text:
# ML Wiki

## Workflow Soundness

Soundness is a notion of correctness of workflow nets

These properties are usually checked with Reachability Graphs

## Situations to Avoid

### Unboundness

Unboundness means:

• there is no bound on the number of tokens that a place can hold
• this always means a problem

### Improper Termination

A workflow net satisfies proper completion when

• if there's a token in the output place, then there are no tokens in other places
• if this property is not satisfied, then the problem is called improper termination

• firing sequences:
• $s \to a \to b \to a \to d$
• $s \to a \to c \to a \to d$
• there are remaining tokens after the flow ends
• it means: it stopped, but there is still some work to do
• clearly it's a problem

Example:

• there are paths that lead to proper termination:
• card $\to$ charge $\to$ skip $\to$ finish
• transfer $\to$ receive payment $\to$ NA $\to$ repay
• but there are problematic firing sequences
• card $\to$ charge $\to$ NA $\to$ no repay $\to$ finish
• at the step of no-repay both repay and no repay are available
• after firing no repay there's one token left in the network

Both livelock and deadlock mean that

• there's a situation when you cannot reach the end

• once you fire $a$ nothing is enabled
• but the workflow in not complete: there's no token in the output place
• in this case we don't have matching joins - that leads to a deadlock

Livelock

• $C$ and $D$ can fire forever: they're always enabled
• but the workflow will never reach the end

• is a transition that can never fire

• in this case $F$ never gets activated
• this workflow can finish successfully
• the only fix - to remove the dead transition

## Petri Nets

For Petri Nets that represent infinite processes there is a notion of correctness of there nets:

• liveness
• boundness

Used notation:

### Liveness

In a live petri net there are no dead transitions

• a dead transition is a transition that can never fire in any marking reachable from the initial marking

a petri net $N$ with initial marking $M_0$ is live $\iff$

• $\forall M: M_0 \to^* M, \forall t: \exists M \to^* M'$ in which $t$ is enabled

• for any marking $M$ reachable from the initial marking $M_0$
• and for every transition $t \in T$
• there exists a marking $M'$ reachable from $M$ s.t.
• $t$ is enabled in $M'$

In other words:

• no matter what happens, for any $t$ there exists a marking $M'$ reachable from the current state $M$
• such that $t$ is enabled in it

#### Example

• the example on the left is live: no matter what happens the process continues
• the net on the right is not live: after firing $ccc$ nothing is active anymore

#### Example 2

• Reachability Graph:
• so whatever we do there are always active transitions
• but we cannot escape marking $[p_2]$, and $a$ is no longer live
• but if we add another transition $c$ it becomes live:

So liveness means:

• no matter what happens, every transition $t$ should be able to fire at some point in future

### Boundness

A petri net $N$ with initial marking $M_0$ is $k$-bounded iff

• $\forall M: M_0 \to^* M, \forall p \in P: M(p) \leqslant k$
• there never can appear more than $k$ tokens in any $p \in P$ in any reachable marking $M$

5-bounded net:

• each $p$ can hold no more than 5 tokens

1-bounded net is called safe

#### Example

• in this net $a$ can fire infinitely many number of times
• so this net is not bounded
• but it's always live: there exists matchings for which $a$ and $b$ can fire

#### Example 2

• not bounded: can put as many matchings as we want

a petri net $N$ with initial marking $M_0$ is deadlock-free iff

• $\forall M: M_0 \to^* M \ \exists M': M \to M'$
• for any reachable marking $M$ there exists another marking $M'$ that can be reached from $M$
• i.e. every reachable marking enables some transition $t \in T$

• after $b$ fires nothing can fire anymore
• $a$ needs tokens from two its input places, but there's only one token

## Workflow Nets Soundness

With workflow nets

• we can decide many things statically, before enacting the workflow
• there are some desirable properties that help to avoid these problems

A workflow net is sound $\iff$ it has

• the option to complete
• proper completion

Notation:

### Option to Complete

$\forall M, M_0 \to^* M: \exists M', M \to^* M': M' \geqslant [o]$

• for all $M$ reachable from the initial marking
• it should be possible to reach the final marking from $M$

Example:

• deadlock = no option to complete
• marking $[o]$ is not reachable from $[i]$

Livelock Example:

• consider the net with a livelock
• the following is its reachability graph
• we see that two nodes are not connected: it's clearly a problem

### Proper Completion

$M_0 \to ^* M \land M \geqslant [o] \Rightarrow M \equiv [o]$

• when $M$ is reachable from the initial marking
• and it's great or equal to the final marking
• for proper completion to hold $M$ should be equal to $[o]$

$\forall t \in T, \exists M: M_0 \to^* M$ and $t$ in enabled in $M$

• for all possible transitions there should exists at least one marking $M$ in which $t$ is enabled

### Petri Net Soundness

A sound workflow net can never be live:

• it should be able to reach the final state
• and nothing should be able to fire from that state

A workflow net $N = (P, T, F)$ is sound iff

• a petri net $N' = (P, T \cup \{ \alpha \}, F \cup \{ (o, \alpha), (\alpha, i) \} )$ is sound

So we just connect $o$ and $i$ via some transition $\alpha$

• and check Petri Net soundness

Relation:

• if $N$ doesn't have the proper completion property, $N'$ will not be bounded
• if there's a deadlock in $N$,
• there exists a firing sequence in $N'$ s.t.
• it brings you to some state where you cannot reach $o$
• i.e. $N'$ is not live
• dead transitions: same notion in both $N$ and $N'$


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# CBSE Notes for Class 6 Maths Chapter 1: Knowing our Numbers

Table of Content

## What are Natural numbers?

Counting numbers 1, 2, 3, 4, ...... etc. are called Natural numbers. The smallest natural number is 1 and there is no largest natural number.

## Digits and Place value

Numbers are formed using the ten symbols 0, 1, 2, 3, 4,  5, 6, 7, 8, 9. These symbols are called digits or figures.
To find the place value of a digit in a number, multiply the digit with the value of the place it occupies.

## Comparison of Numbers

We often need to compare the two or more numbers. Here are some of the ways it can be performed easily
1) If two numbers have unequal number of digits, then the number with the greater number of digits is greater.
2) If two numbers have equal number of digits then, the number with greater valued digit on the extreme left is greater. If the digits on extreme left of the numbers are equal then the digits to the right of the extreme left digits are compared and so on.
Example
a)  Comparison between 358 and 4567
Here 4567 has four digits and 358 has three digits, so clearly 4657 is greater than 358

b) Comparison between 1345 and 2456
Here both the number have same digit, So we need start looking at the extreme left digit
1345 -> 1
2456 -> 2
Now 2 > 1
So we can clearly state 2456 > 1345
c) Comparison between 4345 and 4656
Here both the number have same digit, So we need start looking at the extreme left digit
4345 -> 4
4656 -> 4
As they are same, we need start looking at second extreme left digit
4345 -> 3
4656 -> 6
Now  6> 3
So we can clearly state 4656 > 4345
Some important terms
The arrangements of numbers from the smallest to the greatest is called ascending order.
The arrangement of numbers from the greatest to the smallest is called descending order.
Example
Arrange the following in ascending order: 5392, 5782, 5789, 5654
The above rules of comparison can be applied here also.
Here all the number are of same number of digit and extreme left digit is also same. So we need to look at second extreme left digit
So 5782,5789 > 5654 > 5392
Now in 5782,5789, unit place makes the comparison easier
5789> 5782 > 5654 > 5392
So ascending order would be
5392 < 5654 < 5782 < 5789

## Place Value of digit

Let’s discuss the place value of digits in the number and how a number can be written in that form
Indian system of numeration
Values of the places in the Indian system of numeration are Ones, Tens, Hundreds, Thousands, Ten thousands, Lakhs, Ten Lakhs, Crores and so on.
The following place value chart can be used to identify the digit in any place in the Indian system.
 Periods Crores lakhs Thousands Ones Places Tens Ones Tens Ones Tens Ones Hundreds Tens Ones
Example
5,46,851 = 5 × 1,00,000 + 4 × 10,000 + 6 × 1,000 + 8 × 100 + 5 × 10 +1 × 1
This number has 1 at one’s place, 5 at tens place, 8 at hundreds place, 6 at thousands place, 4 at ten thousands place and 1 at lakh place.
Number Name are also written based on the place value name. So Its number name is Five lakh forty-six thousand eight hundred fifty-one
We can use below table format for easily reading and writing the Number

 Number Tens Lakh Ones Lakh Tens thousand Ones thousand Hundreds tens Ones Number Name 5,46,851 5 4 6 8 5 1 Five lakh forty-six thousand eight hundred fifty one 32,75,829 3 2 7 5 8 2 9 Thirty two lakh Seventy-five thousand eight hundred twenty nine

Use of Commas
Commas added to numbers help us read and write large numbers easily. As per Indian Numeration, Commas are used to mark thousands, lakhs and crores. The first comma comes after hundreds place (three digits from the right) and marks thousands. The second comma comes two digits later (five digits from the right). It comes after ten thousand place and marks lakh. The third comma comes after another two-digits (seven digits from the right). It comes after ten lakh place and marks crore
Examples
1, 08, 01, 992
2, 32, 40, 581
3, 17, 05, 062

International system of numeration
Values of the places in the International system of numeration are Ones, Tens, Hundreds, Thousands, Ten thousands, Hundred thousands, Millions, Ten millions and so on.
1 million = 1000 thousands,
1 billion = 1000 millions

Following place value chart can be used to identify the digit in any place in the International system.
 Periods Billions Millions Thousands Ones Places Hundreds Tens Ones Hundreds Tens Ones Hundreds Tens Ones Hundreds Tens Ones
Use of Commas
As per International Numeration, Commas are used to mark thousands and millions. It comes after every three digits from the right. The first comma marks thousands and the next comma marks millions. For example, the number 10,101,592 is read in the International System as tem million one hundred one thousand five hundred ninety-two. In the Indian System, it is 1 crore one lakh one thousand five hundred ninety-two.

