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36th DPS Meeting, 8-12 November 2004
Session 40 Satellite Formation and Origins
Poster II, Thursday, November 11, 2004, 4:15-7:00pm, Exhibition Hall 1A

## [40.02] Runaway Growth and Isolation of Planetary Embryos

S. J. Weidenschilling (Planetary Science Institute), F. Marzari (Universita di Padova), D. R. Davis (Planetary Science Institute)

It is often assumed that runaway growth yields planetary embryos with the limiting mass imposed by the restricted 3-body problem. A small planetesimal can encounter a massive embryo their initial orbit separation is less than 2 \sqrt{3} Hill radii. If the embryo accretes all particles within this distance, the isolation mass" Miso = 0.0021 R3 \sigma 3/2 MEarth , where R = heliocentric distance (AU) and \sigma = surface density (g cm-2). However, deviations from the idealized problem can alter this outcome, e.g., collisions among small bodies, migration due to gas drag, and mutual scattering and mergers among embryos. We use a multizone accretion code to model embryo growth to 5 My in the region 0.5 - 4 AU. Parameters and processes are varied to determine effects of planetesimal size, surface density profile, fragmentation, gas drag, and tidal interactions of embryos with a gaseous disk. Although Miso increases with R, embryos grow larger inside 2 AU due to more effective drag-induced migration of planetesimals and embryo mergers, producing bodies several times Miso. At larger R these processes are less effective, resulting in embryo masses ~1027 g nearly independent of R. The time of onset of runaway varies inversely with planetesimal size and increases with R; if initial planetesimals are 10 km in size runaway is not complete in the asteroid zone in 5 My. Fragmentation moderately decreases the final masses and spacings of embryos. The standard model of tidal migration results in rapid loss of embryos into the Sun, leaving insufficient mass to produce terrestrial planets. Either the migration rate is slower than predicted, or gas was depleted from the inner nebula on a timescale ~ 1 My. Supported by NASA PGG; computations at NCCS.

Bulletin of the American Astronomical Society, 36 #4
© 2004. The American Astronomical Soceity.

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# Chapter 15: Multiple Integrals - Section 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 900: 8

-6

#### Work Step by Step

$\int^{\sqrt{2}}_0 \int^{3y}_0 \int^{8-x^2-y^2}_{x^2+3y^2}$ dz dx dy =$\int^{\sqrt{2}}_0 \int^{3y}_0(8-2x^2-4y^2)$ dx dy =$\int^{\sqrt{2}}_0 [8x-\frac{2}{3}-4xy^2]^{3y}_0$ dy =$\int^{\sqrt{2}}_0 (24y-18y^3-12y^3)dy$ =$[12y^2-\frac{15}{2}y^4]^{\sqrt{2}}_0$ =24-30 =-6

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# How to set the width of the left border with JavaScript?

HTMLJavascriptProgramming Scripts

To set the width of the left border in JavaScript, use the borderLeftWidth property. Set the width under this property that you want for the border. You can try to run the following code to learn how to set the width of the left border −

## Example

Live Demo
<!DOCTYPE html>

Demo Text

<button type = "button" onclick = "display()">Change left border width</button>

Published on 26-Feb-2018 13:34:48


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# integrating with respect to parameters in beta function

I would like to evaluate an integral: $$\int_t^1\frac{1}{B(1+s\phi,1+\phi(1-s))}p^{s\phi}(1-p)^{\phi(1-s)}ds,$$ where $B(a,b)$ is a beta function and $p\in(0,1)$ and $\phi>0$ are some parameters. Notice that the integration is with respect to $s$, which appears both in the exponent and in the beta function. This is integrating over a family of pdfs of beta distributions, indexed by the parameters in beta.

As an analogy with this question, I would expect the integral to be connected to $B(t\phi,1+\phi(1-t))$ distribution, presumably, equal to the incomplete beta function evaluated at $p$: $$B(p;t\phi,1+\phi(1-t)).$$

However, I have no idea whether this intuition is correct and how to prove it.

Edit: apparently, this integral has sth do to with $\nu$ function. However, I have not found any formula to help me with the problem.

Edit2: since I got no hints, I am thinking of a workaround. In principle, I do not need to caculate the integral, I just need to determine for which $t$ we have: $$\int_t^1\frac{1}{B(1+s\phi,1+\phi(1-s))}p^{s\phi}(1-p)^{\phi(1-s)}ds>c,$$ for some predefined $c\in[0,1]$. Maybe there is a way to derive it without explicit formula for the integral.

## migrated from math.stackexchange.comNov 21 '16 at 13:26

This question came from our site for people studying math at any level and professionals in related fields.


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# External Direct Product Inverses

## Theorem

Let $\struct {S \times T, \circ}$ be the external direct product of the two monoids $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.

Let:

$s^{-1}$ be an inverse of $s \in \struct {S, \circ_1}$

and:

$t^{-1}$ be an inverse of $t \in \struct {T, \circ_2}$.

Then $\tuple {s^{-1}, t^{-1} }$ is an inverse of $\tuple {s, t} \in \struct {S \times T, \circ}$.

### General Result

Let $\displaystyle \left({S, \circ}\right) = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_n, \circ_n}\right)$.

Let $\left({x_1, x_2, \ldots, x_n}\right) \in S$.

Let $y_k$ be an inverse of $x_k$ in $\left({S_k, \circ_k}\right)$ for each of $k \in \N^*_n$.

Then $\left({y_1, y_2, \ldots, y_n}\right)$ is the inverse of $\left({x_1, x_2, \ldots, x_n}\right) \in S$ in $\left({S, \circ}\right)$.

## Proof

Let:

$e_S$ be the identity for $\struct {S, \circ_1}$

and:

$e_T$ be the identity for $\struct {T, \circ_2}$.

Also let:

$s^{-1}$ be the inverse of $s \in \struct {S, \circ_1}$

and

$t^{-1}$ be the inverse of $t \in \struct {T, \circ_2}$.

Then:

 $\displaystyle \tuple {s, t} \circ \tuple {s^{-1}, t^{-1} }$ $=$ $\displaystyle \tuple {s \circ_1 s^{-1}, t \circ_2 t^{-1} }$ $\displaystyle$ $=$ $\displaystyle \tuple {e_S, e_T}$ $\displaystyle$ $=$ $\displaystyle \tuple {s^{-1} \circ_1 s, t^{-1} \circ_2 t}$ $\displaystyle$ $=$ $\displaystyle \tuple {s^{-1}, t^{-1} } \circ \tuple {s, t}$

Thus the inverse of $\tuple {s, t}$ is $\tuple {s^{-1}, t^{-1} }$.

$\blacksquare$


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This functions creates a simple barplot that visually displays how much variance in the outcome (e.g., the regression coefficient) different analytical choices or combinations therefor account for. To use this approach, one needs to estimate a multilevel model that includes all analytical choices as grouping variables (see examples and vignettes). This function uses icc_specs() to compute the intraclass correlation coefficients (ICCs), which provides the data basis for the plot (see examples).

plot_variance(model)

Arguments

model a multilevel model that captures the variances of the specification curve (based on the data frame resulting from run_specs).

a ggplot object.


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By | September 25, 2020

CBSE Class 11 Biology

NCERT Solutions
Chapter 05
Structural organization in plants and animals

1. What is meant by modification of root? What type of modification of root is found in the
(a) Banyan tree
(b) Turnip
(c) Mangrove trees

Ans. Main function of the roots is to absorb water and minerals from the soil. However, roots are also modified to perform various other functions. The roots of some plants act as storage sites for food, some provide support to massive plant structures, while others absorb oxygen from the atmosphere.
Roots and its modifications in various plants:
(a) Banyan tree : Prop roots.
(b) Turnip : Storage of food.
(c) Mangrove tree : Pneumatophores.

2. Justify the following statements on the basis of external features
(i) Underground parts of a plant are not always roots
(ii) Flower is a modified shoot

Ans. (i) Various parts of plants are modified into underground structures to perform various functions such as stems, leaves, and even fruits. The stems in ginger and banana are underground and swollen due to storage of food. They are called rhizomes. Similarly, corm is an underground stem in Colocasia and Zamin-khand. The tips of the underground stem in potato become swollen due to the accumulation of food and forms tuber. Tubers bear eyes, which are subtended by a leaf scar. Basal leaves in onions become fleshy because of the accumulation of food. In peanuts, the flower after fertilization gets pushed inside the soil by growing a flower stalk. The formation of fruits and seeds takes place inside the soil.

(ii) During the flowering season, the apical meristem gives rise to the floral meristem. The axis of the stem gets condensed, while the internodes lie near each other. Instead of leaves, various floral appendages arise from the node. Therefore, it can be said that the flower is a modified shoot.

3. How is pinnately compound leaf different from palmately compound leaf?
Ans.

 Pinnately Compound Leaf Palmately Compound Leaf 1. In a pinnately compound leaf, several leaflets are present on a common axis. 2. Leaflets are borne in pairs on an elongated axis giving a feather-like appearance. 3. The leaflet-bearing axis is the continuation of the petiole or modified mid-rib. It is prominent. Example: Neem leaves 1. In a palmately compound leaf, several leaflets are attached at a common point. 2. Leaflets are attached at the tip of the petiole giving the appearance of the palm of the hand. 3 .The leaflet-bearing axis is very short and represents the tip of the petiole. It is not prominent. Example: Cotton leaves

4. Explain with suitable examples the different types of phyllotaxy?
Ans. Phyllotaxy is of three types, alternate, opposite, and whorled phyllotaxy.
In alternate phyllotaxy, a single leaf arises from the node of a branch. This type of phyllotaxy is observed in the sunflower, mustard, and peepal.

Plants with opposite phyllotaxy have two leaves arising from the node in opposite directions. It is found in guava and jamun plants.

Plants with whorled phyllotaxy have three or more leaves arising from the node. It is found in Alstonia.

5. Define the following terms:
(a) Aestivation
(b) Placentation
(c) Actinomorphic
(d) Zygomorphic
(e) Superior ovary
(f) Perigynous flower
(g) Epipetalous Stamen

Ans. (a) Aestivation: The term ‘aestivation’ refers to the mode in which sepals or petals are arranged in a floral bud with respect to other floral members. There are four types of aestivation in plants i.e., valvate, twisted, imbricate, and vexillary.
(b) Placentation: The term ‘placentation’ refers to the arrangement of ovules within the ovary of a flower. It is primarily of five types, namely marginal, basal, parietal, axile, and free central.
(c) Actinomorphic: Actinomorphic flowers can be divided into two radial halves by any radial plane passing through its centre. Examples of these flowers include chilly and mustard.
(d) Zygomorphic: Zygomorphic flowers are those flowers which can be divided into two similar halves by a single vertical plane. Examples of these flowers include pea and beans.
(e) Superior ovary: Superior ovary flowers are those flowers in which the gynoecium is present at the highest position, while other floral parts are arranged below it. A flower with this arrangement is described as hypogynous. Examples include brinjal and mustard.
(f) Perigynous flower: In perigynous flowers, the gynoecium is present in the centre and the rest of the floral parts are arranged at the rim of the thalamus at the same level. Examples include plum and rose.
(g) Epipetalous Stamen: Epipetalous stamens are stamens attached to the petals. They are found in brinjal.

