odds on each participant in a sporting contest can be listed in one of three different formats.
You may be wondering how we determined which of the two teams was the favorite and which was the underdog. You may also be wondering how much you get paid out for a bet on either side of this game. If you look at the odds above (this is a screenshot from an actual online sportsbook), you’ll see that all of that information is given to you. Before the spread number of 4 ½, you’ll see a plus or minus sign. The plus sign indicates the underdog and the minus sign indicates the favorite.
This is a huge difference. The potential profit on the moneyline wager (\$143) is over 40% greater than that of the point spread wager (\$100). You're a little less likely to win, as there is a chance that Seattle would lose by one or two points, but there's a more than fair chance that if they did cover they would actually win the match. And, of course, if they lost by three or more then you'd have lost either way.
Bets on “Winner of Point”, “Scorer of Goal" and similar offers refer to the participant winning the listed occurrence. For the settlement of these offers, no reference to events happening prior to the listed occurrence will be taken into consideration. Should the listed event not be won within the stipulated time frame, all bets will be declared void, unless otherwise stated.
As you can see, each team is listed, followed by the adjustment or line change for each team, and then the odds that you would be paid out. If you notice, Chelsea has the word scratch next to their name. This is because they are the league favorite to win and all other adjustments are made about them. If your team is in first place at the end of the regular season after the adjustments are made, you will win your bet and be paid the posted odds. As you can see, the odds pay out fairly well on these bets as they are season long and are more difficult to win.
As you may already be assuming, adjustments are made if you want to bet on the Razorbacks. The point spread for the Razorbacks would be set at +7. It will always be the exact opposite of the other team. As the negative sign represents the favorite, the plus sign here represents the underdog. If you were to bet on the Razorbacks, they can actually still lose, and you win your bet.
A quick word on that annoying half point in the point spread – most lines you’ll come across will use half points, but it’s not standard practice across the board. When you see a line with a full number instead of a number with a half point, your wager could end up as a push. In our example, if the line were 7 instead of 7.5 and the final difference in points was exactly 7, your wager is returned to you, and neither you nor the book makes money.
The number-one key to success here (as it is with any type of sports bet) is understanding what value is and knowing when and how to take advantage of it. Value, in a nutshell, is finding sports bets that are paying you at a better rate than you think they should. If you place enough of these bets to overcome variance, you’re going to be a long-term winner.
Sports betting would be easy — or maybe just easier — if all that was required was to correctly pick the winning team. Gambling institutions, sportsbooks and bookies fall back on point spreads to make the process a little more difficult and to create the ultimate wagering challenge. You'll need a solid understanding of the point spread system if you hope to have a profitable season.
The spread on offer will refer to the betting firm’s prediction on the range of a final outcome for a particular occurrence in a sports event e.g. the total number of goals to be scored in a football match, the number of runs to be scored by a team in a cricket match or the number of lengths between the winner and second-placed finisher in a horse race.
### Identify the favorite. Lines with a - before the number (i.e. -200) indicate the favorite. A -200 should be read as: "For every \$200 wagered, I win \$100." When there is a negative sign, the line should always be read with relation to 100. That does not mean you have to bet that much, it's just easiest to understand! When a + sign is present, just reverse the reading, always keeping reference to 100:
The second number in our example (-110 for both teams) tells you how much you have to wager in order to win \$100. It’s an easy way to calculate how much you’ll win if your bet pays off, presented in units of \$100 at a time for simplicity’s sake. Most of the time, these two numbers will be the same, because oddsmakers want to set lines so that they get as much action on the underdog as on the favorite, guaranteeing them a profit. If a book gets a single bet of \$110 (by a customer hoping to win \$100) on the Cowboys and a single bet of \$110 on the Giants, it will have taken in \$220, but will only have to pay back \$210 to whichever customer wins the bet. That’s a guaranteed profit of \$10, and since sportsbooks take far more than a single bet in either direction, they stand to earn that seemingly small amount of profit many times over. The \$10 difference between what you wager and what you win is known as juice or vig in the sports betting industry, and it’s the way books earn their bread and butter.
Another thing to consider is popular winning margins, which are particularly applicable to football. Consider that many tight games may finish with either a three point or a seven point margin. If the point spread is around either of these marks, make every attempt to be the right side. For example, if you were to back a team that is either 2.5 or 3 point favorites, you’d want to back them at the 2.5 mark, as if they were to win by 3 you’d win as opposed to a push.
Doc’s Sports is offering \$60 worth of member’s picks absolutely free – no obligation, no sales people – you don’t even have to enter credit card information. You can use this \$60 credit any way you please for any handicapper and any sport on Doc’s Sports Advisory Board list of expert sports handicappers. Click here for more details and take advantage of this free \$60 picks credit today
What may look like a jumble of words, numbers, and punctuation is actually a precise and easy-to-read breakdown of the various odds and point spread details your book is offering. Here is a breakdown of each unit of information given above. Once you understand each part of the jumbled details above, you’ll be able to read a sports betting line with confidence.
The 2-way moneyline is what most North American bettors would simply refer to as “the moneyline”. This is one of the most common wagering options where the user bets which side will win the game straight up. (A draw or tie results in a push with the 2-way moneyline.) The term is sometimes highlighted during soccer betting to differentiate from the 3-way moneyline - a more popular option with the draw added as a wagering option.
Moneylines are a viable alternative to point spreads when betting on football. If you're one of those bettors who only ever bets on the spread, then you could very well be missing out on some good opportunities to find better value. We don't recommend that you stop placing point spreads and only place moneyline wagers, but you should definitely consider both when betting on a game of football. Try to decide which one offers the better value, and then go with that option.
Obviously, the first three letters on the top two lines of the three-line package of symbols represents a team in the game you’re wagering on; NYG stands for the New York Giants, while DAL stands for the Dallas Cowboys. The number next to each team’s name is known as the spread or the point spread. Wagers on the point spread are among the most popular sports wagers in the world. The reason this wager is popular is that it doesn’t matter which team wins or loses; what matters is the amount of points the teams score, and whether or not the team you place your money on beats the difference in points (the ‘spread’) or not.
If you were correct though but getting paid at the sportsbooks rate, you would lose the bet 55.6 times (-\$5560) and win the bet 44.4 times (44.4 x \$250 = \$11,100). You would profit over \$5,000 for betting on bets that you thought you were going to lose! This is finding value. Value bets are great as a part of a long term winning strategy and are the key to conquering the “simple” moneyline/win bets.
This is a huge difference. The potential profit on the moneyline wager (\$143) is over 40% greater than that of the point spread wager (\$100). You're a little less likely to win, as there is a chance that Seattle would lose by one or two points, but there's a more than fair chance that if they did cover they would actually win the match. And, of course, if they lost by three or more then you'd have lost either way.
```In general, the betting public tends to gravitate towards favorites when betting the games regardless of the actual pointspread. This is especially true with high-profile teams such as Dallas and Green Bay in the NFL and Golden State and Cleveland in the NBA. The sportsbooks are well aware of this phenomenon and often times they will adjust the betting spreads accordingly. This, in turn, actually adds some value to the underdog when you consider that a pointspread is nothing more than a handicapping tool that is designed to even out the match.
```
Actually handicapping games is far from simple though. Or, at least, handicapping them well is. There are all kinds of different factors that can affect the outcome of a game, and you need to take as many of them as possible into account. You also need to assess just how much of an impact those factors will have, and try to accurately evaluate just how likely a team is to win. For more advice on how to do this, please see the following article.
## It is a pretty simple concept once you get the hang of it, and you will also start to see profitable opportunities in football and hoops where wagering on the moneyline makes more sense than betting the point spread. If you really like an 8-point underdog in the NFL and think they will win, you can take the 8 points and hope they cover the spread. Or you can check out the moneyline option where they might be +280 and make more money betting them to win (\$280) than on the point spread (\$100).
The last format we want to look at is fractional odds. Personally, we aren’t a huge fan of fractional odds because they’re the most challenging to work with. The formula is almost the same as with decimal odds, but it gives your profit instead of total money returned. It also requires you to solve a fraction, which may be a nightmare for a lot of people. Regardless, we are going to walk you through how to do it with the same bet we’ve been working with.
###### Let's take a look at a sample Asian handicap bet to make this make more sense. Some things are just better learned through getting your hands dirty. For example, imagine that you choose to bet Manchester United at (-1, -1.5). Half of your bet would be for Manchester United at -1, and a half would be at Manchester United -1.5. Let's say Manchester United wins the game by one goal. You would push on your first bet and lose on your second bet. If you bet \$100 on this, you would receive \$50 back for the push and lose on the other portion of your bet.
This highlights a notable advantage of the moneyline wager. You get to control, to some extent, the risk versus reward. For example, you might be quite certain that the Cardinals are going to win this game, but not convinced that they're going to cover the spread. So a moneyline wager is the safe option. There's less money to be made, but less chance of losing. On the other hand, you might think that the Packers are going to cause an upset. Rather than betting on them to cover the spread, you can bet on them to win outright. There's less chance of winning such a wager, but the potential returns are much greater.
Shopping betting lines is one of the most important things you can do when betting point spreads in basketball. While the majority of sportsbooks will have the same line, it's fairly common that you can find lines that are a half or full point different. This can have a huge impact on your bottom line. If you don't think a half point or one full point makes that big of a difference, just ask anyone who has bet sports for a while. They will inform you that getting an extra half point is like winning the sports betting lottery.
You have three choices for the three betting options: Home, Away, or Draw (tie). The result of the game is decided after regulation play (90 minutes plus injury time). Overtime, the Golden Goal rule, and penalty kicks are not taken into consideration for soccer bets unless otherwise stipulated. You can usually bet on a winner or advancement (including OT & shootouts) but with different odds would be given.
Through our partnership with FanDuel sports book, you will participate in the BetBattle, a private sports betting competition. You will select multiple wagers from a private parlay for games on Saturday and Sunday immediately following the bet camp. The winners will be determined by a point system and will be announced on the Tuesday following the camp.!
We go in depth on this in the advanced guide on understanding value that we referenced earlier, but we will give you a brief intro on it here. If you’re able to calculate the percentage chance that you have to win in order to break even, and you can figure out the percentage chance that you think you’re going to win the bet, you can figure out very quickly if there is value.
For example, let’s say that two players are playing a tennis match and one player is +250. You think that this player has a MUCH better chance than that but still is an underdog. Most people would tell you that you are crazy to make a bet on someone that you think is going to lose. The thing is, though, underdogs do win and if you’re getting paid more than you should when they do, you’re going to be profitable. Here’s a simple math breakdown.
Here in this point spread example for the NFL, the Falcons are playing the Panthers. Atlanta has been set as a three-point favorite on the betting line. That means that for Atlanta to cover the spread that has been set, they will need to win by at least four points. And for Carolina to cover the point spread, they can do so with a loss by two points or less, or obviously a win straight up. If the Falcons win by exactly three points, the bet would result in a push with no payouts.
### Winning at sports betting is challenging. If it were easy, everyone would quit their jobs and do it, and sportsbooks would all be out of business. What makes it so challenging is that the lines are usually set pretty spot on which means it's a bit more challenging to pick the correct side of the bet. That being said, it's definitely not impossible to make money betting basketball point spreads. You'll have to develop a winning strategy and continually tweak it until it's perfect. Here are a few tips and strategies that will help point you in the right direction.
Doc's Sports is offering \$60 worth of handicapper picks absolutely free - no obligation, no sales people - you don't even have to enter credit card information. You can use this \$60 credit any way you please for any handicapper and any sport on Doc's Sports Advisory Board list of expert sports handicappers. Click here for more details and take advantage of this free \$60 picks credit today .
Sometimes a line will move far enough to create a “middle” opportunity. Say the Texas Longhorns end up facing the Wisconsin Badgers in the first round of March Madness. If you have Texas early as a 5-point favorite, and I move the line to Texas –7 later in the week, then you can also place a bet on Wisconsin +7. If Texas happens to win by six points, both your bets cash in. Texas winning by either five or seven gives you a win and a push. Any other result creates a win and a loss, so you’re only risking the vigorish.
If all the money at one sportsbook comes in on Team A and all the money comes into a second sportsbook on Team B, they’re both going to adjust their lines accordingly to what is going on in their book. This means that if you want to bet on Team A, you should go to the second sportsbook where the line will be great. If you want to be on Team B, you should go to the first sportsbook where the line will be better.
You may often notice that the spread is sometimes set at an even number such as 3, 6 , 10, etc. In this case if the favored team won by the exact amount set for the spread the bet would be pushed, and all bets would be returned. For example, if the Patriots were 3 point favorites and they won by a FG (3 points) than this would results in a push, meaning no matter which side you bet on you would get your money returned to you.
One of the biggest mistakes that bettors make is trying to make a judgement on every single game that's taking place. This is especially true of those who only focus on the NFL. There aren't that many games each week, and bettors think they stand the best chance of making money if they can predict the outcomes in all of them. This is not an approach we recommend.
