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http://www2.math.ou.edu/~nbrady/teaching/f09-2513/ | 1,674,999,859,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00736.warc.gz | 95,381,552 | 3,527 | ## Math 2513 - 001 Discrete Mathematics Fall 2009
### Course Handouts and Messages
• Information Sheet. First week handout. Office hours, grade breakdown, exam dates, course policies etc.
• Clicking on the Collaboration Distance tab of the MathSciNet database will give you an author's Erdos number, and enable you to compute collaboration distances between authors.
• The Mathematics Genealogy Project is a wonderful database which locates all your math professors in relation to the greats.
• You can learn a little bit about Paul Erdos here. There is also an entry about Erdos on the OU MathClub blog, complete with an xkcd strip.
• Solutions to Quiz 1, and a blank Quiz I.
• As announced in class, Mid I is now taking place in class on Thursday, September 24. It will cover all of chapter I and will cover proofs by mathematical induction.
• You can look at an old Midterm I exam to help you prepare for our first midterm. Question 5 is on set theory. You should look at some induction problems in place of the set theory material.
• Solutions to Midterm 1, and a blank Midterm I.
• Solutions to Quiz 2, and a blank Quiz 2.
• permutations handout.
• symmetries handout.
• Solutions to Quiz 3, and a blank Quiz 3.
• Midterm II will take place in class on Tuesday, October 27.
There is a review on Monday evening (6:00pm in 809 PHSC) October 26.
Here is a midterm II exam from a previous semester. Try doing this before the review.
Here are some questions which you should also try to do before the review.
• Here is Midterm 2 and some solutions.
• Here is the cardinality handout that we are working through in class. It is updated from the original paper copy.
• Here are solutions to some problems from Hwk 10 and Hwk 11. There is also a write up of an answer to a question that was raised in class today (Thursday, Nov 19).
• Here is a cardinality problem which was asked on a past exam.
• Solutions to Quiz 4, and a blank Quiz 4.
• Midterm III will take place in class on Tuesday, November 24.
There is a review on Monday evening (7:00pm in 809 PHSC) November 23.
• Here are solutions to some problems from the cardinality handout which you were to look at for Monday, Nov 23.
• Here is Midterm 3 and some solutions.
• Here is the number theory handout for those of you who may have missed it in class, or misplaced it.
### Homework
• [01]. Due Tuesday 09/01. Pages 16-21: 17, 18, 26, 28(b), 28(d), 32(e).
Pages 28-30: 8, 18, 24, 25, 32, 42, 43, 44, 45, 55.
• [02]. Visit my office hours sometime in the next week (by Thu Sept 3).
• [03]. Visit the OU MathClub Blog (you can find a link to it from my home page)
• [04]. Due Tuesday 09/08. Pages 46-50: 10, 12, 19, 25, 36, 44, 50
Pages 58-62: 9, 24, 26, 30, 40, 46
• [05]. Due Tuesday 09/15. Pages 85-86: 7, 8, 9, 16, 24, 38, 39, 40. Also...
a. Show where the proof that the square root of 2 is irrational breaks down in the case of the square root of 4.
b. Give a proof that the square root of 5 is irrational.
c. Give a proof that the square root of 6 is irrational.
d. Give a proof that the square root of 8 is irrational.
e. Formulate a conjecture about which square roots (of positive integers) are irrational.
• [XX]. For discussion in class next Thursday. Attempt, write notes, but you do not have to turn these in.
You may be asked to present answers on the board, so do attempt them.
Pages 102-104: 7, 12, 16, 21, 23, 32, 33, 34.
Pages 279-283: 4, 5, 6, 10, 11, 15, 16, 20, 21, 34, 35, 47.
• [06]. Due Tuesday, 10/06. Pages 119-121: 3, 5, 8, 19, 20, 23, 38.
Pages 130-133: 1, 15, 29, 30, 34, 35, 36, 40, 44, 48.
• [07]. Due Tuesday, 10/13. Pages 146-149: 8, 12, 15, 16, 22, 29, 36, 38, 40, 66, 67. (omit 36, 38, 40)
• [08]. Due Tuesday, 10/20. Pages 146-149: 36, 38, 40 and 67 (again!).
Let f : A --> B be a function, S a subset of A, and T a subset of B.
Show that the preimage of the image of S contains S, and that the two are equal if the function is injective.
Show that the image of the preimage of T is contained in T, with equality holding if the function is surjective.
• [XX]. Due Thursday, 10/22. Exercises 3, 4, 5, 10, 11, 12 and 19 from the symmetries handout.
• [09]. Due Tuesday, 11/03. Write up answers to the following questions (many of which were discussed in the review).
• [10]. Due Thursday, 11/12. Do Q1 from the following handout and Q31, Q32 from page 162 of the text.
• [11]. Due Tuesday, 11/17. Do 45, 46, 47 from page 163 of the text, and Q2 from the handout.
• [XX]. Due Monday, 11/23. Do 24, 26, 31, 34, 35, 36 from the cardinality handout. These are not for credit. We will discuss them in the review.
• [12]. Due Tuesday, 12/8. Pages 217-218: 4, 5, 6, 7, 8, 10, 35.
• [13]. Due Thursday, 12/10. Page 230: 24. (turn this in) Also look over these. | 1,519 | 4,792 | {"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 | 0.931447 |
https://www.r-bloggers.com/some-quirks-of-the-r-language/ | 1,477,697,589,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476990033880.51/warc/CC-MAIN-20161020190033-00341-ip-10-142-188-19.ec2.internal.warc.gz | 968,749,787 | 22,833 | # Some Quirks of the R Language
August 14, 2012
By
(This article was first published on librestats » R, and kindly contributed to R-bloggers)
R is my favorite programming language. It’s just so useful for getting work done. Sometimes people will complain that R is a difficult language. To me, this begs the questions: difficult for what? And for whom? I personally think R is just about the easiest thing in the world for prototyping. Meaning if you want to quickly crank out some result, R is king. Now when you get into optimization, interfacing R to foreign languages, parallel computing, or god help you, parallel computing in foreign languages that R interfaces to, things get hard. Really hard. But if you just need to get in, analyse and visualise your data, and get on with your life, it really doesn’t get much easier than R.
Now, that doesn’t mean that R is without its problems. It’s quirky as all hell. You have probably seen Ross Ihaka’s example of a particular function having a variable which is randomly local or global. I emphasise particular function, because I have actually heard people ascribe this particular function’s interesting proprty as something common to all R functions. And, uhh, no.
If you’ve never taken the time to think about his example, I guess I’ll spoil the fun for you. So here’s a quick recreation of his function and some runs of it:
```> x <- 0
> f <- function(.)
{
if (runif(1) > .5)
x <- 1
return(x)
}
> sapply(1:10, f)
[1] 1 0 1 1 0 0 1 1 0 1
> x
[1] 0
```
So just what is that function doing? You can think of the ‘if’ statement as basically being if (coinflip==’tails’). So we call the function, and flip a coin. If ‘heads’, the ‘if’ does nothing and moves on. The only thing left to do in that case is return ‘x’, which wasn’t defined within inside f’s scope. So f shrugs its proverbial shoulders and assumes you must have been talking about that x you declared before you even started talking about f, and so in that case f returns 0. On the other hand, if the coinflip results in a ‘tails’, then the ‘if’ statement declares x to be 1 — but this only happens within f’s scope. So locally, f assigns the value of 1 to x, and in this case it returns 1. Afterwards we can check that the x that had nothing to do with f is still 0.
Beauty really is in the eye of the beholder, friends. Ihaka calls this ‘ugly’, something ‘no sensible language would allow’. I resoundingly, categorically, emphatically disagree. I think this is really beautiful, elegant even. And to me, it quite nicely demonstrates R’s very simple concept of scope. An object is global, unless it isn’t.
Now, say you don’t really care about that. Ok fine, jerk. How about this? Sometimes R’s quirkiness leads to some interesting expressive power. For example, when defining function defaults, you can express argument defaults as parameters that are defined within the function scope:
```f <- function(x=y)
{
y <- 0
return( x )
}
```
Here, a call to f() will return 0, because f() evaluates with the default x=y. It doesn’t matter that y isn’t defined at the time f is called, because R does no calculation before its time, and at the beginning of f’s scope, it doesn’t need to know what x is. So why ask? There’s no reason to be nosy. Now, as we chase through function scope, we define y, then demand for the return of x. At this point, we need to know what x is, because hey, you asked for it. So we remember that we wanted x to be equal to the now-defined y. And so R returns 0 without segfaulting. Of course, if we really wanted, we could call f(1), which will return 1, because x is equal to y, unless of course it isn’t.
This handy trick gets used in core R here and there; off the top of my head, it gets used in svd()/La.svd(). I’m sure I remember seeing it elsewhere, but it’s not extremely common.
Of course, R has other quirks that are less useful…annoying ones, even. Probably everyone has run into this one:
```> x <- factor(c(1,3,5))
> x
[1] 1 3 5
Levels: 1 3 5
> as.numeric(x)
[1] 1 2 3
```
If you haven’t run into this one before, count yourself lucky. One of the ways you can get around this is by first casting x as a character, and then casting it as numeric, like so:
```> as.numeric(as.character(x))
[1] 1 3 5
```
I’ll be honest; I hate this. Like, a lot. At one point I remember reading a convincing argument for this behavior, but the behavior itself enrages me so much that I forced this argument from my brain so that I could once again enjoy maximum rage.
I’ll close with something that makes basically no sense to me. I found this adorable little thing one day when I was fiddling around with strings in R:
```> x <- 1:10
> as.character(x)
[1] "1" "2" "3" "4" "5" "6" "7" "8" "9" "10"
> as.character(list(x))
[1] "1:10"
```
Which strikes me as odd. We can force it to return a vector for us…sort of. So seemingly, when you declare something like “x<-1:10”, R doesn’t allocate that vector until it needs to. This is further suggested by the following:
```> system.time(x <- 1:1e8)
user system elapsed
0.040 0.076 0.116
> system.time(x <- x+0)
user system elapsed
0.192 0.196 0.392
> system.time(x <- x+0)
user system elapsed
0.124 0.156 0.282
```
When you tell R to add 0 to x and store the result in x, one of the things it does is farm off the work of actually adding 0 to x to a C function, which takes some work to perform, even though this operation isn’t mathematically interesting. But that isn’t all that it does; note the sharp dropoff in runtime when running the operation again. The runtime is basically stable after that (i.e., performing the addition “x+0” after the first time takes roughly the same amount of time each time thereafter). Without having dug deeper into the internals, my suspicion is that the time discrepancy is caused by allocation of the vector.
Anyway, this is how we get it to return a vector…sort of:
```> x <- 1:10
> x <- x + 0
> as.character(x)
[1] "1" "2" "3" "4" "5" "6" "7" "8" "9" "10"
> as.character(list(x))
[1] "c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)"
```
Now we basically get a vector back instead of a literal “1:10”. But it’s more like something you would pass to the parse() function than an actual array (like what you get with as.character(x)). And there are probably very good reasons for this behavior, but it’s still quirky.
R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more... | 1,865 | 6,741 | {"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-2016-44 | longest | en | 0.896309 |
https://www.storyofmathematics.com/square-root-function-domain-and-range/ | 1,714,008,861,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00438.warc.gz | 894,930,944 | 39,257 | # Square Root Function Domain and Range – Understanding the Basics
he domain of a function is the set of all possible input values it can accept, and for the square root function $f(x) = \sqrt{x}$, this is all non-negative real numbers, represented as $[0, \infty)$.
This is because the square root of a negative number is not a real number, which is what the function is defined over. Regarding its range, the square root function can only output non-negative numbers, as the square root of any non-negative real number is also non-negative.
Hence, the range is the same as the domain, $[0, \infty)$. As I explore this topic, it’s clear to me how vital understanding these concepts is, not just in mathematics, but in practical scenarios where model real-life situations are required.
Stay tuned as we further explore how the graph of the square root function visibly reflects these mathematical truths.
## Defining Domain and Range of a Square Root Function
In my exploration of square root functions, I’ve found that understanding their domain and range is critical. The domain of a function refers to the set of all possible inputs.
For square root functions, like $f(x) = \sqrt{x}$, the domain includes all non-negative real numbers since taking the square root of a negative number isn’t defined within the real number system. Therefore, the domain of this function is $[0, \infty)$.
The range of a function, on the other hand, represents all possible outputs. As square root functions produce only positive number values for the output (or zero), the range of a basic square root function is also $[0, \infty)$.
This outcome is due to the principle that a square root can only yield a non-negative result when dealing with real numbers.
When sketching the graph of a square root function, it essentially looks like half of a sideways parabola, which is a quadratic function. The critical point here, the vertex, is at the origin (0,0) for the basic function $f(x) = \sqrt{x}$.
FunctionDomainRange
$f(x) = \sqrt{x}$$[0, \infty)$$ [0, \infty)$
Modifications and transformations to the basic formula, such as $f(x) = \sqrt{x – h} + k$, will shift the graph horizontally and vertically, affecting the vertex location and sometimes the domain and range as well.
However, regardless of the horizontal shifts, the domain remains all real numbers greater than or equal to the horizontal shift and the range remains the same.
When dealing with a piecewise function that contains a square root component, the domain might be further confined to match the conditions of other parts of the piecewise definition.
In every case, the domain and range reveal much about the behavior of these interesting mathematical constructs.
## Applications and Examples of Square Root Functions
In my exploration of mathematics, I find square root functions to be particularly useful when dealing with real-world scenarios, and they crop up in various fields such as engineering, physics, and finance.
These functions are defined for all non-negative real numbers, and their domain is typically the set of non-negative numbers, expressed in interval notation as $[0, \infty)$.
For instance, if we look at the function $f(x) = \sqrt{x}$, its domain consists of all real numbers greater than or equal to zero, because the square root of negative numbers is not defined in the real number system.
On the other hand, its range is also $[0, \infty)$ because square root functions only yield non-negative real numbers as a result.
In algebra, square root functions represent a type of radical function. A basic example of this is the function $g(x) = \sqrt{x – 4}$, which has a domain of $[4, \infty)$ because the expression inside the square root must be non-negative to belong to the real numbers. This inequality represents a transformation, shifting the domain 4 units to the right.
FunctionEquationDomain
Square Root$f(x) = \sqrt{x}$$[0, \infty) Transformedg(x) = \sqrt{x-4}$$[4, \infty)$
Additionally, square root functions can serve as the inverse of quadratic or polynomial functions. Take for example $h(x) = x^2$, where to find the inverse, I solve the equation $y = x^2$ for $x$, resulting in $x = \sqrt{y}$.
I often encounter square root functions in linear algebra, where they are used to compute the length of vectors in Euclidean space which is essential for many geometric and physics applications.
This use illustrates the square root‘s role in the broader context of linear functions and systems.
In summary, square root functions are an indispensable tool in various equations, representing solutions to radical equations or modeling scenarios involving growth and decay.
Understanding their domain and range is crucial in correctly applying them to solve practical problems.
## Conclusion
In exploring the square root function, we’ve seen that its domain is quite specific; it includes all non-negative real numbers. Mathematically speaking, this can be denoted as $[0, \infty)$.
I find it intriguing how this function, which effortlessly extracts the roots of numbers, has such a simple yet rigid domain.
The range, which refers to the possible output values, is also the set of non-negative real numbers. For the function $f(x) = \sqrt{x}$, the range is the same as the domain, expressed as $[0, \infty)$.
This symmetry between domain and range is a remarkable feature of the square root function and reflects its underlying principles.
Thinking about the utility of these concepts, they’re crucial when I’m graphing the function or solving equations that involve square roots. They provide a clear boundary for the values that I can plug into the function as well as the ones that I can expect to get out of it.
Moreover, it ensures that the calculations I perform are grounded in the real number system and keeps me from attempting to take the square root of a negative number, which would lead us out of the realm of real numbers.
Remembering the boundaries established by the domain and range can help keep our mathematical work both accurate and meaningful.
When we respect these limitations, we ensure that our work in functions and calculus will remain within the scope of real-world applications. | 1,328 | 6,235 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-18 | latest | en | 0.858243 |
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# Time is what Satan made, look here......
## 11 posts in this topic
I've been doing a lot of thinking and ive concluded as follows;
" Time is not an object. And objects can be altered(refer to dimensions below). Time is the unremovable measurement of cellular and worldwide existance. Luck is a part of time and is a dimension of it's own.
Objects may reside in the past whilst the others ride the future measurement. To visit the past, everything existing must be retained(However, this may prove a problem). The World may be in financial , moral etc chaos
Minds may have control.
EXISTANCE- 2nd dimension
LOCATION IN EXISTANCE- 3rd and 4th Dimension
WHY(mind) 5th Dimension
Location actually is a mix up as it is positioned and also in a point of existance.
TEXT
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That's a very interesting thought process you have there. The quote you have on your profile defines time more accurately.
There is 3 known spacial dimensions and 1 non-spacial dimension. The 3 known are the x, y and z axis where anything within our physical 3 dimensional universe can be charted on a graph. Time is widely believed to be the 4th dimension. Some people don't consider time a dimension because all dimensions have time, but the most accepted view is that time is it's own dimension, and that dimension is apart of all the other dimensions.
Quite literally though time is a non-spatial continuum in which events occur in apparently irreversible succession from the past through the present to the future. So that the x, y and z axis can be used to chart anywhere in the known universe and the time dimension can be used to chart when. Meaning with these 4 dimensions you can chart exactly where in the universe and exactly when in the universe at any point in history or in the future.
Time is nothing more than that, and certainly nothing less
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Thanks,Homer! I'm not really a genius but i just keep trying. It seems anything anybody else says is better! :s03
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What really is interesting about time is it's relationship to mass - plainly put, time is different the nearer you are to a large body of mass.
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AHHHHHHH now I understand :s9 :s2
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That's right PS, although most people seem to understand that time slows down the closer one approaches the speed of light, few people understand that time slows down as it passes through a strong gravitational field, which is caused by dense matter(mass)
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Well, Homer that sure makes time relative. :s6
But is relativity relevant? ???
HOMER IS COOL!
Dalia ;D
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Thanks Dalia, you're cool too
But to answer your question about relativity being relevant, I think the question itself is irrelevant ;D
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Homer,
I suppose the relevance would depend on what is being considered.
Dalia :s6
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[blue]So if I'm sitting beside some big fat slob, the impression that it seems like I've been there forever is actually not an impression but reality?
[/blue] :s9
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LB,
That would be subjective relativity and therefore quite relevant to you. :s2
Now, substitute the big fat slob with a Jaffa Cake!
Dalia
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Brunardot Series
The Brunardot Series is an additive, unending series, of unending sequences, that have a Natural integer for the second term; and, the third term is the square of the first term and, as with successive terms, is the sum of the preceding two terms.
Brunardot Series' sequences:
Second Seq: 1, 0, 1, 1, 2, 3, 5, 8...
Third Seq: Phi, 1, Phi +1, Phi + 2, 2 Phi + 3...
Fourth Seq: 2, 2, 4, 6, 10, 16, 26, 42...
Fifth Seq: (SqRt(13) + 1)/2, 3...
Sixth Seq: (SqRt(17) + 1)/2, 4...
Sixth Series: (SqRt(21) + 1)/2, 5...
Seventh Series: 3, 6, 9, 15, 24, 39, 63, 103...
Eighth Series: (SqRt(29) + 1)/2, 7...
. . .
Thirteenth Series:
4, 12, 16, 28, 44, 72, 116, 188...
. . .
Twenty-first Series: 5, 20, 25, 45, 70, 115, 185, 300...
*
The Second Series can also be written:
(SqRt(5) + 1)/2, 1...
The first Brunardot Series, beginning with the third term, is the Fibonacci series (1,1,2,3,5...). Because the Fibonacci series is commonly found in nature; and, the Brunardot Series is a Natural Series, arguably, it can be said that the two initial terms of the first Brunardot Series (1 and 0) properly belong to the Fibonacci Sequence. The Fibonacci Sequence, so completed: 1, 0, 1, 1, 2, 3, 5..., is referred to as the Revised Fibonacci Sequence.
The first two terms of a Brunardot Series are the Pulser of Brunardot Ellipses.
The first term of the Second Brunardot Series begins with Phi,
F, the Golden Ratio, which is: (SqRt(5) + 1)/2. The Second Brunardot Series is referred to as the Golden Series.
Thus, Brunardot Ellipses, defined by the Brunardot Series, closely relate the Fibonacci Series and Phi, The Golden Ratio.
The formulas for the first and third terms of the Brunardot Series in terms of the second term, s, are as follows:
First Term: (SqRt(4s + 1) + 1)/2
Third Term: (SqRt(4s + 1) + 1)/2 + s
E-mail : Brunardot@Brunardot.com
000101 0:01a | 674 | 1,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-49 | latest | en | 0.854067 |
http://quantummechanics.ucsd.edu/ph130a/130_notes/node73.html | 1,369,193,904,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701281163/warc/CC-MAIN-20130516104801-00067-ip-10-60-113-184.ec2.internal.warc.gz | 209,605,637 | 3,173 | # The Solution: Probability Amplitudes
For EM waves, the intensity, and hence the probability to find a photon, is proportional to the square of the fields. The fields obey the wave equation. The fields from two slits can add constructively or destructively giving interference patterns. The and fields are 90 degrees out of phase and both contribute to the intensity.
We will use the same ideas for electrons, although the details of the field will vary a bit because electrons and photons are somewhat different kinds of particles. For both particles the wavelength is given by
and the frequency by
We will use a complex probability amplitude for the electron. The real and imaginary parts are out of phase like the EM fields. The traveling wave with momentum and energy then is
The probability to find an electron is equal to the absolute square of the complex probability amplitude.
(We will overcome the problem that this probability is 1 everywhere for our simple wavefunction.)
We have just put in most of the physics of Quantum Mechanics. Much of what we do for the rest of the course will be deduced from the paragraph above. Our input came from deBroglie and Plank, with support from experiments.
Lets summarize the physics input again.
• Free particles are represented by complex wave functions with a relationship between their particle properties - energy and momentum, and their wave properties - frequency and wavelength given by Plank and deBroglie.
• The absolute square of the wavefunction gives the probability distribution function. Quantum Mechanics only tells us the probability.
• We can make superpositions of our free particle wave functions to make states that do not have definite momentum. We will find that any state can be made from the superposition of free particle states with different momentum.
We now have a wave-particle duality for all the particles, however, physics now only tells us the probability for some quantum events to occur. We have lost the complete predictive power of classical physics.
Gasiorowicz Chapter 1
Rohlf Chapter 5
Griffiths 1.2, 1.3
Cohen-Tannoudji et al. Chapter
Subsections
Jim Branson 2013-04-22 | 447 | 2,176 | {"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-2013-20 | latest | en | 0.919301 |
https://www.justanswer.com/homework/9nb1c-part-1-the-south-dakota-campus-its-45.html | 1,500,995,332,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425254.88/warc/CC-MAIN-20170725142515-20170725162515-00084.warc.gz | 782,182,579 | 20,298 | • 100% Satisfaction Guarantee
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# PART 1:----------------The South Dakota Campus with its 45
### Customer Question
PART 1:
----------------
The South Dakota Campus with its 45 full-time employees has recently increased its enrollment to 1500 students per night. The increase in student population has had a significant slowdown in the network connectivity in the classroom. Reports have stated that it can take up to twenty minutes, sometimes longer, to get connected to the online classroom or to get to the Internet. The campus is using a Cisco 3500 Router and a Cisco Switch that is wired to the access points in each classroom.
Your team has been tasked to determine the root cause of this problem. The analysis should begin by taking a holistic view of the network diagram to understand the topology. Then, progress through the OSI model to find the problem. Also, look for potential choke points that could cause a significant network slowdown. After locating the potential cause of the problem, design a network with the proposed solution.
Develop a diagram in Microsoft® Visio®, Word, or PowerPoint® based on the description provided above of the current South Dakota network. Assumptions can be made in order complete the diagram. The finished diagram should include the following:
Routers & Switches
Wireless Access points
Potential choke pointsPART 2:
----------
Your team has been tasked to determine the root cause of this problem. The analysis should begin by taking a holistic view of the network diagram to understand the topology. Then, progress through the OSI model to find the problem. Also, look for potential choke points that could cause a significant network slowdown. After locating the potential cause of the problem, design a network with the proposed solution.
Create a 1-page summary of your findings and the team's solution based on the description above and your Week Two diagram. Your summary should include the following:
Methods used to locate potential problems
How choke points were determined
The team's solution
A cost breakdown of the budgetPART 3:
-----------
Using your Week Three summary, develop a diagram in Visio®, Word®, or PowerPoint® of the team's proposed network design solution.
The finished diagram should include the following:
Routers & Switches
Wireless Access points
Any new hardwarePART 4:
----------
Develop a 12- to 14-slide proposal presentation in which you present your solutions to the South Dakota Network dilemma. The presentation should include the following:
Develop a logical and physical network for the South Dakota Campus.
Explain the learning team's recommendations to design the network.
Calculate the cost of the network design.
Explain the implementation process and timeline.
The budget for this project is 50,000 for network design, devices and cabling.
The projected student enrollment is 1,500 in class each night (consider choke points.)
There are 45 full-time employees at the campus. Consider network security requirements between employee access and the student population.
Submitted: 1 year ago.
Category: Homework
Customer: replied 1 year ago.
I am ready to increase it to \$100.
How can I do it?
Expert: David L. replied 1 year ago.
Hello,My name is ***** ***** contact JA customer service and they can help you in changing the amount offered.By when do you need this completed?Thanks,David.
Customer: replied 1 year ago.
Hi David,
I did sent them an email and waiting for a reply.
I will appreciate it if I can have it by Friday or Saturday.Thanks
Expert: David L. replied 1 year ago.
Hello,I noticed that the JA question thread states that you are in France...is that correct?Also, can the due date be extended?Thanks,David.
Customer: replied 1 year ago.
Hi David,
I am back to the USA.
Can you at least send me Part 1 and 2 by Saturday?Thanks
Customer: replied 1 year ago.
is it possible?
Expert: David L. replied 1 year ago.
Hello,My apologies for the delay. Since I am not absolutely certain that I can complete the first two parts by Saturday, I am opening up this question to other experts that are able to complete the project within the required time frame.Thank you,David. | 905 | 4,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-30 | longest | en | 0.929874 |
https://www.physicsforums.com/threads/short-hand-interval-notation.578820/ | 1,511,095,383,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805578.23/warc/CC-MAIN-20171119115102-20171119135102-00366.warc.gz | 854,295,436 | 14,598 | # Short hand interval notation
1. Feb 18, 2012
### rudders93
1. The problem statement, all variables and given/known data
Consider the following intervals:
A = [-3,5), B = (3,8), C = (0,4]
Find: A$\cap$B and A$\cap$C
3. The attempt at a solution
I thought that: A$\cap$B=(3,5) and that A$\cap$C=[0,4] as that is the intersection point, but this book (Schaum's Probability Outlines) says that A$\cap$B=[-3,8) and A$\cap$C=[-3,5)
I'm looking to confirm that the book might be wrong (Amazon reviews indicate alot of typographical errors) and instead maybe their answer refers to $A\cup B$ and $A\cup C$ perhaps? Or am I getting confused?
Thanks!!
2. Feb 18, 2012
### Joffan
The book answers are for union $\cup$, and you are really close to correct with your answers for intersection $\cap$. It could be a typo in either question or answer.
3. Feb 18, 2012
### HallsofIvy
Staff Emeritus
Yes, $A\cap B= (3, 5)$ while $A\cup B= [-3, 8)$ as Joffan says. $A\cup C= [-3, 5)$. But $A\cap C$ is NOT [0, 4] because 0 is not in C.
In fact, C is a subset of A so $A\cap C= C$ and $A\cup C= A$. | 373 | 1,095 | {"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} | 3.890625 | 4 | CC-MAIN-2017-47 | longest | en | 0.900988 |
http://science.howstuffworks.com/game-theory2.htm | 1,500,626,694,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423764.11/warc/CC-MAIN-20170721082219-20170721102219-00558.warc.gz | 286,096,459 | 20,877 | How Game Theory Works
Games
John von Neumann, co-author of "Theory of Games and Economic Behavior," gives an extemporaneous lecture on computing machines before the American Philosophical Society.
Alfred Eisenstaedt//Time Life Pictures/Getty Images
John von Neumann and Oskar Morgenstern introduced game theory to the world in 1943 with "Theory of Games and Economic Behavior." They hoped to find mathematical answers to economic problems.
According to economic theory, producers could make a greater profit by reacting to conditions such as supply and demand. But these theories fail to account for the strategies of other producers, and how the anticipation of those strategies affects each producer's moves. Game theory attempted to account for all of these strategic interactions. It didn't take long for military strategists to see the value in this.
When we discuss game theory, we assume a few things:
• A game is considered any scenario in which two players are able to strategically compete against one another, and the strategy chosen by one player will affect the actions of the other player. Games of pure chance don't count, because there's no freedom of choice, and thus no strategy involved. And one-player games, such as solitaire, aren't considered by game theorists to be games, because they don't require strategic interaction between two players.
• Players in a game know every possible action that any player can make. We also know all possible outcomes. All players have preferences regarding these possible outcomes, and, as players, we know not only our own preferences but also those of the other players.
• Outcomes can be measured by the amount of utility, or value, a player derives from them. If you prefer reaching point A to reaching point B, then point A has higher utility. By knowing that you value A over B, and B over C, a player can anticipate your actions, and plan strategies that account for them.
• All players behave rationally. Even seemingly irrational actions are rational in some way. For instance, if you were to play two games of pool, you wouldn't intentionally lose your money on the first game unless you believed that doing so would bolster your opponent's confidence when he or she was deciding how much to bet on game 2 -- a game you anticipate winning. This is an essential difference between one-shot and repeating games. In a one-shot game, you play once; in a repeating game, you play multiple times. (A little later, we'll look at how rational thinking varies between one-shot and repeating games.)
• If no player can reach a better outcome by switching strategies, the game reaches an impasse called the Nash Equilibrium. Essentially, this boils down to players keeping their current strategies (even if they don't have the highest preference) because switching won't accomplish anything.
In the next section, we'll put this information to use and see what we can learn about strategy by plotting it on a game tree. | 599 | 2,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-30 | longest | en | 0.958794 |
https://zhivarvet.com/veronica-mars-okabgcj/mixed-fraction-formula-0dd669 | 1,620,306,207,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00040.warc.gz | 1,138,478,340 | 8,007 | A mixed numeral (also called a mixed fraction or mixed number) is a traditional denotation of the sum of a non-zero integer and a proper fraction (having the same sign). For example, suppose you want to convert the mixed number . You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. one and three-quarters. A common multiple of 2 and 3 is 6.. This has been a guide to Fraction Number in Excel. Key Terms. A mixed fraction is a whole number and a proper fraction combined, i.e. Reciprocal of 20/5 = 5/20. Short Answer: Quote of the day ... Show me Another Quote! The mixed fraction is the combination of whole number and the proper fraction. Use this calculator to convert your improper fraction to a mixed fraction. Make the … Like Fractions Including: multiplication, division, addition and subtraction problems. Factor the Numbers. : 3/2, 11/3, etc. Mixed fractions are also known as mixed numbers. To change mixed numbers to improper fractions, use the formula: (ac+b)/c, where a is the whole number, b is the numerator, and c is the denominator. Let’s take a look at a few examples of a reciprocal. Convert Improper Fraction to Mixed Fraction Take these printable division worksheets blends with all types of fractions. Reciprocal of X/Y = Y/X. A mixed number is a combination of a whole number and a fraction. Show Step-by-step Solutions. Have you ever really looked at a fraction and a division problem? How to convert between improper fractions and mixed numbers? You use equivalent fractions to make them the same. Divide the numerator by the denominator. Your answer is 6 plus a remainder of 2. Remember to change each mixed number to an improper fraction, then multiply by the reciprocal and simplify where you can. The whole number is the number of times the denominator divides into the numerator. The Formula for Converting Improper Fractions to Mixed Numbers. A mixed fraction is a fraction of the form \$$c {n \over d}\$$, where \$$c\$$ is an integer and \$$n d\$$. When fractions have unlike denominators the first step is to find equivalent fractions so that all of the denominators are the same. Convert from Improper Fraction to Mixed Number An improper fraction can be converted into a mixed number - first divide the numerator by the denominator. Then, add that number to the numerator to find the new numerator. Dividing mixed numbers by fractions and with whole numbers. Reciprocal of 1/2 = 2/1. Our result is the equivalent improper fraction 14/5. One concept to remember is that fractions can also be thought of as a division problem. Greatest Common Divisor (GCD) The biggest or largest integer value which will divide the numerator and denominator without producing a fraction. Fraction Division 1. Proper and Improper Fraction. Divide 402 by 11, which equals 36 with a remainder of 6. Converting between fractions and decimals: Converting from decimals to fractions is straightforward. What Are Fractions? Reciprocal of 7/11 = 11/7. Converting improper fractions to mixed numbers - Part a. Instead of multiplying, this time you divide the top number by the bottom number. Computations With Fractions. As can be see from the equation above, the reciprocal is another word for opposite. The resultant becomes the whole number, and the remainder becomes the numerator of the new fraction. Different types. To keep fractions for such kinds of numbers we have Fractions option, first, select the cell whose value we need to convert in fractions and select Format Cells option from the right-click menu list. Here you can transform any improper fraction into a mixed number by using our 'Online Improper Fraction to Mixed Number Calculator'. When the Denominators are Unlike or Different. Improper Fraction to Mixed Number Calculator. Dirt Bike Proportions . An improper fraction can be converted into a mixed fraction and mixed fraction can be converted into an improper fraction with the help of simple multiplication, division and addition. Fractions Lessons. Mixed Fraction is a sum of a whole number and a proper fraction. An improper fraction can be converted into a mixed fraction. These worksheets are pdf files.. Scientific measurements almost invariably use decimal notation rather than mixed numbers. If the fraction is mixed, the steps to convert to an improper fraction are displayed. If necessary we can simplify the fraction to lowest terms or a mixed number. Please enter your whole number on the left and the fraction on the right then press "Simplify Mixed Number" to simplify it: The following formula is used to calculate the reciprocal of a fraction. Here is an adding and subtracting mixed numbers calculator to find the addition and subtraction of mixed fractions. This lesson starts with tins of paint that hold half a litre each. Mixed fractions are also called mixed numbers. Below are six versions of our grade 5 math worksheet on adding mixed numbers where the fractional part of the numbers have like denominators. It is therefore the sum of a whole number and a proper fraction. (You could do the above example this way, too, if you wanted: convert the whole number 3 to an improper fraction, 3/1. If the divisor had been a mixed number instead of a whole number, you would convert both mixed numbers to improper fractions, invert the second improper fraction, and multiply it by the first improper fraction to get your answer. Fraction Worksheets and Printables. Reduce if possible. Proper Fraction and Mixed Number. Do you see that will work the same way? We can show fractions in two ways, one as 4’s, i.e., ¼, 2/4, ¾, and another one as 8’s, i.e., 2/8, 4/8, 6/8. To enter the fraction 4/7 (four sevenths) into cell A1, select cell A1 and type 0 4/7 (with a 0 or Excel thinks you want to enter a date). Multiple Fractions Addition is a basic arithmetic operation which combines two or more fractions together. Law of Cosines Worksheets and Printables. Improper Fraction: If the fraction is mixed, the values of the final improper fraction. That was easy, but, what about mixed numbers? How to divide mixed numbers? A mixed number is a combination of a whole number and a fraction. However, sometimes the denominators are different. Product of Multiple Fractions is a basic arithmetic operation used to find the product of two, three or more whole numbers, positive and or negative fractions. Dividing Fractions 3. Different combinations. 402 11 = 36 r6. Find the whole number. Multiply any number of fractions with like or unlike denominators, positive and negative fractions, or fractions with whole numbers by using this multiple fractions multiplication formula & calculator. Now we know the different types of fractions, let’s look at some other key terms and phrases: Equivalent fractions – These are fractions that appear different but hold the same value. Math worksheets: Adding mixed numbers with like denominators. Find more fraction resources (and much more!) It is used primarily in measurement: inches, for example. From the category section, choose Fraction under which we have multiple conditions for fractioning type up to one digit, two digits or three digits such as halves, eighths, etc. Worksheets > Math > Grade 5 > Fractions: add / subtract > Adding mixed numbers. Show Step-by-step Solutions. Step-by-Step Solution:... What do 'Improper' Fractions Mean? Get it? The calculator provided returns fraction inputs in both improper fraction form, as well as mixed number form. Although mixed numbers are great for everyday use, it is often easier to work with improper fractions when you want to solve math problems. Snow Sprint. How about this? Different fractions. Here we discuss how to format Fractions in Excel using Home Tab and Format Cell Option along with practical uses of fractions in excel. A mixed fraction is always greater than 1. For example, \$${11 \over 4} = 2 {3 \over 4}\$$. Students apply their learning with fractions and mixed numbers to solve a wide range of calculations with fractions and mixed numbers. Finally, place the new numerator over the original denominator to get the improper fraction. Dividing 101 by 18 gives us 5 with a remainder of 11. Types of Fraction – Mixed Fraction. For example, if you have two whole apples and one half apple, you could describe this as 2 + 1/2 apples, or 21/2 apples.Writing Mixed Numbers as FractionsThis mixed number can also be expressed as a fraction… Step by step look at multiplication of mixed numbers, building on multiplication of proper fractions and addition, to create a simple algorithm. Dividing mixed numbers. This unique tool will simplify your mixed number to its lowest form. In both cases, fractions are presented in their lowest forms by dividing both numerator and denominator by their greatest common factor. As always, there is a “formula” for converting these improper fractions to mixed numbers. The below formula is the mathematical representation to add any number of fractions with like or unlike denominators, positive and negative fractions or fractions with whole or mixed numbers. Both as PDF for ease of showing on muliple devices. 11 (the denominator) divides into 402 (the numerator) 36 times, so 36 is the whole number. Head on over to the next page to continue... 1; 2; 3; Continue > Fractions Games. The calculator evaluates the expression or solves the equation with step-by-step calculation progress information. Clear presentation plus two sided worksheet progressing from multiplication of common fractions to multiplication of mixed numbers. Reciprocal Example. All we have to do is change these to improper fractions... Then we can add them! The Fraction Calculator will reduce a fraction to its simplest form. Fraction Division 2. Invert it to 1/3. Make your calculation easy by this mixed fraction calculator. To convert back to a mixed number perform the division. Examples and practice problems of dividing mixed fractions with different denominators. Now that you've entered a fraction correctly, Excel has applied a Fraction format to cell A1 and you can edit the fraction by simply typing 5/7 (without a 0). It is basically an improper fraction. Reciprocal of 5/6 = 6/5. Reciprocal Formula. Dividing Fractions 4 The 6 becomes your whole number. Improper to Mixed Mixed to Improper. Try the free Mathway calculator and problem solver below to practice various math topics. Enter a improper fraction: Ex. If your answer is an improper fraction, convert it back to a mixed number after simplifying. Prime Number Worksheets. Introduce children to improper fractions and mixed numbers. First, multiply the whole number and the denominator. Note: Mixed fractions can always be converted into a fraction. Puppy Pull. A mixed fraction is the combination of a natural number and fraction. The mixed fraction are very difficult to calculate. To enter the mixed fraction 1 4/7 (one and four sevenths), type 1 4/7. Mixed Review. To convert a mixed number to an improper fraction, follow these steps: Multiply the denominator of the fractional part by the whole number, and add the result to the numerator. Recommended Articles. Solve problems with two or more mixed numbers fractions in one expression. We also offer step by step solutions. Mixed fractions – A mixed fraction is presented as a whole number followed by a fractional number, such as 2⅔, 6⅘ or 25⅝. Multiply Fractions With Common Denominators Worksheets. In the improper fraction 32/5, divide 32 by 5. Equivalent fractions. Sometimes it is useful to convert the solution to a mixed fraction by dividing the numerator by the denominator. Mixed Fraction. This video tutorial explains the process of dividing mixed numbers by fractions and with whole numbers. Teach children how to convert between the two. The improper fraction can be converted to the mixed fraction. Dirt Bike Fractions. Example: Change the improper fraction 402/11 to a mixed number. 9 Worksheets on Simplifying Fractions for 6th Graders . The denominator of the new fraction is the same as the original denominator. | 2,558 | 12,110 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-21 | longest | en | 0.889025 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=SignalProcessing/CosineWindow | 1,462,265,370,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121090.75/warc/CC-MAIN-20160428161521-00114-ip-10-239-7-51.ec2.internal.warc.gz | 658,690,061 | 27,631 | SignalProcessing - Maple Help
Home : Support : Online Help : Science and Engineering : Signal Processing : Windowing Functions : SignalProcessing/CosineWindow
SignalProcessing
CosineWindow
multiply an array of samples by a cosine windowing function
Calling Sequence CosineWindow( A, alpha )
Parameters
A - Array of real or complex numeric values; the signal alpha - real numeric constant
Options
• container : Array, predefined Array for holding results
• inplace : truefalse, specifies that output should overwrite input
Description
• The CosineWindow( A, alpha ) command multiplies the Array A by the cosine windowing function, with parameter $\mathrm{alpha}$, and returns the result in an Array having the same length.
• The cosine windowing function $w\left(k\right)$ with parameter $\mathrm{alpha}$ is defined as follows for a sample with $N$ points.
$w\left(k\right)={\mathrm{sin}\left(\frac{k\mathrm{\pi }}{N}\right)}^{\mathrm{\alpha }}$
• Before the code performing the computation runs, A is converted to datatype float[8] or complex[8] if it does not have one of those datatypes already. For this reason, it is most efficient if A has one of these datatypes beforehand. This does not apply if inplace is true.
• If the container=C option is provided, then the results are put into C and C is returned. With this option, no additional memory is allocated to store the result. The container must be an Array of the same size and datatype as A.
• If the inplace or inplace=true option is provided, then A is overwritten with the results. In this case, the container option is ignored.
• The SignalProcessing[CosineWindow] command is thread-safe as of Maple 18.
Examples
> $\mathrm{with}\left(\mathrm{SignalProcessing}\right):$
> $N≔1024:$
> $a≔\mathrm{GenerateUniform}\left(N,-1,1\right)$
${a}{≔}\left[\begin{array}{c}{\mathrm{1 .. 1024}}{\mathrm{Array}}\\ {\mathrm{Data Type:}}{{\mathrm{float}}}_{{8}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{C_order}}\end{array}\right]$ (1)
> $\mathrm{CosineWindow}\left(a,1.23\right)$
$\left[\begin{array}{c}{\mathrm{1 .. 1024}}{{\mathrm{Vector}}}_{{\mathrm{row}}}\\ {\mathrm{Data Type:}}{{\mathrm{float}}}_{{8}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{C_order}}\end{array}\right]$ (2)
> $c≔\mathrm{Array}\left(1..N,'\mathrm{datatype}'={'\mathrm{float}'}_{8},'\mathrm{order}'='\mathrm{C_order}'\right):$
> $\mathrm{CosineWindow}\left(\mathrm{Array}\left(1..N,'\mathrm{fill}'=1,'\mathrm{datatype}'={'\mathrm{float}'}_{8},'\mathrm{order}'='\mathrm{C_order}'\right),0.72,'\mathrm{container}'=c\right)$
$\left[\begin{array}{c}{\mathrm{1 .. 1024}}{{\mathrm{Vector}}}_{{\mathrm{row}}}\\ {\mathrm{Data Type:}}{{\mathrm{float}}}_{{8}}\\ {\mathrm{Storage:}}{\mathrm{rectangular}}\\ {\mathrm{Order:}}{\mathrm{C_order}}\end{array}\right]$ (3)
> $u≔\mathrm{~}[\mathrm{log}]\left(\mathrm{FFT}\left(c\right)\right):$
> $\mathbf{use}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{plots}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{display}\left(\mathrm{Array}\left(\left[\mathrm{listplot}\left(\mathrm{ℜ}\left(u\right)\right),\mathrm{listplot}\left(\mathrm{ℑ}\left(u\right)\right)\right]\right)\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end use}$
>
Compatibility
• The SignalProcessing[CosineWindow] command was introduced in Maple 18. | 1,045 | 3,426 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "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} | 3.015625 | 3 | CC-MAIN-2016-18 | longest | en | 0.567458 |
https://math.answers.com/Q/What_are_3_consecutive_numbers_when_the_product_of_the_smallest_two_numbers_is_28_less_than_the_square_of_the_largest_number | 1,701,947,802,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00222.warc.gz | 431,449,557 | 45,208 | 0
# What are 3 consecutive numbers when the product of the smallest two numbers is 28 less than the square of the largest number?
Updated: 9/16/2023
Wiki User
14y ago
Suppose the three numbers are x-1, x and x+1
then (x-1)*x + 28 = (x+1)2
or x2 - x + 28 = x2 + 2x + 1
or 3x = 27 so that x = 9 and the three numbers are 8, 9 and 10.
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14y ago
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Q: What are 3 consecutive numbers when the product of the smallest two numbers is 28 less than the square of the largest number?
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### What 3 consecutive composite numbers whose product of 192?
There are no such numbers. The smallest set of three consecutive composite numbers is {8, 9, 10} and the product of these numbers is 720.
### Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.?
Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.
### The product of 2 consecutive odd numbers is 63 what is the largest number?
The largest number is 9.
Yes.
5,6,7
11,12,13
2 x 47 = 94
2 x 47 = 94
8,10,12
### What is the greatest common factor of two consecutive numbers?
The GCF of two consecutive numbers is always 1. The GCF of any set of numbers can't be greater than the smallest of the differences between the numbers.
### What four consecutive numbers have the product of 182?
There is no set of four consecutive numbers with a product of 182. There is a set of four consecutive numbers with a sumof 182: 9, 20, 21 and 22.
1, 3, 5 | 443 | 1,605 | {"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-2023-50 | latest | en | 0.933435 |
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# 9th Class - Math | Chapter: Linear Equations in Two Variables MCQs
Dear candidates you will find MCQ questions of 9th Class - Math | Chapter: Linear Equations in Two Variables here. Learn these questions and prepare yourself for coming examinations. You can check the right answer of any question by clicking the option or by clicking view answer button.
All subjects and chapters of 9th Class are listed below. You can select any subject and chapter below to learn mcq questions of that chapter and subject.
## Q1) The linear equation 3x-11y=10 has
(A) Unique solution
(B) Two solutions
(C) Infinitely many solutions
(D) No solutions
## Q2) 3x+10 = 0 will has
(A) Unique solution
(B) Two solutions
(C) Infinitely many solutions
(D) No solutions
(A) (0,2)
(B) (2,0)
(C) (4,0)
(D) (1,1)
(A) 5
(B) 6
(C) 7
(D) 8
(A) 4/3
(B) 5/3
(C) 3
(D) 7/3
(A) (2,0)
(B) (0,2)
(C) (0,1)
(D) (1,1)
(A) (k, -k)
(B) (0, k)
(C) (k, 0)
(D) (k, k)
## Q8) The graph of x = 3 is a line
(A) Parallel to x-axis at a distance of 3 units from the origin
(B) Parallel to y-axis at a distance of 3 units from the origin
(C) Makes an intercept 3 on x-axis
(D) Makes an intercept 3 on y-axis
## Q9) In equation, y = mx+c, m is
(A) Intercept
(B) Slope of the line
(C) Solution of the equation
(D) None of the above
(A) 5/2
(B) 2/5
(C) 10
(D) 0
(A) x = 5
(B) x + 5 = y
(C) y – 5
(D) x – y = 0
(A) 0
(B) 1/2
(C) 2
(D) 3
(A) y = 0
(B) x + y = z
(C) y = x
(D) x = a
(A) (a, 3a)
(B) (3a, a)
(C) (a, a/3)
(D) (a/3, -a)
## Q16) To which equation does the graph represent?
(A) 3x – 7y = 10
(B) y – 2x = 3
(C) 8y – 6x = 4
(D) 5x + (35/2)y = 25
(A) x = -a
(B) y = a
(C) y = x
(D) x + y = 0
(A) (0, 3)
(B) (3, 0)
(C) (2, 0)
(D) (0, 2)
## Q19) Graph of linear equation ax + by + c = 0, a * 0, 6*0 cuts x-axis and y-axis respectively at the points.
(A) ($$\frac{-c}{a}$$, 0), (0, $$\frac{-c}{b}$$)
(B) (0, $$\frac{-c}{b}$$, 0), ($$\frac{-c}{a}$$, 0)
(C) (-c, 0) (0, -c)
(D) (x, 0) (y, 0)
(A) (2, 4)
(B) (0, 3)
(C) (-4, 1)
(D) (4, -1)
## Q21) How many linear equation in x and y can be satisfied by x = 1 and y = 2?
(A) only one
(B) two
(C) infinitely many
(D) three
(A) (9/2, m)
(B) (n, −9/2)
(C) (0, −9/2)
(D) (−9/2, 0)
(A) 2
(B) 4
(C) 3
(D) 1/2
## Q24) Cost of book (x) exceeds twice the cost of pen (y) by Rs 10. This statement can be expressed as linear equation.
(A) x – 2y – 10 = 0
(B) 2x – y – 10 = 0
(C) 2x + y – 10 = 0
(D) x – 2y + 10 = 0
(A) 6x + y = 5
(B) x = 6y + 5
(C) x + 6y = 5
(D) x – 6 = 5
(A) x + y = 1
(B) x = 2y – 4
(C) x + y = 0
(D) y = x – 1
(A) a ≠ 0, b = 0
(B) b ≠ 0, a = 0
(C) a = 0, b = 0
(D) a ≠ 0, b ≠ 0
## Q28) The maximum number of points that lie on the graph of a linear equation in two variables is.
(A) two
(B) definite
(C) infinitely many
(D) three
(A) y – x
(B) x + y = 0
(C) y = 2x
(D) 2 + 3y = 7x
## Q30) The linear equation 4x – 10y = 14 has
(A) A unique solution
(B) Two solutions
(C) Infinitely many solutions
(D) No solutions
(A) 0
(B) 1
(C) 2
(D) Infinite
(A) 4
(B) 6
(C) 5
(D) 2
(A) (2, 0)
(B) (0, 3)
(C) (3, 0)
(D) (0, 2)
## Q34) The equation y = 5, in two variables, can be written as
(A) 1 .x + 1 .y = 5
(B) 0 .x + 0 .y = 5
(C) 1 .x + 0 .y = 5
(D) 0 .x + 1 .y = 5
(A) (a, –a)
(B) (0, a)
(C) (a, 0)
(D) (a, a)
## Q36) The graph of x = 5 is a line
(A) Parallel to x-axis at a distance 5 units from the origin
(B) Parallel to y-axis at a distance 5 units from the origin
(C) Making an intercept 5 on the x-axis
(D) Making an intercept 5 on the y-axis
(A) 2x + y = 17
(B) x + y = 17
(C) x + 2y = 17
(D) 3x – 2y = 17
(A) (0, y)
(B) (x, 0)
(C) (x, x)
(D) (x, y)
(A) y – x = 0
(B) x + y = 0
(C) –2x + y = 0
(D) –x + 2y = 0
(A) (0, 3)
(B) (3, 0)
(C) (2, 0)
(D) (0, 2)
(A) x = a
(B) y = –a
(C) y = x
(D) x + y = 0
## Q43) The graph of x = 9 is a straight line
(A) Intersecting both the axes
(B) parallel to y-axis
(C) parallel to x-axis
(D) Passing through the origin
(A) x = 6
(B) x = –6
(C) y = 6
(D) y = –6 | 1,813 | 4,032 | {"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": 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} | 4.15625 | 4 | CC-MAIN-2024-18 | latest | en | 0.805976 |
http://cboard.cprogramming.com/cplusplus-programming/115757-random-checkerboard-help-printable-thread.html | 1,419,645,504,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447550218.103/warc/CC-MAIN-20141224185910-00096-ip-10-231-17-201.ec2.internal.warc.gz | 12,916,844 | 3,492 | Random Checkerboard Help
• 05-10-2009
ninety3gd
Random Checkerboard Help
have a C++ book I am currently looking at and trying to make some modifications...
Here is the code I modified from C++ text I have:
Have a checkerboard of random 0s and 1s and check to see if of the same number is found vertically, horizontally or diagonals...
diagonals apparently do not work..
Runs:
Code:
```11101001 01100101 11110110 10000011 11111111 11001010 01001100 11001001 All 1's on row 4 No same numbers on the same column No same numbers on the same diagonal No same numbers on the same subdiagonal 11011011 00000001 00000011 11110001 11100111 10000001 10110101 01001101 All 1's on column 7 All 1's on sub-diagonal No same numbers on the same row No same numbers on the same diagonal```
The above is wrong...
Here is the code...I cant see the problem...thanks for help
Code:
```#include <iostream> #include <ctime> // for time function #include <cstdlib> // for rand and srand functions using namespace std; int main() { srand(time(0)); int board[8][8]; bool isSameOnARow = false, isSameOnAColumn = false, isSameOnADiagonal = false, isSameOnASubdiagonal = false; for (int i = 0; i < 8; i++) { for (int j = 0; j < 8; j++) { board[i] [j] = rand() % 2; cout << board[i] [j]; } cout << endl; } // Check rows for (int i = 0; i < 8; i++) { bool same = true; for (int j = 1; j < 8; j++) { if (board[i][0] != board[i] [j]) { same = false; break; } } if (same) { cout << "All " << board[i] [0] << "'s on row " << i << endl; isSameOnARow = true; } } // Check columns for (int j = 0; j < 8; j++) { bool same = true; for (int i = 1; i < 8; i++) { if (board[0] [j] != board[i] [j]) { same = false; break; } } if (same) { cout << "All " << board[0] [j] << "'s on column " << j << endl; isSameOnAColumn = true; } } // Check major diagonal bool same = true; for (int i = 1; i < 8; i++) { if (board[0] [0] == board[i] [i]) { same = false; break; } } if (same) { cout << "All " << board[0] [0] << "'s on major diagonal" << endl; isSameOnADiagonal = true; } // Check subdiagonal same = true; for (int i = 1; i < 8; i++) { if (board[0] [7] == board[i] [7 - i]) { same = false; break; } } if (same) { cout << "All " << board[0][0] << "'s on sub-diagonal" << endl; isSameOnASubdiagonal = true; } if (!isSameOnARow) { cout << "No same numbers on the same row" << endl; } if (!isSameOnAColumn) { cout << "No same numbers on the same column" << endl; } if (!isSameOnADiagonal) { cout << "No same numbers on the same diagonal" << endl; } if (!isSameOnASubdiagonal) { cout << "No same numbers on the same subdiagonal" << endl; } /* Scaffolding code for testing purposes */ cin.ignore(256, '\n'); cout << "Press ENTER to continue..." << endl; cin.get(); /* End Scaffolding */ return 0; }```
• 05-11-2009
Salem
Compare your if() statement with one which works, with one that doesn't. | 1,044 | 3,148 | {"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-2014-52 | longest | en | 0.448063 |
https://manpages.debian.org/buster/liblapack-doc/single_eig.3 | 1,632,719,070,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00416.warc.gz | 412,884,816 | 7,069 | Scroll to navigation
single_eig(3) LAPACK single_eig(3)
single_eig
# SYNOPSIS¶
## Functions¶
subroutine slarfy (UPLO, N, V, INCV, TAU, C, LDC, WORK)
SLARFY
# Detailed Description¶
This is the group of real LAPACK TESTING EIG routines.
# Function Documentation¶
## subroutine slarfy (character UPLO, integer N, real, dimension( * ) V, integer INCV, real TAU, real, dimension( ldc, * ) C, integer LDC, real, dimension( * ) WORK)¶
SLARFY
Purpose:
``` SLARFY applies an elementary reflector, or Householder matrix, H,
to an n x n symmetric matrix C, from both the left and the right.
H is represented in the form
H = I - tau * v * v'
where tau is a scalar and v is a vector.
If tau is zero, then H is taken to be the unit matrix.
```
Parameters:
UPLO
``` UPLO is CHARACTER*1
Specifies whether the upper or lower triangular part of the
symmetric matrix C is stored.
= 'U': Upper triangle
= 'L': Lower triangle
```
N
``` N is INTEGER
The number of rows and columns of the matrix C. N >= 0.
```
V
``` V is REAL array, dimension
(1 + (N-1)*abs(INCV))
The vector v as described above.
```
INCV
``` INCV is INTEGER
The increment between successive elements of v. INCV must
not be zero.
```
TAU
``` TAU is REAL
The value tau as described above.
```
C
``` C is REAL array, dimension (LDC, N)
On entry, the matrix C.
On exit, C is overwritten by H * C * H'.
```
LDC
``` LDC is INTEGER
The leading dimension of the array C. LDC >= max( 1, N ).
```
WORK
``` WORK is REAL array, dimension (N)
```
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
December 2016
# Author¶
Generated automatically by Doxygen for LAPACK from the source code.
Tue Dec 4 2018 Version 3.8.0 | 537 | 1,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-39 | latest | en | 0.808991 |
https://www.open.edu/openlearn/science-maths-technology/engineering-and-technology/radar-what-happened-next-smaller | 1,638,507,089,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362589.37/warc/CC-MAIN-20211203030522-20211203060522-00522.warc.gz | 986,148,824 | 16,700 | Science, Maths & Technology
Author:
Radar: What happened next – Smaller
Updated Tuesday, 22nd July 2014
The race against time to develop smaller radar systems during the Second World War.
This page was published over five years ago. Please be aware that due to the passage of time, the information provided on this page may be out of date or otherwise inaccurate, and any views or opinions expressed may no longer be relevant. Some technical elements such as audio-visual and interactive media may no longer work. For more detail, see our Archive and Deletion Policy
There were working radar systems before Chain Home, but they were large, power hungry and provided only the most basic information; the direction in which a target might be found. Robert Watson-Watt and his team were the first to develop an effective range finding radar by timing the delay between the outgoing pulse and the return from the target.
Chain Home had to operate at an effective range: far enough to allow fighter aircraft in the air to be notified and ideally to allow planes on the ground to be sent up. The theoretical range of a radar system can be derived from a number of operational parameters to give the Range Equation, a fundamental idea in radar and one which appears on the blackboard several times in “Castle in the Sky”:
Leaving the constants aside, the various symbols are
• R is the range – how far away the target is
• PS is the transmitter power – the more you can send out the more likely you are to get some back
• G is the antenna gain – how efficiently the aerial can convert electrical signals into radio waves and vice versa.
• λ is the wavelength – long wavelengths work better than short ones
• σ is the radar cross section of the target – the bigger it is, the more signal is sent back
• PE is the received power at the antenna – the more sensitive the equipment attached to the antenna is, the smaller the signals which can be processed and the further away the target which can be detected. When the received signal is the smallest possible, the range is the maximum possible.
When Chain Home was being developed, the aerials used had to be large for several reasons:
• to provide enough aerial gain (efficiency)
• to give strongly directional reception, needed to establish the direction of the target
• to allow the use of longer wavelengths. Aerial size is generally proportional to wavelength, and Chain Home used signals of around 12m – ludicrously high by modern standards and high even in comparison with later WW2 systems.
These large antennae had many disadvantages. They were expensive to build, they were obvious to the enemy and they were easy to attack – though, curiously it seems to have taken some time for the Germans to decide that it was worth doing so. They also had to be set up to send signals a long distance and as a result were unable to detect lower, closer aircraft.
Making use of a shorter wavelength
Copyright: RAF Mark's Castle To fill this gap, a second system was added, called Chain Home Low, which used a much shorter wavelength (around 1.5m) and could therefore have much smaller antennae. Some of these were mounted on lorries to make portable radar systems and some used steadily rotating antennae – as we are now familiar with on ships and at airports.
Having smaller antennae was only half the battle, though. The transmitting – and to a lesser extent the receiving – stations also had to be reduced in size, and that meant significant improvements in radio frequency engineering technology. Some of this improvement was underway before the war; Watson Watt and his team were able to make use of transmitter valves which had been developed for the Marconi electronic TV system, but even so radar transmitters were large and power hungry.
The magnetron
HCRS under CC-BY-SA licence under Creative-Commons license The breakthrough needed for smaller radars came at the University of Birmingham in 1940, where John Randall and Harry Boot invented the cavity magnetron, a very small microwave generator which works by passing electrons pass the openings to a series of cavities. This process isn't unlike producing a note by blowing across the mouth of an empty bottle.
The magnetron revolutionized wartime radar allowing not only much more powerful and accurate ground-based systems but also the development of airborne radar, starting with the H2S system which was fitted to Stirling and Halifax bombers in 1943 and still in use on RAF Handley Page Victors fifty years later.
Although magnetrons were powerful and relatively efficient, their output was not particularly stable and they have largely been replaced in radar systems by now. However, they are still very widely in use as the source of microwaves inside microwave ovens, and it has been estimated that over one billion have now been produced for that use.
Next: Further
This article is part of the Radar: What happened next? collection, which looks at different aspects of radar technology since the introduction of Chain Home, the first functioning radar defence system developed during the Second World War. This collection was inspired by the OU/BBC drama Castles in the Sky, featuring Eddie Izzard as Robert Watson-Watt, the father of radar. | 1,063 | 5,272 | {"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-2021-49 | latest | en | 0.958559 |
https://www.seeitmarket.com/discounted-cash-flow-method-company-valuation-tools-14021/ | 1,725,919,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00708.warc.gz | 933,835,382 | 59,706 | # The Discounted Cash Flow Method: Company Valuation Tools
When valuing a company, an analyst has several options for valuation tools. In this valuation series, we have considered the Asset Valuation method, Calculated Intangible Value and the P/E Method. There was also a brief digression to look at the Weighted Average Cost of Capital (WACC) calculation and the PEG Ratio. This time, we will look at the discounted cash flow method (DCF); theoretically speaking this is probably the best way in which to value a company.
As we know, if we discount the future cashflows of an entity we have a value for current shareholder wealth. To do this, we can estimate what the future annual post-tax cashflows will be and discount them with an appropriate cost of capital. Note that we will once again use Caterpillar, Inc (CAT) in our example.
Establishing the cashflows may not be a straightforward task and if possible we should use free cash flow. Free cash flow is what is left to distribute to shareholders after the company has sufficiently invested in non-current assets and working capital to ensure that it is able to continue operating. Post-tax cash flows may be used if the free cash flow cannot be identified. It is worth noting that if a company pays out all of its free cash flow as dividend, the discounted cash flow method will return the same value as the dividend value model valuation (DVM) (subject of my next article), if the free cash flows are expected to grow at a constant rate and this growth rate is also used in the DVM.
We mentioned that an appropriate cost of capital must be used to discount the cash flows, and in this instance, calculate the terminal value. If we use the cost of equity, this can be used on free cash flows (i.e. post-tax cash flows after financing charges) and this will give us the value of the equity in the company. If we use the WACC, this can be used to discount post-tax cashflows before financing charges and give us the debt plus equity value. You may then deduct the value of debt to arrive at the equity value. My preference is to use WACC and as we saw in a previous article, I calculated this to be 11.8% for CAT. For simplicity, we will round this up to 12%.
As mentioned above, it can be difficult to identify free cash flows but fortunately we have them for 2009 to 2013:
  Â
     1. Source: Marketwatch – Caterpillar Free Cash Flow
From these figures, we can calculate the compounded annual growth rate, which removes the volatility in the cash flow:
(4√5.75/4.03)-1 = 9.2%
We have to treat this figure with some caution though. If we use 9.2% in perpetuity, we are in effect saying that CAT will grow at a faster pace than the overall economy for ever. Using historical figures is advisable, 3% is a reasonably cautious estimate:
2014 Â Â Â 5.75 x 1.03 Â Â = Â Â 5.92
2015Â Â Â Â Â 5.92 x 1.03 Â Â = Â Â 6.1
2016 Â Â Â 6.1 x 1.03 Â Â = Â Â 6.28
Alternatively, we could simply have calculated 5.75 x 1.033 = 6.28
Present Values (DCF 12%)
5.92 x 0.893 = 5.29
6.1Â Â x 0.797 = 4.86
6.28 x 0.712 = 4.47
Present Value is 14.62bn
We then have to calculate the terminal value, which assumes that the growth rate and discount rate will remain constant.
Terminal Value (TV)
(6.28 x 1.03) / 12%-3%) = 71.87bn
We then discount this to get the present value:
Present Value of TV:
71.87 x (0.797) = 57.28
Add PV of cash flows and TV: 14.62bn + 57.28bn =71.9bn
At the 2013 year end, CAT had total debt of 37.75m, and cash and short term investments of 6m. We can then subtract 31.75m from our present value and arrive at 71.87bn
Outstanding shares: 605.4m
 Value per share: \$118.71
It is worth remembering though that there are some issues with this method of valuation as well. The most obvious are determining the future cash flows and an appropriate discount rate. In addition, what time period do we calculate in detail and are we sure about our terminal value calculation. Finally, the discounted cash flow method does not consider any future options which may arise and there is a major assumption that discount and tax rates are constant throughout the evaluation period.
Having said that, this method does give you a maximum value for the company and, crucially, it takes into account the time value of money. In my next article, the final in the valuation series, will look at the dividend valuation model. | 1,135 | 4,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.919799 |
https://www.studypool.com/discuss/378793/pleasw-help-me-find-the-domain-of-the-function?free | 1,495,552,168,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607647.16/warc/CC-MAIN-20170523143045-20170523163045-00411.warc.gz | 934,618,061 | 14,295 | Time remaining:
##### Pleasw help me! Find the domain of the function
Algebra Tutor: None Selected Time limit: 0 Hours
Find the domain of the function f(x) = -3x + 2
Feb 9th, 2015
There is no limit on either domain or range. So The domain of the function is negative infinity to positive infinity.
Feb 9th, 2015
...
Feb 9th, 2015
...
Feb 9th, 2015
May 23rd, 2017
check_circle | 119 | 382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-22 | longest | en | 0.823644 |
https://studylib.net/doc/9035303/chemistry-30---4.3---potential-energy-diagrams | 1,596,478,877,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735823.29/warc/CC-MAIN-20200803170210-20200803200210-00560.warc.gz | 520,584,179 | 12,914 | # Chemistry 30 - 4.3 - Potential Energy Diagrams
```4.3 – Potential Energy Diagrams - Worksheet
1. Answer the following questions based on the potential energy diagram
shown here:
a. Does the graph represent an endothermic or exothermic reaction?
b. Label the postion of the reactants, products, and activated complex.
c.
Determine the heat of reaction, ΔH, (enthalpy change) for this
reaction.
d. Determine the activation energy, Ea for this reaction.
e.
How much energy is released or absorbed during the reaction?
f.
How much energy is required for this reaction to occur?
2. Sketch a potential energy curve that is represented by the following values of
ΔH and Ea as large as you can in the space provided. You may make up
appropriate values for the y-axis (potential energy).
ΔH = -100 kJ and Ea = 20 kJ
Is this an endothermic or exothermic reaction?
3. In the next unit we will be discussing reactions that are reversible, and can go
in either the forward or reverse directions. For example, hydrogen gas and
oxygen gas react to form water, but water can also be broken down into
hydrogen and oxygen gas.