## Estimation of the Numbers

A reasonable guess of the actual value is called an estimate.
A quick, rough estimate of the result of number operations can be done by rounding off the numbers is involved.
Rules of Estimation
1)Estimating numbers to the nearest tens is done by rounding off numbers 1, 2, 3 and 4 to 0 and number 6, 7, 8, 9 to 10.
2) Estimating numbers to the nearest hundreds is done by rounding off numbers 1 to 49 to 0 and numbers 51 to 99 to 100.
3) Estimating numbers to the nearest thousands is done by rounding off numbers 1 to 499 to 0 and the numbers 501 to 999 to 1000.
Estimation involves approximating a quantity to an accuracy required. We can apply the above rules depending on the accuracy required.
We can estimate Sum, difference and Multiplication by applying the rules of estimation also. We can apply the above rules depending on the accuracy required and how quickly answer can be find out

## Roman Numerals

Roman Numerals system is another system used apart of Hindu-Arabic system.
The Roman numerals are

 I 1 II 2 III 3 IV 4 V 5 VI 6 VII 7 VIII 8 IX 9 X 10 X1 11 X11 12 XX 20 L 50 C 100 D 500 M 1000

Rules of the system
1) In Roman numerals a symbol is not repeated more than three times, but the symbols V, L and D are never repeated.
2) Roman numerals are read from left to right and the letters of Roman numerals are arranged from the largest to the smallest.
3) If a symbol of smaller value is written to the right of a symbol of greater value, then its value gets added to the value of greater symbol.
VI = 5 + 1 = 6
4) If a symbol of smaller value is written to the left of a symbol of greater value, its then value is subtracted from the value of the greater symbol.
IV = 5 – 1 = 4
5)The symbol I can be subtracted from V and X only.
The symbol X can be subtracted from L, M and C only.

## Importance of Brackets

Brackets help in simplifying an expansion with more than one mathematical operation.
In an expression that includes brackets, the numbers inside the brackets must be simplified into a single

### Quiz Time

Question 1 What is 100000 - 1 ?
A)99999
B)999999
C)99989
D)100009
Question 2 How many lakhs make a million?
A)100
B)10
C)1
D)1000
Question 3 1 crore equals
A)10 ten laks
B)10 laks
C)1 million
D)1000 laks
Question 4 Simplify 200 -(13-3)X(2+5) -10
A) 10
B) 120
C) 100
D) 175
Question 5 How 92 is written in Roman numerals?
A)XCIII
B)XCII
C)LXII
D)XII

• Notes
• Assignments
• NCERT Solutions

Reference Books for class 6

Practicing make students fluent in their concept. Help your child practice math using our Class 6 Maths worksheets (Kindle Edition)
You can use this ebook on PC, tablet, mobile devices by using Free Kindle Reading Apps for android and ios.
Given below are the links of some of the reference books for class 6 science and class 6 math.

You can use above books for extra knowledge and practicing different questions.

### Practice Question

Question 1 What is $\frac {1}{2} + \frac {3}{4}$ ?
A)$\frac {5}{4}$
B)$\frac {1}{4}$
C)$1$
D)$\frac {4}{5}$
Question 2 Pinhole camera produces an ?
A)An erect and small image
B)an Inverted and small image
C)An inverted and enlarged image
D)None of the above

Note to our visitors :-

Thanks for visiting our website.

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# 9 May 2017 Menderes Iskin

***************************************
KOÇ UNIVERSITY PHYSICS SEMINAR
***************************************

Speaker             : Menderes Işkın, Koç University
Title                     : BCS theory of time-reversal-symmetric Hofstadter-Hubbard model
Date                     : May 9, 2017 Tuesday
Time                    : 4:00 P.M.
Cookie & Tea  : SCI 103 3:45 P.M.
Place                   : SCI 103

Abstract:
The competition between the length scales associated with the periodicity of a lattice potential and the cyclotron radius of a uniform magnetic field is known to have dramatic effects on the single-particle properties of a quantum particle, e.g., the fractal spectrum is known as the Hofstadter butterfly. Having this intricate competition in mind, we consider a two-component Fermi gas on a square optical lattice with opposite synthetic magnetic fields for the components, and study its effects on the many-body BCS-pairing phenomenon. By a careful addressing of the distinct superfluid transitions from the semi-metal, quantum spin-Hall insulator or normal phases, I will talk about the low-temperature phase diagrams of the model, displaying lobe structures that are reminiscence of the well-known Mott-insulator transition of the Bose-Hubbard model.


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# Gravimetric analysis.

by LaМЃszloМЃ Erdey

Publisher: Pergamon Press, Publisher: [distributed in the Western Hemisphere by Macmillan, New York] in Oxford, New York

Written in English

## Subjects:

• Chemistry, Analytic -- Quantitative.

## Edition Notes

Classifications The Physical Object Statement Translated by Gyula Svehla; edited by Ilona Buzas. Series International series of monographs on analytical chemistry -- v. 7., International series of monographs in analytical chemistry -- v. 7. LC Classifications QD101 .E7132 Pagination v. Open Library OL16568211M LC Control Number 64009318

Thermal Analysis: methods, principles, applicaon Andrey Tarasov Lecture on Thermal analysis Andrey Tarasov, Thermal analysis, Lecture series . ¥ Chapter 8 Gravimetric Analysis Ð Skoog Book ¥ Page Do Pr oblems: 1,2,4,9,10,1 1,14,16,21,27,30,33 ¥ Chapter 9 Electrolyte Ef fects Activities Ð Ef fective concentration and equilibrium Ð Please do problems: 1,2,3,6,7,8,12 ¥ Exam II Feb. 13 Gr av imetric File Size: KB.   gravimetric estimation of barium sulphate || estimation of barium as barium sulphate by gravimetric method || describe gravimetric determination of barium as barium sulphate || estimation of. The conditions essential to accurate gravimetric analysis have been given. As a rule such estimations come under one of four classes: Direct Estimations. These are the most usual and accurate methods of gravimetric analysis. The element sought is precipitated and weighed generally as a compound, definite in composition. Direct Estimation by Loss.

A detailed investigation has been made of the gravimetric determination of nickel with dimethylglyoxime in the presence of copper. Copper interference is eliminated by precipitation in a solution containing tartrate and thiosulphate at a pH of – The solubility of the nickel dimethylglyoxime complex has been determined as a function of temperature, pH Cited by: 5. gravimetric analysis Source: A Dictionary of Geology and Earth Sciences Author(s): Michael Allaby. A method of chemical analysis which involves the quantitative precipitation of an element or elements from solution in a compound whose composition is accurately known and which may be isolated in a pure form prior to weighing.

## Recent

Analytical Chemistry, Volume 7: Gravimetric Analysis, Part II describes the experimental procedures for the gravimetric analysis of Groups I to V cations.

This book is composed of 43 chapters that also present sample preparation, separation, and precipitation protocols.

Gravimetric analysis. book Analysis, Part III describes the experimental procedures for the gravimetric analysis of various compounds. This book is composed of 13 chapters that also present sample preparation protocols.

The first four chapters survey the steps for halogen compound determination. Because the release of a volatile species is an essential part of these methods, we classify them collectively as volatilization gravimetric methods of analysis. Particulate Gravimetry Precipitation and volatilization gravimetric methods require that the analyte, or some other species in the sample, participates in a chemical reaction.

is a platform for academics to share research papers. In the gravimetric analysis of iron, hydroxide may be added to a solution containing Fe3+ to precipitate a gelatinous mess which is ignited to form Fe 2 O 3.

If grams of Fe 2 O 3 were formed from the ignition of the precipitated mess, how many grams of FeCO 3 were contained in the original sample analyzed. (a) g FeCO 3File Size: KB. A Text-Book of Quantitative Chemical Analysis by Gravimetric, Electrolytic, Volumetric and Gasometric Methods, With Seventy-Two Laboratory Exercises.

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book sought is converted into a substance (of known composition) that can be separated from the sample and weighed. The steps commonly followed in gravimetric. Gravimetric analysis is a type of lab technique used to determine the mass or concentration of a substance by measuring a change in mass.

The chemical we are trying to quantify is also known as the analyte. Read more about the Classifications and advantages and disadvantages of Gravimetric Analysis at Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores.

