text:
# Sets intersection and the axiom of choice?

1. Dec 13, 2011

### Damidami

I'm working on some topology in $\mathbb{R}^n$ problem, and I run across this:
Given $\{F_n\}$ a family of subsets of $\mathbb{R}^n$, then if $x$ is a point in the clausure of the union of the family, then

$x \in \overline{\cup F_n}$
wich means that for every $\delta > 0$ one has
$B(x,\delta) \cap (\cup F_n) \neq \emptyset$

Now, if I say that because the intersection is a nonempty set, I have a point $y \in \mathbb{R}^k$ such that

$d(x,y) < \delta \wedge y \in (\cup F_n)$

did I use the axiom of choice?

Because I think it this way, I choose an element $y$ from an infinite (nonumerable) indexed family parametrized by $\delta$ (for each value of $\delta$ I have another intersection).

So If I don't want to use the AC I can't just say that I can choose an element from that nonempty intersection?
I'm a little confused :?

Thanks!

2. Dec 14, 2011

### zhentil

The way you've written it, you've chosen one y from one set. In any event, you can't be invoking the axiom of choice if you can prove that your collection is a set without invoking it.

3. Dec 14, 2011

### micromass

You did not use the axiom of choice. In fact, you've made only one choice. The axiom of choice is only necessary if you want to make an infinite number of choices (and if you can't specify the choices you make).

4. Dec 14, 2011

### Damidami

Hi zhentil and micromass.

Maybe I wasn't clear, let me explain again.

I have an infinte indexed family of nonempty sets, each one of the kind:

$A_{\delta} = B(x,\delta) \cap (\cup F_n) \neq \emptyset$
$A_{\delta} \subseteq \mathbb{R}^n$
$A_{\delta}$ is closed.

So the family is
$\{ A_{\delta} \}_{\delta \in \mathbb{R}_{>0}}$

and for each set in the family I want to choose an element
$y_{\delta} \in A_{\delta}$
they are not necesarily differents elements (but maybe so)

So I don't think I'm just choosing one element from one set, but that I'm choosing one element from each of the nonumerable family of sets (and I can't specify the choices I make).

Still didn't I use the AC?
Thanks.

5. Dec 14, 2011

### micromass

This is completely different!!!

Now you are indeed making an infinite number of choices. So now you are indeed using the axiom of choice!!

(also, I don't really see why your $A_\delta$ is closed, I don't think this is necessarily the case).

6. Dec 14, 2011

### Damidami

Now that I think about it, I'm not even sure what I mean when I say I 'choose' an element.
I know the family is of nonempty sets, so obviously there is an (at least) an element in each of them, but that didn't mean I 'choose' it, doesn't?

The problem is I want one to use it in the proof. (and the element I choose depends on $\delta$, so I don't just choose one, but one for every $\delta$)

I think I'm now even more confused.

7. Dec 14, 2011

### Damidami

Hi micromass, thanks for your reply (I wrote the next post while you where writing this)
It did sound like I was using the AC, but please read the next post, now I'm not even sure what I mean to 'choose' an element, saying it exists is not enough to have 'choosen it', but using it for something else is?

(my $A_\delta$ is closed because it's part from an exercice where that was one of the hipothesys, and I run across this doubt with the AC). If curious, the exercise is this: to show that if $\{ F_n \}$ is a family of closed subsets of $\mathbb{R}^k$ with $d(F_n, F_m) > 1$ for all $m,n > 0$ with $m \neq n$, then $\cup F_n$ is closed.

Thanks!

8. Dec 14, 2011

### micromass

Saying that something exists is actually equivalent to choosing it. If it exists, then you can choose it.

For example, [0,1] is nonempty. So there exists an element in it. That means that I can choose an element in it. Indeed, I can choose the element 1.

I can also look at $A_i=[0,1]$

All these sets are nonempty, so I can choose an element in each of these sets. However, can I make a simultanious choice?? That is, can I choose something for all i at once?? Yes I can, I can choose the element 1 for each $A_i$ This does not use the axiom of choice.

However, when I cannot specify the choice, then I need to use the axiom of choice.

Maybe you should read my blogs on the axiom of choice: https://www.physicsforums.com/blog.php?u=205308 [Broken] I hope that settles some questions.

Last edited by a moderator: May 5, 2017
9. Dec 14, 2011

### micromass

OK, but if the $A_\delta$ is closed and bounded (=compact), then there might be ways to make the choice not using AC.

For example, if $A_i, i\in I$ are compact subsets of $\mathbb{R}$, then I can choose elements $x_i\in A_i$ without invoking AC. Indeed, I can just take

$$x_i=\max(A_i)$$

the maximum exists because of compactness. This works in $\mathbb{R}^n$, but it can be generalized to $\mathbb{R}^n$.

10. Dec 14, 2011

### Damidami

Hi micromass,
If only saying an element exists in an infinite family is 'choosing it', isn't it obvious the AC has to be true? I mean, if I have an infinite colection of nonempty subsets, each of them is nonempty, so an element 'obviously' has to exist in each of them, doesn't?

I'll surely read your blog on the AC later, I only read the conclusion by now (started backwards)

On the other hand, the sets $F_n$ in the family in my exercice are closed, but doesn't have to be bounded. Even more, the sets I make a choose isn't from them, but from the intersection of them with an open ball, so they aren't even closed, so the choice is quite complicated indeed (I think).

11. Dec 14, 2011

### micromass

The point of the AC is not saying that an element exist. The point is that you can make a simultaneous choice. Surely, if I give you sets $A_1,A_2,A_3,A_4,...$ that are nonempty. Surely, you can pick an element from each of them. You can pick an element from $A_{45345455}$ if you want because it is nonempty. The trick is to pick an element from all of them at once.

The axiom of choice might be obviously true to you (and to many other people), but it cannot be demonstrated from the usual axioms.

12. Dec 14, 2011

### Damidami

Hi micromass,
So, what I'm not sure now, is if the choice I made from the infinite indexed family I have was all at once, or one after another. It is a nonumerable family so there is no "next" set in the family, and I have to chose an $y_\delta$ for each $A_{\delta}$.
So do you think I'm choosing them all at once? How can I realize?
Thanks.

13. Dec 14, 2011

### micromass

You can't choose them one after another. The only way to do that is by using a form of transfinite induction, this is not what happens here.

If you reason: I choose one for $A_0$, then I choose one for $A_1$, then I choose on for $A_2$, etc. then you can only make a finite choice. Once you get to something infinite, this reasoning fails.

So if you want to choose something from an infinite set, then you need to make it all at once.

I urge you to read part 1 of my blog, that will reveal a lot I think.


text:
zgribestika

2021-12-15

Can vectors be inverted in this question?
Enquire if it is possible to solve the below for C.
${B}^{-1}\left(x-\mu \right)=xc$
Here obviously B is an invertible matrix and both c and μ are column vectors.
Would the solution be ${x}^{-1}{B}^{-1}\left(x-\mu \right)=c$ is it possible to invert vectors ?
How about if it was the other way
${B}^{-1}\left(x-\mu \right)=cx$
Is there any other way to do this ?

Bubich13

Expert

Vectors, in general, cant

Melissa Moore

Expert

$q=\mathrm{△}H$, since pressure is constant
$\mathrm{△}H={\int }_{{t}_{1}}^{{t}_{1}}\left(\frac{H}{J}\right)={\int }_{298}^{473}\left\{20.17+0.3665\left(\frac{T}{K}\right)\right\}d\left(\frac{T}{K}\right)$
$=\left(20.17\right)×\left(473-298\right)+\left(\frac{0.3665}{2}\right)×{\left(\frac{T}{K}\right)}^{2}{\mid }_{298}^{473}$
$=\left(3.530×{10}^{3}\right)+\left(2.4725×{10}^{4}\right)=2.83×{10}^{4}$
$q=\mathrm{△}H=2.83×{10}^{4}J=+28.3kJ$
$w=-{p}_{cx}\mathrm{△}V\left[2,8\right]$. where ${p}_{ex}=p$
$w=-p\mathrm{△}V=-\mathrm{△}\left(pV\right)\left[\text{constant pressure}\right]=-\mathrm{△}\left(nRT\right)\left[\text{perfect gas}\right]=-nR\mathrm{△}T$
$=\left(-1.00mol\right)×\left(8.314J{K}^{-1}mo{l}^{-1}\right)×\left(473K-298K\right)=-1.45×{10}^{3}J=-1.45kJ$
$\mathrm{△}U=q+w=\left(28.3kJ\right)-\left(1.45kJ\right)=+26.8kJ$
(b) The energy and enthalpy of a perfect gas depend on temperature alone, hence it does not matter whether the temperature change is brought about at constant volume or constant pressure; are the same.
$\mathrm{△}H=+28.3kJ,\mathrm{△}U=+26.8kJ$
Under constant volume, $w=0$
$q=\mathrm{△}U-w=+26.8kJ$

nick1337

Expert

An inverse rectilinear vector ${\stackrel{\to }{a}}^{\prime }$ is a vector which is co-directed (in the same direction as) a vector $\stackrel{\to }{a}$ and differs from it in magnitude according to:
$|\stackrel{―}{d}|=\frac{1}{|\stackrel{―}{a}|}$
Projections on the coordinate axes of inverse rectilinear vectors are equal according to:
${a}_{x}^{‘}=\frac{{a}_{x}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}};$
${a}_{y}^{‘}=\frac{{a}_{y}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}}$
${a}_{z}^{‘}=\frac{{a}_{z}}{{a}_{x}^{2}+{a}_{y}^{2}+{a}_{z}^{2}};$

Do you have a similar question?

format: plain


text:
# SqlDriverInfo

(Requires Analytica Enterprise or Power Player)

## SqlDriverInfo(driverName)

Returns a list of Attribute-value pairs for the specified driver.

If driverName = "" (an empty text value), returns a list of the names of the drivers.

«driverName» must be a text value—it cannot be a list of text values or an index that is defined as a list of text values. This statement would not normally be used in a model, but may be helpful in understanding the SQL drivers that are available.

## Library

Database Functions


text:
# Selecting terms containing some expression

Imagine I have an expression like

a*k + (a^2)*b*c + b*e

and I would like to obtain the term containing, for example, some power of a. In that case I would extract the two first terms. If I wanted to obtain the term b*e, I would extract the last.

How can I do it? I guess I should use Cases or something similar, but I don't understand its syntax.

-
Maybe Most[CoefficientList[a*k + (a^2)*b*c + b*e, a]] –  Daniel Lichtblau Jan 25 '13 at 16:39

You can use pattern matching. For instance to match a which is multiplied by something else you use

Cases[a*k + (a^2)*b*c + b*e, a*_]

or

Cases[a*k + (a^2)*b*c + b*e, a^_, 2]

The 2 is necessary because a^2 appears deeper in you expression. Furthermore, to prevent trouble, you should be aware of FullForm which shows you the full underlying representation of your structure.

If something does not work, check the FullForm of you pattern and the thing you want to match first. You should start by

Introduction to Patterns

## Update

The first part of my answer was not a ready to use solution. It assumed, that you do some work for yourself: Finding the patterns which help you to extract the parts you want.

