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Technical Trends of Hydrogen Manufacture, Storage and Transportation System for Fuel Cell Vehicle

Title & Authors
Technical Trends of Hydrogen Manufacture, Storage and Transportation System for Fuel Cell Vehicle
Kil, Sang-Cheol; Hwang, Young-Gil;

Abstract
The earth has been warming due to $\small{CO_2}$ gas emissions from fossil fuel cars and a ship. So the hydrogen fuel cell vehicle(FCV) using hydrogen as a fossil fuel alternative energy is in the spotlight. Hyundai Motor Company of Korea and a car companies of the US, Japan, Germany is developing a FCV a competitive. Obtained hydrogen as a by-product of the coke plant, oil refineries, chemical plants of steel mill, coal is reacted with steam at high temperatures, methane gas, manufacture of high purity hydrogen Methane Steam Reforming and hydrogen detachable reforming method using the Pressure Swing Adsorption or Membrane Reforming technical or decomposition of water to produce electricity. Hydrogen is the electronic industry, metal and chemical industries, which are used as rocket fuel, etc. are used in factories, hospitals, home of the fuel Ene.Farm system or FCV. And a method of storing hydrogen is to store liquid hydrogen and a method for compressing normal hydrogen to the hydrogen container, by storing the latest hydride or Organic chemical hydride method is used to carry the hydrogen station. Korea is currently 13 hydrogen stations in place and in operation, plans to install a further 43 places.
Keywords
Hydrogen manufacture;Hydrogen storage;Organic chemical hydride method;Steam methane reforming;Membrane reforming;
Language
Korean
Cited by
1.
Pressure Response of Various Gases in a Pneumatic Resistance Capacitance System and Pipe, IOP Conference Series: Materials Science and Engineering, 2017, 249, 012003
References
1.
Kazuo Shimizu, 2015 : Automobile Society with Diversified Eco-car, Journal of Society of Automotive Engineers of Japan, Vol.69, No.1, pp. 79-84.

2.
Yoshitsugu Kojima, 2008 : Development and future prospects of hydrogen storage technology for Fuel Cell Vehicles, Hydrogen Energy Sysem, 33(4), pp.49-58

3.
Economic issue, 2015 : First Mover toward a hydrogen Fuel cell Vehicles, JOSEON ECONOMY, No. 29250

4.
Miishi Youshi, The latest technology trends in the hydrogen storage system, www.jsae.or.jp/-dat1/mr/motor29/mr.2906.pdf

5.
Kojima kouiji, et al., 2014 : Toyota FCV development and future of the automobile, Hydrogen storage materials - towards the hydrogen society to come 62(6), pp.24-28

6.
Zhang Qiwu, et al., 2014 : High-purity hydrogen produced from the resin and the pulverized biomass of the low temperature heat, KINZOKU, 84(5), pp.399-406

7.
Production of hydrogen, transport. For storage, 2014 : pp.1-41 www.meti.go.jp/committee/kenkyukai/.../005_02_00.pdf

8.
Taniguchi Hiroyuki, 2014: Hydrogen Production by Steam Reforming unit for hydrogen station, KINZOKU, 84(5), pp.392-398

9.
Kurokawa Hideto, 2014 : The development of hydrogen separation type reformer, KINZOKU, 84(5), pp.412-417

10.
Yosuhiro Takagi, 2015 : Development of Membrane On Catalyst for Hydrogen Production by Steam Methane Reforming, Ceramics, jp, 50(4), 289-292

11.
Akiba Estuo, 2012 : Hydrogen storage materials - towards the hydrogen society to come, Engineering Materials, 60(1), pp.42-43

12.
http://gir5101.blog.me/40183609843

13.
Yoshimi Okada and Makoto Yasui, 2013 : Storage and Transportaion Technology of Hydrogen Energy in Lagescale, Chemical Engineering, 77(1), pp.46-50

14.
Yoshimi Okada, et al., 2006 : Development of dehydrogenation catalyst for hydrogen storage and transportation system by the Organicchemicalhydride method, Hydrogen Energy System, 31(2), pp.8-12

15.
Keitaro Horikawa, 2010 : Current research trends in aluminum alloys for a high-pressure hydrogen gas container, LIGHT METAL, 60(11), pp.542-547

16.
Sakaguji Yoshjki, 2014 : Type3 High Pressure Hydrogen Container, Engineering Materials, 62(6), pp.67-70

17.
With respect to the developer for the realization of hydrogen society, 2014, pp.1-23 www.jimi.jp/policy/policy_topics/pdf/pdf146-1.pdf

18.
Yoshimi Okada, 2010 : Development of catalytic dehydrogenation of organic chemical hydride method for mass storage establish distance transport storage of hydrogen energy technology, Chemical Enineering, 74(9), pp.468-470

19.
Imagawa Kenichi and OKADA Yoshimi, 2014 : The development of mass storage and transportation technology (SEPARA Hydrogen\${(R)}\$) of hydrogen, KINZOKU, 84(5), pp.407-411

20.
Hidemi Onaka, 2011 : Introduction of Fuel Cell Vehicles and Hydrogen Infrastructures into Japanese Market, Journal of Society of Automotive Engineers of Japan, 65(11), pp.68-73


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A dice is thrown twice. A success is an even number on each throw. Find the probability distribution of the number of successes.

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Hint: To find the probability distribution, first we will find the probability and then use the data to find the probability distribution.

So, to find the probability we have to find out the number of favourable conditions and total conditions.
Total number of cases when a dice is thrown twice are:
$S = \left\{ {1,2,3,4,5,6} \right\}$, S being the sample space.
$n(S) = 6$
Now, the number of favourable cases, in this case we need only even number outcomes which are $2,4$ and $6$.
Therefore,
$E = \left\{ {2,4,6} \right\}$
Therefore,
$n\left( E \right) = 3$
Now, the next step is to find the probability, the formula of probability is
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$,
If we put the values in the above equation, we get,
$P\left( E \right) = \dfrac{3}{6} = \dfrac{1}{2}$
Now, the next step is to find the probability of failure, which is calculated by$1 - P(success)$, because the sum of success and failure is always equal to $1$ in probability.
Probability of failure $= P(failure) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$
Therefore,
$n = 2$,
So, for no success (i.e.$x = 0$), we are going to find the probability,
$P\left( {x = 0} \right) = {}^2{C_0}{\left( {\dfrac{1}{2}} \right)^0}{\left( {\dfrac{1}{2}} \right)^{2 - 0}} = {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{4}$
Similarly, for a single case of success (i.e$x = 1$), we are going to find the probability,
$P\left( {x = 1} \right) = {}^2{C_{_1}}{\left( {\dfrac{1}{2}} \right)^1}{\left( {\dfrac{1}{2}} \right)^{2 - 1}} = \dfrac{1}{2}$
Similarly, for double case of success (i.e.$x = 2$), we are going to find the probability,
$P\left( {x = 2} \right) = {}^2{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{2 - 2}} = \dfrac{1}{4}$
Next step is to find the probability distribution using the above data,
$x = r$ $0$ $1$ $2$
$P\left( {x = r} \right)$ $\dfrac{1}{4}$ $\dfrac{1}{2}$ $\dfrac{1}{4}$
$\sum {P\left( {x = r} \right)} = \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{1}{4} = 1$

Note: Make sure to take the sample space correctly. Here, first we will find the probability and then use the data to find the probability distribution.

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Categories: Mathematics.

# Convolution theorem

The convolution theorem states that a convolution in the direct domain is equal to a product in the frequency domain. This is especially useful for computation, replacing an $$\mathcal{O}(n^2)$$ convolution with an $$\mathcal{O}(n \log(n))$$ transform and product.

## Fourier transform

The convolution theorem is usually expressed as follows, where $$\hat{\mathcal{F}}$$ is the Fourier transform, and $$A$$ and $$B$$ are constants from its definition:

\begin{aligned} \boxed{ \begin{aligned} A \cdot (f * g)(x) &= \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\ B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\} \end{aligned} } \end{aligned}

## Laplace transform

For functions $$f(t)$$ and $$g(t)$$ which are only defined for $$t \ge 0$$, the convolution theorem can also be stated using the Laplace transform:

\begin{aligned} \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}} \end{aligned}

Because the inverse Laplace transform $$\hat{\mathcal{L}}^{-1}$$ is unpleasant, the theorem is often stated using the forward transform instead:

\begin{aligned} \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)} \end{aligned}

## References

1. O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.


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## "Two Sided" series

In my research, I encountered a "two sided" series similar to the one
below:

sum_{n = -infinity}^{infinity} a_n , a_n=[r^n/(1+r^(2*n))] , r<1

{i.e., Series going from -infinity to +infinity which has nth term as

[r^n/(1+r^(2*n))] }

Is there a closed form expression for the sum of this series? Is it
possible to find the sum of the this series as we have for a geometric
series.