We typically write a reaction that can be reversed this way, using the double
arrow symbol (
or ↔ or ):
2 H2 + O2 ↔ 2 H2O
This reaction is exothermic in the forward direction:
2 H2 + O2 2 H2O + 285 kJ
but endothermic in the reverse direction:
2 H2O + 285 kJ 2 H2 + O2
Consider a general reversible reaction such as:
A+B↔C+D
Given the following potential energy
diagram for this reaction, determine ΔH
and Ea for both the forward and reverse
directions. Is the forward reaction
endothermic or exothermic?
4. Sketch a potential energy diagram for a general reaction A + B↔C + D
Given that ΔHreverse = -10 kJ and Ea forward = +40 kJ
``` | 459 | 1,750 | {"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-2020-34 | latest | en | 0.89917 |
http://www.purplemath.com/learning/viewtopic.php?f=7&t=2677 | 1,498,411,983,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320545.67/warc/CC-MAIN-20170625170634-20170625190634-00543.warc.gz | 629,517,813 | 5,980 | ## simplifiying expressions?
Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
EpidemicBeatz
Posts: 1
Joined: Thu Aug 09, 2012 8:26 pm
Contact:
### simplifiying expressions?
I forgot how to do these and I don't remember where to begin
Help?
(x-5)/(x^2-4x-5)
little_dragon
Posts: 226
Joined: Mon Dec 08, 2008 5:18 pm
Contact:
### Re: simplifiying expressions?
I forgot how to do these and I don't remember where to begin
Help?
(x-5)/(x^2-4x-5)
they show how here: http://www.purplemath.com/modules/rtnldefs2.htm
start by factoring: http://www.purplemath.com/modules/factquad.htm
x^2-4x-5=(x-5)(x+1)
cancel common factors
dont forget x-5=1(x-5) | 230 | 708 | {"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-2017-26 | longest | en | 0.809806 |
https://ru.scribd.com/document/294230109/corporate-finance-practice-exam | 1,606,232,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176864.5/warc/CC-MAIN-20201124140942-20201124170942-00323.warc.gz | 469,913,572 | 90,468 | Вы находитесь на странице: 1из 11
# Fin 221 Fall 2012 Exam 1
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1. Assume that interest rates on 20-year Treasury and corporate bonds are as follows:
T-bond = 7.72%
AAA = 8.72%
A = 9.64%
BBB = 10.18%
## The differences in these rates were probably caused primarily by:
A) Maturity risk differences.
B) Tax effects.
C) Default and liquidity risk differences.
D) Real risk-free rate differences.
E) Inflation differences.
2. A Whole Lotta Dough Pizza (AWLD) trades on the NASDAQ stock market. Below is the current stock price
information. What price would you sell this stock at now if you already own stock in A Whole Lotta Dough
Pizza?
Stock Ticker
Bid
AWLD
\$35.50
\$36.00
A)
B)
C)
D)
E)
\$35.50
\$36.50
\$35.00
\$35.75
\$36.00
3. Astro Investors is interested in purchasing the bonds of the Jetson Company. Jetsons bonds are currently priced at
\$1,100.00 and have 14.5 years to maturity. If the bonds have a 6% coupon rate what is the yield-to-maturity of
these SEMIANNUAL coupon paying bonds?
A) 2.51%
B) 2.50%
C) 5.02%
D) 5.00%
4. Chief Wiggums plans to make 11 semi-annual deposits of \$2000 beginning today and ending 5 years from now
into an account that will pay a 5% nominal annual rate compounded semi-annually. How much will Chief
Wiggums have in this account 5 years from today after he makes his last deposit?
A)
B)
C)
D)
E)
\$22,966.93
\$22,000.00
\$24,966.93
\$22,406.76
\$25,591.11
5. You found your dream house. It will cost you \$175,000 and you will put down \$35,000 as a down payment. For
the rest you get a 30-year 6.25% mortgage. What will be your monthly mortgage payment (assume no early
repayment)?
A) \$729
B) \$605
C) \$389
D) \$862
Statement I:
Statement II:
Statement III:
A)
B)
C)
D)
## As you increase the interest rate, the future value of an investment
increases.
As you increase the length of the investment (to receive some lump sum),
the present value of the investment increases.
The present value of an ordinary annuity is larger than the present value of
an annuity due. (all else equal)
Statements I and II
Statement II only
Statements I and III only
Statement I only
7. After graduating from college with a finance degree, you begin an ambitious plan to retire in 25 years. To build up
your retirement fund, you will make quarterly payments into a mutual fund that on average will pay 12% APR
compounded quarterly. To get you started, a relative gives you a graduation gift of \$5,000. Once retired, you plan
on moving your investment to a money market fund that will pay 6% APR with monthly compounding. As a
young retiree, you believe you will live for 30 more years and will make monthly withdrawals of \$10,000. To meet
your retirement needs, what quarterly payment should you make?
A) \$2,746.50
B) \$2,904.73
C) \$2,221.45
D) \$2,588.27
8. A two-year bond offers a yield of 6% and a three year bond offers a yield of 7.5%. Under the expectations theory
what should be the yield on a one year bond in two years?
A) 3.06%
B) 12.49%
C) 5.95%
D) 10.56%
9. The Springfield Crusaders just signed their quarterback to a 10 year \$50 million contract. Is this contract really
worth \$50 million? (assume r >0)
A) No, it would only be worth \$50 million if it were all paid out today.
B) Yes, because the payments over time add up to \$50 million.
C) Yes, because his agent told him so.
D) No, it is worth more because he can invest the money.
10. Assume that a 3-year Treasury note has no maturity premium, and that the real, risk-free rate of interest is 3
percent. If the T-note carries a yield to maturity of 13 percent, and if the expected average inflation rate over the
next 2 years is 11 percent, what is the implied expected inflation rate during Year 3?
A) 9%
B) 7%
C) 18%
D) 8%
E) 17%
11. As a young graduate, you have plans on buying your dream car in three years. You believe the car will cost
\$50,000. You have two sources of money to reach your goal of \$50,000. First, you will save money for the next
three years in a money market fund that will return 8% annually. You plan on making \$5,000 annual payments to
this fund. You will make yearly investments at the BEGINNING of the year. The second source of money will be a
car loan that you will take out on the day you buy the car. You anticipate the car dealer to offer you a 6% APR loan
with monthly compounding for a term of 60 months. To buy your dream car, what monthly car payment will you
anticipate?
A) \$540.15
B) \$627.73
C) \$483.99
D) \$652.83
## 12. Which of the following transactions takes place in secondary markets?
A) Stock sold in a seasoned equity offering.
B) New stock sold in an initial public offering.
C) Treasury securities auctioned off by the government.
D) Stock sold by an insurance company to adjust its portfolio of assets'.
E) None of the above.
13. You want to buy a new car. The car you picked will cost you \$32,000 and you decide to go with the dealers
financing offer of 5.9% compounded monthly for 60 months. Unfortunately, you can only afford monthly loan
payments of \$300. However, the dealer allows you to pay off the rest of the loan in a one time lump sum payment
at the end of the loan. How much do you have to pay to the dealer when the lump sum is due?
A) \$25,455.37
B) \$14,000.00
C) \$21,890.43
D) \$22,071.75
14. The primary operating goal of a publicly-owned firm interested in serving its stockholders should be to
A) Maximize the stock price on a specific target date.
B) Maximize the stock price per share over the long run, which is the stock's intrinsic value.
C) Maximize its expected total corporate income.
D) Minimize the chances of losses.
E) Maximize its expected EPS.
15. MAD Inc.'s bond rating is downgraded by Standard and Poor's from AAA to BBB. Which of the following would
occur in light of this news?
A) MAD Inc.'s bond price would fall.
B) MAD Inc.'s bond price would increase.
C) MAD Inc.'s bond price would remain the same.
E) Both A and D would occur.
16. Assume that you own an annuity that will pay you \$15,000 per year for 12 years, with the first payment being
made today. You need money today to start a new business, and your uncle offers to give you \$120,000 for the
annuity. If you sell it, what rate of return would your uncle earn on his investment?
A) 8.41%
B) 7.59%
C) 7.99%
D) 6.85%
E) 7.21%
17. What is the future value of cash flows 1-5 AT THE END YEAR 5, assuming a 6% interest rate (compounded
annually)?
End of year
Cash flow
1
\$2,500
2
3,000
3
1,250
4
3,500
5
1,250
6
4,530
7
2,350
A)
B)
C)
D)
\$13,093.74
\$9,7844.40
\$11,548.48
\$13,879.36
18. Other things held constant, if a bond indenture contains a call provision, the yield to maturity that would exist
without such a call provision will generally be ____ the YTM with it.
A) The same as
B) Either higher or lower, depending on the level of call premium, than
C) Unrelated to
D) Lower than
E) Higher than
19. Which answer is FALSE regarding bond prices and interest rates?
A) Interest rate risk can be described as the risk that changes in market interest rates will
cause fluctuations in the bonds price.
B) The price of a bond is the present value of the coupon payments and the face value.
C) Bond prices and interest rates move in opposite directions.
D) The prices of short-term bonds display greater price sensitivity to interest rate changes
than do the prices of long-term bonds.
20. Relaxant Inc. operates as a partnership. Now the partners have decided to convert the business into a corporation.
Which of the following statements is CORRECT?
A) The firm's investors will be exposed to less liability, but they will find it more difficult to
transfer their ownership.
B) The firm will find it more difficult to raise additional capital to support its growth.
C) The company will probably be subject to fewer regulations and required disclosures.
D) Assuming the firm is profitable, none of its income will be subject to federal income taxes.
E) Relaxant's shareholders (the ex-partners) will now be exposed to less liability.
21. Which of the following actions would be most likely to reduce potential conflicts of interest between stockholders
and managers?
A) Change the corporation's formal documents to make it easier for outside investors to
acquire a controlling interest in the firm through a hostile takeover.
B) Beef up the restrictive covenants in the firm's debt agreements.
C) Pay managers large cash salaries and give them no stock options.
D) For a firm that compensates managers with stock options, reduce the time before options
are vested, i.e., the time before options can be exercised and the shares that are received
can be sold.
E) Eliminate a requirement that members of the board of directors must hold a high
percentage of their personal wealth in the firm's stock.
22. If the expectations theory of the term structure of interest rates is correct, and if the other term structure theories
are invalid, and we observe a downward sloping yield curve, which of the following is a true statement?
A) Investors expect short-term rates to increase in the future.
B) Investors expect short-term rates to be constant over time.
C) Investors expect short-term rates to decrease in the future.
D) It is impossible to say unless we know whether investors require a positive or negative
E) The maturity risk premium must be positive.
23. Bavarian Sausage just issued a 10-year 12% coupon bond. The face value of the bond is \$1,000 and the bond
makes ANNUAL coupon payments. If the bond is trading at \$967.25, what is the bonds yield to maturity?
A) 11.26%
B) 12.00%
C) 13.27%
D) 12.59%
24. You want to buy your dream car, but you are \$5,000 short. If you could invest your entire savings of \$2,350 at an
annual interest of 12%, how long would you have to wait until you have accumulated enough money to buy the
car?
A) 9.40 years
B) 3.48 years
C) 6.66 years
D) 7.24 years
25. You are planning your retirement and you come to the conclusion that you need to have saved \$1,250,000 in 30
years. You can invest into an retirement account that guarantees you a 5% annual return. How much do you have
to put into your account at the end of each year to reach your retirement goal?
A) \$81,314.29
B) \$12,382.37
C) \$18,814.30
D) \$23,346.59
26. If interest rates fall from 8 percent to 7 percent, which of the following bonds will have the largest percentage
increase in its value?
A) A 10-year zero-coupon bond.
B) A 10-year bond with a 10 percent semiannual coupon.
C) A 10-year bond with a 10 percent annual coupon.
D) A 5-year zero-coupon bond.
E) A 5-year bond with a 12 percent annual coupon.
## 27. The normal yield curve is upward sloping implying that
A) the return on short-term securities are lower than the return on long-term securities of
similar risk.
B) the return on bonds with a higher default risk is higher than the returns on bonds with
lower default risk.
C) the return on short-term securities are higher than the return on long-term securities of
similar risk.
D) the return on bonds with a lower default risk is higher than the returns on bonds with
higher default risk.
E) the return on long-term securities are equal to the return on short-term securities of similar
risk.
28. Which of the following would be most likely to lead to a higher level of interest rates in the economy?
A) The level of inflation begins to decline.
B) The Federal Reserve decides to try to stimulate the economy.
C) The economy moves from a boom to a recession.
D) Corporations step up their expansion plans and thus increase their demand for capital.
E) Households start saving a larger percentage of their income.
29. Steaks Galore needs to arrange financing for its expansion program. One bank offers to lend the required
\$1,000,000 on a loan which requires interest to be paid at the end of each quarter. The quoted rate is 10 percent,
and the principal must be repaid at the end of the year. A second lender offers 9 percent, daily compounding
(365-day year), with interest and principal due at the end of the year. What is the difference in the effective annual
rates (EFF%) charged by the two banks?
A) 0.96%
B) 0.75%
C) 0.31%
D) 1.25%
E) 0.53%
30. Cold Boxes Ltd. has 100 bonds outstanding (maturity value = \$1,000). The required rate of return on these bonds is
currently 10 percent, and interest is paid semiannually. The bonds mature in 5 years, and their current market value
is \$768 per bond. What is the annual coupon interest rate?
A) 6%
B) 4%
C) 2%
D) 0%
E) 8%
31. Keenan Industries has a bond outstanding with 15 years to maturity, an 8.25% nominal coupon, semiannual
payments, and a \$1,000 par value. The bond has a 6.50% nominal yield to maturity, but it can be called in 6 years
at a price of \$1,120. What is the bond's nominal yield to call?
A) 6.85%
B) 6.53%
C) 6.20%
D) 7.55%
E) 7.20%
32. Kern Corporation's 5-year bonds yield 7.30% and 5-year T-bonds yield 4.10%. The real risk-free rate is r* = 2.5%,
the default risk premium for Kern's bonds is DRP = 1.90% versus zero for T-bonds, the liquidity premium on
Kern's bonds is LP = 1.3%, and the maturity risk premium for all bonds is found with the formula MRP = (t 1)
0.1%, where t = number of years to maturity. What is the inflation premium (IP) on all 5-year bonds?
A) 1.68%
B) 1.20%
C) 1.60%
D) 1.45%
E) 1.32%
33. When you retire you expect to live for another 30 years. During those 30 years you want to be able to withdraw
\$45,000 at the BEGINNING of each year for living expenses. How much money do you have to have in your
retirement account to make this happen. Assume that you can earn 8% on your investments.
A) \$547,128.27
B) \$506,600.25
C) \$723,745.49
D) \$1,350,000.00
34. Bonds issued by BB&C Communications that have a coupon rate of interest equal to 10.65 percent currently have
a yield to maturity (YTM) equal to 15.25 percent. Based on this information, BB&C's bonds must currently be
selling at ____ in the financial markets.
A) par value
C) a discount
D) Not enough information is given to answer this question.
E) None of the above is a correct answer.
35. If you buy a bond that is selling for less than its face, or maturity, value what will happen to the price (value) of the
bond as the maturity date nears if market interest rates do not change during the life of the bond?
A) The price of the bond should decrease even further below the bond's face value because
the rates in the market are too high.
B) The price of the bond will increase as the bond gets closer to its maturity because the
bond's value has to equal its face value at maturity.
C) Because interest rates remain constant, nothing happens to the market value of the bond.
E) None of the above is a correct answer.
## 36. Which of the following would be considered a primary market transaction?
A) The Federal Reserves purchase of mortgage bonds
B) The purchase of the Water Fund, an ETF traded on NASDAQ
C) The purchase of Fidelitys Total Market Fund, an open-ended mutual fund.
D) The purchase of Out of The Rough Fund, a closed-end mutual fund investing in companies
that sponsor golf tournaments and pro golfers.
E) None of the above
37. Bavarian Sausage just issued a 10-year 12% coupon bond. The face value of the bond is \$1,000 and the bond
makes SEMIANNUAL coupon payments. If the required return on the bond is 10%, what is the bonds price?
A) \$1,122.89
B) \$1,000.00
C) \$1,124.62
D) \$815.26
38. You want to buy a house in 4 years and expect to need \$25,000 for a down payment. If you have \$15,000 to invest,
how much interest do you have to earn (compounded annually) to reach your goal?
A) 25.74%
B) 13.62%
C) 16.67%
D) 21.53%
39. If you hold the annual percentage rate constant while increasing the number of compounding periods per year, then
A) the effective interest rate will decrease.
B) the effective interest rate will increase.
C) the effective interest rate will not change.
D) none of the above.
40. Your child's orthodontist offers you two alternative payment plans. The first plan requires a \$4,000 immediate
up-front payment. The second plan requires you to make monthly payments of \$137.41, payable at the end of each
month for 3 years. What nominal annual interest rate is built into the monthly payment plan?
A) 13.64%
B) 12.96%
C) 15.08%
D) 12.31%
E) 14.36%
MULTIPLE CHOICE
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.
C
A
C
C
D
D
D
D
A
D
B
D
D
B
E
A
A
D
D
E
A
C
D
C
C
A
A
D
A
B
B
B
A
C
B
C
C
B
B
E | 4,555 | 16,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-50 | latest | en | 0.885497 |
http://exxamm.com/QuestionSolution19/Class+10/A+swimming+pool+is+40m+long+and+15m+wide+Its+shallow+and+deep+ends+are+1+5m+and+3m+deep+res/3072801736 | 1,553,492,251,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203755.18/warc/CC-MAIN-20190325051359-20190325073359-00443.warc.gz | 71,797,396 | 9,893 | A swimming pool is 40m long and 15m wide. Its shallow and deep ends are 1.5m and 3m deep respectively. If the bo
# Ask a Question
### Question Asked by a Student from EXXAMM.com Team
Q 3072801736. A swimming pool is 40m long and 15m wide. Its shallow and deep ends are 1.5m and 3m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.
A
42,33,000 litres
B
13,50,000 litres
C
22,17,000 litres
D
41,12,000 litres
#### HINT
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#### Access free resources including
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• Syllabus & Pattern Analysis | 226 | 817 | {"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} | 2.71875 | 3 | CC-MAIN-2019-13 | latest | en | 0.866414 |
https://www.iarcs.org.in/inoi/online-study-material/problems/placing-rooks.php | 1,709,286,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00589.warc.gz | 804,441,900 | 2,282 | Training Material
Given a chess board in which some squares are obstacles (#) and other are holes (O). You have to place the maximum number of rooks on this chess board such that no pair of rooks attack each other.
1. A rook cannot be placed on a hole.
2. A rook attacks the row and column on which it is placed. The attacking range of a rook can span a hole, but is blocked by an obstacle.
Here is an example (not maximum!)
``` -------------------------------
| | | | | | | R | |
-------------------------------
| | R | | | | | | |
-------------------------------
| R | # | R | | | | | |
-------------------------------
| | | | | | | O | |
-------------------------------
| | | | | # | | | |
-------------------------------
| | | | | R | | | |
-------------------------------
| | | | | | | | |
-------------------------------
```
Solution | 243 | 951 | {"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-2024-10 | latest | en | 0.792348 |
https://mathbooks.unl.edu/Calculus/sec-8-4-separable.html | 1,721,389,280,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00719.warc.gz | 329,652,035 | 13,913 | $\newcommand{\dollar}{\} \DeclareMathOperator{\erf}{erf} \DeclareMathOperator{\arctanh}{arctanh} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
## Section8.4Separable differential equations
###### Motivating Questions
• What is a separable differential equation?
• How can we find solutions to a separable differential equation?
• Are some of the differential equations that arise in applications separable?
In Sections8.2 and 8.3, we have seen several ways to approximate the solution to an initial value problem. Given the frequency with which differential equations arise in the world around us, we would like to have some techniques for finding explicit algebraic solutions of certain initial value problems. In this section, we focus on a particular class of differential equations (called separable) and develop a method for finding algebraic formulas for their solutions.
###### Separable differential equations
A first-order separable differential equation is a differential equation of the form
\begin{equation*} \frac{dy}{dt} = g(y) h(t) \text{.} \end{equation*}
This structure allows the variables to be separated so that expressions involving $t$ can be collected on one side, and expressions involving $y$ can be collected on the other side, multiplied by $\frac{dy}{dt} \text{.}$
For instance, consider the equation
\begin{equation*} \frac{dy}{dt} = ty\text{.} \end{equation*}
We would like to separate the variables $t$ and $y$ so that all occurrences of $t$ appear on the right-hand side, and all occurrences of $y$ appear on the left, multiplied by $dy/dt\text{.}$ For this example, we divide both sides by $y$ so that
\begin{equation*} \frac1y \frac{dy}{dt} = t\text{.} \end{equation*}
Note that when we attempt to separate the variables in a differential equation, we require that one side is a product in which the derivative $dy/dt$ is one factor and the other factor is solely an expression involving $y\text{.}$
Not every differential equation is separable. For example, if we consider the equation
\begin{equation*} \frac{dy}{dt} = t-y\text{,} \end{equation*}
it may seem natural to separate it by writing
\begin{equation*} y + \frac{dy}{dt} = t\text{.} \end{equation*}
As we will see, this will not be helpful, since the left-hand side is not a product of a function of $y$ with $\frac{dy}{dt}\text{.}$
###### Example8.39
In this example, we explore whether certain differential equations are separable or not, and then revisit some key ideas from earlier work in integral calculus.
1. Which of the following differential equations are separable? If the equation is separable, write the equation in the revised form $g(y) \frac{dy}{dt} = h(t)\text{.}$
1. $\displaystyle \frac{dy}{dt} = -3y\text{.}$
2. $\displaystyle \frac{dy}{dt} = ty - y\text{.}$
3. $\displaystyle \frac{dy}{dt} = t + 1\text{.}$
4. $\displaystyle \frac{dy}{dt} = t^2 - y^2\text{.}$
2. Explain why any autonomous differential equation is guaranteed to be separable.
3. Why do we include the term $+C$ in the expression
\begin{equation*} \int x~dx = \frac{x^2}{2} + C? \end{equation*}
4. Suppose we know that a certain function $f$ satisfies the equation
\begin{equation*} \int f'(x)~dx = \int x~dx\text{.} \end{equation*}
What can you conclude about $f\text{?}$
### SubsectionSolving separable differential equations
Before we discuss a general approach to solving a separable differential equation, it is instructive to consider an example.
###### Example8.40
Find all functions $y$ that are solutions to the differential equation
\begin{equation*} \frac{dy}{dt}= \frac{t}{y^2}\text{.} \end{equation*}
Solution
We begin by separating the variables and writing
\begin{equation*} y^2\frac{dy}{dt} = t\text{.} \end{equation*}
Integrating both sides of the equation with respect to the independent variable $t$ shows that
\begin{equation*} \int y^2\frac{dy}{dt}~dt = \int t~dt\text{.} \end{equation*}
Next, we notice that the left-hand side allows us to change the variable of antidifferentiation3This is why we required that the left-hand side be written as a product in which $dy/dt$ is one of the terms. from $t$ to $y\text{.}$ In particular, $dy = \frac{dy}{dt}~dt\text{,}$ so we now have
\begin{equation*} \int y^2 ~dy = \int t~dt\text{.} \end{equation*}
This equation says that two families of antiderivatives are equal to each other. Therefore, when we find representative antiderivatives of both sides, we know they must differ by an arbitrary constant $C\text{.}$ Antidifferentiating and including the integration constant $C$ on the right, we find that
\begin{equation*} \frac{y^3}{3} = \frac{t^2}{2} + C\text{.} \end{equation*}
It is not necessary to include an arbitrary constant on both sides of the equation; we know that $y^3/3$ and $t^2/2$ are in the same family of antiderivatives and must therefore differ by a single constant.
Finally, we solve the last equation above for $y$ as a function of $t\text{,}$ which gives
\begin{equation*} y(t) = \sqrt[3]{\frac 32 \, t^2 + 3C}\text{.} \end{equation*}
Of course, the term $3C$ on the right-hand side represents 3 times an unknown constant. It is, therefore, still an unknown constant, which we will rewrite as $C\text{.}$ We thus conclude that the function
\begin{equation*} y(t) = \sqrt[3]{\frac 32 \, t^2 + C} \end{equation*}
is a solution to the original differential equation for any value of $C\text{.}$
Notice that because this solution depends on the arbitrary constant $C\text{,}$ we have found an infinite family of solutions. This makes sense because we expect to find a unique solution that corresponds to any given initial value.
For example, if we want to solve the initial value problem
\begin{equation*} \frac{dy}{dt} = \frac{t}{y^2}, \ y(0) = 2\text{,} \end{equation*}
we know that the solution has the form $y(t) = \sqrt[3]{\frac32\, t^2 + C}$ for some constant $C\text{.}$ We therefore must find the appropriate value for $C$ that gives the initial value $y(0)=2\text{.}$ Hence,
\begin{equation*} 2 = y(0) = \sqrt[3]{\frac 32 \, 0^2 + C} = \sqrt[3]{C}\text{,} \end{equation*}
which shows that $C = 2^3 = 8\text{.}$ The solution to the initial value problem is then
\begin{equation*} y(t) = \sqrt[3]{\frac32\, t^2+8}\text{.} \end{equation*}
The strategy of Example8.40 may be applied to any differential equation of the form $\frac{dy}{dt} = g(y) \cdot h(t)\text{,}$ and any differential equation of this form is said to be separable. We work to solve a separable differential equation by writing
\begin{equation*} \frac{1}{g(y)} \frac{dy}{dt} = h(t)\text{,} \end{equation*}
and then integrating both sides with respect to $t\text{.}$ After integrating, we try to solve algebraically for $y$ in order to write $y$ as a function of $t\text{.}$
###### Example8.41
Solve the differential equation
\begin{equation*} \frac{dy}{dt} =3y\text{.} \end{equation*}
Solution
Following the same strategy as in Example8.40, we have
\begin{equation*} \frac 1y \frac{dy}{dt} = 3\text{.} \end{equation*}
Integrating both sides with respect to $t\text{,}$
\begin{equation*} \int \frac 1y\frac{dy}{dt}~dt = \int 3~dt\text{,} \end{equation*}
and thus
\begin{equation*} \int \frac 1y~dy = \int 3~dt\text{.} \end{equation*}
Antidifferentiating and including the integration constant, we find that
\begin{equation*} \ln|y| = 3t + C\text{.} \end{equation*}
Finally, we need to solve for $y\text{.}$ Here, one point deserves careful attention. By the definition of the natural logarithm function, it follows that
\begin{equation*} |y| = e^{3t+C} = e^{3t}e^C\text{.} \end{equation*}
Since $C$ is an unknown constant, $e^C$ is as well, though we do know that it is positive (because $e^x$ is positive for any $x$). When we remove the absolute value in order to solve for $y\text{,}$ however, this constant may be either positive or negative. To account for a possible $+$ or $-\text{,}$ we denote this updated constant by $C$ to obtain
\begin{equation*} y(t) = Ce^{3t}\text{.} \end{equation*}
There is one more technical point to make. Notice that $y=0$ is an equilibrium solution to this differential equation. In solving the equation above, we begin by dividing both sides by $y\text{,}$ which is not allowed if $y=0\text{.}$ To be perfectly careful, therefore, we should consider the equilibrium solutions separately. In this case, notice that the final form of our solution captures the equilibrium solution by allowing $C=0\text{.}$
###### Example8.42
Suppose that the population of a town is growing continuously at an annual rate of 3% per year.
1. Let $P(t)$ be the population of the town in year $t\text{.}$ Write a differential equation that describes the annual growth rate.
2. Find the solutions of this differential equation.
3. If you know that the town's population in year 0 is 10,000, find the population $P(t)\text{.}$
4. How long does it take for the population to double? This time is called the doubling time.
5. Working more generally, find the doubling time if the annual growth rate is $k$ times the population.
Hint
1. Small hints for each of the prompts above.
1. $\frac{dP}{dt} = 0.03 P$
2. $P = Ce^{0.03t}\text{.}$
3. $P = 10000 e^{0.03t}\text{.}$
4. The doubling time is $t = \frac{\ln(2)}{0.03} \approx 23.105$ years.
5. The doubling time is $t = \frac{1}{k} \ln(2)\text{.}$
Solution
1. Since the population $P(t)$ of the town is growing at an annual rate of $3\%$ per year, this means that $\frac{dP}{dt} = 0.03 P$
2. We use separation of variables and write $\frac{1}{P} \frac{dP}{dt} = 1.03\text{.}$ We then have
\begin{equation*} \int \frac{1}{P} \frac{dP}{dt} dt = \int 0.03 dt \end{equation*}
and thus $\int \frac{1}{P} dP = \int 0.03 dt\text{.}$ Evaluating the integrals on each side, we have $\ln|P| = 0.03 t + c$ By definition of the logarithm function, it follows that
\begin{equation*} |P| = e^{0.03t + c} = e^c e^{0.03t} = C e^{0.03t}\text{.} \end{equation*}
Incorporating the $\pm$ that follows from the absolute value sign into the constant $C\text{,}$ we conclude that $P = Ce^{0.03t}\text{.}$
3. Given that $P(0) = 10000\text{,}$ we see that $10000 = C e^0 = C\text{.}$ So
\begin{equation*} P = 10000 e^{0.03t}\text{.} \end{equation*}
4. The population will have doubled when $P = 200000\text{.}$ To determine the time it takes the population to double, we solve the equation
\begin{equation*} 10000 e^{0.03t} = 20000\text{.} \end{equation*}
Thus $e^{0.03t} = 2\text{,}$ so $t = \frac{\ln(2)}{0.03} \approx 23.105\text{.}$ It will take about 23.1 years for the population to double in size.
5. If the annual growth rate is $k\text{,}$ then the differential equation for the population is
\begin{equation*} \frac{dP}{dt} = kP\text{.} \end{equation*}
We can use separation of variables to show that $P = P_0 e^{kt}\text{,}$ where $P_0$ is the population at time $t = 0\text{.}$ We then find the doubling time by solving the equation $P_0 e^{kt} = 2P_0\text{.}$ We first solve for $e^{kt}$ and get $e^{kt} = 2\text{.}$ Taking natural logs, $kt = \ln(2)\text{,}$ and dividing by $k\text{,}$ $t = \frac{1}{k} \ln(2)\text{.}$ The doubling time for the population is thus $\frac{1}{k} \ln(2)\text{.}$
###### Example8.43
Suppose that a cup of coffee is initially at a temperature of $105^\circ$ F and is placed in a $75^\circ$ F room. Newton's law of cooling says that
\begin{equation*} \frac{dT}{dt} = -k(T-75)\text{,} \end{equation*}
where $k$ is a constant of proportionality.
1. Suppose you measure that the coffee is cooling at one degree per minute at the time the coffee is brought into the room. Use the differential equation to determine the value of the constant $k\text{.}$
2. Find all the solutions of this differential equation.
3. What happens to all the solutions as $t\to\infty\text{?}$ Explain how this agrees with your intuition.
4. What is the temperature of the cup of coffee after 20 minutes?
5. How long does it take for the coffee to cool to $80^\circ\text{?}$
Hint
1. Small hints for each of the prompts above.
1. $k = \frac{1}{30}$
2. $T = 75 + Ce^{-t/30}$
3. The temperature of the coffee tends to 75 degrees.
4. $T(20) = 75 + 30e^{-2/3} \approx 90.4^\circ$F.
5. $t = -30 \ln \left( \frac{1}{6} \right) \approx 53.75$ minutes.
Solution
We have
\begin{equation*} \frac{dT}{dt} = -k(T - 75)\text{,} \end{equation*}
where $k$ is a constant. We also have $T(0) = 105\text{.}$
1. Since the coffee is cooling at a rate of one degree per minute at time $0\text{,}$ it follows that $\frac{dT}{dt} = -1$ at this instant. From the given differential equation, we know $-k(T(0) - 75) = -1\text{.}$ Next, $T(0) = 105\text{,}$ so $-k(105 - 75) = -1\text{,}$ and thus $k = \frac{1}{30}\text{.}$
2. We use separation of variables to solve the differential equation $\frac{dT}{dt} = -\frac{1}{30} (T - 75)\text{.}$ Separating variables and integrating both sides with respect to $t\text{,}$
\begin{equation*} \int \frac{1}{T - 75} dT = -\int \frac{1}{30} dt \end{equation*}
Integrating both sides,
\begin{equation*} \ln \left| T - 75 \right| = -\frac{1}{30} t + c \end{equation*}
Using the definition of the logarithm function and writing $e^c = C\text{,}$ we see that
\begin{equation*} |T - 75| = e^{-t/30 + c} = C e^{-t/30} \end{equation*}
Finally, including the absolute value in $C$ and solving for $T\text{,}$ $T - 75 = C e^{-t/30}\text{,}$ so $T = 75 + Ce^{-t/30}\text{.}$
3. As $t \to \infty\text{,}$ $e^{-t/30} \to 0\text{.}$ So
\begin{equation*} \lim_{t \to \infty} \left(75 + Ce^{-t/30} \right) = 75\text{.} \end{equation*}
This means that as $t \to \infty\text{,}$ $T \to 75\text{,}$ or the temperature of the coffee tends to 75 degrees.
4. In order to determine the temperature of the coffee after 20 minutes, we need to first determine the value of the constant $C$ in the solution $T = 75 + Ce^{-t/30}\text{.}$ To do this, we use the fact that the temperature of the coffee at $t = 0$ is $105^\circ$F. Thus, $105 = 75 + Ce^0\text{,}$ so $C = 30\text{.}$ Now we know that $T = 75 + 30e^{-t/30}\text{,}$ and when $t = 20\text{,}$ $T(20) = 75 + 30e^{-2/3}\text{.}$ The temperature of the coffee after 20 minutes is approximately $90.4^\circ$F.
5. To determine how long it will take the coffee to cool to $80^\circ$F, we solve the equation $80 = 75 + 30 e^{-t/30}$ for $t\text{.}$ Solving first for $e^{-t/30}\text{,}$ we subtract $75$ from both sides and divide by $30$ to get $e^{-t/30} = \frac{5}{30}\text{.}$ Taking logarithms and solving for $t\text{,}$ $-\frac{t}{30} = \ln \left( \frac{1}{6} \right)\text{,}$ and thus $t = -30 \ln \left( \frac{1}{6} \right)\text{.}$ We conclude that it will take about 53.75 minutes for the coffee to cool to $80^\circ$F.
###### Example8.44
Solve each of the following differential equations or initial value problems.
1. $\frac{dy}{dt} - (2-t) y = 2-t$
2. $\frac{1}{t}\frac{dy}{dt} = e^{t^2-2y}$
3. $y' = 2y+2\text{,}$ $y(0)=2$
4. $y' = 2y^2\text{,}$ $y(-1) = 2$
5. $\frac{dy}{dt} = \frac{-2ty}{t^2 + 1}\text{,}$ $y(0) = 4$
Hint
1. Small hints for each of the prompts above.
1. $y = -1 + C e^{\left(2t - \frac{t^2}{2} \right)}\text{.}$
2. $y = \frac{1}{2} \ln \left( e^{t^2} + C \right)\text{.}$
3. $y = -1 + 3 e^{2t}\text{.}$
4. $y = -\frac{1}{2t + \frac{3}{2}} = -\frac{2}{4t + 3}\text{.}$
5. $y = \frac{4}{t^2 + 1}\text{.}$
Solution
1. We are given that $\frac{dy}{dt} - (2 - t)y = 2 - t\text{.}$ We first separate the variables by writing $\frac{dy}{dt} = (2 - t)y + (2 - t) = (2-t)(y+1)$ and then dividing both sides by $(y+1)$ to find
\begin{equation*} \frac{1}{1+y} \frac{dy}{dt} = 2 - t\text{.} \end{equation*}
Integrating both sides with respect to $t\text{,}$ we have
\begin{equation*} \int \frac{1}{1+y} dy = \int (2 - t) dt\text{.} \end{equation*}
Evaluating the integrals,
\begin{equation*} \ln|1 + y| = \left( 2t - \frac{1}{2}t^2 \right) + c \end{equation*}
Rewriting without logarithms,
\begin{equation*} |1+y| = e^{\left(2t - \frac{t^2}{2} \right)} e^c \end{equation*}
Writing $e^c = C$ and including the sign from the absolute value in this constant,
\begin{equation*} 1 + y = Ce^{\left(2t - \frac{t^2}{2} \right)}\text{,} \end{equation*}
and therefore $y = -1 + C e^{\left(2t - \frac{t^2}{2} \right)}\text{.}$
2. Given that $\frac{1}{t} \frac{dy}{dt} = e^{t^2 - 2y}\text{,}$ we first observe that $e^{t^2 - 2y} = e^{t^2}e^{ -2y}\text{.}$ Separating the variables,
\begin{equation*} e^{2y} \frac{dy}{dt} = t e^{t^2}\text{.} \end{equation*}
Integrating with respect to $t\text{,}$
\begin{equation*} \int e^{2y} dy = \int t e^{t^2} dt\text{.} \end{equation*}
Using standard substitution techniques on both sides, we find that
\begin{equation*} \frac{1}{2} e^{2y} = \frac{1}{2} e^{t^2} + c \end{equation*}
Writing $2c = C\text{,}$ we then have $e^{2y} = e^{t^2} + C\text{,}$ and taking logarithms to solve for $y\text{,}$ we find $2y = \ln \left( e^{t^2} + C \right)\text{,}$ and thus
\begin{equation*} y = \frac{1}{2} \ln \left( e^{t^2} + C \right)\text{.} \end{equation*}
3. For the initial value problem $y' = 2y + 2\text{,}$ $y(0) = 2\text{,}$ we first solve the differential equation. Noting that $2y + 2 = 2(y+1)\text{,}$ we have
\begin{equation*} \frac{dy}{dt} = 2(y + 1)\text{,} \end{equation*}
so
\begin{equation*} \frac{1}{y + 1} \frac{dy}{dt} = 2 \end{equation*}
Integrating with respect to $t\text{,}$
\begin{equation*} \int \frac{1}{y + 1} dy = \int 2 dt \end{equation*}
and thus evaluating the integrals, $\ln|y+1| = 2t + c\text{.}$ It follows that $|y + 1| = e^{2t + c} = e^c e^{2t}\text{,}$ so in the usual way we can write $y+1 = Ce^{2t}\text{,}$ and therefore
\begin{equation*} y = -1 + Ce^{2t}\text{.} \end{equation*}
Using $y(0) = 2$ to solve the IVP, we obtain $2 = -1 + Ce^0$ and so $C = 3\text{.}$ The solution to the initial value problem is
\begin{equation*} y = -1 + 3 e^{2t}\text{.} \end{equation*}
4. For the initial value problem $y' = 2y^2\text{,}$ $y(-1) = 2\text{,}$ we first separate variables and write
\begin{equation*} \frac{1}{y^2} \frac{dy}{dt} = 2\text{.} \end{equation*}
Next, integrating with respect to $t\text{,}$
\begin{equation*} \int \frac{1}{y^2} dy = \int 2 dt\text{,} \end{equation*}
and thus $-y^{-1} = 2t + c\text{.}$ Solving for $y\text{,}$ it follows
\begin{equation*} y = -\frac{1}{2t + c}\text{.} \end{equation*}
We now use the initial value $y(-1) = 2$ and obtain $2 = -\frac{1}{-2 + c}\text{.}$ Solving for $c$ gives $c = \frac{3}{2}\text{.}$ The solution of the initial value problem is thus
\begin{equation*} y = -\frac{1}{2t + \frac{3}{2}} = -\frac{2}{4t + 3}\text{.} \end{equation*}
5. For the IVP $\frac{dy}{dt} = \frac{-2ty}{t^2 + 1}\text{,}$ $y(0) = 4\text{,}$ we first separate variables and write
\begin{equation*} \frac{1}{y} \frac{dy}{dt} = \frac{-2t}{t^2 + 1}\text{.} \end{equation*}
Integrating with respect to $t$ we find
\begin{equation*} \int \frac{1}{y} dy = -\int \frac{2t}{t^2 + 1} dt\text{.} \end{equation*}
Using the substitution $u = t^2 + 1$ on the right, it follows that
\begin{equation*} \ln|y| = -\ln\left| t^2 + 1 \right| + c\text{.} \end{equation*}
By definition and properties of the logarithm,
\begin{equation*} |y| = e^{-\ln\left| t^2 + 1 \right|}e^c = e^{\ln\left| t^2 + 1 \right|^{-1}}e^c = \left| t^2 + 1 \right|^{-1} e^c\text{.} \end{equation*}
Writing $C = e^c$ and absorbing the absolute value from the left side, we have
\begin{equation*} y = \frac{C}{t^2 + 1}\text{.} \end{equation*}
Using the initial condition $y(0) = 4\text{,}$ we obtain $4 = \frac{C}{1}$ or $C = 4\text{.}$ Thus, the solution to the initial value problem is
\begin{equation*} y = \frac{4}{t^2 + 1}\text{.} \end{equation*}
### SubsectionSummary
• A separable differential equation is one that may be rewritten with all occurrences of the dependent variable multiplying the derivative and all occurrences of the independent variable on the other side of the equation.
• We may find the solutions to certain separable differential equations by separating variables, integrating with respect to $t\text{,}$ and ultimately solving the resulting algebraic equation for $y\text{.}$
• This technique allows us to solve many important differential equations that arise in the world around us. For instance, questions of growth and decay and Newton's Law of Cooling give rise to separable differential equations. Later, we will learn in Section8.6 that the important logistic differential equation is also separable.
### SubsectionExercises
The mass of a radioactive sample decays at a rate that is proportional to its mass.
1. Express this fact as a differential equation for the mass $M(t)$ using $k$ for the constant of proportionality.
2. If the initial mass is $M_0\text{,}$ find an expression for the mass $M(t)\text{.}$
3. The half-life of the sample is the amount of time required for half of the mass to decay. Knowing that the half-life of Carbon-14 is 5730 years, find the value of $k$ for a sample of Carbon-14.
4. How long does it take for a sample of Carbon-14 to be reduced to one-quarter its original mass?
5. Carbon-14 naturally occurs in our environment; any living organism takes in Carbon-14 when it eats and breathes. Upon dying, however, the organism no longer takes in Carbon-14. Suppose that you find remnants of a pre-historic firepit. By analyzing the charred wood in the pit, you determine that the amount of Carbon-14 is only 30% of the amount in living trees. Estimate the age of the firepit.4This approach is the basic idea behind radiocarbon dating.
Consider the initial value problem
\begin{equation*} \frac{dy}{dt} = -\frac ty, \ y(0) = 8 \end{equation*}
1. Find the solution of the initial value problem and sketch its graph.
2. For what values of $t$ is the solution defined?
3. What is the value of $y$ at the last time that the solution is defined?
4. By looking at the differential equation, explain why we should not expect to find solutions with the value of $y$ you noted in (c).
Suppose that a cylindrical water tank with a hole in the bottom is filled with water. The water, of course, will leak out and the height of the water will decrease. Let $h(t)$ denote the height of the water. A physical principle called Torricelli's Law implies that the height decreases at a rate proportional to the square root of the height.
1. Express this fact using $k$ as the constant of proportionality.
2. Suppose you have two tanks, one with $k=-1$ and another with $k=-10\text{.}$ What physical differences would you expect to find?
3. Suppose you have a tank for which the height decreases at $20$ inches per minute when the water is filled to a depth of $100$ inches. Find the value of $k\text{.}$
4. Solve the initial value problem for the tank in part (c), and graph the solution you determine.
5. How long does it take for the water to run out of the tank?
6. Is the solution that you found valid for all time $t\text{?}$ If so, explain how you know this. If not, explain why not.
The Gompertz equation is a model that is used to describe the growth of certain populations. Suppose that $P(t)$ is the population of some organism and that
\begin{equation*} \frac{dP}{dt} = -P\ln\left(\frac P3\right) = -P(\ln P - \ln 3)\text{.} \end{equation*}
1. Sketch a slope field for $P(t)$ over the range $0\leq P\leq 6\text{.}$
2. Identify any equilibrium solutions and determine whether they are stable or unstable.
3. Find the population $P(t)$ assuming that $P(0) = 1$ and sketch its graph. What happens to $P(t)$ after a very long time?
4. Find the population $P(t)$ assuming that $P(0) = 6$ and sketch its graph. What happens to $P(t)$ after a very long time?
5. Verify that the long-term behavior of your solutions agrees with what you predicted by looking at the slope field. | 7,894 | 23,760 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.855127 |
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### Messages - cnc rookie
Pages: 1
1
##### General Mach Discussion / Re: Mach3 and A axis....
« on: November 21, 2016, 11:01:49 AM »
Hi Mike....
Sounds great...I'll have a look at the xml and let you know if I have any questions. I'm behind the 8 ball this week with the holiday so it may be a few days before I can do anything.
I greatly appreciate your time and effort.
Regards Dave
2
##### G-Code, CAD, and CAM discussions / Re: Using M98 and M99
« on: November 20, 2016, 09:13:16 AM »
Gee....I would have never thought to look in the manual that was written for the software.
I'm searching all over the web for the information.
They don't call me rookie for nothing!
Thanks RICH for pointing me in the direction of greatness.
3
##### G-Code, CAD, and CAM discussions / Using M98 and M99
« on: November 20, 2016, 12:55:56 AM »
Would someone be so kind as write a sample of code. I've done several searches and watched numerous videos but still unclear about this one. If I could see a sample of code I believe it would help me understand the commanded machine movements more. I write a lot of simple linear codes for myself but I'm trying to understand more how the M98 function works.
Here is all I need the machine to do. First I'll briefly describe the set up. Mill...XYZ with a rotary A drive with a separate control so it does not need to be included in the code. It just basically operates as a continuous rotating lathe spindle. I need to slice off a row of "buttons" from a rotating piece in the A drive.
So here is what I need the M98 code to do.....from X0 (the end of the part)Y0...Z0 will be top of the part. The Z starts at 1. above Z0 then retract to .050 above Z0 for the clearance.
The X will do a G0 move to -.042 and the Z go -.25 at a G1Fxx from Z0 and just repeat that X and Z incremental move 25 times with the Z retracting to the .050 after each -.25 depth cut.
I need a 1 inch Tool Length offset in the program.
So basically at the start of the program....Z will be 1. above the part at X0Y0 and have the X to move into the minus .042 of an inch and each time the X moves the Z needs to retract -.25....all this 25 times before return back to its start.
I would very much appreciate a sample of what this code should look like and I'm confident I can do my modifications, if needed, from there.
Thank you
Regards Dave aka cnc rookie
4
##### General Mach Discussion / Re: Mach3 and A axis....
« on: November 19, 2016, 09:26:52 AM »
Thanks Mike.....
I have a machine I use in my shop loaded with mach2 that is set up that way. It's a 3 stepper set up. X - Y and a spindle. All 3 steppers are direct connects to the screws. The spindle is programed just as that and you can even adjust the rpm while the program is running and see the rpm readout on the mach screen.
I've messed with it some more but just can't figure it out.
Unless they did a major change in the way the software works from mach2 to 3 I don't see why it cant be done.
But at this point ......who knows?
5
##### General Mach Discussion / Re: Mach3 and A axis....
« on: November 15, 2016, 07:53:18 PM »
Thanks Mike.....I had to pull back from my mill project for a couple days anyway so when ever you have a chance is fine. I think my head was getting ready to explode. Interesting stuff and I'm having fun learning about the software end of it but my real job keeps me overloaded also. I'd enjoy a few more brains in the mix if they're out there and willing.
Regards Dave
6
##### General Mach Discussion / Re: Mach3 and A axis....
« on: November 14, 2016, 11:19:37 AM »
I believe we're in the mix. Forgot I had already named it so I was looking for something different.
I believe this may be the file your looking for.
Regards Dave
7
##### General Mach Discussion / Re: Mach3 and A axis....
« on: November 14, 2016, 09:49:17 AM »
Thanks Mike...
I appreciate the willingness to help me learn.
Let me see what I can figure out with the xml file when I get into the shop today. I did some more reading yesterday but got in over my pay grade. I need to pull back and like you say concentrate on one thing at a time. Trying to understand what all the acronyms (PWM, etc) stand for.
Is there a page somewhere that may have a list of those? For me, it is very confusing to understand what I'm reading about if I run across those and don't understand what they stand for or their basic function. I don't want an electronics engineer degree but I am very interested in this stuff and I'm trying to learn as I go. Eventually I'll get there but I may be a little slow.
I feel fortunate I was able to get thru the computer stuff and have the machine moving on all 4 axis at this point.
Dave
8
##### General Mach Discussion / Re: Mach3 and A axis....
« on: November 13, 2016, 04:54:00 PM »
Ok...so in between football games I messed with it a little. Here's what I have....
I moved the pin numbers from the A to the spindle and enabled that with the same checks clicked
Then I went to the spindle page and enabled the spindle in the top middle of the page and clicked on step/dir
Here's what I get......on the MDI mach tab When I put a value in the spindle rpm (say 120) to test it then I click on F5... the number value goes to zero and the yellow light around the spindle function blinks like it's running but no movement. The part that really has me stumped is I can't turn the spindle off and make the flashing yellow stop. Have to hit the E-stop to make it quit flashing.
What is the spindle box in the bottom left on the spindle page? maybe that's where I'm supposed to be?
So confused...........................
9
##### General Mach Discussion / Re: Mach3 and A axis....
« on: November 13, 2016, 11:07:40 AM »
Thanks Mike....
Yes, I did build it myself. I collected the parts over 10 or so yrs ago and finally just got around to putting them together. I finally got to know enough about my cnc BPT to be able cut the parts needed. Stepper mounts and base and such. You can see the steppers are the outdated round ones instead of square but they are working fine for my needs.
I got the BPT on a song and dance and have taught myself just enough to get into trouble with it. Was able to get it hooked to a computer and transfer files to run parts. I call it Dino..my Brontosaurus. Boss 6 control but still functions fine. At one time there was a guy on cnczone who made boards to be able to run the BPT on mach. But I snoozed on them and they disappeared. Not sure what happened...maybe liability issues or something.
Anyway.................. right now I have two separate issues with my new build.
#1.... Have all the motors turning in the right direction and the incremental distances correct to the DRO. My initial thoughts are I would like to have the A turning as the spindle on the mach read out. I've sat down and wrote a program that will operate the A the way it is set up now for one operation I want to do. But still need the A to turn continually with controlled speed in the spindle function of mach in another program.
So i need the help with pin numbers and box clicks for that.
Issue#2......The control I'm using was originally assembled to run desknc (dos) so I believe some of the pin numbers may be a bit different. It has gecko drives...540 I believe.
I had the controller hooked to another machine years ago and it has a relay wired into it that commands a 120V plug that is attached to the precise spindle I have on this machine. I remember the controller uses m08 and m09 to turn the relay on and off. I see this is the coolant on/off in mach. I haven't changed anything in these settings in the Mach config but......if I plug my spindle into it now....it has a constant flow of 120v. So I assume the relay is closed and it is allowing juice to flow full time. I would also like very much to get this hooked up and be able to turn on/off my precise spindle using the m08/m09 functions in mach if possible. The precise already has it's own speed control 0 to 45K so I don't need the spindle in mach for this function. Just the on/off function. I went back through all of my old papers i have on the desknc pin outs and it appears the m08/9 function was pinned on #16.
Please forgive my greeness......what is a XML file? where would i find it and how would I post it?
Thanks for your willingness to put up with me.
Regards Dave
10
##### General Mach Discussion / Mach3 and A axis....
« on: November 12, 2016, 08:56:39 PM »
Can someone help me with the config on my machine please. I need to configure the A axis run as the spindle in Mach3.
I assume it can done as I have a machine already running like that but its Mach2 and the config sections are different so it was no help. I've tried every combination I can think of to get to work but to no avail.
All 4 axis are running fine and tuned I just need to switch the A.
I included a picture of my newly built machine so you have an idea what I need. I don't care that its running on a stepper and speeds are limited yada yada yada. I would just like it to just turn at a controllable thru the spindle option at a constant speed for the entire length of the programs I write.
Thanks for any and all help.
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Expand Messages
• Gregoire Michel David coined the term relative semi-variogram back in the 70s for what I think you mean by general relative -- that is, each lag is divided
Message 1 of 3 , Apr 5 12:55 PM
Gregoire
Michel David coined the term "relative semi-variogram"
back in the 70s for what I think you mean by general
relative -- that is, each lag is divided by the square
of the mean of the samples used at that lag.
Gary Raymond proposed the pairwise relative soon
after. I used the type you are describing where the
whole semi-variogram is divided by the same
mean-squared in my 1979 paper (Does Geostatistics
Work) because I was analysing a line of samples where
all samples are used at every lag.
The term "standardised" in general statistics usually
means dividing through by the variance or standard
deviation (not a mean). This is the first time I have
seen it in context with a semi-variogram. Seen with no
other information, I would have taken this to imply
standardised to total sill of 1. This would mean
dividing by the variance, not the mean-squared.
proportional effect if you are trying to calculate a
semi-variogram on positively skewed data. Noel Cressie
wrote a paper in Mathematical Geology (early 90s?)
which showed that the David relative semi-variogram
was topologically equivalent to using logarithms. You
data does not have to be lognormal to do this.
Computationally, taking logarithms is faster and more
stable than relative semi-variograms. Probably why
most people don't bother. Gary Raymond provides
software for the pair-wise and Geostat Systems will
have relative semi-variograms. Don't know of any free
stuff.
Isobel
http://geoecosse.bizland.com
• Hi Gregoire, I agree with you regarding the merits of the standardized semivariogram as implemented in variowin software. In one of my last studies, the
Message 2 of 3 , Apr 5 2:50 PM
Hi Gregoire,
I agree with you regarding the merits of the standardized semivariogram as implemented
in variowin software. In one of my last studies, the rescaling by the lag variance helped
correcting the preferential sampling of wells with high arsenic levels, leading to a
susbtantial decrease in random fluctuations of the experimental semivariograms.
While the general relative semivariogram approximates the lag variance by the square
of the lag mean, the standardized semivariogram uses the actual lag variance, hence
makes less assumptions.
Regarding the terminology, I guess we should used a term like "lag-standardized"
to distinguish the global and lag-specific standardization or rescaling of semivariogram
values.
Cheers,
Pierre
-----Original Message-----
From: Gregoire Dubois [mailto:gregoire.dubois@...]
Sent: Tue 4/5/2005 9:48 AM
To: ai-geostats@...
Cc: mueller@...
Subject: [ai-geostats] Definition of standardize variograms
Dear list,
While playing around with different software, I encounter different definitions for standardized variograms.
Surfer (which is using the terminology of Variowin), uses the term "standardized semivariogram" for variograms obtained by dividing the semivariance by the lag variance, while GS+ uses the total variance. While the function obtained in GS+ is only a matter of rescaling variograms, allowing so various variograms to be compared, those proposed in Surfer have the same pupose as the local, pairwise and/or general relative variograms (see Isaaks & Srivastava, page 163-170), that is to reduce the influence of local means. Interestingly enough, one may note that very few software propose relative variograms while I, very personally, consider these functions as essential for detecting spatial structures of many environmental variables.
I have thus here two questions about the use of standardized/relative variogram:
1) What is the correct terminology or definition for standardized variograms? (I personally do not like very much the use of "standardized" when the standardisation is only applied to each lag...)
2) The general relative variogram (lag divided by the mean of the lag) has properties that are very similar to the "standardized" variogram (lag divided by the variance of the lag) but both functions differ. How shall one decide what to use and what are the relative properties of these functions?
Thank you in advance for any feedback.
Gregoire
PS: a few points here good be added to Tom Mueller's FAQ on Geostatistical Software Conventions.
__________________________________________
Gregoire Dubois (Ph.D.)
JRC - European Commission
IES - Emissions and Health Unit | 1,054 | 4,591 | {"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-2017-17 | latest | en | 0.928849 |
https://thispointer.com/check-if-all-elements-in-a-numpy-array-are-equal-to-value/ | 1,708,709,911,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474440.42/warc/CC-MAIN-20240223153350-20240223183350-00724.warc.gz | 590,094,630 | 67,468 | # Check if all elements in a NumPy Array are equal to value
This tutorial will discuss about unique ways to check if all elements in a numpy array are equal to value.
## Technique 1: Using all() method
We can use the numpy.all() method to check if all elements of a NumPy array are equal to a given value.
Compare the NumPy array with the given value, and it will give a NumPy Array of boolean values. A `True` value in this boolean NumPy Array represents that the corresponding value in the original array is equal to the given value. Then to confirm if all the values in this boolean array are True or not, pass this boolean array to the numpy.all() method. If it returns True, then it means that all the values in orignal NumPy array are same, and equal to the given value.
Let’s see the complete example,
```import numpy as np
# Create a NumPy array
arr = np.array([23, 23, 23, 23, 23, 23])
value = 23
# Check if all values in NumPy array are equal to a given value
if np.all(arr == value):
print(f"All elements in Array are equal to the given value")
else:
print(f"All elements in Array are not equal to the given value")
```
Output
```All elements in Array are equal to the given value
```
It should also work with a 2D Numpy Array. Suppose we have 2D NumPy array, and we will use the same technique to check if all the values in this 2D Numpy array are equal.
Let’s see the complete example,
```import numpy as np
# Create a NumPy array
arr = np.array([[16, 16, 16],
[16, 16, 16],
[16, 16, 16]])
value = 16
# Check if all values in 2D NumPy array are equal to a given value
if np.all(arr == value):
print(f"All elements in Array are equal to the given value")
else:
print(f"All elements in Array are not equal to the given value")
```
Output
```All elements in Array are equal to the given value
```
## Technique 2: using array_equal() method
Create an array of equal size but with the repeated values, and this value should be equal to the given value. Then match this new NumPy Array with the original array, using the numpy.array_equal() method. If both the arrays are equal, then it means that all the elements in original NumPy array are equal to the given value.
Let’s see the complete example,
```import numpy as np
# Create a NumPy array
arr = np.array([23, 23, 23, 23, 23, 23])
value = 23
# Create another array of same size but with repeated value
# Check if both the arrays are equal
if np.array_equal(arr, np.full_like(arr, value)):
print(f"All elements in Array are equal to the given value")
else:
print(f"All elements in Array are not equal to the given value")
```
Output
```All elements in Array are equal to the given value
```
## Summary
We learned about different ways to check if all values in a NumPy array are equal to a value in Python. Thanks.
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Topic: Matheology § 038
Replies: 94 Last Post: Jun 19, 2012 12:56 PM
Messages: [ Previous | Next ]
mueckenh@rz.fh-augsburg.de Posts: 12,782 Registered: 1/29/05
Re: Matheology § 038
Posted: Jun 15, 2012 2:08 PM
On 15 Jun., 19:37, PotatoSauce <kiwisqu...@gmail.com> wrote:
> On Friday, June 15, 2012 12:42:39 PM UTC-4, WM wrote:
> > Of course there is a limit involved. Otherwise you could not biject
> > all numbers.
>
> Bijection between Natural numbers and the Even Natural numbers.
>
> f: N ----> 2N,
>
> f(n) = 2n.
>
> What limit?
>
> > Without limit you cannot include all numbers unless you
> > find a last one.
>
> What "last one"? Tell me exactly why I need to find a "last one" to define the map
>
> f: N ---> 2N
>
> f(n) = 2n.
>
You define nothing else, but a bijection between finite initial
segments.
> > Do you think there was a last natural number in an
> > infinite set?
>
> 1) Your question makes no sense.
You are wrong. What does not make sense is to talk about "all" natural
numbers as if there was a last one. What does absolutely not make
sense is finished infinity.
>
> 2) It has absolutely nothing to do with bijections.
>
Look here:
Every initial segment of the even positive numbers
(2, 4, ..., 2n) has cardinality n, i.e., less than 2n.
That can be proved for every finite initial segment and every natural
number n.
|{2, 4, ..., 2n}| = n < 2n
You need the limit in order to breack that rule.
Unless we go to the limit, there is no chance to have a cardinality
aleph_0 of a set that contains only finite numbers 2n.
Regards, WM
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6/19/12 mueckenh@rz.fh-augsburg.de
6/19/12 Uergil
6/15/12 Math | 1,571 | 3,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2013-20 | latest | en | 0.840509 |
http://physics.stackexchange.com/questions/tagged/voltage?page=4&sort=newest&pagesize=15 | 1,469,356,834,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823996.40/warc/CC-MAIN-20160723071023-00010-ip-10-185-27-174.ec2.internal.warc.gz | 188,821,956 | 25,977 | # Tagged Questions
Voltage is the unit of measurement for electronic potential, from one point location to another.
21 views
### Voltmeter across two loops
If one terminal of a voltmeter is connected to a closed loop with current and the other to an open loop, will there be a reading? Theoretically, there is a potential difference across the terminals, ...
16 views
### Could a Kelvin Probe Force Microscope, in principal, be used as a voltmeter?
This question goes into the very nature of the work function that the Kelvin Probe Force Microscope (KPFM) measures. https://en.wikipedia.org/wiki/Kelvin_probe_force_microscope Let's say, you have a ...
48 views
### Voltage of a quadrupole magnet
I have a simple question and it's my first one in this community. :) Does the voltage of a quadrupole magnet depend on the power of the electron beam in a synchrotron? Perhaps someone has a good ...
370 views
### Which power equation to use: $P = I^2 * R$ or $P = V^2 / R$?
Given are ideal max voltage $V = 200\;\mathrm{V}$ and max current $I = 5\;\mathrm{A}$. Therefore: ideal resistance is $$R = \frac VI = \frac{200 \;\mathrm{V}}{5\;\mathrm{A}} = 40 \;\mathrm{\Omega}$$...
4k views
### How can be the neutral wire at 0 volts when current flowing through it?
Voltage is potential difference, and current flows because of voltage. So if the voltage is zero, how can current flow through the neutral wire.
73 views
### Current flow direction, based off the equipotential line?
From the following diagram: Given the wires are only connected to a certain portion of the conductor, will current flow all around the conductor? Or only throughout the equipotential line(diagram ...
21 views
### Does doping and size determine the characteristics of a semi-conductor?
Does the maximum voltage and current that a semiconductor can withstand (without being damaged) depend upon the size and doping of the semiconductor? If so, then please explain, 1.What is the reason ...
249 views
### Voltage - Energy drop
I'm having a hard time understanding the nature of voltage and am hoping you guys can help. The main issue is the concept of the voltage drop. Take the following circuits : In regard to the first, ...
123 views
### Difference in induced current, when magnetic field “span” is reduced?
A conductor of known volume $(V)$ passes a uniform magnetic field$(B)$with a constant velocity $(v)$ the conductor is a source of induced EMF, a power source to a circuit. The induced EMF can be ...
75 views
### What happens if I have a square conducting wire being permeated by a magnetic field and the field suddenly disappears?
Suppose there is a square conducting wire in a magnetic field. The two vertical branches will have the same emf, and the two horizontal branches will have 0 emf, resulting in a net emf of zero and ...
296 views
### Can electromagnets repel?
So I saw a video here https://www.youtube.com/watch?v=wzXRFp0DDrU (See the full video first) I don't understand that how is the electromagnet repelling things in the second half of the video? Also ...
40 views
### Is this diagram in my textbook not constructed to scale?
Or am I having an illusion. Clearly, the distance from 6v to 4v is not the same as 4v to 2v. I know it should be based on: $V = k\frac{q}{r}$.
952 views
### How power lines use high voltages with a low current?
I've read that power lines use high voltages and low currents to reduce power loss due to resistance. Looking at the formula for power - P = VI So to increase P, you increase V rather than I for ... | 844 | 3,562 | {"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} | 3.421875 | 3 | CC-MAIN-2016-30 | latest | en | 0.881081 |
https://quant.stackexchange.com/posts/43761/revisions | 1,561,473,316,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999838.27/warc/CC-MAIN-20190625132645-20190625154645-00484.warc.gz | 564,742,459 | 16,363 | Post Closed as "duplicate" by Lliane, skoestlmeier, amdopt, Helin, Ezy of
3 edited body
I'm trying to calculate my return on investment (ROI) for an options position on margin that has been rolled. I'll give an example:
1. Sell to Open (STO) a naked put position, for which I collect 100 premium, and the margin put up is 1000, for example. I have 1 contract.
2. Roll the position by Buying to Close (BTC) the short, which costs me 50, and then STO another naked put for 150, with a new margin requirement of 1750.
I know each transactional ROI:
1. STO 100/1000 = 10%
2. BTC 50/1000 = -5%
3. STO 150/1750 = 8.6%
My question is, what is the overall positional ROI, calculated from these transactional ROIs? How to weight each ROI?
Is it weighted according to the latest margin, so:
10.1000/1750 - 5.1000/1750 + 8.6.1750/1750 = 1711.16%45%?
I'm trying to calculate my return on investment (ROI) for an options position on margin that has been rolled. I'll give an example:
1. Sell to Open (STO) a naked put position, for which I collect 100 premium, and the margin put up is 1000, for example. I have 1 contract.
2. Roll the position by Buying to Close (BTC) the short, which costs me 50, and then STO another naked put for 150, with a new margin requirement of 1750.
I know each transactional ROI:
1. STO 100/1000 = 10%
2. BTC 50/1000 = -5%
3. STO 150/1750 = 8.6%
My question is, what is the overall positional ROI, calculated from these transactional ROIs? How to weight each ROI?
Is it weighted according to the latest margin, so:
10.1000/1750 - 5.1000/1750 + 8.6.1750/1750 = 17.16%?
I'm trying to calculate my return on investment (ROI) for an options position on margin that has been rolled. I'll give an example:
1. Sell to Open (STO) a naked put position, for which I collect 100 premium, and the margin put up is 1000, for example. I have 1 contract.
2. Roll the position by Buying to Close (BTC) the short, which costs me 50, and then STO another naked put for 150, with a new margin requirement of 1750.
I know each transactional ROI:
1. STO 100/1000 = 10%
2. BTC 50/1000 = -5%
3. STO 150/1750 = 8.6%
My question is, what is the overall positional ROI, calculated from these transactional ROIs? How to weight each ROI?
Is it weighted according to the latest margin, so:
10.1000/1750 - 5.1000/1750 + 8.6.1750/1750 = 11.45%?
2 edited body
I'm trying to calculate my return on investment (ROI) for an options position on margin that has been rolled. I'll give an example:
1. Sell to Open (STO) a naked put position, for which I collect 100 premium, and the margin put up is 1000, for example. I have 1 contract.
2. Roll the position by Buying to Close (BTC) the short, which costs me 50, and then STO another naked put for 150, with a new margin requirement of 1750.
I know each transactional ROI:
1. STO 100/1000 = 10%
2. BTC 50/1000 = -5%
3. STO 150/15001750 = 8.6%
My question is, what is the overall positional ROI, calculated from these transactional ROIs? How to weight each ROI?
Is it weighted according to the latest margin, so:
10.1000/15001750 - 5.1000/15001750 + 8.6.15001750/15001750 = 1817.59%16%?