6. Differentiate between
(a) Racemose and cymose inflorescence
(b) Fibrous roots and adventitious roots
(c) Apocarpous and syncarpous ovary

Ans. (a) Racemose and cymose inflorescences:

 Racemose Cymose 1. The terminal growing point of the inflorescence axis continues to grow indefinitely. 2. The inflorescence axis is monopodial. 3. The number of flowers is not restricted in an inflorescence. 4 .The arrangement of flowers is acropetal and centripetal, i.e. the outer and the basal flowers are old and central or the upper flowers are young. 1. The terminal growing point of the inflorescence axis always terminates into a flower. 2. The inflorescence axis is sympodial or multipodial. 3 .The number of flowers is usually restricted in an inflorescence. 4. The arrangement of flowers is basipetal and centrifugal, i.e. the outer or the basal flowers are young and the central or upper flowers are old.

 Fibrous root Adventitious root In monocotyledonous plants, the primary root is short lived and is replaced by a large number of roots. These roots originate from the base of the stem and constitute the fibrous root system. Example: Wheat In some plants, roots arise from parts of the plant other than the radicle and are called adventitious roots. Examples: Grass,Monstera, banyan tree

(c) Apocarpous and syncarpous ovaries:

 Apocarpous ovary Syncarpous ovary Apocarpous ovary means two or more carpels which are free. Examples: Lotus, rose The syncarpous ovary means two or more carpels which are fused. Examples: Mustard, tomato

7. Draw the labelled diagram of the following:
(i) Gram seed
(ii) V.S. of maize seed

Ans.

8. Describe modifications of stem with suitable examples
Ans. Stems of various plants have undergone modifications to perform different functions.
Underground stems or storage stems: Examples: Rhizomes, Corms, tubers In ginger and banana, the underground stem is called a rhizome. The underground stem in Colocasia (arvi) is known as corm. Rhizomes and corms are underground stems, modified for the storage of food. Also, these stems help in vegetativereproduction of these plants. The tips of the underground stem in potato plants become swollen due to the accumulation of food. The potato is a tuber that helps in the storage of food and bears eyes on it. Subtended by a leaf scar, these eyes bear buds that give rise to new plants.
Supporting stem: The stem in some weak plants bear thin, slender, and spirally-coiled structures called tendrils that help the plant get attached to nearby structures for support. Tendrils are found in cucumbers, melons, and other members of the family Cucurbitaceae.
Protective stems : The stem in bougainvillea and citrus plants (like lemon and orange) bear sharp, pointed structures called thorns, which provide protection to the plant from herbivores.
Photosynthetic stems: The stem in the Opuntia is green. It carries out the process of photosynthesis in the absence of leaves.
Others stem modifications : In some plants, underground stems such as grasses spread in the soil and help in perenation. These stems are called runners.
The short lateral stem called the offset in some aquatic plants (such as Eichhornia) bears leaves and tufts of roots at the node and gives rise to new plants.

9. Take one flower each of families Fabaceae and Solanaceae and write its semi-technical description. Also draw their floral diagrams after studying them.
Ans. (1) Family Fabaceae/Papilionaceae (pea plantFabaceae/Papilionaceae is a sub-family of the Leguminoseae family.
Vegetative features: Habit: Pinnately compound, alternately arranged with leaf tendrils with the pulvinus present at the leaf base along folacious stipules.
Root: Tap root system with root nodules.
Floral features:
Inflorescence: Racemose, generally axial than terminal Flower: Zygomorphic and bisexual flowers are found
Calyx: It contains five sepals which are gamosepalous while aestivation is imbricate. Corolla: It contains five petals (polypetalous) with vexillary aestivation. Androecium: It consists of ten anthers that are diadelphous with dithecous anthers. Gynoecium: Monocarpellary superior ovary which is unilocular with marginal placentation.
Fruit: Legume pod with non-endospermic seeds
Floral formula:

Economic importance: Peas are used as vegetables for making various culinary preparations.

(2) Flowers of Solanum nigrum
Family Solanaceae
Vegetative features:
Habit: Erect, herbaceous plant
Leaves: Simple, exstipulate leaves with reticulate venation Stem: Erect stem with numerous branches.
Floral features:
Inflorescence: Solitary and axillary Flowers: Actinomorphic, bisexual flowers
Calyx: Calyx is composed of five sepals that are united and persistent. Aestivation is valvate.
Corolla: Corolla consists of five united petals with valvate aestivation. Androecium: It consists of five epipetalous stamens.
Gynoecium: It consists of bicarpellary syncarpous superior ovary with axile placentation.
Fruits: Berry
Seeds: Numerous, endospermous
Floral formula:
Economic importance: Peas are used as vegetables for making various culinary preparations.

10. Describe the various types of placentations found in flowering plants.
Ans. Placentation refers to the arrangement of ovules inside the ovary. It is of five basic types.
(A) Marginal placentation:
The ovary in which the placenta forms a ridge along the ventral suture of the ovary and the ovules develop on two separate rows is known to have marginal placentation. This type of placentation is found in peas.

(B) Parietal placentation
When the ovules develop on the inner walls of the ovary, the ovary is said to have parietal placentation.

(C) Axile placentation

In axile placentation, the placenta is axial and ovules are attached to it. Examples include China rose, lemon, and tomato.

(D) Basal placentation
The ovary in which the placenta develops from its base and a single ovule is found attached to the base is said to have basal placentation. It is found in marigold and sunflower.

(E) Free central placentation
In free central placentation, the ovules develop on the central axis while the septa are absent. This type of placentation is found in Dianthus and primrose.

11. What is a flower? Describe the parts of a typical angiosperm flower?
Ans. A flower can be defined as the reproductive unit of any flowering plant. Flowers carry out sexual reproduction in angiosperms. A typical flower is a modified stem with a condensed axis. A flower has four different parts i.e., the calyx, corolla, androecium, and gynoecium. Androecium and gynoecium represent the male and female reproductive organs of a flower (respectively). Bisexual flowers are those which contain both androecium and gynoecium, while unisexual flowers contain either gynoecium or androecium. The corolla and the calyx are generally distinct, but may sometimes be fused (called perianth). A flower that contains all four floral parts is called a complete flower.
Parts of flowers
(A) The calyx forms the outermost whorl of a flower, which contains sepals. They are green, leaf–like structures that cover and protect the flowers during the bud stage. When the sepals of a flower are free, they are called polysepalous, while fused sepals of a flower are called gamosepalous.
(B) The corolla of a flower is a layer that lies inside the calyx. It contains beautifully coloured petals, which help in attracting insects for pollination. When the petals are free, they are called polypetalous, while fused petals are called gamopetalous.
(C) The androecium or the stamen is the male reproductive part of a flower. It consists of two parts, the filament and the bilobed anther. The bilobed anther is the site for meiosis and the generation of pollen grains.
(D) Gynoecium represents the female reproductive part of a flower. It consists of an ovary. The ovary is connected by a long tube (called style) to the stigma. The ovary bears numerous ovules attached to the placenta.

12. How do the various leaf modifications help plants?
Ans. The main function of the leaves is to carry out the process of photosynthesis. However, in a few plants, leaves are modified to perform different functions.
(a) Tendrils: The leaves of a pea plant are modified into tendrils that help the plant in climbing.
(b) Spines: The leaves in cactus are modified into sharp spines that act as an organ of defense.
(c) Phyllode: The leaves of some Australian acacia are short-lived and soon replaced by flattened, green structures called phyllodes that arise from the petiole of the leaves. The petioles in these plants synthesize food.
(d) Pitcher: The leaves of the pitcher plant are modified into pitcher-like structures, which contain digestive juices and help in trapping and digesting insects.

13. Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.
Ans. Inflorescence is the manner in which the flowers are arranged on the flowering axis. During the flowering season, the vegetative apex of the stem gets converted into a floral meristem. Based on whether the floral axis continues to grow or end in a flower, inflorescence is classified into racemose and cymose. In racemose inflorescence, the floral axis continues to grow and produces flowers laterally. On the other hand, in cymose inflorescence, the main axis terminates into a flower. Hence, it is limited in growth.

14. Write the floral formula of an actinomorphic bisexual, hypogynous flower with five united sepals, five free petals. Five free stamens and two united carpals with superior ovary and axile placentation.
The floral formula of the described flower is represented as:

Actinomorphic flowers are represented by the symbol

$\oplus$

.
A bisexual flower is indicated by

$\underset{+}{\stackrel{↗}{O}}$

The calyx contains five united sepals which can be represented as

${K}_{\left(5\right)}$

The androecium consists of five free stamens and is represented by

${A}_{\left(5\right)}$

The gynoecium consists of a superior ovary with two united carpels and axile placentations, which can be represented as

${\underset{_}{G}}_{\left(2\right)}$

The calyx contains five united sepals which can be represented as K(5). The corolla consists of five free petals and it represented as C5.
The androecium consists of five free stamens and is represented by A5.
The gynoecium consists of a superior ovary with two united carpels and axile placentations, which can be represented as

15. Describe the arrangement of floral members in relation to their insertion on thalamus?
Ans. Based on the position of the calyx, corolla, and androecium (with respect to the ovary on the thalamus), the flowers are described as hypognous, perigynous, and epigynous.
In hypogynous flowers, the ovary occupies the highest position on the thalamus while other floral parts are situated below it. In such flowers, the ovary is superior e.g., China rose, mustard etc.
In perigynous flowers, the ovary is situated at the centre and other floral parts are arranged on the rim of the thalamus. The ovary here is said to be half inferior e.g., plum, rose, peach
In epigynous flowers, the thalamus grows around the ovary fusing with its wall. The other floral parts are present above the ovary. Hence, the ovary is said to be inferior e.g., flowers of guava and cucumber.


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# First Derivative

1. Mar 24, 2015

### nazmulislam

Hi,

I have some points say, 100 points which come from a periodic tube profile, i.e., (z,r), where z and r are the axial and radial coordinates, respectively.

Now, I need to calculate the first derivative at each point.

Cheers

2. Mar 24, 2015

### Staff: Mentor

Can you post a sketch of the setup? And you want to calculate the derivatave of what with respect to what? Can you show some sample data?

Is this for schoolwork, or is a data set from your work?

