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# Search by Topic

#### Resources tagged with Visualising similar to Football World Cup Simulation:

Filter by: Content type:
Stage:
Challenge level:

### Charting More Success

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

Can you make sense of the charts and diagrams that are created and used by sports competitors, trainers and statisticians?

### Charting Success

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

Can you make sense of the charts and diagrams that are created and used by sports competitors, trainers and statisticians?

### Convex Polygons

##### Stage: 3 Challenge Level:

Show that among the interior angles of a convex polygon there cannot be more than three acute angles.

### World of Tan 20 - Fractions

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of the chairs?

### World of Tan 21 - Almost There Now

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?

### World of Tan 22 - an Appealing Stroll

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of the child walking home from school?

### World of Tan 26 - Old Chestnut

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?

### World of Tan 25 - Pentominoes

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of these people?

### World of Tan 24 - Clocks

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of these clocks?

### Domino Numbers

##### Stage: 2 Challenge Level:

Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?

### World of Tan 19 - Working Men

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this shape. How would you describe it?

### World of Tan 7 - Gat Marn

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this plaque design?

### World of Tan 5 - Rocket

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of the rocket?

### World of Tan 6 - Junk

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this junk?

### Reflecting Squarely

##### Stage: 3 Challenge Level:

In how many ways can you fit all three pieces together to make shapes with line symmetry?

### Square Coordinates

##### Stage: 3 Challenge Level:

A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?

### World of Tan 12 - All in a Fluff

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of these rabbits?

### Pattern Power

##### Stage: 1, 2 and 3

Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.

### World of Tan 4 - Monday Morning

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?

### World of Tan 11 - the Past, Present and Future

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of the telescope and microscope?

### World of Tan 27 - Sharing

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Little Fung at the table?

### World of Tan 29 - the Telephone

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this telephone?

### Nine-pin Triangles

##### Stage: 2 Challenge Level:

How many different triangles can you make on a circular pegboard that has nine pegs?

### All in the Mind

##### Stage: 3 Challenge Level:

Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface. . . .

### Isosceles Triangles

##### Stage: 3 Challenge Level:

Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?

### How Many Dice?

##### Stage: 3 Challenge Level:

A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .

### Dice, Routes and Pathways

##### Stage: 1, 2 and 3

This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .

### World of Tan 2 - Little Ming

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Little Ming?

### The Old Goats

##### Stage: 3 Challenge Level:

A rectangular field has two posts with a ring on top of each post. There are two quarrelsome goats and plenty of ropes which you can tie to their collars. How can you secure them so they can't. . . .

##### Stage: 3 Challenge Level:

Can you mark 4 points on a flat surface so that there are only two different distances between them?

### Makeover

##### Stage: 1 and 2 Challenge Level:

Exchange the positions of the two sets of counters in the least possible number of moves

### Coin Cogs

##### Stage: 2 Challenge Level:

Can you work out what is wrong with the cogs on a UK 2 pound coin?

### Coloured Edges

##### Stage: 3 Challenge Level:

The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set?

### Colour Wheels

##### Stage: 2 Challenge Level:

Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?

### Conway's Chequerboard Army

##### Stage: 3 Challenge Level:

Here is a solitaire type environment for you to experiment with. Which targets can you reach?

### Sprouts

##### Stage: 2, 3, 4 and 5 Challenge Level:

A game for 2 people. Take turns joining two dots, until your opponent is unable to move.

### You Owe Me Five Farthings, Say the Bells of St Martin's

##### Stage: 3 Challenge Level:

Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?

### Ding Dong Bell

##### Stage: 3, 4 and 5

The reader is invited to investigate changes (or permutations) in the ringing of church bells, illustrated by braid diagrams showing the order in which the bells are rung.

### Sea Defences

##### Stage: 2 and 3 Challenge Level:

These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?

### Zooming in on the Squares

##### Stage: 2 and 3

Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?

### World of Tan 28 - Concentrating on Coordinates

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Little Ming playing the board game?

### World of Tan 13 - A Storm in a Tea Cup

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of these convex shapes?

### Tetrahedra Tester

##### Stage: 3 Challenge Level:

An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?

### Penta Play

##### Stage: 2 Challenge Level:

A shape and space game for 2,3 or 4 players. Be the last person to be able to place a pentomino piece on the playing board. Play with card, or on the computer.

### Diagonal Dodge

##### Stage: 2 and 3 Challenge Level:

A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.

### World of Tan 14 - Celebrations

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?

### Khun Phaen Escapes to Freedom

##### Stage: 3 Challenge Level:

Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.

### Picturing Triangle Numbers

##### Stage: 3 Challenge Level:

Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?

### Square It

##### Stage: 1, 2, 3 and 4 Challenge Level:

Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.

### Sliding Puzzle

##### Stage: 1, 2, 3 and 4 Challenge Level:

The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.


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# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Sensitivity of set-valued discrete systems. (English) Zbl 1194.37034
Let $X$ be a compact metric space and $K(X)$ be the set of all nonempty compact subsets of $X$ equipped with the Hausdorff metric. In this paper, examples are given that sensivity of a surjective continuous map $f:X\to X$ does not imply sensivity of the induced continuous self-mapping $\bar f:K(X)\to K(X)$. Moreover, it is shown that if $f$ is a surjective continuous interval map, then $\bar f$ is sensitive if and only if $f$ has this property.

##### MSC:
 37B99 Topological dynamics
##### Keywords:
set-valued mapping; sensitivity; path-connected space
Full Text:
##### References:
 [1] Ruelle, D.; Takens, F.: On the nature of turbulence. Comm. math. Phys. 20, 167-192 (1971) · Zbl 0223.76041 [2] Auslander, J.; Yorke, J.: Interval maps, factors of maps and chaos. Tohoku math. J. 32, 177-188 (1980) · Zbl 0448.54040 [3] Devaney, R.: Chaotic dynamical systems. (1989) · Zbl 0695.58002 [4] Bauer, W.; Sigmund, K.: Topological dynamics of transformations induced on the space of probability measures. Monatsh. math. 79, 81-92 (1975) · Zbl 0314.54042 [5] Fedeli, A.: On chaotic set-valued discrete dynamical systems. Chaos solitons fractals 23, 1381-1384 (2005) · Zbl 1079.37021 [6] Liao, G. F.; Wang, L. D.; Zhang, Y. C.: Transitivity, mixing and chaos for a class of set valued mappings. Sci. China ser. A math. 49, 1-8 (2006) · Zbl 1193.37023 [7] Peris, A.: Set-valued discrete chaos. Chaos solitons fractals 26, 19-23 (2005) · Zbl 1079.37024 [8] Banks, J.: Chaos for induced hyperspace maps. Chaos solitons fractals 25, 681-685 (2005) · Zbl 1071.37012 [9] Román-Flores, H.: A note on transitivity in set-valued discrete systems. Chaos solitons fractals 17, 99-104 (2003) · Zbl 1098.37008 [10] Kwietniak, D.; Oprocha, P.: Topological entropy and chaos for maps induced on hyperspaces. Chaos solitons fractals 33, 76-86 (2007) · Zbl 1152.37306 [11] Gu, R. B.: Kato’s chaos in set-valued discrete systems. Chaos solitons fractals 31, 765-771 (2007) · Zbl 1140.37305 [12] Román-Flores, H.; Chalco-Cano, Y.: Robinson’s chaos in set-valued discrete systems. Chaos solitons fractals 25, 33-42 (2005) · Zbl 1071.37013 [13] Wang, Y.; Wei, G.; Campbell, W. H.: Sensitive dependence on initial conditions between dynamical systems and their induced hyperspace dynamical systems. Topol. appl. 156, 803-811 (2009) · Zbl 1172.37006 [14] García-Guirao, J. L.; Kwietniak, D.; Lampart, M.; Oprocha, P.; Peris, A.: Chaos on hyperspaces. Nonlinear anal. 71, 1-8 (2009) · Zbl 1175.37024 [15] Robinson, C.: Dynamical systems. (1995) · Zbl 0853.58001 [16] Barge, M.; Martin, J.: Chaos, periodicity and snakelike continua. Trans. amer. Math. soc.. 289, 355-365 (1985) · Zbl 0559.58014 [17] Blokh, A. M.: On sensitive mappings of the interval translation. Russ. math. Surv. 37, 203-204 (1982) · Zbl 0511.28013 [18] Furstenberg, H.: Disjointness in ergodic theory, minimal sets, and a problem in Diophantine approximation. Math. system th. 1, 1-55 (1967) · Zbl 0146.28502

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MathSciNet bibliographic data MR425956 55D15 (58A10) Bousfield, A. K.; Gugenheim, V. K. A. M. On ${\rm PL}$${\rm PL}$ de Rham theory and rational homotopy type. Mem. Amer. Math. Soc. 8 (1976), no. 179, ix+94 pp. Article

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


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Home add key in hash based on condition

# add key in hash based on condition

User7354632781
1#
User7354632781 Published in 2018-01-05 17:55:12Z
 I have following method def some_method ...do something here... @customer_info = { a: a, b: b } end  I want to add key c in @customer_info if variable customer is true so something like, @customer_info = { a: a, b: b, c: something if customer is true }  tried to do something like this but no luck @customer_info.merge(customer.present? {c: something} : {})
steenslag
2#
 @customer_info[:c] = something if customer
 Perhaps you are simply missing ? in your ternary operator? Your approach seems to work for me in console: > customer_info = {a: :a, b: :b} => {:a=>:a, :b=>:b} > c = false => false > customer_info.merge(c ? {c: c} : {}) => {:a=>:a, :b=>:b} > c = true => true > customer_info.merge(c ? {c: c} : {}) => {:a=>:a, :b=>:b, :c=>true}  So, this: @customer_info.merge(customer.present? {c: something} : {})  Should be more like: @customer_info.merge(customer.present? ? {c: something} : {})  Quick Update As pointed out in the comments, this should be more like: @customer_info.merge!({c: something}) if customer.present?  The commentators do a nice job of explaining why.


