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Wiktorówka (ukr. Вікторівка) – wieś na Ukrainie, w obwodzie czerkaskim, w rejonie humańskim.

Przypisy

Bibliografia

Wsie w obwodzie czerkaskim
Rejon chrystyniwski

redpajama_set_name: RedPajamaWikipedia


Outreach has always been one of the major pillars of Grace Bible Church. Though we started with just a few missionaries and a single ministry serving marginalized areas, we have grown to reach our neighborhoods through multiple strategies and have over seventy missionary families in more than twenty countries. There is a place for everyone to serve!
Jesus said to love God first and then love our neighbors. Grace has numerous strategies that empower individuals to build relationships with our neighbors and walk with them as we walk with Christ.
The Global Outreach Ministry exists to promote and support missions and Grace missionaries, and to engage everyone at Grace in the pursuit of fulfilling our mission in our lifetime. But, churches aren't the only ones called to fulfill the Great Commission. We are all called to do the same. Everyone should take part in missions by going, sending, supporting or mobilizing! Being in a church that supports missions is only a first step. The next steps are up to you.
Every year, thousands of men and women come from around the world as undergrad and grad students. Many arrive with very few possessions or friends. We want to mobilize the church to help provide for the wonderful students as they arrive to the Bryan-College Station area.
Outreach has always been one of the major pillars of Grace Bible Church. There is a place for everyone to serve!


× Vashon Chamber Music celebrates its 10th Anniversary season with a 3-concert series in the spring of 2019.
Monday, March 4th at 7:30pm: Cello Virtuosity III features the return of Seattle Symphony Principal Cellist Efe Baltacigil and his regular pianist collaborator Jessica Choe. They will be joined by new Seattle Symphony Concertmaster Noah Geller for a program of masterworks for cello/piano duo and piano trio.
Monday, April 22nd at 7:30pm: The Girsky Quartet with soprano Holly Boaz in a program of Hungarian classics. Tickets for this concert can be purchased at brownpapertickets.com. The concert will be held at the Vashon United Methodist Church.


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(Anonymous) on November 30th, 2012 12:39 pm (UTC)
I just found your posts/FanFic through a different site, and I have to say I'm enjoying them greatly. I am not a member of this site so that's why the anonymous post, but I will continue to follow you recaps/fan-fiction as they come. It's very compelling ;). I really like the way you are showing the characters (all of them) in an (almost) better way than the TV show, but as another commenter has stated, it's easier to get into the heads of the characters in a written format as opposed to a TV format. I think you really get Elena's character and what she's going through and I like that you're playing Damon with a little bit more of an edge, but still keeping all the emotional growth he's shown through the seasons. I also like that you're acknowledging the larger roll he's had in the group as a whole, and legitimizing it by having most of the characters (Matt, Bonnie, Jeremy, Alaric, Meredith … ) recognize it and start to embrace him more. It's humorous that you're keeping Caroline and Stefan (to an extent) on the Damon is bad bandwagon lol. I'm intrigued to see where you go next with the story :).

redpajama_set_name: RedPajamaCommonCrawl


After importing the CCeS XML import file the Incomes are showing as assessed at the incorrect rate, How do I fix this?
Here the list of incomes can be added to, changed, removed and amended.
Step 2. There are two types of Incomes in this list ICS and non-ICS incomes. ICS being an income downloaded from Income Confirmation Services (Centrelink).
In order to import the CCeS file, you must first update the ICS Incomes listed in this table to reflect the correct rates, fixed amount or not and whether or not the income type is assessable relevant to your State, Territory or Organisation.
Step 3. Update each ICS Income Type to reflect the correct rates, fixed amount or not and whether or not the income type is assessable relevant to your State, Territory or Organisation.
Step 4. Exit System Maintenance.
Click the button to find the CCeS response file.
Status Comments will appear in the display window, and the Client incomes will be updated to reflect the changes to the Income Table.


text:
2023 Select
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'21 Molly Terraciano Commits to Lafayette
By Matt Chandik | September 8, 2020 | Comments Off on '21 Molly Terraciano Commits to Lafayette
Molly Terraciano's introduction to the goalie position was… unceremonious, to say the least.
Terraciano's mother, Joanie, was coaching her daughter's youth team in third grade and struggled to find volunteers to get in the cage. As coaches' kids tend to be, the younger Terraciano was told that she had to get in goal for the team's benefit.
"For a while, I didn't like it," said the Team 91 Tristate 2021 and Pelham Memorial senior. "My mom would tell me, 'you're the only girl that doesn't cry.' I always had bruises, which definitely concerned some teachers. But I definitely started liking it around sixth grade, and in eighth grade, I got pulled up to varsity and loved it."
She got good at it, too, becoming a rock in the 91 Tristate cage. When it came to recruiting, though, she admittedly got nervous when she saw schools stockpiling '21 goalies. Her strong interest in a nursing program also whittled down her options, and she wasn't sure what to think when Lafayette took another goalie after showing her significant interest.
"I was pretty stressed when I saw them take another goalie because I didn't think it was looking good for me," Terraciano said. They were still looking to recruit me and I just had to be forward and ask questions. Lafayette was honest with me and told me the truth about the whole process."
That kept Terraciano in the hunt for a spot, and when the Leopards finally offered her the opportunity to be a part of their program, she jumped all over it. She's the fourth Tristate '21 to commit, the second to a Division I program after Sam Spieler (John Jay / Merrimack) and the first player in program history to choose Lafayette.
"I actually got very lucky because Emily Constantine is an assistant coach there and she used to coach with 91," Terraciano said. "I knew her well, which helped. It's a really good academic school, which pushed me even more to the school. When I visited, I felt really close to the girls on the team, who were so welcoming. The coaches are so fun."
Terraciano started playing for Team 91 four years ago, and in that time, has brought her game to that of a Division I recruit. A big part of her improvement was a result of her improving both her athleticism and communication skills. The latter were something she was reluctant to work on at first, but those improvements paid off in spades.
"I'm very good at communicating. I'm very loud and I make sure everyone knows where they should be," said Terraciano, who has also skied since the age of five. "At first, I was afraid that girls would get frustrated and think that I was yelling at them. Emily taught me not to be afraid in goal. I think having a big voice and being a leader is probably the most important thing for a goalie."
"I definitely got faster on my feet and all of my goalie coaches were very helpful with my training," Terraciano continued. "I also did a lot of training with my brother, Tony, who plays club at Vanderbilt."
That athleticism definitely helped Terraciano land on a bunch of radars, and to hear her current coach say it, it's a big reason why her best lacrosse is still ahead of her.
"Molly is a very athletic and explosive goalie, and she's got a ton of potential in the cage," Team 91 Tri-State girls director Rachel Vallarelli said. "I am so excited for her to play at the next level at Lafayette and continue her lacrosse career there. She's going to do great things. She has the talent and work ethic to develop into a top-flight goalie for the Leopards, and coach Katie McConnell and her staff did a great job of recruiting her."
Not bad for someone who didn't even want to be in the cage.
Posted in Tri-State Club News, Tri-State Girls Commits and tagged 2021, College Commits, Lafayette, Molly Terraciano, Pelham Memorial
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very young age depending on their fighting skills. The ones who fail the tests are labeled lower class anoys and they look after construction of villages, etc. The ones who prove their worth at these tests are the ones who take care of the defence of these villages from other annoy colonies or outsiders. Anoy tribes had a very important part to play in the wars. Xoloantra : This planet is the 2nd smallest planet in the universe after Anoyamba and is not located fully on the orbit. It is placed just in place for it to revolve properly. It is not at a very comfortable position on the orbit like others. This planet being small in size has considerably low gravitational force as compared to other. The inhabitants that live here who are called Xoloantrans, they can't fly like Anoys but they don't walk fully on the ground either. As this planet is not in a very comfortable positon on the orbit, the energy received by this planet is the lowest. Due to which the creatures living on this planet are weak and malnourished and have a very short life span of 20-25 year. Because of low life expectancy they reproduce more. This has again resulted in a negative way by giving rise to the population. High population and low energy from Bob makes life difficult in Xoloantra. But the positive result of high population is that the people interact with a large group of people which expands the horizon their knowledge. So people from this planet are very smart and witty.
Marq quietly followed the guard. He just had to tell the Chief…and also, he'd better get answers for his questions. Marq was only seven years old when he had first met Chief Heinz. Heinz was like a fatherly figure to Marq as he was the only one who encouraged Marq's interest and talent at handling technology at any level. Chief Heinz was the only one who knew Marq's parents, it was said. Marq gawped at the huge hall he was standing in. He had never been this far into the 'monastery'. He saw the old man sitting at the other end of the hall shrouded in the smoke from his pipe. The man had a grave expression on his face that left Marq wondering… "Greetings, Marq. It's been a long time…how are you?" "Chief…I need answers. All my friends have been killed during your damned expedition to Bob. Why did you have to send them to such a dangerous place? " "Ah, my boy…it was not my doing nor my wish. I know that sending an expedition to Bob is just being greedy. But alas, power is no longer in my hands. The Elders no longer listen to me. They feel that we should make use of the energy at Bob before any other planet tries." "But that's just asking for war..!" "I know, but as I said, I'm just a namesake leader. Most of the Elders now are banding under Joel Limiz, the Cornel. They are just waiting for me to pass away…" Marq glanced around the hall to see if anyone was in sight before leaning towards the Chief. "I've received a private message from Maresk. He gave me all info required for me to make a conditioner that will help us survive on Bob without any kind of hassle. I've already made a prototype and intend to test it out, myself. Maresk may still be alive somewhere on Bob." The Old Chief grinned at him and said, "I knew you'd come for that, Marq. You're the type who wouldn't let your friends down, no matter what. I have only a few days left.
made our intentions clear. The majority of the elders have decided that Mazinia will put in all its effort to inhabit Bob. We have leads about some device known as the atmospheric conditioner which makes living possible on Bob for us Maz. Therefore, we are replicating this technology and as soon as the conditioners are ready, I will leave for Bob along with entire military." People were talking in hushed noises to each other as they dispersed, seemingly excited by this new prospect. The Cornel turned and went back to his room only to catch the familiar scent of smoke. "Chief Larry...greetings. I presume you heard my speech?" The Chief cast a dry look on him and said, "Of course, I did. It must have resonated throughout Mazinia. But that's not the point. I don't understand why you need the entire military." The Cornel replied curtly, "I have explained this to the other elders. It's because I fear that there are some dangerous forces actually living on the planet and considering the possibilities of the being very strong, we'll need our whole army." "What? You think there are inhabitants on Bob? That's not possible. If it was, why would you let Marq go through?" The Cornel looked taken aback for a moment before retaining his cool. "He should be glad that I didn't arrest him for making such great devices and not even telling us about them. Who knows, he could be developing something against the state. The least I could do was have him watched and that I did." The chief left the room not wanting to talk anymore. Chapter 5: Marq picked up another rock. Something was very delightful about this planet, something that made him feel good about himself. Or maybe, it was the effect of the conditioner which made him feel shiny and brand new, always.
Clank, clank. Marq continued his search for Maresk. He followed a trail of what could possibly be Maz footsteps. He saw it. It was Maresk's body, lying still and with the satellite communicator still in his hand. Marq rushed over and gasped. He was alive. Or at least, that's what he looked like. However, something was wrong. He seemed to be in a coma or something similar, as there was no response from the body no matter how much Marq tried 'waking' him up. He was worse than dead and there was nothing Marq could do about him. And then, it struck him. It was the atmosphere that did this to him. The others died in a crash. What was that all about? Maresk looked like he was the only survivor. All of this didn't make much sense to him, right now. "I need answers…" Marq thought to himself as he spoke into his transmitter. "Chief, this is Maz Z310 from Planet Bob…come in, come in." -All of the Binesians seemed astonished at the sight of a spaceship landing in their main settlement. They were even more surprised when they saw humanoids with leaves, branches and roots sticking out of them coming out from the ship. They took up a defensive stance and asked them to identify themselves.
The most elegant one among the plant men came up front. "I am Mordred, the prince of Contra. Greetings, neighbors. I'll need to speak to your king." One of the larger Binesians asked, "Why do you want to meet him? How come we've never seen you before? Another Binesian pointed to the spaceship and shouted, "And what the hell is that thing behind you?" Mordred smiled calmly and said, "At ease, gentlemen. We mean no harm. We have a proposal and it's very urgent. So, I'll need to speak to your king as soon as possible." The Binesians were still confused and started arguing among themselves. Then, one of the Binesians said, "Okay, we trust you. Follow me; I'll lead you to Binnet's place." Binnet was the fearsome leader of the Binesia. This planet consisted of people who valued strength over brains. The reason that they were backward in technology was simple. They didn't care, really. The ten feet tall and 800 pound leader, Binnet, was someone who was universally feared. Many people dreaded approaching him as Binnet's short temper was also something that was renowned. Mordred calmly bowed in front of Binnet and told the contra to get back to the ship. Taking this as a friendly gesture, Binnet asked, "So, tell me prince. What is that you want from me?" "Okay, let me get to it directly. I'm sure you remember Mazinia, the land of metal men. I have proof that they are up to something sinister and something that endangers the peace between all our planets. I want to do something about it, so maybe; I could use your help too?" Binnet roared with laughter. "So you want back up in a fight and are making up excuses for being weaklings, yourselves?" His expression turned into a frown immediately when he saw Mordred smiling calmly. "You are probably right. But maybe you should go through the evidence we have.." Clank, clank.
A hooded figure comes into the picture, limping. Chapter 7 Binnet came out f his room and blew the mighty Binesian war horn. The horn was used to signify that Binesia was at war and were supposed to assemble. Most of the Binesians looked perplexed as they couldn't understand why they were at war. They hadn't fought any external enemy since a long time now. As soon as they settled, Binnet addressed them. "My dear Binesian folk, we have a problem. The entire armies of Mazinia are making their way to Bob. We can't allow this as they plan to use the energy on Bob to their own advantage. If they don't want peace, fine…we'll give them war." This little speech was followed by the roars and shouts of agreement from the Binesians. "But how do we go and fight them?" asked one of the Binesians still looking confused. Even Binnet looked confused. Mordred with yet another smile, waved at some of the contra standing beside him. They spoke something into their transmitters. The Binesians looked on as the sky seemed to light up. To their astonishment, about a hundred spaceships descended from the sky. Mordred spoke out loud, "Binesians, here are your spaceships; we have a contra for maneuvering each ship. This is a critical situation. I'll need all help you can give me." Having said this, Mordred got into his spaceship and flew away. Binnet roared, "Contra's armies will be meeting us on the battlefield. Let's put on a show the Maz will have to pay a heavy price for." -Cornel Joel Limiz looked at the battalions of Maz spaceships that had just landed on Bob.
"Alright, that's all of them "he thought to himself. "Everyone, set camp" he said. Everything was going according to plan. Despite a delay and their slow spaceships, they had no trouble till now. All the conditioners were working perfectly, so, they could take their time to scan the planet for signs of life. Suddenly, he saw a figure in the distance. It was approaching them at a brisk pace. "Ah", he said recognizing the Maz. It was Marq. He had a strange expression on his face. "What the hell are you doing? Do you know the effect the planet will have on us?" Marq asked. "If you are talking about the energy shifts…yes, I do. That's our purpose; we need the energy at any cost." "Why do you need it, it's not like anyone has attacked us, right?" argued Maz. "How dare you contradict my decision? I ba—" The voice dropped suddenly. All of the Maz looked to the east. "Oh my…who the hell are they?" shouted one Maz soldier. The sky was raining spaceships and out of each spaceship came ten humanoids of immense size and about 8 foot in size. The Binesians were here. The Maz army looked on in fright as the Binesians kept unloading from the spaceships. The largest of them came to the forefront and blew a huge horn. "Binesians…attack!" he shouted. Marq was horrified. Not by the advancing army, but of the expression on Cornel Limiz's face. They had been betrayed.
Mordred sat in the Maz monastery. His plans had gone as smoothly as he wanted them to. There seemed to be no resistance from the remaining Maz on Mazinia and the same went for the Binesians on Binesia. A valet came up to Mordred and whispered something into his ear. Mordred seemed to be mildly amused by that. He waved to his valet to let the man in. It was the first time he was going to see the legendary ace, Annoyer. Annoyer was an Anoy from Anoyamba. Anoyamba was a very small planet inhabited by humanoids who had wings and faces resembling those of birds. The bird-man came in, escorted by two contra soldiers. Mordred asked him to be released. The escorts left the room. Mordred looked at Annoyer and said, "What are you here for, bird-man? Are you anoys actually thinking of coming here to oppose us? Let me tell you that it will be of little or no use." "And why would you say that? What do you actually know about us? Your mother was said to be a madwoman but I knew her well. She actually created peace on Contra. You're just spreading war." "It's all over, Annoyer. My mom did tell about your so called greatness. I'd want to…test that for myself!" he said, as he unsheathed a sword and leaped on Annoyer. Annoyer leaped to the air in a flash and in an instant, Mordred was on the ground with Annoyer bearing down upon him. "I knew you wouldn't agree so easily. In fact, the army of anoys is already wiping the Contras out of Mazinia." Mordred frowned, "Oh, I see…I never counted you in my plans as I thought you wouldn't dare attack Contra. Seems I was wrong, eh. Oh well..." Annoyer was taken aback as vines sprouted from Mordred's body, entangling him. He tried to fly away, but he was trapped. Mordred raised his sword.
from the crowd but Binnet continued. "There is no point in going on like this. We can't afford to lose any more lives." With this, Binnet signaled to both the oncoming armies that he was surrendering. -The Maz army had captured all of the Binesians and had chained Binnet, the monster. The other army that had been approaching identified themselves as the Xoloantrans. Their leader, Phynisis, walked up to Binnet. "You won't be killing anymore Maz, you monster." Binnet protested, "But they were the ones who are going to disturb the balance in the universe on their side." Phynisis replied, "No way, you are lying." A voice among the crowd said, "No, it's true. The Maz were led here to steal energy and disrupt the universe's balance." It was Marq. Coming out of the crowd, he said "The Binesians and the Maz have been cheated by contra and Cornel Limiz. They have led us into this disastrous war while they have already conquered Mazinia and Binesia. We have been fighting like fools…for nothing. " Chapter 11 Mordred couldn't believe it. For once, he was not smiling at all. The anoys had defeated his armies on Binesia and Mazinia. Shaking with rage, he tells his valet to let the person in. Cornel Limiz comes in, hooded. He says, "I have bad news for you, Mordred. The armies of Mazinia and Binesia have stopped fighting. Marq ruined it all." Mordred starts laughing maniacally. "Of course, that was bound to happen. I have bad news for you. You are dead, bro." Saying this, Mordred sliced off Limez' head with a single strike.
"Haha, how do you like that, Limez? You stupid metal head?!! I am—" His voice dropped and he turned back to receive the shock of his life. It was his mother with her blade right through him. "I love you, son. But I have realized that one life cannot be equal to a thousand or much more. " Mordred fell to the ground. He looked at his mother, closed his eyes…and smiled.