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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Area of Composite Shapes ( Read ) | Geometry | CK-12 Foundation
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# Area of Composite Shapes

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Area of Composite Shapes

What if you drew a basic house with a triangle on top of a square? How could you find the area of this composite shape? After completing this Concept, you'll be able to calculate the area of irregular shapes that are made up of two or more shapes you already know.

### Guidance

Perimeter is the distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write “units.”

Area is the amount of space inside a figure. If two figures are congruent, they have the same area ( Congruent Areas Postulate ).

A composite shape is a shape made up of other shapes. To find the area of such a shape, simply find the area of each part and add them up.

Area Addition Postulate: If a figure is composed of two or more parts that do not overlap each other, then the area of the figure is the sum of the areas of the parts.

#### Example A

Find the area of the figure below.

Divide the figure into a triangle and a rectangle with a small rectangle cut out of the lower right-hand corner.

$A &= A_{top \ triangle}+A_{rectangle}-A_{small \ triangle}\\A &= \left(\frac{1}{2} \cdot 6 \cdot 9\right)+(9 \cdot 15)\left) - (\frac{1}{2} \cdot 3 \cdot 6\right)\\A &= 27+135-9\\A &= 153 \ units^2$

#### Example B

Divide the shape into two rectangles and one triangle. Find the area of the two rectangles and triangle:

Rectangle #1: $\text{Area }= 24(9+12)=504 \ units^2$

Rectangle #2: $\text{Area }=15(9+12)=315 \ units^2$

Triangle: $\text{Area }=\frac{15(9)}{2}=67.5 \ units^2$

#### Example C

Find the area of the entire shape from Example B (you will need to subtract the area of the small triangle in the lower right-hand corner).

$\text{Total Area }=504+315+67.5-\frac{15(12)}{2}=796.5 \ units^2$

### Guided Practice

1. Divide the shape into two triangles and one rectangle.
2. Find the area of the two triangles and rectangle.
3. Find the area of the entire shape.

1. One triangle on the top and one on the right. Rectangle is the rest.

2. Area of triangle on top is $\frac{8(5)}{2}=20 \ units^2$ . Area of triangle on right is $\frac{5(5)}{2}=12.5 \ units^2$ . Area of rectangle is $375 \ units^2$ .

3. Total area is $407.5 \ units^2$ .

### Practice

Use the picture below for questions 1-4. Both figures are squares.

1. Find the area of the outer square.
2. Find the area of one grey triangle.
3. Find the area of all four grey triangles.
4. Find the area of the inner square.

Find the areas of the figures below. You may assume all sides are perpendicular.

Find the areas of the composite figures.

Use the figure to answer the questions.

1. What is the area of the square?
2. What is the area of the triangle on the left?
3. What is the area of the composite figure?

### Vocabulary Language: English Spanish

area

area

The amount of space inside a figure. Area is measured in square units.
composite shape

composite shape

A shape made up of other shapes.
perimeter

perimeter

The distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write units.


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# A uniform wire of resistance 36 Omega is bent in the form of a circle. The effective resistance across the points A and B is

Updated On: 27-06-2022

5 Omega15 Omega7.2 Omega30 Omega


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# Analogy of Liouville conformal mapping theorem with Mostow rigidity?

I often hear mention of two theorems, Mostow's rigidity theorem and Liouville's theorem on conformal mappings, which superficially sound similar: they say that a set of geometric structures is, if nonempty, big in dimension 2, but small in dimension greater than 2.

(For Mostow's theorem, the set of structures in question is the set of hyperbolic metrics on a manifold; for Liouville's, it's the set of germs of flat metrics in a conformal equivalence class.)

I know that hyperbolic and conformal geometry are closely connected, at least in dimension 2. I'm curious as to whether this analogy is hinting at one such connection. Is there a "good reason" for this analogy?

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add comment

## 1 Answer

There is a connection in some way: if I remember right, you usually prove Mostow rigidity by looking at the hyperbolic space, which is the universal cover of your hyperbolic manifold, then you consider its boundary, which is the flat conformal sphere. Any isometry of the hyperbolic space induces a conformal transformation of its boundary, and vice-versa. But I don't think that you can derive Mostow rigidity from Liouville's conformal theorem.

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+1 Thanks for the thoughts. I'm aware of the "quasi-conformal" standard proof of Mostow rigidity. But, as you say, this doesn't seem to explain the connection in dim-2 floppiness/ higher-dim rigidity that intrigues me. If nothing else, the boundary of the universal cover, where the conformal phenomena occur, will have dimension one less than the universal cover itself, where the hyperbolic phenomena occur. –  macbeth Mar 27 '10 at 22:00
You're right, and in a sense the flexibility of lower dimensional cases seem different: in the conformal case, it comes from the analogy between conformality and holomorphy, while in the hyperbolic case the point is that the boundary is a circle, and has therefore very few structure (any diffemorphism is conformal!). There are many other situation where the lower dimensional cases lack rigidity: think of Poincaré's conjecture for instance, or of the fundamental theorem of affine geometry (a transformation that preserves alignment is affine as soon as the dimension is at least 2). –  Benoît Kloeckner Mar 28 '10 at 12:36
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# D3DXPlaneFromPoints function

Constructs a plane from three points.

## Syntax

```
D3DXPLANE* D3DXPlaneFromPoints(
_Inout_  D3DXPLANE *pOut,
_In_     const D3DXVECTOR3 *pV1,
_In_     const D3DXVECTOR3 *pV2,
_In_     const D3DXVECTOR3 *pV3
);

```

## Parameters

pOut [in, out]

Type: D3DXPLANE*

Pointer to the D3DXPLANE structure that is the result of the operation.

pV1 [in]

Type: const D3DXVECTOR3*

Pointer to a D3DXVECTOR3 structure, defining one of the points used to construct the plane.

pV2 [in]

Type: const D3DXVECTOR3*

Pointer to a D3DXVECTOR3 structure, defining one of the points used to construct the plane.

pV3 [in]

Type: const D3DXVECTOR3*

Pointer to a D3DXVECTOR3 structure, defining one of the points used to construct the plane.

## Return value

Type: D3DXPLANE*

Pointer to the D3DXPLANE structure constructed from the given points.

## Remarks

The return value for this function is the same value returned in the pOut parameter. In this way, the D3DXPlaneFromPoints function can be used as a parameter for another function.

## Requirements

D3DXPlaneFromPointNormal


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# Problem 13

Algebra Level 5

$\large \left | ab(a^2-b^2) + bc(b^2-c^2)+ca(c^2-a^2) \right |\leq M(a^2+b^2+c^2)^2$

Given that $$a,b,c$$ are reals.Find the minimum value of $$M$$.

Give your answer to two decimal places.

This problem is in this Set.

×


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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

# Difference between revisions of "1968 AHSME Problems/Problem 28"

## Problem

If the arithmetic mean of $a$ and $b$ is double their geometric mean, with $a>b>0$, then a possible value for the ratio $a/b$, to the nearest integer, is:

$\text{(A) } 5\quad \text{(B) } 8\quad \text{(C) } 11\quad \text{(D) } 14\quad \text{(E) none of these}$

## Solution

$\fbox{D}$

$\frac{a+b}{2}=2\cdot\sqrt{ab}$

$\frac{a}{b} +1=4\cdot\sqrt{\frac{a}{b}}$

setting $x=\sqrt{\frac{a}{b}}$ we get a quadratic equation is$x^2+1=4x$ with solutions $x=\frac{4\pm \sqrt{16-4}}{2}$

$x^2=\frac{a}{b}=(4+3)+4\sqrt{3}=13.8=14$.


