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## Sunday, September 14, 2014

### Mixin Constraint on Self-typed Methods

This post illustrates the appropriate use of self-type to restrict the composition of (stackable) traits (mixins) in relation to an existing class or trait.

Overview
Mixin traits with self-type restriction has commonly used in Scala. Dependency injection and the Cake pattern in particular, relies on constraint imposed by a trait that it can be used only with subclass of a predefined types. The same approach can be used to constraint a trait to be used with class that support one or several methods.

Note: For the sake of readability of the implementation of algorithms, all non-essential code such as error checking, comments, exception, validation of class and method arguments, scoping qualifiers or import is omitted

Mixin constraint on self-type
Let's consider the case of a class taxonomy (or hierarchy) that classifies machine learning algorithms as either supervised learning or unsupervised learning.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 trait Learning { def predict(x: Double): Double { } } trait SupervisedLearning { def train(x: Array[Double]): Int = { ... } } trait Validation { self: SupervisedLearning => def validate(x: Array[Double]): Double } class SVM extends SupervisedLearning with Validation { override def train(x: Array[Double]): Int = { ... } }

The support vector machine of type SVM is a type of supervised learning algorithm, and therefore extends theSupervisedLearning trait (line 5 & 115). The code snippet compiles because the class SVM (line 14) complies with the restriction imposed by the trait Validation on sub-types of SupervisedLearning (line 10).

  1 2 3 4 5 6 7 8 9 10 11 12 trait UnsupervisedLearning { def group(x: Array[Double]): int } // Validation: failed self-typed inheritance // from SupervisedLearning trait class EM extends UnupervisedLearning with Validation { override def train(x: Array[Double]): Int = { ... } }

The Scala code snippet does not compile because the expectation-maximization algorithm, EM is an unsupervised learning algorithm and therefore is not a sub-class of SupervisedLearning.

Mixin constraint on self-typed method
Marking a trait to be extended (implemented) with sub-class with predefined method(s) is the same as marking a trait to be extended (implemented) with sub-class with predefined type.
Let's reuse the class hierarchy introduced in the previous section.

trait Validation {
self: { def train(x: Array[Double]): Int } =>
def validate(x: Array[Double]): Double = -1.0
}

This time around the Validation can be mixin with a class that implements the method train
As previously seen, the class SVM complies with the restriction imposed by the Validation. However the declaration of the reinforcement learning algorithm QLearning generated a compilation error because it does not implement the method train asd required.

// No training needed for Q-Learning
class QLearning
extends Learning
with Validation{

def predict(x: Double): Double { }
}

Although brief, this introduction to self-referential condition should help you to start considering this technique to protect you code from unintentional erroneous sub-typing and trait mixing.

References
Scala Cookbook A. Alexander O' Reilly 2013
The Scala Programming Language - M. Odersky, L. Spoon, B.Venners - Artima 2007


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# Testing for ammonia by smelling it?

My chemistry teacher taught us that one of the ways to see if we had produced ammonia gas in a reaction was if it released a pungent smell (Even BBC supported that here! http://www.bbc.co.uk/schools/gcsebitesize/science/triple_edexcel/qualitative_analysis/qualitative_analysis/revision/3/ ). However, as far as I knew, Ammonia was poisonous. Is it okay to breathe in small amounts of Ammonia, or is this a hazardous practice that can potentially harm us?

This question What chemical compounds used in qualitative chemistry can be distinguished by smell? did say that Ammonia could harm you, but it doesn't say how much is needed to cause the damage, which is a big part of my question.

Thanks!

-
Yes, large amounts of ammonia are poisonous. However ammonia is a common household cleaner. Very small amounts are ok. Your teacher should have taught you the technique to use the smell test too. You don't stick a nostril over the test tube and deeply inhale, but rather fan the top of the tube with your hand towards your nose to just get a whiff. – MaxW Feb 15 at 16:22

First, using the smell test for chemistry is dangerous - don't do it!

Second, for this an similar questions, consult a Safety Data Sheet (ex-MSDS) for exposure limits.

The copy that I looked at indicates:

TWA of 25ppm (from the ACGIH) or 50ppm (OSHA) -- these are based on a 8 hour/day, 40 hour/week exposure.

STEL of 35ppm (short term exposure limit). Exceeding the STEL may result in 'upper respiratory tract irritation, eye damage'.

Now, the yield from your lab bench chemistry experiment is unlikely to be very high, and as you may have noticed ammonia is very noticeable (although of the several SDSs that I consulted, none list an odor thershold, surprisingly).

But, the bottom line is, do not inhale your chemistry experiments.

-
This answer is an overreaction. The smell test is fine as a conformation for certain chemicals. Ammonia is sold as a common household cleaner. If getting a whiff was deadly there would be dead folks dropping like flies. Obviously the smell test isn't something that you'd use for nerve gas. – MaxW Feb 15 at 18:02
That, of course, depends on the scale of your experiment/bath/whatnot. It is quite likely that his particular case is not deadly. But, it is not made clear if this is a small test tube, or a 30 liter bucket - I've seen 'simple' scale ups result in really bad outcomes involving rescue folks in SCBA. Getting in to the habit of sniffing, particularly without checking an SDS, is a really dumb idea. And, yes, people do drop dead from mixing common household chemicals. And, yes, ammonia could certainly trigger an asthma attack. Use your smell for cooking, not chemistry. – Jon Custer Feb 15 at 23:11
@JonCuster: If a reaction emits dangerous sufficiently dangerous gases that breathing apparatus is required, removing the apparatus to sniff the gases would obviously be a bad idea. If breathing apparatus would not be required, however, I don't think sniffing is any more dangerous than breathing in fumes without trying to sniff them. – supercat Feb 16 at 0:44
And there is a reason we use fume hoods for real work. Look, I would rather tell people to do things the right way rather than telling random folks with unknown facilities and safety support to go around sniffing their chemistry experiments. Much of the time it might work out OK. Some times it won't and people get hurt or killed. In the real world, in real labs, folks get hurt. – Jon Custer Feb 16 at 1:15

I would advise against placing any importance on smell as a test.

1. Just because you smell ammonia doesn't mean it's necessarily ammonia. If humans had such a great sense of smell, we wouldn't need NMR and all those fancy machines, we would just confirm our products by their smells. Identifying chemicals by smell and taste is something that you would do a century ago and certainly not in (CURRENT YEAR).

2. As you progress in your study of chemistry, you will make more and more dangerous compounds which obviously preclude using smell as a test. For goodness' sake, don't do $\ce{[Mo(CO)6] + 3py -> [Mo(CO)3(py)3] + 3CO}$ and try to smell the reaction to tell when it's done. (Not that you would be able to tell anyway!)

Essentially, if you're making ammonia by heating some ammonium salt, please make sure that it also turns damp red litmus paper blue. I know that in some syllabuses you will be required to write "a colourless, pungent gas was evolved" whenever you heat some ammonium salt. If you have to, by all means, write it. Just don't place so much weight on it in your mind.

Safety-wise, if it was going to kill you, it wouldn't be done in a high school lab. You are not going to die if you sniff some ammonia in a high school lab, unless you continuously and voluntarily breathe it in. You'd also have to be either pretty stupid or masochistic to do so, since it stinks.

With that said, the smartest thing to do is to avoid smelling or tasting any of your chemistry experiments. Things that can be smelled or tasted without harm are the exceptions, not the norm.

-
Thank you. We do also use the litmus test for it. I knew it was not going to kill me, but I was mainly wondering if it was a violation of lab safety rules. It looked like a widespread practice, but it seemed kind of illogical to me. – N A Feb 16 at 2:18
"[I]f it was going to kill you, it wouldn't be done in a high school lab"? Never underestimate the ingenuity of high school students! – SQB Feb 16 at 6:46

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# It is a trigonometry related sum and you have to find the value of the given expression

Arun
25763 Points
2 years ago
sqrt (3)*csc20° - sec20°
= sqrt (3)*(1/sin20°) - 1/cos20°
= (sqrt (3)*cos20° - sin20°)/(sin20°*cos20°)
= 4*[(sqrt (3)/2)*cos20° - (1/2) sin20°]/(2sin20°cos20°)
= 4*(cos30°cos20° - sin30°sin20°)/sin (2*20°)
= 4*cos (30°+20°)/sin40°
= 4*cos50°/sin (90°-50°)
= 4*cos50°/cos50°
= 4.
Soumendu Majumdar
159 Points
2 years ago
Dear Student,
Though I know that your question is wrong as per your class since you have posted it I’ll find the value of the expression and you should also know the question should be $\sqrt3 cosec20^{\circ}-sec20^{\circ}$
Now to find the value of the expression you have posted we need to know the values of $cosec20^{\circ}$ & $sec20^{\circ}$
$cosec20^{\circ} =2.92380440294$(You can use an approximated value)
$sec20^{\circ} =1.06417777336$(You can use an approximated value)
so,
$cosec^220^{\circ} =8.54863218665$
$sec^220^{\circ} =1.13247433331$
so the value of $\sqrt3 cosec20^{\circ}-sec20^{\circ}$$=13.6741909491$ (Take approximate value)
Now to find the value of the correct expression,
First, multiply with ½ to both numerator & denominator
so we get
$\sqrt3(0.5)cosec20^{\circ}-0.5sec20^\circ$
$4(cos30^{\circ}cos20{^\circ}-sin30^{\circ}sin20{^\circ})/(2sin20{^\circ}cos20{^\circ})$
=$4 cos(30+20)^\circ /sin(40)^\circ$
=$4 cos50^\circ/sin(90-50)^\circ$
=$4 cos50^\circ/cos(50)^\circ$
=$4$
Hope it helps!


text:
Reading the wiki article I get confused about matrix norms. My question, is it true that $$\lVert Mx \rVert \leq C(\sup_{ij}{M_{ij}}) \lVert x \rVert$$ where $M$ is a matrix and $x$ is a vector and $C$ is some constant? What does $C$ depend on?
According to wiki, $\lVert Mx \rVert \leq \rho(M) \lVert x \rVert$ holds where $\rho$ denotes the spectral radius.
Yes, you can (the exact value of $C$ will depend on the particular choice of norm in the space and the matrix size). ---According to wiki, $\|Mx\|\le\rho(M)\|x\|$ holds where $\rho$ denotes the spectral radius--- This is nonsense as written. Where have you seen it? –  fedja Nov 15 '12 at 19:56


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# Synchronization network

(Redirected from Synchronization networks)

A synchronization network is a network of coupled dynamical systems. It consists of a network connecting oscillators, where oscillators are nodes that emit a signal with somewhat regular (possibly variable) frequency, and are also capable of receiving a signal.[citation needed]

Particularly interesting is the phase transition where the entire network (or a very large percentage) of oscillators begins pulsing at the same frequency, known as synchronization. The synchronization network then becomes the substrate through which synchronization of these oscillators travels. Since there is no central authority organizing nodes, this is a form of self organizing system.[citation needed]

## Definition

Generally, oscillators can be biological, electronic, or physical. Some examples are fireflies, crickets, heart cells, lasers, microwave oscillators, and neurons. Further example can be found in many domains.