Now, it seems you are asking something different: When you look at your expression as a polynomial in a, then it's worth looking at functions like CoefficientList (as already mentioned by Daniel in the comment), Coefficient, ... Please look here to have an overview

Polynomial algebra

With this you get the expression containing some power (even the one with a^0) easily

CoefficientList[a*k + (a^2)*b*c + b*e, a]

Or you can pick one special power, like the second one

Coefficient[a*k + (a^2)*b*c + b*e, a, 2]

-
Thanks, however the second one only returns a^2 instead of b*c ... Maybe the best thing I could do is taking derivatives and evaluate at 0 ... –  pablo Jan 25 '13 at 16:10
@pablo, please see my update. I maybe misinterpreted your question and hopefully gave now some more insight. –  halirutan Jan 27 '13 at 11:32

I think this is what you want:

Cases[
a*k + (a^2)*b*c + b*e,
term_ /; ! FreeQ[term, a],
{1}
]

{a^2 b c, a k}

Note that the terms are not in the explicit order given as Plus is Orderless. I am assuming this does not matter in practice but other measures can be taken of needed.

As a function:

selectTerms[expr_, pat_] := Cases[expr, term_ /; ! FreeQ[term, pat], {1}]

-

Maybe:

 Select[a*k + (a^2)*b*c + b*e, MemberQ[#, a | a^_] &]

=> a^2 b c + a k

Select[a*k + (a^2)*b*c + b*e, MemberQ[#, b] && FreeQ[#, a] &]

=> b e

Select[a*k + (a^2)*b*c + b*e, MemberQ[#, e] &]

=> b e

-


text:
×

# CALLSCHE - Editorial

Author: Maksym Bevza
Tester: Istvan Nagy
Editorialist: Miguel Oliveira

Medium

# PREREQUISITES:

Graph Theory, Minimum Cut, Maximum Flow

# PROBLEM:

Consider that there are D business days, each with N working hours. The availability of each of the P employees is given for each hour of the business days. The employees also need to have at least 1 hour for lunch in a given time interval.

Given a call center schedule constraints, find if it's possible to make a schedule for all employees so that there are always a given number of employees available to talk with clients, satisfying the constraints.

# EXPLANATION:

N, H, D and P are really small. We can brute force to see if there is a feasible solution.

Same as subtask 3 but with an inefficient implementation.

The problem involves deciding, for each person, what activity they will do in each hour of the day, satisfying some restrictions. This resembles an assignment problem. In fact, we can use a similar approach to assignment problems based on network flow.

We can model this problem as a graph where the vertices represent people, person i talking to clients or having lunch, and the edges contain the given constraints.

A constraint of the form 'person x must not talk to clients more than $L_x$ hours per week' can be modelled as an edge with capacity $L_x$.

Now the tricky part is the lunch time.

There are N hours in each day and we must guarantee there is a free hour during lunch time. To do this, we create 2 nodes per person and per day: "lunch hours at day d" and "normal hours at day d".

Suppose $\Delta_{LT} = {LT}*{end} - {LT}*{begin} - meetings$, the number of lunch hours not filled with meetings. Then, we have $N - \Delta_{LT} - meetings$ normal hours and $\Delta_{LT}$ free lunch hours. In order to have a free lunch hour, $\Delta_{LT}$ must be positive. One of these must not be filled with meetings nor clients, so the actual capacity associated with lunch hours is $\Delta_{LT} - 1$. This way, we ensure that at least 1 of these hours is not assigned work.

Now, we create nodes for each pair (day, hour) and connect these with the nodes associated with each <person, normal="" lunch="" hours=""> if the person does not have meeting during that hour in that day: $F_{k, i, j} = 1$.

Finally, we connect the (day, hour) nodes with a sink vertex with capacity equal to $R_{i, j}$.

Suppose we have 5 days: monday - friday, with H = 5 and lunch hours at hours 3 and 4. A possible resulting graph is the following

If the minimum cut between vertices S and T equals $\sum_{R_{i, j}}$ then we can satisfy all customer requirements.

The minimum cut can be calculated using an efficient maximum flow algorithm such as Dinic's algorithm.

# Sample Solutions

This question is marked "community wiki".

5★mogers
659716
accept rate: 25%

19.8k350498541

 0 So... $O(MAX^2)$ vertices, $O(MAX^3)$ edges, $O(MAX^7)$ Dinic? Ford-Fulkerson is better. The asymptotic complexity is $O(15\cdot MAX^5)$ and in practice, a DFS-based implementation of looking for a path from the source to the sink and stopping the DFS as soon as a path is found is quite fast. answered 16 Feb '16, 03:38 7★xellos0 5.9k●5●42●92 accept rate: 10% The worst case time complexity is a bit misleading in these flow algorithms, especially in these kind of networks. While I did not test thoroughly, from what I've seen, Dinic's algorithm performs faster than Ford-Fulkerson in this problem. (16 Feb '16, 04:34) mogers5★ A sample of 15 tests, some of which can be solved using some necessary condition, isn't good enough to measure the speed. In addition, it depends on how many optimisations are made - I'm sure either could seem faster than the other based on it. But it's still good to note the complexities, if only for a theoretical purpose. Anyway, I have never encountered a flow problem unsolvable by Ford-Fulkerson. (16 Feb '16, 07:10) xellos07★
 0 Please someone tell whether greedy solution would be correct for this problem or not as I got an AC using greedy technique in O(n^3) complexity. AC Solution . answered 16 Feb '16, 04:53 6★likecs 3.7k●23●80 accept rate: 9% Can you describe the idea of your greedy algorithm ? Its painful to look at some code and derive the idea out of it. (17 Feb '16, 22:48)
 toggle preview community wiki:
Preview

By Email:

Markdown Basics

• *italic* or _italic_
• **bold** or __bold__
• image?![alt text](/path/img.jpg "title")
• numbered list: 1. Foo 2. Bar
• to add a line break simply add two spaces to where you would like the new line to be.
• basic HTML tags are also supported
• mathemetical formulas in Latex between \$ symbol

Question tags:

×15,684
×2,598
×1,219
×114
×107
×20

question asked: 16 Feb '16, 00:18

question was seen: 2,381 times

last updated: 18 Feb '16, 22:21


text:
# An Upper Bound on the $$n$$th Prime

by John J. Jaroma (Austin College)

College Mathematics Journal
March, 2005

Subject classification(s): Algebra and Number Theory | Number Theory | Primes
Applicable Course(s): 4.3 Number Theory

Builds on Nagura's theorem saying that there exists at least one prime between $$n$$ and $$6n/5$$ if $$n>24$$

A pdf copy of the article can be viewed by clicking below. Since the copy is a faithful reproduction of the actual journal pages, the article may not begin at the top of the first page.


text:
Hard random lattices

The content I ask in this question is in the following picture in [GPV08][1]. I do not understand the proof of the first claim in $$\bf Lemma$$ 5.2. In the proof, which reprensents the uniform distribution over $$\mathbb{Z}_{q}^{n}$$ and which represents $$\bf{u=Ae} \mod q$$.

Could someone explain it to me? Thanks!

[GPV08] Craig Gentry, Chris Peikert, Vinod Vaikuntanathan, [How to use a short basis: trapdoors for hard lattices and new cryptographic constructions][1], August 25, 2008.

The first claim has three steps to it:

1. Assumption of Primitivity: We want to make sure that the matrix $$A$$ isn't such that the function $$x\mapsto Ax\bmod q$$ (equivalently, $$A$$ viewed as a linear operator $$\mathbb{Z}_q^m\to\mathbb{Z}_q^n$$) "misses" most of its codomain. Imagine that $$A$$ is the all zero's matrix --- then the claim can't possibly be true!

If $$q$$ were prime we could assume that $$A$$ is "full rank" and this issue would disappear. We often don't like to restrict to the case that $$q$$ is prime though (for example, $$q = 2^k$$ is common in practice), so our matrices become defined over rings-that-aren't-fields, and "rank" can be a slightly problematic construct to work with (In particular, consider the lattice with basis matrix $$\mathsf{diag}(2,1,1,\dots,1)$$. If $$q$$ is not prime, this may not be equivalent to the identity matrix, so despite being "full rank" it will not surject).

Instead, we assume that $$A$$ is surjective as a map. This can be described several different ways in the literature (here it is "the columns generate the codomain", other places the term "is a primitive matrix" is used. When $$A$$ is a vector one can specify "the GCD of its entries is one"). This ends up being the right analogue of "$$A$$ is full rank" to use when working over rings-that-aren't-fields like $$\mathbb{Z}/2^k\mathbb{Z}$$. I believe this notion still works over number fields (so in the RLWE case) as everything can be viewed as an integer lattice still, but am less familiar with this area.

2. Near-uniformity of Discrete Gaussians mod the lattice: Lemma 2.8 states the following:

Let $$\Lambda, \Lambda'$$ be $$n$$-dimensional lattices with $$\Lambda'\subseteq\Lambda$$. Then, for any $$\epsilon\in(0,1/2)$$, any $$s\geq \eta_\epsilon(\Lambda')$$, and any $$c\in\mathbb{R}^n$$, the distribution of $$(D_{\Lambda, s, c}\bmod \Lambda')$$ is within statistical distance of at most $$2\epsilon$$ over $$(\Lambda\bmod\Lambda')$$

A useful simplification here is to set $$\Lambda =\mathbb{Z}^n$$. Then, this states that for any $$n$$-dimensional integer lattice $$\Lambda'$$, that for "large enough $$s$$" that $$(D_{\mathbb{Z}^n, s, c}\bmod \Lambda')$$ is within $$2\epsilon$$ being uniformly distributed over $$\mathbb{Z}^n\bmod\Lambda'$$, the fundamental domain of $$\Lambda'$$.

This is "what the smoothing parameter is defined for". The smoothing parameter specifies the "phase transition" between the Discrete Gaussian being non-uniformly distributed over the fundamental domain of $$\Lambda'$$ (when working mod the lattice of course), and it being very close to uniformly distributed (for a graphical demonstration of this, look at this other question). As long as we take $$s$$ larger than the smoothing parameter of the relevant lattice, the uniformity follows easily as a consequence.

3. The quotient group: It's worth mentioning explicitly what we should "expect $$\mathbb{Z}^m/\Lambda^\perp$$ to look like". Let $$q$$ be prime so everything is simpler. Recall that:

$$\Lambda_q^\perp(A) = \{x\in\mathbb{Z}^m \mid Ax \equiv 0 \bmod q\}$$

When $$q$$ is prime, this is just the kernel of the linear map $$A : \mathbb{F}_q^m\to\mathbb{F}_q^n$$. Then the group $$\mathbb{Z}_q^m/\ker(A)\cong \mathsf{im}(A)$$ by the first isomorphism theorem (for abelian groups). Of course we're dealing with $$\mathbb{Z}^m/\ker(A)$$ (no $$\bmod q$$ upstairs), but this doesn't actually matter --- as $$\Lambda_q^\perp(A)$$ is always $$q$$-ary (contains the sublattice $$q\mathbb{Z}^m$$), we have that all points in $$\mathbb{Z}^m$$ which differ by a point in $$q\mathbb{Z}^m$$ will be identified anyway, so we may as well pick representatives for these points and call them $$\mathbb{Z}_q^m$$, and "continue evaluating the quotient".

I'm under the impression this argument extends to the case of composite $$q$$ if one makes the appropriate substitutions of "linear algebraic arguments" to "module-theoretic arguments" along the way, but I haven't made the substitutions myself before, and don't feel like going through with it here. If someone has a reference I'd be interested.