I'm trying to calculate my return on investment (ROI) for an options position on margin that has been rolled. I'll give an example:
1. Sell to Open (STO) a naked put position, for which I collect 100 premium, and the margin put up is 1000, for example. I have 1 contract.
2. Roll the position by Buying to Close (BTC) the short, which costs me 50, and then STO another naked put for 150, with a new margin requirement of 1750.
I know each transactional ROI:
1. STO 100/1000 = 10%
2. BTC 50/1000 = -5%
3. STO 150/1500 = 8.6%
My question is, what is the overall positional ROI, calculated from these transactional ROIs? How to weight each ROI?
Is it weighted according to the latest margin, so:
10.1000/1500 - 5.1000/1500 + 8.6.1500/1500 = 18.59%?
I'm trying to calculate my return on investment (ROI) for an options position on margin that has been rolled. I'll give an example:
1. Sell to Open (STO) a naked put position, for which I collect 100 premium, and the margin put up is 1000, for example. I have 1 contract.
2. Roll the position by Buying to Close (BTC) the short, which costs me 50, and then STO another naked put for 150, with a new margin requirement of 1750.
I know each transactional ROI:
1. STO 100/1000 = 10%
2. BTC 50/1000 = -5%
3. STO 150/1750 = 8.6%
My question is, what is the overall positional ROI, calculated from these transactional ROIs? How to weight each ROI?
Is it weighted according to the latest margin, so:
10.1000/1750 - 5.1000/1750 + 8.6.1750/1750 = 17.16%?
1
Return on Investment for rolled options position on margin
I'm trying to calculate my return on investment (ROI) for an options position on margin that has been rolled. I'll give an example:
1. Sell to Open (STO) a naked put position, for which I collect 100 premium, and the margin put up is 1000, for example. I have 1 contract.
2. Roll the position by Buying to Close (BTC) the short, which costs me 50, and then STO another naked put for 150, with a new margin requirement of 1750.
I know each transactional ROI:
1. STO 100/1000 = 10%
2. BTC 50/1000 = -5%
3. STO 150/1500 = 8.6%
My question is, what is the overall positional ROI, calculated from these transactional ROIs? How to weight each ROI?
Is it weighted according to the latest margin, so:
10.1000/1500 - 5.1000/1500 + 8.6.1500/1500 = 18.59%? | 1,663 | 5,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-26 | latest | en | 0.91389 |
https://examsdb.com/admission/asvab-section-2/after-one-year-what-is-the-interest-owed-on-this-loan-3/ | 1,542,492,544,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743854.48/warc/CC-MAIN-20181117205946-20181117231946-00452.warc.gz | 633,448,765 | 12,373 | Connie loaned Sam \$500 at a yearly interest rate of 3%. After one year what is the interest owed on this loan?
A. \$36
B. \$515
C. \$3
D. \$15
Explanation:
The yearly interest charged on this loan is the annual interest rate (3%) multiplied by the amount borrowed (\$500). To get the interest charged first convert the
interest rate from a percentage to a decimal by dividing by 100 (3 / 100) to get 0.03 then multiply 0.03 by \$500 to get \$15.
NOTE
When you’re dealing with percentages you’ll always need to convert the percent to a decimal by dividing by 100 before doing any calculations with it. | 154 | 607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | latest | en | 0.938347 |
https://forum.freecodecamp.org/t/algorithms-find-the-symmetric-difference/628164 | 1,723,238,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00683.warc.gz | 190,293,888 | 5,731 | # Algorithms - Find the Symmetric Difference
Tell us what’s happening:
Can someone explain what is wrong with my solution
``````function sym(args) {
var currentElement =[]
for(var i =0; i< args.length; i+=2){
if(i ==0){
currentElement = args[i]
}else{
currentElement = checkSym(currentElement, args[i]);
}
if(i !== args.length-1){
currentElement = checkSym(currentElement, args[i+1]);
}
}
return currentElement;
}
function checkSym(a1, a2){
const a = a1.filter(x=> !a2.includes(x));
const b = a2.filter(x=> !a1.includes(x));
const result = a.concat(b);
return result;
}
sym([[1, 2, 3], [5, 2, 1, 4]]);
``````
User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/115.0.0.0 Safari/537.36 Edg/115.0.1901.188`
Challenge: Algorithms - Find the Symmetric Difference
Minor formatting fixes:
``````function sym(args) {
let currentElement = [];
for (let i = 0; i < args.length; i += 2) {
if (i === 0) {
currentElement = args[i];
} else {
currentElement = checkSym(currentElement, args[i]);
}
if (i !== args.length - 1) {
currentElement = checkSym(currentElement, args[i + 1]);
}
}
return currentElement;
}
function checkSym(a1, a2) {
const a = a1.filter(x => !a2.includes(x));
const b = a2.filter(x => !a1.includes(x));
const result = a.concat(b);
return result;
}
sym([[1, 2, 3], [5, 2, 1, 4]]);
``````
It looks like you changed the meaning of the function signature of `sym` - look at the sample function calls.
Also, you aren’t handling duplication inside of a single array.
1 Like
This topic was automatically closed 182 days after the last reply. New replies are no longer allowed. | 483 | 1,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-33 | latest | en | 0.528711 |
https://www.anakpintar.web.id/2010/01/huffman-coding-in-tree-diagram.html | 1,652,715,278,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510138.6/warc/CC-MAIN-20220516140911-20220516170911-00378.warc.gz | 734,400,604 | 28,434 | ### Huffman Coding in a Tree Diagram
Huffman code is basically Initialize_model code prefix (prefix code) which is a set that contains a set of binary code. Prefix code is represented as a labeled binary tree where each side is labeled 0 (left branch) or 1 (right branch). Series of bits that form on each path from the root to leaf is a prefix code for the character that matched.
This code also has a wide variety including:
1. Adaptive Huffman coding
2. Length-Limited Huffman Coding
3. N-Ary Huffman Template Algorithm
4. Huffman With Unequal Letter Costs
1. Variations various Huffman Code
A. Adaptive Huffman Coding
Adaptive methods are used at the time of renewal (update) the new algorithm models both the process of compression and decompression
Basic Concept:
Encoder:
Initialize_model
Repeat for each character
(
Encode character
Update_model
)
Decoder
Initialize_model
Repeat for each character
(
Decode character
Update_model
)
The problem is how to update the model consists of algorithms that increase the number and update the Huffman tree. The trick is to update the tree where it is the compression / nirmampat. Huffman tree initialized with a single node, known as Not-Yet-Transmitted (NYT) Code that is sent each time a new character is found. The algorithm works with a unique numbering on the nodes with different number of leaves.
Lawyer steps update the model
1. If the code was first discovered NYT, then add the two nodes at node NYT. One node as a node NYT and other node as a leaf. Add the number of leaves. If not the NYT, straight to the leaves.
2. If the block does not have the highest rates, exchange with the highest number of blocks.
3. Add the number of node.
4. Check whether the node is the root node. If not go to the parent node.
B. Length-Limited Huffman Coding
Huffman variation is used to obtain the depth of the smallest distance from a symbol, with the restriction that the length of each of which included no less than a given constant value. This method is usually used with GNU gzip.
The steps in Method Length-Limited Huffman are:
1. Selecting two or more symbols to be compressed
2. Combine these symbols and replace them with pseudo-symbol and its frequency.
3. Perform the above steps are iterative until all the nodes that have a single root node.
4. If these nodes have the same frequency, then choose the node with the shortest depth.
C. Binary Huffman Template Algorima
This algorithm is similar to the ordinary Huffman algorithm. The difference, Huffman tree used in this algorithm has more than two roots (0 and 1). While the Huffman template algorithm, allowing to use non-numerical size (the size and frequency in addition to costs).
D. Huffman With Unequal Letter Costs
In this method there is a problem where a set of code that consists of several letters by frequency of occurrence and cost (cost) is different. This method is intended to find a prefix code (prefix code) and calculate the minimum cost (minimum cost). Prefix code is a set of prefix code-free. Cost (cost) of these is the amount of the cost of each letter in the code.
General steps method of Huffman codes with Unequal Letter Cost:
1. Looking for K-prefix code is optimal.
2. Changing K-prefix code into the optimal prefix code.
3. Having obtained the code and then calculate the cost (cost) of his.
Conclusion
Huffman codes proved to have many variations, among them the Adaptive Huffman Code, Code Length-limited Huffman, n-ary Huffman template algorithm, and Huffman codes with unequal letter costs. Various techniques can be used in applications different, but generally used for data compression. This naturally happens because the essence of all the variations of this technique is similar to Huffman codes, namely solving the optimization problem data.
### Komentar
1. Hopefully always succeeded for admin who always give the information and I say thank you a lot | 829 | 3,925 | {"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-2022-21 | latest | en | 0.920925 |
https://groups.yahoo.com/neo/groups/primenumbers/conversations/topics/2526?xm=1&o=1&m=p&tidx=1 | 1,430,086,385,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246656168.61/warc/CC-MAIN-20150417045736-00285-ip-10-235-10-82.ec2.internal.warc.gz | 962,508,337 | 18,135 | ## Wieferich primes
Expand Messages
• This post is about the Wieferich primes, 1093 and 3511. (1) Concerning the two in concert producing a prime P (2) Concerning what I have found to be an
Message 1 of 5 , Sep 4, 2001
• 0 Attachment
This post is about the Wieferich primes, 1093 and
3511.
(1) Concerning the two in concert producing a prime P
(2) Concerning what I have found to be an interesting
composite for P+1 and possibilities for primes close
to it
(3) Concerning 3511 itself, a Wieferich prime.
(1) Let f(x,y) = x^2 - y^2 -2xy.
f(3511,1093) = 7294949 and is prime P.
(2) 7294950 = 2*3*3*5*5*13*29*43 which made it easy to
factorise into its 8 factors, of which 5 are less than
(P+1)^(1/8) and all less than (P+1)^(1/4). This fact
seem to me to make 7294950 fairly interesting as only
6435 previous composites have the property of having 8
prime factors, one in more than a thousand.
(3) f(79,26) = 3511 itself remarkable as 79 = 3*26+1
which makes it a "near" Lucas series with its
characteristic second term 79 = 1mod3 and 1mod(the
first term viz. 26 in this case).
(I have postulated that any prime 1mod10 or -1mod10 is
always expressible with x > 3y in f(x,y), and that x
is the second term of a Fibonacci series T(1) = y,
T(2) = x.)
I have more interesting material but unless this
creates interest, I will spare you it.
John McNamara
__________________________________________________
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• ... The problem is that given two numbers one can find any number of features for them. The important thing is, does any of your message help the search for
Message 2 of 5 , Sep 4, 2001
• 0 Attachment
John McNamara wrote:
> This post is about the Wieferich primes, 1093 and 3511.
The problem is that given two numbers
one can find any number of features for them.
The important thing is, does any of your message help
the search for the next Wieferich prime, above 46 trillion:
http://www.loria.fr/~zimmerma/records/Wieferich.status
• ... There s a distributed hunt for them ongoing. http://catalan.ensor.org/ The client source is available, so anyone with a compiler can join. My alpha got to
Message 3 of 5 , Sep 5, 2001
• 0 Attachment
On Tue, 04 September 2001, d.broadhurst@... wrote:
> John McNamara wrote:
> > This post is about the Wieferich primes, 1093 and 3511.
> The problem is that given two numbers
> one can find any number of features for them.
> The important thing is, does any of your message help
> the search for the next Wieferich prime, above 46 trillion:
> http://www.loria.fr/~zimmerma/records/Wieferich.status
There's a distributed hunt for them ongoing.
http://catalan.ensor.org/
The client source is available, so anyone with a compiler can join. My alpha got to ~#5 on their tables a short while back.
Phil
Mathematics should not have to involve martyrdom;
Support Eric Weisstein, see http://mathworld.wolfram.com
Find the best deals on the web at AltaVista Shopping!
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• ... Notice that (1) should be read f(x,y) = x^2 - y^2 -xy to get f(3511,1093) = 7294949. Assuming this, other prime pairs (x,y) exist that satisfy f(x,y) = P
Message 4 of 5 , Sep 5, 2001
• 0 Attachment
--- In primenumbers@y..., John McNamara <mistermac39@y...> wrote:
> (1) Let f(x,y) = x^2 - y^2 -2xy.
> f(3511,1093) = 7294949 and is prime P.
>
> (2) 7294950 = 2*3*3*5*5*13*29*43 which made it easy to
> factorise into its 8 factors, of which 5 are less than
> (P+1)^(1/8) and all less than (P+1)^(1/4). This fact
> seem to me to make 7294950 fairly interesting as only
> 6435 previous composites have the property of having 8
> prime factors, one in more than a thousand.
Notice that (1) should be read f(x,y) = x^2 - y^2 -xy
to get f(3511,1093) = 7294949.
Assuming this, other prime pairs (x,y) exist that
satisfy f(x,y) = P (prime) and P+1 have the property
of having 8 prime factors.
As an example, the prime pair (47,2) gives
P=f(47,2)=2111 prime and P+1=2*2*2*2*2*2*3*11.
Regards
Flavio Torasso
• ... That s recently gone up to 150 trillion, and climbing fast: http://www.spacefire.com/numbertheory/wieferich.htm
Message 5 of 5 , Sep 5, 2001
• 0 Attachment
I wrote:
> the search for the next Wieferich prime, above 46 trillion
That's recently gone up to 150 trillion, and climbing fast:
http://www.spacefire.com/numbertheory/wieferich.htm
Your message has been successfully submitted and would be delivered to recipients shortly. | 1,389 | 4,492 | {"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} | 3.265625 | 3 | CC-MAIN-2015-18 | latest | en | 0.886689 |
http://mathhelpforum.com/calculus/146093-integrating-linear-polynomial-approx.html | 1,529,380,171,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861752.19/warc/CC-MAIN-20180619021643-20180619041643-00265.warc.gz | 211,382,605 | 9,929 | # Thread: Integrating a linear polynomial approx
1. ## Integrating a linear polynomial approx
hi,
thanks for reading.can anyone help me out?
see the attached file.i am working out for many hours how to integrate P1(t). does anyone know?
how do you get -1/2hf(..... ) + 3/2hf(....) ?
thats all i need to know!
i tried integrating MANY times. it looks easy.so what am i not understanding?
thanks all
2. Hi
The first integral is
$\displaystyle \left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i+1}^2}{2}- t_i t_{i+1}+\frac{t_i^2}{2}$
Substituting $\displaystyle t_{i+1} = t_i + h$
$\displaystyle \left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i}^2}{2} + h t_i + \frac{h^2}{2} - t_i^2 - h t_i + \frac{t_i^2}{2} = \frac{h^2}{2}$
3. Originally Posted by running-gag
Hi
The first integral is
$\displaystyle \left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i+1}^2}{2}- t_i t_{i+1}+\frac{t_i^2}{2}$
Substituting $\displaystyle t_{i+1} = t_i + h$
$\displaystyle \left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i}^2}{2} + h t_i + \frac{h^2}{2} - t_i^2 - h t_i + \frac{t_i^2}{2} = \frac{h^2}{2}$
aha...! thats where confusion happened... h can be anything =S
a difference between one point and another. thanks! =D | 495 | 1,259 | {"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} | 4.28125 | 4 | CC-MAIN-2018-26 | latest | en | 0.66335 |
https://www.diagramelectric.co/how-to-use-schematic-diagram/ | 1,701,233,269,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00065.warc.gz | 841,386,626 | 16,021 | # How To Use Schematic Diagram
## How to Use a Schematic Diagram
A schematic diagram is a simplified representation of an electrical circuit. It uses symbols to represent the components of the circuit, and lines to show how they are connected. Schematic diagrams are used to help engineers and technicians understand how a circuit works, and to troubleshoot problems.
### Parts of a Schematic Diagram
A schematic diagram typically includes the following parts: *
### Components:
The components of the circuit are represented by symbols. Common symbols include resistors, capacitors, inductors, transistors, and diodes. *
### Lines:
The lines show how the components are connected. Lines can be solid or dashed, and they can have arrows to indicate the direction of current flow. *
### Labels:
Labels are used to identify the components of the circuit. Labels can be numbers, letters, or words.
To read a schematic diagram, you first need to identify the components of the circuit. The symbols for the most common components are shown below.
Once you have identified the components, you can follow the lines to see how they are connected. The direction of current flow is indicated by arrows on the lines.
### Troubleshooting a Circuit
Schematic diagrams can be used to troubleshoot problems in a circuit. If a circuit is not working properly, you can use the schematic diagram to identify the components that are not working. To do this, you can start by checking the power supply. If the power supply is working properly, you can then check the components one by one. To check a component, you can use a multimeter to measure its resistance. If the resistance is too high or too low, the component is probably defective. You can also check the component with an ohmmeter to see if it is shorted. If the component is shorted, it will have a resistance of zero ohms.
### Conclusion
Schematic diagrams are a valuable tool for engineers and technicians. They can be used to understand how a circuit works, to troubleshoot problems, and to design new circuits. By learning how to read and use schematic diagrams, you can improve your understanding of electrical circuits. | 441 | 2,183 | {"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} | 3.125 | 3 | CC-MAIN-2023-50 | latest | en | 0.913597 |
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Example35
# Example35 - If however a ≠ 1(that is the first term has a...
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Example 5 Factor x 2 – 5 x – 14. x – 7)( x + 2) Notice that 7 × 2 = 14 and 7 – 2 = 5, the coefficient of the middle term. Also note that the sign of the larger factor, 7, is –, while the other factor, 2, has a + sign. To check,
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Ask a homework question - tutors are online | 268 | 767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-13 | latest | en | 0.281821 |
https://imoneypage.com/rotation-revolution-worksheet/ | 1,568,585,035,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00428.warc.gz | 528,071,980 | 29,981 | # Rotation Revolution Worksheet
In Free Printable Worksheets198 views
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Author: Inna Berg
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
Top | 1,360 | 7,151 | {"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-2019-39 | latest | en | 0.899556 |
https://btechgeeks.com/java-program-to-check-whether-given-four-points-form-pythagorean-quadruple/ | 1,680,023,555,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00247.warc.gz | 197,023,378 | 12,441 | # Java Program to Check Whether Given Four Points Form Pythagorean Quadruple
In the previous article, we have seen Java Program to Find Volume and Surface Area of Cuboid
In this article we are going to see how to check Whether Given Four Points Form Pythagorean Quadruple using Java programming language.
## Java Program to Check Whether Given Four Points Form Pythagorean Quadruple
Before Jumping into the program directly let’s see how we can Check Whether Given Four Points Form Pythagorean Quadruple.
Explanation:
Let
l be the length,
h be the height and
d be the diagonal of the quadruple.
If (l*l)+(b*b)+(h*h)=(d*d) then it is said to be Pythagorean quadruple.
Example:
Suppose we have below values
l = 1
b = 2
h = 2
d = 3
Then according to Pythagorean quadruple principle,
LHS
(l*l)+(b*b)+(h*h) = 1+4+4 = 9
RHS
d*d = 3*3 = 9
Since LHS = RHS
Therefore it is said to be Pythagorean quadruple.
Let’s see different ways to check whether given four points form Pythagorean quadruple or not.
### Method-1: Java Program to Check Whether Given Four Points Form Pythagorean Quadruple By Using Static Value
Approach:
• Declare an integer variable say ‘l’, ‘b’, ‘h’, ‘d’ and assign the value to it, which holds the value of length, breadth, height, diagonal of quadruple respectively.
• Find the LHS and RHS of the formula (l*l)+(b*b)+(h*h)=(d*d)
• If LHS is equal to RHS then four points are from Pythagorean quadruple.
• Print the result.
Program:
class Main
{
public static void main(String[] args)
{
//length, breadth, height, diagonal value declared
int l = 1;
int b = 2;
int h = 2;
int d = 3;
//find sum of (l*l)+(b*b)+(h*h)
int sum = (l*l)+(b*b)+(h*h);
//check the sum is equal to d*d or not
if(sum == d*d)
else
System.out.println("It is not a Pythagorean quadruple");
}
}
Output:
It is a Pythagorean quadruple
### Method-2: Java Program to Check Whether Given Four Points Form Pythagorean Quadruple By Using User Input Value
Approach:
• Declare an integer variable say ‘l’, ‘b’, ‘h’, ‘d’ and take the values as user input by using Scanner class, which holds the value of length, breadth, height and diagonal of quadruple respectively.
• Find the LHS and RHS of the formula (l*l)+(b*b)+(h*h)=(d*d)
• If LHS is equal to RHS then four points are from Pythagorean quadruple.
• Print the result.
Program:
import java.util.*;
class Main
{
public static void main(String[] args)
{
//Scanner class object created
Scanner s = new Scanner(System.in);
System.out.println("Enter the value of l:");
int l = s.nextInt();
System.out.println("Enter the value of b:");
int b = s.nextInt();
System.out.println("Enter the value of h:");
int h = s.nextInt();
System.out.println("Enter the value of d:");
int d = s.nextInt();
//find sum of (l*l)+(b*b)+(h*h)
int sum = (l*l)+(b*b)+(h*h);
//check the sum is equal to d*d or not
if(sum == d*d)
else
System.out.println("It is not a Pythagorean quadruple");
}
}
Output:
Case-1
Enter the value of l:
1
Enter the value of b:
2
Enter the value of h:
2
Enter the value of d:
3
Case-2
Enter the value of l:
3
Enter the value of b:
4
Enter the value of h:
5
Enter the value of d:
6
It is not a Pythagorean quadruple
Have you mastered basic programming topics of java and looking forward to mastering advanced topics in a java programming language? Go with these ultimate Advanced java programs examples with output & achieve your goal in improving java coding skills.
Related Java Articles: | 920 | 3,443 | {"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} | 3.859375 | 4 | CC-MAIN-2023-14 | latest | en | 0.716989 |
https://www.nag.com/numeric/cl/nagdoc_cl26.0/html/f12/f12agc.html | 1,638,726,914,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00049.warc.gz | 985,505,165 | 7,938 | nag_real_banded_sparse_eigensystem_sol (f12agc) (PDF version)
f12 Chapter Contents
f12 Chapter Introduction
NAG Library Manual
# NAG Library Function Documentnag_real_banded_sparse_eigensystem_sol (f12agc)
Note: this function uses optional parameters to define choices in the problem specification. If you wish to use default settings for all of the optional parameters, then the option setting function nag_real_sparse_eigensystem_option (f12adc) need not be called. If, however, you wish to reset some or all of the settings please refer to Section 11 in nag_real_sparse_eigensystem_option (f12adc) for a detailed description of the specification of the optional parameters.
## 1 Purpose
nag_real_banded_sparse_eigensystem_sol (f12agc) is the main solver function in a suite of functions consisting of nag_real_sparse_eigensystem_option (f12adc), nag_real_banded_sparse_eigensystem_init (f12afc) and nag_real_banded_sparse_eigensystem_sol (f12agc). It must be called following an initial call to nag_real_banded_sparse_eigensystem_init (f12afc) and following any calls to nag_real_sparse_eigensystem_option (f12adc).
nag_real_banded_sparse_eigensystem_sol (f12agc) returns approximations to selected eigenvalues, and (optionally) the corresponding eigenvectors, of a standard or generalized eigenvalue problem defined by real banded nonsymmetric matrices. The banded matrix must be stored using the LAPACK column ordered storage format for real banded nonsymmetric (see Section 3.3.4 in the f07 Chapter Introduction).
## 2 Specification
#include #include
void nag_real_banded_sparse_eigensystem_sol (Integer kl, Integer ku, const double ab[], const double mb[], double sigmar, double sigmai, Integer *nconv, double dr[], double di[], double z[], double resid[], double v[], double comm[], Integer icomm[], NagError *fail)
## 3 Description
The suite of functions is designed to calculate some of the eigenvalues, $\lambda$, (and optionally the corresponding eigenvectors, $x$) of a standard eigenvalue problem $Ax=\lambda x$, or of a generalized eigenvalue problem $Ax=\lambda Bx$ of order $n$, where $n$ is large and the coefficient matrices $A$ and $B$ are banded, real and nonsymmetric.
Following a call to the initialization function nag_real_banded_sparse_eigensystem_init (f12afc), nag_real_banded_sparse_eigensystem_sol (f12agc) returns the converged approximations to eigenvalues and (optionally) the corresponding approximate eigenvectors and/or an orthonormal basis for the associated approximate invariant subspace. The eigenvalues (and eigenvectors) are selected from those of a standard or generalized eigenvalue problem defined by real banded nonsymmetric matrices. There is negligible additional computational cost to obtain eigenvectors; an orthonormal basis is always computed, but there is an additional storage cost if both are requested.
The banded matrices $A$ and $B$ must be stored using the LAPACK column ordered storage format for banded nonsymmetric matrices; please refer to Section 3.3.2 in the f07 Chapter Introduction for details on this storage format.
nag_real_banded_sparse_eigensystem_sol (f12agc) is based on the banded driver functions dnbdr1 to dnbdr6 from the ARPACK package, which uses the Implicitly Restarted Arnoldi iteration method. The method is described in Lehoucq and Sorensen (1996) and Lehoucq (2001) while its use within the ARPACK software is described in great detail in Lehoucq et al. (1998). An evaluation of software for computing eigenvalues of sparse nonsymmetric matrices is provided in Lehoucq and Scott (1996). This suite of functions offers the same functionality as the ARPACK banded driver software for real nonsymmetric problems, but the interface design is quite different in order to make the option setting clearer and to combine the different drivers into a general purpose function.
nag_real_banded_sparse_eigensystem_sol (f12agc), is a general purpose function that must be called following initialization by nag_real_banded_sparse_eigensystem_init (f12afc). nag_real_banded_sparse_eigensystem_sol (f12agc) uses options, set either by default or explicitly by calling nag_real_sparse_eigensystem_option (f12adc), to return the converged approximations to selected eigenvalues and (optionally):
– the corresponding approximate eigenvectors; – an orthonormal basis for the associated approximate invariant subspace; – both.
Please note that for ${\mathbf{Generalized}}$ problems, the ${\mathbf{Shifted Inverse Imaginary}}$ and ${\mathbf{Shifted Inverse Real}}$ inverse modes are only appropriate if either $A$ or $B$ is symmetric semidefinite. Otherwise, if $A$ or $B$ is non-singular, the ${\mathbf{Standard}}$ problem can be solved using the matrix ${B}^{-1}A$ (say).
## 4 References
Lehoucq R B (2001) Implicitly restarted Arnoldi methods and subspace iteration SIAM Journal on Matrix Analysis and Applications 23 551–562
Lehoucq R B and Scott J A (1996) An evaluation of software for computing eigenvalues of sparse nonsymmetric matrices Preprint MCS-P547-1195 Argonne National Laboratory
Lehoucq R B and Sorensen D C (1996) Deflation techniques for an implicitly restarted Arnoldi iteration SIAM Journal on Matrix Analysis and Applications 17 789–821
Lehoucq R B, Sorensen D C and Yang C (1998) ARPACK Users' Guide: Solution of Large-scale Eigenvalue Problems with Implicitly Restarted Arnoldi Methods SIAM, Philidelphia
## 5 Arguments
1: $\mathbf{kl}$IntegerInput
On entry: the number of subdiagonals of the matrices $A$ and $B$.
Constraint: ${\mathbf{kl}}\ge 0$.
2: $\mathbf{ku}$IntegerInput
On entry: the number of superdiagonals of the matrices $A$ and $B$.
Constraint: ${\mathbf{ku}}\ge 0$.
3: $\mathbf{ab}\left[\mathit{dim}\right]$const doubleInput
Note: the dimension, dim, of the array ab must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}},×,\left(2×{\mathbf{kl}}+{\mathbf{ku}}+1\right)\right)$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On entry: must contain the matrix $A$ in LAPACK column-ordered banded storage format for nonsymmetric matrices (see Section 3.3.4 in the f07 Chapter Introduction).
4: $\mathbf{mb}\left[\mathit{dim}\right]$const doubleInput
Note: the dimension, dim, of the array mb must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}},×,\left(2×{\mathbf{kl}}+{\mathbf{ku}}+1\right)\right)$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On entry: must contain the matrix $B$ in LAPACK column-ordered banded storage format for nonsymmetric matrices (see Section 3.3.4 in the f07 Chapter Introduction).
5: $\mathbf{sigmar}$doubleInput
On entry: if one of the ${\mathbf{Shifted Inverse Real}}$ modes (see nag_real_sparse_eigensystem_option (f12adc)) have been selected then sigmar must contain the real part of the shift used; otherwise sigmar is not referenced. Section 4.3.4 in the f12 Chapter Introduction describes the use of shift and inverse transformations.
6: $\mathbf{sigmai}$doubleInput
On entry: if one of the ${\mathbf{Shifted Inverse Real}}$ modes (see nag_real_sparse_eigensystem_option (f12adc)) have been selected then sigmai must contain the imaginary part of the shift used; otherwise sigmai is not referenced. Section 4.3.4 in the f12 Chapter Introduction describes the use of shift and inverse transformations.
7: $\mathbf{nconv}$Integer *Output
On exit: the number of converged eigenvalues.
8: $\mathbf{dr}\left[\mathit{dim}\right]$doubleOutput
Note: the dimension, dim, of the array dr must be at least ${\mathbf{nev}}+1$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On exit: the first nconv locations of the array dr contain the real parts of the converged approximate eigenvalues. The number of eigenvalues returned may be one more than the number requested by nev since complex values occur as conjugate pairs and the second in the pair can be returned in position ${\mathbf{nev}}+1$ of the array.
9: $\mathbf{di}\left[\mathit{dim}\right]$doubleOutput
Note: the dimension, dim, of the array di must be at least ${\mathbf{nev}}+1$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On exit: the first nconv locations of the array di contain the imaginary parts of the converged approximate eigenvalues. The number of eigenvalues returned may be one more than the number requested by nev since complex values occur as conjugate pairs and the second in the pair can be returned in position ${\mathbf{nev}}+1$ of the array.
10: $\mathbf{z}\left[\mathit{dim}\right]$doubleOutput
Note: the dimension, dim, of the array z must be at least ${\mathbf{n}}×\left({\mathbf{nev}}+1\right)$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On exit: if the default option ${\mathbf{Vectors}}=\text{Ritz}$ has been selected then z contains the final set of eigenvectors corresponding to the eigenvalues held in dr and di. The complex eigenvector associated with the eigenvalue with positive imaginary part is stored in two consecutive array segments. The first segment holds the real part of the eigenvector and the second holds the imaginary part. The eigenvector associated with the eigenvalue with negative imaginary part is simply the complex conjugate of the eigenvector associated with the positive imaginary part.
For example, if ${\mathbf{di}}\left[0\right]$ is nonzero, the first eigenvector has real parts stored in locations ${\mathbf{z}}\left[\mathit{i}\right]$, for $\mathit{i}=0,1,\dots {\mathbf{n}}-1$ and imaginary parts stored in ${\mathbf{z}}\left[\mathit{i}\right]$, for $\mathit{i}={\mathbf{n}},\dots ,2{\mathbf{n}}-1$.
11: $\mathbf{resid}\left[\mathit{dim}\right]$doubleInput/Output
Note: the dimension, dim, of the array resid must be at least ${\mathbf{n}}$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On entry: need not be set unless the option ${\mathbf{Initial Residual}}$ has been set in a prior call to nag_real_sparse_eigensystem_option (f12adc) in which case resid must contain an initial residual vector.
On exit: contains the final residual vector.
12: $\mathbf{v}\left[\mathit{dim}\right]$doubleOutput
Note: the dimension, dim, of the array v must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}×{\mathbf{ncv}}\right)$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On exit: if the option ${\mathbf{Vectors}}$ (see nag_real_sparse_eigensystem_option (f12adc)) has been set to Schur or Ritz then the first nconv sections of v, of length $n$, will contain approximate Schur vectors that span the desired invariant subspace.
The $i$th Schur vector is stored in locations ${\mathbf{v}}\left[{\mathbf{n}}×\left(\mathit{i}-1\right)+\mathit{j}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{nconv}}$ and $\mathit{j}=1,2,\dots ,n$.
13: $\mathbf{comm}\left[\mathit{dim}\right]$doubleCommunication Array
Note: the dimension, dim, of the array comm must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{lcomm}}\right)$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On exit: contains no useful information.
14: $\mathbf{icomm}\left[\mathit{dim}\right]$IntegerCommunication Array
Note: the dimension, dim, of the array icomm must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{licomm}}\right)$ (see nag_real_banded_sparse_eigensystem_init (f12afc)).
On exit: contains no useful information.
15: $\mathbf{fail}$NagError *Input/Output
The NAG error argument (see Section 2.7 in How to Use the NAG Library and its Documentation).
## 6 Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information.
NE_BAD_PARAM
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_COMP_BAND_FAC
Failure during internal factorization of complex banded matrix. Please contact NAG.
NE_COMP_BAND_SOL
Failure during internal solution of complex banded matrix. Please contact NAG.
NE_INITIALIZATION
Either the initialization function has not been called prior to the first call of this function or a communication array has become corrupted.
NE_INT
On entry, ${\mathbf{kl}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{kl}}\ge 0$.
On entry, ${\mathbf{ku}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ku}}\ge 0$.
The maximum number of iterations $\text{}\le 0$, the option ${\mathbf{Iteration Limit}}$ has been set to $〈\mathit{\text{value}}〉$.
NE_INTERNAL_EIGVAL_FAIL
Error in internal call to compute eigenvalues and corresponding error bounds of the current upper Hessenberg matrix. Please contact NAG.
NE_INTERNAL_EIGVEC_FAIL
Error in internal call to compute eigenvectors. Please contact NAG.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
An unexpected error has been triggered by this function. Please contact NAG.
See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information.
NE_INVALID_OPTION
On entry, ${\mathbf{Vectors}}=\text{Select}$, but this is not yet implemented.
NE_MAX_ITER
The maximum number of iterations has been reached. The maximum number of $\text{iterations}=〈\mathit{\text{value}}〉$. The number of converged eigenvalues $\text{}=〈\mathit{\text{value}}〉$.
NE_NO_ARNOLDI_FAC
Could not build an Arnoldi factorization. The size of the current Arnoldi factorization $=〈\mathit{\text{value}}〉$.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information.
NE_NO_SHIFTS_APPLIED
No shifts could be applied during a cycle of the implicitly restarted Lanczos iteration.
NE_OPT_INCOMPAT
The options ${\mathbf{Generalized}}$ and ${\mathbf{Regular}}$ are incompatible.
NE_REAL_BAND_FAC
Failure during internal factorization of real banded matrix. Please contact NAG.
NE_REAL_BAND_SOL
Failure during internal solution of real banded matrix. Please contact NAG.
NE_SCHUR_EIG_FAIL
During calculation of a real Schur form, there was a failure to compute a number of eigenvalues. Please contact NAG.
NE_SCHUR_REORDER
The computed Schur form could not be reordered by an internal call. Please contact NAG.
NE_TRANSFORM_OVFL
Overflow occurred during transformation of Ritz values to those of the original problem.
NE_ZERO_EIGS_FOUND
The number of eigenvalues found to sufficient accuracy is zero.
NE_ZERO_INIT_RESID
The option ${\mathbf{Initial Residual}}$ was selected but the starting vector held in resid is zero.
NE_ZERO_SHIFT
The option ${\mathbf{Shifted Inverse Imaginary}}$ has been selected and ${\mathbf{sigmai}}=\text{}$ zero on entry; sigmai must be nonzero for this mode of operation.
## 7 Accuracy
The relative accuracy of a Ritz value, $\lambda$, is considered acceptable if its Ritz estimate $\le {\mathbf{Tolerance}}×\left|\lambda \right|$. The default ${\mathbf{Tolerance}}$ used is the machine precision given by nag_machine_precision (X02AJC).
## 8 Parallelism and Performance
nag_real_banded_sparse_eigensystem_sol (f12agc) is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
nag_real_banded_sparse_eigensystem_sol (f12agc) makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the x06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
None.
## 10 Example
This example constructs the matrices $A$ and $B$ using LAPACK band storage format and solves $Ax=\lambda Bx$ in shifted imaginary mode using the complex shift $\sigma$.
### 10.1 Program Text
Program Text (f12agce.c)
### 10.2 Program Data
Program Data (f12agce.d)
### 10.3 Program Results
Program Results (f12agce.r)
nag_real_banded_sparse_eigensystem_sol (f12agc) (PDF version)
f12 Chapter Contents
f12 Chapter Introduction
NAG Library Manual | 4,373 | 16,215 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 92, "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-2021-49 | latest | en | 0.392793 |
http://excel.bigresource.com/Sum-Alternate-Cells-dNvGVWaO.html | 1,526,976,450,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864648.30/warc/CC-MAIN-20180522073245-20180522093245-00232.warc.gz | 102,827,667 | 12,645 | # Sum Alternate Cells
Apr 24, 2008
I need to sum the contents of every fourth cell in row A .... ie sum(A1 + E1 + I1 ... BY1)
Can I do this via the formula bar, or do I need to use a VBA routine?
## Row Range Sum Of Alternate Cells
Apr 18, 2013
I have a row range to sum only alternate cells.
How can i do this job by means of formula.
## Formula To Add Alternate Cells
Jul 17, 2006
I can't remember any formula to do the task. I have a time series with quarterly sales. The dates are in format given below.
01/01/1990 - first quarter of 1990
02/01/1990 - second quarter of 1990
03/01/1990 - Third quarter of 1990
04/01/1990 - fourth quarter of 1990
I need to add the sales of all first quarters , all second quarters..and so on.
Is there any formula that an elegant way or I have to do it manually?
## Countif In Alternate Cells
Jun 19, 2007
I have a table of data that is laid out in multiples of two columns. I have attached a simplified example. Essentially all I need to do is count the contents of every second cell when it equals a specific value. If you look at my attached example it will be clearer. how I can do this using formula? (I don't want to use VBA in this instance).
## How To Protect Alternate Cells Of Column
Apr 16, 2013
Is there any possibility to protect alternate cells of column see below for example. I want to lock cells B1, B6, B11 and B16 only so that no one is able to delete the average rate.
A B
average rate0.0
1
1
1
1
average rate0.0
1
1
1
1
average rate0.0
1
1
1
1
average rate0.0
1
1
1
1
## Macro To Populate Alternate Cells In A Row With A Value...
Mar 17, 2009
I have a spreadsheet for work rosters. Each person on the roster has a row with info on what they're doing for a given week. Split into the am and pm session of each day.
All I want to do is be able to fill the row with the same value if they're doing the same thing for the week - eg if they're on leave, I'd like to hit CTRL-L and have the row populated with 'LEAVE' in each alternate cell, rather than typing it manually. Note that it is each alternate cell, not each cell as the alternate cell has different info.
So a row would look like this:
Leave | blank |Leave | blank |Leave | blank |Leave | blank |Leave | blank |Leave | blank |Leave | blank |Leave | blank |Leave | blank |Leave | blank |
For the am and pm session of the five working days. And just to complicate matters, each 'cell' on my spreadsheet is actually 4 merged cells, and the alternate 'cell' that I want left alone is two merged cells! This is the macro generated when I do a simple record - it does what I want, but obviously jumps to the original row that I recorded it in whenever I run it - I need it to fill the row that I start it from. I recorded starting in cell I133:L134
## Fill Alternate Cells Macro, All Sheets
Dec 22, 2005
Would like a script that will go to each sheet in the workbook, and fill in light grey background, each alternate row that contains data, EXCLUDING the 1st row (header row).
## Find & Copy Cells & Paste To Alternate Columns
Dec 6, 2006
I have a spreadsheet that I would like to loop through column "C" and if criteria is met copy and paste A:C on sheet1 to sheet2 over multiple columns alternately. What I mean by alternately is that I would past the first row in column A6 then the second in E6 then the third in A7 and so on until all items are copied. I purposely left a blank column between both columns of information. I've tried sorting/and advance filtering and couldn't get it to work.
Example:
loop through column "C" If I have the Letter "A" copy data to column "A" and "E" alternately back an forth until I no longer meet the criteria. I start putting data on the 6th row due to header information in rows 1-5.
... If column "C" is the letter "B" copy to column I,M,Q,U
Lastly I could always have less rows of information than I do columns. SO the last column could be empty. I always sort my data by column "C" so data will be sequential.
## Change Colorindex Within A Range Of Multiple Cells [ Alternate Method]
Jan 8, 2010
I have tried to go around the long way to achieve this but came up with pages of pointless code .... I know there is a better way I just dont know enough about VB to do it myself ... And I know this is EASY for many :-)
--------
Cell ranges h11 to as11 are a totals row.
If the total is 0, colorindex is set to vbpatternnone, if >= 1, then colorindex is set to vbpatterngray. Easy right ? I just dont kn ow how to do FROM/IF/DO range loops...
--------
Details:
The code in worksheet_SelectionChange will contain the following:
1: From range h11 to as11, variable1 = application.interior.colorindex of the cell.
2: Check if the cell is >=1 or <=0 ....
3: If >=1 then set application.interior.colorindex = vbpatterngray. Go to #5.
' (This inserts a pattern over the original color of the cell)
4: If <=0 then set application.interior.colorindex = vbpatternNONE
ALSO set application.interior.colorindex = variable1
' (This clears the cell pattern and returns it to original color)
5. Repeat steps to clear cell pattern and restore color / or insert pattern for all cells from range H11:AS11
6. End sub
## Excel 2003 :: Copying Cells Based On Alternate Cell Value?
Sep 26, 2011
I am wondering if it possible to automate the copying of data from particular cells, based on a value in a different cell, into a different format.
So to go from this simplified table:
AB1NameLevel2Arthur2a3Briony3c4Catherine3b5David3a6Edward2a7Felicity3c8George3c
to something like this:
FGHI12a3c3b3a2ArthurBrionyCatherineDavid3EdwardFelicity
4
George
At the moment I do it all manually, and it takes forever. I am sure there must be a simple way of doing it. I am using Excel 2003, but could work in a newer version if required.
## Alternate Is Statement
May 12, 2009
i have attached a copy of an excel file and if you look at the end i am trying to write an IF statement that reads if J = "2-0" then column M = 12-(K) or 12--4 so answer is 16 and then column N = -column M or -16 and then if J = "2-1" then M = 6-(K) or 6--0 = 6 and then N = -M .....
so if J was "2-1" and (K) -4 then the answer given for column M would be 6--4=10 and column N = -10
## Sum Alternate Rows
May 6, 2008
I do an analysis that contains 100 or more rows with 1 of 2 row labels in column A; "existing" or "retrofit." There may be numerous successive rows labeled "existing" in a given place where there is no retrofit. Rows labeled "existing" contain existing equipment details such as area, equip description, operating cost. Rows labeled "retrofit" contain recommended efficient replacements with details such as area operating cost and savings. I have to display total cost and savings. Right now I use the awful method, F1+F3+F4+F5... for existing and F2+F6... for retrofit. This is a terrible method, time consuming, prone to error, etc. Is there a way to total rows with specific labels when the row labels are not consistently alternating? Would like to attach an example spreadsheet but... the permissions say I may not, for some strange reason.
## How To Delete Alternate Rows
Jun 13, 2014
I have data of about 3176 rows,
But after every record one row is blank,
I have to manually delete each blank row.
How to delete it in one Go.
## Reference To Alternate Tables
Jun 2, 2009
I want to reference to a range of tables depending on the value of a separate cell. Problem is that I don't know how to insert this into the range part of the lookup formula.
EG
=vlookup(a12,NAMED RANGE AS PER CELL c9,2,false)
Cell a12 is the lookup reference which is fine.
Named range is set-up and working fine.
Cell c9 is the description of the named range - season_indices_Asda_Cream
I want that to be changeable by he user so they can change the name in cell c9 from a drop down list so that the lookup formula redirects to the alternate named range...
## How To Paste In Alternate Columns
Jul 27, 2003
Is there any way that i can copy a row of data, but paste each individual cell that has been copied into every other column, or every 3rd column, or every x number column.
I want to do this as i have 3 column headers repeated across spreadsheet representing each month. and would like to paste into the relevant column i.e budget actual difference budget actual difference budget actual difference.
So I can paste each value under the actual column.
## HLookup On Alternate Cell?
Mar 5, 2012
Im wondering if their is a formula to return the results in a Hlookup but adjusted for the cell 1 cell to the right?
For instance if Row 1 Column one contains "Tom Jones" that the forumula knows to return the value in row 2 column 2 instead of row 2 column 1.
## One Column Into Two Using Alternate Rows?
Nov 29, 2013
I have a single column of data and need to convert it into 2 columns, by alternating each row:
1
2
3
4
5
want to convert to:
1
2
3
4
5
6
I've seen a previous post from 2007 where someone asked for a formula to do the reverse of this. 'Domenic' provided a formula =INDEX(\$A\$2:\$B\$6,INT((ROWS(D\$2:D2)-1)/2)+1,MOD(ROWS(D\$2:D2)-1,2)+1) which converted
a
1
b
2
c
3
d
4
[code]....
how to reverse this formula?
## How To Delete Alternate Rows
Nov 26, 2007
I have a huge excel file and every row is repeated, e.g.:
john smith 10 23
john smith 10 23
bob jones 11 22
bob jones 11 22
etc..
So I want to delete every second row.
Is there a command to do that?
## Alternate Way Of Displaying Data
Jan 25, 2009
I have been using a lot of sumproducting lately with multiple conditions to extract data. Lately I have noticed that though it is a good way to extract data there is a lot of calculation time involved in it. The Excel workbooks that i make are in a database like format where there is 1 sheet usually a data dump which has data from one column to the 200th column and rows being filled with data points till the 10000th row. Data headers in the columns are usually like Date(ColumnA), Tenure(ColumnB), Person, Type, etc and then from Column Z onwards there are columns which contain Data in the form of numbers like Number of cases, Number of this and number of that.
Now usually when creating a dashboard of this data for performance management I use the sumpoduct formula to retrieve data. It normally has conditions in it like for some given date ranges, Tenure ranges, People ranges extract x data for me. For Example something like this
=SUMPRODUCT((Sheet2!\$A\$4:\$A\$4898>=VALUE(\$E\$3))*(Sheet2!\$A\$4:\$A\$4898
## Select Every Other Row :: Alternate Rows
Apr 5, 2002
Im trying to find a way select several rows at the same time but starting at say row 3 and then alternate rows so rows 3,5,7,9 etc
## Alternate Formula For Sum Indirect
Feb 14, 2007
I have tried to apply '= SUM(INDIRECT("A2:A10"))' formula to do the SUM at cell A11. But, if I add two more Rows, then my formula moves down to cell A13 but numbers in Cell A11 and A12 does not get added to the total. How can I avoid that? I have reserached this site extensively and could not find an archived solution.
## Highlighting Alternate Weeks On Spreadsheet?
Dec 9, 2013
I would like to know how I could highlight Alternate Weeks on my Spreadsheet.
I am currently highlighting Weekends with Conditional Formatting. I need to change this to highlight alternating weeks Starting with the first Sunday to next Saturday, then skipping a Sunday-Sat and so on.
My Dates are in row C2:AH2 and days of the week C3:AH3.
I would like it to Highlight from row 2 to 52 on the appropriate days
## Delete Every Alternate Rows In Excel
Jun 25, 2014
How to delete every alternate rows in excel.
Eg. I need to delete even rows in my spreadsheet, (row 2, 4, 6 etc..)
AND
How to delete rows that contain both text and number but start with text then number.
Eg. My data as below:
12345hello
hello12345
123423hehehe
kekeek11
Result: what i want
12345hello
123423hehehe
## Limitation Of 7 Nested IF Statements - Alternate
Apr 27, 2007
I have a workbook with two worksheets, Sheet1 and Sheet2. Sheet2 contains a table of values that need to be input into a cell on Sheet1, pending the results of comparing two other cells on Sheet1. I have 8 possible variations resulting from that comparison and I cannot make this work as the IF statement limits you to 7 deep.
Example:
Sheet1
A1 (text string value) = LOWER
B1 (text string value) = L1
C1 (currency with no decimals) = Sheet2!Somecell (decision of which cell to use depends on combination of A1 and B1)
A1 can be either the string "LOWER" or "MIDDLE". B1 can be the strings "L1", "L2", "L3", or "L4". The strings in B1 are not cell references, but simple text. This leads to four variations for a row that has "LOWER" in it's A column, and the same for "MIDDLE" - totaling 8 possible combinations.
Depending on the combination, I need to input a number from Sheet2 and that number is different for each unique combination of the eight possibilities. There is no mathematical calculation taking place on Sheet2 - just an "if x and y then z" decision on Sheet1. I will use the value of Sheet1!C1 in other math functions on Sheet1.
## Copy & Paste In Alternate Columns
Jun 23, 2006
I'm trying to copy and paste range in alternate columns from one worksheet to another. I can record this macro, but I believe it'll be really long because I have 21 alternate columns to copy and paste. What I'm trying to do:
-copy range B9:B41 in workbook 'Channel OU template' then paste values only in range BI9:BI41 in workbook 'final'
-copy range D9:D41 to range BK9:BK41
-F9:F41 to BM9:BM41
..and so on until the last column AP9:AP41 to CW9:CW41
Basically it's just simple copying and pasting from alternate columns. This is the really basic code that I have just for one column:
Sub copy()
Windows("Channel OU template").Activate
Sheets("sheet1").Select
Range("b9:b41").copy
Windows("final").Activate
Sheets("ou").Select
Range("bi9").PasteSpecial xlPasteValues
End Sub
## Fill Alternate Rows With A Color
Jul 18, 2006
is there any easy way to fill alternate rows of a worksheet with a particular color? I have a worksheet with 175 rows and alternate rows are to be filled with green!
## Alternate Button Macro & Caption Between 2
Apr 23, 2008
I have a worksheet with 2 buttons labelled "Hide" and "Show". As the names imply, they allow the user to hide or show parts of the worksheet. I would like to combine them into one button and have the button label and the associated macro change with each press of the button. Here is what I have so far;
Rows("10:15").Select
Selection.EntireRow.Hidden = False
ActiveSheet.Shapes("Button 20").Select
Range("A1").Select 'is there a better way to remove the focus from the button than selecting a cell off the button?
End Sub
Rows("10:14").Select
Selection.EntireRow.Hidden = True
ActiveSheet.Shapes("Button 20").Select
Range("A1").Select
End Sub
These macros change the label fine after hiding or unhding the rows but I can't find the proper terms to use to change the macro associated with the button (if there is one?)
## If Row Contains Non Empty Cell Copy To Alternate Worksheet
Sep 11, 2013
When the worksheet is changed;
-For each row between A3 and A5000
-If cell in column L is empty
-Do Nothing and move on to next row
-Otherwise if cell in column L is not empty
-Copy entire row to alternate sheet, and delete row from original sheet.
-After all rows between A3 and A5000 have been checked, sort alternate sheet in ascending order based on the contents of column A.
The issue is that the code keeps skipping some rows that should be copied, possibly due to the 'for each' command not liking how I'm deleting rows (maybe?)
VB:
Private Sub Worksheet_Change(ByVal Target As Range)
Dim rCell As Range
Dim lRow As Long
lRow = Range("A3:A5000").Rows.Count
[Code] .....
## Alternate Colors On Stacked Columns Plot?
Feb 19, 2014
I'm trying to find a way to alternate between two colors (white, grey, for example) in a stacked columns plot. Each stacked column should begin with a grey layer then white, then grey and so on. These plots will be created and changed frequently so I cannot manually do this every time.
## Shading Alternate Unhidden Rows In Excel?
Mar 5, 2014
I have a file that contains multiple rows of data. I built a macro that will then hide rows with a \$0 balance. When the macro is completed, my spreadsheet is only showing the lines with a \$ amount. (The \$0 rows are hidden)
What I'd like to do is then format the unhidden rows so that each alternate row is color coded. This will separate the rows visually so I can more easily follow the rows across to view \$'s per line.
I've attached a sample spreadsheet (it does contain the macro to hide the \$0 rows). How can I then format the remaining rows with alternate shading? | 4,324 | 16,860 | {"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-2018-22 | latest | en | 0.923112 |
https://www.exceldemy.com/tag/excel-min-function/ | 1,723,262,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00042.warc.gz | 604,663,369 | 62,183 | Excel MIN Function
### Excel MIN Function Returns 0: 2 Possible Scenarios
You may use Excel’s MIN function to get the minimum value from a range of cells. In some cases your Excel MIN function returns 0. In this article, ...
### How to Use the MIN Function to Exclude Zero in Excel – 5 Easy Methods
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### How to Find Minimum Value That Is Greater Than 0 in Excel
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### How to Use Combined MIN and IF Functions in Excel (2 Ways)
Here, we have the sales data of three different brands for two different products along the location of their stores. We will apply a single ...
### How to Find Lowest 3 Values in Excel: 5 Easy Ways
Method 1 - Using SMALL Function Step 1: Click on any cell E5. Insert the formula: =SMALL(C5:C10,1) Step 2:Â Press ENTER. The First lowest ...
### How to Find Minimum Value with VLOOKUP in Excel (6 Ways)
We will use the following data table to explain the methods to find minimum value with VLOOKUP in Excel. Method 1 - Minimum Value with ...
### How to Find Minimum Value Based on Multiple Criteria in Excel
Method 1 - Find the Minimum Value Based on Multiple Criteria Using MIN and IF Functions Let’s assume we have a food order dataset with their product ...
### How to Find the Lowest Value with Criteria in Excel – 7 Methods
This is the sample dataset. Method 1 - Combine Excel MIN & IF Functions to Get the Lowest Value The MIN function ...
### How to Find the Minimum Value in Excel – 6 Easy Methods
the MIN function, the SMALL function, the MINIFS function, a Pivot table, and a VBA code can be used. This is the sample datset. ...
Advanced Excel Exercises with Solutions PDF | 441 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.797133 |
https://ghstudents.com/calculate-cut-off-points-for-universities/comment-page-2/ | 1,713,814,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00305.warc.gz | 238,899,175 | 65,725 | # How to Calculate Cut Off Points for Universities in Ghana
Maybe you’ve been wondering how to calculate your cut off points for university admissions. You’re at the right place. In this guide, you’ll learn how to calculate cut off points for all universities in Ghana.
As a policy, tertiary institutions in Ghana use the cut off points to offer candidates admission into various programmes. Most of these universities adopt the WAEC grading system in offering applicants admission.
The cut-off point is calculated based on the grade scored by a student in the West African Examination Council (WAEC). Different grades are assigned a numerical value which students can use to calculate their cut off point.
Contents
## What is a Cut Off Point?
A cut off point is a system adopted by tertiary institutions to grade students into the various programme of study. Different universities have their own specific cut off points for specific courses. That is to say, a cut off point for biochemistry differs from that of veterinary medicine in the same institution.
A student that scores an aggregate cut off point of 24 or lower can apply to any university in the country. However, there are some specific courses that require lower aggregate cut off point.
To calculate your cut off point aggregate, you need to understand the WAEC grading system.
## WASSCE Grading System & Interception
GRADE NUMERIC VALUE INTERPRETATION A1 1 EXCELLENT B2 2 VERY GOOD B3 3 GOOD C4 4 CREDIT C5 5 CREDIT C6 6 CREDIT D7 7 PASS E8 8 PASS F9 9 FAIL
## WAEC Grading and Interpretation in Percentage
Grade Definition Interpretation in Percentage A1 Excellent 80 – 100 B2 Very Good 70 – 79 B3 Good 65 – 69 C4 Credit 60 – 64 C5 Credit 55 – 59 C6 Credit 50 – 54 D7 Pass 49 – 45 E8 Pass 40 – 44 F9 Fail 0 – 39
## Universities Admission Requirements in Ghana
Before a student is offered admission into the university, he/she must, first of all, meet the minimum admission requirements for a specific programme of choice. Admission requirements for science-related courses are quite different for that of non-science related courses.
For instance, an applicant for admission to a degree programme must have at least credits (A1 – C6 in WASSCE and A – D in SSSCE) in English, Core Mathematics and Integrated Science (for Science related programmes) or Social Studies (for non-Science related programmes) and three elective subjects in Science for applicants applying to Science or Agriculture related disciplines or three elective subjects in General Arts/Business for applicants applying to non-Science related disciplines, with the total aggregate not exceeding 24.
In addition, Science applicants should have at least a grade C6 in WASSCE/D in SSSCE in Social Studies/Life Skills and non-Science applicants should also have at least a grade C6 in WASSCE/D IN SSSCE in Integrated Science/Core Science.
Example:
For Bachelor of Medicine and Bachelor of Surgery, an applicant must have credit passes in Core subjects which include English, Core Mathematics, Integrated Science.
In addition, an applicant must also have credit passes in Chemistry and any two from Physics, Biology and Elective Mathematics.
Having said that, the better the grades of applicants in the core courses, the greater their chances of getting admitted. For instance, an applicant might have 6 As in elective subjects and 2 Bs in core subjects and not be offered admission. On the other hand, an applicant who scores 5 As may be way ahead of those who score 7 As, depending on the subject combination.
## How to Calculate Cut Off Points for All Universities in Ghana
Universities use the WAEC grading system and grade students based on their performance on their best six subjects from the core and elective subjects while ensuring that they pass English and Mathematics.
Having said that, let us look at the possible ways of calculating cut off points for university admission in Ghana. In this guide, we’ll be looking at some possible examples of how to calculate cut off points using the WAEC grading system.
Example 1.
An applicant for a Bachelor of Medicine and Bachelor of Surgery has the following grades:
• English language (A1) – 1
• Core Mathematics (A1) – 1
• Integrated Science (B3) – 3
• Social Studies (C5) – 5
• Chemistry (A1) – 1
• Physics (A2) – 2
• Biology (A1) – 2
• Elective Mathematics (B2) – 2
According to the WAEC grading system, each grade has a numerical value, and we’ll be using the values to calculate the aggregate cut-off points.
Using the admission requirements, we’ll be calculating the best six (6) subjects. Going with that, the aggregate cut-off point will be 10 excluding Social Studies and Elective Mathematics.
Example 2.
Another applicant for a Bachelor of Medicine and Bachelor of Surgery has the following grades:
• English language (A1) – 1
• Core Mathematics (A1) – 1
• Integrated Science (A1) – 1
• Social Studies (C6) – 6
• Chemistry (A1) – 1
• Physics (B2) – 2
• Biology (B2) – 2
• Elective Mathematics (B2) – 2
Using the best six (6) subjects, the aggregate cut-off point will be 8, excluding Social Studies and Elective Mathematics.
Furthermore, a candidate who scores 7 As and scores only one B3 in English or Core Mathematics will have an aggregate of eight (8) for his best six (6) subjects.
The one who scores five As (5 As) and three B 2s (3 B2s) in the remaining subjects, depending on his performance in English and Mathematics, can have an aggregate of seven and by WAEC standards, has performed better than the one with 7 As. As such, the candidate with 5 As and three B2s might be offered admission ahead of the one with 7 As and one B3.
Also, the applicant with 5 As and 3 Bs is admitted as a regular, non-fee-paying student and the one with 7 As and a B if admitted becomes a fee-paying student for the same course.
## 50 Comments
1. Danny says:
I had English -B2
integrated science -A1
Social studies -A1
Chemistry -B3
elective maths -C6
Biology -B3
Cmaths-b2
physics-C6
2. Ampomah says:
Please I had social studies A1,core mathematics C6,science c5,English c6,Geography c4,government B3,Economics c6,Elective mathematics c6 Please which course can I offer In the university
3. Isaac amankrah says:
Please I had aggregate 17 can I offer physician assistance
4. Gabi says:
Please I had
English B3
Core MathsA1
ScienceA 1
Social B2
Chemistry A1
Biology B3
Physics C6
Elective MathsB3
Best 6 sum up to 12,and cut off point for Optometry in UCC is 11,can I get admission?
5. Hope says:
Marketing A1
Economics c4
Civic Education B2
English B3
Further mathematics A1
Mathematics A1
Biology B2
Chemistry A1
Physics B2
What my cut off point
Can i go for medicine
6. Gertrude says:
I had
English and science (B3)
Mathematics (C6)
Social (B2)
Government and Ewe (B3)
French (C5)
Literature (C6)
Please will I be admitted in Legon to offer bachelor of arts
7. Esther says:
Please u had grade 24 and I want to offer Business administration(banking and finance) please is that possible
8. Ewormi Hope says:
Please I had
Core math-A1
Social studies-C4
English-C5
Integrated science-B3
Biology-B3
Physics-B2
Chemistry-B2
Elective maths-B3
Can I get an admission for petroleum engineering in KNUST
9. Glory Adjetey says:
Pls can I offer PA , MEDILAB OR HUMANBIOLGY at knust with grade 17
Or if possible which course can I offer . pls I need a reply cuz am a little bit confused and am a student from Nigeria
10. Aikins says:
I got social studies B3,
English language C5,
Mathematics C6,
integrated science c4
crs c6,
economics d7,
geography c6,
government c6
I want to go to university will it be possible?
11. Abdul Malik says:
Here are my results ,I had a1 in science and social, b2 in math, b3 in physics, chemistry and English, c4 in emaths and biology. What is my aggregate. Can I apply for geomatic engineering
12. ALFRED ATIBILA says:
Please l had
Core maths c4
Science c4
English c6
Agric B2
E-maths B3
Animal husbandry c6
Please can apply for information technology or computer science
1. bobby says:
no,this is because you didnt reach the cut off point for that course
13. Obeng Newton says:
I had English C4, Science B2,core math B3, social studies A1, chemistry B3,physics B3 and biology C4..
Can i offer PA or medilab
1. ABDUL-WAHAB KOJO YAHAYA says:
Here are my results, I had b3 in social studies ,B3 in science,c4 in maths,d7 in English,c4 in government, c6 in economics, c6 in geography and d7 in emaths. With this results can I get admission to offer electrical engineering or pharmacy(those who work at polytechnic)
14. Bright says:
I’m Evans can I do applied science with aggregate 24?
15. Bernice Tay says:
I had
Social_ B3
English_C6
Science_D7
Maths_D7
French_B3
Ewe_C5
ICT_B3
Economics_E8
Please would I be able to get into any institutions to offer public administration or ICT. Please help me out
16. Fianyeku favour says:
Pls can I offer nursing at knust and ucc with grade 18
Or if possible which course can I offer . pls I need a reply cuz am a little bit confused and am a student from Nigeria
1. Obeng Newton says:
I had elective math-A1
Int Science-B2
Chemistry- B3
Physics- B3
C.math B3
Social A1
English C4
Biology B3… Can i offer PA, medilab or pharm in your institution
17. Salima says:
English-B3
Social- B3
Coremaths- A1
Science- A1
French- A1
Management in living- A1
Biology- B3
Food and nutrition- C5
Please is it possible to offer general nursing in legon since the cut off point is 10. And calculating my best 6 i had 10.
18. Kwaku Oteng says:
Please I had grade 24 and I want to offer Economics or Geography
Please will it be possible | 2,476 | 9,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.895848 |
https://www.statslife.org.uk/sports/1530-how-well-do-fifa-s-ratings-predict-world-cup-success | 1,561,581,535,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000545.97/warc/CC-MAIN-20190626194744-20190626220744-00310.warc.gz | 908,149,690 | 7,665 | # How well do FIFA’s ratings predict World Cup success?
Written by Ray Stefani on . Posted in Sports
The first match has yet to kickoff, but it’s not too early for the weeping and gnashing of teeth to begin. Fans of the 32 World Cup teams already worry that their group is too hard and their hated rival’s group is too easy, based on FIFA’s team ratings. Just how much of a bellwether have those ratings been? FIFA changed the original 1993 system in 1999 (for the worse as we shall see) and again in 2006 (for the better).
Two WCs (1994 and 1998) were played under the original 1993 system, two (2002 and 2006) under the 1999 system while the 2010 WC was played under the 2006 system, as will the upcoming 2014 WC. Are there other rating systems that might better serve FIFA for the men’s WC? As in life, men should look to women for the answer.
How the FIFA World Ranking system works
The three FIFA systems have the same general form. Match points are calculated for each game using five factors, Match points are summed over a period of years.
match points = result x opponent strength x importance x continent x ageing
Table 1 below shows the history of those five factors, along with a system I suggested in a letter to FIFA in 1996, which is quite close to the eventual 2006 FIFA system. FIFA combined my suggestion with others into a serious of options given to their consultants. In 1993, FIFA’s result points were based on 2,1,0, for wins, draws and losses respectively, even though FIFA mandated 3,1,0 points for all domestic and international matches. Those points were assigned to each team in some mysterious way that they would not publish, based on relative opponent strength.
Table 1: History of the five factors for FIFA rating system Factors FIFA 1993 Stefani 1996 FIFA 1999 FIFA 2006 Result Win = 2, Tie = 1, Loss = 0 Win = 3, Tie = 1, Loss = 0 Win = 2, Tie = 1, Loss = 0 Win = 3, Tie = 1, Loss = 0 Opponent Strength Points shared based on ranking difference Step-down table 10 x 1993 points 100 x Step down equation Importance (Friendly-WC) 1 - 1.5 1 - 3 1 - 2 1 - 4 Continent 0.8 - 1 0.8 - 1 0.85 - 1 0.85 - 1 Ageing over 6 years 4 years 8 years 4 years
In 1996, I suggested 3,1,0 points multiplied by a step-down table, so that progressively fewer points are gained when defeating or tying progressively worse-ranked opponents. In 2006, FIFA changed to 3,1,0 multiplied by 100 times a step-down equation, 2 - rank/100, which is more precise than the table I had suggested and which (coincidentally) went through the points in my 1996 step-down table.
In 1996 I suggested making a WC match count at least 3x that of a friendly, while FIFA undervalued a WC match as being only 1.5x as important as a friendly. They now count a WC match 4x as much as a friendly. The continental factors have been adjusted only slightly.
In 1993, FIFA used a six year window. I suggested using a more logical four-year WC cycle in 1996. FIFA moved to an unusually large eight-year window in 1999, which appears to have resulted in worse predictive accuracy, and then to a four year window.
Table 2 below shows the predictive success of the three FIFA systems over the last five WCs, based on pre-WC ratings. Ties were not counted in the group phase to facilitate comparison with the knockout phase and with other systems. The 1993 system was 69.2% successful in predicting winners in the 1994 and 1998 WCs. The 1999 system only predicted 64.1% of the 2002 and 2006 matches correctly, with an abysmal 37.5% accuracy in the 2006 knockout phase. On the other hand, the 2006 system resulted in 72.5% success in 2010 with an excellent 87.5% success in the knockout phase. The 2006 system is headed in the right direction, but might other systems be a better choice for FIFA?
Table 2: Predictive success of FIFA’s higher ranked teams Group Phase Knockout Phase All WC Games Won Lost Tied % Games Won Lost % % 1994 36 19 8 9 70.3 16 12 4 75.0 72.0 1998 48 23 9 16 71.9 16 9 7 56.2 66.7 2002 48 22 12 14 64.7 16 12 4 75.0 68.0 2006 48 26 11 11 70.3 16 6 10 37.5 60.4 2010 48 23 12 13 65.7 16 14 2 87.5 72.5
Some better rating systems
World-wide attention focuses on the rugby and cricket WCs too; but, the rating systems are more predictive. The International Rugby Board, IRB, and International Cricket Council, ICC, use the same basic scheme for adjusting ratings after each match.