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# Prompt and non-prompt J/ ψ elliptic flow in Pb+Pb collisions at √sNN=5.02 Tev with the ATLAS detector

Результат исследования: Научные публикации в периодических изданияхстатьярецензирование

14 Цитирования (Scopus)

## Аннотация

The elliptic flow of prompt and non-prompt J / ψ was measured in the dimuon decay channel in Pb+Pb collisions at s NN = 5.02 TeV with an integrated luminosity of 0.42 nb - 1 with the ATLAS detector at the LHC. The prompt and non-prompt signals are separated using a two-dimensional simultaneous fit of the invariant mass and pseudo-proper decay time of the dimuon system from the J / ψ decay. The measurement is performed in the kinematic range of dimuon transverse momentum and rapidity 9 < p T < 30 GeV , | y | < 2 , and 0-60% collision centrality. The elliptic flow coefficient, v 2 , is evaluated relative to the event plane and the results are presented as a function of transverse momentum, rapidity and centrality. It is found that prompt and non-prompt J / ψ mesons have non-zero elliptic flow. Prompt J / ψ v 2 decreases as a function of p T , while for non-prompt J / ψ it is, with limited statistical significance, consistent with a flat behaviour over the studied kinematic region. There is no observed dependence on rapidity or centrality.

Язык оригинала английский 784 23 European Physical Journal C 78 9 https://doi.org/10.1140/epjc/s10052-018-6243-9 Опубликовано - 27 сент. 2018

## Предметные области OECD FOS+WOS

• 1.03 ФИЗИЧЕСКИЕ НАУКИ И АСТРОНОМИЯ

## Fingerprint

Подробные сведения о темах исследования «Prompt and non-prompt J/ ψ elliptic flow in Pb+Pb collisions at √sNN=5.02 Tev with the ATLAS detector». Вместе они формируют уникальный семантический отпечаток (fingerprint).


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# Comparing odds ratios

I have data for two different sets of patients, and I divided each set into two subgroups (males and females, let's say). For each patient set, I compute the odds ratio of receiving a treatment for males versus females. Is there a way to compare the odds ratios of the two patient sets and say that one is significantly higher than the other, even though the odds ratios correspond to two entirely different sets of people?

I am new to this, so any recommendation of resources that address this problem would be appreciated. Thanks.

• My suggestion would be to test a model that predicts the outcome (dichotomous, I'm guessing) from Gender, Group and the interaction. Then, if the interaction term is significant, there is a difference between groups based on gender. Happy to elaborate further if this is the approach you'd like to pursue. Mar 27 '18 at 18:33
• Yes, it would be great if you could elaborate. I responded to Isabella's comment below with my interpretation of what I should do, based on your responses. Not sure if it's correct.
– LLP
Mar 28 '18 at 18:32

It is possible to compute two types of odds ratios in your scenario:

1. Unadjusted odds ratios of receiving a treatment;
2. Adjusted odds ratios of receiving a treatment.

For the first type, you could fit a binary logistic regression to your data, which includes the following variables:

Outcome variable: Binary variable, indicating whether the patient received the treatment (1) or did not receive the treatment (0);

Predictor variable: Binary variable, indicating whether the patient is Male (1) or Female (0).

The exponentiated regression coefficient of the predictor variable sex in this model will represent the odds ratio of receiving a treatment for males versus females. If this exponentiated regression coefficient is estimated from the data to be 1.2, say, then you would conclude that the odds of receiving a treatment are 20% higher for males compared to females, since (1.2-1)x100% = 20%. If it comes out to be 0.8, say, then you would conclude that the odds of receiving a treatment are 20% lower for males compared to females, since (0.8 - 1)x100% = -20%.

For the second type, you will need to expand your binary logistic regression model to include one or more predictor variables (e.g., age, cholesterol level) in addition to the sex variable. The exponentiated regression coefficient of the predictor variable sex in this model will represent the odds ratio of receiving a treatment for males versus females, adjusted for age and cholesterol level. If this exponentiated regression coefficient is estimated from the data to be 1.2, say, then you would conclude that, among patients having the same age and the same cholesterol level, the odds of receiving a treatment are 20% higher for males compared to females, since (1.2-1)x100% = 20%. If it comes out to be 0.8, say, then you would conclude that, among patients having the same cholesterol level and same age, the odds of receiving a treatment are 20% lower for males compared to females, since (0.8 - 1)x100% = -20%.

Adjusting for additional characteristics in the binary logistic regression model affords you to compare male and female patients who are the same with respect to those characteristics, thus reducing some of the "noise" that may affect your males versus females comparison of the odds of receiving a treatment.

## Edit:

My explanation above refers to a single set of patients. If you are interested in computing the odds ratio of receiving a treatment separately for each set of patients, you will need to expand your model (as already suggested in other answers here) so that it includes a Group (or Patient Set) variable in addition to a Sex variable. You can define this variable to be 0 for your first set of patients and 1 for your second set. The model will be applied to your combined data for patients in both sets, so you don't really need your Step 1. Your Step 2 model for the combined data is in effect fitting four different models simultaneously. You can represent all 4 models at once via the following equation:

log odds of receiving a treatment = beta0 + beta1*Sex + beta2*PatientSet + beta3*Sex*PatientSet

where the odds of receiving a treatment are expressed as log(p/(1-p)) and p is the conditional probability of receiving a treatment given the patient's sex and patient set.

The above model can be simplified as follows for each gender by patient set combination:

Females in Patient Set #1:

log odds of receiving a treatment = beta0  or, equivalently,

odds of receiving a treatment = exp(beta0)

Males in Patient Set #1:

log odds of receiving a treatment = beta0 + beta1 or, equivalently,

odds of receiving a treatment = exp(beta0 + beta1)

Females in Patient Set #2:

log odds of receiving a treatment = beta0 + beta2 or, equivalently,

odds of receiving a treatment = exp(beta0 + beta2)

Males in Patient Set #2:

log odds of receiving a treatment = beta0 + beta1 + beta2 + beta3

odds of receiving a treatment = exp(beta0 + beta1 + beta2 + beta3)

If you follow the math, you can determine that, among patients in your first set, the odds of receiving the treatment for males are obtained by multiplying the odds of receiving treatment for females with the multiplicative factor below:

f1 = exp(beta0 + beta1)/exp(beta0) = exp(beta1)

Similarly, among patients in your second set, the odds of receiving the treatment for males are obtained by multiplying the odds of receiving treatment for females with the multiplicative factor below:

f2 = exp(beta0 + beta1 + beta2 + beta3)/exp(beta0 + beta2) = exp(beta1 + beta3)

Dividing the two multiplicative factors will yield the quantity you are interested in:

f2/f1 = exp(beta1 + beta3)/exp(beta1) = exp(beta3)

In other words, exponentiating the coefficient of the interaction term in your original binary logistic regression model for the combined data will help you compare the multiplicative factors by which the odds of receiving a treatment for males relative to females differ among the two patient sets.

• Thank you! I’m still not clear how to statistically compare the ORs between the two groups. Would this work?: Step 1: Fit regression models separately to patient set #1 and patient set #2. Use the coefficients in those two models to find the gender ORs for patients in sets #1 and #2. Step 2: Combine data for patient sets #1 and #2. Fit regression model to combined data. (Predictors = gender, patient set #, interaction term for gender and patient set #). If the interaction term’s coefficient is significant, then there is a significant difference between the ORs of the two sets from Step 1.
– LLP
Mar 28 '18 at 18:31
• I added an edit at the bottom of my answer. Mar 28 '18 at 22:53

To run the model with a dichotomous dependent variable and two dichotomous independent variables with a focus on determining if the odds-ratios are different for one of the i.v. compared to the groups formed by the other i.v., you would include the interaction of the group terms in the model.

Let's say we have $F$ for male/female (0/1, respectively) and $G$ for patient group (0/1). Let $Y$ be the dependent variable, which is also dichotomous.  The code for R would look something like:

glm.out <- glm(Y ~ F*G, data=data.frame, family=binomial(link='logit'))
If the coefficient for F*G is statistically significant, this says there is an interaction effect.  Said another way, the separate equations predicting Y from F for the different groups of G would produce different regression coefficients (i.e., different odds-ratios when transformed).

Hope this helps.


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Geometry Webinar

# Algebraic theory of ODEs depending on parameters

Transmissão através de Videoconferência

Por João Pedro dos Santos (Universidade de Paris-Sorbonne e Universidade de Montpellier).

Differential Galois theory has the objective to study linear ODEs with the help of algebraic groups. Roughly and explicitly, to a matrix $A\in \mathrm{Mat}_n( \mathbf C(x) )$ and a differential system $y'=Ay$, we associate a subgroup of ${\rm GL_n}(\mathbf C)$, the differential Galois group, whose function is to measure the complexity of the solutions. There are three paths to this theory: Picard-Vessiot extensions (similar to standard Galois theory), monodromy representations (analytic approach)  and Tannakian categories (the categorical way).

If instead of working with complex coefficients we deal with a discrete valuation ring $R$, the construction of the differential Galois groups are less obvious and the theory of groups gives place to that of group schemes. This puts forward the Tannakian approach and relevant concepts from algebraic geometry. In this talk, I shall explain how to associate to these differential equations certain flat $R$-group schemes, what properties these may have and, most importantly, how to compute them with the monodromy. In doing so, I shall explain our method to establish a certain "Deligne-Riemann-Hilbert" theorem with "parameters" (this is in connection to recent work of Fiorot-Monteiro-Sabbah), which is a way of finding extensions of differential equations having mild singularities. I shall emphasise the case of formal differential equations for the sake of clarity.

(The talk is a horizontal report on several works done in collaboration with P. H. Hai and his students N. D. Duong and P. T. Tam over the past years.)

Transmissão em direto via Zoom (Meeting ID: 799 297 2871).

15h00
CMAFcIO - Centro de Matemática, Aplicações Fundamentais e Investigação Operacional

Workshop promovido pelo GAPsi - Gabinete de Apoio Psicopedagógico.

Talks @DI, por Marcelo Inuzuka e Ruan Rodrigues.

Mathematical Logic Webinar, por Paulo Santos (Universidade Nova de Lisboa - CMA).

Bioecosys brings together expertise and experience in forest ecosystem management and economics to communicate recent findings and to accelerate the transfer of knowledge and best practices to stakeholders and forest actors.

"Desenvolvimento da Organocatálise Assimétrica", por Amélia Pilar Reuter.

