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$$\scriptsize{\overset{{\large{\textbf{Mark Distribution in Previous GATE}}}}{\begin{array}{|c|c|c|c|c|c|c|c|}\hline \textbf{Year}&\textbf{2019}&\textbf{2018}&\textbf{2017-1}&\textbf{2017-2}&\textbf{2016-1}&\textbf{2016-2}&\textbf{Minimum}&\textbf{Average}&\textbf{Maximum} \\\hline\textbf{1 Mark Count}&2&0&2&2&3&3&0&2&3 \\\hline\textbf{2 Marks Count}&2&4&2&3&2&3&2&2.7&4 \\\hline\textbf{Total Marks}&6&8&6&8&7&9&\bf{6}&\bf{7.3}&\bf{9}\\\hline \end{array}}}$$
# Recent questions in Algorithms
1
Let $G$ be a graph with $n$ vertices and $m$ edges.What is the tightest upper bound on the running time of Depth First Search of $G$, when $G$ is represented using adjacency matrix? $O(n)$ $O(m+n)$ $O(n^2)$ $O(mn)$
2
Suppose $T(n)=2T(n/2)+n$, $T(0)=T(1)=1$ which one of the following is false? $T(n)=O(n^2)$ $T(n)=\Theta(n\log n)$ $T(n)=\Omega(n^2)$ $T(n)=O(n\log n)$
1 vote
3
Let $A$ be an array of $31$ numbers consisting of a sequence of $0$’s followed by a sequence of $1$’s. The problem is to find the smallest index $i$ such that $A[i]$ is $1$ by probing the minimum number of locations in $A$. The worst case number of probes performed by an optimal algorithm is $2$ $4$ $3$ $5$
4
Which algorithm has same average, worst case and best case time ? Binary search Maximum of n number Quick sort Fibonacci search
5
Binary search tree is an example of : Divide and conquer technique Greedy algorithm Back tracking Dynamic Programming
6
How much extra space is used by heapsort ? $O (1)$ $O (\log n)$ $O (n)$ $O (n^2)$
7
The asymptotic upper bound solution of the recurrence relation given by $T(n)=2T \left ( \frac{n}{2} \right)+\frac{n}{\lg n}$ is: $O(n^{2})$ $O(n \lg n)$ $O(n \lg \lg n)$ $O(\lg \lg n)$
8
Any decision tree that sorts $n$ elements has height $\Omega (\lg n)$ $\Omega ( n)$ $\Omega (n \lg n)$ $\Omega ( n^{2})$
9
Red-black trees are one of many search tree schemes that are “balanced” in order to guarantee that basic dynamic–set operations take____ time in the worst case. $O(1)$ $O(\lg n)$ $O( n)$ $O(n \lg n)$
10
The minimum number of scalar multiplication required, for parenthesization of a matrix-chain product whose sequence of dimensions for four matrices is $< 5,10,3,12,5>$ is $630$ $580$ $480$ $405$
11
Dijkstra’s algorithm is based on Divide and conquer paradigm Dynamic programming Greedy approach Backtracking paradigm
12
Match the following with respect to algorithm paradigms: ...
13
Which of the following is true for computation time in insertion, deletion and finding maximum and minimum element in a sorted array ? Insertion - $0(1)$, Deletion - $0(1)$, Maximum - $0(1)$, Minimum - $0(1)$ Insertion - $0(1)$, Deletion - $0(1)$, Maximum - $0(n)$, Minimum - $0(n)$ ... , Maximum - $0(1)$, Minimum - $0(1)$ Insertion - $0(n)$, Deletion - $0(n)$, Maximum - $0(n)$, Minimum - $0(n)$
14
Which of the following statements is false? Optimal binary search tree construction can be performed efficiently using dynamic programming. Breadth-first search cannot be used to find connected components of a graph. Given the prefix and postfix walks of a binary tree, the tree cannot be reconstructed uniquely. Depth-first-search can be used to find the connected components of a graph. a b c d
15
For parameters $a$ and $b$, both of which are $\omega(1)$, $T(n) = T(n^{1/a})+1$, and $T(b)=1$. Then $T(n)$ is $\Theta (\log_a \log _b n)$ $\Theta (\log_{ab} n$) $\Theta (\log_{b} \log_{a} \: n$) $\Theta (\log_{2} \log_{2} n$)
16
What is the worst case time complexity of inserting $n^{2}$ elements into an AVL-tree with $n$ elements initially? $\Theta (n^{4})$ $\Theta (n^{2})$ $\Theta (n^{2}\log n)$ $\Theta (n^{3})$
17
Consider a double hashing scheme in which the primary hash function is $h_1(k)= k \text{ mod } 23$, and the secondary hash function is $h_2(k)=1+(k \text{ mod } 19)$. Assume that the table size is $23$. Then the address returned by probe $1$ in the probe sequence (assume that the probe sequence begins at probe $0$) for key value $k=90$ is_____________.
Let $G = (V, G)$ be a weighted undirected graph and let $T$ be a Minimum Spanning Tree (MST) of $G$ maintained using adjacency lists. Suppose a new weighed edge $(u, v) \in V \times V$ is added to $G$. The worst case time complexity of determining if $T$ is still an MST of the resultant graph ... $\Theta (\mid E \mid \mid V \mid) \\$ $\Theta(E \mid \log \mid V \mid) \\$ $\Theta( \mid V \mid)$
Let $G = (V,E)$ be a directed, weighted graph with weight function $w: E \rightarrow \mathbb{R}$. For some function $f: V \rightarrow \mathbb{R}$, for each edge$(u,v)\in E$, define ${w}'(u,v)$ as $w(u,v)+f(u)-f(v)$. Which one of the options completes the ... distance from $s$ to $u$ in the graph obtained by adding a new vertex $s$ to $G$ and edges of zero weight from $s$ to every vertex of $G$
Consider the array representation of a binary min-heap containing $1023$ elements. The minimum number of comparisons required to find the maximum in the heap is ___________.
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# Resampling Algorithms¶
## Introduction¶
In order to restore numerical stability to the sequential Monte Carlo algorithm as the effective sample size is reduced, resampling is used to adaptively move particles so as to better represent the posterior distribution. QInfer allows for such algorithms to be specified in a modular way.
## LiuWestResampler - Liu and West (2000) resampling algorithm¶
### Class Reference¶
class qinfer.resamplers.LiuWestResampler(a=0.98, h=None, maxiter=1000, debug=False, postselect=True, zero_cov_comp=1e-10, kernel=<built-in method randn of mtrand.RandomState object>)[source]
Bases: object
Creates a resampler instance that applies the algorithm of [LW01] to redistribute the particles.
Parameters: a (float) – Value of the parameter $$a$$ of the [LW01] algorithm to use in resampling. h (float) – Value of the parameter $$h$$ to use, or None to use that corresponding to $$a$$. maxiter (int) – Maximum number of times to attempt to resample within the space of valid models before giving up. debug (bool) – Because the resampler can generate large amounts of debug information, nothing is output to the logger, even at DEBUG level, unless this flag is True. postselect (bool) – If True, ensures that models are valid by postselecting. zero_cov_comp (float) – Amount of covariance to be added to every parameter during resampling in the case that the estimated covariance has zero norm. kernel (callable) – Callable function kernel(*shape) that returns samples from a resampling distribution with mean 0 and variance 1.
Warning
The [LW01] algorithm preserves the first two moments of the distribution (in expectation over the random choices made by the resampler) if and only if $$a^2 + h^2 = 1$$, as is set by the h=None keyword argument.
a
## ClusteringResampler - Cluster-based recursive resampler¶
class qinfer.resamplers.ClusteringResampler(eps=0.5, secondary_resampler=None, min_particles=5, metric='euclidean', weighted=False, w_pow=0.5, quiet=True)[source]
Bases: object
Creates a resampler that breaks the particles into clusters, then applies a secondary resampling algorithm to each cluster independently.
Parameters: secondary_resampler – Resampling algorithm to be applied to each cluster. If None, defaults to LiuWestResampler().
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# Multi-Fluids with Collisions (collisionalMultiFluid.pre)¶
Keywords:
Multi-Fluid Collisions
## Problem description¶
This problem shows collisions between three (separate) fluid species in a simple shock problem and allows one to compare it with the single-fluid solution. In the highly collisional regime the multi-fluid problem converges to the single fluid case.
This simulation can be performed with a USimHEDP license.
## Creating the run space¶
The Multi-Fluids with Collisions example is accessed from within USimComposer by the following actions:
• Select the New from Template menu item in the File menu.
• In the resulting New from Template dialog, expand USimHEDP: High Energy Density Plasmas.
• Select Multi-Fluids with Collisions and press the Choose button.
• In the Choose a name for the new runspace dialog, press the Save button to create a copy of this example in your run area.
• Press the Save And Process Setup button in the upper right corner of the Editor pane.
The basic example variables are editable in the Editor pane of the Setup window. After any change is made, the Save and Process Setup button must be pressed again before a new run commences.
## Input file features¶
The following parameters can be varied to look at the effects of collisionality on the shock solution:
• NUMDUMPS - Number of data dumps during the simulation
• XUPPER - Domain size
• PRESSURE - Reference pressure of the gas
• DENSITY - Reference density of the gas
• GAMMA - Gas constant
• PRL - the total pressure on the left half of the domain.
• RHOL - the total density on the left half of the domain.
• PRR - the total pressure on the right half of the domain.
• RHOR - the total density on the right half of the domain.
• FRAC1L - the fraction of gas 1 on the left half initially.
• FRAC2L - the fraction of gas 2 on the left half initially.
• FRAC3L - the fraction of gas 3 on the left half initially.
• FRAC1R - the fraction of gas 1 on the right half initially.
• FRAC2R - the fraction of gas 2 on the right half initially.
• FRAC3R - the fraction of gas 3 on the right half initially.
• MI - Reference mass of ion
• DI - Reference diameter of ion
• MI1 - Mass of ion1
• MI2 - Mass of ion2
• MI3 - Mass of ion3
• DI1 - Diameter of ion1
• DI2 - Diameter of ion2
• DI3 - Diameter of ion3
## Running the simulation¶
After performing the above actions, continue as follows:
• Proceed to the Run window as instructed by pressing the Run icon in the workflow panel.
• To run the simulation, click on the Run button in the upper right corner of the Logs and Output Files pane.
You will also see the engine log output in the Logs and Output Files pane. The run has completed when you see the output, “Engine completed successfully.”
## Visualizing the results¶
After performing the above actions, continue as follows:
• Proceed to the Visualize window as instructed by pressing the Visualize icon in the workflow panel.
• Press the Open button to begin visualizing. The visualization opens with four panels consisting of 1D line plots.
• Drag the slider at the bottom of the Visualization Results pane to see results at the end of the simulation, as shown in Fig. Fig. #collisionalmultifluidvizwin.
The following values can be visualized
• N1,N2,N3 the number densities for species 1, 2 and 3
• T1,T2,T3 the temperatures for species 1, 2 and 3
• V1_0,V1_1,V1_2 the velocity components for species 1
• V2_0,V2_1,V2_2 the velocity components for species 2
• V3_0,V3_1,V3_2 the velocity components for species 3
• collisionMatrix_0 through collisionMatrix_8 the collisional cross frequencies between species
• q1_0,q1_1,q1_2,q1_3,q1_4 the mass density, momentum density and energy of the first species
• q2_0,q2_1,q2_2,q2_3,q2_4 the mass density, momentum density and energy of the second species
• q3_0,q3_1,q3_2,q3_3,q3_4 the mass density, momentum density and energy of the third species
• qTotal_0,qTotal_1,qTotal_2,qTotal_3,qTotal_4 the mass density, momentum density and energy of the sum of the 3 species
Figure 84: Visualization of the densities of each species and the temperature of the first species for the Multi-Fluids with Collisions example
## Further experiments¶
• Increase RHOL and RHOR by a factor of 10 and the fluids will become much more collisional, producing the standard sod shock result.
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# Can I introduce a special environment which will reappear at the end?
I am writing a book-length document, and occasionally I have some text that I would like to automatically repeat at the end. For example, something like
\begin{important_thing}
\medskip
{\bf Lorem Ipsum Principle.} Blah blah blah.
\end{important_thing}
This text should display in the document as if the \begin and \end lines weren't there. In addition I want all the "important things" to appear again at the end of the document, in the order that they appeared.
I feel like it should be possible to define some macro to handle this automatically for me, but I don't know how to do it. (I could define a macro consisting of my Lorem Ipsum Principle, and then refer to it twice -- this comes close to what I'm asking for, although if I reorder things in the document I would then have to manually reorder them at the end.) Am I asking too much?
Minimal changes from my answer at Collect the input of all \TODO commands used in the document at the end. I called the stuff "cross-refs" only because that was a tag in your question. You can call it whatever you want.
MACRO VERSION See later for environment version.
\documentclass{article}
\usepackage{ifthen}
\newcounter{crossrefindex}
\setcounter{crossrefindex}{0}
\newcommand\CROSSREF[1]{%
\expandafter\def\csname crossref\roman{crossrefindex}\endcsname{#1}%
#1%
}
\newcounter{index}
\newcommand\showCROSSREFs{%
\vspace{5ex}%
\rule{10ex}{.5ex}CROSS-REF LIST\rule{10ex}{.5ex}\\%
\setcounter{index}{0}%
\whiledo{\value{index} < \value{crossrefindex}}{%
\arabic{index}): \csname crossref\roman{index}\endcsname\\%
}%
}
\begin{document}
I start hear \CROSSREF{Fix this bug} and do some work.
Then I do thiis \CROSSREF{Get spelling fixed, too} which I have to get back
to
and then I am done
\showCROSSREFs
\end{document}
ENVIRONMENT VERSION
Unfortunately, this eats spaces after the environment, and so one must add an explicit {} after the environment if one wants to preserve it.
\documentclass{article}
\usepackage{ifthen,environ,etoolbox}
\newcounter{crossrefindex}
\setcounter{crossrefindex}{0}
\makeatletter
\NewEnviron{CROSSREF}{%
\expandafter\xdef\csname crossref\romannumeral\value{crossrefindex}\endcsname{%
\expandonce{\BODY}}%
\BODY%
}
\makeatother
\newcounter{index}
\newcommand\showCROSSREFs{%
\vspace{5ex}%
\rule{10ex}{.5ex}CROSS-REF LIST\rule{10ex}{.5ex}\\%
\setcounter{index}{0}%
\whiledo{\value{index} < \value{crossrefindex}}{%
\arabic{index}): \csname crossref\romannumeral\value{index}\endcsname\\%
}%
}
\begin{document}
I start hear \begin{CROSSREF}Fix this bug\end{CROSSREF}{} and do some work.
Then I do thiis \begin{CROSSREF}Get spelling fixed, too\end{CROSSREF}{} which I have to get back
to
and then I am done
\showCROSSREFs
\end{document}
• @Anonymous EDITED to (correctly) provide an environment version. – Steven B. Segletes Sep 15 '16 at 17:34
Several variations are possible, but here's a start.
\documentclass{article}
\usepackage{environ}
\usepackage{lipsum} % just for the example
\newcounter{repeatatend}
\makeatletter
\def\repeat@at@end{} % initialize
\NewEnviron{important}{%
\par
\medskip
\BODY\par
\medskip
\refstepcounter{repeatatend}%
\label{repeatatend@\romannumeral\value{repeatatend}}%
\xdef\repeat@at@end{%
\unexpanded\expandafter{\repeat@at@end}%
\unexpanded{\noindent(p.~\pageref}{repeatatend@\romannumeral\value{repeatatend}}) %
\unexpanded\expandafter{\BODY\par}%
}%
}
\AtEndDocument{\section*{Important}\repeat@at@end}
\begin{document}
Here I state something very important
\begin{important}
\textbf{Lorem Ipsum Principle.} Blah blah blah.
\end{important}
\lipsum % to get to another page
Then another important thing
\begin{important}
Ducks can fly.
\end{important}
which is something to take into account.
And then I am done.
\end{document}
Here's the relevant part at page 1
Here's what happens at the end
A variation of the answer given by the master @egreg, with support for verbatim :) courtesy of the package scontents.
\documentclass{article}
\usepackage[print-env=true,store-env=important]{scontents}
\setlength{\parindent}{0pt}
\pagestyle{empty}
% Uncomment this to automatic repeat ...thanks @egreg :)
%\AtEndDocument{\section*{Important}
%\begin{enumerate}
%\foreachsc[before={\item }]{important}
%\end{enumerate}}
\begin{document}
\section{Important things}
Here I state something very important
\begin{scontents}
\textbf{Lorem Ipsum Principle.} Blah blah blah.
\end{scontents}
Here I state something very important
\begin{scontents}
\textbf{Lorem Ipsum Principle.} Tick Tack Tack.
\end{scontents}
Then another important thing
\begin{scontents}
\verb*|Ducks can fly?|.
\end{scontents}
\section{Repeat manually}
\getstored[1]{important}\\[10pt]
\getstored[2]{important}\\[10pt]
\getstored[3]{important}
\section{Repeat all}
\begin{enumerate}
\foreachsc[before={\item }]{important}
\end{enumerate}
\end{document}
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# How precise to be when describing a Turing machine?
I'm kind of new to the theory of computation and I was working on this problem:
We say that a Turing machine $$M$$ uses $$k$$ squares of tape for an input string $$w$$ if and only if there exists a configuration $$(q, u\underline{a}v)$$ of $$M$$, such that starting with input $$w$$, $$M$$ yields $$(q, u\underline{a}v)$$ and $$|uav| \geq k$$.
Now show that the following problem is solveable: Given a Turing machine $$M$$, an input string $$w$$ and a number $$k$$ does $$M$$ use $$k$$ squares of tape with input $$w$$?
So the solution I came up with is this:
If it is solvable there has to be a Turing machine $$M^*$$ that decides $$U = \{\ll M\gg\ll w \gg \ll k\gg\ :M\ \text{uses}\ k\ \text{squares of tape with input}\ w\}$$
And I describe such a machine:
$$M^*$$ writes in it's tape $$w$$ and starts operating as $$M$$ would. After each step of $$M$$, $$M^*$$ counts the number consecutive squares of tape from the start of the tape until it finds an empty square, and if that number is $$\geq k$$ it goes into an accept state and halts. If not, it continues its operation as M would. If $$M$$ halts without having used at least $$k$$ steps, $$M^*$$ goes into a reject state. So $$M^*$$ decides $$U$$, and thus the problem is solveable.
So my questions are these:
Is my solution correct?, and
Is my description of $$M^*$$ precise, or say, good enough?
You are asking two different questions: whether your solution is correct, and whether the level of description of your Turing machine is appropriate. I will answer them separately.
Correctness Your Turing machine $$M^*$$ is not guaranteed to halt. In other words, your solution is incorrect. The missing observation is that if a Turing machine uses only space $$k$$, then there is an upper bound on the number of steps that it can run without getting into an infinite loop.
Level of description Here you should be asking two questions:
1. What does it mean to describe a Turing machine?
2. How do I prove that a certain language is decidable?
The level of description of Turing machines is a touchy subject. Usually they are described quite informally. The level of description should be such that in principle you could convert the informal description into a formal Turing machine. In practice, I suggest looking at some examples to get the general feel.
As to how to prove that a language is decidable, there is no need to describe a Turing machine. Decidability can be proved using any model of computation. In particular, you can use the "English" model of computation, in which you describe how to decide the problem in the English language (or in any other language). For example, you could say "Simulate $$M$$ on $$w$$, and if it ever uses more than $$k$$ space, immediately reject; otherwise, once $$M$$ halts, accept". (This is just your incorrect solution.)
• Thanks for the answer! I get what you say for the level of description question, but I don't quite understand why my solution to the problem is incorrect. Could you maybe explain it a bit differently? – Da Mike May 5 at 16:58
• Your machine won’t halt if $M$ used the correct amount of space but never halts. – Yuval Filmus May 6 at 0:44
• So basically you say that if $M$ never uses $k$ spaces and runs forever my machine $M^*$ will never halt, right? Then I guess if $M$ is not recurisve, then the problem is not solvable. But the exercise I'm working on says to prove that it is solvable. So maybe they forgot to add that $M$ is recursive in the description of the exercise? – Da Mike May 6 at 8:23
• The problem is perfectly solvable, just not with your solution. – Yuval Filmus May 6 at 11:08
• That's the entire point of the question. I gave one possible option in the answer. Another possible option is to store all configurations of $M$, and to stop if a configuration repeats. Eventually either $M$ will use more than $k$ space, or a configuration will repeat. – Yuval Filmus May 6 at 13:46
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# zbMATH — the first resource for mathematics
## Found 3 Documents (Results 1–3)
100
MathJax
Chuong, N.M. (ed.) et al., Harmonic, wavelet and $$p$$-adic analysis. Based on the summer school, Quy Nhon, Vietnam, June 10–15, 2005. Hackensack, NJ: World Scientific (ISBN 978-981-270-549-5/hbk). 291-309 (2007).
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How to solve the linearized Navier-Stokes equations in L^P? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:08:22Z http://mathoverflow.net/feeds/question/58227 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58227/how-to-solve-the-linearized-navier-stokes-equations-in-lp How to solve the linearized Navier-Stokes equations in L^P? jack 2011-03-12T03:33:17Z 2012-06-06T00:37:00Z <p>Let $\Omega\subset \mathbb{R}^3$ be an open set with smooth boundary $\partial \Omega$. Consider the following linearized Navier-Stokes equations in $Q_T=\Omega\times (0,T)$ for an arbitrarily fixed $T\in (0,\infty)$, $$u_t-\Delta u+a(x,t)u+b\cdot \nabla u+\nabla p=f(x,t),\text{div } u=0$$ with the initial and boundary conditions $u(x,0)=0, \left.u(x,t)\right|_{\partial \Omega\times (0,T)}=0$. Here $u(x,t)=(u^1(x,t),u^2(x,t),u^3(x,t))$ and $p(x,t)$ denote the unknown velocity and pressure respectively, $a(x,t)$ and $b(x,t)$ denote the given coefficients.</p> <p>Question: Suppose that $$a\in L^r(0,T; L^s(\Omega)), b\in L^{r_1}(0,T; L^{s_1}(\Omega)),$$ where $2/r+3/s<2$, $2/r_1+3/s_1<1$, and $f(x,t)\in C_0^\infty(\Omega\times (0,T))$, can we solve the above equations in arbitrary $L^p$? Can we get the estimates such as $$\|u_t\|_{L^p(Q_T)}+\|D^2 u\|_{L^p(Q_T)}+\|u\|_{L^p(Q_T)}\leq \|f\|_{L^p(Q_T)}?$$</p> <p>Solonnikov dealed with this problem in his paper "Estimates for solution of nonstationary Navier-Stokes equations" (http://www.springerlink.com/index/N8374858XNT22P11.pdf). However, I can not verify his proof (Page 487 to Page 489).</p> <p>Who can help me? Any comment will be deeply appreciated.</p> http://mathoverflow.net/questions/58227/how-to-solve-the-linearized-navier-stokes-equations-in-lp/98924#98924 Answer by timur for How to solve the linearized Navier-Stokes equations in L^P? timur 2012-06-06T00:37:00Z 2012-06-06T00:37:00Z <p>Have a look at the following review article and the relevant references therein:</p> <blockquote> <p>Yoshikazu Giga, Weak and strong solutions of the Navier-Stokes initial value problem, <em>Publ. RIMS. Kyoto Univ.</em>, 19:887-910, 1983.</p> </blockquote>
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# How to do State-Based Animation with ECS
I'm making a simple platformer game and was suggested to use ECS from a friend. It looked pretty interested and at least worth a try, so I did. I got ECS working and started work on the rendering system. But I have no idea how to do it, at least correctly, given that each enemy type will have multiple animation sequences that it'll switch to depending on what its doing. A zombie might have a walk, idle, and attack sequences, while a bat would only have the fly animation. This question also extends to AI systems, but rendering is the priority right now.
The way I see it there are 2 possible ways, but neither seem like the right way.
1. create a different component and system for every type, like ZombieSpriteComponent, ZombieSpriteRenderSystem, BatSpriteComponent, BatSpriteRenderSystem, etc.
2. shove all the states from the individual types into a giant component and system which switches depending on whether the given component has the state. hasWalkState, hasAttackState, hasFlyState, etc.
I'm using C++, SDL, and a custom ECS I wrote based on this, in case it helps.
I'd recommend taking a look at how this is handled in other component-based engines (not necessarily full ECS). That will give you some examples of how to attack this problem with reusable components, in a way that you know has successfully shipped other games.
Taking Unity for example, they separate out a reusable Animator component (/system) that just generically plays animation state machines provided to it as data.
The state machine (called an Animator Control Graph) is a collection of Animation Clips (timelines of keyframes & interpolation curves) which serve as nodes/states in the graph, and transitions which describe when to switch/blend between two clips, basically directed edges between nodes. It can also contain a list of properties to use in conditions that enable/disable various transitions or modulate the playback and blending.
An instance of this graph will have variables to keep track of the current state / transition, the playhead position in any playing clips, and something like a map containing the current values of its properties.
This helps decouple the visual/animated representation of a state from the player control / AI behaviour that triggers those states.
A movement component might get a reference to its Animator and send it data like:
• SetFloat("horizontalSpeed", velocity.x)
• SetBool("isGrounded", groundRaycast.hit)
...and the decision about how to translate those properties to animations lives completely in the data of the graph being navigated by a general-purpose animation system, rather than needing one-off code.
I could use the same movement and animation components/systems and their underlying code for a character that walks and a character that slithers, just by putting walking keyframe sequences into one graph and slithering keyframe sequences into the other, but using the same vocabulary of property keys to drive both state machines.
This pattern of "put the differences in data" repeats for the AI behaviour side of the equation too. I could make a movement component take parameters that describe its movement speed and acceleration to get many different movement behaviours out of one set of code. For their AI brains, I could use a reusable BehaviourTree component, and feed it different trees of behaviour states as data.
But to get the difference between flying enemies like bats and ground-based enemies like zombies, you'd probably want different code. To avoid creating new components/systems for every single object type, try to focus on composition: build up a complex behaviour out of reusable modular pieces.
Have a Ground Movement component and a Flying Movement component that can both expose a similar API to the brain logic in the behaviour component(s), and both communicate with the animation component. That way you can mix and match those components freely to get new behaviour combinations out of the same code (like a Vampire enemy that has both a FlyingMovement and GroundMovement, with only one enabled at a time as it switches between bat and humanoid forms...)
Overall, whenever you can, try to rephrase your problems from "is a" to "has a". It's not that a bat enemy "is a" bat. It "has a" flying animation, and a flying movement, etc. Those "has a" relationships clarify how you can represent the particular traits of a game entity with swappable data and reusable building blocks, rather than one-off code.
• I get the whole "put the differences in data" thing, the issue is I'm not sure how to set up the state machine side of it in ECS. Do I just have the attack node on the ai behavior tree tell the AnimationSystem that it's attacking and to switch to the attack animation? (I just decided to use behavior trees >15 minutes ago) Oct 17 at 14:16
• You tell it "set parameter 'isAttacking' to 'true'" (data). The next time the state machine evaluates its transitions from the current state (also data), it will find that the transition with the condition (also data) isAttacking == true is satisfied, and it will execute that transition. Oct 17 at 14:19
• So after the transition, it'll set a value in the SpriteComponent to let the SpriteRenderSystem to use the attack animation/sprite? Oct 17 at 14:41
• The animation clip (more data) will consist of a set of keyframes that each specify a property path to modify when the playhead reaches that frame (like This.SpriteRenderer.CurrentSprite) and a value to set it to (ZombieAttackFrame0) Oct 17 at 14:43
• The more generic you want your animations, the more complex your systems will get. What I've described is how it's done in Unity, which is made to be very generic - for animation 3D characters with skeletal rigs, or material properties, etc. If animations in your game only ever need to set a frame index into a sprite sequence, then you can hard-code that path as something like a direct call to spriteRenderer.SetSpriteIndex() rather than navigating arbitrary property paths. Oct 17 at 15:04
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# zbMATH — the first resource for mathematics
Almost smooth algebras. (English) Zbl 0777.16003
The “almost smooth algebras” (which generalize the formally smooth algebras of Grothendieck) are introduced. An $$A$$-algebra $$B$$ is almost smooth if for every singular $$A$$-extension of $$B$$ by a $$B$$-module $$M$$ the second fundamental exact sequence of $$B$$-modules $0\to M\to\Omega_{E\setminus A} \otimes_ E B\to\Omega_{B\setminus A}\to 0$ is short exact. Several characterizations of almost smooth algebras are given. It is proved that an $$A$$-algebra $$B$$ is almost smooth iff for any $$A$$-algebra $$C$$, any ideal $$I$$ of $$C$$ satisfying $$I^ 2=0$$, and any $$A$$-algebra homomorphism $$g:B\to C/I$$ such that $$I$$ is an injective $$B$$- module via $$g$$, there exists a lifting $$f:B\to C$$ of $$g$$. In the end of the paper an example of an almost smooth algebra which is not formally smooth is given.
##### MSC:
16E30 Homological functors on modules (Tor, Ext, etc.) in associative algebras 16S80 Deformations of associative rings 16D50 Injective modules, self-injective associative rings 16S70 Extensions of associative rings by ideals
Full Text:
##### References:
[1] 1 A. Grothendieck , Eléments de géometrie algébrique IV, Première Partie , Publ. Math. I.H.E.S. , 1964 . Numdam · Zbl 0136.15901 [2] 2 P.J. Hilton & U. Stammbach , A course in homological algebra , Springer , 1971 . MR 346025 | Zbl 0863.18001 · Zbl 0863.18001 [3] 3 S. Lichtenbaum & S. Schlessinger , The cotangent complex of a morphism , Trans. Amer. Math. Soc. 128 ( 1967 ), 41 - 70 . MR 209339 | Zbl 0156.27201 · Zbl 0156.27201 [4] 4 T. Matsuoka , On almost complete intersections , Manuscr. Math. 21 ( 1977 ), 329 - 340 . MR 450258 | Zbl 0356.13009 · Zbl 0356.13009 [5] 5 H. Matsumura , Commutative algebra , Benjamin , New York , 1970 . MR 266911 | Zbl 0211.06501 · Zbl 0211.06501 [6] 6 P. Seibt , Infinitesimal extensions of commutative algebras , J. Pure Appl. Algebra 16 ( 1980 ), 197 - 206 . MR 556160 | Zbl 0431.13014 · Zbl 0431.13014
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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• ### Investigating Galactic supernova remnant candidates with LOFAR(1706.08826)
May 9, 2018 astro-ph.HE
We investigate six supernova remnant (SNR) candidates --- G51.21+0.11, G52.37-0.70, G53.07+0.49, G53.41+0.03, G53.84-0.75, and the possible shell around G54.1-0.3 --- in the Galactic Plane using newly acquired LOw-Frequency ARray (LOFAR) High-Band Antenna (HBA) observations, as well as archival Westerbork Synthesis Radio Telescope (WSRT) and Very Large Array Galactic Plane Survey (VGPS) mosaics. We find that G52.37-0.70, G53.84-0.75, and the possible shell around pulsar wind nebula G54.1+0.3 are unlikely to be SNRs, while G53.07+0.49 remains a candidate SNR. G51.21+0.11 has a spectral index of $\alpha=-0.7\pm0.21$, but lacks X-ray observations and as such requires further investigation to confirm its nature. We confirm one candidate, G53.41+0.03, as a new SNR because it has a shell-like morphology, a radio spectral index of $\alpha=-0.6\pm0.2$ and it has the X-ray spectral characteristics of a 1000-8000 year old SNR. The X-ray analysis was performed using archival XMM-Newton observations, which show that G53.41+0.03 has strong emission lines and is best characterized by a non-equilibrium ionization model, consistent with an SNR interpretation. Deep Arecibo radio telescope searches for a pulsar associated with G53.41+0.03 resulted in no detection, but place stringent upper limits on the flux density of such a source if it is beamed towards Earth.
• ### Asymmetric Type-Ia supernova origin of W49B as revealed from spatially resolved X-ray spectroscopic study(1707.05107)
April 10, 2018 astro-ph.HE
The origin of the asymmetric supernova remnant (SNR) W49B has been a matter of debate: is it produced by a rare jet-driven core-collapse supernova, or by a normal supernova that is strongly shaped by its dense environment? Aiming to uncover the explosion mechanism and origin of the asymmetric, centrally filled X-ray morphology of W49B, we have performed spatially resolved X-ray spectroscopy and a search for potential point sources. We report new candidate point sources inside W49B. The Chandra X-ray spectra from W49B are well-characterized by two-temperature gas components ($\sim 0.27$ keV + 0.6--2.2 keV). The hot component gas shows a large temperature gradient from the northeast to the southwest and is over-ionized in most regions with recombination timescales of 1--$10\times 10^{11}$ cm$^{-3}$ s. The Fe element shows strong lateral distribution in the SNR east, while the distribution of Si, S, Ar, Ca is relatively smooth and nearly axially symmetric. Asymmetric Type-Ia explosion of a Chandrasekhar-mass white dwarf well-explains the abundance ratios and metal distribution of W49B, whereas a jet-driven explosion and normal core-collapse models fail to describe the abundance ratios and large masses of iron-group elements. A model based on a multi-spot ignition of the white dwarf can explain the observed high $M_{\rm Mn}/M_{\rm Cr}$ value (0.8--2.2). The bar-like morphology is mainly due to a density enhancement in the center, given the good spatial correlation between gas density and X-ray brightness. The recombination ages and the Sedov age consistently suggest a revised SNR age of 5--6 kyr. This study suggests that despite the presence of candidate point sources projected within the boundary of this SNR, W49B is likely a Type-Ia SNR, which suggests that Type-Ia supernovae can also result in mixed-morphology SNRs.
• ### Suzaku and Chandra observations of the galaxy cluster RXC J1053.7+5453 with a radio relic(1708.07004)
Aug. 22, 2017 astro-ph.HE
We present the results of Suzaku and Chandra observations of the galaxy cluster RXC J1053.7+5453 ($z=0.0704$), which contains a radio relic. The radio relic is located at the distance of $\sim 540$ kpc from the X-ray peak toward the west. We measured the temperature of this cluster for the first time. The resultant temperature in the center is $\sim 1.3$ keV, which is lower than the value expected from the X-ray luminosity - temperature and the velocity dispersion - temperature relation. Though we did not find a significant temperature jump at the outer edge of the relic, our results suggest that the temperature decreases outward across the relic. Assuming the existence of the shock at the relic, its Mach number becomes $M \simeq 1.4$. A possible spatial variation of Mach number along the relic is suggested. Additionally, a sharp surface brightness edge is found at the distance of $\sim 160$ kpc from the X-ray peak toward the west in the Chandra image. We performed X-ray spectral and surface brightness analyses around the edge with Suzaku and Chandra data, respectively. The obtained surface brightness and temperature profiles suggest that this edge is not a shock but likely a cold front. Alternatively, it cannot be ruled out that thermal pressure is really discontinuous across the edge. In this case, if the pressure across the surface brightness edge is in equilibrium, other forms of pressure sources, such as cosmic-rays, are necessary. We searched for the non-thermal inverse Compton component in the relic region. Assuming the photon index $\Gamma = 2.0$, the resultant upper limit of the flux is $1.9 \times 10^{-14} {\rm erg \ s^{-1} \ cm^{-2}}$ for $4.50 \times 10^{-3} {\rm \ deg^{2}}$ area in the 0.3-10 keV band, which implies that the lower limit of magnetic field strength becomes $0.7 {\rm \ \mu G}$.
• ### CHEERS: The chemical evolution RGS sample(1707.05076)
The chemical yields of supernovae and the metal enrichment of the hot intra-cluster medium (ICM) are not well understood. This paper introduces the CHEmical Enrichment RGS Sample (CHEERS), which is a sample of 44 bright local giant ellipticals, groups and clusters of galaxies observed with XMM-Newton. This paper focuses on the abundance measurements of O and Fe using the reflection grating spectrometer (RGS). The deep exposures and the size of the sample allow us to quantify the intrinsic scatter and the systematic uncertainties in the abundances using spectral modeling techniques. We report the oxygen and iron abundances as measured with RGS in the core regions of all objects in the sample. We do not find a significant trend of O/Fe as a function of cluster temperature, but we do find an intrinsic scatter in the O and Fe abundances from cluster to cluster. The level of systematic uncertainties in the O/Fe ratio is estimated to be around 20-30%, while the systematic uncertainties in the absolute O and Fe abundances can be as high as 50% in extreme cases. We were able to identify and correct a systematic bias in the oxygen abundance determination, which was due to an inaccuracy in the spectral model. The lack of dependence of O/Fe on temperature suggests that the enrichment of the ICM does not depend on cluster mass and that most of the enrichment likely took place before the ICM was formed. We find that the observed scatter in the O/Fe ratio is due to a combination of intrinsic scatter in the source and systematic uncertainties in the spectral fitting, which we are unable to disentangle. The astrophysical source of intrinsic scatter could be due to differences in AGN activity and ongoing star formation in the BCG. The systematic scatter is due to uncertainties in the spatial line broadening, absorption column, multi-temperature structure and the thermal plasma models. (Abbreviated).
• ### Supernova 1604, Kepler's supernova, and its remnant(1612.06905)
Jan. 16, 2017 astro-ph.HE
Supernova 1604 is the last Galactic supernova for which historical records exist. Johannes Kepler's name is attached to it, as he published a detailed account of the observations made by himself and European colleagues. Supernova 1604 was very likely a Type Ia supernova, which exploded 350 pc to 750 pc above the Galactic plane. Its supernova remnant, known as Kepler's supernova remnant, shows clear evidence for interaction with nitrogen-rich material in the north/northwest part of the remnant, which, given the height above the Galactic plane, must find its origin in mass loss from the supernova progenitor system. The combination of a Type Ia supernova and the presence of circumstellar material makes Kepler's supernova remnant a unique object to study the origin of Type Ia supernovae. The evidence suggests that the progenitor binary system of supernova 1604 consisted of a carbon- oxygen white dwarf and an evolved companion star, which most likely was in the (post) asymptotic giant branch of its evolution. A problem with this scenario is that the companion star must have survived the explosion, but no trace of its existence has yet been found, despite a deep search. 1 Introduction; 2 The supernova remnant, its distance and multiwavelength properties; 2.1 Position, distance estimates and SN1604 as a runaway system; 2.2 X-ray imaging spectroscopy and SN1604 as a Type Ia supernova 2.3 The circumstellar medium as studied in the optical and infrared; 3 The dynamics of Kepler's SNR; 3.1 Velocity measurements; 3.2 Hydrodynamical simulations; 4 The progenitor system of SN 1604; 4.1 Elevated circumstellar nitrogen abundances, silicates and a single degenerate scenario for SN1604; 4.2 Problems with a single degenerate Type Ia scenario for SN 1604; 4.3 Was SN 1604 a core-degenerate Type Ia explosion?; 4.4 What can we learn from the historical light curve of SN 1604? ; 5 Conclusions
• The X-ray Integral Field Unit (X-IFU) on board the Advanced Telescope for High-ENergy Astrophysics (Athena) will provide spatially resolved high-resolution X-ray spectroscopy from 0.2 to 12 keV, with 5 arc second pixels over a field of view of 5 arc minute equivalent diameter and a spectral resolution of 2.5 eV up to 7 keV. In this paper, we first review the core scientific objectives of Athena, driving the main performance parameters of the X-IFU, namely the spectral resolution, the field of view, the effective area, the count rate capabilities, the instrumental background. We also illustrate the breakthrough potential of the X-IFU for some observatory science goals. Then we briefly describe the X-IFU design as defined at the time of the mission consolidation review concluded in May 2016, and report on its predicted performance. Finally, we discuss some options to improve the instrument performance while not increasing its complexity and resource demands (e.g. count rate capability, spectral resolution). The X-IFU will be provided by an international consortium led by France, The Netherlands and Italy, with further ESA member state contributions from Belgium, Finland, Germany, Poland, Spain, Switzerland and two international partners from the United States and Japan.
• ### Origin of central abundances in the hot intra-cluster medium - II. Chemical enrichment and supernova yield models(1608.03888)
The hot intra-cluster medium (ICM) is rich in metals, which are synthesised by supernovae (SNe) and accumulate over time into the deep gravitational potential well of clusters of galaxies. Since most of the elements visible in X-rays are formed by type Ia (SNIa) and/or core-collapse (SNcc) supernovae, measuring their abundances gives us direct information on the nucleosynthesis products of billions of SNe since the epoch of the star formation peak (z~2-3). In this study, we compare the most accurate average X/Fe abundance ratios (compiled in a previous work from XMM-Newton EPIC and RGS observations of 44 galaxy clusters, groups, and ellipticals), representative of the chemical enrichment in the nearby ICM, to various SNIa and SNcc nucleosynthesis models found in the literature. The use of a SNcc model combined to any favoured standard SNIa model (deflagration or delayed-detonation) fails to reproduce our abundance pattern. In particular, the Ca/Fe and Ni/Fe ratios are significantly underestimated by the models. We show that the Ca/Fe ratio can be reproduced better, either by taking a SNIa delayed-detonation model that matches the observations of the Tycho supernova remnant, or by adding a contribution from the Ca-rich gap transient SNe, whose material should easily mix into the hot ICM. On the other hand, the Ni/Fe ratio can be reproduced better by assuming that both deflagration and delayed-detonation SNIa contribute in similar proportions to the ICM enrichment. In either case, the fraction of SNIa over the total number of SNe (SNIa+SNcc) contributing to the ICM enrichment ranges within 29-45%. This fraction is found to be systematically higher than the corresponding SNIa/SNe fraction contributing to the enrichment of the proto-solar environnement (15-25%). We also discuss and quantify two useful constraints on both SNIa and SNcc that can be inferred from the ICM abundance ratios.
• ### XMM-Newton Large Program on SN1006 - II: Thermal Emission(1606.08423)
June 27, 2016 astro-ph.HE
Based on the XMM-Newton large program on SN1006 and our newly developed spatially resolved spectroscopy tools (Paper~I), we study the thermal emission from ISM and ejecta of SN1006 by analyzing the spectra extracted from 583 tessellated regions dominated by thermal emission. With some key improvements in spectral analysis as compared to Paper~I, we obtain much better spectral fitting results with less residuals. The spatial distributions of the thermal and ionization states of the ISM and ejecta show different features, which are consistent with a scenario that the ISM (ejecta) is heated and ionized by the forward (reverse) shock propagating outward (inward). Different elements have different spatial distributions and origins, with Ne mostly from the ISM, Si and S from the ejecta, and O and Mg from both ISM and ejecta. Fe L-shell lines are only detected in a small shell-like region SE to the center of SN1006, indicating that most of the Fe-rich ejecta has not yet or just recently been reached by the reverse shock. The overall ejecta abundance patterns for most of the heavy elements, except for Fe and sometimes S, are consistent with typical Type~Ia SN products. The NW half of the SNR interior probably represents a region with turbulently mixed ISM and ejecta, so has enhanced emission from O, Mg, Si, S, lower ejecta temperature, and a large diversity of ionization age. In addition to the asymmetric ISM distribution, an asymmetric explosion of the progenitor star is also needed to explain the asymmetric ejecta distribution.
• ### XMM-Newton Large Program on SN1006 - I: Methods and Initial Results of Spatially-Resolved Spectroscopy(1508.02950)
Aug. 12, 2015 astro-ph.IM, astro-ph.HE
Based on our newly developed methods and the XMM-Newton large program of SN1006, we extract and analyze the spectra from 3596 tessellated regions of this SNR each with 0.3-8 keV counts $>10^4$. For the first time, we map out multiple physical parameters, such as the temperature ($kT$), electron density ($n_e$), ionization parameter ($n_et$), ionization age ($t_{ion}$), metal abundances, as well as the radio-to-X-ray slope ($\alpha$) and cutoff frequency ($\nu_{cutoff}$) of the synchrotron emission. We construct probability distribution functions of $kT$ and $n_et$, and model them with several Gaussians, in order to characterize the average thermal and ionization states of such an extended source. We construct equivalent width (EW) maps based on continuum interpolation with the spectral model of each regions. We then compare the EW maps of OVII, OVIII, OVII K$\delta-\zeta$, Ne, Mg, SiXIII, SiXIV, and S lines constructed with this method to those constructed with linear interpolation. We further extract spectra from larger regions to confirm the features revealed by parameter and EW maps, which are often not directly detectable on X-ray intensity images. For example, O abundance is consistent with solar across the SNR, except for a low-abundance hole in the center. This "O Hole" has enhanced OVII K$\delta-\zeta$ and Fe emissions, indicating recently reverse shocked ejecta, but also has the highest $n_et$, indicating forward shocked ISM. Therefore, a multi-temperature model is needed to decompose these components. The asymmetric metal distributions suggest there is either an asymmetric explosion of the SN or an asymmetric distribution of the ISM.
• ### On the electron-ion temperature ratio established by collisionless shocks(1407.4499)
Astrophysical shocks are often collisionless shocks. An open question about collisionless shocks is whether electrons and ions each establish their own post-shock temperature, or whether they quickly equilibrate in the shock region. Here we provide simple relations for the minimal amount of equilibration to expect. The basic assumption is that the enthalpy-flux of the electrons is conserved separately, but that all particle species should undergo the same density jump across the the shock. This assumption results in an analytic treatment of electron-ion equilibration that agrees with observations of collisionless shocks: at low Mach numbers ($<2$) the electrons and ions are close to equilibration, whereas for Mach numbers above $M \sim 60$ the electron-ion temperature ratio scales with the particle masses $T_e/T_i = m_e/m_i$. In between these two extremes the electron-ion temperature ratio scales as $T_e/T_i \propto 1/M_s^2$. This relation also hold if adiabatic compression of the electrons is taken into account. For magnetised plasmas the compression is governed by the magnetosonic Mach number, whereas the electron-ion temperatures are governed by the sonic Mach number. The derived equations are in agreement with observational data at low Mach numbers, but for supernova remnants the relation requires that the inferred Mach numbers for the observations are over- estimated, perhaps as a result of upstream heating in the cosmic-ray precursor. In addition to predicting a minimal electron/ion temperature ratio, we also heuristically incorporate ion-electron heat exchange at the shock, quantified with a dimensionless parameter ${\xi}$. Comparing the model to existing observations in the solar system and supernova remnants suggests that the data are best described by ${\xi} \sim 5$ percent. (Abridged abstract.)
• ### On cosmic-ray production efficiency at supernova remnant shocks propagating into realistic diffuse interstellar medium(1412.2874)
March 26, 2015 astro-ph.HE
Using three-dimensional magnetohydrodynamics simulations, we show that the efficiency of cosmic-ray (CR) production at supernova remnants (SNRs) is over-predicted if it could be estimated based on proper motion measurements of H$\alpha$ filaments in combination with shock-jump conditions. Density fluctuations of upstream medium make shock waves rippled and oblique almost everywhere. The kinetic energy of the shock wave is transferred into that of downstream turbulence as well as thermal energy which is related to the shock velocity component normal to the shock surface. Our synthetic observation shows that the CR acceleration efficiency as estimated from a lower downstream plasma temperature, is overestimated by 10-40%, because rippled shock does not immediately dissipate all upstream kinetic energy.
• ### The many sides of RCW 86: a type Ia supernova remnant evolving in its progenitor's wind bubble(1404.5434)
April 22, 2014 astro-ph.HE
We present the results of a detailed investigation of the Galactic supernova remnant RCW 86 using the XMM-Newton X-ray telescope. RCW 86 is the probable remnant of SN 185 A.D, a supernova that likely exploded inside a wind-blown cavity. We use the XMM-Newton Reflection Grating Spectrometer (RGS) to derive precise temperatures and ionization ages of the plasma, which are an indication of the interaction history of the remnant with the presumed cavity. We find that the spectra are well fitted by two non-equilibrium ionization models, which enables us to constrain the properties of the ejecta and interstellar matter plasma. Furthermore, we performed a principal component analysis on EPIC MOS and pn data to find regions with particular spectral properties. We present evidence that the shocked ejecta, emitting Fe-K and Si line emission, are confined to a shell of approximately 2 pc width with an oblate spheroidal morphology. Using detailed hydrodynamical simulations, we show that general dynamical and emission properties at different portions of the remnant can be well-reproduced by a type Ia supernova that exploded in a non-spherically symmetric wind-blown cavity. We also show that this cavity can be created using general wind properties for a single degenerate system. Our data and simulations provide further evidence that RCW 86 is indeed the remnant of SN 185, and is the likely result of a type Ia explosion of single degenerate origin.
• ### Early X-ray emission from Type Ia supernovae originating from symbiotic progenitors or recurrent novae(1401.7332)
Jan. 28, 2014 astro-ph.HE
One of the key observables for determining the progenitor nature of Type Ia supernovae is provided by their immediate circumstellar medium, which according to several models should be shaped by the progenitor binary system. So far, X-ray and radio observations indicate that the surroundings are very tenuous, producing severe upper-limits on the mass loss from winds of the progenitors. In this study, we perform numerical hydro-dynamical simulations of the interaction of the SN ejecta with circumstellar structures formed by possible mass outflows from the progenitor systems and we estimate numerically the expected numerical X-ray luminosity. We consider two kinds of circumstellar structures: a) A circumstellar medium formed by the donor star's stellar wind, in case of a symbiotic binary progenitor system; b) A circumstellar medium shaped by the interaction of the slow wind of the donor star with consecutive nova outbursts for the case of a symbiotic recurrent nova progenitor system. For the hydro-simulations we used well-known Type Ia supernova explosion models, as well as an approximation based on a power law model for the density structure of the outer ejecta. We confirm the strict upper limits on stellar wind mass loss, provided by simplified interpretations of X-ray upper limits of Type Ia supernovae. However, we show that supernova explosions going off in the cavities created by repeated nova explosions, provide a possible explanation for the lack of X-ray emission from supernovae originating from symbiotic binaries. Moreover, the velocity structure of circumstellar medium, shaped by a series of nova explosion matches well with the Na absorption features seen in absorption toward several Type Ia supernovae.
• ### The Hot and Energetic Universe: A White Paper presenting the science theme motivating the Athena+ mission(1306.2307)
This White Paper, submitted to the recent ESA call for science themes to define its future large missions, advocates the need for a transformational leap in our understanding of two key questions in astrophysics: 1) How does ordinary matter assemble into the large scale structures that we see today? 2) How do black holes grow and shape the Universe? Hot gas in clusters, groups and the intergalactic medium dominates the baryonic content of the local Universe. To understand the astrophysical processes responsible for the formation and assembly of these large structures, it is necessary to measure their physical properties and evolution. This requires spatially resolved X-ray spectroscopy with a factor 10 increase in both telescope throughput and spatial resolving power compared to currently planned facilities. Feedback from supermassive black holes is an essential ingredient in this process and in most galaxy evolution models, but it is not well understood. X-ray observations can uniquely reveal the mechanisms launching winds close to black holes and determine the coupling of the energy and matter flows on larger scales. Due to the effects of feedback, a complete understanding of galaxy evolution requires knowledge of the obscured growth of supermassive black holes through cosmic time, out to the redshifts where the first galaxies form. X-ray emission is the most reliable way to reveal accreting black holes, but deep survey speed must improve by a factor ~100 over current facilities to perform a full census into the early Universe. The Advanced Telescope for High Energy Astrophysics (Athena+) mission provides the necessary performance (e.g. angular resolution, spectral resolution, survey grasp) to address these questions and revolutionize our understanding of the Hot and Energetic Universe. These capabilities will also provide a powerful observatory to be used in all areas of astrophysics.
• ### The northwestern ejecta knot in SN 1006(1210.7249)
Dec. 14, 2012 astro-ph.HE
Aims: We want to probe the physics of fast collision-less shocks in supernova remnants. In particular, we are interested in the non-equilibration of temperatures and particle acceleration. Specifically, we aim to measure the oxygen temperature with regards to the electron temperature. In addition, we search for synchrotron emission in the northwestern thermal rim. Methods: This study is part of a dedicated deep observational project of SN 1006 using XMM-Newton, which provides us with currently the best resolution spectra of the bright northwestern oxygen knot. We aim to use the reflection grating spectrometer to measure the thermal broadening of the O vii line triplet by convolving the emission profile of the remnant with the response matrix. Results: The line broadening was measured to be {\sigma}_e = 2.4 \pm 0.3 eV, corresponding to an oxygen temperature of 275$^{+72}_{-63}$ keV. From the EPIC spectra we obtain an electron temperature of 1.35 \pm 0.10 keV. The difference in temperature between the species provides further evidence of non-equilibration of temperatures in a shock. In addition, we find evidence for a bow shock that emits X-ray synchrotron radiation, which is at odds with the general idea that due to the magnetic field orientation only in the NE and SW region X-ray synchrotron radiation should be emitted. We find an unusual H{\alpha} and X-ray synchrotron geometry, in that the H{\alpha} emission peaks downstream of the synchrotron emission. This may be an indication for a peculiar H{\alpha} shock, in which the density is lower and neutral fraction are higher than in other supernova remnants, resulting in a peak in H{\alpha} emission further downstream of the shock.
• ### Supernova Remnants as the Sources of Galactic Cosmic Rays(1206.2363)
June 11, 2012 astro-ph.GA, astro-ph.HE
The origin of cosmic rays holds still many mysteries hundred years after they were first discovered. Supernova remnants have for long been the most likely sources of Galactic cosmic rays. I discuss here some recent evidence that suggests that supernova remnants can indeed efficiently accelerate cosmic rays. For this conference devoted to the Astronomical Institute Utrecht I put the emphasis on work that was done in my group, but placed in a broader context: efficient cosmic-ray acceleration and the im- plications for cosmic-ray escape, synchrotron radiation and the evidence for magnetic- field amplification, potential X-ray synchrotron emission from cosmic-ray precursors, and I conclude with the implications of cosmic-ray escape for a Type Ia remnant like Tycho and a core-collapse remnant like Cas A.
• ### The continued spectral and temporal evolution of RX J0720.4-3125(1203.3708)
March 16, 2012 astro-ph.HE
RX J0720.4-3125 is the most peculiar object among a group of seven isolated X-ray pulsars (the so-called "Magnificent Seven"), since it shows long-term variations of its spectral and temporal properties on time scales of years. This behaviour was explained by different authors either by free precession (with a seven or fourteen years period) or possibly a glitch that occurred around $\mathrm{MJD=52866\pm73 days}$. We analysed our most recent XMM-Newton and Chandra observations in order to further monitor the behaviour of this neutron star. With the new data sets, the timing behaviour of RX J0720.4-3125 suggests a single (sudden) event (e.g. a glitch) rather than a cyclic pattern as expected by free precession. The spectral parameters changed significantly around the proposed glitch time, but more gradual variations occurred already before the (putative) event. Since $\mathrm{MJD\approx53000 days}$ the spectra indicate a very slow cooling by $\sim$2 eV over 7 years.
• ### Supernova remnants: the X-ray perspective(1112.0576)
Jan. 3, 2012 astro-ph.HE
Supernova remnants are beautiful astronomical objects that are also of high scientific interest, because they provide insights into supernova explosion mechanisms, and because they are the likely sources of Galactic cosmic rays. X-ray observations are an important means to study these objects.And in particular the advances made in X-ray imaging spectroscopy over the last two decades has greatly increased our knowledge about supernova remnants. It has made it possible to map the products of fresh nucleosynthesis, and resulted in the identification of regions near shock fronts that emit X-ray synchrotron radiation. In this text all the relevant aspects of X-ray emission from supernova remnants are reviewed and put into the context of supernova explosion properties and the physics and evolution of supernova remnants. The first half of this review has a more tutorial style and discusses the basics of supernova remnant physics and thermal and non-thermal X-ray emission. The second half offers a review of the recent advances.The topics addressed there are core collapse and thermonuclear supernova remnants, SN 1987A, mature supernova remnants, mixed-morphology remnants, including a discussion of the recent finding of overionization in some of them, and finally X-ray synchrotron radiation and its consequences for particle acceleration and magnetic fields.
• ### Cooling curves for neutron stars with hadronic matter and quark matter(1112.1880)
The thermal evolution of isothermal neutron stars is studied with matter both in the hadronic phase as well as in the mixed phase of hadronic matter and strange quark matter. In our models, the dominant early-stage cooling process is neutrino emission via the direct Urca process. As a consequence, the cooling curves fall too fast compared to observations. However, when superfluidity is included, the cooling of the neutron stars is significantly slowed down. Furthermore, we find that the cooling curves are not very sensitive to the precise details of the mixing between the hadronic phase and the quark phase and also of the pairing that leads to superfluidity.
• ### Narrow absorption features in the co-added XMM-Newton RGS spectra of isolated Neutron Stars(1109.2506)
Sept. 12, 2011 astro-ph.SR, astro-ph.HE
We co-added the available XMM-Newton RGS spectra for each of the isolated X-ray pulsars RX\,J0720.4$-$3125, RX\,J1308.6+2127 (RBS\,1223), RX\,J1605.3+3249 and RX\,J1856.4$-$3754 (four members of the "Magnificent Seven") and the "Three Musketeers" Geminga, PSR\,B0656+14 and PSR\,B1055-52. We confirm the detection of a narrow absorption feature at 0.57 keV in the co-added RGS spectra of RX\,J0720.4$-$3125 and RX\,J1605.3+3249 (including most recent observations). In addition we found similar absorption features in the spectra of RX\,J1308.6+2127 (at 0.53 keV) and maybe PSR\,B1055-52 (at 0.56 keV). The absorption feature in the spectra of RX\,J1308.6+2127 is broader than the feature e.g. in RX\,J0720.4$-$3125. The narrow absorption features are detected with 2$\sigma$ to 5.6$\sigma$ significance. Although very bright and frequently observed, there are no absorption features visible in the spectra of RX\,J1856.4$-$3754 and PSR\,B0656+14, while the co-added XMM-Newton RGS spectrum of Geminga has not enough counts to detect such a feature. We discuss a possible origin of these absorption features as lines caused by the presence of highly ionised oxygen (in particular OVII and/or OVI at 0.57 keV) in the interstellar medium and absorption in the neutron star atmosphere, namely the absorption features at 0.57 keV as gravitational redshifted ($g_{r}$=1.17) OVIII.
• ### The imprint of a symbiotic binary progenitor on the properties of Kepler's supernova remnant(1103.5487)
March 28, 2011 astro-ph.GA
We present a model for the Type Ia supernova remnant (SNR) of SN 1604, also known as Kepler's SNR. We find that its main features can be explained by a progenitor model of a symbiotic binary consisting of a white dwarf and an AGB donor star with an initial mass of 4-5 M_sun. The slow, nitrogen rich wind emanating from the donor star has partially been accreted by the white dwarf, but has also created a circumstellar bubble. Based on observational evidence, we assume that the system moves with a velocity of 250 km/s. Due to the systemic motion the interaction between the wind and the interstellar medium has resulted in the formation of a bow shock, which can explain the presence of a one-sided, nitrogen rich shell. We present two-dimensional hydrodynamical simulations of both the shell formation and the SNR evolution. The SNR simulations show good agreement with the observed kinematic and morphological properties of Kepler's SNR. Specifically, the model reproduces the observed expansion parameters (m=V/(R/t)) of m=0.35 in the north and m=0.6 in the south of Kepler's SNR. We discuss the variations among our hydrodynamical simulations in light of the observations, and show that part of the blast wave may have traversed through the one-sided shell completely. The simulations suggest a distance to Kepler's SNR of 6 kpc, or otherwise require that SN 1604 was a sub-energetic Type Ia explosion. Finally, we discuss the possible implications of our model for Type Ia supernovae and their remnants in general.
• ### A Decline in the Nonthermal X-ray Emission from Cassiopeia A(1012.0243)
Dec. 1, 2010 astro-ph.HE
We present new Chandra ACIS-S3 observations of Cassiopeia A which, when combined with earlier ACIS-S3 observations, show evidence for a steady ~ 1.5-2%/yr decline in the 4.2-6.0 keV X-ray emission between the years 2000 and 2010. The computed flux from exposure corrected images over the entire remnant showed a 17% decline over the entire remnant and a slightly larger (21%) decline from regions along the remnant's western limb. Spectral fits of the 4.2-6.0 keV emission across the entire remnant, forward shock filaments, and interior filaments indicate the remnant's nonthermal spectral powerlaw index has steepened by about 10%, with interior filaments having steeper powerlaw indices. Since TeV electrons, which give rise to the observed X-ray synchrotron emission, are associated with the exponential cutoff portion of the electron distribution function, we have related our results to a change in the cutoff energy and conclude that the observed decline and steepening of the nonthermal X-ray emission is consistent with a deceleration of the remnant's ~5000 km/s forward shock of ~10--40 km/s/yr
• ### Cold fronts and multi-temperature structures in the core of Abell 2052(1008.3109)
Oct. 5, 2010 astro-ph.CO
The physics of the coolest phases in the hot Intra-Cluster Medium (ICM) of clusters of galaxies is yet to be fully unveiled. X-ray cavities blown by the central Active Galactic Nucleus (AGN) contain enough energy to heat the surrounding gas and stop cooling, but locally blobs or filaments of gas appear to be able to cool to low temperatures of 10^4 K. In X-rays, however, gas with temperatures lower than 0.5 keV is not observed. Using a deep XMM-Newton observation of the cluster of galaxies Abell 2052, we derive 2D maps of the temperature, entropy, and iron abundance in the core region. About 130 kpc South-West of the central galaxy, we discover a discontinuity in the surface brightness of the hot gas which is consistent with a cold front. Interestingly, the iron abundance jumps from ~0.75 to ~0.5 across the front. In a smaller region to the North-West of the central galaxy we find a relatively high contribution of cool 0.5 keV gas, but no X-ray emitting gas is detected below that temperature. However, the region appears to be associated with much cooler H-alpha filaments in the optical waveband. The elliptical shape of the cold front in the SW of the cluster suggests that the front is caused by sloshing of the hot gas in the clusters gravitational potential. This effect is probably an important mechanism to transport metals from the core region to the outer parts of the cluster. The smooth temperature profile across the sharp jump in the metalicity indicates the presence of heat conduction and the lack of mixing across the discontinuity. The cool blob of gas NW of the central galaxy was probably pushed away from the core and squeezed by the adjacent bubble, where it can cool efficiently and relatively undisturbed by the AGN. Shock induced mixing between the two phases may cause the 0.5 keV gas to cool non-radiatively and explain our non-detection of gas below 0.5 keV.
• ### Annihilation emission from young supernova remnants(1006.2537)
June 13, 2010 astro-ph.HE
A promising source of the positrons that contribute through annihilation to the diffuse Galactic 511keV emission is the beta-decay of unstable nuclei like 56Ni and 44Ti synthesised by massive stars and supernovae. Although a large fraction of these positrons annihilate in the ejecta of SNe/SNRs, no point-source of annihilation radiation appears in the INTEGRAL/SPI map of the 511keV emission. We exploit the absence of detectable annihilation emission from young local SNe/SNRs to derive constraints on the transport of MeV positrons inside SN/SNR ejecta and their escape into the CSM/ISM, both aspects being crucial to the understanding of the observed Galactic 511keV emission. We simulated 511keV lightcurves resulting from the annihilation of the decay positrons of 56Ni and 44Ti in SNe/SNRs and their surroundings using a simple model. We computed specific 511keV lightcurves for Cas A, Tycho, Kepler, SN1006, G1.9+0.3 and SN1987A, and compared these to the upper-limits derived from INTEGRAL/SPI observations. The predicted 511keV signals from positrons annihilating in the ejecta are below the sensitivity of the SPI instrument by several orders of magnitude, but the predicted 511keV signals for positrons escaping the ejecta and annihilating in the surrounding medium allowed to derive upper-limits on the positron escape fraction of ~13% for Cas A, ~12% for Tycho, ~30% for Kepler and ~33% for SN1006. The transport of ~MeV positrons inside SNe/SNRs cannot be constrained from current observations of the 511keV emission from these objects, but the limits obtained on their escape fraction are consistent with a nucleosynthesis origin of the positrons that give rise to the diffuse Galactic 511keV emission.
• ### The kinematics and chemical stratification of the Type Ia supernova remnant 0519-69.0(1001.0983)
May 10, 2010 astro-ph.HE
We present an analysis of the XMM-Newton and Chandra X-ray data of the young Type Ia supernova remnant 0519-69.0 in the Large Magellanic Cloud. We used data from both the Chandra ACIS and XMM-Newton EPIC-MOS instruments, and high resolution X-ray spectra obtained with the XMM-Newton Reflection Grating Spectrometer. The Chandra data show that there is a radial stratification of oxygen, intermediate mass elements and iron, with the emission from more massive elements more toward the center. Using a deprojection technique we measure a forward shock radius of 4.0(3) pc and a reverse shock radius of 2.7(4) pc. We took the observed stratification of the shocked ejecta into account in the modeling of the X-ray spectra with multi-component NEI models, with the components corresponding to layers dominated by one or two elements. An additional component was added in order to represent the ISM, which mostly contributed to the continuum emission. This model fits the data well, and was also employed to characterize the spectra of distinct regions extracted from the Chandra data. From our spectral analysis we find that the fractional masses of shocked ejecta for the most abundant elements are: M(O)=32%, M(Si/S)=7%/5%, M(Ar+Ca)=1%, and M(Fe) = 55%. From the continuum component we derive a circumstellar density of nH= 2.4(2)/cm^3. This density, together with the measurements of the forward and reverse shock radii suggest an age of 450+/-200 yr,somewhat lower than, but consistent with the estimate based on the optical light echo (600+/-200 yr). From the RGS spectra we measured a Doppler broadening of sigma=1873+/-50 km/s, from implying a forward shock velocity of vS = 2770+/-500 km/s. We discuss the results in the context of single degenerate explosion models, using semi-analytical and numerical modeling, and compare the characteristics of 0519-69.0 with those of other Type Ia supernova remnants.
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# Using FunctionExpand to evaluate symbolic derivatives
Some symbolic derivatives of certain special functions are not expanded automatically, but FunctionExpand often helps to get a derivative-free closed form expression.
Derivative[1, 0][BesselJ][0, 1]
(* Derivative[1, 0][BesselJ][0, 1] *)
FunctionExpand[%]
(* 1/2 π BesselY[0, x] *)
But for some functions, it takes too much time to evaluate. Possibly, there is even an infinite loop. For example, I left the following expression to evaluate overnight, and it was still running in the morning without any result or messages:
FunctionExpand[Derivative[1, 0][StruveL][0, 1]]
• Is there a workaround that could get an expanded form of the expression Derivative[1, 0][StruveL][0, 1] in reasonable time?
• Is there an infinite-loop bug in the implementation of FunctionExpand or do I just have to wait longer for the results (weeks, months, ...)?
• Is there any public information about what approaches are used by FunctionExpand to expand derivatives?
-
See Low-order differentiation here functions.wolfram.com/NB/StruveH.nb – Dr. belisarius Nov 19 '13 at 1:07
It seems the current version (10.3) is now aware of the Meijer $G$ expressions for the order derivatives (see this math.SE answer as well):
Derivative[1, 0][StruveL][0, z] // FunctionExpand
Nevertheless, FunctionExpand[Derivative[1, 0][StruveL][0, 1]] still takes a ridiculous amount of time (certainly longer than the purely symbolic version); I'm not sure if this is a bug.
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# Approximation and error
1. Feb 8, 2014
### chomool
Triangle ABC is an equilateral triangle with side 4 cm long which is measured corrected to the nearest cm.
Find the percentage error of the perimeter of triangle ABC.
3. The attempt at a solution
Is
[(0.5 x 2 x 3) / 12] x 100% correct?
the '2' here is the measurement errors of the starting pt and ending pt of line segment.
or
it should be:
[(0.5 x 3) / 12] x 100%
plz help~!
2. Feb 8, 2014
### Ray Vickson
There are two distinct possibilities:
(1) The triangle is known to be exactly equilateral, but having (three equal) sides measured with possible errors.
(2) The triangle was measured to have all three sides equal to 4 cm, but the individual sides may have (independent) measurement errors. Therefore, while the "measured" triangle is equilateral, the actual, true, triangle might not be.
I assume you want to go with interpretation (1), which is probably the one meant by the person who set the problem. In that case, it is straightforward: each side is between 3 cm and 5 cm, so the perimeter is between 9 cm and 15 cm, with 12 cm being the measured value. In other words, the perimeter is within the interval $12 \pm 3$ cm. The estimate of 12 cm could be "off" by as much as 3 cm.
3. Feb 8, 2014
### CompuChip
Doesn't "correct to the nearest cm" mean that it would be between 3.5 and 4.5? I.e. the value rounded to whole cm is 4.
4. Feb 8, 2014
### Ray Vickson
Yes, I think you are right.
5. Feb 8, 2014
### haruspex
The percentage error will also be a matter of ± so many %, so you don't need to double up here.
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# Quantization of heat flow in the fractional quantum Hall regime - Mitali Banerjee, Columbia
Tue, Dec 11, 2018, 1:00 pm
Location:
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Discussions of PEA mention that it’s almost useless without a MAOI to pave the way; hence, when I decided to get deprenyl and noticed that deprenyl is a MAOI, I decided to also give PEA a second chance in conjunction with deprenyl. Unfortunately, in part due to my own shenanigans, Nubrain canceled the deprenyl order and so I have 20g of PEA sitting around. Well, it’ll keep until such time as I do get a MAOI.
## For example, a study published in the journal Psychopharmacology in 2000 found that ginkgo improved attention. A 2001 study in the journal Human Psychopharmacology suggested that it improves memory. Nevertheless, in a review of studies on ginkgo in healthy people, researchers found no good evidence that it improved mental abilities, according to a 2002 report in Psychopharmacology Bulletin.
As it happens, these are areas I am distinctly lacking in. When I first began reading about testosterone I had no particular reason to think it might be an issue for me, but it increasingly sounded plausible, an aunt independently suggested I might be deficient, a biological uncle turned out to be severely deficient with levels around 90 ng/dl (where the normal range for 20-49yo males is 249-839), and finally my blood test in August 2013 revealed that my actual level was 305 ng/dl; inasmuch as I was 25 and not 49, this is a tad low.
But Baldino may have been overly modest. In 2002, researchers at Cambridge University gave 60 healthy young male volunteers a battery of standard cognitive tests. One group received modafinil, the other a placebo. The modafinil group performed better on several tasks, such as the "digit span" test, in which subjects are asked to repeat increasingly longer strings of numbers forwards, then backwards. They also did better in recognising repeated visual patterns and at a spatial-planning challenge known as the Tower of London task. (It's not nearly as fun as it sounds.) Writing in the journal Psychopharmacology, the study's authors said the results suggested that "modafinil offers significant potential as a cognitive enhancer".
I can test fish oil for mood, since the other claimed benefits like anti-schizophrenia are too hard to test. The medical student trial (Kiecolt-Glaser et al 2011) did not see changes until visit 3, after 3 weeks of supplementation. (Visit 1, 3 weeks, visit 2, supplementation started for 3 weeks, visit 3, supplementation continued 3 weeks, visit 4 etc.) There were no tests in between the test starting week 1 and starting week 3, so I can’t pin it down any further. This suggests randomizing in 2 or 3 week blocks. (For an explanation of blocking, see the footnote in the Zeo page.)
See Melatonin for information on effects & cost; I regularly use melatonin to sleep (more to induce sleep than prolong or deepen it), and investigating with my Zeo, it does seem to improve & shorten my sleep. Some research suggests that higher doses are not necessarily better and may be overkill, so each time I’ve run out, I’ve been steadily decreasing the dose from 3mg to 1.5mg to 1mg, without apparently compromising the usefulness.
11:30 AM. By 2:30 PM, my hunger is quite strong and I don’t feel especially focused - it’s difficult to get through the tab-explosion of the morning, although one particularly stupid poster on the DNB ML makes me feel irritated like I might on Adderall. I initially figure the probability at perhaps 60% for Adderall, but when I wake up at 2 AM and am completely unable to get back to sleep, eventually racking up a Zeo score of 73 (compared to the usual 100s), there’s no doubt in my mind (95%) that the pill was Adderall. And it was the last Adderall pill indeed.
Clarke and Sokoloff (1998) remarked that although [a] common view equates concentrated mental effort with mental work…there appears to be no increased energy utilization by the brain during such processes (p. 664), and …the areas that participate in the processes of such reasoning represent too small a fraction of the brain for changes in their functional and metabolic activities to be reflected in the energy metabolism of the brain… (p. 675).
Take the synthetic nootropic piracetam, for example. Since piracetam has been shown to improve cell membrane function and cause a host of neuroprotective effects, when combined with other cell membrane stabilizing supplements such as choline and DHA, the brain cells on piracetam can better signal and relay messages to each other for a longer period of time, which improves cognition and brain activity and decreases risk of a crash. So one example of an intelligent “stack” is piracetam taken with choline and DHA.
An unusual intervention is infrared/near-infrared light of particular wavelengths (LLLT), theorized to assist mitochondrial respiration and yielding a variety of therapeutic benefits. Some have suggested it may have cognitive benefits. LLLT sounds strange but it’s simple, easy, cheap, and just plausible enough it might work. I tried out LLLT treatment on a sporadic basis 2013-2014, and statistically, usage correlated strongly & statistically-significantly with increases in my daily self-ratings, and not with any sleep disturbances. Excited by that result, I did a randomized self-experiment 2014-2015 with the same procedure, only to find that the causal effect was weak or non-existent. I have stopped using LLLT as likely not worth the inconvenience.
I’ve been actively benefitting from nootropics since 1997, when I was struggling with cognitive performance and ordered almost $1000 worth of smart drugs from Europe (the only place where you could get them at the time). I remember opening the unmarked brown package and wondering whether the pharmaceuticals and natural substances would really enhance my brain. Last summer, I visited Phillips in the high desert resort town of Bend, Oregon, where he lives with his wife, Kathleen, and their two daughters, Ivy and Ruby. Phillips, who is now 36, took me for coffee at a cheery café called Thump. Wearing shorts, flip-flops and a black T-shirt, he said: "Poker is about sitting in one place, watching your opponents for a long time, and making better observations about them than they make about you." With Provigil, he "could process all the information about what was going on at the table and do something about it". Though there is no question that Phillips became much more successful at poker after taking neuroenhancers, I asked him if his improvement could be explained by a placebo effect, or by coincidence. He doubted it, but allowed that it could. Still, he said, "there's a sort of clarity I get with Provigil. With Adderall, I'd characterise the effect as correction - correction of an underlying condition. Provigil feels like enhancement." And, whereas Adderall made him "jittery", Provigil's effects were "completely limited to my brain". He had "zero difficulty sleeping". # He used to get his edge from Adderall, but after moving from New Jersey to San Francisco, he says, he couldn’t find a doctor who would write him a prescription. Driven to the Internet, he discovered a world of cognition-enhancing drugs known as nootropics — some prescription, some over-the-counter, others available on a worldwide gray market of private sellers — said to improve memory, attention, creativity and motivation. My answer is that this is not a lot of research or very good research (not nearly as good as the research on nicotine, eg.), and assuming it’s true, I don’t value long-term memory that much because LTM is something that is easily assisted or replaced (personal archives, and spaced repetition). For me, my problems tend to be more about akrasia and energy and not getting things done, so even if a stimulant comes with a little cost to long-term memory, it’s still useful for me. I’m going continue to use the caffeine. It’s not so bad in conjunction with tea, is very cheap, and I’m already addicted, so why not? Caffeine is extremely cheap, addictive, has minimal effects on health (and may be beneficial, from the various epidemiological associations with tea/coffee/chocolate & longevity), and costs extra to remove from drinks popular regardless of their caffeine content (coffee and tea again). What would be the point of carefully investigating it? Suppose there was conclusive evidence on the topic, the value of this evidence to me would be roughly$0 or since ignorance is bliss, negative money - because unless the negative effects were drastic (which current studies rule out, although tea has other issues like fluoride or metal contents), I would not change anything about my life. Why? I enjoy my tea too much. My usual tea seller doesn’t even have decaffeinated oolong in general, much less various varieties I might want to drink, apparently because de-caffeinating is so expensive it’s not worthwhile. What am I supposed to do, give up my tea and caffeine just to save on the cost of caffeine? Buy de-caffeinating machines (which I couldn’t even find any prices for, googling)? This also holds true for people who drink coffee or caffeinated soda. (As opposed to a drug like modafinil which is expensive, and so the value of a definitive answer is substantial and would justify some more extensive calculating of cost-benefit.)