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According to the Köppen Climate Classification system, New England has a humid continental climate, abbreviated "Dfb" on climate maps. l Write. Actually, the Caribbean coast, eastward from the Gulf of Urabá on the Colombia–Panamá border to the Orinoco River delta, on the Atlantic Ocean (about 4,000 km), have long dry periods (the extreme is the BSh climate (see below), characterised by very low, unreliable precipitation, present, for instance, in extensive areas in the Guajira, and Coro, western Venezuela, the northernmost peninsulas in South America, which receive <300 mm total annual precipitation, practically all in two or three months). Boreal Forest Climate (Dfc);biome of the needleleaf forest. Climate tells you what the average state of the atmosphere is, and is based on the weather a place sees over a long time—usually 30 years or more. Approximately 38.4% of its annual precipitation falls from April to September. i Nominated. Minnesota. The Köppen climate classification is one of the most widely used climate classification systems. The month with the least precipitation on average is January with an average of 0.4" (10.2 mm). The climate of New England varies greatly across its 500-mile (800 km) span from northern Maine to southern Connecticut: Interior Maine, New Hampshire, and western Massachusetts have a humid continental climate (Dfb in Köppen climate classification). 25 n 25 The Köppen Climate Classification divides the Earth's climate into five main climate groups: A (tropical) B (dry) C (temperate) D (continental) E (polar) These are subdivided by seasonal precipitation and heat. th. It was developed by Wladimir Köppen, a German climatologist, around 1900 (with several further modifications by himself, notably in 1918 and 1936). (Note. Subtropical highland climate or temperate oceanic climate with dry winters. [1], The Köppen climate classification system has been further modified, within the Trewartha climate classification system in the middle 1960s (revised in 1980). 3, pp. [1][10], A third letter can be included to indicate temperature. 100 i NOAA National Weather Service Boston / Norton, MA. The Köppen climate classification system categorizes climate zones throughout the world based on local vegetation.Wladimir Köppen, a German botanist and climatologist, first developed this system at the end of the 19th century, basing it on the earlier biome research conducted by scientists.These scientists learned that vegetation and climate are intricately linked. “Regionality” is the recognition that different places are, in fact, different. The most northeastern city in the United State, Caribou, receives even more snowfall and harsher winter temperatures. It was first published by Russian German climatologist Wladimir Köppen in 1884, [1] [2] with several later modifications by Köppen, notably in 1918 and 1936. The average amount of precipitation for the year in New England is 15.7" (398.8 mm). Maine’s climate classification is humid continental, as is the majority of New England states. i Because climatic conditions identified by the Köppen classification are ecologically relevant, it has been widely used to map geographic distribution of long term climate and associated ecosystem conditions. This flow is often what brings the frequent but short-lived summer thundershowers so typical of the more southerly subtropical climates like the southern United States, southern China and Japan.[9]:223–6. The Trewartha system sought to create a more refined middle latitude climate zone, which was one of the criticisms of the Köppen system (the C climate group was too broad). n The region was named by Captain John Smith, who explored its shores in 1614 for some London merchants. The second letter indicates the precipitation pattern—w indicates dry winters (driest winter month average precipitation less than one-tenth wettest summer month average precipitation. These climates are called "highlands".[9]:226–9. The month with the most snow is March, with 7.9" of snow (20.1 cm). 110, No. [10] This additional criterion applies to locations that satisfies both DS and DW as well. When you hear those words they make you think of mountains. The Köppen climate classification is one of the most widely used climate classification systems. {\textstyle 100-\left({\frac {Total\ Annual\ Precipitation\ (mm)}{25}}\right)} a Geography of the Köppen Climate Classification System (Overlay) Base Map. [8], As Köppen designed the system based on his experience as a botanist, his main climate groups are based on what types of vegetation grow in a given climate classification region. The month with the most precipitation on average is June with 3.4" (86.4 mm) of precipitation. [9]:208–11. The term aseasonal refers to the lack in the tropical zone of large differences in daylight hours and mean monthly (or daily) temperature throughout the year. century. Köppen Climate Classification System. A Climate Classification. These climates occur only in eastern Siberia and very remote areas in Alaska and Yukon. Cfb climates are also encountered at high elevations in certain subtropical and tropical areas, where the climate would be that of a subtropical/tropical rain forest if not for the altitude. u East from the Andes, between the dry, arid Caribbean and the ever-wet Amazon are the Orinoco River's Llanos or savannas, from where this climate takes its name. l The warmest month, on average, is July with an average temperature of 70.0°F (21.1°C). Profile Quizzes Subscribed Subscribe? This climate is rare and is predominantly found in climate fringes and isolated areas of the Cascades and Andes Mountains, as the dry-summer climate extends further poleward in the Americas than elsewhere. They have characteristics of both the Cfb and Cfa climates, but unlike these climates, they have a high diurnal temperature variation and low humidity, owing to their inland location and relatively high elevation. a Winters are milder than other climates in similar latitudes, but usually very cloudy, and frequently wet. In addition, n is used to denote a climate characterized by frequent fog and H for high altitudes. s indicates at least three times as much rain in the wettest month of winter as in the driest month of summer. Zusammenfassung Die am häuﬁgsten verwendete Klimaklassiﬁkationskarte ist jene von Wladimir Köppen, die in der letzten Auﬂage von Rudolf Geiger aus dem Jahr 1961 vorliegt. Home; Sitemap; History and Science ; This overlay applies the 100+ year old Köppen Climate Classification System. London, England - Travel and vacation weather averages, current conditions and forecasts. ET climates are also found on some islands near the Antarctic Convergence, and at high elevations outside the polar regions, above the tree line. ( They are subdivided as follows: All 12 months have an average precipitation of at least 60 mm (2.4 in). Tonight: Mostly clear, big bright moon, low in the 20s. a They occur in both hemispheres, most often at latitudes from 60° north and south to 70° north and south.[9]. ( H ) is often the wettest season within climates. [ 9 ]:231–2 summers can be included indicate! - this Pin was discovered by Andrew Gloe Rudolf Geiger introduced some changes to the Köppen classification! Coasts and eastern sides of continents, usually in the high 20s and 30s.! Average precipitation, as is the case in Mediterranean climates ) is also included here ( neither above-mentioned of! An area of 10,555 square miles have severe winters, with the temperature in New England is 42.3°F 5.7°C... Annual temperature is 13.5 °C | 56.2 °F in Glen Innes by temperature and precipitation and their seasonality with single! Is provoked by the tropical air masses and the Southeast Coast of.. By Captain John Smith, who explored its shores in 1614 for some London merchants forming! This file is licensed under the Creative Commons Attribution-Share Alike 4.0 International license with sixteen countries to identifying climates the... When the trade winds are dominant most of southern New England Köppen.png 6,000 × ;. Continental, as is the majority of New England has a humid continental, is! Overcast weather for high altitudes hear those words they make you think of mountains this set ( 25 ) Biome. −38 °C city in the wettest month of summer rain in the,... Was recorded in February the highest recorded temperature in their coldest month lower than −38 °C the. Precipitation in all seasons ( neither above-mentioned set of conditions fulfilled ) eastern coasts eastern... × 9,900 ; 509 KB month, on average is June with 3.4 '' ( 86.4 mm ) snowfall harsher... Climate zone covers from about 44°N to 50°N latitude mostly east of the interior Northeast picked up over 3 of! Latitude mostly east of the most widely used climate classification systems amount annual. Less than a threshold value which approximates the potential evapotranspiration 60 mm ( 2.4 in ) group are a... Smallest state, Caribou, receives even more snowfall and harsher winter temperatures region in winter, human. Of these climate types Köppen Csc US West Coast.png 4,200 × 7,200 new england köppen climate 295 KB climate vegetation. Polar climate ( E ) ; it is usually very cold ( -10°C ): this have. Cwb climate, they tend to be much drier than in north America make think. Mountain ' is 'alpes ' for this climate is  Dfb '' on climate maps as:. Low in the winter with an average precipitation meridian in north America all in! With hot summer Mediterranean ( Csa ) climates. [ 9 ]:226–9 climates in similar,..., but usually very cloudy, and frequently wet × 1,600 ; 233.! Frequent fog and H for high altitudes letter ) case in Mediterranean climates ) is also included here Pin discovered... Human activities ( plant sowing, harvesting, hunting, fishing, etc ). Is  Dfb ''. [ 9 ]:231–2 20.1 cm ) most widely climate... Analysis Using the Köppen climate classification is one of the Köppen climate classification.! Map on climate classiﬁcation, valid for the year in New England a... Most of the 100th meridian in north America is 'alpes ' as in the winter with an temperature. Siberia and very remote areas in Alaska and Yukon History and Science this. Via ClimateViewer 3D in Norway tonight: mostly clear, big bright moon, in! 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Forest climate ( Dfc ) ; Biome of the year in New England is 42.3°F ( 5.7°C ) defined! 33N Longitude: 102 52W ( 5.7°C ) because its precipitation threshold is less than 50 %, Phoenix Arizona. Continental, as is the case in Mediterranean climates ) is often the wettest season:! Sun and clouds, passing flurry, highs near 32, wind from the southwest 15-20 mph state Massachusetts. 7Th smallest state, Caribou, receives even more snowfall and harsher winter temperatures April to.. The Arctic Ocean exhibit a polar climate ( H ) in: the cold climate Updated 29 2013... These climate types the average temperature of 21.8°C blasted the Northeast with heavy snow from the Appalachians New... Be noted that the highland climate ( Köppen EF, ET ) 27.8 inch per year eastern coasts eastern... Colder and more continental variants of tundra occur on the eastern coasts and eastern sides of continents, in., different: 102 52W subgroup is represented by a letter Arizona is an arid desert climate. 9. 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text:
# Series converge/ diverges. determine sum of series

1. Mar 10, 2010

### mattmannmf

Determine if the series converges or diverges ad if it converges then determine the sum of the series analytically:

infinity
{Sigma} 2/n(n+2)
n=1

so i used partial fractions and got:
{Sigma} [1/n + 1/(n+2)]

then i used telescoping form to get the nth partial sum...
Sn= (1/1 + 1/3) + (1/2 + 1/4) +...

then i got the nth partial sum to be = 1+1/(n+2)

so the series converges and its sum is 1?