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Electoral College meets to elect new President
Sean Douglas
OPPOSITION'S NOMINEE: Israel Khan, SC. -
DAVID LEE, Opposition whip in the House of Representatives, hopes some Government members will not support the Government's nominee Christine Kangaloo for President, but vote for the Opposition's nominee Senior Counsel Israel Khan when the Electoral College meets at 1.30 pm on Friday.
He said it will be a secret ballot, and recalled the Prime Minister's recent admission of once defying his party in electing Arthur NR Robinson as President in 1997.
On Thursday, Lee told Newsday the Opposition regards Kangaloo, the former senate president, as unsuitable.
He questioned her political affiliations and past involvement as acting President in terminating Jwala Rambarran as Central Bank governor.
However, Senate Leader Dr Amery Browne, in a text to Newsday, fully endorsed Kangaloo and, in turn, questioned Opposition members' suitability, citing their conduct in October 2021 in the Electoral College debating the Opposition's unsuccessful call to impeach President Paula-Mae Weekes.
On Friday in the Electoral College, the Government will have 39 votes, namely 23 from the House and 16 from the Senate, including two presiding officers.
Against this, the Opposition will have 25 votes comprised of 19 from the House and six from the Senate, while the Independent Senators have nine votes.
This means even if every Independent Senator voted with the Opposition to total 34 votes, the Government's 39 votes would still carry, provided there were no defections to the Opposition's nominee.
Rowley had hailed Kangaloo as a past opposition senator, government senator, government MP, government minister and senate president.
"I don't disqualify people on the basis of being active politicians. Having been a government minister is part of her pool of experience. She is also a lawyer and has served on both sides of the Parliament floor and both houses."
Rowley said the Constitution specifically envisaged an MP or senator becoming President.
However, the Opposition has feared Balisier House (the headquarters of the ruling party) influencing President's House.
St Augustine MP Khadijah Ameen on Wednesday made an eleventh-hour call for Kangaloo to withdraw her nomination.
Lee told Newsday that Khan was "a very noble and upstanding candidate" with no political affiliation, but someone who has been critical of many governments and even Opposition Leader Kamla Persad-Bissessar.
He said Kangaloo has been a strong PNM activist and former Pointe-a-Pierre MP.
"The electorate then rejected her in favour of Errol Mc Leod in 2010."
GOVT'S NOMINEE: Christine Kangaloo. -
Lee alleged, "She has been left very wanting over the past seven years in her judgement whether it be as senate president against the Opposition, and when she has acted as President."
He reckoned Khan could get elected instead of Kangaloo.
"We feel that tomorrow, if good sense prevails among the Independent Senators along with some of the PNM members, PNM MPs or PNM Senators, our candidate should be chosen."
Lee said Rowley recently admitted supporting Robinson rather than the PNM's candidate retired judge Anthony Lucky.
"I am hoping that this time around the same good sense that four PNM members, including Rowley, showed in the candidature of ANR Robinson, they will see it in the candidature of Israel Khan.
"So we hope and we pray something good will come out tomorrow, because we will be in some dark times if and when Christine Kangaloo wins."
Lee said the Government seemingly has the numbers to win.
"The Opposition just has 19 plus six which is 25, so we need some support from the Independents and also from the PNM. It is done by secret ballot."
Lee said Kangaloo's first act as acting president had been to terminate Jwala Rambarran as Central Bank governor, but a court had recently ruled in favour of Rambarran, saying the process of firing had been unfair.
He opined, "She should have waited until then substantive president (Anthony Carmona) returned from vacation. I don't think an acting president should take that kind of role."
Lee also complained about Kangaloo's partial proclamation of the Data Protection Act to controversially grant certain rights to the police.
"If she is President she has got to appoint nine Independent Senators. Will we get nine independent senators or will they be a carbon copy of the PNM bench?"
On Friday, Browne told Newsday, "What I can say about Mrs Kangaloo is that she is an outstanding citizen, a proven servant of the people of TT, and a lady of immaculate integrity.
"I have absolutely no doubt that she will maintain the highest standards of the Office of the President of the Republic, and will bring her own unique personality and presence to the Presidency.
"As Minister of Foreign and Caricom Affairs, I have been privileged to witness the class and charisma of then acting president Kangaloo during her hosting of state events for visiting heads such as the President of Guyana in 2022, and I was left with a deep sense of admiration and appreciation of the energy, patriotism, and compassion of our nominee for this office."
Browne noted the Opposition's nominee, saying, "Dissent and divergence are important parts of democracy.
"Those who hold an opposing view are entitled so to do, and any such members are empowered under our Constitution to vote as they see fit.
"When this election process is completed, we still have to live together, work together, and improve and develop our land for the next generation. The Prime Minister has made an outstanding choice as the Government's nominee."
Asked how proceedings might go on Friday, he said the Opposition's behaviour was hard to predict, based on their allegedly "appalling behaviour" and "verbal abuse" of Independent Senators at the Electoral College when it sat in relation to Weekes' tenure.
"What I do know is that the members on the Government side will conduct ourselves with dignity and a solemn sense of duty to the people of this land under our Constitution. Whilst we cannot always control the disposition of others, it behoves us to ourselves maintain the highest standards of conduct."
Political analyst Dr Bishnu Ragoonath told Newsday there was an "urgent need for Constitutional reform" in the composition of the Electoral College.
Asked about the Government's majority against a combined Opposition and Independent vote, he said the Electoral College majority resulted from the membership of both houses. However, he said the House and Senate numbers were designed to give a government a majority against any stalemate over finance bills, but were not originally intended for an Electoral College.
Saying Weekes and Carmona had been consensus candidates, Ragoonath viewed Kangaloo as "an overt politician" seeking the Presidency, whose bid the Opposition could not support.
Asked if Independent Senators faced undue pressure to vote for Kangaloo so as to be reappointed, he said they should not have to sing for their supper. Whether they should complete their five-year term or resign upon her election, was entirely a personal choice for each independent senator, Ragoonath said.
"Electoral College meets to elect new President"


