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# Kentico/delivery-sdk-php

### Showing 11 of 16 total issues

#### QueryParams has 23 functions (exceeds 20 allowed). Consider refactoring. Open

class QueryParams implements \ArrayAccess
{
/**
* Filter key-value pairs.
*
Found in src/KenticoCloud/Delivery/QueryParams.php - About 2 hrs to fix

#### Function bindModel has a Cognitive Complexity of 16 (exceeds 5 allowed). Consider refactoring. Open

    public function bindModel($modelType,$data, $linkedItems = null,$processedItems = null)
{
$processedItems =$processedItems ?? array();
$model = new$modelType();
$modelProperties =$this->propertyMapper->getModelProperties($model,$data);
Found in src/KenticoCloud/Delivery/ModelBinder.php - About 2 hrs to fix

# Cognitive Complexity

Cognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.

### A method's cognitive complexity is based on a few simple rules:

• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one
• Code is considered more complex for each "break in the linear flow of the code"
• Code is considered more complex when "flow breaking structures are nested"

#### The class DeliveryClient has 17 fields. Consider redesigning DeliveryClient to keep the number of fields under 15. Open

class DeliveryClient
{
private $sdkVersion = '2.0.0'; private$urlBuilder = null;

Since: PHPMD 0.1

Classes that have too many fields could be redesigned to have fewer fields, possibly through some nested object grouping of some of the information. For example, a class with city/state/zip fields could instead have one Address field.

Example:

class Person {
protected $one; private$two;
private $three; [... many more fields ...] } #### Function bindLinkedItems has a Cognitive Complexity of 12 (exceeds 5 allowed). Consider refactoring. Open  public function bindLinkedItems($element, $linkedItems,$processedItems)
{
$modelLinkedItems = null; if ($linkedItems != null) {
$modelLinkedItems = array(); Found in src/KenticoCloud/Delivery/ModelBinder.php - About 1 hr to fix # Cognitive Complexity Cognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend. ### A method's cognitive complexity is based on a few simple rules: • Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one • Code is considered more complex for each "break in the linear flow of the code" • Code is considered more complex when "flow breaking structures are nested" ### Further reading #### Method bindModel has 28 lines of code (exceeds 25 allowed). Consider refactoring. Open  public function bindModel($modelType, $data,$linkedItems = null, $processedItems = null) {$processedItems = $processedItems ?? array();$model = new $modelType();$modelProperties = $this->propertyMapper->getModelProperties($model, $data); Found in src/KenticoCloud/Delivery/ModelBinder.php - About 1 hr to fix #### Method __construct has 7 arguments (exceeds 4 allowed). Consider refactoring. Open  public function __construct($id = null, $name = null,$codename = null, $lastModified = null,$type = null, $sitemapLocations = null,$language = null)
Found in src/KenticoCloud/Delivery/Models/Items/ContentItemSystem.php - About 50 mins to fix

#### Method __construct has 6 arguments (exceeds 4 allowed). Consider refactoring. Open

    public function __construct(string $projectId, string$previewApiKey = null, string $securedProductionApiKey = null, bool$waitForLoadingNewContent = null, bool $debugRequests = null, int$retryAttempts = null)
Found in src/KenticoCloud/Delivery/DeliveryClient.php - About 45 mins to fix

#### The class DeliveryClient has a coupling between objects value of 21. Consider to reduce the number of dependencies under 13. Open

class DeliveryClient
{
private $sdkVersion = '2.0.0'; private$urlBuilder = null;

Since: PHPMD 1.1.0

A class with too many dependencies has negative impacts on several quality aspects of a class. This includes quality criteria like stability, maintainability and understandability

Example:

class Foo {
/**
* @var \foo\bar\X
*/
private $x = null; /** * @var \foo\bar\Y */ private$y = null;

/**
* @var \foo\bar\Z
*/
private $z = null; public function setFoo(\Foo$foo) {}
public function setBar(\Bar $bar) {} public function setBaz(\Baz$baz) {}

/**
* @return \SplObjectStorage
* @throws \OutOfRangeException
* @throws \InvalidArgumentException
* @throws \ErrorException
*/
public function process(\Iterator $it) {} // ... } #### Function getTypeClass has a Cognitive Complexity of 6 (exceeds 5 allowed). Consider refactoring. Open  public function getTypeClass($typeName)
{
switch ($typeName) { case 'asset':$type = \KenticoCloud\Delivery\Models\Items\Asset::class;
Found in src/KenticoCloud/Delivery/DefaultMapper.php - About 25 mins to fix

# Cognitive Complexity

Cognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.

### A method's cognitive complexity is based on a few simple rules:

• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one
• Code is considered more complex for each "break in the linear flow of the code"
• Code is considered more complex when "flow breaking structures are nested"

#### Avoid unused local variables such as '$element'. Open  foreach ($data[self::ELEMENTS_ATTRIBUTE_NAME] as $elementName =>$element) {

Since: PHPMD 0.2

Detects when a local variable is declared and/or assigned, but not used.

Example:

class Foo {
public function doSomething()
{
$i = 5; // Unused } } #### Refactor this function to reduce its Cognitive Complexity from 16 to the 15 allowed. Open  public function bindModel($modelType, $data,$linkedItems = null, \$processedItems = null)

Cognitive Complexity is a measure of how hard the control flow of a function is to understand. Functions with high Cognitive Complexity will be difficult to maintain.


text:
Algebra 2 (1st Edition)

For starters, the function is not linear, for the slope is not a constant value due to the $x^3$ and $x^2$ terms. We now evaluate $g(-5)$: $$g(-5) = (-5)^3 -2 (-5)^2 +5(-5) - 8 \\ g(-5) = -125 -50 -25 - 8 \\ g(-5) = -208$$

format: plain


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## Thomas' Calculus 13th Edition

a) $\dfrac{4 \pi}{15}$ and b ) $\dfrac{7\pi}{30}$
a) $V= 2 \pi \int_{0}^{1} (y) \cdot (y-y^3) dy=\dfrac{-2y^3 \pi(3y^2-5)}{15}=\dfrac{4 \pi}{15}$ b) We need to use the shell model as follows: $V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx \\= 2 \pi \int_{0}^{1} (1-y)\cdot (y-y^3) dy=\dfrac{2y^2 \pi(12y^3-15y^2-20y+30)}{30}=\dfrac{7\pi}{30}$


text:
# Compact sets in $\mathbb R$ are closed

What is the best way to prove the statement? I know the version of using nested intervals, but is there another way to approach the problem?

• This is also generally true in any metric (or more generally Hausdorff) space. You prove that for a compact set $C$, $\mathbb{R}\setminus C$ is open. You use the open covering definition of compactness to do so. – Moya Oct 3 '15 at 6:45
• So you have three answers, and you selected as the best answer the only one that doesn't help you, (the other two focus on $\mathbb R$) makes sense.. – Gero Oct 4 '15 at 5:36

Let $X \subseteq \mathbb{R}$ be compact, and let $y \in \mathbb{R} \setminus X$. It suffices to prove $\mathbb{R} \setminus X$ is open, i.e. that $\mathbb{R} \setminus X$ contains an open ball about $y$. Since $\mathbb{R}$ is Hausdorff, for each $x\in X$, there exists a neighborhood $U_{x}$ of $x$ and $U_{y}^{x}$ of $y$ such that $U_{x}$ and $U_{y}^{x}$ are disjoint. The set of all $U_{x}$'s for each $x \in X$ form an open cover for $X$, so by the compactness of $X$, there exist finitely many $U_{x_{1}}, \ldots, U_{x_{n}}$ which cover $X$. Can you wrap things up from here?

• Thanks for the suggestion. What I'm looking for is how to view this problem under real analysis? – Quang Dao Oct 3 '15 at 7:13
• Sure. I'm not sure I have a much better answer for you, I'm afraid: what tools did you have in mind, exactly? – Alex Wertheim Oct 3 '15 at 7:18

Yes. Let $A$ be a compact set and let $p\in\mathbb R\setminus A.$ For each $x\in A,$ Let $B_x$ be the open interval $\left(x-\dfrac{|p-x|}{2},x+\dfrac{|p-x|}{2}\right)$ and let $V_x$ be the open interval $\left(p-\dfrac{|p-x|}{2},p+\dfrac{|p-x|}{2}\right).$ Then by the compactness of $A,$ there are $x_1,x_2,\ldots,x_n\in A$ such that $A\subseteq B_{x_1}\cup\cdots\cup B_{x_n}$ and it follows that the set $V_{x_1}\cap\cdots\cap V_{x_n}$ does not intersect $A$ and hence $p$ is an interior point of $\mathbb R\setminus A,$ which implies that $A$ is closed.

Note that a similar argument can be used to prove that every compact subset of a metric space is closed.

If $A \subseteq \mathbb R$ is not closed, take $a \in \overline{A} \setminus A$, and for each $n > 0$ consider $$U_n = ( - \infty , a - \tfrac{1}{n} ) \cup ( a + \tfrac{1}{n} , + \infty ).$$ Then $\bigcup_n U_n = \mathbb{R} \setminus \{ a \} \supseteq A$, however no finite subfamily of $\{ U_n : n > 0 \}$ can cover $A$ since $a$ is a limit point of $A$ (for each $n > 0$ there is an $x \in A$ such that $| x - a | < \frac{1}{n}$).

Therefore $A$ cannot be compact.

(To alter this for a a general metric space $(X,d)$, take $U_n = \{ x \in X : d(x,a) > \frac{1}{n} \}$.)


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## Engage NY Eureka Math 7th Grade Module 3 Lesson 22 Answer Key

### Eureka Math Grade 7 Module 3 Lesson 22 Example Answer Key

Example 1.
The pyramid in the picture has a square base, and its lateral faces are triangles that are exact copies of one another. Find the surface area of the pyramid.

The surface area of the pyramid consists of one square base and four lateral triangular faces.
B = s2
B = (6 cm)2
B = 36 cm2
The pyramid’s base area is 36 cm2.

LA = 4($$\frac{1}{2}$$ bh)
LA = 4 ∙ $$\frac{1}{2}$$ (6 cm ∙ 7 cm)
LA = 2(6 cm ∙ 7 cm)
LA = 2(42 cm2 )
LA = 84 cm2
The pyramid’s lateral area is 84 cm2.
SA = LA + B
SA = 84 cm2 + 36 cm2
SA = 120 cm2
The surface area of the pyramid is 120 cm2.

Example 2: Using Cubes
There are 13 cubes glued together forming the solid in the diagram. The edges of each cube are $$\frac{1}{4}$$ inch in length. Find the surface area of the solid.

The surface area of the solid consists of 46 square faces, all having side lengths of $$\frac{1}{4}$$ inch. The area of a square having sides of length $$\frac{1}{4}$$ inch is $$\frac{1}{16}$$ in2 .
SA = 46 ∙ Asquare
SA = 46 ∙ $$\frac{1}{16}$$ in2
SA = $$\frac{46}{16}$$ in2
SA = 2 $$\frac{14}{16}$$ in2
SA = 2 $$\frac{7}{8}$$ in2 The surface are of the solid is 2 $$\frac{7}{8}$$ in2.

Example 3.
Find the total surface area of the wooden jewelry box. The sides and bottom of the box are all $$\frac{1}{4}$$ inch thick.
What are the faces that make up this box?

The box has a rectangular bottom, rectangular lateral faces, and a rectangular top that has a smaller rectangle removed from it. There are also rectangular faces that make up the inner lateral faces and the inner bottom of the box.

How does this box compare to other objects that you have found the surface area of?
The box is a rectangular prism with a smaller rectangular prism removed from its inside. The total surface area is equal to the surface area of the larger right rectangular prism plus the lateral area of the smaller right rectangular prism.

Large Prism: The surface area of the large right rectangular prism makes up the outside faces of the box, the rim of the box, and the inside bottom face of the box.

SA = LA + 2B
LA = P ∙ h
LA = 32 in. ∙ 4 in.
LA = 128 in2

B = lw
B = 10 in. ∙ 6 in.
B = 60 in2
The lateral area is 128 in2.
The base area is 60 in2.

SA = LA + 2B
SA = 128 in2 + 2(60 in2)
SA = 128 in2 + 120 in2
SA = 248 in2
The surface area of the larger prism is 248 in2.

Small Prism: The smaller prism is $$\frac{1}{2}$$ in. smaller in length and width and $$\frac{1}{4}$$ in. smaller in height due to the thickness of the sides of the box.
SA = LA + 1B
LA = P ∙ h
LA = 2(9 $$\frac{1}{2}$$ in. + 5 $$\frac{1}{2}$$ in.) ∙ 3 $$\frac{3}{4}$$ in.
LA = 2(14 in. + 1 in.) ∙ 3 $$\frac{3}{4}$$ in.
LA = 2(15 in.) ∙ 3 $$\frac{3}{4}$$ in.
LA = 30 in. ∙ 3 $$\frac{3}{4}$$ in.
LA = 90 in2 + $$\frac{90}{4}$$ in2
LA = 90 in2 + 22 $$\frac{1}{2}$$ in2
LA = 112 $$\frac{1}{2}$$ in2
The lateral area is 112 $$\frac{1}{2}$$ in2.