Conversely, you have to consider that the long term effects of Modafinil haven’t been studies very well. It significantly upsets sleep cycles, and 50% of Modafinil users report a number of short term side effects, such as mild to severe headaches, insomnia, nausea, anxiety, nervousness, hypertension, decreased appetite, and weight loss. PET scans show it affects the same areas of the brain that are stimulated by substance abuse.
At this point, I began thinking about what I was doing. Black-market Adderall is fairly expensive; $4-10 a pill vs prescription prices which run more like$60 for 120 20mg pills. It would be a bad idea to become a fan without being quite sure that it is delivering bang for the buck. Now, why the piracetam mix as the placebo as opposed to my other available powder, creatine powder, which has much smaller mental effects? Because the question for me is not whether the Adderall works (I am quite sure that the amphetamines have effects!) but whether it works better for me than my cheap legal standbys (piracetam & caffeine)? (Does Adderall have marginal advantage for me?) Hence, I want to know whether Adderall is better than my piracetam mix. People frequently underestimate the power of placebo effects, so it’s worth testing. (Unfortunately, it seems that there is experimental evidence that people on Adderall know they are on Adderall and also believe they have improved performance, when they do not5. So the blind testing does not buy me as much as it could.)
Fitzgerald 2012 and the general absence of successful experiments suggests not, as does the general historic failure of scores of IQ-related interventions in healthy young adults. Of the 10 studies listed in the original section dealing with iodine in children or adults, only 2 show any benefit; in lieu of a meta-analysis, a rule of thumb would be 20%, but both those studies used a package of dozens of nutrients - and not just iodine - so if the responsible substance were randomly picked, that suggests we ought to give it a chance of 20% \times \frac{1}{\text{dozens}} of being iodine! I may be unduly optimistic if I give this as much as 10%.
And when it comes to your brain, it’s full of benefits, too. Coconut oil works as a natural anti-inflammatory, suppressing cells responsible for inflammation. It can help with memory loss as you age and destroy bad bacteria that hangs out in your gut. (5) Get your dose of coconut oil in this Baked Grouper with Coconut Cilantro Sauce or Coconut Crust Pizza.
Nootropics That is offered through an email showing Ben Carson and Bill O Reily talking about it and they offer a deal the more you buy the cheaper it is with free bottles is a scam. I ordered 3 bottles with 2 free and free shipping which should have been 120.00. They had a offer for 59.99 for cleansing product and I didn’t order it. My total came out to 189.94. I called them and they removed it and then the 5 bottles was still going to be 189.99. I told them about the offer and they would not honor it. I ended up canceling the order. This is a scam!!
I usually use Alpha Brain but have found the delivery times and cost per bottle to be too much .Its great to find a UK based nootropic supplement company giving Alpha Brain a run for their Yankee Dollar . Ultra is a very smooth nootropic - i find my thinking clear and my brain feels more alert and alive - hard to explain but Ultra makes me feel more 'present' and alive . Highly recommended. 5 STARS! MC (Wales)
20 March, 2x 13mg; first time, took around 11:30AM, half-life 3 hours, so halved by 2:30PM. Initial reaction: within 20 minutes, started to feel light-headed, experienced a bit of physical clumsiness while baking bread (dropped things or poured too much thrice); that began to pass in an hour, leaving what felt like a cheerier mood and less anxiety. Seems like it mostly wore off by 6PM. Redosed at 8PM TODO: maybe take a look at the HRV data? looks interestingly like HRV increased thanks to the tianeptine 21 March, 2x17mg; seemed to buffer effects of FBI visit 22 March, 2x 23 March, 2x 24 March, 2x 25 March, 2x 26 March, 2x 27 March, 2x 28 March, 2x 7 April, 2x 8 April, 2x 9 April, 2x 10 April, 2x 11 April, 2x 12 April, 2x 23 April, 2x 24 April, 2x 25 April, 2x 26 April, 2x 27 April, 2x 28 April, 2x 29 April, 2x 7 May, 2x 8 May, 2x 9 May, 2x 10 May, 2x 3 June, 2x 4 June, 2x 5 June, 2x 30 June, 2x 30 July, 1x 31 July, 1x 1 August, 2x 2 August, 2x 3 August, 2x 5 August, 2x 6 August, 2x 8 August, 2x 10 August, 2x 12 August: 2x 14 August: 2x 15 August: 2x 16 August: 1x 18 August: 2x 19 August: 2x 21 August: 2x 23 August: 1x 24 August: 1x 25 August: 1x 26 August: 2x 27 August: 1x 29 August: 2x 30 August: 1x 02 September: 1x 04 September: 1x 07 September: 2x 20 September: 1x 21 September: 2x 24 September: 2x 25 September: 2x 26 September: 2x 28 September: 2x 29 September: 2x 5 October: 2x 6 October: 1x 19 October: 1x 20 October: 1x 27 October: 1x 4 November: 1x 5 November: 1x 8 November: 1x 9 November: 2x 10 November: 1x 11 November: 1x 12 November: 1x 25 November: 1x 26 November: 1x 27 November: 1x 4 December: 2x 27 December: 1x 28 December: 1x 2017 7 January: 1x 8 January: 2x 10 January: 1x 16 January: 1x 17 January: 1x 20 January: 1x 24 January: 1x 25 January: 2x 27 January: 2x 28 January: 2x 1 February: 2x 3 February: 2x 8 February: 1x 16 February: 2x 17 February: 2x 18 February: 1x 22 February: 1x 27 February: 2x 14 March: 1x 15 March: 1x 16 March: 2x 17 March: 2x 18 March: 2x 19 March: 2x 20 March: 2x 21 March: 2x 22 March: 2x 23 March: 1x 24 March: 2x 25 March: 2x 26 March: 2x 27 March: 2x 28 March: 2x 29 March: 2x 30 March: 2x 31 March: 2x 01 April: 2x 02 April: 1x 03 April: 2x 04 April: 2x 05 April: 2x 06 April: 2x 07 April: 2x 08 April: 2x 09 April: 2x 10 April: 2x 11 April: 2x 20 April: 1x 21 April: 1x 22 April: 1x 23 April: 1x 24 April: 1x 25 April: 1x 26 April: 2x 27 April: 2x 28 April: 1x 30 April: 1x 01 May: 2x 02 May: 2x 03 May: 2x 04 May: 2x 05 May: 2x 06 May: 2x 07 May: 2x 08 May: 2x 09 May: 2x 10 May: 2x 11 May: 2x 12 May: 2x 13 May: 2x 14 May: 2x 15 May: 2x 16 May: 2x 17 May: 2x 18 May: 2x 19 May: 2x 20 May: 2x 21 May: 2x 22 May: 2x 23 May: 2x 24 May: 2x 25 May: 2x 26 May: 2x 27 May: 2x 28 May: 2x 29 May: 2x 30 May: 2x 1 June: 2x 2 June: 2x 3 June: 2x 4 June: 2x 5 June: 1x 6 June: 2x 7 June: 2x 8 June: 2x 9 June: 2x 10 June: 2x 11 June: 2x 12 June: 2x 13 June: 2x 14 June: 2x 15 June: 2x 16 June: 2x 17 June: 2x 18 June: 2x 19 June: 2x 20 June: 2x 22 June: 2x 21 June: 2x 02 July: 2x 03 July: 2x 04 July: 2x 05 July: 2x 06 July: 2x 07 July: 2x 08 July: 2x 09 July: 2x 10 July: 2x 11 July: 2x 12 July: 2x 13 July: 2x 14 July: 2x 15 July: 2x 16 July: 2x 17 July: 2x 18 July: 2x 19 July: 2x 20 July: 2x 21 July: 2x 22 July: 2x 23 July: 2x 24 July: 2x 25 July: 2x 26 July: 2x 27 July: 2x 28 July: 2x 29 July: 2x 30 July: 2x 31 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The single most reliable way to protect our brain cells as we age, most researchers agree, is to eat plenty of fruits and vegetables, which are chock-full of antioxidants and nutrients. In a study published in the October 1997 issue of the American Journal of Clinical Nutrition, researchers tested 260 people aged 65 to 90 with a series of mental exercises that involved memorizing words or doing mental arithmetic. The top performers were those who consumed the most fruits and vegetables and ate the least artery-clogging saturated fat.
When I spoke with Sahakian she had just flown from England to Scottsdale, Arizona, to attend a conference, and she was tired. "We may be healthy and high-functioning, and think of ourselves that way," she told me, "but it's very rare that we are actually functioning at our optimal level. Take me. I'm over here and I've got jet lag and I've got to give a talk tonight and perform well in what will be the middle of the night, UK time." She mentioned businessmen who have to fly back and forth across the Atlantic: "The difference between making a deal and not is huge, and they sometimes only have one meeting to try and do it." She added: "We are a society that so wants a quick fix that many people are happy to take drugs."
It looks like the overall picture is that nicotine is absorbed well in the intestines and the colon, but not so well in the stomach; this might be the explanation for the lack of effect, except on the other hand, the specific estimates I see are that 10-20% of the nicotine will be bioavailable in the stomach (as compared to 50%+ for mouth or lungs)… so any of my doses of >5ml should have overcome the poorer bioavailability! But on the gripping hand, these papers are mentioning something about the liver metabolizing nicotine when absorbed through the stomach, so…
Eliminating foggy-headedness seems to be the goal of many users of neuroenhancers. But can today's drugs actually accomplish this? I recently posed this question to Chatterjee's colleague Martha Farah, who is a psychologist at Penn and the director of its Center for Cognitive Neuroscience. She is deeply fascinated by, and mildly critical of, neuroenhancers, but basically in favour - with the important caveat that we need to know much more about how these drugs work. While Farah does not take neuroenhancers, she had just finished a paper in which she reviewed the evidence on prescription stimulants as neuroenhancers from 40 laboratory studies involving healthy subjects. Most of the studies looked at one of three types of cognition: learning, working memory, and cognitive control. A typical learning test asks subjects to memorise a list of paired words; an hour, a few days, or a week later, they are presented with the first words in the pairs and asked to come up with the second. Neuroenhancers did improve retention, especially where subjects had been asked to remember information for several days or longer.
[…] The 7 Best Brain Boosting Supplements | Live in the Now … – While under estimated in the brain health arena, adequate vitamin C is associated with a 20% … If you are looking for a way to maximize brain power I have come across … […] medicines, dietary supplements and organic food products. Justin has also been writing on best brain supplements for … […]
Remembering what Wedrifid told me, I decided to start with a quarter of a piece (~1mg). The gum was pretty tasteless, which ought to make blinding easier. The effects were noticeable around 10 minutes - greater energy verging on jitteriness, much faster typing, and apparent general quickening of thought. Like a more pleasant caffeine. While testing my typing speed in Amphetype, my speed seemed to go up >=5 WPM, even after the time penalties for correcting the increased mistakes; I also did twice the usual number without feeling especially tired. A second dose was similar, and the third dose was at 10 PM before playing Ninja Gaiden II seemed to stop the usual exhaustion I feel after playing through a level or so. (It’s a tough game, which I have yet to master like Ninja Gaiden Black.) Returning to the previous concern about sleep problems, though I went to bed at 11:45 PM, it still took 28 minutes to fall sleep (compared to my more usual 10-20 minute range); the next day I use 2mg from 7-8PM while driving, going to bed at midnight, where my sleep latency is a more reasonable 14 minutes. I then skipped for 3 days to see whether any cravings would pop up (they didn’t). I subsequently used 1mg every few days for driving or Ninja Gaiden II, and while there were no cravings or other side-effects, the stimulation definitely seemed to get weaker - benefits seemed to still exist, but I could no longer describe any considerable energy or jitteriness.
In addition to this, privilege also plays an important role in this epidemic. "Not everyone has access to eat healthily", she mentions. In fact, she recalls an anecdote in which a supermarket owner noticed how people living off food stamps rarely use them to buy fruits and vegetables. Curious about this trend, the owner approached someone with food stamps, to which she admitted she didn't buy them because she didn't know the price prior to weighing them and felt ashamed of asking. His solution? Pre-cutting and packaging fruits in order to make them more accessible to those with lower incomes.
When many of us think of memory enhancers, we think of ginkgo biloba, the herb that now generates more than $240 million in sales a year worldwide. The October 22-29, 1997 issue of the Journal of the American Medical Association reported that Alzheimer's patients who took 120 mg of ginkgo showed small improvements in tests designed to measure mental performance. One fairly powerful nootropic substance that, appropriately, has fallen out of favor is nicotine. It’s the chemical that gives tobacco products their stimulating kick. It isn’t what makes them so deadly, but it does make smoking very addictive. When Europeans learned about tobacco’s use from indigenous tribes they encountered in the Americas in the 15th and 16th centuries, they got hooked on its mood-altering effects right away and even believed it could cure joint pain, epilepsy, and the plague. Recently, researchers have been testing the effects of nicotine that’s been removed from tobacco, and they believe that it might help treat neurological disorders including Parkinson’s disease and schizophrenia; it may also improve attention and focus. But, please, don’t start smoking or vaping. Check out these 14 weird brain exercises that make you smarter. A picture is worth a thousand words, particularly in this case where there seems to be temporal effects, different trends for the conditions, and general confusion. So, I drag up 2.5 years of MP data (for context), plot all the data, color by magnesium/non-magnesium, and fit different LOESS lines to each as a sort of smoothed average (since categorical data is hard to interpret as a bunch of dots), which yields: Hericium erinaceus (Examine.com) was recommended strongly by several on the ImmInst.org forums for its long-term benefits to learning, apparently linked to Nerve growth factor. Highly speculative stuff, and it’s unclear whether the mushroom powder I bought was the right form to take (ImmInst.org discussions seem to universally assume one is taking an alcohol or hotwater extract). It tasted nice, though, and I mixed it into my sleeping pills (which contain melatonin & tryptophan). I’ll probably never know whether the$30 for 0.5lb was well-spent or not.
When it comes to brain power, greens should be on your plate (and cover a lot of that plate) every meal. “Leafy greens are a great base. You swap out a lot of the empty carbohydrates you get from things like pastas or breads, and you can use some leafy greens,” says Psychiatrist Drew Ramsey, MD, author of The Happiness Diet and Eat Complete: The 21 Nutrients That Fuel Brainpower, Boost Weight Loss, and Transform Your Health. “Again, just lots of nutrient density.”
Chocolate or cocoa powder (Examine.com), contains the stimulants caffeine and the caffeine metabolite theobromine, so it’s not necessarily surprising if cocoa powder was a weak stimulant. It’s also a witch’s brew of chemicals such as polyphenols and flavonoids some of which have been fingered as helpful10, which all adds up to an unclear impact on health (once you control for eating a lot of sugar).
We can read off the results from the table or graph: the nicotine days average 1.1% higher, for an effect size of 0.24; however, the 95% credible interval (equivalent of confidence interval) goes all the way from 0.93 to -0.44, so we cannot exclude 0 effect and certainly not claim confidence the effect size must be >0.1. Specifically, the analysis gives a 66% chance that the effect size is >0.1. (One might wonder if any increase is due purely to a training effect - getting better at DNB. Probably not26.)
Be patient. Even though you may notice some improvements right away (sometimes within the first day), you should give your brain supplement at least several months to work. The positive effects are cumulative, and most people do not max out their brain potential on a supplement until they have used it for at least 90 days. That is when the really dramatic effects start kicking in!
In this large population-based cohort, we saw consistent robust associations between cola consumption and low BMD in women. The consistency of pattern across cola types and after adjustment for potential confounding variables, including calcium intake, supports the likelihood that this is not due to displacement of milk or other healthy beverages in the diet. The major differences between cola and other carbonated beverages are caffeine, phosphoric acid, and cola extract. Although caffeine likely contributes to lower BMD, the result also observed for decaffeinated cola, the lack of difference in total caffeine intake across cola intake groups, and the lack of attenuation after adjustment for caffeine content suggest that caffeine does not explain these results. A deleterious effect of phosphoric acid has been proposed (26). Cola beverages contain phosphoric acid, whereas other carbonated soft drinks (with some exceptions) do not.
For illustration, consider amphetamines, Ritalin, and modafinil, all of which have been proposed as cognitive enhancers of attention. These drugs exhibit some positive effects on cognition, especially among individuals with lower baseline abilities. However, individuals of normal or above-average cognitive ability often show negligible improvements or even decrements in performance following drug treatment (for details, see de Jongh, Bolt, Schermer, & Olivier, 2008). For instance, Randall, Shneerson, and File (2005) found that modafinil improved performance only among individuals with lower IQ, not among those with higher IQ. [See also Finke et al 2010 on visual attention.] Farah, Haimm, Sankoorikal, & Chatterjee 2009 found a similar nonlinear relationship of dose to response for amphetamines in a remote-associates task, with low-performing individuals showing enhanced performance but high-performing individuals showing reduced performance. Such ∩-shaped dose-response curves are quite common (see Cools & Robbins, 2004)
I always romecmend I always romecmend doing the best you can. Even acid rain or toxins in the air float onto the food I grow in my garden. I like to look at things as good, better, best. Its best to grow seaweed in a controlled enviorment (farming) and eat it. Of course for most people, in my opinion its far better to eat some seaweed to get some trace minerals than not. Im not saying eat them in MASS quantity but some here and there. Its best to grow your own food in TRACE MINERALS to get them. Was this answer helpful?
The first night I was eating some coconut oil, I did my n-backing past 11 PM; normally that damages my scores, but instead I got 66/66/75/88/77% (▁▁▂▇▃) on D4B and did not feel mentally exhausted by the end. The next day, I performed well on the Cambridge mental rotations test. An anecdote, of course, and it may be due to the vitamin D I simultaneously started. Or another day, I was slumped under apathy after a promising start to the day; a dose of fish & coconut oil, and 1 last vitamin D, and I was back to feeling chipper and optimist. Unfortunately I haven’t been testing out coconut oil & vitamin D separately, so who knows which is to thank. But still interesting.
On the other metric, suppose we removed the creatine? Dropping 4 grams of material means we only need to consume 5.75 grams a day, covered by 8 pills (compared to 13 pills). We save 5,000 pills, which would have cost $45 and also don’t spend the$68 for the creatine; assuming a modafinil formulation, that drops our $1761 down to$1648 or $1.65 a day. Or we could remove both the creatine and modafinil, for a grand total of$848 or \$0.85 a day, which is pretty reasonable.
Second, users are concerned with the possibility of withdrawal if they stop taking the nootropics. They worry that if they stop taking nootropics they won’t be as smart as when they were taking nootropics, and will need to continue taking them to function. Some users report feeling a slight brain fog when discontinuing nootropics, but that isn’t a sign of regression.
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All Solutions of Type: Thermodynamics
0 Click here to jump to the problem! GR8677 #14 Thermodynamics$\Rightarrow$}Exact differentials The key equation is: $PV=nRT$, and its players, $P,V,n,R,T$, are terms one should be able to guess. (A) True, according to the ideal gas law. (This is also the final step in deriving Mayer's Equation, as shown below.) (B) This translates into the statement $\left.\frac{dq}{dT}\right|_V=\left.\frac{dq}{dT}\right|_P\Rightarrow c_V=c_P$.The problem gives away the fact that for an ideal gas $C_P \neq C_V$. B can't be right. (C) According to the ideal gas law, the volume might change. (D) False. An ideal gas's internal energy is dependent only on temperature. More elegantly, $u=u(T)$. (E) Heat needed for what? If one is interested in the formal proof of the relation $c_p=c_v+nR$, read on about Mayer's equation: For thermo, in general, there's an old slacker's pride line that goes like, When in doubt, write a bunch of equations of states and mindlessly begin taking exact differentials. Without exerting much brainpower, one will quickly arrive at a brilliant result." Doing this, $\begin{eqnarray} Q=U+W&\Rightarrow& dQ=dU+PdV\\ U=U(T,V)&\Rightarrow& dU=\left.\partial_T U\right|_T +\left.\partial_V U\right|_V\\ PV=nRT&\Rightarrow&PdV+VdP=nRdT \end{eqnarray}$ Plugging in the first law of thermodynamics into the $U$ equation of state, one gets $dQ=\left.\partial_T U\right|_T+\left(\left.\partial_T U\right|_V+P\right)dV=\left.\partial_T U\right|_T+PdV$, where the last simplification is made by remembering the fact that the internal energy of an ideal gas depends only on temperature. (Taking the derivative with respect to T at constant volume, one gets $\left.\frac{dQ}{dT}\right|_V=\left.\partial_T U\right|_T=C_V$.) Plugging in the simplified result for $dQ=...$ into the third equation of state, the ideal gas equation, one gets: $dQ-C_vdT+VdP=nRdT$. Taking the derivative at constant pressure, one gets: $\left.\frac{dQ}{dT}\right|_P=C_V+nR$ So, one sees that it is the ideal gas equation that makes the final difference. The work of an ideal gas changes when temperature is varied. Click here to jump to the problem!
1 Click here to jump to the problem! GR8677 #66 Thermodynamics$\Rightarrow$}First Law Recall that $dQ=TdS=PdV+dU$, where work done by the system is positive and heat input into the system is positive. For constant volume, the equation becomes $dQ=TdS=dU$. $\Rightarrow T=\left(\frac{\partial U}{\partial S}\right)_V$. The reciprocol gives the right answer. Click here to jump to the problem!
2 Click here to jump to the problem! GR8677 #95 Thermodynamics$\Rightarrow$}Carnot Cycle The Carnot cycle is the cycle of the most efficient engine, which does NOT have $e=1$ (unless $T_1=0$), but rather, $dS=0$---this means that entropy stays constant. Choice (C) is thus false. The efficiency of the Carnot cycle is dependent on the temperature of the hot and cold reservoir. The hot reservoir has decreasing entropy because it gets cooler as the cycle proceeds. From writing down the thermodynamic relations for isothermal and adiabatic paths and matching P-V boundary conditions, one can determine that $Q_1/Q_2 = T_1/T_2$. The efficiency is thus $e=\frac{Q_2-Q_1}{Q_2}=1-Q_1/1_2 = 1-T_1/T_2$. Click here to jump to the problem!
3 Click here to jump to the problem! GR8677 #13 Thermodynamics$\Rightarrow$}Heat Given $P=100$ W and $V=1L=1m^3=1kg$ for water, one can chunk out the specific heat equation for heat, $Q=mc\Delta T=Pt\Rightarrow 4200(1^\circ) = 100t\Rightarrow t \approx 40 s$, as in choice (B). Click here to jump to the problem!
4 Click here to jump to the problem! GR8677 #14 Thermodynamics$\Rightarrow$}Heat The final temperature is $50^\circ C$. The heat exchanged from the hot block to the cool block is $Q=mc\Delta T = 5 kcal$, as in choice (D). Click here to jump to the problem!
5 Click here to jump to the problem! GR8677 #15 Thermodynamics$\Rightarrow$}Phase Diagram Recall that for an ideal gas $U=C_v \Delta T$ and $PV=nRT$. Don't forget the first law of thermodynamics, $Q=W+U$. For $A\rightarrow B$, $U=0$, since the temperature is constant. Thus, $Q=W=RT_H\ln V_2/V_1$. For $B\rightarrow C$, $W=P_2(V_1-V_2)=R(T_c-T_h)$. $U=C_v (T_c - T_h)$, and thus $Q=W+U=C_v (T_c - T_h)-R(T_h-T_c)$. For $C\rightarrow A$, $W=0$, $U=C_v(T_h-T_c)$, thus $Q=U=C_v(T_h-T_c)$. Add up all the Q's from above, cancel the $C_v$ term, to get $Q_tot=RT_h\ln(V_2/V_1)-R(T_h-T_c)$, as in choice (E).s Click here to jump to the problem!
6 Click here to jump to the problem! GR8677 #16 Thermodynamics$\Rightarrow$}Mean Free Path Air is obviously less dense than the atomic radius $10^{-10}$, thus choices (C), (D), and (E) are out. Air is not dilute enough that the distance between particles is actually within human visible range, as in (A)! Thus, the answer must be (B). (Note how this problem exemplifies the usefulness of common sense.) Click here to jump to the problem!
7 Click here to jump to the problem! GR8677 #73 Thermodynamics$\Rightarrow$}Adiabatic Work One should recall the expression for work done by an ideal gas in an adiabatic process. But, if not, one can easily derive it from the condition given in the problem, viz., $PV^\gamma =C \Rightarrow P=C/V^\gamma$. Recall that the definition of work is $W=\int PdV =\int_{V_1}^{V_2} C dV/V^\gamma =\left.-\frac{1}{\gamma-1} C/V^{\gamma -1} \right|_{V_1}^{V_2}$, which when one plugs in the endpoint limits, becomes choice (C). Click here to jump to the problem!
8 Click here to jump to the problem! GR8677 #74 Thermodynamics$\Rightarrow$}Entropy Recall the definition of entropy to be $dS = dQ/T$. The heat is defined here as $dQ = m c dT$, and thus $S = \int mcdT/T$. One is given two bodies of the same mass. One mass is at $T_1=500$ and the other is at $T_2=100$ before they're placed next to each other. When they're put next to each other, one has the net heat transferred being 0, thus $Q_1 = -Q_2 \Rightarrow T_f = (T_1+T_2)/2=300$. The entropy is thus $S = \int^{T_f}_{T_1} mcdT/T + \int^{T_f}_{T_2} mcdT/T = mc \left ( \ln(3/5) + \ln(3) \right) = 2mc \ln 3 - mc \ln 5 = mc(\ln 9 - \ln 5) = mc \ln(9/5)$, as in choice (B). Click here to jump to the problem!
9 Click here to jump to the problem! GR8677 #75 Thermodynamics$\Rightarrow$}Fourier's Law Recall Fourier's Law $q = -k\nabla T$, where $q$ is the heat flux vector (rate of heat flowing through a unit area) and $T$ is the temperature and $k$ is the thermal conductivity. (One can also derive it from dimensional analysis, knowing that the energy flux has dimensions of $J/(s m^2)$) Fourier's Law implies the following simplification: $q = -k \frac{\Delta T}{\Delta l}$ The problem wants the ratio of heat flows $q_A/q_B=\frac{k_A l_B}{k_B l_A}=\frac{0.8 \times 2}{0.025 \times 4}=32/2=16$, as in choice (D). (The problem gives $l_A = 4$, $l_B=2$, and $k_A=0.8$, $k_B=0.025$.) Click here to jump to the problem!
10 Click here to jump to the problem! GR8677 #91 Thermodynamics$\Rightarrow$}Second Law The Second Law of thermodynamics has to do with entropy; that entropy can never decrease in the universe. One form of it states that from hot to cold things flow. A cooler body can thus never heat a hotter body. Since the oven is at a much lower temperature than the wanted sample temperature, the oven can only heat the sample to a maximum of 600K without violating the Second Law. (This solution is due to David Latchman.) (Also, since the exam is presumably written by theorists, one can narrow down the choices to either (D) or (E), since the typical theorist's stereotype of experimenters usually involves experimenters attempting to violate existing laws of physics---usually due to naivity.) Click here to jump to the problem!
11 Click here to jump to the problem! GR8677 #5 Thermodynamics$\Rightarrow$}Degree of Freedom Acording to the Equipartition Theorem, there is a $kt/2$ contribution to the energy from each degree of quadratic freedom in the Hamiltonian. In equation form, the average total energy is $\bar{E} = s (kt/2)$, where s is the degrees of freedom. For a n-dimensional 1-particle system, the Hamiltonian is $H \propto \sum_i p_i^2 + \sum_i q_i^2$, where $p_i$ ($q_i$) refer, respectively, to the $i^{th}$ component of momentum (position). Thus, for a 3-dimensional 1-particle system, one has 6 quadratic terms in the Hamiltonian, as in $H \propto p_1^2+p_2^2 + p_3^2 + q_1^2 + q_2^2 + q_3^2$. Plugging in s=6, one finds that the average energy is 3kT. (This revised solution is due to the GREPhysics.NET user kolndom.) Click here to jump to the problem!
12 Click here to jump to the problem! GR8677 #6 Thermodynamics$\Rightarrow$}Work The work done by an adiabatic process is $W=-\frac{1}{\gamma - 1}(P_2V_2-P_1V_1)$. (One can quickly derive this from noting that $PV^\gamma = const$ in an adiabatic process and $W=\int P dV$.) The work done by an isothermal process is $W=nRT_iln(V_2/V_1)=P_iV_iln(V_2/V_1)$ (One can quickly derive this from noting that $P_2V_2=P_2V_2=nRT_1=nRT_2$ for an isothermal process.) From the above formulae, one can immediately eliminate choice (A). One can calculate the isothermal work to be $W_i = nRT_1 ln 2 = P_1V_1 ln 2$. One can calculate the adiabatic work to be: $W_a = \frac{1}{1-\gamma}(P_1-2P_2)V_1=\frac{1}{1-\gamma}(P_1-2/2^\gamma P_1)V_1 = \frac{1}{1-\gamma}(1-2^{1-\gamma})P_1V_1$. For a monatomic gas, $\gamma=5/3$, and one finds that $0 . Choice (E). (Also, in general, an adiabatic process always does less work than an isothermal process in a closed cycle.) Click here to jump to the problem!
13 Click here to jump to the problem! GR8677 #36 Thermodynamics$\Rightarrow$}Adiabatic In an adiabatic expansion, $dS=0$ (entropy), since $dQ=0$ (heat). No heat flows out, by definition, and thus choice (A) is out, as well as choice (B). Choice (C) is true since by the first law, one has $Q=U+W \Rightarrow U=-W$, and the given integral is just the definition of work. Choice (D) defines work. Choice (E) remains---so take that. Click here to jump to the problem!
14 Click here to jump to the problem! GR8677 #37 Thermodynamics$\Rightarrow$}Cycle Analysis For path C to A, one has $W=0$ for an isochoric (constant volume) process. For path A to B, one has just $W=P\Delta V = 200(V_B-2)$ for an isobaric (constant pressure) process. For path B to C, one has $W=P_c V_c ln(V_C/V_B) = 1000 ln(2/V_b)$. One can figure out $V_B$ from the isothermal condition $P_C V_C = P_B V_B \rightarrow V_B = P_C/P_B V_C = 5/2 \times 2 = 5$. Plug that in above to get, $W_{CA}=0$ $W_{AB}=600$ $W_{BC}=1000 ln(2/5) > -1000$ $\sum W \approx -400$, which is closest to choice (D). Click here to jump to the problem!
15 Click here to jump to the problem! GR8677 #47 Thermodynamics$\Rightarrow$}Entropy Entropy is given as $dS=\int dQ/T$. Since the volume expands by a factor of 2, the work in the isothermal process is $W=nRTln(V_2/V_1)=nRTln2$. But, for an ideal gas, the internal energy change in an isothermal process is 0, thus from the first law of Thermodynamics, one has $dQ=dW+dU \Rightarrow dQ=dW$. The temperature cancels out in the entropy integral, and thus the entropy is just $nR ln2$, as in choice (B). Click here to jump to the problem!
16 Click here to jump to the problem! GR8677 #48 Thermodynamics$\Rightarrow$}RMS Speed In case one forgets the RMS speed, one does not need to go through the formalism of deriving it with the Maxwell-Boltzmann distribution. Instead, one can approximate its dependence on mass and temperature by $3/2 kT = 1/2 m v^2$. One thus has $v \propto \sqrt{kT/m}$. For the ratio of velocities, one has $\frac{v_N}{v_O}=\frac{m_O}{m_N}$. Plug in the given molecular masses for Oxygen and Nitrogen to get choice (C). Incidentally, the trick for memorizing the diatomic gasses is Have No Fear Of Ice Cold Beer (Hydrogen, Nitrogen, Florine, Oxygen, Iodine, Chlorine, Bromine). Click here to jump to the problem!
17 Click here to jump to the problem! GR8677 #16 Thermodynamics$\Rightarrow$}Carnot Engine Recall the common-sense definition of the efficiency $e$ of an engine, $e=\frac{W_{accomplished}}{Q_{input}},$ where one can deduce from the requirements of a Carnot process (i.e., two adiabats and two isotherms), that it simplifies to $e=1-\frac{T_{low}}{T_{high}}$ for Carnot engines, i.e., engines of maximum possible efficiency. ($Q_{input}$ is heat put into the system to get stuff going, $W$is work done by the system and $T_{low}$ ($T_{high}$) is the isotherm of the Carnot cycle at lower (higher) temperature.) The efficiency of the Carnot engine is thus $e=1-\frac{800}{1000}=0.2$, where one needs to convert the given temperatures to Kelvin units. (As a general rule, most engines have efficiencies lower than this.) The heat input in the system is $Q_{input}=2000J$, and thus $W_{accomplished}=400 J$, as in choice (A). Click here to jump to the problem!
18 Click here to jump to the problem! GR8677 #46 Thermodynamics$\Rightarrow$}Critical Isotherm The critical isotherm is the (constant temperature) line that just touches the critical liquid-vapor region, explained in the next question. The condition for the critical isotherm is $\left(\frac{dP}{dV}\right)_c=0$ and $\left(\frac{d^2P}{dV^2}\right)_c=0$, where c denotes the critical point. Click here to jump to the problem!
19 Click here to jump to the problem! GR8677 #47 Thermodynamics$\Rightarrow$}Liquid-Vapor Equilibrium The liquid-vapor region is where the substance can coexist as both a liquid and vapor. (A gas is just a vapor at normal temperatures.) In this region, the liquid and vapor are in equilibrium, hence their coexistence. Equilibrium occurs when $P_v=P_l$ and $\mu_v=\mu_l$, i.e., when the pressure and chemical potential of the liquid and vapor are equal to each other. Since region B shows a constant pressure behavior, despite the volume-decrease, it is the region of liquid-vapor equilibrium. Click here to jump to the problem!
20 Click here to jump to the problem! GR8677 #62 Thermodynamics$\Rightarrow$}Work The work done by a gas in an isothermal expansion is related to the log of the volumes. If one forgets this, one can quickly derive it from recalling the definition of work $W=\int P dV$ and the ideal gas law equation of state $PV=nRT \Rightarrow P=nRT/V$. One has $W=\int nRTdV/V = nRT ln(V_1/V_0)$. For 1 mole, one has $n=1$, which yields choice (E). (And the condition for isothermality $P_1V_1=P_0V_0=nRT_1=nRT_0$ allows one to change the argument in the log.) Click here to jump to the problem!
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Jarvis and Numbers
Tag(s):
## Very-Easy, Very-Easy
Problem
Editorial
Analytics
After defeating Mandarin in Iron man 3, Jarvis is most of time free and now he has started playing with numbers. Jarvis being an AI (Artificial Intelligence) solves almost all the riddle given by Tony Stark very quickly. This time Tony gave him a problem about base conversion’s but the problem statement given by Tony seems to be confusing and Jarvis asked for Help! Problem Statement – “123 when converted in base 16, it consist of two digits 7 and 11 so the sum of the numbers is 18,(For Given N) Find denominator of average(irreducible form) of sum of all the numbers formed on conversions from base 2 to N-1”.
Can you help Jarvis?
Input:
First line will have the number of test case (t) and then t subsequent lines will contain a number $N$.
Output:
For every test case, give answer in new line.
Constraints:
$T < 10$
$3 <= N <= 1000$
SAMPLE INPUT
2
5
7
SAMPLE OUTPUT
3
1
Explanation
1 st test case -> average on converting 5 to different bases is = 7/3 so answer to the problem is 3
2nd test case -> 3/1 so answer is 1
Time Limit: 1.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme: Marks are awarded when all the testcases pass.
Allowed Languages: C, C++, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Scala 2.11.8, Swift, Visual Basic
## CODE EDITOR
Initializing Code Editor...
## This Problem was Asked in
Challenge Name
Beginner Practice Round 1.0
OTHER PROBLEMS OF THIS CHALLENGE
• Basic Programming > Bit Manipulation
• Basic Programming > Implementation
• Math > Basic Math
• Basic Programming > Implementation
• Algorithms > Sorting
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<lift:loc locid="stock.discuss"></lift:loc>
Smooth compactness of self-shrinkers
2009.07.15
http://arxiv.org/abs/0907.2594
We prove a smooth compactness theorem for the space of embedded self-shrinkers in $\RR^3$. Since self-shrinkers model singularities in mean curvature flow, this theorem can be thought of as a compactness result for the space of all singularities and it plays an important role in studying generic mean curvature flow.
Discussions
• Pls. be polite and constructive.
• You can input La|TeX for math formulas. E.g. $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$
• Any attachment files should still be uploaded to arXiv.org
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wx.lib.pubsub.core.listenerbase.ListenerBase¶
Base class for listeners, ie. callables subscribed to pubsub.
Class Hierarchy¶
Inheritance diagram for class ListenerBase:
Methods Summary¶
__init__ Use callable_ as a listener of topicName. The argsInfo is the getCallable Get the listener that was given at initialization. Note that isDead Return True if this listener died (has been garbage collected) module Get the module in which the callable was defined. name Return a human readable name for listener, based on the typeName Get a type name for the listener. This is a class name or wantsAllMessageData True if this listener wants all message data: it has a **kwargs argument wantsTopicObjOnCall True if this listener wants topic object: it has a arg=pub.AUTO_TOPIC
Class API¶
class ListenerBase
Base class for listeners, ie. callables subscribed to pubsub.
Methods¶
__init__(self, callable_, argsInfo, onDead=None)
Use callable_ as a listener of topicName. The argsInfo is the return value from a Validator, ie an instance of callables.CallArgsInfo. If given, the onDead will be called with self as parameter, if/when callable_ gets garbage collected (callable_ is held only by weak reference).
getCallable(self)
Get the listener that was given at initialization. Note that this could be None if it has been garbage collected (e.g. if it was created as a wrapper of some other callable, and not stored locally).
isDead(self)
Return True if this listener died (has been garbage collected)
module(self)
Get the module in which the callable was defined.
name(self)
Return a human readable name for listener, based on the listener’s type name and its id (as obtained from id(listener)). If caller just needs name based on type info, specify instance=False. Note that the listener’s id() was saved at construction time (since it may get garbage collected at any time) so the return value of name() is not necessarily unique if the callable has died (because id’s can be re-used after garbage collection).
typeName(self)
Get a type name for the listener. This is a class name or function name, as appropriate.
wantsAllMessageData(self)
True if this listener wants all message data: it has a **kwargs argument
wantsTopicObjOnCall(self)
True if this listener wants topic object: it has a arg=pub.AUTO_TOPIC
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M00-03
Author Message
TAGS:
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 50730
Show Tags
08 Oct 2017, 07:37
mackhandelwal wrote:
Hi,
If I solve the question by using method :
2 apprentices-2*3/4*5=7.5 hours
2 trainees-2*(1/5)*5=2 hours
1 worker-5 hours
so,(1/7.5 +1/2+1/5) * x=1
so comes out to be 1 hr 12 min.
If 1 apprentices can work $$\frac{3}{4}$$ as fast as a qualified worker and if a qualified worker needs 5 hours, then 1 apprentice will need 4/3 as much time, so 20/3 hours, thus 2 apprentices, will need half that time, 20/6 hours.
If 1 trainee can work $$\frac{1}{5}$$ as fast as a qualified worker and if a qualified worker needs 5 hours, then 1 trainee will need 5 times as much time, so 25 hours, thus 2 trainees, will need half that time, 25/2 hours.
(1/5 + 6/20 + 2/25)*x = 1
x = 50/29.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50730
Show Tags
08 Oct 2017, 07:41
Bunuel wrote:
mackhandelwal wrote:
Hi,
If I solve the question by using method :
2 apprentices-2*3/4*5=7.5 hours
2 trainees-2*(1/5)*5=2 hours
1 worker-5 hours
so,(1/7.5 +1/2+1/5) * x=1
so comes out to be 1 hr 12 min.
If 1 apprentices can work $$\frac{3}{4}$$ as fast as a qualified worker and if a qualified worker needs 5 hours, then 1 apprentice will need 4/3 as much time, so 20/3 hours, thus 2 apprentices, will need half that time, 20/6 hours.
If 1 trainee can work $$\frac{1}{5}$$ as fast as a qualified worker and if a qualified worker needs 5 hours, then 1 trainee will need 5 times as much time, so 25 hours, thus 2 trainees, will need half that time, 25/2 hours.
(1/5 + 6/20 + 2/25)*x = 1
x = 50/29.
17. Work/Rate Problems
For more check:
ALL YOU NEED FOR QUANT ! ! !
Hope it helps.
_________________
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Joined: 08 Oct 2017
Posts: 1
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GPA: 4
Show Tags
08 Apr 2018, 00:56
Bunuel
I solved the question using the below approach
Let A be the qualified worker, B be the apprentice and C be the trainees
A does 20% of the work in 1 hour as he completes the work in 5hours.
B does 15% of work in 1 hour. So 2B does 30% of work in 1 hour
C does 24%of work in 1 hour. So 2C does 48% of work in 1 hour.
Therefore in 1 hour they all combined can complete 20 + 30 + 48 = 98% of the work.
Using this approach the I answer that I would get is nowhere close to the answer choices.
Re: M00-03 &nbs [#permalink] 08 Apr 2018, 00:56
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10 months ago
## Electronic Devices And Circuit Theory Boylestad 10 weihnachtszeit regga
Electronic Devices And Circuit Theory Boylestad 10th Edition
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analysis of transport phenomena william m deen solution manual pdf
Ed Sheeran Plus Deluxe Edition 2011 PLAN9.zip
Al Green - Definitive (Greatest Hits)
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Question: How to change the text size for the names of ColAttributes
0
17 days ago by
researcher0 wrote:
I am using the following command:
> plot(DBdata,cexRow=2,cexCol=2,dendrogram='none',lhei = c(0.5,6),lwid = c(0.5,4),
ColAttributes=c(DBA_TREATMENT,DBA_CONDITION),key=F,
colSideCols=list(c("lightpink","skyblue"),c("chocolate","darkgrey","darkorchid4")))
but unable to control the font size for the ColAttributes in the annotation bar. Any help or suggestion will be highly registration.
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# Applications of Brouwer's fixed point theorem
I'm presenting Brouwer's fixed point theorem to an audience that knows some point-set topology. Does anyone have any zippy / enlightening / cool applications or consequences of it? So far, I have:
• Physical realizations: stuff involving maps of my city, crumpled pieces of graph paper, a stationary gin molecule in a cocktail shaker, etc.
• That every n*n real matrix with all-positive entries has a positive eigenvalue.
Thanks! :)
-
I like the gin thing. Maybe that's something I'll mention next time I try to explain to a relative what topology is. :-) – Saul Glasman Mar 25 '10 at 11:37
Nitpick: there doesn't have to be a stationary gin molecule, just one which ends up at the same point where it starts. – Pete L. Clark Mar 25 '10 at 15:07
I got a negative critics of my book Matrices (Springer GTM 216) because I used Brouwer FPT to prove that every $n\times n$ real matrix with non-negative entries has a non-negative eigenvalue (its spectral radius). Yet I intended only to illustrate the powerness of Brouver's FPT. The reviewer took it too seriously. – Denis Serre Jan 24 '11 at 15:36
I wonder if this should be community wiki. It's asking for a list of answers after all, and it's unclear if there can be a "best" answer. – David White Sep 22 '11 at 13:24
The theorem is equivalent to the determinacy of the Hex game. That's a very famous application'.
The details can be found in [David Gale (1979). "The Game of Hex and Brouwer Fixed-Point Theorem". The American Mathematical Monthly 86: 818–827], a beautiful paper, which JSTOR serves at http://www.jstor.org/stable/2320146.
-
Mariano, do you really mean equivalent here? I would interpret that to mean that over a very weak base theory, as in Reverse Mathematics, you can prove Brouwer's theorem just from the assumption that Hex is determined. This seems unlikely for any finite instantiation of Hex, since the determinacy of finite games amounts to De Morgan's rules of logic. Perhaps you are referring to some kind of continuous analogue of Hex? Could you explain? – Joel David Hamkins Mar 25 '10 at 18:20
It is a much less technical sense of 'equivalent', I guess... If you know the truth of one of the statements, the truth of the other follows easily'. – Mariano Suárez-Alvarez Mar 25 '10 at 19:11
"determinacy" is not the right word to use here; the relevant fact is that Hex cannot end in a draw (a "topological" fact; any way to assign half the cells to each player gives at least one player a winning path). – Sridhar Ramesh Mar 25 '10 at 22:07
@Joel: This is one area where reverse mathematics as it is currently set up does not quite capture the informal sense of "equivalent." Many people feel intuitively that Sperner's lemma and Brouwer's fixed-point theorem are "equivalent," in that the "tricky part" is the same and you can pass from one to the other via "straightforward" reasoning. However, Sperner's lemma is provable in $RCA_0$ while Brouwer's fixed-point theorem requires $WKL_0$. Very roughly speaking, you need greater logical strength to take a limit, but of course logical strength isn't the same as psychological difficulty. – Timothy Chow Jan 24 '11 at 16:09
Both Sperner's lemma and the non-draw in Hex are of course easier than Brouwer fixed point theorem since both posess elementary and short arguments. – Lennart Meier Jan 24 '11 at 22:18
The existence of mixed Nash equilibriums in multiplayer games can be proved using the Brouwer fixed point theorem. If you imagine that each players is trying to improve his results based on the current actions of his opponents, then there is some combination of strategies so that each player can't improve.
-
It has to be noted, though, tha Nash's original proof of the result was based on Kakutani's generalized fixed point theorem. – J. H. S. Mar 25 '10 at 5:37
Well, the first published proof used the Kakutani fixed point theorem, which was suggested to Nash by David Gale. However, he had a proof based on the Brouwer fixed point theorem before and the proof in Nash's thesis uses the Brouwer fixed point theorem. – Michael Greinecker Mar 25 '10 at 9:59
Existence of Nash equilibrium is, in fact, also equivalent to Brower's fixed point theorem. theoryclass.wordpress.com/2012/01/05/… – Henrique de Oliveira Nov 15 '12 at 5:54
In his Annals paper Nash gives a proof based on Brouwer's fixed point theorem. He says that this proof is a considerable improvement of the previous one published in PNAS, which was based on Kakutani's generalized fixed point theorem. – GH from MO Nov 15 '12 at 15:42
Just for fun you could give them the 2-dimensional version of the standing-in-a-train theorem. That theorem says that if you want to go to sleep while standing up in a train that goes along a perfectly straight track, there is some starting angle that will cause you not to fall over. (Proof: if the angle is almost all the way forwards, then you will fall forwards; if it is almost all the way backwards, then you will fall backwards; by the intermediate value theorem there must be an angle that leaves you still standing at the end.) Ian Stewart argues vigorously that the implicit continuity assumption (that your final position depends continuously on your initial position) is wrong, but you can just forget about that.
Now let's suppose that you are standing on a surface that can move horizontally in any direction. (Perhaps you are in a boat, say.) This time if you fall, your head will be somewhere in a circle centre your feet and radius your height. Assuming that the position you end up in depends continuously on your starting position, this defines for you a continuous map of the disc that preserves the boundary. By Brouwer it is not a retraction, so there must be a starting position that stops you falling over.
Even if the continuity assumption is not in the end justified, I think this is a good and amusing illustration of the theorem.
-
As a bonus: this is a good illustration of the importance of the continuity assumption You're free to believe or disbelieve the existence of this magical position depending on whether you buy the continuity of the "map". – Thierry Zell Jan 24 '11 at 15:58
belated link for those curious as to what Stewart has said: dx.doi.org/10.1007/s00283-010-9189-9 (subscription probably required). I don't know if this is the last word on the matter, though – Yemon Choi Mar 9 '11 at 22:58
The Brouwer Theorem can be used to prove that a mapping of ${\bf R}^n$ to itself that has bounded displacement, in the sense that any point is moved at most a fixed amount from its original location, is onto. This seems be a folklore result. I wonder if anyone has a reference for it.
-
In economics and game theory, it's used to prove the existence of equilibrium (eg Nash equilibrium, general equilibrium)
-
This is a really important example: almost the whole of modern macroeconomic theory rests on it. – Tom Smith Mar 25 '10 at 16:23
By the fixed point theorem, one can prove that a polynomial with complex coefficients has at least a root in complex plane. It is not really cool but fits your audience.
-
The Fundamental Theorem of Algebra is cool by any sensible definition of cool! – Mariano Suárez-Alvarez Mar 25 '10 at 19:48
Cool as all get-out! – user4893 Mar 25 '10 at 20:09
@7-adic: I'm sure one can prove the the FTA using Brouwer (or Poincaré-Miranda) somehow, but I have yet to see a (correct) proof (there are famous incorrect derivations) - could you supply a reference ? Kind regards, Stephan F. Kroneck. – Stephan F. Kroneck Jan 5 '12 at 9:44
The following is not exactly an application, but rather a funny picture to illustrate the theorem, precisely in the form of the non-retraction theorem.
Suppose a shark jumps into a shoal of fish (a kind of big ball). The small fishes start escaping in all directions towards the border of the shoal, where the fishes stand still. Yet they escape with a certain disposition to follow a continuous flow, as they usually do, since everybody tends to follow its neighborhoods. But since there is no continuous retraction to the boundary, somebody doesn't know where to go, and stay there for a moment, much to the shark's satisfaction. There is also a 2D version, with a wolf entering into a herd of sheeps. This is just a funny picture, though I like to think that there is some truth in it.
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In many nonlinear equations, the existence of a solution (but not its uniqueness) follows from a topological argument in the vein of BFP Theorem. The BFP is at work especially when the equation is posed in some finite dimensional vector space, and you can establish an a priori estimate of the size of the solution. This means that you know a ball $B_R$ containing all the solutions.
The most important example of this situation is the stationary Navier-Stokes equation, with Dirichlet condition $u=0$ on the boundary of the domain. Of course, the ambient space is infinite dimensional, so you first establish the existence of an approximate solution in a subspace of dimension $n$ (Galerkin procedure); this is where you use the BFP Theorem, or its equivalent form that a continuous vector field over $B_R$ which is outgoing on $\partial B_R$ must vanish somewhere. Then passing to the limit as $n\rightarrow\infty$ is pedestrian.
The BFP Theorem is a consequence of the fact that the Euler-Poincaré characteristic of the ball is non-zero. There are counterparts when you work on a compact manifold (with boundary) whose EPC is non-zero. This happened to me in a very interesting way. I considered the free fall of a rigid body in water filling the entire space. The mathematical problem is a coupling between Navier-Stokes and the Euler equation for the top. I looked at a permanent regime, in which the solid body has a time-independent velocity field, and that of the fluid is time-independent as well, once you consider it in the moving frame attached to the solid. The difficulty is that you don't know a priori the direction of the vertical axis (the direction of gravity) in this frame. After a Galerkin procedure, the problem reduces to the search of a zero of a tangent vector field over $B_R\times S^2$. This vector field is outgoing on the boundary $\partial B_R\times S^2$. Because $$EP(B_R\times S^2)=EP(B_R)\cdot EP(S^2)=1\cdot2\ne0,$$ such a zero exists. Therefore the permanent regime does exist. Remark that because $EP=2$, we even expect an even number of solutions when counting multiplicities, at least at each level of the Galerkin approximation.
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The Schauder fixed point theorem can be proved using the Brouwer fixed point theorem. It says that if $K$ is a convex subset of a Banach space (or more generally: topological vector space) $V$ and $T$ is a continuous map of $K$ into itself such that $T(K)$ is contained in a compact subset of $K$, then $T$ has a fixed point.
The Schauder fixed point theorem in its turn is an important tool for existence proofs in differential equations. One easy application is the Peano existence theorem, but there are also more sophisticated examples.
-
caveat: The Schauder theorem for arbitrary (hausdorff!) topological vector spaces is hard compared to the case of Banach spaces or even locally convex TVS. The general case was only proven in 2002 by Cauty. But of course the locally convex case is the interesting one for applications. – Johannes Hahn Jan 25 '11 at 12:55
One standard consequence of Brouwer's theorem is Borsuk's antipodal mapping theorem, which in turn is used to prove that if $E$, $F$ are subspaces of a normed space and the dimension of $E$ is strictly less than the dimension of $F$, then there is a unit vector in $F$ whose distance to $E$ is one. This result is used often in Banach space theory.
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@Bill Johnson: That Borsuk's antipodal theorem implies Brouwer's is folklore (cp. Granas' monograph), but how does one argue in the other direction ? I'd be very interested to see a proof ! Kind regards, Stephan F. Kroneck. – Stephan F. Kroneck Jan 5 '12 at 9:40
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It can be used to show that a matrix with positive entries has an eigenvector with positive entries: that shows the existence of an invariant measure in the theory of finite space Markov chains.
-
It is worth noting, however, that this can be proven in several more elementary ways. It is a consequence of linear programming duality / the separating hyperplane theorem, and it can also be proven by just iterating the Markov chain if one takes some care dealing with the case of periodic chains. The Brouwer proof is certainly shorter, though. – Noah Stein May 27 '10 at 12:21
Noah, can you (or anyone else) give references for the other proofs? Thanks. – Phil Isett Sep 16 '11 at 17:35
I really like the short proof of Jordan curve theorem using Brouwer’s fixed-point theorem given in these lecture notes.
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The proof of the fundamental theorem of algebra using the Brouwer fixed point theorem is given here:
B.H. Arnold, "A topological proof of the fundamental theorem of algebra" Amer. Math. Monthly , 56 (1949) pp. 465–466
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In fact, this proof is flawed, for it applies Brouwer fixed point theorem to a discontinuous function. For a correct proof of the fundamental theorem of algebra using the Brouwer fixed point theorem, see M. K. Fort, Jr, Some properties of continuous functions, Amer. Math. Monthly 59 (1952), p. 372-375. – ACL Feb 12 '13 at 21:37
Here is a web-link to Fort's article: oldweb.cecm.sfu.ca/personal/jborwein/Expbook/Manuscript/… – Todd Trimble Jun 5 '13 at 13:43
I've been thinking about a very similar issue: I was considering giving a talk about the Brouwer fixed point theorem to some math majors (but not necessarily ones very familiar with point-set topology).
There is an elementary proof which uses Sperner's lemma; see Michael Henle's book A Combinatorial Introduction to Topology for details. You can outline a proof of Sperner's lemma pretty quickly (induction on dimension, and dimension 1 is easy), and from that, you can wave your hands (or be more precise, depending on how much your audience knows about compactness) to get Brouwer's theorem. Since Sperner's lemma holds in higher dimensions, you can prove the fixed point theorem in higher dimensions, also.
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I gave a similar talk last fall. I did not use sperners lemma though, instead I just computed that $\pi_1(D^2)$ has one element and $\pi_1(S^1)$ has more than that. Then you use functoriality, which I had to explain. I also used $\mathbb{C}-0$ to model $S^1$. It worked reasonably well, and was not very formal. – Sean Tilson Jan 25 '11 at 2:25
Simple and example, for which You may perform demonstration here, is to use some piece of rubber balloon and a lamp.If You stretch it and show that it leaves a shadow on the table. Then leave rubber to fall on the table, it should drop within area where the shadow was dropped previously. There should be point which after falling ( and release of stretching) should remain in precisely the same position.
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The Perron-Frobenius theorem (in various degrees of generality) is an easy consequence of Brouwer's fixed-point theorem.
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# fncychap: Remove vertical space between heading and first element in toF and toT
I'm trying to remove the vertical spacing that occurs when using the fncychap package, but how to do this? I've already done it in the toc by writing the following:
\renewcommand\contentsname{Table of Contents}
\tableofcontents
The last line is the one that moves things back to place in toc. Is there a similar way to do this in lof and lot?
Don't use fncychap. All those styles can be generated easily by titlesec
\documentclass{book}
\usepackage{titlesec}
\titleformat{\chapter}[display]
{\normalfont\huge\filleft\bfseries}
{\titlerule[1pt]%
\vspace{1ex}%
\chaptertitlename\ \thechapter}
{20pt}
{\Huge}[\vspace{1ex}{\titlerule[1pt]}]
\titleformat{name=\chapter,numberless}[display]
{\normalfont\huge\filleft\bfseries}
{}
{0pt}
{\titlerule[1pt]
\vspace{1ex}%
\Huge}[\vspace{1ex}{\titlerule[1pt]}]
\titlespacing*{\chapter} {0pt}{20pt}{20pt} %% adjust these numbers
\titlespacing*{name=\chapter,numberless} {0pt}{20pt}{20pt} %% adjust these numbers
\begin{document}
\tableofcontents
\chapter{Introduction}
\end{document}
• Thank you for your answer, however I will not accept this for solving my problem, as it don't. I asked how to do it while using fncychap. :) – Nicolai Anton Lynnerup May 3 '15 at 19:15
• @NicolaiAntonLynnerup You are welcome :-) – user11232 May 3 '15 at 22:59
I will have to admit that I now agree with @HarishKumar. The solution to my problem was to skip the fncychap package and use titlesec instead.
Here's the code I've generated from @HarishKumar's example.
\usepackage{titlesec}
\titleformat{\chapter}[display]
{\normalfont\Large\filleft}
{\sc\chaptertitlename\ \Huge{\thechapter}\\%
\vspace{1.5cm}
\titlerule[1pt]}
{-20pt}
{\Large}[\vspace{2ex}{\titlerule[1pt]}]
\titleformat{name=\chapter,numberless}[display]
{\normalfont\Large\filleft}
{}
{0pt}
{\titlerule[1pt]
\vspace{2ex}%
\Large}[\vspace{2ex}{\titlerule[1pt]}]
\titlespacing*{\chapter} {0pt}{0pt}{40pt} %% adjust these numbers
\titlespacing*{name=\chapter,numberless} {0pt}{0pt}{40pt} %% adjust these numbers
|
{}
|
## ==> Sticky: User Created 70 Sample Questions with answers
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### ==> Sticky: User Created 70 Sample Questions with answers
Questions 1-70 in this thread are questions that imitate those of the real PGRE test. For those of you who want to practise only on ETS official questions there are directions on how to do this (the 5 practice tests etc.). The questions I post here periodically are for those who want to practise on PGRE questions beyond the official material. The questions in this thread are updated when possible!
----------------------------------------------------------------------------------------------------------------------------------------------------------------------
Notice:
ALL ANSWERS TO THE FOLLOWING SAMPLE QUESTIONS ARE GIVEN BELOW IN THIS THREAD (TBA answers will be given soon!)
----------------------------------------------------------------------------------------------------------------------------------------------------------------------
Time alloted: Use as much as you need for practice!
SAMPLE QUESTION 1:
Suppose $$\psi_1$$ and $$\psi_2$$ are the ground state wavefunctions of two single (non-interacting) potential wells #1 and #2 respectively. Consider the following wavefunctions representing the ground state of the system of the two aforementioned wells (now seen as a double well):
I. $$\frac{1}{\sqrt{2}}(\psi_1+\psi_2)$$
II. $$\frac{1}{2}(\psi_1 + \sqrt{3} \psi_2)$$
III. $$\frac{1}{\sqrt{2}}(\psi_1-\psi_2)$$
Given the above information which of the following statements is FALSE?
(A) Wavefunction (I) can represent the ground state of a system of two identical wells.
(B) Wavefunction (II) can represent the ground state of an asymmetric double well.
(C) Wavefunction (III) cannot represent the ground state of a symmetric double well.
(D) If the double well is described by wavefunction (II), then potential well #2 is deeper than potential well #1.
(E) If the double well is described by wavefunction (I) its fundamental energy level lies higher than the fundamental energy levels of the non-interacting wells.
SAMPLE QUESTION 2:
A beam of radioactive particles is measured as it shoots through a laboratory. It is found that, on average, each particle "lives" for a time of 20 ns. When at rest in the laboratory, the same particles "live" 7 ns on average. How fast do the particles in the beam move?
(A) 0.50c
(B) 0.88c
(C) 0.12c
(D) 0.65c
(E) 0.93c
{c = speed of light}
SAMPLE QUESTION 3:
A diffraction grating with a width of 2.0 cm contains 1000 lines/cm across that width. For an incident wavelength of 500 nm, what is the smallest wavelength difference this grating can resolve in the second order?
(A) 0.125 nm
(B) 0.25 nm
(C) 0.50 nm
(D) 8 nm
(E) 2 nm
SAMPLE QUESTION 4:
If the temperature $$T$$ of an ideal gas is increased at constant pressure, what happens to the mean free path of its molecules/atoms?
(A) It decreases in proportion to $$1/T$$.
(B) It decreases in proportion to $$1/T^2$$.
(C) It increases in proportion to $$T$$.
(D) It increases in proportion to $$T^2$$.
(E) It is not affected by the temperature change.
Last edited by physics_auth on Tue Jan 16, 2018 4:03 pm, edited 68 times in total.
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### SAMPLE QUESTIONS 5 - 11 AND HOW TO FIND ETS RELEASED SAMPLE
SAMPLE QUESTION 5:
A homogeneous disk of mass M = 2 kg and radius R is rotated about an axis perpendicular to its plane that passes through its center. Initially, the disk rotates at an angular speed of 40 rad/sec. At time t = 0, sand starts dropping uniformly onto the area of the disk at a rate of 0.2 kg/sec. By how much did the angular speed of the system (disk + sand) change in the time interval from 10 to 30 secs?
(A) It decreased by 30 rad/s.
(B) It decreased by 20 rad/s.
(C) It decreased by 10 rad/s.
(D) It increased by 10 rad/s.
(E) It increased by 20 rad/s.
SAMPLE QUESTION 6:
The work function for sodium is 2.28 eV. A portion of sodium is irradiated separately and successively by the following three types of electromagnetic radiation
I. ultraviolet radiation
II. blue optical radiation
III. infrared radiation
For which radiation or radiations does the photoelectric effect take place?
(A) I and II only
(B) I only
(C) II and III only
(D) I, II and III
(E) III only
SAMPLE QUESTION 7:
An atom transits from an excited state to the ground state emitting a photon of energy 4.7 eV. The lifetime of the excited state is 0.1 picosec. What is approximately (i.e. the order of magnitude of) the spectral line width Δν/ν of the photon? v stands for frequency.
(A) $$10^{-21}$$
(B) 0.001
(C) 1
(D) 10
(E) 100
SAMPLE QUESTION 8:
A block of mass M is attached to the free end of a horizontal spring of stiffness constant k (the other end of the spring is fixed to a vertical wall). A projectile of mass m (m < M) moves horizontally and directs ahead to the block approaching it at a speed u. The projectile is embedded into the block. Friction from the floor is negligible. What is the amplitude A of the oscillation of the system projectile-block?
(A) $$A= \frac{mu}{M+m} \cdot \sqrt{\frac{M}{k}}$$
(B) $$A = \frac{mu}{M-m} \cdot \sqrt{\frac{M+m}{k}}$$
(C) $$A = \frac{Mu}{\sqrt{k(M+m)}}$$
(D) $$A = \frac{(M-m)u}{\sqrt{k(M+m)}}$$
(E) $$A = \frac{mu}{\sqrt{k(M+m)}}$$
SAMPLE QUESTION 9:
In some experiment, we have a velocity selector whose disks are separated by 0.5 m and its transmission axes (slits) make a relative angle of π rad. It is found that molecules pass through the selector when it turns at a rate of 600 rev/sec. What is the maximum speed of a molecule that passes the selector?
(A) 300/π m/s
(B) 100/π m/s
(C) 600 m/s
(D) 300 m/s
(E) 600/π m/s
SAMPLE QUESTION 10:
If the internal electrons of lithium atom (Z = 3) are considered to screen fully the nucleus from the outer (valence) electron, then the ionization work of this atom will be most nearly equal to:
(A) 13.6 eV
(B) 1.5 eV
(C) 54.4 eV
(D) 3.4 eV
(E) 17 eV
SAMPLE QUESTION 11:
In a series of experiments, three unpolarized beams of neutral atoms in their ground state are sent consecutively through a Stern-Gerlach apparatus. The first beam consists of sodium atoms, and it is known that sodium belongs to the third row and the first column of the periodic table of elements. The second beam consists of magnesium atoms, and it is known that magnesium is next to sodium in the periodic table. Finally, the third beam consists of nitrogen atoms, and it is known that the ground state electron configuration of nitrogen is $${}^4\text{S}_{3/2}$$. What are the results that a detector in the output would record for each one of the three experiments?
______(Na atoms)________(Mg atoms)__________(N atoms)
(A)____ 1 beam __________ 1 beam ___________ 3 beams
(B)____ 2 beams _________ 2 beams __________ 3 beams
(C)____ 2 beams _________ 1 beam ___________ 4 beams
(D)____ 2 beams _________ 3 beams __________ 4 beams
(E)____ 2 beams _________ 1 beam ___________ 1 beam
Last edited by physics_auth on Tue Jan 16, 2018 4:50 am, edited 45 times in total.
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### PGRE SAMPLE QUESTIONS 12-15
SAMPLE QUESTION 12:
An object is 20 cm to the left of a lens of focal length +10 cm. A second lens, of focal length +12.5 cm, is 30 cm to the right of the first lens. What is the distance between the original object and the final image?
(A) 28 cm
(B) 50 cm
(C) 100 cm
(D) 0 cm
(E) $$\infty$$
SAMPLE QUESTION 13:
In a piece of metal at an equilibrium temperature of 0 Kelvin, what does the Fermi energy represent?
(A) The energy of the top of the valence band.
(B) The energy of the bottom of the conduction band.
(C) The mean thermal energy of the electrons.
(D) The highest energy that an electron of the metal can have.
(E) The energy gap between the top of the valence band and the bottom of the conduction band.
SAMPLE QUESTION 14:
A laboratory source of electromagnetic waves is placed opposite a perfect conducting reflecting surface of large dimensions. Between the source and the reflecting surface is placed a small receiver which can move along a straight line that is perpendicular to the reflecting surface. When the receiver is displaced by 15 cm, exactly ten rises and falls of the intensity are registered. What is the frequency of the electromagnetic waves?
(A) $$5 \cdot 10^9$$ Hz
(B) $$2 \cdot 10^9$$ HZ
(C) $$1 \cdot 10^{10}$$ Hz
(D) $$5\cdot 10^8$$ Hz
(E) $$1\cdot10^8$$ Hz
SAMPLE QUESTION 15:
The potential energy of quantum harmonic oscillator is given by $$V(x) = m\omega^2 x^2/2$$, where m is the mass and $$\omega$$ the angular frequency of the oscillator. What is the position uncertainty of a quantum harmonic oscillator in its first excited state?
(A) $$\Delta x = \sqrt\frac{3\hbar}{2m\omega}}{$$
(B) $$\Delta x = 0$$
(C) $$\Delta x = \sqrt\frac{\hbar}{m\omega}}{$$
(D) $$\Delta x = \sqrt\frac{\hbar}{2m\omega}}{$$
(Ε) $$\Delta x = \sqrt\frac{3\hbar}{4m\omega}}{$$
Last edited by physics_auth on Wed Jan 17, 2018 2:36 am, edited 15 times in total.
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### PGRE SAMPLE QUESTIONS 16 - 19
SAMPLE QUESTION 16:
A particle is in the second excited state of an infinite square well of width $$L$$. The probability of NOT finding the particle in the region $$\left[ \frac{L}{2},\frac{5L}{6}\right]$$ of the well is equal to:
(A) 2/3
(B) 3/4
(C) 1/3
(D) 1/2
(E) 1
SAMPLE QUESTION 17:
The number of particles emitted each minute by a radioactive source is recorded for a period of 10 hours. A total of 60 counts are registered. During how many 1-minute intervals, approximately, should we expect to observe no particles?
(A) 60 1-minute intervals
(B) 540 1-minute intervals
(C) 54 1-minute intervals
(D) 600 1-minute intervals
(E) 90 1-minute intervals
SAMPLE QUESTION 18:
A sodium atom (Na) consists of 11 electrons. What is the ground state electron configuration of doubly ionized sodium, i.e. of $$Na^{2+}$$?
(A) $$1s^2 2s^2 2p^5$$
(B) $$1s^2 2s^2 2p^6$$
(C) $$1s^2 2s^2 2p^6 3s^1$$
(D) $$1s^2 2s^2 2p^5 3s^2$$
(E) $$1s^2 2s^2 2p^6 3s^2$$
SAMPLE QUESTION 19:
After traveling a distance of 10cm in a material medium, the intensity of a beam of photons has decreased by 75%. The mean free path of the photons in this medium is most nearly equal to which of the following? (ln2 = 0.69)
(A) 0.029 cm
(B) 0.14 cm
(C) 7.2 cm
(D) 13.6 cm
(E) 34.8 cm
Last edited by physics_auth on Mon Jul 11, 2011 7:11 pm, edited 11 times in total.
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### PGRE SAMPLE QUESTIONS 20 -23
SAMPLE QUESTION 20:
Operator $$\hat{H}$$ which operates on a space of two states $$|1 \rangle$$ and $$|2\rangle$$ is given by the formula: $$\hat{H} = |1\rangle \langle 1| + |2\rangle \langle 2|- |1\rangle \langle 2|-|2\rangle \langle 1|$$. Assuming that $$\langle i | j \rangle = \delta_{ij}$$, where i,j = 1, 2, what are the eigenvalues of operator $$\hat{H}$$?
(A) $$\lambda_1= 0,\, \lambda_2 = 1$$
(B) $$\lambda_1= 0,\, \lambda_2 = 2$$
(C) $$\lambda_1= 1,\, \lambda_2 = 2$$
(D) $$\lambda_1= \lambda_2 = 0$$
(E) $$\lambda_1= \lambda_2 = 1$$
SAMPLE QUESTION 21:
The one end of a pendulum of length $$l$$ is fixed on the ceiling of an elevator (the elevator moves within the gravitational field of the Earth). The elevator moves upwards with an acceleration $$a=g/2$$ (where $$g$$ is the gravitational acceleration on the surface of the Earth). If the pendulum's motion is simple harmonic, what is the frequency of oscillation $$f$$?
(A) $$\frac{1}{2\pi}\sqrt{\frac{3g}{2l}}$$
(B) $$\frac{1}{2\pi}\sqrt{\frac{2g}{3l}}$$
(C) $$\frac{1}{2\pi}\sqrt{\frac{g}{l}}$$
(D) $$\frac{1}{2\pi}\sqrt{\frac{g}{2l}}$$
(E) $$\frac{1}{2\pi}\sqrt{\frac{2g}{l}}$$
SAMPLE QUESTION 22:
A point particle with charge +q is to be brought from far away to a point near an electric dipole. Suppose that the dipole is along the x-axis of an Oxyz coordinate system and that the positions of the charges of the dipole, -Q and +Q, are (-s/2,0,0) and (+s/2,0,0) respectively (i.e. s is the distance between the two charges). Where should the final position of the point particle be so that the net work done for its transference is equal to zero?
(A) On the axis of the dipole, on the segment from -s/2 to +s/2 (i.e. between the two charges).
(B) On the axis of the dipole, on the segment from -s/2 to -infinity.
(C) On the axis of the dipole, on the segment from +s/2 to +infinity.
(D) On a line that is perpendicular to the dipole moment and passes through the midpoint.
(E) On a line that makes an angle of 45 degrees with the dipole moment.
SAMPLE QUESTION 23:
When we keep a circular copper (ohmic) conductor at a constant temperature $$\theta_1$$, thermal energy is produced along it at a rate $$P_1$$ and the magnetic field at its center has magnitude $$B_1$$. Then, we heat the conductor to a higher constant temperature $$\theta_2$$ (i.e.$$\theta_2 > \theta_1$$). At temperature $$\theta_2$$, let $$P_2$$ and $$B_2$$ be the rate of dissipation of energy and the magnitude of the magnetic field at its center respectively. Which of the following is TRUE, if ALL other factors remain the same?
(A) It holds that $$B_2>B_1$$ and $$P_2 = P_1$$.
(B) It holds that $$B_2<B_1$$ and $$P_2 < P_1$$.
(C) It holds that $$B_2=B_1$$ and $$P_2 = P_1$$.
(D) It holds that $$B_2>B_1$$ and $$P_2 >P_1$$.
(E) It holds that $$B_2=B_1$$ and $$P_2 > P_1$$.
Last edited by physics_auth on Mon Jul 11, 2011 7:08 pm, edited 13 times in total.
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### 8 NEW SAMPLE PHYSICS GRE QUES. WITH THEIR ANSWERS (24-42)
SAMPLE QUESTION 24:
A neutral particle is at rest in a uniform magnetic field of magnitude B. At time t = 0 it decays into two charged particles each of mass m. The two particles move off in separate orbits, both of which lie in a plane perpendicular to the magnetic field. The charge of one of the particles is +q. What is going to happen at a future time t > 0?
(A) The particles will collide after a time interval equal to 2πm/qB.
(B) The particles will collide after a time interval equal to πm/qB.
(C) The particles will collide after a time interval equal to πm/2qB.
(D) The particles will recede from each other until they are infinitely apart.
(E) The particles will recede from each other, moving along a straight line and subsequently approach each other (moving along the same line) under the action of the mutual Coulomb attraction.
SAMPLE QUESTION 25:
A point charge +Q is placed at the vertex of a cube. What is the electric flux through the cube?
(A) 0
(B) +$$\frac{Q}{\epsilon_0}$$
(C) +$$\frac{Q}{2\epsilon_0}$$
(D) +$$\frac{Q}{6\epsilon_0}$$
(E) +$$\frac{Q}{8\epsilon_0}$$
SAMPLE QUESTION 26:
The mass density of a certain planet has spherical symmetry but varies in such a way that the mass inside every spherical surface with center at the center of the planet is proportional to the radius of the surface. If r is the distance from the center of the planet to a point mass inside the planet, the gravitational force on this mass is:
(A) not dependent on r
(B) proportional to $$r^2$$
(C) proportional to r
(D) proportional to $$1/r$$
(E) proportional to $$1/r^2$$
SAMPLE QUESTION 27: (This is a tough question - a more analytical answer is provided in the thread below)
A soap film immersed in air is illuminated by white light almost normal to its surface. The index of refraction of the film is 1.50. Wavelengths of 480 nm and 800 nm are ONLY intensified in the reflected beam. (No other wavelengths between them are intensified.) The thickness of the film is:
(A) 150 nm
(B) 240 nm
(C) 360 nm
(D) 400 nm
(E) 600 nm
SAMPLE QUESTION 28:
A certain nucleus, after absorbing a neutron, undergoes beta minus decay and then splits into two alpha particles. Which of the following can be the A and Z of the original nucleus?
(A) A = 6 _____ Z = 2
(B) A = 6 _____ Z = 3
(C) A = 7 _____ Z = 3
(D) A = 7 _____ Z = 2
(E) A = 8 _____ Z = 4
SAMPLE QUESTION 29:
The energy supplied by a thermal neutron in an induced fission event is essentially equal to:
(A) its rest energy
(B) its kinetic energy
(C) its binding energy to the nucleus which undergoes fission
(D) the total energy of the initial fission fragments
(E) the energy released during the fission process
SAMPLE QUESTION 30:
A baryon with strangeness 0 decays into two particles, one of which is a baryon with strangeness +1. Which of the following could be the other particle?
(A) a baryon with strangeness 0
(B) a baryon with strangeness -1
(C) a lepton
(D) a meson with strangeness +1
(E) a meson with strangeness -1
SAMPLE QUESTION 31:
Two bodies move with speeds 0.8c and 0.6c respectively. If the two bodies have the same rest mass, what is the ratio of their relativistic kinetic energies?
(A) 16/9
(B) 4/3
(C) 3/2
(D) 8/3
(E) 32/27
SAMPLE QUESTION 32:
A large transparent slab of uniform thickness and index of refraction $$n$$ is initially immersed in the air. A monochromatic beam of light is incident on the slab at an angle $$\theta$$ such that the reflected and the refracted beam emerge on mutually orthogonal directions. The previous experiment is then repeated (i.e. same angle of incidence for the aforementioned monochromatic beam), this time with the slab immersed in a liquid with index of refraction $$n_l$$. In terms of the angle of incidence $$\theta$$ and the refraction indices $$n$$ and $$n_l$$, what is the angle between the emergent reflected and refracted beams in the latter case?
(A) They are still orthogonal to each other.
(B) It is equal to $$$$\arcsin \left( {\frac{{{n_l}}}{{{n^2}}} \cdot \sqrt {1 + {n^2}} } \right) - \theta$$$$.
(C) It is equal to $$$$\arcsin \left( {{n_l }/\sqrt {1 + {n^2}} } \right) - \theta$$$$.