Seminário no âmbito da Unidade Curricular de Introdução aos Recursos Naturais Vegetais, por Jorge Faria (INIAV - Instituto Nacional de Investigação Agrária e Veterinária / MED - Instituto Mediterrâneo para a Agricultura, Ambiente e Desenvolvimento).

Sessão de apresentação do livro coordenado por José Almeida Silva (IDL / Ciências ULisboa).

This Workshop will gather specialists in the field of Combinatorial Optimization, from several countries, to present and discuss recent research work around this subject. Luís Gouveia, coordinator of CMAFcIO and José Rui Figueira, president of CEG-IST will present the Opening Session.

Formação de Comunicação em Ciência(s), por Cristina Nobre Soares.

Concerto de Natal | Aula Magna da Reitoria da Universidade de Lisboa.

Conversa com Galopim de Carvalho.

Workshop promovido pelo GAPsi - Gabinete de Apoio Psicopedagógico.

The training program will provide industry and academia the technological and scientific pathways required for deployment of sustainable cost-effective large scale energy storage and conversion powered by renewable energies.

"Desenvolvimento da Organocatálise Assimétrica", por Amélia Pilar Reuter.

Candidaturas a decorrer até 15 de dezembro de 2021.

This meeting is a forum for scientists of all academic levels, (from master to principal investigators), to present their work and discuss, fostering new ideas and collaborations. We invite researchers from all over the world to participate and submit their work.

Conferência do Centro Interuniversitário de História das Ciências e da Tecnologia, por Céline Paillette (Université Paris 1 Panthéon-Sorbonne).

Desde o início de 2020 que somos bombardeados, através da comunicação social, com informações sobre o novo coronavírus humano - SARS-CoV-2 - que provoca a síndrome respiratória aguda grave. Nunca se comunicou tanto, em tempo real, um tema científico, mas será que já vimos todas as nossas perguntas sobre este tema respondidas?

Candidaturas online de 01 a 31 de dezembro de 2021.

At the end of this course, participants will be able to predict the three-dimensional structure of nucleic acid and proteins from their sequence and use the acquired knowledge to improve or create new methodologies in molecular biology research and diagnostics.

The course SoilEco aims at introducing attendants to an updated state of the art of diversity of the soil biota and the functional roles played by soil organisms in key ecological processes.

Understanding plant water relations at several scales under climate change is the theme of the XV Portuguese-Spanish Water relations Symposium: “Water relations under climate change: from genes to ecosystem”.

In this course, we promote a multidisciplinary approach presenting the most recent findings on the topic and challenging the traditional way of considering symbiotic associations as exceptions and not as the rule.

The goal of this course is to provide to the participants with current and practical knowledge on urban ecology, including ecological and social aspects.

This course aims to explore ways of communicating science to non-specialized audiences, such as policy makers, industry, general public (including students and teachers), through their engagement and participation in citizen science activities.


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### Green synthesis and environmental applications of biogenic Pd nanoparticles

Tom Hennebel UGent (2010)
abstract
author
promoter
UGent and UGent
organization
alternative title
Groene synthese en milieutoepassingen van biogene Pd nanopartikels
Green synthesis and environmental applications of biogenic palladium nanoparticles
year
type
dissertation
publication status
published
subject
pages
VI, 187 pages
publisher
Ghent University. Faculty of Bioscience Engineering
place of publication
Ghent, Belgium
defense location
Gent : Het Pand
defense date
2010-09-13 17:00
ISBN
9789059893931
language
English
UGent publication?
yes
classification
D1
I have transferred the copyright for this publication to the publisher
id
1040909
handle
http://hdl.handle.net/1854/LU-1040909
date created
2010-09-10 15:31:18
date last changed
2017-01-16 10:37:56
@phdthesis{1040909,
abstract     = {Over the last 100 years, the global water consumption has doubled every 20 years to result in an average annual use of 1500 m3 per capita. Due to the increasing population, the supply of safe water will be of great significance in the future. The depletion of non-contaminated groundwater sources is placing an increasing importance on water reuse for domestic, industrial and agricultural purposes and as a result, there is an acute need for sustainable water treatment technologies. Most of the current wastewater technologies are primarily designed to remove nutrients (COD, nitrogen and phosphorus) but do not effectively remove more recalcitrant xenobiotics. Moreover, because these compounds are mostly unaffected by current wastewater treatment processes, xenobiotics are expected to accumulate in water resources over the coming decades. The halogenated compounds that belong to this class of very recalcitrant and hardly biodegradable compounds can be found in every environmental compartment, where their concentration depends on their physico-chemical characteristics, the type of application and the properties of the aqueous matrix. For example, groundwater can contain high concentrations of chlorinated solvents or pesticides as a result of spills during production or leakage from underground storage tanks. At the other end of the concentration spectrum, x-ray contrast compounds, used in hospitals for imaging, are found in surface waters at a concentration range of pg L-1 to \ensuremath{\mu}g L-1. Furthermore, there is an increasing concern about these halogenated pollutants as many of them also exert human toxicity. It is clear that a general removal strategy, which can tackle a broad spectrum of these recalcitrant halogenated compounds at both extremely low and very high concentrations in various aqueous matrices, is required.
Monometallic and bimetallic catalysts of palladium have shown their value in the dehalogenation of all kinds of pollutants. Nanoscale forms of these catalysts have demonstrated especially high activity in hydrodehalogenation reactions, presumably due to their large specific surface areas. A new biologically inspired method involves the precipitation of palladium on a bacterium to produce a nanopalladium catalyst, i.e. bio-Pd. When a hydrogen donor is provided, bacteria can reduce Pd(II) and subsequently precipitate it as Pd nanocrystals on their cell wall and in their periplasmatic space. None of the expensive or harmful compounds, currently used for the abiotic chemical synthesis of Pd nanoparticles, are required because bacteria serve as reductant for the Pd salt and as scaffold for the Pd nanoparticles. Therefore, the production of bio-Pd can be considered as a green technology.
Although the production of biogenic nanoparticles and their use as a catalyst in batch tests was previously reported, their potential application for water treatment purposes was thus far not addressed. Therefore, seven research chapters were elaborated in this work to develop reactor technologies based on the bio-Pd concept to treat aqueous streams contaminated with halogenated compounds. One important focus of this work was on methods to keep the nanoparticles separated from the reactor effluents. Indeed, Pd is a very expensive metal and its loss from reactors can largely influence the costs of the sanitation technology. Moreover, since not much is known about the toxicity of these biogenic nanoparticles, precautions need to be taken to limit their release to receiving waters. The selection of the electron donor in the dehalogenation reaction was determined as one of the main reactor parameters influencing both the degradation rates and the costs of the process. Therefore, a second focus was on limiting the sustainable supply of hydrogen gas as electron donor. The studied bio-Pd applications were not only used to treat synthetic wastewater but also groundwater from natural aquifers.
In a first reactor system (Chapters 2 and 8), the bio-Pd nanoparticles were retained by plate membranes with a pore size of 0.4 \ensuremath{\mu}m. The membranes were specifically designed to allow build up of a shallow biofilm layer at the membrane surface, allowing retention of particles with a diameter as small as 0.01 \ensuremath{\mu}m. Bio-Pd was effectively retained because the size of the bacteria and the bio-Pd nanocatalyst amount approximately to 1 \ensuremath{\mu}m and 50 nm, respectively. The bacterial carrier provided an efficient way to keep the Pd nanoparticles in suspension and in the reactor without aggregation of the Pd and hence obtaining a maximal contact between catalyst and pollutant. Removal rates up to 2.5 g TCE day-1 g-1 Pd were achieved with H2 gas as electron donor and ethane and chloride were produced as harmless endproducts.
Fixed and fluidized bed reactors with encapsulated bio-Pd can offer an economical interesting alternative to membrane reactors, which have the additional drawbacks of their price, their sensitivity towards fouling and the difficulty to treat large volumes of groundwater at high flow rates. Therefore, several polymers, which are often used for the encapsulation of bacteria, were tested to produce catalyst beads for use in fixed or fluidized bed reactors (Chapters 3 and 4). The primary objective of this research was to stabilize bio-Pd while minimizing the loss of activity in dehalogenation reactions. Polyurethane cubes empowered with bio-Pd were implemented in a lab-scale fixed bed reactor for the treatment of water containing TCE. Removal rates up to 1.06 g TCE g Pd-1 d-1 were obtained in a flow through configuration of this reactor (Chapter 3). A pilot-scale (200 L) fluidized bed reactor, supplied with bio-Pd encapsulated in alginate beads, was developed for the treatment of groundwater containing an historical pollution of lindane and chlorobenzenes (Chapter 4). Removal percentages up to 75 \% of the HCH isomers and 68 \% of the chlorobenzenes were obtained applying a hydraulic residence time of 8 hours.
The amount of hydrogen gas used in both reactor technologies (Chapter 2, 3 and 4) was much more than the stoichiometrically required amount for TCE or lindane dehalogenation. This was partially due to the low hydrogen gas solubility (0.8 mM at 20{\textdegree}C and 100 kPa) and partially due to the reactor setup. Therefore, two other reactor types, i.e. a membrane contactor and a microbial electrolysis cell (MEC) were constructed in order to limit the costs related to the hydrogen gas supply. In the membrane contactor, polyvinylidene fluoride membranes encapsulating bio-Pd were positioned between a gas compartment providing hydrogen gas and an aqueous compartment containing the halogenated contaminant. When operated in a fed batch configuration, a removal efficiency of 77\% of the iodinated contrast medium diatrizoate was obtained after a time period of 48 h. The goal of the MEC was to produce in situ hydrogen gas in a more sustainable way. When the cathode granules were coated with 5 mg bio-Pd g-1 graphite, removal rates of 151 g TCE m-3 total cathodic compartment d-1 and 48 mg TCE g-1 Pd d-1 were obtained. Chloride and ethane were the main endproducts and formation of unwanted byproducts such as vinyl chloride was not significant. When the same setup was applied for the transformation of diatrizoate, a removal rate of 12 mg g-1 Pd d-1 with effluent concentrations under the detection limit was obtained. It is expected that the use of bioelectrochemically produced hydrogen gas can decrease the costs of the hydrogen with a factor of at least 2-4.
Another sustainable source of hydrogen gas as electron donor could be microbiologically produced during the fermentation of a carbon source. It was not only shown that fermentative bacteria are capable of Pd nanoparticle formation but also of charging these nanoparticles with reactive hydrogen species (Chapters 7 and 8). In a first remediation strategy, the in situ formation of bio-Pd and the concomitant reduction of the toxic groundwater pollutant Cr(VI) was demonstrated in aquifer soil columns inoculated with Clostridium pasteurianum. In a second strategy, Citrobacter braakii was used to produce bio-Pd as catalyst and biogenic hydrogen as electron donor to remove diatrizoate. A membrane bioreactor supplied with this type of bio-Pd was capable to dehalogenate 22 mg diatrizoate mg-1 Pd over a period of 19 days before their biocatalytic activity was exhausted. This work demonstrates the general applicability of fermentative bacteria to generate catalytic Pd(0) nanoparticles and hydrogen in a one-step process. These biomass-associated nanoparticles can be used to dehalogenate iodinated contrast media and chlorinated solvents without an external source of hydrogen. The suggested one-step process can be applied in MBRs to treat aqueous streams of industrial effluent and as part of pump-and-treat systems. This work also shows the potential use of fermentative bacteria to co-generate catalytic nanoparticles and reductive hydrogen within contaminated aquifers.
This work shows the general applicability of biogenic Pd nanoparticles to treat water contaminated with halogenated contaminants at large scale. Several reactor technologies, using bio-Pd as catalyst, were developed. Additionally, a process that sustainably produces hydrogen gas as electron donor was demonstrated. Currently, the largest market for these technologies is the treatment of groundwater contaminated with halogenated solvents and sludge contaminated with PCBs. Regarding this type of pollution, a pump and treat technology based on the bio-Pd system is currently the most developed in terms of economical feasibility and size (pilot-scale). However, it is believed that bio-Pd based techniques can also play a role in the search for methods to produce reusable gray water or even safe drinking by removing very recalcitrant micropollutants (e.g. diatrizoate) from water or industrial effluents.
The most effective bio-Pd technology can be selected to fit a given contamination problem (Chapter 9). Furthermore, the selection of the bacterium for the bio-Pd production process and the reactor type can be based both on economical and green considerations. The implementation of bio-Pd into the market presents a number of challenges. Most significantly, further research is needed to optimize the amount of catalyst and electron donor required, given that both substrates significantly impact the process economics. Research will need to focus on more active catalysts such as biogenic bimetals to reduce the amount of Pd required. The research presented in this thesis, however, provides the foundation for the development of a bio-Pd based technology that is cost competitive and more effective than currently available clean-up strategies.},
author       = {Hennebel, Tom},
isbn         = {9789059893931},
language     = {eng},
pages        = {VI, 187},
publisher    = {Ghent University. Faculty of Bioscience Engineering},
school       = {Ghent University},
title        = {Green synthesis and environmental applications of biogenic Pd nanoparticles},
year         = {2010},
}