# If a fight is scheduled for more than four rounds and, after four rounds, an accidental foul occurs which causes an injury (further to which the referee stops the fight), the fight will be deemed to have resulted in a technical decision in favor of the boxer who is ahead on the scorecards at the time the fight is stopped (and all markets on the fight will stand).
Understand that negative odds indicate how much money your must spend to make \$100. When betting on the favorite, you take less risk, and thus earn less. When betting on a favorite, the moneyline is the amount of money you need to spend to make \$100 profit. In the previous example, in order to make \$100 of profit betting for the Cowboys, you would need to spend \$135. Like positive odds, you earn back your bet when winning.
If the Cowboys are 6-point favorites, their odds are -6. If the Giants are 6-point underdogs, their odds are +6. From the oddsmakers' perspective, the Giants are starting the game with a 6-0 lead, while from the Dallas side, the Cowboys are starting with a 0-6 deficit. If you bet on the Cowboys and they win 34-30, they failed to cover the spread by two points. If you bet on the Giants, they beat the spread by two points.
With moneyline bets, there is no point spread to manipulate. Instead, the sportsbook will alter the payouts you’ll receive for a correct pick. The bigger the favorite, the less you’ll get paid. The bigger the underdog, the more you’ll get paid for a correct wager. This line will fluctuate as the sportsbook needs it to in order to encourage or discourage bets on either side.
###### Used in high-scoring sports like NFL and NBA, the point spread is a handicap that is placed on the favorite team in terms of points for betting purposes. If 10 points favor the Broncos over the Seahawks the point spread is 10. The Broncos must win the game by 11 or more points for you to win your wager. If you’ve made your bet on the Seahawks, you’ll win your bet if your team wins the game or losses by nine points or less. If the Broncos manage to win by exactly 10 points then the bet will be a tie.
The team that has the minus sign, which is the favorite, has points deducted from its final score, while the dog, with the plus sign, has points added. The favorite must beat the spread, which means they have to win by more than the negative number to pay off. The underdog pays off in two instances—if they win outright or if they lose by less than the spread.
The last types of bets that you should be aware of when betting on the NBA are proposition bets. A prop bet is a wager in which you bet on whether or not something is going to happen in a particular game. For example, will a certain player make more than four 3-pointers in a game? Which team will score first? Which team will win the opening tip-off? Will a certain team shoot over X % from the field? All of these are examples of prop bets that you can make on a particular NBA game.
```We're often asked a question along the lines of "why would I place moneyline wagers rather than point spread wagers?" There's no simple answer to this question really, as point spreads and moneylines shouldn't be viewed as "either/or" options as such. You don't have to decide that you're always going to bet on the spread, or that you're always going to bet moneylines. These are two different wager types that have their own merits, and any bettor should have them both in their arsenal.
```
Betting “against the spread” (ATS) just means you’re betting on the point spread in a particular matchup as opposed to the moneyline, or some other type of wager. Bettors often use a team’s ATS record to gauge its performance against the spread. For example, the New England Patriots were 11-5 ATS in the 2017 regular season, meaning they covered the posted point spread 11 times, and failed to cover five times. | 5,310 | 24,251 | {"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} | 2.796875 | 3 | CC-MAIN-2019-39 | latest | en | 0.945082 |
https://math.answers.com/algebra/What_does_x_equals_in_3x_plus_11_plus_5x_equals_75 | 1,679,925,139,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00035.warc.gz | 445,485,430 | 49,802 | 0
What does x equals in 3x plus 11 plus 5x equals 75?
Wiki User
2010-06-22 17:27:30
8x+11=75(-11)
8x=64(/ 8)
Wiki User
2010-06-22 17:27:30
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2256 Reviews | 92 | 203 | {"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} | 2.703125 | 3 | CC-MAIN-2023-14 | longest | en | 0.549859 |
https://de.mathworks.com/matlabcentral/cody/problems/1636-zigzag-matrix-with-reflected-format/solutions/2042851 | 1,582,403,108,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145713.39/warc/CC-MAIN-20200222180557-20200222210557-00202.warc.gz | 349,281,417 | 15,657 | Cody
# Problem 1636. ZigZag matrix with reflected format
Solution 2042851
Submitted on 2 Dec 2019 by Asif Newaz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 5; y_correct = [ 2 4 6 8 10; 1 3 5 7 9; 1 3 5 7 9; 2 4 6 8 10]; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = 7; y_correct = [ 2 4 6 8 10 12 14; 1 3 5 7 9 11 13; 1 3 5 7 9 11 13; 2 4 6 8 10 12 14]; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = 11; y_correct = [ 2 4 6 8 10 12 14 16 18 20 22; 1 3 5 7 9 11 13 15 17 19 21; 1 3 5 7 9 11 13 15 17 19 21; 2 4 6 8 10 12 14 16 18 20 22]; assert(isequal(your_fcn_name(x),y_correct)) | 349 | 735 | {"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-2020-10 | latest | en | 0.544966 |
https://www.printablemultiplication.com/worksheets/Multiplication-Problems-4th-Grade-Worksheet | 1,606,901,262,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141706569.64/warc/CC-MAIN-20201202083021-20201202113021-00205.warc.gz | 813,928,919 | 76,530 | ## Multiplication Chart X30
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…fundamentals of multiplication. Also, assess the fundamentals how to use a multiplication table. Let us evaluation a multiplication illustration. Employing a Multiplication Table, flourish 4 times a few and get… | 480 | 2,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-50 | longest | en | 0.765534 |
https://arpadt.com/articles/memoization-in-javascript-with-examples | 1,603,149,799,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00589.warc.gz | 228,249,173 | 7,571 | # Memoization in JavaScript with examples
Memoization is a great technique which helps developers write more efficient code. In this post, I'll present solutions to two popular problems with the use of memoization.
## What is memoization?
Memoization is a programming technique which helps expensive operations become less expensive by storing (caching) the results of previous calculations.
What is an expensive operation?
If you take a look at the big-O chart, everything in red should be avoided, if possible.
I wrote a post on the big-O notation, so I’m not going to go into details here. Basically, big-O (or time complexity) shows the running time of an algorithm.
The more calculation a function needs to do in order to complete its task, the more expensive (and time and resource consuming) the operation is.
The classic example is the calculation of Fibonacci numbers, which usually happens with recursion, i.e. a function calls itself for a number of times to calculate the next member of the sequence.
In the Fibonacci-example above, the time complexity is horrible (O(2n)), and if you ever tried to submit a solution similar to this to Hackerrank or freeCodeCamp, you noticed that these platforms wouldn’t accept it, although the algorithms returns the correct result. Instead, they want you to think about less expensive solutions, which may be more complex in terms of code quantity, but take less time to run.
## Examples
Storing the result of the previous calculation helps decrease time complexity, which means that the algorithm becomes more efficient (and your solution will pass the tests).
Let’s see how memoization works by looking at two examples.
### Find factorials
The task is the following: Calculate the factorial of a given number.
Factorial is when we multiply the positive whole numbers together from 1 to the given number. The notation of the factorial is the `!` (exclamation mark). For example, 5! = 120, because 1 * 2 * 3 * 4 * 5 = 120.
The function will be called `findFactorial` and will accept an argument, the number we want to calculate the factorial of.
The code can look like this:
``````function findFactorial(num) {
if (num < 0) return 'Please enter a non-negative number.'
if (!Number.isInteger(num)) return 'Please enter an integer.'
if (num === 0) return 1;
let factorial = 1;
let i = 1;
while (i <= num) {
factorial *= i;
i++;
}
return factorial;
}
``````
Let’s go over it step-by-step.
First, we can do some validation. Factorials only make sense for positive whole numbers, so we can return the relevant messages if the input is other than that. For the input of zero, we’ll return `1`, because 0! = 1 by convention.
The interesting part is the `factorial` variable, which has an initial value of `1`. We’ll store the results of previous calculations in this variable.
The `while` statement does the lion share of the work. The value of `i` is increased (`i++`) after each multiplication until it becomes greater than `num`, which is our input.
For each iteration, we perform a multiplication, where the already stored product of numbers (`factorial`) is multiplied by the next `i` value.
This solution is simple and `findFactorial` is a pure function, meaning that its output only depends on the input and it has no side effects. If `num` is changed, the return value of `findFactorial` will also change.
This algorithm has a time complexity of `O(n)` (because of the iteration) and this is OK.
### Find the n-th Fibonacci number
The classic Fibonacci-solution involves recursion, but we can do better here.
The task is to find the greatest Fibonacci-number that is not greater than a given upper limit.
We can create the next term of the Fibonacci-sequence by adding the two previous terms together. The first and second terms are 1 and 1. The third term (F3) is 2 (1 + 1), then 3, 5, 8, 13, 21, 34, and so on. The terms (members) of the Fibonacci-sequence are called Fibonacci numbers.
Our function can be called `findFib` and it’ll have one argument, which is the upper limit (`num`).
In the factorial example, we stored the already calculated product in a variable (`factorial`) of number type.
In case of `findFib`, we’ll need to add two numbers and one of these numbers will be different in each addition. A single variable won’t be enough here, so it might be a good idea to store the temporary results in an array.
As we go along and calculate the next Fibonacci number, we can insert the new numbers to the array by adding the last two elements.
The code can look like this:
``````function findFib(num) {
if (num <= 0) return 'Please enter a positive number';
if (num === 1) return 1;
const fibs = [1, 1];
let nextFib = 0;
while ((nextFib = fibs[0] + fibs[1]) <= num) {
fibs.unshift(nextFib);
}
return fibs[0];
}
``````
We can do a simple validation first. Fibonacci numbers are positive integers, so it doesn’t make sense to have a negative upper limit. If `num` is 1, we can return `1` because both the first and second Fibonacci numbers are equal to 1, so `1` meets the condition of not being greater than the upper limit.
The `fibs` array will store the already calculated Fibonacci numbers. The first two terms of the sequence are always defined, otherwise we couldn’t move on to the next member. We need to list them as the first two elements of the array (`1` and `1`).
It’s shocking but `nextFib` will be the next Fibonacci number in the sequence. Because it’s always recalculated, we’ll need to declare it with the `let` keyword.
The engine of the function is - again - the `while` statement.
The new Fibonacci number will be inserted to the beginning of the array, because it’ll be easier to calculate the next term this way. If we `push`ed the new element to the end of the `fibs` array, we would need to mess with the length. No big deal, it’s doable, but it’s probably a bit easier and cleaner to do it this way.
We can use the `unshift` array method to add elements to the beginning of the array. It mutates `fibs`, but it’s not an issue here, because `fibs` need to continually change so that we can calculate the next term.
In the condition of the `while` statement, we add the first and the second elements of `fibs` (which are the largest and the second largest Fibonacci numbers before the next term is calculated), thanks to the `unshift` method. We’ll keep adding the first two elements until we reach the upper limit (`num`).
Finally, the first element of the array is returned.
With this, we could reduce the original O(2n) complexity with the recursion to O(n).
### Bonus: sum of the first n Fibonacci numbers
We can also add up the Fibonacci numbers up to the upper limit.
Because the numbers are stored in an array, we can use the `reduce` method to add the numbers together.
Let’s rename the function to `sumOfFibs`, and the new, modified function looks like this:
``````function sumOfFibs(num) {
if (num <= 0) return 'Input must be a positive number';
if (num === 1) return 1;
const fibs = [1, 1];
let nextFib = 0;
while ((nextFib = fibs[0] + fibs[1]) <= num) {
fibs.unshift(nextFib);
}
return fibs.reduce((acc, val) => {
return acc + val;
}, 0);
}
``````
The initial value is `0` and we keep adding up the numbers until we get to the end of the array.
## Conclusion
As we have seen, the key is that we store (cache) the already calculated values, so that they don’t need to be recalculated later on. The hardest part is to find the best way to store the values.
I hope this post was useful. If so, see you next time. | 1,815 | 7,566 | {"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} | 3.578125 | 4 | CC-MAIN-2020-45 | longest | en | 0.925211 |
https://exercism.io/tracks/elm/exercises/sum-of-multiples/solutions/816e89f6b2c84431b65527ec74b19140 | 1,624,570,409,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488559139.95/warc/CC-MAIN-20210624202437-20210624232437-00271.warc.gz | 222,447,045 | 6,331 | 🎉 Exercism Research is now launched. Help Exercism, help science and have some fun at research.exercism.io 🎉
jjunho's solution
to Sum Of Multiples in the Elm Track
Published at Dec 27 2020 · 0 comments
Instructions
Test suite
Solution
Given a number, find the sum of all the unique multiples of particular numbers up to but not including that number.
If we list all the natural numbers below 20 that are multiples of 3 or 5, we get 3, 5, 6, 9, 10, 12, 15, and 18.
The sum of these multiples is 78.
Elm Installation
Refer to the Installing Elm page for information about installing elm.
Writing the Code
The code you have to write is located inside the `src/` directory of the exercise. Elm automatically installs packages dependencies the first time you run the tests so we can start by running the tests from the exercise directory with:
``````\$ elm-test
``````
To automatically run tests again when you save changes:
``````\$ elm-test --watch
``````
As you work your way through the tests suite in the file `tests/Tests.elm`, be sure to remove the `skip <|` calls from each test until you get them all passing!
