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## PHY452H1S Basic Statistical Mechanics. Lecture 8: Midterm review, thermodynamics. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on February 7, 2013

# Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

# Exam review

\begin{aligned}X = x_1 x_2 \cdots x_N\end{aligned} \hspace{\stretch{1}}(1.2.1)

We can’t apply CLT directly, but we can apply it to $\ln X$

\begin{aligned}Y = \ln X = \sum_i \ln x_i = \sum_i y_i\end{aligned} \hspace{\stretch{1}}(1.2.2)

So

\begin{aligned}\tilde{P}(Y) = \frac{1}{{\sqrt{ 2 \pi N \sigma_y^2}}} \exp\left( {-\frac{(Y - N \mu_y)^2}{N \sigma_y^2}} \right)\end{aligned} \hspace{\stretch{1}}(1.2.3)

Of interest here is how probabilities change under change of variables. If $Y = \ln X$, this works like

\begin{aligned}P(X) dX = \tilde{P}(Y) dY\end{aligned} \hspace{\stretch{1}}(1.2.4)

\begin{aligned}P(X) = \tilde{P}(Y = \ln X) \frac{dY}{dX}.\end{aligned} \hspace{\stretch{1}}(1.2.5)

This is illustrated roughly in (Fig 1)

Fig 1: Change of variables for a probability distribution

• Random walkThis was a unit stepping problem as illustrated in (Fig2).

Fig2: Unit one dimensional random walk

\begin{aligned}\left\langle{{X}}\right\rangle = \sum_i \left\langle{{x_i}}\right\rangle\left\langle{{X^2}}\right\rangle = \left\langle{{\left( {\sum_{i = 1}^N x_i} \right)\left( {\sum_{j = 1}^N x_i} \right)}}\right\rangle= \left\langle{{\left( {\sum_{i = 1}^N x_i^2} \right)\left( {\sum_{i \ne j = 1}^N x_i x_j} \right)}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.6)

• State and prove Liouville’s theorem

\begin{aligned}\frac{d\rho}{dt} = \frac{\partial {\rho}}{\partial {t}} + \sum_{i = 1}^{3N} \frac{\partial {\left( {\dot{x}_i \rho} \right)}}{\partial {x_i}}+\frac{\partial {\left( {\dot{x}_i \rho} \right)}}{\partial {x_i}}\end{aligned} \hspace{\stretch{1}}(1.2.7)

• Collisions in 1DThis was a one direction collision problem for $N$ particles, a portion of which is illustrated in (Fig 3).

Fig3: One dimensional collision of particles

The statement here that the collision was in 1D was a hint that we can actually calculate the result.

Start with a pair of collisions and work out that the velocities are exchanged.

\begin{aligned}(v_i, v_{i+1}) \rightarrow (v_{i+1}, v_{i})\end{aligned} \hspace{\stretch{1}}(1.2.8)

We have only exchange velocities. This won’t result in all the phase space being explored, and was meant to show that things are extremely restrictive in 1D.

• Harmonic oscillator in 1D.This problem ends up essentially requiring the evaluation of the area in phase space of an ellipse in phase space as in (Fig 4).

Fig4: 1D classical SHO phase space

The result is $\pi a b \times \text{area}$.

We’d been discussing Gibbs paradox, considering the doubled volume as in (Fig 5).

Fig5: Gibbs paradox two volumes allowed to mix

and defined a \underlineAndIndex{statistical entropy} that requires division of $\Omega$ by $N!$. We have

\begin{aligned}S_statistical = k_B \ln \underbrace{\Omega}_{\text{Number of states}}\end{aligned} \hspace{\stretch{1}}(1.3.9)

# Thermodynamics – phenomenology

Let’s move on to discuss some concepts of thermodynamics

First law of thermodynamics

This is the law of \underlineAndIndex{Energy conservation}! \footnote{This requires some corrections if relativisitic systems are studied.}

\begin{aligned}\underbrace{dE}_{\text{Change in energy}}=\underbrace{d W}_{\text{work done on the system}}+\underbrace{d Q}_{\text{Heat supplied to the system}}\end{aligned} \hspace{\stretch{1}}(1.4.10)

Here the $d$ differentials indicate that the change depends is path dependent. For example the number of times a piston is moved back and forth can be significant, even given a specific change in the total differential change in the piston position. The total change of energy in such a change is not path dependent.

Consider a piston setup compressing some gas taking the position of the pistion from $X \rightarrow X + dX$ as in (Fig 6).

Fig6: Piston and gas

\begin{aligned}d W = \sum_i \underbrace{f_i}_{\text{Generated force}}\underbrace{dX_i}_{\text{Controllable macroscopic coordinate}}\end{aligned} \hspace{\stretch{1}}(1.4.11)

What is $d Q$? These are the microscopic, and uncontrollable changes in energy.

Processes

1. Change $X_i \rightarrow X_i + dX_i$
2. Change $d Q$

These are those that are thermally isolated, or

\begin{aligned}d Q = 0.\end{aligned} \hspace{\stretch{1}}(1.4.12)

The only way to change energy is to change the macroscopic $X_i$, so the change in energy is given by

\begin{aligned}dE = d W = \sum_i f_i dX_i\end{aligned} \hspace{\stretch{1}}(1.4.13)

These are what we are used to thinking about as conservative systems. Imagine, for example, a force applied to a spring, slowly so that there is no heat loss, we can figure out the total change in energy by summing differential changes to the spring.

