I actually have very few negative things to say about this hike (shocking, I know). My expectations were misaligned (meaning, I get stressed out when I KNOW we should cover 14 miles in a day and we only cover 10) but overall I found this hike to be thoroughly enjoyable (all complaining aside).
We parked at Timberline Lodge and started our hike from there. Hindsight being 20/20, we probably would have preferred to start the hike somewhere else and have one of our nights along the trail be a relaxing stay at the lodge. Would it be hard to get on the road again after that? Yes. Would it have been totally worth it to sleep in a real bed and eat delicious food and drink wine by the fire? Definitely.
The road home is a loooooong one. There are several horrifying river crossings along the way (beware if hiking this with a puppy or a little one) but the last river crossing is a wide and swift one. Wide meaning you are DEFINITELY going to get wet-foot and swift meaning that rocks will likely belt your calves along your wet-foot way. By the time you emerge you have to climb up a steep, sandy slope (meaning wet, sandy feet) and then you can see the lodge far off in the distance for several miles before you actually arrive there. Which is a slow torture. You can picture a juicy burger in one hand and a cold beer in the other but you are still HOURS and a wet, sandy climb away from that reality. Keep on keeping on – you'll get there. Refer to my section on "fighting with your camping buddy" for tips to get through this difficult time.
To say I was stressed out about the Elliot Crossing from what I'd read would be an understatement. We had read that you could shimmy down a rope on a steep slope or cross a very sketchy glacier. We opted for the rope and hiked up a steep slope only to turn around thinking we had missed it somewhere along the way. We turned around and took a few unlikely side trails (no luck) and then continued up the way we came. We realized to our exhausted horror that we had turned around about 100 yards from the rope spot. So don't turn back blog-readers! It's pretty clear where the rope is (but it's a much steeper climb than you would anticipate).
The rope climb down into the canyon and then the subsequent river crossing and climb back up were SO NOT THAT BAD! A little sketchy, sure. But on a scale of 1 to deadly I'd give this a 4. Don't fall but you would likely not even be doing this hike if you were clumsy. Point is, Elliot Crossing is not to be feared.
Take the time to take your time. Because the hike was so long, I had it set in my mind that we would hike more miles in the first 2 days and then take it easier on the 3rd day and be done with plenty of time for a burger and a beer at Skyline Restaurant down the way. So when Nick "Bearshield" wanted to take a side hike I FREAKED OUT (internally at least). Turns out, the side hike was one of my favorite experiences of the entire hike. We took some funny nudie pics and a butterfly landed on my finger which made me THE HAPPIEST PERSON IN THE WHOLE WORLD. If I had freaked out externally as much as I freaked out internally, we would've missed the whole experience.
A much needed cold foot soak!


The recent days leading up to Stout Burgers & Beer opening their third location just off the Promenade in Santa Monica, it felt akin to the days before Christmas. I was giddy with excitement. I could not wait to unwrap my presents. A good friend forewarned me though: "Your burger's going to arrive at your table and upon first impression, you may be a tad underwhelmed. It won't look like it's packed and stacked with flavor. But, trust me, it is. It'll be one of the best burgers you've ever eaten."
My buddy was absolutely correct with this assessment. Unlike the food pornographic photo you see above generously provided by the restaurant (mine came out way too dark), the burgers at Stout arrive hidden underneath their golden shell of a brioche bun. Yet, upon first bite (or first peek under the hood), you are welcomed with a dizzying concordance of flavor. That's right, a dizzying concordance. It's almost like there's a whole new, undiscovered world under that bun. Like "Legends of the Hidden Temple" or something. A delicious "Legends of the Hidden Temple." Especially for their namesake burger, the Stout Burger. We're talking about a nicely browned/crusted patty with blue cheese, gruyere, rosemary bacon, carmelized onion, horseradish cream and roasted tomatoes. Despite all these varying flavor profiles, the Stout Burger really works and succeeds unlike many LA burgers. There's so much going on, yet it all works rather harmonically.
I can very comfortably place the Stout Burger, on first taste alone, in the upper Unemployed Eater Santa Monica Burger echelon. Obviously an incredibly coveted place to be, it certainly belongs near the top of the list, amongst the Father's Office Burger, Rustic Canyon Burger and Hole in the Wall burger. Only problem: it will be hard to ever order a different burger there - and the others sound mighty damn fine as well.


14th of Malin's Welcome, Year 1708 Marie would come up to Ashley in the town square both could be seeing talking very few words were spoken then hugging, soon Marie would walk off and leave Ashley waiting by the fountain... Soon after coming back with her Five year old Philip and her her one year old that has still been with out a name. Ashley would take the sleeping one year old from Marie, going to now be know as Cole until given a better name by his father when he was well once again. Marie says hugs Philip once again and kisses Cole on the head then leaves with again saying only a few words as a goodbye.... Ashley would then start to walk to the Fernández Manor where she was staying until there place at the docs had been set up. Letting Philip in and then herself, Philip still holding on to her hand and Cole still asleep in her arm, shed sigh then lean upon the door and look down and both boys with a smile... hiding the fear,worried and doubt that she had in her head thinking she didn't have what is took to help out her friend,her sister, her family out, but never the less she had the navy of Curon to help her out if needed. Taken a deep breath shed calm herself knowing that she has everything and more that she needs to help out her friend.


This story of Fiona, the blind dog living in a trash pile in Los Angeles, who gets an amazing rescue, has become wildly popular.
Watch this video, and you'll see why. It's a really touching portrayal, accompanied by April Lavigne's "I'm With You," and a great story.
By the end Fiona can see again, for the first time in a long time.


Born in Sydney, Graham's first life was in the corporate game. However, it was not long before his desire to run a farm emerged and got the better of him. Moving to the Hunter Valley in 1980 following the purchase of Valley View Vineyard he jumped right into the deep end.
Over the last 30+ years Graham has been active in the Hunter Valley wine industry working for Hungerford Hill, Rothbury Estate, Bimbadgen Estate, McWilliam's Mount Pleasant and currently Ivanhoe Wines.
In between working at Cellar Doors and raising a family, Graham has managed to be at the frontline of vineyard operations and cuts no corners when it comes to quality.