Surface Area of the Box
SAbox = SA + LA
SAbox = 248 in2 + 112 $$\frac{1}{2}$$ in2
SAbox = 360 $$\frac{1}{2}$$ in2
The total surface area of the box is 360 $$\frac{1}{2}$$ in2.

### Eureka Math Grade 7 Module 3 Lesson 22 Exercise Answer Key

Opening Exercise
What is the area of the composite figure in the diagram? Is the diagram a net for a three-dimensional image? If so, sketch the image. If not, explain why.

There are four unit squares in each square of the figure. There are 18 total squares that make up the figure, so the total area of the composite figure is
A = 18 ∙ 4 units2 = 72 units2.
The composite figure does represent the net of a three-dimensional figure. The figure is shown below.

### Eureka Math Grade 7 Module 3 Lesson 22 Problem Set Answer Key

Question 1.
For each of the following nets, draw (or describe) the solid represented by the net and find its surface area.
a. The equilateral triangles are exact copies.

The net represents a triangular pyramid where the three lateral faces are identical to each other and the triangular base.
SA = 4B since the faces are all the same size and shape.

SA = 4B
SA = 4(35 $$\frac{1}{10}$$ mm2 )
SA = 140 mm2 + $$\frac{4}{10}$$ mm2
SA = 140 $$\frac{2}{5}$$ mm2

B = $$\frac{1}{2}$$ bh
B = $$\frac{1}{2}$$ ∙ 9 mm ∙ 7 $$\frac{4}{5}$$ mm
B = $$\frac{9}{2}$$ mm ∙ 7 $$\frac{4}{5}$$ mm
B = $$\frac{63}{2}$$ mm2 + $$\frac{36}{10}$$ mm2
B = $$\frac{315}{10}$$ mm2 + $$\frac{36}{10}$$ mm2
B = $$\frac{351}{10}$$ mm2 The surface area of the triangular pyramid is 140 $$\frac{2}{5}$$ mm2.
B = 35 $$\frac{1}{10}$$ mm2

b.

The net represents a square pyramid that has four identical lateral faces that are triangles. The base is a square.
B = s2
B = (12 in.)2
B = 144 in2

LA = 2(12 in. ∙ 14 $$\frac{3}{4}$$ in.)
SA = LA + B
LA = 4 ∙ $$\frac{1}{2}$$ (bh)
LA = 4 ∙ $$\frac{1}{2}$$ (12 in. ∙ 14 $$\frac{3}{4}$$ in.)
LA = 2(168 in2 + 9 in2 )
LA = 336 in2 + 18 in2
LA = 354 in2
SA = LA + B
SA = 354 in2 + 144 in2
SA = 498 in2
The surface area of the square pyramid is 498 in2.

Question 2.
Find the surface area of the following prism.

SA = LA + 2B
LA = P ∙ h
LA = (4 cm + 6 $$\frac{1}{2}$$ cm + 4 $$\frac{1}{5}$$ cm + 5 $$\frac{1}{4}$$ cm) ∙ 9 cm
LA = (19 cm + $$\frac{1}{2}$$ cm + $$\frac{1}{5}$$ cm + $$\frac{1}{4}$$ cm) ∙ 9 cm
LA = (19 cm + $$\frac{10}{20}$$ cm + $$\frac{4}{20}$$ cm + $$\frac{5}{20}$$ cm) ∙ 9 cm
LA = (19 cm + $$\frac{19}{20}$$ cm) ∙ 9 cm
LA = 171 cm2 + $$\frac{171}{20}$$ cm2
LA = 171 cm2 + 8 $$\frac{11}{20}$$ cm2
LA = 179 $$\frac{11}{20}$$ cm2

B = Arectangle + Atriangle
B = (5 $$\frac{1}{4}$$ cm ∙ 4 cm) + $$\frac{1}{2}$$(4 cm ∙ 1 $$\frac{1}{4}$$ cm)
B = (20 cm2 + 1 cm2 ) + (2 cm ∙ 1 $$\frac{1}{4}$$ cm)
B = 21 cm2 + 2 $$\frac{1}{2}$$ cm2
B = 23 $$\frac{1}{2}$$ cm2

SA = LA + 2B
SA = 179 $$\frac{11}{20}$$ cm2 + 2(23 $$\frac{1}{2}$$ cm2 )
SA = 179 $$\frac{11}{20}$$ cm2 + 47 cm2
SA = 226 $$\frac{11}{20}$$ cm2
The surface area of the prism is 226 $$\frac{11}{20}$$ cm2.

Question 3.
The net below is for a specific object. The measurements shown are in meters. Sketch (or describe) the object, and then find its surface area.

SA = LA + 2B
LA = P ∙ h
LA = 6 cm ∙ $$\frac{1}{2}$$ cm
LA = 3 cm2

B = ($$\frac{1}{2}$$ cm ∙ $$\frac{1}{2}$$ cm) + ($$\frac{1}{2}$$ cm ∙ 1 cm) + ($$\frac{1}{2}$$ cm ∙ 1 $$\frac{1}{2}$$ cm)
B = ($$\frac{1}{4}$$ cm2) + ($$\frac{1}{2}$$ cm2 ) + ($$\frac{3}{4}$$ cm2 )
B = ($$\frac{1}{4}$$cm2) + ($$\frac{2}{4}$$ cm2) + ($$\frac{3}{4}$$ cm2)
B = $$\frac{6}{4}$$ cm2
B = 1 $$\frac{1}{2}$$ cm2

SA = LA + 2B
SA = 3 cm2 + 2(1 $$\frac{1}{2}$$ cm2 )
SA = 3 cm2 + 3 cm2
SA = 6 cm2

The surface area of the object is 6 cm2.

Question 4.
In the diagram, there are 14 cubes glued together to form a solid. Each cube has a volume of $$\frac{1}{8}$$ in3. Find the surface area of the solid.

The volume of a cube is s3, and 1/8 in3 is the same as ($$\frac{1}{2}$$ in.)3, so the cubes have edges that are $$\frac{1}{2}$$ in. long. The cube faces have area s2, or ($$\frac{1}{2}$$ in.)2, or $$\frac{1}{4}$$ in2. There are 42 cube faces that make up the surface of the solid.
SA = $$\frac{1}{4}$$ in2 ∙ 42
SA = 10 $$\frac{1}{2}$$ in2
The surface area of the solid is 10 $$\frac{1}{2}$$ in2.

Question 5.
The nets below represent three solids. Sketch (or describe) each solid, and find its surface area.
a.

SA = LA + 2B
LA = P ∙ h
LA = 12 ∙ 3
LA = 36 cm2

B = s2
B = (3 cm)2
B = 9 cm2

SA =
36 cm2 + 2(9 cm2)
SA = 36 cm2 + 18 cm2
SA = 54 cm2

b.

SA = 3Asquare + 3Art triangle + Aequ triangle
Asquare = s2
Asquare = (3 cm)2
Asquare = 9 cm2
Art triangle = $$\frac{1}{2}$$ bh
Art triangle = $$\frac{1}{2}$$ ∙ 3 cm ∙ 3 cm
Art triangle = $$\frac{9}{2}$$
Art triangle = 4 $$\frac{1}{2}$$ cm2
Aequ triangle = $$\frac{1}{2}$$ bh
Aequ triangle = $$\frac{1}{2}$$ ∙ (4 $$\frac{1}{5}$$ cm) ∙ (3 $$\frac{7}{10}$$ cm)
Aequ triangle = 2 $$\frac{1}{10}$$ cm ∙ 3 $$\frac{7}{10}$$ cm
Aequ triangle = $$\frac{21}{10}$$ cm ∙ $$\frac{37}{10}$$ cm
Aequ triangle = 7 $$\frac{77}{100}$$ cm2
Aequ triangle = 7 $$\frac{77}{100}$$ cm2
SA = 3(9 cm2 ) + 3(4 $$\frac{1}{2}$$ cm2 ) + 7 $$\frac{77}{100}$$ cm2
SA = 27 cm2 + (12 + $$\frac{3}{2}$$) cm2 + 7 $$\frac{77}{100}$$ cm2
SA = 47 cm2 + $$\frac{1}{2}$$ cm2 + $$\frac{77}{100}$$ cm2
SA = 47 cm2 + $$\frac{50}{100}$$ cm2 + $$\frac{77}{100}$$ cm2
SA = 47 cm2 + $$\frac{127}{100}$$ cm2
SA = 47 cm2 + 1 cm2 + $$\frac{27}{100}$$ cm2
SA = 48 $$\frac{27}{100}$$ cm2

c.

SA = 3Art triangle + Aequ triangle
SA = 3(4 $$\frac{1}{2}$$ ) cm2 + 7 $$\frac{77}{100}$$ cm2
SA = 12 cm2 + $$\frac{3}{2}$$ cm2 + 7 cm2 + $$\frac{77}{100}$$ cm2
SA = 20 cm2 + $$\frac{1}{2}$$ cm2 + $$\frac{77}{100}$$ cm2
SA = 20 cm2 + 1 cm2 + $$\frac{27}{100}$$ cm2
SA = 21 $$\frac{27}{100}$$ cm2

d. How are figures (b) and (c) related to figure (a)?
If the equilateral triangular faces of figures (b) and (c) were matched together, they would form the cube in part (a).

Question 6.
Find the surface area of the solid shown in the diagram. The solid is a right triangular prism (with right triangular bases) with a smaller right triangular prism removed from it.

SA = LA + 2B
LA = P ∙ h
LA = (4 in. + 4 in. + 5 $$\frac{13}{20}$$ in.) ∙ 2 in.
LA = (13 $$\frac{13}{20}$$ in.) ∙ 2 in.
LA = 26 in2 + $$\frac{13}{10}$$ in2
LA = 26 in2 + 1 in2 + $$\frac{3}{10}$$ in2
LA = 27 $$\frac{3}{10}$$ in2
The $$\frac{1}{4}$$in. by 4 $$\frac{19}{20}$$ in. rectangle has to be taken away from the lateral area:
A = lw
A = 4 $$\frac{19}{20}$$ in ∙ $$\frac{1}{4}$$ in
A = 1 in2 + 19/80 in2
A = 1 19/80 in2

LA = 27 $$\frac{3}{10}$$ in2-1 $$\frac{19}{80}$$ in2
LA = 27 $$\frac{24}{80}$$ in2-1 $$\frac{19}{80}$$ in2
LA = 26 $$\frac{5}{80}$$ in2
LA = 26 $$\frac{1}{16}$$ in2

Two bases of the larger and smaller triangular prisms must be added:
SA = 26 $$\frac{1}{16}$$ in2 + 2(3 $$\frac{1}{2}$$ in ∙ $$\frac{1}{4}$$ in) + 2($$\frac{1}{2}$$ ∙ 4 in ∙ 4 in)
SA = 26 $$\frac{1}{16}$$ in2 + 2 ∙ $$\frac{1}{4}$$ in ∙ 3 $$\frac{1}{2}$$ in + 16 in2
SA = 26 $$\frac{1}{16}$$ in2 + $$\frac{1}{2}$$ in ∙ 3 $$\frac{1}{2}$$ in + 16 in2
SA = 26 $$\frac{1}{16}$$ in2 + ($$\frac{3}{2}$$ in2 + $$\frac{1}{4}$$ in2 ) + 16 in2
SA = 26 $$\frac{1}{16}$$ in2 + 1 in2 + $$\frac{8}{16}$$ in2 + $$\frac{4}{16}$$ in2 + 16 in2
SA = 43 $$\frac{13}{16}$$ in2
The surface area of the solid is 43 $$\frac{13}{16}$$ in2.

Question 7.
The diagram shows a cubic meter that has had three square holes punched completely through the cube on three perpendicular axes. Find the surface area of the remaining solid.

Exterior surfaces of the cube (SA1):
SA1 = 6(1 m)2-6($$\frac{1}{2}$$ m)2
SA1 = 6(1 m2 )-6($$\frac{1}{4}$$ m2)
SA1 = 6 m2–$$\frac{6}{4}$$ m2
SA1 = 6 m2-(1 $$\frac{1}{2}$$ m2)
SA1 = 4 $$\frac{1}{2}$$ m2
Just inside each square hole are four intermediate surfaces that can be treated as the lateral area of a rectangular prism. Each has a height of $$\frac{1}{4}$$ m and perimeter of $$\frac{1}{2}$$ m + $$\frac{1}{2}$$ m + $$\frac{1}{2}$$ m + $$\frac{1}{2}$$ m or 2 m.
SA2 = 6(LA)
SA2 = 6(2 m ∙ $$\frac{1}{4}$$ m)
SA2 = 6 ∙ $$\frac{1}{2}$$ m2
SA2 = 3 m2

The total surface area of the remaining solid is the sum of these two areas:
SAT = SA1 + SA2.
SAT = 4 $$\frac{1}{2}$$ m2 + 3 m2
SAT = 7 $$\frac{1}{2}$$ m2
The surface area of the remaining solid is 7 $$\frac{1}{2}$$ m2.