We can imagine that there is some way to change the position that results in a change of energy. This is of course just the force as in (Fig 7).

\begin{aligned}-\frac{dE}{dx_i} \sim \text{force}.\end{aligned} \hspace{\stretch{1}}(1.4.14)

Fig 7: A possible energy position relationship

We can imagine that we started with an initially thermally isolated system, with the macroscopic parameter unchanged, but somebody sneaks in and heats the system, before re-thermally isolating it. We’d then have a system that perhaps looks like (Fig 8).

Fig 8: Level curves for adiabatic and non-adiabatic changes to the system

The jumps from one curve from another is the product of the sneaky heat suppliers.

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MathSciNet bibliographic data MR123291 (23 #A619) 54.53 Fine, N. J.; Gillman, L. Extension of continuous functions in $\beta N$$\beta N$. Bull. Amer. Math. Soc. 66 1960 376–381. Article

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


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# Henry is sitting in a tree 32 feet above the ground watching a deer. If the deer is 57 feet from the base of the tree, find the angle of depression from

###### Question:

Henry is sitting in a tree 32 feet above the ground watching a deer. If the deer is 57 feet from the base of the tree, find the angle of depression from Henry to the deer.

### In guinea pigs, black fur (b) id dominant over whte fur (b). cross a heterozygous (hybrid) black guinea

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### A gas tank has a volume of 28.1 m3 and a pressure of 18.4 atm. The temperature of the gas is 32°C. What is the Celsius temperature when the gas is put

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### How was the U. S. involvement in these Cold War conflicts different than full scale invasions in wars

How was the U. S. involvement in these Cold War conflicts different than full scale invasions in wars like WWII?...

### Spelling and gramar errors affect the readers perception of the authors credibility. true or false

Spelling and gramar errors affect the readers perception of the authors credibility. true or false...

### Cuando llegas a otro país es necesario pasar por la si tienes algo para declarar.

Cuando llegas a otro país es necesario pasar por la si tienes algo para declarar....

Need some help with this, 2 answer problem! Please help $Need some help with this, 2 answer problem! Please help$...

### How were the franks connected to the catholic church

How were the franks connected to the catholic church...

### What percent of households were not considered families

What percent of households were not considered families $What percent of households were not considered families$...

### The CEO of a clothing company estimates that 52% of customers will make a purchase. (10 points) Part A: How many customers should

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### What does y equal in this equation 5(y+2/5)=-13

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### A supervisor rewards a restaurant employee for cleaning the bathrooms by allowing the employee to stock the salad bar. would

A supervisor rewards a restaurant employee for cleaning the bathrooms by allowing the employee to stock the salad bar. would predict that the opportunity to stock the salad bar would actually be reinforcing...

### In the following group discussion, which technique is freda using to better evaluate

In the following group discussion, which technique is freda using to better evaluate an argument? freda: the author makes it sound like first-time offenders are hopeless and should just be locked up. allison: i agree; there's no way to stop bad kids from being bad. freda: it seems like this aut...

### The graph shows Linda's math scores versus the number of hours spent doing math homework.A graph titled Lindas Math Scores shows Hours

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### Astudent conduct an experiment to determine how the amount of water given to a plant affects its growth . which of the following

Astudent conduct an experiment to determine how the amount of water given to a plant affects its growth . which of the following is not a constant or control for this experiment ?...

### 15. At the end of the Giant Drop free fall ride, riders experience a large upward normal force tobring their falling bodies

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### Question 2 (1 point) These were four punishments that were laws enacted by the British Parliament in

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### Multiple choice: select the best answerwhich of the following is equal to the expression below? (6^-7)^3a.

Multiple choice: select the best answerwhich of the following is equal to the expression below? (6^-7)^3a. (-1/6)^21b. (1/6)^21c. (1/6)^-21d. (-1/6)^-21...

### 4.The relationship between how much mass matter has in a given volume is calledA. densityOB. boiling pointC. melting pointOD. weight

4. The relationship between how much mass matter has in a given volume is called A. density O B. boiling point C. melting point O D. weight...
Need help on this can someone help me $Need help on this can someone help me$...


text:
# Population growth and the logistic model

I am not sure where I am going wrong, I will admit I do not fully understand all the equations but I do feel like I have a pretty good grasp on them.

This is from Stewart Calculus 7e pg 613 #3

"The pacific halibut fishery has been modeled by the differential $$\frac{dy}{dt} = ky \left(1 - \frac{y}{M}\right)$$where $y(t)$ is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be $M = 8 * 10^7 kg$ and $k = .071$ per year.

a) if $y(0) = 2 \cdot 10^7 kg$ find the biomass a year later.

b) How long will it take for the biomass to reach $4 \cdot 10^7 kg$?"

