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# Quantifier elimination for theory of equivalence relations

Let $\mathcal{L}=\{\sim\}$ and $\Sigma_\infty$ be the set of axioms stating that:

(i) $\sim$ is an equivalence relation
(ii) Every equivalence class is infinite
(iii) there are infinitely many equivalence classes

It suffices to show that every $\mathcal{L}$ formula $\exists y\varphi(x,y)$ is $\Sigma_\infty$-equivalent to a quantifier free $\psi(x)$, where $x=(x_1,\ldots,x_n)$ are distinct variables, and $\varphi(x,y)$ is a conjunction of literals. Since it's fairly trivial when $\varphi(x,y)$ has a conjunct of the form $x_i=y$, we can assume we need to reduce $$\exists y (\bigwedge_{i\in I} x_i\not=y\wedge\bigwedge_{j\in J}x_j\sim y\wedge\bigwedge_{h\in H}\neg (x_j\sim y))$$ My question is whether or not the following condition holds:

$$\Sigma_\infty\vdash\exists y (\bigwedge_{i\in I} x_i\not=y\wedge\bigwedge_{j\in J}x_j\sim y\wedge\bigwedge_{h\in H}\neg (x_j\sim y))\leftrightarrow(\bigwedge_{j\in J,h\in H}\neg(x_i\sim x_j)\wedge\bigwedge_{j,j'\in J}x_j\sim x_{j'})$$ I can't think of a reason why it would not, but I'm a bit worried about ignoring the information about non-identical elements in the unreduced formula.

• I think that if you have infinitely many different equivalence classes, you have also infinitely many different elements, since the same element cannot belong to two different classes. – Taroccoesbrocco Nov 26 '15 at 5:44

The equivalence holds because the equivalence classes are infinite. If you have $(x_1,…,x_n)$ such that the RHS holds, the set $\{x_j \mid j\in J\}$ is finite. If it is nonempty, there is an element $y$ such that $y\sim x_j$ for $j\in J$ and $y\neq x_j$. If the set is empty, the existence of $y$ is implied by the fact that there are infinitely many equivalence classes, so you can pick a $y$ that is not $\sim$-equivalent to any $x_h,h\in H$.


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+0

# Why does it say invalid argument? sqrt(11, 0)

0
95
2

Why does it say invalid argument? sqrt(11, 0)

Feb 22, 2022

#1
0

I don't get what you want your argument to be.

If you are taking the square root of a coordinate point, you have do individually root it for the x argument and the y argument.

If you mean 11.0, just type 11.

In general, your system is interpreting your answer as the sqaure root of a coordinate point, which is obviously invalid.

Feb 22, 2022
#2
0

What you have entered    sqrt ( 11,0)      is    this :   $$\sqrt[0]{11}$$

which would be     111/0       and we all know you cannot divide by 0

Feb 22, 2022

format: plain


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Browse Questions

# The activation energy for a certain reaction is 76.7 kJ mol$^{-1}$. How many times faster will the reaction occur at 50$^\circ$C compared to 0$^\circ$C?

$\begin{array}{1 1} 186 \\ 93 \\ 47.5 \\ 33 \end{array}$

From Arrhenius equation, $k=Ae^{\large\frac{-E_a}{RT}}$, we get:
$\Rightarrow \large\frac{k_2}{k_1} $$= \large\frac{Ae^{\large\frac{-E_a}{RT_1}}}{Ae^{\large\frac{-E_a}{RT_2}}} Using the fact that \large\frac{e^x}{e^y}$$ = e^{x-y}$ and taking the natural log on both sides and simplifying, we arrive at:
$ln\large\frac{k_2}{k_1}$$= \large\frac{E_a}{R}$$ (\large\frac{1}{T_1}$$- \large\frac{1}{T_2}$$)$
Substituting the values (note: convert $^\circ$C to $^\circ$K) , we get:
$\Rightarrow ln\large\frac{k_2}{k_1}$$= \large\frac{76700\; Jmol^{-1}}{8.314 JK.mol^{-1}}$$ (\large\frac{1}{273 ^\circ K}$$- \large\frac{1}{323 ^\circ K}$$)$
$\Rightarrow ln\large\frac{k_2}{k_1}$$= \large\frac{76700 \times 0.0005670}{8.314}$$=5.231$


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# Damped Oscillations: Incoherence between a general solution and a specific one

In my 'Classical Dynamics of Particles and Systems, THORNTON/MARION, 5th Edition' book of classical mechanics it is given the following general solution for a damped oscillation solving $\ddot{x}+2\beta\dot{x}+w_o^2x=0$:

$$x(t)=e^{-\beta t}[A_{1}\exp(\sqrt{\beta^2-w_o^2}t)+A_{2}\exp(-\sqrt{\beta^2-w_o^2}t)]$$

as $A_1$ and $A_2$ some arbitrary constants, $\beta$ the damping parameter and $w_0$ the the characteristic angular frequency.

So, if I want to find the $x(t)$ expression for the critical damped case, I have to consider that $$w_0^2=\beta^2$$

so my critical damped motion is described by:

$$x(t)=(A_1+A_2)(e^{-\beta t})$$

But, it we solve the specific differential equation for the critical damped case, the result is (as equal roots exists):

$$x(t)=(A+B t)e^{-\beta t}$$

So, my question is: What am I missing?

The first solution is not correct since it implies a strange connection $\dot x (0)=-\beta x(0)$. Check the general solution for its region of validity.


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# Additive inverse of a number explained

Learn and know what is additive inverse in mathematics and know how to find additive inverse of a any number.

In the lower classes of mathematics, additive identity, multiplicative identity, multiplicative inverse (reciprocal), additive inverse are most important concepts. All these four are very simple concepts which we can understand easily. Now we will learn about what is additive inverse? How to find additive inverse for any number, fractions and for a rational number.

## What is meant by the additive inverse in math?

First we will know the meaning of additive inverse. We have a very simple meaning i.e. for any number if we add another number such that the result should become zero. Each of these numbers will be additive inverse of other number.

For example, consider number 8. So now we need to add another number such that result should become zero. Can you say what is that number we need to add? We need to add -8 because 8 + (-8) =0. So now we say that the additive inverse of 8 is -8 and also the additive inverse of -8 is 8.

## How do we write the additive inverse for numbers?

To find additive inverse of a number there exists 2 methods. In first method, just writing the opposite sign to the given number.

For example, the additive inverse of 9 will be -9.

In the second method, for the given number add a any variable and make it equal to zero (0). Then find variable value. Here variable value will be additive inverse of the given number.

For example, let us take a number. Am considering -2. So add a variable to -2 and make it equal to zero and simplify for variable.

-2 + y = 0

Y = 2

Now we say that additive inverse of -2 is 2 or additive inverse of 2 is -2.

## How do we find Additive inverse of any given fraction?

The method for finding additive inverse of a fraction is same as the methods which we applied in the above. Just change the sign of a fraction or add a variable to the given fraction and make it equal to zero then find variable value.

For example, if $\frac { 2 }{ 3 }$ is a fraction then the $\frac { -2 }{ 3 }$ will be the additive inverse of that fraction.

## How do we find additive inverse of any given rational numbers?

Additive inverse of rational numbers also follows the above given methods only. Just changing the sign will give you additive inverse of rational numbers.

For example, $\frac { -5 }{ 7 }$ is a rational number then $\frac { 5 }{ 7 }$ will be additive inverse of that rational number.


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Volume 421 - Neutrino Oscillation Workshop (NOW2022) - Session III. Multimessenger astrophysics
Analytical treatment of neutrino oscillation and decay in matter
S. Goswami*, D.S. Chattopadhyay, K. Chakraborty and A. Dighe
Full text: pdf
Pre-published on: January 25, 2023
Published on:
Abstract
We discuss the propagation of decaying neutrinos in matter and outline a procedure to obtain the analytic
oscillation probabilities in two and three generation framework. For unstable neutrinos,
the Hermitian and anti-Hermitian components of the effective Hamiltonian
do not commute and cannot be simultaneously diagonalised by unitary transformations for all
matter densities. We present the compact analytic expressions for neutrino probabilities in matter for the two flavour case,
obtained using a re-summation of the inverse Baker-Campbell-Hausdorff (BCH) or Zassenhaus
expansion,
We also outline briefly how the approximate probabilities for the three flavour case can be obtained assuming
only the vacuum mass eigenstate $\nu_3$ decays. We show the baselines and energies where
the different approximations give good matching with the numerical probabilities
DOI: https://doi.org/10.22323/1.421.0059
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
Fehler gefunden?
Erweiterte Suche

# Die Anschauung des Werdens

## Zu Husserls Theorie des Zeitbewusstseins

Zurück zum Heft: Phänomenologische Forschungen 2006
DOI: https://doi.org/10.28937/1000107930
EUR 14,90

Husserl discovered that the phenomenon of time cannot be analyzed in terms of objective temporality. The intuitively given succession is not one of worldly objective time, but rather of the time of consciousness itself. Here it seems that the successive phases can exist together, such as they are given together in intuition. This poses the problem of how the time of consciousness can appear in its own course, that is, how to understand the relation between the time that appears and the process of its appearing. Can time perception be understood as a temporal self-perception of consciousness?


text:
• # question_answer The kinetic energy of one gramme molecule of a gas, at normal temperature and pressure will be $(R=8.31\,J\,mole\,K)$: A)  $3.4\times {{10}^{3}}J$         B)  $2.7\times {{10}^{2}}J$C)  $1.2\times {{10}^{2}}J$          D)  none of these

Solution :

We know that the kinetic energy of $1g$ molecule of an ideal gas is $E=\frac{3}{2}\,RT$ Here     $R=8.31\,J/mol\,K$   and        $~T={{273}^{o}}K$ $\therefore$  $E=\frac{3}{2}\times 8.31\times 273=3.4\times {{10}^{3}}J$

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# The reduced row echelon form of the augmented matrix of a system of three linear equations in 3 variables must be of the form where a,b,c are real nmb

The reduced row echelon form of the augmented matrix of a system of three linear equations in three variables must be of the form $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ a& b& c\end{array}\right]$ where a, b, c, are real numbers. Justify your answer.

You can still ask an expert for help

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Reduced row echelon form consists of the first row starting off with a 1 followed by 0s


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# How do you implicitly differentiate -y=xy-xe^y ?

Jul 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - {e}^{y}}{x {e}^{y} - x - 1}$

#### Explanation:

I'm finding $\frac{\mathrm{dy}}{\mathrm{dx}}$. Differentiate term by term, the derivative of y wrt x is $\frac{\mathrm{dy}}{\mathrm{dx}}$ while remembering to be careful with our product rule for both terms on the right hand side.