3. Mar 24, 2015

### nazmulislam

Hi, I have given the profile below

I want to calculate dh/dz. Some data are below:

z h
-1 1
-0.8 0.823664
-0.6 0.714683
-0.4 0.714683
-0.2 0.823664
0 1
0.2 1.176336
0.4 1.285317
0.6 1.285317
0.8 1.176336
1 1

The data set from my work.

Thanks

#### Attached Files:

• ###### Profile.pdf
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86.6 KB
Views:
71
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2.6 KB
Views:
84
4. Mar 24, 2015

### Staff: Mentor

It looks to me like your data do not match your graph...

But in any case, are you aware of the definition of the derivative? Or implementations in discrete datasets?

One way to do it is:

$$h'(z) =\frac{h(z+1)-h(z-1)}{2z}$$

Okay, LaTeX isn't working for me tonight... In clear text:
Note from Mark44: It's fixed now @Berkman -- there was an extra brace that I removed.

h'(z) = ( h(z+1) - h(z-1) ) / 2z

Where 2z is the distance between the 2 bracketing datapoints (in the units of z). Does that make sense?

Last edited by a moderator: Mar 25, 2015
5. Mar 24, 2015

### nazmulislam

Yes, I understand the facts. Many thanks.

Cheers

6. Mar 25, 2015

### Integral

Staff Emeritus
Numeric derivatives are notoriously bad, so take your results with a grain of salt. A better approach would be to attempt to fit your data to a sinusoidal function. Then get your derivatives from that.


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• # question_answer 9) A bar magnet of length 10 cm and having pole strength equal to ${{10}^{-3}}$ A x m is kept in a magnetic field of $4\pi \,\,\times \,\,{{10}^{-}}^{3}$ Tesla. It makes an angle $30{}^\circ$ with the direction of the field. The torque acting on the magnet is $\left( newton\text{ }\times \text{ }metre \right)$- A) $2\pi \,\,\times \,\,{{10}^{-}}^{7}$                 B) $2\pi \,\,\times \,\,{{10}^{-}}^{5}$C) $0.5$                           D) $\,50$

$M=mL$ $={{10}^{-}}^{3}\,\,\times \,\,0.1$ $={{10}^{-4}}$ $Torque\,\tau \,\,=\,\,MB\text{ }\sin 30{}^\circ$ $={{10}^{-4}}\times 4\pi \times {{10}^{-3}}\times \frac{1}{2}$ $=\,\,\,2\pi \times {{10}^{-}}^{7}\,N-m$


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# Icy tensions slow glacial collateralisation talks

The stand-off over the rules for non-US collateralisation continues. Marc Jones looks at the situation

The deadlock continues in the ongoing debate over collateralisation rules for foreign companies that wish to participate in the US (re)insurance market.

At the moment, as US law now stands, foreign (re)insurers cannot trade in the US market without posting 100% collateral to cover their liabilities, a rule that has been attacked in Europe as being unfair.

Although the National Association of Insurance Commissioners (NAIC) has released the results of the last meeting into the debate over this issue, progress towards any kind of a resolution of the debate remains glacial, with no sign of an agreement on the subject any time soon.

Unfortunately, agencies and companies in Europe and the UK have been protesting against the law for some years now without making much progress and substantial obstacles remain in the path of progress.

The biggest supporter of the current state of US law is the US (re)insurance industry itself. According to the Reinsurance Association of America (RAA), there are a number of reasons why the present system should stay in place.

The RAA, which along with the American Insurance Association and the National Association of Mutual Insurance Companies (NAMIC) is a member of the US Coalition to Protest Unauthorised Reinsurance Collateral, has even listed a number of these reasons on its website, citing just why the 100% collateral requirement is so important.

The first and biggest reason for the Coalition is a claim by NAIC that there is a $26bn 'credit risk' affect for US insurers and their customers. It points out that the NAIC insolvency committee says that without the collateral: "Insurers or receivers would first have to litigate in the US; and then seek to enforce the judgements abroad to collect the money. No treaty to enforce judgments exists between the US and other jurisdictions. The current US collateral requirements ensure that reinsurance can be collected when a US judgement is entered. NAIC's insolvency committee has unanimously voted to keep the collateral requirements and has documented problems in collecting judgements abroad." It also claims that collateral protects consumers and prevents larger guarantee-fund assessments, ensures reinsurance collectibility since judgements in foreign countries may not be honoured and is consistent with free trade since the US collateral requirement is not a barrier to entry into the market. In addition, the Coalition claims that state law allows non-US reinsurers a choice of how to do business with US insurers, with many of the former taking advantage of those choices. Business can be transacted in three ways: through a US subsidiary or branch, through a multiple-beneficiary trust fund or by posting collateral for each reinsurance transaction. In fact, according to the Coalition, more not less collateral is likely needed as the amount owed by reinsurers is subject to wide variations over time. It points out that 100% funding does not allow any margin for error, so collateral requirements should not be reduced. It adds that reducing collateral will not reduce reinsurance rates and that reducing collateral does not guarantee increased capacity. The Coalition's position is supported by further pressure from the US domestic market. The Alliance of American Insurers, the American Association of State Compensation Insurance Funds, the National Association of Independent Insurers and others including the RAA and NAMIC, which represent approximately 85% of the property and casualty premium written by more than 2,000 primary insurers in the United States, have taken a broad-front approach to this issue and have openly appealed to NAIC to keep the existing rules. In a joint letter sent to NAIC, the above organisations said that: "We believe that security for primary insurers is the foremost issue in the Working Group's consideration of the alien reinsurers' proposal to diminish collateral requirements. Departure from full collateralisation of risks written by alien reinsurers brings potential disadvantage to primary insurers, who are promised no reciprocal advantage in compensation of alien reinsurers and who may additionally become liable to make whole the estates of failed companies in which uncollateralised reinsurance was uncollectible." On the other hand, the London-based International Underwriting Association (IUA) disagrees with these fears. Although in 2003, Stephen Cane, chairman of the IUA said that: "Collateral requirements can represent an extraordinary burden. As an industry, we must be extremely careful about imposing them without first thinking through all possible implications," the IUA has been pressing for reform of the US collateralisation rules for almost four years. Two other organisations that have been pressuring for a relaxation of the rules are the Comite Europeen des Assurances (CEA) and Lloyd's of London. The CEA is especially keen for some form of resolution on this point as it points out that the US is the European Union's major trading partner and that an agreement is crucial as the two combined make up some 70% of the world insurance market. The CEA's stance is almost directly opposite that of the US Coalition. In a statement released by the CEA in December 2004, it said that the current US reinsurance rules treat non-US reinsurers proving reinsurance protection to US cedants less favourably than US reinsurers and therefore places non-US reinsurers "at a significant competitive disadvantage". It added that: "Under the current US rules, the funds of non-US reinsurers must be reserved at 100% of the liabilities assumed from US insurers (together with a surplus of$20m, or in the case of Lloyd's, £100m). These funds have to be localised in the US. Non-US reinsurers' liabilities are therefore funded 'gross' and no credit may be taken for the retrocession protection that they purchase, even if this is from US-based reinsurers and even when such assets are accorded full-value on their own balance sheets.

"The US rules also require non-US reinsurers to hold their funds as regulatory deposits in semi-static trust accounts in the US. US reinsurers, on the other hand, are permitted to reserve their liabilities on a 'net' basis, i.e. taking credit for their own reinsurance protection. US reinsurers have no obligation to hold funds in trust and, paradoxically, are permitted to hold their statutory assets in foreign accounts."

In direct contrast to the Coalition, the CEA claims that the current rules drive reinsurance rates up as they inhibit the efficient use of capital to support the efficient spread of risk and have a significant frictional cost impact on pricing, with the costs involved ultimately being borne by the consumer. According to the CEA, it is estimated that US collateral requirements amount to £45bn for non-US reinsurers and cost around £500m a year, resulting from the loss on yield of 20-25 basis points, bank charges of 75 basis points and overheads of between 10-15 basis points.

The CEA therefore wants the NAIC credit for reinsurance laws "radically and quickly reformed" to at least substantially reduce or, better still, abolish all collateral requirements for non-US reinsurers that are both financially strong and subject to high-quality domestic regulatory regimes.

It also feels that the European Union should bring more pressure onto the US on this issue.

Lloyd's of London has also been pressing hard on this issue. Chairman of Lloyd's Lord Levene has spoken out freely about the importance of changes to the current system. In one speech, made in 2003 he said: "Currently, we have around \$9bn tied up in funds in the US. We are required to deposit cash regardless of the amount or quality of reinsurance protection we have bought from other companies. This represents massive and needless over-funding of our liabilities.

"This funding burden cannot be sustained without a cost. Ultimately, that cost is passed on in the price insurance companies are charged for reinsurance cover. Insurance companies in turn must charge policyholders, so the price of these requirements are often borne by the ordinary policyholder."

"Demanding that all foreign reinsurers, whatever their size, strength or claims-paying record, lodge 100% collateral in the US, does not account for the fact that carriers like Lloyd's ... are already subject to detailed regulatory oversight in their home country."

In a counter-offensive, the IUA has joined forces with the CEA to propose a compromise, by which foreign reinsurers would be able to voluntarily apply to be included on a list of reinsurers that would be maintained by NAIC. In order to qualify for this list, reinsurers would have to prove to US regulators that they had sufficient capital resource, management expertise and integrity. This would involve making detailed financial filings, being submitted to meaningful US regulatory scrutiny and coming under the jurisdiction of the US courts. Those able to make it onto the list would be able to fund their liabilities to US cedants at less that the 100% demanded at the moment, down to a minimum of 50%.

The IUA/CEA proposals have so far met with a cool reception and the Coalition has urged the US market to turn down the plan. It claims that the proposals lack specific details, such as which standards are to be used to measure the adequacy of assets, what the required level of regulation is, which financial reporting requirements are used, who ensures compliance and which provisions provide for fair competition. As a result of US opposition, the proposals appear to be stuck firmly in neutral.

However, a little progress has been made. The National Conference of Insurance Legislators (NCOIL) has voted to consider the proposals, while NAIC's reinsurance task force is also looking at the matter. Although neither the IUA/CEA nor the Coalition has changed their positions in the meantime, the two are still debating the matter.

