text:
cultural city in Mandalay Region, Myanmar
Mandalay (/ˌmændəˈleɪ/ or /ˈmændəleɪ/; Burmese: ᝫᝨᝧᝮᝡ; MLCTS: manta.le: [màɴdəlé]) is a city in Myanmar. There is only one city in Myanmar that has more people than Mandalay. It is 716 km (445 mi) north of Yangon. Mandalay is on the east side of the Irrawaddy River. In 2014, there were 1,225,553 people living there.
ᝫᝨᝧᝮᝡ
Location of Mandalay, Myanmar
Coordinates: 21°58′59″N 96°05′04″E / 21.9831°N 96.0844°E / 21.9831; 96.0844
Mandalay Region
Mandalay District
Aung Maung[1]
• City
163.84 km2 (63.26 sq mi)
(2014 census)[3][4]
7,500/km2 (19,000/sq mi)
• Ethnic groups
Bamars Indian Chinese Shan
• Religions
Theravada Buddhism Christianity Hinduism Islam
UTC+6:30 (MST)
Area code(s)
2 (mobile: 20,69, 90)[5]
2.1 Early history
2.2 Mandalay in British Burma (1885-1948)
2.3 Mandalay today (1948 - now)
3 Important places
3.1 Mandalay Palace
3.2 Atumashi Monastery
3.3 Mandalay Hill
3.4 Kuthowdaw Pagoda
3.5 Mahamuni Buddha Temple
3.6 Other places
6 Weather
7 Mandalay in writing
8 Twin towns
The name "Mandalay" comes from the nearby hill called Mandalay Hill. "Mandalay" may come from the Pali language. Some people think that the name "Mandalay" comes from the Pali word mandala.[4] Other people think that the name comes from Mount Mindara. Mindara is a mountain in Hindu mythology.[6]
The original name of the city was Yadanabon (ရတနာပုံ, [jədənàbòʊɴ]). This name came from the Pali word Ratanapūra (ရတနပူရ). "Ratanapūra" means "City of Gems."[7][8] The city also had another name: Lay Kyun Aung Myei (လေးကျွန်းအောင်မြေ, [lé dʑʊ́ɴ àʊɴ mjè]). This means "Victorious Land over the Four Islands."
Early historyEdit
The Mandalay Palace grounds.
Mandalay was started by King Mindon. He started the city on 13 February 1857. He built Mandalay as the new capital city of the Kingdom of Burma.[9]
Mindon's city was 66 km2 (25.5 sq mi) in size. There were rivers on each side. It had 144 square blocks in a grid. In the middle was a palace that was 16 square blocks in size.[10]
A tower in the Mandalay Palace walls.
The palace had four walls that were each 203 m (666 ft) long. Around the walls was a moat that was 64 m (210 ft) wide and 4.6 m (15 ft) deep. Each wall had turrets (towers) for watchmen. These turrets were placed every 169 m (555 ft). Each turret had a spire made of gold.[11] Each side had three gates and five bridges across the moat.[10]
The king also built some other buildings in the city. These buildings were the Kuthodaw Pagoda, the Pahtan-haw Shwe Thein upasampada hall, the Thudamma "Good Dharma" zayats (IPA: [zəjaʔ]), and a library for the Pāli Canon. A zayat is a building where Buddhism is taught.
In June 1857, the king moved the royal palace to Mandalay. The palace had been in Amarapura. The building was taken apart. Each piece was moved by elephants to Mandalay. The palace was put back together and finished on 23 May 1859. The palace is now at the bottom of Mandalay Hill.[9]
Mandalay was the capital for 26 years. It was the last royal capital of the Konbaung Dynasty. (The Konbaung Dynasty were the last rulers of the Kingdom of Burma before it was taken over by the British Empire.) On 28 November 1885, the British conquered Burma. This ended the Third Anglo-Burmese War. King Thibaw Min and Queen Supayalat were sent into exile.
Mandalay in British Burma (1885-1948)Edit
Map of Mandalay, 1911
Mandalay was still an important city in British Burma. The city of Rangoon became the capital of Burma.
The British wanted to make Mandalay an important commercial city. They did not want it to be important for politics, education, or other reasons. Mandalay was connected to Yangon by railroad in 1889.[12] The first college in Mandalay wasn't built until 1925.[13]
The royal palace was looted by the British. All of the treasures were stolen. Some treasures can be seen in the Victoria and Albert Museum.[14] The British changed the name of the palace to Fort Dufferin. British soldiers used the palace as a barracks (a place to camp.)
A street scene in Mandalay during British colonial times. This street is in Mandalay's Chinatown neighborhood.
Mandalay was the most important city for Burmese culture during the time the British were in charge. It was also the most important place for Buddhist learning. The Burmese people saw Mandalay as a symbol. Between World War I and World War II, there were many protests in Mandalay against the British.[4]
During the time the British were in charge, many people moved to Mandalay from India. (India was also a British colony.) From 1904 to 1905, a plague happened. About one third of the people of Mandalay left to get away from the disease.[4]
During World War II, Mandalay was bombed by the Japanese. The British moved all of their airplanes to India. Mandalay had no defenses against the Japanese air raids. On 3 April 1942, the Imperial Japanese Army Air Service bombed the city. Three-fifths (3/5) of the houses were destroyed. More than 2,000 civilians (people who are not military) were killed.[15][16][17]
The Japanese soon invaded Burma. The Imperial Japanese Army took over the city of Mandalay. They were in charge from May 1942 until March 1945. The palace citadel was made into a supply depot. (A supply depot is a building that stores supplies like food and machines.)
The Allied forces bombed the city after the Japanese took it over. Much of the palace was destroyed. Only the royal mint and watch tower were not destroyed. The palace was rebuild in the 1990s.
Mandalay today (1948 - now)Edit
In 1948, the British left Burma. Mandalay was still the most important city for culture, education, and commerce in Upper Burma.
Until the 1990s, most students went to university in Mandalay. Until 1991, there were only three universities in Upper Burma: Mandalay University, The University of Medicine, Mandalay, and the Defense Services Academy. All of these universities are in Mandalay. Only some other cities had schools called "Degree Colleges." These were small schools that were part of Mandalay University. These colleges had classes in a small number of subjects.
From 1962 to 1988, Ne Win was in charge of the government. He was very isolationist. This means that he did not want Myanmar to be part of global trade or tourism. In the 1980s, Mandalay was the second biggest city in Myanmar. It looked like a town with low buildings, dusty streets, and more bicycles than cars.
During the 1980s, there were two big fires. In May 1981, a fire destroyed more than 6,000 buildings. 36,000 people were left homeless. On 24 March 1984, there was another fire. This fire destroyed 2,700 buildings. 23,000 more people were made homeless.[18][19]
Mandalay metropolitan area seen from satellite
Fires are still a big problem. In February 2008, there was another big fire in Mandalay. It destroyed the Yadanabon Market. This was the second-biggest market in the city. In February 2009, a fire destroyed 320 houses. More than 1,600 people were made homeless.[20][21]
The fires during the 1980s destroyed a lot of buildings. This made a lot of land empty. Many Chinese people in Burma bought this land. A lot of these people had come from Yunnan in China.[22] Many Chinese people came to Upper Burma and Mandalay from Yunnan and Sichuan in the 1990s.[23] 250,000 to 350,000 people from Yunnan came to Mandalay in the 1990s.[24] Today, about 30% to 40% of the people of Mandalay are Chinese.[24] The native Burmese people complain that Mandalay is now like a part of China.[25]
The Chinese have rebuilt the parts of the city that burned down. They built apartment buildings, hotels, and shopping centers. They helped make Mandalay again an important city for trade and commerce. It now connects trade for Upper Burma, Lower Burma, China, and India.
Mandalay has grown much bigger over time. It now includes what used to be Amarapura. (Amarapura was where King Mindon's capital city used to be before he moved it to Mandalay.) Mandalay celebrated its 150th birthday on 15 May 2009 at 4:31:36 am.[4]
Important placesEdit
There are many important places and buildings in Mandalay.
Mandalay PalaceEdit
The Mandalay Palace.
The royal palace in Mandalay is a very important building. A lot of it was burned down during World War II. The royal mint and watch tower did not burn down. The rest of the buildings were rebuilt.
The Mandalay Palace's name in Burmese is Mya Nan San Kyaw (မြနန်းစံကျော် [mja̰ nán sàn tɕɔ̀]; "The Famed Royal Emerald Palace"). It is also called ရွှေနန်းတော်ကြီး ([ʃwè nán dɔ̀ dʑí]), or the "Great Golden Royal Palace".
Today the palace is an important symbol of Mandalay. Many tourists and visitors come to see it.
Atumashi MonasteryEdit
The Atumashi Monastery was built again after it burned down.
The Atumashi Monastery is a Buddhist monastery. A monastery is a building where monks (holy men) live, work, and pray. This place is a kyaung, a special kind of monastery. The name means "inimitable vihara."
The monastery burned down in 1890. Only the plinth (bottom platform) did not burn. The government decided to rebuild the monastery. They started building on 2 May 1995. It was finished being built in June of 1996.
Mandalay HillEdit
There are many important religious places on Mandalay Hill.
Mandalay Hill is a big hill in Mandalay. It is a holy hill. There is a legend about the hill. The Buddha visited the hill and made a prophecy. He said that some day a great city would be built at the bottom of the hill.
The hill is 230 metres (750 ft) tall. There is a road to the top. There are many temples, monasteries, and pagodas built on the hill.
Kuthowdaw PagodaEdit
Kuthodaw Pagoda. This place holds the world's largest book.
The Kuthodaw Pagoda is a very large pagoda. It was built by King Mindon in 1857. It was designed to look like the Shwezigon Pagoda in Nyaung U.
There are 729 rocks around the pagoda. Each rock is smooth and has writing on it. The writing are the words of the Tipiṭaka. (The Tipiṭaka are the Buhhdist holy books.) These stone rocks are the "world's largest book."
Mahamuni Buddha TempleEdit
Mahamuni Temple
The Mahamuni Buddha Temple is one of the most important Buddhist temples. It has an important Buddha statue. The statue was made during the life of the Gautama Buddha. The Gauthama Buddha hugged the statue seven times. This caused the statue to come to life.
Buddhists believe that the statue is alive because of this. They call it the Mahamuni Sacred Living Image. This makes the Mahamuni Temple the most holy pagoda in Mandalay.
The temple was built by King Bodawpaya in 1784. The Buddha is sitting. It is 384 centimetres (12 ft 7 in) tall.
Other placesEdit
The Shwenandaw Monastery.
Near the bottom of Mandalay Hill is the Kyauktawgyi Pagoda. It was built by King Mindon from 1853 to 1878. It was carved out of one big block of marble. Around it are 80 arahants.
The Yadanabon Zoological Garden is a small zoo in Mandalay. It is in between the palace and Mandalay Hill. It is the only zoo in the world that has Burmese roofed turtles.
The Shwenandaw Monastery (Burmese: ရွှေနန်းတော်ကျောင်း; MLCTS: hrwe. nan: taw kyaung:, IPA: [ʃwènándɔ̀ tʃáun]; lit. "Golden Palace Monastery") is an important Buddhist monastery. It was built by King Thibaw Min in 1880.[26] It is made from teak, a kind of wood. Inside of it are many carvings on the walls and roofs. The carvings show Buddhist myths.[27]
In 2007, the United Nations (UN) guessed that there were about 1 million people living in Mandalay.[3] In 2014, the Myanmar government counted all of the people in the country for a census. They found that there were 1,225,133 people living in Mandalay then.[4] The UN thinks that there will be 1.5 million people in Mandalay in 2025.[3]
Mandalay has always been an important city for the Bamar people. Most of the people who have lived in Mandalay are Burman people. Lots of Chinese people have been moving to Mandalay. The Chinese have been coming for the past 20 years. Now there are more Chinese people than Burman people in Mandalay.[23][28][29]
Today, 40-50% of the people in Mandalay are Chinese. About 30% of Mandalay's people are Yunnanese. The rest of the people in Mandalay are probably Burman.[30] There are also a lot of Burmese Indians in Mandalay. (These are Indian people who live in Myanmar.)
People in Mandalay speak the Burmese language. A lot of people also speak the Standard Chinese language. The Chinese language is used a lot in markets like Zegyo Market. People who have a lot of money and education also speak the English language.
PlaceEdit
Mandalay is in the country of Myanmar. It is by the side of the Irrawaddy river. Its coordinates are 21.98° North, 96.08° East.
The city is 79 metres (260 ft) above sea level. It is in the same time zone as the rest of Mandalay: Myanmar Standard Time (MST), UTC+6:30.
Mandalay is on a tectonic fault. This means that there are two tectonic plates that come together underneath the city. (A tectonic plate is a big piece of earth.) The tectonic fault underneath Mandalay is called the Sagaing Fault. It is where the India Plate and the Sunda Plate meet.
A lot of earthquakes happen on tectonic faults. The biggest earthquake in Mandalay happened in 1956. That one was a magnitude 7 earthquake. That earthquake did a lot more damage to the city of Sagaing, which is nearby. It is now called the "Great Sagaing Quake."
WeatherEdit
Mandalay has a tropical savanna climate. This means that the city has a wet season and a dry season. There is a lot of rain during the wet season. There is very little rain during the dry season. The wet season is from May to October. The dry season is from November to April.
The coldest month is January. The average temperature is 21 °C (70 °F). The hottest month is April. Then, the average temperature is 31 °C (88 °F).
It is very hot in April and May. Most days it is hotter than 35 °C (95 °F). Some days it is hotter than 40 °C (104 °F).
The hottest temperature ever in Mandalay was 45.6 °C (114.1 °F). The coldest temperature ever was 5.6 °C (42.1 °F).[31]
Climate data for Mandalay (1961–1990)
Average high °C (°F)
(101.1) 36.8
Daily mean °C (°F)
Average low °C (°F)
Average rainfall mm (inches)
(0.2) 2
(1.6) 138
Average rainy days
Average relative humidity (%)
Mean monthly sunshine hours
309 280 301 291 267 208 182 168 215 223 269 278 2,991
Source 1: World Meteoroglogical Organization,[32] Weatherbase (record highs and lows).[33]
Source 2: Danish Meteorological Institute (sun and relative humidity)[34]
Mandalay in writingEdit
There are some important books and poems that talk about Mandalay. Some of these are famous.
Rudyard Kipling wrote a poem called "Mandalay." He wrote it in 1890. This poem has a famous saying in it: "on the road to Mandalay."[35]
In 1907, Oley Speaks wrote a song using the words of the poem. He called the song "On the Road to Mandalay." The song is very famous. Frank Sinatra sang the song and called it Come Fly with Me.
George Orwell wrote a novel called Burmese Days. He wrote it in 1934. Orwell was part of the Indian Imperial Police. He was stationed in Mandalay. The novel is about his experiences (what he did and saw.) He also wrote some non-fiction essays and short stories. Some of these are "A Hanging" (1931) and "Shooting an Elephant" (1936).
Twin townsEdit
Mandalay has three twin towns:
Cirebon, Indonesia
Kunming, China
↑ Thein Sein (28 February 2012). "Mandalay Mayor appointed Republic of the Union of Myanmar". New Light of Myanmar. Government of Myanmar. Archived from the original on 13 November 2012. Retrieved 28 February 2012.
↑ "Water Purification Plant No. 8 in Aungmyethazan Township 60% Coomplete". Bi-Weekly Eleven (in Burmese). Eleven Media Group. 28 April 2011.
↑ 3.0 3.1 3.2 "United Nations World Urbanization Prospects, 2007 revision". The United Nations Population Division. Archived from the original on 18 December 2008. Retrieved 24 January 2009.
↑ 4.0 4.1 4.2 4.3 4.4 4.5 Zon Pann Pwint, Minh Zaw and Khin Su Wai (18–24 May 2009). "Mandalay marks 150th birthday". The Myanmar Times. Archived from the original on 2 June 2009. Retrieved 5 March 2016.
↑ "Myanmar Area Codes". Archived from the original on 2009-12-01. Retrieved 2016-03-05.
↑ Issac Taylor (1898). Names and Their Histories: A Handbook of Historical Geography and Topographical Nomenclature (2nd ed.). Rivingtons. p. 186.
↑ ဦးဟုတ်စိန်. "Entry for ratana". ပါဠိမြန်မာ အဘိဓာန် (Pāḷi-Myanmar Dictionary) (in Burmese). Pali Canon E-Dictionary Version 1.94. Retrieved 15 February 2015.
↑ ဦးဟုတ်စိန်. "Entry for pūra". ပါဠိမြန်မာ အဘိဓာန် (Pāḷi-Myanmar Dictionary) (in Burmese). Pali Canon E-Dictionary Version 1.94. Retrieved 15 February 2015.
↑ 9.0 9.1 "Mandalay Palace" (PDF). Directorate of Archaeological Survey, Burma. 1963. Retrieved 22 August 2006.
↑ 10.0 10.1 Kyaw Thein (1996). The Management of Secondary Cities in south-east Asia. Case Study: Mandalay. United Nations Centre for Human Settlements. UN-Habitat. ISBN 9789211313130.
↑ Vincent Clarence Scott O'Connor (1907). Mandalay: And Other Cities of the Past in Burma. Hutchinson & Co. pp. 6–9. ISBN 9789748495170.
↑ Herbert Thirkell White (1913). A Civil Servant in Burma. London: E. Arnold.
↑ Ko Yin Aung (23 December 1999). "Prospects of Education in Myanmar". The New Light of Myanmar.
↑ Bird, George W (1897). Wanderings in Burma. London: F J Bright & Son. p. 254.
↑ "The Axis Conquers the Philippines: January 1942-July 1942". HowStuffWorks. 11 September 2007.
↑ "World War 2 Timelines 1939-1945 - Asian Mainland 1942 - Worldwar-2.net". www.worldwar-2.net.
↑ Jackson, Ashley (May 10, 2006). The British Empire and the Second World War. Bloomsbury Academic. p. 41. ISBN 9780826440495.
↑ "23,000 Homeless in Burma Fire". New York Times via Reuters. 26 March 1984. Retrieved 25 January 2009.
↑ "Myanmar Fire Mar 1984 UNDRO Information Reports 1 – 2". ReliefWeb. 27 March 1984. Archived from the original on 29 April 2011. Retrieved 25 January 2009.
↑ "Huge fire rages major market in Myanmar second largest city". China View via Xinhua. 25 February 2008. Retrieved 25 January 2009.
↑ Ne Nwe Moe Aung and Sithu Naing (2 March 2009). "Dry weather brings upsurge in outbreaks of fire in Myanmar". The Myanmar Times. Archived from the original on 16 December 2010. Retrieved 5 March 2016. {{cite journal}}: Cite journal requires |journal= (help)
↑ Min Lwin (April 2009). "The Chinese Road to Mandalay". The Irrawaddy. {{cite journal}}: Cite journal requires |journal= (help)
↑ 23.0 23.1 Stephen Mansfield (13 May 1999). "Myanmar's Chinese connection". Japan Times.
↑ 24.0 24.1 Poon Kim Shee (2002). "The Political Economy of China–Myanmar Relations: Strategic and Economic Dimensions" (PDF). Ritsumeikan Annual Review of International Studies. Ritsumeikan University. Archived from the original (PDF) on 2016-03-04. Retrieved 2016-03-05.
↑ Larkin 2006, p. 12.
↑ List of Ancient Monuments in Burma 1910, p. 4.
↑ Fiala 2002.
↑ China's Ambitions in Myanmar (July 2000). "China's Ambitions in Myanmar". IISS Strategic Comments.
↑ Forbes, Andrew ; Henley, David (2011). Traders of the Golden Triangle. Chiang Mai: Cognoscenti Books. ASIN: B006GMID5
↑ Rieffel, Lex (2010). Myanmar/Burma: inside challenges, outside interests. Brookings Institution Press. pp. 95–97. ISBN 978-0-8157-0505-5.
↑ "Extreme Temperatures Around the World". Retrieved 11 January 2013.
↑ "World Weather Information Service – Mandalay". World Meteorological Organization. Retrieved 23 February 2013.
↑ "Weatherbase: Historical Weather for Mandalay, Myanmar". Weatherbase. Retrieved 23 February 2013.
↑ Cappelen, John; Jensen, Jens. "Myanmar – Mandalay" (PDF). Climate Data for Selected Stations (1931–1960) (in Danish). Danish Meteorological Institute. p. 188. Archived from the original (PDF) on April 27, 2013. Retrieved 23 February 2013.
↑ "Rudyard Kipling's Poem Mandalay". Archived from the original on 2008-12-05. Retrieved 2016-03-05.
BooksEdit
List of Ancient Monuments in Burma (I. Mandalay Division). Vol. 1. Rangoon: Office of the Superintendent, Govt. Printing, Burma. 1910.
Fiala, Robert D. (2002). "Shwenandaw Kyaung Temple, Mandalay, Myanmar". Asian Historical Architecture. Retrieved 11 November 2013.
Larkin, Emma (6 March 2006). Finding George Orwell in Burma. Penguin Publishing Group. ISBN 978-1-101-11812-2.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Mandalay&oldid=8337042"