Source
A variation on Problem 1 at Project Euler http://projecteuler.net/problem=1
Submitting Incomplete Solutions
It is possible to submit an incomplete solution so you can see how others have completed the exercise.
Tests.elm
``````module Tests exposing (tests)
import Expect
import SumOfMultiples exposing (sumOfMultiples)
import Test exposing (..)
tests : Test
tests =
describe "Sum Of Multiples"
[ test "[3, 5] 15" <|
\() -> Expect.equal 45 (sumOfMultiples [ 3, 5 ] 15)
, skip <|
test "[7, 13, 17] 20" <|
\() -> Expect.equal 51 (sumOfMultiples [ 7, 13, 17 ] 20)
, skip <|
test "[4, 6] 15" <|
\() -> Expect.equal 30 (sumOfMultiples [ 4, 6 ] 15)
, skip <|
test "[5, 6, 8] 150" <|
\() -> Expect.equal 4419 (sumOfMultiples [ 5, 6, 8 ] 150)
, skip <|
test "[43, 47] 10000" <|
\() -> Expect.equal 2203160 (sumOfMultiples [ 43, 47 ] 10000)
, skip <|
test "[5, 25] 51" <|
\() -> Expect.equal 275 (sumOfMultiples [ 5, 25 ] 51)
]``````
``````module SumOfMultiples exposing (multiples, sumOfMultiples)
import Set
sumOfMultiples : List Int -> Int -> Int
sumOfMultiples divisors limit =
divisors
|> List.map (\d -> multiples d limit)
|> List.concat
|> Set.fromList
|> Set.toList
|> List.sum
multiples : Int -> Int -> List Int
multiples number limit =
List.range 1 (limit - 1)
|> List.filter (\element -> modBy number element == 0)`````` | 746 | 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} | 3.828125 | 4 | CC-MAIN-2021-25 | latest | en | 0.767827 |
https://www.doubtnut.com/question-answer/find-the-sum-of-first-10-even-natural-numbers-32537219 | 1,632,814,716,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00635.warc.gz | 728,024,320 | 71,598 | Home
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Find the sum of first 10 even ...
# Find the sum of first 10 even natural numbers.
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 1-10-2020
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## Latest Questions
Class 10th
Arithmetic Progression
Class 10th
Arithmetic Progression
Class 10th
Arithmetic Progression
Class 10th
Arithmetic Progression
Class 10th
Arithmetic Progression
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Class 10th
Arithmetic Progression
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Arithmetic Progression
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Arithmetic Progression
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Arithmetic Progression | 458 | 1,253 | {"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} | 2.59375 | 3 | CC-MAIN-2021-39 | latest | en | 0.678852 |
https://simplywall.st/stocks/cl/materials/snse-cementos/cementos-bio-bio-shares/news/heres-why-cementos-bio-bio-s-a-s-snsecementos-returns-on-capital-matters-so-much/ | 1,597,478,811,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740733.1/warc/CC-MAIN-20200815065105-20200815095105-00219.warc.gz | 479,446,465 | 15,216 | # Here’s why Cementos Bio Bio S.A.’s (SNSE:CEMENTOS) Returns On Capital Matters So Much
Today we’ll evaluate Cementos Bio Bio S.A. (SNSE:CEMENTOS) to determine whether it could have potential as an investment idea. Specifically, we’re going to calculate its Return On Capital Employed (ROCE), in the hopes of getting some insight into the business.
First of all, we’ll work out how to calculate ROCE. Then we’ll compare its ROCE to similar companies. Finally, we’ll look at how its current liabilities affect its ROCE.
### What is Return On Capital Employed (ROCE)?
ROCE is a measure of a company’s yearly pre-tax profit (its return), relative to the capital employed in the business. All else being equal, a better business will have a higher ROCE. Overall, it is a valuable metric that has its flaws. Renowned investment researcher Michael Mauboussin has suggested that a high ROCE can indicate that ‘one dollar invested in the company generates value of more than one dollar’.
### So, How Do We Calculate ROCE?
Analysts use this formula to calculate return on capital employed:
Return on Capital Employed = Earnings Before Interest and Tax (EBIT) ÷ (Total Assets – Current Liabilities)
Or for Cementos Bio Bio:
0.044 = CL\$17b ÷ (CL\$428b – CL\$54b) (Based on the trailing twelve months to September 2019.)
So, Cementos Bio Bio has an ROCE of 4.4%.
See our latest analysis for Cementos Bio Bio
### Is Cementos Bio Bio’s ROCE Good?
One way to assess ROCE is to compare similar companies. Using our data, Cementos Bio Bio’s ROCE appears to be significantly below the 8.0% average in the Basic Materials industry. This performance is not ideal, as it suggests the company may not be deploying its capital as effectively as some competitors. Putting aside Cementos Bio Bio’s performance relative to its industry, its ROCE in absolute terms is poor – considering the risk of owning stocks compared to government bonds. Readers may wish to look for more rewarding investments.
We can see that, Cementos Bio Bio currently has an ROCE of 4.4%, less than the 8.8% it reported 3 years ago. So investors might consider if it has had issues recently. You can click on the image below to see (in greater detail) how Cementos Bio Bio’s past growth compares to other companies.
It is important to remember that ROCE shows past performance, and is not necessarily predictive. Companies in cyclical industries can be difficult to understand using ROCE, as returns typically look high during boom times, and low during busts. ROCE is, after all, simply a snap shot of a single year. If Cementos Bio Bio is cyclical, it could make sense to check out this free graph of past earnings, revenue and cash flow.
### What Are Current Liabilities, And How Do They Affect Cementos Bio Bio’s ROCE?
Current liabilities include invoices, such as supplier payments, short-term debt, or a tax bill, that need to be paid within 12 months. Due to the way ROCE is calculated, a high level of current liabilities makes a company look as though it has less capital employed, and thus can (sometimes unfairly) boost the ROCE. To counter this, investors can check if a company has high current liabilities relative to total assets.
Cementos Bio Bio has current liabilities of CL\$54b and total assets of CL\$428b. As a result, its current liabilities are equal to approximately 13% of its total assets. This is not a high level of current liabilities, which would not boost the ROCE by much.
### The Bottom Line On Cementos Bio Bio’s ROCE
Cementos Bio Bio has a poor ROCE, and there may be better investment prospects out there. But note: make sure you look for a great company, not just the first idea you come across. So take a peek at this free list of interesting companies with strong recent earnings growth (and a P/E ratio below 20).
If you like to buy stocks alongside management, then you might just love this free list of companies. (Hint: insiders have been buying them).
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material. Thank you for reading. | 1,013 | 4,550 | {"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} | 2.59375 | 3 | CC-MAIN-2020-34 | latest | en | 0.935054 |
https://thefloatingschoolid.org/and-pdf/3624-pdf-and-cdf-probability-examples-in-the-real-world-724-505.php | 1,653,204,986,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00745.warc.gz | 648,471,521 | 9,741 | # Pdf And Cdf Probability Examples In The Real World
File Name: and cdf probability examples in the real world.zip
Size: 22915Kb
Published: 17.05.2021
## Using the cumulative distribution function (CDF)
Random Variables play a vital role in probability distributions and also serve as the base for Probability distributions. Before we start I would highly recommend you to go through the blog — understanding of random variables for understanding the basics. Today, this blog post will help you to get the basics and need of probability distributions. What is Probability Distribution? Probability Distribution is a statistical function which links or lists all the possible outcomes a random variable can take, in any random process, with its corresponding probability of occurrence.
Values o f random variable changes, based on the underlying probability distribution. It gives the idea about the underlying probability distribution by showing all possible values which a random variable can take along with the likelihood of those values. Let X be the number of heads that result from the toss of 2 coins. Here X can take values 0,1, or 2.
X is a discrete random variable. The table below shows the probabilities associated with the different possible values of X. The probability of getting 0 heads is 0. Simple example of probability distribution for a discrete random variable. Need of Probability Distribution. However, it lacks the capability to capture the probability of getting those different values. So, probability distribution helps to create a clear picture of all the possible set of values with their respective probability of occurrence in any random process.
Different Probability Distributions. Probability Distribution of discrete random variable is the list of values of different outcomes and their respective probabilities. In other words, for a discrete random variable X, the value of the Probability Mass Function P x is given as,.
If X, discrete random variable takes different values x1, x2, x3……. Example: Rolling of a Dice. If X is a random variable associated with the rolling of a six-sided fair dice then, PMF of X is given as:. Unlike discrete random variable, continuous random variable holds different values from an interval of real numbers.
Hence its difficult to sum these uncountable values like discrete random variables and therefore integral over those set of values is done. Probability distribution of continuous random variable is called as Probability Density function or PDF. Given the probability function P x for a random variable X, the probability that X belongs to A, where A is some interval is calculated by integrating p x over the set A i.
Example: A clock stops at any random time during the day. Let X be the time Hours plus fractions of hours at which the clock stops. The PDF for X is. And the density curve is given by. Cumulative Distribution Function. All random variables, discrete and continuous have a cumulative distribution function CDF.
Similarly if x is a continuous random variable and f x is the PDF of x then,. I hope this post helped you with random variables and their probability distributions. Probability distributions makes work simpler by modeling and predicting different outcomes of various events in real life.
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## What is Probability Density Function (PDF)?
Actively scan device characteristics for identification. Use precise geolocation data. Select personalised content. Create a personalised content profile. Measure ad performance.
The probability for a continuous random variable can be summarized with a continuous probability distribution. Continuous probability distributions are encountered in machine learning, most notably in the distribution of numerical input and output variables for models and in the distribution of errors made by models. Knowledge of the normal continuous probability distribution is also required more generally in the density and parameter estimation performed by many machine learning models. As such, continuous probability distributions play an important role in applied machine learning and there are a few distributions that a practitioner must know about. In this tutorial, you will discover continuous probability distributions used in machine learning. The relationship between the events for a continuous random variable and their probabilities is called the continuous probability distribution and is summarized by a probability density function , or PDF for short.
Cumulative Distribution Functions (CDF); Probability Density Function (PDF) of course, are often lacking even a mentionable fraction of such knowledge of the world. Such a function, x, would be an example of a discrete random variable. take on a infinite number of values within the continuous range of real numbers.
## Continuous Probability Distributions for Machine Learning
Say you were to take a coin from your pocket and toss it into the air. While it flips through space, what could you possibly say about its future? Will it land heads up? More than that, how long will it remain in the air? How many times will it bounce?
Random Variables play a vital role in probability distributions and also serve as the base for Probability distributions. Before we start I would highly recommend you to go through the blog — understanding of random variables for understanding the basics. Today, this blog post will help you to get the basics and need of probability distributions.
### Probability Distributions: Discrete and Continuous
These ideas are unified in the concept of a random variable which is a numerical summary of random outcomes. Random variables can be discrete or continuous. A basic function to draw random samples from a specified set of elements is the function sample , see?
a way to represent the probability distribution of such continuous variables, and the pur- There are many such examples in everyday life and also in research.
Recall that continuous random variables have uncountably many possible values think of intervals of real numbers. Just as for discrete random variables, we can talk about probabilities for continuous random variables using density functions. The first three conditions in the definition state the properties necessary for a function to be a valid pdf for a continuous random variable. So, if we wish to calculate the probability that a person waits less than 30 seconds or 0.
However, for some PDFs e. Even if the PDF f x takes on values greater than 1, i f the domain that it integrates over is less than 1 , it can add up to only 1. As you can see, even if a PDF is greater than 1 , because it integrates over the domain that is less than 1 , it can add up to 1. Because f x can be greater than 1. Check it out here.
The cumulative distribution function CDF calculates the cumulative probability for a given x-value. Use the CDF to determine the probability that a random observation that is taken from the population will be less than or equal to a certain value. You can also use this information to determine the probability that an observation will be greater than a certain value, or between two values.
Городские огни сияли, как звезды в ночном небе. Он направил мотоцикл через кустарник и, спрыгнув на нем с бордюрного камня, оказался на асфальте. Веспа внезапно взбодрилась. Под колесами быстро побежала авеню Луис Монтоно. Слева остался футбольный стадион, впереди не было ни одной машины.
- Включи свет. - Прочитаешь за дверью. А теперь выходи.
### Related Posts
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However, continuous models often approximate real-world situations very Definition. Let X be a continuous r.v. Then a probability distribution or probability. | 1,636 | 8,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-21 | latest | en | 0.862258 |
https://openturns.github.io/openturns/1.15/examples/reliability_sensitivity/gauss_product_experiment.html | 1,632,494,973,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00232.warc.gz | 476,865,311 | 4,697 | Create a Gauss product designΒΆ
In this example we are going to create a deterministic weighted design experiment using Gauss product.