However, there are a few signs that times might be changing. The issue of collateralisation is too important to fade away and, sooner or later, it will have to be resolved. Furthermore, recent moves towards a harmonised reinsurance regulatory regime in Europe may strengthen the European negotiating position and give it a little more leverage in the debate.


text: To a degree, that statement is correct, but it is not true that $$\sqrt{x^2} = x$$. Subtract the similar radicals, and subtract also the numbers without radical symbols. Any exponents in the radicand can have no factors in common with the index. No, you wouldn't include a "times" symbol in the final answer. Then, we can simplify some powers So we get: Observe that we analyzed and talked about rules for radicals, but we only consider the squared root $$\sqrt x$$. This theorem allows us to use our method of simplifying radicals. Radicals ( or roots ) are the opposite of exponents. Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. Julie. Since I have two copies of 5, I can take 5 out front. Simplifying square roots review. Just to have a complete discussion about radicals, we need to define radicals in general, using the following definition: With this definition, we have the following rules: Rule 1.1:    $$\large \displaystyle \sqrt[n]{x^n} = x$$, when $$n$$ is odd. Is the 5 included in the square root, or not? Get your calculator and check if you want: they are both the same value! 1. root(24) Factor 24 so that one factor is a square number. Quotient Rule . The expression " katex.render("\\sqrt{9\\,}", rad001); " is read as "root nine", "radical nine", or "the square root of nine". This website uses cookies to improve your experience. One rule is that you can't leave a square root in the denominator of a fraction. These date back to the days (daze) before calculators. Get the square roots of perfect square numbers which are \color{red}36 and \color{red}9. If you notice a way to factor out a perfect square, it can save you time and effort. You'll usually start with 2, which is the … This theorem allows us to use our method of simplifying radicals. 72 36 2 36 2 6 2 16 3 16 3 48 4 3 A. How could a square root of fraction have a negative root? Oftentimes the argument of a radical is not a perfect square, but it may "contain" a square amongst its factors. Free Radicals Calculator - Simplify radical expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. To indicate some root other than a square root when writing, we use the same radical symbol as for the square root, but we insert a number into the front of the radical, writing the number small and tucking it into the "check mark" part of the radical symbol. Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify. Rule 2:    $$\large\displaystyle \sqrt[n]{xy} = \sqrt[n]{x} \sqrt[n]{y}$$, Rule 3:    $$\large\displaystyle \sqrt[n]{\frac{x}{y}} = \frac{\sqrt[n]{x}}{\sqrt[n]{y}}$$. Just as the square root undoes squaring, so also the cube root undoes cubing, the fourth root undoes raising things to the fourth power, et cetera. It's a little similar to how you would estimate square roots without a calculator. 0. Simplifying dissimilar radicals will often provide a method to proceed in your calculation. Another way to do the above simplification would be to remember our squares. You don't want your handwriting to cause the reader to think you mean something other than what you'd intended. For instance, 3 squared equals 9, but if you take the square root of nine it is 3. 1. (Technically, just the "check mark" part of the symbol is the radical; the line across the top is called the "vinculum".) In other words, we can use the fact that radicals can be manipulated similarly to powers: There are various ways I can approach this simplification. + 1) type (r2 - 1) (r2 + 1). Simplifying simple radical expressions IntroSimplify / MultiplyAdd / SubtractConjugates / DividingRationalizingHigher IndicesEt cetera. Special care must be taken when simplifying radicals containing variables. The radicand contains no fractions. To simplify radical expressions, we will also use some properties of roots. We use the fact that the product of two radicals is the same as the radical of the product, and vice versa. Simplify the following radicals. Example 1. After taking the terms out from radical sign, we have to simplify the fraction. Step 2 : If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. Often times, you will see (or even your instructor will tell you) that $$\sqrt{x^2} = x$$, with the argument that the "root annihilates the square". 1. root(24) Factor 24 so that one factor is a square number. When doing this, it can be helpful to use the fact that we can switch between the multiplication of roots and the root of a multiplication. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. Quotient Rule . I can simplify those radicals right down to whole numbers: Don't worry if you don't see a simplification right away. Break it down as a product of square roots. Take a look at the following radical expressions. Let us start with $$\sqrt x$$ first: So why we should be excited about the fact that radicals can be put in terms of powers?? The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. Generally speaking, it is the process of simplifying expressions applied to radicals. No radicals appear in the denominator. Learn How to Simplify Square Roots. The index is as small as possible. Step 3 : Simplify each of the following. All right reserved. Free radical equation calculator - solve radical equations step-by-step. 1. That is, the definition of the square root says that the square root will spit out only the positive root. 1. (Much like a fungus or a bad house guest.) Here are some tips: √50 = √(25 x 2) = 5√2. ANSWER: This fraction will be in simplified form when the radical is removed from the denominator. How to simplify radicals . Special care must be taken when simplifying radicals containing variables. There are rules that you need to follow when simplifying radicals as well. There are five main things you’ll have to do to simplify exponents and radicals. Generally speaking, it is the process of simplifying expressions applied to radicals. Here’s the function defined by the defining formula you see. I used regular formatting for my hand-in answer. ... Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify. On the other hand, we may be solving a plain old math exercise, something having no "practical" application. Since I have only the one copy of 3, it'll have to stay behind in the radical. Some radicals do not have exact values. Concretely, we can take the $$y^{-2}$$ in the denominator to the numerator as $$y^2$$. This calculator simplifies ANY radical expressions. By using this website, you agree to our Cookie Policy. Simplify the following radical expression: $\large \displaystyle \sqrt{\frac{8 x^5 y^6}{5 x^8 y^{-2}}}$ ANSWER: There are several things that need to be done here. There are lots of things in math that aren't really necessary anymore. One rule is that you can't leave a square root in the denominator of a fraction. And take care to write neatly, because "katex.render("5\\,\\sqrt{3\\,}", rad017);" is not the same as "katex.render("\\sqrt[5]{3\\,}", rad018);". 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text:
AIMS Mathematics, 2017, 2(4): 586-609. doi: 10.3934/Math.2017.4.586

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• Citation and Abstract

Permutational behavior of reversed Dickson polynomials over finite fields II

School of Mathematics and Information, China West Normal University, Nanchong 637009, P.R.China

## Abstract    Full Text(HTML)    Figure/Table

In this paper, we study the special reversed Dickson polynomial of theform $D_{p^{e_1}+...+p^{e_s}+\ell, k}(1,x)$, where $s,e_1, ..., e_s$are positive integers, $\ell$ is an integer with $0 ≤ \ell < p$. In fact, by using Hermite criterion we first give an answer to the questionthat the reversed Dickson polynomials of the forms $D_{p^{s}+1, k}(1,x)$,$D_{p^{s}+2, k}(1,x)$, $D_{p^{s}+3, k}(1,x)$, $D_{p^{s}+4, k}(1,x)$,$D_{p^{s}+p^{t}, k}(1,x)$ and $D_{p^{s}+p^{t}+1, k}(1,x)$ are permutationpolynomials of ${\mathbb F}_{q}$ or not. Finally, utilizing the recursiveformula of the reversed Dickson polynomials, we represent$D_{p^{e_1}+...+p^{e_s}+\ell, k}(1,x)$ as the linear combinationof the elementary symmetric polynomials with the power of $1-4x$being the variables. From this, we present a necessary and sufficient conditionfor $D_{p^{e_1}+...+p^{e_s}+\ell, k}(1,x)$ to be a permutation polynomialof ${\mathbb F}_{q}$.
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# References

1. K. Cheng, Permutational Behavior of Reversed Dickson Polynomials over Finite Fields, AIMS Math., 2 (2017), 244-259.

2. R. Coulter, Explicit evaluation of some Weil sums, Acta Arith., 83 (1998), 241-251.

3. S. Hong, X. Qin andW. Zhao, Necessary conditions for reversed Dickson polynomials of the second kind to be permutational, Finite Fields Appl., 37 (2016), 54-71.

4. X. Hou, G. L. Mullen, J.A. Sellers and J.L. Yucas, Reversed Dickson polynomials over finite fields, Finite Fields Appl., 15 (2009), 748-773.

5. R. Lidl and H. Niederreiter, Finite Fields, second ed., Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge, 20, 1997.

6. X. Qin and S. Hong, Constructing permutation polynomials over finite fields, Bull. Aust. Math. Soc., 89 (2014), 420-430.

7. X. Qin, G. Qian and S. Hong, New results on permutation polynomials over finite fields, Int. J. Number Theory, 11 (2015), 437-449.

8. Q. Wang and J. Yucas, Dickson polynomials over finite fields, Finite Fields Appl., 18 (2012), 814-831.


text:
# 4.10: Formulas

The last kind of variable that I want to introduce before finally being able to start talking about statistics is the formula. Formulas were originally introduced into R as a convenient way to specify a particular type of statistical model (see Chapter15) but they’re such handy things that they’ve spread. Formulas are now used in a lot of different contexts, so it makes sense to introduce them early.

Stated simply, a formula object is a variable, but it’s a special type of variable that specifies a relationship between other variables. A formula is specified using the “tilde operator” #~#. A very simple example of a formula is shown below:62

formula1 <- out ~ pred
formula1
## out ~ pred

The precise meaning of this formula depends on exactly what you want to do with it, but in broad terms it means “the out (outcome) variable, analysed in terms of the pred (predictor) variable”. That said, although the simplest and most common form of a formula uses the “one variable on the left, one variable on the right” format, there are others. For instance, the following examples are all reasonably common

formula2 <-  out ~ pred1 + pred2   # more than one variable on the right
formula3 <-  out ~ pred1 * pred2   # different relationship between predictors
formula4 <-  ~ var1 + var2         # a 'one-sided' formula

and there are many more variants besides. Formulas are pretty flexible things, and so different functions will make use of different formats, depending on what the function is intended to do.

This page titled 4.10: Formulas is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Danielle Navarro via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.


text:
# How do you write an equation of a line with slope -15/11 and y intercept (0,-9)#?

Dec 26, 2016

$y = - \frac{15}{11} x - 9$

#### Explanation:

To write this equation we will use the slope-intercept formula.

The slope-intercept form of a linear equation is:

$y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

In this problem we are give:

$\textcolor{red}{m} = - \frac{15}{11}$

and

$\textcolor{b l u e}{b} = - 9$

Substituting these into the formula gives:

$y = \textcolor{red}{- \frac{15}{11}} x + \textcolor{b l u e}{- 9}$

$y = \textcolor{red}{- \frac{15}{11}} x - \textcolor{b l u e}{9}$


text:
qscqesze
Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

# B. Tokitsukaze and Mahjong

time limit per test1 second
memory limit per test256 megabytes

Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from 1 to 9). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, …, 9m, 1p, 2p, …, 9p, 1s, 2s, …, 9s.

In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand.

Do you know the minimum number of extra suited tiles she needs to draw so that she can win?

Here are some useful definitions in this game:

A mentsu, also known as meld, is formed by a koutsu or a shuntsu;
A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu;
A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu.
Some examples:

[2m, 3p, 2s, 4m, 1s, 2s, 4s] — it contains no koutsu or shuntsu, so it includes no mentsu;
[4s, 3m, 3p, 4s, 5p, 4s, 5p] — it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu;
[5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] — it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu.
Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite.

## Input

The only line contains three strings — the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from 1 to 9 and the second character is m, p or s.