In a particular system, oscillators may be identical or non-identical. That is, either the network is made up of homogeneous or heterogeneous nodes.

Properties of oscillators include: frequency, phase and natural frequency.

Network edges describe couplings between oscillators. Couplings may be physical attachment, or consist of some proximity measure through a medium such as air or space.

Networks have several properties, including: number of nodes (oscillators), network topology, and coupling strength between oscillators.

## Kuramoto model

Kuramoto developed a major analytical framework for coupled dynamical systems, as follows: [1][2][3][4][5]

A network of oscillators with varied natural frequencies will be incoherent while the coupling strength is weak.

Letting ${\displaystyle \theta _{i}(t)}$ be the phase of the ${\displaystyle i}$th oscillator and ${\displaystyle \omega _{i}}$ be its natural frequency, randomly selected from a Cauchy-Lorentz distribution as follows,

${\displaystyle g(\omega )={\frac {\gamma }{\pi [\gamma ^{2}+(\omega -\omega _{0})^{2})}}}$, having width ${\displaystyle \gamma }$ and central value ${\displaystyle \omega _{0}}$,

we obtain a description of collective synchronization:

${\displaystyle {\frac {d\theta _{i}}{dt}}=\omega _{i}+{\frac {1}{N}}\sum _{j=1}^{N}K_{ij}\sin(\theta _{j}-\theta _{i}),i=1,...,N}$,

where ${\displaystyle N}$ is the number of nodes (oscillators), and ${\displaystyle K_{ij}}$ is the coupling strength between nodes ${\displaystyle i}$ and ${\displaystyle j}$.

Kuramoto has also developed an "order parameter", which measures synchronization between nodes:

${\displaystyle r(t)={\bigg |}{\frac {1}{N}}\sum _{j=1}^{N}e^{i\theta _{j}(t)}{\bigg |}}$

This leads to the asymptotic definition of ${\displaystyle K_{c}}$, the critical coupling strength, as ${\displaystyle N\to \infty }$ and ${\displaystyle t\to \infty }$

${\displaystyle r={\begin{cases}0,&K

with ${\displaystyle K_{c}=2\gamma }$.

Note that ${\displaystyle r=0\Rightarrow }$ no synchronization, and ${\displaystyle r=|e^{i\theta }|=1\Rightarrow }$ perfect synchronization.

Beyond ${\displaystyle K_{c}}$, each oscillator will belong to one of two groups:

• a group that is synchronized.
• a group that will never synchronize, since their natural frequencies vary too greatly from the synchronization frequency.

## Network topology

Synchronization networks may have many topologies. Topology may have a great deal of influence over the spread of dynamics.[6]

Some major topologies are listed below:

• Regular networks: This describes networks where every node has the same number of links. Lattices, rings, and fully connected networks are some examples of this topology.
• Random graphs: Developed by Erdős and Rényi, these graphs are characterized by a constant probability of a link existing between any two nodes.
• Small world networks: These networks are the result of rewiring a certain number of edges in regular lattice networks. The resulting networks have much smaller average path length than the original networks.
• Scale-free networks: Found ubiquitously in naturally occurring systems, scale free networks are characterized by a large number of high-degree nodes. In particular, the degree distribution follows a power-law.

## History

Coupled oscillators have been studied for many years, at least since the Wilberforce pendulum in 1896. In particular, pulse coupled oscillators were pioneered by Peskin in 1975 with his study of cardiac cells.[7] Winfree developed a mean-field approach to synchronization in 1967, which was developed further in the Kuramoto model in the 1970s and 1980s to describe large systems of coupled oscillators.[8] Crawford brought the tools of manifold theory and bifurcation theory to bear on the stability of synchronization with his work in the mid-1990s.[9] These works coincided with the development of a more general theory of coupled dynamical systems and popularization by Strogatz et al. in 1990, continuing through the early 2000s.


text:
2019
Том 71
№ 8

Gorbachuk M. L.

Articles: 51
Article (Ukrainian)

Spaces of smooth and generalized vectors of the generator of an analytic semigroup and their applications

Ukr. Mat. Zh. - 2017. - 69, № 4. - pp. 478-509

For a strongly continuous analytic semigroup $\{ e^{tA}\}_{t\geq 0}$ of linear operators in a Banach space $B$ we investigate some locally convex spaces of smooth and generalized vectors of its generator $A$, as well as the extensions and restrictions of this semigroup to these spaces. We extend Lagrange’s result on the representation of a translation group in the form of exponential series to the case of these semigroups and solve the Hille problem on description of the set of all vectors $x \in B$ for which there exists $$\mathrm{l}\mathrm{i}\mathrm{m}_{n\rightarrow \infty }\biggl( I + \frac{tA}n \biggr)^n x$$ and this limit coincides with etAx. Moreover, we present a short survey of particular problems whose solutions are necessary for the introduction of the above-mentioned spaces, namely, the description of all maximal dissipative (self-adjoint) extensions of a dissipative (symmetric) operator; the representation of solutions to operator-differential equations on an open interval and the analysis of their boundary values, and the existence of solutions to an abstract Cauchy problem in various classes of analytic vector-valued functions.

Anniversaries (Ukrainian)

Mykola Oleksiiovych Perestyuk (on his 70th birthday)

Ukr. Mat. Zh. - 2016. - 68, № 1. - pp. 142-144

Article (Ukrainian)

Representations of a Group of Linear Operators in a Banach Space on the Set of Entire Vectors of its Generator

Ukr. Mat. Zh. - 2015. - 67, № 5. - pp. 592-601

For a strongly continuous one-parameter group $\{U(t)\} t ∈(−∞,∞)$ of linear operators in a Banach space $\mathfrak{B}$ with generator $A$, we prove the existence of a set $\mathfrak{B}_1$ dense in $\mathfrak{B}_1$ on the elements $x$ of which the function $U(t)x$ admits an extension to an entire B$\mathfrak{B}$-valued vector function. The description of the vectors from $\mathfrak{B}_1$ for which this extension has a finite order of growth and a finite type is presented. It is also established that the inclusion $x ∈ \mathfrak{B}_1$ is a necessary and sufficient condition for the existence of the limit ${ \lim}_{n\to 1}{\left(I+\frac{tA}{n}\right)}^nx$ and this limit is equal to $U(t)x$.

Article (Ukrainian)

History of the Appearance of Infinite-Dimensional Analysis and its Development in Ukraine

Ukr. Mat. Zh. - 2014. - 66, № 8. - pp. 1058–1073

We present a brief survey of the development of functional analysis in Ukraine and the problems of infinite-dimensional analysis posed and solved for thousands of years, which laid the foundations of this branch of mathematics.

Anniversaries (Ukrainian)

Yurii Stephanovych Samoilenko (on his 70th birthday)

Ukr. Mat. Zh. - 2013. - 65, № 10. - pp. 1408-1409

Anniversaries (Ukrainian)

Anatolii Mykhailovych Samoilenko (on his 75th birthday)

Ukr. Mat. Zh. - 2013. - 65, № 1. - pp. 3 - 6

Anniversaries (Ukrainian)

Oleksandr Ivanovych Stepanets’ (on the 70 th anniversary of his birthday)

Ukr. Mat. Zh. - 2012. - 64, № 5. - pp. 579-581

Anniversaries (Ukrainian)

Yuri Yurievich Trokhimchuk (on his 80th birthday)

Ukr. Mat. Zh. - 2008. - 60, № 5. - pp. 701 – 703

Anniversaries (Ukrainian)

Fifty years devoted to science (on the 70th birthday of Anatolii Mykhailovych Samoilenko)

Ukr. Mat. Zh. - 2008. - 60, № 1. - pp. 3–7

Obituaries (Ukrainian)

Alexander Ivanovich Stepanets

Ukr. Mat. Zh. - 2007. - 59, № 12. - pp. 1722-1724

Anniversaries (Ukrainian)

Mark Grigorievich Krein (to the centenary of his birth)

Ukr. Mat. Zh. - 2007. - 59, № 5. - pp. 579-587

Article (Ukrainian)

Phragmén-Lindelöf theorem for solutions of elliptic differential equations in a banach space

Ukr. Mat. Zh. - 2007. - 59, № 5. - pp. 650–657

For a second-order elliptic differential equation considered on a semiaxis in a Banach space, we show that if the order of growth of its solution at infinity is not higher than the exponential order, then this solution tends exponentially to zero at infinity.

Anniversaries (Ukrainian)

Evgen Yakovich Khruslov (on his 75 th birthday)

Ukr. Mat. Zh. - 2007. - 59, № 4. - pp. 549-550

Anniversaries (Ukrainian)

On the 90th birthday of Yurii Alekseevich Mitropol’skii

Ukr. Mat. Zh. - 2007. - 59, № 2. - pp. 147–151

Anniversaries (Ukrainian)

Yaroslav Borisovich Lopatinsky (09.11.1906 - 03.10.1981)

Ukr. Mat. Zh. - 2006. - 58, № 11. - pp. 1443-1445

Article (Ukrainian)

On the correct solvability of the Dirichlet problem for operator differential equations in a Banach space

Ukr. Mat. Zh. - 2006. - 58, № 11. - pp. 1462–1476

We investigate the structure of solutions of an equation $y″(t) = By(t)$, where $B$ is a weakly positive operator in a Banach space B, on the interval $(0, \infty)$ and establish the existence of their limit values as $t → 0$ in a broader locally convex space containing $B$ as a dense set. The analyticity of these solutions on $(0, \infty)$ is proved and their behavior at infinity is studied. We give conditions for the correct solvability of the Dirichlet problem for this equation and substantiate the applicability of power series to the determination of its approximate solutions.

