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## Monte Carlo approximation certificates for k-means clustering

This week, I visited Afonso Bandeira at NYU to give a talk in the MaD seminar on the semidefinite relaxation of k-means. Here are the slides. The last part of the talk is very new; I worked it out with Soledad Villar while she visited me a couple weeks ago, and our paper just hit the arXiv. In this blog entry, I’ll briefly summarize the main idea of the paper.

Suppose you are given data points $\{x_i\}_{i\in T}\subseteq\mathbb{R}^m$, and you are tasked with finding the partition $C_1\sqcup\cdots\sqcup C_k=T$ that minimizes the k-means objective

$\displaystyle{\frac{1}{|T|}\sum_{t\in[k]}\sum_{i\in C_t}\bigg\|x_i-\frac{1}{|C_t|}\sum_{j\in C_t}x_j\bigg\|^2\qquad(T\text{-IP})}$

(Here, we normalize the objective by $|T|$ for convenience later.) To do this, you will likely run MATLAB’s built-in implementation of k-means++, which randomly selects $k$ of the data points (with an intelligent choice of random distribution), and then uses these data points as proto-centroids to initialize Lloyd’s algorithm. In practice, this works very well: After running it a few times, you generally get a very nice clustering. But when do you know to stop looking for an even better clustering?

## Talks from the Summer of ’17

This summer, I participated in several interesting conferences. This entry documents my slides and describes a few of my favorite talks from the summer. Here are links to my talks:

UPDATE: SIAM AG17 just posted a video of my talk.

Now for my favorite talks from FoCM, ILAS, SIAM AG17 and SPIE:

Ben RechtUnderstanding deep learning requires rethinking generalization

In machine learning, you hope to fit a model so as to be good at prediction. To do this, you fit to a training set and then evaluate with a test set. In general, if a simple model fits a large training set pretty well, you can expect the fit to generalize, meaning it will also fit the test set. By conventional wisdom, if the model happens to fit the training set exactly, then your model is probably not simple enough, meaning it will not fit the test set very well. According to Ben, this conventional wisdom is wrong. He demonstrates this by presenting some observations he made while training neural nets. In particular, he allowed the number of parameters to far exceed the size of the training set, and in doing so, he fit the training set exactly, and yet he still managed to fit the test set well. He suggested that generalization was successful here because stochastic gradient descent implicitly regularizes. For reference, in the linear case, stochastic gradient descent (aka the randomized Kaczmarz method) finds the solution of minimal 2-norm, and it converges faster when the optimal solution has smaller 2-norm. Along these lines, Ben has some work to demonstrate that even in the nonlinear case, fast convergence implies generalization.

## Global Guarantees for Enforcing Deep Generative Priors by Empirical Risk

Vlad Voroninski recently posted an arXiv preprint with Paul Hand that provides compressed sensing guarantees using a neural net-based generative signal model. This offers some theoretical justification for the shocking empirical results presented in the “Compressed sensing using generative models” paper, which demonstrates compressed sensing with 10 times fewer measurements than conventional compressed sensing (the source code is available here). I was especially excited to see this paper, having recently read Michael Elad’s editorial on deep learning. To learn more, I interviewed Vlad (see below); I’ve lightly edited his responses for formatting and hyperlinks:

DGM: What is the origin story of this project? Were you and Paul inspired by the “Compressed sensing using generative models” paper?

VV: I have been working extensively with applied deep learning for the last year or so, and have been inspired by recent applications of deep generative image priors to classical inverse problems, such as the super resolution work by Fei Fei Li et al. Moreover, recent work on regularizing with deep generative priors for synthesizing the preferred inputs to neural activations, by Yosinski et al., made me optimistic that GAN-based generative priors are capturing sophisticated natural image structure (the synthetic images obtained in this paper look incredibly realistic).

## A polynomial-time relaxation of the Gromov-Hausdorff distance

Soledad Villar recently posted her latest paper on the arXiv (joint work with Afonso Bandeira, Andrew Blumberg and Rachel Ward). This paper reduces an instance of cutting-edge data science (specifically, shape matching and point-cloud comparison) to a semidefinite program, and then investigates fast solvers using non-convex local methods. (Check out her blog for an interactive illustration of the results.) Soledad is on the job market this year, and I read about this paper in her research statement. I wanted to learn more, so I decided to interview her. I’ve lightly edited her responses for formatting and hyperlinks:

## The Voronoi Means Conjecture

UPDATE (July 26, 2016): Boris Alexeev recently disproved the Voronoi Means Conjecture! In particular, he found that certain stable isogons fail to exhibit the conjectured behavior, and his solution suggests a certain refinement of the conjecture. I asked him to write a guest blog entry about his solution, so expect to hear more in the coming weeks.

Suppose you’re given a sample from an unknown balanced mixture of $k$ spherical Gaussians of equal variance in dimension $d$:

In the above example, $k=3$ and $d=2$. How do you estimate the centers $\{\gamma_i\}_{i=1}^k$ of each Gaussian from the data? In this paper, Dasgupta provides an algorithm in which you project the data onto a randomly drawn subspace of some carefully selected dimension so as to concentrate the data points towards their respective centers. After doing so, there will be $k$ extremely popular regions of the subspace, and for each region, you can average the corresponding points in the original dataset to estimate the corresponding Gaussian center. With this algorithm, Dasgupta proved that

$\displaystyle{\mathrm{MSE}:=\frac{1}{k}\sum_{i=1}^k\|\hat{\gamma}_i-\gamma_i\|^2\lesssim d\sigma^2 \qquad\text{whp}}$

provided $\mathrm{SNR}:=\frac{1}{\sigma^2}\min_{i\neq j}\|\gamma_i-\gamma_j\|^2\gtrsim d$.

## On the low-rank approach for semidefinite programs arising in synchronization and community detection

From MaxCut to PhaseLift, semidefinite programming has proven to be rather powerful, especially for convex relaxation. SDP solvers take polynomial time, but the exponent is large, and anyone who’s run an SDP on CVX has experienced some frustration with the runtime. In practice, the SDP-optimal matrix tends to have extremely low rank, and so one may apply a rank constraint to facilitate the search for the SDP’s solution. This heuristic was first introduced by Burer and Monteiro, and it works well in practice, but the rank-constrained program is nonconvex and the theory is scant. Recently, the theory gap started to close with this paper:

On the low-rank approach for semidefinite programs arising in synchronization and community detection

Afonso S. Bandeira, Nicolas Boumal, Vladislav Voroninski

As the title suggests, this paper provides strong performance guarantees for the Burer-Monteiro heuristic in the particular cases of $\mathbb{Z}_2$ synchronization and community detection. I was very excited to see this paper, and so I interviewed one of the authors (Nicolas Boumal). I’ve lightly edited his responses for formatting and hyperlinks:

## Clustering noisy data with semidefinite relaxations

Soledad Villar recently posted our latest paper on the arXiv (this one coauthored by her advisor, Rachel Ward). The paper provides guarantees for the k-means SDP when the points are drawn from a subgaussian mixture model. This blog entry will discuss one of the main ideas in our analysis, which we borrowed from Guedon and Vershynin’s recent paper.

The first application comes from graph clustering. Consider the stochastic block model, in which the $n$ vertices are secretly partitioned into two communities, each of size $n/2$, and edges between vertices of a common community are drawn iid with some probability $p$, and all other edges are drawn with probability $q. The goal of community estimation is to estimate the communities given a random draw of the graph. For this task, you might be inclined to find the maximum likelihood estimator for this model, but this results in an integer program. Relaxing the program leads to a semidefinite program, and amazingly, this program is tight and recovers the true communities with high probability when $p=(\alpha\log n)/n$ and $q=(\beta\log n)/n$ for good choices of $(\alpha,\beta)$. (See this paper.) These edge probabilities scale like the threshold for connected Erdos-Renyi graphs, and this makes sense since we wouldn’t know how to assign vertices in isolated components. If instead, the probabilities were to scale like $1/n$, then we would be in the “giant component” regime, so we’d still expect enough signal to correctly assign a good fraction of the vertices, but the SDP is not tight in this regime.


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## College Algebra (10th Edition)

Yes, $\overline{\overline{z}}=z$
We are given $z=a+bi$. We solve for $\overline{\overline{z}}$: $\overline{\overline{z}}=\overline{\overline{a+bi}}=\overline{a-bi}=a+bi=z$

format: plain


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The Importance of Moral Information

This article will argue that information is heterogeneous. It will further argue that a particular kind of information is moral information and that information is a key driver of consumer and producer preference. These preferential changes can result in significant shifts of entire supply and demand curves.

Let’s say you are told something. You are told that apples cost $.50 per pound and bananas cost$.40 per pound. These statements are information. In themselves they do not determine if you will make a purchase. Let’s say we know that your budget is \$4.00. We can now create a budget line and combine that with your preference curves to determine exactly what purchase, if any, that you will make. This is neoclassical economics as usual.

The determination of these preference curves is not neoclassical economics as usual. Economists today realize that these curves will not form a reverse L-shape, but that’s about all they understand. The way these curves are calculated is usually by presuming that the past patterns will hold into the future but this is not always true, and it is further problematic when dealing with a double innovation or something that has no past precedent.

Image credit: http://ingrimayne.com/econ/MaximizingBeha/DerivingDemand.html

Preference curves derive from moral information. Information about what should be done. Moral information is equivalent to preferential information. Let me be clear about the fact that moral information is usually not correct moral information. ‘Moral’ here is simply referring to what should be done. In this view a moral belief is the same as accepted moral information, which is also the same as a preference. The less complicated idea here is that people want to do the things that they think they should do. This is in some ways similar to and in other ways different from the rational actor model of economics.

Remember the information on fruit cost. Now let’s propose that the consumer’s belief is that a balanced diet between the two is preferable, or that the most calories per dollar is preferable, or that bananas are clearly always preferable because they have potassium, or that the consumer is allergic to bananas and should always buy only apples and so on. Each of these scenarios will result in some particular indifference curve.

This method is useful for at least four reasons. First, this method can be used as a independent check on the validity of the implications of the neoclassical method. Second, this method can sometimes be used for double innovations, or things which have no precedent. Third, this method is sometimes easier to analyze, although it is also sometimes harder. For example, research for my method can be conducted through survey of beliefs rather than observation of action. Lastly, my method implies that a major way to change supply and demand curves in a whole economy is by affecting preferences through related moral information rather than any kind of information.

Spreading information through an advertisement on low prices, for example, will not change consumer behavior unless the consumer behavior is contingent on prices. Usually it is, but sometimes it is not. If you are allergic to bananas it doesn’t matter how cheap they get! If you are opposed to guns or abortions, would a discount really make you change your mind? Here we discover something like an price elasticity of behavior where some actions are more or less contingent on price, while others are virtually unresponsive to changes in price. These critical points can be addressed through moral and preferential analysis but not through neoclassical mechanisms.