text:
\section{Introduction}

Let $G=(V,E)$ be a simple connected graph, where $V = \{v_1, v_2, \ldots, v_n\}$.
The adjacency matrix of $G$ is a 0-1 matrix $A(G)=(a_{ij})_{n \times n}$ with $a_{ij}=1$ iff $v_i v_j \in E$.
Let $D(G)=diag(d_1, d_2, \ldots, d_n)$ be the degree matrix of $G$, where $d_i$ is the degree of vertex $v_i$.
Then $L(G)=D(G)-A(G)$ is termed as the Laplacian matrix,
and $\mathcal{L}(G) =D(G)^{-\frac{1}{2}} L(G) D(G)^{-\frac{1}{2}}$
the normalized Laplacian matrix of graph $G$.
Let $0= \mu_1 <\mu_2 \leq \cdots \leq \mu_n$ be the eigenvalues of $L(G)$,
and $0= \nu_1 <\nu_2 \leq \cdots \leq \nu_n$ the eigenvalues of $\mathcal{L}(G)$.
The sets $Sp(L(G)) = \{\mu_1, \mu_2, \ldots, \mu_n\}$
and $Sp(\mathcal{L}(G)) = \{\nu_1, \nu_2, \ldots, \nu_n\}$ are called
the Laplacian spectrum and normalized Laplacian spectrum of $G$, respectively.

Let $d_{ij}$ denotes the distance between vertices $v_i$ and $v_j$ in $G$, i.e.,
the length of a shortest path connecting them.
The Wiener index \cite{b1} and Gutman index \cite{b2},
defined as $W(G) =\sum_{i<j} d_{ij}$ and $Gut(G) =\sum_{i<j} d_i d_j d_{ij}$,
are two famous distance-based topological indices,
which have successful applications in chemistry and communication networks.

By regarding graph $G$ as an electronic network, in which a unit resistor is placed on each edge of $G$,
Klein and Randi\'{c} \cite{b3} proposed the concept of resistance distance.
The resistance distance $r_{ij}$ between vertex $v_i$ and $v_j$ is defined
as the effective resistance between them.
Similar to the Wiener index and Gutman index, the Kirchhoff index \cite{b3} and
degree-Kirchhoff index \cite{b4} of $G$ are defined as
$Kf(G)=\sum_{i<j} r_{ij}$ and $Kf^*(G)= \sum_{i <j} d_i d_j r_{ij}$, respectively.
Furthermore, $Kf(G)$ (resp. $Kf^*(G)$) was shown to have close relationship
with $Sp(L(G))$ (resp. $Sp(\mathcal{L}(G))$).

\noindent \textbf{Lemma 1.1} \cite{b5,b6}\textbf{.} Let $G$ be a simple graph of order $n \geq 2$.
Then
$$Kf(G) = n \sum_{i=2}^n \frac{1}{\mu_i}.$$

\noindent \textbf{Lemma 1.2} \cite{b4}\textbf{.}
Let $G$ be a simple connected graph of order $n \geq 2$ and size $m$. Then
$$Kf^*(G) = 2m \sum_{i=2}^n \frac{1}{\nu_i}.$$

Due to their practical applications in complex networks, physics, and chemistry, the Kirchhoff index and degree-Kirchhoff index have attracted a lot of attentions.
Up to now, the closed-form formulas of (degree-)Kirchhoff index have been established
for cycles \cite{cycle}, circulant graphs \cite{circulant}, regular graphs \cite{regular}, composite graphs \cite{composite}, complete multipartite graphs \cite{complete-mult}, flower graphs \cite{flower},
the strong prism of a star \cite{strong}, the Cartesian product of $S_n$ and $K_2$ \cite{Cartesian},
zigzag polyhex nanotube TUHC$[2n,2]$ \cite{nano}, linear phenylenes \cite{LinearPheny,LinearPheny2},
cyclic phenylenes \cite{CyclicPheny}, M\"{o}bius phenylenes chain and cylinder phenylenes chain \cite{MobiusPheny,CylinderPheny}, linear [n] phenylenes \cite{Linear-n-Pheny},
generalized phenylenes \cite{GeneralPheny,GeneralPheny2}, periodic linear chains \cite{Periodic},
linear polyomino chain \cite{Linear4}, linear crossed polyomino chain \cite{LinearCrossed4},
linear pentagonal chain \cite{Linear5}, linear hexagonal chain \cite{Linear6,Linear6-2},
linear crossed hexagonal chain \cite{LinearCrossed6},
M\"{o}bius hexagonal chain \cite{Mobius6}, linear octagonal chain \cite{Linear8},
linear crossed octagonal chain \cite{LinearCrossed8}, M\"{o}bius and cylinder octagonal chains \cite{Mobius8}, linear octagonal-quadrilateral chain \cite{Linear8-4},
and M\"{o}bius octagonal-quadrilateral networks \cite{Mobius8-4}.