Article (Ukrainian)

On the behavior of orbits of uniformly stable semigroups at infinity

Ukr. Mat. Zh. - 2006. - 58, № 2. - pp. 148–159

For uniformly stable bounded analytic $C_0$-semigroups $\{T(t)\} t ≥ 0$ of linear operators in a Banach space $B$, we study the behavior of their orbits $T (t)x, x ∈ B$, at infinity. We also analyze the relationship between the order of approaching the orbit $T (t)x$ to zero as $t → ∞$ and the degree of smoothness of the vector $x$ with respect to the operator $A^{-1}$ inverse to the generator A of the semigroup $\{T(t)\}_{t \geq 0}$. In particular, it is shown that, for this semigroup, there exist orbits approaching zero at infinity not slower than $e^{-at^{\alpha}}$, where $a > 0,\; 0 < \alpha < \pi/(2 (\pi - 0 )),\; \theta$ is the angle of analyticity of $\{T(t)\}_{t \geq 0}$, and the collection of these orbits is dense in the set of all orbits.

Anniversaries (Ukrainian)

Leonid Pavlovych Nyzhnyk (on his 70-th birthday)

Ukr. Mat. Zh. - 2005. - 57, № 8. - pp. 1120-1122

Anniversaries (Ukrainian)

Yurij Makarovich Berezansky (the 80th anniversary of his birth)

Ukr. Mat. Zh. - 2005. - 57, № 5. - pp. 3-11

Article (Ukrainian)

Functional Analysis in the Institute of Mathematics of the National Academy of Sciences of Ukraine

Ukr. Mat. Zh. - 2005. - 57, № 5. - pp. 582–600

We give a brief survey of results on functional analysis obtained at the Institute of Mathematics of the Ukrainian National Academy of Sciences from the day of its foundation.

Article (Ukrainian)

Direct and Inverse Theorems in the Theory of Approximation by the Ritz Method

Ukr. Mat. Zh. - 2005. - 57, № 5. - pp. 633–643

For an arbitrary self-adjoint operator B in a Hilbert space $\mathfrak{H}$, we present direct and inverse theorems establishing the relationship between the degree of smoothness of a vector $x \in \mathfrak{H}$ with respect to the operator B, the rate of convergence to zero of its best approximation by exponential-type entire vectors of the operator B, and the k-modulus of continuity of the vector x with respect to the operator B. The results are used for finding a priori estimates for the Ritz approximate solutions of operator equations in a Hilbert space.

Anniversaries (Ukrainian)

Dmytro Yakovych Petryna (on his 70 th birthday)

Ukr. Mat. Zh. - 2004. - 56, № 3. - pp. 291-292

Anniversaries (Ukrainian)

Mykhailo Vasyl'ovych Ostrograds'kyi and his role in the development of mathematics

Ukr. Mat. Zh. - 2001. - 53, № 8. - pp. 1011-1023

Anniversaries (Ukrainian)

Yurii Makarovich Berezanskii

Ukr. Mat. Zh. - 2000. - 52, № 5. - pp. 579-581

Article (Ukrainian)

On analytic solutions of operator differential equations

Ukr. Mat. Zh. - 2000. - 52, № 5. - pp. 596-607

We find conditions on a closed operator A in a Banach space that are necessary and sufficient for the existence of solutions of a differential equation y′(t) = Ay(t), t ∈[0,∞),in the classes of entire vector functions with given order of growth and type. We present criteria for the denseness of classes of this sort in the set of all solutions. These criteria enable one to prove the existence of a solution of the Cauchy problem for the equation under consideration in the class of analytic vector functions and to justify the convergence of the approximate method of power series. In the special case where A is a differential operator, the problem of applicability of this method was first formulated by Weierstrass. Conditions under which this method is applicable were found by Kovalevskaya.

Article (Ukrainian)

On one generalization of the Berezanskii evolution criterion for the self-adjointness of operators

Ukr. Mat. Zh. - 2000. - 52, № 5. - pp. 608-615

We describe all weak solutions of a first-order differential equation in a Banach space on (0, ∞) and investigate their behavior in the neighborhood of zero. We use the results obtained to establish necessary and sufficient conditions for the essential maximal dissipativity of a dissipative operator in a Hilbert space.

Anniversaries (Ukrainian)

Bohdan Iosypovych Ptashnyk

Ukr. Mat. Zh. - 1997. - 49, № 9. - pp. 1155–1156

Article (Ukrainian)

Behavior of the extended spectral measure of a self-adjoint operator at infinity

Ukr. Mat. Zh. - 1997. - 49, № 4. - pp. 510–516

We study the dependence of the rate of growth of the extended spectral measure of a self-adjoint operator at infinity on the order of singularity of the vector on which this measure is considered.

Anniversaries (Ukrainian)

Yurii Dmitrievich Sokolov (on his 100th birthday)

Ukr. Mat. Zh. - 1996. - 48, № 11. - pp. 1443-1445

Article (Ukrainian)

Estimate of error of an approximated solution by the method of moments of an operator equation

Ukr. Mat. Zh. - 1996. - 48, № 11. - pp. 1477-1483

For an equationAu = f whereA is a closed densely defined operator in a Hilbert spaceH, f εH, we estimate the deviation of its approximated solution obtained by the moment method from the exact solution. All presented theorems are of direct and inverse character. The paper refers to direct methods of mathematical physics, the development of which was promoted by Yu. D. Sokolov, the well-known Ukrainian mathematician and mechanic, a great humanitarian and righteous man. We dedicate this paper to his blessed memory.

Anniversaries (Ukrainian)

On the 60th birthday of Leonid Pavlovich Nizhnik

Ukr. Mat. Zh. - 1995. - 47, № 10. - pp. 1416–1417

Article (Ukrainian)

Theory of self-adjoint extensions of symmetric operators. entire operators and boundary-value problems

Ukr. Mat. Zh. - 1994. - 46, № 1-2. - pp. 55–62

Article (Ukrainian)

Inverse problem for a sturm-liouville elliptic equation in a Hilbert space

Ukr. Mat. Zh. - 1991. - 43, № 11. - pp. 1537–1540

Anniversaries (Russian)

Samuil Davidovich Eidelman (On his sixtieth birthday)

Ukr. Mat. Zh. - 1991. - 43, № 5. - pp. 578

Article (Ukrainian)

Extension theory for symmetric operators and boundary value problems for differential equations

Ukr. Mat. Zh. - 1989. - 41, № 10. - pp. 1299–1313

Article (Ukrainian)

Convergence of series of independent Gaussian operators

Ukr. Mat. Zh. - 1984. - 36, № 4. - pp. 500 – 502

Article (Ukrainian)

Behavior at infinity of solutions of operator differential equations

Ukr. Mat. Zh. - 1983. - 35, № 4. - pp. 489—493

Article (Ukrainian)

Weak solutions of differential equations in Hilbert space

Ukr. Mat. Zh. - 1981. - 33, № 4. - pp. 513–518

Article (Ukrainian)

Scattering problem for first-order differential equations with operator coefficients

Ukr. Mat. Zh. - 1978. - 30, № 4. - pp. 452–461

Article (Ukrainian)

Some questions of the spectral theory of differential equations of elliptic type in the space of vector-functions

Ukr. Mat. Zh. - 1976. - 28, № 3. - pp. 313–324

Article (Ukrainian)

Some questions of the spectral theory of differential equations of elliptic type in the space of vector functions on a finite interval

Ukr. Mat. Zh. - 1976. - 28, № 1. - pp. 12–26

Article (Ukrainian)

The spectrum of self-adjoint extensions of the minimal operator generated by a Sturm-Liouville equation with operator potential

Ukr. Mat. Zh. - 1972. - 24, № 6. - pp. 726—734

Article (Ukrainian)

Classes of boundary-value problems for the sturm — Liouville equation with an operator potential

Ukr. Mat. Zh. - 1972. - 24, № 3. - pp. 291—304

Article (Ukrainian)

Problems of the spectral theory of the second order linear differential equation with unbounded operator coefficients

Ukr. Mat. Zh. - 1971. - 23, № 1. - pp. 3–14

Article (Ukrainian)

On self-adjoint semibounded abstract differential operators

Ukr. Mat. Zh. - 1970. - 22, № 6. - pp. 806—808

Article (Ukrainian)

The stability of motion of a solid body containing a liquid-filled cavity suspended by a cord

Ukr. Mat. Zh. - 1968. - 20, № 5. - pp. 586–602

Article (Ukrainian)

Basic results of studies at the mathematical-analysis department of the Institute of Mathematics of the Academy of Sciences of the Ukrainian SSR

Ukr. Mat. Zh. - 1967. - 19, № 6. - pp. 73–92

Brief Communications (Russian)

On the continuation of positively definite functions of two variables

Ukr. Mat. Zh. - 1965. - 17, № 5. - pp. 96-102

Brief Communications (Russian)

On the description of continuations of a positively definite operator function

Ukr. Mat. Zh. - 1965. - 17, № 5. - pp. 102-110

Article (Russian)

On the representation of positively determined operator functions

Ukr. Mat. Zh. - 1965. - 17, № 2. - pp. 29-46

Brief Communications (Russian)

Spectral expansions of Infinitesimal operators of the fourth dimensional group of translations

Ukr. Mat. Zh. - 1964. - 16, № 4. - pp. 528-534


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• # Terence TAO in U4/8, il 22.06.2018 alle ore 14:30

Lezioni LEONARDESCHE

Venerdì, 22 Giugno 2018, ore 14.30,

in Aula "Luisella Sironi", Edificio U4, Piazza della Scienza, 4 - Milano

il Professor TERENCE TAO

(University  of California, Los Angeles)

terrà un seminario dal titolo: VAPORIZING AND FREEZING THE RIEMANN ZETA FUNCTION

Abstract:  In 1950, de Bruijn studied the effect of evolving the Riemann zeta function (or more precisely, a closely related function known as the Riemann xi function) by the (backwards) heat equation. His analysis, together with later work by Newman, showed that there existed a finite constant Lambda, at most 1/2 in value, such that the Riemann hypothesis for this evolved function was true at times greater than or equal to Lambda, and false below that threshold. Thus the Riemann hypothesis for the zeta function is equivalent to Lambda being non-positive. Recently, in joint work with Brad Rodgers, I was able to establish the complementary estimate that Lambda is non-negative, confirming a conjecture of Newman; thus, the Riemann hypothesis for zeta, if true, is only “barely so”. The proof relies on an analysis of the dynamics of zeroes of entire functions under heat flow; it turns out that as one evolves forward in time, the zeroes “freeze” into approximate arithmetic progressions, while if one evolves backwards, the zeroes “vaporize” to leave the critical line. In followup work in an online collaborative “Polymath” project, the upper bound on Lambda has also been improved. We describe these results and their proofs in this talk.