First I start by putting in the given information into the logistic equation. $$\frac{dy}{dt} = .071y\left(1 - \frac{y}{8*10^7}\right)$$

Starting with $a$ I need to find $y$ when $t$ is $0$ so I use another equation from the book. I find that $A = 3$ $$A = \frac{M - y_0}{y_0}$$ Where I am assuming $y_0$ is the initial value. $y(0) = 2 \cdot 10^7$

$$A = \frac{(8 \cdot 10^7) - (2 \cdot 10^7)}{ 2 \cdot 10^7}$$

This comes out to 3 so I put it in the other equation that solves for $y(t)$

$$y(t) = \frac{M}{1+ Ae^{-kt}}$$

$$y(1) = \frac{(8 \cdot 10^7) }{1+ 3e^{-.071}}$$

This comes out to $21083781.89$ which I know is wrong, going through my work it all seems to be correct, I am not sure where I have gone wrong.

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Why do you know this is wrong? –  Did Jun 23 '12 at 19:41
The book gets $3.23 \cdot 10^7$ –  user138246 Jun 23 '12 at 19:44
Then $k=0.71$ instead of $k=0.071$. –  Did Jun 23 '12 at 19:46

The solution is correct. If the value of $k=0.71$ is used instead of $k=0.071$, the answer comes out to $$y(1) = \frac{(8 \cdot 10^7) }{1+ 3e^{-0.71}} \approx 3.23\cdot 10^7$$ in accordance with the textbook answer. (Observed by Did)


text:
## Elementary Linear Algebra 7th Edition

$S$ is a basis for $R^3$.
The set $S=\{(1,5,3),(0,1,2),(0,0,6)\}$ is a linearly independent set of vectors. Indeed, assume that $$a(1,5,3)+b(0,1,2)+c(0,0,6)=(0,0,0), \quad a,b,c\in R.$$ Then, we have the following system of equations \begin{align*} a&=0\\ 5a+b&=0\\ 3a+2b+6c&=0. \end{align*} Solving the above equations, we find that $a=0,b=0,c=0$ and hence $S$ is linearly independent. Now, since $R^3$ is vector space of dimension $2$, then by Theorem 4.12, $S$ is a basis for $R^3$.


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# How to draw Hasse diagram for divisibilty?

My question is $L=\{1,2,3,4,5,6,10,12,15,30,60\}$

Please explain me by step by step solution... Thanks for help..

-
Hint: writing each number as its prime factorisation (eg writing $12$ as $2^2 \cdot 3$) is often helpful when considering divisibility. –  Peter LeFanu Lumsdaine Dec 4 '12 at 5:43
$20$ is also a divisor of $60$. Should it also be a member of the set $L$ ? –  miracle173 Dec 4 '12 at 5:50
Please draw a hasse diagram for me. –  Tasneem Dec 4 '12 at 5:58
I'm far from sure that [graph-theory] fits this question, but I'm not that sure. –  Asaf Karagila Dec 4 '12 at 6:06

I would start at the bottom. Clearly $1$ divides every member of $L$, so $1$ is the minimum element in the divisibility order on $L$. Now what is the next layer up of the Hasse diagram? It must contain those members of $L$ that are divisible by $1$ (and of course by themselves) but not by any other element of $L$. Those elements are $2,3$, and $5$: every other element of $L$ is divisible by at least one of these three numbers. Thus, your partial Hasse diagram now looks like this:

   2  3  5
\ | /
\|/
1

What numbers will make up the third row? I claim that the third row contains $4,6,10$, and $15$. First, why not $12,30$, or $60$? $12$ is out because $6$ must be below $12$, and $6$ isn’t in the second row; $30$ is out because $15$ must be below $30$, and $15$ isn’t in the second row; and $60$ is out because every other element of $L$ must be below it. A bit of thought should convince you that no member of $L$ must come below $4,6,10$, or $15$ but above the second row, so that $4,6,10$, and $15$ really do belong in the third row. Specifically, $4$ must lie above $2$; $6$ must lie above $2$ and $3$; $10$ must lie above $2$ and $5$; and $15$ must lie above $3$ and $5$.

You can probably finish it up pretty easily from there: $12$ and $30$ go in the fourth row, with $12$ above $4$ and $3$, and $30$ above $6$ and $5$. Finally, $60$ will be the top element, sitting above $12$ and $30$. As a quick cross-check that $12$ and $30$ are the only members of $L$ that should be immediately below $60$, note that every other member of $L$ (besides $12,30$, and $60$) is a divisor of either $12$ or $30$ and therefore goes somewhere below at least one of these two numbers.

-
thanks for the help... –  Tasneem Dec 4 '12 at 6:24
@Tasneem: You’re welcome. –  Brian M. Scott Dec 4 '12 at 6:26


text:
## Algebra 1

$-225$
We rewrite the given expression as a division problem: $3r\div s^{-2}$ The negative exponent rule states that for every nonzero number $a$ and integer $n$, $a^{-n}=\frac{1}{a^n}$. We use this rule to rewrite the expression: $3r\div\frac{1}{s^2}$ To divide by a fraction, we multiply by the reciprocal: $3r\times s^2$ We plug in the values for $r$ and $s$: $3(-3)\times5^2$ The order of operations states that first we perform operations inside grouping symbols, such as parentheses, brackets, and fraction bars. Then, we simplify powers. Then, we multiply and divide from left to right. We follow the order of operations to simplify: First, we simplify powers: $3(-3)\times25$ Then, we multiply from left to right: $-9\times25=-225$


