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# Math Help - Factoring Help
1. ## Factoring Help
Hi,
For my FST project, I am supposed to factor n^3 - 8 = 0
So,
I turn it into the difference of cubes:
(n)^3 - (2)^3
Which ends up being:
(n - 2) (n^2 + 2n + 4)
So, (n-2) is one factor. However, I don't know how to find the other zeros. I need to factor (n^2 + 2n + 4), but I can't seem to figure it out.
Thanks!
Ethan
2. Originally Posted by EthanDavis
Hi,
For my FST project, I am supposed to factor n^3 - 8 = 0
So,
I turn it into the difference of cubes:
(n)^3 - (2)^3
Which ends up being:
(n - 2) (n^2 + 2n + 4)
So, (n-2) is one factor. However, I don't know how to find the other zeros. I need to factor (n^2 + 2n + 4), but I can't seem to figure it out.
Thanks!
Ethan
That's good. There are no real factors. You'll have to use the quadratic formula to get the two complex roots.
-Dan
3. ## Re
Alright, let me try this:
(n^2 + 2n + 4) =
-2 +/- root{2^2 - 4(1)(4)
______________________
2
This equals:
-2 +/- root{-12
_____________
2
This equals:
-2 +/- i root{12
_____________
2
This equals:
-2 +/1 2i root {3
______________
2
Question: Now is it okay to divide everything by 2?
-1 +/- i root{3
So, would those be my factors? (-1+ i root{3)(-1 - i root{3)
I have a strong feeling I went wrong somewhere along the line...
(By the way, I apologize for having to write out "+/-" and "root{". How do you draw out the math problem like so many on this forum do?)
Thanks,
Ethan
4. Originally Posted by EthanDavis
Alright, let me try this:
(n^2 + 2n + 4) =
-2 +/- root{2^2 - 4(1)(4)
______________________
2
This equals:
-2 +/- root{-12
_____________
2
This equals:
-2 +/- i root{12
_____________
2
This equals:
-2 +/1 2i root {3
______________
2
Question: Now is it okay to divide everything by 2?
-1 +/- i root{3
So, would those be my factors? (-1+ i root{3)(-1 - i root{3)
I have a strong feeling I went wrong somewhere along the line...
(By the way, I apologize for having to write out "+/-" and "root{". How do you draw out the math problem like so many on this forum do?)
Thanks,
Ethan
Those are your roots. Your factors, if you choose to write them out, will be
$n^3 - 8 = (n - 2)(n - (-1 + i \sqrt{3} ))(n - (-1 - i \sqrt{3}))$
-Dan
|
The Community for Technology Leaders
ABSTRACT
<p>A new rank-based distributed deadlock avoidance algorithm for the <it>AND</it> resource request model is presented. Deadlocks are avoided by dynamically maintaining an invariant <tmath>$Con(WFG)$</tmath>: For each pair of processes <tmath>$p_i$</tmath> and <tmath>$p_j$</tmath>, <tmath>$p_i$</tmath> is allowed to wait for process <tmath>$p_j$</tmath> iff the rank of <tmath>$p_j$</tmath> is greater than that of <tmath>$p_i$</tmath> for the <it>WFG</it> (<it>Wait-For Graph</it>). Our algorithm neither restricts the order of resource requests nor needs a priori information about resource requests nor causes unnecessary abortion of processes. Multidimensional ranks, which are partially ordered and dynamically modified, are used to drastically reduce the cost of maintaining <tmath>$Con(WFG)$</tmath>. Our simulation results show that the performance of our algorithm is better than that of existing algorithms.</p>
INDEX TERMS
distributed deadlock avoidance, partially ordered rank, wait-for graph, concurrency control
CITATION
W. Chin, H. Wu and J. Jaffar, "An Efficient Distributed Deadlock Avoidance Algorithm for the AND Model," in IEEE Transactions on Software Engineering, vol. 28, no. , pp. 18-29, 2002.
doi:10.1109/32.979987
|
Finite complement topology over $\Bbb{R}$ is not second-countable under ZF?
Under $\mathsf{ZF+AC_\omega}$, the space $\Bbb{R}$ with finite complement topology is not second-countable, since a countable union of countable subsets of $\Bbb{R}$. However, such argument doesn't work if we does not assume any form of choice.
My question is, $\Bbb{R}$ with finite complement topology is not second-countable under ZF (Zermelo-Fraenkel set theory without choice), or it is consistently second-countable? I would appreciate your answer.
Proof of "$\Bbb{R}$ with finite complement topology is not second-countable" under choice: Let assume $\{V_n\}_{n<\omega}$ be a countable basis of the space. We know that $\Bbb{R}-\{x\}$ is open for each $x\in \Bbb{R}$ so for each $x$ there is $n<\omega$ s.t. $V_n\subset \Bbb{R}-\{x\}$ so $$\bigcap_{n<\omega} V_n\subseteq \bigcap_{x\in\Bbb{R}} \Bbb{R}-\{x\} = \varnothing.$$ However, $\bigcap_{n<\omega} V_n = \Bbb{R} - \bigcup_{n<\omega} (\Bbb{R}-V_n)$ and $(\Bbb{R}-V_n)$ is finite set. Since countable union of at most countable sets is at most countable, $\bigcup_{n<\omega} (\Bbb{R}-V_n)$ is countable so $\bigcap_{n<\omega} V_n$ is not empty, a contradiction.
Here's the thing about $\Bbb R$. It's linearly ordered. And this means that finite subsets are well-ordered canonically by that linear ordering.
In particular, this means that the countable union of finite sets of real numbers is always countable. So the usual proof works, with the minor modification that now we need to appeal to the properties of $\Bbb R$ to claim that $\bigcup(\Bbb R-V_n)$ is countable.
|
# Tracking / calculating dependencies of nodes
I have a problem regarding dependencies. I'm making a circuit simulator thingy and I stumbled upon the problem of determining which circuit component should be updated first (since updates in my case mean that the output is transformed based on its inputs according to some specified behaviour, the behaviour being irrelevant). How does one efficiently tackle this problem and how does he approach the problem of circular dependencies? (e.g. in counters)
Sources on the topic are much appreciated!
If there are no circular dependencies you can just use a topological order of the associated directed graph (that contains edge $$(u,v)$$ if $$v$$ uses the output of $$u$$ as an input).
|
# Is the classifying space of a symmetric monoidal category an infinite loop space?
Wikipedia states:
The classifying space (geometric realization of the nerve) of a symmetric monoidal category is an infinite loop space.
If my mind is correct, Segals delooping machine gives a functor $Sp$ from topological symmetric monoidal categories $C$ to prespectra (In this question, a presepectrum is a sequence of based spaces $X_0,X_1$,... with structure maps $\Sigma X_i\rightarrow X_{i+1}$) with $Sp(C)_0=BC$ the classifying space of the category.
If I remember correctly, one of Segals results is that the adjoints of the structure maps $Sp(C)_i\rightarrow \Omega Sp(C)_{i+1}$ are weak homotopy equivalences for $i\ge1$ and for $i=1$ this is the case iff $\pi_0(Sp(C)_0)=\pi_0(BC)$ is an abelian group with the induced multiplication which comes from the fact that $BC$ is a commutative monoid up to homotopy.
So the classifying space of a topological symmetric monoidal category $C$ can in this sense infinitely delooped only if $\pi_0(BC)$ is a group.
1.) Did I say anything stupid or wrong?
2.) Is Wikipedia wrong?
• What you say is correct, so Wikipedia was wrong. I have added a minimal correction to the statement in Wikipedia, but it would be nice if someone could expand on it. Dec 3, 2014 at 17:44
• Wouldn't this be a more appropriate discussion for the Wikipedia talk page? Dec 3, 2014 at 17:44
• in general $BC$ is an $E_{\infty}$-space and in the case where $\pi_{0}BC$ is a group (abelian in this case) then $BC$ is an infinite loop space.
– Max
Dec 3, 2014 at 18:24
• @RyanBudney People are a lot more likely to notice a question here on MO than on the Wikipedia page, I think. There have been a number of similar questions about issues with Wikipedia articles, and I think they are reasonable questions. Usually they get answered quickly and have the positive effect of improving Wikipedia. Dec 3, 2014 at 18:39
• Hi Dan, it looks to me like this question was asked and answered on the talk page a few months ago. Dec 3, 2014 at 18:43
|
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## Triangular Patches
I understand the principles of barycentric coordinates and I also understand what Bernstein Polynomials are in the sense of the coordinate plane. Can someone please explain to me in a constructive way from these ideas how to get to triangular bernstein bezier patches, please?
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# Markov Chain
## Homework Statement
-A bus is moving along an infinite route and has stops at n = 0, 1, 2, 3, ......
-The bus can hold up to B people
-The number of people waiting at stop n, Yn, is distributed Poisson(10) and is independent from the number waiting at all the other stops.
-At any given stop each passenger has probability p of getting off and the passengers are mutually independent.
-Assume after the passengers get off at stop n, everyone waiting at that stop gets on as long as there are spots remaining
$$\{X_n; n \geq 0\}$$ is a Markov Chain representing the number of passengers on the bus after it leaves stop n.
1) What are the transition probabilities, pij?
2) What is $$E[X_n]$$ for n = 0, 1, 2, ..., 20 if p = 0.4 and B = 20
## The Attempt at a Solution
(1)
$$X_n = \min\{X_{n-1} - L_n + Y_n, B\}$$
Where L_n ~ Binomial(X_{n-1}, p) is the number of people who got off at stop n
The transition probabilities need to be broken up into cases. For i > 0, j < i $$p_{i,j} = \sum_{k=i-j}^i {i \choose k} p^k(1-p)^{i-k} \left( \frac{10^{j-i+k} e^{-10}}{(j-i+k)!}\right)$$
for i>0, B > j >= i $$p_{i,j} = \sum_{k=0}^i {i \choose k} p^k(1-p)^{i-k} \left( \frac{10^{j-i+k} e^{-10}}{(j-i+k)!}\right)$$
for i>=0 j = B$$p_{i,j} = \sum_{k=0}^i {i \choose k} p^k(1-p)^{i-k} \left(\sum_{h=j-i+k}^{\infty}\frac{10^{h} e^{-10}}{(h)!}\right)$$
(2)
I feel reasonably okay about question (1) although maybe it's completely wrong. My main concern was part 2. If I want to use $$X_n = \min\{X_{n-1} - L_n + Y_n, B\}$$ to find the expected value, I get this $$E[X_n] =\min\{E[X_{n-1}] - E[L_n] + E[Y_n], B\}$$
which seems like it should be doable Except L_n ~ binomial (X_{n-1}, p) so the parameter is a random variable. Am I allowed to say$$E[L_n] = E[X_{n-1}]p$$
We haven't learned anything about having random variables in sum boundaries so I'm a bit clueless. Thank you.
Last edited:
Related Calculus and Beyond Homework Help News on Phys.org
Ray Vickson
Homework Helper
Dearly Missed
## Homework Statement
-A bus is moving along an infinite route and has stops at n = 0, 1, 2, 3, ......
-The bus can hold up to B people
-The number of people waiting at stop n, Yn, is distributed Poisson(10) and is independent from the number waiting at all the other stops.
-At any given stop each passenger has probability p of getting off and the passengers are mutually independent.
-Assume after the passengers get off at stop n, everyone waiting at that stop gets on as long as there are spots remaining
$$\{X_n; n \geq 0\}$$ is a Markov Chain representing the number of passengers on the bus after it leaves stop n.
1) What are the transition probabilities, pij?
2) What is $$E[X_n]$$ for n = 0, 1, 2, ..., 20 if p = 0.4 and B = 20
## The Attempt at a Solution
(1)
$$X_n = \min\{X_{n-1} - L_n + Y_n, B\}$$
Where L_n ~ Binomial(X_{n-1}, p) is the number of people who got off at stop n
The transition probabilities need to be broken up into cases. For i > 0, j < i $$p_{i,j} = \sum_{k=i-j}^i {i \choose k} p^k(1-p)^{i-k} \left( \frac{10^{j-i+k} e^{-10}}{(j-i+k)!}\right)$$
for i>0, B > j >= i $$p_{i,j} = \sum_{k=0}^i {i \choose k} p^k(1-p)^{i-k} \left( \frac{10^{j-i+k} e^{-10}}{(j-i+k)!}\right)$$
for i>=0 j = B$$p_{i,j} = \sum_{k=0}^i {i \choose k} p^k(1-p)^{i-k} \left(\sum_{h=j-i+k}^{\infty}\frac{10^{h} e^{-10}}{(h)!}\right)$$
(2)
I feel reasonably okay about question (1) although maybe it's completely wrong. My main concern was part 2. If I want to use $$X_n = \min\{X_{n-1} - L_n + Y_n, B\}$$ to find the expected value, I get this $$E[X_n] =\min\{E[X_{n-1}] - E[L_n] + E[Y_n], B\}$$
which seems like it should be doable Except L_n ~ binomial (X_{n-1}, p) so the parameter is a random variable. Am I allowed to say$$E[L_n] = E[X_{n-1}]p$$
We haven't learned anything about having random variables in sum boundaries so I'm a bit clueless. Thank you.
I guess in part (2) you are supposed to use the numerical values of B and p to get a numerical ##21 \times 21## one-step transition matrix ##P = \left( p_{i,j} \right) ##. Then, if you are given a starting state ##X_0 \in \{ 0, \ldots, 20 \}##, say ##X_0 = i_0##, the distribution of ##X_n## is just row ##i_0## of the matrix ##P^n##, so the expected value is readily computable. This is more easily obtained by starting with initial state probability (row) vector ##v_0 = (0,0, \ldots,1,0 \ldots, 0)##, where the 1 is in position ##i_0##. Then the row vector ##v_n## of state probabilities at time ##n## is given by ##v_n = v_{n-1} P##, so we can get ##v_1, v_2, \ldots, v_{20}## recursively, without the need to compute the whole matrix ##P^n## for each ##n##. In general, the values of ##EX_n = \sum_{k=0}^{20} k v_n(k)## will depend on the starting state ##i_0##.
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edr1c 2 years ago in fourier series cos(npi) is (-1)^n and sin(npi)=0, what is cos(npi/2) and sin(npi/2)? and when the answer is 0 when n is odd i change n into 2n, while if answer is 0 when n is even i change n into 2n-1. do i apply the same thing for npi/2?
1. edr1c
im always stucked here when im doing the cosine and sine series in the PDE, especially when its stepfunction with pi as interval.
2. Herp_Derp
$\cos(\pi/2n) = \begin{cases} (-1)^{n/2} &\;\;\;\;\;\;\text{if }n\text{ is even}\\ 0 &\;\;\;\;\;\;\text{if }n\text{ is odd} \end{cases}$$\sin(\pi/2n) = \begin{cases} 0 &\;\text{if }n\text{ is even}\\ (-1)^{(n-1)/2} &\;\text{if }n\text{ is odd} \end{cases}$
3. edr1c
oh. so if i want n to represent for all n, change n into 2n if n is odd gives 0, and change n into 2n-1 if n is even gives 0?
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# Quantum Computation via Sparse Distributed Representation
15 Jul 2017Gerard J. Rinkus
Quantum superposition says that any physical system simultaneously exists in all of its possible states, the number of which is exponential in the number of entities composing the system. The strength of presence of each possible state in the superposition, i.e., its probability of being observed, is represented by its probability amplitude coefficient... (read more)
PDF Abstract
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# How do I organize chapters into unnumbered supra chapters?
I'm working on a book that has two parts. Each part is, of course, composed of several chapters which are composed of sections. Great, right? Sure!
But....
For logistical reasons we want to organize the chapters into "chunks" within a part. So:
• Part 1: DESSERTS
• [intro to Part 1]
• ICE CREAM
• [ intro]
• Chapter 1: Chocolate ice cream
• Chapter 2: Vanilla ice cream
• PIE
• [ intro]
• Chapter 3: Apple pie
• Chapter 4: Blueberry pie
But of course, "ice cream" and "pie" aren't chapters in the same way that "vanilla ice cream" and "blueberry pie" are chapters.
How can I organize these two "supra chapters" above the level of the other chapters, but without giving them a chapter number?
• \chapter*{ICE CREAM} perhaps, slightly modified to get rid of the clearpage (if that is an issue?) – user31729 Sep 6 '16 at 13:57
• It lacks Gosky patties and Amblongus pies ;o) – Bernard Sep 6 '16 at 15:49
Perhaps this is the simplest method: Using a wrapper command that performs \chapter*{} and adds the line to the ToC automatically.
\frontmatter\chapter{foo}\mainmatter would be possible too, but this changes the pagenumbering as well, which is not desired, most likely.
I provide two variants, both having the same effect. I prefer the 2nd one with no explicit \addcontentsline, however!
The 2nd variant uses a temporary change of the secnumdepth coutner which controls whether a sectioning unit is numbered in display or not. Setting it to -1 will only leave \part countered, restoring after the suprachapter has been used will provide the usual behaviour.
\documentclass{book}
\usepackage{xparse}
\usepackage{hyperref}
\newcounter{oldsecnumdepth}
\NewDocumentCommand{\suprachapter}{som}{%
\IfBooleanTF{#1}{%
\chapter*{#1}%
}{%
\chapter*{#3}%
\IfValueTF{#2}{%
}{%
}%
}%
}
\NewDocumentCommand{\suprachapterother}{som}{%
\IfBooleanTF{#1}{%
\chapter*{#1}%
}{%
\setcounter{oldsecnumdepth}{\value{secnumdepth}}%
\setcounter{secnumdepth}{-1}%
\IfValueTF{#2}{%
\chapter[#2]{#3}%
}{%
\chapter{#3}%
}%
\setcounter{secnumdepth}{\value{oldsecnumdepth}}%
}%
}
\begin{document}
\tableofcontents
\part{Foo}
\suprachapter{Ice scream}
\chapter{Vanilla flavour}
\chapter{Albatross flavour}
\suprachapterother{PIE}
\chapter{Apple Pie}
\chapter{Black Forest Cherry Tarte}
\end{document}
• you, my friend, are an all-star genius. THANK YOU – Teusz Sep 8 '16 at 14:10
• @Teusz: You're welcome. Happy TeXing! – user31729 Sep 8 '16 at 14:13
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## JDCC '17 Contest 3 S3 - Interview Progression
View as PDF
Points: 7 (partial)
Time limit: 2.0s
Memory limit: 512M
Problem type
Atharva is currently in an interview. He has memorized a list of statements that he is prepared to say during the interview. The interviewer is marking him by points. He starts the interview with one point. Every statement he says affects his points in some way – as a function of his current points. Find a way to maximize the amount of points he gets in the interview.
The statements change his points in one of three ways (as a function of p-number of points):
1.+c where the new number of points is $$p + c$$
2.^c where the new number of points is
3.*c where the new number of points is
#### Input Specification
The first line contains , the number of statements that Atharva has prepared.
The next lines contain a statement in the form stated above, describing how the statement affects the points.
$$1 \le S \le 10^5$$
$$0 \le c \le 10^9$$ (whole number)
#### Output Specification
On the first line, output the number of statements he should say.
On the second line, output a number of indices (0-indexed), representing the order Atharva should say the statements so that he can maximize his points.
If there are multiple ways to do this, pick the way with the least number of statements and the least lexicographical order.
#### Sample Input 1
3
+0
^1
*1
#### Sample Output 1
0
#### Sample Input 2
3
^2
+5
*3
#### Sample Output 2
3
1 2 0
#### Explanation for Sample Outputs
For the first sample, no matter what Atharva says, his points stay at .
For the second sample Atharva can achieve a maximum score of $$2^{3 \times (1+5)} = 262144$$.
|
# Logical Form and Formal Validity
Often, the validity of an argument is connected to its logical form. Both of our sample arguments from the previous section—the one about the cars and the one about the platypus—have the same logical form. Consider:
1. No are .
2. All are .
3. Therefore, all are not .
Each of the boxes above is actually a text input box: you can type in anything you want. Try replacing with ‘mammals’, with ‘egg-layers’, and with ‘platypus’.
Play around with it: plug in any terms—that is, any common nouns or noun phrases—into the form. Each time, note whether or not the premises and conclusion are true.
Now for a more focused challenge. Try plugging in terms that make the premises all true. Once you’ve done that, the conclusion will be specified. Is it true or false? Can you find terms that make all the premises true but the conclusion false?
Now try plugging in terms that make the conclusion false. Once you’ve done that, all but one of the terms will be specified. Can you find a value for that third term that makes one of the premises false?
This argument is formally valid: there is no argument of this form—no argument that you can get by plugging in different terms—that has actually true premises and an actually false conclusion.
Here is another argument form that involves three terms:
1. Some are
2. Some are
3. ∴ Some are
This argument is not formally valid. Can you find values for , , and that make both of the premises true and the conclusion false?
Here is one way to do it: let ‘As’ be ‘cats’; let ‘Bs’ be ‘animals’; let ‘Cs’ be ‘dogs’.
In the previous section, we saw that validity had to do with what is or isn’t possible: is it possible that all the premises be true but the conclusion false? Figuring out what is or isn’t possible requires a lot of imagination, and it isn’t always clear what we should say.1
But formal validity is more straightforward.
Formal Validity
An argument is formally valid just in case there is no argument with the same logical form that has all (actually) true premises and a(n actually) false conclusion.
We can test the formal validity of an argument by trying to find an argument that has the same logical form whose premises are actually true, and whose conclusion is actually false.
Let’s practice.
The following form is invalid. Show this by finding values for ‘A’, ‘B’, and ‘C’ that make the premises true and the conclusion false.
1. Some are
2. All are not
3. ∴ All are not
Here is one way to do it: let ‘A’ be ‘numbers’, ‘B’ be ‘odd numbers’, and ‘C’ be ‘divisible by two’. Then the premises are both true, but the conclusion is false.
Here is another way to do it: let ‘A’ be ‘animals’, ‘B’ be ‘dogs’, and ‘C’ be ‘cats’.
Is this form valid? Try to find values for ‘A’, ‘B’, and ‘C’ that make the premises true and the conclusion false.
1. All are
2. All are
3. ∴ All are
The form is valid. No matter what nouns you plug in for ‘A’, ‘B’, and ‘C’, if the premises are true, the conclusion will be true too. Aristotle called this argument form “Barbara”.
Notice that we cannot really show that the argument is valid. All we can do is try to come up with an example that shows that it is invalid. When we cannot come up with any such examples, how can we be sure that this is not simply due to our lack of imagination?
Is this form valid?
1. Some are
2. ∴ Some are
Yes, it is.
# Sentential Logic: ‘if’, ‘not’, ‘and’, ‘or’
In the previous section, we took the common nouns that occurred in an argument—words like ‘platypus’ and ‘police car’—to be the non-logical content of an argument—and we took the words that surrounded them—words like ‘all’, ‘some’, ‘no’, ‘not’, and ‘are’—to indicate the logical form of the argument.
The resulting logic is called Term Logic (or sometimes Aristotelean Logic, because this was the sort of logic that Aristotle developed). It is also called Traditional Logic, because it was the logic used throughout the medieval and early modern periods. But this is not the only way one might think about logical form.
Consider the following argument:
1. If , then .
2. .
This argument is valid: any possible situation that makes both premises true makes the conclusion true too.
Further, the argument is formally valid. Replace with any sentence, and with any sentence, and the resulting argument will also be valid. Try it, by typing in other sentences into the box and the circle.
Here is another argument, similar but slightly different in form:
1. If capital punishment deters crimes, then it is justified.
2. Capital punishment does not deter crimes.
3. ∴ Capital punishment is not justified.
Is this argument valid? The answer may not be immediately obvious. We can represent the form of the argument as,
1. If , then
2. It is not the case that
3. ∴ It is not the case that
Can you find sentences that, when plugged into the box and circle, make the premises true but the conclusion false? (Note that the box and circle are still interactive: you can type sentences into them.)
Try plugging in ‘Hilary Clinton is President’ into the box and ‘a Democrat is President’ into the circle. Then the premises are both true, but the conclusion is false.
This invalid form of argument is a fairly common logical fallacy—that is, logical mistake people make when reasoning. It is common enough that it has a name: denying the antecedent.
To understand that name, you need to know a few more technical terms. We call an ‘if … then …’ sentence a conditional. We call the ‘if’-part of a conditional the antecedent, and we call the ‘then’-part the consequent. So, for example, here is a conditional:
• If Bonzo is an ape, then he should go to bed,
The antecedent is ‘Bonzo is an ape’. The consequent is ‘Bonzo should go to bed’. So, to deny the antecedent of this conditional would be to say that the Bonzo is not an ape.
Here are four common forms of argument that involve conditionals. Which are valid and which are invalid?
Modus Ponens:
If , then . $$∴$$
Denying the Antecedent
If , then . It is not the case that $$∴$$ It is not the case that
Affirming the Consequent
If , then . $$∴$$
Modus Tollens: If P, then Q. It is not the case that Q. ∴ It is not the case that P.
If , then . It is not the case that $$∴$$ It is not the case that
Which of these four forms are valid, and which invalid? Try to work out your answer before clicking the box below.
The first and last forms, with the weird Latin names, are valid. The other two are invalid. For the invalid forms, can you come up with sentences that make the premises true but the conclusion false?
There are other ways to combine sentences to form new more complicated sentences. For example, given the sentences and , we can form sentences like,
• and .
• or .
• just in case .
These examples all share an important feature: they are truth-functional. That is, the truth or falsehood of the complex sentence is a function of the truth or falsehood of its component parts.
This isn’t true of all connectives in English. Consider,
• because .
Both of the component sentences are true. The NRA gave Senator Jim Renacci \$9,900 in support of his 2012 election campaign, and Renacci voted against a gun control bill. But that is not enough to tell us whether or not the complex sentence is true, whether the contribution caused his vote. So ‘because’ is not truth-functional.
Can you think of other connectives in English that are not truth-functional?
There are lots of examples. Here are two:
• ‘while’, as in: “It rained while I walked.” Suppose ‘it rained’ and ‘I walked’ are both true. That does not settle whether or not ‘it rained while I walked’ is true.
• ‘necessarily’, as in: “Necessarily, 2+2=4,” or “Necessarily, there are cows”. The former is true and the second is false, but ‘2+2=4’ and ‘there are cows’ are both true.
# Review
The aim of chapter 0 was to provide you a first introduction to logic. Key terms that were defined include logic, argument, premise, conclusion, valid, sound, formally valid, conditional, antecedent, and consequent. You were introduced to two kinds of logic form—the sort studied by term logic and the sort studied by sentential logic. In the next chapter, we will begin developing sentential logic.
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# What is the motivation behind the arbitrary union topological axiom?
1. Why is the arbitrary union axiom in the definition of topology necessary?
2. Why is it useful? Why might we expect ("intuitively") that it should be useful?
3. What is the (historical) motivation for the axiom?
Please do not read any of the below unless you are really bored or have a lot of time to kill; the essence of my question is the above.
In what follows, I will follow Dellacherie and Meyer in denoting:
$\underset{f}{\cup}$ = finite unions
$\underset{c}{\cup}$ = countable unions
$\underset{u}{\cup}$ = uncountable unions
$\underset{a}{\cup}$ = arbitrary unions
Similar notation applies for intersections.
In what follows I will take closure under (relative) complementation for granted, simply because almost all set systems which I know of (besides $\pi$ systems) have this property, its motivation is obvious, and because it is involutory it is easily added to the axioms of a set system without changing the included sets very much.
Most spaces which are commonly studied have at most the continuity of the continuum (not that there aren't counterexamples, but at least I don't run into them in my work).
Thus, it seems we could have at most the following types of set systems for a space which has at most the cardinality of the continuum.
1. $\underset{f}{\cup}$, $\underset{f}{\cap}$ (Boolean) algebra
2. $\underset{c}{\cup}$, $\underset{f}{\cap}$/$\underset{c}{\cap}$, $\underset{f}{\cup}$ (slightly stronger than a $\lambda$ system)
3. $\underset{c}{\cup}$, $\underset{c}{\cup}$ $\sigma-$algebra
4. $\underset{u/a}{\cup}$, $\underset{c}{\cup}$/ $\underset{u/a}{\cap}$, $\underset{c}{\cup}$ ????? Related: Is there a difference between allowing only countable unions/intersections, and allowing arbitrary (possibly uncountable) unions/intersections?
5. $\underset{u/a}{\cup}$, $\underset{f}{\cap}$/ $\underset{u/a}{\cap}$, $\underset{f}{\cup}$ systems of open sets (topologies)/systems of closed sets resepctively
6. $\underset{u/a}{\cup}$, $\underset{u/a}{\cap}$ Alexandroff topologies
The motivation for the set axioms of a Boolean algebra are obvious, because they basically constitute "closure under elementary logical operations". For example, all truth functions are well-defined given a Boolean algebra, as is propositional logic.
The motivation for a $\sigma-$algebra is also fairly clear to me -- it is essentially the "countable closure/extension" of a regular algebra, and this property allows us to define countably additive set functions, and along with that obtain closure under pointwise limits of functions as well as the monotone convergence theorem, the two foundations on which integration theory is based.
It is also clear that such additivity could not be extended to the cardinality of the continuum, since the uncountable sum of any non-zero numbers is infinite.
But why do open/closed set systems turn out to be useful? How are these uncountable axioms applied fruitfully in topology?
As far as I am aware, for example with stochastic processes, we usually rely on some separability/first/second-countability properties in order to be able to apply most topologies fruitfully -- i.e. some reduction of the "uncountable" axiom to an essentially countable property. So why not study only metric spaces?
Moreover, the study of limits is only defined/sensible once we have the T1 and T2 separation axioms, so the claim that we study set systems of type 5. in order to study limits seems unjustified to me; at the very least we need that in combination with some separation axioms.
Also, why do we study set systems of type 5., but not of types 4. or 6.? What is the essential property that type 5. set systems capture (in the way that type 1. and type 3. systems capture)?
The best discussion of this topic which I have found so far: https://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets
I agree with the OP of that question however, that the top-voted answer is more of a justification for the notion of metric space (which are always separable in some sense due to the density of the rationals in the reals).
The discussion of how notions of limit really seem to be more of a consequence of a notion of "Hausdorfness" is here: Why not just study the consequences of Hausdorff axiom? What do statements like, "The arbitrary union of open sets is open," gain us? I would also like to point out that the axiomatization of topology in terms of neighborhood systems, plus the Hausdorff axiom, seems to be more relevant to understanding the concept of "limit" than the axioms of a general topological space. And again, I don't see why we would need uncountably many neighborhoods of a given point to define it as the limit of something, since countably many neighborhoods always suffices (at least in the metric space setting).
Motivation:
The motivation for this question came from some rumination about stochastic processes. Namely, the reason why their study can be so difficult is that we are confined to using set systems of type 3., which are only countably closed, whereas we are studying uncountably many random variables.
In order to get any degree of meaningful results, we therefore rely on having some degree of "separability/regularity" properties that essentially allow us to reduce to the countable case where it counts (for instance with regards to the existence of regular conditional distributions).
So as frustrating as it may be trying to study uncountable objects with set systems of type 3., it is clear to me why we can not replace such set systems with any of type 4, 5, or 6, again so that integration and probability are tractable.
So then I thought "oh so we must use topologies because we need uncountability for some reason". But then I thought about how almost all topologies which I am aware of/use are in some way "separable" or "first/second-countable", which basically constitutes reducing back to the countable case, the same way we do with stochastic processes. I.e. there is no fundamental gain in generality, since the only cases of uncountable objects we can/do study are those special cases which can be reduced back to the countable case in some meaningful way.
I'm really at a loss as to why an uncountable closure axiom would be at all useful, and why we would want only finite closure for the complementary operation, rather than some type of infinite closure.
I feel like it has to have something to do with the distinction between "discrete" and "continuous" spaces, which leads us back again to the study of stochastic processes, and the distinction between "discrete" and "continuous" time.
Namely, for a set with countable cardinality, like the natural numbers, the power set coincides with the reasonable choices for sigma algebra and topology.
In other words, both concepts are clearly only needed when moving to the uncountable case, where they don't necessarily coincide with the power set, i.e. we can gain a concept of "continuousness", rather than the "discreteness" which we are necessarily confined to with set systems of countable sets. Again, the purpose of $\sigma-$algebra in this case is clear to me, since it is actually only closed under countable set operations, but now I really don't understand why we are able to use the uncountable set axiom profitably in open/closed set systems, especially since in all cases which I am aware of we have some sort of "separability" which reduces the concept of topology to a countable one. So why not define the concept which we actually use, which is countable, not uncountable in nature?
• I think the first nonmetrizable space that we encounter as analysts is the weak star topology on an infinite dimensional Banach space which is the dual of some other infinite dimensional Banach space. For example, $L^p(I)$ with $1<p<\infty$ and $I$ an interval in $\mathbb{R}$. We wind up needing this basically because the strong topology doesn't have enough compact sets for us to be able to extract convergent subsequences from certain approximation procedures. – Ian Jun 9 '16 at 17:30
• This also comes up in stochastic process theory (there are whole books about proving things in probability theory using weak convergence methods; for example I've looked at one that looks at large deviation theory from this perspective). Is this enough motivation? – Ian Jun 9 '16 at 17:33
• Well, my point was that as you've said, the uncountable union axiom basically doesn't come up in separable metric spaces, which is to say essentially all metric spaces of interest in analysis. So I went for the first nonmetrizable space I could think of. As it happens that $L^p$ space in particular is separable, and the dual of a separable space is separable in its weak star topology, so you are right. – Ian Jun 9 '16 at 19:17
• I guess this is another illustration case: if open sets only allow countable unions, then closed sets must only allow countable intersections. This could be bad: for example, if you have a mapping that sends each point $x$ in the Cantor set to a closed interval $I(x)$, the intersection of $\{ f \in C[0,1] : f(x) \in I(x) \}$ should be closed (if a sequence of functions is in that intersection and converges uniformly, then the limit is in that intersection, certainly), but you would not be able to infer that from the topology axioms. – Ian Jun 9 '16 at 19:17
• Aside: you do have arbitrary meets; it's just that they're not given by intersection. $$\bigwedge_{i \in I} U_i = \text{interior}\left( \bigcap_{i \in I} U_i \right)$$ (also, finite meets distribute over arbitrary unions, but infinite meets don't have to) – Hurkyl Jun 12 '16 at 8:33
Here are three characterizations that have nothing to do with arbitrary unions
A base on $X$ is a collection $N \subseteq \mathcal{P}(X)$ of neighborhoods satisfying
• Every point is contained in at least one neighborhood
• If $A$ and $B$ are neighborhoods and $P \in A \cap B$, then there is a neighborhood $C$ such that $P \in C \subseteq A \cap B$.
You can develop topology in terms of bases. e.g.
$f : X \to Y$ is continuous if and only if, for every neighborhood $V \subseteq Y$ and every point $P \in X$ such that $f(P) \in V$, there is a neighborhood $U \subseteq X$ such that $P \in U$ and $f(P) \subseteq V$.
Let $O(N)$ denote the set of all arbitrary unions of neighborhoods from $N$. Then $O(N)$ satisfies the open set axioms. Furthermore:
If $N$ and $N'$ are two different bases on $X$, then the identity map defines a homeomorphism $(X,N) \to (X,N')$ if and only if $O(N) = O(N')$.
Conversely, the collection of open sets in a topological space satisfies the neighborhood axioms.
A closure operation on $X$ is an operation on subsets of $X$ satisfying
• $\overline{\varnothing} = \varnothing$
• $A \subseteq \overline{A}$
• $\overline{A \cup B} = \overline{A} \cup \overline{B}$
• $\overline{\overline{A}} = \overline{A}$
The collection of sets satisfying $A = \overline{A}$ satisfies the closed set axioms.
Conversely, in a topological space, closure is a closure operation.
A nearness relation on $X$ is a relation between elements of $X$ and subsets of $X$
• No point is near $\varnothing$
• If $P \in A$, then $P$ is near $A$
• $P$ is near $A \cup B$ if and only if $P$ is near $A$ or $P$ is near $B$ (or both)
• If $P$ is near $A$ and every point of $A$ is near $B$, then $P$ is near $B$
The set of near points to $A$ defines a closure relation, so we again get a topological space.
Conversely, $P \in \bar{A}$ is a nearness relation.
• The axiomatization of "nearness" is mine and maybe has errors, although I'm sure I've seen it elsewhere. Also, I didn't look up the phrasing of continuity for bases, so it may have an error too. The other two formulations are well-known. – Hurkyl Jun 12 '16 at 8:30
• I guess what I am confused about is why we need the arbitrary union of base sets to be open, when in all cases I know of, countable unions suffice. Do you know a (non-discrete) case where arbitrary unions of base sets are necessary as opposed to just countable unions? – Chill2Macht Jun 12 '16 at 14:08
• @William: If every arbitrary union can be written as a countable union, then you still have arbitrary unions. If you don't have that property, then all three of the characterizations I gave imply that requiring arbitrary unions is the "right" definition and merely countable unions is "wrong". – Hurkyl Jun 12 '16 at 14:18
• Two simple counterexamples I can think of are the discrete topology on an uncountable set (take a base of singletons) and the long line (take a base consisting of covers of the individual intervals constructing it). Maybe there are things involving function spaces? I don't know this sort of detail about those. – Hurkyl Jun 12 '16 at 14:34
• @William: Oh, I should also add that typical arguments use arbitrary unions (e.g. for every point having some property choosing a neighborhood with some property); so even if you had a space where arbitrary unions could be written as countable unions, you would still prefer to have theorems formulated in terms of the arbitrary unions. Otherwise, arguments would have to take unnecessary detours through working with countable dense subsets or countable subcovers. – Hurkyl Jun 12 '16 at 14:48
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Top
# Parallel optimization
Iaroslav Shcherbatyi, May 2017.
Reviewed by Manoj Kumar and Tim Head.
## Introduction
For many practical black box optimization problems expensive objective can be evaluated in parallel at multiple points. This allows to get more objective evaluations per unit of time, which reduces the time necessary to reach good objective values when appropriate optimization algorithms are used, see for example results in [1] and the references therein.
One such example task is a selection of number and activation function of a neural network which results in highest accuracy for some machine learning problem. For such task, multiple neural networks with different combinations of number of neurons and activation function type can be evaluated at the same time in parallel on different cpu cores / computational nodes.
The “ask and tell” API of scikit-optimize exposes functionality that allows to obtain multiple points for evaluation in parallel. Intended usage of this interface is as follows:
1. Initialize instance of the Optimizer class from skopt
2. Obtain n points for evaluation in parallel by calling the ask method of an optimizer instance with the n_points argument set to n > 0
3. Evaluate points
4. Provide points and corresponding objectives using the tell method of an optimizer instance
5. Continue from step 2 until eg maximum number of evaluations reached
## Example
A minimalistic example that uses joblib to parallelize evaluation of the objective function is given below.
from skopt import Optimizer
from skopt.learning import GaussianProcessRegressor
from skopt.space import Real
from sklearn.externals.joblib import Parallel, delayed
# example objective taken from skopt
from skopt.benchmarks import branin
optimizer = Optimizer(
dimensions=[Real(-5.0, 10.0), Real(0.0, 15.0)],
random_state=1
)
for i in range(10):
x = optimizer.ask(n_points=4) # x is a list of n_points points
y = Parallel()(delayed(branin)(v) for v in x) # evaluate points in parallel
optimizer.tell(x, y)
# takes ~ 20 sec to get here
print(min(optimizer.yi)) # print the best objective found
0.398019152257
Note that if n_points is set to some integer > 0 for the ask method, the result will be a list of points, even for n_points=1. If the argument is set to None (default value) then a single point (but not a list of points) will be returned.
The default "minimum constant liar" [1] parallelization strategy is used in the example, which allows to obtain multiple points for evaluation with a single call to the ask method with any surrogate or acquisition function. Paralellization strategy can be set using the "strategy" argument of ask. For supported parallelization strategies see the documentation of scikit-optimize.
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## The problem of the extension of a parametric family of diophantine triples.(English)Zbl 0903.11010
A set of positive integers $$\{a_1,a_2,\ldots,a_m\}$$ is said to have the property of Diophantus if $$a_ia_j+1$$ is a perfect square for all $$1\leq i<j\leq m$$. Such a set is called a diophantine $$m$$-tuple. There is a well-known diophantine quadruple for all integers $$k\geq 2$$: the set $$\{k-1,k+1,4k,16k^3-4k\}$$.
The author proves the following statement. Theorem. Let $$k\geq 2$$ be an integer. If the set $$\{k-1,k+1,4k,d\}$$ has the property of Diophantus, then $$d$$ has to be $$16k^3-4k$$. It is worth mentioning that this result was known only in the cases $$k=2$$ and $$k=3$$. To prove the mentioned theorem, the author applies a system of Pellian equations, a result of Rickert on simultaneous rational approximations, linear forms of logarithms and the Grinstead method.
### MSC:
11D09 Quadratic and bilinear Diophantine equations 11D25 Cubic and quartic Diophantine equations 11J13 Simultaneous homogeneous approximation, linear forms 11J86 Linear forms in logarithms; Baker’s method
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# International Standards - Matrix Algebra - 2.4
IS MX 2-4
Understand that the transpose of an m \times n matrix is an n \times m matrix.
Page last modified on September 16, 2008, at 10:23 AM
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### Bounds on information exchange for Byzantine agreementBounds on information exchange for Byzantine agreement
Access Restriction
Subscribed
Author Dolev, Danny ♦ Reischuk, Rdiger Source ACM Digital Library Content type Text Publisher Association for Computing Machinery (ACM) File Format PDF Copyright Year ©1985 Language English
Subject Domain (in DDC) Computer science, information & general works ♦ Data processing & computer science Abstract Byzantine Agreement has become increasingly important in establishing distributed properties when errors may exist in the systems. Recent polynomial algorithms for reaching Byzantine Agreement provide us with feasible solutions for obtaining coordination and synchronization in distributed systems. In this paper the amount of information exchange necessary to ensure Byzantine Agreement is studied. This is measured by the total number of messages the participating processors have to send in the worst case. In algorithms that use a signature scheme, the number of signatures appended to messages are also counted.First it is shown that $&OHgr;(\textit{nt})$ is a lower bound for the number of signatures for any algorithm using authentication, where $\textit{n}$ denotes the number of processors and $\textit{t}$ the upper bound on the number of faults the algorithm is supposed to handle. For algorithms that reach Byzantine Agreement without using authentication this is even a lower bound for the total number of messages. If $\textit{n}$ is large compared to $\textit{t},$ these bounds match the upper bounds from previously known algorithms. For the number of messages in the authenticated case we prove the lower bound $&OHgr;(\textit{n}$ + $\textit{t}2).$ Finally algorithms that achieve this bound are presented. ISSN 00045411 Age Range 18 to 22 years ♦ above 22 year Educational Use Research Education Level UG and PG Learning Resource Type Article Publisher Date 1985-01-01 Publisher Place New York e-ISSN 1557735X Journal Journal of the ACM (JACM) Volume Number 32 Issue Number 1 Page Count 14 Starting Page 191 Ending Page 204
#### Open content in new tab
Source: ACM Digital Library
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# I get an error every time I try to run my Latex document in Texmaker, using a chunk code from R
I'm following the following guide: https://joshldavis.com/2014/04/12/beginners-tutorial-for-knitr/
I've installed the knitr package and run it in R, and I'm trying to copy and paste the example from the link above:
From ISLR: Chapter 3, Problem 14.
Using a created simulated data, answer the questions regarding simple
linear regression.
<<>>=
# Ensure consistent values
set.seed(1)
# Create uniform distribution for first input
x1 <- runif(100)
# Normal distribution for second input
x2 <- 0.5 * x1 + rnorm(100) / 10
# Our Linear Model
y <- 2 + (2 * x1) + (.3 * x2) + rnorm(100)
@
However, when I try to run it, I get the following error in Latex:
! You can't use macro parameter character #' in horizontal mode.
l.30 #
Ensure consistent values
?
I'm not quite sure what I'm doing wrong here? Is there any package I'm supposed to run in Latex as well?
• The file should be a normal and complete LaTeX file (i.e., starting with \documentclass ... and ending with \end{document}`) but with the .Rnw extension, and configure Texmaker properly. Please add a minimal working example (MWE) of a .Rnw and the configuration of the command that you run. I you are new with nitr files, check first that your first .Rnw file can be compiled with RStudio before to deal with configuration of Texmaker. – Fran Aug 30 '18 at 20:06
• Duplicate on SO – Troy Aug 30 '18 at 23:59
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Converting CTI and XML input files to YAML¶
If you want to convert an existing, legacy CTI or XML input file to the YAML format, this section will help.
cti2yaml¶
Cantera comes with a converter utility cti2yaml (or cti2yaml.py) that converts legacy CTI format mechanisms into the new YAML format introduced in Cantera 2.5. This program can be run from the command line to convert files to the YAML format. (New in Cantera 2.5)
Usage:
cti2yaml [-h] input [output]
The input argument is required, and specifies the name of the input file to be converted. The optional output argument specifies the name of the new output file. If output is not specified, then the output file will have the same name as the input file, with the extension replaced with .yaml.
Example:
cti2yaml mymech.cti
will generate the output file mymech.yaml.
If the cti2yaml script is not on your path, but the Cantera Python module is, cti2yaml can be used by running:
python -m cantera.cti2yaml mymech.cti
It is not necessary to use cti2yaml to convert any of the CTI input files included with Cantera. YAML versions of these files are already included with Cantera.
For input files where you have both the CTI and XML versions, cti2yaml is recommended over ctml2yaml. In cases where the mechanism was originally converted from a CK-format mechanism, it is recommended to use ck2yaml if the original input files are available.
ctml2yaml¶
Cantera comes with a converter utility ctml2yaml (or ctml2yaml.py) that converts legacy XML (CTML) format mechanisms into the new YAML format introduced in Cantera 2.5. This program can be run from the command line to convert files to the YAML format. (New in Cantera 2.5)
Usage:
ctml2yaml [-h] input [output]
The input argument is required, and specifies the name of the input file to be converted. The optional output argument specifies the name of the new output file. If output is not specified, then the output file will have the same name as the input file, with the extension replaced with .yaml.
Example:
ctml2yaml mymech.xml
will generate the output file mymech.yaml.
If the ctml2yaml script is not on your path, but the Cantera Python module is, ctml2yaml can be used by running:
python -m cantera.cti2yaml mymech.xml
It is not necessary to use ctml2yaml to convert any of the XML input files included with Cantera. YAML versions of these files are already included with Cantera.
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# Proofs of AM-GM
This pages lists some proofs of the weighted AM-GM Inequality. The inequality's statement is as follows: for all nonnegative reals $a_1, \dotsc, a_n$ and nonnegative reals $\lambda_1, \dotsc, \lambda_n$ such that $\sum_{i=1}^n \lambda_i = 1$, then $$\sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i},$$ with equality if and only if $a_i = a_j$ for all $i,j$ such that $\lambda_i, \lambda_j \neq 0$.
We first note that we may disregard any $a_i$ for which $\lambda_i= 0$, as they contribute to neither side of the desired inequality. We also note that if $a_i= 0$ and $\lambda_i \neq 0$, for some $i$, then the right-hand side of the inequality is zero and the left hand of the inequality is greater or equal to zero, with equality if and only if $a_j = 0 = a_i$ whenever $\lambda_j\neq 0$. Thus we may henceforth assume that all $a_i$ and $\lambda_i$ are strictly positive.
## Proofs of Unweighted AM-GM
These proofs use the assumption that $\lambda_i = 1/n$, for all integers $1 \le i \le n$.
### Proof by Cauchy Induction
We use Cauchy Induction, a variant of induction in which one proves a result for $2$, all powers of $2$, and then that $n$ implies $n-1$.
Base Case: The smallest nontrivial case of AM-GM is in two variables. By the properties of perfect squares (or by the Trivial Inequality), $(x-y)^2 \geq 0,$ with equality if and only if $x-y=0$, or $x=y$. Then because $x$ and $y$ are nonnegative, we can perform the following manipulations: $$x^2 - 2xy + y^2 > 0$$ $$x^2 + 2xy + y^2 > 4xy$$ $$(x+y)^2 = 4xy$$ $$\frac{(x+y)^2}{4} \geq xy$$ $$\frac{x+y}{2} \geq \sqrt{xy},$$ with equality if and only if $x=y$, just as before. This completes the proof of the base case.
Powers of Two: We use induction. Suppose that AM-GM is true for $n$ variables; we will then prove that the inequality is true for $2n$. Let $x_1, x_2, \ldots x_{2n}$ be any list of nonnegative reals. Then, because the two lists $x_1, x_2 \ldots x_n$ and $x_{n+1}, x_{n+2}, \ldots x_{2n}$, each have $n$ variables, $$\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n} \textrm{ and } \frac{x_{n+1} + x_{n+2} + \cdots + x_{2n}}{n} \geq \sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}.$$ Adding these two inequalities together and dividing by two yields $$\frac{x_1 + x_2 + \cdots + x_{2n}}{2n} \geq \frac{\sqrt[n]{x_1 x_2 \cdots x_n} + \sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}}{2}.$$ From here, we perform AM-GM in two variables on $\sqrt[n]{x_1 x_2 \cdots x_n}$ and $\sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}$ to get $$\frac{\sqrt[n]{x_1 x_2 \cdots x_n} + \sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}}{2} \geq \sqrt[2n]{x_1 x_2 \ldots x_{2n}}.$$ Combining this inequality with the previous one yields AM-GM in $2n$ variables, with one exception — equality.
For equality, note that every AM-GM application mentioned must have equality as well; thus, inequality holds if and only if all the numbers in $x_1, x_2, \ldots x_n$ are the same, all the numbers in $x_{n+1}, x_{n+2}, \ldots x_{2n}$ are the same, and $\sqrt[n]{x_1 x_2 \cdots x_n} = \sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}$. From here, it is trivial to show that this implies $x_1 = x_2 = \cdots x_{2n}$, which is the equality condition for AM-GM in $2n$ variables.
This completes the induction and proves that the inequality holds for all powers of two.
Backward Step: Assume that AM-GM holds for $n$ variables. We will then use a substitution to derive AM-GM for $n-1$ variables. Letting $x_n = \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}$, we have that $$\frac{x_1 + x_2 + \cdots + x_{n-1} + \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_{n-1} \left(\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}\right)}.$$ Because we assumed AM-GM in $n$ variables, equality holds if and only if $x_1 = x_2 = \cdots x_{n-1} = \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}$. However, note that the last equality is implied if all the numbers of $x_1, x_2 \ldots x_{n-1}$ are the same; thus, equality holds if and only if $x_1 = x_2 = \cdots x_{n-1}$.
We first simplify the lefthand side. Multiplying both sides of the fraction by $n-1$ and combining like terms, we get that $$\frac{x_1 + x_2 + \cdots + x_{n-1} + \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}}{n} = \frac{nx_1 + nx_2 + \cdots + nx_{n-1}}{n(n-1)} = \frac{x_1 + x_2 + \cdots x_{n-1}}{n-1}.$$ Plugging this into the earlier inequality yields $$\frac{x_1 + x_2 + \cdots x_{n-1}}{n-1} \geq \sqrt[n]{x_1 x_2 \cdots x_{n-1} \left(\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1} \right)}.$$ Raising both sides to the $n$th power yields $$\left( \frac{x_1 + x_2 + \cdots x_{n-1}}{n-1}\right)^n \geq x_1 x_2 \cdots x_{n-1}\left(\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}\right).$$ From here, we divide by $\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}$ and take the $(n-1)^{\textrm{th}}$ root to get that $$\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1} \geq \sqrt[n-1]{x_1 x_2 \cdots x_{n-1}}.$$ This is the inequality in $n-1$ variables. Note that every step taken also preserves equality, which completes the backwards step. Then by Cauchy Induction, the AM-GM inequality holds. $\square$
### Proof by Rearrangement
Define the $n$ sequence $\{ r_{i,j}\}_{i=1}^{n}$ as $r_{i,j} = \sqrt[n]{a_i}$, for all integers $1 \le i,j \le n$. Evidently these sequences are similarly sorted. Then by the Rearrangement Inequality, $$\sum_i a_i = \sum_i \prod_j r_{i,j} \ge \sum_i \prod_j r_{i+j,j} = n \prod_i \sqrt[n]{a_i} ,$$ where we take our indices modulo $n$, with equality exactly when all the $r_{i,j}$, and therefore all the $a_i$, are equal. Dividing both sides by $n$ gives the desired inequality. $\blacksquare$
### Proof by Calculus
We will start the proof by considering the function $f(x) = \ln x - x$. We will now find the maximum of this function. We can do this simply using calculus. We need to find the critical points of $f(x)$, we can do that by finding $f'(x)$ and setting it equal to $0$. Using the linearity of the derivative $f'(x) = \frac{1}{x} - 1$. We need $f'(x) = 0$ $$f'(x) = \frac{1}{x} - 1 = 0$$ $$\frac{1}{x} = 1$$ $$x = 1$$ Note that this is the only critical point of $f(x)$. We can confirm it is the maximum by finding it's second derivative and making sure it is negative. $f''(x) = -\frac{1}{x^2}$ letting x = 1 we get $f''(1) = -\frac{1}{1} < 0$. Since the second derivative $f''(x) < 0$, $f(1)$ is a maximum. $f(1) = \ln 1 - 1 = -1$. Now that we have that $f(1) = -1$ is a maximum of $f(x)$, we can safely say that $\ln x - x \leq -1$ or in other words $\ln x \leq x - 1$. We will now define a few more things and do some manipulations with them. Let $A = \frac{\sum_{i=1}^{n} a_i}{n}$, with this notice that $nA = \sum_{i=1}^{n} a_i$. This fact will come into play later. now we can do the following, let $x = \frac{a_i}{A}$ and plug this into $\ln x < x - 1$, we get $$\ln\left(\frac{a_1}{A}\right) \leq \frac{a_1}{A} - 1$$ $$\ln\left(\frac{a_2}{A}\right) \leq \frac{a_2}{A} - 1$$ $$\vdotswithin{=}$$ $$\ln\left(\frac{a_n}{A}\right) \leq \frac{a_n}{A} - 1$$ Adding all these results together we get $$\ln\left(\frac{a_1}{A}\right) + \ln\left(\frac{a_2}{A}\right) + \dots + \ln\left(\frac{a_n}{A}\right) \leq \frac{a_1}{A} - 1 + \frac{a_2}{A} - 1 + \dots + \frac{a_n}{A} - 1$$ $$\ln\left(\frac{\prod_{i=1}^{n} a_i}{A^n}\right) \leq \frac{\sum_{i=1}^{n}a_i}{A} - n$$ $$\ln\left(\frac{\prod_{i=1}^{n} a_i}{A^n}\right) \leq \frac{nA}{A} - n$$ $$\ln\left(\frac{\prod_{i=1}^{n} a_i}{A^n}\right) \leq n - n$$ $$\ln\left(\frac{\prod_{i=1}^{n} a_i}{A^n}\right) \leq 0$$ Now exponentiating both sides we get $$e^{\ln\left(\frac{\prod_{i=1}^{n} a_i}{A^n}\right)} \leq e^{0}$$ $$\frac{\prod_{i=1}^{n} a_i}{A^n} \leq 1$$ $$\prod_{i=1}^{n} a_i \leq A^n$$ $$\sqrt[\leftroot{2}\uproot{3}n]{\prod_{i=1}^{n} a_i} \leq \sqrt[\leftroot{2}\uproot{3}n]{A^n}$$ $$\sqrt[\leftroot{2}\uproot{3}n]{\prod_{i=1}^{n} a_i} \leq A$$ $$\sqrt[\leftroot{2}\uproot{3}n]{\prod_{i=1}^{n} a_i} \leq \frac{\sum_{i=1}^{n} a_i}{n}$$ This proves the AM-GM inequality. $\blacksquare$
## Proof of Weighted AM-GM
### Proof by Convexity
We note that the function $x \mapsto \ln x$ is strictly concave. Then by Jensen's Inequality, $$\ln \sum_i \lambda_i a_i \ge \sum_i \lambda_i \ln a_i = \ln \prod_i a_i^{\lambda_i} ,$$ with equality if and only if all the $a_i$ are equal. Since $x \mapsto \ln x$ is a strictly increasing function, it then follows that $$\sum_i \lambda_i a_i \ge \prod_i a_i^{\lambda_i},$$ with equality if and only if all the $a_i$ are equal, as desired. $\blacksquare$
### Alternate Proof by Convexity
This proof is due to G. Pólya.
Note that the function $f:x \mapsto e^x$ is strictly convex. Let $g(x)$ be the line tangent to $f$ at $(0,1)$; then $g(x) = x+1$. Since $f$ is also a continuous, differentiable function, it follows that $f(x) \ge g(x)$ for all $x$, with equality exactly when $x=0$, i.e., $$1+x \le e^x,$$ with equality exactly when $x=0$.
Now, set $$r_i = a_i/\biggl( \sum_{j=1}^n \lambda_j a_j \biggr) - 1,$$ for all integers $1\le i \le n$. Our earlier bound tells us that $$r_i +1 \le \exp(r_i),$$ so $$(r_i +1)^{\lambda_i} \le \exp(\lambda_i r_i) .$$ Multiplying $n$ such inequalities gives us
$$\prod_{i=1}^n (r_i + 1)^{\lambda_{i}} \le \prod_{i=1}^n \exp \lambda_i r_i$$
Evaluating the left hand side:
$$\prod_{i=1}^n (r_i + 1)^{\lambda_{i}} = \frac{\prod_{i=1}^n a_i^{\lambda_i} }{ (\sum_{j=1}^n \lambda_j a_j)^{\sum_{j=1}^n \lambda_i} } = \frac{\prod_{i=1}^n a_i^{\lambda_i} }{ \sum_{j=1}^n \lambda_j a_j } ,$$
for
$$\sum_{j=1}^n \lambda_i = 1 .$$
Evaluating the right hand side:
$$\prod_{i=1}^n \exp \lambda_i r_i = \prod_{i=1}^n \exp \left(\frac{a_i \lambda_i}{ \sum_{i=1}^n \lambda_j a_j} - \lambda_i\right) = \exp \left(\frac{\sum_{i=1}^n a_i \lambda_i}{ \sum_{i=1}^n \lambda_j a_j} - \sum_{i=1}^n \lambda_i\right) = \exp 0 = 1 .$$
Substituting the results for the left and right sides:
$$\frac{\prod_{i=1}^n a_i^{\lambda_i} }{ \sum_{j=1}^n \lambda_j a_j } \le 1$$
$$\prod_{i=1}^n a_i^{\lambda_i} \le \sum_{j=1}^n \lambda_j a_j = \sum_{i=1}^n \lambda_i a_i ,$$
as desired. $\blacksquare$
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# torch.manual_seed¶
torch.manual_seed(seed) → torch._C.Generator[source]
Sets the seed for generating random numbers. Returns a torch.Generator object.
Parameters
seed (int) – The desired seed. Value must be within the inclusive range [-0x8000_0000_0000_0000, 0xffff_ffff_ffff_ffff]. Otherwise, a RuntimeError is raised. Negative inputs are remapped to positive values with the formula 0xffff_ffff_ffff_ffff + seed.
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## anonymous one year ago The function f(x) = −x2 + 44x − 384 models the daily profit, in dollars, a shop makes for selling donut combos, where x is the number of combos sold and f(x) is the amount of profit. Part A: Determine the vertex. What does this calculation mean in the context of the problem? (5 points) Part B: Determine the x-intercepts. What do these values mean in the context of the problem? (5 points) CAN YOU GVM THE ANSWERS PLEASE
1. anonymous
@jim_thompson5910
2. anonymous
@jim_thompson5910 ARE STILL THERE?
3. jim_thompson5910
what do you have so far?
4. anonymous
The function f(x) = −x2 + 44x − 384 models the daily profit, in dollars, a shop makes for selling donut combos, where x is the number of combos sold and f(x) is the amount of profit. Part A: Determine the vertex. What does this calculation mean in the context of the problem? (5 points) Part B: Determine the x-intercepts. What do these values mean in the context of the problem? (5 points)
5. jim_thompson5910
6. anonymous
yea can you help me its my last question?
7. jim_thompson5910
|dw:1436488305187:dw|
8. jim_thompson5910
|dw:1436488324499:dw|
9. jim_thompson5910
|dw:1436488354162:dw|
10. jim_thompson5910
oops I meant to say "a = -1"
11. anonymous
Part A: -1? Part B: 44? Part C: -384?
12. jim_thompson5910
no
13. anonymous
14. jim_thompson5910
a,b,c that i listed are the coefficients not the answers
15. jim_thompson5910
have you seen $\Large y = ax^2 + bx + c$ before?
16. jim_thompson5910
btw there is no part c
17. anonymous
OH YEA MY Bad, so i need to solve that problem?
18. jim_thompson5910
if a = -1 and b = 44 then what is $$\LARGE \frac{-b}{2a}$$ equal to?
19. anonymous
a=2 and b=2 ?
20. jim_thompson5910
$\Large \frac{-b}{2a} = \frac{-44}{2*(-1)} = ??$
21. anonymous
22?
22. jim_thompson5910
yes, so x = 22 is where the vertex is plug x = 22 into the original equation to find y
23. anonymous
so whats nexts? i need to find y?
24. jim_thompson5910
yes
25. jim_thompson5910
|dw:1436488910837:dw|
26. anonymous
-342?
27. jim_thompson5910
incorrect
28. jim_thompson5910
you should have gotten -1*(22)^2+44(22) - 384 = 100
29. jim_thompson5910
when x = 22, y = 100
30. jim_thompson5910
|dw:1436489230679:dw|
31. anonymous
oh ok, so whats next?
32. jim_thompson5910
What does this calculation mean in the context of the problem? what does the vertex tell us?
33. anonymous
(22, 100)?
34. jim_thompson5910
yeah what does it mean in the real world? how does it translate into english?
35. jim_thompson5910
if you're stuck, what does x represent in this problem?
36. anonymous
twenty-two
37. jim_thompson5910
go back up at the top what does it say x represents? where x is the...
38. anonymous
x = 22 ?
39. jim_thompson5910
look at the instructions. It tells you what x means
40. jim_thompson5910
I don't want the number
41. jim_thompson5910
The function f(x) = −x2 + 44x − 384 models the daily profit, in dollars, a shop makes for selling donut combos, where x is _________ and f(x) is _____________.
42. anonymous
vertex?
43. jim_thompson5910
do you see what I mean? by asking to look at the instructions
44. jim_thompson5910
Fill in the blanks The function f(x) = −x2 + 44x − 384 models the daily profit, in dollars, a shop makes for selling donut combos, where x is _________ and f(x) is _____________.
45. jim_thompson5910
scroll to the very top of the page. You posted these instructions up there
46. anonymous
where x is the "number of combos sold" and f(x) is the amount of profit
47. jim_thompson5910
good
48. jim_thompson5910
when we say x = 22 and y = 100 that means when 22 combos are sold, the profit is 100 dollars
49. anonymous
oh
50. jim_thompson5910
(22,100) is the vertex at the highest point, so the most profit we can get is $100 and that happens when 22 combos are sold 51. jim_thompson5910 this is usually where managers want to aim 52. anonymous so that the answer for part A? 53. jim_thompson5910 yeah all so far for A 54. anonymous OR THAT only one part? 55. jim_thompson5910 we found the vertex and then gave an explanation of what it really means 56. anonymous ok thank you and for part B? 57. jim_thompson5910 solve 0 = -x^2 + 44x - 384 for x you can use technology (graphing) or the quadratic formula 58. anonymous Part A:(22,100) is the vertex at the highest point, so the most profit we can get is$100 and that happens when 22 combos are sold Part B:?
59. jim_thompson5910
part b) solve 0 = -x^2 + 44x - 384 for x you can use technology (graphing) or the quadratic formula
60. jim_thompson5910
desmos is a good graphing calculator https://www.desmos.com/calculator
61. jim_thompson5910
type in -x^2 + 44x - 384 then look for the x intercepts
62. anonymous
TYPE IN THE CALCULATOR?
63. jim_thompson5910
you see where the input box is?
64. anonymous
NO :/
65. jim_thompson5910
66. anonymous
OH OKAY WAIT LET ME SEE
67. anonymous
1?
68. jim_thompson5910
69. anonymous
OH IN THAT CALCULATOR I DONT KNOW HOW TO PUT X^2
70. jim_thompson5910
do you see the points (12,0) and (32,0) ?
71. anonymous
yes
72. jim_thompson5910
you just put in x^2 like you typed
73. jim_thompson5910
it understands that notation
74. jim_thompson5910
(12,0) and (32,0) are the x-intercepts x-intercept (12,0) when x = 12, the value of y is 0 so when 12 combos are sold, the profit is $0 x-intercept (32,0) when x = 32, the value of y is 0 so when 32 combos are sold, the profit is$0
75. jim_thompson5910
anything in between x = 12 and x = 32 means you'll have positive profits otherwise you will have 0 profit or negative profit (you lose money or owe it)
76. anonymous
lose?
77. jim_thompson5910
negative profit = you lost money ie you spent more than what you made
78. jim_thompson5910
example you send $10 but you only make$6 so you lost $4 (6-10 = -4) 79. anonymous oh yea 80. anonymous so? 81. jim_thompson5910 we're practically done with the problem 82. jim_thompson5910 Part B: Determine the x-intercepts. What do these values mean in the context of the problem? the x-intercepts are 12 and 32. As ordered pairs they are (12,0) and (32,0). They represent the points where the profit is$0. This is where you break even. You don't gain any money and you don't lose any money.
83. anonymous
so thats the asnwer for part B?
84. jim_thompson5910
more or less
85. anonymous
86. anonymous
LESS?
87. jim_thompson5910
huh? I don't know what you mean? I'm basically saying "yes"
88. anonymous
OH I THINK YOU ARE ASK ME IS MORE OR LESS
89. jim_thompson5910
ignore that portion
90. anonymous
SO WHAT WITH NEED TO DO NOW?
91. anonymous
WE*
92. jim_thompson5910
you're done, the problem is over
93. anonymous
SO WAIT THE ANSWER FOR PART B ITS: MORE OR LESS?
94. jim_thompson5910
no
95. jim_thompson5910
I already gave it to you. Just go back and re-read it
96. anonymous
the x-intercepts are 12 and 32. As ordered pairs they are (12,0) and (32,0). They represent the points where the profit is \$0. This is where you break even. You don't gain any money and you don't lose any money.
97. anonymous
???
98. jim_thompson5910
yeah just put that into your own words. Hopefully you understand what I mean?
99. anonymous
IN my own words? i dont know
100. anonymous
@jim_thompson5910 ???
101. jim_thompson5910
you can do it
102. anonymous
103. anonymous
:(
104. jim_thompson5910
Give it your best shot. Just be confident.
105. anonymous
so wait i need to write the answer in my own?
106. anonymous
with my own letters?
107. jim_thompson5910
yes
108. anonymous
thankss for you help
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1. ## [Matrix] Q about Trivial&NonTrivial System
I just like to check
Trivial solution = linearly independent
= unique = when n x m matrix, n=m=rank
NonTrivial solution = linearly independent
= infinite = row of zeros = n>rank
Also. (True of False)
(a) If the system is homogeneous, every solution is trival.
- False, since if the matrix contains a row of zeros (0 0 ... 0|0), then the system is non-trival.
(b) If the system has a nontrivial solution, it cannot be homogeneous.
- False, since homogeneous system is either unique(trival) or infinite(nontrival). Therefore, nontrivial solution must be a homogenous system.
(c) If there exists a trivial solution, the system is homogeneous.
- True, since homogenous system is either unique(trival) or infinite(nontrival). Therefore, trivial solution must be a homogeneous system.
(d) If the system is consistent, it must be homogenous.
- True, since any system that is not inconsistent is called consistent and consistent either has unique or infinite solutions which is a property of a homogeneous system.
Now assume that the system is homogeneous:
(e) If there exists a nontrivial solution, there is no trivial solution.
- True. A matrix with a nontrivial solution must contain a row of all zeros but no trivial solution must not contain a row of zeros in RREF.(What is no trival soln?)
(f) If there exists a solution, there are infinitely many solutions.
- False. If the system has basic solutions (systems without a row of zeros), the system contains unique solutions.
(g) If there exist nontrivial solutions, the row-echelon form of A has a row of zeros.
- True, since linearly dependent systems requires a row of zeros for parameters, which results infinite solutions.
(h) If the row-echelon form of A has a row of zeros, there exist nontrival solutions.
((((((((((((((((((((I think it is possible. I believe what my Professor said... ??? but y??)))))))))))))))))))
(i) If a row operation is applied to the system, the new system is also homogeneous.
-True. If the starting matrix is a homogenous system then the applied row operation system is always a homogenous system because it does not change property of original matrix.
SORRY, IT IS BIT LONG...
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# Convincing pdfbook to assume letterpaper
I have a small booklet in half-letterpaper size:
\documentclass{book}
\usepackage[paperheight=8.5in,paperwidth=5.5in]{geometry}
\usepackage{mwe}
\begin{document}
\lipsum[1-13]
\end{document}
But when I run pdfbook on the resulting PDF, I invariably receive—no matter what the original page size is—a PDF that is 11.7 by 8.27 inches. How can I use pdfbook (or pdfjam) to simply (and appropriately) mash these pages together without trying to alter the paper size?
### Ideal
After compiling the above MWE, I have a PDF that is exactly half of the size of a sheet of letter paper. I would like to run pdfbook (or a custom call to pdfjam) such that these pages are set two-up on the long edge in the correct configuration to bind it as a book.
$pdfbook -p letter test.pdf out.pdf ---- pdfjam: This is pdfjam version 2.08. pdfjam: Reading any site-wide or user-specific defaults... (none found) pdfjam ERROR: no PDF/JPG/PNG file found at -p • Not knowing what arguments you're using with pdfbook, it appears its default paper size is a4. Have you tried pdfbook -p letter in.pdf out.pdf already? – Mike Renfro Apr 22 '14 at 1:48 • @MikeRenfro I have not, but I've edited in its output. – Sean Allred Apr 22 '14 at 2:01 • As far as I can tell, pdfbook passes --landscape to pdfjam no matter what. Since you want portrait, I am not sure pdfbook is a good choice. Perhaps pdfjam can do this if invoked directly? pdfbook is a fairly simple wrapper for pdfjam - it isn't adding any functionality of its own to speak of. pdfjam should be able to do it given that it is a much more powerful interface to pdfpages. Failing that, I guess you could use pdfpages itself. – cfr Apr 22 '14 at 2:14 • OK. Try this. I still wouldn't let this script fetch me a jammy dodger but it does seem to work. pdfbook --paper letter --outfile out.pdf -- in.pdf. Note that the -- before the input pdf is essential. This is not at all normal behaviour for a shell script. It is parsing options extremely weirdly. I wouldn't like to say it is dangerous but I wouldn't trust that it is not. However, as long as you don't pass it anything dangerous, it should be fine ;). – cfr Apr 22 '14 at 2:51 • There is an alternative pdfbook program available at: ctan.org/tex-archive/support/pdfbook. You might be interested to have a look at that. – Jaap Eldering Oct 1 '14 at 13:01 ## 1 Answer The basic problem here is that pdfjam exhibits non-standard behaviour and is rather poorly documented. Since pdfbook is basically a minimal wrapper for pdfjam, it inherits these deficiencies while further obscuring the source of problems. In this case, you can workaround the problem using pdfbook --paper letter --outfile out.pdf -- in.pdf Note that the -- before the input pdf is essential. This is not at all normal behaviour for a shell script. pdfjam is parsing options extremely weirdly. I wouldn't like to say it is dangerous but I wouldn't trust that it is not. However, as long as you don't pass it anything dangerous, it should (hopefully) be reasonably safe. Figuring this out takes a little detective work. The manual pages for pdfbook and pdfjam are not terribly informative. Essentially, I started by reading the script pdfbook which shows how it invokes pdfjam. This is relatively straightforward. I then switched to testing pdfjam directly to try to narrow down the problem. pdfjam --help is the most useful source of documentation I found. Part of this includes the following: Usage: pdfjam [OPTIONS] [--] [FILE1 [SEL1]] [FILE2 [SEL2]]... and --paper PAPERSPEC (or simply --PAPERSPEC) Specify a LaTeX paper size, for example '--paper a4paper' or simply '--a4paper' for ISO A4 paper. If the LaTeX 'geometry' package is installed, a wider range of paper sizes is available. For details see documentation for LaTeX and/or the 'geometry' package. [Default for you at this site: paper=a4paper] --papersize '{WIDTH,HEIGHT}' Specify a custom paper size, e.g., --papersize '{10in,18cm}' (Note the braces, and the comma!) If the 'geometry' package is not found, this has no effect. and * '--' can be used to signal that there are no more options to come. What this does not tell you is that the script is parsing options non-standardly and that -- is essential if certain options are passed. Normally, you only need -- if one of the main arguments passed to a shell script might otherwise be mistaken for an option (e.g. a filename begins with -) or if you wish to prevent misuse (e.g. when a shell is invoked). In the case of pdfjam, though, if you write pdfjam --paper letter --outfile out.pdf in.pdf you will get errors. Debugging with --no-try shows that the script is passing something like paper letter--outfile as an optional argument when invoking the PDF merge command offered by pdfpages. It then tries to merge out.pdf and in.pdf. None of which works, obviously. Turning to pdfjam itself, you can see how it parses options: for arg do case$arg in
--quiet | -q | --configpath)
verbose=false ;
;;
--version | -V)
echo "\$version"
exit 0 ;
;;
--batch)
batch=true ;
;;
--help | -u | -h)
help=true ;
;;
--vanilla)
vanilla=true ;
;;
*)
;;
esac
done
It then parses any remaining options separately later in the file. I'm not entirely sure why it does this but the upshot is that it is essentially unable to tell the difference between an option and an argument. If you don't tell it explicitly that the options are finished, it fails to parse any remaining options correctly. This is actually pretty bad as it is letting arbitrary stuff get passed onto pdfpages even when it makes no sense. Although I don't know if this could be exploited, it is the kind of thing which can make scripts vulnerable to exploitation.