### Eureka Math Grade 7 Module 3 Lesson 22 Exit Ticket Answer Key

Question 1.
The right hexagonal pyramid has a hexagon base with equal-length sides. The lateral faces of the pyramid are all triangles (that are exact copies of one another) with heights of 15 ft. Find the surface area of the pyramid.

SA = LA + 1B
LA = 6 ∙ $$\frac{1}{2}$$ (bh)
LA = 6 ∙ $$\frac{1}{2}$$ (5 ft. ∙ 15 ft.)
LA = 3 ∙ 75 ft2
LA = 225 ft2

B = Arectangle + 2Atriangle
B = (8 ft. ∙ 5 ft.) + 2 ∙ $$\frac{1}{2}$$ (8 ft. ∙ 3 ft.)
B = 40 ft2 + (8 ft. ∙ 3 ft.)
B = 40 ft2 + 24 ft2

B = 64 ft2
SA = LA + 1B
SA = 225 ft2 + 64 ft2
SA = 289 ft2
The surface area of the pyramid is 289 ft2.

Question 2.
Six cubes are glued together to form the solid shown in the diagram. If the edges of each cube measure 1 $$\frac{1}{2}$$ inches in length, what is the surface area of the solid?

There are 26 square cube faces showing on the surface area of the solid (5 each from the top and bottom view, 4 each from the front and back view, 3 each from the left and right side views, and 2 from the “inside” of the front).
A = s2
A = (1 $$\frac{1}{2}$$ in.)2
A = (1 $$\frac{1}{2}$$ in.)(1 $$\frac{1}{2}$$ in.)
A = 1 $$\frac{1}{2}$$ in.(1 in. + $$\frac{1}{2}$$ in.)
A = (1 $$\frac{1}{2}$$ in. ∙ 1 in.) + (1 $$\frac{1}{2}$$ in. ∙ $$\frac{1}{2}$$ in.)
A = 1 $$\frac{1}{2}$$ in2 + $$\frac{3}{4}$$ in2
A = 1 $$\frac{2}{4}$$ in2 + $$\frac{3}{4}$$ in2
A = 1 $$\frac{5}{4}$$ in2 = 2 $$\frac{1}{4}$$ in2

SA = 26 ∙ (2 $$\frac{1}{4}$$ in2)
SA = 52 in2 + $$\frac{26}{4}$$ in2
SA = 52 in2 + 6 in2 + $$\frac{1}{2}$$ in2
SA = 58 $$\frac{1}{2}$$ in2
The surface area of the solid is 58 $$\frac{1}{2}$$ in2.

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# On refined volatility smile expansion in the Heston model

## Author Info

• P. Friz
• S. Gerhold
• A. Gulisashvili
• S. Sturm
Registered author(s):

## Abstract

It is known that Heston's stochastic volatility model exhibits moment explosion, and that the critical moment $s_+$ can be obtained by solving (numerically) a simple equation. This yields a leading order expansion for the implied volatility at large strikes: $\sigma_{BS}( k,T)^{2}T\sim \Psi (s_+-1) \times k$ (Roger Lee's moment formula). Motivated by recent "tail-wing" refinements of this moment formula, we first derive a novel tail expansion for the Heston density, sharpening previous work of Dragulescu and Yakovenko [Quant. Finance 2, 6 (2002), 443--453], and then show the validity of a refined expansion of the type $\sigma_{BS}( k,T) ^{2}T=( \beta_{1}k^{1/2}+\beta_{2}+...)^{2}$, where all constants are explicitly known as functions of $s_+$, the Heston model parameters, spot vol and maturity $T$. In the case of the "zero-correlation" Heston model such an expansion was derived by Gulisashvili and Stein [Appl. Math. Optim. 61, 3 (2010), 287--315]. Our methods and results may prove useful beyond the Heston model: the entire quantitative analysis is based on affine principles: at no point do we need knowledge of the (explicit, but cumbersome) closed form expression of the Fourier transform of $\log S_{T}$\ (equivalently: Mellin transform of $S_{T}$ ); what matters is that these transforms satisfy ordinary differential equations of Riccati type. Secondly, our analysis reveals a new parameter ("critical slope"), defined in a model free manner, which drives the second and higher order terms in tail- and implied volatility expansions.

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File URL: http://arxiv.org/pdf/1001.3003

## Bibliographic Info

Paper provided by arXiv.org in its series Papers with number 1001.3003.

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 Length: Date of creation: Jan 2010 Date of revision: Nov 2010 Handle: RePEc:arx:papers:1001.3003 Contact details of provider: Web page: http://arxiv.org/

## References

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1. S. Benaim & P. Friz, 2009. "Regular Variation And Smile Asymptotics," Mathematical Finance, Wiley Blackwell, vol. 19(1), pages 1-12.
2. Leif Andersen & Vladimir Piterbarg, 2007. "Moment explosions in stochastic volatility models," Finance and Stochastics, Springer, vol. 11(1), pages 29-50, January.
3. A. Gulisashvili & E. M. Stein, 2009. "Asymptotic Behavior of the Stock Price Distribution Density and Implied Volatility in Stochastic Volatility Models," Papers 0906.0392, arXiv.org.
4. A. Gulisashvili, 2009. "Asymptotic Formulas with Error Estimates for Call Pricing Functions and the Implied Volatility at Extreme Strikes," Papers 0906.0394, arXiv.org.
5. Adrian A. Dragulescu & Victor M. Yakovenko, 2002. "Probability distribution of returns in the Heston model with stochastic volatility," Papers cond-mat/0203046, arXiv.org, revised Nov 2002.
6. Adrian Dragulescu & Victor Yakovenko, 2002. "Probability distribution of returns in the Heston model with stochastic volatility," Quantitative Finance, Taylor & Francis Journals, vol. 2(6), pages 443-453.
7. A. Dragulescu & V. M. Yakovenko, 2002. "Probability distribution of returns in the Heston model with stochastic volatility," Computing in Economics and Finance 2002 127, Society for Computational Economics.
8. Martin Forde & Antoine Jacquier & Aleksandar Mijatovic, 2009. "Asymptotic formulae for implied volatility in the Heston model," Papers 0911.2992, arXiv.org, revised May 2010.
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text:
# Challenge Math Challenge - July 2019

• Featured

#### QuantumQuest

Gold Member
I may have overthought about problem 2.
Yes indeed. The solution you give in post #19 using simple calculus is much faster and simpler. Well done.

I noticed that P4 required the use of calculus? What, precisely, does that mean? If my approach doesn't qualify, it's fine.
There are various ways that can be used but I ask for the use of calculus as it is a fast and simple approach. Your first solution for question $4$ gets unnecessarily complex but the second is fine.

#### QuantumQuest

Gold Member
Let $f(x) = arctan(x).$

Since $arctan(x)$ is differentiable for all x, by the Mean Value Theorem we know
$$\frac{f(b) - f(a)}{b - a} = f'(c)$$
for some c on the interval $a < c < b$.

$f'(x) = \frac{1}{1 + x^2}$, which is decreasing for $x > 0$, so for $0 < a < c < b$, $f'(b) < f'(c) < f'(a)$.

Substituting, we have
$$f'(b) < \frac{f(b) - f(a)}{b - a} < f'(a)$$
$$\frac{1}{1 + b^2} < \frac{arctan(b) - arctan(a)}{b - a} < \frac{1}{1 + a^2}$$
$$\frac{b - a}{1 + b^2} < arctan(b) - arctan(a) < \frac{b - a}{1 + a^2}$$

Let $a = 1, b = 4/3$.

$$\frac{\frac{4}{3} - 1}{1 + (\frac{4}{3})^2} < arctan(4/3) - arctan(1) < \frac{\frac{4}{3} - 1}{1 + 1^2}$$
$$\frac{3}{25} < arctan(4/3) - \frac{\pi}{4} < 1/6$$
$$\frac{\pi}{4} + \frac{3}{25} < arctan(4/3) < \frac{\pi}{4} + \frac{1}{6}$$
Well done @Flatlanderr

#### LagrangeEuler

My solution to 11.
View attachment 246120

By the way i wanted to know whether can we solve the same integral but indefinite.If we can please provide a solution.
How you change $x \to \frac{\pi}{2}-x$ without changing interval of integration?

#### Math_QED

Homework Helper
How you change $x \to \frac{\pi}{2}-x$ without changing interval of integration?
The new integration bounds are from $\pi/2$ to $0$ and we get a minus sign for the integral as a result of the substitution. Switching the integral bounds removes the minus sign.

#### Physics lover

How you change $x \to \frac{\pi}{2}-x$ without changing interval of integration?
This is a propertie of definite integral
$\int_0^a f(x) \, dx =$ $\int_0^a f(0+a-x) \, dx$

#### fresh_42

Mentor
2018 Award
Oh, come on ...!

No graphics, handwritings and even more, no uploads on external servers please!
Presentation is part of what should be learnt with these riddles. To know a solution is far from being able to convince others that your solution is one.

#### Pi-is-3

I am having a lot of problem with latex. I am not that bad at latex in AoPS or Stack exchange but I am not able to use it properly here :(. Example, my first line of answer should be this-

By AM-GM

$$\frac{(u+(1-v)+v+(1-w)+w+(1-u))}{6} \geq (u(1-v)v(1-w)w(1-u))^{\frac{1}{6}$$

but the Latex is not working. What am I doing wrong?

#### fresh_42

Mentor
2018 Award
I am having a lot of problem with latex. I am not that bad at latex in AoPS or Stack exchange but I am not able to use it properly here :(. Example, my first line of answer should be this-

By AM-GM

$$\frac{(u+(1-v)+v+(1-w)+w+(1-u))}{6} \geq (u(1-v)v(1-w)w(1-u))^{\frac{1}{6}$$

but the Latex is not working. What am I doing wrong?
This is due to your laziness. If you copy and paste instead of actually writing it, you also copy hidden control sequences as colors or fonts. This doesn't work in LaTeX here.

If it helps you, you can download a script program, e.g. AutoHotkey, which allows you to abbreviate certain key sequences by keyboard shortcuts. For example I have \frac{}{} on Ctrl+F or \ begin{bmatrix} \ end{bmatrix} on Alt+M (without the blanks).

#### Pi-is-3

@fresh_42 Thanks for telling my mistake! Now I can latex again
By AM-GM

$$\frac {(u)+(1-v)+(v)+(1-w)+(w)+(1-u)} {6} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{6}}$$

Implies

$$\frac {1} {2} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{6}}$$

Squaring both sides (works in this inequality as both sides are positive)

$$\frac {1} {4} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{3}}$$

Now using GM-HM

$$\frac {1} {4} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{3}} \geq \frac {3} {\frac{1}{u(1-v)}+\frac{1}{v(1-w)} + \frac{1}{w(1-u)}}$$

Implies

$$\frac{1}{u(1-v)} + \frac{1}{v(1-w)} + \frac{1}{w(1-u)} \geq 12$$

W.L.O.G assume $u(1-v) \leq v(1-w) \leq w(1-u)$

then $\frac{3}{u(1-v)} \geq 12$

implies $$\frac{1}{4} \geq u(1-v)$$

Hence Proved

#### fresh_42

Mentor
2018 Award
@fresh_42 Thanks for telling my mistake! Now I can latex again
By AM-GM

$$\frac {(u)+(1-v)+(v)+(1-w)+(w)+(1-u)} {6} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{6}}$$

Implies

$$\frac {1} {2} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{6}}$$

Squaring both sides (works in this inequality as both sides are positive)

$$\frac {1} {4} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{3}}$$

Now using GM-HM

$$\frac {1} {4} \geq [(u)(1-v)(v)(1-w)(w)(1-u)]^{\frac{1}{3}} \geq \frac {3} {\frac{1}{u(1-v)}+\frac{1}{v(1-w)} + \frac{1}{w(1-u)}}$$

Implies

$$\frac{1}{u(1-v)} + \frac{1}{v(1-w)} + \frac{1}{w(1-u)} \geq 12$$

W.L.O.G assume $u(1-v) \leq v(1-w) \leq w(1-u)$

then $\frac{3}{u(1-v)} \geq 12$

implies $$\frac{1}{4} \geq u(1-v)$$

Hence Proved
Well done and far better written than your first attempts, and I do not mean LaTeX, I mean the structure of your proof!

You should consider to download this little helper. It saves so much time, that I can almost type at a normal speed without all these special characters. I even use keys for \alpha , \beta, \omega etc.

If anyone still wants to try: there is another proof using a binomial formula.

Last edited:

#### BWV

#7
Isnt #7 just the variance of W?
Var(W)=E[W^2]-E^2[W]
E[W]=E^2[W]=0
So
var(w)=E[W^2]=t

Last edited:

#### fresh_42

Mentor
2018 Award
Well, from science advisors and such experienced members of PF as you are we can certainly expect that they can solve problems meant for the kids still in school.