Every couple should go on a romantic vacation every now and then. It can be very fun and exciting. There is nothing like experiencing a new place for the first time with someone who is special to you. Internet travel tools have made it very easy to find vacation package deals for two. Do you already have a place in mind? Is there a city or beach the two of you have your hearts set on already? Or are you open to suggestions? Either way, you should be able to find exactly what you are looking for.
If you've only started dating you might not feel comfortable enough to go very far, so a private day tour of the closest city might be a good place to start. When you are feeling adventurous, you could check and see what kind of fun activities are available in your state or neighboring state. If you've been together long enough that you feel comfortable going to another country, you have plenty of options, from exploring ancient Mayan ruins in Cozumel or going all the way to the top of the Eiffel Tower.
There is no shortage of romantic getaways for couples. Whether you want to celebrate your anniversary, honeymoon, or simply spend some time alone at a cozy resort, you should look over vacation package deals for two to find out what is available.
Keep in mind that a lot of travel packages rates that you will come across are only the price per person. Even if the hotel or resort allows two people, you do not want to forget that other plane ticket. Also, make sure that admissions to concerts or attractions that an all inclusive package might offer cover two people.
There are also walking and trekking adventures to consider. Simply walking side by side through amazing places can be both romantic and adventurous. If you're not the sporty type and do not want to anything thrilling or scary, a walking tour probably looks safe and comfortable. You can go on a walking tour through an exciting city, take a mountain path, walk along an exotic beach, walk around Everest Base Camp, etc.
You and your special one might want to go on a cruise for a few days. There are romantic suites and cabins lined toward couples. While at sea, watch for dolphins. When docked, head to shore for a fun excursion. There are so many cruise destinations to choose from: Hawaiian Islands, Caribbean, Bahamian, Alaskan, Mediterranean, South American, and more.
Just get together and make a list of places you would like to go and activities you both will enjoy and see what kind of vacation package deals for two you can find.
There are many reasons to use online websites to find vacation package deals for two . For starters, you are guaranteed the lowest price. You can browse through the offers and read reviews on the hotels, cruise ships, resorts, inns, and more. With online promotional codes, you'll likely be able to get a very good price on a trip.


It is time for Aarhus Beach Party 2017 at Strandbaren. NB: The party has been postponed to next week!
Skilled workers can have their certificate assessed and recognized, which might improve their job search or employment.
Enjoy food from Kødbyens Mad & Marked at Officerspladsen during Aarhus Jazz Festival 2017.


text:
The Electronic Communications Code and what local authorities need to know
Emma Barnfield and Jeni Hanstock set out some of the issues that local authorities need to be aware of when it comes to the Electronic Communications Code.
Local authorities (LAs) are in a unique position to help deliver the government's 5G connectivity targets and to reassure the public as to the misinformation around 5G, in particular health and safety concerns regarding electro-magnetic radiation emitted from 5G masts.
However, as well as delivering the local planning function, LAs are also extensive landowners in their own right and need to balance competing interests, including the Department for Levelling Up, Housing and Communities (DLUHC) 27 August 2020 guidance, with obligations to achieve "best value" when entering into land agreements and the provisions of the National Planning Policy Framework.
Code issues for LAs to consider
There has been a raft of recent cases under the Code resulting in issues that LAs should consider:
Access to LA sites for MSVs
Since the DLUHC guidance calling on LAs to help boost gigabit broadband rollout and 5G mobile coverage to assist in the UK's coronavirus recovery, we have seen a substantial increase in operator requests seeking access to survey LA sites (so called multi-skilled visits, or MSVs).
LA buildings in urban areas may be attractive if they meet an operator's height criteria. CTIL v University of London established that rooftop access surveys constitute a Code right. Often, any prejudice to the landowner is considered capable of being compensated in money. In EE Ltd and another v London Underground Ltd, the landowner's security concerns, and the building being designated as critical national infrastructure was not enough to convince the Tribunal that the operator's Code access rights should be cast aside, so it is difficult to see what reasons could be given to prevent access for MSVs.
The best course is likely to be early engagement with operators, and advice from a telecoms surveyor, to ensure that an LA's landowner position is protected. If terms cannot be agreed, the operator may apply to the Tribunal for rights to carry out an MSV and, given the recent case law, will likely succeed.
Negotiating Code agreements
The Code has broadly increased operator rights and, as it sets out the terms a court would impose, operators seek to include them by agreement. From a landowner's perspective, the key consideration in negotiating a Code agreement is asset protection, which subject to the particular attributes of the site may include:
Safeguarding the interests of residents whilst the operator accesses the building to install and thereafter inspect, maintain and repair their apparatus. This may include supervised access rights, a pre-approved list of operator personnel able to access the building, an operator obligation to cause the least possible inconvenience when undertaking works (for example, agreeing set down areas for equipment and the positioning of cranes etc.) and an indemnity covering damage the operator may cause during visits and works.
A requirement for the operator to give reasonable prior notice before visits to the building and/or works may take place, to enable the landowner to communicate with affected parties in the building and to minimise the risk of any undue interference or disruption to the day-to-day running of the building.
Securing a costs recovery provision for the landowner's benefit, to include taking professional advice, instructing a telecoms surveyor and negotiating and entering into a Code agreement.
Considering the ICNIRP public exposure guidelines with the operator and seeking input from a telecoms surveyor on the technical aspect of any health risk posed to the landowner or its asset by the operator's proposed apparatus. This is particularly relevant to LAs, in light of the MHCLG guidance to quell concerns over the risk posed by 5G technology, if installation is to take place in proximity to residential buildings.
Making provision for any maintenance programmes or redevelopment works planned by the landowner, and who bears the costs of switching off the apparatus, or if necessary, relocating the apparatus to enable plans to be realised.
It is highly unlikely that a landowner's interests will be aligned with that of an operator. Each request for a Code agreement, and the appropriate terms to sufficiently safeguard a landowner's asset, must be considered on a case-by-case basis and where there is a high level of prejudice to a landowner for which money is inadequate compensation, a landowner might justifiably refuse a Code agreement without inclusion of a particular asset-protecting measure.
Compensation and rent payable to LAs
Under the Code, the basis for calculating payments to be made to the landowner by an operator disregards the operator's use of the site for its apparatus and any uniqueness the site has to that operator.
There have been a number of decisions which give an indication of the approach the Tribunal may take, if asked, to the valuation of a particular type of site. In CTIL v London and Quadrant Housing Trust, the Tribunal held that an annual rent of £5,000 is a good indication of the market value of a Code agreement for a rooftop site on a residential building, and whilst there may be features of a particular building to justify a modest range, the Tribunal would not expect significant variations one way or the other. In On Tower UK Ltd v JH and FW Green Ltd, the Tribunal considered a rural greenfield site and determined that absent of particular special attributes an annual rent of £750 would be appropriate. Whether these decisions are creating a valuation benchmark under the Code remains to be seen.
Emma Barnfield is a partner and Jeni Hancock is an Associate at Trowers & Hamlins.
The Colston 4 and the listed building
High Court rejects appeals over business rates and sham transactions findings
Decision by council leader to put brakes on housing development flawed and left resident "in limbo", says Ombudsman
'Grasping the nettle' - the role of previous inconsistent decisions in planning - Cornerstone Barristers
Cost: Free Speakers: Josef Cannon, Dr Ashley Bowes, Dr Alex Williams To what extent does a planning decision-maker have to have regard to previous decisions about the same or similar development? When do you have to 'grasp the nettle' of any disagreement with such a decision? What kind of prev…
Environmental Information Regulations - Act Now
Cost: £399 + VAT Speaker: Lynn Wyeth This interactive workshop explores the Environmental Information Regulations and provides a complete overview of how to handle requests under the EIR. After briefly considering the background to the EIR, it explores the definition of environmental information…
FOI: Contracts and Commercial Confidentiality - Act Now
Cost £399 + VAT Speaker Susan Wolf Public authorities often receive FOI requests for information about contracts awarded, tenders or other information that might be considered to be commercially sensitive. This interactive workshop examines the legal and practical issues that arise when dealing …
New certificated planning enforcement training course - Ivy Legal
Until Sun, 31 December 2023
Free Webinars for Local Authority Lawyers - IQ Legal Training
Until Thu, 10 March 2022
Diversions and Stopping Up Orders - IQ Legal Training
Legal Leaders Programme - A lawyer at the table: being influential in your organisation - The Centre for Legal Leadership
On Wed, 23 February 2022
Managing Difficult People - IQ Legal Training
NSIPs - Recent Legal Challenges - Francis Taylor Building
On Mon, 28 February 2022
Commercial Property Transactions for Support Staff 2022 - IQ Legal Training
From Mon, 11 April 2022 Until Mon, 27 June 2022