Moreover, including
* '--' can be used to signal that there are no more options to come.
does not constitute proper documentation of this behaviour because this is a standard behaviour for shell scripts. What is not standard is that this is required even if your first argument proper has a non-weird name i.e. even if in.pdf has a filename which does not start with -.
This is buggy behaviour in my view. Shell scripts ought not behave this way. Moreover, it uses /var/tmp for temporary files. This is also non-standard and problematic. It ought to use /tmp so that the temporary files it creates are removed on reboot, for example. Although these remain around only if --no-tidy is passed, the behaviour is still unexpected and, I think, undesirable for a script of this kind.
So I would be careful in using pdfjam and wrappers such as pdfbook. For example, I would avoid piping the output of other processes to pdfjam and only invoke the script or its wrappers manually.
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# Pattern matching using Cases: why does this fail? [closed]
I have objects that look like, for example:
F2 = 2 d[x] ^ d[y] + 3 d[x] ^ d[z]
where ^ is Wedge (and not exponentiation). The variables inside of d[...] are not known beforehand and the number of terms is also not known; I would like to extract the coefficients that lie in front of these terms (here 2 and 3). The real objects that I have to work with are enormous, so I would like to do this as efficiently as possible, so I try to extract what differentials I have like this:
Cases[ Level[F2, 1] , _ d[a_] ^ d[b_] -> d[a] ^ d[b] ]
This returns: { d[x] ^ d[y], d[x] ^ d[z] }. So, now I can use Coefficient and all is well.
However, if I have an object like this:
F2 = 2 d[x] ^ d[y] + 3 e[1] ^ e[2]
I would like to be returned the list: { d[x] ^ d[y], e[1] ^ e[2] }; so, I try:
Cases[ Level[F2, 1] , _ d1_[a_] ^ d2_[b_] -> d1[a] ^ d2[b] ]
But this does not work. It returns an empty list. Why? And is there a way to make this work?
Edit
So, contrary to my claim, this does actually work as intended, except that an expression like:
F2 = d[x] ^ d[y]
returns an empty list. I naively expected the code I wrote above to handle an implicit coefficient as well as explicit ones. I needed to modify my code to:
Cases[ Level[F2, 1] , _ d1_[a_] ^ d2_[b_] | d1_[a_] ^ d2_[b_] -> d1[a] ^ d2[b] ]
Thanks for the help! I'll probably delete the question in due course...
• Your example works on my machine. Try to either remove any definitions you have for d1, d2, a and b, or use :> instead of ->. – Marius Ladegård Meyer Mar 2 '16 at 19:53
• It works for me, too. – march Mar 2 '16 at 19:54
• Hang on. Let me see if I'm being dumb... – Zorawar Mar 2 '16 at 19:58
• Still works here :p – Marius Ladegård Meyer Mar 2 '16 at 20:25
• Did you, by any chance, use exponentiation in your last line of code (in Cases)? Because if I use exponentiation I do get an empty list, while using Wedge returns the desired terms. – Berg Mar 2 '16 at 21:13
Is it not much easier and faster to simply use the following?
List@@F2/._Wedge->1
A possible solution for the more general problem described in the comments would (perhabs) be:
Replace[F2, a_*w_Wedge -> CustomSimplify[a]*w, {1}]
which applies some custom simplifications to the coefficients of the Wedge-products. One can change the pattern to a_. if the coefficient 1 should also be transformed.
• Yes, but my question is a simplified version of my real needs :) I need to simplify the expressions but Mathematica finds simplifying p-forms very difficult to do, so my strategy is to extract coefficients, simplify them individually and then recombine them; so, I need to track what the coefficients are in front of also. Probably I'm still doing it inefficiently, though... – Zorawar Mar 2 '16 at 22:05
• In effect, for F2 above, what I need is something like: Simplify[2] d[x] ^ d[y] + Simplify[3] d[x] ^ d[z]. But where the actual coefficients are functions of the variables x, y, z, ... and not just 2 and 3 :) – Zorawar Mar 2 '16 at 22:08
• I changed my reply a bit. Those patterns with default values are neat way to simplify code, see the tutorial on Patterns and Transformation Rules. – Berg Mar 2 '16 at 23:19
• Interesting. Thanks! I'm going to delete the question soon, but thanks for the suggestion! – Zorawar Mar 3 '16 at 20:31
|
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Mat. Zametki: Year: Volume: Issue: Page: Find
Mat. Zametki, 1977, Volume 22, Issue 5, Pages 671–678 (Mi mz8091)
Diameters of a class of smooth functions in the space $L_2$
R. S. Ismagilova, Kh. Nasyrova
a Moscow Institute of Electronic Engineering
Abstract: The class $V_\psi$, consisting of the smooth functions $f(t)$, $0\le t\le1$, satisfying the condition $\int_0^1\psi[f^{(r)}(t)] dt\le1$, where the function $\psi(t)$ is nonnegative and $r$ is a natural number, is studied. Under certain restrictions on the function $\psi(t)$ ensuring the compactness of the class $V_\psi$, the order of decrease of the Kolmogorov diameters $d_n(V_\psi)$ is computed. The analogous problem for the case $r=1$ is solved also for functions of several variables.
Full text: PDF file (494 kB)
English version:
Mathematical Notes, 1977, 22:5, 865–870
Bibliographic databases:
UDC: 517
Citation: R. S. Ismagilov, Kh. Nasyrova, “Diameters of a class of smooth functions in the space $L_2$”, Mat. Zametki, 22:5 (1977), 671–678; Math. Notes, 22:5 (1977), 865–870
Citation in format AMSBIB
\Bibitem{IsmNas77}
\by R.~S.~Ismagilov, Kh.~Nasyrova
\paper Diameters of a~class of smooth functions in the space $L_2$
\jour Mat. Zametki
\yr 1977
\vol 22
\issue 5
\pages 671--678
\mathnet{http://mi.mathnet.ru/mz8091}
\mathscinet{http://www.ams.org/mathscinet-getitem?mr=473653}
\zmath{https://zbmath.org/?q=an:0372.41013}
\transl
\jour Math. Notes
\yr 1977
\vol 22
\issue 5
\pages 865--870
\crossref{https://doi.org/10.1007/BF01098351}
|
# Generalization of Cauchy's eigenvalue interlacing theorem?
Cauchy's Interlacing Theorem says that given an $$n \times n$$ symmetric matrix $$A$$, let $$B$$ be an $$(n-1) \times (n-1)$$ principal submatrix of it, then the eigenvalues of $$A$$ and those of $$B$$ interlace.
Using this property, one can obtain a lower bound on the $$k$$-th largest eigenvalue of a $$t \times t$$ principal submatrix of $$A$$, using the $$(k+n-t)$$-th largest eigenvalue of $$A$$. This lower bound is best possible, for example when $$A$$ is diagonal. But for many interesting (fixed) matrices, such bound is usually far from being optimal. For example, let $$A=\begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1\\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}$$
The eigenvalues of $$A$$ are $$2, 0, 0, -2$$. So if we would like to bound from below the largest eigenvalue of its $$3 \times 3$$ principal submatrix using Cauchy's Theorem, we only get a lower bound of $$0$$. However it is straightforward to check that it is always at least $$\sqrt{2}$$.
I am wondering if there is a more "quantitative" Interlacing Theorem, say if your matrix satisfies some additional properties (non-negative, binary, etc.), then one can obtain a better lower bound on the $$k$$-th largest eigenvalue of a $$t \times t$$ principal submatrix of $$A$$?
If $$r_i = r_i(A) := \sum_{k=1}^n a_{ik}$$ and $$\rho = \rho(A) := \max_{\lambda \in \sigma(A)}\{|\lambda|\}$$, then $$\min_{1 \le i \le n} r_i \le \rho \le \max_{1 \le i \le n} r_i.$$ This result is due to Frobenius [MR0235974].
For example, for the $$3$$-by-$$3$$ leading principal sub-matrix $$B:= \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix},$$ we obtain $$1 \le \rho(B) \le 2$$.
One way to try and improve these bounds is by applying the same bounds to $$D^{-1} A D$$, where $$D$$ is a diagonal matrix with positive diagonal entries.
There are many other works in the literature that are dedicated to improving the Frobenius bounds above (too many to list here, but I would be happy to email you a list if you'd like), but many are quite complicated (it seems to me that you want to get $$\sqrt{2}$$ as a lower bound).
|
# Vagrant Getting Started
## Why Vagrant?
Remember all those other projects you wanted to work on, but you had to install about a thousand things, have exactly the right path variables setup and had to perform some voodoo ritual to get started and your object reference was still not set to the instance of an object? Vagrant makes that pain happen only once! That’s right, if you can sucker someone else into building your vagrant environment for you, you don’t feel any pain at all.
Vagrant lets you build a custom virtual machine from scripts and set it up so your can work from your normal directories and IDEs. Then, when someone else wants to work in the same environment you are, they just download your Vagrantfile and “vagrant up” their own machine.
## Starting From Scratch
1. Install Vagrant and Virtual Box
2. Clone whatever repository you want to setup your image under
• Skip this step if you already have it
3. Open a command line to your repository
• In windows, if you have the explorer window open you can hold shift and right click to get an “Open command window here” option in the context menu
4. Type vagrant init hashicorp/precise32
• This creates a file named Vagrantfile that contains the options used to setup the box
5. Type vagrant up
• The image gets stored in your user profile so you don’t have to download it more than once, incase you setup another vagrant box later for another repository.
6. Type vagrant ssh
• This requires some SSH tools. If it complains, make sure git is installed.
7. Hooray! You’re ssh’d into your own VM!
1. Install Vagrant and Virtual Box as normal
2. Clone some repository that already has a Vagrantfile
3. Open the command line to your repository
4. Type vagrant up
5. Type vagrant ssh
6. Hooray! You’re ssh’d into a VM someone else setup for you!
## Vagrantfile
There are many options you can set in your vagrant file to customize your VM. Here are some that are extra common or useful.
#### config.vm.box
This tells the virtual machine what image to build from. It can be an image you made yourself somewhere else, or it can be an image pulled from a website somewhere.
#### config.vm.network
This config parameter lets you forward ports from your host machine to your VM. For example, if you setup your VM to have a web server hosted on port 80, you could set port 8080 on your real machine to point directly to port 80 on the VM.
Example: config.vm.network "forwarded_port", guest: 80, host: 8080
#### config.vm.synced_folder
While vagrant automatically syncs /vagrant on the VM to the folder you ran “vagrant up” in, you may want to sync up other folders in different spots. This command lets you do it.
Example: config.vm.synced_folder "../data", "/vagrant_data"
The example is attaching a folder named data on the host machine to the directory /vagrant_data on the VM
#### config.vm.provision
This configuration is a lot of the bread and butter of Vagrant. This lets you run commands to setup your VM as it boots. Take the example below
Example: config.vm.provision :shell, path: "bootstrap.sh"
Here we are telling vagrant to run the file bootstrap.sh, which is located in the same directory as the Vagrantfile, to effectively set up anything else on the machine. This is a bash script that is used to run commands to install other programs.
Ok, so now you have a VM. If you didn’t use the cheat codes, that means you got suckered into setting up the VM. Vagrant has quite a few options that will make your life easier.
The VM we setup is a Ubuntu Linux image which uses a program called apt-get to install/update programs used in the system. To setup our machine, we can do whats called “provisioning the system.” This uses the config.vm.provision function of the Vagrantfile. To provision the system we will write a bash script that gets run once the machine is booted that will install all our needs. Lets say we need python, pip (an easy python library installer), and a few useful python libraries as a use case.
In the Vagrantfile, add the line config.vm.provision :shell, path: "bootstrap.sh" This line is telling Vagrant to run the bootstrap.sh file using bash when we tell it to provision the system.
Next, in your repository directory create a bootstrap.sh file. This file will get run to provision the system.
Since this file is a bash file, it will run every line of the file as if we typed it into the command line ourselves. Note that lines starting with # are comments. Our example file might look something like this:
#!/usr/bin/env bash
#Get Packages
#the -y tells the apt-get program to
#respond with "yes" to every "are you sure" question
apt-get -y update
apt-get install -y python2.7-dev
apt-get install -y curl
apt-get install -y python-pip
#Get python libraries using pip
pip install beautifulsoup4
pip install mechanize
pip install selenium
Now, if your vagrant isn’t already running, when you run the command vagrant up the system will automatically be provisioned. If your vagrant machine is already running, that’s ok, you can just run vagrant provision to start the provisioning process.
Finally your system is setup. All that’s left to do is to commit your Vagrantfile and bootstrap.sh file back to your repository so that others can use it and start developing quickly and painlessly. Make sure they thank you for taking one for the team and setting everything up!
## Shutting Down
When you’re ready to shut down your vagrant VM you have a couple different options.
vagrant suspend
This command is a “point in time” shutdown for your VM. This allows you to resume your VM quickly but requires more disk space as it copies the entire RAM of the VM. It bring it out of a suspended state, use the vagrant resume command.
vagrant halt
This is the same as shutting any other PC. All data in the image is saved and it can be brought back up with a simple vagrant up command again.
vagrant destroy
This command wipes everything. It will clean up all the VM space and it will be like you never created a VM in the first place. Don’t worry about having to download the base image again though, the box itself gets stored elsewhere so you can load up other environments more quickly.
## IDE Integration
By now you’re probably wondering, “This is all well and good, but what about my IDE? Doesn’t it need to know where python or (insert other language here) lives to work?” If you’re using a magical IDE, like almost anything from JetBrains, you can have it point to the interpreters living on your VM so you don’t need anything installed locally other than the IDE, vagrant and virtual box. Unfortunately, remote interpreters only comes with the paid version of JetBrain’s IDEs, but they are well worth the cost. I’ll leave the setting up the interpreter as an exercise for the reader. But, I will leave a hint.
|
An ideal gas occupying a volume of 2 dm^3 at a pressure of 5 bar undergoes isothermal and irreversible expansion
### Question Asked by a Student from EXXAMM.com Team
Q 1157756684. An ideal gas occupying a volume of 2 dm^3 at a pressure of 5 bar undergoes isothermal and irreversible expansion against external pressure of 1 bar. Find the final volume of the system and the work involved in the process.
A
10 dm^3, –500J
B
10 dm^3, –800 J
C
8 dm^3, 800 J
D
8 dm^3, 500 J
#### HINT
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)
#### Access free resources including
• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis
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## div
'div' |
<defdiv>|
<line> [<division>]|
<set> [<division>]|
['mult'|'div' <factor-div> <factor-bias>]|
['auto' <node-dist> <angle> <elem-ratio>]
This keyword can be used to re-define the default division of lines:
div 4
The div keyword works on a line or a set of lines (see qadd). The division controls the number of nodes created when the geometry is meshed (see elty and mesh). For example,
div all 4
attaches the division of 4 to all lines. With the keyword mult or div in combination with a value, it is possible to multiply or divide already assigned divisions:
div all mult 2.
Or in case you need a starting-point for the individual divisions you can use the option auto with the optional parameters node-dist and angle. Node-dist is the maximum allowed distance between nodes and angle is the maximum allowed angle defined by three sequential nodes. If one parameter is not fulfilled then the division is halved until the requirements are fulfilled. Default values are defined in the file cgx.h and can be listed with
div
without parameters
div all auto
uses the defaults. The following example sets them expicidly:
div all auto 2. 10. 0.5
will use a maximum element lenght of 2., the angle between successive nodes is less than 10 degree and the minimum element is only half of the maximum-length as long as the length of the line is sufficient. It should be noted that it could make sense to use different values for different sets.
|
main-content
## Über dieses Buch
This book results from various lectures given in recent years. Early drafts were used for several single semester courses on singular perturbation meth ods given at Rensselaer, and a more complete version was used for a one year course at the Technische Universitat Wien. Some portions have been used for short lecture series at Universidad Central de Venezuela, West Vir ginia University, the University of Southern California, the University of California at Davis, East China Normal University, the University of Texas at Arlington, Universita di Padova, and the University of New Hampshire, among other places. As a result, I've obtained lots of valuable feedback from students and listeners, for which I am grateful. This writing continues a pattern. Earlier lectures at Bell Laboratories, at the University of Edin burgh and New York University, and at the Australian National University led to my earlier works (1968, 1974, and 1978). All seem to have been useful for the study of singular perturbations, and I hope the same will be true of this monograph. I've personally learned much from reading and analyzing the works of others, so I would especially encourage readers to treat this book as an introduction to a diverse and exciting literature. The topic coverage selected is personal and reflects my current opin ions. An attempt has been made to encourage a consistent method of ap proaching problems, largely through correcting outer limits in regions of rapid change. Formal proofs of correctness are not emphasized.
## Inhaltsverzeichnis
### Chapter 1. Examples Illustrating Regular and Singular Perturbation Concepts
Abstract
Consider a linear spring-mass system with forcing, but without damping, and with a small spring constant. This yields the differential equation
$$y + \in y = f\left( x \right)$$
for the displacement y(x) as a function of time x, with the small positive parameter ε being the ratio of the spring constant to the mass of the spring. This should be solved on the semi-infinite interval x ≥ 0, with both the initial displacement y(0) and the initial velocity y’(0) prescribed. The traditional approach to solving such initial value problems [cf. Boyce and DiPrima (1986)] is to note that for ε small the homogeneous equation has the slowly varying solutions $$cos\left( {\sqrt \in x} \right)$$ and $$sin\left( {\sqrt \in x} \right)$$ and to look for a solution through variation of parameters. Specifically, one sets $$y\left( x \right) = v_1 \left( x \right)\cos \left( {\sqrt \in x} \right) + v_2 \left( x \right)\sin \left( {\sqrt \in x} \right)$$, where $$v'_1 \cos \left( {\sqrt \in x} \right) + v'_2 \sin \left( {\sqrt \in x} \right) = 0$$ and $$- \sqrt \in v'_1 \sin \left( {\sqrt \in x} \right) + \sqrt \in v'_2 \cos \left( {\sqrt \in x} \right) = f\left( x \right).$$ Since $$y\left( 0 \right) = v_1 \left( 0 \right)$$ and $$y'\left( 0 \right) = \sqrt \in v_2 \left( 0 \right)$$, solving for v’1 and v’2 and integrating provides the unique solution
$$\begin{array}{*{20}c} {y(x, \in ) = y(0)\cos \left( {\sqrt \in x} \right) + \frac{1} {{\sqrt \in }}y'(0)\sin (\sqrt \in x)} \\ { - \frac{1} {{\sqrt \in }}\cos (\sqrt { \in x} )\int_0^x {\sin } (\sqrt \in t)f(t)dt} \\ { + \frac{1} {{\sqrt \in }}\sin (\sqrt \in x)\int_0^x {\cos (\sqrt \in t)f(t)dt.} } \\ \end{array}$$
Robert E. O’Malley
### Chapter 2. Singularly Perturbed Initial Value Problems
Abstract
Readers should refer to Murray (1977) and to earlier chemical engineering literature [especially Bowen et al. (1963) and Heinekin et al. (1967)] for experts’ explanations of the significance of the pseudo-steady-state hypothesis in biochemistry. The theory of Michaelis and Menton (1913) and Briggs and Haldane (1925) concerns a substrate S being converted irreversibly by a single enzyme E into a product P. There is also an intermediate substrate-enzyme complex SE. Since the back reaction is negligible, we shall systematically write
$$S + E\begin{array}{*{20}{c}} {{{k}_{1}}} \\ \to \\ \leftarrow \\ {{{k}_{{ - 1}}}} \\ \end{array} SE\xrightarrow{{{{k}_{2}}}}P + E.$$
Robert E. O’Malley
### Chapter 3. Singularly Perturbed Boundary Value Problems
Abstract
Consider the two-point problem εy+a(x)y+b(x)y=f(x) on 0≤x≤1 where a(x)>0 and with the boundary values y(0) and y(1) prescribed. We shall suppose that a, b, and f are arbitrarily smooth, and we shall prove that the asymptotic solution will exist, be unique, and have the form
$$y\left( {x,\varepsilon } \right) = Y\left( {x,\varepsilon } \right) + \xi \left( {x/\varepsilon,\varepsilon } \right)$$
where the outer expansion Y(x, c) has a power series expansion
$$Y\left( {x,\varepsilon } \right) \sim \sum\limits_{j = 0}^\infty {{Y_j}\left( x \right){\varepsilon ^j}}$$
and the initial layer correction $$\xi \left( {\tau,\varepsilon } \right)$$ has an expansion
$$\xi \left( {\tau , \in } \right)\sim \sum\limits_{{j = 0}}^{\infty } {{{\xi }_{j}}\left( \tau \right){{ \in }^{j}}}$$
such that each ξj(and its derivatives) will tend to zero as the stretched variable $$\tau = x/\varepsilon$$ tends to infinity. We shall base our proof on the existence of both a smooth asymptotic solution of the differential equation and of a (linearly independent) asymptotic solution which features an initial layer of nonuniform convergence near x = O. Recognizing that the asymptotic solution is an additive composite function of the slow “time” x and the fast time x/E generalizes to multitime expansions in many asymptotic contexts [cf. Nayfeh (1973) and, especially, Kevorkian and Cole (1981).
Robert E. O’Malley
### Backmatter
Weitere Informationen
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# Floating point precission issues.
This topic is 766 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I ran into problem when two vertices that are on diffrent side of a plane (n,d) do not produce intersection by float point precission issue:
template <class T>
bool SegmentPlaneIntersection(t3dpoint<T> n, T d, t3dpoint<T> rA, t3dpoint<T> rB, t3dpoint<T> & res)
{
t3dpoint<T> PointOnPlane;
long double originDistance = (long double)d;
long double distance1 = ((n.x * rA.x) + (n.y * rA.y) + (n.z * rA.z)) + originDistance;
long double distance2 = ((n.x * rB.x) + (n.y * rB.y) + (n.z * rB.z)) + originDistance;
if(distance1 * distance2 > 0.0) return false; //here
t3dpoint<T> vLineDir = Normalize(rB - rA);
long double Numerator = -1.0 * (n.x * rA.x + n.y * rA.y + n.z * rA.z + originDistance);
long double Denominator = ( (n.x * vLineDir.x) + (n.y * vLineDir.y) + (n.z * vLineDir.z) );
if( Denominator == 0.0) PointOnPlane = rA;
else
{
long double dist = Numerator / Denominator;
PointOnPlane = (rA + (vLineDir * T(dist)));
}
res = PointOnPlane;
return true;
}
code that is responsible for this is here:
if(distance1 * distance2 > 0.0) return false;
even when one point is lets say 50, 1.1, 50 and second is 50, -0.8, 50 it produces positive dot product (that was odd really
exact points of triangle:
------------------
ADDED WHOLE FACE AS BELOW WATERLINE:
TESTING POINT A: 62.9247131347656 -0.0228385925292969 47.2107238769531 dot: -0.0228385925292969
TESTING POINT B: 40.1480102539063 1.10841751098633 44.3472938537598 dot: 1.10841751098633
TESTING POINT C: 40.1480102539063 -0.918651580810547 49.4782066345215 dot: -0.918651580810547
----------------------
note AB, BC they are clearly between that plane but dot in this function failed gratefully (please do not tell me that its the resaon of c casting i made a function like this: and the result was the same no interesection
[spoiler]
bool SegmentPlaneIntersectionf(vec3 n, float d, vec3 rA, vec3 rB, vec3 & res)
{
vec3 PointOnPlane;
float distance1 = ((n.x * rA.x) +
(n.y * rA.y) +
(n.z * rA.z)) + d;
float distance2 = ((n.x * rB.x) +
(n.y * rB.y) +
(n.z * rB.z)) + d;
if (distance1 * distance2 > 0.0) return false;
vec3 vLineDir = Normalize(rB - rA);
float Numerator = -( dot(n, rA) + d);
float Denominator = dot(n, vLineDir);
if( Denominator == 0.0)
PointOnPlane = rA;
else
{
float dist = Numerator / Denominator;
PointOnPlane = (rA + (vLineDir * dist));
}
res = PointOnPlane;
return true;
}
[/spoiler]
Now to see maybe why is this happening i went to function that calls it: (its a long function so i put parts of it)
full
[spoiler]
void SliceHullByWater(TWaterWaves * ocean)
{
int arri = -1;
TPolygon<float> sliced_area;
float surface_area = 0.0;
t3dpoint<float> A;
t3dpoint<float> B;
t3dpoint<float> output;
for (int f=0; f < hull->VBO_BE.Length; f++)
{
float dst;
vec3 pop;
vec3 normal = vec3(0.0, 1.0, 0.0);//FindOceanTriangleNormal(pos + ROTATION_MAT * hull->FACE_CENTER_POINT[f], ocean, 0.0, dst, pop);
dst = getplaneD(normal, vec3(0.0,0.0,0.0));
int ahue = hull->VBO_BE[f].INDEX_START;
int flen = hull->VBO_BE[f].length;
for (int i=ahue; i < ahue+flen; i++) //through all vertices in that poly..., 3 only
{
sliced_area.Count = 0;
A = pos + ROTATION_MAT * hull->VBO_V[i];
int next = i + 1;
if (next >= ahue+flen) next = ahue;
B = pos + ROTATION_MAT * hull->VBO_V[next];
if ( SegmentPlaneIntersectionf(normal, dst, A, B, output) )
}
//~~~~~~~~~~ Test if Area is fully submerged or not at all
if (sliced_area.Count == 0)
{
if (dot( pos + ROTATION_MAT * hull->VBO_V[ahue], normal) + dst < 0.0) //fully submerged triangle
{
arri = arri + 1;
ACTUAL_FRAME[arri].V[0] = pos + ROTATION_MAT * hull->VBO_V[ahue];
ACTUAL_FRAME[arri].V[1] = pos + ROTATION_MAT * hull->VBO_V[ahue+1];
ACTUAL_FRAME[arri].V[2] = pos + ROTATION_MAT * hull->VBO_V[ahue+2];
ACTUAL_FRAME[arri].normal = normal;
ACTUAL_FRAME[arri].dst = dst;
ACTUAL_FRAME[arri].pop = pop;
ALOG(""); ALOG("");ALOG("------------------");
ALOG("ADDED WHOLE FACE AS BELOW WATERLINE:");
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[0]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[0], normal) + dst));
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[1]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[1], normal) + dst));
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[2]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[2], normal) + dst));
ALOG("----------------------");
ALOG("");
}
continue; //go to next face
}
//**************************if partially submerged
//see how many vertices are behind to know what type of cut we have
int backfaces = 0;
ALOG("");ALOG("");
for (int i=ahue; i < ahue + flen; i++)
{
ALOG("TESTING POINT: "+POINT_TO_TEXT(pos + ROTATION_MAT * hull->VBO_V[i]) + " dot: " +FloatToStr(dot( pos + ROTATION_MAT * hull->VBO_V[i], normal) + dst));
if (dot( pos + ROTATION_MAT * hull->VBO_V[i], normal) + dst < 0.0)
{
// if (classify_point_plane( pos + ROTATION_MAT * hull->VBO_V[i], normal, dst) == isBack)
backfaces = backfaces + 1;
}
}
//see how many vertices are behind to know what type of cut we have
sliced_area.Count = 0;
for (int i=ahue; i < ahue + flen; i++)
{
A = pos + ROTATION_MAT * hull->VBO_V[i];
if (dot( A, normal ) + dst < 0.0)
{
}
int next = i + 1;
if (next >= ahue+flen) next = ahue;
B = pos + ROTATION_MAT * hull->VBO_V[next];
if ( SegmentPlaneIntersection(normal, dst, A, B, output) )
{
}
else
{
float a = dot( A, normal ) + dst;
float b = dot( B, normal ) + dst;
if (!( ( (a <= 0) && (b <= 0) ) || ( (a > 0) && (b > 0) ) ))
ShowMessage("Definetly a hit but i cant see it");
}
}
vec3 cp;
if (backfaces == 1)
{
arri = arri + 1;
ACTUAL_FRAME[arri].V[0] = sliced_area.V[0];
ACTUAL_FRAME[arri].V[1] = sliced_area.V[1];
ACTUAL_FRAME[arri].V[2] = sliced_area.V[2];
ACTUAL_FRAME[arri].normal = normal;
ACTUAL_FRAME[arri].dst = dst;
ACTUAL_FRAME[arri].pop = pop;
ALOG("FIRST TRI: "+IntToStr(sliced_area.Count));
if (dot(ACTUAL_FRAME[arri].V[0],normal) + dst > 0) ALOG("Vert 0: above waterline");
if (dot(ACTUAL_FRAME[arri].V[1],normal) + dst > 0) ALOG("Vert 1: above waterline");
if (dot(ACTUAL_FRAME[arri].V[2],normal) + dst > 0) ALOG("Vert 2: above waterline");
} else //two triangle vertices under the water
{
if (backfaces < 2)
{
ShowMessage("DUD KURWA WTF!");
}
arri = arri + 1;
ACTUAL_FRAME[arri].V[0] = sliced_area.V[0];
ACTUAL_FRAME[arri].V[1] = sliced_area.V[1];
ACTUAL_FRAME[arri].V[2] = sliced_area.V[2];
ACTUAL_FRAME[arri].normal = normal;
ACTUAL_FRAME[arri].dst = dst;
ACTUAL_FRAME[arri].pop = pop;
ALOG("FIRST TRI");
if (dot(ACTUAL_FRAME[arri].V[0],normal) + dst > 0) ALOG("Vert 0: above waterline");
if (dot(ACTUAL_FRAME[arri].V[1],normal) + dst > 0) ALOG("Vert 1: above waterline");
if (dot(ACTUAL_FRAME[arri].V[2],normal) + dst > 0) ALOG("Vert 2: above waterline");
arri = arri + 1;
ACTUAL_FRAME[arri].V[0] = sliced_area.V[2];
ACTUAL_FRAME[arri].V[1] = sliced_area.V[3];
ACTUAL_FRAME[arri].V[2] = sliced_area.V[0];
ACTUAL_FRAME[arri].normal = normal;
ACTUAL_FRAME[arri].dst = dst;
ACTUAL_FRAME[arri].pop = pop;
ALOG("SECOND TRI");
if (dot(ACTUAL_FRAME[arri].V[0],normal) + dst > 0) ALOG("Vert 0: above waterline");
if (dot(ACTUAL_FRAME[arri].V[1],normal) + dst > 0) ALOG("Vert 1: above waterline");
if (dot(ACTUAL_FRAME[arri].V[2],normal) + dst > 0) ALOG("Vert 2: above waterline");
}
}
submerged_tri_count = arri + 1;
}
[/spoiler]
now to the code that was failing:
click for description:
[spoiler]
Function name is: void SliceHullByWater(TWaterWaves * ocean) so it slices hull model with water planes (but for simplification i used one plane for all triangles
- Ok so first of all I loop through all faces in the model
- then I go through all vertices in that model and check whenever they are on different sides of plane: if they are all on one side of plane we check if first vertex of the face is below that plane (water plane) if it is then we add this triangle to the stack. (ACTUAL_FRAME), then skip this face iteration. If not then we do nothing and skip this face iteration (because triangle is above water and we return only submerged part)
[/spoiler]
Ahue is index in vertex array (index of first vertex of that face, flen is face length)
for (int i=ahue; i < ahue+flen; i++) //through all vertices in that poly..., 3 only
{
sliced_area.Count = 0;
A = pos + ROTATION_MAT * hull->VBO_V[i];
int next = i + 1;
if (next >= ahue+flen) next = ahue;
B = pos + ROTATION_MAT * hull->VBO_V[next];
//here we get vector AB (for each side of triangle)
//Then we test if theres intersection if there is add it to some temp polygon
//Like I am trying to explain it fails here like crap
if ( SegmentPlaneIntersectionf(normal, dst, A, B, output) )
}
Now If there was no intersections at all
//~~~~~~~~~~ Test if Area is fully submerged or not at all
if (sliced_area.Count == 0)
{
if (dot( pos + ROTATION_MAT * hull->VBO_V[ahue], normal) + dst < 0.0) //fully submerged triangle it test first vertex of face (if its below waterline)
{
arri = arri + 1;
ACTUAL_FRAME[arri].V[0] = pos + ROTATION_MAT * hull->VBO_V[ahue];
ACTUAL_FRAME[arri].V[1] = pos + ROTATION_MAT * hull->VBO_V[ahue+1];
ACTUAL_FRAME[arri].V[2] = pos + ROTATION_MAT * hull->VBO_V[ahue+2];
ACTUAL_FRAME[arri].normal = normal;
ACTUAL_FRAME[arri].dst = dst;
ACTUAL_FRAME[arri].pop = pop;
Log which you saw above tells that first vertex in this trinagle
hull->VBO_V[ahue] is below waterline so that’s why i posted the log
ALOG(""); ALOG("");ALOG("------------------");
ALOG("ADDED WHOLE FACE AS BELOW WATERLINE:");
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[0]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[0], normal) + dst));
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[1]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[1], normal) + dst));
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[2]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[2], normal) + dst));
ALOG("----------------------");
ALOG("");
}
continue; //go to next face
}
and i got this note that triangle 0,1,2 is actually this triangle:
and the result:
SegmentPlaneIntersection doesnt see anything for 0,1,2 vertices so sliced_area.Count == 0
then it tests dot(vertex 0, waterline_normal) + waterline_dst < 0 since it is below waterline it adds whole triangle to submerged triangle list (idea was good but SegmentPlaneIntersection fails in this part of code or maybe the problem is somewhere else) but then SegmentPlaneIntersection should work...
the hting is in the second part of full function o do the same loop with same code but second one works but not this first. D:!
after thinking why the hell (dot(a,n)+d ) * (dot(b,n)+d) fails i come with different solution to the problem (note that this dot * dot thing is very important here because i willl show you that it may work in another part of the function.
here is fixed part of code:
bool fully_submerged = true;
for (int i=ahue; i < ahue+flen; i++) //through all vertices in that poly..., 3 only
{
A = pos + ROTATION_MAT * hull->VBO_V[i];
int next = i + 1;
if (next >= ahue+flen) next = ahue;
B = pos + ROTATION_MAT * hull->VBO_V[next];
int aside = int(PLUSORMINUS(dot(A, normal) + dst));
int bside = int(PLUSORMINUS(dot(B, normal) + dst));
if ((aside*bside) < 0) fully_submerged = false;
}
//~~~~~~~~~~ Test if Area is fully submerged or not at all
if (fully_submerged) //fully submerged triangle
if ((dot(pos + ROTATION_MAT * hull->VBO_V[ahue], normal) + dst) < 0)
{
arri = arri + 1;
ACTUAL_FRAME[arri].V[0] = pos + ROTATION_MAT * hull->VBO_V[ahue];
ACTUAL_FRAME[arri].V[1] = pos + ROTATION_MAT * hull->VBO_V[ahue+1];
ACTUAL_FRAME[arri].V[2] = pos + ROTATION_MAT * hull->VBO_V[ahue+2];
ACTUAL_FRAME[arri].normal = normal;
ACTUAL_FRAME[arri].dst = dst;
ACTUAL_FRAME[arri].pop = pop;
ALOG(""); ALOG("");ALOG("------------------");
ALOG("ADDED WHOLE FACE AS BELOW WATERLINE:");
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[0]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[0], normal) + dst));
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[1]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[1], normal) + dst));
ALOG("TESTING POINT: "+POINT_TO_TEXT(ACTUAL_FRAME[arri].V[2]) + " dot: " +FloatToStr(dot( ACTUAL_FRAME[arri].V[2], normal) + dst));
ALOG("----------------------");
ALOG("");
continue; //go to next face
} else continue;
where
template <class type> type PLUSORMINUS(type k)
{
if (k >= 0) return type(1.5); else return -type(1.5);
}
i basically checked dots and whenever dot was 0 or greater i returned integer = 1
if it was below i returned integer -1
so multiplying these two will always give me proper positive / negative values
it gave me this:
now theres important part of code i didint mention earlier the one that uses SegmentPlaneIntersection with crappy dot*dot thing.
but before i post part of the code and whole function code i want to explain what whole function does and what does that important second part of code:
Function name is: void SliceHullByWater(TWaterWaves * ocean) so it slices hull model with water planes (but for simplification i used one plane for all triangles
- Ok so first of all I loop through all faces in the model
- then I go through all vertices in that model and check whenever they are on different sides of plane: if they are all on one side of plane we check if first vertex of the face is below that plane (water plane) if it is then we add this triangle to the stack. (ACTUAL_FRAME), then skip this face iteration. If not then we do nothing and skip this face iteration (because triangle is above water and we return only submerged part)
But whenever we have points on both sides of water plane then we need to perform cut.
And here is the second part I use that same crappy dot*dot function that determines whenever theres intersection or not.
sliced_area.Count = 0;
for (int i=ahue; i < ahue + flen; i++) //throug all face vertices
{
A = pos + ROTATION_MAT * hull->VBO_V[i];
if (dot( A, normal ) + dst < 0.0) // add to temp polygon as new underwater triangle vertex
int next = i + 1;
if (next >= ahue+flen) next = ahue;
B = pos + ROTATION_MAT * hull->VBO_V[next];
//here check for intersection if theres intersection then put that to output var and add that output to as another vertex to temp polygon
if ( SegmentPlaneIntersection(normal, dst, A, B, output) )
}
Like you see I am again using SegmentPlaneIntersection but now it works even when I test same sides, it just works....
now i am really confused coz i spent last 3 days thinking why the hell in first loop SegmentPlaneIntersection function didin't work and in this part of code is working. (my first thoughts are because of some floating point precision issues - since i dont see any bugs in original function (that whole code i posted in spoiler) i am not quite sure what is going on and this is my question is it really a FP issue or i just simply wrote (first part) it wrong? and i dont see some statement that causes magic to happen..
Im confused because it may work on windows machine but when i put that into android phone - depending on phone it can slice properly or wrongly liek in the first picture.
Edited by WiredCat
##### Share on other sites
Did not read through this all
The line segement plane intersection seems more complex than necessary (e.g. normalization).
Below the code i'm using without sqrt.
You wanna do Buoyancy simulation and you clip the whole detailed ship by the waterplane?
Why don't you use a low res approximization, like a box? Should be good enough?
inline float IntersectRayPlane (qVec3 &rO, qVec3 &rD, qVec3 &pO, qVec3 &pN)
{
// rD does not need to be unit length
float d = pN.Dot(rD);
float n = pN.Dot(pO - rO);
if (fabs(d) < FP_EPSILON) // ray parallel to plane
{
if (fabs(n) < FP_EPSILON) return 0; // ray in plane
else return FLT_MAX; // no intersection
}
float t = n / d;
//qVec3 intersectionPoint = rP + rD * t;
return t;
}
##### Share on other sites
when using boxes i cant calculate drag acting on a boat properly without using some imaginary coefficients that act on specified angle at specified power, additionally it was flipping the boat during turns(but thats not related to box thing) any way slicing box and hull uses the same algorithm
##### Share on other sites
if( Denominator == 0.0) PointOnPlane = rA;
else
That's very very bad - you compare a floating point number against zero.
You should test against a small epsilon instead, like i do here: if (fabs(d) < FP_EPSILON)
(FP_EPSILON = 1.0e-5f for single precision float)
The problem is, a tiny number larger than zero but smaller than a reasonable small number like 0.00001
may result in garbage after the division - testing for exactly zero is not enough.
There is almost never a reason to compere floating point against a exact value.
You need to use epsilons everywhere.
##### Share on other sites
main reason: if(distance1 * distance2 > 0.0) return false;
##### Share on other sites
main reason: if(distance1 * distance2 > 0.0) return false;
That's the same bad practice. First you should classify if the points are ON the plane (fabs(distance) < epsilon).
If one point is on the plane you can return, and afterwards your test would be ok, because the signs are guaranteed to make sense.
For robust polygon clipping you need to handle special cases like point on plane, using doubles won't fix it.
I can't tell if this causes your actual problem because you forgot to give numbers for the plane.
However - seems you're not willing to listen anyways.
##### Share on other sites
You should also look up Sutherland Hodgman clipping inside of Christer Ericson's "Real-Time Collision Detection" book. He has some text that explains planes, thick planes, clipping and how to avoid common bugs/problems like you are running into. It seems you are missing some fundamentals of using floats for geometry stuff on the CPU, and these links and this book will help clear that up. Like Joe was saying if we deal with planes with floats we must think in terms of "close enough" with error tolerances. These resources are great reads and directly help with what you are working on
Bouyancy demo (exactly what you are doing): http://box2d.org/files/Misc/Buoyancy.zip
Float tolerances: http://realtimecollisiondetection.net/blog/?p=89
Edited by Randy Gaul
##### Share on other sites
i saw that buoyancy.zip some time ago, it actually isn't the same thing because i clip whole model then make tetrahedrons where all of them share same apex in center of mass and then i clip these tetrahedrons then i compute center of buoyancy for each tetrahedron (or a clipped solid) (by adding vertices of each shape and dividing by vert amount) not quite sure if thats the way of doing it but, from what i understand i need a mass of every tetrahedron (even when sliced to another shape - basically had to compute a mass at wich i apply a force thats a bit stupid from now since i began to understand that after computing inertia matrix i would only add moments as: Moments = moments + vector_buo_center_2_center_of_mass x force, so i wont have to know the mass of each sliced tetrahedron/solid. but i figoured that after making that ;x
now i am a bit confused if i should put apex of all tetrahedrons in a water plane and calc each tetrahedroon buoyancy force or to calc whole buoyant force for them and somehow apply torque without specyfying point of contact (where force is applied like in this demo but it uses the mass of the submerged part)
or do it like i do: every tetrahedron for each submerged triangle share on apex (at hulls cetner of mass) then i slice again all tetrahedrons to properly calculate volume of submerged part of a hull. after having list of solids that are inside ship hull and they are below waterline i compute buoyant force for each one (and applying torque (after specifying buoyancy center as each solid geometric center point) i will also haveto add form drag to each submerged triangle of a hull that i slice but ill post that after i finish calculation of inertia matrix and fixing whole rotational part of the 'rigid body of mine' - basically i would love to know how to calculate form drag (as far as i remember normalized(-(linear vel + ang vel)) was the initial dragforce direction, since i have a shape i could do that even when it would be a really retarded approximation but at least in that case i won't have to use some hardcoded drag coefficients.
Edited by WiredCat
##### Share on other sites
I think you're overlooking the Catto bouyancy demo. Did you read his article in game programming gems 6? He's doing the same thing you are, and speaks about where to place the apex. He also covers drag.
And, if you need mass you can calculate the volume of your mesh and multiply by an appropriate density value, or just assign a custom mass yourself.
|
#### Vol. 6, No. 1, 2012
Recent Issues
The Journal Cover Editorial Board Editors' Addresses Editors' Interests About the Journal Scientific Advantages Submission Guidelines Submission Form Subscriptions Editorial Login Contacts Author Index To Appear ISSN: 1944-7833 (e-only) ISSN: 1937-0652 (print)
$L$-series of Artin stacks over finite fields
### Shenghao Sun
Vol. 6 (2012), No. 1, 47–122
##### Abstract
We develop the notion of stratifiability in the context of derived categories and the six operations for stacks. Then we reprove the Lefschetz trace formula for stacks, and give the meromorphic continuation of $L$-series (in particular, zeta functions) of ${\mathbb{F}}_{q}$-stacks. We also give an upper bound for the weights of the cohomology groups of stacks, and an “independence of $\ell$” result for a certain class of quotient stacks.
##### Keywords
$l$-adic cohomology, algebraic stack, Lefschetz trace formula, $L$-function
##### Mathematical Subject Classification 2010
Primary: 14F20
Secondary: 14F05, 19F27
|
Creating a table with super columns and rows
After attempts with little success, I can't seem to achieve the table depicted by the image below. The image demonstrates the format I wish to achieve, to which I will add more columns and rows to eventually. If someone can point me in the right direction, that would be highly appreciated!
• Pretty simple, because all text is on one horitontal line. Use \multicolumn, that's all ...
– user2478
Nov 8, 2018 at 12:24
Enjoy (though all those rules don't look good):
\documentclass[]{article}
\newcommand\mc{\multicolumn}
\begin{document}
\begin{tabular}[]{|ll|*6{c|}}
\cline{3-8}
\mc{1}{l}{}&&\mc{3}{|c|}{Col 1}&\mc{3}{c|}{Col 2}\\
\cline{3-8}
\mc{1}{l}{}&& \mc{1}{c}{C} & \mc{1}{c}{NC} & \mc{1}{c|}{SR}
& \mc{1}{c}{C} & \mc{1}{c}{NC} & \mc{1}{c|}{SR} \\
\hline
& Ex 1 &&&&&&\\
\cline{3-8}
Row 1& Ex 2 &&&&&&\\
\cline{3-8}
& Ex 3 &&&&&&\\
\hline
& Ex 1 &&&&&&\\
\cline{3-8}
Row 2& Ex 2 &&&&&&\\
\cline{3-8}
& Ex 3 &&&&&&\\
\hline
& Ex 1 &&&&&&\\
\cline{3-8}
Row 3& Ex 2 &&&&&&\\
\cline{3-8}
& Ex 3 &&&&&&\\
\hline
\end{tabular}
\end{document}
|
Réunion d'hiver SMC 2013
Université d'Ottawa, 6 - 9 décembre 2013
Algèbres d'opérateurs
Org: Benoit Collins (University of Ottawa), Thierry Giordano (University of Ottawa), Mehrdad Kalantar (Carleton University) et Matthew Kennedy (Carleton University)
[PDF]
Quantum unique ergodicity for generalized Wigner matrices [PDF]
For large generalized Wigner matrices, we prove a probabilistic version of quantum unique ergodicity at any scale, and gaussianity of the eigenvectors entries. The proof relies on analyzing the effect of the Dyson Brownian motion on eigenstates. Relaxation to equilibrium of the eigenvectors is related to a new multi-particle random walk in a random environment, the eigenvector moment flow. This is joint work with H.T. Yau.
RAPHAEL CLOUATRE, University of Waterloo
Unitary equivalence to Jordan models for weak contractions of class $C_0$ [PDF]
We study the problem of classification of Hilbert space contractions belonging to the class $C_0$ through the properties of some associated operator algebras. More precisely, we obtain results on the unitary equivalence of weak contractions of class $C_0$ to their Jordan models under an assumption on their commutants. In particular, our work addresses the case of arbitrary finite multiplicity. The main tool is the theory of boundary representations due to Arveson. Our approach also raises some interesting questions about general non-self-adjoint operator algebras and possible refinements of Paulsen's theorem on completely bounded homomorphisms.
KENNETH R. DAVIDSON, University of Waterloo
Semicrossed products over lattice semigroups [PDF]
We study semicrossed products by semigroups, and in this talk I will specialize to the case of lattice ordered Ore semigroups. For these semigroups, one can restrict attention to representations of the covariance relations which are Nica covariant, a doubly commuting condition that applies to elements with meet zero. In the abelian case, we show that the C*-envelope of the Nica-covariant semicrossed product for injective systems on C*-algebras is a full crossed product. For the special case of $\mathbb{Z}_+^n$, we embed an arbitrary C* dynamical system into a canonical injective system. We can then deduce that the C*-envelope of the Nica-covariant semicrossed product is a full corner of a full crossed product. In the non-abelian case, we consider right covariance relations, which automatically reduce to the injective case, and again the C*-envelope is a full crossed product.
GEORGE ELLIOTT, University of Toronto
A short essay on the Pimsner-Voiculescu embedding [PDF]
In some sense, all embeddings of the irrational rotation C*-algebra in an AF algebra are the same---they are approximately unitarily equivalent (since the domain algebra is simple AT with unique trace). In another sense, the original Pimsner-Voiculescu embedding would appear to have special properties, as it would seem to be somewhat rigid, with unusually rapid (summable) convergence of the finite-dimensional approximate embeddings, with these being rigidly specified in terms of the continued fraction expansion. An interesting question, perhaps, is whether this map (although it is not quite uniquely determined) is conjugate to another AF-embedding discovered recently by Zhuang Niu and me, obtained simply by cutting the spectrum of each of the canonical unitary generators (at a single point---in some representation of the algebra). Certainly, not all embeddings are conjugate, whether one looks at the mapping or even just the subalgebra (as for subfactors).
DOUG FARENICK, University of Regina
New characterisations of the weak expectation property [PDF]
It is known, by a theorem of Kirchberg, that every separable C$^*$-algebra is a C$^*$-subalgebra of a separable C$^*$-algebra with the weak expectation property. Furthermore, Kirchberg has shown that the assertion that every separable C$^*$-algebra is a quotient of a C$^*$-algebra with the weak expectation property is logically equivalent to the assertion that the Connes Embedding Problem has an affirmative solution. Therefore, it is not surprising that it is rather difficult to ascertain whether a given C$^*$-algebra has the weak expectation property or not. In this lecture I will report on joint work with A. Kavruk, V. Paulsen, and I.G. Todorov in which unital C$^*$-algebras with the weak expectation property are characterised by matrix completion properties and by certain tensorial identities involving group C$^*$-algebras and operator systems.
Semicrossed Products over Semigroups [PDF]
Group dynamical systems and the crossed product algebras they generate have long been a source of interesting operator algebras. The natural generalization, inspired by concrete examples of operator algebras, is to consider semigroup dynamical systems. That is, $A$ is an operator algebra and $S$ is a semigroup acting on $A$ by endomorphisms. The goal is to construct a larger algebra, containing $A$, that also encodes the information of the action of $S$ on $A$. The non-self-adjoint versions of these algebras are called semicrossed product algebras.
Recent work has shown that the C$^*$-envelope of a semicrossed algebra can be useful discovering properties of the underlying dynamical system. However, whilst in the C$^*$-literature a wide class of semigroups are considered for crossed-products, the majority of the work on semicrossed products has been carried out in the case when the semigroup $S$ is the non-negative integers.
In this talk we present some recent results on semicrossed products arising from a wider class of positive cones. This is joint work with Kenneth R. Davidson and Evgenios T.A. Kakariadis.
DANIEL GONÇALVES, UFSC - Universidade Federal de Santa Catarina
Simplicity of partial skew group rings [PDF]
Let R be a commutative ring, G a group and $\alpha$ a partial action by ideals that contain local units. In this talk we will describe a simplicity criterion for the partial skew group ring R*G in terms of maximal commutativity and G-simplicity of R. If time permits we will sketch how to use this result to give a new proof of the simplicity criterion for Leavitt path algebras and how to apply it in partial topological dynamical systems.
MICHAEL HARTZ, University of Waterloo
DAVID KERR, Texas A\&M University
Borel complexity and automorphisms of operator algebras [PDF]
We show that, for the standard examples of strongly self-absorbing $C^*$-algebras, the action of the automorphism group on itself by conjugation is generically turbulent. Using this we then prove that, for all separable $C^*$-algebras which tensorially absorb the Jiang-Su algebra, the relation of conjugacy on automorphisms is not classifiable by countable structures. This includes the $C^*$-algebras that fall under the scope of classification results based on the Elliott invariant. The main novelty of our turbulence argument is the use of the malleability of the tensor product shift over a strongly self-absorbing $C^*$-algebra. This is joint work with M. Lupini, N.C. Phillips, and W. Winter.
JAMIE MINGO, Queen's University
Asymptotic Freeness and the Transpose [PDF]
We show a surprising connection between asymptotic freeness of certain ensembles of random matrices and the transpose.
JONATHAN NOVAK, MIT
Asymptotics of unitary multimatrix models [PDF]
A fundamental result of Voiculescu gives algebraic machinery for computing the large $N$ limit of the normalized trace of a polynomial in several independent Haar-distributed random unitary matrices. This result has previously been generalized in two ways: first, to the large $N$ limit of several interacting random matrices (Collins, Guionnet, Maurel-Segala); second, to global fluctuations of several independent random matrices (Mingo, Sniady, Speicher). I will present a unification and extension of these results to the complete asymptotics of all cumulants of several interacting random unitary matrices. The approach rests on the systematic analysis of a hierarchy of noncommutative partial differential equations known as the "Schwinger-Dyson lattice." This is joint work with Alice Guionnet.
ALEXEY POPOV, University of Waterloo
Abelian, amenable operator algebras are similar to C*-algebras [PDF]
Suppose that $H$ is a complex Hilbert space and that $B(H)$ denotes the bounded linear operators on $H$. We show that every abelian, amenable operator algebra is similar to a $C^*$-algebra. We do this by showing that if $A \subseteq B(H)$ is an abelian algebra with the property that given any bounded representation $\varrho: A \to B(H_\varrho)$ of $A$ on a Hilbert space $H_\varrho$, every invariant subspace of $\varrho(A)$ is topologically complemented by another invariant subspace of $\varrho(A)$, then $A$ is similar to an abelian $C^*$-algebra.
This is a joint work with Laurent W. Marcoux.
CHRIS RAMSEY, University of Virginia
Triangular algebras vs. tensor algebras [PDF]
In 1959 Kadison and Singer defined an operator algebra T to be triangular if $T \cap T^*$ is abelian. The tensor algebra of a C$^*$-correspondence over a commutative C$^*$-algebra is then triangular, which begs the question whether all triangular algebras are tensor algebras. We will look at this question in the case of triangular UHF algebras.
LUIS SANTIAGO, University of Oregon
PAUL SKOUFRANIS, University of California - Los Angeles
Freely Independent Random Variables with Non-Atomic Distributions [PDF]
One central concept in the study of free probability is to describe the spectral distributions of non-commutative polynomials of freely independent random variables. In this talk, we will examine the spectral distributions of non-commutative polynomials of non-atomic, freely independent random variables. In particular, the construction of an analogue to the Strong Atiyah Conjecture for free groups implies that the measure of each atom of any $n \times n$ matricial polynomial of non-atomic, freely independent random variables is an integer multiple of $n^{-1}$. In addition, we will show that the distribution of any matricial polynomial of freely independent semicircular variables has an algebraic Cauchy transform and thus is real-analytic except at a finite number of points. This is joint work with D. Shlyakhtenko.
A central sequence in a C*-algebra is a sequence that asympotically commutes in norm with every element of the algebra. Similarly a we have a definition of central sequences for II$_1$ factors, where the convergence is now in the L$^2$-norm associated with the trace. It is well known that the group von Neumann algebra of the free group on two generators have only trivial central sequences. On the other hand, the reduced C*-algebra of the free group on two generators has an abundance of central sequences. To solve this dichotomy we introduce a new notion of central sequences, the tracially central sequences. We believe that the notion of tracially central sequence should replace the notion of central sequences in a C*-algebra. In fact we show that if A is a simple, stably finite, unital, separable C*-algebra, which has strict comparison of positive elements and a unique tracial state, and if in addition A satisfies an extra condition, then the tracially central algebra of A coincides with the central algebra of the von Neumann algebra associated to the Gelfand-Naimark-Segal representation of A.
Tracial Rokhlin property for finite group actions on C*-algebras was introduced by Chris Phillips to study the structure of the crossed product, which could be viewed as a finite group analogue of the classical Rokhlin lemma. In this talk, we will give a characterization of product-type actions with tracial Rokhlin property. Then we will show that, if $\alpha\colon G \rightarrow Aut(A)$ is a finite group action with the weak tracial Rokhlin property, where $A$ is a $\alpha$-simple unital C*-algebra with either tracial rank zero, or real rank zero, or stable rank one, the same property on $A$ passes to the crossed product $C^*(G,A,\alpha)$.
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## What is Photon
The photon is a type of elementary particle, the quantum of the electromagnetic field including electromagnetic radiation such as light, and the force carrier for the electromagnetic force (even when static via virtual particles). The photon has zero rest mass and always moves at the speed of light within a vacuum.
Like all elementary particles, photons are currently best explained by quantum mechanics and exhibit wave-particle duality, exhibiting properties of both waves and particles. For example, a single photon may be refracted by a lens and exhibit wave interference with itself, and it can behave as a particle with definite and finite measurable position or momentum, though not both at the same time. The photon’s wave and quantum qualities are two observable aspects of a single phenomenon – they cannot be described by any mechanical model; a representation of this dual property of light that assumes certain points on the wavefront to be the seat of the energy is not possible. The quanta in a light wave are not spatially localized.
The modern concept of the photon was developed gradually by Albert Einstein in the early 20th century to explain experimental observations that did not fit the classical wave model of light. The benefit of the photon model was that it accounted for the frequency dependence of light’s energy, and explained the ability of matter and electromagnetic radiation to be in thermal equilibrium. The photon model accounted for anomalous observations, including the properties of black-body radiation, that others (notably Max Planck) had tried to explain using semiclassical models. In that model, the light was described by Maxwell’s equations, but material objects emitted and absorbed light in quantized amounts (i.e., they change energy only by certain particular discrete amounts). Although these semiclassical models contributed to the development of quantum mechanics, many further experiments beginning with the phenomenon of Compton scattering of single photons by electrons, validated Einstein’s hypothesis that light itself is quantized In 1926 the optical physicist Frithiof Wolfers and the chemist Gilbert N. Lewis coined the name photon for these particles After Arthur H. Compton won the Nobel Prize in 1927 for his scattering studies, most scientists accepted that light quanta have an independent existence, and the term photon was accepted.
In the Standard Model of particle physics, photons and other elementary particles are described as a necessary consequence of physical laws having a certain symmetry at every point in spacetime. The intrinsic properties of particles, such as charge, mass and spin, are determined by this gauge symmetry. The photon concept has led to momentous advances in experimental and theoretical physics, including lasers, Bose-Einstein condensation, quantum field theory, and the probabilistic interpretation of quantum mechanics. It has been applied to photochemistry, high-resolution microscopy, and measurements of molecular distances. Recently, photons have been studied as elements of quantum computers, and for applications in optical imaging and optical communication such as quantum cryptography.
## Properties of Photon
The cone shows possible values of wave 4-vector of a photon. The “time” axis gives the angular frequency and the “space” axis represents the angular wavenumber. Green and indigo represent left and right polarization
A photon is massless, has no electric charge, and is a stable particle. A photon has two possible polarization states. In the momentum representation of the photon, which is preferred in quantum field theory, a photon is described by its wave vector, which determines its wavelength λ and its direction of propagation. A photon’s wave vector may not be zero and can be represented either as a spatial 3-vector or as a (relativistic) four-vector; in the latter case, it belongs to the light cone (pictured). Different signs of the four-vector denote different circular polarizations, but in the 3-vector representation one should account for the polarization state separately; it actually is a spin quantum number. In both cases the space of possible wave vectors is three-dimensional.
The photon is the gauge boson for electromagnetism, and therefore all other quantum numbers of the photon (such as lepton number, baryon number, and flavor quantum numbers) are zero. Also, the photon does not obey the Pauli exclusion principle.
Photons are emitted in many natural processes. For example, when a charge is accelerated it emits synchrotron radiation. During a molecular, atomic or nuclear transition to a lower energy level, photons of various energy will be emitted, ranging from radio waves to gamma rays. Photons can also be emitted when a particle and its corresponding antiparticle are annihilated (for example, electron-positron annihilation).
In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the magnitude of the momentum vector p. This derives from the following relativistic relation, with m = 0:
{\displaystyle E^{2}=p^{2}c^{2}+m^{2}c^{4}.} E^{2}=p^{2} c^{2} + m^{2} c^{4}.
The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ):
{\displaystyle E=\hbar \omega =h\nu ={\frac {hc}{\lambda }}} E=\hbar\omega=h\nu=\frac{hc}{\lambda}
{\displaystyle {\boldsymbol {p}}=\hbar {\boldsymbol {k}},} \boldsymbol{p}=\hbar\boldsymbol{k},
where k is the wave vector (where the wave number k = |k| = 2π/λ), ω = 2πν is the angular frequency, and ħ = h/2π is the reduced Planck constant.
Since p points in the direction of the photon’s propagation, the magnitude of the momentum is
{\displaystyle p=\hbar k={\frac {h\nu }{c}}={\frac {h}{\lambda }}.} p=\hbar k=\frac{h\nu}{c}=\frac{h}{\lambda}.
The photon also carries a quantity called spin angular momentum that does not depend on its frequency.[23] The magnitude of its spin is {\displaystyle \scriptstyle {{\sqrt {2}}\hbar }} \scriptstyle{\sqrt{2} \hbar} and the component measured along its direction of motion, its helicity, must be ±ħ. These two possible helicities, called right-handed and left-handed, correspond to the two possible circular polarization states of the photon.[24]
To illustrate the significance of these formulae, the annihilation of a particle with its antiparticle in free space must result in the creation of at least two photons for the following reason. In the center of momentum frame, the colliding antiparticles have no net momentum, whereas a single photon always has momentum (since, as we have seen, it is determined by the photon’s frequency or wavelength, which cannot be zero). Hence, conservation of momentum (or equivalently, translational invariance) requires that at least two photons are created, with zero net momentum. (However, it is possible if the system interacts with another particle or field for the annihilation to produce one photon, as when a positron annihilates with a bound atomic electron, it is possible for only one photon to be emitted, as the nuclear Coulomb field breaks translational symmetry.) The energy of the two photons, or, equivalently, their frequency, may be determined from conservation of four-momentum. Seen another way, the photon can be considered as its own antiparticle. The reverse process, pair production, is the dominant mechanism by which high-energy photons such as gamma rays lose energy while passing through matter. That process is the reverse of “annihilation to one photon” allowed in the electric field of an atomic nucleus.
The classical formulae for the energy and momentum of electromagnetic radiation can be re-expressed in terms of photon events. For example, the pressure of electromagnetic radiation on an object derives from the transfer of photon momentum per unit time and unit area to that object, since the pressure is force per unit area and force is the change in momentum per unit time.
Each photon carries two distinct and independent forms of angular momentum of light. The spin angular momentum of light of a particular photon is always either {\displaystyle +\hbar } {\displaystyle +\hbar } or {\displaystyle -\hbar } {\displaystyle -\hbar }. The light orbital angular momentum of a particular photon can be any integer N, including zero
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# Math Help - Convergence problem
1. ## Convergence problem
Hello
Let f be a function [0 ,1] ->[0,1] and
|f(x)-f(z)| <= |x-z| x,z in [0,1].
I need to proove that the sequence
x_n+1 = 1/2 * [x_n + f(x_n)], n in N
converges to a fixed point of f.
Thanks
/Mark
2. Is it sufficient to prove that $|x_{n+1} - x_{n}| \le |x_{n} - x_{n-1}|$? What do you think? The successive absolute differences are non-increasing.
Or will we also have to prove that it's bounded? (Which seems pretty obvious.)
3. Originally Posted by TKHunny
Is it sufficient to prove that $|x_{n+1} - x_{n}| \le |x_{n} - x_{n-1}|$? What do you think? The successive absolute differences are non-increasing.
Or will we also have to prove that it's bounded? (Which seems pretty obvious.)
Consider the sequence of partial sums of the harmonic series:
$
S_n=\sum_{r=1}^n \frac{1}{r}
$
This sequences satisfies:
$|S_{n+1} - S_{n}| < |S_{n} - S_{n-1}|$,
but does not converge.
RonL
4. I believe TKHunny was referring to something called a "contractive sequence" which is $|x_{n+1}-x_n|\leq k|x_n-x_{n-1}|$ where $k\in (0,1)$. (Note TKHunny did not fullfil this condtion). The theorem says that contractive sequences converges. See Here.
This is Mine 68th Post!!!
5. I wasn't quite seeing it. I was hoping I could get one of you'all to talk me into it.
It was rather weak to rely on "non-increasing". Since the sequence would quikly leave the Domain if it had to stick with equality, it would be stronger to state "decreasing", but that remains only as convincing as the Harmonic example.
Well, Mark84, what are your thoughts? What is it about that slope that ensures convergence?
$\frac{|f(x) - f(z)|}{|x - z|} < 1$
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brechtel2019farranging
Far-ranging generalist top predators enhance the stability of meta-foodwebs
Andreas Brechtel, Thilo Gross and Barbara Drossel
Sci Rep 9, 12268, 2019
Identifying stabilizing factors in foodwebs is a long standing challenge with wide implications for community ecology and conservation. Here, we investigate the stability of spatially resolved meta-foodwebs with far-ranging super-predators for whom the whole meta-foodwebs appears to be a single habitat. By using a combination of generalised modeling with a master stability function approach, we are able to efficiently explore the asymptotic stability of large classes of realistic many-patch meta-foodwebs. We show that meta-foodwebs with far-ranging top predators are more stable than those with localized top predators. Moreover, adding far-ranging generalist top predators to a system can have a net stabilizing effect, despite increasing the food web size. These results highlight the importance of top predator conservation.
Figure 1: Proportion of stable webs for a 10-patch Holling type 2 system with $$S = 10$$ local species and a global predator for the different numbers of prey species of the global predator. The dashed line represents the proportion of stable webs of the system when the global species is removed. The system’s stability increases when the global species has more prey species. For more than 4 prey species the system with the global predator exceeds the stability of the system without it.
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## Pre-Algebra Review
1.
Premix concrete uses a mixture of $5 2/3$ pounds of gravel for every $3 1/3$ pounds of cement. If the plant operator needs to make a batch of premix using 4,300 pounds of gravel, how much cement will be needed?
1. 2,529 pounds
2. 6,318 pounds
3. 4,516 pounds
4. 2,689 pounds
2.
Which are equivalent ratios?
1. 4:8 and 1:2
2. 8:6 and 6:8
3. 3:7 and 5:14
4. 5:3 and 12:15
3.
Tina drives 40 miles per hour. At this rate, how long will it take her to travel 120 miles?
1. 2 hrs
2. 4 hrs
3. 3 hrs
4. 5 hrs
4.
A factory is producing smart phones. In one hour, they can produce 14,000 smart phones ($y$). Write the equation that represents how many smart phones they can produce per hour ($t$).
1. $y=14000t$
2. $t=14000y$
3. $y=1/14000t$
4. $t=1/14000y$
5.
In a class of 50 students, 36% own a bicycle.
How many students in the class own a bicycle?
1. 18
2. 15
3. 20
4. 45
6.
A clothing shop has a markup of 45% on the wholesale price of shirts. If the wholesale price of men's shirts is $35.00, what will be the retail price? 1.$80.00
2. $52.50 3.$57.50
4. $50.75 5.$45.00
7.
If you were to place a point at $+2$ on the number line, add $-5$ to that number, and move the point to the location of the new number, how many units and in what direction would you have to move the new point in order to bring it back to where it originally was?
1. 3 units to the right
2. 5 units to the right
3. 0 units in either direction
4. 2 units to the left
8.
A plane takes off from an airport that is 1,500 feet above sea level. If the plane ascends 6,000 feet from the ground, how many feet will the plane have to descend if it wants to land at another airport that is also 1,500 feet above sea level?
1. 9,000 feet
2. 7,500 feet
3. 4,500 feet
4. 6,000 feet
9.
Which of the following has a sum of 0?
1. $3 + | -3|$
2. $-3 + | 3|$
3. $|3| + |3|$
4. $| -3| + | -3|$
10.
What number should be placed in the square to make the equation true?
$5.32 - square = 5.32 + (-7.89)$
1. 2.57
2. -2.57
3. -7.89
4. 7.89
11.
If you add $-3/4$ and $-2/8$, where would your answer be on the number line?
1. A
2. B
3. C
4. D
12.
All non-terminating decimals are irrational numbers.
1. True
2. False
13.
$pi$ is a rational number.
1. True
2. False
14.
Which of these numbers is irrational?
1. $sqrt49$
2. $16/3$
3. $24/3$
4. $sqrt5$
15.
$sqrt2$ is what type of number?
1. integer
2. irrational
3. rational
4. whole
16.
$sqrt(20)$ is between which two numbers?
1. 2 and 3
2. 3 and 4
3. 4 and 5
4. 5 and 6
17.
Expand.
$4(5x+3)$
1. $20x+7$
2. $20x+1$
3. $20x-1$
4. $20x+12$
18.
Factor.
$48x+8$
1. $8(6x+1)$
2. $8(x+6)$
3. $40x$
4. $6(x+8)$
19.
Combine like terms.
$3x-2y+4z-x+5y+z$
1. $12x+7y+4z$
2. $2x-3y+5z$
3. $2x+3y+5z$
4. $5xy+4xz+5yz$
20.
Tristan wants to replace his computer screen. Which of these equations would represent Tristan's scenario if the screen costs $87 plus 16% in taxes, and x represents the total amount he must pay? 1. $x=87xx0.16$ 2. $x=87xx1.16$ 3. $x=0.16/87$ 4. $x=87/1.16$ 21. Solve. $a-2+3=-2$ 1. $a=-3$ 2. $a= -1$ 3. $a=7$ 4. $a=0$ 22. During the third quarter of a basketball game between the Tigers and the Bears, four different players scored all of the points for the Tigers. Player A scored $1//2$ of the Tigers points. Player B scored $1//3$ of the Tigers points. Player C made one three-point basket. Player D scored the last point by sinking a free throw. How many points did the Tigers score in the third quarter? 1. 24 points 2. 21 points 3. 18 points 4. 12 points 23. Solve for x. $7x+5=61$ 1. x=7 2. x=8 3. x=9 4. x=10 24. The price of tickets in a group when purchased in bulk can be found with the equation $C=px+24$ where $C$ is the cost, $p$ is the number of people, and $x$ is the price per ticket. What is the price of each ticket if it costs$189 to buy tickets for 15 people?
1. $8 2.$24
3. $9 4.$11
25.
Which graph shows the solution to the inequality?
$x + 4 <= 4$
26.
$1/32$ can also be written which way?
1. $2^-3$
2. $2^-4$
3. $2^-5$
4. $2^-6$
27.
Which of these is equivalent to $4^-3?$
1. $1/4 xx 1/4 xx 1/4$
2. $-4 xx -4 xx -4$
3. $3^4$
4. $4 xx 3$
28.
Simplify the following expression.
$3x^5*2x^12$
1. $5x^17$
2. $5x^60$
3. $6x^17$
4. $6x^60$
29.
Find the square root.
$sqrt144$
1. 72
2. 12
3. 6
4. 3
30.
Simplify. $root3 27$
1. 9
2. 3
3. 2
4. 4
31.
Which of the following is a correct notation for 58,000?
1. $5.8xx10^4$
2. $5.8xx10^3$
3. $5.8xx10^2$
4. $5.8xx10^5$
32.
Which of the following is a correct notation for 0.0000032?
1. $3.2xx10^-6$
2. $3.2xx10^-5$
3. $3.2xx10^-7$
4. $3.2xx10^-4$
33.
$(4.9xx10^9)/(1.6xx10^3)$
1. $3.02xx10^6$
2. $3.2xx10^6$
3. $3.6xx10^6$
4. $3.06xx10^6$
34.
A teacher grades 5 tests in 30 minutes. If a graph was made to show the number of tests per hour that the teacher can grade, what would be the slope of the graph?
1. $6$
2. $1/6$
3. $60$
4. $10$
35.
An airplane is gaining 1,000 meters of altitude for every 2 kilometers that it travels. This graph
1. has a smaller rate of change than the airplane.