(D) It is equal to $$$$\arcsin \left( {\frac{{{n_l } \cdot n}}{{1 + {n^2}}}} \right) - \theta$$$$.
(E) It is equal to $$$$\arcsin \left( {\frac{{{n_l }}}{n} \cdot \sqrt {1 - {n^2}} } \right) - \theta$$$$.
SAMPLE QUESTION 33:
The largest number of beats per second will be heard from which of the following pairs of tuning forks?
(A) 200 and 201 Hz
(B) 256 and 260 Hz
(C) 534 and 540 Hz
(D) 763 and 774 Hz
(E) 8420 and 8422 Hz
SAMPLE QUESTION 34:
A beam of particles of energy E = 9 eV strikes on a step potential and 25% of the beam's particles is reflected back. What is the height of the step potential?
(A) 4 eV
(B) 8 eV
(C) 3 eV
(D) 1 eV
(E) 5 eV
SAMPLE QUESTION 35:
The magnetic dipole moment of a current-carrying loop of wire is in the positive z direction. The magnetic dipole is placed in space where there is a magnetic field $$\vec{B}=B_0\hat {\vec{i}}+B_0 \hat{\vec{j}}$$, where $$B_0$$ is some positive constant and $$\hat{\vec{i}}$$, $$\hat{\vec{j}}$$ the unit vectors along directions x and y respectively. What is the direction of the magnetic torque on the loop?
(A) It is along the negative z direction.
(B) It is along the line y = -x in the forth quadrant.
(C) It is along the line y = x, in the third quadrant.
(D) It is along the line y = -x, in the second quadrant.
(E) It is along the line x = y = z towards positive x,y and z.
SAMPLE QUESTION 36:
A system is composed of 4000 non-interacting particles distributed among three possible energy states $$E_1 = 0$$, $$E_2 = \epsilon$$ and $$E_2 = 2\epsilon$$. A particular partition corresponds to the occupation numbers $$n_1 = 2000$$ and $$n_2 = 1500$$ for the first two energy states. The total number of particles of the system remains constant. What is approximately the average energy of this configuration?
(A) $$2500 \epsilon$$
(Β) $$0.63 \epsilon$$
(C) $$5000 \epsilon$$
(D) $$1.25 \epsilon$$
(E) $$0.38 \epsilon$$
SAMPLE QUESTION 37:
The positive terminals of two batteries with emf's of $$E_1$$ and $$E_2$$, respectively, are connected together. It is $$E_2 > E_1$$. The circuit is completed by connecting the negative terminals. If each battery has an internal resistance r, what is the rate with which electrical energy is converted to chemical energy in the smaller battery?
(A) $$\frac{E_1^2}{r}$$
(B) $$\frac{E_1^2}{2r}$$
(C) $$\frac{(E_2-E_1)E_1}{r}$$
(D)$$\frac{(E_2-E_1)E_1}{2r}$$
(E) $$\frac{E_2^2}{2r}$$
SAMPLE QUESTION 38:
The Lagrangian for a mechanical system is $$L(q,\dot{q})=\frac{\dot{q}^3 + q^3}{3}$$, where $$q$$ is a generalized coordinate and $$\dot{q}=dq/dt>0$$. What is the Hamiltonian $$H$$ of this system?
(A) $$H=pq-\frac{p^{\frac{3}{2}}+q^3}{3}$$
(B) $$H=\frac{2p^{\frac{1}{2}}-q^3}{3}$$
(C) $$H=\frac{2p^{\frac{3}{2}}-q^3}{3}$$
(D) $$H=\frac{2 \dot{q}^3-q^3}{3}$$
(E) $$H=\frac{p^{\frac{3}{2}}+q^3}{3}$$
SAMPLE QUESTION 39:
Massless and inextensible string is wrapped around the periphery of a homogeneous cylinder of radius R = 0.5 m and mass m = 2 kg. The string is pulled straight away from the upper part of the periphery of the cylinder, without relative slipping. The cylinder moves on a horizontal floor, for which the friction coefficient (μ) is 0.4. What is most nearly the maximum force $$F_{max}$$ that can be exerted on the free end of the string so that the cylinder rolls without sliding?
(A) $$F_{max}$$ = 24 N
(B) $$F_{max}$$= 12 N
(C) $$F_{max}$$ = 8 N
(D) $$F_{max}$$ = 6 N
(E) $$F_{max}$$ = 8/3 N
SAMPLE QUESTION 40:
Which of the following facts could be a remnant of the Big Bang theory for the evolution of the universe from a space-time singularity?
I. Uniform distribution of microwave background radiation.
II. Uniform distribution of background electrons.
III. Uniform distribution of background neutrinos.
IV. Uniform distribution of gluons and quarks.
(A) I and II only
(B) I, II and III only
(C) I and III only
(D) I, III and IV only
(E) II and III only
SAMPLE QUESTION 41:
Six identical point charges +q are placed on the vertices of a regular hexagon of side of length l. There is one charge on each vertex. The hexagon is rotated about an axis that passes through its center of symmetry and is perpendicular to its plane. If the frequency of revolution is f, what is the magnitude of the induced magnetic field B at the center of the hexagon?
(A) $$B = \frac{\mu_0 q f}{2l}$$
(B) $$B = \frac{\mu_0 q f}{l}$$
(C) $$B = \frac{\mu_0 q f}{6l}$$
(D) $$B = \frac{3 \mu_0 q f}{l}$$
(E) $$B = \frac{3 \mu_0 q f}{2l}$$
$$\mu_0$$= magnetic susceptibility of free space
SAMPLE QUESTION 42:
A Carnot engine operates using a monoatomic gas as its working substance. Which of the following procedures could lead to the greatest possible increase of the efficiency of the Carnot thermal engine?
(A) The increase of the temperature of the hot reservoir by 40 K.
(B) The lowering of the temperature of the cold reservoir down to 0 K.
(C) The substitution of the monoatomic gas by a diatomic gas.
(D) The increase of the temperature of the hot reservoir by 20 K and the simultaneous decrease of the temperature of the cold reservoir by 20 K.
(E) The substitution of the reversible mode of operation of the engine by an irreversible one.
Last edited by physics_auth on Tue Jan 30, 2018 5:41 am, edited 79 times in total.
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### Questions 43-70
SAMPLE QUESTION 43:
A particle is initially in the second excited level of an infinite square well and makes a transition to the first excited level. The transition is accompanied by the emission of a photon of wavelength 1200 A ($$A = 10^{-10} m$$). What is the minimum possible energy that the particle can have in the well?
(A) 1 eV
(B) 2 eV
(C) 4 eV
(D) 8 eV
(E) 10 eV
SAMPLE QUESTION 44:
Consider a system of four non-identical spin 1/2 particles. What are the possible values of the total spin $$S_{tot}$$ of the four-particle system and their corresponding degeneracy?
(A) $$S_{tot} = 0$$ which is two-fold degenerate, and $$S_{tot} = 1$$ which is two-fold degenerate
(B) $$S_{tot} = 0$$ which is two-fold degenerate, $$S_{tot} = 1$$ which is three-fold degenerate, and $$S_{tot} = 2$$ which
is non-degenerate
(C) $$S_{tot} = 0$$ which is two-fold degenerate, and $$S_{tot} = 2$$ which is non-degenerate
(D) $$S_{tot} = 0$$ which is two-fold degenerate, $$S_{tot} = 1$$ which is nine-fold degenerate, and $$S_{tot} = 2$$ which
is five-fold degenerate
(E) $$S_{tot} = 0$$ which is non-degenerate, $$S_{tot} = 1$$ which is three-fold degenerate, and $$S_{tot} = 2$$ which
is five-fold degenerate
-------------------------------------------------------------------------------------------------------------------------------
SAMPLE QUESTIONS 45 & 46:
In a region of space where there are no charges ($$\rho = 0$$), the electric field is given by $$\vec{E} = E_0 x^2 \,\hat{i} + E_y \,\hat{j} -2 E_0 (x+y)z \,\hat{k}$$, where $$\hat{i},\,\hat{j},\,\hat{k}$$ are unit vectors along x-, y- and z-directions respectively. The y-component of the electric field was found to vary only along y-direction (i.e. it doesn't change along x- or z-direction).
SAMPLE QUESTION 45:
What is the value of the $$E_y$$ component?
(A) $$E_y = E_0 \,y$$
(B) $$E_y = 2 E_0 \,y$$
(C) $$E_y = E_0 \,y^2$$
(D) $$E_y = \frac{E_0}{2} \,y^2$$
(E) $$E_y = 2 E_0 \,y^2$$
SAMPLE QUESTION 46:
Which of the following statements is TRUE?
(A) The aforementioned electric field is electrostatic.
(B) The displacement current is nonzero.
(C) If we place a small circular loop with its surface perpendicular to the z-direction, induced current will appear.
(D) The aforementioned electric field is due to the presence of a time-varying magnetic field.
(E) There is a net conductivity current along z-direction.
----------------------------------------------------------------------------------------------------------------------------------
SAMPLE QUESTION 47:
Suppose that the temperature of an initially very hot gas of hydrogen atoms is continually lowered until room temperature is reached, and that the intensity of the a-Lyman spectral line is observed with the help of a high resolution spectroscope. Which of the following statements is/are true for the process described above?
I. The width of the a-Lyman line continually decreases until it becomes vanishing.
II. The number of atoms that the ground level accomodates increases during the process.
III. The image of the a-Lyman line becomes more and more pronounced.
IV. Resonant fluoresence stops taking place by the time the temperature of the gas reaches room temperature.
(A) I and II only
(B) II and IV only
(C) II only
(D) I, II and III only
(E) II, III and IV only
SAMPLE QUESTION 48:
A cube has proper volume $$10^{-3}\,\, m^3$$. What volume is calculated by an observer O' who moves at a velocity of 0.8c relative to the cube, in a direction parallel to one edge (of the cube)?
(A) $$1000 \,\,cm^3$$
(B) $$800 \,\,cm^3$$
(C) $$600 \,\,cm^3$$
(D) $$500 \,\,cm^3$$
(E) $$400 \,\,cm^3$$
SAMPLE QUESTION 49:
Two metalic spheres of radii $$R_1$$ and $$R_2$$ respectively, with $$R_2>R_1$$ are placed at a distance $$l$$from each other, so that $$l >> R_1+R_2$$. There is a very thin conducting wire connecting the two far-apart spheres. A total charge $$Q$$ is distributed between them. How much charge does the larger sphere carry (i.e. the sphere of radius $$R_2$$) ?
(A) $$\frac{R_1}{R_2}\,Q$$
(B) $$\frac{R_2}{R_1}\,Q$$
(C) $$\frac{R_1}{R_1+ R_2}\,Q$$
(D) $$\frac{R_2}{R_1 + R_2}\,Q$$
(E) $$\frac{R_2-R_1}{R_1+R_2}\,Q$$
SAMPLE QUESTION 50:
Solid A, with mass M is at its melting point $$T_A$$. It is placed in thermal contact with solid B, with heat capacity $$C_B$$ and initially at temperature $$T_B$$, where $$T_B>T_A$$. The combination is thermally isolated. A has latent heat of fusion L and when it has melted has heat capacity $$C_A$$. Supposing that A completely melts, what is the final common temperature of both A and B?
(A) $$\frac{C_A T_A + C_B T_B - ML}{C_A + C_B}$$
(B) $$\frac{C_A T_A - C_B T_B + ML}{C_A + C_B}$$
(C) $$\frac{C_A T_A - C_B T_B - ML}{C_A + C_B}$$
(D) $$\frac{C_A T_A + C_B T_B + ML}{C_A + C_B}$$
(E) $$\frac{C_A T_A + C_B T_B}{C_A + C_B}$$
SAMPLE QUESTION 51:
Two idential disks with mass M and radius R roll without sliding across a horizontal floor with the same speed and then up inclines. The two inclines are identical. Disk A rolls up its incline without sliding whereas disk B rolls up a frictionless incline. Disk A reaches a height of 12 cm above the horizontal floor before rolling down again. At what height does disk B reach above the horizontal floor before rolling down again?
(A) 24 cm
(B) 18 cm
(C) 12 cm
(D) 8 cm
(E) 6 cm
SAMPLE QUESTION 52:
Pi mesons at rest have a half-life of $$T$$. If a beam of pi mesons is travelling at a speed of $$u=\beta c$$ over what distance is the intensity of the beam halved? {$$|\beta|<1$$}
(A) $$c\beta T \frac{1}{\sqrt{1-\beta^2}}$$
(B) $$c\beta T \sqrt{\frac{1+\beta}{1-\beta}}$$
(C) $$c\beta T \frac{\ln2}{\sqrt{1-\beta^2}}$$
(D) $$c\beta T \sqrt{1-\beta^2}$$
(E) $$c\beta T \ln2 \sqrt{1-\beta^2}$$
SAMPLE QUESTION 53:
What is the ratio of the wavelength of the $$K_{a}$$ x-ray line for Nb (Z=41) to that of Ga (Z=31)?
(A) 9/16
(B) 16/9
(C) 3/4
(D) 4/3
(E) 41/31
SAMPLE QUESTION 54:
Which of the following is most essential for laser action to occur between two energy levels of an atom?
(A) The upper level should be a rapidly decaying state.
(B) The lasing material should be a gas.
(C) The lower level should be the ground state.
(D) There should be more atoms in the lower level than in the upper level.
(E) The upper level should be metastable.
SAMPLE QUESTION 55:
A light emitting diode (LED) emits light when:
(A) electrons are excited from the valence band to the conduction band.
(B) electrons collide with atoms.
(C) electrons are accelerated by the electric field in the depletion region.
(D) electrons from the conduction band recombine with holes from the valence band.
(E) the temperature of the junction has significantly increased.
SAMPLE QUESTION 56:
Suppose the operator $$\hat{A}=\kappa \frac{d}{dx}- \lambda x$$ where $$\kappa,\,\lambda$$ could in general be complex numbers. Under what conditions is $$\hat{A}$$ a hermitian operator?
(A) Both $$\kappa$$ and $$\lambda$$ should be real numbers.
(B) Both $$\kappa$$ and $$\lambda$$ should be imaginary.
(C) $$\kappa$$ should be imaginary and $$\lambda$$ should be real number.
(D) $$\kappa$$ should be real and $$\lambda$$ should be imaginary.
(E) $$\kappa$$ and $$\lambda$$ should be complex conjugates of each other.
SAMPLE QUESTION 57:
Two ideal monatomic gases are in thermal equilibrium with each other. Gas A is composed of molecules with mass $$m$$, while gas B is composed of molecules with mass $$4m$$. What is the ratio of the average molecular speeds $$u_A/u_B$$?
(A) 1/4
(B) 1/2
(C) 1
(D) 2
(E) 4
SAMPLE QUESTION 58:
In a certain mass spectrometer, an ion beam passes firstly through a velocity filter consisting of mutually perpendicular fields $$\vec{E}$$ and $$\vec{B}$$. Afterwards, the beam enters a region of another magnetic field $$\vec{B'}$$ perpendicular to the beam. The radius of curvature of the resulting ion beam is proportional to which of the following?
(A) $$EB'/B$$
(B) $$EB/B'$$
(C) $$BB'/E$$
(D) $$B/EB'$$
(E) $$E/BB'$$
SAMPLE QUESTION 59:
A vibrating tuning fork is held over a water column with one end closed and the other open. As the water level is allowed to fall, a loud sound is heard for water levels separated by 17 cm. If the speed of sound in air is 340 m/s, what is the frequency of the tuning fork?
(A) 250 Hz
(B) 500 Hz
(C) 1000 Hz
(D) 2000 Hz
(E) 5780 Hz
SAMPLE QUESTION 60:
A long straight cylindrical shell has inner radius $$R_{i}$$ and outer radius $$R_{o}$$. It carries a current $$i$$, uniformly distributed over its cross section. A wire is parallel to the cylinder axis, in the hollow region ($$r<R_{i}$$). The magnetic field is zero everywhere in the hollow region. Which of the following statements is TRUE?
(A) The wire is on the cylinder axis and carries current $$i$$ in the same direction as the current in the shell.
(B) The wire may be anywhere in the hollow region but must be carrying current $$i$$ in the direction opposite to that of the current in the shell.
(C) The wire may be anywhere in the hollow region but must be carrying current $$i$$ in the same direction as the current in the shell.
(D) The wire is on the cylinder axis and carries current $$i$$ in the direction opposite to that of the current in the shell.
(E) The wire doesn't carry any current.
SAMPLE QUESTION 61:
Suppose that the Hamiltonian of the (valence) electron of a triply ionized Ti atom (Z = 22) is given by $$\hat{H}_{ion} = \epsilon \hat{1} + \lambda \hat{L} \cdot \hat{S}$$, where $$\epsilon$$ and $$\lambda > 0$$ are real numbers (parameters of the problem), and $$\hat{1}$$ denotes the unit operator. In terms of the given parameters, what are the possible energy eigenstates and their corresponding denegeracy?
(A) $$\varepsilon + \lambda {\hbar ^2}$$ which is ten-fold degenerate
(B) $$\varepsilon - {\textstyle{3 \over 2}}\lambda {\hbar ^2}$$ which is six-fold degenerate and $$\varepsilon + \lambda {\hbar ^2}$$ which is four-fold degenerate
(C) $$\varepsilon - \lambda {\hbar ^2}$$ which is five-fold degenerate and $$\varepsilon + \lambda {\hbar ^2}$$ which is five-fold degenerate
(D) $$\varepsilon - \lambda {\hbar ^2}$$ which is four-fold degenerate and $$\varepsilon + \lambda {\hbar ^2}$$ which is six-fold degenerate
(E) $$\varepsilon - {\textstyle{3 \over 2}}\lambda {\hbar ^2}$$ which is four-fold degenerate and $$\varepsilon + \lambda {\hbar ^2}$$ which is six-fold degenerate
SAMPLE QUESTION 62:
Consider a system with two energy levels $$E_{1}$$ and $$E_{2}$$ with $$E_{1}<E_{2}$$ and total number of particles $$N=n_{1}+n_{2}$$ where $$n_{1}$$ and $$n_{2}$$ is the number of particles accomodated by the first and second energy level respectively. The system is in contact with a heat reservoir at temperature T. At some moment one particle decays from the upper energy level to the lower one (i.e. we have the transition $$E_{2} \rightarrow E_{1}$$). What is the change in the entropy of the system of the two energy levels? ($$k$$ denotes the Boltzmann constant)
(A) $$\Delta S= k\ln \left((n_{1}+1)(n_{2}-1)\right)$$
(B) $$\Delta S= k\ln \left(\frac{n_{2}}{n_{1}+1}\right)$$
(C) $$\Delta S= k\ln \left(\frac{n_{2}-1}{n_{1}+1}\right)$$
(D) $$\Delta S= k\ln \left(\frac{n_{2}-1}{n_{1}}\right)$$
(E) $$\Delta S= k\ln \left(\frac{n_{2}}{n_{1}}\right)$$
SAMPLE QUESTION 63:
Two thermal engines are connected so that the heat rejected by the first thermal engine withefficiency $$e_1$$ be absorbed by a second thermal engine with efficiency $$e_2$$. The efficiency of the combined system of the two thermal engines is
(A) $$\left| e_1 - e_2 \right|$$
(B) $$e_1 \cdot e_2$$
(C) $$e_1 + e_2$$
(D) $$e_1 + e_2 - e_1 \cdot e_2$$
(E) $$e_1/e_2$$
SAMPLE QUESTION 64:
Suppose that $$\hat{A}$$ is a Hermitian operator, and let us further define the operator $$\hat{U} = e^{i\hat{A}}$$. Which of the following statements is/are FALSE?
I. The operator $$\hat{U}$$ is unitary.
II. The determinant of the operator $$\hat{U}$$ is invariant under a similarity transformation.
III. The determinant of the operator $$\hat{U}$$ is given by $$\det (\hat U) = {e^{i\det(\hat A)}}$$.
IV. The eigenvalues of the operator $$\hat{U}$$ are real.
(A) Statements I and III only
(B) Statements II and III only
(C) Statement IV only
(D) Statement III only
(E) Statements III and IV only
SAMPLE QUESTION 65:
In a neutron-induced fission process, what is the origin of the delayed neutrons?
(A) They are produced by the moderator material.
(B) They are produced by the original nucleus after it absorbs a neutron.
(C) They are components of the cosmic background radiation.
(D) They are produced by the fission fragments.
(E) They are produced by the control rods of the fission reactor.
SAMPLE QUESTION 66:
In an RLC series circuit the capacity C is variable. The RLC series circuit is driven by an external source of time-varying voltage. By changing the capacity C of the circuit in a continuous manner over a sufficiently wide range two well defined discrete resonances are observed with the help of an oscilloscope, for $$C = C_1$$ and $$C = C_2 > C_1$$. Which of the following statements is/are always TRUE?
I. The external time-varying voltage consists of at least two sinusoidal components of different frequencies.
II. The bandwidth of each resonance is small compared to the difference between the two observed resonant frequencies.
III. If $$\omega_1^{res}$$ and $$\omega_2^{res}$$ are the two observed resonant frequencies, then it is $$\omega_1^{res} < \omega_2^{res}$$.
(A) III only
(B) All of the above statements
(C) II and III only
(D) II only
(E) I and II only
SAMPLE QUESTION 67:
Which of the following statements is NOT true?
(A) A continuous spectrum is one that contains all the wavelengths of visible light, such as that emitted by an incandescent material.
(B) Band spectra are characteristic emissions produced by excited molecules. The bands are actually groups of lines very close together.
(C) Fluorescence is the process by which bodies absorb shorter wavelengths of light and subsequently emit light of longer wavelengths.
(D) Fraunhofer lines are dark lines in the atomic spectrum of helium.
(E) A bright line spectrum consists of bright lines at wavelengths characteristic of the elements emitting them when excited in the gaseous state.
SAMPLE QUESTION 68:
Two point sources, oscillating in phase, produce an interference pattern over the surface of the water of a tank. If the frequency of oscillation of the two point sources is increased by 20%, then the number of the dark fringes (fringes due to destructive interference)
(A) is roughly increased by 40%.
(B) is roughly increased by 20%.
(C) is roughly decreased by 20%
(D) is roughly decreased by 40%.
(E) remains invariant.
SAMPLE QUESTION 69:
A homogeneous uniform rod is initially at rest on a frictionless surface. At some moment, forces of equal magnitude and opposite direction are applied at the two edges of the rod. Each force is applied in a direction normal to the length of the rod, and parallel to the surface on which the latter lies. Which of the following quantities does NOT change after the application of the aforementioned forces?
(A) The angular momentum of the rod with respect to an axis normal to the frictionless surface.
(B) The net torque with respect to the center of mass of the rod.
(C) The total kinetic energy of the rod.
(D) The momentum of the center of mass of the rod.
(E) The total energy of the rod.
SAMPLE QUESTION 70:
A thin uniform and homogeneous rod is suspended on a ceiling and is initially at rest (in a vertical position). When the previous rod is partially immersed into a liquid of density $$\rho_l$$, it comes into equilibrium when half of it is immersed into the liquid, at an angle with the vertical orientation (i.e. within the liquid the new equilibrium position of the rod is not the vertical one any more). What is the density $$\rho_r$$of the rod in terms of the density of the liquid?
(A) It is $$\rho_r = \rho_l/2$$.
(B) It is $$\rho_r = \rho_l/4$$.
(C) It is $$\rho_r = \rho_l$$.
(D) It is $$\rho_r = 3\rho_l/4$$.
(E) It is $$\rho_r = 2\rho_l$$.
Last edited by physics_auth on Wed Jan 17, 2018 3:26 pm, edited 106 times in total.
blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am
### Re: All 31 posted sample PGRE quest. & ETS-released sample quest
Really nice pool of questions.. i really appreciate it.. can you please post the answers too.. i am planning to take them as a short test..
by the way.. i might be a bit nosy but you don't seem to be a student.. who are you exactly? .
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### Re: Answers to all posted sample questions
ANSWERS TO ALL POSTED SAMPLE QUESTIONS:
SAMPLE QUESTION 1: (E)
SAMPLE QUESTION 2: (E)
SAMPLE QUESTION 3: (A)
SAMPLE QUESTION 4: (C)
SAMPLE QUESTION 5: (C)
SAMPLE QUESTION 6: (A)
SAMPLE QUESTION 7: (B)
SAMPLE QUESTION 8: (E)
SAMPLE QUESTION 9: (C)
SAMPLE QUESTION 10: (D)
SAMPLE QUESTION 11: (C)
SAMPLE QUESTION 12: (D)
SAMPLE QUESTION 13: (D)
SAMPLE QUESTION 14: (C)
SAMPLE QUESTION 15: (A)
SAMPLE QUESTION 16: (A)
SAMPLE QUESTION 17: (B)
SAMPLE QUESTION 18: (A)
SAMPLE QUESTION 19: (C)
SAMPLE QUESTION 20: (B)
SAMPLE QUESTION 21: (Α)
SAMPLE QUESTION 22: (D)
SAMPLE QUESTION 23: (B)
SAMPLE QUESTION 24: TBA
SAMPLE QUESTION 25: (E)
SAMPLE QUESTION 26: (D)
SAMPLE QUESTION 27: (D)
SAMPLE QUESTION 28: (C)
SAMPLE QUESTION 29: (C)
SAMPLE QUESTION 30: (E)
SAMPLE QUESTION 31: (D)
SAMPLE QUESTION 32: TBA
SAMPLE QUESTION 33: (D)
SAMPLE QUESTION 34: (B)
SAMPLE QUESTION 35: (D)
SAMPLE QUESTION 36: TBA
SAMPLE QUESTION 37: (C)
SAMPLE QUESTION 38: (C)
SAMPLE QUESTION 39: (A)
SAMPLE QUESTION 40: (C)
SAMPLE QUESTION 41: (D)
SAMPLE QUESTION 42: TBA
SAMPLE QUESTION 43: (B)
SAMPLE QUESTION 44: TBA
SAMPLE QUESTION 45: (C)
SAMPLE QUESTION 46: (D)
SAMPLE QUESTION 47: (C)
SAMPLE QUESTION 48: (C)
SAMPLE QUESTION 49: (D)
SAMPLE QUESTION 50: (A)
SAMPLE QUESTION 51: (D)
SAMPLE QUESTION 52: (A)
SAMPLE QUESTION 53: (A)
SAMPLE QUESTION 54: (E)
SAMPLE QUESTION 55: (D)
SAMPLE QUESTION 56: (C)
SAMPLE QUESTION 57: (D)
SAMPLE QUESTION 58: (B)
SAMPLE QUESTION 59: (B)
SAMPLE QUESTION 60: (E)
SAMPLE QUESTION 61: (E)
SAMPLE QUESTION 62: TBA
SAMPLE QUESTION 63: TBA
SAMPLE QUESTION 64: E
SAMPLE QUESTION 65: TBA
SAMPLE QUESTION 66: TBA
SAMPLE QUESTION 67: TBA
SAMPLE QUESTION 68: TBA
SAMPLE QUESTION 69: TBA
SAMPLE QUESTION 70: TBA
Last edited by physics_auth on Tue Jan 16, 2018 5:58 am, edited 54 times in total.
blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am
### Re: All 31 posted sample PGRE quest. & ETS-released sample quest
physics_auth wrote:
blackcat007 wrote:Really nice pool of questions.. i really appreciate it.. can you please post the answers too.. i am planning to take them as a short test..
by the way.. i might be a bit nosy but you don't seem to be a student.. who are you exactly? .
No, I am a graduate student who has finished with all tests (not from US), but due to the fact that I missed the November deadlines for PGRE (I sat it in spring and) I have to wait some more time in order to be able to apply for the fall 2010 or the spring 2009, though I have limited opportunities in the latter case. Anyway, I plan to ask advise when is better to do so... .
P.S.: If I knew how to post images taken from pdfs i would have posted even more questions. Some questions require a figure or else they will tend to be very lengthy and probably appalling for many of you. The good thing about all this is that when I constructed some questions, I "bumped into" them in the real test, but the irony is that I didn't pay much attention to them before! One such question was about Coriolis effect.
Physics_auth
AFAIK spring sems are poorly funded, and since you are an intl student, may be it will be exorbitant for you. i think fall 2010 will be better
well if you have your homepage you can load them there and give the link here. or you may even mail them to the interested members.
hmm coriolis effect?? i thought they never ask such questions.. what else did you encounter new?
oh and yes.. please post the answers for these questions
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### Analytic solution of question 27 (soap films)
physics_auth wrote:Question 27:
This is a hard PGRE question. At first, I would like to say that I omitted to say that the soap film is surrounded by air. Anyway. I corrected it. The reflected beam includes the ray that is reflected by the upper surface of the film and that that is reflected by the lower surface of it. The phase difference is in general:
Δφ = phase difference due to different optical path + phase shift due to reflection
In our case it is:
phase difference due to different optical path = (2π/λ) *(2nd) = 4πnd/λ
phase shift due to the reflection off the upper surface of the film = π
thus Δφ = 4πnd/λ + π and for constructive interference it should be Δφ = 2mπ, where m = integer. For constructive interference it is therefore
4πnd/λ + π = 2mπ => {constant quantity} = 4nd = (2m - 1)λ (1)
From (1) it is that as λ increases, the order of interference should decrease so that their product is constant (since d and n are given, constant quantities).
Let's assume that the order of interference that corresponds to λ1 = 480 nm is m1, then since no other wavelength between 480 nm and 800 nm is intensified in the reflected beam, it should be that
m2 = m1 - 1 (2),
where m2 the order of interference of wv λ2 = 800 nm, since according to (1) for λ2 > λ1 it should be m2 < m1 (since the left hand product of (1) is constant quantity). This explains why in (2) we got minus 1 instead of plus 1.
Applying (1) for m1 and m2 and equating the righ-hand members of the equations (i.e eliminating the constant product) we find a relationship between m1 and m2. If in this last equation we substitute m2 from (2), then we have an equation of only one variable -> m1. After some algebra it is found that m1 = 3. By substituting m1 = 3 into (1) and solving for d we end up to answer (D).
Last edited by physics_auth on Mon Aug 17, 2009 6:09 pm, edited 5 times in total.
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### Re: All 31 posted sample PGRE quest. & ETS-released sample quest
blackcat007 wrote:
physics_auth wrote:
blackcat007 wrote:Really nice pool of questions.. i really appreciate it.. can you please post the answers too.. i am planning to take them as a short test..
by the way.. i might be a bit nosy but you don't seem to be a student.. who are you exactly? .
No, I am a graduate student who has finished with all tests (not from US), but due to the fact that I missed the November deadlines for PGRE (I sat it in spring and) I have to wait some more time in order to be able to apply for the fall 2010 or the spring 2009, though I have limited opportunities in the latter case. Anyway, I plan to ask for advice when is better to do so... .
P.S.: If I knew how to post images taken from pdfs i would have posted even more questions. Some questions require a figure or else they will tend to be very lengthy and probably appalling for many of you. The good thing about all this is that when I constructed some questions, I "bumped into" them in the real test, but the irony is that I didn't pay much attention to them before! One such question was about Coriolis effect.
Physics_auth
AFAIK spring sems are poorly funded, and since you are an intl student, may be it will be exorbitant for you. i think fall 2010 will be better
well if you have your homepage you can load them there and give the link here. or you may even mail them to the interested members.
hmm coriolis effect?? i thought they never ask such questions.. what else did you encounter new?
oh and yes.. please post the answers for these questions
blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am
### Re: Analytic solution of question 27 (soap films)
physics_auth wrote:
physics_auth wrote:Question 27:
This is a hard PGRE question. At first, I would like to say that I omitted to say that the soap film is surrounded by air. Anyway. I corrected it. The reflected beam includes the ray that is reflected by the upper surface of the film and that that is reflected by the lower surface of it. The phase difference is in general:
Δφ = phase difference due to different optical path + phase shift due to reflection
In our case it is:
phase difference due to different optical path = (2π/λ) *(2nd) = 4πnd/λ
phase shift due to the reflection off the upper surface of the film = π
thus Δφ = 4πnd/λ + π and for constructive interference it should be Δφ = 2mπ, where m = integer. For constructive interference it is therefore
4πnd/λ + π = 2mπ => {constant quantity} = 4nd = (2m - 1)λ (1)
From (1) it is that as λ increases, the order of interference should decrease so that their product is constant (since d and n are given, constant quantities).
Let's assume that the order of interference that corresponds to λ1 = 480 nm is m1, then since no other wavelength between 480 nm and 800 nm is intensified in the reflected beam, it should be that
m2 = m1 - 1 (2),
where m2 the order of interference of wv λ2 = 800 nm, since according to (1) for λ2 > λ1 it should be m2 < m1 (since the left hand product of (1) is constant quantity). This explains why in (2) we got minus 1 instead of plus 1.
Applying (1) for m1 and m2 and equating the righ-hand members of the equations (i.e eliminating the constant product) we find a relationship between m1 and m2. If in this last equation we substitute m2 from (2), then we have an equation of only one variable -> m1. After some algebra it is found that m1 = 3. By substituting m1 = 3 into (1) and solving for d we end up to answer (D).
this was really a nice and conceptual question. thanks!!
blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am
### Re: Three NEW questions for futher practice
physics_auth wrote:
EXTRA QUESTION 3:
Infrared electromagnetic radiation of broad range (it covers near and far infrared) is sent to pass through a gaseous mixture which consists amongst others of the following diatomic molecules
H_2, O_2, CO, HCl, N_2, HF, HI
Which of the above molecules will absorb the incident radiation?
(A) All of them
(B) HCl, HF, HI and O_2 only
(C) H_2, O_2 and N_2 only
(D) CO, HCl, HF and HI only
(E) None of the above
what is the principle used in this question?
physics_auth
Posts: 163
Joined: Sat Jul 18, 2009 7:24 pm
### Re: Three NEW questions for futher practice
what is the principle used in this question?
Brief answer: As you may have noticed, the question pertains to diatomic molecules. The difference between these molecules is that some of them are homonuclear molecules whereas the rest of them aren't. The theory for diatomic molecules (if we ignore electronic transitions) says that the spectrum is composed of two components: (1) the vibrational component, whose spectrum looks like the energy levels of a quantum mechanical harmonic oscillator, and which obeys the selection rule Δn = {+1, -1}, where n as in equation E_n = (n + 1/2) * hbar*ω, and (2) the rotational component, whose spectrum is composed of a series of very closely situated -but non-equidistant- energy levels, and which obeys the selection rule Δl = {+1, -1}, where l is the quantum number of angular momentum. For in general the rotational energy is much smaller than the vibrational energy (about 1000 times smaller -in eV-) it comes out that to each vibrational level there correspond several rotational levels. In a typical diatomic spectrum, for instance, we have a vibrational level upon which sits a sequence of rotational levels, then at a greater energy height there is the next vibrational level upon which sits another sequence of rotational levels and so on. Under normal cisrcumstances, a molecule lies in the lowest vibrational level (though, it can occupy any rotational level of the sequence that sits upon the fundamental vibrational level). Keep in mind that allowed transitions are only those which conform with both of the selection rules mentioned above (one for each component). The above behavior is valid for the lower energy states, for higher enough states the situation is different.
For diatomic molecules, the rotational spectrum corresponds usually to the far infrared region, the vibrational spectrum corresponds usually to the near infrared region and the electronic spectrum is further, corresponding to the unltraviolet region or beyond. Since, the radiation employed covers infrared region (near to far), electronic transitions are indifferent to our case. Now, the essential point of the questions is this: For either a vibrational or a rotational transition to occur, either in emission or absorption, the diatomic molecule must have a permanent electric dipole moment so that it can behave as a rotating or oscillating electric dipole which, according to the classical electromagnetic theory, will radiate (and amongst others be able to come back to its fundamental energy level after it has been excited in some way). Homonuclear molecules H_2, O_2, N_2 do NOT have permanent electric dipole moment and do not show the spectrum discussed above, i.e. they do not absorb the incident infrared radiation. On the other hand, heteronuclear molecules such as CO, HCl, HF and HI have permanent electric dipole moment and show strong such spectrums, which means that they absorb the incident infrared radiation. To see why these molecules have permanent electric moment remember from basic chemistry how the electronegativity series develops ... with the hydrogen being the least electronegative element ... . Thus, the correct answer is (D).
sravanskarri
Posts: 58
Joined: Sat Jun 14, 2008 10:19 pm
### Re: ==> Sticky: More ETS and User-Created Sample Questions
Can you pls explain the solution for extra question 5. I think we need to apply rocket eq but I could not get to the answer. Was this on the sample GRE Qs ?
Thanks for the previous posts
blackcat007
Posts: 378
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### Re: ==> Sticky: More ETS and User-Created Sample Questions
sravanskarri wrote:Can you pls explain the solution for extra question 5. I think we need to apply rocket eq but I could not get to the answer. Was this on the sample GRE Qs ?
Thanks for the previous posts
I did it in this way:
since the thread is pulled at constant rate 10cm/s, thus the spool is rotating at a constant rate of v/r (since there is no slipping) any element of the thread in contact with the spool is also rotating at this constant angular velocity, thus it has only centrifugal acceleration (in non inertial frame) v^2/r , now as each segment leaves the cylinder, the only change in acceleration is from v^2/r to 0 (0 acceleration when it leaves the spool contact) thus the answer is v^2/r=0.01/0.05 = 0.2 m/s^2.
sravanskarri
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### Re: ==> Sticky: More ETS and User-Created Sample Questions
Thanks for the reply. But isn't the change in magnitude of acceleration of the "string" zero; when it is in contact with the cylinder it has a_radial = v^2/r and a_tangential =0 => total magnitude is a_radial and when the string comes off the cylinder it is a_tangential = v^2/r and a_radial = 0 leaving the magnitude constant.
I thought there will be some change in velocity due to some mass leaving the cylinder + string system...giving it some acceleration.May be I overdid it.
blackcat007
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### Re: ==> Sticky: More ETS and User-Created Sample Questions
sravanskarri wrote: when the string comes off the cylinder it is a_tangential = v^2/r and a_radial = 0 leaving the magnitude constant.
no since the thread is being pulled at a constant rate ie 10cm/s there is no tangential acceleration.
physics_auth
Posts: 163
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### Re: ==> Sticky: More ETS and User-Created Sample Questions
sravanskarri wrote:Thanks for the reply. But isn't the change in magnitude of acceleration of the "string" zero; when it is in contact with the cylinder it has a_radial = v^2/r and a_tangential =0 => total magnitude is a_radial and when the string comes off the cylinder it is a_tangential = v^2/r and a_radial = 0 leaving the magnitude constant.
I thought there will be some change in velocity due to some mass leaving the cylinder + string system...giving it some acceleration.May be I overdid it.
The solution given by blackcat007 is absolutely correct. It is a simple exercise in kinematics. Simply focus on what is going with the acceleration, which changes from υ^2 / R (centripetal only) to zero because the thread leaves the cylinder with a contant velocity thus zero acceleration. Please, do not make very complex thoughts, since the questions must be answered in at most 2 to 3 minutes. The questions I post in this thread are not ETS questions, but noone knows if they have ever been included in the real physics test (or even if a similar concept has appeared in the real test). Anyway, the underlying intention of my posted questions is simply to emphasize some details that may be helpful somehow for one's preparation. Furthermore, though it is highly improbable for you to meet a question in the real test that will be facsimile of the above questions, it is very probable to apply some of the strategies you used to solve some of the sample questions in the solution of your real test questions. This is also an aspect of success. I think so. By the way, if anyone of you want to see harder questions, let me know (send a message or sth, but i am not going to provide harder questions than those appeared in the 9677 real test!)
p.s.: ETS sample questions are in a pdf released archive. I describe how one can find it in some previous response in this thread. Know, in general that the questions in the real test are going to be a bit harder than the sample questions that ETS releases!
Physics_auth
sravanskarri
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### Re: ==> Sticky: More ETS and User-Created Sample Questions
My bad. Agree with the solution. As I mentioned earlier, I kind of mis-understood the Q, at this part
>>"As each small segment of string leaves the cylinder, by what amount does its acceleration change"
I never thought "its acceleration" implied "string acceleration" but "cylinder acceleration" which is a foolish mistake on my part. I did not mean to say the questions you had need to be from ETS. Please keep posting ,the questions are really good and helping me identify my shortcomings.
physics_auth
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### Re: ==> Sticky: More ETS and User-Created Sample Questions
sravanskarri wrote:My bad. Agree with the solution. As I mentioned earlier, I kind of mis-understood the Q, at this part
>>"As each small segment of string leaves the cylinder, by what amount does its acceleration change"
I never thought "its acceleration" implied "string acceleration" but "cylinder acceleration" which is a foolish mistake on my part. I did not mean to say the questions you had need to be from ETS. Please keep posting ,the questions are really good and helping me identify my shortcomings.
As you correctly found, it refers to the string acceleration. Keep in mind this: if in a question you make a more complex ratiocination than necessary -> by taking a quick look at the answer choices it may help you to "harness" your ratiocination. Keep also in mind that: (as I have observed) though ETS scarcely provides more information in a question than necessary to find its solution (maybe in a way to cause confusion to the candidate), they never provide questions deficient of data. All data necessary for the solution are given. If you think that sth is missed then you probably made a more complex ratiocination than required.
Sorry, if i misunderstood your answer but I usually visit the site deep in the night and sometimes it is possible to miss sth ... . You know.
Physics_auth
blackcat007
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### Re: ==> Sticky: More ETS and User-Created Sample Questions
yeah physics_auth please keep posting your questions they are very helpful.. the only prob i am facing is knowing the the topics widely. i have found that the topics which i know like EM, mechanics etc ;i am able to answer almost all, but topics like particle physics and stuffs, really are keeping me low
sravanskarri
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### Re: ==> Sticky: More ETS and User-Created Sample Questions
Thanks for the Questions.. they are pretty clear ..I am feeling a little better now
betelgeuse1
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
physics_auth
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
betelgeuse1 wrote:small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
Try to solve the problem using single geometry related to the probability density. For example the prob. density partitions into three sections from [0, L/3], [L/3, 2L/3] ... . See where the midpoints of these intervals are and use simple geometric thoughts to find the asked probability (don't forget to subtract your result from 1). For example the probability of finding the particle in [L/3, 2L/3] = 1/3, use that kind of geometrical manipulations. This method is quicker ... .
blackcat007
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
betelgeuse1 wrote:small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
apart from the method mentioned by Physics_auth here is another one: 2nd excited means n=3, (n=1 is ground state) and the probability of not finding the particle in between L/2 and 5L/6 is (1- Probability of finding the particle between L/2 and 5L/6 )
betelgeuse1
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
UPS! It's a bad idea to start solving problems when you are ILL like me now...
so the second EXCITED state, not the second state: n=3 ok!
betelgeuse1
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
physics_auth: Nice method! any other nice tricks you may write about?
betelgeuse1
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### problem 30, particle alert!
what if the interaction is weak? As I know you should specify something about the life time of the stable initial state, otherwise you could choose D too if strangeness is not conserved... I agree with the conservation of barion number though
betelgeuse1
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
me again : problem 34: I don't have any idea about it now... some hlp! I guess quantum tunneling is not the solution...
physics_auth
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### Re: problem 30, particle alert!
betelgeuse1 wrote:what if the interaction is weak? As I know you should specify something about the life time of the stable initial state, otherwise you could choose D too if strangeness is not conserved... I agree with the conservation of barion number though
My intention in this question is to check if you know selection rules for strangeness. If the interaction is strong or electromagnetic then ΔS = 0, whereas if it is weak it holds that ΔS = +1 or -1 (but not zero). The answers are carefully selected so that there is only one answer that can cover all three cases. This is why I don't mention sth about the kind of interaction. Did you get my point?
physics_auth
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
betelgeuse1 wrote:me again : problem 34: I don't have any idea about it now... some hlp! I guess quantum tunneling is not the solution...
Simply use the formula about the reflection coefficient for the one-dimensional scattering in a potential step. If you don't remember it by heart, then try to work out this formula analytically. Then, check if you can make a mnemonics so as to remember it for the exam or invent a more quick way that will help you find the asked formula in the quickest possible time. Use your imagination in the latter case ... . You expect the formula for the reflection coefficient to depend on E and the step height amongst other possible things. And be carefull -> it is a step not a barrier.
physics_auth
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
blackcat007 wrote:
betelgeuse1 wrote:small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
apart from the method mentioned by Physics_auth here is another one: 2nd excited means n=3, (n=1 is ground state) and the probability of not finding the particle in between L/2 and 5L/6 is (1- Probability of finding the particle between L/2 and 5L/6 )[/quote
Yeah man. It is impossible that ETS will ask you to use Bragg's formula for a non-simple cubic lattice, since in this case one needs to know things about Miller indices, and the edge of the cube also, in order to use a formula for the distance between lattice planes. At least me I have never met such question in PGRE. By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
physics_auth
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
physics_auth wrote:
blackcat007 wrote:
betelgeuse1 wrote:small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
apart from the method mentioned by Physics_auth here is another one: 2nd excited means n=3, (n=1 is ground state) and the probability of not finding the particle in between L/2 and 5L/6 is (1- Probability of finding the particle between L/2 and 5L/6 )[/quote
Yeah man. It is impossible that ETS will ask you to use Bragg's formula for a non-simple cubic lattice, since in this case one needs to know things about Miller indices, and the edge of the cube also, in order to use a formula for the distance between lattice planes. At least me I have never met such question in PGRE.
Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
blackcat007
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
well i think it should be monochromatic, because if its a continuous spectrum, then for a given lattice spacing, the final pattern of diffraction will be very cumbersome.. there will resolution problem (Rayleigh's Criterion), ie difficult to resolve, since dispersion d(theta)/d(lambda) will be very large and thus overlapping will prevent proper identification of the angle diffreacted.
The wavelength is continuous, so in 2*d*sin(theta)=n*(lambda), since lambda is continuous we will get a continuous pattern and thus difficult to differentiate.
physics_auth
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
blackcat007 wrote:
Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
well i think it should be monochromatic, because if its a continuous spectrum, then for a given lattice spacing, the final pattern of diffraction will be very cumbersome.. there will resolution problem (Rayleigh's Criterion), ie difficult to resolve, since dispersion d(theta)/d(lambda) will be very large and thus overlapping will prevent proper identification of the angle diffreacted.
The wavelength is continuous, so in 2*d*sin(theta)=n*(lambda), since lambda is continuous we will get a continuous pattern and thus difficult to differentiate.
Good attempt, but we use radiation of continuous spectrum. In fact, it is difficult to find the correct direction of the incident ray -assuming it is monochromatic- for which it can lead to Bragg diffraction. Besides the detector assumes only a small solid angle in space - detectors of solid angle of 4π are only very rarely used due their high cost (and probably other factors that I miss... for the time being). By using a continuous radiation, Bragg diffraction picks automatically out all these wavelengths that satisfy Bragg's law and in the pattern in the output we see well resolved peaks sitting upon an almost flat background. The peaks do correspond to Bragg reflections (or diffractions) and then we try to find out which possible groups of lattice planes could give a specific peak (this is done by the formula that connects d -the distance between consecutive planes in a group- with the Miller indices and the dimensions of the cell.) The point is that we don't know in advance which group of planes (i.e. the d) has given a specific Bragg reflection ... .
blackcat007
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
physics_auth wrote:
blackcat007 wrote:
Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
well i think it should be monochromatic, because if its a continuous spectrum, then for a given lattice spacing, the final pattern of diffraction will be very cumbersome.. there will resolution problem (Rayleigh's Criterion), ie difficult to resolve, since dispersion d(theta)/d(lambda) will be very large and thus overlapping will prevent proper identification of the angle diffreacted.
The wavelength is continuous, so in 2*d*sin(theta)=n*(lambda), since lambda is continuous we will get a continuous pattern and thus difficult to differentiate.
Good attempt, but we use radiation of continuous spectrum. In fact, it is difficult to find the correct direction of the incident ray -assuming it is monochromatic- for which it can lead to Bragg diffraction. Besides the detector assumes only a small solid angle in space - detectors of solid angle of 4π are only very rarely used due their high cost (and probably other factors that I miss... for the time being). By using a continuous radiation, Bragg diffraction picks automatically out all these wavelengths that satisfy Bragg's law and in the pattern in the output we see well resolved peaks sitting upon an almost flat background. The peaks do correspond to Bragg reflections (or diffractions) and then we try to find out which possible groups of lattice planes could give a specific peak (this is done by the formula that connects d -the distance between consecutive planes in a group- with the Miller indices and the dimensions of the cell.) The point is that we don't know in advance which group of planes (i.e. the d) has given a specific Bragg reflection ... .
but then how do we know which was the wavelength that actually created the diffraction pattern that the detector is detecting? suppose i use lambda varying from say 5-6 nm, and we want to find the plane spacing d, we know theta, and as you said, we try to see which plane was responsible for the diffraction.. then which value of wavelength we use?
blackcat007
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### Re: ==> Sticky: More ETS and 41 "active" Sample Questions
blackcat007 wrote:
physics_auth wrote: Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
well i think it should be monochromatic, because if its a continuous spectrum, then for a given lattice spacing, the final pattern of diffraction will be very cumbersome.. there will resolution problem (Rayleigh's Criterion), ie difficult to resolve, since dispersion d(theta)/d(lambda) will be very large and thus overlapping will prevent proper identification of the angle diffreacted.
The wavelength is continuous, so in 2*d*sin(theta)=n*(lambda), since lambda is continuous we will get a continuous pattern and thus difficult to differentiate.
Good attempt, but we use radiation of continuous spectrum. In fact, it is difficult to find the correct direction of the incident ray -assuming it is monochromatic- for which it can lead to Bragg diffraction. Besides the detector assumes only a small solid angle in space - detectors of solid angle of 4π are only very rarely used due their high cost (and probably other factors that I miss... for the time being). By using a continuous radiation, Bragg diffraction picks automatically out all these wavelengths that satisfy Bragg's law and in the pattern in the output we see well resolved peaks sitting upon an almost flat background. The peaks do correspond to Bragg reflections (or diffractions) and then we try to find out which possible groups of lattice planes could give a specific peak (this is done by the formula that connects d -the distance between consecutive planes in a group- with the Miller indices and the dimensions of the cell.) The point is that we don't know in advance which group of planes (i.e. the d) has given a specific Bragg reflection ... .
hehehe somehow in experimental physics things turn out the opposite of what i think.. !!! thats a clue.. i will answer whatever seems wrong to me in experimental physics questions
betelgeuse1
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
Now that I jumped into the pool and came out in a poor condition but still alive I would like to see the other pool... you said something about harder problems. Where are they?
physics_auth
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
betelgeuse1 wrote:Now that I jumped into the pool and came out in a poor condition but still alive I would like to see the other pool... you said something about harder problems. Where are they?
I will NOT post this other pool that is mentioned in this thread. However, I can send it via an e-mail upon "private" request (it is a pdf archive and a pdf with solutions). It is a small pool of 60 questions with brief answers only. I have already sent it to blackcat007. You can ask for his opinion as well.
Don't worry if my pool "brought to light" possible weaknesses... it is your responsibility to find out your weak points and re-read the pertinent piece of material to see if you missed any essential detail. I will be glad to hear that this pool helped you -even a bit- to answer correctly one, two or more questions in the real test.
betelgeuse1
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
First about problem 34: I don't understand how is it possible to solve this problem using R=(k1-k2)^2/(k1+k2)^2, k1~sqrt(E) and k2~sqrt(E-V) in 2 minutes. I am sure there may be a quicker solution but I don't get it.
physics_auth
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
betelgeuse1 wrote:First about problem 34: I don't understand how is it possible to solve this problem using R=(k1-k2)^2/(k1+k2)^2, k1~sqrt(E) and k2~sqrt(E-V) in 2 minutes. I am sure there may be a quicker solution but I don't get it.
Who told you that all questions in PGRE are designed to be solved in 2 minutes? There are always questions that require >2 minutes. If you pinpoint such questions in the test you can avoid them in a first browsing, and return to them at a later time, given that there is sufficient time to do so! If you stick to such "time-consuming" questions then you will probably miss time to solve easier questions. From my experience, when the answers to a question are in the form of pure numbers then there is no trick to zero in on the correct result. You can only eliminate answers that are false according to your physics knowledge or intuition. But in such questions the test comittee want you to do some rough calculation and for this reason 2 or 3 answers can not be eliminated unless you have done some calculations in your scratch paper.
colonblow
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
hi, shouldn't the answer to #1 problem be letter D, which is 3g/L? If 3g/2L is the correct answer, can someone please tell me how it was solved? thanks thanks!
blackcat007
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
colonblow wrote:hi, shouldn't the answer to #1 problem be letter D, which is 3g/L? If 3g/2L is the correct answer, can someone please tell me how it was solved? thanks thanks!
the only torque that is acting is that due to the weight of the beam. so mgl/2 =Ia
I=(ml^2)/3
equate and solve..
colonblow
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
I was doing mgl=Ia... thanks again!
blackcat007
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
QUESTION 35: (Magnetism - CR question)
The magnetic dipole moment of a current-carrying loop of wire is in the positive z direction. The magnetic dipole is placed in space where there is a magnetic field B = B0*i + B0*j, where B0 = const. and i , j the unit vectors along directions x and y respectively. What is the direction of the magnetic torque on the loop?
(A) It is along the negative z direction.
(B) It is along the line y = -x in the forth quadrant.
(C) It is along the line y = x, in the third quadrant.
(D) It is along the line y = -x, in the second quadrant.
(E) It is along the line x = y = z towards positive x,y and z.
I think B0>0 should be mentioned for avoiding any ambiguity between B and D option
QUESTION 40: (Cosmology - CR question)
Which of the following facts could be a remnant of the Big Bang theory for the evolution of the universe from a space-time singularity?
I. Uniform distribution of microwave background radiation.
II. Uniform distribution of background electrons.
III. Uniform distribution of background neutrinos.
IV. Uniform distribution of gluons and quarks.
(A) I and II only
(B) I, II and III only
(C) I and III only
(D) I, III and IV only
(E) II and III only
but the recent WMAP found anisotropy in the cosmic background of the order of micro kelvins and astrophysicists (esp Smoot and Fire) have made some prediction of structure formations from these anisotropy.
physics_auth
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
blackcat007 wrote:
QUESTION 35: (Magnetism - CR question)
The magnetic dipole moment of a current-carrying loop of wire is in the positive z direction. The magnetic dipole is placed in space where there is a magnetic field B = B0*i + B0*j, where B0 = const. and i , j the unit vectors along directions x and y respectively. What is the direction of the magnetic torque on the loop?
(A) It is along the negative z direction.
(B) It is along the line y = -x in the forth quadrant.
(C) It is along the line y = x, in the third quadrant.
(D) It is along the line y = -x, in the second quadrant.
(E) It is along the line x = y = z towards positive x,y and z.
I think B0>0 should be mentioned for avoiding any ambiguity between B and D option.
B0 is supposed to represent the magnitude of magnetic field component which is always a non-zero quantity. To avoid confusion with the direction in space I used the unit vectors i and j. If I said B0 *(-i) this means that the i-th component of the magnetic induction is towards negative x-direction. Anyway, I corrected it to avoid ambiguity as you said ... . Thanks!
QUESTION 40: (Cosmology - CR question)
Which of the following facts could be a remnant of the Big Bang theory for the evolution of the universe from a space-time singularity?
I. Uniform distribution of microwave background radiation.
II. Uniform distribution of background electrons.
III. Uniform distribution of background neutrinos.
IV. Uniform distribution of gluons and quarks.
(A) I and II only
(B) I, II and III only
(C) I and III only
(D) I, III and IV only
(E) II and III only
but the recent WMAP found anisotropy in the cosmic background of the order of micro kelvins and astrophysicists (esp Smoot and Fire) have made some prediction of structure formations from these anisotropy.
Well, PGRE candidates are not supposed to have such specific knowledge ... answer according to your undergraduate school knowledge.
noospace
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
Where did there questions come from? Are they really official ETS? If so how come they're not available on the website?
What's going on here?
physics_auth
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### Re: ==> Sticky: More ETS and 42 "active" Sample Questions
noospace wrote:Where did there questions come from? Are they really official ETS? If so how come they're not available on the website?
What's going on here?
It is a bit irritating to have to apologize again ... but I will do it once and for all ... . Questions 1-42 in this post are not ETS official questions, but questions that imitate those of the real PGRE test. Above all, to post ETS questions is plagiarism! Furthermore, I give directions (in green color somewhere in the 1st page) in this post how to find a pdf archive with ETS questions (about 30 questions with their brief solutions). The pdf mentioned contains ETS official questions. Is that clear? For those of you who want to practise only on ETS official questions there are directions on how to do this (the pdf I mentioned... the 4 practice tests etc.), the questions I post here periodically are for those who want to practise on PGRE questions beyond the official material. Ok?
Last edited by physics_auth on Sun Sep 27, 2009 12:46 pm, edited 1 time in total.
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## Finding the Right Words
At Warby Parker, we get a lot of tweets. Our social media team does a great job of responding to each one and recording various metadata that the team uses for reporting purposes. One type of post that we often see from our Home Try-On customers is an informal poll to Warby Parker, friends, and family regarding Home Try-On choices. Here’s an example of one that we posted ourselves, featuring Chris Becker from our Tech team:
The Social Media team recently wanted to track this type of post in order to better engage customers and help them with their Home Try-Ons, as well as gauge their involvement in our Home Try-On program. By providing meaningful responses to these posts, Warby Parker can both assist customers through the process and provide a better customer experience. They approached the Data Science team with the task of automating this process for tracking incoming tweets and historic data.
#### The Problem
From a data science perspective, this is simply a text classification problem. Given the tweet text, the task is to classify whether it is a Home Try-On poll or not. In particular, we need to map the contents of the post to one of two classes: “Positive” if the post is a Home Try-On poll and “Negative” otherwise. The issues faced were the following:
1. Dealing with natural language is always an issue. People can sometimes use the exact same words and mean entirely different things. For example, consider the following sentences: “Meatloaf for dinner, great!” and “Meatloaf for dinner? Great…”. The same words were used, but one implied that meatloaf for dinner was a good thing, and the other implied the opposite.
2. There is an inherent class imbalance. Warby Parker gets a lot of tweets, and only a small fraction of them are the type of posts we would label as positive. In fact, a classifier that simply labels all tweets as negative would have an accuracy of about 99%.
3. There were many unlabeled data points. When our Social Media team approached us with the problem, Warby Parker had received well over 100,000 tweets to date but had fewer than 100 positive-labeled examples.
#### Feature Selection
Selecting numeric features from a text post usually involves text pre-processing, followed by text vectorization. In our case, minimal text pre-processing was used. Punctuation was removed, and words from a given stopword list were filtered out (words such as the, and, and of).
For text vectorization, we used Term Frequency Inverse Document Frequency (tf-idf) as a means of counting word frequencies in a text. For a given term (or word) $t$ and a given text document $d$, $tf-idf$ for a given text is calculated by
$tfidf(t, d) = tf(t,d) \times idf(t, d)$,
where $tf(t,d)$, the term frequency is defined by
$tf(t, d) = \frac{f(t, d)}{N_{d}}$,
where $tf(t,d)$ is the number of times $t$ appears in $d$, and $N$ is the number of terms in document $d$. The inverse document frequency is the inverse of the ratio of documents containing term $t$ to the total number of documents $idf(t,d)$, and is defined as
$idf(t,d) = log\left ( \frac{N}{\left | \left \{ d \in D; t\in d \right \} \right |} \right )$,
where $N$ is the number of documents in the corpus.
Using the Scikit-Learn Python package, we can do both stopword filtering and tf-idf all at once.
In the above code snippet, we initialize a TfidfVectorizer object, where we specify the stopword list. We then call .fit_transform(corpus), which creates the $tf-idf$ model of the corpus and then transforms the corpus. The max_df parameter specifies the maximum term frequency that we will consider. Words appearing more frequently than this in a given document will be ignored. The sublinear_tf parameter, when True, transforms the term frequency calculation to:
$tf_{new} = 1 + log(tf)$.
This transformation, as the name implies, is sublinear, which in practice lessens the effect of frequently occurring words. Since this calculation is only for the term frequency—and thus only affects the numerator of the $tf-idf$ calculation—having a sublinear transform is a good way to reduce the effect of very common words.
#### Hand-made features
Because of the nature of tweets and the nature of tweets about our Home Try-On program, some features can be exploited. We created a few hand-made features that performed well. These features were:
1. True/False value indicating whether the tweet contained a URL
2. The domain of the URL, or None if no URL was present
3. How far into the tweet the URL appears, expressed as the starting index divided by tweet length
In general, the tweets polling for Home Try-On help included URLs, so feature #1 proved to be a good feature to filter out inputs. The remaining features were helpful in providing insight into what type of possible content the URL will contain.
#### Class Imbalances and Unlabeled Data
Upon receiving the data from our Social Media team, we had a large corpus of tweets, but very few labeled points. Worse yet, the natural proportion of positive-to-negative examples was low (i.e., most tweets, although unlabeled, were not positive examples). The main way we dealt with this problem was:
1. Build classifier on data that we have, inferring the label of unlabeled points (using semi-supervised learning methods)
2. Examining the performance of the classifier, fixing any obviously misclassified examples (using good old fashioned elbow grease)
3. Repeat
To infer labels of unlabeled points, we utilized the LabelSpreading class from Sci-kit learn. This is a semi-supervised learning technique that infers labels on unlabeled points given a small number of labeled points. Using it is rather straightforward:
and voila! The variable y_inferred now contains inferred labels for each points.
#### Model Selection
After feature selection, we made both a multinomial Naive Bayes and an SVM model for classification. The SVM model outperformed the Naive Bayes model and is what I will discuss here.
SVMs come out of the box with a few hyperparameters. The C parameter specifies how sensitive the model is to errors (effectively acting as a regularization parameter), and the radial basis kernel comes with a gamma parameter, specifying the width of the kernel.
In general, optimizing hyperparameters is a difficult problem, but we can approximate the optimization through sampling. It has been shown that random sampling can be more effective than more heuristic approaches like grid search[1]. We can easily perform random hyperparameter search in Scikit-learn by the following:
When specifying hyperparameters, we input a scipy distribution object by which to sample. The RandomizedSearchCV object takes care of sampling from this distribution. In this case, we used the scipy.stats.expon distribution, an exponential distribution. The n_iter parameter specifies how many models to build and how many hyperparameter configurations to sample. The n_jobs parameter specifies how many jobs to run concurrently.
#### Conclusion
Binary text classification can always be messy, but by taking a few precautions, you can avoid many headaches. In particular, keeping in mind the types of behaviors present in the domain and exploiting these behaviors through hand-made features, noting the proportion of positive to negative samples, and searching for more optimal hyperparameters all help reduce the ambiguities.
You might be asking yourself, how well did our classifier perform? Well on a corpus with 74,016 examples, 10,520 of which were labeled and 63,496 were unlabeled, the classifier had a cross-validation F-score of 0.92 (0.88 for positive and 0.96 for negative). It turns out that people tend to use a lot of the same words when making a Home Try-On poll post! For example, the word ‘pick’ appeared in about 34% of positive examples and only about 2% of negative examples, and the word ‘help’ appeared in about 47% of positive versus about 10% of negative examples.
Oh, and by the way, Chris ended up choosing the Beckett, and we think it was a good choice.
[1] : Bergstra, James, and Yoshua Bengio. “Random search for hyper-parameter optimization.” The Journal of Machine Learning Research 13 (2012): 281-305.
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# Math Help - Help on multiplication algebra problem.
1. ## Help on multiplication algebra problem.
The problem is "A worker at a baking company packs rolls twice as fast as he can pack muffins. He packs crackers 2.5 times paster than rolls. He can pack 250 boxes of crackers per hour. How many packs of muffins can he pack in one hour?" Thanks.
2. ## I like muffins
Originally Posted by KanyeWest
The problem is "A worker at a baking company packs rolls twice as fast as he can pack muffins. He packs crackers 2.5 times paster than rolls. He can pack 250 boxes of crackers per hour. How many packs of muffins can he pack in one hour?" Thanks.
Well since he packs rolls twice as fast as muffins we get
$r=2m$ Does this make sense?? lets check
if he packs one muffin he would then have packed two rolls.
He packs crackers 2.5 times faster than rolls
$c=2.5r$ Does this make sense? (yes)
Now we can use the above info to solve 250 crackers in one hr so we solve
$250=2.5r \iff 100 =r$
now using r=100 in the 1st equation we get
$100=2m \iff 50 = m$
so he can pack 50 muffins in one hour. yeah!!
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Mathematics
Straight Lines and Pair of Straight Lines
Previous Years Questions
## Numerical
Consider the lines L1 and L2 defined by $${L_1}:x\sqrt 2 + y - 1 = 0$$ and $${L_2}:x\sqrt 2 - y + 1 = 0$$For a fixed constant $$\lambda$$, let C be ...
Consider the lines L1 and L2 defined by $${L_1}:x\sqrt 2 + y - 1 = 0$$ and $${L_2}:x\sqrt 2 - y + 1 = 0$$For a fixed constant $$\lambda$$, let C be ...
For a point $$P$$ in the plane, Let $${d_1}\left( P \right)$$ and $${d_2}\left( P \right)$$ be the distance of the point $$P$$ from the lines $$x - y ... ## MCQ (Single Correct Answer) For$$a > b > c > 0,$$the distance between$$(1, 1)$$and the point of intersection of the lines$$ax + by + c = 0$$and$$bx + ay + c = 0$$... A straight line$$L$$through the point$$(3, -2)$$is inclined at an angle$${60^ \circ }$$to the line$$\sqrt {3x} + y = 1.$$If$$L$$also inters... Let$$O\left( {0,0} \right),P\left( {3,4} \right),Q\left( {6,0} \right)$$be the vertices of the triangles$$OPQ$$. The point$$R$$inside the triangl... The lines$${L_1}:y - x = 0$$and$${L_2}:2x + y = 0$$intersect the line$${L_3}:y + 2 = 0$$at$$P$$and$$Q$$respectively. The bisector of the acu... Area of the triangle formed by the line$$x + y = 3$$and angle bisectors of the pair of straight line$${x^2} - {y^2} + 2y = 1$$is The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices$$\l...
Orthocentre of triangle with vertices $$\left( {0,0} \right),\left( {3,4} \right)$$ and $$\left( {4,0} \right)$$ is
A triangle with vertices $$(4, 0), (-1, -1), (3, 5)$$is
Locus of mid point of the portion between the axes of $$x$$ $$\cos \alpha + y\sin \alpha = p$$ where $$p$$ is constant is
If the pair of lines $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$ intersect on the $$y$$ axis then
The pair of lines represented by $$3a{x^2} + 5xy + \left( {{a^2} - 2} \right){y^2} = 0$$ are perpendicular to each other for
Let $$0 < \alpha < {\pi \over 2}$$ be fixed angle. If $$P = \left( {\cos \theta ,\,\sin \theta } \right)$$ and $$Q = \left( {\cos \left( {\alp... Let$$P = \left( { - 1,\,0} \right),\,Q = \left( {0,\,0} \right)$$and$$R = \left( {3,\,3\sqrt 3 } \right)$$be three points. Then the equation of t... A straight line through the origin$$O$$meets the parallel lines$$4x+2y=9$$and$$2x+y+6=0$$at points$$P$$and$$Q$$respectively. Then the point ... The number of integer values of$$m$$, for which the$$x$$-coordinate of the point of intersection of the lines$$3x + 4y = 9$$and$$y = mx + 1$$is... Area of the parallelogram formed by the lines$$y = mx$$,$$y = mx + 1$$,$$y = nx$$and$$y = nx + 1$$equals Let$$PS$$be the median of the triangle with vertices$$P(2, 2),Q(6, -1)$$and$$R(7, 3).$$The equation of the line passing through$$(1, -1)$$... The incentre of the triangle with vertices$$\left( {1,\,\sqrt 3 } \right),\left( {0,\,0} \right)$$and$$\left( {2,\,0} \right)$$is Lt$$PQR$$be a right angled isosceles triangle, right angled at$$P(2, 1)$$. If the equation of the line$$QR$$is$$2x + y = 3,$$then the equation ... If$${x_1},\,{x_2},\,{x_3}$$as well as$${y_1},\,{y_2},\,{y_3}$$, are in G.P. with the same common ratio, then the points$$\left( {{x_1},\,{y_1}} \r...
The diagonals of a parralleogram $$PQRS$$ are along the lines $$x + 3y = 4$$ and $$6x - 2y = 7$$. Then $$PQRS$$ must be a.
If $$\left( {P\left( {1,2} \right),\,Q\left( {4,6} \right),\,R\left( {5,7} \right)} \right)$$ and $$S\left( {a,b} \right)$$ are the vertices of a parr...
The orthocentre of the triangle formed by the lines $$xy=0$$ and $$x+y=1$$ is
The locus of a variable point whose distance from $$\left( { - 2,\,0} \right)$$ is $$2/3$$ times its distance from the line $$x = - {9 \over 2}$$ is...
The equations to a pair of opposites sides of parallelogram are $${x^2} - 5x + 6 = 0$$ and $${y^2} - 6y + 5 = 0,$$ the equations to its diagonals are
If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is
Line $$L$$ has intercepts $$a$$ and $$b$$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same ...
If $$P=(1, 0),$$ $$Q=(-1, 0)$$ and $$R=(2, 0)$$ are three given points, then locus of the point $$S$$ satisfying the relation $$S{Q^2} + S{R^2} = 2S{P... The points$$\left( {0,{8 \over 3}} \right),\,\,\left( {1,\,3} \right)$$and$$\left( {82,\,30} \right)$$are vertices of A vector$$\overline a $$has components$$2p$$and$$1$$with respect to a rectangular cartesian system. This system is rotated through a certain ang... The straight lines$$x + y = 0,\,3x + y - 4 = 0,\,x + 3y - 4 = 0$$form a triangle which is The point$$\,\left( {4,\,1} \right)$$undergoes the following three transformations successively. Reflection about the line$$y=x$$. Translation thro... The points$$\left( { - a,\, - b} \right),\,\left( {0,\,0} \right),\,\left( {a,\,b} \right)$$and$$\left( {{a^2},\,ab} \right)$$are : ## Subjective The area of the triangle formed by intersection of a line parallel to$$x$$-axis and passing through$$P (h, k)$$with the lines$$y = x $$and$$x + ...
A straight line $$L$$ through the origin meets the lines $$x + y = 1$$ and $$x + y = 3$$ at $$P$$ and $$Q$$ respectively. Through $$P$$ and $$Q$$ two...
A straight line $$L$$ with negative slope passes through the point $$(8, 2)$$ and cuts the positive coordinate axes at points $$P$$ and $$Q$$. Find th...
Let $$a, b, c$$ be real numbers with $${a^2} + {b^2} + {c^2} = 1.$$ Show that the equation $$\left| {\matrix{ {ax - by - c} & {bx + ay} &... For points$$P\,\,\, = \left( {{x_1},\,{y_1}} \right)$$and$$Q\,\,\, = \left( {{x_2},\,{y_2}} \right)$$of the co-ordinate plane, a new distance$$d\...
Let $$ABC$$ and $$PQR$$ be any two triangles in the same plane. Assume that the prependiculars from the points $$A, B, C$$ to the sides $$QR, RP, PQ$$...
Using co-ordinate geometry, prove that the three altitudes of any triangle are concurrent.
A rectangle $$PQRS$$ has its side $$PQ$$ parallel to the line $$y = mx$$ and vertices $$P, Q$$ and $$S$$ on the lines $$y = a, x = b$$ and $$x = -b,$$...
A line through $$A (-5, -4)$$ meets the line $$x + 3y + 2 = 0,$$ $$2x + y + 4 = 0$$ and $$x - y - 5 = 0$$ at the points $$B, C$$ and $$D$$ respectiv...
Tagent at a point $${P_1}$$ {other than $$(0, 0)$$} on the curve $$y = {x^3}$$ meets the curve again at $${P_2}$$. The tangent at $${P_2}$$ meets the ...
Determine all values of $$\alpha$$ for which the point $$\left( {\alpha ,\,{\alpha ^2}} \right)$$ lies insides the triangle formed by the lines $$\... Show that all chords of the curve$$3{x^2} - {y^2} - 2x + 4y = 0,$$which subtend a right angle at the origin, pass through a fixed point. Find the co... Find the equation of the line passing through the point$$(2, 3)$$and making intercept of length 2 units between the lines$$y + 2x = 3$$and$$y + 2...
Straight lines $$3x + 4y = 5$$ and $$4x - 3y = 15$$ intersect at the point $$A$$. Points $$B$$ and $$C$$ are choosen on these two lines such that $$AB... A line cuts the$$x$$-axis at$$A (7, 0)$$and the$$y$$-axis at$$B (0, -5)$$. A variable line$$PQ$$is drawn perpendicular to$$AB$$cutting the$$...
Let $$ABC$$ be a triangle with $$AB = AC$$. If $$D$$ is the midpoint of $$BC, E$$ is the foot of the perpendicular drawn from $$D$$ to $$AC$$ and $$F... Lines$${L_1} = ax + by + c = 0$$and$${L_2} = lx + my + n = 0$$intersect at the point$$P$$and make an angle$$\theta $$with each other. Find the ... One of the diameters of the circle circumscribing the rectangle$$ABCD$$is$$4y = x + 7$$. If$$A$$and$$B$$are the points$$(-3, 4)$$and$$(5, 4)...
Two sides of rhombus $$ABCD$$ are parallel to the lines $$y = x + 2$$ and $$y = 7x + 3$$. If the diagonals of the rhombus intersect at the point $$(1,... Two equal sides of an isosceles triangle are given by the equations$$7x - y + 3 = 0$$and$$x + y - 3 = 0$$and its thirds side passes through the po... The vertices of a triangle are$$\left[ {a{t_1}{t_2},\,\,a\left( {{t_1} + {t_2}} \right)} \right],\,\,\left[ {a{t_2}{t_3},a\left( {{t_2} + {t_3}} \rig...
The coordinates of $$A, B, C$$ are $$(6, 3), (-3, 5), (4, -2)$$ respectively, and $$P$$ is any point $$(x, y)$$. Show that the ratio of the area of ...
The end $$A, B$$ of a straight line segment of constant length $$c$$ slide upon the fixed rectangular axes $$OX, OY$$ respectively. If the rectangle \$...
A straight line $$L$$ is perpendicular to the line $$5x - y = 1.$$ The area of the triangle formed by the line $$L$$ and the coordinate axes is $$5$$....
(a) Two vertices of a triangle are $$(5, -1)$$ and $$(-2, 3).$$ If the orthocentre of the triangle is the origin, find the coordinates of the third po...
A straight line segment of length $$\ell$$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the lin...
The area of a triangle is $$5$$. Two of its vertices are $$A\left( {2,1} \right)$$ and $$B\left( {3, - 2} \right)$$. The third vertex $$C$$ lies on $$... One side of rectangle lies along the line$$4x + 7y + 5 = 0.$$Two of its vertices are$$(-3, 1)$$and$$(1, 1).$$Find the equations of the other thr... ## MCQ (More than One Correct Answer) Let$${L_1}$$be a straight line passing through the origin and$${L_2}$$be the straight line$$x + y = 1$$. If the intercepts made by the circle$${...
If the vertices $$P, Q, R$$ of a triangle $$PQR$$ are rational points, which of the following points of the triangle $$PQR$$ is (are) always rational ...
All points lying inside the triangle formed by the points $$\left( {1,\,3} \right),\,\left( {5,\,0} \right)$$ and $$\left( { - 1,\,2} \right)$$ satisf...
Three lines $$px + qy + r = 0$$, $$qx + ry + p = 0$$ and $$rx + py + q = 0$$ are concurrent if
## Fill in the Blanks
The vertices of a triangle are $$A\left( { - 1, - 7} \right)B\left( {5,\,1} \right)$$ and $$C\left( {1,\,4} \right).$$ The equation of the bisector of...
Let the algebraic sum of the perpendicular distances from the points $$\left( {2,0} \right),\,\left( {0,\,2} \right)$$ $$\left( {1,\,1} \right)$$ to a...
The orthocentre of the triangle formed by the lines $$x + y = 1,\,2x + 3y = 6$$ and $$4x - y + 4 = 0$$ lies in quadrant number .............
If $$a,\,b$$ and $$c$$ are in A.P., then the straight line $$ax + by + c = 0$$ will always pass through a fixed point whose coordinates are .............