redpajama_set_name: RedPajamaCommonCrawl


Valmet will supply Valmet DNA automation systems and Valmet IQ quality control systems to Jiangxi Taison Paper's four new tissue machines, TM7, TM8, TM9 and TM10, at its greenfield mill in Jiujiang, Jiangxi Province, China.
By installing Valmet's advanced automation technology in these tissue machines, the mill will be able to optimize machine performance and end-product quality.
The order is included in Valmet's first quarter 2017 orders received. The value of the order will not be disclosed. Typically, the order value of these kinds of automation system deliveries ranges from below one million to three million euros.
"Valmet is one of the top suppliers for the Chinese pulp and paper industry. The company has very good tissue machine and automation technologies, wide experience and many successful references in China. It's also very important for us that Valmet has a professional local service team close to us. We trust in Valmet," says Wu Jianxiong, General Manager, Jiangxi Taison.
"Jiangxi Taison can count on highly efficient project management, on-time delivery and a successful startup when working with Valmet. Our proven technology for tissue making, such as non-nuclear measurements and the reliable Valmet DNA platform, will enable the mill to succeed and move its performance forward," says Andy Zhang, Area Sales Manager, Valmet.


text:
Dorner Manhunt, Transparency and the Ethics of Law Enforcement
Posted on February 15, 2013 by Kendall Coffey
In the 10-day manhunt for ex-Los Angeles Police Department officer Christopher Dorner, reporters and the general public hoped for more clues in a deadly rampage involving former LAPD colleagues and others. The San Bernardino Sheriff's Department ended its press conference on Wednesday with dozens of unanswered questions.
How did the final hideout cabin burn?
How was the suspect able to hide in a condo 20 yards away from the sheriff's mountain command post?
In the Internet age, the lines between law enforcement, public information and media begin to blur. A report in the Christian Science Monitor raises the question: "Is it fair to expect all these issues to be addressed?"
As the media aims to inform the public with accurate and reliable information, experts in the ethics of law enforcement want to present transparency but also have a duty to protect their investigation. LAPD Chief Charlie said, "They cannot put out information when they have barely begun their own investigation into what actually happened at a crime scene," he says. "This is an active, on-going investigation into many crime scenes by multiple jurisdictions…"
On the subject of media-strong messages, former U.S. Attorney from Miami, Kendall Coffey said, "Some messages may never reach the court of law but still register big time with public opinion."
Although some information may never make it to the jury, "…they can be fed routinely to the press because some themes that never appear in trial transcripts will splash hugely in newsprint transcript."
The public is focused on its demands for transparency and accountability, says Donald Tibbs, professor at the Earle Mack School of Law at Drexel University in Philadelphia. He says the police, "… have civil liability to think about. Anything they say could be used in a civil lawsuit that might be brought by the Dorner family, for instance."
In the report, former Los Angeles deputy sheriff and FBI agent Frank Scafidi says the LAPD is in an unfortunate position of having to defend against charges of a "coverup" in its handling of the investigation and Dorner's termination. He added, "…that's why it is extremely important for officials to get as much detailed information as possible out as soon as possible before the conspiracies have time to wind up."
Posted in Kendall Coffey | Tagged Dorner, Kendall Coffey, law, public information | Leave a reply
Kendall Coffey on the Silent Majority
Posted on February 1, 2013 by Kendall Coffey
Earlier this month, Richard Nixon would have celebrated his 100th birthday.
In Kendall Coffey's latest tweet, he discusses changing demographics from 1972 to 2012, social issues and the fall of the silent majority in the latest presidential election. He discussed how Nixon was usually demonized by Baby Boomers but how he has nevertheless had a profound role in shaping America's political landscape.
In his post, he refers to the current Republican foundation as a "Silent Majority." The term encompasses value issues, including "…faith, patriotism, and law and order, as well as a shrewd manipulation of attitudes concerning intellectualism."
He refers to the Nixonian constituents as a "Vocal Minority." He states, "…the Nixon coalition of values may be backfiring with Hispanics, women, and gays, who are underwhelmed by the anti-immigration, anti-choice, and anti-gay marriage positions of the GOP."
Possibly by 2016, the Republican party may reconfigure their strategy to gain support from Hispanics and younger voters.
Posted in Election, Kendall Coffey | Leave a reply
Kendall Coffey's comment: Trump passed documents to Special Counsel Robert Mueller
Singleton & Serino Testify in Zimmerman Case
OJ Simpson- blasted by the court of public opinion
Boston Bombing Suspect to be Prosecuted Through the Civilian Process
Dzhokar Tsarnaev
Kendall Coffey
Massey Energy
Pat Santeramo
Peter Madoff
PoliticKING
spinning the law