2. has a greater rate of change than the airplane.
3. has the same rate of change as the airplane.
4. has no relation than the airplane.
36.
What is the y-intercept of the equation $y=3x-2?$
1. $-3$
2. $3$
3. $2$
4. $-2$
37.
What is the slope of the graph?
1. $-2$
2. $-1$
3. $1/2$
4. $2$
38.
What is the slope of the equation $y=4x-5$?
1. 5
2. -5
3. 4
4. -4
39.
The lines $y=2x+5$ and $3y-2x=2y+5$ have
1. no solutions.
2. 1 solution.
3. infinitely many solutions.
4. 2 solutions.
40.
The lines $4y-5=3x$ and $y=5x-1$ have
1. no solutions.
2. 2 solutions.
3. infinite solutions.
4. 1 solution.
41.
Simplify. $2y-3x+2=5x+12$
1. $y=4x+5$
2. $y=-4x+14$
3. $y=x+5$
4. $y=x+10$
42.
The equation $-3y+8=2y-6x$ is equivalent to which of the following equations?
1. $y=(6x)/5+8$
2. $y=6x+5$
3. $y=(6x)/5+8/5$
4. $y=6x+8$
43.
What is the solution to $y=5x+5$ and $y=x+5 ?$
1. $(0,0)$
2. $(0,5)$
3. $(5,0)$
4. No solution
44.
What is the solution to $y=-3x+2$ and $y=-3x-1 ?$
1. $(-3,6)$
2. $(3,-6)$
3. $(2/3,-3)$
4. No solution
45.
Melody and Brigitta open savings accounts at the bank at the same time. Melody deposits $100 and then deposits$20 per week. Brigitta deposits $50 and then$30 per week. Will Melody and Brigitta ever have the same amount of money in their savings accounts? If so when?
1. After 15 weeks
2. After 2 weeks
3. After 5 weeks
4. Never
46.
This is a function.
1. True
2. False
3. Sometimes
4. Not enough information
47.
Which equation has a smaller rate of change than this graph?
1. $y=-1/3x$
2. $y=-3x$
3. $y=-5/2x-3$
4. $y=-8x$
48.
The equation $y=-2x^2+2$ is
1. linear.
2. nonlinear.
3. not a function.
4. both b and c.
49.
Is this graph linear?
1. Yes
2. No
3. Not enough information
4. Sometimes
50.
Given the table and equation below, what is the value of $square$ that makes the equation true?
$y = squarex$
xy
28
312
416
520
1. 2
2. 4
3. 6
4. 8
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# Quadratic Function Formula – How To Find The Vertex Of A Quadratic Function?
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In elementary algebra, the quadratic formula would be the solution for quadratic equation. There are two possible techniques for solving the quadratic equation instead you put values in quadratic equation directly. These techniques are factoring, completing the square or graphing etc. Out of all these, working with quadratic formula is always the most convenient option.
The general form of quadratic equation could be given as –
$\large ax^{2} + bx + c =0$
Here x is the unknown variable, while a, b, and c are constants that could not be equal to the zero.
This is easy to verify the quadratic formula by satisfying the quadratic equations and inserting the former values to the latter. Hence, the standard quadratic formula in mathematics is given as below –
$\large x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
Here, the solution given y the quadratic formula is called as roots of quadratic equation. Geometrically, roots are used to give values on the parabola where y crosses the x-axis. If there is some formula whose yield is calculated equal to zero then this is easy to find immediately how many real zeros a particular quadratic equation has.
$\large x=\frac{\sqrt{b^{2}-4ac} -b }{2a}$
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OBJECTIVE
To examine whether lower serum levels of serum 25-hydroxyvitamin (OH) D [25(OH)D] are associated with increased risk of developing type 2 diabetes.
RESEARCH DESIGN AND METHODS
A post hoc analysis of three nested case-control studies of fractures, colon cancer, and breast cancer that measured serum 25(OH)D levels in women participating in the Women’s Health Initiative (WHI) Clinical Trials and Observational Study who were free of prevalent diabetes at baseline. Diabetes was defined as self-report of physician diagnosis or receiving insulin or oral hypoglycemic medication. We used inverse probability weighting to make the study population representative of the WHI population as a whole. Weighted logistic regression models compared 25(OH)D levels (divided into quartiles, clinical cut points [<50, 50–<75, ≥75 nmol/L], or as a continuous variable) using the distribution of control subjects and adjusted for multiple confounding factors.
RESULTS
Of 5,140 women (mean age 66 years) followed for an average of 7.3 years, 317 (6.2%) developed diabetes. Regardless of the cut points used or as a continuous variable, 25(OH)D levels were not associated with diabetes incidence in either age or fully adjusted models. Nor was any relationship found between 25(OH)D and incident diabetes when evaluated by strata of BMI, race/ethnicity, or randomization status in the Calcium Vitamin D trial.
CONCLUSIONS
Lower serum 25(OH)D levels were not associated with increased risk of developing type 2 diabetes in this racially and ethnically diverse population of postmenopausal women.
Vitamin D has been shown to have numerous nonskeletal effects, including an important role in pancreatic insulin secretion and insulin action (1). Although several studies have reported a protective relationship between vitamin D and the risk of developing diabetes, the data are not consistent. A recent meta-analysis found that three of six observational studies (OS) found an association between low vitamin D status and increased risk of incident type 2 diabetes or metabolic syndrome (2). In contrast, eight clinical trials found vitamin D supplementation had no effect on glycemia or incident diabetes (1). It may be that higher doses of vitamin D than those tested in clinical trials may be required to affect diabetes risk. Alternatively, the associations of calcium and vitamin D intake with improved glucose metabolism reported in OS may be the result of confounding by other components of foods containing these nutrients (3), outdoor exercise associated with solar radiation, or other factors (4,5).
We undertook this analysis to further evaluate the relationship between 25-hydroxyvitamin D [25(OH)D] levels and diabetes risk in a large cohort of postmenopausal women participating in the Women’s Health Initiative (WHI). Several nested case-control studies within the WHI measured 25(OH)D levels in over 5,000 women, providing a unique opportunity to evaluate an older, multiethnic population at high risk for both vitamin D insufficiency and diabetes (6,7). Our objectives were to examine whether decreased serum levels of 25(OH)D were associated with increased risk of incident type 2 diabetes and whether ethnicity or BMI modified the relationship.
This was a post hoc analysis of data collected from three nested case-control studies that measured 25(OH)D levels among women participating in the WHI clinical trials (CT) and OS who did not have prevalent diabetes at baseline (Supplementary Data).
Between 1993 and 1998, the 40 WHI clinical centers throughout the United States recruited postmenopausal women aged 50–79 years for participation in trials of postmenopausal hormone therapy (HT), dietary modification (DM), and calcium and vitamin D (CaD) supplementation (8). Women who were not eligible for or chose not to participate in the clinical trials were enrolled in the WHI OS. At 1–2 years after joining the hormone or diet intervention trials, 36,282 women were enrolled in the CaD supplementation trial to be randomly assigned to either placebo or calcium carbonate 1,000 mg combined with 25(OH) vitamin D3 400 IU daily. Women were allowed to continue their personal use of calcium and vitamin D as long as vitamin D intake did not exceed 600 IU (and later 1,000 IU) daily. Within the CaD trial, three nested case-control studies were conducted to analyze associations between serum concentrations of 25(OH)D and incidence of colorectal cancer, breast cancer, or hip, spine, or lower wrist fracture; control subjects were matched on age, race/ethnicity, blood draw date, and clinic center at CaD randomization. The CaD breast cancer nested case-control study was also matched on HT and DM trial arm. In the WHI OS, incident hip fracture cases were identified through August 2004; control subjects were matched to case subjects on age, race/ethnicity, and blood draw date.
We analyzed baseline and semiannual visits and annual questionnaires as of the termination dates for the clinical trials (912). Diabetes was defined as self-report of physician diagnosis of “sugar diabetes” treated with insulin or oral medications, or use of an oral hypoglycemic agent or insulin on a medication inventory (13). Physical activity was measured in MET-hrs/wk spent on recreational physical activity. Cardiovascular disease (CVD) was defined as myocardial infarction (MI), coronary revascularization, stroke, and peripheral arterial disease other than abdominal aortic aneurysm.
Serum 25(OH)D concentrations, which reflect total body stores of vitamin D (14), were obtained from fasting serum samples drawn at the baseline (OS) or year 1 (CT) visit that was processed and stored at −80°C. Serum 25(OH)D concentrations (nmol/L) were determined using the DiaSorin LIAISON chemiluminescence method (DiaSorin, Stillwater, MN), and the coefficient of variation (CV) determined using blinded controls was 11.8%, which was similar the CV found in other cohort studies (15). Serum 25(OH)D concentrations varied by month of blood draw, thus month of blood draw was adjusted for in the statistical analyses.
### Statistical analysis
We combined three separate nested case-control studies that included measurement of 25(OH)D concentrations and disease outcomes (CaD/fracture, colorectal cancer; CaD/breast cancer; OS/hip fracture) into one population (Supplementary Data) to evaluate the association between serum 25(OH)D on incident diabetes. The CaD case-control studies supplied 86% of participants in this analysis. Because fractures and breast and colon cancer are not considered along the causal pathway of diabetes, both case and control subjects were included in the analysis; indeed, diabetes incidence was similar across case-control status. Because individuals in these studies were matched for other conditions, we used inverse probability weighting so that our study population would be representative of the WHI population as a whole. The explanatory variables in the models used for weighting included age, ethnicity, latitude of clinical center, and month of blood draw. Previous case-control outcomes (hip fracture, spine fracture, lower arm/wrist fracture, breast cancer, and colorectal cancer) were included in the models as well. For example, because fracture case and control subjects were a large part of our sample, we have a significantly older population in our sample (mean age at blood draw was 66.3 years) than in the full CaD/OS cohort (mean age 63.6 years). Therefore, the highest weights were assigned to those participants least likely to be sampled (for example, younger participants with no fractures). Overall, the weights increased the influence of the control subjects in the analysis while still preserving information from the case subjects. This method better represents the whole population than unrestricted analysis of case and control subjects or a control subjects–only analysis. Nonetheless, we conducted sensitivity analyses to evaluate control subject–only specimens from the above case-control subject pairs (i.e., those serving as control subjects for the fracture cases, colorectal cancer cases, and breast cancer cases), as well as using unweighted data.
We used weighted logistic regression models to compare exposure levels (divided into quartiles or clinical cut points <50, 50–<75, ≥75 nmol/L), using the distribution of the controls (and simultaneously adjusting for multiple confounding factors and effect modifiers), or used as a continuous variable. Models were adjusted for known diabetes risk factors as well as sun exposure (based on latitude of clinical center), which may be a potential effect modifier of vitamin D level that may also influence diabetes risk. To examine the independent relationship of 25(OH)D to risk of diabetes, we evaluated 25(OH)D as a predictor of diabetes after progressively adjusting for potential confounders; model 1: age and ethnicity; model 2: model 1 covariates + latitude of clinical center, month of blood draw, and WHI study indicators (clinical trial, randomization assignment, and case-control status); and model 3: model 2 covariates + BMI, hypertension, fiber intake, magnesium intake, and physical activity. The variables in model 3 were selected by taking an initial full model using the covariates in model 3, along with Langley units, smoking, history of CVD, intake of protein, fruits and vegetables, calcium, fat, alcohol, total glycemic load, multivitamin use, hormone use, waist circumference, education level, and skin cancer, then applying backward selection techniques (P value criteria for removal from model = 0.10) to get a reduced model with BMI (P < 0.0001), hypertension (P = 0.003), fiber (P = 0.010), and magnesium (P = 0.024) (Supplementary Data). In addition to these four variables, all of the adjustments from model 2 and total physical activity (which may reflect sun exposure and is a risk factor for diabetes) were forced into the model. Logistic regression models using model 3 were used to evaluate associations between 25(OH)D and diabetes in women stratified by BMI (normal or underweight <25 kg/m2, overweight 25–<30, and obese ≥30), CaD trial enrollment (active, placebo), or race/ethnicity. Because logistic regression is a complete case subject analysis and we wanted to avoid excluding additional participants as a result of missing physical activity data (∼10% of the sample), an indicator variable for missing physical activity was included in all models. Sensitivity analyses were performed on unweighted data and without physical activity. At all levels of analysis, we tested the assumptions and examined the goodness-of-fit of logistic models. SAS 9.1 (SAS Institute, Cary, NC) was used.
Of 5,140 women, 1,263 (25%) had a 25(OH)D level <34.7 nmol/L, 2,741 (53%) had a 25(OH)D level <50 nmol/L, and 317 (6.2%) developed incident diabetes over a mean follow-up of 7.3 years. Mean age at blood draw was 66 years in women with and without diabetes, with a similar proportion (∼38%) of those aged 70–79 years in both groups (Table 1). Nonwhite women were more likely to develop diabetes than white women. Women who developed diabetes had a higher prevalence of hypertension, obesity, and increased waist circumference, lower education and income levels, lower levels of physical activity, a history of CVD, and were more likely to report a family history of diabetes or premature coronary heart disease. Women who developed diabetes were also more likely to be past or never alcohol users and to report consuming fewer servings of fruits and vegetables and more total, fat, and protein servings. Total vitamin D and calcium intakes were similar in women with and without diabetes, as were U.S. region of residence, sun exposure, season of blood draw, and history of skin cancer. Participants with a hysterectomy were more likely to develop diabetes than those without. Women who developed diabetes were no more likely to be in the CaD intervention group than the placebo group.
Table 1
Baseline characteristics of study participants (CaD and OS) by incident diabetes status
No diabetesDiabetesP
Age at blood draw (years), mean (SD) 66.26 (7.30) 66.29 (7.18) 0.597
50–59 1,044 (21.6) 61 (19.2)
60–69 1,949 (40.4) 133 (42.0)
70–79 1,830 (37.9) 123 (38.8)
Ethnicity <0.001
White 4,400 (91.2) 261 (82.3)
Black 207 (4.3) 21 (6.6)
Hispanic 96 (2.0) 19 (6.0)
Other/Unknown 120 (2.5) 16 (5.0)
U.S. region 0.383
Northeast 1,208 (25.0) 72 (22.7)
South 1,009 (20.9) 79 (24.9)
Midwest 1,199 (24.9) 76 (24.0)
West 1,407 (29.2) 90 (28.4)
Season of blood draw 0.227
Winter 1,064 (22.1) 57 (18.0)
Spring 1,317 (27.3) 82 (25.9)
Summer 1,238 (25.7) 92 (29.0)
Fall 1,204 (25.0) 86 (27.1)
Region by solar irradiance in Langley categories 0.237
475–500 975 (20.2) 68 (21.5)
400–430 777 (16.1) 46 (14.5)
375–380 464 (9.6) 40 (12.6)
350 1,099 (22.8) 80 (25.2)
300–325 1,506 (31.2) 83 (26.2)
Smoking status 0.324
Never 2,541 (52.7) 176 (55.5)
Past 1,895 (39.3) 114 (36.0)
Current 337 (7.0) 26 (8.2)
Hypertension at blood draw <0.001
No 2,569 (53.3) 116 (36.6)
Yes 2,254 (46.7) 201 (63.4)
BMI at blood draw (kg/m2), mean (SD) 27.86 (5.45) 31.27 (5.86) <0.001
≤25 1,628 (33.8) 46 (14.5)
25–<30 1,725 (35.8) 95 (30.0)
≥30 1,470 (30.5) 176 (55.5)
Waist ≥88 cm at blood draw <0.001
No 2,755 (57.1) 97 (30.6)
Yes 2,067 (42.9) 219 (69.1)
Physical activity at blood draw (MET h/week), mean (SD) 11.63 (13.28) 8.98 (10.79) 0.003
<2 1,075 (22.3) 92 (29.0)
2–<7.5 1,011 (21.0) 76 (24.0)
7.5–17 1,218 (25.3) 60 (18.9)
≥17 1,051 (21.8) 53 (16.7)
Hysterectomy 3,017 (62.6) 165 (52.1) <0.001
No 1,806 (37.4) 152 (47.9)
Yes 1,011 (21.0) 76 (24.0)
Not randomized 658 (13.6) 38 (12.0)
Intervention 2,108 (43.7) 141 (44.5)
Comparison 2,057 (42.6) 138 (43.5)
HT arm 0.066
Not randomized 2,979 (61.8) 175 (55.2)
Placebo 949 (19.7) 72 (22.7)
Active 895 (18.6) 70 (22.1)
DM arm 0.953
Not randomized 1,906 (39.5) 123 (38.8)
Intervention 1,808 (37.5) 119 (37.5)
Comparison 1,109 (23.0) 75 (23.7)
OS flag 0.404
No 4,165 (86.4) 279 (88.0)
Yes 658 (13.6) 38 (12.0)
HT use at blood draw 0.054
Never used 1,924 (39.9) 114 (36.0)
Past user 803 (16.6) 69 (21.8)
Current user 2,096 (43.5) 134 (42.3)
Education 0.001
≤High school diploma/GED 1,131 (23.5) 86 (27.1)
School after high school 1,843 (38.2) 143 (45.1)
College degree or higher 1,823 (37.8) 85 (26.8)
Family income <0.001
<$20,000 820 (17.0) 83 (26.2)$20,000–$49,999 2,262 (46.9) 135 (42.6)$50,000–$74,999 861 (17.9) 44 (13.9)$75,000 + 627 (13.0) 34 (10.7)
Alcohol use 0.007
Never drinker 512 (10.6) 45 (14.2)
Past drinker 756 (15.7) 66 (20.8)
Current drinker 3,522 (73.0) 203 (64.0)
Cancer ever at blood draw 0.455
No 4,223 (87.6) 285 (89.9)
Yes 534 (11.1) 28 (8.8)
Skin cancer ever at blood draw 0.159
No 4,363 (90.5) 297 (93.7)
Yes 396 (8.2) 17 (5.4)
CVD ever at blood draw 0.014
No 4,497 (93.2) 282 (89.0)
Yes 254 (5.3) 28 (8.8)
Family history of adult diabetes <0.001
No 3,248 (67.3) 168 (53.0)
Yes 1,349 (28.0) 132 (41.6)
Family hx premature MI (<55 male, <65 female) 0.055
No 3,521 (73.0) 213 (67.2)
Yes 800 (16.6) 68 (21.5)
Multivitamin use at blood draw 0.879
No 2,946 (61.1) 195 (61.5)
Yes 1,877 (38.9) 122 (38.5)
Total vitamin D intake at blood draw (IU), mean (SD) 374.09 (276.51) 374.57 (263.06) 0.332
<200 1,781 (36.9) 122 (38.5)
200–<400 928 (19.2) 51 (16.1)
400–<600 1,150 (23.8) 78 (24.6)
≥600 955 (19.8) 64 (20.2)
Total calcium intake (mg), mean (SD) 828.63 (438.66) 858.84 (458.64) 0.247
<513 1,212 (25.1) 71 (22.4)
513–<740 1,214 (25.2) 71 (22.4)
740–<1053 1,195 (24.8) 92 (29.0)
≥1053 1,202 (24.9) 83 (26.2)
Total magnesium intake (mg), mean (SD) 255.21 (95.32) 253.36 (98.01) 0.858
<187 1,209 (25.1) 80 (25.2)
187–< 242 1,193 (24.7) 78 (24.6)
242–< 311 1,208 (25.0) 85 (26.8)
≥311 1,213 (25.2) 74 (23.3)
Total fat intake (g), mean (SD) 55.54 (29.74) 59.97 (32.37) 0.177
<34.6 1,220 (25.3) 66 (20.8)
34.6–< 49.5 1,201 (24.9) 79 (24.9)
49.5–< 69.2 1,210 (25.1) 79 (24.9)
≥69.2 1,192 (24.7) 93 (29.3)
Total protein intake (g), mean (SD) 66.82 (26.73) 70.24 (29.83) 0.503
<47.6 1,218 (25.3) 69 (21.8)
47.6–<63.2 1,203 (24.9) 78 (24.6)
63.2–<81.8 1,202 (24.9) 86 (27.1)
≥81.8 1,200 (24.9) 84 (26.5)
Total glycemic load (total carbohydrates), mean (SD) 104.96 (41.74) 109.33 (52.32) 0.480
<75.1 1,201 (24.9) 82 (25.9)
75.1–< 98.8 1,218 (25.3) 68 (21.5)
98.8–<127.2 1,204 (25.0) 81 (25.6)
≥127.2 1,200 (24.9) 86 (27.1)
Total fiber intake (g) at blood draw, mean (SD) 16.26 (6.95) 15.71 (7.22) 0.423
<11.2 1,200 (24.9) 92 (29.0)
11.2–<15.2 1,206 (25.0) 77 (24.3)
15.2–<20.1 1,214 (25.2) 75 (23.7)
≥20.1 1,203 (24.9) 73 (23.0)
Fruit/vegetable servings at blood draw, mean (SD) 4.32 (2.12) 4.12 (2.35) 0.016
<2.7 1,192 (24.7) 98 (30.9)
2.7–<4.0 1,156 (24.0) 82 (25.9)
4.0–<5.6 1,262 (26.2) 62 (19.6)
≥5.6 1,213 (25.2) 75 (23.7)
No diabetesDiabetesP
Age at blood draw (years), mean (SD) 66.26 (7.30) 66.29 (7.18) 0.597
50–59 1,044 (21.6) 61 (19.2)
60–69 1,949 (40.4) 133 (42.0)
70–79 1,830 (37.9) 123 (38.8)
Ethnicity <0.001
White 4,400 (91.2) 261 (82.3)
Black 207 (4.3) 21 (6.6)
Hispanic 96 (2.0) 19 (6.0)
Other/Unknown 120 (2.5) 16 (5.0)
U.S. region 0.383
Northeast 1,208 (25.0) 72 (22.7)
South 1,009 (20.9) 79 (24.9)
Midwest 1,199 (24.9) 76 (24.0)
West 1,407 (29.2) 90 (28.4)
Season of blood draw 0.227
Winter 1,064 (22.1) 57 (18.0)
Spring 1,317 (27.3) 82 (25.9)
Summer 1,238 (25.7) 92 (29.0)
Fall 1,204 (25.0) 86 (27.1)
Region by solar irradiance in Langley categories 0.237
475–500 975 (20.2) 68 (21.5)
400–430 777 (16.1) 46 (14.5)
375–380 464 (9.6) 40 (12.6)
350 1,099 (22.8) 80 (25.2)
300–325 1,506 (31.2) 83 (26.2)
Smoking status 0.324
Never 2,541 (52.7) 176 (55.5)
Past 1,895 (39.3) 114 (36.0)
Current 337 (7.0) 26 (8.2)
Hypertension at blood draw <0.001
No 2,569 (53.3) 116 (36.6)
Yes 2,254 (46.7) 201 (63.4)
BMI at blood draw (kg/m2), mean (SD) 27.86 (5.45) 31.27 (5.86) <0.001
≤25 1,628 (33.8) 46 (14.5)
25–<30 1,725 (35.8) 95 (30.0)
≥30 1,470 (30.5) 176 (55.5)
Waist ≥88 cm at blood draw <0.001
No 2,755 (57.1) 97 (30.6)
Yes 2,067 (42.9) 219 (69.1)
Physical activity at blood draw (MET h/week), mean (SD) 11.63 (13.28) 8.98 (10.79) 0.003
<2 1,075 (22.3) 92 (29.0)
2–<7.5 1,011 (21.0) 76 (24.0)
7.5–17 1,218 (25.3) 60 (18.9)
≥17 1,051 (21.8) 53 (16.7)
Hysterectomy 3,017 (62.6) 165 (52.1) <0.001
No 1,806 (37.4) 152 (47.9)
Yes 1,011 (21.0) 76 (24.0)
Not randomized 658 (13.6) 38 (12.0)
Intervention 2,108 (43.7) 141 (44.5)
Comparison 2,057 (42.6) 138 (43.5)
HT arm 0.066
Not randomized 2,979 (61.8) 175 (55.2)
Placebo 949 (19.7) 72 (22.7)
Active 895 (18.6) 70 (22.1)
DM arm 0.953
Not randomized 1,906 (39.5) 123 (38.8)
Intervention 1,808 (37.5) 119 (37.5)
Comparison 1,109 (23.0) 75 (23.7)
OS flag 0.404
No 4,165 (86.4) 279 (88.0)
Yes 658 (13.6) 38 (12.0)
HT use at blood draw 0.054
Never used 1,924 (39.9) 114 (36.0)
Past user 803 (16.6) 69 (21.8)
Current user 2,096 (43.5) 134 (42.3)
Education 0.001
≤High school diploma/GED 1,131 (23.5) 86 (27.1)
School after high school 1,843 (38.2) 143 (45.1)
College degree or higher 1,823 (37.8) 85 (26.8)
Family income <0.001
<$20,000 820 (17.0) 83 (26.2)$20,000–$49,999 2,262 (46.9) 135 (42.6)$50,000–$74,999 861 (17.9) 44 (13.9)$75,000 + 627 (13.0) 34 (10.7)
Alcohol use 0.007
Never drinker 512 (10.6) 45 (14.2)
Past drinker 756 (15.7) 66 (20.8)
Current drinker 3,522 (73.0) 203 (64.0)
Cancer ever at blood draw 0.455
No 4,223 (87.6) 285 (89.9)
Yes 534 (11.1) 28 (8.8)
Skin cancer ever at blood draw 0.159
No 4,363 (90.5) 297 (93.7)
Yes 396 (8.2) 17 (5.4)
CVD ever at blood draw 0.014
No 4,497 (93.2) 282 (89.0)
Yes 254 (5.3) 28 (8.8)
Family history of adult diabetes <0.001
No 3,248 (67.3) 168 (53.0)
Yes 1,349 (28.0) 132 (41.6)
Family hx premature MI (<55 male, <65 female) 0.055
No 3,521 (73.0) 213 (67.2)
Yes 800 (16.6) 68 (21.5)
Multivitamin use at blood draw 0.879
No 2,946 (61.1) 195 (61.5)
Yes 1,877 (38.9) 122 (38.5)
Total vitamin D intake at blood draw (IU), mean (SD) 374.09 (276.51) 374.57 (263.06) 0.332
<200 1,781 (36.9) 122 (38.5)
200–<400 928 (19.2) 51 (16.1)
400–<600 1,150 (23.8) 78 (24.6)
≥600 955 (19.8) 64 (20.2)
Total calcium intake (mg), mean (SD) 828.63 (438.66) 858.84 (458.64) 0.247
<513 1,212 (25.1) 71 (22.4)
513–<740 1,214 (25.2) 71 (22.4)
740–<1053 1,195 (24.8) 92 (29.0)
≥1053 1,202 (24.9) 83 (26.2)
Total magnesium intake (mg), mean (SD) 255.21 (95.32) 253.36 (98.01) 0.858
<187 1,209 (25.1) 80 (25.2)
187–< 242 1,193 (24.7) 78 (24.6)
242–< 311 1,208 (25.0) 85 (26.8)
≥311 1,213 (25.2) 74 (23.3)
Total fat intake (g), mean (SD) 55.54 (29.74) 59.97 (32.37) 0.177
<34.6 1,220 (25.3) 66 (20.8)
34.6–< 49.5 1,201 (24.9) 79 (24.9)
49.5–< 69.2 1,210 (25.1) 79 (24.9)
≥69.2 1,192 (24.7) 93 (29.3)
Total protein intake (g), mean (SD) 66.82 (26.73) 70.24 (29.83) 0.503
<47.6 1,218 (25.3) 69 (21.8)
47.6–<63.2 1,203 (24.9) 78 (24.6)
63.2–<81.8 1,202 (24.9) 86 (27.1)
≥81.8 1,200 (24.9) 84 (26.5)
Total glycemic load (total carbohydrates), mean (SD) 104.96 (41.74) 109.33 (52.32) 0.480
<75.1 1,201 (24.9) 82 (25.9)
75.1–< 98.8 1,218 (25.3) 68 (21.5)
98.8–<127.2 1,204 (25.0) 81 (25.6)
≥127.2 1,200 (24.9) 86 (27.1)
Total fiber intake (g) at blood draw, mean (SD) 16.26 (6.95) 15.71 (7.22) 0.423
<11.2 1,200 (24.9) 92 (29.0)
11.2–<15.2 1,206 (25.0) 77 (24.3)
15.2–<20.1 1,214 (25.2) 75 (23.7)
≥20.1 1,203 (24.9) 73 (23.0)
Fruit/vegetable servings at blood draw, mean (SD) 4.32 (2.12) 4.12 (2.35) 0.016
<2.7 1,192 (24.7) 98 (30.9)
2.7–<4.0 1,156 (24.0) 82 (25.9)
4.0–<5.6 1,262 (26.2) 62 (19.6)
≥5.6 1,213 (25.2) 75 (23.7)
Data are n (%) unless otherwise indicated.
Increasing quartiles of serum 25(OH)D levels were associated with a suggestion of a possible trend toward lower diabetes incidence in the model adjusted for only age and ethnicity and in the model adjusted for age, ethnicity, latitude, month of blood draw, and trial enrollment (P for trend = 0.20; Table 2). However, after more complete adjustment for risk factors for BMI, hypertension, fiber, magnesium intake, and physical activity there was no association between 25(OH)D quartile and incident diabetes (odds ratio 1.01, P for trend = 0.94), nor was there evidence of a linear association between 25(OH)D level and diabetes risk using clinical cut points or a continuous variable (in 5 nmol/L increments). Although there was a suggestion of a U-shaped relationship between 25(OH)D level and diabetes risk, with the 50–<75 nmol/L group having the lowest odds ratio, the 95% CIs were broad and included an odds ratio of 1.
Table 2
Relationship of serum 25(OH)D levels to incident type 2 diabetes among postmenopausal women in the Women’s Health Initiative (N = 5,262)
25(OH)D quartiles (nmol/L)
25(OH)D levels
<34.734.7–47.847.9–64.2>64.2P*Continuous<50 nmol/L50–< 75 nmol/L≥75 nmol/LP*
N 1,263 1,295 1,295 1,287 5,140 2,741 1,668 731
Model 1 Reference 1.15 (0.73–1.79) 0.85
(0.55–1.32) 0.78
(0.48–1.26) 0.195 0.98
(0.95–1.02) Reference 0.79
(0.55–1.13) 0.86
(0.53–1.40) 0.355
Model 2§ Reference 1.15
(0.73–1.80) 0.87
(0.56–1.35) 0.78
(0.47–1.29) 0.195 0.98
(0.95–1.02) Reference 0.80
(0.56–1.15) 0.86
(0.52–1.42) 0.355
Model 3 Reference 1.25
(0.78–1.99) 1.00
(0.64–1.57) 1.05
(0.62–1.76) 0.935 1.01
(0.97–1.05) Reference 0.89
(0.61–1.30) 1.14
(0.68–1.90) 0.873
25(OH)D quartiles (nmol/L)
25(OH)D levels
<34.734.7–47.847.9–64.2>64.2P*Continuous<50 nmol/L50–< 75 nmol/L≥75 nmol/LP*
N 1,263 1,295 1,295 1,287 5,140 2,741 1,668 731
Model 1 Reference 1.15 (0.73–1.79) 0.85
(0.55–1.32) 0.78
(0.48–1.26) 0.195 0.98
(0.95–1.02) Reference 0.79
(0.55–1.13) 0.86
(0.53–1.40) 0.355
Model 2§ Reference 1.15
(0.73–1.80) 0.87
(0.56–1.35) 0.78
(0.47–1.29) 0.195 0.98
(0.95–1.02) Reference 0.80
(0.56–1.15) 0.86
(0.52–1.42) 0.355
Model 3 Reference 1.25
(0.78–1.99) 1.00
(0.64–1.57) 1.05
(0.62–1.76) 0.935 1.01
(0.97–1.05) Reference 0.89
(0.61–1.30) 1.14
(0.68–1.90) 0.873
Data are odds ratio (95% CI) unless otherwise indicated.
*P for linear trend across quartiles.
†5 nmol/L increase.
§Adjusted for age, ethnicity, latitude of clinical center, month of blood draw, and WHI study indicators.
∥Adjusted for age, ethnicity, latitude of clinical center, month of blood draw, WHI study indicators, BMI, hypertension, fiber intake, magnesium intake, and physical activity.
In the fully adjusted model, the P value for interaction between 25(OH)D quartile and BMI was of borderline significance (P = 0.045) (Table 3). All BMI–25(OH)D combinations had 95% CIs that included 1, although different patterns of diabetes risk were suggested for each BMI category: normal weight women in the highest 25(OH)D quartile (>64 nmol/L) had the lowest risk of diabetes whereas obese women had no difference in diabetes risk across 25(OH)D quartiles. No consistent pattern was observed for overweight women. On the other hand, there was no evidence of a BMI interaction with 25(OH)D clinical cut points on diabetes risk (P = 0.48). Nor was there evidence of increased diabetes risk by race/ethnicity or randomization status in the CaD trial for any 25(OH)D categorization (quartiles, clinical cut points, or as a continuous variable; Table 3). Similar relationships between 25(OH)D levels and diabetes risk were observed in fully adjusted analyses of controls only, unweighted participants, and without physical activity in model 3 (data not shown; Supplementary Data).
Table 3
Relationship of serum 25(OH)D levels (categorical) to type 2 diabetes incidence by subgroups
25(OH)D quartiles (nmol/L)
25(OH)D levels (nmol/L)
25(OH)D (nmol/L)
<34.734.7–47.847.9–64.2≥64.2P*<5050–75≥75P*5-unit increasesP*
BMI
< 25 Ref. 0.70 (0.22–2.20) 1.26 (0.46–3.49) 0.37 (0.10–1.33) 0.04 Ref. 1.21 (0.52–2.82) 0.70 (0.20–2.50) 0.48 0.96 (0.89–1.03) 0.29
25–< 30 Ref. 2.10 (0.91–4.85) 0.60 (0.22–1.65) 1.72 (0.74–4.03) Ref. 0.61 (0.29–1.29) 1.15 (0.51–2.60) 1.02 (0.96–1.08)
≥30 Ref. 1.05 (0.59–1.89) 1.12 (0.63–1.99) 1.04 (0.54–2.03) Ref. 0.99 (0.60–1.61) 1.37 (0.67–2.82) 1.02 (0.97–1.07)
Ethnicity
White Ref. 1.05 (0.65–1.69) 0.99 (0.62–1.60) 0.83 (0.49–1.41) 0.22 Ref. 0.90 (0.61–1.34) 0.99 (0.58–1.70) 0.20 0.99 (0.95–1.03) 0.43
Black Ref. 4.31 (1.12–16.54) 0.05 (0.01–0.49) 0.27 (0.02–3.09) Ref. 0.03 (0.00–0.24) 0.33 (0.03–4.08) 0.99 (0.87–1.13)
Hispanic Ref. 1.86 (0.25–13.85) 1.02 (0.25–4.28) 1.21 (0.12–12.65) Ref. 0.75 (0.20–2.86) 2.18 (0.23–20.33) 1.04 (0.91–1.18)
Other Ref. 0.34 (0.04–2.74) 0.19 (0.03–1.24) 3.21 (0.58–17.83) Ref. 1.90 (0.35–10.42) 8.65 (1.54–48.44) 1.18 (1.00–1.40)
Placebo Ref. 1.08 (0.56–2.08) 0.71 (0.36–1.41) 1.10 (0.53–2.26) 0.77 Ref. 0.78 (0.43–1.41) 1.23 (0.59–2.56) 0.82 1.01 (0.96–1.07) 0.99
Active Ref. 1.25 (0.64–2.42) 1.16 (0.60–2.24) 1.16 (0.57–2.33) Ref. 0.99 (0.57–1.73) 1.19 (0.55–2.56) 1.01 (0.96–1.06)
25(OH)D quartiles (nmol/L)
25(OH)D levels (nmol/L)
25(OH)D (nmol/L)
<34.734.7–47.847.9–64.2≥64.2P*<5050–75≥75P*5-unit increasesP*
BMI
< 25 Ref. 0.70 (0.22–2.20) 1.26 (0.46–3.49) 0.37 (0.10–1.33) 0.04 Ref. 1.21 (0.52–2.82) 0.70 (0.20–2.50) 0.48 0.96 (0.89–1.03) 0.29
25–< 30 Ref. 2.10 (0.91–4.85) 0.60 (0.22–1.65) 1.72 (0.74–4.03) Ref. 0.61 (0.29–1.29) 1.15 (0.51–2.60) 1.02 (0.96–1.08)
≥30 Ref. 1.05 (0.59–1.89) 1.12 (0.63–1.99) 1.04 (0.54–2.03) Ref. 0.99 (0.60–1.61) 1.37 (0.67–2.82) 1.02 (0.97–1.07)
Ethnicity
White Ref. 1.05 (0.65–1.69) 0.99 (0.62–1.60) 0.83 (0.49–1.41) 0.22 Ref. 0.90 (0.61–1.34) 0.99 (0.58–1.70) 0.20 0.99 (0.95–1.03) 0.43
Black Ref. 4.31 (1.12–16.54) 0.05 (0.01–0.49) 0.27 (0.02–3.09) Ref. 0.03 (0.00–0.24) 0.33 (0.03–4.08) 0.99 (0.87–1.13)
Hispanic Ref. 1.86 (0.25–13.85) 1.02 (0.25–4.28) 1.21 (0.12–12.65) Ref. 0.75 (0.20–2.86) 2.18 (0.23–20.33) 1.04 (0.91–1.18)
Other Ref. 0.34 (0.04–2.74) 0.19 (0.03–1.24) 3.21 (0.58–17.83) Ref. 1.90 (0.35–10.42) 8.65 (1.54–48.44) 1.18 (1.00–1.40)
Placebo Ref. 1.08 (0.56–2.08) 0.71 (0.36–1.41) 1.10 (0.53–2.26) 0.77 Ref. 0.78 (0.43–1.41) 1.23 (0.59–2.56) 0.82 1.01 (0.96–1.07) 0.99
Active Ref. 1.25 (0.64–2.42) 1.16 (0.60–2.24) 1.16 (0.57–2.33) Ref. 0.99 (0.57–1.73) 1.19 (0.55–2.56) 1.01 (0.96–1.06)
Data are odds ratio (95% CI) unless otherwise indicated.
Model adjusted for age, ethnicity, latitude of clinical center, month of blood draw, WHI study indicators, BMI, hypertension, fiber intake, magnesium intake, and physical activity.
*P value for interaction.
†Combined black, Hispanic, and other race/ethnicity because some cells contained zero.
After adjustment for BMI and other risk factors, we found no relationship between serum 25(OH)D concentration and diabetes incidence over 7 years in this multiethnic cohort of over 5,000 older women. Nor was a relationship between 25(OH)D status and diabetes risk found upon further evaluation by race/ethnicity or CVD status. In the analysis stratified by BMI, there was a suggestion that normal weight women with the highest level of serum 25(OH)D (>64 nmol/L) may have had a lower risk of diabetes, but this relationship was not present using the clinical cut point of ≥75 nmol/L. Although this finding may merit further investigation in other prospective cohorts, the lack of a trend for a protective relationship for the similarly-sized group of overweight women suggests it may have been a spurious finding. No evidence of a relationship between 25(OH)D and diabetes was found in obese women—the group at greatest risk of developing diabetes.
Our observational findings are consistent with those of the WHI CaD intervention trial, which found that supplementation with elemental calcium 1,000 mg plus vitamin D3 400 IU did not reduce the risk of developing diabetes over 7 years of follow-up (12). The hazard ratio for incident diabetes associated with CaD treatment was 1.01 (95% CI 0.94–1.10) with no association found in subgroup analyses, or efficacy analyses accounting for nonadherence. A higher dose of 2,000 IU/d of vitamin D supplementation is undergoing evaluation in the VITamin D and OmegA-3 TriaL (VITAL) of 20,000 women aged ≥65 years and men aged ≥60 years who will be followed for 5 years (16). The primary end points are cancer, coronary heart disease, and stroke, with diabetes as one of the secondary end points.
The few small studies that have evaluated the association between circulating 25(OH)D and diabetes incidence have been conflicting, as have the larger epidemiologic cohort studies of dietary vitamin D intake (1). The Women’s Health Study found an inverse relationship between vitamin D intake and incident diabetes risk when adjusted for age (17). In contrast, the even larger Nurses’ Health Study found no relationship between vitamin D intake from supplements and incident type 2 diabetes after further adjustment for confounders—similar to the findings of our study (18). The Nurses’ Health Study investigators did find one group with an inverse association between vitamin D and diabetes: those with a combined intake of >1,200 mg calcium and >800 IU vitamin D (1.3% of the cohort) had a significantly lower risk of diabetes when compared with intakes <600 mg calcium and <400 IU vitamin D.
It is possible that in men or other racial or ethnic groups, more severe vitamin D deficiency, or vitamin D deficiency earlier in life could contribute to diabetes risk (19,20). Two small Finnish nested case-control studies evaluating serum 25(OH)D concentrations, rather than dietary intake, found no association between mean 25(OH)D level and diabetes in women, but did find an inverse association in men (21). On the other hand, an analysis of the Framingham Offspring study found an association between a higher prediction score for vitamin D status and diabetes risk in both men and women (22).
Vitamin D may influence the development of diabetes via a number of potential mechanisms (1). Vitamin D receptors for the biologically active form of vitamin D are present on the pancreatic β-cells, and severe vitamin D deficiency can inhibit insulin secretion in some animal models. Administration of 1,25(OH)2D or its metabolites has improved insulin sensitivity and insulin secretion in some human in vitro and in vivo studies. Insulin sensitivity, β-cell function, and oral glucose tolerance have been shown to be inversely related to 25(OH)D concentrations in healthy normoglycemic individuals, elderly men, and east Asians. On the other hand, some studies have found that in the absence of vitamin D deficiency, vitamin D supplementation does not improve insulin sensitivity or glucose tolerance. Treatment with 100,000 IU vitamin D by intramuscular injection has been shown to improve C-peptide and insulin levels, although abnormal glucose tolerance by oral glucose tolerance test was unchanged (23).
The WHI was a well-characterized, large, ethnically diverse cohort of women. Although we used women from several case-control studies, it does not appear that case-control status influenced our findings. Limitations of the study include reliance on a self-report of treated diabetes. Although we evaluated numerous potential confounders, residual confounding cannot be excluded. Because lipid levels were not measured in >85% of case and control subjects, we were unable to adjust for other fat soluble vitamins, such as vitamin E. We also had low power to detect race/ethnicity specific associations.
In conclusion, serum concentrations of 25(OH)D were not associated with the incidence of type 2 diabetes in this cohort of postmenopausal women after adjusting for BMI and other risk factors for diabetes. A number of plausible mechanisms have been proposed for a protective effect of 25(OH)D on insulin and glucose metabolism, and it may be that evaluation of other at-risk population groups or cohorts with higher serum levels of 25(OH)D may reveal a protective association between vitamin D and diabetes. The results of ongoing trials of higher doses of vitamin D supplementation are needed before recommendations for vitamin D supplementation for the prevention of type 2 diabetes or other nonskeletal chronic diseases could be justified.
The WHI program is funded by the National Heart, Lung, and Blood Institute, National Institutes of Health through contracts N01WH22110, 24152, 32100-2, 32105-6, 32108-9, 32111-13, 32115, 32118-32119, 32122, 42107-26, 42129-32, and 44221. J.E.M. is Principal Investigator of VITAL, which is being conducted at Brigham and Women’s Hospital, Harvard Medical School. VITAL is funded by the National Institutes of Health (National Cancer Institute; National Heart, Lung, and Blood Institute; and other institutes and agencies are cosponsors).
J.G.R. received grants to her institution from Abbott, Bristol-Myers Squibb, Daiichi-Sankyo, GlaxoSmithKline, Hoffmann-La Roche, Merck, and Merck Schering-Plough. For VITAL, vitamin D pills are donated by Pharmavite (Northridge, CA) (vitamin D) and omega-3 pills are donated by Pronova BioPharma (Sandefjord, Norway). N.W. received support from Amgen, Eli Lilly and Company, Merck, and NPS Pharmaceuticals. N.W. has received honoraria from Amgen, Novartis, Procter & Gamble, sanofi-aventis, and Warner-Chilcott. Consulting fees have been received from Amgen, Baxter Healthcare, InteKrin, Johnson & Johnson, MannKind, MedPace, NPS, Pfizer, Procter & Gamble, sanofi-aventis, Takeda Pharmaceuticals, Vivus, and Warner Chilcott. No other potential conflicts of interest relevant to this article were reported.
J.G.R. conceived of, designed, oversaw statistical analysis, and wrote the manuscript in addition to collecting data for WHI. J.E.M., S.L., B.V.H., Y.S., J.D.C., J.M.S., M.A., K.C.J., L.P., and N.W. contributed to the statistical analysis and critical revisions to the manuscript in addition to collecting data for WHI. J.L. performed the statistical analysis and contributed critical revisions to the manuscript.
The short list for investigator acknowledgments can be found at http://www.whiscience.org/publications/write_paper.php.
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# Tag Info
6
I don't quite understand what you mean by "share that goes into capital", but the common interpretation is that $\alpha$ is the share of income/output spent on capital. You can show that the following way: Since the factors will be compensated according to their marginal products, under the assumption of competitive markets, we have (for capital): $$\... 4 For completeness, let me illustrate this in the continuous time framework. The Solow equation, in the simplest of cases, is \dot{k} = s f(k) - \delta k = \phi(k) Then we have \frac{\partial \phi}{\partial k} = s f'(k) - \delta = \frac{sf'(k)k - \delta k }{k}. In steady state (i.e., \dot{k} = \phi(k^{\ast}) = 0), we have \delta k = s f(k), hence ... 4 An answer along macro textbook lines is given by @1muflon1. A shorter answer is as follows. Consider an investor who borrows capital from household (who owns the capital, in growth models) to invest in the firm with production technology f(k). The rate of return r on capital for the household is the interest rate of borrowing for the investor. Each ... 4 This is not proven in Romer but it is a well known result. To derive it mathematically you need to take the following steps: First, the capital as in Romer depreciates so the evolution of capital will be given:$$k_t = k_{t-1} + i_t- \delta k_{t-1} \tag{1}$$where k_t is the present stock of capital, k_{t-1} previous stock of capital, i_t is ... 4 I will assume that you are talking about the Solow, R. M. (1956). A contribution to the theory of economic growth. The quarterly journal of economics, 70(1), 65-94. as to my best knowledge Solow did not published paper entitled the "Theory of Economic Growth" in 1956 and the first equation you use in your question is from the paper. Moreover, I also think ... 4 Since, I was urged to present an answer for didactic reasons to which I totally agree, I will provide the full set of corrections to avoid any ambiguity. % Model parameters alpha = 1/3; % s = 0.2; % Investment rate delta = 0.2; % Depreciation rate n = 0.02; % Growth of labor force g = 0.01; % technological progress eps = 1; k(1) = ... 4 These two assumptions are not necessarily contradictory. Just check whether the assumptions are satisfied by any candidate function. For example, take F(K,N) = K^{\alpha}N^{1-\alpha}, with \alpha \in (0,1). Constant returns to scale: For any scaling factor c \in (0, \infty):$$F(cK,cN) = (cK)^{\alpha}(cN)^{1-\alpha} = c^{\alpha}c^{1-\alpha} K^{\alpha}N^...
4
Yes, there are DSGE models that can be used for forecasting. These models typically have a particular kind of steady-state, which is, more precisely, called balanced growth path (BGP). On the BGP (in the absence of shocks), key indicators growth at the same constant rate. For example, GDP, household consumption, investment all grow at 2% a year. This is ...
3
Take a look at the dynamics of the capital: $k_{t+1}=sA_ty_t+(1-\delta-n)k_t$. A sudden positive shock to TFP in period $t$ increases the capital stock of the next period $k_{t+1}$. So, there is no contemporaneous effect on $k$, convergence to the new steady state will be gradual. The other variables have contemporaneous relationship with TFP. EDIT: A ...
3
Have you seen the GitHub Project Replicating Mankiw, Romer and Weil 1992? It seems to have both the data and a replication of the original results. For the curious, the paper is A Contribution to the Empirics of Economic Growth. Abstract: This paper examines whether the Solow growth model is consistent with the international variation in the standard ...
3
Yes the answer should be C. I have attached an image showing the variation with time of the variables $y$(per capita output), $c$(per capita consumption) and $i$(per capita investment). I am assuming at $t=t_0$ the savings rate is increased. The consumption per capita initially will fall because the savings rate has increased. Eventually it must go above ...
3
Let $$Q = F(K,L)$$ Assume a) $F(K,L)$ exhibits consant returns to scale. We need this to aggregate from the individual firms to the total. b) Price taking behavior and c) Profit maximizing behavior from the part of the firms. Then, throughout the dynamic process, $$w = \frac {\partial F(K,L)}{\partial L} \equiv F_L$$ i.e. the wage is equal to ...
3
Below please find a portion of a lecture slide a professor of mine used last year. Please note that $\gamma_{\tilde{y}}$ denotes per-capita output growth, $\gamma_{\tilde{k}}$ denotes per-capita capital growth and $\alpha(t)$ denotes the income share of capital. $R(t)$ - the Solow residual - can then be easily computed. The production function you have ...
3
It represents two things: (1) the elasticity of output with respect to capital, and (2) capital's share of output. To show (1), just take the natural log of the production equation, and then take the derivative of the logged equation with respect to time. $$\ln{Y} = \alpha \ln{K} + (1 - \alpha)(\ln{A} + \ln{L})$$ $$\frac{\dot{Y}}{Y} = \alpha \frac{\dot{K}}{... 3 Following Yorgos's alternative interpretation, (about \alpha which shows the percentage change of Y at 1\% change in K), one intuition may also be to log-linearize your production function. As follows \ln Y = \alpha \ln K + (1-\alpha) \ln L. Then recalling that log may be used to compute continuous variations, you could substitute, e.g K for \... 3 We want to prove that$$\frac{n+g+\delta}{s} > f'(k^*)$$Replace the left hand side with the equivalent from the expression sf(k^*)=(n+\delta+g)k^*, and you get:$$ \frac{f(k^*)}{k^*} > f'(k^*) $$Cobb-Douglas case Without loss of generality, assume$$f(k^*) = {k^*}^{\alpha}$$Then, the above inequality is:$$ {k^*}^{\alpha-1} > \alpha{k^*}...
3
For stability, we want $$\frac{\partial k_{t+1}}{\partial k_t}\Big|_{\bar k} <1 \implies \frac{(1-\delta) + \sigma A_0 f'(\bar k)}{1+n} <1$$ $$\implies f'(\bar k) < \frac {\delta+n}{\sigma A_0 } = \frac {f(\bar k)}{\bar k}$$ So we need the marginal product of capital to be smaller than the average product at the steady state. Equivalently,...
3
The story in the other answer is not fundamentally wrong but incomplete and bit inaccurate. Saving does actually affect capital stock through investment in Solow model (assuming based on the Solow tag that is the model you want intuition for), but there is more complexity to it. What is saving Investment is actually equal not just to private saving but both ...
2
To simplistically answer your question, use the following: $Y = K^\alpha (AL)^{1 - \alpha}$ In order to prove all three can be equal: We will assume that technological progress is labor augmenting (Harrod Neutral) or "labor saving", while holding the following true: Returns to capital are roughly constant over time. Capital share of income is roughly ...
2
In the model with technological progress the capital per effective worker remains constant, implies that capital per worker grows at the rate of exogenous rate of technological progress. See Barro and Martin book, Chapter 1.
2
The steady-state value $k^*$ must be a fixed point : $(1+g)(1+n)k^* = s(k^*)^{\alpha} +(1 - \delta) k^*$ Taking the difference between this equation and the dynamic one : $(1+g)(1+n)(k_{t+1} - k^*) = s((k_{t+1})^{\alpha} - (k^*)^{\alpha}) +(1 - \delta) (k_t - k^*)$ Now if you denote by $d_t = k_t - k^*$ the distance to steady-state, this gives : $(1+g)(... 2 if the population growth rate grows, why could the capital and income per capita decreases This is basically asking "why does higher population growth lower the steady state capital per worker"? Mathematics Let us assume a Cobb-Douglas production function with constant returns to scale. This is, $$Y=K^{\alpha}L^{1-\alpha}$$ It can be shown that in the ... 2 The following graph shows a point at which the total amount of depreciation is greater than savings. This decreases the amount of capital per effective worker, over time, shown by the arrow pointing downwards and to the left. 2 Yes, for any A and B $$\text{A}\cdot(g+n) + \text{B}\cdot(g+n) = \text{(A+B)}\cdot(g+n)$$ does indeed hold. 2 Business cycle fluctuations can be studied in many ways including also growth models. An example of using growth model to study business cycle would be this 1995 paper by Cho & Cooley. This being said growth models are not as common for examining business cycles as for example DSGE models so saying they are usually used for that purpose is correct. To my ... 2 Savings in macro are equal to investment$(S=I)$. Investment is how you get new capital stock. When people save more they also by definition invest more and when they invest more there is more capital. 2 I don't know how this can be answered without any recourse to at least basic mathematics, since the question itself refers to movement along some curves which is mathematical concept. Hence I will go over some math and then provide intuition at the end. Explanation Including Math The investment curve in Solow model is defined as$sf(k)$where$f(k) = Y\$ and ...
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# P and Q are each circular regions. What is the radius of P, if the are
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P and Q are each circular regions. What is the radius of P, if the are [#permalink]
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04 Sep 2018, 23:26
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P and Q are each circular regions. What is the radius of P, if the area of P minus the area of Q is 15π and P has a circumference that is 4 times that of Q?
A. 1
B. 2
C. 4
D. 5
E. 20
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Re: P and Q are each circular regions. What is the radius of P, if the are [#permalink]
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04 Sep 2018, 23:37
Bunuel wrote:
P and Q are each circular regions. What is the radius of P, if the area of P minus the area of Q is 15π and P has a circumference that is 4 times that of Q?
A. 1
B. 2
C. 4
D. 5
E. 20
Let p and q are the radius of circular regions P and Q respectively.
Given, area of P minus the area of Q is 15π
Or, $$p^2-q^2=15$$
Or, (p+q)(p-q)=15------------(1)
Given, P has a circumference that is 4 times that of Q
or, p=4q-------------(2)
from(1), $$(p+\frac{p}{4})(p-\frac{p}{4})=15$$
Or, $$\frac{5p}{4} * \frac{3p}{4}=15$$
Or, $$p^2=16$$
or, p=4
Ans. (C)
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Re: P and Q are each circular regions. What is the radius of P, if the are &nbs [#permalink] 04 Sep 2018, 23:37
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# Ross (Michigan) Class of 2016- Calling all applicants!
Author Message
Current Student
Joined: 07 Apr 2012
Posts: 33
Concentration: Real Estate, Entrepreneurship
GMAT 1: 690 Q44 V41
WE: General Management (Real Estate)
Followers: 1
Kudos [?]: 11 [0], given: 10
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 03 Sep 2013, 09:12
Anyone else having trouble with Ross' word limits? 150 words per question in Essay 2 is just too little!!
Current Student
Joined: 31 Oct 2011
Posts: 521
Schools: Johnson '16 (M)
GMAT 1: 690 Q45 V40
WE: Asset Management (Mutual Funds and Brokerage)
Followers: 36
Kudos [?]: 211 [0], given: 57
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 04 Sep 2013, 16:27
Stephen Ross donates again! $200M this time, super impressed. http://dealbook.nytimes.com/2013/09/04/ ... igan/?_r=0 What is he trying to do, get the school to be the University of Ross? _________________ My Applicant Blog: http://hamm0.wordpress.com/ Moderator Joined: 02 Jul 2012 Posts: 1228 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Followers: 89 Kudos [?]: 975 [1] , given: 116 Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 04 Sep 2013, 22:27 1 This post received KUDOS hamm0 wrote: Stephen Ross donates again!$200M this time, super impressed.
http://dealbook.nytimes.com/2013/09/04/ ... igan/?_r=0
What is he trying to do, get the school to be the University of Ross?
Haha.. Might have happened if it weren't a public university....
Harvard, Stanford, Yale and Columbia are the only ones remaining in the top 20 who are yet to name their b-schools.. Unfortunately I'm not applying to any of them.. And it doesn't make sense to donate to a school which wasn't my alma mater.. Guess I wont be having a top 20 b-school named after me after all..
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Joined: 23 Mar 2011
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Schools: Duke '16 (M)
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Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 05 Sep 2013, 04:04
MacFauz wrote:
Harvard, Stanford, Yale and Columbia are the only ones remaining in the top 20 who are yet to name their b-schools.. Unfortunately I'm not applying to any of them.. And it doesn't make sense to donate to a school which wasn't my alma mater.. Guess I wont be having a top 20 b-school named after me after all..
That, my friend, is a good deduction.
Manager
Status: Hell of a job to do now !
Joined: 31 Dec 2012
Posts: 69
Location: India
GMAT 1: 710 Q49 V37
GPA: 3.2
WE: Web Development (Computer Software)
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Kudos [?]: 16 [1] , given: 34
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 13 Sep 2013, 21:37
1
KUDOS
stephenleet wrote:
Hey applicants,
What do you all think about Ross's location? I know Ross is in Ann Arbor, but Detroit will inevitably have some hegemony. Wanted to know what you all thought about the role Detroit's situation would play regarding access to recruiting and even entertainment.
Hi,
I had similar doubts about Ross, but go through these links. The information helps to decide -
and
http://poetsandquants.com/2010/07/24/mi ... s-kellogg/
There are different pages on this link. Make sure you go through all the pages of information.
Manager
Status: Hell of a job to do now !
Joined: 31 Dec 2012
Posts: 69
Location: India
GMAT 1: 710 Q49 V37
GPA: 3.2
WE: Web Development (Computer Software)
Followers: 0
Kudos [?]: 16 [0], given: 34
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 13 Sep 2013, 21:41
saidba wrote:
Anyone else having trouble with Ross' word limits? 150 words per question in Essay 2 is just too little!!
I think that is the best part about Ross essays. It doesn't eat up a lot of time to complete your essays. And you get to say what you really want to say. You don't beat around the bush.
Current Student
Status: Duke!
Joined: 20 Jan 2012
Posts: 132
GMAT 1: 710 Q49 V38
Followers: 3
Kudos [?]: 40 [0], given: 10
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 16 Sep 2013, 09:33
nalin1 wrote:
saidba wrote:
Anyone else having trouble with Ross' word limits? 150 words per question in Essay 2 is just too little!!
I think that is the best part about Ross essays. It doesn't eat up a lot of time to complete your essays. And you get to say what you really want to say. You don't beat around the bush.
100% agree. Short and to the point, without making you write long, never-ending stories. And by the way, there are other schools that took this to the extreme (Duke offers about 40 words for you ST goals and another 40 words for LT. That's super short).
BTW, does anyone know whether the recommendations has this "ranking" element across multiple personality traits?
Intern
Joined: 26 Jun 2010
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Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 20 Sep 2013, 03:32
amitkpandey wrote:
arpanpatnaik wrote:
amitkpandey wrote:
Non-availability of co-signor free loans to international students worries me a lot..but still i'm in for R1.
I feel the same Amit, but Ross has a good tradition in terms of Financial Aid being offered. As per Business Week, 75% of the students requested aid, and they have received some. Of course, how much of what they expected is unclear, but something is truly better than nothing when it comes to scholarships
I am getting ready for my GMAT. Hope I'd have enough time for the apps. Really looking forward to applying and scoring a college this year!
Regards,
A
I'm not sure about the how much will the scholarships help..Considering the free fall of Rupee these days..paying in dollars after taking an Indian loan is not gonna be fine. Anyways funding obviously comes much later.. its the GMAT first and then the apps..I'm in for R1
How are you planning to finance if admitted? Have you looked for any options? I guess as an international applicant one should take into account no cosigner loans before applying rather than spending efforts in application. What if one is admitted but can't attend because of the financial situation. I have seen some of such cases in forums from last few years. Credila offers loans upto CoA but the interest rate is insane. It is almost 13-14 %.
I saw on last year's forum that there was a link for a loan option for internationals. But now that link is not valid anymore.
Current Student
Joined: 01 Mar 2010
Posts: 106
Location: United States (FL)
Concentration: Finance, Nonprofit
GMAT 1: 640 Q36 V41
GMAT 2: 630 Q42 V34
GMAT 3: 640 Q42 V35
GMAT 4: 620 Q42 V34
GMAT 5: 700 Q45 V40
GPA: 2.71
WE: Underwriter (Other)
Followers: 1
Kudos [?]: 13 [0], given: 20
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 20 Sep 2013, 18:15
Anyone notice that on the employment section of the application it only asks for two employers? At first I thought it may be an issue with my browser, but that doesn't seem to be the case. Is there a way to add additional employers or is Ross only interested in the most recent two ?
Intern
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Schools: Mays '16
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Kudos [?]: 0 [0], given: 3
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 22 Sep 2013, 14:37
Hi Everyone,
I have a quick doubt. Do I need to send my GMAT scores to Ross via GMAC? Or is submitting the scanned copy of my report enough?
Current Student
Status: Duke!
Joined: 20 Jan 2012
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Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 23 Sep 2013, 00:42
kirncha wrote:
Hi Everyone,
I have a quick doubt. Do I need to send my GMAT scores to Ross via GMAC? Or is submitting the scanned copy of my report enough?
From Ross' FAQ page:
Quote:
Q. How long will it take for my scores to be sent to Ross after I take the test?
A. We recommend that you allow for at least four weeks for Ross to receive your GMAT or GRE scores. Ross’ GRE reporting code is 1839. If you are uncertain as to whether we will receive your official scores by the deadline, you can upload your unofficial score via your online application, fax your unofficial score report to our office at 734-763-7804, or send it to 701 Tappan St., Room E2540, Ann Arbor, MI 48109-1234.
So it seems like while you SHOULD send your official reports to them, it is OK to upload an unofficial score till the the official ones arrive to them.
Current Student
Status: Duke!
Joined: 20 Jan 2012
Posts: 132
GMAT 1: 710 Q49 V38
Followers: 3
Kudos [?]: 40 [0], given: 10
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 24 Sep 2013, 08:27
doggdetroit wrote:
Anyone notice that on the employment section of the application it only asks for two employers? At first I thought it may be an issue with my browser, but that doesn't seem to be the case. Is there a way to add additional employers or is Ross only interested in the most recent two ?
I think you are correct. I didn't find a way to add a third employer as well.
Current Student
Status: Duke!
Joined: 20 Jan 2012
Posts: 132
GMAT 1: 710 Q49 V38
Followers: 3
Kudos [?]: 40 [0], given: 10
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 28 Sep 2013, 09:01
Application submitted!
Good luck to us all!
Current Student
Joined: 31 Oct 2011
Posts: 521
Schools: Johnson '16 (M)
GMAT 1: 690 Q45 V40
WE: Asset Management (Mutual Funds and Brokerage)
Followers: 36
Kudos [?]: 211 [0], given: 57
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 29 Sep 2013, 17:45
Joining the Round 1 submitted club! Seems like just yesterday I was trying to figure out how to write a 100 word introduction...
Good luck everyone!
_________________
My Applicant Blog: http://hamm0.wordpress.com/
Intern
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Concentration: Entrepreneurship, Strategy
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Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 29 Sep 2013, 18:34
targetting for 2nd round
Current Student
Joined: 13 Jan 2013
Posts: 148
Followers: 1
Kudos [?]: 34 [0], given: 43
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 29 Sep 2013, 19:06
Submitted my R1 app earlier this afternoon. Good luck everyone!
Intern
Joined: 30 May 2013
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GMAT 1: 700 Q50 V35
GPA: 3.28
Followers: 0
Kudos [?]: 0 [0], given: 9
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 30 Sep 2013, 09:54
Hi guys,
I'm also working on the R1 application.
But I'm having a problem.
I submitted wrong email ID of my supervisor and the system is not allowing me to update it
Does anyone have any solution?
Can we provide offline recommendation?
Regards
Manager
Status: Hell of a job to do now !
Joined: 31 Dec 2012
Posts: 69
Location: India
GMAT 1: 710 Q49 V37
GPA: 3.2
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Followers: 0
Kudos [?]: 16 [0], given: 34
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 30 Sep 2013, 10:10
goan wrote:
Hi guys,
I'm also working on the R1 application.
But I'm having a problem.
I submitted wrong email ID of my supervisor and the system is not allowing me to update it
Does anyone have any solution?
Can we provide offline recommendation?
Regards
You can try removing that recommender and try adding him again !
Intern
Joined: 30 May 2013
Posts: 11
GMAT 1: 700 Q50 V35
GPA: 3.28
Followers: 0
Kudos [?]: 0 [0], given: 9
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 30 Sep 2013, 10:12
not happening dude Ross people are not responding also
nalin1 wrote:
goan wrote:
Hi guys,
I'm also working on the R1 application.
But I'm having a problem.
I submitted wrong email ID of my supervisor and the system is not allowing me to update it
Does anyone have any solution?
Can we provide offline recommendation?
Regards
You can try removing that recommender and try adding him again !
Current Student
Joined: 14 Jan 2013
Posts: 23
GMAT 1: 720 Q49 V39
GRE 1: 1400 Q790 V610
WE: Engineering (Manufacturing)
Followers: 0
Kudos [?]: 7 [0], given: 5
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 30 Sep 2013, 14:39
goan wrote:
not happening dude Ross people are not responding also
nalin1 wrote:
goan wrote:
Hi guys,
I'm also working on the R1 application.
But I'm having a problem.
I submitted wrong email ID of my supervisor and the system is not allowing me to update it
Does anyone have any solution?
Can we provide offline recommendation?
Regards
You can try removing that recommender and try adding him again !
Check this out
Below content is quoted from Ross's website
Q: How do I submit supplemental information (e.g., transcripts or a letter of recommendation) if I am unable to upload it with my online application?
A: We do not accept updated documentation after application submission. This includes new resumes, essays or additional recommendations. The only exception is updated test scores.
If there was a problem submitting transcripts or a recommendation via the online application, you can scan these items and attach in an email to [email protected]. Please include your name and date of birth in the email, and let us know this is supplemental to your application.
Re: Ross (Michigan) Class of 2016- Calling all applicants! [#permalink] 30 Sep 2013, 14:39
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# Brilliant Sub Junior Calculus Contest (Season-1) Note-3
This is the third note to the season of Brilliant Sub Junior Calculus Contest (Season-1).
Click here for First Note (if you want to visit it)
Click here for Second Note (if you want to visit it)
Hi everyone! This is the first season of Brilliant Sub Junior Calculus Contest. This contest is for beginners or intermediate ones who want to sharpen their skills of problem-solving in overall calculus.
The aim of this contest is to improve the skill of in the computation in all sorts of problem (of basic level) in calculus like integrals (both definite and indefinite), differentiation, limits, ecetra by learning from each other and of course to have fun!
Eligibility:- People should fulfill either of the 2 following
• 17 years or below
• Level 4 or below in Calculus
Eligible people here may participate in this contest.
The rules are as follows:
• I will start by posting the first problem. If there is a user solves it, then they must post a new one.
• You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.
• Only make substantial comment that will contribute to the discussion.
• Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.
• If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.
• The scope of questions is only computation of basic level problems in calculus.
• It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.
• You are also NOT allowed to post a solution using a contour integration or residue method.
Answer shouldn't contain any Special Function.
Please post your solution and your proposed problem in a single new thread.
Format your post is as follows:
1 2 3 4 5 6 7 **SOLUTION OF PROBLEM xxx (number of problem) :** **[Post your solution here]** **PROBLEM xxx (number of problem) :** **[Post your problem here]**
The comments will be easiest to follow if you sort by "Newest":
Note by Akshay Yadav
5 years, 1 month ago
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## Comments
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Problem 40 :
Evaluate $\int_{0}^{1} \dfrac{5x^3+3x-1}{(x^3+3x+1)^3} dx$
- 5 years, 1 month ago
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$I=\displaystyle \int { \frac { -({ x }^{ 3 }+3x+1)+6{ x }^{ 3 }+6x }{ { ({ x }^{ 3 }+3x+1) }^{ 3 } } dx }$
This can be further written as,
$I=\displaystyle \int { \frac { -{ ({ x }^{ 3 }+3x+1) }^{ 2 }+{ 2x({ x }^{ 3 }+3x+1) }({ 3x }^{ 2 }+3) }{ { ({ x }^{ 3 }+3x+1) }^{ 4 } } dx }$
This is like the derivative form of $\displaystyle \frac { -x }{ { ({ x }^{ 3 } }+3x+1)^{ 2 }}$
The integral simplifies to,
$I=\displaystyle \frac { -x }{ { ({ x }^{ 3 } }+3x+1)^{ 2 } } +constant$
Now putting the limits $0$ to $1$ we get the value as $-1/25$
- 5 years, 1 month ago
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Just for enlightening, I'm posting my method too.
\begin{aligned}\int\frac{5x^3 + 3x - 1}{\left(x^3 + 3x + 1\right)^3}\, dx &=\int\frac{5x^3 + 3x-1}{\left(\sqrt{x}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)\right)^3}\,dx\\&=\int\frac{5x^3 + 3x-1}{x^{3/2}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\, dx\\&=\int\frac{5x^{3/2}+3x^{-1/2}-x^{-3/2}}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\, dx\\&=\int\frac{2\,d(x^{5/2}+3\sqrt{x}+x^{-1/2})}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\\& -\frac{1}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^2}+C\\& \end{aligned}
$\implies \int_{0}^{1} \frac{5x^3 + 3x - 1}{\left(x^3 + 3x + 1\right)^3} dx = \boxed{-\dfrac{1}{25}}$
- 5 years, 1 month ago
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Nice method.
- 5 years, 1 month ago
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Problem 41:
Evaluate
$\displaystyle \int _{0}^{\pi} \frac{1}{ (5+\cos (x))^{\frac{1}{3}}} dx$
- 5 years, 1 month ago
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Can you verify the problem?
- 5 years, 1 month ago
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@Aditya Kumar The problem is correct.
- 5 years, 1 month ago
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I think the solution should be posted
- 5 years, 1 month ago
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@Akshay Yadav @Nihar Mahajan @Samuel Jones what you guyz suggest about continuing the contest as @Saarthak Marathe hasn't posted the solution yet . I think someone should post a new problem and keep it going !
- 5 years, 1 month ago
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Since Sarthak Marathe has exceeded the time limit, from this moment onwards anyone can post the new question.
- 5 years, 1 month ago
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@Saarthak Marathe Why did you change the problem? I solved the original problem and came here to post the solution, only to find that it has been changed.
- 5 years, 1 month ago
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Hmm was it changed? I mean he gave a problem which was there in brilliant and he posted another one which is this one.
- 5 years, 1 month ago
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@Aditya Sharma @Saarthak Marathe Earlier problem was to evaluate $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\sin^2 x +k^2 \cos^2 x) \ dx$
- 5 years, 1 month ago
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Ya. This was pointed out to be a problem on brilliant with a name "the one with log and trigonometry" (maybe). So he changed the problem
- 5 years, 1 month ago
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Oh ok. Btw, do you know why alternate solutions have been deleted in JEE contest?
- 5 years, 1 month ago
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The contest which sandeep sir started? I thnik only particular solutions are kept for each problem.
- 5 years, 1 month ago
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Why is that so? I think if a solution is entirely different from the accepted one, it shouldn't be deleted.
- 5 years, 1 month ago
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I do agree with u personally in that case, maybe the moderator wasn't too much careful while removing the solution. If sandeep sir or Calvin sir is informed they will rectify it
- 5 years, 1 month ago
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I have informed them. Hope they'll resolve the issue.
- 5 years, 1 month ago
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Yeah and @Saarthak Marathe should post the solution as the allotted time is over.
- 5 years, 1 month ago
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@Saarthak Marathe Can you post your solution to this question? Thanks.
- 5 years ago
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To be honest,I don't know how to solve this one. When I first saw it,I thought it would be easy. But found out later that it is quite tough. It uses some kind of special function to solve. I managed to bring it to an algebraic form,
$I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { x } \sqrt { 1-x } { (2+x) }^{ 1/3 } } }$
I have posted this question on Brilliant hoping to get a less complicated solution to this question.
- 5 years ago
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@Saarthak Marathe That goes against the rules. Please post questions that you know the solution to from next time.