Given the points $$A\left( {0,4} \right)$$ and $$B\left( {0, - 4} \right)$$, the equation of the locus of the point $$P\left( {x,y} \right)$$ such tha...
$$y = {10^x}$$ is the reflection of $${\log _{10}}\,x$$ in the line whose equation is ...........
The set of lines $$ax + by + c = 0,$$ where $$3a + 2b + 4c = 0$$ is concurrent at the point ..........
The area enclosed within the curve $$\left| x \right| + \left| y \right| = 1$$ is .................
## True or False
The lines $$2x + 3y + 19 = 0$$ and $$9x + 6y - 17 = 0$$ cut the coordinates axes in concyclic points.
The straight line $$5x + 4y = 0$$ passes through the point of intersection of the straight lines $$x + 2y - 10 = 0$$ and $$2x + y + 5 = 0.$$
EXAM MAP
Joint Entrance Examination
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# 数学代写|变分法代写variational methods代考|Math595
## 数学代写|变分法代写variational methods代考|Black Box Variational Inference
As seen in Sect. 3.2.1.5, the SVI computes the distribution updates in a closed form, which requires model-specific knowledge and implementation. Moreover, the gradient of the ELBO must have a closed-form analytical formula. Black Box Variational Inference (BBVI) [25] avoids these problems by estimating the gradient instead of actually computing it.
BBVI uses the score function estimator [34]
$$\nabla_\phi \mathbb{E}{q(\mathbf{z} ; \phi)}[f(\mathbf{z} ; \theta)]=\mathbb{E}{q(\mathbf{z} ; \phi)}\left[f(\mathbf{z} ; \theta) \nabla_\phi \log q(\mathbf{z} ; \phi)\right]$$
where the approximating distribution $q(\mathbf{z} ; \phi)$ is a continuous function of $\phi$ (see Appendix A.1). Using this estimator to compute the gradient of the ELBO in Eq. (3.7) gives us
$$\nabla_\phi \mathrm{ELBO}=\mathbb{E}q\left[\left(\nabla\phi \log q(\mathbf{z} ; \phi)\right)(\log p(\mathbf{x}, \mathbf{z})-\log q(\mathbf{z} ; \phi))\right] .$$
The expectation in $\mathrm{Fq}$. (3.77) is approximated by a Monte Carlo integration.
The sole assumption of the gradient estimator in Eq. (3.77) about the model is the feasibility of computing the log of the joint $p\left(\mathbf{x}, \mathbf{z}_s\right)$. The sampling method and the gradient of the log both rely on the variational distribution $q$. Thus, we can derive them only once for each approximating family $q$ and reuse them for different models $p\left(\mathbf{x}, \mathbf{z}_s\right)$. Hence the name black box: we just need to specify the model $p\left(\mathbf{x}, \mathbf{z}_s\right)$ and can directly perform VI on it. Actually, $p\left(\mathbf{x}, \mathbf{z}_s\right)$ does not even need to be normalized, since the log of the normalization constant does not contribute to the gradient in Eq. (3.77).
## 数学代写|变分法代写variational methods代考|Black Box α Minimization
Black Box $\alpha$ minimization [9] (BB- $\alpha$ ) optimizes an approximation of the power EP energy function $[19,20]$. Instead of considering $i$ different local compatibility functions $\widetilde{f}_i$, it ties them together so that all $\widetilde{f}_i$ are equal, that is, $\widetilde{f}_i=\widetilde{f}$. We may view it as an average factor approximation, which we use to approximate the average effect of the original $f_i[9]$.
Further restricting these factors to belong to the exponential family amounts to tying their natural parameters. As a consequence, BB- $\alpha$ no longer needs to store an approximating site per likelihood factor, which leads to significant memory savings in large data sets. The fixed points differ from power EP, though they become equal in the limit of infinite data.
$\mathrm{BB}-\alpha$ dispenses with the need for double-loop algorithms to directly minimize the energy and employs gradient-descent methods for this matter. This contrasts with the iterative update scheme of Sect. 3.2.3. As other modern methods designed for large-scale learning, it employs stochastic optimization to avoid cycling through the whole data set. Besides, it estimates the expectation over the approximating distribution $q$ present in the energy function by Monte Carlo sampling.
Differently from BBVI [25], the BB- $\alpha$ uses the pathwise derivative estimator [24] to estimate the gradient (see Appendix A.1). We must be able to express the random variable $\mathbf{z} \sim q(\mathbf{z}, \phi)$ as an invertible deterministic transformation $g(\cdot ; \phi)$ of a base random variable $\epsilon \sim p(\epsilon)$, so we can write
$$\nabla_\phi \mathbb{E}{q(\mathbf{z} ; \phi)}[f(\mathbf{z} ; \theta)]=\mathbb{E}{p(\epsilon)}\left[\nabla_\phi f(g(\epsilon ; \phi) ; \theta)\right]$$
The approach requires not only the distribution $q(\mathbf{z} ; \phi)$ to be reparameterizable but also $f(\mathbf{z} ; \theta)$ to be known and a continuous function of $\phi$ for all values of $\mathbf{z}$. Note that it requires, in addition to the likelihood function, its gradients. Still, we can readily obtain them with automatic differentiation tools if the likelihood is analytically defined and differentiable.
As observed in Sect. 3.2.3, the parameter $\alpha$ in Eq. (3.68) controls the divergence function. Hence, the method is able to interpolate between VI $(\alpha \rightarrow-1)$ and an algorithm similar to EP $(\alpha \rightarrow 1)$. Interestingly, the authors [9] claim to usually obtain the best results by setting $\alpha=0$, halfway through VI and EP. This value corresponds to the so-called Hellinger distance, the sole member of the $\alpha$-family that is symmetric.
## 数学代写|变分法代写variational methods代考|Black Box Variational Inference
BBVI 使用评分函数估计器 [34]
$$\nabla_\phi \mathbb{E} q(\mathbf{z} ; \phi)[f(\mathbf{z} ; \theta)]=\mathbb{E} q(\mathbf{z} ; \phi)\left[f(\mathbf{z} ; \theta) \nabla_\phi \log q(\mathbf{z} ; \phi)\right]$$
$$\nabla_\phi \mathrm{ELBO}=\mathbb{E} q[(\nabla \phi \log q(\mathbf{z} ; \phi))(\log p(\mathbf{x}, \mathbf{z})-\log q(\mathbf{z} ; \phi))]$$
## 数学代写|变分法代写variational methods代考|Black Box α Minimization
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# statsmodels.stats.diagnostic.linear_rainbow¶
statsmodels.stats.diagnostic.linear_rainbow(res, frac=0.5, order_by=None, use_distance=False, center=None)[source]
Rainbow test for linearity
The null hypothesis is the fit of the model using full sample is the same as using a central subset. The alternative is that the fits are difference. The rainbow test has power against many different forms of nonlinearity.
Parameters
resRegressionResults
A results instance from a linear regression.
fracfloat, default 0.5
The fraction of the data to include in the center model.
order_by{ndarray, str, List[str]}, default None
If an ndarray, the values in the array are used to sort the observations. If a string or a list of strings, these are interpreted as column name(s) which are then used to lexicographically sort the data.
use_distancebool, default False
Flag indicating whether data should be ordered by the Mahalanobis distance to the center.
center{float, int}, default None
If a float, the value must be in [0, 1] and the center is center * nobs of the ordered data. If an integer, must be in [0, nobs) and is interpreted as the observation of the ordered data to use.
Returns
fstatfloat
The test statistic based on the F test.
pvaluefloat
The pvalue of the test.
Notes
This test assumes residuals are homoskedastic and may reject a correct linear specification if the residuals are heteroskedastic.
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# Into a square with side K is inscribed a circle with radius
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Into a square with side K is inscribed a circle with radius [#permalink]
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Updated on: 25 Feb 2014, 10:19
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Into a square with side K is inscribed a circle with radius r. If the ratio of area of square to the area of circle is P and the ratio of perimeter of the square to that of the circle is Q. Which of the following must be true?
(A) P/Q > 1
(B) P/Q = 1
(C) 1 > P/Q > 1/2
(D) P/Q = 1/2
(E) P/Q < 1/2
Originally posted by eyunni on 18 Oct 2007, 13:47.
Last edited by Bunuel on 25 Feb 2014, 10:19, edited 1 time in total.
Edited the question and added the OA.
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Re: Into a square with side K is inscribed a circle with radius [#permalink]
### Show Tags
25 Feb 2014, 22:17
4
1
honchos wrote:
Bunuel,
Can you look into this solution.
Inscribed means circle is touching all the 2 sides of a square?
OR
2R<K
According to me this should be the condition:
2R<=K
Final solution will be P/Q >=1 NOT P/Q = 1
Also, check out the following posts on regular polygons inscribed in circles and circles inscribed in regular polygons:
http://www.veritasprep.com/blog/2013/07 ... relations/
http://www.veritasprep.com/blog/2013/07 ... other-way/
http://www.veritasprep.com/blog/2013/07 ... n-circles/
http://www.veritasprep.com/blog/2013/07 ... -polygons/
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##### General Discussion
SVP
Joined: 01 May 2006
Posts: 1777
### Show Tags
18 Oct 2007, 14:14
1
(B) for me
We have:
o k= 2*r
o Area of the circle = pi * r^2
o Area of the square = k^2 = 4 * r^2
o P = k^2 / (pi * r^2) = 4 / pi
o Perimeter of the circle = 2*pi*r = pi * k
o Perimeter of the square = 4*k
o Q = 4*k / pi * k = 4 / pi = P
Finally,
P = Q
<=> P/Q = 1
Director
Joined: 11 Jun 2007
Posts: 576
### Show Tags
18 Oct 2007, 14:17
Fig wrote:
(B) for me
We have:
o k= 2*r
o Area of the circle = pi * r^2
o Area of the square = k^2 = 4 * r^2
o P = k^2 / (pi * r^2) = 4 / pi
o Perimeter of the circle = 2*pi*r = pi * k
o Perimeter of the square = 4*k
o Q = 4*k / pi * k = 4 / pi = P
Finally,
P = Q
<=> P/Q = 1
oops...OA is not unconvincing anymore..
I missed the equation k = 2r. Thanks.
OA is B indeed.
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Re: Into a square with side K is inscribed a circle with radius [#permalink]
### Show Tags
25 Feb 2014, 10:13
Bunuel,
Can you look into this solution.
Inscribed means circle is touching all the 2 sides of a square?
OR
2R<K
According to me this should be the condition:
2R<=K
Final solution will be P/Q >=1 NOT P/Q = 1
_________________
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Posts: 50002
Re: Into a square with side K is inscribed a circle with radius [#permalink]
### Show Tags
25 Feb 2014, 10:40
2
1
honchos wrote:
Into a square with side K is inscribed a circle with radius r. If the ratio of area of square to the area of circle is P and the ratio of perimeter of the square to that of the circle is Q. Which of the following must be true?
(A) P/Q > 1
(B) P/Q = 1
(C) 1 > P/Q > 1/2
(D) P/Q = 1/2
(E) P/Q < 1/2
Bunuel,
Can you look into this solution.
Inscribed means circle is touching all the 2 sides of a square?
OR
2R<K
According to me this should be the condition:
2R<=K
Final solution will be P/Q >=1 NOT P/Q = 1
When a circle is inscribed in a square it touches all 4 sides of the square:
Attachment:
Untitled.png [ 3.85 KiB | Viewed 4925 times ]
Thus when a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square.
Into a square with side K is inscribed a circle with radius r. If the ratio of area of square to the area of circle is P and the ratio of perimeter of the square to that of the circle is Q. Which of the following must be true?
(A) P/Q > 1
(B) P/Q = 1
(C) 1 > P/Q > 1/2
(D) P/Q = 1/2
(E) P/Q < 1/2
Into a square with side K is inscribed a circle with radius r --> k = 2r --> ;
The ratio of area of square to the area of circle is P --> $$\frac{k^2}{\pi{r^2}}=\frac{(2r)^2}{\pi{r^2}}=\frac{4}{\pi}=P$$.
The ratio of perimeter of the square to that of the circle is Q --> $$\frac{4k}{2\pi{r}}=\frac{4(2r)}{2\pi{r}}=\frac{4}{\pi}=Q$$.
Thus we have that P=Q --> P/Q=1.
Hope it's clear.
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Re: Into a square with side K is inscribed a circle with radius [#permalink]
### Show Tags
21 Jul 2016, 10:58
eyunni wrote:
Into a square with side K is inscribed a circle with radius r. If the ratio of area of square to the area of circle is P and the ratio of perimeter of the square to that of the circle is Q. Which of the following must be true?
(A) P/Q > 1
(B) P/Q = 1
(C) 1 > P/Q > 1/2
(D) P/Q = 1/2
(E) P/Q < 1/2
Area of square=K^2
Radius of Circle=K/2(as shown in fig.)
Area of circle=pi*(K/2)^2---->pi*K^2/4
Ratio(P)=K^2/(pi*K^2/4)----->4/pi
Again Perimeter of square=4K
Perimeter of circle=2*pi*K/2---->pi*K
Ratio (Q)= 4K/pi*K-------4/pi
So P/Q=(4/pi)/(4/pi)=1
Ans B
Attachments
Untitled (1).png [ 7.81 KiB | Viewed 2321 times ]
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Joined: 04 Jun 2016
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Re: Into a square with side K is inscribed a circle with radius [#permalink]
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29 Jul 2016, 22:58
eyunni wrote:
Into a square with side K is inscribed a circle with radius r. If the ratio of area of square to the area of circle is P and the ratio of perimeter of the square to that of the circle is Q. Which of the following must be true?
(A) P/Q > 1
(B) P/Q = 1
(C) 1 > P/Q > 1/2
(D) P/Q = 1/2
(E) P/Q < 1/2
Area of the square = $$k^2$$
Area of the circle = $$Pi * r^2$$; Diameter of the circle is k ; so the radius $$r=\frac{k}{2}$$
Area of the circle = $$pi * \frac{k^2}{4}$$
Ratio $$P = \frac{K^2}{Pi * k^2/4} = \frac{4}{Pi}$$
Perimeter of square = $$4k$$
Perimeter of circle = $$2pi*r; 2pi$$ is diameter $$= k$$
Perimeter of the circle = $$pi* k$$
Ratio $$Q= \frac{4k}{pi*k} = \frac{4}{Pi}$$
Now P/Q become $$\frac{4}{pi}$$ divided by $$\frac{4}{pi}$$ = 1
B
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Re: Into a square with side K is inscribed a circle with radius [#permalink]
### Show Tags
29 Jun 2017, 00:39
1
honchos wrote:
Bunuel,
Can you look into this solution.
Inscribed means circle is touching all the 2 sides of a square?
OR
2R<K
According to me this should be the condition:
2R<=K
Final solution will be P/Q >=1 NOT P/Q = 1
Inscribed means(In geometry) :draw (a figure) within another so that their boundaries touch but do not intersect.
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# Into a square with side K is inscribed a circle with radius
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# Diff. paths of a Force in xy-plane
1. Nov 17, 2007
### KvnBushi
[SOLVED] diff. paths of a Force in xy-plane
This was an accidental incomplete post.
(Thanks for the reply HallsofIvy, I have posted the complete post)
Last edited: Nov 17, 2007
2. Nov 17, 2007
### HallsofIvy
Staff Emeritus
I almost hate to have to say this, but: WHAT are A, O, B, and C? I assume they are points in the plane but obviously we have to know exactly what points they are before we can answer this. Even then are we to assume that "moving along OAC" means moving along the straight line from O to A, the the straight line from A to C?
I am inclined to guess that O is (0,0), A is (0.5, 0), B is (0, 0.5) and C is (0.5, 0.5) but don't you see that we can't help you if you don't TELL us that?
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# Math Help - Chinese Remainder Theorem
1. ## Chinese Remainder Theorem
Suppose (a,b) = 1. As x runs through a complete residue system (mod b) and y runs through a complete residue system (mod a), then ax + by runs through a complete residue system (mod ab).
I'm curious as to whether this is true. I've tried many examples but have failed to prove the above statement. Any help would be great. Thanks!
2. Suppose $ax_0+by_0 \equiv ax_1+by_1 \mod ab$. We want to show that $x_0\equiv x_1 \mod b$ and $y_0 \equiv y_1 \mod a$. We have $a(x_0-x_1)+b(y_0-y_1)\equiv 0 \mod ab$. Reducing $\mod a$, we have $a(x_0-x_1)+b(y_0-y_1) \equiv b(y_0-y_1) \equiv 0 \mod a$. Since $(a,b)=1$, this implies $y_0-y_1 \equiv 0 \mod a$. The other part follows similarily.
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# Caltech-UCLA Logic Seminar
Friday, October 9, 2020
12:00pm to 12:50pm
Online Event
Part 1 of Martin's conjecture and measure-preserving functions
Patrick Lutz, Department of Mathematics, UC Berkeley,
Martin's conjecture is an attempt to make precise the idea that the only natural functions on the Turing degrees are the constant functions, the identity, and transfinite iterates of the Turing jump. The conjecture is typically divided into two parts. Very roughly, the first part states that every natural function on the Turing degrees is either eventually constant or eventually increasing and the second part states that the natural functions which are increasing form a well-order under eventual domination, where the successor operation in this well-order is the Turing jump.
In joint work with Benny Siskind, we prove part 1 of Martin's conjecture for a class of functions that we call measure-preserving. This has a couple of consequences. First, it allows us to connect part 1 of Martin's conjecture to the structure of ultrafilters on the Turing degrees. Second, we also show that every order-preserving function on the Turing degrees is either eventually constant or measure-preserving and therefore part 1 of Martin's conjecture holds for order-preserving functions. This complements a result of Slaman and Steel from the 1980s showing that part 2 of Martin's conjecture holds for order-preserving Borel functions.
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# How few terms may appear in a polynomial with given (cyclotomic) roots and nonnegative coefficients?
Given $W \subset \mathbb C$, let $S_W$ be the set of polynomials in $\mathbb R[x]$ that vanish on $W$ and have only nonnegative coefficients.
Warm-up question: It's clear that if $W$ contains a positive real number, then $S_W = \{0\}$. Is the converse true?
I'm pretty sure the answer is "yes" but I don't know if I'd quite call what I have a proof.
More generally, how do the combinatorics of $W$ affect what you can say about $S_W$? In particular, I'd like a handle on the following problem.
Problem: Find (bounds on) the smallest number of nonzero terms in a polynomial belonging to $S_W$.
What conditions on $W$ would allow us to do this? As I alluded to above, I'm especially interested in combinatorial conditions, e.g., on where the numbers in $W$ lie on the complex plane.
Maybe the following special case is more tractable. Actually, it's the only one that really matters to me anyway.
Special case: What if $W$ is a set of complex $n$th roots of unity?
In this case, we'll want to assume $1 \not\in W$. Then the polynomial $x^{n-1}+x^{n-2}+\cdots+x+1$ gives an upper bound of $n$. If every number in $W$ lies in the left half-plane and $W$ is a self-conjugate set, then an upper bound of $|W|+1$ can be obtained by taking the product of the corresponding quadratic factors.
Meanwhile, I'm not sure I know how to derive a lower bound better than $2$ under any conditions!
Any thoughts or theory out there that might help shed light on these questions?
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As G.Myerson already noted we must assume $W$ is finite. Then a different approach for the "warmup" is to observe that if $\omega \in W$ is not a positive real then some power $\omega^N$ ($N \geq 1$) has non-positive real part. Therefore $\omega$ is a root of the polynomial $(X^N-\omega^N)(X^N-\overline\omega^N)$ which has only nonnegative coefficients. The product over all $\omega\in W$ gives a nonnegative polynomial of the desired kind. It also gives an upper bound $3^{\left|W\right|}$ on the number of monomials.
If $\{{\rm Im}\log\omega : \omega \in W, \omega \neq 0\} \cup \{2\pi\}$ is ${\bf Q}$-linearly independent then a single $N$ will do for all $\omega\in W$, because as $N$ varies the $\left|W\right|$-tuples of angles formed by the $N$-th powers are dense in $({\bf R}/2\pi{\bf Z})^{\left|W\right|}$. In that case a polynomial of degree $2\left|W\right|$ in $X^N$ will do, and we get an upper bound $2\left|W\right|+1$.
For a lower bound, clearly $2$ is enough iff there is some $N$ such that all nonzero elements of $W$ have the same $N$-th power and that power is negative; else we need at least $3$ monomials. If $W$ contains all $n-1$ nontrivial $n$-th roots of unity then we need at least $n$ monomials, because a polynomial $\sum_k a(k) x^k$ is a multiple of $(x^n-1)\big/(x-1)$ iff the $n$ sums $\sum_j a(k_0+nj)$ ($k_0=0,1,2,\ldots,n-1$) are all equal, and at least one of them must be positive if the $a(k)$ are nonnegative and not all zero.
[added later] Another sharp bound is obtained if $W$ consists entirely of negative numbers: then $\prod_{\omega\in W} (X-\omega)$ has $\left|W\right|+1$ monomials, and this is best possible by Descartes' rule of signs. Come to think of it, this also gives a sharp bound if $W$ consists entirely of pure imaginaries: we may assume $W = -W$, and then $P_0(X) = \prod_{\omega\in W} (X-\omega)$ is again nonnegative, and if $P$ is any multiple of $P_0$ we can apply Descartes' rule to the even and odd parts of $P$ separately.
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For the warm-up question, if $W$ is an infinite set, then $S_W=\{0\}$ whether $W$ contains a positive real or not, so let's assume it was intended that $W$ be finite.
If $\alpha$, not a positive real, is in $W$, then $p(x)=(x-\alpha)(x-\overline\alpha)$ vanishes at $\alpha$ and not at any positive real. More generally, if $W=\{\alpha_1,\dots,\alpha_m\}$ then $p(x)=\prod_j(x-\alpha_j)(x-\overline\alpha_j)$ vanishes on $W$ and not at any positive real.
I believe that if $p(x)$ has no positive real root then for $n$ sufficiently large $(x+1)^np(x)$ has no negative coefficients. This follows from a result cited in the accepted answer to Application of polynomials with non-negative coefficients. So the answer to the warm-up question (if I have interpreted that earlier answer correctly) is, yes, the converse is true.
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$\def\ZZ{\mathbb{Z}}$ $\def\RR{\mathbb{R}}$ $\def\QQ{\mathbb{Q}}$ $\def\Re{\mathrm{Re}}$ $\def\Im{\mathrm{Im}}$Let $|W|=n$. I will show that $2n+1$ monomials are always sufficient, and are generically necessary. (Noam Elkies has already shown that $2n+1$ is generically sufficient.) If $W$ is closed under complex conjugation and has no real points, then this argument shows that $n+1$ is enough, since we could replace $W$ by $W' := \{ \omega \in W : \Im(\omega) > 0 \}$ and any real polynomial which vanishes on $W'$ will also vanish on $W$.
Let the elements of $W$ be $\omega_j = x_j + i y_j = r_j e^{i \theta_j}$. For any nonnegative integer $m$, let $v_m$ be the vector $$v_m := (\Re(\omega_1^m), \Im(\omega_1^m), \ldots, \Re(\omega_n^{m}), \Im(\omega_n^{m}))$$ So the $v_i$ are vectors in $\RR^{2n}$, and our goal is to find a positive linear relation between them.
Generic necessity: Suppose that we had a linear relation $\sum a_i v_{m_i}=0$ using only $2n$ terms. Then the vectors $v_{m_1}$, ..., $v_{m_{2n}}$ would be linearly dependent, so the $2n \times 2n$ matrix they form would have determinant zero. This is a nontrivial polynomial relation between the $x_j$ and $y_j$ with integer coefficients. If the $\omega_j$ are chosen generically then the $x_j$ and $y_j$ will be algebraically independent over $\QQ$, and no such relation will exist.
Sufficiency: This is essentially Noam's proof when the $\theta_j$'s are linearly independent over $\mathbb{Q}$, but with a lot more checking of degenerate cases. Suppose, for the sake of contradiction, that we cannot write $0$ as $\sum_{i=1}^{2n+1} a_i v_{m_i}$ with the $a_i \geq 0$ and not all $0$. Equivalently, assume that, for any $(2n+1)$-tuple of vectors of the form $v_m$, the origin is not in the convex hull of the tuple.
By the contrapositive of Carathéodory's theorem, we conclude that $0$ is not in the convex hull of the vectors $v_m$. Let $K$ be the closure of the convex hull of the $v$'s. We conclude that $0$ is not in the interior of $K$. By Farkas's lemma, there is a linear function $\lambda : \RR^{2n} \to \RR$ with is $\geq 0$ on $K$ and not identically $0$ on $K$. Equivalently, $\lambda(v_m)$ is nonnegative for all $m$ and is positive for some $m$.
We can write $\lambda$ in the form $$(f_1, g_1, f_2, g_2, \ldots, f_n, g_n) \mapsto \Re \left( \sum (a_j+i b_j) (f_j+i g_j) \right)$$ for some $a_j$ and $b_j$. Set $\zeta_j = a_j+i b_j$. Our hypothesis now is that $$\phi(m) : = \Re \left( \sum_j \zeta_j \omega_j^m \right)$$ is nonnegative for all $m$ and positive for some $m$; our goal is to deduce a contradiction.
Let $R$ be the set of distinct values of $|\omega_j|$, and let the elements of $R$ be $r_1 > r_2 > \cdots > r_p$. Let the elements of $R$ with norm $r_j$ be $r_j \exp(i \theta^j_1)$, $r_j \exp(i \theta^j_2)$, ..., $r_j \exp(i \theta^j_{k(j)})$. Reindex the $\zeta$'s accordingly as $\zeta^1_1$, $\zeta^1_2$, ...., $\zeta^1_{k(1)}$, .... $\zeta^p_1$, ...., $\zeta^p_{k(p)}$. Put $$\phi_j(m) := \Re \left( \sum_{\ell=1}^{k(j)} \zeta^j_{\ell} \exp(i m \theta^j_{\ell}) \right)$$ so $$\phi(m) = \sum_j r_j^m \phi_j(m). \quad (\ast)$$ Since $\phi(m)$ is not identically zero, not all of the $\phi_j(m)$ are identically zero. Let $j_0$ be minimal such that $\phi_{j_0}(m)$ takes nonzero values.
Lemma There is $c>0$ so that $\phi_{j_0}(m)$ is infinitely often less than $-c$.
Proof By assumption, $\phi_{j_0}$ is not identically zero. Since none of the $\theta$'s are $0 \bmod 2 \pi \ZZ$, the Cesaro limit $\lim_{M \to \infty} \frac{1}{M} \sum_{1 \leq m \leq M} \phi_{j_0}(m)$ is $0$. (This is where we use that the $\omega$'s are not positive reals. Note that this is true even if some of the $\theta$'s are in $2 \pi \QQ$.) So $\phi_{j_0}$ is negative for some $m$, say $m_0$. Let $\phi_{j_0}(m_0)=-2c$.
Consider any $\delta>0$. By a basic pigeonhole argument, we can find infinitely many $N$ such that $| N \theta^{j_0}_{\ell} \bmod 2 \pi \ZZ| < \delta$. (Note that this is true even if the $\theta$'s are not linearly independent over $\QQ$.) Choosing $\delta$ small enough, for such $N$'s, we have $\phi_{j_0}(m_0+N) < -c$. $\square$.
Therefore, there are infinitely many $m$ for which the leading term of $(\ast)$ is as negative as $-c r_{j_0}^m$, and all the other terms are exponentially less. Thus, we have found an $m$ for which $\phi(m)<0$. This is a contradiction and the theorem follows.
Remark: Carathéodory + Farkas + (something clever) is a general proof technique. Chapter 2 of Barvinok's A Course in Convexity has many nontrivial exercises of this form.
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• February 5th 2009, 10:42 PM
reshma
ABCD is a cyclic quadrilateral. Diagonals of ABCD (ie.,AC & BD) meet at right angles at P.PL is a perpendicular bisector to chord AB. If PL is extended to meet chord DC at M, show that PM bisects DC.
• February 6th 2009, 08:07 AM
red_dog
$AL=LB, \ PL\perp AB\Rightarrow\Delta APB$ isosceles $\Rightarrow\widehat{PAB}=\widehat{PBA}$
But $\widehat{PAB}=\widehat{PDC}, \ \widehat{PBA}=\widehat{PCD}\Rightarrow\widehat{PDC }=\widehat{PCD}\Rightarrow\Delta PCD$ isosceles.
We have $\widehat{ACD}=\widehat{CAB}\Rightarrow AB\parallel DC\Rightarrow PM\perp CD\Rightarrow DM=MC$
• February 9th 2009, 11:26 AM
fardeen_gen
1)angle APB and DPC are vertically opposite,hence equal
2)opposite sides are similar in ABCD because the diagonals are perpendicular to each other, hence have equal length
3)angle PLA and angle PMD are alternate angles, hence equal
Therefore, by ASA(angle-side-angle) postulate, triangles ABP and DCP are congruent.
Also,
PL=PM
Therefore, if PL bisects AB then, PM bisects DC.
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## The DSL, MDA, UML thing again...
Simon Johnston (IBM/Rational) on Domain Specific Languages and refinement:
My position is that the creation of domain specific languages that do not seamlessly support the ability to transform information along the refinement scale are not helpful to us. So, for example, a component designer that is a stand alone tool unconnected to the class designer that provides the next logical level of refinement (classes being used to construct components) is a pot hole in the road from concept to actual implementation. Now, this is not as I have said to indicate that domain specific languages are bad, just that many of us in this industry love to create new languages be they graphical, textual or conceptual. We have to beware of the tendency to build these disjoint languages that force the user to keep stopping and jumping across another gap.
I am not sure I want to go back to the argument between IBM and Microsoft about this stuff, but I think the notion of refinement is important from a linguistic point of view (e.g., embedding, language evolution, type systems, reasoning etc.)
But can refinement of the sort discussed here work in practice, or does it depend on language-design savvy architects? Maybe the difference between IBM and Microsoft is that the IBM approach assumes all modelling knowledge will come pre-packaged, being designed by modelling professionals and embedded in tools, where as the Microsoft approach assumes more design expertise from users?
Feel free to correct me, I am really unsure where all this is heading anyway...
## Comment viewing options
### Specialization?
Was wondering whether DSL is meant to promote specialization in the software process?
(see related article on Separating job functions).
The problem with software, to paraphrase Hawking, is that "It's software all the way down".
### paraphrase Hawking?
What's he got to do with it? The anecdote isn't about him, or is there more than one?
### Just gisting
The original quote is about it being Turtles all the way down. For some reason that reminds me of the difficulty of seperating functions within software development - no matter where you draw the lines, it's still software on both sides of the division.
### Of course it's about turtles, it 's just not from Hawking
Eddington's the guy, IIRC. Maybe Russell (but I think that's the one with the punch line "Oh alright, I thought you said 100 million years"). Definitely not Hawking.
### Hawking popularized it
The opening of "A Brief History of Time" starts with the turtle story.
### Darn it...
Ok, right. Now I see it.
But to make things even more frustrating Hawking just writes "a wellknown scientist (some say it was Bertrand Russell)". No wonder I wasn't sure who it was.
### russel or feynman
I want to remember it was feynman, perhaps recounted in his "pleasure of finding things out".
However, tha intarweb is siding with the russel hypothesis.
### Similar problem
no matter where you draw the lines, it's still software on both sides of the division.
This reminds me of a coworker who likes to pull my chain saying that "it's all syntax." If you formally define semantics, you are simply using syntax again etc. I am sure everyone here understands what we mean when we say semantics aren't the same as syntax, but try arguing with someone who is smart and tries to resist the distinction. It's quite frustrating :-)
### Syntax vs. semantics
This reminds me of a coworker who likes to pull my chain saying that "it's all syntax."... try arguing with someone who is smart and tries to resist the distinction. It's quite frustrating :-)
If he can argue that, he could probably equally convincingly argue that "it's all semantics." A little cleverness is a dangerous thing! :)
I think most researchers understand that the barrier between syntax and semantics is porous. Like rationals, an object typically admits many ways of factoring itself into syntactic and semantic parts, including ways in which one or the other is trivial. But (and this is very vague but) I think you can only find factorizations where the semantic part is trivial if the object in question is computable'. A way to attack the "it's all syntax" argument is to point out the chicken-egg problem between, for example, set theory and logic; the same argument works for "it's all semantics." (We don't know that set theory is consistent, for instance.)
In programming languages, there is a very strong tendency to factor a language so the syntactic portion is context-free, both for historical reasons and because we have decent tools for parsing context-free grammars. I think this fact confuses a lot of programmers into thinking the syntax-semantics barrier is immovable.
In the calculus I'm working on now, my terms have a grammar which is not context-free, and I think it's quite nice that way, but I am considering refactoring it so that the context-sensitive conditions turn out as typing rules, just because people find context-sensitive grammars weird'.
### Semantics vs. Syntax
I read that semantics is the meaning while syntax is the representation.
To communicate meaning though, you have to represent it. So to specify the semantics, you have to write something which would have a syntax.
I'm guessing a simple example of this is prefix/postfix/mixfix syntax. Underneath the parameter ordering is a function with some semantics. For example, integer addition is integer addition whether you're on a RPN HP calculator, in Scheme, or doing the usual infix representation on paper (excluding space/representation limitations).
Last week or so I castigated myself for not clearly distinguishing between semantics in the usual PLT sense and what I called "intentional PL semantics", that is, the meaning of the program from the point of view of the programmer/user,
Now I don't feel so bad, as I've since noticed that this is quite common. ;-)
For example, integer addition is integer addition whether you're on a RPN HP calculator, in Scheme, or doing the usual infix representation on paper
I think this is a nice example of the distinction I want to draw. The intentional semantics of all of these are the same, though the PLT semantics may not be.
The reason is that the latter type of semantics must have well-defined formal properties to be considered adequate, and formal details matter.
Some of the different cases you mention might in fact have different formal properties.
For example, the various computer or calculator implementation are actually going to do addition modulo some value (depending on the width integer storage).
This value will vary from case to case and therefore they are not equivalent formal systems.
On paper we normally don't assume any roll-over, which makes it formally different again.
Part of the reason that it is hard to distinguish between formal syntax and formal semantics is that they both must be computational in nature, so they are the "same type of thing" in a sense.
Intentional semantics (and general semantics) has no such limitation (that we know of yet), and finding ways to map it onto more formal semantics is 99% of the challenge of software development.
Last week or so I castigated myself for not clearly distinguishing between
semantics in the usual PLT sense and what I called "intentional PL semantics" ...
Good lord, man!! Intellectual honesty is important and all, but I think you're being way to hard on yourself!
Oh, wait ...
You said castigated?
Uhhh ..... never mind ...
: )
[With apologies to Gilda Radner, as Emily Litella, may she rest in peace.]
### Uh, why not?
If thou didst put this soure cold habit on
To castigate thy pride, 'twere well.
### Can of worms
I'm guessing a simple example of this is prefix/postfix/mixfix syntax. Underneath the parameter ordering is a function with some semantics. For example, integer addition is integer addition whether you're on a RPN HP calculator, in Scheme, or doing the usual infix representation on paper (excluding space/representation limitations).
Nice try. :-) This is a good example of exactly what's being discussed above. Lambda calculus provides a simple example of a case where integer addition (along with anything else that's computable) can be implemented as nothing but a series of (arguably) purely syntactic transformations.
If you're not familiar with this, just imagine how the first cavemen might have done addition: using a "syntax" involving a number of stones. To add one to a number, represented by a pile of stones, you perform the syntactic transformation of adding a syntactic token (a stone) to the pile. You can perform any addition this way, if you have enough time, and enough stones. For such cavemen, the actual act of addition would have been entirely syntactic. Assigning a meaning to the resulting pile of stones introduces semantics, but that's only after the answer has been arrived at, and only because it involves translating between systems.
I can't tell if you're demonstrating me wrong. You've just done two implementations of the natural numbers. How is it that you know they are the same? They share the same semantics!
Have you ever written a program with perfect syntax and bad semantics? Lots of people do. That's why they crash.
If you're pulling the article in, then it sounds very much like that people are making a syntax and have no way to map to another syntax while keeping the semantics of programs. It seems that most programmers are so tied up in syntax and representation (rocks vs. numerals) that they aren't thinking about the meaning of what they are writing.
I can't tell if you're demonstrating me wrong. You've just done two implementations of the natural numbers. How is it that you know they are the same? They share the same semantics!
As Marc Hamann pointed out, the word "semantics" is being used in two different senses here. What Marc called the "usual PLT sense" refers to an aspect of a language's definition that is (usually) necessary in order to determine the result that any given program will produce. However, as Frank pointed out, there are typically many ways of factoring something into syntactic and semantic parts. The point of the examples I gave was to show cases where evaluation operations are entirely syntactic, which illustrates an aspect of the point that Ehud was referring to. Incidentally, it also demonstrates that the PL semantics of integer addition may be different between different systems.
The sort of semantics you seem to be referring to, which Marc called "intentional PL semantics", are another matter. That's what I referred to as "translating between systems". Re-reading your original post, you mentioned "communicating meaning", and seem to be focusing on that communication aspect, as opposed to the way in which a result is arrived at. However, communicating meaning can't be done without first dealing with the "internal" PL theory semantics of the languages in question.
To relate this to your prefix/postfix/mixfix example, the "function with some semantics" that lies beneath the syntax first has a semantics in the PL theory sense. This semantics may be different for each of the different systems that implement integer addition, if the systems work in different ways. If the semantics were the same for each of them, they wouldn't really be different systems. To demonstrate that two different systems implement the "same" operation, you'd have to prove the necessary correspondence between their respective semantics.
In your example, you may have intended to focus on a case where the PL semantics were in fact identical in each case, and only the syntax differed. But the fact that any given semantics can support multiple syntaxes doesn't tell us much about the syntactic/semantic divide that Ehud referred to.
If you're pulling the article in, then it sounds very much like that people are making a syntax and have no way to map to another syntax while keeping the semantics of programs. It seems that most programmers are so tied up in syntax and representation (rocks vs. numerals) that they aren't thinking about the meaning of what they are writing.
Again, you're not talking about PL semantics here, which makes things sound a lot easier than they really are. You'd be absolutely right if all languages were simply syntactic layers over some universal language with a single common semantics. Alas, that's not the case. The problem is that languages have (PL) semantics which are non-trivial to map to other languages with different semantics.
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# Discrete LTI filter impulse response
If I have the unit impulse response function for a discrete-time LTI system (Unit sequence response?), h[n], how can I calculate the time taken for the output to fall below 1% of its initial value, after a unit impulse is applied to the input?
In particular, I have:
$$h[n]=(\alpha ^{-1}-\alpha )u[n-1]-\alpha \delta [n-1]$$
## Answers and Replies
berkeman
Mentor
If I have the unit impulse response function for a discrete-time LTI system (Unit sequence response?), h[n], how can I calculate the time taken for the output to fall below 1% of its initial value, after a unit impulse is applied to the input?
In particular, I have:
$$h[n]=(\alpha ^{-1}-\alpha )u[n-1]-\alpha \delta [n-1]$$
Can you just run a simulation? I've used Excel for that before.
Can you just run a simulation? I've used Excel for that before.
I could do, but I would like to know the method for determining it algebraically.
That h[n] doesn't seem to decay, so it will never fall bellow 1% of its initial value.
But, for others h[n] which do decay, I will solve for n this simple inequality:
h[n] < 0.01 · h[0]
That h[n] doesn't seem to decay, so it will never fall bellow 1% of its initial value.
But, for others h[n] which do decay, I will solve for n this simple inequality:
h[n] < 0.01 · h[0]
Thank you, it was meant to be:
$$h[n]=(1-\alpha ^2)\alpha ^{n-1}u[n-1]-\alpha \delta [n-1]$$
In which case it will converge for $$\alpha<1$$
I managed to get the inequality:
$$n < \frac{ln\left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right)}{ln\alpha}$$
Is this right?
Edit this could also be written as:
$$n < \log_\alpha \left( \frac{0.01\alpha(1-\alpha-\alpha^2)}{1-\alpha^2} \right)$$
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# Cut Pieces
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 69 Accepted Submission(s): 27
Problem Description
Suppose we have a sequence of n blocks. Then we paint the blocks. Each block should be painted a single color and block i can have color 1 to color ai. So there are a total of prod(ai) different ways to color the blocks.
Consider one way to color the blocks. We call a consecutive sequence of blocks with the same color a "piece". For example, sequence "Yellow Yellow Red" has two pieces and sequence "Yellow Red Blue Blue Yellow" has four pieces. What is S, the total number of pieces of all possible ways to color the blocks?
This is not your task. Your task is to permute the blocks (together with its corresponding ai) so that S is maximized.
Input
First line, number of test cases, T.
Following are 2*T lines. For every two lines, the first line is n, length of sequence; the second line contains n numbers, a1, ..., an
Sum of all n <= 106
All numbers in the input are positive integers no larger than 109.
Output
Output contains T lines.
Each line contains one number, the answer to the corresponding test case.
Since the answers can be very large, you should output them modulo 109+7.
Sample Input
1 3 1 2 3
Sample Output
14
Hint
Both sequence 1 3 2 and sequence 2 3 1 result in an S of 14.
Source
Recommend
zhuyuanchen520
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
//
const int V = 1000000 + 50;
const int MaxN = 80 + 5;
const int mod = 1000000000 + 7;
const __int64 INF = 0x7FFFFFFFFFFFFFFFLL;
const int inf = 0x7fffffff;
int T, n, num[V], ans[V];
__int64 dp[V], sum[V];
int main() {
int i, j;
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
for(i = 0; i < n; ++i)
scanf("%d", &num[i]);
sort(num, num + n);
int ii = 0, jj = n - 1;
for(i = 0; i < n; ++i) {
if(i % 2 == 0) {
ans[i] = num[ii];
ii++;
}
else {
ans[i] = num[jj];
jj--;
}
}
sum[n] = 1;
sum[n - 1] = ans[n - 1];
dp[n - 1] = ans[n - 1];
for(i = n - 2; i >= 0; --i) {
if(ans[i] >= ans[i + 1])
dp[i] = (ans[i] - ans[i + 1]) * (dp[i + 1] + sum[i + 1]) + (sum[i + 1] - sum[i + 2] + dp[i + 1]) * ans[i + 1];
else
dp[i] = ans[i] * (dp[i + 1] + sum[i + 1] - sum[i + 2]);
dp[i] %= mod;
dp[i] = (dp[i] + mod) % mod;
sum[i] = (sum[i + 1] * ans[i]) % mod;
}
printf("%d\n", dp[0]);
}
}
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# Adjust design of algorithm when using the algorithmic package
I am using overleaf and I have inserted an algorithm into an template for a conference. Unfortunately, the algorithm is not displayed in the way I like: -the caption should be above the steps -there should be two lines before the caption "Algorithm" above - there should be a line at the end of the algorithm
Basically it should look similar to this examples (line numbers are not necessary): https://math-linux.com/latex-26/faq/latex-faq/article/how-to-write-algorithm-and-pseudocode-in-latex-usepackage-algorithm-usepackage-algorithmic
I tried to do this like they did, but it did not work out and I had to use the "algorithmic" \usepackage{algorithmic}
EDIT: Here is the Latex Code
\documentclass[conference]{IEEEtran}
\IEEEoverridecommandlockouts
% The preceding line is only needed to identify funding in the first footnote. If that is unneeded, please comment it out.
\usepackage{cite}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{algorithmic}
\usepackage{graphicx}
\usepackage{textcomp}
\usepackage{gensymb}
\usepackage{xcolor}
\usepackage[options ]{algorithm2e}
\def\BibTeX{{\rm B\kern-.05em{\sc i\kern-.025em b}\kern-.08em
T\kern-.1667em\lower.7ex\hbox{E}\kern-.125emX}}
\begin{document}
\title{Title \\
{\footnotesize \textsuperscript{}}
}
\author{\IEEEauthorblockN{}
\IEEEauthorblockA{\textit{} \\
\textit{}\\
\\
}
\and
\IEEEauthorblockN{}
\IEEEauthorblockA{\textit{} \\
\textit{}\\
\\
}
\and
\IEEEauthorblockN{}
\IEEEauthorblockA{\textit{} \\
\textit{}\\
\\
}
}
\maketitle
\begin{abstract}
Abstract
\end{abstract}
\begin{IEEEkeywords}
component, formatting, style, styling, insert
\end{IEEEkeywords}
\section{Introduction}
\section{Optimization Problem}
\section{Methodology}
\label{section: Methodology}
\subsection{Method 1}
\begin{algorithm}[h]
\SetAlgoLined
\While{$t < Z$}{
\If{$T^{min} \leq T_t^{BS} \leq T^{max}$}{
Set $x_t = x_t^S$\;
}
\If{$T_t^{BS} > T^{max}$}{
Set $x_t = mDeg^{min}$\;
}
\If{$T_t^{BS} < T^{min}$}{
Set $x_t = 1$ \;
}
Set $t=t+1$\;
}
\caption{Algorithm 1}
\label{algo:BS}
\end{algorithm}
\subsection{Method 2}
\bibliography{bibtex}
\end{document}
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# Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/Practice Problems and Solutions
## Introduction
This is a compilation of problems and solutions from past numerical methods qualifying exams at the University of Maryland.
## August 2008
### Problem 1
Consider the system ${\displaystyle \displaystyle Ax=b}$ . The GMRES method starts with a point ${\displaystyle \displaystyle x_{0}}$ and normalizes the residual ${\displaystyle \displaystyle r_{0}=b-Ax_{0}}$ so that ${\displaystyle \textstyle v_{1}={\frac {r_{0}}{\nu }}}$ has 2-norm one. It then constructs orthonormal Krylov bases ${\displaystyle \scriptstyle V_{k}=(v_{1}\,v_{2}\,\cdots v_{m})}$ satisfying
${\displaystyle \displaystyle AV_{k}=V_{k+1}H_{k}}$
where ${\displaystyle \displaystyle H_{k}}$ is a ${\displaystyle \textstyle (k+1)\times k}$ upper Hessenberg matrix. One then looks for an approximation to ${\displaystyle \displaystyle x}$ of the form
${\displaystyle \displaystyle x(c)=x_{0}+V_{k}c}$
choosing ${\displaystyle \displaystyle c_{k}}$ so that ${\displaystyle \textstyle \|r(c)\|=\|b-Ax(c)\|}$ is minimized, where ${\displaystyle \textstyle \|\cdot \|}$ is the usual Euclidean norm.
#### Part 1a
Show that ${\displaystyle \displaystyle c_{k}}$ minimizes ${\displaystyle \|\nu e_{1}-H_{k}c\|}$ .
#### Solution 1a
We wish to show that
${\displaystyle \displaystyle \|b-Ax(c)\|=\|\nu e_{1}-H_{k}c\|}$
{\displaystyle {\begin{aligned}\|b-Ax(c)\|&=\|b-A(x_{0}+V_{k}c)\|\\&=\|b-Ax_{0}-AV_{k}c\|\\&=\|r_{0}-AV_{k}c\|\\&=\|r_{0}-V_{k+1}H_{k}c\|\\&=\|\nu v_{1}-V_{k+1}H_{k}c\|\\&=\|V_{k+1}\underbrace {(\nu e_{1}-H_{k}c)} _{h_{c}}\|\\&=(V_{k+1}h_{c},V_{k+1}h_{c})^{\frac {1}{2}}\\&=((V_{k+1}h_{c})^{T}V_{k+1}h_{c})^{\frac {1}{2}}\\&=(h_{c}^{T}V_{k+1}^{T}V_{k+1}h_{c})^{\frac {1}{2}}\\&=(h_{c}^{T}h_{c})^{\frac {1}{2}}\\&=\|h_{c}\|\\&=\|\nu e_{1}-H_{k}c\|\end{aligned}}}
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rnodeDAGWishart {BCDAG} R Documentation
## Draw one observation from a Normal-Inverse-Gamma distribution (internal function)
### Description
This function performs one draw from the Multivariate-Normal-Inverse-Gamma (prior/posterior) distribution of the parameters of a Normal linear regression model. Response variable is node and covariates are given by the parents of node in DAG. It is implemented node-by-node in rDAGWishart to obtain draws from a compatible (prior/posterior) DAG-Wishart distribution.
### Usage
rnodeDAGWishart(node, DAG, aj, U)
### Arguments
node numerical label of the node in DAG DAG (q,q) adjacency matrix of the DAG aj common shape hyperparameter of the compatible DAG-Wishart, a > q - 1 U position hyperparameter of the compatible DAG-Wishart, a (q, q) s.p.d. matrix
### Value
A list with two elements; a vector with one draw for the (vector) regression coefficient and a scalar with one draw for the conditional variance
[Package BCDAG version 1.0.0 Index]
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Warning: pg_query(): Query failed: ERROR: missing chunk number 0 for toast value 29512337 in pg_toast_2619 in /dati/webiit-old/includes/database.pgsql.inc on line 138 Warning: ERROR: missing chunk number 0 for toast value 29512337 in pg_toast_2619 query: SELECT data, created, headers, expire, serialized FROM cache_page WHERE cid = 'https://www-old.iit.cnr.it/node/927' in /dati/webiit-old/includes/database.pgsql.inc on line 159 Warning: pg_query(): Query failed: ERROR: missing chunk number 0 for toast value 29512337 in pg_toast_2619 in /dati/webiit-old/includes/database.pgsql.inc on line 138 Warning: ERROR: missing chunk number 0 for toast value 29512337 in pg_toast_2619 query: SELECT data, created, headers, expire, serialized FROM cache_page WHERE cid = 'https://www-old.iit.cnr.it/node/927' in /dati/webiit-old/includes/database.pgsql.inc on line 159 Packet Classification via Improved Space Decomposition Techniques | IIT - CNR - Istituto di Informatica e Telematica
## Packet Classification via Improved Space Decomposition Techniques
Packet Classification is a common task in modern Internet routers.
The goal is to classify packets into classes'' or
flows'' according to some ruleset that looks at multiple fields
of each packet. Differentiated actions can then be applied to the
traffic depending on the result of the classification.
Even though rulesets can be expressed in a relatively compact way
by using high level languages, the resulting decision trees can
partition the search space (the set of possible attribute
values) in a potentially very large (10^6 and more)
number of regions. This calls for methods that scale to
such large problem sizes, though the only scalable proposal
in the literature so far is the one based on a Fat Inverted Segment
Tree.
In this paper we propose a new geometric technique called {\em G-filter}
for packet classification on d dimensions.
G-filter is based on an improved space decomposition
technique. In addition to a theoretical analysis showing that
classification in G-filter has O(1) time complexity and
slightly super-linear space in the number of rules, we provide
thorough experiments showing that the constants
involved are extremely small on a wide range of problem sizes,
and that G-filter improve the best results in the literature
for large problem sizes, and is competitive for small sizes as well.
IEEE Infocom 2005, Miami, 2005
Autori: F. Geraci, M. Pellegrini, P. Pisati and L. Rizzo
Autori IIT:
Tipo: Articolo in Atti di convegno internazionale con referee
Area di disciplina: Information Technology and Communication Systems
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# An object's two dimensional velocity is given by v(t) = ( t-2 , 5t^2-3t). What is the object's rate and direction of acceleration at t=1 ?
##### 1 Answer
Feb 18, 2017
"acceleration"=dotv(1))=(1,7)ms^(-2)
" direction tan^(-1)(7)
#### Explanation:
if velocity is given a s a function of time the acceleration is found by differentiating the velocity function.
thus.
$v \left(t\right) = \left(t - 2 , 5 {t}^{2} - 3 t\right)$
$a = \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = \dot{v} \left(t\right) = \left(1 , 10 t - 3\right)$
the acceleration is then (assuming the measurements all are SI
$\dot{v} \left(1\right) = = \left(1 , 10 - 3\right) = \left(1 , 7\right) m {s}^{- 2}$
direction is ${\tan}^{- 1} \left(\frac{y}{x}\right)$
" direction tan^(-1)(7)
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# The fundamental group of the von Neumann algebra of a free group of infinite rank
It is well-known that that the fundamental group (in the sense of Murray and von Neumann) of the factor $L(F_{\mathbb{N}})$ is $\mathbb{R} \smallsetminus \{0\}$. I think that by the cutting and pasting technique and random matrix model, it is not hard to show that the fundamental group of $L(F_{S})$ is $\mathbb{R} \smallsetminus \{0\}$, where $S$ is an infinite set (not countable) and $F_S$ is the free group over $S$. Can anyone point me to some reference includes this kind of result? Thank you in advance!
• I don't know about others, but I, for one, have no clue what you are talking about. Can you please define your terms? Sep 1 '17 at 14:42
• $L(G)$ denotes the von Neumann algebra of a group $G$. This is standard terminology for von Neumann algebraists. "Factor" is in the sense of von Neumann. I agree that I was initially as stuck as Igor, given that not even the word "von Neumann algebra" was written by the OP...
– YCor
Mar 15 '18 at 22:04
• To a von Neumann algebra (at least those called $II_1$-factors) is associated a certain subgroup of $\mathbf{R}_{>0}$ called its "fundamental group". This was defined by Murray and von Neumann. Probably they had no more imagination this very day and now this is widespread terminology. It's unrelated to topologists's fundamental group, it's not even a group (but a subgroup of $\mathbf{R}_{>0}$ - sure it's a group but what's important is to remember which subgroup it is, not just the isomorphism class).
– YCor
Mar 15 '18 at 22:54
• For uses of "fundamental group" that is not Poincaré's original topological one, I tend (especially out of context) to use an additional name. For instance "Bass-Serre fundamental group". I'd see nothing against using, at least out of context (e.g., in the title of such a post), "Murray-von Neumann fundamental group".
– YCor
Mar 16 '18 at 6:13
F. Rădulescu. The fundamental group of the von Neumann algebra of a free group with infinitely many generators is $\mathbb{R}\smallsetminus\{0\}$. J. Amer. Math. Soc. 5(3) (1992), 517-532.
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Glossary
/
Financials
/
Loan to Value (LTV)
# Loan to Value (LTV)
### What is Loan to Value?
You might have heard your lender talk about Loan to Value and that they're unwilling to lend if the loan to value ratio is too high. What is loan to value anyways?
Loan to Value is the ratio of the amount you're borrowing versus how much the underlying property is worth. If you have a high loan to value (LTV) ratio, it means you're borrowing more money and investing less cash into the property. If you have a low loan to value (LTV) ratio, it means you're borrowing less money and investing more cash into the property yourself.
### How to calculate the Loan to Value Ratio?
Property investors can simply calculate the Loan to Value ratio using this formula:
${ Total \, amount \, borrowed \over Total \, appraised \, value }$
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# How do you simplify 2/x + 3/(2x^3)?
Jul 29, 2016
=$\frac{4 {x}^{2} + 3}{2 {x}^{3}}$
#### Explanation:
Treat algebraic fractions the same as arithmetic fractions.
Find the LCD first and make equivalent fractions.
$\frac{2}{x} + \frac{3}{2 {x}^{3}} \text{ } L C D = 2 {x}^{3}$
$\frac{2}{x} \times \frac{2 {x}^{2}}{2 {x}^{2}} + \frac{3}{2 {x}^{3}}$
=$\frac{4 {x}^{2} + 3}{2 {x}^{3}}$
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# Two likelihoods dependent on missing predictors
Hi,
I am trying to solve a silly indexing issue and realize I’m not sure I understand pymc3 syntax. Suppose I want to model a y that is normally distributed, and have predictors x0 and x1. x1 has only sometimes been measured, but I do not wish to impute it. Instead, I want to define two likelihoods for y. The two likelihoods would have different linear models, the less complex one where x1 is missing has, instead of x1 as a predictor, an additional residual uncertainty (sigma). Thus, when complete data is available, I want to model y like so:
y \sim \mathcal{N}(\mu, \sigma) \\ \mu = a+\beta_0 * x_0 + \beta_1 * x_1
and in the case I only know x0:
y \sim \mathcal{N}(\mu, \sigma + \sigma_{x1}) \\ \mu = a+\beta_0 * x_0
In stan I can do this like so (with N observations, and x1_measured an indicator vector):
// likelihood, which we only evaluate conditionally
if(run_estimation==1){
for (i in 1:N) {
if (x1_measured[i]==1) y[i] ~ normal(mu[i] , sigma ) ;
if (x1_measured[i]==0) y[i] ~ normal(mu[i] , sigma+sigma_x1 ) ; // more uncertainty when non-measured
}
I tried this approach in pymc3 (after defining priors etc, idx_without_m is an array of indeces):
with pm.Model() as model:
# define priors...(not shown)
mu_with = a + beta_x0*x0 + beta_x1*x1
mu_without = a + beta_x0 * x0
y_likelihood_with = pm.Normal("y_with",mu_with, sigma, observed=np.array(dw)[idx_with_m])
y_likelihood_without = pm.Normal("y_lik_without",mu_without, sigma+sigma_m, observed=np.array(dw)[idx_without_m])
Is this “canonical”? Should I used Theano shared variables etc?
That looks fine. (At least pymc syntax wise, I really don’t know about the imputation/non-imputation).
There is no need to use theano shared variables, that is only something you can use to switch out datasets.
Shouldn’t the second sigma be something like tt.sqrt(sigma**2 + sigma_m**2)?
1 Like
Thanks! And you are right.
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# Time-Delay Beamforming of Microphone ULA Array
This example shows how to perform wideband conventional time-delay beamforming with a microphone array of omnidirectional elements. Create an acoustic (pressure wave) chirp signal. The chirp signal has a bandwidth of 1 kHz and propagates at a speed of 340 m/s at ground level.
Note: This example runs only in R2016b or later. If you are using an earlier release, replace each call to the function with the equivalent step syntax. For example, replace myObject(x) with step(myObject,x).
c = 340;
t = linspace(0,1,50e3)';
sig = chirp(t,0,1,1000);
Collect the acoustic chirp with a ten-element ULA. Use omnidirectional microphone elements spaced less than one-half the wavelength at the 50 kHz sampling frequency. The chirp is incident on the ULA with an angle of $6{0}^{\circ }$ azimuth and ${0}^{\circ }$ elevation. Add random noise to the signal.
microphone = phased.OmnidirectionalMicrophoneElement(...
'FrequencyRange',[20 20e3]);
array = phased.ULA('Element',microphone,'NumElements',10,...
'ElementSpacing',0.01);
collector = phased.WidebandCollector('Sensor',array,'SampleRate',5e4,...
'PropagationSpeed',c,'ModulatedInput',false);
sigang = [60;0];
rsig = collector(sig,sigang);
rsig = rsig + 0.1*randn(size(rsig));
Apply a wideband conventional time-delay beamformer to improve the SNR of the received signal.
beamformer = phased.TimeDelayBeamformer('SensorArray',array,...
'SampleRate',5e4,'PropagationSpeed',c,'Direction',sigang);
y = beamformer(rsig);
subplot(2,1,1)
plot(t(1:5000),real(rsig(1:5e3,5)))
axis([0,t(5000),-0.5,1])
title('Signal (real part) at the 5th element of the ULA')
subplot(2,1,2)
plot(t(1:5000),real(y(1:5e3)))
axis([0,t(5000),-0.5,1])
title('Signal (real part) with time-delay beamforming')
xlabel('Seconds')
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Dual Spaces of Sobolev Spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:25:43Z http://mathoverflow.net/feeds/question/25293 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25293/dual-spaces-of-sobolev-spaces Dual Spaces of Sobolev Spaces euklid345 2010-05-19T22:38:27Z 2010-05-20T15:35:25Z <p>I will consider Sobolev spaces with $p=2$, only, so that they are Hilbert spaces. Hence the Sobolev inner product identifies each Sobolev space with its dual. In other words, I have an isomorphism $W_m\to (W_m)^\ast$ given by $x\mapsto \langle x,\cdot\rangle_m$</p> <p>Now, if $\sigma>0$, then I have an embedding $W_{m+\sigma}\hookrightarrow W_m$. Under the above isomorphism, how can I describe the image of $W_{m+\sigma}$ inside $W_m^\ast$? In particular, is there a $\tau>0$ such that $W_{m+\sigma}$ it identified with $(W_{m-\tau})^\ast$? </p> http://mathoverflow.net/questions/25293/dual-spaces-of-sobolev-spaces/25300#25300 Answer by Harald Hanche-Olsen for Dual Spaces of Sobolev Spaces Harald Hanche-Olsen 2010-05-20T00:24:05Z 2010-05-20T15:35:25Z <p>I am going to stick with the standard terminology $H^m$ here. Taking Fourier transforms one finds that <code>$$\langle u,v\rangle_m=\int\hat u(\xi)\bar{\hat v}(\xi)(1+|\xi|^2)^m\,d\xi$$</code> (give or take the odd multiplicative constant), where $H^m$ consists precisely of those $u\in L^2$ for which <code>$\langle u,u\rangle<\infty$</code>. This works even for <code>$m<0$</code>, if you allow distributions whose Fourier transforms are functions. Everything follows from this, including the fact that $H^{-m}$ acts as the dual of $H^m$ simply by the distribution $u$ acting on the function $v$, which corresponds to the integral <code>$$\langle u,v\rangle=\int \hat u(\xi)\bar{\hat v}(\xi)\,d\xi=\int \hat u(\xi)(1+|\xi|^2)^{-m/2}\cdot\hat{\bar v}(\xi)(1+|\xi|^2)^{m/2}\,d\xi$$</code> where I have split up the integrand into a product of two $L^2$ functions.</p> <p>For this reason, it seems more natural to identify $H^{-m}$ with the dual of $H^m$ than to identify $H^m$ with its own dual. However, you can go ahead and identify any Sobolev space with the dual of any other just by inserting a suitable power of $1+|\xi|^2$ in the integral defining the pairing between the two.</p> <p>Rather than coming straight out and answering your question, I'll leave it to you to ponder the consequences of the above. In particular, note that you we embed and identify you have to keep careful track of what space you have identified with whose dual, or you will be endlessly befuddled.</p> <p><strong>Addendum:</strong> To spell out a more direct answer to your question, <code>$\langle\cdot,\cdot\rangle_m$</code> can identify $H^{m+\sigma}$ with the dual of $H^{m-\sigma}$, since we can write <code>$$\langle u,v\rangle_m=\int \hat u(\xi)(1+|\xi|^2)^{(m-\sigma)/2}\cdot\bar{\hat v}(\xi)(1+|\xi|^2)^{(m+\sigma)/2}\,d\xi$$</code> where I have split the integrand into a product of two $L^2$ functions.</p> <p><strong>Edit:</strong> Changed a couple $\hat{\bar v}$ into $\bar{\hat v}$.</p>
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# Linear State Space Models¶
## Contents¶
“We may regard the present state of the universe as the effect of its past and the cause of its future” – Marquis de Laplace
## Overview¶
This lecture introduces the linear state space dynamic system
This model is a workhorse that carries a powerful theory of prediction
Its many applications include:
• representing dynamics of higher-order linear systems
• predicting the position of a system $j$ steps into the future
• predicting a geometric sum of future values of a variable like
• non financial income
• dividends on a stock
• the money supply
• a government deficit or surplus, etc.
• key ingredient of useful models
• Friedman’s permanent income model of consumption smoothing
• Barro’s model of smoothing total tax collections
• Rational expectations version of Cagan’s model of hyperinflation
• Sargent and Wallace’s “unpleasant monetarist arithmetic,” etc.
### Setup¶
In [1]:
using InstantiateFromURL
activate_github("QuantEcon/QuantEconLecturePackages", tag = "v0.9.7");
In [2]:
using LinearAlgebra, Statistics, Compat
## The Linear State Space Model¶
The objects in play are:
• An $n \times 1$ vector $x_t$ denoting the state at time $t = 0, 1, 2, \ldots$
• An iid sequence of $m \times 1$ random vectors $w_t \sim N(0,I)$
• A $k \times 1$ vector $y_t$ of observations at time $t = 0, 1, 2, \ldots$
• An $n \times n$ matrix $A$ called the transition matrix
• An $n \times m$ matrix $C$ called the volatility matrix
• A $k \times n$ matrix $G$ sometimes called the output matrix
Here is the linear state-space system
\begin{aligned} x_{t+1} & = A x_t + C w_{t+1} \\ y_t & = G x_t \nonumber \\ x_0 & \sim N(\mu_0, \Sigma_0) \nonumber \end{aligned} \tag{1}
### Primitives¶
The primitives of the model are
1. the matrices $A, C, G$
2. shock distribution, which we have specialized to $N(0,I)$
3. the distribution of the initial condition $x_0$, which we have set to $N(\mu_0, \Sigma_0)$
Given $A, C, G$ and draws of $x_0$ and $w_1, w_2, \ldots$, the model (1) pins down the values of the sequences $\{x_t\}$ and $\{y_t\}$
Even without these draws, the primitives 1–3 pin down the probability distributions of $\{x_t\}$ and $\{y_t\}$
Later we’ll see how to compute these distributions and their moments
#### Martingale difference shocks¶
We’ve made the common assumption that the shocks are independent standardized normal vectors
But some of what we say will be valid under the assumption that $\{w_{t+1}\}$ is a martingale difference sequence
A martingale difference sequence is a sequence that is zero mean when conditioned on past information
In the present case, since $\{x_t\}$ is our state sequence, this means that it satisfies
$$\mathbb{E} [w_{t+1} | x_t, x_{t-1}, \ldots ] = 0$$
This is a weaker condition than that $\{w_t\}$ is iid with $w_{t+1} \sim N(0,I)$
### Examples¶
By appropriate choice of the primitives, a variety of dynamics can be represented in terms of the linear state space model
The following examples help to highlight this point
They also illustrate the wise dictum finding the state is an art
#### Second-order difference equation¶
Let $\{y_t\}$ be a deterministic sequence that satisfies
$$y_{t+1} = \phi_0 + \phi_1 y_t + \phi_2 y_{t-1} \quad \text{s.t.} \quad y_0, y_{-1} \text{ given} \tag{2}$$
To map (2) into our state space system (1), we set
$$x_t= \begin{bmatrix} 1 \\ y_t \\ y_{t-1} \end{bmatrix} \qquad A = \begin{bmatrix} 1 & 0 & 0 \\ \phi_0 & \phi_1 & \phi_2 \\ 0 & 1 & 0 \end{bmatrix} \qquad C= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \qquad G = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix}$$
You can confirm that under these definitions, (1) and (2) agree
The next figure shows dynamics of this process when $\phi_0 = 1.1, \phi_1=0.8, \phi_2 = -0.8, y_0 = y_{-1} = 1$
Later you’ll be asked to recreate this figure
#### Univariate Autoregressive Processes¶
We can use (1) to represent the model
$$y_{t+1} = \phi_1 y_{t} + \phi_2 y_{t-1} + \phi_3 y_{t-2} + \phi_4 y_{t-3} + \sigma w_{t+1} \tag{3}$$
where $\{w_t\}$ is iid and standard normal
To put this in the linear state space format we take $x_t = \begin{bmatrix} y_t & y_{t-1} & y_{t-2} & y_{t-3} \end{bmatrix}'$ and
$$A = \begin{bmatrix} \phi_1 & \phi_2 & \phi_3 & \phi_4 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \qquad C = \begin{bmatrix} \sigma \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad G = \begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}$$
The matrix $A$ has the form of the companion matrix to the vector $\begin{bmatrix}\phi_1 & \phi_2 & \phi_3 & \phi_4 \end{bmatrix}$.
The next figure shows dynamics of this process when
$$\phi_1 = 0.5, \phi_2 = -0.2, \phi_3 = 0, \phi_4 = 0.5, \sigma = 0.2, y_0 = y_{-1} = y_{-2} = y_{-3} = 1$$
#### Vector Autoregressions¶
Now suppose that
• $y_t$ is a $k \times 1$ vector
• $\phi_j$ is a $k \times k$ matrix and
• $w_t$ is $k \times 1$
Then (3) is termed a vector autoregression
To map this into (1), we set
$$x_t = \begin{bmatrix} y_t \\ y_{t-1} \\ y_{t-2} \\ y_{t-3} \end{bmatrix} \quad A = \begin{bmatrix} \phi_1 & \phi_2 & \phi_3 & \phi_4 \\ I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & I & 0 \end{bmatrix} \quad C = \begin{bmatrix} \sigma \\ 0 \\ 0 \\ 0 \end{bmatrix} \quad G = \begin{bmatrix} I & 0 & 0 & 0 \end{bmatrix}$$
where $I$ is the $k \times k$ identity matrix and $\sigma$ is a $k \times k$ matrix
#### Seasonals¶
We can use (1) to represent
1. the deterministic seasonal $y_t = y_{t-4}$
2. the indeterministic seasonal $y_t = \phi_4 y_{t-4} + w_t$
In fact both are special cases of (3)
With the deterministic seasonal, the transition matrix becomes
$$A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$
It is easy to check that $A^4 = I$, which implies that $x_t$ is strictly periodic with period 4:[1]
$$x_{t+4} = x_t$$
Such an $x_t$ process can be used to model deterministic seasonals in quarterly time series.
The indeterministic seasonal produces recurrent, but aperiodic, seasonal fluctuations.
The model $y_t = a t + b$ is known as a linear time trend
We can represent this model in the linear state space form by taking
$$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \qquad C = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \qquad G = \begin{bmatrix} a & b \end{bmatrix} \tag{4}$$
and starting at initial condition $x_0 = \begin{bmatrix} 0 & 1\end{bmatrix}'$
In fact it’s possible to use the state-space system to represent polynomial trends of any order
For instance, let
$$x_0 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \qquad A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \qquad C = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
It follows that
$$A^t = \begin{bmatrix} 1 & t & t(t-1)/2 \\ 0 & 1 & t \\ 0 & 0 & 1 \end{bmatrix}$$
Then $x_t^\prime = \begin{bmatrix} t(t-1)/2 &t & 1 \end{bmatrix}$, so that $x_t$ contains linear and quadratic time trends
### Moving Average Representations¶
A nonrecursive expression for $x_t$ as a function of $x_0, w_1, w_2, \ldots, w_t$ can be found by using (1) repeatedly to obtain
\begin{aligned} x_t & = Ax_{t-1} + Cw_t \\ & = A^2 x_{t-2} + ACw_{t-1} + Cw_t \nonumber \\ & \qquad \vdots \nonumber \\ & = \sum_{j=0}^{t-1} A^j Cw_{t-j} + A^t x_0 \nonumber \end{aligned} \tag{5}
Representation (5) is a moving average representation
It expresses $\{x_t\}$ as a linear function of
1. current and past values of the process $\{w_t\}$ and
2. the initial condition $x_0$
As an example of a moving average representation, let the model be
$$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \qquad C = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$
You will be able to show that $A^t = \begin{bmatrix} 1 & t \cr 0 & 1 \end{bmatrix}$ and $A^j C = \begin{bmatrix} 1 & 0 \end{bmatrix}'$
Substituting into the moving average representation (5), we obtain
$$x_{1t} = \sum_{j=0}^{t-1} w_{t-j} + \begin{bmatrix} 1 & t \end{bmatrix} x_0$$
where $x_{1t}$ is the first entry of $x_t$
The first term on the right is a cumulated sum of martingale differences, and is therefore a martingale
The second term is a translated linear function of time
For this reason, $x_{1t}$ is called a martingale with drift
## Distributions and Moments¶
### Unconditional Moments¶
Using (1), it’s easy to obtain expressions for the (unconditional) means of $x_t$ and $y_t$
We’ll explain what unconditional and conditional mean soon
Letting $\mu_t := \mathbb{E} [x_t]$ and using linearity of expectations, we find that
$$\mu_{t+1} = A \mu_t \quad \text{with} \quad \mu_0 \text{ given} \tag{6}$$
Here $\mu_0$ is a primitive given in (1)
The variance-covariance matrix of $x_t$ is $\Sigma_t := \mathbb{E} [ (x_t - \mu_t) (x_t - \mu_t)']$
Using $x_{t+1} - \mu_{t+1} = A (x_t - \mu_t) + C w_{t+1}$, we can determine this matrix recursively via
$$\Sigma_{t+1} = A \Sigma_t A' + C C' \quad \text{with} \quad \Sigma_0 \text{ given} \tag{7}$$
As with $\mu_0$, the matrix $\Sigma_0$ is a primitive given in (1)
As a matter of terminology, we will sometimes call
• $\mu_t$ the unconditional mean of $x_t$
• $\Sigma_t$ the unconditional variance-convariance matrix of $x_t$
This is to distinguish $\mu_t$ and $\Sigma_t$ from related objects that use conditioning information, to be defined below
However, you should be aware that these “unconditional” moments do depend on the initial distribution $N(\mu_0, \Sigma_0)$
#### Moments of the Observations¶
Using linearity of expectations again we have
$$\mathbb{E} [y_t] = \mathbb{E} [G x_t] = G \mu_t \tag{8}$$
The variance-covariance matrix of $y_t$ is easily shown to be
$$\textrm{Var} [y_t] = \textrm{Var} [G x_t] = G \Sigma_t G' \tag{9}$$
### Distributions¶
In general, knowing the mean and variance-covariance matrix of a random vector is not quite as good as knowing the full distribution
However, there are some situations where these moments alone tell us all we need to know
These are situations in which the mean vector and covariance matrix are sufficient statistics for the population distribution
(Sufficient statistics form a list of objects that characterize a population distribution)
One such situation is when the vector in question is Gaussian (i.e., normally distributed)
This is the case here, given
1. our Gaussian assumptions on the primitives
2. the fact that normality is preserved under linear operations
In fact, it’s well-known that
$$u \sim N(\bar u, S) \quad \text{and} \quad v = a + B u \implies v \sim N(a + B \bar u, B S B') \tag{10}$$
In particular, given our Gaussian assumptions on the primitives and the linearity of (1) we can see immediately that both $x_t$ and $y_t$ are Gaussian for all $t \geq 0$ [2]
Since $x_t$ is Gaussian, to find the distribution, all we need to do is find its mean and variance-covariance matrix
But in fact we’ve already done this, in (6) and (7)
Letting $\mu_t$ and $\Sigma_t$ be as defined by these equations, we have
$$x_t \sim N(\mu_t, \Sigma_t) \tag{11}$$
By similar reasoning combined with (8) and (9),
$$y_t \sim N(G \mu_t, G \Sigma_t G') \tag{12}$$
### Ensemble Interpretations¶
How should we interpret the distributions defined by (11)(12)?
Intuitively, the probabilities in a distribution correspond to relative frequencies in a large population drawn from that distribution
Let’s apply this idea to our setting, focusing on the distribution of $y_T$ for fixed $T$
We can generate independent draws of $y_T$ by repeatedly simulating the evolution of the system up to time $T$, using an independent set of shocks each time
The next figure shows 20 simulations, producing 20 time series for $\{y_t\}$, and hence 20 draws of $y_T$
The system in question is the univariate autoregressive model (3)
The values of $y_T$ are represented by black dots in the left-hand figure
In the right-hand figure, these values are converted into a rotated histogram that shows relative frequencies from our sample of 20 $y_T$‘s
(The parameters and source code for the figures can be found in file linear_models/paths_and_hist.jl)
Here is another figure, this time with 100 observations
Let’s now try with 500,000 observations, showing only the histogram (without rotation)
The black line is the population density of $y_T$ calculated from (12)
The histogram and population distribution are close, as expected
By looking at the figures and experimenting with parameters, you will gain a feel for how the population distribution depends on the model primitives listed above, as intermediated by the distribution’s sufficient statistics
#### Ensemble means¶
In the preceding figure we approximated the population distribution of $y_T$ by
1. generating $I$ sample paths (i.e., time series) where $I$ is a large number
2. recording each observation $y^i_T$
3. histogramming this sample
Just as the histogram approximates the population distribution, the ensemble or cross-sectional average
$$\bar y_T := \frac{1}{I} \sum_{i=1}^I y_T^i$$
approximates the expectation $\mathbb{E} [y_T] = G \mu_T$ (as implied by the law of large numbers)
Here’s a simulation comparing the ensemble averages and population means at time points $t=0,\ldots,50$
The parameters are the same as for the preceding figures, and the sample size is relatively small ($I=20$)
The ensemble mean for $x_t$ is
$$\bar x_T := \frac{1}{I} \sum_{i=1}^I x_T^i \to \mu_T \qquad (I \to \infty)$$
The limit $\mu_T$ is a “long-run average”
(By long-run average we mean the average for an infinite ($I = \infty$) number of sample $x_T$‘s)
Another application of the law of large numbers assures us that
$$\frac{1}{I} \sum_{i=1}^I (x_T^i - \bar x_T) (x_T^i - \bar x_T)' \to \Sigma_T \qquad (I \to \infty)$$
### Joint Distributions¶
In the preceding discussion we looked at the distributions of $x_t$ and $y_t$ in isolation
This gives us useful information, but doesn’t allow us to answer questions like
• what’s the probability that $x_t \geq 0$ for all $t$?