$- \frac{\mathrm{dy}}{\mathrm{dx}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} - x \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y}$

Collect like terms

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[x {e}^{y} - x - 1\right] = y - {e}^{y}$

and solve:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - {e}^{y}}{x {e}^{y} - x - 1}$


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# Hiding trivial conditions when using Reduce

I use Reduce to find conditions on some parameters such that certain inequalities are fulfilled. An additional condition is that all parameters are positive. The first problem is that I have to mention all those conditions when calling Reduce, and the second is that they appear in the result and make it less readable.

I tried to use Assumptions, but it doesn't change the result.

I wonder if it is possible to hide these trivial conditions or at least suppress them in the output?

$Assumptions = Flatten[{Thread[{r, p, h, K} > 0], Element[{r, p, h, K}, Reals]}] Reduce[K*(r - h)/r > 0 && (p*h)/(r - h) < p && K >0 && p>0 && h>0 && r>0,h]  • MemberQ[Keys[Options[Reduce]], Assumptions] returns False, so it is inappropriate to use $Assumptions with Reduce[]. – J. M.'s technical difficulties Jan 22 at 9:23
• Dmitry, apply Simplify to the result of Reduce. – Henrik Schumacher Jan 22 at 9:24
• @HenrikSchumacher, this seems to be the most simple and efficient solution so far. Thank you! – Dmitry Jan 22 at 9:38
• You're welcome! – Henrik Schumacher Jan 22 at 9:45

conds = K > 0 && p > 0 && h > 0 && r > 0;

• Then you can use Simplify[], which does take assumptions: With[{conds = K > 0 && p > 0 && h > 0 && r > 0 && ρ > 0}, Assuming[conds, Simplify[Reduce[K*(r - h)/r > 0 && (p*h)/(r - ρ - h) < p && conds, h, Reals]]]]. – J. M.'s technical difficulties Jan 22 at 9:42


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Contents:

### 3.1.1 The review of Axiom 1 ( measurement: $\S$2.7)

In the previous chapter, we introduced Axiom 1 (measurement ) , which is mainly due to M. Born.
3.1.1 The review of Axiom 1 ( measurement: $\S$2.7)
With any system $S$, a basic structure $[ {\mathcal A} \subseteq \overline{\mathcal A}]_{B(H)}$ can be associated in which measurement theory of that system can be formulated. In $[ {\mathcal A} \subseteq \overline{\mathcal A}]_{B(H)}$, consider a $W^*$-measurement ${\mathsf M}_{\overline{\mathcal A}} \big({\mathsf O}{{=}} (X, {\cal F} , F{}), S_{[{}\rho] } \big)$ $\Big($ or, $C^*$-measurement ${\mathsf M}_{{\mathcal A}} \big({\mathsf O}{{=}} (X, {\cal F} , F{}), S_{[{}\rho] } \big)$ $\Big)$. That is, consider
 $\bullet$ a $W^*$-measurement ${\mathsf M}_{\overline{\mathcal A}} \bigl({\mathsf O} , S_{[{}\rho{}] } \bigl)$ $\Big($ or, $C^*$-measurement ${\mathsf M}_{{\mathcal A}} \big({\mathsf O}{{=}} (X, {\cal F} , F{}), S_{[{}\rho] } \big)$ $\Big)$ of an observable ${\mathsf O}{{=}} (X, {\cal F} , F{})$ for a state $\rho(\in {\frak S}^p({\cal A}^*):\mbox{state space})$
Then, the probability that a measured value $x$ $( \in X )$ obtained by the $W^*$-measurement ${\mathsf M}_{\overline{\mathcal A}} \bigl({\mathsf O} , S_{[{}\rho{}] } \bigl)$ $\Big($ or, {$C^*$-measurement} ${\mathsf M}_{{\mathcal A}} \big({\mathsf O}{{=}} (X, {\cal F} , F{}), S_{[{}\rho] } \big)$ $\Big)$ belongs to $\Xi$ $(\in {\cal F}{})$ is given by \begin{align*} \rho( F(\Xi)) (\equiv {}_{{{\mathcal A}^*}}(\rho, F(\Xi) )_{\overline{\mathcal A}} ) \end{align*} (if $F(\Xi)$ is essentially continuous at $\rho$, or see (2.56) in Definition 2.19 ).
Here, note that
 $(A):$ the above axiom is a kind of spell (i.e., incantation, magic words, metaphysical statement), and thus, it is impossible to verify them experimentally.
In this sense, the above axiom corresponds to "a priori synthetic judgment" in Kant's philosophy. And thus, we say:
 $(B):$ After we learn the spell (= Axiom 1) by rote, we have to exercise and lesson the spell (= Axiom 1). Since quantum language is a language, it may be unable to use well at first. It will make progress gradually, while applying a trial-and-error method.
However,
 $(C_1):$ if we would like to make speed of acquisition of a quantum language as quick as possible, we may want the good manual on how to use the axioms.
Here, we think that
 $(C_2):$ the linguistic interpretation= the manual on how to use the spells (Axiom 1 and 2)
After all, I want to assert that \begin{align} \underset{\mbox{[dualistic idealism]}}{\mbox{Descartes philosophy}}\longrightarrow \left\{\begin{array}{ll} \color{blue}{\underset{\mbox{[Axioms]}}{\mbox{Continental Rationalism}}} \\ \\ \color{red}{\underset{\mbox{[Linguistic interpretation]}}{\mbox{British empiricism}}} \end{array}\right\} \longrightarrow \underset{\mbox{[quantum language]}}{\mbox{Kant philosophy}} \end{align}

### 3.1.2: Descartes figure (in the linguistic interpretation)

In what follows, let us explain the linguistic interpretation. The concept of "measurement" can be, for the first time, un erstood@in dualism.@Let us explain it.@The image@of@"measurement"@is as shown in@Figure 3.1.
In the above,
 $(D_1):$ measurement = "① interfere" + "② perceive the reaction"
That is, "measurement" is characterized as the interaction between "observer" and "measuring object". However, it can not be represented by a kinetic equation. (If the kinetic equation exists, we do not need dualism but monism.) Thus,
 $(D_2):$ In measurement theory, we do not use the term "interaction" but "measurement".

Therefore, in order to avoid confusion, it might better to omit the interaction "① and ①" in Figure 3.1. After all, we think that:
 $(D_3):$ It is clear that there is no measured value without observer (i.e., brain). Thus, we consider that measurement theory is composed of three key-words (cf. Note 3.1 later): \begin{align} \underset{\mbox{ (observer,brain, mind)}}{\fbox{measured value}}, \quad \underset{\mbox{ (thermometer, eye, ear, body, polar star)}}{\fbox{observable (= measuring instrument )}}, \quad \underset{\mbox{ (matter)}}{\fbox{state}}, \end{align} and thus, it might be called "trialism" (and not "dualism"). But, according to the custom, it is called "dualism" in this note.

### 3.1.3: The linguistic interpretation $(E_1)-(E_7)$

The linguistic interpretation is "the manual on how to sue Axiom1 and 2". Thus, there are various explanations for the linguistic interpretations. However, it is usual to consider that the linguistic interpretation is characterized as the following (E). And the most important is
Only one measurement is permitted
(E):The linguistic interpretation (=quantum language interpretation)
With Descartes figure Figure 3.1 (and (E$_1$)-(E$_7$)) in mind, describe every phenomenon in terms of Axioms 1 and 2. Here,
 $(E_1):$ Consider the dualism composed of "observer" and "system( =measuring object)". And therefore, "observer" and "system" must be absolutely separated. If it says for a metaphor, we say "Audience should not be up to the stage". $(E_2)$: Of course, "matter(=measuring object)" has the space-time. On the other hand, the observer does not have the space-time. Thus, the question: "When and where is a measured value obtained?" is out of measurement theory, Thus, there is no tense in measurement theory. This implies that there is no tense in science. $(E_3)$: In measurement theory, "interaction" must not be emphasized. $(E_4)$: Only one measurement is permitted. Thus, the state after measurement (or, the influence of measurement) is meaningless. $(E_5)$: There is no probability without@measurement. $(E_6)$: State never moves,
and so on. Also, since our assertion is
quantum language is the final goal of@dualistic idealism (="Descartes=Kant philosophy")
we have to assert that
$(E_7)$: Many of maxims of the philosophers@(particularly, the dualistic idealism ) can be regarded as a part of the linguistic interpretation.
Some may think that@the (E$_7$) is unbelievable.@However, note that
 $(F_1):$ @the purpose of Descartes=Kant philosophy and@that of quantum language are the same,@that is,@to establish @the world view $(F_2)$: @the methodology of Descartes=Kant philosophy and@that of quantum language are the same,@that is,@the non-realistic dualism.
Therefore, it is natual to consider
@maxims of philosophers@$\approx$@the linguistic interpretation
Recall the following figure ( in Chapter 1 ):
Fig.1.1: the location of "quantum language" in the world-views
In the above, we regard
as a genealogy of the dualistic idealism.@Talking cynically,@we say that
 $\bullet$ @Philosophers continued investigating@"linguistic interpretation"@(="how to use Axioms 1 and 2")@without@Axioms 1 and 2.

For example, "Only one measurement is permitted" and "State never moves" may be related to Parmenides' words;

$\left\{\begin{array}{ll} \mbox{ There is no "plurality", but only "one". } \\ \\ \mbox{ And therefore, there is no movement.} \end{array}\right.$

Thus, we want to assert that@Parmenides (born around BC. 515) is the oldest discoverer of the linguistic interpretation. Also, we propose the following table:

Although I am not familiar with ontology, I want to consider "key-words exist in each world-view" or "Key-word = existence ".
 $\fbox{Note 3.1}$ In the above table, Newtonian mechanics may be the most understandable.@Also, we want to identify "Plato idea" with "absolute standard".@If the identification is permitted, we can understand that Table 3.1 implies Whitehead' word: the development of Western philosophy is a series of footnotes to Plato since each world-view including "observable" is regarded as dualistic idealism. And, we want to understand that@Newton is similar to Aristotle,@since their assertions belong to the realistic world view@(cf.@Figure 1.1).@Also,@recall the formula (3.1),@that is,@"observable"="measuring instrument"="body".@Thus,@as the examples of "observable",@we think: $$\mbox{ eyes, ears, glasses, telescope, compass, etc. }$$ If "compass" is accepted,@"the polar star" should be also accepted@as the example of the observable.@In the same sense,@"the jet stream to an airplane" is a kind of observable.@It is clear that there is no measured value without@brain ( $\approx$ mind, brain ). Thus, we consider the correspondence@:"measured value $\longleftrightarrow$ I". @Also, if it is certain that Descartes is the first discoverer of "I",@I have to retract our understanding of@Scholasticism in Table 3.1.@Although I have no confidence about Scholasticism,@the discover of three words@("post rem", "ante rem", "in re")@should be remarkable.