- 5 years ago
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Hey, Can you lift the bar on the level. I would also like to solve a few good problems
- 5 years, 1 month ago
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$\mathbf{Problem }$ $\mathbf{ 42}$
$\text{Integrate : } \large \int_{0}^{1} \frac{x^4(1-x)^4}{1+x^2}dx$
- 5 years, 1 month ago
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Using partial fractions,
$\displaystyle \int_{0}^{1} \dfrac{x^4(1-x)^4}{1+x^2} dx = \int_{0}^{1} \left( x^6 -4x^5 +5x^4 -4x^2 - \dfrac{4}{x^2+1} +4 \right) dx$
$= \boxed{\dfrac{22}{7} - \pi}$
Interesting Note : Since the integrand is positive, the integral must also be positive, giving us $\pi < \dfrac{22}{7}$ which is the widely used approximation for $\pi$ .
- 5 years, 1 month ago
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It is a JEE problem yeah?(This came in the year my sister wrote IIT :P)
- 5 years, 1 month ago
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Oh you identified :p no need to mention the year, yes it's a Jee Advanced problem
- 5 years, 1 month ago
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haha I had seen it some time before!That is why I was able to identify,i won't mention don't worry!
- 5 years, 1 month ago
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Haha it is worth sharing IIT JEE problems
- 5 years, 1 month ago
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Yeah.
- 5 years, 1 month ago
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I just completed reading the solution from a book!:P So,i guess i would wait sometime,in case somebody else is posting.
- 5 years, 1 month ago
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Is the solution very hard?
- 5 years, 1 month ago
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Nah,just requires a small trick!
- 5 years, 1 month ago
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Yeah basically the method is to decompose and expand smartly
- 5 years, 1 month ago
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Exactly,I will post the solution by night,if nobody else does,then we will solve whole night!If you are free !
- 5 years, 1 month ago
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Ya ofcourse !! If it's about maths and brilliant nothing comes more imp than it @mention532456:Adarsh Kumar . we will have a night maths party :-)
- 5 years, 1 month ago
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Problem 43:
Evaluate
$\huge \int_{0}^{\frac{\pi}{2}} \dfrac{\ln (\cos x)}{\sin x} dx$
- 5 years, 1 month ago
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Substitute $\cos x = t$, the integral becomes
$\displaystyle \text{I} = \int_{0}^{1} \dfrac{\log t}{1-t^2} \ dt$
$\displaystyle = \sum_{n=0}^{\infty} \int_{0}^{1} t^{2n} \log t \ dt \quad \left(\text{Using infinite G.P.} \right)$
$\displaystyle \int_{0}^{1} t^{2n} \log t \ dt = - \dfrac{1}{(2n+1)^2}$ which can be shown using integration by parts many times.
$\displaystyle \implies \text{I} = - \sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2}$
$\displaystyle = - \left(\sum_{n=1}^{\infty} \dfrac{1}{n^2} - \sum_{n=1}^{\infty} \dfrac{1}{(2n)^2} \right)$
$\displaystyle = - \dfrac{3}{4} \sum_{n=1}^{\infty} \dfrac{1}{n^2}$
$\displaystyle = \boxed{-\dfrac{\pi^2}{8}}$
- 5 years ago
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Nice observation samuels (+1)!!
- 5 years ago
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Hey,some of my friends told me that it is either quite hard or not possible to solve this without using special functions,does your solution also use any such functions? @Samuel Jones
- 5 years, 1 month ago
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Yeah I tried it , maybe i m wrong but are special functions required? I too have the same question as adarsh
- 5 years, 1 month ago
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I saw your solution, but in the step $\frac {ln(t)}{1-t^2}$, you have forgotten a negative sign when substituting. Therefore your answer is positive rather than negative. I will give a solution without special functions though so just wait for a while.
- 5 years ago
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Thanks! I will fix it, I wasn't sure where I have missed the sign.
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I have solved without special functions.
- 5 years ago
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Is it the answer ?
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No, the answer is $-\dfrac{\pi^2}{8}$
- 5 years ago
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@Aditya Sharma @Adarsh Kumar Do you want me to write the solution, or do I wait until you guys figure out your mistake ?
- 5 years ago
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Let the alotted time be over, then post your solution. Lets see if we can make it in that time @Samuel Jones
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Sure.
- 5 years ago
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$\text{Let the integral be J = } \int_{0}^{\frac{\pi}{2}} \frac{log(cosx)}{sinx}dx$
$\text{Put cosx = t }$
$-sinx dx=dt\implies dx = -\frac{dt}{\sqrt{1-t^2}}$
$\text{We have } J = -\int_{0}^{1}\frac{ln(t)}{1-t^2}$
$\text{Put ln(t) = u }\implies \frac{1}{t}dt=du\implies dt=tdu=e^u du$
$J = -\int_{\infty}^{0} \frac{xe^x}{1-e^{2x}}dx = -\frac{1}{2}\int_{\infty}^{0} \frac{x[(e^x+1)+(e^x-1)]}{1-e^{2x}}dx$
$2J = -\int_{\infty}^{0} [\frac{x}{1-e^x} - \frac{x}{e^x+1}]dx$
$2J = -\color{#D61F06}{\underbrace{\int_{0}^{\infty} [\frac{x}{e^x-1}}} + \color{#3D99F6}{\underbrace{\frac{x}{e^x+1}]dx}}$
$\text{In this step we are going to implement some of the special functions , their definitions are as below : }$
$\text{Polygarithm Function :}$
$\large Li_{s}(z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s-1}}{\frac{e^x}{z}-1}dx$
$\large -Li_{s}(-z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s-1}}{\frac{e^x}{z}+1}dx$
$\text{The special values where s=2 \& z=1,-1 are : } Li_{2}(1) = \frac{(\pi)^2}{6} , Li_{2}(-1) = -\frac{(\pi)^2}{12}$
$-2J = \color{#D61F06}{\underbrace{Li_{2}(1)}} + \color{#3D99F6}{\underbrace{-Li_{2}(-1)}}\implies -2J = \frac{(\pi)^2}{6}+\frac{(\pi)^2}{12}\implies \boxed{J=-\frac{\pi^2}{8}}$
$\text{There might be an method to do this by elementary functions . Please post if anyone gets this solved. }$
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You should post the solution now.
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@Aditya Sharma I've posted the solution, but you post the next problem, since you had solved it first.
- 5 years ago
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Problem 46:
$\displaystyle \int_{-\pi}^{\pi} \dfrac{2x(1+sinx)}{1+{cos}^{2}x}dx$
- 5 years ago
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$\text{Problem 44 :}$
$\large \int \frac{dx}{\sec^2x+\tan^2x}$
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Take $tanx=t$
Then the integral becomes,
$I=\displaystyle \int { \frac { dt }{ (1+{ t }^{ 2 })(1+{ 2t }^{ 2 }) } }$
Now separate the terms,
$I=\displaystyle 2\int { \frac { dt }{ (1+2{ t }^{ 2 }) } } -\int { \frac { dt }{ ({ t }^{ 2 }+1) } }$
which on integration and resubstituting the value $t= \tan x$ comes out as,
$\sqrt { 2 } { \tan }^{ -1 }(\sqrt { 2 } \tan x)-x+C$
- 5 years ago
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Absolutely , Post the next problem
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Problem 45:
$\displaystyle \int { \frac { x }{ { (7x-10-{ x }^{ 2 }) }^{ 3/2 } } dx }$
- 5 years ago
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Solution:
$I=\displaystyle \int { \frac { x }{ ({ (2-x)(x-5) })^{ 3/2 } } dx }$
$I= \displaystyle \int { \frac { x }{ \frac { { (x-5) }^{ 3/2 } }{ { (2-x) }^{ 3/2 } } { (2-x) }^{ 3 } } dx }$
Take $t=\frac{x-5}{2-x}$ which gives $x=\frac{2t+5}{1+t}$
$I=\displaystyle \frac { 1 }{ 9 } \int { \frac { \frac { (2t+5) }{ (1+t) } }{ { t }^{ 3/2 }\frac { 1 }{ (1+t) } } } dt$
This simplifies to ,
$I=\displaystyle \frac { 2 }{ 9 } \int { { t }^{ -1/2 }dt+\frac { 5 }{ 9 } \int { { t }^{ -3/2 }dt } }$
This on integration gives,
$I=\displaystyle \frac { 4 }{ 9 } { t }^{ 1/2 }-\frac { 10 }{ 9 } { t }^{ -1/2 }$
Resubstituting $t=\frac{x-5}{2-x}$ we get our answer.
- 5 years ago
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@Akshay Yadav Should I post the solution to this problem ,as no one has posted one yet ?
- 5 years ago
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Go ahead, post the solution. The time limit has been already exceeded for anyone to solve it.
- 5 years ago
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@Akshay Yadav Should I post the next question?
- 5 years ago
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# Math Help - How to solve this trig equation!
1. ## How to solve this trig equation!
1. The problem statement, all variables and given/known data
Solve for θ.
2sinθcosθ + 1 - 2sin^2θ = 0
2. Relevant equations
No equations, just identities.
3. The attempt at a solution
I tried factoring, but I don't know how to continue with 2 diff ID's in one equation.
I don't know how to simplify it so that everything is in terms of one similar Identity.
:\
2. $\displaystyle 2sinxcosx=sin2x$
Make the sub:
$\displaystyle sin2x+1-2sin2x=1-sin2x\rightarrow .....$
3. Originally Posted by dwsmith
$\displaystyle 2sinxcosx=sin2x$
Make the sub:
$\displaystyle sin2x+1-2sin2x=1-sin2x\rightarrow .....$
I think you might have misread that I meant -2sin(square)x instead of -2sin2x, lol.
If I subbed that
I would get:
-2sin^2(x) + sin2(x) + 1 = 0
I wanted to factor, but I got stuck at the sin2(x) part. I only know how to factor if the 2 was in front...?
4. Originally Posted by TN17
1. The problem statement, all variables and given/known data
Solve for θ.
2sinθcosθ + 1 - 2sin^2θ = 0
2. Relevant equations
No equations, just identities.
3. The attempt at a solution
I tried factoring, but I don't know how to continue with 2 diff ID's in one equation.
I don't know how to simplify it so that everything is in terms of one similar Identity.
:\
First substitute $\displaystyle 2 \sin(\theta) \cos(\theta) = \sin(2 \theta)$ and $\displaystyle \sin^2 (\theta) = \frac{1 - \cos(2 \theta)}{2}$ and simplify.
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# Projectile with linear drag [closed]
A particle with mass $$m$$ travels with speed $$v_0$$ and angle $$\theta$$ compared to the horizontal ground. Air resistance given as $$F_R = -m \alpha v$$, where $$v$$ is the particle's speed, is working against the projectile.
The particle's motion is described as
$$\frac{{dv_x} }{dt} + \alpha v_x = 0$$
$$\frac{{dv_y} }{dt} + \alpha v_y = -g$$
where
$$v_x = \frac{{dx} }{dt}$$
$$v_y = \frac{{dy} }{dt}$$
Show that the particles position is given as
$$x(t) = \frac{{1} }{\alpha}(v_0 \cos \theta)(1-e^{-\alpha t})$$ $$y(t) = \frac{{1} }{\alpha}(v_0 \sin \theta + \frac{g}{\alpha})(1-e^{-\alpha t}) - \frac{g}{\alpha}t$$
I understand how to set up the equations and do the integrals (at least a part of it) and get the following result:
$$x(t) = \frac{{1} }{\alpha}(1-e^{-\alpha t})$$
What I don't understand is where the $$v_0 cos \theta$$ is coming from. I'm thinking it has something to do with decomposing forces in $$x$$ and $$y$$ direction, but I can't seem to work it out and where it goes in the equation.
(I haven't started with $$y(t)$$ yet).
## closed as off-topic by John Rennie, Kyle Kanos, Aaron Stevens, Thomas Fritsch, stafusaAug 30 at 7:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Kyle Kanos, Aaron Stevens, Thomas Fritsch, stafusa
If this question can be reworded to fit the rules in the help center, please edit the question.
• What I don't understand is where the $v_0\cos\theta$ is coming from. From the initial conditions. – Gert Aug 29 at 14:27
• $v_{x,0}=v_0\cos\theta$, $v_{y,0}=v_0\sin\theta$ – Gert Aug 29 at 15:50
I think, you forgot to add the integration constant. Integration of $${dv_x \over dt}+\alpha v_x=0$$ gives $$ln {v_x}=\alpha t+c_0$$. After some algebraic manipulation we can write $$v_x=Ae^{\alpha t}$$, where $$A=e^{c_0}$$, and applying the initial condition gives $$A=v_{x0}$$ and from component decomposition we know $$v_{x0}=v_0 cos \theta$$. So now we're left with the following differential $$eq^n$$-$${dx \over dt}=v_0 cos \theta e^{ \alpha t}$$ whose $$sol^n$$ is $$x(t)={1 \over \alpha} v_0 cos \theta e^{\alpha t}+c_1$$. Applying the initial condition $$x=0$$ at $$t=0$$ , we get $$c_1= -{1 \over \alpha}v_0 cos \theta$$. Substituting it in the $$x(t)$$ expression you'll get your desired expression for $$x(t)$$.
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# 8.20. Geospatial Functions
Presto Geospatial functions that begin with the ST_ prefix support the SQL/MM specification and are compliant with the Open Geospatial Consortium’s (OGC) OpenGIS Specifications. As such, many Presto Geospatial functions require, or more accurately, assume that geometries that are operated on are both simple and valid. For example, it does not make sense to calculate the area of a polygon that has a hole defined outside of the polygon, or to construct a polygon from a non-simple boundary line.
Presto Geospatial functions support the Well-Known Text (WKT) and Well-Known Binary (WKB) form of spatial objects:
• POINT (0 0)
• LINESTRING (0 0, 1 1, 1 2)
• POLYGON ((0 0, 4 0, 4 4, 0 4, 0 0), (1 1, 2 1, 2 2, 1 2, 1 1))
• MULTIPOINT (0 0, 1 2)
• MULTILINESTRING ((0 0, 1 1, 1 2), (2 3, 3 2, 5 4))
• MULTIPOLYGON (((0 0, 4 0, 4 4, 0 4, 0 0), (1 1, 2 1, 2 2, 1 2, 1 1)), ((-1 -1, -1 -2, -2 -2, -2 -1, -1 -1)))
• GEOMETRYCOLLECTION (POINT(2 3), LINESTRING (2 3, 3 4))
Use ST_GeometryFromText and ST_GeomFromBinary functions to create geometry objects from WKT or WKB. In WKT/WKB, the coordinate order is (x, y). For spherical/geospatial uses, this implies (longitude, latitude) instead of (latitude, longitude).
The basis for the Geometry type is a plane. The shortest path between two points on the plane is a straight line. That means calculations on geometries (areas, distances, lengths, intersections, etc) can be calculated using cartesian mathematics and straight line vectors.
The SphericalGeography type provides native support for spatial features represented on “geographic” coordinates (sometimes called “geodetic” coordinates, or “lat/lon”, or “lon/lat”). Geographic coordinates are spherical coordinates expressed in angular units (degrees).
The basis for the SphericalGeography type is a sphere. The shortest path between two points on the sphere is a great circle arc. That means that calculations on geographies (areas, distances, lengths, intersections, etc) must be calculated on the sphere, using more complicated mathematics. More accurate measurements that take the actual spheroidal shape of the world into account are not supported.
For SphericalGeography objects, values returned by the measurement functions ST_Distance and ST_Length are in the unit of meters; values returned by ST_Area are in square meters.
Use to_spherical_geography() function to convert a geometry object to geography object. For example, ST_Distance(ST_Point(-71.0882, 42.3607), ST_Point(-74.1197, 40.6976)) returns 3.4577 in the unit of the passed-in values on the euclidean plane, while ST_Distance(to_spherical_geography(ST_Point(-71.0882, 42.3607)), to_spherical_geography(ST_Point(-74.1197, 40.6976))) returns 312822.179 in meters.
## Constructors
ST_AsBinary(Geometry) → varbinary
Returns the WKB representation of the geometry.
ST_AsText(Geometry) → varchar
Returns the WKT representation of the geometry. For empty geometries, ST_AsText(ST_LineFromText('LINESTRING EMPTY')) will produce 'MULTILINESTRING EMPTY' and ST_AsText(ST_Polygon('POLYGON EMPTY')) will produce 'MULTIPOLYGON EMPTY'.
ST_GeometryFromText(varchar) → Geometry
Returns a geometry type object from WKT representation.
ST_GeomFromBinary(varbinary) → Geometry
Returns a geometry type object from WKB representation.
ST_LineFromText(varchar) → LineString
Returns a geometry type linestring object from WKT representation.
ST_LineString(array(Point)) → LineString
Returns a LineString formed from an array of points. If there are fewer than two non-empty points in the input array, an empty LineString will be returned. Throws an exception if any element in the array is null or empty or same as the previous one. The returned geometry may not be simple, e.g. may self-intersect or may contain duplicate vertexes depending on the input.
ST_MultiPoint(array(Point)) → MultiPoint
Returns a MultiPoint geometry object formed from the specified points. Return null if input array is empty. Throws an exception if any element in the array is null or empty. The returned geometry may not be simple and may contain duplicate points if input array has duplicates.
ST_Point(x, y) → Point
Returns a geometry type point object with the given coordinate values.
ST_Polygon(varchar) → Polygon
Returns a geometry type polygon object from WKT representation.
to_spherical_geography(Geometry) → SphericalGeography
Converts a Geometry object to a SphericalGeography object on the sphere of the Earth’s radius. This function is only applicable to POINT, MULTIPOINT, LINESTRING, MULTILINESTRING, POLYGON, MULTIPOLYGON geometries defined in 2D space, or GEOMETRYCOLLECTION of such geometries. For each point of the input geometry, it verifies that point.x is within [-180.0, 180.0] and point.y is within [-90.0, 90.0], and uses them as (longitude, latitude) degrees to construct the shape of the SphericalGeography result.
to_geometry(SphericalGeography) → Geometry
Converts a SphericalGeography object to a Geometry object.
## Relationship Tests
ST_Contains(Geometry, Geometry) → boolean
Returns true if and only if no points of the second geometry lie in the exterior of the first geometry, and at least one point of the interior of the first geometry lies in the interior of the second geometry.
ST_Crosses(Geometry, Geometry) → boolean
Returns true if the supplied geometries have some, but not all, interior points in common.
ST_Disjoint(Geometry, Geometry) → boolean
Returns true if the give geometries do not spatially intersect – if they do not share any space together.
ST_Equals(Geometry, Geometry) → boolean
Returns true if the given geometries represent the same geometry.
ST_Intersects(Geometry, Geometry) → boolean
Returns true if the given geometries spatially intersect in two dimensions (share any portion of space) and false if they do not (they are disjoint).
ST_Overlaps(Geometry, Geometry) → boolean
Returns true if the given geometries share space, are of the same dimension, but are not completely contained by each other.
ST_Relate(Geometry, Geometry) → boolean
Returns true if first geometry is spatially related to second geometry.
ST_Touches(Geometry, Geometry) → boolean
Returns true if the given geometries have at least one point in common, but their interiors do not intersect.
ST_Within(Geometry, Geometry) → boolean
Returns true if first geometry is completely inside second geometry.
## Operations
geometry_union(array(Geometry)) → Geometry
Returns a geometry that represents the point set union of the input geometries. Performance of this function, in conjunction with array_agg() to first aggregate the input geometries, may be better than geometry_union_agg(), at the expense of higher memory utilization.
ST_Boundary(Geometry) → Geometry
Returns the closure of the combinatorial boundary of this geometry.
ST_Buffer(Geometry, distance) → Geometry
Returns the geometry that represents all points whose distance from the specified geometry is less than or equal to the specified distance. If the points of the geometry are extremely close together (delta < 1e-8), this might return an empty geometry.
ST_Difference(Geometry, Geometry) → Geometry
Returns the geometry value that represents the point set difference of the given geometries.
ST_Envelope(Geometry) → Geometry
Returns the bounding rectangular polygon of a geometry.
ST_EnvelopeAsPts(Geometry) -> array(Geometry)
Returns an array of two points: the lower left and upper right corners of the bounding rectangular polygon of a geometry. Returns null if input geometry is empty.
expand_envelope(Geometry, double) → Geometry
Returns the bounding rectangular polygon of a geometry, expanded by a distance. Empty geometries will return an empty polygon. Negative or NaN distances will return an error. Positive infinity distances may lead to undefined results.
ST_ExteriorRing(Geometry) → Geometry
Returns a line string representing the exterior ring of the input polygon.
ST_Intersection(Geometry, Geometry) → Geometry
Returns the geometry value that represents the point set intersection of two geometries.
ST_SymDifference(Geometry, Geometry) → Geometry
Returns the geometry value that represents the point set symmetric difference of two geometries.
ST_Union(Geometry, Geometry) → Geometry
Returns a geometry that represents the point set union of the input geometries.
## Accessors
ST_Area(Geometry) → double
Returns the 2D Euclidean area of a geometry.
For Point and LineString types, returns 0.0. For GeometryCollection types, returns the sum of the areas of the individual geometries.
ST_Area(SphericalGeography) → double
Returns the area of a polygon or multi-polygon in square meters using a spherical model for Earth.
ST_Centroid(Geometry) → Point
Returns the point value that is the mathematical centroid of a geometry.
ST_Centroid(SphericalGeography) → Point
Returns the point value that is the mathematical centroid of a spherical geometry.
It supports Points and MultiPoints as input and returns the three-dimensional centroid projected onto the surface of the (spherical) Earth e.g. MULTIPOINT (0 -45, 0 45, 30 0, -30 0) returns Point(0, 0) Note: In the case that the three-dimensional centroid is at (0, 0, 0), the spherical centroid is undefined and an arbitrary point will be returned e.g. MULTIPOINT (0 0, -180 0) returns Point(-90, 45)
ST_ConvexHull(Geometry) → Geometry
Returns the minimum convex geometry that encloses all input geometries.
ST_CoordDim(Geometry) → bigint
Return the coordinate dimension of the geometry.
ST_Dimension(Geometry) → bigint
Returns the inherent dimension of this geometry object, which must be less than or equal to the coordinate dimension.
ST_Distance(Geometry, Geometry) → double
Returns the 2-dimensional cartesian minimum distance (based on spatial ref) between two geometries in projected units.
ST_Distance(SphericalGeography, SphericalGeography) → double
Returns the great-circle distance in meters between two SphericalGeography points.
ST_GeometryN(Geometry, index) → Geometry
Returns the geometry element at a given index (indices start at 1). If the geometry is a collection of geometries (e.g., GEOMETRYCOLLECTION or MULTI*), returns the geometry at a given index. If the given index is less than 1 or greater than the total number of elements in the collection, returns NULL. Use :func:ST_NumGeometries to find out the total number of elements. Singular geometries (e.g., POINT, LINESTRING, POLYGON), are treated as collections of one element. Empty geometries are treated as empty collections.
ST_InteriorRingN(Geometry, index) → Geometry
Returns the interior ring element at the specified index (indices start at 1). If the given index is less than 1 or greater than the total number of interior rings in the input geometry, returns NULL. Throws an error if the input geometry is not a polygon. Use :func:ST_NumInteriorRing to find out the total number of elements.
ST_GeometryType(Geometry) → varchar
Returns the type of the geometry.
ST_IsClosed(Geometry) → boolean
Returns true if the linestring’s start and end points are coincident.
ST_IsEmpty(Geometry) → boolean
Returns true if this Geometry is an empty geometrycollection, polygon, point etc.
ST_IsSimple(Geometry) → boolean
Returns true if this Geometry has no anomalous geometric points, such as self intersection or self tangency. Use geometry_invalid_reason() to determine why the geometry is not simple.
ST_IsRing(Geometry) → boolean
Returns true if and only if the line is closed and simple.
ST_IsValid(Geometry) → boolean
Returns true if and only if the input geometry is well formed. Use geometry_invalid_reason() to determine why the geometry is not well formed.
ST_Length(Geometry) → double
Returns the length of a linestring or multi-linestring using Euclidean measurement on a two dimensional plane (based on spatial ref) in projected units.
ST_Length(SphericalGeography) → double
Returns the length of a linestring or multi-linestring on a spherical model of the Earth. This is equivalent to the sum of great-circle distances between adjacent points on the linestring.
ST_PointN(LineString, index) → Point
Returns the vertex of a linestring at a given index (indices start at 1). If the given index is less than 1 or greater than the total number of elements in the collection, returns NULL. Use :func:ST_NumPoints to find out the total number of elements.
ST_Points(Geometry) -> array(Point)
Returns an array of points in a linestring.
ST_XMax(Geometry) → double
Returns the X maximum of the geometry’s bounding box.
ST_YMax(Geometry) → double
Returns the Y maximum of the geometry’s bounding box.
ST_XMin(Geometry) → double
Returns the X minimum of the geometry’s bounding box.
ST_YMin(Geometry) → double
Returns the Y minimum of the geometry’s bounding box.
ST_StartPoint(Geometry) → point
Returns the first point of a LineString geometry as a Point. This is a shortcut for ST_PointN(geometry, 1).
ST_EndPoint(Geometry) → point
Returns the last point of a LineString geometry as a Point. This is a shortcut for ST_PointN(geometry, ST_NumPoints(geometry)).
ST_X(Point) → double
Return the X coordinate of the point.
ST_Y(Point) → double
Return the Y coordinate of the point.
ST_InteriorRings(Geometry) -> array(Geometry)
Returns an array of all interior rings found in the input geometry, or an empty array if the polygon has no interior rings. Returns null if the input geometry is empty. Throws an error if the input geometry is not a polygon.
ST_NumGeometries(Geometry) → bigint
Returns the number of geometries in the collection. If the geometry is a collection of geometries (e.g., GEOMETRYCOLLECTION or MULTI*), returns the number of geometries, for single geometries returns 1, for empty geometries returns 0. Note that empty geometries inside of a GEOMETRYCOLLECTION will count as a geometry; eg ST_NumGeometries(ST_GeometryFromText('GEOMETRYCOLLECTION(MULTIPOINT EMPTY)')) will evaluate to 1.
ST_Geometries(Geometry) -> array(Geometry)
Returns an array of geometries in the specified collection. Returns a one-element array if the input geometry is not a multi-geometry. Returns null if input geometry is empty.
For example, a MultiLineString will create an array of LineStrings. A GeometryCollection will produce an un-flattened array of its constituents: GEOMETRYCOLLECTION(MULTIPOINT(0 0, 1 1), GEOMETRYCOLLECTION(MULTILINESTRING((2 2, 3 3)))) would produce array[MULTIPOINT(0 0, 1 1), GEOMETRYCOLLECTION(MULTILINESTRING((2 2, 3 3)))].
flatten_geometry_collections(Geometry) -> array(Geometry)
Recursively flattens any GeometryCollections in Geometry, returning an array of constituent non-GeometryCollection geometries. The order of the array is arbitrary and should not be relied upon. Examples:
POINT (0 0) -> [POINT (0 0)], MULTIPOINT (0 0, 1 1) -> [MULTIPOINT (0 0, 1 1)], GEOMETRYCOLLECTION (POINT (0 0), GEOMETRYCOLLECTION (POINT (1 1))) -> [POINT (0 0), POINT (1 1)], GEOMETRYCOLLECTION EMPTY -> [].
ST_NumPoints(Geometry) → bigint
Returns the number of points in a geometry. This is an extension to the SQL/MM ST_NumPoints function which only applies to point and linestring.
ST_NumInteriorRing(Geometry) → bigint
Returns the cardinality of the collection of interior rings of a polygon.
simplify_geometry(Geometry, double) → Geometry
Returns a “simplified” version of the input geometry using the Douglas-Peucker algorithm. Will avoid creating derived geometries (polygons in particular) that are invalid.
line_locate_point(LineString, Point) → double
Returns a float between 0 and 1 representing the location of the closest point on the LineString to the given Point, as a fraction of total 2d line length.
Returns null if a LineString or a Point is empty or null.
line_interpolate_point(LineString, double) → Geometry
Returns the Point on the LineString at a fractional distance given by the double argument. Throws an exception if the distance is not between 0 and 1.
Returns an empty Point if the LineString is empty. Returns null if either the LineString or double is null.
geometry_invalid_reason(Geometry) → varchar
Returns the reason for why the input geometry is not valid or not simple. If the geometry is neither valid no simple, it will only give the reason for invalidity. Returns null if the input is valid and simple.
great_circle_distance(latitude1, longitude1, latitude2, longitude2) → double
Returns the great-circle distance between two points on Earth’s surface in kilometers.
## Aggregations
convex_hull_agg(Geometry) → Geometry
Returns the minimum convex geometry that encloses all input geometries.
geometry_union_agg(Geometry) → Geometry
Returns a geometry that represents the point set union of all input geometries.
## Bing Tiles
These functions convert between geometries and Bing tiles. For Bing tiles, x and y refer to tile_x and tile_y. Bing Tiles can be cast to and from BigInts, using an internal representation that encodes the zoom, x, and y efficiently:
cast(cast(tile AS BIGINT) AS BINGTILE)
While every tile can be cast to a bigint, casting from a bigint that does not represent a valid tile will raise an exception.
bing_tile(x, y, zoom_level) → BingTile
Creates a Bing tile object from XY coordinates and a zoom level. Zoom levels from 1 to 23 are supported.
bing_tile(quadKey) → BingTile
Creates a Bing tile object from a quadkey.
bing_tile_at(latitude, longitude, zoom_level) → BingTile
Returns a Bing tile at a given zoom level containing a point at a given latitude and longitude. Latitude must be within [-85.05112878, 85.05112878] range. Longitude must be within [-180, 180] range. Zoom levels from 1 to 23 are supported.
bing_tiles_around(latitude, longitude, zoom_level) -> array(BingTile)
Returns a collection of Bing tiles that surround the point specified by the latitude and longitude arguments at a given zoom level.
bing_tiles_around(latitude, longitude, zoom_level, radius_in_km) -> array(BingTile)
Returns a minimum set of Bing tiles at specified zoom level that cover a circle of specified radius in km around a specified (latitude, longitude) point.
bing_tile_coordinates(tile) → row<x, y>
Returns the XY coordinates of a given Bing tile.
bing_tile_polygon(tile) → Geometry
Returns the polygon representation of a given Bing tile.
bing_tile_quadkey(tile) → varchar
Returns the quadkey of a given Bing tile.
bing_tile_zoom_level(tile) → tinyint
Returns the zoom level of a given Bing tile.
geometry_to_bing_tiles(geometry, zoom_level) -> array(BingTile)
Returns the minimum set of Bing tiles that fully covers a given geometry at a given zoom level. Zoom levels from 1 to 23 are supported.
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# Classical Nusselt Evaporation
Schematic of Nusselt evaporation.
The simplest evaporation heat transfer coefficient calculation uses the Nusselt treatment of falling film evaporation, as shown in the figure on the right. A flat plate with constant temperature Tw is placed vertically and a liquid film flows downward due to gravity. Evaporation takes place on the surface of the liquid film because the temperature of the vertical plate is above the saturation temperature of the vapor corresponding to the partial pressure of the vapor in the mixture. Since the heat must be transferred from the heat source (vertical wall) to the liquid-vapor interface, where evaporation takes place, this is an example of heterogeneous evaporation. The following assumptions are made:
1. The vapor is quiescent.
2. Inertia in the laminar liquid film is negligible.
3. Heat transfer through the liquid film is by conduction only, i.e., convective terms in the energy equation are negligible.
4. Constant wall temperature.
5. The process occurs at steady state.
6. A nonslip condition exists at the wall.
Proceeding from the assumptions of negligible inertia and constant fluid properties, the momentum balance is
${{\mu }_{\ell }}\frac{{{d}^{2}}{{u}_{\ell }}}{d{{y}^{2}}}=-\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g$ (1)
Integrating eq. (1) twice, with respect to y and using the boundary conditions of non-slip at the wall (u=0 at y=0) and zero shear at the liquid-vapor interface ($\partial u/\partial y=0$ at y = δ), the following velocity profile is obtained:
${{u}_{\ell }}=\frac{\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g}{{{\mu }_{\ell }}}\left( \delta y-\frac{{{y}^{2}}}{2} \right)$ (2)
The mass flow rate per unit width is defined by integrating the velocity profile from the wall to the interface.
$\Gamma =\int_{0}^{\delta }{{{\rho }_{\ell }}udy}$ (3)
The liquid velocity profile, eq. (2), is substituted into eq. (3) and the result is integrated from the wall to the interface. The result is then rearranged to obtain liquid film thickness:
$\delta ={{\left[ \frac{3{{\mu }_{\ell }}\Gamma }{{{\rho }_{\ell }}\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g} \right]}^{{1}/{3}\;}}$ (4)
Since heat transfer through the film is by conduction only, the heat transfer coefficient is ${{h}_{x}}={{k}_{\ell }}/\delta$. Considering eq. (4), the local heat transfer coefficient becomes
${{h}_{x}}={{k}_{\ell }}{{\left[ \frac{{{\rho }_{\ell }}\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g}{3{{\mu }_{\ell }}\Gamma } \right]}^{{1}/{3}\;}}$ (5)
Note that the film thickness δ drops out of the equation, and the heat transfer coefficient h is now a function of the mass flow rate per unit width, Γ.
The local energy balance across the liquid film is
${{h}_{\ell v}}\frac{d\Gamma }{dx}=-{{h}_{x}}\left( {{T}_{w}}-{{T}_{v}} \right)$ (6)
Combining eqs. (5) and (6), one obtains,
${{\Gamma }^{{1}/{3}\;}}d\Gamma =-\frac{{{k}_{\ell }}}{{{h}_{\ell v}}}{{\left[ \frac{{{\rho }_{\ell }}\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g}{3{{\mu }_{\ell }}} \right]}^{{1}/{3}\;}}\left( {{T}_{w}}-{{T}_{v}} \right)dx$ (7)
Integrating from the inlet to some arbitrary length L,
$\int_{{{\Gamma }_{o}}}^{{{\Gamma }_{L}}}{{{\Gamma }^{{1}/{3}\;}}}d\Gamma =-\int_{0}^{L}{\frac{{{k}_{\ell }}}{{{h}_{\ell v}}}}{{\left[ \frac{{{\rho }_{\ell }}\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g}{3{{\mu }_{\ell }}} \right]}^{{1}/{3}\;}}\left( {{T}_{w}}-{{T}_{v}} \right)dx$ (8)
A relation for Γ at outlet is obtained in terms of the inlet mass flow rate, Γ0, for the case of constant wall and vapor temperatures:
$\int_{{{\Gamma }_{o}}}^{{{\Gamma }_{L}}}{{{\Gamma }^{{1}/{3}\;}}}d\Gamma =-\int_{0}^{L}{\frac{{{k}_{\ell }}}{{{h}_{\ell v}}}}{{\left[ \frac{{{\rho }_{\ell }}\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g}{3{{\mu }_{\ell }}} \right]}^{{1}/{3}\;}}\left( {{T}_{w}}-{{T}_{v}} \right)dx$ (9)
Now the mass flow rate per unit width is related to the Reynolds number by
$\operatorname{Re}=\frac{4\Gamma }{{{\mu }_{\ell }}}$ (10)
The local heat transfer coefficient from eq. (5) can be expressed in terms of Reynolds number as follows:
$\frac{{{h}_{x}}}{{{k}_{\ell }}}{{\left[ \frac{\mu _{\ell }^{2}}{{{\rho }_{\ell }}\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g} \right]}^{{1}/{3}\;}}={{\left( \frac{4}{3} \right)}^{{1}/{3}\;}}\frac{1}{{{\operatorname{Re}}^{{1}/{3}\;}}}=1.10{{\operatorname{Re}}^{-1/3}}$ (11)
Equation (11) for local heat transfer coefficient is valid for both laminar condensation and evaporation at both constant wall temperature and constant heat flux at wall.
To calculate the average heat transfer coefficient, $\bar{h}$, the energy flux over the entire length 0 < x < L is balanced with the mass flux from the interface:
$\overset{\_}{\mathop{h}}\,=\frac{{{h}_{\ell v}}\left( {{\Gamma }_{0}}-{{\Gamma }_{L}} \right)}{L\left( {{T}_{w}}-{{T}_{v}} \right)}=\frac{{{\mu }_{\ell }}{{h}_{\ell v}}\left( {{\operatorname{Re}}_{0}}-{{\operatorname{Re}}_{L}} \right)}{4L\left( {{T}_{w}}-{{T}_{v}} \right)}$ (12)
Equation (12) is valid for laminar, wavy laminar, and turbulent with constant wall temperature.
Equation (6), which reflects the local energy balance across the liquid film, can be written in terms of Reynolds number as
$\frac{d\operatorname{Re}}{{{h}_{x}}}=-\frac{4\left( {{T}_{w}}-{{T}_{v}} \right)}{{{\mu }_{\ell }}{{h}_{\ell v}}}dx$ (13)
Integrating eq. (13) in the interval of (0, L), one obtains
$\int_{{{\operatorname{Re}}_{0}}}^{{{\operatorname{Re}}_{L}}}{\frac{d\operatorname{Re}}{{{h}_{x}}}}=-\frac{4\left( {{T}_{w}}-{{T}_{v}} \right)L}{{{\mu }_{\ell }}{{h}_{\ell v}}}$ (14)
Combining eqs. (12) and (14), the average heat transfer coefficient is expressed as
$\overset{\_}{\mathop{h}}\,=-\frac{\left( {{\operatorname{Re}}_{0}}-{{\operatorname{Re}}_{L}} \right)}{\int_{{{\operatorname{Re}}_{0}}}^{{{\operatorname{Re}}_{L}}}{\frac{1}{{{h}_{x}}}d\operatorname{Re}}}$ (15)
Equation (15) for $\bar{h}$ is valid for laminar, wavy laminar, and turbulent flow.
Substituting eq. (11) into eq. (15), an expression for the average heat transfer coefficient for laminar flow in terms of Reynolds number is obtained:
$\frac{\overset{\_}{\mathop{h}}\,}{{{k}_{\ell }}}{{\left[ \frac{\mu _{\ell }^{2}}{{{\rho }_{\ell }}\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g} \right]}^{{1}/{3}\;}}={{\left( \frac{4}{3} \right)}^{{4}/{3}\;}}\frac{\left( {{\operatorname{Re}}_{o}}-{{\operatorname{Re}}_{L}} \right)}{\left( \operatorname{Re}_{o}^{{4}/{3}\;}-\operatorname{Re}_{L}^{{4}/{3}\;} \right)}$ (16)
In order to use eq. (16) to obtain the average heat transfer coefficient, it is necessary to know Reynolds number at x=L, ${{\operatorname{Re}}_{L}}$. It can be obtained by rewriting the relation for Γ, eq. (9), in terms of Reynolds number, as follows:
$\operatorname{Re}_{L}^{{4}/{3}\;}=\operatorname{Re}_{o}^{{4}/{3}\;}-4{{\left( \frac{4}{3} \right)}^{{4}/{3}\;}}\frac{{{k}_{\ell }}\left( {{T}_{w}}-{{T}_{v}} \right)L}{\mu _{\ell }^{{4}/{3}\;}{{h}_{\ell v}}}{{\left[ \frac{{{\rho }_{\ell }}\left( {{\rho }_{\ell }}-{{\rho }_{v}} \right)g}{3{{\mu }_{\ell }}} \right]}^{{1}/{3}\;}}$ (17)
The above analysis is valid for falling film evaporation on a vertical wall maintained at a constant temperature.
## References
Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA
Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.
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• ### Gaia Data Release 2: Summary of the variability processing & analysis results(1804.09373)
May 8, 2018 astro-ph.SR, astro-ph.IM
The Gaia Data Release 2 (DR2): we summarise the processing and results of the identification of variable source candidates of RR Lyrae stars, Cepheids, long period variables (LPVs), rotation modulation (BY Dra-type) stars, delta Scuti & SX Phoenicis stars, and short-timescale variables. In this release we aim to provide useful but not necessarily complete samples of candidates. The processed Gaia data consist of the G, BP, and RP photometry during the first 22 months of operations as well as positions and parallaxes. Various methods from classical statistics, data mining and time series analysis were applied and tailored to the specific properties of Gaia data, as well as various visualisation tools. The DR2 variability release contains: 228'904 RR Lyrae stars, 11'438 Cepheids, 151'761 LPVs, 147'535 stars with rotation modulation, 8'882 delta Scuti & SX Phoenicis stars, and 3'018 short-timescale variables. These results are distributed over a classification and various Specific Object Studies (SOS) tables in the Gaia archive, along with the three-band time series and associated statistics for the underlying 550'737 unique sources. We estimate that about half of them are newly identified variables. The variability type completeness varies strongly as function of sky position due to the non-uniform sky coverage and intermediate calibration level of this data. The probabilistic and automated nature of this work implies certain completeness and contamination rates which are quantified so that users can anticipate their effects. This means that even well-known variable sources can be missed or misidentified in the published data. The DR2 variability release only represents a small subset of the processed data. Future releases will include more variable sources and data products; however, DR2 shows the (already) very high quality of the data and great promise for variability studies.
• ### OGLE-2015-BLG-1459L: The Challenges of Exo-Moon Microlensing(1711.09651)
May 2, 2018 astro-ph.EP
We show that dense OGLE and KMTNet $I$-band survey data require four bodies (sources plus lenses) to explain the microlensing light curve of OGLE-2015-BLG-1459. However, these can equally well consist of three lenses and one source (3L1S), two lenses and two sources (2L2S) or one lens and three sources (1L3S). In the 3L1S and 2L2S interpretations, the host is a brown dwarf and the dominant companion is a Neptune-class planet, with the third body (in the 3L1S case) being a Mars-class object that could have been a moon of the planet. In the 1L3S solution, the light curve anomalies are explained by a tight (five stellar radii) low-luminosity binary source that is offset from the principal source of the event by $\sim 0.17\,\au$. These degeneracies are resolved in favor of the 1L3S solution by color effects derived from comparison to MOA data, which are taken in a slightly different ($R/I$) passband. To enable current and future ($WFIRST$) surveys to routinely characterize exomoons and distinguish among such exotic systems requires an observing strategy that includes both a cadence faster than 9 min$^{-1}$ and observations in a second band on a similar timescale.
• ### Spitzer Opens New Path to Break Classic Degeneracy for Jupiter-Mass Microlensing Planet OGLE-2017-BLG-1140Lb(1803.04437)
April 30, 2018 astro-ph.SR, astro-ph.EP
We analyze the combined Spitzer and ground-based data for OGLE-2017-BLG-1140 and show that the event was generated by a Jupiter-class $(m_p\simeq 1.6\,M_{\rm jup})$ planet orbiting a mid-late M dwarf $(M\simeq 0.2\,M_\odot)$ that lies $D_{LS}\simeq 1.0\,\mathrm{kpc}$ in the foreground of the microlensed, Galactic-bar, source star. The planet-host projected separation is $a_\perp \simeq 1.0\,\mathrm{au}$, i.e., well-beyond the snow line. By measuring the source proper motion ${\mathbf{\mu}}_s$ from ongoing, long-term OGLE imaging, and combining this with the lens-source relative proper motion ${\mathbf{\mu}}_\mathrm{rel}$ derived from the microlensing solution, we show that the lens proper motion ${\mathbf{\mu}}_l={\mathbf{\mu}}_\mathrm{rel} + {\mathbf{\mu}}_s$ is consistent with the lens lying in the Galactic disk, although a bulge lens is not ruled out. We show that while the Spitzer and ground-based data are comparably well fitted by planetary (i.e., binary-lens, 2L1S) models and by binary-source (1L2S) models, the combination of Spitzer and ground-based data decisively favor the planetary model. This is a new channel to resolve the 2L1S/1L2S degeneracy, which can be difficult to break in some cases.
• We highlight the power of the Gaia DR2 in studying many fine structures of the Hertzsprung-Russell diagram (HRD). Gaia allows us to present many different HRDs, depending in particular on stellar population selections. We do not aim here for completeness in terms of types of stars or stellar evolutionary aspects. Instead, we have chosen several illustrative examples. We describe some of the selections that can be made in Gaia DR2 to highlight the main structures of the Gaia HRDs. We select both field and cluster (open and globular) stars, compare the observations with previous classifications and with stellar evolutionary tracks, and we present variations of the Gaia HRD with age, metallicity, and kinematics. Late stages of stellar evolution such as hot subdwarfs, post-AGB stars, planetary nebulae, and white dwarfs are also analysed, as well as low-mass brown dwarf objects. The Gaia HRDs are unprecedented in both precision and coverage of the various Milky Way stellar populations and stellar evolutionary phases. Many fine structures of the HRDs are presented. The clear split of the white dwarf sequence into hydrogen and helium white dwarfs is presented for the first time in an HRD. The relation between kinematics and the HRD is nicely illustrated. Two different populations in a classical kinematic selection of the halo are unambiguously identified in the HRD. Membership and mean parameters for a selected list of open clusters are provided. They allow drawing very detailed cluster sequences, highlighting fine structures, and providing extremely precise empirical isochrones that will lead to more insight in stellar physics. Gaia DR2 demonstrates the potential of combining precise astrometry and photometry for large samples for studies in stellar evolution and stellar population and opens an entire new area for HRD-based studies.
• ### OGLE-2017-BLG-0373Lb: A Jovian Mass-Ratio Planet Exposes A New Accidental Microlensing Degeneracy(1802.10067)
April 5, 2018 astro-ph.EP
We report the discovery of microlensing planet OGLE-2017-BLG-0373Lb. We show that while the planet-host system has an unambiguous microlens topology, there are two geometries within this topology that fit the data equally well, which leads to a factor 2.5 difference in planet-host mass ratio, i.e., $q=1.5\times 10^{-3}$ vs. $q=0.6\times 10^{-3}$. We show that this is an "accidental degeneracy" in the sense that it is due to a gap in the data. We dub it "the caustic-chirality degeneracy". We trace the mathematical origins of this degeneracy, which should enable similar degenerate solutions to be easily located in the future. A Bayesian estimate, based on a Galactic model, yields a host mass $M=0.25^{+0.30}_{-0.15} M_\odot$ at a distance $D_L=5.9^{+1.3}_{-1.95}$ kpc. The lens-source relative proper motion is relatively fast, $\mu=9$ mas/yr, which implies that the host mass and distance can be determined by high-resolution imaging after about 10 years. The same observations could in principle resolve the discrete degeneracy in $q$, but this will be more challenging.
• ### MOA-2015-BLG-337: A Planetary System with a Low-mass Brown Dwarf/Planetary Boundary Host, or a Brown Dwarf Binary(1804.00830)
April 3, 2018 astro-ph.EP
We report the discovery and the analysis of the short timescale binary-lens microlensing event, MOA-2015-BLG-337. The lens system could be a planetary system with a very low mass host, around the brown dwarf/planetary mass boundary, or a brown dwarf binary. We found two competing models that explain the observed light curves with companion/host mass ratios of q~0.01 and ~0.17, respectively. From the measurement of finite source effects in the best-fit planetary model, we find a relatively small angular Einstein radius of theta_E ~ 0.03 mas which favors a low mass lens. We conduct a Bayesian analysis to obtain the probability distribution of the lens properties. The results for the planetary models strongly depend on the minimum mass, M_min, in the assumed mass function. In summary, there are two solutions of the lens system: (1) a brown dwarf/planetary mass boundary object orbited by a super-Neptune (the planetary model with M_min=0.001 M_sun) and (2) a brown dwarf binary (the binary model). If the planetary models is correct, this system can be one of a new class of planetary system, having a low host mass and also a planetary mass ratio (q <0.03) between the host and its companion. The discovery of the event is important for the study of planetary formation in very low mass objects. In addition, it is important to consider all viable solutions in these kinds of ambiguous events in order for the future comprehensive statistical analyses of planetary/binary microlensing events.
• ### OGLE-2017-BLG-0482Lb: A Microlensing Super-Earth Orbiting a Low-mass Host Star(1803.10830)
March 28, 2018 astro-ph.EP
We report the discovery of a planetary system in which a super-earth orbits a late M-dwarf host. The planetary system was found from the analysis of the microlensing event OGLE-2017-BLG-0482, wherein the planet signal appears as a short-term anomaly to the smooth lensing light curve produced by the host. Despite its weak signal and short duration, the planetary signal was firmly detected from the dense and continuous coverage by three microlensing surveys. We find a planet/host mass ratio of $q\sim 1.4\times 10^{-4}$. We measure the microlens parallax $\pi_{\rm E}$ from the long-term deviation in the observed lensing light curve, but the angular Einstein radius $\theta_{\rm E}$ cannot be measured because the source trajectory did not cross the planet-induced caustic. Using the measured event timescale and the microlens parallax, we find that the masses of the planet and the host are $M_{\rm p}=9.0_{-4.5}^{+9.0}\ M_\oplus$ and $M_{\rm host}=0.20_{-0.10}^{+0.20}\ M_\odot$, respectively, and the projected separation between them is $a_\perp=1.8_{-0.7}^{+0.6}$ au. The estimated distance to the lens is $D_{\rm L}=5.8_{-2.1}^{+1.8}$ kpc. The discovery of the planetary system demonstrates that microlensing provides an important method to detect low-mass planets orbiting low-mass stars.
• ### OGLE-2017-BLG-1522: A giant planet around a brown dwarf located in the Galactic bulge(1803.05095)
March 14, 2018 astro-ph.SR, astro-ph.EP
We report the discovery of a giant planet in the OGLE-2017-BLG-1522 microlensing event. The planetary perturbations were clearly identified by high-cadence survey experiments despite the relatively short event timescale of $t_{\rm E} \sim 7.5$ days. The Einstein radius is unusually small, $\theta_{\rm E} = 0.065\,$mas, implying that the lens system either has very low mass or lies much closer to the microlensed source than the Sun, or both. A Bayesian analysis yields component masses $(M_{\rm host}, M_{\rm planet})=(46_{-25}^{+79}, 0.75_{-0.40}^{+1.26})~M_{\rm J}$ and source-lens distance $D_{\rm LS} = 0.99_{-0.54}^{+0.91}~{\rm kpc}$, implying that this is a brown-dwarf/Jupiter system that probably lies in the Galactic bulge, a location that is also consistent with the relatively low lens-source relative proper motion $\mu = 3.2 \pm 0.5~{\rm mas}~{\rm yr^{-1}}$. The projected companion-host separation is $0.59_{-0.11}^{+0.12}~{\rm AU}$, indicating that the planet is placed beyond the snow line of the host, i.e., $a_{sl} \sim 0.12~{\rm AU}$. Planet formation scenarios combined with the small companion-host mass ratio $q \sim 0.016$ and separation suggest that the companion could be the first discovery of a giant planet that formed in a protoplanetary disk around a brown dwarf host.
• ### The ASAS-SN Catalog of Variable Stars I: The Serendipitous Survey(1803.01001)
March 2, 2018 astro-ph.SR
The All-Sky Automated Survey for Supernovae (ASAS-SN) is the first optical survey to routinely monitor the whole sky with a cadence of $\sim2-3$ days down to V$\lesssim17$ mag. ASAS-SN has monitored the whole sky since 2014, collecting $\sim100-500$ epochs of observations per field. The V-band light curves for candidate variables identified during the search for supernovae are classified using a random forest classifier and visually verified. We present a catalog of 66,533 bright, new variable stars discovered during our search for supernovae, including 27,753 periodic variables and 38,780 irregular variables. V-band light curves for the ASAS-SN variables are available through the ASAS-SN variable stars database (https://asas-sn.osu.edu/variables). The database will begin to include the light curves of known variable stars in the near future along with the results for a systematic, all-sky variability survey.
• ### OGLE-2017-BLG-0329L: A Microlensing Binary Characterized with Dramatically Enhanced Precision Using Data from Space-based Observations(1802.10196)
Feb. 27, 2018 astro-ph.SR
Mass measurements of gravitational microlenses require one to determine the microlens parallax $\pie$, but precise $\pie$ measurement, in many cases, is hampered due to the subtlety of the microlens-parallax signal combined with the difficulty of distinguishing the signal from those induced by other higher-order effects. In this work, we present the analysis of the binary-lens event OGLE-2017-BLG-0329, for which $\pie$ is measured with a dramatically improved precision using additional data from space-based $Spitzer$ observations. We find that while the parallax model based on the ground-based data cannot be distinguished from a zero-$\pie$ model at 2$\sigma$ level, the addition of the $Spitzer$ data enables us to identify 2 classes of solutions, each composed of a pair of solutions according to the well-known ecliptic degeneracy. It is found that the space-based data reduce the measurement uncertainties of the north and east components of the microlens-parallax vector $\pivec_{\rm E}$ by factors $\sim 18$ and $\sim 4$, respectively. With the measured microlens parallax combined with the angular Einstein radius measured from the resolved caustic crossings, we find that the lens is composed of a binary with components masses of either $(M_1,M_2)\sim (1.1,0.8)\ M_\odot$ or $\sim (0.4,0.3)\ M_\odot$ according to the two solution classes. The first solution is significantly favored but the second cannot be securely ruled out based on the microlensing data alone. However, the degeneracy can be resolved from adaptive optics observations taken $\sim 10$ years after the event.
• ### OGLE-2017-BLG-1130: The First Binary Gravitational Microlens Detected From Spitzer Only(1802.09023)
Feb. 27, 2018 astro-ph.SR, astro-ph.EP
We analyze the binary gravitational microlensing event OGLE-2017-BLG-1130 (mass ratio q~0.45), the first published case in which the binary anomaly was only detected by the Spitzer Space Telescope. This event provides strong evidence that some binary signals can be missed by observations from the ground alone but detected by Spitzer. We therefore invert the normal procedure, first finding the lens parameters by fitting the space-based data and then measuring the microlensing parallax using ground-based observations. We also show that the normal four-fold space-based degeneracy in the single-lens case can become a weak eight-fold degeneracy in binary-lens events. Although this degeneracy is resolved in event OGLE-2017-BLG-1130, it might persist in other events.
• ### OGLE-2016-BLG-1266: A Probable Brown-Dwarf/Planet Binary at the Deuterium Fusion Limit(1802.09563)
Feb. 26, 2018 astro-ph.SR, astro-ph.EP
We report the discovery, via the microlensing method, of a new very-low-mass binary system. By combining measurements from Earth and from the Spitzer telescope in Earth-trailing orbit, we are able to measure the microlensing parallax of the event, and find that the lens likely consists of an $(12.0 \pm 0.6) M_{\rm J}$ + $(15.7 \pm 1.5) M_{\rm J}$ super-Jupiter / brown-dwarf pair. The binary is located at a distance of $(3.08 \pm 0.18)$ kpc in the Galactic Plane, and the components have a projected separation of $(0.43 \pm 0.03)$ AU. Two alternative solutions with much lower likelihoods are also discussed, an 8- and 6-$M_{\rm J}$ model and a 90- and 70-$M_{\rm J}$ model. Although disfavored at the 3-$\sigma$ and 5-$\sigma$ levels, these alternatives cannot be rejected entirely. We show how the more-massive of these models could be tested with future direct imaging.
• ### A Neptune-mass Free-floating Planet Candidate Discovered by Microlensing Surveys(1712.01042)
Feb. 13, 2018 astro-ph.EP
Current microlensing surveys are sensitive to free-floating planets down to Earth-mass objects. All published microlensing events attributed to unbound planets were identified based on their short timescale (below two days), but lacked an angular Einstein radius measurement (and hence lacked a significant constraint on the lens mass). Here, we present the discovery of a Neptune-mass free-floating planet candidate in the ultrashort ($t_{\rm E}=0.320\pm0.003$ days) microlensing event OGLE-2016-BLG-1540. The event exhibited strong finite-source effects, which allowed us to measure its angular Einstein radius of $\theta_{\rm E}=9.2\pm0.5\,\mu$as. There remains, however, a degeneracy between the lens mass and distance. The combination of the source proper motion and source-lens relative proper motion measurements favors a Neptune-mass lens located in the Galactic disk. However, we cannot rule out that the lens is a Saturn-mass object belonging to the bulge population. We exclude stellar companions up to 15 au.
• ### OGLE-2017-BLG-1434Lb: Eighth q < 1 * 10^-4 Mass-Ratio Microlens Planet Confirms Turnover in Planet Mass-Ratio Function(1802.02582)
Feb. 7, 2018 astro-ph.EP
We report the discovery of a cold Super-Earth planet (m_p=4.4 +/- 0.5 M_Earth) orbiting a low-mass (M=0.23 +/- 0.03 M_Sun) M dwarf at projected separation a_perp = 1.18 +/- 0.10 AU, i.e., about 1.9 times the snow line. The system is quite nearby for a microlensing planet, D_Lens = 0.86 +/- 0.09 kpc. Indeed, it was the large lens-source relative parallax pi_rel=1.0 mas (combined with the low mass M) that gave rise to the large, and thus well-measured, "microlens parallax" that enabled these precise measurements. OGLE-2017-BLG-1434Lb is the eighth microlensing planet with planet-host mass ratio q < 1 * 10^-4. We apply a new planet-detection sensitivity method, which is a variant of "V/V_max", to seven of these eight planets to derive the mass-ratio function in this regime. We find dN/d(ln q) ~ q^p, with p = 1.05 (+0.78,-0.68), which confirms the "turnover" in the mass function found by Suzuki et al. relative to the power law of opposite sign n = -0.93 +/- 0.13 at higher mass ratios q >~ 2 * 10^-4. We combine our result with that of Suzuki et al. to obtain p = 0.73 (+0.42,-0.34).
• ### A Gravitationally Lensed Quasar Discovered in OGLE(1801.08481)
Jan. 25, 2018 astro-ph.GA
We report the discovery of a new gravitationally lensed quasar (double) from the Optical Gravitational Lensing Experiment (OGLE) identified inside the $\sim$670 sq. deg area encompassing the Magellanic Clouds. The source was selected as one of $\sim$60 "red W1-W2" mid-IR objects from WISE and having a significant amount of variability in OGLE for both two (or more) nearby sources. This is the first detection of a gravitational lens, where the discovery is made "the other way around", meaning we first measured the time delay between the two lensed quasar images of $-132<t_{\rm AB}<-76$ days (90% CL), with the median $t_{\rm AB}\approx-102$ days (in the observer frame), and where the fainter image B lags image A. The system consists of the two quasar images separated by 1.5" on the sky, with $I\approx20.0$ mag and $I\approx19.6$ mag, respectively, and a lensing galaxy that becomes detectable as $I \approx 21.5$ mag source, 1.0" from image A, after subtracting the two lensed images. Both quasar images show clear AGN broad emission lines at $z=2.16$ in the NTT spectra. The SED fitting with the fixed source redshift provided the estimate of the lensing galaxy redshift of $z \approx 0.9 \pm 0.2$ (90% CL), while its type is more likely to be elliptical (the SED-inferred and lens-model stellar mass is more likely present in ellipticals) than spiral (preferred redshift by the lens model).
• ### The OGLE Collection of Variable Stars. Classical, Type II, and Anomalous Cepheids toward the Galactic Center(1712.01307)
Jan. 23, 2018 astro-ph.SR
We present a collection of classical, type II, and anomalous Cepheids detected in the OGLE fields toward the Galactic center. The sample contains 87 classical Cepheids pulsating in one, two or three radial modes, 924 type II Cepheids divided into BL Her, W Vir, peculiar W Vir, and RV Tau stars, and 20 anomalous Cepheids - first such objects found in the Galactic bulge. Additionally, we upgrade the OGLE Collection of RR Lyr stars in the Galactic bulge by adding 828 newly identified variables. For all Cepheids and RR Lyr stars, we publish time-series VI photometry obtained during the OGLE-IV project, from 2010 through 2017. We discuss basic properties of our classical pulsators: their spatial distribution, light curve morphology, period-luminosity relations, and position in the Petersen diagram. We present the most interesting individual objects in our collection: a type II Cepheid with additional eclipsing modulation, W Vir stars with the period doubling effect and the RVb phenomenon, a mode-switching RR Lyr star, and a triple-mode anomalous RRd star.
• ### Spitzer Microlensing Parallax for OGLE-2016-BLG-1067: a sub-Jupiter Orbiting an M-dwarf in the Disk(1801.05806)
Jan. 17, 2018 astro-ph.EP
We report the discovery of a sub-Jupiter mass planet orbiting beyond the snow line of an M-dwarf most likely in the Galactic disk as part of the joint Spitzer and ground-based monitoring of microlensing planetary anomalies toward the Galactic bulge. The microlensing parameters are strongly constrained by the light curve modeling and in particular by the Spitzer-based measurement of the microlens parallax, $\pi_\mathrm{E}$. However, in contrast to many planetary microlensing events, there are no caustic crossings, so the angular Einstein radius, $\theta_\mathrm{E}$ has only an upper limit based on the light curve modeling alone. Additionally, the analysis leads us to identify 8 degenerate configurations: the four-fold microlensing parallax degeneracy being doubled by a degeneracy in the caustic structure present at the level of the ground-based solutions. To pinpoint the physical parameters, and at the same time to break the parallax degeneracy, we make use of a series of arguments: the $\chi^2$ hierarchy, the Rich argument, and a prior Galactic model. The preferred configuration is for a host at $D_L=3.73_{-0.67}^{+0.66}~\mathrm{kpc}$ with mass $M_\mathrm{L}=0.30_{-0.12}^{+0.15}~\mathrm{M_\odot}$, orbited by a Saturn-like planet with $M_\mathrm{planet}=0.43_{-0.17}^{+0.21}~\mathrm{M_\mathrm{Jup}}$ at projected separation $a_\perp = 1.70_{-0.39}^{+0.38}~\mathrm{au}$, about 2.1 times beyond the system snow line. Therefore, it adds to the growing population of sub-Jupiter planets orbiting near or beyond the snow line of M-dwarfs discovered by microlensing. Based on the rules of the real-time protocol for the selection of events to be followed up with Spitzer, this planet will not enter the sample for measuring the Galactic distribution of planets.
• ### OGLE-2016-BLG-1045: A Test of Cheap Space-Based Microlens Parallaxes(1801.00169)
Dec. 30, 2017 astro-ph.SR, astro-ph.IM
Microlensing is a powerful and unique technique to probe isolated objects in the Galaxy. To study the characteristics of these interesting objects based on the microlensing method, measurement of the microlens parallax is required to determine the properties of the lens. Of the various methods to measure microlens parallax, the most robust way is to make simultaneous ground- and space-based observations, i.e., by measuring the space-based microlens parallax. However, space-based campaigns usually require "expensive" resources. Gould & Yee (2012) proposed an idea called the "cheap space-based microlens parallax" that can measure the lens-parallax using only $2$ or $3$ space-based observations of high-magnification events. This cost-effective observation strategy to measure microlens parallaxes could be used by space-borne telescopes to build a complete sample for studying isolated objects. This would enable a direct measurement of the mass function including both extremely low-mass objects and high-mass stellar remnants. However, to adopt this idea requires a test to check how it would work in actual situations. Thus, we present the first practical test of the idea using the high-magnification microlensing event OGLE-2016-BLG-1045, for which a subset of Spitzer observations fortuitously duplicate the prescription of Gould & Yee (2012). From the test, we confirm that the measurement of the lens-parallax adopting this idea has sufficient accuracy to determine the physical properties of the isolated lens.
• ### OGLE-2016-BLG-1190Lb: First Spitzer Bulge Planet Lies Near the Planet/Brown-Dwarf Boundary(1710.09974)
Nov. 20, 2017 astro-ph.EP
We report the discovery of OGLE-2016-BLG-1190Lb, which is likely to be the first Spitzer microlensing planet in the Galactic bulge/bar, an assignation that can be confirmed by two epochs of high-resolution imaging of the combined source-lens baseline object. The planet's mass M_p= 13.4+-0.9 M_J places it right at the deuterium burning limit, i.e., the conventional boundary between "planets" and "brown dwarfs". Its existence raises the question of whether such objects are really "planets" (formed within the disks of their hosts) or "failed stars" (low mass objects formed by gas fragmentation). This question may ultimately be addressed by comparing disk and bulge/bar planets, which is a goal of the Spitzer microlens program. The host is a G dwarf M_host = 0.89+-0.07 M_sun and the planet has a semi-major axis a~2.0 AU. We use Kepler K2 Campaign 9 microlensing data to break the lens-mass degeneracy that generically impacts parallax solutions from Earth-Spitzer observations alone, which is the first successful application of this approach. The microlensing data, derived primarily from near-continuous, ultra-dense survey observations from OGLE, MOA, and three KMTNet telescopes, contain more orbital information than for any previous microlensing planet, but not quite enough to accurately specify the full orbit. However, these data do permit the first rigorous test of microlensing orbital-motion measurements, which are typically derived from data taken over <1% of an orbital period.
• ### OGLE-2017-BLG-0173Lb: Low Mass-Ratio Planet in a "Hollywood" Microlensing Event(1709.08476)
Nov. 8, 2017 astro-ph.EP
We present microlensing planet OGLE-2017-BLG-0173Lb, with planet-host mass ratio either $q\simeq 2.5\times 10^{-5}$ or $q\simeq 6.5\times 10^{-5}$, the lowest or among the lowest ever detected. The planetary perturbation is strongly detected, $\Delta\chi^2\sim 10,000$, because it arises from a bright (therefore, large) source passing over and enveloping the planetary caustic: a so-called "Hollywood" event. The factor $\sim 2.5$ offset in $q$ arises because of a previously unrecognized discrete degeneracy between Hollywood events in which the caustic is fully enveloped and those in which only one flank is enveloped, which we dub "Cannae" and "von Schlieffen", respectively. This degeneracy is "accidental" in that it arises from gaps in the data. Nevertheless, the fact that it appears in a $\Delta\chi^2=10,000$ planetary anomaly is striking. We present a simple formalism to estimate the sensitivity of other Hollywood events to planets and show that they can lead to detections close to, but perhaps not quite reaching, the Earth/Sun mass ratio of $3\times 10^{-6}$. This formalism also enables an analytic understanding of the factor $\sim 2.5$ offset in $q$ between the Cannae and von Schlieffen solutions. The Bayesian estimates for the host-mass, system distance, and planet-host projected separation are $M=0.39^{+0.40}_{-0.24}\,M_\odot$, $D_L=4.8^{+1.5}_{-1.8}\,\kpc$, and $a_\perp=3.8\pm 1.6\,\au$. The two estimates of the planet mass are $m_p=3.3^{+3.8}_{-2.1}\,M_\oplus$ and $m_p=8^{+11}_{-6}\,M_\oplus$. The measured lens-source relative proper motion $\mu=6\,\masyr$ will permit imaging of the lens in about 15 years or at first light on adaptive-optics imagers on next-generation telescopes. These will allow to measure the host mass but probably cannot resolve the planet-host mass-ratio degeneracy.
• ### Hydrogen-rich supernovae beyond the neutrino-driven core-collapse paradigm(1709.10475)
Sept. 29, 2017 astro-ph.SR
We present our study of OGLE-2014-SN-073, one of the brightest Type II SN ever discovered, with an unusually broad lightcurve combined with high ejecta velocities. From our hydrodynamical modelling we infer a remarkable ejecta mass of $60^{+42}_{-16}$~M$_\odot$, and a relatively high explosion energy of $12.4^{+13.0}_{-5.9} \times10^{51}$~erg. We show that this object belongs, with a very small number of other hydrogen-rich SNe, to an energy regime that is not explained by standard core-collapse (CC) neutrino-driven explosions. We compare the quantities inferred by the hydrodynamical modelling with the expectations of various exploding scenarios, trying to explain the high energy and luminosity released. We find some qualitative similarities with pair-instabilities SNe, although a prompt injection of energy by a magnetar seems also a viable alternative to explain such extreme event.
• ### OGLE-2016-BLG-0693LB: Probing the Brown Dwarf Desert with Microlensing(1707.01222)
Sept. 27, 2017 astro-ph.SR, astro-ph.EP
We present an analysis of microlensing event OGLE-2016-BLG-0693, based on the survey-only microlensing observations by the OGLE and KMTNet groups. In order to analyze the light curve, we consider the effects of parallax, orbital motion, and baseline slope, and also refine the result using a Galactic model prior. From the microlensing analysis, we find that the event is a binary composed of a low-mass brown dwarf 49+-20 M_J companion and a K- or G-dwarf host, which lies at a distance 5.0+-0.6 kpc toward the Galactic bulge. The projected separation between the brown dwarf and its host star is less than 5 AU, and thus it is likely that the brown dwarf companion is located in the brown dwarf desert.
• ### An Isolated Microlens Observed from K2, Spitzer and Earth(1709.09959)
We present the result of microlensing event MOA-2016-BLG-290, which received observations from the two-wheel Kepler (K2), Spitzer, as well as ground-based observatories. A joint analysis of data from K2 and the ground leads to two degenerate solutions of the lens mass and distance. This degeneracy is effectively broken once the (partial) Spitzer light curve is included. Altogether, the lens is found to be an extremely low-mass star located in the Galactic bulge. MOA-2016-BLG-290 is the first microlensing event for which we have signals from three well-separated ($\sim1$ AU) locations. It demonstrates the power of two-satellite microlensing experiment in reducing the ambiguity of lens properties, as pointed out independently by S. Refsdal and A. Gould several decades ago.
• ### Extracting Microlensing Signals from K2 Campaign 9(1704.08692)
Aug. 20, 2017 astro-ph.EP, astro-ph.IM
The reduction of the \emph{K2}'s Campaign 9 (K2C9) microlensing data is challenging mostly because of the very crowded field and the unstable pointing of the spacecraft. In this work, we present the first method that can extract microlensing signals from this K2C9 data product. The raw light curves and the astrometric solutions are first derived, using the techniques from Soares-Furtado et al. and Huang et al. for \emph{K2} dense field photometry. We then minimize and remove the systematic effect by performing simultaneous modeling with the microlensing signal. We also derive precise $(K_p-I)$ vs. $(V-I)$ color-color relations that can predict the microlensing source flux in the \emph{Kepler} bandpass. By implementing the color-color relation in the light curve modeling, we show that the microlensing parameters can be better constrained. In the end, we use two example microlensing events, OGLE-2016-BLG-0980 and OGLE-2016-BLG-0940, to test our method.
• ### OGLE-2016-BLG-0263L\lowercase{b}: Microlensing Detection of a Very Low-mass Binary Companion Through a Repeating Event Channel(1708.02727)
Aug. 9, 2017 astro-ph.EP
We report the discovery of a planet-mass companion to the microlens OGLE-2016-BLG-0263L. Unlike most low-mass companions that were detected through perturbations to the smooth and symmetric light curves produced by the primary, the companion was discovered through the channel of a repeating event, in which the companion itself produced its own single-mass light curve after the event produced by the primary had ended. Thanks to the continuous coverage of the second peak by high-cadence surveys, the possibility of the repeating nature due to source binarity is excluded with a $96\%$ confidence level. The mass of the companion estimated by a Bayesian analysis is $M_{\rm p}=4.1_{-2.5}^{+6.5}\ M_{\rm J}$. The projected primary-companion separation is $a_\perp = 6.5^{+1.3}_{-1.9}$ au. The ratio of the separation to the snow-line distance of $a_\perp/a_{\rm sl}\sim 15.4$ corresponds to the region beyond Neptune, the outermost planet of the solar system. We discuss the importance of high-cadence surveys in expanding the range of microlensing detections of low-mass companions and future space-based microlensing surveys.
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## Convergence, Divergence, and Comparison
In this section we discuss using comparison to determine if an improper integrals converges or diverges. Recall that if and are continuous functions on an interval and , then
This observation can be incredibly useful in determining whether or not an improper integral converges.
Not only does this technique help in determing whether integrals converge, but it also gives you some information about their values, which is often much easier to obtain than computing the exact integral.
Theorem 5.7.8 (Comparison Theorem (special case)) Let and be continuous functions with for .
1. If converges, then converges.
2. If diverges then diverges.
Proof. Since for all , the function
is a non-decreasing function. If converges to some value , then for any we have
Thus in this case is a non-decreasing function bounded above, hence the limit exists. This proves the first statement.
Likewise, the function
is also a non-decreasing function. If diverges then the function defined above is still non-decreasing and does not exist, so is not bounded. Since we have for all , hence is also unbounded, which proves the second statement.
The theorem is very intuitive if you think about areas under a graph. If the bigger integral converges then so does the smaller one, and if the smaller one diverges so does the bigger ones.''
Example 5.7.9 Does converge? Answer: YES.
Since , we really do have
as illustrated in Figure 5.7.5. Thus
so converges.