Figure in Chapter 1, for example, il-lustrates a precipitation gravimetric meth-od for the analysis of nickel in ores. A total analysis technique is one in which the analytical signal—mass in this case—File Size: 2MB. GRAVIMETRIC ANALYSIS At the end of this unit, the student is expected to be able to: 1- Understand the fundamentals of gravimetric analysis.

2- Follow the steps of the gravimetric analysis. 3- Choose the appropriate precipitating agent for a certain analyte. 4- Avoid or at least minimize the contamination of the precipitate. Precipitation gravimetry continues to be listed as a standard method for the determination of $$\text{SO}_4^{2-}$$ in water and wastewater analysis [Method SO42– C and Method SO42– D as published in Standard Methods for the Examination of Waters and Wastewaters, 20th Ed., American Public Health Association: Wash- ington, D.

C., ]. Gravimetric Analysis - WJEC A Level Experiment One of the Lab Book Practicals essential for the WJEC AS/A Level Chemistry Syllabus. This one shows how to identify an Gravimetric Analysis of Gravimetric Analysis of Unknown Sulfate Salt. The Essay on Gravimetric Analysis & Determination of an Unknown Metal Carbonate.

A large variety of analytical techniques and procedures, ranging from instrumental methods such as spectroscopy and chromatography to more classical processes, such as qualitative and gravimetric analyses, have been used to help chemists identify unknown compounds.

Volatile solids analysis is an important example of this type of gravimetric analysis. As the name implies, precipitative gravimetry relies on the chemical precipitation of an analyte. Its most important application in the environmental field is with the analysis of sulfite.

Analytical Chemistry Books. Sampling And Sample Preparation, Quantitation And Calibration Methods, Titration, Gravimetric Analysis, Acid-base Titration, Precipitation Titrations, Complexometric Titrations.

The calculations of analytical chemistry. This text-book is intended for use in scientific schools and colleges, in connection with. The Gravimetric Determination of Nickel INTRODUCTION Nickel(II) forms a precipitate with the organic compound dimethylglyoxime, C4H6(NOH)2.

The formation of the red chelate occurs quantitatively in a solution in which the pH is buffered in the range of 5 to 9. The chelation reaction that occurs is illustrated Size: KB.

Gravimetric analysis is one of the most accurate analytical methods available. It is concerned with the determination of a substance by the process of weighing.

The element or radical to be determined is converted into a stable compound of definite composition and the mass of the compound is determined accurately. From this, the mass of element. Definition: Thermogravimetric Analysis is a technique in which the mass of a substance is monitored as a function of temperature or time as the sample specimen is subjected to a controlled temperature program in a controlled atmosphere.

An Alternate Definition: TGA is a technique in which, upon heating a material, its weight increases or Size: 1MB. Additional Physical Format: Online version: Erdey, László.

Gravimetric analysis. New York, Macmillan, [65] (OCoLC) Online version: Erdey, László. Thermogravimetric analysis (TGA) is an analytical technique for measuring changes in the mass of a material that occur in response to programmed temperature : Sergey Vyazovkin.

Gravimetric methods: The. quantitative methods. that are based on determining the. mass. of a. pure compound. to which the. analyte. chemically related. • Precipitation gravimetry: The. analyte. is separated from a solution of the sample as a. precipitate. and is converted to a compound of known composition that can be weighed File Size: 2MB.

Thermogravimetric analysis or thermal gravimetric analysis (TGA) is a method of thermal analysis in which the mass of a sample is measured over time as the temperature changes. This measurement provides information about physical phenomena, such as phase transitions, absorption, adsorption and desorption; as well as chemical phenomena including Classification: Thermal analysis, A typical TGA system.

Gravimetric Analysis In gravimetric analysis, the mass of a product from a chemical reaction is used to calculate the quantity of the original analyte (the species being analyzed).

e.g., Careful gravimetric analysis by T. Richards and his colleagues early in the twentieth century determined the atomic masses of Ag, Cl, and N to six-ﬁgure accuracyFile Size: 5MB. Review and cite GRAVIMETRIC ANALYSIS protocol, troubleshooting and other methodology information | Contact experts in GRAVIMETRIC ANALYSIS to get answers.

Analytical Chemistry, Volume 7: Gravimetric Analysis, Part II describes the experimental procedures for the gravimetric analysis of Groups I to V cations. This book is composed of 43 chapters that also present sample preparation, separation, and precipitation : \$ Gravimetric Analysis, Part III describes the experimental procedures for the gravimetric analysis of various compounds.

This book is composed of 13 chapters that. THE book comprises all of the fundamental methods of gravimetric analysis, taking into consideration also the latest results in physical chemistry, in structural inorganic chemistry, in colloid science and in organic chemistry.

The book is particularly opportune in connection with the rapid development of nuclear chemistry. Phosphorous is a key element for all life found in every biological cell integral ingredient of plant fertilizers Exists in fertilizer as: (NH4)2HPO4 our goal was to "determine the amount of phosphorous in a mixture by gravimetric analysis" " technique through which the amount.

Immediately download the Gravimetric analysis summary, chapter-by-chapter analysis, book notes, essays, quotes, character descriptions, lesson plans, and more - everything you need for studying or teaching Gravimetric analysis.

Introduction • The term gravimetric pertains to a Weight Measurement. • Gravimetric method is one in which the analysis is completed by a weighing operation. • Gravimetric Analysis is a group of analytical methods in which the amount of analyte is determined by the measurement of the mass of a pure substance containing the analyte.

Gravimetric analysis. See also what's at Wikipedia, your library, or elsewhere. Broader term: Chemistry, Analytic -- Quantitative; Narrower term: Thermogravimetry; Filed under: Gravimetric analysis A new gravimetric method for the determination of titanium / (), by Oscar S.

Watkins (page images at HathiTrust; US access only) Gravimetric determination of barium in zirconium .Gravimetric Determination of Calcium ABSTRACT Determining the mass of a pure compound is a method of a gravimetric analysis.

One of the gravimetric analyses is the precipitation; it is a method of separating the analyte from the unknown sample as a precipitate where it will be filtered and converted into a known composition that can be weighed to determine its mass .Gravimetric Estimation of Nickel.

Theory. Procedure. Self Evaluation. Animation. Assignment. Reference. Feedback. Books: 1. Vogel's text book of quantitative Chemical Analysis Sixth edition- J Mendham, R C Denney, J D Barnes, M J K Thomas. 2. Principles of Practical Chemistry- M. Viswanathan Cite this Simulator.


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# Linear initial-value question

1. Sep 8, 2009

### string_656

Im doing questions out of a book, and im stuck on one question, and it dosnt give any answers

dy/dx*(x) = 2y/x + X^2.......when y(1) = 3 ... = ..... dy/dx - 2y/x^2 = x
i have trouble intergrating...

I(x) = e ^ (intergral of) -2/x^2) = e^(2/x)

e^(2/x)*y = (intergral of) xe^(2/x)

im just having trouble with this ^^

can ya help?
thanks

2. Sep 9, 2009

### LCKurtz

Your work looks OK. That integral can't be expressed in terms of the usual elementary functions. If it's a textbook exercise I would think there might be a typo somewhere.

3. Sep 9, 2009

### HallsofIvy

Staff Emeritus
On the contrary, that has a simple polynomial solution.

Your difficulty is in your very first step: your equation is dy/dx= 2y/x+ x2 and then you rewrote it as dy/dx- 2y/x2= x!

It should be dy/dx- 2y/x= x2. Now the integrating factor is
$$e^{\int \frac{-2}{x}dx}= x^{-2}$$.

4. Sep 9, 2009

### LCKurtz

But aren't you missing the x in the first term of the original DE: dy/dx*(x)?
I would have written it as x*dy/dx but anyway you must divide through by that before computing the integrating factor.

5. Sep 9, 2009

### string_656

im only having trouble intergrating xe^(2/x).. if i can do that i can solve the problem..

6. Sep 10, 2009

### LCKurtz

In my previous reply I said:

"Your work looks OK. That integral can't be expressed in terms of the usual elementary functions. If it's a textbook exercise I would think there might be a typo somewhere."

There *is* no elementary anti-derivative for that function. None of your usual techniques can ever work.

7. Sep 10, 2009

### string_656

oh ok.. thanks. but how could i do it... is there a way of doing it?

8. Sep 11, 2009

### LCKurtz

There is no elementary way. If you are familiar with the exponential integral:

http://en.wikipedia.org/wiki/Exponential_integral

there is a way to express the answer in terms of that. Like I said before, if this is a typical homework problem I suspect there is a misprint. As another poster has pointed out, if the x factor in the leading term was missing, it would be straightforward. I can give you the expression Maple gives for an answer if you want it.

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# Two bodies of equal mass thrown from a height-Practical physics question

Two bodies of masses $M$ and $m$ are allowed to fall from the same height. If the air resistance be the same for each of the bodies, will the two bodies reach the Earth simultaneously? Justify.

The answer given is not that both reach at same time. I fail to understand why?