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GATE CSE
Theory of Computation
Finite Automata and Regular Language
Previous Years Questions

## Marks 1

Which one of the following regular expressions correctly represents the language of the finite automation given below? ...
Which one of the following regular expressions represents the set of all binary strings with an odd number of 1’s?
Consider the following statements. I. If L1 $$\cup$$ L2 is regular, then both L1 and L2 must be regular. II. The class of regular languages is close...
If L is a regular language over Σ = {a,b}, which one of the following languages is NOT regular?
For Σ = {a, b}, let us consider the regular language L = {x | x = a2+3k or x = b10+12k, k ≥ 0}. Which one of the following can be a pumping length (th...
The number of states in the minimum sized $$DFA$$ that accepts the language defined by the regular expression $${\left( {0 + 1} \right)^ * }\left( {... Language$${L_1}$$is defined by the grammar:$$S{}_1 \to a{S_1}b|\varepsilon $$Language$${L_2}$$is defined by the grammar:$$S{}_2 \to ab{S_2}|\va...
Which one of the following regular expressions represents the language: the set of all binary strings having two consecutive $$0s$$ and two consecutiv...
Let $$L$$ be the language represented by the regular expression $$\sum {^ * 0011\sum {^ * } }$$ where $$\sum { = \left\{ {0,1} \right\}} .$$ What is ...
If $${L_1} = \left\{ {{a^n}\left| {n \ge \left. 0 \right\}} \right.} \right.$$ and $${L_2} = \left\{ {{b^n}\left| {n \ge \left. 0 \right\}} \right.} \... The length of the shortest string NOT in the language (over$$\sum { = \left\{ {a,\,\,b} \right\}} $$) of the following regular expression is ________... Which one of the following is TRUE? Consider the finite automation in the following figure. What is the set of reachable states for the input string$$0011?$$... Consider the languages$${L_1} = \phi $$and$${L_2} = \left\{ a \right\}.$$Which one of the following represents$${L_1}\,L_2^ * UL_1^ * ?$$What is the complement of the language accepted by the$$NFA$$shown below? Assume$$\sum { = \left\{ a \right\}\,\,} $$and$$\varepsilon $$is the ... Let$${L_1}$$recursive language. Let$${L_2}$$and$${L_3}$$be languages that are recursively enumerable but not recursive. Which of the following s... Which one of the following languages over the alphabet$$\left\{ {0,\left. 1 \right)} \right.$$is described by the regular expression$${\left( {0 + ...
The regular expression $${0^ * }\left( {{{10}^ * }} \right){}^ *$$denotes the same set as
Consider the set $$\sum {^ * }$$ of all strings over the alphabet $$\,\sum { = \,\,\,\left\{ {0,\,\,\,1} \right\}.\sum {^ * } }$$ with the concatena...
Consider the following two statements; $${S_1}\,\,:\,\,\left\{ {{0^{2n}}\left| {n \ge 1} \right.} \right\}$$ is a regular language $${S_2}\,\,:\,\,\le... Given an arbitrary non-deterministic finite automaton$$(NFA)$$with$$N$$states, the maximum number of states in an equivalent minimized$$DFA$$is ... Let$$S$$and$$T$$be languages over$$\sum { = \left\{ {a,b} \right\}} $$represented by the regular expressions$${\left( {a + {b^ * }} \right)^ * ...
Let $$L$$ denote the language generated by the grammar $$S \to 0S\left. 0 \right|00.$$ Which one of the following is true?
Consider the regular expression $$(0+1)(0+1).......n$$ times. The minimum state finite automation that recognizes the language represented by this reg...
If the regular set $$A$$ is represented by $$A = {\left( {01 + 1} \right)^ * }$$ and the regular set $$'B'$$ is represented by $$B = \left( {{{\left( ... Which of the following sets can be recognized by a Deterministic Finite-state Automation? The string$$1101$$does not belong to the set represented by How many substrings of different lengths (non-zero) can be formed from a character string of length$$n$$?$$\sum { = \left\{ {a,b} \right\},\,\,} $$which one of the following sets is not countable. Which two of the following four regular expressions are equivalent? (i)$${\left( {00} \right)^ * }\left( {\varepsilon + 0} \right)$$(ii)$${\left(...
Let $$L \subseteq \sum {^{^ * }\,}$$ where $$\,\sum { = \,\,\left\{ {a,b} \right\}\,\,}$$ which of the following is true?
State True or False with one line explanation: A FSM (Finite State Machine) can be designed to add two integers of any arbitrary length (arbitrary num...

## Marks 2

Consider the following language. L = { w ∈ {0, 1}* | w ends with the substring 011} Which one of the following deterministic finite automata ac...
Consider the following language. L = {x $$\in$$ {a, b}* | number of a’s in x is divisible by 2 but not divisible by 3} The minimum number of states ...
Let $$N$$ be an $$NFA$$ with $$n$$ states. Let $$k$$ be the number of states of a minimal $$DFA$$ which is equivalent to $$N.$$ Which one of the follo...
Given a language $$𝐿,$$ define $${L^i}$$ as follows: $${L^0} = \left\{ \varepsilon \right\}$$ $${L^i} = {L^{i - 1}}.\,\,L$$ for all $$i > 0$$ ...
Consider the following two statements: $$\,\,\,\,\,\,\,{\rm I}.\,\,\,\,\,$$ If all states of an $$NFA$$ are accepting states then the language accepte...
Consider the following languages: \eqalign{ & {L_1} = \left\{ {{a^n}{b^m}{c^{n + m}}:m,n \ge 1} \right\} \cr & {L_2} = \left\{ {{a^n... Consider the transition diagram of aPDA$$given below with input alphabet$$\sum {\, = \left\{ {a,b} \right\}} $$and stack alphabet$$\Gamma = \l...
The number of states in the minimal deterministic finite automaton corresponding to the regular expression $${\left( {0 + 1} \right)^{\,\, * }}\left( ... Which of the following languages is/are regular?$${L_1}:\left\{ {wx{w^R}|w,x\, \in \left\{ {a,b} \right\}{}^ * } \right.$$and$$\left. {\left| w \ri...
Consider the alphabet $$\sum { = \left\{ {0,1} \right\},}$$ the null/empty string $$\lambda$$ and the sets of strings $${X_0},\,{X_1},$$ and $${X_2}... Consider the DFAs M and N given above. The number of states in a minimal DFA that accepts the language L(M) ∩ L(N) is___________.... Consider the NPDA$$\left\langle {Q = \left\{ {{q_0},{q_1},{q_2}} \right\}} \right.,\Sigma = \left \{ 0, 1 \right \},\Gamma = \left \{ 0, 1,...
Let $${L_1} = \left\{ {w \in \left\{ {0,1} \right\}{}^ * \left| w \right.} \right.$$ has at least as many occurrences of $$(110)'s$$ as $$(011)'s$$$$... Let$${L_1} = \left\{ {w \in \left\{ {0,1} \right\}{}^ * \left| w \right.} \right.$$has at least as many occurrences of$$(110)'s$$as$$(011)'s...
Which of the regular expression given below represent the following $$DFA?$$ ...
Consider the following languages $${L_1} = \left\{ {{0^p}{1^q}{0^r}\left| {p,q,r \ge 0} \right.} \right\}$$ $${L_2} = \left\{ {{0^p}{1^q}{0^r}\left| ... Consider the set of strings on$$\left\{ {0,1} \right\}$$in which, every substring of$$3$$symbols has at most two zeros. For example,$$001110$$an... Definition of the language$$L$$with alphabet$$\left\{ a \right\}$$is given as following.$$L = \left\{ {{a^{nk}}} \right.\left| {k > 0,\,n} \ri...
Let $$L = \left\{ {w \in {{\left( {0 + 1} \right)}^ * }\left| {\,w} \right.} \right.$$ has even number of $$\,\left. {1's} \right\},$$ i.e $$L$$ is th...
Let $$w$$ be any string of length $$n$$ in $${\left\{ {0,1} \right\}^ * }$$. Let $$L$$ be the set of all substrings of $$w.$$ What is the minimum numb...
$$L = {L_1} \cap {L_2}$$ where $${L_1}$$ and $${L_2}$$ are languages defined as follows. $${L_1} = \left\{ {{a^m}{b^m}\,c\,{a^n}{b^n}\left| {m,n \ge... The above$$DFA$$accepts the set of all strings over$$\left\{ {0,\,\,1} \right\}$$that... Given below are two finite state automata ($$ \to $$indicates the start state and$$F$$indicates a final state) Which of the following represen... Match the following$$NFAs$$with the regular expressions they correspond to ... Which of the following are regular sets? ... A minimum state deterministic finite automation accepting the language$$L = \left\{ {w\left| {w \in } \right.\,\,{{\left\{ {0,1} \right\}}^ * },\,\,}...
Which of the following languages is regular?
Consider the following finite state automation The language accepted by this automation is given by the regular expression...
Consider the following finite state automation The minimum state automation equivalent to the above $$FSA$$ has the following number of states...
If $$s$$ is a string over $${\left( {0 + 1} \right)^ * }$$ then let $${n_0}\left( s \right)$$ denote the number of $$0'$$ s in $$s$$ and $${n_1}\left(... Consider the regular language$$L = {\left( {111 + 11111} \right)^ * }.$$The minimum number of states in any$$DFA$$accepting this language is Consider the machine$$M:$$The language recognized by$$M$$is: ... The following finite state machine accepts all those binary strings in which the number of$$1's$$and$$0's$$are respectively ... Consider the$$NFAM$$shown below. Let the language accepted by$$M$$be$$L.$$Let$${L_1}$$be the language accepted by the$$NFA$$,$${M_1}\$...
Consider the following deterministic finite state automation $$M.$$ Let $$S$$ denote the set of seven bit binary strings in which the first, the fo...
The smallest finite automaton which accepts the language $$L = \left. {\left\{ x \right.} \right|$$ length of $$x$$ is divisible by $$\left. 3 \righ... The Finite state machine described by the following state diagram with$$A$$as starting state, where an arc label is$$x/y$$and$$x$$stands for$$1...
Consider a $$DFA$$ over $$\sum { = \left\{ {a,\,\,b} \right\}}$$ accepting all strings which have number of $$a'$$s divisible by $$6$$ and number of ...
Consider the following languages: $${L_1} = \left\{ {w\,w\left| {w \in {{\left\{ {a,\,b} \right\}}^ * }} \right.} \right\}$$ $${L_2} = \left\{ {w\,{w^... What can be said about a regular language$$L$$over$$\left\{ a \right\}$$whose minimal finite state automation has two states? Let$$L$$be the set of all binary strings whose last two symbols are the same. The number of states in the minimum state deterministic finite-state a... Which one of the following regular expressions over$$\left\{ {0,\,\,1} \right\}$$denotes the set of all strings not containing$$100$$as substring? A finite state machine with the following state table has a single input$$X$$and a single out$$Z$$. If the initial state is unknown, then the sh... Which of the following definitions below generates the same language as$$L$$Where$$L = {\left\{ x \right.^n}{y^n}\left| {n \ge \left. 1 \right\}} ...
The number of sub-strings (of all lengths inclusive) that can be formed from a character string of length $$n$$ is
The regular expression for the language recognized by the finite state automation of is _________. ...
Which of the following regular expression identities are true?
If $$G$$ is a context-free grammar and $$w$$ is a string of length $$n$$ in $$L(G),$$ how long is a derivation of $$w$$ in $$G,$$ if $$G$$ is Chomsky ...
Let $$r = 1\,{\left( {1 + 0} \right)^ * },s = {11^ * }\,0$$ and $$\,t = {1^ * }\,0$$ be three regular expressions. Which one of the following is true?
Let $${R_1}$$ and $${R_2}$$ be regular sets defined over the alphabet $$\sum \,$$ then:
How many substrings (of all lengths inclusive ) can be formed from a character string of length $$n$$? Assume all characters to be distinct. Prove you...