But why did we use ? It's a guess that turned out to work. You could have used something else, e.g., for some constant . This is an illustration of how in mathematics sometimes you have to use your imagination or guess and see what happens. Don't get anxious--instead, relax, take a deep breath and explore.
For example, alternatively we could have done the following:
and this works just as well, since converges (as is continuous).
Example 5.7.10 Consider . Does it converge or diverge? For large values of , the term very quickly goes to 0, so we expect this to diverge, since diverges. For , we have , so for all we have
(verify by cross multiplying)
But
Thus must also diverge.
Note that there is a natural analogue of Theorem 5.7.8 for integrals of functions that blow up'' at a point, but we will not state it formally.
Example 5.7.11 Consider
We have
(Coming up with this comparison might take some work, imagination, and trial and error.) We have
thus converges, even though we haven't figured out its value. We just know that it is . (In fact, it is .)
What if we found a function that is bigger than and its integral diverges?? So what! This does nothing for you. Bzzzt. Try again.
Example 5.7.12 Consider the integral
This is an improper integral since has a pole at . Does it converge? NO.
On the interal we have . Thus
Thus diverges.
William Stein 2006-03-15
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# 16: Vector Calculus
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## Twenty Tips For Interpreting Scientific Claims
It’s science!
Title of today’s post is taken from article of the same name in Nature by William Sutherland, David Spiegelhalter, Mark Burgman. Several readers asked me to comment.
I’ll assume you’ve read the original. I kept the same order and wording as their points, and try not to repeat any of their good points.
Differences and chance cause variation. Chance can’t and doesn’t cause anything. Chance isn’t a thing, therefore it can never be a cause. Differences don’t cause things per se: things do (sizes of differences can certainly change rates of change). We cannot always identify causal agents, just correlates of change.
No measurement is exact. Well, not quite, but I take their point. Measurement error is vastly more prevalent than acknowledged and almost never accounted for. Leading to…can you guess? Over-certainty.
Bias is rife. Amen and amen. But, just like admonishing the public by reminding them they look ugly in jeans, they always think it’s the other guy and not them. Yes, you, even you, are biased. Even you. And you. Even if you’re part of a team that won prizes.
Bigger is usually better for sample size. Indeed, except for cost and the possibility of being overwhelmed or misled by errata, bigger is always better.
Correlation does not imply causation. But the opposite is true: causation causes correlation. People often forget the distinction between ontology and epistemology. This is also my fault for not making this distinction clearer more often. Most probability models are epistemological, meaning they say what are the changes in probability of some outcome given changes in input variables. The problem comes when people interpret the changing probabilities of the outcome as being caused by the input variables, which is usually not true.
Regression to the mean can mislead. See this on the so-called Sports Illustrated curse.
Extrapolating beyond the data is risky. The reason is probability models are usually not causative and even when they are few check them for accuracy (everybody checks them for fit, via p-values, posteriors and the like).
Beware the base-rate fallacy. Think of it this way. If you’re forecasting “No rain” for Tucson each day, you’re likely to be right most of the time. But your boast carries little importance. Try the same forecast for Norman, Oklahoma and you’re accuracy heads south. This is why we should speak of skill—the improvement over naive predictions—instead of accuracy rates. Right, climatologists?
Controls are important. The only thing wrong with this point is that important should read everything. The more you can control, the closer your model comes to causality. Problem is that controlling “everything” in human interactions or in anything contingent is impossible. It will always—as in always—be possible that something other than what we thought caused the outcome.
Randomization avoids bias. No. Randomization is not a property. It “gives” nothing to your results. “Randomization” belongs to the old days of magical thinking. Rather, assigning control of an experiment to persons without a financial or emotional interest reduces but cannot avoid bias. That residual bias exists is why there are always calls for replication.
Seek replication, not pseudoreplication. And speaking of replication… Listen up sociologists, psychologists, and so on: It is not a replication unless the experiment is repeated in exactly the same way where they only differences are those things you could not control in the first experiment. “More or less” the same way is not exactly the same way and is therefore not a replication. A mass of published literature on the same subject is only a weak indicator of truth. Who remembers frontal lobotomies, etc., etc., etc.?
Scientists are human. And because they are typically in positions commanding money and people, they fall prey more often to the standard sins.
Significance is significant. No, it is not, or at least not necessarily. “Significance” means attaining a wee p-value, one less than the magic number. And this result may not have and usually does not have practical bearing on questions of interest about the thing at hand. Finding a wee p-value is child’s play. Finding something useful to say is far harder.
Separate no effect from non-significance. Here I must quote: “The lack of a statistically significant result (say a P-value > 0.05) does not mean that there was no underlying effect: it means that no effect was detected.” This is only partially true. Lack of a wee p-value might mean the effect was there but undetected. On the other hand, the effect might be there and detectable, too. It’s just the p-values are terrible at discovering which situation we’re in. An effect without a wee p-value may still be important. If instead we looked at probability models as they should be looked at, as predictive statements, we could say more.
Effect size matters. Wee p-values alone mean nothing. Repeat that until you get sick of repeating it. This is another call for predictive analytics.
Study relevance limits generalizations. It’s funny how many reporters never read the papers they report on.
Feelings influence risk perception. And this is because feelings are part of what we risk! Money, after all, is only a crude device to measure our feelings. And just because you hate fat people eating transfats does not mean the risk of disease from eating transfats is high. And just because you hate smoking does not mean that “second-hand” smoke is perilously dangerous, etc., etc.
Dependencies change the risks. Try not to look at anything in isolation, unless the thing is amenable to isolation. Dice come to mind. The changes that await us when global warming finally strikes (soon, soon) do not.
Data can be dredged or cherry picked. “Big data” anyone? One thing Big Data guarantees is shocked looks on the face of managers who were certain sure they picked up a “significant” signal in their gleaming, massive datasets.
Extreme measurements may mislead. Like I always say, any survey or result is true conditional on the set of premises belonging to the experiment. Vary any of these premises, the result no longer holds. The more premises, i.e. conditions, there are, the greater the chance the results are not meaningful beyond the realm of this single experiment. Journalists often change this premises in their reporting; but to be fair, so do many scientists when summarizing their work. Memorize this.
## Friedrich A. Hayek’s Lecture “The Pretense of Knowledge”
Somehow Hayek wrote his multitudinous works wearing a suit.
This was reprinted in the Wall Street Journal over the weekend. I’ve chopped it into parts for commenting.
To act on the belief that we possess the knowledge and the power which enable us to shape the processes of society entirely to our liking, knowledge which in fact we do not possess, is likely to make us do much harm. In the physical sciences there may be little objection to trying to do the impossible; one might even feel that one ought not to discourage the over-confident because their experiments may after all produce some new insights. But in the social field the erroneous belief that the exercise of some power would have beneficial consequences is likely to lead to a new power to coerce other men being conferred on some authority.
If a physics experiment goes south, all that happens is some money is wasted but with the reasonable chance something is learned. Not always, of course. Think of the “cold” fusion hiccup twenty years ago. But when an, as Mill put it, “experiment in living” sours people suffer. And nothing is learned.
The love of theory is far too strong in economics, sociology, education and the like for observation to wound belief. In physics, what has gone wrong is usually identifiable, but in the “nudging” sciences the evidence is always somewhat ambiguous. There is aways wiggle room in whatever happens that the theory which drove the experiment might be true. And this is enough.
Human society is so unfathomably complex that all experiments should be approached with trepidation and fear and with an “out,” a way to revert, if at all possible, to the old ways. The essence of at least one definition of conservatism it that “change we can believe in” just for the sake of change is more likely to lead to grief than to happiness. There is a vast amount of wisdom packed into tradition which should not be overthrown lightly.
Even if such power is not in itself bad, its exercise is likely to impede the functioning of those spontaneous ordering forces by which, without understanding them, man is in fact so largely assisted in the pursuit of his aims. We are only beginning to understand on how subtle a communication system the functioning of an advanced industrial society is based — a communications system which we call the market and which turns out to be a more efficient mechanism for digesting dispersed information than any that man has deliberately designed.
The (highly degreed, well-placed) bureaucrat looks at the market and says to himself, “There’s no way to understand this, therefore it is not understandable. Therefore I must direct it, then I will understand it.” The conclusion has elements of truth—it is easier to claim to understand what one directs—but it does not follow from the premise. And anyway, we’re right back at the beginning. The directions (regulations, laws, taxes) cause changes nobody could have foreseen. The urge to perfection is never humiliated by history.
If man is not to do more harm than good in his efforts to improve the social order, he will have to learn that in this, as in all other fields where essential complexity of an organized kind prevails, he cannot acquire the full knowledge which would make mastery of the events possible. He will therefore have to use what knowledge he can achieve, not to shape the results as the craftsman shapes his handiwork, but rather to cultivate a growth by providing the appropriate environment, in the manner in which the gardener does this for his plants.
Hayek follows Burke and says change should be small, incremental, with the acknowledgement, like in gardening, that despite our best efforts, a poor harvest is more likely to result from enthusiasm than from conservatism.
There is danger in the exuberant feeling of ever growing power which the advance of the physical sciences has engendered and which tempts man to try, “dizzy with success,” to use a characteristic phrase of early communism, to subject not only our natural but also our human environment to the control of a human will. The recognition of the insuperable limits to his knowledge ought indeed to teach the student of society a lesson of humility which should guard him against becoming an accomplice in men’s fatal striving to control society — a striving which makes him not only a tyrant over his fellows, but which may well make him the destroyer of a civilization which no brain has designed but which has grown from the free efforts of millions of individuals.
That cannot be said better, but just as a for instance, note that the Expertism embraced by all modern liberal democratic societies has gone so far that government social services feels it can judge the mental health of tourists, as in Britain, where an Italian woman having a panic attack was forcibly sedated and cut open, her baby removed from her womb, taken away, and not given back. The experts feel they would do a better job raising this non-citizen’s baby than its mother. Link.
## On Dressing Well
Make that, “On Dressing Not So Well.” Here is a picture from a mall in Santa Monica, first appearing in the Los Angeles Times.
Tis always the season for looking poorly.
The article which surrounded this picture is titled “The season of excess begins”, and the caption of the photo aptly begins “Black Friday shoppers are seen…”
Truly it was a black day, and not just because of the fights, riots, and nauseating avarice displayed by many shoppers. No. It was the visual assault that cast dark shadows.
Let’s engage in some amateur photo reconnaissance. The location? A mall with top-end stores, meaning that to shop in them one has to have piles of the green stuff at arm’s reach. These are not poor people. Further proof: nobody looks like they’re going hungry. There are no beggars. The accoutrement of the mall—the tree, the giant present—are of superior quality. The buildings themselves are clean, sharp.
So money isn’t a problem. What is?
What first struck me were the two gentleman in the center. One is wearing what looks like a modified logo of an oil company and a hat advertising his preference in sports, which is evidently something he wished others to know about, the depth of his ardency so strong that he put it where everyone must see it. His neighbor wears a hipster hat and blue rumpled t-shirt on which is imprinted a message thankfully too small for us to read, but one which this man thought of such important that he should emblazon it on his chest.
Both men are wearing jeans, which is no story. Therefore, that they look sloppy is a given. The jeans are too long and gather in folds around their feet, which are shod with tennis shoes the shape of which would have been familiar to Frankenstein’s monster. We can bet that, if asked, both would say they are “comfortable.”
Next are the two older women leading them on either side. Both are wearing jeans. The pair of the leftmost lady’s are faded, as if she could not afford a newer pair. The rightmost lady has done herself no favor and has chosen a pair which are about as unflattering as is possible. But then jeans only flatter bodies which don’t need praise. Otherwise, they are ugly and make the wearers ugly. Yes, that means you, too. (Yes, even you, who thought the “you” of the previous sentence wasn’t really you. It was.)
The blouse of the rightmost lady resembles one of those hidden art pictures which if stared at reveal a rocketship or a man riding a horse. The blouse of the other lady looks to be a smock, such as one might wear while casting pottery.
There is no need to continue. Every person, without exception, looks terrible. No one picked something which fit his or her body. Everybody’s clothes are ugly, sad, unkempt. Everybody did their damnedest to appear like an unthinking high school sophomore. In this and only this did they succeed.
The clothes are the visual equivalent of popular, which is to say, bad music. There is no pleasure to be had in looking at anybody.
## What To Do On Black Friday
This, one of our most sacred days of the year, can be exhausting, especially in the choices one must make. Should one wake at 3am or should one even sleep? Should one take part in a riot, tumbling headlong into a store to be the first to secure the new iWhatsit, which is rumored to be 0.00132″ slimmer than last month’s model? Or should one circle the mall’s outer limits for hours spying for a spot to park?
Just what form should the joy of the upcoming and unnamed and soon to be unnameable holiday take? Well, here are two “black” ideas.
1. Go see a priest. Confessions heard daily.
Truly, confession is good for the soul.
2. Read Wordsworth, the old curmudgeon. Nothing black here, but the sheer absence of it can put one in mind of it.
England, 1802
O FRIEND! I know not which way I must look
For comfort, being, as I am, opprest,
To think that now our life is only drest
For show; mean handy-work of craftsman, cook,
Or groom!—We must run glittering like a brook
In the open sunshine, or we are unblest:
The wealthiest man among us is the best:
No grandeur now in nature or in book
Delights us. Rapine, avarice, expense,
This is idolatry; and these we adore:
Plain living and high thinking are no more:
The homely beauty of the good old cause
Is gone; our peace, our fearful innocence,
And pure religion breathing household laws.
— William Wordsworth. 1770-1850
3. Or chat with a nun! Develop the habit.
Calling all souls!
Update Our first story. Violence flares as shoppers slug it out for the best Black Friday deals.
## Old Lodge Skins’ Prayer Of Thanksgiving
A real Little Big Man existed. Bottom row, second from right.
From Little Big Man, Thomas Berger. Eschew the movie, which shares only the title and the names of a few characters in the book, which is the moral and historical opposite of the (more or less) politically correct film.
Also highly recommended (as orientation) is The Fighting Cheyennes by George Bird Grinnell, who was born in 1849 and who wrote the book in 1915 (it’s still in print). It is a non-patronizing, non-romantic look at the battles the Cheyenne fought in, as much as possible, their own words.
The Cheyenne are Human Beings. They call themselves that because they are and were, and because they act and acted just like human beings everywhere, including those white and black versions with which they had and have many strange interactions. Little Big Man was born white and named Jack Crabbe, but through a series of curious incidents was raised by the Cheyenne during the time in which that great nation was going into eclipse.
Berger wrote Little Big Man at a time (1964) when white boys still wanted to run off and be Indians. (Nearly twenty years later, Grizzly Adams fulfilled the same function.) Some call Berger’s work “comic”, which is the most inapt description which could be imagined.
Old Lodge Skins was Little Big Man’s adoptive grandfather. The scene takes place shortly after the Battle of Little Big Horn slash Battle at the Greasy Grass. There is much in this prayer that still works.
Then he commenced to pray to the Everywhere Spirit in the same stentorian voice, never sniveling but bold and free.
“Thank you for making me a Human Being! Thank you for helping me become a warrior! Thank you for all my victories and for all my defeats. Thank you for my vision, and for the blindness in which I saw further.
“I have killed many men and loved many women and eaten much meat. I have also been hungry, and I thank you for that and for the added sweetness that food has when you receive it after such a time.
“You make all things and direct them in their ways, O Grandfather, and now you have decided that the Human Beings will soon have to walk a new road. Thank you for letting us win once before that happened. Even if my people must eventually pass from the face of the earth, they will live on in whatever men are fierce and strong. So that when women see a man who is proud and brave and vengeful, even if he has a white face, they will cry: ‘That is a Human Being!’…”
I stood there in awe and Old Lodge Skins started to sing, and when the cloud arrived overhead, the rain started to patter across his uplifted face, mixing with the tears of joy there.
It might have been ten minutes or an hour, and when it stopped and the sun’s setting rays cut through, he give his final thanks and last request.
“Take care of my son here,” he says, “and see that he does not go crazy.”
He laid down then on the damp rocks and died right away. I descended to the treeline, fetched back some poles, and built him a scaffold. Wrapped him in the red blanket and laid him thereon. Then after a while I started down the mountain in the fading light.
Posted in Fun
## What Random Means In Random Number Generation
Dice throws appear random because their outcomes cannot be predicted naively. But knowing the physics and initial conditions, they can be predicted, and so are not truly random.
It’s simple, really. A “random” number generator spits out a string of numbers or characters from some set, say c1, c2,…, cp, where each of these is a character out of p total. Example: c1 = ‘a’, c2 = ‘b’, etc.
Before the generator starts, using only the premises that the generator can produce only these p characters, we deduce the probability that the first character is ci (for any i in 1 to p) as 1/p.
Start the generator. The characters roll out one at a time. To be be considered “random”, this condition must hold:
Pr( Ct = ci | C1, C2, Ct-1, generator premises) = 1/p for all t and all i,
where I mean by this notation that the probability the next character (Ct) will be one of the set is 1/p no matter how far along in the series we are (short of infinity) and no matter what character we consider. The “generator premises” are those which state only the characters c1 through cp are available, etc.
In other words, if there is any information in the series to date that allows us to calculate any probability other than 1/p for Ct, then the series is not “random.” Be clear that “allows us” means “logically possible” and not necessarily practical or in-practice possible. There may be some existence proof which says something like “Given this generator and an output series like this-and-such, the probability of Ct is x, which does not equal 1/p”. We may never see the this-and-such series, but still if the output series is possible, then the output is not “random.”
Now don’t start going all frequentist on me and say, “Look here, Briggs, you fool. I’ve ran the generator a finite number of times and the relative frequencies of the observed Ct don’t match 1/p.” I’d reply, “Get back to me when you reach an infinite number of times.”
Run the generator for just one character. The observed relative frequencies will be 0, 0, …, 1, …, 0, where it’s 0s for all ci except the 1 character which showed. What does this prove? Next to nothing. Probability just isn’t relative frequency, but relative frequency can match the probability. Probabilities predict relative frequencies. In that spirit, we know the series is not “random” if we can do a better job predicting the series than saying the probability of the next character is 1/p (for any character).
But I see the idea of relative frequency is still alluring. Perhaps this is why. There is in math the idea of a “normal” number, unfortunately named because “normal” in probability means something else. A normal string or number is one in which the digits/characters repeat equally often, yea even unto infinity. Examples: 0.111222333444555666777888999000111222… and ‘abcabcabcabc…’ (we know the numbers are limited to digits 0-9, but here I limit the characters to a,b,c).
These normal numbers are in no sense “random”, because if you know where you are you know with certainty what the next digit or character is. Plus, there are some technical ideas, where a number may be normal in one base (say 10 or 2) but not normal in another base. Here is an example of a number, Stoneham’s constant, which is normal in one base but not another. So “normal” does not imply random, but we have the sense that “random” implies “normal”, which it does.
Truly “random” numbers are probably (I don’t believe there is a proof) normal in any base. Another way to put it, in terms on information, is to say that the number cannot be compressed (in any base); that is, speaking loosely, that it takes just as many characters to compress the number as to display it. The above-linked article gives some hints on the “randomness” of π, which appears normal in many (all?) bases and which cannot be compressed. So where do the digits of π and other transcendental numbers come from? Only God knows.
Last point: many “random” number generators—where by now we see that “random” merely means unknown or unpredictable in some sense—are wholly predictable, they are fully deterministic. Start at a known output and the path can be predicted with certainty. These are called “pseudo-random” generators because the numbers only appear unpredictable.
And what does that mean? Appear unpredictable? Well, it means not-random, that we can prove that a set of premises exist which predict the series perfectly. The difference between a “random” generator is that we can prove no such set of premises exist.
## Winner Announced in What Do You Call A Believer In Scientism Contest
At least they’ll subsidize your cane.
Last week we asked what is the best word or term for a die-hard believer in scientism? The response was overwhelming!
More than 50 people entered some 100 words. I grouped these into three-plus-one categories: Honorable Mentions, Runner Ups, and Top 10. But there can be only one! Winner, that is, who will receive a Kindle copy of Iain Murray’s not-to-be-missed Stealing You Blind.
All complaints or suggestions about the entries, rules, or my judging may be entered at this site.
Special Mention
Thinkologist (davebowne).
Love it, but I was saving this for a special use on certified Experts. So it was not in the running for the contest.
Honorable Mentions
In no particular order:
Evidencist (Francsois). Inscienaty, Narrow-Minder, Rationer, Reasoner, Sciencerapist (Pedro Erik). Science Zombie (JohnK; Davebowne). Bright (Thinkling; Mariner). Supercilioust (HowardW). Scimoron (GaryL). Positivist (Philip Neal). Psysciphilliac (BradH). Sheldonists, Optimists, Triverist (Paul Murphy). Knowlatan, Pedantocrat, Patronisiac (Hamish McCallum). Unscientific Hucksterists, Philistine Scientists, Scientific Excrescence, Vulgar Scientists (Jim Fedako). Scatomancer, Weepees (Bruce Foutch). Scisyphist, Scisyphean (Jonathan S). Scienterrific (An Engineer). Scienzealot (Charles Boncelet), Scientomancer (hmi). Anthropodogmatists (Pangloss). Scinsist (Jeffrey).
Runner Ups
In no particular order:
Scientite (Adrian Hilton). Fatuotist (as in “fatuous”; Bob Mrotek). Scientizer, Scientaster, Scifollogist, Sciphist (Jester). Spockists (Toby Young). Scientician (TheRealAaron). Sciencista (Mike Anderson). Scientocracy (K). Sciphist (John Baglien). Technogogue, Techogogue, Scientologue (Mark Webster). Scientificant, Saintificist, Saintologist, Scientificist (Bob). Aoristicist (Don Jackson). Scienticist (Edmund Kartoffel; David). Scientipher, Scientifie (Mike B). Scientient (Andy). Scientismion, Scientismidel, (Aloysius Hogan, who had 42 generated entries, such as Scientismafuego derived from Cacafuego, a ship all fans of Patrick O’Brian will recognize; only the top are shown).
Top 10
In winning order:
1. Scientificalist (Ye Olde Statisician)
2. Obfuscator Scienista (Bruce Foutch)
4. Sciphiliac (Rich)
5. Scientophile (Andy)
6. Scientiscubus (Aloysius Hogan)
7. Sighintists (John M)
8. Scientocrat (K)
It is not only that Science cannot answer these questions now, but that it never can. All these and many more are forever beyond the reach of empiricism. There is no observation in the universe, nor can there be, nor will there ever be, which proves $e^{i\pi} = -1$. It is impossible to peer at the Unmoved Mover, yet He must be there or, quite literally, nothing would happen.
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1
JEE Main 2022 (Online) 27th June Evening Shift
Numerical
+4
-1
The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 $$\Omega$$ is __________ mA.
2
JEE Main 2022 (Online) 26th June Morning Shift
Numerical
+4
-1
As per the given circuit, the value of current through the battery will be ____________ A.
3
JEE Main 2022 (Online) 25th June Evening Shift
Numerical
+4
-1
In an experiment of CE configuration of n-p-n transistor, the transfer characteristics are observed as given in figure.
If the input resistance is 200 $$\Omega$$ and output resistance is 60 $$\Omega$$, the voltage gain in this experiment will be ____________.
4
JEE Main 2022 (Online) 24th June Evening Shift
Numerical
+4
-1
In the given circuit, the value of current IL will be ____________ mA. (When RL = 1k$$\Omega$$)
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
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EXAM MAP
Joint Entrance Examination
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# Given an infinite set of events, prove that the probability of an event is smaller than $1$
I have two infinite sets of events $$A$$ and $$B$$ with the following probabilities:
• $$P(A_n)=\frac{2}{6n-1}$$
• $$P(B_n)=\frac{2}{6n+1}$$
Note that I have divided them into two sets only because it is easier to depict the probabilities; other than that, these two sets are essentially "indistinguishable".
I want to prove that the probability that any one of the events will take place is smaller than $$1$$.
I've figured I can use Inclusion-Exclusion principle, but the problem, of course, is that I have an infinite amount of events so I'm not quite sure how to accomplish that.
I tested it using the following Python script, so I believe that the probability of one of the events taking place is indeed smaller than $$1$$:
from decimal import Decimal
from decimal import getcontext
getcontext().prec = 100
def getBits(num):
bit = 0
bits = []
while num > 0:
if num & 1:
bits.append(bit)
bit += 1
num >>= 1
return bits
def prod(arr):
res = Decimal(1)
for val in arr:
res *= val
return res
SIZE = 20
sums = [Decimal(0) for k in range(SIZE)]
A = [Decimal(2)/(6*k-1) for k in range(1,SIZE)]
B = [Decimal(2)/(6*k+1) for k in range(1,SIZE)]
probabilities = [p for pair in zip(A,B) for p in pair]
for n in range(1,1<<SIZE):
bits = getBits(n)
sums[len(bits)-1] += prod([probabilities[bit] for bit in bits])
print(sum([sums[k]*(-1)**k for k in range(SIZE)]))
After a minute or two, the printout shows a probability of approximately $$89\%$$.
Does anyone see a way for me to achieve my purpose mathematically?
Again - my primary objective is to prove that the probability of one of the events taking place is smaller than $$1$$. I don't mind calculating the exact value of this probability along the way, but if there's another way which only proves that it is smaller than $$1$$ then I'm fine with that.
Thank you very much!
• So you want to prove that, for all $n \in \Bbb Z^+$, that $P(A_n)$ and $P(B_n)$ (individually) are always less than $1$? Just want to clarify. – Eevee Trainer May 11 at 6:53
• @EeveeTrainer: No, that one is obvious from the definition. – goodvibration May 11 at 6:54
• @EeveeTrainer: I want to prove that there is a chance that none of the events will take place (or equivalently, that the probability that either one of them will take place is smaller than $1$). – goodvibration May 11 at 6:56
Without further information about the relation between any two of $$\{A_n\}_{n\ge 1} \cup \{B_n\}_{n\ge 1}$$, like independence or exclusivity, you cannot prove what you want to prove, as there are counterexamples:
Take a uniformly random continuous variable $$u$$ on $$[0,1]$$. With $$0 \le a \le b \le1$$, the probability for the event "$$u \in [a,b]$$" is $$b-a$$.
Define
$$a_0:=0, \text{ and }a_{n}:=a_{n-1}+\frac2{6n-1} \text{ for }n \ge 1.$$
This means
$$a_k=\sum_{n=1}^k\frac2{6n-1}.$$
Now that sum in the above forumula is (asymptotically, for large $$n$$) equivalent to the harmomic series, so it diverges. That means at some point $$k_1$$ we get $$a_{k_1} \ge 1$$ for the first time. According to Wolfram Alpha we have $$k_1=8$$, but the exact value is unimportant.
$$A_n=u \in [a_{n-1},a_n], \text{ for }n=1,2,\ldots,k_1-1$$
and
$$A_{k_1} = u \in [1-\frac2{6k_1-1},1]$$
We have $$P(A_n)=\frac2{6n-1}$$ for $$n=1,2,\ldots,k_1$$. Also note that by definition of $$k_1$$, we have $$1-\frac2{6k_1-1} \ge a_{k_1-1}$$.
In the end that means that
$$P(A_1 \cup A_2 \cup \ldots A_{k_1-1} \cup A_{k_1})=1$$, because the intervals used to define each $$A_n$$ cover the complete interval $$[0,1]$$.
In your python code (I have no knowledge of python) it looks like you are multiplying probabilites, which suggests that there is some kind of independence. If you can specify more about that, maybe what you want to prove is correct. But as my example proves, if you can't assume anything, your desired conclusion is incorrect.
• Thank you. I will rephrase the question then. – goodvibration May 13 at 7:09
• @goodvibration Make a new question, if possible, and add a link to it in this question's text. – Ingix May 13 at 8:37
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## Got, got, need: how much does a 2014 World Cup sticker collection cost, or why is swapping so important?
The year 2014 brings with it a World Cup, and with that an activity that is familiar to many: the collection of stickers in the attempt to complete Panini’s World Cup sticker album. The pastime has not changed much over the years since Panini’s first collection for the 1970 World Cup. The Italian company still prints the same multilingual booklet, with the pages for most, but not all, of the teams headed with the name of that country in its official language – for Belgium they split the difference and go with ‘Belgique/België’ (sorry German-speaking Belgians!), but rather than entertain a four-way tie they plump with ‘Switzerland’. The major change though is that the number of stickers to collect has increased, largely due to the World Cup itself having expanded to 32 teams in 1998. This year’s collection comprises 640 stickers: 19 for each team (17 players, the team and the logo) with the remaining stickers being the stadia, trophy, official ball etc. So how much does this cost?
The good news is that the empty album can be picked up for free, and furthermore comes with an insert of 6 stickers: Yaya Touré (Côte d’Ivoire), Cristiano Ronaldo (Portugal), Hugo Lloris (France), Gonzalo Higuaín (Argentina), Daniele De Rossi (Italy) and Roman Shirokov (Russia). This leaves 634 stickers to buy. Stickers can be bought in packs of 5 for 50p, but you get multipacks of 5 packs for £2 in Asda. So 25 multipacks and 2 extra packs would get you 640 stickers at a cost of £51 (just buying individual packs this cost would be £63.50.)
But, of course, among these stickers are likely to be duplicates. If I were to hand you 636 randomly selected stickers the chances you’d have a complete collection would be about 1 in $10^{276}$. So how much would we expect to spend before we have a complete collection. We’ll assume that each sticker is equally likely to appear (even the ones that come in the insert), so each sticker has a $\frac{1}{640}$ probability of being Yaya Touré, or of being Wayne Rooney, or of being the Colombian logo.
So far we already have 6 of our album stickers (as stickers stuck in our album will be referred to hereon in), so when we open the first pack the first sticker we see could be one of 634 we don’t already have or one of the 6 we do, so the probability it is our 7th album sticker is $\frac{634}{640}$, while the probability it is a duplicate is $\frac{6}{640}$. If it is a duplicate then the next sticker still has a probability of $\frac{634}{640}$ of being our 7th album sticker, and this remains the case until we find our 7th album sticker. The number of stickers we see until we have our 7th album sticker is equivalent to the number of success/failure (i.e. Bernoulli) trials until the first success. The number of Bernoulli trials of success probability $p$ until success follows a Geometric distribution and its mean value is simply given by $\frac{1}{p}$. So we expect to see $\frac{640}{634}$ stickers until we have our 7th album sticker.
Once we have our 7th album sticker, we have now 633 stickers left to collect. So the probability of a sticker being our 8th album sticker is $\frac{633}{640}$. After our 7th album sticker, the number of stickers we see until we have our 8th album sticker again follows a Geometric distribution, so we expect to see $\frac{640}{633}$ more stickers until we have our 8th album sticker. We then expect to see $\frac{640}{632}$ stickers until we see our 9th album sticker, and $\frac{640}{631}$ until we see our 10th and so on.
So, once we have $latex i-1$ stickers, the number of stickers we expect to see before we find our $i$th album sticker is $\frac{640}{641-i}$. All told then the number of stickers we’d expect to see to complete our album of 640 stickers is
$\sum_{i=7}^{640}\frac{640}{641-i}=\frac{640}{634}+\frac{640}{633}+\frac{640}{632}+\cdots+\frac{640}{2}+\frac{640}{1}=4500,$
which would be 180 multipacks at a total cost of £360! (Or 900 individual packs at a cost of £450.)
The main reason for this high cost is that as you fill up your sticker album, it becomes harder and harder to add to it. Once you have 639 stickers, you’d expect to have to see 640 stickers before you get your final sticker, at a cost of £51.20 (at the 8p per sticker multipack price) or £64 (at the 10p per sticker individual pack price), which is a lot of money to get one sticker! You can see the expected number of stickers and cost at each rate for each album sticker in the spreadsheet below (the would-be cost of album stickers is included – their being free doesn’t offer a huge saving.)
Thankfully Panini have mercy and allow you a one-off order of 50 specified stickers to help you complete your collection, at the price of 14p per sticker plus £1 for postage. This works out at 16p per sticker if you order 50, which reduces to 14.6p with a 10% discount on the stickers if you order online using a code in your album. While this is more expensive than the cost of a sticker it’s still much cheaper than the cost of later album stickers, in fact from the spreadsheet we can see that even with multipacks each new album sticker is expected to cost over 14.6p when you only have 290 album stickers – not even half full! Ordering these stickers is a no-brainer and requires us then to only collect 590 album stickers beforehand, the number of stickers we’d expect to see to reach this amount is
$\sum_{i=7}^{590}\frac{640}{641-i}=1619,$
which requires 65 multipacks at a total cost of £130 or 324 individual packs at a total cost of £162. Add in your ordered album stickers at the reduced rate and you’re expecting to pay £137.30 with multipacks or £169.30 with individual packs, a significant reduction thanks to being able to order those last 50 otherwise incredibly expensive stickers. It is still however a good deal more than the £51 minimum cost – which is what makes the ability to make good swaps so vital, to reduce the expected cost closer towards that lower value.
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# Remainder theorem of factorials and powers
If $$7^n$$ divides $$68!$$ Then what is the greatest value of $$n$$?
Can someone tell me how to find the answer.
• Basically, this is asking you, how many factors of 7 are there in 68!. DonAntonio's answer gives you a formula to compute this, but you should try to come up with an answer without the formula first to understand how it works. – Don Thousand Apr 21 at 19:48
$$\sum_{k=1}^\infty\left\lfloor\frac{68}{7^k}\right\rfloor$$
• You should add the detail that it uses Legendre's formula for the $p$-valuation of $n!$, $p$ being a prime number. – Bernard Apr 21 at 19:55
• I believe it should be the sum from $k=1$ – Will Jagy Apr 21 at 20:02
• for example $7!$ should give exponent $1,$ while $6!$ should give exponent $0$ – Will Jagy Apr 21 at 20:06
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# All Questions
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## Return to Answer
1 [made Community Wiki]
The canonical example to introduce this idea early to students is the Fundamental Theorem of Calculus. In order to figure out one area $\int_{a}^b f(t)dt$, you must come to grips with the generalized problem $x \mapsto \int_{a}^x f(t)dt$
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# The Unapologetic Mathematician
## Differentiable Convex Functions
We showed that all convex functions are continuous. Now let’s assume that we’ve got one that’s differentiable too. Actually, this isn’t a very big imposition. It turns out that a result called Rademacher’s Theorem will tell us that any Lipschitz function is differentiable “almost everywhere”.
Okay, so what does differentiability mean? Remember our secant-slope function:
$\displaystyle s(\left[a,b\right])=\frac{f(b)-f(a)}{b-a}$
Differentiability says that as we shrink the interval $\left[a,b\right]$ down to a single point $c$ the function has a limit, and that limit is $f'(c)$.
So now take $a. We can pick a $c$ between them and points $x$ and $y$ so that $a. Now we compare slopes to find
$s(\left[a,x\right])\leq s(\left[a,c\right])\leq s(\left[c,b\right])\leq s(\left[y,b\right])$
so as we let $x$ approach $a$ and $y$ approach $b$ we find
$f'(a)\leq s(\left[a,c\right])\leq s(\left[c,b\right])\leq f'(b)$
And so the derivative of $f$ must be nondecreasing.
Let’s look at the statement $f'(a)\leq s(\left[a,x\right])$ a little more closely. We can expand this out to say
$\displaystyle f'(a)\leq\frac{f(x)-f(a)}{x-a}$
which we can rewrite as $f(a)+f'(x)(x-a)\leq f(x)$. That is, while the function lies below any of its secants it lies above any of its tangents. In particular, if we have a local minimum where $f'(a)=0$ then $f(a)\leq f(x)$, and the point is also a global minimum.
If the derivative $f'(x)$ is itself differentiable, then the differential mean-value theorem tells us that $f''(x)\geq0$ since $f'(x)$ is nondecreasing. This leads us back to the second derivative test to distinguish maxima and minima, since a function is convex near a local minimum.
April 16, 2008 Posted by | Analysis, Calculus | 4 Comments
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# Infinite set on finite closed topology
I'm having trouble making sense of certain terminology. So the question asks me to determine whether a finite closed topology on an infinite set is a $T_1$ space. Now before I get into that, I'm having trouble dissecting the statement of a finite closed topology on an infinite set.
So to start, the definition of a finite closed topology on a set $X$ is the closed subsets of $X$ are $X$ itself and all finite subsets of $X$; that is the open sets are the empty set and all subsets of $X$ which have finite complements.
Here is my problem. If this is the definition of the finite complement topology, then what topological space are the open sets on? So a closed set on a topological space is defined to be closed if its complement in $X$ is open in the topology put upon it, but if that is so then how are these open sets even in the finite complement topology if they are open?
Also I was trying to imagine what the finite complement topology would look like on an infinite set I came up with this:
$$X = \{X,\emptyset, \{a_1\},\,\{a_2\},\,\dots\{a_i\},\,\dots\{a_1,a_2\},\dots,\{a_1,a_2,\dots,a_n\}\}$$ and the sets would continue to get larger.
The "finite closed topology" describes the closed sets. Which gives you an exact definition of the open sets, they are the complements of closed sets. So indeed this is the co-finite topology (or finite complement topology).
A subset $U$ is open if and only if $U=\varnothing$ or $X\setminus U$ is finite.
Imagining the entire topology is a bit tricky, since infinite sets can be very large, and therefore have many finite sets. So it's best to think about this in terms of definitions. There is a definition when a set is open, and when it is closed. Now check if these definitions meet the requirement of being $T_1$.
And for that matter, allow me to remind you that $X$ is a $T_1$ space if and only if every singleton is closed.
• the open sets are the complements of closed sets. So in this specific topology the "closed" sets are the open sets because they are in the topology, meanwhile the "open" sets are the closed sets because they are the complement to what are the "open" sets on the topological space? I'm really confused... – dc3rd May 1 '15 at 20:57
• You can define the topology by giving the open sets; or the closed sets. Note that if all singletons are open, then every set is open. So it is impossible that all finite sets are open, but not all sets are open. – Asaf Karagila May 1 '15 at 21:02
• hmmm ok, well if we use the closed definition of a topology, then all of the singleton sets would be members of the finite closed topology, but the complements of those finite singleton sets would be open, but also infinite, which would mean that this is not a $T_1$ space. – dc3rd May 1 '15 at 21:34
• I don't get your argument. Finite closed means finite sets are closed, and complement of finite sets are open. Not that finite complements of finite sets are open, because that makes no sense, in an infinite set the complement of size finite set is never finite itself. – Asaf Karagila May 1 '15 at 21:38
• what I am saying is that since all of the singleton sets are in the topology and since each member of this topology is closed, that if I take the complement of any singleton set that is a member of the finite closed topology, the complement of that singleton set is infinite. – dc3rd May 1 '15 at 21:47
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Algebra Level 5
$$\displaystyle x^{4} + ax^{3} + bx^{2} + cx + d = 0$$
With $$a,b,c,d \in \mathbb R$$
Has 4 non real roots, two with the sum $$3 + 4i$$ and the other two with product $$13 + i$$.
Then find the value of b.
Also try JEE Quadratic
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# As part of a game, four people each must secretly choose an
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As part of a game, four people each must secretly choose an [#permalink]
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As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?
A. 9%
B. 12%
C. 16%
D. 20%
E. 25%
[Reveal] Spoiler: OA
Originally posted by tarek99 on 30 Nov 2007, 08:57.
Last edited by Bunuel on 02 Mar 2012, 23:50, edited 1 time in total.
Edited the question
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30 Nov 2007, 09:14
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The first guy can choose anything
chance of success 1/1
Second guy can choose any three of the four numbers
chance of success 3/4
Third guy can choose any two of the four numbers
chance of success 2/4
Last guy can choose only one of the four numbers
chance of success 1/4
Prob: 1 x 3/4 x 2/4 x 1/4 = 6/64 = 9%
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01 Dec 2007, 03:02
out of curiosity guys, sometimes there are questions that we will have to look for the opposite and then subtract that answer from 1. but how come it doesn't work with this problem? for example:
what is the probability that they will choose the same number instead of different numbers:
4/4 * 1/4 * 1/4 * 1/4 = 1/64
so 1 - 1/64 = 63/64 which is not even close to 9%. how come this doesn't work? cause i always confuse between when to use the opposite and when not. anybody?
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01 Dec 2007, 16:33
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tarek99 wrote:
out of curiosity guys, sometimes there are questions that we will have to look for the opposite and then subtract that answer from 1. but how come it doesn't work with this problem? for example:
what is the probability that they will choose the same number instead of different numbers:
4/4 * 1/4 * 1/4 * 1/4 = 1/64
so 1 - 1/64 = 63/64 which is not even close to 9%. how come this doesn't work? cause i always confuse between when to use the opposite and when not. anybody?
Because
(1 - everyone chooses same number) "Not equal to" (everyone chooses different number)
Infact , (1 - everyone chooses same number) "equal to" (everyone chooses not the same number)
But it still means 2 or 3 people can choose the same number, just not all 4.
Hope it helps..
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Updated on: 24 Dec 2007, 09:12
Possible combinations of each choosing a diffrent number (favorable events):
4*3*2*1=24
Total possible combinations (total events): 4*4*4*4=256
Favorable events/total events = 24/256 approx 10%
Is this a correct approach?
Originally posted by cstefanita on 24 Dec 2007, 07:28.
Last edited by cstefanita on 24 Dec 2007, 09:12, edited 1 time in total.
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24 Dec 2007, 08:08
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Raffie wrote:
The first guy can choose anything
chance of success 1/1
Second guy can choose any three of the four numbers
chance of success 3/4
Third guy can choose any two of the four numbers
chance of success 2/4
Last guy can choose only one of the four numbers
chance of success 1/4
Prob: 1 x 3/4 x 2/4 x 1/4 = 6/64 = 9%
Excellent and quite fast.
Quote:
(1 - everyone chooses same number) "Not equal to" (everyone chooses different number)
Infact , (1 - everyone chooses same number) "equal to" (everyone chooses not the same number)
But it still means 2 or 3 people can choose the same number, just not all 4.
This is also possible, if we can write the combinations for the cases in which same number is selected by 4 people and then by 3 people and then by 2 people, and subtract the sum total from 1.
People are A,B,C,and D.
Number are 1,2,3,and 4
If 4 people select the same number:
number selected is 1:
A can choose it in 1/4 ways
B can choose it in 1/4 ways
C can choose it in 1/4 ways
D can choose it in 1/4 ways
Therefore: (1/4)*(1/4)*(1/4)*(1/4) = (1/4)^4
Other numbers which can be selected are 2,3,and 4
Therefore total number of ways = 4* [ (1/4)^4] = (1/4)^3 = 1/64
If 3 people select the same number:
number selected is 1:
A can choose it in 1/4 ways
B can choose it in 1/4 ways
C can choose it in 1/4 ways
D can choose any other numbers in 3/4 ways
Therefore: (1/4)*(1/4)*(1/4)*(3/4) = (3/4^4)
Other numbers which can be selected are 2,3,and 4
Therefore total number of ways = 4*(3/4^4) = 3/(4)^3 =3/64
If 2 people select the same number:
number selected is 1:
A can choose it in 1/4 ways
B can choose it in 3/4 ways
C can choose it in 2/4 ways
D can choose any other numbers in 3/4 ways
Therefore: (1/4)*(1/4)*(3/4)*(2/4) = (6/4^4)
Other numbers which can be selected are 2,3,and 4
Therefore total number of ways = 4*(6/4^4) = 6/(4)^3 =6/64
Ohhhh I think I messed up somewhere!!!!
but I guess it's possible this way too!
[/u]
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10 Jan 2008, 06:03
very quickly: total area of probability equals 4*4*4*4=4^4
at the numerator: 4! (if the first choose one number, there are 3 possible alternatives remaining...)
so, 4!/4^4=0.093....
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10 Jan 2008, 10:09
favourable outcomes=4!
total outcomes=4^4
Probality=4!/4^4=3/32= aprox 9%
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01 Sep 2008, 01:09
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arjtryarjtry wrote:
suppose the question is modified to the probability that all of them select the same no.???
give detailed working...
p for all of them select 1 = 1/4^4
p for all of them select 1 or 2 or 3 or 4 = 4*1/4^4= 1/4^3
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07 Sep 2008, 01:03
arjtryarjtry wrote:
suppose the question is modified to the probability that all of them select the same no.???
give detailed working...
(1/4*1/4*1/4*1/4)*4 = 4/256 = 1/64
Guys let me know if I am missing something here.
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27 Sep 2009, 01:59
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?
a) 9%
b) 12%
c) 16%
d) 20%
e) 25%
Soln:
Chances that all four choose different numbers is
Assuming first person chooses any of 4 numbers, second can choose the 1 from the left over 3, third can choose one number from the left over two and the last person chooses the last number. Hence
4 * 3 * 2 * 1 = 24 ways
Total possible chances is 4 * 4 * 4 * 4 = 256
Thus probability is = (24/256) * 100 = 9%
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22 Apr 2010, 00:18
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As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?
a) 9%
b) 12%
c) 16%
d) 20%
e) 25%
Ans:a)
1st person has option no's- (1,2,3,4) - there fore probability of getting a no = 4c1/4c1 = 1
2nd person has option no's any three ,
he has to choose a no from three no's - there fore probability of getting a no = 3c1/4c1 = 3/4
3rd person has option no's any two ,
he has to choose a no from two no's -there fore probability of getting a no = 2c1/4c1 = 1/2
4th person has only one option - there fore probability of getting a no
= 1c1/4c1 = 1/4
=1*3/4*1/2*1/4 = 3/32 = 9%
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As part of a game, four people each must secretly choose an [#permalink]
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08 Sep 2010, 04:48
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tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?
a) 9%
b) 12%
c) 16%
d) 20%
e) 25%
SOLUTIONS FOR ALL SCENARIOS
When four people choose an integer between 1 and 4, inclusive 5 cases are possible:
A. All choose different numbers - {a,b,c,d};
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c};
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b};
D. 3 people choose same number - {a,a,a,b};
E. All choose same number - {a,a,a,a}.
Some notes before solving:
As only these 5 cases are possible then the sum of their individual probabilities must be 1: $$P(A)+P(B)+P(C)+P(D)+P(E)=1$$
$$Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}$$
As each person has 4 options, integers from 1 to 4, inclusive, thus denominator, total # of outcomes would be 4^4 for all cases.
A. All choose different numbers - {a,b,c,d}:
$$P(A)=\frac{4!}{4^4}=\frac{24}{256}$$.
# of ways to "assign" four different objects (numbers 1, 2, 3, and 4) to 4 persons is 4!.
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:
$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.
$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b}:
$$P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}$$.
$$C^2_4$$ - # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b};
$$\frac{4!}{2!2!}$$ - # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons;
D. 3 people choose same number - {a,a,a,b}:
$$P(D)=\frac{C^3_4*4*3}{4^4}=\frac{48}{256}$$.
$$C^3_4$$ - # of ways to choose which 3 persons out of 4 will have same number;
$$4$$ - # of ways to choose which number it will be;
$$3$$ - options for 4th person.
E. All choose same number - {a,a,a,a}:
$$P(E)=\frac{4}{4^4}=\frac{4}{256}$$.
$$4$$ - options for the number which will be the same.
Checking: $$P(A)+P(B)+P(C)+P(D)+P(E)=\frac{24}{256}+\frac{144}{256}+\frac{36}{256}+\frac{48}{256}+\frac{4}{256}=1$$.
Hope it's clear.
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10 Sep 2010, 12:33
Hi Bunuel,
For the case C, I am going wrong somewhere. Could you point out where please?
# of ways to chose 2 people from the group = 4C2
# of ways for this group to select one number out of 4 numbers = 4
# of ways to select 2 people out of remaining 2 = 2C2
# of ways for this group to select a number from the remaining 3 = 3
Hence total number of ways = 4C2*4*2C2*3 = 72.
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10 Sep 2010, 13:13
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jainsaurabh wrote:
Hi Bunuel,
For the case C, I am going wrong somewhere. Could you point out where please?
# of ways to chose 2 people from the group = 4C2
# of ways for this group to select one number out of 4 numbers = 4
# of ways to select 2 people out of remaining 2 = 2C2
# of ways for this group to select a number from the remaining 3 = 3
Hence total number of ways = 4C2*4*2C2*3 = 72.
# of ways to divide group of 4 into two groups of 2 when order of the groups does not matter is $$\frac{C^2_4*C^2_2}{2!}$$ and then you can do 4*3 for numbers;
But if you do just $$C^2_4*C^2_2$$ then you get the # of divisions of group of 4 into two groups of 2 when the order of the groups matters (you'll have group XY and also group YX with this fromula) then for numbers you should use $$C^2_4$$.
So your formula needs to be divided by 2! in any case to get rid of the duplications.
Check the following links for more on this issue:
combination-anthony-and-michael-sit-on-the-six-member-87081.html?hilit=dividing#p767453
probability-88685.html?hilit=factorial%20teams
ways-to-divide-99053.html?hilit=factorial%20teams
combinations-problems-95344.html?hilit=factorial%20teams
Hope it's clear.
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10 Sep 2010, 13:14
Bunuel wrote:
jainsaurabh wrote:
Hi Bunuel,
For the case C, I am going wrong somewhere. Could you point out where please?
# of ways to chose 2 people from the group = 4C2
# of ways for this group to select one number out of 4 numbers = 4
# of ways to select 2 people out of remaining 2 = 2C2
# of ways for this group to select a number from the remaining 3 = 3
Hence total number of ways = 4C2*4*2C2*3 = 72.
# of ways to divide group of 4 into two groups of 2 when order of the groups does not matter is $$\frac{C^2_4*C^2_2}{2!}$$ and then you can do 4*3 for numbers;
But if you do just $$C^2_4*C^2_2$$ then you get the # of divisions of group of 4 into two groups of 2 when the order of the groups matters (you'll have group XY and also group YX with this fromula) then for numbers you should use $$C^2_4$$.
So your formula needs to be divided by 2! in any case to get rid of the duplications.
Check the following links for more on this issue:
combination-anthony-and-michael-sit-on-the-six-member-87081.html?hilit=dividing#p767453
probability-88685.html?hilit=factorial%20teams
ways-to-divide-99053.html?hilit=factorial%20teams
combinations-problems-95344.html?hilit=factorial%20teams
Hope it's clear.
Thanks, makes a lot more sense looking at it this way.
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10 Sep 2010, 13:26
Thanks for the reply Bunuel !
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Bunuel wrote:
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?
a) 9%
b) 12%
c) 16%
d) 20%
e) 25%
SOLUTIONS FOR ALL SCENARIOS
When four people choose an integer between 1 and 4, inclusive 5 cases are possible:
A. All choose different numbers - {a,b,c,d};
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c};
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b};
D. 3 people choose same number - {a,a,a,b};
E. All choose same number - {a,a,a,a}.
Some notes before solving:
As only these 5 cases are possible then the sum of their individual probabilities must be 1: $$P(A)+P(B)+P(C)+P(D)+P(E)=1$$
$$Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}$$
As each person has 4 options, integers from 1 to 4, inclusive, thus denominator, total # of outcomes would be 4^4 for all cases.
A. All choose different numbers - {a,b,c,d}:
$$P(A)=\frac{4!}{4^4}=\frac{24}{256}$$.
# of ways to "assign" four different objects (numbers 1, 2, 3, and 4) to 4 persons is 4!.
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:
$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.
$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b}:
$$P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}$$.
$$C^2_4$$ - # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b};
$$\frac{4!}{2!2!}$$ - # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons;
D. 3 people choose same number - {a,a,a,b}:
$$P(D)=\frac{C^3_4*4*3}{4^4}=\frac{48}{256}$$.
$$C^3_4$$ - # of ways to choose which 3 persons out of 4 will have same number;
$$4$$ - # of ways to choose which number it will be;
$$3$$ - options for 4th person.
E. All choose same number - {a,a,a,a}:
$$P(E)=\frac{4}{4^4}=\frac{4}{256}$$.
$$4$$ - options for the number which will be the same.
Checking: $$P(A)+P(B)+P(C)+P(D)+P(E)=\frac{24}{256}+\frac{144}{256}+\frac{36}{256}+\frac{48}{256}+\frac{4}{256}=1$$.
Hope it's clear.
Way cool. Wish I could bookmark just this response. Many times, I bookmark a topic, but forget that it was actually a constituent post way way down that really prompted me to do so...
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Re: As part of a game, four people each must secretly choose an [#permalink]
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28 Apr 2012, 12:38
Exactly 2 people choose the same number and other 2 choose different numbers - {a,a,b,c}: 4*4*3*2/4^4 what's wrong with it, plz explain?
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Posts: 125
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28 Apr 2012, 18:34
Bunuel wrote:
tarek99 wrote:
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?
a) 9%
b) 12%
c) 16%
d) 20%
e) 25%
SOLUTIONS FOR ALL SCENARIOS
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:
$$P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}$$.
$$C^2_4$$ - # of ways to choose which 2 persons will have the same number;
$$4$$ - # of ways to choose which number it will be;
$$P^2_3$$ - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b}:
$$P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}$$.
$$C^2_4$$ - # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b};
$$\frac{4!}{2!2!}$$ - # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons;
Bunuel,
I am teribbly sorry to revive this old post but I am still trying to find out to out compute options (B) and (C) that you have laid out.
For (B) this was my thought process- - {a,a,b,c}:
(4C1)(4C2)(3C1)(2C1)(2C1)(1C1)=288 ( You got 144, which is half my answer, why do I need to divide by 2?)
(# of ways to pick the paired number)(# of ways to place the pair within the 4 slots) (# of ways to pick the first non-pair number)(# of ways to place first non-pair number in the remaining two slots)(# of ways to pick the second non-pair number)(# of ways to place the second non-pair in the last slot)
For (C) this was my thought process - {a,a,b,b}:
(4C1)(4C2)(3C1)(2C2)=72
(# of ways to pick the first pair)(# of ways to place first pair number in the 4 slots)(# of ways to pick the second pair )(# of ways to place second pair in the remaining 2 slots)
What am I doing wrong here? Help Bunuel!
Thank you!
Re: PS: Probability II [#permalink] 28 Apr 2012, 18:34
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Could anyone please point to one or more websites where is possible to download a working implementation of a #SAT solver? I'm interested in those returning the exact solution count, not an approximation.
• Hi Walter, your question is close to the border of what would be officially "on-topic" for this site. However, if you have nowhere else to ask this question and we can answer it, perhaps it's not that bad... (Since the site is still under development, I think we are being more open than other sites may be.) Rest assured that the point of this comment is not to "scold" or "warn", it is just a friendly notice. – Ryan Williams Sep 14 '10 at 19:57
• Hi Ryan, thanks for your notice. I'm sorry if this question is close to the border. I've searched on the web and I didn't find anything: only some SAT solvers, but no #SAT solvers. That's why I've asked here. Of course I know that I can write my own code which uses a SAT solver as an engine to count solutions, but I was looking for something already made and ready to use. – Giorgio Camerani Sep 15 '10 at 7:48
• I'd like to disagree. I think such questions are within scope, and should be ! – Suresh Venkat Sep 15 '10 at 8:05
• agree its in scope. fyi/imho its not too practical to build a #SAT solver from a SAT solver unless one has the source code and even in that case, not so practical, except for very small formulas, because of a very bad exponential blowup. usually special techniques unique to #SAT and not SAT would be required... – vzn Mar 3 '13 at 18:40
You can do this with SAT4J, simply by iterating over all models: http://www.sat4j.org/howto.php#models. I imagine that most SAT solvers have this ability.
• Hi supercooldave, thanks for your pointer. I didn't know that SAT4J had this ability. – Giorgio Camerani Sep 15 '10 at 7:51
You can also try the #SAT solver "sharpSAT" (website, github) for counting the number of satisfying assignments of CNF formulas.
One option is to use a BDD library, such as JavaBDD. All such libraries either have a function that counts solutions fast or, at least, they make it easy to write such a function. The disadvantage, however, is that constructing the BDD will be slow in many cases and may require much memory.
In case your input is in CNF, a simple heuristic that speeds up the construction of the BDD is the following. First, build a small BDD for each clause and put them into a priority queue whose root is the smallest BDD. Second, pop two BDDs, compute AND between them and push the result to the priority queue. Here's the idea: Since computing AND between BDDs of size $m$ and $n$ takes $O(mn)$ in theory but $\sim m+n$ in practice, minimizing the runtime is the same as finding a Huffman code.
Related topic: Best SAT Solver.
• Thanks Sadeq. The topic you indicated seems to be theoretical-oriented. It lists several papers on decreasing the upper bound. It's very interesting, but I was looking for a ready-to-use working implementation. – Giorgio Camerani Sep 15 '10 at 7:56
• You are most welcome. Among the links cited there, there was one which was purely practical: satcompetition.org. I think you can find very good implementations there. – M.S. Dousti Sep 15 '10 at 14:07
The best I found is "c2d compiler". http://reasoning.cs.ucla.edu/c2d/
It uses d-DNNF and you need the -count option.
• c2d solves much more CNFs than sharpsat. For toy purposes the "relsat" sat solver will do too. – Leon Leon Dec 31 '10 at 12:47
The MBound Solver given here http://www.cs.cornell.edu/~sabhar/ can give model counts with probabilistic guarantees. It's much faster than enumerating all solutions.
I wrote a small model/prime implicant enumerator. This can already be used for model counting with the model enumeration but that's not very practical. If anybody's interested, I can extend it so it counts models from prime implicants.
Here is one called tensorCSP and based on a tool called tensor networks. It is explained in this paper.
The website BeyondNP contains a good inventory of the existing tools to solve #SAT (and other related hard problems on CNF formulas). You may also find a list of tools for approximate model counting and knowledge compilation (the task of transforming the CNF into a hopefully succinct data structure that often supports polynomial time model counting).
You may also find a list of tools for preprocessing CNF formulas which may be useful to improve the performances of model counters and various benchmarks.
Glucose is a very efficient SAT solver developed at university of Bordeaux.
https://www.labri.fr/perso/lsimon/glucose/
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# Parts XXIX – END
## Part XXIX
The journey to The Stith where the stealth-scout picked them up and led them to the hidden jumpgate was fairly uneventful, although long. Most pilots were either staying docked to avoid the cross-fire or were in the fighting themselves, so there was little traffic to notice them and their hurried journey into unregulated space. There had been little conversation along the way, Arouin and Yyanis both lost in thought, Arouin maintaining a steady gaze out of the cockpit and into the depths of space beyond.
They jumped through into the hidden sector with it´s secret TRI base, the glow from the jump fading as Arouin smoothly accelerated the Phoenix out of the gate, then swung it around to head for the station. His eyes flickered away from the HUD, looking out across the void, and he lost his breath.
“What the fuck are those?” he said, eyes wide. Yyanis had been lying down on the floor, trying to get some rest. She pulled herself to her feet, rubbed her eyes, then walked to the front of the cockpit, yawning with her hand to her mouth as she looked out. She froze, mid-yawn.
“I don´t have a clue… I´ve never seen ships like them before.” she said, hand slowly falling back to her side. One of the ships sped over to them, trailing three drive trails, and blazed past them with a flash of darkness. As they flew slowly towards the station, it doubled-back and flew up alongside them, holding formation, and they got a good look at it. Matte-black painted, all angles, no curves, a faint shimmer where there might have been plexi-glass for a cockpit canopy. It was a double-pronged sliver of darkness, meanacing, shaped like a delta-wing, with two leading points, vicious-looking.
“I want one.” murmured Arouin, staring sideways out of the cockpit at the object of his desire.
“Arouin?” said Yyanis, tapping him on the shoulder.
“What?” he asked, somewhat dreamily.
“We´re going to crash into the station unless you do something about it.” she pointed out. Arouin snapped out of it, turned, and sat back down at the pilot´s chair. The dark spacecraft peeled off, three drive trails flaring brightly as it headed away from them.
“This is the _Lithe Shadow_ to docking control, requesting clearance to land.”
“This is docking control, you´re cleared. Welcome back, Lithe Shadow
The docking rings sprang into life as they coasted towards the station. Arouin bursted the thrust a few times to steady them, then pushed the throttle gently forwards, easing them into the darkness.
The Director was waiting for them as the lift brought them up into the hangar bay, standing perfectly still amongst a group of heavily-armed security guards who looked around the area with unease. Arouin hit the canopy release before the lift had stopped moving, then climbed down the ladder quickly, giving Yyanis a hand as she took the last step down to the floor, bundle under one arm.
“You have it?” said the Director, smiling.
“We have it” said Arouin, nodding. Yyanis walked forward and passed the bundle to the Director, who took off the jacket which it was wrapped in and dropped it distastefully on the floor before gazing, eyes seemingly alight, at his new prize. Yyanis took a step forward and picked her jacket off of the floor, annoyed.
“I am indebted to you. We will organise your respective payment methods over the next few days, during which you are invited to stay onboard the station as our guests. We will, of course, also need to debrief you, lay out what you cannot say about your time doing work for… TRI.” Yyanis nodded, Arouin remained silent for a moment before speaking up,
“What are those ships out there?” The Director looked uncomfortable for a moment before replying,
“I´m sorry that´s classified information. Need-to-know basis. And you don´t need to know.”
Arouin looked pained.
“Can I buy one?” The Director laughed,
“I´m afraid not, Mr Simmel, no matter how much you wish for it to happen it´s not going to, sorry to shatter any illusions you may have been harbouring about receiving one of those spacecraft in payment.” The Director turned and walked away, followed by his security contingent. After a few moments, one of the men who had stayed behind walked up to them and nodded to each in turn.
“If you will follow me, I´ll show you to your quarters.”
“Yes, of course” said Yyanis, smiling. Arouin grunted his approval then followed the man as he walked away. Yyanis fell into step beside him.
“You don´t seem too happy” she remarked.
“Why should I?” asked Arouin, glumly, “Back to being hunted after a few days of operations, no doubt. There´s a ship out there which can beat mine into a bloody pulp by the looks of it, at least. It´s been exciting, made a change from my normal way of life, doing something for good instead of just myself. Meeting new people.” He smiled at Yyanis, then looked away slowly.
“You don´t have to stop ´meeting new people´, you know” she said as they walked through the corridors, a smile on her lips.
“Sooner or later I´m going to get killed. By what or who I don´t know, but sooner or later I will. I´ve gone too far along this road now, I can´t just turn back. Even if I stop my various illegal activities, there are some people who won´t care that I no longer have a criminal record. They want me dead, it just so happened that before the law was on their side. Now it´s against them, but that won´t stop them, I´ve made too many enemies.” Their guide stopped in front of a pair of doors.
“Here are your rooms.” he said in a neutral tone of voice.
“Thanks” said Arouin, nodding to the man and opening the door in front of him. “I need sleep. Lots of sleep. I´ll… see you later.” He walked inside, sat down on the bed, and took off his boots, setting them on the floor, then swung his legs up onto the matress. Arouin closed his eyes, and let himself drift into unconcious bliss.
Arouin awoke and lay there in comforting darkness for a few moments, eyes shut against the world. An electronic chime from the door sounded, faint and insignificant in his still-fading dreamscape of dark nothingness. It sounded again, and Arouin opened his eyes slowly, seeing the ceiling, ghostly white, barely iluminated by the dimmed lighting in the room. He slowly rolled himself out of the bed and onto his feet, then padded softly over to the door where he rubbed his eyes with one hand before opening it. Yyanis was stood there.
“They thought it was about time to wake you up, they were going to send someone else, but I figured I should do it.” she said, by way of explanation. Arouin ran a hand through his hair then rubbed his face again.
“A good choice. I feel lousy. Anyone else and I might have bitten their head off, verbally or otherwise.” he said, chuckling quietly to himself, “How long have I been out for?”
“Thirty nine hours” said Yyanis, with a wry smile. Arouin yawned, putting the back of his hand to his mouth.
“Thirty nine hours? And they didn´t drug me or anything?”
“Apparently not”
“Wow.” he said, looking impressed, “That´s pretty damned good for me.”
“Making up for lost time I think.”
“I guess so” he said, nodding, “what have we got in store for us today? Whenever today might be, that is.”
“The Director wants to debrief us, let us know when we´re getting our goodies.”
“Oh joy. Sounds like such fun. Well, I guess we´d better move.”
Arouin stepped outside the door, brushing past Yyanis, and closed it behind him before stopping and turning around again. He opened the door, stepped inside, then put on his boots.
“That´s better” he said to Yyanis, who took his arm and started off down the corridor. A couple of security guards appeared behind them and kept pace.
“What´s with the shadow?” he whispered in her ear.
“I guess there are some things they don´t want us seeing” she suggested as they rounded the last corner and stopped in front of the door into The Director´s office. Arouin knocked smartly on the door, then took a step back until the reply came from within,
Arouin and Yyanis stood and walked out of the office, Yyanis closing the door behind them.
“Heavy stuff…” muttered Arouin as they walked away.
“Yes… but I get my operation in twelve hours” she said, smiling. Arouin looked back at her, grinning,
“Free at last, hey?”
“Yeah, I suppose so.”
“The Director´s quite a guy, you know… the only time I´ve actually seen him happy is when he got his hands on that positron coupling. I wish I knew more about him…” said Arouin, musing quietly. Yyanis looked at him, surprised,
“You didn´t check out any information on him before you took the job?”
“I was strapped to a bed surrounded by armed guards and if I _didn´t_ take the job I was going to be shot.”
“Oh,” said Yyanis, looking slightly put out, “okay. Well, I had the leisure of a slightly more relaxed recruitment, and I did quite a bit of research into him.”
“And?” asked Arouin as they walked into a cafeteria.
“It was hard to find much on him, mainly bits from old news transmissions, databases and the like. Probably the most interesting thing about him is that he´s from Hyperial.” At this Arouin turned around to stare at her, eyes wide.
“Hyperial?!” he said in surprise, pausing half-way through sitting down at a bench.
“So it seems. Since the war between Quantar and Hyperial, the Hyperions have had little or no spacefaring activity, while remaining fairly advanced. The war´s over now, of course, and they needed someone to head up TRI. It was thought that since a Hyperion would have no factionalist tendancies, since there weren´t any of his faction´s ships in space, so they would be fair to all the other factions. Obviously they might be a little covertly hostile to the Quantar, but nothing that couldn´t be spotted and easily patched over.” Arouin rested his elbows on the table and put one hand to each temple, slumping forwards slightly, lost in thought.
“Yyanis… could you do something for me?” Yyanis looked at him dubiously.
“I guess so,” she said cautiously, smiling slightly, “what?”
“I need you to try and get access to the station´s files and find out what those ships are.”
“Oh.” she said, deflating slightly, “Okay, I guess so.”
“Want something to eat?” asked Arouin.
“Sure, just get me whatever you´re having.
“Soy-beef soup sound okay to you?”
“Yeah I guess so.”
Arouin walked over to the vending machine, then returned a few moments later with their meals, which they sat and ate together in silence.
A couple of hours later, they were sat in Yyanis´ quarters, looking at her terminal screen as her hands flew over the keyboard.
“Have you got any idea what we´re looking for?” she asked.
“Not really… check anything you can relating to hangar bay operations, refueling, I don´t know… something that would store ID codes or ship specifications. I want to find out what that ship can do… I might have to fight one someday.” Lists of files and directories scrolled past the screen, illuminating Yyanis´ face in the dimmed lighting of the room.
“Hmm… they´ve got an automated factory set up here.”
“What´s it making?”
“What´s that got to do with the ships?”
“Call it natural inquisitiveness” said Arouin, grinning. Yyanis´ hands tapped out a sequence of keys, and a batch production listing came up on the screen.
“´X-Type Power System´, got five hundred counts. Nothing else.” she shrugged.
“The new system… the one we retrieved the technology for?” mused Arouin.
“Probably right… they must have worked like demons. The batch started a day ago… so yeah, it could be the new system.”
“Is it linked with any others files?”
“Let´s see… there´s a file for the X-Type specs, I´ll pull that up.” There as a pause as the data loaded to the terminal, then Yyanis nodded, “Very good guess, dead on. You can see from the blueprints, there are the three pieces of tech that we picked up.” She pointed to each in turn on the screen..
“Why only five hundred?” asked Arouin, to no-one in particular. Yyanis shrugged,
“I don´t know, but you wanted to find out about those ships. I´ll take a look in the hangar bay log files. Hangar bay one or two?” Arouin looked up in alarm,
“There´s a hangar bay two?”
“I tell no lies, look for yourself.”
“Hangar bay two then, I guess.” As Yyanis´ hands flickered over the keyboard, there was a knock at the door.
“Go away.” said Arouin loudly, still staring at the screen.
“There´s four hundred and eighty-three ships docked there!” she said in amazement, “It must be huge!”
The knock at the door came again, considerably louder.
Go. Away.” said Arouin, louder.
“Open the door now, damnit.” came a muffled reply. Yyanis was staring at the screen, opening another file.
“They´re… they´re…”
“What, damnit?” asked Arouin, still looking at the door.
“Hyperial, Xavier-class Heavy Interceptors. Five hundred of them are currently stationed here.” she lapsed into silence. The knocking at the door stopped, then came a thump on it as someone hit it hard. And again.
“Five hundred…” breathed Arouin, looking down at Yyanis, “…Hyperial Ships…”
“…with a Hyperion as director of TRI…” she gulped, “…a fleet that no-one knows about…”
The hammering at the door intensified, the metal buckled slightly.
“…equipped with brand new power plants that are god-knows how many times more powerful than anything before, and smaller to boot… …more shields…. …more guns… …more engine power…” said Arouin, mouth agape.
The door buckled further.
“…and after two hundred years of playing second fiddle… …they might just feel it´s time for them to take the limelight.”
Ladies and Gentlemen!” cried Arouin theatrically, springing to his feet and pulling out a pair of Bullroarers from his trenchcoat as a pair of shots were fired through the door and it was kicked in, “We are officially in shit!
## Part XXX
Arouin didn´t hesitate for a moment longer, the gun blazed in his hand and punched gaping holes in the metal and plastic, light from the corridor outside streaming them into the darkened room within. There was a burst of automatic fire from outside and a line of bullet-holes stitched themselves along the wall, Arouin dropping to one knee as the blind firing from outside went high, sending bullets screaming through the air right above his head.
“Get out of here!” he shouted at Yyanis, face a grimace as he carried on pounding at the trigger, the muzzle flashes lighting up his features with a firey yellow glow. Yyanis looked around desperately for a moment searching for anything that could help them, before her eyes travelled upwards and came to rest on one of the ceiling panels. Arouin fired another two rounds into the door then scrambled on his hands and knees over the the desk where Yyanis was sheltering, dropping a clip out of his pistol and quickly replacing it as he did so.
“We can´t stay in here,” he said, the click of the new clip slamming home punctuating the sentance, “they can just throw a grneade in or something and we´re dead and,” Arouin popped up and took a quick look over the desk, “I can´t get us out of here with brute force.”
“The ceiling, look”, said Yyanis, pointing up at the roof. Arouin looked up at the panelling above, then back down at Yyanis.
“I´ll keep them busy, you get up there and get ready to help me up.” he said, looking unsettled. Yyanis nodded quickly. “Three, two, one…”
Arouin lept to his feet, his other hand drawing out another pistol, then moved across the width of the room, guns alternately belching flame and shaking from the recoil, gunshots ringing out and mingling with the scream of someone outside the room. As he ran across the room a pair of assault rifles opened up outside and the lines of bullet-holes rushed across the wall as he moved, trying to catch up with his fleeting form. Yyanis jumped up onto the desk and struck out at the ceiling panel above her as Arouin slammed boldily into the far wall, fired another couple of shots from his pistols as the lines of bullet-holes closed the distance to him, then flung himself into a roll under the twinned streams of death that flew over his head. Yyanis hoisted herself up into the pitch black space above the ceiling as Arouin crawled the remaining few paces to the desk, the roar from outside not lessening as the pair of assault rifles carried on with a steady barrage of ammunition at the point which Arouin had last shot from. Larger bits of the wall were being flung into the room as the bullets tore into it, then the pieces were hit again as they fell sending them jinking around as sucessive bullets smashed into them, light streaming through the increasing number of holes that they left. Arouin lept onto the desk, then jumped up and caught hold of the edge of the hole before hauling himself up into the space above with Yyanis´ help. It was hot and dark, the ceiling panels visible only as glowing outlines where light was let through the thinner edges of the panels and imperfect joins. The corridors were clearly visible from above, the tops of them poking up through the ceiling to make miniature walls about half a meter high; the result was a low-height carbon copy of the floor below. Yyanis was standing balanced on a pair of struts which kept the ceiling panels up, holding onto a metal pipe that ran overhead.
“Are these safe to walk on?” asked Arouin, quietly, pointing at the ceiling tiles.