• what’s the probability that the process $\{y_t\}$ exceeds some value $a$ before falling below $b$?
• etc., etc.
Such questions concern the joint distributions of these sequences
To compute the joint distribution of $x_0, x_1, \ldots, x_T$, recall that joint and conditional densities are linked by the rule
$$p(x, y) = p(y \, | \, x) p(x) \qquad \text{(joint }=\text{ conditional }\times\text{ marginal)}$$
From this rule we get $p(x_0, x_1) = p(x_1 \,|\, x_0) p(x_0)$
The Markov property $p(x_t \,|\, x_{t-1}, \ldots, x_0) = p(x_t \,|\, x_{t-1})$ and repeated applications of the preceding rule lead us to
$$p(x_0, x_1, \ldots, x_T) = p(x_0) \prod_{t=0}^{T-1} p(x_{t+1} \,|\, x_t)$$
The marginal $p(x_0)$ is just the primitive $N(\mu_0, \Sigma_0)$
In view of (1), the conditional densities are
$$p(x_{t+1} \,|\, x_t) = N(Ax_t, C C')$$
#### Autocovariance functions¶
An important object related to the joint distribution is the autocovariance function
$$\Sigma_{t+j, t} := \mathbb{E} [ (x_{t+j} - \mu_{t+j})(x_t - \mu_t)' ] \tag{13}$$
Elementary calculations show that
$$\Sigma_{t+j,t} = A^j \Sigma_t \tag{14}$$
Notice that $\Sigma_{t+j,t}$ in general depends on both $j$, the gap between the two dates, and $t$, the earlier date
## Stationarity and Ergodicity¶
Stationarity and ergodicity are two properties that, when they hold, greatly aid analysis of linear state space models
### Visualizing Stability¶
Let’s look at some more time series from the same model that we analyzed above
This picture shows cross-sectional distributions for $y$ at times $T, T', T''$
Note how the time series “settle down” in the sense that the distributions at $T'$ and $T''$ are relatively similar to each other — but unlike the distribution at $T$
Apparently, the distributions of $y_t$ converge to a fixed long-run distribution as $t \to \infty$
When such a distribution exists it is called a stationary distribution
### Stationary Distributions¶
In our setting, a distribution $\psi_{\infty}$ is said to be stationary for $x_t$ if
$$x_t \sim \psi_{\infty} \quad \text{and} \quad x_{t+1} = A x_t + C w_{t+1} \quad \implies \quad x_{t+1} \sim \psi_{\infty}$$
Since
1. in the present case all distributions are Gaussian
2. a Gaussian distribution is pinned down by its mean and variance-covariance matrix
we can restate the definition as follows: $\psi_{\infty}$ is stationary for $x_t$ if
$$\psi_{\infty} = N(\mu_{\infty}, \Sigma_{\infty})$$
where $\mu_{\infty}$ and $\Sigma_{\infty}$ are fixed points of (6) and (7) respectively
### Covariance Stationary Processes¶
Let’s see what happens to the preceding figure if we start $x_0$ at the stationary distribution
Now the differences in the observed distributions at $T, T'$ and $T''$ come entirely from random fluctuations due to the finite sample size
By
• our choosing $x_0 \sim N(\mu_{\infty}, \Sigma_{\infty})$
• the definitions of $\mu_{\infty}$ and $\Sigma_{\infty}$ as fixed points of (6) and (7) respectively
we’ve ensured that
$$\mu_t = \mu_{\infty} \quad \text{and} \quad \Sigma_t = \Sigma_{\infty} \quad \text{for all } t$$
Moreover, in view of (14), the autocovariance function takes the form $\Sigma_{t+j,t} = A^j \Sigma_\infty$, which depends on $j$ but not on $t$
This motivates the following definition
A process $\{x_t\}$ is said to be covariance stationary if
• both $\mu_t$ and $\Sigma_t$ are constant in $t$
• $\Sigma_{t+j,t}$ depends on the time gap $j$ but not on time $t$
In our setting, $\{x_t\}$ will be covariance stationary if $\mu_0, \Sigma_0, A, C$ assume values that imply that none of $\mu_t, \Sigma_t, \Sigma_{t+j,t}$ depends on $t$
### Conditions for Stationarity¶
#### The globally stable case¶
The difference equation $\mu_{t+1} = A \mu_t$ is known to have unique fixed point $\mu_{\infty} = 0$ if all eigenvalues of $A$ have moduli strictly less than unity
That is, if all(abs(eigvals(A)) .< 1) == true
The difference equation (7) also has a unique fixed point in this case, and, moreover
$$\mu_t \to \mu_{\infty} = 0 \quad \text{and} \quad \Sigma_t \to \Sigma_{\infty} \quad \text{as} \quad t \to \infty$$
regardless of the initial conditions $\mu_0$ and $\Sigma_0$
This is the globally stable case — see these notes for more a theoretical treatment
However, global stability is more than we need for stationary solutions, and often more than we want
To illustrate, consider our second order difference equation example
Here the state is $x_t = \begin{bmatrix} 1 & y_t & y_{t-1} \end{bmatrix}'$
Because of the constant first component in the state vector, we will never have $\mu_t \to 0$
How can we find stationary solutions that respect a constant state component?
#### Processes with a constant state component¶
To investigate such a process, suppose that $A$ and $C$ take the form
$$A = \begin{bmatrix} A_1 & a \\ 0 & 1 \end{bmatrix} \qquad C = \begin{bmatrix} C_1 \\ 0 \end{bmatrix}$$
where
• $A_1$ is an $(n-1) \times (n-1)$ matrix
• $a$ is an $(n-1) \times 1$ column vector
Let $x_t = \begin{bmatrix} x_{1t}' & 1 \end{bmatrix}'$ where $x_{1t}$ is $(n-1) \times 1$
It follows that
\begin{aligned} x_{1,t+1} & = A_1 x_{1t} + a + C_1 w_{t+1} \\ \end{aligned}
Let $\mu_{1t} = \mathbb{E} [x_{1t}]$ and take expectations on both sides of this expression to get
$$\mu_{1,t+1} = A_1 \mu_{1,t} + a \tag{15}$$
Assume now that the moduli of the eigenvalues of $A_1$ are all strictly less than one
Then (15) has a unique stationary solution, namely,
$$\mu_{1\infty} = (I-A_1)^{-1} a$$
The stationary value of $\mu_t$ itself is then $\mu_\infty := \begin{bmatrix} \mu_{1\infty}' & 1 \end{bmatrix}'$
The stationary values of $\Sigma_t$ and $\Sigma_{t+j,t}$ satisfy
\begin{aligned} \Sigma_\infty & = A \Sigma_\infty A' + C C' \\ \Sigma_{t+j,t} & = A^j \Sigma_\infty \nonumber \end{aligned} \tag{16}
Notice that here $\Sigma_{t+j,t}$ depends on the time gap $j$ but not on calendar time $t$
In conclusion, if
• $x_0 \sim N(\mu_{\infty}, \Sigma_{\infty})$ and
• the moduli of the eigenvalues of $A_1$ are all strictly less than unity
then the $\{x_t\}$ process is covariance stationary, with constant state component
Note
If the eigenvalues of $A_1$ are less than unity in modulus, then (a) starting from any initial value, the mean and variance-covariance matrix both converge to their stationary values; and (b) iterations on (7) converge to the fixed point of the discrete Lyapunov equation in the first line of (16)
### Ergodicity¶
Let’s suppose that we’re working with a covariance stationary process
In this case we know that the ensemble mean will converge to $\mu_{\infty}$ as the sample size $I$ approaches infinity
#### Averages over time¶
Ensemble averages across simulations are interesting theoretically, but in real life we usually observe only a single realization $\{x_t, y_t\}_{t=0}^T$
So now let’s take a single realization and form the time series averages
$$\bar x := \frac{1}{T} \sum_{t=1}^T x_t \quad \text{and} \quad \bar y := \frac{1}{T} \sum_{t=1}^T y_t$$
Do these time series averages converge to something interpretable in terms of our basic state-space representation?
The answer depends on something called ergodicity
Ergodicity is the property that time series and ensemble averages coincide
More formally, ergodicity implies that time series sample averages converge to their expectation under the stationary distribution
In particular,
• $\frac{1}{T} \sum_{t=1}^T x_t \to \mu_{\infty}$
• $\frac{1}{T} \sum_{t=1}^T (x_t -\bar x_T) (x_t - \bar x_T)' \to \Sigma_\infty$
• $\frac{1}{T} \sum_{t=1}^T (x_{t+j} -\bar x_T) (x_t - \bar x_T)' \to A^j \Sigma_\infty$
In our linear Gaussian setting, any covariance stationary process is also ergodic
## Noisy Observations¶
In some settings the observation equation $y_t = Gx_t$ is modified to include an error term
Often this error term represents the idea that the true state can only be observed imperfectly
To include an error term in the observation we introduce
• An iid sequence of $\ell \times 1$ random vectors $v_t \sim N(0,I)$
• A $k \times \ell$ matrix $H$
and extend the linear state-space system to
\begin{aligned} x_{t+1} & = A x_t + C w_{t+1} \\ y_t & = G x_t + H v_t \nonumber \\ x_0 & \sim N(\mu_0, \Sigma_0) \nonumber \end{aligned} \tag{17}
The sequence $\{v_t\}$ is assumed to be independent of $\{w_t\}$
The process $\{x_t\}$ is not modified by noise in the observation equation and its moments, distributions and stability properties remain the same
The unconditional moments of $y_t$ from (8) and (9) now become
$$\mathbb{E} [y_t] = \mathbb{E} [G x_t + H v_t] = G \mu_t \tag{18}$$
The variance-covariance matrix of $y_t$ is easily shown to be
$$\textrm{Var} [y_t] = \textrm{Var} [G x_t + H v_t] = G \Sigma_t G' + HH' \tag{19}$$
The distribution of $y_t$ is therefore
$$y_t \sim N(G \mu_t, G \Sigma_t G' + HH')$$
## Prediction¶
The theory of prediction for linear state space systems is elegant and simple
### Forecasting Formulas – Conditional Means¶
The natural way to predict variables is to use conditional distributions
For example, the optimal forecast of $x_{t+1}$ given information known at time $t$ is
$$\mathbb{E}_t [x_{t+1}] := \mathbb{E} [x_{t+1} \mid x_t, x_{t-1}, \ldots, x_0 ] = Ax_t$$
The right-hand side follows from $x_{t+1} = A x_t + C w_{t+1}$ and the fact that $w_{t+1}$ is zero mean and independent of $x_t, x_{t-1}, \ldots, x_0$
That $\mathbb{E}_t [x_{t+1}] = \mathbb{E}[x_{t+1} \mid x_t]$ is an implication of $\{x_t\}$ having the Markov property
$$x_{t+1} - \mathbb{E}_t [x_{t+1}] = Cw_{t+1}$$
The covariance matrix of the forecast error is
$$\mathbb{E} [ (x_{t+1} - \mathbb{E}_t [ x_{t+1}] ) (x_{t+1} - \mathbb{E}_t [ x_{t+1}])'] = CC'$$
More generally, we’d like to compute the $j$-step ahead forecasts $\mathbb{E}_t [x_{t+j}]$ and $\mathbb{E}_t [y_{t+j}]$
With a bit of algebra we obtain
$$x_{t+j} = A^j x_t + A^{j-1} C w_{t+1} + A^{j-2} C w_{t+2} + \cdots + A^0 C w_{t+j}$$
In view of the iid property, current and past state values provide no information about future values of the shock
Hence $\mathbb{E}_t[w_{t+k}] = \mathbb{E}[w_{t+k}] = 0$
It now follows from linearity of expectations that the $j$-step ahead forecast of $x$ is
$$\mathbb{E}_t [x_{t+j}] = A^j x_t$$
The $j$-step ahead forecast of $y$ is therefore
$$\mathbb{E}_t [y_{t+j}] = \mathbb{E}_t [G x_{t+j} + H v_{t+j}] = G A^j x_t$$
### Covariance of Prediction Errors¶
It is useful to obtain the covariance matrix of the vector of $j$-step-ahead prediction errors
$$x_{t+j} - \mathbb{E}_t [ x_{t+j}] = \sum^{j-1}_{s=0} A^s C w_{t-s+j} \tag{20}$$
Evidently,
$$V_j := \mathbb{E}_t [ (x_{t+j} - \mathbb{E}_t [x_{t+j}] ) (x_{t+j} - \mathbb{E}_t [x_{t+j}] )^\prime ] = \sum^{j-1}_{k=0} A^k C C^\prime A^{k^\prime} \tag{21}$$
$V_j$ defined in (21) can be calculated recursively via $V_1 = CC'$ and
$$V_j = CC^\prime + A V_{j-1} A^\prime, \quad j \geq 2 \tag{22}$$
$V_j$ is the conditional covariance matrix of the errors in forecasting $x_{t+j}$, conditioned on time $t$ information $x_t$
Under particular conditions, $V_j$ converges to
$$V_\infty = CC' + A V_\infty A' \tag{23}$$
Equation (23) is an example of a discrete Lyapunov equation in the covariance matrix $V_\infty$
A sufficient condition for $V_j$ to converge is that the eigenvalues of $A$ be strictly less than one in modulus.
Weaker sufficient conditions for convergence associate eigenvalues equaling or exceeding one in modulus with elements of $C$ that equal $0$
### Forecasts of Geometric Sums¶
In several contexts, we want to compute forecasts of geometric sums of future random variables governed by the linear state-space system (1)
We want the following objects
• Forecast of a geometric sum of future $x$‘s, or $\mathbb{E}_t \left[ \sum_{j=0}^\infty \beta^j x_{t+j} \right]$
• Forecast of a geometric sum of future $y$‘s, or $\mathbb{E}_t \left[\sum_{j=0}^\infty \beta^j y_{t+j} \right]$
These objects are important components of some famous and interesting dynamic models
For example,
• if $\{y_t\}$ is a stream of dividends, then $\mathbb{E} \left[\sum_{j=0}^\infty \beta^j y_{t+j} | x_t \right]$ is a model of a stock price
• if $\{y_t\}$ is the money supply, then $\mathbb{E} \left[\sum_{j=0}^\infty \beta^j y_{t+j} | x_t \right]$ is a model of the price level
#### Formulas¶
Fortunately, it is easy to use a little matrix algebra to compute these objects
Suppose that every eigenvalue of $A$ has modulus strictly less than $\frac{1}{\beta}$
It then follows that $I + \beta A + \beta^2 A^2 + \cdots = \left[I - \beta A \right]^{-1}$
• Forecast of a geometric sum of future $x$‘s
$$\mathbb{E}_t \left[\sum_{j=0}^\infty \beta^j x_{t+j} \right] = [I + \beta A + \beta^2 A^2 + \cdots \ ] x_t = [I - \beta A]^{-1} x_t$$
• Forecast of a geometric sum of future $y$‘s
$$\mathbb{E}_t \left[\sum_{j=0}^\infty \beta^j y_{t+j} \right] = G [I + \beta A + \beta^2 A^2 + \cdots \ ] x_t = G[I - \beta A]^{-1} x_t$$
## Code¶
Our preceding simulations and calculations are based on code in the file lss.jl from the QuantEcon.jl package
The code implements a type which the linear state space models can act on directly through specific methods (for simulations, calculating moments, etc.)
Examples of usage are given in the solutions to the exercises
## Exercises¶
### Exercise 1¶
Replicate this figure using the LSS type from lss.jl
### Exercise 2¶
Replicate this figure modulo randomness using the same type
### Exercise 3¶
Replicate this figure modulo randomness using the same type
The state space model and parameters are the same as for the preceding exercise
### Exercise 4¶
Replicate this figure modulo randomness using the same type
The state space model and parameters are the same as for the preceding exercise, except that the initial condition is the stationary distribution
Hint: You can use the stationary_distributions method to get the initial conditions
The number of sample paths is 80, and the time horizon in the figure is 100
Producing the vertical bars and dots is optional, but if you wish to try, the bars are at dates 10, 50 and 75
## Solutions¶
In [3]:
using QuantEcon, Plots
gr(fmt=:png);
### Exercise 1¶
In [4]:
ϕ0, ϕ1, ϕ2 = 1.1, 0.8, -0.8
A = [1.0 0.0 0
ϕ0 ϕ1 ϕ2
0.0 1.0 0.0]
C = zeros(3, 1)
G = [0.0 1.0 0.0]
μ_0 = ones(3)
lss = LSS(A, C, G; mu_0=μ_0)
x, y = simulate(lss, 50)
plot(dropdims(y, dims = 1), color = :blue, linewidth = 2, alpha = 0.7)
plot!(xlabel="time", ylabel = "y_t", legend = :none)
Out[4]:
### Exercise 2¶
In [5]:
using Random
Random.seed!(42) # For deterministic results.
ϕ1, ϕ2, ϕ3, ϕ4 = 0.5, -0.2, 0, 0.5
σ = 0.2
A = [ϕ1 ϕ2 ϕ3 ϕ4
1.0 0.0 0.0 0.0
0.0 1.0 0.0 0.0
0.0 0.0 1.0 0.0]
C = [σ
0.0
0.0
0.0]''
G = [1.0 0.0 0.0 0.0]
ar = LSS(A, C, G; mu_0 = ones(4))
x, y = simulate(ar, 200)
plot(dropdims(y, dims = 1), color = :blue, linewidth = 2, alpha = 0.7)
plot!(xlabel="time", ylabel = "y_t", legend = :none)
Out[5]:
### Exercise 3¶
In [6]:
ϕ1, ϕ2, ϕ3, ϕ4 = 0.5, -0.2, 0, 0.5
σ = 0.1
A = [ ϕ1 ϕ2 ϕ3 ϕ4
1.0 0.0 0.0 0.0
0.0 1.0 0.0 0.0
0.0 0.0 1.0 0.0]
C = [σ
0.0
0.0
0.0]
G = [1.0 0.0 0.0 0.0]
I = 20
T = 50
ar = LSS(A, C, G; mu_0 = ones(4))
ymin, ymax = -0.5, 1.15
ensemble_mean = zeros(T)
ys = []
for i ∈ 1:I
x, y = simulate(ar, T)
y = dropdims(y, dims = 1)
push!(ys, y)
ensemble_mean .+= y
end
ensemble_mean = ensemble_mean ./ I
plot(ys, color = :blue, alpha = 0.2, linewidth = 0.8, label = "")
plot!(ensemble_mean, color = :blue, linewidth = 2, label = "y_t_bar")
m = moment_sequence(ar)
pop_means = zeros(0)
for (i, t) ∈ enumerate(m)
(μ_x, μ_y, Σ_x, Σ_y) = t
push!(pop_means, μ_y[1])
i == 50 && break
end
plot!(pop_means, color = :green, linewidth = 2, label = "G mu_t")
plot!(ylims=(ymin, ymax), xlabel = "time", ylabel = "y_t", legendfont = font(12))
Out[6]:
### Exercise 4¶
In [7]:
ϕ1, ϕ2, ϕ3, ϕ4 = 0.5, -0.2, 0, 0.5
σ = 0.1
A = [ϕ1 ϕ2 ϕ3 ϕ4
1.0 0.0 0.0 0.0
0.0 1.0 0.0 0.0
0.0 0.0 1.0 0.0]
C = [σ
0.0
0.0
0.0]''
G = [1.0 0.0 0.0 0.0]
T0 = 10
T1 = 50
T2 = 75
T4 = 100
ar = LSS(A, C, G; mu_0 = ones(4))
ymin, ymax = -0.6, 0.6
μ_x, μ_y, Σ_x, Σ_y = stationary_distributions(ar)
ar = LSS(A, C, G; mu_0=μ_x, Sigma_0=Σ_x)
colors = ["c", "g", "b"]
ys = []
x_scatter = []
y_scatter = []
for i ∈ 1:80
rcolor = colors[rand(1:3)]
x, y = simulate(ar, T4)
y = dropdims(y, dims = 1)
push!(ys, y)
x_scatter = [x_scatter; T0; T1; T2]
y_scatter = [y_scatter; y[T0]; y[T1]; y[T2]]
end
plot(ys, linewidth = 0.8, alpha = 0.5)
plot!([T0 T1 T2; T0 T1 T2], [-1 -1 -1; 1 1 1], color = :black, legend = :none)
scatter!(x_scatter, y_scatter, color = :black, alpha = 0.5)
plot!(ylims=(ymin, ymax), ylabel = "y_t", xticks =[], yticks = ymin:0.2:ymax)
plot!(annotations = [(T0+1, -0.55, "T");(T1+1, -0.55, "T'");(T2+1, -0.55, "T''")])
Out[7]:
Footnotes
[1] The eigenvalues of $A$ are $(1,-1, i,-i)$.
[2] The correct way to argue this is by induction. Suppose that $x_t$ is Gaussian. Then (1) and (10) imply that $x_{t+1}$ is Gaussian. Since $x_0$ is assumed to be Gaussian, it follows that every $x_t$ is Gaussian. Evidently this implies that each $y_t$ is Gaussian.
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# What is the isosceles triangle theorem?
Jan 2, 2016
If two sides of a triangle are congruent, the angles opposite them are congruent.
#### Explanation:
If...
$\overline{\text{AB")congbar("AC}}$
then...
$\angle \text{B"congangle"C}$
If two sides of a triangle are congruent, the angles opposite them are congruent.
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# Local-measurement-based quantum state tomography via neural networks
## Abstract
Quantum state tomography is a daunting challenge of experimental quantum computing, even in moderate system size. One way to boost the efficiency of state tomography is via local measurements on reduced density matrices, but the reconstruction of the full state thereafter is hard. Here, we present a machine-learning method to recover the ground states of $$k$$-local Hamiltonians from just the local information, where a fully connected neural network is built to fulfill the task with up to seven qubits. In particular, we test the neural network model with a practical dataset, that in a 4-qubit nuclear magnetic resonance system our method yields global states via the 2-local information with high accuracy. Our work paves the way towards scalable state tomography in large quantum systems.
## Introduction
Quantum state tomography (QST) plays a vital role in validating and benchmarking quantum devices,1,2,3,4,5 because it can completely capture properties of an arbitrary quantum state. However, QST is not feasible for large systems because of its need for exponential resources. In recent years, there has been extensive research on methods for boosting the efficiency of QST.6,7,8,9,10,11,12 One of the promising candidates among these methods is QST via reduced density matrices (RDMs);13,14,15,16,17,18,19 because local measurements are convenient and accurate on many experimental platforms.
QST via RDMs is also a useful tool for characterizing ground states of local Hamiltonians. A many-body Hamiltonian $$H$$ is $$k$$-local if $$H={\sum }_{i}{H}_{i}^{(k)}$$, where each term $${H}_{i}^{(k)}$$ acts non-trivially on at most $$k$$ particles. For $$k$$-local Hamiltonians, only polynomial number of parameters are needed to characterize the whole system. Moreover, generally, a single eigenstate of such $$k$$-local Hamiltonian can encode the information of the system.18,20,21 Therefore, for these ground states, one only needs $$k$$-local measurements for state tomography. Although local measurements are efficient and even if $$\left|\psi \right\rangle$$ is uniquely determined by its $$k$$-local measurements, reconstructing $$\left|\psi \right\rangle$$ from its $$k$$-local measurements is computationally hard.22 We remark that this is not due to the problem that $$\left|\psi \right\rangle$$ needs to be described by exponentially many parameters. In fact, in many cases, ground states of $$k$$-local Hamiltonians can be effectively represented by tensor product states.18,23
The state reconstruction problem naturally connects to the regression problem in supervised learning. Regression analysis, in general, seeks to discover the relation between inputs and outputs, i.e., to recover the underlying mathematical model. Unsupervised learning techniques have been applied to QST in various cases, such as in refs. 24,25 In our case, as shown in Fig. 1, by knowing the Hamiltonian $$H$$, it is relatively easy to get the ground state $$\left|{\psi }_{H}\right\rangle$$ since the ground state is nothing but the eigenvector corresponding to the smallest eigenvalue. And then we could naturally achieve the $$k$$-local measurements $${\bf{M}}$$ of $$\left|{\psi }_{H}\right\rangle$$. Therefore, the data for tuning our reverse engineering model is accessible, which allows us to realize QST through supervised learning practically. Additionally, artificial neural networks are often noise tolerable,26,27,28 so they are favorable for working with experimental data.
In this work, we propose a local-measurement-based QST by fully connected feedforward neural network, in which every neuron connects to every neuron in the next layer and information only passes forward (i.e., there is no loop in the network). We first build a fully connected feedforward neural network for $$4$$-qubit ground states of fully connected $$2$$-local Hamiltonians. Our trained $$4$$-qubit network not only analyzes the test dataset with high fidelity but also reconstruct $$4$$-qubit nuclear magnetic resonance (NMR) experimental states accurately. We use the $$4$$-qubit case to demonstrate the potential of using neural networks to realize QST via $$k$$-local measurements. The versatile framework of neural networks for recovering ground states of $$k$$-local Hamiltonians could be extended to more qubits and various interaction structures; we then apply our methods to the ground states of seven-qubit 2-local Hamiltonians with nearest-neighbor couplings. In both cases, neural networks give accurate estimates with high fidelities. We observe our framework yields higher efficiency and better noise tolerance compared with the least-squares tomography (the approximated maximum likelihood estimation (MLE)) when the added noise is >5%.
## Results
### Theory
The universal approximation theorem29 states that every continuous function on the compact subsets of $${{\mathbb{R}}}^{n}$$ can be approximated by a multi-layer feedforward neural network with a finite number of neurons, i.e., computational units. And by observing the relation between $$k$$-local Hamiltonian and local measurements of its ground state, as shown in Fig. 1, we are empowered to turn the tomography problem to a regression problem which fits perfectly into the neural network framework.
In particular, we first construct a deep neural network for $$4$$-qubit ground states of full $$2$$-local Hamiltonians as follows:
$$H=\sum _{i=1}^{4}\sum _{1\le k\le 3}{\omega }_{k}^{(i)}{\sigma }_{k}^{(i)}+\sum _{1\le i<j\le 4}\sum _{1\le n,m\le 3}{J}_{nm}^{(ij)}{\sigma }_{n}^{(i)}\otimes {\sigma }_{m}^{(j)},$$
(1)
where $${\sigma }_{k},{\sigma }_{n},{\sigma }_{m}\in \Delta$$, and $$\Delta =\{{\sigma }_{1}={\sigma }_{x},{\sigma }_{2}={\sigma }_{y},{\sigma }_{3}={\sigma }_{z},{\sigma }_{4}=I\}$$.
We denote the set of Hamiltonian coefficients as $$\overrightarrow{h}=\{{\omega }_{k}^{(i)},{J}_{nm}^{(ij)}\}$$. The coefficient vector $$\overrightarrow{h}$$ is the vector representation of $$H$$ according to the basis set $${\bf{B}}=\{{\sigma }_{m}\otimes {\sigma }_{n}:n+m\,\ne \,8,{\sigma }_{m},{\sigma }_{n}\in \Delta \}$$. The configuration of the ground states is illustrated in Fig. 2a.
The number of parameters of the local observables of ground states determines the amount of input units of the neural network. Concretely, $${\bf{M}}=\{{s}_{m,n}^{(i,j)}:{s}_{m,n}^{(i,j)}={\rm{Tr}}({{\rm{Tr}}}_{(i,j)}\rho \cdot {B}_{(m,n)}),{B}_{(m,n)}\in {\bf{B}},1\le i\;<\;j\le 4,1\le n,m\le 4\}$$, where $${\sigma }_{n},{\sigma }_{m}\in \Delta$$ and $$\rho$$ is the density matrix of the ground state. $${\bf{M}}$$ is a set of true expectation values $${s}_{m,n}^{(i,j)}$$ of the local observables $${B}_{(m,n)}$$ in the ground states $$\rho$$. Notice that we are using the true values of expectation values instead of their estimations (which contain statistical fluctuations), since we are theoretically generating all the training and testing data. The input layer has $$66$$ neurons since the cardinality of the set of measurement results is $$66$$. Our network then contains two fully connected hidden layers, in which every neuron in the previous layer is connected to every neuron in the next layer. The number of output units equals to the number of parameters of our $$2$$-local Hamiltonian, which is $$66$$ in our $$4$$-qubit case. More details of our neural network can be found in “Methods” section.
Our training data consist of the 120,000 randomly generated $$2$$-local Hamiltonians as output and the local measurements of their corresponding ground states. The test data include 5000 pairs of Hamiltonians and local measurement results $$({H}_{i},{{\bf{M}}}_{i})$$.
We train the network by a popular optimizer in the machine-learning community called Adam (adaptive moment estimation).30,31 For loss function, we choose cosine proximity $$\cos (\theta )=({\overrightarrow{h}}_{{\rm{pred}}}\cdot \overrightarrow{h})/(\parallel {\overrightarrow{h}}_{{\rm{pred}}}\parallel \cdot \parallel \overrightarrow{h}\parallel )$$, where $${\overrightarrow{h}}_{{\rm{pred}}}$$ is the estimate of the neural network and $$\overrightarrow{h}$$ is the desired output. Generally speaking, the role of loss functions in supervised learning is to efficiently measure the distance between the true value and the estimated outcome. (In our case, it is the distance between $$\overrightarrow{h}$$ and $${\overrightarrow{h}}_{{\rm{pred}}}$$). And the training procedure seeks to minimize this distance. We find the cosine proximity function fits our scenario better than the more commonly chosen loss functions, such as mean square error or mean absolute error.32 The reason can be understood as follows. Because the parameter vector $$\overrightarrow{h}$$ is a representation of the corresponding Hamiltonian in the Hilbert space expanded by the local operators $${\bf{B}}$$, the angle $$\theta$$ between the two vectors $$\overrightarrow{h}$$ and $${\overrightarrow{h}}_{{\rm{pred}}}$$ is a “directional distance measure” between two corresponding Hamiltonians.20 Notice that the Hamiltonian corresponding to the parameter $$\overrightarrow{h}$$ has the same eigenvectors as those of the Hamiltonian of $$c\cdot \overrightarrow{h}$$, where $$c\in {\mathbb{R}}$$ is a constant. In other words, we only care about the “directional distance”. Instead of forcing every single element close to its true value (as mean squared error or mean absolute error does), the cosine loss function tends to train the angle $$\theta$$ towards zero, which is more adapted to our situation.
As illustrated in Fig. 1, after getting estimated Hamiltonian from the neural network, we calculate the ground state $${\rho }_{{\rm{nn}}}$$ of the estimated Hamiltonian and take the result as the estimate of the ground state that we attempt to recover. We remark that our estimated Hamiltonian is not necessarily exactly the same as the original Hamiltonian; Even if that happens, our numeric results suggest their ground states are still close.
There are two different fidelity functions that we can use to measure the distance between the randomly generated states $${\rho }_{{\rm{rd}}}$$ and our neural network estimated states $${\rho }_{{\rm{nn}}}$$, namley:
$$f({\rho }_{1},{\rho }_{2})\equiv {\rm{Tr}}\sqrt{\sqrt{{\rho }_{1}}{\rho }_{2}\sqrt{{\rho }_{1}}},$$
(2)
$$C({\rho }_{1},{\rho }_{2})\equiv \frac{{\rm{Tr}}({\rho }_{1}{\rho }_{2})}{\sqrt{{\rm{Tr}}({\rho }_{1}^{2})}\cdot \sqrt{{\rm{Tr}}({\rho }_{2}^{2})}}.$$
(3)
The fidelity measure $$f$$ defined in Eq. (2) is standard,33 which requires the matrix $${\rho }_{1}$$ and $${\rho }_{2}$$ to be positive semi-definite. Considering that the density matrix obtained directly from the raw data of a state tomography experiment may possibly not be positive definite, we usually adopt the definition of $$C$$ for processing the raw data in NMR.34 In this work, there are not negative matrices after constraining the raw density matrices to be positive semi-definite. The values of the fidelities are calculated by $$f$$ if there is no additional explanation in the following.
After supervised learning on the training data, our neural network is capable of estimating the 4-qubit output of the test set with high performance. The fidelity averaged over the whole test set is 98.7%. The maximum, minimum, standard deviations of fidelities for the test set are shown in Table 1. Figure 2c illustrates the fidelities between 100 random states $${\rho }_{{\rm{rd}}}$$ and our neural network estimates $${\rho }_{{\rm{nn}}}$$.
Our framework generalizes directly to more qubits and different interaction patterns. We apply our framework to recover 7-qubit ground states of $$2$$-local Hamiltonians with nearest-neighbor interaction. The configuration of our $$7$$-qubit states is shown in Fig. 2b. The Hamiltonian of this 7-qubit case is
$$H=\sum _{i=1}^{7}\sum _{1\le k\le 3}{\omega }_{k}^{(i)}{\sigma }_{k}^{(i)}+\sum _{i=1}^{6}\sum _{1\le n,m\le 3}{J}_{nm}^{(i)}{\sigma }_{n}^{(i)}\otimes {\sigma }_{m}^{(i+1)},$$
(4)
where $${\sigma }_{k},{\sigma }_{n},{\sigma }_{m}\in \Delta$$, $${\omega }_{k}^{(i)}$$ and $${J}_{nm}^{(i)}$$ are coefficients. We trained a similar neural network with 250,000 pairs of randomly generated Hamiltonians and $$2$$-local measurements of the corresponding ground states. For the 5000 randomly generated test sets, the network estimates have an average fidelity of 97.9%. More statistical performance are shown in Table 1 and fidelity results of 100 random generated states are shown in Fig. 2d.
Due to the variance inherent to this method, it is natural to ask how to determine whether a neural network estimate $${\rho }_{{\rm{nn}}}$$ is acceptable without knowing the true state $${\rho }_{{\rm{rd}}}$$. This problem can be easily solved by calculating the measurement estimate $${{\bf{M}}}_{{\rm{pred}}}$$, i.e., using the estimate $${\rho }_{{\rm{nn}}}$$ to measure the set of local operators $${\bf{B}}$$. By setting an acceptable error bound and comparing $${{\bf{M}}}_{{\rm{pred}}}$$ with the true measurements $${\bf{M}}$$, one can decide whether to accept $${\rho }_{{\rm{nn}}}$$ or not. Please see the “Methods” section for details.
Our neural-network-based framework is also significantly faster than the approximated MLE method. Once the network is trained sufficiently well, it can be used to deal with thousands of datasets without much effort on a regular computer. Calculating $${\rho }_{{\rm{nn}}}$$ from $${\overrightarrow{h}}_{{\rm{pred}}}$$, which is essentially the computation of the eigenvector corresponding to the smallest eigenvalue, is the only part that may take some time. Detailed discussions could be found in the “Methods” section.
### Experiment
So far, our theoretical model is noise-free. To demonstrate that our trained machine-learning model is resilient to experimental noises, we experimentally prepare the ground states of the random Hamiltonians and then try to reconstruct the final quantum states from 2-local measurements using a four-qubit NMR platform.35,36,37,38 The four-qubit sample is 13C-labeled trans-crotonic acid dissolved in d6-acetone, where C1–C4 are encoded as the four work qubits, and the rest spin-half-nuclei are decoupled throughout all experiments. Figure 3 describes the parameters and structure of this molecule. Under the weak-coupling approximation, the Hamiltonian of the system writes
$${{\mathcal{H}}}_{{\rm{int}}}=\sum _{j=1}^{4}\pi ({\nu }_{j}-{\nu }_{0}){\sigma }_{z}^{j}+\sum _{j {<} k=1}^{4}\frac{\pi }{2}{J}_{jk}{\sigma }_{z}^{j}{\sigma }_{z}^{k},$$
(5)
where $${\nu }_{j}$$ are the chemical shifts, $${J}_{jk}$$ are the J-coupling strengths, and $${\nu }_{0}$$ is the reference frequency of 13C channel in the NMR platform. All experiments were carried out on a Bruker AVANCE 400 MHz spectrometer at room temperature. We briefly describe our three experimental steps here and leave the details in the “Methods” section: (i) Initialization: The pseudo-pure state39,40,41 for being the input of quantum computation $$\left|0000\right\rangle$$ is prepared. (More details are provided in the “Methods” section). (ii) Evolution: Starting from the state $$\left|0000\right\rangle$$, we create the ground state of the random two-body Hamiltonian by applying the optimized shaped pulses. (iii) Measurement: In NMR experiments, the expectation values of all 2-qubit Pauli products can be measured by the ensemble measurement. From them, we can directly obtain all 2-local measurements, and perform four-qubit QST to estimate the quality of our implementations, which is accomplished by the least-squares tomography from the experimental data. More details about the least-squares tomography can be found in “Methods” section.
In experiments, we created the ground states of 20 random Hamiltonians of the form in Eq. (1) and performed 4-qubit QST for them after the state preparations. It is worth emphasizing that the experimental raw density matrices obtained from ensemble measurements on NMR are usually negative. First, we further performed the least-squares QST from the raw density matrices in experiments as $${\rho }_{\exp }$$, and estimated that the fidelities between the experimental states $${\rho }_{\exp }$$ and the target ground state $${\rho }_{{\rm{th}}}$$ are over $$99.2$$%. It is noted that the purpose of reconstructing the states $${\rho }_{\exp }$$ is to use them to compare with the results estimated by our neural network. We collected the expectation values of all 2-qubit Pauli product operators, such as $$\left\langle {\sigma }_{x}\otimes I\otimes I\otimes I\right\rangle$$ and $$\left\langle {\sigma }_{x}\otimes {\sigma }_{y}\otimes I\otimes I\right\rangle$$, which were directly obtained by measuring the expectation values of these Pauli strings in NMR. Then we fed them into our neural-network-based framework to reconstruct the 4-qubit states, obtaining an average fidelity of 98.8% between $${\rho }_{\exp }$$ and $${\rho }_{{\rm{nn}}}$$, where $${\rho }_{{\rm{nn}}}$$ is the neural network estimated state. Figure 4 shows the fidelity details of these density matrices. The results indicate that the original 4-qubit state can be efficiently reconstructed by our trained neural network using only 2-local measurements, instead of the traditional full QST.
## Discussion
As a famous double-edged sword in experimental quantum computing, QST captures full information of quantum states on the one hand, while on the other hand, its implementation consumes a tremendous amount of resources. Unlike traditional QST that requires exponentially many experiments with the growth of system size, the recent approach by measuring RDMs and reconstructing the full state thereafter opens up a new avenue to efficiently realize experimental QST. However, there is still an obstacle in this approach, that it is in general computationally hard to construct the full quantum state from its local information.
This is a typical problem empowered by machine learning. In this work, we apply the neural network model to solve this problem and demonstrate the feasibility of our method with up to seven qubits in the simulation. It should be noticed that 7-qubit QST in experiments is already a significant challenge in many platforms—the largest QST to date is of 10 qubits in superconducting circuits, where the theoretical state is a GHZ state with rather simple mathematical form.61 We further demonstrate that our method works well in a 4-qubit NMR experiment, thus validating its usefulness in practice. We anticipate this method to be a powerful tool in future QST tasks of many qubits due to its accuracy and convenience. Comparing with the MLE, our method has acceptable fidelities, better noise tolerance and also has a significant advantage in terms of speed.
Our framework can be extended in several ways. First, we can consider excited states. As stated in the “Results” section, the Hamiltonian recovered by our neural network is not necessarily the original Hamiltonian, but their ground states are fairly close. We preliminarily examined the eigenstates of estimated Hamiltonians. Although the ground states have considerable overlap, the excited states are not close to each other. It means, in this reverse engineering problem, ground states are numerically more stable than excited states. To recover excited states using our method, one may need to use more sophisticated neural networks, such as convolutional neural network62 (CNN) or residual neural network63 (ResNet). Second, although we have not included noise in the training and test data, our network estimates the experimental 4-qubit fully connected 2-local states with high fidelities. This indicates our method has certain error tolerant ability. For future study, one can add different noise to the training and test data. Third, one can also study how to incorporate the current method into the existing quantum tomography methods, such as compressive sensing techniques.9,64,65
## Methods
### Machine learning
In this subsection, we discuss our training/test dataset generation procedure, the structure, and hyperparameters of our neural network, and the required number of training data during training. And we also provide a criterion for determining whether the neural network estimate is acceptable without knowing the true states.
The training and test data sets are formed by random $$k$$-local Hamiltonians and $$k$$-local measurements of corresponding ground states. For our 4-qubit case, the 2-local Hamiltonians are defined in Eq. (1). The parameter vector $$\overrightarrow{h}$$ of random Hamiltonians are uniformly drawn from random normal distributions without uniform mean values and standard deviations. It is realized by applying function np.random.normal in Python. Similarly, for the 7-qubit case, Hamiltonian is defined in Eq. (4), and the corresponding parameter vector $$\overrightarrow{h}$$ is generated by the same method. As the dashed lines in Fig. 1 shows, after getting random Hamiltonians $$H$$, we calculate the ground states $$\left|{\psi }_{H}\right\rangle$$ (the eigenvector corresponds to the smallest eigenvalue of $$H$$) and then get the 2-local measurements $${\bf{M}}$$.
In this work, we use a fully connected feedforward neural network, which is famous as the first and most simple type of neural network.42 By fully connected, it means every neuron is connected to every other neuron in the next layer. Feedforward or acyclic, as the word indicated, means information only passes forward; the network has no cycle. Our machine-learning process is implemented using Keras,43 which is a high-level deep learning library running on top of the popular machine-learning framework: Tensorflow.44
As mentioned in the “Results” section, the true value of local measurements have been used as input to our neural network. The input is $${\bf{M}}=\{{s}_{m,n}^{(i,j)}:{s}_{m,n}^{(i,j)}={\rm{Tr}}({{\rm{Tr}}}_{(i,j)}\rho \cdot {B}_{(m,n)}),{B}_{(m,n)}\in {\bf{B}},1\le i<j\le 4,1\le n,m\le 4\}$$. For the 4-qubit case, it is easy to see that $${\bf{M}}$$ has $$3\,\times\, 4=12$$ single body terms and $${C}_{4}^{2}\times 9=54$$ 2-body terms. By arranging these $$66$$ elements in $${\bf{M}}$$ into a row, we set it as the input of our neural network.
The output set to be the vector representation of the Hamiltonian $$\overrightarrow{h}$$, which also has 66 entries. For the 7-qubit 2-local case, where 2-body terms only appear on nearest qubits, the network takes 2-local measurements as input, and the number of neurons in the input layer is $$7\,\times\, 3+6\,\times\, 3\,\times\, 3=75$$. The number of neurons in the output layer is also 75.
The physical aspect of our problem fixes the input and output layers. The principle for choosing the number of hidden layers is efficiency. While training networks, inspired by Occam’s Razor principle, we choose fewer layers and neurons when increasing them do not significantly increase the performance but increases the required training epochs. In our 4-qubit case, two hidden layers of 300 neurons have been inserted between the input layer and the output layer. In the 7-qubit case, we use four fully connected hidden layers with the following number of hidden neurons: 150-300-300-150. The activate function for each layer is rectified linear unit (ReLU),45 which is a widely used non-linear activation function. We also choose the optimizer having the best performance in our problem over almost all the built-in optimizers in Tensorflow: AdamOptimizer (adaptive moment estimation).30 The learning rate is set to be 0.001.
The whole training dataset has been split into two parts, 80% used for training, and 20% used for validation after each epoch. A new data set of 5000 data was used as the test set after training. The initial batch size was chosen as 512. As the amount of training data increases, the average fidelity of estimated states and the true test states goes up. The neural network reaches a certain performance after we fed sufficient training data. More training data requires more training epochs; however, replete epochs ebb the neural network performance due to over-fitting. Table 2 shows the average fidelities of using different training data and epochs. The first round of training locks down the optimal amount of training data, then we change the batch size and find the optimal epoch. We report the results for the second round training in Table 3. For the 4-qubit case, appropriate increase in the batch size can benefit the stability of training process, thus improves the performance of the neural network. Though, by choosing the batch size as 512 and 2048, the network can also reach the same performance with larger epochs, we chose the batch size as 1028, since more epochs require more training time. After the same attempting for the 7-qubit case, we find 512 a promising batch size.
The time cost for preparing the network involves two parts—generating training and testing data, and training the networks. Most of the time spending on data generating is to solve the ground states (eigenvector corresponding to the smallest eigenvalue) of randomly generated Hamiltonians. It takes roughly 5 min (2.2 h) to generate the whole data set for 4-qubit (7-qubit) by implementing eigs in MATLAB. With sufficient data in hand, the network-training procedure takes about 12 min (49 min) for 4-qubit (7-qubit).
As reported in the “Results” section, the fidelities of neural network estimates have a slight variation, for example, the fidelity of the 4-qubit case is range from 91.4% to 99.8%. One who uses this framework might wonder how precise the neural network outcome is compared to the true state. In contrast of the scenario when we are testing our framework theoretically, we do not have the true state in hand. Now it is natural to ask that how to determine whether the estimate is precise enough. Providentially, we could solve this question in a straightforward way.
Based on $${\rho }_{{\rm{nn}}}$$, we compute $${{\bf{M}}}_{{\rm{pred}}}=\{{s}_{m,n}^{(i,j)}:{s}_{m,n}^{(i,j)}={\rm{Tr}}({{\rm{Tr}}}_{(i,j)}{\rho }_{{\rm{nn}}}\cdot {B}_{(m,n)}),{B}_{(m,n)}\in {\bf{B}},1\le i<j\le 4,1\le n,m\le 4\}$$ and compare with the original $${\bf{M}}$$. Root-mean-square-error (RMSE) between two variables $$\overrightarrow{x}$$ and $$\overrightarrow{y}$$, defined as $${\rm{rmse}}(\overrightarrow{x},\overrightarrow{y})=\sqrt{\frac{1}{d}{\sum }_{i=1}^{d}{({x}_{i}-{y}_{i})}^{2}}$$, is a frequently used quantity to measure the closeness of $$\overrightarrow{x}$$ and $$\overrightarrow{y}$$. In reality, how bad an error is also depends on the magnitude of the true value. That means, with the same RMSE, the larger the magnitude of the true value is the better the accuracy. A measure referring to the real value reveals more about how precise between an estimation and a expected outcome. We, therefore, define a quantity called relative RMSE, namely $${\rm{rrmse}}(\overrightarrow{x},\overrightarrow{y})=\sqrt{\frac{1}{d}{\sum }_{i=1}^{d}{({x}_{i}-{y}_{i})}^{2}}/| | \overrightarrow{y}| | ={\rm{rmse}}(\overrightarrow{x},\overrightarrow{y})/| | \overrightarrow{y}| |$$, where $$y$$ is the true value and $$| | \overrightarrow{y}| |$$ is its $${l}^{2}$$-norm. The relative RMSE between $${{\bf{M}}}_{{\rm{pred}}}$$ and $${\bf{M}}$$ is $${\rm{rmse}}({{\bf{M}}}_{{\rm{pred}}},{\bf{M}})/| | {\bf{M}}| |$$. By bounding the relative RMSE <0.2%, 4692 out of 5000 (93.8%) estimations of our 4-qubit network are acceptable, and the probability of these estimations having fidelities higher than 97% is 99.8%.
### Comparison with the approximated MLE
The standard MLE46,47,48,49 is usually adopted to reconstruct a legal and full quantum state whose local information is closest to the measured results. It technically maximizes the likelihood of the estimate by the given data. When we make the Gaussian distribution and assume that all measurements have the same standard deviation, the MLE is approximately the least-squares tomography which minimizes the distance between the searched results and the measurement outcomes.50 In this section, we make a comparison of efficiency, accuracy, and noise tolerance between the approximated MLE and our method.
With a personal computer,51 every single 4-qubit state takes about 1 min to compute for the approximated MLE. The estimating procedure of our method analyzed 5000 data in 2 min (about 0.024 s per data set) using the same computer. For the 7-qubit case, the approximated MLE requires about 168 min to converge for each single data point. Remarkably, our method can process 5000 data sets within <6 min (about 0.070 s per data set). This suggests that our method is substantially faster than the approximated MLE. We can reasonably expect that when the system size gets even larger, our computation time advantage will become more impressive.
In the 4-qubit cases, the approximated MLE can yield estimates with an average fidelity of 99.9%. In the 7-qubit cases, it can still achieve an average fidelity of 99.9%. Therefore in terms of accuracy, the approximated MLE slightly outperforms our method.
We also analyze the noise tolerance of the two methods by adding noise to the input measurements. The set of unbiased noise $$\overrightarrow{n}$$ was generated according to the normal distribution with mean value $$0$$ and standard deviation $$1$$. The percentile noise vector $$\alpha \overrightarrow{n}$$ is formed by multiplying the factor $$\alpha \in \{5 \% ,10 \% ,15 \% ,20 \% ,25 \% ,30 \% \}$$ to the unbiased noise $$\overrightarrow{n}$$. By adding $$\alpha\overrightarrow{n}$$ to the true measurements $${\bf{M}}$$, we formed the noisy input $${\bf{M}}+\alpha\overrightarrow{n}$$. Suppose the approximated MLE or our neural network estimates the noisy output $${\rho }_{{\rm{noise}}}$$. We calculate the fidelities of the estimate $${\rho }_{{\rm{noise}}}$$ with the true state $$\rho$$ for 100 pairs of 4-qubit data. As depicted in Fig. 5, our method has better noise tolerance than the approximated MLE with the pure state constraint when noise >5% is added to the measurements of a pure state.
### NMR states preparation
Our experimental procedure consists of three steps: initialization, evolution, and measurement. In this subsection, we discuss these three steps in details.
1. (i)
Initialization: The computational basis state $${\left|0\right\rangle }^{\otimes n}$$ is usually chosen as the input state for quantum computation. Most of the quantum systems do not start from such an input state, so a proper initialization processing is necessary before applying quantum circuits. In NMR, the sample initially stays in the Boltzmann distribution at room temperature,
$${\rho }_{{\rm{thermal}}}={\mathcal{I}}/16+\epsilon ({\sigma }_{z}^{1}+{\sigma }_{z}^{2}+{\sigma }_{z}^{3}+{\sigma }_{z}^{4}),$$
where $${\mathcal{I}}$$ is the $$16\,\times\, 16$$ identity matrix and $$\epsilon \approx 1{0}^{-5}$$ is the polarization. We cannot directly use it as the input state for quantum computation, because such a thermal state is a highly mixed state.39,52 We instead create a so-called pseudo-pure state (PPS) from this thermal state by using the spatial averaging technique,39,40,41 which consists of applying local unitary rotations and using $$z$$-gradient fields to destroy the unwanted coherence. The form of the 4-qubit PPS can be written as
$${\rho }_{0000}=(1-\epsilon ^{\prime} ){\mathcal{I}}/16+\epsilon ^{\prime} \left|0000\right\rangle \left\langle 0000\right|.$$
Here, although the PPS $${\rho }_{0000}$$ is also a highly mixed state, the identity part $${\mathcal{I}}$$ neither changes under any unitary operations nor contributes to observable NMR signal. It means that we can focus on the deviated part $$\left|0000\right\rangle \left\langle 0000\right|$$ and consider $$\left|0000\right\rangle \left\langle 0000\right|$$ as the initial state of our quantum system. Finally, 4-qubit QST was performed to evaluate the quality of our PPS. We found that the fidelity between the perfect pure state $$\left|0000\right\rangle$$ and the experimentally measured PPS is about 98.7% by the definition of $$C$$ in Eq. (3), where the raw PPS density matrix obtained directly from the experiment is negative. This sets a solid ground for the subsequent experiments.
2. (ii)
Evolution: In this step, we prepared the ground states of the given Hamiltonians using optimized pulses. The form of the considered Hamiltonian is chosen as Eq. (1).
Here, the parameters $${\omega }_{k}^{(i)}$$ and $${J}_{nm}^{(ij)}$$ mean the chemical shift and the J-coupling strength, respectively. In experiments, we create the ground states of different Hamiltonians by randomly changing the parameter set $$({\omega }_{k}^{(i)},{J}_{nm}^{(ij)})$$. For the given Hamiltonian, the gradient ascent pulse engineering (GRAPE) algorithm53,54,55,56 is adapted to optimize a radio-frequency (RF) pulse to realize the dynamical evolution from the initial state $$\left|0000\right\rangle$$ to the target ground state. The GRAPE pulses are designed to be robust to the static field distributions and RF inhomogeneity, and the simulated fidelity is over $$0.99$$ for each dynamical evolution.
3. (iii)
Measurement: In principle, we only need to measure the 2-local measurements to determine the original 4-qubit Hamiltonian through our trained network. Experimentally, we performed 4-qubit QST, which naturally includes the 2-local measurements after preparing these states,57,58,59 to evaluate the performance of our implementations. Hence, we can estimate the quality of the experimental implementations by computing the fidelity between the target ground state $${\rho }_{{\rm{th}}}=\left|{\psi }_{{\rm{th}}}\right\rangle \left\langle {\psi }_{{\rm{th}}}\right|$$ and the experimentally reconstructed density matrix $${\rho }_{\exp }$$.60 By reconstructing states $${\rho }_{{\rm{nn}}}$$ merely based on the experimental 2-local measurements, the performance of our trained neural network can be evaluated by comparing the experimental states $${\rho }_{\exp }$$ with the states $${\rho }_{{\rm{nn}}}$$.
Finally, we attempt to evaluate the confidence of the expected results by analyzing the potential error sources in experiments. The infidelity of the experimental density matrix is mainly caused by some unavoidable factors in experiments, including decoherence effects, imperfections of the PPS preparation, and imprecision of the optimized pulses. From a theoretical perspective, we numerically simulate the influence of the optimized pulses and the decoherence effect of our qubits. Then we compare the fidelity computed in this manner with the ideal case to evaluate the quality of the final density matrix. As a numerical result, about 0.2% infidelity was created on average and the 1.2% error is related to the infidelity of the initial state preparation. Additionally, other errors can also contribute to the infidelity, such as imperfections in the readout pulses and spectral fitting.
### The approximated MLE
We briefly describe the approximated MLE we used in the numerical simulation. The standard MLE48,49 is also a method to produce satisfactory results in recovering the full states from the experimental measurements. In general, the standard MLE can be divided into three steps.
1. (i)
Parameterize a density matrix in a legal way. Here, we describe a pure density matrix by
$$\rho (\overrightarrow{x})=V(\overrightarrow{x}){V}^{\dagger }(\overrightarrow{x})/{\rm{Tr}}(V(\overrightarrow{x}){V}^{\dagger }(\overrightarrow{x})).$$
$$V$$ is a $${2}^{N}$$-dimensional vector with the parameters $$\overrightarrow{x}$$ and the number of qubits $$N$$. $$\rho (\overrightarrow{x})$$ is a normalized and non-negative definite Hermitian density matrix under such a parameterization.
2. (ii)
Construct a likelihood function to be maximized. The measurements calculated from the parameterized density matrix $$\rho (\overrightarrow{x})$$ are $${\rm{Tr}}({{\rm{Tr}}}_{(i,j)}\rho (\overrightarrow{x})\cdot {B}_{(m,n)})$$ with $${B}_{(m,n)}\in {\bf{B}},1\le i<j\le 4$$ and $$1\le n,m\le 4$$, and the total probability of $$\rho (\overrightarrow{x})$$ yielding the results closely to the true measurements $${\bf{M}}$$ can be written as
$$P(\overrightarrow{x})=\frac{1}{{\mathcal{N}}}\prod _{i,j,m,n}\exp \left[-\frac{{\{{\rm{Tr}}({{\rm{Tr}}}_{(i,j)}\rho (\overrightarrow{x})\cdot {B}_{(m,n)})-{s}_{m,n}^{(i,j)}\}}^{2}}{2{({\chi }_{m,n}^{(i,j)})}^{2}}\right],$$
where $${\chi }_{m,n}^{(i,j)}$$ is the standard deviation of each measurement $${s}_{m,n}^{(i,j)}$$ and $${\mathcal{N}}$$ is the normalization (Gaussian model). $$P(\overrightarrow{x})$$ is the likelihood function we need to maximize. If we assume the standard deviation is the same, the standard MLE is approximately least-squares tomography.50 It is equivalent to minimize the following function:
$${\mathcal{F}}(\overrightarrow{x})=\sum _{i,j,m,n}{\left[{\rm{Tr}}({{\rm{Tr}}}_{(i,j)}\rho (\overrightarrow{x})\cdot {B}_{(m,n)})-{s}_{m,n}^{(i,j)}\right]}^{2}.$$
Here, we ignore some constants which do not influence the optimization, e.g., the normalization factor $${\mathcal{N}}$$. $${\mathcal{F}}(\overrightarrow{x})$$ is the cost function that we minimize with the least-squares tomography.
3. (iii)
Minimize the cost function using some techniques. We use the function lsqnonlin of MATLAB with an initial guess and a default setting. It takes a while to optimize a sum of squares like $${\mathcal{F}}(\overrightarrow{x})$$. Finally, quantum state $$\rho (\overrightarrow{x})$$ can be recovered from the parameters $$\overrightarrow{x}$$ when the optimization is finished.
## Data availability
The experimental data and the source code that support the findings of this study are available from the corresponding author on reasonable request.
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Hradil, Z. & Řeháček, J. Efficiency of maximum-likelihood reconstruction of quantum states. Fortschr. Phys. 49, 1083–1088 (2001).
48. 48.
Řeháček, J., Hradil, Z., Knill, E. & Lvovsky, A. Diluted maximum-likelihood algorithm for quantum tomography. Phys. Rev. A 75, 042108 (2007).
49. 49.
Paris, M. & Rehacek, J. Quantum State Estimation, Vol. 649 (Springer Science & Business Media, 2004).
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Acharya, A., Kypraios, T. and Guţă, M. A comparative study of estimation methods in quantum tomography. J. Phys. A: Math Theor. 52(23), 234001, (2019). https://iopscience.iop.org/article/10.1088/1751-8121/ab1958.
51. 51.
MacBook Pro, Processor: 2.3 GHz Intel Core i5, Memory: 8 GB.
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Gershenfeld, N. A. & Chuang, I. L. Bulk spin-resonance quantum computation. Science 275, 350–356 (1997).
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Boulant, N., Edmonds, K., Yang, J., Pravia, M. & Cory, D. Experimental demonstration of an entanglement swapping operation and improved control in NMR quantum-information processing. Phys. Rev. A 68, 032305 (2003).
54. 54.
Khaneja, N., Reiss, T., Kehlet, C., Schulte-Herbrüggen, T. & Glaser, S. J. Optimal control of coupled spin dynamics: design of nmr pulse sequences by gradient ascent algorithms. J. Magn. Reson. 172, 296–305 (2005).
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Ryan, C., Negrevergne, C., Laforest, M., Knill, E. & Laflamme, R. Liquid-state nuclear magnetic resonance as a testbed for developing quantum control methods. Phys. Rev. A 78, 012328 (2008).
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Lu, D. et al. Enhancing quantum control by bootstrapping a quantum processor of 12 qubits. npj Quantum Inf. 3, 45 (2017).
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Leskowitz, G. M. & Mueller, L. J. State interrogation in nuclear magnetic resonance quantum-information processing. Phys. Rev. A 69, 052302 (2004).
58. 58.
Lee, J.-S. The quantum state tomography on an NMR system. Phys. Lett. A 305, 349–353 (2002).
59. 59.
Li, J. et al. Optimal design of measurement settings for quantum-state-tomography experiments. Phys. Rev. A 96, 032307 (2017).
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Altepeter, J. B., Jeffrey, E. R. & Kwiat, P. G. Photonic state tomography. Adv. Atom. Mol. Opt. Phys. 52, 105–159 (2005).
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Song, C. et al. 10-qubit entanglement and parallel logic operations with a superconducting circuit. Phys. Rev. Lett. 119, 180511 (2017).
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Krizhevsky, A., Sutskever, I. & Hinton, G.E. Advances in Neural Information Processing Systems 25. In Advances in Neural Information Processing Systems (eds Pereira, F., Burges, C. J. C., Bottou, L. & Weinberger, K. Q.) 1097–1105 (Curran Associates, Inc., 2012). https://dl.acm.org/citation.cfm?id=3065386.
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He, K., Zhang, X., Ren, S. & Sun, J. Deep residual learning for image recognition. In Proc. IEEE Conference on Computer Vision and Pattern Recognition, 770–778 (2016).
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Flammia, S. T., Gross, D., Liu, Y.-K. & Eisert, J. Quantum tomography via compressed sensing: error bounds, sample complexity and efficient estimators. New J. Phys. 14, 095022 (2012).
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Riofrío, C. et al. Experimental quantum compressed sensing for a seven-qubit system. Nat. Commun. 8, 15305 (2017).
## Acknowledgements
We thank Yi Shen for helpful discussions. G.L. is grateful to the following funding sources: the National Natural Science Foundation of China (11175094); National Basic Research Program of China (2015CB921002). T.X., D.L., and J.L. are supported by the National Natural Science Foundation of China (Grants nos. 11605153, 11605005, 11875159, U1801661, 11905099, and 11975117), Science, Technology and Innovation Commission of Shenzhen Municipality (Grants nos. ZDSYS20170303165926217 and JCYJ20170412152620376), Guangdong Innovative and Entrepreneurial Research Team Program (Grant no. 2016ZT06D348). T.X. is also supported by Guangdong Basic and Applied Basic Research Foundation (No. 2019A1515011383). N.C. and B.Z. acknowledge the Natural Sciences and Engineering Research Council of Canada (NSERC), Canadian Institute for Advanced Research (CIFAR) and Chinese Ministry of Education (20173080024).
## Author information
B.Z. conceived the idea of this paper. S.L. and N.C. wrote and implemented the computer code and simulations. T.X. accomplished the NMR experiments. B.Z. supervised the project. T.X., S.L., and N.C. wrote the manuscript with feedback from all authors.
Correspondence to Jun Li or Bei Zeng.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
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2015
05-24
Aeroplane chess
Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are 1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.
There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid.
There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).
The input end with N=0, M=0.
There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).
The input end with N=0, M=0.
2 0
8 3
2 4
4 5
7 8
0 0
1.1667
2.3441
5，如果5和8连到一起了，那你还可以继续跳到8，最后问跳到n时平均置几次骰子。也就是求期望。
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int N=100005;
struct node
{
int y,next;
};
bool vis[N];
node path[N];
int first[N],t;
double dp[N];
{
path[t].y=y;
path[t].next=first[x];
first[x]=t++;
}
int main()
{
double s;
int n,m,v;
while(cin>>n>>m)
{
if(m==0&&n==0) break;
memset(dp,0,sizeof(dp));
memset(vis,0,sizeof(vis));
memset(first,0,sizeof(first));
int x,y;
t=1;
while(m--)
{
cin>>x>>y;
}
dp[n]=-1;
for(int i=n; i>=0; i--)
{
if(!vis[i])
{
vis[i]=true;
s=0;
for(int k=1; k<=6; k++)
s+=dp[i+k];
s/=6;
dp[i]+=(s+1);
}
for(int k=first[i]; k; k=path[k].next)
{
v=path[k].y;
dp[v]=dp[i];
vis[v]=true;
}
}
printf("%.4lf\n",dp[0]);
}
return 0;
}
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# View-Based
In addition to the other partitioning methods, SGL provides a view-based distribution mechanism. For example, one can create a graph that is distributed block-cyclically:
using spec_type = stapl::distribution_spec<>;
using graph_type = stapl::dynamic_graph<
stapl::DIRECTED,
stapl::MULTIEDGES,
int,
int,
stapl::view_based_partition<spec_type>,
stapl::view_based_mapper<spec_type>
>;
const std::size_t n = 100;
auto cyclic_spec = stapl::block_cyclic(n, 4);
graph_type g(cyclic_spec);
Using this code, g will be distributed cyclically across the locations in block sizes of 4.
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# Hackerrank – Compute the Perimeter of a Polygon
## Hackerrank – Problem Statement
A description of the problem can be found on Hackerrank.
## Solution
Calculate a length of a line between each pair of following points p, defined with x and y coordinate:
• (p0, p1), (p1, p2), ..., (pn-1, p0)
Length of a line between two points:
$$l = \sqrt{{(x_1 – x_2)}^2 + {(y_1 – y_2)}^2}$$
Make sum of the lengths.
I created solution in:
All solutions are also available on my GitHub.
Scala
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# Tale of sentient phone exchanges [duplicate]
Years ago I recall reading a story by, I think, Isaac Asimov. Its central theme was phone exchanges becoming sentient.
Does anyone know what I am thinking about or am I dreaming and should quickly write this story?
## marked as duplicate by user14111 story-identification StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 11 '15 at 19:47
• Could it be Heinlein's The Moon is a Harsh Mistress? Though that novel would be hard to confuse with Asimov. – b_jonas Jun 5 '14 at 18:06
• Please remember to click on the check to accept whichever answer you find correct. Thanks – Jim Green Jun 6 '14 at 19:07
Dial F For Frankenstein by Arthur C Clarke is probably what you are looking for. I found it in The Wind from the Sun
A tech-crew discuss the strange happenings since they have linked the world's telecommunications system with a satellite network.
• That's the one! Thanks all and sounds like some other good tales to read here. – user27993 Jun 6 '14 at 18:20
• @user27993 -- If it's the right answer, please be sure to click the check box below the voting arrows and 'accept' it! – K-H-W Oct 16 '14 at 13:30
John Varley's Hugo-winning 1984 novella "Press Enter ■" wasn't written by Asimov, but it was published in Isaac Asimov's Science Fiction Magazine (May 1984 issue, available at the Internet Archive); could that be what you were thinking of?
The story sort of fits the meager description you gave. As I seem to recall from reading it 30 years ago, it's not exactly the phone exchanges per se that become sentient, it's all the computers in the world, communicating over the telephone lines. Or something like that.
I couldn't find a useful review, but I found my copy of the story (in Varley's collection Blue Champagne), so I'll copy out a few quotations. You should be able to tell from them if this is the story you're looking for.
The beginning:
"This is a recording. Please do not hang up until—"
I slammed the phone down so hard it fell onto the floor. Then I stood there, dripping wet and shaking with anger. Eventually, the phone started to make that buzzing noise they make when a receiver is off the hook. It's twenty times as loud as any sound a phone can normally make, and I always wondered why. As though it was such a terrible disaster: "Emergency! Your telephone is off the hook!!!"
Phone answering machines are one of the small annoyances of life. Confess, do you really like to talk to a machine? But what had just happened to me was more than a petty irritation. I had just been called by an automatic dialing machine.
Some plot explanation:
"The connections. Again, it's different, but the concept of networking is the same. A neuron is connected to a lot of others. There are trillions of them, and the way messages pulse through them determine what we are and what we think and what we remember. And with that computer I can reach a million others. It's bigger than the human brain, really, because the information in that network is more than all humanity could cope with in a million years. It reaches from Pioneer Ten, out beyond the orbit of Pluto, right into every living room that has a telephone in it. With that computer you can tap tons of data that has been collected but nobody's even had the time to look at it.
"That's what Kluge was interested in. The old 'critical mass computer' idea, the computer that becomes aware, but with a new angle. Maybe it wouldn't be the size of the computer, but the number of computers. There used to be thousands of them. Now there's millions. They're putting them in cars. In wristwatches. Every home has several, from the simple timer on a microwave oven up to a video game or home terminal. Kluge was trying to find out if critical mass could be reached that way.
The ending:
I live by candlelight, and kerosene lamp. I grow most of what I eat.
It took a long time to taper off the Tranxene and the Dilantin, but I did it, and now take the seizures as they come. I've usually got bruises to show for it.
In the middle of a vast city I have cut myself off. I am not part of the network growing faster than I can conceive. I don't even know if it's dangerous to ordinary people. It noticed me, and Kluge, and Osborne. And Lisa. It brushed against our minds like I would brush away a mosquito, never noticing I had crushed it. Only I survived.
But I wonder.
It would be very hard . . . Lisa told me how it can get in through the wiring. There's something called a carrier wave that can move over wires carrying household current. That's why the electricity had to go.
I need water for my garden. There's just not enough rain here in southern California, and I don't now how else I could get the water.
Do you think it could come through the pipes?
Might be Alfred Bester's "Something Up There Like Me". Not quite a sentient phone system, rather a sentient satellite that takes control of earth's communication systems.
Primo Levi's "For a Good Purpose" in Sixth Day and other tales
"For a Good Purpose": the telephone network develops intelligence when it is connected to the French and German networks. Slowly it experiments with its powers
• Thank you so very much. I knew there was a story by Primo Levi along these lines that I had read once, but all I could find was the Clarke story. – SQB Feb 9 '17 at 15:51
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## 邪魔的笔记 · · · ( 50篇 )
### 战国策（全二册） (20)
• ##### 第403页
秦之有贤相也，非楚国之利也。
• ##### 第394页
以财交者，财尽而交绝；以色交者，华落而爱渝。是以嬖女不敝席，宠臣不敝轩。
• ##### 第393页
然则且有子杀其父，臣弑其主者，而王终己不知者，何也？以王好闻人之美而恶闻人之恶也！
### 名著典藏 (11)
• ##### 第56页
I had treated seeing Catherine very lightly,Ihad gotten somewhat drunk and had nearly forgotten to come but when I could not see her there I was feeling lonely and hollow.
• ##### 第48页
It was a long time since I had written to the States and I knew I should write but I had let it go so long that it was almost impossible to write now.
• ##### 第34页
I leaned forward in the dark to kiss her and there was a sharp stinging flash.She had slapped my face hard.Her hand had hit my nose and eyes,and tears came in my eyes from the reflex.
### 儒林外史 (1)
• ##### 第209页
蒋刑房等他说完了，慢慢提起来，说：“潘三哥在监里，前日再三和我说，听见尊驾回来了，意思要会一会，叙叙苦情。不知先生你意下何如？”匡超人道：“潘三哥是个豪杰，他不曾遇事时，会着我们，到酒店里坐坐，鸭...
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# [texhax] New Bibtex style needed for citing datasets
Philip G. Ratcliffe philip.ratcliffe at fastwebnet.it
Mon Nov 2 14:17:22 CET 2009
> I'm attempting to create a new bibtex style that includes
> support for data. I think I can create the style but am
> having difficulty figuring out how to have the *.bst file
> found when I refer to it in a \bibliographystyle(cca.bst) statement.
>
> Can anyone tell me how to refer to a non-standard *.bst file?
> For example, I tried
>
> \bibliographystyle('/Users/hellyj/cca.bst')
Well, it's hardly a complete path, is it?
> and that didn't seem to have any effect: positive or negative.
First of all, this is something of an FAQ. And it is dealt with in the FAQ
linked at the bottom of the page.
Anyway, in general, you should construct a parallel TeX Directory Structure
(TDS).
Under Unix-like systems this is usually under /usr/local/share/texmf, for
MiKTeX it might me under C:...\localtexmf.
For BibTeX style files, the TDS should contain the subdirectory
...texmf/bibtex/bst or ...texmf\bibtex\bst.
In either case you must use the wizard provided (I'm afraid I can't remember
what it's called under Linux) to make sure the new files added are known to
the system - and then update the filename database (this last step is
important).
Cheers, Phil
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de.mpg.escidoc.pubman.appbase.FacesBean
Deutsch
Hilfe Wegweiser Datenschutzhinweis Impressum Kontakt
# Datensatz
DATENSATZ AKTIONENEXPORT
Freigegeben
Konferenzbeitrag
#### Heaps are better than Buckets: Parallel Shortest Paths on Unbalanced Graphs
##### MPG-Autoren
http://pubman.mpdl.mpg.de/cone/persons/resource/persons45038
Meyer, Ulrich
Algorithms and Complexity, MPI for Informatics, Max Planck Society;
##### Externe Ressourcen
Es sind keine Externen Ressourcen verfügbar
##### Volltexte (frei zugänglich)
Es sind keine frei zugänglichen Volltexte verfügbar
##### Ergänzendes Material (frei zugänglich)
Es sind keine frei zugänglichen Ergänzenden Materialien verfügbar
##### Zitation
Meyer, U. (2001). Heaps are better than Buckets: Parallel Shortest Paths on Unbalanced Graphs. In R. Sakellariou, J. Keane, J. Gurd, & L. Freeman (Eds.), Proceedings of the 7th International Euro-Par Conference on Parallel Processing (Euro-Par-01): (pp. 343-351). Berlin, Germany: Springer.
We propose a new parallel algorithm for the single-source shortest-path problem (SSSP). Its heap data structure is particularly advantageous on graphs with a moderate number of high degree nodes. On arbitrary directed graphs with $n$ nodes, $m$ edges and independent random edge weights uniformly distributed in the range $[0,1]$ and maximum shortest path weight $\Diam$ the PRAM version of our algorithm runs in ${\cal O}(\log^2 n \cdot \min_{i} \{2^i \cdot \Diam \cdot \log n+ |V_i| \})$ average-case time using ${\cal O}(n \cdot \log n +m )$ operations where $|V_i|$ is the number of graph vertices with degree at least $2^i$. For power-law graph models of the Internet or call graphs this results in the first work-efficient $o(n^{1/4})$ average-case time algorithm.
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# Spark
The 2001 The Spark of Life, by Christopher Wills[1], a biology professor, and Jeffrey Bada[2], marine chemistry professor, then director of the NASA exobiology center at Scripps Institution of Oceanography, wherein the argue that life arose on the earth’s surface in the form of a genetic material containing proto-virus.[3]
In terms, spark (TR:91) (LH:3) (TL:94) refers to []
## Quotes
The following are quotes:
“As such, if we are to naively believe that the 5-element RNA was the ‘first form of life’, then we would also have to believe the following backwards logic”
${\displaystyle {\ce {C4H7O4N}}}$ (aspartic acid) = not alive [?]
${\displaystyle {\ce {C10H12O6N5P}}}$ (ribonucleic acid) = alive!
${\displaystyle {\ce {C21H36O16N7P3S}}}$ (coenzyme A) = more alive [?]
This type of reasoning, in which small 4-element molecules, such as aspartic acid, a crystalline amino acid found especially in plants, are ‘not alive’, whereas 5-element molecules, such as RNA, are ‘alive’, is clearly ridiculous. The hypothesis put forward herein, to reconcile these areas of theoretical inconsistency, is that the human organism is a 26-element molecule and that it as well as all other large-element molecules are dynamic atomic structures found within a 92-element, heat-fluxed, environment, which together react, form, and break bonds, evolve, and reproduce according to the four laws of thermodynamics. Moreover, there is NO such reality as there being a specific energy-filled ‘spark day’ in the earth’s past in which molecules suddenly became lifelike, alive, or imbued with life, etc., as is currently believed.”
Libb Thims (2007), Human Chemistry, Volume One (§Molecule Evolution, pgs. 130-31) [4]
“Why is defining life so frustratingly difficult? Why have scientists and philosophers failed for centuries to find a specific physical property or set of properties that clearly separates the living from the inanimate? Because such a property does not exist. Life is a concept that we invented. On the most fundamental level, all matter that exists is an arrangement of atoms and their constituent particles. These arrangements fall onto an immense spectrum of complexity, from a single hydrogen atom to something as intricate as a brain. In trying to define life, we have drawn a line at an arbitrary level of complexity and declared that everything above that border is alive and everything below it is not. In truth, this division does not exist outside the mind. There is no threshold at which a collection of atoms suddenly becomes alive, no categorical distinction between the living and inanimate, NO Frankensteinian spark. We have failed to define life because there was never anything to define in the first place.”
Ferris Jabr (2013), “Why Life Does Not Really Exist”, Dec 2 [5]
## End matter
### References
1. Christopher Wills – Wikipedia.
2. Jeffrey Bada – NASA Astrobiology Institute.
3. Willis, Christopher; Bada, Jeffrey. (2001). The Spark of Life: Darwin and the Primeval Soup. Oxford.
4. (a) Thims, Libb. (2007). Human Chemistry, Volume One. LuLu.
(b) Thims, Libb. (2007). Human Chemistry, Volume Two. LuLu.