## Marks 5

Given that language $${L_1}$$ is regular and that the language $${L_1} \cap {L_2}$$ is regular is the language $${L_2}$$ is always regular?
Is the class of regular sets closed under infinite union? Explain.
Give the regular expression over $${\left\{ {0,\,\,1} \right\}}$$ to denote the set of proper non-null substrings of the string $$0110$$.
Give minimal $$DFA$$ that performs as a Mod-$$3$$ $$1's$$ counter, i.e., outputs a $$1$$ each time the number of $$1's$$ in the input sequence is a se...
EXAM MAP
Joint Entrance Examination


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# How do you solve the system x + 4y = 28 and - x + 4y = 20?

Jun 14, 2015

$\left(x , y\right) = \left(4 , 6\right)$

#### Explanation:

[1]$\textcolor{w h i t e}{\text{XXXX}}$$x + 4 y = 28$
[2]$\textcolor{w h i t e}{\text{XXXX}}$$- x + 4 y = 20$

[3]$\textcolor{w h i t e}{\text{XXXX}}$$8 y = 48$
dividing by 8
[4]$\textcolor{w h i t e}{\text{XXXX}}$$y = 6$

substituting $6$ for $y$ (from [4]) in [1]
[5]$\textcolor{w h i t e}{\text{XXXX}}$$x + 4 \left(6\right) = 28$
simplifying
[6]$\textcolor{w h i t e}{\text{XXXX}}$$x = 4$


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# zbMATH — the first resource for mathematics

Classification of unitary representations in irreducible representations of general linear group (non-Archimedean case). (English) Zbl 0614.22005
The author classifies the irreducible unitary representations of the general linear group GL(n) over a local, non-Archimedean field. He constructs a set B so that every unitary representation is either in B or is unitarily induced from a representation in B. The representations in B are either irreducible quotients of representations induced from square integrable representations or complementary series representations. The classifications of the unitary dual is given in Langlands as well as in Zelevinski parameters.
A similar classification of the unitary dual for Archimedean fields was obtained by D. Vogan.
Reviewer: B.Speh

##### MSC:
 2.2e+51 Representations of Lie and linear algebraic groups over local fields
Full Text:


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# The Mattress Group

Dr. Strogatz

Steven Strogatz is a professor of applied math at Cornell University.  In addition to doing lots of interesting math, he writes a column for the New York Times about mathematics.

His most recent article is about group theory in the bedroom (which is a lot less racy than it sounds!).  As you know, you are supposed to periodically rotate and/or flip your mattress that it doesn’t get squished flat from people sleeping in the same spot each night.  Dr. Strogatz discusses the fact that if you have a standard rectangular mattress, then there are a total of four possible positions that the mattress can be in (either upside-up or upside-down and either your head on one end or your head on the other end).  But, as Dr. Strogatz discusses, what is actually even more interesting is the actual operations which move the mattress from position to position.  The basic ones are to either rotate the mattress or to flip the mattress, and all the rest are some combination of these two.

Since if you do one of these operations and then another one, you get a third such operation, it turns out that the collection of all four of these forms a group.  Just by checking the various possibilities, it turns out that there are two groups with four elements:

$\mathbb{Z}_{4} \text{ and } \mathbb{Z}_2 \times \mathbb{Z}_2$

The first is the integers with addition modulo 4 (you might have learned it in Discrete Math as modular arithmetic) and the second one is the Klein Four Group.  So which one is the Mattress Group?  Well, in $\mathbb{Z}_4$ there is an element (the number $\bar{1}$) which you can do with itself (ie. add to itself) once, twice, thrice, or four times and each time get a different answer.  But it’s not too hard to see that in the Mattress Group, if you do any operation twice, the mattress is back to where you started!

via the NYTimes

So, by process of elimination, the Mattress Group is the Klein Four Group! (P.S.  Technically, it’s isomorphic to the Klein Four Group.)

You should definitely go to the NYTimes and read the whole article and Dr. Strogatz’s other interesting essay!  For example, check out this one on the Pythagorean Theorem which is closely related to the Primitive Integral Triangles (PITs) that Dr. Miller talked about last month in Math Club.

Thanks to Dr. Ravi Shankar for pointing out the article by Dr. Strogatz.  Dr. Shankar also pointed out that if you have a square mattress or a pillow top, then you get other interesting Mattress Groups.  Which makes us think that Dr. Shankar has a nicer bed than we do!


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• anonymous
You decide to put $100 in a savings account to save for a$3,000 down payment on a new car. If the account has an interest rate of 2% per year and is compounded monthly, how long does it take you to earn \$3,000 without depositing any additional funds? A. 170.202 years B. 14.3129 years C. 171.755 years D. 168.354 years
Mathematics

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# Séminaires : Séminaire Théorie des Nombres

 Equipe(s) : fa, tn, tga, Responsables : Marc Hindry, Bruno Kahn, Wieslawa Niziol, Cathy Swaenepoel Email des responsables : cathy.swaenepoel@imj-prg.fr Salle : Adresse : Description http://www.imj-prg.fr/tn/STN/stnj.html

 Orateur(s) Adebisi Agboola - , Titre Restricted Selmer groups and special values of p-adic L-functions Date 01/02/2010 Horaire 14:00 à 16:00 Diffusion Résume Let $E/Q$ be an elliptic curve with complex multiplication by thering of integers of an imaginary quadratic field $K$. In 1991, bystudying a certain special value of the Katz two-variable $p$-adic $L$-function lying outside the range of $p$-adic interpolation, K. Rubin formulated a $p$-adic variant of the Birch and Swinnerton-Dyer conjecture when $E(K)$ is infinite, and he proved that his conjecture is true for $E(K)$ of rank one. When $E(K)$ is finite, however, the statement of Rubin's original conjecture no longer applies, and the relevant special value of the appropriate $p$-adic $L$-function is equal to zero. We shall explain what happens in this case. We shall also describe what happens in an analogous situation associated to CM modular forms of higher weight. Salle Adresse


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# The equation 3 x 2 + 3 y 2 + 6 x − y = 0 represents a circle and evaluate the center and radius of the circle.

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.8, Problem 108E
To determine

## To show: The equation 3x2+3y2+6x−y=0 represents a circle and evaluate the center and radius of the circle.

Expert Solution

The equation 3x2+3y2+6xy=0 represents a circle. The center of the circle is (1,16) and radius is 3736 units.

### Explanation of Solution

Given information:

The equation 3x2+3y2+6xy=0 .

Formula used:

In order to solve a quadratic equation, completing the square method is used that transforms the equation in the form of square trinomial.

Step1: Divide the equation by coefficient of x2 if it is not equal to 1.

Step 2: Take the square of the half of the coefficient of x and add it to both sides of the equation.

Step 3: Factor the equation.

The standard form of the equation of the circle is (xh)2+(yk)2=r2 , where (h,k) denote the center of the circle and r denote the radius.

Calculation:

Consider the equation 3x2+3y2+6xy=0 .

Recall that in order to solve a quadratic equation, completing the square method is used that transforms the equation in the form of square trinomial.

Step1: Divide the equation by coefficient of x2 if it is not equal to 1.

Step 2: Take the square of the half of the coefficient of x and add it to both sides of the equation.

Step 3: Factor the equation.

In the provided equation, divide the entire equation by 3,

x2+y2+2xy3=0

Add 1 to both the sides of the equation,

x2+y2+2xy3+1=0+1

Again add 136 to both the sides of the equation,

x2+y2+2xy3+1+136=0+1+136

Group the terms,

x2+y2+2xy3+1+136=0+1+136(x2+2x+1)+(y2y3+136)=3736

Factor out the trinomial, recall that (a+b)2=a2+2ab+b2 and (ab)2=a22ab+b2 .

Apply it,

x2+y2+2xy3+1+136=0+1+136(x2+2x+1)+(y2y3+136)=3736(x+1)2+(y16)2=3736

Recall that the standard form of the equation of the circle is (xh)2+(yk)2=r2 , where (h,k) denote the center of the circle and r denote the radius.

Convert the equation obtained above in standard form,

x2+y2+2xy3+1+136=0+1+136(x2+2x+1)+(y2y3+136)=3736(x+1)2+(y16)2=3736(x(1))2+(y16)2=(376)2

Compare, (xh)2+(yk)2=r2 and (x(1))2+(y16)2=(376)2 .

Here, h=1,k=16 and r=3736 .

Therefore, center of circle is (1,16) and radius is 3736 .

Thus, the equation 3x2+3y2+6xy=0 represents a circle. The center of the circle is (1,16) and radius is 3736 units.

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text:
This is a line editing library for UNIX. It can be linked into almost any program to provide command line editing and history. It is call compatible with the FSF readline library, but is a fraction of the size (and offers fewer features).

The small size (<30k), lack of dependencies (no ncurses needed!) and the free license should make this library interesting to many embedded developers seeking a replacement for the GNU readline library.

Configuration is made by supplying different options to the GNU configure script. In the examples/ directory you can find some small code snippets used for testing.

Issue tracker and GIT repository available at GitHub:

See also the OpenHub page, or the old Free(code) page.

## Origin & References

The editline library was created by Simmule Turner and Rich Salz back in 1992. At the time they chose to distribute the code under a “C News-like” copyright, see the file LICENSE for details. Today Rich disitributes the original sources under the Apache 2.0 license. The version distributed here, however, continues to use the original license.

This version of the editline library is the continuation of the Minix3 sources, where it is called libedit. Other known versions, often based on the original comp.sources.unix posting are:

The most intersting patches and bug fixes from each fork have been merged here. Outstanding issues are listed in the TODO file.


text:
# How do you integrate  (e^(2x-1))-1?

$= \frac{1}{2} {e}^{2 x - 1} - x + C$
$\int \setminus \left({e}^{2 x - 1}\right) - 1 \setminus \mathrm{dx}$
$= \int \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{2} {e}^{2 x - 1}\right) - 1 \setminus \mathrm{dx}$
$= \frac{1}{2} {e}^{2 x - 1} - x + C$


text:
# A bathtub contains 45 gallons of water and the total weight of the tub and water is approximately...

## Question:

A bathtub contains 45 gallons of water and the total weight of the tub and water is approximately 760.725 pounds. You pull the plug and the water begins to drain. Let v represent the number of gallons of waters that has drained from the tub since the plus was pulled. Noe that water weighs 8.345 pounds per gallon.

a. Write an expression in terms of v that represents the weight of the water that has drained from the tub (in pounds).

b. Write an expression in terms of v that represents the total weight of the tub and water (in pounds).

c. How much does the tub weight when there is no water in it?

d. If the weight of the tub and water is 618.41 pounds, how many gallons of water are in the tub?