Tea Break: ore 15.30

ABOUT THE SPEAKER: Terence Tao is nothing short of a mathematical superstar whose breadth, depth and innovative spirit is rapidly changing the mathematical landscape. Gifted with a prodigious talent which propelled him to a gold medal at the International Mathematical Olympiad at age 13 and a Full Professorship at UCLA at the age of 24, Tao has combined this talent with his stated vision that “mathematics is a unified subject and I am particularly happy when I get the opportunity to work on a project that involves several fields at once”. Tao’s spectacular problem solving skills in a have earned him a list of prizes and awards too long to mention, but it is telling that the citation for his 2006 Fields Medal reads “for his contributions to partial differential equations, combinatorics, harmonic analysis and additive number theory” which shows the unprecedented range of his activity that continues to expand. Among his most well-known achievements, Tao together with various collaborators has established: existence of arithmetic sequences of prime numbers of arbitrary length, the first entropy estimates in analytic number theory, wide ranging results for nonlinear Schrödinger equations, L^p harmonic analysis and critical Sobolev regularity for wave maps, the solution of the Kakeya problem in geometric measure theory and the Horn conjecture on eigenvalue inequalities for Hermitian matrices. Tao is also a pioneer in mathematical exposition and interaction in the internet age with an unprecedented success in mathematical blogs which have stimulated countless textbooks and collaborations worldwide. About him, fellow Fields medalist Timothy Gowers speculated that Terence Tao may be the modern day David Hilbert who comes to know all of mathematics. Tao is currently holds the James and Carol Collins Chair in Mathematics at UCLA.


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## VC++ is confused by registry/drives

I had an extra hard drive present in my PC which had drive allocation D:. I installed MS Visual Studio to drive E:. I then had to remove the Drive D: which meant that the drive that was E: is now D:. Consequently all the registry entries for MS Visual Studio are now pointing to the wrong drive. There are too many entries for me to correct easily and I've now lost the facility to correct it myself by writing a program! Can somebody help me out? Given that all registry entries of "E:\" should be "D:\"... Is there another way?? Please help! It's very frustrating....


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# Values of $\alpha$, $\beta$ and $\gamma$ in ets in forecast package

I am using the forecast package in R. I wanted to know how the ets() function finds the value of $\alpha$, $\beta$ and $\gamma$?

Is it by minimizing the SSE or some other criteria?

-
The question is clearly answered by the help for the function - ?ets –  Glen_b Nov 21 '12 at 0:17


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# Why do accelerated charges emits a photon?

Is it only when the acceleration is negative? If yes, when it is positive it absorbs a photon?

Demystifier
Gold Member
Perhaps you should first understand why classical (not quantum) charge emits electromagnetic radiation.

ZapperZ
Staff Emeritus
This thread has been moved from QM sub-forum to the classical physics sub-forum. I agree that the OP needs to first understand this from the classical E&M perspective.

Zz.

Andrew Mason
Homework Helper
This is an important question and, remarkably, one that has no definitive answer as far as I can tell.

A charged particle that is accelerated by gravity does not emit an em wave/photon. The only other way to accelerate a charged particle is to apply a mechanical or electro-magnetic force to it over some distance. And a mechanical force is nothing but an em force. Thus, the observational evidence amounts to:
a charged particle does not emit an electro-magnetic wave or photon except when it receives electro-magnetic energy.

So this leads to a reformulation of the question: does the charged particle emit a photon because it accelerates? Or does it emit a photon in response to receiving a photon, which incidentally causes it to accelerate?

The question seems to be still unresolved: See this site, for example.

AM

olgranpappy
Homework Helper
This is an important question and, remarkably, one that has no definitive answer as far as I can tell.

This very cute problem is addressed to some extent by Peierls in his little book "Surprises in Theoretical Physics".

A charged particle that is accelerated by gravity does not emit an em wave/photon.
What is your basis for that assertion?

olgranpappy
Homework Helper
What is your basis for that assertion?

Apparently, the principle of equivalence...

Consider a man standing on a scale on earth. The scale reads, say, 170 pounds.

Consider a man in completely empty space in a rocketship accelerating at 9.8m/s^2. The man is standing on a scale. The scale reads 170 pounds.

Now, consider the exact same situation but now with a point charge stapled to the man's nose.

If the principle of equivalence holds then... well, shouldn't the stationary point charge in a gravitational field be radiating? Regardless of whether or not it "should", it does not.

Andrew Mason
Homework Helper
What is your basis for that assertion?
Further to what olegranpappy has said, according to the principle of equivalence, a charged particle accelerating in freefall in a gravitational field is locally equivalent to a charged particle at rest in an inertial frame of reference. Since a charged particle at rest in an inertial frame does not radiate, the same must hold true for the charged particle in freefall. This is consistent with all observation.

A charged particle that is stationary in a gravitional field is locally equivalent to a charged particle accelerating in a gravity-free space. So it should radiate. This is a bit of a problem, because as olegranpappy points out, it is not observed. Here is a paper that explains why this occurs - it suggests that the radiation is a relative phenomenon that depends on the relative acceleration between the observer and the charge:

From "Radiation from a Charge in a Gravitational Field" said:
It is found that the “naive” conclusion from the principle of equivalence - that a freely falling charge does not radiate, and a charge supported at rest in a gravitational field does radiate - is a correct conclusion, and one should look for rdiation whenever a relative acceleration exists between an electric charge and its electric field. The electric field which falls freely in the gravitational field is accelerated relative to the static charge. The field is curved, and the work done in overcoming the stress force created in the curved field, is the source of the energy carried by the radiation. This work is done by the gravitational field on the electric field, and the energy carried by the radiation is created in the expence of the gravitational energy of the system.

AM

Last edited:
Further to what olegranpappy has said, according to the principle of equivalence, a charged particle accelerating in freefall in a gravitational field is locally equivalent to a charged particle at rest in an inertial frame of reference. Since a charged particle at rest in an inertial frame does not radiate, the same must hold true for the charged particle in freefall. This is consistent with all observation.
AM
But I thought that principle of equivalence contradicts special relativity, and by consequence, classical electromagnetic theory as well.

olgranpappy
Homework Helper
But I thought that principle of equivalence contradicts special relativity, and by consequence, classical electromagnetic theory as well.

Hopefully not too much, since special relativity and classical electromagnetic theory are correct.

Andrew Mason
Homework Helper
But I thought that principle of equivalence contradicts special relativity, and by consequence, classical electromagnetic theory as well.
The General Theory and the Special Theory of Relativity are perfectly consistent. The principle of equivalence is the cornerstone of GR. Maxwell's equations are consistent with both - certainly with SR. In fact, the belief that Maxwell's equations are valid in all inertial frames of reference was one of the things that drove Einstein to develop SR.

AM

Andrew Mason
Homework Helper
A charged particle that is stationary in a gravitional field is locally equivalent to a charged particle accelerating in a gravity-free space. So it should radiate.
I should have added after: "So it should radiate"... "IF accelerating charges radiate because they are accelerating.

AM

olgranpappy
Homework Helper
Oh, I finally found that book I was talking about earlier. Here's some highlights:

Although the formula for the rate at which a small object loses energy via radiation is
$$\frac{dW}{dt}=\frac{2e^2}{3c^3}a^2$$
where a is acceleration. And this gives the right answer for total loss. It can not be the right expression for instantaneous rate of loss of energy.

blah blah blah. For periodic or other types of motion we can use instead the expression
$$\frac{dW}{dt}=\frac{-2e^2}{3c^2}v\dot a$$
where v is velocity and $$\dot a$$ is "jerk".

Unfortunately, further thought shows that this rewriting does not resolve the paradox... actually, it gets a bit complicated. A careful discussion was given by Rohrlich and Fulton in Ann. Phys. 9, 499 (1960). But the paradox was not truly resolved until arguments by D. Boulware came along (unpublished as of 1979, but given by Peierls in his book).

Hi,

I was reading this thread and I have an additional question to ask. If an electron and a proton were accelerating at the same rate, would they emit the same photon?

Last edited:
Andrew Mason
Homework Helper
Hi,

I was reading this thread and I have an additional question to ask. If an electron and a proton were accelerating at the same rate, would they emit the same photon?
The force on, or rate of momentum change of, the proton would be 1800 times greater than that for the electron. This means that the electromagnetic interaction needed to cause this (ie. the momentum of the photon) would have to be 1800 times greater.

AM

So, it would take really high frequencies of light to accelerate a proton? And would the proton re-emit this frequency of light as well?

A related question:
If we were to accelerate a proton using a very weak magnetic field, would it emit radiation?

This is a bit of a problem, because as olegranpappy points out, it is not observed.

You're saying that if an apparatus stands stationary to the Earth, it can not detect radiation from co-supported charges (this much seems sensible, else where would the energy come from: no other part of the system has given up change). But just to clarify:

If a charge is released to fall freely past the apparatus (and vice-versa), is radiation then detected? If the apparatus is placed on (say) an accelerating train, will it continue not to detect radiation from co-accelerated charges?

Actually, my understanding was that the field lines of a constantly accelerated charge merely "droop". This isn't radiation if you can stay stationary with respect to charge and the direction of acceleration. I think the equivalence principle here is unscathed (until you add boundary conditions, but that feels like cheating).

Last edited:
olgranpappy
Homework Helper
Hi,

I was reading this thread and I have an additional question to ask. If an electron and a proton were accelerating at the same rate, would they emit the same photon?

classically you would use the same formula to determine the power radiated, there's no reference to mass in that formula.

Andrew Mason
Homework Helper
If a charge is released to fall freely past the apparatus (and vice-versa), is radiation then detected? If the apparatus is placed on (say) an accelerating train, will it continue not to detect radiation from co-accelerated charges?
There is a difference between a charge in free fall in a gravitational field and a charge accelerating due to a non-gravitational force (ie. an electromagnetic force). In the first case it does not radiate. In the second, it does.

Actually, my understanding was that the field lines of a constantly accelerated charge merely "droop". This isn't radiation if you can stay stationary with respect to charge and the direction of acceleration. I think the equivalence principle here is unscathed (until you add boundary conditions, but that feels like cheating).
I am not sure I understand what you mean here.

AM

There is a difference between a charge in free fall in a gravitational field and a charge accelerating due to a non-gravitational force (ie. an electromagnetic force). In the first case it does not radiate. In the second, it does.
Do you have evidence of that? How do you define "radiate"?
I am not sure I understand what you mean here.
I say that to a classical detector-apparatus, "radiation" means any time-varying field at the apparatus.