5. Jabr, Ferris. (2013). “Why Life Does Not Really Exist”, Scientific American, Brainwaves Blog, Dec 2.
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## Introduction to Fraction Calculations
Often we want to express a relationship between two numbers that measure different things. For example, you may want to track the number of customers who pass through your store each day. Or maybe you want to narrow that number down to customers per hour to determine the busiest time of day, so you can staff the store appropriately. These rates can all be expressed as fractions, or one measurement over another: $\frac{2253\text{ customers}}{2\text{ days}}$, or comparing $\frac{88\text{ customers}}{1\text{ hour}}$ at 11am to $\frac{356\text{ customers}}{1\text{ hour}}$ at 3pm on Fridays. We can then use these numerical fractions to make further calculations or decisions.
## Contribute!
Did you have an idea for improving this content? We’d love your input.
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# tex input or include another tex file
In a tex file, I need to draw a system diagram, so I put it in a separate tex like diagram.tex:
\documentclass[tikz]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\draw (0,0) -- (1,1);
\end{tikzpicture}
\end{document}
and then in main.tex:
\documentclass[10pt, a4paper]{article}
\usepackage[UTF8]{ctex}
\begin{document}
\section{sectionA}
\include{diagram} % I also tried input the same error. I wish diagram.tex figure could be inserted at this place.
\subsection{subsectionA}
\section{sectionB}
\end{document}
besides: running diagram.tex alone in latex gives:
! LaTeX Error: File standalone.cls' not found.Type X to quit or <RETURN> to proceed,or enter new name. (Default extension: cls)Enter file name:! Emergency stop.<read > \usepackage
So the question would be how to insert a standalone tex file? And is it possible to run a standalone tex alone? I am also studying tikz, while demos I found are mostly using standalone.
• I think the answer is in post : tex.stackexchange.com/questions/32127/standalone-tikz-pictures – Stan Feb 7 '18 at 7:18
• it is possible (but unnecessarily complicated) to make that work but it is much simpler to just have the tikzpicture in a separate file (no \documentclass etc) then you can simply \input it. – David Carlisle Feb 7 '18 at 7:48
• @DavidCarlisle make what work but complicated? – Tiina Feb 7 '18 at 8:16
• \documentclass[tikz]{standalone} load tikz, so it is not need to load it again with \usepackage{tikz}.
• in main document you need to load
• standalone for stripping out preamble in your diagra file
• tikz with necessary tikz libraries
i.e. all packages used in included document.
\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}
\draw (0,0) -- (1,1);
\end{tikzpicture}
\end{document}
and
\documentclass[10pt, a4paper]{article}
\usepackage[UTF8]{ctex}
• are you suggesting using input instead of include – Tiina Feb 7 '18 at 8:38
• @Tiina, yes. i assume, that included file should not start on new page (this happens with \include`). – Zarko Feb 7 '18 at 8:43
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Home Eclipses
# Eclipses
### Solar Eclipse in Pisces February 2017 – Deep Waters
The Solar Eclipse and New Moon occurs at 14:58 (UT) on the 26th February 2017 at 08°Pi12' This eclipse is an annular eclipse. Instead of the Moon entirely blocking out the light of the Sun which occurs at a total eclipse, the smaller apparent diameter...
### Solar Saros 138
We are gearing up to the solar eclipse so I thought I would take a look at Saros Cycle 138 as the eclipse on the 10th May will fall under this series. The Saros Cycle is rather like the main story arc of a...
### Solar Eclipse in Capricorn January 2019 – Reframe
The solar eclipse occurs at 01:28 (UT) on January 6, 2019 at 15°Cp25'. This is a partial eclipse which belongs to Saros cycle 122. This series of eclipses began way back in 991AD taking us back to the times of King Ethelred the Unready. There's...
### Solar Eclipse July 2019 – Karmic Clowns
The Solar Eclipse occurs at 20:16 (BST) on July 2, 2019 at 10°Cn37'. This is a total solar eclipse, visible from parts of New Zealand, Chile and Argentina. The eclipse belongs to Saros 127, a family of eclipses that began back in Viking times in...
### Lunar Eclipse in Leo February 2017 – Cue
The lunar eclipse occurs at 00:32 (UT) on the 11th February 2017 at 22°Le28'. This is the first of the Leo/Aquarius eclipses that will correspond to the Nodes moving into these signs in May. As Leo often represents the entertainer, I feel like this eclipse...
### Solar Eclipse in Virgo September 2015 – Chronic
The Solar Eclipse occurs at 06:41 (UT) on the 13th September 2015 at 20°VIR10'. This partial eclipse will be visible only from South Africa, Antarctica and places in the Indian and Atlantic ocean. The eclipse belongs to Saros cycle 125 which began way back in...
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# Floating leg of a standard swap still has a value at par when we use the OIS as discount factor?
Does a bond paying floating coupon LIBOR, still has the value at par when we use the OIS as discount factor? It seems only when the Identity: $$B(t,T_2)(1+(T_2-T_1)F(t,T_1,T_2))=B(t,T_1)$$ still holds, the proposition above will be true. Here $B(t,T)$ is the value of zero coupon bond, $F(t,T_1,T_2)$ is the forward LIBOR.
In John Hull's book Options, Futures and Other Derivatives 9th page 205 ,shows the way to calculate the forward LIBOR implied in Swap rate under OIS discounting. But it's the case we know $B(t,T_1),$ but don't know $B(t,T_2).$
If we know both $B(t,T_1)$ and $B(t,T_2).$ Can we calculate the forward LIBOR still as above identity?
Denote
$D_{ois}(t):$ the discounted factor of OIS
$B(t,T):$ Bond price
$E_t[]:$ Conditional expectation at time $t$ under OIS-risk neutral measure which makes $D_{ois}(t)B(t,T)$ martingale for all $T.$
Use $N(t) = D_{ois}(t)B(t,T_1)$ as a numeraire to change the measure into OIS $T_1$-forward measure $E^{T_1}_t[]$(simply use expectation represent new measure).
Then $$\dfrac{B(t,T)}{B(t,T_1)} = \dfrac{D_{ois}(t)B(t,T)}{D_{ois}(t)B(t,T_1)}$$ should be martingale under $E^{T_1}_t[].$ Then use the definition of the forward LIBOR $F(t, T, T_{1})$ we can prove that $$\dfrac{1}{D_{ois}(T)}E_{T}\left[D_{ois}(T_1)\Big((T_1-T) \cdot F(T, T, T_{1})+1\Big)\right] = 1.$$
In a dual curve settings discounting is done at $B_{OIS}(t, T)$, whereas the forward libors are computed on the projection curve as $F(t, T_1, T_2) = (B_{libor}(t, T_1)/B_{libor}(t, T_2) - 1)/(T_2 - T_1)$, where $B_{OIS}(t, T)$ is the discount factor on the OIS curve and $B_{libor}(t, T)$ is the discount factor on the libor curve.
• actually I am not much understand what's the $B_{OIS}$ and $B_{LIBOR}?$ Why is discounted factor not D(t) and B(t,T)is the bond price. As you say, the forward LIBOR $F(t,T,T_1)$ will not change under the OIS discounting? – A.Oreo Oct 18 '17 at 16:01
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The current GATK version is 3.8-0
Examples: Monday, today, last week, Mar 26, 3/26/04
#### Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
You can opt in to receive email notifications, for example when your questions get answered or when there are new announcements, by following the instructions given here.
#### ☞ Got a problem?
1. Search using the upper-right search box, e.g. using the error message.
3. Include tool and Java versions.
4. Tell us whether you are following GATK Best Practices.
5. Include relevant details, e.g. platform, DNA- or RNA-Seq, WES (+capture kit) or WGS (PCR-free or PCR+), paired- or single-end, read length, expected average coverage, somatic data, etc.
6. For tool errors, include the error stacktrace as well as the exact command.
7. For format issues, include the result of running ValidateSamFile for BAMs or ValidateVariants for VCFs.
8. For weird results, include an illustrative example, e.g. attach IGV screenshots according to Article#5484.
9. For a seeming variant that is uncalled, include results of following Article#1235.
#### ☞ Formatting tip!
Wrap blocks of code, error messages and BAM/VCF snippets--especially content with hashes (#)--with lines with three backticks ( ` ) each to make a code block as demonstrated here.
GATK version 4.beta.3 (i.e. the third beta release) is out. See the GATK4 beta page for download and details.
# "Raw" VCF FILTER field, previous GATK builds vs Current
Member
edited September 2012
Hello,
Did the UnifiedGenotyper of previous builds use to place "PASS" in the filter fields of vcf files? I have re-ran some data using current best practices, including the HaplotypeCaller. My first snps.indels.raw.vcf file has all . in the FILTER field, which I remember was not a good thing. Im compared the vcf produced at this step with my previous UnifiedGenotyper counterpart and the FILTER field was populated with "PASS". I am bit concerned that is there were no PASS then no LowQual filter applied by the caller either.
Besides using HaplotypeCaller, the only other argument I had changed was returning the -stand_emit_conf to the default. Is this perhaps the root of my concern? Does having both stand_call_conf and stand_emit_conf at defaults ie. equal not apply any FILTER?
Thank you.
The genotyper doesn't apply any filters other than LowQual - which is discussed in the documentation.
The behavior should not have changed. "." in the filter field is not bad (it means that no filtering was applied). Please see the VCF spec for more information.
• Member
edited September 2012
Understood. I guess Im just curious how I applied a filter previously when running UnifiedGenotyper for the first time. I apologize if this is a bit frivolous, but its driving me a bit crazy.
My current vcf header has ##FILTER=<ID=LowQual,Description="Low quality">, but I do not see any LowQual or PASS, so I am inclined to believe this was not applied (as you suggested) and I would like to change that, especially if it can be applied automatically during initial variant calling (before I apply other hard filters). I will have to change my -stand_emit_conf again. Im guessing that is applied to the QUAL? Thank you.
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# Challenge
Given a list of integers, return the list of these integers after repeatedly removing all pairs of adjacent equal items.
Note that if you have an odd-length run of equal numbers, one of them will remain, not being part of a pair.
### Example:
[0, 0, 0, 1, 2, 4, 4, 2, 1, 1, 0]
First, you should remove 0, 0, 4, 4, and 1, 1 to get:
[0, 1, 2, 2, 0]
Now, you should remove 2, 2:
[0, 1, 0]
And this is the final result.
# Test Cases
[] -> []
[1] -> [1]
[1, 1] -> []
[1, 2] -> [1, 2]
[11, 11, 11] -> [11]
[1, 22, 1] -> [1, 22, 1]
[-31, 46, -31, 46] -> [-31, 46, -31, 46]
[1, 0, 0, 1] -> []
[5, 3, 10, 10, 5] -> [5, 3, 5]
[5, 3, 3, 3, 5] -> [5, 3, 5]
[0, -2, 4, 4, -2, 0] -> []
[0, 2, -14, -14, 2, 0, -1] -> [-1]
[0, 0, 0, 1, 2, 4, 4, 2, 1, 1, 0] -> [0, 1, 0]
[3, 5, 4, 4, 8, 26, 26, 8, 5] -> [3]
[-89, 89, -87, -8, 8, 88] -> [-89, 89, -87, -8, 8, 88]
# Scoring
This is , so the shortest answer in each language wins!
• Sandbox for those who can see deleted posts – musicman523 Jul 21 '17 at 20:53
• It doesn't matter, they are all equal. The meaning of this phrase is that [14, 14, 14] collapses to [14] – musicman523 Jul 21 '17 at 21:01
• Misread the challenge, sorry. Thought you had to remove all pairs of numbers increasing by 1 (1,2, 11,12, etc.) – Stephen Jul 21 '17 at 21:02
• Can we take input as a delimited string? – Shaggy Jul 21 '17 at 21:18
• Could you add a test case such as -89,89,-87,-8,-88? Both my (unposted) Japt solution and Fry's Retina solution fail there, outputting --87,8. – Shaggy Jul 21 '17 at 21:55
# Jelly, 10 bytes
Œgœ^/€FµÐL
Try it online!
### How it works
Œgœ^/€FµÐL Main link. Argument: A (array)
µ Combine all links to the left into a chain.
Œg Group all adjacent equal items.
/€ Reduce each group by...
œ^ symmetric multiset difference.
In each step, this maps ([], n) to [n] and ([n], n) to [], so the
group is left with a single item if its length is odd, and no items
at all if its length if even.
F Flatten the resulting array of singleton and empty arrays.
ÐL Apply the chain until the results are no longer unique. Return the last
unique result.
• Using Ẏ instead of F would make you support lists in your list too. – Erik the Outgolfer Jul 22 '17 at 11:37
• No, œ^ relies on integer-to-array promotion here. Since 1D arrays don't get promoted to 2D arrays, it won't work for anything except an array of numbers. – Dennis Jul 22 '17 at 17:53
• Heh...I mean you could've just used ŒgWẎ$œ^/$€ẎµÐL...oh wait that's too naive. :P – Erik the Outgolfer Jul 22 '17 at 17:59
# Retina, 17 15 bytes
+m^(.+)¶\1$¶? Try it online! Saved 2 bytes thanks to Neil and Martin! Replaces each pair of numbers with nothing. This process loops until no changes are made. • Worked up an identical solution in Japt before spotting this. Unfortunately, we both fail on inputs such as -89 89 -87 -88 -88, which outputs --87. – Shaggy Jul 21 '17 at 21:46 • @Shaggy Thanks, I corrected it by adding a boundary check and using _ to denote negatives, as is common in some languages. – FryAmTheEggman Jul 21 '17 at 21:55 • I've since discovered that this'll also fail on _89 89 _87 _8 _88, outputting _89 89 _87 8. Sorry :\ – Shaggy Jul 21 '17 at 21:58 • @Shaggy Don't be sorry! Thanks for finding the problem! I added another boundary check to fix that case. – FryAmTheEggman Jul 21 '17 at 22:45 • @FryAmTheEggman Not sure whether that's what Neil meant but you could then also use m to turn the \bs into ^ and $. – Martin Ender Jul 22 '17 at 7:40
# Mathematica 29 bytes
This repeatedly removes pairs of equal adjacent elements, a_,a_ until there are none left.
#//.{b___,a_,a_,c___}:>{b,c}&
# Python 2, 57 bytes
r=[]
for x in input():r+=x,;r[-2:]*=r[-2:-1]!=[x]
print r
Try it online!
Iteratively constructs the output list by appending the next element, then chopping off the end if the appending element equals the one before it. Checking the second-to-last element r[-2:-1]!=[x] turns out awkward because it's possible the list has length only 1.
• Awesome answer, well done :) – musicman523 Jul 23 '17 at 2:19
Œr;ṪḂ$$€x/€FµÐL Try it online! # Explanation Œr;ṪḂ$$€x/€FµÐL Main Link
Œr Run-length encode
; Concatenate (?)
€ For each element
ṪḂ$$Is the last element odd? € For each element // Non-breaking alternative x/ Reduce by repeating // for run-length decode F Flatten µ (New monadic link) ÐL Repeat until results are no longer unique -1 byte thanks to miles, and fixed :) • @FryAmTheEggman Fixed; thanks! – HyperNeutrino Jul 21 '17 at 21:26 • I'm not sure if throwing an error and leaving the output empty counts as a correct solution. You program throws ValueError: not enough values to unpack (expected 2, got 0) for test case [1,2,2,1]. Also note that empty output is different from [] and 2 is different from [2]. – user72349 Jul 21 '17 at 21:56 • 13 bytes with Œr;ṪḂ$$€ŒṙµÐL. To avoid the error, replace Œṙ with x/€F since run-length decode is throwing an error when given an empty list. To see the output as a list, tacking ŒṘ will show it. – miles Jul 21 '17 at 23:10
• @ThePirateBay Jelly's representation of an empty list is - empty, of one item - just that item, and of multiple items - a bracketed and comma separated list. The submission is of a link (function) not a full program (much like a lambda would be in Python) - to see a more "normal" view place ÇŒṘ in the footer to call the last link (Ç) and print a Python representation (ŒṘ). The error might not be acceptable however. – Jonathan Allan Jul 21 '17 at 23:43
• @JonathanAllan. Ok, I realized that Jelly's string representation of a list is acceptable. The main point of my first comment is to mention that the error is thrown when the list become empty. – user72349 Jul 21 '17 at 23:56
## JavaScript (ES6), 54 53 bytes
Saved 1 byte thanks to @ThePirateBay
f=a=>1/a.find(q=>q==a[++i],i=-2)?f(a,a.splice(i,2)):a
Naive recursive solution, may be improvable.
• You can check current and previous element instead of current and next one, so you can replace i=0 with i=-2 and i-1 with i which is -1 byte in total. – user72349 Jul 21 '17 at 21:18
• @guest44851 Thanks, but... wouldn't that mean I'd need to change it to i+1? (I tried this before with moving the ++ as well and couldn't figure it out, though I only had about a minute to do so) – ETHproductions Jul 21 '17 at 22:19
• You can see that it works properly. – user72349 Jul 21 '17 at 22:41
• @ThePirateBay By golly, you're right! But how? – ETHproductions Jul 22 '17 at 0:32
# Python 2, 73 bytes
Since I do not have enough reputation to comment: I just changed @officialaimm 's answer to use r!=[] instead of len(r) to save a byte. Very clever solution to you, @officialaimm !
r=[] # create list that will hold final results. A new list is important because it needs to be removable.
for i in input():
if r!=[]and r[-1]==i:r.pop() # Ensure that we have at least 1 char added to the list (r!=[])... or that the last character of our final result isn't the current character being scanned. If that is, well, remove it from the final list because we do not want it anymore
else:r+=[i] # Shorthand for r.append(i). This adds i to the final result
print r
Try it online!
It is, again, way too late... why am I even still up?
# Python, 60 58 bytes
f=lambda a:a and(a[:1]+f(a[1:]))[2*(a[:1]==f(a[1:])[:1]):]
Try it online!
• [a[0]] is a[:1] – xnor Jul 23 '17 at 19:18
• @xnor So it is. Thanks! – Anders Kaseorg Jul 23 '17 at 20:14
# MATL, 7 bytes
t"Y'oY"
For some of the test cases where the result is empty the program exits with an error, but in any case it produces the correct (empty) output.
### Explanation
t % Implicit input. Duplicate
" % For each (i.e. do as many times as input size)
Y' % Run-length encode. Gives array of values and array of run lengths
o % Parity, element-wise. Reduces run-lengths to either 0 or 1
Y" % Run-length decode. Gives array of values appearing 0 or 1 times;
% that is, removes pairs of consecutive values
% Implicit end. Implicit display
Consider input
0 0 0 1 2 4 4 2 1 1 0
Each iteration removes pairs of consecutive pairs. The first iteration reduces the array to
0 1 2 2 0
The two values 2 that are now adjacent were not adjacent in the initial array. That's why a second iteration is needed, which gives:
0 1 0
Further iterations will leave this unchanged. The number of required iterations is upper-bounded by the input size.
An empty intermediate result causes the run-length decoding function (Y") to error in the current version of the language; but the ouput is empty as required.
• Could you add an explanation? I'd like to understand how you beat me so soundly. :P – Dennis Jul 25 '17 at 15:11
• @Dennis Sure! I had forgotten. Done :-) – Luis Mendo Jul 25 '17 at 15:33
• Ah, RLE pushes two arrays. That's useful. – Dennis Jul 25 '17 at 15:46
# x86 Machine Code (32-bit protected mode), 36 bytes
52
8B 12
8D 44 91 FC
8B F9
8D 71 04
3B F0
77 10
A7
75 F9
83 EF 04
4A
4A
A5
3B F8
75 FB
97
EB E7
58
89 10
C3
The above bytes of machine code define a function that takes an array as input, collapses adjacent duplicates in-place, and returns to the caller without returning a result. It follows the __fastcall calling convention, passing the two parameters in the ECX and EDX registers, respectively.
The first parameter (ECX) is a pointer to the first element in the array of 32-bit integers (if the array is empty, it can point anywhere in memory). The second parameter (EDX) is a pointer to a 32-bit integer that contains the length of the array.
The function will modify the elements of the array in-place, if necessary, and also update the length to indicate the new length of the collapsed array. This is a bit of an unusual method for taking input and returning output, but you really have no other choice in assembly language. As in C, arrays are actually represented in the language as a pointer to the first element and a length. The only thing a bit weird here is taking the length by reference, but if we didn't do that, there would be no way to shorten the array. The code would work fine, but the output would contain garbage, because the caller wouldn't know where to stop printing elements from the collapsed array.
Ungolfed assembly mnemonics:
; void __fastcall CollapseAdjacentDuplicates(int * ptrArray, int * ptrLength);
; ECX = ptrArray ; ECX = fixed ptr to first element
; EDX = ptrLength
push edx ; save pointer to the length
mov edx, [edx] ; EDX = actual length of the array
lea eax, [ecx+edx*4-4] ; EAX = fixed ptr to last element
mov edi, ecx ; EDI = ptr to element A
lea esi, [ecx+4] ; ESI = ptr to element B
FindNext:
cmp esi, eax ; is ptr to element B at end?
ja Finished ; if we've reached the end, we're finished
cmpsd ; compare DWORDs at ESI and EDI, set flags, and increment both by 4
jne FindNext ; keep looping if this is not a pair
; Found an adjacent pair, so remove it from the array.
sub edi, 4 ; undo increment of EDI so it points at element A
dec edx ; decrease length of the array by 2
dec edx ; (two 1-byte DECs are shorter than one 3-byte SUB)
movsd ; move DWORD at ESI to EDI, and increment both by 4
cmp edi, eax ; have we reached the end?
jne RemoveAdjacentPair ; keep going until we've reached the end
xchg eax, edi ; set new end by updating fixed ptr to last element
Finished:
pop eax ; retrieve pointer to the length
mov [eax], edx ; update length for caller
ret
The implementation was inspired by my C++11 answer, but meticulously rewritten in assembly, optimizing for size. Assembly is a much better golfing language. :-)
Note: Because this code uses the string instructions, is does assume that the direction flag is clear (DF == 0). This is a reasonable assumption in most operating environments, as the ABI typically requires that DF is clear. If this cannot be guaranteed, then a 1-byte CLD instruction (0xFC) needs to be inserted at the top of the code.
It also, as noted, assumes 32-bit protected mode—specifically, a "flat" memory model, where the extra segment (ES) is the same as the data segment (DS).
## Batch, 133 bytes
@set s=.
:l
@if "%1"=="%2" (shift/1)else set s=%s% %1
@shift/1
@if not "%1"=="" goto l
@if not "%s:~2%"=="%*" %0%s:~1%
@echo(%*
I set s to . because Batch gets confused if there are only duplicates. I also have to use shift/1 so that I can use %0%s:~1% to set the argument list to the new array and loop.
• I have to ask ... why? Good answer ... but why? – Zacharý Jul 22 '17 at 0:05
• @Zacharý Because it's there. – Neil Jul 22 '17 at 0:23
• @Zacharý In part, a good reason to golf in non-golfing languages is because this might actually be useful. No one is going to fire up a Jelly interpreter in real life to do this, but they might need to do it in a batch file! – Cody Gray Jul 25 '17 at 16:04
• Oh. that makes sense. – Zacharý Jul 25 '17 at 17:21
# Jelly, 12 bytes
ŒgṁLḂ$$€ẎµÐL A monadic link taking and returning lists of numbers. Try it online! or see a test suite ### How? ŒgṁLḂ$$€ẎµÐL - Link: list
µÐL - perform the chain to the left until no changes occur:
Œg - group runs (yield a list of lists of non-zero-length equal runs)
$€ - last two links as a monad for €ach run:$ - last two links as a monad:
L - length (of the run)
Ḃ - modulo 2 (1 if odd, 0 if even)
ṁ - mould (the run) like (1 or 0) (yields a list of length 1 or 0 lists)
Ẏ - tighten (make the list of lists into a single list)
• ṁLḂ$$€ is equivalent to ḣLḂ$$€ which is equivalent to ṫḊ¿€3$ which you can replace with ṫḊ¿€3 here to form a dyad/nilad pair. – Erik the Outgolfer Jul 22 '17 at 11:40 • That does not work with, for example, an input with a run of length 4. What is the input to the dequeue at each iteration of the while loop? – Jonathan Allan Jul 22 '17 at 12:43 • You are supposed to be left with a list with 0 or 1 elements. If len(x) == 1, then Ḋ will return [] while if len(x) == 0 Ḋ will return 0, both being falsy values. The input to Ḋ is of course the current value, and ṫ will have the current value as left argument and 3 as the right. If len(x) == 4, then it would be the same as ṫ3ṫ3 or ṫ5 leaving you with []. – Erik the Outgolfer Jul 22 '17 at 12:46 • I can see what it is supposed to do, but is x in your description there really the current value? Try this out for size. – Jonathan Allan Jul 22 '17 at 12:49 • To be honest I do not know if that is the code or a bug :) – Jonathan Allan Jul 22 '17 at 12:49 # Japt, 34 bytes ó¥ k_l vîò k_l É}Ãc ó¥ l ¥Ul ?U:ß Recursively removes pairs of equal numbers until none exist. Try it online! with the -Q flag to format the output array. Run all test cases using my WIP CodePen. # 05AB1E, 15 bytes [γʒgÉ}€нÐγ‚€gË# Try it online! ### Explanation [γʒgÉ}€нÐγ‚€gË# [ # Start infinite loop γ # Group Array into consecutive equal elements ʒgÉ} # Keep the subarrays with an uneven amount of elements €н # Keep only the first element of each subarray Ð # Triplicate the result on the stack γ # Group the top element into consecutive equal elements ‚ # Wrap the top two items of the stack in an array €g # Get the length of each subarray Ë# # Break if they are equal # Implicit print # 05AB1E, 13 bytes [DγʒgÉ}€нDŠQ# Try it online! Explanation: [DγʒgÉ}€нDŠQ# Implicit input [ Start infinite loop D Duplicate γ Split into chunks of equal elements ʒ } Filter by g Length É Odd? (0=falsy 1=truthy) € Foreach command н Head D Duplicate Š Push c, a, b Q Equal? (0=falsy 1=truthy) # Break if true (i.e. equal to 1) # Haskell, 33 bytes a!(b:c)|a==b=c a!b=a:b foldr(!)[] Try it online! # Python 2, 74 70 66 bytes • Thanks @SteamyRoot for 4 bytes: r instead of len(r) is enough to check emptiness of the list/stack. • Thanks @ovs for 4 bytes: better if condition [i]==r[-1:] # Python 2, 66 bytes r=[] for i in input(): if[i]==r[-1:]:r.pop() else:r+=[i] print r Try it online! • If the purpose of len(r) is just to check whether or not the list is empty, you should be able to replace it by just r, I think? – sTertooy Jul 22 '17 at 13:47 • Oh yes, Thanks. – officialaimm Jul 22 '17 at 13:55 • 66 bytes – ovs Jul 23 '17 at 13:49 • @ovs Thanks a lot, that is awesome! (y) – officialaimm Jul 23 '17 at 13:58 • Alternative 66 bytes long version, though only requiring three lines. – Jonathan Frech Nov 9 '17 at 17:21 ## Clojure, 100 bytes #(loop[i % j[]](if(= i j)i(recur(mapcat(fn[p](repeat(mod(count p)2)(last p)))(partition-by + i))i))) Not sure if this is the shortest possible. ## Bash, 82 bytes cat>b while cat b>a perl -pe 's/(\d+) \1( |$)//g' a>b
! diff a b>c
do :
done
cat a
There's probably a way out of all those cats, but I don't know it.
## Husk, 10 bytes
ωoṁS↑o%2Lg
Try it online!
## Explanation
ωoṁS↑o%2Lg
ω Repeat until fixed point
o the following two functions:
ṁ b) map over groups and concatenate:
L length of group
o%2 mod 2
S↑ take that many elements of group
# PHP, 81 bytes
function f(&$a){for($i=count($a);--$i;)$a[$i]-$a[$i-1]||array_splice($a,$i-1,2);}
function, call by reference or try it online.
fails for empty input; insert $i&& or $a&& before --$i to fix. # V, 10 bytes òͨ.«©î±î* Try it online! Compressed Regex: :%s/$$.\+$$\n\1\n*. The optional newline is so that it works at the end of the file also. If I assume that there is a newline after the end it would be 8 bytes... but that seems like a stretch # dc, 84 78 bytes [L.ly1-dsy0<A]sA[LtLtS.ly1+sy]sP[dStrdStr!=Pz1<O]sO[0syzdsz1<Oly0<Azlz>M]dsMxf Try it online! Unpacking it a bit, out of order in some attempt at clarity: • [0syzdsz1<Olydsx0<Alx1+lz>M]dsMxf The main macro M resets counter y to 0, retrieves the number of items on the stack, stores this in register z, then runs macro O if there are at least two items on the stack. Once O finishes, it loads counter y and copies it into register x before checking to make sure y is nonzero (meaning stack . has data). If this is the case, it runs macro A. Finally it checks whether the original stack size is larger than the current stack size and reruns itself if so. Once it has finished, it prints the stack with f. • [dStrdStr!=Pz1<O]sO Macro O temporarily stores the top two items on the stack into stack t. It then compares the top two items and runs macro P if they are not equal. Finally it checks whether or not there are at least two items on the stack, and runs itself if so. • [LtLtS.ly1+sy]sP Macro P takes the two items from stack t, pushes the top one back onto the main stack, and pushes the following one onto stack .. It then increments counter y. • [L.ly1-dsy0<A]sA Macro A takes stack . and turns it back into the primary stack. It does that, decrementing counter y until there's nothing left to push. Edited for explanation, and to golf off 6 bytes as I was needlessly storing the size of the stack. # C++11, 161 bytes #include<vector> #include<algorithm> using V=std::vector<int>;void f(V&v){V::iterator i;while((i=std::adjacent_find(v.begin(),v.end()))!=v.end())v.erase(i,i+2);} The above code defines a function, f, that takes a std::vector<int> by reference, modifies it in place to collapse adjacent duplicates according to the specification, and then returns. Try it online! Before I checked the byte count, I thought this was pretty svelte code. Over 150 bytes is, however, not so good! Either I'm not very good at golfing, or C++ is not a very good golfing language… Ungolfed: #include <vector> #include <algorithm> using V = std::vector<int>; void f(V& v) { V::iterator i; // Use std::adjacent_find to search the entire vector for adjacent duplicate elements. // If an adjacent pair is found, this returns an iterator to the first element of the // pair so that we can erase it. Otherwise, it returns v.end(), and we stop. while ((i=std::adjacent_find(v.begin(), v.end())) != v.end()) { v.erase(i, i+2); // erase this adjacent pair } } • C++ isn't the best golfing language. Nice use of std::adjacent_find! I wonder if you implemented this function yourself if it would be shorter, since you can remove #include <algorithm> as well – musicman523 Jul 26 '17 at 17:55 • @musicman523 My first attempt did implement it by hand, although I used a little bit different algorithm. I was adapting the implementation of std::unique to do what I needed. But it takes a lot of code to do all the logic, and when I happened across std::adjacent_find, it was pretty obvious that that was a winner in terms of code size. – Cody Gray Jul 27 '17 at 9:35 # PHP, 74 bytes function c(&$a){foreach($a as$k=>$v)$a[$k+1]===$v&&array_splice($a,$k,2);}
Function c calls by reference to reduce array. Try it online.
Interestingly this works in Php5.6 but not 7.
l=rle(scan());while(any(x<-!l$l%%2))l=rle(l$v[!x]);l$v Try it online! uses a run-length encoding to remove pairs. # J, 38 bytes ;@(<@($~2|#)/.~0+/\@,}.~:}:)^:(0<#)^:_
Try it online!
# GNU sed, 19 + 1 = 20 bytes
+1 byte for -r flag.
:
s/\b(\S+ )\1//g
t
Try it online!
# Pyth, 10 bytes
Bit late to the party.
ueMf%hT2r8
`
Test Suite.
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{}
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# How do you simplify 4.2/0.05?
$\frac{4.2}{0.05}$ can be rewritten as:
$4.2 \div 0.05$
$4.2 \div 0.05 = 84$
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# Two persons start walking on a road that diverge at an angle of 120°.
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Manager
Joined: 01 Jun 2015
Posts: 194
Location: India
GMAT 1: 620 Q48 V26
Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
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Updated on: 24 Oct 2018, 02:18
1
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Difficulty:
55% (hard)
Question Stats:
68% (02:46) correct 32% (02:42) wrong based on 118 sessions
### HideShow timer Statistics
Two persons start walking on a road that diverge at an angle of 120°. If they walk at the rate of 3 km/h and 2 km/h respectively. Find the distance between them after 4 hours.
A. 5
B. 4√19
C. 7
D. 8√19
E. √19
Attachment:
2018-10-24_1415.png [ 11.16 KiB | Viewed 3205 times ]
Originally posted by techiesam on 12 May 2016, 04:01.
Last edited by Bunuel on 24 Oct 2018, 02:18, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Aug 2009
Posts: 8741
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
12 May 2016, 05:16
2
1
1
techiesam wrote:
Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5
B. 4√19
C.7
D. 8√19
E. √19
hi,
Trignometry is NOT tested in GMAT so there has to be another solution..
SEE att image..
Draw an altitude from A to BC extende at D..
Triangle ACD is a right angle triangle, whose HYP is 2*4..
rest sides are CD-4, opp 30 degree and AD - $$4\sqrt{3}$$, opp 60 degree..
c is hyp..
BD = 3*4 + 4 = 16 and AD = $$4\sqrt{3}$$..
HYp = AC = $$\sqrt{16^2+(4*Sq Root 3})^2$$ = $$\sqrt{304}$$ = $$4\sqrt{19}$$
B
Attachments
IMG_5618.JPG [ 1.75 MiB | Viewed 4260 times ]
_________________
Math Expert
Joined: 02 Aug 2009
Posts: 8741
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
12 May 2016, 05:37
2
ronny123 wrote:
chetan2u wrote:
techiesam wrote:
Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5
B. 4√19
C.7
D. 8√19
E. √19
hi,
Trignometry is NOT tested in GMAT so there has to be another solution..
SEE att image..
Draw an altitude from A to BC extende at D..
Triangle ACD is a right angle triangle, whose HYP is 2*4..
rest sides are CD-4, opp 30 degree and AD - $$4\sqrt{3}$$, opp 60 degree..
c is hyp..
BD = 3*4 + 4 = 16 and AD = $$4\sqrt{3}$$..
HYp = AC = $$\sqrt{16^2+(4*Sq Root 3})^2$$ = $$\sqrt{304}$$ = $$4\sqrt{19}$$
B
hi,
Don't be offended but isn't the maths inherent to subject correct ?
if you use c^2 = a^2 + b^2 - 2abcos(120)
= a^2 + b^2 - 2ab(-1/2)
= a^2 + b^2 + ab
a = 2 x 4 = 8
b = 3 x 4 = 12
Putting these values,
c^2 = 8^2 + 12^2 + (8)(12)
= 64 +144 + 96
= 304
c = 4√‾19
Saying Trigo isn't tested on the GMAT isn't changing the solution, my friend
Again, no offense meant.
Hi,
It is not meant for you ..
It is for people who are preparing for GMAT so that they do not get into the area where it is NOT required..
You are most welcome to learn anything being tested or not,
BUT people who are interested in just GMAT should not go away with thought that they have to do trignometry to do this Q..
I don't require to take offence or offend you, I am giving the WAY one should think about this solution
_________________
##### General Discussion
Manager
Joined: 12 Jun 2015
Posts: 74
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
Updated on: 12 May 2016, 05:32
hi techiesam,
can you please check if the question is correct, specifically whether one of the speeds is 3 kmph or 33 kmph
Also,
you can simply use the following formula for cos x to get your answer :
c^2 = a^2 + b^2 - 2abcos(120)
= a^2 + b^2 - 2ab(-1/2)
= a^2 + b^2 + ab
Originally posted by ronny123 on 12 May 2016, 04:52.
Last edited by ronny123 on 12 May 2016, 05:32, edited 1 time in total.
Manager
Joined: 12 Jun 2015
Posts: 74
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
12 May 2016, 05:31
chetan2u wrote:
techiesam wrote:
Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5
B. 4√19
C.7
D. 8√19
E. √19
hi,
Trignometry is NOT tested in GMAT so there has to be another solution..
SEE att image..
Draw an altitude from A to BC extende at D..
Triangle ACD is a right angle triangle, whose HYP is 2*4..
rest sides are CD-4, opp 30 degree and AD - $$4\sqrt{3}$$, opp 60 degree..
c is hyp..
BD = 3*4 + 4 = 16 and AD = $$4\sqrt{3}$$..
HYp = AC = $$\sqrt{16^2+(4*Sq Root 3})^2$$ = $$\sqrt{304}$$ = $$4\sqrt{19}$$
B
hi,
Don't be offended but isn't the maths inherent to subject correct ?
if you use c^2 = a^2 + b^2 - 2abcos(120)
= a^2 + b^2 - 2ab(-1/2)
= a^2 + b^2 + ab
a = 2 x 4 = 8
b = 3 x 4 = 12
Putting these values,
c^2 = 8^2 + 12^2 + (8)(12)
= 64 +144 + 96
= 304
c = 4√‾19
Saying Trigo isn't tested on the GMAT isn't changing the solution, my friend
Again, no offense meant.
VP
Joined: 27 May 2012
Posts: 1070
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
19 Oct 2018, 05:31
1
techiesam wrote:
Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5
B. 4√19
C.7
D. 8√19
E. √19
Dear Moderator,
There is a typo in this question, the latter speed seems to be 3 kmph rather than 33 kmph, hope you will do the needful. Thank you.
_________________
- Stne
Math Expert
Joined: 02 Sep 2009
Posts: 64939
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
24 Oct 2018, 02:19
stne wrote:
techiesam wrote:
Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5
B. 4√19
C.7
D. 8√19
E. √19
Dear Moderator,
There is a typo in this question, the latter speed seems to be 3 kmph rather than 33 kmph, hope you will do the needful. Thank you.
________________
Edited. Thank you.
_________________
Senior Manager
Joined: 21 Jun 2017
Posts: 432
Location: India
Concentration: Finance, Economics
Schools: IIM
GMAT 1: 620 Q47 V30
GPA: 3
WE: Corporate Finance (Commercial Banking)
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
24 Oct 2018, 05:07
1
Longest side will be opposite to angle C. This means c should be more than 12.
Also by triangle property
4<c<20
Combining,
12<c<20.
PS: sqrt 19 is almost 4.5
Posted from my mobile device
VP
Joined: 27 May 2012
Posts: 1070
Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
20 Jul 2019, 08:34
chetan2u wrote:
techiesam wrote:
Two person start walking on a road that diverge at angle of 120 degree.If they walk at the rate of 2kmph and 3 3kmph relatively, then find the distance between them after 4 hours or find the value of c as given in the image attached below.
A.5
B. 4√19
C.7
D. 8√19
E. √19
hi,
Trignometry is NOT tested in GMAT so there has to be another solution..
SEE att image..
Draw an altitude from A to BC extende at D..
Triangle ACD is a right angle triangle, whose HYP is 2*4..
rest sides are CD-4, opp 30 degree and AD - $$4\sqrt{3}$$, opp 60 degree..
c is hyp..
BD = 3*4 + 4 = 16 and AD = $$4\sqrt{3}$$..
HYp = AC = $$\sqrt{16^2+(4*Sq Root 3})^2$$ = $$\sqrt{304}$$ = $$4\sqrt{19}$$
B
Hi chetan2u,
Just a small clarification required here, how can we conclude that the Hypotenuse is 8, why can't it be 12.
How do we deduce that the one going along side b is the one walking at 2mph , the one walking along side b could also be walking at 3mph then the Hypotenuse could be 12.
Am I missing anything? Thank you.
_________________
- Stne
Manager
Joined: 30 Jun 2019
Posts: 223
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink]
### Show Tags
22 Jul 2019, 21:12
You should modify the answers to be in the correct order of either increasing or decreasing value as per GMAT guidelines.
Fastest way to do this problem is to use pythagorean theorum and find c = ~14.
Since the angle is greater than 90, we know the answer must be greater than 14.
4sqrt(19) and 8sqrt(19) are your only two choices.
Besides the fact that 8sqrt(19) is way too huge, if you multiply 14 by 1.3 (~30% larger since 120 is about 30% larger) you get the the answer should be somewhere between 14 and 18, with a number closer 18. 4sqrt(19) = 17.4 = ~17.
This kind of math obviously wouldn't fly on a straight math test, but this is the fastest way to approximate an answer. And unless the GMAT decides to be extremely unforgiving and interested in testing math skills rather than reasoning - then there should only be 1 answer that fits the condition of between 14 and 18, closer to 18.
Re: Two persons start walking on a road that diverge at an angle of 120°. [#permalink] 22 Jul 2019, 21:12
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1. ## Continuity
f(x) = cx/2 if x<2 and 4(c^2+2)/6x if x>=2
continuous at x=2.
find the values for c.
I'm not sure how to go about calculating this.. Any help would be greatly appreciated.
2. Originally Posted by mmfoxall
f(x) = cx/2 if x<2 and 4(c^2+2)/6x if x>=2
continuous at x=2.
find the values for c.
I'm not sure how to go about calculating this.. Any help would be greatly appreciated.
You require $\displaystyle \frac{c(2)}{2} = \frac{4(c^2 + 2)}{6(2)}$. Simplify and solve for c.
3. In other words, you want the two parts to "match up" at x= 2.
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## Zhou, Jun
Compute Distance To:
Author ID: zhou.jun.1 Published as: Zhou, Jun External Links: ORCID · dblp
Documents Indexed: 177 Publications since 2005 Co-Authors: 23 Co-Authors with 79 Joint Publications 1,013 Co-Co-Authors
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### Co-Authors
54 single-authored 31 Mu, Chunlai 15 Ding, Hang 11 Xu, Guangyu 7 Xu, Da 5 Shi, Junping 4 Wang, Xiongrui 3 Deng, Xiumei 3 Hao, Aijing 3 Kim, Chan-Gyun 3 Qiu, Wenlin 3 Zhang, Huan 2 Chen, Hongbin 2 Dong, Zhihua 2 Duan, Zhaoxia 2 Gao, Ketian 2 Guo, Jianguo 2 Guo, Jing 2 Guo, Zongyi 2 Li, Yuhuan 2 Li, Zhongping 2 Liang, Xiaosong 2 Liao, Wudai 2 Liu, Jinxing 2 Liu, Xu 2 Liu, Zongsheng 2 Wang, Renhai 1 Büsze, Ben 1 Cai, Li 1 Cao, Jiliang 1 Cao, Zhenfu 1 Chang, Jing 1 Chen, Bokui 1 Chen, Hui 1 Chen, Jinhuan 1 Chen, Peng 1 Chen, Qiaoyu 1 Cheng, Qiansheng 1 Cheng, Shihong 1 Cheng, Yingying 1 Da, Liexiong 1 Ding, Zhongjun 1 Dong, Xiaolei 1 Du, Haibo 1 Fan, Mingshu 1 Feng, Jianhu 1 Feng, Min 1 Gao, Wubin 1 Guan, Nan 1 Guo, Boling 1 He, Yong 1 Huang, Haoqian 1 Huang, Li 1 Huang, Wei 1 Jayapal, Senthil 1 Je, Minkyu 1 Jiang, Huifa 1 Jiang, Ronghua 1 Jiang, Yanqun 1 Kim, Tony Tae-Hyoung 1 Li, Mengquan 1 Li, Mingjun 1 Li, Xiaolan 1 Lin, Xiangze 1 Liu, Weichen 1 Lou, De Bin 1 Lu, Feng 1 Lu, Xiangyang 1 Lu, Xinbiao 1 Luo, Lirong 1 Ma, Xin 1 Ohsawa, Yasuharu 1 Peng, Jingmei 1 Qian, Yuntao 1 Qiao, Leijie 1 She, Hong Wei 1 Shen, Jian 1 Song, Xiaojun 1 Stuyt, Jan 1 Tang, Bo 1 Tian, Xin 1 Tong, Dongbing 1 Wang, Shi-yao 1 Wang, Yong 1 Wei, Guoliang 1 Wu, Di 1 Wu, Peng 1 Xiao, Chunhua 1 Xie, Yiyuan 1 Xiong, Fengchao 1 Xue, Guoqing 1 Yang, Jingping 1 Yang, Wenhua 1 Yuan, Shujuan 1 Zeinolabedin, Seyed Mohammad Ali 1 Zhang, Lina 1 Zhao, Jinlong 1 Zhao, Zhibing 1 Zheng, Meng 1 Zhou, Wuneng 1 Zhu, Chungang ...and 1 more Co-Authors
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### Serials
15 Journal of Mathematical Analysis and Applications 9 Applied Mathematics Letters 8 ZAMP. Zeitschrift für angewandte Mathematik und Physik 8 Communications on Pure and Applied Analysis 7 Applicable Analysis 7 Computers & Mathematics with Applications 5 Bulletin of the Korean Mathematical Society 5 International Journal of Systems Science. Principles and Applications of Systems and Integration 4 Applied Mathematics and Optimization 4 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 4 Nonlinear Analysis. Real World Applications 4 IEEE Transactions on Circuits and Systems I: Regular Papers 3 Applied Mathematics and Computation 3 Journal of Dynamical and Control Systems 2 International Journal of Control 2 Mathematical Methods in the Applied Sciences 2 Nonlinearity 2 Annales Polonici Mathematici 2 Glasgow Mathematical Journal 2 Mathematics in Practice and Theory 2 Topological Methods in Nonlinear Analysis 2 Bulletin of the Belgian Mathematical Society - Simon Stevin 2 NoDEA. Nonlinear Differential Equations and Applications 2 Nonlinear Dynamics 2 Acta Mathematica Scientia. Series A. (Chinese Edition) 2 Boundary Value Problems 2 Discrete and Continuous Dynamical Systems. Series S 2 Electronic Research Archive 1 Acta Mechanica 1 Analysis Mathematica 1 IMA Journal of Applied Mathematics 1 Journal of the Franklin Institute 1 Physica A 1 Rocky Mountain Journal of Mathematics 1 Chaos, Solitons and Fractals 1 Annali di Matematica Pura ed Applicata. Serie Quarta 1 IEEE Transactions on Computers 1 Journal of Computational and Applied Mathematics 1 Journal of Differential Equations 1 Journal of the Korean Mathematical Society 1 Mathematics and Computers in Simulation 1 Proceedings of the Edinburgh Mathematical Society. Series II 1 Rendiconti del Circolo Matemàtico di Palermo. Serie II 1 Studies in Applied Mathematics 1 Mathematica Numerica Sinica 1 Journal of Sichuan University. Natural Science Edition 1 Advances in Mathematics 1 Computer Aided Geometric Design 1 IMA Journal of Mathematical Control and Information 1 Multidimensional Systems and Signal Processing 1 Numerical Algorithms 1 Journal of Contemporary Mathematical Analysis. Armenian Academy of Sciences 1 International Journal of Computer Mathematics 1 Proceedings of the Royal Society of Edinburgh. Section A. Mathematics 1 International Journal of Bifurcation and Chaos in Applied Sciences and Engineering 1 Journal of Nonlinear Science 1 Electronic Journal of Differential Equations (EJDE) 1 Computational and Applied Mathematics 1 Discrete and Continuous Dynamical Systems 1 Mathematical Problems in Engineering 1 Electronic Journal of Qualitative Theory of Differential Equations 1 Mathematical and Computer Modelling of Dynamical Systems 1 Communications of the Korean Mathematical Society 1 Communications in Nonlinear Science and Numerical Simulation 1 The ANZIAM Journal 1 IEEE Transactions on Image Processing 1 Nonlinear Analysis. Modelling and Control 1 Applied Mathematics E-Notes 1 Journal of Applied Mathematics 1 Acta Mathematica Scientia. Series B. (English Edition) 1 Journal of University of Science and Technology of China 1 Analysis and Applications (Singapore) 1 Journal of Hefei University of Technology. Natural Science 1 Mediterranean Journal of Mathematics 1 Global Journal of Pure and Applied Mathematics 1 Mathematical Biosciences and Engineering 1 International Journal of Evolution Equations 1 Acta Mathematica Sinica. Chinese Series 1 Surveys in Mathematics and its Applications 1 Journal of Nonlinear Science and Applications 1 Differential Equations and Applications 1 Advances in Applied Mathematics and Mechanics 1 Science China. Mathematics 1 Scientia Sinica. Mathematica 1 ISRN Mathematical Analysis 1 Analysis and Mathematical Physics 1 Advances in Nonlinear Analysis 1 East Asian Journal on Applied Mathematics
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### Fields
135 Partial differential equations (35-XX) 28 Biology and other natural sciences (92-XX) 14 Systems theory; control (93-XX) 12 Fluid mechanics (76-XX) 11 Numerical analysis (65-XX) 8 Information and communication theory, circuits (94-XX) 7 Mechanics of deformable solids (74-XX) 6 Dynamical systems and ergodic theory (37-XX) 6 Operator theory (47-XX) 6 Computer science (68-XX) 4 Ordinary differential equations (34-XX) 4 Integral equations (45-XX) 3 Real functions (26-XX) 2 Harmonic analysis on Euclidean spaces (42-XX) 2 Functional analysis (46-XX) 2 Mechanics of particles and systems (70-XX) 2 Statistical mechanics, structure of matter (82-XX) 2 Operations research, mathematical programming (90-XX) 1 Associative rings and algebras (16-XX) 1 Potential theory (31-XX) 1 Several complex variables and analytic spaces (32-XX) 1 Difference and functional equations (39-XX) 1 Statistics (62-XX) 1 Optics, electromagnetic theory (78-XX) 1 Geophysics (86-XX) 1 Game theory, economics, finance, and other social and behavioral sciences (91-XX)
### Citations contained in zbMATH Open
102 Publications have been cited 664 times in 379 Documents Cited by Year
Global existence and blow-up for a mixed pseudo-parabolic $$p$$-Laplacian type equation with logarithmic nonlinearity. Zbl 1447.35202
Ding, Hang; Zhou, Jun
2019
Coexistence states of a Holling type-II predator-prey system. Zbl 1196.35217
Zhou, Jun; Mu, Chunlai
2010
Pattern formation of a coupled two-cell Brusselator model. Zbl 1195.35292
Zhou, Jun; Mu, Chunlai
2010
The existence, bifurcation and stability of positive stationary solutions of a diffusive Leslie-Gower predator-prey model with Holling-type II functional responses. Zbl 1306.92054
Zhou, Jun; Shi, Junping
2013
Critical blow-up and extinction exponents for non-Newton polytropic filtration equation with source. Zbl 1180.35314
Zhou, Jun; Mu, Chunlai
2009
Blow-up for a thin-film equation with positive initial energy. Zbl 1353.35078
Zhou, Jun
2017
Positive solutions of a diffusive predator-prey model with modified Leslie-Gower and Holling-type II schemes. Zbl 1234.35288
Zhou, Jun
2012
Local existence, global existence and blow-up of solutions to a nonlocal Kirchhoff diffusion problem. Zbl 1430.35124
Ding, Hang; Zhou, Jun
2020
Positive solutions for a three-trophic food chain model with diffusion and Beddington-DeAngelis functional response. Zbl 1206.35239
Zhou, Jun; Mu, Chunlai
2011
Global existence and blow-up of solutions for a Kirchhoff type plate equation with damping. Zbl 1410.35237
Zhou, Jun
2015
The critical curve for a non-Newtonian polytropic filtration system coupled via nonlinear boundary flux. Zbl 1173.35348
Zhou, Jun; Mu, Chunlai
2008
Lower bounds for blow-up time of two nonlinear wave equations. Zbl 1316.35055
Zhou, Jun
2015
A multi-dimension blow-up problem to a porous medium diffusion equation with special medium void. Zbl 1320.35102
Zhou, Jun
2014
Positive steady state solutions of a diffusive Leslie-Gower predator-prey model with Holling type II functional response and cross-diffusion. Zbl 1304.35275
Zhou, Jun; Kim, Chan-Gyun; Shi, Junping
2014
Global asymptotical behavior of solutions to a class of fourth order parabolic equation modeling epitaxial growth. Zbl 1460.74061
Zhou, Jun
2019
Global existence and finite time blow-up for a class of thin-film equation. Zbl 1379.35161
Dong, Zhihua; Zhou, Jun
2017
Global existence and blow-up of solutions to a nonlocal parabolic equation with singular potential. Zbl 06877206
Feng, Min; Zhou, Jun
2018
On the critical Fujita exponent for a degenerate parabolic system coupled via nonlinear boundary flux. Zbl 1172.35007
Zhou, Jun; Mu, Chunlai
2008
Bifurcation analysis of a diffusive predator-prey model with ratio-dependent Holling type III functional response. Zbl 1348.92139
Zhou, Jun
2015
Lifespan for a semilinear pseudo-parabolic equation. Zbl 1387.35058
Xu, Guangyu; Zhou, Jun
2018
Global existence and finite time blow-up of the solution for a thin-film equation with high initial energy. Zbl 1375.35058
Xu, Guangyu; Zhou, Jun
2018
$$L^{2}$$-norm blow-up of solutions to a fourth order parabolic PDE involving the Hessian. Zbl 1405.35074
Zhou, Jun
2018
Global existence and blow-up for a fourth order parabolic equation involving the Hessian. Zbl 1375.35240
Xu, Guangyu; Zhou, Jun
2017
A time two-grid algorithm based on finite difference method for the two-dimensional nonlinear time-fractional mobile/immobile transport model. Zbl 1452.65175
Qiu, Wenlin; Xu, Da; Guo, Jing; Zhou, Jun
2020
Global existence and blow-up for a parabolic problem of Kirchhoff type with logarithmic nonlinearity. Zbl 1469.35122
Ding, Hang; Zhou, Jun
2021
Global asymptotical behavior and some new blow-up conditions of solutions to a thin-film equation. Zbl 1394.35059
Zhou, Jun
2018
Positive steady state solutions of a Leslie-Gower predator-prey model with Holling type II functional response and density-dependent diffusion. Zbl 1318.92045
Zhou, Jun
2013
Global existence and blow-up for a non-Newton polytropic filtration system with nonlocal source. Zbl 1178.35203
Zhou, Jun; Mu, Chunlai
2008
Positive solutions of a diffusive Leslie-Gower predator-prey model with Bazykin functional response. Zbl 1293.35349
Zhou, Jun
2014
Ground state solution for a fourth-order elliptic equation with logarithmic nonlinearity modeling epitaxial growth. Zbl 1442.35109
Zhou, Jun
2019
Global existence and blow-up of solutions to a nonlocal Kirchhoff diffusion problem. Zbl 1452.35073
Ding, Hang; Zhou, Jun
2020
Blow-up and global existence of solutions to a parabolic equation associated with the fraction $$p$$-Laplacian. Zbl 1409.35219
Jiang, Ronghua; Zhou, Jun
2019
Coexistence of a diffusive predator-prey model with Holling type-II functional response and density dependent mortality. Zbl 1254.35226
Zhou, Jun; Mu, Chunlai
2012
Pattern formation in a general glycolysis reaction-diffusion system. Zbl 1338.35445
Zhou, Jun; Shi, Junping
2015
Blowup, extinction and non-extinction for a nonlocal $$p$$-biharmonic parabolic equation. Zbl 1355.35089
Hao, Aijing; Zhou, Jun
2017
Qualitative analysis of a modified Leslie-Gower predator-prey model with Crowley-Martin functional responses. Zbl 1312.35015
Zhou, Jun
2015
Blowup for degenerate and singular parabolic system with nonlocal source. Zbl 1146.35378
Zhou, Jun; Mu, Chunlai; Li, Zhongping
2006
An ADI compact difference scheme for the two-dimensional semilinear time-fractional mobile-immobile equation. Zbl 1476.35150
Jiang, Huifa; Xu, Da; Qiu, Wenlin; Zhou, Jun
2020
Initial boundary value problem for a inhomogeneous pseudo-parabolic equation. Zbl 1439.35301
Zhou, Jun
2020
A weak Galerkin finite element method for multi-term time-fractional diffusion equations. Zbl 1468.65157
Zhou, Jun; Xu, Da; Chen, Hongbin
2018
Positive solutions for a Lotka-Volterra prey-predator model with cross-diffusion and Holling type-II functional response. Zbl 1315.35089
Zhou, Jun; Kim, Chan-Gyun
2014
Fujita exponent for an inhomogeneous pseudoparabolic equation. Zbl 1445.35044
Zhou, Jun
2020
Time periodic solutions of porous medium equation. Zbl 1203.35019
Zhou, Jun; Mu, Chunlai
2010
Blowup for a degenerate and singular parabolic equation with non-local source and absorption. Zbl 1206.35153
Zhou, Jun; Mu, Chunlai
2010
Incomplete quenching of heat equations with absorption. Zbl 1165.35394
Zhou, Jun; He, Yong; Mu, Chunlai
2008
Blow-up and global existence to a degenerate reaction-diffusion equation with nonlinear memory. Zbl 1139.35068
Zhou, Jun; Mu, Chunlai; Lu, Feng
2007
Upper bounds of blow-up time and blow-up rate for a semi-linear edge-degenerate parabolic equation. Zbl 1378.35047
Xu, Guangyu; Zhou, Jun
2017
A new blow-up condition for semi-linear edge degenerate parabolic equation with singular potentials. Zbl 1366.35077
Hao, Aijing; Zhou, Jun
2017
Qualitative analysis of an autocatalytic chemical reaction model with decay. Zbl 1292.35150
Zhou, Jun; Shi, Junping
2014
Initial-boundary value problem for a fourth-order plate equation with Hardy-Hénon potential and polynomial nonlinearity. Zbl 1442.35247
Liu, Xu; Zhou, Jun
2020
Upper bound estimate for the blow-up time of an evolution $$m$$-Laplace equation involving variable source and positive initial energy. Zbl 1443.35082
Zhou, Jun; Yang, Di
2015
Global existence and blow-up of solutions to a singular non-Newton polytropic filtration equation with critical and supercritical initial energy. Zbl 1395.35049
Xu, Guangyu; Zhou, Jun
2018
Global existence and blow-up to a degenerate reaction-diffusion system with nonlinear memory. Zbl 1154.35391
Zhou, Jun; Mu, Chunlai; Fan, Mingshu
2008
Uniform blow-up profiles and boundary layer for a parabolic system with localized sources. Zbl 1146.35377
Zhou, Jun; Mu, Chunlai
2008
Blow-up and lifespan of solutions to a nonlocal parabolic equation at arbitrary initial energy level. Zbl 06832832
Zhou, Jun
2018
Alternating direction implicit difference scheme for the multi-term time-fractional integro-differential equation with a weakly singular kernel. Zbl 1443.65153
Zhou, Jun; Xu, Da
2020
Global existence and blow-up of solutions for a non-Newton polytropic filtration system with special volumetric moisture content. Zbl 1443.35081
Zhou, Jun
2016
Quenching for a parabolic equation with variable coefficient modeling MEMS technology. Zbl 1426.35141
Zhou, Jun
2017
Global existence and blow-up for non-Newton polytropic filtration system coupled with local source. Zbl 1151.35381
Zhou, Jun; Mu, Chunlai
2009
Buckling analysis of a plate with built-in rectangular delamination by strip distributed transfer function method. Zbl 1071.74022
Li, D.; Tang, G.; Zhou, J.; Lei, Y.
2005
Global existence and blowup of solutions for a class of nonlinear higher-order wave equations. Zbl 1262.35172
Zhou, Jun; Wang, Xiongrui; Song, Xiaojun; Mu, Chunlai
2012
Corrigendum to “Coexistence states of a Holling type-II predator-prey system” [J. Math. Anal. Appl. 369 (2) (2010) 555-563]. Zbl 1235.35277
Zhou, Jun; Mu, Chunlai
2011
Global existence and blow-up for weakly coupled degenerate and singular parabolic equations with localized source. Zbl 1228.35127
Zhou, Jun; Mu, Chunlai
2011
Blowup for a degenerate and singular parabolic equation with nonlocal source and nonlocal boundary. Zbl 1338.35262
Zhou, Jun; Yang, Di
2015
Bifurcation analysis of the Oregonator model. Zbl 1330.35030
Zhou, Jun
2016
A new blow-up condition for a parabolic equation with singular potential. Zbl 1356.35058
Hao, Aijing; Zhou, Jun
2017
Algebraic criteria for global existence or blow-up for a boundary coupled system of nonlinear diffusion equations. Zbl 1132.35404
Zhou, Jun; Mu, Chunlai
2007
Global existence and blowup for a degenerate and singular parabolic system with nonlocal source and absorptions. Zbl 1296.35082
Zhou, Jun
2014
Global existence, extinction, and non-extinction of solutions to a fast diffusion $$p$$-Laplace evolution equation with singular potential. Zbl 1444.35108
Deng, Xiumei; Zhou, Jun
2020
Global existence and blow-up for a mixed pseudo-parabolic $$p$$-Laplacian type equation with logarithmic nonlinearity. II. Zbl 1474.35416
Ding, Hang; Zhou, Jun
2021
Global existence and blow-up of solutions to a semilinear heat equation with logarithmic nonlinearity. Zbl 1475.35079
Peng, Jingmei; Zhou, Jun
2021
Infinite time blow-up of solutions to a class of wave equations with weak and strong damping terms and logarithmic nonlinearity. Zbl 1476.35144
Ding, Hang; Wang, Renhai; Zhou, Jun
2021
Blow-up and exponential decay of solutions to a class of pseudo-parabolic equation. Zbl 1430.35146
Zhou, Jun
2019
Behavior of solutions to a fourth-order nonlinear parabolic equation with logarithmic nonlinearity. Zbl 1470.35174
Zhou, Jun
2021
Well-posedness of solutions for the sixth-order Boussinesq equation with linear strong damping and nonlinear source. Zbl 1471.35004
Zhou, Jun; Zhang, Huan
2021
Analysis of a pseudo-parabolic equation by potential wells. Zbl 1473.35348
Zhou, Jun; Xu, Guangyu; Mu, Chunlai
2021
Spatiotemporal pattern formation of a diffusive bimolecular model with autocatalysis and saturation law. Zbl 1345.92170
Zhou, Jun
2013
The second critical exponent for a nonlocal porous medium equation in $$\mathbb{R}^N$$. Zbl 1328.35113
Zhou, Jun
2014
Coexistence of a three species predator-prey model with diffusion and density dependent mortality. Zbl 1231.35276
Zhou, Jun; Mu, Chunlai
2011
Positive solutions for a modified Leslie-Gower prey-predator model with Crowley-Martin functional responses. Zbl 1316.92078
Zhou, Jun
2014
Asymptotic behavior for a fourth-order parabolic equation involving the Hessian. Zbl 1406.35145
Xu, Guangyu; Zhou, Jun
2018
Stability analysis for complicated sampled-data systems via descriptor remodelling. Zbl 1475.93085
Zhou, Jun; Gao, Ketian; Lu, Xinbiao
2019
Two new blow-up conditions for a pseudo-parabolic equation with logarithmic nonlinearity. Zbl 1423.35215
Ding, Hang; Zhou, Jun
2019
Global existence, finite time blow-up, and vacuum isolating phenomenon for a class of thin-film equation. Zbl 1439.76012
Xu, Guangyu; Zhou, Jun; Mu, Chunlai
2020
Well-posedness of solutions for the dissipative Boussinesq equation with logarithmic nonlinearity. Zbl 1491.35035
Ding, Hang; Zhou, Jun
2022
Global existence and blow-up of solutions to a semilinear heat equation with singular potential and logarithmic nonlinearity. Zbl 1430.35139
Deng, Xiumei; Zhou, Jun
2020
Asymptotic behaviors of solutions to a sixth-order Boussinesq equation with logarithmic nonlinearity. Zbl 1466.35049
Zhang, Huan; Zhou, Jun
2021
Blow-up of solutions to a parabolic system with nonlocal source. Zbl 1391.35216
Dong, Zhihua; Zhou, Jun
2018
The lifespan for 3D quasilinear wave equations with nonlinear damping terms. Zbl 1230.35062
Zhou, Jun; Mu, Chunlai
2011
Stability analysis for a predator-prey system with nonlocal delayed reaction-diffusion equations. Zbl 1274.35191
Li, Yuhuan; Zhou, Jun; Mu, Chunlai
2012
Blow-up for a non-Newton polytropic filtration system with nonlinear nonlocal source. Zbl 1168.35379
Zhou, Jun; Mu, Chunlai
2008
Uniqueness of the positive solution for a non-cooperative model of nuclear reactors. Zbl 06421005
Zhou, Jun; Shi, Junping
2013
Non-simultaneous blow-up for a semilinear parabolic system with nonlinear memory. Zbl 1128.35356
Zhou, Jun
2007
Blow-up rate for a porous medium equation with convection. Zbl 1144.35424
Zhou, Jun; Mu, Chunlai
2007
Turing instability and Hopf bifurcation of a bimolecular model with autocatalysis and saturation law. Zbl 1389.35069
Zhou, Jun
2017
Pattern formation in a general Degn-Harrison reaction model. Zbl 1373.35044
Zhou, Jun
2017
Global existence and blow-up to the solutions of a singular porous medium equation with critical initial energy. Zbl 1381.35096
Luo, Lirong; Zhou, Jun
2016
On the Cauchy problem for a reaction-diffusion system with singular nonlinearity. Zbl 1299.35040
Zhou, Jun
2013
Qualitative analysis for a degenerate Kirchhoff-type diffusion equation involving the fractional $$p$$-Laplacian. Zbl 1476.35056
Xu, Guangyu; Zhou, Jun
2021
Global existence and blow-up of solutions to a class of nonlocal parabolic equations. Zbl 1442.35222
Xu, Guangyu; Zhou, Jun
2019
Well-posedness of solutions for the dissipative Boussinesq equation with logarithmic nonlinearity. Zbl 1491.35035
Ding, Hang; Zhou, Jun
2022
Global existence and blow-up for a parabolic problem of Kirchhoff type with logarithmic nonlinearity. Zbl 1469.35122
Ding, Hang; Zhou, Jun
2021
Global existence and blow-up for a mixed pseudo-parabolic $$p$$-Laplacian type equation with logarithmic nonlinearity. II. Zbl 1474.35416
Ding, Hang; Zhou, Jun
2021
Global existence and blow-up of solutions to a semilinear heat equation with logarithmic nonlinearity. Zbl 1475.35079
Peng, Jingmei; Zhou, Jun
2021
Infinite time blow-up of solutions to a class of wave equations with weak and strong damping terms and logarithmic nonlinearity. Zbl 1476.35144
Ding, Hang; Wang, Renhai; Zhou, Jun
2021
Behavior of solutions to a fourth-order nonlinear parabolic equation with logarithmic nonlinearity. Zbl 1470.35174
Zhou, Jun
2021
Well-posedness of solutions for the sixth-order Boussinesq equation with linear strong damping and nonlinear source. Zbl 1471.35004
Zhou, Jun; Zhang, Huan
2021
Analysis of a pseudo-parabolic equation by potential wells. Zbl 1473.35348
Zhou, Jun; Xu, Guangyu; Mu, Chunlai
2021
Asymptotic behaviors of solutions to a sixth-order Boussinesq equation with logarithmic nonlinearity. Zbl 1466.35049
Zhang, Huan; Zhou, Jun
2021
Qualitative analysis for a degenerate Kirchhoff-type diffusion equation involving the fractional $$p$$-Laplacian. Zbl 1476.35056
Xu, Guangyu; Zhou, Jun
2021
Local existence, global existence and blow-up of solutions to a nonlocal Kirchhoff diffusion problem. Zbl 1430.35124
Ding, Hang; Zhou, Jun
2020
A time two-grid algorithm based on finite difference method for the two-dimensional nonlinear time-fractional mobile/immobile transport model. Zbl 1452.65175
Qiu, Wenlin; Xu, Da; Guo, Jing; Zhou, Jun
2020
Global existence and blow-up of solutions to a nonlocal Kirchhoff diffusion problem. Zbl 1452.35073
Ding, Hang; Zhou, Jun
2020
An ADI compact difference scheme for the two-dimensional semilinear time-fractional mobile-immobile equation. Zbl 1476.35150
Jiang, Huifa; Xu, Da; Qiu, Wenlin; Zhou, Jun
2020
Initial boundary value problem for a inhomogeneous pseudo-parabolic equation. Zbl 1439.35301
Zhou, Jun
2020
Fujita exponent for an inhomogeneous pseudoparabolic equation. Zbl 1445.35044
Zhou, Jun
2020
Initial-boundary value problem for a fourth-order plate equation with Hardy-Hénon potential and polynomial nonlinearity. Zbl 1442.35247
Liu, Xu; Zhou, Jun
2020
Alternating direction implicit difference scheme for the multi-term time-fractional integro-differential equation with a weakly singular kernel. Zbl 1443.65153
Zhou, Jun; Xu, Da
2020
Global existence, extinction, and non-extinction of solutions to a fast diffusion $$p$$-Laplace evolution equation with singular potential. Zbl 1444.35108
Deng, Xiumei; Zhou, Jun
2020
Global existence, finite time blow-up, and vacuum isolating phenomenon for a class of thin-film equation. Zbl 1439.76012
Xu, Guangyu; Zhou, Jun; Mu, Chunlai
2020
Global existence and blow-up of solutions to a semilinear heat equation with singular potential and logarithmic nonlinearity. Zbl 1430.35139
Deng, Xiumei; Zhou, Jun
2020
Active finite-time disturbance rejection control for attitude tracking of quad-rotor under input saturation. Zbl 1450.93039
Zhou, Jun; Cheng, Yingying; Du, Haibo; Wu, Di; Zhu, Min; Lin, Xiangze
2020
Stability analysis and stabilisation in linear continuous-time periodic systems by complex scaling. Zbl 1453.93196
Zhou, J.