## Algebraic Expressions

In this question, we form algebraic expressions at each step to solve the question. When an expression is equated to a value, we have an equation as well.

a) The weight of the water that must have drained out will be the product of the quantity of water that has drained out (v gallons) and the weight of water per gallon (8.345 pounds). This is:

$$8.345v$$

b) The total weight when v gallons have drained out will be the difference between the initial weight and the weight that has reduced. So,

$$760.725-8.345v$$

c) When there is not water, the amount of water that must have been drained is v=45 gallons. So, the weight of the tub is:

$$760.725-8.345*45=385.2\text{ pounds}$$

d) First, let's find the amount of water that must have drained out by solving the following equation.

\begin{align} 760.725-8.345v&=618.41\\ -8.345v&=618.41-760.725\\ v&=\frac{142.315}{8.345}\\ &=17.05 \end{align}

If 17.05 gallons has drained out, the amount remaining is {eq}45-17.05=27.95\text{ gallons} {/eq}.


text:
## Building a Student Intervention Model

As education has grown to rely more on technology, vast amounts of data has become available for examination and prediction. Logs of student activities, grades, interactions with teachers and fellow students, and more, are now captured in real time through learning management systems like Canvas and Edmodo. Within all levels of education, there exists a push to help increase the likelihood of student success, without watering down the education or engaging in behaviors that fail to improve the underlying issues. Graduation rates are often the criteria of A local school district has a goal to reach a 95% graduation rate by the end of the decade by identifying students who need intervention before they drop out of school. We will build a model that predicts how likely a student is to pass their high school final exam: **the model must be effective while using the least amount of computation costs.

# Classification vs. Regression

This is a binary classification problem with a labeled dataset. Our model must output either “1” if the student passed the final exam or “0” otherwise.

# Exploring the Data

Let’s begin by investigating the dataset to determine how many students we have information on, and learn about the graduation rate among these students.

[+] Total number of students: 395
[+] Number of features: 30
[+] Number of students who passed: 265
[+] Number of students who failed: 130
[+] Graduation rate of the class: 67.09%

[+] There is more Female than Male students in the dataset:
F: 208 | M: 187 [+] The number of grad Female students is about the same as that of Male students:
[+] The Male students have a higher graduation rate: F: 0.64|M: 0.71
[+] The ratio of students attending G. Pereira versus Mousinho is 7 : 1
[+] Graduation rate at Gabriel Pereira: 0.68 | Mousinho da Silveira 0.63
[+] The graduation rate is higher for students attending Pereira: 0.05 points higher

## Visualization of a few data

More observations: Most students with age 15 to 21 pass the final exam, with the exception of the 19 yo students (grad rate < 0.5. Note that there is only 1 student of age 21 and 1 student of age 22. There are more Urban than Rural students. The graduation rate of Urban students is higher than that of Rural Students

# Preparing the Data

In this section, we will prepare the data for modeling, training and testing. It is often the case that the data you obtain contains non-numeric features. This can be a problem, as most machine learning algorithms expect numeric data to perform computations with.

[+] Feature columns
[‘school’, ‘sex’, ‘age’, ‘address’, ‘famsize’, ‘Pstatus’, ‘Medu’, ‘Fedu’, ‘Mjob’, ‘Fjob’, ‘reason’,
‘guardian’, ‘traveltime’, ‘studytime’, ‘failures’, ‘schoolsup’, ‘famsup’, ‘paid’, ‘activities’, ‘nursery’,
‘higher’, ‘internet’, ‘romantic’, ‘famrel’, ‘freetime’, ‘goout’, ‘Dalc’, ‘Walc’, ‘health’, ‘absences’]

 school sex age address famsize Pstatus Medu Fedu Mjob Fjob higher internet romantic famrel freetime goout Dalc Walc health absences 0 GP F 18 U GT3 A 4 4 at_home teacher yes no no 4 3 4 1 1 3 6 1 GP F 17 U GT3 T 1 1 at_home other yes yes no 5 3 3 1 1 3 4 2 GP F 15 U LE3 T 1 1 at_home other yes yes no 4 3 2 2 3 3 10 3 GP F 15 U GT3 T 4 2 health services yes yes yes 3 2 2 1 1 5 2 4 GP F 16 U GT3 T 3 3 other other yes no no 4 3 2 1 2 5 4

[5 rows x 30 columns]
There are several non-numeric columns that need to be converted! Many of them are simply ‘yes’/’no’, e.g. ‘internet’. These can be reasonably converted into ‘1’/’0’ (binary) values. Other columns, like ‘Mjob’ and ‘Fjob’, have more than two values, and are known as categorical variables. The recommended way to handle such a column is to create as many columns as possible values (e.g. ‘Fjob_teacher’, ‘Fjob_other’, ‘Fjob_services’, etc.), and assign a ‘1’ to one of them and ‘0’ to all others. These generated columns are sometimes called dummy variables, and we will use the pandas.get_dummies() function to perform this transformation.

Processed feature columns (48 total features): [‘school_GP’, ‘school_MS’, ‘sex_F’, ‘sex_M’, ‘age’, ‘address_R’, ‘address_U’, ‘famsize_GT3’, ‘famsize_LE3’, ‘Pstatus_A’, ‘Pstatus_T’, ‘Medu’, ‘Fedu’, ‘Mjob_at_home’, ‘Mjob_health’, ‘Mjob_other’, ‘Mjob_services’, ‘Mjob_teacher’, ‘Fjob_at_home’, ‘Fjob_health’, ‘Fjob_other’, ‘Fjob_services’, ‘Fjob_teacher’, ‘reason_course’, ‘reason_home’, ‘reason_other’, ‘reason_reputation’, ‘guardian_father’, ‘guardian_mother’, ‘guardian_other’, ‘traveltime’, ‘studytime’, ‘failures’, ‘schoolsup’, ‘famsup’, ‘paid’, ‘activities’, ‘nursery’, ‘higher’, ‘internet’, ‘romantic’, ‘famrel’, ‘freetime’, ‘goout’, ‘Dalc’, ‘Walc’, ‘health’, ‘absences’]

 school sex_M sex_F age address_R address_U famsize_GT3 famsize_LE3 Pstatus_A higher internet romantic famrel freetime goout Dalc Walc health absences 0.0 1.0 1.0 0.0 18 0.0 1.0 1.0 0.0 1.0 1.0 0.0 0.0 4.0 3.0 4.0 1.0 1.0 3.0 6.0 1.0 1.0 1.0 0.0 17 0.0 1.0 1.0 0.0 0.0 1.0 1.0 0.0 5.0 3.0 3.0 1.0 1.0 3.0 4.0 2.0 1.0 1.0 0.0 15 0.0 1.0 0.0 1.0 0.0 1.0 1.0 0.0 4.0 3.0 2.0 2.0 3.0 3.0 10.0 3.0 1. 1.0 0.0 15.0 0.0 1.0 1.0 0.0 0.0 1.0 1.0 1.0 3.0 2.0 2.0 1.0 1.0 5.0 2.0 4.0 1.0 1.0 0.0 16.0 0.0 1.0 1.0 0.0 0.0 1.0 0.0 0.0 4.0 3.0 2.0 1.0 2.0 5.0 4.0

[5 rows x 48 columns]

# Implementation: Training and Testing Data Split

Now, we split the data (both features and corresponding labels) into training and test sets. In the following code cell below, you will need to implement the following:
[+] Randomly shuffle and split the data (‘X_all’, ‘y_all’) into training and testing subsets.
[+] Use 300 training points (approximately 75%) and 95 testing points (approximately 25%).
[+] Set a ‘random_state’ for the function(s) you use, if provided.
[+] Store the results in ‘X_train’, ‘X_test’, ‘y_train’, and ‘y_test’.

This “dummy” model predicts that all the students passed the final exam (i.e a graduation rate of 100%).

[+] Number of True Positive: 205
[+] Number of False Positive: 95
[+] The F1-score on the training set is 0.81 and on the test set 0.77. Our supervised Learning Model must perform better if using F1-score as a metric of model performance.

The confusion matrix of the naive model for the training set is shown below (in parenthesis is the count):

 Positive predict Negative predict Positive Truth tp (=205) fn (=0) Negative Truth fp(=95) tn(=0)

The fp are false positive, i.e the examples where the model wrongly classified the students as passing the tests.

1. Recall = tp/(tp+fn) = 205/(205+0) = 1
2. Precision = tp/(tp + fp) = 205/(205+95) = 0.68

The F1-score might not be the best metric to evaluate the performance of the model in our problem. $F1Score = 2 \times \frac{Precision \times Recall}{Precision + Recall}$

We will express F1-Score in terms of tp, tn, fp, fn instead: $F1Score = 2 \times \frac{\frac{tp}{tp + fp} \times \frac{tp}{tp+fn}}{\frac{tp}{tp+fp}+\frac{tp}{tp+fn}}$

The equation can be simplified to: $F1Score =\frac{2 tp}{2tp + fn + fp}$

Let’s now look at some cases:

1. Case1: fn=fp=0 (i.e no misclassification), then F1-score=1 as expected
2. Case2: fn=0.5 but fp=0 i.e we have 50% of the students who passed the exam were predicted to fail the exam. $F1Score =\frac{2\times 0.5}{2 \times 0.5 + 0 + 0.5} = 0.67$
3. Case3: fn=0 but fp=0.5, i.e the student that failed the exam were predicted to pass the exam. $F1Score =\frac{2\times 0.5}{2 \times 0.5 + 0.5 + 0} = 0.67$

Case2 and Case3 result in the same F1-score. However, a classifier that misclassifies failing students as graduating students can be considered to be worst than a classifier that misclassifies graduating students as failing students. We might want to use a metric that put more weights on Case3 misclassifications

# Training and Evaluating Models

We choose 3 supervised learning models that are appropriate for this problem.

Boosted Decision Tree Classifier(modelA)
A Boosted Decision Tree Classifier (DTC)is a supervised learning algorithm using Ensemble method to improve the accuracy of the learning algorithm. The real-world application of Boosted DTC includes star galaxy classification, assess quality of Financial Products, diagnosis of desease, object recognition and text processing. One of the advantages of the Boosted DTC is that it can be easily visualized and explained: given a new example, its classification can be explained by Boolean Logic. A Boosted DTC has few parameters to tune, and it requires little data preparation such as normalization of the features. One of the disadvantages of a Boosted DTC is that if the weak learners are too complex, it can result in high variance. Typically, Boosted Decision Tree Classifiers use an ensemble of DTC of depth=1, i.e Decision Stumps.

Naive Bayes Classification (modelB)
The Naive Bayes Classification (NBC) algorithm is a classification technique based on Bayes Theorem and can be used for binary and multi-class problems. It is used in spam detection, recognizing letters from handwritten texts as well as facial analysis problems. The Naive Bayes Classifier assumes that the features are conditionally independent, that is to say that given the class variable, the value of a particular feature is unrelated to that of any other features. The NBC presents the advantage to have no tuning parameters. It is easy to implement and fast to predict classes of datasets. It is also useful for models that require transparency. NBC performs well in the case of categorical input variables compared to numerical variables. For the later case, one needs to assume normal distribution which is a strong assumption. Another disadvantage is that: if a category that is not present in the training data is found in the test data, the model will not be able to make a prediction. The assumption of independent predictors adds to the bias of the model. NBC tends to perform poorly due to their simplicity.