Let $L_{n}$ be the linear octagonal-quadrilateral network with $n$ quadrilaterals and $n$ octagons, as shown in Figure 1. The crossed octagonal-quadrilateral network $Q_{n}$ (shown in Figure 2) is obtained from $L_n$ by connecting vertex $i$ to vertex $(i+1)'$ and $i'$ to vertex $i+1$, $1 \leq i \leq 4n$.
Inspired by the above works, in this paper we consider the resistance distance-based invariants of linear crossed octagonal-quadrilateral networks.
Explicit closed-form formulas of $Kf(Q_n)$, $Kf^*(Q_n)$, and $\tau(Q_n)$ are obtained, where $\tau(G)$ denotes the number of spanning trees of graph $G$.
Moreover, it is shown that $\lim_{n\rightarrow\infty} Kf(Q_{n})/W(Q_{n})= \lim_{n\rightarrow\infty} Kf^*(Q_{n})/Gut(Q_n) =1/4$.

\begin{figure}[htbp]
\centering\includegraphics[width=15cm,height=4cm]{1.jpg}
\caption{The linear octagonal-quadrilateral network $L_n$.}
\end{figure}

\begin{figure}[htbp]
\centering\includegraphics[width=15cm,height=4cm]{2.jpg}
\caption{The linear crossed octagonal-quadrilateral network $Q_n$.}
\end{figure}

\section{Preliminary}
It is easily seen that $|V(Q_{n})|=8n+2,|E(Q_{n})|=18n+1$, and $\pi=(1,1')(2,2')\cdots\big(4n+1,(4n+1)'\big)$ is an automorphism of $Q_{n}$. Let $ V_{1}=\{1,2,\ldots,4n+1\}$ and $V_2=\{1',2',\ldots,(4n+1)'\}$.
Then we have
\begin{equation*}
  L(G)=\begin{pmatrix}
    L_{V_1V_1} &L_{V_1V_2}\\
    L_{V_2V_1} &L_{V_2V_2}
  \end{pmatrix} ~\text{and}~
  \mathcal{L}(G)=\begin{pmatrix}
    \mathcal L_{V_1V_1} &\mathcal L_{V_1V_2}\\
    \mathcal L_{V_2V_1} &\mathcal L_{V_2V_2}
  \end{pmatrix},
  \end{equation*}
\noindent where $L_{V_i V_j}$ (resp. $\mathcal L_{V_i V_j}$) is the matrix formed by the rows and columns of $L(G)$ (resp. $\mathcal{L}(G)$) corresponding to the vertices in $V_i \bigcup V_j$, $1 \leq i, j \leq 2$.
By symmetry, it holds that $L_{V_1 V_2}=L_{V_2 V_1}$, $L_{V_1 V_1}=L_{V_2 V_2}$, $\mathcal L_{V_1 V_2}=\mathcal L_{V_2 V_1}$, and $\mathcal L_{V_1 V_1}=\mathcal L_{V_2 V_2}$.

Let
\begin{equation*}
  T=\begin{pmatrix}
    \frac{1}{\sqrt{2}}I_{4n+1} &\frac{1}{\sqrt{2}}I_{4n+1} \\
  \frac{1}{\sqrt{2}}I_{4n+1} &-\frac{1}{\sqrt{2}}I_{4n+1}
  \end{pmatrix}.
\end{equation*}

\noindent Then we have

\begin{equation*}
  TL(G)T'=\begin{pmatrix}
    L_A &0\\
    0 &L_S\\
  \end{pmatrix} ~\text{and}~
   T\mathcal{L}(G)T'=\begin{pmatrix}
    \mathcal L_{A} &0\\
    0 &\mathcal L_{S}\\
  \end{pmatrix},
\end{equation*}
\noindent where $L_A= L_{V_{11}}+L_{V_{22}}$, $L_S=L_{V_1V_1}-L_{V_1V_2}$,
$\mathcal L_{A}= \mathcal L_{V_{11}}+\mathcal L_{V_{22}}$, and $\mathcal L_{S}=\mathcal L_{V_1V_1}-\mathcal L_{V_1V_2}$.

Based on the above arguments, by applying the decomposition theorem obtained in \cite{Decom},
we immediately have the following result, where $\Phi_B(x) = det(xI-B)$ refers
the characteristic polynomial of a square matrix $B$.

\begin{lem}
Let $L_{A},L_{S},\mathcal L_{A}, and \mathcal L_{S}$ be defined as above. Then
\begin{eqnarray*}
\Phi_{L(Q_{n})}(x)=\Phi_{L_{A}}(x) \Phi_{L_{S}}(x) ~\text{and}~
\Phi_{\mathcal{L}(Q_{n})}(x)=\Phi_{\mathcal L_{A}}(x)\Phi_{\mathcal L_{S}}(x).
\end{eqnarray*}
\end{lem}

\begin{lem}\textup{\cite{TreeNum}}
Assume that $G$ is a connected graph with $n \geq 2$ vertices,
then $$\tau(G) = \frac{1}{n}\prod_{i=2}^{n} \mu_i.$$
\end{lem}

\section{The Kirchhoff index and Wiener index of $Q_{n}$}\label{sct3}
It is easily seen that
\begin{eqnarray*}
 L_{V_1 V_1}&=&
\left(
  \begin{array}{ccccccccccc}
    3 & -1 & & & & & & & & &\\
    -1 & 4 & -1 & & & & & & & &\\
    & -1 & 4 & -1 & & & & & & &\\
    & & -1 & 5 & -1 & & & & & &\\
    & & & -1 & 5 & -1 & & & & &\\
    & & & & -1 & 4 & -1 & & & &\\
    & & & & & & \ddots & & & &\\
    & & & & & & -1 & 4 & -1 & &\\
    & & & & & & & -1 & 4 & -1 &\\
    & & & & & & & & -1 & 5 & -1\\
    & & & & & & & & & -1 & 3\\
  \end{array}
\right)_{(4n+1)\times(4n+1)}
\end{eqnarray*}
and
\begin{eqnarray*}
 L_{V_1 V_2}&=&
\left(
  \begin{array}{ccccccccccc}
    -1 & -1 & & & & & & & & &\\
    -1 & 0 & -1 & & & & & & & &\\
    & -1 & 0 & -1 & & & & & & &\\
    & & -1 & -1 & -1 & & & & & &\\
    & & & -1 & -1 & -1 & & & & &\\
    & & & & -1 & 0 & -1 & & & &\\
    & & & & & & \ddots & & & &\\
    & & & & & & -1 & 0 & -1 & &\\
    & & & & & & & -1 & 0 & -1 &\\
    & & & & & & & & -1 & -1 & -1\\
    & & & & & & & & & -1 & -1\\
  \end{array}
\right)_{(4n+1)\times(4n+1)}.
\end{eqnarray*}

\noindent Thus,
\begin{eqnarray*}
 L_{A}&=&
\left(
  \begin{array}{ccccccccccc}
    2 & -2 & & & & & & & & &\\
    -2 & 4 & -2 & & & & & & & &\\
    & -2 & 4 & -2 & & & & & & &\\
    & & -2 & 4 & -2 & & & & & &\\
    & & & -2 & 4 & -2 & & & & &\\
    & & & & -2 & 4 & -2 & & & &\\
    & & & & & & \ddots & & & &\\
    & & & & & & -2 & 4 & -2 & &\\
    & & & & & & & -2 & 4 & -2 &\\
    & & & & & & & & -2 & 4 & -2\\
    & & & & & & & & & -2 & 2\\
  \end{array}
\right)_{(4n+1)\times(4n+1)}
\end{eqnarray*}
and $L_S=diag(4,4,4,6,6,4,\ldots,4,4,6,4)$.

Suppose $Sp(L_{A}) = \{0=\rho_{1}, \rho_{2}, \ldots, \rho_{4n+1}\}$
and $Sp(L_S) = \{\sigma_{1}, \sigma_{2}, \ldots, \sigma_{4n+1}\}$. From Lemma 1.1 we have
\begin{equation}
  Kf(Q_{n})=(8n+2)\Big(\sum_{i=2}^{4n+1}\frac{1}{\rho_{i}}+\sum_{i=1}^{4n+1}\frac{1}{\sigma_{i}}\Big).
\end{equation}

Obviously,
\begin{equation}
\sum_{i=1}^{4n+1}\frac{1}{\sigma_{i}}=\frac{1}{4}\times(2n+2)+\frac{1}{6}\times(2n-1)=\frac{5n+2}{6}.
\end{equation}

To get the value of $\sum_{i=2}^{4n+1}\frac{1}{\rho_{i}}$, suppose $\Phi_{L_{A}}(x)=det(xI-L_{A})=x^{4n+1}+a_{1}x^{4n}+\cdots+a_{4n}x$.
Then $1/\rho_2, 1/\rho_3, \ldots, 1/\rho_{4n+1}$ are the roots of the polynomial $a_{4n}x^{4n}+a_{4n-1}x^{4n-1}+\cdots+a_{1}x+1$. From the Vieta theorem we have
\begin{equation}
\sum_{i=2}^{4n+1}\frac{1}{\rho_{i}}=\frac{(-1)^{4n-1}a_{4n-1}}{(-1)^{4n}a_{4n}}.
\end{equation}

Let $W_{i}$ (resp. $U_{i}$) be the $i$-th order principal submatrix of $L_{A}$ formed by its first (resp. last) $i$ rows and $i$ columns, $w_i = det W_i$, and $u_i = det U_i$, $i=1,2,\ldots,4n$.
By direct calculations we have $w_1=2$, $w_2=4$, and $w_i=4w_{i-1}-4w_{i-2}$ if $3\leq i\leq4n$. By solving the recurrence equation we get $w_{i}=2^{i}$, $i=1,2,\ldots,4n$. Furthermore, from the structure of $L_A$, obviously it holds that $u_i = w_i$. Also, let\\
\begin{equation*}
  N_i = \begin{pmatrix}
     4 &-2 & & & & & & & & \\
     -2& 4 & -2 & & & & & & &\\
     &-2 & 4 &-2 & & & & & & \\
     & & -2 &4 & & & & & & \\
     & & & & & \ddots& & & &\\
     & & & & & &  4& -2 & &\\
     & & & & & &  -2& 4 & -2&\\
     & & & & & & &-2 & 4 &-2 \\
     & & & & & & & & -2 & 4
  \end{pmatrix}_{i \times i}
\end{equation*}
\noindent and $n_i=detN_i$, $i = 1,2,\ldots,4n-1$. By direct calculations, we have $n_i=(i+1)2^i$.
It is both consistent and convenient to define $w_0 = u_0 =n_0 =1$.

\begin{lem} $\sum\nolimits_{i=2}^{4n+1}\frac{1}{\rho_{i}}= n(4n+2)/3$.
\end{lem}

\noindent{\bf Proof.} Let $L_A[i]$ denote the principal submatrix of $L_A[i]$ by deleting its $i$-th row and column. Then
$$(-1)^{4n}a_{4n} = \sum_{i=1}^{4n+1}det L_A[i]= \sum_{i=1}^{4n+1}det \begin{pmatrix} W_{i-1} &0\\ 0 & U_{4n+1-i}\end{pmatrix} =\sum_{i=1}^{4n+1}w_{i-1} w_{4n+1-i}=(4n+1)2^{4n}.$$

On the other hand, let $L_A[i,j]$ denote the submatrix of $L_A[i]$ by deleting its $i$-th and $j$-th rows and columns. Then
\begin{align*}
(-1)^{4n-1}a_{4n-1} &=\sum_{1\leq{i}<j\leq4n+1}det L_A[i,j] \\
   &=\sum_{1\leq{i}<j\leq 4n+1}det\begin{pmatrix} W_{i-1}& & \\& N_{j-i-1} &\\& &U_{4n+1-j}\end{pmatrix}\\
   &=\sum_{1\leq{i}<j\leq 4n+1} n_{j-i-1} w_{i-1} u_{4n+1-j}\\
   &=\sum_{1\leq{i}<j\leq 4n+1} (j-i)2^{j-i-1} 2^{i-1} 2^{4n+1-j}\\
   &=\frac{4n(4n+1)(4n+2)2^{4n-2} }{3},
\end{align*}
\noindent and the result follows from Equation (3.3). \hfill\rule{1ex}{1ex}

Now, from Equations (3.1) -- (3.2) and Lemma 3.1, we immediately have the following result.

\begin{thm} $Kf(Q_{n}) = (32n^{3}+44n^{2}+17n+2) /3$.
\end{thm}

The values of $Kf(Q_{n})$ for $1 \leq n \leq 15$ are listed in Table 1.
\begin{table}[htbp]
\setlength{\abovecaptionskip}{0.05cm} \centering\vspace{.3cm}
\caption{The values of $Kf(Q_{n})$, $1 \leq n \leq 15$.}
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
  \hline

  $n$ & $Kf(Q_{n})$ & $n$ & $Kf(Q_{n})$ & $n$ & $Kf(Q_{n})$ & $n$ & $Kf(Q_{n})$ & $n$ & $Kf(Q_{n})$ \\
  \hline
  $1$ & $31.67$ & $4$ & $940.67$ & $7$ & $4417.67$ & $10$ & $12190.70$ & $13$ & $25987.70$ \\
  $2$ & $156.00$ & $5$ & $1729.00$ & $8$ & $6446.00$ & $11$ & $16035.00$ & $14$ & $32224.00$\\
  $3$ & $437.67$ & $6$ & $2866.67$ & $9$ & $9015.67$ & $12$ & $20612.70$ & $15$ & $39385.70$\\
  \hline
\end{tabular}
\end{table}

\begin{thm} $\lim\nolimits_{n\rightarrow\infty} Kf(Q_{n})/W(Q_{n})=1/4$.
\end{thm}

\noindent{\bf Proof.} For a subset $U \subseteq V$, let $W(U) = \Sigma_{i \in U}\Sigma_{j \in V} d_{ij}$, i.e., the sum of the sum of the distances between each vertex in $U$ and other vertices in $Q_n$.