[9]:
from __future__ import print_function
import openturns as ot
[10]:
# Define the underlying distribution, degrees
distribution = ot.ComposedDistribution([ot.Exponential(), ot.Triangular(-1.0, -0.5, 1.0)])
marginalDegrees = [15, 8]
[11]:
# Create the design
experiment = ot.GaussProductExperiment(distribution, marginalDegrees)
sample = experiment.generate()
[12]:
# Plot the design
graph = ot.Graph("GP design", "x1", "x2", True, "")
cloud = ot.Cloud(sample, "blue", "fsquare", "")
[12]: | 167 | 636 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-39 | latest | en | 0.641863 |
https://sunlimingbit.wordpress.com/2012/08/04/template/ | 1,529,939,845,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00403.warc.gz | 728,413,609 | 19,185 | $\mathbf{Problem:}$ Let stripe $S=\{z\in \mathbb{C}|0<\Re(z)<1\}$, $f$ is analytic in $S$ and continuous to the boundary of $S$. Suppose $|f(z)|\leq e^{C|z|^{2-\delta}}$ for some $C>0$ and $\delta>0$. Define $\displaystyle M(x)=\sup\limits_{y}|f(x+iy)|$. Then $M(t)\leq M(0)^{1-t}M(1)^{t}$ for $0\leq t\leq 1$, which means $logM(t)$ is a convex function.
$\mathbf{Proof:}$ First let us prove if $M(0)\leq 1$ and $M(1)\leq 1$, then $|f|\leq 1$.
Fix $\epsilon>0$, define $F(z)=e^{\epsilon z^2}f(z)$ on $S$, then $|F(z)|\to 0$ as $z\to \infty$. By the maximum principle, $\sup_S|F(z)|$ must occur on the edge of $S$.
$|F(z)|\leq e^{\epsilon z^2}$ for $z\in S$
Let $\epsilon \to 0$, we get $|f(z)|\leq 1$ for $z\in S$.
For the general case, define $F(z)=f(z)M(0)^{z-1}M(1)^{-z}$, then $|F(z)|\leq 1$ on $\partial S$. And $F$ has growth rate no more than some $e^{C|z|^{2-\delta}}$, so apply the above result, we know that $\sup_S |F(z)|\leq 1$, which means $\sup_S |f(z)|\leq M(0)^{1-z}M(1)^{z}$.
Let $z=t\in [0,1]$, we get the conclusion.
$\text{Q.E.D}\hfill \square$
$\mathbf{Remark:}$ $f$ can not grow too fast, otherwise this theorem fails.
Consider $\displaystyle f(z)=e^{e^{\pi(z-\frac 12)i}}$ on $S$. Easy to verify that $|f(z)|\leq 1$ on $\partial S$. But on $\Re(z)=\frac 12$, $f$ grows extremely fast. | 551 | 1,316 | {"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": 45, "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} | 3.984375 | 4 | CC-MAIN-2018-26 | latest | en | 0.672905 |
https://www.math-only-math.com/worksheet-on-fundamental-concepts-of-geometry.html | 1,713,379,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00537.warc.gz | 810,152,795 | 14,122 | # Worksheet on Fundamental Concepts of Geometry
Practice the questions given in the worksheet on fundamental concepts of geometry. The questions are related to the basic geometrical shapes with which we are already familiar with.
1. Fill in the blanks:
(i) is at point ____ on the line AB. (ii) is at point ____ on the line AB. (iii) is at point ____ on the line AB. (iv) is at point ____ on the line AB.
2. Draw line segments EF and GH which cut each other. Label the point which is on both line segments as point K.
3. Draw figures by joining the line segments named below:
(i) AB, BC and CA, (example is shown)
(ii) ZY, YX, XW and WZ
(iii) PQ, QR and RP
(iv) AB, BC, CD, DE and EA
4. Look for at least 3 objects of different shapes in your surroundings like classroom, garden, your home and prepare a table writing the names of objects against each shape.
Shape Name of objects
Rectangle
Square
Circle
5. Count how many circles, squares, rectangles and triangles are there in the given picture. Color different shapes in different color:
Circle – Blue Square – Red Rectangle – Green Triangle – Yellow
6. Name the points:(i) inside the circle (i) inside the triangle (i) inside the rectangle
7. Measure the lengths of AB and BC in centimeters.
(i) AB = _____ cm (ii) BC = _____ cm (iii) AB + BC = _____ cm
8. The figure is made up of:(i) _____ curved lines (ii) _____ straight lines
Answers for the worksheet on fundamental concepts of geometry are given below to check the exact answers of the basic questions on geometry.
1. (i) P
(ii) Q
(iii) R
(iv) S
4. Shape Name of objects
Rectangle pencil box, book and bench
Square chess board, square stamps, tiles on the floor
6. (i) the points inside the circle are A, B, D and F
(i) the points inside the triangle are D, E and F
(i) the points inside the rectangle are B, C and F
8. (i) 5 curved lines
(ii) 7 straight lines
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http://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-10-section-10-1-radical-expressions-and-radical-functions-exercise-set-page-686/27 | 1,524,311,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945143.71/warc/CC-MAIN-20180421110245-20180421130245-00316.warc.gz | 423,490,853 | 14,751 | ## Algebra: A Combined Approach (4th Edition)
$\frac{1}{2}$
$\sqrt[3]\frac{1}{8}=\frac{1}{2}$ because, $(\frac{1}{2})^3=\frac{1}{8}$ | 57 | 133 | {"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} | 3.640625 | 4 | CC-MAIN-2018-17 | latest | en | 0.524338 |
https://socratic.org/questions/given-point-6-8-how-do-you-find-the-distance-of-the-point-from-the-origin-then-f#624536 | 1,638,179,328,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00037.warc.gz | 580,783,482 | 5,650 | # Given point (6,8) how do you find the distance of the point from the origin, then find the measure of the angle in standard position whose terminal side contains the point?
The distance is given by Pythagoras: $r = \setminus \sqrt{{6}^{2} + {8}^{2}} = 10$
The angle is $\theta = \arctan \left(\frac{8}{6}\right) \approx {53.13}^{\circ}$ | 105 | 339 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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} | 3.765625 | 4 | CC-MAIN-2021-49 | latest | en | 0.761709 |
http://math.stackexchange.com/questions/189945/finding-the-probability-that-the-population-becomes-extinct-for-the-first-time-i | 1,455,082,050,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701158609.98/warc/CC-MAIN-20160205193918-00201-ip-10-236-182-209.ec2.internal.warc.gz | 145,542,256 | 17,197 | # Finding the probability that the population becomes extinct for the first time in the third generation.
I am trying to solve the following problem and am wondering firstly if my solution is correct and alternatively if there is a shorter way to compute this.
In a branching process the number offspring per individual has a binomial distribution with parameters 2, p. Starting with a single individual, what is the probability that the population becomes extinct for the first time in the third generation.
The PGF of the binomial distribution is $G(s) = (ps + q)^n$
The probability that the population becomes extinct for the first time in the third generation can be evaluated by the function $G(G(G(0))) - G(G(0))$ and more precisely it is meant to be the smallest non-negative solution to the above equation.
$$G(G(G(s))) - G(G(s)) = (p(p(ps + q)^2+ q)^2)^2 - (p(ps + q)^2+ q)^2$$ Solving and substituting $n = 2$, $q= 1-p$ and $s=0$ I got $$G(G(G(0))) - G(G(0)) = p^2(p^3 - 2p^2 + 1)^4 - (p(p - 1)^2 - p + 1)^2$$ which expands out
$$= p^{14} - 8p^{13} + 24p^{12} - 28p^{11} - 8p^{10} + 48p^9 - 26p^8 - 24p^7 + 23p^6 + 8p^5 - 12p^4 - 2p^3 + 5p^2 - 1$$
According to Matlab other than 1 most of the other roots are either greater than, negative or complex.
I am unsure how to interpret these results and i suspect there to be a mistake in my solution.
Any help would be much appreciated.
-
Your solution is almost corrrect. I get different coefficients for $p^8$ and below. Apparently, you left out a $+q$ in the outermost parentheses of the first summand. Plotting the graph of the correct expression $p^{14}-8p^{13} \pm\ldots$ for $0\le p\le1$ shows a nice curve that starts at 0, ends at 0 and has a single max inbetween. Why are you looking for roots? That would be values of $p$ such that extinction in exactly 3 generations is impossible and this is obviously the case only for $p=0$ or $p=1$. | 561 | 1,914 | {"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} | 3.921875 | 4 | CC-MAIN-2016-07 | longest | en | 0.899276 |
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Mathematics 2,754 rep 11031 Physics 165 rep 8 Stack Overflow 123 rep 3 Theoretical Computer Science 101 rep 1 MathOverflow 101 rep 13 | 497 | 1,530 | {"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} | 2.96875 | 3 | CC-MAIN-2016-30 | latest | en | 0.692525 |
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## Travel Time & Distance
Estimate distance and duration of a road trip.
Travel between two distant cities.
In[1]:=
```cities = {Entity["City", {"Lisbon", "Lisboa", "Portugal"}], Entity["City", {"Beijing", "Beijing", "China"}]};```
This is the geodesic distance between them.
In[2]:=
`GeoDistance[cities]`
Out[2]=
This is the length of the computed road trip.
In[3]:=
`TravelDistance[cities]`
Out[3]=
And this is the estimated driving time, assuming continuous driving with no stops.
In[4]:=
`TravelTime[cities]`
Out[4]=
This object contains the actual set of travel instructions.
In[5]:=
```td = TravelDirections[{Entity[ "City", {"Lisbon", "Lisboa", "Portugal"}], Entity["City", {"Beijing", "Beijing", "China"}]}]```
Out[5]=
Represent the trajectory (in red) on a Mercator map and compare with the geodesic trajectory (in blue), which is shorter, as you saw before.
In[6]:=
```GeoGraphics[{Thick, Red, GeoPath[td], Blue, GeoPath[{Entity["City", {"Lisbon", "Lisboa", "Portugal"}], Entity["City", {"Beijing", "Beijing", "China"}]}]}, GeoProjection -> "Mercator", GeoGridLines -> Automatic]```
Out[6]=
An azimuthal projection shows more clearly that the geodesic is shorter than the travel path.
In[7]:=
```GeoGraphics[{Thick, Red, GeoPath[td], Blue, GeoPath[{Entity["City", {"Lisbon", "Lisboa", "Portugal"}], Entity["City", {"Beijing", "Beijing", "China"}]}]}, GeoProjection -> "Mercator", GeoGridLines -> Automatic]; Show[%, GeoProjection -> "LambertAzimuthal", GeoZoomLevel -> 4]```
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http://steamcommunity.com/id/God_ownerOfTheUniverse/?l=schinese | 1,488,019,196,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171706.94/warc/CC-MAIN-20170219104611-00532-ip-10-171-10-108.ec2.internal.warc.gz | 235,381,750 | 13,144 | God, owner of the Universe
God Germany
Starting November 2016, I'm using a review score system that scales with my certainty. Imgur album with two images (exported by SlaloM ) showing the concept: https://imgur.com/a/CYdpn
Examples of what this entails (some of which are obviously ridiculous):
[*0*] = ~50% (0%-100%)
[*0* | 1] = ~25% (0%-50%)
[0 | *1*] = ~75% (50%-100%)
[*0* | 1 | 2] = ~17% (0%-33%)
[0 | *1* | 2] = ~50% (33%-67%)
[0 | 1 | *2*] = ~83% (67%-100%)
[*0* 1 | 2 3] = ~12% (0%-25%)
[0 *1* | 2 3] = ~38% (25%-50%)
[0 1 | *2* 3] = ~62% (50%-75%)
[0 1 | 2 *3*] = ~88% (75%-100%)
[*0* 1 | 2 | 3 4] = ~10% (0%-20%)
[0 *1* | 2 | 3 4] = ~30% (20%-40%)
[0 1 | *2* | 3 4] = ~50% (40%-60%)
[0 1 | 2 | *3* 4] = ~70% (60%-80%)
[0 1 | 2 | 3 *4*] = ~90% (80%-100%)
etc.
(The "|" separates the thumbs-down from the thumbs-up half. If there are two "|", that's because there's middle ground (which there isn't when it comes to Steam's either-or reviews).
The *stars* surround the score that was chosen.
The percentage range is a consequence of the coarseness: If we'd grow the scale to [0 ... 100] and mark the 14, the text after the equals would read "~14% (14%-14%)".
The "~30%" is just the middle of (20%-40%). I mean, if I could give a more decisive score here, like e.g. 34%, then this would imply that I could increase the scale in the first place, which I didn't so the ~30% are all I can say and the"~" is very important. "ROUGHLY." Sometimes "VERY ROUGHLY."
It's a coarse scale with variable certainty, and it must also be read as such. It's not useless or whatever, it's instead closer to the truth than just throwing out a "85%!" statement. It also makes it easier for the reviewer, because you don't have to ponder until the end of time until you have the answer that's required for the bottom of the article, you can make and refine the statement as you go.
............................................................................................................