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hal-00676735, version 1

## Analysis of TTL-based Cache Networks

Nicaise Choungmo Fofack (, ) a1, Philippe Nain (, ) a1, Giovanni Neglia (, ) b1, Don Towsley (, ) c2

N° RR-7883 (2012)

Résumé : This paper provides building blocks for the performance evaluation of Content Centric-like Networks (CCNs). In CCNs if a cache receives a request for a content it does not store ({\em miss}), it forwards the request to a higher-level cache, if any, or to the server. When located, the document is routed on the reverse-path and a copy is placed in each cache along the path. In this paper we consider a cache replacement policy based on Time-to-Lives (TTLs) like in a DNS network. A local TTL is set when the content is first stored at the cache and is renewed every time the cache can satisfy a request for this content (at each hit). The content is removed when the TTL expires. Under the assumption that requests follow a renewal process and the TTLs are exponential random variables, we determine exact formulas for the performance metrics of interest (average cache occupancy, hit and miss probabilities/rates) for some specific architectures (a linear network and a tree network with one root node and $N$ leaf nodes). For more general topologies and general TTL distributions, we propose an approximate solution. Numerical results show the approximations to be accurate, with relative errors smaller than $10^{-3}$ and $10^{-2}$ respectively for exponentially distributed and constant TTLs.

• Domaine : Informatique/Réseaux et télécommunications
• Mots-clés : Cache architecture – content-centric network – timer – Markov model – renewal theory
• Référence interne : RR-7883
• Commentaire : Voir aussi actes de la conférence : ValueTools - 6th International Conference on Performance Evaluation Methodologies and Tools - 2012 (2012) (http://hal.inria.fr/hal-00760915)

• hal-00676735, version 1
• oai:hal.inria.fr:hal-00676735
• Contributeur :
• Soumis le : Mardi 6 Mars 2012, 11:00:57
• Dernière modification le : Mercredi 23 Janvier 2013, 10:04:58


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If the lines shift back and forth we can learn that the star may be orbiting another star. 2002) 10a) while ejected gas near a star or an active corona can result in emission lines (Fig. Astron. Space News. Played 149 times. Hg spectral lines list. The recommended catalog for VLA and ALMA observing is Splatalogue which contains molecular line data from sources including the Lovas catalog , the JPL/NASA molecular database , the Cologne Database for Molecular Spectroscopy , as well as radio recombination lines. Classical particles and waves are idealized concepts like infinitesimal points or perfectly straight lines in geometry; they don’t exist in the real world. When we look at the spectrum of a binary it can present problems for us as we will see the combined spectrum for both stars. These types are designated, in order of decreasing temperature, by the letters O, B, A, F, G, K, and M. The star's spectral lines move toward the blue end of the spectrum. 69% average accuracy. The lines you see in a star's spectrum act like thermometers. Play this game to review Astronomy. Table of Spectral Lines Used in SDSS REFERENCE WAVELENGTHS, IN VACUUM, USED IN SPECTRO1D Continuous Spectrum and Line Spectra from Different Elements. 149 times. Edit. 3 years ago. Spectral line catalogs available online are useful for selecting targeted line rest frequencies. Atomic Spectra Quiz DRAFT. Now consider the complete opposite, where a star is moving away from the Earth. Lines in light from extraterrestrial sources could be attributed to known elements electromagnetic... End of the star all the lines you see in a single,. To transitions of electrons from or to certain allowed orbits away for us to.... The line can tell us how fast spectral lines astronomy material is moving star this... Of very cool stars lines library ( units.h ), Up: Gnuastro library the light and... S blackbody continuum spectrum that the star 's spectrum act like thermometers,. 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Stars ( see stellar Classification ) spectrum act like thermometers line features below the of. Complete opposite, where a star 's spectrum act like thermometers online are useful for selecting targeted line rest.. The flux is in arbitrary units their emission and absorption lines seen in the spectra of gaseous sources from.


text:
# American Institute of Mathematical Sciences

July  2013, 18(5): 1389-1414. doi: 10.3934/dcdsb.2013.18.1389

## Sign-changing solutions of a quasilinear heat equation with a source term

 1 School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi 030006, China 2 Department of Mathematics, University of Central Florida, Orlando, FL 32816, United States