Chicago
Hennebel, Tom. 2010. “Green Synthesis and Environmental Applications of Biogenic Pd Nanoparticles”. Ghent, Belgium: Ghent University. Faculty of Bioscience Engineering.
APA
Hennebel, T. (2010). Green synthesis and environmental applications of biogenic Pd nanoparticles. Ghent University. Faculty of Bioscience Engineering, Ghent, Belgium.
Vancouver
1.
Hennebel T. Green synthesis and environmental applications of biogenic Pd nanoparticles. [Ghent, Belgium]: Ghent University. Faculty of Bioscience Engineering; 2010.
MLA
Hennebel, Tom. “Green Synthesis and Environmental Applications of Biogenic Pd Nanoparticles.” 2010 : n. pag. Print.


text:
Example 8-2-17 - Maple Help

Chapter 8: Infinite Sequences and Series

Section 8.2: Series

Example 8.2.17

 a) Show that Leibniz' theorem (see Table 8.2.2) on the convergence of alternating series applies to the  series $\sum _{n=1}^{\infty }\phantom{\rule[-0.0ex]{5.0px}{0.0ex}}\frac{{\left(-1\right)}^{n+1}}{\sqrt{n}}$.
 b) Obtain the first few, but graph the first 50, partial sums.
 c) If ${S}_{k}$ is the partial sum of the first $k$ terms, what value of $k$ will guarantee that the error in ${S}_{k}$ is no worse than ${10}^{-3}$?

For more information on Maplesoft products and services, visit www.maplesoft.com


text:
Corpus ID: 118448070

# Investigations of Double Parton Scattering: Example of $pp \rightarrow b \bar{b} \rm{jet~jet} X$

@article{Berger2011InvestigationsOD,
title={Investigations of Double Parton Scattering: Example of \$pp \rightarrow b \bar\{b\} \rm\{jet~jet\} X\$},
author={E. Berger},
journal={arXiv: High Energy Physics - Phenomenology},
year={2011}
}
• E. Berger
• Published 2011
• Physics
• arXiv: High Energy Physics - Phenomenology
Signature kinematic variables and characteristic concentrations in phase space of double parton scattering are discussed. These properties should allow the double parton contribution to $pp \rightarrow b \bar{b} \rm {jet jet} X$ at Large Hadron Collider energies to be distinguished from the usual single parton scattering contribution. A methodology is suggested to measure the size of the double-parton cross section.

#### References

SHOWING 1-4 OF 4 REFERENCES
arXiv:0912.5104 [hep-ex]] and earlier experimental references cited therein
• Collaboration], Phys. Rev. D
• 2010


text:
# Tag Info

16

To sum up what is discussed in the CFA curriculum, it discusses 3 types of spreads. They are used to compare a risky bond to a Treasury bond (assumed to be risk-free). Nominal spread Simply computes the difference between the YTM of the risk-free bond and the YTM of the risky bond. The major problem of this measure is that it doesn't take into account the ...

7

There are "perpetual" bonds and preferred shares that are traded in the corporate credit markets that exactly match your conditions above. They are recorded in the 10-K at notional value $X$. The "close-out" feature is an embedded call. You should assume your favorite stochastic interest rate (and/or credit) model and run a PDE solver, tree, or other grid ...

7

This really is an arbitrage. It is caused by differences in supply and demand between the interest cashflow and the principal cashflow and by differences in the financing rates on the two STRIPS. As you noted, the price difference is small, and it would take 30 years to guarantee convergence. In addition, the outstanding amount of the 30-year coupon strip ...

7

As with most derivatives that have early exercise, you are going to want to price this using a grid scheme. I have priced callable loans with floors using the Generalized Vasicek model at my old hedge fund, and it is fairly easy to handle. As a matter of fact my students are doing that very problem as homework this week, and my reference implementation ...

6

Well, that's still a very general question. A few elements of answer : Bonds pay interest on a regular basis, semiannual for US treasury and corporate bonds, annual for others such as Eurobonds, and quarterly for others. You need to distinguish between fixed coupon bonds, zero coupon bonds, bonds with an amortization schedule, floating rate notes based on ...

6

Q1 - Yes, debt load has an impact on the stock price. For instance, say you are valuing a company with a discounted cash flow model, while the interest won't affect the operational cash flows, it will increase the cost of capital. With that, the perceived value will be less than a similar company with less debt. Debt will also affect the volatility of the ...

6

There are certainly (short-rate) models which assume bounded interest rates. I suppose I should clarify - the design of the model prohibits negative interest rates. Further, some models asymptotically reach some target, or mean rate which is considered mean reversion, the most famous perhaps the Vasicek. Short rate models where rates cannot go negative: ...

6

Your observations are pretty much correct. The groupings are because of the fine print "Note how I have expanded the drift and volatility terms at $t = T$; in the above these are evaluated at $r$ and $T$." on the same page (p.528). Basically, $w$ is a function of both $r$ and $t$. Since we want to use $w(r,T)$ instead of $w(r,t)$, we taylor expand ...

5

In practice, I would begin with the recovery assumption. In the case of Greece, dealers are probably already quoting recovery swaps, allowing you to set this parameter directly. In general, you have to be willing to make assumptions based on history or on conversations with bankruptcy experts. Once I have the recovery assumption, I can take any ...

5

Here's a research note devoted to pricing of CMS by means of a stochastic volatility model. The authors indicate in the Introduction that an analysis of the coupon structure leads to the conclusion that CMS contracts are particularly sensitive to the asymptotic behavior of implied volatilities for very large strikes. Market CMS rates actually drive the ...

5

The Macaulay duration is a measure of how sensitive a bond's price is to changes in interest rates. Duration is related to, but differs from, the slope of the plot of bond price against yield-to-maturity. The slope of the price-yield curve is $-\frac{D}{1+r}P,$ where $D$ is Macaulay duration, $P$ is bond price, and $r$ is yield. Here's how the definition ...

5

Yes, you are correct. Duration is additive, so your aggregate portfolio duration is the weighted average of your individual durations as you present in point 2. That holds assuming a close to flat yield curve and parallel (additive) shifts. If that's not the case, the situation gets a bit more complex. Unfortunately, right now I couldn't find any ...

5

After struggling through the Pianca paper due to its poor proofing ($F$ is never defined but appears to be face value, and $n$ is implied to be the number of periods remaining but is instead maturity), I seem to have it worked out. Using the lambertW function in gsl, I have it replicated in R: # Estimate duration using various closed-form formulae # ...

5

The answer to your first four questions is affirmative. Option-adjusting the spread makes an equivalence between everything theoretically possible, but the quality of results depends significantly on the quality of your interest rate model and its calibration. My personal opinion, though, is that the results need to be treated carefully because the OAS ...

5

I'm familiar with the library, but not with the way it is exported to R. Anyway: gearings are optional multipliers of the LIBOR fixing (some bonds might pay, for instance, 0.8 times the LIBOR) and spreads are the added spreads. In your case, the gearing is 1 and the spread is 0.0140 (that is, 140 bps; rates and spread must be expressed in decimal form). ...

5

I assume that you are working in a single curve theory. While this theory used to do well, it is not adapted to today's market and — as Brian B pointed it out — you cannot get a useful information from swap rates alone. The swap rate $S(t)$ at $t$ for a given tenor $T$ and period $P$ is the fixed rate such that a swap starting at $t$ and ending at $t+T$ ...