text:
Rockies' Tom Murphy changes approach, looks to…
SportsRockies
Rockies' Tom Murphy changes approach, looks to land on 25-man roster
John Leyba / The Denver Post
Rockies catcher Tom Murphy performs drills during the team's first full squad workout on Feb. 20 during last season's spring training at Salt River Fields at Talking Stick in Scottsdale, Ariz.
By Patrick Saunders and Denver Post |
SCOTTSDALE, Ariz. — Tom Murphy has fielded the questions before. Indeed, he has faced them every spring training since 2014.
Is this the season you finally break through? The season you stay healthy? The season you become a full-time big-league catcher? The answer has always been "no," primarily because injuries have stalled his career.
Last year, for example, an optimistic Murphy arrived at Salt River Fields looking to share catching duties with Tony Wolters. Hopes were sky high.
"This spring is a great opportunity for Tom to make a big impact on this ball club," catching coach Mike Redmond said at the time. "He's got the toughness and the skill set. We think he's ready to take that next step."
But Murphy never got the chance. His dream was shattered during a March 11 Cactus League game when the Cubs' Anthony Rizzo swung his bat and hit Murphy's arm while Murphy was throwing to second base on a steal attempt. Rizzo's swing cracked a bone in Murphy's right wrist and forearm. In a blink, Murphy's camp was over. He ended up playing in only 12 games for the Rockies, over two brief call-ups, with one hit in 24 at-bats.
"Last year seems like a blur, and I really don't have any details in my mind about it," said Murphy, 26. "This year, I've come in to work out every day in good spirits, and I've set out to accomplish my goals. I'm going to keep it that way."
As the Rockies officially opening camp Wednesday with workouts for pitchers and catchers, Murphy sits third on the depth chart behind veteran Chris Iannetta, who returned to Colorado with a two-year, $8.5 million contract, and Wolters, who began last season strong but then spent most of August at Triple-A after the Rockies acquired veteran Jonathan Lucroy at the trade deadline.
The Rockies have been waiting for Murphy to blossom ever since drafting him in the third round in 2012 out of the University at Buffalo. They aren't about to give up on him now. This winter, general manager Jeff Bridich applauded Murphy's efforts at Triple-A Albuquerque last summer, when Murphy batted hit .255, socked four homers and drove in 19 RBIs in 38 games. Bridich also complimented Murphy's improved dexterity behind the plate and noted his improved ability to block pitches in the dirt.
"That was a tough year for him — high hopes, high expectations that he probably put on himself," Bridich said. "It sort of went the other way for Tom. But he never gave in. He kept battling. These things will all be beneficial as he tries to bounce back this season."
In past years, Murphy spent too much time worrying about things outside of his control, especially when it came to competition at his position. Now he realizes that all of that was just wasted energy and became counterproductive.
"If I think about somebody else, that's not being in the moment, that's not trying to accomplish what I want to accomplish," he said. "I can't worry about what somebody else does, or if they had a good BP (batting practice) or whatever. Now, it's nonexistent to me."
In other words, Murphy is trying to take outside pressures off his broad shoulders.
"I have seen more bad than good from the other way of thinking," he said. "Trying to compete with everybody else? Trying to please the coach? That's not something I'm wholeheartedly into. I've seen the downside of that, and that's not somebody I want to be."


Home » Cycling » Travel » Cycling to Spiti! Day 0, Noida.
This week has been a torture, the last minute shopping spree that doesn't seem to end, and money bleeding here and there in the family. A sudden shakeup at work.
A bad bout of anxiety yesterday I have no idea how and why this happened. But it just happened, had I not had a few precious people around, I am pretty sure I would have cancelled this tour, I have thought through it multiple time. " Cancel it! So what if you've spent so much time and money already! I am definitely cancelling it". And today I was double checking my list, optimistically emptying my refrigerator of stuff that could spoil, packed, unpacked, packed, and unpacked again.
I had to leave at 5:30, and I didn't realize till 5:45 that it was way beyond 5:30. I left at 6 finally, after a hurried shower and a milkshake of the last packet of milk.
6 easy kilometers and a thousand stares later, I kep hearing a clicking noise of the bike. "Definitely stones skidding of the back wheel" and I kept moving, a Kilometer later I figured that the noise just wouldn't go. I stopped and checked only to find a nail on a cable fastener entirely inside the tyre. I pulled it out immediately only to hear the tyre go flat in an instant.
It was 6:30 already and my bus was at 8:30, I had to decide against fixing the flat myself. I unpacked the bags, put them up in an auto along with the cycle and reached the nearest bike shop at 6:50.
Fixing the flat took another 20 minutes as the mechanic kept stopping to make a sale to other customers. By the time I was packed and moving again it was 7:20 already and the bus stand was 15 kilometers away.
I rushed through flyovers, asking directions while speeding alongside autos, and reach the bus stand at 8:10, After a help from the conductor in unpacking and loading the bike, I successfully got onto the bus profusely sweating and barely breathing.
Picture Courtesy: Precious People I mentioned above.


text:
\section{Introduction}

Several different methods to construct minimal polynomials of Salem numbers have been investigated in the literature (see e.g.\ \cite{BDGPS,boyd,grossmcmullen,smyth}). Various authors associate Salem numbers to Coxeter polynomials and use this relation in order to construct Salem numbers (cf.\ for instance \cite{cannwagr,floydplotnick,grohirmcm,lakatos,mckrowsmy}). In this paper we follow the very explicit approach of Gross et al.~\cite{grohirmcm} and provide precise information on the decomposition of Coxeter polynomials of certain star-like trees into irreducible factors, thereby giving estimates on the degree of the occurring Salem factor.

To be more precise, let $r, a_0,\ldots, a_r \in \NN$ such that $a_0\ge 2,\ldots, a_r \ge 2$. We consider the star-like tree
 $T=T (a_0,\ldots, a_r)$ with $r+1$ arms of  $a_0-1,\ldots, a_r-1$ edges, respectively. According to \cite[Lemma 5]{mckrowsmy}
the Coxeter polynomial of $T (a_0,\ldots, a_r)$ is given by
$$
R_{T (a_0,\ldots, a_r)} (z) =\prod_{i=0}^{r} \Bigl( \frac{z^{a_i}-1}{z-1}\Bigr)\Bigl(z+1- z \sum_{i=0}^{r}  \frac{z^{a_i-1}-1}{z^{a_i}-1}\Bigr).
$$
Note that $R_T$ can be written as
\begin{equation}\label{eq:decomp}
R_T(z)=C(z)S(z),
\end{equation}
where $C$ is a product of cyclotomic polynomials and $S$ is the minimal polynomial of a Salem number or of a quadratic Pisot number. Indeed, by the results of \cite{pena},  the zeros of $R_T$ are either real and positive or  have modulus 1. The decomposition \eqref{eq:decomp} now follows from \cite[Corollaries~7 and~9, together with the remark after the latter]{mckrowsmy}, as these results imply that $R_T$ has exactly one irrational real positive root of modulus greater than $1$.