Putting it together:

From here everything is fairly mechanical. We want to analyze the distribution of $$(D_{\mathbb{Z}^m, s, c}\bmod\Lambda_q^\perp(A))$$. By the second point this is within $$2\epsilon$$ (in statistical distance) of $$\mathsf{Unif}(\mathbb{Z}^m/\Lambda_q^\perp(A))$$. By the third point the underlying set of this is (assuming $$q$$ prime for simplicity) isomorphic to $$\mathsf{im}(A)$$, so the uniform distribution on the first set is identical to $$\mathsf{Unif}(\mathsf{im}(A))$$. Then, by the first point we have that $$\mathsf{im}(A)\cong \mathbb{Z}_q^n$$, so this distribution is simply $$\mathsf{Unif}(\mathbb{Z}_q^n)$$. In total, we end up with $$(D_{\mathbb{Z}^m, s, c}\bmod\Lambda_q^\perp(A))$$ being within $$2\epsilon$$ in statistical distance of $$\mathsf{Unif}(\mathbb{Z}_q^n)$$, which is the first claim.

• Thanks！ Your answer is useful. According to your resaoning, we still need that $D_{\mathbb{Z^{m}}, s, c}$ mod $\wedge^{\bot}_{q}(A)$ is isomophic to the distribution of the syndrome $u=Ae$ mod q since the first claim needs to prove that the distribution of the syndrome $u=Ae$ mod q is within statistical distance $2\varepsilon$ of uniform over $\mathbb{Z}^{n}_{q}$. The reason is about this (crypto.stackexchange.com/questions/80522/random-lattices). Is it right?@Mark – Alex Ideal May 8 '20 at 13:44
• The distribution of the syndrome ends up being uniform over the codomain due to a combination of $A$ being surjective (mentioned above), and the fact that linear maps are "regular" in the sense that all of their preimages are the same size (this is simply because the preimage of $u$ under $A$ can be written as $e + \ker(A)$ for any distinguished element $e$ in the preimage). This property ends up being precisely what you want to prove that the distribution of $A(\mathsf{Unif}(\mathbb{Z}_q^m)) = \mathsf{Unif}(\mathbb{Z}_q^n)$, which is a simple direct proof once one has the regularity property. – Mark May 8 '20 at 19:09


text:
# VNM Theorem

Starting with some set of outcomes, gambles (or lotteries) are defined recursively. An outcome is a gamble, and for any countablefinite set of gambles, a probability distribution over those gambles is a gamble.

The VNM theorem is one of the classic results of Bayesian decision theory. It establishes that, under four assumptions,assumptions known as the VNM axioms, a preference relation must be representable by maximum-expectation decision making over some real-valued utility function. (In other words, rational decision making is best-average-case decision making.)

Starting with some set of outcomes, gambles (or lotteries) are defined recursively. An outcome is a gamble, and for any countable set of gambles, a probability distribution over those gambles is a gamble.

Preferences are then expressed over gambles via a preference relation. if A is preferred to B, this is written B">A>B. We also have indifference, written AB. If A is either preferred to B or indifferent with B, this can be written AB.

The four VNM axioms are:

1. Completeness. For any gambles A and B, either B">A>BA">B>A, or AB.
2. Transitivity. If A<B and B<C, then A<C.
3. Continuity. If ABC, then there exists a probability p[0,1] such that  pA+(1p)CB. In other words, there is a probability which hits any point between two gambles.
4. Independence. For any C and p[0,1], we have AB if and only if pA+(1p)CpB+(1p)C. In other words, substituting A for B in any gamble can't make that gamble worth less.

In contrast to Utility Functions, this tag focuses specifically on posts which discuss the VNM theorem itself.

The VNM theorem is one of the classic results of Bayesian decision theory. It establishes that, under four assumptions, a preference relation must be representable by maximum-expectation decision making over some real-valued utility function. (In other words, rational decision making is best-average-case decision making.)

Created by abramdemski at 1y


text:
# Math Help - definition of a limit

1. ## definition of a limit

Suppose $g:[0,\infty)\rightarrow\mathbb{R}$ and $L\in[-\infty,\infty]$. Prove that $lim_{t\rightarrow\infty}g(t)=L \iff$ for every increasing sequence $(a_n)$ in $[0,\infty)$, if $a_n\rightarrow\infty$, then $g(a_n) \rightarrow L$.

I'm trying to prove [<==] when $L=\infty$ by contradiction. But not sure about the definition:
$lim_{t\rightarrow\infty}g(t)=\infty$ means that $\forall M>0$, there is $x\in[0,\infty)$ such that $\forall t>x, g(t)>M.$ Is this the correct definition?
So if $lim_{t\rightarrow\infty}g(t)\not=\infty$, then there exists $M>0$ such that $\forall x\in[0,\infty)$ there exists $t>x$ such that $g(t) \leq M.$ Right?

2. Hello.

Your definition seems to be correct.

Now, you can try for a contradiction/contrapositive by constructing an increasing sequence $\{a_n\}$ which diverges to infinity but $\{g(a_n)\}$ does not.


text:
# Finding the matrix of this linear transformation

We're given $V$, which is an $n$ dimensional vector space. $T : V \to V$ is a linear transformation. There is a vector $v \in V$ such that $T^n(v) = 0$. We're also told that the vectors $T^{n-1}(v), T^{n-2}(v), \ldots, T(v), v$ form a basis for $V$.

The questions are:

1. If $n = 4$, calculate the matrix of $T$ w.r.t the basis.
2. If $n = 4$, calculate the matrix for $T^n$ for $n = 2,3,4$.

From a previous question, I know that if you want to form the matrix for a transformation, you simply compute the value for $T$ at the basis, and express your answer as a matrix w.r.t the basis. But, we can't really "compute" in this case because we don't know what the actual transformation is.

Also, when they say $T^2(v)$, do they just mean $T(T(v))$? If so, I suppose $T(T^{n-1}(v)) = 0$, but I'm not sure what else we can figure out or even how to construct a matrix.

As for the second question, I don't really see what they're asking.

Lastly, say we do find a matrix, call it $M$. Let's say I have a vector $u$, and say $T(u) = s$, where $s$ is some other vector. Does the relationship $Mu = s$ always hold in this case?

Thanks a bunch for all your help!

-
Starting with your last question, yes, "$M$ is the matrix of $T$" means "for all $u$, $T(u)=Mu$." –  Gerry Myerson Dec 14 '11 at 3:30

Yes, $T^2(v)$ means $T(T(v))$ in general $T^k(v)$ is composition of $T$ with itself $k$ times. And you are correct about how we should go about this problem, compute how the transformation acts on the basis. I will describe the process for how to do the first part below, and to do the second part you will do the exact same thing, but replace the transformation $T$ with the transformations $T^n$ for each $n$.

For $n = 4$ we have the basis as $$v_1 = T^3(v), v_2 = T^2(v), v_3 = T(v), v_4=v$$

To compute the matrix of a transformation with respect to an ordered basis we simply compute how the transformation acts on that basis. In this case we have $$T(v_1) = T(T^3(v)) = T^4(v) = 0$$ by our assumption that $T^n(v) = 0$ for each $n$. Then we have $$T(v_2) = T(T^2(V)) = T^3(V) = v_1,$$ $$T(v_3) = T(T(v)) = T^2(v) = v_2$$ and $$T(v_4) = T(v) = v_3$$

So now we compose these into a matrix and get

$$[T]_{(v_1, v_2, v_3, v_4)} = \left(\begin{smallmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{smallmatrix}\right)$$

You should be able to preform the same process for the transformation $T^n$ for each $n$ required in part (b).

-
Makes total sense, thanks a lot! –  John Doe Dec 14 '11 at 4:14


text:
# Getting started with BSM Models¶

## Basics¶

Herwig uses a Model class which:

• stores the input parameters of model and calculates derived quantities;
• provides access to the Feynman rules for the model to allow the calculation of hard processes and decays.

Obviously by default the StandardModel class which implements the parameters and Feynman rules of the Standard Model is used. In general setting up all the parameters and Vertex classes which implement the Feynman rules is complicated and a .model file is supplied which does all of this and sets replaces the StandardModel object used by default by that for the new physics model. In addition it creates the ParticleData objects for any new particles needed in the model, sets their branchings in the shower so that QCD radiation is generated and ensures they are decayed during the parton shower.

The first step of any BSM simulation is therefore reading the new model. We will use the built-in MSSM model as an example. There is an example input file LHC-MSSM.in provided with the release. In this case we start with

read MSSM.model

Often, particularly in SUSY models, many of the parameters of the model and decay modes of the new particles are read from a spectrum file calculated by an external spectrum generator. For example in the MSSM these are read using

setup [name of model]/Model [name of SLHA file]

Furthermore as Herwig includes the generation of additional radiation in many BSM decays using the POWHEG scheme it is useful to switch the generation of the parton shower to use these:

# Set emission to POWHEG for radiation in decays
set /Herwig/Shower/ShowerHandler:HardEmission POWHEG

Following this we need to specify which hard processes and decays to include in the simulation.

## Calculation of Hard Processes¶

Herwig includes a built-in mechanism for the calculation of most hard scattering processes of interest in BSM models. Rather than implement matrix elements for specific scattering processes we include a mechanism where a range of HardProcessConstructor classes construct the scattering processes using the Feynman rules.

In general $$2\to2$$ scattering processes are the most important. There are two mechanisms for generating these processes in Herwig

1. The TwoToTwoProcessConstructor class constructs general $$2\to2$$ scattering processes based on the Feynman rules of the model. The default object created for this class is called HPConstructor. We therefore need to specify the incoming and outgoing particles for the scattering processes we wish to consider, for example for up and down left-handed squark production in the MSSM with the incoming proton only containing gluons, up and down quarks and antiquarks:

# Example hard process: Incoming proton, outgoing squarks

insert HPConstructor:Incoming 0 /Herwig/Particles/g
insert HPConstructor:Incoming 0 /Herwig/Particles/u
insert HPConstructor:Incoming 0 /Herwig/Particles/ubar
insert HPConstructor:Incoming 0 /Herwig/Particles/d
insert HPConstructor:Incoming 0 /Herwig/Particles/dbar

insert HPConstructor:Outgoing 0 /Herwig/Particles/~u_L
insert HPConstructor:Outgoing 0 /Herwig/Particles/~u_Lbar
insert HPConstructor:Outgoing 0 /Herwig/Particles/~d_L
insert HPConstructor:Outgoing 0 /Herwig/Particles/~d_Lbar

By default all processes with only the incoming particles, but any of the outgoing particles in association with any other particle are generated. There are a number of switches which control which specific processes are generated:

• Processes controls the generation of the hard processes and specifically how the list of outgoing particles is interpreted in order to construct the relevant scattering processes. There are three options:

1. SingleParticleInclusive the default value which requires at least one particle from the list of outgoing particles as an outgoing particle in the hard process;

2. TwoParticleInclusive which requires that both the outgoing particles in the hard processes are in the list of outgoing particles;

3. Exclusive which require that there are only two particles in the list of outgoing particles and that they are the outgoing particles in every hard process generated.

• IncludeEW whether to include both electroweak (EW) and strong processes. By default it is set Yes in include both EW and strong processes although in many models it is reset to No so that only the dominant strong processes are generated to increase the speed of the calculations.

• Excluded particles which are not allowed as s- or t-channel intermediates in the hard processes

• ExcludedExternal particles which are not allowed external particles in the scattering processes.

• ExcludedVertices vertices which are not to be used in the generation of the hard scattering processes.