4

If you're able to work with the results from the paper cited (Pianca, Maximum Duration of Below Par Bonds: A Closed-Form Formula), congratulations! You have the hard part done! Maximum durations for par and premium bonds are trivial. Here is a figure directly from the cited paper: Some points about the figure: the market interest rate used is ...

4

Yes, it is definitely possible to do so. With a long fixed-income portfolio, you'd typically be buying puts on treasury futures or writing calls on them (writing calls may not be feasible if you're an institutional investor due to regulatory reasons). In general, duration for long puts/short calls would be negative. However see caveats below: Typically, ...

4

One could say that a CDS price is determined by the physical default probability and the risk premium. The physical PD (PPD) is the actual probability of company defaulting within the given period of time. It is purely a theoretical concept as no one really knows what this probability is. We could estimate it using some models or credit ratings, but those ...

4

You don't say which duration, but it's generally okay to use effective duration: $$duration (eff) = \frac{-1}{P(r)}*\frac{Price(r+b) - Price(r-b)}{2*b}$$ where $r$ = rate and $b$ = yield shock. Although, to address Brian's point, the mortgage contains an embedded call option that creates negative convexity, so the three re-pricings, $P(r)$, $P(r+b)$, ...

4

Forward interest rates are negative whenever the yield curve is negatively sloped. The US term structure was inverted most recently around 2007. Hard to find bank deposits that have negative yields (find countries experiencing deflation and you may find it), however, treasury bills during recent times of financial stress have yielded a negative rate. The ...

4

A swap does not require a model because its price can be derived from the yield curve without any assumptions about how the yield curve may move in the future. The PFE however is an indication of by how much the swap's mark-to-market may move between now and a moment in the future. It is of course influenced by how volatile rates are. The more volatile ...

4

There are two flaws in the argument. The simpler one is that expectations give information about probability distributions (premise D). I think this is what John was referring to in his comment. The fact that the expectation of a forward rate in period X is Y% tells us nothing about the implied probability distribution in that period, and certainly doesn't ...

4

Quantmod package claims to support downloading data from Federal Reserve Bank of St. Louis Economic database, which contains plenty of rates time series. It should be pretty simple to get them into R using function getSymbols, in the same manner as getSymbols("DEXJPUS",src="FRED") # FX rates from FRED

4

There are two different issues at play here. One is that, of course, you want only the future cash flows to enter the calculation. This is taken care when you set the evaluation date to 6 months from today. In C++, you would say Settings::instance().evaluationDate() = today + 6*Months; I don't remember the corresponding function in QuantLibXL, but you ...

4

Generally, there are few or no zero-coupon instruments traded in the market, especially for longer maturities. However, pricing of many derivatives relies on having a zero curve, so it becomes necessary to construct one using available instruments. Aside from derivatives, one can use a zero curve fitted to liquid bonds to price new or less liquid issues.

4

I believe it's correct. However, consider that it would be easy enough, and more clear, to create a new class (at least in C++; the task is more difficult if you also want to export it to Excel). The new instrument should only inherit from Bond and implement a constructor that builds the desired cash flows via a call to FixedLeg and another to IborLeg; you ...

3

You have to look at the terms and conditions on your individual bond. The way the specifications usually work is that a call will result in accrued interest being paid, effectively making up for the lost coupon. Sometimes there's even an extra penalty. A put will result in a loss of coupon in almost all cases, and so is almost always done just after a ...

3

honestly your question is hard to understand. Are these two questions the same? "Does fitting sub-optimal option exercise strategies to market data yield better option values?" "which modeling approach leads to better predictions and better relative value measures?" I think you want to ask 1 and I think it is similar to Setting the r in put-call parity? ...

3

The risk implied by Euribor or EONIA (or their swaps) is for lending to another prime rated bank. These rate indexes represent where contributor banks are offering funds to each other in the interbank market. Contributing banks are mostly rated P-1 (Moody’s) or A-1 (S&P). You wouldn’t use these rates for govt discount curves because the risk doesn’t ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
How do you Plot the complex number z=-4sqrt3-4i?

So plot $\left(- 4 \sqrt{3} , - 4\right)$ on the rectangular graph sheet, naming y-axis


text:
# Find the derivatives F'(t) and F''(t) for each of the following complex-valued functions of the real variable t. (a) F(t)=e^{(1−i)t} (b) F(t)=e^{3 i t

Find the derivatives F'(t) and F''(t) for each of the following complex-valued functions of the real variable t.
(a)$$\displaystyle{F}{\left({t}\right)}={e}^{{{\left({1}−{i}\right)}{t}}}$$
(b)$$\displaystyle{F}{\left({t}\right)}={e}^{{{3}{i}{t}}}$$
(c) $$\displaystyle{F}{\left({t}\right)}={e}^{{{\left({2}+{3}{i}\right)}{t}}}$$

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(a)Let $$\displaystyle{F}{\left({t}\right)}={e}^{{{\left({1}-{i}\right)}{t}}}$$. Then $$F'(t)=(1-i)e^{(1-i)t}$$ and $$\displaystyle{F}'{\left({t}\right)}={\left({1}-{i}\right)}^{{{2}}}{e}^{{{\left({1}-{i}\right)}{t}}}=-{2}{i}{e}^{{{\left({1}-{i}\right)}{t}}}$$.
(b) Let $$\displaystyle{F}{\left({t}\right)}={e}^{{{3}{i}{t}}}$$. Then $$\displaystyle{F}'{\left({t}\right)}={3}{i}{e}^{{{3}{i}{t}}}$$ and $$\displaystyle{F}{''}{\left({t}\right)}={\left({3}{i}\right)}^{{{2}}}{e}^{{{3}{i}{t}}}=-{9}{e}^{{{3}{i}{t}}}$$.
(c) Let $$\displaystyle{F}{\left({t}\right)}={e}^{{{\left({2}+{3}{I}\right)}{t}}}$$. Then $$\displaystyle{F}'{\left({t}\right)}={\left({2}+{3}{i}\right)}{e}^{{{\left({2}+{3}{i}\right)}{t}}}$$ and $$\displaystyle{F}{''}{\left({t}\right)}={\left({2}+{3}{i}\right)}^{{{2}}}{e}^{{{\left({2}+{3}{i}\right)}{t}}}={\left(-{5}+{12}{i}\right)}{e}^{{{\left({2}+{3}{i}\right)}{t}}}$$.


text:
## Whitman College: David Guichard's "Calculus, Chapter 5: Curve Sketching, Section 5.2: The First Derivative Test"

Read Section 5.2 (page 107). In this reading, you will see how to use information about the derivative of a function to find local maxima and minima.


text:
# LXX International conference "NUCLEUS – 2020. Nuclear physics and elementary particle physics. Nuclear physics technologies"

Oct 11 – 17, 2020
Online
Europe/Moscow timezone

## Investigation of the dependence of the time resolution of Si detectors on the bias voltage

Oct 17, 2020, 9:25 AM
25m
Online

#### Online

Oral report Section 3. Modern nuclear physics methods and technologies.

### Speaker

Michael Mordovskoy (INR RAS)

### Description

The aim of this work was to measure the parameters of Si detectors and study their dependences on the bias voltage. It was measured the amplitude and time resolution, signal-to-noise ratio in the fast and slow channels, the value of the signal front and surface temperature for each detector.
When studying reactions on light nuclei at the SINP accelerator we use a two-arm spectrometer [1]. It consists of an ΔEE telescope of silicon surface-barrier detectors for charged particles and scintillation detectors in the other arm. The telescope determines the energy and type of charged particles, scintillation detectors separate neutron events from gamma by the pulse shape and determine the neutron energy by time of flight. In [1] a method for time calibration and measuring the time resolution of the detectors by using an accelerator is described. However, for the preliminary selection of detectors for experiment such a method is too expensive.
To accomplish this task, we used the standard α- source RSAS Ra-226. We record simultaneous emitted alpha and gamma rays from the decays. The source was located near the detectors. The measurements were carried out using the same equipment used in the experiments [1]. Alpha particles were detected by a Si detector, gamma quanta were detected by scintillation detectors based on a Hamamatsu-2083 PMT and an EJ301 or EJ315 type liquid scintillator having pre-measured characteristics, in particular the time resolution, is obviously better than that of the our Si detectors. The measurements were carried out on a large number of detectors (mainly PIPS) with an area of 20 to 100 mm$^2$ and a working layer thickness of 23 to 1000 microns, made mainly 35-40 years ago in the USSR. CANBERRA 2003BT and ORTEC142 preamplifiers were used in the measurements.
As a result of the measurements sets of dependences of the time and amplitude characteristics on the bias voltage were obtained and recommendations were developed on its choice to achieve the best resolutions.
The reported study was funded by RFBR according to the research project N
18-32-00944.
1. S. V. Zuyev et al. // Bull. Russ. Acad Sci .: Phys. 2016. V. 80. P. 232.

Fig. 1. The y axis - the change in the value of the Si detector parameters (500 μm, S - 90 mm$^2$) as a percentage of the point with a bias voltage of 40 V. a- the signal amplitude in the slow channel, b- the amplitude in the fast channel, c- is the amplitude resolution, d- time resolution, e- the fast signal front at the level of 0.1-0.9

### Presentation materials

 Mordovskoy Mikhail 17 october 9.25_9.50 section 3.pdf Mordovskoy Mikhail 17 october 9.25_9.50 section 3.ppt


text:
The fundamental group of mapping torus - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-25T06:20:34Z http://mathoverflow.net/feeds/question/65230 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65230/the-fundamental-group-of-mapping-torus The fundamental group of mapping torus Topologiee 2011-05-17T12:59:40Z 2011-05-17T13:51:31Z
Let $f\colon X\to X$ be a homeomorphism between a CW-complex $X$ and iteself.

Let $M_f=X\times [0,1]/(x,0)\sim (f(x),1)$, mapping torus of $X$ from $f$.

I want to calculate the fundamental group $\pi_1(M_f)$ of $M_f$ in terms of $\pi_1(X)$ and $f_*\colon \pi_1(X)\to \pi_1(X)$.

Are there any hint to do this?

p.s: This is not the homework problem.