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# Isometry on metric

1. Dec 1, 2007

### jamiemmt

How do I prove every isometry from R->R is of the form f(x)=a+-x , regardless of the metric?

I know it has to do with considering d_1(0,x_1)=d_2(0, f(x_1)), but beyond that I am lost.

2. Dec 2, 2007

### HallsofIvy

Staff Emeritus
Start by writing out the definition of "isometry"!


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Cryptology ePrint Archive: Report 2021/1128

Continuously Non-Malleable Secret Sharing: Joint Tampering, Plain Model and Capacity

Gianluca Brian and Antonio Faonio and Daniele Venturi

Abstract: We study non-malleable secret sharing against joint leakage and joint tampering attacks. Our main result is the first threshold secret sharing scheme in the plain model achieving resilience to noisy-leakage and continuous tampering. The above holds under (necessary) minimal computational assumptions (i.e., the existence of one-to-one one-way functions), and in a model where the adversary commits to a fixed partition of all the shares into non-overlapping subsets of at most $t-1$ shares (where $t$ is the reconstruction threshold), and subsequently jointly leaks from and tampers with the shares within each partition.

We also study the capacity (i.e., the maximum achievable asymptotic information rate) of continuously non-malleable secret sharing against joint continuous tampering attacks. In particular, we prove that whenever the attacker can tamper jointly with $k > t/2$ shares, the capacity is at most $t - k$. The rate of our construction matches this upper bound.

An important corollary of our results is the first non-malleable secret sharing scheme against independent tampering attacks breaking the rate-one barrier (under the same computational assumptions as above).

Category / Keywords: foundations / secret sharing, non-malleability, leakage resilience.

Original Publication (with minor differences): IACR-TCC-2021


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## DMOPC '18 Contest 5 P2 - A Photography Problem

View as PDF

Points: 7
Time limit: 1.0s
Memory limit: 128M

Author:
Problem type

When taking digital photos, it is important to get the proper exposure (brightness). A photo can be represented as a grid of by pixels, and the pixel in row and column has a brightness , which can be any real number from to inclusive. If you average the brightnesses of all the pixels in a typical image, the result is . Images whose average is lower than that are underexposed, and those whose average is higher are overexposed.

Most digital cameras have an exposure correction feature. By choosing a correction constant and multiplying all the pixel brightnesses in an image by , a darker or brighter image can be obtained. When applying a correction constant, if any pixel brightnesses become greater than 1, those values are "clipped" and reduced to 1.

Given an image, first determine if it is underexposed, overexposed, or correctly exposed. If the image is not correctly exposed, please find the that would correct its average brightness to . You may consider the image correctly exposed if the required is within of .

#### Constraints

It is guaranteed that no underexposed or overexposed image has a required within of .

#### Input Specification

The first line will contain two space-separated integers, and .
The next lines will each contain space-separated real numbers, the pixel brightnesses.

#### Output Specification

On one line, output the string underexposed, overexposed, or correctly exposed. If the image is not correctly exposed, output an additional line containing one positive real number, the required . Your answer will be considered correct if it has an absolute error of no more than .

#### Sample Input 1

2 3
0.36 0.304 0.12
0.408 0.312 0.96

#### Sample Output 1

underexposed
1.25

#### Sample Input 2

2 3
0.48 0.58 0.56
0.38 0.40 0.48

#### Sample Output 2

correctly exposed


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Identifiability

In statistics, identifiability is a property which a model must satisfy in order for precise inference to be possible. A model is identifiable if it is theoretically possible to learn the true values of this model's underlying parameters after obtaining an infinite number of observations from it. Mathematically, this is equivalent to saying that different values of the parameters must generate different probability distributions of the observable variables. Usually the model is identifiable only under certain technical restrictions, in which case the set of these requirements is called the identification conditions.

For a less technical treatment, see Parameter identification problem.

A model that fails to be identifiable is said to be non-identifiable or unidentifiable: two or more parametrizations are observationally equivalent. In some cases, even though a model is non-identifiable, it is still possible to learn the true values of a certain subset of the model parameters. In this case we say that the model is partially identifiable. In other cases it may be possible to learn the location of the true parameter up to a certain finite region of the parameter space, in which case the model is set identifiable.

Aside from strictly theoretical exploration of the model properties, identifiability can be referred to in a wider scope when a model is tested with experimental data sets, using identifiability analysis.[1]

Definition

Let ${\displaystyle {\mathcal {P}}=\{P_{\theta }:\theta \in \Theta \}}$ be a statistical model where the parameter space ${\displaystyle \Theta }$ is either finite- or infinite-dimensional. We say that ${\displaystyle {\mathcal {P}}}$ is identifiable if the mapping ${\displaystyle \theta \mapsto P_{\theta }}$ is one-to-one:[2]

${\displaystyle P_{\theta _{1}}=P_{\theta _{2}}\quad \Rightarrow \quad \theta _{1}=\theta _{2}\quad \ {\text{for all }}\theta _{1},\theta _{2}\in \Theta .}$

This definition means that distinct values of θ should correspond to distinct probability distributions: if θ1θ2, then also Pθ1Pθ2.[3] If the distributions are defined in terms of the probability density functions (pdfs), then two pdfs should be considered distinct only if they differ on a set of non-zero measure (for example two functions ƒ1(x) = 10  x < 1 and ƒ2(x) = 10  x  1 differ only at a single point x = 1 — a set of measure zero — and thus cannot be considered as distinct pdfs).

Identifiability of the model in the sense of invertibility of the map ${\displaystyle \theta \mapsto P_{\theta }}$ is equivalent to being able to learn the model's true parameter if the model can be observed indefinitely long. Indeed, if {Xt}  S is the sequence of observations from the model, then by the strong law of large numbers,

${\displaystyle {\frac {1}{T}}\sum _{t=1}^{T}\mathbf {1} _{\{X_{t}\in A\}}\ {\xrightarrow {\text{a.s.}}}\ \Pr[X_{t}\in A],}$

for every measurable set A  S (here 1{...} is the indicator function). Thus, with an infinite number of observations we will be able to find the true probability distribution P0 in the model, and since the identifiability condition above requires that the map ${\displaystyle \theta \mapsto P_{\theta }}$ be invertible, we will also be able to find the true value of the parameter which generated given distribution P0.

Examples

Example 1

Let ${\displaystyle {\mathcal {P}}}$ be the normal location-scale family:

${\displaystyle {\mathcal {P}}={\Big \{}\ f_{\theta }(x)={\tfrac {1}{{\sqrt {2\pi }}\sigma }}e^{-{\frac {1}{2\sigma ^{2}}}(x-\mu )^{2}}\ {\Big |}\ \theta =(\mu ,\sigma ):\mu \in \mathbb {R} ,\,\sigma \!>0\ {\Big \}}.}$

Then

{\displaystyle {\begin{aligned}&f_{\theta _{1}}=f_{\theta _{2}}\\[6pt]\Longleftrightarrow {}&{\frac {1}{{\sqrt {2\pi }}\sigma _{1}}}\exp \left(-{\frac {1}{2\sigma _{1}^{2}}}(x-\mu _{1})^{2}\right)={\frac {1}{{\sqrt {2\pi }}\sigma _{2}}}\exp \left(-{\frac {1}{2\sigma _{2}^{2}}}(x-\mu _{2})^{2}\right)\\[6pt]\Longleftrightarrow {}&{\frac {1}{\sigma _{1}^{2}}}(x-\mu _{1})^{2}+\ln \sigma _{1}={\frac {1}{\sigma _{2}^{2}}}(x-\mu _{2})^{2}+\ln \sigma _{2}\\[6pt]\Longleftrightarrow {}&x^{2}\left({\frac {1}{\sigma _{1}^{2}}}-{\frac {1}{\sigma _{2}^{2}}}\right)-2x\left({\frac {\mu _{1}}{\sigma _{1}^{2}}}-{\frac {\mu _{2}}{\sigma _{2}^{2}}}\right)+\left({\frac {\mu _{1}^{2}}{\sigma _{1}^{2}}}-{\frac {\mu _{2}^{2}}{\sigma _{2}^{2}}}+\ln \sigma _{1}-\ln \sigma _{2}\right)=0\end{aligned}}}

This expression is equal to zero for almost all x only when all its coefficients are equal to zero, which is only possible when |σ1| = |σ2| and μ1 = μ2. Since in the scale parameter σ is restricted to be greater than zero, we conclude that the model is identifiable: ƒθ1 = ƒθ2θ1 = θ2.

Example 2

Let ${\displaystyle {\mathcal {P}}}$ be the standard linear regression model:

${\displaystyle y=\beta 'x+\varepsilon ,\quad \mathrm {E} [\,\varepsilon \mid x\,]=0}$

(where ′ denotes matrix transpose). Then the parameter β is identifiable if and only if the matrix ${\displaystyle \mathrm {E} [xx']}$ is invertible. Thus, this is the identification condition in the model.

Example 3

Suppose ${\displaystyle {\mathcal {P}}}$ is the classical errors-in-variables linear model:

${\displaystyle {\begin{cases}y=\beta x^{*}+\varepsilon ,\\x=x^{*}+\eta ,\end{cases}}}$

where (ε,η,x*) are jointly normal independent random variables with zero expected value and unknown variances, and only the variables (x,y) are observed. Then this model is not identifiable,[4] only the product βσ² is (where σ² is the variance of the latent regressor x*). This is also an example of a set identifiable model: although the exact value of β cannot be learned, we can guarantee that it must lie somewhere in the interval (βyx, 1÷βxy), where βyx is the coefficient in OLS regression of y on x, and βxy is the coefficient in OLS regression of x on y.[5]

If we abandon the normality assumption and require that x* were not normally distributed, retaining only the independence condition ε  η  x*, then the model becomes identifiable.[4]

Software

In the case of parameter estimation in partially observed dynamical systems, the profile likelihood can be also used for structural and practical identifiability analysis.[6] An implementation of the is available in the MATLAB Toolbox PottersWheel.

References

Citations

1. Raue, A.; Kreutz, C.; Maiwald, T.; Bachmann, J.; Schilling, M.; Klingmuller, U.; Timmer, J. (2009-08-01). "Structural and practical identifiability analysis of partially observed dynamical models by exploiting the profile likelihood". Bioinformatics. 25 (15): 1923–1929. doi:10.1093/bioinformatics/btp358. PMID 19505944.
2. Lehmann & Casella 1998, Definition 1.5.2
3. van der Vaart 1998, p. 62
4. Reiersøl 1950
5. Casella & Berger 2001, p. 583
6. Raue, A; Kreutz, C; Maiwald, T; Bachmann, J; Schilling, M; Klingmüller, U; Timmer, J (2009), "Structural and practical identifiability analysis of partially observed dynamical models by exploiting the profile likelihood", Bioinformatics, 25 (15): 1923–9, doi:10.1093/bioinformatics/btp358, PMID 19505944, archived from the original on 2013-01-13.