## Output

Print a single integer — the minimum number of extra suited tiles she needs to draw.

input
1s 2s 3s
output
0
input
9m 9m 9m
outputCopy
0
inputCopy
3p 9m 2p
outputCopy
1

## Note

In the first example, Tokitsukaze already has a shuntsu.

In the second example, Tokitsukaze already has a koutsu.

In the third example, Tokitsukaze can get a shuntsu by drawing one suited tile — 1p or 4p. The resulting tiles will be [3p, 9m, 2p, 1p] or [3p, 9m, 2p, 4p].

## 代码

#include<bits/stdc++.h>
using namespace std;

pair<int,int> form(string s){
pair<int,int> p;
p.first=int(s[0]-'0');
if(s[1]=='m'){
p.second=0;
}else if(s[1]=='p'){
p.second=1;
}else{
p.second=2;
}
return p;
}
bool check(vector<pair<int,int> >v){
sort(v.begin(),v.end());
for(int i=0;i<v.size();i++){
for(int j=i+1;j<v.size();j++){
for(int k=j+1;k<v.size();k++){
if(v[i].second==v[j].second&&v[j].second==v[k].second){
if(v[i].first==v[j].first&&v[j].first==v[k].first){
return true;
}
if(v[i].first+1==v[j].first&&v[j].first+1==v[k].first){
return true;
}
}
}
}
}
return false;
}
int main(){
vector<pair<int,int> >v;
for(int i=0;i<3;i++){
string s;
cin>>s;
v.push_back(form(s));
}
if(check(v)){
cout<<"0"<<endl;
return 0;
}
for(int i=1;i<10;i++){
for(int j=0;j<3;j++){
vector<pair<int,int> >d = v;
d.push_back(make_pair(i,j));
if(check(d)){
cout<<"1"<<endl;
return 0;
}
}
}
cout<<"2"<<endl;
return 0;
}
posted @ 2019-07-29 01:29  qscqesze  阅读(207)  评论(0编辑  收藏


text:
# Design Contracts for Cyber-Physical Systems: Making Timing Assumptions Explicit

### http://www.eecs.berkeley.edu/Pubs/TechRpts/2012/EECS-2012-191.pdf

Building Cyber-Physical Systems (CPS) involves several teams of engineers with different backgrounds. We focus on interactions between control engineers and embedded software engineers. Lack of rigorous methodologies for communication and interaction between the two groups has been recognized as an obstacle to the development of dependable and cost-effective CPS. We advocate the use of {\em design contracts} as a step towards bridging this gap. Design contracts act as a medium for communication and interaction between the control and computation/communication design problems in a CPS. The contracts provide a focal point for negotiation and for decision making. Once design contracts have been established, they help to separate the global CPS design problem into two sub-problems, each of which can be tackled separately by the control-engineering and the embedded software teams, thus making the design more tractable and efficient. We propose a framework of design contracts encompassing (i) concepts relevant to timing constraints and functionality; (ii) a formulation of some popular design approaches as design contracts; and (iii) a process and guidelines on how to choose, derive and employ design contracts. The guidelines place specific emphasis on modeling and simulation support.

BibTeX citation:

@techreport{Toerngren:EECS-2012-191,
Author = {Toerngren, Martin and Tripakis, Stavros and Derler, Patricia and Lee, Edward A.},
Title = {Design Contracts for Cyber-Physical Systems: Making Timing Assumptions Explicit},
Institution = {EECS Department, University of California, Berkeley},
Year = {2012},
Month = {Aug},
URL = {http://www.eecs.berkeley.edu/Pubs/TechRpts/2012/EECS-2012-191.html},
Number = {UCB/EECS-2012-191},
Abstract = {Building Cyber-Physical Systems (CPS) involves several  teams of engineers with different backgrounds. We focus on interactions between control engineers and embedded software engineers. Lack of rigorous methodologies for communication and interaction between the two groups has been recognized as an obstacle to the development of dependable and cost-effective CPS. We advocate the use of {\em design contracts} as a step towards bridging this gap. Design contracts act as a medium for communication and interaction between the control and computation/communication design problems in a CPS. The contracts provide a focal point for negotiation and for decision making. Once design contracts have been established, they help to separate the global CPS design problem into two sub-problems, each of which can be tackled separately by the control-engineering and the embedded software teams, thus making the design more tractable and efficient.

We propose a framework of design contracts encompassing
(i) concepts relevant to timing constraints and functionality;
(ii) a formulation of some popular design approaches as design contracts; and
(iii) a process and guidelines on how to choose, derive and employ design contracts. The guidelines place specific emphasis on modeling and simulation support.}
}

EndNote citation:

%0 Report
%A Toerngren, Martin
%A Tripakis, Stavros
%A Derler, Patricia
%A Lee, Edward A.
%T Design Contracts for Cyber-Physical Systems: Making Timing Assumptions Explicit
%I EECS Department, University of California, Berkeley
%D 2012
%8 August 21
%@ UCB/EECS-2012-191
%U http://www.eecs.berkeley.edu/Pubs/TechRpts/2012/EECS-2012-191.html
%F Toerngren:EECS-2012-191


text:
I recently started working on a managed wrapper for the Terminal Services API, and as my C++/CLI is a bit rusty I ran into some issues which I'm sure are common when trying to handle the impedance mismatch between the managed and unmanaged worlds.

I'm going to take a look at one of those issues here, and that is using System::String with native functions. The Win32 API is one such body of native functions and they fairly consistently take LPTCHAR parameters for strings. This type is a typedef to TCHAR*, TCHAR in turn is a typedef to wchar_t for Unicode builds and char otherwise.

## Toll-free bridge with CString

One easy and automatic way to do this conversion is to bridge it through CString, this is a type that is part of Microsoft's ATL. I believe it was first part of MFC but it's since been divorced from depending on the rest of MFC and even the ATL proper or it's server classes; include atlstr.h instead of cstringt.h.

The char type of CString is internally based on TCHAR as well, so depending on if you are doing a Unicode build the internal representation of the CString will be either wchar_t or char. The CLI has no such type differentiation, String is always a wide character string. This means that in Unicode builds we don't need to do a conversion, but non-Unicode builds will, and CString takes care of this for us through it's helpful conversion constructor that accepts a System::String^.

 template <class SystemString> CStringT( SystemString^ pString ) : CThisSimpleString( StringTraits::GetDefaultManager() ) { cli::pinptr<const System::Char> pChar = PtrToStringChars( pString ); const wchart *psz = pChar; *this = psz; }

PtrToStringChars retrieves a pointer to the String's internal memory buffer, no copy here. The pointer returned is then pinned so that the Garbage Collector will not move it while we use it. It then uses an implicit conversion to go from pin_ptr<const System::Char> to const wchar_t*. Finally it uses the copy assignment operator of CString (which delegates to a base class operator) to copy the contents of the String buffer into itself. In a Unicode build this copy assignment operator decays to a basic memory copy, otherwise a different copy assignment operator is invoked that uses WideCharToMultiByte to convert to the CStrings internal char type.

 // This gets called when the right-hand-side pszSrc is the same char type as the internal storage // It simply delegates to it's base class to copy the source buffer into itself CStringT& operator=( Inoptz PCXSTR pszSrc ) { CThisSimpleString::operator=( pszSrc );

 return( *this ); } // This gets called when the right-hand-side pszSrc has a different char type than the internal storage // It converts to the internal storage char type directly into it's internal buffer CStringT& operator=( Inoptz PCYSTR pszSrc ) { // nDestLength is in XCHARs int nDestLength = (pszSrc != NULL) ? StringTraits::GetBaseTypeLength( pszSrc ) : 0; if( nDestLength > 0 ) { PXSTR pszBuffer = GetBuffer( nDestLength ); StringTraits::ConvertToBaseType( pszBuffer, nDestLength, pszSrc); ReleaseBufferSetLength( nDestLength ); } else { Empty(); } return( *this ); }

static void ConvertToBaseType(Outcap(nDestLength) _CharType* pszDest, _In int nDestLength, Incount(nSrcLength) const wchart* pszSrc, In int nSrcLength = -1) throw() { // nLen is in XCHARs ::WideCharToMultiByte(_AtlGetConversionACP(), 0, pszSrc, nSrcLength, pszDest, nDestLength, NULL, NULL); }

Now we can use the CString in place of LPCTSTR parameter because it has a user-defined conversion operator that simply returns the guts of the CString. This automatically does-the-right-thing when calling functions in Unicode builds and non-Unicode builds, all with no #ifdefs.

 operator PCXSTR() const throw() { return( m_pszData ); }

PCXSTR is a chain of typedefs that eventually, through TCHAR, finds it's way to a const wchar_t*. PCXSTR means "pointer to a const null-terminated string of the same char type I am", it also has PXSTR and XCHAR typedefs. In addition to the X typedefs it also has a set of Y typedefs (PCYSTR, PYSTR, YCHAR) that map to the opposite of the X typedefs, if the X is wchar_t then the Y is char. It uses the set of Y typedefs to create a number of copy assignment operators and conversion constructors that convert from either character type to the internal type.

## Conversion to Mutlibyte UTF8

There is one problem that this doesn't cover however, and that is how to convert your Unicode String into a const char* for use as parameters to functions that don't have Unicode counterparts. Stan Lippman has a blog post from back in 2004 where he presents a couple of functions to handle this conversion to char* and to std::string. We'll see some similarities with the CString constructors.

 bool ToCharStar( String^ source, char*& target ) { pinptr<const wchar_t> wch = PtrToStringChars( source ); int len = (( source->Length+1) * 2); target = new char[ len ]; return wcstombs( target, wch, len ) != -1; }

bool Tostring( String^ source, string &target ) { pinptr<const wchar_t> wch = PtrToStringChars( source ); int len = (( source->Length+1) * 2); char *ch = new char[ len ]; bool result = wcstombs( ch, wch, len ) != -1; target = ch; delete ch; return result; }

There are a couple problems with these functions. First, they are not reentrant and hence not thread safe, wcstombs keeps a global internal state during the conversion of a string. Second, they only work with UTF-16, which String always is, but it over allocates for simple single-byte character sets like ASCII. Third, they create an unecissary temporary buffer in the case of the std::string converter. And in the case of the char* converter it put the onus of freeing the buffer on the caller, which is doubly dangerous here because since the function uses new[] to allocate the buffer, the caller needs to know to call delete[].