Starting with "Deus Ex: Mankind Divided" (August 2016), every game whose FieldOfView is too low will get a negative review from me. This game itself is a 3 of 4 in my book (1 or 2 of 4 would be negative), so the reason is not quite the game itself.
See my relatively short article here on why exactly.
Too low FOV has been a problem in too many PC games and is unacceptable in light of the fact that it's just a little number that needs to be increased to make the experience of tens of thousands of players significantly better: Via Cheat Engine hack [pcgamingwiki.com], I have been playing DX:MD with proper FOV without crashes, look-through-walls side-effects, low performance, or weird geometry. It just works. The decision to give us low FOV without hard reason is either that of a malicious person , or it is a sign that the developers do not understand the medium they are developing for. It's definitely one of the two.
And due to the fact that large screens become more and more affordable (else I wouldn't be playing on a 16:9 49" at 23" distance) and therefore more prevalent, starting with this game, I draw the vote-line at FOV, period.
............................................................................................................
It's pretty hard to lead by example while invisible.
Btw., just take my name literally, ok.
SlaloM users - 公共组
Manage your game library like a pro!
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about FOV (field of view) in computer games
my "low FieldOfView = thumbs down" policy
Compendium for the perfectionist game developer
5
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93 小时（记录在案的）
200 点经验值
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## #1 2006-02-20 15:09:14
ineedhelp
Novice
Offline
### Rational Expressions -- Homework :(
1. Find all values of x that make the rational expression undefined:
(xsquared -9)
----------------
(xsquared +13x +36)
(x+3) (x-3)
---------------
(x+4) (x-4)
2. Simplify:
4
(3r-1 )
---------------
4
----- +4
(3r-1)
4 (3r-1)
----- -4 times ------- +4
(3r-1 ) -4
12r-4
------- -16
12r-4
3. Simplify:
6x to the 8th - 10x to the 6th
----------------------------------
-2x to the 8th
2x to the 6th (3xsquared -5)
----------------------------------
2x to the 8th
Answer: x to the 3rd (3xsquared -5)
4. The average number of vehicles waiting to pay a toll at the toll booth of a super highway is modeled by the function
n(x) = xsquared
--------------------
.5(1-x)
where x is a quantity between 0 and 1 known as traffic intensity. What happens to the average number of vehicles waiting as the value representing the traffic intensity increases? Explain your answer.
A) The average number of vehicles waiting decreases at first, but then increases.
B) The average number of vehicles waiting remains constant.
Answer: C) The average number of vehicles waiting increases.
D) The average number of vehicles waiting decreases.
5. Perform the indicated operation and simplify.
4p-4 5p-5
------ divided by ------
p 3psquared
4p-4 3squared
----- times -----------
p 5p-5
3psquared +4p -4
---------------------
5psquared - 5p
(3p+2) (3p-2)
----------------
p
I have a feeling I got all of these wrong. Any help would be greatly appreciated. Thanks.
Also, how do you all get the complex formula's and exponents and all to show up in the forums? I can't seem to figure that one out.
## #2 2006-02-21 03:16:59
kempos
Full Member
Offline
### Re: Rational Expressions -- Homework :(
1. Denominator: (x+4)(x+9)
kempos
Full Member
Offline
3. (3x^2-5)/x^2 | 643 | 2,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2014-10 | longest | en | 0.754757 |
https://www.physicsforums.com/threads/trig-question.464399/ | 1,656,748,627,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00359.warc.gz | 977,731,741 | 15,028 | # Trig Question
## Homework Statement
Prove:( (1 - cos A)/ (1+cos A) ) ^(1/2) = cosec A - cot A
Then they have.....
((1 - cos A)/ (1+cos A) ) ^(1/2) = ( (1 - cos A)^2 )/ (1-cos^2 A) ) ^(1/2)
= (1 - cos A) / (1-cos^2 A) ^(1/2)/
I have never done anything like this... I have just been studying trig for the past two days.....
Why did they square the entire numerator but only square the cos in the denominator?
... Can some help me understand this the easiest way possible? thank you.
## The Attempt at a Solution
Last edited by a moderator:
Nope lol sorry... However I did figure it out on my own by just looking at it! So I was proud about that lol....
One important question that I would like answered......
Lets say..... I have.... (1 - cos A)/ (1+cos A) ) ^(1/2)
Should I leave it as so... or try and simplify it down to cosec A - cot A?
Because to be honest I dont think I could have looked at the square root portion and though " Ohh... maybe this can be simplified..."
Last edited:
HallsofIvy
Homework Helper
## Homework Statement
Prove:( (1 - cos A)/ (1+cos A) ) ^(1/2) = cosec A - cot A
Then they have.....
((1 - cos A)/ (1+cos A) ) ^(1/2) = ( (1 - cos A)^2 )/ (1-cos^2 A) ) ^(1/2)
= (1 - cos A) / (1-cos^2 A) ^(1/2)
Once you are here, use the fact that $1- cos^2(A)= sin^2(A)$
(From $sin^2(A)+ cos^2(A)= 1$. Do you know that identity?)
$\left(\frac{1- cos(A)}{1- cos^2(A)}\right)^{1/2}= \frac{1- cos(A)}{sin(A)}= \frac{1}{sin(A)}- \frac{cos(A)}{sin(A)}$
Do you know the definition of "cosec(A)" and "cot(A)"?
I have never done anything like this... I have just been studying trig for the past two days.....
Why did they square the entire numerator but only square the cos in the denominator?
They didn't. What they did is multiply both numerator and denominator by $1- cos(A)$. Since there was already $1- cos(A)$, that becomes $(1- cos(A))^2$. The denominator was $1+ cos(A)$ so it becomes $(1- cos(A))(1+ cos(A))= 1+ 1(cos(A))- cos(A)(1)- cos(A)cos(A)= 1- cos^2(A)$ because the "$1(cos(A))$"and "[mat]-cos(A)(1)[/itex] cancel.
... Can some help me understand this the easiest way possible? thank you.
## The Attempt at a Solution
Last edited by a moderator:
Ive been reading in my book.. and it says,
sin^2 A + cos^2 A = 1
Sec^2A = 1 + tan^2A
.
.
. Should I also rearange sin^2 A + cos^2 A = 1 ; cos^2 A = 1 -sin^2 A = sin^2 A =1 -cos^2..
and the same with the other one?
Should I not only memorize the two but also memorize their rearangements? | 807 | 2,482 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.919 |
https://www.themagiccafe.com/forums/printtopic.php?topic=163110&forum=99 | 1,624,536,243,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00384.warc.gz | 913,892,517 | 2,165 | Topic: How does this work ?
Message: Posted by: DavidF (May 17, 2006 11:32AM)
Has anyone seen this web site ? http://digicc.com/fido/ Im sure there is a simple answer to this puzzle but I cannot work it out.I don't want the answer I would just like to know if anyone knows how it works
Message: Posted by: landmark (May 17, 2006 03:08PM)
Cute website. It's a very old math effect. I don't think anyone will mind if I tell you that when you do that kind of subtraction, the digits must add up to a multiple of nine. Lots of other fun math stuff in this forum, check it out.
Jack Shalom
Message: Posted by: Doctor Whoston (May 17, 2006 03:10PM)
I do!
DW
Message: Posted by: Slim King (May 17, 2006 07:46PM)
Relax there Jack! That's my favorite secret...LOL
Message: Posted by: TomasB (May 18, 2006 02:38AM)
That is definitely my favourite way to force a number to be a multiple of nine.
Here's a hint to why it works (n>m): a*10^n-a*10^m=a*10^m(10^(n-m)-1)=9*a*10^m*111...111 where the last number has n-m ones in it.
/Tomas
Message: Posted by: landmark (May 18, 2006 02:58PM)
[quote]Here's a hint to why it works (n>m): a*10^n-a*10^m=a*10^m(10^(n-m)-1)=9*a*10^m*111...111 where the last number has n-m ones in it. [/quote]
That's easy for you to say . . . :)
Message: Posted by: roi_tau (Jun 23, 2006 10:16AM)
What is it about the nine that so many tricks involve him.
will it work with 7 in a octali basis?
Have fun
ROi
Message: Posted by: Nir Dahan (Jun 23, 2006 11:03AM)
[quote]
On 2006-06-23 11:16, roi_tau wrote:
What is it about the nine that so many tricks involve him.
will it work with 7 in a octali basis?
Have fun
ROi
[/quote]
yes
Gardner (who else) has written about it.
N | 553 | 1,701 | {"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} | 3.75 | 4 | CC-MAIN-2021-25 | latest | en | 0.942861 |
https://www.peruzinasi.com/form-six-mathematics-study-notes-vector-analysis-1/ | 1,669,554,465,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00027.warc.gz | 1,010,395,063 | 26,678 | # FORM SIX MATHEMATICS STUDY NOTES VECTOR ANALYSIS-1
FORM SIX MATHEMATICS STUDY NOTES VECTOR ANALYSIS-1
UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU:0787237719
## FORM SIX MATHEMATICS STUDY NOTES VECTOR ANALYSIS-1
Definition
Vectors are any quantity that possess both magnitude and direction .
Example
(i) Displacement
(ii) Velocity
(iii) Acceleration
REPRESENTATION OF A VECTOR
The vector quantity is always described by using two capital letters with respect to arrow on top or small letters with respect to bar at the bottom.
n
A B
or
COMPONENTS OF VECTORS.
This depends on the dimension of vector as follows;
(i) For two dimensions namely as and
Where =x- value
= y – value
e.g. = (x,y) coordinate form
component form
Diagram
For three dimensions
Re:
This involves three components namely as , and
Where,
= x – value
= y – value
TERMINOLOGIES APPLIED IN VECTOR ANALYSIS
1. PARALLEL VECTORS
These are vectors having the same direction.
e.g
2. EQUAL VECTORS
These are vectors having the same magnitude and direction.
e.g.
= 5N
= 5N
3. NEGATIVE (OPPOSITE) VECTORS
These are vectors having the same magnitude but opposite direction.
Hence
(i) =- (vector in opposite direction
(ii) = = b vector on the same direction
FREE VECTORS
These are vectors which originate from different points.
POSITION VECTORS
These are vectors which originate from the same points.
NB:
– Position vector of = –
– Position vector of = –
– Position vector of = –
– Position vector of = –
NULL VECTORS
These are vectors which have a magnitude of zero ( length of zero) or
These are vectors which contain zero point
Eg.
= (0,0)
= (0,0,0)
COLLINEAR VECTORS
– These are vectors which lie on the same line.
i.e.
COPLANAR VECTORS
These are vectors which lie on the same plane.
, and are coplanar vector
NB:
Consider the vector diagram below;
Where
= initial (starting) point
B = Final (terminal) point
OPERATION IN VECTORS
These are
(ii) Subtraction
(iii) Multiplication
Suppose two dimensional vectors
Suppose three dimensional vectors
+
k
RESULTANT VECTOR
Is the single vector which represents the effect of all vectors acting at a point.
Consider are acting at a point
Hence
Where F is the result force/ vector
A: TRIANGULAR LAW OF VECTORS APPLICATION
Consider the vector diagram below;
+ – = 0
= +
= +
Where r is the resultant vector
B. PARALLELOGRAM LAW OF VECTORS ADDITION
– Consider the vector diagram below
+ = …………………………(i)
+ – = 0
+ = ……………………………(ii)
From (i) and (ii) above
Proved
(ii) Addition of vectors is associative for any three vectors a, b and c
Proof
Consider the vector diagram below;
Individual but not considered
=
=
=
……(i)
…………………..(ii)
From (i) and (ii) above
=
Proved
For every vector a, we have;
Where;
0 The null (zero) vector
For every vector a we have
Where
→ is the positive of vector
→ is the null vector
ii. SUBTRACTION OF VECTORS
Suppose two dimensional vectors
Hence
=
=
– Suppose three dimensional vectors
Hence
Question 1
1. If
(a Find (i)
(ii)
Comment of the results in (a) above
Question 2
Given that
(i) Find
(ii)
(b) Comment on the results in a above
MAGNITUDE OF A VECTOR
– The magnitude of a vector is a measure of length of the vector.
– – This is denoted by the symbol
(a) Consider two dimensional vector
By using Pythagoras theorem
Recall;
Where
– is the magnitude/ module of the vector r
(b) Consider three dimensional vector
RECTANGULAR RESOLUTION OF A VECTOR
Let: be three rectangular axes and be three unit vectors parallel to axes respectively.
Consider
+
+
Also consider the right angled OFP
Using Pythagoras theorem
i.e a2 + b2 = c2
Where
= is the magnitude of the vector
Question
Given that
Find
DIRECTION RATIO AND DIRECTION COSINES
I. DIRECTION RATIO
Suppose the vector
The direction ratio is given by
II. DIRECTION COSINE
Consider the vector
From three dimension plane.
Makes angles with direction respectively
Hence
Therefore the direction cosines are
FACT IN DIRECTION COSINES
– Suppose the vector
Also the direction cosines are
Hence
++= 1
– The sum of square of the direction cosines is one.