Sources

• Walter, É.; Pronzato, L. (1997), Identification of Parametric Models from Experimental Data, Springer


text:
#### Volume 13, issue 4 (2009)

 Recent Issues
 The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1364-0380 ISSN (print): 1465-3060
Lefschetz fibrations and symplectic homology

### Mark McLean

Geometry & Topology 13 (2009) 1877–1944
##### Abstract

We show that for each $k>3$ there are infinitely many finite type Stein manifolds diffeomorphic to Euclidean space ${ℝ}^{2k}$ which are pairwise distinct as symplectic manifolds.

##### Keywords
Stein manifold, symplectic homology, Lefschetz fibration
Primary: 53D35
Secondary: 53D40


text:
# energy balance equation chemical engineering

0000008569 00000 n \left(\frac{\text{mass}}{\text{time}}\right) This type of situation is also known as chemical equilibrium. ? Chemical and Bio-Process Control, Texas Tech … \left(\frac{P}{\rho} + The Elements of Chemical Reaction Engineering, Prentice Hall. 0000005484 00000 n 0000189615 00000 n 0000004059 00000 n The mass balance for a steady state system is: A batch system is one in which all of the materials of the system are placed into a fixed vessel and left to react for an amount of time. 0000229628 00000 n What IS accumulation? g \, z \right)_{1} = \frac{\dot m \, w_{s}}{\epsilon}\], \begin{align}\begin{aligned}\SI{1}{hp} = \SI{550}{ft.lb_{f}/s}\\\SI{1}{hp} = \SI{0.746}{kW}\end{aligned}\end{align}, generalized form of the conservation of energy equation. An overall mass balance is a mass balance which models the entire mass of the system, as opposed to the mass of one component of the system. 0000000016 00000 n 0000004441 00000 n w_{s} - w_{f}\], $\left(\frac{P}{\rho} + \left(\frac{\text{work}}{\text{mass}}\right) = https://www.conservapedia.com/index.php?title=General_balance_equation&oldid=1256382. w_{s} - w_{f}$, \begin{align}\begin{aligned}\text{Power} = \frac{\text{work}}{\text{time}} 0000007933 00000 n \left(\frac{P}{\rho} + 30 0 obj <> endobj 0000215173 00000 n This means that the in, out, generation, and consumption terms are constants, and the accumulation term is 0. 0000229826 00000 n g \, z \right)_{2} - The general balance equation is a fundamental concept of chemical engineering which is based on the principles of conservation of mass and conservation of energy. 0000191025 00000 n \frac {1}{2} \, v^{2} + This page was last modified on 13 July 2016, at 13:20. \dot m \, w_{s} = \rho \, \dot V \, w_{s}\\\epsilon = \text{Efficiency} = \frac{\text{power delivered to the fluid}} xref Energy Balances - Leon Grado ... Chemical engineers do a mass balance to account for what happens to each of the compounds used in a chemical process. \left(\frac{P}{\rho} + \frac {1}{2} \, \alpha \, v_{avg}^{2} + g \, z \right)_{1} = 0000002110 00000 n trailer \[\left(\frac{P}{\rho} + ��.����3��c�{�Nt ���^��;�վgA0Y\A�"P�^���@zXHݓ˵�Q��V�a����0�,�>��Gg~m -��\�? Introduction to Chemical Engineering Processes/Unsteady state energy and mass balances. The balance equation then becomes. g \, (z_{2} - z_{1})= 0, $\left(\frac{P}{\rho} + \frac {1}{2} \, \alpha \, v_{avg}^{2} + g \, z \right)_{\text{out}} - Mathematics in Chemical Engineering" within 3rd year of study. \frac{1}{2} \left(v_{2}^{2} - v_{1}^{2} \right) + \frac {1}{2} \, v^{2} + 0000008611 00000 n Assuming an efficiency of $$\epsilon = 0.8 \, (80 \text{%})$$, 0000229356 00000 n <<845EAA49F2A2214AB2A85E904FBD3CC9>]>> ISBN 0-13-047394-4; Kravaris, Costas. 2. Chapter 7 A general Strategy for Solving Material Balance Problems The strategy outlined below is designed to focus your attention on the main path rather than the detours: 1. Jump to navigation Jump to search. 0000002397 00000 n 0000003900 00000 n %PDF-1.6 %���� 0000001407 00000 n 0000005533 00000 n what size pump (in units of $$\si{hp}$$) would be required? 0000214980 00000 n For such a purpose, the general principle of balancing is reduced to the bookkeeping procedure. • Only ONE energy balance equation is written for any system (or sub-system) irrespective of the number of components in the product(s) Note: Once mass and energy balance equations are written, make sure that the number of equations and number of unknowns are equal. Another way the general balance equation is used is the mole balance, that is, a balance on the number of moles of one component of the system. 0000189852 00000 n startxref 0000004944 00000 n 0000005201 00000 n 63 0 obj<>stream 0000004696 00000 n Since each mole of any given chemical must weigh the same, a mole balance is simply a direct product of the mass balance. Thus the general balance equation applied to a batch system is: If the system is at steady state, then the accumulation term is 0. Below is a general description of how to use mass balances for different situations. :Grt�������e���c�tiP=��߲� �g��a0o-�a�{1X���*M��d7e��E�!ӄ�!޸���y�bt�&�&+�e6F�n0��k������2 ��U��HJ8͟Ekr�UTq�hVd��~�ӘeP3S�ˊ։�c��2hJ�}W����Y��g��/������>N�ź���@\7�� � 0000003466 00000 n 0000001543 00000 n 0 \frac {1}{2} \, v^{2} + 0000214715 00000 n Mathematical formula derived from Eq. From Wikibooks, open books for an open world < Introduction to Chemical Engineering Processes. Read and understand the problem statement This mean read the problem carefully so that you know what is given and what is to be accomplished. 0000200827 00000 n Unreviewed. One of the ways chemical engineers ensure their calculations on linked process units have been performed correctly is to perform a mass balance over the entire system. You may assume that the velocity adjustment factor is unity If the system is at steady state then there are no terms which vary over time. The energy equation is often used for incompressible flow problems and is called the Mechanical Energy Equation or the Extended Bernoulli Equation. 0000001955 00000 n Since there is no flow in or out, the in and out terms are both 0. 0000005609 00000 n The general balance equation states that the total or component mass or energy of any system can be modeled by: The equation is used to describe and design almost every process unit within chemical engineering. Thus the overall balance is shown by: If the system is also at steady state, then the equation reduces to: Note that an overall mass balance can not be performed on a batch system, as all terms would equal 0. 0000008077 00000 n \frac{W}{\epsilon} = g \, z \right)_{2} - The mechanical energy equation for a turbine - where power is produced - can be written as: p in / ρ + v in 2 / 2 + g h in = p out / … 0000002938 00000 n the internal energy consists of ‘rate of flow’ work only ($$\dot W_{S} = 0$$) Under these conditions, we arrive at the Bernoulli Equation, which is often employed in engineering to relate velocities and pressures in various types of flows. 30 34 endstream endobj 31 0 obj<. g \, z \right)_{1} = 0$, \[\frac{P_{2} - P_{1}}{\rho} + 0000004198 00000 n However, mole balances still have advantages over mass balances in some situations (for example, if the flow rates are given in moles per time). \left(\frac{P}{\rho} + \frac {1}{2} \, \alpha \, v_{avg}^{2} + g \, z \right)_{\text{in}} =


text:
# the bug re-appears (Delay when rendering TeX)

The bug that I reported in this question is back today.

${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

• Like Arturo (from the linked question) this is how it has always been for me. I usually compose in notepad and copy it over periodically to make sure it renders. As it is, the textbox for typing an answer is barely functional. – Antonio Vargas Apr 1 '12 at 18:19
• @Antonio The point is that the preview used to work splendidly. It was broken recently - so much so that it makes the site almost unusable for me, since it crashes every browser I have, causing me to lose work, etc. I've had to stop using the site because it is far too painful with all the crashes. – Bill Dubuque Apr 1 '12 at 21:59
• Still there hours later. This is horrible. Last time it got cleared up shortly after the problem began. Now if I've typed part of an answer, I have to stop typing and wait for TeX to get rendered, and then as soon as I start typing again, I just see unrendered code throught the whole answer, and it it doesn't get rendered again until 10 seconds or so after I stop typing. Stackexchange has never been like this (for me...) except during that brief period when I reported this bug earlier. – Michael Hardy Apr 1 '12 at 22:01
• The problem wasn't here yesterday. – Michael Hardy Apr 1 '12 at 22:01
• @Michael Are you talking about the preview crash/freeze bug (as linked), or the quality standard block (or both)? If you refer to the crashes then you should explicitly mention such. Otherwise, readers who don't chase the link may think that you refer only to the far less serious quality standard issue. – Bill Dubuque Apr 1 '12 at 22:06
• @BillDubuque : You wrote "as linked". I don't see any link. – Michael Hardy Apr 1 '12 at 22:20
• @Michael The link you gave as "this question". – Bill Dubuque Apr 1 '12 at 22:28
• @MichaelHardy I think Bill is referring to the link in your question. The way it is written right now it looks like you are talking about the quality standard block that comes up on short questions. (That is what I thought when I read this earlier, and didn't click on the link in your question. I didn't realize what you were actually complaining about until reading the comments.) – user23784 Apr 1 '12 at 22:29
• @Michael Not to mention the paper-clip unicorn that's romping across all of my idle stackexhange screens today... – Bill Dubuque Apr 1 '12 at 22:31
• Using MSE is much more unpleasant than it used to be, with long delays while typing answers. – André Nicolas Apr 2 '12 at 16:06
• The problem continued for a couple of days, and now seems to be gone. – André Nicolas Apr 3 '12 at 17:24

The current preview code in use on MSE no longer has a delay before it starts typesetting the preview, so I do not see how you are having the same issue as you previously reported. Are you saying that there is a significant delay between when you type and when the math starts to be displayed?

The current algorithm is to replace the preview with the new text and immediately start typesetting it, but if new characters are typed, to interrupt the typesetting, replace the preview and start typesetting again.

MathJax does finish the current equation before it can be interrupted, so there is a possible delay there (though it should be small for most equations), and some equations may cause MathJax to load external files (for example, if there is no mathematics on the page already, the first equation to be typeset will cause MathJax to load the input and output processors), and that can cause a delay. But other than that, typesetting should proceed immediately.