Logistic Regression (modelC)
Logistic Regression is a Supervised Learning Algorithm that can be used for binary classification, but also for multi-class when applying the oneVsAll method. Logistic Regression is used in Marketing to predict customer rentention, in Finance to predit stock price, and text classification. Unlike DTC or Naive Bayes Algorithm, Logistic Regression can include interaction terms to investigate potential combined effects of the variables. The model coefficients can be interpreted to understand the direction and strength of the relationships between the features and the class. A Logistic regression without polynomial terms can only represent linear/planar boundary. In order to represent more complex boundaries, high order feature terms are required. This comes with the disadvantage of increasing the number of features, with the risk of high variance, and also increases the computing time. Though, regularization can be used to prevent overfitting. Another disadvantage of the Logistic Regression is that it requires additional data processing such as feature normalization.

We will try out Logistic Regression because it is a simple model that can be used as a benchmark. Because we are dealing with a small dataset, we also chose to experiment with the Naive Bayes Classifier because it is a high Bias classifier. Even if the features are not entirely independent, Naive Bayes can still perform well. The Model interpretability is also an important factor in the choice of the model. Indeed, those 3 models can be interpreted by the school board and reveal correlation between the student performance and the features. Those models are also cheap and fast to compute. Finally, it is also important that the models achieve high accuracy.

# Setup

We run the code cell below to initialize three helper functions which you can use for training and testing the three supervised learning models. The functions are as follows:

• “train_classifier” - takes as input a classifier and training data and fits the classifier to the data.
• “predict_labels” - takes as input a fit classifier, features, and a target labeling and makes predictions using the F1 score.
• “train_predict” - takes as input a classifier, and the training and testing data, and performs train_clasifier and predict_labels.
• This function will report the F1 score for both the training and testing data separately.

# Implementation: Model Performance Metrics

[+] model A: Decision Tree Classifier with Boosting
Training a AdaBoostClassifier using a training set size of 100. . .
Trained model in 0.0072 seconds
Total observed 64 | predicted Success 64
F1 score for training set: 1.0000.
Total observed 60 | predicted Success 56
F1 score for test set: 0.6552.

[+] model B: Bernouilli Naive Bayes
Training a BernoulliNB using a training set size of 100. . .
Trained model in 0.0034 seconds
Total observed 64 | predicted Success 76
F1 score for training set: 0.8286.
Total observed 60 | predicted Success 81
F1 score for test set: 0.7943.

[+] model C: Linear Regression
Training a LogisticRegression using a training set size of 100. . .
Trained model in 0.0026 seconds
Total observed 64 | predicted Success 70
F1 score for training set: 0.9104.
Total observed 60 | predicted Success 67
F1 score for test set: 0.7087.

[+] model A: Decision Tree Classifier with Boosting
Training a AdaBoostClassifier using a training set size of 200. . .
Trained model in 0.0064 seconds
Total observed 138 | predicted Success 138
F1 score for training set: 1.0000.
Total observed 60 | predicted Success 67
F1 score for test set: 0.7402.

[+] model B: Bernouilli Naive Bayes
Training a BernoulliNB using a training set size of 200. . .
Trained model in 0.0015 seconds
Total observed 138 | predicted Success 162
F1 score for training set: 0.8267.
Total observed 60 | predicted Success 76
F1 score for test set: 0.7794.

[+] model C: Linear Regression
Training a LogisticRegression using a training set size of 200. . .
Trained model in 0.0091 seconds
Total observed 138 | predicted Success 147
F1 score for training set: 0.8421.
Total observed 60 | predicted Success 77
F1 score for test set: 0.7883.

[+] model A: Decision Tree Classifier with Boosting
Training a AdaBoostClassifier using a training set size of 300. . .
Trained model in 0.0034 seconds
Total observed 205 | predicted Success 205
F1 score for training set: 1.0000.
Total observed 60 | predicted Success 58
F1 score for test set: 0.6102.

[+] model B: Bernouilli Naive Bayes
Training a BernoulliNB using a training set size of 300. . .
Trained model in 0.0025 seconds
Total observed 205 | predicted Success 228
F1 score for training set: 0.8037.
Total observed 60 | predicted Success 75
F1 score for test set: 0.7556.

[+] model C: Linear Regression
Training a LogisticRegression using a training set size of 300. . .
Trained model in 0.0120 seconds
Total observed 205 | predicted Success 236
F1 score for training set: 0.8435.
Total observed 60 | predicted Success 73
F1 score for test set: 0.7820.

# Tabular Results

** Classifer 1 - Decision Tree Classifier **

 Training Set Size Prediction Time (test) Training Time F1 Score (train) F1 Score (test) 100 0.0059406 0.0006511 1.0 0.6207 200 0.0061325 0.0005101 1.0 0.7360 300 0.0065699 0.0005306 1.0 0.6281

** Classifer 2 - Naive Bayes **

 Training Set Size Prediction Time (test) Training Time F1 Score (train) F1 Score (test) 100 0.001072 0.0004 0.8206 0.7943 200 0.001585 0.0007 0.8267 0.7794 300 0.001323 0.0004 0.8037 0.7556

** Classifer 3 - Logistic Regression**

 Training Set Size Prediction Time (test) Training Time F1 Score (train) F1 Score (test) 100 0.002504 0.000122 0.9104 0.7087 200 0.002925 0.000122 0.8421 0.7883 300 0.003895 0.000122 0.8435 0.7820

The Boosted Decision Tree Classifier (DTC) is the best performing model, although it is the slowest of the 3 models. The Boosted DTC when using the default settings does not perform better than the Naive Bayes nor the Logistic Regression Classifier. However, the settings can be tweaked (as it will be done later) in order to improve the performance. In particular, it is clear that, as is (i.e with the default settings), our Boosted DTC suffers from overfitting as demonstrated by a F1-score=1 for the training set (300 examples), and a poor F1-score of 0.63 for the test set.

The goal of the model is to predict whether a student will pass the final exam. In order to build our model, we take a sample of students whose the result at the final exam is known. Each student is “defined” by a set of characteristics/attributes: gender, address, the school attended, etc…

The model consists in generating a number of simple rules on the students attributes, with a classification error below better than 0.5, i.e better than . For example, one possible rule could be to split the students depending on the area: Urban or Rural. If all the students with the Urban attribute had passed the exam, and all the students with Rural attribute had failed, then this rule alone would be sufficient to classify students, and predict outcome for future students. But such a simple rule is very likely to make errors. A new rule is then generated and it will pay more attention to the examples that were misclassified by the previous rule. As new rules are generated, they will continue to pay more attention to the examples that have been misclassified by the preceding rule.

After many iterations of generating new rules, the prediction from all those simple rules are combined in the form of a weighted average, which will have much more accuracy than the prediction from a single rule alone.

# Implementation: Model Tuning

The final model has a F1-score of 0.82 for the training set and 0.80 for the test set. The F1-score of training set is lower for the tuned model: 0.82 versus 1.0 for the untuned model. And the F1-score of the test set is improved after tuning the model. The results clearly indicate that there is less overfitting in the tuned model. Indeed, ($\Delta$ F1-score), i.e “F1-score of training set” minus “F1-score of test set”, is lower for the tuned model.


text:
# How to work with two encodings?

I'll try to explain my problem here. I know that there are many posts dealing with this but I'm still confused.

We have many files from different machines with different systems. The important is:

1. the tex file will be compiled on a Linux machine with utf-8.
2. the tex files comes from many Windows machines with iso.

Without converting, is there some good preamble to solve my problem?

For example, on Windows Vista, using Winedt to produce a tex file encoded in iso, is there some package which allows me to run pdflatex on a Linux machine with Kile and utf-8?

This will be a big project with a lot of people sending many tex files to me and I'd like to create a good template with necessary packages to avoid confusions.

-
It doesn't matter what machine a file comes from; if it's Latin-1 encoded, just having \usepackage[latin1]{inputenc} in it is sufficient. – egreg Jan 8 '13 at 13:40
I believe that Kile is able to detect the encoding of a file. If you open the Windows file, check that the characters are correct, select all, copy and paste in the window where the UTF-8 file is opened. The machine should do the conversion. Or just decide for one encoding. I believe that recent versions of WinEdt are able to cope with UTF-8, so this would be the best choice. – egreg Jan 8 '13 at 13:48
@egreg can we have multiple encodings,say \include{iso-latin1.tex} and \include{utf-8.tex} in a single project.tex like this ? – texenthusiast Jan 8 '13 at 13:55
Yes, just use \inputencoding{latin1}\input{fileinlatin1}\inputencoding{utf8}. But working with multiple encodings is better avoided. – egreg Jan 8 '13 at 13:57
Better standardize in UTF-8, it is easier in the long run (and I believe current Windows uses it by default now). You can use recode (or a similar command) to change the encoding of the files you get. – vonbrand Jan 19 '13 at 2:33

In collaborative projects with source files being edited on a variety of platforms, editors, and locales, encoding issues can become a real PITA.

In such setting, it is best to define one "right", mandatory encoding (UTF-8), communicate this clearly and stick to it. However, in many cases the collaborators, especially if they stem from the Windows or Mac world, do not even know what an encoding is. What, however, works reasonably well in practice, is the following.

1. Prepare templates that already are in the right encoding, including (if appropriate) the \usepackage[utf8]{inputenc} line. The template also should (at least in some comment), contain some UTF-8 characters: Many modern editors have reasonably good heuristics to detect the encoding "automagically" if non-ASCII characters are present.

2. Provide editor modelines for as many editors as possible. Many editors, including WinEdt and TeXShop, provide a modeline syntax that can also be used to specify the encoding explicitly. Many other support either the vim or emacs modeline syntax, so those should always be present. But some put additional constraints in this respect: WinEdt, for instance, supports the emacs modeline syntax, but each setting (mode and coding, below) has to be placed in a different line. (Thanks to karlkoeller for correcting this!)

The following document header includes some UTF-8 characters and modeline commands to set the file type (for syntax coloring) to TeX/LaTeX and the encoding to UTF-8 for, respectively: vim, WinEdt, TexShop, Emacs, and Kate:

% vim: set fenc=utf-8 ft=latex
% !Mode:: "TeX:UTF-8"
% !TEX encoding = UTF-8 Unicode
% -*- mode: latex -*-
% -*- coding: UTF-8 -*-
% kate: encoding utf-8;
%
% IMPORTANT Note: This file is UTF-8 encoded!
%                 You should see some umlauts here: ÄÖÜäöü

The Kate modeline might also work for Texmaker, as both are based on the generic Qt editor component, but I have not checked this. For most editors, modelines have to appear in the first 10 lines of a document, some (Vim, Emacs) also support them at the end of the document (within the last 10 lines).

-
If anybody knows the respective modelines of other editors than the mentioned: Please tell (or edit it into the answer :-) – Daniel Jan 23 '13 at 20:41
OK. It seems that this one was a forgotten (but interesting) question, seeing all people who are upvoting now! – karlkoeller Jun 26 '13 at 9:08
off the (tex-related) topic ... to refer to a coding as “iso” is essentially meaningless -- i regularly work with three iso encodings (iso 646 ~= ascii, iso 88591-1 = iso latin1, and iso 10646 = unicode [with utf-8 encoding]). one might guess that your “iso” means “iso 8859-1”, but the usage is ambiguous and therefore unhelpful. – wasteofspace Jun 26 '13 at 9:23
@wasteofspace: I totally agree and, thus, don't use the term "iso". Maybe your comment was intended for the question? – Daniel Jun 26 '13 at 10:09


text:
12 Fluid Dynamics and Its Biological and Medical Applications

# 88 12.4 Viscosity and Laminar Flow; Poiseuille’s Law

### Summary

• Define laminar flow and turbulent flow.
• Explain what viscosity is.
• Calculate flow and resistance with Poiseuille’s law.
• Explain how pressure drops due to resistance.