“I don´t actually know” said Yyanis, looking dubiously at them, “I haven´t tried yet.” There was a shout from below and a crash as the door was kicked off of it´s hinges.
“Now seems like a good time to find out” said Arouin, grabbing Yyanis firmly by the arm and starting to run across the ceiling tiles, which buckled as their feet hit them, but held. They ran through the hot darkness, making low, fast jumps over the tops of the walls, ducking under the occasional pipe and girder, until Arouin stopped suddenly. Yyanis almost fell over, but regained her footing, and stood there breathing heavily, exhasted. She brushed some sweat-matted hair out of her eyes before looking at Arouin.
“What now?”
“Oh nothing much,” said Arouin, checking the clips in each of the pistols he was holding, “we´re just going to pay someone a visit.”
He pointed one pistol at the ceiling panel he was standing on, and pulled the trigger twice. The panel disintegrated and Arouin dropped through the gap, landing heavily on the floor in the room below, bits of panelling cascading down around him. He sprang up from the crouch he had dropped into, and span, a gun in each hand, to face the grey-haired man behind the desk. The Director sat there, alone in his office, looking defenceless, yet unafraid. He was smiling slightly, in a distant way, and held Arouin´s gaze a moment before saying anything.
“I don´t know what you hope to achieve by coming here.” said a quiet, tired voice, “it´s not as if you can make any difference any more.”
Arouin remained impassive, staring him down, taking in the view on a pair of raised terminal screens, security forces running through the corridors that they had run over the top of, guns still trained on the older man
“You´ve played me for a fool…” said Arouin, shaking his head slowly, pistols raised, eyebrows knotted.
“It was necessary,” said the Director, in a not unkind tone, “I didn´t enjoy it, but you were the only man for the job. I needed you, and I don´t think anything could have convinced you to help me if I hadn´t tricked you.”
“And what about everyone else?” said Arouin, gesturing around with a pistol, “what about everyone else on this damn station?”
“Quite a few of them are in on the plot,” said the Director in a neutral tone, smiling slightly, “the rest, well, they will have to co-operate.”
Arouin shook his head, and slowly raised his right hand, knuckles white around the cold metal of the pistols.
“It won´t help.” said the Director, patiently, “You might as well accept what´s going to happen, it´s not as if you can stop it. There are five hundred ships with the new power system docked in the hangars, nothing can stand up against that, not even you…” the Director smiled thinly at this, then continued, “Aren´t you interested in the larger picture, don´t you want to know what´s going to happen next? We´ve got untold power, Hyperial can unite TRI like nothing else has managed to… You must admit that the whole plan is far in excess of anything that´s ever been attempted before, you must, at the very least, want to know the details of the scheme.”
“Not really, I think I know enough.”
The blast from the pistol sent the Director and the chair he was sitting on careening back into the wall behind him, where it bounced off and toppled over onto one side. The Director looked down in surprise at the red stain spreading over the front of his shirt, touched it with his finger, then drew it away quickly, as though the pain from the wound was nothing, yet the sensation of the blood on his fingers was a burning agony. A pair of booted feet slowly made their way across this horizontal vision of the floor, and he looked up along them, up to the face of an angered man.
“You´re not supposed… to do that…” he managed, an incredulous look on his face as he tried to push himself up off the debris-strewn carpet.
“If you think I´m going to sit around all day why you cackle madly over the fine details of your plot for galaxy-wide domination, you´re sorely mistaken.” Arouin said grimly, bringing the gun to bear on the Director´s forehead, “I´ve got work to do.”
The crack of the gunshot rang out, echoed quietly, there was a barely-audible thump as the body hit the floor for one last time, then silence flooded back into the room.
Arouin turned and looked up into the space above the ceiling at Yyanis´ white face.
“Let´s go”
Yyanis nodded slowly, then lowered herself down from the ceiling above.
“Where first?” she asked
There was a knock at the door, and a moment´s pause, then Arouin´s hand whipped back up and put three rounds through the metal and plastic panelling. There was a split-second of silence, then the tinkle of the cartridge cases hitting the floor and the thud of a body hitting the ground outside the door. Arouin walked over to it, and kicked the door open, a sickening fleshy noise coming from the body of the security guard who had been outside the door as it was crushed against the wall. Arouin strode out of the door without giving it a second glance, but Yyanis gave the cooling body a wide berth as she walked by.
“We have a few of our own objectives now,” explained Arouin, striding forcefully down the corridor, one pistol held out in front of him as he walked, “we need to take out 500 fighters of a completely unknown type and power, then get the hell out of here, preferably alive.” They went around a corner in the corridor and came face to face with a pair of running security guards, Arouin barely blinking as he pumped at the trigger, sending the two men sprawling backwards to the floor screaming, then walked past them without breaking his stride.
“Any suggestions?” he asked as they approached a T-junction and the sound of running feet approached.
“Uh, none right now” said Yyanis, drawing a pistol from her pocket. Arouin dropped to one knee as the first security guard sprinted around the corner, his eyes widening when he saw them, then he tried to slow down and bring his weapon to bear, but too late. Arouin´s pistols barked, flame gouting from their barrels, and the guard jerked spasmodically as the bullets tore through him before succumbing to gravity and falling to the floor. Yyanis heard something behind them and span suddenly, just in time to see two guards coming around the corner of the corridor.
“Behind us!” she shouted, bringing the pistol to bear and opening fire. One of the guards dropped to the floor just in time to avoid the first shot, but his colleague behind him took the bullet in the face, sending him sprawling into the wall behind him. Yyanis shifted her aim downwards and dropped into a crouch as a burst of automatic rifle fire zipped over her shoulder, squeezing off three quick shots which tore into the torso of the man, spinning him like a top before he hit the floor. Arouin was still firing as more guards came around the corner, firing their weapons, the heat and smoke in the corridor building up rapidly and the clamour of battle ringing down the hallway. Through the haze Arouin saw a rifle swing around the corner, the owner out of sight. He reached back to Yyanis who was franticly reloading, and shouted a warning,
Get down!“, he screamed, pushing her out the way before diving forward as the rifle opened up. Bullets rained down the corrdior, hitting the metal walls and sending sparks flying off them which arced through the air and rained down on Yyanis´ prone form. Arouin hit the ground hard and slid along the corridor floor, slick with blood, into the corridor at the end of the T-junction. The guard blind-firing the assault rifle looked down with surprise as Arouin slid into view, then was blown up into the air by a flurry of rounds from the blazing akimbo pistols. Another guard appeared through the haze from the other side, looked down to see Arouin lying on the floor, mid-reload, and brought his rifle up, only for Yyanis´ blazing pistol to send bullets smashing through him, shaking his body and spraying blood on the wall.
“Nice one,” said Arouin, rolling back onto his feet and wiping a fleck of someone else´s blood from his brow, then looking around warily. He finished reloading, then prodded one of the nearby bodies with one foot. It rolled over, revealing the glassy stare of a body with no soul left in it, bereft of life, then rolled back when Arouin took his foot away. Yyanis was flat against one side of the corridor, back against the wall, looking down the length of the hallway, pistol held tightly in one hand and breathing heavily.
“I don´t think I can take much more of this” she said, glancing up and down the hallway again. Arouin, sweat dripping from his brow, looked at her through the haze,
“It´s not a question of what you can and cannot take, it´s a simple question of whether we´re going to survive this or not.”
“I just want to get out of here” she said, shaking her head slowly.
“Until those ships are destroyed,” said Arouin, starting to walk down the corridor again, “we aren´t going anywhere.”
“How _are_ you going to destroy five hundred ships?” said Yyanis, still feeling jumpy, glancing nervously from side to side.
“I´m not to sure on that point myself,” he admitted, “I´m hoping something will present itself.”
“So… you´re _relying_ on good luck?” she said incredulously as they took a turn into the corridor which led to the second hangar.
“Hasn´t let me down so far, has it?” he replied, closing the distance to the door into the hangar with purposful strides.
“Luck runs out.”
“Sometimes you have to make your own luck.”
“Doesn´t sound like an exact science to me, not something I´d like to stake my life on.”
Arouin stopped just before the door, and turned to face Yyanis.
“It´s not like we have a choice… how do you fancy your chances trying to outrun five hundred ships?”
“I suppose so, it´s just that sometimes, just occasionally, it´s nice to maintain the illusion that we´re working to some kind of plan here.”
“Never mind eh?” he said, turning back to the door, pushing it open, and walking through into the hangar. Arouin stood still and gazed around the cavernous space, ceiling veiled in darkness, the floor a stark grey, and the ships, stretching off into the distance, in neat, eye-straining rows. Black, venomous, deadly, they looked although they were the essence of speed and destruction condensed into material form. Yyanis walked up behind Arouin and put one hand on his shoulder, the other still holding her pistol.
“You´re busy wondering if they´ve got a self-destruct mechanism?” she said, airily.
“Uh… something like that.”
“I think not.”
“Neither do I,” he admitted, “but I do have an idea.”
“Is it a good idea?”
“No, it´s a stupid idea.” he said, gazing out over the rows of gleaming black fighters.
“But it might just work?”
“Yeah…” he said, “it´d better…”
## Part XXXI
Arouin ran. The avenue of glistening black fighters stretched away ahead of him, gleaming where the widely spaced overhead lights illuminated them, reflected light contrasting sharply with the darkness around. Lights flashed past above him as he ran, Yyanis following, feet hammering on the hangar floor.
“Any hints on what this idea is?” she managed between laboured breaths.
“We´re headed… for Docking… Control… for… this… hangar bay.” he said slowly, pausing for breath every couple of words. As the wall ahead of them neared, details revealed themselves in the half-light of the gloomy hangar, rivets studding the wall, fire extinguisers, safety signs, a door… The door slowly swung away from the wall, light trickling out through the narrow crack, becoming a cascade of illumination as the door opened fully. The shadow silhouetted resolved into a uniformed figure, passing through the portal, turning to close the door behind him, hearing the footsteps. He turned.
“Who´s th…” he managed before the crack of Arouin´s pistol sounded out across the hangar, echoing and reverberating amongst the insectile hulks of the black spacecraft. The man fell backwards, Arouin sprinting over his falling body without breaking his stride, crashing through the door and into the room beyond. Yyanis lept over the man´s corpse and looked up to see Arouin swing left, then right. There was a scream, Arouin´s right hand tracked a movement that was out of sight, then the pistol kicked in his hand once, twice, three times. Another movement, another brief blaze of fire as Arouin emptied the rest of the clip. Yyanis ran past him as he dropped to one knee and reloaded his pistol, pulling another one out of a pocket and repeating the process. Gun in hand, movements cautious and near-silent, she crept crab-wise into the next room, not sparing a glance for the blood-stained wall behind the table where the two technicians had been eating their meal. She sprang through into the next room, dropped into a crouch, quickly swung right, left, then looked back to Arouin.
“It´s clear – there´s a stairway here” she said, tersely. Arouin walked through the doorway, glancing around and taking in the contents of the room; a few storage lockers, a vending machine and a rack of coat-hooks. Yyanis moved closer to the bottom of the stairs, pistol grasped in front of her with two hands, covering the top.
“That should be the way up to the control deck” said Arouin, nodding, “Shouldn´t be too many people up there – it´s not in an active state.”
Arouin walked past Yyanis, guns held out in front of him, her pistol trained over his shoulder. He took the stairs carefully, one step at a time, keeping his gaze fixed at the top of the stairs. At the last step he paused for a moment, then leapt through into the room beyond the doorway at the top. There was a pregnant pause, complete silence for a few seconds and Yyanis held her breath, waiting for the blasts of firearms, but none came. Arouin´s head appeared again at the top of the stairs. Yyanis let out her breath and her gun dropped slightly.
“It´s clear, come on up. Looks like there´s no-one here.”
“let´s hope it stays that way,” said Yyanis, “I´m running out of ammo if nothing else.”
Arouin stepped through into the control room, Yyanis running up the last few stairs to catch up. The room was dimly lit, the better to see the glowing multitude of screens covering three walls and a control console in front of the wide window looking out over the hangar bay that made up the fourth. Five swivel chairs were scattered in front of the control console, head-sets hooked over the seat-backs, consoles before them blinking. Arouin relaxed after a few more seconds and looked around at the monitors showing a vast array of figures, diagrams and graphs. Then he turned to Yyanis. Yyanis, who had been following his gaze, took a step backwards and held out a hand.
“Don´t look at me!” she said hastily, “I don´t know what that lot means.”
“I don´t care what it means, I just want you to launch the fighters.” replied Arouin, calmly.
What?!” she said incredulously, but before Arouin had time to reply there was a scream from down below. Arouin turned to look at the stairs, then looked back at Yyanis. There was a pause of three or four seconds, then the alarm went off. The speakers situated in the corners of the room hummed into life, whined with feedback for a moment, then started blaring out,
“Hangar Bay Two is being sabotaged! All pilots and security personnel report to Hangar Bay Two!”
“Oh dear” murmured Arouin, putting one hand to his forehead. There was the sound of running feet; someone coming up the stairs. Yyanis looked at him, face a mask of horror.
All?! Isn´t that a _tad_ excessive?!”
Arouin gave her a withering stare for a moment as one hand came up with a pistol in it, pointing at the doorway. The other pointed at one of the consoles.
“Just do it. We haven´t got time to argue.”
There was a sudden intake of breath from behind Arouin as he looked at Yyanis, and without looking he pumped at the trigger, bullets slamming into the worker who´d found the bodies and raised the alarm.
“Now.”
Yyanis paused for a second, there was a thump as the body behind Arouin hit the floor, out of sight. She looked at him.
“Okay,” she conceeded, “tell me the plan as I go.”
Yyanis walked over to one of the consoles and sat herself down at it, peering intently at the slightly flickering screen.
“Pretty simplistic,” she said as her fingers danced across the keyboard, Arouin standing silently behind her, gaze alternating from the doorway to the window. “Of course you want the system to be simple,” she commented after a few seconds, “It wouldn´t do to have anyone too smart around here for extended periods of time. They might go digging, find out a few uncomfortable home truths. Get a little too nosy…” Arouin wasn´t moving any more; wasn´t paying attention to Yyanis´ idle chatter, just stood, staring out of the window.
“We may have a problem” he said slowly and deliberately. Yyanis looked up from the command console and stared through the glass. In the distance men ran from entrances in the hangar wall towards the ships.
“Launch the ones furthest from us first,” Arouin said urgently, “and fast.”
The line of ships furthest from them sank into the hangar floor, lift doors closing over them, sealing them away from the pilots who ran to try and board their ships. Yyanis looked up brightly.
“Oh, I get it! You launch the ships…”
“…with no pilots in them, then destroy them at will as they have to take time to ship pilots out to them. That takes about an hour a ship.” Arouin finished for her, a satisfied tone in his voice.
“That long?” said Yyanis, looking up at him, surprised, hands still tapping away at the keys.
“Oh yes,” he said, noticable relish in his voice, “modern ships don´t have airlocks and things like that – just a seal around the escape pod. Absolute hell to get a pilot into a drifting ship, risky too.”
Yyanis looked out of the window as the men, ant-like in their distant movements, ran for the ships which disappeared before them.
“You have also, I take it, thought up some subtle and cunning ruse to get us out of the sitation we´re about to find ourselves in?” she said, after a few more moments. Arouin looked slightly uncomfortable.
“Not as such, no.” he admitted after a few seconds.
“Start thinking?”
Arouin was still staring out of the window.
“The ships aren´t launching fast enough – the system can get the ships ready for launch faster than the launching system is sending them into space… I´m going to try and buy us some time. We can´t let any of these get away, and some of the pilots´ll get to their ships if they carry on at the rate they are.”
Yyanis looked unsure.
“Uh, are you sure that´s absol…” she started, but Arouin was gone, feet thundering down the stairs. As she looked out of the window, fingers a blur as she send ships, pilotless, out into the void, she saw Arouin running out towards the nearing line of pilots.
The line of grey-fatigued pilots ran with all their might towards their ships which were disappearing before them, a row at a time. The ships, the pride of the rebuilt Hyperian Navy, slipped away from them just as they approached, but they gained ground, slowly, surely. Sweat ran in trickles down their anguished faces, screwed up with the effort of the continued strain. Feet blistered, boots blurred, breath laboured. The row before them vanished, revealing the row behind, which vanished, revealing the row behind, which vanished, revealing the row behind, which vanished, revealing a black-trenchcoat wearing man, akimbo pistols gripped in white-knuckled fists, eyes glinting in the grey light of the hangar.
Feet planted firmly apart, a sleek, black weapon in either hand, trenchcoat billowing behind him in the wind created by the vacuum of space greedily accepting the ships of his enemies, Arouin opened fire. His hands tracked across the width of the hangar, trigger fingers spasming rhythmically, guns blazing with fire and noise, cartridge cases fountaining up in a shower of metal. Pilots twitched, knocked this way and that as bullets slammed home, falling like sickening dominos of flesh and blood and bone, striking the ground and lying there, immobile, as their life leaked away from them. The guns ran dry, the scream and whine of bullets ceased, there was a moment´s silence as the stream of pilots stared around them at the carnage wrought by Arouin, their friends and comrades scattered on the hard ground before them, bleeding, screaming, dying. Arouin span, coat swirling around him, and broke into a run, dropping the clips out of the pistols, sprinting for all his worth towards the disappearing line of ships. Bullets whined past him as his feet pounded on the grey metal floor, driving him ever onward. Of the next line of ships to sink into the floor before him, one did not, a safe haven of darkness. Arouin ran past it, then squatted on it´s far side, back against the cold metal, gasping for breath.
“Thank you Yyanis” me muttered, relieved. After a few moments of rest he hauled himself to his feet once more and started running again, reloading his pistols as he went. There was a shout behind him of an unfriendly nature, and he redoubled his pace, heading straight for the door into docking control. The welcoming rectangle of light drew nearer, erratic weapon fire starting anew behind him, bullets striking the wall and floor before him. With a quick glance back, Arouin squeezed off a couple of shots and was rewarded with a scream of pain and one of the pilots collapsing to the floor, screaming, clutching at his leg. He turned back again to see a silhouette taking up most of the doorway, raising a gun in one hand, bringing it to bear…
Get out the way!” screaned Yyanis, Arouin quickly zipping off to one side as the gun in her hand flamed, snapped around to cover another target, blazed again and again as bullets screamed out and ripped into Arouin´s pursuers, knocking some from their feet in sprays of blood. Arouin almost knocked Yyanis to the floor as he sprinted through the door, then snapped around and put his back to the wall, reloading the single pistol he had used as Yyanis continued firing out of the doorway.
“How many ships got away?” Arouin shouted at her over the bark of the pistol. It ran dry, the trigger clicking twice before Yyanis sprang out of the way, changed the clip, then ran for the stairs.
“Only a couple, come on!” she yelled, running up the stairs, “The good news is it doesnt matter…”
Arouin reached the top of the stairs and ran into the control room.
“What?” he said, puzzled and out of breath.
“After the first three rows the ships started colliding into the ones that had already been launched. They´re one big lump of twisted, cooling metal outside the launch tubes.”
Arouin smiled wordlessly and turned back to the stairs.
“Any work on the getting-out-of-here-alive front?” she asked, scathingly.
“Getting out, no. Staying alive, yes.” said Arouin, grabbing her by the arm and dragging her over to the console.
## Part XXXII
They moved swiftly with practiced co-ordination and grace, cautious yet confident, weapons held tightly, aim unshaking. The fire team moved up the stairs, leap frogging their positions; one man running forward as his partner covered him, the reversing the roles, the first man standing frozen, alert, as his buddy ran past. There was a pause, a silence laden with the promise of violence, then a short hand signal from the lead. Afer a brief flurry of movement a stun grenade ricocheted off the wall and into the control room. Light blazed out for a split second, white, intense, throwing the room into sharp relief, then a thump as the secondary explosive went off, deafening and disorientating those inside. A thunder of feet preceeded the fire team as they stormed into the room, weapons sweeping theeir assigned sectors with geometric prescision. But silence fell; no clatter of automatic weapons, no screams of the dying. The fire team stood stock still, eyes flickering around the room, looking for a trap, a hidden figure, a suggestion of movement. The room seemed untouched, uninhabited, five seats were all tucked underneath the overhanging control console, monitor command prompts blinking balefully in the darkness. Scattered over the floor were four small screws, dull in the grey, dim light that the monitors cast over the room. Then, in the corner of the room, a glimmer of reflected light danced over polished metal. Instantly, as one man, the fire team turned on the glint, guns raised. The leader made a quick gesture with one hand, and one of the other troops moved cautiously forward, one careful step after another. When he was a meter or two away he crouched down and leant forward for a better look.
“It´s a grate,” he said, voice seeming unnaturally loud in the hushed silence of the room, “they´re in the ventilation shafts.”
The leader of the team shook his head slightly and grimaced,
“Cunning bastards… Hoskins and Turunae, stay here, make sure they don´t double back. Everyone else with me, seal off all the exits then we´ll move in, closing off sections as we sweep them. Call in the medic teams to clear up this mess.”
And with that, all save two of the troops left the room…
…The medic teams arrived less than a minute after the radio call to tell them that the area was secure, moving amongst the dead and dying, sorting those who might be saved from those who would surely die. Over the next twenty minutes they moved across the hangar floor, until they finally reached docking control. One medic checked the body nearest to the door, fingers to the throat. Was that a pulse? There it was again…
“We´ve got a live one! Give me a hand over here!” he shouted, standing and waving, Off to his right one of his comrades lept to her feet.
“Another one here!” she yelled, looking around and then kneeling back down beside the young woman with the blonde hair as two pairs of men ran over, bringing stretchers with them…
…Hoskins lay on the floor, gun still trained on the vent, Turunae kneeling beside him, still, quiet. They stayed there for ten minutes before Turunae broke the silence.
Hoskins shifted his weight around slightly before replying,
“He took out the Director.”
“Shit… why?” said Turunae, concern and surprise in his voice.
“He wasn´t briefed on the… Hyperian aspect of the project.”
“He wasn´t told? He wasn´t with us?”
“Nope, all behind his back. They bought him off with some equipment and didn´t give him the full story”
“…And when he went off the rails no-one tried to stop him?” asked Turunae, shocked.
“You obviously haven´t glanced over this guy´s CV…”
…”Okay, we´ve sealed all of the vent exits wuth regular security detatchments,” said Jarren, the fire team leader, pacing the briefing room, “Now it´s up to us. We need to move through the vent system and flush them out. They´re heavily armed and very, _very_ dangerous, so we´ll be moving in pairs. We can´t afford to let him get away. We´ve already lost the existing ships because of his actions, and the Director´s been killed. It´s up to us to avenge his death and stop this madman from doing any more damage. Command says there´s another one hundred ships en route that they had been saving for follow-up operations. When they get here we´ll re-start production of the new power systems, and then proceed with the plan. You know your places, we have work to do.” He stood up, and looked over the assembled faces of his men, “Move out troops…”
…The medics ran through the corridors towards the med-lab, pusing the two injured base personnel on trollies in front of them.
“He seems stable, no obvious blood loss or physical injuries save for a few bruises here and there. Get an oxygen mask on him anyway, won´t do him any harm, might do some good.”
The medic that had been looking into the eyes of the blankly staring man ran ahead of the trolley and further down the corridor to the next one, where they woman lay under the white sheets.
“What about her?” he enquired, a concerned expression on his face.
“Stable, dilated pupils, again no obvious signs of physical injury. Probably psychological trauma.”
As the woman stared, unfocused at the ceiling above, lights passed overhead, glaringly bright. An oxygen mask from somewhere out of sight moved across her line of vision and settled over her face. Back down the corridor a medic meant over the man in the trolley.
“He´s snapping out of it!”
The man´s eyes focused on the medic´s concerned face.
“You are very lucky to be alive my friend.” said the medic, with a smile. The face of the man looking back up at him broke into a grin.
“I guess I am…”
“…This is Droden in sector A5. No movement.” The trooper stared ahead, tense, listening for the replies of the rest of the fire team.
“Pinero, sector A6. Nada.”
“Chaz, backing up Pinero. Fuck all at A6, over.”
“Murver, B9, wth Genin. We´ve got nothin´ here.”
Droden moved forward a little more, bent almost double in the cramped vent. Behind him his partner, a sleight man by the name of Shelton, or ´Shorty´ to his friends, moved up as well.
“Bet you´re loving this Shorty.” said the taller man, grinning.
“Not small enough, Dro´” he said in a high-pitched, whiney voice, “I´m just short enough that I think I can stand up. Just tall enough to crack my head on the bloody ceiling.”
Droden laughed to himself quietly, but stopped suddenly when his radio crackled into life.
“This is Pinero, A7, we´ve got movement, we´ve got morement!”
Droden gripped his weapon more tightly and started moving forwards…
“…and then he blasted his way out of there, took on three Typhoons by himself in a Phoenix.” finished Hoskins with relish. Turunae was wide-eyed.
“No-one seems to know her name”, said Hoskins with a grin, “but from the camera footage I´ve seen she´s a nice piece of work.”
Turunae laughed heartily at this, but didn´t relax his unwavering aim on the enterance into the vent system.
“Well, I really don´t think that taking her alive is going to be an option.” he said with a smile. Hoskins grinned a stupid grin,
“You never know, she might try and get us to spare her if…”
There was a metallick clunk from the vent, and both men shut up immediately, concentrating on the vent, their weapons, their breathing slowing, measured. Another clunking noise came from the vent. And another…
…The medics moved the two injured personnel over to normal beds, kept their oxygen masks on, brought them water, and monitored them. After five or ten minutes of nervously glancing around the man lying in the bed beckoned to a nurse.
“Could I… have my gun with me… I feel so unsafe… so vunerable…”
A pleading look entered his eyes as the nurse frowned, looking dubiously at the man under the bedsheets.
“I´ll have to ask Doctor Phaar,” he said, guardedly, then turned slowly, looking over his shoulder at the man in the bed, and then walked off. A few minutes later the nurse found Phaar sitting behind his desk behind a pile of patient charts, looking them over. After the man´s request was put to him, Phaar looked at the nurse with a quizzical stare.
“Well, it would no doubt help his psychological state, but it´s a bit dangerous…” The nurse nodded in agreement with the older man as he spoke.
“I think we can take care of it though.”
“Hmmm… okay. I trust you know what to do with the weapon before handing it over.” The doctor grinned.
“Of course,” said the nurse, looking slightly put out, “I´m not stupid you know.”
A few minutes later, the nurse handed the cut-down rifle to the man who sat upright in his bed to accept the weapon. He backed off and gave a somewhat distasteful look at it before turning and walking off.
“Thanks,” said the man to the retreating back of the nurse, smiling, “I feel better already…”
“…It´s close, it´s fucking close!” came the voice over the radio.
“Stay cool, stay cool, we´re on our way…” came Droden´s steady voice.
“Two traces… they´re closing!”
Droden ran down the vent, doubled over, the metal sides a blur as he rushed through the enclosed space, Shelton moving fast behind him, staying close, weapon raised.
“Stay cool, we´re almost there!” he said, running.
“They´re right on-fucking-top of us!” came a paniced cry from his radio…
“…well, he´s asleep now. The gun wasn´t loaded anyway, not that it mattered – he dropped off minutes after he had it”
The medic looked at the sleeping man.
“And what about the woman?”
“Diz Therran, a technician from sector V6, never seen combat before; scared out of her wits…”
…Droden stormed around the corner, Shelton close behind, to see two muzzles pointing directly at him. His mind immediately went into overdrive, assessing the situation, his reactions kicked in instantaneously, hurling himself to the ground. The guns before him blazed, fire licking out of the muzzles, bullets ripping overhead, smashing into the sides of the vent, sparks dancing in the air, falling to the ground in a shower of light. Shelton jerked spasmodically as the lumps of metal tore through his body, twitching, his finger jerked on the trigger as he fell and the limply-held gun blazed away, recoil spraying it around the inside of the vent, light and sound merging into a maelstrom of death.
Cease fire, cease fire!!” screamed Droden, looking up into the smoking barrels before him, “Pinero… you fucking idiot man… you´ve killed Shelton… you… you´ve been reading us on the tracker… you… fuck…”
…Turunae winced at the carnage and curses over the radio, and turned to look at Hoskins. His face was pale, eyes wide, aghast at what had happened. There was another clunk from the vent, and both men looked back at it, both shaking slightly. Turunae put a hand up to his ear.
“Is anyone in sector C3? Sector C3?” he asked the squad, anxiously, gun in one hand, wavering slightly, still aimed at the vent.
“Uh… yeah man… It´s Garris… we´ve, uh… taken a wrong turn somewhere. We´re gonna head back now. Shit, man, Shelton´s dead…”
Turunae shook his head, and visibly relaxed a little. Hoskins sighed with relief, clambered to his feet, then stretched, arms held high above his head as he yawned.
“That was fucking close buddy,” he said, quietly, “That was just too fucking close… almost ended up wasting Garris and his number two… Shelton dead… it´s all screwed up.” Hoskins nodded in silent agreement.
Behind the two soldiers, in the dark space between the swivel chairs and the command console, there was a hint of movement, and gunmetal gleamed coldly in the grey light.
## Part XXXIII
In the silence of the docking control room, in the darkness under the control console, metal scraped on metal; slowly, delicately, the sound barely audible from a few centimeters. As the thread turned the black, sleek cylinder edged down to eventually connect with the grey-silver gleaming metal of the gun, joining firmly, mated perfectly. Arouin smiled, teeth an ivory row of small, pale slabs seeming to glow softly in the darkness for a moment before disappearing again. His heand moved slowly, swinging around in the shadows to bring the deadly burden it carried to bear on the closer of the two troops who still watched the vent before them intently. The gun came to rest, deadly snout aiming for the back of the kneeling man´s head. He sawyed back and forth slightly, fiddling with the strap that his weapon was slung on. The hand with the pistol shifted back and forth with the man´s movement. Arouin relaxed, breathing in time with his unsuspecting target. The finger resting on the trigger shifted, tensed, squeezed. Despite the silencer the sound of the pistol firing seemed loud in the still air of the room, the sharp report muffled and reduced to a dull thump which seemed to carry abnormally well. The man gave a quiet grunt as the back of his head exploded, throwing him forwards and onto the back of his comrade who lay on the floor, closer to the vent. Blood spilt over the floor and sprayed up into the air before arcing back down and splattering onto the ground making a sound like rain.
“What the fuck?!” shouted the man on the floor, writhing around to look back over his shoulder and into the face of his friend, blood running over the man´s frozen features. His eyes went wide, face a mask of panic, incomprehension and fear. Out of one corner of his eye he saw a movement within the darkness, a steely glint of light. Reactions honed through ten years of training and simulated combat instantaneously took over, legs thrust out to try and get him away from the body of his partner, weapon arm already swinging around to combat the threat, finger jamming down on the trigger as the gun swang around. The assault rifle blazed, cartridge cases spewing up into the air, muzzle flash describing an arc of yellow flame as it curved around. Arouin´s pistol snapped off two quick shots, the first shot slapping wetly into the top of the man´s forehead, sending his head snapping back as if he´d been hit in the face with a metal pole, blood and spittle spraying out. The body arced backwards, the second shot catching it in the shoulder and knocking it back into the air, spinning it around one and half times before it crashed back to the floor with a thud, face down. There was a moment´s pause before Arouin rolled off of the pair of seats he had been lying on underneath the console. Yyanis rolled off the three swivel chairs she´d been on and landed awkwardly, winding herself slightly. Arouin crawled out from the cramped space and watched Yyanis critically as she did likewise, then picked herself up off the floor.
“Why did you get three chairs?” he said, sullenly.
“It was your idea, I would have suggested something more comfortable,” she said, grinning, “besides, I´m a lady. You´re meant to be gracious and polite to me remember?”
“Uh… yeah. Plan worked though didn´t it? One hundred percent success in my opinion; we´re both alive.”
“Not the most original I´ve heard; ´Hide under the bed and hope they won´t notice us´”
“Don´t knock it,” he said dryly, “it worked.”
“Great, we´re still alive. We´re also still trapped in this damn station, so let´s work on the next bit; getting the hell out of here.”
“I´m taking out the production facility first.” he said firmly, checking his pockets to see how much ammo he had left.
“You´re taking out the production facility.” she said levelly.
“Yeah.”
“Well, I haven´t worked out the technicalities yet, but we can burn that bridge when we come to it.”
“You´re the only person I´ve ever met who actually relies upon luck. Couldn´t we just get rid of the blueprints of the power system? It would have the same overall effect; they couldn´t make anyway.” she pointed out.
“Yeah,” said Arouin nodding slowly, “I guess so… but how can we be sure that all trace of the thing´s been removed?”
“Just a matter of getting high level access then deleting all copies of the blueprints on the database. Sod it,” she said thoughtfully, “we could just nuke the whole d-base.”
“You´d better rephrase that ´we´,” said Arouin, glancing around, “if you think I can do anything to do with breaking into a secure system you´re woefully mistaken.”
“Well yes, I can do it. I´ll need core access though.”
“What? Can´t you do it from here?” Arouin pointed offhandedly towards the command console with a pistol.
“No; they´ve got better security than that. I need to physically get to the core to do anything as far-reaching as deleting the whole database. Even then they´ll have backups and safeguards.”
“A self destruct mechanism, perhaps?” said Arouin looking hopeful. Yyanis looked at him blankly.
“This is not a holo series. People don´t build self destruct mechanisms into highly expensive computer systems.”
“Okay, okay,” he said, holding his hands up and spreading them apart, “just a suggestion.” He looked down at the bodies again. “Whatever we do, we´ve got to do it fast; someone´ll notice they haven´t reported in sooner or later. Judging by today, probably sooner.” Arouin stepped over the bodies and walked out of the room, Yyanis followed him out of the room and down the stairs, checking the clip in her gun as she did so. He paused just inside the door which led out into the hangar, stuck his head out into the cavernous space, then looked left and right. The hangar stretched away from them, seeming to have grown by several orders of magnitude since the last time they had run through it. It was more brightly lit and completely devoid of cover; the floor featureless and flat save for cartridge cases and bloodstains scattered over it´s enormous expanse.
“Looks like the clean-up crew´s gone… no-one on the hangar floor.” He withdrew and looked back at Yyanis who was staring over his shoulder, “What do you think?”
“Eh?”
“Run or walk?” he said, rubbing the stuble on his chin. He hadn´t shaved for what seemed like a week, and thinking about it, yes it was probably a week.
“Running will attract attention if someone sees us… but, of course, we´re less likely to be seen on security cameras because we´ll get across it faster.”
“If no-one´s meant to be in the area at all then it won´t matter if we run or walk.”
“Hmmm…” said Yyanis, still gazing off into the distance.
“Fuck it, let´s run.”
“Okay” she said brightly, and walked over to the door. Arouin took one last all-encompassing look around the hangar, then started off across the expanse at a run. Side by side they ran, the distance before them steadily reducing, Arouin´s coat whipping around behind him, billowing in the wind. Arms and legs a blur they ran on. It took the best part of two minutes to cross the hangar; they arrived at the far side breathless, getting out of the hangar as fast as they could and into the comparitive safety of a camera-less corridor. They stood against the walls of the hallway, one on each side, backs against the cool walls, silently regarding one another for a few moments before letting their gazes drop in exhastion.
“No alarms.” said Yyanis eventually. Arouin looked up from the patch of floor he had been staring dumbly at.
“No.” he said simply, then a couple of moments later; “Not yet, at least.”
“I don´t think we were seen.”
“I hope we weren´t.” said Arouin, pulling himself upright and away from the wall. “We´ve got to move”. He loked back at Yyanis, who nodded slowly and pushed herself away from the wall. He drew a brace of pistols from his coat, and weighed them thoughfully in his hands before turning and starting to walk off down the corridor. After a few paces he stopped, then turned around and looked over his shoulder to Yyanis who was still standing there, checking over her pistol. She looked up and met his gaze.
“All right; I´m coming.” She said, slamming a clip home and re-cocking the pistol before setting off after Arouin. Their pace was measured and cautious, checking corners with quick, stealthy glances before rounding them; sneaking past rooms which sounds came from behind their doors.
“Where are we headed?” asked Yyanis after a couple of minutes of tension-laden pacing.
“The core system´s at the central station core; the main module.” replied Arouin after a few moments, “From what you say, that´s where you should be able to access everywhere from and get rid of all trace of this powerplant.”
“You have touching faith in my abilities.”
“You haven´t failed me yet.” he said, smiling distractedly.
“Any hint of a backup plan forming in that head of yours, just in case I can´t do it?”
“If you can´t do it electronically I´d have to find out a way of nuking the entire production facility using two kilos of ´tex, which is all I´ve got on me. There´s some more in the ship, but I really don´t fancy our chances of making it to the ship, to the core, then back to the ship again.”
Yyanis nodded silently, and they moved on.
The station seemed to be in a controlled state of panic, the occasional alarm going off, personnel in the corridors unsure of what to do, rushing around trying to find out what was going on. Arouin walked with hands in his pockets, guns grasped firmly in balled fists. They moved slightly faster when they started to find crowds of people rushing around; ducking and weaving amongst the bodies, keeping on the move lest someone recognise them. At one point Arouin lept to one side of the corridor and ducked into a shadowy alcove, dragging Yyanis with him. They paused in the confined space for a few seconds as a troop of security personnel thundered past, weapons readied for action, then moved on again again as soon as the coast was clear.
“Not far now” said Arouin steadily, glancing around warily. After another minute or so of walking they saw, suspended above their heads, a sign which read “Central Core”.
“Wonderful; signposts.” said Arouin with a grin.
“I guess we´re there” murmured Yyanis as they followed the direction the sign indicated, into a short corridor of a few metres in length. At the end of the corrdior there sat a forbiddingly large door, painted an off-white colour. Warning stripes covered a strip at the bottom of the door and a red-painted sign with white type gave a list of unpleasant things which would happen to anyone who wasn´t authorised to enter and tried to. A swipe card slot sat to one side of the door, a single orange light on it blinking slowly. Arouin regarded the door´s steely countenance for a moment before turning to Yyanis.
“Any ideas?”
Yyanis studied the door for a couple of seconds before looking sideways at Arouin.
“What about the ´tex?”
“We don´t know how thick the door is. Plus, of course, it´s probably a ´plug´ defence door – a blast from the outside will wedge it backwards and jam it there, preventing access.”
“So not the ´tex.”
“No, not the ´tex.”
Arouin took a couple of steps forwards and ran his hands along the seal around the outside of the door, fingertips seeking a purchase on the smooth metal. Seeminly satisfied there was nothing of use he stepped back again and cast his gaze over a variet of pipes, wires and conduits that ran over the walls and floor, and into the surround of the door. Yyanis looked at them.
“They wouldn´t be stupid enough to put the power for the door on the outside” she said.
Arouin deflated slightly.
“Any other ideas?” he said glumly.
Arouin continued staring blankly at the door. Outside in the corridor there was the sound of running feet. Arouin looked up, then pushed his back against the wall as a technician rounded the corner and ran down towards the door. The man looked at Arouin strangely for a moment, then leant past him and hammered on the door with the base of his fist. There was a click, a whine, and the door started opening. Arouin looked past the back of the technician´s head at Yyanis, who pulled a face and shrugged. The door finished opening with a chunky metallic thump and the technician stepped through, and turned to his left.
“Hi, sorry, forgot my card again.” he said smiling, then glanced back past the doorway at Arouin, who had turned to face him. “What´s with the new guards?”
“Guards?” said a voice. Arouin was already moving, lashing out with a fist at the technician´s face as he ran past him into the room. The blow sent the man sprawling backwards into the room and sliding along the floor. Arouin span to his left, gun in hand, and fired into the face of a burly security guard who sat there, mouth open, coffee in hand.
“Wha..?” he managed before the gun went off, the blast knocking him back off the chair, his flailing hands dragging the coffee mug and a magazine he had been reading with him to the floor. Arouin glanced around the room quickly. It was octagonal and double-layered; the centre of the room was sunk into the floor, filled with grey-black boxes with status lights playing over their surfaces. The door Arouin had entered by was on the upper level, the walls of which were covered by terminals and other technical-looking equipment. Yyanis stepped through the doorway and looked around the room as the door gave a faint click and then hummed shut behind them. Arouin´s gaze fell to the man who lay at his feet, eyes closed, body slack, white face, and jaw set at a funny angle. Yyanis brought her pistol around to bear on the unconcious technician and looked at Arouin, face questioning. Arouin shook his head.
“No need.”
Yyanis nodded absent-mindedly, then turned her attention to the wall consoles. She walked over to the nearest one and ran one finger down the screen, stopping at the bottom and holding it there, tapping her fingertip against the glass. Arouin grabbed a chair and dragged it over to Yyanis, who sat down on it without looking, then grabbed another two and dragged them over to the door. The wheels shreiked and skittered over the blue-grey floor tiles. Arouin knocked one chair over, then moved it so that the seat back was underneath the back of the other chair. Then he carefully knealt down and aimed his pistols at the door, resting his arms on the seat, body covered by the backs of the two chairs. In the background Yyanis tapped away a the keys, abserbed in the task at hand.
“None of the terminals are logged in with high enough clearance to delete the backup files.” she said, after a minute or so.
“What does this mean?” called Arouin, gaze not shifting from the door.
“I can delete all the other copies – this terminal does have a high enough access level for that – but I can´t delete the remaining copy of it in backup unless I get can get complete control. It looks like the only way I can actually do that is try and bypass the security measures they´ve got installed…”
“You don´t sound very hopeful.”
“I´m not.”
Minutes passed, the silence of the room only broken by the tapping of keys and the whirring of fans. Suddenly Yyanis shouted with frustration and pushed herself away from the console with both hands, sending the chair squealing backwards across the floor.
“It´s secure! I can´t find any loopholes, no little bugs that can be exploited… doesn´t look like anyone´s been stupid enough to leave a log of entering the core system either… I can´t get into it.” she said, exasperated.
“Well… what else can we do?” said Arouin, looking around.
“I can try and crack it using brute force by feeding it a dictionary, if we can guess or find out a username to find a password for.”
“Sounds good, how long would that take?”
“Uh… could take a few weeks.”
Arouin closed his eyes, and rubbed his forehead with one hand.
“Yes. Very good. Yet not particularly useful.”
“No… I suppose not.” she conceeded. Arouin carried on looking around, then stopped suddenly and turned back to face Yyanis.
“The _only_ place it is stored is in the backup.”
“Yeah, everything else I can delete but they´ve got it set up so that you can´t delete the backup and if it´s needed it´ll just re-propogate through the system.”
“Propawhat?”
“Don´t worry,” said Yyanis in a tired tone, “yes this is the only place the plans are stored.”
“The only place…” he said, slowly, “in the entire universe… that these plans are stored… are in those boxes?”
Yes, dammit!” shouted Yyanis at him, “And I can´t do anything about it! Okay? Lay off for fuck´s sake!”
“Calm down… I can do it.”
“Oh you can, can you?” she said scathingly, turning around and grabbing the swivel chair from behind her. She pushed it towards Arouin. “Pray, have a seat and tell us mere mortals how it might be done.”
Arouin laughed at her angry expression, then dug a hand in one pocket, fished around in it for a moment, then drew out a large off-white slab.
“Do you really wish to delete this file?” he said with a grin and a glint in his eye. Yyanis held his gaze for a moment, then burst out laughing. She reached over to take the ´tex from his hand.
“Hell yeah.”
Arouin held onto the explosive for a moment, and a worried frown crossed his face.
“Will there be any debris left they could retrieve the data from?”
“No,” said Yyanis, shaking her head slightly, “The data´s stored on high density crystals. They´re fragile enough normally, never mind what´ll happen to them when a pile of ´tex goes off.”
She plucked the slab from his outstretched hand, then walked over to the gap in the railings which circled the drop to the floor below, turned around, then climbed down the ladder. Arouin turned back to face the door, waiting. Behind and below him there sounded the slow and measured footsteps of Yyanis packing round the storage stacks. Every few footsteps there was a pause and a quiet, plasticy noise, as of putty being slapped against a metal wall. The work took less than a minute, then there was the metallic ringing of Yyanis climbing back up the rungs of the ladder to the upper floor.
“Done.”
“Okay, let´s get out of here.” said Arouin, starting to get to his feet.
“Aren´t you forgetting something?” asked Yyanis, head cocked to one side.
“Uh…” Arouin paused and glanced around uncertainly.
“Detonators?”
“Ah, yeah, good point.”
Arouin fished around in a pocket for a second, then stuffed his hand into another, then patted down the length of the coat. Yyanis looked increasingly concerned.
“You have got the detonators, haven´t you.”
“Yeah… somewhere.”
Arouin checked a pocket inside the trenchcoat, and the worried expression that had been building slowly on his face vanished as he grinned.
“Got ´em.” he said, triumphantly, pulling a pair of metal cylinders out and tossing them one at a time over to Yyanis who caught them deftly. Arpuin turned his attention back to the door as she walked back over to the ladder and climbed down it. A few moments later Yyanis was back on the upper level.
“Looks clear.” he said, turning back to face Yyanis, who stepped around a console and into view. Arouin saw her tense, the gun in her hand leap up to aim at his face, and, as his mind raced, his body moved to avoid the shot, but to late. The gun flared, yellow-white flame licked out from the barrel, the edges tinged red-brow. He felt a ripple of air past one cheek and a stray strand of hair tugged at his hair as the noise of the shot washed over him. Then he felt a fine spray on the back of his neck. He froze, stooped slightly, watching smoke rise in whisps from the end of Yyanis´ gun, and slowly lifted one hand to the nape of his neck. He drew it back, and looked down at it, sticky and red with someone else´s blood. He looked back up at Yyanis, then turned around and looked down at a body with half it´s throat and lower jaw missing.
“Thanks” he said with feeling, staring down at the multilated body. Yyanis slowly untensed and stood up from the slight crouch she´d dropped into.
“Can we actually leave now?” she said, keeping a close eye on the door.
“Actually, while we´re here, there´s one more thing I´d like you to do.”
Yyanis looked at him warily.
“What?”
“Tell the hangar bay 1 systems to take the _Lithe Shadow_ into a refitting bay, then get the automatic systems to install item ref number…” he closed his eyes, “….3778359-A.”
Yyanis looked at him blankly for a moment, then snapped out of it.
“Okay.” she said with a shrug, then walked over to the closest console. She tapped out a quick squence of commands then took a step back.
“Done”
“_Now_ let´s get out of here.”
## Part XXXIV
The station shook; a rumble passed through it´s structure rattling those inside. Coffee mugs bounced along desks and desk lamps fell to the floor. People stopped in corridors and looked around uncertainly, wondering what caused the tremor. Inside the central operations complex, Jarren looked up.
“What the fuck was that?” he said, calmly.
The men gathered around him and the table he stood in front of glanced around at one another, faces carefully blank. The fire team had pulled back to the ops complex to try and find out where their prey had gone; for the last twenty minutes they had been studying reports overlaid onto the large electronic map of the station that was set into the central table in the room. After no-one moved for a couple of seconds one of the men at the table turned away and walked quickly over to a terminal. The tapping sounds of his fingers hitting the keys rang out curiously loud in the room, and although no-one turned to watch him it was quite obvious where everyone´s attention lay. The screen of the terminal that the man was typing at lit up blue with a smattering of white writing and an unintelligible error message. He paused for a moment, his face changing expression to one of increasing concern, before hammering out another sequence. The screen lit up blue again. He turned and walked back to the table, the colour draining from his face.
“The… uh… system core is offline.”
Jarren´s jam dropped a little way open and then hung there slackly.
What?!
“It´s down! It´s not responding… that blast must have been…”
“Oh shit.” murmured one of the other men, covering his eyes with his hands and rubbing his face.
“Check on the plans for the power system.” said Jarren hurriedly, walking around the table and following the man who turned once more and headed for the terminal again.
“Nothing in the database… backup d-base is down…” there was more insistent tapping, “…and all traces of Project Divinity have been removed from local d-bases.”
That bastard!” shouted Jarren, casting his eyes to the ground, balling one fist and hammering it into the open palm of the other hand. He looked up again, teeth gritted. “What… about… the prototype?” he said slowly.
“Uh… unknown sir… the logs show that it was removed from storage about three minutes ago…”
“He´s going.” said Jarren, lips forming a thin line. “He´s going and he´s taking it with him.” He swung around to face the officers gathered around the table. “Well don´t just stand there…” he said, arms stretching wide, hands splayed, “Get him!
Arouin and Yyanis burst into the refitting bay from one of the doors on the seventh level in the bay´s back wall. The catwalk they stood on was three meters wide and ran along all four of the interior walls of the bay, ladders leading down to the six floors below and up to the two floors above them. The lowest five levels of catwalks only covered three of the walls; the wall furthest from them had a massive five-story door set into it which led to the hangar proper. Arouin walked to the edge, boots scraping on the metal grille of the catwalk, then leant forward bracing himself against the rusting metal railing with both hands. Heavy chains stretched past them from the ceiling and the floors above, lifting equipment attached to the ends below them, magno-clamps and hooks swaying gently and casting their moving shadows on the form of the _Lithe Shadow_ which sat five stories below Arouin. A cradle held the Phoenix class fighter in the air, supported by two massive oil-covered metal cylinders which disappeared into a small, dark pit in the floor. Fluro-strips hung from the ceiling high above, chains holding them up with power cables twisted around them like snakes around a bough of a tree, disappearing off into the darkness above where their light was not cast. Water fell from on high, distant cooling units condenstating the moisture out of the air, the droplets falling down, glinting in the light of the fluro-strips, before splattering on the floor twenty meters below and the hulking mass of Arouin´s fighter. Droplets formed, ran together, then flowed over the brown-painted armour of the ships in rivulets of silver that glittered like rivers of mercury in the glare from the overhead lights. Dust motes glittered, catching the light as they rose and fell and swirled around, contrasting sharply with the dark walls beyond.
“Beautiful.” whispered Arouin under his breath, looking down upon the sleek lines of his ship. There were footsteps on metal behind him as Yyanis walked up to the railing.
“Let´s move.”
“Yup.” he grinned, still looking down, “Time to get out of here.” Arouin pushed himself upright, away from the railing, turned, and walked along the catwalk towards the nearest ladder downwards. He wove through the stacks of equipment and afterburner fuel, stepping over the occasional bit of piping or roll of cable that had been left lying around, and was only a few meters away from the ladder when there was a whine of servos and then the metallic thump of a door locking open. Arouin stopped slowly, and turned. Yyanis was already looking out across the refitting bay, at a now-open doorway one level above their own, set into the far bay wall. Two lines of troops ran out of the door, wearing black battle dress and plasteel combat jackets. They ran out along the catwalk, split evenly to each side of the door; twenty troops, ten each side. Another man walked out, hair greying slightly, grey eyes, an impassive expression on his face, a sub-machinegun hanging loosely from his right hand, a shotgun slung across his back. Slowly, his left arm came up, and one bony digit extended from his hand to point at Arouin.
“You have too much luck for one man, Mr.Simmel.” he shouted across the bay.
“Doesn´t look like it from here.” Arouin called back, then turned his head to Yyanis “Get ready to make a dash for the ship” he whispered urgently. The man on the far side of the refitting bay chuckled loudly.
“Very observant, Mr.Simmel, very observant. You don´t have, you´ve had, past tense, too much luck for one man. That luck has now run out.”
The sub-machinegun started to rise up in Jarren´s right hand, his left coming around to cradle the barrel just forward of the magazine. Arouin turned and pushed Yyanis out the way with both hands, throwing himself backwards behind a large packing crate who´s top was off revealing an array of expensive engine parts. Yyanis stumbled sideways, falling to the metal floor, bouncing, and then scrambling under the cover of a half stripped-down _Dream_ engine. The guns let rip. The plasteel crate Arouin was behind shuddered and quivered as bullets ploughed into it, shards of the material springing into the air and scattering across the metal grille of the catwalk, some pieces spilling over the edge and tumbling end over end down to the floor below. Arouin grabbed a pair of pistols, then squirmed around and up onto his feet, squatting on his haunches to keep his head down. Inside his head a counter was running.
25… 26… 27… 28… 29… 30 shots… 31… 32… 33… 34… 35
Phew, empty…
Four levels to go…
He looked across to the left side of the bay and saw Yyanis already a floor below him, her nearest pursuer two floors above, then turned back, grabbed the ladder before him, and quickly climbed down it. Arouin couldn´t see most of the troops any more; they were hidden by multiple catwalk layers,
But they´re there… and they´re still coming for me…
He glanced around the catwalk, then looked up suddenly when there was a rattling somewhere above him. In the distance there was a burst of machinegun fire, two pistol shots, and then a scream of pain. There was a rattle again from above him, and Arouin moved closer to the edge of the catwalk. He was about to lean out over the drop when out of the corner of one eye he saw a movement of something black and sleek. The darkness exploded into light again as someone three levels above at the back wall of the bay opened fire, flame from the muzzle flash lancing towards him. Arouin backpedaled quickly, firing his pistols as bullets ricocheted off of the handrail which hummed as the shots struck it. Arouin´s shots kicked up showers of sparks from the railing beside the black-clad figure, who quickly dropped back down out of view again. Arouin looked around, breathing laboured, and dropped to the ground beside a _Sport_ powerplant which was half-dismantled on the floor to reload. He dropped the clips out of both pistols, slammed a fresh clip into one, and was about to do the same with the second when there was a loud rattling noise right beside him. Arouin looked up to see one of the heavy chains shaking violently, water flying from it, and was half way to his feet again when a figure holding a sub-machinegun slid into view, holding onto the chain with one hand. Arouin launched himself out the way to his left, trying to put the dismantled powerplant between him and the other man, but he moved too slowly; the gun flamed, sweeping across to cover Arouin´s movement. In mid-air, Arouin felt the bullet smash into his left shoulder, spinning him around as he fell, his whole body shuddering with the impact. He hit the ground, shocked, and froze for a moment as the clamour of battle continued, face down to the deck, watching his own blood mingle with the water and drip down to the level below. There was a thump of someone landing on the catwalk, then the tread of heavy boots approaching. He felt a prod in his ribs, then a kick rolled him over onto his back, the pain adding to the cacophony of torment from his wounded shoulder. He watched the gun, hanging below from it´s owner´s hand, swing up, steady, and he looked down the darkness of the barrel which seemed to fill his vision. Arouin blinked, and focused on the face of the man who stood over him.
Bugger.
Slowly, it seemed, the man brought the gun up, nestling the stock in his shoulder, and sighted on Arouin´s forehead. Arouin looked on, watching water and sweat mingle and run down the mans face; he could see the stubble where he hadn´t shaved, a scar on one cheek from who knows what injury in the past, and brown, hate-filled eyes. There was a shout from the far side of the bay, and the man standing over him turned his head, water spraying off of his hair as his head snapped around. Arouin dreamily brought up his right hand, the pistol still in it, and sighted on the man´s temple. He snarled something at someone unseen across the bay, then turned back to face Arouin, who fired. The shot punched a hole in the middle of the man´s face, blood and bone shrapnel spraying out over the catwalk as his body was launched backwards into the air. Arouin threw himself forwards off the ground, still firing as the trooper´s body shuddered again and again before landing heavily on the deck. Arouin spat a bit of bone out of his mouth, dropped his pistol, then stooped over the body and grabbed the sub-machinegun, slinging it´s strap over his neck before reloading it as quickly as he could, one-handed. The _Lithe Shadow_ was now only two stories below. Arouin made a stumbling run to the next ladder and clambered awkwardly down it, gun clattering on the metal of the rungs as it bounced between the ladder and his chest. He looked out across the bay.
“Yyanis!” he shouted, straining his voice to it´s limit, shoulder screaming in agony, “We… are… leaving!”
He took a few steps back, then ran forward and leapt over the railing, fell almost three meters, and landed heavily on the slope of the Phoenix´s right wing. The surface was slippery, and he fought to keep on it, feet scrabbling over the wet metal. There was a metal series of high-pitched thunking noises and a row of small indentations in the metal sprang up half a metre from Arouin, followed by a shout and the echoing crackle of a carbine. He stuck one foot in the crack by an inspection hatch to brace himself, then twisted around and loosed off a one-handed burst from the sub towards the figure high above before scrambling over the upper surface of the Phoenix to it´s cockpit. He glanced up to see someone sighting at him along a laser carbine from four stories above, leaning far out over the railing to get a clear shot. As he started to shift his weight to try and get into a position where he could aim at the man there was a flurry of shots which split the man´s head open like an over-ripe melon. Arouin slapped his hand down on the canopy controls and the black glass raised up, water running off of it, to let Arouin scramble over the edge of the cockpit and into the pilot´s chair.
“Come on!” he screamed hoarsely to Yyanis who was two levels above and franticly scrabling down a ladder. He slumped back in the chair to try and use the control panel in front of him as cover, and rested the barrel of the sub-machinegun on the top edge of the radar display. One level above Yyanis, someone leant out over the edge of the catwalk to steal a quick glance at the situation, but moved too slowly; a burst of fire from Arouin sent them toppling backwards missing most of their neck. Yyanis made it to the level of the wing, ran, and jumped onto it. She almost slipped and fell off, but, with a great deal of frantic arm-windmilling, managed to regain her balance and run the handful of meters to the cockpit before falling over the side and landing heavily on the floor. Arouin slapped the canopy close button and sprayed gunfire from the sub at whatever he could see until it ran dry, before throwing it out of the cockpit just before the canopy clicked shut. There was a succession of muffled thuds as small arms fire struck the cockpit glass and the armour around it. Yyanis lay on the floor, watching blood drip from Arouin´s shoulder and pool on the decking. Arouin wasn´t hesitating; one-handed he powered up the ship´s systems, bringing everything online as fast as possible.
“Disengage the cradle clamps… and we´re free.” he announced, voice tired. Below the Phoenix the yellow and black striped holding cradle dropped away, the metal cylinders supporting it sliding smoothly into the floor. Inside the cockpit Arouin grabbed the control stick. The Phoenix tilted slightly as the cradle lowered away from her, thrusters keeping her hovering in the artificial gravity of the refitting bay. Arouin hammered at the control to remotely open the door, and waited for a few seconds to see the result. Nothing happened. After a couple of seconds of glancing over the controls, Arouin shrugged and pulled the trigger. Quad Barraks flamed; two lances of blue fire leapt out from either side of the _Lithe Shadow´s_ cockpit and hammered into the door which screened off the main hangar. Giant holes were punched in the thin metal, then Arouin nudged the ship forward and accelerated out of the bay through the remanants of the doors, the Phoenix shuddering as it smashed through them, sending a horrific shriek of tortured metal through the air. The bay stretched out away in front of them and on both sides. The occasional ship sat here and there on the featureless grey floor, equipment and refueling nozzles lying around them. Arouin turned around in his seat, wincing a little as he did so.
“Which side of the bay faces outward?”
Yyanis closed her eyes, remembering a schematic seen half an hour but seemingly two decades ago.
“Should be the one to our right” she said eventually. Arouin nodded slowly. Behind the Phoenix Jarren ran out through the shattered door and into the hangar, then stared up at the Octavian fighter which hung above him, shaking with rage and frustration.
“Exit stage right” Arouin said, smiling faintly and thumbing the missile launch button. Four streaks of flame and smoke screamed away from under the wings of the Lithe Shadow towards the wall, growing smaller with distance before blossoming into flame. The flame broiled out from the impact point of the four, almost instantaneous, detonations, and a shockwave rippled through the bay, tearing up the floor and sending metal plates meters wide flying up into the air. The ceiling rippled like waves on an ocean, mirroring the reaction of the floor below, girders screaming and bending, snapping, hawsers whipping around. Pieces of equipment and bits of the structure started falling onto the floor of the bay, bouncing around before lying still, and above it all, above the crashing of metal on metal and the death throes of metal, rose the howl of the wind. The smoke blossomed out towards the Phoenix, paused, then, like a film being rewound, sucked back again to reveal the blackness of space beyond. Bits of debris started sliding along the floor, gaining speed and momentum, crashing into one another and bouncing up, heading for the oblivion of the void. Around the gaping hole in the side of the station were white contrails as the moisture in the air froze, rendering it briefly visible before the vacuum of space stole it away from sight. Arouin, fearful of debris hitting his fighter, glanced at the rear view screen just in time to see a grey haired, black suited body crash into the camera, then go spinning off past the cockpit before tumbling into the depths of space. After twenty seconds the storm became a gale, became a breeze, and died. Arouin accelerated the ship forwards towards the gap in the outer hull, the fused metal edges of which seemed to reach towards them before they rushed past and into the soothing darkness of space. He quickly targetted the only Jumpgate in the sector, and headed towards it, sighing with relief.
“I hate to say it, but I wasn´t _absolutely_ sure we were going to fit through that.”
Arouin smiled a satisfied smile, despite the pain, and slipped back into the comfort of the padded pilot´s seat.
_Well,_ he reflected, _that wasn´t so bad. Got the powerplant. Got the ship. Still alive._ Arouin grinned broadly to himself. Hell, even got the girl…
He turned to look at Yyanis. She was already on her feet, and looking at him in a slightly different way. The in her hand gun crashed, flame licking out from it´s barrel. Arouin shuddered, once, then collapsed back into the chair, blood running down his neck from the gaping wound in his forehead. Bits of red and grey flesh dropped away from the inside slope of the cockpit, splattering on the control panel. Yyanis leant past Arouin´s body and it´s incredulous, unbelieving expression, and tapped at the communications controls.
“This is Agent Ulevi-09. I have the prototype…”
## Epilogue
In the cold of space, there was only the darkness, and the Phoenix.
##Hope this is the stuff you wanted Arouin, I've tested it, and it shouldn't##
##malfunction on you; not that I expect you to come back and complain if##
##it does :) - Raquel##
mort=false
WHILE (mort=false) DO[
t=0
Body(get_hrate)
WHILE (hrate < 30) DO[ ##3 minutes should give anyone time to revive you##
t=t+1
delay(1000) ##(Time in ms)##
Body(get_hrate)
IF (t=180) THEN mort=true
]
]
Shipsys(DefMech(Destruct))
And then there was light.
THE END
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# TeX capacity exceeded, sorry [input stack size=5000]. \begin{figure}[h]
I have found similar errors posted but none related to the figure command. I am trying to show a figure in my Latex template, if I remove the \begin{figure}[h] block, my file runs OK, but currently I cannot show any figure. The files I have are:
Main.tex
\documentclass[12pt,PhD,twoside]{muthesis}
\usepackage{CJKutf8}
\usepackage[doublespacing]{setspace}
\usepackage{tocloft}
\renewcommand\cftsecafterpnum{\vskip\baselineskip}
\renewcommand\cftsubsecafterpnum{\vskip\baselineskip}
\renewcommand\cftsubsubsecafterpnum{\vskip\baselineskip}
\begin{document}
\title{Title}
\author{Author}
\school{School name}
\faculty{Faculty}
\def\wordcount{xxxxx}
\beforeabstract
\input{abstract}
\afterabstract
\prefacesection{Acknowledgements}
\afterpreface
\chapter{Introduction}
\input{chapter1}
\chapter{Background}
\chapter{Conclusions}
\begin{thebibliography}{999}
\bibitem{ANO} A.N.~Other, ....
\end{thebibliography}
\appendix
\chapter{A Long Proof}
\end{document}
abstract.tex
\begin{singlespace}
This is the abstract.
\end{singlespace}
chapter 1.tex
This is a ref to Figure 1 below.
\begin{figure}[h]
\centering
\includegraphics[width=0.7\linewidth]{figure1}
\caption{Caption}
\label{fig:figure1}
\end{figure}
Other errors:
chapter1 - TeX capacity exceeded, sorry [input stack size=5000]. \begin{figure}[h]
abstract - Underfull \hbox (badness 10000) in paragraph
chapter 1 - Underfull \vbox (badness 10000) has occurred while \output is active []
The errors above disappear if I remove the \begin{figure}[h] block but I need to put figures.
Thank you for any help.
The class I am using is here.
• The files are shown there :) Files are "chapter1" and "abstract" – User2130 Apr 13 '18 at 11:52
• If you copy and past the files as they are shown there. The error appears. The contents shown in chpater1.tex and abstract.tex is all I have inside those. – User2130 Apr 13 '18 at 11:54
• Sorry, I will edit to include the class file which is the only missing thing – User2130 Apr 13 '18 at 11:59
• oh that is a different muthesis to the one in texlive (look at line 25:-):-) – David Carlisle Apr 13 '18 at 12:09
• the one in texlive was edited by Graham 2011 %% Time-stamp: <2011-08-25 16:41:19 grahamgough> The one you are using is dated 2013 but based on an ancient version of Graham's %% Time-stamp: <July 23, 1997 11:33:00 BST graham@zzzzz> and seems to be basically broken. I would use the muthesis that is in texlive and add graphicx package and hyperref. – David Carlisle Apr 13 '18 at 12:15
%% Time-stamp: <2011-08-25 16:41:19 grahamgough>
%% Time-stamp: <July 23, 1997 11:33:00 BST graham@zzzzz>
and seems to be basically broken. I would use the muthesis.cls that is in texlive and add graphicx package and hyperref in your preamble.
|
# American Institute of Mathematical Sciences
January 2020, 19(1): 31-46. doi: 10.3934/cpaa.2020003
## Infinitely many solutions and Morse index for non-autonomous elliptic problems
Department of Mathematical Sciences, University of Cincinnati, Cincinnati Ohio 45221-0025 USA
Received February 2018 Revised April 2019 Published July 2019
This paper deals with changes of variables, and the exact bifurcation diagrams for a class of self-similar equations. Our first result is a change of variables which transforms radial $k$-Hessian equations into radial $p$-Laplace equations. Then, in another direction, we generalize the classical results of D.D. Joseph and T.S. Lundgren [10] by using the method we developed in [13] and [14]. We provide a considerably simpler approach, which yields additional information on the Morse index of solutions.
Citation: Philip Korman. Infinitely many solutions and Morse index for non-autonomous elliptic problems. Communications on Pure & Applied Analysis, 2020, 19 (1) : 31-46. doi: 10.3934/cpaa.2020003
##### References:
[1] I. Bihari, A generalization of a lemma of Bellman and its application to uniqueness problems of differential equations, Acta Math. Acad. Sci. Hungar., 7 (1956), 81-94. doi: 10.1007/BF02022967. Google Scholar [2] C. Budd and J. Norbury, Semilinear elliptic equations and supercritical growth, J. Differential Equations, 68 (1987), 169-197. doi: 10.1016/0022-0396(87)90190-2. Google Scholar [3] L. Caffarelli, L. Nirenberg and J. Spruck, The Dirichlet problem for nonlinear second-order elliptic equations Ⅲ. Functions of the eigenvalues of the Hessian, Acta Math., 155 (1985), 261-301. doi: 10.1007/BF02392544. Google Scholar [4] S. Chandrasekhar, An Introduction to the Study of Stellar Structure, Dover Publications Inc., New York, 1957. Google Scholar [5] J. Dávila, M. del Pino, M. Musso and J. Wei, Fast and slow decay solutions for supercritical elliptic problems in exterior domains, Calc. Var. Partial Differential Equations, 32 (2008), 453-480. doi: 10.1007/s00526-007-0154-1. Google Scholar [6] B. Gidas, W.-M. Ni and L. Nirenberg, Symmetry and related properties via the maximum principle, Commun. Math. Phys., 68 (1979), 209-243. Google Scholar [7] E. Hopf, On Emden's differential equation, Monthly Notices of the Royal Astronomical Society, 91 (1931), 653-663. Google Scholar [8] J. Jacobsen, Global bifurcation problems associated with k-Hessian operators, Topol. Methods Nonlinear Anal., 14 (1999), 81-130. doi: 10.12775/TMNA.1999.023. Google Scholar [9] J. Jacobsen and K. Schmitt, The Liouville-Bratu-Gelfand problem for radial operators, J. Differential Equations, 184 (2002), 283-298. doi: 10.1006/jdeq.2001.4151. Google Scholar [10] D. D. Joseph and T. S. Lundgren, Quasilinear Dirichlet problems driven by positive sources, Arch. Rational Mech. Anal., 49 (1972/73), 241-269. doi: 10.1007/BF00250508. Google Scholar [11] P. Korman, Uniqueness and exact multiplicity of solutions for a class of Dirichlet problems, J. Differential Equations, 244 (2008), 2602-2613. doi: 10.1016/j.jde.2008.02.014. Google Scholar [12] P. Korman, Global Solution Curves for Semilinear Elliptic Equations, World Scientific, Hackensack, NJ, 2012. doi: 10.1142/8308. Google Scholar [13] P. Korman, Global solution curves for self-similar equations, J. Differential Equations, 257 (2014), 2543-2564. doi: 10.1016/j.jde.2014.05.045. Google Scholar [14] P. Korman, Infinitely many solutions for three classes of self-similar equations, with the $p$-Laplace operator, Proc. Roy. Soc. Edinburgh Sect. A, 147A (2017), 1-16. doi: 10.1017/S0308210517000038. Google Scholar [15] P. Korman, Explicit solutions and multiplicity results for some equations with the $p$-Laplacian, Quart. Appl. Math., 75 (2017), 635-647. doi: 10.1090/qam/1471. Google Scholar [16] C. S. Lin and W.-M. Ni, A counterexample to the nodal domain conjecture and related semilinear equation, Proc. Amer. Math. Soc., 102 (1988), 271-277. doi: 10.2307/2045874. Google Scholar [17] F. Merle and L. A. Peletier, Positive solutions of elliptic equations involving supercritical growth, Proc. Roy. Soc. Edinburgh Sect. A, 118 (1991), 49-62. doi: 10.1017/S0308210500028882. Google Scholar [18] K. Nagasaki and T. Suzuki, Spectral and related properties about the Emden-Fowler equation $-\Delta u=\lambda e^ u$ on circular domains, Math. Ann., 299 (1994), 1-15. doi: 10.1007/BF01459770. Google Scholar [19] L. A. Peletier and J. Serrin, Uniqueness of positive solutions of semilinear equations in $R^{n}$, Arch. Rational Mech. Anal., 81 (1983), 181-197. doi: 10.1007/BF00250651. Google Scholar [20] J. A. Pelesko, Mathematical modeling of electrostatic MEMS with tailored dielectric properties, SIAM J. Appl. Math., 62 (2002), 888-908. doi: 10.1137/S0036139900381079. Google Scholar [21] J. Sánchez and V. Vergara, Bounded solutions of a $k$-Hessian equation in a ball, J. Differential Equations, 261 (2016), 797-820. doi: 10.1016/j.jde.2016.03.021. Google Scholar [22] N. S. Trudinger and X.-J. Wang, Hessian measures Ⅰ, Topol. Methods Nonlinear Anal., 10 (1997), 225-239. doi: 10.12775/TMNA.1997.030. Google Scholar
show all references
##### References:
[1] I. Bihari, A generalization of a lemma of Bellman and its application to uniqueness problems of differential equations, Acta Math. Acad. Sci. Hungar., 7 (1956), 81-94. doi: 10.1007/BF02022967. Google Scholar [2] C. Budd and J. Norbury, Semilinear elliptic equations and supercritical growth, J. Differential Equations, 68 (1987), 169-197. doi: 10.1016/0022-0396(87)90190-2. Google Scholar [3] L. Caffarelli, L. Nirenberg and J. Spruck, The Dirichlet problem for nonlinear second-order elliptic equations Ⅲ. Functions of the eigenvalues of the Hessian, Acta Math., 155 (1985), 261-301. doi: 10.1007/BF02392544. Google Scholar [4] S. Chandrasekhar, An Introduction to the Study of Stellar Structure, Dover Publications Inc., New York, 1957. Google Scholar [5] J. Dávila, M. del Pino, M. Musso and J. Wei, Fast and slow decay solutions for supercritical elliptic problems in exterior domains, Calc. Var. Partial Differential Equations, 32 (2008), 453-480. doi: 10.1007/s00526-007-0154-1. Google Scholar [6] B. Gidas, W.-M. Ni and L. Nirenberg, Symmetry and related properties via the maximum principle, Commun. Math. Phys., 68 (1979), 209-243. Google Scholar [7] E. Hopf, On Emden's differential equation, Monthly Notices of the Royal Astronomical Society, 91 (1931), 653-663. Google Scholar [8] J. Jacobsen, Global bifurcation problems associated with k-Hessian operators, Topol. Methods Nonlinear Anal., 14 (1999), 81-130. doi: 10.12775/TMNA.1999.023. Google Scholar [9] J. Jacobsen and K. Schmitt, The Liouville-Bratu-Gelfand problem for radial operators, J. Differential Equations, 184 (2002), 283-298. doi: 10.1006/jdeq.2001.4151. Google Scholar [10] D. D. Joseph and T. S. Lundgren, Quasilinear Dirichlet problems driven by positive sources, Arch. Rational Mech. Anal., 49 (1972/73), 241-269. doi: 10.1007/BF00250508. Google Scholar [11] P. Korman, Uniqueness and exact multiplicity of solutions for a class of Dirichlet problems, J. Differential Equations, 244 (2008), 2602-2613. doi: 10.1016/j.jde.2008.02.014. Google Scholar [12] P. Korman, Global Solution Curves for Semilinear Elliptic Equations, World Scientific, Hackensack, NJ, 2012. doi: 10.1142/8308. Google Scholar [13] P. Korman, Global solution curves for self-similar equations, J. Differential Equations, 257 (2014), 2543-2564. doi: 10.1016/j.jde.2014.05.045. Google Scholar [14] P. Korman, Infinitely many solutions for three classes of self-similar equations, with the $p$-Laplace operator, Proc. Roy. Soc. Edinburgh Sect. A, 147A (2017), 1-16. doi: 10.1017/S0308210517000038. Google Scholar [15] P. Korman, Explicit solutions and multiplicity results for some equations with the $p$-Laplacian, Quart. Appl. Math., 75 (2017), 635-647. doi: 10.1090/qam/1471. Google Scholar [16] C. S. Lin and W.-M. Ni, A counterexample to the nodal domain conjecture and related semilinear equation, Proc. Amer. Math. Soc., 102 (1988), 271-277. doi: 10.2307/2045874. Google Scholar [17] F. Merle and L. A. Peletier, Positive solutions of elliptic equations involving supercritical growth, Proc. Roy. Soc. Edinburgh Sect. A, 118 (1991), 49-62. doi: 10.1017/S0308210500028882. Google Scholar [18] K. Nagasaki and T. Suzuki, Spectral and related properties about the Emden-Fowler equation $-\Delta u=\lambda e^ u$ on circular domains, Math. Ann., 299 (1994), 1-15. doi: 10.1007/BF01459770. Google Scholar [19] L. A. Peletier and J. Serrin, Uniqueness of positive solutions of semilinear equations in $R^{n}$, Arch. Rational Mech. Anal., 81 (1983), 181-197. doi: 10.1007/BF00250651. Google Scholar [20] J. A. Pelesko, Mathematical modeling of electrostatic MEMS with tailored dielectric properties, SIAM J. Appl. Math., 62 (2002), 888-908. doi: 10.1137/S0036139900381079. Google Scholar [21] J. Sánchez and V. Vergara, Bounded solutions of a $k$-Hessian equation in a ball, J. Differential Equations, 261 (2016), 797-820. doi: 10.1016/j.jde.2016.03.021. Google Scholar [22] N. S. Trudinger and X.-J. Wang, Hessian measures Ⅰ, Topol. Methods Nonlinear Anal., 10 (1997), 225-239. doi: 10.12775/TMNA.1997.030. Google Scholar
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2019 Impact Factor: 1.105
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# Grew • Command Line Interface
The command to run Grew is: grew <subcommand> [<args>]
Main subcommands are:
• 🔗 transform: application of a rewriting system to a set of graphs
• 🔗grep: search for a pattern in a corpus
• 🔗 compile: compile a set of corpora
• 🔗clean: clean a set of corpora
• 🔗 count: compute stats of a set of patterns in a set of corpora)
• OBSOLETE 🔗 gui: run the GTK interface
Other subcommands:
• version: Print version numbers of the Grew Ocaml library and of the Grew tool
• help: Print general help
• help <subcommand>: Print help for the given subcommand
# Transform
In this mode, Grew apply a Graph Rewriting System to a graph or a set of graphs.