#### hilbert2

Gold Member
For anyone learning proof writing and presentation skills in mathematics, it's important to remember that a proof never becomes any better by making it look like something esoteric is taking place there. Most "normal" people won't understand it anyway, even when it's made "as simple as possible but no simpler" (a quote from Einstein). When a professional mathematician writes a new long proof of some theorem, it's difficult enough to follow the reasoning in it even when it contains all the necessary detail and the reader is a professional of equal standing.

#### nuuskur

Great, I am not normal

#### Pi-is-3

Set $A=1+i, B=1-i, C=-1-i, D=-1+i$ as the vertices of the square.
Our circle passes through $A, D$ and $\frac{C+B}{2}=M=-i$
The center lies on the intersection of y-axis and perpendicular bisector of $DM$.
equating slopes (or equations lines, depends on you), we get the center $K= \frac{i}{4}$
Using distance formula we find circle intersects the side DC at point $L=(-1,-0.5)$.
So we get, the circle divides the side DC into the ratio 3:1

#### Math_QED

Homework Helper
#7
Isnt #7 just the variance of W?
Var(W)=E[W^2]-E^2[W]
E[W]=E^2[W]=0
So
var(w)=E[W^2]=t
It is not. Why do you think so?

Note that the question is changed to calculating $\mathbb{E}[X]$, which is a lot easier.

#### nuuskur

I know P13 was solved, but I like the exercise so I tried something, too.
For every $0<x<1$ it holds $x(1-x)\leq 1/4$ (find the vertex of the parabola). Assume $u(1-v) > 1/4$ and $v(1-w) > 1/4$. Need to show $w(1-u) \leq 1/4$.

If $1-u \leq 1-w$, then
$$w(1-u) \leq w(1-w) \leq 1/4.$$

Assume $1-u>1-w$. From
$$v(1-u) > v(1-w) > 1/4 \geq w(1-w)$$
we further obtain $u<w<v$. But now
$$1/4 < u(1-v) < u(1-u) \leq 1/4,$$
which is impossible. Thus $1-u \leq 1-w$ must hold.

#### lpetrich

Problem 9
I recognize this problem as Cholesky decomposition - Wikipedia

The proof of uniqueness is by construction, and this construction uses the Cholesky-Banachiewicz-Crout algorithm, as it might be called. One must find lower-triangular matrix L for matrix A such that they suffer $\mathbf{A} = \mathbf{L} \mathbf{L}^T$. It is easy to show that taking the transpose of both sides yields this equation again. Here is that algorithm, for A having size n*n:

For j = 1 to n do:
$$L_{jj} = \sqrt{ A_{jj} - \sum_{k=1}^{j-1} L_{jk}^2 }$$
For i = (j+1) to n do:
$$L_{ij} = \frac{1}{L_{jj}} \left( A_{ij} - \sum_{k=1}^{j-1} L_{ik} L_{jk} \right)$$
Next i
Next j

Each new L component depends only on an A component and on previously-calculated L components, so after one pass, the calculation is complete.

I must also calculate L for
$$A = \begin{pmatrix} 4 & 2 & 4 & 4 \\ 2 & 10 & 17 & 11 \\ 4 & 17 & 33 & 29 \\ 4 & 11 & 29 & 39 \end{pmatrix}$$

I used Mathematica's CholeskyDecomposition[] function and took the result's transpose, since that function calculates an upper triangular matrix. I then verified that that result is correct. It is
$$L = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 \\ 2 & 5 & 2 & 0 \\ 2 & 3 & 5 & 1 \end{pmatrix}$$

#### lpetrich

Problem 8, partial solution
It is necessary to find the Galois group for the splitting field of $x^4 - 2x^2 - 2$ over Q.

The roots of that equation are $\pm \sqrt{1 \pm \sqrt{3} }$. Their four symmetry operations are (identity), (reverse outer square-root sign), (reverse inner square-root sign), and (reverse both square-root signs). The group of these operations is rather obviously $Z_2 \times Z_2$.

#### Math_QED

Homework Helper
Problem 8, partial solution
It is necessary to find the Galois group for the splitting field of $x^4 - 2x^2 - 2$ over Q.

The roots of that equation are $\pm \sqrt{1 \pm \sqrt{3} }$. Their four symmetry operations are (identity), (reverse outer square-root sign), (reverse inner square-root sign), and (reverse both square-root signs). The group of these operations is rather obviously $Z_2 \times Z_2$.
You are right about the roots, but that's about it. The Galoisgroup is not the Klein-group, nor has it order 4.

#### ArcHorizon

HI! I just came across those questions and thought if it would be okay if I could send you some math questions for the August math challenge 2019!

#### nuuskur

Not really my area, but here goes nothing. Based on Brownian motion characterisation under section titled "Mathematics"
Put $X := \int _0^t W_s^2 ds$. Based on conditions 1 and 4 of the characterisation we have $W_s \sim \mathcal N(0,s)$, where $s\geq 0$. Therefore
$$s = \mbox{var}(W_s) = \mathbb E (W_s - \mathbb EW_s)^2 = \mathbb E(W_s^2).$$
Fubini allows us to change order of integration, so we get
$$\mathbb EX = \mathbb E \left ( \int _0^t W_s^2ds\right ) = \int _0^t \mathbb E(W_s^2)ds = \int_0^t sds = \frac{t^2}{2}$$
Initially, P7 asked for variance, which turns out to be
$$\mbox{var}(X) = \mathbb E (X - \mathbb EX)^2 = \mathbb E\left (X - \frac{t^2}{2}\right )^2 = \mathbb E(X^2) - \frac{t^4}{4}$$
So we need to calculate the second moment of $X$. I'm not sure at the moment what's happening here.

Last edited:

#### Math_QED

Homework Helper
Not really my area, but here goes nothing. Based on Brownian motion characterisation under section titled "Mathematics"
Put $X := \int _0^t W_s^2 ds$. Based on conditions 1 and 4 of the characterisation we have $W_s \sim \mathcal N(0,s)$, where $s\geq 0$. Therefore
$$s = \mbox{var}(W_s) = \mathbb E (W_s - \mathbb EW_s)^2 = \mathbb E(W_s^2).$$
Fubini allows us to change order of integration, so we get
$$\mathbb EX = \mathbb E \left ( \int _0^t W_s^2ds\right ) = \int _0^t \mathbb E(W_s^2)ds = \int_0^t sds = \frac{t^2}{2}$$
Initially, P7 asked for variance, which turns out to be
$$\mbox{var}(X) = \mathbb E (X - \mathbb EX)^2 = \mathbb E\left (X - \frac{t^2}{2}\right )^2 = \mathbb E(X^2) - \frac{t^4}{4}$$
So we need to calculate the second moment of $X$. I'm not sure at the moment what's happening here.
Correct! It should be noted that is non-trivial that the map $(t,\omega) \mapsto W_t(\omega)$ is jointly measurable (i.e. measurable w.r.t. the product sigma algebra), which is necessary to apply Fubini. But you could use this without a proof.

The calculation of $\mathbb{E}[X^2]$ is harder. If you want to attempt it anyway, here is a hint to get you (or someone else who wants to try) started:

$$X^2 = \left(\int_0^t W_u^2 du\right) \left(\int_0^t W_v^2 dv\right)=\int_0^t \int_0^t W_u^2 W_v^2 dudv$$

Now, taking expectations on both sides you can use Fubini on a triple integral and the problem reduces to finding $\mathbb{E}[W_u^2 W_v^2]$.

Last edited:

#### nuuskur

It gets weird, there has to be some kind of algebraic trick involved, which I can't think of.
We could try
$$\mathbb E(W_uW_v)^2 = \mbox{var}(W_uW_v) + \mathbb E^2 (W_uW_v)$$
This brings more complications, though. By linearity we get
$$v\leq u \implies u-v=\mbox{var}(W_u-W_v) = \mathbb E(W_u-W_v)^2 = u - 2\mathbb E(W_uW_v) + v$$
from which
$$v\leq u \implies \mathbb E(W_uW_v) = v$$
not sure how that's helpful, though.

#### Math_QED

Homework Helper
It gets weird, there has to be some kind of algebraic trick involved, which I can't think of.
We could try
$$\mathbb E(W_uW_v)^2 = \mbox{var}(W_uW_v) + \mathbb E^2 (W_uW_v)$$
This brings more complications, though. By linearity we get
$$v\leq u \implies u-v=\mbox{var}(W_u-W_v) = \mathbb E(W_u-W_v)^2 = u - 2\mathbb E(W_uW_v) + v$$
from which
$$v\leq u \implies \mathbb E(W_uW_v) = v$$
not sure how that's helpful, though.
Your attempt shows in fact that $E(W_u W_v) = \min\{u,v\}$, which is correct. But you don't need it here. What you need is $E(W_u^2 W_v^2)$. Here is a hint:

Suppose $u \geq v$. Then

$$E(W_u^2 W_v^2) = E[(\{W_u-W_v\}+W_v)^2 W_v^2]$$

and $W_u- W_v, W_v$ are independent variables. What do you know about the expectation of the product of two independent random variables?

"Math Challenge - July 2019"

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text:
# Double Angle Formulas for Sine and Cosine

• PRACTICE (online exercises and printable worksheets)

The double-angle formulas for sine and cosine tell how to find the sine and cosine of twice an angle ($\,2x\,$),
in terms of the sine and cosine of the original angle ($\,x\,$):

 double-angle formula for the sine: $$\sin 2x = 2\sin x\,\cos x$$ double-angle formulas for the cosine: three versions: \begin{align} \cos 2x &=\cos^2 x - \sin^2 x\cr\cr \cos 2x &= 1 - 2\sin^2 x\cr\cr \cos 2x &= 2\cos^2 x - 1 \end{align}

## Notes on the Double-Angle Formulas:

• The double-angle formulas are identities. That is, they are always true.
They could each be prefaced with:   ‘For all real numbers $\,x\,$ ... ’
• Recall the convention regarding the use of function notation with multi-letter function names (like ‘$\,\sin\,$’ and ‘$\,\cos\,$’):
parentheses can be dropped, as long as there is no confusion about order of operations.

Thus, you typically see ‘$\,\sin 2x\,$’ (no parentheses) instead of ‘$\,\sin(2x)\,$’ (with parentheses).

Both ‘$\,\sin 2x\,$’ and ‘$\,\sin(2x)\,$’ represent the same thing:
• the function name is ‘$\,\sin\,$’ (pronounced ‘sine’, long i)
• the input to the function is $\,2x\,$
• both ‘$\,\sin 2x\,$’ and ‘$\,\sin(2x)\,$’ represent the output from the function ‘$\,\sin\,$’ when the input is $\,2x\,$
• Recall that ‘$\,\sin^2 x\,$’ is a shorthand for $\,(\sin x)^2\,$, which equals $\,(\sin x)(\sin x)\,$.
Similarly, ‘$\,\cos^2 x\,$’ is a shorthand for $\,(\cos x)^2 = (\cos x)(\cos x)\,$.
• Here is a verbalization of the double-angle formula for the sine: $$\overbrace{\sin\strut}^{\text{ the sine }} \overbrace{(2x)\strut}^{\text{ of twice an angle }} \overbrace{=\strut}^{\text{ is }} \overbrace{\strut 2}^{\text{ twice }} \overbrace{\strut \sin x}^{\text{ the sine of the original angle }} \overbrace{\cdot\strut}^{\text{ times }} \overbrace{\strut \cos x}^{\text{ the cosine of the original angle }}$$
• Here is a verbalization of a double-angle formula for the cosine.
The other two versions can be similarly verbalized. $$\overbrace{\cos\strut}^{\text{ the cosine }} \overbrace{(2x)\strut}^{\text{ of twice an angle }} \overbrace{=\strut}^{\text{ is }} \overbrace{\strut 1}^{\text{ one }} \overbrace{-\strut}^{\text{ minus }} \overbrace{2\strut}^{\text{ twice }} \overbrace{\sin^2\strut}^{\text{ the sine squared }} \overbrace{(x)\strut}^{\text{ of the original angle}}$$
• In this section, the three versions of the double-angle formula for the cosine are referred to as follows:  $\cos 2x = \cos^2 x - \sin^2 x\,$ the version involving both sine and cosine $\cos 2x = 1 - 2\sin^2 x\,$ the version involving only the sine $\cos 2x = 2\cos^2 x - 1\,$ the version involving only the cosine
Note that these descriptions refer to what is happening on the right-hand side of the formulas.