(1) Let $U_1=\{1,1',4n+1,(4n+1)'\}$, then $W(U_1) =4(16n^{2}+4n+1)$;

(2) Let $U_2 =\{4i-2,(4i-2)'|i = 1,2,\ldots,n\}$, then
$$W(U_2)=2\sum_{i=1}^{n}(1+\sum_{j=1}^{4i-3}j+\sum_{j=1}^{4n-4i+3}j)=\frac{4}{3}(8n^{3}+3n^{2}+n);$$

(3) Let $U_3 = \{4i-1,(4i-1)'|i = 1,2,\ldots,n\}$, then
$$W(U_3) =2\sum_{i=1}^{n}(1+\sum_{j=1}^{4i-2}j+\sum_{j=1}^{4n-4i+2}j)=\frac{2}{3}(16n^{3}+6n^{2}-n);$$

(4) Let $U_4 = \{4i,(4i)'|i = 1,2,\ldots,n\}$, then
$$W(U_4) =2\sum_{i=1}^{n}(1+\sum_{j=1}^{4i-1}j+\sum_{j=1}^{4n-4i+1}j)=\frac{4}{3}(8n^{3}+3n^{2}+n);$$

(5) Let $U_5 = \{4i+1,(4i+1)'|i = 1,2,\ldots,n-1\}$, then
$$W(U_5) =2\sum_{i=1}^{n-1}(1+\sum_{j=1}^{4i}j+\sum_{j=1}^{4n-4i}j)=\frac{2}{3}(16n^{3}-18n^{2}+5n-3).$$

\noindent Hence $W(Q_{n}) =(W(U_1)+W(U_2)+W(U_3)+W(U_4)+W(U_5))/2=(128n^{3}+96n^{2}+40n)/3$, and from Theorem 3.2 we immediately have
$\lim_{n\rightarrow\infty} Kf(Q_{n})/W(Q_{n})=1/4$. \hfill\rule{1ex}{1ex}

\begin{thm} $\tau(Q_n)=2^{10n+2}\cdot3^{2n-1}$.
\end{thm}

\noindent{\bf Proof.} From Lemma 2.2 and the proof of Lemma 3.1, we have
\begin{align*}
\tau(Q_{n}) &=\frac{1}{8n+2} \cdot \prod_{i=2}^{4n+1}\rho_i \cdot \prod_{i=1}^{4n+1}\sigma_i\\
            &=\frac{1}{8n+2} \cdot (-1)^{4n}a^{4n} \cdot 4^{2n+2} \cdot 6^{2n-1} \\\
            &=2^{10n+2}\cdot3^{2n-1}.
\end{align*}

The proof is completed. \hfill\rule{1ex}{1ex}

The values of $\tau(Q_{n})$, $1 \leq n \leq 8$, are shown in Table 3.
\begin{table}[htbp]
\setlength{\abovecaptionskip}{0.05cm}
 \centering \vspace{.3cm}
\caption{The number of spanning trees of $Q_n$, $1 \leq n \leq 8$.}
\begin{tabular}{c|c|c|c}
  \hline

  $n$ & $\tau(Q_{n})$ & $n$ & $\tau(Q_{n})$ \\
  \hline
  $1$ & $12288$ & $5$ & $88644351465533472768$ \\
  $2$ & $113246208$ & $6$ & $816946343106356485029888$ \\
  $3$ & $1043677052928$ & $7$ & $7528977498068181366035447808$ \\
  $4$ & $9618527719784448$ & $8$ & $69387056622196359469382686998528$ \\
  \hline
\end{tabular}
\end{table}

\section{Degree-Kirchhoff index and the complexity of $Q_n$}

By the same method used in Section 3, we will determine the degree-Kirchhoff index of $Q_n$ in this section.
It is easily seen that
\begin{eqnarray*}
\mathcal{L}_{V_1 V_1}&=&
\left(
  \begin{array}{ccccccccccc}
    1 & \frac{-1}{\sqrt{12}} & & & & & & & & &\\
    \frac{-1}{\sqrt{12}} & 1 & \frac{-1}{4} & & & & & & & &\\
    & \frac{-1}{4} & 1 & \frac{-1}{\sqrt{20}} & & & & & & &\\
    & & \frac{-1}{\sqrt{20}} & 1 & \frac{-1}{5} & & & & & &\\
    & & & \frac{-1}{5} & 1 & \frac{-1}{\sqrt{20}} & & & & &\\
    & & & & \frac{-1}{\sqrt{20}} & 1 & \frac{-1}{4} & & & &\\
    & & & & & & \ddots & & & &\\
    & & & & & & \frac{-1}{\sqrt{20}} & 1 & \frac{-1}{4} & &\\
    & & & & & & & \frac{-1}{4} & 1 & \frac{-1}{\sqrt{20}} &\\
    & & & & & & & & \frac{-1}{\sqrt{20}} & 1 & \frac{-1}{\sqrt{15}}\\
    & & & & & & & & & \frac{-1}{\sqrt{15}} & 1\\
  \end{array}
\right)_{(4n+1)\times(4n+1)}
\end{eqnarray*}

\noindent and

\begin{eqnarray*}
\mathcal{L}_{V_1 V_2}&=&
\left(
  \begin{array}{ccccccccccc}
    \frac{-1}{3} & \frac{-1}{\sqrt{12}} & & & & & & & & &\\
    \frac{-1}{\sqrt{12}} & 0 & \frac{-1}{4} & & & & & & & &\\
    & \frac{-1}{4} & 0 & \frac{-1}{\sqrt{20}} & & & & & & &\\
    & & \frac{-1}{\sqrt{20}} & \frac{-1}{5} & \frac{-1}{5} & & & & & &\\
    & & & \frac{-1}{5} & \frac{-1}{5} & \frac{-1}{\sqrt{20}} & & & & &\\
    & & & & \frac{-1}{\sqrt{20}} & 0 & \frac{-1}{4} & & & &\\
    & & & & & & \ddots & & & &\\
    & & & & & & \frac{-1}{\sqrt{20}} & 0 & \frac{-1}{4} & &\\
    & & & & & & & \frac{-1}{4} & 0 & \frac{-1}{\sqrt{20}} &\\
    & & & & & & & & \frac{-1}{\sqrt{20}} & \frac{-1}{5} & \frac{-1}{\sqrt{15}}\\
    & & & & & & & & & \frac{-1}{\sqrt{15}} & \frac{-1}{3}\\
  \end{array}
\right)_{(4n+1)\times(4n+1)}.
\end{eqnarray*}
Thus
\begin{eqnarray*}
\mathcal L_A&=&
\left(
  \begin{array}{ccccccccccc}
    \frac{2}{3} & \frac{-1}{\sqrt{3}} & & & & & & & & &\\
    \frac{-1}{\sqrt{3}} & 1 & \frac{-1}{2} & & & & & & & &\\
    & \frac{-1}{2} & 1 & \frac{-1}{\sqrt{5}} & & & & & & &\\
    & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{5} & & & & & &\\
    & & & \frac{-2}{5} & \frac{4}{5} & \frac{-1}{\sqrt{5}} & & & & &\\
    & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{2} & & & &\\
    & & & & & & \ddots & & & &\\
    & & & & & & \frac{-1}{\sqrt{5}} & 1 & \frac{-1}{2} & &\\
    & & & & & & & \frac{-1}{2} & 1 & \frac{-1}{\sqrt{5}} &\\
    & & & & & & & & \frac{-1}{\sqrt{5}} & \frac{4}{5} & \frac{-2}{\sqrt{15}}\\
    & & & & & & & & & \frac{-2}{\sqrt{15}} & \frac{2}{3}\\
  \end{array}
\right)_{(4n+1)\times(4n+1)}
\end{eqnarray*}

\noindent and $\mathcal L_S = diag(\frac{4}{3},1,1,\frac{6}{5},\frac{6}{5},1,\ldots,1,1,\frac{6}{5},\frac{4}{3})$.

Suppose $Sp(\mathcal L_A) = \{0=\lambda_{1}, \lambda_{2},\ldots,\lambda_{4n+1}\}$ and
$Sp(\mathcal L_S) = \{\varphi_1, \varphi_2, \ldots, \varphi_{4n+1}\}$.
Then from Lemma 1.2 we have

\begin{equation}
Kf^{*}(Q_{n})=2(18n+1)\Big(\sum_{i=2}^{4n+1}\frac{1}{\lambda_{i}}+\sum_{i=1}^{4n+1}\frac{1}{\varphi_{i}}\Big).
\end{equation}

Obviously, \begin{equation}
\sum_{i=1}^{4n+1}\frac{1}{\varphi_{i}}=1\times{2n}+\frac{5}{6}\times(2n-1)+\frac{3}{4}\times2=\frac{11n+2}{3}.
\end{equation}

To get the value of $\sum_{i=2}^{4n+1}\frac{1}{\lambda_{i}}$,
suppose $\Phi_{\mathcal L_A}(x) =det(xI-\mathcal L_A) =x^{4n+1}+b_{1}x^{3n}+\cdots+b_{4n}x$.
Similarly to Equation (3.3),
we have
\begin{equation}
\sum_{i=2}^{4n+1}\frac{1}{\lambda_{i}}=\frac{(-1)^{4n-1}b_{4n-1}}{(-1)^{4n}b_{4n}}.
\end{equation}

Let $X_{i}$ be the $i$-th order principal submatrix of $\mathcal L_{A}$ formed by its first $i$ rows and $i$ columns, and $x_i = det X_i$, $i=1,2,\ldots,4n$.
By direct calculations we have $x_{1}=\frac{2}{3}$, $x_{2}=\frac{1}{3}$, $x_{3}=\frac{1}{6}$, $x_{4}=\frac{1}{15}$, $x_{5}=\frac{2}{75}$, $x_{6}=\frac{1}{75}$, $x_{4n}=\frac{5}{3} \cdot (\frac{1}{25})^n$,
and for $1\leq k \leq n-1$, it holds that
\begin{eqnarray*}
\begin{cases}
x_{4k}=\frac{4}{5}x_{4k-1}-\frac{1}{5}x_{4k-2}\\
x_{4k+1}=\frac{4}{5}x_{4k}-\frac{4}{25}x_{4k-1}\\
x_{4k+2}=x_{4k+1}-\frac{1}{5}x_{4k}\\
x_{4k+3}=x_{4k+2}-\frac{1}{4}x_{4k+1}
\end{cases}.
\end{eqnarray*}
By solving the above recursive equations we have
\begin{eqnarray*}
\begin{cases}
x_{4k}=\frac{5}{3}\cdot(\frac{1}{25})^{k}\\
x_{4k+1}=\frac{2}{3}\cdot(\frac{1}{25})^{k}\\
x_{4k+2}=\frac{1}{3}\cdot(\frac{1}{25})^{k}\\
x_{4k+3}=\frac{1}{6}\cdot(\frac{1}{25})^{k}\\
\end{cases},~1\leq k \leq n-1.
\end{eqnarray*}

Similarly, let $Y_{i}$ be the $i$-th order principal submatrix of $\mathcal L_{A}$ formed by its last $i$ rows and $i$ columns, and $y_i = det Y_i$, $i=1,2,\ldots,4n$. Then $y_{1}=\frac{2}{3}$, $y_{2}=\frac{4}{15}$, $y_{3}=\frac{2}{15}$, $y_{4}=\frac{1}{15}$, $y_{5}=\frac{2}{75}$, $y_{6}=\frac{4}{375}$, $y_{4n}=\frac{5}{3}\cdot(\frac{1}{25})^{n}$, and for $1\leq k \leq n-1$, it holds that
\begin{eqnarray*}
\begin{cases}
y_{4k}=y_{4k-1}-\frac{1}{4}y_{4k-2}\\
y_{4k+1}=\frac{4}{5}y_{4k}-\frac{1}{5}y_{4k-1}\\
y_{4k+2}=\frac{4}{5}y_{4k+1}-\frac{4}{25}y_{4k}\\
y_{4k+3}=y_{4k+2}-\frac{1}{5}y_{4k+1}
\end{cases},
\end{eqnarray*}
and so
\begin{eqnarray*}
\begin{cases}
y_{4k}=\frac{5}{3}\cdot(\frac{1}{25})^{k}\\
y_{4k+1}=\frac{2}{3}\cdot(\frac{1}{25})^{k}\\
y_{4k+2}=\frac{4}{15}\cdot(\frac{1}{25})^{k}\\
y_{4k+3}=\frac{2}{15}\cdot(\frac{1}{25})^{k}
\end{cases}.
\end{eqnarray*}

Define $x_0=y_0=1$.

\begin{lem} $\sum_{i=2}^{4n+1}\frac{1}{\lambda_i}= \frac{2n(54n^{2}+9n+4)}{18n+1}$.
\end{lem}

\noindent{\bf Proof.} Let $\mathcal L_A[i]$ denote the principal submatrix of $\mathcal L_A[i]$ by deleting its $i$-th row and column. Then
\begin{align*}
(-1)^{4n}b_{4n}&=\sum_{j=1}^{4n+1}det\mathcal L_A[j] =x_{4n}+y_{4n}+\sum_{j=2}^{4n}det\mathcal L_A[j]\\
&=x_{4n}+y_{4n}+\sum_{k=1}^{n}x_{4(k-1)+3}y_{4(n-k)+1}+\sum_{k=1}^{n-1}x_{4k}y_{4(n-k)}\\
&~~+\sum_{k=0}^{n-1}x_{4k+1}y_{4(n-k-1)+3}+\sum_{k=0}^{n-1}x_{4k+2}y_{4(n-k-1)+2}\\
&=\frac{10}{3}\cdot(\frac{1}{25})^{n}+\sum_{k=1}^{n}\frac{1}{6}\cdot(\frac{1}{25})^{k-1}\cdot\frac{2}{3}\cdot(\frac{1}{25})^{n-k}
+\sum_{k=1}^{n-1}\frac{5}{3}\cdot(\frac{1}{25})^{k}\cdot\frac{5}{3}\cdot(\frac{1}{25})^{n-k}\\
&~~+\sum_{k=0}^{n-1}\frac{2}{3}\cdot\left(\frac{1}{25}\right)^{k}\cdot\frac{2}{15}\cdot(\frac{1}{25})^{n-k-1}+\sum_{k=0}^{n-1}\frac{1}{3}\cdot(\frac{1}{25})^{k}\cdot\frac{4}{15}\cdot(\frac{1}{25})^{n-k-1}\\
&=\frac{18n+1}{45}(\frac{1}{25})^{n-1}.
\end{align*}

From Equation (4.3), it suffices to show
$(-1)^{4n-1}b_{4n-1}=\frac{2(54n^{3}+9n^{2}+4n)}{45}(\frac{1}{25})^{n-1}
$.