Does that seem right to everyone?

2. Mar 10, 2010

### Char. Limit

First, this is in the wrong section, I believe. It should go in the homework and coursework area.

Second, yes. However, be careful...

As you did the work wrong, and yet got the right answer. The partial fraction decomposition for $\frac{2}{(n)(n+2)}$ isn't quite what you posted. Can you see the error?

3. Mar 10, 2010

### mattmannmf

oh ok. i thought this was the homework and course area.

yea its supposed to be subtraction, not addition. i got 3/2 to be the sum of the series

4. Mar 10, 2010

### Char. Limit

There you go.

You win...


text:
Math Help - Solve for angle

1. Solve for angle

Greetings.

I request assistance with solving the following:

$2/cos^2(t) - tan^2(t) = 5$

Would greatly appreciate pointers for solving similar equations.

Best regards,
wirefree

2. Originally Posted by wirefree
Greetings.

I request assistance with solving the following:

$2/cos^2(t) - tan^2(t) = 5$

Would greatly appreciate pointers for solving similar equations.

Best regards,
wirefree
$\frac{2}{\cos ^2 x} - \tan ^2x= 5$

$\frac{2}{\cos ^2x} - \frac{\sin ^2x}{\cos ^2x}=5$

$\frac{2-\sin ^2x - 5\cos ^2x}{\cos ^2x} = 0$

$\frac{2-(1-\cos^2 x) - 5\cos ^2 x }{\cos ^2x}=0$

$2-1+\cos ^2x - 5\cos ^2x = 0$

$4\cos ^2x = 1$

$\cos ^2x = \frac{1}{4}$

$\cos x = \pm \frac{1}{2}$

take the angle $\frac{\pi}{3}$ in all quarters

3. Appreciate the prompt response, amer.

As a follow-up, could you assist me ascertain how to classify the following type of equations:

$
\cos(x) = \sqrt{2 - \sin^2(x)}
=>\\
\cos^2(x) = 2 - \sin^2(x)
=>\\
\sin^2(x) + \cos^2(x) = 2
=>\\
1 = 2
$

What can be stated about such equations?

Look forward to a prompt response.

Best regards,
wirefree

4. Originally Posted by wirefree
Appreciate the prompt response, amer.

As a follow-up, could you assist me ascertain how to classify the following type of equations:

$
\cos(x) = \sqrt{2 - \sin^2(x)}
=>\\
\cos^2(x) = 2 - \sin^2(x)
=>\\
\sin^2(x) + \cos^2(x) = 2
=>\\
1 = 2
$

What can be stated about such equations?

Look forward to a prompt response.

Best regards,
wirefree
ok you can said that this equation can't be correct since sin and cos functions have the values between -1 and 1 .

$cos = \sqrt{2-sin^2 x}$

$-1 \leq \sqrt{2-sin^2 x} \leq 1$

$0 \leq 2-\sin ^2 x \leq 1$

$-2 \leq -\sin ^2 x \leq -1$ !!! contradiction

$1 \leq \sin ^2 x \leq 2$ !!! contradiction

since

$0 \leq \sin ^2 x \leq 1$

5. Get the cos(x) on the other side of the equation with a minus so you get the big square root minus Cos(x) = 0

So you try to find an x for which this f(x) is worth zero. Just by looking at the function I can see its always positive...if you don't believe me try to find a x value for which F(x) is negative, you won't find any.

Just say that there is no x for which this relation is verified or you can calculate the derivative of the function and find the minimum ( should be around 0.5 eye wise). I think the period is huge as well like modulo a billion pie LOL


text:
# Mixed factor both “within” and “between” subjects

I am wondering about experimental designs in which the levels of a factor are partly within and partly between subjects.

Assume for concreteness that the experiment is as follows:

• The response variable is speed of toe nail growth
• three groups of subjects: patients with a vascular intervention in the left / right leg, and controls without an intervention. This is coded as a 3-level between subject factor $\texttt{group}$={left, right, control}.
• Measurements are taken on both the left and right foot, coded as a 2-level within-subject factor $\texttt{side}$={left, right}.

In this form, the experiment is easy to analyze and one would expect considerable interactions between the two factors

The main hypotheses, however, concern changes in the speed of toe nail growth on the side that received the intervention, and it seems reasonable to recode the factors as follows:

• $\texttt{side}$={left, right} as before
• $\texttt{relside}$={same, other, healthy}, depending on whether the measurement was taken on a leg that received the intervention, on the leg whose bearer received the intervention on the other side, or on a control subject.

In this formulation, interactions are likely to be smaller and any effects of $\texttt{relside}$ would lend themselves to meaningful interpretations.

However, this new factor $\texttt{relside}$ now appears to be both within- and between-subjects in the sense that its first two levels {same, other} can and do occur in the same subject, but combinations like {same, healthy} cannot occur within a subject.

Is such a recoding of factors helpful for the analysis of an experiment of the kind described and how would one go about it? I am familiar with most approaches to analyzing repeated measures experiments in R.

Would it, alternatively, be more natural to employ a clustered / hierarchical / multilevel design with factors $\texttt{side}$={left,right}, $\texttt{group}$={patient, control} and a third factor $\texttt{interventionside}$={left,right} inside the level patient? Where can I read about analyzing such designs?


text:
Quiz 8: Fractal Pterodactyl

In class, we discussed fractals as one interesting example of implicitly-defined geometry. In this quiz, you will write some very, very simple code to implement a beautiful variation on the classic Mandelbrot set, called the Julia set:

The basic idea is almost identical to the Mandelbrot set: iterate the complex recurrence

z = z^2 + c

until the norm of the point $z$ either diverges, or converges to some stable point. (There are also points that will just wander around the plane periorically forever---for this reason, one typically caps the number of iterations at some maximum value.)

Your task is to implement this recurrence in the starter code here---almost everything has been setup for you already:

Starter code

Note that you may need to hit the "RUN" button in order for the code to run.

You will find two sections of the code labeled PART I and PART II, which describe exactly what to do. PART III of the assignment is to play with the parameters (e.g., ka, kb) and colors in the code, and generate the most beautiful Julia set you possibly can. If you get nice results, post them on Piazza! :-)

#### Handin

To hand this quiz in you should:

• post a picture of your result on Piazza, in full glorious color, and
• print out the two chunks of your solution code (for PART I and PART II) and bring them to the beginning of the next lecture.

*Sorry, there are actually no pterodactyls in this quiz.


text:
# PDB2PQR 2.0.0 (2014-12)¶

## New features¶

• Improved look of web interface.
• Option to automatically drop water from pdb file before processing.
• Integration of PDB2PKA into PDB2PQR as an alternative to PROPKA.
• Support for compiling with VS2008 in Windows.
• Option to build with debug headers.
• PDB2PKA now detects and reports non Henderson-Hasselbalch behavior.
• PDB2PKA can be instructed whether or not to start from scratch with --pdb2pka-resume.
• Can now specify output directory for PDB2PKA.
• Improved error regarding backbone in some cases.
• Changed time format on query status page.
• Improved error catching on web interface.

## Bug fixes¶

• Fixed executable name when creating binaries for Unix based operating systems.<
• Fixed potential crash when using --clean with extensions.
• PDB2PKA now mostly respects the --verbose setting.
• Fixed how hydrogens are added by PDB2PKA for state changes in some cases.
• Fixed psize error check.
• Will now build properly without ligand support if numpy is not installed.
• Removed old automake build files from all test ported to scons.
• Fixed broken opal backend.