But typesetting isn't instantaneous, and (depending on how fast you are typing and how long the answer is), you may interrupt MathJax before any equations are completed, so you may not see typeset equations while you type. But they should show more quickly (and more quickly after you stop typing) than when there was a delay before the typesetting was started. With a short answer having few equations, things should move fast enough that you don't see the untypeset mathematics.

Note that this is how the preview has always worked. In the past, the TeX was re-typeset every time you typed a new character, so your claim that "I could keep typing and still see the displayed $\rm\TeX$ as I typed" is based on the fact that the typesetting completed fast enough for you not to see it happening. When the delay was added (the source for your original complaint), that was no longer the case, and you did have to wait for the math to show up. The difference between the current approach and the older one that you liked is that typing new material now interrupts the typesetting and restarts it on the new material (rather than having to wait for everything to typeset). That should make the preview more responsive to your typing, especially for longer answers, but it does mean that mathematics later in the text might not get typeset as you continue to type.

Just to be clear, it was never the case that previously typeset math was being kept unchanged while you type. It has always been re-typeset for each character.

A better solution would be to update only the portion of the message that has actually changed, and only re-typeset the math in that portion of the text, but that would require a much more sophisticated editor than the one currently in use on SE.

• Hi Davide, thanks for your answer. Am I understanding you correctly that it is normal behavior for TeX to rerender with every new character typed, slowing down text input? When composing an answer, if there is TeX anywhere earlier in the answer I can usually type (non-TeX) faster than it displays. Sometimes it chugs so much that it stops accepting input at all for a split second and ignores a character or two. – Antonio Vargas Apr 2 '12 at 22:05
• Yes, the TeX is retested each time the input changes. Prior to a week ago, all the TeX would be typeset before responding to te next character typed. That is no longer the case. Now typing will interrupt the typesetting, and can even do so before any typesetting occurs, so the preview should be more responsive than it used to be, and you should no longer have to wait as long as you did, or lose characters. Has that happened in the last week? – Davide Cervone Apr 3 '12 at 12:01


text:
hbeLabs.com

## Weyl Counting Function and the Plots

Bob Jones
Summer 2016
I believe that original material on this page includes the NB, NC, DB, and DC Weyl counting function; and the trivial observation that "cumsum(d)" plot is far more (visually) sensitive to mistakes than the "d" plot, and the related observation that "cumsum(d)" appears to widen linearly.
Weyl counting plot for all of the lowest 164300+ regular pentagon eigenvalues. Includes both Dirichlet and Neumann eigenvalues. These plots have about a quarter of a million data points and are 20000x600 pixels, but file-size is less than 1MB. The very last handful of dots may not be consecutive, and although all eigenvalues are included, some may be misplaced but very slightly.

### Weyl's eigenvalue counting formula

Weyl's counting formulas are well known for pure Dirichlet and pure Neumann polygons, but often not (well known) for the polygonal shapes with the relevant boundary/symmetry conditions for the problem at hand.

Of course, Weyl's counting function is approximate and the usual practice is to append an order term, which I will disregard for now. Instead, I will consider only three terms of the form: $$N_w(\lambda)= c_1\lambda + c_2\sqrt{\lambda}+c_3$$ where the $c_i$ are three constants. This is called the "two-term" Weyl counting function (because two terms involve $\lambda$), and is the form appropriate to our polygons. This is a smooth function which treats $\lambda\ge 0$ as a continuous eigenparameter. It is supposed to approximate the actual eigenvalue counting function $\#(\lambda)$, which is a step-like function, starting at zero and increasing by $g_i$ (degeneracy of $\lambda_i$) as $\lambda$ increases past the eigenvalue $\lambda_i$.

The function $\delta(\lambda)=\#(\lambda)-N_w(\lambda)$ is usually displayed to demonstrate that $N_w(\lambda)$ is correct (for example), and it can be made to have a mean value of zero by appropriately choosing $c_3$.

In our problem, some formulas (specifically, those for DA,NA,DS,NS) can be trivially derived (see Appendix below), but the others (specifically, those for DB,DC,NB,NC) are not obvious to me. Despite that, with a little experience, this is what I came up with: $$N_w(\lambda) = C+\frac{1}{4\pi}\times \begin{cases} \left[ A \lambda + P \sqrt{\lambda} \right] &\enspace \text{NS} \\ \left[ A \lambda - P \sqrt{\lambda} \right] &\enspace \text{DA} \\ \left[ 2 A \lambda + \sqrt{\lambda} \right] &\enspace \text{NB or NC} \\ \left[ 2 A \lambda - \sqrt{\lambda} \right] &\enspace \text{DB or DC} \\ \left[ A \lambda + (P-1) \sqrt{\lambda} \right] &\enspace \text{DS}\\ \left[ A \lambda - (P-1) \sqrt{\lambda} \right] &\enspace \text{NA} \end{cases}$$ where $P$ is the perimeter and $A$ is the area of the $\pi/2$-$\pi/5$-$3\pi/10$ right triangle with shortest edge equal to 1/2. This triangle is the fundamental domain of the unit-edged regular pentagon according to its dihedral symmetry of order 10. Specifically, those constants are $$\begin{eqnarray} H &=& \frac{1}{2\tan(\pi/5)} \\ P &=& \frac{1}{2} + H + \sqrt{\frac{1}{4}+H^2} \\ A &=& \frac{1}{8\tan(\pi/5)} \end{eqnarray}$$

Note that the "$(P-1)$" factor scales with the linear dimensions: So if it is doubled so that the new pentagon perimieter doubles and its area becomes four times bigger, then that factor would be "$2(P-1)$" where this $P$ is still the fundamental triangle perimeter of the unit-edged regular pentagon. Similar words apply to the factor of one (before $\sqrt{\lambda}$) in the DB, DC, NB, and NC formulas.

In the Weyl counting function, $C$ is a (possibly) different constant for each of the eight cases. In some cases, it seems easy to calculate exactly, but we don't need to know what it is to make this work.

Adding up all the Dirichlet modes and Neumann modes for the pentagon, respectively, the well-known result is $$N_w(\lambda) = C + \frac{1}{4\pi}\times \begin{cases} \left[ 10 A \lambda - 5 \sqrt{\lambda} \right] &\enspace \text{Dirichlet }\unicode{x2b20} \\ \left[ 10 A \lambda +5 \sqrt{\lambda} \right] &\enspace \text{Neumann }\unicode{x2b20} \end{cases}$$ where one must not forget to double up the B and C modes since they are doubly degenerate in the pentagon. Note $10A$ is the area and 5 is the the perimeter of the unit-edged regular pentagon, as they should be.

The B and C modes have a sort of periodic-type symmetry condition on some of their symmetry lines, and I don't know how to figure out the Weyl counting function from first principles. However, I'm sure it is possible since I apparently found the answer. Perhaps a starting point is to assume the B and C counting functions are equal to each other (to within an additive constant) for a given pentagon boundary condition, then it is easy to solve.

### The Algorithm

To describe in more detail the plots, I use the octave language, which is similar to matlab, but octave is free software (GNU octave). I first give the strategy, then an octave script file, with its graphic output.

First note that there are no accidental degeneracies within the eight towers: The smallest eigenvalue gap happens to be $\delta\lambda\approx 0.164$ (at DB $\lambda\approx 907100$). So, I don't have to deal with such degeneracies right now, but the B and C eigenvalues are actually doubly degenerate in the pentagon.

Put all the eigenvalues of a given tower in an array L. These are indexed beginning from 1 so that 0<L(1)<L(2)<...<L(end).

Define next an index array $\fbox{I=1:length(L);}$, which is simply the vector of numbers [1,2,3,4,...,length(L)]. Notice that I(end)=length(L).

Define the actual counting function $\#(\lambda)$, which is the number of eigenvalues less than (or equal to) $\lambda$. Here $\lambda$ is a continuous parameter starting at zero. For L(n)<$\lambda$<L(n+1), this counting function is equal to n for valid values of n=1,2,3,...,I(end)-1.

What happens exactly at an eigenvalue for $\#(\lambda)$ is not important for this procedure. What we really want to do is adjust our constant $C$ so that the Weyl counting function effectively passes through (on average) the middle of the risers of the steps in $\#(\lambda)$. Below, I impose this convention by subtracting the mean of d0 so that the mean of d is zero.

Now, observe that I(n)=n is also the value of $\#(\lambda)$ for L(n)<$\lambda$<L(n+1). This vector I behaves very much like the actual counting function $\#(\lambda)$, although we have to watch out for n=0 or for n>length(L).

Let Nw0(L) be the Weyl counting vector (where L is our octave vector of eigenvalues), where the constant $C$ is set equal to zero. Then, I can form the difference vector, $\fbox{d0 = I - Nw0(L);}$ where, for clarity, d0(1)=1-Nw0(L(1)), d0(2)=2-Nw0(L(2)), and so on.

If there are no missing eigenvalues (below some specified number), and the Weyl counting function (with $C=0$) is correct, then this vector d0 will appear to fluctuate around a constant (the mean of d0). With this in mind, subtract this constant off so that $\fbox{d = d0 - mean(d0);}$ This vector d has mean of zero.

Finally, when d is plotted versus L, we obtain our "Weyl Plots". If everything worked, they should fluctuate around zero, and not vary very much, even for tens of thousands of eigenvalues. One interesting feature of these plots is that, even though the Weyl counting formula is supposed to be an asymptotic formula (for large $\lambda$), it seems to offer a nice relationship all the way down to $\lambda=0$.

Assume the DB eigenvalues are listed in the file L.DB.dat, which start out like

 27.786 89.167 111.120 172.736 200.818

Because of the script's simplicity, the data file must not have any extra blank lines or extraneous characters.

The octave script might look like the following (for the DB eigenvalues):

 #!/usr/local/bin/octave-cli --persist format long; # octave specific for displaying more digits H = 1/(2*tan(pi/5)); # set up our constants P = (1/2) + H + sqrt((1/2)^2 + H^2); # perimeter of fundamental triangle A = 1/(8*tan(pi/5)); # area of fundamental triangle FH=fopen("L.DB.dat","r"); # open and read in the data from the file L = textscan(FH,"%f"){1}; fclose(FH); I = 1:length(L); # Take transpose, so I and Nw0 I = I'; # will have the same size. Nw0 = @(L) (2*A*L-sqrt(L))/(4*pi); # inline function for DB modes d0 = I - Nw0(L); # Difference between actual counting # function and zero-Weyl # counting function d = d0 - mean(d0); # Here is the difference d that # has zero mean. graphics_toolkit gnuplot; # gnuplot toolkit hold "on"; # useful to make more complicated plots axis([0,L(end)]); # specify the horizontal interval of the graph plot(L,d,".b"); # plot blue dots at (L(1),d(1)), (L(2),d(2)), etc. xlabel("eigenvalue"); # label the x-axis title("DB d=#(L)-Nw(L)"); # add a title saveas (gca(),"DB.png", 'png'); #print it out as a png bitmap graphics file

Plotting the cummulative sum of d (cumsum(d)) is also very useful, as illustrated below.