Am linken Drauufer inmitten von Weinbergen in der Ortschaft Kanischa der Pfarrei Jahring, wo der Vater Michael Lorber seine Bergholdengründe bewirtschaftete, wurde Jakob Lorber am 22. Juli 1800 geboren. Und nicht zufällig wuchs er in einem armen, jedoch für Kunst und Religion offenen Haus in dieser ländlichen Umgebung heran. Sein Diplom als Hauptschullehrer erwarb er 1829 in Graz, der steiermärkischen Landeshauptstadt, wo er aber keine entsprechende Anstellung fand, so daß er weiterhin intensive Musikstudien betrieb, komponierte, Gesangs und Violin-Unterricht gab und gelegentlich auch öffentliche Konzerte.
Untergang der katholischen Kirche durch neue Propheten?
Werden die Propheten und Warnzeichen beachtet werden?
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text:
One-Shots, Shmups
Waifu Uncovered: Return of the Strip 'Em Up
April 13, 2020 Pete Davison 1 Comment
So it seems that "strip 'em up" is a thing now. I am neither surprised nor particularly upset about this, given that Kaneko's classic erotic Qix-alike Gals Panic also spawned its own subgenre; it's just amusing to see this sort of thing happen in the modern day.
For the unfamiliar, the strip 'em up, previously seen here on MoeGamer in the form of Deep Space Waifu (and in a tangentially different form in Crawlco Block Knockers) is a take on the shoot 'em up — or, more broadly, arcade game — formula in which you not only blast enemies, you also, through engaging with the game's specific mechanics, find some means of disrobing the (usually anime-style) young lady in the background. Success provides titties; failure provides frustration.
Which brings us to Waifu Uncovered, product of the delightfully named One Hand Free Studios, and a game that, rather pleasingly, isn't just a clone of Deep Space Waifu. Let's take a closer look!
Mild NSFW stuff ahead!
In Waifu Uncovered, you take on the role of K. Vaio, a horse-headed ninja piloting an unsubtly phallic spaceship who is tasked with rescuing a series of cute girls from their infected clothing that will, over time, turn them into "hot but ugly aliens", whatever that means.
Anyway, in order to fix this terrible issue, K. Vaio needs to travel around the world in one of several selectable (universally phallic) spaceships, blast away hordes of enemies and collect the ninja stars they drop, which will subsequently home in on the infected clothing and eventually destroy it when a counter reaches zero. Once all the infected clothing has been successfully removed, the boss for the stage appears, at which point you must blast it into oblivion; the first time you achieve this it generally unlocks something new such as an alternative ship to pilot or perhaps one of the two "uncensor" modes.
The mechanics of the game are actually pretty interesting, as this isn't just about blasting enemies. For one thing, the essential ninja stars that foes drop fall down the screen and can be missed, so you need to position yourself in order to catch them, and collecting more in succession adds to a combo and provides the potential for higher scores.
The stages aren't just randomly generated hordes of enemies, either; there are distinct patterns and groups of enemies you'll encounter, each with their own unique ways of attacking. Particular highlights come during "asteroid" sequences, where large, spinning, "rage face"-inspired obstacles float down the screen at varying speeds, forcing you to spot and quickly plan out a route to safety while still blowing enemies away.
Of special note is the game's power-up system, which has a fun amount of depth to it. Rather than just powering up your ship bit by bit, each ship can be upgraded in various ways by collecting the appropriate icons. You can increase the ship's movement speed, its spread of shots, its bullet power and even add different types of orbiting options. The twist is that each ship has its own "cap" on each of these upgradeable statistics, though collecting "level up" markers allows you to bust through these limits and continue to improve the ship, with the sprite changing with each level increase for a nice bit of visual flair.
These "caps" on various areas of performance are the main way in which the different ships are made to feel distinct from one another. The one you start with, for example, is balanced, allowing for a decent spread shot with good power and a nice bit of speed. Elsewhere there's a ship that starts as speedy to move but has no capacity for spread shots until a level up; a ship with powerful shots but slow speed; and various other combinations. Each feels nicely different from the others, making multiple playthroughs of the game something well worth exploring.
The game is divided into three main modes, which all essentially unfold in a similar manner with varying difficulty. The "Normal" mode is extremely easy and can probably be cleared by most people with a passing interest in shoot 'em ups on their first try, which will unlock most of the ships for subsequent playthroughs; the "Arcade" mode provides a much stiffer challenge through denser bullet patterns and more enemies, but allows access to a character not present in the "Normal" mode if you can clear it; and "One-Hand" mode allows you to play with the mouse, leaving your other free for masturbating furiously, should you feel the need. Oh, and there's also a ship you can only unlock in "One-Hand" mode, too.
The game can be played with one or two players, and the developers are rigid and unflinching on their desire to make this an arcade-style experience: you have one life, no continues, and if you mess up, you have to go right back to the beginning and start again. Some players on Steam have expressed a certain degree of upset about this — likely since Deep Space Waifu allows you to tackle a single level at a time and saves your progress as you go — but with the score-attack structure in place here, the merciless, no-nonsense arcade structure works perfectly. If you want to see everything the game has to offer, you better get good; that "Arcade" mode takes no prisoners!
Presentation of the game is very nice. The artwork for the girls is good quality and has a distinctive sense of style about it; I particularly like the way it "posterises" itself with bright colours and high contrast under certain circumstances, and the film grain effect applied over the background during the pre-stage Galaga-style enemy swarm is pleasing to the eye. The girls run a nice gamut, too, including a rocker chick, a soccer player, a cheerleader, a cowgirl and more; something for most tastes!
The enemies are a creative and fun bunch, ranging from Robotron-style brains to monstrous, dark takes on Hello Kitty via what appear to be sentient assholes, a boss that is a literal pile of shit and the aliens from Mars Attacks! There's an obvious Japanese influence going on in more than just the artwork for the girls, too; the "ghost"-type enemies in particular are strongly reminiscent of comedic anime and manga depictions of spirits from beyond. Each enemy type has its own bullet and movement patterns, meaning you can quickly recognise them by sight and know how to deal with them after a playthrough or two.
The music is pretty great, too. Rather than attempting to ape the popular '80s-inspired vaporwave style heard in Deep Space Waifu (and numerous other indie games right now), Waifu Uncovered instead opts for a more '90s trance and techno-inspired soundscape, featuring some cuts that sound like they were put together on Dance eJay. And I mean that as an absolute compliment, believe me. It's a bit of a shame that at the time of writing there isn't completely unique music for every stage, but the game has been undergoing continual development since it first appeared on Steam a while back.
In fact, as I type this, the game has just left Steam's Early Access programme, and judging by forum posts and reviews, it seems the developers have been serious about listening to feedback from players, implementing lots of tweaks and fixes to make the overall package really rather solid.
Ultimately, Waifu Uncovered is a bit of cheesy, silly fun that was almost certainly never intended to keep you super-invested over the long term — but it will doubtless be a lot of fun to return to every so often in the pursuit of high scores.
Also, at just £2.99, this is the modern equivalent of those straight-to-cassette budget releases gamers of a certain age used to be able to get for their Spectrum, Commodore 64 or Atari 8-bit computers: cheap, disposable fun that will bring a smile to your face every time you boot it up. That said, it is to the developers' credit that they haven't skimped on actually making a good game in their attempts to get more budget-price tits on Steam; like Deep Space Waifu, this is a solid, enjoyable shoot 'em up in its own right that just happens to incorporate anime tiddies into its overall aesthetic.
And I am 100% cool with that!
More about Waifu Uncovered
A review copy was provided by the developer.
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text:
\section{Introduction}\label{section-intro}
Lauricella's hypergeometric series $F_C$ of $m$ variables $x_1 ,\ldots ,x_m$
with complex parameters $a$, $b$, $c_1$, $\ldots$, $c_m$ is defined by
\begin{align*}
 F_C (a,b,c ;x )
 =\sum_{n_1 ,\ldots ,n_m =0} ^{\infty }
 \frac{(a,n_1 +\cdots +n_m )(b,n_1 +\cdots +n_m )}
 {(c_1 ,n_1 )\cdots (c_m ,n_m ) n_1 ! \cdots n_m !} x_1 ^{n_1} \cdots x_m ^{n_m} ,
\end{align*}
where $x=(x_1 ,\ldots ,x_m),\ c=(c_1 ,\ldots ,c_m)$,
$c_1 ,\ldots ,c_m \not\in \{ 0,-1,-2,\ldots \}$, and $(c_1 ,n_1)=\Gamma (c_1+n_1)/\Gamma (c_1)$.
This series converges in the domain
$$
D_C =\left\{ (x_1 ,\ldots ,x_m ) \in \C ^m  \ \middle| \ \sum _{k=1} ^{m} \sqrt{|x_k|} <1  \right\} .
$$
It is shown in \cite{HT} that the hypergeometric system $E_C=E_C (a,b,c)$
of differential equations satisfied by  $F_C (a,b,c;x)$ is a holonomic system
of rank $2^m$ with the singular locus
\begin{align*}

  &S= \Big( \prod_{k=1}^m x_{k} \cdot R(x)=0 \Big) \subset \C^m ,\\

  &R(x_1 ,\ldots ,x_m)=\prod_{\vep_1 ,\ldots ,\vep_m =\pm 1}
  \Big( 1+\sum_{k=1}^m \vep_k \sqrt{x_k} \Big),
\end{align*}
and that the system $E_C$ is
irreducible (in the sense of $D$-modules)
if and only if
\begin{align}
  \label{irred-1}
  a-\sum_{k=1}^m i_k c_k ,\quad b-\sum_{k=1}^m i_k c_k \not\in \Z ,\qquad
  \forall I=(i_1,\dots,i_m)\in \ztwom.
\end{align}
It is classically known that
there are $2^m$ solutions to $E_C (a,b,c)$
expressed in terms of $F_C$ with different parameters(see (\ref{series-sol})).
If parameters satisfy (\ref{irred-1}) and $c_1 ,\ldots ,c_m \not\in \Z$
then they form a fundamental system of solutions to $E_C (a,b,c)$
in a simply connected domain in $D_C -S$.

Let $X$ be the complement of the singular locus $S$.
The fundamental group of $X$ is generated by $m+1$ loops
$\rho_0 ,\ \rho_1 ,\ldots ,\ \rho_m$ (see \S \ref{section-MR}).
In \cite{G-FC-monodromy},
we express the circuit transformations
$\CM_i$ along $\rho_i$ $(i=0,\dots , m)$
by using the $2^m$ solutions and the intersection form on
twisted homology groups associated with
Euler-type integrals of solutions to $E_C$.
These expressions are independent of the choice of a basis
of the twisted homology group. The circuit transformations $\CM_i$ are
also studied in \cite{M} by the specification of the intersection form
regarded as indeterminate.

In this paper, we show the following.
\begin{Th}[Main theorem]
  The monodromy representation
  $$
  \CM : \pi_1 (X,\dot{x}) \to GL(\sol )
  $$
  is irreducible under the assumption (\ref{irred-1}),
  where $\sol =\sol (a,b,c)$ is the local solution space to $E_C(a,b,c)$
  around a point $\dot{x}\in D_C -S$.
\end{Th}
We also show that the monodromy representation is reducible if
one of the assumption (\ref{irred-1}) is not satisfied.
We specify an invariant subspace in $\sol$
under the monodromy representation $\CM$ in this case.

Note that under the assumption (\ref{irred-1}), for example,
$c_k$ may be an integer.
In such a case, the solutions (\ref{series-sol}) expressed by $F_C$
do not form a basis of
the local solution space.
We give a linear transformation of them
so that the transformed solutions are valid even in cases where any of
$c_k$'s are integers.
We construct it inductively on $m$ using tensor products of matrices.

We remark that the irreducibility of the monodromy representation $\CM$
is implied from that of the system $E_C$ under the assumption (\ref{irred-1}).
We prove it explicitly by using properties of the circuit transformations in
\cite{G-FC-monodromy}, not applying results of $D$-modules.
We here briefly explain our idea of the proof of the main theorem.
It is shown in \cite{G-FC-monodromy} that the $1$-eigenspace $V$
of $\CM_0$ is
$(2^m-1)$-dimensional. Let $f_0\in \sol$ (corresponding to $\ev$ in \S \ref{section-rep-mat})
be a non-zero vector in
its orthogonal complement with respect to the intersection form.
It is quite easy to give a basis of the whole space $\sol$
by actions $\CM_1,\dots,\CM_m$ on $f_0$.
Let $W$ be an invariant subspace of $\sol$ under the monodromy
representation $\CM$.
If $W\not\subset V$ then we can show $f_0 \in W$, which yields
that $W$ becomes the whole space $\sol$ by the previous fact.
Otherwise, we can show that $W$ becomes the zero space
by the perfectness of the intersection form.

\section{Preliminaries}
Except in \S \ref{section-red}, we assume the conditions
for parameters $a,b,c_1,\dots,c_m$ in (\ref{irred-1})
(it is equivalent to (\ref{irred-2}) or (\ref{irred-3}) mentioned below).