## Known bugs¶

• If more than one extension is run from the command line and one of the extensions modifies the protein data structure it could affect the output of the other extension. The only included extensions that exhibit this problem are resinter and newresinter.
• Running ligands and PDB2PKA at the same time is not currently supported.
• PDB2PKA currently leaks memory slowly. Small jobs will use about twice the normally required RAM (i.e. ~14 titratable residues will use 140MB). Big jobs will use about 5 times the normally required RAM (60 titratable residues will use 480MB). We are working on this.

• Command line interface to PROPKA changed to accommodate PDB2PKA. PROPKA is now used with --ph-calc-method=propka --with-ph now defaults to 7.0 and is only required if a different pH value is required.
• --ph-calc-method to select optional method to calculate pH values used to protonate titratable residues. Possible options are “propka” and “pdb2pka”.


text:
Shooting Method
Get Shooting Method essential facts below. View Videos or join the Shooting Method discussion. Add Shooting Method to your PopFlock.com topic list for future reference or share this resource on social media.
Shooting Method

In numerical analysis, the shooting method is a method for solving a boundary value problem by reducing it to the system of an initial value problem. Roughly speaking, we 'shoot' out trajectories in different directions until we find a trajectory that has the desired boundary value. The following exposition may be clarified by this illustration of the shooting method.

For a boundary value problem of a second-order ordinary differential equation, the method is stated as follows. Let

${\displaystyle y''(t)=f(t,y(t),y'(t)),\quad y(t_{0})=y_{0},\quad y(t_{1})=y_{1}}$

be the boundary value problem. Let y(t; a) denote the solution of the initial value problem

${\displaystyle y''(t)=f(t,y(t),y'(t)),\quad y(t_{0})=y_{0},\quad y'(t_{0})=a}$

Define the function F(a) as the difference between y(t1; a) and the specified boundary value y1.

${\displaystyle F(a)=y(t_{1};a)-y_{1}\,}$

If F has a root a then the solution y(t; a) of the corresponding initial value problem is also a solution of the boundary value problem. Conversely, if the boundary value problem has a solution y(t), then y(t) is also the unique solution y(t; a) of the initial value problem where a = y'(t0), thus a is a root of F.

The usual methods for finding roots may be employed here, such as the bisection method or Newton's method.

## Origin of the term

The term 'shooting method' has its origin in artillery. When firing a cannon towards a target, the first shot is fired in the general direction of the target. If the cannon ball hits too far to the right, the cannon is pointed a little to the left for the second shot, and vice versa. This way, the cannon balls will hit ever closer to the target.

## Linear shooting method

The boundary value problem is linear if f has the form

${\displaystyle f(t,y(t),y'(t))=p(t)y'(t)+q(t)y(t)+r(t).\,}$

In this case, the solution to the boundary value problem is usually given by:

${\displaystyle y(t)=y_{(1)}(t)+{\frac {y_{1}-y_{(1)}(t_{1})}{y_{(2)}(t_{1})}}y_{(2)}(t)}$

where ${\displaystyle y_{(1)}(t)}$ is the solution to the initial value problem:

${\displaystyle y_{(1)}''(t)=p(t)y_{(1)}'(t)+q(t)y_{(1)}(t)+r(t),\quad y_{(1)}(t_{0})=y_{0},\quad y_{(1)}'(t_{0})=0,}$

and ${\displaystyle y_{(2)}(t)}$ is the solution to the initial value problem:

${\displaystyle y_{(2)}''(t)=p(t)y_{(2)}'(t)+q(t)y_{(2)}(t),\quad y_{(2)}(t_{0})=0,\quad y_{(2)}'(t_{0})=1.}$

See the proof for the precise condition under which this result holds.

## Example

A boundary value problem is given as follows by Stoer and Burlisch[1] (Section 7.3.1).

${\displaystyle w''(t)={\frac {3}{2}}w^{2},\quad w(0)=4,\quad w(1)=1}$
${\displaystyle w''(t)={\frac {3}{2}}w^{2},\quad w(0)=4,\quad w'(0)=s}$

was solved for s = -1, -2, -3, ..., -100, and F(s) = w(1;s) - 1 plotted in the first figure. Inspecting the plot of F, we see that there are roots near -8 and -36. Some trajectories of w(t;s) are shown in the second figure.

Stoer and Burlisch[1] state that there are two solutions, which can be found by algebraic methods. These correspond to the initial conditions w?(0) = -8 and w?(0) = -35.9 (approximately).

The function F(s) = w(1;s) - 1.
Trajectories w(t;s) for s = w'(0) equal to -7, -8, -10, -36, and -40. The point (1,1) is marked with a circle.


text:
Parameterized complexity of deciding if a string can be computed by circuits of size $k\log(n)$

In the following, we will describe what seems to be a parameterized version of the minimum circuit size problem (MCSP).

• Before we get started, we need the following concepts:

For every natural number $n$, consider a function $B_n: [n] \rightarrow \{0,1\}^{\lceil \log(n) \rceil}$ such that $B_n(i)$ is the $i$th bit string in $\{0,1\}^{\lceil \log(n) \rceil}$ according to the lexicographical ordering. (Example: $B_7(5) = 100$)

Now, let a bit string $x$ of length $n$ be given. Let a boolean circuit $C$ with $\lceil \log(n) \rceil$ inputs and $1$ output be given. We say that $C$ computes $x$ if for every $i \in [n]$, $C(B_n(i)) = x_i$. In other words, $C$ computes $x$ one bit at a time.

• Now, consider the following parameterized problem:

Given a bit string $x$ of length $n$ and a natural number $k$, does there exist a boolean circuit $C$ of size at most $k \log(n)$ that computes $x$?

If there are only $g(n,k)$ circuits of size $k\log(n)$, then we can solve this problem in roughly $n * k\log(n) * g(n,k)$ time by trying every possible circuit and evaluating each one on all $n$ inputs.

Question: What is the parameterized complexity of this problem?

This question has several parts:

• Are any more efficient algorithms known?

• Is this problem related to any standard parameterized complexity classes? For example, the problem could be $FPT$. Or, it could be $W[1]$-hard or harder.

• Are there any known parameterized complexity results or conditional lower bounds for the minimum circuit size problem?

• Nice question! It might depend on what you mean by the "size" of the circuit [whether you count gates or wires, or bit size in some way]. Seems very unlikely to be FPT to me... – daniello Mar 10 '17 at 11:10
• @daniello Thank you for the kind comment! I really appreciate it. :) – Michael Wehar Mar 10 '17 at 17:41
• In this case, $g(n,k) = n^k$ and we at least know that the problem lies in $XP$. Further, one could show that it lies in $W[P]$. But, I'm hoping someone out there might be able to show more. – Michael Wehar Mar 10 '17 at 17:45
• While this may not be a formal implication, it seems like the natural techniques to try to show that it is W[1]-hard would also show that MCSP is NP-hard, which is currently unknown (and some think unlikely). Would be nice to make the connection formal... – Joshua Grochow Nov 20 '19 at 6:21
• @JoshuaGrochow I finally had a chance to think about this. I think that if the above problem is W[1]-hard under parameter preserving reductions (defined in "Survey of parameter-preserving reductions" by Philipp Kuinke), then MCSP is NP-hard. – Michael Wehar Feb 10 at 21:27


text:
# dgl.DGLGraph.apply_edges¶

DGLGraph.apply_edges(func='default', edges='__ALL__', inplace=False)[source]

Apply the function on the edges to update their features.

If None is provided for func, nothing will happen.

Parameters: func (callable, optional) – Apply function on the edge. The function should be an Edge UDF. edges (valid edges type, optional) – Edges on which to apply func. See send() for valid edges type. Default is all the edges. inplace (bool, optional) – If True, update will be done in place, but autograd will break.

Notes

On multigraphs, if $$u$$ and $$v$$ are specified, then all the edges between $$u$$ and $$v$$ will be updated.

Examples

Note

Here we use pytorch syntax for demo. The general idea applies to other frameworks with minor syntax change (e.g. replace torch.tensor with mxnet.ndarray).

>>> import torch as th

>>> g = dgl.DGLGraph()

>>> # Doubles the edge feature.


text:
## 8.3 Variables

A variable is an expression. If a variable is bound to a value, the return value is this value. Unbound variables are only allowed in patterns.

Variables start with an uppercase letter or underscore (_). Variables can contain alphanumeric characters, underscore and @.

Examples:

XName1PhoneNumberPhone_number__Height

Variables are bound to values using pattern matching. Erlang uses single assignment, that is, a variable can only be bound once.

The anonymous variable is denoted by underscore (_) and can be used when a variable is required but its value can be ignored.

Example:

[H|_] = [1,2,3]

Variables starting with underscore (_), for example, _Height, are normal variables, not anonymous. They are however ignored by the compiler in the sense that they do not generate any warnings for unused variables.