Proof
i.e = x+y+z
=
Also
—————-(i)
———————-(ii)
—————————-(iii)
Adding the equation (i) , (ii) and (iii)
+= +
=
But
+=
+
UNIT VECTOR
-Is the vector whose magnitude (modules) is one line a unit
-The unit vector in the direction of vector a is donated by read as “ a cap” thus
NOTE:
Any vector can be compressed as the product of it’s magnitude and it’s unit vector
i.e
QUESTIONS
1. Find a vector in the direction of vector which has a magnitude of 8 units
2. Find the direction ratio and direction cosines of the vector where p is the point (2, 3, -6)
THE FORMULA OF DISTANCE BETWEEN TWO POINTS
Suppose the line joining the points and whose position vectors are a and b respectively
HENCE
=
=
Hence
-Formula distance between two point
MID POINT OF A LINE
Suppose M is the point which divide the line joining the points and whose position vectors are respectively a and b into two equal parts
i.e
a =
b =
Hence
Therefore
The co-ordinate of M is
INTERNAL AND EXTERNAL DIVISION OF A LINE (RATIO THEOREM)
I. INTERNAL DIVISION OF A LINE
– Suppose M- is the point which divides the line joining the points and whose point vectors are a and b respectively internally in the ratio X:ee
a = =
b = =
……………(i)
……………(ii)
By using ratio theorem
By using multiplication
The ordinate form of M is
II. EXTERNAL DIVISION OF A LINE
Suppose M- is the point which divides the line joining the points and where position vectors are respectively , externally is the ration
=
……………(i)
……………(ii)
By using ratio theorem
i.e
BY CROSSING MULTIPLICATION
Therefore
The co-ordinate of M
External division of a line where
QUESTIONS
1.Find the length of the line of
2. Find the position vector which divides line having point into two equal points.
3. A and B are two points whose vectors are 3 + and respectively. Find the position vector of the points dividing AB.
(a) Internally in the ratio 1:3
(b) Externally in the ratio 3:1
III. MULTIPLICATION OF A VECTOR
(A) SCALAR MULTIPLICATION OF A VECTOR
In this case a vector is multiplied by a certain constant called scalar
Let
THEREFORE
QUESTIONS
If
(a) Find (i)
(ii).
(b) Comment on results in above
DEFINITION OF DOT PRODUCT
For vectors
Where
is the above between
cos Q
cos Q
CHARACTERISTICS
1. 1. PARALLEL VECTOR
Two vector are said to be parallel of the angle between them is zero
Mathematically
From
= cos Q
But Q =
= cos Q
This is one among the
2. Orthogonal vectors
Two vectors are said to be orthogonal of the angle between them is 90
Mathematically
From
But Q = 90 (orthogonal or perpendicular vector)
= cos 90
This is conditional for the orthogonal vector
THEOREM.
(a)For the definition
cos
=(1,0)
=(1,0)
Therefore
From the definition
cos
cos 90
x 0
Therefore
– Suppose the vector
=
= + ++
=
From the definition
= cos
= cos
– Consider the vector
QUESTIONS
1. If + 2+2
Find the angle between and
2. Show that the vectors
=
are orthogonal
3. If =2, =3
, find
4. The vectors and
where k are such that
and are orthogonal find k
5. If and
=
Find the value of if and are orthogonal
6.If =2, =3,
Find
APPLICATION OF DOT PRODUCT
1. TO VERIFY COSINE RULE
Consider the vector diagram below
+=
…………………..(i)
Dot by
. =
=
= + | 2,159 | 7,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-49 | latest | en | 0.708999 |
http://www.theanalysisfactor.com/missing-data-two-recommended-solutions/ | 1,506,119,672,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689373.65/warc/CC-MAIN-20170922220838-20170923000838-00588.warc.gz | 579,648,428 | 15,736 | # Two Recommended Solutions for Missing Data: Multiple Imputation and Maximum Likelihood
by
Two methods for dealing with missing data,vast improvements over traditional approaches, have become available in mainstream statistical software in the last few years.
Both of the methods discussed here require that the data are missing at random–not related to the missing values. If this assumption holds, resulting estimates (i.e., regression coefficients and standard errors) will be unbiased with no loss of power.
The first method is Multiple Imputation (MI). Just like the old-fashioned imputation methods, Multiple Imputation fills in estimates for the missing data. But to capture the uncertainty in those estimates, MI estimates the values multiple times. Because it uses an imputation method with error built in, the multiple estimates should be similar, but not identical.
The result is multiple data sets with identical values for all of the non-missing values and slightly different values for the imputed values in each data set. The statistical analysis of interest, such as ANOVA or logistic regression, is performed separately on each data set, and the results are then combined. Because of the variation in the imputed values, there should also be variation in the parameter estimates, leading to appropriate estimates of standard errors and appropriate p-values.
Multiple Imputation is available in SAS, S-Plus, R, and now SPSS 17.0 (but you need the Missing Values Analysis add-on module).
The second method is to analyze the full, incomplete data set using maximum likelihood estimation. This method does not impute any data, but rather uses each cases available data to compute maximum likelihood estimates. The maximum likelihood estimate of a parameter is the value of the parameter that is most likely to have resulted in the observed data.
When data are missing, we can factor the likelihood function. The likelihood is computed separately for those cases with complete data on some variables and those with complete data on all variables. These two likelihoods are then maximized together to find the estimates. Like multiple imputation, this method gives unbiased parameter estimates and standard errors. One advantage is that it does not require the careful selection of variables used to impute values that Multiple Imputation requires. It is, however, limited to linear models.
Analysis of the full, incomplete data set using maximum likelihood estimation is available in AMOS. AMOS is a structural equation modeling package, but it can run multiple linear regression models. AMOS is easy to use and is now integrated into SPSS, but it will not produce residual plots, influence statistics, and other typical output from regression packages.
References:
Schafer, J. Software for Multiple Imputation
Hox, J.J. (1999) A Review of Current Software for Handling Missing Data, Kwantitatieve Methoden, 62, 123-138.
Allison, P. (2000). Multiple Imputation for Missing Data: A Cautionary Tale, Sociological Methods and Research, 28, 301-309.
Want to learn more about how to handle Missing Data? In this 5-part workshop, you’ll learn all about two fabulous modern missing data techniques: multiple imputation and maximum likelihood.
Lau August 25, 2017 at 12:21 pm
Hi Karen,
I have the same problem as LF. I’m doing an Exploratory Factor Analysis and just 27 of all 198 participants completed every item. So I did a multiple imputation. But now I’m struggeling how to run the factor analysis.
For any advice I would be very very thankful!!
hosein April 26, 2016 at 1:45 pm
hello – i am working in mineral exploration field -Do Cohen likelihood maximum Method for censored (missing) data replacement use for Geochemical data Now?
Emily Stone July 15, 2015 at 9:27 am
Hello! I am doing Asymptotically distribution free estimation in AMOS due to a data set that is not normal and has ordinal data. I am trying to determine how to handle missing data with this type of estimation in AMOS. Can you do multiple imputation in AMOS? Thank you so much!
Karen July 18, 2015 at 9:54 am
Hi Emily,
AMOS doesn’t do multiple imputation, but you don’t need it to. It does maximum likelihood. You might find this helpful, though it’s not exactly what you’re doing:
How to Use Full Information Maximum Likelihood in AMOS to Analyze Regression Models with Missing Data
LF April 7, 2015 at 12:24 pm
Thanks Karen. Any help to the above question about the difference in MPlus and AMOS is much appreciated.
I am struggling with dealing with missing data and doing an Exploratory Factor Analysis with a complete dataset. I thought perhaps I could do Multiple Imputation in SPSS and do the EFA there but I don’t think it is one of the supported analyses for pooled data. Any suggestions how to use MI in an EFA in SPSS or do I have to switch to another software? Any help is much appreciated.
Thank you.
LF March 10, 2015 at 7:37 am
Hello Karen,
In AMOS, when you use ML estimation with missing data, it says that the full sample is used. I’ve recently tried using MPlus and when it runs there, it says it takes out those cases from the analysis that doesn’t have any data on those variables. If it’s the same estimation method for missing data between the two packages, then why would it come out different. Is AMOS doing the same just not telling us it’s based on part of the sample?
Thank you.
Karen March 23, 2015 at 12:28 pm
Hi LF,
I don’t know MPlus, so I’m not sure what it is doing. AMOS isn’t dropping cases for having some missing data. I would suggest looking into the defaults in MPlus. Perhaps you just need to change an option.
Any Mplus users want to chime in?
Patrick Onyeneho January 18, 2015 at 3:19 am
How do i implement the add on of missing data using the .ML method in spss
Thank you
Michael December 25, 2014 at 8:25 pm
Thanks Karen for the R free resource website..
Hi Peng, If you are looking for some case studies in R with real world proven examples you can try for some free classes at http://my-classes.com/
there are practice tests also available to self assess your knowledge.
kaushal Chaudhary March 12, 2014 at 2:14 pm
Hi Karen,
SAS also used ml (maximum likelihood) or reml (restricted maximum likelihood) method for parameter estimation. Does this mean it also impute missing values in the data? So, if there are missing observartions, we do not have to impute. Thanks for your clarification.
Karen April 4, 2014 at 9:56 am
Hi Kaushal,
ML isn’t imputing. But yes, you can use SAS proc calis for missing predictors in a linear model or proc mixed for missing outcome values in a multilevel model.
Dong November 1, 2013 at 7:28 pm
I am looking into how to run an MLE. Can SPSS 20 run an MLE in it’s easy-to-use pull-down menus or can this only be done via syntax? Thank you!
Karen November 8, 2013 at 11:38 am
It can. It is based on the analysis, however. What kind of model are you looking for?
peng January 30, 2010 at 1:49 am
hi friends,
I am new to R.I would like to know R-PLUS.Does any know where can I get the free training for R-PLUS.
Regards,
Peng.
Karen September 18, 2012 at 9:08 am
Hi Peng,
If you need free, I would suggest: http://www.ats.ucla.edu/stat/r/
Karen | 1,680 | 7,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-39 | longest | en | 0.835332 |
https://brainmass.com/physics/partial-differential-equation/partial-differential-equations-heat-equation-579736 | 1,656,786,020,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00396.warc.gz | 216,015,136 | 75,444 | Explore BrainMass
# Partial Differential Equations - Heat Equation
Not what you're looking for? Search our solutions OR ask your own Custom question.
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The causal solution f(x t) of the one-dimensional diffusion equation with source term f
See attached
https://brainmass.com/physics/partial-differential-equation/partial-differential-equations-heat-equation-579736
## SOLUTION This solution is FREE courtesy of BrainMass!
The causal solution of the one-dimensional diffusion equation with source term f
(1)
is given by
(2)
where is the causal Green function satisfying
(3)
a) Change variables in (3) to , . Letting with the Heaviside step function , show (3) is solved if satisfies
(4)
Solve by Fourier transforming with respect to the position variable. Hence obtain an expression for
b) The temperature of a long uniform insulating metal rod satisfies
A heating element is used to create an initial temperature distribution Writing obtain an equation for , and combining (1), (2) and the answer to a) identify the temperature variation at the center of the rod, .
Solution:
a) After changing the variables, equation (3) becomes:
(5)
and putting
(6)
one yields:
(7)
where the following property of Dirac delta distribution was used:
(7')
The other partial derivative will be:
(8)
By replacing (7) and (8) in (5), we will have:
(9)
which can be written also as
(10)
This expression is equivalent to the following equations simultaneously satisfied:
(11)
(proof of statement (4))
One applies now the Fourier transform on the homogeneous equation (11) with respect to variable "z", using the formula:
(12)
and the property of derivative:
(13)
It follows that
(14)
and (15)
Applying the Fourier transform to the initial condition, we will get:
(16)
where the Fourier transform applied to Dirac delta distribution yields "1".
The equation (11) becomes a simple ordinary differential equation of 1'st order:
(17)
whose general solution is:
(18)
By applying the initial condition, it follows that C0 = 1, so that
(19)
The inverse of Fourier transform (12) is
(20)
and, if applied to (19), yields:
In the last integral, the variable change:
(21)
yields
(22)
Substituting in (6), the expression of Green function will be found:
(23)
or
(24)
b) It is required to solve the p.d. equation:
(25)
with the initial condition
(26)
One considers the function:
(27)
whose partial derivatives are:
(28)
(29)
It follows that
Since the last parenthesis is null (according to (25)), the equation to be solved is:
(30)
which is a inhomogeneous p.d. equation of general form
(31)
The solution of this kind of equation is expressed by formula:
(32)
where the Green function G is given by (24) and
(33)
Replacing with the appropriate expressions, the integral (32) becomes:
(34)
Using again the property (7') of Dirac distribution as well as the property
(35)
the integral (34) becomes:
(36)
Comparing to (27), one deduces that
(37)
The temperature in the center of the rod will be given by:
(38)
Using a new variable change, the Poisson integral will be again obtained:
(39)
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! | 791 | 3,286 | {"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} | 3.9375 | 4 | CC-MAIN-2022-27 | longest | en | 0.873304 |
https://www.kidpid.com/combine-tens-ones-worksheets-for-grade-1/ | 1,685,328,615,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00473.warc.gz | 917,939,196 | 29,082 | # Combine tens & ones worksheets for Grade 1
In the abstract science of number and quantity – mathematics, addition is one of the four basic operations. The addition of two numbers results in the sum of those values combined. The addition is usually signified by the “+” symbol, hence the sum of two numbers ‘a’ and ‘b’ would be represented as a+b.