2. The ResonantProcessConstructor class constructs $$2\to2$$ scattering processes where a specific s-channel intermediate resonance is required. The default object created for this class is called ResConstructor. As before we need to specify the incoming and outgoing particles for the scattering processes as well as the possible s-channel resonances. For example from the built in model for Randell-Sundrum, from LHC-RS.in, if we wished to consider the production of the graviton decaying to $$e^+, W^+, Z^0,\gamma$$ with the incoming proton only containing gluons, up and down quarks and antiquarks

# Example hard process: Incoming proton, outgoing graviton

insert ResConstructor:Incoming 0 /Herwig/Particles/g
insert ResConstructor:Incoming 0 /Herwig/Particles/u
insert ResConstructor:Incoming 0 /Herwig/Particles/ubar
insert ResConstructor:Incoming 0 /Herwig/Particles/d
insert ResConstructor:Incoming 0 /Herwig/Particles/dbar

insert ResConstructor:Intermediates 0 /Herwig/Particles/Graviton

insert ResConstructor:Outgoing 0 /Herwig/Particles/e+
insert ResConstructor:Outgoing 1 /Herwig/Particles/W+
insert ResConstructor:Outgoing 2 /Herwig/Particles/Z0
insert ResConstructor:Outgoing 3 /Herwig/Particles/gamma

By default all processes with only the incoming particles, but any of the outgoing particles in association with any other particle are generated. There are a number of switches which control which specific processes are generated:

• Processes The setting described above are supported together with the additional Inclusive option where instead of using the list of outgoing particles all the processes are generated where the outgoing particles are those allowed for the decay of the on-shell intermediate.

In addition to the options described above it is possible to change the choice of the scale used for the hard processes using ScaleChoice and include a prefactor for the scale using ScaleFactor.

As well as the general mechanism for generating $$2\to2$$ scattering processes we provide av small number of more specialised HardProcessConstructor classes for the generation of processes involving the Higgs bosons of BSM models:

• HiggsVectorBosonProcessConstructor, the default object is HVConstructor, generates hard process for Higgs boson production in association with a vector boson;
• HiggsVBFProcessConstructor, the default object is HiggsVBFConstructor, generates hard processes for Higgs boson production via vector boson fusion;
• QQHiggsProcessConstructor, the default object is QQHiggsConstructor, generates hard processes for the production of the Higgs boson in association with a heavy quark-antiquark pair.

## Calculation of Decays¶

Most particle in BSM models are unstable and decay. We therefore include the calculation of the allowed decay modes and their partial widths which allows us to calculate the total width and branching ratios for the various modes. In order to do this we include a number of classes inheriting from the NBodyDecayConstructorBase class to calculate different types of decay mode.

By default we use:

• TwoBodyDecayConstructor, with an instance called TwoBodyDC, to generate $$1\to2$$ decays;
• ThreeBodyDecayConstructor, with an instance called ThreeBodyDC, to generate $$1\to3$$ decays.

These classes construct all the $$1\to2,3$$ decays consistent with the Feynman rules for the model and are switched on by default. In addition we have two more specific classes:

• WeakCurrentDecayConstructor, with an instance called WeakDecayConstructor, for generating decays of the form $$a\to bW^*$$ with a small mass difference between a and b so that the $$W^\pm$$ boson is highly off-shell and decays to a small number of hadrons, the production of which is then described using the same weak currents as the decay of the $$\tau^\pm$$ lepton;
• FourBodyDecayConstructor, with an instance called FourBodyDC, for generating decays of the scalars to four fermions as this can be important for the Higgs boson in extended models and the decay of the stop in some SUSY scenarios;

which generate more specific types of decay which can be important in specific models but aren’t relevant in most cases and are switched off by default. N.B. as both of these simulate higher multiplicity decays the phase-space integration of the matrix elements has to be performed using the Monte Carlo technique and can therefore be slow if there are a lot of decay modes.


text:
# How to determine the annual harmonics of temperature and salinity data?

This is a question on methodology. But I have 4d temperature and salinity gridded data (time, depth, lat and lon). The time is in monthly time steps. How would I get the annual harmonics of temperature and salinity using a periodicity of 12 months:

The equation is:

$$Var(t,z,y,x) = A (z,y,x) * \cos [ (2 \pi t/P + \phi(z,y,x)]$$

Where $$A$$ is the amplitudes of the annual component, where $$var$$ is either temp or salinity. $$\phi$$ is the phase angle which determine the time when the maximum of the annual harmonic occurs. And $$t$$ varies from 0 - n months (however long the time series is).

We can isolate $$A(z,y,x)$$ just with algebra, but the issue is finding the phase angle where the maximum of the annual harmonic occurs.

Do you you need to take a fourier transform of monthly means (January - December), or do you take the FT of the entire time series but just look at the power spectrum at 12 months... I am using Python, and taking the fourier transform is no problem. I just don't know how to treat the data to obtain the phase angle where the maximum of the annual harmonic occurs. What might be the steps to find the annual harmonic amplitude given 4D temp and salinity (time in months, depth, lat and lon)?

With a little bit of math:

$$A * \cos [ (2 \pi t/P + \phi] = a\sin(2\pi t/P)+b\cos(2\pi t/P)$$

Then, the amplitude is:

$$A=\sqrt{a^2+b^2}$$ and the phase is $$\phi=\arctan{\frac{b}{a}}$$

https://en.wikibooks.org/wiki/Trigonometry/Simplifying_a_sin(x)_%2B_b_cos(x)

• yeah, thank you! My issue is applying this to the 4D data... and to what extent, like monthly means (January - December) or just the trended, monthly data itself Nov 5 '21 at 18:11
• I would recommend that you use the monthly data. There is also the possibility that the A(x,y,z) could have a change in time. You have to do it point by point of the grid. You could try to do a regional average and then get the seasonal cycle in each region, but that may alias the magnitude of the cycle Nov 5 '21 at 18:49
• Oh ok, thank you! Sorry one last question... to get the amplitude of the annual harmonic, could I just get the amplitude from the FFT just at the period of 12 months (or frequency of 1/12)? Would that constitute as the annual harmonic? Nov 5 '21 at 19:23
• You don't even need to do the FFT, you could just do a fit to that function exclusively. The issue of the FFt is that you might get some of the energy of the annual cycle into other frequencies. You can difinitely try the 1/12 frequency and see if it makes sense Nov 5 '21 at 21:39


text:
• 简体中文
• ScienceCloud
•  welcome!

2019-01-17 10:11
Catalog：
1099

# Introduction

Large Sky Area Multi-object Fiber Spectroscopic Telescope (LAMOST) is a quasi-meridian reflecting Schmidt telescope with an effective light-collecting aperture of 3.6 m to 4.9 m (depending on pointing direction) and a 5$$^\circ$$ field of view in diameter (Wang et al. 1996; Su & Cui 2004; Cui et al. 2012), located in Xinglong Station of National Astronomical Observatory, China (NAOC). It has 4000 robotic fibers, with 3.3 arcsec diameter, mounted on its focal plane and connected to 16 spectrographs. Each spectrum of the target is split into two channels, red and blue, and then recorded on red and blue cameras, respectively. The blue channel is optimized for 3700–5900 Å, and the red channel for 5700–9000 Å, with 200 Å overlaps between the two channels. The spectral resolution reaches $$R\sim1800$$ over the entire wavelength range (see Cui et al. 2012).

Left: Primary mirror of LAMOST. Upper right: The focal plane with 4000 fiber positioning units. Lower right: the fiber positioning units. (http://www.lamost.org)

LAMOST regular survey, carried out from September 2012, was designed to have two major components: the LAMOST ExtraGAlactic Survey (LEGAS) and the LAMOST Experiment for Galactic Understanding and Exploration (LEGUE) survey of Milky Way stellar structure (Zhao et al. 2012). The LAMOST quasar survey was conducted under LEGAS, which covers the high Galactic latitude area in the northern sky. This survey aims to make census of quasars at differenct redshifts for better understanding the growth of supermassive black holes and the co-evolution with their host galaxies.

### Target Selection

The details of the method used to select the targets for LAMOST quasar survey can be found in Wu & Jia (2010), Wu et al. (2012) and Peng et al. (2012). Most of the quasar candidates for spectroscopic followup are selected based on the photometry data of the Sloan Digital Sky Survey (SDSS, e.g., Ahn et al. 2012). The targets are required to be point sources in SDSS images to avoid large numbers of galaxies. A faint limit of $$i = 20$$ is adopted to avoid the too low signal-to-noise ratio, and a bright limit of $$i = 16$$ is adopted to avoid the saturation and the cross-talk of the neighboring fibers. The selection is mainly based on the following methords.

• Optical-infrared colors: The distinct optical-infrared colors could be an useful tool in selecting quasar candidates. The photometric point sources are cross-matched with the dataset from UKIDSS/Large Area Survey (Lawrence et al. 2007) and WISE all-sky data release (Wright et al. 2010). Then, the selection is made according to the location of sources in the multi-dimensional SDSS and UKIDSS/WISE color space (Wu & Jia 2010, Wu et al. 2012).
• Data-mining algorithms: A mixture of heterogeneous data-mining algorithm, such as support vector machine (SVM, Peng et al. 2012), extreme deconvolution (XD, Bovy et al. 2011) and kernel-density-estimation technique (KDE, Richards et al. 2009), are also adopted in combination with the optical-infrared color selection method.
• Multi-wavelength data matching: The final input catalog of the targets are supplemented by cross-matching the SDSS photometry with sources detected in X-ray surveys (XMM-Newton, Chandra, ROSAT) and radio survey (FIRST, NVSS).

### Quasar Identification

After the observations, the spectra were automatically classified by LAMOST 1D pipeline into four categories according to their object types, namely STAR, GALAXY, QSO and Unknown (Luo et al. 2012). Then visual inspection was performed to identify not only the spectra of targets in the input quasar candidate list but also the spectra that are classified as QSO by the pipeline but not in the candidate list. A Java program ASERA (Yuan et al. 2013) was used for the visual inspection and scuring the estimation of redshifts. The misclassified spectra are rejected or re-classified by eye. Each spectrum is inspected by two or three classifiers to check if the features of spectrum can match the quasar template. A flag of ZWARNING=1 is given when there is only one emission line visible in the spectra. After the quasars are identified, we performed spectral modeling of both narrow and broad components of H$$\alpha$$, H$$\beta$$, MgII and CIV emission lines for each spectrum when possible. The full width at half maximum and equivalent width for these lines are measured.

### Overview

After 5 years regular observations, LAMOST quasar survey has observed spectra for about 42,500 quasars (Ai et al. 2016; Dong et al. 2018; Yao et al. 2019). Among all LAMOST quasars, $$\sim$$24,000 quasars are discovered by LAMOST, $$\sim$$7000 of which were reported in SDSS DR12/14 quasar catalogs (Pâris et al. 2017,2018) after the LAMOST quasar survey started, while the other $$\sim$$17,000 are newly discovered. Thus, 56% are discovered independently by LAMOST quasar survey. In catalogs, we also provide our measurements of the broad emission lines and the estimations of black hole masses. LAMOST quasar catalog provides not only a great supplment to the new quasar discoveries, but also a large database for investigating quasar spectral variability and discovering unusual AGNs, including changing-look quasars (Yang et al. 2018)

Left: The absolute magnitude and redshift distribution for the visually confirmed quasars observed in five-year LAMOST quasar survey (black) and in the fourth and fifth year (blue). Right: The pie plot shows the fraction of new discoveries among the LAMOST quasars observed between 2015 September and 2017 June as an example.