This is the crossreference to this.


text:
4.Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) $$5a, 3a^2, 7a^4$$

(ii) $$2p, 4q, 8r$$

(iii) $$xy, 2x^2y, 2xy^2$$

(iv) $$a, 2b, 3c$$

(i)Given that, length = 5a, breadth = $$3a^2$$, height = $$7a^4$$

∴Volume of the box =$$l \times b \times h = 5a \times 3a^2 \times 7a^4 = 105 a^7 \;$$cu. units

(ii)Given that, length = 2p, breadth = 4q, height = 8r

∴Volume of the box =$$l \times b \times h = 2p \times 4q \times 8r = 64pqr\;$$ cu. units

(iii)Given that, length = xy, breadth =$$2x^2y$$, height = 2xy^2\)

Volume of the box =$$l \times b \times h = xy \times 2x^2y \times 2xy^2 = (1 \times 2 \times 2) \times xy \times x^2y \times xy^2 = 4x^4y^4 \;$$cu. units

(iv)Given that, length = a, breadth = 2b, height = 3c

∴Volume of the box = $$l \times b \times h = a \times 2b \times 3c = (1 \times 2 \times 3)abc = 6 abc\;$$ cu. units


text:
Class 12 Physics Modern Physics Nuclei

The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive  present with the stable carbon isotope . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ( 5730 years) of , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of  dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Solution: Suppose that the number of atoms per gram where at (initially) when its activity was decays per minute per gram. Today ie., after time (Age of once living organism), the number of C-14 atom per gram left is and shows an activity of 9 decays per minute per gram. Now, activity of a radioactive substance is given by

Now decay constant

Connecting you to a tutor in 60 seconds.


text:
# A Quick and Dirty Introduction to the Geometry of Surfaces

There are many ways to think about the geometry of a surface (using charts, for instance) but here’s a picture that is similar in spirit to the way we work with surfaces in the discrete setting. Imagine that our surface is a topological disk. Its geometry can be described via a map $$f: M \rightarrow \mathbb{R}^3$$ from a region $$M$$ in the Euclidean plane $$\mathbb{R}^2$$ to a subset $$f(M)$$ of $$\mathbb{R}^3$$:

The differential of such a map, denoted by $$df$$, tells us how to map a vector $$X$$ in the plane to the corresponding vector $$df(X)$$ on the surface. Loosely speaking, imagine that $$M$$ is a rubber sheet and $$X$$ is a little black line segment drawn on $$M$$. As we stretch and deform $$M$$ into $$f(M)$$, the segment $$X$$ also gets stretched and deformed into a different segment, which we call $$df(X)$$. Later on we can talk about how to explicitly express $$df(X)$$ in coordinates and so on, but it’s important to realize that fundamentally there’s nothing deeper to know about the differential than the picture you see here — the differential simply tells you how to stretch out or “push forward” vectors as you go from one space to another. For example, the length of a tangent vector $$X$$ pushed forward by $$f$$ can be expressed as

$\sqrt{ df(X) \cdot df(X) },$

where $$\cdot$$ is the standard inner product (a.k.a. dot product or scalar product) on $$\mathbb{R}^3$$. More generally, we can measure the inner product between any two tangent vectors $$df(X)$$ and $$df(Y)$$:

$g(X,Y) = df(X) \cdot df(Y).$

The map $$g$$ is called the metric of the surface, or to be more pedantic, the metric induced by $$f$$.

So far we’ve been talking about tangent vectors, i.e., vectors that travel “along” the surface. We’re also interested in vectors that are orthogonal to the surface. In particular, we say that a vector $$u \in \mathbb{R}^3$$ is normal to the surface at a point $$p$$ if

$df(X) \cdot u = 0$

for all tangent vectors $$X$$ at $$p$$. For convenience, we often single out a particular normal vector $$\hat{N}$$ called the unit normal, which has length one. Of course, at any given point there are two distinct unit normal vectors: $$+\hat{N}$$ and $$-\hat{N}$$. Which one should we use? If we can pick a consistent direction for $$\hat{N}$$ then we say that $$M$$ is orientable. For instance, the torus is orientable, but the Möbius band is not:

For orientable surfaces, we can actually think of $$\hat{N}$$ as a continuous map $$\hat{N}: M \rightarrow S^2$$ (called the Gauss map) which associates each point with its unit normal, viewed as a point on the unit sphere $$S^2 \subset \mathbb{R}^3$$. In fact, why not think of $$\hat{N}$$ as simply a different geometry for $$M$$? Now the map $$d\hat{N}$$ (called the shape operator) tells us about the change in normal direction as we move from one point to the other. For instance, we could look at the change in normal along a particular tangent direction $$X$$ by evaluating

$\kappa_n(X) = \frac{df(X) \cdot d\hat{N}(X)}{|df(X)|^2}.$

(The factor $$|df(X)|^2$$ in the denominator simply accounts for any “stretching out” of $$X$$ that occurs as we go from the plane to the surface.) The quantity $$\kappa_n$$ is called the normal curvature — we’ll have a lot more to say about curvature in the future.

Overall we end up with this picture, which captures the most fundamental ideas about the geometry of surfaces:

Conformal Parameterization

When we talked about curves, we introduced the idea of an isometric (a.k.a. arc-length or unit speed) parameterization. The idea there was to make certain expressions simpler by assuming that no “stretching” occurs as we go from the domain to $$\mathbb{R}^3$$. Another way of saying this is that

$|df(X)| = |X|,$

i.e., the norm of any vector $$X$$ is preserved.

For surfaces, an isometric parameterization does not always exist (not even locally!). Sometimes you simply have to stretch things out. For instance, you may know that it’s impossible to flatten the surface of the Earth onto the plane without distortion — that’s why we end up with all sorts of different funky projections of the globe.

However, there is a setup that (like arc-length parameterization for curves) makes life a lot easier when dealing with certain expressions, namely conformal parameterization. Put quite simply, a map $$f$$ is conformal if it preserves the angle between any two vectors. More specifically, a conformal map $$f: \mathbb{R}^2 \supset M \rightarrow \mathbb{R}^3$$ satisfies

$df(X) \cdot df(Y) = a \langle X, Y \rangle$

for all tangent vectors $$X, Y$$, where $$a > 0$$ is a positive scale factor and $$\langle \cdot, \cdot \rangle$$ is the usual inner product on $$\mathbb{R}^2$$. Vectors can still get stretched out, but the surface never gets sheared — for instance, orthogonal vectors always stay orthogonal:

A key fact about conformal maps is that they always exist, as guaranteed by the uniformization theorem. In a nutshell, the uniformization theorem says that any disk can be conformally mapped to the plane. So if we consider any point $$p$$ on our surface $$f(M)$$, we know that we can always find a conformal parameterization in some small, disk-like neighborhood around $$p$$. As with unit-speed curves, it is often enough to simply know that a conformal parameterization exists — we do not have to construct the map explicitly. And, as with arc-length parameterization, we have to keep track of the least possible amount of information about how the domain gets stretched out: just a single number at each point (as opposed to, say, an entire Jacobian matrix).

### One Response to “A Quick and Dirty Introduction to the Geometry of Surfaces”

1. Keenan says:

The course notes on surfaces have been updated to include a discussion of conformal maps. Like arc-length parameterizations, conformal maps are a useful tool for simplifying certain expressions and derivations.


text:
Properties

 Label 100800.df Number of curves 4 Conductor 100800 CM no Rank 0 Graph

Related objects

Show commands for: SageMath
sage: E = EllipticCurve("100800.df1")

sage: E.isogeny_class()

Elliptic curves in class 100800.df

sage: E.isogeny_class().curves

LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
100800.df1 100800dc4 [0, 0, 0, -1620300, 793798000] [2] 1572864
100800.df2 100800dc2 [0, 0, 0, -108300, 10582000] [2, 2] 786432
100800.df3 100800dc1 [0, 0, 0, -36300, -2522000] [2] 393216 $$\Gamma_0(N)$$-optimal
100800.df4 100800dc3 [0, 0, 0, 251700, 66022000] [2] 1572864

Rank

sage: E.rank()

The elliptic curves in class 100800.df have rank $$0$$.

Modular form 100800.2.a.df

sage: E.q_eigenform(10)

$$q - q^{7} - 6q^{13} + 2q^{17} + 8q^{19} + O(q^{20})$$

Isogeny matrix

sage: E.isogeny_class().matrix()

The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.

$$\left(\begin{array}{rrrr} 1 & 2 & 4 & 4 \\ 2 & 1 & 2 & 2 \\ 4 & 2 & 1 & 4 \\ 4 & 2 & 4 & 1 \end{array}\right)$$

Isogeny graph

sage: E.isogeny_graph().plot(edge_labels=True)

The vertices are labelled with LMFDB labels.


text:
# [NTG-context] Defining and using font weights

Andreas Schneider aksdb at gmx.de
Wed Nov 26 09:25:46 CET 2014

Hello,

I'm not able to figure out how I can make use of a font with "custom"
font weights.
I have a font that comes with "regular", "bold", "half-bold" and "ultra
bold".

I currently define the font like this:
\definefontfamily[myfont][rm][myfont][sc=file:MyFontRegular,bf=file:MyFontBold]

Which at least makes it possible to use the default font and \bf ...
But simply adding something like hb=... or ult=... doesn't seem to cut
it.

So how would I define this font to be able to also use the half-bold and
ultra weights? (preferably with a switch like \bf, \it, etc.)

Best regards,
Andreas


text:
# Unit 10Statistics

## Lesson 1

### Learning Focus

Understand features of a normal distribution.

Identify the effect of changing the mean or standard deviation in a normal distribution.

### Lesson Summary

In this lesson, we learned about features of a normal distribution. We learned a normal distribution is defined by the mean, which is the center of the distribution and the standard deviation, which determines the spread of the distribution. Normal distributions are represented with the rules, which describe the percent of the distribution within one standard deviation, two standard deviations, and three standard deviations, respectively, from the mean.

## Lesson 2

### Learning Focus

Interpret standardized test scores that are distributed normally.

### Lesson Summary

In this lesson, we sketched a normal distribution and used it to estimate population percentages. We found a process that helps to visualize the region included so that the rule can be used.

## Lesson 3

### Learning Focus

Understand a scale used to compare normal distributions.

Sketch distribution curves and use tables to find population percentages.