New rating = old rating + K [result – prediction]
The IRB and ICC both score a result 1, 0.5, 0 for a win, tie and loss respectively. The prediction (the probability of winning) is also limited to a 1-0 basis depending on rating difference, using a simple equation (0.5 + 0.5 diff /10 for rugby and 0.5 + 0.5 diff /50 for the David Kendix system used in cricket). K depends on importance and score difference in rugby and on importance and games played in cricket. The rugby system includes an adjustment for home advantage. The higher-rated team won an average of 83% of all matches in the 2007 and 2011 rugby WCs, with 90% success in 2011. David Kendix’ system was 79% predictive in the 2007 and 2011 One Day International WCs and 76% predictive in the 2014 T20 WC. Both systems outperformed FIFA’s men’s system.
FIFA need only look to their women’s Elo system for improvement. The form is the same as for rugby and cricket. FIFA cleverly converts game score to a 1-0 scale, instead of making every win the same and every loss the same. The prediction (again the probability of winning) is calculated from rating difference using the well-known Elo equation from chess ratings. The Elo system, which adjusts for home advantage, was 82% successful for the women’s WCs of 2007 and 2011. The rugby, cricket and women’s FIFA systems more precisely adjust after each win, tie or loss and each is more strongly weighted toward pre-WC strength. FIFA would do well to use the women’s Elo system for men’s competition as well.
The four top-ranked teams
Table 3 contains the success of the four top-ranked teams in the last five men’s football WCs using pre-WC rankings. Also shown are the top four as of 8 May, 2014. One more set of ratings will be published prior to the 2014 WC. Of the 20 teams, 75% reached the knockout phase, meaning that 25% of those supposedly dominant teams were eliminated in the group phase. 55% reached the quarter finals and 30% (6) reached the semi-finals. Three of those six won the WC and three were second. No top rated team won, presenting a challenge for Spain in 2014. The second-rated teams won twice, good news for Germany. The fourth-rated teams were second twice, with host nation Brazil currently being fourth. Brazil is particularly well placed to win, given a host-nation advantage. 18th ranked France won as the host nation in 1998. The average winner was ranked 8th, four positions worse than Brazil.
Table 3: Success of the four top-ranked teams World Cup Host FIFA ranked 1 Finish FIFA ranked 2 Finish FIFA ranked 3 Finish FIFA ranked 4 Finish 1994 USA Germany QF Neth. QF Brazil 1 Italy 2 1998 France Brazil 2 Germany QF Spain Group England Round of 16 2002 SK-Japan France Group Brazil 1 Argentina Group Portugal Group 2006 Germany Brazil QF Czech Group Neth. Round of 16 Mexico Round of 16 2010 South Africa Brazil QF Spain 1 Portugal Round of 16 Neth. 2 2014 Brazil Spain TBC Germany TBC Portugal TBC Brazil TBC
Please excuse me as I return to gnashing my teeth. Landon Donavan was not chosen for the US team. Silly me: I thought that loyalty, dedication and experience were reasons to be selected. Let the serious gnashing now begin. | 1,995 | 7,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-26 | latest | en | 0.961169 |
http://www.top-law-schools.com/forums/viewtopic.php?f=6&t=195535 | 1,529,837,620,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866932.69/warc/CC-MAIN-20180624102433-20180624122433-00634.warc.gz | 511,927,984 | 10,510 | ## predict my lsat score
Prepare for the LSAT or discuss it with others in this forum.
jess1921
Posts: 15
Joined: Sat Aug 25, 2012 1:29 pm
### predict my lsat score
Expected shit
LR: -2/3
LR: -2/3
LG: -0
RC; -7 to -12
pedestrian
Posts: 461
Joined: Sat Jun 18, 2011 9:38 pm
### Re: predict my lsat score
Sure. Curve will probably be between -10 and-12. Based on past 100 question tests with those curves and your estimated minimum/maximum wrong, I'd say:
Best: 171 (-11, -12 curve)
Worst: 165 (-18, -10 curve)
Of course, that's only as good as the assumptions - if the curve is -8 or you missed a bunch you didn't expect, things could go differently.
Poo-T
Posts: 266
Joined: Fri Jul 06, 2012 10:32 am
### Re: predict my lsat score
best: - 11
worst: - 25
pedestrian
Posts: 461
Joined: Sat Jun 18, 2011 9:38 pm
### Re: predict my lsat score
Poo-T wrote:best: - 11
worst: - 25
Well, using the same assumptions as above:
Best: 171
Worst: 159
Edited: sorry, I initially used the -12 curve for both. You can get an idea from past scores using this table: http://lsatblog.blogspot.com/2010/03/ls ... rsion.html. The curve is the number you can get wrong and still get a 170.
SumStalwart
Posts: 201
Joined: Wed Aug 01, 2012 2:37 am
### Re: predict my lsat score
Best: -8
Worst: -20
Huge range, but RC is the bulk of that volatility :/
CalAlumni
Posts: 205
Joined: Fri May 11, 2012 11:58 am
### Re: predict my lsat score
jess1921 wrote:Expected shit
LR: -2/3
LR: -2/3
LG: -0
RC; -7 to -12
I have a feeling you did better than you give yourself credit for in RC.
jess1921
Posts: 15
Joined: Sat Aug 25, 2012 1:29 pm
### Re: predict my lsat score
I hope so.
wwrhodes
Posts: 15
Joined: Mon Oct 08, 2012 3:19 pm
RC: -2
LR: -2
LG: -7 | 614 | 1,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-26 | latest | en | 0.882808 |
https://www.clutchprep.com/chemistry/practice-problems/144045/methane-when-liquefied-has-a-density-of-0-466-g-ml-the-density-of-methanol-is-0- | 1,638,744,178,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363226.68/warc/CC-MAIN-20211205221915-20211206011915-00229.warc.gz | 765,266,503 | 31,121 | Thermochemical Equations Video Lessons
Concept:
# Problem: Part C. Methane, when liquefied, has a density of 0.466 g/ mL, the density of methanol is 0.791 g/ mL at 25°C. Calculate the enthalpy of combustion in kilojoules of 1.00 L of liquid methane and liquid methanol, respectively.Enter the energy change for methane followed by the energy change for methanol in kilojoules separated by a comma.Natural gas (mostly methane, CH4) is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which has a boiling point of − 164°C. One possible strategy is to oxidize the methane (in a 1:1 mole ratio) to methanol, CH3OH, which has a boiling point of 65°C and can therefore be shipped more readily.
###### FREE Expert Solution
92% (497 ratings)
###### Problem Details
Part C. Methane, when liquefied, has a density of 0.466 g/ mL, the density of methanol is 0.791 g/ mL at 25°C.
Calculate the enthalpy of combustion in kilojoules of 1.00 L of liquid methane and liquid methanol, respectively.
Enter the energy change for methane followed by the energy change for methanol in kilojoules separated by a comma.
Natural gas (mostly methane, CH4) is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which has a boiling point of − 164°C. One possible strategy is to oxidize the methane (in a 1:1 mole ratio) to methanol, CH3OH, which has a boiling point of 65°C and can therefore be shipped more readily.
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Thermochemical Equations concept. You can view video lessons to learn Thermochemical Equations. Or if you need more Thermochemical Equations practice, you can also practice Thermochemical Equations practice problems. | 493 | 2,031 | {"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} | 3.109375 | 3 | CC-MAIN-2021-49 | latest | en | 0.903337 |
http://www.qsl.net/wb6tpu/si-list6/0133.html | 1,531,786,432,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589536.40/warc/CC-MAIN-20180716232549-20180717012549-00013.warc.gz | 523,638,504 | 3,262 | # RE: [SI-LIST] : skin effect
From: Ray Anderson ([email protected])
Date: Thu Aug 31 2000 - 12:56:58 PDT
As a couple of kind souls have pointed out offline,
I made a math error in my first posting today
regarding skin effect.
.7 mils is .0007 inch not .007 inch DUH!
So the calculated skin depth of .002 mm
turns out to be of more significance since
the line is real .018 mm thick, not .18 mm.
----------------
| ----------- |
| | | |
| | a |b|
| | | |
| ----------- |
----------------
However, despite my math error (of which I retain copyright)
I think my original argument holds.
Consider the above diagram. The annular region "b" is of
skin depth "t" in thickness.
The interior region "a" is effectively out of the picture
at high frequencies as the current for the most part
concentrates in region "b".
The total loss of the conductor is a comprised of
the copper losses (simply a resistance based on the
cross section area of the conductor) and frequency dependent
losses (skin effect) (these are per unit length losses)
and dielectric losses (and radiation losses to I suppose).
To complicate matters further, the simple skin effect
equation found in most texts doesn't include corrections
for geometry dependent "proximity effect", which due to
current crowding, caused by adjacent current return paths
in non-symmetrical geometries, causes the frequency dependent
losses to increase because the entire area of region "b" in the
diagram is not utilized 100% by the current in some situations.
We know that skin effect loss is supposed to scale with
sqrt(f) and it does except when the proximity effect is
significant (due to non-symmetrical geometries) in which
case the sqrt(f) relation ceases to be the controlling function.
This is a major cause of error in most simulator t-line models.
Note that the proximity effect is most noticeable in on-chip
t-line situations due to the aspect ratios and spacings involved,
and usually isn't too much of a concern in most pcb level situations.
Anyway, the point I was trying to make was that as long
as the "skin depths" on both sides of the conductor
are not thick enough to merge together and consume the
entire conductor (effectively removing region "a") then
increasing the conductor thickness will have little effect
on the skin-effect losses, though it will have marked effect
on the DC losses. Increasing the width of the conductor
ameliorates the effects of the skin-effect losses as well
as reducing the copper losses. (but can increase dielectric
losses somewhat in some situations.....)
-Ray Anderson
Sun Microsystems
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# Georgetown University MATH 137 Multivariable Calculus Solutions to Midterm 2.
### Document Content and Description Below
Georgetown University MATH 137 Multivariable Calculus Fall 2018 Solutions to Midterm 2 1. Find the critical points of f(x; y) = -x3 + 12x - 4y2 and classify each point as a local maximum, loc... al minimum, or saddle point. To solve for the critical points, we set fx = -3x2 + 12 = 0 fy = -8y = 0 The first equation gives x2 = 4, whence x = ±2. The second equation gives y = 0. Thus there are two critical points, (2; 0) and (-2; 0). We now use the Hessian H = fxxfyy - (fxy)2 to determine the nature of each critical point. We have ) - 02 = 48x. We have Therefore f(2; 0) is a local maximum. Also, H(-2; 0) = 48(-2) = -96 < 0 So (-2; 0; f(-2; 0)) is a saddle point. 2. Use the Lagrange multiplier method to find the maximum and minimum values of subject to the constraint We have We may deduce from the first four equations of this system that x = y = z = t. Hence x2 + y2 + z2 + t2 = 4x2 = 1, which implies x2 = 1 4, or x = ±12. Therefore there are two critical points to consider, namely 1 2; 1 2; 12; 12 and -1 2; -1 2; -1 2; -1 2 . We have Therefore, on the hypersphere x2 + y2 + z2 + t2 = 1, the absolute maximum of f is 2 and the absolute minimum of f is -2. 3. Consider the surface defined by the equation x3 + x2y2 + z2 = 0. Find an equation for the plane tangent to the surface at the point P(-2; 1; 2). This surface is a level surface of the function f(x; y; z) = x3 +x2y2 +z2. Consequently an equation of the tangent plane at P(-2; 1; 2) is Therefore the tangent plane is given by (a) Find the directional derivative of f at P (1; 1) in the direction of Q(2; 3). (b) Determine the direction of maximum increase of f at P (1; 1). This direction is specified by the gradient vector rf(1; 1) = h2; 6i. 5. Show that the level curves f(x; y) = x2 - y2 = k1 and g(x; y) = xy = k2, where k1 and k2 are constants, intersect orthogonally at any point of intersection P (a; b). Let P (a; b) be any point of intersection of two level curves. We have Therefore the level curves intersect orthogonally. 6. Let f(x; y) satisfy rf(x; y) = chx; yi for some constant c and all (x; y) 2 R2. Show that f is constant on any circle of radius a > 0 centered at the origin. [Show More]
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https://de.mathworks.com/matlabcentral/cody/problems/1702-maximum-value-in-a-matrix/solutions/1856685 | 1,600,912,550,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213006.47/warc/CC-MAIN-20200924002749-20200924032749-00481.warc.gz | 356,604,919 | 16,381 | Cody
# Problem 1702. Maximum value in a matrix
Solution 1856685
Submitted on 24 Jun 2019 by Johan Cuperus
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 3; 4 5 6; 7 8 9]; y_correct = 9; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = -10:0; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = 17; y_correct = 17; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
x = magic(6); y_correct = 36; assert(isequal(your_fcn_name(x),y_correct))
5 Pass
x = [5 23 6 2 9 0 -1]'; y_correct = 23; assert(isequal(your_fcn_name(x),y_correct)) | 249 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-40 | latest | en | 0.60762 |
http://math.stackexchange.com/questions/579072/sequences-not-strictly-increasing | 1,469,591,443,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825365.1/warc/CC-MAIN-20160723071025-00043-ip-10-185-27-174.ec2.internal.warc.gz | 157,465,428 | 17,832 | # Sequences not strictly increasing.
Let $d_n=a_{n+1}-a_n$ be the difference of two consecutive terms of a sequence of natural numbers.
We can easily construct sequences of natural numbers $a_n$ using trigonometric functions or the floor function which have the property:
*(1)*For infinitely many $n, d_{n+1}<d_n$ .
For example, let $a_n=cos(\frac{n\pi}{2})$ or $a_n=\lfloor \frac{n}{3}\rfloor$ and $n\geq 1$ for both cases.
• My question is
Can anybody give me an example of a sequence having the property (1) but which has a closed formula not containing trigonometric functions,or the floor function?
• Note
I would like to see a closed formula of such a sequence (if possible) and not something general like the primes or a sequence whose sum of reciprosals diverges.
Thank you very much in advance.
EDIT:$(-1)^n=cos(n\pi)$ so don't try something like this. So what i would like to see is something without trigonometric functions ,floor functions or $(-1)^n$.
Everything else is acceptable.
-
I’m probably missing something, but doesn’t $a_n=42+(-1)^n$ already work?
EDIT: Excluding $(-1)^n$ still leaves arbitrarily many sequences, e.g., A027642, (the absolute values of) A124449, or even A067029, but it all depends on your definition of “closed formula”. To me, something like
$$a_n = \min\left\{v_p(n) | v_p(n)>0\right\}$$
(which encodes A067029) is a closed formula, but you may of course disagree.
-
No. $(-1)^n=cos(n\pi)$ – Konstantinos Gaitanas Nov 24 '13 at 11:40
I.e., you are looking for something that not only does not contain trigonometric functions, but additionally, cannot be rewritten in terms of them? Or not be rewritten in fewer than a thousand trig terms? Or … I don’t know. To me, $(-1)^n$ does not contain trig functions. – Christopher Creutzig Nov 24 '13 at 11:50
i made an edit i think that it looks much more precise now. – Konstantinos Gaitanas Nov 24 '13 at 11:56
The denominator of Bernoulli numbers looks like an answer. – Konstantinos Gaitanas Nov 24 '13 at 12:33 | 563 | 2,014 | {"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} | 3.5 | 4 | CC-MAIN-2016-30 | latest | en | 0.8963 |
https://nalinkpithwa.com/2021/11/03/topological-spaces-and-groups-part-2-fast-review/ | 1,638,578,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00211.warc.gz | 476,155,727 | 36,620 | # Topological Spaces and Groups: Part 2: Fast Review
Reference: Chapter 1, Topological Transformation Groups by Deane Montgomerry and Leo Zippin
1.10 Sequential Convergence:
The proof of the principal theorem of this section illustrates a standard technique. It will involve choosing an infinite sequence, then an infinite subsequence, then again an infinite subsequence and so on repeating this construction a countably infinite number of times. A special form of this method is called the Cantor diagonalization procedure. To facilitate the working of this technique we shall sometimes use the following notation:
1.10.1
The letter I will denote the sequence of natural numbers 1, 2, 3, …When subsequences of I need to be chosen, they will be labelled in some systematic way: $I_{1}, I_{2}, \ldots$ or $I^{'}, I^{''}, \ldots$ or $I^{*}, I^{**}, \ldots$ and so on. Then given a sequence of elements $x_{n}$, $n \in I$, we can refer to a subsequence as : $x_{n}$, $n \in I_{1}$, or : $x_{n}, n \in I^{*}$ and so on.
DEFINITION:
Let S denote a metric space and let $K_{n}, n \in I$ be a sequence of subsets of S. The sequence $K_{n}$ is said to converge to a set K if for every $\epsilon >0$
1.10.3 $K_{n} \subset S_{\epsilon}(K)$ and $K \subset S_{\epsilon}(K_{n})$
for n sufficiently large (depending only on $\epsilon$)
If $K_{n}$ is given and if a K exists satisfying relation 1.10.3 then $\overline{K}$ also satisfies 1.10.3. If two closed sets $K^{'}$ and $K^{''}$ satisfy 1.10.3 for the same sequence $K_{n}$ then $K^{'}=K^{''}$. In the special case that the sets $K_{n}$ are single points, the set K if it exists is a point and is unique.
1.10.4 THEOREM
Every sequence of non-empty subsets of a compact metric space S has a convergent subsequence.
Proof:
Let $K_{n}, n \in I$ be an arbitrary sequence of subsets of S and let $W_{n}, n\in I$ be a basis for open sets in S. (Theorem 1.9)
Let I be called $I_{0}$ and suppose a sequence $I_{m-1}$ has been defined. Consider $W_{m} \bigcap K_{n}$, $n \in I_{m-1}$ m fixed. Then either $W_{m} \bigcap K_{n}$ is not empty for an infinite subsequence of integers $n \in I_{m-1}$ or on the contrary $W_{m} \bigcap K_{n}$ is empty for almost all $n \in I_{n-1}$. In the first of these cases define $I_{m}$ as the set of indices n in $I_{m-1}$. In the first of these cases define $I_{m}$ as the set of indices n in $I_{m-1}$ such that $W_{m} \bigcap K_{n}$ is not empty for $n \in I_{m}$. Then in all possible cases $I_{m} \subset I_{m-1}$ is uniquely defined. We now consider $I_{m}$ to be defined by induction for all $m \in I$.
We can now specify what subsequence of $K_{n}$ we may take as convergent. Let $I^{*} \subset I$ denote the diagonal sequence of the sequences $I_{m}$, that is $I^{*}$ contains the m-th element of $I_{m}$ for each m. We shall show that $K_{n}, n \in I^{*}$ is convergent. It follows from the definition of I^{*} that for each $m \in I$, if we except at most the first m integers in I^{*},
$W_{m} \bigcap K_{n}, n \in I^{*}$,
is always empty or is never empty depending on m.
We next define the set K to which the sets $K_{n}, n \in I^{*}$ will be shown to converge. Let W denote the union of those sets $W_{m}, m \in I$ for which $W_{m} \bigcap K_{n}, n \in I^{*}$ is almost empty. Let $K = S - W$ Since each $W_{m}$ forming W meets at most a finite number of the sets $K_{n}$ no finite number of these $W_{m}$ can cover S. Therefore W cannot cover S, since S is compact, and hence K is not empty. The set K is closed and therefore compact.
Let $\epsilon >0$ be given. There is a covering of K by sets $W_{k_{1}}, W_{k_{2}}, \ldots, W_{k_{s}}, k_{i} \in I$
each meeting K and each of diameter less than $\epsilon$. None of the sets $W_{k_{i}}$ can belong to W and each must intersect almost all the $K_{n}, n \in I^{*}$. Therefore for sufficiently large n, $n \in I^{*}$
$W_{k_{i}} \bigcap K_{n} \neq \Phi$
It follows that $K \subset S_{\epsilon}(K_{n})$, $n \in I^{*}$ for all n sufficiently large. Finally it can be seen that the closed set $S - S_{\epsilon}(K) \subset W$. It follows that the complement of $S_{\epsilon}(K)$ is covered by sets $W_{j_{1}}, W_{j_{2}}, \ldots, W_{j_{t}}$ each of which is an element in the union defining W. Therefore $W_{j_{k}} \bigcap K_{n}$, $n \in I^{*}$ $k=1,2, \ldots t$
is almost always empty. Therefore for sufficiently large $n \in I^{*}$
$K_{n} \subset S_{\epsilon}(K)$.
This completes the proof of the theorem. QED.
1.10.5
Let X and Y be closed subsets of a compact metric space S. Define Hausdorff metric: $d(X,Y)$ as the greatest lower bound of all $\epsilon$ such that symmetrically $X \subset S_{\epsilon}(Y), Y \subset S_{\epsilon(Y)}$
This is a metric for the collection of closed subsets of S.
THEOREM:
The set F of all closed subsets of a compact metric space S is a compact metric space in the metric defined above.
Proof.
The set F is a metric space in the metric defined above. It $A_{n}$ is in F then $A_{n}$ has a subsequence converging to a set A in the sense of convergence defined above and the set A may be assumed closed. It follows that the subsequence also converges to A in the sense of the metric of F. Hence, every sequence in F has a convergent subsequence.
Let $W_{n}$ where $n \in \mathcal{N}$ be a basis for open sets in S as in the preceding Theorem. For each n and each choice of integers $k_{1}, k_{2}, \ldots, k_{n}$ let $W(k_{1}, k_{2}, \ldots, k_{n})$ denote the union of the sets $W_{k_{1}}, W_{k_{2}}, \ldots, W_{k_{n}}$. Now in F let $W^{*}(k_{1}, \ldots, W_{k_{n}})$ consist of all the compact sets in S which belong to $W(k_{1}, \ldots, k_{n})$ and meet each $W_{k_{i}}$. This gives a countable collection of subsets of F. The proof of the preceding theorem shows that this collection is a basis for open sets in F.
Finally, let $\{ O_{n}\}$ where n=1, 2, …be a countable collection of open sets of F which cover F. We have to find an integer m such that $\bigcup_{1}^{m}O_{i}$ covers F. If no such integer existed we could find a sequence of points $x_{n}$ where $x_{n} \in F - \bigcup_{1}^{n}U_{i}$. This sequence would have to have a susbsequence converging to some point x. Since $x \in O_{m}$ for some m, it follows that infinitely many of the $x_{n}$ belong to $O_{m}$; this contradiction proves the theorem.
QED.
1.10.6
Note that separability implies that every collection of covering sets has a countable covering subcollection. It also implies that there exists a countable set of points which is everywhere dense in the space.
THEOREM.
A metric space S is compact if and only if every infinite sequence of points has a convergent subsequence.
Proof:
Suppose that every infinite subsequence of points of S has a convergent subsequence. We shall prove that S is separable. The last paragraph of the preceding section then shows that S is compact. The converse is shown in 1.10.4
For each positive integer n construct a set $P_{n}$ such that (1) every point of $P_{n}$ is at a distance at least $\frac{1}{n}$ from every other point of $P_{n}$ (2) every point of S not in $P_{n}$ is at a distance less than 1/n from some point of $P_{n}$. It is easy to see that no sequence of points in any one $P_{n}$ can be convergent and it follows that $P_{n}$ is a finite point set. Let $P= \bigcup P_{n}$. Then P is countable and every point of S is a limit point of P. For each rational $r >0$ and each point of P construct the “sphere” with that point as centre and radius r. The set of these spheres is countable and is a basis for open sets. This concludes the proof.
QED.
EXAMPLE.
Let S be a compact metric space let H be the space of real continuous functions defined on S with values in $R_{1}$. Each continuous function $f(x)$ determines a closed subset of $S \times R_{1}$ namely the graph consisting of the pairs $(x, f(x))$, $x \in S$. Hence H is a subset of a compact metric space (see examples 2 and 3 in Sec 1.9).
Example 2 of Section 1.9 reproduced below:
The set F of continuous functions defined on a compact space S with values in a metric space M becomes a metric space by defining for f, g in F $d(f,g) = lub (x \in S) [d_{M}(f(x), g(x))]$ where $d_{M}$ is the metric in M. Recall also the following here in this example: Theorem: Let S be a compact space and $\{ D_{a}\}$ a collection of closed subsets such that $\bigcap_{a}D_{a}$ is empty. Then there is some finite set $D_{a_{1}}, \ldots, D_{a_{n}}$ such that $\bigcap_{i}D_{a_{i}}$ is empty. From this theorem it follows that: A lower semi-continuous (upper semi-continuous) real valued function on a compact space has finite g.l.b. (and L.u.b) and always attains these bounds at some points of space.
Example 3 of Section 1.9 reproduced below:
If S is a compact metric space and $E_{1}$ denotes the real line, then $S \times E_{1}$ is a metrizable locally compact space with a countable basis for open sets.
Remarks: Notice that the metric defined for H in example 2 of 1.9 and the metric which it gets from $S \times H_{1}$ are topologically equivalent. This proves that H itself is a separable metric space.
QED.
1.11 TOPOLOGICAL GROUPS
Topological groups were first considered by Lie, who was concerned with groups defined by analytic relations (to be discussed later). Around 1900-1910 Hilbert and others were interested in more general topological groups. Brouwer showed that the Cantor middle third set can be made into an abelian topological group. Later Schreier and Leja gave a definition in terms of topological spaces whose theory had been developed in the intervening time.
A topological group is a topological space whose points are elements of an abstract group, the operations of the group being continuous in the topology of the space. A detailed definition containing some redundancies is as follows:
DEFINITION:
A topological group G is a space in which for $x, y \in G$ there is a unique product $xy \in G$, and
i) there is a unique identity element e in G such that $xe = ex =x$ for all $x \in G$
ii) to each $x \in G$ there is an inverse $x^{-1} \in G$ such that $xx^{-1} = x^{-1}x=e$
iii) $x(yz) = (xy)z$ for $x, y, z \in G$
iv) the function $x^{-1}$ is continuous on G and $xy$ is continuous on $G \times G$
Familiar examples are the real or complex numbers under addition with the usual topologies for $E_{1}$ and $E_{2}$ respectively, and the complex numbers of absolute value one under multiplication with their usual topology as a subset of $E_{2}$. The space of this last group is homeomorphic to the circumference of a circle.
1.11.1
Of course, properties 1, 2 and 3 define a group in the customary sense and a topological group may be thought of as a set of elements which is both an abstract group and a space, the two concepts being united through 4. Wnen a subset H of G is itself a group we shall call H a subgroup of G, but we shall understand that H is to be given the relative topology. It is easy to see that then H becomes a topological group.
If H is a subgroup of G and if x and y are points of G belonging to the closure of H, then every neighbourhood of the product xy contains points of H. For, let U be a neighbourhood of xy. Then by property 4, there exists neighbourhoods V of x and W of y such that every product of an element of V and an element of W is contained in U. We see from this that xy belongs to the closure of H. Similarly, $x^{-1}$ belongs to the closure. Thus, $\overline{H}$ is a group, and we shall call it a closed subgroup.
1.11.2
If $G_{a}$ where $a \in \{ a\}$ is a collection of topological groups and if G denotes the product space defined in 1.6, (previous blog), then G can be regarded as a group (the product of two elements of G being defined by the product of their components in each factor $G_{a}$). Because each neighbourhood of the PROD $G_{a}$ depends on only a finite number of the factors it is easy to see that G becomes a topological group. We shall call it the topological group of the factors $G_{a}$.
By way of examples, note that the product of an arbitrary number of groups each isomorphic to $C_{1}$: the group of reals modulo one (isomorphic to the complex numbers of modulus one under multiplication) is a compact topological group. The product of an arbitrary number of factors each isomorphic to the group of reals is a topological group which is locally compact if the number of factors is finite.
A finite group with the discrete topology is compact and the topological product of any collection of finite groups is therefore compact.
1.11.3
If x and y are in topological group, $x \neq y$, then as will be seen (section 1.16) we may choose a neighbourhood W of e such that
$y \notin WW^{-1}$
Hence, yW and xW are disjoined, and thus two distinct points of a topological group are in disjoined open sets. This is called the Hausdorff property (see section 1.10); it implies that a point is a closed set.
It is easy to see that if H is an abelian subgroup of a topological group G then $\overline{H}$ is also abelian. Thus, if x, y $\in \overline{H}$ and $xy \neq yx$ there are neighbourhoods $U_{1}$ of xy and $U_{2}$ of yx with $U_{1} \bigcap U_{2} = \Phi$. There exist neighbourhoods V of x and W of y such that for every $v \in V$ and $w \in W$ $vw \in U_{1}$ and $wv \in U_{2}$. However, if v, w $\in H$ then $wv = vw$ and we are led to a contradiction proving that the closure of H is abelian.
QED.
1.11.4
Important examples of topological groups are given below:
EXAMPLE 1.
The sets $M_{n}(R)$ and $M_{n}(C)$ of all $n \times n$ matrices of real and complex elements under addition with the distance of $A = (a_{ij})$, $B = (b_{ij})$ defined by
$d(A, B) = max_{i,j}|a_{ij} - b_{ij}|$
The spaces of these two groups are homeomorphic to $E_{n^{2}}$ and $E_{2n^{2}}$. They are in fact the sets of real or complex vectors with $n^{2}$ co-ordinates, and hence are vector spaces as well as groups. Another example is the set H of continuous real valued functions on a metric space under addition.
EXAMPLE 2.
The sets of non-singular real or complex $n \times n$ matrices $GL(n,R), GL(n,C)$ under multiplication; these are subsets of $M_{n}(R)$ and $M_{n}(C)$ respectively and are given the induced topology. They are open subsets and are therefore locally compact and locally euclidean. (see 1.27 later)
EXAMPLE 3.
Let S be a compact metric space and let G be the group of all homeomorphisms of S onto itself topologized as a subspace of the space of continuous maps of S into itself. (Section 1.9 previous blog, example 2)
1.12 ISOMORPHISM of TOPOLOGICAL GROUPS
The spaces associated with two topological groups may be homeomorphic but the groups essentially different; for example, one abelian and the other not.
EXAMPLE 1.
The matrices $\left |\begin{array}{cc}a & 0 \\ 0 & b \end{array}\right |$ where a and b are real numbers under addition. This is an abelian group with $E_{2}$ as space.
EXAMPLE 2.
The matrices $\left| \begin{array}{cc} e^{a} & b \\ 0 & e^{-a} \end{array} \right |$ where a and b are real under multiplication. This is a non-abelian group with $E_{2}$ as space.
If we give the space in this example (or example 1) the discrete topology we obtain a new topological group with the same algebraic structure.
EXAMPLE 3.
In the additive group of integers, for each pair of integers h and k, $k \neq 0$ let the set $\{ h \pm nk\}$ where n=0, 1,2, …, be called an open set and let the collection of all these sets be taken as a basis for open sets.
EXAMPLE 4.
Introduce a metric into the additive gorup of integers, depending on the prime number p, defined thus:
$d(a,b) = \frac{1}{p^{n}}$
if $a \neq b$ and $p^{n}$ is the highest power of p which is a factor of $a-b$.
EXAMPLE 5.
Let G be the integers under addition with any set called open if it is the complement of a finite set (or is the whole space or the null set). Algebraically G is a group and it is also a space. However, it is not a topological group because addition is now not simultaneously continuous. It is true however that addition is continuous in each variable separately.
For some types of group spaces separate continuity implies simultaneous continuity. It is not known whether this is true for a compact Hausdorff group space.
DEFINITION.
Two topological groups will be called isomorphic if there is a one-one correspondence between their elements which is a group isomorphism (preserves products and inverses) and a space homeomorphism (preserves open sets).
An isomorphic map of G onto G is called an automorphism.
In examples (3) and (4) the abstract group structure of the additve group of integers is embodied in infinitely many non-isomorphic topological groups.
1.13 SET PRODUCTS
If G is a group $A \subset G$ let $A^{-1}$ denote the inverse set $\{ a^{-1}\}$ where $a \in A$. Clearly, $(A^{-1})^{-1}=A$. If $B \subset G$ let AB denote the set $\{ ab\}$ where $a \in A, b \in B$. It is understood that the product set is empty if either factor is empty. We shall write $AA = A^{2}$ and so on. It can be seen that $(AB)C = A (BC)$ and $(AB)^{-1} = B^{-1}A^{-1}$. The set $AA^{-1}$ satisfies
$(AA^{-1})^{-1}=AA^{-1}$
that is, it is symmetric. Similarly, the set intersection $A \bigcap A^{-1}$ is symmetric. The intersection of symmetric sets is symmetric.
A set H in G is called invariant if $gH = Hg$ for every $g \in G$ equivalently if $gHg^{-1}=H$.
THEOREM 1.
Let G be a topological group and let $A \subset G$ be an open set. Then $A^{-1}$ is open.
Proof:
Let $a^{-1}$ be in A. By the continuity of the inverse there exists an open set B containing a such that $b \in B$ implies $b^{-1} \in A$. This means that $B^{-1} \subset A$ and therefore $B \subset A^{-1}$. Thus $A^{-1}$ is a union of open sets and is open.
COROLLARY:
The map $x \rightarrow x^{-1}$ is a homeomorphism.
LEMMA.
Let G be a topological group, A an open subset, b an element. Then Ab and bA are open.
Proof:
Let $a \in A$ and let $c=ab$. Then $a = cb^{-1}$. Because “a” regarded as a product is continuous in c there must exist an open set C containing c such that if $c^{'} \in C$ then $c^{'}b^{-1} \in A$. But then $c^{'} \in Ab$ and it follows that $C \subset Ab$. Therefore Ab is a union of open sets and is open. The proof that bA is open is similar.
QED.
COROLLARY.
For each $a \in G$, the left and right translations : $a \rightarrow ax$ and $x \rightarrow xa$ are homeomorphisms.
THEOREM 2.
Let G be a topological group and let A and B be subsets. If A or B is open then $AB$ is open.
Proof.
Since AB is a union of sets of the form Ab, $b \in B$, it is open if A is open. SimilarlyAB is a union of sets aB, $a \in A$ and is open if B is open.
COROLLARY.
Let A be a closed subset of a topological group. Then Ab and bA are closed.
Proof:
This is true because left and right translations by the constant b are homeomorphisms of G onto G.
Now, a function $f(x)$ taking a group $G_{1}$ into another group $G_{2}$ will be called a homomorphism if
*) $f(x)f(y) = f(xy)$ where $x, y \in G_{1}$
When $G_{1}$ and $G_{2}$ are topological we shall ordinarily require that f be continuous. The most useful case is where f is open as well as continuous. In many situations (see 1.26.4 and 2.13 in later blogs) continuity implies openness but this is not true in general. The set of elements going into e is a subgroup and if f is continuous it is a closed subgroup (a point in a topological group is a closed set). This is the kernel of the homomorphism.
EXAMPLE.
Let $V_{1}$ be the additive group of real numbers and G be a topological group. A continuous homomorphism $h(t)$ of $V_{1}$ into G is called a one-parameter group in G. If h is defined only on an open interval around zero satisyfying the definition of homomorphism so far as it has meaning, then $h(t)$ is called a local one-parameter group in G. If $h(t)$ is a one-parameter group, the image of $V_{1}$ may consist of e alone and then $h(t)$ is a trivial one-parameter group. If this is not the case and if for some $t_{1} \neq 0$ and $h(t_{1})=e$, then the image of $V_{1}$ is homeomorphic to a circumference. In case $h(t)=e$ only for $t =0$ the image of $V_{1}$ is a one-one image of the line which may be homeomorphism of the line or a very complicated imbedding of the line. To illustrate this let G be a torus which we obtain from the plane vector group $V_{2}$ by reducing mod one in both the x and y directions. In $V_{2}$ any line through the origin is a subgroup isomorphic to $V_{1}$ and after reduction the line $y=ax$ is mapped onto the torus G thus giving a one-parameter group in G. If a is rational the image is a simple closed curve but if a is irrational the image is everywhere dense on the torus.
1.14 PRODUCTS of Closed Sets
If A and B are subgroups of a group G, AB is not necessarily a subgroup. However if A is an invariant subgroup (that is, $g^{-1}Ag = A$ where $g \in G$) and B is a subgroup then $AB$ is a subgroup.
The product of closed subsets, even if they are subgroups, need not be closed. As an example let G be the additive group of real numbers, $H_{1}$ the subgroup of integers $\{ \pm n\}$ and $H_{2}$ the subgroup $\{ \pm n\sqrt{2}\}$. The product $H_{1}H_{2}$ is countable and a subgroup but it is not a closed set.
It will be shown later that if A is a compact invariant subgroup and B is a closed subgroup then AB is a closed subgroup (corollary of 2.1 in later blogs). (Remark: I think in I N Herstein’s language of Topics in Algebra, an “invariant subgroup” is a normal subgroup. Kindly correct me if I am mistaken).
1.15 Neighbourhoods of the identity
Let G be a topological group and U an open subset containing the identity e. We showed in 1.13 that $xU$ is open and clearly $x \in xU$. Conversely, if $x \in O$, O is open, then $U = x^{-1}O$ is an open set containing e.
If a collection of open sets $\{ U_{a}\}$ is a basis for open sets at e then every open set of G is a union of open sets of the form $x_{a}U_{a}$, where $x_{a} \in G$, $U_{a} \in \{ U_{a}\}$ and the topology of G is completely determined by the basis at e. In particular the collection $\{ xU_{a}\}$ is a basis for open sets at x so also is $\{ U_{a}x\}$
If U is a neighbourhood of e, $U^{-1}$ is a neighbourhood of e and $U \bigcap U^{-1}$ is a symmetric neighbourhood of e.
THEOREM
Let G be a topological group and U a neighbourhood of e. There exists a symmetric neighbourhood W of e such that $W^{2} \subset U$.
Proof:
Since $e.e = e$ and the product is simultaneously continuous in x and y there must exist neighbourhoods $V_{1}$ and $V_{2}$ of e such that $V_{1}V_{2} \subset U$. Define $W = V_{1} \bigcap V_{1}^{-1} \bigcap V_{2} \bigcap V_{2}^{-1}$. Then $W^{2} \subset U$ and this completes the proof.
QED.
COROLLARY.
Let G be a topological group. If $x \neq e$ there exists a neighbourhood W of e such that $W \bigcap xW$ is empty.
Proof.
Since G is a topological space there is a neighbourhood $V_{1}$ of e not containing x or there is a neighbourhood $V_{2}$ of e not containing x. In either case there is a neighbourhood of e not containing x. Let W be a symmetric neighbourhood of e, $W^{2} \subset U$. If $W \bigcap xW$ were not vacuous there would exist $w, w^{'} \in W$ with $w^{'}=w$. But this gives $x = w^{'}w^{-1} \in W^{2} \subset U$ which is false.
QED.
1.15.1
If G is a topological group then a set S of open neighbourhoods $\{ V\}$ which forms a basis at the identity e has the following properties:
(a) the intersection of all V in S is $\{ e\}$
(b) the intersection of two sets of S contains a third set of S
(c) given U in S there is a V in S such that $VV^{-1} \subset U$
(d) if U is in S and $a \in U$ then there is a V in S such that $Va \subset U$
(e) If U is in S and a is in G there is a V in S such that $aVa^{-1} \subset U$.
Conversely, a system of subsets of an abstract group having these properties may be used to determine a topology in G as will be formulated in the following theorem, the proof of which is contained in main part in remarks already made.
THEOREM.
Let G be an abstract group in which there is given a system S of subsets satisfying (a) to (e) above. If open sets in G are defined as unions of sets of the form Va, $a \in G$ then G becomes topological with S a basis for open sets at e. This is the only topology making G a topological group with S a basis at e.
1.16 COSET Spaces
Let G be a group and H a subgroup. The sets $xH$ and $yH$ where $x, y \in G$ either coincide or are mutually exclusive; and $xH = yH$ if and only if $x^{-1}y \in H$. Each set xH is called a coset of H, more specifically a left coset. Right cosets Hx, Hy will be used infrequently. We use the notation G/H for the set of all left cosets. When G is a (topological space) G/H will be made into a space (see below 1.16.2) but we speak of it as a space in the present case also, although it carries no topology at present.
If H is an invariant subgroup, that is if $x^{-1}Hx=H$ equivalently if $xH=Hx$ for any $x \in G$, then
$xHyH = xyH$ where $x, y \in G$
is a true equation in sets of elements of G. In this case, the coset space becomes a group, the factor group $G/H$.
1.16.1
DEFINITION
By the natural map T of a group G onto the coset space G/H, H being a subgroup of G, we mean the map
$T : x \rightarrow xH$, where $x \in G, xH \in G/H$
For any subset $U \subset G$ we have
$T^{-1}(T(U)) = UH \subset G$
Let G be a topological group, H a subgroup. It is useful to topologize the coset space and to do this so that the natural map T is continuous. It will become clear as we proceed that unless the group H is a closed subgroup of G, it will not be possible in general to have T continuous and G/H a topological space; for this reason only the case where H is closed will be considered.
1.16.2
DEFINITION.
Let G be a topological group and let H be a closed subgroup of G, that is a subgroup which is a closed set. By an open set in G/H we mean a set whose inverse under the natural map I is an open set in G/H.
THEOREM.
With open sets defined as above, G/H is a topological space and the map T is continuous and open. If $xH \neq yH$, there exist neighbourhoods $W_{1}$ and $W_{2}$ of xH and yH respectively such that $W_{1} \bigcap W_{2}$ is empty.
Proof:
Let U denote an arbitrary open set in G. Then U/H is open (1.13) and is the inverse of $T(U)$. Since T(U) is open in G/H, T is an open map.
Let $x, y \in G$ and $xH \neq yH$. Then $x \in yH$ and yH is closed because H is closed. There exists a neighbourhood U of e such that $Ux \bigcap yH$ is empty. Let W be open, $e \in W$ and $W^{*} \subset U$ (Theorem in 1.15 above). Then if \$latex $WxH \bigcap WyH$ is not empty we can find $w, w_{1} \in W$ and $h, h_{1} \in H$ so that $w_{1}xh = wy h_{1}$. But this leads to $w^{-1}w_{1}x = yh_{1}h^{-1}$ and implies that Ux meets yH. Therefore $WxH \bigcap WyH$ is empty. The sets Wx and Wy are open. Therefore $W_{1}=T(Wx)$ and $W_{2} = T(Wy)$ are open in G/H; $xH \in W_{1}$ $yH \in W_{2}$ and $W_{1} \bigcap W_{2}$ is empty. This is the main part of the proof and depends on the fact that H is closed. We have proved more than condition (4) of 1.1 (The T_{0} separation axiom) The remaining three conditions of 1.1 are easy to verify. The fact that T is a continuous map is stated in the definition of open set in G/H. The fact that T is open was proved in the last paragraph.
QED.
COROLLARY.
If G is a topological group, $x, y \in G$ where $x \neq y$, then there exist neighbourhoods $W_{1}$ of x and $W_{2}$ of y such that $W_{1} \bigcap W_{2}$ is empty.
Proof:
To see this it is only necessary to take $H=e$. A space in which every pair of distinct points belong to mutually exclusive open sets is called a Hausdorff space. Therefore it has been shown that a topological group G and a coset space G/H, H closed in G, are Hausdorff spaces.
COROLLARY.
Suppose that G is a topological group and H a closed invariant subgroup. Then with the customary definition of product : $(xH)(yH) = xyH$,, G/H becomes a topological group. The natural map of G onto G/H is a continuous and open homomorphism.
1.17 A FAMILY OF NEIGHBOURHOODS
Suppose that we are given a topological group G and a sequence of neighbourhoods of e: $Q_{0}, Q_{1}, \ldots$ By repeated of the following Theorem 1.15: ( Let G be a topological group and U a neighbourhood of e. There exists a symmetric neighbourhood W of e such that $W^{2} \subset U$) .We can choose a sequence of symmetric open neighbourhoods of e : $U_{0}, U_{1}, \ldots$ with $U_{0}=Q_{0}$ such that
(1) $U_{n+1}^{2} \subset U_{n}\bigcap Q_{n}$ where n=0, 1, ….
In this section we shall show how to imbed the sets $U_{n}$ in a larger family of neighbourhoods possessing a multiplicative property which generalizes (1). We shall use this family in the next section to construct a real and non-constant function which is continuous on G. In 1.22 we shall use a similar family in order to construct a metric in a metrizable group. We remark in passing that in groups which do not satisfy the first countability axiom the set: $\bigcap U_{n}$ may be of considerable interest (to be discussed in a future blog section 2.6); it is a closed group (if $x, y \in \bigcap U_{n}$ then for every n, $xy \in U_{n+1}^{2} \subset U_{n}$).
Now, for each dyadic rational $r = \frac{k}{2^{n}}$, for n=0, 1, …and $k=1, \ldots, 2^{n}$ we define an open neighbourhood $V_{r}$ of e as follows:
(2) $V_{1/2^{n}}=U_{n}$ for all n
and then using (3) and (4) alternately by induction on k.
(3) $V_{2k/2^{n+1}}=V_{k/2^{n}}$
(4) $V_{(2k+1)/2^{n+1}} = V_{1/2^{n+1}}V_{k/2^{n}}$
Each $V_{n}$ depends on the dyadic rational r, and not on the particular representation by $k/2^{n}$. The entire family has the property:
(5) $V_{1/2^{n}}V_{m/2^{n}} \subset V_{m+1/2^{n}}$ where $m+1 \leq 2^{n}$
For $m=2k$, (5) is an immediate consequence of (3) and (4). For $m=2k+1$ the left side of (5) becomes: $V_{1/2^{n}}(V_{1/2^{n}}V_{k/2^{n-1}}) \subset V_{1/2^{n-1}}V_{k/2^{n-1}}$
The right side of (5) becomes $V_{(k+1)/2^{n-1}}$. This sets up an induction on n and since (5) holds for $n=1$, we have proved that (5) is true for all n. It follows from (5) and also more directly that:
(6) $V_{r} \subset V_{r^{'}}$ if $r
1.18 COMPLETE REGULARITY
THEOREM.
Suppose that G is a topological group and that F is a closed subset of G not containing e. Then one can define on G a continuous real function f, $0 \leq f(x) \leq 1$, $f(e) =0$, $f(x)=1$ if $x \in F$.
Proof:
In virtue of the property described in the theorem, G is called a completely regular space at the point e. The theorem is due to Pontrjagin.
Set $Q_{a}=G-F$ and setting $Q_{n}=Q_{0}$ for every n, construct a family of sets $V_{r}$ as in the preceding section.
Define f(x), $x \in G$ as follows:
(1) f(x)=0 if $x \in V_{r}$ for every r
(2) f(x)=1 if $x \in V_{1}$
and in all other cases
(3) $f(x)=lub_{r}$ where $\{ r \leq 1, x \in V_{r}\}$
It is clear that e belongs to every $V_{r}$ and that $F = G - V_{1}$ and does not meet $V_{1}$ so that there remains only to prove that f is continuous; let $\epsilon>0$ be given and let n be a positive integer such that $1/2^{n}<\epsilon$.
Now suppose that $f(x) <1$ at some point $x \in G$. Then there is a pair of integers m and k such that $k >n$ (same n as above), $m < 2^{n}$ and (interpreting $V_{0}$, if it occurs, as the null set),
$x \in V_{m/2^{k}}-V_{m-1/2^{k}}$
Let y be an arbitrary element in the neighbourhood $V_{1/2^{k}}x$. Then,
$y \in V_{1/2^{k}}V_{m/2^{k}} \subset V_{(m+1)/2^{k}}$
By the choice of y, $yx^{-1} \in V_{1/2^{k}}$; therefore $xy^{-1} \in V_{1/2^{k}}$ and $x \in V_{1/2^{k}}y$. It follows from this that $y$ cannot belong to $V_{(m-2)/2^{k}}$ and this shows that
$(m-2)/2^{k} \leq f(y) \leq (m+1)2^{k}$
Concerning x we know that
$(m-1)/2^{2^{k}} \leq f(x) \leq m/2^{k}$ and we may conclude from both inequalities that
$|f(x)-f(y)| \leq 2/2^{k} \leq 1/2^{n}<\epsilon$
Suppose next that $f(y)=1$ and choose $k>n$ as before. Let y be an arbitrary element in $V_{1/2^{k}}x$. Now y cannot belong to $V_{m/2^{k}}$ with $m < 2^{k}-2$ without implying that $f(x)<1$. It follows that
$1-2/2^{k} \leq f(y) \leq 1$
and again we get $|f(x) - f(y)| \leq e$
This concludes the proof and $f(x)$ is continuous on G.
QED.
COROLLARY.
A topological group is completely regular at every point.
1.19 HOMOGENEOUS SPACES
THEOREM.
Let G be a topological group, H a closed subgroup. Each element of G determines a homeomorphism of the coset space G/H onto itself, and G becomes a group of homeomorphisms of this space (topological space); furthermore G is transitive on the space: that is, each point may be carried to any other by an element of G.
Proof
Let $a \in G$. Associate to a the mapping $T_{a}: xH \rightarrow axH$.
This is a one-one transformation of G/H onto itself with $T_{a^{-1}}$ as inverse. These transformations are open and each transformation is a homeomorphism.
Since $T_{b}T_{a}$ is given by
$T_{b}T_{a}(xH) = T_{b}(axH)=baxH = T_{ba}(xH)$.
the association of $a \in G$ and $T_{a}$ makes a group of transformations of G/H. It is clear that xH is carried to yH by $T_{a}$ with $a = yx^{-1}$. This completes the proof.
QED.
A space is called HOMOGENEOUS when a group of homeomorphisms is transitive on it. We have shown that G/H is homogeneous with G being the transitive group of homeomorphisms. This implies that the unit segment $R_{1}$, for example, cannot be the underlying space of a group or even a coset space G/H since an end point of $R_{1}$ cannot go to an interior point to a homeomorphism of $R_{1}$.
It follows from the simultaneous continuity of $ax \in G$ in the factors a and x that the image point $axH$ of $xH$ under $T_{a}$ is continuous simultaneously in the counter point xH and the element $T_{a}$ of G. This makes G an instance of what is called a topological transformation group of a space M which will be defined below.
However, we shall be principally concerned with transformation groups which are locally compact and separable, acting on spaces which are topologically locally euclidean.
1.20 LOCAL GROUPS
An open neighbourhood of the identity of a topological group when it is regarded as a space in the relative topology has some of the properties of a group. There will usually be pairs of elements for which no product element exists in the neighbourhood. A structure of this kind is called a local group and will be defined below. Local groups often arise in a natural way, especially in the case of analytic group (Lie groups of transformations) and they have been intensively studied in that form
DEFINITION.
A space G is called a local group if a product xy is defined as an element in G for some pairs x, y in G and the following conditions are satisfied:
i) there is a unique element e in G such that ex and xe are defined for each x in G and $ex = xe=x$.
ii) If x, y are in G and xy exists then there is a neighbourhood U of x and a neighbourhood V of y such that if $x^{'} \in U$ and $y^{'} \in V$ then $x^{'}y^{'}$ exists. The product xy is continuous wherever defined.
iii) The associative law holds whenever it has meaning.
iv) If $ab=e$ then $ba=e$. An element b satisfying this relation is called an inverse and is denoted by $a^{-1}$. We assume that $a^{-1}$ is unique and continuous where defined and that if it exists for an element a it exists for all elements in some neighbourhood of a. Note that $a^{-1}$ always exists in some neighbourhood of e. In fact, there exists a symmetric open neighbourhood U of e such that $U^{2}$ is defined.
The above definition is somewhat redundant.
EXAMPLE.
Any neighbourhood O of the identity oa topological group is a local group if the neighbourhood is open.
We shall call two local groups isomorphic if there is a homeomorphism between their elements which carries inverse to inverse and product to product in so far as they are defined. However, in some applications, it is natural to regard two local groups as equivalent if they belong to the same local equivalence class, that is, a neighbourhood of e in one is isomorphic to a neighbourhood of e in the other. In this book an isomorphism and preserves group operations so far as they are defined.
LEMMA.
Let G be a local group with U the symmetric open neighbourhood of e described in the definition. Given any neighbourhood V of e, $V \subset U$, there is a symmetric neighbourhood W of e, $W^{3} \subset V$. The product sets AB, BC, (AB)C, A(BC) exist for $A, B, C \subset W$ and $A(BC) = (AB)C$. The set AB is open if either A or B is open. The sets A, bA, and Ab are homeomorphic for $b \in W$. Any two points of W have homeomorphic neighbourhoods.
The proof is omitted, the details being as in 1.13, 1.14 and 1.15.
Regards,
Nalin Pithwa
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 10,829 | 36,879 | {"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": 426, "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} | 3.6875 | 4 | CC-MAIN-2021-49 | latest | en | 0.822755 |
https://puzzling.stackexchange.com/questions/33879/what-happened-there-between-alice-and-bob/33884 | 1,582,837,989,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146809.98/warc/CC-MAIN-20200227191150-20200227221150-00059.warc.gz | 515,861,990 | 32,344 | # What happened there between Alice and Bob
So our favourite couple, Alice and Bob finally decided to get married ;)
Bob was already working in a well reputed company as Information Security Analyst. Alice, used to get bored at home, so she decided to be a chef. Both were happy, it all worked out well.
On weekdays, Bob's routine is:
Bob wakes up at 8, eats lovely and healthy breakfast prepared by Alice, goes to his office (which is nearby their home). He works hard all the day (as his work is highly confidential, we don't know much).
Bob arrives at the metro station from work at 9 pm. Alice leaves home in her car to meet him there at exactly 9 pm, and drives him home.
On one fine Friday, Bob got to the station an hour early, and started walking home, until Alice met him on the road. They got home 20 minutes earlier than usual.
The questions is: For how many minutes, was he walking?
Assumptions:
All the speeds are constant!
He was walking for
50 minutes
Explanation :
Let's call $d_w$ the distance he walked, $S_w$ his speed while walking, $d_c$ the distance in the car and $S_c$ the speed of the car.
Usually the travel takes $\frac{d_c + d_w}{S_c}$ but this time it takes $\frac{d_c}{S_c} + \frac{d_w}{S_w}$ and it was 40 minutes longer so :
$\frac{d_c + d_w}{S_c} = \frac{d_c}{S_c} + \frac{d_w}{S_w} +40$
so $\frac{d_w}{S_c} = \frac{d_w}{S_w} +40$
The wife drove 20 minutes less so she was at 10 minutes of the station ( it would have taken 10 minutes to drive the distance he walked) :
$\frac{d_w}{S_c} = 10$
When we combine the two equations : $\frac{d_w}{S_w} = 50$
• Correct and good explanation, accepted :) – ABcDexter Jun 4 '16 at 12:57
He was walking for
50 minutes
Explanation :
Alice usually drives from home to the station back home. Today she drove from home to the point where she met Bob back home. She saved 20 minutes. Because both ways are the same distance, she saved 10 minutes each way. Normally she meets Bob at 9pm. But today she met him 10 minutes earlier, at 8:50pm. Since Bob started walking at 8pm, he walked for 50 minutes.
• Way easier than my solution – Fabich Jun 4 '16 at 15:20
• That's simply brilliant :) – ABcDexter Jun 4 '16 at 17:44
Simple.
Let's assume that it takes an hour to drive to the station. Alice would leave home at 8 thinking she'd arrive at the station at 9. Bob would also be leaving the station on foot at 8. Alice would encounter Bob after driving X minutes, where X < 60 and X is the time Bob was walking. Then, Alice would return home, driving X minutes on the return trip. They arrive home 20 minutes early, meaning (as per assumption) they arrive at 9:40, since it would have normally taken an hour to get home from the station.
The total trip time (2X) is 1 hour and 40 minutes, or X = 50 minutes.
For further proof, let's now assume that it takes the minimum of 20 minutes to drive to the station. If Alice leaves the house at 8:40 expecting to be at the station at 9, then she would instead need to arrive home at 9 in order to get home twenty minutes early. She would spend ten minutes getting to Bob, and ten minutes getting home. She would encounter Bob on the road at 8:50. Bob started walking at 8. Thus, Bob has been walking for 50 minutes in this scenario as well.
• You made it sound very simple, kudos :D – ABcDexter Jun 4 '16 at 12:57 | 895 | 3,344 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.40625 | 4 | CC-MAIN-2020-10 | latest | en | 0.974062 |
http://en.wikibooks.org/wiki/Discrete_Mathematics/Logic/Page_2 | 1,432,699,965,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928869.6/warc/CC-MAIN-20150521113208-00189-ip-10-180-206-219.ec2.internal.warc.gz | 73,870,370 | 14,602 | # Discrete Mathematics/Logic/Page 2
## Conditional Propositions
Example 8
Discuss what Andy means when he says to Bernard: "If you want some more coffee, there’s some in the pot".
What Andy probably means is simply "There's some coffee in the pot, and if you want some, help yourself". Almost certainly, he doesn't really mean to imply that the presence (or otherwise) of the coffee in the pot is somehow dependent upon Bernard's desire for some. As you may have realised, we are sometimes very sloppy in our use of English! One of the things we need to do if we want to represent a sentence using logic symbols, is to work out what it really means.
### The ⇒ IMPLICATION
Example 9
Suppose that
p is "The weather is warm"; and
q is "I go swimming at lunchtime"
Then we can represent the conditional proposition "If the weather is warm, then I go swimming at lunchtime" by the symbols:
pq
So ⇒ means: "if ... then", or "implies (or means) that".
Other ways of saying the same thing are:
• "The weather is warm means that I go swimming at lunchtime"
• "Whenever the weather is warm, I go swimming at lunchtime"
• "The weather is warm only if I go swimming at lunchtime"
Be careful with this last option, using the words "only if". It can be a bit tricky. What it means is that if on a particular day I didn't go swimming at lunchtime, then the weather cannot have been warm on that day. It doesn't, of course, mean that my going swimming (or otherwise) determines the weather on that day!
### Necessary and Sufficient Conditions
Suppose that last Tuesday I went swimming at lunchtime. Given p and q as above, and given also that pq, can you be sure that the weather was warm last Tuesday?
The answer is that, no, you can't. I might have gone swimming anyway, for some other reason.
Note, then, that pq means that p is a sufficient condition for q. It is not, however, a necessary condition: q can still be true without p.
### The truth table for ⇒
Note that pq is itself a proposition; i.e. it has a truth value - it may or may not be the case that if the weather is warm, I go swimming at lunchtime.
Now the value of the proposition pq depends upon the combination of the values of p and q. So we can construct its truth table:
p q p ⇒ q T T T T F F F T T F F T
Note in particular the (perhaps surprising) third line of this table! The fact is that pq means that we cannot draw any conclusions about q if p is false. That is to say, if p is false, q may equally well be true or false without contradicting the truth of pq.
To put this another way, pq is false only when p is T and q is F; i.e. a true statement cannot imply a false one.
To clarify this further, consider the statement above: "If the weather is warm, then I go swimming at lunchtime". This says nothing about what happens when the weather is not warm: I may go swimming or I may not. The only time this statement is untrue is when the weather is warm, but I don't go swimming.
## Biconditional Propositions
As we have already noted, we often use English in a very imprecise way. Using Example 9, suppose what I really mean to say is:
If the weather is warm I go swimming at lunchtime, and if it’s not I don’t.
In this case p and q are either both true or both false: you can’t have one without the other. We could re-phrase this and say:
I go swimming at lunchtime if and only if the weather is warm.
The phrase "if and only if" is represented by the symbol ⇔, and so we can say in this case:
pq
In such a case as this, p is a necessary and sufficient condition for q.
Example 10
p is "x2 = 9". Find a suitable statement q about x (rather than x2) for which pq is true.
Solution
If x = 3, then certainly x2 = 9. So if q is "x = 3", then qp is true, and this would make q a sufficient condition.
But is it necessary and sufficient? No, because 3 is not the only number whose square is 9. x = -3 would also make x2 = 9.
To ensure a necessary and sufficient q, then, we would have to say:
q is "x = ±3"
### Logically Equivalent
We have said that pq means that p and q are either both true, or both false, and we can therefore say that in such a case they are logically equivalent.
### The truth table for ⇔
Since pq means that p and q are either both true or both false, the truth table for ⇔ is:
p q p ⇔ q T T T T F F F T F F F T
Example 11
By drawing up a truth table show that
pq ≡ (pq) $\scriptstyle \wedge$ (qp)
Solution
The truth table for (pq) $\scriptstyle \wedge$ (qp) is:
p q (p ⇒ q) $\scriptstyle \wedge$ (q ⇒ p) T T T T T T F F F T F T T F F F F T T T 1 3 output 2
The output is T, F, F, T which is the same as the output for pq, and therefore
pq ≡ (pq) $\scriptstyle \wedge$ (qp)
### The ⇐ Notation
We sometimes use ⇐ to mean "is implied by". So qp is an alternative way of writing pq, and we could have written Example 11 as:
pq ≡ (pq) $\scriptstyle \wedge$ (pq)
## Logic Exercise 4
Click the link for Logic Exercise 4.
## Predicate Logic
So far, we have considered only Propositional Logic, with statements like:
p is "All people with red hair have fiery tempers"
q is "Joe has red hair"
Given p and q as above and r is "Joe has a fiery temper", we can write:
p $\scriptstyle \wedge$ qr
If we want to make a statement about Brenda having red hair, and therefore a fiery temper, we should need further propositions, like this:
s is "Brenda has red hair"
t is "Brenda has a fiery temper"
and so:
p $\scriptstyle \wedge$ st
… and so on.
Each time we want to make a statement about another person having red hair, and therefore a fiery temper (which may or may not be true!), we shall need further propositions. A much better way of representing these ideas is to use predicates like this:
redHair is the phrase "... has red hair"
(Notice that redHair is not a proposition. Why not?)
We can now use this predicate to form statements about anyone who has red hair; like this:
redHair(Joe) is "Joe has red hair"
redHair(Brenda) is"‘Brenda has red hair"
... and so on.
In the same way, we can define the predicate fieryTemper to stand for the phrase:
"... has a fiery temper"
So the statement "If Joe has red hair, then he has a fiery temper" can be represented by:
redHair(Joe) ⇒ fieryTemper(Joe)
and the statement "If Brenda has red hair, then she has a fiery temper" by:
redHair(Brenda) ⇒ fieryTemper(Brenda)
### Notation
We shall use single words or several words joined together, using upper- and lower-case letters as shown, to denote predicates.
The "object" to which the predicate applies - a person, number, or whatever - will be written in parenthesis following the predicate.
### Negation
With red Hair defined as above:
¬with red Hair is "it is not the case that with red hair"
## Logic Exercise 5
Click the link for Logic Exercise 5.
## Propositional Functions
If we want to talk about general, undefined, predicates, we shall use upper-case letters: P, Q, ..., and if we want a general, undefined, object, we shall use a lower-case letter: x, y, ...
So if P is "... has property P", then P(x) is "x has property P".
• P(x) can then be described as a propositional function whose predicate is P.
A function has the property that it returns a unique value when we know the value(s) of any parameter(s) supplied to it. P(x) is therefore a function since it returns a truth value which depends upon the value of its parameter, x.
For example, if American(x) is the propositional function "x is an American", then American(x) will return the value
T if x = Bill Clinton; and the value
F if x = HM The Queen
## Quantifiers
We now extend the ideas in Exercise 5 above. Suppose we want to make statements like:
(a) All of my friends are wealthy.
(b) Some of my friends are boring.
The problem here is that we are making statements about my friends in general, without referring to a particular individual. So we need to define propositional functions as follows:
friend(x) is "x is a friend of mine"
wealthy(x) is "x is wealthy"
boring(x) is "x is boring"
We can then write the two statements above as:
(a) For all x, friend(x) ⇒ wealthy(x)
(b) For some x, friend(x) $\scriptstyle \wedge$ boring(x)
### Notation: ∀ and ∃
The symbol ∀ (called the universal quantifier) stands for the phrase "For all …"
So we can write (a) above as:
(a) ∀ x, friend(x) ⇒ wealthy(x)
The symbol ∃ (called the existential quantifier) stands for the phrase "For some …"
So we can write (b) above as:
(b) ∃ x, friend(x) $\scriptstyle \wedge$ boring(x)
#### Plural or singular?
Note that, although statements (a) and (b) above use plural words – all, are, some, friends – when we symbolise them, the predicates and the variables are singular: x is wealthy, x is boring, etc. It is important to realise, then, that x can stand for just one value at a time.
So the symbolic statement:
x, friend(x) ⇒ wealthy(x)
would be literally translated using singular words as:
"For each value of x, if x is a friend of mine, then x is wealthy".
and
x, friend(x) $\scriptstyle \wedge$ boring(x)
is more literally translated:
"For at least one value of x, x is a friend of mine and x is boring".
To emphasise this, you might find it helpful to use the following as translations of the symbols:
∀ means "For each (value of) ..."
and
∃ means "For at least one (value of) ..."
We can now make our earlier statement that "All people with red hair have fiery tempers" using Propositional Function notation as follows:
redHair(x) is: "x has red hair"
fieryTemper(x) is: "x has a fiery temper"
Now "All people with red hair have fiery tempers" is re-written in the singular as:
For each value of x, if x has red hair, then x has a fiery temper.
In symbols, then:
x, redHair(x) ⇒ fieryTemper(x)
Example 12
Define suitable propositional functions and then express in symbols:
(a) Some cats understand French.
(b) No footballers can sing.
(c) At least one lecturer is not boring.
(d) I go swimming every sunny day.
Solutions
(a) Re-write in the singular: "At least one cat understands French". So we shall need to define propositional functions as:
cat(x) is "x is a cat"
French(x) is "x understands French"
So there is at least one x that is a cat and understands French; or, in symbols:
x, cat(x) $\scriptstyle \wedge$ French(x)
(b) Re-write in the singular: "It is not true that at least one footballer can sing". So:
footballer(x) is "x is a footballer"
sing(x) is "x can sing"
In symbols, then:
¬ (∃x, footballer(x) $\scriptstyle \wedge$ sing(x))
Alternatively, we might re-write "No footballers can sing" as "For each x, if x is a footballer, then x cannot sing". In symbols, then, this gives the equally valid solution:
x, footballer(x) ⇒ ¬ sing(x)
(c) This is already in the singular; so:
lecturer(x) is "x is a lecturer"
boring(x) is "x is boring"
So:
x, lecturer(x) $\scriptstyle \wedge$ ¬ boring(x)
(d) In this example, it is important to realise what the variable represents. In (a) to (c) the variable x has denoted an animal or a person. In the sentence "I go swimming every sunny day" it is the day that changes, and, with it, the weather and my going swimming. So we must form our propositional functions around a variable x that stands for a day. Thus:
sunny(x) is "x is a sunny day"
swimming(x) is "x is a day when I go swimming"
Then, re-writing "I go swimming every sunny day" in the singular, we get:
"For each day, if it is a sunny day then it is a day when I go swimming"
So, in symbols:
x, sunny(x) ⇒ swimming(x)
## Logic Exercise 6
Click the link for Logic Exercise 6.