The addition is one of the most fundamental concepts of mathematics. 1+1 is 2. This is the very first addition that we learn. It is a concept that is not only important in mathematics but is essential for a basic understanding of our life. Take for instance a person who goes to a grocery shop to buy some groceries. He’ll then have to use addition to calculate the price he has to pay for the things he bought.
The worksheet allows for practicing the addition of whole numbers. It is meant for beginners who are just getting started with it.
## Combine Tens & Ones Practice Worksheets
Hence none of the sums in the worksheet use the concept of regrouping since we’ll be adding tens and ones in the following worksheets.
This is the second sheet. All of the worksheets consist of 12 questions. Allow your students ample time to learn in an easy manner. Leaning is a process and it isn’t complete without making mistakes. Mistakes often result in a better understanding of what is being taught.
It is a general assumption that mathematics is hard. Many kids fear and hesitate to try solving sums for the fear of making mistakes. It is necessary that they understand that to err is human. The third sheet allows them to make mistakes and learn from them.
If you think your student is getting good at doing sums, challenge him by asking him to perform the calculations in a limited amount of time. It will challenge him and allow for better concentration and learning as kids by nature are competitive and keen on learning.
This is the final sheet of the worksheet. By now your student must have a grasp of the fundamentals of addition. Test his understanding by this worksheet.
## Colorful Rainbow Crafts for Kids
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+ | 612 | 2,916 | {"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} | 3.75 | 4 | CC-MAIN-2023-23 | longest | en | 0.969277 |
https://www.storyofmathematics.com/glossary/concave/ | 1,716,844,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00630.warc.gz | 876,321,429 | 41,994 | # Concave|Definition & Meaning
## Definition
Concave is defined as a shape that has an inward curve. Any shape with an internal angle greater than 180° has a concave surface. It has the word “cave” that tells that a concave shape will be like the entrance of a cave.
For example, in a hexagon, if any two sides are curved inwards, then it is called a concave hexagon. Figure 1 shows a regular hexagon and a concave hexagon.
Figure 1 – A Regular Hexagon and a Concave Hexagon
## Lens
A lens is defined as a transparent optical device that is used to refract light rays. They focus at a point to form an image of an object.
There are two types of lenses; simple and compound lenses.
A concave lens is a simple lens as it only consists of a single lens whereas a compound lens consists of more than one lens in its geometry.
## Concave Lens
A concave lens has a thin middle area and is thicker in the outer edges. Figure 2 shows a concave lens.
Figure 2 – A Concave Lens
It refracts light rays to disperse them in all directions. It has a variety of uses including the treatment of myopia that is shortsightedness.
## A Diverging Lens
The concave lens is also known as the diverging lens. This is because it diverges the light rays that are refracted through it.
Figure 3 shows light rays passing through the concave lens.
Figure 3 – Light Rays Diverging Through a Concave Lens
In figure 3, F is the concave lens’s focal point or focus. It is the point where all the light rays coincide when traced backward. Thus, the concave lens focuses the light rays at point F.
The principle axis is the axis that passes through the focal point and the optical center of the concave lens.
## Image Formed by a Concave Lens
The image of the object formed by the concave lens is real or virtual(diminished) and erect(upright). Figure 4 shows a virtual image formed by a concave lens.
Figure 4 – A Virtual Upright Image Formed by a Concave Lens
The image is formed between the focal point F and the optical center C of the concave lens.
The image is upright and inward. It is also virtual or diminished, meaning the image appears smaller than the object’s size.
## Focal Length of a Concave Lens
The distance from the center to the focal point F of a lens is called the focal length denoted by “f”.
It can be calculated by using the formula given below:
$\frac{1}{f} = \frac{1}{i} \ – \ \frac{1}{o}$
Where i and o are the distances of the image and object respectively from the center of the lens.
Figure 5 shows the focal length, object distance, and image distance when light rays are refracted through the concave lens.
Figure 5 – Demonstration of Focal Length, Image and Object Distance of a Concave Lens
The concave lens’s focal length is always negative by sign convention. It is because the concave lens diverges the light rays. It is on the lens’s left side, which also makes it negative.
That is why it is also called a negative lens.
## Magnification
Magnification is the ability to enlarge the object’s apparent size from its original size. It is calculated by using a ratio thus magnification has no units.
A magnification of one means that the object’s size and the image size are the same.
A magnification number greater than one means that the image is enlarged and a magnification less than one means that the image is diminished or de-magnified.
A concave lens always has a magnification of less than one. This is because the image formed is always smaller than the object’s size.
Magnification can be calculated by using the formula given below:
$m = \frac{ h_{i} }{ h_{o} }$
Where,
$h_i$ = height of the image
$h_o$ = height of the object.
Height is the vertical distance. Another formula for magnification is:
$m = \frac{i}{o}$
Where i is the image distance and o is the object distance.
## Applications
There are many applications of concave lenses one of which the most common are eyeglasses for myopia correction.
The eyeball with myopia is elongated which can’t focus light from far away objects. The concave lens diverges the light rays coming to the eyeball which helps it make a clearer image of distant objects.
Concave lenses are also used in binoculars, compound microscopes, cameras, spy holes, mobile phones, telescopes, and flashlights.
## Types of Concave Lens
Different variations and additions to concave lenses produce the following types.
### Plano-concave Lens
A Plano-concave lens has a plane surface on one side and a concave surface on the other. It has a negative focal length like the concave lens.
### Biconcave Lens
A biconcave lens is a compound lens and has two concave lenses joined together. Both lenses have the same optical center.
### Convexo-concave Lens
A Convexo-concave lens has one concave surface and one convex surface. The concave side is more curved as compared to the convex side.
Figure 6 shows these three lenses.
Figure 6 – Types of Concave Lens
## A Solved Problem Involving a Concave Lens
A concave lens forms an image at 7 cm from the optical center. It has a focal length of 14 cm. Find the distance of the object from the lens. Also, calculate the magnification.
### Solution
The formula for focal length is:
$\frac{1}{f} = \frac{1}{i} \ – \ \frac{1}{o}$
The focal length and the image distance of a concave lens will be negative as it is on the left of the lens. So, the given values are:
f = -14 cm , i = -7 cm
Calculating the object distance o, the formula becomes:
$\frac{1}{o} = \frac{1}{i} \ – \ \frac{1}{f}$
Putting the values gives:
$\frac{1}{o} = \frac{1}{-7} \ – \ \frac{1}{-14}$
$\frac{1}{o} = – \frac{1}{7} + \frac{1}{14}$
$\frac{1}{o} = \frac{ – 2 + 1 }{14}$
$\frac{1}{o} = \frac{ -1 }{14}$
$o = – \ 14$
So, the object distance is 14 cm. The negative sign indicates that the image formed was virtual.
The magnification is calculated as follows:
$m = \frac{i}{o}$
$m = \frac{-7}{-14}$
$m = \frac{1}{2}$
So, the magnification is 0.5.
All the images are created using GeoGebra. | 1,517 | 6,024 | {"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.940842 |
https://www.physicsforums.com/threads/impulse-and-momentum.16888/ | 1,542,756,226,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746847.97/warc/CC-MAIN-20181120231755-20181121013755-00324.warc.gz | 946,655,838 | 12,838 | # Homework Help: Impulse and Momentum
1. Mar 23, 2004
### Draygon_Phly
Hi! I'm a sixteen year old grade eleven physics student,and I'm having trouble understanding impulse and momentum.I would like you to clear up a few things for me.
"suppose two identical cars traveling at the same speed are brought to a stop: one by its braks, the other by a concrete abutment. In both cases the changes in momentum is the same. the cars have equal masses and experience an equal change in velocity. The impulse to stop each car must be equal too, because impulse and momentum are equivalent quantities.
This is what it says in my text. I don't really understand it, could somone explane this to me? Thank you.
Help!
2. Mar 23, 2004
### Chen
For starters, the book isn't very precise. Impluse doesn't equal momentum, in magnitude or direction. Impluse is the change in momentum, and momentum is defined as the mass of the object times its velocity, i.e $$m\vec v$$.
So to stop a car traveling at $$v$$, you need to change its momentum from $$m\vec v$$ to 0 (the momentum of a resting body is zero since its velocity is zero):
$$\vec F\Delta t = m\Delta \vec v = m\vec v - 0$$
To change the momentum you need to exert a force $$F$$ for $$\Delta t$$ time. You can either exert a very strong force for a short period of time (like the concrete wall would), or you can exert a weaker force for a longer period of time (like the friction would). But no matter what, the product of the force and time must be the same.
You can probably find a lot of sites online that will explain this even better:
Although if you have any specific questions I (and others I'm sure) will be happy to answer them.
Last edited: Mar 23, 2004
3. Mar 25, 2004
### Draygon_Phly
Thanx your lots of help. I don't think I could figure it out with out you.
4. Mar 28, 2004
### expscv
hey thx too that help me recall something | 486 | 1,898 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-47 | latest | en | 0.946881 |
http://de.metamath.org/mpeuni/seqfveq.html | 1,718,229,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00399.warc.gz | 11,215,370 | 6,494 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > seqfveq Structured version Visualization version GIF version
Theorem seqfveq 12687
Description: Equality of sequences. (Contributed by NM, 17-Mar-2005.) (Revised by Mario Carneiro, 27-May-2014.)
Hypotheses
Ref Expression
seqfveq.1 (𝜑𝑁 ∈ (ℤ𝑀))
seqfveq.2 ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐹𝑘) = (𝐺𝑘))
Assertion
Ref Expression
seqfveq (𝜑 → (seq𝑀( + , 𝐹)‘𝑁) = (seq𝑀( + , 𝐺)‘𝑁))
Distinct variable groups: 𝑘,𝐹 𝑘,𝐺 𝑘,𝑀 𝑘,𝑁 𝜑,𝑘
Allowed substitution hint: + (𝑘)
Proof of Theorem seqfveq
StepHypRef Expression
1 seqfveq.1 . . . 4 (𝜑𝑁 ∈ (ℤ𝑀))
2 eluzel2 11568 . . . 4 (𝑁 ∈ (ℤ𝑀) → 𝑀 ∈ ℤ)
31, 2syl 17 . . 3 (𝜑𝑀 ∈ ℤ)
4 uzid 11578 . . 3 (𝑀 ∈ ℤ → 𝑀 ∈ (ℤ𝑀))
53, 4syl 17 . 2 (𝜑𝑀 ∈ (ℤ𝑀))
6 seq1 12676 . . . 4 (𝑀 ∈ ℤ → (seq𝑀( + , 𝐹)‘𝑀) = (𝐹𝑀))
73, 6syl 17 . . 3 (𝜑 → (seq𝑀( + , 𝐹)‘𝑀) = (𝐹𝑀))
8 eluzfz1 12219 . . . . 5 (𝑁 ∈ (ℤ𝑀) → 𝑀 ∈ (𝑀...𝑁))
91, 8syl 17 . . . 4 (𝜑𝑀 ∈ (𝑀...𝑁))
10 seqfveq.2 . . . . 5 ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐹𝑘) = (𝐺𝑘))
1110ralrimiva 2949 . . . 4 (𝜑 → ∀𝑘 ∈ (𝑀...𝑁)(𝐹𝑘) = (𝐺𝑘))
12 fveq2 6103 . . . . . 6 (𝑘 = 𝑀 → (𝐹𝑘) = (𝐹𝑀))
13 fveq2 6103 . . . . . 6 (𝑘 = 𝑀 → (𝐺𝑘) = (𝐺𝑀))
1412, 13eqeq12d 2625 . . . . 5 (𝑘 = 𝑀 → ((𝐹𝑘) = (𝐺𝑘) ↔ (𝐹𝑀) = (𝐺𝑀)))
1514rspcv 3278 . . . 4 (𝑀 ∈ (𝑀...𝑁) → (∀𝑘 ∈ (𝑀...𝑁)(𝐹𝑘) = (𝐺𝑘) → (𝐹𝑀) = (𝐺𝑀)))
169, 11, 15sylc 63 . . 3 (𝜑 → (𝐹𝑀) = (𝐺𝑀))
177, 16eqtrd 2644 . 2 (𝜑 → (seq𝑀( + , 𝐹)‘𝑀) = (𝐺𝑀))
18 fzp1ss 12262 . . . . 5 (𝑀 ∈ ℤ → ((𝑀 + 1)...𝑁) ⊆ (𝑀...𝑁))
193, 18syl 17 . . . 4 (𝜑 → ((𝑀 + 1)...𝑁) ⊆ (𝑀...𝑁))
2019sselda 3568 . . 3 ((𝜑𝑘 ∈ ((𝑀 + 1)...𝑁)) → 𝑘 ∈ (𝑀...𝑁))
2120, 10syldan 486 . 2 ((𝜑𝑘 ∈ ((𝑀 + 1)...𝑁)) → (𝐹𝑘) = (𝐺𝑘))
225, 17, 1, 21seqfveq2 12685 1 (𝜑 → (seq𝑀( + , 𝐹)‘𝑁) = (seq𝑀( + , 𝐺)‘𝑁))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ∀wral 2896 ⊆ wss 3540 ‘cfv 5804 (class class class)co 6549 1c1 9816 + caddc 9818 ℤcz 11254 ℤ≥cuz 11563 ...cfz 12197 seqcseq 12663 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-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-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-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-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 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-n0 11170 df-z 11255 df-uz 11564 df-fz 12198 df-seq 12664 This theorem is referenced by: seqfeq 12688 seqf1olem2 12703 seqf1o 12704 sumeq2ii 14271 fsum 14298 fsumser 14308 prodeq2ii 14482 fprod 14510 fprodntriv 14511 gsumccat 17201 gsumzaddlem 18144 gsumconst 18157 wilthlem3 24596 gsumnunsn 29942 mblfinlem2 32617
Copyright terms: Public domain W3C validator | 2,643 | 4,174 | {"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-2024-26 | latest | en | 0.225739 |
https://oeis.org/A112506/internal | 1,670,191,677,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00043.warc.gz | 456,203,417 | 3,656 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A112506 Number of counties per state of the USA in alphabetical order by state. 2
%I
%S 67,27,15,75,58,63,8,3,67,159,5,44,102,92,99,105,120,64,16,23,14,83,
%T 87,82,114,56,93,16,10,21,33,62,100,53,88,77,36,67,5,46,66,95,254,29,
%U 14,95,39,55,72,23
%N Number of counties per state of the USA in alphabetical order by state.