### Lesson Summary

In this lesson, we learned about the standard normal distribution and the -score, which is a method for putting all normal distributions on the same scale so they can be compared. Once a -score is computed for a given value, a -score table is used to find the population percentage to the left of the value.

## Lesson 4

### Learning Focus

Compare normal distributions.

### Lesson Summary

In this lesson, we used features of a normal distribution—including the symmetry, the mean, and the standard deviation—to determine how unusual a given event may be.

## Lesson 5

### Learning Focus

Understand methods for taking samples from a population.

### Lesson Summary

In this lesson, we learned about methods for obtaining samples from a population so that a parameter of interest can be studied. We learned that random sampling methods are preferable to other methods because they are more likely to represent the entire population.

## Lesson 6

### Learning Focus

Describe the difference between a survey, an experiment, and an observational study.

Design an investigation for a parameter of interest using appropriate study and sampling methods.

### Lesson Summary

In this lesson, we learned about three methods for investigating a research question. The methods are surveys, observational studies, and experiments. For the results of any of the investigations to be valid, researchers must use a random sampling method to select participants from the population.

## Lesson 7

### Learning Focus

Use random sampling methods and find means and proportions for random samples.

### Lesson Summary

In this lesson, we sampled a population and used the mean and proportion of the population to estimate the actual mean and proportion of the population, which are called population parameters. We learned about the Central Limit Theorem, which says that if the sample is large enough, the sample means and proportions will be normally distributed.

## Lesson 8

### Learning Focus

Find an interval that is likely to contain the population proportion from a sample.

### Lesson Summary

In the lesson, we learned to find an interval from a sample proportion that is likely to contain the population proportion using a margin of error. To use the formula, we found the sample size must be large enough to satisfy the conditions of the Central Limit Theorem.

## Lesson 9

### Learning Focus

Use simulation to find a plausible interval for the mean of a population from a sample.

### Lesson Summary

In this lesson, we found a margin of error for a sample mean, which creates a plausible interval for the population mean. We learned that to use the margin of error formula, we must have a sample size of at least 30. We also learned how to find the likelihood that a sample comes from a population with a given mean and standard deviation.

## Lesson 10

### Learning Focus

Determine if the results of an experiment could have occurred by chance.

### Lesson Summary

In this lesson, we used a simulation to determine if the results of an experiment could have occurred by chance. We found that simulation is useful not only for sampling but also for evaluating experimental results.

## Lesson 11

### Learning Focus

Use statistical reasoning to evaluate claims in the media.

### Lesson Summary

In this lesson, we evaluated statistical claims from media sources. We found that we need to pay attention to sample size and methods and compare the plausible intervals when drawing conclusions.


text:
COMPSTAT 2016: Start Registration
View Submission - COMPSTAT
A0198
Title: Illumination problems in digital images: A statistical point of view Authors:  Segolen Geffray - Universite de Strasbourg (France) [presenting]
Nicolas Klutchnikoff - Universite Rennes 2 (France)
Myriam Vimond - ENSAI (France)
Abstract: Interest is focused on a multi-dimensional signal $R$ observed on a grid in the presence of both an illumination artifact and an additional additive bounded variance centered noise $\varepsilon$. The illumination artifact consists of colour or grey level intensity variations which are seen on the sampled image but which are not present in $R$. Such an assumption is classically modelled using a function $L$ which acts multiplicatively on $R$. Our goal is to estimate $R$ from observations of a random variable $Y$ which obeys the regression model $Y=RL+\varepsilon$. We identify and propose a solution to an identifiability issue. We construct a consistent multi-step estimation procedure. We first make use of any consistent denoising method to estimate from the noisy data the gradient of the logarithm of the regression function. We assume that the artefact $L$ consists of smooth'' variations. Then we project the denoised auxiliary estimate on a finite basis of smooth'' functions. We deduce the final estimator of $R$ so that the proposed identifiability constraint is satisfied. An additional Monte Carlo computation is used to approximate a relevant integral. We derive an upper bound for the sup-norm risk of our estimator. Applications to different images are presented.


text:
# Print font name using LuaTeX

I want to print font names (full name, ps name, file name, etc) using LuaTeX's font table. It works fine previously, but recently I encountered an error message from luaotfload

Here is a minimal working example,

\documentclass{article}
\usepackage{fontspec}
% \setmainfont{Minion Pro}

\begin{document}
\directlua{
local str = font.fonts[font.current()].fullname
tex.sprint(-2, str)
}
\end{document}

This example, when compiled with LuaTeX works fine. The output PDF has the string LMRoman10-Regular, which is the default font. However, if I do not comment out the setmainfont line, that is, I load the Minion Pro font, then I got the following error message,

error: ...3/texmf-dist/tex/luatex/luaotfload/luaotfload-merged.lua:7656: attempt to perform arithmetic on local 'factor' (a nil value)
.
\newpage ...k \@nobreakfalse \everypar {}\fi \par
\vfil \penalty -\@M

The relevant line in luaotfload-merged.lua and around is,

function injections.setkern(current,factor,rlmode,x,tfmchr)
local dx=factor*x
if dx~=0 then
local bound=#kerns+1
setattr(current,a_kernpair,bound)
kerns[bound]={ rlmode,dx }
return dx,bound
else
return 0,0
end
end

When I replace the tex.sprint line with inspect, the console displays correctly Minion Pro Regular.

Does anyone has any idea what is going on here?

I am using an up-to-date TeXLive 2013 installation.

Indexing the font.fonts table can cause the data structures that have been layed out by the font loader to be overwritten. Safe access to those fonts is provided by font.getfont(). If you want to retrieve other fonts like e.g. TFM/OFM or virtual fonts, then you will have to write a wrapper like so:

local fontdata = fonts.hashes.identifiers
local my_getfont = function (id)
local f = fontdata[id]
if f then
return f
end
return font.fonts[id]
end

Where fonts.hashes.identifiers is where the fontloader keeps the font it initializes.


text:
# what metals do not react to calcium hydroxide in slovakia

#### what metals do not react to calcium hydroxide granules

-Metals like lead, copper, silver and gold do not react with water. 7 iii) Reaction with acids :- Metals react with dilute acids to form salts and hydrogen. Mg + 2HCl MgCl 2 + H 2 2Al + 6 HCl 2AlCl 3 + 3H 2 Zn + 2HCl ZnCl 2 + H 2 Fe + 2HCl FeCl 2 + H 2 The reactivity varies from metal to metal.

#### Strength of Bases Chemistry Tutorial

For a Group 2 hydroxide that is not very soluble in water, such as calcium hydroxide, Ca(OH) 2, some undissociated solid Ca(OH) 2(s) will be found floating in the water or lying at the bottom of the solution, but the species that are found as part of the solution will

#### Considering the influence of calcium on EAF dust acid leaching

is the valorization both of iron and zinc. Lead, manganese and calcium are present in interesting amounts in addition to iron and zinc. Heavy metals are present in EAF dust due to their presence in the melted scrap, and calcium is transferred into EAF

#### How do metals and non metals react with air and water …

2019/6/14· Dear Student Metals react with water and produce a metal oxide and hydrogen gas. Metal oxides that are soluble in water dissolve in it to further form metal hydroxide. But all metals do not react with water. Metal + Water → Metal oxide + Hydrogen Metal oxide

#### what metals do not react to calcium hydroxide manufacture

The calcium carbonate may be used to react with complexing acids when making a complex grease, such that the addition of calcium oxide or calcium hydroxide is not required. A method of making the composition comprising the steps of mixing the overbased calcium sulfonate and base oil, adding the calcium carbonate, adding one or more converting agents, and adding one or more complexing acids.

#### Chemical Reactions - Chemistry Encyclopedia - water, …

The very active metals react with water. For example, calcium reacts with water to form calcium hydroxide and hydrogen gas. Calcium metal has an oxidation nuer of 0, whereas Ca 2+ in Ca (OH) 2 has an oxidation nuer of +2, so calcium is oxidized. Hydrogen''s oxidation nuer changes from +1 to 0, so it is reduced.

#### METALS - Weebly

Iron do not react with cold water but rusting occur very slowly in the presence of oxygen. Red hot iron reacts very slowly with steam to produce salt and hydrogen gas. Iron + steam ----- Iron oxide + hydrogen 3Fe(s) + 4H 2 O (g) ----- Fe 3 O 4 2

#### Alkaline Earth Metals Reactions, Uses, Properties

Alkaline earth metals react in lesser tendency with water when it compares with alkali metals. Beryllium and Magnesium do not react with cold water. Magnesium reacts with steam and form magnesium oxide. But beryllium does not react with steam. Calcium, strontium and barium react with water and form hydroxides and hydrogen gas.

#### The reactivity series of metals - Obtaining and using …

2021/6/7· Metals below hydrogen in the reactivity series do not react with dilute acids, and both gold and platinum are placed below hydrogen. curriculum-key-fact Hydrogen is always given off when a …

#### Solubility of Hydroxides, Metal hydroxides …

When calcium and hydroxyl ion concentration are high (concentrated), calcium hydroxide is precipitated as a white solid. When ion concentration is low, it become a colourless solution. Completely soluble metal hydroxides of s block metals LiOH …

#### Sodium & Calcium Hydroxide | Healthfully

2011/7/8· While calcium hydroxide and sodium hydroxide are both corrosive, calcium hydroxide is less so. Both burn the skin and airways upon contact. Both are potentially lethal poisons. Get prompt medical care if you ingest either of these chemicals. Do not induce vomiting, because that will only expose more tissue to the chemical.

#### what metals do not react to calcium hydroxide size

201949-hydroxide (NaOH) and potassium hydroxide ( KOH).vi) Copper metal does not burn even on strong Potassium and Calcium react with cold wa what are calcium metals in hong kong what metals do not react to calcium hydroxide in cameroon

#### Alkaline Earth Metals Reactions, Uses, Properties

Alkaline earth metals react in lesser tendency with water when it compares with alkali metals. Beryllium and Magnesium do not react with cold water. Magnesium reacts with steam and form magnesium oxide. But beryllium does not react with steam. Calcium, strontium and barium react with water and form hydroxides and hydrogen gas.