## Derivations of the Double-Angle Formulas

The double-angle formulas are simple to prove, once the Addition Formulas for Sine and Cosine are in place.
For all real numbers $\,x\,$:

 $\sin 2x$ $=$ $\sin (x + x)$ (rename $\,2x\,$ as $\,x + x\,$) $=$ $\sin x\,\cos x + \cos x\,\sin x$ (the sine addition formula) $=$ $2\sin x\,\cos x$ (add like terms) $\cos 2x$ $=$ $\cos (x + x)$ (rename $\,2x\,$ as $\,x + x\,$) $=$ $\cos x\,\cos x - \sin x\,\sin x$ (the cosine addition formula) $=$ $(\cos x)^2 - (\sin x)^2$ (exponent notation) $=$ $\cos^2 x - \sin^2 x$ (shorthand notation)

By the Pythagorean Identity,   $\,\cos^2 x = 1 - \sin^2 x\,$   and   $\,\sin^2 x = 1 - \cos^2 x\,$.
Thus, we get two alternate versions of the cosine double angle formula: $$\cos 2x \quad =\quad \cos^2 x - \sin^2 x \quad =\quad (1 - \sin^2 x) - \sin^2 x \quad =\quad 1 - 2\sin^2 x$$ and $$\cos 2x \quad =\quad \cos^2 x - \sin^2 x \quad =\quad \cos^2 x - (1 - \cos^2 x) \quad =\quad \cos^2 x - 1 + \cos^2 x \quad =\quad 2\cos^2 x - 1$$

## A Geometric Proof of the Double-Angle Formulas, for Small Angles

For the sketch and derivation below, assume $\,x\,$ is measured in degrees and $\,2x < 90^\circ\,$.
This is a different ‘stacking’ of triangles than in the prior lesson, to show alternative proof approaches.

 Start with right triangle $\,\triangle ABC\,$ (yellow). It has acute angle $\,x\,$ and hypotenuse of length $\,1\,$. Thus: bottom leg: $\,\cos x\,$ side leg: $\,\sin x\,$ Stack a right triangle $\,\triangle ACD\,$ (mostly green) on the hypotenuse of the yellow triangle. It also has acute angle $\,x\,$. The bottom leg has length $\,1\,$. Since $\,\cos x = \frac{\text{ADJ}}{\text{HYP}} = \frac{1}{\text{HYP}}\,$, the hypoteuse has length $\,\frac{1}{\cos x}\,$. Thus, the side leg is $\,\frac{1}{\cos x}\cdot \sin x = \frac{\sin x}{\cos x}\,$. the blue right triangle: since $\,\overline{EC}\, ||\, \overline{AB}\,$, $\,\angle ACE = x\,$ (alternate interior angles) thus, $\,\angle ECD = 90^\circ - x\,$ thus, $\,\angle EDC = x\,$ Thus: bottom leg: $\,\frac{\sin x}{\cos x}(\sin x) = \frac{\sin^2 x}{\cos x}\,$ side leg: $\,\frac{\sin x}{\cos x}(\cos x) = \sin x\,$ Using right triangle $\,\triangle AFD\,$, we get the cosine double-angle formula: \begin{align} \cos 2x &= \frac{\text{ADJ}}{\text{HYP}}\cr &= \frac{\cos x - \frac{\sin^2 x}{\cos x}}{1/\cos x}\cr\cr &= \left(\cos x - \frac{\sin^2 x}{\cos x}\right)(\cos x)\cr\cr &= \cos^2 x - \sin^2 x\cr\cr \end{align} Again using right $\,\triangle AFD\,$, we get the sine double-angle formula: $$\sin 2x \ =\ \frac{\text{OPP}}{\text{HYP}} \ =\ \frac{\sin x +\sin x}{1/\cos x} \ =\ 2\sin x\,\cos x$$

## Review:   Does $\,\cos x\,$ Uniquely Determine $\,\sin x\,$?   No!

Recall that every nonzero real number is uniquely identified by knowing two things:

• its size (distance from zero)
• its sign (plus or minus)
This fact is often crucial when working with trigonometric expressions.

Often, knowing the value of a trigonometric expression (like $\,\cos x\,$)
allows you to correctly determine the size of other trigonometric expressions, but not the sign.
Additional information must be provided to get the correct sign.

For example, suppose you know that $\,\cos x = \frac 45\,$, and nothing else.
Does this allow us to uniquely determine $\,\sin x\,$?
No!
It gives the size of $\,\sin x\,$, but not the sign, as follows:
• Knowing $\,\cos x = \frac 45\,$ puts the terminal point for $\,x\,$ in either quadrant I (point $\,A\,$)
or quadrant IV (point $\,B\,$), as shown at right.
• Finding the size of $\,\sin x\,$ is routine.
For example, you can use the Pythagorean Identity: $$\sin^2 x \ =\ 1 - \cos^2 x \ =\ 1 - \left(\frac 45\right)^2 \ =\ \frac{25}{25} - \frac{16}{25}\ =\ \frac{9}{25}$$ Thus, $\sin x = \pm \frac{3}{5}\,$.
The size of $\,\sin x\,$ is $\,\frac 35\,$. The sign is unknown (without additional information).
There is a variety of additional information that might be given to determine the sign of $\,\sin x\,$.
Here are some examples:

 information: consequence: information: consequence: $\,x\,$ is in Quadrant I $\,\sin x\,$ is positive $\,x\,$ is in Quadrant IV $\,\sin x\,$ is negative $\csc x > 0$ cosecant is the reciprocal of the sine;if cosecant is positive, so is sine;$\,\sin x\,$ is positive $\csc x < 0$ cosecant is the reciprocal of the sine;if cosecant is negative, so is sine;$\,\sin x\,$ is negative $\tan x > 0$ tangent is sine over cosine;if cosine is positiveand tangent is positive, then:$\,\sin x\,$ is positive $\tan x < 0$ tangent is sine over cosine;if cosine is positiveand tangent is negative, then:$\,\sin x\,$ is negative

## Example: Using the Double-Angle Formulas

Suppose that $\displaystyle\,\cos x = \frac 45\,$ and $\,\csc x < 0\,$.
Find $\,\sin 2x\,$, $\,\cos 2x\,$, and $\,\tan 2x\,$.

Solution:

• By the Pythagorean Identity,
$\displaystyle\sin^2 x \ =\ 1 - \cos^2 x \ =\ 1 - \left(\frac 45\right)^2 \ =\ \frac{25}{25} - \frac{16}{25}\ =\ \frac{9}{25}\,,$
so $\displaystyle\, \sin x = \pm \frac{3}{5}\,$.
• Since $\displaystyle\,\csc x := \frac{1}{\sin x} < 0\,$, we have $\,\sin x < 0\,$.
• Together, $\,\displaystyle\sin x = -\frac{3}{5}\,$.
• $\displaystyle\sin 2x \ =\ 2\sin x\,\cos x \ =\ 2\bigl(-\frac 35\bigr)\bigl(\frac 45\bigr) \ =\ -\frac{24}{25}\,$
• $\displaystyle\cos 2x \ =\ 2\cos^2 x - 1\ =\ 2\left(\frac 45\right)^2 - 1 \ =\ 2\cdot\frac{16}{25} - \frac{25}{25}\ =\ \frac{7}{25}$

(Any of the three cosine double-angle formulas can be used.
The one chosen here is ‘safest’ since it uses only given—not computed—information.)
• $\displaystyle\tan 2x \ :=\ \frac{\sin 2x}{\cos 2x}\ =\ \frac{-\frac{24}{25}}{\frac{7}{25}}\cdot \frac{25}{25}\ =\ -\frac{24}{7}$

A complex fraction is an expression that involves ‘fractions within fractions’, like $\,\displaystyle\frac{-\frac{24}{25}}{\frac{7}{25}}\,$.
Here is a technique used to transform a complex fraction to a simple fraction:
• Find the least common denominator (lcd) of all the simple fractions involved.
• Multiply the complex fraction by $\,1\,$ in the form of $\,\frac{\text{lcd}}{\text{lcd}}\,$.
This is a beautiful example of ‘multiplying by $\,1\,$ in an appropriate form’!

Here's how this technique was applied above:
• The top fraction ($\,-\frac{24}{25}\,$) has denominator $\,25\,$.
• The bottom fraction ($\,\frac{7}{25}\,$) also has denominator $\,25\,$.
• The least common multiple of $\,25\,$ and $\,25\,$ is $\,25\,$.
Thus, the lcd of all the simple fractions involved is $\,25\,$.
• The complex fraction was multipled by $\,\frac{25}{25}\,$, and all the ‘little’ fractions disappeared!

Master the ideas from this section


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# Algebraic topology

The branch of mathematics in which one studies such properties of geometrical figures (in a wider sense, of all objects for which one can speak of continuity), and their mappings into each other, which remain unchanged under continuous deformations (homotopies). In principle, the objective of algebraic topology is a complete listing of such properties. The very name of algebraic topology originates from the decisive role of algebraic notions and algebraic methods in solving problems in this field. The most important classes of objects whose properties are studied in algebraic topology include complexes (polyhedra, cf. Complex), which may be simplicial, cellular, etc.; manifolds (cf. Manifold), which may be open, closed or with boundary, and may in turn be divided into smooth (differentiable), analytic, complex-analytic, piecewise-linear or topological; and fibre bundles (fibrations, cf. Fibration) and their sections. The principal types of mappings considered in algebraic topology are arbitrary continuous, piecewise-linear and smooth mappings, and their most important subclasses are: homeomorphisms, in particular, continuous, piecewise-linear or smooth (diffeomorphisms); imbeddings of one object into another and also immersions (local imbeddings). (cf. Homeomorphism; Diffeomorphism.)

A very important notion in algebraic topology is that of a deformation. The principal types of deformations include: a homotopy, i.e. an arbitrary continuous (smooth, piecewise-linear) deformation of continuous mappings; an isotopy (continuous, smooth, piecewise-linear), i.e. a deformation of homeomorphisms, imbeddings or immersions where in the process of deformation the mapping is a homomorphism, an imbedding or an immersion at any moment of time.

The principal internal problems of algebraic topology include the problem of the classification of manifolds by homeomorphisms (continuous, smooth, piecewise-linear), the classification of imbeddings (or immersions) with respect to isotopies (regular homotopies), and the classification of general continuous mappings up to homotopy. An important intermediate role in the solution of these problems is played by the problem of the classification of complexes or manifolds by the so-called homotopy equivalence or homotopy type.

The following, somewhat special, problems played an important role in the development of algebraic topology.

1) The problem of imbedding is usually considered not in its general form, but rather as imbedding into a Euclidean space. A very important special case is knot theory (and the theory of links) in three-dimensional space, which was one of the origins of algebraic topology; braid theory is a related case.

2) An important role in the history of algebraic topology was played by the theory of homology invariants of the position of various sets in a Euclidean space and the duality laws (cf. Duality in algebraic topology), connecting the homology of a set and its complement.

3) A number of fundamental results concerned the computation of the algebraic number of fixed points of a mapping of a manifold into itself. Many fundamental facts were discovered for the fixed points of compact smooth groups of transformations, including cyclic groups of finite order.

4) Of major technical importance in the development of algebraic topology were methods developed to solve the problems on so-called cobordism: Does a manifold with boundary (cobordism) with a given closed manifold as boundary exist? Problems of this kind originally arose in the context of the computation of the homotopy groups of spheres (cf. Homotopy group). Important cases of the cobordism problem are solved in the language of characteristic classes (cf. Characteristic class).

5) Many facts are now available concerning the homology invariants of the singularities of vector fields, frame fields and tensor fields, as well as of the singularities of smooth mappings of manifolds, in particular into Euclidean space. The solution of this problem leads, in particular, to characteristic classes. An especially important case is that of stationary points of smooth functions on manifolds or of various functionals on path spaces (extremals); their connection with homology theory is important in the clarification of the geometrical structure of manifolds and in obtaining lower bounds for the number of extremals.

6) The study of the algebraic-topological properties of an important kind of special spaces — Lie groups — is closely connected with their algebraic structure, their representations and the calculus of variations on Lie groups (cf. Lie group). The results obtained for the topological structure of Lie groups form the base of numerous methods and facts of algebraic topology which apply to arbitrary manifolds. Algebraic topology of homogeneous manifolds is closely related to methods of Lie groups.

7) A major role in algebraic topology is played by special invariants connected with various algebraic structures over the fundamental group. The simplest invariants of this type appeared in knot theory and in the theory of three-dimensional manifolds; the development of their algebraic theory subsequently made substantial progress, and eventually formed a separate discipline — stable algebra or algebraic $K$-theory.

8) The analysis of the geometrical structure of a very large number of examples of the simplest and most frequently encountered manifolds (e.g. Lie groups, homogeneous spaces, manifolds of line elements and manifolds with discrete groups of motions), in conjunction with the fundamental principles of Riemannian geometry, led to the notion of a fibre bundle, which consists of a space (the total or fibre space), a base space, a projection of the total space onto the base space, a fibre (homeomorphic to a "fibre" of the projection), and the structure group of fibre transformations. In this context, one of the central problems in algebraic topology is the problem of the classification with respect to homotopy of fibre bundles and their sections. Principal fibre bundles and vector bundles are especially important (cf. Principal fibre bundle; Vector bundle).

A universal method for the solution of all principal problems in algebraic topology involves the construction of algebraic invariants which are effectively calculable in actual examples and which assume some discrete set of values; the value of the invariant may not vary during deformations in the corresponding class of mappings for the study of which the invariant has been constructed. A large number of essential invariants, the abundance of algebraic links between them and the difficulty in their calculation have determined the features of modern algebraic topology.