Now, if an apparatus inside an elevator is stationary with respect to some charge, regardless of whether the elevator is "standing in a planet's gravitational field" or "accelerating constantly in space" the apparatus will detect no radiation (because in both cases the field at the apparatus is time-independent). Hence I disagree with your statement that there is some problem with the equivalence principle.

(As for the rough shape of the field inside the elevator, it happens to droop -- in both cases as per the equivalence principle -- as illustrated here. Actually if you're interested in the exact shape, you should be able to distinguish the two cases due to the differing global geometry.)

Andrew Mason
Homework Helper
Do you have evidence of that?
Sure. An electron whipping around a curved path radiates (synchrotron radiation). Microgravity experiments have shown that a falling electron does not radiate: .
Emit a photon.

I say that to a classical detector-apparatus, "radiation" means any time-varying field at the apparatus.

Now, if an apparatus inside an elevator is stationary with respect to some charge, regardless of whether the elevator is "standing in a planet's gravitational field" or "accelerating constantly in space" the apparatus will detect no radiation (because in both cases the field at the apparatus is time-independent). Hence I disagree with your statement that there is some problem with the equivalence principle.
I never said that there was a problem with the principle of equivalence. There is no problem. The problem may be with the theory that a charge radiates because it accelerates. Outside a gravitational field, a charge cannot accelerate unless it interacts with a an em field ie. a photon. Perhaps a charge radiates because it interacts with a photon which, because it has momentum, causes the charge (which has mass) to accelerate.

AM

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I'm trying to avoided muddying the issue with photons, since this thread was tossed to the classical forum.

So you're saying you think charges only radiate if they are driven by an electromagnetic force (not allowed to fall)? So then, if the moon was given an electric charge, you think it also would not radiate (presuming one could detect light-month wavelengths)?

As I understand it, you have to be *very* careful about things. We're talking about 3 different things here -- the charged particle, the gravitational field, and the observing equipment. They can be all moving relative to each other. Assuming that we've got a (quasi-)uniform gravitational field, there are a few interesting cases:

1. The charge is not moving relative to the measuring equipment, but both are free-falling in the gravitational field. In this case, you get the same answer as in flat space, not moving.

2. The charge is not moving relative to the measuring equipment. They are also stationary relative to the gravitational field. In this case, you get the same answer as if you were all accelerated, as required by the equivalence principle. Thus, you get the "drooping" referred to in the papers so far cited.

3. The measuring instrument is stationary wrt gravitational field. The charge free-falls. We get the same pattern as if a uniformly accelerating charge moves past.

I think that's what the equations say... Corrections anyone?

The measuring instrument is stationary wrt gravitational field
I presume you mean 'supported in a grav. field' as in
charges only radiate if they are driven by an electromagnetic force (not allowed to fall)?

Andrew Mason


text:
MathSciNet bibliographic data MR1134992 (92m:43004) 43A22 (43A15) Furuta, Koji Algebras \$A\sb p\$$A\sb p$ and \$B\sb p\$$B\sb p$ and amenability of locally compact groups. Hokkaido Math. J. 20 (1991), no. 3, 579–591. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
My Math Forum Infinite geometric series application

 Calculus Calculus Math Forum

 May 12th, 2011, 10:55 AM #1 Newbie   Joined: Dec 2010 Posts: 9 Thanks: 1 Infinite geometric series application A doctor prescribes a 240 milligram (mg), pain-reducing drug to a patient who has chronic pain. The medical instructions read that this drug should be taken every 4 hours. After 4 hours, 60% of the original dose leaves the body. Under these conditions, the amount of drug remaining in the body, at 4-hour intervals, forms a geometric series. 1. Supposing that the patient takes just one dose of the medicine write an equation for the amount of the drug in the patient's blood stream t hours after taking the medicine Qe^-(ct) where c is a positive constant? 2. How many mgs of the drug are present in the body after 4 hours? (just after second dose?) 3. Graph the amount of medicine in the blood stream for the first 24 hour period. 4. Show that the amount of medicine in the patient's bloodstream after the Nth dose can be expressed by a geometric series. Use sigma notation to express the series. 5. The minimum lethal dosage of this pain-reducing drug is 600mg. if the patient is continuously (infinitely) taking the drug as prescribed, will the patient ever have this much of the drug in their body? 6. If the patient decides to abuse the drug and take it every 2 hours, against the doctor’s orders, then will the patient meet reach the minimal lethal dosage? Thanks from Elizabeth Thomas Last edited by skipjack; December 12th, 2018 at 08:41 PM.
 May 12th, 2011, 11:05 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Infinite geometric series application 1. Since we are told the concentration of the drug decays exponentially, we know the time rate of change of the amount of the drug in the patient's bloodstream is proportional to the amount A(t). We may assume initially at time t = 0, A(t) = 240 mg. This gives rise to the IVP: $\displaystyle \frac{dA}{dt}=-kA$, (k > 0), $\displaystyle A(0)=A_0$ Separate variables: $\displaystyle \frac{1}{A}\,dA=-k\,dt$ Integrate, using boundaries as limits of integration: $\displaystyle \int_{A_0}^A\frac{1}{A}\,dA=-k\int_0^t\,dt$ $\displaystyle \ln\left(\frac{A}{A_0}\right)=-kt$ Convert from logarithmic to exponential form: $\displaystyle \frac{A}{A_0}=e^{-kt}$ $\displaystyle A=A_0e^{-kt}$ We are told it takes 4 hours for 60% of the dose to leave the body. This means A(4) = 0.4A(0)...this gives: $\displaystyle \ln\left(\frac{\frac{2}{5}A_0}{A_0}\right)=-k(4)$ $\displaystyle \ln\left(\frac{2}{5}\right)=-4k$ $\displaystyle \frac{\ln\left(\frac{2}{5}\right)}{4}=-k$ $\displaystyle -k=\frac{1}{4}\ln\left(\frac{2}{5}\right)=\ln\left( \left(\frac{2}{5}\right)^{\frac{1}{4}}\right)$ Now, substituting for -k into the expression for A gives: $\displaystyle A = A_0e^{\ln\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)t}$ $\displaystyle A = A_0\left(e^{\ln\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)}\right)^t$ $\displaystyle A = A_0\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)^t$ $\displaystyle A = A_0\left(\frac{2}{5}\right)^{\frac{t}{4}}$ So, with $\displaystyle A_0=240\text{ mg}$ we have: $\displaystyle A(t) = 240\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ 2. Just after the second dose, we would then have: $\displaystyle A(t)=\left(240\left(\frac{2}{5}\right)^{\frac{4}{4 }} + 240\right)\text{ mg} = \frac{7}{5}240\text{ mg} = 336\text{ mg}$ 3.) I don't own any graphing software, so we'll have to piece this together. As we found in part 1, the first 4 hours (up until the 2nd dose is taken) is given by: $\displaystyle A(t) = 240\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ Then, for the 2nd four hour period, using the result from part 2, we have: $\displaystyle A(t) = 336\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ For the 3rd four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}336\right) \left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 374.4\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text {mg}$ For the 4th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}374.4\right)\left(\frac{2}{5} \right)^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 389.76\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ For the 5th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}389.76\right)\left(\frac{2}{5 }\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 395.904\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ For the 6th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}395.904\right)\left(\frac{2}{ 5}\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 398.3616\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ Graph these six pieces for each successive 4 hour period. 4. As we saw in part 3, the amount A after the doses is: dose...amount (mg) 1..........240 2..........240 + 0.4�240 = 1.4�240 = 336 3..........240 + 0.4(240 + 0.4�240) = 240(1+0.4+0.4�) = 374.4 We see then that the amount after the nth dose will be given by: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\frac{2}{5}\right)^k$ 5. Using the formula for the first n terms of a geometric series, we get: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\frac{2}{5}\right)^k=240\frac{1-\left(\frac{2}{5}\right)^n}{1-\frac{2}{5}} = 400\left(1-\left(\frac{2}{5}\right)^n\right)$ Thus, we find: $\displaystyle \lim_{n\to\infty}A_n = 400\text{ mg}$ So we see the patient will never reach the lethal dose of 600 mg. 6. To answer this, we need to know what portion of the drug remains in the body after 2 hours. Using the formula from part 1: $\displaystyle A = A_0\left(\frac{2}{5}\right)^{\frac{2}{4}} = A_0\sqrt{\frac{2}{5}}$ So now, we have: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\sqrt{\frac{2}{5}}\right)^k = 240\frac{1-\left(\sqrt{\frac{2}{5}}\right)^n}{1-\sqrt{\frac{2}{5}}} = \left(400+80\sqrt{10}\right)\left(1-\left(\sqrt{\frac{2}{5}}\right)^n\right)$ $\displaystyle \lim_{n\to\infty}A_n = 400+80\sqrt{10}\text{ mg}\approx652.98\text{ mg}$ So we see the patient will eventually reach the lethal dosage. To find after which dosage, we may use: $\displaystyle \left(400+80\sqrt{10}\right)\left(1-\left(\sqrt{\frac{2}{5}}\right)^n\right) = 600$ Now we solve for n: $\displaystyle 1-\left(\sqrt{\frac{2}{5}}\right)^n = \frac{600}{400+80\sqrt{10}}=\frac{5-\sqrt{10}}{2}$ $\displaystyle \left(\frac{2}{5}\right)^{\frac{n}{2}} = 1-\frac{5-\sqrt{10}}{2}=\frac{\sqrt{10}-3}{2}$ Take the natural logarithm of both sides: $\displaystyle \ln\left(\left(\frac{2}{5}\right) ^{\frac{n}{2}}\right) = \ln\left(\frac{\sqrt{10}-3}{2}\right)$ $\displaystyle \frac{n}{2}\ln\left(\frac{2}{5}\right) = \ln\left(\frac{\sqrt{10}-3}{2}\right)$ $\displaystyle n = 2\frac{\ln\left(\frac{\sqrt{10}-3}{2}\right)}{\ln\left(\frac{2}{5}\right)}\approx5 .48$ Thus, the patient will reach a lethal amount after the 6th dose. Last edited by skipjack; December 12th, 2018 at 08:34 PM.
 May 12th, 2011, 04:53 PM #3 Newbie   Joined: Dec 2010 Posts: 9 Thanks: 1 Re: Infinite geometric series application Thanks a ton for the solution! I'll post back if I come up with any questions. Last edited by skipjack; December 12th, 2018 at 08:36 PM.
 May 12th, 2011, 04:57 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Infinite geometric series application You're welcome!
 September 16th, 2015, 10:37 AM #5 Newbie   Joined: Sep 2015 From: malaysia Posts: 1 Thanks: 0 I would like to ask... where did 4c/t come from (the solution of fourth four hour period)? The second question is why I did not get the value of lethal dose of paracetamol when I use the infinite limit geometric series? Last edited by skipjack; December 12th, 2018 at 08:37 PM.
December 11th, 2018, 12:45 AM   #6
Newbie