# Laminar Flow and Viscosity

When you pour yourself a glass of juice, the liquid flows freely and quickly. But when you pour syrup on your pancakes, that liquid flows slowly and sticks to the pitcher. The difference is fluid friction, both within the fluid itself and between the fluid and its surroundings. We call this property of fluids viscosity. Juice has low viscosity, whereas syrup has high viscosity. In the previous sections we have considered ideal fluids with little or no viscosity. In this section, we will investigate what factors, including viscosity, affect the rate of fluid flow.

The precise definition of viscosity is based on laminar, or nonturbulent, flow. Before we can define viscosity, then, we need to define laminar flow and turbulent flow. Figure 1 shows both types of flow. Laminar flow is characterized by the smooth flow of the fluid in layers that do not mix. Turbulent flow, or turbulence, is characterized by eddies and swirls that mix layers of fluid together.

Figure 2 shows schematically how laminar and turbulent flow differ. Layers flow without mixing when flow is laminar. When there is turbulence, the layers mix, and there are significant velocities in directions other than the overall direction of flow. The lines that are shown in many illustrations are the paths followed by small volumes of fluids. These are called streamlines. Streamlines are smooth and continuous when flow is laminar, but break up and mix when flow is turbulent. Turbulence has two main causes. First, any obstruction or sharp corner, such as in a faucet, creates turbulence by imparting velocities perpendicular to the flow. Second, high speeds cause turbulence. The drag both between adjacent layers of fluid and between the fluid and its surroundings forms swirls and eddies, if the speed is great enough. We shall concentrate on laminar flow for the remainder of this section, leaving certain aspects of turbulence for later sections.

### MAKING CONNECTIONS: TAKE-HOME EXPERIMENT: GO DOWN TO THE RIVER

Try dropping simultaneously two sticks into a flowing river, one near the edge of the river and one near the middle. Which one travels faster? Why?

Figure 3 shows how viscosity is measured for a fluid. Two parallel plates have the specific fluid between them. The bottom plate is held fixed, while the top plate is moved to the right, dragging fluid with it. The layer (or lamina) of fluid in contact with either plate does not move relative to the plate, and so the top layer moves at$\boldsymbol{v}$while the bottom layer remains at rest. Each successive layer from the top down exerts a force on the one below it, trying to drag it along, producing a continuous variation in speed from$\boldsymbol{v}$to 0 as shown. Care is taken to insure that the flow is laminar; that is, the layers do not mix. The motion in Figure 3 is like a continuous shearing motion. Fluids have zero shear strength, but the rate at which they are sheared is related to the same geometrical factors$\boldsymbol{A}$and$\boldsymbol{L}$as is shear deformation for solids.

A force$\boldsymbol{F}$is required to keep the top plate in Figure 3 moving at a constant velocity$\boldsymbol{v},$and experiments have shown that this force depends on four factors. First,$\boldsymbol{F}$is directly proportional to$\boldsymbol{v}$(until the speed is so high that turbulence occurs—then a much larger force is needed, and it has a more complicated dependence on$\boldsymbol{v}$). Second,$\boldsymbol{F}$is proportional to the area$\boldsymbol{A}$of the plate. This relationship seems reasonable, since$\boldsymbol{A}$is directly proportional to the amount of fluid being moved. Third,$\boldsymbol{F}$is inversely proportional to the distance between the plates$\boldsymbol{L}.$This relationship is also reasonable;$\boldsymbol{L}$is like a lever arm, and the greater the lever arm, the less force that is needed. Fourth,$\boldsymbol{F}$is directly proportional to the coefficient of viscosity,$\boldsymbol{\eta}.$The greater the viscosity, the greater the force required. These dependencies are combined into the equation

$\boldsymbol{F=\eta}$$\boldsymbol{\frac{vA}{L}},$

which gives us a working definition of fluid viscosity$\boldsymbol{\eta}.$Solving for$\boldsymbol{\eta}$gives

$\boldsymbol{\eta\:=}$$\boldsymbol{\frac{FL}{vA}},$

which defines viscosity in terms of how it is measured. The SI unit of viscosity is$\boldsymbol{\textbf{N}\cdotp\textbf{m}/[(\textbf{m/s})\textbf{m}^2]=(\textbf{N/m}^2)\textbf{s or Pa}\cdotp\textbf{s}}.$Table 1 lists the coefficients of viscosity for various fluids.

Viscosity varies from one fluid to another by several orders of magnitude. As you might expect, the viscosities of gases are much less than those of liquids, and these viscosities are often temperature dependent. The viscosity of blood can be reduced by aspirin consumption, allowing it to flow more easily around the body. (When used over the long term in low doses, aspirin can help prevent heart attacks, and reduce the risk of blood clotting.)

# Laminar Flow Confined to Tubes—Poiseuille’s Law

What causes flow? The answer, not surprisingly, is pressure difference. In fact, there is a very simple relationship between horizontal flow and pressure. Flow rate$\boldsymbol{Q}$is in the direction from high to low pressure. The greater the pressure differential between two points, the greater the flow rate. This relationship can be stated as

$\boldsymbol{Q\:=}$$\boldsymbol{\frac{P_2-P_1}{R}},$

where$\boldsymbol{P_1}$and$\boldsymbol{P_2}$are the pressures at two points, such as at either end of a tube, and$\boldsymbol{R}$is the resistance to flow. The resistance$\boldsymbol{R}$includes everything, except pressure, that affects flow rate. For example,$\boldsymbol{R}$is greater for a long tube than for a short one. The greater the viscosity of a fluid, the greater the value of$\boldsymbol{R}.$Turbulence greatly increases$\boldsymbol{R},$whereas increasing the diameter of a tube decreases$\boldsymbol{R}.$

If viscosity is zero, the fluid is frictionless and the resistance to flow is also zero. Comparing frictionless flow in a tube to viscous flow, as in Figure 4, we see that for a viscous fluid, speed is greatest at midstream because of drag at the boundaries. We can see the effect of viscosity in a Bunsen burner flame, even though the viscosity of natural gas is small.

The resistance$\boldsymbol{R}$to laminar flow of an incompressible fluid having viscosity$\boldsymbol{\eta}$through a horizontal tube of uniform radius$\boldsymbol{r}$and length$\boldsymbol{\ell},$such as the one in Figure 5, is given by

$\boldsymbol{R\:=}$$\boldsymbol{\frac{8\eta\ell}{\pi{r}^4}}.$

This equation is called Poiseuille’s law for resistance after the French scientist J. L. Poiseuille (1799–1869), who derived it in an attempt to understand the flow of blood, an often turbulent fluid.

Let us examine Poiseuille’s expression for$\boldsymbol{R}$to see if it makes good intuitive sense. We see that resistance is directly proportional to both fluid viscosity$\boldsymbol{\eta}$and the length$\boldsymbol{\ell}$of a tube. After all, both of these directly affect the amount of friction encountered—the greater either is, the greater the resistance and the smaller the flow. The radius$\boldsymbol{r}$of a tube affects the resistance, which again makes sense, because the greater the radius, the greater the flow (all other factors remaining the same). But it is surprising that$\boldsymbol{r}$is raised to the fourth power in Poiseuille’s law. This exponent means that any change in the radius of a tube has a very large effect on resistance. For example, doubling the radius of a tube decreases resistance by a factor of$\boldsymbol{2^4=16}.$

Taken together,$\boldsymbol{Q=\frac{P_2-P_1}{R}}$and$\boldsymbol{R=\frac{8\eta\ell}{\pi{r}^4}}$give the following expression for flow rate:

$\boldsymbol{Q\:=}$$\boldsymbol{\frac{(P_2-P_1)\pi{r}^4}{8\eta\ell}}.$

This equation describes laminar flow through a tube. It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

### Example 1: Using Flow Rate: Plaque Deposits Reduce Blood Flow

Suppose the flow rate of blood in a coronary artery has been reduced to half its normal value by plaque deposits. By what factor has the radius of the artery been reduced, assuming no turbulence occurs?

Strategy

Assuming laminar flow, Poiseuille’s law states that

$\boldsymbol{Q\:=}$$\boldsymbol{\frac{(P_2-P_1)\pi{r}^4}{8\eta\ell}}.$

We need to compare the artery radius before and after the flow rate reduction.

Solution

With a constant pressure difference assumed and the same length and viscosity, along the artery we have

$\boldsymbol{\frac{Q_1}{r_1^4}}$$\boldsymbol{=}$$\boldsymbol{\frac{Q_2}{r_2^4}}.$

So, given that$\boldsymbol{Q_2=0.5Q_1},$we find that$\boldsymbol{r_2^4=0.5r_1^4}.$

Therefore,$\boldsymbol{r_2=(0.5)^{0.25}r_1=0.841r_1},$a decrease in the artery radius of 16%.

Discussion

This decrease in radius is surprisingly small for this situation. To restore the blood flow in spite of this buildup would require an increase in the pressure difference$\boldsymbol{(P_2-P_1)}$of a factor of two, with subsequent strain on the heart.

Fluid Temperature (ºC) Viscosity
η(mPa·s)
Gases
Air 0 0.0171
20 0.0181
40 0.0190
100 0.0218
Ammonia 20 0.00974
Carbon dioxide 20 0.0147
Helium 20 0.0196
Hydrogen 0 0.0090
Mercury 20 0.0450
Oxygen 20 0.0203
Steam 100 0.0130
Liquids
Water 0 1.792
20 1.002
37 0.6947
40 0.653
100 0.282
Whole blood1 20 3.015
37 2.084
Blood plasma2 20 1.810
37 1.257
Ethyl alcohol 20 1.20
Methanol 20 0.584
Oil (heavy machine) 20 660
Oil (motor, SAE 10) 30 200
Oil (olive) 20 138
Glycerin 20 1500
Honey 20 2000–10000
Maple Syrup 20 2000–3000
Milk 20 3.0
Oil (Corn) 20 65
Table 1. Coefficients of Viscosity of Various Fluids

The circulatory system provides many examples of Poiseuille’s law in action—with blood flow regulated by changes in vessel size and blood pressure. Blood vessels are not rigid but elastic. Adjustments to blood flow are primarily made by varying the size of the vessels, since the resistance is so sensitive to the radius. During vigorous exercise, blood vessels are selectively dilated to important muscles and organs and blood pressure increases. This creates both greater overall blood flow and increased flow to specific areas. Conversely, decreases in vessel radii, perhaps from plaques in the arteries, can greatly reduce blood flow. If a vessel’s radius is reduced by only 5% (to 0.95 of its original value), the flow rate is reduced to about$\boldsymbol{(0.95)^4=0.81}$of its original value. A 19% decrease in flow is caused by a 5% decrease in radius. The body may compensate by increasing blood pressure by 19%, but this presents hazards to the heart and any vessel that has weakened walls. Another example comes from automobile engine oil. If you have a car with an oil pressure gauge, you may notice that oil pressure is high when the engine is cold. Motor oil has greater viscosity when cold than when warm, and so pressure must be greater to pump the same amount of cold oil.