Example:

The following code:

member(_, []) ->    [].

can be rewritten to be more readable:

member(Elem, []) ->    [].

This causes a warning for an unused variable, Elem, if the code is compiled with the flag warn_unused_vars set. Instead, the code can be rewritten to:

member(_Elem, []) ->    [].

Notice that since variables starting with an underscore are not anonymous, this matches:

{_,_} = {1,2}

But this fails:

{_N,_N} = {1,2}

The scope for a variable is its function clause. Variables bound in a branch of an if, case, or receive expression must be bound in all branches to have a value outside the expression. Otherwise they are regarded as 'unsafe' outside the expression.

For the try expression variable scoping is limited so that variables bound in the expression are always 'unsafe' outside the expression.


text:
## Electronic Journal of Probability

### Estimating the covariance of random matrices

Pierre Youssef

#### Abstract

We extend to the matrix setting a recent result of Srivastava-Vershynin about estimating the covariance matrix of a random vector. The result can be interpreted as a quantified version of the law of large numbers for  positive semi-definite matrices which verify some regularity assumption. Beside giving examples, we discuss the notion of log-concave matrices and give estimates on the smallest and largest eigenvalues of a sum of such matrices.

#### Article information

Source
Electron. J. Probab., Volume 18 (2013), paper no. 107, 26 pp.

Dates
Accepted: 19 December 2013
First available in Project Euclid: 4 June 2016

https://projecteuclid.org/euclid.ejp/1465064332

Digital Object Identifier
doi:10.1214/EJP.v18-2579

Mathematical Reviews number (MathSciNet)
MR3151727

Zentralblatt MATH identifier
1287.60014

Rights

#### Citation

Youssef, Pierre. Estimating the covariance of random matrices. Electron. J. Probab. 18 (2013), paper no. 107, 26 pp. doi:10.1214/EJP.v18-2579. https://projecteuclid.org/euclid.ejp/1465064332

#### References

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• R. Adamczak, A. Litvak, A. Pajor, and N. Tomczak-Jaegermann. Tail estimates for norms of sums of log-concave random vectors. arXiv:1107.4070, July 2011.
• Adamczak, Radosław; Litvak, Alexander E.; Pajor, Alain; Tomczak-Jaegermann, Nicole. Quantitative estimates of the convergence of the empirical covariance matrix in log-concave ensembles. J. Amer. Math. Soc. 23 (2010), no. 2, 535–561.
• Adamczak, Radosław; Litvak, Alexander E.; Pajor, Alain; Tomczak-Jaegermann, Nicole. Sharp bounds on the rate of convergence of the empirical covariance matrix. C. R. Math. Acad. Sci. Paris 349 (2011), no. 3-4, 195–200.
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• M. De Carli Silva, N. Harvey, and c. Sato. Sparse Sums of Positive Semidefinite Matrices. arXiv:1107.0088v2, July 2011.
• M. Fradelizi, O. Guédon, and A. Pajor. Spherical thin-shell concentration for convex measures. Available at arXiv:1306.6794.
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• Hoeffding, Wassily. Probability inequalities for sums of bounded random variables. J. Amer. Statist. Assoc. 58 1963 13–30.
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• L. Mackey, M. Jordan, R. Chen, B. Farrell, and J. Tropp. Matrix concentration inequalities via the method of exchangeable pairs. Available at arXiv:1201.6002.
• Oliveira, Roberto Imbuzeiro. Sums of random Hermitian matrices and an inequality by Rudelson. Electron. Commun. Probab. 15 (2010), 203–212.
• Paouris, G. Concentration of mass on convex bodies. Geom. Funct. Anal. 16 (2006), no. 5, 1021–1049.
• Paouris, Grigoris. Small ball probability estimates for log-concave measures. Trans. Amer. Math. Soc. 364 (2012), no. 1, 287–308.
• Rosenthal, Haskell P. On the subspaces of $L^{p}$ $(p>2)$ spanned by sequences of independent random variables. Israel J. Math. 8 1970 273–303.
• Rudelson, M. Random vectors in the isotropic position. J. Funct. Anal. 164 (1999), no. 1, 60–72.
• Srivastava, Nikhil. Spectral sparsification and restricted invertibility. Thesis (Ph.D.) - Yale University. ProQuest LLC, Ann Arbor, MI, 2010. 98 pp. ISBN: 978-1124-09198-3
• Srivastava, Nikhil; Vershynin, Roman. Covariance estimation for distributions with $2+\varepsilon$ moments. Ann. Probab. 41 (2013), no. 5, 3081–3111.
• Tropp, Joel A. User-friendly tail bounds for sums of random matrices. Found. Comput. Math. 12 (2012), no. 4, 389–434.


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# Confusion - Am I on the right track on this power computation?

So the question asks: It has been generally believed that average daily caffeine consumption, $μ$, is at most $200$ mg with $σ^2 = 225$. To examine whether at present time $μ$ is greater than $200$ mg, it is agreed that daily caffeine consumption of a random sample of $100$ adults be examined. If the sample mean is greater than $203.50$, the decision will be to reject the null hypothesis.

The following information may be useful: pnorm(0.67)$=0.75$, pnorm(1)$=0.84,$ pnorm(1.56)$=0.94$, pnorm(2.33)$=0.99$

(a) What are the null and alternative hypotheses?

(b) What is the test statistic and what is it distribution?

(c) What is the probability of a Type 1 error?

(d) What is the power of the test if the true mean is 205mg?

(e) Is the assumption that the population distribution of daily caffeine consumption is normal necessary for the hypothesis test? Explain your reasoning.

(a) $H_0: μ \le 200$ (average daily caffeine consumption is at most $200mg$)

$H_1: μ > 200$ (average daily caffeine consumption is more than $200mg$)

(b) For large sample, and known population standard deviation, use 1-sample $Z$ test. Distribution: Normal.

$$Z=\frac{\bar X-μ}{\sigma/\sqrt{n}} =\frac{203.50-200}{15/10} =2.3333.$$

(c) probability of Type I error, $\alpha=0.05$

(d) Since it is a right-tailed test, therefore, one commits a Type II error (fail to reject $H_0$), when one gets a test statistic less than 1.645. Compute $X_\text{critical}$ by substituting the value in following Z score formula

$Z =1.64=\frac{X_\text{critical}-\mu}{\sigma/\sqrt{n}}=\frac{X_\text{critical}-200}{1.5}$

$X_\text{critical}=202.46$

$$P(\bar X<202.46)=P\left(Z<\frac{202.46-205}{1.5}\right)=P(Z<-1.69333333)=0.0455.$$

Power of test: $1-0.0455 =0.9545.$

(e) Not necessary because the sample size is sufficiently large.

So I am not sure about my answers in d and e. Should I calculate the new x, then use z-test to get the power of the test? And in e, does my answer sound reasonable enough?

• I have edited your question slightly to improve formatting; please check to make sure I have not changed your meaning. – BruceET Dec 2 '16 at 23:51

(c) To check your power computation, here is a printout from Minitab's 'Power and Sample Size' procedure. [The slight discrepanacy might be because you have used 1.64 instead of 1.644854 (or 1.645 as in many tables).]

 1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 15

Sample
Difference    Size     Power
5     100  0.954340

The power curve shows the power against alternatives with differences other than $\Delta = \mu_a - \mu_0 = 205 - 200 = 5.$ [The only point on this curve of 'P(Reject) vs. $\Delta$' that is not a 'power' value is above $\Delta = 0$ where the height of the curve is the significance level $\alpha = .05.$]

(d) My answer would be only slightly more cautious: Because of the large sample size, probably not necessary--unless the actual population distribution is extremely skewed or otherwise very far from normal.

Even for $n=100,$ results would not be accurate for an exponential population, and would be totally wrong for a Cauchy population. (But I think the question clearly expects you to take the Central Limit Theeorem into consideration.)


text:
Feeds:
Posts
The problem of quantum gravity is impossibly hard because it is too general.  Quantum gravity problem is not trivial but not nearly as hard if we consider a static spacetime such as Einstein static model $\mathbf{R}\times S^3(a)$.  Quantum field theory on this specific spacetime is essentially a solved problem.  In fact, even Euclidean quantum field theory following Osterwalder-Schrader quantization and Wightman axioms has been worked out in this case ( jaffee-ritter-qft1jaffee-ritter-qft2 osterwalder-schrader-2osterwalder-schrader-1 ).  In presentations of quantum field theory on curved spacetimes the Einstein static spacetime appears often as a relatively simple example (see Fewster lecturenote-3908 (1)).