Lets take a whack at implementing these functions in the way of the C++ standard library while solving these deficiencies (thanks to Kniht on freenode ##C++ for helping me distill this).

 include <limits.h> include <wchar.h> include <algorithm> include <stdexcept> struct ConversionError : std::runtimeerror { ConversionError() : std::runtimeerror("ConversionError") {} explicit ConversionError(std::string const& what) : std::runtime_error("ConversionError: " + what) {} protected: struct NoPrefix {}; ConversionError(NoPrefix, std::string const& what) : std::runtime_error(what) {} }; template<class OutIter> struct mboutputt : std::iterator<wchart,void,void,void,void> { mboutput_t(OutIter out) : _mbstate(), _out(out) {} mboutputt& operator++() { return *this; } mboutputt& operator++(int) { return this; } mboutput_t& operator () { return *this; } void operator=(wchart wc) { char buf[MBLENMAX]; int len = ::wcrtomb(buf, wc, &mbstate); if (len == -1) { throw ConversionError("wcrtomb"); } _out = std::copy(buf, buf + len, _out); } mbstate_t _mbstate; OutIter _out; }; template<class OutIter> mboutputt<OutIter> mboutput(OutIter out) { return mboutputt<OutIter>(out); } template<class Cont> mboutputt<std::backinsertiterator<Cont> > mbbackinserter(Cont& c) { return mboutput(std::backinserter(c)); }

template<class Cont> void wcstomb(Cont& c, wchart const* s) { if (s) { std::copy(s, s + wcslen(s), mbback_inserter(c)); } }

This code is completely standards compliant so it will work perfectly well on any standards compliant compiler/OS. In our examples however our source is a System::String and there are two targets we're interested in, std::string and char*. We can accomplish each of these targets easily using std::copy with our custom mbbackinserter.

 template<class Cont> void Stringtomb(Cont& c, System::String^ source) { pinptr<const wchart> wch = ::PtrToStringChars( source ); wcstomb(c, wch); }

 void ctest(const char* str) { std::cout << str << std::endl; } void stest(const std::string& str) { std::cout << str << std::endl; } void convtest() { String^ sstr = L"Hello!"; std::vector<char> charStar; String_to_mb(charStar, sstr); // vector<char> can be used as a char* by passing &charStar[0] to a function taking char* ctest(&charStar[0]); std::string s; String_to_mb(s, sstr); // the std::string can be used as-is or it can also produce a char* by calling string::c_str() stest(s); ctest(s.c_str()); // the vector and string are automatically deallocated leaving this scope

}

Making effective use of containers, iterators, and wcrtomb(...) we've created a solution that doesn't require caller deallocation and that is reentrant.


text:
# $\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}=1$

How can I prove it without using Taylor series, L'Hopital's rule, or integrals,or series expansion? I don't know how. Please help.

First I have tried to use epsilon-delta, I also tried to use sequences but I failed.

• Are you allowed to use L'Hopital's Rule? – Kavi Rama Murthy Dec 14 '19 at 5:42
• No. @KaboMurphy – New2Math Dec 14 '19 at 5:43
• What is your definition of the logarithm? – Martin R Dec 14 '19 at 6:09
• Then how do you want to prove it without using Taylor series? Note that the definition of $\exp$ as a Taylor series easily implies $\lim_{y \to 0}\frac{e^y-1}{y} = 0$, and that implies your desired limit. – Martin R Dec 14 '19 at 6:19
• I suggest that you clarify your question with the necessary information (such as how log/exp are defined). Requiring “without using Taylor series” does not make much sense if the definitions are based on Taylor series. – Also note that your question should be clear without the title. – Martin R Dec 14 '19 at 6:45

The Inequalities section of Wikipedia's "List of Logarithmic Identities" page states

$$\frac{x}{\sqrt{1+x+x^2/12}} \le \ln(1 + x) \le \frac{x}{\sqrt{1+x}} \tag{1}\label{eq1A}$$

for $$0 \le x$$, and the reverse for $$-1 \lt x \le 0$$. Dividing by $$x$$ and then having $$x \to 0^{+}$$ as shown, and with the reverse inequalities for $$x \to 0^{-}$$, gives $$1$$ on either side in both cases. Thus, by the Squeeze Theorem, you have

$$\lim_{x \to 0}\frac{\log(1+x)}{x} = 1 \tag{2}\label{eq2A}$$

• Does not help me – New2Math Dec 14 '19 at 6:10
• @New2Math This answers your question as you've written it. Since it doesn't help you, I suggest you reword your question to be more explicit about what you are willing to, and not willing to, accept for an answer. Thanks. – John Omielan Dec 14 '19 at 6:14
• It is too difficult – New2Math Dec 14 '19 at 6:16

Since $$\mathop {\lim }\limits_{x \to \pm \infty } \left( {1 + \frac{1} {x}} \right)^x = e$$ you have that $$\mathop {\lim }\limits_{x \to \pm \infty } x\log \left( {1 + \frac{1} {x}} \right) = 1$$ Now, let be $$t=\frac{1}{x}$$. You have that $$\mathop {\lim }\limits_{t \to 0} \frac{1} {t}\log \left( {1 + t} \right) = 1$$

• We have not talked about this identity of e – New2Math Dec 14 '19 at 6:17

Still calculus approach, just no integrals or explicit L'Hopital:

Let $$\varphi(x)=\log(x+1)-x$$ for $$x>0$$, then $$\varphi'(x)<0$$ for $$x>0$$, so $$\varphi(x)\leq\varphi(0)=0$$, we get $$\dfrac{\log(1+x)}{x}\leq 1$$ for $$x>0$$.

On the other hand, let $$\varphi(x)=\log(x+1)-x+x^{2}/2$$ for $$0, then $$\varphi'(x)>0$$ for $$0, so $$\dfrac{\log(x+1)}{x}\geq 1-\dfrac{1}{2}x$$ for $$0.

Finally, one concludes by Squeeze Theorem.

For negative part, we exploit the trick like the following: \begin{align*} \lim_{x\rightarrow 0^{-}}\dfrac{\log(x+1)}{x}&=\lim_{y\rightarrow 0^{+}}\dfrac{\log(1-y)}{-y}\\ &=\lim_{y\rightarrow 0^{+}}\dfrac{-\log(1-y)}{y}\\ &=\lim_{y\rightarrow 0^{+}}\dfrac{\log\left(\dfrac{1}{1-y}\right)}{y}\\ &=\lim_{y\rightarrow 0^{+}}\dfrac{\log\left(1+\dfrac{y}{1-y}\right)}{y}\\ &=\lim_{y\rightarrow 0^{+}}\dfrac{\log\left(1+\dfrac{y}{1-y}\right)}{\dfrac{y}{1-y}}\dfrac{1}{1-y}\\ &=1\cdot 1\\ &=1. \end{align*}

• What is $\varphi'$ I also don't know that $e= \lim (n+1/n)^n$ – New2Math Dec 14 '19 at 6:13
• I didn't use the limit fact. And $\varphi'$ is the derivative of $\varphi$, I use the derivative criterion to show the monotonicity of the function $\varphi$. – user284331 Dec 14 '19 at 6:27
• But we haven't talked about derivatives yet – New2Math Dec 14 '19 at 6:29
• Then I really don't know how to help, sorry. – user284331 Dec 14 '19 at 6:29

Since you have stated you know that exp(x) = $$\Sigma_{n=0}^{\infty}\frac{x^n}{n!}$$

Let $$\log(1+x) = a \implies e^a-1=x$$

as $$x \to 0 \ , a\to0$$

Now, the question rephrases to $$\lim_{a \to 0}\frac{a}{e^a-1}$$ , Now using the expansion of exp(a) the limit evaluates to 1.

Hope this helps.

In the comments you appear to accept that $$\lim_{y\to0}\frac{e^y-1}y=1.$$ As $$x\to0$$, $$1+x\to1$$ and so $$\ln(1+x)\to0$$. Therefore setting $$y=\ln(1+x)$$ gives $$1=\lim_{y\to0}\frac{e^y-1}y=\lim_{x\to0}\frac{(1+x)-1}{\ln(1+x)} =\lim_{x\to0}\frac{x}{\ln(1+x)}$$ etc.

Hint For $$x >0$$ whenever when $$n \leq \frac{1}{x} you have $$\log (1+\frac{1}{n+1})^n\leq \frac{\log(x+1)}{x} \leq \log (1+\frac{1}{n})^{n+1}$$

Use the fact that

$$\lim_n (1+\frac{1}{n+1})^n= \lim_n (1+\frac{1}{n})^{n+1}=e$$

Edit With your definition of $$e$$.

Hint 1: Use your definition of $$e$$ to show that $$\lim_{x \to 0} \frac{e^x-1}{x} =e$$

Hint 2 If you meet it, apply the formula for $$(f^{-1})'(0)$$. If not, calculate $$\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}$$ via the substitution $$x=e^y-1$$. Then, you get $$\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}=\lim_{y\rightarrow 0}\frac{\log(e^y-1+1)}{e^y-1}$$

• The last fact is not written in the book. We have defined e as exp(1) where exp(x)= $\sum x^n/n!$ we also know that exp(x)= $\sum^n x^n/n!+ R_{n+1}$ with $|R_{n+1}|\leq 2|x^{n+1}/(n+1!)|$ with $x\leq 1+0.5n$ now help me please. I had to write down everything on mobile phone because my computer is broken – New2Math Dec 14 '19 at 6:01
• @New2Math The usual definition of $e$ is the one in my post. I will correct the answer to use your definition. – N. S. Dec 14 '19 at 18:27


text:
# Posts by Tags

## Simulating Poisson Networks Efficiently

Published:

When I first started my Ph.D. I learned about modeling large wireless network using point processes. They allow you to obtain tractable mathematical models that scale well with the size of the network. The most commonly used model in wireless networks is the homogeneous Poisson point proces (PPP), because it has a reasonable assumption about the network geometry and leads to tractable models. To briefly introduce the concept, say we have a set of two dimensional points $\phi = \{u_1, \cdots\}$ randomly distributed in $\mathbb{R}^2$, such that for every compact set $B \in \mathbb{R}^2$. The set of points $\phi$ is a HPPP if:

## Simulating Poisson Networks Efficiently

Published:

When I first started my Ph.D. I learned about modeling large wireless network using point processes. They allow you to obtain tractable mathematical models that scale well with the size of the network. The most commonly used model in wireless networks is the homogeneous Poisson point proces (PPP), because it has a reasonable assumption about the network geometry and leads to tractable models. To briefly introduce the concept, say we have a set of two dimensional points $\phi = \{u_1, \cdots\}$ randomly distributed in $\mathbb{R}^2$, such that for every compact set $B \in \mathbb{R}^2$. The set of points $\phi$ is a HPPP if:

## Simulating Poisson Networks Efficiently

When I first started my Ph.D. I learned about modeling large wireless network using point processes. They allow you to obtain tractable mathematical models that scale well with the size of the network. The most commonly used model in wireless networks is the homogeneous Poisson point proces (PPP), because it has a reasonable assumption about the network geometry and leads to tractable models. To briefly introduce the concept, say we have a set of two dimensional points $\phi = \{u_1, \cdots\}$ randomly distributed in $\mathbb{R}^2$, such that for every compact set $B \in \mathbb{R}^2$. The set of points $\phi$ is a HPPP if:


text:
# zbMATH — the first resource for mathematics

Control of nonlinear variable structure systems. (English) Zbl 0622.93026
This paper is devoted to the research of different nonlinear control problems with variable structure described by $$\dot x=f(t,x,u)$$, $$x(t)\in R^ n$$, $$u(t)\subset V\in R^ m$$ with sliding manifold $$S(x)=0,S(x)\in R^ m$$. Controllability conditions based on Filipov’s definition of the solution of ordinary differential equations with discontinuous right-hand sides are derived. The derived conditions are illustrated via particular control problems with variable structure.


text:
## Prerequisite

To start this tutorial be sure to have the version 3.1 of the TAO platform. You can see your version at bottom left of the screen

If you don't have a TAO platform version 3.1 running please have a look to [installation guide](../../forge/Installation and Upgrading/installubuntuapachemysql)

Copy files from the newly downloaded release 3.2 to the directory where is stored the release 3.1 on your server. You should replace everything except data and config directories. Make sure that the user that will access the directories (www-data) have the rights on them.