The calculation of algebraic-topological invariants of the simplest and most frequently encountered manifolds is not always a simple matter. Thus, the calculation of homology invariants of Lie groups and of many homogeneous spaces required much effort and involved the use of complicated methods. The calculation of homotypy groups is even more difficult. A number of the most important homotypy groups for Lie groups was calculated, against expectation, using the variational theory of geodesics; the knowledge of the table of these homotypy groups for Lie groups made the classification of certain vector bundles possible.

Most algebraic-topological invariants are a so-called functor on the category of topological spaces of the type studied. This means, roughly speaking, that the values of the invariant undergo natural transformations when the spaces are mapped into each other. For example, the fundamental groups of arbitrary spaces or their homology (cohomology) groups (rings) are connected by homomorphisms induced by continuous mappings; the characteristic classes (distinguished elements in (co)homology groups) are mapped into each other by the homomorphism corresponding to a morphism of manifolds; the result of cohomology operations performed on an element of the homology (cohomology) is converted, after a continuous mapping of the space, into the result of this operation performed on the image of this element, etc.

One of the principal properties, on which the study and the application of almost-all algebraic-topological invariants are based, is the fact that their effective construction is usually connected with an essential complementary geometrical structure. For instance, the construction of all the principal invariants up to homeomorphism for complexes involves a subdivision into simplices or cells, whereas the result of such a construction has to be invariant with respect to all continuous homeomorphisms, and even with respect to homotopy equivalences; examples are the fundamental group, the Euler characteristic, the homology group (Betti group), the cohomology ring and the cohomology operations (cf. Cohomology operation). In a similar manner, the structure of all the basic homeomorphism invariants for smooth manifolds involves their preliminary triangulation, i.e. reduction to complexes, or else to a significant extent the use of analytical tools, such as the construction of a homology ring by way of differential forms (skew-symmetric tensors) and differential operations on them, or the construction of characteristic classes using the singularities of vector fields, frame fields or tensor fields. Moreover, it may also be necessary to use tools of Riemannian geometry in certain cases, e.g. the definitions of the characteristic classes of a manifold or of a fibre bundle in terms of the Riemannian curvature play an important role, even though the result is invariant wih respect to all continuous homeomorphisms. In this context, the appearance of basic, effectively calculable invariants in the history of topology involved the difficult problem of proving the invariance of these quantities. This is another type of problem dealt with by algebraic topology. Thus, rational characteristic classes (or integrals of classes over cycles) proved to be topologically invariant and homotopically non-invariant. The totality of characteristic classes over the integers proved to be non-invariant with respect to continuous or even piecewise-linear homeomorphisms.

On the contrary, if an invariant is so constructed that it is invariant (by definition) with respect to continuous homeomorphisms (or some wider class of transformations — homotopy equivalences), then the calculation of such an invariant is usually difficult. The most important invariants of this type are the homotopy groups of classes of homotopic mappings of a sphere into the space under study. The calculation of even the simplest examples of homotopy groups involves major difficulties. Thus, the homotopy groups of the spheres themselves are only incompletely known, despite the large number of methods which were proposed for such calculations in the course of development of algebraic topology (cf. Homotopy group).

The method of classification of classes of homotopic mappings is based on homology theory in conjunction with the theory of fibre bundles with its apparatus of spectral sequences, as well as cohomology operations and their generalizations. The algebraic discipline which arose on the basis of the complicated computational tools of algebraic topology is known as homological algebra.

All the basic primary constructions of homology theory for complexes and smooth manifolds by way of triangulation or differential forms are effectively combinatorial — algebraic or analytic. However, it was noted in the course of development of algebraic topology that homology theory can be completely defined by a small number of formal properties (axioms) on which its computational tools are based:

1) Homotopic invariance of homology groups.

2) The homology groups are homomorphically mapped into each other under continuous mappings of spaces, and products of mappings correspond to the products of these homomorphisms (functoriality).

3) A certain form of interconnection is postulated between the cohomology of a complex, a subcomplex and a quotient complex (exactness axiom).

4) The so-called excision axiom.

5) Normalization: Only the homology groups for a point need be known, and they must be known to vanish in non-zero dimensions.

It was subsequently noted that objects of an altogether different geometrical nature may display all of the above properties, except for normalization. Such objects were called generalized or extra-ordinary homology theories. They were used to improve the calculation methods of algebraic topology. The most important example is $K$-theory, which is based on vector bundles over the space studied instead of their cycles or differential forms which served for the construction of the ordinary homology groups. Another important example is cobordism (bordism) theory, in which only the mappings of closed manifolds into the space under study are taken instead of arbitrary cycles, and only mappings of manifolds with boundary of some classes are taken instead of arbitrary boundaries.

The combination of the methods of algebraic topology with a small number of pure geometric facts applied to the structure of manifolds and their homeomorphism groups eventually resulted in the solution (which, from the modern point of view, may or may not be considered complete) of classification problems of manifolds with respect to all kinds of homeomorphisms (smooth, piecewise-linear, continuous), and in the solution of fundamental problems in the classification of imbeddings and immersions. Strangely enough, problems connected with three-dimensional and four-dimensional manifolds, where the fundamental problems are still (1977) unsolved, represent an exception.

Since most of the fundamental internal problems have now been solved, modern efforts in algebraic topology tend to concentrate on the application of these ideas in other fields.

#### References

 [1] M.F. Atiyah, "$K$-theory: lectures" , Benjamin (1967) MR224083 [2] H. Seifert, W. Threlfall, "A textbook of topology" , Acad. Press (1980) (Translated from German) MR0575168 Zbl 0469.55001 [3] J.W. Milnor, "Morse theory" , Princeton Univ. Press (1963) MR0163331 Zbl 0108.10401 [4] J. Milnor, "Lectures on the $h$-cobordism theorem" , Princeton Univ. Press (1965) MR190942 [5] J. Milnor, "Singular points of complex hypersurfaces" , Princeton Univ. Press (1968) MR0239612 Zbl 0184.48405 [6] J.W. Milnor, J. Stasheff, "Characteristic classes" , Ann. of Math. Studies , Princeton Univ. Press (1974) MR0440554 Zbl 0298.57008 [7] R.M. Switzer, "Algebraic topology - homotopy and homology" , Springer (1975) MR0385836 Zbl 0305.55001 [8] S.-T. Hu, "Homotopy theory" , Acad. Press (1959) MR0106454 Zbl 0088.38803 [9] N.E. Steenrod, "The topology of fibre bundles" , Princeton Univ. Press (1951) MR0039258 Zbl 0054.07103 [10] K. Nomizu, "Lie groups and differential geometry" , Math. Soc. Japan (1956) MR0084166 Zbl 0071.15402 [11] J. Leray, "Le calcule différentiel et intégrale sur une variété analytique complexe. (Problème de Cauchy III)" Bull. Soc. Math. France , 87 (1959) pp. 81–180 [12] S.S. Chern, "Complex manifolds without potential theory" , Springer (1979) MR0533884 Zbl 0444.32004 [13] G. Springer, "Introduction to Riemann surfaces" , Addison-Wesley (1957) MR0092855 Zbl 0078.06602 [14] P.S. Aleksandrov, "Combinatorial topology" , Graylock , Rochester (1956) (Translated from Russian) MR0076324 Zbl 0441.55002 Zbl 0097.15903 Zbl 0024.08404 Zbl 66.0947.03 [15] L.S. Pontryagin, "Grundzüge der kombinatorischen Topologie" , Deutsch. Verlag Wissenschaft. (1956) (Translated from Russian) [16] L.S. Pontryagin, "Smooth manifolds and their applications in homology theory" , Moscow (1976) (In Russian) [17] E.H. Spanier, "Algebraic topology" , McGraw-Hill (1966) MR0210112 MR1325242 Zbl 0145.43303 [18] S.P. Novikov, "Topology" R.V. Gamkrelidze (ed.) , Modern mathematics, fundamental directions , 1 (To appear) (Translated from Russian) ((Original: Moscow, 1986)) MR0895592 MR0887530 MR0876295 Zbl 0668.55001 Zbl 0639.00031 Zbl 0635.55002 Zbl 0839.55001 [19] S.P. Novikov, "The Cartan–Serre theorem and intrinsic homology" Russian Math. Surveys , 21 : 5 (1966) pp. 209–224 Uspekhi Mat. Nauk , 21 : 5 (1966) [20] F. Hirzebruch, "Topological methods in algebraic geometry" , Springer (1978) (Translated from German) MR1335917 MR0202713 Zbl 0376.14001 [21] B.A. Dubrovin, A.T. Fomenko, S.P. Novikov, "Methods and applications" , Modern geometry , 1–2 , Springer (1985) (Translated from Russian) MR0822730 MR0822729 MR0807945 Zbl 0565.57001 [22] B.A. Dubrovin, A.T. Fomenko, S.P. Novikov, "Methods of homology theory" , Modern geometry , 3 , Springer (To appear) (Translated from Russian)

 [a1] J.F. Adams, "Stable homotopy and generalized homology" , Univ. Chicago Press (1974) [a2] R. Bott, L.W. Tu, "Differential forms in algebraic topology" , Springer (1982) MR0658304 Zbl 0496.55001 [a3] S. Eilenberg, N.E. Steenrod, "Foundations of algebraic topology" , Princeton Univ. Press (1952) MR0050886 Zbl 0047.41402 [a4] M. Karoubi, "$K$-theory. An introduction" , Springer (1978) MR0488029 Zbl 0382.55002 [a5] J. Vaisman, "Cohomology and differential forms" , M. Dekker (1973) MR0341344 Zbl 0267.58001 [a6] G.W. Whitehead, "Elements of homotopy theory" , Springer (1978) pp. 23; 415–455 MR0516508 Zbl 0406.55001


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×

# raise to power zero?

Why is it that anything raise to power 0 is 1?

Note by Raideep Singh
3 years, 1 month ago

Sort by:

$$0 = k-k$$ , Thus $$n^0 = n^{k-k}$$ (say k is non-zero as we want to prove for 0) = $$\frac{n^k}{n^k} = 1$$

But this won't be true for $$n=0$$ as the denominator becomes $$0^k = 0$$ for non zero k , and hence the value is not defined (this is why we say that $$0^0$$ is not defined . ) Thus ANYTHING raised to 0 is not 1 .... any "non-zero" thing raised to 0 is 1....
;) · 3 years, 1 month ago

as u said that , 0 power 0 is not defined.... bt as far i know.. 0 to the 0 is e whose value is 2.17828(approx.) · 3 years, 1 month ago

Take example as 2^0. It can be expessed as 2^(x-x) = (2^x)/(2^x)=1 · 3 years, 1 month ago

Let a number $$a$$ be raised to $$x$$.

We have $$a^x$$.

Let's divide it by itself-

$$\frac{a^x}{a^x}$$.

Surely, these two quantities are the same-

$$\frac{a^x}{a^x} = \frac{a^x}{a^x}$$

Let's apply the law of indices to L.H.S & normal division to the R.H.S. Now, we have,

$$a^{x-x} = 1$$

$$\Rightarrow \boxed{a^{0} = 1}$$. · 3 years, 1 month ago


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# 15.1.1.2.1.12. cobra.flux_analysis.variability¶

## 15.1.1.2.1.12.1. Module Contents¶

cobra.flux_analysis.variability.flux_variability_analysis(model, reaction_list=None, loopless=False, fraction_of_optimum=1.0, pfba_factor=None)[source]

Determine the minimum and maximum possible flux value for each reaction.

Parameters: model (cobra.Model) – The model for which to run the analysis. It will not be modified. reaction_list (list of cobra.Reaction or str, optional) – The reactions for which to obtain min/max fluxes. If None will use all reactions in the model (default). loopless (boolean, optional) – Whether to return only loopless solutions. This is significantly slower. Please also refer to the notes. fraction_of_optimum (float, optional) – Must be <= 1.0. Requires that the objective value is at least the fraction times maximum objective value. A value of 0.85 for instance means that the objective has to be at least at 85% percent of its maximum. pfba_factor (float, optional) – Add an additional constraint to the model that requires the total sum of absolute fluxes must not be larger than this value times the smallest possible sum of absolute fluxes, i.e., by setting the value to 1.1 the total sum of absolute fluxes must not be more than 10% larger than the pFBA solution. Since the pFBA solution is the one that optimally minimizes the total flux sum, the pfba_factor should, if set, be larger than one. Setting this value may lead to more realistic predictions of the effective flux bounds. A data frame with reaction identifiers as the index and two columns: - maximum: indicating the highest possible flux - minimum: indicating the lowest possible flux pandas.DataFrame

Notes

This implements the fast version as described in [1]. Please note that the flux distribution containing all minimal/maximal fluxes does not have to be a feasible solution for the model. Fluxes are minimized/maximized individually and a single minimal flux might require all others to be suboptimal.

Using the loopless option will lead to a significant increase in computation time (about a factor of 100 for large models). However, the algorithm used here (see [2]) is still more than 1000x faster than the “naive” version using add_loopless(model). Also note that if you have included constraints that force a loop (for instance by setting all fluxes in a loop to be non-zero) this loop will be included in the solution.