%C Two states do not have counties so a(2)=27 for the 27 divisions of Alaska (whose names each include "Borough" or "Census Area") and a(18)=64 for the 64 parishes of Louisiana. No term includes the District of Columbia (Washington, D.C.) nor the independent cities found in Maryland (1), Missouri (1), Nevada (1) and Virginia (40). The part of Yellowstone National Park in Montana is not part of any county and is also not counted here. Sum_{n=1..50} a(n) = 3097. This World Almanac source indicates that its sources include the U.S. Bureau of the Census and the U.S. Department of Commerce.
%D "Populations and Areas of Counties and States", The World Almanac and Book of Facts (2000 edition). New Jersey: Primedia Reference Inc., 1999, pp. 427-445.
%e Alphabetically, Alabama is the first state of the United States of America so a(1)=67 because that state has 67 counties. Wyoming, the 50th state alphabetically, has a(50)=23 counties.
%Y Cf. A112507 (same but in increasing order).
%K fini,full,nonn
%O 1,1
%A _Rick L. Shepherd_, Sep 08 2005
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Last modified December 4 16:41 EST 2022. Contains 358563 sequences. (Running on oeis4.) | 624 | 2,153 | {"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-2022-49 | latest | en | 0.868706 |
https://quant.stackexchange.com/questions/61372/what-is-the-best-way-to-interpret-changes-in-treasury-yields | 1,702,241,233,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102637.84/warc/CC-MAIN-20231210190744-20231210220744-00537.warc.gz | 508,619,999 | 43,076 | # What is the best way to interpret changes in Treasury yields?
My question is quick and simple. However, I would like to use this answer to further my understanding of bonds and yields.
If the YTM on a 10yr Note yesterday was 1.00% and the YTM on the same 10yr Note today is 1.10%, did the yield increase by:
1. 0.10%, i.e. (1.10% - 1.00% = 0.10% ?
2. 10.0%, i.e. (1.10%/1.00% - 1) = 10% ?
What I am looking to get out of an answer to this question:
1. What does yield actually represent? We look at stock returns during a holding period as in terms of the current price and purchase price (basis). How can we look at stock returns in terms of yield?
2. THE THING I DO NOT UNDERSTAND ABOUT BONDS (chicken or egg). Do traders who trade bonds trade the yield or the face value? Does the yield go up because bond prices go down or do bond prices go down because yields go up
Thank you for taking the time to answer this seemingly trivial question. However, this is a major blockage point in my studies for the CFA and Finance in general.
TL;DR: If the YTM on a 10yr Note yesterday was 1.00% Yesterday and the YTM on the same 10yr Note Today is 1.10%, did the yield increase by:
1. 0.10%, i.e. (1.10% - 1.00% = 0.10% ?
2. 10.0%, i.e. (1.10%/1.00% - 1) = 10% ?
• Yield is more intuitive (especially for people who are not professional bond traders), but your doubts about how to measure a yield change are precisely why traders sometimes measure market changes by $\Delta P/P$, especially when dealing with bonds of very different maturities. So both $P$ and $y$ are useful. Feb 25, 2021 at 21:12
• Usually people quote nominal changes in yield, i.e. 1. You may be interested in total return calculation where you have your change in price plus accrued interest to get you a percent that can compare you to another asset (say, stock). Keep in mind that yield cuts are selling (ie higher yields) and bumps are buying (lower yields). I would say 10s sold off 10 bps in that scenario.
– Kch
Feb 25, 2021 at 23:10
This is a very profound question, actually.
Definitely not 2. This breaks when rates become zero and negative. This works for prices, not for rates. (I wrote more on it here.)
In practice, most people use 1: the yield changed by 0.10% (10 basis points).
This is not ideal either, because what if you're trying to compare this change to a historical change when the yield changed from 10% to 11% - which of these was a bigger change?
Some people compare the changes in log(1 + rate), or some variant of it. This does not seem as intuitive, but gets around some of these problems.
Edit: you asked what a yield is. A yield is one number that aims to summarize multiple numbers - the price you pay for the bond now, and the cash flows that you are promised to be paid in the future. You lose some details but gain the convenience of seeing just one number. If you are trying to decide whether one bond is rich or cheap in comparison with another, then you probably need to look at a lot more than their yield.
For dividend-paying stocks, there is a similar concept of dividend yield - the amount of dividend you expect divided by the price. | 818 | 3,157 | {"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} | 2.953125 | 3 | CC-MAIN-2023-50 | longest | en | 0.943336 |
renfreeshawlab.science.unimelb.edu.au | 1,726,607,454,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00719.warc.gz | 436,769,289 | 12,268 | # Organising data in Excel for analysis
From time to time I get asked to help with a statistical analysis. When I ask for the data I often get a spreadsheet like this. OK, I can read this and sometimes make sense of it, but there are a number of important issues including:
• how much work is involved in laying out these tables? For example, calculating averages and SEM requires multiply entering the relevant functions for each column for each group, or a lot of cutting and pasting of formulae and double checking that it worked as expected.
• is this format easy to import into a statistics program? The answer to this is clearly NO!
• How easy is it to regroup the data, filter the data, graph the data? Not very easy at all!
With a little bit of learning, this can all be done so much easier. Here is a suggestion.
First, enter the data in columns with one row for each data value, like this.
The individual measurements are in column D in this example.
The corresponding treatment, animal code and time are recorded in columns A-C.
There is a bit of repetition (eg you need to copy the treatment codes over, in this case, 12 rows each, but excel makes replicating down or across easy (enter the value in the first cell of a range, select the range, press ctrl-D (copy down) or ctrl-R ( copy right). There is repetition of the animal numbers – ditto. Time, in this case, can be duplicated from the block of control to the Treatment1 block of rows. Easy. (or use Excel’s Power Query Editor to automate these steps and make it really easy – see Organising data in Excel for analysis part 2. There is a bit of learning to do, but it will save heaps of time in the long run.)
Now to analyse the data – Say you want basic means etc by treatments and times. Select the columns with the relevant data (in this case click in the A header for column A to select column A, then hold down shift and click the D header for column D. Now you will have A-D selected.
From the Insert menu select Pivot Table. Click OK on the create pivot table dialog that pops up. This will create a pivot table with the selected data on a new worksheet.
The new worksheet will look like this. On the left you have the Pivot Table template that will be populated when you fill in the The Pivot table fields form on the right which appears whenever you select a cell within the pivot table.
In this case we wanted averages for each treatment and time. This is easy. In the the Pivot table fields form click and drag “treatment” from the top and release it in the “rows” area. The distinct values of the treatment column appear on the pivot table. Repeat this process to add “time” below treatment in the rows field. Now drag “measurement” down to the “Σ Values” area.
In column A we now have a breakdown by treatment and time.
In Column B we have sum of Measurement. OOPS! we wanted average not sum. Easy. in the The Pivot table fields form, in the “Σ Values” area, Sum of measurement has a drop down icon.
Click this and select Value field settings. Here you can choose the desired function. Choose average, and the pivot table now shows the averages.
So far, so good. Now for sem. Sadly excel lacks a build in SEM function so we have to calculate it from StDev / sqrt(count). So, drag measurement to the “Σ Values” area and set it to show StDev from the Value Field Settings. And again drag measurement to the “Σ Values” area and set it to show Count from the Value Field Settings. Now we have this:
Now you have a pivot table with data grouped by treatment and time, with average, stdDev and count … select the table and Paste-Special :: Values to a new worksheet and you can then calculate the SEM easily. Enter the formula on the first row then copy the formula down and all the SEMs are done in one go.
As it is, the pivot table is a bit untidy. There are lots of things you can do. You can remove unneeded values – for example the “Blank” treatment group… Click on the “row Labels” header dropdown icon and unselect “blank”.
Don’t want subtotals etc… right click in the pivot table and select Pivot Table Options from the context menu.
Do you want a different layout. Try dragging “treatment” from rows to columns… have a play. Google tutorials on Pivot tables… they are very powerful ways to arrange and collate your data.
Want to visualise your collated data with a graph? Easy, Select your raw data columns and instead of choosing pivot table, choose Pivot Chart or Pivot Table and Pivot Chart. WIth a couple of key presses you can generate a quick graph to see what your data looks like. For example:
Now consider if you have data for, say, 10 genes at each treatment/time… Add a new column called Gene, and add 9 more blocks of data, one for each gene…
Now you can make pivot tables and pivot graphs with ease and with just a few clicks you can collate for each gene, filter the data, change graph types etc etc.
I have put a couple of examples below with a sample set of 3 genes.
Or how about a line chart showing differences between genes for the different treatments
Or how about just gene1 vs gene 3 looking at changes over time … it is all very easy if the data is organised the right way.
And did I mention that when the data is organised nicely, you can simply export the data as, say, a CSV file ready for import into your favourite statistics program for more detailed statistical analysis. Statistics programs like data in columns.
Want to find out more about pivot tables and graphs – look on the web for endless information. Here is one example that looks reasonably clear and easy https://www.excel-easy.com/data-analysis/pivot-tables.html
Have fun. | 1,248 | 5,679 | {"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-2024-38 | latest | en | 0.913746 |
https://discourse.vtk.org/t/use-of-tessellation-shader-in-vtk/1181 | 1,653,008,488,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00755.warc.gz | 263,241,622 | 4,985 | Use of Tessellation Shader in VTK
I have a triangular mesh(polygons as cells in vtkPolyData). Each point has
different scalar value. There is a lookup table for coloring based on scalars. On using Point Color data as scalars, triangles
have graded color. But, what I want is like in the image below
I know that subdividing the triangular mesh into triangles having same
color data vertices can give result like this by using Cell data as scalars.
Now, our workflow is similar to example
https://vtk.org/gitweb?p=VTK.git;a=blob;f=Examples/VisualizationAlgorithms/Cxx/FilledContours.cxx
and I have the desired output. Meshes we deal with have 10s of millions of
triangles and the process is quite slow.
http://ogldev.atspace.co.uk/www/tutorial30/tutorial30.html , and they
mentioned about “Primitive Generator” part of Tessellation where we can just
mention the tessellation levels of each triangle by gl_TessLevelOuter and
gl_TessLevelInner. I believe, specifying gl_TessLevelOuter for every
triangle of original mesh based on Vertices Scalar values for all
triangles solves my problem with performance.
Can anybody please let me know how to do this in VTK?
As noted by @Dave_DeMarle on the mailing list, VTK only provides support for vertex, geometry, and fragment shaders at this time.
You might be able to get a similar look if you disable color interpolation across the triangles. If you pass your own per-vertex color information into the shader pipeline, you can disable interpolation by declaring the variable with the `flat` qualifier. There is some useful information in this Stack Overflow post. I haven’t actually tried this approach, but it might be worth a shot.
Thank you for the response.
The problem is solved now. I have used the in-built function itslef. Colors are actually generated from scalar values of vertices from a lookup table of color bar. Without dividing into triangles, I am using InterpolateScalarsBeforeMappingOn() as explained here and it’s working perfectly.
That’s a much better solution. Glad you got it working.
1 Like | 464 | 2,076 | {"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-2022-21 | latest | en | 0.892456 |
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