### Fix taoWorkspace

If you had taoWorkspace installed on your installation you should manually add this extension. To do so you should first get the latest release using composer require "oat-sa/extension-tao-workspace":~1.1 or by downloading the archive and placing files in the correct directory.

### Update extensions

On your web server launch the command

# Linux{#linux}

sudo -u www-data php tao/scripts/taoUpdate.php

# Windows{#windows}

php tao/scripts/taoUpdate.php

You should now see something like the following image.

### Fix Tests

This version change will break some part of the application to fix them you will need to change some config files. First go into the file config\generis\filesystem.conf.php. You should see an array and something like

'http://tao-31.dev/tao31.rdf#i1514541812949527' => array(
'class' => 'Local',
'options' => array(
'root' => '/var/www/tao-31/data/taoQtiTest/'
)
),

Add testData and the end of the root so you will end up with :

'http://tao-31.dev/tao31.rdf#i1514541812949527' => array(
'class' => 'Local',
'options' => array(
'root' => '/var/www/tao-31/data/taoQtiTest/testData'
)
),

### Update server

It is highly recommended to update PHP from 5.6 to 7.0 after TAO is upgraded to TAO 3.2. Unlike TAO 3.2, TAO 3.1 is not fully compatible with PHP 7.0 so please update the TAO software prior to updating PHP.


text:
# Genetic Variation of the β2-Adrenoceptor

@article{Taylor2001GeneticVO,
title={Genetic Variation of the $\beta$2-Adrenoceptor},
author={D. Taylor and M. Kennedy},
journal={American Journal of Pharmacogenomics},
year={2001},
volume={1},
pages={165-174}
}
• Published 2001
• Biology
• American Journal of Pharmacogenomics
Asthma is a polygenic disease for which no clear genotype-phenotype relationships have emerged. In contrast, although not associated with the diagnosis of asthma per se, variant forms of the β2-adrenoceptor (β2-AR) gene (ADRB2) display functional effects that may be clinically relevant. Single nucleotide polymorphisms (SNPs) of ADBR2 are common and result in amino acid substitutions at positions 16, 27, and 164 of the receptor as well as position 19 of its 5′ upstream peptide. These SNPs… Expand
1 Citations

#### References

SHOWING 1-10 OF 64 REFERENCES


text:
5

# [) For f(x) - * -* 3x + find the following and then sketch the graph. DomainFirst DerivativeIncreasing Intervals Decreasing Intervals Relative Max Relative Min_ Sec...

## Question

###### [) For f(x) - * -* 3x + find the following and then sketch the graph. DomainFirst DerivativeIncreasing Intervals Decreasing Intervals Relative Max Relative Min_ Second Derivative Concave Down Intervals Concave Up Intervals _ Points of Inflection _2) For cos X - 2sin X, 0 sX s 21 find the following and then sketch the graph DomainFirst DerivativeIncreasing Intervals Decreasing IntervalsRelative Max

[) For f(x) - * -* 3x + find the following and then sketch the graph. Domain First Derivative Increasing Intervals Decreasing Intervals Relative Max Relative Min_ Second Derivative Concave Down Intervals Concave Up Intervals _ Points of Inflection _ 2) For cos X - 2sin X, 0 sX s 21 find the following and then sketch the graph Domain First Derivative Increasing Intervals Decreasing Intervals Relative Max

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(a) Find an approximation to the integral 4x) dx using Riemann sum with right endpoints and n = 8 5.8125 (b) If f is integrable on [a, b]_ then f(x) dx lim n =0 b _ f(x;) Ax, where Ax = and xi = a + Ax: Use this to evaluate (x2 _ 4x) dx:...
##### Point) Substitution in the Definite IntegralsinS dx using the substitution; cos?Suppose we want to evaluate the definite integral,cos(x)Part 1.Re-write the definite integral in terms of the variable and remember t0 use the limits of integration for the function u = f(x) Then, input the antiderivative of the integrand and the limits of integration you found:NOTE: enter an answer in which the lower Iimit of integration is less than the upper Iimit of integration.Part 2.Finally; evaluate the origin
point) Substitution in the Definite Integral sinS dx using the substitution; cos? Suppose we want to evaluate the definite integral, cos(x) Part 1. Re-write the definite integral in terms of the variable and remember t0 use the limits of integration for the function u = f(x) Then, input the antideri...
##### Derivatives of products and quotients Find the derivative of the following functions by first expanding or simplifying the expression. Simplify your answers. $$f(x)=(\sqrt{x}+1)(\sqrt{x}-1)$$
Derivatives of products and quotients Find the derivative of the following functions by first expanding or simplifying the expression. Simplify your answers. $$f(x)=(\sqrt{x}+1)(\sqrt{x}-1)$$...


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# MATLAB: Question about bvp4c solution

bvp4cdifferential equationsnumerical calculation

I used bvp4c to solve second order differential equation: A*y'' = f(y), now I want to get some special values from the solutions. How to get the values of y'' and y' from each point?

y1'=y2y2'=f(y)/A,
yp=deval(sol.yp,sol.x);plot(sol.x,yp(1,:),sol.x,yp(2,:));


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# Algebraic and geometric classification of nilpotent anticommutative algebras

#### Preprint

We give algebraic and geometric classifications of 6-dimensional complex nilpotent anticommutative algebras. Specifically, we find that, up to isomorphism, there are 14 one-parameter families of 6-dimensional nilpotent anticommutative algebras, complemented by 130 additional isomorphism classes. The corresponding geometric variety is irreducible and determined by the Zariski closure of a one-parameter family of algebras. In particular, there are no rigid 6-dimensional complex nilpotent anticommutative algebras.

Ivan Kaygorodov

Mykola Khrypchenko

### Publication

Year of publication: 2019

### File

main_3.pdf377.36 KB


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# Distribution of the variate Y

• May 4th 2009, 05:26 AM
Stats
Distribution of the variate Y
It X is binomially distributed with parameters n and p , what can you say about the distribution of variate
$
Y_n = \frac{X-np}{\sqrt{npq}} \; \; as \qquad n \to \infty
$
• May 4th 2009, 03:46 PM
mr fantastic
Quote:

Originally Posted by Stats
It X is binomially distributed with parameters n and p , what can you say about the distribution of variate
$
Y_n = \frac{X-np}{\sqrt{npq}} \; \; as \qquad n \to \infty
$

Have you heard of the normal approximation to the binomial distribution?


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# How do you solve 5( 9+ 7) - 14\cdot 2?

Sep 23, 2016

The value of this expression is $52$
$5 \times \left(9 + 7\right) - 14 \times 2 = 5 \times 16 - 14 \times 2 = 80 - 14 \times 2 = 80 - 28 = 52$


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# Conversion Funnel

## Introduction

Since 7th December 2020, we've applied a redesigned version of Conversion Funnel (see our announcement post here). If you are finding information about the legacy Conversion Funnel, please refer to this document.

Conversion Funnel refers to a "journey" that a subject goes through in a particular business context. For example:

• A lead's progress in a sales funnel: lead sourcing → lead qualified with a sales call → follow-ups → lead becomes customers
• A customer's progress in an e-commerce's funnel: visits website → browses for products → adds products to cart → purchases → becomes repeated customers

The Conversion Funnel Chart helps you visualize such journeys and the conversion rates between the steps.

## When to use Conversion Funnel?

Conversion Funnel is useful when you want to have full visibility on a process, identify opportunities to improve a drop between steps in the funnel.

## Create a Conversion Funnel

The Conversion Funnel Chart has two components to visualize your funnel: the chart and the table.

The chart will display the overall funnel, while the table shows a detailed break-down of the funnel over multiple legends.

• Breakdown: help you break your funnel down by one dimension. In our example, we want to see this conversion funnel by country so we dragged the Country field in.
• Values: each value field you drag in here will represent a step in the funnel. In our example, we used 4 values to build that funnel: Number of people who visited the page (Sum of Visited), Number of people who signed in (Sum of Signed In), number of people who viewed listings (Sum of Viewed Listings), number of people who booked a hotel (Sum of Booked Hotel).

Note that you can change the steps' order by re-arranging the fields' order in visualization settings.

## Styling options

• Columns color: choose the color for your the columns
• Circle color: choose the color for the circle containing the overall conversion rate


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## Lecture 14: Spectral sparsifiers

1. Spectral Sparsifiers

1.1. Graph Laplacians

Let ${G=(V,E)}$ be an unweighted graph. For notational simplicity, we will think of the vertex set as ${V = \{1,\ldots,n\}}$. Let ${e_i \in {\mathbb R}^n}$ be the ${i}$th standard basis vector, meaning that ${e_i}$ has a ${1}$ in the ${i}$th coordinate and ${0}$s in all other coordinates. For an edge ${uv \in E}$, define the vector ${x_{uv}}$ and the matrix ${X_{uv}}$ as follows:

$\displaystyle \begin{array}{rcl} x_{uv} &:=& e_u - u_v \\ X_{uv} &:=& x_{uv} x_{uv} ^T \end{array}$

In the definition of ${x_{uv}}$ it does not matter which vertex gets the ${+1}$ and which gets the ${-1}$ because the matrix ${X_{uv}}$ is the same either way.

Definition 1 The Laplacian matrix of ${G}$ is the matrix

$\displaystyle L_G ~:=~ \sum_{uv \in E} X_{uv}$

Let us consider an example.

Note that each matrix ${X_{uv}}$ has only four non-zero entries: we have ${X_{uu} = X_{vv} = 1}$ and ${X_{uv} = X_{vu} = -1}$. Consequently, the ${u}$th diagonal entry of ${L_G}$ is simply the degree of vertex ${u}$. Moreover, we have the following fact.