Joined: Dec 2018
From: United States

Posts: 1
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Quote:
 Originally Posted by MarkFL 1. Since we are told the concentration of the drug decays exponentially, we know the time rate of change of the amount of the drug in the patient's bloodstream is proportional to the amount A(t). We may assume initially at time t = 0, A(t) = 240 mg. This gives rise to the IVP: $\displaystyle \frac{dA}{dt}=-kA$, (k > 0), $\displaystyle A(0)=A_0$ Separate variables: $\displaystyle \frac{1}{A}\,dA=-k\,dt$ Integrate, using boundaries as limits of integration: $\displaystyle \int_{A_0}^A\frac{1}{A}\,dA=-k\int_0^t\,dt$ $\displaystyle \ln\left(\frac{A}{A_0}\right)=-kt$ Convert from logarithmic to exponential form: $\displaystyle \frac{A}{A_0}=e^{-kt}$ $\displaystyle A=A_0e^{-kt}$ We are told it takes 4 hours for 60% of the dose to leave the body. This means A(4) = 0.4A(0)...this gives: $\displaystyle \ln\left(\frac{\frac{2}{5}A_0}{A_0}\right)=-k(4)$ $\displaystyle \ln\left(\frac{2}{5}\right)=-4k$ $\displaystyle \frac{\ln\left(\frac{2}{5}\right)}{4}=-k$ $\displaystyle -k=\frac{1}{4}\ln\left(\frac{2}{5}\right)=\ln\left( \left(\frac{2}{5}\right)^{\frac{1}{4}}\right)$ Now, substituting for -k into the expression for A gives: $\displaystyle A = A_0e^{\ln\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)t}$ $\displaystyle A = A_0\left(e^{\ln\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)}\right)^t$ $\displaystyle A = A_0\left(\left(\frac{2}{5}\right) ^{\frac{1}{4}}\right)^t$ $\displaystyle A = A_0\left(\frac{2}{5}\right)^{\frac{t}{4}}$ So, with $\displaystyle A_0=240\text{ mg}$ we have: $\displaystyle A(t) = 240\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ 2. Just after the second dose, we would then have: $\displaystyle A(t)=\left(240\left(\frac{2}{5}\right)^{\frac{4}{4 }} + 240\right)\text{ mg} = \frac{7}{5}240\text{ mg} = 336\text{ mg}$ 3.) I don't own any graphing software, so we'll have to piece this together. As we found in part 1, the first 4 hours (up until the 2nd dose is taken) is given by: $\displaystyle A(t) = 240\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ Then, for the 2nd four hour period, using the result from part 2, we have: $\displaystyle A(t) = 336\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text{m g}$ For the 3rd four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}336\right) \left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 374.4\left(\frac{2}{5}\right)^{\frac{t}{4}}\:\text {mg}$ For the 4th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}374.4\right)\left(\frac{2}{5} \right)^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 389.76\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ For the 5th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}389.76\right)\left(\frac{2}{5 }\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 395.904\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ For the 6th four hour period: $\displaystyle A(t) = \left(240+\frac{2}{5}395.904\right)\left(\frac{2}{ 5}\right) ^{\frac{t}{4}}\:\text{mg}$ $\displaystyle A(t) = 398.3616\left(\frac{2}{5}\right) ^{\frac{t}{4}}\:\text{mg}$ Graph these six pieces for each successive 4 hour period. 4. As we saw in part 3, the amount A after the doses is: dose...amount (mg) 1..........240 2..........240 + 0.4�240 = 1.4�240 = 336 3..........240 + 0.4(240 + 0.4�240) = 240(1+0.4+0.4�) = 374.4 We see then that the amount after the nth dose will be given by: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\frac{2}{5}\right)^k$ 5. Using the formula for the first n terms of a geometric series, we get: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\frac{2}{5}\right)^k=240\frac{1-\left(\frac{2}{5}\right)^n}{1-\frac{2}{5}} = 400\left(1-\left(\frac{2}{5}\right)^n\right)$ Thus, we find: $\displaystyle \lim_{n\to\infty}A_n = 400\text{ mg}$ So we see the patient will never reach the lethal dose of 600 mg. 6. To answer this, we need to know what portion of the drug remains in the body after 2 hours. Using the formula from part 1: $\displaystyle A = A_0\left(\frac{2}{5}\right)^{\frac{2}{4}} = A_0\sqrt{\frac{2}{5}}$ So now, we have: $\displaystyle A_n = 240\sum_{k=0}^{n-1}\left(\sqrt{\frac{2}{5}}\right)^k = 240\frac{1-\left(\sqrt{\frac{2}{5}}\right)^n}{1-\sqrt{\frac{2}{5}}} = \left(400+80\sqrt{10}\right)\left(1-\left(\sqrt{\frac{2}{5}}\right)^n\right)$ $\displaystyle \lim_{n\to\infty}A_n = 400+80\sqrt{10}\text{ mg}\approx652.98\text{ mg}$ So we see the patient will eventually reach the lethal dosage. To find after which dosage, we may use: $\displaystyle \left(400+80\sqrt{10}\right)\left(1-\left(\sqrt{\frac{2}{5}}\right)^n\right) = 600$ Now we solve for n: $\displaystyle 1-\left(\sqrt{\frac{2}{5}}\right)^n = \frac{600}{400+80\sqrt{10}}=\frac{5-\sqrt{10}}{2}$ $\displaystyle \left(\frac{2}{5}\right)^{\frac{n}{2}} = 1-\frac{5-\sqrt{10}}{2}=\frac{\sqrt{10}-3}{2}$ Take the natural logarithm of both sides: $\displaystyle \ln\left(\left(\frac{2}{5}\right) ^{\frac{n}{2}}\right) = \ln\left(\frac{\sqrt{10}-3}{2}\right)$ $\displaystyle \frac{n}{2}\ln\left(\frac{2}{5}\right) = \ln\left(\frac{\sqrt{10}-3}{2}\right)$ $\displaystyle n = 2\frac{\ln\left(\frac{\sqrt{10}-3}{2}\right)}{\ln\left(\frac{2}{5}\right)}\approx5 .48$ Thus, the patient will reach a lethal amount after the 6th dose.
Was wondering whether you still had the photos of the work you did for this problem. Unfortunately, I am not seeing them and they would help me immensely. Thank you!

Last edited by skipjack; December 12th, 2018 at 08:38 PM.

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text:
Tick (check mark)

A tick (known as a check mark or check in North American English) is a mark (, , , etc.) used to indicate the concept "yes", for example "yes; this has been verified", "yes; that is the correct answer", "yes; this has been completed", or "yes; this [item or option] applies to me". The x mark is also sometimes used for this purpose (most notably on election ballot papers), but otherwise usually indicates "no", incorrectness, or failure.

As a verb, to tick (off) or to check (off) means to add such a mark. It is quite common, especially on printed forms, printed documents, and computers (see check box), for there to be squares in which to place ticks.

International differences

The tick mark is a predominant affirmative symbol of convenience in the English-speaking world because of its instant and facile composition. In other countries, however, the tick mark is more complicated. In some countries (e.g., Finland, Norway, Sweden, Korea, and Japan)[citation needed], the tick can be used as an error mark and indicates "no" rather than "yes". For example, it is common in Swedish schools for a to indicate that an answer is incorrect (while 'R', from the Swedish rätt i.e. "correct", is used to indicate that an answer is correct). This often causes confusion as to whether something that has been ticked is correct or incorrect, especially when reading English media. In Finnish, stands for väärin i.e. "wrong". (The opposite, "right", is marked with $\cdot \! / \! \cdot$, a slanted vertical line emphasized with two dots).

Asian replacement

In Japan, an "O Mark" (in the appearance of a circle, Unicode symbol "◯"), also known as "丸印" marujirushi, is used instead of a tick to mean "yes"; this symbol is also used in Korea.[citation needed] In China, tick is used for "correct". A tick placed in brackets can mean a previously used or acceptable fact or definition is being looked into, usually for the purpose of expanding academic research.[citation needed]

Unicode

The "Symbol, Other" category of Unicode provides various check marks:

• U+237B not check mark
• U+2610 ballot box
• U+2611 ballot box with check
• U+2705 white heavy check mark
• U+2713 check mark
• U+2714 heavy check mark


text:
### Asymptotically Tight Bounds for Composing ORAM with PIR

Ittai Abraham, Christopher W. Fletcher, Kartik Nayak, Benny Pinkas, and Ling Ren

##### Abstract

Oblivious RAM (ORAM) is a cryptographic primitive that allows a trusted client to outsource storage to an untrusted server while hiding the client's memory access patterns to the server. The last three decades of research on ORAMs have reduced the bandwidth blowup of ORAM schemes from $O(\sqrt{N})$ to $O(1)$. However, all schemes that achieve a bandwidth blowup smaller than $O(\log N)$ use expensive computations such as homomorphic encryptions. In this paper, we achieve a sub-logarithmic bandwidth blowup of $O(\log_d N)$ (where $d$ is a free parameter) without using expensive computation. We do so by using a $d$-ary tree and a two server private information retrieval (PIR) protocol based on inexpensive XOR operations at the servers. We also show a $\Omega(\log_{cD} N)$ lower bound on bandwidth blowup in the modified model involving PIR operations. Here, $c$ is the number of blocks stored by the client and $D$ is the number blocks on which PIR operations are performed. Our construction matches this lower bound implying that the lower bound is tight for certain parameter ranges. Finally, we show that C-ORAM (CCS'15) and CHf-ORAM violate the lower bound. Combined with concrete attacks on C-ORAM/CHf-ORAM, we claim that there exist security flaws in these constructions.

Note: This extends our original submission with a lower bound on ORAMs using PIR and shows two attacks on C-ORAM/CHf-ORAM.