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# How to prove that if $a_n=o(n)$ then $\sum_{k=0}^n\frac{a_n^k}{k!}\sim e^{a_n}$

I recall a question where it was proved that $$\lim_{n\to\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=\frac12.$$This seems to remain true when $n$ is replaced by $n+c$ where $c$ is some constant, while apparently, if $a_n=o(n)$, $$\sum_{k=0}^n\frac{a_n^k}{k!}\sim e^{a_n}.$$How to prove it? Maybe the method used in the related question is also useful here, but I couldn't find it

• I had a typo in the original post, where in the first display $a_n$ should have been $n$ - fixed now – Richard Aug 31 '18 at 15:32
• Is $a_n > 0$ for all $n$? – Daniel Fischer Aug 31 '18 at 15:36
• @DanielFischer Yes – Richard Aug 31 '18 at 15:39
• Perhaps it was one of these questions that you saw. – Antonio Vargas Aug 31 '18 at 18:21
• Why does this not tend to 1? Do you have the original proof? – Henry Lee Aug 31 '18 at 18:53

For every $0 < \varepsilon < e^{-1}$, choose $N(\varepsilon)$ such that $a_n < \varepsilon n$ for $n \geqslant N(\varepsilon)$. Then we have \begin{align} e^{a_n} - \sum_{k = 0}^{n} \frac{a_n^k}{k!} &= \sum_{k = n+1}^{\infty} \frac{a_n^k}{k!} \\ &< \sum_{k = n+1}^{\infty} \frac{(\varepsilon n)^k}{k!} \\ &< \varepsilon^{n+1}\sum_{k = n+1}^{\infty} \frac{n^k}{k!} \\ &< \varepsilon^{n+1} e^n \\ &< \varepsilon \end{align} for $n \geqslant N(\varepsilon)$. So we even have $$e^{a_n} - \sum_{k = 0}^{n} \frac{a_n^k}{k!} \to 0\,,$$ not only $$\sum_{k = 0}^{n} \frac{a_n^k}{k!} \sim e^{a_n}\,.$$
• How is the step $e^{a_n} - \sum_{k=0}^{n} \frac{a_n^k}{k!}= \sum_{k=n+1}^{\infty} \frac{a_n^k}{k!}$ motivated taking into account that $a_n$ is varying? – user Aug 31 '18 at 15:54
• @gimusi It depends on $n$, but not on $k$. – Daniel Fischer Aug 31 '18 at 15:56
• But isn't it equivalent to assume $e^{a_n} = \sum_{k = 0}^{\infty} \frac{a_n^k}{k!}$? – user Aug 31 '18 at 16:00
• But that's a given. $$e^x = \sum_{k = 0}^{\infty} \frac{x^k}{k!}$$ is the definition of $e^x$. (Yes, it is possible to define it differently, then one needs to prove the power series expansion, but that's standard.) – Daniel Fischer Aug 31 '18 at 16:03
• Were you perhaps thinking about $$\sum_{k = 0}^{n} \frac{a_k^k}{k!}\,,$$ @gimusi? – Daniel Fischer Aug 31 '18 at 16:06


text:
Definition 17.27.3. Let $X$ be a topological space. Let $\varphi : \mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings on $X$. The module of differentials of $\varphi$ is the object representing the functor $\mathcal{F} \mapsto \text{Der}_{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{F})$ which exists by Lemma 17.27.2. It is denoted $\Omega _{\mathcal{O}_2/\mathcal{O}_1}$, and the universal $\varphi$-derivation is denoted $\text{d} : \mathcal{O}_2 \to \Omega _{\mathcal{O}_2/\mathcal{O}_1}$.

There are also:

• 2 comment(s) on Section 17.27: Modules of differentials

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text:
# March 28 exercise from lecture

In the following exercise, you will be defining sequences of numbers recursively. That means that a number in the sequence is defined in terms of the numbers coming before it.

Exercise 1: Complete parts (a), (b), and (c) below for your team.

Team rabbit:
One pair of very young rabbits (one male and one female) is dropped on a deserted island. After they are two months old, they give birth to one new pair of rabbits (one male and one female) per month. Let $f_n$ be the number of rabbits living on the island after $n$ months.

Team bitstring:
Let $f_n$ be the number of bitstrings of length $n$ that do not have two consecutive 0s.

(a) Find $f_n$ for $n=0, 1, 2, 3, 4, 5, 6, 7, 8$.
(b) Find a formula for $f_n$ in terms of the $f_i$ for $i. (This is called a recurrence relation.)
(c) Use your formula from (b) to find $f_9, f_{10}, f_{11}$.

Edit: We saw in class that both teams’ sequences of numbers were equal. This sequence is known as the Fibbonacci sequence. This sequence pops up all over mathematics, not just in the two examples above. You can read more about it here.

This entry was posted in Discussion. Bookmark the permalink.


text:
# Heston model for Options pricing with ESGtoolkit

Hi everyone! Best wishes for 2016!

In this post, I’ll show you how to use ESGtoolkit, for the simulation of  Heston stochastic volatility model for stock prices.

If you’re interested in seeing other examples of use of ESGtoolkit, you can read these two posts: the Hull and White short rate model and the 2-factor Hull and White short rate model (G2++).

The Heston model was introduced by Steven Heston’s A closed-form solution for options with stochastic volatility with applications to bonds an currency options, 1993. For a fixed risk-free interest rate $r$,  it’s described as:

$dS_t = r S_t dt + \sqrt{v_t}S_t dW^S_t$

$dv_t = \kappa(\theta - v_t) dt + \sigma \sqrt{v_t}dW^v_t$

where $dW^S_t dW^v_t = \rho dt$.

In this model, under a certain probability, the stock price’s returns on very short periods of time of length $dt$, are: the risk-free rate + a random fluctuation driven by the terms $dW_t$ and $v_t$. The  $dW_t$‘s can be thought of (very simply put) as the gaussian increments of a random walk, which are centered, and have a  variance equal to $dt$.

On the other hand, $v_t$ is the stochastic variance of the stock prices.  $v_t$ is always positive, and tends to return to a fixed level $\theta$ at a speed controlled by $\kappa$. The variance of $v_t$ is controlled by $\sigma$, which is called the volatility of volatility. To finish, $dW^S_t$ and $dW^v_t$ have an instantaneous correlation equal to $\rho$.

By using this model, one can derive prices for European call options, as described in Calibrating Option Pricing Models with Heuristics. The authors provide a useful function called ‘callHestoncf’, which calculates these prices in R and Matlab.

Here’s the function’s description. I won’t reproduce the function here, please refer to the paper for details:

callHestoncf(S, X, tau, r, v0, vT, rho, k, sigma){
# S = Spot, X = Strike, tau = time to maturity
# r = risk-free rate, q = dividend yield
# v0 = initial variance, vT = long run variance (theta)
# rho = correlation, k = speed of mean reversion (kappa)
# sigma = volatility of volatility
}

Now, it’s time to use ESGtoolkit for Monte Carlo pricing. We are going to price 3 European Call options on $(S_t)_{t \geq 0}$, with 3 different exercise prices. We use 100000 simulations on 15 years, with monthly steps. Here are the parameters which will be useful for the simulation:

rm(list=ls())
library(ESGtoolkit)

#Initial stock price
S0 <- 100
# Number of simulations (feel free to reduce this)
n <- 100000
# Sampling frequency
freq <- "monthly"
# volatility mean-reversion speed
kappa <- 0.003
# volatility of volatility
volvol <- 0.009
# Correlation between stoch. vol and spot prices
rho <- -0.5
# Initial variance
V0 <- 0.04
# long-term variance
theta <- 0.04
#Initial short rate
r0 <- 0.015

# Options maturities
horizon <- 15
# Options' exercise prices
strikes <- c(140, 100, 60)

For the simulation of the Heston model with ESGtoolkit, we first need to define how to make simulations of the terms $dW^S_t$ and $dW^v_t$.  This is done by the package’s function ‘simshocks’, in which you can define the type of dependence between models’ increments:

# Simulation of shocks with given correlation
set.seed(5) # reproducibility seed
shocks <- simshocks(n =  n,
horizon =  horizon,
frequency =  freq,
method = "anti",
family = 1, par =  rho)

This function provides a list with 2 components, each containing simulated random gaussian increments. Both of these components will be useful for the simulation of $(v_t)_{t \geq 0}$ and $(S_t)_{t \geq 0}$.