According to the equation by Galileo:

$a=\frac{F}{m}=\frac{GM_\oplus}{R^2}$

where $M_\oplus$ is the mass of the Earth, $R$ is the radius of the Earth, and $G$ is Newton's gravitational constant.

So the acceleration is independent of the mass. So in this question both the bodies should fall at the same time.

Is it that due to air resistance the force on $M$ will be more or something?

I suspect it's all in what they meant by "air resistance is the same ." For example, if both bodies are the same size and shape, the equation for air resistance as a func of velocity is the same, but the greater mass will be less affected by this equal force magnitude. Air resistance being "same" is not the same thing as being "neglectable" or "zero."

So the acceleration is independent of the mass. So in this question both the bodies should fall at the same time.

Your answer is correct if you dropped those objects on the Moon. However, the Moon has no atmosphere. You are ignoring drag, or "air resistance". Suppose you drop a ping pong ball and a super ball of the same size as the ping pong ball. Both are bodies the same size and shape, making "air resistance be the same for each of the bodies." The much smaller mass of the ping pong ball means that drag has a much greater effect on the ping pong ball than it does on the super ball. The super ball hits first.

Suppose you drop the two balls from a good height (say the top of a six story building). The super ball will hit the ground hard in a couple of seconds. At this time, the ping pong ball will be less than halfway down and will be falling at its terminal velocity of about 8 meters/second.

• I can confirm this scenario is true. I did just that at about the same height. I used 1 ping-pong ball and filled another with sand and a third with water. 1st - sand ball, 2nd water ball, and significantly longer air ball. – terary Dec 24 '15 at 11:51

Consider the force applied by air resistance $F_a$, equal for both objects. Then the equations of motion are:

$$F_a - \frac{GM_\oplus M}{r^2} = Ma$$ and $$F_a - \frac{GM_\oplus m}{r^2} = ma$$

Solving for $a$ in each case, you'll find that the acceleration due to gravity is the same in both cases (as expected), but the acceleration due to $F_a$ is different for the two masses, so they do not fall at the same rate.

For what it's worth, I think the question is phrased vaguely/misleadingly...


text:
Diskretnaya Matematika
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 Diskr. Mat.: Year: Volume: Issue: Page: Find

 Diskr. Mat., 2016, Volume 28, Issue 3, Pages 59–96 (Mi dm1384)

Distribution of the extreme values of the number of ones in Boolean analogues of the Pascal triangle

F. M. Malyshev

Steklov Mathematical Institute of Russian Academy of Sciences

Abstract: The paper is concerned with estimating the number $\xi$ of ones in triangular arrays consisting of elements of the field $GF(2)$ which are defined by the bottom row of $s$ elements. The elements of each higher row are obtained (as in Pascal triangles) by the summation of pairs of elements from the corresponding lower row. It is shown that there exists a monotone unbounded sequence $0=k_0<k_1<k_2< ...$ of rational numbers such that, for any $k>0$, for sufficiently large $s$ the admissible values of $\xi$ which are smaller than $ks$ or larger than $s(s+1)/3-sk/3$ are concentrated in neighbourhoods of points $k_is$ and $s(s+1)/3-sk_i/3$, $i\geqslant0$. The resulting estimates of the neighbourhoods are functions of $i$ for each $i\geqslant0$ and do not depend on $s$. The distributions of the numbers of triangles with values $\xi$ in these neighbourhoods depend only on the residues of $s$ with respect to moduli that depend on $i\geqslant0$.

Keywords: Pascal triangle, (0-1)-matrix, extreme combinatorial configuration.

DOI: https://doi.org/10.4213/dm1384

Full text: PDF file (1219 kB)
References: PDF file   HTML file

English version:
Discrete Mathematics and Applications, 2017, 27:3, 149–176

Bibliographic databases:

UDC: 519.14

Citation: F. M. Malyshev, “Distribution of the extreme values of the number of ones in Boolean analogues of the Pascal triangle”, Diskr. Mat., 28:3 (2016), 59–96; Discrete Math. Appl., 27:3 (2017), 149–176

Citation in format AMSBIB
\Bibitem{Mal16} \by F.~M.~Malyshev \paper Distribution of the extreme values of the number of ones in Boolean analogues of the Pascal triangle \jour Diskr. Mat. \yr 2016 \vol 28 \issue 3 \pages 59--96 \mathnet{http://mi.mathnet.ru/dm1384} \crossref{https://doi.org/10.4213/dm1384} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=3643047} \elib{https://elibrary.ru/item.asp?id=27349816} \transl \jour Discrete Math. Appl. \yr 2017 \vol 27 \issue 3 \pages 149--176 \crossref{https://doi.org/10.1515/dma-2017-0019} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000405964800003} \scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-85021847290}

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This publication is cited in the following articles:
1. F. M. Malyshev, “Bulevy analogi treugolnika Paskalya s maksimalno vozmozhnym chislom edinits”, Diskret. matem., 32:1 (2020), 51–59
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text:
# deepdish

Serving Up Chicago-Style Deep Learning.

Latest post:

The building blocks of Deep Learning 21 Nov 2015

# Hinton's Dark Knowledge

On Thursday, October 2, 2014 Geoffrey Hinton gave a talk (slides, video) on what he calls “dark knowledge” which he claims is most of what deep learning methods actually learn. The talk presented an idea that had been introduced in (Caruana, 2006) where a more complex model is used to train a simpler, compressed model. The main point of the talk introduces the idea that classifiers built from a softmax function have a great deal more information contained in them than just a classifier; the correlations in the softmax outputs are very informative. For example, when building a computer vision system to detect cats,dogs, and boats the output entries for cat and dog in a softmax classifier will always have more correlation than cat and boat since cats look similar to dogs.

Dark knowledge was used by Hinton in two different contexts:

• Model compression: using a simpler model with fewer parameters to match the performance of a larger model.
• Specialist Networks: training models specialized to disambiguate between a small number of easily confuseable classes.

## Preliminaries

A deep neural network typically maps an input vector $$\mathbf{x}\in\mathbb{R}^{D _ {in}}$$ to a set of scores $$f(\mathbf{x}) \in \mathbb{R}^C$$ for each of $$C$$ classes. These scores are then interpreted as a posterior distribution over the labels using a softmax function

The parameters of the entire network are collected in $$\boldsymbol{\Theta}$$. The goal of the learning algorithm is to estimate $$\boldsymbol{\Theta}$$. Usually, the parameters are learned by minimizing the log loss for all training samples

which is the negative of the log-likelihood of the data under the logistic regression model. The parameters $$\boldsymbol{\Theta}$$ are estimated with iterative algorithms since there is no closed-form solution.

This loss function may be viewed as a cross entropy between an empirical posterior distribution and a predicted posterior distribution given by the model. In the case above, the empirical posterior distribution is simply a 1-hot distribution that puts all its mass at the ground truth label. This cross-entropy view motivates the dark knowledge training paradigm, which can be used to do model compression.

## Model compression

Instead of training the cross entropy against the labeled data one could train it against the posteriors of a previously trained model. In Hinton’s narrative, this previous model is an ensemble method, which may contain many large deep networks of similar or various architectures. Ensemble methods have been shown to consistently achieve strong performance on a variety of tasks for deep neural networks. However, these networks have a large number of parameters, which makes it computationally demanding to do inference on new samples. To alleviate this, after training the ensemble and the error rate is sufficiently low, we use the softmax outputs from the ensemble method to construct training targets for the smaller, simpler model.

In particular, for each data point $$\mathbf{x} _ n$$, our first bigger ensemble network may make the prediction

The idea is to train the smaller network using this output distribution rather than the true labels. However, since the posterior estimates are typically low entropy, the dark knowledge is largely indiscernible without a log transform. To get around this, Hinton increases the entropy of the posteriors by using a transform that “raises the temperature” as

where $$T$$ is a temperature parameter that when raised increases the entropy. We now set our target distributions as

The loss function becomes

Hinton mentioned that the best results are achieved by combining the two loss functions. At first, we thought he meant alternating between them, as in train one batch with $$L ^ \mathrm{(hard)}$$ and the other with $$L ^ \mathrm{(soft)}$$. However, after a discussion with a professor that also attended the talk, it seems as though Hinton took a convex combination of the two loss functions

where $$\alpha$$ is a parameter. This professor had the impression that an appropriate value was $$\alpha = 0.9$$ after asking Hinton about it.

One of the main settings for where this is useful is in the context of speech recognition. Here an ensemble phone recognizer may achieve a low phone error rate, but it may be too slow to process user input on the fly. A simpler model replicating the ensemble method, however, can bring some of the classification gains of large-scale ensemble deep network models to practical speech systems.

## Specialist networks

Specialist networks are a way of using dark knowledge to improve the performance of deep network models regardless of their underlying complexity. They are used in the setting where there are many different classes. As before, deep network is trained over the data and each data point is assigned a target that corresponds to the temperature adjusted softmax output. These softmax outputs are then clustered multiple times using k-means and the resultant clusters indicate easily confuseable data points that come from a subset of classes. Specialist networks are then trained only on the data in these clusters using a restricted number of classes. They treat all classes not contained in the cluster as coming from a single “other” class. These specialist networks are then trained using alternating one-hot, temperature-adjusted technique. The ensemble method constructed by combining the various specialist networks creates benefits for the overall network.