## Universe of Discourse
Many of the propositions in Exercises 5 and 6 referred to 'my friends'. For example, consider the proposition: "All of my friends are either wealthy or clever."
Using predicates, we can symbolise this as:
x, friend(x) ⇒ (wealthy(x) ∨ clever(x))
However, if we agreed that the variable x can only stand for one of my friends, then we could symbolise this more simply as:
x, wealthy(x) ∨ clever(x)
For a given propositional function P(x), the universe of discourse is the set from which the value of x may be chosen. Defining a universe of discourse can simplify the symbolisation of propositional functions. If a universe of discourse is not defined, then we shall have to assume that any object or individual may be substituted for x.
Example 13
Define, in each case, a suitable universe of discourse and predicates to symbolise:
(a) Some students are hard-working or drink too much.
(b) Everybody was hot and someone fainted.
Solutions
(a) Define the following:
Universe of discourse = {students}
hardWorking(x) is "x is hard-working"
drink(x) is "x drinks too much
Writing in the singular: "For at least one x, x is hard-working or x drinks too much", we get:
x, hardWorking(x) ∨ drink(x)
(b) In the given sentence, the word "everybody" clearly doesn't mean everybody in the whole world, simply everybody in the story being recounted. So we can define as follows:
Universe of discourse = {people in the story}
hot(x) is "x was hot"
fainted(x) is "x fainted"
Then, in the singular, we have "For each x, x was hot and for at least one x, x fainted". So:
x, hot(x) $\scriptstyle \wedge$x, fainted(x)
## Two-place Predicates
The predicates we have looked at so far have been one-place predicates. To convert each predicate into a proposition, we have had to supply a single object or individual - the value of a single parameter, if you like.
We can create predicates which require two objects (or parameter values) to be supplied to convert them to propositions. Such predicates are called two-place predicates.
Example 14
Consider the following predicates:
greaterThan is "… is greater than …"
loves is "… loves …"
belongsTo is "… belongs to …"
Then the following are two-variable propositional functions:
greaterThan(x, y) is "x is greater than y"
loves(x, y) is "x loves y"
belongsTo(x, y) is "x belongs to y"
So, for example, the following are propositions:
greaterThan(5, 2) is "5 is greater than 2"
loves(Bob, Lizzie) is "Bob loves Lizzie"
belongsTo(This coat, Harry) is "This coat belongs to Harry"
### Two-place Predicates and Quantifiers
With the predicates above, we can quantify over one variable:
x, belongsTo(x, me) is "Everything belongs to me"
x, greaterThan(2, x) is "2 is greater than something"
x, loves(Mary, x) is "Mary loves everyone"
... or both variables:
x, ∃y, loves(x, y) is "Everybody loves somebody"
x, ∀ y, loves(x, y) is "Somebody loves everybody"
x, ∃ y, loves(x, y) is "Somebody loves somebody"
x, ∀ y, loves(x, y) is "Everybody loves everybody"
## Negation of Quantified Propositional Functions
In Question 5 of Exercise 1, we had to say which of several options represented the negation of a proposition. Let’s look at this again, using our Quantified Propositional Function notation.
Example 15
Consider the negation of the proposition
"All sheep are black".
The negation is:
"It is not true that all sheep are black".
which is equivalent to:
"At least one sheep is not black".
If we define the universe of discourse as {sheep} and isBlack in the obvious way, then we can symbolise all this as follows:
¬ (∀ x, isBlack(x)) ≡ ∃ x, ¬isBlack(x)
Now consider the proposition
"Some sheep are black".
The negation of this is:
"It is not true that some sheep are black"
which is equivalent to:
"All sheep are not black"
This can then be symbolised as:
¬ (∃ x, isBlack(x)) ≡ ∀ x, ¬isBlack(x)
We can generalise what we have found here to any propositional function P(x), as follows:
¬(∀ x, P(x)) ≡ ∃ x, ¬ P(x)
¬(∃ x, P(x)) ≡ ∀ x, ¬P(x)
## Logic Exercise 7
Click the link for Logic Exercise 7.
Previous topic:Discrete Mathematics/Number theory|Contents:Discrete Mathematics|Next topic:Discrete Mathematics/Enumeration | 4,040 | 15,802 | {"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": 14, "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-2015-22 | latest | en | 0.970543 |
http://www.stata.com/statalist/archive/2005-10/msg00000.html | 1,438,400,353,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988458.74/warc/CC-MAIN-20150728002308-00125-ip-10-236-191-2.ec2.internal.warc.gz | 730,371,690 | 2,761 | # Re: st: number of unique values within an observation
From "Svend Juul" To Subject Re: st: number of unique values within an observation Date Fri, 30 Sep 2005 22:30:33 +0200
```Scott Imberman wrote:
I was wondering if there is s a simple way to count the number of
unique values in a variable list for each observation. For example, i
am looking at the id numbers for colleges that a set of student attends.
A student may list having attended a college in multiple years like
this:
year id
1995 3
1996 2
1997 3
1998 3
I want to know how many unique schools the student attended (in this
-------------------
In Stata terminology the information for the four years for one student
is not organized in one, but four observations (long format). A solution
follows here (but I guess there are simpler ways):
------------------------------------------------------------------------
clear
input student year school
17 1995 3
17 1996 2
17 1997 3
17 1998 3
18 1995 1
18 1996 1
18 1997 1
18 1998 1
19 1995 1
19 1996 2
19 1997 3
19 1998 4
end
gen x=1
sort student school
list
replace x=x[_n-1]+1 if student==student[_n-1] & school != school[_n-1]
replace x=x[_n-1] if student==student[_n-1] & school == school[_n-1]
by student: gen nschools=x[_N]
list
+----------------------------------------+
| student year school x nschools |
|----------------------------------------|
1. | 17 1996 2 1 2 |
2. | 17 1997 3 2 2 |
3. | 17 1995 3 2 2 |
4. | 17 1998 3 2 2 |
5. | 18 1997 1 1 1 |
|----------------------------------------|
6. | 18 1996 1 1 1 |
7. | 18 1995 1 1 1 |
8. | 18 1998 1 1 1 |
9. | 19 1995 1 1 4 |
10. | 19 1996 2 2 4 |
|----------------------------------------|
11. | 19 1997 3 3 4 |
12. | 19 1998 4 4 4 |
+----------------------------------------+
------------------------------------------------------------------------
Hope this helps,
Svend
________________________________________________________
Svend Juul
Institut for Folkesundhed, Afdeling for Epidemiologi
(Institute of Public Health, Department of Epidemiology)
Vennelyst Boulevard 6
DK-8000 Aarhus C, Denmark
Phone, work: +45 8942 6090
Phone, home: +45 8693 7796
Fax: +45 8613 1580
E-mail: sj@soci.au.dk
_________________________________________________________
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 829 | 2,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2015-32 | longest | en | 0.676598 |
https://support.nag.com/numeric/nl/nagdoc_28.6/flhtml/f01/f01lef.html | 1,702,124,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100909.82/warc/CC-MAIN-20231209103523-20231209133523-00723.warc.gz | 600,954,127 | 7,630 | # NAG FL Interfacef01lef (real_gen_tridiag_lu)
## ▸▿ Contents
Settings help
FL Name Style:
FL Specification Language:
## 1Purpose
f01lef computes an $LU$ factorization of a real tridiagonal matrix, using Gaussian elimination with partial pivoting.
## 2Specification
Fortran Interface
Subroutine f01lef ( n, a, b, c, tol, d, ipiv,
Integer, Intent (In) :: n Integer, Intent (Inout) :: ifail Integer, Intent (Out) :: ipiv(n) Real (Kind=nag_wp), Intent (In) :: lambda, tol Real (Kind=nag_wp), Intent (Inout) :: a(n), b(n), c(n) Real (Kind=nag_wp), Intent (Out) :: d(n)
#include <nag.h>
void f01lef_ (const Integer *n, double a[], const double *lambda, double b[], double c[], const double *tol, double d[], Integer ipiv[], Integer *ifail)
The routine may be called by the names f01lef or nagf_matop_real_gen_tridiag_lu.
## 3Description
The matrix $T-\lambda I$, where $T$ is a real $n×n$ tridiagonal matrix, is factorized as
$T-λI=PLU,$
where $P$ is a permutation matrix, $L$ is a unit lower triangular matrix with at most one nonzero subdiagonal element per column, and $U$ is an upper triangular matrix with at most two nonzero superdiagonal elements per column.
The factorization is obtained by Gaussian elimination with partial pivoting and implicit row scaling.
An indication of whether or not the matrix $T-\lambda I$ is nearly singular is returned in the $n$th element of the array ipiv. If it is important that $T-\lambda I$ is nonsingular, as is usually the case when solving a system of tridiagonal equations, then it is strongly recommended that ${\mathbf{ipiv}}\left(n\right)$ is inspected on return from f01lef. (See the argument ipiv and Section 9 for further details.)
The argument $\lambda$ is included in the routine so that f01lef may be used, in conjunction with f04lef, to obtain eigenvectors of $T$ by inverse iteration.
## 4References
Wilkinson J H (1965) The Algebraic Eigenvalue Problem Oxford University Press, Oxford
Wilkinson J H and Reinsch C (1971) Handbook for Automatic Computation II, Linear Algebra Springer–Verlag
## 5Arguments
1: $\mathbf{n}$Integer Input
On entry: $n$, the order of the matrix $T$.
Constraint: ${\mathbf{n}}\ge 1$.
2: $\mathbf{a}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Input/Output
On entry: the diagonal elements of $T$.
On exit: the diagonal elements of the upper triangular matrix $U$.
3: $\mathbf{lambda}$Real (Kind=nag_wp) Input
On entry: the scalar $\lambda$. f01lef factorizes $T-\lambda I$.
4: $\mathbf{b}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Input/Output
On entry: the superdiagonal elements of $T$, stored in ${\mathbf{b}}\left(2\right)$ to ${\mathbf{b}}\left(n\right)$; ${\mathbf{b}}\left(1\right)$ is not used.
On exit: the elements of the first superdiagonal of $U$, stored in ${\mathbf{b}}\left(2\right)$ to ${\mathbf{b}}\left(n\right)$.
5: $\mathbf{c}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Input/Output
On entry: the subdiagonal elements of $T$, stored in ${\mathbf{c}}\left(2\right)$ to ${\mathbf{c}}\left(n\right)$; ${\mathbf{c}}\left(1\right)$ is not used.
On exit: the subdiagonal elements of $L$, stored in ${\mathbf{c}}\left(2\right)$ to ${\mathbf{c}}\left(n\right)$.
6: $\mathbf{tol}$Real (Kind=nag_wp) Input
On entry: a relative tolerance used to indicate whether or not the matrix ($T-\lambda I$) is nearly singular. tol should normally be chosen as approximately the largest relative error in the elements of $T$. For example, if the elements of $T$ are correct to about $4$ significant figures, then tol should be set to about $5×{10}^{-4}$. See Section 9 for further details on how tol is used. If tol is supplied as less than $\epsilon$, where $\epsilon$ is the machine precision, then the value $\epsilon$ is used in place of tol.
7: $\mathbf{d}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Output
On exit: the elements of the second superdiagonal of $U$, stored in ${\mathbf{d}}\left(3\right)$ to ${\mathbf{d}}\left(n\right)$; ${\mathbf{d}}\left(1\right)$ and ${\mathbf{d}}\left(2\right)$ are not used.
8: $\mathbf{ipiv}\left({\mathbf{n}}\right)$Integer array Output
On exit: details of the permutation matrix $P$. If an interchange occurred at the $k$th step of the elimination, then ${\mathbf{ipiv}}\left(k\right)=1$, otherwise ${\mathbf{ipiv}}\left(k\right)=0$. If a diagonal element of $U$ is small, indicating that $\left(T-\lambda I\right)$ is nearly singular, then the element ${\mathbf{ipiv}}\left(n\right)$ is returned as positive. Otherwise ${\mathbf{ipiv}}\left(n\right)$ is returned as $0$. See Section 9 for further details. If the application is such that it is important that $\left(T-\lambda I\right)$ is not nearly singular, then it is strongly recommended that ${\mathbf{ipiv}}\left(n\right)$ is inspected on return.
9: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, $-1$ or $1$ to set behaviour on detection of an error; these values have no effect when no error is detected.
A value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not.
If halting is not appropriate, the value $-1$ or $1$ is recommended. If message printing is undesirable, then the value $1$ is recommended. Otherwise, the value $0$ is recommended. When the value $-\mathbf{1}$ or $\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 1$.
${\mathbf{ifail}}=-99$
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
## 7Accuracy
The computed factorization will satisfy the equation
$PLU=(T-λI)+E,$
where
$‖E‖1≤ 9×maxi≥j (|lij|,|lij|2) ε ‖T-λI‖1$
where $\epsilon$ is the machine precision.
## 8Parallelism and Performance
f01lef makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The time taken by f01lef is approximately proportional to $n$.
The factorization of a tridiagonal matrix proceeds in $\left(n-1\right)$ steps, each step eliminating one subdiagonal element of the tridiagonal matrix. In order to avoid small pivot elements and to prevent growth in the size of the elements of $L$, rows $k$ and ($k+1$) will, if necessary, be interchanged at the $k$th step prior to the elimination.
The element ${\mathbf{ipiv}}\left(n\right)$ returns the smallest integer, $j$, for which
$|ujj|≤‖(T-λI)j‖1×tol,$
where ${‖{\left(T-\lambda I\right)}_{j}‖}_{1}$ denotes the sum of the absolute values of the $j$th row of the matrix ($T-\lambda I$). If no such $j$ exists, then ${\mathbf{ipiv}}\left(n\right)$ is returned as zero. If such a $j$ exists, then $|{u}_{jj}|$ is small and hence ($T-\lambda I$) is singular or nearly singular.
This routine may be followed by f04lef to solve systems of tridiagonal equations. If you wish to solve single systems of tridiagonal equations you should be aware of f07caf, which solves tridiagonal systems with a single call. f07caf requires less storage and will generally be faster than the combination of f01lef and f04lef, but no test for near singularity is included in f07caf and so it should only be used when the equations are known to be nonsingular.
## 10Example
This example factorizes the tridiagonal matrix $T$ where
$T= ( 3.0 2.1 0 0 0 3.4 2.3 -1.0 0 0 0 3.6 -5.0 1.9 0 0 0 7.0 -0.9 8.0 0 0 0 -6.0 7.1 )$
and then to solve the equations $Tx=y$, where
$y= ( 2.7 -0.5 2.6 0.6 2.7 )$
by a call to f04lef. The example program sets ${\mathbf{tol}}=5×{10}^{-5}$ and, of course, sets ${\mathbf{lambda}}=0$.
### 10.1Program Text
Program Text (f01lefe.f90)
### 10.2Program Data
Program Data (f01lefe.d)
### 10.3Program Results
Program Results (f01lefe.r) | 2,635 | 8,871 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 115, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-50 | latest | en | 0.712344 |
http://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-7-section-7-2-multiplying-and-dividing-rational-expressions-exercise-set-page-500/33 | 1,524,173,728,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937045.30/warc/CC-MAIN-20180419204415-20180419224415-00528.warc.gz | 423,401,618 | 14,564 | ## Algebra: A Combined Approach (4th Edition)
$\dfrac{x^{2}+5x}{8}\cdot\dfrac{9}{3x+15}=\dfrac{3x}{8}$
$\dfrac{x^{2}+5x}{8}\cdot\dfrac{9}{3x+15}$ Take out common factor $x$ from the numerator of the first fraction and common factor $3$ from the denominator of the second fraction: $\dfrac{x^{2}+5x}{8}\cdot\dfrac{9}{3x+15}=\dfrac{x(x+5)}{8}\cdot\dfrac{9}{3(x+5)}=...$ Evaluate the product of the two rational expressions and then simplify by removing repeated factors in the numerator and the denominator: $...=\dfrac{9x(x+5)}{24(x+5)}=\dfrac{9x}{24}=\dfrac{3x}{8}$ | 205 | 566 | {"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} | 4.375 | 4 | CC-MAIN-2018-17 | latest | en | 0.719059 |
https://documen.tv/question/39-when-you-heat-a-flask-of-water-how-are-you-changing-the-ordered-kinetic-energy-of-the-water-m-15940394-32/ | 1,638,274,344,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00043.warc.gz | 282,508,240 | 15,988 | ## 39. When you heat a flask of water, how are you changing the ordered kinetic energy of the water molecules (if at all)? Likewis
Question
39. When you heat a flask of water, how are you changing the ordered kinetic energy of the
water molecules (if at all)? Likewise, how are you changing the random kinetic energy of the
molecules (if at all)?
in progress 0
3 months 2021-07-30T16:54:45+00:00 1 Answers 5 views 0
The average kinetic energy of the molecules increases
Explanation:
The temperature of a substance is proportional to the average kinetic energy of the particles in the substance.
In fact, for an ideal gas for instance, there is the following relationship:
where
KE is the average kinetic energy of the particles
k is the Boltzmann’s constant
T is the absolute temperature of the gas
When we heat a substance (such as the flask of water in this problem), we are giving thermal energy to the particles of the substance; therefore, these particles will move faster on average, so their kinetic energy will increase (and the temperature of the substance will increase as well). | 248 | 1,101 | {"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-2021-49 | latest | en | 0.897555 |
https://dlmf.nist.gov/search/search?q=generators | 1,726,502,157,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00854.warc.gz | 190,176,488 | 5,295 | # generators
(0.001 seconds)
## 1—10 of 395 matching pages
##### 2: 16.2 Definition and Analytic Properties
###### Polynomials
Note also that any partial sum of the generalized hypergeometric series can be represented as a generalized hypergeometric function via …
##### 4: 1.16 Distributions
$\Lambda:\mathcal{D}(I)\rightarrow\mathbb{C}$ is called a distribution, or generalized function, if it is a continuous linear functional on $\mathcal{D}(I)$, that is, it is a linear functional and for every $\phi_{n}\to\phi$ in $\mathcal{D}(I)$, … More generally, for $\alpha\colon[a,b]\to[-\infty,\infty]$ a nondecreasing function the corresponding Lebesgue–Stieltjes measure $\mu_{\alpha}$ (see §1.4(v)) can be considered as a distribution: … More generally, if $\alpha(x)$ is an infinitely differentiable function, then … Friedman (1990) gives an overview of generalized functions and their relation to distributions. …
##### 7: 8.16 Generalizations
###### §8.16 Generalizations
For a generalization of the incomplete gamma function, including asymptotic approximations, see Chaudhry and Zubair (1994, 2001) and Chaudhry et al. (1996). Other generalizations are considered in Guthmann (1991) and Paris (2003).
##### 9: 7.16 Generalized Error Functions
###### §7.16 Generalized Error Functions
Generalizations of the error function and Dawson’s integral are $\int_{0}^{x}e^{-t^{p}}\,\mathrm{d}t$ and $\int_{0}^{x}e^{t^{p}}\,\mathrm{d}t$. …
##### 10: 16.26 Approximations
###### §16.26 Approximations
For discussions of the approximation of generalized hypergeometric functions and the Meijer $G$-function in terms of polynomials, rational functions, and Chebyshev polynomials see Luke (1975, §§5.12 - 5.13) and Luke (1977b, Chapters 1 and 9). | 489 | 1,740 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 10, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.746697 |
https://ogrecave.github.io/ogre/api/latest/tut__static_geom.html | 1,723,066,371,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00404.warc.gz | 356,869,595 | 7,455 | OGRE 14.2 Object-Oriented Graphics Rendering Engine
Static Geometry
If we have a collection of entities in our scene that will not be moved, then Ogre can perform an optimization by rendering them in "batches". Modern GPUs are designed to render an enormous amount of triangles at once. We are able to take advantage of this through the batching techniques used by a Ogre::StaticGeometry object. Static geometry is a bit of a misnomer in this case, because we can use some tricks to accomplish things like grass waving in the wind, but the general idea is that this is an object that will not be manipulated a great deal. Good examples include rocks, trees, and buildings.
The full source for this tutorial can be found in samples directory Samples/Simple/include/Grass.h.
The first thing we will do is create the grass mesh we will be rendering. We will use a pattern you've probably seen to create the illusion of grass. We will render three square quads that have a grass texture applied to them. We will create one, then place another rotated 60 degrees, and then place a third rotated 120 degrees. This will create a simple illusion of 3D grass.
The first step will be to define some variables. We will define the width and height of our quad, then we will initialize a vector that will be used to define the four corners of our quad. We will again be using quaternions to handle rotations. Our plan is to use the vector to represent the orientation of the base of our quad.
// planes intersect along the Y axis with 60 degrees between them
Vector3 vec = Quaternion(Degree(i * 60), Vector3::UNIT_Y) * Vector3(width / 2, 0, 0);
Vector< 3, Real > Vector3
Definition: OgrePrerequisites.h:252
This should look somewhat familiar. We have created a quaternion that rotates by multiples of 60 degree around the y-axis.
The vector we are using starts out pointing down the x-axis with a length that is half the width of our quad. This may be a little hard to visualize. Here is a picture to help:
Remember that x and z are in the plane of the floor. So our vector keeps track of where the foundation of our quad is, we build everything from that.
We will now begin defining our manual object. We set the render operation to be OT_TRIANGLE_LIST. This means that after we define our vertices with the position method, we then have to let Ogre know how to set up the index buffer by giving it a list of triangles made from the vertices.
ManualObject obj("GrassObject");
For each quad we are going to define four vertices representing the corners. We will also specify a texture coordinates. These are normalized coordinates that tell Ogre how to map the texture on to our mesh. In our case, these coordinates are very simple since we are creating a solid square.
for (unsigned int j = 0; j < 4; j++) // each plane has 4 vertices
{
vec.y = j % 2 ? 0 : height;
obj.position(j < 2 ? Vector3(-1, 1, -1) * vec : vec);
obj.textureCoord(j < 2 ? 0 : 1, j % 2);
// all normals point straight up
obj.normal(0, 1, 0);
}
We've also labeled the four corners with the order they were created to help with creating the triangles. The count starts with the 0th corner.
To ensure that both triangles face the same direction, we need to provide the points in counter-clockwise order. The triangle method does not directly take positions, instead it takes three numbers that represent the order in which the points were created. This is why we labeled the four corners in our image.
unsigned int off = i * 4;
// each plane consists of 2 triangles
obj.triangle(off + 0, off + 3, off + 1);
obj.triangle(off + 0, off + 2, off + 3);
First, ignore the offset value and look at the numbers we are adding. They match the numbers we assigned in the image. The first triangle connects the 0th, 3rd, and 1st corners. The second triangle connects the 0th, 2nd, and 3rd corners. You can look at the image to see these are in counter-clockwise order. The purpose of the offset is because we are creating three different quads, but they are all going to be a part of one manual object. So the second quad's corners will be numbered 4, 5, 6, 7. Adding the offset accounts for this.
After we've created all three quads, the loop finishes and we call end to finalize the object.
obj.end();
obj.convertToMesh("grass");
The last line converts our manual object into an actual mesh. Meshes require less storage compared to directly using the manual object for rendering.
We are now finished creating the grass mesh. If you create a complex mesh, then you may save it to a file instead of rebuilding the mesh each time. To do this, you would save the mesh pointer that is returned by convertToMesh. Then you would use a mesh serializer to export the mesh to a file.
Here is an example. Do not add this code to our current project.
ser.exportMesh(mesh, "my_grass.mesh");
Class for serialising mesh data to/from an OGRE .mesh file.
Definition: OgreMeshSerializer.h:92
void exportMesh(const Mesh *pMesh, const String &filename, MeshVersion version, Endian endianMode=ENDIAN_NATIVE)
Exports a mesh to the file specified, in a specific version format.
Pre-transforms and batches up meshes for efficient use as static geometry in a scene.
Definition: OgreStaticGeometry.h:121
# A field of grass
Now we get to the creation of our static geometry. The first thing we do is create an entity from the grass mesh we constructed, then we ask the scene manager to give us a pointer to a new StaticGeometry object.
// create our grass mesh, and create a grass entity from it
createGrassMesh();
Entity* grass = mSceneMgr->createEntity("Grass", "grass");
// create a static geometry field, which we will populate with grass
mField = mSceneMgr->createStaticGeometry("Field");
mField->setRegionDimensions(Vector3(140, 140, 140));
mField->setOrigin(Vector3(70, 70, 70));
Here, RegionDimensions refer to the physical size of one batch. All grass patches located within the region will be treated as one. Patches outside the region will be in a separate batch. This allows you to trade culling for batching effectiveness.
Now we will prepare the actual build. We are going to loop through points on the floor of our region and place a grass patch with a random offset at each point.
// add grass uniformly throughout the field, with some random variations
for (int x = -280; x < 280; x += 20)
{
for (int z = -280; z < 280; z += 20)
{
Vector3 pos(x + Math::RangeRandom(-7, 7), 0, z + Math::RangeRandom(-7, 7));
Quaternion ori(Degree(Math::RangeRandom(0, 359)), Vector3::UNIT_Y);
Vector3 scale(1, Math::RangeRandom(0.85, 1.15), 1);
}
}
mField->build(); // build our static geometry (bake the grass into it)
We've partitioned the floor of our region into enough sections to fit all of our grass patches. The first thing we do is calculate an offset that will be used to randomly nudge each grass patch. This will help it look a little more natural. We use this offset to define a position vector for our object. To do this, we use the Ogre::Math::RangeRandom. We then use the same method to create a randomized scale vector to add some more variety to our grass. Our three quads were already rotated 60 degrees from each other, but now we are randomly rotating the entire grass patch as a whole. Always remember to play around with numbers like these to create surprising effects in your scene. Some great game mechanics have been discovered by doing exactly this.
The last thing we do is add a new grass entity to our scene using all of the information we've just set up. We then call the build method to construct our StaticGeometry object.
Note
When defining static geometry, you will either use Ogre::StaticGeometry::addEntity or Ogre::StaticGeometry::addSceneNode. The latter method adds all of the entities attached to that scene node to the static geometry. It uses the positions, orientations, and scales of the child nodes instead of requiring you to specify them manually. When using addSceneNode, be sure to remove the scene node from its previous parent. If you do not, Ogre will render them both.
# Animating StaticGeometry
Once you have created a StaticGeometry object, you are not supposed to do much more with it. After all, that's the entire point behind static geometry - it's supposed to be static. As mentioned before, you can use certain tricks to do things like grass waving in the wind. If you are interested in how to do this, then take a look at the Examples/GrassBladesWaver material, which adds wave-like behavior to our grass meshes through a vertex-shader. | 1,990 | 8,549 | {"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} | 3.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.90213 |
https://ch.mathworks.com/matlabcentral/answers/375525-how-do-i-use-parameters-in-the-upper-and-lower-limits-of-an-intergral?s_tid=prof_contriblnk | 1,642,439,002,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00451.warc.gz | 238,694,782 | 24,444 | # How do I use parameters in the upper and lower limits of an intergral?
2 views (last 30 days)
Sergio Manzetti on 4 Jan 2018
Commented: Sergio Manzetti on 4 Jan 2018
Hi, I have the following code
if true
% code
end
syms h g x C n L p e E C m
h = 1.0545718E-34
g = 5.344285879E-28
m = 9.10938356E-31
E = h^2/8*m*L^2
y = - (exp(-(x*(g*1i + (- 2*g^2 + E)^(1/2)))/h)*(g*1i - (- 2*g^2 + E)^(1/2)))/(2*(E - 2*g^2)^(1/2)) + (exp(-(x*(g*1i - (- 2*g^2 + E)^(1/2)))/h)*(g*1i + (- 2*g^2 + E)^(1/2)))/(2*(E - 2*g^2)^(1/2))
z = - (exp(-(x*(g*(-i) + (- 2*g^2 + E)^(1/2)))/h)*(g*(-i) - (- 2*g^2 + E)^(1/2)))/(2*(E - 2*g^2)^(1/2)) + (exp(-(x*(g*(-i) - (- 2*g^2 + E)^(1/2)))/h)*(g*(-i) + (- 2*g^2 + E)^(1/2)))/(2*(E - 2*g^2)^(1/2))
v = i*h*diff(z, x)
S = y*z
W = [int(S, 0, L)] % Operator check hermiticity
vpa(W) % vpa= Variable precision arithmetic, simplifies the large fractions to nunmbers
however the upper limit set to L is not accepted by MATLAB. Is there any chance of telling MATLAB to treat this as a constant?
Thanks
##### 2 CommentsShowHide 1 older comment
Sergio Manzetti on 4 Jan 2018
Sorry. There was an error, but related to vpa(W). Removing this makes it work, although it looks horribly long.
Birdman on 4 Jan 2018
The same code works me without any error, including vpa(W). To decide number of digits, use the following:
vpa(W,3)
##### 1 CommentShowHide None
Sergio Manzetti on 4 Jan 2018
Thanks Birdman! | 592 | 1,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-05 | latest | en | 0.669402 |
https://studylib.net/doc/8964774/grade-7-math--1.5-chap.1-factors-and-exponents-%E2%80%93-lesson-%23... | 1,618,721,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038468066.58/warc/CC-MAIN-20210418043500-20210418073500-00628.warc.gz | 637,402,094 | 11,188 | # Grade 7 Math: 1.5 Chap.1 Factors and Exponents – Lesson #3 P. 16
```Grade 7 Math: 1.5
Chap.1 Factors and Exponents – Lesson #3
P. 16-17 Powers
*Would you rather take \$100 000 cash OR take one dollar the first day and triple it everyday
for 11 days days? (Re-visit at end of class)
Power: a numerical __numerical expression_____ that shows __repeated_____ multiplication.
(also known as a number written in exponential form
Eg. 5 x 5 x 5 x 5 = 54
Exponent
Base
Base: the _factor_____ you multiply.
Exponent: the number of factors you multiply. (the number of times you repeat the base as a
factor)
Practice: Write each repeated multiplication as a power.
a) 3 x 3 = ____32_
b) 9 x 9 x 9 = __93
Write the following power as repeated multiplication: 63 = __6 x 6 x 6_____
Standard Form: the solution to a power. Eg : 63 = _216____
Naming powers: 45 is a power, which includes a base and an exponent, and can be named:
four to the fifth power
four to the exponent five
the fifth power of four.
45 can be written as repeated multiplication: 4 x 4 x 4 x 4 x 4 and then evaluated or solved and
written in standard form: 1024.
Try this: Write 74 in words, in three ways: _Seven to the fourth, seven to the exponent four and
the fourth power of seven.
Square Number: the ____product________ of ___two_______ equal factors.
Eg. 16 is the square of 4, since 4 x 4 = 16 or 42 = 16
42 is read as four squared.
Area of a square: A = s2
Cubic Number: the ___product______ of __three___________ equal factors.
Eg. 8 is the cube of 2, since 2 x 2 x 2 = 8 or 23 = 8
23 is read as two cubed.
Volume of a cube: V = s3
Writing a number as a power of a given base factor:
If you are given a number and then are asked to write it as a power of another number:
Eg. Write 243 as a power of 3 (243 is the solution to the power, the base must be 3, raised to
some exponent value which must be calculated)
1. Use repeated division method:
1
3 3
Divide 3 into 243, write the answer above 243, then divide 3 into 81,
3 9
write the answer above 81, then divide 3 into 27 ….. until the answer
3 27
becomes 1. The number of 3’s used as the divisor becomes the
3 81
exponent to be placed on the base of 3.
3 243
So 243 written as a power of 3: 243 = 35
2. Use repeated multiplication method:
243 = 3 x 3 x 3 x 3 x 3
= 9x3x3x3
Multiply the base factor “3” by itself as many times as needed
to arrive at the product of 243.
= 27 x 3 x 3
= 81 x 3
= 243
Try This: Write 256 as a power of 4: 44
*Would you rather take \$100 000 cash OR take one dollar the first day and triple it everyday
for 11 days? (Re-visit at end of class)
Day
Repeated
Power
Prize (\$)
Multiplication
1
3
31
3
2
2
3x3
3
9
3
3
3x3x3
3
27
4
4
3x3x3x3
3
81
5
5
3x3x3x3x3
3
243
6
3x3x3x3x3x3
36
729
7
7
3x3x3x3x3x3x3
3
2187
8
8
3x3x3x3x3x3x3x3
3
6561
9
9
3x3x3x3x3x3x3x3x3
3
19683
10
3x3x3x3x3x3x3x3x3x3
310
59049
11
11
3x3x3x3x3x3x3x3x3x3x3 3
177147 | 1,043 | 2,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-17 | latest | en | 0.909463 |
https://www.slideshare.net/rlzijdeman/introduction-into-r-for-historians-part-3-examine-and-import-data | 1,576,132,539,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00130.warc.gz | 882,119,658 | 40,206 | Successfully reported this slideshow.
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# Introduction into R for historians (part 3: examine and import data)
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Introduction into R for the European Historical Population Sample summerschool, Cluj-Napoca, Romana, 2015. Aimed at a public of historians with little quantitative skills
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### Introduction into R for historians (part 3: examine and import data)
1. 1. Recap Getting data in R Do it yourself! Plotting using ggplot2 Examining data and importing data in R Richard L. Zijdeman May 29, 2015 Richard L. Zijdeman Examining data and importing data in R
2. 2. Recap Getting data in R Do it yourself! Plotting using ggplot2 1 Recap 2 Getting data in R 3 Do it yourself! 4 Plotting using ggplot2 Richard L. Zijdeman Examining data and importing data in R
3. 3. Recap Getting data in R Do it yourself! Plotting using ggplot2 Recap Richard L. Zijdeman Examining data and importing data in R
4. 4. Recap Getting data in R Do it yourself! Plotting using ggplot2 The structure of objects Store just about anything in R: numbers, sentences, datasets Objects Study the structure of objects: str() type of object features of object ships <- data.frame(year = c(1850, 1860, 1870, 1880), inbound = c(215, 237, 237, NA), outbound = c(212, 239, 260, 265)) Richard L. Zijdeman Examining data and importing data in R
5. 5. Recap Getting data in R Do it yourself! Plotting using ggplot2 Study the structure of object “ships”" str(ships) ## 'data.frame': 4 obs. of 3 variables: ## \$ year : num 1850 1860 1870 1880 ## \$ inbound : num 215 237 237 NA ## \$ outbound: num 212 239 260 265 Richard L. Zijdeman Examining data and importing data in R
6. 6. Recap Getting data in R Do it yourself! Plotting using ggplot2 Characteristics of objects Class: class() Length: length() Dimensions: dim() class(ships) ## [1] "data.frame" length(ships) ## [1] 3 dim(ships) # rows, columns ## [1] 4 3 Richard L. Zijdeman Examining data and importing data in R
7. 7. Recap Getting data in R Do it yourself! Plotting using ggplot2 Closer inspection of data.frames names of columns (variables): names() top/bottom rows: head(), tail() missing data: is.na() names(ships) ## [1] "year" "inbound" "outbound" is.na(ships) ## year inbound outbound ## [1,] FALSE FALSE FALSE ## [2,] FALSE FALSE FALSE ## [3,] FALSE FALSE FALSE ## [4,] FALSE TRUE FALSE Richard L. Zijdeman Examining data and importing data in R
8. 8. Recap Getting data in R Do it yourself! Plotting using ggplot2 Summarizing data in data.frames descriptive statistics: summary() calculus: e.g. min(), mean(), sum() results table format: table() summary(ships) ## year inbound outbound ## Min. :1850 Min. :215.0 Min. :212.0 ## 1st Qu.:1858 1st Qu.:226.0 1st Qu.:232.2 ## Median :1865 Median :237.0 Median :249.5 ## Mean :1865 Mean :229.7 Mean :244.0 ## 3rd Qu.:1872 3rd Qu.:237.0 3rd Qu.:261.2 ## Max. :1880 Max. :237.0 Max. :265.0 ## NA's :1 Richard L. Zijdeman Examining data and importing data in R
9. 9. Recap Getting data in R Do it yourself! Plotting using ggplot2 is.na(ships) ## year inbound outbound ## [1,] FALSE FALSE FALSE ## [2,] FALSE FALSE FALSE ## [3,] FALSE FALSE FALSE ## [4,] FALSE TRUE FALSE table(is.na(ships)) ## ## FALSE TRUE ## 11 1 Richard L. Zijdeman Examining data and importing data in R
10. 10. Recap Getting data in R Do it yourself! Plotting using ggplot2 Visualizing your data Not just for analyses! Data quality representativeness missing data Richard L. Zijdeman Examining data and importing data in R
11. 11. Recap Getting data in R Do it yourself! Plotting using ggplot2 plot(ships) year 215 220 225 230 235 1850186018701880 215220225230235 inbound 1850 1855 1860 1865 1870 1875 1880 210 220 230 240 250 260 210220230240250260 outbound Richard L. Zijdeman Examining data and importing data in R
12. 12. Recap Getting data in R Do it yourself! Plotting using ggplot2 Getting data in R Richard L. Zijdeman Examining data and importing data in R
13. 13. Recap Getting data in R Do it yourself! Plotting using ggplot2 Data already in R The “datasets” package very slim datasets specific example data To obtain list of datasets, type: library(help = "datasets") To obtain information on a specific dataset, type: help(swiss) # thus: help(name_of_package) or to just see the data: help(swiss) Richard L. Zijdeman Examining data and importing data in R
14. 14. Recap Getting data in R Do it yourself! Plotting using ggplot2 Reading in data Different functions for different files: Base R: read.table() (read.csv()) foreign package: read.spss(), read.dta(), read.dbf() openxlsx package: read.xlsx() alternatives packages: xlsx(Java required) gdata (perl-based) Richard L. Zijdeman Examining data and importing data in R
15. 15. Recap Getting data in R Do it yourself! Plotting using ggplot2 read.xlsx() from openxlsx package file: your file, including directory sheet: name of sheet Richard L. Zijdeman Examining data and importing data in R
16. 16. Recap Getting data in R Do it yourself! Plotting using ggplot2 read.csv() file: your file, including directory header: variable names or not? sep: seperator read.csv default: “,” read.csv2 default: “;” skip: number of rows to skip nrows: total number of rows to read stringsAsFactors encoding (e.g. “latin1” or “UTF-8”) Richard L. Zijdeman Examining data and importing data in R
17. 17. Recap Getting data in R Do it yourself! Plotting using ggplot2 Do it yourself! Richard L. Zijdeman Examining data and importing data in R
18. 18. Recap Getting data in R Do it yourself! Plotting using ggplot2 Read in the following files as data.frames: HSN_basic.xlsx check the data.frame: using dim(), length() check the variables: using summary(), min(), table() Repeat for HSN_marriages.csv: read in only 100 lines Richard L. Zijdeman Examining data and importing data in R
19. 19. Recap Getting data in R Do it yourself! Plotting using ggplot2 Plotting using ggplot2 Richard L. Zijdeman Examining data and importing data in R
20. 20. Recap Getting data in R Do it yourself! Plotting using ggplot2 ggplot2 Package by Hadley Wickham Generic plotting for a great range of plots ggplot2 website: http://ggplot2.org excellent tutorial: https://jofrhwld.github.io/avml2012/#Section_1.1 Richard L. Zijdeman Examining data and importing data in R
21. 21. Recap Getting data in R Do it yourself! Plotting using ggplot2 Building your graph Each plot consists of multiple layers Think of a canvas on which you ‘paint’ data layer geometries layer statistics layer Richard L. Zijdeman Examining data and importing data in R
22. 22. Recap Getting data in R Do it yourself! Plotting using ggplot2 Data layer data.frame and aesthetics ggplot(data.frame, aes(x= ..., y = ...)) geometries layer ggplot(..., aes(x= ..., y = ...)) + geom_...() # e.g. geom_line statistics layer ggplot(..., aes(x= ..., y = ...)) + geom_...() + stat_...() # e.g. stat_smooth Richard L. Zijdeman Examining data and importing data in R
23. 23. Recap Getting data in R Do it yourself! Plotting using ggplot2 an example Reading in the data hmar <- read.csv("./../data/derived/HSN_marriages.csv", stringsAsFactors = FALSE, encoding = "latin1", header = TRUE, nrows = 100) Richard L. Zijdeman Examining data and importing data in R
24. 24. Recap Getting data in R Do it yourself! Plotting using ggplot2 Plotting the data install.packages(ggplot2) library(ggplot2) ggplot(hmar, aes(x= M_year, y = Age_bride)) + geom_point() Richard L. Zijdeman Examining data and importing data in R
25. 25. Recap Getting data in R Do it yourself! Plotting using ggplot2 20 30 40 50 1830 1840 1850 1860 1870 M_year Age_bride Richard L. Zijdeman Examining data and importing data in R
26. 26. Recap Getting data in R Do it yourself! Plotting using ggplot2 Improving the plot Specify characteristics of the geom_layer ggplot(hmar, aes(x= M_year, y = Age_bride)) + geom_point(colour = "blue", size = 3, shape = 18) See http: //www.cookbook-r.com/Graphs/Shapes_and_line_types/ Richard L. Zijdeman Examining data and importing data in R
27. 27. Recap Getting data in R Do it yourself! Plotting using ggplot2 Specify characteristics of the geom_layer 20 30 40 50 1830 1840 1850 1860 1870 M_year Age_bride Richard L. Zijdeman Examining data and importing data in R
28. 28. Recap Getting data in R Do it yourself! Plotting using ggplot2 A PTE example Does age at marriage depend on educational attainment? To marry you need resources the more attainment the longer it takes to acquire resources ergo: brides with edu attainment marry later in life Not a statistical test: but let’s graph this Richard L. Zijdeman Examining data and importing data in R
29. 29. Recap Getting data in R Do it yourself! Plotting using ggplot2 A request from yesterday Can I plot labels? ggplot(hmar, aes(x= M_year, y = Age_bride, label = SIgn_bride)) + geom_text() Richard L. Zijdeman Examining data and importing data in R
30. 30. Recap Getting data in R Do it yourself! Plotting using ggplot2 Yes you can! Not really useful though. . . h a h h h a h a h a a a a h a a h h h h h h h a a h h a a h a a a hh h hh a a a a h a h a h h a a h hh h a h h h h h h h a h a h h a h a h h a hh a h h h h h h a a h h h h h h h h h a h a a h a h 20 30 40 50 1830 1840 1850 1860 1870 M_year Age_bride Richard L. Zijdeman Examining data and importing data in R
31. 31. Recap Getting data in R Do it yourself! Plotting using ggplot2 Let’s try with colours. . . ggplot(hmar, aes(x= M_year, y = Age_bride)) + geom_point(aes(colour = factor(SIgn_bride)), size = 3, shape = 18) Richard L. Zijdeman Examining data and importing data in R
32. 32. Recap Getting data in R Do it yourself! Plotting using ggplot2 20 30 40 50 1830 1840 1850 1860 1870 M_year Age_bride factor(SIgn_bride) a h No real pattern, though. . . Richard L. Zijdeman Examining data and importing data in R
33. 33. Recap Getting data in R Do it yourself! Plotting using ggplot2 Finalizing the graph ggplot(hmar, aes(x= M_year, y = Age_bride)) + geom_point(aes(colour = factor(SIgn_bride)), size = 3, shape = 18) + labs(list(title = "Age of marriage over time", x = "time (years since A.D.)", y = "age of bride (years)", colour = "Signature")) # here we use colour since legend shows colour Richard L. Zijdeman Examining data and importing data in R
34. 34. Recap Getting data in R Do it yourself! Plotting using ggplot2 20 30 40 50 1830 1840 1850 1860 1870 time (years since A.D.) ageofbride(years) Signature a h Age of marriage over time Richard L. Zijdeman Examining data and importing data in R
35. 35. Recap Getting data in R Do it yourself! Plotting using ggplot2 Satisfied? Richard L. Zijdeman Examining data and importing data in R
36. 36. Recap Getting data in R Do it yourself! Plotting using ggplot2 Actually not. . . the points are plotted on top of each other. . . Solution: geom_jitter ggplot(hmar, aes(x= M_year, y = Age_bride)) + geom_jitter(aes(colour = factor(SIgn_bride)), size = 3, shape = 18) + labs(list(title = "Age of marriage over time", x = "time (years since A.D.)", y = "age of bride (years)", colour = "Signature")) # here we use colour since legend shows colour Richard L. Zijdeman Examining data and importing data in R
37. 37. Recap Getting data in R Do it yourself! Plotting using ggplot2 20 30 40 50 1830 1840 1850 1860 1870 time (years since A.D.) ageofbride(years) Signature a h Age of marriage over time Richard L. Zijdeman Examining data and importing data in R
38. 38. Recap Getting data in R Do it yourself! Plotting using ggplot2 Final remarks on ggplot2 We have just scratched the surface of ggplot2 Build your graph slowly start with the basics add complexity step-wise Now it’s your turn! Richard L. Zijdeman Examining data and importing data in R
39. 39. Recap Getting data in R Do it yourself! Plotting using ggplot2 A small PTE project Look at the variables in the HSN files Think of a research question Provide a general mechanism and hypothesis Plot your results Richard L. Zijdeman Examining data and importing data in R | 3,495 | 12,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-51 | latest | en | 0.689901 |
https://www.physicsforums.com/threads/energy-heat-and-speed.224193/ | 1,480,899,709,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541517.94/warc/CC-MAIN-20161202170901-00083-ip-10-31-129-80.ec2.internal.warc.gz | 997,078,146 | 18,829 | # Energy, heat and speed
1. Mar 25, 2008
### GRB 080319B
My question is two-fold:
1). I understand that according to Einstein's theories that as one approaches the speed of light, e.g. interstellar flight in a spaceship, the mass of the moving object approaches infinity. E=mc^2 the increasing mass correspond to an increasing amount of energy. Is this energy in the form of potential energy, i.e. could be released by a collision with another object, or some other type of energy?
2). In reference to my first question, is there any relationship between the energy of a moving object and thermal energy? I know that c is the upper limit for speed of an object, but does mean it has the maximum amount of energy it can store as potential energy? Is there an upper bound for thermal energy? I have heard that thermal energy is the kinetic energy of particles in a system. I also have heard that all bodies emit thermal radiation in the form of black-body radiation, but I've never heard of any object emitting energy because of the speed its traveling at. Is there any correlation between the thermal energy of a hot object, e.g. a hot cloud of plasma, and the energy of a moving object,e.g. a fast moving comet?
I have no real knowledge of physics so feel free to correct me if I have missed anything or made absurd assumptions.
2. Mar 25, 2008
### GRB 080319B
To clarify the second question, is there correlation between the energy added by linear motion, e.g. a spaceship where all of its particles are moving with the same velocity, and heat added to a system causing the particles to move faster?
3. Mar 25, 2008
### yuiop
Your question touches on the relationship of relativity and thermodynamics. The relationship is not clear. I have Tolman's book on relativity and thermodymanics that concludes the temperature of a moving body cools (T' = T/y). This is the classic interpretation but a number of experts since have claimed the relativistic temperature is covariant (T'=Ty) and yet others that claim temperature is invariant (T'=T). To that you can add experts that claim there is no simple relationship between temperature and relative velocity and recent paper that introduces a 4 inverse temperature. I have been comparing these papers on and off for a number of weeks to try and see how and why they differ and hope to start a thread on that soon.
I can comment on some of the issues you raise.
Yes, the energy would be released in some form in a collision.
An infinite amount of energy would be required to accelerate a massive body to the speed of light so that observation does not put any finite upper bound to the kinetic energy or temperature of a body.
Yes, all bodies that are not at absolute zero temperature (which is supposed to be impossible) emit radiation which is proportional (for an ideal black body) to their temperature. We can note that the peak frequency of the radiation from a black body is proportional to its temperature. For a body with relative motion we can also note that the frequency of the radiation from the body reduces by the Lorentz factor (after we have deducted the doppler shift due the the body comeing towards us or going away) hinting at the drop in relativistic temperature due to relative motion. Not all bodies are exact ideal black bodies but most cosmologists/ astronomers aproximate stars as black bodies when trying to figure out their temperaures.
While it is not absolutely clear if a body emits energy because of the speed it is traveling at, it is certain that particles emit radiation while accelerating. This is known as synchrotron radiation. This is regularly observed in synchroton accelerators and particles spiralling down towards a black hole in the acretion disk are thought to give off large quantities of radiation in the form of x-rays.
The assumption that all the particles of a rocket are moving at the same velocity is an aproximation because a large object like a rocket can not be at absolute zero temperature. Thermal vibrations of atoms mean some of the atome will be be moving faster than the average velocity of the rocket and some will be moving slower. However, it is a reasonable aproximation at low temperatures. On the understanding of that aproximation, we used 10,000 Joules of energy to accelerate brick A to a given velocity and another 10,000 Joules of energy to heat brick B to a given temperature then the average kinetic energy of the atoms in both bricks will be the same. The only difference is that motion of individual atoms in the first brick are are all in the same direction (coherant), while the motion of the vibrating atoms of the second brick are in random directions (incoherant).
I hope at least some of that helps.
4. Mar 26, 2008
### GRB 080319B
Yes, the explanation about relativity and temperature and the brick experiment did help, but also led me to some additional questions:
As far as the quote above, does this mean that there is no bound to the amount of thermal energy that can exist in a system? I thought there was a bound at the amount of energy/mass that can be in one place before a black hole is created?
Does one have to add more energy to an incoherent system than a coherent system to arrive at the same average speed of each particle? I would think that the collisions taking place in an incoherent system would cause it to radiate more energy as heat, and subsequently take longer for individual particles to speed up, than a coherent system. Therefore, would it take less time to accelerate the particles in a coherent system to a certain speed than heat up particles in an incoherent one? Do incoherent systems radiate more energy as heat than coherent systems? I've also heard the reason why a object would require an infinite amount of energy to accelerate to the speed of light was because the mass of increases as the speed of the object increases. Is it true that particles in an incoherent system become more massive as their thermal energy is increased?
I though as an object approached the speed of light, its mass increased, not its temperature. I would think though that adding energy to a system in the form of speeding it up would make it hotter? Why would it lose energy as heat? Would this cooling reduce the objects speed in any way, or make the object less massive? I assume by the last sentence of this quote that the relation between relativity and thermodynamics is still not completely understood, so if these questions are unanswerable at this time I understand. Thank you.
5. Mar 26, 2008
### yuiop
You are assuming that the particles in the sytem stay within the same volume ("in one place") as it is heated. A star with just over ten solar masses has enough gravitational energy to form a black hole. While the star still has nuclear fuel to burn, the thermal energy of its gas particles is enough to resist it collapsing to a black hole. If we somehow added even more energy to the star it would expand, increasing its volume and reducing its density preventing it becomeing a black hole. When the star runs out of nuclear fuel, it collapses rapidly, releasing a great deal of gravitational potential energy which creates a supernova explosion ejecting matter all over the place but enough mass remains to form a black hole.
In the brick example you would have to insulate the brick that you are heating up to make it a fair comparison.
Yes
This is a difficult subject. Classically mass increases with relative velocity. The formal modern view is that rest mass is invariant and we do not discuss relativistic mass. One reason for this is that an object moveing at a very high relative speed to the observer might be considered to have enough mass within a given volume to become a black hole. To an observer comoving with the object, it is not a black hole. An object that is a black hole to one one observer while not a black hole to another observer is clearly a contadiction. Einstein said there is no clear definition of mass. Like I said, it is not an easy subject.
Consider a particle fired horizontally from a tower at 0.8c relative to an observer on the tower. Say that observer measures 10 seconds for that particle to fall to the ground. (It is a very long flat planet :P) To an observer comoving with the particle it takes 6 seconds for the particle to fall to the ground. Assuming the the height of the tower and mass of the planet are such that the difference in gravitational time dilation between the top of the tower and the ground is negligable then why the difference?
Once again this is difficult to answer, if we can not talk about relativistic mass. What we can note, is that if we have a very hot brick and a very cold brick that it will take more energy to accelerate the the hot brick to a given velocity than the cold brick. The hot brick is harder to accelerate. Some people interpret that as the hot brick having more inertial mass (more resistance to acceleration) than the cold brick.
As I said before, the formal aproach is that its mass does not increase, but it does resist acceleration more. I think the more formal version is that the momentum energy of the system increases. As for the temperature, the experts are not agreed on whether it increases or decreases or something in between.
Well, at least some of the energy has to spent on speeding up the system and increasing the overall (coherent) momentum of the system). This usually involves ejecting mass in the opposite direction so that momentum of the universe is conserved. The difficulty comes when analysing the motions of individual particles in a moving system in deciding how much of the motion is coherent and how much is incoherent. I have been looking at it for a while and it is not easy, and papers on the subject disagree.
See all the above.
Here are some papers on relativistic thermodynamics if you want to sift through them:
P' and T' are the transformed pressure and temperature.
P' = Py^2 , T' = Ty http://arxiv.org/PS_cache/arxiv/pdf/0712/0712.3793v2.pdf
P' = ? , T' = ? http://arxiv.org/PS_cache/physics/pdf/0303/0303091v3.pdf
P' = ? , T' = ? http://arxiv.org/PS_cache/gr-qc/pdf/9803/9803007v2.pdf
P' = P , T' = Ty http://arxiv.org/PS_cache/arxiv/pdf/0801/0801.2639v1.pdf
P' = P , T= T/y http://arxiv.org/PS_cache/physics/pdf/0505/0505004v2.pdf
P' = P, T= T/y is my interpretation of the last paper but they do not spell it out. Tolman's book on relativity and thermodynamics also concludes P' = P, T= T/y but the book is rather old.
Last edited: Mar 26, 2008
6. Mar 27, 2008
### GRB 080319B
Thank you for the help. | 2,342 | 10,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2016-50 | longest | en | 0.941669 |
https://blog.finxter.com/numpy-arange/ | 1,726,206,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00073.warc.gz | 117,523,810 | 24,336 | # NumPy arange(): A Simple Illustrated Guide
The `np.arange()` function appears in 21% of the 35 million Github repositories that use the NumPy library! This illustrated tutorial shows you the ins and outs of the NumPy arange function. So let’s get started!
## What’s the NumPy Arange Function?
The `np.arange([start,] stop[, step])` function creates a new NumPy array with evenly-spaced integers between `start` (inclusive) and `stop` (exclusive). The `step` size defines the difference between subsequent values. For example, `np.arange(1, 6, 2)` creates the NumPy array `[1, 3, 5]`.
Have a look at the following graphic:
## Video np.arange()
[Reading time: 4 minutes] – Or watch the video.
## Syntax and Arguments of np.arange()
The NumPy `arange()` function (commonly misspelled: NumPy arrange) creates a NumPy array of evenly spaced numbers within a fixed interval.
Let’s explore these examples in the following code snippet that shows the four most important uses of the `np.arange()` function:
First, import the NumPy library:
`import numpy as np`
Now, you can use the `np.arange()` function to create sequences with equal step sizes in various ways. Go over the table and study the examples thoroughly:
To master the NumPy `arange` function, read over the following basic function calls with different sets of arguments.
### np.arange(stop)
Here’s the most basic example of the NumPy `arange` function. You only specify the `stop` argument. Like many other sequence operations, `np.arange()` starts at index 0. If you specify one `stop` argument, it is the same as setting the start argument to 0.
Let’s say, you want to represent the seven weekdays Monday-Sunday with seven sequential numbers 0-7.
```>>> np.arange(7)
array([0, 1, 2, 3, 4, 5, 6])```
The result is a NumPy array with seven elements starting from the implicitly chosen index 0 (inclusive) and ending in the explicitly chosen index 7 (exclusive).
But what if we want to define a `start` index? Let’s say you feel anxious every Sunday evening because you hate going to work on Mondays for a big accountancy firm. First, you should get a new job as a programmer! Second, to stop working Mondays, we need to skip the start index 0.
### np.arange(start, stop)
Your boss at the big accountancy firm has agreed to let you take every Monday off. Yay! Now we need to update the company records.
If you add a second argument to `np.arange(start, stop)`, Python interprets the first one as the `start` index and the second one as the `stop` index.
```>>> np.arange(1, 7)
array([1, 2, 3, 4, 5, 6])```
But what about the three-argument version.?
### np.arange(start, stop, step)
Your new boss has asked you to fill in the days you want to work. You’d love to start your week on Tuesdays, but you’d also like more time off. So you say you’ll work every second day. Luckily, you have a solid salary as a programmer, so this isn’t a problem!
To do this, add a third argument to `np.arange` to set the step size. It is 2 in this case because you work every second day.
```>>> np.arange(1, 7, 2)
array([1, 3, 5])```
Well done, now you only work on Tuesdays, Thursdays, and Saturdays! All programmers are lazy and you are lazy! It’s a perfect job for you.
But wait, there’s a problem… let’s move on to the four-argument function call to solve it.
### np.arange(start, stop, step, dtype)
As you put your working days `[1, 3, 5]` into your company’s tracking system, it complains. Your NumPy array is formatted incorrectly. It expects all NumPy array values to be floats rather than integers. “What a design flaw for weekday data!”.
Thankfully, there is a simple fix. We need to use the fourth argument of `np.arange()` to set the data type (`dtype`) of the output array. The `dtype` argument accepts two kinds of data types. First, traditional language-specific data types such as float and integer. Second, NumPy-specific data types such as` np.int16` or `np.float32`.
For instance, the NumPy-specific data types `np.int16` or `np.float32` allow for an integer value with 16 bits (=2 bytes) or a float value with 32 bits (=4 bytes). Keep in mind that more bits leads to higher overheads. But it gives you a greater range of numbers to work with (or greater precision in the case of floats). Here is a collection of `dtypes` you can use (check out this excellent post if you need more information about NumPy `dtypes`):
• `bool`: The default boolean data type in Python (1 Byte).
• `int`: The default Python integer data type in Python (4 or 8 Bytes).
• `float`: The default float data type in Python (8 Bytes).
• `complex`: The default complex data type in Python (16 Bytes).
• `np.int8`: Integer (1 Byte).
• `np.int16`: Integer (2 Bytes).
• `np.int32`: Integer (4 Bytes).
• `np.int64`: Integer (8 Bytes).
• `np.float16`: Float (2 Bytes).
• `np.float32`: Float (4 Bytes).
• `np.float64`: Float (8 Bytes).
By default, NumPy chooses `np.float64` and `np.int64` for floats and integers. So only specify a different `dtype` if you want something other than those. At the start of your Python journey, it’s unlikely you will need to deeply understand the different `dtypes`. Once you start working on more complex problems, you will need this knowledge though.
Lastly, note that you can only spot differences once numbers get large. The largest `np.int8` is 127 but the largest `np.int16` is 32767. Yet, if you compare `np.int8` 127 and `np.int16` 127, they are the same.
```>>> np.int8(127) == np.int16(127)
True```
Here is an example of the 4-argument version of `np.arange()`:
```>>> np.arange(1, 7, 2, dtype=np.float32)
array([1., 3., 5.], dtype=float32)```
You’ve completed the first part of the NumPy arange tutorial! But the one who prepares best wins. Let’s dive into some practice examples and attack the highest level of NumPy arange expertise!
## Interactive Shell NumPy Arange
Can you solve the following basic puzzle about NumPy arange?
Exercise: Guess the output of this code snippet. Then, run the code snippet to check your result!
## Examples
You have mastered all four different uses of the NumPy `arange()` function. To wrap things up, make sure to work through this more comprehensive list of examples:
```>>> np.arange(10)
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> np.arange(2, 10, 3)
array([2, 5, 8])
# Non-integer step sizes
np.arange(1.5, 10., 1.5)
array([1.5, 3., 4.5, 6., 7.5, 9. ])
# Create np.arange in reverse with a negative step size
# Start is inclusive, end is exclusive so it stops at 2
>>> np.arange(10, 1, -1)
array([10, 9, 8, 7, 6, 5, 4, 3, 2])
# Reversed and select every second value
>>> np.arange(10, 1, -2)
array([10, 8, 6, 4, 2])
# Reversed non-integer step size
>>> np.arange(9.7, 9.2, -0.1)
array([9.7, 9.6, 9.5, 9.4, 9.3])
# End being exclusive beats the start being inclusive
>>> np.arange(1, 1, 1)
array([], dtype=int64)
>>> np.arange(1, 5, dtype=np.int16)
array([1, 2, 3, 4], dtype=int16)
>>> np.arange(1, 5, dtype=np.float32)
array([1., 2., 3., 4.], dtype=float32)```
I‘ve also added several NumPy `arange` puzzles to my puzzle-based learning app Finxter.com. Check it out to train yourself and become a master coder.
Finxter App: Test your skills now!
## Questions np.arange()
Congratulations, you now know the most important details about the NumPy arange function. But you may still have a few questions. Let’s answer them one by one!
### np.arange vs np.linspace – When Should I Use Which One?
Use `np.arange()` if you want to create integer sequences with evenly distributed integer values within a fixed interval.
Use `np.linspace()` if you have a non-integer step size. The `np.linspace` function handles the endpoints better. This prevents you from introducing unnecessary bugs into your code. One such bug is when you assume an endpoint is not in the NumPy array but it is because of floating-point arithmetic.
If you want to understand the NumPy `linspace()` function in detail, check out our blog article.
### What Are Some np.arange() Use Cases?
NumPy is mostly about multi-dimensional matrices. It is common to create a 1D NumPy array with the NumPy arange function and to transform it immediately into a 2D array using the `np.reshape()` function. Below we create a 2D array with three rows and two columns from a 1D array.
```np.arange(6).reshape((3, 2))
# array([[0, 1],
# [2, 3],
# [4, 5]])```
### np.arange vs range – What’s the Difference?
You’re probably familiar with the built-in `range()` function. We use it all the time to write for loops and list comprehensions.
```>>> for i in range(5):
print(i)
0
1
2
3
4
# Square numbers from 0-4
>>> squares = [i**2 for i in range(5)]
>>> squares
[0, 1, 4, 9, 16]```
So can we do the same with `np.arange`? And even if we can, should we?
Let’s first see if it’s possible.
```# Works the same as range()
>>> for i in np.arange(5):
print(i)
0
1
2
3
4
# Also works the same as range()
>>> np_squares = [i**2 for i in np.arange(5)]
>>> np_squares
[0, 1, 4, 9, 16]
# They're even identical!
>>> np_squares == squares
True```
So it seems like we can do the same with both functions. But we’re working with a tiny amount of data. What happens if we work with much larger numbers?
### np.arange vs range – How to Work with Big Data?
We’ll use iPython’s magic function `%timeit` to see how long our for loops take when using `np.arange` or `range`.
```# Only works in iPython
In [1]: %timeit for i in np.arange(1000000): pass
78.1 ms ± 4.9 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [2]: %timeit for i in range(1000000): pass
32.8 ms ± 899 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
# range is more than twice as fast!
In [3]: 78.1 / 32.8
Out[3]: 2.381...```
This happens because the two functions work differently. The function `range()` is a generator function. It creates the next value when it needs to and stops once it has run out. This makes it very efficient. But `np.arange` creates an array of length 1,000,000 and stores it in memory. This is computationally expensive for Python and slows it down. Thus, if you are ever looping, you should use `range()`. It’s much faster.
However, there is a time when you should not use range.
### How to Iterate Over Large NumPy Arrays
Let’s say you have a NumPy array, A, containing 1 million values. You want to create a new array, B, by performing a calculation on each element of A. For this example, we will add 1 to every element of A to get B.
First, we create an array of 1 million random numbers using the `random` module and a list comprehension. Check out our article to learn more about this built-in library.
```>>> import random
# Set seed so we can reproduce our results
>>> random.seed(1)
# Use list comprehension to generate random numbers
>>> A = [random.random() for i in range(1000000)]
# Convert to a numpy array
>>> A = np.array(A)
# Check first 5 values (you should have the same as me)
>>> A[:5]
[0.13436424411240122, 0.8474337369372327, 0.763774618976614, 0.2550690257394217, 0.49543508709194095]```
Let’s first create B using a `for` loop:
```>>> B = np.array([])
>>> for i in A:
B.append(i + 1)
# Check first 5 values - looks good
>>> B[:5]
[1.134364244112401, 1.8474337369372327, 1.7637746189766141, 1.2550690257394217, 1.4954350870919408]```
Let’s time it:
```# Only works in iPython
In [1]: B = np.array([])
# NumPy arrays don't have an .append() method so we use np.append()
In [2]: %timeit for i in A: np.append(B, i+1)
6.91 s ± 628 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)```
That is rather slow. Can we make this any faster? As NumPy arrays are made up of lists, we can do this using a list comprehension. Let’s see if this improves the speed.
```In [3]: %timeit B = np.array([i+1 for i in A])
600 ms ± 19.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)```
Wow! That’s 11.5x faster than using a for loop! But can we get even faster? Yes, we can and the answer is: vectorized computation.
Most of the time, the operation you want to perform can be done using functions. You should always use these as they have been optimised by the NumPy developers. Check out this Stack Overflow answer for an introduction. We can, and probably will, write a full article on this topic soon. But for now, check out Chapter 4 from Python for Data Analysis by Wes McKinney.
The fastest way to solve this is to use NumPy’s broadcasting property i.e. if we +1 to a NumPy array it ‘broadcasts’ this to all the elements of the array.
```In [4]: %timeit B = A + 1
1.99 ms ± 386 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)```
This is 3,455x faster than using a for loop. It will take some time to get used to using vectorized computations. But once you get used to it, you will save a lot of time.
🌍 Recommended Finxter Tutorial: How to Iterate over a NumPy Array?
## Summary
The `np.arange([start,] stop[, step])` function creates a new NumPy array with evenly-spaced integers between `start` (inclusive) and `stop` (exclusive). The `step` size defines the difference between subsequent values.
Consider the following examples:
```import numpy as np
# np.arange(stop)
>>> np.arange(10)
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
# np.arange(start, stop)
>>> np.arange(2, 10)
array([2, 3, 4, 5, 6, 7, 8, 9])
# np.arange(start, stop, step)
>>> np.arange(2, 10, 2)
array([2, 4, 6, 8])
# np.arange(start, stop, step, dtype)
>>> np.arange(2, 10, 2, float)
array([2., 4., 6., 8.])```
The examples show all four variants of using `np.arange()` with one, two, three, or four arguments.
## Where to Go From Here?
How to join the top earners in any field? Read more books!
Because I always find it difficult to find time to learn, I have written a new NumPy book that can be entirely consumed in small doses — for example as you drink your daily morning coffee.
Don’t miss out on this exciting new way of learning to code — it’s so much more fun! 🙂
## 3 thoughts on “NumPy arange(): A Simple Illustrated Guide”
1. Thanks for your nice post. However, I am prefering numpy linspace for the reasons given above. | 3,950 | 14,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-38 | latest | en | 0.675545 |
https://goforexams.com/railways-mcq-question-and-answers/page/11/ | 1,569,111,501,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574710.66/warc/CC-MAIN-20190921231814-20190922013814-00278.warc.gz | 496,982,912 | 7,982 | # Go For Exams
> > Page 11
###### RPF Constables
Q 101.
Nine persons went to a hotel for taking their meals. Eight of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine. What was the total money spent by them.
A) Rs. 115
B) Rs. 116
C) Rs. 117
D) Rs. 118
Explanation:
Q 102.