References

 [1] Computationally efficient flux variability analysis. Gudmundsson S, Thiele I. BMC Bioinformatics. 2010 Sep 29;11:489. doi: 10.1186/1471-2105-11-489, PMID: 20920235
 [2] CycleFreeFlux: efficient removal of thermodynamically infeasible loops from flux distributions. Desouki AA, Jarre F, Gelius-Dietrich G, Lercher MJ. Bioinformatics. 2015 Jul 1;31(13):2159-65. doi: 10.1093/bioinformatics/btv096.
cobra.flux_analysis.variability.find_blocked_reactions(model, reaction_list=None, zero_cutoff=1e-09, open_exchanges=False)[source]

Finds reactions that cannot carry a flux with the current exchange reaction settings for a cobra model, using flux variability analysis.

Parameters: model (cobra.Model) – The model to analyze reaction_list (list) – List of reactions to consider, use all if left missing zero_cutoff (float) – Flux value which is considered to effectively be zero. open_exchanges (bool) – If true, set bounds on exchange reactions to very high values to avoid that being the bottle-neck. List with the blocked reactions list
cobra.flux_analysis.variability.find_essential_genes(model, threshold=None, processes=None)[source]

Return a set of essential genes.

A gene is considered essential if restricting the flux of all reactions that depend on it to zero causes the objective, e.g., the growth rate, to also be zero, below the threshold, or infeasible.

Parameters: model (cobra.Model) – The model to find the essential genes for. threshold (float, optional) – Minimal objective flux to be considered viable. By default this is 1% of the maximal objective. processes (int, optional) – The number of parallel processes to run. Can speed up the computations if the number of knockouts to perform is large. If not passed, will be set to the number of CPUs found. Set of essential genes set
cobra.flux_analysis.variability.find_essential_reactions(model, threshold=None, processes=None)[source]

Return a set of essential reactions.

A reaction is considered essential if restricting its flux to zero causes the objective, e.g., the growth rate, to also be zero, below the threshold, or infeasible.

Parameters: model (cobra.Model) – The model to find the essential reactions for. threshold (float, optional) – Minimal objective flux to be considered viable. By default this is 1% of the maximal objective. processes (int, optional) – The number of parallel processes to run. Can speed up the computations if the number of knockouts to perform is large. If not passed, will be set to the number of CPUs found. Set of essential reactions set


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# 0: Review of Concepts and Notation

Modular Arithmetic

We write the set of integers and the set of natural numbers as:

$Z\stackrel{\text{def}}{=}\{..., -2, -1, 0, 1, 2, ...\}\nonumber$

$N\stackrel{\text{def}}{=}\{0, 1, 2, ...\}\nonumber$

0: Review of Concepts and Notation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mike Rosulek.


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# zbMATH — the first resource for mathematics

## Mitrea, Dorina

Compute Distance To:
 Author ID: mitrea.dorina Published as: Mitrea, D.; Mitrea, Dorina
 Documents Indexed: 65 Publications since 1994, including 8 Books
all top 5

#### Co-Authors

 16 single-authored 43 Mitrea, Marius 18 Mitrea, Irina 4 Hofmann, Steve 4 Martell, José María (Chema) 4 Yan, Lixin 3 Monniaux, Sylvie 2 Brewster, Kevin 2 Marmolejo-Olea, Emilio 2 Morris, Andrew Joel 2 Taylor, Michael Eugene 1 Brigham, Dan 1 Duduchava, Roland 1 Duong, Xuan Thinh 1 Gesztesy, Fritz 1 Lu, Guozhen 1 McIntosh, Alan 1 Pipher, Jill Catherine 1 Rosenblatt, Heather 1 Schmutzler, Bernd 1 Shaw, Mei-Chi 1 Shi, Qiang 1 Verdera, Joan 1 Ziadé, Elia
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#### Serials

 5 Journal of Functional Analysis 3 Mathematical Methods in the Applied Sciences 3 Memoirs of the American Mathematical Society 2 Integral Equations and Operator Theory 2 Mathematische Nachrichten 2 Revista Matemática Iberoamericana 2 Differential and Integral Equations 2 Communications in Partial Differential Equations 2 Journal of Mathematical Sciences (New York) 2 Communications on Pure and Applied Analysis 2 Analysis & PDE 2 Universitext 1 American Mathematical Monthly 1 Indiana University Mathematics Journal 1 Mathematische Annalen 1 Numerical Functional Analysis and Optimization 1 Transactions of the American Mathematical Society 1 Journal of Integral Equations and Applications 1 Atti della Accademia Nazionale dei Lincei. Classe di Scienze Fisiche, Matematiche e Naturali. Serie IX. Rendiconti Lincei. Matematica e Applicazioni 1 M$^3$AS. Mathematical Models & Methods in Applied Sciences 1 Potential Analysis 1 Memoirs on Differential Equations and Mathematical Physics 1 Advances in Applied Clifford Algebras 1 Advances in Differential Equations 1 Electronic Research Announcements of the American Mathematical Society 1 The Journal of Fourier Analysis and Applications 1 Comptes Rendus. Mathématique. Académie des Sciences, Paris 1 Complex Variables and Elliptic Equations 1 De Gruyter Studies in Mathematics 1 Proceedings of Symposia in Pure Mathematics 1 Electronic Research Announcements in Mathematical Sciences 1 Applied and Numerical Harmonic Analysis
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#### Fields

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text:
## Finding Centroid

1. The problem statement, all variables and given/known data
Let U be the solid region in the first octant bounded by the ellipsoid (x^2)/4 + (y^2)/9 + (z^2)/4 = 1. Find the centroid of U.

2. Relevant equations

3. The attempt at a solution

I tried to do this problem but I'm not sure if my answer is right. First, I find the mass and I got 2pi. Then I find the moment, and divided it by mass, and I got the centroid to be like (1,9/8,1). But I'm not sure if I did it right or not. Can anyone help me?

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Gold Member Science Advisor Staff Emeritus That can't be right. Since the "y-length" is 3, and the other lengths 2, the y coordinate of the centroid must be 3/2 the other coordinates. I did this by changing to "elliptic coordinates". More precisely, I let $x= 2\rho cos(\theta) sin(\phi)$, $y= 3\rho sin(\theta) sin(\phi)$ and $z= 2\rho cos(\phi)$- spherical coordinates 'altered' to fit the ellipse. Calculating the Jacobian gives $12\rho^2 sin(\phi)d\rho d\phi d\theta$ as the differential. The ellipse in the first octant take $\rho$ from 0 to 1, $\phi$ from 0 to $\pi/2$ and $\theta$ from 0 to $\pi/2$. I get (3/8, 9/16, 3/8) as the centroid. (By the way, $2\pi$ is the volume of the ellipse, not the "mass". "Centroid" is a purely geometrical concept and geometric figures do not have "mass".)


text:
# How to create a pivot table in Google BigQuery

on
jun 27, 2021
in

To calculate the elapsed times of Terraform Cloud run stages, I needed to create a pivot table in Google BigQuery. As the classic SQL query was very expensive, I changed it to use the BigQuery PIVOT operator which was 750 times cheaper.

Terraform Cloud allows you to manage your infrastructure at an enterprise scale. It is a managed service, and the pricing is based on the number of users and the number of concurrent runs you need. By default you get a single agent or runner. Extra runners cost a few thousand dollar per year. So, you need to balance throughput and cost. To do this, you need data on the utilization of the runs. Terraform Cloud does not provide any data out of the box, so I activated the Terraform run notifications webhook and streamed the data into BigQuery. To get useful information I had to pivot the data.

The Terraform run notification webhook, emits a notification whenever a run changes state. These notifications look as follows:

{
"run_url": "https://app.terraform.io/app/binx-io/my-ws/runs/run-BbUuHJv3w1kztNDi",
"run_id": "run-BbUuHJv3w1kztNDi",
"run_message": "Queued manually using Terraform",
"run_created_at": "2021-06-12T12:11:20.000Z",
"run_created_by": "mjvanholsteijn",
"workspace_id": "ws-tgHKyu5U8a3dDUGC",
"workspace_name": "my-ws",
"organization_name": "binx-io",
{
"message": "Run Created",
"trigger": "run:created",
"run_status": "pending",
"run_updated_at": "2021-06-12T12:11:20.000Z",
"run_updated_by": "mjvanholsteijn"
}
]
}

The notification only reports the time of the state change. It does not report the elapsed time since the previous state change.

As I loaded all of these events into BigQuery, I had to find a smart way to collect all of the timestamps of the appropriate state changes in a single row.

As you can see from the sample, the notifications is modeled as an array. So Terraform potentially sends multiple state changes in a single notification. To make it a simple table, I use the unnest operator:

SELECT
e.workspace_id,
e.run_id,
n.run_updated_at,
n.run_status
FROM
UNNEST(notifications) AS n

Which delivers a table with a row for each of the notifications in the array:

|workspace_id       |run_id              |run_updated_at     |run_status|
|-------------------|--------------------|----------|--------|---|------|
|ws-tgHKyu5U8a3dDUGC|run-ubJZBXt4aT2DprjJ|2021-06-13 20:49:02|applied|
|ws-tgHKyu5U8a3dDUGC|run-EMYsYn6oZd36Bk2o|2021-06-13 19:47:53|pending|
|ws-tgHKyu5U8a3dDUGC|run-ubJZBXt4aT2DprjJ|2021-06-13 20:48:53|planned|
|ws-tgHKyu5U8a3dDUGC|run-jLcoV2rTiWpncAzp|2021-06-13 19:43:37|applying|
|ws-tgHKyu5U8a3dDUGC|run-jLcoV2rTiWpncAzp|2021-06-13 19:43:27|planning|
|ws-tgHKyu5U8a3dDUGC|run-BbUuHJv3w1kztNDi|2021-06-12 12:11:20|pending|

To be able to calculate the duration of each stage, I needed to pivot the table.

## Using plain-ol’ SQL

Unaware of the PIVOT operator, my first solution was to use plain old SQL to solve this problem. The following statement uses a sub-select for each of the event notification times:

SELECT
workspace_id,
run_id,
(SELECT ANY_VALUE(n.run_updated_at)
WHERE workspace_id = e.workspace_id AND run_id = e.run_id AND n.run_status = 'pending') as pending,
(SELECT ANY_VALUE(n.run_updated_at)
WHERE workspace_id = e.workspace_id AND run_id = e.run_id AND n.run_status = 'planning') as planning,
(SELECT ANY_VALUE(n.run_updated_at)
WHERE workspace_id = e.workspace_id AND run_id = e.run_id AND n.run_status = 'planned') as planned,
(SELECT ANY_VALUE(n.run_updated_at)
WHERE workspace_id = e.workspace_id AND run_id = e.run_id AND n.run_status = 'applying') as applying,
(SELECT ANY_VALUE(n.run_updated_at)
WHERE workspace_id = e.workspace_id AND run_id = e.run_id AND n.run_status = 'applied') as applied,
FROM
( SELECT DISTINCT workspace_id, run_id
) e

It produces the expected result:

workspace_id run_id pending planning planned applying applied
ws-tgHKyu5U8a3dDUGC run-ubJZBXt4aT2DprjJ 2021-06-13 20:48:45 2021-06-13 20:48:47 2021-06-13 20:48:53 2021-06-13 20:48:55 2021-06-13 20:49:02
ws-tgHKyu5U8a3dDUGC run-jLcoV2rTiWpncAzp 2021-06-13 19:43:25 2021-06-13 19:43:27 2021-06-13 19:43:34 2021-06-13 19:43:37 2021-06-13 19:43:44
ws-tgHKyu5U8a3dDUGC run-EMYsYn6oZd36Bk2o 2021-06-13 19:47:53 2021-06-13 19:47:53 2021-06-13 19:48:00 2021-06-13 19:48:02 2021-06-13 19:48:06

But the cost were horrendous! The query returned in 1.5 seconds and spend almost 30 seconds of slot time. For a table with just a handful of rows, that is clearly not acceptable.

## pivot table in Google BigQuery

After some googling, I found that the PIVOT operator does exactly what I want. The following statement will add the columns pending, planning, planned, applying, applied with
the column’s value set to max(run_updated_at).

SELECT
*
FROM (
SELECT
e.workspace_id,
e.run_id,
n.run_updated_at,
n.run_status
FROM
)
PIVOT (
MAX(run_updated_at)
FOR run_status IN ('pending','planning','planned', 'applying','applied')
)

It produces the following result:

workspace_id run_id pending planning planned applying applied
ws-tgHKyu5U8a3dDUGC run-ubJZBXt4aT2DprjJ 2021-06-13 20:48:45 2021-06-13 20:48:47 2021-06-13 20:48:53 2021-06-13 20:48:55 2021-06-13 20:49:02
ws-tgHKyu5U8a3dDUGC run-jLcoV2rTiWpncAzp 2021-06-13 19:43:25 2021-06-13 19:43:27 2021-06-13 19:43:34 2021-06-13 19:43:37 2021-06-13 19:43:44
ws-tgHKyu5U8a3dDUGC run-EMYsYn6oZd36Bk2o 2021-06-13 19:47:53 2021-06-13 19:47:53 2021-06-13 19:48:00 2021-06-13 19:48:02 2021-06-13 19:48:06

The data was returned in 0.2 seconds and only consumed 0.04 seconds of slot time. That is 750 times cheaper. How awesome is that!