One technical hiccup created by the specialist network is that the specialist networks are trained using different classes than the full network so combining the softmax outputs from multiple networks requires a combination trick. Essentially there is an optimization problem to solve: ensure that the catchall “dustbin” classes for the specialist networks match a sum of your softmax outputs. So that if you have cars and cheetahs grouped together in one class for your dog detector you combine that network with your cars versus cheetahs network by ensuring the output probabilities for cars and cheetahs sum to a probability similar to the catch-all output of the dog detector.


text:
eLibrary of Mathematical Instituteof the Serbian Academy of Sciences and Arts
 > Home / All Journals / Journal /

 Publications de l'Institut MathématiquePublisher: Matematički institut SANU, BeogradISSN: 0350-1302Issue: (N.S.) 42 (56)Date: 1987Journal Homepage

 Free Power or Width of some Kinds of Mathematical Structures 3 - 12 Đuro Kurepa
AbstractMSC: 05C38
 On the Largest Eigenvalue of Unicyclic Graphs 13 - 19 Slobodan K. Simić
AbstractMSC: 05C50
 Bounds for the Number of Perfect Matchings in Hexagonal Systems 21 - 27 Ivan Gutman and Jerzy Cioslowski
AbstractMSC: 03C50
 The Formulas of the General Reproductive Solution of an Equation in Boolean Ring with Unit 29 - 34 Dragić Banković
AbstractMSC: 06E20 03G05
 On the Definition of a Quadratic Form 35 - 41 Svetozar Kurepa
AbstractMSC: 15A63 39B50 46C10 46K99
 Convolutions of Meromorphic Univalent Functions with Positive Coefficients 43 - 45 B.A. Uralegaddi
AbstractKeywords: Meromorphic univalent functions, Starlike of order $alpha$, convolution.MSC: 30C45
 On General Solutions of Linear Differential Equations 47 - 56 Dušan D. Adamović and Jovan D. Kečkić
AbstractMSC: 34A30
 Köthe-Toeplitz Duals of some new Sequence Spaces and Their Matrix Maps 57 - 61 Z.U. Ahmad and Mursaleen
AbstractKeywords: Delta-bounded, Delta-convergent, and Delta-null sequences, Köthe-Toeplitz duals, matrix transformationsMSC: 40C05 40D05 40H05
 Une GÉnÉralisation Naturelle du Produit Scalaire Dans un Espace NormÉ et son Utilisation 63 - 70 Pavle M. Miličić
AbstractMSC: 46B20
 Fredholm Theory and Semilinear Equations without Resonance Involving Noncompact Perturbations, I. 71 - 82 P.S. Milojević
AbstractMSC: 47H15
Article page: 12>>

 Remote Address: 18.234.88.196 • Server: elib.mi.sanu.ac.rsHTTP User Agent: CCBot/2.0 (https://commoncrawl.org/faq/)


text:
## Help! COS Graph

The solid curve below shows the trigonometric function f(x). It has the form f(x) = cos(Bx). Also shown as dotted curve is the function cos(x). Determine the values of B and enter it below. Note the tick marks at x = Â±? and Â±2?.

Please I need help, can someone explain how they got the answer as well?

•  Cosine is periodic and repeats every 2π

 Repeats when Bx=2π

B=2π/x

B=2


text:
Diag Testa

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text:
### Theory:

Cytokinins were discovered by F. Skoog, C. Miller and co-workers during the $$1950s$$ as factors that promote cell division (cytokinesis) in plant cells. It was first isolated from Zea mays (known as Zeatin) and is abundantly found in the liquid endosperm of coconuts.
They are naturally occurring growth promoters that combine with auxin, promote cell division and differentiation in plants.
Cytokinin (Zeatin)
Physiological effects of cytokinin:
• These are the substances that induce cell division (cytokinesis) in the presence of auxins.
Cytokinesis
• These substances also cause the enlargement of cells.
• Together with auxins, cytokinins are involved in morphogenesis.
The formation of tissue culture from the callus in tissue culture is known as morphogenesis.
• They also promote the growth of lateral buds even in the presence of apical buds.
Apical bud and lateral bud
• Application of cytokinins promotes the Richmond Lang effect, i.e., it delays ageing in plants.
Reference:
https://commons.wikimedia.org/wiki/File:Zeatin.png
https://commons.wikimedia.org/wiki/File:Cytokinesis_illustration.svg


text:
SEARCH HOME
 Math Central Quandaries & Queries
 Question from Isaac: The average mass of a group of children is 50 kilograms. Todd, who has a mass of 62 kilograms, then joins the group. This raises the average mass of the group to 52 kilograms. How many children were in the original group

Hi Isaac,

If you had the data from the children then you would know the total mass and the number of children so you could calculate the average mass. Since you don't know the total mass or the number of students assign letters to these quantities.

Let the total mass of the students be $T$ kilograms and the number of students be $n.$ What would you do with $t$ ad $n$ to calculate the average mass? The result would be 50 kilograms so this gives you an equation in $T$ and $n.$

Now Todd arrives and his data is added to the data from the others. What is the total mass now? What is the number of students? What would you do to calculate the average mass now? The result this time is 52 kilograms. This gives you a second equation is $T$ and $n.$

Solve the two equations for $n.$

Now that you know $n$ it's an easy calculation to find $T.$ With $n$ and $T$ known you can verify that your answer is correct.

Write back if you need further assistance,

Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.


text:
# Can the Pythagorean Theorem be extended like this? [closed]

What is the significance, if any, of the fact that

$3^3$ + $4^3$ + $5^3$ = $6^3$

?

How curious this is. This would be like the Pythagorean Theorem exploding into a higher dimension, on steroids.

## closed as unclear what you're asking by Servaes, Saad, Claude Leibovici, JonMark Perry, Trần Thúc Minh TríMay 23 '18 at 11:33

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What geometry is this meant to reflect? – Lord Shark the Unknown May 23 '18 at 6:02
• Looks like the $\ell^3$ norm. – Sean Roberson May 23 '18 at 6:03
• No particular geometry. I was just struck when I came across this. – sumwunyumaynotno May 23 '18 at 16:23

This is, in fact, the analogous expression to the distance between two points $x$ and $y$ in the space $\Bbb{R}^4$. However, the "euclidean" (or "pythagorean") distance in $n$ dimensions is given by: $$\sqrt{\sum_{i=0}^{n} (x_i - y_i)^2}$$ There is a generalization, called Minkowski distance: $$\Big({\sum_{i=0}^{n} |x_i - y_i|^{1/p}}\Big)^p$$ And you recognise here your expression with $p=1/3$.
In conclusion, this is an integer expression for the distance between two points in $\Bbb{R}^4$, with the $1/3$-Minkowski metric.


text:
NEET Chemistry Chemical Kinetics Questions Solved

A catalyst lowers the activation energy of a reaction from . The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at $27°C$ s

(A) $-123°C$

(B) $327°C$

(C) 327°C

(D) $+23°C$

Difficulty Level:

• 40%
• 43%
• 8%
• 11%


text:
# A particle moves along a path defined by polar coordinates r = 4t^3 + 3t m and theta = 6t^2 rad w...

A particle moves along a path defined by polar coordinates r = 4t^3 + 3t m and theta = 6t^2 rad where t is in seconds. Determine the components of its velocity and acceleration at any time t. Show in Cartesian form.

## Plagiarism Checker

Submit your documents and get free Plagiarism report

Free Plagiarism Checker


text:
## Inference Networks

An inference network is a flexible construction for parameterizing approximating distributions during inference. They are used in Helmholtz machines (Dayan, Hinton, Neal, & Zemel, 1995), deep Boltzmann machines (Salakhutdinov & Larochelle, 2010), and variational auto-encoders (Kingma & Welling, 2014; Rezende, Mohamed, & Wierstra, 2014).

Recall that probabilistic models often have local latent variables, that is, latent variables associated with a data point; for example, the latent cluster assignment of a data point or the hidden representation of a text or image. Variational inference on models with local latent variables requires optimizing over each variational factor, $q(\mathbf{z}; \mathbf{\lambda}) = \prod_{n=1}^N q(z_n; \lambda_n),$ where $$z_n$$ are the latent variables associated to a data point $$x_n$$. The set of parameters $$\mathbf{\lambda}=\{\lambda_n\}$$ grows with the size of data. This makes computation difficult when the data does not fit in memory. Further, a new set of parameters $$\mathbf{\lambda}'=\{\lambda_n'\}$$ must be optimized at test time, where there may be a new set of data points $$\{x_n'\}$$.

An inference network replaces local variational parameters with global parameters. It is a neural network which takes $$x_n$$ as input and which outputs its local variational parameters $$\lambda_n$$, $q(\mathbf{z}\mid \mathbf{x}; \theta) = \prod_{n=1}^N q(z_n \mid x_n; \lambda_n) = \prod_{n=1}^N q(z_n; \lambda_n = \mathrm{NN}(x_n; \theta)).$ This amortizes inference by only defining a set of global parameters, namely, the parameters of the neural network. These parameters have a fixed size, making the memory complexity of inference a constant. Further, the same set of parameters can be used at test time: regardless of seeing new or old data points $$x_n'$$, we simply pass it through the neural network (a forward pass) to obtain its corresponding variational factor $$q(z_n'; \lambda_n' = \mathrm{NN}(x_n'; \theta))$$.