The average of 50 numbers is 30. If two numbers, 35 and 40 are discarded, then the average of the remaining numbers is nearly:
A) 29.68
B) 28.78
C) 28.32
D) 29.27
Explanation:
Q 103.
Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km per hour. Find the average speed of train during the whole journey.
A) 30.5 km/hr
B) 60 km/hr
C) 57 km/hr
D) 67.2 km/hr
Explanation:
Q 104.
There are two sections A and B of a class, consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class.
A) 35 kg
B) 30 kg
C) 37.25 kg
D) 42.5 kg
Explanation:
Q 105.
The average of runs of a cricket player of 10 innings was 32. How many runes must be made in his next innings so as to increase his average of runs by 4 ?
A) 70
B) 72
C) 74
D) 76
Explanation:
Average after 11 innings = 36. Required number of runs, = ( 36 * 11) - (32 * 10) = 396-320. = 76.
Q 106.
David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, mathematics, physics, chemistry and biology. What are his average marks ?
A) 69
B) 65
C) 75
D) None of these
Q 107.
Nine persons went to a hotel for taking their meals. Eight of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine. What was the total money spent by them.
A) Rs. 116
B) Rs. 115
C) Rs. 118
D) | 614 | 1,863 | {"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} | 4.0625 | 4 | CC-MAIN-2019-39 | latest | en | 0.931967 |
https://stats.stackexchange.com/questions/505813/estimate-covariance-matrix-using-r-without-knowing-the-joint-distribution | 1,726,237,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00166.warc.gz | 500,830,120 | 40,514 | # Estimate covariance matrix using R without knowing the joint distribution
n <- 10000
X <- rnorm(n, mean = 3, sd = 3)
Y <- rexp(n, rate = 2)
Z <- rnorm(n, mean = -2, sd = 0.5)
cov(X, Y)
cov(X, Z)
cov(Y, Z)
Suppose I know the marginal distributions of the random variables $$X \sim N(3, 9)$$, $$Y \sim Exp(2)$$, and $$Z \sim N(-2, 0.025)$$. However, I do not have any information on the joint distribution of $$(X, Y, Z)$$ and I do not know whether these r.v.s are independent or not.
I'm interested in estimating the covariance matrix $$Cov\begin{bmatrix}X\\ Y \\ Z\end{bmatrix}$$. Is it appropriate to do what I did above: generate observations from each marginal distribution and simply use the sample covariance as an estimate? i.e., cov(X, Y), cov(X, Z), cov(Y, Z) in R?
• @stans thanks. would you like to add your comment as an answer? Commented Jan 21, 2021 at 16:01
• Sure. Have done it. Commented Jan 21, 2021 at 16:14
No. What you did above is simulated $$(X,Y,Z)$$ where $$X$$, $$Y$$ and $$Z$$ are independent. The sample covariances will simply converge to 0 as $$n$$ converges to infinity. To simulate any random variables jointly, you have to have an idea how they are related | 362 | 1,197 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-38 | latest | en | 0.880533 |
https://gmatclub.com/forum/headway-is-now-targeted-at-a-wider-readership-to-include-92233.html?sort_by_oldest=true | 1,496,000,384,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610374.3/warc/CC-MAIN-20170528181608-20170528201608-00407.warc.gz | 942,844,998 | 66,676 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# Headway is now targeted at a wider readership - to include
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03 Apr 2010, 05:10
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Headway is now targeted at a wider readership - to include all NUT members who are on the leadership salary spine. This year's NUT membership credential gave members the opportunity to identify themselves as teachers on the leadership spine and Headway will, in future, be mailed directly to all members so identified.
From the text above, which of the following can we reasonably conclude?
a) Headway is a free newspaper.
c) Teachers can only be present on the leadership spine thanks to the NUT.
I am not clear on the answer provided. Please provide some pointers.
Thanks.
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03 Apr 2010, 12:40
is it D ?
all other options can be easily proved wrong.
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03 Apr 2010, 13:22
It should be B......
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04 Apr 2010, 03:38
between A and B, I think B is a better answer.
Headway is sending the magazine to a new group which has been not identified before. I recognize that my argument little shaky, but this was best one among given.
source? OA?
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09 Apr 2010, 13:04
IMO B....Whats the OA !!!
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09 Apr 2010, 19:35
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I'm stuck between B and D here.
Headway is now targeted at a wider readership - to include all NUT members who are on the leadership salary spine. This year's NUT membership credential gave members the opportunity to identify themselves as teachers on the leadership spine and Headway will, in future, be mailed directly to all members so identified.
From the text above, which of the following can we reasonably conclude?
a) Headway is a free newspaper. - (don't know)
b) Formerly, the readership of Headway did not include members on the leadership salary spine. (possibly hence now they are including ALL NUT members who are on the leadership salary spine) - (the tricky part here is the word "ALL" - B says that they did not include members, meaning no members were included before or some were included but not all, not sure here)
c) Teachers can only be present on the leadership spine thanks to the NUT. ( clearly not something we can conclude)
d) The readership for Headway will increase in future years. ( well their target is wider readership and "Headway will, in future, be mailed directly to all members so identified." so it's safe to assume that readership will increase )
e) Teachers cannot get access to Headway unless it is mailed to them. ( there could be other ways to get access we don't know)
I'm leaning towards D on this one, but B is very close. I'm not sure if my logic in B makes sense though.
Please let us know the OA. Thanks.
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10 Apr 2010, 00:41
OA is B.
Even I was stumped by the choice and selected D at first. But if you check closely by using Process of Elimination, PoE, then B is the best choice.
I would request you all to write some explanation why you chose some option. Writing just the answer as "OA is B" or "IMO it is B" would not be of much use because people are interested in the food for thought and that this forum is all about. This explanation will be helpful for others to understand the problem from various perspectives leading to same conclusion.
Give me KUDOS if I am right.
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10 Apr 2010, 11:32
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B
It is an inference based question and the inference must be true taking the argument to be true.
The evidence says that:
Headway would target all Nut members on the leadership spine. (maybe it was targeting some but not all earlier)
Headway would mail magazine directly to teachers who are NUT members.
Using negation method in the options now:
a. Headway is a free newspaper. (If it is not a free newspaper, would it effect the argument. NO)
c) Teachers can only be present on the leadership spine thanks to the NUT. (No. Extreme case. This time nut members credentials gave an opportunity to the teachers to be identified. Does not mean that they can only be present because of nut.)
d) The readership for Headway will increase in future years. (Headway is targeted at all NUT members but does not mean that they would accept or it can also happen that some existing Headway readers would not continue with Headway. So No.)
e) Teachers cannot get access to Headway unless it is mailed to them. (No. headway mails to teachers does not mean that this is the only source for teachers to access headway.)
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12 Apr 2010, 14:26
I selected D at first but now I am totally agree with you
ankithakhu wrote:
B
It is an inference based question and the inference must be true taking the argument to be true.
The evidence says that:
Headway would target all Nut members on the leadership spine. (maybe it was targeting some but not all earlier)
Headway would mail magazine directly to teachers who are NUT members.
Using negation method in the options now:
a. Headway is a free newspaper. (If it is not a free newspaper, would it effect the argument. NO)
c) Teachers can only be present on the leadership spine thanks to the NUT. (No. Extreme case. This time nut members credentials gave an opportunity to the teachers to be identified. Does not mean that they can only be present because of nut.)
d) The readership for Headway will increase in future years. (Headway is targeted at all NUT members but does not mean that they would accept or it can also happen that some existing Headway readers would not continue with Headway. So No.)
e) Teachers cannot get access to Headway unless it is mailed to them. (No. headway mails to teachers does not mean that this is the only source for teachers to access headway.)
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30 Mar 2012, 23:16
Guys, It is a word game.
One can also interpret above statement as follows:
Headway did not include all members on the salary spine, but it included few members. Now headway wants to expand readership by including remaining members.
Therefore one cannot conclusively say that readership of headway did not include members on the leadership salary spine. Headway might have included few members but not all.
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16 May 2012, 08:09
Agree with Cmplkj123. Headway could be read by people who belong to leadership salary spine group. We will include more such people does not mean that there was no representation/ readership earlier.
D is also suspect. Existing readers may discontinue or teachers may not sign up leadership salary spine. Any of the given option can take place. Difficult to make an inference that the readership will increase.
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10 Jul 2012, 03:26
I think "Headway is now targeted at a wider readership - to include all NUT members who ...... " makes B the answer.
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31 Jul 2012, 09:01
wow.. i could come down to b and d only.. beyond that unable to move an inch ! as much as b could be wrong because the word "all" in premise, D could be wrong as well because needs an assumption - that it will not decrease in other areas.
How to resolve?
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31 Jul 2012, 09:15
Desperate123 wrote:
I think "Headway is now targeted at a wider readership - to include all NUT members who ...... " makes B the answer.
Thanks for highlighting.. Yup, because of NOW, option D with future is wrong!
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02 Aug 2012, 06:51
OA seems to be a BS to me.
Headway is targeting a wider readership, to include ALL NUT members. This states that the company is plans to 'register' all members of NUT, however, it does not necessarily imply that there is not one of NUT members currently reading Headway - it is simply going after ALL members.
d) The readership for Headway will increase in future years. - EXTRA assumption
Via POE, I chose A.
B & D cannot be inferred from the stimulus.
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02 Aug 2012, 07:15
Like others, i am also stuck between (B) & (D), can someone tell me that conclusion usually presents the the action following a certain process, i.e. the future not the past.(D) tells the future but he answer is (B) indeed.
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19 Feb 2013, 11:56
1
KUDOS
Expert's post
ykaiim wrote:
Headway is now targeted at a wider readership - to include all NUT members who are on the leadership salary spine. This year's NUT membership credential gave members the opportunity to identify themselves as teachers on the leadership spine and Headway will, in future, be mailed directly to all members so identified.
From the text above, which of the following can we reasonably conclude?
(A) Headway is a free newspaper.
(C) Teachers can only be present on the leadership spine thanks to the NUT.
fameatop wrote:
Hi Mike
I would like to know your take on this question. I think both options B & D are doubtful. As per me D is best of the worst. Can you kindly explain the logic applied in this question. Waiting eagerly for your valuable inputs.
Fame
I am responding to Fame's p.m.
This question reaches new depths of poor quality. If I set out to compose a CR question of the poorest possible quality, I truly doubt whether I would be able to create one as spectacularly bad as this one.
Flaw #1: The grammar & diction --- "is now targeted at" --- really? Everything about the questions and the answer choices suggest an incredibly poor grasp of the English language.
Flaw #2: What is NUT? How is the general read suppose to know that this stands for "National Union of Teachers"? The GMAT never gives an acronym without explicitly defining it. Cf. OG13, CR #7 (BTW, "Headway" is a real magazine, published for members of NUT.)
Flaw #3: How does a "membership credential" give anyone anything? Does it mean a membership credential program?
Flaw #4: What in tarnation is a "leadership salary spine"? That is not a standard term in any particularly well known discipline or profession. The GMAT might occasionally use a term, such as "tenure track" or "middle management", a term well recognized and understood in particular contexts, but if it uses a term invented from whole cloth, such as the one here, it would explain this new term quite clearly.
This question is not worth any consideration or analysis.
As always, some of the longest thread and most involved discussions on this forum are debates about exceedingly poorly written questions. A well written questions entails little discussion --- some folks get it right, and the folks who get it wrong understand quickly why the right answer has to be right and why the wrong answers have to be wrong. There's not much to debate. It's only these bottom-of-the-barrel questions that entails pages of arguments back and forth --- that's a sure-fire sign that it's a poorly written question.
Here's a relatively strong GMAT CR practice question:
http://gmat.magoosh.com/questions/3137
When you submit your answer, the following page will have a complete video explanation.
Let me know if anyone reading this has any questions.
Mike
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Mike McGarry
Magoosh Test Prep
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27 Aug 2015, 03:41
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: Headway is now targeted at a wider readership - to include [#permalink] 27 Aug 2015, 03:41
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Display posts from previous: Sort by | 4,299 | 16,633 | {"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-2017-22 | latest | en | 0.90144 |
http://www.w3resource.com/c-programming-exercises/input-output/c-input-output-statement-exercises-2.php | 1,500,626,622,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423764.11/warc/CC-MAIN-20170721082219-20170721102219-00604.warc.gz | 576,831,168 | 7,633 | C exercises: Calculates the volume of a sphere - w3resource
# C Exercises: Calculates the volume of a sphere
## C Input Output statement and Expressions: Exercise-2 with Solution
Write a C program that calculates the volume of a sphere.
C Programming : Volume of a sphere
A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a completely round ball.
In three dimensions, the volume inside a sphere is derived to be V = 4/3*π*r3 where r is the radius of the sphere
Sample Solution:
C Code:
``````#include <stdio.h>
float myvolume; /* volume of the sphere (to be computed) */
char line_text[50]; /* a line from the keyboard */
/* the value of pi to 50 places, from wikipedia */
const float PI = 3.14159265358979323846264338327950288419716939937510;
int main() {
printf("Input the radius of the sphere : ");
fgets(line_text, sizeof(line_text), stdin);
myvolume = (4.0 / 3.0) * PI * (myradius * myradius * myradius); /* volumn=(4/3) * pi * r^3*/
printf("The volume of sphere is %f.\n", myvolume);
return(0);
}
```
```
Sample Output:
```Input the radius of the sphere : 2.56
The volume of sphere is 70.276237.
```
C Programming Code Editor:
```#include <stdio.h>
float myvolume; /* volume of the sphere (to be computed) */
char line_text[50]; /* a line from the keyboard */
/* the value of pi to 50 places, from wikipedia */
const float PI = 3.14159265358979323846264338327950288419716939937510;
int main() {
printf("Input the radius of the sphere : ");
fgets(line_text, sizeof(line_text), stdin); | 423 | 1,586 | {"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} | 3.453125 | 3 | CC-MAIN-2017-30 | latest | en | 0.676002 |
https://devel.isa-afp.org/browser_info/current/HOL/HOL-Analysis/Binary_Product_Measure.html | 1,702,294,794,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679511159.96/warc/CC-MAIN-20231211112008-20231211142008-00614.warc.gz | 229,002,564 | 45,121 | # Theory Binary_Product_Measure
```(* Title: HOL/Analysis/Binary_Product_Measure.thy
Author: Johannes Hölzl, TU München
*)
section ‹Binary Product Measure›
theory Binary_Product_Measure
imports Nonnegative_Lebesgue_Integration
begin
lemma Pair_vimage_times[simp]: "Pair x -` (A × B) = (if x ∈ A then B else {})"
by auto
lemma rev_Pair_vimage_times[simp]: "(λx. (x, y)) -` (A × B) = (if y ∈ B then A else {})"
by auto
subsection "Binary products"
definition✐‹tag important› pair_measure (infixr "⨂⇩M" 80) where
"A ⨂⇩M B = measure_of (space A × space B)
{a × b | a b. a ∈ sets A ∧ b ∈ sets B}
(λX. ∫⇧+x. (∫⇧+y. indicator X (x,y) ∂B) ∂A)"
lemma pair_measure_closed: "{a × b | a b. a ∈ sets A ∧ b ∈ sets B} ⊆ Pow (space A × space B)"
using sets.space_closed[of A] sets.space_closed[of B] by auto
lemma space_pair_measure:
"space (A ⨂⇩M B) = space A × space B"
unfolding pair_measure_def using pair_measure_closed[of A B]
by (rule space_measure_of)
lemma SIGMA_Collect_eq: "(SIGMA x:space M. {y∈space N. P x y}) = {x∈space (M ⨂⇩M N). P (fst x) (snd x)}"
by (auto simp: space_pair_measure)
lemma sets_pair_measure:
"sets (A ⨂⇩M B) = sigma_sets (space A × space B) {a × b | a b. a ∈ sets A ∧ b ∈ sets B}"
unfolding pair_measure_def using pair_measure_closed[of A B]
by (rule sets_measure_of)
lemma sets_pair_measure_cong[measurable_cong, cong]:
"sets M1 = sets M1' ⟹ sets M2 = sets M2' ⟹ sets (M1 ⨂⇩M M2) = sets (M1' ⨂⇩M M2')"
unfolding sets_pair_measure by (simp cong: sets_eq_imp_space_eq)
lemma pair_measureI[intro, simp, measurable]:
"x ∈ sets A ⟹ y ∈ sets B ⟹ x × y ∈ sets (A ⨂⇩M B)"
by (auto simp: sets_pair_measure)
lemma sets_Pair: "{x} ∈ sets M1 ⟹ {y} ∈ sets M2 ⟹ {(x, y)} ∈ sets (M1 ⨂⇩M M2)"
using pair_measureI[of "{x}" M1 "{y}" M2] by simp
lemma measurable_pair_measureI:
assumes 1: "f ∈ space M → space M1 × space M2"
assumes 2: "⋀A B. A ∈ sets M1 ⟹ B ∈ sets M2 ⟹ f -` (A × B) ∩ space M ∈ sets M"
shows "f ∈ measurable M (M1 ⨂⇩M M2)"
unfolding pair_measure_def using 1 2
by (intro measurable_measure_of) (auto dest: sets.sets_into_space)
lemma measurable_split_replace[measurable (raw)]:
"(λx. f x (fst (g x)) (snd (g x))) ∈ measurable M N ⟹ (λx. case_prod (f x) (g x)) ∈ measurable M N"
unfolding split_beta' .
lemma measurable_Pair[measurable (raw)]:
assumes f: "f ∈ measurable M M1" and g: "g ∈ measurable M M2"
shows "(λx. (f x, g x)) ∈ measurable M (M1 ⨂⇩M M2)"
proof (rule measurable_pair_measureI)
show "(λx. (f x, g x)) ∈ space M → space M1 × space M2"
using f g by (auto simp: measurable_def)
fix A B assume *: "A ∈ sets M1" "B ∈ sets M2"
have "(λx. (f x, g x)) -` (A × B) ∩ space M = (f -` A ∩ space M) ∩ (g -` B ∩ space M)"
by auto
also have "… ∈ sets M"
by (rule sets.Int) (auto intro!: measurable_sets * f g)
finally show "(λx. (f x, g x)) -` (A × B) ∩ space M ∈ sets M" .
qed
lemma measurable_fst[intro!, simp, measurable]: "fst ∈ measurable (M1 ⨂⇩M M2) M1"
by (auto simp: fst_vimage_eq_Times space_pair_measure sets.sets_into_space Times_Int_Times
measurable_def)
lemma measurable_snd[intro!, simp, measurable]: "snd ∈ measurable (M1 ⨂⇩M M2) M2"
by (auto simp: snd_vimage_eq_Times space_pair_measure sets.sets_into_space Times_Int_Times
measurable_def)
lemma measurable_Pair_compose_split[measurable_dest]:
assumes f: "case_prod f ∈ measurable (M1 ⨂⇩M M2) N"
assumes g: "g ∈ measurable M M1" and h: "h ∈ measurable M M2"
shows "(λx. f (g x) (h x)) ∈ measurable M N"
using measurable_compose[OF measurable_Pair f, OF g h] by simp
lemma measurable_Pair1_compose[measurable_dest]:
assumes f: "(λx. (f x, g x)) ∈ measurable M (M1 ⨂⇩M M2)"
assumes [measurable]: "h ∈ measurable N M"
shows "(λx. f (h x)) ∈ measurable N M1"
using measurable_compose[OF f measurable_fst] by simp
lemma measurable_Pair2_compose[measurable_dest]:
assumes f: "(λx. (f x, g x)) ∈ measurable M (M1 ⨂⇩M M2)"
assumes [measurable]: "h ∈ measurable N M"
shows "(λx. g (h x)) ∈ measurable N M2"
using measurable_compose[OF f measurable_snd] by simp
lemma measurable_pair:
assumes "(fst ∘ f) ∈ measurable M M1" "(snd ∘ f) ∈ measurable M M2"
shows "f ∈ measurable M (M1 ⨂⇩M M2)"
using measurable_Pair[OF assms] by simp
lemma
assumes f[measurable]: "f ∈ measurable M (N ⨂⇩M P)"
shows measurable_fst': "(λx. fst (f x)) ∈ measurable M N"
and measurable_snd': "(λx. snd (f x)) ∈ measurable M P"
by simp_all
lemma
assumes f[measurable]: "f ∈ measurable M N"
shows measurable_fst'': "(λx. f (fst x)) ∈ measurable (M ⨂⇩M P) N"
and measurable_snd'': "(λx. f (snd x)) ∈ measurable (P ⨂⇩M M) N"
by simp_all
lemma sets_pair_in_sets:
assumes "⋀a b. a ∈ sets A ⟹ b ∈ sets B ⟹ a × b ∈ sets N"
shows "sets (A ⨂⇩M B) ⊆ sets N"
unfolding sets_pair_measure
by (intro sets.sigma_sets_subset') (auto intro!: assms)
lemma sets_pair_eq_sets_fst_snd:
"sets (A ⨂⇩M B) = sets (Sup {vimage_algebra (space A × space B) fst A, vimage_algebra (space A × space B) snd B})"
(is "?P = sets (Sup {?fst, ?snd})")
proof -
{ fix a b assume ab: "a ∈ sets A" "b ∈ sets B"
then have "a × b = (fst -` a ∩ (space A × space B)) ∩ (snd -` b ∩ (space A × space B))"
by (auto dest: sets.sets_into_space)
also have "… ∈ sets (Sup {?fst, ?snd})"
apply (rule sets.Int)
apply (rule in_sets_Sup)
apply auto []
apply (rule insertI1)
apply (auto intro: ab in_vimage_algebra) []
apply (rule in_sets_Sup)
apply auto []
apply (rule insertI2)
apply (auto intro: ab in_vimage_algebra)
done
finally have "a × b ∈ sets (Sup {?fst, ?snd})" . }
moreover have "sets ?fst ⊆ sets (A ⨂⇩M B)"
by (rule sets_image_in_sets) (auto simp: space_pair_measure[symmetric])
moreover have "sets ?snd ⊆ sets (A ⨂⇩M B)"
by (rule sets_image_in_sets) (auto simp: space_pair_measure)
ultimately show ?thesis
apply (intro antisym[of "sets A" for A] sets_Sup_in_sets sets_pair_in_sets)
apply simp
apply simp
apply simp
apply (elim disjE)
apply (simp add: space_pair_measure)
apply (simp add: space_pair_measure)
apply (auto simp add: space_pair_measure)
done
qed
lemma measurable_pair_iff:
"f ∈ measurable M (M1 ⨂⇩M M2) ⟷ (fst ∘ f) ∈ measurable M M1 ∧ (snd ∘ f) ∈ measurable M M2"
by (auto intro: measurable_pair[of f M M1 M2])
lemma measurable_split_conv:
"(λ(x, y). f x y) ∈ measurable A B ⟷ (λx. f (fst x) (snd x)) ∈ measurable A B"
by (intro arg_cong2[where f="(∈)"]) auto
lemma measurable_pair_swap': "(λ(x,y). (y, x)) ∈ measurable (M1 ⨂⇩M M2) (M2 ⨂⇩M M1)"
by (auto intro!: measurable_Pair simp: measurable_split_conv)
lemma measurable_pair_swap:
assumes f: "f ∈ measurable (M1 ⨂⇩M M2) M" shows "(λ(x,y). f (y, x)) ∈ measurable (M2 ⨂⇩M M1) M"
using measurable_comp[OF measurable_Pair f] by (auto simp: measurable_split_conv comp_def)
lemma measurable_pair_swap_iff:
"f ∈ measurable (M2 ⨂⇩M M1) M ⟷ (λ(x,y). f (y,x)) ∈ measurable (M1 ⨂⇩M M2) M"
by (auto dest: measurable_pair_swap)
lemma measurable_Pair1': "x ∈ space M1 ⟹ Pair x ∈ measurable M2 (M1 ⨂⇩M M2)"
by simp
lemma sets_Pair1[measurable (raw)]:
assumes A: "A ∈ sets (M1 ⨂⇩M M2)" shows "Pair x -` A ∈ sets M2"
proof -
have "Pair x -` A = (if x ∈ space M1 then Pair x -` A ∩ space M2 else {})"
using A[THEN sets.sets_into_space] by (auto simp: space_pair_measure)
also have "… ∈ sets M2"
using A by (auto simp add: measurable_Pair1' intro!: measurable_sets split: if_split_asm)
finally show ?thesis .
qed
lemma measurable_Pair2': "y ∈ space M2 ⟹ (λx. (x, y)) ∈ measurable M1 (M1 ⨂⇩M M2)"
by (auto intro!: measurable_Pair)
lemma sets_Pair2: assumes A: "A ∈ sets (M1 ⨂⇩M M2)" shows "(λx. (x, y)) -` A ∈ sets M1"
proof -
have "(λx. (x, y)) -` A = (if y ∈ space M2 then (λx. (x, y)) -` A ∩ space M1 else {})"
using A[THEN sets.sets_into_space] by (auto simp: space_pair_measure)
also have "… ∈ sets M1"
using A by (auto simp add: measurable_Pair2' intro!: measurable_sets split: if_split_asm)
finally show ?thesis .
qed
lemma measurable_Pair2:
assumes f: "f ∈ measurable (M1 ⨂⇩M M2) M" and x: "x ∈ space M1"
shows "(λy. f (x, y)) ∈ measurable M2 M"
using measurable_comp[OF measurable_Pair1' f, OF x]
by (simp add: comp_def)
lemma measurable_Pair1:
assumes f: "f ∈ measurable (M1 ⨂⇩M M2) M" and y: "y ∈ space M2"
shows "(λx. f (x, y)) ∈ measurable M1 M"
using measurable_comp[OF measurable_Pair2' f, OF y]
by (simp add: comp_def)
lemma Int_stable_pair_measure_generator: "Int_stable {a × b | a b. a ∈ sets A ∧ b ∈ sets B}"
unfolding Int_stable_def
by safe (auto simp add: Times_Int_Times)
lemma (in finite_measure) finite_measure_cut_measurable:
assumes [measurable]: "Q ∈ sets (N ⨂⇩M M)"
shows "(λx. emeasure M (Pair x -` Q)) ∈ borel_measurable N"
(is "?s Q ∈ _")
using Int_stable_pair_measure_generator pair_measure_closed assms
unfolding sets_pair_measure
proof (induct rule: sigma_sets_induct_disjoint)
case (compl A)
with sets.sets_into_space have "⋀x. emeasure M (Pair x -` ((space N × space M) - A)) =
(if x ∈ space N then emeasure M (space M) - ?s A x else 0)"
unfolding sets_pair_measure[symmetric]
by (auto intro!: emeasure_compl simp: vimage_Diff sets_Pair1)
with compl sets.top show ?case
by (auto intro!: measurable_If simp: space_pair_measure)
next
case (union F)
then have "⋀x. emeasure M (Pair x -` (⋃i. F i)) = (∑i. ?s (F i) x)"
by (simp add: suminf_emeasure disjoint_family_on_vimageI subset_eq vimage_UN sets_pair_measure[symmetric])
with union show ?case
unfolding sets_pair_measure[symmetric] by simp
qed (auto simp add: if_distrib Int_def[symmetric] intro!: measurable_If)
lemma (in sigma_finite_measure) measurable_emeasure_Pair:
assumes Q: "Q ∈ sets (N ⨂⇩M M)" shows "(λx. emeasure M (Pair x -` Q)) ∈ borel_measurable N" (is "?s Q ∈ _")
proof -
obtain F :: "nat ⇒ 'a set" where F:
"range F ⊆ sets M"
"⋃ (range F) = space M"
"⋀i. emeasure M (F i) ≠ ∞"
"disjoint_family F" by (blast intro: sigma_finite_disjoint)
then have F_sets: "⋀i. F i ∈ sets M" by auto
let ?C = "λx i. F i ∩ Pair x -` Q"
{ fix i
have [simp]: "space N × F i ∩ space N × space M = space N × F i"
using F sets.sets_into_space by auto
let ?R = "density M (indicator (F i))"
have "finite_measure ?R"
using F by (intro finite_measureI) (auto simp: emeasure_restricted subset_eq)
then have "(λx. emeasure ?R (Pair x -` (space N × space ?R ∩ Q))) ∈ borel_measurable N"
by (rule finite_measure.finite_measure_cut_measurable) (auto intro: Q)
moreover have "⋀x. emeasure ?R (Pair x -` (space N × space ?R ∩ Q))
= emeasure M (F i ∩ Pair x -` (space N × space ?R ∩ Q))"
using Q F_sets by (intro emeasure_restricted) (auto intro: sets_Pair1)
moreover have "⋀x. F i ∩ Pair x -` (space N × space ?R ∩ Q) = ?C x i"
using sets.sets_into_space[OF Q] by (auto simp: space_pair_measure)
ultimately have "(λx. emeasure M (?C x i)) ∈ borel_measurable N"
by simp }
moreover
{ fix x
have "(∑i. emeasure M (?C x i)) = emeasure M (⋃i. ?C x i)"
proof (intro suminf_emeasure)
show "range (?C x) ⊆ sets M"
using F ‹Q ∈ sets (N ⨂⇩M M)› by (auto intro!: sets_Pair1)
have "disjoint_family F" using F by auto
show "disjoint_family (?C x)"
by (rule disjoint_family_on_bisimulation[OF ‹disjoint_family F›]) auto
qed
also have "(⋃i. ?C x i) = Pair x -` Q"
using F sets.sets_into_space[OF ‹Q ∈ sets (N ⨂⇩M M)›]
by (auto simp: space_pair_measure)
finally have "emeasure M (Pair x -` Q) = (∑i. emeasure M (?C x i))"
by simp }
ultimately show ?thesis using ‹Q ∈ sets (N ⨂⇩M M)› F_sets
by auto
qed
lemma (in sigma_finite_measure) measurable_emeasure[measurable (raw)]:
assumes space: "⋀x. x ∈ space N ⟹ A x ⊆ space M"
assumes A: "{x∈space (N ⨂⇩M M). snd x ∈ A (fst x)} ∈ sets (N ⨂⇩M M)"
shows "(λx. emeasure M (A x)) ∈ borel_measurable N"
proof -
from space have "⋀x. x ∈ space N ⟹ Pair x -` {x ∈ space (N ⨂⇩M M). snd x ∈ A (fst x)} = A x"
by (auto simp: space_pair_measure)
with measurable_emeasure_Pair[OF A] show ?thesis
by (auto cong: measurable_cong)
qed
lemma (in sigma_finite_measure) emeasure_pair_measure:
assumes "X ∈ sets (N ⨂⇩M M)"
shows "emeasure (N ⨂⇩M M) X = (∫⇧+ x. ∫⇧+ y. indicator X (x, y) ∂M ∂N)" (is "_ = ?μ X")
proof (rule emeasure_measure_of[OF pair_measure_def])
show "positive (sets (N ⨂⇩M M)) ?μ"
by (auto simp: positive_def)
have eq[simp]: "⋀A x y. indicator A (x, y) = indicator (Pair x -` A) y"
by (auto simp: indicator_def)
show "countably_additive (sets (N ⨂⇩M M)) ?μ"
proof (rule countably_additiveI)
fix F :: "nat ⇒ ('b × 'a) set" assume F: "range F ⊆ sets (N ⨂⇩M M)" "disjoint_family F"
from F have *: "⋀i. F i ∈ sets (N ⨂⇩M M)" by auto
moreover have "⋀x. disjoint_family (λi. Pair x -` F i)"
by (intro disjoint_family_on_bisimulation[OF F(2)]) auto
moreover have "⋀x. range (λi. Pair x -` F i) ⊆ sets M"
using F by (auto simp: sets_Pair1)
ultimately show "(∑n. ?μ (F n)) = ?μ (⋃i. F i)"
by (auto simp add: nn_integral_suminf[symmetric] vimage_UN suminf_emeasure
intro!: nn_integral_cong nn_integral_indicator[symmetric])
qed
show "{a × b |a b. a ∈ sets N ∧ b ∈ sets M} ⊆ Pow (space N × space M)"
using sets.space_closed[of N] sets.space_closed[of M] by auto
qed fact
lemma (in sigma_finite_measure) emeasure_pair_measure_alt:
assumes X: "X ∈ sets (N ⨂⇩M M)"
shows "emeasure (N ⨂⇩M M) X = (∫⇧+x. emeasure M (Pair x -` X) ∂N)"
proof -
have [simp]: "⋀x y. indicator X (x, y) = indicator (Pair x -` X) y"
by (auto simp: indicator_def)
show ?thesis
using X by (auto intro!: nn_integral_cong simp: emeasure_pair_measure sets_Pair1)
qed
proposition (in sigma_finite_measure) emeasure_pair_measure_Times:
assumes A: "A ∈ sets N" and B: "B ∈ sets M"
shows "emeasure (N ⨂⇩M M) (A × B) = emeasure N A * emeasure M B"
proof -
have "emeasure (N ⨂⇩M M) (A × B) = (∫⇧+x. emeasure M B * indicator A x ∂N)"
using A B by (auto intro!: nn_integral_cong simp: emeasure_pair_measure_alt)
also have "… = emeasure M B * emeasure N A"
using A by (simp add: nn_integral_cmult_indicator)
finally show ?thesis
by (simp add: ac_simps)
qed
lemma (in sigma_finite_measure) times_in_null_sets1 [intro]:
assumes "A ∈ null_sets N" "B ∈ sets M"
shows "A × B ∈ null_sets (N ⨂⇩M M)"
using assms by (simp add: null_sets_def emeasure_pair_measure_Times)
lemma (in sigma_finite_measure) times_in_null_sets2 [intro]:
assumes "A ∈ sets N" "B ∈ null_sets M"
shows "A × B ∈ null_sets (N ⨂⇩M M)"
using assms by (simp add: null_sets_def emeasure_pair_measure_Times)
subsection ‹Binary products of ‹σ›-finite emeasure spaces›
locale✐‹tag unimportant› pair_sigma_finite = M1?: sigma_finite_measure M1 + M2?: sigma_finite_measure M2
for M1 :: "'a measure" and M2 :: "'b measure"
lemma (in pair_sigma_finite) measurable_emeasure_Pair1:
"Q ∈ sets (M1 ⨂⇩M M2) ⟹ (λx. emeasure M2 (Pair x -` Q)) ∈ borel_measurable M1"
using M2.measurable_emeasure_Pair .
lemma (in pair_sigma_finite) measurable_emeasure_Pair2:
assumes Q: "Q ∈ sets (M1 ⨂⇩M M2)" shows "(λy. emeasure M1 ((λx. (x, y)) -` Q)) ∈ borel_measurable M2"
proof -
have "(λ(x, y). (y, x)) -` Q ∩ space (M2 ⨂⇩M M1) ∈ sets (M2 ⨂⇩M M1)"
using Q measurable_pair_swap' by (auto intro: measurable_sets)
note M1.measurable_emeasure_Pair[OF this]
moreover have "⋀y. Pair y -` ((λ(x, y). (y, x)) -` Q ∩ space (M2 ⨂⇩M M1)) = (λx. (x, y)) -` Q"
using Q[THEN sets.sets_into_space] by (auto simp: space_pair_measure)
ultimately show ?thesis by simp
qed
proposition (in pair_sigma_finite) sigma_finite_up_in_pair_measure_generator:
defines "E ≡ {A × B | A B. A ∈ sets M1 ∧ B ∈ sets M2}"
shows "∃F::nat ⇒ ('a × 'b) set. range F ⊆ E ∧ incseq F ∧ (⋃i. F i) = space M1 × space M2 ∧
(∀i. emeasure (M1 ⨂⇩M M2) (F i) ≠ ∞)"
proof -
obtain F1 where F1: "range F1 ⊆ sets M1"
"⋃ (range F1) = space M1"
"⋀i. emeasure M1 (F1 i) ≠ ∞"
"incseq F1"
by (rule M1.sigma_finite_incseq) blast
obtain F2 where F2: "range F2 ⊆ sets M2"
"⋃ (range F2) = space M2"
"⋀i. emeasure M2 (F2 i) ≠ ∞"
"incseq F2"
by (rule M2.sigma_finite_incseq) blast
from F1 F2 have space: "space M1 = (⋃i. F1 i)" "space M2 = (⋃i. F2 i)" by auto
let ?F = "λi. F1 i × F2 i"
show ?thesis
proof (intro exI[of _ ?F] conjI allI)
show "range ?F ⊆ E" using F1 F2 by (auto simp: E_def) (metis range_subsetD)
next
have "space M1 × space M2 ⊆ (⋃i. ?F i)"
proof (intro subsetI)
fix x assume "x ∈ space M1 × space M2"
then obtain i j where "fst x ∈ F1 i" "snd x ∈ F2 j"
by (auto simp: space)
then have "fst x ∈ F1 (max i j)" "snd x ∈ F2 (max j i)"
using ‹incseq F1› ‹incseq F2› unfolding incseq_def
by (force split: split_max)+
then have "(fst x, snd x) ∈ F1 (max i j) × F2 (max i j)"
by (intro SigmaI) (auto simp add: max.commute)
then show "x ∈ (⋃i. ?F i)" by auto
qed
then show "(⋃i. ?F i) = space M1 × space M2"
using space by (auto simp: space)
next
fix i show "incseq (λi. F1 i × F2 i)"
using ‹incseq F1› ‹incseq F2› unfolding incseq_Suc_iff by auto
next
fix i
from F1 F2 have "F1 i ∈ sets M1" "F2 i ∈ sets M2" by auto
with F1 F2 show "emeasure (M1 ⨂⇩M M2) (F1 i × F2 i) ≠ ∞"
by (auto simp add: emeasure_pair_measure_Times ennreal_mult_eq_top_iff)
qed
qed
sublocale✐‹tag unimportant› pair_sigma_finite ⊆ P?: sigma_finite_measure "M1 ⨂⇩M M2"
proof
obtain F1 :: "'a set set" and F2 :: "'b set set" where
"countable F1 ∧ F1 ⊆ sets M1 ∧ ⋃ F1 = space M1 ∧ (∀a∈F1. emeasure M1 a ≠ ∞)"
"countable F2 ∧ F2 ⊆ sets M2 ∧ ⋃ F2 = space M2 ∧ (∀a∈F2. emeasure M2 a ≠ ∞)"
using M1.sigma_finite_countable M2.sigma_finite_countable by auto
then show
"∃A. countable A ∧ A ⊆ sets (M1 ⨂⇩M M2) ∧ ⋃A = space (M1 ⨂⇩M M2) ∧ (∀a∈A. emeasure (M1 ⨂⇩M M2) a ≠ ∞)"
by (intro exI[of _ "(λ(a, b). a × b) ` (F1 × F2)"] conjI)
(auto simp: M2.emeasure_pair_measure_Times space_pair_measure set_eq_iff subset_eq ennreal_mult_eq_top_iff)
qed
lemma sigma_finite_pair_measure:
assumes A: "sigma_finite_measure A" and B: "sigma_finite_measure B"
shows "sigma_finite_measure (A ⨂⇩M B)"
proof -
interpret A: sigma_finite_measure A by fact
interpret B: sigma_finite_measure B by fact
interpret AB: pair_sigma_finite A B ..
show ?thesis ..
qed
lemma sets_pair_swap:
assumes "A ∈ sets (M1 ⨂⇩M M2)"
shows "(λ(x, y). (y, x)) -` A ∩ space (M2 ⨂⇩M M1) ∈ sets (M2 ⨂⇩M M1)"
using measurable_pair_swap' assms by (rule measurable_sets)
lemma (in pair_sigma_finite) distr_pair_swap:
"M1 ⨂⇩M M2 = distr (M2 ⨂⇩M M1) (M1 ⨂⇩M M2) (λ(x, y). (y, x))" (is "?P = ?D")
proof -
let ?E = "{a × b |a b. a ∈ sets M1 ∧ b ∈ sets M2}"
obtain F :: "nat ⇒ ('a × 'b) set" where F: "range F ⊆ ?E"
"incseq F" "⋃ (range F) = space M1 × space M2" "∀i. emeasure (M1 ⨂⇩M M2) (F i) ≠ ∞"
using sigma_finite_up_in_pair_measure_generator by auto
show ?thesis
proof (rule measure_eqI_generator_eq[OF Int_stable_pair_measure_generator[of M1 M2]])
show "?E ⊆ Pow (space ?P)"
using sets.space_closed[of M1] sets.space_closed[of M2] by (auto simp: space_pair_measure)
show "sets ?P = sigma_sets (space ?P) ?E"
by (simp add: sets_pair_measure space_pair_measure)
then show "sets ?D = sigma_sets (space ?P) ?E"
by simp
from F show "range F ⊆ ?E" "(⋃i. F i) = space ?P" "⋀i. emeasure ?P (F i) ≠ ∞"
by (auto simp: space_pair_measure)
next
fix X assume "X ∈ ?E"
then obtain A B where X[simp]: "X = A × B" and A: "A ∈ sets M1" and B: "B ∈ sets M2" by auto
have "(λ(y, x). (x, y)) -` X ∩ space (M2 ⨂⇩M M1) = B × A"
using sets.sets_into_space[OF A] sets.sets_into_space[OF B] by (auto simp: space_pair_measure)
with A B show "emeasure (M1 ⨂⇩M M2) X = emeasure ?D X"
by (simp add: M2.emeasure_pair_measure_Times M1.emeasure_pair_measure_Times emeasure_distr
measurable_pair_swap' ac_simps)
qed
qed
lemma (in pair_sigma_finite) emeasure_pair_measure_alt2:
assumes A: "A ∈ sets (M1 ⨂⇩M M2)"
shows "emeasure (M1 ⨂⇩M M2) A = (∫⇧+y. emeasure M1 ((λx. (x, y)) -` A) ∂M2)"
(is "_ = ?ν A")
proof -
have [simp]: "⋀y. (Pair y -` ((λ(x, y). (y, x)) -` A ∩ space (M2 ⨂⇩M M1))) = (λx. (x, y)) -` A"
using sets.sets_into_space[OF A] by (auto simp: space_pair_measure)
show ?thesis using A
by (subst distr_pair_swap)
(simp_all del: vimage_Int add: measurable_sets[OF measurable_pair_swap']
M1.emeasure_pair_measure_alt emeasure_distr[OF measurable_pair_swap' A])
qed
lemma (in pair_sigma_finite) AE_pair:
assumes "AE x in (M1 ⨂⇩M M2). Q x"
shows "AE x in M1. (AE y in M2. Q (x, y))"
proof -
obtain N where N: "N ∈ sets (M1 ⨂⇩M M2)" "emeasure (M1 ⨂⇩M M2) N = 0" "{x∈space (M1 ⨂⇩M M2). ¬ Q x} ⊆ N"
using assms unfolding eventually_ae_filter by auto
show ?thesis
proof (rule AE_I)
from N measurable_emeasure_Pair1[OF ‹N ∈ sets (M1 ⨂⇩M M2)›]
show "emeasure M1 {x∈space M1. emeasure M2 (Pair x -` N) ≠ 0} = 0"
by (auto simp: M2.emeasure_pair_measure_alt nn_integral_0_iff)
show "{x ∈ space M1. emeasure M2 (Pair x -` N) ≠ 0} ∈ sets M1"
by (intro borel_measurable_eq measurable_emeasure_Pair1 N sets.sets_Collect_neg N) simp
{ fix x assume "x ∈ space M1" "emeasure M2 (Pair x -` N) = 0"
have "AE y in M2. Q (x, y)"
proof (rule AE_I)
show "emeasure M2 (Pair x -` N) = 0" by fact
show "Pair x -` N ∈ sets M2" using N(1) by (rule sets_Pair1)
show "{y ∈ space M2. ¬ Q (x, y)} ⊆ Pair x -` N"
using N ‹x ∈ space M1› unfolding space_pair_measure by auto
qed }
then show "{x ∈ space M1. ¬ (AE y in M2. Q (x, y))} ⊆ {x ∈ space M1. emeasure M2 (Pair x -` N) ≠ 0}"
by auto
qed
qed
lemma (in pair_sigma_finite) AE_pair_measure:
assumes "{x∈space (M1 ⨂⇩M M2). P x} ∈ sets (M1 ⨂⇩M M2)"
assumes ae: "AE x in M1. AE y in M2. P (x, y)"
shows "AE x in M1 ⨂⇩M M2. P x"
proof (subst AE_iff_measurable[OF _ refl])
show "{x∈space (M1 ⨂⇩M M2). ¬ P x} ∈ sets (M1 ⨂⇩M M2)"
by (rule sets.sets_Collect) fact
then have "emeasure (M1 ⨂⇩M M2) {x ∈ space (M1 ⨂⇩M M2). ¬ P x} =
(∫⇧+ x. ∫⇧+ y. indicator {x ∈ space (M1 ⨂⇩M M2). ¬ P x} (x, y) ∂M2 ∂M1)"
by (simp add: M2.emeasure_pair_measure)
also have "… = (∫⇧+ x. ∫⇧+ y. 0 ∂M2 ∂M1)"
using ae
apply (safe intro!: nn_integral_cong_AE)
apply (intro AE_I2)
apply (safe intro!: nn_integral_cong_AE)
apply auto
done
finally show "emeasure (M1 ⨂⇩M M2) {x ∈ space (M1 ⨂⇩M M2). ¬ P x} = 0" by simp
qed
lemma (in pair_sigma_finite) AE_pair_iff:
"{x∈space (M1 ⨂⇩M M2). P (fst x) (snd x)} ∈ sets (M1 ⨂⇩M M2) ⟹
(AE x in M1. AE y in M2. P x y) ⟷ (AE x in (M1 ⨂⇩M M2). P (fst x) (snd x))"
using AE_pair[of "λx. P (fst x) (snd x)"] AE_pair_measure[of "λx. P (fst x) (snd x)"] by auto
lemma (in pair_sigma_finite) AE_commute:
assumes P: "{x∈space (M1 ⨂⇩M M2). P (fst x) (snd x)} ∈ sets (M1 ⨂⇩M M2)"
shows "(AE x in M1. AE y in M2. P x y) ⟷ (AE y in M2. AE x in M1. P x y)"
proof -
interpret Q: pair_sigma_finite M2 M1 ..
have [simp]: "⋀x. (fst (case x of (x, y) ⇒ (y, x))) = snd x" "⋀x. (snd (case x of (x, y) ⇒ (y, x))) = fst x"
by auto
have "{x ∈ space (M2 ⨂⇩M M1). P (snd x) (fst x)} =
(λ(x, y). (y, x)) -` {x ∈ space (M1 ⨂⇩M M2). P (fst x) (snd x)} ∩ space (M2 ⨂⇩M M1)"
by (auto simp: space_pair_measure)
also have "… ∈ sets (M2 ⨂⇩M M1)"
by (intro sets_pair_swap P)
finally show ?thesis
apply (subst AE_pair_iff[OF P])
apply (subst distr_pair_swap)
apply (subst AE_distr_iff[OF measurable_pair_swap' P])
apply (subst Q.AE_pair_iff)
apply simp_all
done
qed
subsection "Fubinis theorem"
lemma measurable_compose_Pair1:
"x ∈ space M1 ⟹ g ∈ measurable (M1 ⨂⇩M M2) L ⟹ (λy. g (x, y)) ∈ measurable M2 L"
by simp
lemma (in sigma_finite_measure) borel_measurable_nn_integral_fst:
assumes f: "f ∈ borel_measurable (M1 ⨂⇩M M)"
shows "(λx. ∫⇧+ y. f (x, y) ∂M) ∈ borel_measurable M1"
using f proof induct
case (cong u v)
then have "⋀w x. w ∈ space M1 ⟹ x ∈ space M ⟹ u (w, x) = v (w, x)"
by (auto simp: space_pair_measure)
show ?case
apply (subst measurable_cong)
apply (rule nn_integral_cong)
apply fact+
done
next
case (set Q)
have [simp]: "⋀x y. indicator Q (x, y) = indicator (Pair x -` Q) y"
by (auto simp: indicator_def)
have "⋀x. x ∈ space M1 ⟹ emeasure M (Pair x -` Q) = ∫⇧+ y. indicator Q (x, y) ∂M"
by (simp add: sets_Pair1[OF set])
from this measurable_emeasure_Pair[OF set] show ?case
by (rule measurable_cong[THEN iffD1])
qed (simp_all add: nn_integral_add nn_integral_cmult measurable_compose_Pair1
nn_integral_monotone_convergence_SUP incseq_def le_fun_def image_comp
cong: measurable_cong)
lemma (in sigma_finite_measure) nn_integral_fst:
assumes f: "f ∈ borel_measurable (M1 ⨂⇩M M)"
shows "(∫⇧+ x. ∫⇧+ y. f (x, y) ∂M ∂M1) = integral⇧N (M1 ⨂⇩M M) f" (is "?I f = _")
using f proof induct
case (cong u v)
then have "?I u = ?I v"
by (intro nn_integral_cong) (auto simp: space_pair_measure)
with cong show ?case
by (simp cong: nn_integral_cong)
qed (simp_all add: emeasure_pair_measure nn_integral_cmult nn_integral_add
nn_integral_monotone_convergence_SUP measurable_compose_Pair1
borel_measurable_nn_integral_fst nn_integral_mono incseq_def le_fun_def image_comp
cong: nn_integral_cong)
lemma (in sigma_finite_measure) borel_measurable_nn_integral[measurable (raw)]:
"case_prod f ∈ borel_measurable (N ⨂⇩M M) ⟹ (λx. ∫⇧+ y. f x y ∂M) ∈ borel_measurable N"
using borel_measurable_nn_integral_fst[of "case_prod f" N] by simp
proposition (in pair_sigma_finite) nn_integral_snd:
assumes f[measurable]: "f ∈ borel_measurable (M1 ⨂⇩M M2)"
shows "(∫⇧+ y. (∫⇧+ x. f (x, y) ∂M1) ∂M2) = integral⇧N (M1 ⨂⇩M M2) f"
proof -
note measurable_pair_swap[OF f]
from M1.nn_integral_fst[OF this]
have "(∫⇧+ y. (∫⇧+ x. f (x, y) ∂M1) ∂M2) = (∫⇧+ (x, y). f (y, x) ∂(M2 ⨂⇩M M1))"
by simp
also have "(∫⇧+ (x, y). f (y, x) ∂(M2 ⨂⇩M M1)) = integral⇧N (M1 ⨂⇩M M2) f"
by (subst distr_pair_swap) (auto simp add: nn_integral_distr intro!: nn_integral_cong)
finally show ?thesis .
qed
theorem (in pair_sigma_finite) Fubini:
assumes f: "f ∈ borel_measurable (M1 ⨂⇩M M2)"
shows "(∫⇧+ y. (∫⇧+ x. f (x, y) ∂M1) ∂M2) = (∫⇧+ x. (∫⇧+ y. f (x, y) ∂M2) ∂M1)"
unfolding nn_integral_snd[OF assms] M2.nn_integral_fst[OF assms] ..
theorem (in pair_sigma_finite) Fubini':
assumes f: "case_prod f ∈ borel_measurable (M1 ⨂⇩M M2)"
shows "(∫⇧+ y. (∫⇧+ x. f x y ∂M1) ∂M2) = (∫⇧+ x. (∫⇧+ y. f x y ∂M2) ∂M1)"
using Fubini[OF f] by simp
subsection ‹Products on counting spaces, densities and distributions›
proposition sigma_prod:
assumes X_cover: "∃E⊆A. countable E ∧ X = ⋃E" and A: "A ⊆ Pow X"
assumes Y_cover: "∃E⊆B. countable E ∧ Y = ⋃E" and B: "B ⊆ Pow Y"
shows "sigma X A ⨂⇩M sigma Y B = sigma (X × Y) {a × b | a b. a ∈ A ∧ b ∈ B}"
(is "?P = ?S")
proof (rule measure_eqI)
have [simp]: "snd ∈ X × Y → Y" "fst ∈ X × Y → X"
by auto
let ?XY = "{{fst -` a ∩ X × Y | a. a ∈ A}, {snd -` b ∩ X × Y | b. b ∈ B}}"
have "sets ?P = sets (SUP xy∈?XY. sigma (X × Y) xy)"
by (simp add: vimage_algebra_sigma sets_pair_eq_sets_fst_snd A B)
also have "… = sets (sigma (X × Y) (⋃?XY))"
by (intro Sup_sigma arg_cong[where f=sets]) auto
also have "… = sets ?S"
proof (intro arg_cong[where f=sets] sigma_eqI sigma_sets_eqI)
show "⋃?XY ⊆ Pow (X × Y)" "{a × b |a b. a ∈ A ∧ b ∈ B} ⊆ Pow (X × Y)"
using A B by auto
next
interpret XY: sigma_algebra "X × Y" "sigma_sets (X × Y) {a × b |a b. a ∈ A ∧ b ∈ B}"
using A B by (intro sigma_algebra_sigma_sets) auto
fix Z assume "Z ∈ ⋃?XY"
then show "Z ∈ sigma_sets (X × Y) {a × b |a b. a ∈ A ∧ b ∈ B}"
proof safe
fix a assume "a ∈ A"
from Y_cover obtain E where E: "E ⊆ B" "countable E" and "Y = ⋃E"
by auto
with ‹a ∈ A› A have eq: "fst -` a ∩ X × Y = (⋃e∈E. a × e)"
by auto
show "fst -` a ∩ X × Y ∈ sigma_sets (X × Y) {a × b |a b. a ∈ A ∧ b ∈ B}"
using ‹a ∈ A› E unfolding eq by (auto intro!: XY.countable_UN')
next
fix b assume "b ∈ B"
from X_cover obtain E where E: "E ⊆ A" "countable E" and "X = ⋃E"
by auto
with ‹b ∈ B› B have eq: "snd -` b ∩ X × Y = (⋃e∈E. e × b)"
by auto
show "snd -` b ∩ X × Y ∈ sigma_sets (X × Y) {a × b |a b. a ∈ A ∧ b ∈ B}"
using ‹b ∈ B› E unfolding eq by (auto intro!: XY.countable_UN')
qed
next
fix Z assume "Z ∈ {a × b |a b. a ∈ A ∧ b ∈ B}"
then obtain a b where "Z = a × b" and ab: "a ∈ A" "b ∈ B"
by auto
then have Z: "Z = (fst -` a ∩ X × Y) ∩ (snd -` b ∩ X × Y)"
using A B by auto
interpret XY: sigma_algebra "X × Y" "sigma_sets (X × Y) (⋃?XY)"
by (intro sigma_algebra_sigma_sets) auto
show "Z ∈ sigma_sets (X × Y) (⋃?XY)"
unfolding Z by (rule XY.Int) (blast intro: ab)+
qed
finally show "sets ?P = sets ?S" .
next
interpret finite_measure "sigma X A" for X A
proof qed (simp add: emeasure_sigma)
fix A assume "A ∈ sets ?P" then show "emeasure ?P A = emeasure ?S A"
by (simp add: emeasure_pair_measure_alt emeasure_sigma)
qed
lemma sigma_sets_pair_measure_generator_finite:
assumes "finite A" and "finite B"
shows "sigma_sets (A × B) { a × b | a b. a ⊆ A ∧ b ⊆ B} = Pow (A × B)"
(is "sigma_sets ?prod ?sets = _")
proof safe
have fin: "finite (A × B)" using assms by (rule finite_cartesian_product)
fix x assume subset: "x ⊆ A × B"
hence "finite x" using fin by (rule finite_subset)
from this subset show "x ∈ sigma_sets ?prod ?sets"
proof (induct x)
case empty show ?case by (rule sigma_sets.Empty)
next
case (insert a x)
hence "{a} ∈ sigma_sets ?prod ?sets" by auto
moreover have "x ∈ sigma_sets ?prod ?sets" using insert by auto
ultimately show ?case unfolding insert_is_Un[of a x] by (rule sigma_sets_Un)
qed
next
fix x a b
assume "x ∈ sigma_sets ?prod ?sets" and "(a, b) ∈ x"
from sigma_sets_into_sp[OF _ this(1)] this(2)
show "a ∈ A" and "b ∈ B" by auto
qed
proposition sets_pair_eq:
assumes Ea: "Ea ⊆ Pow (space A)" "sets A = sigma_sets (space A) Ea"
and Ca: "countable Ca" "Ca ⊆ Ea" "⋃Ca = space A"
and Eb: "Eb ⊆ Pow (space B)" "sets B = sigma_sets (space B) Eb"
and Cb: "countable Cb" "Cb ⊆ Eb" "⋃Cb = space B"
shows "sets (A ⨂⇩M B) = sets (sigma (space A × space B) { a × b | a b. a ∈ Ea ∧ b ∈ Eb })"
(is "_ = sets (sigma ?Ω ?E)")
proof
show "sets (sigma ?Ω ?E) ⊆ sets (A ⨂⇩M B)"
using Ea(1) Eb(1) by (subst sigma_le_sets) (auto simp: Ea(2) Eb(2))
have "?E ⊆ Pow ?Ω"
using Ea(1) Eb(1) by auto
then have E: "a ∈ Ea ⟹ b ∈ Eb ⟹ a × b ∈ sets (sigma ?Ω ?E)" for a b
by auto
have "sets (A ⨂⇩M B) ⊆ sets (Sup {vimage_algebra ?Ω fst A, vimage_algebra ?Ω snd B})"
unfolding sets_pair_eq_sets_fst_snd ..
also have "vimage_algebra ?Ω fst A = vimage_algebra ?Ω fst (sigma (space A) Ea)"
by (intro vimage_algebra_cong[OF refl refl]) (simp add: Ea)
also have "… = sigma ?Ω {fst -` A ∩ ?Ω |A. A ∈ Ea}"
by (intro Ea vimage_algebra_sigma) auto
also have "vimage_algebra ?Ω snd B = vimage_algebra ?Ω snd (sigma (space B) Eb)"
by (intro vimage_algebra_cong[OF refl refl]) (simp add: Eb)
also have "… = sigma ?Ω {snd -` A ∩ ?Ω |A. A ∈ Eb}"
by (intro Eb vimage_algebra_sigma) auto
also have "{sigma ?Ω {fst -` Aa ∩ ?Ω |Aa. Aa ∈ Ea}, sigma ?Ω {snd -` Aa ∩ ?Ω |Aa. Aa ∈ Eb}} =
sigma ?Ω ` {{fst -` Aa ∩ ?Ω |Aa. Aa ∈ Ea}, {snd -` Aa ∩ ?Ω |Aa. Aa ∈ Eb}}"
by auto
also have "sets (SUP S∈{{fst -` Aa ∩ ?Ω |Aa. Aa ∈ Ea}, {snd -` Aa ∩ ?Ω |Aa. Aa ∈ Eb}}. sigma ?Ω S) =
sets (sigma ?Ω (⋃{{fst -` Aa ∩ ?Ω |Aa. Aa ∈ Ea}, {snd -` Aa ∩ ?Ω |Aa. Aa ∈ Eb}}))"
using Ea(1) Eb(1) by (intro sets_Sup_sigma) auto
also have "… ⊆ sets (sigma ?Ω ?E)"
proof (subst sigma_le_sets, safe intro!: space_in_measure_of)
fix a assume "a ∈ Ea"
then have "fst -` a ∩ ?Ω = (⋃b∈Cb. a × b)"
using Cb(3)[symmetric] Ea(1) by auto
then show "fst -` a ∩ ?Ω ∈ sets (sigma ?Ω ?E)"
using Cb ‹a ∈ Ea› by (auto intro!: sets.countable_UN' E)
next
fix b assume "b ∈ Eb"
then have "snd -` b ∩ ?Ω = (⋃a∈Ca. a × b)"
using Ca(3)[symmetric] Eb(1) by auto
then show "snd -` b ∩ ?Ω ∈ sets (sigma ?Ω ?E)"
using Ca ‹b ∈ Eb› by (auto intro!: sets.countable_UN' E)
qed
finally show "sets (A ⨂⇩M B) ⊆ sets (sigma ?Ω ?E)" .
qed
proposition borel_prod:
"(borel ⨂⇩M borel) = (borel :: ('a::second_countable_topology × 'b::second_countable_topology) measure)"
(is "?P = ?B")
proof -
have "?B = sigma UNIV {A × B | A B. open A ∧ open B}"
by (rule second_countable_borel_measurable[OF open_prod_generated])
also have "… = ?P"
unfolding borel_def
by (subst sigma_prod) (auto intro!: exI[of _ "{UNIV}"])
finally show ?thesis ..
qed
proposition pair_measure_count_space:
assumes A: "finite A" and B: "finite B"
shows "count_space A ⨂⇩M count_space B = count_space (A × B)" (is "?P = ?C")
proof (rule measure_eqI)
interpret A: finite_measure "count_space A" by (rule finite_measure_count_space) fact
interpret B: finite_measure "count_space B" by (rule finite_measure_count_space) fact
interpret P: pair_sigma_finite "count_space A" "count_space B" ..
show eq: "sets ?P = sets ?C"
by (simp add: sets_pair_measure sigma_sets_pair_measure_generator_finite A B)
fix X assume X: "X ∈ sets ?P"
with eq have X_subset: "X ⊆ A × B" by simp
with A B have fin_Pair: "⋀x. finite (Pair x -` X)"
by (intro finite_subset[OF _ B]) auto
have fin_X: "finite X" using X_subset by (rule finite_subset) (auto simp: A B)
have card: "0 < card (Pair a -` X)" if "(a, b) ∈ X" for a b
using card_gt_0_iff fin_Pair that by auto
then have "emeasure ?P X = ∫⇧+ x. emeasure (count_space B) (Pair x -` X)
∂count_space A"
by (simp add: B.emeasure_pair_measure_alt X)
also have "... = emeasure ?C X"
apply (subst emeasure_count_space)
using card X_subset A fin_Pair fin_X
apply (auto simp add: nn_integral_count_space
of_nat_sum[symmetric] card_SigmaI[symmetric]
simp del: card_SigmaI
intro!: arg_cong[where f=card])
done
finally show "emeasure ?P X = emeasure ?C X" .
qed
lemma emeasure_prod_count_space:
assumes A: "A ∈ sets (count_space UNIV ⨂⇩M M)" (is "A ∈ sets (?A ⨂⇩M ?B)")
shows "emeasure (?A ⨂⇩M ?B) A = (∫⇧+ x. ∫⇧+ y. indicator A (x, y) ∂?B ∂?A)"
by (rule emeasure_measure_of[OF pair_measure_def])
(auto simp: countably_additive_def positive_def suminf_indicator A
nn_integral_suminf[symmetric] dest: sets.sets_into_space)
lemma emeasure_prod_count_space_single[simp]: "emeasure (count_space UNIV ⨂⇩M count_space UNIV) {x} = 1"
proof -
have [simp]: "⋀a b x y. indicator {(a, b)} (x, y) = (indicator {a} x * indicator {b} y::ennreal)"
by (auto split: split_indicator)
show ?thesis
by (cases x) (auto simp: emeasure_prod_count_space nn_integral_cmult sets_Pair)
qed
lemma emeasure_count_space_prod_eq:
fixes A :: "('a × 'b) set"
assumes A: "A ∈ sets (count_space UNIV ⨂⇩M count_space UNIV)" (is "A ∈ sets (?A ⨂⇩M ?B)")
shows "emeasure (?A ⨂⇩M ?B) A = emeasure (count_space UNIV) A"
proof -
{ fix A :: "('a × 'b) set" assume "countable A"
then have "emeasure (?A ⨂⇩M ?B) (⋃a∈A. {a}) = (∫⇧+a. emeasure (?A ⨂⇩M ?B) {a} ∂count_space A)"
by (intro emeasure_UN_countable) (auto simp: sets_Pair disjoint_family_on_def)
also have "… = (∫⇧+a. indicator A a ∂count_space UNIV)"
by (subst nn_integral_count_space_indicator) auto
finally have "emeasure (?A ⨂⇩M ?B) A = emeasure (count_space UNIV) A"
by simp }
note * = this
show ?thesis
proof cases
assume "finite A" then show ?thesis
by (intro * countable_finite)
next
assume "infinite A"
then obtain C where "countable C" and "infinite C" and "C ⊆ A"
by (auto dest: infinite_countable_subset')
with A have "emeasure (?A ⨂⇩M ?B) C ≤ emeasure (?A ⨂⇩M ?B) A"
by (intro emeasure_mono) auto
also have "emeasure (?A ⨂⇩M ?B) C = emeasure (count_space UNIV) C"
using ‹countable C› by (rule *)
finally show ?thesis
using ‹infinite C› ‹infinite A› by (simp add: top_unique)
qed
qed
lemma nn_integral_count_space_prod_eq:
"nn_integral (count_space UNIV ⨂⇩M count_space UNIV) f = nn_integral (count_space UNIV) f"
(is "nn_integral ?P f = _")
proof cases
assume cntbl: "countable {x. f x ≠ 0}"
have [simp]: "⋀x. card ({x} ∩ {x. f x ≠ 0}) = (indicator {x. f x ≠ 0} x::ennreal)"
by (auto split: split_indicator)
have [measurable]: "⋀y. (λx. indicator {y} x) ∈ borel_measurable ?P"
by (rule measurable_discrete_difference[of "λx. 0" _ borel "{y}" "λx. indicator {y} x" for y])
(auto intro: sets_Pair)
have "(∫⇧+x. f x ∂?P) = (∫⇧+x. ∫⇧+x'. f x * indicator {x} x' ∂count_space {x. f x ≠ 0} ∂?P)"
by (auto simp add: nn_integral_cmult nn_integral_indicator' intro!: nn_integral_cong split: split_indicator)
also have "… = (∫⇧+x. ∫⇧+x'. f x' * indicator {x'} x ∂count_space {x. f x ≠ 0} ∂?P)"
by (auto intro!: nn_integral_cong split: split_indicator)
also have "… = (∫⇧+x'. ∫⇧+x. f x' * indicator {x'} x ∂?P ∂count_space {x. f x ≠ 0})"
by (intro nn_integral_count_space_nn_integral cntbl) auto
also have "… = (∫⇧+x'. f x' ∂count_space {x. f x ≠ 0})"
by (intro nn_integral_cong) (auto simp: nn_integral_cmult sets_Pair)
finally show ?thesis
by (auto simp add: nn_integral_count_space_indicator intro!: nn_integral_cong split: split_indicator)
next
{ fix x assume "f x ≠ 0"
then have "(∃r≥0. 0 < r ∧ f x = ennreal r) ∨ f x = ∞"
by (cases "f x" rule: ennreal_cases) (auto simp: less_le)
then have "∃n. ennreal (1 / real (Suc n)) ≤ f x"
by (auto elim!: nat_approx_posE intro!: less_imp_le) }
note * = this
assume cntbl: "uncountable {x. f x ≠ 0}"
also have "{x. f x ≠ 0} = (⋃n. {x. 1/Suc n ≤ f x})"
using * by auto
finally obtain n where "infinite {x. 1/Suc n ≤ f x}"
by (meson countableI_type countable_UN uncountable_infinite)
then obtain C where C: "C ⊆ {x. 1/Suc n ≤ f x}" and "countable C" "infinite C"
by (metis infinite_countable_subset')
have [measurable]: "C ∈ sets ?P"
using sets.countable[OF _ ‹countable C›, of ?P] by (auto simp: sets_Pair)
have "(∫⇧+x. ennreal (1/Suc n) * indicator C x ∂?P) ≤ nn_integral ?P f"
using C by (intro nn_integral_mono) (auto split: split_indicator simp: zero_ereal_def[symmetric])
moreover have "(∫⇧+x. ennreal (1/Suc n) * indicator C x ∂?P) = ∞"
using ‹infinite C› by (simp add: nn_integral_cmult emeasure_count_space_prod_eq ennreal_mult_top)
moreover have "(∫⇧+x. ennreal (1/Suc n) * indicator C x ∂count_space UNIV) ≤ nn_integral (count_space UNIV) f"
using C by (intro nn_integral_mono) (auto split: split_indicator simp: zero_ereal_def[symmetric])
moreover have "(∫⇧+x. ennreal (1/Suc n) * indicator C x ∂count_space UNIV) = ∞"
using ‹infinite C› by (simp add: nn_integral_cmult ennreal_mult_top)
ultimately show ?thesis
by (simp add: top_unique)
qed
theorem pair_measure_density:
assumes f: "f ∈ borel_measurable M1"
assumes g: "g ∈ borel_measurable M2"
assumes "sigma_finite_measure M2" "sigma_finite_measure (density M2 g)"
shows "density M1 f ⨂⇩M density M2 g = density (M1 ⨂⇩M M2) (λ(x,y). f x * g y)" (is "?L = ?R")
proof (rule measure_eqI)
interpret M2: sigma_finite_measure M2 by fact
interpret D2: sigma_finite_measure "density M2 g" by fact
fix A assume A: "A ∈ sets ?L"
with f g have "(∫⇧+ x. f x * ∫⇧+ y. g y * indicator A (x, y) ∂M2 ∂M1) =
(∫⇧+ x. ∫⇧+ y. f x * g y * indicator A (x, y) ∂M2 ∂M1)"
by (intro nn_integral_cong_AE)
(auto simp add: nn_integral_cmult[symmetric] ac_simps)
with A f g show "emeasure ?L A = emeasure ?R A"
by (simp add: D2.emeasure_pair_measure emeasure_density nn_integral_density
M2.nn_integral_fst[symmetric]
cong: nn_integral_cong)
qed simp
lemma sigma_finite_measure_distr:
assumes "sigma_finite_measure (distr M N f)" and f: "f ∈ measurable M N"
shows "sigma_finite_measure M"
proof -
interpret sigma_finite_measure "distr M N f" by fact
obtain A where A: "countable A" "A ⊆ sets (distr M N f)"
"⋃ A = space (distr M N f)" "∀a∈A. emeasure (distr M N f) a ≠ ∞"
using sigma_finite_countable by auto
show ?thesis
proof
show "∃A. countable A ∧ A ⊆ sets M ∧ ⋃A = space M ∧ (∀a∈A. emeasure M a ≠ ∞)"
using A f
by (intro exI[of _ "(λa. f -` a ∩ space M) ` A"])
(auto simp: emeasure_distr set_eq_iff subset_eq intro: measurable_space)
qed
qed
lemma pair_measure_distr:
assumes f: "f ∈ measurable M S" and g: "g ∈ measurable N T"
assumes "sigma_finite_measure (distr N T g)"
shows "distr M S f ⨂⇩M distr N T g = distr (M ⨂⇩M N) (S ⨂⇩M T) (λ(x, y). (f x, g y))" (is "?P = ?D")
proof (rule measure_eqI)
interpret T: sigma_finite_measure "distr N T g" by fact
interpret N: sigma_finite_measure N by (rule sigma_finite_measure_distr) fact+
fix A assume A: "A ∈ sets ?P"
with f g show "emeasure ?P A = emeasure ?D A"
by (auto simp add: N.emeasure_pair_measure_alt space_pair_measure emeasure_distr
T.emeasure_pair_measure_alt nn_integral_distr
intro!: nn_integral_cong arg_cong[where f="emeasure N"])
qed simp
lemma pair_measure_eqI:
assumes "sigma_finite_measure M1" "sigma_finite_measure M2"
assumes sets: "sets (M1 ⨂⇩M M2) = sets M"
assumes emeasure: "⋀A B. A ∈ sets M1 ⟹ B ∈ sets M2 ⟹ emeasure M1 A * emeasure M2 B = emeasure M (A × B)"
shows "M1 ⨂⇩M M2 = M"
proof -
interpret M1: sigma_finite_measure M1 by fact
interpret M2: sigma_finite_measure M2 by fact
interpret pair_sigma_finite M1 M2 ..