Fact 2 Let ${D}$ be the diagonal matrix with ${D_{u,u}}$ equal to the degree of vertex ${u}$. Let ${A}$ be the adjacency matrix of ${G}$. Then ${L_G = D - A}$.

If ${G}$ had weights ${w : E \rightarrow {\mathbb R}}$ on the edges we could define the weighted Laplacian as follows:

$\displaystyle L_G ~=~ \sum_{uv \in E} w_{uv} \cdot X_{uv}.$

Claim 3 Let ${G=(V,E)}$ be a graph with non-negative weights ${w : E \rightarrow {\mathbb R}}$. Then the weighted Laplacian ${L_G}$ is positive semi-definite.

Proof: Since ${X_{uv} = x_{uv} x_{uv} ^T}$, it is positive semi-definite. So ${L_G}$ is a weighted sum of positive semi-definite matrices with non-negative coefficients. Fact 5 in the Notes on Symmetric Matrices implies ${L_G}$ is positive semi-definite. $\Box$

The Laplacian can tell us many interesting things about the graph. For example:

Claim 4 Let ${G=(V,E)}$ be a graph with Laplacian ${L_G}$. For any ${U \subseteq V}$, let ${\chi(U) \in {\mathbb R}^n}$ be the characteristic vector of ${U}$, i.e., the vector with ${\chi(U)_v}$ equal to ${1}$ if ${v \in U}$ and equal to ${0}$ otherwise. Then ${\chi(U) ^T \, L_G \, \chi(U) = | \delta(U) |}$.

Proof: For any edge ${uv}$ we have ${ \chi(U) ^T \, X_{uv} \, \chi(U) = ( \chi(U) ^T \, x_{uv} )^2 }$. But ${| \chi(U) ^T \, x_{uv} |}$ is ${1}$ if exactly one of ${u}$ or ${v}$ is in ${U}$, and otherwise it is ${0}$. So ${\chi(U) ^T \, X_{uv} \, \chi(U) = 1}$ if ${uv \in \delta(U)}$, and otherwise it is ${0}$. Summing over all edges proves the claim. $\Box$

Similarly, if ${G=(V,E)}$ is a graph with edge weights ${w : E \rightarrow {\mathbb R}}$ and ${L_G}$ is the weighted Laplacian, then then ${\chi(U) ^T \, L_G \, \chi(U) = w( \delta(U) )}$.

Fact 5 If ${G}$ is connected then ${\mathrm{image}(L_G) ~=~ \{\: x \::\: \sum_i x_i = 0 \:\} }$, which is an ${(n-1)}$-dimensional subspace.

1.2. Main Theorem

Theorem 6 Let ${G=(V,E)}$ be a graph with ${n = |V|}$. There is a randomized algorithm to compute weights ${w : E \rightarrow {\mathbb R}}$ such that:

• only ${O(n \log n / \epsilon^2)}$ of the weights are non-zero, and
• with probability at least ${1-2/n}$,

$\displaystyle (1-\epsilon) \cdot L_G ~\preceq~ L_w ~\preceq~ (1+\epsilon) \cdot L_G,$

where ${L_w}$ denotes the weighted Laplacian of ${G}$ with weights ${w}$. By Fact 4 in Notes on Symmetric Matrices, this is equivalent to

$\displaystyle (1-\epsilon) x ^T L_G x ~\leq~ x ^T L_w x ~\leq~ (1+\epsilon) x ^T L_G x \qquad\forall x \in {\mathbb R}^n. \ \ \ \ \ (1)$

By (1) and Claim 4, the resulting weights are a graph sparsifier of ${G}$:

$\displaystyle (1-\epsilon) \cdot |\delta(U)| ~\leq~ w(\delta(U)) ~\leq~ (1+\epsilon) \cdot |\delta(U)| \qquad\forall U \subseteq V.$

The algorithm that proves Theorem 6 is as follows.

• Initially ${w = 0}$.
• Set ${k=8 n \log(n) / \epsilon^2}$.
• For every edge ${e \in E}$ compute ${r_e = \mathop{\mathrm{tr}}\,( X_e L_G^+ )}$.
• For ${i=1,\ldots,k}$
• ${\quad}$ Let ${e}$ be a random edge chosen with probability ${r_e/(n-1)}$.
• ${\quad}$ Increase ${w_e}$ by ${\frac{n-1}{r_e \, k}}$.

Claim 7 The values ${\{ r_e/(n-1) \::\: e \in E \}}$ indeed form a probability distribution.

Proof: (of Theorem 6). How does the matrix ${L_w}$ change during the ${i}$th iteration? The edge ${e}$ is chosen with probability ${\frac{r_e}{n-1}}$ and then ${L_w}$ increases by ${\frac{n-1}{r_e \cdot k} X_e}$. Let ${Z_i}$ be this random change in ${L_w}$ during the ${i}$th iteration. So ${Z_i}$ equals ${\frac{n-1}{r_e \cdot k} X_e}$ with probability ${\frac{r_e}{n-1}}$. The random matrices ${Z_1,\ldots,Z_k}$ are mutually independent and they all have this same distribution. Note that

$\displaystyle {\mathrm E}[Z_i] ~=~ \sum_{e \in E} \frac{r_e}{n-1} \cdot \frac{n-1}{r_e \cdot k} X_e ~=~ \frac{1}{k} \sum_e X_e ~=~ \frac{L_G}{k}.$

The final matrix ${L_w}$ is simply ${\sum_{i=1}^k Z_i}$. To analyze this final matrix, we will use the Ahlswede-Winter inequality. All that we require is the following claim, which we prove later.

Claim 8 ${Z_i \preceq (n-1) \cdot {\mathrm E}[Z_i]}$.

We apply Corollary 2 from the previous lecture with ${R=n-1}$, obtaining

$\displaystyle \begin{array}{rcl} {\mathrm{Pr}}\big[ (1-\epsilon) L_G \:\preceq\: L_w \:\preceq\: (1+\epsilon) L_G \big] &=& {\mathrm{Pr}}\Bigg[ (1-\epsilon) \frac{L_G}{k} \:\preceq\: \frac{1}{k} \sum_{i=1}^k Z_i \:\preceq\: (1+\epsilon) \frac{L_G}{k} \Bigg] \\ &\leq& 2n \cdot \exp\big( - \epsilon^2 k / 4 (n-1) \big) \\ &\leq& 2n \cdot \exp\big( - 2 \ln n \big) ~<~ 2/n. \end{array}$

$\Box$

Proof: (of Claim 7) First we check that the ${r_e}$ values are non-negative. By the cyclic property of trace

$\displaystyle \mathop{\mathrm{tr}}\,( X_e L_G^+ ) ~=~ \mathop{\mathrm{tr}}\,( x_e ^T L_G^+ x_e ) ~=~ x_e ^T L_G^+ x_e,$

This is non-negative since ${L_G^+ \succeq 0}$ because ${L_G \succeq 0}$. Thus ${r_e \geq 0}$. Next, note that

$\displaystyle \sum_e \mathop{\mathrm{tr}}\,( X_e L_G^+ ) ~=~ \mathop{\mathrm{tr}}\,( \sum_e X_e L_G^+ ) ~=~ \mathop{\mathrm{tr}}\,( L_G L_G^+ ) ~=~ \mathop{\mathrm{tr}}\,( I_{\mathrm{im}~L_G} ),$

where ${I_{\mathrm{im}~L_G}}$ is the orthogonal projection onto the image of ${L_G}$. The image has dimension ${n-1}$ by Fact 5, and so

$\displaystyle \sum_e r_e ~=~ \frac{1}{n-1} \sum_e \mathop{\mathrm{tr}}\,( X_e L_G^+ ) ~=~ \frac{1}{n-1} \mathop{\mathrm{tr}}\,( I_{\mathrm{im}~L_G} ) ~=~ 1.$

$\Box$

Proof: (of Claim 8). The maximum eigenvalue of a positive semi-definite matrix never exceeds its trace, so

$\displaystyle \lambda_{\mathrm{max}}( L_G^{+/2} \cdot X_e \cdot L_G^{+/2} ) ~\leq~ \mathop{\mathrm{tr}}\,( L_G^{+/2} \cdot X_e \cdot L_G^{+/2} ) ~=~ r_e.$

By Fact 8 in the Notes on Symmetric Matrices,

$\displaystyle L_G^{+/2} \cdot X_e \cdot L_G^{+/2} ~\preceq~ r_e \cdot I.$

So, by Fact 4 in the Notes on Symmetric Matrices, for every vector ${v}$,

$\displaystyle v ^T \frac{L_G^{+/2} \cdot X_e \cdot L_G^{+/2}}{r_e} v ~\leq~ v ^T v.$

Now let us write ${v=v_1 + v_2}$ where ${v_1 = I_{\mathrm{im}~L_G} \, v}$ is the projection onto the image of ${L_G}$ and ${v_2 = I_{\mathrm{ker}~L_G} \, v}$ is the projection onto the kernel of ${L_G}$. Then ${L_G \, v = 0}$ and ${L_G^{+/2} \, v = 0}$. So

$\displaystyle \begin{array}{rcl} v ^T \frac{L_G^{+/2} \cdot X_e \cdot L_G^{+/2}}{r_e} v &=& v_1 ^T \frac{L_G^{+/2} \cdot X_e \cdot L_G^{+/2}}{r_e} v_1 ~+~ \underbrace{v_2 ^T \frac{L_G^{+/2} \cdot X_e \cdot L_G^{+/2}}{r_e} v_2}_{=0} \\ &=& v_1 ^T \frac{L_G^{+/2} \cdot X_e \cdot L_G^{+/2}}{r_e} v_1 \\ &\leq& v_1 ^T v_1 ~=~ v ^T I_{\mathrm{im}~L_G} v. \end{array}$

Since this holds for every vector ${v}$, Fact 4 in the Notes on Symmetric Matrices again implies

$\displaystyle \frac{L_G^{+/2} \cdot X_e \cdot L_G^{+/2}}{r_e} ~\preceq~ I_{\mathrm{im}~L_G}.$

Since ${\mathrm{im}~X_e \subseteq \mathrm{im}~L_G}$, Claim 16 in the Notes on Symmetric Matrices shows this is equivalent to

$\displaystyle \frac{n-1}{r_e \cdot k} X_e ~\preceq~ \frac{n-1}{k} L_G.$

This completes the proof of the claim. $\Box$


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# Which of the following reagents can be used to oxidised primary alcohol to aldehydes ?

$\begin{array}{1 1} \text{CrO_3 in anhydrous medium } \\ KMnO_4 \text {in acidic medium } \\ \text{pyridinium chlorochromate} \\ \text{Heat in the presence of Cu at 573 k} \end{array}$

$\text{pyridinium chlorochromate}$