For Coxeter polynomials corresponding to star-like trees with three arms we are able to say much more about the factors of the decomposition \eqref{eq:decomp}. In particular, we shall prove the following result.

\begin{theorem}\label{thm1}
Let  $a_0,  a_1, a_2 \in \ZZ$  such that $a_2 > a_1 >a_0>1$ and
$(a_0, a_1, a_2)\ne (2,3,t)$ for all $t\in \set{4, 5, 6}$. Further, let $T:=T (a_0, a_1, a_2)$ be the star-like tree with three arms of  $a_0-1, a_1-1, a_2-1$ edges, and let $\lambda$ be its largest eigenvalue. Then $\tau >1$ defined by
$$
\sqrt {\tau}+ 1/\sqrt {\tau}=  \lambda
$$
is a Salem or a quadratic Pisot root of the Coxeter polynomial $R_{T}$ of  $T$.
If $S$ is the minimal polynomial of $\tau$  then we can write
\begin{equation}\label{eq:clear}
R_T(x)=S(x)C(x),
\end{equation}
where $C$ is a product of cyclotomic polynomials of orders bounded by $420 (a_2-a_1+a_0-1)$ whose roots have multiplicity bounded by an effectively computable constant $m(a_0,a_2-a_1)$. Thus
\begin{equation}\label{SalemLowerBound}
\deg S \ge \deg R_T - m(a_0,a_2-a_1)\sum_{k\le 420 (a_2-a_1+a_0-1)}\varphi(k),
\end{equation}
where $\varphi$ denotes Euler's $\varphi$-function.
\end{theorem}

\begin{remark}[Periodicity properties of cyclotomic factors]\label{rem:period}
Gross et al.~\cite{grohirmcm} study certain Coxeter polynomials and prove periodicity properties of their cyclotomic factors. Contrary to their case, our Coxeter polynomials $R_T$ do not have the same strong separability properties (cf.\ Lemma~\ref{lem:mult}). For this reason, we could not exhibit analogous results for $C(x$), however, we obtain weaker periodicity properties in the following way.

In the setting of Theorem~\ref{thm1} assume that $a_0$ as well as $a_2-a_1$ are constant. For convenience set $S_{a_1}=R_{T(a_0,a_1,a_2)}$ and let $\zeta_k$ be a root of unity of order $k$. It follows from \eqref{expzdarst} below (see \eqref{pzdef} for the definition of $P$) that $S_{a_1}(\zeta_k)=0$ if and only if  $S_{a_1+k}(\zeta_k)=0$, i.e., the
fact that the $k$-th cyclotomic polynomial divides $S_{a_1}$ depends only on the residue class of $a_1 \pmod k$.
Therefore, setting $K:=\lcm\{1,2,\ldots,  420 (a_2-a_1+a_0-1)\}$,
the set of all cyclotomic polynomials dividing $S_{a_1}$ is determined by the residue class of $a_1 \pmod K$.

If we determine the set $\{k \,:\, k \le 420 (a_2-a_1+a_0-1),\, S_{a_1}(\zeta_k) = 0 \}$ for all $a_1 \le K$ we thus know exactly which cyclotomic factor divides which of the polynomials $S_{a_1}$ for $a_1\in\mathbb{N}$. Obviously, this knowledge would allow to improve the bound \eqref{SalemLowerBound}.
\end{remark}

\begin{remark}[Degrees of the Salem numbers]\label{rem:degree}
Theorem~\ref{thm1} enables us to exhibit Salem numbers of arbitrarily large degree.
 Indeed, if $a_0$ and the difference $a_2-a_1$ are kept small and  $a_1\to \infty$ then \eqref{SalemLowerBound} assures that $\deg S \to \infty$. We also mention here that Gross and McMullen \cite[Theorem~1.6]{grossmcmullen} showed that for any odd integer
$n\ge 3$ there exist infinitely many {\it unramified} Salem numbers of degree $2 n$;
recall that a Salem polynomial $f$ is said to be unramified if it satisfies $|f(-1)|=|f(1)|=1$.
The construction pursued in this work substantially differs from ours: it is proved that every unramified Salem polynomial arises from
an  automorphism of an indefinite lattice.
\end{remark}

If  two of the arms of the star-like tree under consideration get longer and longer, the associated Salem numbers converge to the $m$-bonacci number $\varphi_m$, where $m$ is the (fixed) length of the third arm. This is made precise in the following theorem.

\begin{theorem}\label{th:m}
Let $a_1 > a_0 \ge 2$ and $\eta \ge 1$ be given and set $a_2 = a_1+\eta$.  Then, for $a_1\to\infty$, the Salem root $\tau(a_0, a_1, a_2)$ of the  Coxeter polynomial associated with $T (a_0, a_1, a_2)$ converges to $\varphi_{a_0}$, where the degree of $\tau(a_0, a_1, a_2)$ is bounded from below by \eqref{SalemLowerBound}.
\end{theorem}

Besides that, we are able to give the following result which is valid for more general star-like trees.

\begin{theorem}\label{th:general}
Let $r\ge 1$, $a_r>  \cdots > a_1 > a_0 \ge 2$, and choose $k\in\{1,\ldots,r-1\}$.  Then, for fixed $a_0,\ldots,a_k$ and $a_{k+1},\ldots, a_r \to \infty$, the Salem root $\tau(a_0, \ldots , a_r)$ of the Coxeter polynomial associated with $T (a_0, \ldots, a_r)$ converges to the dominant Pisot root of
\begin{equation}\label{eq:qz}
Q(z)=(z+1-r+k)\prod_{i=0}^k(z^{a_i}-1)-z\sum_{i=0}^k(z^{a_i-1}-1)
\prod_{\begin{subarray}{c} j=0 \\ j\not=i \end{subarray}}^k (z^{a_j}-1).
\end{equation}
\end{theorem}

\section{Salem numbers generated by Coxeter polynomials of star-like trees}

For convenience, we introduce the polynomial
\begin{equation}\label{pzdef}
\begin{array}{rrl}
\displaystyle P(z)&:=&\displaystyle (z-1)^{r +1} R_{T (a_0,\ldots, a_r)} (z) \\[3pt]
&=&\displaystyle \Bigl( \prod_{i=0}^{r} (z^{a_i}-1)\Bigr) (z+1)- z \sum_{i=0}^{r}  \Bigl( (z^{a_i-1}-1) \prod_{j=0, j\ne i}^{r} (z^{a_j}-1)\Bigr).
\end{array}
\end{equation}
Of course, like $R_T$,  the polynomial $P$ can be decomposed as a product of a Salem (or quadratic Pisot) factor times a factor containing only cyclotomic polynomials.