2020
Global existence and blow-up for a mixed pseudo-parabolic $$p$$-Laplacian type equation with logarithmic nonlinearity. Zbl 1447.35202
Ding, Hang; Zhou, Jun
2019
Global asymptotical behavior of solutions to a class of fourth order parabolic equation modeling epitaxial growth. Zbl 1460.74061
Zhou, Jun
2019
Ground state solution for a fourth-order elliptic equation with logarithmic nonlinearity modeling epitaxial growth. Zbl 1442.35109
Zhou, Jun
2019
Blow-up and global existence of solutions to a parabolic equation associated with the fraction $$p$$-Laplacian. Zbl 1409.35219
Jiang, Ronghua; Zhou, Jun
2019
Blow-up and exponential decay of solutions to a class of pseudo-parabolic equation. Zbl 1430.35146
Zhou, Jun
2019
Stability analysis for complicated sampled-data systems via descriptor remodelling. Zbl 1475.93085
Zhou, Jun; Gao, Ketian; Lu, Xinbiao
2019
Two new blow-up conditions for a pseudo-parabolic equation with logarithmic nonlinearity. Zbl 1423.35215
Ding, Hang; Zhou, Jun
2019
Global existence and blow-up of solutions to a class of nonlocal parabolic equations. Zbl 1442.35222
Xu, Guangyu; Zhou, Jun
2019
Global existence and blow-up of solutions to a nonlocal parabolic equation with singular potential. Zbl 06877206
Feng, Min; Zhou, Jun
2018
Lifespan for a semilinear pseudo-parabolic equation. Zbl 1387.35058
Xu, Guangyu; Zhou, Jun
2018
Global existence and finite time blow-up of the solution for a thin-film equation with high initial energy. Zbl 1375.35058
Xu, Guangyu; Zhou, Jun
2018
$$L^{2}$$-norm blow-up of solutions to a fourth order parabolic PDE involving the Hessian. Zbl 1405.35074
Zhou, Jun
2018
Global asymptotical behavior and some new blow-up conditions of solutions to a thin-film equation. Zbl 1394.35059
Zhou, Jun
2018
A weak Galerkin finite element method for multi-term time-fractional diffusion equations. Zbl 1468.65157
Zhou, Jun; Xu, Da; Chen, Hongbin
2018
Global existence and blow-up of solutions to a singular non-Newton polytropic filtration equation with critical and supercritical initial energy. Zbl 1395.35049
Xu, Guangyu; Zhou, Jun
2018
Blow-up and lifespan of solutions to a nonlocal parabolic equation at arbitrary initial energy level. Zbl 06832832
Zhou, Jun
2018
Asymptotic behavior for a fourth-order parabolic equation involving the Hessian. Zbl 1406.35145
Xu, Guangyu; Zhou, Jun
2018
Blow-up of solutions to a parabolic system with nonlocal source. Zbl 1391.35216
Dong, Zhihua; Zhou, Jun
2018
Blow-up for a thin-film equation with positive initial energy. Zbl 1353.35078
Zhou, Jun
2017
Global existence and finite time blow-up for a class of thin-film equation. Zbl 1379.35161
Dong, Zhihua; Zhou, Jun
2017
Global existence and blow-up for a fourth order parabolic equation involving the Hessian. Zbl 1375.35240
Xu, Guangyu; Zhou, Jun
2017
Blowup, extinction and non-extinction for a nonlocal $$p$$-biharmonic parabolic equation. Zbl 1355.35089
Hao, Aijing; Zhou, Jun
2017
Upper bounds of blow-up time and blow-up rate for a semi-linear edge-degenerate parabolic equation. Zbl 1378.35047
Xu, Guangyu; Zhou, Jun
2017
A new blow-up condition for semi-linear edge degenerate parabolic equation with singular potentials. Zbl 1366.35077
Hao, Aijing; Zhou, Jun
2017
Quenching for a parabolic equation with variable coefficient modeling MEMS technology. Zbl 1426.35141
Zhou, Jun
2017
A new blow-up condition for a parabolic equation with singular potential. Zbl 1356.35058
Hao, Aijing; Zhou, Jun
2017
Turing instability and Hopf bifurcation of a bimolecular model with autocatalysis and saturation law. Zbl 1389.35069
Zhou, Jun
2017
Pattern formation in a general Degn-Harrison reaction model. Zbl 1373.35044
Zhou, Jun
2017
Global existence and blow-up of solutions for a non-Newton polytropic filtration system with special volumetric moisture content. Zbl 1443.35081
Zhou, Jun
2016
Bifurcation analysis of the Oregonator model. Zbl 1330.35030
Zhou, Jun
2016
Global existence and blow-up to the solutions of a singular porous medium equation with critical initial energy. Zbl 1381.35096
Luo, Lirong; Zhou, Jun
2016
Global existence and blow-up of solutions for a Kirchhoff type plate equation with damping. Zbl 1410.35237
Zhou, Jun
2015
Lower bounds for blow-up time of two nonlinear wave equations. Zbl 1316.35055
Zhou, Jun
2015
Bifurcation analysis of a diffusive predator-prey model with ratio-dependent Holling type III functional response. Zbl 1348.92139
Zhou, Jun
2015
Pattern formation in a general glycolysis reaction-diffusion system. Zbl 1338.35445
Zhou, Jun; Shi, Junping
2015
Qualitative analysis of a modified Leslie-Gower predator-prey model with Crowley-Martin functional responses. Zbl 1312.35015
Zhou, Jun
2015
Upper bound estimate for the blow-up time of an evolution $$m$$-Laplace equation involving variable source and positive initial energy. Zbl 1443.35082
Zhou, Jun; Yang, Di
2015
Blowup for a degenerate and singular parabolic equation with nonlocal source and nonlocal boundary. Zbl 1338.35262
Zhou, Jun; Yang, Di
2015
A multi-dimension blow-up problem to a porous medium diffusion equation with special medium void. Zbl 1320.35102
Zhou, Jun
2014
Positive steady state solutions of a diffusive Leslie-Gower predator-prey model with Holling type II functional response and cross-diffusion. Zbl 1304.35275
Zhou, Jun; Kim, Chan-Gyun; Shi, Junping
2014
Positive solutions of a diffusive Leslie-Gower predator-prey model with Bazykin functional response. Zbl 1293.35349
Zhou, Jun
2014
Positive solutions for a Lotka-Volterra prey-predator model with cross-diffusion and Holling type-II functional response. Zbl 1315.35089
Zhou, Jun; Kim, Chan-Gyun
2014
Qualitative analysis of an autocatalytic chemical reaction model with decay. Zbl 1292.35150
Zhou, Jun; Shi, Junping
2014
Global existence and blowup for a degenerate and singular parabolic system with nonlocal source and absorptions. Zbl 1296.35082
Zhou, Jun
2014
The second critical exponent for a nonlocal porous medium equation in $$\mathbb{R}^N$$. Zbl 1328.35113
Zhou, Jun
2014
Positive solutions for a modified Leslie-Gower prey-predator model with Crowley-Martin functional responses. Zbl 1316.92078
Zhou, Jun
2014
The existence, bifurcation and stability of positive stationary solutions of a diffusive Leslie-Gower predator-prey model with Holling-type II functional responses. Zbl 1306.92054
Zhou, Jun; Shi, Junping
2013
Positive steady state solutions of a Leslie-Gower predator-prey model with Holling type II functional response and density-dependent diffusion. Zbl 1318.92045
Zhou, Jun
2013
Spatiotemporal pattern formation of a diffusive bimolecular model with autocatalysis and saturation law. Zbl 1345.92170
Zhou, Jun
2013
Uniqueness of the positive solution for a non-cooperative model of nuclear reactors. Zbl 06421005
Zhou, Jun; Shi, Junping
2013
On the Cauchy problem for a reaction-diffusion system with singular nonlinearity. Zbl 1299.35040
Zhou, Jun
2013
Positive solutions of a diffusive predator-prey model with modified Leslie-Gower and Holling-type II schemes. Zbl 1234.35288
Zhou, Jun
2012
Coexistence of a diffusive predator-prey model with Holling type-II functional response and density dependent mortality. Zbl 1254.35226
Zhou, Jun; Mu, Chunlai
2012
Global existence and blowup of solutions for a class of nonlinear higher-order wave equations. Zbl 1262.35172
Zhou, Jun; Wang, Xiongrui; Song, Xiaojun; Mu, Chunlai
2012
Stability analysis for a predator-prey system with nonlocal delayed reaction-diffusion equations. Zbl 1274.35191
Li, Yuhuan; Zhou, Jun; Mu, Chunlai
2012
Positive solutions for a three-trophic food chain model with diffusion and Beddington-DeAngelis functional response. Zbl 1206.35239
Zhou, Jun; Mu, Chunlai
2011
Corrigendum to “Coexistence states of a Holling type-II predator-prey system” [J. Math. Anal. Appl. 369 (2) (2010) 555-563]. Zbl 1235.35277
Zhou, Jun; Mu, Chunlai
2011
Global existence and blow-up for weakly coupled degenerate and singular parabolic equations with localized source. Zbl 1228.35127
Zhou, Jun; Mu, Chunlai
2011
Coexistence of a three species predator-prey model with diffusion and density dependent mortality. Zbl 1231.35276
Zhou, Jun; Mu, Chunlai
2011
The lifespan for 3D quasilinear wave equations with nonlinear damping terms. Zbl 1230.35062
Zhou, Jun; Mu, Chunlai
2011
Coexistence states of a Holling type-II predator-prey system. Zbl 1196.35217
Zhou, Jun; Mu, Chunlai
2010
Pattern formation of a coupled two-cell Brusselator model. Zbl 1195.35292
Zhou, Jun; Mu, Chunlai
2010
Time periodic solutions of porous medium equation. Zbl 1203.35019
Zhou, Jun; Mu, Chunlai
2010
Blowup for a degenerate and singular parabolic equation with non-local source and absorption. Zbl 1206.35153
Zhou, Jun; Mu, Chunlai
2010
Critical blow-up and extinction exponents for non-Newton polytropic filtration equation with source. Zbl 1180.35314
Zhou, Jun; Mu, Chunlai
2009
Global existence and blow-up for non-Newton polytropic filtration system coupled with local source. Zbl 1151.35381
Zhou, Jun; Mu, Chunlai
2009
The critical curve for a non-Newtonian polytropic filtration system coupled via nonlinear boundary flux. Zbl 1173.35348
Zhou, Jun; Mu, Chunlai
2008
On the critical Fujita exponent for a degenerate parabolic system coupled via nonlinear boundary flux. Zbl 1172.35007
Zhou, Jun; Mu, Chunlai
2008
Global existence and blow-up for a non-Newton polytropic filtration system with nonlocal source. Zbl 1178.35203
Zhou, Jun; Mu, Chunlai
2008
Incomplete quenching of heat equations with absorption. Zbl 1165.35394
Zhou, Jun; He, Yong; Mu, Chunlai
2008
Global existence and blow-up to a degenerate reaction-diffusion system with nonlinear memory. Zbl 1154.35391
Zhou, Jun; Mu, Chunlai; Fan, Mingshu
2008
Uniform blow-up profiles and boundary layer for a parabolic system with localized sources. Zbl 1146.35377
Zhou, Jun; Mu, Chunlai
2008
Blow-up for a non-Newton polytropic filtration system with nonlinear nonlocal source. Zbl 1168.35379
Zhou, Jun; Mu, Chunlai
2008
Blow-up and global existence to a degenerate reaction-diffusion equation with nonlinear memory. Zbl 1139.35068
Zhou, Jun; Mu, Chunlai; Lu, Feng
2007
Algebraic criteria for global existence or blow-up for a boundary coupled system of nonlinear diffusion equations. Zbl 1132.35404
Zhou, Jun; Mu, Chunlai
2007
Non-simultaneous blow-up for a semilinear parabolic system with nonlinear memory. Zbl 1128.35356
Zhou, Jun
2007
Blow-up rate for a porous medium equation with convection. Zbl 1144.35424
Zhou, Jun; Mu, Chunlai
2007
...and 2 more Documents
all top 5
### Cited by 497 Authors
63 Zhou, Jun 34 Mu, Chunlai 14 Xu, Guangyu 12 Ding, Hang 11 Nguyen Huy Tuan 9 Han, Yuzhu 8 Liu, Dengming 8 Wei, Junjie 7 Mi, Yongsheng 6 Chen, Shanshan 6 Liu, Chein-Shan 5 Guo, Shangjiang 5 Li, Haixia 5 Liu, Bingchen 4 Boudjeriou, Tahir 4 Fang, Zhongbo 4 Gao, Wenjie 4 Thach, Tran Ngoc 4 Wang, Xiaoli 4 Wang, Yifu 4 Wu, Jianhua 4 Xu, Runzhang 4 Yang, Wenbin 4 Yao, Xiaobin 4 Zhang, Guohong 4 Zheng, Pan 4 Zhou, Shouming 3 Au, Vo Van 3 Balachandran, Krishnan 3 Bie, Qunyi 3 Chen, Botao 3 Chen, Mengxin 3 Deng, Xiumei 3 Di, Huafei 3 Jia, Yunfeng 3 Li, Yanling 3 Liu, Lishan 3 Liu, Wenjun 3 Ngoc, Tran Bao 3 Qi, Yuanwei 3 Shao, Xiangkun 3 Shi, Junping 3 Sun, Fenglong 3 Wang, Jinfeng 3 Wang, Mingxin 3 Wang, Qiru 3 Wang, Xingchang 3 Wu, Ranchao 3 Wu, Yonghong 3 Xu, XiangHui 3 Yin, Jingxue 3 Yuan, Hailong 3 Zeng, Rong 2 Ahmad, Bashir 2 Al-saedi, Ahmed Eid Salem 2 Bai, Yuzhen 2 Can, Nguyen Huu 2 Cao, Yang 2 Caraballo Garrido, Tomás 2 Chen, Xinfu 2 Chu, Ying 2 Chuong, Quach Van 2 Dong, Mengzhen 2 Dong, Zhihua 2 Fang, Zhong 2 Fu, Xiaoxue 2 Guo, Bin 2 Hammouch, Zakia 2 Han, Jiangbo 2 Jiang, Hongling 2 Kim, Chan-Gyun 2 Kumari, Nitu 2 Le Xuan Truong 2 Li, Chenglin 2 Li, Fengjie 2 Li, Shanbing 2 Li, Shangzhi 2 Li, Zhongping 2 Liao, Menglan 2 Liu, Gongwei 2 Liu, Hongxia 2 Liu, Meng 2 Ma, Qiaozhen 2 Min, Na 2 Misra, Om Prakash 2 Mohan, Nishith 2 Nhan, Le Cong 2 O’Regan, Donal 2 Peng, Congming 2 Piṣkin, Erhan 2 Polat, Mustafa 2 Shang, Yadong 2 Tang, Guoji 2 Tri, Vo Viet 2 Wang, Jian 2 Wang, Weiming 2 Wang, Xiongrui 2 Wang, Yulan 2 Wang, Yuxia 2 Wei, Xin ...and 397 more Authors
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### Cited in 97 Serials
25 Journal of Mathematical Analysis and Applications 22 Nonlinear Analysis. Real World Applications 15 Computers & Mathematics with Applications 14 Applicable Analysis 14 Boundary Value Problems 13 Applied Mathematics and Computation 13 International Journal of Bifurcation and Chaos in Applied Sciences and Engineering 13 Discrete and Continuous Dynamical Systems. Series B 11 Applied Mathematics Letters 10 Discrete and Continuous Dynamical Systems. Series S 9 ZAMP. Zeitschrift für angewandte Mathematik und Physik 9 Journal of Differential Equations 9 Communications on Pure and Applied Analysis 8 Nonlinear Dynamics 7 International Journal of Biomathematics 7 Electronic Research Archive 6 Journal of Dynamical and Control Systems 5 Mathematical Methods in the Applied Sciences 5 Applied Mathematics and Optimization 5 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 5 Acta Applicandae Mathematicae 5 Discrete and Continuous Dynamical Systems 5 Taiwanese Journal of Mathematics 5 Advances in Difference Equations 4 Chaos, Solitons and Fractals 4 Electronic Journal of Differential Equations (EJDE) 4 Journal of Mathematical Chemistry 4 Journal of Inequalities and Applications 4 Nonlinear Analysis. Modelling and Control 4 Journal of Applied Mathematics and Computing 4 Advances in Nonlinear Analysis 4 Journal of Function Spaces 3 Journal of Mathematical Physics 3 Nonlinearity 3 Rocky Mountain Journal of Mathematics 3 Annales Polonici Mathematici 3 Glasgow Mathematical Journal 3 Journal of Dynamics and Differential Equations 3 Journal of Nonlinear Science 3 Turkish Journal of Mathematics 3 Abstract and Applied Analysis 3 Discrete Dynamics in Nature and Society 3 Acta Mathematica Scientia. Series A. (Chinese Edition) 3 Bulletin of the Malaysian Mathematical Sciences Society. Second Series 3 Mediterranean Journal of Mathematics 3 Open Mathematics 2 Bulletin of the Iranian Mathematical Society 2 International Journal of Computer Mathematics 2 Filomat 2 NoDEA. Nonlinear Differential Equations and Applications 2 Communications in Nonlinear Science and Numerical Simulation 2 Qualitative Theory of Dynamical Systems 2 Mathematical Biosciences and Engineering 2 Journal of Biological Dynamics 2 Advances in Mathematical Physics 2 Journal of Applied Analysis and Computation 2 Evolution Equations and Control Theory 2 AIMS Mathematics 2 SN Partial Differential Equations and Applications 1 Journal of Mathematical Biology 1 Lithuanian Mathematical Journal 1 Mathematical Biosciences 1 Mathematical Notes 1 Annali di Matematica Pura ed Applicata. Serie Quarta 1 Proceedings of the American Mathematical Society 1 Bulletin of the Korean Mathematical Society 1 Applied Mathematics and Mechanics. (English Edition) 1 Chinese Annals of Mathematics. Series B 1 Stochastic Analysis and Applications 1 Applied Numerical Mathematics 1 Applied Mathematical Modelling 1 Journal of Partial Differential Equations 1 Applied Mathematics. Series B (English Edition) 1 Bulletin of the Belgian Mathematical Society - Simon Stevin 1 Opuscula Mathematica 1 Journal of Applied Analysis 1 European Journal of Control 1 Chaos 1 Acta Mathematica Sinica. English Series 1 The ANZIAM Journal 1 Stochastics and Dynamics 1 Analysis and Applications (Singapore) 1 Journal of Function Spaces and Applications 1 Cubo 1 Complex Variables and Elliptic Equations 1 Applications and Applied Mathematics 1 Journal of Fixed Point Theory and Applications 1 Frontiers of Mathematics in China 1 Journal of Nonlinear Science and Applications 1 Science China. Mathematics 1 International Journal of Numerical Methods and Applications 1 Analysis and Mathematical Physics 1 Journal of Applied Nonlinear Dynamics 1 Journal of Elliptic and Parabolic Equations 1 Philosophical Transactions of the Royal Society of London. A. Mathematical, Physical and Engineering Sciences 1 Results in Applied Mathematics 1 Advanced Studies: Euro-Tbilisi Mathematical Journal
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### Cited in 23 Fields
328 Partial differential equations (35-XX) 131 Biology and other natural sciences (92-XX) 34 Ordinary differential equations (34-XX) 25 Dynamical systems and ergodic theory (37-XX) 14 Fluid mechanics (76-XX) 12 Mechanics of deformable solids (74-XX) 9 Operator theory (47-XX) 9 Probability theory and stochastic processes (60-XX) 7 Real functions (26-XX) 7 Numerical analysis (65-XX) 6 Integral equations (45-XX) 5 Systems theory; control (93-XX) 2 Difference and functional equations (39-XX) 2 Statistical mechanics, structure of matter (82-XX) 2 Information and communication theory, circuits (94-XX) 1 Special functions (33-XX) 1 Integral transforms, operational calculus (44-XX) 1 Functional analysis (46-XX) 1 Calculus of variations and optimal control; optimization (49-XX) 1 Mechanics of particles and systems (70-XX) 1 Optics, electromagnetic theory (78-XX) 1 Classical thermodynamics, heat transfer (80-XX) 1 Operations research, mathematical programming (90-XX)
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100, rue des maths 38610 Gières / GPS : 45.193055, 5.772076 / Directeur : Thierry Gallay
# An algebraic study of unitary one dimensional quantum cellular automata.
Mardi, 24 Janvier, 2006 - 15:00
Prénom de l'orateur :
Pablo
Nom de l'orateur :
ARRIGHI
Résumé :
One dimensional quantum cellular automata (1QCA) consist of a row of
identical, finite dimensional, quantum systems. These evolve in
discrete time steps according to a global evolution G -- which itself
arises from the application of a local transition function delta,
homogeneously and synchronously across space. But in order to grant
them the status of physically acceptable models, one must ensure that
the global evolution G is physically acceptable in a quantum
theoretical setting, i.e. one must ensure that $Delta$ is unitary.
Unfortunately this global property is non-trivially related to the
description of the local transition function $delta$ -- witness of
this the abundant literature on reversible cellular automata (RCA). We
provide algebraic characterizations of unitary one dimensional quantum
cellular automata. We do so both by algebraizing existing decision
procedures, and by adding constraints into the model which do not
change the quantum cellular automata's computational power. The
configurations we consider have finite but unbounded size.
Institution de l'orateur :
Leibniz-IMAG
Thème de recherche :
Physique mathématique
Salle :
1 tour Irma
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# Recurring Decimal Expansion
For any natural number $$n>1$$, we write the infinite decimal expansion of $$\frac 1n$$ (for example, $$\frac 14$$ is written as $$0.24999$$... instead of $$0.25$$). We need to determine the length of the non-periodic part of the infinite decimal expansion of $$\frac 1n$$.
I tried many methods, a somewhat promising one was to assume $$\frac 1n$$ to be some $$0.abbbbb$$..., where ‘$$a$$’ denotes the non-recurring part which has $$r$$ digits including zero, while ‘$$b$$’ is the recurring part. But I get stuck at deciding the lower and upper bounds for $$r$$. Please help.
(Please note: this is my first post on this website. So if I have to improve the way I should post the question in, please let me know how to correct the errors in my post. Thanks.)
• Leading question: can you prove that if $n$ is divisible by neither $2$ nor $5$, then the decimal fraction is immediately periodic (that is, the length of the non-periodic part is $0$)? (By the way, the standard name for that part is the "pre-periodic" part.) – Greg Martin Dec 31 '18 at 18:51
• Take the examples of $\frac{1}{3}=0.3333333...$, f $\frac{1}{7}=0.142857142857...$, and $\frac{1}{11}=0.09090909090909...$. If $n$ is not divisible by $2$ or $5$, the pre-periodic length will always be zero. – poetasis Dec 31 '18 at 19:33
• @GregMartin, It actually seemed intuitive for me, but I’m not able to come up with a rigorous proof for that. (Well, actually, every step of the solution seems very intuitive, but I don’t know how to write a rigorous solution by giving proofs for them :/ ) – Yellow Jan 1 '19 at 3:50
• By the way, is there any way I can rigorously prove that the the length of the pre-periodic part is related to powers of $2$ and $5$? Because, if powers of any other prime are not going to affect the length of the pre- periodic part, then powers of $2$ or $5$ might have some relation with its length, right? – Yellow Jan 1 '19 at 15:30
• Yes, the length of the pre-periodic part is definitely going to be determined by the powers of $2$ and $5$! Do you know modular arithmetic? Do you know what the "order of $a$ modulo $n$" is? Because knowing that the period of $1/n$ is actually equal to the order of $10$ modulo $n$ (when $n$ is not divisible by $2$ or $5$) makes the periodicity easier to prove. – Greg Martin Jan 1 '19 at 20:01
Lemma:
For every number $$n\in N$$ that is not divisible by 2 and 5, there exists $$k\in N$$ such that $$n\mid10^k-1$$
Proof: Suppose that the statement is not true, i.e. $$10^k-1\not\equiv0 \pmod n$$ for all values of $$k$$. There are infinitely many values of $$k$$ and just $$n-1$$ possible values ($$1\dots n-1$$) for $$10^k-1\pmod n$$. So by pidgeon hole principle there are two different values $$k_1, k_2$$ such that:
$$10^{k_1}-1\equiv10^{k_2}-1\pmod n,\quad (k_1>k_2)$$
This simply means that:
$$10^{k_1}-10^{k_2}\equiv0\pmod n$$
$$10^{k_2}(10^{k1-k2}-1)\equiv0\pmod n$$
Number $$n$$ has no factors 2 and 5 so obviously $$n\nmid 10^{k_2}$$ which implies that $$n\mid(10^{k_1-k_2}-1)$$ or:
$$n\mid10^k-1$$
...where $$k=k_1-k_2$$.
End of lemma proof.
Part 1:
Let us now show that:
For every number $$n$$ such that $$2\nmid n$$ and $$5\nmid n$$, decimal representation of $$1/n$$ has no pre-periodic part. In other words, $$1/n$$ can be written as: $$\frac 1n=0.aaa\dots=0.\bar{a}\tag{1}$$
...where $$a$$ stands for a group of repeating digits (possibly starting with zero) of length $$l_a$$. For example for $$n=7$$: $$1/7=0.\overline{142857}$$, so $$a=142857$$ and $$l_a=6$$.
One can easily show that (1) can be rewritten in the following way:
$${\frac1n}=\frac{a}{10^{l_a}-1}$$
$$a=\frac{10^{l_a}-1}{n}$$
According to our lemma, it's guaranted that there exists $$l_a$$ such that $$n\mid 10^{l_a}-1$$ so it's is possible to find $$a$$ for every $$n$$ such that $$1/n=0.\bar{a}$$, without a pre-periodic part.
Part 2
If $$2\mid n$$ or $$5\mid n$$, decimal representation of $$1/n$$ has a pre-periodic part: $$\frac1n=0.b\overline {a}\tag{2}$$
...with the lenght of pre-periodic group of digits $$b$$ equal to $$l_b$$ and length of periodic group of digits $$a$$ equal to $$l_a$$.
Suppose the pposite, that there is some number $$n$$ divisible by either 2 or 5 such that:
$$\frac1n=0.\bar{a}=\frac{a}{10^{l_a}-1}$$
$$na=10^{l_a}-1$$
...which is impossible because the LHS is divisible by 2 or 5 and the RHS is clearly not.
Based on part 1 and 2 we now know that:
Decimal representation of $$1/n$$ has pre-periodic part if and only if $$2\mid n$$ or $$5\mid n$$.
Part 3
For a number $$n$$ of the form $$n=2^p5^qm$$ and $$2,5\nmid m$$ the length of pre-periodic part is exactly $$\max(p,q)$$.
It can be easily proved that any number of the form $$0.b\bar{a}$$ can be written as:
$$0.b\bar{a}=\frac{b}{10^{l_b}}+\frac{a}{10^{l_b}(10^{l_a}-1)}\tag{3}$$
Because $$m$$ is not divisible by 2 or 5, we can write $$1/m$$ as:
$$\frac1m=\frac{a}{10^{l_a}-1}$$
which means that:
$$\frac1n=\frac1{2^p5^q} \cdot \frac1m$$
If we introduce:
$$r=\max(p,q)$$
we get:
$$\frac1n=\frac{2^{r-p}5^{r-q}}{10^r} \cdot \frac1m=\frac{2^{r-p}5^{r-q}a}{10^r(10^{l_a}-1)}\tag{4}$$
Now look at (4) carefully.
Case 1:
$$2^{r-p}5^{r-q}a<10^{l_a}-1$$
By comparing (3) and (4), the length of pre-periodic part is $$r$$, and the pre-periodic part is made of zeroes $$b=0$$. Periodic part is equal to $$2^{r-p}5^{r-q}a$$ and the length of the periodic part is $$l_a$$.
Case 2:
$$2^{r-p}5^{r-q}a>10^{l_a}-1$$
In that case you can write:
$$2^{r-p}5^{r-q}a=s(10^{l_a}-1)+a_1$$
...and (4) becomes:
$$\frac1n=\frac{s(10^{l_a}-1)+a_1}{10^r(10^{l_a}-1)}=\frac{s}{10^r}+\frac{a_1}{10^r(10^{l_a}-1)}$$
By comparing the last expression with (4), the length of the pre-periodic part $$s$$ is again $$r$$ and the length of repeating sequence $$a_1$$ is again $$l_a$$.
Conclusion
1. The length of the periodic part in the decimal representation of $$1/n$$ is determined by the length of periodic part in $$1/m$$ with $$m$$ being the greatest divisor of $$n$$ such that $$2\nmid m$$ and $$5\nmid m$$.
2. Pre-periodic part exists only if $$n$$ is of the form $$2^p5^qm$$.
3. The length of the pre-periodic part is $$\max(p,q)$$
Interesting example
$$\frac{1}{19}=0.\overline{052631578947368421}$$
Periodic part has 18 digits. Now take a look at:
$$\frac{1}{760}=\frac{1}{2^3\cdot5\cdot19}=0.001\overline{315789473684210526}$$
Pre-periodic part has length 3 (because the biggest power of 2 or 5 in $$n=760$$ is 3). And the periodic part has length 18, same length as in $$1/19$$.
• I have a question: Are part $1$ and $2$ really necessary? Because stating that the length of the non-periodic part of $1/n$ if $n = 2^m5^n p$ is $max(m,n)$ also takes care of the situations where $n$ is not at all divisible by $2$ and $5$, in which case $max(m,n)=0$ and so no non-periodic part. – Yellow Jan 7 '19 at 18:52
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Measurement of charged particle spectra in minimum-bias events from proton–proton collisions at $\sqrt s$ = 13 TeV
CMS Collaboration; Canelli, Florencia; Kilminster, Benjamin; Aarestad, Thea; Brzhechko, Danyyl; Caminada, Lea; De Cosa, Annapaoloa; Del Burgo, Riccardo; Donato, Silvio; Galloni, Camilla; Hreus, Tomas; Leontsinis, Stefanos; Mikuni, Vinicius Massami; Neutelings, Izaak; Rauco, Giorgia; Robmann, Peter; Salerno, Daniel; Schweiger, Korbinian; Seitz, Claudia; Takahashi, Yuta; Wertz, Sebastien; Zucchetta, Alberto; et al (2018). Measurement of charged particle spectra in minimum-bias events from proton–proton collisions at $\sqrt s$ = 13 TeV. European Physical Journal B: Condensed Matter and Complex Systems, C78(9):697.
Abstract
Pseudorapidity, transverse momentum, and multiplicity distributions are measured in the pseudorapidity range $|\eta | < 2.4$ for charged particles with transverse momenta satisfying $p_{\mathrm {T}} > 0.5\,\text {GeV}$ in proton–proton collisions at a center-of-mass energy of $\sqrt{s} = 13\,\text {TeV}$. Measurements are presented in three different event categories. The most inclusive of the categories corresponds to an inelastic $\mathrm {p} \mathrm {p}$ data set, while the other two categories are exclusive subsets of the inelastic sample that are either enhanced or depleted in single diffractive dissociation events. The measurements are compared to predictions from Monte Carlo event generators used to describe high-energy hadronic interactions in collider and cosmic-ray physics.
Abstract
Pseudorapidity, transverse momentum, and multiplicity distributions are measured in the pseudorapidity range $|\eta | < 2.4$ for charged particles with transverse momenta satisfying $p_{\mathrm {T}} > 0.5\,\text {GeV}$ in proton–proton collisions at a center-of-mass energy of $\sqrt{s} = 13\,\text {TeV}$. Measurements are presented in three different event categories. The most inclusive of the categories corresponds to an inelastic $\mathrm {p} \mathrm {p}$ data set, while the other two categories are exclusive subsets of the inelastic sample that are either enhanced or depleted in single diffractive dissociation events. The measurements are compared to predictions from Monte Carlo event generators used to describe high-energy hadronic interactions in collider and cosmic-ray physics.
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## integral priors for binomial regression
Diego Salmerón and Juan Antonio Cano from Murcia, Spain (check the movie linked to the above photograph!), kindly included me in their recent integral prior paper, even though I mainly provided (constructive) criticism. The paper has just been arXived.
A few years ago (2008 to be precise), we wrote together an integral prior paper, published in TEST, where we exploited the implicit equation defining those priors (Pérez and Berger, 2002), to construct a Markov chain providing simulations from both integral priors. This time, we consider the case of a binomial regression model and the problem of variable selection. The integral equations are similarly defined and a Markov chain can again be used to simulate from the integral priors. However, the difficulty therein follows from the regression structure, which makes selecting training datasets more elaborate, and whose posterior is not standard. Most fortunately, because the training dataset is exactly the right dimension, a re-parameterisation allows for a simulation of Bernoulli probabilities, provided a Jeffreys prior is used on those. (This obviously makes the “prior” dependent on the selected training dataset, but it should not overly impact the resulting inference.)
### 10 Responses to “integral priors for binomial regression”
1. Dan Simpson Says:
A better way of asking the question (I always [?] get it in the end) is as follows.
How far from a “bad” prior (aka a prior that gives bad results) are the integral priors?
And, to answer my own question, I think that they’re quite far away, in the sense that you’re solving a well-posed problem (an integral equation of the second kind) to get the prior, so a “nearby” prior should be the solution of a “nearby” integral equation.
• Uh?! A prior that gives “bad” results?! Whazat?!
We define “objective” [testing] priors as a result of an information minimisation goal. The principle was laid by Pérez & Berger (2002) and we follow it in this less manageable setting of binomial regression.. I kind of like it for the reason that it allows for the ‘improper prior sin’ in testing, offering a way out or rather a way in for improper priors. The implementation issue is not part of this question.
Now, I agree with you [?] that we could have conducted experiments where we knew the “truth” and had the possibility of finding the error rate of a model selection principle based on integral priors. A nice proposal for a summer project.
2. Dan Simpson Says:
Could you also expand upon the procedure for generating training data? I’m clearly missing something. But in my mind training data begets pi^N, but step 2 requires pi^N, so I fail to see how to avoid the circular definition.
A different thing: drawing linearly independent columns isn’t, to my knowlege, trivial, especially in the big data context. Isn’t that part of why g-priors exist? (The X^T X but deals with the approximate colinearity) is there a similar trick here? I imagine drawing independent but almost colinear columns would be a bad thing…
• Training data: this is a sample with the smallest possible size so that the posterior $\pi_i(\theta_i|z_i)$ is proper. The posterior associated with the reference prior $\pi_i^N$ associated with model i. Which usually is improper. If you look at the four steps on page 5, each sounds clear enough to me. Mind that the reference prior $\pi_i^N$ is an objective Bayes prior associated with the model $M_i$ per se, not the reference prior we are seeking. Maybe this explains for your confusion…
• linear independence and near colinearity:I had not though of this problem indeed. In the paper, we pick the column indices at random. This is, I think, related with the overall debate as to whether or not we should condition on X (as opposed to modelling X as well). I am of the “condition on X” school.
• Dan Simpson Says:
I think I’m of the “condition on X until it doesn’t work and then panic” school….
3. Dan Simpson Says:
I wonder if those integral equation could be solved numerically (at least for the dimensions of theta considered here (if I counted right that’s 5 and 12). I suspect it would be faster, cheaper, and more accurate than MCMC (although that’s not much of a bar to clear in moderate dimensions… ). They seem to just be second kind integral equations…
Then I’d probably stick the resulting approximate priors into INLA (but that’s personal preference :p)
Did you look at how the MC error upsets the balance ? (i.e. are the priors still neutral ) Because 10k chains will (if you’re lucky) give you 1 significant figure (maaaaybe 2).
(NB – I’ve only read the start and the end… Apologies if this was addressed in the middle (pp 5-10)… I’m getting to it presently)
A more general question: is this the sort of things scientists want? As opposed to designing objective priors on the whole of 2^X (X=set of covariates) and then leaping around the model space with gay abandon? Or is it more common/practical/useful to test given groupwise in/out hypotheses?
• Interesting suggestion about the numerical or even analytical resolution: I would think the answer is highly dependent on the model, although it may be that Beta/Bernoulli models can be handled rather easily… I am less sure I get the remark about MC(MC) error vs. numerical approximation error: it seems to apply every time MCMC is used!
• Dan Simpson Says:
It’s a little different in this case. It is (moderately well) understood what MCMC does for posterior inference, but here you’re propagating this through another layer of inference machinery. So I guess it’s worth thinking about.
It’s probably just an awkward way of framing a “prior sensitivity” question. The prior that you’re actually using is a perturbation of a theoretically motivated prior, so it’s worth checking how good/bad that is.
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# Physics307L:People/Meyers/Electron Diffraction Lab Summary
Steve Koch 21:06, 21 December 2010 (EST):Nice data and presentation.
## Purpose
Low Voltage Correction Device
The purpose of this lab is to become more familiar with the DeBroglie relationship. We calculated the wavelengths of the primary and secondary scattering of a beam of electrons. We showed that the length of the diameters is related to the voltage the electrons are accelerated through. We also compared the wavelengths we calculated with the accepted values.
## Procedure
High Voltage and Heater Supply
Along with the Lab Manual the procedure for this experiment is primarily just adjusting the high voltage and measuring the associated diameters of the projected diffraction pattern. We used a micrometer to measure the diameters of the rings of the diffraction pattern. The wiring diagram for the set up is shown Here.
## Data
The Diffraction Bulb
This is the Raw data from which we calculated the wavelengths:
After correcting the diameters for going from a sphere to a plane by these equations:
$y=R-\sqrt{R^2-\frac{D_{observed}^2}{4}}\,\!$
$tan(\theta)=\frac\frac{D_{observed}}{2}{L-y}\,\!$
$D_{corrected}=2Ltan(\theta)\,\!$
We get this graph of the inverse square root of the Voltage versus the Diameters.
SJK 21:05, 21 December 2010 (EST)
21:05, 21 December 2010 (EST)
This is an excellent way of plotting the data. Wondering what accounts for the large spread between runs...
Using the experimentally found slopes and h being Plank's constant, e being the elementary charge, m being the mass of an electron and L and R being defined in the manual as 13cm and 66mm respectively. we can use the following equation:
$d=\frac\frac{2hL}{\sqrt{2me}}{slope}\,\!$
to calculate d as:
$d1=2.59(5)*10^{-10}m=0.259(5)nm\,\!$
$d2=1.59(3)*10^{-10}m=0.159(3)nm\,\!$
## Error
The error I calculated using STDEV in EXCEL is 2.272 and 3.696 for the inner and outer ring respectively. From there I put these into the above equations to get the reported diameters. To calculate the percent error see below:
$%=\frac{0.259-0.213}{0.213}*100=21.59%\,\!$
$%=\frac{0.159-0.123}{0.123}*100=29.27%\,\!$
For how unsteady we were at taking the data with the calipers I think to have 30% error is fair.
## Conclusion
We showed that the wavelength follow the DeBroglie relation. We also found the wavelengths close to the accepted values. The percent errors of 21 and 29 percent are a bit troublesome but because the measuring style is primitive this is acceptable. An interesting way to get better measures would be to set up a stationary camera and take images to do a visual comparison in MATLAB. This would theoretically give better results. This was an interesting lab but the painstakingly boring measuring process takes some of the life out of the process.
## Thanks
1)To exce2wiki.net for the conversion of an excel doc to wiki code converter. It saved me for hours of menial data input.
2)To Kirstin from who I got the relations for the diameter correction.
3)To Nathan for being a great lab partner.
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## College Physics (4th Edition)
The maximum compression distance of the spring is $8.12~m$
We can use conservation of momentum to find the speed of the block just after the collision with the dart: $m_f~v_f= m_0~v_0$ $v_f= \frac{m_0~v_0}{m_f}$ $v_f= \frac{(0.122~kg)~(132~m/s)}{5.00~kg+0.122~kg}$ $v_f = 3.144~m/s$ We can use work and energy to find the maximum compression distance of the spring: $U_s+Work = KE_0$ $\frac{1}{2}kx^2 -(m_fg~\mu_k~x) = \frac{1}{2}m_f~v_f^2$ $kx^2 -(2m_fg~\mu_k~x) - m_f~v_f^2 = 0$ $(8.56~N/m)x^2 -(2)(5.122~kg)(9.80~m/s^2)(0.630)~x - (5.122~kg)(3.144~m/s)^2 = 0$ $(8.56~N/m)x^2 -(63.25~N)~x - (50.63~J) = 0$ We can use the quadratic formula to find $x$: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{63.25 \pm\sqrt{(-63.25)^2 - 4(8.56)(-50.63)}}{(2)(8.56)}$ $x = -0.729~m, 8.12~m$ Since $x$ must be positive, $x = 8.12~m$. The maximum compression distance of the spring is $8.12~m$
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# if the
Question:
If $4 \sin ^{-1} x+\cos ^{-1} x=\pi$, then what is the value of $x$ ?
Solution:
We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$\therefore 4 \sin ^{-1} x+\cos ^{-1} x=\pi$
$\Rightarrow 4 \sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} x=\pi \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$
$\Rightarrow 3 \sin ^{-1} x=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{6}$
$\Rightarrow x=\sin \frac{\pi}{6}$
$\Rightarrow x=\frac{1}{2}$
$\therefore x=\frac{1}{2}$
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# Scale-free networks
Post-publication activity
Curator: Albert-Laszlo Barabasi
A network that has a power-law degree distribution, regardless of any other structure, is called a scale-free network.
## Power-Law degree distribution
The degree of a node is the number of links adjacent to it. If we call the degree of a node $$k\ ,$$ a scale-free network is defined by a power-law degree distribution, which can be expressed mathematically as $$P(k)\sim k^{-\gamma}$$ From the form of the distribution it is clear that when:
• $$\gamma<2$$ the average degree diverges.
• $$\gamma<3$$ the standard deviation of the degree diverges.
It has been found that most scale-free networks have exponents between 2 and 3. Thus, they lack a characteristic degree or scale, and therefore their name. High degree nodes are called hubs.
## Scale-free network models
There are several models that are able to create a scale-free network. But most of them introduce in one way or another two main ingredients, growth and preferential attachment. By growth we mean that the number of nodes in the network increases in time. Preferential attachment refers to the fact that new nodes tend to connect to nodes with large degree. One can naively argue that for large networks this is nonsense because it requires a global knowledge of the network, i.e., knowing which are the high degree nodes, but this is not the case. There are several local mechanisms that introduce preferential attachment (see below).
In mathematical terms, preferential attachment means that the probability that a node with degree $$k_i$$ acquires a link goes as $$P(k_i)=\frac{k_i}{\sum_{i}k_i}$$
## The Barabasi-Albert model
The Barabasi-Albert model (a.k.a. BA model) introduced in 1998 explains the power-law degree distribution of networks by considering two main ingredients: growth and preferential attachment (Barabasi and Albert 1999). The algorithm used in the BA model goes as follows.
1. Growth: Starting with a small number ($$m_0$$) of connected nodes, at every time step, we add a new node with $$m(<m_0)$$ edges that link the new node to $$m$$ different nodes already present in the network.
2. Preferential attachment: When choosing the nodes to which the new node connects, we assume that the probability $$P$$ that a new node will be connected to node $$i$$ depends on the degree $$k_i$$ of node $$i\ ,$$ such that
$P\sim \frac{k_i}{\sum_{i}k_{i}}$
Numerical simulations and analytic results indicate that this network evolves into a scale-invariant state with the probability that a node has $$k$$ edges following a power law with an exponent $$\gamma=3\ .$$ The scaling exponent is independent of $$m\ ,$$ the only parameter in the model.
## Analytical solution for the BA model
This model can be solved analytically by setting up a differential equation in which the rate at which a node acquires links is equal to the number of links added times the probability of acquiring a link $\frac{dk_i}{dt}= m\frac{k_i}{\sum_{j}k_j}$
This equation can be simplified by realizing that at each time step $$m$$ links are added, thus $\sum_j k_j = 2mt$ and $\frac{dk_i}{dt}=\frac{k_i}{2t}$ $\ln{k_i}=\frac{1}{2}\ln{t} + C$
where $$C$$ is an integration constant that can be determined by using the fact that the $$i^{th}$$ node arrived to the network at time $$t^i$$ having degree $$m\ .$$ Thus $k_i = m(\frac{t}{t_i})^{1/2}\ .$ The degree distribution can be calculated by finding the probability that a node has a degree smaller than $$k$$ $P(k_i(t) > k) = P(t_i < \frac{mt}{k^2}) = 1 - P(t_i > \frac{mt}{k^2})$ Without loss of generality we can assume that nodes are added at a constant rate thus $P(t_i) =\frac{1}{m_0 + t}$ where $$m_0$$ is the total degree of the nodes that got the network started. Using this distribution $P(k_{i} - k) = 1 -\frac{mt}{k^2}\frac{1}{m_0 + t}$ Finally we get the degree distribution by differentiating and conclude that $\frac{d}{dk}P(k_i < k) = P(ki = k) = \frac{2m^2 t}{k^3}\frac{1}{m_0 + t}$
This mechanism was first introduced by Yule in the early 20th century to explain the distribution of different Taxa and was later generalized by Price in the 70’s and coined as cumulative advantage. The example shown here is not the most general version of the Price model that can be found in the original paper as well as in Newman (2005). In any case, the lesson that should be learned from this is that whenever we found a system in which the probability of increasing is proportional to the actual value, we should expect its distribution to follow a power-law.
## Local alternatives for preferential attachment
We could imagine that when nodes join the network they follow a link. For example, you move to a new town and an old friend tells you that you should visit a friend of him. If we consider that link to be a randomly chosen one, the probability $$\pi$$ that you were referred to a person with degree $$k$$ is $\pi(k)= k P(k)$ which is the precise definition of preferential attachment. This is because $$k$$ links end up in an agent of degree $$k\ .$$
### Duplication and divergence model
In a biological context the "duplication and divergence" model has been proposed as an explanation for the scale-free nature of protein-protein interaction networks. Let us consider a protein that interacts with $$k_p$$ other proteins. Eventually some of the genes that encode a protein get duplicated and now 2 copies become available. This redundancy allows one copy of the gene to mutate without changing the fitness of the organism. This process ends up with different proteins which are likely to share some interactions. If these duplication and divergence process occurs at random, proteins that have a high degree are more likely to have one of its neighbors change, and again, the probability of winning a neighbor through this process is proportional to the actual number of neighbors. This is another example of linear preferential attachment.
### Limited information
A different scenario that one can imagine is the one in which a node incorporates to the network with a limited or local information about it. If linear preferential attachment is used as the rule to create links in the local information context, a power-law degree distribution is also recovered, regardless of having limited information.
## Properties of scale-free networks
Scale-free networks have qualitatively different properties from strictly random, Erdos and Renyi, networks. These are:
• Scale-free networks are more robust against failure. By this we mean that the network is more likely to stay connected than a random network after the removal of randomly chosen nodes.
• Scale-free networks are more vulnerable against non-random attacks. This means that the network quickly disintegrates when nodes are removed according to their degree.
• Scale-free networks have short average path lengths. In fact the average path length goes as $$L\sim \log{N}/\log{\log{k}}$$
## Scale-free networks in nature
Scale-free networks have been observed in social, technological and biological systems. These include the citation and co-author scientific networks, the internet and world-wide web, and protein-protein interaction and gene regulatory networks (Albert and Barabasi 2002).
## References
• R. Albert, A.-L. Barabási (2002) Statistical mechanics of complex networks. Reviews of Modern Physics 74, 47-97.
• A.-L. Barabási and R. Albert (1999) Emergence of scaling in random networks. Science 286, 509-512
• M. E. J. Newman (2005) Contemporary Physics 46, 323-351.
Internal references
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I Does this go to 0 for large enough x?
1. Aug 3, 2016
ChrisVer
I have one question... In general I always thought that the exponential function was "dying" out faster than any other polynomial function, such that:
$e^{-x} x^a \rightarrow 0$ for $x \rightarrow \infty$.
[eg this is was used quiet commonly and so I got it as a rule-of-thumb, when deriving wavefunctions for a simple example for the Hydrogen atom]
However recently I read in a paper that this is not true, and as an illustration of how can that be, they logarithm-ized the function like:
$\ln (e^{-x} x^n) = -x + n \ln x$ which goes to infinity for $x,n\rightarrow \infty$.
http://arxiv.org/pdf/1108.4270v5.pdf
in Sec4 (the new paragraph after Eq4.1)
This has confused me, can someone shred some light?
2. Aug 3, 2016
micromass
$n$ is variable. In your example, $a$ is fixed.
3. Aug 3, 2016
ChrisVer
well even if $n$ was not a variable, then the quantity I wrote $-x + n \ln x$ (fixed n), is not really going to zero for large x...
oops sorry... you want the logarithm to go to infinity.
4. Aug 3, 2016
Stephen Tashi
which is on page 10 of the PDF.
However, there is a distinction between limits taken with respect to one variable and limits taken with respect to two variables.
$\lim_{s\rightarrow \infty} g(s)$ is defined differently than $\lim_{s\rightarrow\infty,\ n\rightarrow\infty} g(s,n)$.
(The wording in the paper is "when both $n$ and $s$ go to $\infty$".)
There is a further distinction between the definition of a "double limit" $\lim_{s\rightarrow\infty,\ n\rightarrow\infty} g(s,n)$ and the definition of the two "iterated limits":
$\lim_{s\rightarrow\infty}( \lim_{n\rightarrow\infty} g(s,n))$
and
$\lim_{n\rightarrow\infty}( \lim_{s\rightarrow\infty} g(s,n))$.
5. Aug 4, 2016
vanhees71
The first limit can be evaluated by repeated application of de L'Hospital's rule,
$$\lim_{x \rightarrow \infty} \frac{x^{a}}{\exp x}=\lim_{x \rightarrow \infty} \frac{a x^{a-1}}{\exp x}=\ldots = \lim_{x \rightarrow \infty} \frac{a(a-1) \ldots (a-n+1) x^{a-n}}{\exp x}=0,$$
where I made $n>a-1$.
6. Aug 4, 2016
Stephen Tashi
L'hospitals rule doesn't apply once the numerator becomes constant, so we should stop when that happens.
7. Aug 5, 2016
vanhees71
If $a \in \mathbb{N}$ then for $n=a$ the numerator becomes constant, and then the limit is also shown to be 0. So there's nothing wrong arguing with de L'Hospital in all cases.
8. Aug 5, 2016
Stephen Tashi
I'm just saying L'Hospital's rule does not apply to a case like $lim_{x\rightarrow\infty} \frac{6}{f(x)}$ since L'Hospital's rule only applies when both the numerator and denominator both have a limit of zero or both have infinite limits. Once you get a constant like 6 in the numerator, you have to use a different justification for finding the limit.
So we shouldn't write a chain of equalities that implies L'Hospital's rule is being applied to the case where the numerator is constant because it is not, in general, true that $lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = lim_{x\rightarrow\infty} \frac{f'(x)}{g'(x)} =lim_{x\rightarrow\infty} \frac{f"(x)}{g"(x)} = .... = lim_{x\rightarrow\infty} \frac{\frac{d^n}{dx} f(x)}{\frac{d^n}{dx}g(x)}$, running through an arbitary number $n$ of differentiations.
9. Aug 5, 2016
vanhees71
Of course it works only for limits of the type "$0/0$" or $\infty/\infty$. Perhaps I was not strict enough in my formulation. I did not mean to use it arbitrarily many times of course but only as long as it is applicable.
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# Mechanical Vibrations problem
1. Mar 2, 2009
### naggy
1. The problem statement, all variables and given/known data
A mass m is attached to a spring(massless) that is located inside a massless box. The box is falling under gravity. When the box starts to fall the spring is in it's equilibrium position and the box sticks to the ground when it hits it.
-The box is a distance H from the ground
-Spring has spring constant k
-The mass on the spring is m
Find the equation of motion (and initial conditions) when
a)the box is falling and
b)when the box has landed.
Variables
x is movement from equilibrium position of spring
y is distance from ground to mass
2. Relevant equations
$$L=KE - PE$$
or
$$F=m\ddot{x}$$
3. The attempt at a solution
I prefer using Lagrangian equations. When the box is falling:
$$KE= \frac{1}{2}m\dot{x^2}$$
$$PE= mgy +\frac{1}{2}kx^2$$
Now can I connect y(distance from the ground to m) and x(movement from equilibrium position of mass) with y=constant + x and use the Euler lagrange equations?
I'm also not sure on intial conditions, it would be x(0)=0 and x'(0)=0 for the first eq. of motion
when the box lands, maybe x(tH)=H and x'(tH)=sqrt(2gH) ??
File size:
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Views:
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2. Mar 3, 2009
### Imperitor
Just think what happens when the box hits the ground. It will stop but the mass on the spring will still have same velocity because nothing is stopping it. The only contribution of the fall on the system is an initial velocity. Hope that helps.
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# 3. Player Characters¶
In the previous lesson about rules and dice rolling we made some assumptions about the “Player Character” entity:
• It should store Abilities on itself as character.strength, character.constitution etc.
• It should have a .heal(amount) method.
So we have some guidelines of how it should look! A Character is a database entity with values that should be able to be changed over time. It makes sense to base it off Evennia’s DefaultCharacter Typeclass. The Character class is like a ‘character sheet’ in a tabletop RPG, it will hold everything relevant to that PC.
## 3.1. Inheritance structure¶
Player Characters (PCs) are not the only “living” things in our world. We also have NPCs (like shopkeepers and other friendlies) as well as monsters (mobs) that can attack us.
In code, there are a few ways we could structure this. If NPCs/monsters were just special cases of PCs, we could use a class inheritance like this:
from evennia import DefaultCharacter
# stuff
# more stuff
# more stuff
All code we put on the Character class would now be inherited to NPC and Mob automatically.
However, in Knave, NPCs and particularly monsters are not using the same rules as PCs - they are simplified to use a Hit-Die (HD) concept. So while still character-like, NPCs should be separate from PCs like this:
from evennia import DefaultCharacter
# stuff
# separate stuff
# more separate stuff
Nevertheless, there are some things that should be common for all ‘living things’:
• All can take damage.
• All can die.
• All can heal
• All can hold and lose coins
• All can loot their fallen foes.
• All can get looted when defeated.
We don’t want to code this separately for every class but we no longer have a common parent class to put it on. So instead we’ll use the concept of a mixin class:
from evennia import DefaultCharacter
class LivingMixin:
# stuff common for all living things
# stuff
# stuff
# more stuff
Above, the LivingMixin class cannot work on its own - it just ‘patches’ the other classes with some extra functionality all living things should be able to do. This is an example of multiple inheritance. It’s useful to know about, but one should not over-do multiple inheritance since it can also get confusing to follow the code.
## 3.2. Living mixin class¶
Create a new module mygame/evadventure/characters.py
Let’s get some useful common methods all living things should have in our game.
# in mygame/evadventure/characters.py
from .rules import dice
class LivingMixin:
# makes it easy for mobs to know to attack PCs
is_pc = False
def heal(self, hp):
"""
Heal hp amount of health, not allowing to exceed our max hp
"""
damage = self.hp_max - self.hp
healed = min(damage, hp)
self.hp += healed
self.msg(f"You heal for {healed} HP.")
def at_pay(self, amount):
"""When paying coins, make sure to never detract more than we have"""
amount = min(amount, self.coins)
self.coins -= amount
return amount
def at_damage(self, damage, attacker=None):
"""Called when attacked and taking damage."""
self.hp -= damage
def at_defeat(self):
"""Called when defeated. By default this means death."""
self.at_death()
def at_death(self):
"""Called when this thing dies."""
# this will mean different things for different living things
pass
def at_do_loot(self, looted):
"""Called when looting another entity"""
looted.at_looted(self)
def at_looted(self, looter):
"""Called when looted by another entity"""
# default to stealing some coins
max_steal = dice.roll("1d10")
stolen = self.at_pay(max_steal)
looter.coins += stolen
Most of these are empty since they will behave differently for characters and npcs. But having them in the mixin means we can expect these methods to be available for all living things.
## 3.3. Character class¶
We will now start making the basic Character class, based on what we need from Knave.
# in mygame/evadventure/characters.py
from evennia import DefaultCharacter, AttributeProperty
from .rules import dice
class LivingMixin:
# ...
"""
A character to use for EvAdventure.
"""
is_pc = True
strength = AttributeProperty(1)
dexterity = AttributeProperty(1)
constitution = AttributeProperty(1)
intelligence = AttributeProperty(1)
wisdom = AttributeProperty(1)
charisma = AttributeProperty(1)
hp = AttributeProperty(8)
hp_max = AttributeProperty(8)
level = AttributeProperty(1)
xp = AttributeProperty(0)
coins = AttributeProperty(0)
def at_defeat(self):
"""Characters roll on the death table"""
if self.location.allow_death:
# this allow rooms to have non-lethal battles
dice.roll_death(self)
else:
self.location.msg_contents(
"$You()$conj(collapse) in a heap, alive but beaten.",
from_obj=self)
self.heal(self.hp_max)
def at_death(self):
"""We rolled 'dead' on the death table."""
self.location.msg_contents(
### 3.3.2. Backtracking¶
We make our first use of the rules.dice roller to roll on the death table! As you may recall, in the previous lesson, we didn’t know just what to do when rolling ‘dead’ on this table. Now we know - we should be calling at_death on the character. So let’s add that where we had TODOs before:
# mygame/evadventure/rules.py
# ...
def roll_death(self, character):
ability_name = self.roll_random_table("1d8", death_table)
# kill the character!
character.at_death() # <------ TODO no more
else:
# ...
if current_ability < -10:
# kill the character!
character.at_death() # <------- TODO no more
else:
# ...
## 3.4. Connecting the Character with Evennia¶
You can easily make yourself an EvAdventureCharacter in-game by using the type command:
type self = evadventure.characters.EvAdventureCharacter
You can now do examine self to check your type updated.
If you want all new Characters to be of this type you need to tell Evennia about it. Evennia uses a global setting BASE_CHARACTER_TYPECLASS to know which typeclass to use when creating Characters (when logging in, for example). This defaults to typeclasses.characters.Character (that is, the Character class in mygame/typeclasses/characters.py).
There are thus two ways to weave your new Character class into Evennia:
1. Change mygame/server/conf/settings.py and add BASE_CHARACTER_CLASS = "evadventure.characters.EvAdventureCharacter".
2. Or, change typeclasses.characters.Character to inherit from EvAdventureCharacter.
You must always reload the server for changes like this to take effect.
Important
In this tutorial we are making all changes in a folder mygame/evadventure/. This means we can isolate our code but means we need to do some extra steps to tie the character (and other objects) into Evennia. For your own game it would be just fine to start editing mygame/typeclasses/characters.py directly instead.
## 3.5. Unit Testing¶
Create a new module mygame/evadventure/tests/test_characters.py
For testing, we just need to create a new EvAdventure character and check that calling the methods on it doesn’t error out.
# mygame/evadventure/tests/test_characters.py
from evennia.utils import create
from evennia.utils.test_resources import BaseEvenniaTest
class TestCharacters(BaseEvenniaTest):
def setUp(self):
super().setUp()
def test_heal(self):
self.character.hp = 0
self.character.hp_max = 8
self.character.heal(1)
self.assertEqual(self.character.hp, 1)
# make sure we can't heal more than max
self.character.heal(100)
self.assertEqual(self.character.hp, 8)
def test_at_pay(self):
self.character.coins = 100
result = self.character.at_pay(60)
self.assertEqual(result, 60)
self.assertEqual(self.character.coins, 40)
# can't get more coins than we have
result = self.character.at_pay(100)
self.assertEqual(result, 40)
self.assertEqual(self.character.coins, 0)
# tests for other methods ...
If you followed the previous lessons, these tests should look familiar. Consider adding tests for other methods as practice. Refer to previous lessons for details.
For running the tests you do:
evennia test --settings settings.py .evadventure.tests.test_character
## 3.6. About races and classes¶
Knave doesn’t have any D&D-style classes (like Thief, Fighter etc). It also does not bother with races (like dwarves, elves etc). This makes the tutorial shorter, but you may ask yourself how you’d add these functions.
In the framework we have sketched out for Knave, it would be simple - you’d add your race/class as an Attribute on your Character:
# mygame/evadventure/characters.py
from evennia import DefaultCharacter, AttributeProperty
# ...
# ...
charclass = AttributeProperty("Fighter")
charrace = AttributeProperty("Human")
We use charclass rather than class here, because class is a reserved Python keyword. Naming race as charrace thus matches in style.
We’d then need to expand our rules module (and later character generation to check and include what these classes mean.
## 3.7. Summary¶
With the EvAdventureCharacter class in place, we have a better understanding of how our PCs will look like under Knave.
For now, we only have bits and pieces and haven’t been testing this code in-game. But if you want you can swap yourself into EvAdventureCharacter right now. Log into your game and run the command
type self = evadventure.characters.EvAdventureCharacter
If all went well, ex self will now show your typeclass as being EvAdventureCharacter. Check out your strength with
py self.strength = 3
Important
When doing ex self you will not see all your Abilities listed yet. That’s because Attributes added with AttributeProperty are not available until they have been accessed at least once. So once you set (or look at) .strength above, strength will show in examine from then on.
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# Cannot reach Internet from instances [closed]
Hi Everyone, I am having issues in reaching the internet from any of the instacnes created on the open-stack dashboard.
I have created 2 instances with following IP's.
Instance1 - 10.0.0.2(Floating IP: 192.168.2.2)
Instance2- 10.0.0.3(Floating IP: 192.168.2.3)
Host IP: 192.168.2.50
I can ping the GW 192.168.2.1 or the floating IP's 192.168.2.2/3 from the instances but i cant ping the host IP or the internet from instances. Same applies from host side as i cant ping any of the floating IP's from the host.
I currently do not have any FW rules on host and on the instances i am allowing all traffic.
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Vectors
### Vectors
The concept of a vector is one of the most important in physics. Vectors represent all kinds of important quantities like velocity, acceleration, momentum and force.
A vector is a directed line segment. It is drawn as an arrow (right), and has only two important aspects, its length and its direction.
The length or magnitude of a vector represents the size of a physical quantity, like a force or speed (speed is the magnitude of velocity).
The direction of a vector (where the arrow points) is the direction of action of the physical quantity. For example, it might be the direction of an applied pushing force.
The only two features of a vector that are important are length (which captures the magnitude or size of the quantity) and direction. As long as length and direction are preserved, a vector can be moved anywhere in a coordinate system, purely for convenience.
### Vector translation
Above we noted that the only important things about a vectors are its length and direction. It doesn't matter where it is located on the plane (or in space). In fact, we are free to move vectors where ever we'd like, just for the sake of convenience, without changing their meaning.
Vectors A, B and C on the left could be force vectors, velocity vectors, acceleration vectors ... you name it. Often it's convenient to translate vectors to the origin.
Play the animation to translate all three vectors to the origin. None of the meaning of the vectors is altered in any way.
Vectors are of little use unless we can add them. Below are the two principal methods of vector addition.
#### Method 1: Tip to tail
The easiest way to add vectors is the tip-to-tail (or head-to-tail) method. Remember that the only two important things about vectors are length and direction. Therefore we can move any vector to any location in the plane as we like and, as long as we don't change the length or direction, it remains the same vector
Adding by the tip-to-tail method means to move one vector so that its tail lies on the tip of the first vector. The resultant vector, A+B - the sum of the two - is simply the new vector drawn from the origin of the first vector to the arrow of the second.
Run this animation to add vectors A and B to find the resultant vector A+B. The resultant vector (just a vector sum) is often labeled "R".
Any number of vectors can be added in this way by just chaining them together, arrow of the current vector to origin of the next, and drawing in the vector R.
#### Method 2: Parallelogram
The parallelogram method of vector addition is shown on below. Notice that it's the same thing as tip-to-tail, but in this case vector B is moved down so that the vectors are tail-to-tail or origin-to-origin.
The resultant vector is the diagonal of the parallelogram formed by two copies of each vector.
Notice that the head-to-tail method of vector addition is embedded within the parallelogram method (twice). Look for it.
#### Which is better?
Once in a while the parallelogram method is more convenient, but the head-to-tail method is usually the place to start. It's much more useful for adding more than two vectors, and it's the method we'll almost always use to program computers to do vector addition.
### Adding more than two vectors
Using the head-to-tail method to add more than two vectors is very easy, graphically or numerically (we'll look at numerical addition below). One thing that often confuses students is whether, during the addition process, vectors can cross. They can.
Below is an example of a four-vector addition done in head-to-tail fashion, in which vectors cross. It's fine, no problem.
Adding more than two vectors with the parallelogram method is more cumbersome. You'd have to do the first addition to come up with a resultant vector, then add that to the next vector to find the new resultant, and so on.
Another thing that the head to tail method shows is that vector addition is commutative. Look at the six ways to add three vectors in the panel below. All six yield, graphically, the same resultant vector R.
X
### Commutative property
The commutative property applies to both addition and multiplication. It says that order of pure addition or pure multiplication doesn't matter:
#### a · b = b · a
Notice that regardless of the order of addition of vectors, the resultant or sum vector (in magenta) is the same – vector addition is commutative.
### A thought experiment ...
Here's a thought experiment. Imagine a bowling ball suspended in the middle of a room by 1000 bungee cords, each attached somewhere on a wall, the ceiling or the floor. Each cord, of course, is exerting a force on the ball, and each can be represented by a vector - only in three dimensions instead of two - that's legal. The length of the vector is proportional to the strength of the force and the direction is the direction of the pull.
Now the ball isn't moving, so there can't be any net force on it, otherwise it would be accelerating.
The net force is the sum of all force vectors. That means that the vector sum of all 1000 vectors has to be zero. So if we join all of those vectors together, tip-to-tail on a 3-D grid, the tip of the last vector will touch the tail of the first - no matter what order of addition, and the resultant vector will have a length of zero. That's cool.
The four scenarios below might help you to visualize vector addition. In all cases, the airplane has a forward velocity vector of 150 Km/h (the plane is traveling – or trying to travel – at 150 Km/h in the forward direction). As we would expect, a tailwind adds to the overall velocity of an airplane and a headwind subtracts. You can experience this by flying to the west coast and back on a commercial plane.
The westbound trip can take 6 hours or more while the eastbound flight often takes substantially less time.
A cross wind (coming in at 90˚ to the direction of travel) shifts the course of the plane away from the wind. In the case of wind blowing at an odd angle to the forward velocity vector, use the law of cosines to solve for the resultant velocity vector.
### Resolving vectors into components
We know that we can add two vectors to get another. Very often (quite often, really) we need to find two convenient vectors that add to a vector of interest. This is called resolving a vector into components.
The example at right shows a vector drawn on a Cartesian plane. By drawing the dashed lines from the tip of the vector to the axes, at right angles, we come up with the vectors Vx and Vy, which sum to vector V.
As we'll see, resolving V into two vectors that lie along our coordinate axes will be a big help in solving some problems.
### Using trigonometry to resolve vectors
When the length of a vector and the angle it makes with either axis is known, we can use trigonometry to find the lengths fo the components along each axis.
Using either of the right triangles drawn by the dashed line (left), it's easy to see that
$$sin(\theta) = \frac{V_y}{V} \phantom{000} \text{and} \phantom{000} cos(\theta) = \frac{V_x}{V},$$
so we can derive the relationships shown.
It might be helpful at this point to review your trig. and the special triangles, the 45-45-90 triangle and the 30-60-90 triangle, for the angles of which we can find convenient exact solutions for the sine and cosine functions.
### Making the coordinate system work for you
There are many problems in physics and other fields where changing from one coordinate system to a more convenient one makes solving a problem simpler and more intuitive. One example is a mass on an inclined plane. The problem is shown in panel 1. The only force that makes the ball roll down the ramp is the force of gravity, Fg. The trouble is that Fg is at an odd angle to the ramp. But we can impose a different coordinate system upon the problem, a more convenient one in which the x-axis is classed along the ramp and the y-axis perpendicular to it (panel 2). In this case, we can resolve the Fg vector into its Fgx and Fgy components (panel 3).
Panels 4, 5 & 6 show ramps of increasing steepness. Notice that the component of Fg pointing down the ramp in each increases with the steepness. The larger force accounts for the fact that the ball will roll faster as the ramp steepens.
By the way, no matter what kind of crazy coordinate system you impose on this problem, the ball will still roll down the ramp like it always did. Nature couldn't give one whit about your coordinate system. Your choice of coordinate system is made to make your mathematical modeling of the situation easier, or even just possible.
### Practice problems
1. Sketch head-to-tail additions of the following vector pairs. Roll over or tap the images to see the solution.
1. A boat travels at 5 knots across a river with a current of 1 knot (a knot is one nautical mile per hour). If the intended direction of the boat is due north (the river runs east-west), find the actual course (in degrees from north) and the speed of the boat as it moves.
2. Use trigonometry to find the x- and y-components of these vectors (roll over for solutions):
1. The figure below shows an airplane flying due west (270˚ compass bearing — see the compass "rose" on the right). Find the actual speed and direction of the plane if it encounters a wind of (a) 30 mi./h from the north (0˚), (b) 30 mi./h from the southwest (225˚).
Solution
Part (a)
\begin{align} R &= \sqrt{165^2 + 30^2} \\ &= 167.7 \frac{mi}{h} \; \text{ (a little faster)} \\ \\ \theta &= tan^{-1} \left( \frac{130}{165} \right) = 10.3˚ \\ \\ \text{course } &= 270˚ - 10.3˚ = 259.7˚ \end{align}
Part (b) Begin with the law of cosines
\begin{align} R^2 &= 30^2 + 165^2 - 2(30)(165) cos(45˚) \\ &= 30^2 + 165^2 - 2(30)(165) \frac{\sqrt{2}}{2} \\ \\ &= 23,175 \\ \\ R &= \sqrt{23175} = 152.2 \; \frac{mi}{h} \end{align}
The plane is a little slower into a headwind. For the angle, we use the law of sines:
$$\frac{sin(\theta)}{30} = \frac{sin(45)}{152.2}$$
\begin{align} \theta &= sin^{-1}\left( \frac{30}{152.2} sin(45˚) \right) \\ \\ &= 8.01˚ \\ \\ \text{course } &= 270˚ + 8˚ = 278˚ \end{align}
### Special triangles can help
Often we work with angles that are special fractions of a circle. It's a good thing to memorize the dimensions of two special triangles with hypotenuses of lenght 1, the 30-60-90 triangle and the 45-45-90 triangle. I can't overemphasize how much knowing these comes in handy later in math and physics.
Doing math with vectors graphically really helps with learning, but in order to really do any serious computations with vectors, we need to learn how to manipulate them numerically.
We'll begin by noticing that there are two ways we can describe a vector on the plane:
• specify the beginning and end points of the vector
• translate the vector to the origin [where one endpoint is (0, 0)], then the vector is specified only be giving the endpoint.
Take the vectors A and B (black) in the figure below. It's easy to translate them to the origin by simply subtracting the coordinates of the beginning of the vector (the dot) from both coordinates. For example, to translate vector A to the origin, we subtract the coordinate (-4, -2) from each coordinate. Then we get
• start: (-4,-2) - (-4,-2) = ((-4+4), (-2+2)) = (0, 0)
• end: (-2,4) - (-4,-2) = ((-2+4),(4+2)) = (2, 6)
This result is shown in the magenta vector A. You should work your way through the same translation for the B vector.
We can think of vector translation another way, too. Suppose we specify the two ends of some vector A (below) with two vectors from the origin, v1 & v2. Now if we subtract v1 from v2 (or add the negative of v1 to v2) we will get vector A moved to the origin. Check out the graph below to see how it works. You should be familiar with both ways of translating vectors to the origin mathematically.
We can also flip vectors by 180˚ numerically. This just involves flipping the signs of all coordinates. For example, in the graph below, vector A is flipped around to -A by transforming its beginning coordinate (-1, 3) to (1, -3) and its end coordinate (6, 6) to (-6, -6).
These vectors can be translated to the orign graphically or numerically to show that they lie on the same line but point in opposite directions. They also have the same magnitude because the Pythagorean theorem, which we use to calculate vector lengths, depends only on the squares of the coordinates, therefore sign is unimportant.
Finally, we can add vectors numerically. To do so we simply add coordinates. For a 2-dimensional vector, that means
(x1, y1) + (x2, y2) = (x1+x2, y1+y2)
Vectors can be added and then translated to the origin, or translated to the origin first, then added.
Vector translation and addition of vectors are commutative (can be done in either order).
#### Vector translation
A vector with beginning at $(x_1, \, y_1)$ and end at $(x_2, \, y_2)$ can be translated to the origin by subtracting $(x_1, \, y_1)$ from each coordinate.
Vectors $\vec{A} = (a, \, b)$ and $\vec{B} = (c, \, d)$ are added by adding respective coordinates: $\vec{A} + \vec{B} = (a+c, \, b+d)$
Vector subtraction is just the same as adding the negative of a vector.
The negative of vector $(a, \, b)$ is $(-a, \, -b).$
### Example 1
Add vectors $\vec{A} = (-3, \, 3)$ and $\vec{B} = (1, \, 6).$ These vectors both originate from $(0, 0).$
Solution: Adding the vectors is straightforward:
$$\vec{A} + \vec{B} = (-3 + 1, \, 3 + 6) = (-2, \, 9)$$
All of these vectors originate from $(0, 0).$ The graphical result is shown. A parallelogram is drawn in to help you see the addition.
### Example 2
Add vector A with endpoints (1, 1) & (-2, 3) to vector B with endpoints (-2, -2) & (3, 4). Translate the result to the origin.
Solution: The coordinates of the endpoints of our sum vector are
start = (1, 1) + (-2, -2) = (-1, -1)
end = (-2, 3) + (3, 4) = (1, 7)
So the coordinates of the two ends of our vector are (-1, -1) and (1, 7). We can translate this vector to the origin by subracting (-1, -1) from each to get:
start = (-1, -1) - (-1, -1) = (0, 0)
end = (1, 7) - (-1, -1) = (2, 8)
Now we can show that we get the same result by first translating each vector to the origin and then adding the resulting vectors:
A = (-2, 3) - (1, 1) = (-3, 2), where the start coordinate just turns into the origin: (1, 1) - (1, 1) = (0, 0).
and
B = (3, 4) - (-2, -2) = (5, 6), where the start coordinate is again (-2, -2) - (-2, -2) = (0, 0)
The sum of our two translated vectors is
(-3, 2) + (5, 6) = (2, 8), just what we got on the first try. Easy peasy.
Below: vectors A & B add to the magenta vector.
In the graph below, A & B are translated to the origin. The sum vector is easier to visualize in that view. The original vectors are in green for comparison.
### Practice problems
Calculate the sum of the vectors and translate to the origin if necessary. Problems 1-3 give vectors from the origin (0, 0). Problems 4-6 include the start and endpoints (respectively) of the vectors.
1 (-2, -5) & (5, 4) Solution \begin{align} \bar{v}_1 + \bar{v}_2 &= (-2 + 5, -5 + 4) \\ &= \bf (3, -1) \end{align} 2 (2, -3) & (-2, -6) Solution \begin{align} \bar{v}_1 + \bar{v}_2 &= (2 + (-2), -3 + (-6)) \\ &= (2 - 2, -3 - 6) \\ &= \bf (0, -9) \end{align} 3 (7, 1) & (1, 5) Solution \begin{align} \bar{v}_1 + \bar{v}_2 &= (7 + 1, 1 + 5) \\ &= \bf (8, 6) \end{align}
4 (-2, -2) to (3, 4) and (-1, 7) to (2, 2) Solution First translate the vectors to the origin. We need to "zero-out" the origin of each vector, by adding (2, 2) to the start and end of the first, and (1, -7) to the start and end of the second: \begin{align} \bar{v}_1 &= (2 + 3, 2 + 4) = (5, 6) \\ \bar{v}_2 &= (2 + 1, 2 - 7) = (3, -5) \end{align} Now add the translated vectors: \begin{align} \bar{v}_1 + \bar{v}_2 &= (5 + 3, 6 - 5) \\ &= \bf (8, 1) \end{align} 5 (4, 5) to (-4, -4) and (-2, -4) to (2, 6) Solution First translate the vectors to the origin. We need to "zero-out" the origin of each vector, by adding (-4, -5) to the start and end of the first, and (2, 4) to the start and end of the second: \begin{align} \bar{v}_1 &= (-4 - 4, -4 - 5) = (-8, -9) \\ \bar{v}_2 &= (2 + 2, 6 + 4) = (4, 10) \end{align} Now add the translated vectors: \begin{align} \bar{v}_1 + \bar{v}_2 &= (-8 + 4, -9 + 10) \\ &= \bf (-4, 1) \end{align} 6 (7, -1) to (2, 5) and (4, 4) to (6, -1) Solution First translate the vectors to the origin. We need to "zero-out" the origin of each vector, by adding (-7, 1) to the start and end of the first, and (-2, -5) to the start and end of the second: \begin{align} \bar{v}_1 &= (2 - 7, 5 + 1) = (-5, 6) \\ \bar{v}_2 &= (6 - 4, -1 - 4) = (2, -5) \end{align} Now add the translated vectors: \begin{align} \bar{v}_1 + \bar{v}_2 &= (-5 + 2, 6 - 5) \\ &= \bf (-3, 1) \end{align}
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• Post:Stuart Morrison
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# Are 200 level classes harder than 100?
Looking for an answer to the question: Are 200 level classes harder than 100? On this page, we have gathered for you the most accurate and comprehensive information that will fully answer the question: Are 200 level classes harder than 100?
They may have either three digits or four digit classification like 100 or 1000 depending on the school. 100- 600 or 1000 -6000 level : Typically, this slab of 100 to 600 corresponds to Undergrad level classes.
There's a recurring debate on my campus about mandating a certain number of 200 level classes for a degree. Advocates tend to frame the argument around academic rigor. 200 level classes are more rigorous than 100, the argument goes, so we should require some 200 levels in every program to ensure that students are appropriately challenged.
Some 400-level classes include first-year graduate students who are preparing to take 500- and 600-level classes offered through graduate schools. 100- and 200-Level Course Expectations 100-level courses come with no prerequisites regarding knowledge of a disciplines concepts and terminology.
In my classes, the 100 level usually was the course that would be an intro type and the 200 levels were taken after. Not that they were "harder" rather they built on 100 levels. A 200 level course would, in theory, have the expectation that you had some background knowledge- or perhaps even a prereq from the same dept at the 100 level.
## What is the most failed subject in high school?
Algebra 1 (CBS4) – Algebra 1 is the most failed class in high schools across the country.
## What is the difference between a 100 level and a 200 level course?
Upper-Level. Lower-level courses are those at the 100-level and 200-level. Upper-level courses are those at the 300-level and 400-level. In addition, a 200-level course may be proposed to count as an upper-level course, particularly if it has a university-level prerequisite.
## What does a 200 level course mean?
200-level course designation Courses of intermediate college-level difficulty; courses with 100-level course(s) as prerequisite(s); or survey courses devoted to particular areas or fields within a discipline. Assumptions: 1. Students will have completed expository writing (ENG 102) or the equivalent; 2.
## Are 500 level classes hard?
Master-level graduate courses numbered 500-600 require a bachelor's degree and admission to a graduate program. 500 level course are more rigorous than undergraduate courses.
## Are 200 classes for sophomores?
200-299: Lower division courses of freshman and sophomore level. ... 500-599: Courses intended primarily for graduate students that may be taken by advanced undergraduate students for baccalaureate credit. Content requires significant independent thinking on the part of the student and offers opportunity for research.
## How much harder are college classes?
In summary, college classes are definitely harder than high school classes: the topics are more complicated, the learning is more fast-paced, and the expectations for self-teaching are much higher. HOWEVER, college classes are not necessarily harder to do well in.
## Are 200 level classes hard?
200 level - Very difficult, often weedout courses. These were usually the "foundation" courses for your field of study.
## What are 700 level courses?
700—900 or 7000—9000 level : This classes with this numbering correspond to Graduate level classes for MS, MBA or PhD. Masters classes are ideally in the range of 700 to 800. 900 level classes correspond to PhD and Thesis or research level classes and much advanced.
## What are 400 level classes?
400-level course designation Advanced upper-division courses, seminars, practicums, or internships for majors and upper- division students.
## Which subject is easy in 11th?
Physical Education. This is one of the easiest and scoring disciplines in different commerce subjects in Class 11. Physical Education is ideal for those interested in sports, yoga, physical fitness, physiology, etc.
## What is the easiest major in college?
CollegeVine's Top Easiest MajorsBusiness Administration. Average GPA: 3.2.Psychology. Average GPA: 3.3. ... Education. Average GPA: 3.6. ... Social Work. Average GPA: 3.4. ... Public Relations & Advertising. Average GPA: 3.0. ... Criminal Justice. Average GPA: 3.1. ... Journalism. Average GPA: 3.2. ... Economics. Average GPA: 3.0. ...
## What is level 200 in the university?
A 200 level course code indicates the course is expanding on introductory knowledge and skills. You may need to have completed a pre-requisite course to study a 200 level course. These courses are normally studied in your second year of full-time study.
## What are considered upper level courses?
Suggestions for the differentiation between lower and upper division courses are as follows: Lower-division courses comprise all 100-level courses and all 200-level courses. Upper-division courses comprise all 300- and 400-level courses.
## Which subject is most easy?
What are the 12 easiest A-Level subjects?Geography. ... Textiles. ... Film Studies. ... Sociology. ... Information Technology. ... Health and Social Care. ... Media Studies. With a pass rate of 100% in 2019, Media Studies is definitely one of the easier A-Levels. ... Law. A-Level Law is surprisingly easy, especially compared to degree-level Law.
## What are 400 level courses?
400-level course designation Advanced upper-division courses; and/or seminars, tutorials and honor courses for majors and upper-division students. Assumptions: 1. that students have completed a substantial amount of work on the 300 level, and, for seminars, tutorials and honor courses , 2.
## Are 200 level classes harder than 100? Expert Answers
none
none
### In college, how hard are 200 level courses compared to 100 ...
Assuming you are talking standard. 100 intros. 200 Specifics. 300 Majors/Minors (for the most part) 400+ Majors only (For the most part) It really depends on the class. 200s generally aren't that ...
### 100 and 200 Level Courses | Confessions of a …
Are 200-level courses more rigorous than 100 level? There's a recurring debate on my campus about mandating a certain number of 200 level classes for a degree. Advocates tend to frame the argument around academic rigor. 200 level classes are more rigorous than 100, the argument goes, so we should require some 200 levels in every program to ensure that students …
### Differences between levels of courses... 100, 200, etc ...
In my classes, the 100 level usually was the course that would be an intro type and the 200 levels were taken after. Not that they were "harder" rather they built on 100 levels. A 200 level course would, in theory, have the expectation that you had some background knowledge- or perhaps even a prereq from the same dept at the 100 level.
none
### What is really the difference between 100/200/300 etc ...
100 are generally intro level courses. The courses above those tend to expand on that information or teach about more specific aspect of it. 100 classes are often prerequisites for 200-level classes, which are often prereqs for 300-level classes. How much time you need to devote to them is going to vary a lot.
### "Difficulty" of 200/300/400 level undergrad classes? | GBCN
It will be harder than your lower level classes and will build on your 200 level classes. I feel you on the summer classes too. I had to take my 5 semester credit hour calc class over summer session so I could apply to get into the business school before my junior year (I had transferred and changed my major a few times).