In this section, we collect some facts about Lauricella's $F_C$
mentioned in \cite{G-FC}, \cite{G-FC-monodromy}, \cite{HT} and \cite{L}.
\begin{Nota}
  We put
  \begin{align*}
    \al =\exp (2\pi \sqrt{-1} a),\quad
    \be =\exp (2\pi \sqrt{-1} b),\quad
    \ga_k =\exp (2\pi \sqrt{-1} c_k)\ (k=1,\ldots ,m).
  \end{align*}
  We often regard $\al$, $\be$ and $\ga_k$ as indeterminants,
  and consider the rational function field
  $\C(\al ,\be ,\ga)=\C(\al ,\be ,\ga_1 ,\ldots ,\ga_m)$.
  For a rational function
  $g(\al ,\be ,\ga_1 ,\ldots ,\ga_m) \in \C(\al ,\be ,\ga)$,
  we denote $g(\al ,\be ,\ga_1 ,\ldots ,\ga_m)^{\vee}
  =g(\al^{-1} ,\be^{-1} ,\ga_1^{-1} ,\ldots ,\ga_m^{-1})$.
\end{Nota}
Under these notations, the condition (\ref{irred-1}) is equivalent to
\begin{align}
  \label{irred-2}
  \al-\prod_{k=1}^m \ga_k^{i_k} ,\quad \be-\prod_{k=1}^m \ga_k^{i_k} \neq 0 ,\qquad
  \forall I=(i_1,\dots,i_m) \in\ztwom.
\end{align}
For example, $\ga_k =1$ or $\al \be -(-1)^{m-1}\prod_{k=1}^m \ga_k =0$ are allowed.

Note that though in \cite{G-FC-monodromy} the indices $I$ run the subsets of $\{ 1,\dots ,m \}$,
in this paper we use $\ztwom$ as a set of indices.
The correspondence is given by
$$
\{ 1,\dots ,m \} \supset \{ i_1 ,\dots ,i_r \} \longleftrightarrow
e_{i_1}+\dots+e_{i_r}\in \ztwom ,
$$
where $e_k$ is the $k$-th unit vector of size $m$.
We put $|I|=\sum_{k=1}^m i_k$.

\subsection{System of differential equations}
Let $\pa_k \ (k=1 ,\dots , m)$ be the partial differential operator with respect to $x_k$.
We set $\theta_k =x_k \pa_k$, $\theta =\sum_{k=1}^m \theta_k$.
Lauricella's $ F_C (a,b,c;x)$ satisfies differential equations
$$
\left[ \theta_k (\theta_k+c_k-1)-x_k(\theta +a)(\theta +b)  \right] f(x)=0,
\quad k=1,\dots , m.
$$
The system generated by them is
called Lauricella's hypergeometric system $E_C (a,b,c)$ of
differential equations.
The system $E_{C} (a,b,c)$ is a holonomic system of rank $2^m$
with the singular locus $S$.
It is shown in \cite{HT} that
the system $E_C(a,b,c)$ is irreducible,
that is, the system $E_{C} (a,b,c)$ defines a maximal ideal in the ring of
differential operators with rational function coefficients,
if and only if the parameters $a,b,c_1,\dots,c_m$ satisfy
(\ref{irred-1}).

For an element $I=(i_1,\dots,i_m)$ of $\ztwom$, we set
\begin{align}
\label{series-sol}
F_I(x)=\frac{\prod_{k=1}^m\Ga((-1)^{i_k}(1-c_k))}
{\Ga(1-a^I)\Ga(1-b^I)}\cdot \prod_{k=1}^m x_k^{i_k(1-c_k)}
\cdot F_C(a^I,b^I,c^I;x),
\end{align}
where
\begin{align*}
\label{eq:abI}
  &a^I=a+\sum_{k=1}^m i_k(1-c_k),\quad
  b^I=b+\sum_{k=1}^m i_k(1-c_k), \\
  &c^I=(c_1+2i_1(1-c_1),\dots,c_m+2i_m(1-c_m)).
\end{align*}
Note that the assumption (\ref{irred-1}) is equivalent to
\begin{equation}
\label{irred-3}
a^I,b^I\notin \Z,
\qquad
  \forall I=(i_1,\dots,i_m)\in \ztwom,
\end{equation}
and that
$$
c_k+2i_k(1-c_k)=\left\{
\begin{matrix} c_k &\textrm{if}& i_k=0,\\
2-c_k &\textrm{if}& i_k=1.\\
\end{matrix}
\right.
$$
\begin{Ex}\label{ex-series-1}
  We give examples for $m=1$ (we put $c_1 =c$, $x_1=x$):
  \begin{align*}
    &F_{0}(x)=\frac{\Ga(1-c)}{\Ga(1-a)\Ga(1-b)}F_C(a,b,c;x)\\
    &=\frac{\Ga(1-c)\Ga(c)}{\Ga(1-a)\Ga(1-b)\Ga(a)\Ga(b)}\sum_{n=0}^\infty
    \dfrac{\Ga(a+n)\Ga(b+n)}{\Ga(c+n)\Ga(1+n)}x^n\\
    &=\frac{\sin(\pi a)\sin(\pi b)}{\pi\sin(\pi c)}\sum_{n=0}^\infty
    \dfrac{\Ga(a+n)\Ga(b+n)}{\Ga(c+n)\Ga(1+n)}x^n,\\

    &F_{1}(x)=\frac{\Ga(c-1)}{\Ga(c-a)\Ga(c-b)}
    x^{1-c}F_C(a+1-c,b+1-c,2-c;x)\\
    &=\frac{\Ga(c\!-\!1)\Ga(1\!-\!(c\!-\!1))}
    {\Ga(c\!-\! a)\Ga(c\!-\! b)\Ga(1\!-\! c\!+\! a)\Ga(1\!-\! c\!+\! b)}
    \sum_{n=0}^\infty \dfrac{\Ga(a\!-\! c\!+\!1\!+\!n)\Ga(b\!-\! c\!+\!1\!+\! n)}
    {\Ga(2\!-\! c\!+\! n)\Ga(1\!+\! n)}
    x^{n+1-c},\\
    &=-\frac{\sin(\pi(a-c))\sin(\pi(b-c))}
    {\pi\sin(\pi c)}
    \sum_{n'=1-c}^\infty \dfrac{\Ga(a\!+\!n')\Ga(b+\! n')}
    {\Ga(c\!+\! n')\Ga(1\!+\! n')}
    x^{n'} ,
  \end{align*}
  where $n'=n+1-c$ and
  $\sum_{n'=1-c}^\infty$ means the sum of $n'$ running
  over the set $1-c+\N=\{1-c+n\mid n\in \N\}$.
\end{Ex}
The functions $\{ F_{I}(x)\}_{I\in \ztwom}$
form a basis of the local solution space $\sol =\sol (a,b,c)$ to the system $E_C(a,b,c)$
around a point $\dot x$ in $D_C -S$
under conditions (\ref{irred-1}) and
\begin{align}
  \label{c-cond}
  c_1,\dots,c_m\notin \Z.
\end{align}
We set a (row) vector valued function
$$\bfF(x)=(\dots, F_{I}(x),\dots),$$
where $I\in \ztwom$ are aligned by the pure lexicographic order as
$$
(0,\dots,0),\ (1,0,\dots,0), \ (0,1,\dots,0),\  (1,1,\dots,0),\
(0,0,1,\dots,0),\
\dots, (1,\dots,1).
$$
Note that its entries have factors
$$1 ,\ x_1^{1-c_1},\ x_2^{1-c_2},\ x_1^{1-c_1}x_2^{1-c_2},\  x_3^{1-c_3},\
\dots,\  x_1^{1-c_1}x_2^{1-c_2}\cdots x_m^{1-c_m},$$
respectively.

\subsection{Monodromy representation}
\label{section-MR}
Put $\dot{x}=\left( \frac{1}{2m^2},\ldots ,\frac{1}{2m^2} \right) \in X$.
For $\rho \in \pi_1 (X,\dot{x})$ and $g \in \sol$,
let $\rho_* g$ be the analytic continuation of $g$ along $\rho$.
Since $\rho_* g$ is also a solution to $E_C (a,b,c)$,
the map $\rho_* :\sol \to \sol ;\ g\mapsto \rho_* g$ is
a $\C$-linear automorphism which satisfies
$(\rho \cdot \rho')_* =\rho'_* \circ \rho_*$ for $\rho,\rho' \in \pi_1 (X,\dot{x})$.
Here, the composition $\rho \cdot \rho'$ of loops $\rho$ and $\rho'$ is
defined as the loop going first along $\rho$,
and then along $\rho'$.
We thus obtain a representation
$$
\CM :\pi_1 (X,\dot{x}) \to GL(\sol)
$$
of $\pi_1 (X,\dot{x})$, where $GL(V)$ is
the general linear group on a $\C$-vector space $V$.
This representation $\CM$ is called the monodromy representation of $E_C(a,b,c)$.

Let $\rho_0, \rho_1 ,\ldots ,\rho_m$ be loops in $X$ so that
\begin{itemize}
\item $\rho_0$ turns the hypersurface $(R(x)=0)$ around the point
  $\left( \frac{1}{m^2},\ldots, \frac{1}{m^2} \right)$, positively,
\item $\rho_k \ (k=1,\dots , m)$ turns the hyperplane $(x_k=0)$, positively.
\end{itemize}
For explicit definitions of them, see \cite{G-FC-monodromy}.
\begin{Fact}[\cite{G-FC-monodromy}]\label{pi1}
  The loops $\rho_0 ,\rho_1 ,\ldots ,\rho_m$ generate
  the fundamental group $\pi_1 (X,\dot{x})$.
  Moreover, if $m\geq 2$, then they satisfy the following relations:
  \begin{align*}
    \rho_i \rho_j =\rho_j \rho_i \quad (i,j=1,\dots , m) ,\quad
    (\rho_0 \rho_k)^2 =(\rho_k \rho_0)^2 \quad (k=1,\dots , m).
  \end{align*}
\end{Fact}
In \cite{G-FC-monodromy}, $m+1$ linear maps
$\CM_i =\CM (\rho_i) \ (i=0,\dots , m)$ are investigated in terms of
twisted homology groups and the intersection form.
In this paper, we do not explain them.
What we need is the following fact.
\begin{Fact}[\cite{G-FC}]\label{solution-cycle}
  Suppose (\ref{irred-1}) and (\ref{c-cond}).
  \begin{enumerate}[(i)]
  \item By integration, the twisted homology group is isomorphic
    to the solution space $\sol$.
  \item We can construct twisted cycles $\{\De_I\}_I$ that correspond to $\{F_I\}_I$.
  \item The intersection matrix $H=\left( H_{I,I'} \right)_{I,I'}$
    with respect to the basis $\{\De_I\}_I$ is
    diagonal, and its $(I,I)$-entry is
  \begin{align*}
    H_{I,I}
    =\prod_{k=1}^m \frac{(-1)^{i_k} \ga_k^{1-i_k}}{\ga_k-1}
    \cdot \frac{( \al -\prod_{k=1}^m \ga_k^{i_k}) (\be -\prod_{k=1}^m \ga_k^{i_k} ) }
    {( \al -\prod_{k=1}^m  \ga_k ) (\be -1)} .
  \end{align*}  \end{enumerate}
\end{Fact}
By using this fact and properties of the intersection form,
we can induce the intersection form on $\sol$.

\begin{Def}
  \label{def-intersection-form}
  We assume (\ref{irred-1}) and (\ref{c-cond}).
  We define a bilinear form
 (called the intersection form)
  \begin{align*}
    \CI (\cdot ,\cdot ):\sol \times \sol \to \C (\al ,\be ,\ga)
  \end{align*}
  as follows.
  For any $F(x),G(x) \in \sol$, we express them as
  linear combinations of the basis $\{ F_I \}_I$:
  \begin{align*}
    F(x)=\bfF(x)\cdot \bff ,\quad G(x)=\bfF(x)\cdot \bfg,
    \qquad \bff,\bfg \in \C(\al,\be,\ga)^{2^m} ,
  \end{align*}
  and define
  \begin{align*}
    \CI(F(x),G(x))=\tp \bff \cdot H \cdot \bfg^{\vee} .
  \end{align*}
\end{Def}

\begin{Rem}
\label{rem-zure}
Let $\CI_H$ be the intersection form  on $\sol$ induced from that on
the twisted homology group by the isomorphism
in Fact \ref{solution-cycle} (i).
The intersection form $\CI$ coincides with $\CI_H$
modulo a constant multiple which never vanishes
under the conditions (\ref{irred-1}) and (\ref{c-cond}).
\end{Rem}

\begin{Cor}
  \label{cor-intersection-form}
  Under the conditions (\ref{irred-1}) and (\ref{c-cond}),
  the intersection form  $\CI$ is a monodromy invariant form, that is,
  for any loops $\rho \in \pi_1 (X,\dot{x})$, we have
  \begin{align*}
    \CI(\CM(\rho)(F(x)),\CM(\rho)(G(x)))=\CI(F(x),G(x)) .
  \end{align*}
  In other words, $H$ satisfies
  \begin{align*}
    \tp M_{\rho} \cdot H \cdot M_{\rho}^{\vee} =H ,
  \end{align*}
  where $M_{\rho}$ is the representation matrix of $\CM(\rho)$
  with respect to the basis $\{ F_I\}_I$.
\end{Cor}