Now, we concentrate on star-like trees with three arms, i.e., we assume that $r=2$.

\begin{lemma} \label{pblock}
Let $a_2 > a_1 > a_0$. Then for $T(a_0,a_1,a_2)$ the polynomial $P(z)$ reads
\begin{equation}\label{expzdarst}
  P(z)= z^{a_1 + a_2} Q(z) + z^{a_1 + 1} R(z)+ S(z)
\end{equation}
with
\begin{eqnarray*}
 Q(z) &=& z^{a_0+1} - 2 z^{a_0}+ 1, \\
  R(z) &=& z^{a_2-a_1+a_0-1} - z^{a_2-a_1} + z^{a_0-1} -1,\\
  S(z) &=& -z^{a_0 +1} + 2z -1.
  \end{eqnarray*}
Moreover,
$$
\max\{ \deg(Q), \deg(R), \deg(S) \} = a_2-a_1+a_0-1, \quad \deg (P)= a_0 + a_1 +a_2 + 1,
$$
and the (naive) height of $P$ equals $2$.
\end{lemma}
\begin{proof}
This can easily be verified by direct computation.
\end{proof}

\begin{lemma}\label{lem:mult}
Let $a_2 > a_1 > a_0$ and let $P$ as in \eqref{pzdef} be associated to $T(a_0,a_1,a_2)$. Then there exists an effectively computable constant $m=m(a_0,a_2-a_1)$ which bounds the multiplicity of every root $z$ of $P$ with $|z|=1$.
\end{lemma}

\begin{proof}
Observe, that $1$ is a root of $Q$, $R$, and $S$. Thus, for technical reasons, we work with $\tilde P(z)=P(z)/(z-1)$ and, defining $\tilde Q(z)$, $\tilde R(z)$, and $\tilde S(z)$ analogously, we write
\[
\tilde P(z)= z^{a_1 + a_2} \tilde Q(z) + z^{a_1 + 1} \tilde R(z)+ \tilde S(z).
\]
Our first goal is to bound the $n$-th derivatives $|\tilde P^{(n)}(z)|$ with $|z|=1 $ away from zero. To this end we define the quantities
\begin{align*}
\eta(a_0)&:=\min\{|\tilde Q(z)| \,:\, |z|=1 \} > 0, \\
E_n=E_n(a_0)&:=\max\{|\tilde Q^{(k)}(z)| \,:\, 1\le k\le n,\, |z|=1 \},  \\
F_0=F_0(a_0,a_2-a_1)&:=\max\{|\tilde R(z)| \,:\,  |z|=1 \}, \\
F_n=F_n(a_0,a_2-a_1,n)&:=\max\{|\tilde R^{(k)}(z)| \,:\, 1\le k\le n,\, |z|=1 \},  \\
G_n=G(a_0,n)&:=\max\{|\tilde S^{(n)}(z)| \,:\, |z|=1 \}.
\end{align*}
For $n\ge 1$ one easily computes that (note that $(x)_{n}=x(x-1)\cdots(x-n+1)$ denotes the Pochhammer symbol)
\begin{align*}
\tilde P^{(n)}(z) =& (a_1+a_2)_{(n)}\tilde Q(z)z^{a_1+a_2-n} +
(a_1+1)_{(n)}\tilde R(z)z^{a_1+1-n} \\
&+ \sum_{k=0}^{n-1}{n\choose k}(a_1+a_2)_{(k)}\tilde Q^{(n-k)}(z)z^{a_1+a_2-k} \\
&+ \sum_{k=0}^{n-1}{n\choose k}(a_1+1)_{(k)}\tilde R^{(n-k)}(z)z^{a_1+1-k}
+ \tilde S^{(n)}(z).
\end{align*}
Now for  $|z|=1$ we estimate
\begin{align}
|\tilde P^{(n)}(z)| \ge & (a_1+a_2)_{(n)}\eta(a_0) - 2^{-n+1}(a_1+a_2)_{(n)}F_0 \nonumber \\
&- 2^{n-1}(a_1+a_2)_{(n-1)}E_n - 2^{n-1}(a_1+1)_{(n-1)}F_n - G_n\label{eq:betrag}\\
\ge& (a_1+a_2)_{(n)} \Big(\eta(a_0) - 2^{-n+1}F_0 - \frac{2^{n-1}(E_n+F_n)}{a_1+a_2-n+1} - \frac{G_n}{(a_1+a_2)_{(n)}}\Big). \nonumber
\end{align}
Now we fix $a_0$ and the difference $a_2-a_1$. Then we choose $n_0=n_0(a_0,a_2-a_1)$ such that
$$
\eta(a_0) - 2^{-n_0+1}F_0 >0.
$$
In view of \eqref{eq:betrag} there exists a constant $c=c(a_0,a_2-a_1)$ such that for all $a_1,a_2$ with $a_1+a_2 > c$ (with our fixed difference) we have $|\tilde P^{(n_0)}(z)|  > 0$ for all $z$ with $|z|=1$. If, on the other hand, $a_1+a_2 \le c$, then we have $\deg \tilde P \le c+a_0$. Therefore, in any case, the multiplicity of a root of $\tilde P$ on the unit circle is bounded by $\max(n_0,c+a_0)$ and the result follows by taking $m=\max(n_0,c+a_0)+1$.
\end{proof}

The following lemma is a simple special case of Mann's theorem.

\begin{lemma}\label{abcz}
Let   $a, b, c, p, q  \in \ZZ$ such that  $(p,q)\ne (0,0)$ and $a,b,c$ nonzero.
If $\zeta$ is  a  root of unity such that
$$a \zeta^p + b \zeta^q  +c= 0$$
then  the order of $\zeta$ divides $\; 6 \, \gcd (p, q).$
\end{lemma}

\begin{proof}
This is a special case of \cite[Theorem 1]{mann}.
\end{proof}

For subsequent use we recall some notation and facts (used in a similar context in \cite{grohirmcm}).
A  {\em divisor} on the complex plane is a finite sum
$$  D=\sum_{j  \in J}  a_j \cdot z_j$$
where $a_j \in \ZZ\setminus \set{0} $ and $${\rm supp \,} (D):= \set{z_j \in  \CC\;:\; j \in J}$$
is the support of $D$; $D$ is said to be  {\em effective} if all its coefficients are positive.