### Courses Numbering System in US :100—900 Level? BS or MS ...
Some schools have more advanced classes that are around 500 and 600 level, they are also Undergrad level classes but more advanced. 700—900 or 7000—9000 level : This classes with this numbering correspond to Graduate level classes for MS, MBA or PhD. Masters classes are ideally in the range of 700 to 800. 900 level classes correspond to PhD ...
### Are classes numbered dependent on how difficult they …
At the three schools I've attended/worked at, numbers beyond the first digit are meaningless (with the sole exception of the intro class, generally 100 or 101 or 111, which is usually a little easier than any following in-depth course like 116 or 140). 201/202 can be harder than 260, 302s are insane, whereas 360, 378, 380 are kind like easy electives, etc.
### Water Hardness Scale Chart: Do You Have Hard Water ...
Example of white stains hard water causes. Limescale build-up in internal piping is extremely problematic. According to the water hardness scale, more than 85% of US households have hard water in their piping. How to know if you have hard water? Simple: The water hardness scale is a benchmark on how hard our water is.
### What is a Cleanroom? Cleanroom Classifications, Class 1 ...
Large numbers like "class 100" or "class 1000" refer to FED_STD-209E, and denote the number of particles of size 0.5 µm or larger permitted per cubic foot of air. The standard also allows interpolation, so it is possible to describe e.g. "class 2000."
### How many 300 level classes is too much a semester ...
I typically take 5 courses a semester. Second semester sophomore year, 4 out of my 5 classes were 300-level and the one that was 200-level (German) was one of the harder courses (certainly harder than music composition and dance history, both of …
### 300/400 level classes. : college
By the time you're taking 300/400 level courses, you should have the basics down and be more experienced with college, so it's about the same difficulty as a 100/200 level course as a freshman. Obviously different teachers have different teaching styles, so you could just request a copy of last semesters' syllabus from a few professors to get a ...
### Difference Between Entry Level College Course Vs. Upper ...
One of the first things you need to learn is how classes are structured and the differences between lower and upper level courses. Commonly, lower division courses are numbered as 100 or 200 level courses and upper division courses are 300 to 400 level courses. Understanding the difference between the two will help you plan and prepare in advance.
### Course Levels and Numbering | Registrar & Academic Systems ...
The course number indicates the level of the course, with the exception of the first-year seminars, all of which are open only to first-year students and considered to be at the 100 level. Fall and Winter offerings: 100 – 199 – Generally courses numbered 100 to 199 are introductory and open to first-year students. They do not have ...
### Clean Room Classifications & ISO Standards | Quotes - 48 Hours
ISO 5 is a super clean cleanroom classification. A cleanroom must have less than 3,520 particles >0.5 micron per cubic meter and 250-300 HEPA filtered air changes per hour. The equivalent FED standard is class 100 or 100 particles per cubic foot. Common applications are semiconductor manufacturing and pharmaceutical filling rooms.
### The 5 Easiest and 5 Hardest College Classes | CollegeVine Blog
5 Hardest College Classes. 1. Organic Chemistry. The notorious requirement for pre-meds is known for separating the future doctors from those who might not make the cut. Not only are the stakes extremely high, but the coursework itself is grueling, and students often study incorrectly for it. Many students think that Orgo is about memorization ...
### What is the difference between 100, 200, 300, 400, and 500 ...
All 100 level classes focus on skills and core communication competencies. 200 level classes introduce key terms and topics from the mission. 300 level classes provide students with a deeper exploration of key topics, skills, and issues in communication through the use of theoretically grounded research methods. 400 level courses are either internships or “case …
### What is the difference between a 400 level course and a ...
Answer (1 of 32): It depends on the college. At the university I attended, classes with codes from 000–099 were substitutes for high-school prerequisite classes that students might have missed (such as year 12 calculus, physics or chemistry). Classes with codes from 100–199 had no …
### Choosing Courses at Laurier | Students - Wilfrid Laurier ...
Understanding Course Numbering and Levels. Our courses are numbered by academic year. 100-level courses are first year, 200-level courses are second year, 300-level courses are third year and 400-level courses are fourth year. You are not permitted to take 300- or 400-level courses in a subject that is not your major of study.
none
### Introductory (100 Level), Intermediate (200/300 Level ...
The 300/400 Level assignment involves students’ developing a research proposal, a project that would be appropriate for content-specific courses as well as senior seminar, research methods, and capstone courses. Type: Annotated bibliography Article critique Paper. Level: 100, 200/300, 300/400. Block Plan Context: Level 100
### What Does Course Level Mean for Transferring College Credit?
Higher-level courses, like those in the 200 and 300 ranges, may be easier to transfer. Many universities view these as being more focused and more in …
### Hardest A Level Subjects List | Acrosophy
At Acrosophy students often ask us for help in deciding their a level choices – expecting that we have a magical list of A level subjects in order of difficulty. As you can imagine the truth is a bit more complicated than each subject being clearly harder or easier than another.
### 200 or 300? | Confessions of a Community College Dean
Third, and the point of today’s piece, is that there is no industry-wide standard in many fields for which courses should fall at the 200 level and which should fall at the 300 level. In states in which community colleges are limited to the first two years, such as Massachusetts, the distinction matters. If we teach a class that a receiving ...
### 200-Hour Standards | Yoga Alliance
New standards underlying the foundational-level RYS 200 credential took effect February 27, 2020. To view the new RYS 200 standards, visit our Common Core Curriculum for RYS™ 200 page . These Standards describe Yoga Alliance’s requirements for a Registered Yoga School that offers a 200-hour training.
### The Top 10 Hardest A-levels
Students doing their A Levels always like to claim that their subjects are the hardest, even if they may seem easy to other people. Whilst it is true that all A Levels are hard, there are still some which seem to be much harder than the rest. Hopefully this post will settle once and for all which ten A Levels are the hardest of them all. 1.
### 400 Level Course Difficulty — College Confidential
Yeah, I definitely disagree w/ the "higher number courses are harder" belief. In many cases, they are just specialized major-driven classes that build from the very broad(++ courseload) 100-level classes. Dont go by the numbers. For example, I had to take a 220 level math class before I could take a 160-level math class, lol. Anyway, I'm sure ...
### What is the difference between a 100-level class and a 300 ...
Most of the time 100 level classes are introductory, 200 level are intermediate, and 300 and 400 level are upper level. Always check the prerequisites and descriptions of courses you are interested in before registering for the course. 0 …
### Level Scale - The London School of English
Your English Level. You can discover your level of English on a scale from 1 (Beginner) to 9 (Very advanced). Check the table below to see which level you have, or take a 20 minute free Online English Level Test which will help you understand your …
### What’s the difference between 500- and 800-level courses?
The LDT Program does not design 400-level courses to be “easier” than other courses, rather we designate certain topics as the strong and sturdy base upon which future courses can build. There is a limit on the number of 400-level courses that can count toward a master’s degree from Penn State. Students enrolled in the 30-credit Master of ...
### OFFICE OF THE HUNTER COLLEGE SENATE
200-level course designation Courses of intermediate college-level difficulty; courses with 100-level course(s) as prerequi-site(s); or survey courses devoted to particular areas or fields within a discipline. Assumptions: 1. that students will have completed expository writing …
### Which Matters More—High Grades or Challenging Courses?
That said, students with some B grades in difficult courses will still have plenty of college options. A "B" in AP Chemistry shows that you are able to succeed in a challenging college-level class. Indeed, an unweighted "B" in an AP class is a better measure of your ability to succeed in college than an "A" in band or woodworking.
### 100-Level Math Courses | U-M LSA Mathematics
4 Credits. No credit after completeing any 200+ level amth course, except 385, 485, 489, and 497. Background and Goals: This course is intended for students who want to engage in mathematical reasoning without having to take calculus first. It is particularly well-suited for non-science majors or those who are thoroughly undecided.
### Classroom Briefing - 100 200 level - For 100 200-Level ...
View Notes - Classroom Briefing - 100 200 level from CON 200 at Defense Acquisition University. For 100- & 200-Level Courses Need a photo (or collage?) or something that transitions to
### Taking on the challenge: Freshmen enroll in 3000 level courses
“A student with a foreign language AP credit can probably handle the first 3000 level foreign language class, but some would find it harder than they expect,” she said.
### College Classes in High School: Is AP or Community College ...
A rigorous high school course load is very important to selective colleges, and AP courses may be considered stronger indicators of your academic abilities than community college classes. With community college classes, the difficulty of the class and your mastery of the material are harder for colleges to judge.
### Samsung
Through innovative, reliable products and services, and a responsible approach to business, Samsung is taking the world in imaginative new directions.
### Paragraph on Importance of Education 50, 100, 150, 200 ...
Paragraph on Importance of Education – 200 Words for Classes 6, 7, 8 Children Every kid has his own vision of doing something unique in life. Sometimes parents also dream of their kids to be at high professions like doctors, engineers, IAS or PCS officers, or any other high-level professions.
### 200-Level Math Courses | U-M LSA Mathematics
Prerequisites: Math 116, 156, 176, 186, or 296. Credit: 4 Credits. Credit is granted for only one course among Math 215 and 285. Background and Goals: The sequence Math 115-116-215 is the standard complete introduction to the concepts and methods of calculus. It is taken by the majority of students intending to concentrate in mathematics ...
### 100 or 200 Ton Master Class - THE PRACTICAL NAVIGATOR
Upon successful completion, mariners qualify for near coastal or inland commercial captain's licenses up to 100 or 200 gross registered tons. Topics include navigation, chart plotting, deck seamanship, deck safety, vessel administration, and rules of the road. The level of license is commensurate with sea time and experience. License Requirements:
Search the world's information, including webpages, images, videos and more. Google has many special features to help you find exactly what you're looking for.
### Eight things I wish I'd known before starting my A-levels ...
There’s more to free periods than social gatherings I wasted my first month of free periods sat around a table eating Doritos, struggling to get past level 33 on Candy Crush.
### terminology - What does $200\%$ faster mean? (How can ...
$100\%$ faster than $0.2$ batteries per minute would be $0.4$ batteries per minute, or charging a single battery in $2.5$ minutes. $200\%$ faster than $0.2$ batteries per minute would be $0.6$ batteries per minute, or charging a single battery in $1\frac{2}{3}$ minutes, or …
### Strawverry Supreme's Blog :)title> var a1 = ;var a2 =;var ...
Strawverry Supreme's Blog :)title> var a1 = ;var a2 =;var ...
### Degree Requirements | L&S Advising
36 upper division units (courses numbered between 100-199). Graduate-level courses numbered between 200-299 may also apply those units to the Upper Division Unit requirement. 6 upper division units outside your major department or department cluster. This can be included in your 36 upper division total.
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# How do you convert the following equation from standard to vertex form by completing the square: y=3x^2+12x+5?
Nov 15, 2016
#### Explanation:
The given equation is the equation of a parabola that opens upward (or downward). The vertex form of the equation of a parabola that opens upward (or downward) is:
$y = a {\left(x - h\right)}^{2} + k$
where $\left(h , k\right)$ is the vertex and "a" is the coefficient of the ${x}^{2}$ term.
Given: $y = 3 {x}^{2} + 12 x + 5$
$a = 3$, therefore, add 0 in the form $3 {h}^{2} - 3 {h}^{2}$ to the equation:
$y = 3 {x}^{2} + 12 x + 3 {h}^{2} - 3 {h}^{2} + 5$
Factor out 3 from the first 3 terms:
$y = 3 \left({x}^{2} + 4 x + {h}^{2}\right) - 3 {h}^{2} + 5$
Set the middle term in right side of the pattern, ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, equal to the middle term in the equation:
$- 2 h x = 4 x$
Solve for h:
$h = - 2$
Substitute the left side of the pattern into the equation:
$y = 3 {\left(x - h\right)}^{2} - 3 {h}^{2} + 5$
Substitute -2 for h:
$y = 3 {\left(x - - 2\right)}^{2} - 3 {\left(- 2\right)}^{2} + 5$
Combine the constant terms:
$y = 3 {\left(x - - 2\right)}^{2} - 7$
The above is the vertex form.
The vertex can be read directly from the equation; it is at:
$\left(- 2 , - 7\right)$
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## Energy calculation precision
$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$
$\Delta G^{\circ}= -RT\ln K$
$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$
JennyCKim1J
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am
### Energy calculation precision
Out of three methods for calculating free energy, which one is the most accurate/least accurate and why?
Deap Bhandal L1 S1J
Posts: 77
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 2 times
### Re: Energy calculation precision
There is actually a fourth equation too which derives delta g from a standard state cell potential: delta G= (-)moles of electrons*faraday's constant*cell potential. Depending on what information is given will tell you which equation to use. The accuracy of them depends on how accurate and precise the variables are in them. Otherwise, they should give very similar delta Gs.
Justin Chang 2K
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am
### Re: Energy calculation precision
I don't think Dr. Lavelle has told us which one is most/least accurate of the three methods, so I wouldn't worry about it unless he says it in lecture sometime. You will just have to calculate deltaG based on the information that is given to you (if you have deltaH, deltaS, and a temperature, then use G=H-TS, and etc)
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English
# Item
ITEM ACTIONSEXPORT
Released
Paper
#### Note on higher-point correlation functions of the TbarT or JbarT deformed CFTs
##### MPS-Authors
/persons/resource/persons2717
He, Song
Canonical and Covariant Dynamics of Quantum Gravity, AEI Golm, MPI for Gravitational Physics, Max Planck Society;
##### External Ressource
No external resources are shared
##### Fulltext (public)
2012.06202.pdf
(Preprint), 322KB
##### Supplementary Material (public)
There is no public supplementary material available
##### Citation
He, S. (in preparation). Note on higher-point correlation functions of the TbarT or JbarT deformed CFTs.
Cite as: http://hdl.handle.net/21.11116/0000-0007-A48E-0
##### Abstract
We investigate the generic n-point correlation functions of the conformal field theories (CFTs) with the $T\bar{T}$ and $J\bar{T}$ deformations in terms of perturbative CFT approach. We systematically obtain the first order correction to the generic correlation functions of the CFTs with $T\bar{T}$ or $J\bar{T}$ deformation. As applications, we compute the out of time ordered correlation function (OTOC) in Ising model with $T\bar{T}$ or $J\bar{T}$ deformation which confirm that these deformations do not change the integrable property up to the first order level.
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MathSciNet bibliographic data MR304699 32C35 Hill, C. Denson A Kontinuitätssatz for $\bar \partial \sb{M}$$\bar \partial \sb{M}$ and Lewy extendibility. Indiana Univ. Math. J. 22 (1972/73), 339–353. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
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# Question 1 of 10
Helen uses some toothpicks to form the pattern below.
\begin{array}{|c|c|} \hline \mbox{Pattern} & \mbox{Number of toothpick} \\ \hline 1 & 6\\ \hline 2 & 15\\ \hline 3 & 25\\ \hline 4 & 35\\ \hline 5 & \\ \hline \end{array}
(a) How many toothpicks will she need to form Pattern 5?
(b) How many toothpick will she need to form Pattern 40?
(c) Helen uses 4955 toothpicks to form a Pattern. Which Pattern is it?
A
(a) 45
(b) 40
(c) 496
B
(a) 47
(b) 43
(c) 498
C
(a) 49
(b) 45
(c) 500
D
(a) 50
(b) 49
(c) 502
E
None of the above
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# Sets A, B and C are shown below. If number 100 is included
Author Message
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### Hide Tags
Manager
Joined: 10 Nov 2010
Posts: 164
Followers: 5
Kudos [?]: 192 [0], given: 6
Sets A, B and C are shown below. If number 100 is included [#permalink]
### Show Tags
25 Mar 2011, 04:24
5
This post was
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Difficulty:
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Question Stats:
71% (01:41) correct 29% (01:04) wrong based on 326 sessions
### HideShow timer Statictics
Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from largest to smallest?
A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50}
(A) A, C, B
(B) A, B, C
(C) C, A, B
(D) B, A, C
(E) B, C, A
As per me answer should be
[Reveal] Spoiler:
E
.
Wants to confirm or reject the OA.
[Reveal] Spoiler: OA
Last edited by Bunuel on 22 Jul 2013, 05:37, edited 1 time in total.
Edited the OA.
Director
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
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Re: standard deviation after including a number [#permalink]
### Show Tags
25 Mar 2011, 08:13
A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50}
Initially SD(A) > SD(B) > SD(C). When 100 is added. The range of A is unchanged, so least change. But to calculate any useful relationship between the modified A Vs B we have to know the fact that B contains negative numbers. So we will get new sds as follows SD(B) > SD(C) > SD(A). Pls verify this reasoning.
To prove this inference let me calculate change in mean for sets B and C -
m(B) changes by (100 - 0)/6 = 50/3 = 16.67 hence the new mean of set B is 16.67 + Old mean = 16.67
m(C) changes by (100 - 40)/6 = 10. Hence the new mean of set C is 10 + Old mean = 10 + 40 = 50
Now the new distances from their respective means of set B (mean 16.67) and set C (mean 50)
B = {36.67, 26.67,16.67, 6.67, 3.33, 83.33}
C = {20,15,10,5,0,50}
Hence SD(B) > SD(C) > SD (A). Answer E. So this is not a 120 sec question. How to save time ?
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Re: standard deviation after including a number [#permalink]
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14 Jun 2011, 20:36
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vjsharma25 wrote:
Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following
represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from
largest to smallest?
A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50}
(A) A, C, B
(B) A, B, C
(C) C, A, B
(D) B, A, C
(E) B, C, A
As per me answer should be
[Reveal] Spoiler:
E
.
Wants to confirm or reject the OA.
It is a Veritas Prep Book X question for which the OA given is (E). The explanation clearly explains you why the answer is E.
You don't have to calculate anything. SD measures the distance between each element and mean. If a new element is added which is far away from the mean, it will distort the mean more than if it were added close to the mean.
The means of the 3 sets are 70, 0 and 40.
100 is farthest from 0 so it will change the SD of set B the most (in terms of absolute increase). It is closest to 70 so it will change the SD of set A the least. Hence answer is B, C, A
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 30 Oct 2011 Posts: 48 Followers: 0 Kudos [?]: 8 [0], given: 13 Re: standard deviation after including a number [#permalink] ### Show Tags 14 Nov 2012, 11:34 VeritasPrepKarishma wrote: vjsharma25 wrote: Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from largest to smallest? A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50} (A) A, C, B (B) A, B, C (C) C, A, B (D) B, A, C (E) B, C, A As per me answer should be [Reveal] Spoiler: E . Wants to confirm or reject the OA. Can you please post your source of this question? It is a Veritas Prep Book X question for which the OA given is (E). The explanation clearly explains you why the answer is E. You don't have to calculate anything. SD measures the distance between each element and mean. If a new element is added which is far away from the mean, it will distort the mean more than if it were added close to the mean. The means of the 3 sets are 70, 0 and 40. 100 is farthest from 0 so it will change the SD of set B the most (in terms of absolute increase). It is closest to 70 so it will change the SD of set A the least. Hence answer is B, C, A Thanks Karishma. You saved a lot of time and effort !! Manager Joined: 05 Nov 2012 Posts: 172 Followers: 1 Kudos [?]: 26 [0], given: 57 Re: standard deviation after including a number [#permalink] ### Show Tags 14 Nov 2012, 14:57 VeritasPrepKarishma wrote: vjsharma25 wrote: Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following Hey Karishma.... I am stuck.... I dealt with it in another way..... for any series.... If we draw Gaussian curve.... http://upload.wikimedia.org/wikipedia/c ... am.svg.png the min and the max are 8sigma interval difference... where sigma is standard deviation.... in other terms... min and max are 4sigma intervals away from mean.... So if we calculate sigma for the sets as max-min/8.... Set A will be 110-30/8 which is 80/8 Set B will be 100-(-20)/8 which is 120/8 Set C will be 100-30/8 which is 70/8 So in order of highest to lowest wouldn't it be B,A,C? Where am I going wrong? thank you Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6480 Location: Pune, India Followers: 1759 Kudos [?]: 10494 [0], given: 206 Re: standard deviation after including a number [#permalink] ### Show Tags 14 Nov 2012, 20:10 Expert's post Amateur wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following Hey Karishma.... I am stuck.... I dealt with it in another way..... for any series.... If we draw Gaussian curve.... http://upload.wikimedia.org/wikipedia/c ... am.svg.png the min and the max are 8sigma interval difference... where sigma is standard deviation.... in other terms... min and max are 4sigma intervals away from mean.... So if we calculate sigma for the sets as max-min/8.... Set A will be 110-30/8 which is 80/8 Set B will be 100-(-20)/8 which is 120/8 Set C will be 100-30/8 which is 70/8 So in order of highest to lowest wouldn't it be B,A,C? Where am I going wrong? thank you First of all, this is not a normal distribution. In a normal distribution, the values are concentrated around the mean (as is obvious from the normal distribution curve). You cannot calculate the SD of these sets based on the ND curve. Secondly, you have to order them in terms of the absoluteincrease in their standard deviation, not in terms of their new SD. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: standard deviation after including a number [#permalink]
### Show Tags
21 Nov 2012, 11:54
VeritasPrepKarishma wrote:
vjsharma25 wrote:
Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following
represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from
largest to smallest?
A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50}
(A) A, C, B
(B) A, B, C
(C) C, A, B
(D) B, A, C
(E) B, C, A
As per me answer should be
[Reveal] Spoiler:
E
.
Wants to confirm or reject the OA.
It is a Veritas Prep Book X question for which the OA given is (E). The explanation clearly explains you why the answer is E.
You don't have to calculate anything. SD measures the distance between each element and mean. If a new element is added which is far away from the mean, it will distort the mean more than if it were added close to the mean.
The means of the 3 sets are 70, 0 and 40.
100 is farthest from 0 so it will change the SD of set B the most (in terms of absolute increase). It is closest to 70 so it will change the SD of set A the least. Hence answer is B, C, A
Hi Karishma,
I can understand that by adding 100 to the three sets the extent to which the S.D changes is based on the absolute difference b/w the mean and 100. But based on this, how can you conclude that the new SD will be in B, C & A order??
Thanks.
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Re: standard deviation after including a number [#permalink]
### Show Tags
23 Nov 2012, 13:05
1
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Expert's post
aalriy wrote:
VeritasPrepKarishma wrote:
vjsharma25 wrote:
Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following
represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from
largest to smallest?
A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50}
(A) A, C, B
(B) A, B, C
(C) C, A, B
(D) B, A, C
(E) B, C, A
As per me answer should be
[Reveal] Spoiler:
E
.
Wants to confirm or reject the OA.
It is a Veritas Prep Book X question for which the OA given is (E). The explanation clearly explains you why the answer is E.
You don't have to calculate anything. SD measures the distance between each element and mean. If a new element is added which is far away from the mean, it will distort the mean more than if it were added close to the mean.
The means of the 3 sets are 70, 0 and 40.
100 is farthest from 0 so it will change the SD of set B the most (in terms of absolute increase). It is closest to 70 so it will change the SD of set A the least. Hence answer is B, C, A
Hi Karishma,
I can understand that by adding 100 to the three sets the extent to which the S.D changes is based on the absolute difference b/w the mean and 100. But based on this, how can you conclude that the new SD will be in B, C & A order??
Thanks.
Notice that the denominator in the calculation of SD will be the same in the case of all the 3 sets (since they all have 5 elements each). When you add 100 to each one of them, they will have 6 elements each and hence the denominator will still stay the same.
In case of set B, the numerator increases by 100^2 (before you take the root)
In case of set C, the numerator increases by 60^2 (before you take the root)
In case of set A, the numerator increases by 30^2 (before you take the root)
So in absolute terms, B will see the most effect and A will see the least. You can look at the actual calculation to understand exactly why this happens. The formula for SD is discussed in the first post below.
For more on SD, check out these posts:
http://www.veritasprep.com/blog/2012/06 ... deviation/
http://www.veritasprep.com/blog/2012/06 ... n-part-ii/
http://www.veritasprep.com/blog/2012/06 ... questions/
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Re: standard deviation after including a number [#permalink]
### Show Tags
24 Jan 2013, 22:34
VeritasPrepKarishma wrote:
Sachin9 wrote:
Hi Karishma,
From what I can infer, it seems that the order of 'increase in the SD' and the order of 'new SD' will always be the same ..
Please correct me if its incorrect.
Regards,
Sach
Actually no, that may not be the case.
The increase in SD depends on the distance between the number added (100 here) and the mean. Set B has the smallest mean (0) so it is farthest from 100 hence it will see maximum increase.
The 'new SD' depends on the difference between all the elements (including the new one) and the mean. If the rest of the numbers are very close to the mean, it is certainly possible that the new SD does not have the same ordering.
e.g.
Set A = {-1, 0, 1}
Set B = (0, 20, 40}
If you add another number say, 30, the increase in SD of set A will be substantial because 30 is far from mean
but increase in SD of set B will not be very much.
Nevertheless, new SD of set A will be less than the new SD of set B.
I guess the new SD of A will be more than the new SD of B..
Numbers in A would be more dispersed than those in B..
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Re: standard deviation after including a number [#permalink]
### Show Tags
24 Jan 2013, 23:21
1
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Sachin9 wrote:
I guess the new SD of A will be more than the new SD of B..
Numbers in A would be more dispersed than those in B..
No. You can use a fin calc to find that the SD of A is 13 and that of B is 14.8. The difference isn't much but still the new SD of A is less than the new SD of B. As I said, what matters is that how far apart are all the elements from the mean in case of new SD. One element can have a huge impact but it still may not be sufficient. So you cannot infer that the new SD will be in the same order.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Status: Gonna rock this time!!! Joined: 22 Jul 2012 Posts: 547 Location: India GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29 WE: Information Technology (Computer Software) Followers: 3 Kudos [?]: 49 [0], given: 562 Re: standard deviation after including a number [#permalink] ### Show Tags 25 Jan 2013, 04:03 VeritasPrepKarishma wrote: Sachin9 wrote: I guess the new SD of A will be more than the new SD of B.. Numbers in A would be more dispersed than those in B.. No. You can use a fin calc to find that the SD of A is 13 and that of B is 14.8. The difference isn't much but still the new SD of A is less than the new SD of B. As I said, what matters is that how far apart are all the elements from the mean in case of new SD. One element can have a huge impact but it still may not be sufficient. So you cannot infer that the new SD will be in the same order. Thanks a lot, Karishma.. From what I understand,highest effect would depend on how far is the new no. from the mean in all the sets.. and Actual order of SD among sets would depend on how dispersed all elements are from the mean.. _________________ hope is a good thing, maybe the best of things. And no good thing ever dies. Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595 My GMAT Journey : end-of-my-gmat-journey-149328.html#p1197992 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6480 Location: Pune, India Followers: 1759 Kudos [?]: 10494 [1] , given: 206 Re: standard deviation after including a number [#permalink] ### Show Tags 25 Jan 2013, 19:34 1 This post received KUDOS Expert's post Sachin9 wrote: VeritasPrepKarishma wrote: Sachin9 wrote: I guess the new SD of A will be more than the new SD of B.. Numbers in A would be more dispersed than those in B.. No. You can use a fin calc to find that the SD of A is 13 and that of B is 14.8. The difference isn't much but still the new SD of A is less than the new SD of B. As I said, what matters is that how far apart are all the elements from the mean in case of new SD. One element can have a huge impact but it still may not be sufficient. So you cannot infer that the new SD will be in the same order. Thanks a lot, Karishma.. From what I understand,highest effect would depend on how far is the new no. from the mean in all the sets.. and Actual order of SD among sets would depend on how dispersed all elements are from the mean.. Yes, that's correct. 'Change' in SD depends on how far the new no is from the mean. If the new no is close to the mean, the change in SD is very little because it adds very little dispersion to the scenario. If the new no is far from the mean, the change in SD is significant because it adds a lot more dispersion. The mean changes, it becomes farther than the previous mean and hence overall dispersion in a lot higher. Actual SD depends in big part on how the previous numbers were dispersed around the mean. So it is hard to say what the new order will be based on just the new no. and the previous mean. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: standard deviation after including a number [#permalink]
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25 Jan 2013, 22:04
Thanks a lot.. U rock, Karishma!!
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
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01 Feb 2014, 11:34
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
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03 Feb 2014, 11:37
5 sec. approach:
SD is the deviation of the mean to the smallest and biggest number in the set. So if a number is added inside the boundaries of that set, there will be no changes in the SD of that set. So no absolute increase will occur in set A.
A has to be the set with the smallest increase, what's bigger than 0 right? And hence E is the only right answer.
Correct me if I'm wrong
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
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27 Jul 2014, 04:44
vjsharma25 wrote:
Sets A, B and C are shown below. If number 100 is included in each of these sets, which of the following represents the correct ordering of the sets in terms of the absolute increase in their standard deviation, from largest to smallest?
A {30, 50, 70, 90, 110}, B {-20, -10, 0, 10, 20}, C {30, 35, 40, 45, 50}
(A) A, C, B
(B) A, B, C
(C) C, A, B
(D) B, A, C
(E) B, C, A
As per me answer should be
[Reveal] Spoiler:
E
.
Wants to confirm or reject the OA.
For evenly spaced sets, median = mean
so mean of 3 sets are 70, 0 & 40.
see for yourself 100 is farthest from which mean
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
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15 Aug 2015, 12:08
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink]
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18 Mar 2016, 04:30
of all sets the SD of set A will either see no change or decrease because 100 is less than 110, whereas B and C will definitely see the increase in SD.
the only answer that has A at the end is E
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Re: Sets A, B and C are shown below. If number 100 is included [#permalink] 18 Mar 2016, 04:30
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## Cryptology ePrint Archive: Report 2020/111
Adaptively Secure Constrained Pseudorandom Functions in the Standard Model
Alex Davidson and Shuichi Katsumata and Ryo Nishimaki and Shota Yamada and Takashi Yamakawa
Abstract: Constrained pseudorandom functions (CPRFs) allow learning constrained'' PRF keys that can evaluate the PRF on a subset of the input space, or based on some predicate. First introduced by Boneh and Waters [AC’13], Kiayias et al. [CCS’13] and Boyle et al. [PKC’14], they have shown to be a useful cryptographic primitive with many applications. These applications often require CPRFs to be adaptively secure, which allows the adversary to learn PRF values and constrained keys in an arbitrary order. However, there is no known construction of adaptively secure CPRFs based on a standard assumption in the standard model for any non-trivial class of predicates. Moreover, even if we rely on strong tools such as indistinguishability obfuscation (IO), the state-of-the-art construction of adaptively secure CPRFs in the standard model only supports the limited class of NC1 predicates.
In this work, we develop new adaptively secure CPRFs for various predicates from different types of assumptions in the standard model. Our results are summarized below.
- We construct adaptively secure and $O(1)$-collusion-resistant CPRFs for $t$-conjunctive normal form ($t$-CNF) predicates from one-way functions (OWFs) where $t$ is a constant. Here, $O(1)$-collusion-resistance means that we can allow the adversary to obtain a constant number of constrained keys. Note that $t$-CNF includes bit-fixing predicates as a special case.
- We construct adaptively secure and single-key CPRFs for inner-product predicates from the learning with errors (LWE) assumption. Here, single-key security means that we only allow the adversary to learn one constrained key. Note that inner-product predicates include $t$-CNF predicates for a constant $t$ as a special case. Thus, this construction supports more expressive class of predicates than that supported by the first construction though it loses the collusion-resistance and relies on a stronger assumption.
- We construct adaptively secure and $O(1)$-collusion-resistant CPRFs for all circuits from the LWE assumption and indistinguishability obfuscation (IO).
The first and second constructions are the first CPRFs for any non-trivial predicates to achieve adaptive security outside of the random oracle model or relying on strong cryptographic assumptions. Moreover, the first construction is also the first to achieve any notion of collusion-resistance in this setting. Besides, we prove that the first and second constructions satisfy weak $1$-key privacy, which roughly means that a constrained key does not reveal the corresponding constraint. The third construction is an improvement over previous adaptively secure CPRFs for less expressive predicates based on IO in the standard model.
Category / Keywords: foundations / constrained PRF, collusion resistance, adpative security
Date: received 4 Feb 2020, last revised 6 Feb 2020
Contact author: ryo nishimaki at gmail com, ryo nishimaki zk@hco ntt co jp, takashi yamakawa ga@hco ntt co jp, shuichi katsumata@aist go jp, shuichi katsumata000@gmail com, alex davidson92@gmail com, yamada-shota@aist go jp, shota yamada enc@gmail com
Available format(s): PDF | BibTeX Citation
Note: This is a major update version of https://eprint.iacr.org/2018/982 with many new results.
Short URL: ia.cr/2020/111
[ Cryptology ePrint archive ]
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# Production possibility frontier and stochastic programming
### Citation:
Chovanec P. Production possibility frontier and stochastic programming. Proceedings of the 14th Annual Conference of Doctoral Students -- WDS 2005. 2005:108--113.
### Abstract:
By its nature, Data Envelopment Analysis (DEA) leaves no room for uncertainty in data such as measurement errors. To improve this fact, we consider $\alpha$-stochastic efficiency concept, and we relate this problem to the stochastic programming problem. Two types of probability inequalities are employed for introducing new criteria for efficiency.
wds05.pdf 147.28 KB
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Solution to tensor differential equations
1. May 30, 2010
jfy4
hello all,
I need two solutions to two different tensor diffeqs. I think I may have the solution to the sourceless equation, however I am in the dark about the one with the source.
$$\left(\partial_{\gamma}\partial_{\alpha}+\imath k^{\beta}g_{\alpha\beta}\partial_{\gamma}\right) \phi=T_{\gamma\alpha}\phi$$
and
$$\left(\partial_{\gamma}\partial_{\alpha}+\imath k^{\beta}g_{\alpha\beta}\partial_{\gamma}\right) \phi=0$$.
Any help would be appreciated.
2. May 30, 2010
jfy4
here is my solution for the source less equation, feel free to check it please.
$$\phi^{\gamma\alpha}=Ae^{-\imath\left(\delta^{\gamma}_{\alpha}k_{\gamma}x^{\alpha}\right)}+Be^{-\imath\left(k_{\alpha}x^{\alpha}-k_{\gamma}x^{\gamma}\right)}$$
thanks.
3. May 30, 2010
jfy4
I also made the replacement $$k_{\beta}=k^{\alpha}g_{\alpha\beta}$$
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# À¬»ø±éµØµÄ¿ÆѧÑо¿
ÒÑÓÐ 3764 ´ÎÔĶÁ 2013-10-19 21:25 |¸öÈË·ÖÀà:¿Æѧ°¼ÏÂÈ¥|ϵͳ·ÖÀà:¿ÆÑбʼÇ|¹Ø¼ü´Ê:ѧÕß
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How science goes wrong
Scientific research has changed the world.Now it needs to change itself
A SIMPLE idea underpins science: ¡°trust,but verify¡±. Results should always be subject to challenge from experiment.That simple but powerful idea has generated a vast body of knowledge. Since itsbirth in the 17th century, modern science has changed the world beyondrecognition, and overwhelmingly for the better.
But success can breed complacency. Modernscientists are doing too much trusting and not enough verifying¡ªto thedetriment of the whole of science, and of humanity.
Too many of the findings that fill theacademic ether are the result of shoddy experiments or poor analysis (seearticle). A rule of thumb among biotechnology venture-capitalists is that halfof published research cannot be replicated. Even that may be optimistic. Lastyear researchers at one biotech firm, Amgen, found they could reproduce justsix of 53 ¡°landmark¡± studies in cancer research. Earlier, a group at Bayer, adrug company, managed to repeat just a quarter of 67 similarly importantpapers. A leading computer scientist frets that three-quarters of papers in hissubfield are bunk. In 2000-10 roughly 80,000 patients took part in clinicaltrials based on research that was later retracted because of mistakes orimproprieties.
Even when flawed research does not putpeople¡¯s lives at risk¡ªand much of it is too far from the market to do so¡ªitsquanders money and the efforts of some of the world¡¯s best minds. Theopportunity costs of stymied progress are hard to quantify, but they are likelyto be vast. And they could be rising.
One reason is the competitiveness ofscience. In the 1950s, when modern academic research took shape after itssuccesses in the second world war, it was still a rarefied pastime. The entireclub of scientists numbered a few hundred thousand. As their ranks haveswelled, to 6m-7m active researchers on the latest reckoning, scientists havelost their taste for self-policing and quality control. The obligation to¡°publish or perish¡± has come to rule over academic life. Competition for jobsis cut-throat. Full professors in America earned on average $135,000 in2012¡ªmore than judges did. Every year six freshly minted PhDs vie for everyacademic post. Nowadays verification (the replication of other people¡¯sresults) does little to advance a researcher¡¯s career. And withoutverification, dubious findings live on to mislead. Careerism also encourages exaggeration andthe cherry-picking of results. In order to safeguard their exclusivity, theleading journals impose high rejection rates: in excess of 90% of submittedmanuscripts. The most striking findings have the greatest chance of making itonto the page. Little wonder that one in three researchers knows of a colleaguewho has pepped up a paper by, say, excluding inconvenient data from results¡°based on a gut feeling¡±. And as more research teams around the world work on aproblem, the odds shorten that at least one will fall prey to an honestconfusion between the sweet signal of a genuine discovery and a freak of thestatistical noise. Such spurious correlations are often recorded in journalseager for startling papers. If they touch on drinking wine, going senile orletting children play video games, they may well command the front pages ofnewspapers, too. Conversely, failures to prove a hypothesisare rarely even offered for publication, let alone accepted. ¡°Negative results¡±now account for only 14% of published papers, down from 30% in 1990. Yetknowing what is false is as important to science as knowing what is true. Thefailure to report failures means that researchers waste money and effortexploring blind alleys already investigated by other scientists. The hallowed process of peer review is notall it is cracked up to be, either. When a prominent medical journal ranresearch past other experts in the field, it found that most of the reviewersfailed to spot mistakes it had deliberately inserted into papers, even afterbeing told they were being tested. If it¡¯s broke, fix it All this makes a shaky foundation for anenterprise dedicated to discovering the truth about the world. What might bedone to shore it up? One priority should be for all disciplines to follow theexample of those that have done most to tighten standards. A start would begetting to grips with statistics, especially in the growing number of fieldsthat sift through untold oodles of data looking for patterns. Geneticists havedone this, and turned an early torrent of specious results from genomesequencing into a trickle of truly significant ones. Ideally, research protocols should be registeredin advance and monitored in virtual notebooks. This would curb the temptationto fiddle with the experiment¡¯s design midstream so as to make the results lookmore substantial than they are. (It is already meant to happen in clinicaltrials of drugs, but compliance is patchy.) Where possible, trial data alsoshould be open for other researchers to inspect and test. The most enlightened journals are alreadybecoming less averse to humdrum papers. Some government funding agencies,including America¡¯s National Institutes of Health, which dish out$30 billionon research each year, are working out how best to encourage replication. Andgrowing numbers of scientists, especially young ones, understand statistics.But these trends need to go much further. Journals should allocate space for¡°uninteresting¡± work, and grant-givers should set aside money to pay for it.Peer review should be tightened¡ªor perhaps dispensed with altogether, in favourof post-publication evaluation in the form of appended comments. That systemhas worked well in recent years in physics and mathematics. Lastly,policymakers should ensure that institutions using public money also respectthe rules.
Science still commands enormous¡ªifsometimes bemused¡ªrespect. But its privileged status is founded on the capacityto be right most of the time and to correct its mistakes when it gets thingswrong. And it is not as if the universe is short of genuine mysteries to keepgenerations of scientists hard at work. The false trails laid down by shoddyresearch are an unforgivable barrier to understanding.
http://blog.sciencenet.cn/blog-71685-734270.html
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## SRN – about the magical 0.234 acceptance rate
Sunday Reading Notes series is back : Let’s understand the magical rule of ‘tuning your MH algorithm so that the acceptance rate is roughly 25%’ together!
‘Tune your MH algorithm so that the acceptance rate is roughly 25%’ has been general advice given to students in Bayesian statistics classes. It has been almost 4 years since I first read about it from the book Bayesian Data Analysis, but I never read the original paper where this result first appeared. This Christmas, I decided to read the paper ‘Weak Convergence and Optimal Scaling of Random Walk Metropolis Algorithms’ by Roberts, Gelman and Gilks and to resume by Sunday Reading Notes series with a short exposition of this paper.
In Roberts, Gelman and Gilk (1997), the authors obtain a weak convergence result for the sequence of algorithms targeting the sequence of distributions ${\pi_d(x^d) = \prod_{i=1}^{d} f(x_i^d)}$ converging to a Langevin diffusion. The asymptotic optimal scaling problem becomes a matter optimizing the speed of the Langevin diffusion, and it is related to the asymptotic acceptance rate of proposed moves.
A one-sentence summary of the paper would be
if you have a d-dimensional target that is independent in each coordinate, then choose the step size of random walk kernel to be 2.38 / sqrt(d) or tune your acceptance rate to be around 1/4.
Unfortunately, in practice the ‘if’ condition is often overlooked and people are tuning the acceptance rate to be 0.25 as long as the proposal is random walk, no matter what the target distribution is. It has been 20 years since the publication of the 0.234 result and we are witnessing the use of MCMC algorithms on more complicated target distributions, for example parameter inference for state-space models. I feel that this is good time that we revisit and appreciate the classical results while re-educating ourselves on their limitations.
Reference:
Roberts, G. O., Gelman, A., & Gilks, W. R. (1997). Weak convergence and optimal scaling of random walk Metropolis algorithms. The annals of applied probability7(1), 110-120.
——–TECHNICAL EXPOSITION——-
Assumption 1 The marginal density of each component $f$ is such that ${f'/f}$ is Lipschitz continuous and
$\displaystyle \mathbb{E}_f\left[\left(\frac{f'(X)}{f(X)}\right)^8\right] = M < \infty, \ \ \ \ \ (1)$
$\displaystyle \mathbb{E}_f\left[\left(\frac{f''(X)}{f(X)}\right)^4\right] < \infty. \ \ \ \ \ (2)$
Roberts et al. (1997) considers random walk proposal ${y^d - x^d \sim \mathcal{N}(0,\sigma_d I_d)}$ where ${\sigma_d^2 = l^2 / (d-1).}$ We use ${X^d = (X_0^d,X_1^d,\ldots)}$ to denote the Markov chain and define another Markov process ${(Z^d)}$ with ${Z_t^d = X_{[dt]}^d}$, which is the speed-up version of ${X^d}$. Let ${[a]}$ denote the floor of ${a \in \mathbb{R}}$. Define ${U^d_t= X^d_{[dt],1}}$, the first component of ${X_{[dt]}^d = Z^d_t}$.
Theorem 1 (diffusion limit of first component) Suppose ${f}$ is positive and in ${\mathbb{C}^2}$ and that (1)-(2) hold. Let ${X_0^{\infty} = (X^1_{0,1},X^{2}_{0,2},\ldots)}$ be such that all components are distributed according to ${f}$ and assume ${X^{i}_{0,j} = X^{j}_{0,j}}$ for all ${i \le j}$. Then as ${d \to \infty}$,
$\displaystyle U^d \to U.$
The ${U_0 \sim f}$ and ${U}$ satisfies the Langevin SDE
$\displaystyle dU_t = (h(l))^{1/2}dB_t + h(l)\frac{f'(U_t)}{2f(U_t)}dt \ \ \ \ \ (3)$
and
$\displaystyle h(l) = 2 l^2 \Phi(-l\sqrt{I}/2)$
with ${\Phi}$ being the standard normal cdf and
$\displaystyle I = \mathbb{E}_f\left[\left(f'(X)/ f(X)\right)^2\right].$
Here ${h(l)}$ is the speed measure of the diffusion process and the most efficient’ asymptotic diffusion has the largest speed measure. ${I}$ measures the roughness’ of ${f}$.
Example 1 If ${f}$ is normal, then ${f(x) = (2\pi\sigma^2_f)^{-1/2}\exp(-x^2/(2\sigma_f^2)).}$
$\displaystyle I = \mathbb{E}_f\left[\left(f'(x) / f(x) \right)^2\right] = (\sigma_f)^{-4}\mathbb{E}_f\left[x^2\right] = 1/\sigma^2_f.$
So when the target density ${f}$ is normal, then the optimal value of ${l}$ is scaled by ${1 / \sqrt{I}}$, which coincides with the standard deviation of ${f}$.
Proof: (of Theorem 1.1) This is a proof sketch. The strategy is to prove that the generator of ${Z^n}$, defined by
$\displaystyle G_n V(x^n) = n \mathbb{E}\left[\left(V(Y^n) - V(x^n)\right) \left( 1 \wedge \frac{\pi_n(Y^n)}{\pi_n(x^n)}\right)\right].$
converges to the generator of the limiting Langevin diffusion, defined by
$\displaystyle GV(x) = h(l) \left[\frac{1}{2} V''(x) + \frac{1}{2} \frac{d}{dx}(\log f)(x) V'(x)\right].$
Here the function ${V}$ is a function of the first component only.
First define a set
$\displaystyle F_d = \{|R_d(x_2,\ldots,x_d) - I| < d^{-1/8}\} \cap \{|S_d(x_2,\ldots,x_d) - I| < d^{-1/8}\},$
where
$\displaystyle R_d(x_2,\ldots,x_d) = (d-1)^{-1} \sum_{i=2}^d \left[(\log f(x_i))'\right]^2$
and
$\displaystyle S_d(x_2,\ldots,x_d) = - (d-1)^{-1} \sum_{i=2}^d \left[(\log f(x_i))''\right].$
For fixed ${t}$, one can show that ${\mathbb{P}\left(Z^d_s \in F_d , 0 \le s \le t\right)}$ goes to 1 as ${d \to \infty}$. On these sets ${\{F_d\}}$, we have
$\displaystyle \sup_{x^d \in F_d} |G_d V(x^d) - G V(x_1)| \to 0 \quad \text{as } d \to \infty ,$
which essentially says ${G_d \to G}$, because we have uniform convergence for vectors contained in a set of limiting probability 1.
$\Box$
Corollary 2 (heuristics for RWMH) Let
$\displaystyle a_d(l) = \int \int \pi_d(x^d)\alpha(x^d,y^d)q_d(x^d,y^d)dx^d dy^d$
be the average acceptance rate of the random walk MH in ${d}$ dimensions.
We must have ${lim_{d\to\infty} a_d(l) \to a(l)}$ where ${a(l) = 2 \Phi(-l\sqrt{I}/2)}$.
${h(l)}$ is maximized by ${l = \hat{l} = 2.38 / \sqrt{I}}$ and ${a(\hat{l}) = 0.23}$ and ${h(\hat{l}) = 1.3 / I.}$
The authors consider two extensions of the target density ${\pi_d}$, where the convergence and optimal scaling properties will still hold. The first extension concerns the case where ${f_i}$‘s are different, but there is an law of large numbers on these density functions. Another extension concerns the case ${\pi_d(x^d) = f_1(x_1) \prod_{i=2}^{d} P(x_{i-1}, x_{i})}$, with some conditions on ${P}$.
## SRN – A Geometric Interpretation of the Metropolis-Hastings Algorithm by Billera and Diaconis
Coming back to the Sunday Reading Notes, this week I discuss the paper ‘A Geometric Interpretation of the Metropolis-Hastings Algorithm’ by Louis J. Billera and Persi Diaconis from Statistical Science. This paper is suggested to me by Joe Blitzstein.
In Section 4 of ‘Informed proposals for local MCMC in discrete spaces’ by Giacomo Zanella (see my SRN Part I and II), Zanella mentions that the Metropolis-Hasting acceptance probability function(APF) $\min\left(1,\frac{\pi(y)p(x,y)}{\pi(x)p(y,x)}\right)$ is not the only APF that makes the resulting kernel $\pi$-reversible as long as detailed-balance is satisfied. This comes first as a ‘surprise’ to me as I have never seen another APF in practice. But very quickly I realize that this fact was mentioned in both Stat 213 & Stat 220 at Harvard and I have read about it from Section 5.3 – ‘Why Does the Metropolis Algorithm Work?‘ of ‘Monte Carlo Strategies in Scientific Computing‘ by Jun S. Liu. Unfortunately, I did not pay enough attention. Joe suggested this article to me after I posted on Facebook about being upset with not knowing such a basic fact.
In this Billera and Diaconis paper, the authors focus on the finite state space case $X$ and considers the MH kernel as the projection of stochastic matrices (row sums are all 1 and all entries are non-negative, denoted by$\mathcal{s}(X)$) onto the set of $\pi$-reversible Markov chains (stochastic matrices that satisfy detailed balance $\pi(x)M(x,y) = \pi(y)M(y,x)$, denoted by $R(\pi)).$ If we introduce a metric on the stochastic matrices: $d(K,K') = \sum_{x} \sum_{x\not=y} \pi(x) |K(x,y)-K'(x,y)|$.
The key result in this paper is Theorem 1. The authors prove that the Metropolis maps $M := M(K)(x,y) = \min\left( K(x,y), \frac{\pi(y}{\pi(x)}K(y,x)\right)$ minimizes the distance $d$ from the proposal kernel $K$ to $R(\pi).$ This means that $M(K)$ is the unique closest element in $R(\pi)$ that is coordinate-wise smaller than $K$ on its off-diagonal entries. So $M$ is in a sense the closest reversible kernel to the original kernel $K$.
I think this geometric interpretation offers great intuition about how the MH algorithm works: we start with a kernel $K$ and change it to another kernel with stationary distribution $\pi$. And the change must occur as follows:
from $x$, choose $y$ from $K(x,y)$ and decide to accept $x$ or stay at $y$; this last choice may be stochastic with acceptance probabilty $F(x,y) \in [0,1]$. This gives the new chain with transition probabilities: $K(x,y) F(x,y)$, x \not =y\$. The diagonal entries are changed so that each row sums to 1.
Indeed the above procedure describes how the MH algorithm works. If we insist on $\pi$-reversibility, we must have $0 \leq 0 \leq \min(1,R(x,y)$ where $R(x,y) = \frac{\pi(y)K(y,x)}{\pi(x)K(x,y)}.$ So the MH choice of APF is one that maximizes the chance of moving from $x$ to $y$. The resulting MH kernel $M$ has the largest spectral gap (1 – second largest eigenvalue) and by Peksun’s theorem must have the minimum asymptotic variance estimating additive functionals.
In Remark 3.2, the authors point out if we consider only APF that are functions of $R(x,y)$, then the function must satisfy $g(x) = x g(1/x)$ which is the characteristic of balancing functions in Zanella’s ‘informed proposals’ paper.
This paper allows me to study Metropolis-Hastings algorithm from another angle and review facts I have neglected in my coursework.
References:
• Billera, L. J., & Diaconis, P. (2001). A geometric interpretation of the Metropolis-Hastings algorithm. Statistical Science, 335-339.
• Zanella, G. (2017). Informed proposals for local MCMC in discrete spaces. arXiv preprint arXiv:1711.07424.
• Liu, J. S. (2008). Monte Carlo strategies in scientific computing. Springer Science & Business Media.
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## Fixed points
1. Recall that ω ≡ \x. x x, and Ω ≡ ω ω. Is Ω a fixed point for ω? Find a fixed point for ω, and prove that it is a fixed point.
ANSWER: Most of you just stated that Ω a fixed point for ω, but didn't explain why. The mere fact that ω Ω doesn't immediately reduce to Ω doesn't show that the two terms are not convertible. However, one can argue that Ω syntactically has two lambdas, and that upon reduction it always yields itself. So it suffices to show that neither ω Ω nor anything it can reduces to can have two lambdas. And indeed it has three lambdas, and anything it reduces to will either have three or four lambdas. For the last part of the question, Y X is a fixed point for any X, as we've already demonstrated in the notes.
2. Consider Ω ξ for an arbitrary term ξ. Ω is so busy reducing itself (the eternal narcissist) that it never gets around to noticing whether it has an argument, let alone doing anything with that argument. If so, how could Ω have a fixed point? That is, how could there be an ξ such that Ω ξ <~~> ξ? To answer this question, begin by constructing Y Ω. Prove that Y Ω is a fixed point for Ω.
ANSWER: Already demonstrated in the notes. We don't need Ω ξ to reduce to ξ, because Y Ω is a ξ that can do its own reducing.
3. Find two different terms that have the same fixed point. That is, find terms F, G, and ξ such that F ξ <~~> ξ and G ξ <~~> ξ. (If you need a hint, reread the notes on fixed points.)
ANSWER: Everything is a fixed point of I, so in particular I is. I is also a fixed point for K I. There are many other examples.
4. Assume that Ψ is some fixed point combinator; we're not telling you which one. (You can just write Psi in your homework if you don't know how to generate the symbol Ψ.) Prove that Ψ Ψ is a fixed point of itself, that is, that Ψ Ψ <~~> Ψ Ψ (Ψ Ψ).
ANSWER: By the definition of a fixed point operator, Ψ Ψ is a fixed point for Ψ; and Ψ (Ψ Ψ) is a fixed point for Ψ Ψ. That is: (a) Ψ (Ψ Ψ) <~~> Ψ Ψ; and (b) Ψ Ψ (Ψ (Ψ Ψ)) <~~> Ψ (Ψ Ψ). Now a fact we did not discuss in class, but which Hankin and other readings do discuss, is that substitution of convertible subterms preserves convertibility (and as part of that, <~~> is transitive). Hence in the lhs of (b), we can substitute Ψ Ψ for the subterm Ψ (Ψ Ψ), because of (a), getting (c) Ψ Ψ (Ψ Ψ) <~~> Ψ (Ψ Ψ). But then by (a) and transitivity of <~~>, we get (d) Ψ Ψ (Ψ Ψ) <~~> Ψ Ψ. Which states that Ψ Ψ is a fixed point for itself.
## Writing recursive functions
1. Helping yourself to the functions given below, write a recursive function called fact that computes the factorial. The factorial n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1. For instance, fact 0 ~~> 1, fact 1 ~~> 1, fact 2 ~~> 2, fact 3 ~~> 6, and fact 4 ~~> 24.
let true = \y n. y in
let false = \y n. n in
let pair = \a b. \v. v a b in
let fst = \a b. a in ; aka true
let snd = \a b. b in ; aka false
let zero = \s z. z in
let succ = \n s z. s (n s z) in
let zero? = \n. n (\p. false) true in
let pred = \n. n (\p. p (\a b. pair (succ a) a)) (pair zero zero) snd in
let add = \l r. r succ l in
let mult = \l r. r (add l) 0 in
let Y = \h. (\u. h (u u)) (\u. h (u u)) in
let fact = ... in
fact 4
ANSWER: let fact = Y (\fact n. (zero? n) 1 (mult n (fact (pred n)))).
2. For this question, we want to implement sets of numbers in terms of lists of numbers, where we make sure as we construct those lists that they never contain a single number more than once. (It would be even more efficient if we made sure that the lists were always sorted, but we won't try to implement that refinement here.) To enforce the idea of modularity, let's suppose you don't know the details of how the lists are implemented. You just are given the functions defined below for them (but pretend you don't see the actual definitions). These define lists in terms of one of the new encodings discussed last week.
; all functions from the previous question, plus
; num_cmp x y lt eq gt returns lt when x<y, eq when x==y, gt when x>y
let num_cmp = (\base build consume. \l r. r consume (l build base) fst)
; where base is
(pair (\a b c. b) (K (\a b c. a)))
; and build is
(\p. pair (\a b c. c) p)
; and consume is
(\p. p fst p (p snd) (p snd)) in
let num_equal? = \x y. num_cmp x y false true false in
let neg = \b y n. b n y in
let empty = \f n. n in
let cons = \x xs. \f n. f x xs in
let empty? = \xs. xs (\y ys. false) true in
let head = \xs. xs (\y ys. y) err in
let tail = \xs. xs (\y ys. ys) empty in
let append = Y (\append. \xs zs. xs (\y ys. (cons y (append ys zs))) zs) in
let take_while = Y (\take_while. \p xs. xs (\y ys. (p y) (cons y (take_while p ys)) empty) empty) in
let drop_while = Y (\drop_while. \p xs. xs (\y ys. (p y) (drop_while p ys) xs) empty) in
...
The functions take_while and drop_while work as described in Week 1's homework.
Using those resources, define a set_cons and a set_equal? function. The first should take a number argument x and a set argument xs (implemented as a list of numbers assumed to have no repeating elements), and return a (possibly new) set argument which contains x. (But make sure x doesn't appear in the result twice!) The set_equal? function should take two set arguments xs and ys and say whether they represent the same set. (Be careful, the lists [1, 2] and [2, 1] are different lists but do represent the same set. Hence, you can't just use the list_equal? function you defined in last week's homework.)
Here are some tips for getting started. Use drop_while, num_equal?, and empty? to define a mem? function that returns true if number x is a member of a list of numbers xs, else returns false. Also use take_while, drop_while, num_equal?, tail and append to define a without function that returns a copy of a list of numbers xs that omits the first occurrence of a number x, if there be such. You may find these functions mem? and without useful in defining set_cons and set_equal?. Also, for set_equal?, you are probably going to want to define the function recursively... as now you know how to do.
Some comments comparing this exercise to The Little Schemer, and Scheme more generally:
• The set_equal? you're trying to define here is like eqset? in Chapter 7 of The Little Schemer, and set_cons x xs would be like (makeset (cons x xs)), from that same chapter.
• mem? and without are like the member? and rember functions defined in Chapter 2 and 3 of The Little Schemer, though those functions are defined for lists of symbolic atoms, and here you are instead defining them for lists of numbers. The Little Schemer also defines multirember, which removes all occurrences of a match rather than just the first; and member* and rember* in Chapter 5, that operate on lists that may contain other, embedded lists.
• The native Scheme function that most resembles the mem? you're defining is memv, though that is defined for more than just numbers, and when that memv finds a match it returns a list starting with the match, rather than #t.
let not_equal? = \n m. neg (num_equal? n m) in
let mem? = \n xs. neg (empty? (drop_while (not_equal? n) xs)) in
let without = \n xs. append (take_while (not_equal? n) xs) (tail (drop_while (not_equal? n) xs)) in
let set_cons = \x xs. (mem? x xs) xs (cons x xs) in
let set_equal? = Y (\set_equal?. \xs ys. (empty? xs)
(empty? ys)
; else when xs aren't empty
(set_equal? (tail xs) (without (head xs) ys))
; else when head xs not in ys
false))
3. Linguists often analyze natural language expressions into trees. We'll need trees in future weeks, and tree structures provide good opportunities for learning how to write recursive functions. Making use of our current resources, we might approximate trees as follows. Instead of words or syntactic categories, we'll have the nodes of the tree labeled with Church numbers. We'll think of a tree as a list in which each element is itself a tree. For simplicity, we'll adopt the convention that a tree of length 1 must contain a number as its only element.
Then we have the following representations:
.
/|\
/ | \
1 2 3
[[1], [2], [3]]
.
/ \
/\ 3
1 2
[[[1], [2]], [3]]
.
/ \
1 /\
2 3
[[1], [[2], [3]]]
Some limitations of this scheme: there is no easy way to label an inner, branching node (for example with a syntactic category like VP), and there is no way to represent a tree in which a mother node has a single daughter.
When processing a tree, you can test for whether the tree is a leaf node (that is, contains only a single number), by testing whether the length of the list is 1. This will be your base case for your recursive definitions that work on these trees. (You'll probably want to write a function leaf? that encapsulates this check.)
Your assignment is to write a Lambda Calculus function that expects a tree, encoded in the way just described, as an argument, and returns the sum of its leaves as a result. So for all of the trees listed above, it should return 1 + 2 + 3, namely 6. You can use any Lambda Calculus implementation of lists you like.
The tricky thing about defining these functions is that it's easy to unwittingly violate the conventions about how we're encoding trees. Thus Leaf1 ≡ [1] is a tree, and [Leaf1, Leaf1, Leaf1] is also a tree, but [Leaf1] is not a tree. (It's a singleton whose content is not a number, but rather a leaf, a list containing a number.) Thus when we are recursively processing [Leaf1, Leaf1, Leaf1], we have to be careful not to just blindly call our function with the tail of our input, since when we get to the end the tail [Leaf1] is not itself a tree, on the conventions we've adopted. And our function is likely to rely on those conventions in such a way that it chokes when they are violated.
That understood, here is a recursive implementation of sum_leaves:
let singleton? = \xs. num_equal? one (length xs) in
let singleton = \x. cons x empty in
let doubleton = \x y. cons x (singleton y) in
let second = \xs. head (tail xs) in
let sum_leaves = Y (\sum_leaves. \t.
(singleton? t)
; else if t is a tree, it contains two or more subtrees
; don't recurse if (tail t) is a singleton
((singleton? (tail t))
(sum_leaves (second t))
; else it's ok to recurse
(sum_leaves (tail t))))) in
...
4. The fringe of a leaf-labeled tree is the list of values at its leaves, ordered from left-to-right. For example, the fringe of all three trees displayed above is the same list, [1, 2, 3]. We are going to return to the question of how to tell whether trees have the same fringe several times this course. We'll discover more interesting and more efficient ways to do it as our conceptual toolboxes get fuller. For now, we're going to explore the straightforward strategy. Write a function that expects a tree as an argument, and returns the list which is its fringe. Next write a function that expects two trees as arguments, converts each of them into their fringes, and then determines whether the two lists so produced are equal. (Convert your list_equal? function from last week's homework into the Lambda Calculus for this last step.)
; are xs strictly longer than ys?
let longer? = \xs ys. neg (leq? (length xs) (length ys)) in
; uncons xs f ~~> f (head xs) (tail xs)
let uncons = \xs f. f (head xs) (tail xs) in
let check = \x p. p (\bool ys. uncons ys (\y ys. pair (and (num_equal? x y) bool) ys)) in
let finish = \bool ys. (empty? ys) bool false in
let list_equal? = \xs ys. (longer? xs ys) false (xs check (pair true (rev ys)) finish) in
let get_fringe = Y (\get_fringe. \t.
; this uses a similar pattern to previous problem
(singleton? t)
t
; else if t is a tree, it contains two or more subtrees
(append
; don't recurse if (tail t) is a singleton
((singleton? (tail t))
(get_fringe (second t))
; else it's ok to recurse
(get_fringe (tail t))))) in
Here is some test data:
let leaf1 = singleton 1 in
let leaf2 = singleton 2 in
let leaf3 = singleton 3 in
let t12 = doubleton leaf1 leaf2 in
let t23 = doubleton leaf2 leaf3 in
let alpha = cons leaf1 t23 in
let beta = doubleton t12 leaf3 in
let gamma = doubleton leaf1 t23 in
list_equal? (get_fringe gamma) (get_fringe alpha)
And here are some cleverer implementations of some of the functions used above:
let box = \a. \v. v a in
let singleton? = \xs. xs (\x b. box (b (K true))) (K false) not in
; this function works by first converting [x1,x2,x3] into (true,(true,(true,(K false))))
; then each element of ys unpacks that stack by applying its fst to its snd and itself
; so long as we've not gotten to the end, this will have the result of selecting the snd each time
; when we get to the end of the stack, ((K false) fst) ((K false) snd) (K false) ~~> K false
; after ys are done iterating, we apply the result to fst, which will give us either true or ((K false) fst) ~~> false
let longer? = \xs ys. ys (\y p. (p fst) (p snd) p) (xs (\x. pair true) (K false)) fst in
let shift = \x t. t (\a b c. triple (cons x a) a (pair x))) in
let uncons = \xs. xs shift (triple empty empty (K err_head)) (\a b c. c b) in
...
## Arithmetic infinity?
The next few questions involve reasoning about Church arithmetic and infinity. Let's choose some arithmetic functions:
succ = \n s z. s (n s z)
add = \l r. r succ l in
mult = \l r. r (add l) 0 in
exp = \base r. r (mult base) 1 in
There is a pleasing pattern here: addition is defined in terms of the successor function, multiplication is defined in terms of addition, and exponentiation is defined in terms of multiplication.
1. Find a fixed point ξ for the successor function. Prove it's a fixed point, i.e., demonstrate that succ ξ <~~> ξ.
We've had surprising success embedding normal arithmetic in the Lambda Calculus, modeling the natural numbers, addition, multiplication, and so on. But one thing that some versions of arithmetic supply is a notion of infinity, which we'll write as inf. This object sometimes satisfies the following constraints, for any finite natural number n:
n + inf == inf
n * inf == inf
n ^ inf == inf
leq n inf == true
(Note, though, that with some notions of infinite numbers, like ordinal numbers, operations like + are defined in such a way that inf + n is different from n + inf, and does exceed inf; similarly for * and ^. With other notions of infinite numbers, like the cardinal numbers, even less familiar arithmetic operations are employed.)
ANSWER: Let H ≡ \u. succ (u u), and X ≡ H H ≡ (\u. succ (u u)) H. Note that X ≡ H H ~~> succ (H H). Hence X is a fixed point for succ.
2. Prove that add ξ 1 <~~> ξ, where ξ is the fixed point you found in (1). What about add ξ 2 <~~> ξ?
Comment: a fixed point for the successor function is an object such that it is unchanged after adding 1 to it. It makes a certain amount of sense to use this object to model arithmetic infinity. For instance, depending on implementation details, it might happen that leq n ξ is true for all (finite) natural numbers n. However, the fixed point you found for succ and (+n) (recall this is shorthand for \x. add x n) may not be a fixed point for (*n), for example.
ANSWER: Prove that add X 1 <~~> X:
add X 1 == (\m n. n succ m) X 1
~~> 1 succ X
== (\s z. s z) succ X
~~> succ X
Which by the previous problem is convertible with X. (In particular, X ~~> succ X.) What about add X 2?
add X 2 == (\m n. n succ m) X 2
~~> 2 succ X
== (\s z. s (s z)) succ X
~~> succ (succ X)
And we know the inner term will be convertible with X, and hence we get that the whole result is convertible with succ X. Which we already said is convertible with X. We can readily see that add X n <~~> X for all (finite) natural numbers n.
## Mutually-recursive functions
1. (Challenging.) One way to define the function even? is to have it hand off part of the work to another function odd?:
let even? = \x. (zero? x)
; if x == 0 then result is
true
; else result turns on whether x-1 is odd
(odd? (pred x))
At the same tme, though, it's natural to define odd? in such a way that it hands off part of the work to even?:
let odd? = \x. (zero? x)
; if x == 0 then result is
false
; else result turns on whether x-1 is even
(even? (pred x))
Such a definition of even? and odd? is called mutually recursive. If you trace through the evaluation of some sample numerical arguments, you can see that eventually we'll always reach a base step. So the recursion should be perfectly well-grounded:
even? 3
~~> (zero? 3) true (odd? (pred 3))
~~> odd? 2
~~> (zero? 2) false (even? (pred 2))
~~> even? 1
~~> (zero? 1) true (odd? (pred 1))
~~> odd? 0
~~> (zero? 0) false (even? (pred 0))
~~> false
But we don't yet know how to implement this kind of recursion in the Lambda Calculus.
The fixed point operators we've been working with so far worked like this:
let ξ = Y h in
ξ <~~> h ξ
Suppose we had a pair of fixed point operators, Y1 and Y2, that operated on a pair of functions h and g, as follows:
let ξ1 = Y1 h g in
let ξ2 = Y2 h g in
ξ1 <~~> h ξ1 ξ2 and
ξ2 <~~> g ξ1 ξ2
If we gave you such a Y1 and Y2, how would you implement the above definitions of even? and odd??
let proto_even = \even odd. \n. (zero? n) true (odd (pred n)) in
let proto_odd = \even odd. \n. (zero? n) false (even (pred n)) in
let even = Y1 proto_even proto_odd in
let odd = Y2 proto_even proto_odd in
...
By the definitions of Y1 and Y2, we know that even <~~> proto_even even odd, and odd <~~> proto_odd even odd. Hence the bound variables even and odd inside the proto_... functions have the values we want. even will be bound to (something convertible with) the underlined portion of proto_even, and odd will be bound to (something convertible with) the underlined portion of proto_odd.
2. (More challenging.) Using our derivation of Y from this week's notes as a model, construct a pair Y1 and Y2 that behave in the way described above.
Here is one hint to get you started: remember that in the notes, we constructed a fixed point for h by evolving it into H and using H H as h's fixed point. We suggested the thought exercise, how might you instead evolve h into some T and then use T T T as h's fixed point. Try solving this problem first. It may help give you the insights you need to define a Y1 and Y2. Here are some hints.
ANSWER: One solution is given in the hint. Here is another:
let Y1 = \f g . (\u v . f (u u v)(v v u))
(\u v . f (u u v)(v v u))
(\v u . g (v v u)(u u v)) in
let Y2 = \f g . Y1 g f in
...
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Rust by Example
18.2 Channels
Rust provides asynchronous channels for communication between threads. Channels allow a unidirectional flow of information between two end-points: the Sender and the Receiver.
use std::sync::mpsc::{Sender, Receiver};
use std::sync::mpsc;
fn main() {
// Channels have two endpoints: the Sender<T> and the Receiver<T>,
// where T is the type of the message to be transferred
// (type annotation is superfluous)
let (tx, rx): (Sender<i32>, Receiver<i32>) = mpsc::channel();
// The sender endpoint can be copied
// Each thread will send its id via the channel
// The thread takes ownership over thread_tx
// Each thread queues a message in the channel
// Sending is a non-blocking operation, the thread will continue
// immediately after sending its message
});
}
// Here, all the messages are collected
let mut ids = Vec::with_capacity(NTHREADS as usize);
// The recv method picks a message from the channel
// recv will block the current thread if there no messages available
}
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Calc based Kinetics problem (Very easy to solve with physics- confused with calc0
1. Dec 15, 2007
lLovePhysics
1. The problem statement, all variables and given/known data
With what intial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (approx 550ft)?
2. Relevant equations
Here's what I know:
s(0)=0
a(t)=-32ft/s^2
$$s(t_{max})=550\\ s'(t_{max})=0$$
3. The attempt at a solution
Actually I got the correct answer but I don't understand something. How do you know whether the constants "C" given by the indefinite integrals are the same?
For example, when you integrate a(t)=s''(t) you get:
s'(t)=-32+C
When you integrate the velocity or s'(t) you get:
$$s(t)=-16t^2+Ct+C$$
So are those C's the same or are they different? How do you know? When I treated them the same I got the correct answer, but when I didn't the answer turned out to be wrong.
Can someone please explain the "constants dilemma?" Thanks.
Last edited: Dec 15, 2007
2. Dec 15, 2007
Dick
They are different. And s'(t)=-32t+C, you left out the t. You fix the constants by making sure that s(0)=0 and the max of s(t) is 550ft.
3. Dec 15, 2007
dynamicsolo
Since you integrated the (constant) acceleration function, you now have the velocity function. The velocity at time t = 0 would be
v(0) = s'(0) = -32·0 + C = C ,
so the "arbitrary constant" becomes your initial velocity v(0). Now,
The position function is now
s(t) = -16(t^2) + v(0)·t + D ,
which at time t = 0 becomes
s(0) = -16·(0^2) + v(0)·0 + D .
So the second arbitrary constant is D = s(0), the initial position. This is where the textbooks get the formulas for constant acceleration (a) kinematics
v(t) = v(0) + at ,
x(t) = x(0) + v(0)·t + (1/2)·a·(t^2) .
Last edited: Dec 15, 2007
4. Dec 15, 2007
lLovePhysics
WOW That's so cool!!! Okay thanks guys. First time seeing things in both the calculus and physics perspectives. lol
5. Dec 15, 2007
dynamicsolo
I really wish the two were simply taught together, since they grew up together. A great deal of mathematical technique and theory for millenia, but particularly over the last four hundred years, was in aid of solving increasingly sophisticated problems in physics and engineering...
6. Dec 15, 2007
lLovePhysics
Yeah, me too. Unfortunately I learned physics before learning calculus, which I'm currently taking. I wish I had learned both together. I'm glad I took physics though, or else I wouldn't understand the physics-based calculus problems as much.
7. Dec 15, 2007
lLovePhysics
Also, I can't believe I never knew how to actually derive these formulas.. I guess I couldn't anyways since I did not know any calculus to begin with.
8. Dec 15, 2007
dynamicsolo
Actually, the constant acceleration kinematic equations were already known in medieval Europe. If you make a graph of a constant acceleration function and ask how the area under it increases as time on the graph advances, you get the velocity equation given above. If you do the same with that function, you get the position equation. (They were doing what we would now called "graphical integration".) Newton developed calculus in order to grapple with problems involving non-constant accelerations.
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# Simple linear algebra proof
1. Feb 10, 2014
### U.Renko
1. The problem statement, all variables and given/known data
find a formula for $\begin{bmatrix} 1 & 1& 1\\ 0& 1& 1\\ 0& 0 & 1 \end{bmatrix} ^n$
and prove it by induction
the induction part is ok.
I'm just having trouble finding a pattern
I may have figured it out but it looks too cumbersome
2. Relevant equations
3. The attempt at a solution
Lets call that matrix A
I computed A^2 through A^5 and noticed a pattern:
$A^2 = \begin{bmatrix} 1 & 2&3\\ 0& 1& 2\\ 0& 0 & 1 \end{bmatrix}$
$A^3 = \begin{bmatrix} 1 & 3& 6\\ 0& 1& 3\\ 0& 0 & 1 \end{bmatrix}$
$a^4 = \begin{bmatrix} 1 & 4& 10\\ 0& 1& 4\\ 0& 0 & 1 \end{bmatrix}$
so the pattern is :
below the diagonal is always 0
the diagonal is always 1
$a_12 = a_23 = n$
$a_13 = some number$ thats where I had trouble figuring the pattern
I noticed that, it is also the sum of the elements in the first row of A^(n-1) but that is a bit awkward to generalize.
Last edited: Feb 10, 2014
2. Feb 10, 2014
### kduna
This is a fun little problem, just do the computation for a couple small n and the the pattern should be easy to pick out.
3. Feb 10, 2014
### U.Renko
ok, here is what I've done and why I said it looked cumbersome
I thought about how $a_{1,3}$ came up in the matrices:
following the multiplicattion of matrices procedure.
it is the sum of $1*1 + 1*(n-1)$ plus 1 times the $a_{1,3}$ element of the $A^{n-1}$ matrix.
thus
if n=2
if n=3
if n=4
so, the element $a_{13}$ of $A^n$ is always $1 + (n-1) + something$
then I took as an example n =4
in this case we have
1+ (4-1) + [1+(4-2) +[1 +(4-3) +[ 1 +[4-4] ] ] ]
in other words
1+ 3+ 1 + 2 + 1+1+1
which is:
4 + (1+2+3)
which I expressed as
$n + \sigma$ where $\sigma = \sum_{i=1}^{n-1}i$
the formula asked then becomes: $\begin{bmatrix} 1 & n & n+ \sigma\\ 0 & 1 & n\\ 0& 0 & 1 \end{bmatrix}$
that is where I thought was too cumbersome and was wondering if there is a simpler way
4. Feb 10, 2014
### kduna
5. Feb 10, 2014
### U.Renko
well, indeed it is.
so now the formula becomes $A^n = \begin{bmatrix} 1 & n & \frac{n(n+1)}{2} \\ 0& 1& n\\ 0&0 & 1 \end{bmatrix}$
and then is just using induction
thanks a lot!
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Reduced graph divisors
Let $G$ be a graph and $v_0 \in V(G)$. A divisor $D$ of $G$ is called $v_0$-reduced if:
1. $f (v) \geq 0$ for all $v ∈ V(G)- v_0$;
2. and for every non-empty set $S \subseteq V (G) − v_0$, there exists a vertex $v \in S$ such that $f (v) \lt d^-_S (v)$ (where $d^-_S (v)$ denotes the number of edges of $v$ leaving $S$).
Remarks
• The definition makes sense for directed graphs but is only interesting in the case of undirected (multi)graphs.
• The following theorem plays a fundamental role in graph divisor theory:
Theorem[1] Let $G$ be an undirected graph and let $v_0 \in V(G)$ be a fixed base vertex. Then for every divisor $D$ of $G$, there exists a unique $v_0$-reduced divisor $D$ such that $D \sim D'$ (here $\sim$ denotes the linear equivalence of graph divisors).
• As a consequence, elements of the Picard group of a graph can be represented by reduced divisors.
References
1. M. Baker, S. Norine, Riemann--Roch and Abel--Jacobi theory on a finite graph, Advances in Mathematics (2007) DOI link, ArXiv Link
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