Let $\mathcal L_A[i,j]$ denote the submatrix of $\mathcal L_A$ by deleting its $i$-th and $j$-th rows and columns, $1 \leq i < j \leq 4n+1$. Then
\begin{align}\label{sum}
(-1)^{4n-1}b_{4n-1} &=\sum_{1\leq i < j \leq 4n+1}det \mathcal L_A[i,j]
   =\sum_{1\leq i<j \leq 4n+1}det\begin{pmatrix} X_{i-1}& & \\& Z_{i,j} &\\& &Y_{4n+1-j}\end{pmatrix}\\ \nonumber
   &= \sum_{1\leq i < j \leq 4n+1}x_{i-1} y_{4n+1-j} z_{i,j},
\end{align}
\noindent where $Z_{i,j}$ is the $(j-i-1)$-th order principal submatrix of $\mathcal L_A$ formed by its $i+1, i+2, \ldots, j-1$ rows and columns. Let $z_{i,j} = detZ_{i,j}$,
and defined $z_{i,j}=1$ if $j=i+1$.
It is an easy but tedious task to calculate the values of $z_{i,j}$ for different $i$'s and $j$'s.
In fact, in the Appendix, we calculate the values of $z_{i,j}$ in 16 cases, and have the following result.

\begin{equation}\label{z_ij}
z_{i,j}=
  \begin{cases}
\frac{2}{5}(b-a)(\frac{1}{25})^{b-a-1},&i=4a, j=4b, 1\leq a < b \leq n;\\
(4b-4a+1)(\frac{1}{25})^{b-a}, &i=4a, j=4b+1, 1\leq{a}\leq b\leq{n};\\
\frac{2}{5}(4b-4a+2)(\frac{1}{25})^{b-a}, &i=4a, j=4b+2, 1\leq{a}\leq b\leq{n-1};\\
\frac{1}{5}(4b-4a+3)(\frac{1}{25})^{b-a}, &i=4a, j=4b+3, 1\leq{a}\leq b\leq{n-1};\\
\frac{1}{4}(4b-4a-1)(\frac{1}{25})^{b-a-1}, &i=4a+1, j=4b, 0\leq{a}< b\leq{n};\\
\frac{2}{5}(b-a)(\frac{1}{25})^{b-a-1}, &i=4a+1, j=4b+1, 0\leq{a}< b\leq{n};\\
(4b-4a+1)(\frac{1}{25})^{b-a}, &i=4a+1, j=4b+2, 0\leq{a}\leq b\leq{n-1};\\
(2b-2a+1)(\frac{1}{25})^{b-a}, &i=4a+1, j=4b+3, 0\leq{a}\leq b\leq{n-1};\\
(2b-2a-1)(\frac{1}{25})^{b-a-1}, &i=4a+2, j=4b, 0\leq{a}< b\leq{n};\\
\frac{1}{5}(4b-4a-1)(\frac{1}{25})^{b-a-1}, &i=4a+2, j=4b+1, 0\leq{a}< b\leq{n};\\
\frac{2}{25}(4b-4a)(\frac{1}{25})^{b-a-1}, &i=4a+2,j=4b+2, 0\leq{a}< b\leq{n-1};\\
(4b-4a+1)(\frac{1}{25})^{b-a}, &i=4a+2, j=4b+3, 0\leq{a}\leq b\leq{n-1};\\
(4b-4a-3)(\frac{1}{25})^{b-a-1}, &i=4a+3, j=4b, 0\leq{a}<b\leq{n};\\
\frac{2}{5}(4b-4a-2)(\frac{1}{25})^{b-a-1}, &i=4a+3, j=4b+1, 0\leq{a}<b\leq{n};\\
\frac{4}{25}(4b-4a-1)(\frac{1}{25})^{b-a-1}, &i=4a+3, j=4b+2, 0\leq{a}<b\leq{n-1};\\
\frac{2}{25}(4b-4a)(\frac{1}{25})^{b-a-1}, &i=4a+3, j=4b+3, 0\leq{a}<b\leq{n-1}.
\end{cases}
\end{equation}

Accordingly, in the Appendix, we obtain the values of $K_{i,j}$, $0\leq i,j\leq 3$, as follows.
\begin{equation}\label{K_ij}
  \begin{cases}
K_{0,0}=\sum_{1\leq{a}<b\leq{n}} det\mathcal L_A[4a,4b]=
  \frac{5(n^3-n)}{27}\left(\frac{1}{25}\right)^{n-1};\\
K_{0,1}=\sum_{1\leq{a}\leq b\leq n} det\mathcal L_A[4a,4b+1]
  =\frac{20n^3-9n^2+7n}{108}\left(\frac{1}{25}\right)^{n-1};\\
K_{0,2}=\sum_{1\leq{a}\leq b\leq{n-1}} det\mathcal L_A[4a,4b+2]
  =\frac{2(2n^3-3n^{2}+n)}{27}\left(\frac{1}{25}\right)^{n-1};\\
K_{0,3}=\sum_{1\leq{a}\leq b\leq{n-1}} det\mathcal L_A[4a,4b+3]
  =\frac{4n^3-3n^{2}-n}{27}\left(\frac{1}{25}\right)^{n-1};\\
K_{1,0}=\sum_{0\leq{a}<b\leq{n}} det\mathcal L_A[4a+1,4b]
  =\frac{20n^3+21n^2+13n}{108} \left(\frac{1}{25}\right)^{n-1};\\
K_{1,1}=\sum_{0\leq{a}<b\leq{n}} det\mathcal L_A[4a+1,4b+1]
  =\frac{25n^3+15n^2+14n}{135}\left(\frac{1}{25}\right)^{n-1};\\
K_{1,2}=\sum_{0\leq{a}\leq b\leq{n-1}} det\mathcal L_A[4a+1,4b+2]
  =\frac{20n^3-9n^2+7n}{135}\left(\frac{1}{25}\right)^{n-1};\\
K_{1,3}=\sum_{0\leq{a}\leq b\leq{n-1}} det\mathcal L_A[4a+1,4b+3]=
  \frac{20n^3+6n^2+10n}{135}\left(\frac{1}{25}\right)^{n-1};\\
K_{2,0}=\sum_{0\leq{a}<b\leq{n}} det\mathcal L_A[4a+2,4b]=
  \frac{2(2n^3+3n^{2}+n)}{27}\left(\frac{1}{25}\right)^{n-1};\\
K_{2,1}=\sum_{0\leq{a}<b\leq n} det\mathcal L_A[4a+2,4b+1]=
  \frac{20n^3+21n^2+13n}{135}\left(\frac{1}{25}\right)^{n-1};\\
K_{2,2}=\sum_{0\leq{a}<b\leq{n-1}} det\mathcal L_A[4a+2,4b+2]=
  \frac{16(n^3-n)}{135}\left(\frac{1}{25}\right)^{n-1};\\
K_{2,3}=\sum_{0\leq{a}<b\leq{n-1}} det\mathcal L_A[4a+2,4b+3]=
  \frac{4(4n^3+3n^{2}-n)}{135}\left(\frac{1}{25}\right)^{n-1};\\
K_{3,0}=\sum_{0\leq{a}<b\leq{n}} det\mathcal L_A[4a+3,4b]=
  \frac{4n^3+3n^2-n}{27}\left(\frac{1}{25}\right)^{n-1};\\
K_{3,1}=\sum_{0\leq{a}<b\leq{n}} det\mathcal L_A[4a+3,4b+1]=
  \frac{20n^3+6n^2+10n}{135}\left(\frac{1}{25}\right)^{n-1};\\
K_{3,2}=\sum_{0\leq{a}<b\leq{n-1}} det\mathcal L_A[4a+3,4b+2]=
   \frac{4(4n^3-3n^{2}-n)}{135}\left(\frac{1}{25}\right)^{n-1};\\
K_{3,3}=\sum_{0\leq{a}<b\leq{n-1}} det\mathcal L_A[4a+3,4b+3]=
  \frac{16(n^3-n)}{135}\left(\frac{1}{25}\right)^{n-1}.
\end{cases}
\end{equation}
Therefore, $
(-1)^{4n-1}b_{4n-1}=\sum_{0\leq i,j\leq 3}K_{i,j} =\frac{2n(54n^{2}+9n+4)}{45}(\frac{1}{25})^{n-1}. \hfill\rule{1ex}{1ex}$

From Equations (4.1)--(4.2) and Lemma 4.1, we immediately have the following result.
\begin{thm} $Kf^{*}(Q_{n})= 2(324n^{3}+252n^{2}+71n+2)/3$.
\end{thm}

\begin{thm} $\lim\nolimits_{n\rightarrow\infty} Kf^{*}(Q_{n})/Gut(Q_{n})=1/4$.
\end{thm}

\noindent{\bf Proof.} For a subset $U \subseteq V$,
let $Gut(U) = \Sigma_{i \in U}\Sigma_{j \in V} d_i d_j d_{ij}$.

(1) Let $U_1=\{1,1'\}$, then
\begin{align*}
Gut(U_1)&=2\left[3\cdot3\cdot(1+4n+4n)+\sum_{i=1}^{n}2\cdot3\cdot5\cdot(4i-1)
  +\sum_{i=1}^{n-1}2\cdot3\cdot5\cdot(4i)\right.\\
&~~~\left.+\sum_{i=0}^{n-1}2\cdot3\cdot4\cdot(4i+1)+\sum_{i=0}^{n-1}2\cdot3\cdot4\cdot(4i+2)\right]\\
&=18(24n^{2}+2n+1).
\end{align*}

(2) Let $U_2=\{4n+1,(4n+1)'\}$, then
\begin{align*}
Gut(U_2)&=2\left[3\cdot3\cdot(1+4n+4n)+\sum_{i=1}^{n}2\cdot3\cdot5\cdot(4n-4i+1)+\sum_{i=1}^{n-1}2\cdot3\cdot4\cdot(4n-4i+2)\right.\\
&~~~\left.+\sum_{i=1}^{n}2\cdot3\cdot4\cdot(4n-4i+3)+\sum_{i=1}^{n}2\cdot3\cdot5\cdot(4n-4i+4)\right]\\
&=6(72n^{2}+42n-13).
\end{align*}

(3) Let $U_3 =\{4i-2,(4i-2)'|i = 1,2,\ldots,n\}$, then
\begin{align*}
Gut(U_3)&=2\sum_{i=1}^{n}\left[4\cdot4\cdot2+2\cdot3\cdot4\cdot(4i-3)+2\cdot3\cdot4\cdot(4n-4i+3)+\sum_{k=2}^{i}2\cdot4\cdot5\cdot(4i-4k+1)\right.\\
&~~~+\sum_{k=i+1}^{n}2\cdot4\cdot5\cdot(4k-4i-1)+\sum_{k=1}^{i-1}2\cdot4\cdot4\cdot(4i-4k-1)+\sum_{k=i}^{n}2\cdot4\cdot4\cdot(4k-4i+1)\\
&~~~+\sum_{k=1}^{i-1}2\cdot4\cdot4\cdot(4i-4k)+\sum_{k=i+1}^{n}2\cdot4\cdot4\cdot(4k-4i)+\sum_{k=1}^{i-1}2\cdot4\cdot5\cdot(4i-4k-2)\\
&~~~\left.+\sum_{k=i}^{n}2\cdot4\cdot5\cdot(4k-4i+2)\right]\\
&=32(12n^{3}+n^{2}+2n).
\end{align*}