Available format(s)
Publication info
A minor revision of an IACR publication in PKC 2017
Keywords
Oblivious RAMPIRlower bound
Contact author(s)
kartik @ cs umd edu
History
2017-01-17: last of 2 revisions
See all versions
Short URL
https://ia.cr/2016/849

CC BY

BibTeX

@misc{cryptoeprint:2016/849,
author = {Ittai Abraham and Christopher W.  Fletcher and Kartik Nayak and Benny Pinkas and Ling Ren},
title = {Asymptotically Tight Bounds for Composing ORAM with PIR},
howpublished = {Cryptology ePrint Archive, Paper 2016/849},
year = {2016},
note = {\url{https://eprint.iacr.org/2016/849}},
url = {https://eprint.iacr.org/2016/849}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# regular tiling of a surface of genus 2 by heptagons

Let $$S$$ be a Riemann Surface of genus 2. Is there a picture in the literature for a regular tiling of $$S$$ by 12 heptagons (where three heptagons meet at each vertex)? Also, apart from the obvious restriction given by the Euler characteristic $$2-2g=f-nf/2+nf/v$$ (where $$g$$ is the genus, $$f$$ is the number of faces, $$v$$ is the number of faces meeting at each vertex) are there any obstructions for the existence of a tiling of a surface of genus $$g$$ by $$n$$-gons (where the same number $$v$$ of $$n$$-gons meet at each vertex).

I know that such a tiling exists (for a surface of genus 2 by heptagons), but I am unable to make a drawing. The construction goes like this (thanks to Mladen Bestvina): Start with a sphere. View this as a octahedron so that you have 8 triangles. Pick a hexagon (start from the top go down, left, down come back up from the opposite side) We fatten up these 6 sides to look like circles with one ray coming from the middle. You should think of these as the quotient of 2 triangles attached along one edge <|> via rotation by 180 degrees (so you get two edges meeting in a point A and 2 edges plus a ray meeting in the other point B). We call the center of the circle C so that the ray goes from C to B.
You have to view the point $$B_i$$ (of the $$i$$-th side, $$i=1,2,3,4,5,6$$) as being attached to $$A_{i+1}$$ for $$i=1,2,3,4,5,6$$ (of course mod 6) so each of these vertices has valence $$5+2=7$$. Now you take the double cover with fixed points $$C_1,...,C_6$$ and you get a surface of genus 2 with 28 triangles (28=2x8+2x6 where 8 is the number of triangles and 6 are the circles with a ray that unwrap in to two triangles). The vertices still have valence 7 (because the cover is étale here). Take the dual tiling and you are done!

This is just a curiosity. I tried asking the first part of this on stackexchange with no luck. I am also aware of the beautiful pictures of the famous tiling of a surface of genus 3 by 24 heptagons but you can not use that to obtain the tiling of a genus two surface in an obvious way.

Addressing the second question positively, this is the Main Theorem of this paper:

Edmonds, Allan L.; Ewing, John H.; Kulkarni, Ravi S., Regular tessellations of surfaces and (p,q,2)-triangle groups, Ann. Math. (2) 116, 113-132 (1982). ZBL0497.57001. MR0662119

In fact, they show this also works for non-orientable surfaces, except for the non-existence of a degree 3 triangulation of the projective plane.

A picture is in the comments to John Baez' Blog Post. EDIT For the second part of the question, no there is no obstruction, as long as the implied angle ($2\pi/v$) is smaller than the angle of a Euclidean regular $n$-gon. This is true by an obvious argument I had originally heard from Bill Thurston: a very small hyperbolic regular $n$-gon has angles close to Euclidean, a very large regular hyperbolic $n$-gon has angles $0$ so, by continuity, somewhere in between you will have the angle you want.

• I could see lot's of beautiful pictures of tilings of a genus 3 surface, but I couldn't find the genus 2, but there are many comments so maybe I missed it; could you tell me the date of the comment so I can find the picture? Also, do you know the answer to the second part of the question? – Hacon Feb 20 '15 at 16:02
• @Hacon Sure, it's 29 May 2013. – Igor Rivin Feb 20 '15 at 16:28
• @Hacon And I edited to answer the second part of the question. – Igor Rivin Feb 20 '15 at 16:31
• I see it now. Very nice! thank you. – Hacon Feb 20 '15 at 16:41
• Still that picture is somewhat incomplete in that the surface as shown is in fact one of infinitely many fundamental domains for a group action on an infinite (noncompact) covering of the needed surface. I wonder if there is a satisfactory picture of the surface itself somewhere. – მამუკა ჯიბლაძე Jun 19 '15 at 6:19


text:
# Metonymy in mathematics

Metonymy is a figure of speech where a word or another expression is used to mean something other than its literal meaning. This phenomenon is not restricted to the "usual human languages" (such as English or Finnish) but also appears in the symbolic language of mathematics. I am not interested in drawing boundaries between English and mathematics in a mathematical presentation; I am interested in metonymy in all mathematical contexts.

Examples of metonyms:

• When we discuss a function $f$, we often actually write or say $f(x)$, although $f(x)$ is the value of the function $f$ at $x$. That is, the value of the function at a "generic point" is used as a metonym for the function itself. (An extended example of this metonymy is the statement "The Fourier transform of $f(x)$ is $\hat f(\xi)$".)
• A continuous function $\gamma:I\to\mathbb R^n$ from a real interval $I$ is a curve. "The curve $\gamma$" occasionally also refers to the image $\gamma(I)$, the trace of the curve, by metonymy. Different authors use different conventions here, but the word "curve" is used in various different meanings.
• A group is a set together with a (certain kind of) binary operation. The set alone is often used to denote the group by metonymy, the operation being understood implicitly.

Metonyms are always about using the same expression (word, symbol…) in several different meanings. Metonymy is often convenient, but it can also be a source of confusion if it is not introduced explicitly. Sloppy notation and abuse of notation are typically metonymous.

Questions:

1. How should I take metonymy into account when teaching mathematics? Should I try to explicitly state whenever this linguistic phenomenon is present? In my experience this is rarely done, and I have a feeling that some students have difficulties because they do not recognize when an expression is used metonymously.
2. What are the most important metonyms that I should make my students aware of? (I am thinking about university level education, but answers on any level are welcome.)
3. Are there some metonyms that I should avoid or instruct my students to avoid? Do students commonly have misconceptions of a metonymous origin? This is the kind of problem that I had in mind: If a banana costs $3\$$and an apple 2\$$, a student may be tempted to write$\text{banana}=3\$$and \text{apple}=2\$$ and conclude $2\times\text{banana}=3\times\text{apple}$ which is not true. This is because of using the symbol "$=$" metonymously, not actually referring to two things being equal. If one introduces a price function $p$ and lets $p(\text{apple})=2\$$and so on, then$p(2\times\text{banana})=p(3\times\text{apple})$is a reasonable statement. 4. To what extent should I delay exposition to metonyms? Metonymy is a part of standard mathematical communication, and mathematical educations should prepare students for it. But this does not necessarily mean that I should express myself metonymously from day one when teaching a new topic. 5. Can metonymy be used as a tool to help learn mathematics? That is, are there useful metonyms that I should introduce? • I found post focusing on the "$f$or$f(x)$" example. It is relevant but different. matheducators.stackexchange.com/q/604/2074 – Joonas Ilmavirta Oct 1 '14 at 18:19 • I guess you meant$2\times p(banana) = 3\times p(apple)$– Jonas Gomes Oct 9 '14 at 21:10 • @JonasGomes, the price functions is linear, so that is the same thing. I wanted to emphasize that it's not$a=b$but$p(a)=p(b)\$. – Joonas Ilmavirta Oct 10 '14 at 4:45
• What about bulk discounts in non-competitive markets?!! (I kid, I kid.) – Vandermonde Nov 11 '15 at 4:13

Metonymy and its relatives, metaphor, polysemy, synecdoche occur all over the place in mathematical writing, and sometimes cause students problems and sometimes don't, because those thought processes are basic to our understanding of everything.

Where mathematics comes from, by George Lakoff and Rafael E. Núñez. Basic Books, 2000.

Exploring the role of metonymy in mathematical understanding and reasoning: The concept of derivative as an example, by Michelle J. Zandieh and Jessica Knapp. The Journal of Mathematical Behavior, Volume 25, Issue 1, Pages 1-17. This is owned by (boo hiss) Elsevier who charges an exorbitant amount to get a copy, so I have not read it. I have not checked if it is on Jstor.

Your Metaphor, or My Metonymy?, by John Mason, 2010.

Metonymy and metaphor in mathematics, by Tony Phillips, 1999.

Handbook of mathematical discourse, by Charles Wells.

Images and metaphors, article in abstractmath.org.

Incomplete notation, section of an article in abstractmath.org.

Embedding, section of an article in abstractmath.org.

One of my favorite examples of this sort of thing is when a mathematician says something like "x^2-1 vanishes at x=1".


text:
Lemma 10.50.10. Let $A$ be a valuation ring. Then $A$ is a normal domain.

Proof. Suppose $x$ is in the field of fractions of $A$ and integral over $A$, say $x^{d + 1} + \sum _{i \leq d} a_ i x^ i = 0$. By Lemma 10.50.4 either $x \in A$ (and we're done) or $x^{-1} \in A$. In the second case we see that $x = - \sum a_ i x^{i - d} \in A$ as well. $\square$

There are also:

• 3 comment(s) on Section 10.50: Valuation rings

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text:
We want to evaluate limits for which the Limit Laws do not apply.

In the last section we computed limits using continuity and the limit laws. What about limits that cannot be directly computed using these methods? Consider the following limit, Here in light of this, you may think that the limit is one or zero. Not so fast. This limit is of an indeterminate form. What does this mean? Read on, young mathematician.

Which of the following limits are of the form $\relax \boldsymbol {\tfrac {0}{0}}$?
$\lim _{x\to 0}\frac {\sin (x)}{x}$ $\lim _{x\to 0}\frac {\cos (x)}{x}$ $\lim _{x\to 0}\frac {x^2-3x+2}{x-2}$ $\lim _{x\to 2}\frac {x^2-3x+2}{x-2}$ $\lim _{x\to 3}\frac {x^2-3x+2}{x-3}$

Let’s finish the example with the function above.

Let’s consider a few more examples of the form $\relax \boldsymbol {\tfrac {0}{0}}$.

We will look at one more example.

All of the examples in this section are limits of the form $\relax \boldsymbol {\tfrac {0}{0}}$. When you come across a limit of the form $\relax \boldsymbol {\tfrac {0}{0}}$, you should try to use algebraic techniques to come up with a continuous function whose limit you can evaluate.

Notice that we solved multiple examples of limits of the form $\relax \boldsymbol {\tfrac {0}{0}}$ and we got different answers each time. This tells us that just knowing that the form of the limit is $\relax \boldsymbol {\tfrac {0}{0}}$ is not enough to compute the limit. The moral of the story is

Limits of the form $\relax \boldsymbol {\tfrac {0}{0}}$ can take any value.