### Example 2: What pressure Produces This Flow Rate?

An intravenous (IV) system is supplying saline solution to a patient at the rate of$\boldsymbol{0.120\textbf{ cm}^3\textbf{/s}}$through a needle of radius 0.150 mm and length 2.50 cm. What pressure is needed at the entrance of the needle to cause this flow, assuming the viscosity of the saline solution to be the same as that of water? The gauge pressure of the blood in the patient’s vein is 8.00 mm Hg. (Assume that the temperature is$\boldsymbol{20^0\textbf{C}}$.)

Strategy

Assuming laminar flow, Poiseuille’s law applies. This is given by

$\boldsymbol{Q\:=}$$\boldsymbol{\frac{(P_2-P_1)\pi{r}^4}{8\eta\ell},}$

where$\boldsymbol{P_2}$is the pressure at the entrance of the needle and$\boldsymbol{P_1}$is the pressure in the vein. The only unknown is$\boldsymbol{P_2}.$

Solution

Solving for$\boldsymbol{P_2}$yields

$\boldsymbol{P_2\:=}$$\boldsymbol{\frac{8\eta\ell}{\pi{r}^4}}$$\boldsymbol{Q+P_1}.$

$\boldsymbol{P_1}$is given as 8.00 mm Hg, which converts to$\boldsymbol{1.066\times10^3\textbf{ N/m}^2}.$Substituting this and the other known values yields

$\begin{array}{lcl} \boldsymbol{P_2} & \boldsymbol{=} & \boldsymbol{[\frac{8(1.00\times10^{-3}\textbf{ N}\cdotp\textbf{s/m}^2)(2.50\times10^{-2}\textbf{ m})}{\pi(0.150\times10^{-3}\textbf{ m})^4}](1.20\times10^{-7}\textbf{ m}^3\textbf{/s})+1.066\times10^3\textbf{ N/m}^2} \\ {} & \boldsymbol{=} & \boldsymbol{1.62\times10^4\textbf{ N/m}^2.} \end{array}$

Discussion

This pressure could be supplied by an IV bottle with the surface of the saline solution 1.61 m above the entrance to the needle (this is left for you to solve in this chapter’s Problems and Exercises), assuming that there is negligible pressure drop in the tubing leading to the needle.

# Flow and Resistance as Causes of Pressure Drops

You may have noticed that water pressure in your home might be lower than normal on hot summer days when there is more use. This pressure drop occurs in the water main before it reaches your home. Let us consider flow through the water main as illustrated in Figure 6. We can understand why the pressure$\boldsymbol{P_1}$to the home drops during times of heavy use by rearranging

$\boldsymbol{Q\:=}$$\boldsymbol{\frac{P_2-P_1}{R}}$

to

$\boldsymbol{P_2-P_1=RQ},$

where, in this case,$\boldsymbol{P_2}$is the pressure at the water works and$\boldsymbol{R}$is the resistance of the water main. During times of heavy use, the flow rate$\boldsymbol{Q}$is large. This means that$\boldsymbol{P_2-P_1}$must also be large. Thus$\boldsymbol{P_1}$must decrease. It is correct to think of flow and resistance as causing the pressure to drop from$\boldsymbol{P_2}$to$\boldsymbol{P_1}.\:\boldsymbol{P_2-P_1=RQ}$is valid for both laminar and turbulent flows.

We can use$\boldsymbol{P_2-P_1=RQ}$to analyze pressure drops occurring in more complex systems in which the tube radius is not the same everywhere. Resistance will be much greater in narrow places, such as an obstructed coronary artery. For a given flow rate$\boldsymbol{Q},$the pressure drop will be greatest where the tube is most narrow. This is how water faucets control flow. Additionally,$\boldsymbol{R}$is greatly increased by turbulence, and a constriction that creates turbulence greatly reduces the pressure downstream. Plaque in an artery reduces pressure and hence flow, both by its resistance and by the turbulence it creates.

Figure 7 is a schematic of the human circulatory system, showing average blood pressures in its major parts for an adult at rest. Pressure created by the heart’s two pumps, the right and left ventricles, is reduced by the resistance of the blood vessels as the blood flows through them. The left ventricle increases arterial blood pressure that drives the flow of blood through all parts of the body except the lungs. The right ventricle receives the lower pressure blood from two major veins and pumps it through the lungs for gas exchange with atmospheric gases – the disposal of carbon dioxide from the blood and the replenishment of oxygen. Only one major organ is shown schematically, with typical branching of arteries to ever smaller vessels, the smallest of which are the capillaries, and rejoining of small veins into larger ones. Similar branching takes place in a variety of organs in the body, and the circulatory system has considerable flexibility in flow regulation to these organs by the dilation and constriction of the arteries leading to them and the capillaries within them. The sensitivity of flow to tube radius makes this flexibility possible over a large range of flow rates.

Each branching of larger vessels into smaller vessels increases the total cross-sectional area of the tubes through which the blood flows. For example, an artery with a cross section of$\boldsymbol{1\textbf{ cm}^2}$may branch into 20 smaller arteries, each with cross sections of$\boldsymbol{0.5\textbf{ cm}^2},$with a total of$\boldsymbol{10\textbf{ cm}^2}.$In that manner, the resistance of the branchings is reduced so that pressure is not entirely lost. Moreover, because$\boldsymbol{Q=A\bar{v}}$and$\boldsymbol{A}$increases through branching, the average velocity of the blood in the smaller vessels is reduced. The blood velocity in the aorta ($\boldsymbol{\textbf{diameter }=1\textbf{ cm}}$) is about 25 cm/s, while in the capillaries ($\boldsymbol{20\:\mu\textbf{m}}$in diameter) the velocity is about 1 mm/s. This reduced velocity allows the blood to exchange substances with the cells in the capillaries and alveoli in particular.

# Section Summary

• Laminar flow is characterized by smooth flow of the fluid in layers that do not mix.
• Turbulence is characterized by eddies and swirls that mix layers of fluid together.
• Fluid viscosity$\boldsymbol{\eta}$is due to friction within a fluid. Representative values are given in Table 1. Viscosity has units of$\boldsymbol{(\textbf{N/m}^2)\textbf{s}}$or$\boldsymbol{\textbf{Pa}\cdotp\textbf{s}}.$
• Flow is proportional to pressure difference and inversely proportional to resistance:
$\boldsymbol{Q\:=}$$\boldsymbol{\frac{P_2-P_1}{R}}.$
• For laminar flow in a tube, Poiseuille’s law for resistance states that
$\boldsymbol{R\:=}$$\boldsymbol{\frac{8\eta\ell}{\pi{r}^4}.}$
• Poiseuille’s law for flow in a tube is
$\boldsymbol{Q\:=}$$\boldsymbol{\frac{(P_2-P_1)\pi{r}^4}{8\eta\ell}.}$
• The pressure drop caused by flow and resistance is given by
$\boldsymbol{P_2-P_1=RQ}.$

### Conceptual Questions

1: Explain why the viscosity of a liquid decreases with temperature—that is, how might increased temperature reduce the effects of cohesive forces in a liquid? Also explain why the viscosity of a gas increases with temperature—that is, how does increased gas temperature create more collisions between atoms and molecules?

2: When paddling a canoe upstream, it is wisest to travel as near to the shore as possible. When canoeing downstream, it may be best to stay near the middle. Explain why.

3: Why does flow decrease in your shower when someone flushes the toilet?

4: Plumbing usually includes air-filled tubes near water faucets, as shown in Figure 8. Explain why they are needed and how they work.

### Problems & Exercises

1: (a) Calculate the retarding force due to the viscosity of the air layer between a cart and a level air track given the following information—air temperature is$\boldsymbol{20^0\textbf{C}},$the cart is moving at 0.400 m/s, its surface area is$\boldsymbol{2.50\times10^{-2}\textbf{ m}^2},$and the thickness of the air layer is$\boldsymbol{6.00\times10^{-5}\textbf{ m}}.$(b) What is the ratio of this force to the weight of the 0.300-kg cart?

2: What force is needed to pull one microscope slide over another at a speed of 1.00 cm/s, if there is a 0.500-mm-thick layer of$\boldsymbol{20^0\textbf{C}}$water between them and the contact area is$\boldsymbol{8.00\textbf{ cm}^2}?$

3: A glucose solution being administered with an IV has a flow rate of$\boldsymbol{4.00\textbf{ cm}^3\textbf{/min}}.$What will the new flow rate be if the glucose is replaced by whole blood having the same density but a viscosity 2.50 times that of the glucose? All other factors remain constant.

4: The pressure drop along a length of artery is 100 Pa, the radius is 10 mm, and the flow is laminar. The average speed of the blood is 15 mm/s. (a) What is the net force on the blood in this section of artery? (b) What is the power expended maintaining the flow?

5: A small artery has a length of$\boldsymbol{1.1\times10^{-3}\textbf{ m}}$and a radius of$\boldsymbol{2.5\times10^{-5}\textbf{ m}}.$If the pressure drop across the artery is 1.3 kPa, what is the flow rate through the artery? (Assume that the temperature is $\boldsymbol{37^0\textbf{C}}$.)

6: Fluid originally flows through a tube at a rate of$\boldsymbol{100\textbf{ cm}^3\textbf{/s}}.$To illustrate the sensitivity of flow rate to various factors, calculate the new flow rate for the following changes with all other factors remaining the same as in the original conditions. (a) Pressure difference increases by a factor of 1.50. (b) A new fluid with 3.00 times greater viscosity is substituted. (c) The tube is replaced by one having 4.00 times the length. (d) Another tube is used with a radius 0.100 times the original. (e) Yet another tube is substituted with a radius 0.100 times the original and half the length, and the pressure difference is increased by a factor of 1.50.

7: The arterioles (small arteries) leading to an organ, constrict in order to decrease flow to the organ. To shut down an organ, blood flow is reduced naturally to 1.00% of its original value. By what factor did the radii of the arterioles constrict? Penguins do this when they stand on ice to reduce the blood flow to their feet.

8: Angioplasty is a technique in which arteries partially blocked with plaque are dilated to increase blood flow. By what factor must the radius of an artery be increased in order to increase blood flow by a factor of 10?

9: (a) Suppose a blood vessel’s radius is decreased to 90.0% of its original value by plaque deposits and the body compensates by increasing the pressure difference along the vessel to keep the flow rate constant. By what factor must the pressure difference increase? (b) If turbulence is created by the obstruction, what additional effect would it have on the flow rate?

10: A spherical particle falling at a terminal speed in a liquid must have the gravitational force balanced by the drag force and the buoyant force. The buoyant force is equal to the weight of the displaced fluid, while the drag force is assumed to be given by Stokes Law,$\boldsymbol{F_{\textbf{s}}=6\pi{r}\eta{v}}.$Show that the terminal speed is given by

$\boldsymbol{v\:=}$$\boldsymbol{\frac{2R^2g}{9\eta}}$$\boldsymbol{(\rho_{\textbf{s}}-\rho_1),}$

where$\boldsymbol{R}$is the radius of the sphere,$\boldsymbol{\rho_{\textbf{s}}}$is its density, and$\boldsymbol{\rho_1}$is the density of the fluid and$\boldsymbol{\eta}$the coefficient of viscosity.

11: Using the equation of the previous problem, find the viscosity of motor oil in which a steel ball of radius 0.8 mm falls with a terminal speed of 4.32 cm/s. The densities of the ball and the oil are 7.86 and 0.88 g/mL, respectively.