(4) If $U_4 = \{4i-1,(4i-1)'|i = 1,2,\ldots,n\}$, then
\begin{align*}
Gut(U_4)&=2\sum_{i=1}^{n}\left[2\cdot4\cdot4\cdot2
+2\cdot3\cdot4\cdot(4i-2)+2\cdot3\cdot4\cdot(4n-4i+2)+\sum_{k=1}^{i-1}2\cdot4\cdot4\cdot(4i-4k)\right.\\
&~~~+\sum_{k=i+1}^{n}2\cdot4\cdot4\cdot(4k-4i)+\sum_{k=1}^{i}2\cdot4\cdot4\cdot(4i-4k+1)+\sum_{k=i+1}^{n}2\cdot4\cdot4\cdot(4k-4i-1)\\
&~~~+\sum_{k=2}^{i}2\cdot4\cdot5\cdot(4i-4k+2)+\sum_{k=i+1}^{n}2\cdot4\cdot5\cdot(4k-4i-2)+\sum_{k=1}^{i-1}2\cdot4\cdot5\cdot(4i-4k-1)\\
&~~~\left.+\sum_{k=i}^{n}2\cdot4\cdot5\cdot(4k-4i+1)\right]\\
&=16(24n^{3}+2n^{2}-n).
\end{align*}

(5) If $U_5 = \{4i,(4i)'|i = 1,2,\ldots,n\}$, then
\begin{align*}
Gut(U_5)&=2\sum_{i=1}^{n}\left[5\cdot5\cdot1+2\cdot3\cdot5\cdot(4i-1)
+2\cdot3\cdot5\cdot(4n-4i+1)+\sum_{k=1}^{i}2\cdot4\cdot5\cdot(4i-4k+1)\right.\\
&~~~+\sum_{k=i+1}^{n}2\cdot4\cdot5\cdot(4k-4i-1)+\sum_{k=1}^{i}2\cdot4\cdot5\cdot(4i-4k+2)+\sum_{k=i+1}^{n}2\cdot4\cdot5\cdot(4k-4i-2)\\
&~~~+\sum_{k=2}^{i}2\cdot5\cdot5\cdot(4i-4k+3)+\sum_{k=i+1}^{n}2\cdot5\cdot5\cdot(4k-4i-3)+\sum_{k=1}^{i-1}2\cdot5\cdot5\cdot(4i-4k)\\
&~~~\left.+\sum_{k=i+1}^{n}2\cdot5\cdot5\cdot(4k-4i)\right]\\
&=10(48n^{3}+4n^{2}+n).
\end{align*}

(6) If $U_6 = \{4i-3,(4i-3)'|i = 2,\ldots,n\}$, then
\begin{align*}
Gut(U_6)&=2\sum_{i=2}^{n}\left[5\cdot5\cdot1+2\cdot3\cdot5\cdot(4i-4)+2\cdot3\cdot5\cdot(4n-4i+4)+\sum_{k=1}^{i-1}2\cdot4\cdot5\cdot(4i-4k-2)\right.\\
&~~~+\sum_{k=i}^{n}2\cdot4\cdot5\cdot(4k-4i+2)+\sum_{k=2}^{i-1}2\cdot4\cdot5\cdot(4i-4k-1)+\sum_{k=i}^{n}2\cdot4\cdot5\cdot(4k-4i+1)\\
&~~~+\sum_{k=2}^{i-1}2\cdot5\cdot5\cdot(4i-4k)+\sum_{k=i+1}^{n}2\cdot5\cdot5\cdot(4k-4i)+\sum_{k=1}^{i-1}2\cdot5\cdot5\cdot(4i-4k-3)\\
&~~~\left.+\sum_{k=i}^{n}2\cdot5\cdot5\cdot(4k-4i+3)\right]\\
&=10(48n^{3}-84n^{2}+49n-13).
\end{align*}

Hence $Gut(Q_{n}) =(Gut(U_1)+Gut(U_2)+ \cdots +Gut(U_6))/2=864n^{3}+64n^{2}+418n-95$, and from Theorem 4.2 we immediately have
$\lim_{n\rightarrow\infty} Kf^{*}(Q_{n})/Gut(Q_{n})=1/4$. \hfill\rule{1ex}{1ex}

The values of $Kf^{*}(Q_{n})$ for $1 \leq n \leq 15$ are listed in Table 2.
\begin{table}[htbp]
\setlength{\abovecaptionskip}{0.05cm} \centering\vspace{.3cm}
\caption{The values of $Kf^*(Q_n)$, $1 \leq n \leq 15$.}
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
  \hline

  $n$ & $Kf^{*}(Q_{n})$ & $Q_{n}$ & $Kf^{*}(Q_{n})$ & $n$ & $Kf^{*}(Q_{n})$ & $n$ & $Kf^{*}(Q_{n})$ & $Q_{n}$ & $Kf^{*}(Q_{n}))$ \\
  \hline
  $1$ & $432.67$ & $4$ & $16702.70$ & $7$ & $82652.70$ & $10$ & $233274.67$ & $13$ & $503560.67$ \\
  $2$ & $2496.00$ & $5$ & $31438.00$ & $8$ & $121724.00$ & $11$ & $308316.00$ & $14$ & $626296.00$
  \\
  $3$ & $7487.33$ & $6$ & $52989.30$ & $9$ & $171499.34$ & $12$ & $398009.34$ & $15$ & $767511.34$
  \\
  \hline
\end{tabular}
\end{table}

redpajama_set_name: RedPajamaArXiv


text:
Q: My If statement isn't returning a value based on cell being blank or has data My If statement says that if the cell it is looking at is null, then return x, if it's got a value in it, then Y... The logger shows data in the cell, but it still returns x!
I've tried changing to isblank, is null, is "". I can see that it is logging data, but it is still returning the answer for a blank cell
function trackingSheetUpdate() {

  var now = new Date();

  var ss = SpreadsheetApp.getActiveSpreadsheet();

  var sourceSheet = ss.getSheetByName('Active Campaign');

  var Tracking = ss.getSheetByName('Tracking')

  var getUpdate = sourceSheet.getRange(2, 5).getValue();

  var dateResult = now - sourceSheet.getRange(2, 6).getValue();

  var openSS = SpreadsheetApp.openByUrl(getUpdate).getSheetByName('Sheet1')

  var weeklyUpdate = openSS.getRange(2, 1).getValue();

  if (dateResult > 1 && dateResult <= 7); {

    if (weeklyUpdate == null, Tracking.getRange(2, 2).setValue("x"));

    else

      ("Y");
  }
}

I want x if the cell is blank and Y if cell is not blank, but it only returns x! Eventually I am going to loop this through a dynamic range, but want to get the basics right first!

A: I created some test sheet data of my own and the below code seemed to work as you're expecting it to.
function trackingSheetUpdate() {
  var now = new Date();
  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var sourceSheet = ss.getSheetByName('Active Campaign');
  var trackSheet = ss.getSheetByName('Tracking');
  var getUpdate = sourceSheet.getRange(2, 5).getValue();
  var dateResult = now - sourceSheet.getRange(2, 6).getValue();
  var openSheet = SpreadsheetApp.openByUrl(getUpdate).getSheetByName('Sheet1');
  var weeklyUpdate = openSheet.getRange(2, 1).getValue();

  if (dateResult > 1 && dateResult <= 7); {
    if (weeklyUpdate == '') {
      trackSheet.getRange(2, 2).setValue("x");
    } else {
      trackSheet.getRange(2, 2).setValue("y");
    }
  }
}

Of course this relies on your sheet data being correct, for example you need var getUpdate to be a URL. I have also renamed some variables for consistency with the rest of the script, along with some grammatical changes like making sure there are semicolons to end lines etc.

redpajama_set_name: RedPajamaStackExchange


Note: Each tool is uniquely identified and has "UP" coil pipe indicated in RED. Inside is threaded to be used with Handle/Rod EDC Item #221295.
Each tool is uniquely identified and has "UP" coil pipe indicated in RED. Inside is threaded to be used with Handle/Rod EDC Item #221295.


text:
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We have awarded a grant to assist Rainbow Trust in delivering their services to Island Families! Having identified a need for better support in the home, we contacted Rainbow Trust back in 2012 to ask them to offer their services on the island. Now, we are working alongside each other with Layla's Trust gifting £12,000 last year and a further £14,000 this year to enable Rainbow Trust to support families in their own homes, offering practical and emotional support where they can. We feel strongly that by utilising 'Island' funds to support this national charity, we ensure that the money remains on the Island or the benefit of our Island community.
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Possible values for $L_x$ I've a physical system with $l=1$ and I have to calculate the values I can obtain if I measure $L_x$ and their probability.
I know that:

*

*the values I can obtain are $\ m=0, \pm 1$

*$\displaystyle L_x=\frac{L_+ + L_-}{2}$

*$L_x|1, m>_x=\hbar m |1, m>_x$

But I can't understand, for example, why I should obtain
$|1, 1>_x=\frac{1}{2}[|1,1>+\sqrt{2}|1,0>+|1,-1>] $
(I have obtained $|1, 1>_x=\frac{\sqrt{2}}{2} |1,0>$)
Can I have some hints?

A: The simplest way is to exploit the symmetries here. So, instead of mindlessly going through the algebra, solving equations etc. you just use the fact that you are free to call any direction the x-direction, and set up a right handed coordinate system. In particular, this means that you are free to cyclically permute x, y, z in the equations. The problem of expressing $|1,1\rangle_x$ in terms of the eigenvectors of $L_z$, can be solved as follows. If you apply the raising operator for the x-component of the angular momentum, $L_x^{+}$ to $|1,1\rangle_x$ then this should yield zero. But to do this we need to express $L_x^{+}$ in terms of $L_z$ and $L_z^{\pm}$. You can do that as follows. We have:
$$L_z^{+} = L_x + i L_y$$
If you cyclically permute x,y, and z, you get:
$${L_{x}}^{+} = L_y + i L_z$$
We can write this as:
$$L_x^{+} =\frac{1}{2 i}\left({L_{z}}^{+} - {L_{z}}^{-}\right) + i L_z$$
This means that the state $|1,1\rangle_x$ satisfies the equation
$$\left[L_z^{+} - {L_z}^{-} - 2 L_z\right]|1,1\rangle_x = 0$$
If we take inner product with $|1,1\rangle$, we get the equation:
$$\sqrt{2}\langle 1,0|1,1\rangle_x - 2 \langle 1,1|1,1\rangle_x = 0$$
Taking the inner product with $|1,0\rangle$ gives:
$$\langle 1,-1|1,1\rangle_x = \langle 1,1|1,1\rangle_x$$
So, we see that the $|1,0\rangle$ component of the state is $\sqrt{2}$ times that of the other two components, and by normalizing the state you obtain the overall factor of 1/2 in your equation.