 Full catalog Download (19MB) Documentation containing quasars observed during the first five-year LAMOST regular survey. DR4&5 Download (15MB) containing quasars observed between 2015.09.12 and 2017.06.16. Full content and detailed description can be found in Yao et al. (2018) DR2&3 Download (10MB) containing quasars observed between 2013.09.10 and 2015.05.30. Full content and detailed description can be found in Dong et al. (2018) DR1 containing quasars observed between 2012.09.28 and 2015.06.03. DR1 quasar catalogs are provided seperately into new discoveries and known ones. Full content and detailed description can be found in Ai et al. (2016)

### References

1. “The Large Sky Area Multi-object Fiber Spectroscopic Telescope Quasar Survey: Quasar Properties from the First Data Release ( 2016, AJ, 151, 24)” Ai, Y. L.; Wu, Xue-Bing; Yang, Jinyi; Yang, Qian; Wang, Feige; Guo, Rui; Zuo, Wenwen; Dong, Xiaoyi; Zhang, Y.-X.; Yuan, H.-L.; Song, Y.-H.; Wang, Jianguo; Dong, Xiaobo; Yang, M.; -Wu, H.; Shen, S.-Y.; Shi, J.-R.; He, B.-L.; Lei, Y.-J.; Li, Y.-B.; Luo, A.-L.; Zhao, Y.-H.; Zhang, H.-T.

2. “The Large Sky Area Multi-Object Fibre Spectroscopic Telescope (LAMOST) Quasar Survey: Quasar Properties from Data Release Two and Three (2018, AJ, 155, 189)” Dong, X. Y.; Wu, Xue-Bing; Ai, Y. L.; Yang, J. Y.; Yang, Q.; Wang, F.; Zhang, Y. X.; Luo, A. L.; Xu, H.; Yuan, H. L.; Zhang, J. N.; Wang, M. X.; Wang, L. L.; Li, Y. B.; Zuo, F.; Hou, W.; Guo, Y. X.; Kong, X.; Chen, X. Y.; Wu, Y.; Yang, H. F.; Yang, M.

3. “The Large Sky Area Multi-Object Fiber Spectroscopic Telescope (LAMOST) Quasar Survey: the 4th and 5th Data Release (2019, ApJS, 240, 1)” Yao, Su; Wu, Xue-Bing; Ai, Y. L.; Yang, Jinyi; Yang, Qian; Dong, Xiaoyi; Joshi, Ravi; Wang, Feige; Feng, Xiaotong; Fu, Yuming; Hou, Wen; Luo, A.-L.; Kong, Xiao; Liu, Yuanqi; Zhao, Y.-H.; Zhang, Y.-X.; Yuan, H.-L.; Shen, Shiyin

### Acknowledgments

If the LAMOST quasar catalogs have been helpful for your research the following acknowledgment would be appreciated:

This research has made use of data obtained from LAMOST quasar survey. LAMOST is a National Major Scientific Project built by the Chinese Academy of Sciences. Funding for the project has been provided by the National Development and Reform Commission. LAMOST is operated and managed by the National Astronomical Observatories, Chinese Academy of Sciences.

### Contact

Prof. Xue-Bing Wu (PI of the LAMOST Quasar Survey), Department of Astronomy & Kavli Institute for Astronmy and Astrophysics, Peking University. Email: wuxb(at)pku.edu.cn


text:
# A Study of NK Landscapes' Basins and Local Optima Networks

Abstract : We propose a network characterization of combinatorial fitness landscapes by adapting the notion of inherent networks proposed for energy surfaces (Doye, 2002). We use the well-known family of $NK$ landscapes as an example. In our case the inherent network is the graph where the vertices are all the local maxima and edges mean basin adjacency between two maxima. We exhaustively extract such networks on representative small NK landscape instances, and show that they are 'small-worlds'. However, the maxima graphs are not random, since their clustering coefficients are much larger than those of corresponding random graphs. Furthermore, the degree distributions are close to exponential instead of Poissonian. We also describe the nature of the basins of attraction and their relationship with the local maxima network.
Keywords :
Type de document :
Communication dans un congrès
Genetic And Evolutionary Computation Conference, Jul 2008, Atlanta, United States. ACM New York, NY, USA, pp.555--562, 2008, 〈10.1145/1389095.1389204〉
Domaine :

https://hal.archives-ouvertes.fr/hal-00331868
Contributeur : Sébastien Verel <>
Soumis le : dimanche 19 octobre 2008 - 00:01:26
Dernière modification le : mercredi 31 janvier 2018 - 10:24:05
Document(s) archivé(s) le : lundi 7 juin 2010 - 20:45:15

### Fichiers

t07pap218-ochoa.pdf
Fichiers produits par l'(les) auteur(s)

### Citation

Gabriela Ochoa, Marco Tomassini, Sébastien Verel, Christian Darabos. A Study of NK Landscapes' Basins and Local Optima Networks. Genetic And Evolutionary Computation Conference, Jul 2008, Atlanta, United States. ACM New York, NY, USA, pp.555--562, 2008, 〈10.1145/1389095.1389204〉. 〈hal-00331868〉

### Métriques

Consultations de la notice

## 361

Téléchargements de fichiers


text:
# B8.2 Continuous Martingales and Stochastic Calculus - Material for the year 2019-2020

## Primary tabs

2019-2020
Lecturer(s):
Prof. Samuel Cohen
General Prerequisites:

B8.1 Martingales through Measure Theory is a prerequisite. Consequently, Part A Integration and Part A Probability are also prerequisites.

Course Term:
Hilary
Course Lecture Information:

16 lectures

Course Weight:
1.00 unit(s)
Course Level:
H

### Assessment type:

Course Overview:

Stochastic processes - random phenomena evolving in time - are encountered in many disciplines from biology, through geology to finance. This course focuses on mathematics needed to describe stochastic processes evolving continuously in time and introduces the basic tools of stochastic calculus which are a cornerstone of modern probability theory. The motivating example of a stochastic process is Brownian motion, also called the Wiener process - a mathematical object initially proposed by Bachelier and Einstein, which originally modelled displacement of a pollen particle in a fluid. The paths of Brownian motion, or of any continuous martingale, are of infinite variation (they are in fact nowhere differentiable and have non-zero quadratic variation) and one of the aims of the course is to define a theory of integration along such paths equipped with a suitable integration by parts formula (Itô formula).

Learning Outcomes:

The students will develop an understanding of Brownian motion and continuous martingales in continuous time. They will became familiar with stochastic calculus and in particular be able to use Itô's formula.

Course Synopsis:

An introduction to stochastic processes in continuous time. Brownian motion - definition, construction and basic properties, regularity of paths. Filtrations and stopping times, first hitting
times. Brownian motion - martingale and strong Markov properties, reflection principle. Martingales - definitions, regularisation and convergence theorems, optional sampling theorem, maximal and Doob's $L^p$ inequalities. Quadratic variation, local martingales, semimartingales. Recall of Stieltjes integral. Stochastic integration and Itô's formula with applications.

1. D. Revuz and M. Yor, Continuous martingales and Brownian motion, Springer (Revised $3^{rd}$ ed.), 2001, Chapters 0-4.
2. I. Karatzas and S. Shreve, Brownian motion and stochastic calculus, Springer ($2^{nd}$ ed.), 1991, Chapters 1-3.


text:
## anonymous one year ago What is the length of stack M N with bar on top ? Round to the nearest tenth of a unit. units

1. anonymous

Graph

2. anonymous

WILL FAN AND MEDAL!!

3. anonymous

@Hero

4. anonymous

Can you help me @Hero?

5. Hero

The quickest way to do it is to find the length of the unit segment. Then multiply by the total number of unit segments that span segment MN.

6. anonymous

What??

7. anonymous

@Vocaloid

8. Hero

The unit segment. If you notice, the entire length of MN can be broken down into six equal segments. Of those six smaller segments, if you take one of them, it would represent the unit segment.

9. anonymous

Can you walk me through using numbers I learn better that way.

10. Hero

11. Hero

When you have understood that, let me know.

12. anonymous

When I said numbers I meant equations... like @Vocaloid does.

13. Hero

We'll get to that. But first, do you see that there are six unit segments?

14. anonymous

yes

15. Hero

Okay, now all we have to do is take just one of those segments create a right triangle and apply pythagorean theorem. Notice the length of the legs of the triangle is just one unit. The pythagorean theorem is a^2 + b^2 = c^2. If we let a and b represent the length of the legs, then $$1^2 + 1^2 = c^2$$ or $$1 + 1 = c^2$$ or simply $$2 = c^2$$. So $$\sqrt{2} = c$$

16. Hero

$$c$$ is the length of one unit segment. Multiply that by 6 and you'll have the length of $$MN$$

17. UnkleRhaukus

surely it is simpler to apply pythag to the entire interval

18. anonymous

I am SO lost.

19. Hero

In theory, yes, but if properly applied, this is the simpler method.

20. Hero

Or the more intuitive method.

21. anonymous

I am NOT good at math at ALL... Sorry.

22. UnkleRhaukus

|dw:1434682942799:dw|

23. UnkleRhaukus

|dw:1434683007008:dw|

24. UnkleRhaukus

${MN}^2 = (4-(-2))^2 + (8-2)^2\\ {MN} = \sqrt{(4-(-2))^2 + (8-2)^2}$

25. anonymous

ok hold on.

26. anonymous

4?

27. UnkleRhaukus

?

28. anonymous

the answer for the equation... is it 4?

29. UnkleRhaukus

no

30. UnkleRhaukus

what is 8 - 2 ?

31. anonymous

oh 6

32. UnkleRhaukus

and what is 4 - (-2) ?

33. anonymous

6

34. UnkleRhaukus

so you have a right angled triangle, with two sides equal to 6, and you want to find the hypotenuse pythag. says: 6^2 + 6^2 = MN^2

35. UnkleRhaukus

ie MN = √ ( 6^2+6^2 )

36. anonymous

72

37. UnkleRhaukus

dont forget to take the square root

38. anonymous

8.49?

39. UnkleRhaukus

yes!, but careful, the question wants the answer accurate to the nearest tenth of a unit

40. anonymous

ok I have 3 more can you help me @UnkleRhaukus ?

41. UnkleRhaukus

I want to check your final answer for this one before we move on, what is 8.49 with one decimal place values represented only

42. anonymous

yeah I came to 8.5.

43. UnkleRhaukus

Good work!

44. anonymous

So can you help me with the rest?

45. UnkleRhaukus

perhaps, (close this post and open a new question so the page doesn't get too long )

46. anonymous

ok I'll tag you.