12: A skydiver will reach a terminal velocity when the air drag equals their weight. For a skydiver with high speed and a large body, turbulence is a factor. The drag force then is approximately proportional to the square of the velocity. Taking the drag force to be$\boldsymbol{F_{\textbf{D}}=\frac{1}{2}\rho{A}v^2}$and setting this equal to the person’s weight, find the terminal speed for a person falling “spread eagle.” Find both a formula and a number for$\boldsymbol{v_{\textbf{t}}},$with assumptions as to size.

13: A layer of oil 1.50 mm thick is placed between two microscope slides. Researchers find that a force of$\boldsymbol{5.50\times10^{-4}\textbf{ N}}$is required to glide one over the other at a speed of 1.00 cm/s when their contact area is$\boldsymbol{6.00\textbf{ cm}^2}.$What is the oil’s viscosity? What type of oil might it be?

14: (a) Verify that a 19.0% decrease in laminar flow through a tube is caused by a 5.00% decrease in radius, assuming that all other factors remain constant, as stated in the text. (b) What increase in flow is obtained from a 5.00% increase in radius, again assuming all other factors remain constant?

15: Example 2 dealt with the flow of saline solution in an IV system. (a) Verify that a pressure of$\boldsymbol{1.62\times10^4\textbf{ N/m}^2}$is created at a depth of 1.61 m in a saline solution, assuming its density to be that of sea water. (b) Calculate the new flow rate if the height of the saline solution is decreased to 1.50 m. (c) At what height would the direction of flow be reversed? (This reversal can be a problem when patients stand up.)

16: When physicians diagnose arterial blockages, they quote the reduction in flow rate. If the flow rate in an artery has been reduced to 10.0% of its normal value by a blood clot and the average pressure difference has increased by 20.0%, by what factor has the clot reduced the radius of the artery?

17: During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.0% of its normal value, and the blood pressure difference across the circulatory system has increased by 50.0%. By what factor has the average radii of her blood vessels increased?

18: Water supplied to a house by a water main has a pressure of$\boldsymbol{3.00\times10^5\textbf{ N/m}^2}$early on a summer day when neighborhood use is low. This pressure produces a flow of 20.0 L/min through a garden hose. Later in the day, pressure at the exit of the water main and entrance to the house drops, and a flow of only 8.00 L/min is obtained through the same hose. (a) What pressure is now being supplied to the house, assuming resistance is constant? (b) By what factor did the flow rate in the water main increase in order to cause this decrease in delivered pressure? The pressure at the entrance of the water main is$\boldsymbol{5.00\times10^5\textbf{ N/m}^2},$and the original flow rate was 200 L/min. (c) How many more users are there, assuming each would consume 20.0 L/min in the morning?

19: An oil gusher shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter. Neglecting air resistance but not the resistance of the pipe, and assuming laminar flow, calculate the gauge pressure at the entrance of the 50.0-m-long vertical pipe. Take the density of the oil to be$\boldsymbol{900\textbf{ kg/m}^3}$and its viscosity to be$\boldsymbol{1.00(\textbf{N/m}^2)\cdotp\textbf{s}}$(or$\boldsymbol{1.00\textbf{ Pa}\cdotp\textbf{s}}$). Note that you must take into account the pressure due to the 50.0-m column of oil in the pipe.

20: Concrete is pumped from a cement mixer to the place it is being laid, instead of being carried in wheelbarrows. The flow rate is 200.0 L/min through a 50.0-m-long, 8.00-cm-diameter hose, and the pressure at the pump is$\boldsymbol{8.00\times10^6\textbf{ N/m}^2}.$(a) Calculate the resistance of the hose. (b) What is the viscosity of the concrete, assuming the flow is laminar? (c) How much power is being supplied, assuming the point of use is at the same level as the pump? You may neglect the power supplied to increase the concrete’s velocity.

Consider a coronary artery constricted by arteriosclerosis. Construct a problem in which you calculate the amount by which the diameter of the artery is decreased, based on an assessment of the decrease in flow rate.

22: Consider a river that spreads out in a delta region on its way to the sea. Construct a problem in which you calculate the average speed at which water moves in the delta region, based on the speed at which it was moving up river. Among the things to consider are the size and flow rate of the river before it spreads out and its size once it has spread out. You can construct the problem for the river spreading out into one large river or into multiple smaller rivers.

## Footnotes

1. 1 The ratios of the viscosities of blood to water are nearly constant between 0°C and 37°C.
2. 2 See note on Whole Blood.

## Glossary

laminar
a type of fluid flow in which layers do not mix
turbulence
fluid flow in which layers mix together via eddies and swirls
viscosity
the friction in a fluid, defined in terms of the friction between layers
Poiseuille’s law for resistance
the resistance to laminar flow of an incompressible fluid in a tube: R = 8ηl/πr4
Poiseuille’s law
the rate of laminar flow of an incompressible fluid in a tube: Q = (P2P1)πr4/8ηl

### Solutions

Problems & Exercises

1:

(a)$\boldsymbol{3.02\times10^{-3}\textbf{ N}}$

(b)$\boldsymbol{1.03\times10^{-3}}$

3:

$\boldsymbol{1.60\textbf{ cm}^3\textbf{/min}}$

5:

$\boldsymbol{8.7\times10^{-11}\textbf{ m}^3\textbf{/s}}$

7:

0.316

9:

(a) 1.52

(b) Turbulence will decrease the flow rate of the blood, which would require an even larger increase in the pressure difference, leading to higher blood pressure.

11:

$\boldsymbol{225\textbf{ mPa}\cdotp\textbf{s}}$

13:

$\boldsymbol{0.138\textbf{ Pa}\cdotp\textbf{s}},$

or

Olive oil.

15:

(a)$\boldsymbol{1.62\times10^4\textbf{ N/m}^2}$

(b)$\boldsymbol{0.111\textbf{ cm}^3\textbf{/s}}$

(c)10.6 cm

17:

1.59

19:

$\boldsymbol{2.95\times10^6\textbf{ N/m}^2}$(gauge pressure)


text:
# On the bias of the distance–redshift relation from gravitational lensing

@inproceedings{Kaiser2016OnTB,
title={On the bias of the distance–redshift relation from gravitational lensing},
author={Nick Kaiser and John A. Peacock},
year={2016}
}
• Published 2016
• Physics
• A long standing question in cosmology is whether gravitational lensing changes the distance-redshift relation $D(z)$ or the mean flux density of sources. Interest in this has been rekindled by recent studies in non-linear relativistic perturbation theory that find biases in both the area of a surface of constant redshift and in the mean distance to this surface, with a fractional bias in both cases on the order of the mean squared convergence $\langle \kappa^2 \rangle$. Any such area bias could… CONTINUE READING

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#### References

##### Publications referenced by this paper.
SHOWING 1-3 OF 3 REFERENCES

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• C LBennett
• Bertotti B.,
• 1996
VIEW 7 EXCERPTS
HIGHLY INFLUENTIAL

## Narlikar 2010 for derivations) are particularly important in the present context since, as we have discussed, they are the basis for the ‘focusing theorem

• Schneider, Ehlers, Falco
• 1992


text:
# Why do some molecules form T shapes instead of trigonal planar shapes [duplicate]

Consider a molecule such as $\ce{ClF_3}$.
Shouldn't the electron clouds (which are more negative than the Chlorine atom) want to repel each other and so the $\ce{ClF_3}$molecule should arrange into a trigonal planar shape where the electron clouds are as far away as possible?

I think at a deeper level than VSEPR, Chlorine is either partially ionized to a hybrid of $\ce{F^- + Cl^+F_2}$ configurations or has some electrons promoted up to d shaped orbitals but I don't know how to calculate the resulting shape. I think the main thing throwing me off is that the $\ce{F^- + Cl^+F_2}$ shapes should have the $\ce{F^-}$ ion repulsed from the lone pair electrons.

• There is a 3-center 4-electron bond between the chlorine and two fluorine atoms. This bond is linear, so the fluorines are necessarily on opposite sides of the chlorine. – f'' Jul 16 '16 at 20:23
• A more general question is $$\mathrm{Why\ does\ VSEPR\ predict\ a\ T-shaped\ geometry\ for AX_3 E_2 \ and\ not\ trigonal\ planar?}$$ We also have to remember that VSEPR was developed to rationalize what we were learning about electrons, bonding, and quantum mechanics withe known geometries. – Ben Norris Jul 16 '16 at 21:08
• – ron Jul 16 '16 at 21:52


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# An object with a mass of 12 kg is on a surface with a kinetic friction coefficient of  1 . How much force is necessary to accelerate the object horizontally at 7  ms^-2?

Feb 13, 2016

The total force is composed of two pieces: the force required to accelerate the mass and the force required to overcome friction. $F = m a + \mu m g = 12 \cdot 7 + 1 \cdot 12 \cdot 9.8 = 201.6$ $N$.

#### Explanation:

The total force is made up of two different forces.

If there were no friction, the force required to accelerate a mass of $12$ $k g$ at an acceleration of $7$ $m {s}^{-} 2$ is given by Newton's Second Law :

$F = m a = 12 \cdot 7 = 84$ $N$

The second part is the frictional force:

${F}_{f} = \mu {F}_{N}$ where $\mu$ is the frictional coefficient and ${F}_{N}$ is the normal force, which in turn is $m g$, the mass times the acceleration due to gravity.

${F}_{f} = \mu m g = 1 \cdot 12 \cdot 9.8 = 117.6$ $N$

To find the total force, just add these two forces:

$F = m a + \mu m g = 12 \cdot 7 + 1 \cdot 12 \cdot 9.8 = 201.6$ $N$.


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1. ## Checking my work

i need to use the product rule

f(x)= (sgaure root of x)(sinx)

and what I have is

sqaure root of x (cosx) + (sinx)(1/2 x ^ -1/2)

Is that right?? Doesn't seem like it as I am stuck.

2. $\displaystyle (\sqrt{x}\sin x )' = \frac{1}{2\sqrt{x}}\sin x + \sqrt{x}\cos x$

3. i have no clue how you got that.....

4. Originally Posted by MZTakara
i have no clue how you got that.....
pickslides just cleaned up the algebra ...

note that $\dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}$


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Physics
OpenStudy (anonymous):

How does increasing the speed of a wave travels affect its wavelength?

OpenStudy (anonymous):

anyone????????

OpenStudy (anonymous):

www.physicsclassroom.com

OpenStudy (anonymous):

what.... http://www.physicsclassroom.com is too general

OpenStudy (anonymous):

i need specific info

OpenStudy (anonymous):

Is it an electromagnetic wave or mechanical wave?

OpenStudy (anonymous):

electromagnetic wave

OpenStudy (jamesj):

And now you see why linking to www.physicsclassroom.com is not useful

OpenStudy (anonymous):

why?

OpenStudy (anonymous):

ohhhhhhh

OpenStudy (anonymous):

???

OpenStudy (anonymous):

Electromagnetic waves are considered to be traveling at the speed of light for simple analysis, but$\lambda = {v \over f}$From the equation we see a direct correlation between velocity and wavelength. At the same frequency, if we increase velocity, we increase wavelength.

OpenStudy (anonymous):

$wavelength = v \div f$?????

OpenStudy (anonymous):

yes, lambda $$\lambda$$ is the symbol for wavelength.