In Ada, a floating point literal must be written with both an integral and decimal part. 10 is not a valid literal for a floating point value, while 10.0 is.
The principal scalar types predefined by Ada are Integer, Float, Boolean, and Character. These correspond to int, float, bool/boolean, and char, respectively. The names for these types are not reserved words; they are regular identifiers.
Ada's type system encourages programmers to think about data at a high level of abstraction. The compiler will at times output a simple efficient machine instruction for a full line of source code (and some instructions can be eliminated entirely). The careful programmer's concern that the operation really makes sense in the real world would be satisfied, and so would the programmer's concern about performance.
The next example below defines two different metrics: area and distance. Mixing these two metrics must be done with great care, as certain operations do not make sense, like adding an area to a distance. Others require knowledge of the expected semantics; for example, multiplying two distances. To help avoid errors, Ada requires that each of the binary operators "+", "-", "*", and "/" for integer and floating-point types take operands of the same type and return a value of that type.
Even though the Distance and Area types above are just Floats, the compiler does not allow arbitrary mixing of values of these different types. An explicit conversion (which does not necessarily mean any additional object code) is necessary.
The predefined Ada rules are not perfect; they admit some problematic cases (for example multiplying two Distances yields a Distance) and prohibit some useful cases (for example multiplying two Distances should deliver an Area). These situations can be handled through other mechanisms. A predefined operation can be identified as abstract to make it unavailable; overloading can be used to give new interpretations to existing operator symbols, for example allowing an operator to return a value from a type different from its operands; and more generally, GNAT has introduced a facility that helps perform dimensionality checking.
Ada enumerations work similarly to C++ and Java's enums.
In the above example, Grade is a new integer type associated with a range check. Range checks are dynamic and are meant to enforce the property that no object of the given type can have a value outside the specified range. In this example, the first assignment to G1 is correct and will not raise a run-time exceprion. Assigning N to G1 is illegal since Grade is a different type than Integer. Converting N to Grade makes the assignment legal, and a range check on the conversion confirms that the value is within 0 .. 100. Assigning G1+10 to G2 is legal since + for Grade returns a Grade (note that the literal 10 is interpreted as a Grade value in this context), and again there is a range check.
Range checks are useful for detecting errors as early as possible. However, there may be some impact on performance. Modern compilers do know how to remove redundant checks, and you can deactivate these checks altogether if you have sufficient confidence that your code will function correctly.
These declarations don't create new types, just new names for constrained ranges of their base types.
The clause beginning with with introduces an Ada aspect, which is additional information provided for declared entities such as types and subtypes. The Dynamic_Predicate aspect is the most general form. Within the predicate expression, the name of the (sub)type refers to the current value of the (sub)type. The predicate is checked on assignment, parameter passing, and in several other contexts. There is a "Static_Predicate" form which introduce some optimization and constrains on the form of these predicates, outside of the scope of this document.
Attributes start with a single apostrophe ("tick"), and they allow you to query properties of, and perform certain actions on, declared entities such as types, objects, and subprograms. For example, you can determine the first and last bounds of scalar types, get the sizes of objects and types, and convert values to and from strings. This section provides an overview of how attributes work. For more information on the many attributes defined by the language, you can refer directly to the Ada Language Reference Manual.
Other interesting examples are the 'First and 'Last attributes which, respectively, return the first and last values of a scalar type. Using 32-bit integers, for instance, Integer'First returns -231 and Integer'Last returns 231 - 1.
C++ arrays are pointers with offsets, but the same is not the case for Ada and Java. Arrays in the latter two languages are not interchangable with operations on pointers, and array types are considered first-class citizens. Arrays in Ada have dedicated semantics such as the availability of the array's boundaries at run-time. Therefore, unhandled array overflows are impossible unless checks are suppressed. Any discrete type can serve as an array index, and you can specify both the starting and ending bounds---the lower bound doesn't necessarily have to be 0. Most of the time, array types need to be explicitly declared prior to the declaration of an object of that array type.
In C++ and Java, only the size of the array is given during declaration. In Ada, array index ranges are specified using two values of a discrete type. In this example, the array type declaration specifies the use of Integer as the index type, but does not provide any constraints (use <>, pronounced box, to specify "no constraints"). The constraints are defined in the object declaration to be 1 to 26, inclusive. Arrays have an attribute called 'Range. In our example, Arr'Range can also be expressed as Arr'First .. Arr'Last; both expressions will resolve to 1 .. 26. So the 'Range attribute supplies the bounds for our for loop. There is no risk of stating either of the bounds incorrectly, as one might do in C++ where "I <= 26" may be specified as the end-of-loop condition.
As in C++, Ada Strings are arrays of Characters. The C++ or Java String class is the equivalent of the Ada type Ada.Strings.Unbounded_String which offers additional capabilities in exchange for some overhead. Ada strings, importantly, are not delimited with the special character '\0' like they are in C++. It is not necessary because Ada uses the array's bounds to determine where the string starts and stops.
You see here that no constraints are necessary for this variable definition. The initial value given allows the automatic determination of My_String's bounds.
Ada offers high-level operations for copying, slicing, and assigning values to arrays. We'll start with assignment. In C++ or Java, the assignment operator doesn't make a copy of the value of an array, but only copies the address or reference to the target variable. In Ada, the actual array contents are duplicated. To get the above behavior, actual pointer types would have to be defined and used.
This assigns the 4th, 5th, and 6th elements of A1 into the 1st, 2nd, and 3rd elements of A2. Note that only the length matters here: the values of the indexes don't have to be equal; they slide automatically.
Pointers, references, and access types differ in significant ways across the languages that we are examining. In C++, pointers are integral to a basic understanding of the language, from array manipulation to proper declaration and use of function parameters. Java goes a step further: everything is a reference, except for primitive types like scalars. Ada's design goes in the other direction: it makes more features available without requiring the explicit use of pointers.
There's a fundamental difference between the Ada and C++ semantics above and the semantics for Java. In Ada and C++, objects are allocated on the stack and are directly accessed. V1 and V2 are two different objects and the assignment statement copies the value of V1 into V2. In Java, V1 and V2 are two references to objects of class R. Note that when V1 and V2 are declared, no actual object of class R yet exists in memory: it has to be allocated later with the new allocator operator. After the assignment V2 = V1, there's only one R object in memory: the assignment is a reference assignment, not a value assignment. At the end of the Java code, V1 and V2 are two references to the same objects and the V2.A = 1 statement changes the field of that one object, while in the Ada and the C++ case V1 and V2 are two distinct objects.
For those coming from the Java world: there's no garbage collector in Ada, so objects allocated by the new operator need to be expressly freed.
Dereferencing is performed automatically in certain situations, for instance when it is clear that the type required is the dereferenced object rather than the pointer itself, or when accessing record members via a pointer. To explicitly dereference an access variable, append .all. The equivalent of V1->A in C++ can be written either as V1.A or V1.all.A.


Surveillance of healthcare associated infections is an overlooked parameter of good clinical practice in most healthcare institutions, due to the workload demanded in the absence of adequate IT-systems. The aim of the present study was to investigate whether a simple protocol could be used to estimate the burden of healthcare associated infections in three university hospitals in Huddinge in Sweden, Riga in Latvia and Vilnius in Lithuania and form the basis for initiating a long term follow up system. The medical records of all patients receiving antibiotics were reviewed according to a standardised protocol, focusing on the indications for the drugs and on the frequency of hospital acquired infection (HAI) in a point-prevalence survey. Only comparable specialities were included. The proportion of patients treated with antibiotics (prophylaxis not included) were 63/280 (22%) in Huddinge, 73/649 (11%) in Riga and 99/682 (15%) in Vilnius. The proportion of admitted patients treated for a HAI were 15%, 3% and 4%, respectively, (both comparisons Huddinge versus other centres P<0.001). Surgical site infections were most common, followed by infections with an onset more than 2 days after admission without any of the other registered risk factors present. Our inexpensive and simple method showed that healthcare associated infections were a significant problem among patients admitted to Huddinge. The figures obtained can be used for further discussion and form a baseline for follow up at the local level. The comparison of figures between centres was far less relevant than the process the study created.


text:
Events Film & Culture Night: Live in Between 此岸彼岸
Film & Culture Night: Live in Between 此岸彼岸
Martijn van Veen Saturday November 30th, 2019
Date: December 17, 2019 8:00 pm – 10:00 pm
Venue: Huis de Pinto
Categories: CinemAsia Film Festival Events
3 dialogues x 3 filmmakers
Louis Hothothot
He works as a multimedia artist combining video art, performance and graphic design. His research deals with the subject of memory, identity, emotion and pain. Louis Hothothot receives support from institutions such as Netherland Film Fonds, Amsterdams Fonds voor de Kunst, Eye Film Museum in Amsterdam, International Documentary Filmfestival Amsterdam and Netherlands Film Academy.
Mr. Hu
When the spring came, I was positive about the US-China trade war and thought it would soon end with China making a compromise. I started to gather Dutch-Chinese' stories, first in Bos en Lommer, then in Den Haag, Haarlem, and Diemen.
Though a Chinese restaurant owner would not tell me her name, she truly opened up through telling me her story. She was disappointed about her role as both a mom and a daughter, having to take care of both her parents in China and her daughters in Holland. Mr Hu wished his paintings and antique clock collection could be a benefit to his family and kids, but they were against his ideas. René Liem's family had moved to Indonesia from China 150 years ago, but his family and other Chinese Indonesians were discrimi- nated against by the Suharto government, etc. I enjoyed hearing their warm, sad family stories all through the spring.
Will there be a new wave of immigration from Asia? The idea made me nervous, and I realised that all of the characters I had been working with had moved to Holland dur- ing the first Cold War. Some of them had left China during the Cultural Revolution, one had moved after the Tiananmen Massacre, one family immigrated during the period of Suharto government, and so on. I was surprised that these histories, hidden by official Chinese media, had begun to be reveal themselves. Not in China, but here in a foreign land. Each story featured a soul that was hurt by politics, but still holding on to dignity. They had all spent more than half of their lives building up a new sense of belonging in a foreign place.
Hangfeng Chen
His work includes drawing & painting, paper-cuts, photo, video, installation and performance. His practice deals with issues of surrounding commercial- ization, environmentalism, globalization and cultural transmutation, often realized through playful metaphors when he merges concepts and mediums like an alchemist. Chen Hangfeng sees artists as present-day incarnations of the ancient philosopher, artisans and literati.
Tai Bai Yu Tou
This is a popular dish in China – a soupy dish cooks with the Big Head Carp and Tofu. The legendary was the great poet from Tang Dynasty – Li Bai, who loved this dish, and name was given after. This piece shows normally how people here will eat the fish, from buying the fish at the wet market, the process of killing, preparing and cooking the fish… It focuses on the eating technic which is picking the bones with teeth, tongue and lips, avoiding the fish bone stuck in the throat! The Big Head Carp is the exact fish that is highly invasive in the US at the moment. It is seen as a dangerous invasive spe- cies that threatens the ecosystem. But it's considered as a delicious dish in China…
Minhong Yu
Minhong Yu stands simultaneously in two worlds and two systems, trying to start a dialogue between them. How can they find a relationship, how can they find a balance? Her work is concerned with visibilities and invisi- bilities relating to social issues in this digital world and ranges from prints to video installations which explore representations of power. She depicts existing systems of power, and questions symbols of identity. Minhong Yu graduated in 2013 from the China Central Academy of Fine Art in Bei- jing, and obtained her masters degree in design at the Sandberg Instituut, Rietveld Academy in 2016.
Breathe confronts opposing forces, and proposes a dialogue between poetry and rigid socio-political structures, between opacity and transparency. Breathe shows the polluting factories of Minhong's hometown and her newly overseas residence as contributors of a beautiful landscape, where air, smoke and water are the links for transportation between these two continents. Smog is in both installations the invisible medium that connects these otherwise distant worlds softly together. Living in an unfamiliar environment, coping with a language barrier, and forced to explore the meaning of identity, Minhong wants to be- come invisible: to wear a smog-mask and disappear into the herd. Because in order to be stronger, one first has to be weaker. The voice of silence is the highest state of bonding. Transparency can be the perfect camouflage for any existence in this world.


Bone Morphogenetic Proteins are growth factors that were originally found to help induce the formation of bone and cartilage but also regulate tissue architecture throughout the body. The absence or misregulation of BMP signaling has been linked in the progression of various cancer types, including bone and colon. Noggin proteins are signal transducers which are integral inhibitors for many proteins in the TGF (Transforming Growth Factor) family, including BMP4, and are crucial in cognition, bone development, joint formation and nuerulation. GoldBio's high quality recombinant proteins have a purity level exceeding 95% in most cases and also have excellent bioactivity with low endotoxin levels.


text:
William Hague to be News of the World columnist
The News of the World has completed its line-up of new columnists by signing up former Conservative leader William Hague.
The MP joins TV presenter Ulrika Johnson, celebrity gossip writers Rav Singh and Polly Graham and former royal press secretary Mark Bolland, who have all been added to the line-up of columnists in the past six months.
Film director Michael Winner will continue to write monthly features for the paper.
NoW editor Andy Coulson said: "William Hague was famed for his sharp, no-nonsense delivery at the dispatch box and we're sure that will now translate into his column. Being a Tory won't stop him targeting politicians of all parties."
The paper has also revealed a wideranging reshuffle of its senior team following the announcement this week that news editor Ricky Sutton has handed in his notice and is moving abroad.
Picture editor Clare Wood is also leaving to pursue a career outside journalism.
Gary Thompson has been promoted from assistant editor (features) to associate editor (news and pictures). His first job will be to recruit replacements for the two leavers.
In other changes at the paper, Paul Ashton has been promoted to chief photographer and Jules Stenson has been promoted to assistant editor (features).
By Dominic Ponsford


A small team of fourteen swimmers competed at Newmarket Pool for the 2013 Suffolk Development Championships and produced some excellent results.
Many thanks to the volunteers who helped: on the door, in the hall and on poolside and to the Coaching Team who all played a part in the swimmers successes.
Minnie Wayman (9yrs) – 50BF – 44.47 – a 10's PB!
Megan Claydon also came third overall on a points system of 6 points for first place, 5 points for etc which is a fantastic achievement out of 164 female swimmers all ages! Well done Megan!
The results are attached and you will see from the improvements (a minus equals a PB) there were some brilliant swims.
Out of thirteen clubs Deben SC came 9th and considering we were one of the smallest in terms of numbers competing I think this result really shows the strength and depth that these swimmers had.


text:
The airline says it will start serving the coffee on a flight from Toronto to Calgary and expand service across its fleet.
WestJet has signed a deal to serve McDonald's coffee aboard its flights. The airline says it will start serving the coffee on a flight from Toronto to Calgary and expand service across its fleet. By December 1st WestJet says it also plans to offer McDonald's coffee across its fleet of 737NG and 767NG aircraft with its regional airline, WestJet Encore to follow.
McDonald's has been working to improve its coffee in recent years with its McCafé brand. The deal is similar to other arrangements airlines have struck with coffee companies.
With more than 140 aircraft, the partnership will service approximately 30,000 cups of McCafé Premium Roast Coffee every day.
Coffee on board WestJet aircraft will have the same popular on-cup rewards collector card and peelable sticker, redeemable on completion at participating Canadian McDonald's restaurants.
Complimentary beverages are available on most WestJet flights and may include water, soft drinks, juice, tea, and McCafé Premium Roast Coffee.


hope u r allays continue herd work in our dhakamobile forum.
Must do hard work and keep posting help other member...than we will decide for you.
Im just curious, but was there ever a time when this whole process was more unpredictable?
Last edited by M@SUM; 02-02-17 at 04:22 PM.