The resulting graphics files are:

 plot(L,d,".b"); plot(L,cumsum(d),".r");

These plots, at first, appear to be simply "noise", which might not seem useful. However, these plots can be very useful to spot mistakes in our data. To demonstrate, I deliberately edit the eigenvalue file to

• remove one eigenvalue, which is 199,963.842, and
• insert one extra (fake) eigenvalue at 400,000.
The resulting plots look like this:
 plot(L,d,".b"); plot(L,cumsum(d),".r");

Those two errors show up in both plots, but far more dramatically in the cumsum plot. By examining the knees in the cumsum plot, it is easy to see where to look for mistakes in the sweep process, here at 200,000 and 400,000. In this example, only two mistakes were deliberately induced, but there are 18,373 eigenvalues in that file: It is extremely senstive to even one or two mistakes. (Using my sweep method, I would typically make about two or three errors like this in each sweep interval of $\delta\lambda=100,000$.)

For these reasons, when doing an eigenvalue calculation like this, it is rather important to make these plots. They don't prove we are right, but they can easily prove we are wrong ... and show us were to find the mistakes.

Another observation worth pursuing is the relatively simple linear widening of the cumsum (for valid data). This is to compare with the much more well known d function, which grows wider, but in a more complicated fashion.

As one final note, if the lowest $N$ eigenvalues are omitted, this process won't detect those omissions (it is absorbed in $C$), but can be used to see that the list that is used is complete. Thus, to be complete, this means we must make sure the lowest eigenvalue is included.

### Appendix - Elementary derivation of DA,DS,NA,NS Weyl counting functions

Note: If you understand this without drawing triangles and pentagons, good for you. I need to include some pictures.

For a pure Dirichlet or pure Neumann polygon, we have, to within an additive constant, $$4\pi N_w(\lambda) = \text{(area)}\lambda \mp \text{(perimeter)}\sqrt{\lambda}$$ where "$-$" is for Dirichlet and "$+$" is for Neumann.

For this section, I adopt the notation that the ordered triple (abc) will represent the even or odd nature of the edges of the $\pi/5$-$\pi/2$-$3\pi/10$ fundamental triangle. They are arranged in order from shortest to longest edges.

I use "e" for even (Neumann) and "o" for odd (Dirichlet). For example, the DS pentagon mode will have (oee), specifically, the shortest edge is odd (Dirichlet boundary condition of the pentagon) and the other two longer edges are even (even symmetry lines of the apothem and radius of pentagon).

The area of this triangle is $A$ and the edges are $\{1/2,H,S\}$ where $S=\sqrt{1/4+H^2}$ and $H$ is given in the octave script above. The perimeter of that fundamental region triangle is $P=1/2+H+S$.

With this notation, and using the fundamental region of that unit-edged regular pentagon, $$\begin{eqnarray} && \text{NS}\unicode{x2b20}:\enspace 4\pi N_w^\text{(eee)}(\lambda) = A\lambda + (1/2+H+S)\sqrt{\lambda}\\ && \text{DA}\unicode{x2b20}:\enspace 4\pi N_w^\text{(ooo)}(\lambda) = A\lambda - (1/2+H+S)\sqrt{\lambda} \end{eqnarray}$$

To get the DS and NS counting functions, imagine two fundamental region triangles joined together on their shortest edges forming a $\pi/5$-$\pi/5$-$3\pi/5$ isoceles triangle T. The eigenmodes of T are either even or odd due to its simple reflection symmetry. The Dirichlet T modes are the (ooo)+(eoo) modes, and the Neumann T modes are the (oee)+(eee) modes. The (e**) modes are the even T modes and the (o**) are its odd modes.

We know the T counting functions since T is a polygon (triangle) with area $2A$, and perimeter $(2H+2S)$. Therefore, $$\begin{eqnarray} & \text{Neumann T}:\enspace & 4\pi \left( N_w^\text{(oee)}(\lambda)+N_w^\text{(eee)}(\lambda)\right) = (2A)\lambda + (2H+2S)\sqrt{\lambda} \\ & \text{Dirichlet T}:\enspace & 4\pi \left(N_w^\text{(ooo)}(\lambda)+N_w^\text{(eoo)}(\lambda)\right) = (2A)\lambda - (2H+2S)\sqrt{\lambda} \end{eqnarray}$$

When one realizes that the DS$\unicode{x2b20}$ modes are the (oee) modes, and the NA$\unicode{x2b20}$ modes are the (eoo) modes, the above four equations can be used to get $$\begin{eqnarray} && \text{NA}\unicode{x2b20}:\enspace 4\pi N_w^\text{(eoo)}(\lambda) = A\lambda + ([1/2+H+S]-1)\sqrt{\lambda}\\ && \text{DS}\unicode{x2b20}:\enspace 4\pi N_w^\text{(oee)}(\lambda) = A\lambda - ([1/2+H+S]-1)\sqrt{\lambda} \end{eqnarray}$$ Using $P=1/2+H+S$, we get the four equations for DA,DS,NA,NS given at the beginning of this page.

Mon May 30 16:09:55 UTC 2016
rsjones7 at yahoo dot com


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# parasys.net

Home > Error Propagation > Error Propagation Exponential

# Error Propagation Exponential

## Contents

Pearson: Boston, 2011,2004,2000. Your cache administrator is webmaster. soerp package, a python program/library for transparently performing *second-order* calculations with uncertainties (and error correlations). Anmelden Wird geladen... news

Learn more You're viewing YouTube in German. doi:10.1287/mnsc.21.11.1338. error-analysis share|cite|improve this question edited Jan 25 '14 at 20:01 Chris Mueller 4,72711444 asked Jan 25 '14 at 18:31 Just_a_fool 3341413 add a comment| 2 Answers 2 active oldest votes up Melde dich bei YouTube an, damit dein Feedback gezählt wird. http://chem.libretexts.org/Core/Analytical_Chemistry/Quantifying_Nature/Significant_Digits/Propagation_of_Error

## Error Propagation

The relative error is . Hints help you try the next step on your own. How to handle a senior developer diva who seems unaware that his skills are obsolete? This ratio is called the fractional error.

In other classes, like chemistry, there are particular ways to calculate uncertainties. Two numbers with uncertainties can not provide an answer with absolute certainty! doi:10.1016/j.jsv.2012.12.009. ^ Lecomte, Christophe (May 2013). "Exact statistics of systems with uncertainties: an analytical theory of rank-one stochastic dynamic systems". Error Propagation Natural Log p.2.

You will sometimes encounter calculations with trig functions, logarithms, square roots, and other operations, for which these rules are not sufficient. Error Propagation Exponential Function Bitte versuche es später erneut. Wolfram Language» Knowledge-based programming for everyone. Consider a length-measuring tool that gives an uncertainty of 1 cm.

One immediately noticeable effect of this is that error bars in a log plot become asymmetric, particularly for data that slope downwards towards zero. Error Propagation Calculator Since at least two of the variables have an uncertainty based on the equipment used, a propagation of error formula must be applied to measure a more exact uncertainty of the Raising to a power was a special case of multiplication. Reciprocal In the special case of the inverse or reciprocal 1 / B {\displaystyle 1/B} , where B = N ( 0 , 1 ) {\displaystyle B=N(0,1)} , the distribution is

## Error Propagation Exponential Function

National Bureau of Standards. 70C (4): 262. Wenn du bei YouTube angemeldet bist, kannst du dieses Video zu einer Playlist hinzufügen. Error Propagation Management Science. 21 (11): 1338–1341. Exponent Error Propagation Function Variance Standard Deviation f = a A {\displaystyle f=aA\,} σ f 2 = a 2 σ A 2 {\displaystyle \sigma _{f}^{2}=a^{2}\sigma _{A}^{2}} σ f = | a | σ A

Starting with a simple equation: $x = a \times \dfrac{b}{c} \tag{15}$ where $$x$$ is the desired results with a given standard deviation, and $$a$$, $$b$$, and $$c$$ are experimental variables, each http://parasys.net/error-propagation/error-propagation-exponential-function.php The derivative, dv/dt = -x/t2. It will be interesting to see how this additional uncertainty will affect the result! Journal of Sound and Vibrations. 332 (11). Error Propagation Rules

Keith (2002), Data Reduction and Error Analysis for the Physical Sciences (3rd ed.), McGraw-Hill, ISBN0-07-119926-8 Meyer, Stuart L. (1975), Data Analysis for Scientists and Engineers, Wiley, ISBN0-471-59995-6 Taylor, J. Wird geladen... Given the measured variables with uncertainties, I ± σI and V ± σV, and neglecting their possible correlation, the uncertainty in the computed quantity, σR is σ R ≈ σ V More about the author It may be defined by the absolute error Δx.

Why are there no BGA chips with triangular tessellation of circular pads (a "hexagonal grid")? Error Propagation Physics This is the most general expression for the propagation of error from one set of variables onto another. Since $$\frac{\text{d}\ln(x)}{\text{d}x} = \frac{1}{x}$$ the error would be $$\Delta \ln(x) \approx \frac{\Delta x}{x}$$ For arbitraty logarithms we can use the change of the logarithm base:  \log_b

## If you know that there is some specific probability of $x$ being in the interval $[x-\Delta x,x+\Delta x]$, then obviously $y$ will be in $[y_-,y_+]$ with that same probability.

In problems, the uncertainty is usually given as a percent. Peralta, M, 2012: Propagation Of Errors: How To Mathematically Predict Measurement Errors, CreateSpace. When propagating error through an operation, the maximum error in a result is found by determining how much change occurs in the result when the maximum errors in the data combine Error Propagation Chemistry Now a repeated run of the cart would be expected to give a result between 36.1 and 39.7 cm/s.

Hinzufügen Playlists werden geladen... If you like us, please shareon social media or tell your professor! However, we want to consider the ratio of the uncertainty to the measured number itself. http://parasys.net/error-propagation/error-propagation-rules-exponential.php Why does the material for space elevators have to be really strong?

The derivative with respect to x is dv/dx = 1/t. If you measure the length of a pencil, the ratio will be very high. If you are converting between unit systems, then you are probably multiplying your value by a constant.


text:
## anonymous one year ago medal and fan An artist is making a square stained glass window in which a green g lass circle is surrounded by blue glass. The side length of the window is 20x, and the area of the green piece is 64πx^2 . What i s the area of the blue glass? Write your answer in factored form. so far i have Area 1= 400x^2 and Area 2= π(64πx^2)^2 and the GCF is 16x^2, am i right? can someone help me finish?