## conclusion

After first trying some good old fashioned nested SQL to pivot tables, I found the PIVOT operator very simple to use. it is really sweet.

Image by Harry Strauss from Pixabay

Mark van Holsteijn is a senior software systems architect, and CTO of binx.io. He is passionate about removing waste in the software delivery process and keeping things clear and simple.


text:
# Distribution of the magnitude of samples from the multivariate gaussian distribution

Let $\mathbf{X}\sim \mathcal{N}(0,I)$ be an $n-$dimensional standard Gaussian vector. We know that \begin{align} ||\mathbf{X}||^2&=\sum_{i=1}^nX_i^2\sim\chi^2_n\\ ||\mathbf{X}||&=\sqrt{\sum_{i=1}^nX_i^2}\sim\chi_n\\ \end{align}

Suppose the components are correlated, i.e., we consider $\mathbf{Y}\sim \mathcal{N}(0,\Sigma)$ where $\Sigma$ is positive definite. We know that $||\mathbf{Y}||^2$ has the generalized chi-squared distribution. Is there also a corresponding generalized chi distribution for $||\mathbf{Y}||$? If so, is there some asymptotic result on the mean of such a distribution for $n\to\infty$? This is related to

Expectation of square root of sum of independent squared uniform random variables

Central Limit Theorem for square roots of sums of i.i.d. random variables

But here the components of the random vector are neither independent or identically distributed.

• The method I used in the (first) linked post do not really depend on independence, it can be used directly. You just need to find the mgf of $\| \mathbf{ Y} \|^2$ – kjetil b halvorsen Dec 11 '17 at 17:16
• @kjetilbhalvorsen good point, but unfortunately cfs and mgfs weren't part of my (only) academic course on Probability, so I'm not too familiar with them. I know that the mgf of the sum of $n$ iid RVs with common mgf $f(t)$ is just $f(t)^n$ (because convolution becomes multiplication under a Fourier transform), but I know little else. Can you point me to some references which would help me calculate the mgf of correlated RVs? – DeltaIV Dec 12 '17 at 14:20
• You can start from my answer here: math.stackexchange.com/questions/442472/… or maybe cpaindex.naturalspublishing.com/files/published/…. Also, the book referenced in the first link do have the results. – kjetil b halvorsen Dec 12 '17 at 20:28
• Now I have written out the mgf as an answer here: stats.stackexchange.com/questions/262604/… – kjetil b halvorsen Dec 14 '17 at 20:48
• There is much more freedom here than in your former question, which only depended on $n$. Here there are $n$ eigenvalues, which might be all equal (easy case), or all different, or something in between. If you could indicate something about how $n$ tends to infinity (how do $\Sigma$ look like? – kjetil b halvorsen Dec 19 '17 at 21:36

This is not a complete answer so far, I will come back completing it when time. Using results and notation from What is the moment generating function of the generalized (multivariate) chi-square distribution?, we can write $$\| Y \|^2 = Y^T Y$$ which we can write as a sum of $n$ independent scaled chisquare random variables (here it is central chisquare since the expectation of $Y$ is zero) $Y^T Y = \sum_1^n \lambda_j U_j^2$ with the following definitions taken from the link above. Use the spectral theorem to write $\Sigma ^{1/2} A \Sigma^{1/2} = P^T \Lambda P$ with $P$ orthogonal and $\Lambda$ diagonal with diagonal elements $\lambda_j$, $U=P Y$ then have independent standard normal components. Then we find the mgf of $Y^T Y$ as $$M_{Y^T Y}(t) = \exp\left(-\frac12 \sum_1^n \log(1-2\lambda_j t) \right)$$ Then one can continue finding the expectation using the method from the link in comments. The expectation of $\sqrt{Y^T Y}$ is given by $$\DeclareMathOperator{\E}{\mathbb{E}} D^{1/2} M (0) = \Gamma(1/2)^{-1}\int_{-\infty}^0 |z|^{-1/2} M'(z) \; dz$$ where $M(t)$ is the mgf above and $'$ denotes differentiation. Using maple we can calculate

M := (t,lambda1,lambda2,lambda3) -> exp( -(1/2)*( log(1-2*t*lambda1)+log(1-2*t*lambda2)+log(1-2*t*lambda3) ) )

which is the case with $n=3$. As an example, if the covariance matrix $\Sigma$ is the equicorrelation matrix with offdiagonal elements $1/2$, that is, $$\Sigma=\begin{pmatrix} 1 & 1/2 & 1/2 \\ 1/2 & 1 & 1/2 \\ 1/2 & 1/2 & 1 \end{pmatrix}$$ then the eigenvalues $\lambda_j$ can be calculated as $2, 1/2, 1/2$. Then we can calculate

int( diff(M(z,2,1/2,1/2),z)/sqrt(abs(z)),z=-infinity..0 )/GAMMA(1/2)

obtaining the result. $$1/3\,{\frac {\sqrt {3}{\rm arctanh} \left(\sqrt {3}/2\right)+6}{ \sqrt {\pi}}}$$ It is probably better to just use the option numeric=true, we will not find some simple general symbolic answer. An example with other eigenvalues gives another symbolic form of the result:

int( diff(M(z,5,1,1/2),z)/sqrt(abs(z)),z=-infinity..0 )/GAMMA(1/2)

$$-1/15\,{\frac {\sqrt {5} \left( 10\,i\sqrt {5}{\it EllipticF} \left( 3 \,i,1/3 \right) -9\,i\sqrt {5}{\it EllipticE} \left( 3\,i,1/3 \right) -30 \right) }{\sqrt {\pi}}}$$ (I now also suspect that at least one of this results returned by maple is in error, see https://www.mapleprimes.com/questions/223567-Error-In-A-Complicated-Integralcomplex?sq=223567)

Now I will turn to numerical methods, programmed in R. First, one idea is to use the saddlepoint approximation or one of the other methods from the answer here: Generic sum of Gamma random variables combined with the relation $f_{\text{chi}}(x)=2xf_{\text{chi}^2}(x^2)$, but first I will look into use of the R package CompQuadForm, which implements several methods for the cumulative distribution function, and then numerical differentiation with the package numDeriv, code follows:

library(CompQuadForm)
library(numDeriv)

make_pchi2   <-  function(lambda, h=rep(1, length(lambda)), delta=rep(0, length(lambda)), sigma=0, ...)  {
function(q) {
vals <- rep(0, length(q))
warning_indicator <- 0
for (j in seq(along=q)) {
res <-  davies(q[j], lambda=lambda, h=h, delta=delta, sigma=sigma, ...)
if (res$ifault > 0) warning_indicator <- warning_indicator+1 vals[j] <- res$Qq
}
if (warning_indicator > 0) warning("warnings reported", warning_indicator)
vals  <-  1-vals  # converting to cumulative probabilities
return(vals)
}
}

pchi2  <-  make_pchi2(c(2, 0.5), c(1, 2))

# Construction a density function via numerical differentiation:

make_dchi  <-  function(pchi) {
function(x, method="simple")  {
side  <-  ifelse(x==0, +1, NA)
vals  <-  rep(0,  length(x))
for (j in seq(along=x)) {
vals[j]  <-  grad(pchi,  x[j]^2, method=method, side=side[j])
}
vals  <-  2*x*vals
return(vals)
}
}

dchi  <-  make_dchi(pchi2)

plot( function(x) dchi(x, method="Richardson"),  from=0,  to=7, main="density of generalized chi distribution", xlab="x", ylab="density", col="blue")

which gives this result:

and we can test with:

integrate(function(x) dchi(x,  method="Richardson"), lower=0,  upper=Inf)
0.9999565 with absolute error < 0.00011

Back to (approximation) for the expected value. I will try to get some approximation for the moment generating function of $Y^T Y$ for large $n$. Start with the expression for the mgf $$M_{Y^T Y}(t) = \exp\left(-\frac12 \sum_1^n \log(1-2\lambda_j t) \right)$$ found above. We concentrate on its logarithm, the cumulant generating function (cgf). Assume that when $n$ increases, then the eigenvalues will approximatly come from some distribution with density $g(\lambda)$ on $(0,\infty)$ (in the numerical examples we will just use the uniform distribution on $(0,1)$). Then the sum above divided into $n$ will be a Riemann sum for the integral $$I(t) = \int_0^\infty g(\lambda) \log(1-2\lambda t)\; d\lambda$$ for the uniform distribution example we will find $$I(t) = \frac{2t\ln(1-2t)-\ln(1-2t)-2t}{2t}$$ Then the mgf will be (approximately) $M_n(t)=e^{-\frac{n}2 I(t)}$ and the expected value of $\| Y \|$ is $$\Gamma(1/2)^{-1} \int_{-\infty}^0 |z|^{-1/2} M_n'(z) \; dz$$ This seems to be a difficult integral, some work seems to show that the integrand has a vertical asymptote at zero, for instance maple refuses to do a numerical integration (but uses abnormally long time). So it might be that a better strategy is to use numerical integration using the density approximations above. Nevertheless, this approach seems to indicate that the dependence on $n$ should not be very dependent on the eigenvalue distribution.

Going for numerical integration in R, that also shows to be difficult. First some code:

E  <-  function(lambda, h=rep(1, length(lambda)), lower, upper, ...) {
pchi2  <-  make_pchi2(lambda, h, acc=0.001, lim=50000)
dchi  <-   make_dchi(pchi2)
val  <-   integrate(function(x) x*dchi(x), lower=lower,   upper=upper, subdivisions=10000L, stop.on.error=FALSE)$value return(val) }  Some experimentation with this code, using the integration limits lower=-Inf, upper=0, shows that for a large number of eigenvalues, the result is 0! The reason is that the integrand is zero "almost everywhere", and the integration routine misses the dump. But plotting the integrand first, and choosing a sensible interval, we can get reasonable results. For the example I am using uniformly distributed eigenvalues between 1 and 5, with varying$n$. So for$n=5000$the following code: E(seq(from=1, to=5, length.out=5000), lower=-130, upper=-110)  works. This way we can compute the following results:  n Expected value Approximation [1,] 5 3.656979 3.890758 [2,] 15 6.581486 6.738991 [3,] 25 8.559964 8.700000 [4,] 35 10.161615 10.293979 [5,] 45 11.544113 11.672275 [6,] 55 12.777101 12.904185 [7,] 65 13.901465 14.028328 [8,] 75 14.941275 15.068842 [9,] 85 15.915722 16.042007 [10,] 95 16.830085 16.959422 [11,] 105 17.697715 17.829694 [12,] 500 38.702620 38.907583 [13,] 1000 54.749000 55.023631 [14,] 5000 122.449200 123.036580  Where the last column is calculated as$1.74 \sqrt{n}$, and is found by regression analysis. We show this as a plot: The fit is quite good, indicating that a reasonable guess is that the expectation grows with the square root of$n$, with a prefactor which probably depends on the eigenvalue distribution. • Nice! I look forward to reading the rest of the answer. Thanks a lot! The complete answer will definitely deserve a bounty :-) – DeltaIV Dec 15 '17 at 10:37 • Thanks! In the meantime, the mgf can be used to approximate the density function of$Y^T Y$via the saddflepoint approximation, see stats.stackexchange.com/questions/72479/… (since chisquare is a special case of gamma). If that idea can be extended to work also for its squareroot, I do not know. – kjetil b halvorsen Dec 15 '17 at 16:35 • Could we not simplify your moment generating function into the following? $$M_{Y^2}(t)=\prod_1^n (1-2\lambda_jt)^{-\frac{1}{2}}$$ – Sextus Empiricus Dec 15 '17 at 21:41 • Surely, but I am not sure that is a simplification---for instance, I had better results with maple using the exponential form. The exponential form is also often better for numerical computation – kjetil b halvorsen Dec 15 '17 at 22:03 • @kjetilbhalvorsen ok, if I understand correctly a generic answer is not possible, it depends on the variance-covariance matrix. That's fine. Given that we cannot find an explicit expression, do you think it could be possible at least to get an estimate for the growth of$||\mathbf{Y}||$with$n$? If I could at least say that, even with correlated components, the magnitude of the multivariate Gaussian vector goes to$\infty$as fast as, say,$\sqrt{n}$, or maybe$\Gamma(n+1/2)/\Gamma(n)\$, that would be a nice result. – DeltaIV Dec 18 '17 at 11:59


text:
# Monthly Archives: April 2012

## Another Combinatorial Result

Here is Chris Eagle’s presentation from Stevo Todorcevic’s class “Combinatorial Set Theory”. From the abstract: We prove that MA + $\mathfrak{c} = \aleph_2$ implies $\mathfrak{c} \not\rightarrow (\mathfrak{c}, \omega+2)^2$ . The exposition is based on hand-written notes provided by S. Todorcevic. The result itself is due to R. Laver. This is the analogous result to “MA […]