\medskip

The set of all divisors on $\CC$ forms the abelian group ${\rm Div \,} (\CC)$, and the natural
 evaluation map $\sigma: {\rm Div \,} (\CC) \to \CC $ is given by
$$
\sigma ( D) =\sum_{j  \in J}  a_j z_j\,.
$$

\medskip

A  {\em polar rational polygon}  (prp) is an effective divisor  $D=\sum_{j  \in J}  a_j \cdot z_j$ such that each $z_j$ is a root of unity and $ \sigma ( D) =0\,.$ In this case the order ${\rm o } (D) $ is the cardinality of the subgroup of
$ \CC \setminus \set{0} $ generated by the roots of unity $\set{z_j / z_k \;:\; j,k\in J}$.
The prp $D$ is called  {\em primitive} if there do not exist non-zero prp's $D'$ and $D''$ such that $D=D' + D''$.
In particular, the coefficients of $D', D''$ are positive, thus each prp can be expressed as a sum of primitive prp's.

\bigskip

Every polynomial  $f \in \ZZ[X] \setminus \set{0}$ can be uniquely written in the form
\begin{equation}\label{dcf}
  f=\sum_{j \in J} \varepsilon_j a_j X^j
\end{equation}
with $J \subseteq \set{0, \ldots, \deg  (f)}$, $\varepsilon_j=\pm 1$ and $a_j > 0$. We call
$$\ell (f):= \Card (J)$$
the length of $f$. For $\zeta \in \CC$ with $f(\zeta)=0$ we define the effective divisor of $f$ (w.r.t. $\zeta$) by
$$D f (\zeta):= \sum_{j \in J}  a_j (\varepsilon_j \zeta^j) \,.$$

\bigskip

\begin{proposition}\label{farc}
Let $a_2 > a_1 > a_0$ and let $P$ as in \eqref{pzdef} be associated to $T(a_0,a_1,a_2)$. If   $\zeta$ is  a  root of unity such that  $P(\zeta) = 0$ then
the order of  $\zeta$ satisfies
$$
\mathrm{ord}(\zeta) \le 420 (a_2-a_1+a_0-1).
$$
\end{proposition}

\begin{proof}
We follow the proof of \cite[Theorem 2.1]{grohirmcm} and write the polynomials $Q, R, S$
in the form
$$Q (X) =\sum Q_j (X), \qquad R (X) =\sum R_j (X), \qquad S (X) =\sum S_j (X)$$
with finite sums of integer polynomials such that
$$D P (\zeta)= \sum_{j}  D P_j (\zeta)= \sum_{j} \bigl( \zeta^{a_1 + a_2} D Q_j(\zeta) +  \zeta^{a_1 + 1} D R_j(\zeta)+  DS_j(\zeta)\bigr)$$
is a decomposition of the divisor  $ D P (\zeta)$ into primitive polar rational polygons $ D P_j (\zeta)$, thus for every $j$ the sum
$$\zeta^{a_1 + a_2} Q_j(\zeta) + \zeta^{a_1 + 1} R_j(\zeta)+ S_j(\zeta)$$
is the evaluation of the primitive prp $ D P_j (\zeta)$.
Observe that in view of Lemma \ref{pblock} the (naive) height of the polynomials $Q_j, R_j, S_j$ does not exceed  $2$ since the coefficients cannot increase when performing the decomposition of a prp into primitive prp's.

\medskip

Case 1: \qquad $\max \set{\ell (Q_j), \ell (R_j), \ell (S_j)} > 1$ for some $j$

\medskip

 Let us first assume $\ell (Q_j)> 1$ for some $j$. The ratio of any two roots of unity occurring in $D Q_j (\zeta)$
 can be written in the form $\pm \zeta^e$ with $1 \le e \le \deg (Q_j) \le \deg (Q)$. Therefore we have
 $$\frac{\mathrm{ord}(\zeta)}{2 \deg (Q) } \le {\rm o \,} (D P_j (\zeta)) .$$
 By Mann's Theorem \cite{mann}, $ {\rm o \,} (D P_j (\zeta))$ is bounded by the product of primes
 at most equal to
 $$\ell (P_j)\le \ell (Q_j) +\ell (R_j) + \ell (S_j)\le \ell (Q) +\ell (R) + \ell (S)\le 3+4+3=10\,.$$
 The product of the respective primes is at most $2\cdot 3 \cdot 5\cdot 7=210$. Therefore by Lemma \ref{pblock} we find
 $$\frac{\mathrm{ord}(\zeta)}{2 (a_0+1)}=\frac{\mathrm{ord}(\zeta)}{2 \deg (Q) } \le 210,$$
 which yields
 $$\mathrm{ord}(\zeta) \le 420 \; (a_0+1).$$
 Analogously, the other two cases yield
 $$\mathrm{ord}(\zeta) \le 420 \; (a_2-a_1+a_0-1) \qquad \text{ or } \qquad\mathrm{ord}(\zeta)\le 420 \; (a_0+1),$$
 and we conclude
\begin{equation}\label{resge1}
\mathrm{ord}(\zeta) \le 420 \; \max \set{a_0+1, \, a_2-a_1+a_0-1 } =420 \; (a_2-a_1+a_0-1).
\end{equation}

\medskip

Case 2:  \qquad $\max \set{\ell (Q_j), \ell (R_j), \ell (S_j)} \le 1$ for  all $j$

\medskip

In this case, $D P_j (\zeta)$ is either of the form
\begin{equation}\label{Pdouble}
D P_j (\zeta) = c_{j1}\zeta^{b_{j1}} + c_{j2}\zeta^{b_{j2}}
\end{equation}
or of the form
\begin{equation}\label{Ptriple}
D P_j (\zeta) = c_{j1}\zeta^{b_{j1}} + c_{j2}\zeta^{b_{j2}}  + c_{j3}\zeta^{b_{j3}} ,
\end{equation}
where $c_{ji}\in \{-2,\ldots, 2\}$ by Lemma~\ref{lem:mult}. We distinguish two subcases.

\medskip

Case 2.1: There exists $j$ such that $D P_j (\zeta)$ is of the form \eqref{Ptriple}.