Finally, you may find it distressing that we introduced a form, namely $\relax \boldsymbol {\tfrac {0}{0}}$, only to end up saying they give no information on the value of the limit. But this is precisely what makes indeterminate forms interesting… they’re a mystery!


text:
# Discussion regarding work done? [closed]

Suppose I take a box with some masses and pull it along the table at a constant speed. In this case, there is a force pulling to the right (the string) and a frictional force pulling to the left.

Since the net force is zero, there is no change in momentum. Everything is fine.

. . . .◆Work-Energy Principle◆. . . . The work-energy principle says that the work (W) done on a point mass is equal to its change in kinetic energy. Work is done by a force moving a certain distance. Actually, it's only the force in the direction of motion that matters. As an equation, it looks like this $$W=F d \cos (\theta)$$

Here θ is the angle between the force and the displacement. If the force is "pushing backwards" you can have negative work. For the kinetic energy, it depends on the mass and the velocity $$KE=\frac{1}{2}m v^2$$

For the block being pulled along the table with friction the two forces are the force from the string pulling to the right and friction pulling to the left. The total work on the block would be zero, and it would move at a constant speed.

BUT WAIT! There is a problem. What if you measure the temperature of this block before and after you pull it? It's not a huge increase in temperature, but it did indeed warm up. If I slide the block over a larger distance (or back and forth), That is an area where the table increases in temperature—the block also gets hotter.

But if the block gets warmer, that means it increases in energy. In this case, it would be an increase in thermal energy. So, how can the block increase in energy if there is zero work done on the object? How is it possible for there to be zero work AND an increase in energy?

• Friendly comment: your post is too long. I skipped down to the last two paragraphs. Hint: what exactly is $s$ in the definition of work? – garyp Sep 22 at 17:45
• Yup, it is long but I've included each and every minute thing i can include, you can read it fast ! Here s is the distance simply covered by my example – Vishal Kumar Sep 22 at 17:47
• Please mark up your math using mathjax. The question is pretty incoherent. You present lots of numerical data, but it's not clear what the point is. – Ben Crowell Sep 22 at 17:52
• what do you think friction energy does, the table does not move, yet it absorbs friction energy as ____? – Adrian Howard Sep 22 at 18:09
• So what is your question? StackExchange is a question/answer site, not a discussion site. – Jon Custer Sep 23 at 13:21

But if the block gets warmer, that means it increases in energy. In this case, it would be an increase in thermal energy. So, how can the block increase in energy if there is zero work done on the object? That is indeed a mystery. How is it possible for there to be zero work AND an increase in energy. Thanks friends for reading it out , i would like to hear your thoughts about this ,

The work energy principle states: the net work done on an object equals its change in kinetic energy.

The key term is net work. Work can be positive (force in same direction as displacement) or negative (force in opposite direction of displacement). Friction work is always negative because the friction force opposes the motion of an object.

So lets say a constant external force $$F$$ moves a box a distance $$d$$ on a surface with friction. This work is positive. But the friction force $$f$$ acts in the opposite direction so its work is negative. By the work energy principle

$$W_{net}=Fd-fd=\Delta KE$$

If the applied force equals the friction force, the net work done on the object is zero (no change in velocity) and $$Fd-fd=0$$, or $$Fd=fd$$.

The friction force is $$f=μ_{k}mg$$ so the friction work is $$μ_{k}mgd$$

This tells us that the magnitude of the work done by the external force equals the magnitude of the work done by the friction force. The energy of the work done by friction elevates the temperature of the contacting surfaces and eventually transfers to the surroundings as heat.

Bottom line: There is no increase in energy. Although the net work done is zero, meaning there is no change in kinetic energy (velocity is constant) the work done by friction is not zero. It is simply negative work. Friction work take the energy supplied to the object by the positive work done by the external force and performs an equal amount of negative work dissipating the energy done by the external force as heat. Energy is conserved.

Jake pulls a box with a mass of 22 kg. The rope makes an angle of 25 degrees with respect to the horizontal. The coefficient of kinetic friction is 0.1. Find the work done by Jake and the work done by friction for the case where the box moves along the ground a distance of 144 meters.

Bad. Bad question. You could indeed calculate the force of friction, but you can't calculate the work done (unless you also know some stuff about the changes in thermal energy).

It is indeed a bad question, but not for the reasons you are citing. You say you "could indeed calculate the force of friction". But the fact is you can't. You need to know the magnitude of the force applied by Jake to calculate the friction force, the friction work, and the work done by Jake. Here's why, starting with the friction force.

The kinetic friction force, $$f$$, is given by

$$f=μ_{k}N$$

Where $$μ_{k}$$ is the coefficient of kinetic friction and $$N$$ is the normal force exerted by the contact surface. That normal force is usually $$mg$$, the weight of the box, but in this case it is not. That's because Jake is pulling the box at an angle with respect to the horizontal. The vertical component of Jake's force reduces the normal force. So the friction force becomes:

$$f=μ_{k}(mg-Fsin25)$$

where $$F$$ is the magnitude of the force applied by Jake. Without the friction force, you cannot determine the friction work which is.

$$W_{friction}=μ_{k}(mg-Fsin25)d$$

where $$d$$ is the horizontal distance moved.

The only work done by Jake is due to the horizontal component of his force (there is no vertical movement of the box so no vertical work done. So the work done by Jake is

$$W_{Jake}=Fdcos25$$

Since Jake's work is positive and friction work is negative, the net work done is

$$W_{net}=W_{Jake}-W_{friction}=d(Fcos25-μ_{k}(mg-Fsin25))=\Delta KE$$

Everything is known except $$F$$.

If you calculated the work done by friction as the frictional force multiplied by the distance the block moves, how would you account for the increase in thermal energy of the block (and floor)?

If you knew the magnitude of the force applied by Jake, you can calculate the friction work. If the net work done by Jake and friction is zero, all that means is that friction has taken the energy provided by Jake due to his work and converted it into heat. Energy is conserved.

SUMMARY:

I think the main problem you are having is thinking that net work equal to zero means that no work has been done. That is simply not the case. Jake did work in moving the box. Just ask Jake. Jake's work is positive since his force is in the same direction as the movement of the box. Friction did work, as evidenced by the increase in temperature of the contact surfaces. Friction's work is negative since the friction force is in the opposite direction of the movement of the box. If the magnitude of Jake's work equals the friction work, the net work done is zero. That only means there is no change in kinetic energy of the box. It doesn't mean no work was done.

Hope this helps.

• Yup, that's what I'm saying, if the frictinal force and applied force are just equal and hence opposite the net work done would came out to be zero and then by using the work energy theorem the velocity would simply came out to be constant/zero but because since we've considered the system consisiting of block and table and as i pull it along it has no change in kinetic energy but however the thermal energy increases. – Vishal Kumar Sep 23 at 17:50
• And therefore my friend came with an analogy which I've mentioned below please read that out, it means it doesn't necessarily means fiction does the same amout of work using his example. – Vishal Kumar Sep 23 at 17:51
• @VishalKumar I'm afraid I can't visualize what is going on in your example of the brushes, so I can't comment on the validity of you and your friend's conclusion. All I know is that total energy must be conserved. If the hand does work moving the brush and the brush begins and ends at rest, there is no change in kinetic energy of the brush. If all the movement is horizontal there is no change in potential energy. – Bob D Sep 23 at 19:31
• The bottom line is the work done by the hand has to be accounted form. It has to be dissipated as heat due to friction. How else will you account for the energy expended by the hand? At this point I have no more time to invest in this discussion, so I wish you good luck in resolving your issues! – Bob D Sep 23 at 19:31

Here's what my friend texted me today . . . Suppose that I rub two brushes together instead of a block and a table. Watch what happens. Notice that as the brush is pulled, there are two forces that do work. My hand does work (positive work), and the brushes do work (negative work). But look closely. Notice that as the brush (and my hand) move to the left a certain distance, the brushes bend. This means that the force the bottom brush exerts on the top brush moves over a shorter distance than the hand moves. Even if the force of the brush is the same magnitude as the force of my hand, the brush does less work because it moves over a shorter distance. That means the total work done on the brush is NOT zero joules but some positive amount. Of course the brush is an analogy for friction. We like to think of friction as this nice and simple interaction, but it's not. For the block sliding on a table, the frictional force is an interaction between the surface atoms in the block and the surface atoms on the table. It's not so simple. Physics textbooks like to treat a block as a point object—but it is not a point object. It's a complicated object made of countless atoms. In the case of friction, you can't forget that and just treat a block as a point object. It doesn't work.

but you could do this problem with the momentum principle and it wouldn't be a problem. Remember that the momentum principle deals with forces and time, not distance. So even though the frictional force acts over a different distance, the time is the same for both the frictional force and the force pulling the string.

Well, here is the problem. The main goal in physics is to build models that agree with real-life experiences. These models could be a big idea like the work-energy principle—and that's great. Let's consider an example with another model. What about a globe? It's a model of the Earth. It even shows the location of the continents and everything. But what if I want to use this globe and measure its mass and volume so I can determine the density of the real (full-sized) Earth? That wouldn't work, because the globe isn't actually Earth. The same is true with the work-energy principle. It's great for some things, but you can't just use it wherever you like.

• but we could do this problem with the momentum principle and it wouldn't be a problem. Remember that the momentum principle deals with forces and time, not distance. So even though the frictional force acts over a different distance, the time is the same for both the frictional force and the force pulling the string. – Vishal Kumar Sep 23 at 2:13
• First of fall, you should ask your question in short way secondly, you should mention what condition are you assuming like in high school physics in India, while solving Newtonian mechanics problem they assume there is no loss of energy due to friction – yuvraj singh Sep 23 at 4:25
• If I'd have assumed any thing I've already mentioned it in . Secondly I've mentioned all things and makes things crystal clear , that's why my discussion is long but worth to read !and yes that's what I'm trying to focus on that we all are finding it wrong in all these years , no one ever questioned it also! – Vishal Kumar Sep 23 at 6:41
• @VishalKumar See the update to my answer in response to your own answer. Hope it helps. – Bob D Sep 23 at 14:21
• @VishalKumar I've added a summary to my answer that I think gets to the core of the problem you are having. – Bob D Sep 23 at 15:06


text:
# Definition of Deep Learning

There are many conflicting definitions of Deep Learning, here’s some definitions by various people:

Yann LeCun
DL is constructing networks of parameterized functional modules & training them from examples using gradient-based optimization (Source)
• Definition is orthogonal to the learning paradigm: paradigm, reinforcement, supervised, or self-supervised