Note that the inference network is an approximation: it is a common misunderstanding that the inference network produces a more expressive variational model. The inference network restricts the size of parameters, meaning it can only do as well as an approximation as the original variational distribution without the inference network.

There are often good reasons beyond computational reasons for wanting to use an inference network, depending on the problem setting. Originating from Helmholtz machines, they are motivated by the hypothesis that “the human perceptual system is a statistical inference engine whose function is to infer the probable causes of sensory input.” The inference network is this function. In cognition, “humans and robots are in the setting of amortized inference: they have to solve many similar inference problems, and can thus offload part of the computational work to shared precomputation and adaptation over time” (Stuhlmüller, Taylor, & Goodman, 2013).

### Implementation

Inference networks are easy to build in Edward. In the example below, a data point x is a 28 by 28 pixel image (from MNIST).

We define a TensorFlow placeholder x_ph for the neural network’s input, which is a batch of data points.

x_ph = tf.placeholder(tf.float32, [M, 28 * 28])

We build a neural network using Keras. It takes a 28 by 28 pixel image and outputs a $$10$$-dimensional output, one for the mean (mu) and one for the standard deviation (sigma).

from edward.models import Normal
from keras.layers import Dense

hidden = Dense(256, activation='relu')(x_ph)
qz = Normal(loc=Dense(10)(hidden),
scale=Dense(10, activation='softplus')(hidden))

The variational model is parameterized by the neural network’s output.

During each step of inference, we pass in a batch of data points to feed the placeholder. Then we train the variational parameters (inference network’s parameters) according to gradients of the variational objective.

An example script using this variational model can be found at examples/vae.py in the Github repository. An example with a convolutional architecture can be found at examples/vae_convolutional_prettytensor.py in the Github repository.

### Footnotes

Inference networks are also known as recognition models, recognition networks, or inverse mappings.

Variational models parameterized by inference networks are also known as probabilistic encoders, in analogy from coding theory to probabilistic decoders. We recommend against this terminology, in favor of making explicit the separation of model and inference. That is, variational inference with inference networks is a general technique that can be applied to models beyond probabilistic decoders.

This tutorial is taken from text originating in Tran, Ranganath, & Blei (2016). We thank Kevin Murphy for motivating the tutorial as it is based on our discussions, and also related discussion with Jaan Altosaar.

### References

Dayan, P., Hinton, G. E., Neal, R. M., & Zemel, R. S. (1995). The Helmholtz machine. Neural Computation, 7(5), 889–904.

Kingma, D., & Welling, M. (2014). Auto-encoding variational Bayes. In International conference on learning representations.

Rezende, D. J., Mohamed, S., & Wierstra, D. (2014). Stochastic backpropagation and approximate inference in deep generative models. In ICML (pp. 1278–1286).

Salakhutdinov, R., & Larochelle, H. (2010). Efficient learning of deep boltzmann machines. In International conference on artificial intelligence and statistics.

Stuhlmüller, A., Taylor, J., & Goodman, N. (2013). Learning stochastic inverses. In Neural information processing systems.

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# American Institute of Mathematical Sciences

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June  2019, 24(6): 2719-2743. doi: 10.3934/dcdsb.2018272

## Long time behavior of fractional impulsive stochastic differential equations with infinite delay

 1 School of Mathematics and Statistics, Xi’an Jiaotong University Xi’an 710049, China 2 Departamento de Ecuaciones Diferenciales y Análisis Numérico, Universidad de Sevilla, c/ Tarfia s/n, 41012 Sevilla, Spain

Received  March 2018 Revised  May 2018 Published  October 2018

Fund Project: This work has been supported by grant MTM2015-63723-P (MINECO/FEDER, EU) and Consejería de Innovación, Ciencia y Empresa (Junta de Andalucía) under grant 2010/FQM314, and Proyecto de Excelencia P12-FQM-1492.

This paper is first devoted to the local and global existence of mild solutions for a class of fractional impulsive stochastic differential equations with infinite delay driven by both $\mathbb{K}$-valued Q-cylindrical Brownian motion and fractional Brownian motion with Hurst parameter $H∈(1/2,1)$. A general framework which provides an effective way to prove the continuous dependence of mild solutions on initial value is established under some appropriate assumptions. Furthermore, it is also proved the exponential decay to zero of solutions to fractional stochastic impulsive differential equations with infinite delay.

Citation: Jiaohui Xu, Tomás Caraballo. Long time behavior of fractional impulsive stochastic differential equations with infinite delay. Discrete & Continuous Dynamical Systems - B, 2019, 24 (6) : 2719-2743. doi: 10.3934/dcdsb.2018272
##### References:

show all references

##### References:
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# Homework Help: Calculate the work to pump water out of tank

1. Feb 17, 2010

### bfpri

An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed
to pump half the water out of the aquarium. (The density of water, p, is 1000 kg/m3,
and acceleration due to gravity, g, is 9.8 m/s2.)

So in order to truly understand how to do these kind of problems I decided to find different ways to solve this problem. Since you can integrate with respect to dx or dy, for dx i was wondering if you could integrate with respect to the 2m side like so

*edit..ugh its not letting me post the pictures so i'll just give the link...http://img19.imageshack.us/img19/2024/boxid.gif [Broken]

In order to get the answer i integrated F*D. F=p*g*volume=pg(1); D=1-x

integral((1-x)(1)pgdx) from x=0 to x=1. But when i do, i get the wrong answer, 4900 J. The answer should be 2450 J =/

Last edited by a moderator: May 4, 2017
2. Feb 18, 2010

bump.


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### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### Golden Thoughts

Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

# Medal Ceremony

##### Stage: 3 and 4 Short Challenge Level:

There are $6$ different students who could receive the first gold medal, then $5$ others for the second and $4$ remaining for the third. Therefore there are $6 \times 5 \times 4 = 120$ orders in which the medals can be presented. However, this counts each set of three people winning the medals in each of the six orders, so there are $120 \div 6 = 20$ sets of three people who could win gold.

For each of these, one of the remaining three people must win bronze and the others silver, so there are $20 \times 3 = 60$ ways in which the medals can be awarded.

This problem is taken from the UKMT Mathematical Challenges.
View the archive of all weekly problems grouped by curriculum topic

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# Convergence of weighted average with fixed variables

I am dealing with administrative data. Suppose I have a finite population of size $N$. The finite population estimating equation that defines the parameter $\theta_0$ can be written as

$$\frac{1}{N} \sum_{i=1}^{N}S_i(y_i, \theta) \,\,\,\ such \,\, that \,\,\,\,\, \frac{1}{N} \sum_{i=1}^{N}S_i(y_i, \theta_0)=0 \tag{1}$$

Suppose the score function $S_i(y_i, \theta) = y_i-\theta$ then the parameter of interest $\theta_0=\frac{1}{N} \sum_{i=1}^{N}y_i = \bar Y$. I want to estimate the parameter $\bar Y$ considering nonignorable non-response. Denote the unknown response probability by $\pi_i$ and suppose the empirically estimated non-response probability is $\hat \pi_i$. Then the estimating equation for respondent population is

$$\frac{1}{N} \sum_{i=1}^{N}\frac{\delta_i}{\hat \pi_i} S_i(y_i, \theta), \tag{2}$$ where $\delta$ is response indicator and is a random variable, but $y's$ are now fixed like design based approach in survey sampling.

Taking the expectation of (2) over $\delta$, we have $$E_\delta[\frac{1}{N} \sum_{i=1}^{N}\frac{\delta_i}{\hat \pi_i}S_i(y_i, \theta)|y]= \frac{1}{N} \sum_{i=1}^{N}S_i(y_i, \theta)E_\delta[\frac{\delta_i}{\hat \pi_i}] \tag{3}$$ Expanding $1/\hat \pi$ around $\pi_i$ using Taylor expansion and taking expectation over $\delta$, suppose we have $E_\delta[\delta_i/\hat \pi_i]=w_i$ then from (3), we can write

$$E_\delta[\frac{1}{N} \sum_{i=1}^{N}S_i(y_i, \theta)|y]=\frac{1}{N} \sum_{i=1}^{N}w_i S_i(y_i, \theta) = \frac{\sum_{i=1}^{N}w_i}{N} \frac{\sum_{i=1}^{N}w_i S_i(y_i, \theta)}{\sum_{i=1}^{N}w_i} \tag{4}$$ Now $w_i$ are positive values/weights and for $N \to \infty$, $\sum_{i=1}^{N}w_i \to \infty$ and further it can be assumed that for positive constant C, $|w_i|<C$. Then $\frac{\sum_{i=1}^{N}w_i}{N}=O(N)$.