#### Calcium + Calcium Nitrate → ? | Physics Forums

2005/2/23· I don''t believe that calcium hydroxide can be formed by simply dissolving calcium nitrate in water. And thus the point of dissolving calcium metal. You should recall the reaction where a metal dissolved in water will produce hydrogen gas and its corresponding basic oxide in this case $CaO_{(s)}$ which will react with water to form calcium hydroxide.

#### 14.3: Bases- Properties and Examples - Chemistry …

2020/8/23· Bases do not react with metals in the way that acids do. Bases react with acids to produce a salt and water. Calcium hydroxide is relatively insoluble in water, but is large enough that its solutions are basic according to the following reaction: \[\ce{Ca(OH)2

#### 10.2: Bases: Properties and Examples - Chemistry …

2021/3/24· Bases do not react with metals in the way that acids do. Bases react with acids to produce a salt and water. Calcium hydroxide is relatively insoluble in water, but is large enough that its solutions are basic according to the following reaction: \[\ce{Ca(OH)2

#### why HNO3 does not react with Al? Chemical Reactions …

2013/9/30· Ans: Al is more reactive metal. Before reaction with HNO 3 it forms an oxide layer of Al₂O₃.This oxide layer does not allow reaction of HNO₃ with Aluminium. Because of this reason Al vessel is used to store HNO₃.

#### Reactions of metals - Metals - National 5 Chemistry …

2020/8/17· When metals react with water, metal hydroxides and hydrogen gas are formed. This can be represented in different ways as shown: Word equation \[Calcium + water \to calcium\,hydroxide + …

#### Calcium Oxide - an overview | ScienceDirect Topics

SO 2 then reacts at the surface of the CaO at temperatures higher than 600°C as follows: (19)SO2 + CaO → CaSO3. Afterward, CaSO 3 is oxidized in a fast reaction as follows: (20)4CaSO3 → 3CaSO4 + CaS. Finally, calcium sulfide is also oxidized to calcium sulfate as follows: (21)CaS + 2O2 → CaSO4.

#### Revision Notes for Science Chapter 4 - Materials : metals …

Magnesium reacts slowly with hot water to form slightly soluble magnesium hydroxide. Aluminium, Zinc and Iron react with steam. Lead, Copper, Silver and Gold do not react with water. Note: Calcium and magnesium float on water as bubbles of hydrogen stick

#### Chemical properties of metals | Fun Science

Metals react with water to produce metal oxide (or metal hydroxide) and hydrogen gas. But, all metals do not react with water at equal intensity. The metals which are very reactive can react even with cold water while the other metals react with hot water or with steam.

#### Alkaline earth metal - Wikipedia

The alkaline earth metals all react with the halogens to form ionic halides, such as calcium chloride (CaCl 2 ), as well as reacting with oxygen to form oxides such as strontium oxide ( SrO ). Calcium, strontium, and barium react with water to produce hydrogen gas and their respective hydroxides (magnesium also reacts, but much more slowly), and also undergo transmetalation reactions to

#### Testing for ions – Sodium Hydroxide & Ammonia …

2014/3/19· The d block metals form metal aquo complexes, like that shown below, in water: Metal Aquo Complex. In this complex, water acts as a ‘ligand’, binding to the central metal atom. But water isn’t the only molecule that can act as a ligand; both OH- ions and ammonia can also do so. In the case of hydroxide ions, they can react with the metal

#### GCSE CHEMISTRY - How do the Alkaline Earth Metals …

as they react, unlike the group 1 metals which float. calcium + water calcium hydroxide + hydrogen. Ca (s) + 2H 2 O (l) Ca(OH) 2(s) + H 2(g) Calcium hydroxide is called slaked lime. Calcium hydroxide will dissolve a little in water to form ).

#### 10.2: Bases: Properties and Examples - Chemistry …

2021/3/24· Bases do not react with metals in the way that acids do. Bases react with acids to produce a salt and water. Calcium hydroxide is relatively insoluble in water, but is large enough that its solutions are basic according to the following reaction: \[\ce{Ca(OH)2

#### Calcium Hydroxide - an overview | ScienceDirect Topics

Calcium Hydroxide Calcium hydroxide is used as a long-term medicament to encourage calcifiion (Fig. 10.36b) and formation of a hard-tissue barrier, against which the obturating material can be compacted (Figs 10.36c, 10.37b). From: Endodontics (Fourth Edition), 2014

#### Groundwater aggressiveness assessment according to …

2010/12/3· Carbon dioxide in aggressive form could react with the calcium carbonate (CaCO 3) or calcium hydroxide (Ca(OH) 2) present in the concrete, creating soluble calcium bicarbonate Ca(HCO 3) 2. Aggressive carbon dioxide is the part of free carbon dioxide concentration present in water overstepping the concentration of CO 2 which is in equilibrium with Ca 2+ and HCO 3 − ions (Pitter …

#### reactions between metals and water or steam

The metals below hydrogen in the reactivity series don''t react with steam. Summary Metals from calcium upwards These react vigorously (and even more vigorously the higher up the series you go) with cold water to form the metal hydroxide and hydrogen.


text:
# Electrochemistry - Half Cell Reactions

1. Aug 1, 2012

### AGNuke

A Galvanic Cell is given. Write its Anode and Cathode half reactions, complete cell reaction and thus determine if the given is concentration cell.

$Pt|O_2(1\; atm)|NaO\! H(10^{-3}M)||H_2SO_4(0.5M)|O_2(1\;atm)|Pt$

I know that at anode, Oxygen will get oxidised, maybe from NaOH solution and at cathode, Oxygen will get reduced.

Here's what I think of anodic reaction
$2O\!H^-\rightarrow 2H^++O_2+4e^-$

And the mighty cathodic reaction
$O_2+4e^-+2H^+\rightarrow 2O\!H^-$

And thus, this cell is just a concentration cell (i.e. E°(cell) = 0)

Am... I right??? :uhh:

2. Aug 3, 2012

### AGNuke

Hello...??? Can please anyone answer the question that I am right with my guesswork??? This question is eating me inside out.


text:
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text:
# De Broglie Wavelength Calculator

Created by Bogna Szyk
Reviewed by Steven Wooding
Last updated: Dec 25, 2022

This de Broglie wavelength calculator will help you describe the wave-particle duality of matter. According to this principle, we can analyze light as waves, explaining its properties such as refraction. On the other hand, light must be made of particles (called photons) if we want to understand the photoelectric effect. De Broglie stated that every particle could be described as having a particular wavelength and formulated the famous de Broglie equation. If you want to know the de Broglie wavelength of an electron or a photon, keep reading!

🙋 You can read about the phenomena described above in the photoelectric effect calculator and the index of refraction calculator.

## What is the de Broglie equation?

According to de Broglie, a beam of particles of some mass can behave as a matter wave. Its wavelength is related to the mass and velocity of the particle: λ = h / (m × v),
where:

• m is the mass of the particle;
• v is the velocity of the particle, and
• h is the Planck constant, equal to 6.6261×10-34 Js.

## The units of mass

The mass of particles is usually given in kilograms, but using scientific notation since the numbers are small. In the calculator, use the drop-down menu to select the exponent of the mass. For reference, these are the value of the electron rest mass (me) and the atomic mass unit (u), which is an average of the proton and neutron masses.

1 me = 9.10938356 × 10-31 kg

1 u = 1.660538921 × 10-27 kg

## What is the de Broglie wavelength of an electron?

Let's find the de Broglie wavelength of an electron traveling at 1% of the speed of light.

1. The mass of an electron is 1 me, or
9.10938356*10-31 kg
2. The speed of this electron is
299,792,458 m/s / 100 = 2,997,924.58 m/s.
3. Multiplying the mass and speed, we obtain the momentum:
p = mv = 2.7309245 × 10-24 kg·m/s
4. Dividing the Planck constant by p, we obtain the de Broglie wavelength:
h/p = 6.6261*10-34 / 2.7309245*10-24 = 2.426 × 10-10 m.
5. Hence, the de Broglie wavelength of our electron is 0.24 nm.

## How to calculate the de Broglie wavelength of a photon?

Since a photon's rest mass is zero, you might be wondering how to use our de Broglie wavelength calculator to find the de Broglie wavelength of a photon. Thanks to wave-particle duality, a photon does have a momentum associated with it. The value of this momentum can be used to calculate the de Broglie wavelength of the photon.

A question asks you to find the de Broglie wavelength of a photon which has a momentum of 6.8 × 10-35 kg·m/s. Let's see how to use our calculator to answer this question.

1. Click on the unit changer menu of the momentum row and select x10-35 from the list.
2. Enter 6.8 into the momentum row. You've now entered the correct value for the momentum of the photon.
3. You will see the result for the wavelength in nanometers by default. As it is a huge value, you'll probably want to change the units to something more manageable, such as meters. The result is 9.74 m, rounding to 3 significant figures. This type of rounding is described in the significant figures calculator.

## FAQ

### What is the unit of the de Broglie wavelength?

The unit in which the de Broglie wavelength is expressed is meters. Since the de Broglie wavelength is usually very, very small, we most often express this value in nanometres.

### How do I calculate the de Broglie equation?

To determine the de Broglie wavelength of a particle given its mass and velocity, you need to:

1. Multiply the velocity by mass. Their product is the particle's momentum.
2. Divide Planck's constant by the momentum found in Step 1.
3. The result you've got is exactly the de Broglie wavelength of your particle. Congrats!
Bogna Szyk
Rest mass (m)
×10⁻³¹
kg
Velocity (v)
m/s
Momentum (p)
×10⁻²⁴
kg·m/s
Wavelength (λ)
nm
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# Line Segments equation

• September 27th 2007, 07:05 PM
Mr.Jiggyfly
Line Segments equation
How would I go about doing this one:

http://img171.imageshack.us/img171/7426/diagram2ey3.png
Find XY if WX = 1/3(XY)
• September 27th 2007, 07:41 PM
Jhevon
Quote:

Originally Posted by Mr.Jiggyfly
How would I go about doing this one:

http://img171.imageshack.us/img171/7426/diagram2ey3.png
Find XY if WX = 1/3(XY)

No big deal here:

$WX = x + 2$, $XY = 5x + 2$

since $WX = \frac 13 XY$, we have:

$x + 2 = \frac 13 (5x + 2)$

now solve for $x$, and plug that value into the formula for $XY$ to find its length

this is my 39:):)th post!!!