Let $M_i$ be the representation matrix of $\CM_i$ $(i=0,\dots , m)$
with respect to the basis $\{ F_I\}_I$.
We give explicit expressions of them.
\begin{Fact}[\cite{G-FC-monodromy}]\label{fact:rep-mat-1}
  We assume (\ref{irred-1}), (\ref{c-cond}) and
  $\la=(-1)^{m-1} \al^{-1} \be^{-1} \prod_{k=1}^m \ga_k \neq 1$.
  For $k=1,\dots , m$,
  the representation matrix $M_k$ is diagonal,
  and its $(I,I)$-entry is $\ga_k^{-i_k}$.
  The representation matrix $M_0$ is expressed as
$$
  M_0 =E_{2^m} -\frac{1-\la}{\tp \one \cdot H  \cdot\one}
  \cdot \one \cdot \tp \one \cdot H
  =E_{2^m} -\frac{(\be-1)(\al -\prod_{k=1}^m \ga_k )}{\al \be}
  \cdot \one \cdot \tp \one \cdot H ,
$$
  where $E_{2^m}$ is the unit matrix of size $2^m$,
  $\one$ is the column vector of size $2^m$ with all entries $1$, and
  $H$ is the intersection matrix given in Fact \ref{solution-cycle}.
\end{Fact}
\begin{Rem}\label{rem-eigen}
  These expressions are obtained from consideration to eigenvectors of each
  $\CM_i \in GL(\sol )$.
  \begin{enumerate}[(i)]
  \item $\CM_k$ ($k=1,\ldots,m$);
    $F_I$ is an eigenvector of eigenvalue $\ga_k^{-1}$ (resp. $1$)
    if $i_k=1$ (resp. $i_k=0$), where $I=(i_1,\ldots ,i_m)$.
  \item $\CM_0$;
    the eigenvalues of $\CM_0$ are
    $\la$ and $1$.
    The eigenspace of eigenvalue
   $\la$ is one-dimensional and spanned by
    $$
    f_0 = \sum_{I \in \ztwom} F_I ,
    $$
    which corresponds to $\one$ when we take the basis $\{F_I\}_I$.
    The eigenspace of $\CM_0$ of eigenvalue $1$ is characterized as
    $\{ g \in \sol \mid \CI (g,f_0 )=0 \}$.

  \item The first expression of $M_0$ is stable under the non-zero scalar
    multiple to $H$.
  \end{enumerate}
\end{Rem}

\section{Another basis}
In fact, $\{F_I \}$ does not form a basis of $\sol$ when $c_i$'s are integers.
In this section, we introduce another basis $\{ \tF_I \}_I$
which is a well-defined basis even if $c_i$'s are integers,
and we give the circuit matrices with respect to this basis.
Note that these do not coincide with
solutions obtained by integrating the twisted cycles defined in \cite[\S 6]{G-FC-monodromy}

\subsection{Basis of $\sol$}
First, we construct a basis of $\sol$.
\begin{Lem}
\label{lem:scalar-mul}
Let $I=(i_1,i_2\dots,i_m)$ be any element of $\ztwom$ and
$p=(p_1,p_2,\dots,p_m)$ be any element of $\Z^m$.
Then the limit function
$$\lim_{c\to p}(\ga_1-1)(\ga_2-1)\cdots(\ga_m-1)F_{I}(x)$$
is well-defined and not identically zero.
\end{Lem}
\begin{proof}
We have only to note that $F_{I}(x)$ has the factor
$$\prod_{k=1}^m \Ga(c_k-i_k)\Ga(1-(c_k-i_k))
=\prod_{k=1}^m\frac{(-1)^{i_k}\pi}{\sin(\pi c_k)},
$$
which cancels out $(\ga_1-1)(\ga_2-1)\cdots(\ga_m-1)$.
\end{proof}

\begin{Lem}
\label{lem:wa}
Let $I=(i_1,\dots,\overset{k\textrm{-th}}{0},\dots,i_m)$ and
$I'=(i_1,\dots,\overset{k\textrm{-th}}{1},\dots,i_m)$
be elements of $\ztwom$ and $p_k$ be an integer.
Then the limit function
$$\lim_{c_k\to p_k}
(\ga_m-1)\cdots(\ga_{k+1}-1)(\ga_{k-1}-1)\cdots(\ga_1-1)
(F_{I}(x)+F_{I'}(x))$$
is well-defined and not identically zero.
\end{Lem}

\begin{proof}
We show the case $m=1$ (we put $p_1=p$). Firstly, we assume $p=1$.
By Example \ref{ex-series-1},
\begin{align*}
F_{0}(x)
&=\frac{s_0}{\sin(\pi c)}\sum_{n=0}^\infty
\dfrac{\Ga(a+n)\Ga(b+n)}{\Ga(c+n)\Ga(1+n)}x^n,\\
F_{1}(x)
&=-\frac{s_1(c)}{\sin(\pi c)}
\sum_{n'=1-c}^\infty \dfrac{\Ga(a+n')\Ga(b+ n')}
{\Ga(c+ n')\Ga(1+ n')}x^{n'},
\end{align*}
where $n'=n+1-c$, $s_0=\dfrac{\sin(\pi a)\sin(\pi b)}{\pi}$,
$s_1(c)=\dfrac{\sin(\pi(a-c))\sin(\pi(b-c))}{\pi}$.
Both functions $\sin (\pi c) F_0(x)$ and $-\sin (\pi c) F_1(x)$
converge to
$$
\sum_{n=0}^\infty
s_0\dfrac{\Ga(a+n)\Ga(b+n)}{\Ga(1+n)\Ga(1+n)}x^n
=\sum_{n=0}^\infty A_nx^n$$
as $c\to p=1$.
Apply l'H\^opital's rule to the function
\begin{align*}
F_0(x)+F_1(x)
=\dfrac{1}{\sin(\pi c)}\cdot
&\Bigg[\sum_{n=0}^\infty
\big(s_0\dfrac{\Ga(a+ n)\Ga(b+ n)}{\Ga(c+ n)\Ga(1+ n)}- A_n \big)x^n\\
&
-\sum_{n'=1-c}^\infty
\Big\{\big(
s_1(c)\dfrac{\Ga(a+n')\Ga(b+ n')}
{\Ga(c+ n')\Ga(1+ n')}- A_n\big)x^{n'}
- A_n(x^n-x^{n'}\big)\Big\}
\Bigg]
\end{align*}
to verify that its limit as $c\to p=1$ exists, where $n=n'+c-1$
in the second sum. Note that
$\lim\limits_{c\to 1} \frac{A_n(x^n-x^{n'})}{\sin(\pi c)}$
yields the factor $\log x$.

Secondly, we assume $p\ge 2$.
In this case, the sum $\sum\limits_{n'=1-c}$ has negative terms
for $n'=1-p,\dots,-1$ as $c\to p$.
Since $\lim\limits_{c\to p} \dfrac{1}{\sin(\pi c)}\cdot\dfrac{1}{\Ga(n'+1)}$
converges to a non-zero value, these negative terms are well-defined.
Thus $F_0(x)+F_1(x)$ consists of these finite terms and
the infinite sum considered in the case $p=1$.

Thirdly, we assume $p\le 0$.
By regarding $n$ as $n+p$ in the sums of $F_0(x)$ and $F_1(x)$,
we can show that
$F_0(x)+F_1(x)$ is well-defined and not identically zero as in the previous
consideration.

For a general $m$, use a similar argument
by regrading the variables except $x_k$ as constants.
Note that the limit function has the factor $\log x_k$  coming from
$\lim\limits_{n_k\to n'_k} \dfrac{x_k^{n_k}-x_k^{n'_k}}{n_k-n_k'}$.
\end{proof}

We define the tensor product $A\otimes B$ of matrices
$A$ and $B=(b_{ij})_{\substack{1\le i\le r\\ 1\le j\le s}}$ as
$$A\otimes B=\begin{pmatrix}
A\, b_{11} & A\, b_{12} &\cdots & A\, b_{1s} \\
A\, b_{21} & A\, b_{22} &\cdots & A\, b_{2s} \\
\vdots  &\vdots   &\ddots & \vdots  \\
A\, b_{r1} & A\, b_{r2} &\cdots & A\, b_{rs} \\
\end{pmatrix}.
$$
We remark that this is different from the usual definition.
We fix the number $m$ of variables.
We set
$$
G_k=\begin{pmatrix}
1 & 0 \\
0 & \ga_k^{-1}
\end{pmatrix},\quad
Q_k=
\begin{pmatrix}
1- \ga_k& 1 \\
0 & 1
\end{pmatrix},
$$
for $k=1,\dots, m$ and
\begin{align*}
  P_m=Q_1\otimes Q_2\otimes \cdots \otimes  Q_m.
\end{align*}
By using these notations,
the matrices $M_1,\dots ,M_m$ given in Fact \ref{fact:rep-mat-1}
is expressed as
\begin{align*}
  M_k=E_2 \otimes \cdots \otimes E_2 \otimes \underset{k\textrm{-th}}{G_k}
  \otimes E_2 \otimes \cdots \otimes E_2
  \quad (k=1,\ldots ,m)
\end{align*}
where $E_2$ is the unit matrix of size $2$.
For example,
we have
$$
M_1=G_1=\begin{pmatrix}
1 & 0 \\
0 & \ga_1^{-1}
\end{pmatrix},\quad
P_1=Q_1=
\begin{pmatrix}
1- \ga_1& 1 \\
0 & 1
\end{pmatrix}
$$
in the case $m=1$, and
\begin{align*}
  M_1=\begin{pmatrix}
    1 & 0 & 0 &0 \\
    0 & \ga_1^{-1}& 0 &0\\
    0 & 0 &1 & 0 \\
    0 & 0 &0 &\ga_1^{-1}
  \end{pmatrix},\quad
  M_2=\begin{pmatrix}
    1 & 0 & 0 &0 \\
    0 & 1 & 0 &0\\
    0 & 0 &\ga_2^{-1} & 0 \\
    0 & 0 &0 &\ga_2^{-1}
  \end{pmatrix},
  \\
  P_2=Q_1\otimes Q_2=\begin{pmatrix}
    (1-\ga_1)(1-\ga_2)& 1-\ga_2 & 1-\ga_1 & 1 \\
    0                & 1-\ga_2 & 0       & 1 \\
    0                & 0       & 1-\ga_1 & 1 \\
    0                & 0       & 0       & 1
  \end{pmatrix}
\end{align*}
in the case $m=2$.
Note that
$$
P_m=\begin{pmatrix}
P_{m-1}(1-\ga_m) & P_{m-1}\\
O & P_{m-1}\\
\end{pmatrix},
$$
where $O$ is the square zero matrix of size $2^{m-1}$.
We have
\begin{align}
  \label{eq:detPm}
  \det(P_m)=\prod_{k=1}^m(1-\ga_k)^{2^{m-1}},
\end{align}
since $\det(P_1)=1-\ga_1$ and
$$\det(P_m)=\det((1-\ga_m)P_{m-1})\det(P_{m-1})
=(1-\ga_m)^{2^{m-1}}\det(P_{m-1})^2.$$
We use a new basis given by
$$
\tbfF(x)=(\dots, \tF_I(x),\dots)=\bfF(x)\cdot P_m.
$$
The vector-valued function $\tbfF(x)$ takes the form
$$
\Big( (1-\ga_1)F_0(x),F_0(x)+F_1(x) \Big)
$$
for $m=1$, and the form
\begin{align*}
  \Big( &(1-\ga_1)(1-\ga_2)F_{00}(x),
  (1-\ga_2)(F_{00}(x)+F_{10}(x)), \\
  &(1-\ga_1)(F_{00}(x)+F_{01}(x)),
  F_{00}(x)+F_{10}(x)+F_{01}(x)+F_{11}(x) \Big)
\end{align*}
for $m=2$.
\begin{Th}\label{th-sol-basis}
The vector-valued function $\tbfF(x)$ gives a basis of the space $\sol$ of the local solutions
to $E_C(a,b,c)$ around $\dot x$ even in cases $c_k\in \Z$ $(k=1,\dots,m)$.
\end{Th}
\begin{proof}
The entries of $\tbfF(x)$ consist of
\begin{align*}
\Big[\prod_{k=1}^m (1-\ga_k)\Big] &F_{0\dots 0}, \\
\Big[\prod_{1\le k\le m}^{k\ne i}(1-\ga_k)\Big]
&(F_{0\dots 0}+F_{e_i}), \\
\Big[\prod_{1\le k\le m}^{k\ne i,j} (1-\ga_k)\Big]
&(F_{0\dots 0}+F_{e_i}+F_{e_j}+F_{e_i+e_j}),
\\
& \vdots \\
(1-\ga_k)&\sum_{I\in \ztwom} (1-i_k)F_{I}, \\
&\sum_{I\in \ztwom}F_{I},
\end{align*}
where $e_k$ is the $k$-th unit vector of size $m$
and $I=(i_1,\dots,i_m)$.
The functions in the first and second lines are well-defined by Lemmas
\ref{lem:scalar-mul} and \ref{lem:wa}.
Since the functions
\begin{align*}
&\Big[\prod_{1\le k\le m}^{k\ne i}(1-\ga_k)\Big]
\big[
(F_{0\dots 0}+F_{e_i})+(F_{e_j}+F_{e_i+e_j})\big],\\
&\Big[\prod_{1\le k\le m}^{k\ne j}(1-\ga_k)\Big]\big[
(F_{0\dots 0}+F_{e_j})+(F_{e_j}+F_{e_i+e_j})\big]
\end{align*}
are well-defined by Lemma \ref{lem:wa}, the function
$$
\Big[\prod_{1\le k\le m}^{k\ne i,j}(1-\ga_k)\Big]
(F_{0\dots 0}+F_{e_i}+F_{e_j}+F_{e_i+e_j})$$
is also well-defined. In this way, we can show that
the entries of $\tbfF(x)$ are well-defined
even in cases $c_k\in \Z$. In the case $c_k\in \Z$, the functions $\tF_I(x)$
has the factor $\log x_k$,
if $i_k=1$.
This implies that $\tF_I(x)$'s are also linearly independent
in such a case.
\end{proof}