The full command for this mode:
grew transform [<args>]
All arguments are optional:
• -grs <grs_file>: the main file which describes the Graph Rewriting System. If no GRS is given, the empty GRS is loaded: strat main {Seq ()}
• -i <input_file>: describes the input data (CoNLL file of gr file). If no input file is given, Grew reads from stdin
• -o <output_file>: is the name of the output file (CoNLL file). If no output file is given, Grew writes to stdout
• -strat <name>: the strategy used in transformation (default value: main)
• -safe_commands: flag. It makes rewriting process fail in case of ineffective command
# Grep
This mode corresponds to the command line version of the Grew-match tool. The command is:
grew grep -pattern <pattern_file> -i <corpus_file>
where:
• <pattern_file> is a file which describes a pattern
• <corpus_file> is the corpus in which the search is done
The output is given in JSON format.
## Example
With the following files:
• The dev part of the corpus UD_French-GSD version 2.6: fr_gsd-ud-dev.conllu🔗
• A pattern file with the code below: bleu.pat🔗
pattern { e: M -> N; N [lemma="bleu"] }
The command:
grew grep -pattern bleu.pat -i fr_gsd-ud-dev.conllu
produces the following JSON output:
[
{
"sent_id": "fr-ud-dev_00623",
"matching": {
"nodes": { "N": "39", "M": "38" },
"edges": {
"e": { "source": "38", "label": { "1": "amod" }, "target": "39" }
}
}
},
{
"sent_id": "fr-ud-dev_00217",
"matching": {
"nodes": { "N": "17", "M": "16" },
"edges": {
"e": { "source": "16", "label": { "1": "amod" }, "target": "17" }
}
}
}
]
This means that the pattern described in the file bleu.pat was found twice in the corpus, each item gives the sentence identifier and the position of the nodes and the edges matched by the pattern.
Note that two other options exist:
• -html: produces a new html field in each JSON item with the sentence where words impacted by the pattern are in a special HTML span with class highlight
• -dep_dir <directory>: produces a new file in the folder directory with the representation of the sentence with highlighted part (as in Grew-match tool) and a new field in each JSON item with the filename; the output is in dep format (usable with Dep2pict).
# Compile
For the Grew-match server (grew_daemon) or for the command grew count, it is required to first compile corpora. For these two usages, sets of corpora are described in a JSON file.
For compilation, the command is:
grew compile -i <corpora.json>
Note that this produces, for each corpus, a new file with the marshal extension stored in the corpus directory. The marshal is computed only if the corpus has changed since the last compilation.
# Clean
The commands below removes the marshal files produced by the grew compile command for the set of corpora described in the JSON file corpora.json.
grew clean -i <corpora.json>
# Count
This mode computes corpus statistics. Given a set of patterns and a set of corpora, a TSV table is built with the number of occurrences for each pattern in each corpus.
The set of corpora is described in a JSON file and must be compiled before running grew count.
Each pattern is described in a separate file.
## Example
With the two following 1-line files:
• ADJ_NOUN.pat 🔗
pattern { A[upos=ADJ]; N[upos=NOUN]; N -[amod]-> A; A << N }
• NOUN_ADJ.pat 🔗
pattern { A[upos=ADJ]; N[upos=NOUN]; N -[amod]-> A; N << A }
and the example file en_fr_zh.json 🔗
{ "corpora": [
{ "id": "[email protected]",
"directory": "/Users/guillaum/resources/ud-treebanks-v2.6/UD_English-EWT/",
"files": ["en_ewt-ud-dev.conllu", "en_ewt-ud-test.conllu", "en_ewt-ud-train.conllu"]
},
{ "id": "[email protected]",
"directory": "/Users/guillaum/resources/ud-treebanks-v2.6/UD_French-Sequoia/"
},
{ "id": "[email protected]",
"directory": "/Users/guillaum/resources/ud-treebanks-v2.6/UD_Chinese-GSD/"
} ]
}
1. Compile the corpora: grew compile -i en_fr_zh.json
2. Build stat table: grew count -patterns "ADJ_NOUN.pat NOUN_ADJ.pat" -i en_fr_zh.json
The output is given as TSV data:
Corpus # sentences ADJ_NOUN NOUN_ADJ
UD_English-EWT 16622 9835 163
UD_French-Sequoia 3099 891 2779
UD_Chinese-GSD 4997 1505 0
which corresponds to the table:
UD_English-EWT 16622 9835 163
UD_French-Sequoia 3099 891 2779
UD_Chinese-GSD 4997 1505 0
We can then observe that in the annotations of the 3 corpora in use:
• in English, there is a strong preference for adjective position before the noun (98.4%)
• in French, there is a weak preference for adjective position after the noun (75,7%)
• in Chinese, there is a very strong preference for adjective position before the noun (100%)
## Remarks
• The TSV table also contains a column with the size of corpora (in number of sentences), this can be useful to make cross-corpora analysis and to compute ratios instead of raw numbers.
• Pattern syntax can be learned here or with the online Grew-match tool, first with the tutorial and then with snippets given on the right of the text area.
• If some corpus is updated, it is necessary to run again the compilation step.
• Some patterns may take a long time to be searched in corpora.
# GUI (Obsolete)
The command to run the GTK interface: grew gui [<args>]. It supposes that you have installed the grew_gui opam package (see GUI installation page).
Optional arguments:
• -grs <grs_file>: load the given file
• -i <input_file>: input data (graph or corpus) loaded in GUI
• -strat <name>: the strategy selected in the interface (default: main)
• -main_feat <feat_name_list> set the list of feature names used ad the main feat in graph visualization
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## Tag: rational-points
96 Estimating the size of solutions of a diophantine equation 2016-01-05T16:55:40.637
29 Patterns among integer-distance points 2013-07-28T00:35:27.390
22 Why is there a $\sqrt{5}$ in Hurwitz's Theorem? 2015-07-07T19:10:41.480
19 Does $(x^2 - 1)(y^2 - 1) = c z^4$ have a rational point, with z non-zero, for any given rational c? 2012-05-19T08:14:41.457
19 Rational points on the "quintic circle" $x^5 + y^5 = 7$ 2015-11-22T07:43:21.043
18 Integer-distance sets 2013-07-17T00:18:42.993
16 Rational points on a sphere in $\mathbb{R}^d$ 2013-03-22T01:34:24.433
16 Identifying Ramanujan's integer solutions of x^3+y^3+z^3=1 among Elkies' rational solutions 2013-06-28T16:00:12.553
15 Is the perimeter of an ellipse with integer axes irrational? 2013-11-23T02:53:04.860
15 Number of height-limited rational points on a circle 2015-09-28T13:03:56.307
15 Determining the Mordell-Weil group of a universal elliptic curve 2017-10-10T06:59:46.017
14 Totally rational polytopes 2011-08-03T01:29:14.860
14 Rational solutions of the Fermat equation $X^n+Y^n+Z^n=1$ 2015-12-14T18:37:20.517
13 Existence of points on varieties which avoid a given number field. 2012-06-11T19:51:48.333
13 Rational Points on $y^2=x^3-86069^5$ 2012-08-26T21:58:35.453
13 Elliptic curves and connected components 2015-01-23T21:24:45.070
12 Closed vs Rational Points on Schemes 2010-02-02T22:51:59.597
12 Rational points à la Chabauty-Coleman 2011-03-25T23:52:55.560
12 what is the maximum number of rational points of a curve of genus 2 over the rationals 2012-07-27T16:32:14.273
12 Existence of local sections 2014-04-06T02:31:36.743
11 Can a harmonic number be a rational number for non-integer rational argument? 2013-05-07T01:01:27.260
11 Geometrically unirational varieties that are not unirational 2013-06-17T12:46:02.173
11 What evidence is there that $\mathbb{Q}^{ab}$ is ample? 2013-06-23T19:38:12.007
11 What is our current knowledge on the structure of J_0(N)(Q) and J_1(N)(Q) 2014-05-19T13:17:40.027
11 Galois Representations and Rational Points 2017-01-18T06:12:42.297
10 Circles avoiding rational points of height $\le h$ 2014-07-31T23:02:10.470
10 Schoenberg's Rational Polygon Problem 2014-10-12T01:33:31.943
10 rational points of a hyperelliptic curve 2014-12-02T03:28:10.850
10 How much do I need to learn algebraic geometry to understand arithmetics over number fields 2016-04-07T22:01:53.383
10 Possible groups of K-rational points for elliptic curves over arbitrary fields 2017-09-06T09:00:39.760
9 Existence of hyperelliptic curve with specific number of points in a family 2009-11-10T17:29:32.807
9 Rational points on surfaces of general type 2013-03-07T12:40:00.257
9 Rational points on circular spirals 2013-09-08T00:55:29.703
8 Shortest irrational path 2011-07-26T12:13:52.230
8 Smart elliptic curve rational point search given Reg*#Sha 2012-02-04T06:26:27.373
8 Varieties with infinitely many etale covers and rational points 2012-11-03T01:19:50.017
8 Distribution of Mordell–Weil ranks of higher genus curves 2015-06-06T20:51:04.920
8 An elliptic curve for Ramanujan-type cubic identities? 2016-07-04T21:42:50.913
8 Scaling a set of reals to be nearly integers 2017-08-15T11:45:05.773
8 Artin representations appearing in Mordell-Weil groups of elliptic curves 2017-09-07T13:10:42.520
7 Sums of cubes and more 2009-10-13T19:52:05.320
7 Lattice radial-step (ratchet) spirals 2013-10-26T18:46:30.013
7 Endomorphism algebras of abelian surfaces with real multiplication 2015-05-29T11:42:03.320
7 Does Chabauty-Coleman method give an algorithm for finding rational points? 2016-09-11T15:52:08.523
7 Rational points on varieties over local fields 2017-10-07T08:26:59.843
7 Why study unirational and rational varieties? 2017-11-30T06:05:25.897
6 How can you find small denominators inside triangles? 2009-10-04T20:19:43.440
6 Rational points on smooth compactifications 2012-05-03T00:18:56.200
6 Simple field extension and rational points 2013-03-31T05:58:54.857
6 Pick's Theorem for rational points of bounded height 2013-07-01T11:45:59.817
6 Is the following consequence of the Lang conjecture known? 2014-01-27T15:35:47.673
6 Rational points techniques on curves not using their Jacobian 2014-08-23T23:01:51.980
6 Average height of rational points on a curve 2015-04-25T01:45:21.160
6 Why are some solutions of these diophantine equations off the usual patterns? 2015-12-07T09:22:31.633
6 rational points and a local perturbation of an elliptic curve 2017-11-06T20:45:16.260
6 What is the smallest sphere whose surface includes 100 integer points? 2018-01-30T14:32:52.307
5 Existence of a non-trivial zero (in the rational cyclotomic field) of a form 2011-10-11T06:52:50.227
5 12 descent scripts for pari/gp 2012-09-13T04:05:16.270
5 Solving for special rational triangles 2013-06-25T12:07:53.343
5 Maximum number of general-position points with mutual rational distances? 2014-10-05T00:42:00.940
5 Transcendental distance sets 2015-07-26T14:15:41.687
5 A relative version of Hensel's lemma? 2015-09-14T18:28:08.067
5 Counting square zero forms over finite fields 2015-09-21T09:49:28.080
5 Rational $d$-simplices 2016-04-23T00:42:44.000
4 Rational points on $X_0(15)$ 2013-07-02T16:12:31.333
4 What is the complexity of finding an integral point on an elliptic curve? 2014-09-28T08:11:10.800
4 Find all possible rational values of a parametric quartic such that it is reducible 2015-04-06T20:44:45.353
4 Find all rational solutions of this diophantine-equation? 2015-06-07T14:34:25.970
4 Conics, rational points and probability 2015-12-23T17:26:21.080
4 Rational points on the unit circle 2016-08-02T23:24:23.143
4 Understanding Siegel's Theorem on integral points 2016-12-16T16:34:00.507
4 Action of the Picard Scheme of an Elliptic Fibration 2018-02-06T12:05:36.987
4 Integral complete 4-partite graphs 2018-02-17T21:29:00.193
3 k rational points and base change 2011-05-25T12:28:15.813
3 Brauer-Manin obstruction and Hasse principle 2012-01-03T17:13:40.113
3 How many points are there on an elliptic curve reduced at a bad prime? 2012-05-31T07:15:20.070
3 Rational points and Tesla cards 2012-12-12T01:39:39.580
3 Closed points of field extension of k-scheme under projection 2013-09-02T18:45:54.157
3 Lattice-point-free buffers around circles 2013-10-30T12:15:39.503
3 cohomological obstructions and rational points 2015-01-28T19:53:19.803
3 Existence of lattices whose circles have bounded number of points 2016-07-11T14:48:37.603
2 Rational subspaces 2012-10-03T04:03:59.227
2 Question on x coordinates of Mordell Curves where $y^2=x^3+k$ and $k^2 = 1$ mod $24$ 2013-08-03T10:37:55.767
2 The uniform boundedness of rational torsion for traceless abelian surfaces over a function field 2014-12-23T14:50:38.967
2 Rank of the Jacobian of twists of hyperelliptic curves 2015-08-23T05:45:33.907
2 F-points of product of closed subgroups vs. product of F-points, F a local field, reference? 2015-11-25T23:51:59.747
2 Rational points on towers of curves 2015-12-27T20:05:42.590
2 the least point on a variety over a finite field 2016-03-20T07:59:07.333
2 Counting algebraic points of bounded height 2017-11-11T18:12:30.947
2 Number of rational points of a singular cubic surface over a finite field 2017-11-12T07:07:51.430
1 are p-limits scales dense in the infinite musical scale of all rational frequencies? 2012-07-14T22:40:22.283
1 Tricks to produce examples of hypersurfaces with index greater than $1$ 2013-04-06T18:51:35.120
1 Infinite residue field extensions and algebraic closure of residue fields 2013-09-05T16:28:48.527
1 Existence of a curve with no points over finite separable field extensions 2013-09-11T09:58:39.000
1 Benchmark problems for computing rational points on varieties 2014-01-08T20:23:35.893
1 Coarse moduli spaces and rational points 2014-06-07T21:50:06.877
|
# Abstract Harmonic Analysis
@inproceedings{Hewitt1963AbstractHA,
title={Abstract Harmonic Analysis},
author={Edwin Shields Hewitt and Kenneth A. Ross},
year={1963}
}
• Published 1963
• Mathematics
The first € price and the £ and $price are net prices, subject to local VAT. Prices indicated with * include VAT for books; the €(D) includes 7% for Germany, the €(A) includes 10% for Austria. Prices indicated with ** include VAT for electronic products; 19% for Germany, 20% for Austria. All prices exclusive of carriage charges. Prices and other details are subject to change without notice. All errors and omissions excepted. E. Hewitt, K.A. Ross Abstract Harmonic Analysis 3,587 Citations THE PARSEVAL FORMULA IN THE HARMONIC ANALYSIS FOR OPERATOR AND THE SUPPORT FOR OPERATORS Abstract In this paper a harmonic analysis for operators on a homogeneous Banach space on the additional group of real numbers is discussed. Main results are as following: 1. The Parseval formulaExpand Classical Themes of Fourier Analysis This article is devoted to “Harmonic Analysis for itself” (a quotation from the preceding article), more precisely, to specific problems of the classical theory of Fourier series (and, to a certainExpand Operator theory in harmonic analysis The purpose of this paper is to give an illustration of how operator theoretic results in Hilbert space can be applied to obtain results in classical and abstract harmonic analysis. An inequality forExpand Harmonic Analysis and Functional Equations Functional equations occur in many parts of mathematics, also in harmonic analysis. As an example we mention that the complex exponential function 7 :x ? exp (?x) for any ? ∈ R is a solution ofExpand Projections in$L^1(G)$; the unimodular case • Mathematics • 2015 We consider the issue of describing all self-adjoint idempotents (projections) in$L^1(G)$when$G$is a unimodular locally compact group. The approach is to take advantage of known facts concerningExpand Fractional differential equations: alpha-entire solutions, regular and irregular singularities We consider fractional differential equations of order$\alpha \in (0,1)$for functions of one independent variable$t\in (0,\infty)$with the Riemann-Liouville and Caputo-Dzhrbashyan fractionalExpand Harmonic analysis and applications • M. Filali • Mathematics, Computer Science • Kybernetes • 2012 This paper surveys briefly how harmonic analyis started and developed throughout the centuries to reach its modern status and its surprisingly wide range of applications. Expand Fourier multipliers for certain spaces of functions with compact supports • Mathematics • 1977 This is perhaps the final word on how large overall the Fourier coefficients of continuous periodic functions can be. (See the remarks in Paley [,11] on this point.) Helgason [--53, Theorem 2,Expand Positive operator measures, generalised imprimitivity theorem, and their applications Given a topological group$G$and a unitary representation$U$of$G$, we consider the problem of classifying the positive operator measures which are based on a$G$-homogeneous space$X\$ andExpand
Harmonic operators: the dual perspective
• Mathematics
• 2005
The study of harmonic functions on a locally compact group G has recently been transferred to a “non-commutative” setting in two different directions: Chu and Lau replaced the algebra L∞(G) by theExpand
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k nearest neighbors method, temporal trend in error
I have this set of data that looks like this
I was asked o build a $k$-nearest neighbors algorithm for it which I just finished building. I have this question in regards to the data that I do not understand: Do you notice any spatial or temporal trends in error?
I am not sure how to proceed in answering that question. Any suggestions would be appreciated.
|
SN 2014ab: an aspherical Type IIn supernova with low polarization
ABSTRACT We present photometry, spectra, and spectropolarimetry of supernova (SN) 2014ab, obtained through ∼200 d after peak brightness. SN 2014ab was a luminous Type IIn SN (MV < −19.14 mag) discovered after peak brightness near the nucleus of its host galaxy, VV 306c. Pre-discovery upper limits constrain the time of explosion to within 200 d prior to discovery. While SN 2014ab declined by ∼1 mag over the course of our observations, the observed spectrum remained remarkably unchanged. Spectra exhibit an asymmetric emission-line profile with a consistently stronger blueshifted component, suggesting the presence of dust or a lack of symmetry between the far side and near side of the SN. The Pa β emission line shows a profile very similar to that of H α, implying that this stronger blueshifted component is caused either through obscuration by large dust grains, occultation by optically thick material, or a lack of symmetry between the far side and near side of the interaction region. Despite these asymmetric line profiles, our spectropolarimetric data show that SN 2014ab has little detected polarization after accounting for the interstellar polarization. We are likely seeing emission from a photosphere that has only small deviation from circular symmetry in the plane normal to our line of sight, but with either more »
Authors:
; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; more »
Award ID(s):
Publication Date:
NSF-PAR ID:
10281975
Journal Name:
Monthly Notices of the Royal Astronomical Society
Volume:
498
Issue:
3
Page Range or eLocation-ID:
3835 to 3851
ISSN:
0035-8711
2. ABSTRACT We observed the Brackett α emission line (4.05 μm) within the nuclear starburst of NGC 253 to measure the kinematics of ionized gas, and distinguish motions driven by star formation feedback from gravitational motions induced by the central mass structure. Using NIRSPEC on Keck II, we obtained 30 spectra through a $0^{\prime \prime }_{.}5$ slit stepped across the central ∼5 arcsec × 25 arcsec (85 × 425 pc) region to produce a spectral cube. The Br α emission resolves into four nuclear sources: S1 at the infrared core (IRC), N1 at the radio core, and the fainter sources N2 and N3 in the northeast. The line profile is characterized by a primary component with Δvprimary ∼90–130 $\rm km\, s^{-1}$ (full width at half-maximum) on top of a broad blue 2wing with Δvbroad ∼300–350 $\rm km\, s^{-1}$, and an additional redshifted narrow component in the west. The velocity field generated from our cube reveals several distinct patterns. A mean NE–SW velocity gradient of +10 $\rm km\, s^{-1}$ arcsec−1 along the major axis traces the solid-body rotation curve of the nuclear disc. At the radio core, isovelocity contours become S-shaped, indicating the presence of secondary nuclear bar of total extent ∼5 arcsec (90 pc). The symmetry of the bar places the galactic centre, and potential supermassivemore »
5. ABSTRACT After correcting for their light-curve shape and colour, Type Ia supernovae (SNe Ia) are precise cosmological distance indicators. However, there remains a non-zero intrinsic scatter in the differences between measured distance and that inferred from a cosmological model (i.e. Hubble residuals or HRs), indicating that SN Ia distances can potentially be further improved. We use the open-source relational data base kaepora to generate composite spectra with desired average properties of phase, light-curve shape, and HR. At many phases, the composite spectra from two subsamples with positive and negative average HRs are significantly different. In particular, in all spectra from 9 d before to 15 d after peak brightness, we find that SNe with negative HRs have, on average, higher ejecta velocities (as seen in nearly every optical spectral feature) than SNe with positive HRs. At +4 d relative to B-band maximum, using a sample of 62 SNe Ia, we measure a 0.091 ± 0.035 mag (2.7σ) HR step between SNe with Si ii λ6355 line velocities ($v_{Si\, rm{\small II}}$) higher/lower than −11 000 km s−1 (the median velocity). After light-curve shape and colour correction, SNe with higher velocities tend to have underestimated distance moduli relative to a cosmological model. The intrinsic scatter in our sample reduces from 0.094 to 0.082 mag after making thismore »
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Creates or validates a subclass of a tibble. These function is mostly useful for package authors that implement subclasses of a tibble, like sf or tsibble.
new_tibble() creates a new object as a subclass of tbl_df, tbl and data.frame. This function is optimized for performance, checks are reduced to a minimum.
validate_tibble() checks a tibble for internal consistency. Correct behavior can be guaranteed only if this function runs without raising an error.
new_tibble(x, ..., nrow, class = NULL, subclass = NULL)
validate_tibble(x)
## Arguments
x A tibble-like object. Name-value pairs of additional attributes. The number of rows, required. Subclasses to assign to the new object, default: none. Deprecated, retained for compatibility. Please use the class argument.
## Construction
For new_tibble(), x must be a list. The ... argument allows adding more attributes to the subclass. An nrow argument is required. This should be an integer of length 1, and every element of the list x should have vctrs::vec_size() equal to this value. (But this is not checked by the constructor). This takes the place of the "row.names" attribute in a data frame. x must have names (or be empty), but the names are not checked for correctness.
## Validation
validate_tibble() checks for "minimal" names and that all columns are vectors, data frames or matrices. It also makes sure that all columns have the same length, and that vctrs::vec_size() is consistent with the data.
tibble() and as_tibble() for ways to construct a tibble with recycling of scalars and automatic name repair.
## Examples
# The nrow argument is always required:
new_tibble(list(a = 1:3, b = 4:6), nrow = 3)
#> # A tibble: 3 x 2
#> a b
#> <int> <int>
#> 1 1 4
#> 2 2 5
#> 3 3 6
# Existing row.names attributes are ignored:
try(validate_tibble(new_tibble(iris, nrow = 3)))
#> Error : Tibble columns must have compatible sizes.
#> * Size 3: Requested with nrow argument.
#> * Size 150: Columns Sepal.Length, Sepal.Width, Petal.Length, Petal.Width, and Species.
#> ℹ Only values of size one are recycled.
# The length of all columns must be compatible with the nrow argument:
try(validate_tibble(new_tibble(list(a = 1:3, b = 4:6), nrow = 2)))
#> Error : Tibble columns must have compatible sizes.
#> * Size 2: Requested with nrow argument.
#> * Size 3: Columns a and b.
#> ℹ Only values of size one are recycled.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
The given expression, $a^2-12ab+49b^2 ,$ is not a perfect square trinomial because the middle term is not twice the product of the square roots of the first and third terms. Hence, it is $\text{ prime .}$
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## Debugging Ajax’ed JavaScript and jQuery val() calls
June 8, 2017
I develop a web application which displays data in a read-only form, and loads the edit form upon pressing a button
$(function () {$(".btnEdit").click(function () {
$.ajax({ url: '@Url.Action("Form", new { id = Model.ID })', type: 'post' }); }); }); Now I needed to debug the JavaScript code loaded by the call to$.ajax(), but Chrome does not seem to display the loaded response in its tree of Sources.
An answer on SO provided me with the solution: Simply add the line
//# [email protected]("Form", new { id = Model.ID })
inside a <script> block in the AJAX-loaded HTML. This will add the requested URL under the (no domain) node
Now that I have access to the source file, I needed to find all invocations of the jQuery val() function, since I was tracking down a wrong value in an <input>.
Again SO provided a solution, which I added to my code
(function($){ var originalVal =$.fn.val;
$.fn.val = function(){ var selectorPath =$(this).parents().map(function () {return this.tagName;}).get().reverse().join("->");
console.log("val( #" + $(this).attr("id") + " " + selectorPath + " , " + JSON.stringify(arguments) + ")"); var result =originalVal.apply(this,arguments); return result; }; })(jQuery); Now that the calls to val() were logged to the Console, it was easy to find where the wrong value was set. Advertisements ## Adding SSL Wildcard Certificates to IIS Webs March 21, 2017 As web browsers start to issue warnings on plain http websites if you are asked to input username/password, it’s time to add SSL certificates even on dev/test servers. We can expect more aggressive warnings in the future 😉 Apparently there is a way to create a self-signed certificate built into IIS (screenshot from Windows Server 2008) but this seems to create cerficates only for the host name, not for any domain hosted on the machine. Back to square one, start up a current Linux machine, and make sure your openssl is newer than version 1.0.1f. (Remember Heartbeed?). The instructions I found to create self-signed certificates are nearly identical (source, source, source) openssl genrsa 2048 > my-host.key openssl req -new -x509 -nodes -sha1 -days 3650 -key my-host.key > my-host.cert # make sure Common Name starts with "*.", e.g. *.my-host.com openssl x509 -noout -fingerprint -text < my-host.cert > my-host.info cat my-host.cert my-host.key > my-host.pem For use in IIS, you need to create a .pfx from these certificate files: openssl pkcs12 -inkey my-host.pem -in my-host.cert -export -out my-host.pfx Copy the .pfx to your IIS machine. In IIS Manager, select “Server Certificates” on the server node, click “Import…” to import the .pfx certificate. Start up mmc, “File”, “Add/Remove Snap-in”, select “Certificates”, “Add”, “Computer account”, “Finish”, “OK”, (this click orgy shows you how important certificates were in 2008, as compared to Start/Administrative Tools/Data Sources (ODBC) 😉 ) and find the imported certificate(s) under Console Root\Certificates\Personal\Certificates Right-click each of them, select Properties, and make sure that the Friendly Name starts with “*.” for wild-card certificates. Otherwise, you cannot assign a host name for https web sites. Back in IIS Manager, select each site you want to add https support, click Bindings, Add, select Type: https and select the wild-card SSL certificate. Only if the friendly name starts with *, you can/must set the site’s Host name. Click OK and you are done. If you want your sites to redirect http to https automatically, make sure the Require SSL box is not checked in the site’s SSL Settings. The minimal web.config to perform these redirects looks like this (source, source) <?xml version="1.0" encoding="UTF-8"?> <configuration> <system.webServer> <rewrite> <rules> <rule name="Redirect-HTTP-HTTPS-IIS"> <match url="(.*)" /> <conditions> <add input="{HTTPS}" pattern="^OFF$" ignoreCase="true" />
</conditions>
<action type="Redirect" url="https://{HTTP_HOST}/{R:1}"
redirectType="Permanent" />
</rule>
</rules>
</rewrite>
</system.webServer>
</configuration>
Be aware that while these steps enable https for your IIS sites, self-signed certificates still require the users to explicitly accept the certificates in their browsers, which will raise an “Unknown issuer” warning at their first visit.
Update: There also seems to be a Powershell way to do it 😉
## Installing AWStats on Windows Server 2012
March 7, 2017
To install AWStats on Windows, first download the current version from awstats.org. If you don’t have Perl on your machine, get Strawberry Perl for Windows, as ActivePerl requires an annual Business License for production use.
On the server, create a web directory and a data directory for awstats. Follow the steps in the AWStats Setup Guide.
To access the log files of a remote IIS, I created a read-only share on c:\inetpub\LogFiles, and had to run
icacls c:\inetpub\LogFiles /reset /t
To get Strawberry Perl to run on IIS, follow this Installation Guide:
• In the Web Server role, you need to have the CGI feature installed.
• In IIS Administrator, create a web site or application hosting AWStats. In the site or application, you need to add a Script Map for *.pl executing
C:\path\to\perl.exe "%s" %s
Things should be running by now if you browse to
http://myHost/awstats/cgi-bin/awstats.pl?config=mySite
I noticed that the stats only included data from the installation date (IIS logs are configured to daily log files).
Answers on the internetz suggest to merge old log files using logresolvemerg.pl, a script that ships with awstats.
C:\awstats\tools>perl logresolvemerge.pl [all my log files] > merged.log
Replace the LogFile entry in your config file(s) to point to the merged log file
LogFile="C:\awstats-data\merged.log"
#LogFile="\\path\to\LogFiles\W3SVC1\u_ex%YY-1%MM-1%DD-1.log"
and run
perl awstats.pl -config=mySite
again after deleting the previously generated data files.
Unfortunately, the merged log only resulted in “dropped” and “corrupted” records:
Phase 1 : First bypass old records, searching new record...
Searching new records from beginning of log file...
Jumped lines in file: 0
Parsed lines in file: 30376
Found 16100 dropped records,
Found 0 blank records,
Found 14276 corrupted records,
Found 0 old records,
Found 0 new qualified records.
This may be caused by a number of reasons, but it turned out that the merged log requires a specific LogFormat:
LogFormat="%time2 %other %method %url %query %other %logname %host %ua %code %other %other %other"
Finally, I created a batch file awstats.cmd to update all my statistics
net use z: \\host\LogFiles awstats /user:awstats
d:
cd D:\wwwroot\awstats\wwwroot\cgi-bin
perl awstats.pl -update -config=mySite1
perl awstats.pl -update -config=mySite2
...
net use z: /delete
and created a scheduled task to automatically execute the script every day.
## Finding Spammers in hMailServer Log Files
February 4, 2016
hMailServer has a couple of spam protection measures built in, such as DNS blacklists and SURBL support. Among other features, you can also ban single IP adresses or IP ranges from connecting to your mail server.
While recently browsing through the log files, I noticed a couple of IP addresses which repeatedly connected to the mail server to log in, but kept their rate over the default 30 minutes auto-ban timer.
Interestingly those addresses chose to authenticate via AUTH LOGIN, but failed every time to provide a valid password. This results in a
535 Authentication failed
answer by the server, thus closing the conversation.
In the log file, the status code 535 looks like this
"SMTPD" 3228 21101 "2016-02-03 00:05:19.743" "xxx.xx.xx.xxx"
"SENT: 535 Authentication failed. Too many invalid logon attempts."
To find the conversations ending in status code 535, we can simply grep or findstr the relevant log files
grep "SENT: 535" *.log
In the log files, IP address is logged in the sixth column, so we can iterate over the resulting lines with the shell’s for command with option /f “tokens=6”.
Then we sort and count
(for /f "tokens=6" %i in ('grep "SENT: 535" *.log') do @echo %i)
| sort | uniq -c
To count the resulting IP addresses, I use my tool uniq, implemented after the Unix command uniq.
Similarly, one could also search for “550 Unknown user”.
## ASCII(), UNICODE() and the Replacement Character
June 3, 2013
This is a follow-up article on my previous post about string equality and collations.
We know, simply from looking at them, that the characters 'ܐ' and 'አ' are different, but the collation you use may treat them as equal.
Well, we can still compare their code point value by using the UNICODE() function, as in
select unicode(N'ܐ'), unicode(N'አ')
returning 1808 and 4768.
The reason I write this is because I discovered a fallacy in a comment on SO, resulting from mixing Unicode and non-Unicode literals and functions.
Take the statement
select ascii('ܐ'), ascii('አ')
Note that the Unicode characters are not given as Unicode strings (N” notation), but as non-Unicode strings.
Since both characters cannot be mapped onto Latin ASCII (or whatever your collation is), they are replaced by a Replacement Character, which is the question mark ‘?’ in ASCII.
Wikipedia tells us
The question mark character is also often used in place of missing or unknown data. In Unicode, it is encoded at U+003F ? question mark (HTML: ?).
and
In many web browsers and other computer programs, “?” is used to show a character not found in the program’s character set. […] Some fonts will instead use the Unicode Replacement Glyph (U+FFFD, �), which is commonly rendered as a white question mark in a black diamond (see replacement character).
So we can see where the question mark comes from, and thus both functions return 63.
In a similar, but nonetheless different case
select ascii(N'ܐ'), ascii(N'አ')
the characters are defined as Unicode strings, but passed to a function that only accepts non-Unicode strings. In this case, the mapping according to the current collation is performed by the ASCII() function, again resulting in the value 63.
As for the Unicode Replacement Character, you’ll encounter them if you decode a byte array to Unicode, and the decoder encounters a byte sequence that cannot be converted to Unicode given the selected encoding.
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Conference paper Open Access
# Energy Cooperation for Sustainable IoT Services within Smart Cities
Angel Fernandez Gambin; Elvina Gindullina; Leonardo Badia; Michele Rossi
### Dublin Core Export
<?xml version='1.0' encoding='utf-8'?>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:creator>Angel Fernandez Gambin</dc:creator>
<dc:creator>Elvina Gindullina</dc:creator>
<dc:creator>Michele Rossi</dc:creator>
<dc:date>2018-06-11</dc:date>
<dc:description>In this paper, we consider energy cooperation in an Internet of Things (IoT) smart city scenario. We assume the presence of interconnecting energy harvesting IoT gateways (GWs), that are endowed with energy harvesting capabilities and whose role is to collect and aggregate data from field sensing devices. Energy cooperation complements and balances the energetic needs to those devices that are neither connected to the power grid, nor satisfactorily served by energy harvesting due to the instability of ambient energy arrivals. The proposed solution entail energy transfers from energy rich gateways to energy scarce ones, i.e., those which are not connected to the power grid. To identify the optimal energy transfer/allocation scheme, we formulate a convex optimization problem that finds the optimal solution for heterogeneous smart systems. With this energy allocation technique, the gateways are unlikely to run out of energy during operation and the gap between energy offer and demand among interconnected gateways is kept to a minimum. We also quantify the performance of the proposed energy transfer policies as a function of network parameters, including: the amount of traffic generated by sensing devices, the number of smart services in the system, and the number of gateways that are connected to the power grid.</dc:description>
<dc:identifier>https://zenodo.org/record/1445358</dc:identifier>
<dc:identifier>10.1109/WCNC.2018.8377450</dc:identifier>
<dc:identifier>oai:zenodo.org:1445358</dc:identifier>
<dc:relation>info:eu-repo/grantAgreement/EC/H2020/675891/</dc:relation>
<dc:rights>info:eu-repo/semantics/openAccess</dc:rights>
<dc:subject>Energy harvesting; energy cooperation; energy consumption; Smart City; Smart services; Internet of Things</dc:subject>
<dc:title>Energy Cooperation for Sustainable IoT Services within Smart Cities</dc:title>
<dc:type>info:eu-repo/semantics/conferencePaper</dc:type>
<dc:type>publication-conferencepaper</dc:type>
</oai_dc:dc>
70
63
views
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# Pre GED Math Practice Test
1. Which of these numbers cannot be divided by $$2$$?
$$30,\; 22,\; 16,\; 13$$
A.
B.
C.
D.
2. The following chart shows you estimated wild tiger populations, by region. What's the total population of wild tigers worldwide according to this chart?
$$\begin{array}{|l|cr|} \hline \text{ Region } & Tiger\; population\: \: & \; \text{} \\ \hline \text{ India, Nepal and Bhutan } & 1650 & \text{} \\ \text{ China and Russia} & 450 & \text{} \\ \text{ Bangladesh} & 250 & \text{} \\ \text{ Sumatra (Indonesia) } & 400 & \text{} \\ \text{ Malaysia } & 500 & \text{} \\ \text{ other SE Asia } & 350 & \text{} \\ \hline \end{array}$$
A.
B.
C.
D.
3. Which of these numbers cannot be divided by $$2$$?
$$84,\; 34,\; 31,\; 58$$
A.
B.
C.
D.
4. Simplify the following expression:
$$0\div11$$
A.
B.
C.
D.
5. If a math tutor worked for $$47$$ hours and was paid $$\15$$ each hour, how much money would she have made?
A.
B.
C.
D.
6. Simplify the given expression.
$$10 \cdot 0 = 0$$
A.
B.
C.
D.
7. One boat from an island can take $$6$$ people. How many boat trips are required to ferry $$45$$ people from the island?
A.
B.
C.
D.
8. The swimming pool has a length of $$25$$ yards. One lap is swimming up and then back again. How many yards did John swim doing $$27\;$$ laps?
A.
B.
C.
D.
9. Compute the exact value of this exponential expression:
$$5^{2} \cdot 4^{1}$$
A.
B.
C.
D.
10. A square has the given side of $$28$$ inches. Find the area.
A.
B.
C.
D.
Try the Pre GED Science Test
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# Chapter R - Elementary Algebra Review - R.4 Polynomials - R.4 Exercise Set: 12
$x$
#### Work Step by Step
Using $\dfrac{1}{a^{-n}}=a^{n}$, the given expression, $\dfrac{1}{x^{-1}} ,$ simplifies to \begin{array}{l}\require{cancel} x^1 \\\\= x .\end{array}
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Decorative fence
The rectangular garden is 57 m long and 42 m wide. Calculate how much m² it will reduce if it encloses with a 60 cm wide decorative fence.
Correct result:
S = 117.36 m2
Solution:
$a=57 \ \text{m} \ \\ b=42 \ \text{m} \ \\ p=60 \ cm \rightarrow m=60 / 100 \ m=0.6 \ m \ \\ \ \\ S_{1}=a \cdot \ b=57 \cdot \ 42=2394 \ \text{m}^2 \ \\ S_{2}=(a-2 \cdot \ p) \cdot \ (b-2 \cdot \ p)=(57-2 \cdot \ 0.6) \cdot \ (42-2 \cdot \ 0.6)=\dfrac{ 56916 }{ 25 }=2276.64 \ \text{m}^2 \ \\ \ \\ S=S_{1}-S_{2}=2394-2276.64=\dfrac{ 2934 }{ 25 }=117.36 \ \text{m}^2$
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## How would you produce and detect
1. Plane
2. Circularly polarized and
3. Elliptically polarized light
## In a michelson interferomete, When 100 fringes were shifted, The final reading of the screw was found to be 10.735 mm.
If the wavelength of the light was $5.92×{10}^{-7}m,$
What was the initial reading of the screw?
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## Introduction
The Antarctic Ice Sheet feeds ice into the ocean through fast-flowing outlet glaciers and ice streams1. The glaciers and ice streams form floating ice tongues and ice shelves, accounting for 74% of the coastal margins of the ice sheet2. Because approximately half of the ablation of the Antarctic Ice Sheet occurs due to basal melting of floating ice3,4, processes at the ice–ocean interface are very important for the ice sheet mass budget. Furthermore, thinning or loss of an ice shelf causes reduction in the buttressing force acting on inland ice. This accelerates ice discharge into the ocean, as observed in the Antarctic Peninsula region5. The most rapidly melting ice shelves are observed in the Bellingshausen–Amundsen seas sector of West Antarctica6. There, warm and dense circumpolar deep water (CDW) with a temperature of ~1°C (~3°C above the in situ freezing point) flows into subshelf cavities, supplying heat that enables rapid melting7. Observations and numerical modeling have indicated strong temporal and spatial variability in the basal-melt rates of ice shelves in the Bellingshausen and Amundsen seas8,9, where mass loss of the ice sheet has recently been increasing10. Recent observations at Totten Glacier demonstrated that rapid ice shelf base melting occurs in East Antarctica as well11,12. Heat for the basal melting in the Totten Glacier region is carried by modified CDW, which is slightly colder than CDW but still sufficiently warm for melting.
Circulation beneath these rapidly melting ice shelves is driven by an out-flowing buoyant plume generated by basal melting. This density-driven circulation allows for the advection of warmer, saltier, and denser water to the grounding line. The basal-melt rate $$\dot{m}$$ is commonly estimated from ocean temperature T, salinity S, and current speed U in the boundary layer beneath an ice shelf13 (see full equations in “Methods”)
$$\dot{m}=f(U,\ T-{T}_{\mathrm{f}},\ S),$$
(1)
where Tf is the freezing point of the seawater in the mixed layer beneath the ice shelf, but calculated using the pressure of the ice shelf base. Therefore, direct measurements of temperature, salinity, and water current underneath an ice shelf are crucial to estimate the melt rate based on physical processes at the ice–ocean boundary. In addition, in situ oceanic data from a subshelf cavity provides critical information for furthering our understanding of basal melting and water circulation. The equation for estimating melt rate assumes the transfer of ocean heat from a fully turbulent boundary layer to a hydraulically smooth ice shelf base13. However, turbulence in the boundary layer and the roughness of the ice-surface topography can only be investigated using direct measurements.
Subshelf density-driven circulation is categorized into three major modes, which are associated with the magnitude of ocean heat transport into a subshelf cavity14. In Mode 1, salty water at freezing points is generated by brine rejection during sea ice production and flows into the subshelf cavity. Since the freezing point drops as pressure increases, heat is available in the water to melt ice at the depth of the ice shelf base and a part of the upwelling cooled water might refreeze before it exits the cavity. In Mode 2, warm CDW flows onto the continental shelf and into the cavity. CDW causes rapid basal melting since it is warmer than the freezing point at the ice shelf base, typically by ~3°C. In Mode 3, Antarctic Surface Water (ASW) carries heat supplied by the atmosphere into the cavity, causing a significant amount of basal melting near the ice shelf front during the summer15,16. Although these modes are widely accepted in the community, in situ oceanic data with coverage of an entire cavity is required for validation of the proposed processes.
Studies have shown melting near the ice shelf grounding lines to be a key for the future evolution of the ice sheet17. However, understanding of mixing processes in the vicinity of grounding lines is limited due to the difficulty of obtaining measurements. Theoretical studies suggest that tidal currents erode stratification and induce vertical mixing near the grounding line18. On the other hand, recent observations found a lack of tidal mixing in the grounding zone of Ross Ice Shelf, where vertical mixing was instead driven by relatively inefficient diffusive convection19. To better understand the processes controlling basal melting, in situ observations of thermohaline structure, and circulation regimes will be required over extensive areas of a subshelf cavity.
In this study, we made direct measurements of ocean properties beneath the ice shelf of Langhovde Glacier, East Antarctica in January 2018, using boreholes drilled with a hot-water system through ice 234–412-m thick. Four boreholes were located along a glacier flowline from the ice shelf front to the grounding line. Water temperature, salinity, and currents were measured in the subshelf ocean cavity, which had thicknesses between 302 and 12 m. Our observations confirmed the advection of warm water into the cavity, and revealed the circulation regime under the ice shelf of Langhovde Glacier. Based on these data, we quantify the rate of basal melting and the stability of the water column, providing an overview of the ice–ocean interaction of the ice shelf of a small outlet glacier in East Antarctica.
## Results
### Study site
Langhovde Glacier is a 3-km wide outlet glacier draining into Lützow-Holm Bay in East Antarctica (Fig. 1). The lower 10 km of the glacier flows in a channel, which is bounded by bedrock to the west and by slowly moving ice to the east (Fig. 1). In this channelized region, ice near the ice shelf front flows at a rate of ~150 m a−1 (ref. 20), modulated by tides21. Surface topography indicates that the glacier has a several kilometer long ice shelf with a grounding line protruding toward the ocean in its central part22 (Fig. 1b). The grounding zone of the glacier was drilled in a previous field campaign, which confirmed a 10-m-thick ocean-water layer below 430-m-thick ice at 3.1 km from the ice shelf front23. The grounding line location was estimated to be several hundred meters upglacier from the drilling site.
The bathymetry of Lützow-Holm Bay was also surveyed during previous expeditions24. A ≥1-km deep ocean trough is incised into the continental shelf and continues to Shirase, Skallen, and Telen Glaciers, which are major outlet glaciers terminating in the bay (Fig. 1a). Another trough extends to Langhovde Glacier along the Soya Coast through a passage between the ice sheet and Ongul Islands, where Syowa Station, a Japanese research base, is located (Fig. 1a). The depth of the ocean in front of the glacier is ~600 m, but ocean bed topography below the Langhovde Glacier ice shelf is unknown.
Lützow-Holm Bay is located near the eastern boundary of the Weddell Gyre, where the Antarctic Circumpolar Current approaches the coast25 (Fig. 1). The clockwise circulation of the gyre and coastal current carry Warm Deep Water (WDW) along the continental slope, which supplies heat for the melting of the ice sheet’s margins15,26. WDW flows into the main trough of Lützow-Holm Bay27, resulting in significant basal melting of Shirase Glacier28,29. A mooring measurement under Fimbul Ice Shelf in Dronning Maud Land showed that the cavity was occasionally accessed by Modified Warm Deep Water (MWDW), which is a mixture of WDW and Winter Water (WW)15.
### Ice and bed geometries
Borehole drilling revealed ice and ocean bed geometries of the Langhovde Glacier ice shelf at the drill sites (Fig. 1c and Table 1). At the northernmost borehole BH1801 (0.5 km from the front), 234-m thick ice was underlain by a 302-m-deep seawater column. The ice thickness increased toward the grounding zone, i.e., 294 m at BH1802, 389 m at BH1803, and 412 m at the uppermost borehole BH1804 (2.4 km from the front). Bed elevation increases and the water layer thins toward the grounding zone (Fig. 1c). At BH1804, the bed elevation was 364 m below sea level and the subshelf water layer was 12-m thick.
Ice thickness measured at boreholes BH1801, 02, and 04 agrees with those calculations based on the hydrostatic equilibrium assumption to within 3% (Fig. 1c and Table 1). Nevertheless, ice thickness at BH1803 was greater than the estimation by 40 m, showing that the buoyant force at the drill site exceeded the gravitational force by 10%. The ice was being depressed by stresses acting from ice surrounding the site.
An upward-looking video camera was lowered into the borehole at BH1802 to inspect the lower surface of the ice shelf (Supplementary movie). The surface was covered with dimples with a diameter of 0.2–0.3 m and depth of 0.05–0.1 m (Fig. 2).
### Subshelf ocean properties
Oceanographic measurements were repeated 2–3 times at each borehole, so that conditions were observed during both high and low tides (Supplementary Fig. 2). Conservative temperature (Θ) in the cavity ranged between −1.4 and −1.15°C (Fig. 3a). Conservative temperature and absolute salinity (SA) increased from the ice to the seafloor (Fig. 3a). In the vicinity of the ice–ocean boundary, the water temperature was above the in situ freezing point by 0.65–0.95°C (Fig. 3a and Supplementary Fig. 4). Water properties were nearly homogeneous below 420 m. The average conservative temperature and absolute salinity within the layer below 420 m were −1.18°C and 34.48 g kg−1, respectively. Near the grounding line at BH1804, water was colder and fresher than that observed at the same depth in the other profiles by 0.05°C and 0.04 g kg−1 (Fig. 3a).
Brunt-Väisälä frequency (stability of water stratification, N) showed local maxima at 18–48 m below the ice base (Fig. 3b, Table 2, and Supplementary Fig. 3), which we assume to mark the lower boundary of the ice–ocean boundary layer. Salinity within the boundary layer was similar at the four sites, whereas temperature near the grounding line (BH1803–04) was more than 0.1°C higher than closer to the front (BH1801–02) (Table 2).
At BH1801, water flowed toward the open ocean within ~50 m of the ice base, whereas a relatively strong current (40 mm s−1) toward the grounding line was observed at a depth of 260–420 m. The flow below 420 m was relatively weak (10–30 mm s−1) and changed its direction between high and low tides (Fig. 4). At BH1802, water in the boundary layer flowed downglacier at a rate of 20 mm s−1. Below the boundary layer, current deviated eastward at a rate of 10 mm s−1 (Fig. 4). At BH1803 and 04, the magnitude and direction of the water current was substantially different between the casts performed at high and low tides. For example, at BH1803, the entire water column flowed toward the glacier front during high tide (Fig. 4a), whereas flow in the upper layer switched toward the grounding line during low tide (Fig. 4b). Water current generally increased toward the front, except for the relatively weak flow observed at BH1802 (Table 2). The observed range of water current (10–40 mm s−1) corresponds to a travel time of 3–0.7 days between the ice shelf front and the grounding zone.
## Discussion
The water temperature in the ice–ocean boundary layer exceeded the in situ freezing point at all the boreholes (Fig. 3), indicating that the ice was subjected to basal melting from the grounding line to the front. Water properties within the ice–ocean boundary layer generally lie along the meltwater-mixing line except for a thin 2–3-m thick water layer immediately below the ice (Fig. 5c and Supplementary Fig. 4b). Basal-melt rates were estimated by solving the heat and salt conservation equation (Eqs. (3)–(6))13 using previously proposed parameters30 (Supplementary Table 1). The calculation assumes a fully turbulent boundary layer and a hydraulically smooth ice–water boundary, which we discuss based on our borehole observations in the Supplementary Note. Melt rates computed for BH1801–04 varied from 0.56 to 2.46 m a−1 with a mean rate of 1.42 m a−1. We tested the sensitivity of the calculated melt rates to temperature, salinity, and water speed using the observed ranges. The melt rate is highly dependent on the water current (1.3 m a−1/10 mm s−1 at T = −1.25°C and SA = 34.45 g kg−1) and water temperature (0.3 m a−1/0.1°C at SA = 34.45 g kg−1 and U = 20 mm s−1). The melt rate at BH1803 (2.46 m a−1) was 40% greater than the mean over the four locations (Fig. 4). This rapid melting at BH1803 was a result of the presence at the ice base of relatively warm water (−1.2°C) at a greater depth (~360 m, Tf = −2.16°C) and a relatively strong current (Fig. 3a and Table 2).
The melt rates obtained in this study were greater than those observed using borehole-derived oceanographic data at other ice shelves in East Antarctica15,31. Under Fimbul Ice Shelf, the melt rate is low because WW occupies the cavity for most of the year, with MWDW only occasionally entering the cavity15. A borehole measurement near the front of Amery Ice Shelf in Prydz Bay revealed that basal melting at the drill site is limited year-round since basal ice is protected from melting by a layer of super-cooled water31. Relatively rapid basal melting obtained at the ice shelf of Langhovde Glacier is due to subshelf water temperatures of ~−1.2°C, which are warmer than those observed under other ice shelves in East Antarctica. Because the ice shelf is relatively small, subshelf water exits the cavity before the heat is fully utilized for basal melting. Consequently, the cavity is filled with water above freezing, resulting in melting along the entire ice shelf (Fig. 4). In addition, a long-term water temperature measurement near the grounding line of Langhovde Glacier showed that water temperature remains above the freezing point year-round23, suggesting that basal melt occurs throughout the year. These results imply that the subshelf environment of Langhovde Glacier is more directly connected to surrounding ocean conditions, and thus is more vulnerable to oceanic warming than other ice shelves in the region.
Subshelf circulation observed under the Langhovde Glacier ice shelf was characterized by the outflow of a buoyant plume along the ice base and inflow of salty warm water at depth (Figs. 3 and 4). Thermal forcing (water temperature above the in situ freezing point) at depths between 350 and 450 m was 0.69–0.77°C (Fig. 3a). This value is approximately half that observed in the Bellingshausen–Amundsen Seas6 and ~0.3°C higher than observed in front of Totten Glacier, East Antarctica11 in the same depth ranges. Ocean-water properties from previous observations indicated the existence of WDW, MWDW, ASW, and WW in Lützow-Holm Bay (Fig. 5a, b). Subshelf water temperature and salinity obtained in this study are similar to those of WW (Fig. 5b). Furthermore, water properties at BH1801–03 below ~350 m lie along a WDW-WW mixing line on the Θ-SA diagram (Fig. 5c). These observations suggest that MWDW, that originated in the deep ocean, mixes with WW over the course of being transported through a seafloor trough into the subshelf cavity of Langhovde Glacier. The Θ-SA diagram indicates that deep water in the Lützow-Holm Bay is warmer than that under the Langhovde Glacier ice shelf. It is likely that similar or even warmer water reaches other outlet glaciers in the region (e.g., Skallen and Telen Glaciers) (Fig. 1a), supplying heat for melting. Indeed, beneath the floating tongue of Shirase Glacier, recent measurements indicate rapid melting due to inflow of WDW29.
We compared the water properties at BH1804 with those obtained at the same location in the previous field campaign in January 2012 (BH1201) (Fig. 1b)23. Interestingly, subshelf water at BH1804 in January 2018 was ~0.25°C warmer and ~0.02 g kg−1 saltier than the January 2012 observation (Fig. 5)23. We hypothesize that a change in sea ice conditions in the bay affected the subshelf conditions in 2018. Lützow-Holm Bay is usually covered by thick land-fast ice for most of the year32, which was the case for the period from 2006 to 2016. In summer 2016, the sea ice broke off and drifted away from the region, including the forefront of Langhovde Glacier33. The loss of the thick multiyear ice resulted in open water conditions near the glacier front during the following summers in 2017 and 2018 (Fig. 6). Offshore pack-ice cover was also relatively sparse during those summers. We assume that the higher temperature near the grounding line in 2018 was the result of the interaction of the open water with the atmosphere. Several studies suggest that interannual variability in the thickness of subsurface warm water is caused by wind stress27,34 and sea–ice production35. Observations near the front of Ross Ice Shelf indicated that basal melting is enhanced by inflow of warm surface water, generated in front of the ice shelf by summer insolation16,36,37. We are not able to specify the mechanism linking the sea ice condition and water properties in the cavity. Nevertheless, it is likely that the recent change in sea ice in Lützow-Holm Bay is at least partly affecting the ice shelf of Langhovde Glacier and other glaciers in this region.
Within the ice–ocean boundary layer at BH1801–02, water flowed toward the front, which we attribute to a buoyant plume generated by meltwater mixing with MWDW (Figs. 3 and 4). Previous observations in 2012 showed similarly stratified subshelf water near the grounding line and implied outflow along the ice base15. Our new data obtained in 2018 revealed the details of the stratification and flow under the entire ice shelf. A buoyant plume developed under the influence of freshwater generated by basal melting, and a buoyancy-driven current was enhanced by the relatively steep basal slope between BH1801 and 03 (Fig. 4). As a consequence of the interaction with the ice, the buoyant plume layer thickens and water flows faster as it approaches the ice shelf front (Figs. 3 and 4). This observation is consistent with a previously proposed plume theory. Basal melting near the grounding line generates a buoyant plume, which upwells alongside the ice–ocean interface toward the open ocean38. Such subshelf circulation is strongly affected by ocean temperature change39, therefore long-term monitoring is required both in the ocean and in the cavity to understand the response of the subshelf conditions to the oceanic forcing.
Near the grounding line, a complex water circulation was observed (Fig. 4). Within the same vertical distance from the ice, water near the grounding line at BH1804 was fresher and colder than that at BH1803 (Fig. 3a). As a result of the water property distributions, potential density (σΘ) at 370-m depth increased downglacier from 27.58 kg m−3 at BH1804 to 27.61 kg m−3 at BH1803 (Fig. 3b). By assuming that the circulation in the narrow subshelf cavity near the grounding line is non-geostrophic, this horizontal density gradient has the potential to drive water flow near the grounding line, i.e., cold and freshwater flows downglacier as a buoyant plume, whereas warm and dense water fills the grounding zone. In contrast to our observations, a 10-m-thick water column in the grounding zone of Ross Ice Shelf was vertically stratified under conditions of relatively weak current (~10 mm s−1) and low thermal forcing (<0.1°C)19. The stratification was attributed to the absence of tidal mixing, and diffusive convection was suggested as a key process for driving basal melting. Thermal forcing of subshelf water at Langhovde Glacier was an order of magnitude higher than that under Ross Ice Shelf. Presumably, more rapid melting and a steeper basal ice slope generates an upslope-flowing plume, resulting in shear instabilities in the water column in the grounding zone of Langhovde Glacier. We conclude that the mixing conditions near the grounding line are controlled by the strength of basal melting, basal ice topography, and the magnitude of the tidal current, all of which will differ for different ice shelves.
The thermohaline structure and circulation presented in this study are consistent with those proposed by observation and theoretical analysis of Antarctic ice shelves7,40. Observations that cover the entire field of a cavity are sparse7, thus our data serve as an important validation of theories and models used for estimation of basal melting. Our results indicate warm water inflow into the cavity and relatively high melt rates at the base of the Langhovde Glacier ice shelf. Similar processes are expected at other Antarctic outlet glaciers in similar settings. Despite the abundance of small ice shelves distributed around Antarctica, their size means that they are not resolved in numerical models used for predicting future sea-level change40. Further investigation and long-term monitoring of such ice shelves is crucial for an understanding of the response of the Antarctic Ice Sheet to a changing climate.
## Methods
### Hot-water drilling and ice geometry measurements
In December 2017 and January 2018, we drilled four boreholes along a flowline of the ice shelf of Langhovde Glacier. Boreholes were drilled 0.55, 1, 1.6, and 2.5 km from the ice shelf front and referred to as BH1801, BH1802, BH1803, and BH1804 (Fig. 1). The drilling was performed at a mean rate of 42 m h−1 with a hot-water drilling system, which had been previously used on Langhovde Glacier23. The system consists of three high-pressure, hot-water machines (Kärcher HDS1000BE), a 1/2-inch diameter hose, and a winch system41. Meltwater was supplied to the system from a supraglacial stream or pond. Ice thicknesses and ocean bed elevation were determined to an accuracy of ±1 m, using water temperature, salinity, and pressure profiles described in the next section. At each borehole site, ice-surface elevation was measured using a dual-frequency GNSS (Leica Geosystems System 1200) with the static method using data from a reference station (GNSS Technologies, GEM-1) located on the western flank of the glacier. In order to observe lower surface ice topography, an upward-looking video camera (GoPro, HERO5) enclosed in a pressure vessel was lowered into the borehole at BH1802. Roughness of the ice surface was measured by visual inspection of the imagery.
### Oceanographic measurements and data processing
We used a CTD (conductivity, temperature, and depth) profiler (IDRONAUT Ocean Seven 304, Supplementary Fig. 1) to measure water temperature, conductivity, and depth to accuracies of 0.005°C, 7 × 10−3 mS cm−1, and 0.5 m, respectively. The conductivity was used to derive salinity to an accuracy of ~5 × 10−3 psu (ref. 42). The sample rate was 1 Hz. During the subshelf measurements the profiler, suspended on a Kevlar rope, was pulled upwards at a rate of 1 m s−1. To achieve a higher vertical resolution near the ice–ocean boundary, the profiler was moved more slowly, at a rate of 0.1 m s−1. Water depth was derived from pressure and water density computed from salinity and temperature. The magnitude and direction of the water current were measured with a current meter (JFE Advantec, AEM-USB; accuracy 10 mm s−1 and 2°) combined with the CTD profiler (Supplementary Fig. 1). We recorded data for 1 min with 1-s sampling intervals every 10 m in boreholes BH1801–02 and every 5 m at BH1803–04. Mean current and direction were calculated by averaging the 1-min data.
The CTD and current meter measurements were performed a day after drilling to avoid any influence from the drilling. The measurements were repeated two or three times in each borehole during periods of high and low tides to reduce the influence of tidal currents, with the exception of BH1803. As a result of a severe storm, two measurements at BH1803 took place several hours after high tide (Supplementary Fig. 2). Because of the timing of the measurements, the observed speed may be weaker than its long-term mean. However, we speculate that the magnitude of tidal currents is small because of the limited volume of the subshelf cavity particularly near the grounding line.
Water temperature and salinity were converted into conservative temperature (Θ) and absolute salinity (SA) using the standard equations43. To investigate mixing of meltwater and MWDW, we used a Θ-SA diagram. As MWDW mixes with meltwater, temperature and salinity show variations on the diagram along the Gade line or meltwater-mixing line44, which connects water properties of MWDW and meltwater. The meltwater-mixing line was obtained using equations defined ratio of SA to Θ changes when ice melts into seawater44,45, using the Gibbs SeaWater Oceanographic Toolbox of TEOS-1043. MWDW is a mixture of WDW and WW, thus MWDW is identified on the Θ-SA diagram along a line connecting properties of WDW and WW. To find the lower boundary of the ice–ocean boundary layer, we analyzed the Brunt-Väisälä frequency (N)
$${N}^{2}=\frac{g}{\rho }\frac{{\rm{d}}\rho }{{\rm{d}}z},$$
(2)
where ρ is water density, g is the gravitational acceleration, and z is the depth below the ocean surface. Negative or zero N2 values indicate that the water column is unstable, and the stability increases as N2 takes on a larger value. In order to minimize the influence of small-scale instabilities, water density was smoothed using a Gaussian filter with a half-width of 50-m depth. A half-width of 20 m was used at BH1804 because the water column was thin. The local maximum in the calculated N2 was used to define the lower limit of the boundary layer at BH1801–03. At BH1804, we assumed the entire water column to be a boundary layer.
In addition to the data obtained in this study, we used temperature and salinity profiles available for Lützow-Holm Bay and the nearby open ocean (Fig. 5a). These measurements were obtained by the Japanese Antarctic Research Expedition between 1957 and 2018.
### Basal-melt rate
Current, temperature, and salinity measurements were used to calculate the basal-melt rate, $$\dot{m}$$, by solving the balance equation of heat and salt at the ice–water boundary13,46.
$${\rho }_{{\mathrm{fw}}}\dot{m}{L}_{\mathrm{i}}-{K}_{\mathrm{i}}{\left(\frac{\partial T}{\partial z}\right)}_{\mathrm{b}}=\rho {c}_{\mathrm{w}}{\gamma }_{\mathrm{T}}(T-{T}_{{\rm{0}}})$$
(3)
$${\rho }_{\mathrm{fw}}\dot{m}{S}_{{\rm{0}}}=\rho {c}_{\mathrm{w}}{\gamma }_{\mathrm{S}}({S}_{\mathrm{A}}-{S}_{{\rm{0}}})$$
(4)
where T, SA, ρ, and cw are the temperature, absolute salinity, density, and specific heat capacity of water in the boundary layer, Li and Ki are the latent heat of fusion and thermal conductivity of ice, and ρfw is the density of freshwater. The temperature gradient at the base of the ice (∂T/∂z)b was estimated to be 0.015°C m−1 from borehole temperature data obtained in a previous campaign23. The freezing point at the ice–ocean interface T0 was given by absolute salinity at the interface S0 and pressure P as defined by the “gsw_t_freezing()” function43.
The heat and salt transfer coefficients, γT and γS, for a fully developed turbulent flow over a hydraulically smooth boundary are given by
$${\gamma }_{\mathrm{T}}=\frac{{K}^{1/2}U}{2.12{\rm{ln}}({K}^{1/2}Re)+12.5P{r}^{2/3}}$$
(5)
and
$${\gamma }_{\mathrm{S}}=\frac{{K}^{1/2}U}{2.12{\rm{ln}}({K}^{1/2}Re)+12.5S{c}^{2/3}},$$
(6)
where K is the ice-surface drag coefficient, Pr and Sc are the Prandtl and Schmidt numbers of seawater, and Re is the Reynolds number defined by Re = UD/v, with v being the kinematic viscosity13. Measured T, SA, P, and U were averaged over the ice–ocean boundary layer D (Table 2) to numerically solve the equations (Eqs. (3) and (4)) with the function of T0 for $$\dot{m}$$, T0 and S0 by an iterative method. All of the parameter values used in the calculation were obtained from a previous study30 (Supplementary Table 1).
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# 5.4 Summary of the uniform and exponential probability distributions
Page 1 / 1
This module provides a summary of formulas and definitions related to Continuous Random Variables.
Formula
## Uniform
$X$ = a real number between $a$ and $b$ (in some instances, $X$ can take on the values $a$ and $b$ ). $a$ = smallest $X$ ; $b$ = largest $X$
$X$ ~ $U\left(\mathrm{a,}b\right)$
The mean is $(\mu , \frac{a+b}{2})$
The standard deviation is $(\sigma , \sqrt{\frac{\left(b-a{\right)}^{2}}{12}})$
Probability density function: $f(X)=\frac{1}{b-a}$ for $((a, X), b)$
Area to the Left of x: $((P\left(X, x\right)), \text{(base)}\text{(height)})$
Area to the Right of x: $((P\left(X, x\right)), \text{(base)}\text{(height)})$
Area Between c and d: $((((P\left(c, X), d\right)), \left(\text{base}\right)\left(\text{height}\right)), \left(d-c\right)\left(\text{height}\right))$ .
Formula
## Exponential
$X$ ~ $\mathrm{Exp}\left(m\right)$
$X$ = a real number, 0 or larger. $m$ = the parameter that controls the rate of decay or decline
The mean and standard deviation are the same.
$\mu =\sigma =\frac{1}{m}$ and $m=\frac{1}{\mu }=\frac{1}{\sigma }$
The probability density function: $f\left(X\right)=m\cdot {e}^{\mathrm{-m\cdot X}}$ , $(X, 0)$
Area to the Left of x: $((P\left(X, x\right)), 1-{e}^{\mathrm{-m\cdot x}})$
Area to the Right of x: $((P\left(X, x\right)), {e}^{\mathrm{-m\cdot x}})$
Area Between c and d: $((P\left(c, X), d\right))=(P\left(X, d\right))-(P\left(X, c\right))=\left(1-{e}^{\mathrm{- m\cdot d}}\right)-\left(1-{e}^{\mathrm{- m\cdot c}}\right)={e}^{\mathrm{- m\cdot c}}-{e}^{\mathrm{- m\cdot d}}$
Percentile, k: $k=\frac{\text{LN(1-AreaToTheLeft)}}{\mathrm{-m}}$
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid.
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## Is Mathematica's lexical scoping broken?
14
5
Why do these two lines output different things (b and 0, respectively)? Is it a bug?
ReleaseHold@With[{a=0},FlattenAt[Map[List,FlattenAt[Hold@With[{b=0,b=b},b+0a],{1,1}],{2}],{1,-1}]]
ReleaseHold@With[{a=0},FlattenAt[Map[List,FlattenAt[Hold@With[{b=0,b=b},b+0b],{1,1}],{2}],{1,-1}]]
(Sorry, I couldn't find a shorter example that didn't give an error.)
5
This is not a bug, see for instance tutorial/LocalConstants (in particular the last line).
– None – 2016-03-08T13:20:25.433
1
The fact that renaming can occur is documented e.g. in Details section of With documentation: "With constructs can be nested in any way, with inner variables being renamed if necessary.", there are examples of it in Properties & Relations and Possible Issues sections, but it's not clearly stated when renaming is "necessary".
– jkuczm – 2016-03-08T13:53:41.757
1
Related: Enforcing correct variable bindings and avoiding renamings for conflicting variables in nested scoping constructs. @Mehrdad If you think that simulating scoping by means of variable renaming is debatable design decision then you're not alone.
– jkuczm – 2016-03-08T13:54:48.770
1
Please don't continue discussions here, there are already two chats for this topic: jkuczm-and-mehrdad chat and the newest one: on-answer-by-daniel-lichtblau
– Kuba – 2019-08-21T06:22:23.230
## Answers
9
Just wanted to post back 3 years later to dissect what's happening.
### tl;dr: It's a broken (sorta-lexical, sorta-dynamic, sorta-syntactic) substitution. If you really want lexical scoping, Module may be closer to what you want, although it's not clear to me that even Module (or anything else) could possibly do it perfectly.
To see what's happening, here's the same code, echoing every step:
(Edit: Renamed the second b to c to avoid a duplicate variable as suggested in the comments.)
In[1]:= Module[{x},
With[{a = 0},
x = Echo[Hold[With[{b = 0, c = b}, c + 0 a]]];
x = Echo[FlattenAt[x, {1, 1}]];
x = Echo[Map[List, x, {2}]];
x = Echo[FlattenAt[x, {1, -1}]];
ReleaseHold[x]]]
>> Hold[With[{b$$=0,c$$=b},c$+0 0]] >> Hold[With[b$$=0,c$$=b,c$+0 0]]
>> Hold[With[{b$$=0},{c=b},{c$$+0 0}]]
>> Hold[With[{b$$=0},{c=b},c$$+0 0]]
Out[1]= b
Here's what happens if we change 0 a to 0 b (or in fact, 0 followed by any other variable):
In[2]:= Module[{x},
With[{a = 0},
x = Echo[Hold[With[{b = 0, c = b}, c + 0 b]]];
x = Echo[FlattenAt[x, {1, 1}]];
x = Echo[Map[List, x, {2}]];
x = Echo[FlattenAt[x, {1, -1}]];
ReleaseHold[x]]]
>> Hold[With[{b=0,c=b},0 b+c]]
>> Hold[With[b=0,c=b,0 b+c]]
>> Hold[With[{b=0},{c=b},{0 b+c}]]
>> Hold[With[{b=0},{c=b},0 b+c]]
Out[2]= 0
What seems to be happening here is that Mathematica only attempts to figure out the scope of a variable once it is forced to actually manipulate the expression containing it semantically. In this case, that means the semantic substitution of a=0 in the inner expression is triggering variable renaming, which is how Mathematica resolves naming conflicts. If that doesn't ever happen, then the engine doesn't bother to resolve name conflicts at all.