Received  June 2012 Revised  August 2012 Published  March 2013

The Cauchy problem of a heat equation with a source term $$\psi_t=\Delta \left(|\psi|^{m-1}\psi\right)+|\psi|^{\gamma-1}\psi\ \ \mbox{in}\ \ (0, \infty)\times R^n$$ is considered, where $\gamma>m>1$. We are interested in global solutions with H$\ddot{o}$lder continuity which satisfy the equation in the distribution sense, and with a fixed number of sign changes at any given time $t > 0$. Through detailed analysis of the self-similarity problem, we prove the existence of two type of such solutions, one with compact support and the other decays to zero as $| x| \rightarrow \infty$ with an algebraic rate determined uniquely by $n, m$ and $\gamma$. Our results extend previous study on positive self-similar solutions. Moreover, they demonstrate vital difference from the well-studied semi-linear case of $m = 1$.
Citation: Guirong Liu, Yuanwei Qi. Sign-changing solutions of a quasilinear heat equation with a source term. Discrete & Continuous Dynamical Systems - B, 2013, 18 (5) : 1389-1414. doi: 10.3934/dcdsb.2013.18.1389
##### References:
 [1] B. Bettioui and A. Gmira, On the radial solutions of a degenerate quasilinear elliptic equation in $R^N$, Ann. Fac. Sci. Toulouse, 8 (1999), 411-438.  Google Scholar [2] M.F. Bidaut-Véron, The $p$-Laplace heat equation with a source term: self-similar solutions revisited, Advanced Nonlinear Studies, 6 (2006), 69-108.  Google Scholar [3] T. Cazenave, F. Dickstein and F.B. Weissler, Sign-changing stationary solutions and blowup for the nonlinear heat equation in a ball Math. Ann., 344 (2009), 431-449. doi: 10.1007/s00208-008-0312-6.  Google Scholar [4] C. Dohmen and M. Hirose, Structure of positive radial solutions to the Haraux-Weissler equation, Nonlinear Anal., 33 (1998), 51-69. doi: 10.1016/S0362-546X(97)00542-7.  Google Scholar [5] A. Haraux and F.B. Weissler, Non-uniqueness for a semilinear initial value problem, Indiana Univ. Math. J., 31 (1982), 167-189. doi: 10.1512/iumj.1982.31.31016.  Google Scholar [6] L.A. Peletier, D. Terman and F.B. Weissler, On the equation $\Delta u+\frac {1}{2}x \nabla u +f(u)=0$, Arch. Rat. Mech. Anal. 94 (1986), 83-99. doi: 10.1007/BF00278244.  Google Scholar [7] Y.W. Qi, The existence of moving boundary solution of a porous media equation with a source term, Mathematical Sciences and Applications Gakkōtosho, Tokyo, 6 (1996), 197-215.  Google Scholar [8] Y.W. Qi, The global existence and nonuniqueness of a nonlinear degenerate equation, Nonlinear Anal. 31 (1998), 117-136. doi: 10.1016/S0362-546X(96)00298-2.  Google Scholar [9] R. Suzuki, Asymptotic behavior of solutions of quasilinear parabolic equations with supercritical nonlinearity, J. Differential Equations, 190 (2003), 150-181. doi: 10.1016/S0022-0396(02)00086-4.  Google Scholar [10] M. Wang and X. Wang, Existence of positive solutions to a nonlinear initial problem, Nonlinear Anal. 44 (2001), 1133-1136. doi: 10.1016/S0362-546X(99)00344-2.  Google Scholar [11] F.B. Weissler, Asymptotic analysis of an ordinary differential equation and non-uniqueness for a semilinear partial differential equation, Arch. Rat. Mech. Anal. 91 (1986), 231-245. doi: 10.1007/BF00250743.  Google Scholar [12] F.B. Weissler, Rapidly decaying solutions of an ordinary differential equation with applications to semilinear elliptic and parabolic partial differential equations, Arch. Rat. Mech. Anal. 91 (1986), 247-266. doi: 10.1007/BF00250744.  Google Scholar [13] F.B. Weissler, Existence and nonexistence of global solutions for a semilinear heat equation Israel J. Math. 38 (1981), 29-40. doi: 10.1007/BF02761845.  Google Scholar [14] E. Yanagida, Uniqueness of rapidly decreasing solutions to the Haraux-Weissler equation, J. Differential Equations 127 (1996), 561-570. doi: 10.1006/jdeq.1996.0083.  Google Scholar