\subsection{Representation matrices and the intersection matrix}
\label{section-rep-mat}
Next, we consider the representation matrices of $\CM_i$'s and
the intersection matrix with respect to the new basis $\{\tF_I \}_I$.

In the below discussion, we often use the following equality
which is shown by a straightforward calculation:
\begin{align}
  \label{eq:sum}
  &\sum_{J\leq I} (-1)^{|J|}\left( \alpha \beta \prod_{k=1}^m \ga_k^{1-j_k}
    + \prod_{k=1}^m \ga_k^{1+j_k} \right) \\
  \nonumber
  &=\left( \alpha \beta +(-1)^{|I|}\prod_{k=1}^m \ga_k^{i_k} \right)
  \prod_{k=1}^m \left( \ga_k^{1-i_k}(\ga_k-1)^{i_k} \right) ,
\end{align}
where $I=(i_1,\dots ,i_m)$, $J=(j_1,\dots ,j_m)$ and
we define a partial order $\leq$ on $\ztwom$ by
$$
J\leq I \Longleftrightarrow j_k \leq i_k, \quad k=1,\dots , m.
$$
\begin{Cor}\label{cor:rep-mat}
  Let $\tM_i$ be the representation matrix of $\CM_i$ $(i=0, \dots ,m)$
  with respect to the basis $\{ \tF_I\}_I$.
  For $k=1,\dots, m$, we have
  \begin{align*}
    \tM_k =E_2 \ot \cdots \ot E_2 \ot \underset{k\textrm{-th}}{\tG_k} \ot E_2 \ot \cdots \ot E_2 ,
    \quad
    \tG_k =\begin{pmatrix}1 & -\ga_k^{-1} \\ 0 & \ga_k^{-1} \end{pmatrix}.
  \end{align*}
  $\tM_0$ is written as
  \begin{align*}
    \tM_0 =E_{2^m} -N_0 ,\quad
    N_0 =\tp (\zero, \dots ,\zero ,\bfv) ,
  \end{align*}
  where $\bfv \in \C^{2^m}$ is a column vector whose $I$-th entry is
  \begin{align*}
    \left\{
      \begin{array}{ll}
        (-1)^m \frac{(\al-1)(\be-1)\prod_{k=1}^m \ga_k}{\al \be}&
       (I=(0,\dots ,0)) ,\\
        (-1)^{m+|I|} \frac{(\al \be +(-1)^{|I|} \prod_{k=1}^m \ga_k^{i_k})
          \prod_{k=1}^m \ga_k^{1-i_k}}{\al \be}& (I\neq(0,\dots ,0)) .
      \end{array}
    \right.
  \end{align*}
Under the condition (\ref{irred-1}) without assuming  (\ref{c-cond}),
$\tM_0 ,\dots ,\tM_m$ are valid.
\end{Cor}

\begin{proof}
  By the definition of $\{ \tF_I\}_I$, we have $\tM_i =P_m^{-1} M_i P_m$.
  The first claim follows from $Q_k^{-1} G_k Q_k =\tG_k$ and
  the expressions of $M_k$ and $P_k$ as tensor products.

  We show the second claim.
  Note that
  since all of the entries of the $2^m$-th column
  (we also say the $(1,\dots ,1)$-th column)
  of $P_m$ are $1$, we have $P_m \ev =\one$
  and hence $\ev$ is an eigenvector of $\tM_0$ of eigenvalue
  $(-1)^{m-1} \al^{-1} \be^{-1} \prod_k \ga_k$,
  where $\ev =\tp (0,\dots ,0 ,1)\in \C^{2^m}$.
  By $\tM_0 =P_m^{-1} M_0 P_m$ and Fact \ref{fact:rep-mat-1},
  $\tM_0 -E_{2^m}$ should be
  \begin{align*}
    &\frac{(\be-1)(\al -\prod_{k=1}^m \ga_k )}{\al \be} P_m^{-1}
    \cdot \one \cdot \tp \one \cdot H P_m
    = \frac{(\be-1)(\al -\prod_{k=1}^m \ga_k )}{\al \be} \ev \tp \one H P_m \\
    &=\frac{(\be-1)(\al -\prod_{k=1}^m \ga_k )}{\al \be}
    \tp (\zero, \dots ,\zero ,\bfh) \cdot P_m ,
  \end{align*}
  where $\bfh \in \C^{2^m}$ is a column vector whose $I$-th entry is $H_{I,I}$.
  It is sufficient to show that
  $$
  \frac{(\be-1)(\al -\prod_{k=1}^m \ga_k )}{\al \be}
  \tp \bfh \cdot P_m
  =\tp \bfv .
  $$
  The $I$-th entry of the left-hand side is equal to
  \begin{align*}
    &\frac{(\be-1)(\al -\prod_{k=1}^m \ga_k )}{\al \be}
    \sum_{J\leq I} \left( H_{J,J} \cdot \prod_{k=1}^m (1-\ga_k)^{1-i_k} \right) \\

    &=\frac{(-1)^{m}}{\al \be \prod_{k=1}^m (1-\ga_k)^{i_k}}
    \sum_{J\leq I} \left( (-1)^{|J|}
      \cdot \prod_{k=1}^m  \ga_k^{1-j_k}
      \cdot \Big( \al -\prod_{k=1}^m \ga_k^{j_k} \Big)
      \Big(\be -\prod_{k=1}^m \ga_k^{j_k} \Big)  \right) .
  \end{align*}
  If $I=(0,\dots ,0)$, then this is the $(0,\dots ,0)$-th entry of $\bfv$.
  If we assume $I\neq (0,\dots ,0)$,
  then it equals to
  \begin{align*}
    &\frac{(-1)^{m}}{\al \be \prod_{k=1}^m (1-\ga_k)^{i_k}}
    \sum_{J\leq I} (-1)^{|J|}
    \left( \alpha \beta \prod_{k=1}^m \ga_k^{1-j_k}
      + \prod_{k=1}^m \ga_k^{1+j_k} \right) \\
    &=\frac{(-1)^{m}}{\al \be \prod_{k=1}^m (1-\ga_k)^{i_k}}
    \left( \alpha \beta +(-1)^{|I|}\prod_{k=1}^m \ga_k^{i_k} \right)
    \prod_{k=1}^m \left( \ga_k^{1-i_k}(\ga_k-1)^{i_k} \right)
  \end{align*}
  by (\ref{eq:sum}),
  and this coincides with the $I$-th entry of $\bfv$.
\end{proof}

\begin{Lem}
  \label{lem-indep}
  $2^m$ vectors
  \begin{align*}
    \left( \prod_{k=1}^m \tM_k^{i_k} \right) \cdot \ev= \tM_1^{i_1} \tM_2^{i_2} \cdots \tM_m^{i_m}\ev
    \quad (I= ( i_1 ,\ldots i_m ) \in \ztwom )
  \end{align*}
  are linearly independent.
  In other words, actions $\CM_1,\dots,\CM_m$ on $f_0$ give a basis of the whole space $\sol$.
\end{Lem}
\begin{proof}
  It is sufficient to show that the $2^m \times 2^m$ matrix
  $$
  (\ev, \tM_1 \ev, \tM_2 \ev , \tM_1 \tM_2 \ev ,\tM_3 \ev ,\ldots , \tM_1 \cdots \tM_m \ev )
  $$
  is invertible.
  We calculate its determinant.
  Because of $\tM_k=P_m^{-1}M_k P_m$ and $\ev=P_m^{-1}\one$, this matrix equals to
  \begin{align}
    \label{eq-basis-matrix}
    P_m^{-1} \cdot (\one, M_1 \one, M_2 \one, M_1 M_2 \one ,M_3 \one,\ldots ,M_1 \cdots M_m \one ) .
  \end{align}
  By the alignment of the indices set, the right side of this product is
  \begin{align*}
    \begin{pmatrix}
      1 & 1 \\ 1 & \ga_1^{-1}
    \end{pmatrix}
    \ot
    \begin{pmatrix}
      1 & 1 \\ 1 & \ga_2^{-1}
    \end{pmatrix}
    \ot \cdots \ot
    \begin{pmatrix}
      1 & 1 \\ 1 & \ga_m^{-1}
    \end{pmatrix} ,
  \end{align*}
  and its determinant is $\prod_{k=1}^m (\ga_k^{-1}-1)^{2^{m-1}}$.

  By (\ref{eq:detPm}), the determinant of (\ref{eq-basis-matrix}) is equal to
  $\prod_{k=1}^m \ga_k^{-2^{m-1}}$,
  which is not zero.

\end{proof}

\begin{Prop}
  \label{lem-intersection}
  Let $\tH=\tp P_m H P_m^{\vee}$, which represents
  the intersection form $\CI$ with respect to the basis $\{ \tF_I\}_I$.
  Then $\tH$
  is well-defined, and its determinant is
  \begin{align*}
    \det(\tH)=\frac{1}{(\al -\prod _k  \ga_k )^{2^m}(\be -1)^{2^m}}
    \cdot \prod_{I\in \ztwom}\left(\al-\prod_{k=1}^m \ga_k^{i_k}\right)
    \left(\be-\prod_{k=1}^m \ga_k^{i_k}\right).
  \end{align*}
  In particular, $\tH$ is non-degenerate
  even in cases $c_k\in \Z$ $(k=1,\dots,m)$.