Blushing Rose porcelain tray for tea service or vanity use. Pale pink roses and marbled green painted edge with 14kt Gold hand applied trim .
Porcelain Treasures has just announced that they are retiring! These were exactly that, porcelain treasures .
We will certaintly miss the beautiful teaware they produced.


If cancelled within 24 hours of check-in, a cancellation fee of one night will be charged. 24-72 hours before expected arrival, 50% of the total amount will be charged. If cancelled 3-5 days before expected arrival, an $80.00 cancellation fee will be charged. Our cancellation policy for group bookings differs from this. If you cancel within 1 month of your chosen dates we will charge you for any cabins/trains that we cannot re-book for the period of your stay.


text:
Пловдивска област () се налази у јужном делу Бугарске. Ова област заузима површину од 5,972 -{km²}- и има 712.702 становника. Административни центар Пловдивске области је град Пловдив.

Списак насељених места у Пловдивској области
Градови су подебљани

Општина Асеновград
Асеновград,
Бачково,
Бојанци,
Бор,
Врата,
Горни Воден,
Горнослав,
Добростан,
Долни Воден,
Долнослав,
Жалт Камак,
Златоврх,
Избеглии,
Нови извор,
Изворово,
Козаново,
Конуш,
Косово,
Леново,
Љасково,
Мостово,
Мулдава,
Нареченски бањи,
Новаково,
Орешец,
Патриарх Евтимово,
Стоево,
Сини врх,
Тополово,
Три могили,
Узуново,
Червен

Општина Брезово
Бабек,
Борец,
Брезово,
Врбен,
Дрангово,
Зелениково,
Златосел,
Отец Кирилово,
Падарско,
Розовец,
Свежен,
Стрелци,
Сарнегор,
Тјуркмен,
Чехларе,
Чоба

Општина Калојаново
Бегово,
Главатар,
Горна Махала,
Долна Махала,
Дуванлии,
Далго поље,
Житница,
Иван Вазово,
Калојаново,
Отец Паисиево,
Песнопој,
Ражево,
Ражево Конаре,
Сухозем,
Черноземен

Општина Карлово
Бања,
Бегунци,
Богдан,
Васил Левски,
Ведраре,
Војњагово,
Горни Домљан,
Домљан,
Дабене,
Иганово,
Калофер,
Каравелово,
Карлово,
Климент,
Клисура,
Куртово,
Карнаре,
Марино Поље,
Московец,
Мраченик,
Певците,
Пролом,
Розино,
Слатина,
Соколица,
Столетово,
Христо Даново

Општина Кричим
Кричим

Општина Куклен
Галабово,
Добралак,
Куклен,
Руен,
Цар Калојан,
Јаврово

Општина Лаки
Балкан махала,
Белица,
Борово,
Џурково,
Дрјаново,
Здравец ,
Лакавица,
Лаки,
Манастир,
Четрока,
Југово

Општина Марица
Бенковски,
Војводиново,
Војсил,
Граф Игнатиево,
Динк,
Жељазно,
Калековец,
Костиево,
Крислово,
Маноле,
Манолско Конаре,
Радиново,
Рогош,
Скутаре,
Строево,
Трилистник,
Труд,
Царацово,
Јасно поље

Општина Пловдив
Пловдив

Општина Првомај
Брјагово,
Буково,
Бјала река,
Виница,
Воден,
Градина,
Добри дол,
Драгојново,
Далбок извор,
Езерово,
Искра,
Караџалово,
Крушево,
Поројна,
Православен,
Првомај,
Татарево

Општина Раковски
Белозем,
Бољарино,
Момино село,
Раковски,
Стрјама,
Чалакови,
Шишманци

Општина Родопи
Белаштица,
Бојково,
Бранипоље,
Брестник,
Брестовица,
Дедово,
Златитрап,
Извор,
Кадиево,
Крумово,
Лилково,
Марково,
Оризари,
Првенец,
Ситово,
Скобелево,
Устина,
Храбрино,
Цалапица,
Чурен,
Јагодово

Општина Садово
Ахматово,
Богданица,
Бољарци,
Караџово,
Катуница,
Кочево,
Милево,
Моминско,
Поповица,
Садово,
Селци,
Чешнегирово

Општина Сјединеније
Гољам Чардак,
Драгомир,
Љубен,
Мали Чардак,
Најден Герово,
Недељево,
Правиште,
Сјединеније,
Царимир,
Церетељево

Општина Хисарја
Беловица,
Красново,
Крстевич,
Мало Крушево,
Михилци,
Матеница,
Ново Железаре,
Паничери,
Старо Железаре,
Старосел,
Хисарја,
Черничево

Општина Перуштица
Перуштица

Општина Стамболијски
Јоаким Груево,
Куртово Конаре,
Стамболијски,
Ново село,
Триводици

Општина Сопот
Анево,
Сопот

Спољашње везе

 Обласна администрација Пловдивска

Области Бугарске

redpajama_set_name: RedPajamaWikipedia


If there is one thing that will fill up your heart meter, it is cupcakes! Especially Legend of Zelda themed 8-bit heart cupcakes!
I'm not sure who made these delicious looking cupcakes with 8-bit hearts on top but they look delicious! This would be the perfect thing to snack on while having a marathon session of running around Hyrule as Link. Unfortunately, they won't do anything for filling up Link's heart meter.
The only question I have is why nobody ever brings me any cupcakes like this? I like cupcakes too you know!


text:
Latest News Releases 14 October
Summaries Titles
Noninfectious versions of SARS-CoV-2 provide powerful research tools
A new system of "replicons" will facilitate efforts to study different coronavirus variants and develop new drugs for COVID.
Using Disney movies to help with child development
Worried your children are getting too much screen time during the COVID-19 pandemic? A University of Houston study found animated movies can serve as valuable tools for parents and counselors alike to improve communication with children about tough issues.
Child and Adolescent Social Work Journal
Artificial intelligence helps to find new natural substances
More than a third of all medicines available today are based on active substances from nature and a research team from the University of Jena (Germany) has developed a procedure to identify small active substance molecules much more quickly and easily. The researchers present their method, called COSMIC (Confidence Of Small Molecule IdentifiCations), in the current issue of the renowned journal Nature Biotechnology.
Packard Foundation backs Rice bioengineer
Grant and Award Announcement
Rice University bioengineer Jerzy Szablowski wins a prestigious Packard Fellowship to pursue noninvasive brain research.
New Governing Board elected to provide scientific and strategic leadership to the International Science Council
International Science Council
A new Governing Board has been appointed to lead the International Science Council, an international non-governmental organization that works to provide a global voice for science. The Governing Board is a central decision-making body of the Council, which was formed in 2018 by the merger of the International Council of Science (ICSU), representing the natural sciences, and the International Social Science Council (ISSC).
Clues emerge: How harmless bacteria go rogue turning into deadly flesh-eating variants
A new study from the University of Central Florida found that the environmental lifestyle that bacteria possess reveal why some go rogue and turn deadly while others remain harmless to humans. The findings, which published recently in the journal Proceedings of the National Academy of Sciences, focus on Vibrio vulnificus, better known as the flesh-eating bacteria. However, what the scientists found could help create a model that may well extend to other human pathogens.
OU researchers contribute to study on possible alternative treatment for Lyme disease
Researchers at the University of Oklahoma contributed to a study published in the journal, Cell, that describes a new antibiotic with the potential to cure Lyme disease.
Bridging optics and electronics
Harvard John A. Paulson School of Engineering and Applied Sciences
Researchers at the Harvard John A. Paulson School of Engineering and Applied Sciences, in collaboration with a team from Washington University, have developed a simple spatial light modulator made from gold electrodes covered by a thin film of electro-optical material that changes its optical properties in response to electric signals.
Climate change threatens hydropower energy security in the Amazon basin
Cary Institute of Ecosystem Studies
Hydropower is the dominant source of energy in the Amazon region, the world's largest river basin and a hotspot for future hydropower development. However, a new Global Environmental Change study warns that in the coming decades, climate change-driven reductions in precipitation and river discharge will diminish the Amazon's hydropower capacity.
Cornell Atkinson Center for Sustainability, Amazon Dams Computational Sustainability Working Group
Evidence of superionic ice provides new insights into the unusual magnetic fields of Uranus and Neptune
GFZ GeoForschungsZentrum Potsdam, Helmholtz Centre
Not all ice is the same. The solid form of water comes in more than a dozen different - sometimes more, sometimes less crystalline - structures, depending on the conditions of pressure and temperature in the environment. Superionic ice is a special crystalline form, half solid, half liquid - and electrically conductive. Its existence has been predicted on the basis of various models and has already been observed on several occasions under - very extreme - laboratory conditions. However, the exact conditions at which superionic ices are stable remain controversial. A team of scientists led by Vitali Prakapenka from the University of Chicago, which also includes Sergey Lobanov from the German Research Center for Geosciences GFZ Potsdam, has now measured the structure and properties of two superionic ice phases (ice XVIII and ice XX). They brought water to extremely high pressures and temperatures in a laser-heated diamond anvil cell. At the same time, the samples were examined with regard to structure and electrical conductivity. The results were published today in the renowned journal Nature Physics. They provide another piece of the puzzle in the spectrum of the manifestations of water. And they may also help to explain the unusual magnetic fields of the planets Uranus and Neptune, which contain a lot of water.
Nature Physics
Helmholtz Young Investigators Program CLEAR
The inside story of England COVID pandemic described in new study
Wellcome Trust Sanger Institute
The Covid-19 crisis that gripped the UK between September 2020 and June 2021 can be thought of as a series of overlapping epidemics, rather than a single event, say researchers at the Wellcome Sanger Institute, EMBL's European Bioinformatics Institute (EMBL-EBI) and their collaborators.
, National Institute of Health Research (NIHR), Genome Research Limited
HKUST research shows growing dominance of diatom algae in the Pearl River estuary and paves the way for future study on diatom blooms
Hong Kong University of Science and Technology
It is a common perception that waters close to population would be more polluted than those offshore or at higher latitudes. However, researchers from The Hong Kong University of Science and Technology (HKUST) found that the ratio between two common microalgae diatom and dinoflagellate (dino) – a common benchmark of water quality, has been nearly doubled in the Pearl River Estuary (PRE), one of the world's most urbanized subtropical coastal waters, over the past two decades.
Hong Kong Branch of Southern Laboratory of Ocean Science and Engineering Guangdong Laboratory,
Conservatives tweet louder
SwRI expands valve testing capabilities to aid carbon capture, storage
Surface chemistry reveals corrosive secrets
Reports and Proceedings
Meeting Announcement
Book Announcement
Multi-Language News Releases
日本語のニュース


* If you rent does the HOA or management have breed restrictions?
What Boundaries are you planning on setting for your new pet?
Where will your pet be kept while you are away?
Where will your pet be kept when you are home?
How many hours will your pet be home alone during the day?
Have you ever given away, sold or surrendered a pet?
What brand food do you plan on feeding your pet?
What heartworm and flea and tick preventives do you plan on using?
What method of training and discipline do you plan on using?
Are you prepared to hire a dog trainer or take classes if necessary?
How many years do you believe you will have this pet?
Where will you keep your pet when you go out of town?
What will happen to your pet when you move?
Under what circumstances would you return a pet?
How much annually are you prepared to spend on your pet with food, toys, veterinary visits, flea and tick preventative and heartworm preventive?
What would you do if you are unable to care for this pet?
Have you ever been convicted animal neglect or abuse?
Many times even house trained dogs can be stressed with the changes and have accidents in the home for a period of time after adopting. Are you prepared to deal with accidents?
We offer on our dogs over 1 year of age a 2 week foster to adopt to ensure the placement is a good fit. Should you wish to foster to adopt we will provide the food and crate during this period and any medications the dog may be on if any. But we will not pay for any vet visits due to negligence on the fosters behalf. During this time you must contact us first to get permission for the dog to go to vet otherwise you will be responsible for the bill. There is a $50 non refundable fee for this option. However should you proceed with the adoption the fee will be applied to the adoption donation. Do you agree to the above terms?
donation to PP2CR for help with the expenses incurred and vet bills by saving this pet from the shelter.
agree to provide care from this day forward to my new pet. This includes adequate food, water, exercise, shelter, veterinary care and lots of love. The pet will be an inside pet and not kept in the yard or tethered outside. I agree never to sell, trade or to surrender the pet to any agency without notifying PP2CR and receiving PP2CR's approval and/or having first option of reclaiming the pet. I agree that the adoption fee is non refundable and there is no reimbursement if the adoption does not work out. I understand that these dogs came from shelters and that PP2CR does not know their history except what may have been provided from the shelters. I hereby understand that PP2CR has no prior information of age, breed, training, registration, health issues, or any other on the pets other than what is provided at time of adoption per what has been our experience with the pet. PP2cr makes no warranties of any kind concerning the pet. I agree that all expenses incurred after I take possession of the pet will be my sole responsibility and PP2CR is released of any and all liabilities whether financial or other.
I (the adopter) have read this contract fully and understand by signing this contract I am agreeing to the terms of the contract.
We strongly encourage positive reinforcement training.
We do not recommend Kibble n Bits, Purina Beneful, Alpo or Dog Chow, Pedigree and Ol Roy.