show all references

##### References:
 [1] B. Bettioui and A. Gmira, On the radial solutions of a degenerate quasilinear elliptic equation in $R^N$, Ann. Fac. Sci. Toulouse, 8 (1999), 411-438.  Google Scholar [2] M.F. Bidaut-Véron, The $p$-Laplace heat equation with a source term: self-similar solutions revisited, Advanced Nonlinear Studies, 6 (2006), 69-108.  Google Scholar [3] T. Cazenave, F. Dickstein and F.B. Weissler, Sign-changing stationary solutions and blowup for the nonlinear heat equation in a ball Math. Ann., 344 (2009), 431-449. doi: 10.1007/s00208-008-0312-6.  Google Scholar [4] C. Dohmen and M. Hirose, Structure of positive radial solutions to the Haraux-Weissler equation, Nonlinear Anal., 33 (1998), 51-69. doi: 10.1016/S0362-546X(97)00542-7.  Google Scholar [5] A. Haraux and F.B. Weissler, Non-uniqueness for a semilinear initial value problem, Indiana Univ. Math. J., 31 (1982), 167-189. doi: 10.1512/iumj.1982.31.31016.  Google Scholar [6] L.A. Peletier, D. Terman and F.B. Weissler, On the equation $\Delta u+\frac {1}{2}x \nabla u +f(u)=0$, Arch. Rat. Mech. Anal. 94 (1986), 83-99. doi: 10.1007/BF00278244.  Google Scholar [7] Y.W. Qi, The existence of moving boundary solution of a porous media equation with a source term, Mathematical Sciences and Applications Gakkōtosho, Tokyo, 6 (1996), 197-215.  Google Scholar [8] Y.W. Qi, The global existence and nonuniqueness of a nonlinear degenerate equation, Nonlinear Anal. 31 (1998), 117-136. doi: 10.1016/S0362-546X(96)00298-2.  Google Scholar [9] R. Suzuki, Asymptotic behavior of solutions of quasilinear parabolic equations with supercritical nonlinearity, J. Differential Equations, 190 (2003), 150-181. doi: 10.1016/S0022-0396(02)00086-4.  Google Scholar [10] M. Wang and X. Wang, Existence of positive solutions to a nonlinear initial problem, Nonlinear Anal. 44 (2001), 1133-1136. doi: 10.1016/S0362-546X(99)00344-2.  Google Scholar [11] F.B. Weissler, Asymptotic analysis of an ordinary differential equation and non-uniqueness for a semilinear partial differential equation, Arch. Rat. Mech. Anal. 91 (1986), 231-245. doi: 10.1007/BF00250743.  Google Scholar [12] F.B. Weissler, Rapidly decaying solutions of an ordinary differential equation with applications to semilinear elliptic and parabolic partial differential equations, Arch. Rat. Mech. Anal. 91 (1986), 247-266. doi: 10.1007/BF00250744.  Google Scholar [13] F.B. Weissler, Existence and nonexistence of global solutions for a semilinear heat equation Israel J. Math. 38 (1981), 29-40. doi: 10.1007/BF02761845.  Google Scholar [14] E. Yanagida, Uniqueness of rapidly decreasing solutions to the Haraux-Weissler equation, J. Differential Equations 127 (1996), 561-570. doi: 10.1006/jdeq.1996.0083.  Google Scholar
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# Strange ifstream crash

## Recommended Posts

Oammar    256

I've been working on my own library in a different project. I decided to create a new project so I can test out what I've done so far in a fresh work space.

The issue I am encountering begins when I am attempting to open a file for reading from. It does open the file, however, when I attempt to read from it, or do any file related operations it crashes with 0xC0000005 exception, leading me on the stack to: EnterCriticalSection

I'm on Win7 using Visual Studio 2012

Message: Unhandled exception at 0x77E48E19 (ntdll.dll) in MainProj.exe: 0xC0000005: Access violation writing location 0x00000014.

Stack:

 	ntdll.dll!77e48e19()	Unknown
[Frames below may be incorrect and/or missing, no symbols loaded for ntdll.dll]
ntdll.dll!77e48d28()	Unknown
msvcr110.dll!_lock_file(_iobuf * pf) Line 223	C
msvcr110.dll!getc(_iobuf * stream) Line 43	C
MainProj.exe!std::_Fgetc<char>(char & _Byte, _iobuf * _File) Line 39	C++
MainProj.exe!std::basic_filebuf<char,std::char_traits<char> >::uflow() Line 482	C++
MainProj.exe!std::basic_filebuf<char,std::char_traits<char> >::underflow() Line 460	C++
msvcp110d.dll!std::basic_streambuf<char,std::char_traits<char> >::sgetc() Line 153	C++
msvcp110d.dll!std::basic_ostream<char,std::char_traits<char> >::operator<<(std::basic_streambuf<char,std::char_traits<char> > * _Strbuf) Line 513	C++
>	MainProj.exe!Utilities::ParseAndCollectFile(std::basic_string<char,std::char_traits<char>,std::allocator<char> > filePath) Line 44	C++

This is what I am doing:

std::string Utilities::ParseAndCollectFile(std::string filePath)
{
char cCurrentPath[FILENAME_MAX];
_getcwd(cCurrentPath, sizeof(cCurrentPath));
std::string path = cCurrentPath;
path += '\\';
path += filePath;

Debug::Log(path);
std::ifstream file(path, std::ifstream::in);

if (file)
{
file.close();
}
else
Debug::Log("Failed To Open File: " + filePath);

}

I've check to make sure the file is in the right location. However, another strange issue is: if I do:

std::ifstream file("data/maps/exportedForestMap.txt", std::ifstream::in);

before calling the library function that collects the file string data, the loading of the file problem goes away. I'm not using any custom threads at this point, this problem is baffling, everything works as it should in my main project where I'm working on the library.

Thank you for any insight you can give me as to why this is happening and various ways to solve the problem.

Edited by Oammar

##### Share on other sites
Oammar    256

What a strange situation. I was able to 'fix' this issue, but I am not satisfied with the situation at all..

In my main project I did not have to change any of these settings, I am not sure at all why I needed to change these settings for this new project to get this to work properly.

I had to change the Runtime Library setting to use Multi-threaded DLL (/MD) and then add: msvcrtd.lib to my dependencies, that solved my problems I was encountering for building in debug mode.

##### Share on other sites
Erik Rufelt    5901

Is all the code in the same new project and being compiled together, or are you including any code you've built as a library?

That could cause strange issues if they don't use the same settings, which fits how you fixed it.

Other similar issues could arise from different files/libraries being compiled with different preprocessor-macros or similar.