47. UnkleRhaukus

ok!


text:
# Recent posts in Others

1
PGEE 2020 was conducted on 24th June 2020. I took it, my center was Rajkot, Gujarat. As they do not publish previous year papers, I thought some immediate memory-based topics which were most asked in the paper would be useful in coming years. First paper was ... get admission in IIITH. I request all the students who took PGEE 2020 (CSE) to add their experiences and topics which were asked.
2
Hi, We just had a major server upgrade and so some bugs are expected. Please list down any issue you are facing since today. You can also add any other bug which was already present. Also, there are some new themes to try out if you are interested. Muffin Theme which has both Light and Dark Versions Frapuchino – for Desktop Aven
3
I am sharing my email coversation with IITGN academic department. I have asked a bunch Question, since little to no information is provided by the IITGN on their official website. MY MAIL ----> Hello, I am considering to apply for the PGDIIT, ... whereas PGDIIT is a diploma. With Regards, Saswati Roy Assistant Registrar (Academic) Indian Institute of Technology Gandhinagar 079-2395-2051/2532
4
5
COAP http://coap.iitm.ac.in/ open
6
BARC 2020 Paper : Every Year students search BARC previous year paper.Here is this year exam paper. Hope this blog will helpful for future aspirants.not all the questions but most of the questions are covered here. Slot 1: Date 14/03/2020 (Taken From Nupur ( ... .com/How-was-your-BARC-Computer-Science-Exam-2019 please shear your BARC exam experience so that it will helpful for the future aspirant.
7
Hi, I am a 2019 Btech - CSE graduate with a CGPA of 7.49. My 2019 Gate score is 655. I heard that course is new, and not much information is provided in their site. Do I have a chance with my current score ? Will there be an interview ? Thank You
8
Why ‘Upcoming Exams & Job openings’ related section is removed from Gate Overflow home page? It was very useful section ,Admin Please look into it.
9
National Institute of Electronics and Information Technology (NIELIT) on behalf of NIC invites an Application for the post of Scientist - 'B' and Scientific/Technical Assistant - 'A' in NIC [Advt No. NIELIT/NIC/2020/1]. Start date for submission of ... delhi/content/previous-question-papers-recruitment-exam (previous year paper) 3) http://beta.nielit.gov.in/content/old-question-papers-0
10
Admission CGPA Required Expected GATE Rank Cutoff Expected GATE Score Cutoff IISc. M.Tech. TA CSA Direct Gen$:60$ OBC: NIL SC: NIL ST: NIL Gen$:894(2018)$ OBC$:806(2018)$ SC$:632(2018)$ ST$:509(2018)$ IISc. M.Tech. RA CSA Interview Gen: NIL ... those planning to attend CSA Interviews or join TA course, it is highly recommended to attend CSA Open Day: https://www.facebook.com/csaopenday/?ref=br_rs
11
function top_avg(students){ var len = students.length; var top01 = 40; if(len > 10000){ top01 = 80; } var sum = 0.0; for (var i = 0; i < top01; i++) { sum += students[i].marks; } return sum/top01; } The above code means, everyone rank will ... the $top01$ value dynamically based on the length of students.length, just multiply by $0.0001$ Unless its already done. I just checked the $github$.
12
Another big opportunity for aspirants. IIIT HYDERABAD PORTAL IS LIVE NOW. Last Date: 31st Mar 2020 Entrance Exam - 26th Apr 2020 Refer here: http://pgadmissions.iiit.ac.in/pgee.html https://ims.iiit.ac.in/admissions.php
13
PLEASE PROVIDE OVERVIEW,ANALYSE AND EXPERIENCE OF ISRO WRITTEN EXAM .LEVEL OF EXAM ,GOOD NUMBER OF ATTEMPTS? OFFICIAL ANSWER KEYS :- https://www.isro.gov.in/career?field_isro_centre_value=icrb RESULT:- https://www.isro.gov.in/sites/default/ ... pdf View Obtained Marks: https://apps.isac.gov.in/CentralBE-2019/ShowStatus.jsp?postno=03/2019_BE003&postname=Scientist/Engineer-SC(Computer%20Science)
14


text:
# What is the difference between Quaternions and Bicomplex Numbers?

So, I know Quaternions are basically 4 dimensional Complex numbers, and the dimensions can double forever to Octonions, Sedinions, etc. I recently heard about bicomplex numbers, which are also sort of 4 dimensional complex numbers. I looked into it a bit, and I found that bicomplex numbers are a commutative version of quaternions. But I wanted to know how exactly that works. How is the multiplication defined? Also can that be extended to 8-Dimensional biquaternions, 16-D bioctonions, etc? And how are those defined exactly? I know with regular hypercomplex numbers, you lose properties as you go up in dimensions, so does the same thing happen with the bi- ones? If so does it just lose the property that it would normally lose one dimension lower? Or do they not lose properties at all? Any answers would be greatly appreciated.

• Quaternions are basically 4 dimensional real numbers? – Curiosity Mar 12 '15 at 18:38
• Well, however you want to say I guess. I was more just saying that I knew what Quaternions were, not trying to give an in-depth description. Really they are their own thing, 4-Dimensional Hypercomplex Numbers. – RothX Mar 12 '15 at 18:45

I'd never heard of a bicomplex number, but I like algebra, so let's go for it. From what I understand, reading Wikipedia and playing around:

We assume that, in addition to $i$, our favorite square root of $-1$, we have another, $h$, so that $h^2 = -1 = i^2$. Further, we force $i$ and $h$ to commute, so that $ih = hi$.

You can see for yourself that $(ih)^2 = 1$, since $h$ and $i$ commute. So, $ih$ is a square root of $1$. We know of some other square roots of $1$, let's see what happens when we multiply something by $ih$. At this point, we'll assume our numbers are associative (just like we assumed $h$ and $i$ commute).

$(ih)h = ih^2 = i(-1) = -i$. Thus, multiplication by $ih$ sent $h$ to $-i$. This means $ih$ can't be $1$ or $-1$, the square roots of $1$ we already know and love, since $i$ and $h$ are distinct (that is, $h \neq -i$).

It turns out the set $\{1, i, h, ih\}$ forms a basis for these numbers. Thus, a generic "bicomplex number" looks like $a + bi + ch + dih$.

But, these bicomplex numbers are certainly don't act like quaternions. For one, they form a commutative algebra; the order of multiplication never matters! But more seriously, we have zero-divisors:

$$(1 + ih)(-1 + ih) = -1 + ih - ih +(ih)^2 = -1 +ih - ih - 1 = 0.$$

This is a pretty serious departure from the quaternions.

It turns out that the biquaternions are isomorphic to the direct sum $\mathbb{C} \oplus \mathbb{C}$ of the field of complex numbers with itself (here, addition and multiplication are performed component-wise, like vector addition). This is not obvious to me, but it's believable.

Direct sums are fairly uninteresting: Just slap two copies of something together, and there you go! It explains why we have zero-divisors, and commutativity. We just do everything we normally would, but do it twice; business as usual.

The quaternions on the other hand - now that's an algebra! Associative, but not commutative; no zero divisors; we can define a norm. Generally, life with the quaternions is great.

To build the complex numbers, quaternions, or octonians from the reals, the situation is a lot more subtle; the construction is an example of a "twisted group algebra." The complex numbers, quaternions, and octonions are all examples of twisted group algebras over the real numbers. In particular, we get the quaternions by introducing a "twisted" way of multiplying elements of the group algebra $\mathbb{R}[\mathbb{Z}_2^2]$.

Unfortunately, I'm still processing all the details myself, so it's up to you to figure out exactly what "twisting" something entails. Essentially, it's like a direct sum, but with the terms interacting multiplicatively more than just "component-wise," we need some extra magic thrown in.

Sources:

John Baez' write up on octonions, where I first saw the term "twisted group algebra." Lacking on details of the construction, lots of attention to the big picture.

John Bales' write up on "Properly Twisted Group Algebras", which is way more detail than I'd hoped for, but could fill in some of the gaps.

• Ok, thanks, that explains it to an extant. They are more different than I expected. Also, I'm glad my question introduced you to something new as well! – RothX Mar 14 '15 at 0:53
• Also, you said $−1+ih−ih+(ih)^2 = −1+ih−ih−1$, when it should be $−1+ih−ih+(ih)^2 = −1+ih−ih+1$ – RothX May 18 '17 at 13:46
• @RothX Correct you are, and thank you! I'll hold off on changing anything at the moment -- while it is certainly an error, maybe in a day or two I'll be able to improve the post in a more significant way (although probably not :)). – pjs36 May 18 '17 at 14:09


text:
# Understanding Quaternions

### 1. Introduction

Attitude and Heading Sensors from CH Robotics can provide orientation information using both Euler Angles and Quaternions.  Compared to quaternions, Euler Angles are simple and intuitive and they lend themselves well to simple analysis and control.  On the other hand, Euler Angles are limited by a phenomenon called "gimbal lock," which prevents them from measuring orientation when the pitch angle approaches +/- 90 degrees.

Quaternions provide an alternative measurement technique that does not suffer from gimbal lock.  Quaternions are less intuitive than Euler Angles and the math can be a little more complicated.  This application note covers the basic mathematical concepts needed to understand and use the quaternion outputs of CH Robotics orientation sensors.

Sensors from CH Robotics that can provide quaternion orientation outputs include the UM6 Orientation Sensor and the UM6-LT Orientation Sensor.

### 2. Quaternion Basics

A quaternion is a four-element vector that can be used to encode any rotation in a 3D coordinate system.  Technically, a quaternion is composed of one real element and three complex elements, and it can be used for much more than rotations.  In this application note we'll be ignoring the theoretical details about quaternions and providing only the information that is needed to use them for representing the attitude of an orientation sensor.

The attitude quaternion estimated by CH Robotics orientation sensors encodes rotation from the "inertial frame" to the sensor "body frame."  The inertial frame is an Earth-fixed coordinate frame defined so that the x-axis points north, the y-axis points east, and the z-axis points down as shown in Figure 1.  The sensor body-frame is a coordinate frame that remains aligned with the sensor at all times.  Unlike Euler Angle estimation, only the body frame and the inertial frame are needed when quaternions are used for estimation (Understanding Euler Angles provides more details about using Euler Angles for attitude estimation).

Figure 1 - The Inertial Frame

Let the vector $\mathbf{q}_i^b$ be defined as the unit-vector quaternion encoding rotation from the inertial frame to the body frame of the sensor:

$\mathbf{q}_i^b = \begin{pmatrix} a & b & c & d \end{pmatrix}^T.$

where $T$ is the vector transpose operator.  The elements b, c, and d are the "vector part" of the quaternion, and can be thought of as a vector about which rotation should be performed.  The element is the "scalar part" that specifies the amount of rotation that should be performed about the vector part.  Specifically, if $\theta$ is the angle of rotation and the vector $\begin{pmatrix} v_x & v_y & v_z \end{pmatrix}^T$ is a unit vector representing the axis of rotation, then the quaternion elements are defined as

$\begin{pmatrix} a\\ b\\ c\\ d \end{pmatrix} = \begin{pmatrix} \cos(0.5\theta) \\ v_x\sin(0.5\theta)\\ v_y\sin(0.5\theta)\\ v_z\sin(0.5\theta) \end{pmatrix}.$

In practice, this definition needn't be used explicitly, but it is included here because it provides an intuitive description of what the quaternion represents.  CH Robotics sensors output the quaternion $\mathbf{q}_i^b$ when quaternions are used for attitude estimation.

### 3. Rotating Vectors Using Quaternions

The attitude quaternion $\mathbf{q}_i^b$ can be used to rotate an arbitrary 3-element vector from the inertial frame to the body frame using the operation

$\mathbf{v}_B = \mathbf{q}_i^b\begin{pmatrix} 0\\ \mathbf{v}_I \end{pmatrix} (\mathbf{q}_i^b)^{-1}.$

That is, a vector can rotated by treating it like a quaternion with zero real-part and multiplying it by the attitude quaternion and its inverse.  The inverse of a quaternion is equivalent to its conjugate, which means that all the vector elements (the last three elements in the vector) are negated.  The rotation also uses quaternion multiplication, which has its own definition.