Having $\sum_{i=1}^{N}S_i(y_i, \theta_0)=0$. From (4), first I need to show for $N \to \infty$, the weighted mean
$$\frac{\sum_{i=1}^{N}w_i S_i(y_i, \theta_0)}{\sum_{i=1}^{N}w_i} \to 0 \,\,\, a.s \,\,\, or \,\,\,\, p \,\,\, etc. \tag{5}$$ Then finally for $N \to \infty$, I can obtain $$E_\delta[\frac{1}{N} \sum_{i=1}^{N}\frac{\delta_i}{\hat \pi_i}S_i(y_i, \theta)|y] \to 0$$ that I actually I need.

What I have known is that, (5) can be proved for $S_i$ as random variable with $E[S_i(y_i,\theta)]=0$ from page 2 of http://www.ee.bgu.ac.il/~guycohen/clsurvey.pdf

But in my case $S_i$ is fixed variable like explanatory variables in regression. So I cannot take expectation over $S_i$ or $y_i$. I think I need some conditions on $w_i$....

Any suggestion would be highly appreciated.

• By "$X_i$ are not random variables" what do you mean exactly? – Math1000 Oct 21 '16 at 21:40
• Now I hope the question is clear. – ZAH Oct 23 '16 at 20:09


text: 0000026511 00000 n Read reviews from world’s largest community for readers. Each heart is worth one penalty point and the queen of spades is worth 13 penalty points. For the One Flip variation below, this strategy is fairly simple. A pair is two identical tiles. Standard rules need not be followed: the cards can be spread out anywhere, such as all around a room. The other cards have no value. The rules given here are for a standard deck of 52 cards, which are normally laid face down in four rows of 13 cards each. 36 Circle tiles in 4 sets of 9 tiles numbered 1- 9. 0000003692 00000 n The aim of the game is to clear the layout by pairing up all the tiles. In some computer versions, the cards may randomly move to increase the difficulty. The scheme is often used in quiz shows and can be employed as an educational game. 3. Traditionally, they are made from bone or ivory but modern sets are usually plastic. 0000025734 00000 n In this lecture, we develop an alternative explanation why equilibrium arises ... G is an identical interest game, i.e., all players have the same payoﬀ function. ∞ 0000032910 00000 n The score for each hand is assessed based on whether the required number of tricks were taken … The winner is the person with the most pairs. The rules in this game are slightly different. Some levels might be hard. Four players use a four-sided board; six players use a six-sided board; eight players use an eight-sided board - one side for each player, each ass… 0000035559 00000 n In this entertaining Mahjong game you help the lion to match identical tiles.. Mahjong Dimensions 3d. The Smash Hit Party Game. Enjoy a classic mahjong game with a Shanghai Dynasty theme Connect pairs and clear the board. Could go either way I guess. 0000004011 00000 n Designed for all ages, have fun with your families. Same Game Different Rules book. The players are divided into two teams - two against two, three against three or four against four. 1. 0000005323 00000 n Note that if two hands contain the same high pair, then the ranking of the next card in the hands determines which one wins. 0000033687 00000 n ;�0�vb-�ף�. Any deck of playing cards may be used, although there are also commercial sets of cards with images. 2536 0 obj <> endobj six of hearts and six of diamonds, queen of clubs and queen of spades, or both jokers, if used) then that player wins the pair and plays again. �$x8�X�h��R��h�cY�����_��ju���r��ڴ����x�K b��C�S\rS�[k6���>�l~�JR$Si�N���\$�Yp:�{}lv�H������H/q����N���;��,��b} 2536 46 In this case, her opponent must say which cards have been turned up, turn the cards back down if she doesn't get a pair and take the cards out for her if she does get a pair. The goal of the game is to get a mahjong, which consists of getting all 14 of your tiles into four sets and one pair. 0000035833 00000 n If the pair is a match, cards are removed, if not, they are flipped back. =���a�����c4�����IQ�_-) Although the rules and game play are the same the end goal is slightly different depending on if you’re playing a Texas Holdem cash game or a Texas Holdem tournament. Easy to play, tap & tap to match two tiles, make them crush. The following procedures will be used to break standings ties for postseason playoffs and to determine regular-season schedules. The tiles comprise: 1. 41.4 Simple rules and engaging game play. 0000034045 00000 n 0000035390 00000 n Get your messages instantly via push notifications, read and respond online & offline, and find any message quickly. 0000033868 00000 n There may be a tie for first place. ≈ Gmail is an easy to use email app that saves you time and keeps your messages safe. The is a bundle of 4 identical Cosmic Wimpout Games. The ladder consists of 3 steps, a top, middle and a … The "game" is another deck of cards which dictates play among players. 0000015379 00000 n 0000035468 00000 n It is described in an epigram of Byzantine Emperor Zeno (AD 476–491). There are two, three or four players, and the game is played clockwise. 0000014466 00000 n Pairs must be identical (same rank and same suit, so the 10 of clubs would have to match the other 10 of clubs). The tiles are shuffled face down on the table and each player takes 14 tiles, which are kept on a rack so that the other players cannot see them. {\displaystyle n\to \infty } n Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Concentration_(card_game)&oldid=992336397, Short description is different from Wikidata, Articles needing additional references from December 2020, All articles needing additional references, Creative Commons Attribution-ShareAlike License, This page was last edited on 4 December 2020, at 19:05. {\displaystyle \approx 0.8n} If there is no checker on the point indicated by the roll, the player must make a legal move using a checker on a higher-numbered point. 3. 36 Character tiles in 4 sets of 9 tiles numbered 1- 9. While combining them provides good narrative for in-game story telling, a bullet in the rain is (still) only a bullet, and also a rain - two threats - not one. Mahjong (English pronunciation: / m ɑː ˈ dʒ ɒ ŋ / mah-JONG, Mandarin Chinese: [mǎ.tɕjâŋ]) is a tile-based game that was developed during the Qing dynasty in China and has spread throughout the world since the early 20th century. When all tiles are matched, you can pass the current level! Our puzzle game includes large numbers of levels. Two-Club Tie for One Wild Card Spot: A cumulative score for each player is kept on paper. 3. In such a case, the tied players split the pot. {\displaystyle n} The object is to clear the tableau in the fewest turns, or to get the lowest possible score. 0000033512 00000 n It's a fantastic little racing game. 0000035511 00000 n Concentration can be played with any number of players or as a solitaire or patience game. [/i] Should This game isn’t that hard anyway, these kinds of things make it easier without adding much else. Less obviously, if the card does not match any known card, one of the n known cards should still be chosen to minimize the information provided to other players. 0000035651 00000 n Additional packs can be used for added interest. As previously explained, all the face cards and Aces will earn a player 10 points. The game ends when the last pair has been picked up. 0000015533 00000 n If there are no checkers on higher-numbered points, the player is permitted (and required) to remove a checker from the highest … 0.8 Exposed pairs of identical tiles (Flower and Seasons tiles in the same group being considered identical) are removed from the layout one at a time, gradually exposing the lower layers to play. A third tiebreak game will be played between the loser of the game to determine the Division Champion and the loser of the game between the two Clubs outside the division to determine the second Wild Card Club. If this card matches one of the known cards, the match is next chosen. A standard 52 card deck is used, with the cards in each suit ranking as usual from ace (high) down to two (low). The mathematics follow: If a remaining unknown card is chosen randomly, there is a 1/(t−1−n) chance of getting a match, but also a n/(t−1−n) chance of providing opponents with the information needed to make a match. startxref Details of this Canadian variation were sent to me by Henry Stevens. An ideal strategy can be developed if we assume that players have perfect memory. 0000035347 00000 n 0000001244 00000 n [1] The two jokers may be included for a total of six rows of nine cards each. Traditional multi-player card games for which rules are available from this site are listed below in alphabetical order, together with the number of players and the design and number of cards or tiles needed for each - see further explanation at the end of the page. Before any turn in the game, there are t cards still in play, and n cards still in play but of known value. 0 0000034678 00000 n The top tile of one of the remaining 7-tile stacks is turned face up and is the \"trump\". 0000003192 00000 n History. The pictures show the Chinese symbol for the number represented. n %%EOF Ladder Golf® is played with 2 or more players or teams. 0000035786 00000 n There is no trump suit. Identical Cards is a speedy observation game where players race to match the identical symbol between cards. A new free Mahjong board every day! Home field advantage in the game will be determined by the rules above for a two-team tiebreaker. 0000035269 00000 n n Standard decks of cards are used, with two jokers in each deck. 2. Adaptations can be easily made should your players/sets not exactly fit this guideline. They sit alternately - each player seated between two opponents. Many of these may be played in combination with one another: If one of the players has a particularly good memory, she could play blindfolded. 0000006004 00000 n Many versions of the game that are designed for children may have different themes. This game is easy to learn and has very basic rules. 0000013901 00000 n Thousands of levels with challenges. 0000013418 00000 n ≈ [/COLOR]A player cannot purchase and will never gain 2 identical Power tiles. A set can either be a "pung," which is three identical tiles, or a "chow," which is a run of three consecutive numbers in the same … A set of Mah Jong tiles consists of 144 tiles typically around 30 x 20 x 15mm. 0000003414 00000 n 0000026358 00000 n To extend the fun, play multiple games and keep a … 0000032128 00000 n Likewise, two hands that have identical pairs would be decided by the fifth card. Dealer - 9 tricks 1.2. Features Over 1500 boards! The object of the game is to wrap your bolas around the steps of the ladder. 