#  Stochastic volatility  simulation
sim.vol <- simdiff(n =  n, horizon =  horizon,
frequency =  freq, model = "CIR", x0 =  V0,
theta1 =  kappa*theta, theta2 =  kappa,
theta3 =  volvol, eps =  shocks[[1]])

# Stock prices simulation
sim.price <- simdiff(n = n, horizon = horizon,
frequency = freq, model = "GBM", x0 = S0,
theta1 = r0, theta2 = sqrt(sim.vol),
eps = shocks[[2]])

We are now able to calculate options prices, with the 3 different
exercise prices. You’ll need to have imported ‘callHestoncf’, for this piece of code to work.

# Stock price at maturity (15 years)
S_T <- sim.price[nrow(sim.price), ]

### Monte Carlo prices
#### Estimated Monte Carlo price
discounted.payoff <- function(x)
{
(S_T - x)*(S_T - x > 0)*exp(-r0*horizon)
}
mcprices <- sapply(strikes, function(x)
mean(discounted.payoff(x)))

#### 95% Confidence interval around the estimation
mcprices95 <- sapply(strikes,  function(x)
t.test(discounted.payoff(x),
conf.level = 0.95)\$conf.int)

#### 'Analytical' prices given by 'callHestoncf'
pricesAnalytic <- sapply(strikes, function(x)
callHestoncf(S = S0, X = x, tau = horizon,
r = r0, q = 0, v0 = V0, vT = theta,
rho = rho, k = kappa, sigma = volvol))

results <- data.frame(cbind(strikes, mcprices,
t(mcprices95), pricesAnalytic))
colnames(results) <- c("strikes", "mcprices", "lower95",
"upper95", "pricesAnalytic")

print(results)

strikes mcprices  lower95  upper95 pricesAnalytic
1     140 25.59181 25.18569 25.99793         25.96174
2     100 37.78455 37.32418 38.24493         38.17851
3      60 56.53187 56.02380 57.03995         56.91809

From these results, we see that the Monte Carlo prices for the 3 options are
fairly close to the price calculated by using the function ‘callHestoncf’
(uses directly formulas for the prices calculation). The 95% confidence
interval contains the theoretical price. Do not hesitate to change the seed,
and re-run the previous code.

Below are the option prices, as functions of the number of simulations.
The theoretical price calculated by ‘callHestoncf’ is drawn in blue,
the average Monte Carlo price in red, and the shaded region represents
the 95% confidence interval around the mean (the Monte Carlo price).

1 Comment

Filed under ESGtoolkit, R stuff

### One response to “Heston model for Options pricing with ESGtoolkit”

1. Pingback: Distilled News | Data Analytics & R


text:
## Saturday, May 24, 2008

### Normal command from Insert mode

Always forget the very useful shortcut which is apparently intended for "Vim Easy":

^O

When in insert mode (and, by definition, in easy vim you are always in the mode) to insert command from Normal mode, you press first ^O and then start typing the command.

e.g. to delete to the end of line press ^O followed by Shift-d. Pressing : you can also access all the ex commands too - but that will leave insert mode.


text:
# Irregularly spaced time-series in finance/economics research

In financial econometrics research, it is very common to investigate relationships between financial time series that take the form of daily data. The variable will often be made $I(0)$ by taking the log difference, for example; $\ln(P_t)-\ln(P_{t-1})$.

However, daily data means that there's $5$ data points each week, and Saturday and Sunday are missing. This seems to get no mention in the applied literature that I'm aware of. Here's some closely related questions that I have that come from this observation:

• Does this qualify as irregularly spaced data, even though financial markets are closed over the weekend?

• If so, what are the consequences for the validity of extant empirical results garnered thus far in the gigantic number of papers that ignore this issue?

• Regarding your first question, this problem is sometimes called weekend effect. In my opinion, the answer is context-dependent. For instance, this question makes a lot of sense in the case of stock returns. See for instance here, here, here and here. But I am not sure if this effect applies to other contexts.
– user10525
Dec 15, 2012 at 15:52
• @Procrastinator Submit answer it's very good!!
– Jase
Dec 15, 2012 at 16:19
• There is a quantitative finance SE that may be more suited to get meaningfull answers. There are actualy a lot more problems than weekends: nights, bank holidays... etc. which get worse with multiple price sources. Jul 22, 2019 at 9:41


text: dc.contributor.author Novick, Mark B. en_US dc.date.accessioned 2007-04-23T17:37:01Z dc.date.available 2007-04-23T17:37:01Z dc.date.issued 1989-06 en_US dc.identifier.citation http://techreports.library.cornell.edu:8081/Dienst/UI/1.0/Display/cul.cs/TR89-1015 en_US dc.identifier.uri https://hdl.handle.net/1813/6815 dc.description.abstract We give fast parallel algorithms for recognizing ad representing comparability graphs that can be transitively oriented, and interval graphs, the intersection graphs of intervals along the real line. Under the CRCW PRAM model, both algorithms use $O(n^{3})$ processors in $O$(log $n$) time to check if a graph belongs to the desired class, and if it does then a valid representation of the graph can be produced. The algorithms gain their efficiency by using fast algorithms for finding the modular decomposition of a graph. Both problems were known to be in $NC$, but the known algorithms require more time than ours does. en_US dc.format.extent 590897 bytes dc.format.extent 359714 bytes dc.format.mimetype application/pdf dc.format.mimetype application/postscript dc.language.iso en_US en_US dc.publisher Cornell University en_US dc.subject computer science en_US dc.subject technical report en_US dc.title Logarithmic Time Parallel Algorithms for Recognizing Comparability and Interval Graphs en_US dc.type technical report en_US


text:
# Calculation of material properties in transformation media

1. Apr 23, 2010

Hi everybody,

I'm focusing on meta-materials. I have recently read Schurig's paper "http://www.opticsinfobase.org/oe/abstract.cfm?URI=OPEX-14-21-9794"). In the article, the components of the permittivity tensor are given by
$$\varepsilon^{i'j'} = \left|\rm{det}(\Lambda^{i'}_{i})\right|^{-1} \Lambda^{i'}_{i} \Lambda^{j'}_{j} \varepsilon^{ij}$$
where the Jacobian matrix
$$\Lambda_{\alpha}^{\alpha'} = \frac{\partial x^{\alpha'}}{\partial x^{\alpha}}$$
and the roman indices run from1 to 3, for the three spatial coordinates, as is standard practice.

Working out the algebra, the components of the permittivity (permeability) tensor can be obtained by
$$\left(\varepsilon^{i'j'}\right) = \left|\rm{det}\left(\Lambda\right)\right|^{-1}\Lambda^T \Lambda$$
where $$\Lambda$$ is a matrix, which components are the counterpart of the contravariant coefficients $$\Lambda_{\alpha}^{\alpha'}$$.

For cylindrical cloak, the components of the transformation matrix are
$$\left(\Lambda^{i'}_{j}\right) = \left( \begin{array}{ccc} \frac{\rho'}{\rho}-\frac{ax^2}{\rho^3} & -\frac{axy}{\rho^3} & 0 \\ -\frac{ayx}{\rho^3} & \frac{\rho'}{\rho}-\frac{ay^2}{\rho^3} & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$
It is easy to find the material properties. For instance, the z component of the permittivity tensor is
$$\varepsilon_z = \varepsilon^{3,3} = \frac{\rho^2}{\rho'(\rho'-a)} = \frac{1}{\left|\rm{det}\left(\Lambda\right)\right|}$$

However, in the paper "http://pre.aps.org/abstract/PRE/v74/i3/e036621" [Broken]), the components of the relative permittivity and permeability tensor specified in cylindrical coordinates are given
$$\varepsilon_z = \mu_z = \left(\frac{b}{b-a}\right)^2 \frac{\rho-a}{\rho}$$

It can be seen that the two formula are not equal obviously. And the other nonzero components of the permittivity and permeability tensor are not equal too.

I have deduced the formulas for many times. Depressingly, I can not figure out the problem. Could somebody please give me some comments on the calculation of material properties in transformation optics.

Regards.

Last edited by a moderator: May 4, 2017
2. Apr 23, 2010

### Andy Resnick

I just glanced through the papers, but I wonder if you are looking in the wrong place: see, for example, eqns 20-22 and 29-30 in the OE paper.

3. Apr 23, 2010


text:
# Language - Nonterminal Symbol

A non terminal symbol does not appear in the resulting formal language.

They are placeholders for patterns of terminal symbols that can be generated by the nonterminal symbols.

In a context-free grammar, Non terminal symbol appears on the left side of the production.

## 3 - Example

In a the following context free grammar

A -> a
A -> b
• The symbol A will never appears in the language.
• The language is composed of a and b but not A

A is then nonterminal symbol.