Define quaternions $\mathbf{q}_1 = \begin{pmatrix} a_1 & b_1 & c_1 & d_1 \end{pmatrix}^T$ and $\mathbf{q}_2 = \begin{pmatrix} a_2 & b_2 & c_2 & d_2 \end{pmatrix}^T$.  Then the quaternion product $\mathbf{q}_1 \mathbf{q}_2$ is given by

$\mathbf{q}_1 \mathbf{q}_2 = \begin{pmatrix} a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2 \\ a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2 \\ a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2 \\ a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2 \end{pmatrix}.$

To rotate a vector from the body frame to the inertial frame, two quaternion multiplies as defined above are required.  Alternatively, the attitude quaternion can be used to construct a 3x3 rotation matrix to perform the rotation in a single matrix multiply operation.  The rotation matrix from the inertial frame to the body frame using quaternion elements is defined as

$R_i^b(\mathbf{q}_i^b)=\begin{pmatrix} a^2 + b^2 - c^2 - d^2 & 2bc - 2ad & 2bd + 2ac \\ 2bc + 2ad & a^2 - b^2 + c^2 - d^2 & 2cd - 2ab \\ 2bd - 2ac & 2cd + 2ab & a^2 - b^2 - c^2 + d^2 \end{pmatrix}.$

Then the rotation from the inertial frame to the body frame can be performed using the matrix multiplication

$\mathbf{v}_B = R_i^b(\mathbf{q}_i^b)\mathbf{v}_I.$

Regardless of whether quaternion multiplication or matrix multiplication is used to perform the rotation, the rotation can be reversed by simply inverting the attitude quaternion before performing the rotation.  By negating the vector part of the quaternion vector, the operation is reversed.

### 4. Converting Quaternions to Euler Angles

CH Robotics sensors automatically convert the quaternion attitude estimate to Euler Angles even when in quaternion estimation mode.  This means that the convenience of Euler Angle estimation is made available even when more robust quaternion estimation is being used.

If the user doesn't want to have the sensor transmit both Euler Angle and Quaternion data (for example, to reduce communication bandwidth requirements), then the quaternion data can be converted to Euler Angles on the receiving end.

The exact equations for converting from quaternions to Euler Angles depends on the order of rotations.  CH Robotics sensors move from the inertial frame to the body frame using first yaw, then pitch, and finally roll.  This results in the following conversion equations:

$\phi = \arctan\left( \frac{2(ab+cd)}{a^2 - b^2 - c^2 + d^2} \right ),$

$\theta = -\arcsin\left( 2(bd - ac) \right ),$ and

$\psi = \arctan\left( \frac{2(ad+bc)}{a^2 + b^2 - c^2 - d^2} \right ).$

See the chapter on Understanding Euler Angles for more details about the meaning and application of Euler Angles.  When converting from quaternions to Euler Angles, the atan2 function should be used instead of atan so that the output range is correct.  Note that when converting from quaternions to Euler Angles, the gimbal lock problem still manifests itself.  The difference is that since the estimator is not using Euler Angles, it will continue running without problems even though the Euler Angle output is temporarily unavailable.  When the estimator runs on Euler Angles instead of quaternions, gimbal lock can cause the filter to fail entirely if special precautions aren't taken.


text:
PS_4.pdf - Econometrics II Fall 2019 Problem Set 4 due by 10:56 Monday November 18 in the box labeled Econometria II next to office 20.141(For each

# PS_4.pdf - Econometrics II Fall 2019 Problem Set 4 due by...

• Test Prep
• 2

This preview shows page 1 - 2 out of 2 pages.

Econometrics II. Fall 2019 Problem Set 4 due by 10:56, Monday, November 18 in the box labeled Econometria II next to office 20.141 (For each person of your problem set group, please put their seminar group number next to their name.) Problem 1 . Suppose Y i are random variables with mean μ and variance σ 2 = 30. We have a sample of n = 10 observations, and would like to predict the value of an out-of-sample Y OOS . 1. Consider predictor Y / 2. Calculate its MSPE. [Hint: it may depend on μ .] 2. Consider predictor Y . Calculate its MSPE. 3. Suppose μ = 1. Calculate MSPE’s of both predictors. Which one is smaller? 4. Suppose μ = 10. Calculate MSPE’s of both predictors. Which one is smaller? Problem 2 Prove the following results about conditional means, forecasts, and forecast errors: 1. Let W be a random variable with mean μ w and variance σ 2 w . Let c be a constant and show that E[( W - c ) 2 ] = σ 2 w + ( μ w - c ) 2 2. Suppose we are interested in forecasting a random variable Y t and to do this we have random variables

Subscribe to view the full document.

Unformatted text preview: Y t-1 , Y t-2 , . . . . Let f t-1 denote some forecast for Y t that is based on Y t-1 , Y t-2 , . . . . We judge the quality of the forecast by the conditional mean squared forecast error E[( Y t-f t-1 ) 2 | Y t-1 , Y t-2 , . . . ]. Show that taking f t-1 = E[ Y t | Y t-1 , Y t-2 , . . . ] mini-mizes the conditional mean squared forecast error. 3. Consider the AR(p) model under the assumption E[ u t | Y t-1 , . . . , Y t-p ] = 0. Show that this assumption implies that Cov[ u t , u t-j ] = 0 for all j 6 = 0. 1 Problem 3 Consider the stationary AR(1) model Y t = 2 . 7 + 0 . 7 Y t-1 + u t , where u t is iid with E[ u t ] = 0 and variance Var[ u t ] = 9. 1. Compute the mean and variance of Y t . 2. Compute the first two autocovariances of Y t . 3. Compute the first two autocorrelations of Y t . 4. Suppose Y T = 102 . 3.Compute Y T +1 | T = E[ Y T +1 | Y T , Y t-1 , . . . ]. 2...
View Full Document

• Fall '18

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern


text:
# How to evaluate the sum for definite integrals using limit definition?

by rockettj   Last Updated December 06, 2018 15:20 PM - source

So I'm able to solve the first part of this question correctly with getting $$(1-2(2+\frac{2i}{n}))(\frac{2}{n})$$. But for the second part which is 'evaluating the sum', I'm not able to figure out what the question is exactly asking for and how to calculate it.

It's not the same as evaluating the limit so this is confusing me a bit.

It will help if someone's able to help me understand this 'evaluate the sum' and how to solve this part of the question.

Thank you!

Tags :

Without checking the correctness of your representation, you have $$\frac{2}{n} \left[1 - 2\left(2+\frac{2i}{n}\right)\right] = \frac{2}{n} \left[-3 - \frac{4i}{n}\right] = -\frac{6}{n} - \frac{8i}{n}$$ Therefore, $$\sum_{i=0}^{n-1} \frac{2}{n} \left[1 - 2\left(2+\frac{2i}{n}\right)\right] = -\sum_{i=0}^{n-1} \frac{6}{n}-\sum_{i=0}^{n-1} \frac{8i}{n} = -6-\frac{8}{n}\sum_{i=0}^{n-1} i$$ Can you finish this?


text:
parasys.net

Home > Error Propagation > Error Propagation Calculation Examples

Error Propagation Calculation Examples

Contents

The result is most simply expressed using summation notation, designating each measurement by Qi and its fractional error by fi. © 1996, 2004 by Donald E. This is why we could safely make approximations during the calculations of the errors. The derivative with respect to x is dv/dx = 1/t. For example, if some number A has a positive uncertainty and some other number B has a negative uncertainty, then simply adding the uncertainties of A and B together could give news

Uncertainty in measurement comes about in a variety of ways: instrument variability, different observers, sample differences, time of day, etc. But here the two numbers multiplied together are identical and therefore not inde- pendent. Simanek. ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.8/ Connection to 0.0.0.8 failed. Now consider multiplication: R = AB.

Standard Error Propagation

X = 38.2 ± 0.3 and Y = 12.1 ± 0.2. If you're measuring the height of a skyscraper, the ratio will be very low. Since the variables used to calculate this, V and T, could have different uncertainties in measurements, we use partial derivatives to give us a good number for the final absolute uncertainty. Sometimes, these terms are omitted from the formula.

Rules for exponentials may also be derived. Then our data table is: Q ± fQ 1 1 Q ± fQ 2 2 .... Using Beer's Law, ε = 0.012614 L moles-1 cm-1 Therefore, the $$\sigma_{\epsilon}$$ for this example would be 10.237% of ε, which is 0.001291. How To Calculate Error Propagation In Excel Wird geladen...

Similarly, fg will represent the fractional error in g. The error propagation methods presented in this guide are a set of general rules that will be consistently used for all levels of physics classes in this department. All rights reserved.

Since at least two of the variables have an uncertainty based on the equipment used, a propagation of error formula must be applied to measure a more exact uncertainty of the

Uncertainty components are estimated from direct repetitions of the measurement result. How To Calculate Error Propagation Physics WiedergabelisteWarteschlangeWiedergabelisteWarteschlange Alle entfernenBeenden Wird geladen... Nächstes Video Propagation of Errors - Dauer: 7:04 paulcolor 29.438 Aufrufe 7:04 Calculating Uncertainties - Dauer: 12:15 Colin Killmer 11.475 Aufrufe 12:15 Propagation of Uncertainty, Parts 1 and 2 - Dauer: When two quantities are added (or subtracted), their determinate errors add (or subtract).

Propagation Of Error Explained

Now a repeated run of the cart would be expected to give a result between 36.1 and 39.7 cm/s. http://www.itl.nist.gov/div898/handbook/mpc/section5/mpc55.htm So the result is: Quotient rule. Standard Error Propagation This ratio is very important because it relates the uncertainty to the measured value itself. Uncertainty Propagation Formula In summary, maximum indeterminate errors propagate according to the following rules: Addition and subtraction rule.

We'd have achieved the elusive "true" value! 3.11 EXERCISES (3.13) Derive an expression for the fractional and absolute error in an average of n measurements of a quantity Q when navigate to this website This result is the same whether the errors are determinate or indeterminate, since no negative terms appeared in the determinate error equation. (2) A quantity Q is calculated from the law: SOLUTION To actually use this percentage to calculate unknown uncertainties of other variables, we must first define what uncertainty is. A final comment for those who wish to use standard deviations as indeterminate error measures: Since the standard deviation is obtained from the average of squared deviations, Eq. 3-7 must be Error Propagation Equation Calculator

These instruments each have different variability in their measurements. Adding or subtracting a constant doesn't change the SE Adding (or subtracting) an exactly known numerical constant (that has no SE at all) doesn't affect the SE of a number. Später erinnern Jetzt lesen Datenschutzhinweis für YouTube, ein Google-Unternehmen Navigation überspringen DEHochladenAnmeldenSuchen Wird geladen... More about the author So if the angle is one half degree too large the sine becomes 0.008 larger, and if it were half a degree too small the sine becomes 0.008 smaller. (The change

Starting with a simple equation: $x = a \times \dfrac{b}{c} \tag{15}$ where $$x$$ is the desired results with a given standard deviation, and $$a$$, $$b$$, and $$c$$ are experimental variables, each How To Calculate Propagation Of Error In Chemistry The problem might state that there is a 5% uncertainty when measuring this radius. When the error a is small relative to A and ΔB is small relative to B, then (ΔA)(ΔB) is certainly small relative to AB.

Answer: we can calculate the time as (g = 9.81 m/s2 is assumed to be known exactly) t = - v / g = 3.8 m/s / 9.81 m/s2 = 0.387

Laboratory experiments often take the form of verifying a physical law by measuring each quantity in the law. Likewise, if x = 38 ± 2, then x - 15 = 23 ± 2. The relative indeterminate errors add. Error Propagation Example Problems Note that this fraction converges to zero with large n, suggesting that zero error would be obtained only if an infinite number of measurements were averaged!

Table 1: Arithmetic Calculations of Error Propagation Type1 Example Standard Deviation ($$\sigma_x$$) Addition or Subtraction $$x = a + b - c$$ $$\sigma_x= \sqrt{ {\sigma_a}^2+{\sigma_b}^2+{\sigma_c}^2}$$ (10) Multiplication or Division \(x = When we are only concerned with limits of error (or maximum error) we assume a "worst-case" combination of signs. First you calculate the relative SE of the ke value as SE(ke )/ke, which is 0.01644/0.1633 = 0.1007, or about 10 percent. click site You can calculate that t1/2 = 0.693/0.1633 = 4.244 hours.