1. Michele_Laino

i think that the are of blue glass is given by the subsequent difference: $\Large 400{x^2} - 64\pi {x^2}$

2. Michele_Laino

now, please try to factor out 16x^2, what do you get?

3. anonymous

16x^2(25-4pi)?

4. Michele_Laino

that's right!

5. anonymous

thanks again I'm getting better but still suck at math

6. Michele_Laino

:)


text:
# LOG#152. Bohrlogy (II).

An interesting but relatively unknown variation of the Bohr model is to use a logarithmic potential energy. In that case, we have

(1)

(2)

(3)

Bohr quantization rules impose

and that

so

(4)

and then

(5)

(6)

(7)

The momentum is NOT quantized in the logarithmic potential Bohr model. By the other hand,

implies that

so

(8)

The velocities are not quantized either

(9)

The energies are easily calculated to be

(10)

or equivalently

(11)

The forces and accelerations are quantized

(12)

(13)

Areas are also quantized

(14)

The angular frequencies and the periods are quantized as well

(15)

(16)

Interestingly, we can modify and enlarge the Bohr quantization rules. The modified or enhanced Bohr rules imply the addition of the quantization of the area/length via an extra condition

(17)

and thus

(18)

Now, we proceed as Bohr himself

Therefore, momentum and velocity are again quantized (unlike the usual logarithmic potential with the normal Bohr conditions)

(19)

(20)

Forces and accelerations are quantized in a different form

(21)

(22)

Energies are also quantized but modified too

(23)

This last equation is something more complicated that the first logarithmic potential. We can play with it a bit. Introduce the areas

and suppose that

where

The energies are

and algebra provides

(24)

Finally, angular frequencies and periods are also quantized

(25)

(26)

As you can observe, some magnitudes change as we modify the quantization rules. This logarithmic model is useful in some interesting problems in theoretical physics and mathematics.

See you in my next blog post!

View ratings


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# Tag Info

0

Yes, lower base doping: Increase emitter injection efficiency (ratio of emitter:base doping) Increase base transport factor as lower doping means less recombination for minority carriers injected from the emitter Increase early voltage - because the base-collector junction depletion region will deplete more into the collector than the base- therefor the base ...

1

Here are two answers to your questions: 1.) Sallen&Key topology: When an RC network is combined with a (positive) finite-gain amplifier (gain of "one" or "two" or any other value <3) and if there is single-element signal feedback. 2.) Center frequency fo: For finding the center frequency (for maximum output) you must not ...

-2

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course). Well, we are trying to analyze the following opamp-circuit: simulate this circuit – Schematic created using CircuitLab When we use and apply KCL, we can write the following ...

2

Nodal analysis is the way to go. You can work on s-domain to make the life easier. According to your schematic, $\mathrm{v_b=V_o}$. Using the properties of a closed loop operational amplifier, you'll get $$\mathrm{ V_i+(1+s\ RC)V_o=(2+s\ RC)V_a }$$ where $$\mathrm{ V_a=V_o\frac{1+2s\ RC}{sRC} }$$ So the gain function can be written as  \mathrm{ G=\...

0

$CMRR = (A_{dm}-A_{cm})[dB]= 80~ dB ~, A_{dm}= 40 ~dB,~ thus ~A_{cm}=-40~dB$ $V_{IN_{cm}}= 79 mV ~RF???$ This CMRR only applies to DC to < 10 Hz and reduces 20 dB per decade to some small value like < 20dB Considering the question says RF noise, we can safely assume they're is not much differential or CM gain so the output is likely only ...

0

If the RF noise on the leads from the thermocouple sensor to the data logger is 79 mV, what will the noise level be on the amplified signal in mV? Here's where the question isn't tied down enough - does it mean that the RF noise on both leads is simultaneously 79 mV i.e. the common-mode interfering noise is truly 79mV or is it trying to tease out something ...

0

I suspect that your notes have two standard formulas, one for a circuit where RLC are all in series and one where they are all in parallel. This circuit is neither of those cases, so you may need to work a little harder. Notice that the series RL and the capacitor are in parallel. This means that the voltage across the RL and the capacitor must always be the ...

0

First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is way complicated than this one. This makes my answer valuable in my opinion. Well, we are trying to analyze the following circuit: simulate this circuit – Schematic ...

1

The datasheet includes a complete circuit diagram for a monostable (oneshot) circuit. Directly underneath the schematic is a section that describes how to calculate the length of the output pulse: The value for $R_A$ is in ohms. Use farads for the value of $C$. The value of $t_W$ is in seconds. You are given a requirement of 10 seconds - that's $... 1 Here's the circuit with the mesh currents drawn as you described them, all clockwise: simulate this circuit – Schematic created using CircuitLab These are the correct equations applying KVL with mesh currents: Bottom left mesh:  24 (I_1 - I_2) + 4 (I_1 - I_3) - 22 = 0  Top Mesh:  15 \cdot I_3 + 4 (I_3 - I_1) + 3 (I_3 - I_2) = 0  Some points ... 0 You can do a star->delta transformation to simplify first. Then try again by resolving parallel resistors etc. should become straightforward this way ... 0 It is not the resitor that is negative. It is the polarity of the voltage drop expressed as V=IR. It might be more intuitive to write: Loop 4: 0= 5.6(ı3-ı4)+10ı4 0 This kind of circuit is called a PD3 (parallel double, 3 phases -> q=3, so n=6). (unless "writing" errors). General assumption made is that Current load Ic is assumed constant =$Ic$. It has some formulas to describe it. (for 2 diodes current only allowed) Mean value of voltage on load : (Vm is peak voltage);$ Uco=2 * q /Pi * Vm * sin(Pi/q)\...

0

1)You can apply KCL at the negative terminal of V1 to find the current flowing through R2. Finding Vx then is just applying Ohm's law. 2)You can then apply KCL at the negative terminal of V2 to find the third current(two of them are known). 3)Finally you can apply KVL through the loop consisting of the two voltage sources to find out the current through R3. ...

0

Here is the basic approach: Look at the information you have. If you know the voltage across or current through a resistor, use Ohm's Law to find any missing value. Use KVL around any loops. If you have an equation with just one unknown, solve for it. Use KCL at all nodes. If you have an equation with just one unknown, solve for it. Repeat this process ...

3

The 741 can't get even close to the voltage rails: Source: TI 741 datasheet You need another opamp or another voltage source to power your circuit. The op-amps im allowed to use are TL074CN,CA3140EZ, LM358AP and the LM741CN The LM358A is slightly better but none of them have rail-to-rail inputs and outputs: Source: TI LM358 datasheet Since it is just a ...

2

You can't really use an LM741 op amp at low voltage, like 4.5V. The output is not rail-to-rail, so at low voltage you will only get a couple volts swing on the output.

1

The general strategy for problems like this is to combine resistors until you are left with a single resistor in parallel with the voltage source. Then you can use Ohm's Law to find the current through that single resistor. Now that you know the current, take that value back to your next-to-last circuit simplification. Use the newly found current to find the ...

-1

The left side with the power supply and Ra & R8 can be simplified by creating the thevenin equivalent circuit. Ra & R8 are really in series. And you can apply thevenin equivalent circuit again on Ra8 and Rb7, which are really on parallel. That should be able to give you just resistors in series. If you have issues seeing this check that your Thevenin ...

Top 50 recent answers are included


text:
# Introduction to Natural Language Processing (NLP) with TensorFlow

In this article we'll look at how we can use TensorFlow to analyze and predict text using natural language processing (NLP).

Whereas images often come in regular shaped tensors of pixel values, text data is typically messier and more difficult to deal with. Below we'll discuss how to handle  the inherent complexity that comes with language including dealing with short vs. long sentences and processing individual characters, words, and sentences.

Firstly, this involves processing text into numbers that we can use for calculating the weights and biases that we can feed into a neural network. Then, we'll look at more advanced topics such as sentence padding and vocabulary tokens that are key to natural language processing.

1. Word-Based Encodings
2. Text to Sequence
4. Word Embeddings
5. Sequence Models for Natural Language Processing
6. Sequence Models for Text Generation

## 1. Word-Based Encodings

As mentioned in our previous article on convolutional neural networks, computer vision problems use convolutions to identify features in images and use them for classification.

In the case of natural language processing, we're focused on building a classifier based on text models. First off, we'll look at sentiment in text and then learn how to build models that can classify new text based on the training data it has seen.

In order to encode text to numbers, we could take character encodings such as the ASCII values, although the issue is that the semantics of the word aren't encoded in the letters. For example, the words "silent" and "listen" have the exact same characters but a completely different meaning. That said, training a neural network with just characters can be a challenging task.

Instead, we can give words a value and use these to train the network. The value of each word doesn't exactly matter, it just needs to be the same value for the same word each time. For example, we can encode the sentence "I love my dog" as 001, 002, 003, 004. If we have the sentence "I love my cat" we can create a new token for "cat", so it could be encoded as 001, 002, 003, 005.

This is a starting point for training a neural network, although fortunately TensorFlow and Keras have already done this word-based encoding for us and made it available through several APIs, one of which is called Tokenizer.

The Tokenizer generates the dictionary of word encodings and creates vectors from sentences. Here's how we can encode the two sentences we just saw with the Tokenizer API:

• We first put the sentences into an array
• We then create an instance of the Tokenizer and set the parameter num_words to 100. This parameter takes the top 100 words by volume and encodes those, although we only have 5 in this basic example.
• The fit_on_texts method on tokenizer takes in the data and encodes it
• The tokenizer provides a word_index property that returns a dictionary with key-value pairs where the key is the word and the value is the to
import tensorflow as tf
from tensorflow import keras
from tensorflow.keras.preprocessing.text import Tokenizer
sentences = [
'I love my dog',
'I love my cat'
]

tokenizer = Tokenizer(num_words = 100)
tokenizer.fit_on_texts(sentences)
word_index = tokenizer.word_index
print(word_index)

You'll notice that the "I" has been stripped of its capitalization and if we add punctuation this will also be stripped away.

## 2. Text to Sequence

We just looked at how to tokenize words and sentences in order to build up a dictionary of all the words that make up a corpus. The next step is to turn sentences into lists of values based on these tokens.

We'll then also likely need to manipulate these lists, for example making every sentence the same length, in order to use them for training a neural network. We saw the same concept in our guide to CNNs with TensorFlow—when images were sized differently, we needed to resize them to fit the network.

Once again, TensorFlow has APIs to deal with resizing for text data. For example, we can update our code as follows:

• Below we've added one sentence that is longer
• We also call on the tokenizer to get text_to_sequences, which will turn our sentences into a set of sequences
sentences = [
'I love my dog',
'I love my cat',
'You love my dog!',
'Do you think my dog is amazing?'
]
tokenizer = Tokenizer(num_words = 100)
tokenizer.fit_on_texts(sentences)
word_index = tokenizer.word_index

sequences = tokenizer.texts_to_sequences(sentences)

print(word_index)
print(sequences)

At the top we have our new dictionary with new tokens for new words. Below that, we have a list of sentences that have been encoded into integer lists with tokens instead of words.