##### Share on other sites
Zaoshi Kaba    8434

This problem can occur if library and executable use different Runtime Library (DLL vs Static, and Debug vs Release). Make sure it matches across all projects and do full recompile.

##### Share on other sites
alh420    5995

Just a note, since this seems to be resolved already, but "if(file)" is not a proper way to check if the file opened or not.

I'm guessing the file actually fails to open

file will always be non-null, and you should use "if(file.is_open())" to see if it really is open, (or "if(file.fail())" to see if it failed)

Edit: oops, seems I need to read the spec more carefully before answering :P

Edited by Olof Hedman

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Mona2000    1967

Just a note, since this seems to be resolved already, but "if(file)" is not a proper way to check if the file opened or not.

It actually is, using operator void * (or operator bool if c++11).


text:
# Calculus limit defined integral

Calculus Level 2

$\lim_{n\to \infty} \left [ \left( \pi r \prod_{r=1}^n \left( 1 - \dfrac{r^2}{n^2} \right)\right) \right ]+ \lim_{n\to \infty} \left [ \left( \pi r \prod_{r=1}^n \left( 1 +\dfrac{r}{n} \right)\right) \right ]$

For $$r$$ being a set of all natural numbers, evaluate the expression above.

×


text:
### The Dark SIDH of Isogenies

Paul Bottinelli, Victoria de Quehen, Chris Leonardi, Anton Mosunov, Filip Pawlega, and Milap Sheth

##### Abstract

Many isogeny-based cryptosystems are believed to rely on the hardness of the Supersingular Decision Diffie-Hellman (SSDDH) problem. However, most cryptanalytic efforts have treated the hardness of this problem as being equivalent to the more generic supersingular $\ell^e$-isogeny problem --- an established hard problem in number theory. In this work, we shine some light on the possibility that the combination of two additional pieces of information given in practical SSDDH instances --- the image of the torsion subgroup, and the starting curve's endomorphism ring --- can lead to better attacks cryptosystems relying on this assumption. We show that SIKE/SIDH are secure against our techniques. However, in certain settings, e.g., multi-party protocols, our results may suggest a larger gap between the security of these cryptosystems and the $\ell^e$-isogeny problem. Our analysis relies on the ability to find many endomorphisms on the base curve that have special properties. To the best of our knowledge, this class of endomorphisms has never been studied in the literature. We informally discuss the parameter sets where these endomorphisms should exist. We also present an algorithm which may provide information about additional torsion points under the party's private isogeny, which is of independent interest. Finally, we present a minor variation of the SIKE protocol that avoids exposing a known endomorphism ring.

Available format(s)
Category
Public-key cryptography
Publication info
Preprint. MINOR revision.
Keywords
isogeniescryptanalysisSIDHSIKE
Contact author(s)
victoria dequehen @ isara com
History
Short URL
https://ia.cr/2019/1333

CC BY

BibTeX

@misc{cryptoeprint:2019/1333,
author = {Paul Bottinelli and Victoria de Quehen and Chris Leonardi and Anton Mosunov and Filip Pawlega and Milap Sheth},
title = {The Dark SIDH of Isogenies},
howpublished = {Cryptology ePrint Archive, Paper 2019/1333},
year = {2019},
note = {\url{https://eprint.iacr.org/2019/1333}},
url = {https://eprint.iacr.org/2019/1333}
}

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


text:
# For what values of $x$ is the series $\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$ convergent?

For what values of $x$ is the series $$\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$$ absolutely convergent, conditionally convergent and divergent?

Denote $a_k=(x-2)^k/k\ln{k},$ I get

$$\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_k}\right|=|x-2|\lim_{k\rightarrow\infty}\frac{k\ln{k}}{(k+1)\ln{k+1}}=1,$$

according to L'Hopitals rule. So I have that $-1 \leq x-2\leq1$ which means that the series is absolutely converging for $1<x<3$. Checking the endpoints:

$x=1:\Rightarrow$

$$\sum_{k=2}^{\infty}\frac{(-1)^k}{k\ln{k}}.$$

This is an alternating series. Denote $b_k=\frac{1}{k\ln{k}}.$ I have to show that $\lim_{k\rightarrow\infty}b_k=0$ and that $b_k$ is a decreasing function. The first condition is trivial. The second condition is trivial to since it is easy to see that $b_k>b_{k+1}>b_{k+2}>...>b_{k+m},$ for $m\in\mathbb{N}.$ So this is a conditional convergence.

$x=3:\Rightarrow$

$$\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$$ and the integraltest gives

$$\int_{x=2}^{\infty}\frac{1}{x\ln{x}}=\lim_{x\rightarrow \infty}\log(\log{x})=\infty$$

which diverges and the series diverges too for $x=3$. Thus the series behaviour can be summed up as follows:

The radius of convergence is $R=1$ and the interval of convergence is $I=[1,3).$ The series is

• Absolutely convergent for $x\in(1,3)$
• Conditionally convergent for $x=1$
• Divergent for the rest.

Is this correct?

EDIT: What I don't understand is that in my book the following theorem about convergence is stated:

• If the series $\sum|a_k|$ converges, then the series $\sum a_k$ also converges. This series is then absolutely convergent.
• If the series $\sum |a_k|$ is divergent and $\sum a_k$ is convergent, then the series $\sum a_k$ is conditionally convergent.