\end{Prop}
\begin{proof}
  First, we show the well-definedness.
  For $I=(i_1,\dots ,i_m)$, $I'=(i'_1,\dots ,i'_m)$, we put
  $$
  I\cdot I'=(i_1 i'_1 ,\dots ,i_m i'_m )\in \ztwom .
  $$
  Since $H$ is diagonal, the $(I,I')$-entry of $\tH$ is
  \begin{align*}
    &\tH_{I,I'} \\
    &=\prod_{k=1}^m (1-\ga_k)^{1-i_k} (1-\ga_k^{-1})^{1-i'_k}
    \sum_{J\leq I, J\leq I'} \left( \prod_{k=1}^m \frac{(-1)^{j_k} \ga_k^{1-j_k}}{\ga_k-1}
      \cdot \frac{( \al -\prod_{k=1}^m \ga_k^{j_k}) (\be -\prod_{k=1}^m \ga_k^{j_k} ) }
      {( \al -\prod_{k=1}^m  \ga_k ) (\be -1)} \right) \\
    &=\frac{\prod_{k=1}^m (-1)^{i'_k} \ga_k^{i'_k-1}(1-\ga_k)^{1-i_k-i'_k}}
    {( \al -\prod_{k=1}^m  \ga_k ) (\be -1)}
    \sum_{J\leq I \cdot I'}  \left( (-1)^{|J|} \prod_{k=1}^m \ga_k^{1-j_k}
      \cdot \Big( \al -\prod_{k=1}^m \ga_k^{j_k} \Big) \Big(\be -\prod_{k=1}^m \ga_k^{j_k} \Big) \right) .
  \end{align*}
  If $I\cdot I'=(0,\dots ,0)$, then $1-i_k-i'_k \geq 0\ (k=1,\dots ,m)$, and hence
  \begin{align*}
    \tH_{I,I'}
    =\prod_{k=1}^m (-\ga_k)^{i'_k}(1-\ga_k)^{1-i_k-i'_k}
    \cdot \frac{\al -1}{\al -\prod_{k=1}^m  \ga_k}
  \end{align*}
  is well-defined.
  If $I\cdot I' \neq (0,\dots ,0)$, the same calculation as the proof of
  Corollary \ref{cor:rep-mat} shows
  \begin{align*}
    &\tH_{I,I'}\\
    &=\frac{\prod_{k=1}^m (-1)^{i'_k} \ga_k^{i'_k-1}(1-\ga_k)^{1-i_k-i'_k}}
    {( \al -\prod_{k=1}^m  \ga_k ) (\be -1)}
    \left( \alpha \beta +(-1)^{|I\cdot I'|}\prod_{k=1}^m \ga_k^{i_k i'_k} \right)
    \prod_{k=1}^m \left( \ga_k^{1-i_k i'_k}(\ga_k-1)^{i_k i'_k} \right) \\
    &=\frac{\alpha \beta +(-1)^{|I\cdot I'|}\prod_{k=1}^m \ga_k^{i_k i'_k}}
    {( \al -\prod_{k=1}^m  \ga_k ) (\be -1)}
    \cdot \prod_{k=1}^m (-\ga_k)^{i'_k(1-i_k)} (1-\ga_k)^{(1-i_k)(1-i'_k)},
  \end{align*}
  and we can see that
  its denominator does not vanish.

  Next, we evaluate $\det (\tH)$.
  Straightforward calculation and (\ref{eq:detPm}) show
  \begin{align*}
    \det (H) &= (-1)^{m 2^{m-1}} \cdot
    \frac{\prod_{k=1}^m \ga_k^{2^{m-1}}
      \cdot \prod_I( \al -\prod_{k=1}^m \ga_i^{i_k}) (\be -\prod_{k=1}^m \ga_i^{i_k} ) }
    {\prod_{k=1}^m (\ga_k -1)^{2^m} \cdot ( \al -\prod_{k=1}^m \ga_k )^{2^m} (\be -1)^{2^m}} , \\
    \det (P_m^{\vee})
    &=\prod_{k=1}^m \left( \ga_k^{-2^{m-1}}(\ga_k-1)^{2^{m-1}} \right) .
  \end{align*}
  We thus have
  \begin{align*}
    \det (\tH )=\frac{1}{(\al -\prod_{k=1}^m \ga_k )^{2^m}(\be -1)^{2^m}}
    \cdot \prod_{I\in \ztwom}\left(\al-\prod_{k=1}^m \ga_k^{i_k}\right)
    \left(\be-\prod_{k=1}^m \ga_k^{i_k}\right) ,
  \end{align*}
  and it is not zero under the condition (\ref{irred-2}).
\end{proof}
By this proposition, we can relax the condition to define
the intersection from on $\sol$.
\begin{Cor}
  By using $\{ \tF_I\}_I$ and $\tH$, the intersection form
  $\CI :\sol \times \sol \to \C (\al ,\be ,\ga)$
  in Definition \ref{def-intersection-form}
  can be extended even in cases $c_k\in \Z$ $(k=1,\dots,m)$.
\end{Cor}

\begin{Lem}
  \label{lem-eigenspace}
  The eigenspace of $\tM_0$ with eigenvalue $1$ is expressed as
  \begin{align*}
    \ker N_0= \ker \tp \bfv .
  \end{align*}
\end{Lem}
\begin{proof}
  This is obvious because of the expression
  \begin{align*}
    \tM_0 =E_{2^m}-N_0=E_{2^m}- \tp (\zero,\ldots ,\zero ,\bfv ) .
  \end{align*}
\end{proof}

\begin{Rem}
  If the eigenvalue $(-1)^{m-1}\al^{-1}\be^{-1}\prod_{k=1}^m\ga_k$ of $\tM_0$ coincides with $1$,
  then the $(1,\dots ,1)$-th entry of $\bfv$ is zero and
  $\ev$ also belongs to this eigenspace $\ker N_0$.
\end{Rem}

\begin{Lem}
  \label{lem-orthogonal}
  $$
  \ker N_0

  =\{ \bfw\in \C^{2^m} \mid \tp \bfw \tH \ev =0\}.
  $$
\end{Lem}
\begin{proof}
  This is also obvious because of the orthogonality of the eigenspaces (Remark \ref{rem-eigen} (ii))
  and the definition of $\tH$.
\end{proof}

\section{Irreducibility}
We restate the main theorem and give its proof.
\begin{Th}
  The monodromy representation
  $$
  \CM : \pi_1 (X,\dot{x}) \to GL(\sol )
  $$
  is irreducible under the condition (\ref{irred-1}).
\end{Th}
\begin{proof}
  By Theorem \ref{th-sol-basis},
  it is sufficient to consider the matrix representation by $\tM_i$ under
  the isomorphism $\sol \simeq \C^{2^m}$.
  Let $W\subset \C^{2^m}$ be an invariant subspace.
  \begin{enumerate}
    \item[(i)] First, we suppose $W \not\subset \ker N_0$.
      We take $\bfw\in W$ such that $N_0 \bfw \neq \zero$.
      By the definition of $N_0$, the image of $N_0$ is spanned by $\ev$.
      Thus $N_0 \bfw \neq \zero$ implies that
      there exists $\mu \neq 0$ such that $N_0 \bfw =\mu \ev$.
      We obtain
      \begin{align*}
        \ev=\frac{1}{\mu}N_0 \bfw =\frac{1}{\mu} (\bfw-\tM_0 \bfw) \in W.
      \end{align*}
    By Lemma \ref{lem-indep}, the $2^m$ vectors
    \begin{align*}
      \left( \prod_{k=1}^m \tM_k^{i_k} \right) \cdot \ev\in W
      \qquad (I= ( i_1 ,\ldots i_m ) \in \ztwom )
    \end{align*}
    are linearly independent.
    This implies $W=\C^{2^m}$.
    \item[(ii)] Next, we suppose $W \subset \ker N_0$.
      We fix an arbitrary $\bfw\in W$.
      Since $W$ is an invariant subspace, we have
      \begin{align*}
        (\tM_1^{i_1})^{-1}(\tM_2^{i_2})^{-1}\cdots(\tM_m^{i_m})^{-1}\bfw \in W \subset \ker N_0
      \end{align*}
      for any $I=(i_1 ,\ldots i_m ) \in \ztwom$.
      By the monodromy invariance $\tp \tM_i \tH \tM_i^{\vee}=\tH$ of
      the intersection matrix $\tH$,
      commutativity between $\tM_1, \tM_2,\ldots ,\tM_m$, and
      Lemma \ref{lem-orthogonal}, we obtain
      \begin{align*}
        &\tp \bfw \tH (\tM_1^{i_1} \tM_2^{i_2} \cdots \tM_m^{i_m} \ev)^{\vee} \\
        &=\tp \bfw \tp ((\tM_1^{i_1})^{-1}(\tM_2^{i_2})^{-1}\cdots(\tM_m^{i_m})^{-1}) \tH \ev^{\vee} \\
        &=\tp ((\tM_1^{i_1})^{-1}(\tM_2^{i_2})^{-1}\cdots(\tM_m^{i_m})^{-1} \bfw) \tH \ev^{\vee} =0 .
      \end{align*}
      The linear independence of $\{\tM_1^{i_1} \tM_2^{i_2} \cdots \tM_m^{i_m}\ev\}_I$ and
      $\det (\tH ) \neq 0$ (Proposition \ref{lem-intersection}) means that $\bfw=0$.
      We thus have $W=0$.
    \end{enumerate}
  Therefore, the invariant subspaces should be the trivial ones.
\end{proof}

\section{Reducibility}\label{section-red}
Recall that our irreducibility assumption (\ref{irred-1}) consists of
$2^{m+1}$ conditions for parameters.
In this section, we show that if one of them is not satisfied
then the monodromy representation $\CM$ of $E_C(a,b,c)$ is reducible.
More precisely, we have the following theorem.

\begin{Th}
\label{th-reducible}
Suppose that there exists $I=(i_1,\dots,i_m)\in \ztwom$ such that
$a^I\in \Z$, $b^I\notin \Z$ or  $a^I\notin \Z$, $b^I\in \Z$.
If $a^{I'},b^{I'}\notin \Z$ for any $I'\in \ztwom$ different from $I$,
then the monodromy representation $\CM$ of $E_C(a,b,c)$ is reducible, that is,
there exists a non-trivial subspace in $\sol$ invariant under $\CM$.
\end{Th}

\begin{proof}
We fix $I=(i_1,\dots,i_m)\in \ztwom$ and
assume that $a^I\in \Z,\ b^I\notin \Z.$
If there exists $j$ such that $c_j\in \Z$ then
$$
\left\{
\begin{matrix}
  a^{I+e_j}=a^I+ (1-c_j)\in \Z& \textrm{if} & i_j=0,\\
  a^{I-e_j}=a^I- (1-c_j)\in \Z& \textrm{if} & i_j=1,
\end{matrix}
\right.
$$
which contradicts to the assumption.
Thus we have $c_1,\dots,c_m\notin \Z$ and
the solutions $F_{I'}(x)$ in (\ref{series-sol}) for $I'(\ne I)$ are valid.
Note that these $2^m-1$ solutions are linearly independent.

Hereafter, we regard $a^I$ as an indeterminant,
and consider two cases:
\begin{enumerate}[(i)]
\item $a^I$ approaches to a negative integer $-L$;
\item $a^I$ approaches to a non-negative integer $L'$.
\end{enumerate}
To prove the reducibility, we find a non-trivial
invariant subspace in each case.
\begin{enumerate}[(i)]
\item
Note that the solution $F_{I}(x)$ in (\ref{series-sol}) for this $I$
is expressed as a non-zero constant multiple of
$$
\lim_{a^I\to -L} \sin(\pi a^I)\sum_{(n_1',\dots,n_m')}
\frac{\Ga(a+n'_1+\cdots+ n'_m)\Ga(b+n'_1+\cdots+ n'_m)}
{\Ga(c_1\!+\!n_1')\cdots \Ga(c_m\!+\!n_m')\Ga(1\!+\!n_1')\cdots
\Ga(1\!+\!n_m')}x_1^{n_1'}\cdots x_m^{n_m'},
$$
where $n_k'$ $(1\le k\le m)$ runs over the set
$$
\left\{
\begin{array}{lcc}
\N=\{n_k\mid n_k\in \N\}=\{0,1,2,\dots\} & \textrm{if} & k\notin I,\\
1-c_k+\N=\{1-c_k+n_k\mid n_k\in \N\} & \textrm{if} & k\in I.
\end{array}
\right.
$$
Since
$$a+n'_1+\cdots+ n'_m=a^I+n_1+\cdots+n_m=-L+n_1+\cdots+n_m\le 0
$$
for $n_1+\cdots +n_m\le L$ when $a^I\to -L$,
some finite terms of $\Ga(a+n'_1+\cdots+ n'_m)$ diverge.
However, the poles of the Gamma function are simple,
these poles are canceled by the limit of $\sin(\pi a^I)$
as $a^I\to -L$.
Hence the solution $F_I(x)$ is reduced to the sum of finite terms
by the limit $a^I\to -L$.  Since $F_I(x)$ is a polynomial times
$\prod_{k=1}^m x_k^{i_k(1-c_k)}$,
the $1$-dimensional span of $F_I(x)$ in $Sol_{\dot x}$
is invariant under $\CM$. Therefore the monodromy representation $\CM$ is
reducible in this case.
Note that the representation matrices $M_i$ $(i=0,1,\dots,m)$ in
Fact \ref{fact:rep-mat-1} are
valid under the limit and the $I$-th column of $M_0$ is
the $I$-th unit column vector of size $2^m$. We can also see that
the $1$-dimensional span of $F_I(x)$ is invariant under $\CM$
by these representation matrices.
\item
If the parameter $a^I$ goes to a non-negative integer $L'$,
then the solution $F_{I}(x)$ in (\ref{series-sol}) for this $I$
reduces to the identically zero.
Thus we use the fundamental system
$(\dots, F_I'(x),\dots)=(\dots, F_I,\dots)H^{-1}$,
where the diagonal matrix $H$ is
given in Fact \ref{solution-cycle}.
By the explicit form of $H$, we can easily see that
$F_{I'}'(x)$ for $I'\ne I$ are valid under the limit $a^I \to L'$.
The limit $\lim\limits_{a^I \to L'} F_I'(x)$
is a non-zero constant multiple of
$$
\lim_{a^I\to L'} \sum_{(n_1',\dots,n_m')}
\frac{\Ga(a+n'_1+\cdots+ n'_m)\Ga(b+n'_1+\cdots+ n'_m)}
{\Ga(c_1+n_1')\cdots \Ga(c_m+n_m')\Ga(1+n_1')\cdots \Ga(1+n_m')}
x_1^{n_1'}\cdots x_m^{n_m'},
$$
where $n_k'$ $(1\le k\le m)$ runs over the same set as (i).
Since each term of this series converges as $a^I\to L'$,
this limit is a solution to $E_C(a,b,c)$ with
the factor $\prod_{k=1}^m x_k^{i_k(1-c_k)}$.
Thus the fundamental system $(\dots, F_I'(x),\dots)$ is valid
under the limit $a^I \to L'$.
By this change of fundamental systems, the representation matrices
$M_i$ $(i=0,1,\dots,m)$ are transformed into
$$M'_i=H M_i H^{-1}.
$$
For $k=1,\dots,m$,
since $M_k$ and $H$ are diagonal, we have
$M'_k=H M_k H^{-1}=M_k=\tp M_k$.
By Fact \ref{fact:rep-mat-1}, $M_0'$ is given as
\begin{align*}
M'_0=HM_0H^{-1}&=
H\Big(E_{2^m} -\frac{(\be-1)(\al -\prod_{k=1}^m \ga_k )}{\al \be}
  \cdot \one \cdot \tp \one \cdot H\Big)H^{-1}\\
&=E_{2^m} -\frac{(\be-1)(\al -\prod_{k=1}^m \ga_k )}{\al \be}
  \cdot H\cdot \one \cdot \tp \one = \tp M_0.
\end{align*}
These representation matrices are valid under the limit $a^I\to L'$.
By this limit, the $I$-th row of $M'_0$ is the $I$-th
unit row vector of size $2^m$.
Hence the $(2^m-1)$-dimensional space spanned by $F'_{I'}$ $(I'\ne I)$
is invariant under $\CM$.
\end{enumerate}
Therefore, we obtain non-trivial invariant subspaces, and
complete the proof.
\end{proof}