Keep in mind that text_to_sequences can take any set of sentences, so it can encode them based on the word set that we passed to fit_on_texts. This means that if we train a neural network on a corpus of text and a word index is generated, in order to perform inference with the trained model we'll have to encode the text we want to infer on with the same word index, otherwise it will be meaningless.

In other words, we need a lot of training data in order to get a broad vocabulary or we could end up with lists of encoded sentences that don't make any sense.

In many cases, instead of just ignoring unseen words it's a good idea to use a special value in when one is encountered. We can do this using a property on the tokenizer, oov_token, which stands for "out of vocabulary":

tokenizer = Tokenizer(num_words = 100, oov_token="<00V>")

As mentioned, just like how image data that needs to be in uniform size, text data has similar requirements of uniformity, and one way we can do this is with sequence padding.

In order to use padding functions from TensorFlow we need to import the following:

from tensorflow.keras.preprocessing.sequence import pad_sequences

We can then pass our sequences to pad_sequences as follows:

We can now see the list of sentences has been padded out into a matrix and that each row in the matrix has the same length. It does this by putting the required number of 0's before the sentence.

If we want to padding to occur after the sentence we can add padding='post'. If we want to set a maximum padding length we can use maxlen=5, for example.

## 4. Word Embeddings

Let's now look at how to use TensorFlow and Keras to implement word embeddings, which is one of the most important ideas in natural language processing.

If we have a corpus of 10,000 words, for example, the basic idea of word embeddings is to represent these words in a better way than just using the numbers 1-10,000. As MachineLearningMastery describes:

Word embeddings are a type of word representation that allows words with similar meaning to have a similar representation.

Similar to how we extract features from images using convolutions, the process of extracting sentiment from words can be done through embedding.

The idea of embedding is that closely related words will be clustered together as vectors in a multi-dimensional space.

### IMDB Dataset

Another one of the datasets included in TensorFlow is the IMDB reviews dataset, which contains 50,000 movie reviews that are categorized as positive or negative.

We can import the dataset as follows:

import tensorflow_datasets as tfds
imdb, info = tfds.load("imdb_reviews", with_info=True, as_supervised=True)

The data is split into 25,000 images for training and 25,000 for testing, which we can split out like this:

train_data, test_data = imdb['train'], imdb['test']

Each of these are iterables with 25,000 sentences and labels as tensors, so we can iterate over the training and testing data and extract the sentences and labels with for loops as follows:

import numpy as np

training_sentences = []
training_labels = []

testing_sentences = []
testing_labels = []

for s, l in train_data:
training_sentences.append(str(s.numpy()))
training_labels.append(l.numpy())

for s, l in test_data:
testing_sentences.append(str(s.numpy()))
testing_labels.append(l.numpy())

When training the labels are expected to be NumPy arrays so we can convert the list of labels we just created as follows:

training_labels_final = np.array(training_labels)
testing_labels_final = np.array(testing_labels)

Next we can tokenize and pad the sentences with the following hyperparameters:

vocab_size = 10000
embedding_dim = 16
max_length = 120
trunc_type='post'
oov_tok = "<OOV>"

from tensorflow.keras.preprocessing.text import Tokenizer

tokenizer = Tokenizer(num_words = vocab_size, oov_token=oov_tok)
tokenizer.fit_on_texts(training_sentences)
word_index = tokenizer.word_index
sequences = tokenizer.texts_to_sequences(training_sentences)

testing_sequences = tokenizer.texts_to_sequences(testing_sentences)
testing_padded = pad_sequences(testing_sequences,maxlen=max_length)

Now we can define our neural network as a Sequential model with the first layer set as an Embedding layer. The result of the embedding will be a 2D array with the length of the sentence and embedding dimension as the size, so we need to flatten it with a Flatten layer. We then feed that into two Dense layers for classification:

model = tf.keras.Sequential([
tf.keras.layers.Embedding(vocab_size, embedding_dim, input_length=max_length),
tf.keras.layers.Flatten(),
tf.keras.layers.Dense(6, activation='relu'),
tf.keras.layers.Dense(1, activation='sigmoid')
])

We can compile the model and print out the summary with the code below:

model.compile(loss='binary_crossentropy',optimizer='adam',metrics=['accuracy'])
model.summary()

Training can be done by passing padded and training_labels_final as our training set, specifying the number of epochs, and passing the testing_padded and testing_labels_final as our test set:

num_epochs = 10
model.fit(padded, training_labels_final, epochs=num_epochs, validation_data=(testing_padded, testing_labels_final))

## 5. Sequence Models for Natural Language Processing

Let's now look at how to implement sequence models such as recurrent neural networks (RNNs) for natural language processing in TensorFlow.

Sequence models are important in NLP since the relative order of words provide an extra layer of meaning in sentences. One reason that sequence models such as RNNs and LSTMs are useful for NLP is that the context is preserved from timestamp to timestamp.

We won't cover RNNs in detail here, but here's an overview from our Introduction to RNNs and LSTMs:

Recurrent neural networks are designed to learn from sequences of data by passing the hidden state from one step in the sequence, to the next step, and combining this with the input.

LSTMs, or Long Short-Term Memory Networks, are an update to RNNs which, in addition to the context being passed from RNNs, pass another pipeline of context called the cell state.

The cell state helps keep context from earlier tokens so that they can be accessed later on. Cell states can also be bidirectional, which means that later contexts can impact earlier ones.

You can also learn more about LSTMs from Andrew Ng's course on Sequence Models. Below is an example of a model that uses an LSTM in the second layer:

model = tf.keras.Sequential([
tf.keras.layers.Embedding(tokenizer.vocab_size, 64),
tf.keras.layers.Bidirectional(tf.keras.layers.LSTM(64)),
tf.keras.layers.Dense(64, activation='relu'),
tf.keras.layers.Dense(1, activation='sigmoid')
])

## 6. Sequences Models for Text Generation

Now that we've reviewed NLP for text classification, let's look at how we can use TensorFlow to predict text, which leads us to natural language generation.

Text prediction is similar to what we've already discussed—we take a body of text, extract the full vocabulary from it, and then create datasets where we make the phrase the $x$'s and next word we're predicting to be the $y$'s.

Given enough words in a corpus, a neural network can predict and generate relatively sophisticated text. For example, the largest language generation model is currently GPT-3, which was trained on 175 billion parameters.

We won't cover the code for text generation in this article, although you can find the full code on Github here. You can also find a tutorial from TensorFlow on generating text with an RNN here.

## Summary: Natural Language Processing with TensorFlow

In this article, we introduced how to use TensorFlow and Keras for natural language processing.

The first principles of NLP include tokenization and padding in order to prepare our data for a neural network. We also discussed word embeddings and mapping words to vectors. Finally, we reviewed how sequence models such as RNN and LSTMs can be used for classification by not just looking at words in isolation, but instead accessing information previously learned in the sentence.

This is simply an introduction to natural language processing with TensorFlow—below you'll find additional resources on the topic.

### Resources

Join the conversation.


text:
# Multiplying Decimals by Powers of Ten (D)

##### This Multiplying Decimals by Powers of Ten (D) worksheet also includes:

Fifth graders enhance their multiplication skills by solving the 45 decimal problems that range by powers of ten.

Concepts

Resource Details


text:
21 questions linked to/from Tag management 2020
2k views

### Tag management 2019 [duplicate]

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Often mathematics is also a matter of aesthetics and elegance. Sometimes there appear questions asking for a "more elegant" way to do this or that, or a "nice" proof of some fact. I'd like to know ...
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# Not positive definite variance-covariance matrix in meta-regression using 'metafor'

I'm having a strange problem running a meta-regression using the function rma.mv() in the 'metafor' package in R.

Since some of my data are from multiple-endpoint studies, I have calculated the variance-covariance matrix so that correlations between outcomes are taken into account. I'm also using random effects at study and treatment level. As far as I'm aware, I have now covered all issues with regard to dependent effect sizes.

The model looks like this:

cov_mod <- rma.mv(Hedges_g, cov, mods = ~ days, random = ~ treatment | study, data = rev)

When running the code, it gives this error message:

Error in rma.mv(Hedges_g, cov9, mods = ~days, random = ~1 | treatment/study,  :
Error during optimization.
In rma.mv(Hedges_g, cov9, mods = ~days, random = ~1 | treatment/study,  :
V appears to be not positive definite.

I have discovered that the problem lies with one particular study (9 effect sizes in total, coming from 3 treatment groups that were each tested at 3 moments in time). When I remove this study from the data set, the code runs without problem.

Thus, apparently this particular study causes the matrix to be 'not positive definite'. I have read that this likely means that "at least one of [the] variables can be expressed as a linear combination of the others" (source).

However, here comes the strange thing: I have replaced all values in the variance-covariance matrix relating to this particular study with random numbers between 0-1 (maintaining the symmetry), and the error message remains unchanged. I am puzzled, because the matrix can no longer be linearly predictable if it contains random numbers.

What could be the issue?

• A random covariance matrix is unlikely to be positive definite! – whuber Aug 4 '16 at 13:33
• Could you post the 9x9 var-cov matrix for that study? (ideally so that it can be directly read into R, so using dput()). – Wolfgang Aug 4 '16 at 15:59
• @Wolfgang Here you go: structure(c(0.029, 0.0421, 0.0363, 0, 0, 0, 0, 0, 0, 0.0421, 0.0297, 0.037, 0, 0, 0, 0, 0, 0, 0.0363, 0.037, 0.0418, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.0342, 0.0472, 0.0383, 0, 0, 0, 0, 0, 0, 0.0472, 0.0345, 0.0431, 0, 0, 0, 0, 0, 0, 0.0383, 0.0431, 0.0465, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.0231, 0.0368, 0.0208, 0, 0, 0, 0, 0, 0, 0.0368, 0.0231, 0.023, 0, 0, 0, 0, 0, 0, 0.0208, 0.023, 0.0337), .Dim = c(9L, 9L), .Dimnames = list(NULL, c("V40", "V41", "V42", "V43", "V44", "V45", "V46", "V47", "V48"))) – Johanna Aug 5 '16 at 8:04
• That's not a valid var-cov matrix. For example, the first two variances are 0.0290 and 0.0297. The covariance is 0.0421. This would imply a correlation of 1.434514, which is not possible. – Wolfgang Aug 5 '16 at 8:13
• Thank you for the insight - I will work on it and then get back to this thread. – Johanna Aug 5 '16 at 11:58


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