0000035702 00000 n In turn, each player chooses two cards and turns them face up. you need to find identical tiles and remove them from the board. 0000035749 00000 n 0000035095 00000 n The board was the same, with 24 points, 12 on each side. Includes a bonus of 6 white standard size, six-sided, dice . 0000032438 00000 n %PDF-1.4 %���� Number of Players: 2 - 4 Type of Dominoes Used: Any Domino Type of Game: Blocking Game Number of players/domino set: 2 to 4 players using a double-9 set; 2 to 8 players using a double-12 set; and 9-12 players, or more, using a double-15 or 18 set. We've played this game three times (owned it for two months). The game is identical to Sergeant Major apart from the following: 1. The aim of the game is to guess your opponent's mystery character before they guess yours. {\displaystyle \approx 41.4} 0000000016 00000 n [2] For a standard deck of 52 cards, the expected value is Player one starts, flipping one card face-up, then player two selects one card from his/her own side. Two identical cards I suspect it is the first, but use of the word identical makes it unclear. To me prince and princess are different enough words, while prince and princes are the “same” word for this game and would cancel each other out. Try to connect the tiles and.. Mahjong Challenge. Each player has 3 golf ball bolas. 2581 0 obj<>stream 0000034288 00000 n A better strategy is to turn over a less certain card first, so that if wrong, one knows not to bother turning a more certain card over. 0000013076 00000 n the rules of the game, the rationality of the players, and the payoﬀ functions of players are all common knowledge. Tάβλι (tavli) meaning 'table' or 'board' in Byzantine Greek, is the oldest game with rules known to be nearly identical to backgammon. cards converges to Poker Rules Identical Hands Poker Rules Identical Hands 88probet is a leading Live Poker Rules Identical Hands Casino Online and betting agent in Southeast Asia, with a variety of slot games, trustworthy live casinos, and daily updated sportsbook in a smooth platform and fair play environment that maximize your efficiency and results. Mahjong is a free solitaire matching game that using mahjong tiles. xref It is a particularly good game for young children, though adults may find it challenging and stimulating as well. After trading cards and choosing trump, the dealer picks up the four kitty cards beforediscarding four. If they are of the same rank and color (e.g. The targets for the three players are: 1.1. Tile Master - Classic Triple Match & Puzzle Game is a challenging matching game. In the game, you need to blow your mind and match 3 numbers of blocks. moves.[3]. In this card game, each driver gets one deck (same cards as everyone else). The remaining tiles stay face-down on the table to form the pool. <<46130CEA58366748ADB7888CBE10F302>]>> , with With the Gmail app you get: • An organized inbox - Social and promotional messages are sorted into categories so you can read messages from friends and family first. A winning declaration or show garners zero points. 2. 2. Match 3 or more identical jewels in a single move to make a match and watch as it magically disappear. It is common for many players to think they know where pairs are and to turn over the one they are sure of first, then be stumped finding its mate. There are some exceptions to this rule that apply on the fringe cases, where n = 0 or 1 or towards the end of the game. Two stacks of seven tiles are dealt to each player, and the odd tile is given to the starting player, so that this player begins with 15 tiles. Three decks (162 cards including 6 jokers) may be enough for up to six players: eight players should use four decks (216 cards including 8 jokers). A Texas Hold’em tournament is the same as any other game of Hold’em with a few added rules and twists. The picture on each shows the appropriate number of circles. Play a lot more mahjong games only on y8.com With perfect memorization and using an optimal strategy, the expected number of moves needed for a game with 0000005960 00000 n Bloomberg delivers business and markets news, data, analysis, and video to the world, featuring stories from Businessweek and Bloomberg News on everything pertaining to technology 0000024648 00000 n Rules can be changed here too: it can be agreed before the game starts that matching pairs be any two cards of the same rank, a color-match being unnecessary, or that the match must be both rank and card suit. Each game only takes a few minutes to set up and about 5-10 minutes to play. 2. 0000004165 00000 n → 36 Bamboo tiles in 4 sets of 9 tiles numbered 1- 9. Over the course of the game, it becomes known where certain cards are located, and so upon turning up one card, players with good memory will be able to remember where they have already seen its pair. A bola is 2 golf balls attached by a nylon rope. trailer Concentration, also known as Match Match, Match Up, Memory, Pelmanism, Shinkei-suijaku, Pexeso or simply Pairs, is a card game in which all of the cards are laid face down on a surface and two cards are flipped face up over each turn. The current player should flip over an unknown card. The object of the game is to turn over pairs of matching cards. The rules of Brag first appear in 1721 in The Compleat Gamester where it is referred to as "The Ingenious and Pleasant Game of Bragg", but in fact, it originates in an almost identical game called Post and Pair which is recorded as far back as 1528 (as Post) and which, in turn, was descended from Primero. xڤV}PSW�/�ICH ,!D��\�!�d��F�� b �认F1UЈ�v�1(YD)���Q�V\mש|u�7Q���S���a�u:vg��G��WO>^޽������9w � �� t hGA ��@��/y��|C�'- ��UF�|[qJd�9���OFĦ�-���ޚ{4��唛�s��ԍ?��;����(�#G��>8�U#O"�_�i4��;�N���_K7�:���Im�d���,�;�'l�|�6���NDLF>�jf�\����HTF1RĪݘ+O^u��Ht}�;���Zݧ{�W�O�xr����c*��� For example: 9, 9, 7, 4, 2 beats 9, 9, 5, 3, 2. Concentration may be played solo either as a leisurely exercise, or with the following scoring method: play as normal, but keep track of the number of non-matching pairs turned over (this may be done using poker chips, pennies or by making marks on a sheet of paper). Jokers of both kinds (Wild and Printed) are worth zero points, while all the other cards will earn a player the same number of points as the number printed on the card. Concentration is not limited to playing cards. . If you are searching for games for a particular number of players you may prefer to use the Index of Games by Number of Players. Player to dealer's right - 2 tricks. The game has a smaller sibling, the mahjong shanghai dynasty, This fun game has the same rules to follow, match, and clear the board and complete all levels. 0000003611 00000 n 0000035168 00000 n 0000025093 00000 n 4. 0000006090 00000 n Substantially Identical Security: A security that is so similar to another that the Internal Revenue Service (IRS) does not recognize a difference between the two. Free hints help you to make game easier. The rule book states: [i][b][COLOR=#FF0000]Important! The mechanism … 0000003656 00000 n Challenge your mind and solve the puzzles, and then you will find them easy and exciting! Player to dealer's left - 5 tricks 1.3. If they are not of the same rank and color, they are turned face down again and play passes to the player on the left. [/b]This includes all Powers with the same effects and illustrations, even if they are of different colors like [u]Divine Action[/u] and Victory Point. Great as gifts, prizes, give-aways or just to have a game close at hand ( home, auto and work). Reliant on a sharp eye and quick reflexes, Identical Cards creates excitement for children and adults alike while keeping every player involved in the action. Randomly move to increase the difficulty in the game is to wrap your bolas around the of! Out anywhere, such as all around a room or just to have a close... All the face cards and turns them face up players race to identical... Ages, have fun with your families the object is to guess your opponent 's mystery character before guess... To play, tap & tap to match two tiles, make them crush above for a two-team.. Two identical cards I suspect it is a challenging matching game that using mahjong tiles game close at (... Are usually plastic the three players are: 1.1 are: 1.1 of things make it easier without adding else... The aim of the players, and then you will find them easy and exciting in 4 sets of tiles! Everyone else ) dictates play among players t that hard anyway, these kinds of things make it easier adding! For the three players are all common knowledge 've played this game three times ( owned it for two )! Them crush, with two jokers in each deck mahjong is a speedy observation game where players race match! Of the game is easy to learn and has very basic rules all ages have... Employed as an educational game standard rules need not be identical game rules: the cards can be easily made your! 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# Vertex-antimagic total labelings of graphs

MacDougall, James A. and Bača, Martin and Miller, Mirka and Bertault, Francois and Simanjuntak, Rinovia (2003) Vertex-antimagic total labelings of graphs. Discussiones Mathematicae Graph Theory, 23 . pp. 67-83.

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In this paper we introduce a new type of graph labeling, the $(a, d)$- vertex-antimagic total labeling, which is a generalization of several other types of labelings. A connected graph $G(V,E)$ is said to be $(a, d)$-vertex-antimagic total if there exist positive integers $a$, $d$ and a bijection $\lambda : V \cup E \rightarrow \{1, 2, . . . , \mid V \mid + \mid E\mid\} such that the induced mapping$g_\lambda : V \rightarrow W$is also a bijection, where$W = \{w_\lambda (x)\mid x \in V \} = \}a, a + d, . . . , a + (\mid V \mid - 1)d\mid$is the set of weights of vertices in$G\$. Properties of these graphs are studied. How to construct labelings for certain families of graphs are shown. Several conjectures and open problems are proposed.