Guidance on when this is acceptable practice is given below: If the measurements of a and b are independent, the associated covariance term is zero. Example: We have measured a displacement of x = 5.1+-0.4 m during a time of t = 0.4+-0.1 s. Then vo = 0 and the entire first term on the right side of the equation drops out, leaving: [3-10] 1 2 s = — g t 2 The student will, For this discussion we'll use ΔA and ΔB to represent the errors in A and B respectively.

Introduction Every measurement has an air of uncertainty about it, and not all uncertainties are equal. Does it follow from the above rules? The system returned: (22) Invalid argument The remote host or network may be down. References Skoog, D., Holler, J., Crouch, S.

For example, a body falling straight downward in the absence of frictional forces is said to obey the law: [3-9] 1 2 s = v t + — a t o Then we'll modify and extend the rules to other error measures and also to indeterminate errors. We quote the result in standard form: Q = 0.340 ± 0.006. How can you state your answer for the combined result of these measurements and their uncertainties scientifically?


text:
Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 11 Feb 2016, 00:53

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# GMAT Data Sufficiency (DS)

 Question banks Downloads My Bookmarks Reviews Important topics Go to page Previous    1  ...  9   10   11   12   13  ...  148    Next Search for:
Topics Author Replies   Views Last post

Announcements

65
 150 Hardest and easiest questions for DS   Tags:
Bunuel

3

8828

07 Dec 2015, 08:09

348
 DS Question Directory by Topic and Difficulty
bb

0

105109

07 Mar 2012, 07:58

Topics

2
 Last month the Green grocery store sold 1,750 pieces of fruit consisti
Bunuel

4

241

28 Oct 2015, 13:43

1
 If n is an integer, xy does not equal zero, and x^n = y^n, what is the
skylimit

3

242

10 Sep 2015, 10:01

 If m and n are integers and m not equal to n. Does n=0?
harish1986

1

242

14 Oct 2015, 15:03

2
 Maria wants to send a package overseas. She can either pay shipping co
Bunuel

4

242

05 Oct 2015, 05:02

2
 What is the perimeter of triangle MNO?
Bunuel

6

243

20 Jan 2016, 21:10

1
 If w, x, y, and z are all positive numbers, then what is the value of
reto

1

243

25 Aug 2015, 11:30

 What is the value of 6y + 2x^2 ?
Bunuel

5

243

01 Nov 2015, 21:46

1
 If d is an integer, is d = d^3?
hideyoshi

1

244

03 Sep 2015, 18:09

2
 Is x > 3?
Bunuel

3

245

11 Oct 2015, 06:47

 Penny does consulting work part-time and is paid on an hourly basis.
Bunuel

5

245

06 Oct 2015, 07:13

1
 For a certain concert, the price of balcony tickets was exactly half
Bunuel

4

246

27 Sep 2015, 23:33

 In year X, 12.2 percent of the Gross Domestic Product (GDP) came ...
Gmatdecoder

1

246

23 Jun 2015, 10:21

2
 If 0 < a < b, is c < 0? 1.cb < ca 2.|c| > 1
GMATinsight

2

246

12 Nov 2015, 08:35

 For integer n, if n is positive and b = 6*10^2 + 4*10^n + 5 , what is
Bunuel

3

246

04 Nov 2015, 20:16

1
 If x ≠ -y, is (x-y)/(x+y)> 1 ? (1) x > 0 (2) y < 0
NoHalfMeasures

4

246

13 Jan 2016, 20:17

1
 Does p^2 = q if p is a prime number?
Bunuel

3

246

15 Nov 2015, 00:50

1
 Jane and Tom spent a total of $160 on clothes. How much did Tom spend? Bunuel 5 246 17 Nov 2015, 09:15 3  Does there exist an integer d such that x > d > 1 & x/d is an integer? noyaljoseph 9 246 03 Feb 2016, 09:29 3  Did the sum of the prices of three shirts exceed$60?
Bunuel

2

247

26 Oct 2015, 08:55

2
 Is the range of them greater than 3?
gmat2015p

2

247

13 Dec 2015, 17:18

5
 d = 1/(2^k*5^m), m > k (m and k are positive integers) is a terminatin
Mascarfi

1

247

18 Jul 2015, 17:04

 Last year, 'Crazy Matt's Electronics' earned 90% more than did 'Sane D
reto

1

247

26 Sep 2015, 08:10

7
 If j > 1, is integer j a prime number? (1) When j is divided by 3, th

4

248

11 Jan 2016, 15:43

 If represents one of the operations +, -, and x, is k(l+m)=(kl)+(k   Tags:
nuttyaks

0

248

23 Aug 2015, 04:06

1
 If abc ≠ 0, is a > 0? (1) 3*a/b > 0 (2) b/c^2 < 0
Bunuel

3

249

16 Sep 2015, 11:38

 If z=(m + m/3)(n+2/(n^(-1)) and mn#0, what is the value of Z?
harish1986

4

249

17 Jan 2016, 21:55

3
 If x and y are distinct positive integers, what is the value of ....
Prajat

1

249

10 Sep 2015, 21:04

3
 A total of 2,000 t-shirts was divided among a soccer team, two basebal
Bunuel

3

249

05 Oct 2015, 01:25

1
 All of the employees of company Q received bonuses. If Naomi was an
Bunuel

2

250

16 Nov 2015, 23:31

2
 If r, a and b are positive integers; is r a product of two consecutive
Mascarfi

2

252

19 Jul 2015, 03:52

3
 If b is an integer, is a|a| < 2^b
harish1986

3

252

14 Oct 2015, 08:48

 If a and b are positive integers, is a/b = 5/8 ?
Bunuel

4

252

31 Oct 2015, 23:35

1
 All members of the Travelers Club have been to France or India or both
Bunuel

4

252

22 Dec 2015, 11:28

3
 Tyler sells two types of milkshakes: chocolate and Oreo. Chocolate mil
Bunuel

4

252

22 Sep 2015, 19:33

1
 What was the final score in a game between Team X and Team Y?
Bunuel

2

252

05 Oct 2015, 02:21

3
 A candy manufacturer decided to decrease the weight of each candy bar,
Bunuel

8

252

31 Jan 2016, 21:07

 What is the remainder when b is divided by a, if a and b are consecuti
apple08

3

253

07 Oct 2015, 18:10

6
 There are 5 women and 4 men including 2 couples,
jyotibrata

4

253

31 Dec 2015, 06:28

 An insurance company sells only one type of health and one type of
Bunuel

5

254

03 Nov 2015, 21:28

2
 In the fall semester, Mr. Rodriguez’s, Ms. Vangala’s, and Mr. Hollande
Bunuel

1

254

01 Sep 2015, 03:25

2
 If x, y, b and t are all positive integers, x = y/5 + 2
hideyoshi

2

254

03 Sep 2015, 22:14

2
 Does line L pass through Quadrant II?
GMATinsight

2

254

23 Jul 2015, 06:25

3
 Is AB=10 the shortest side of triangle ABC? (1) The perimeter of ΔABC
reto

3

255

12 Oct 2015, 14:06

1
 If t is the sum of three consecutive positive integers, is t a
Bunuel

5

255

22 Dec 2015, 23:59

1
 In a certain store, the price of five pens is equal to the price of a
Bunuel

3

255

17 Sep 2015, 08:31

3
 Mr. X paid for a painting using currency notes of denominations 1$, 2$
VenoMfTw

2

255

05 Aug 2015, 11:17

3
 If a > b, how much greater than b is a ?
AbdurRakib

2

256

28 Aug 2015, 07:59

3
 If a + b = 20, then what is the value of c − d?
Bunuel

7

257

15 Dec 2015, 09:09

1
 If a, b, and c are distinct positive integers where a < b < c and
Bunuel

3

257

11 Oct 2015, 06:51

 If j and k are integers, is j + k an odd integer?
Bunuel

3

258

12 Jan 2016, 15:17

 Question banks Downloads My Bookmarks Reviews Important topics Go to page Previous    1  ...  9   10   11   12   13  ...  148    Next Search for:
 Who is online In total there is 1 user online :: 0 registered, 0 hidden and 1 guest (based on users active over the past 15 minutes) Users browsing this forum: No registered users and 1 guest Statistics Total posts 1447752 | Total topics 176225 | Active members 432246 | Our newest member edocicch

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.


text:
# Three Two One... Wait, Add Two Thousand

$\LARGE 2003^{2002^{2001}}$

What are the last three digits of the above number?

×


text:
Chapter 7

• Special statistical tests are needed when more than two groups are studied or when a group is measured on several variables.
• Analysis of variance, or ANOVA, is a statistical method that divides the variance in an observation into the variance among groups and the rest of the variance, called the within-group or error variance.
• The F test used to compare two variances in Chapter 6 is used to compare the variance among groups to the error.
• An example of the way ANOVA is calculated from the definitional formulas is helpful in understanding the logic behind the test.
• The terms used in ANOVA are important, but the details of the computations are given for illustration only, and computer programs are used for all ANOVA procedures.
• One-way ANOVA is the appropriate method when more than two groups are studied on one variable.
• As with the t test, certain assumptions must be made to use ANOVA, and equal variances is one of the most important.
• Making many comparisons among groups increases the chances of a type I error, that a difference is concluded when there is none.
• Investigators can decide ahead of time what specific comparisons they want to make.
• The Bonferroni procedure is a common way to compensate for making many comparisons among groups; it works by reducing the size of α for each comparison, essentially increasing the difference needed to be significant.
• Some multiple comparison methods, called post hoc, are done only if the ANOVA results are statistically significant.
• Tukey’s test is one of the most highly recommended post hoc tests for comparing mean differences.
• The Scheffé post hoc test is the most conservative (requiring a larger difference to be significant), but it is also the most versatile.
• The Newman–Kuels post hoc test is used frequently in basic science research.
• Dunnett’s procedure is the test of choice if the only comparisons being made are between the mean in a control group and the means in other groups.
• Two-way ANOVA analyzes two factors instead of just one, as in one-way ANOVA. It also permits the analysis of the interaction between two factors.
• ANOVA designs involving more that two factors are possible, generally called factorial designs.
• Confounding variables can be accommodated by the ANOVA randomized block design.
• Repeated-measures ANOVA is a common procedure in medical research; it is analogous to the paired t test with more than two groups and is also called the split-plot design.
• Nonparametric ANOVA methods include Kruskal–Wallis for one-way designs and Friedman two-way ANOVA for repeated measures. These methods are analogous to the Wilcoxon procedures and are used when the assumptions for ANOVA are not met.
• The chi-square test can be used to compare more than two proportions and to determine if there is an association between two factors, each of which can have two or more levels. It is a simple extension of the chi-square test we discussed in Chapter 6.
• As with research questions involving one or two groups, power analysis is ...

Sign in to your MyAccess profile while you are actively authenticated on this site via your institution (you will be able to verify this by looking at the top right corner of the screen - if you see your institution's name, you are authenticated). Once logged in to your MyAccess profile, you will be able to access your institution's subscription for 90 days from any location. You must be logged in while authenticated at least once every 90 days to maintain this remote access.

Ok

## Subscription Options

### AccessPhysiotherapy Full Site: One-Year Subscription

Connect to the full suite of AccessPhysiotherapy content and resources including interactive NPTE review, more than 500 videos, Anatomy & Physiology Revealed, 20+ leading textbooks, and more.