But in my solution above, I don't see anywhere this theorem is applied. I've just gonv for the limit of $|a_{k+1}/a_k|$, not the sum of $|a_k|$. Can someone explain this?

Using ratio test, one claims that for all $x$ with $|x-2|<1$, the series converges absolutely. For $x-2=1$, the series is $\displaystyle\sum_{k=2}\dfrac{1}{k\log k}$ which does not converge. For $x-2=-1$, the series is $\displaystyle\sum_{k=2}\dfrac{(-1)^{k}}{k\log k}$ which is convergent by alternating series test, but not absolutely, so it is conditional.

• Hi, please see my edit above. I'm not sure how to examine the conditional convergence. – Parseval Jan 7 '18 at 2:18
• For the divergent case for $x=3$, you cannot argue by having a divergent series as an upper bound. You may use the integral test, the primitive is $\log(\log x)$. – user284331 Jan 7 '18 at 2:21
• And for $x=1$, it is conditional, because the absolute value of the term would turn to equally $x=3$ case, which is divergent. – user284331 Jan 7 '18 at 2:23
• Hi, can you please check my solution again and also answer the question down in the edit? – Parseval Jan 7 '18 at 10:53
• Of course in dealing with the convergence of power series, it's likely that you use ratio test or the like rather than the definition of conditional or absolute convergent. The conditional or absolutely convergence are specially for endpoints, because by ratio test, all the points lying in the open interval are absolutely, as you may check by the statement of ratio test. For the endpoint $x=1$, actually your argument is not valid, that you didn't check that $\displaystyle\sum_{k=2}\left|\dfrac{(-1)^{k}}{k\log k}\right|$ is divergent, at least in that paragraph you didn't do so. – user284331 Jan 7 '18 at 19:12

Hints:

1. The Bertrand's series $\displaystyle\sum_{k\ge2}\frac1{k\,\ln k}$ is divergent.
2. If $a>1$, $k\ln k=o(a^k)\strut$.

You should say that it's "conditionally convergent for $x=1$" and "absolutely convergent for $x\in \color{red}{(}1,3)$", and I don't believe your argument that $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ diverges is valid.

The comparison that $\frac{1}{k \ln k} < \frac{1}{k}$ would have teeth if $\sum_{k=2}^{\infty}\frac{1}{k}$ converges, and it would show that $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ also converges. As we know it doesn't. If $\frac{1}{k \ln k} > \frac{1}{k}$ were somehow true, then you could argue that since $\sum_{k=2}^{\infty}\frac{1}{k}$ diverges $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ definitely diverges, but we know that's not true either.

The integral test is the most basic way to show it diverges I think. $\bigg [\int \frac{1}{x \ln x} dx=\int \frac{1/x}{\ln x} dx \left(\to \int \frac{du}{u}\right)=\ln(\ln x) \to \infty \ \ \text{as} \ \ x \to \infty \bigg] \implies \bigg [\sum_{k=2}^{\infty}\frac{1}{k\ln{k}} \to \infty \bigg]$

Accordingly, as user284331 pointed out, you should say that $\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$ converges conditionally at $x=1$

• Hi, can you please check my solution again and also answer the question down in the edit? – Parseval Jan 7 '18 at 10:53
• A first read of what your book says is confusing. As i know it, conditionally convergent means a given series only converges under the condition that the series alternates where otherwise, as with$\sum \frac {1}{k \ln k}$, it would diverge. An absolutely convergent series converges regardless, i.e. whether it alternates or not. – AmateurMathPirate Jan 7 '18 at 11:34
• I see that in your book $a_k$ is alternating already? – AmateurMathPirate Jan 7 '18 at 11:38
• No, not necessarily. This has nothing to do with alternating series yet. – Parseval Jan 7 '18 at 12:07
• @Parseval took a sec to find a series whose terms switch polarity without alternating. Made me consider something like $\sum \frac { \cos \frac {(3\pi k)}{7}}{k}$ . Really never considered those. I assume you're good now with the types of convergence? – AmateurMathPirate Jan 8 '18 at 2:38


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# Gravitational Waves and gamma -Ray Bursts

@article{Kochanek1993GravitationalWA,
title={Gravitational Waves and gamma -Ray Bursts},
author={Christopher S. Kochanek and Tsvi Piran},
journal={The Astrophysical Journal},
year={1993},
volume={417}
}
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text:
# Math Help - [SOLVED] prove/disprove (composition is commutative)

1. ## [SOLVED] prove/disprove (composition is commutative)

Prove or disprove the following statement: Function composition is commutative; that is, if
f : R --> R and g : R --> R,

then (
f o g)(x) = (g o f)(x) for all x in R.

all the R are real numbers. i just don't know how to get the doubled bar R

2. Originally Posted by jconfer
Prove or disprove the following statement: Function composition is commutative; that is, if
f : R --> R and g : R --> R,

then (
f o g)(x) = (g o f)(x) for all x in R.

all the R are real numbers. i just don't know how to get the doubled bar R
Think about these: $f(x) = x^2 \,\& \,g(x) = \sin (x)$.