let ?E = "{a × b |a b. a ∈ sets M1 ∧ b ∈ sets M2}"
let ?P = "M1 ⨂⇩M M2"
obtain F :: "nat ⇒ ('a × 'b) set" where F:
"range F ⊆ ?E" "incseq F" "⋃ (range F) = space M1 × space M2" "∀i. emeasure ?P (F i) ≠ ∞"
using sigma_finite_up_in_pair_measure_generator
by blast
show ?thesis
proof (rule measure_eqI_generator_eq[OF Int_stable_pair_measure_generator[of M1 M2]])
show "?E ⊆ Pow (space ?P)"
using sets.space_closed[of M1] sets.space_closed[of M2] by (auto simp: space_pair_measure)
show "sets ?P = sigma_sets (space ?P) ?E"
by (simp add: sets_pair_measure space_pair_measure)
then show "sets M = sigma_sets (space ?P) ?E"
using sets[symmetric] by simp
next
show "range F ⊆ ?E" "(⋃i. F i) = space ?P" "⋀i. emeasure ?P (F i) ≠ ∞"
using F by (auto simp: space_pair_measure)
next
fix X assume "X ∈ ?E"
then obtain A B where X[simp]: "X = A × B" and A: "A ∈ sets M1" and B: "B ∈ sets M2" by auto
then have "emeasure ?P X = emeasure M1 A * emeasure M2 B"
by (simp add: M2.emeasure_pair_measure_Times)
also have "… = emeasure M (A × B)"
using A B emeasure by auto
finally show "emeasure ?P X = emeasure M X"
by simp
qed
qed
lemma sets_pair_countable:
assumes "countable S1" "countable S2"
assumes M: "sets M = Pow S1" and N: "sets N = Pow S2"
shows "sets (M ⨂⇩M N) = Pow (S1 × S2)"
proof auto
fix x a b assume x: "x ∈ sets (M ⨂⇩M N)" "(a, b) ∈ x"
from sets.sets_into_space[OF x(1)] x(2)
sets_eq_imp_space_eq[of N "count_space S2"] sets_eq_imp_space_eq[of M "count_space S1"] M N
show "a ∈ S1" "b ∈ S2"
by (auto simp: space_pair_measure)
next
fix X assume X: "X ⊆ S1 × S2"
then have "countable X"
by (metis countable_subset ‹countable S1› ‹countable S2› countable_SIGMA)
have "X = (⋃(a, b)∈X. {a} × {b})" by auto
also have "… ∈ sets (M ⨂⇩M N)"
using X
by (safe intro!: sets.countable_UN' ‹countable X› subsetI pair_measureI) (auto simp: M N)
finally show "X ∈ sets (M ⨂⇩M N)" .
qed
lemma pair_measure_countable:
assumes "countable S1" "countable S2"
shows "count_space S1 ⨂⇩M count_space S2 = count_space (S1 × S2)"
proof (rule pair_measure_eqI)
show "sigma_finite_measure (count_space S1)" "sigma_finite_measure (count_space S2)"
using assms by (auto intro!: sigma_finite_measure_count_space_countable)
show "sets (count_space S1 ⨂⇩M count_space S2) = sets (count_space (S1 × S2))"
by (subst sets_pair_countable[OF assms]) auto
next
fix A B assume "A ∈ sets (count_space S1)" "B ∈ sets (count_space S2)"
then show "emeasure (count_space S1) A * emeasure (count_space S2) B =
emeasure (count_space (S1 × S2)) (A × B)"
by (subst (1 2 3) emeasure_count_space) (auto simp: finite_cartesian_product_iff ennreal_mult_top ennreal_top_mult)
qed
proposition nn_integral_fst_count_space:
"(∫⇧+ x. ∫⇧+ y. f (x, y) ∂count_space UNIV ∂count_space UNIV) = integral⇧N (count_space UNIV) f"
(is "?lhs = ?rhs")
proof(cases)
assume *: "countable {xy. f xy ≠ 0}"
let ?A = "fst ` {xy. f xy ≠ 0}"
let ?B = "snd ` {xy. f xy ≠ 0}"
from * have [simp]: "countable ?A" "countable ?B" by(rule countable_image)+
have "?lhs = (∫⇧+ x. ∫⇧+ y. f (x, y) ∂count_space UNIV ∂count_space ?A)"
by(rule nn_integral_count_space_eq)
(auto simp add: nn_integral_0_iff_AE AE_count_space not_le intro: rev_image_eqI)
also have "… = (∫⇧+ x. ∫⇧+ y. f (x, y) ∂count_space ?B ∂count_space ?A)"
by(intro nn_integral_count_space_eq nn_integral_cong)(auto intro: rev_image_eqI)
also have "… = (∫⇧+ xy. f xy ∂count_space (?A × ?B))"
by(subst sigma_finite_measure.nn_integral_fst)
(simp_all add: sigma_finite_measure_count_space_countable pair_measure_countable)
also have "… = ?rhs"
by(rule nn_integral_count_space_eq)(auto intro: rev_image_eqI)
finally show ?thesis .
next
{ fix xy assume "f xy ≠ 0"
then have "(∃r≥0. 0 < r ∧ f xy = ennreal r) ∨ f xy = ∞"
by (cases "f xy" rule: ennreal_cases) (auto simp: less_le)
then have "∃n. ennreal (1 / real (Suc n)) ≤ f xy"
by (auto elim!: nat_approx_posE intro!: less_imp_le) }
note * = this
assume cntbl: "uncountable {xy. f xy ≠ 0}"
also have "{xy. f xy ≠ 0} = (⋃n. {xy. 1/Suc n ≤ f xy})"
using * by auto
finally obtain n where "infinite {xy. 1/Suc n ≤ f xy}"
by (meson countableI_type countable_UN uncountable_infinite)
then obtain C where C: "C ⊆ {xy. 1/Suc n ≤ f xy}" and "countable C" "infinite C"
by (metis infinite_countable_subset')
have "∞ = (∫⇧+ xy. ennreal (1 / Suc n) * indicator C xy ∂count_space UNIV)"
using ‹infinite C› by(simp add: nn_integral_cmult ennreal_mult_top)
also have "… ≤ ?rhs" using C
by(intro nn_integral_mono)(auto split: split_indicator)
finally have "?rhs = ∞" by (simp add: top_unique)
moreover have "?lhs = ∞"
proof(cases "finite (fst ` C)")
case True
then obtain x C' where x: "x ∈ fst ` C"
and C': "C' = fst -` {x} ∩ C"
and "infinite C'"
using ‹infinite C› by(auto elim!: inf_img_fin_domE')
from x C C' have **: "C' ⊆ {xy. 1 / Suc n ≤ f xy}" by auto
from C' ‹infinite C'› have "infinite (snd ` C')"
by(auto dest!: finite_imageD simp add: inj_on_def)
then have "∞ = (∫⇧+ y. ennreal (1 / Suc n) * indicator (snd ` C') y ∂count_space UNIV)"
by(simp add: nn_integral_cmult ennreal_mult_top)
also have "… = (∫⇧+ y. ennreal (1 / Suc n) * indicator C' (x, y) ∂count_space UNIV)"
by(rule nn_integral_cong)(force split: split_indicator intro: rev_image_eqI simp add: C')
also have "… = (∫⇧+ x'. (∫⇧+ y. ennreal (1 / Suc n) * indicator C' (x, y) ∂count_space UNIV) * indicator {x} x' ∂count_space UNIV)"
by(simp add: one_ereal_def[symmetric])
also have "… ≤ (∫⇧+ x. ∫⇧+ y. ennreal (1 / Suc n) * indicator C' (x, y) ∂count_space UNIV ∂count_space UNIV)"
by(rule nn_integral_mono)(simp split: split_indicator)
also have "… ≤ ?lhs" using **
by(intro nn_integral_mono)(auto split: split_indicator)
finally show ?thesis by (simp add: top_unique)
next
case False
define C' where "C' = fst ` C"
have "∞ = ∫⇧+ x. ennreal (1 / Suc n) * indicator C' x ∂count_space UNIV"
using C'_def False by(simp add: nn_integral_cmult ennreal_mult_top)
also have "… = ∫⇧+ x. ∫⇧+ y. ennreal (1 / Suc n) * indicator C' x * indicator {SOME y. (x, y) ∈ C} y ∂count_space UNIV ∂count_space UNIV"
by(auto simp add: one_ereal_def[symmetric] max_def intro: nn_integral_cong)
also have "… ≤ ∫⇧+ x. ∫⇧+ y. ennreal (1 / Suc n) * indicator C (x, y) ∂count_space UNIV ∂count_space UNIV"
by(intro nn_integral_mono)(auto simp add: C'_def split: split_indicator intro: someI)
also have "… ≤ ?lhs" using C
by(intro nn_integral_mono)(auto split: split_indicator)
finally show ?thesis by (simp add: top_unique)
qed
ultimately show ?thesis by simp
qed
proposition nn_integral_snd_count_space:
"(∫⇧+ y. ∫⇧+ x. f (x, y) ∂count_space UNIV ∂count_space UNIV) = integral⇧N (count_space UNIV) f"
(is "?lhs = ?rhs")
proof -
have "?lhs = (∫⇧+ y. ∫⇧+ x. (λ(y, x). f (x, y)) (y, x) ∂count_space UNIV ∂count_space UNIV)"
by(simp)
also have "… = ∫⇧+ yx. (λ(y, x). f (x, y)) yx ∂count_space UNIV"
by(rule nn_integral_fst_count_space)
also have "… = ∫⇧+ xy. f xy ∂count_space ((λ(x, y). (y, x)) ` UNIV)"
by(subst nn_integral_bij_count_space[OF inj_on_imp_bij_betw, symmetric])
(simp_all add: inj_on_def split_def)
also have "… = ?rhs" by(rule nn_integral_count_space_eq) auto
finally show ?thesis .
qed
lemma measurable_pair_measure_countable1:
assumes "countable A"
and [measurable]: "⋀x. x ∈ A ⟹ (λy. f (x, y)) ∈ measurable N K"
shows "f ∈ measurable (count_space A ⨂⇩M N) K"
using _ _ assms(1)
by(rule measurable_compose_countable'[where f="λa b. f (a, snd b)" and g=fst and I=A, simplified])simp_all
subsection ‹Product of Borel spaces›
theorem borel_Times:
fixes A :: "'a::topological_space set" and B :: "'b::topological_space set"
assumes A: "A ∈ sets borel" and B: "B ∈ sets borel"
shows "A × B ∈ sets borel"
proof -
have "A × B = (A×UNIV) ∩ (UNIV × B)"
by auto
moreover
{ have "A ∈ sigma_sets UNIV {S. open S}" using A by (simp add: sets_borel)
then have "A×UNIV ∈ sets borel"
proof (induct A)
case (Basic S) then show ?case
by (auto intro!: borel_open open_Times)
next
case (Compl A)
moreover have *: "(UNIV - A) × UNIV = UNIV - (A × UNIV)"
by auto
ultimately show ?case
unfolding * by auto
next
case (Union A)
moreover have *: "(⋃(A ` UNIV)) × UNIV = ⋃((λi. A i × UNIV) ` UNIV)"
by auto
ultimately show ?case
unfolding * by auto
qed simp }
moreover
{ have "B ∈ sigma_sets UNIV {S. open S}" using B by (simp add: sets_borel)
then have "UNIV×B ∈ sets borel"
proof (induct B)
case (Basic S) then show ?case
by (auto intro!: borel_open open_Times)
next
case (Compl B)
moreover have *: "UNIV × (UNIV - B) = UNIV - (UNIV × B)"
by auto
ultimately show ?case
unfolding * by auto
next
case (Union B)
moreover have *: "UNIV × (⋃(B ` UNIV)) = ⋃((λi. UNIV × B i) ` UNIV)"
by auto
ultimately show ?case
unfolding * by auto
qed simp }
ultimately show ?thesis
by auto
qed
lemma finite_measure_pair_measure:
assumes "finite_measure M" "finite_measure N"
shows "finite_measure (N ⨂⇩M M)"
proof (rule finite_measureI)
interpret M: finite_measure M by fact
interpret N: finite_measure N by fact
show "emeasure (N ⨂⇩M M) (space (N ⨂⇩M M)) ≠ ∞"
by (auto simp: space_pair_measure M.emeasure_pair_measure_Times ennreal_mult_eq_top_iff)
qed
end
``` | 18,356 | 48,467 | {"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-2023-50 | latest | en | 0.601711 |
http://www.wowinterface.com/forums/showthread.php?t=41795&page=2 | 1,455,362,952,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701166570.91/warc/CC-MAIN-20160205193926-00211-ip-10-236-182-209.ec2.internal.warc.gz | 750,591,434 | 22,723 | Distance formula to calculate speed between two points - Page 2 - WoWInterface
11-15-11, 08:44 PM #21
A Molten Giant
Join Date: Dec 2005
Posts: 772
It's not quite on topic, but let me ask you smart guys;
This is the data:
A -> B
B -> C
C -> D
E -> C
A -> E
B -> A
C -> B
C -> A
A -> C
You see, if you wish to go from A to D you can go:
(1) A->B->C->D
(2) A->C->D
(3) many others involving cycles like A->C->B->A->C->D and so forth...
In any case, I have this system in MySQL:
{PrimKey, X, Y} where PrimKey is an integer, X and Y are the node values. For example A->B is {1, A, B}, then B->C becomes {2, B, C} and so forth.
This is graph theory, traversal of graphs but considering there are cycles this makes it a bit tricky.
I have to make a class i.e. "Map" to help me keep order of who comes and goes to who. This way if I want to go from A to D I can return a list where I get various paths leading to D without the backtracking (A->C->B->A->C->D e.g.) and it shouldn't stuck looping indefinitely.
Do you have some ideas how I should structure this class?
I was thinking something along the lines of:
class Map {
nodes = array();
addChild(getParent(node::letter), node) // find A in nodes then add this node as that parents child, if not then add the child as a new root-node with node::letter as index key
}
getRoutes(start, end) {
// traverse the nodes, find any paths starting at start and ending at end, paths not leading to the end are ignored, this I reckon will be a recursive function
}
}
Very rough, maybe someone has a better idea or a tip like example code they know of on the net. :P
11-16-11, 01:52 AM #22
SDPhantom
A Pyroguard Emberseer
Join Date: Jul 2006
Posts: 1,339
A method I could think of is to have it scan out, branching off every time it has a choice based on direction from the origin point. I've had experience in writing functions that run recursively though data without the need to call itself. This usually involves building and managing your own data in a stack-like data structure. You end up using more memory instead of increasing your call stack, which is the best way to avoid stack overflows. It'll take a while to write it all from scratch. I'm assuming this would be in PHP?
__________________
"All I want is a pretty girl, a decent meal, and the right to shoot lightning at fools." -Anders (Dragon Age: Origins - Awakening)
11-16-11, 02:30 AM #23
A Molten Giant
Join Date: Dec 2005
Posts: 772
Indeed, php.
11-16-11, 03:34 PM #24
SDPhantom
A Pyroguard Emberseer
Join Date: Jul 2006
Posts: 1,339
I'm looking into writing some code that'll allow you to add paths with a cost metric and the pathfinding AI will choose the lowest cost route. To conserve processing time, it'll "give up" on a path if its cost exceeds that of a route it already discovered.
This method will allow you to customize what you use for a path's cost metric, whether it's the copper cost, distance, or even number of hops. The first two should be easy to figure out, but to do hops, you'll need to set the cost of all paths to a single number greater than zero, like 1.
I may have the pathfinding AI accept a max cost argument that it'll automatically "give up" on if the cost is exceeded.
__________________
"All I want is a pretty girl, a decent meal, and the right to shoot lightning at fools." -Anders (Dragon Age: Origins - Awakening)
Last edited by SDPhantom : 11-16-11 at 03:42 PM.
11-16-11, 06:03 PM #25
SDPhantom
A Pyroguard Emberseer
Join Date: Jul 2006
Posts: 1,339
I have a working prototype running and here's the debug info that it's printing.
The following calls were made to generate this.
Code:
```addPath("A","B",1);
findRoute("B","E");```
Path Data:
Code:
```Create: A>B(1)
Create: B>A(1)
Create: C>A(1)
Create: E>C(1)
Array
(
[A] => Array
(
[0] => Array
(
[0] => B
[1] => 1
)
[1] => Array
(
[0] => C
[1] => 1
)
[2] => Array
(
[0] => E
[1] => 1
)
)
[b] => Array
(
[0] => Array
(
[0] => A
[1] => 1
)
[1] => Array
(
[0] => C
[1] => 1
)
)
[C] => Array
(
[0] => Array
(
[0] => A
[1] => 1
)
[1] => Array
(
[0] => B
[1] => 1
)
[2] => Array
(
[0] => D
[1] => 1
)
)
[E] => Array
(
[0] => Array
(
[0] => C
[1] => 1
)
)
)```
Routing Steps:
Code:
```Routing B>E
Iterating Stack: 0@0=B>A 0+1=1
Push Stack 1
Iterating Stack: 1@0=A>B 1+1=2
Iterating Stack: 1@1=A>C 1+1=2
Push Stack 2
Iterating Stack: 2@0=C>A 2+1=3
Iterating Stack: 2@1=C>B 2+1=3
Iterating Stack: 2@2=C>D 2+1=3
Pop Stack 2
Iterating Stack: 1@2=A>E 1+1=2
Route Found
Array
(
[0] => 2
[1] => B
[2] => A
[3] => E
)
Pop Stack 1
Iterating Stack: 0@1=B>C 0+1=1
Push Stack 1
Iterating Stack: 1@0=C>A 1+1=2
Cost Exceeded
Iterating Stack: 1@1=C>B 1+1=2
Cost Exceeded
Iterating Stack: 1@2=C>D 1+1=2
Cost Exceeded
Pop Stack 1
Pop Stack 0```
Returned Route:
Code:
```Array
(
[0] => B
[1] => A
[2] => E
)```
__________________
"All I want is a pretty girl, a decent meal, and the right to shoot lightning at fools." -Anders (Dragon Age: Origins - Awakening)
Last edited by SDPhantom : 11-16-11 at 06:09 PM.
11-16-11, 06:25 PM #26
SDPhantom
A Pyroguard Emberseer
Join Date: Jul 2006
Posts: 1,339
Polished off a few checks and removed the debug code. Here's the prototype.
PHP Code:
``` \$paths=array();function addPath(\$src,\$dst,\$cost=1){ global \$paths; if(\$src==\$dst || \$cost<0) return NULL; if(\$paths[\$src]){ foreach(\$paths[\$src] as \$i) if(\$i[0]==\$dst) return false; }else \$paths[\$src]=array(); \$paths[\$src][]=array(\$dst,\$cost); return true;}function findRoute(\$src,\$dst,\$maxcost=NULL,\$maxhops=NULL){ global \$paths; if(!\$paths[\$src]) return NULL; \$found=false; foreach(\$paths as \$i) foreach(\$i as \$j) if(\$j[0]==\$dst){\$found=true;break;} if(!\$found) return false; \$pstack=array(array(\$src,0,0)); while((\$id=count(\$pstack)-1)>=0){ \$ptr=&\$pstack[\$id]; if(\$node=\$paths[\$ptr[0]][\$ptr[1]++]){ \$cost=\$ptr[2]+\$node[1]; if(\$maxcost===NULL || \$cost<=\$maxcost){ if(\$node[0]==\$dst){ \$maxcost=\$cost; \$route=array(); foreach(\$pstack as \$i) \$route[]=\$i[0]; \$route[]=\$dst; array_pop(\$pstack); }elseif(\$paths[\$node[0]] && (\$maxhops===NULL || (\$id+1)<\$maxhops)){ \$unique=true; foreach(\$pstack as \$i) if(\$i[0]==\$node[0]){\$unique=false;break;} if(\$unique) array_push(\$pstack,array(\$node[0],0,\$cost)); } } }else array_pop(\$pstack); } return \$route?\$route:false;} ```
Notes:
addPath() returns true on success, false on duplicates, and NULL on error.
findRoute() returns an array on success, false if no path found, and NULL on error.
__________________
"All I want is a pretty girl, a decent meal, and the right to shoot lightning at fools." -Anders (Dragon Age: Origins - Awakening)
Last edited by SDPhantom : 11-17-11 at 03:39 PM.
11-16-11, 08:19 PM #27
A Molten Giant
Join Date: Dec 2005
Posts: 772
Wow SDPhantom, you really put effort into this code, thanks!
I got the basic idea, just trying to sort off cycles now, as it gets stuck looping in a circle thinking it goes forward. I need to add a "visited" check to skip going where we've already been I reckon.
Last edited by Vlad : 11-16-11 at 08:21 PM.
11-16-11, 08:52 PM #28
SDPhantom
A Pyroguard Emberseer
Join Date: Jul 2006
Posts: 1,339
Strange, can you post a copy of the paths you're registering? When it checks a path, it looks back into the stack to see if the node isn't already in there. With a bigger data set, it would often recheck paths again when going back up the stack. You can try setting the maxcost to limit how far it runs away with this.
__________________
"All I want is a pretty girl, a decent meal, and the right to shoot lightning at fools." -Anders (Dragon Age: Origins - Awakening)
11-16-11, 10:35 PM #29
A Molten Giant
Join Date: Dec 2005
Posts: 772
http://pastebin.com/e0LfewY1
For example if I do this:
findRoute(2, 15);
The output of the id is this:
Code:
`0>1>1>2>2>2>3>3>3>3>4>4>4>4>5>5>5>5>5>6>6>7>7>7>6>5>6>6>6>6>5>4>5>6>7>7>7>7>7>7>8>8>8>8>7>6>6>5>5>4>5>6>6>6>7>7>8>9>9>10>10>11>11>11>12>12>12>12>13>13>14>14>14>14>15>15>16>16>16>16>16>17>17>18>19>20>20>21>21>22>22>23>24>25>25>26>26>26>26>27>27>27>28>28>28>29>29>29>30>31>31>31>31>30>30>29>28>27>27>28>29>29>30>31>31>31>31>30>30>30>29>28>28>28>27>...`
It keeps repeating after the dots... going back to 26 then up to 31 and back to 26, e.g. It's a bit odd indeed, are you getting that too?
11-17-11, 03:56 AM #30
SDPhantom
A Pyroguard Emberseer
Join Date: Jul 2006
Posts: 1,339
The stack ID only reflects the size of the stack at the time, it'll always be going up and down until it's all done. It's pretty much equal to how many hops it's at in its current path.
__________________
"All I want is a pretty girl, a decent meal, and the right to shoot lightning at fools." -Anders (Dragon Age: Origins - Awakening)
11-17-11, 03:41 PM #31
SDPhantom
A Pyroguard Emberseer
Join Date: Jul 2006
Posts: 1,339
I looked at the DBC data over and discovered that no paths lead to #15. I added in a check for this before it starts pathing and an additional maxhops argument. I would suggest using it to cut down on processing time. You may use NULL to bypass the maxcost argument.
For example, this'll trace an alliance route from Booty Bay to Light's Hope Chapel with a max of 10 hops.
Code:
`findNode(19,67,NULL,10);`
The code in my previous post has been updated with these changes.
PS: You should only load in paths accessible to a specific faction to help in processing if by chance it's queried for a route cross faction. It can also wander over into cross faction routes if it hits shared paths. It seems the value following the unknown horde/alliance IDs is indeed a 3-bit faction flag. Bit 1 is alliance and bit 2 is horde. I see bit 3 set in a few paths, but it's unknown what that means at this point.
__________________
"All I want is a pretty girl, a decent meal, and the right to shoot lightning at fools." -Anders (Dragon Age: Origins - Awakening)
Last edited by SDPhantom : 11-17-11 at 04:00 PM.
11-17-11, 05:02 PM #32
A Molten Giant
Join Date: Dec 2005
Posts: 772
Hmm, really? Maybe I used a node no longer accessible, should have used the lighthope chapel one, sorry for my mistake too, kind of set you on the wrong path.
Yes, you are right about the faction thing, should not mix them up.
*Edit*
Very nice mate, couldn't have done this all by myself -tough I should, had graph theory in Java a year ago.
Anyway, got it working on my end as well, made a class for it and it only reads in paths that are accessible for the faction when I am flying around in my virtual WoW.
*Edit 2*
I ran some calculations for 10 paths from Stormwind, the results were (time in HH:MM:SS):
Code:
```Stormwind, Elwynn -> Sentinel Hill, Westfall takes 00:01:11
Stormwind, Elwynn -> Lakeshire, Redridge takes 00:01:50
Stormwind, Elwynn -> Ironforge, Dun Morogh takes 00:03:32
Stormwind, Elwynn -> Menethil Harbor, Wetlands takes 00:05:28
Stormwind, Elwynn -> Thelsamar, Loch Modan takes 00:05:13
Stormwind, Elwynn -> Darkshire, Duskwood takes 00:01:53
Stormwind, Elwynn -> Refuge Pointe, Arathi takes 00:06:53
Stormwind, Elwynn -> Booty Bay, Stranglethorn takes 00:03:15
Stormwind, Elwynn -> Aerie Peak, The Hinterlands takes 00:07:29
Stormwind, Elwynn -> Nethergarde Keep, Blasted Lands takes 00:02:53```
Checking the InFlight data that contains the proper times (+-1/2 seconds difference, since latency has something to say on the arrival time as well), in any case these were the differences in seconds:
Code:
```Stormwind, Elwynn -> Sentinel Hill, Westfall takes (71-73) -2
Stormwind, Elwynn -> Lakeshire, Redridge takes (110-113) = -3
Stormwind, Elwynn -> Ironforge, Dun Morogh takes (212-215) = -3
Stormwind, Elwynn -> Menethil Harbor, Wetlands takes (328-233) = 95
Stormwind, Elwynn -> Thelsamar, Loch Modan takes (313-212) = 101
Stormwind, Elwynn -> Darkshire, Duskwood takes (113-116) = -3
Stormwind, Elwynn -> Refuge Pointe, Arathi takes (413-375) = 38
Stormwind, Elwynn -> Booty Bay, Stranglethorn takes (195-198) = -3
Stormwind, Elwynn -> Aerie Peak, The Hinterlands takes (449-412) = 37
Stormwind, Elwynn -> Nethergarde Keep, Blasted Lands takes (173-176) = -3```
Those that are heavily out of mark like Menethil Harbor, Thelsamar, Refuge Pointe, Aerie Peak, those I think are because of my coding of the faction check, if not then something else must have gone wrong, since the others are up to 3 seconds quicker than what InFlight says the times are, 3 seconds is not bad at all considering it's all calculated trough formulas and not stopwatch.
With some fine tuning like adding delay because of the "wobble" effect (large angles, high distances, create a slight deviation in the distance calculation, this is not yet addressed in the code), and proper faction checks to allow neutral paths when flying as any faction but restrict the path finding to either Alliance or Horde and avoid mixing those.
Perhaps these huge deviations may be due to result of flights going quicker than the constant speed I calculate with (30+1/3), maybe those paths used quicker flight speeds, so this also has to be added in the code.
This was my "report" on the matter for now, when this is done I'll share my code for anyone to use that needs to calculate taxi flight times. :P
Last edited by Vlad : 11-17-11 at 07:00 PM.
11-17-11, 08:44 PM #33
Seerah
Fishing Trainer
Join Date: Oct 2006
Posts: 10,032
Do your calculations account for when the taxi circles before landing?
__________________
"You'd be surprised how many people violate this simple principle every day of their lives and try to fit square pegs into round holes, ignoring the clear reality that Things Are As They Are." -Benjamin Hoff, The Tao of Pooh
11-17-11, 09:16 PM #34
A Molten Giant
Join Date: Dec 2005
Posts: 772
Yes it should, hard to tell atm because I need to implement the various speeds on some flightpaths, hehe.
For many regular gryphon based flightpaths with regular speeds the calculations return answers within 3 seconds of the InFlight data, I guess the last 3 seconds are due to latency; it takes a up to a second to start flying and landing and some minor delays due to wobble effect or when jumping from a node to node in general.
I think Ironforge may be the cause of the weird timers on my end, probably because the game does not fly in the city when flying on a connected path, it skips some nodes that I include in the calculations most likely, need to research more.
11-17-11, 10:21 PM #35
Seerah
Fishing Trainer
Join Date: Oct 2006
Posts: 10,032
Ah, yes, that would definitely be a contributing factor. SW, too.
__________________
"You'd be surprised how many people violate this simple principle every day of their lives and try to fit square pegs into round holes, ignoring the clear reality that Things Are As They Are." -Benjamin Hoff, The Tao of Pooh
11-19-11, 02:22 PM #36
A Molten Giant
Join Date: Dec 2005
Posts: 772
The API returns different values but very small changes, decimals only.
Still, at long paths this may change the time up to 10 seconds at worst, maybe this has something to do with lag compensation?
I.e. if the average player ms is 500 then the game would make up for the time lost in between node connections (i.e. sw->goldshire->redrige), each "->" jump would need to be confirmed with the server so if the mount visually flies quicker and the client comes at the last node and requests the next path, it has accounted for the slight delay ocuring so you wouldn't loose time.
Somehow it doesn't make any sense because the location of the player is anyway decided by the server, that explains why I've sometimes seen me land earlier than I should by having my flight just teleport me the last 40 yards instead of flying and landing like it should have.
The gryphons (creature 541) flies between 30.1 to 30.4 so any decimals in between, very weird and hard to account for when flying, thus even tracking the time with a stopwatch would always return 1-3 seconds differences between flights due to this randomness...
Still, some seconds off the mark is not a big deal considering you still know you got over 13 minutes of flying if you want to fly from Booty Bay to Eastern Plaguelands.
11-25-11, 07:21 PM #37
Saiket
A Chromatic Dragonspawn
Join Date: Jul 2008
Posts: 152
I spent a weekend trying to figure out the flight-path mechanics, and while I did come pretty close to accurately modeling them, I'm stumped by a few issues. Since I doubt I'll make anymore progress without a big time commitment, here's what I've found so far:
1. WoW smooths the paths between TaxiPathNode points using splines—specifically, Catmull-Rom splines. I wrote a test addon so I could experimentally compare expected and measured flightpaths, and Catmull-Rom seems 100% accurate.
Cyan line is the expected path, red line is the measured path, alternating yellow and purple sets of dots mark joined path splines' control points, yellow gridlines are map chunk boundaries, and the red crosshair centers on the player.
2. For multi-part flights, WoW takes the shortest combination of paths, unless a direct route is available. The "length" of a path is the length of the Catmull-Rom spline in 3D space, found through integration.
3. Where two paths join together, WoW seems to dynamically strip away spline control points between them that would make too sharp a turn. I didn't attempt to reverse engineer the algorithm, since it would just be a bunch of guesswork and tweaking without WoW's source code available. My test addon defines a function GetSkippedNodes in case anyone wants to attempt to implement this properly. For now, it returns placeholder values 1,1 to skip one innermost point from each joined path. Here are a few examples where paths join incorrectly:
While I was taking screenshots for this post, I noticed that last example where WoW actually makes a tight turn. Who knows.
4. I couldn't find any way to estimate flight-path speeds. It's definitely not constant across all paths, and instead seems to be determined by the node you depart from. The NPC ID of your flight-path mount (from TaxiNodes) works the same way, but I don't know if your mount affects your speed here. Even if it does, I couldn't find a lookup for NPC speed in my cache files or the DBCs.
I have noticed that flight-path speed appears to vary, but GetUnitSpeed doesn't reflect it. I assume any visual speed change is just rubber-banding when you periodically synchronize your flight-path progress with the server.
Overall it's still pretty inaccurate because of lag, unknown speed, and the path joint issue. Oh well, it was still a fun project.
My test mod is attached if anyone wants to play with it. "/taxi" opens the grid, and you can navigate around sort of like Google Maps. If you talk with a flight master while the grid is open, hovering over nodes will plot their expected paths.
Attached Thumbnails
Attached Files
TaxiGraph.zip (1.00 MB, 556 views)
11-25-11, 07:34 PM #38
A Molten Giant
Join Date: Dec 2005
Posts: 772
Very informative post, thanks for sharing your findings!
Very interesting, catmull-rom splines, haven't heard of that until now, gotta look into this!
WoWInterface » Distance formula to calculate speed between two points
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts vB code is On Smilies are On [IMG] code is On HTML code is Off | 5,784 | 19,822 | {"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} | 2.96875 | 3 | CC-MAIN-2016-07 | longest | en | 0.92066 |
http://jnrm.srbiau.ac.ir/article_10898.html | 1,586,415,501,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371830894.88/warc/CC-MAIN-20200409055849-20200409090349-00010.warc.gz | 84,436,404 | 9,782 | # Partial second-order subdifferentials of -prox-regular functions
Document Type: research paper
Authors
1 Ph.D. student, Department of Mathematics, Sahand University of Technology, Tabriz, Iran
2 Department of Applied Mathematics, School of Mathematics Science, University of Tabriz, Tabriz, Iran
Abstract
Although prox-regular functions in general are nonconvex, they possess properties that one would expect to find in convex or lowerC2 functions. The class of prox-regular functions covers all convex functions, lower C2 functions and strongly amenable functions. At first, these functions have been identified in finite dimension using proximal subdifferential. Then, the definition of prox-regular functions have been developed in Banach and Hilbert spaces. In this paper, the parametric prox-regular functions are defined using limiting subdifferentials. Also, a partial second-order subdifferential is defined here for extended real valued functions of two variables corresponding to its variables through coderivatives of first-order partial subdifferential mappings. Then, relations between maximal monotonicity of the partial first-order subdifferentials of these functions and the positive-semidefiniteness of the coderivatives of partial first order subdifferential mapping are investigated. Finally, we present necessary and sufficient conditions for ∂ -prox-regular functions to be convex based on positive-semidefiniteness of the partial second-order subdifferentials mappings.
Keywords
Article Title [Persian]
### زیرمشتقات جزئی مرتبه دوم توابع -تقریباً-منظم
Authors [Persian]
• سمیه نادی 1
• جواد وکیلی 2
1 دانشجوی دکتری، گروه ریاضی، دانشگاه صنعتی سهند، تبریز، ایران.
2 استادیار، گروه ریاضی کاربردی، دانشگاه تبریز، تبریز، ایران.
Abstract [Persian]
با اینکه توابع تقریباً-منظم در حالت کلی محدب نیستند، ولی خصوصیاتی را دارا هستند که انتظار میرود در توابع محدب و یا C2- پایینی یافت شود. کلاس توابع تقریباً-منظم، شامل توابع محدب، C2- پایینی، قویا متمایل و... است. این توابع در ابتدا روی فضاهای با بعد متناهی و با استفاده از زیرمشتق تقریبی تعریف شدند و سپس تعریف این توابع به روی فضاهای باناخ و هیلبرت گسترش داده شد. در این مقاله، توابع تقریباً-منظم پارامتری با استفاده از زیرمشتق حدی تعریف میشوند. همچنین به تعریف زیرمشقات جزئی مرتبه دوم توابع دو متغیره نسبت به متغییرهایشان با استفاده از هممشتق نگاشت زیرمشتق مرتبه اول میپردازیم. سپس ارتباط بین یکنوایی ماکسیمال زیرمشتقات جزئی مرتبه اول این توابع با نیممعین مثبت بودن نگاشت هممشتق زیرمشتق جزئی مرتبه اول بررسی میشود. سرانجام شرایط لازم و کافی برای محدب بودن توابع ∂-تقریباً-منظم برحسب نیممعین مثبت بودن نگاشت زیرمشتقات جزئی مرتبه دوم ارائه میدهیم.
Keywords [Persian]
• توابع تقریباً-منظم
• نگاشت یکنوای ماکسیمال
• زیرمشتقات جزئی مرتبه دوم
• هممشتق
• آنالیز تغییرات | 764 | 2,760 | {"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-2020-16 | latest | en | 0.341778 |
http://www.markedbyteachers.com/gcse/maths/i-am-going-to-investigate-the-relationship-between-height-and-weight-for-both-male-and-females-attending-mayfield-high-school-in-ks4-and-ks3.html | 1,526,971,566,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864626.37/warc/CC-MAIN-20180522053839-20180522073839-00162.warc.gz | 433,093,090 | 20,263 | • Join over 1.2 million students every month
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• Level: GCSE
• Subject: Maths
• Word count: 3132
# I am going to investigate the relationship between height and weight for both male and females attending Mayfield high school in ks4 and Ks3.
Extracts from this document...
Introduction
Math's Coursework Aim: I am going to investigate the relationship between height and weight for both male and females attending Mayfield high school in ks4 and Ks3. Introduction: For this investigation I am using the data from a school called Mayfield High School. This is a secondary school consisting of both male and female with an overall population of 1183 students, having 579 girls and 604 boys. This school starts from yr7 up to yr11 and contains a variety of records such as, surname, family name, height, weight, age, IQ, ks3 and ks4 results and so on. The reason I've decided to choose height and weight is because I thought it is most probable to influence each other. I will undertake the average between height and weight for female and male. Out of the 1183 students I will pick 180, 90 boys and 90 girls from Mayfield High School. An important factor is having my data sample and unbiased so I will also collect the data for gender too and than do a random stratified sample which would give me reports from each gender year group, in correlation to their percentage of the school size. After I have gathered the information I will put it all in tallies and frequency tables, which I would than, find the mode, mean, median and range for height and weight. Finally I will put everything into different graphs and tables. I will make sure to do this for each set so that my data collected will be understandable. ...read more.
Middle
Tally Frequency 130h-140 3 20w-30 3 140h-150 8 30w-40 10 150h-160 23 40w-50 18 160h-170 28 50w-60 29 170h-180 18 60w-70 21 180h-190 8 70w-80 8 190h-200 2 80w-90 1 Below is the frequency table of the weight and height of girls: Height (m) Tally Frequency Weight (w) Tally Frequency 130h-140 2 20w-30 0 140h-150 12 30w-40 5 150h-160 18 40w-50 31 160h-170 40 50w-60 44 170h-180 14 60w-70 9 180h-190 4 70w-80 1 190h-200 0 80w-90 0 Graphs to show height and weight: Since the data I have produced is grouped into class intervals I think it would be a good idea to produce and record it into leaf and stem diagrams. The main advantage of this is it would make it easier to read off the median data. I will be using stem leaf diagrams and frequency density tables below: Boys Height: Stem Leaf Frequency 1.3 2,6,6 3 1.4 1,2,2,6,7,7,8,8 8 1.5 0,0,2,2,2,2,2,2,2,4,4,4,4,5,5,5,5,5,5,6,7,8,9 23 1.6 0,0,0,0,0,0,0,1,2,2,2,2,2,2,3,3,3,5,5,5,6,6,6,7,7,7,7,8 28 1.7 0,0,0,1,1,1,1,2,2,2,3,5,5,5,5,5,7,7 18 1.8 0,0,0,1,5,6,8,8 8 1.9 0,1 2 Height and weight are continuous data so I can easily analyze the height of boys the same way as weight, so recording it on a histogram would be a good idea. Histogram of boys height: By drawing frequency polygons on the same graph it can be useful to compare continuous data. Frequency polygon of boys height: Boys weight: Stem Leaf Frequency 20 5,6,6 3 30 2,5,5,6,7,7,8,8,8,8 10 40 0,0,0,0,0,1,1,4,4,4,5,5,5,5,5,8,9,9 18 50 0,0,0,0,0,0,1,1,2,2,2,2,2,2,2,3,6,6,6,6,7,7,7,7,7,7,8,8,9 29 60 0,0,0,0,0,0,0,0,0,0,0,3,4,4,5,6,6,6,8,8,9 21 70 0,0,0,2,3,5,5,7 8 80 2 1 Histogram of boys weight: Again I will draw a frequency ...read more.
Conclusion
You can calculate the percentage of students that will have a height with a given range. For example to find the % of students that will have a height with a given range. For example to find the % of boys who have a height between 150-160 cm. Using cumulative frequency curve we know that in the sample-----boys have a height up to 150 cm and ......boys have a height up to 160cm. So...-....=....boys in between the range 150cm to 160cm. So we can say that..../.... =.....% of boys in the school will have a height between,,,. Next I calculated the standard deviation. Standard deviation is a type of statistic that shows you how closely all the various examples are clustered around the data. It shows how spread out a data is from the mean. Below I have outlined how you can calculate standard deviation: For the value of x (midpoint) of the class interval subtract the overall average x" from x, and then square the result and divide it by the frequency. Add up all the values and than divide that result by the sum of all the frequencies. Finally square root the last number. A formula for this is shown below: Now I will calculate the standard deviation of the boys' height: Standard deviation= Standard deviation for boys' height = Now I will calculate the standard deviation of the girls height. Standard deviation= Standard deviation for girls' height = Now I will calculate the standard deviation of the boys' weight: Standard deviation= Standard deviation for boys' weight= Standard deviation= Standard deviation for girls' weight= ...read more.
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# Related GCSE Height and Weight of Pupils and other Mayfield High School investigations essays
1. ## mayfield high statistics coursework
there are 10 Both Handed so these student have been left out of my investigation for this hypothesis. As I have found Right Handed Student is common for most student. Right Handed Students= 922 Divided By 1173 multiplied by 20 = 16 Left Handed Students= 251 Divided by 1173 Multiplied
2. ## Mayfield. HYPOTHESIS 1: Boys at Mayfield School are Taller and Weigh more on ...
between 130- 139 cm, which shows that my data may be misleading or there is a lapse in growth. Both Girls and Boys have average heights and are fairly balanced and of equal size. Although my Sample shows that boys on some occasion grow to above average height such as
1. ## Maths Statistics Coursework - relationship between the weight and height
the girls' heights are all around the mean and are all about the same. I haven't done any calculations for the weights for the students because it is not relevant to what I am interpreting to help me answer my hypotheses.
2. ## Conduct an investigation comparing height and weight from pupils in Mayfield School.
(Please refer to the box-and-whisker diagram on page 20). The box-and-whisker shows that the boys interquartile range is 2cm less than the girls interquartile range. This suggests that the girls height were more spread out than the boys heights. The median for the girls is 154cm.
1. ## Determine the relationship between the range of the jump achieved by the ski jumper ...
Analysing the graphs: From my results, I have done a couple of graphs that can, help and provide equations that may help prove the success of my work. Using the theoretical equations, it became clear to me that the relationship between the dropping height on slope and the range is a square root relationship.
2. ## The relationship between height and weight for students of Mayfield High School.
1,1,2,2,2,2,5,7,7,7,8,9 12 1.70 0,0,1,2,2,3,5 7 1.80 0,0,0,4,5,56 7 1.90 2.00 0 1 Boys Weight Stem Leaf Frequency 30 5, 8 2 40 2,7,8 3 50 0,0,0,0,0,1,4,6,7 9 60 0,0,2,2,3,4,6,8 8 70 2,2,3,3,6 5 80 0,6 2 90 2, 1 Girls Height Stem Leaf Frequency 1.30 1.40 1.50 2,2,4,5,6,7,8 7 1.60
1. ## I would like to know whether there is a link between ability in Maths ...
8 Male 5 5 8 Male 4 4 7 Male 3 3 7 Female 5 4 7 Female 5 5 8 Male 4 4 9 Female 3 3 7 Male 4 5 8 Female 4 5 8 Female 4 5 8 Female 4 4 7 Male 5 4 7 Female
2. ## we can see if there is any correlation between a person's height and weight ...
I will be doing this via selective sampling. I chose to do selective sampling because, stratified sampling is used to fairly work out how many certain random samples from (e.g.) a year you take from a piece of data. In this case there is already an equal and fair amount
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• Ideas and feedback to | 2,632 | 8,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-22 | latest | en | 0.931598 |
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Discussion in 'Electronic Design' started by Mike, Oct 17, 2004.
1. ### MikeGuest
Hello everybody!
I'm seeking help in building a little device that will allow me to
control speed on high current fans from a low current fan header found
on motherboards.
Preface: The fan header on motherboards was designed to run small
fans, typically around 5W. Most motherboards nowadays support fan
speed control via pulse width modulation. Speed is controlled by
software which reads various thermistors (ex. CPU, hard drives,
ambient…)
What I basically want to do is have a high current 12v input from the
power supply, pulse width modulated low current 12v input from the
motherboard header, and a high current pulse width modulated output.
(Preferably with parts available from the neighborhood RatShack.)
2. ### Al BorowskiGuest
Since its just a fan, I suppose a simple transistor would work. It
depends on what the motherboard actually modulates - the 12V supply, or
the fan earth return. If the 12V is modulated an NPN transistor should
do the trick
12V
|
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|
FAN +
FAN -
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___ |/
+12V from MB------|___|-----|
|>
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0V
created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de
cheers,
Al
4. ### MikeGuest
If the 12V is modulated an NPN transistor should
Al, yes it is 12V modulated on 3 different motherboards I tested. I
also did find one interesting bit of information while testing this
out.
It seems there is a capacitor before the header, as when I was
measuring voltage on the header after turning it off via software, the
voltage took a good 3 minutes to settle at around 0.3V.
So I have 3 questions:
(please be patient as I have no formal education in electronics
engineering or prior experience :-] )
1.) I'm assuming that the capacitor before the header will make the
single transistor approach impossible, unless a load is presented on
the header. Resistor? Is that a good approach? And how would I
calculate watts from ohms? Given voltage.
2.) I found an old NPN TIP120 transistor laying around and tested it.
I'm getting 420 ohms between the emitter ad the collector and 1.6k
ohms between collector and emitter. Is it dead?
3.) If it is, would a new TIP120 be good for what I'm trying to do?
How many watts (ballpark) would it be able to handle on a small
heatsink inside a sealed project box?
5. ### Al BorowskiGuest
Hello,
If you are just measuring with a multimeter, you might just be measuring
the stray capacitance?
Theres a fair chance the loading from the transistor will be enough.
Resistor? Is that a good approach? And how would I
I'm tired, so theres likely to be a mistake in this, but here's what I
would do (note: The following are _very_ rough 'ballpark' figures that
should be OK for a one-off).
1) Assume that the 'capacitor' is just capacitance, and thus has a very
small value. Assume that 100ma current flow is enough to discharge the
capacitence quickly enough
2) Since I = .1, and V = 12, use ohm's law to get a resistor to use. V =
I*R, R = V / I, R = 120 ohms.
3) check the resistors power rating. P = V * I = 12 * .1 = 1.2W.
However, considering that the PWM probably won't always be at 100%, you
can probably get away with 1W. Or play it safe and use 2 1W resistors.
4) hook up the circuit as follows
12V
|
|
|
FAN +
FAN -
|
|
|
|
|
120 ohms | Collector
___ |/
+12V from MB------|___|-----|
Base |>
| Emitter
|
|
|
0V
No idea sorry.
It looks like it should be OK.
What kind of box? Metal?
No problem, but as I said I'm very tired right now, and may be speaking
crap!
cheers,
Al
6. ### Tim WescottGuest
Is the circuit one that drives the motor straight off of a PWM
transistor, or does it make a switching amplifier out of it with a
filter coil and capacitor to spare the motor? If it's filtered (which
the presence of a cap would indicate) then the simple transistor circuit
we're giving you won't suffice; you'll need something to turn the analog
voltage back into a PWM duty cycle.
You should check: load the output (with a resistor or a real fan). If
you have an o-scope then look at the voltage to the motor -- this will
be good because the simple circuit given is also sensitive to frequency;
above 1kHz you need to actually worry about the transistor speed. If
all you have is a DVM then all you can do is check to see if there's
significant AC on the pin, and hope that it's slow enough for your
transistor circuit.
7. ### MikeGuest
It seems there is a capacitor before the header, as when I was
I thought it would be impossible to find the actual capacitor, but the
are fairly large near the header, and can be traced with the naked
eye. There
is indeed one sitting right before each header, and it is 47uf @ 25v.
Unfortunately no. I tried another small NPN transistor that I found
and the
fan went into full power as soon as the header was turned on. It made
no
difference in fan RPMs if the header was set to 5% or 100%.
And given that the capacitance is now known ([email protected][email protected] here
right?)
based on that how would one calculate the speed at which it discharges
with
a given resistor value @ 12v?
And in this scenario, I guess we also need to know how fast the signal
is
pulsated? As I'm assuming if the capacitor is not discharged fast
enough,
and the transistor has a certain threshold before it switches off, the
pulses coming off the transistor will be longer. Logic correct?
Plastic, 1" x 1" x 2".
Please bear with me for a second, and this might sound silly, but
shouldn’t the resistor connect with one end to the 12v MB and
the other end to 0V? (I think 0v is common ground in computers, that
is the case and every other component shares that ground.)
Power Supply
| 12v+
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|
|
|
|
12v+ ----------------------| NPN
| |> Transistor
.-. |
| | |
| | |
'-' |
Resistor | |
| |
| | +
| |
| .-.
| (FAN)
| '-'
| | -
| |
| |
___ ___
_ Ground _
| |
Would this schematic work? (Modulating (+) instead of ground?)
Thanks,
Mike
8. ### Al BorowskiGuest
Hi,
[...]
Rats. That makes thing harder.
[...]
It doesn't work like that I'm afraid. The '25V' on the capacitor means
the capacitor is rated for a maximum of 25V. It still has the valur of 47uf.
"rc discharge equation". Seeing there is a real capacitor there, the
calculations become a bit trickier. I'd crunch the numbers myself but i
don't have time right now.
Yes.
As I'm assuming if the capacitor is not discharged fast
Yes. One danger is that the transistor may operate in the linear region,
where it is neither fully on nor fully off. This means it will get much
hotter then if it was switched fully on or fully off. In a sealed
plastic box, this could cause problems.
I'm not expert, but it sounds like that could be asking for trouble.
Maybe a cheap metal box would be the way to go?
It effectivly does!
There is a (small) vltage drop accross the Base-Emitter pins, but
otherwise its connected to 0V through the transistor.
Possibly, but its asking for trouble:
1) Too much base current may flow. The resistor in the first circuit
served to limit the bse current, but also to discharge the capacitor.
2) since you just want to use the NPN transistor as a switch, you are
better off placing the fan between 12V and the collector. I don't have
time to explain why right now, but google "common emitter" "emitter
follower" and "transistor switch".
I can give more details later if needed.
cheers,
Al
9. ### MikeGuest
That I do not know, and slowly trying to figure out. The chances of me
getting to an engineer at any motherboard manufacturer are slim to
none.
From what I can see on the board with my own two eyes, there are only
two transistors, a few resistors and a capacitor in the final stage of
the PWD fan control.
The first transistor is small, with the base connected to the central
microprocessor, and the base gets a 0 to 5.38v+ DC, linearly
proportional to how many % the fan header is set to. The second
transistor is larger, with the base feeding from the smaller
transistor's emitter, collector a constant 12v+ DC and emitter going
directly to the capacitor and then the actual fan header.
23.61V AC. The strange part is, if I set the multi-meter to AC, use
the positive probe on the DC (+) I get the 23.61V but if I use the
negative probe on the DC (+) I get 0V AC. Should it not be showing
same voltage either way?
10. ### MikeGuest
That's all very good stuff. I was just looking to be pointed in the
right direction. With some google, and maybe *gasp* a book that is not
too big I'll figure this out, hopefully without burning up all fan | 2,241 | 8,592 | {"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-2020-16 | latest | en | 0.916378 |
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# midterm_spr06 - EE 562a Midterm Exam 1 March 2006 WRITE...
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EE 562a Midterm Exam, 1 March 2006 WRITE YOUR NAME on the exam paper, and indicate if you are an ON-CAMPUS or DEN student. You may use a simple calculator (i.e., not a scientific calculator), and one letter-sized sheet of notes (both sides). Do all problems. Note that these problems have different point values, and differ widely in difficulty. You should look over all the problems before starting your work. Write all answers on the test paper. You may use scratch paper to do the problem, but the answer must be written with the problem. If you do work on scratch paper, and wish that work to be considered for partial credit, hand in the scratch paper along with the exam. Part I. Basic Concepts. For each question in this section, circle the best answer from the choices listed. Each problem has only one best answer. There is no partial credit on this section. (3 points each.) 1. Let x ( u ) and y ( u ) be random variables such that E { x ( u ) } = E { y ( u ) } = E { x ( u ) y ( u ) } = 0. Then we can say that x ( u ) and y ( u ) are: (a) Uncorrelated (b) Orthogonal (c) Independent (d) All of the above (e) Both (a) and (b)
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Jill Tulane University ‘16, Course Hero Intern | 592 | 2,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-22 | latest | en | 0.903669 |
http://maths.crusoecollege.vic.edu.au/s2t4i/ | 1,709,360,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00464.warc.gz | 19,444,469 | 8,074 | # Semester 1 Topic 8 Level I
Task Name : Cat and Dog Task Level : I (Year 8) Semester : 1 Topic : Chance VC Strand : Chance Web Address : http://maths.crusoecollege.vic.edu.au/chanceI Equipment Needed : A selection of six-sided, eight-sided, ten-sided and twelve-sided dice Victorian Curriculum outcome : Identify complementary events and use the sum of probabilities to solve problems. Describe events using language of ‘at least’, exclusive ‘or’ (A or B but not both), inclusive ‘or’ (A or B or both) and ‘and’. Represent events in two-way tables and Venn diagrams and solve related problems. Task description : Students play “Cat and Dog” as an intro to this task. The rules are that two six-sided dice are rolled. If the difference between the two dice are 0, 1 or 2, the “Cat” (one of the players) gets a point. If the difference between the two dice are 3, 4 or 5, the “Dog” gets a point. They then use this unfair game to create their own game which is fair. The new game must have three options – cat gets a point, dog gets a point, and either neither gets a point or both get a point. They use a two way table to show the game is fair, and use a Venn Diagram to show the probability for all possible events (cat gets a point, cat doesn’t get a point, dog gets a point, dog doesn’t get a point, both get a point, neither get a point). Encourage students to use something other than two six-sided die for extra challenge. Assessment options : Photograph of game with two way tables explaining the probability of each event as well as Venn Diagrams. Teacher notes : Extension: Students design a game of chance Intro presentation: Design a game of chance Design a game of chance rubric Extra task: Play the game Higher or Lower: What is your strategy? What was your highest score? Record the probabilities for one full game. For example: Pr(<7) = 24/51 (A, 2, 3, 4, 5, 6 are all lower: 6 x 4 suits = 24 out of 51 cards left) Pr(>4) = 35/50 (5, 6, 7, 8, 9, 10, J, Q, K are all higher: 9 x 4 suits ( – 1) = 35 out of 50 cards left) Pr( | 542 | 2,040 | {"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} | 4.375 | 4 | CC-MAIN-2024-10 | latest | en | 0.905263 |
https://www.jiskha.com/questions/300422/What-is-the-greatest-fraction-in-the-group-1-6-9-6-4-6-5-6 | 1,571,455,722,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986688674.52/warc/CC-MAIN-20191019013909-20191019041409-00018.warc.gz | 950,323,488 | 5,448 | Math
What is the greatest fraction in the group? 1/6, 9/6, 4/6, 5/6
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1. Isn't 9/6 greater than one?
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2. 5/6
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https://rdrr.io/cran/inflection/f/inst/doc/inflectionDevelopingMethods.Rmd | 1,686,350,881,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00176.warc.gz | 546,250,389 | 9,299 | # Developing methods for identifying the inflection point of a convex/concave curve" In inflection: Finds the Inflection Point of a Curve
{css, echo=FALSE} body .main-container { max-width: 1024px !important; width: 1024px !important; } body { max-width: 1024px !important; }
r
library(inflection)
knitr::opts_chunk$set(echo = TRUE) options(max.width = 1000) options(max.print = 100000) ## The Fisher-Pry sigmoid curve with total symmetry (not noisy) Let’ s take the function: $$f (x) = 5 + 5\, tanh (x − 5)$$ after [3], which has p = 5, L = 10, x1 = 2.7024, x99 = 7.2976 and examine it at the interval [2, 8] in order to have data symmetry w.r.t. inflection point. The function is also symmetrical around inflection point, i.e. we have total symmetry. From Corollary 1.1 of [1] we compute$x_l = 5.970315941, x_r = 4.029684059$,$x_{F1} = 3.850750196, x_{F2} = 6.149249804$, all inside [2, 8], thus all methods are theoretically applicable. We first take n = 500 sub-intervals equal spaced without error just for checking our estimators. The results are presented at Table 1 of [1], while here we also present the BESE iterations done by 'bese()'. library(inflection) data("table_01") x=table_01$x
y=table_01$y plot(x,y,cex=0.3,pch=19) grid() bb=ese(x,y,0);bb pese=bb[,3];pese abline(v=pese) cc=bese(x,y,0) cc$iplast
abline(v=cc$iplast,col='blue') knitr::kable(cc$iters, caption = 'BESE')
We observe that $\chi_l = 5.9720, \chi_r = 4.0280$, $\chi_{F1} = 3.8480, \chi_{F2} =6.1520$ are very close to the theoretically expected values, so we are on the results of Lemma 1.3 of [1]. The absolutely accuracy from the first apply of all methods confirms our theoretical analysis.
## The Fisher-Pry sigmoid curve with total symmetry (noisy)
We next add the error term $\epsilon_i\sim\,U(−0.05, 0.05)$ via the process 14 of [1] and run our algorithms again.The results are presented at Table 2 of [1] and here we also present the BESE iterations done by 'bese()'.
library(inflection)
data("table_02")
x=table_01$x y=table_01$y
plot(x,y,cex=0.3,pch=19)
grid()
bb=ese(x,y,0);bb
pese=bb[,3];pese
abline(v=pese)
cc=bese(x,y,0)
cc$iplast abline(v=cc$iplast,col='blue')
knitr::kable(cc$iters, caption = 'BESE') ## The Fisher-Pry sigmoid curve with data left asymmetry (not noisy) We continue with the same sigmoid function, but now we choose a proper [a, b] to show data asymmetry w.r.t. inflection point. Let’ s take for example [4.2, 8]. If we do our theoretical computations we find$x_l = 5.974322740, x_r = 4.029684059$,$x_{F1} = 4.025677260$,$x_{F2} = 5.974322740$. We have that$x_r < a$, so$\chi_r$has to estimate a = 4.2 and$\chi_S$must be close to 4.703504993. Additionally,$x_{F1} < a$, so$\chi_{F1}$must be also an estimation of a, thus$\chi_{D}$must lie near the value 5.087161370. It' s time to see if our theoretical predictions will be confirmed by experiment. We use for comparability the same Standard Partition as before and have the output presented at Table 3 and 4 of [1]. data("table_03_04") x=table_03_04$x
y=table_03_04$y tese=ese(x,y,0);tese pese=tese[,3] tede=ede(x,y,0);tede pede=tede[,3] cc=bese(x,y,0) cc$iplast
dd=bede(x,y,0)
dd$iplast plot(x,y,cex=0.3,pch=19) grid() abline(v=pese) abline(v=cc$iplast,col='blue')
abline(v=dd$iplast,col='red') knitr::kable(cc$iters, caption = 'BESE')
knitr::kable(dd$iters, caption = 'BEDE') ## The Fisher-Pry sigmoid curve with data left asymmetry (noisy) Let’ s add again an same error term$\epsilon_i\sim\,U(−0.05, 0.05)$and run our algorithms. The results at Table 5 of [1] clearly are close enough to the theoretical expectations. Since ESE method did not estimate the inflection point with acceptable accuracy, after running BESE and BEDE iterative methods we find Table 6 of [1] which is a clear improvement of both estimations. data("table_05_06") x=table_05_06$x
y=table_05_06$y tese=ese(x,y,0);tese pese=tese[,3] tede=ede(x,y,0);tede pede=tede[,3] cc=bese(x,y,0) cc$iplast
dd=bede(x,y,0)
dd$iplast plot(x,y,cex=0.3,pch=19) grid() abline(v=pese) abline(v=cc$iplast,col='blue')
abline(v=dd$iplast,col='red') knitr::kable(cc$iters, caption = 'BESE')
knitr::kable(dd$iters, caption = 'BEDE') ## The Gompertz non symmetric sigmoid curve (not noisy) Let’ s examine the function: $$f (x) = 10 e^{-e^{5}e^{−x}}$$ after [4], in the interval [3.5, 8]. It is easy to prove that f is (0.224, 1.0)-asymptotically symmetric around inflection point, so we can handle it similar to a symmetric sigmoid only for a distance of ±1 from p = 5. We use, for comparison reasons, the same SP with 500 sub-intervals without error and obtain the Table 8 of [1] which is absolutely compatible with theoretical predictions. The ESE & EDE iterations are showed at Table 9 of [1] where we observe convergence to the real p for both two methods. data("table_08_09") x=table_08_09$x
y=table_08_09$y tese=ese(x,y,0);tese pese=tese[,3] tede=ede(x,y,0);tede pede=tede[,3] cc=bese(x,y,0) cc$iplast
dd=bede(x,y,0)
dd$iplast plot(x,y,cex=0.3,pch=19) grid() abline(v=pese) abline(v=cc$iplast,col='blue')
abline(v=dd$iplast,col='red') knitr::kable(cc$iters, caption = 'BESE')
knitr::kable(dd$iters, caption = 'BEDE') ## The Gompertz non symmetric sigmoid curve (noisy) We continue with our familiar SP by adding error uniformly distributed by U(−0.05, 0.05) and the results are given at Table 10 of [1] while ESE & EDE iterations are shown at Table 11 of [1]. data("table_10_11") x=table_08_09$x
y=table_08_09$y tese=ese(x,y,0);tese pese=tese[,3] tede=ede(x,y,0);tede pede=tede[,3] cc=bese(x,y,0) cc$iplast
dd=bede(x,y,0)
dd$iplast plot(x,y,cex=0.3,pch=19) grid() abline(v=pese) abline(v=cc$iplast,col='blue')
abline(v=dd$iplast,col='red') knitr::kable(cc$iters, caption = 'BESE')
knitr::kable(dd$iters, caption = 'BEDE') From these Tables we conclude that convergence to the true value of inflection point p = 5 occurs from the iterative application of ESE and EDE methods in one or two steps only. ## A symmetric 3rd order polynomial with total symmetry Let the polynomial function: $$f(x)=-\frac{1}{3}\,x^3+\frac{5}{2}\,x^2-4x+\frac{1}{2}$$ We study it at [-2, 7], it has inflection point at p = 2.5 and we have total symmetry. The SP with 500 sub-intervals without error gives Table 13 of [1] which is absolutely compatible with theoretical predictions. There is no need for any kind of iteration, because both methods agree with the true value. data("table_13") x=table_13$x
y=table_13$y plot(x,y,cex=0.3,pch=19) grid() bb=ese(x,y,0);bb pese=bb[,3];pese abline(v=pese) The same SP with uniform error distributed by U(−2, 2) gives the results of Table 14 of [1] and two ESE iterations are presented at Table 15 of [1]. data("table_14_15") x=table_14_15$x
y=table_14_15$y plot(x,y,cex=0.3,pch=19) grid() bb=ese(x,y,0);bb pese=bb[,3];pese abline(v=pese) cc=bese(x,y,0) cc$iplast
abline(v=cc$iplast,col='blue') knitr::kable(cc$iters, caption = 'BESE')
## A symmetric 3rd order polynomial with data right asymmetry
For the same symmetric 3rd order polynomial as above we change the interval to [-2, 8], thus we have data right asymmetry now. The case of SP with 500 sub-intervals and no error gives Table 17 of [1], while ESE and EDE iterations are presented at Table 18 of [1]. First results are absolutely compatible with theoretical predictions for ESE method.
data("table_17_18")
x=table_17_18$x y=table_17_18$y
bb=ese(x,y,0);bb
pese=bb[,3];pese
plot(x,y,cex=0.3,pch=19)
grid()
cc=bese(x,y,0)
cc$iplast dd=bede(x,y,0) dd$iplast
abline(v=pese)
abline(v=cc$iplast,col='blue') abline(v=dd$iplast,col='red')
knitr::kable(cc$iters, caption = 'BESE') knitr::kable(dd$iters, caption = 'BEDE')
We add uniform error distributed by U(-2, 2) and we have the results of Table 19 of [1], while one ESE & one EDE iteration are given at Table 20 of [1].
data("table_19_20")
x=table_19_20$x y=table_19_20$y
bb=ese(x,y,0);bb
pese=bb[,3];pese
plot(x,y,cex=0.3,pch=19)
grid()
cc=bese(x,y,0)
cc$iplast dd=bede(x,y,0) dd$iplast
abline(v=pese)
abline(v=cc$iplast,col='blue') abline(v=dd$iplast,col='red')
knitr::kable(cc$iters, caption = 'BESE') knitr::kable(dd$iters, caption = 'BEDE')
There exist a problem here. Although we have a symmetric polyno- mial, the TESE is not equal to the true inflection point. A remedy for this problem for the class of 3rd order polynomials is given with Lemma 2.1 of [1]. Lets apply it here. We have that a = −2, b = 8 and from Table 19 of [1] is $\chi_{r} = −0.26, \chi_{l} = 4.74$, so we have that: $$\hat{p}=\frac{1}{3}\,\chi_{l} + \frac{1}{3}\,\chi_{r}+\frac{1}{6}\,a+\frac{1}{6}\,b=2.493333333$$ which is much closer to the true value of 2.5.