\begin{Rem}
Even in the case of $m=1$, we need detailed case analysis
to give a fundamental system of solutions to $E_C(a,b,c)$
in terms of the series (\ref{series-sol})
without the condition (\ref{irred-1}), refer to \cite{KS} and \cite{MS}.
\end{Rem}

redpajama_set_name: RedPajamaArXiv


This factsheet considers the options available to micro entities when preparing their accounts.	At Alexander James & Co, we can provide guidance on the options available and their implications in the Surrey area.
If you are in the Surrey area please do contact us for guidance on the best options for your micro entity accounts.


I usaully oil my hair 3x a week (my hair can be quiet dry and prone to flaky scalp) with oiling ive noticed I dont have this problem anymore which is great!
Im due to start a new job soon, so my hair care routine needs to change (i cant go to work with greasy hair after oiling) so ive read an alternative to provide moisture to my dry hair : "leave in conditioner!"
how many days do you leave it in before washing out?
can it be applied to the scalp too?
I think the amount of oil you use determines how greasy your hair will be. I use lots of coconut/almond/whatever oil on my ends and then spray a leave in conditioner (you can even use normal conditioner if you can figure out the right amount) on top of it. My hair is super thirsty though, it never looks overly greasy unless I overdo it.
You could always alter your regimen a little, and oil when you get home from work, and wash it out before you go to sleep or when you wake up in the morning. It's an option, at the least.
Suggestion - heavy oiling once per week before washing, choose a day off work.
I use my regular conditioner as a leave-in. Currently it's Save Your Hair/Save Your Rainforest, but I've used other kinds, and I like most of them. My ends are a long distance from my scalp and need the extra moisture. You probably won't have to buy something special to use as a leave-in. Start with a pea-sized amount on ends and increase/decrease according to results.
I mix Suave coconut conditioner with water and keep it in a spray bottle. I spritz it on the ends when they start to feel dry. My son gets tangle in the back of his hair and I spray it on his hair to brush the tangles out every day.
Following on from your advice I've brought some almond and coconut leave in conditioner today!
Hopefully this new hair regime works & stops too many splits and breakage occuring :) I shall keep you all updated. Many thanks for your advice.


text:
Two newcomers, one incumbent on D91 board
Cotton Yancy captures the greatest share of votes
by Nona Tepper April 10, 2019
Forest Park District 91 schools will have two new members and retain one who was up for re-election.
Voters chose incumbent Shannon Wood and newcomers Monique Cotton Yancy and Katherine Valleau to represent them on the school board in the D91's first contested race in recent history. Greg Mitchell, who was appointed to the board in September 2018, finished out of the running.
With all 11 precincts reporting — the final precinct's totals did not come in until after 10 p.m. on April 2 — Cotton Yancy received the most votes, with nearly 24 percent of ballots selecting her name. Vote totals will change slightly as mail and provisional ballots are tabulated in the coming days.
"It doesn't even seem real yet," Cotton Yancy said.
She has a particular interest in special education — since she's had two children with individualized education plans (IEPs) — as well as the district's equity imperative, curriculum among other issues.
"I think people see me involved in things already so it's kind of a natural progression to run, and just having my face and name out there doing other things," Cotton Yancy said. She is involved in the Citizens Advisory Council, the Park District of Forest Park, the Diversity Commission and more.
She credits the success of Proviso Together — the incumbent slate for Proviso Township High Schools District 209 school board — for inspiring her to run.
"If they wouldn't have run four years ago, I wouldn't have any idea how this process works," she said.
Wood, meanwhile, who received about 17 percent of the vote, said she felt relieved to be through with campaigning.
"It was fun, I got to meet some new people in Forest Park that I don't know, and I got to hear their questions and concerns, [but] it's hard. It's very out of my comfort zone to just go and talk to people randomly. It was a good experience for sure," she said, adding that she is looking forward to continuing the district's equity work. She hopes to drive greater engagement between the board and community.
"I want parents to feel heard, whether that's an email or at a meeting. I want them to feel they're respectfully heard," she said. "I would like someone to follow up with them about any question they ask at a board meeting. Our main goal is what's best for the kids. Parents want what we want and so we have to listen to parents."
Valleau received about 16 percent of the vote. She is currently a Forest Park business owner but worked as a D91 teacher and union head for many years. She was unable to comment on her win.
"My heart lies in education and I am ecstatic to get back to the work and vocation I care so deeply about," Valleau posted on social media. "I'm also thrilled to share this outcome with such excellent ladies and look forward to working with Monique, Shannon and the rest of the D91 school board."
CONTACT: ntepper@wjinc.com


Cooper is a natural storyteller, and all five novels grip the reader tightly.
The first in the sequence, Over Sea, Under Stone, follows the adventures of Simon, Barney, and Jane on holiday in Cornwall. It is there, in the attic of the grey house, that they find an ancient clue that will point them the way to recovering one of the legendary Arthurian relics, the Holy Grail. Against them are arrayed the powers of the Dark, determined to prevent its discovery. The second book, perhaps the best known, The Dark is Rising, is the best of the sequence, a masterful tale of Will Stanton's rise to becoming a warrior for the Light. It was made into an absolutely abysmal film that should be avoided at all costs. The third and shortest book, Greenwitch, brings Jane, Barney and Simon together with Will against the dark. Book four, The Grey King, set in Wales, is as grey and atmospheric as is Wales itself, pitting Will and his friend Bran against the powers of the Grey King. The final book in the sequence, Silver on the Tree, transcends epic fantasy into the realm of grandiose myth and is one of the stranger - and better - books in the sequence.
Cooper is a natural storyteller, and all five novels grip the reader tightly, helped in this with copious amounts of mythology and spectacular prose. The prose of the second book in the series, The Dark is Rising, is some of the best I've read in its genre. The sequence is an absolute classic, and should be required reading for children between the ages of seven and fifteen. Those who are older who haven't read them yet are really missing out on something wonderful. Highly recommended.
Have you read The Dark is Rising Sequence?


With all my logo designs, I usually start with a blank piece of paper and start to write down words that relate to what I am trying to create. From these words come images that are eventually interpreted and created as logos.
This logo was designed for a small home based business that creates and sells salted caramels. I found a reference of an entire tree looking like the leaves were falling off. I placed it in Illustrator and converted it to vectors. I then edited it, recreating all the leaves as well. The font used was what the client already was using and wanted to keep.
Every year for over 10 years, Forest City Gymnastics has hosted their Emerald Cup. If is a 2 day event having over 300 athletes attend from all over Ontario. One year there was actually a team from Iqaluit. I chose the round shape because this logo was also used for the medal design for the medals that were given to the athletes.
As well as designing and maintaining Conrad Home's website, I created their logo. The colour green was used to convey a earth friendly component and the gold with serifed text was to show the following of tried and true design and construction.
I created this logo in a request to make give the club a cleaner design than their previous logo. The integration of the athletes in the lettering shows the strong connection that the athletes and staff feel to the club.
This logo utilized two themes that are centred in this community which are the Wye Creek and Thorndale Ivy.


text:
Courtney Love Enthusiastically Approved the Use of that Nirvana Song on this Week's 'Succession'
By Dustin Rowles | TV | November 3, 2021 |
During a pivotal scene in this week's episode of Succession, Shiv Roy takes the stage as President of Waystar. It's a big moment for her because it will be the first time she'll be seen by the shareholders — and the public — as having some real power inside the media conglomerate.
Before she can barely get a word out, however, a set of speakers set up by her brother, Kendall Roy, begins loudly playing Nirvana's "Rape Me," over the speech of Shiv. For a few bars, I thought it was a well-targeted troll job. Shiv, after all, was on stage in an attempt to paper over the sexual misconduct allegations leveled against the company, so the use of this particular song was appropriate. But as the song continued, and as a humiliated Shiv shrank in embarrassment, it began to feel less like a pointed attack on the company's values and more of a personal attack against Shiv by her brother. It was blunt and uncomfortable, and any sympathy that Kendall might have incurred for "doing the right thing" (ha!) completely evaporated in that moment. When Shiv walked off the stage, made a beeline for Kendall's office, and spit in his datebook, I wanted nothing more than to see her exact her revenge.
She did exactly that, and by the end of the episode, Kendall was a broken man, completely alienated from his family, and eviscerated by the media by a letter Shiv had written about Ken's fragile mental health.
It's oddly fitting, then, that "Rape Me" was used in this context, not just in reference to the scandal inside Waystar. Cobain's song was literally an anti-rape song, but figuratively it was also a screed against the media for the way it had treated Cobain and Love. While understanding that it's not a term that we use in the metaphorical sense anymore, it very much befit the situation on Succession as the song was understood 28 years ago. "You'll never kill me. I'll survive this and I'm gonna f—king rape you one of these days and you won't even know it," is how Darcey Steinke of Spin described the defiant nature of the song in 1993.
No one appreciated that more than Courtney Love, who took to Instagram to commend the show's use of the song.
"I was in the room watching (agog) as this song was written in about an hour, and I've never been so proud of approving one of Kurt's songs," Courtney Love wrote. "This cue in Succession is as if they truly understood KC. What he was screaming his heart out … without specifics … about. I'm sure wherever he is he's proud of this. #rapeme #kurtdcobain #nirvana #disruption #myfavoriteinsidesource."
At this point, I would love nothing more than to see Shiv destroy Kendall's career and lock away the burnt remnants of his dignity in a heart-shaped box.
via Uproxx
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Header Image Source: HBO
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VAFF 2021 Review: 'Last Night in Soho' Is Immersive, Muddled, Reconfirms Anya Taylor-Joy's Dazzlingness, and Stumbles Over Itself
Review: In 'Found,' Three Adoptees Search for Their Birth Parents But Find Something Better
Kal Penn Comes Out as a NASCAR Fan (Also Gay)
Steve Buscemi Won This Year's Halloween (But There Were Some Decent Celebrity Runner-Ups)
Dave Chappelle Breaks His Promise


text:
Sweet Home é uma cidade localizada no estado norte-americano de Oregon, no Condado de Linn.

Demografia
Segundo o censo norte-americano de 2000, a sua população era de 8016 habitantes.
Em 2006, foi estimada uma população de 8576, um aumento de 560 (7.0%).

Geografia
De acordo com o United States Census Bureau tem uma área de
14,9 km², dos quais 13,8 km² cobertos por terra e 1,1 km² cobertos por água. Sweet Home localiza-se a aproximadamente 149 m acima do nível do mar.

Localidades na vizinhança
O diagrama seguinte representa as localidades num raio de 24 km ao redor de Sweet Home.

Ligações externas

Cidades do Oregon
Condado de Linn (Oregon)