So after the ReleaseHold, what we're left with are these two expressions to evaluate:
With[{b$$= 0}, {c = b}, c$$]
With[{b = 0}, {c = b}, c ]
This is actually a "sequential With" that is highlighted poorly, but equivalent to the following:
With[{b$$= 0}, With[{c = b}, c$$]]
With[{b = 0}, With[{c = b}, c ]]
In the first case, the outer c$= 0 has no effect on the innermost expression, since it is hidden ("shadowed") by the inner one. In the second case, this is also true, but the 0 value of b is substituted into c before b is hidden. Thus in the first case we're left with the unbound symbol b, whereas in the second case it simplifies down to 0. The thing is, this is wildly broken behavior, because lexical scope is a static property, whereas Mathematica triggers it dynamically. Note that whether Mathematica decides to resolve naming conflicts via renaming or via tracking them internally (like most other languages) isn't relevant to this—they're just implementation details. The important thing is that expressions maintain their semantics, and that requires that a variable like c refers to the same thing regardless of whether it is later added to 0 a or 0 b. ## Then why does Mathematica do this? Because the language specification is self-contradictory. Specifically (not sure if pun intended): • With is a scoping construct that implements read-only lexical variables. With replaces symbols in expr only when they do not occur as local variables inside scoping constructs. • With[{x=x0,y=y0,…},expr] specifies that all occurrences of the symbols x, y, … in expr should be replaced by x0, y0, …. At first glance these seem fine, but in fact these two requirements point to a deep inconsistency in how Mathematica treats expressions, stemming from a lack of a clear separation between static and dynamic properties of code. Namely, the first specification is a semantic requirement, and thus requires static semantic analysis of expr. For Mathematica to implement "lexical scope" and understand that it should not replace a nested local variable, it needs to be able to identify each With, Module, etc. construct in the first place. However, this is impossible inside constructs like Hold, because Mathematica is so dynamic that it allows you to construct expressions at runtime. For example, compare the following two statements: ReleaseHold[With[{x = 0}, Hold[With [{x = 1}, x]]]] (* OK?!?! *) With2 = With; ReleaseHold[With[{x = 0}, Hold[With2[{x = 1}, x]]]] (* error! *) You'd expect them to be semantically equivalent, except that it is impossible for Mathematica to know whether a given expression is a With construct without actually evaluating it. (A With might be something else initially, like Symbol["With"], or, conversely, the user might have wanted to later replace a With with something else.) The second specification acknowledges as much. It admits that in fact the language is performing a symbolic (syntactic) substitution, not a semantic one. And that implies, by its definition, that With does not implement lexical scoping. Instead, what With really does is that it fakes a semantic pre-analysis: it pretends that the language isn't dynamic at all, assumes that any With expression will look like With[...] (rather than, say, With2[...] above), and then performs the substitution hoping that you don't notice any discrepancy. Admittedly, in practice, it works fine, because people rarely write expressions whose meanings are extremely dynamic, and thus preserve the static meanings of symbols—otherwise, it becomes harder for people to read such code, too. But it's a fundamentally conflicting language requirement to attempt to statically substitute inside what can be a completely dynamic expression, and there isn't really a true fix for it. The best they can do, therefore, is to fix the documentation and explain precisely what really happens (a half-baked mishmash of symbolic/static/dynamic substitution, with a trigger of variable renaming that can suddenly change the semantics of the inner expression unexpectedly!), as opposed to what they're trying to fake (lexical scoping). 2I fail to see how this is anything other than undefined behavior.If there is a bug, it is that With did not give an error when a variable appeared twice in the lhs of the assignment list. – Daniel Lichtblau – 2019-08-18T13:38:12.223 1This variant might be a better example. In that inner With change the second variable. With[{b = 0, c = b}, c + 0 b] vs. With[{b = 0, c = b}, c + 0 a] – Daniel Lichtblau – 2019-08-18T13:44:29.727 @DanielLichtblau: Thanks, I changed it to clarify. But I disagree that it's undefined behavior. There's a Hold outside which prevents its evaluation, and undefined behavior only occurs when a behavior is demanded of the expression -- that is, during evaluation. That makes it equivalent to Hold[With[{a=1,a=2},a]], which doesn't (and shouldn't) give an error. – user541686 – 2019-08-18T16:06:28.153 @Daniel: actually.... this is giving me second thoughts. Maybe that should also be an error? It's weird because Mathematica is both trying to analyze the expression and also avoid doing so. It might be outright impossible to give it reasonable semantics at all -- I'll write more when I have a chance (maybe in another 3 years? haha). – user541686 – 2019-08-18T16:32:28.400 With the change to use c instead of b it could well be a scoping bug, I'm not sure. Possibly the lexical scoping emulation should always rewrite. The business of using b both as lhs and rhs of With assignments makes it a bit hard for me to determine what to expect for when rewriting will or will not happen. – Daniel Lichtblau – 2019-08-19T02:47:00.400 @DanielLichtblau: I think I fleshed it out -- see my update. It seems to me that it's not really so much a bug as it is an attempt to satisfy an impossible requirement. – user541686 – 2019-08-19T02:54:37.333 I would replace the last three transformation steps with x = Echo[x /. HoldPattern@With[{first_, second_}, expr_] :> With[{first}, With[{second}, expr]]];. It's a lot clearer (at least in my opinion), shorter, does not rely on undocumented behavior, and still shows the same behavior – Lukas Lang – 2019-08-19T08:02:51.390 Also, I don't really see an issue with what's happening - Mathematica knows only one type of variables - global ones. All the different scoping constructs do is perform static analysis on their bodies, renaming and assigning variables as needed. This means that nested scoping constructs are also only recognized statically. If you now deliberately change the semantics of nested scoping constructs after the outer one is evaluated, it's clear that you're breaking things. Admittedly, it a bit strange that renaming is triggered in only one of both cases, but for well behaved cases you don't care. – Lukas Lang – 2019-08-19T08:11:29.873 @LukasLang: You genuinely "don't really see an issue" with the binding of c suddenly changing merely because 0a was added to it in lieu of 0b? Props to you I guess, because to me (and I dare say I would expect to most people) it looks like completely nonsensical behavior. – user541686 – 2019-08-19T08:19:25.340 @Mehrdad As I said, the trigger for renaming seems a bit unpredictable. What I meant is that you're only seeing these "nonsensical" effects because you're deliberately tricking the outer With by changing the semantics of its body after the fact. If you accept that Mathematica is an expression rewriting language with only global symbols, then I think the current implementation is the best thing you can do to implement scoping. The only thing you could do is make the weirdness more predictable by renaming only as needed, but that wouldn't remove the true issue with your examples I think – Lukas Lang – 2019-08-19T08:26:04.573 @LukasLang: The problem with "accepting that Mathematica only uses global symbols" is that *it's both counterintuitive AND in direct contradiction with their own documentation*. And it has absolutely nothing to do with me "tricking the outer With" or anything like that; that's utter nonsense. You don't even need With at all to see this problem. It can manifest itself a million other ways; all I did was just show you 1 example. Here's another example: **The following code is nondeterministic (!):** f = Function[{}, {x$1, x$2, ..., x$2000}]; Module[{x=1}, ContainsAny[f[] - x, {0}]] – user541686 – 2019-08-19T09:39:48.073
@LukasLang: (cont'd...) This is despite the documentation claiming that Module creates a local variable. Clearly it doesn't, but in what world should a user already know the documentation is lying??? "A bit unpredictable" is quite a disingenuous way to portray what's really active lying and complete nonsense in the language... – user541686 – 2019-08-19T09:48:17.533
– Lukas Lang – 2019-08-19T10:25:51.030
7
[Too long for a comment but not a full response.]
I am starting to see this differently. Sequential With assignment lists (which I realize are not yet documented) can behave differently from the flat counterpart, when a symbol appears both as an lhs and rhs. In particular these will behave differently.
With[{b = 0}, {c = b}, c + 0 b] // Trace // InputForm
(* Out[2713]//InputForm=
{HoldForm[With[{b = 0}, {c = b}, c + 0*b]], HoldForm[With[{c$$= 0}, c$$ + 0*0]],
HoldForm[0 + 0*0], {HoldForm[0*0], HoldForm[0]}, HoldForm[0 + 0], HoldForm[0]} *)
I think there is no controversy here.
With[{b = 0, c = b}, c + 0 b] // Trace // InputForm
(* Out[2714]//InputForm=
{HoldForm[With[{b = 0, c = b}, c + 0*b]], HoldForm[b + 0*0],
{HoldForm[0*0], HoldForm[0]}, HoldForm[b + 0], HoldForm[b]} *)
This much is as designed: With assignments only apply to later arguments so the b in that c=b assignment does not get lexically replaced.
Using Hold as in the original examples delays some of the evaluation, in a way that I think is also not controversial. Since the original With argument list has been split at that point, I am inclined to see the eventual behavior as also being according to design.
Now for the renaming, I am again not sure anything is amiss. Here are two examples to show what I have in mind. In the first we see the inner With variables get renamed.
Trace[With[{a = 0}, x = With[{b = 0, c = b}, c + 0 a]]] // InputForm
(* Out[2723]//InputForm=
{HoldForm[With[{a = 0}, x = With[{b = 0, c = b}, c + 0*a]]],
HoldForm[x = With[{b$$= 0, c$$ = b}, c$$+ 0*0]], {HoldForm[With[{b = 0, c = b}, c$$ + 0*0]], HoldForm[b + 0*0],
{HoldForm[0*0], HoldForm[0]}, HoldForm[b + 0], HoldForm[b]}, HoldForm[x = b],
HoldForm[b]} *)
In this next case we do not rename With variables.
Trace[With[{a = 0}, x = With[{b = 0, c = b}, c + 0 b]]] // InputForm
(* Out[2724]//InputForm=
{HoldForm[With[{a = 0}, x = With[{b = 0, c = b}, c + 0*b]]],
HoldForm[x = With[{b = 0, c = b}, c + 0*b]],
{HoldForm[With[{b = 0, c = b}, c + 0*b]], HoldForm[b + 0*0],
{HoldForm[0*0], HoldForm[0]}, HoldForm[b + 0], HoldForm[b]}, HoldForm[x = b],
HoldForm[b]} *)
The salient difference between these and the held examples is that the held ones gave rise to a split in the With` assignment lists, and that does allow for different semantics than the unsplit case. It happened in such a way that it was affected by whether or not renaming took place.
As I mentioned in a comment, I suppose one way to address this would be to force variable renaming to always happen. Not sure how popular such a change would be though.
Comments are not for extended discussion; this conversation has been moved to chat.
– Kuba – 2019-08-21T06:20:33.303
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Explore a number pattern which has the same symmetries in different bases.
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A circle has centre O and angle POR = angle QOR. Construct tangents at P and Q meeting at T. Draw a circle with diameter OT. Do P and Q lie inside, or on, or outside this circle?
### Circle Box
##### Stage: 4 Challenge Level:
It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?
### A Computer Program to Find Magic Squares
##### Stage: 5
This follows up the 'magic Squares for Special Occasions' article which tells you you to create a 4by4 magicsquare with a special date on the top line using no negative numbers and no repeats.
### Unit Interval
##### Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### Euler's Formula and Topology
##### Stage: 5
Here is a proof of Euler's formula in the plane and on a sphere together with projects to explore cases of the formula for a polygon with holes, for the torus and other solids with holes and the. . . .
|
### Solution of n-th order interval fuzzy differential IAL equations using the backstepping method
Document Type : Research Paper
Authors
1 Department of Accounting, Al-Esraa University College, Baghdad, Iraq
2 Department of Mathematics and Computer Applications, College of Science, Al-Nahrain University, Baghdad, Iraq
Abstract
There are two key points in this work as the main objectives. The first is how to convert $n^{th}$ order fuzzy differential equation into a first-order system of fuzzy differential equations using the notion of upper and lower bounds of the fuzzy solution to constitute the so-called interval fuzzy solution. The second is to solve the obtained system from the first step using a powerful method (the backstepping method) to provide an asymptotically stable solution by applying direct methods of stability (Lyapunov direct method).
Keywords
|
×
Minimum Value
I recently solved a trigonometric question which asked for the minimum value of the $$(\sin\theta + \csc\theta)^{2}+(\cos\theta +\sec\theta)^{2}$$.I found there ar two answers, which one of them is correct?
A)8 B)9
Note by Puneet Pinku
1 year, 2 months ago
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Hint: Expand and simplify the expression by using $$\sin^2 A + \cos^2 A = 1$$. Then apply double angle formula $$\sin(2A) = 2\sin A \cos A$$.
- 1 year, 2 months ago
I believe that you didn't check the conditions under which we could apply arithmetic mean - geometric mean.
Staff - 1 year, 2 months ago
|
# Simple Question on Polynomial Rings
#### Peter
##### Well-known member
MHB Site Helper
When we write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] where F is, say, a field, do we necessarily mean the set of all possible polynomials in x_1, x_2, ... ... x_n with coefficients in F? [In this case, essentially all that is required to determine whether a polynomial belongs to [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] is to check that the co-efficients belong to F and the indeterminates only contain [TEX] x_1, x_2, ... ... , x_n [/TEX].]
OR
when e write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] do we mean to include possible cases such as the set of polynomials with even coefficients - that is we may be talking about the set of polynomials with even co-efficients - so we cannot be sure what ring of polynomials we are talking about when we write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] until we specify the exact nature of ring of polynomials we are talking about further.
If the latter is the case when given [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] we can not reason about whether particular polynomials belong to [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] until you know the exact nature of the ring [TEX] F[x_1, x_2, ... ... , x_n] [/TEX]
I very much suspect that the former is the case but ... ... Can someone please confirm or clarify this?
Peter
[This is also posted on MHF]
Last edited:
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Re: Simple question on polynomial ringsWhen we write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] where F
When we write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] where F is, say, a field, do we necessarily mean the set of all possible polynomials in x_1, x_2, ... ... x_n with coefficients in F? [In this case, essentially all that is required to determine whether a polynomial belongs to [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] is to check that the co-efficients belong to F and the indeterminates only contain [TEX] x_1, x_2, ... ... , x_n [/TEX].]
OR
when e write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] do we mean to include possible cases such as the set of polynomials with even coefficients - that is we may be talking about the set of polynomials with even co-efficients - so we cannot be sure what ring of polynomials we are talking about when we write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] until we specify the exact nature of ring of polynomials we are talking about further.
If the latter is the case when given [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] we can not reason about whether particular polynomials belong to [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] until you know the exact nature of the ring [TEX] F[x_1, x_2, ... ... , x_n] [/TEX]
I very much suspect that the former is the case but ... ... Can someone please confirm or clarify this?
Peter
[This is also posted on MHF]
Hey Peter!
I am pretty sure that the former is the case.
Lets take a very simple non-polynomial ring example. When we write $\mathbb R$ we mean the set of all reals, not some specific type of them, like say irrationals or something. There is no reason that mathematicians would choose to use a different and quite ambiguous convention for more complicated structures.
#### Peter
##### Well-known member
MHB Site Helper
Re: Simple question on polynomial ringsWhen we write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] where F
Hey Peter!
I am pretty sure that the former is the case.
Lets take a very simple non-polynomial ring example. When we write $\mathbb R$ we mean the set of all reals, not some specific type of them, like say irrationals or something. There is no reason that mathematicians would choose to use a different and quite ambiguous convention for more complicated structures.
Thanks caffeinemachine,
You write "There is no reason that mathematicians would choose to use a different and quite ambiguous convention for more complicated structures."
I was more thinking that maybe [TEX] F[x_1, x_2, ... ... , x_n] [/TEX]would stand for a set of possible structures in the same way that when we say, a ring R., it can stand for many structures ... in the same way, I was thinking that maybe [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] could stand for a number of different polynomial rings.
Mind you, I think you are correct anyway
Peter
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Re: Simple question on polynomial ringsWhen we write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] where F
I was more thinking that maybe [TEX] F[x_1, x_2, ... ... , x_n] [/TEX]would stand for a set of possible structures in the same way that when we say, a ring R., it can stand for many structures ... in the same way, I was thinking that maybe [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] could stand for a number of different polynomial rings.
I don't quite understand you here. Can you please elaborate?
#### hmmm16
##### Member
Re: Simple question on polynomial ringsWhen we write [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] where F
Thanks caffeinemachine,
You write "There is no reason that mathematicians would choose to use a different and quite ambiguous convention for more complicated structures."
I was more thinking that maybe [TEX] F[x_1, x_2, ... ... , x_n] [/TEX]would stand for a set of possible structures in the same way that when we say, a ring R., it can stand for many structures ... in the same way, I was thinking that maybe [TEX] F[x_1, x_2, ... ... , x_n] [/TEX] could stand for a number of different polynomial rings.
Mind you, I think you are correct anyway
Peter
It does stand for a number of different structures in the same way that $R$ stands for different structures but that is because the $F$ can represent different fields.
So for example the polynomial ring $\mathbb{Q}[x_1,.....,x_n]$ has co-efficients from the rationals and would be analogous to the ring $\mathbb{Q}$
And $F[x_1,.....,x_n]$ has co-efficients from the field $F$ whatever that may be in the same way that $R$ has elements from $R$ whatever that may be.
However once we specify this field it does not then change
|
# Twisting holomorphic vector bundles and Euler characteristics
Given a holomorphic vector bundle $$\mathcal{V}$$ over a compact complex manifold $$M$$, it seems that even if $$\mathcal{V}$$ is non-trivial, then it can still have trivial Euler characteristic, that it, $$\sum_{k=0}^{\mathrm{dim}(M)} (-1)^k H^{(0,k)}(\mathcal{V}) = 0.$$ Is it true that for a positive line bundle $$\mathcal{L}$$, one can always find a large enough $$n$$, such that the twisted vector bundle $$\mathcal{V} \otimes \mathcal{L}^{\otimes n}$$ will have non-trivial Euler characteristic?
• If $M$ is projective, then for $n$ large enough one has that $\mathcal V\otimes \mathcal L^{\otimes n}$ is globally generated and that $H^q(M, V\otimes \mathcal L^{\otimes n})=0$ for $q>0$. Therefore, the holomorphic Euler characteristic of $V\otimes \mathcal L^{\otimes n}$ is positive. Oct 10, 2019 at 12:57
• ... and since $M$ carries a positive line bundle (by assumption), it is in fact projective.
– Ben
Oct 15, 2019 at 17:49
|
# Event Display using Druid
## Introduction
• Druid is used to visualize event information and detector geometry.
• We can better understand event and reconstruction algorithm performance.
• With the gdml file, detector geometry could be visualized to simulation level.
## Event Display
• Firstly, you need to download the file named with *.slcio and add it to the directory geometryfile in the Druid.
• Secondly, using command “Druid *.slcio”, the following window will display:
• From left side, finding the “Eve,Files,Options” three buttons, and choosing “Options”.
• Showing different event by operating “Event Navigation”.
• For more interesting event graph, setting the energy cut and other variables.
• For more information collecting in sub-detector, clicking “Eve” and moving to left interface. You can click whatever you like.
• When you point to one line in graph with mouse, there will show the information about the line left by particle.
## Geometry Display
• The following picture is my operation, you can follow.
• After operation as showed above, we can get following window.
### Shortcut
• Command+W: change the graph to line structure
• Command+E: change the background
• Command+R: back to original graph
### Dynamical Graph
• Double click “Edit Object”. From left side, finding the “Eve, Files, Options” three buttons, and choosing “Eve”. You will see the following window:
• We can see four buttons “Style Guides Clipping Extras” on a row in left, select “Extras” button and click start,you will see dynamical detector’s geometry.
• You can turn the values showed on the left side to control the concrete display.
## Usage
• Flexible parameter managing beside the steering file
Druid # print the instruction for the input format
• Separate geometry & data display
Druid *.slcio # display the first event in given slcio file
Druid *.gdml(*.xml or *.root) # display detector geometry
• Together with other arguments:
Druid *.slcio $EventNumber # given event in giver slcio file Druid *.slcio *.gdml(*.xml) # first event & geometry Druid *.slcio *.gdml(*.xml)$EventNumber
Druid *.slcio *.gdml(*.xml) $RunNumber$EventNumber
If you have any problem, contact [email protected]. You have to install LCIO and ROOT before installing Druid. I had installed Druid on macbook, macos version is 10.13.1, LCIO version is v02-04-03, ROOT version is root_v5.34.36.macosx64-10.11-clang70.tar. Have a good time.
|
# AQA/Edexcel GCSE Pythagoras Theorem
hi guys,
i found this useful so i thought i post it up to share! Feel free to leave feedback!
Thanks :]
HideShow resource information
## Other slides in this set
### Slide 2
Here's a taster:
The Theorem
· The Pythagorean theorem is named after
the Greek mathematician; Pythagoras.
· Who by tradition is credited with its
discovery and proof.
· His theorem is written as a formula.
### Slide 3
Here's a taster:
What Is The Hypotenuse?
· C² represents the hypotenuse multiplied
by itself (squared).
· The hypotenuse is the side of a right
triangle opposite the right angle.
TOP TIP:
Hypotenuse is
the longest
side in a
### Slide 4
Here's a taster:
Working It Out
STEPS:
· Write out the formula:
a² + b² = c²
· Put the numbers in the write place, and label unknown
sides as a letter.
5² + 12² = C²
· Find the squares of the known numbers:
25 + 144 = 169
· As the hyp. is being squared, to find the real value we
must square root it:
### Slide 5
Here's a taster:
Sure...?
· Let's see if it really works using an example. A
"3,4,5" triangle has a right angle in it, so the
formula should work.
· Let's check if the areas are the same:
3² + 4² = 5²
· Calculating this becomes:
9 + 16 = 25
### Slide 6
Here's a taster:
To Be Clear...
1. a² + b² = c²
2. 5² + 12² = c²
3. 25 + 144 = 169
4. C² = 169
5. c = 169
|
Enter the accuracy percentage and the full scale pressure of the gauge into the calculator to determine the full scale accuracy.
Full Scale Accuracy Formula
The following equation is used to calculate the Full Scale Accuracy.
FSA = (A/100) * FSP
• Where FSA is the full scale accuracy
• A is the percentage accuracy (%)
• FSP is the full scale pressure.
To calculate a full scale accuracy, multiply the percentage accuracy by the full scale pressure.
What is a Full Scale Accuracy?
Definition:
A full scale accuracy is a type of accuracy that is dependent on the reading of a gauge. For example, for a flow rate device, the accuracy would be measured as a percentage of the maximum flow rate.
How to Calculate Full Scale Accuracy?
Example Problem:
The following example outlines the steps and information needed to calculate Full Scale Accuracy.
First, determine the accuracy percentage. For this example, the accuracy percentage is 2%.
Next, determine the full scale pressure. In this example, the full scale pressure is 500 N/m^2.
Finally, calculate the full scale accuracy using the formula above:
FSA = (A%/100) * FSP
FSA = (2/100) * 500
FSA = 10 N/m^2
|
## anonymous one year ago Which graph represents the function of f(x) = the quantity of 4 x squared minus 4 x minus 8, all over 2 x plus 65?
1. anonymous
@jdoe0001
2. anonymous
where the graph
3. anonymous
i meant using graphing technology, and then tell where the point of discontinuity is on the graph
4. jdoe0001
or $$\bf \cfrac{4x^2-4x-8}{2x+2}\implies \cfrac{4(x^2-x-2)}{2(x+1)}\implies \cfrac{4\cancel{(x+1)}(x-2)}{2\cancel{(x+1)}}$$
5. anonymous
yes so it would be 2(x-2)
6. jdoe0001
7. jdoe0001
btw, you can zoom in/out on those graphs, by using the middle-mouse button
8. anonymous
would the point go on (-1,-6) or (1,-2)??
9. jdoe0001
yes
10. anonymous
graph of 2 x minus 4, with discontinuity at negative 1, negative 6 OR graph of 2 x minus 4, with discontinuity at 1, negative 2
11. jdoe0001
well... hold the mayo... notice, if x = -1 the denominator turns 0
12. anonymous
Confused
13. jdoe0001
$$\bf \cfrac{4x^2-4x-8}{2x+2}\qquad {\color{brown}{ x=-1}}\implies \cfrac{4x^2-4x-8}{2({\color{brown}{ -1}})+2}\implies \cfrac{4x^2-4x-8}{0}$$
14. jdoe0001
so -1,-6 is not a point on the line
15. anonymous
okay thanks :)
16. jdoe0001
yw
17. anonymous
If I ever have another question, I'll make sure to tag you :) Bye!!
18. jdoe0001
:)
|
# What are the specifications for media?
### Article sections
Media supplied must be original or have permission to use from the originator
• Images/photos (optional) at least 1000 pixels wide
• Video format accepted (MOV. .MPEG4MP4. .AVI. .WMV. .MPEGPS. .FLV. 3GPP)
• Videos will be uploaded onto the TMC YOUTUBE channel (Launch date TBC)
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# Calling a method from a string
I am trying to execute a line of code, such as: bpy.ops.object.modifier_add(type='SUBSURF') from a string of text.
My code is:
import bpy
import math
print("~~~Separator~~~")
obj_list = []
class ExportSomeData(bpy.types.Operator):
"""Test exporter which just writes hello world"""
bl_idname = "export.some_data"
bl_label = "Export Some Data"
filepath = bpy.props.StringProperty(subtype="FILE_PATH")
@classmethod
def poll(cls, context):
if not obj_list:
return False
else:
return True
def callMethod(o, name):
getattr(o, name)()
def execute(self, context):
file = open(self.filepath, 'r+')
for line in file:
callMethod("I don't know what to put here", line)
file.close()
return {'FINISHED'}
def invoke(self, context, event):
return {'RUNNING_MODAL'}
class SelectOperator(bpy.types.Operator):
bl_idname = "scene.select_option"
bl_label = "Multi Select Operator"
@classmethod
def poll(cls, context):
if not bpy.context.selected_objects:
return False
else:
return True
def execute(self, context):
global obj_list
obj_list = []
for obj in bpy.context.selected_objects:
obj_list.append(obj)
print(str(obj))
print("Objects in the active list: " + str(obj_list))
return {'FINISHED'}
"""Subdivide and scale a mesh."""
@classmethod
def poll(cls, context):
global obj_list
if not obj_list:
return False
else:
return True
def execute(self, context):
global obj_list
return {'FINISHED'}
class SelectionPanel(bpy.types.Panel):
bl_label = "Multi Selection Options"
bl_space_type = 'VIEW_3D'
bl_region_type = 'TOOLS'
def draw(self, context):
layout = self.layout
row = layout.row()
split = layout.split()
col = split.column(align=True)
col.operator("scene.select_option", text = "Set Objects", icon = "MESH_CUBE")
col.operator("export.some_data", text = "Open And Do Task", icon = "FILE")
def register():
bpy.utils.register_class(SelectionPanel)
bpy.utils.register_class(SelectOperator)
bpy.utils.register_class(ExportSomeData)
def unregister():
bpy.utils.unregister_class(SelectionPanel)
bpy.utils.unregister_class(SelectOperator)
bpy.utils.unregister_class(ExportSomeData)
if __name__ == "__main__":
register()
Below is what is causing me trouble:
def callMethod(o, name):
getattr(o, name)()
def execute(self, context):
file = open(self.filepath, 'r+')
for line in file:
callMethod("I don't know what to put here", line)
file.close()
return {'FINISHED'}
line is bpy.ops.object.modifier_add(type='SUBSURF'), but I don't know the object to put for the first parameter. If I am going the wrong direction with this, please correct me.
It looks like you are passing a whole function call to the getattr() function as a string.
getattr() takes an object as first argument and a string a second argument.
What you should do is isolate the object, the function and the params
Once you isolated your object string (ie. str="bpy.ops.object") you can isolate the object with a quick but unsafe myObj = eval(str) or use a safer
import sys
def str_to_class(str):
return getattr(sys.modules[__name__], str)
myObj = str_to_class("bpy.ops.object") #line added from original code
After that you should call getattr in this fashion:
getattr(myObj, 'myFunction')(params)
In the specific case you presented it would look something like this in explicit form
getattr(bpy.ops.object, "modifier_add")(type='SUBSURF')
Note: eval() is a very unsafe to use if there's some kind of user input, since the user could inject malicious code in eval()'s parameter string.
Edit: if security is not a concern, for any reason, you could also just eval() the whole string. Just make sure not to share your code without a proper warning.
If security is a concern, in both of the cases described above, you might want to apply a filter on the strings you pass to the converting function. Something like if str[:3]=="bpy": str_to_class(str) #or eval(). It makes things a little safer.
• I apologize, but I am still having an error. getattr(bpy.ops.object, "modifier_add")(type='SUBSURF') works fine. However, import sys def str_to_class(str): return getattr(sys.modules[__name__], str) myObj = str_to_class("bpy.ops.object") #line added from original code gives an error of "AttributeError: 'module' object has no attribute 'bpy.ops.object'" – Tritofic Aug 31 '15 at 14:17
• Yup: Didn't test the code coming from the page I linked, assuming it was working, but it isn't for me either. I've been playing around a bit and worked out something like this: m = __import__("bpy") \ ops = getattr(m, "ops") \ object = getattr(ops, "object") You can easily loop that process with the aid of "my.test.string".split(".") Give a shout out if you need any more help :) – torels Sep 1 '15 at 15:33
• Maybe m = locals().get("bpy", None) , fair to assume bpy would be in locals for a blender script. – batFINGER Jan 1 '17 at 11:41
|
# Almost Sure
## 6 March 17
### The Projection Theorems
In this post, I introduce the concept of optional and predictable projections of jointly measurable processes. Optional projections of right-continuous processes and predictable projections of left-continuous processes were constructed in earlier posts, with the respective continuity conditions used to define the projection. These are, however, just special cases of the general theory. For arbitrary measurable processes, the projections cannot be expected to satisfy any such pathwise regularity conditions. Instead, we use the measurability criteria that the projections should be, respectively, optional and predictable.
The projection theorems are a relatively straightforward consequence of optional and predictable section. However, due to the difficulty of proving the section theorems, optional and predictable projection is generally considered to be an advanced or hard part of stochastic calculus. Here, I will make use of the section theorems as stated in an earlier post, but leave the proof of those until after developing the theory of projection.
As usual, we work with respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}\}_{t\ge0},{\mathbb P})}$, and only consider real-valued processes. Any two processes are considered to be the same if they are equal up to evanescence. The optional projection is then defined (up to evanescence) by the following.
Theorem 1 (Optional Projection) Let X be a measurable process such that ${{\mathbb E}[1_{\{\tau < \infty\}}\lvert X_\tau\rvert\;\vert\mathcal{F}_\tau]}$ is almost surely finite for each stopping time ${\tau}$. Then, there exists a unique optional process ${{}^{\rm o}\!X}$, referred to as the optional projection of X, satisfying
$\displaystyle 1_{\{\tau < \infty\}}{}^{\rm o}\!X_\tau={\mathbb E}[1_{\{\tau < \infty\}}X_\tau\,\vert\mathcal{F}_\tau]$ (1)
almost surely, for each stopping time ${\tau}$.
Predictable projection is defined similarly.
Theorem 2 (Predictable Projection) Let X be a measurable process such that ${{\mathbb E}[1_{\{\tau < \infty\}}\lvert X_\tau\rvert\;\vert\mathcal{F}_{\tau-}]}$ is almost surely finite for each predictable stopping time ${\tau}$. Then, there exists a unique predictable process ${{}^{\rm p}\!X}$, referred to as the predictable projection of X, satisfying
$\displaystyle 1_{\{\tau < \infty\}}{}^{\rm p}\!X_\tau={\mathbb E}[1_{\{\tau < \infty\}}X_\tau\,\vert\mathcal{F}_{\tau-}]$ (2)
almost surely, for each predictable stopping time ${\tau}$.
## 28 February 17
### Pathwise Regularity of Optional and Predictable Processes
As I have mentioned before in these notes, when working with processes in continuous time, it is important to select a good modification. Typically, this means that we work with processes which are left or right continuous. However, in general, it can be difficult to show that the paths of a process satisfy such pathwise regularity. In this post I show that for optional and predictable processes, the section theorems introduced in the previous post can be used to considerably simplify the situation. Although they are interesting results in their own right, the main application in these notes will be to optional and predictable projection. Once the projections are defined, the results from this post will imply that they preserve certain continuity properties of the process paths.
Suppose, for example, that we have a continuous-time process X which we want to show to be right-continuous. It is certainly necessary that, for any sequence of times ${t_n\in{\mathbb R}_+}$ decreasing to a limit ${t}$, ${X_{t_n}}$ almost-surely tends to ${X_t}$. However, even if we can prove this for every possible decreasing sequence ${t_n}$, it does not follow that X is right-continuous. As a counterexample, if ${\tau\colon\Omega\rightarrow{\mathbb R}}$ is any continuously distributed random time, then the process ${X_t=1_{\{t\le \tau\}}}$ is not right-continuous. However, so long as the distribution of ${\tau}$ has no atoms, X is almost-surely continuous at each fixed time t. It is remarkable, then, that if we generalise to look at sequences of stopping times, then convergence in probability along decreasing sequences of stopping times is enough to guarantee everywhere right-continuity of the process. At least, it is enough so long as we restrict consideration to optional processes.
As usual, we work with respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge0},{\mathbb P})}$. Two processes are considered to be the same if they are equal up to evanescence, and any pathwise property is said to hold if it holds up to evanescence. That is, a process is right-continuous if and only is it is everywhere right-continuous on a set of probability 1. All processes will be taken to be real-valued, and a process is said to have left (or right) limits if its left (or right) limits exist everywhere, up to evanescence, and are finite.
Theorem 1 Let X be an optional process. Then,
1. X is right-continuous if and only if ${X_{\tau_n}\rightarrow X_\tau}$ in probability, for each uniformly bounded sequence ${\tau_n}$ of stopping times decreasing to a limit ${\tau}$.
2. X has right limits if and only if ${X_{\tau_n}}$ converges in probability, for each uniformly bounded decreasing sequence ${\tau_n}$ of stopping times.
3. X has left limits if and only if ${X_{\tau_n}}$ converges in probability, for each uniformly bounded increasing sequence ${\tau_n}$ of stopping times.
The only if’ parts of these statements is immediate, since convergence everywhere trivially implies convergence in probability. The importance of this theorem is in the if’ directions. That is, it gives sufficient conditions to guarantee that the sample paths satisfy the respective regularity properties.
Note that conditions for left-continuity are absent from the statements of Theorem 1. In fact, left-continuity does not follow from the corresponding property along sequences of stopping times. Consider, for example, a Poisson process, X. This is right-continuous but not left-continuous. However, its jumps occur at totally inaccessible times. This implies that, for any sequence ${\tau_n}$ of stopping times increasing to a finite limit ${\tau}$, it is true that ${X_{\tau_n}}$ converges almost surely to ${X_\tau}$. In light of such examples, it is even more remarkable that right-continuity and the existence of left and right limits can be determined by just looking at convergence in probability along monotonic sequences of stopping times. Theorem 1 will be proven below, using the optional section theorem.
For predictable processes, we can restrict attention to predictable stopping times. In this case, we obtain a condition for left-continuity as well as for right-continuity.
Theorem 2 Let X be a predictable process. Then,
1. X is right-continuous if and only if ${X_{\tau_n}\rightarrow X_\tau}$ in probability, for each uniformly bounded sequence ${\tau_n}$ of predictable stopping times decreasing to a limit ${\tau}$.
2. X is left-continuous if and only if ${X_{\tau_n}\rightarrow X_\tau}$ in probability, for each uniformly bounded sequence ${\tau_n}$ of predictable stopping times increasing to a limit ${\tau}$.
3. X has right limits if and only if ${X_{\tau_n}}$ converges in probability, for each uniformly bounded decreasing sequence ${\tau_n}$ of predictable stopping times.
4. X has left limits if and only if ${X_{\tau_n}}$ converges in probability, for each uniformly bounded increasing sequence ${\tau_n}$ of predictable stopping times.
Again, the proof is given below, and relies on the predictable section theorem. (more…)
## 29 November 16
### The Section Theorems
Consider a probability space ${(\Omega,\mathcal{F},{\mathbb P})}$ and a subset S of ${{\mathbb R}_+\times\Omega}$. The projection ${\pi_\Omega(S)}$ is the set of ${\omega\in\Omega}$ such that there exists a ${t\in{\mathbb R}_+}$ with ${(t,\omega)\in S}$. We can ask whether there exists a map
$\displaystyle \tau\colon\pi_\Omega(S)\rightarrow{\mathbb R}_+$
such that ${(\tau(\omega),\omega)\in S}$. From the definition of the projection, values of ${\tau(\omega)}$ satisfying this exist for each individual ${\omega}$. By invoking the axiom of choice, then, we see that functions ${\tau}$ with the required property do exist. However, to be of use for probability theory, it is important that ${\tau}$ should be measurable. Whether or not there are measurable functions with the required properties is a much more difficult problem, and is answered affirmatively by the measurable selection theorem. For the question to have any hope of having a positive answer, we require S to be measurable, so that it lies in the product sigma-algebra ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}}$, with ${\mathcal{B}({\mathbb R}_+)}$ denoting the Borel sigma-algebra on ${{\mathbb R}_+}$. Also, less obviously, the underlying probability space should be complete. Throughout this post, ${(\Omega,\mathcal{F},{\mathbb P})}$ will be assumed to be a complete probability space.
It is convenient to extend ${\tau}$ to the whole of ${\Omega}$ by setting ${\tau(\omega)=\infty}$ for ${\omega}$ outside of ${\pi_\Omega(S)}$. Then, ${\tau}$ is a map to the extended nonnegative reals ${\bar{\mathbb R}_+={\mathbb R}_+\cup\{\infty\}}$ for which ${\tau(\omega) < \infty}$ precisely when ${\omega}$ is in ${\pi_\Omega(S)}$. Next, the graph of ${\tau}$, denoted by ${[\tau]}$, is defined to be the set of ${(t,\omega)\in{\mathbb R}_+\times\Omega}$ with ${t=\tau(\omega)}$. The property that ${(\tau(\omega),\omega)\in S}$ whenever ${\tau(\omega) < \infty}$ is expressed succinctly by the inclusion ${[\tau]\subseteq S}$. With this notation, the measurable selection theorem is as follows.
Theorem 1 (Measurable Selection) For any ${S\in\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}}$, there exists a measurable ${\tau\colon\Omega\rightarrow\bar{\mathbb R}_+}$ such that ${[\tau]\subseteq S}$ and
$\displaystyle \left\{\tau < \infty\right\}=\pi_\Omega(S).$ (1)
As noted above, if it wasn’t for the measurability requirement then this theorem would just be a simple application of the axiom of choice. Requiring ${\tau}$ to be measurable, on the other hand, makes the theorem much more difficult to prove. For instance, it would not hold if the underlying probability space was not required to be complete. Note also that, stated as above, measurable selection implies that the projection of S is equal to a measurable set ${\{\tau < \infty\}}$, so the measurable projection theorem is an immediate corollary. I will leave the proof of Theorem 1 for a later post, together with the proofs of the section theorems stated below.
A closely related problem is the following. Given a measurable space ${(X,\mathcal{E})}$ and a measurable function, ${f\colon X\rightarrow\Omega}$, does there exist a measurable right-inverse on the image of ${f}$? This is asking for a measurable function, ${g}$, from ${f(X)}$ to ${X}$ such that ${f(g(\omega))=\omega}$. In the case where ${(X,\mathcal{E})}$ is the Borel space ${({\mathbb R}_+,\mathcal{B}({\mathbb R}_+))}$, Theorem 1 says that it does exist. If S is the graph ${\{(t,f(t))\colon t\in{\mathbb R}_+\}}$ then ${\tau}$ will be the required right-inverse. In fact, as all uncountable Polish spaces are Borel-isomorphic to each other and, hence, to ${{\mathbb R}_+}$, this result applies whenever ${(X,\mathcal{E})}$ is a Polish space together with its Borel sigma-algebra. (more…)
## 22 November 16
### Predictable Processes
In contrast to optional processes, the class of predictable processes was used extensively in the development of stochastic integration in these notes. They appeared as integrands in stochastic integrals then, later on, as compensators and in the Doob-Meyer decomposition. Since they are also central to the theory of predictable section and projection, I will revisit the basic properties of predictable processes now. In particular, any of the collections of sets and processes in the following theorem can equivalently be used to define the predictable sigma-algebra. As usual, we work with respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in{\mathbb R}_+},{\mathbb P})}$. However, completeness is not actually required for the following result. All processes are assumed to be real valued, or take values in the extended reals ${\bar{\mathbb R}={\mathbb R}\cup\{\pm\infty\}}$.
Theorem 1 The following collections of sets and processes each generate the same sigma-algebra on ${{\mathbb R}_+\times\Omega}$.
1. {${[\tau,\infty)}$: ${\tau}$ is a predictable stopping time}.
2. ${Z1_{[\tau,\infty)}}$ as ${\tau}$ ranges over the predictable stopping times and Z over the ${\mathcal{F}_{\tau-}}$-measurable random variables.
3. {$A\times(t,\infty)\colon t\in{\mathbb R}_+,A\in\mathcal{F}_t$}$\cup${$A\times\{0\}\colon A\in\mathcal{F}_0$}.
4. The elementary predictable processes.
5. {${(\tau,\infty)}$: ${\tau}$ is a stopping time}${\cup}${${A\times\{0\}\colon A\in\mathcal{F}_0}$}.
6. The left-continuous adapted processes.
7. The continuous adapted processes.
Compare this with the analogous result for sets/processes generating the optional sigma-algebra given in the previous post. The proof of Theorem 1 is given further below. First, recall that the predictable sigma-algebra was previously defined to be generated by the left-continuous adapted processes. However, it can equivalently be defined by any of the collections stated in Theorem 1. To make this clear, I now restate the definition making use if this equivalence.
Definition 2 The predictable sigma-algebra, ${\mathcal{P}}$, is the sigma-algebra on ${{\mathbb R}_+\times\Omega}$ generated by any of the collections of sets/processes in Theorem 1.
A stochastic process is predictable iff it is ${\mathcal{P}}$-measurable.
## 15 November 16
### Optional Processes
The optional sigma-algebra, ${\mathcal{O}}$, was defined earlier in these notes as the sigma-algebra generated by the adapted and right-continuous processes. Then, a stochastic process is optional if it is ${\mathcal{O}}$-measurable. However, beyond the definition, very little use was made of this concept. While right-continuous adapted processes are optional by construction, and were used throughout the development of stochastic calculus, there was no need to make use of the general definition. On the other hand, optional processes are central to the theory of optional section and projection. So, I will now look at such processes in more detail, starting with the following alternative, but equivalent, ways of defining the optional sigma-algebra. Throughout this post we work with respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in{\mathbb R}_+},{\mathbb P})}$, and all stochastic processes will be assumed to be either real-valued or to take values in the extended reals ${\bar{\mathbb R}={\mathbb R}\cup\{\pm\infty\}}$.
Theorem 1 The following collections of sets and processes each generate the same sigma-algebra on ${{\mathbb R}_+\times\Omega}$.
{${[\tau,\infty)}$: ${\tau}$ is a stopping time}.
• ${Z1_{[\tau,\infty)}}$ as ${\tau}$ ranges over the stopping times and Z over the ${\mathcal{F}_\tau}$-measurable random variables.
• The cadlag adapted processes.
• The right-continuous adapted processes.
• The optional-sigma algebra was previously defined to be generated by the right-continuous adapted processes. However, any of the four collections of sets and processes stated in Theorem 1 can equivalently be used, and the definitions given in the literature do vary. So, I will restate the definition making use of this equivalence.
Definition 2 The optional sigma-algebra, ${\mathcal{O}}$, is the sigma-algebra on ${{\mathbb R}_+\times\Omega}$ generated by any of the collections of sets/processes in Theorem 1.
A stochastic process is optional iff it is ${\mathcal{O}}$-measurable.
## 8 November 16
### Measurable Projection and the Debut Theorem
I will discuss some of the immediate consequences of the following deceptively simple looking result.
Theorem 1 (Measurable Projection) If ${(\Omega,\mathcal{F},{\mathbb P})}$ is a complete probability space and ${A\in\mathcal{B}({\mathbb R})\otimes\mathcal{F}}$ then ${\pi_\Omega(A)\in\mathcal{F}}$.
The notation ${\pi_B}$ is used to denote the projection from the cartesian product ${A\times B}$ of sets A and B onto B. That is, ${\pi_B((a,b)) = b}$. As is standard, ${\mathcal{B}({\mathbb R})}$ is the Borel sigma-algebra on the reals, and ${\mathcal{A}\otimes\mathcal{B}}$ denotes the product of sigma-algebras.
Theorem 1 seems almost obvious. Projection is a very simple map and we may well expect the projection of, say, a Borel subset of ${{\mathbb R}^2}$ onto ${{\mathbb R}}$ to be Borel. In order to formalise this, we could start by noting that sets of the form ${A\times B}$ for Borel A and B have an easily described, and measurable, projection, and the Borel sigma-algebra is the closure of the collection such sets under countable unions and under intersections of decreasing sequences of sets. Furthermore, the projection operator commutes with taking the union of sequences of sets. Unfortunately, this method of proof falls down when looking at the limit of decreasing sequences of sets, which does not commute with projection. For example, the decreasing sequence of sets ${S_n=(0,1/n)\times{\mathbb R}\subseteq{\mathbb R}^2}$ all project onto the whole of ${{\mathbb R}}$, but their limit is empty and has empty projection.
There is an interesting history behind Theorem 1, as mentioned by Gerald Edgar on MathOverflow (1) in answer to The most interesting mathematics mistake? In a 1905 paper, Henri Lebesgue asserted that the projection of a Borel subset of the plane onto the line is again a Borel set (Lebesgue, (3), pp 191–192). This was based on the erroneous assumption that projection commutes with the limit of a decreasing sequence of sets. The mistake was spotted, in 1916, by Mikhail Suslin, and led to his investigation of analytic sets and to begin the study of what is now known as descriptive set theory. See Kanamori, (2), for more details. In fact, as was shown by Suslin, projections of Borel sets need not be Borel. So, by considering the case where ${\Omega={\mathbb R}}$ and ${\mathcal{F}=\mathcal{B}({\mathbb R})}$, Theorem 1 is false if the completeness assumption is dropped. I will give a proof of Theorem 1 but, as it is a bit involved, this is left for a later post.
For now, I will state some consequences of the measurable projection theorem which are important to the theory of continuous-time stochastic processes, starting with the following. Throughout this post, the underlying probability space ${(\Omega,\mathcal{F})}$ is assumed to be complete, and stochastic processes are taken to be real-valued, or take values in the extended reals ${\bar{\mathbb R}={\mathbb R}\cup\{\pm\infty\}}$, with time index ranging over ${{\mathbb R}_+}$. For a first application of measurable projection, it allows us to show that the supremum of a jointly measurable processes is measurable.
Lemma 2 If X is a jointly measurable process and ${S\in\mathcal{B}(\mathbb{R}_+)}$ then ${\sup_{s\in S}X_s}$ is measurable.
Proof: Setting ${U=\sup_{s\in S}X_s}$ then, for each real K, ${U > K}$ if and only if ${X_s > K}$ for some ${s\in S}$. Hence,
$\displaystyle U^{-1}\left((K,\infty]\right)=\pi_\Omega\left((S\times\Omega)\cap X^{-1}\left((K,\infty]\right)\right).$
By the measurable projection theorem, this is in ${\mathcal{F}}$ and, as sets of the form ${(K,\infty]}$ generate the Borel sigma-algebra on ${\mathbb{\bar R}}$, U is ${\mathcal{F}}$-measurable. ⬜
Next, the running maximum of a jointly measurable process is again jointly measurable.
Lemma 3 If X is a jointly measurable process then ${X^*_t\equiv\sup_{s\le t}X_s}$ is also jointly measurable.
## 1 November 16
### Predictable Projection For Left-Continuous Processes
In the previous post, I looked at optional projection. Given a non-adapted process X we construct a new, adapted, process Y by taking the expected value of ${X_t}$ conditional on the information available up until time t. I will now concentrate on predictable projection. This is a very similar concept, except that we now condition on the information available strictly before time t.
It will be assumed, throughout this post, that the underlying filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in{\mathbb R}_+},{\mathbb P})}$ satisfies the usual conditions, meaning that it is complete and right-continuous. This is just for convenience, as most of the results stated here extend easily to non-right-continuous filtrations. The sigma-algebra
$\displaystyle \mathcal{F}_{t-} = \sigma\left(\mathcal{F}_s\colon s < t\right)$
represents the collection of events which are observable before time t and, by convention, we take ${\mathcal{F}_{0-}=\mathcal{F}_0}$. Then, the conditional expectation of X is written as,
$\displaystyle Y_t={\mathbb E}[X_t\;\vert\mathcal{F}_{t-}]{\rm\ \ (a.s.)}$ (1)
By definition, Y is adapted. However, at each time, (1) only defines Y up to a zero probability set. It does not determine the paths of Y, which requires specifying its values simultaneously at the uncountable set of times in ${{\mathbb R}_+}$. So, (1) does not tell us the distribution of Y at random times, and it is necessary to specify an appropriate version for Y. Predictable projection gives a uniquely defined modification satisfying (1). The full theory of predictable projection for jointly measurable processes requires the predictable section theorem. However, as I demonstrate here, in the case where X is left-continuous, predictable projection can be done by more elementary methods. The statements and most of the proofs in this post will follow very closely those given previously for optional projection. The main difference is that left and right limits are exchanged, predictable stopping times are used in place of general stopping times, and the sigma algebra ${\mathcal{F}_{t-}}$ is used in place of ${\mathcal{F}_t}$.
Stochastic processes will be defined up to evanescence, so two processes are considered to be the same if they are equal up to evanescence. In order to apply (1), some integrability requirements need to imposed. I will use local integrability. Recall that, in these notes, a process X is locally integrable if there exists a sequence of stopping times ${\tau_n}$ increasing to infinity and such that
$\displaystyle 1_{\{\tau_n > 0\}}\sup_{t \le \tau_n}\lvert X_t\rvert$ (2)
is integrable. This is a strong enough condition for the conditional expectation (1) to exist, not just at each fixed time, but also whenever t is a stopping time. The main result of this post can now be stated.
Theorem 1 (Predictable Projection) Let X be a left-continuous and locally integrable process. Then, there exists a unique left-continuous process Y satisfying (1).
As it is left-continuous, the fact that Y is specified, almost surely, at any time t by (1) means that it is uniquely determined up to evanescence. The main content of Theorem 1 is the existence of Y, and the proof of this is left until later in this post.
The process defined by Theorem 1 is called the predictable projection of X, and is denoted by ${{}^{\rm p}\!X}$. So, ${{}^{\rm p}\!X}$ is the unique left-continuous process satisfying
$\displaystyle {}^{\rm p}\!X_t={\mathbb E}[X_t\;\vert\mathcal{F}_{t-}]{\rm\ \ (a.s.)}$ (3)
for all times t. In practice, X will usually not just be left-continuous, but will also have right limits everywhere. That is, it is caglad (“continu à gauche, limites à droite”).
Theorem 2 Let X be a caglad and locally integrable process. Then, its predictable projection is caglad.
The simplest non-trivial example of predictable projection is where ${X_t}$ is constant in t and equal to an integrable random variable U. Then, ${{}^{\rm p}\!X_t=M_{t-}}$ is the left-limits of the cadlag martingale ${M_t={\mathbb E}[U\;\vert\mathcal{F}_t]}$, so ${{}^{\rm p}\!X}$ is easily seen to be a caglad process. (more…)
## 25 October 16
### Optional Projection For Right-Continuous Processes
In filtering theory, we have a filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},{\mathbb P})}$ and a signal process ${\{X_t\}_{t\in{\mathbb R}_+}}$. The sigma-algebra ${\mathcal{F}_t}$ represents the collection of events which are observable up to and including time t. The process X is not assumed to be adapted, so need not be directly observable. For example, we may only be able to measure an observation process ${Z_t=X_t+\epsilon_t}$, which incorporates some noise ${\epsilon_t}$, and generates the filtration ${\mathcal{F}_t}$, so is adapted. The problem, then, is to compute an estimate for ${X_t}$ based on the observable data at time t. Looking at the expected value of X conditional on the observable data, we obtain the following estimate for X at each time ${t\in{\mathbb R}_+}$,
$\displaystyle Y_t={\mathbb E}[X_t\;\vert\mathcal{F}_t]{\rm\ \ (a.s.)}$ (1)
The process Y is adapted. However, as (1) only defines Y up to a zero probability set, it does not give us the paths of Y, which requires specifying its values simultaneously at the uncountable set of times in ${{\mathbb R}_+}$. Consequently, (1) does not tell us the distribution of Y at random times. So, it is necessary to specify a good version for Y.
Optional projection gives a uniquely defined process which satisfies (1), not just at every time t in ${{\mathbb R}_+}$, but also at all stopping times. The full theory of optional projection for jointly measurable processes requires the optional section theorem. As I will demonstrate, in the case where X is right-continuous, optional projection can be done by more elementary methods.
Throughout this post, it will be assumed that the underlying filtered probability space satisfies the usual conditions, meaning that it is complete and right-continuous, ${\mathcal{F}_{t+}=\mathcal{F}_t}$. Stochastic processes are considered to be defined up to evanescence. That is, two processes are considered to be the same if they are equal up to evanescence. In order to apply (1), some integrability requirements need to imposed on X. Often, to avoid such issues, optional projection is defined for uniformly bounded processes. For a bit more generality, I will relax this requirement a bit and use prelocal integrability. Recall that, in these notes, a process X is prelocally integrable if there exists a sequence of stopping times ${\tau_n}$ increasing to infinity and such that
$\displaystyle 1_{\{\tau_n > 0\}}\sup_{t < \tau_n}\lvert X_t\rvert$ (2)
is integrable. This is a strong enough condition for the conditional expectation (1) to exist, not just at each fixed time, but also whenever t is a stopping time. The main result of this post can now be stated.
Theorem 1 (Optional Projection) Let X be a right-continuous and prelocally integrable process. Then, there exists a unique right-continuous process Y satisfying (1).
Uniqueness is immediate, as (1) determines Y, almost-surely, at each fixed time, and this is enough to uniquely determine right-continuous processes up to evanescence. Existence of Y is the important part of the statement, and the proof will be left until further down in this post.
The process defined by Theorem 1 is called the optional projection of X, and is denoted by ${{}^{\rm o}\!X}$. That is, ${{}^{\rm o}\!X}$ is the unique right-continuous process satisfying
$\displaystyle {}^{\rm o}\!X_t={\mathbb E}[X_t\;\vert\mathcal{F}_t]{\rm\ \ (a.s.)}$ (3)
for all times t. In practise, the process X will usually not just be right-continuous, but will also have left limits everywhere. That is, it is cadlag.
Theorem 2 Let X be a cadlag and prelocally integrable process. Then, its optional projection is cadlag.
A simple example of optional projection is where ${X_t}$ is constant in t and equal to an integrable random variable U. Then, ${{}^{\rm o}\!X_t}$ is the cadlag version of the martingale ${{\mathbb E}[U\;\vert\mathcal{F}_t]}$. (more…)
## 21 October 16
### The Projection Theorems
Back when I first started this series of posts on stochastic calculus, the aim was to write up the notes which I began writing while learning the subject myself. The idea behind these notes was to give a more intuitive and natural, yet fully rigorous, approach to stochastic integration and semimartingales than the traditional method. The stochastic integral and related concepts were developed without requiring advanced results such as optional and predictable projection or the Doob-Meyer decomposition which are often used in traditional approaches. Then, the more advanced theory of semimartingales was developed after stochastic integration had already been established. This now complete! The list of subjects from my original post have now all been posted. Of course, there are still many important areas of stochastic calculus which are not adequately covered in these notes, such as local times, stochastic differential equations, excursion theory, etc. I will now focus on the projection theorems and related results. Although these are not required for the development of the stochastic integral and the theory of semimartingales, as demonstrated by these notes, they are still very important and powerful results invaluable to much of the more advanced theory of continuous-time stochastic processes. Optional and predictable projection are often regarded as quite advanced topics beyond the scope of many textbooks on stochastic calculus. This is because they require some descriptive set theory and, in particular, some understanding of analytic sets. The level of knowledge required for applications to stochastic calculus is not too great though, and I aim to give complete proofs of the projection theorems in these notes. However, the proofs of these theorems do require ideas which are not particularly intuitive from the viewpoint of stochastic calculus, and hence the desire to avoid them in the initial development of the stochastic integral. The theory of semimartingales and stochastic integration will not used at all in the series of posts on the projection theorems, and all that will be required from these stochastic calculus notes are the initial posts on filtrations and processes. I will also mention quasimartingales, although only the definition and very basic properties will be required.
The subjects related to the projection theorems which I will cover are,
• The Debut Theorem. I have already covered the debut theorem for right-continuous processes. This is a special case of the more general result which applies to arbitrary progressively measurable processes.
• The Optional and Predictable Section Theorems. These very powerful results state that optional processes are determined, up to evanescence, by their values at stopping times and, similarly, predictable processes are determined by their values at predictable stopping times.
• Optional and Predictable Projection. This forms the core of these sequence of posts, and follows in a straightforward way from the section theorems. As the section theorems are required to prove them, the projection theorems are also regarded as an advanced topic. However, for right-continuous and left-continuous processes it is possible to construct respectively the optional and predictable projections in a more elementary and natural way, without involving the section theorems.
• Dual Optional and Predictable Projection. The dual projections are, as the name suggests, dual to the optional and predictable projections mentioned above. These apply to increasing integrable processes or, more generally, to processes with integrable variation. For a process X, the dual projections can be thought of as the optional and predictable projections applied to the differential ${dX}$.
• The Doléans Measure. The Doléans measure can be defined for class (D) submartingales and, applied to the square of a martingale, can be used to construct the stochastic integral for square integrable martingales. Although this does not involve the projection theorems, the Doléans measure in conjunction with dual predictable projection gives a slick proof of the Doob-Meyer decomposition. The Doléans measure also exists for quasimartingales and, similarly, the Doob-Meyer decomposition can be extended to such processes.
## 12 October 16
### Do Convex and Decreasing Functions Preserve the Semimartingale Property — A Possible Counterexample
Figure 1: The function f, convex in x and decreasing in t
Here, I attempt to construct a counterexample to the hypotheses of the earlier post, Do convex and decreasing functions preserve the semimartingale property? There, it was asked, for any semimartingale X and function ${f\colon{\mathbb R}_+\times{\mathbb R}\rightarrow{\mathbb R}}$ such that ${f(t,x)}$ is convex in x and right-continuous and decreasing in t, is ${f(t,X_t)}$ necessarily a semimartingale? It was explained how this is equivalent to the hypothesis: for any function ${f\colon[0,1]^2\rightarrow{\mathbb R}}$ such that ${f(t,x)}$ is convex and Lipschitz continuous in x and decreasing in t, does it decompose as ${f=g-h}$ where ${g(t,x)}$ and ${h(t,x)}$ are convex in x and increasing in t. This is the form of the hypothesis which this post will be concerned with, so the example will only involve simple real analysis and no stochastic calculus. I will give some numerical calculations suggesting that the construction below is a counterexample, but do not have any proof of this. So, the hypothesis is still open.
Although the construction given here will be self-contained, it is worth noting that it is connected to the example of a martingale which moves along a deterministic path. If ${\{M_t\}_{t\in[0,1]}}$ is the martingale constructed there, then
$\displaystyle C(t,x)={\mathbb E}[(M_t-x)_+]$
defines a function from ${[0,1]\times[-1,1]}$ to ${{\mathbb R}}$ which is convex in x and increasing in t. The question is then whether C can be expressed as the difference of functions which are convex in x and decreasing in t. The example constructed in this post will be the same as C with the time direction reversed, and with a linear function of x added so that it is zero at ${x=\pm1}$. (more…)
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## divide and conquer – Clarification on the algorithm for finding a majority element
When the size of your problem reduced to 2, it’s the base case of your
recursion, it’s false that we pair up two remaining elements. so it’s
sufficiently, first count number of occurrence of two remaining
elements in $$A$$.
for your problem we act as follow:
we check the whole array to count number of occurrence $$A(i)=3$$, and
number of occurrence of $$A(i)=1$$ and then if number of occurrence
$$3geq frac{n}{2}+1$$ or if number of occurrence $$1geq frac{n}{2}+1$$, we print it.
but the main problem is we must prove the correctness and running time of algorithm.
Suppose $$mathcal{X}$$ is majority element, because of, $$mathcal{X}$$ is majority element, then the number of occurrence of it, is grater than
$$frac{n}{2}$$, so after pair up, at least there is one pair that elements are $$mathcal{X}$$, otherwise it contradict with this fact that $$mathcal{X}$$ is majority element.
On the other hand, at most all elements of array $$A$$ be $$mathcal{X}$$, so both elements of a pair be equal, as a result at most half of the elements remain in each recursion.
Therefore one of the at most $$frac{n}{2}$$ elements be $$mathcal{X}$$. And we do our divide an conquer the problem until remain only two element.
Our statement holds only if $$A$$ contains $$mathcal{X}$$. So after we get base that $$n=2$$, it’s sufficient to for each of two remaining elements we check number of occurrence it.
Suppose the remaining elements is a,b:
`````` A,B=0
for i = 1 to n
if (a==A(i))
A++
if (b==A(i))
B++
if(a> n/2)
print a
else if (b> n/2)
print b
else
print "there is no majority in the array"
``````
Because at each iteration at most half of the elements are remain
$$T(n)=T(frac{n}{2})+O(n)$$
$$=O(n)$$
Note that at each iteration pairing up of $$n$$ elements need $$O(n)$$ time.
## floating point – Unit conversion – Better to divide by an integer or multiply by a double?
I currently have a `long` timestamp measured in units of 100ns elapsed since January 1st, 1900. I need to convert it to milliseconds.
I have the choice of either multiplying by `0.0001` or dividing by `10_000`. Although at first glance they sound the same, the former would actually cause an implicit cast to a `double` – the latter would of course result in another `long` with the remainder truncated.
Which would yield a better result? Obviously `double` is an imprecise type that introduces errors due to the use of a mantissa, floating radix, and exponent, but would that error be less or more than the error from performing the simple integer division? Or would the error be demonstrably negligible?
To give an example of one of my values, here is one of the timestamps: `38348440316924872` .
I’m specifically referring to C#, but this question should be general to computer science.
## How does one divide massive distances into a floating-point cooperative value system?
I am making a space game, and the large problem I am having is distances, positions, and floating point errors.
Imagine I currently have an galaxy of 100,000 X 2,000 X 100,000 light-years. I need to place objects in this space, and because of floating point limitation, I cannot simply put something at, say, 247014781, 181487104710, 17218383000 (actual distances even larger).
My current idea was to make a coordinate system of incrementally smaller cubes.
For this idea, we divide the galaxy into 4 quadrants of 50k ly. We also say that negative means lower half, positive means upper half. We now essentially have 8 octants of 50k x 50k x 50k light-year length.
using this idea, we could theoretically divide this large area into something reasonable, but the resultant coordinates are a bit odd. Take a 2, 4, 5, 10, 10 division (so divide into quadrants, then those quads into 4, etc) and we get a coordinate that is represented as:
-+1-4, 1-64, 1-125, 1-1000, 1-1000, meters (ei -1, 31, 23, 420, 69, any value between 0 and 2.365e+17)
this DOES make the possible x, y, z position way smaller (a change from 9.46073e+20 to 2.365e+17 meters), but it doesn’t make it small enough, it looks a bit odd, and it doesn’t seem practical from a programmers standpoint.
Let’s look at this on a smaller scale, say a cube of 1000 meters. Let’s use a half, quadrant, sector, meters system, in which each sector is 1/2 of a quadrant (8 sectors a quadrant))
This changes the highest possible value from 1000 to 250, as 1000 / 2 / 2 = 250. A position of 230, 272, -71 would be represented in our new system as (index from topleftback corner) (+3.1.230, +3.1.22, +3.1.71)
We have also confined all value to positive, as we know their direction of each value due to the quadrant and sector, and we add in that direction. That is why z=-71 is now +3.1.71, as 3 tells me I am in a negative quadrant.
Let’s say that we clamp the player position at 0,0,0, and handle movement by procedurally generating using our new coordinate value. The player is, in unity, always at 0, but in the class we designate a value coords as that stated above. We also have a “star” who’s coordinate is (+3.1.70, +3.1.245, +3.1.11). We draw the star (in unity coordinates now) at a position relative to the player (in this case -160, 223, 60)
How can I do something like this on a reasonable scale? Is there a better way to do this?
## combinatorics – For which \$n\$ we can divide set \$M= {1,2,3,…,3n}\$ in to \$n\$ subsets each with \$3\$ elements such that in each subset \${x,y,z}\$ we have \$x+y=3z\$?
For which $$n$$ we can divide set $$M= {1,2,3,…,3n}$$ in to $$n$$ $$3$$ element subsets such that in each subset $${x,y,z}$$ we have $$x+y=3z$$?
Since $$x_i+y_i=3z_i$$ for each subset $$A_i={x_i,y_i,z_i}$$, we have $$4sum _{i=1}^n z_i=sum _{i=1}^{3n}i = {3n(3n+1)over 2} implies 8mid n(3n+1)$$
so $$n=8k$$ or $$n=8k-3$$. Now it is not difficult to see that if $$k=1$$ we have such partition.
Say $$n=5$$ we have: $$A_1= {9,12,15}, A_2= {4,6,14}, A_3= {2,5,13}, A_4= {10,7,11}$$ and $$A_5= {1,3,8}$$
What about for $$kgeq 2$$? Some clever induction step? Or some ”well” known configuration?
## java – Divide BigDecimal by int
I am trying to divide the change owed by 1 to figure out how many dollars are owned. So if 3.60 is owed I want to divide that by 1 but I used BigDecimal to collect the change variable and can’t figure out how to divide it by 1.
This is the code I have tried but it’s not working. I needed to use BigDecimal for round off errors but I also need to use that value with int values.
``````BigDecimal Dollars = (Change.divide(1));
``````
## number theory – if the gcd of (a,b) = 1, and a,b divide an integer x, prove that ab ≤x
A simple doubt regarding GCDs:
if gcd(a,b) = 0, and a/x, b/x, how do we prove that ab≤x ?
I was attempting the proof to this theorem:
if gcd(a,b) = 1, and a/x, b/x, prove that ab/x.
And proving the above mentioned inequality will complete my proof.
## plotting – Cannot divide a list of numbers by their total from uploaded data
I have a list of numbers in a CSV file which my code uploads to Mathematica. I upload the file, then flatten the list and sort it to arrange the numbers in ascending order.
After doing that, I would like to divide each number in the rearranged list by the total of the list and then plot it. I am using the code shown below, but it says it is ‘not a list of numbers or pair of numbers’. What am I doing wrong? What do I need to change?
Thanks,
Sid
``````FormPage[{"data" -> "CSV"},
a = NumericalSort[Flatten[#data]] &;
b = a/Total[a] &;
ListPlot[b]
]
``````
## Divide string only at line breaks
I have a very simple problem but don’tk now how to solve it.
I have a string that looks like this:
string=”a b c
d e f”
I want to break it ONLY at the line breaks, i.e. I want to get a list of the following two strings in return:
“a b c” and “d e f”.
I.e. the answer should be {“a b c”, “d e f”}.
However, if I use StringSplit[string], I instead get {“a”,”b”,”c”,”d”,”e”,”f”}.
How can I fix this? Why does Mathematica treat line breaks and whitespaces the same way?
## java – How do I divide Dao and Dao-Cache into different modules?
I want to improve performance for api. I design cache in dao layer. I use caffine cache. Basic code is:
CacheConfig.java
``````@Configuration
public class CacheConfig {
@Autowired
private AppDao appDao;
@Bean(value = "appCache")
return Caffeine.newBuilder()
.maximumSize(1000)
.refreshAfterWrite(5, TimeUnit.MINUTES)
.build(key -> appDao.getByKeyFromDB(key));
}
}
``````
AppDao.java
``````@Repository
public class AppDao {
@Autowired
@Qualifier(value = "appCache")
@Autowired
private AppMapper appMapper;
// service call getByAppKey
public App getByKey(String appKey) {
App app = appCache.get(appKey);
return app;
}
// appCache load data by getByKey
public App getByKeyFromDB(String appKey) {
LambdaQueryWrapper<App> queryWrapper = Wrappers.lambdaQuery();
queryWrapper.eq(App::getAppKey, appKey);
return appMapper.selectOne(queryWrapper);
}
}
``````
I maybe divide getByKey and getByKeyFromDB to two file, but I have no idea.
How do I divide getByKey and getByKeyFromDB to different modules?
## google sheets – Divide data evenly in row, place in additional rows
Okay, so what I have presented is a random divisor for every number. If you want all the rows in a column to add up, it will not be a random number. for instance, if you want to have 4 rows, the number in which you will use as a divisor will be 4. This is unless you are referring to the average of the random divisors to equal 4… or however that really works. The point is, choosing a random number between 4 and 7 will not have the summation results in which you are looking for.
`````` function myFunction() {
var data = ss.getDataRange().getValues();
var newData = ((,,,,,));
var i = 0;
var j = 0;
var k = 1;
var random = ();
for (var i = 0, dataLength = data.length; i < dataLength; i++) {
for (k = 1; k < 6; k++) {
random(k) = Math.floor(Math.random() * 7) + 4;
}
newData(j) = (data(i)(0) / random(1),data(i)(1) / random(2),data(i)(2) / random(3),data(i)(3) / random(4),data(i)(4) / random(5),data(i)(5));
for (k = 1; k < 6; k++) {
random(k) = Math.floor(Math.random() * 7) + 4;
}
newData(j+1) = (data(i)(0) / random(1),data(i)(1) / random(2),data(i)(2) / random(3),data(i)(3) / random(4),data(i)(4) / random(5),data(i)(5));
for (k = 1; k < 6; k++) {
random(k) = Math.floor(Math.random() * 7) + 4;
}
newData(j+2) = (data(i)(0) / random(1),data(i)(1) / random(2),data(i)(2) / random(3),data(i)(3) / random(4),data(i)(4) / random(5),data(i)(5));
for (k = 1; k < 6; k++) {
random(k) = Math.floor(Math.random() * 7) + 4;
}
newData(j+3) = (data(i)(0) / random(1),data(i)(1) / random(2),data(i)(2) / random(3),data(i)(3) / random(4),data(i)(4) / random(5),data(i)(5));
j = j + 4
}
ss.getRange(1,1, newData.length,6).setValues(newData);
}
``````
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# 2) (15) Evaluale the following integrals.Jx? cosh(x' )dxfsinh " cosh Adtcsc hicothJ= 2I-1 . 4}
###### Question:
2) (15) Evaluale the following integrals. Jx? cosh(x' )dx fsinh " cosh Adt csc hi coth J= 2I-1 . 4}
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# Timer with uncapped frame rate
I have been struggling a ton to get a basic time trial type timer that includes milliseconds.
The kicker is, I REALLY want to have my frame rate uncapped. With that in mind, this solution I found counts WAY too fast (it counts approx 30 seconds every actual second): Timers and animation frame rate
I'm thinking there may be something to using datetime, but I can't figure it out. If I could just somehow start a timer at 0, then add to it the time between the previous and current frame, I'd be golden. I just can't figure out how to do that.
EDIT: I think I am close with bge.logic.getRealTime() ...I just can't display it in the text object for some reason. I get "error return without exception set". It displays in the console just fine if I print it instead.
## 1 Answer
Turns out I've answered my own question. Here's a quick script for anyone who finds this later down the road. Just attach this to a text property and you should be good.
import bge
from datetime import datetime
textObject = bge.logic.getCurrentController().owner
timePassed = bge.logic.getRealTime()
timePassed = str(round(timePassed,2))
textObject.text = timePassed
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