## References
[1] Demetris T. Christopoulos (2014), Developing methods for identifying the inflection point of a convex/concave curve. arXiv:1206.5478v2 [math.NA]. URL: https://doi.org/10.48550/arXiv.1206.5478
[2] Demetris T. Christopoulos (2016), On the Efficient Identification of an Inflection Point, International Journal of Mathematics and Scientific Computing , Volume 6 (1), June 2016, Pages 13-20, ISSN: 2231-5330. URL: https://veltech.edu.in/wp-content/uploads/2016/04/Paper-04-2016.pdf
[3] J.C. Fisher and R.H. Pry (1971), A Simple Substitution Model of Technological Change, Technological Forecasting and Social Change, 3, pp. 5–88. URL: https://doi.org/10.1016/S0040-1625(71)80005-7
[4] B. Gompertz (1825), On the Nature of the Function Expressive of the Law of Human Mortality, and on a New Mode of Determining the Value of Life Contingencies, Philosophical Transactions of the Royal Society of London, 115, pp. 513–585.
## Try the inflection package in your browser
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inflection documentation built on June 15, 2022, 5:07 p.m. | 3,340 | 9,719 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-23 | latest | en | 0.765218 |
https://www.jiskha.com/questions/518648/how-can-you-make-density-of-1g-ml-for-a-600-ml-bottle-whose-mass-is-27-gms | 1,600,816,861,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00583.warc.gz | 924,695,292 | 4,796 | # Science
How can you make density of 1g/ml, for a 600 ml bottle whose mass is 27 gms
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1. Fill it with enough water to make it stop floating, but not sink either.
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# Answer to Question #196369 in General Chemistry for Aven Lewis
Question #196369
flourine gas exerts pressure of 297 torr when the volume changes to 250 mL at STP.what was the original volume?
1
2021-05-21T02:34:00-0400
Given:
Initial Pressure P1=297torr
Let initial volume be V1
According to the question ,Final Volume =250mL
In final condition gas is at STP. it means T=273K,and P2=760torr
So, at constant temperature :-
P1V1=P2V2(Boyle's Law)
297×V1=760×250
V1="\\frac{760\u00d7250}{297}=639.73mL"
So, Original Volume ="639.73mL"
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# What are the common factor of 28 and 35?
Updated: 8/30/2023
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12y ago
The greatest common factor (GCF) is often also called the greatest common divisor (GCD) or highest common factor (HCF). Keep in mind that these different terms all refer to the same thing: the largest integer which evenly divides two or more numbers.
The greatest common factor of 35 and 48 is 1
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13y ago
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Wiki User
15y ago
Factors of 48: 1 2 3 4 6 8 12 16 24 36 Factors of 35: 1 5 7 35 1 is the only common factor of 48 and 35
Wiki User
12y ago
The factors of 21 are: 1, 3, 7, 21
The factors of 35 are: 1, 5, 7, 35
The common factors are: 1, 7
Wiki User
8y ago
The factors of 35 are:
1, 5, 7, 35
The factors of 48 are:
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Wiki User
11y ago
1 and 7 are both common factors of 35 and 49.
Wiki User
8y ago
The GCF is 1.
Wiki User
12y ago
The common factors are: 1, 7
Wiki User
6y ago
The GCF is 1.
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13y ago
1
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8y ago
1 and 7
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Q: What are the common factor of 28 and 35?
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Related questions
### What is the greatest common factor of 28 35 63?
The greatest common factor of the numbers 28, 35 and 63 is 7.
### Highest common factor of 28 and 35?
The greatest common factor or the highest common factor is the highest number that divides exactly into two or more numbers. 28: 1, 2, 4, 7, 14, 28 35: 1, 5, 7, 35 The GCF of 28 and 35 is 7.
The GCF is 7.
### What is the common factors of 49 35 28 20?
The only common factor of 49, 35, 28, and 20 is ' 1 '.
It is 7
The GCF is 7.
### What is the highest common factor of 28 and 35 and 70?
The highest common factor (HCF) is 7.
1 and 7.
The GCF is 7.
The GCF is 7.
7
It is: 1 | 684 | 1,964 | {"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 | 4 | CC-MAIN-2023-40 | latest | en | 0.91825 |
http://www.ocw.titech.ac.jp/index.php?module=General&action=T0300&GakubuCD=100&GakkaCD=12&KeiCD=&KougiCD=201600379&Nendo=2016&vid=03&lang=EN | 1,716,434,815,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058588.75/warc/CC-MAIN-20240523015422-20240523045422-00208.warc.gz | 43,240,281 | 11,495 | ### 2016 Exercises in Classical Mechanics
Font size SML
Academic unit or major
Physics
Instructor(s)
Nishida Yusuke Adachi Satoshi
Class Format
Exercise
Media-enhanced courses
Day/Period(Room No.)
Mon5-6(H102) Thr7-8(H103)
Group
a
Course number
ZUB.Q212
Credits
2
2016
Offered quarter
2Q
Syllabus updated
2016/4/27
Lecture notes updated
-
Language used
Japanese
Access Index
### Course description and aims
The variational method to rewrite a given equation is explained. In analytical mechanics, the minimum action principle, which is a variational method, derives Euler-Lagrange equation. The Euler-Lagrange equation is invariant under general coordinate transformations. When a physical system has a symmetry, the corresponding conserved quantity exists. By expressing analytical mechanics in Hamilton formalism, the invariance of equation of motion under a larger class of transformations is obtained.
Both of understanding logic of analytical mechanics and developing ability of calculation in concrete examples are aimed.
### Student learning outcomes
Newtonian mechanics is reconstructed on basis of the minimum action principle. The equation of motion is derived from Lagrangian expressed in general coordinate system and is analyzed in several examples. The canonical formalism of analytical mechanics is also explained.
### Keywords
principle of variation, minimum action principle, Euler-Lagrange formalism, Lagrange function, general coordinate transformation, symmetry and conserved quantity, Noether's theorem, Hamilton formalism, Hamilton function, canonical transformation, Hamilton=Jacobi formalism, action function, separation of variables
### Competencies that will be developed
✔ Specialist skills Intercultural skills Communication skills Critical thinking skills ✔ Practical and/or problem-solving skills
### Class flow
Materials for exercises are distributed in advance. Students are expected to solve exercises at home and present answers on blackboard. Answers are discussed.
### Course schedule/Required learning
Course schedule Required learning
Class 1 importance and role of analytical mechanics in physics The method of analytical mechanics is used everywhere in physics, for example, electromagnetism, quantum mechanics, field theory and so on.
Class 2 variational method By representing a given equation to fit the variational principle, an invariance with certain range of transformation is obtained automatically.
Class 3 minimum action principle (Euler=Lagrange equation) The variational method for action function is the minimum action principle. It derives Euler=Lagrange equation as equation of motion.
Class 4 system with constraint A physical system that has a holonomic constraint can be analyzed by Euler=Lagrange formalism.
Class 5 method of Lagrange multipliers A physical system with holonomic constrained can be treated by method of Lagrange multipliers, which increases variables to erase the constraint.
Class 6 symmetry and conserved quantity When a physical system has a continuous symmetry, the corresponding conserved quantity exists according to Noether's theorem.
Class 7 arbitrariness of Lagrange function Lagrange function has an arbitrariness to include any complete differential with respect to time.
Class 8 small oscillation Motion around a stable equilibrium is called small oscillation. It is expressed by linear superposition of normal mode oscillations.
Class 9 rigid body motion Motion of direction of undeformable body in three dimensional space is analyzed.
Class 10 canonical formalism Hamilton mechanics is the another central formalism of analytical mechanics than Euler=Lagrange mechanics.
Class 11 canonical transformation Hamilton equation is invariant under the set of canonical transformations, which is larger than that of general coordinate transformations.
Class 12 Poisson bracket Hamilton equation can be expressed in terms of Poisson bracket. Thereby, the symmetry within the equation becomes more clear.
Class 13 Hamilton=Jacobi equation A physical system of several degree of freedom that has some symmetry may be solved by separation of variables based on Hamilton=Jacobi equation.
Class 14 phase space and Liouville theorem State space of dynamical system in the canonical formalism is called phase space. Phase space volume is conserved in time evolution according to Liouvelle theorem.
Class 15 adiabatic invariants In a slow variation of parameters for a physical system, if a quantity is conserved,, it is called an adiabatic invariant. Action integrals supply adiabatic invariants.
not yet decided
### Reference books, course materials, etc.
Landau=Lifshitz,``Mechanics'',(Tokyo tosho)
Yoshirou Oonuki,``Analytical Mechanics'',(Iwanami shoten)
### Assessment criteria and methods
Based on blackboard presentation, report and examination.
### Related courses
• ZUB.E202 ： Electromagnetism I
• ZUB.E211 ： Exercises in Electromagnetism
• ZUB.Q204 ： Quantum Mechanics I
• ZUB.Q215 ： Exercises in Quantum Mechanics I
### Prerequisites (i.e., required knowledge, skills, courses, etc.)
Students are expected to have learned real analysis and linear algebra in the first school year. | 1,056 | 5,194 | {"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} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.843722 |
https://www.coursehero.com/file/6708373/Homework-6-USING-EXCEL/ | 1,493,168,337,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121000.17/warc/CC-MAIN-20170423031201-00472-ip-10-145-167-34.ec2.internal.warc.gz | 875,664,026 | 98,089 | Homework 6 USING EXCEL
# Homework 6 USING EXCEL - HW 6 Page 1 of 2 Homework 6. ANOVA...
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HW 6 Page 1 of 2 Homework 6. ANOVA Analysis USING EXCEL Conduct data management & analyses 1. Generate pivot tables & pivot charts for the following sets of variables in your dataset: PHYS/MENTHLTH by (1) CHECKUP, (2) MSCODE, (3) SMOKESTAT, & (4) RACE5 To create Pivot Tables & Charts Put cursor in cell A1 Go to “Insert” & click on “PivotTable” Make sure the whole set of data is in range From the Pivot Table field list Check PHYS/MENTHLTH & put in “Values” box Choose the BY variable (e.g., CHECKUP) & put in the “Column Labels” box From the table Right click on first value in the “Total” row & choose “Summarize values by” & “Average” Change the value labels for HLTHPLAN Go to “Options” & click “Pivot Chart” Figure 1. Mean Days of Good Mental Health Related to Time Since Last Checkup Choose “Bar graph” & copy the graph & paste it in a Word document & save Insert a title & labels for the X axis & Y axis o Go to “Chart Tools,” “Layout,” “Chart Title,” & “Axis Title” Change the Y-axis values to 0 to 30 o Go to “Chart Tools,” “Layout,” “Axes,” “Primary Vertical Axis,” “More Primary Vertical Axis Options,” & 2. Separate the data for PHYS/MENTHLTH FOR EACH LEVEL of the following variables to make new variables: (1) CHECKUP, (2) MSCODE, (3) SMOKESTAT, & (4) RACE5 To create new variables : Sort the dataset by the variable of interest (e.g., CHECKUP) from the “Home” page by clicking on “Sort & Filter,”
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## This note was uploaded on 01/18/2012 for the course SSC 304 taught by Professor Stacy during the Fall '09 term at University of Texas.
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Homework 6 USING EXCEL - HW 6 Page 1 of 2 Homework 6. ANOVA...
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Ask a homework question - tutors are online | 572 | 2,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-17 | longest | en | 0.812207 |
https://math.stackexchange.com/questions/3401482/multiplicative-closed-subsets-s-subset-mathbbz-such-that-varphi-mathrms | 1,713,832,199,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00187.warc.gz | 350,017,077 | 35,914 | # Multiplicative closed subsets $S\subset \mathbb{Z}$ such that $\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z}))$ is open in $\mathrm{Spec}(\mathbb Z)$.
I want to find all multiplicatively closed subsets $$S\subset \mathbb{Z}$$ such that $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z}))$$ is open in $$\mathrm{Spec}(\mathbb{Z})$$, where $$\varphi(\mathfrak{q}) = f^{-1}(\mathfrak{q})$$, where $$f$$ is the natural homomorphism between $$\mathbb{Z}$$ and $$S^{-1}\mathbb{Z}$$.
My attempt: Notice that $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \{\mathfrak{p}\in \mathrm{Spec}(\mathbb{Z})\mid \mathfrak{p}\cap S = \emptyset\}$$. The case when $$0\in S$$ is clear, since then $$S^{-1}\mathbb{Z} = \{0\}$$ and thus $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \varphi(\emptyset) = \emptyset$$ which is clearly open.
Now suppose that $$0\notin S$$, notice that since $$\mathbb{Z}$$ is a PID every closed set is of the form $$V((n))$$ for some $$n\in \mathbb{Z}$$. Thus for $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z}))$$ to be open we need that $$X_{S} := \mathrm{Spec}(\mathbb{Z})\setminus\varphi(\mathrm{Spec}(S^{-1}(\mathbb{Z})) = V((n))$$ for some $$n\in\mathbb{Z}$$.
If $$n=0$$, then we have $$V((0)) = \mathrm{Spec}(\mathbb{Z})$$. And this is the case if and only if $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \emptyset$$. Thus we need $$\mathrm{Spec}(S^{-1}\mathbb{Z}) = \emptyset$$. Now my claim is that the only possibility is that $$S = \mathbb{Z}\setminus\{0\}$$, but I have problems proving this.
If $$n\neq 0$$, then $$n$$ has finitely many prime divisors, say $$p_{1},...,p_{r}$$. Then we have $$V((n)) = \bigcup_{i=1}^{r}\{(p_{i})\}$$. Then we have $$X_{S} = \bigcup_{i=1}^{r}\{(p_{i})\}$$ if and only if $$(p_{i})\cap S \neq \emptyset$$ and $$\forall \pi$$ prime in $$\mathbb{Z}$$ with $$\pi\neq p_{i}$$ $$\forall i$$ we have $$(\pi)\cap S = \emptyset$$. From here I also have problems completing this case.
New attempt: Using the cofiniteness of the topology on $$\mathrm{Spec}(\mathbb{Z})$$ we have, assuming that $$\varphi(\mathrm{Spec}(\mathbb{Z}))$$ is open that:
$$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \emptyset$$, which implies that $$\mathrm{Spec}(S^{-1}\mathbb{Z}) = \emptyset$$.
$$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \mathrm{Spec}(\mathbb{Z})$$, which implies that $$S=\{1\}$$, or $$S=\{-1,1\}$$.
Or we have a finite set of primes $$\{p_{1},...,p_{r}\}$$ such that $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \{\mathfrak{p}\in\mathrm{Spec}(\mathbb{Z})\rvert p_{1},...,p_{r}\notin \mathfrak{p}\}$$. Rewriting this last equality gives $$\{\mathfrak{p}\in\mathrm{Spec}(\mathbb{Z})\rvert \mathfrak{p}\cap S=\emptyset\} = \{(\pi)\rvert \pi\in\mathbb{Z} \ \text{prime such that} \ \pi\neq p_{i} \ \forall i\in\{1,...,r\}\}\cup\{(0)\}$$. This implies that $$S\subset \cup_{i=1}^{r}\{(p_{i})\}\cup \{-1,1\}$$ and $$S\cap (\pi)=\emptyset = S\cap(0)$$ for every $$\pi\in\mathbb{Z}$$ prime with $$\pi\neq p_{i}$$ for all $$i$$.
In all the cases, assuming that you have a multiplicatively closed $$S$$ satisfying one of the conditions above it is easy to show that in those cases $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z}))$$ is indeed open.
Only question left: Does the condition $$\mathrm{Spec}(S^{-1}\mathbb{Z}) = \emptyset$$ imply something stronger, like $$S = \{0\}$$, or $$S = \mathbb{Z}\backslash\{0\}$$?
• For your "only question left": $\operatorname{Spec} R =\emptyset$ is equivalent to $R=0$. It is easy to check that if we localize at a multiplicative subset not containing a zero divisor then the original ring embeds in to this localization and if we localize at a multiplicative set containing a zero divisor, then we collapse our ring to zero. Thus $\operatorname{Spec} S^{-1}\Bbb Z=\emptyset$ is equivalent to $0\in S$. Oct 21, 2019 at 22:29
Note that the topology on $$Spec(\mathbb{Z})$$ is cofinite.
Let $$S$$ be a multiplicative closed sub set of $$\mathbb{Z}$$. Then $$Spec(S^{-1}\mathbb{Z})$$ is an open subset of $$Spec(\mathbb{Z})$$ if and only if there exists a subset $$A=\{p_1,...,p_n\}$$ of prime number ( $$A$$ may be empty) such that $$S\subseteq \mathbb{Z}-\cup_i\langle p_i\rangle$$ and for any other subset $$\{q_j\}$$ of prime numbers we have $$S\not\subseteq \mathbb{Z}-\cup_j\langle q_j\rangle$$.
Proof: If $$Spec(S^{-1}\mathbb{Z})$$ is an open subset of $$Spec(\mathbb{Z})$$, then either $$Spec(S^{-1}\mathbb{Z})=\{\}$$ or there are prime number $$p_1,...,p_n$$ such that $$Spec(S^{-1}\mathbb{Z})=\{P\in Spec(\mathbb{Z})\mid S\cap P=\{\}\}=\{P\in Spec(\mathbb{Z})\mid p_1,...,p_n\not\in P\}$$. Thus either $$A$$ is empty or $$A=\{p_1,...,p_n\}$$ such that $$S\subseteq \mathbb{Z}-\cup_i\langle p_i\rangle$$ and for any other subset $$\{q_j\}$$ of prime numbers we have $$S\not\subseteq \mathbb{Z}-\cup_j\langle q_j\rangle$$. The converse is trivial.
• If you have a $P\in\mathrm{Spec}(\mathbb{Z})$ with $S\cap P=\emptyset$ and $p_{1},...,p_{n}\notin P$ one could have that $p_{i}\in S$ for some $i$ right? So this means that we might have $S\not\subset \mathbb{Z}-\cup_{i=1}^{n}(p_{i})$. | 1,858 | 5,048 | {"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": 74, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-18 | latest | en | 0.72769 |
https://math.stackexchange.com/questions/3251571/confusion-about-sobolev-spaces | 1,563,448,585,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525627.38/warc/CC-MAIN-20190718104512-20190718130512-00113.warc.gz | 465,158,400 | 35,941 | Reading through Chapter 8 of Brezis. We see that $$C_{c}^{\infty}(\mathbb{R})$$ is dense in $$W^{1,\,p}(I)$$. Later on in the chapter we define $$W^{1,\,p}_{0}(I)$$ to be the closure of $$C_{c}^{1}(I)$$ in $$W^{1,\,p}(I)$$. This $$W^{1,\,p}_{0}(I)$$ space has really confused me.
1. What happens when look at $$C_{c}^{\infty}(I)$$ which causes us to lose density in $$W^{1,\,p}(I)$$?
2. This question is linked to the first question. Suppose $$(u_{n})\in C_{c}^{1}(I)$$, then $$(u_{n}')\in C_{c}(I)$$. Both $$C_{c}^{1}(I)$$ and $$C_{c}(I)$$ are dense in $$L^{p}(I)$$, so $$u_{n}\rightarrow u$$ and $$u_{n}'\rightarrow g$$ in $$L^{p}(I)$$. Hence why cannot we not say that $$u_{n}\rightarrow u$$ in $$W^{1,\,p}$$ by taking $$u'=g$$?
3. By definition $$\overline{C_{c}^{1}(I)}=W^{1,\,p}_{0}(I)$$ in the $$W^{1,\,p}$$ norm. So how does one show density of $$C_{c}^{\infty}(I)$$ in $$W^{1,\,p}_{0}(I)$$? Is it sufficient to show that $$C_{c}^{\infty}(I)$$ is dense in $$C_{c}^{1}(I)$$ with respect to the supremum norm?
It's not true that $$\mathcal C_c^\infty (I)$$ is dense in $$W^{1,p}(I)$$ when $$I\neq \mathbb R$$. What is true is $$\mathcal C_c^\infty (\mathbb R)$$ is dense in $$W^{1,p}(\mathbb R)$$. And indeed, if $$I\neq \mathbb R$$ is an interval, we define $$W_0^{1,p}(I)$$ as the closure of $$\mathcal C_c^\infty (I)$$ in $$W^{1,p}(I)$$.
• Yes I know this. I am asking why is $C_{c}^{\infty}(I)$ not dense in $W^{1,\,p}(I)$? What do we lose when we go from $C_{c}^{\infty}(\mathbb{R})$ to $C_{c}^{\infty}(I)$ that makes us sacrifice density? – Zeta-Squared Jun 5 at 7:40
• Just take any sequence in $\mathcal C_c^\infty (0,1)$ that converges to $f(x)=1\in W^{1,p}(0,1)$. The gradient will explose. (with a draw it's very easy to see). – Surb Jun 5 at 7:49
• I see what you are saying. But $C_{c}^{\infty}((0,1))$ is dense in $L^{p}((0,1))$. So if $f_{n}\rightarrow f$ in $L^{p}((0,1))$ and likewise since $(f_{n}')\in C_{c}((0,1))$ then $f_{n}'\rightarrow f'$ in $L^{p}((0,1))$ wont we have that $f_{n}\rightarrow f$ in the $W^{1,\,p}$ norm? – Zeta-Squared Jun 5 at 7:59
• The thing is in $W^{1,p}(0,1)$ you cannot define a function on the boundary as you want. In $W^{1,p}(0,1)$, the function $f(x)=1$ must have $f(0)=f(1)=1$. You'll find more information with the trace operator – Surb Jun 5 at 8:04
• I agree. I am clearly not understand something properly here. I think I may have identified what I am mistaken about. Please clarify for me. We can show that a function $f$ belongs to $L^{p}$ simply by showing $\|f\|_{p}<\infty$. However, I have been thinking that it is sufficient to show $f$ belongs to $W^{1,\,p}$ if $\|f\|_{W^{1,\,p}}<\infty$. This is incorrect, that is, $\|f\|_{W^{1,\,p}}\nRightarrow f\in W^{1,\,p}$ – Zeta-Squared Jun 5 at 8:09 | 1,088 | 2,768 | {"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-30 | latest | en | 0.755395 |
https://www.mathworks.com/matlabcentral/cody/problems/2520-append-two-matrix-as-shown-below-example/solutions/1685401 | 1,606,532,699,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194982.45/warc/CC-MAIN-20201128011115-20201128041115-00221.warc.gz | 760,427,186 | 18,768 | Cody
# Problem 2520. Append two matrix as shown below example
Solution 1685401
Submitted on 3 Dec 2018 by fatma uysal
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### Test Suite
Test Status Code Input and Output
1 Pass
A=[1 2; 3 4] B=[5 6;7 8] y_correct= [ 1 2 5 6; 3 4 7 8] assert(isequal(addMatrix(A,B),y_correct))
A = 1 2 3 4 B = 5 6 7 8 y_correct = 1 2 5 6 3 4 7 8
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https://www.jiskha.com/display.cgi?id=1218167250 | 1,502,928,284,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102757.45/warc/CC-MAIN-20170816231829-20170817011829-00618.warc.gz | 917,914,491 | 3,956 | # math
posted by .
algebra properties and proportions
(x+1) 1
----- = ---
(x+2) 3
How do you come to this answer?
• math -
You posted this before with (x+1)/(x+2) = 2/3.
Which is it?
Use the rule that when a/b = c/d, ad = bc. Then solve for x
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What is true about microscopic properties of an object?
More Similar Questions | 558 | 2,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-34 | latest | en | 0.91763 |
https://jenisesexton.com/2016/02/12/creating-a-remediation-they-want-to-come-to/?replytocom=133 | 1,621,089,429,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991370.50/warc/CC-MAIN-20210515131024-20210515161024-00444.warc.gz | 330,712,673 | 26,341 | # Creating a Remediation They Want to Come to!
At the end of every 1st semester, students who failed a course are invited to participate in Comet Academy, my schools grade recovery/remediation sessions. This year, those attending Comet Academy come for 2 1/2 hours for 6 consecutive Saturday mornings. During our first session on January 30th, students were given a 30 questions, multiple choice pretest covering 5 key standards from 1st semester. I was assigned the 8th graders who failed to which I was very excited. 8th grade is the one grade level I haven’t been able to infiltrate with the use of manipulatives and building conceptual understanding.
Students completed the pretest in about an hours time, followed by checking the answers of a partner. Stunned by the results, I knew immediately, 6 sessions, even at 150 minutes a pop, wouldn’t be enough to get the students to pass the post test. The highest score was 11 out of 30 correct, ugh!
After proposing to the math AP about the need for a math boot camp for students who failed to stabilize their foundation for high school, I got busy planning. 8th graders who failed 1st semester, whether they attend Comet Academy or not, would be pulled twice a week for 30 minutes each time for 5 weeks. This would occur during their Connections classes (electives, specials, not the core classes). This boot camp would only offer additional support and would not be rewarded with a grade, extra credit or anything tangible outside of a better understanding of the content. For students who are often motivated by outside factors, I needed to ensure the activities were enticing enough to get them to come week after week.
Meaningful Practice
The standards are not new for students, therefore a mix of meaningful practice and concept development is necessary. In their regular classroom, students are subjected to worksheet after worksheet or textbook page after textbook page for practice. In boot camp we used a Solving Equations Bingo game to practice solving equations. Students filled in the answers: x=3, x=11.5, x= 1 1/2, x=5, x= 17, x=4, x=17 and -72 =x into the Bingo board. Then I read off equations in which students needed to solve in order to cover the correct answer.
We also played a game of Knockout! This idea was taken from the basketball game Knockout! In the basketball version, participates line up behind the free throw line, the first two people in line have a basketball. The goal is for the second person in line to make a shot before the person in front of them in order to knock the player out of the game. In boot camp, students sat in a straight line and the second person tried to correctly answer the math problem before the person in front of them did in order to knock them out of the game. Here’s the PowerPoint with the game: Boot Camp 2-9 and 2-11.
To encourage collaboration, students were grouped (they chose girls against boys) and given whiteboards to record the answers to problems. The group representatives would hold up their whiteboard showing their answers. In order to receive a point, teams had to get the correct answer, but also shot a tiny basketball into a toy hoop.
Concept Development
To help to continue to build student understanding of solving equations and integer rules, I used lessons from Hands-On Standards and incorporated color tiles and algeblocks with lessons like this Exponent Activity.
Formative Assessment
I have to know where they are from day to day because I don’t have a lot of time to cover the material. Therefore, each session has a connected formative assessment. This helps me to plan differentiated lessons even within the small group of students I see on the different days. I’ve used a two question quiz, a portion of a FAL and a Ticket out the Door pictured below. Students were able to choose what type of question they wanted to answer which is a formative assessment within a formative assessment :-).
1. Pam White says:
Jenise, I have said it before and I will say it again…you are awesome! We also have a 30 min support time for students in math and reading at our school. The students who do not need support have either band or choir or a enrichment class like designing, art, education computer games, etc. We also offer a stem class during this 30 minute class. I love your ideas and would like to hear more about what you do in class. I would love to incorporate more manipulatives and games to this support time.
2. Michelle says:
Love this! Thanks for sharing friend! Love you!
3. Lauren Cooke says:
Hello! I have read all of your posts and you have helped me tremendously. Thank you. I think the Boot Camp is a great component to your school. I pitched it to my school leader and he is on board. Is there any way that you would be able to share more resources? games? practices? assessments? concept development? I teach 7th grade math. Any help would be GREATLY appreciated and I most certainly gave you credit! Let me know if there is another way to do that besides a footer on the materials that I wrote to send home. Thank you for taking the time to share your knowledge with all of us. You are truly kid-focused and we are definitely going to benefit.
• Lauren, I’m glad you found this information helpful! Of course I can share resources with you. Send me your email address and I can share what I have. Also consider taking a look at the resources on the CCGPS 6-8 wiki. Google that and it should come up. My email address is JeniseSexton12@gmail.com. | 1,258 | 5,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-21 | latest | en | 0.968604 |
https://www.khanacademy.org/math/in-class-8-math-foundation/x5ee0e3519fe698ad:simple-equations/x5ee0e3519fe698ad:untitled-99/v/two-step-equations | 1,701,346,655,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00826.warc.gz | 941,422,985 | 81,419 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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## Class 8 (Foundation)
### Course: Class 8 (Foundation)>Unit 8
Lesson 2: Word problems
# Two-step equations with decimals and fractions
Let's practice some two step equations, some of which require merging terms and using the distributive property. Created by Sal Khan and CK-12 Foundation.
## Want to join the conversation?
• Math needs to solve its own problems. I'm not a therapist.
• words of a true man
• At , Sal kept 0.6 as a decimal. Could you also make that into a fraction (6/10) and then multiply both sides by 10 (getting 6x = 120), then dividing both sides by 6, getting x = 20? It seems much easier that way.
• Gauri, the way you described is a completely correct way of doing it, and I agree: it is easier. Usually, if you can covert decimals to fractions because fractions are easier to work with, that's a good way to do it.
Hope that helps!
• At , why is it necessary to do the division? Couldn't you just do "Well, I know 6 goes into 12 2 times, but this is a 0.6, which is 10 times less that 6, so just multiply that quotient by ten and boom: 20 is the answer."?
• Because even if this certain division problem was easy, he was demonstrating it so that you could understand and do harder ones more easily… I know- this reply is coming a little to far in the future. Outdated but- hope I helped.
• Why do us students have to learn all subjects, but teachers can't teach all subjects?
• They specialize in one topic. Thats why each teacher is so good at one subject cause they excelled in that area and are able to teach us well.
• Dear math,
She is gone.
• Thank You for these tips for two step equations. Is there other ways to do this?
• In two steps equations one need to solve one part of the equation first to solve the whole equation . This is going to be a good way to do one way equation.
• Why did he change 3s/8 to (3/8)S?
• What is 13=-4x+9
• I will demonstrate the steps towards solving the equation:
``Original equation:13 = -4x + 9Algebraic manipulations:13 - 9 = -4xx = (13 - 9) / -4x = -1Verification:13 = -4(-1) + 9, passesAnswer:.________.| x = -1 |.________.`` | 604 | 2,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-50 | latest | en | 0.963332 |
https://math.stackexchange.com/questions/2479861/semidirect-products-of-v-4-ltimes-alpha-c-3-and-isomorphies | 1,558,952,806,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262311.80/warc/CC-MAIN-20190527085702-20190527111702-00530.warc.gz | 543,143,181 | 33,932 | # Semidirect products of $V_4 \ltimes_\alpha C_3$ and isomorphies
My task is to find all semidirect products $V_4 \ltimes_\alpha C_3$ and to find those, who are isomorphic.
First of all, I've got to find the automorphism group of $C_3$. I know, that it is isomorphic to $(\mathbb Z / 3 \mathbb Z)^*$, so it has to have $3-1=2$ elements. The $C_3$ has got three elements $a,a^2,e$, where $e=a^3$.
For an automorphism of $C_3$ I know that it has to map $e \mapsto e$, so I only have to look how the other two elements can be mapped to eachother.
The first possibility would be $\alpha_1=Id_{C_3}$ with $e \mapsto e$ $\;$, $a \mapsto a$ and $a^2 \mapsto a^2$.
The second one would be $\alpha_2$ with $e \mapsto e$ $\;$, $a \mapsto a^2$ and $a^2 \mapsto a$.
One thing I know is, if $\alpha$ is the identity then the semidirect product is the normal direct product, so my first one would deliver $V_4 \ltimes_{\alpha_1} C_3$ = $V_4 \times C_3$. Is this one isomorphic to a known group like $C_n$ or $S_n$?
And for $\alpha_2$ I don't know how I have to go on.
• Is $V_4$ the Klein $4$-group? Direct and semidirect products are generally considered "known groups" and often have no simpler description, the point here is just to identify how many there are and when they're isomorphic. – Jim Oct 19 '17 at 11:18
• Also, the map you really need to look at is $V_4 \to \mathrm{Aut}(C_3)$. You get a direct product when this map is trivial. So now you know that $\mathrm{Aut}(C_3) \simeq C_2$. You need to start by finding all the homomorphisms $V_4 \to C_2$. – Jim Oct 19 '17 at 11:21
• Yes, it is the Klein-4-group. Is their a theorem that gives me the number of homomorphisms betwenn two groups? So I know how many I have to find. – Myrkuls JayKay Oct 19 '17 at 11:31
• I know that for a homomorphism $\phi : V_4 \to Aut(C_3)$, the neutral element of $V_4$ has to be mapped to $Id_{C_3}$. – Myrkuls JayKay Oct 19 '17 at 11:38
• A hint regarding $\alpha_2$: What can you say in that case about (a) the set of Sylow-$3$ subgroups of $G$, (b) the action of $G$ on that set and (c) the kernel of that action? – jpvee Oct 19 '17 at 12:51
Each semidirect products $( \mathbb{Z}_2\times \mathbb{Z}_2 )\ltimes_{\alpha} \mathbb{Z}_3$ is characterised by a homomorphism $\alpha : \mathbb{Z}_2\times \mathbb{Z}_2 \to \operatorname{Aut}(\mathbb{Z}_3)$. As you noted there are two automorphisms of $\mathbb{Z}_3$: \begin{align*} \operatorname{id}: \mathbb{Z}_3 &\to \mathbb{Z}_3 & f : \mathbb{Z}_3 &\to \mathbb{Z}_3 \\ 1& \mapsto 1 & 1 &\mapsto 2 \end{align*}
Next we determine the homomorphisms $\alpha :\mathbb{Z}_2 \times \mathbb{Z}_2 \to \operatorname{Aut}(\mathbb{Z}_3)$. Each $\alpha$ is determined by its effecton on $(1,0),(0,1) \in \mathbb{Z}_2 \times \mathbb{Z}_2$. Since $\operatorname{Aut}(\mathbb{Z}_3)$ has two elements, there are $2\cdot 2 = 4$ such homorphisms. First is the trivial map
\begin{align*} \alpha_1: \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \operatorname{Aut}(\mathbb{Z}_3) \\ x &\mapsto \operatorname{id}, \end{align*}
in which case $\left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_1} \mathbb{Z}_3 \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3$. There are also the three homomorphisms: \begin{align*} \alpha_2: \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \operatorname{Aut}(\mathbb{Z}_3) & \alpha_3 : \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \operatorname{Aut}(\mathbb{Z}_3) & \alpha_4 : \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \operatorname{Aut}(\mathbb{Z}_3) \\ (0,0) & \mapsto \operatorname{id} & (0,0) & \mapsto \operatorname{id} & (0,0) & \mapsto \operatorname{id} \\ (1,0) & \mapsto \operatorname{id} & (1,0) & \mapsto f & (1,0) & \mapsto f \\ (0,1) & \mapsto f & (0,1) & \mapsto \operatorname{id} & (0,1) & \mapsto f \\ (1,1) & \mapsto f & (1,1) & \mapsto f & (1,1) & \mapsto \operatorname{id} \\ \end{align*}
So there are four possible semidirect products $( \mathbb{Z}_2\times \mathbb{Z}_2 )\ltimes_{\alpha} \mathbb{Z}_3$. But are they all distinct groups?
The answer is no. The homomorphisms $\alpha_2,\alpha_3$ and $\alpha_4$ all induce isomorphic groups. To see this, observe that $\alpha_3 = \alpha_2 \circ \phi$ and $\alpha_4 = \alpha_2 \circ \psi$, where $\phi$ and $\psi$ are the automorphisms: \begin{align*} \phi: \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \mathbb{Z}_2 \times \mathbb{Z}_2 & \psi: \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \mathbb{Z}_2 \times \mathbb{Z}_2 \\ (0,0) & \mapsto (0,0) & (0,0) & \mapsto (0,0) \\ (1,0) & \mapsto (0,1) & (1,0) & \mapsto (1,1) \\ (0,1) & \mapsto (1,0) & (0,1) & \mapsto (0,1) \\ (1,1) & \mapsto (1,1) & (1,1) & \mapsto (1,0) \\ \end{align*}
It follows that $\left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_2} \mathbb{Z}_3 \cong \left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_3} \mathbb{Z}_3$ via the map: \begin{align*} \Theta:\left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_2} \mathbb{Z}_3 &\to \left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_3} \mathbb{Z}_3 \\ ((a,b),c) &\mapsto \left( \phi^{-1}(a,b),c \right) \end{align*}
A similar calculation yields, $\left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_2} \mathbb{Z}_3 \cong \left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_4} \mathbb{Z}_3$.
Thus there are two distict semidirect products $( \mathbb{Z}_2\times \mathbb{Z}_2 )\ltimes_{\alpha} \mathbb{Z}_3$.
• I really thank you for this solution! In your definition of $\alpha_4$ are mistakes in the last two mappings. You have there another two times the element $(0,0)$. These should be $(0,1)$ and $(1,1)$. – Myrkuls JayKay Oct 19 '17 at 19:00 | 2,196 | 5,640 | {"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} | 3.421875 | 3 | CC-MAIN-2019-22 | latest | en | 0.900865 |
https://byjus.com/maths/maths-questions/ | 1,656,322,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103329963.19/warc/CC-MAIN-20220627073417-20220627103417-00195.warc.gz | 198,639,347 | 156,116 | Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Maths Questions
Maths questions (Mathematics questions) are a valuable source for students who are preparing for various academic and competitive exams. Solving multiple maths questions on different math concepts will help you boost your problem-solving and logical thinking skills. It is advised for the students to practice questions on different maths concepts to understand, analyse and apply various strategies and maths formulas to get the correct answers. This way of studying maths concepts will enhance your analytical skills and strengthen your mathematics knowledge.
Mathematics is a subject that needs a lot of practice to understand the concepts thoroughly. Thus, it is not enough to simply study the definition and formula for any maths concept. Along with theoretical knowledge, one should acquire practical knowledge of all maths concepts. This can be achieved by solving several maths questions, i.e., questions on various maths topics.
Here, you will get the list of multiple maths concepts and questions on those particular concepts.
Also, access:
## List of Concept wise Maths Questions
Algebra Questions Fractions Questions Parabola Questions Algebraic Expressions Questions Frequency Polygon Questions Percentage Questions Applications Of Derivatives Questions Functions Questions Perimeter And Area Questions Applications Of Integrals Questions Geometric Progression Questions Permutations And Combinations Questions Area Of Parallelogram Questions Geometry Questions Pie Chart Questions Area Of Triangles Questions Hcf Questions Polynomials Questions Areas Related To Circles Questions Heights And Distances Questions Prime Numbers Questions Arithmetic Progression Questions Heron’s Formula Questions Probability Questions Average Questions Histogram Questions Pythagoras Theorem Questions Bar Chart Questions Hyperbola Questions Quadratic Equation Questions Bayes’ Theorem Questions Infinite Series Questions Quadrilaterals Questions Binomial Theorem Questions Integrals Questions Ratio And Proportion Questions Circles Questions Integration Questions Real Numbers Questions Complex Numbers Questions Inverse Trigonometric Functions Questions Rectangle Questions Compound Interest Questions Lcm Questions Relations And Functions Questions Conditional Probability Questions Limits And Derivatives Questions Sequence Questions Cone Questions Line Graph Questions Series Questions Constructions Questions Linear Equations Questions Sets Questions Continuity And Differentiability Questions Linear Inequalities Questions Simple Interest Questions Coordinate Geometry Questions Linear Programming Questions Square Questions Cube Root Questions Lines And Angles Questions Square Root Questions Cubes And Cube Roots Questions Logarithm Questions Squares And Square Roots Questions Cylinder Questions Mathematical Induction Questions Squence And Series Questions Data Handling Questions Mathematical Reasoning Questions Statistics Questions Determinants Questions Matrices Questions Straight Lines Questions Differential Equations Questions Mean Questions Surface Area And Volume Questions Differentiation Questions Median Questions Three Dimensional Geometry Questions Ellipse Questions Mensuration Questions Triangles Questions Euclid’S Geometry Questions Mode Questions Trigonometric Functions Questions Exponents And Powers Questions Multiples Questions Trigonometry Questions Factorial Questions Number System Questions Types Of Angles Questions Factorisation Questions Numbers To Words Questions Vector Algebra Questions Factors Questions Pair Of Linear Equations In Two Variables Questions Venn Diagram Questions
Some of the major advantages of learning and practising maths questions and answers are listed below:
• Solving maths questions will boost your confidence and cognitive skills.
• Helps in scoring good marks in the examination.
• A better understanding of concepts and in-depth understanding of applications of various mathematical concepts.
• Helps in solving real-world problems.
• Practising maths questions will strengthen your logical thinking.
• Improves your decision-making skills, etc.
### Interesting Questions on Maths
Who is the father of maths?
Archimedes is known as the father of maths.
What are the main branches of math?
We generally consider the following four are the main branches of maths.
Algebra
Geometry
Calculus
Statistics & Probability
Why is 1729 called Ramanujan number?
The number 1729 is called the Hardy-Ramanujan Number because he discovered that this is the smallest number that can be written as the sum of two distinct cubes in two various ways.
What is the meaning of 1/0?
The meaning of 1/0 is undefined. Mathematically, we express the value of 1/0 as infinity, i.e., ∞.
What is the largest number?
A Googolplex is the largest number in the world and is written as 10googol. The exponential form of this number is 1010^100.
Visit BYJU’S – The Learning App to get more Maths-related concepts and download the app today! | 887 | 5,119 | {"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-2022-27 | longest | en | 0.863409 |
https://www.tutorialspoint.com/print-first-n-fibonacci-numbers-using-direct-formula | 1,685,563,033,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00075.warc.gz | 1,174,623,339 | 11,413 | # Print first n Fibonacci Numbers using Direct Formula
In this article, we are going to solve the problem of printing first n Fibonacci Numbers using a direct formula.
In mathematics, the fibonacci numbers often denoted by Fn (which indicates nth fibonacci number), form a series in which each number is equal to the sum of the preceding two numbers. The nth fibonacci number can be indicates as below −
$$\mathrm{Fn\:=\:F_{n-1}\:+\:F_{n-2}}$$
The series begins with 0 and 1. The first few values in the fibonacci sequence, starting with 0 and 1 are −
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.
So, in this problem we will be given a number N and we need to print first N fibonacci numbers using a direct formula.
## Example
INPUT: 4
OUTPUT: 0 1 1 2
INPUT: 8
OUTPUT: 0 1 1 2 3 5 8 13
For this problem we need to know the concept of Binet’s formula which gives the direct formula to get the nth fibonacci number which is discussed in the algorithm section in detail.
## Algorithm
According to the formula,$\mathrm{Fn\:=\:F_{n-1}\:+\:F_{n-2}}$ we need (n-1)th term and (n-2)th to get the nth term by adding them. Since in this problem we are supposed to print the first n fibonacci number using a direct formula to get the nth fibonacci number.
To get the nth fibonacci number in the fibonacci sequence, one can apply the explicit formula known as Binet’s formula. It was created by mathematician Jacques Philippe Marie Binet.
### Formula
If $\mathrm{Fn}$ denotes the nth fibonacci number in the fibonacci sequence, then it can be expressed as
$$\mathrm{F_n\:=\:\frac{1}{\sqrt5}((\frac{1+{\sqrt5}}{2})^n\:-\:(\frac{1-{\sqrt5}}{2})^n)}$$
NOTE − This formula gives the fibonacci sequence starting from 1 and 1. To get the fibonacci sequence starting from 0 and 1, use n-1 to get the nth fibonacci number.
We can derive this formula using concepts of quadratic equations. We will be using this formula to print every fibonacci number until nth fibonacci number to print first n fibonacci numbers.
## Approach
• We will use a for loop to print all N fibonacci numbers iterating from 0 to n since we are considering the fibonacci sequence starting from 0 and 1.
• Initialise a variable as fibonacci and store the ith fibonacci number using the above formula for every iteration until i<N.
• Keep printing fibonacci at every iteration which will give us the first N fibonacci numbers.
### Example
Below is the implementation of the above approach in C++ −
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
void fibonacci(long long int N){ //function to print first N fibonacci numbers
long long int fibonacci; //to store ith fibonacci number
for(int i=0;i<N;i++) { //using for loop to print all N fibonacci numbers
//using direct formula to print every fibonacci number until n
fibonacci = 1/sqrt(5)*(pow((1+sqrt(5))/2,i) - (pow((1-sqrt(5))/2,i)));
cout<<fibonacci<<" ";
}
cout<<endl;
}
//Driver Code
int main(){
long long int N=10;
fibonacci(N);
N=6;
fibonacci(N);
return 0;
}
### Output
0 1 1 2 3 5 8 13 21 34
0 1 1 2 3 5
Time Complexity: O(n), since for loop runs until i is less than n.
Space Complexity: O(1), since it uses no extra space.
## Conclusion
In this article, we learned to print the first N fibonacci numbers using direct formula rather than using recursion. We have also learned about the Binet’s formula to directly get the nth fibonacci number in the fibonacci sequence. | 902 | 3,440 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-23 | latest | en | 0.852278 |
https://astarmathsandphysics.com/ib-maths-notes/matrices/1007-simultaneous-equations-with-non-unique-solutions.html | 1,713,890,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00802.warc.gz | 92,865,642 | 11,617 | ## Simultaneous Equations With Non Unique Solutions
The set of simultaneous equations
has the solutionsandandInfact many values ofandsatisfyboth equations.
Notice that if the second equation is divided by 2 throughout, weobtain the first equation, so that these two equations are in factjust one independednt equation. In general, we need two independentlinear equations to solve simultaneous equations with two unknowns tofind unique solutions. If we only have one independent linearequation with two unknowns, then we will be able to find an infinitenumber of solutions.
In general, if we have n equations with n unknowns, we can onlyfind unique solutions if the equations are independent, so that noneof the equations can be expressed in terms of the others.
Consider the equations
The third equation is the sum of the first two, so this system ofequations is not independent and does not have a unique solution.
In fact, we can ignore one of the equations, and still solve thesystem. If we ignore the third equation, we have the system
Settogive
(1)
(2)
(1)-(2) gives(3)
Substitute this expression into (1) to give(4)
Then the set of solutions is given by
These are parametric coordinates for a line. We can make thesolution more obvious.
Eliminate t between (3) and (4) by finding(3)*6+4:
isobvuously the equation of a line.
If we have three equations in three unknowns and each is amultiple of the other, then we can discard two and the result will beone of the equations which must be the equation of a plane.
Example:
The second is twice the first, and the third is three times thefirst, so two are redundant.
If we discard the last two, the solution set is the set ofsolutions toandthis is a plane. | 396 | 1,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-18 | latest | en | 0.922945 |
https://www.unitconverters.net/energy/pound-force-foot-to-inch-ounce.htm | 1,726,540,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00665.warc.gz | 957,880,025 | 3,510 | Home / Energy Conversion / Convert Pound-force Foot to Inch-ounce
# Convert Pound-force Foot to Inch-ounce
Please provide values below to convert pound-force foot [lbf*ft] to inch-ounce [in*ozf], or vice versa.
From: pound-force foot To: inch-ounce
### Pound-force Foot to Inch-ounce Conversion Table
Pound-force Foot [lbf*ft]Inch-ounce [in*ozf]
0.01 lbf*ft1.92 in*ozf
0.1 lbf*ft19.2 in*ozf
1 lbf*ft192 in*ozf
2 lbf*ft384 in*ozf
3 lbf*ft576 in*ozf
5 lbf*ft960 in*ozf
10 lbf*ft1920 in*ozf
20 lbf*ft3840 in*ozf
50 lbf*ft9600 in*ozf
100 lbf*ft19200 in*ozf
1000 lbf*ft192000 in*ozf
### How to Convert Pound-force Foot to Inch-ounce
1 lbf*ft = 192 in*ozf
1 in*ozf = 0.0052083333 lbf*ft
Example: convert 15 lbf*ft to in*ozf:
15 lbf*ft = 15 × 192 in*ozf = 2880 in*ozf | 295 | 769 | {"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} | 2.671875 | 3 | CC-MAIN-2024-38 | latest | en | 0.256699 |
rareenterprise.in | 1,653,342,787,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00358.warc.gz | 494,215,345 | 8,425 | • formula for critical speed of ball millinter komis
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is the mass of steel ball and V is velocity variation ofsteelballwhenthisballruns time Accordingtothevibrationtheory theradialdisplacement of the shell meets the Poisson equation which is expressed by 2 2 2 = where is the propagation velocity of vibration in the mill shell which can be calculated as follows = = 2.0 10 11
Chat Online | 2,614 | 12,167 | {"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-2022-21 | longest | en | 0.890546 |
https://forum.wilmott.com/viewtopic.php?f=3&p=865864&sid=a096904fdb83ef54076c2b19cf84b2b3 | 1,624,489,487,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488544264.91/warc/CC-MAIN-20210623225535-20210624015535-00209.warc.gz | 255,629,405 | 13,471 | Serving the Quantitative Finance Community
Jan_Roman
Topic Author
Posts: 2
Joined: April 13th, 2021, 1:04 pm
### FRN in a multiple curve framework
Hi,
Im dealing with valuation of Floating Rate Notes and calculate:
1.) The Present Value
2.) Discount Margin
3.) Modified Duration
In a multiple curve framework. I.e., when you generate the cash-flows with a Swap Curve (or any floating rate curve) and discount with e.g., a Government curve + Discount Margin.
I guess the best discount curve for bonds and FRN's are Government curves + spread due to bonds vs. bonds.
I have seen similar questions before but where the generating curve and the discount curve are the same (the old way of thinking?).
Senior Risk Analyst, Senior Financial Engineer & Senior Lecturer. Published two books, Analytical Finance I & II.
Hymn
Posts: 8
Joined: February 27th, 2012, 1:52 pm
### Re: FRN in a multiple curve framework
Hi,
the Government curve could be use as base curve, then you have to add a credit spread that capture the issuer riskeness over the Government, for that you can use the Z-spread over the government bond of a listed bond with the same issuer and characteristics of the bond that you are valuing. The Z-spread include also the liquidity spread because it's computed starting from market price.
Jan_Roman
Topic Author
Posts: 2
Joined: April 13th, 2021, 1:04 pm
### Re: FRN in a multiple curve framework
Hi, I have solved the problem by defining the spread between the index curve (LIBOR) and the treasury curve. Then using the total spread above the index curve as the Discount Margin. In that way I found the price at a reset as 100 + an annuity paining the difference between between the Reset Margin (the Quoted Margin) and the Discount Margin. Then I also found that the Modified Duration = Spread Duration + Index Duration. Also the Index Duration can be positive and negative depending on the value Reset Margin minus Discount Margin.
Senior Risk Analyst, Senior Financial Engineer & Senior Lecturer. Published two books, Analytical Finance I & II. | 481 | 2,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-25 | latest | en | 0.925871 |
https://numberworld.info/1013122 | 1,600,840,516,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209999.57/warc/CC-MAIN-20200923050545-20200923080545-00292.warc.gz | 559,065,536 | 3,838 | # Number 1013122
### Properties of number 1013122
Cross Sum:
Factorization:
2 * 11 * 46051
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
f7582
Base 32:
utc2
sin(1013122)
0.71040831914987
cos(1013122)
-0.70378975559655
tan(1013122)
-1.0094041771718
ln(1013122)
13.828547210333
lg(1013122)
6.0056617461861
sqrt(1013122)
1006.5396167067
Square(1013122)
### Number Look Up
Look Up
1013122 (one million thirteen thousand one hundred twenty-two) is a very unique figure. The cross sum of 1013122 is 10. If you factorisate the number 1013122 you will get these result 2 * 11 * 46051. The figure 1013122 has 8 divisors ( 1, 2, 11, 22, 46051, 92102, 506561, 1013122 ) whith a sum of 1657872. 1013122 is not a prime number. The figure 1013122 is not a fibonacci number. 1013122 is not a Bell Number. The number 1013122 is not a Catalan Number. The convertion of 1013122 to base 2 (Binary) is 11110111010110000010. The convertion of 1013122 to base 3 (Ternary) is 1220110202001. The convertion of 1013122 to base 4 (Quaternary) is 3313112002. The convertion of 1013122 to base 5 (Quintal) is 224404442. The convertion of 1013122 to base 8 (Octal) is 3672602. The convertion of 1013122 to base 16 (Hexadecimal) is f7582. The convertion of 1013122 to base 32 is utc2. The sine of 1013122 is 0.71040831914987. The cosine of 1013122 is -0.70378975559655. The tangent of the figure 1013122 is -1.0094041771718. The square root of 1013122 is 1006.5396167067.
If you square 1013122 you will get the following result 1026416186884. The natural logarithm of 1013122 is 13.828547210333 and the decimal logarithm is 6.0056617461861. that 1013122 is very unique figure! | 657 | 1,814 | {"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} | 3.265625 | 3 | CC-MAIN-2020-40 | latest | en | 0.738579 |
https://discuss.codecademy.com/c/python/practice-makes-perfect | 1,558,692,831,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00004.warc.gz | 460,037,206 | 5,107 | # Python Practice Makes Perfect
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I don't get the Remove Duplicates lesson? 5 August 1, 2018 | 537 | 1,622 | {"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-2019-22 | latest | en | 0.898068 |
http://math.stackexchange.com/questions/107706/the-space-c-omega-mathbbr-has-a-predual?answertab=votes | 1,462,452,299,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860127407.71/warc/CC-MAIN-20160428161527-00204-ip-10-239-7-51.ec2.internal.warc.gz | 181,883,695 | 19,904 | # The Space $C(\Omega,\mathbb{R})$ has a Predual?
Let $\Omega$ be a compact metric space and $C(\Omega,\mathbb{R})$ the space of borel continuous real valued functions. I would like to know if there is any real Banach space $V$ such that its dual space (topological)is exactly $C(\Omega,\mathbb{R})$.
My main interest is when $\Omega$ is an infinite cartesian product as for example, $\Omega=E^{\mathbb{Z}^d}$, where $E$ is a compact metric space. If the answer for this case is also negative are we in a better situation if $E$ is a finite set ?
Thanks for any comment or reference.
Edition. Remove the superfluous hypothesis pointed out by Philip.
-
In the infinite Cartesian product example you mention, I don't think you have a separable metric space. Or are you taking some kind of restricted direct product? – user16299 Feb 10 '12 at 3:00
What do you mean by "Borel continuous"? Are they continuous, or only Borel? If you mean continuous, then e.g. if $\Omega=[0,1]$ or another connected space, then the only extreme points of the unit ball of $C(\Omega,\mathbb R)$ will be the constant functions $f=1$ and $f=-1$, and by the Krein-Milman theorem this implies that $C(\Omega,\mathbb R)$ is not a dual space. If $\Omega$ is the one point compactification of the integers then you get a space isomorphic to the space of convergent sequences, which I believe is not a dual space but don't know how to show it. – Jonas Meyer Feb 10 '12 at 3:02
In fact: in the complex-valued case, C(X) for X compact Hausdorff is a dual Banach space precisely when it is an abelian von Neumann algebra, which happens if and only if X is extremely disconnected. I have a feeling that the only Von Neumann algebras that are separable in norm topology are the finite dimensional ones. Now all this should carry over to the real-valued case, implying that $\Omega$ is finite... – user16299 Feb 10 '12 at 3:03
@Yemon: Your feeling about separability of von Neumann algebras is correct. There's maybe an easier way to show it than this, but every infinite dimensional C*-algebra contains an element with infinite spectrum, and with Borel functional calculus of a self-adjoint element with infinite spectrum it isn't hard to show that the von Neumann algebra generated by such an element is nonseparable. – Jonas Meyer Feb 10 '12 at 3:09
@Yemon: your "if and only if" is not exactly right. There are extremely disconnected compact spaces $X$ such that $C(X)$ is not a von Neumann algebra. The right iff is given by hyperstonean spaces; these are extremely disconnected but they have the additional property that $C(X)$ has normal states. – Martin Argerami Feb 22 '12 at 7:41
Firstly, let me just point out that every compact metric space is automatically separable.
Secondly, note that $c_0$ does not embed into any separable dual space, hence neither does any Banach space containing a subspace isomorphic to $c_0$. Since every $C(K)$ space ($K$ compact Hausdorff) contains a subspace isomorphic to $c_0$, no $C(K)$ space embeds isomorphically into a separable dual space. In particular, since metrizability of $K$ is equivalent to $C(K)$ being norm separable, the answer to your question is always no.
For a reference for all of the above claims, look up $C(K)$ and $c_0$ in the index of Albiac and Kalton's book Topics in Banach Space Theory.
-
Hi Philip thanks a lot for your help. I will have a look at the reference you provided. – Leandro Feb 10 '12 at 3:20
The result is false in general. For example, consider $\Omega=\{\frac 1 n: n\in \mathbb N\}\cup \{0\}$, then $C(\Omega,\mathbb R)=c_0$ which is not a dual space. Edit: I'm adding some references: My claim follows from Proposition 4.4.1 of Albion and Kalton's book.
-
hi azarel, you mean that $c_0$ has no a predual space ? could you elaborate more or give me a reference ? – Leandro Feb 10 '12 at 3:00
$c_0$ is a pre-dual space (it is the predual of its dual), however it is not a dual space (which is what you mean to say). – Martin Wanvik Feb 10 '12 at 3:01
In your example, $C(\Omega,\mathbb R)\cong c$, the space of real convergent sequences, not $c_0$. It is easier to see that $c_0$ is not a dual space, because its unit ball has no extreme points. I don't know how to show that $c$ is not a dual space. – Jonas Meyer Feb 10 '12 at 3:04
Note that $c_0=c_0\oplus \mathbb R=c$ – azarel Feb 10 '12 at 3:45
@azarel: No, $c_0\neq c$. – Jonas Meyer Feb 10 '12 at 4:16 | 1,237 | 4,421 | {"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} | 2.65625 | 3 | CC-MAIN-2016-18 | latest | en | 0.891629 |
http://www.theschoolrun.com/what-place-value | 1,490,757,479,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190181.34/warc/CC-MAIN-20170322212950-00513-ip-10-233-31-227.ec2.internal.warc.gz | 700,976,216 | 19,147 | # What is place value?
A good understanding of place value (the value of each digit in a number) is vital in primary-school maths. Our parents' guide explains how your child will be taught about units, tens, hundreds and thousands with number lines, arrow cards and more, as well as outlining how place value is used to help children visualise calculations.
## What is place value?
Place value is the value of each digit in a number. It means understanding that 582 is made up of 500, 80 and 2, rather than 5, 8 and 2.
## How are children taught to understand place value in KS1?
In school two maths aids are used to help make place value clear to children.
Deines blocks are blocks in which cubes represent units / ones, rods of ten cubes represent tens, flats of 100 cubes represent hundreds and blocks of 1000 cubes represent thousands:
In Key Stage 1, a child might be given some ten and units (ones) Deines blocks and asked to make a number such as 43. They would need to select 4 tens rods and 3 ones blocks. This makes it very clear to them that a two-digit number it made up of tens and ones. It also helps them to practise counting in tens.
Arrow cards look like this:
A child might be asked to make the number 34 using arrow cards. They would need to take the 30 and the 4 and put them together so that the arrows were lined up. This again helps to make clear that a two-digit number is made up of tens and ones.
It is absolutely vital that children understand place value before they can go onto adding and subtracting two-digit numbers.
## Place value in KS2
In Year 3, children might be given Deines blocks and arrow cards to help them with adding a pair of two-digit numbers. This would enable them to partition the numbers, so that they could add the tens first and then the units. A Year 3 child would also be expected to know the place value of digits in three-digit numbers.
In Year 3, children also need to know what happens to a number when it is multiplied by 10 or 100. It is really important here that they are aware that the number moves to the left, rather than talking about 'adding zeros' (when children move onto dividing by 10 to find a decimal answer, they will not be able to use the strategy of adding or removing zeros).
This is a common method used by teachers to show children how to multiply by ten:
When multiplying 6 by ten, the number moves to the left and a zero is put in, so the number becomes 60.
If you multiply 6 by 100, the number moves two places to the left and zeros are put in, so the number would become 600.
In Year 4, children would need to use their knowledge of place value to work out sums and difference of pairs of multiples of 10, 100 or 1000. They might be asked to mentally work out 80 + 40 in which case they would need to 'cross 100' to find the answer 120. They might be asked to work out 700 + 600 or 8000 + 3000, in which case they would be crossing 1000 and 10,000. A number line can be helpful in these instances.
Year 4 children also need to start to understand the place value of decimals. A blank hundred square is a good way of demonstrating this:
With the above two decimals, you can make it clear that 0.41 is 41/100 and 0.02 is 2/100.
Sometimes teachers use small whiteboards marked like this:
onto which children have to write a number that the teacher has said out loud. For example: the child might be asked to write the number 5.72. They might then be asked to add 0.1 to this number, at which point they would need to change the number to 5.82. They might be asked to subtract 0.01, at which point they would need to change the number to 5.81.
In Year 4 children need to learn how to divide numbers by 10 and 100. Teachers usually teach this using this method:
When dividing 6 by ten, the number moves one place to the right and becomes 0.6.
If you divide 6 by 100, the number moves two places to the right and becomes 0.06.
In Year 5, children need to use their knowledge of place value to work out sums and differences of decimals. For example, they would need to know that adding 0.8 and 0.4 results in 'crossing 1' because the answer is 1.2. A number line can be helpful in demonstrating this:
In Year 6, children need to know their knowledge of place value to work out multiplication and division of decimals. For example, when working out 0.6 x 4, they would need to know that this is the same as working out 6 x 4 = 24, but then the answer needs to be divided by 10 (because 6 divided by 10 is 0.6) to make the answer 2.4. | 1,100 | 4,537 | {"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} | 4.84375 | 5 | CC-MAIN-2017-13 | latest | en | 0.960538 |
https://morethingsjapanese.com/what-are-some-facts-about-the-number-100/ | 1,726,815,739,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00380.warc.gz | 367,380,533 | 40,858 | # What are some facts about the number 100?
## What are some facts about the number 100?
There are some well-known facts about 100:
• There are 100 years in a century.
• On the Celsius scale, 100 degrees is the boiling temperature of water.
• The United States Senate has 100 Senators.
• There are 100 yards in an American football field.
• 100 is a perfect square number and its square root is 10.
Did u know facts about numbers?
5 Interesting Number Facts
• There are 86,400 seconds in a day.
• The only prime numbers that end with a “2” or “5” are two and five.
• In a room of 23 people, there’s a 50 percent chance of two people having the same birthday.
• Π is an irrational number discovered more than a thousand years ago.
What are some cool facts about math?
Interesting and Amazing Math Facts
• Pi and pizzas are linked.
• Nature loves Fibonacci sequences.
• In a crowded room, two people probably share a birthday.
• Multiplying ones always gives you palindromic numbers.
• The universe isn’t big enough for Googolplex.
• Seven is the favorite number.
• Prime numbers help Cicadas survive.
### What is no 100 called?
100 or one hundred (Roman numeral: C) is the natural number following 99 and preceding 101….100.
← 99 100 101 →
← 100 101 102 103 104 105 106 107 108 109 → List of numbers — Integers ← 0 100 200 300 400 500 600 700 800 900 →
Cardinal one hundred
Ordinal 100th (one hundredth)
Factorization 22 × 52
Why 100 is a special number?
number symbolism Because our notational system for numbers is decimal (base 10), the number 100 takes on a significance that it would probably not possess if we employed other systems of notation. It is a round number and holds hints of perfection.
What are number facts math?
Number facts are simple addition, subtraction, multiplication and division facts which are sometimes referred to as number bonds. Children learn these basic facts between the ages of 4 and 10.
#### How many numbers less than 100 have exactly 3 factors?
Simply count the primes less than 10 and you have your answer. Generate the integers from 1 to 100, factor each one, find the integers in the list that have exactly 3 prime factors, extract them & store them in b, and count them: So the answer is that there are 22 integers from 1–100 that have exactly 3 prime factors:.
Are there any fun facts about the number 100?
In Belgium, 100 is the ambulance and firefighter telephone number Take this quiz to see what 100 calories looks like! as of September 2010, Japan had approximately 44,449 centenarians (people living to the age of 100); the most in the world
What are interesting facts about numbers from 0 to 10?
People often are spellbound by numbers, basically used for counting, measuring and ranking. They are not just symbols. They hold some sort of uniqueness. It is in fact interesting to know some interesting facts about numbers from 0 to 10. Check out the article below and know how fascinating and funny are these numbers.
## Where can I find all factors of numbers 1-100?
You may use this resource to quickly find all the factors of the first one hundred (100) elements of the set of counting numbers. If you find any errors or typos regarding the factors of the numbers below, please email me at [email protected] so that I can fix them immediately. | 785 | 3,320 | {"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} | 3.453125 | 3 | CC-MAIN-2024-38 | latest | en | 0.91664 |
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