\medskip

 In this situation $D P_j (\zeta)$ can be written more explicitly as
\begin{equation}\label{Pt1}
D P_j (\zeta) = c_{j1}\zeta^{a_1+a_2+\eta_1} + c_{j2}\zeta^{a_1+\eta_2}  + c_{j3}\zeta^{\eta_{3}}
\end{equation}
or
\begin{equation}\label{Pt2}
D P_j (\zeta) = c_{j1}\zeta^{a_1+a_2+\eta_1} + c_{j2}\zeta^{a_2+\eta_2}  + c_{j3}\zeta^{\eta_{3}},
\end{equation}
where $\eta_1\in\{0,a_0,a_0+1\}$, $\eta_2\in\{1,a_0\}$, and $\eta_3\in\{0,1,a_0+1\}$. If $D P_j (\zeta)$ is as in \eqref{Pt1} then $P_j (\zeta)=0$ implies that
\[
c_{j1}\zeta^{a_1+a_2+\eta_1-\eta_3} + c_{j2}\zeta^{a_1+\eta_2-\eta_3}  + c_{j3} =0.
\]
Now, using Lemma~\ref{abcz} we gain
\[
\mathrm{ord}(\zeta) \mid 6 \mathrm{gcd}(a_1+a_2+\eta_1-\eta_3, a_1+\eta_2-\eta_3)
\]
which yields
\[
\mathrm{ord}(\zeta) \mid 6(a_2-a_1+\eta_1-2\eta_2+\eta_3),
\]
hence,
\begin{equation}\label{tripleresult}
\mathrm{ord}(\zeta) \le 6(2 a_0 + a_2 - a_1).
\end{equation}
If $D P_j (\zeta)$  is as in \eqref{Pt2}, by analogous arguments we again obtain \eqref{tripleresult}.

\medskip

Case 2.2:  For all $j$ the divisor $D P_j (\zeta)$ is of the form \eqref{Pdouble}.

\medskip

In this case we have to form pairs of the 10 summands of $D P(\zeta)$ to obtain the divisors $D P_j (\zeta)$. As $\ell(R)=\ell(S)=4$ there must exist $j_1,j_2$ such that $\ell(R_{j_1})=0$ and $\ell(S_{j_2})=0$.  In what follows, $c_{ij}\in\{-2,\ldots,2\}$, and  $\eta_1,\eta_1'\in\{0,a_0,a_0+1\}$, $\eta_2,\eta_2'\in\{1,a_0\}$, and $\eta_3,\eta_3'\in\{0,1,a_0+1\}$.
Then $D P_{j_1} (\zeta)$ is of the form
\begin{equation}\label{pairj1}
D P_{j_1} (\zeta) = c_{j_11}\zeta^{a_1+a_2+\eta_1} + c_{j_13}\zeta^{\eta_{3}}
\end{equation}
which yields
\begin{equation*}\label{pairj1eq}
c_{j_11}\zeta^{a_1+a_2+\eta_1-\eta_3} + c_{j_13} = 0,
\end{equation*}
and, hence,
\begin{equation}\label{pairj1ord}
\mathrm{ord}(\zeta)\mid 2(a_1+a_2+\eta_1-\eta_3).
\end{equation}
For $D P_{j_2} (\zeta)$ we have two possibilities.  Either we have
\begin{equation}\label{pairj21}
D P_{j_2} (\zeta) = c_{j_21}\zeta^{a_1+a_2+\eta'_1} + c_{j_22}\zeta^{a_1+\eta'_2} .
\end{equation}
This yields
\begin{equation*}\label{pairj21eq}
c_{j_21}\zeta^{a_2+\eta'_1-\eta_2'} + c_{j_22} = 0,
\end{equation*}
and, hence,
\begin{equation}\label{pairj21ord}
\mathrm{ord}(\zeta)\mid 2(a_2+\eta_1'-\eta_2').
\end{equation}
The second alternative for $D P_{j_2} (\zeta)$ reads
\begin{equation}\label{pairj22}
D P_{j_2} (\zeta) = c_{j_21}\zeta^{a_1+a_2+\eta'_1} + c_{j_22}\zeta^{a_2+\eta'_2}.
\end{equation}
This yields
\begin{equation*}\label{pairj22eq}
c_{j_21}\zeta^{a_1+\eta'_1-\eta_2'} + c_{j_22} = 0,
\end{equation*}
and, hence,
\begin{equation}\label{pairj22ord}
\mathrm{ord}(\zeta)\mid 2(a_1+\eta_1'-\eta_2').
\end{equation}
If $P_{j_2}$ is of the form \eqref{pairj21}, then \eqref{pairj1ord} and \eqref{pairj21ord} yield
that
\[
\mathrm{ord}(\zeta)\mid 2 \mathrm{gcd}(a_1+a_2+\eta_1-\eta_3, a_2+\eta_1'-\eta_2'),
\]
hence, $\mathrm{ord}(\zeta)\mid 2(a_1-a_2+\eta_1-2\eta_1'+2\eta_2'-\eta_3)$ and therefore
\begin{equation}\label{endres2}
\mathrm{ord}(\zeta)\le 2(a_2-a_1+3a_0+1).
\end{equation}
If $P_{j_2}$ is of the form \eqref{pairj22}, then \eqref{pairj1ord} and \eqref{pairj22ord} yield
\[
\mathrm{ord}(\zeta)\mid 2 \mathrm{gcd}(a_1+a_2+\eta_1-\eta_3, a_1+\eta_1'-\eta_2'),
\]
hence, \eqref{endres2} follows again.

Summing up, the proposition is proved by combining  \eqref{resge1}, \eqref{tripleresult}, and \eqref{endres2}.

 \end{proof}

Combining results from \cite{mckrowsmy} with our previous considerations we can now prove the main theorem of this paper.

\begin{proof}[Proof of Theorem~\ref{thm1}]
The fact that $\tau$ is either a Salem number or a quadratic Pisot number  as well as the decomposition of $R_T$ given in \eqref{eq:clear} follows immediately from~\eqref{eq:decomp}. The bound on the orders of the roots of the cyclotomic polynomials is a consequence of Proposition~\ref{farc}. Together with Lemma~\ref{lem:mult} this proposition yields the estimate $\eqref{SalemLowerBound}$ on the degree of $S$, where the explicitly computable constant $m(a_0,a_2-a_1)$ is the one stated in Lemma~\ref{lem:mult}.
 \end{proof}

\section{Convergence properties of Salem numbers generated by star-like trees}

In this section we prove Theorems~\ref{th:m} and~\ref{th:general}.
In the following proof of Theorem~\ref{th:m} we denote by $M_m(x)=x^m-x^{m-1}-\dots-x-1$ the minimal polynomial of the $m$-bonacci number $\varphi_m$.

\begin{proof}[Proof of Theorem~\ref{th:m}]
Note that $Q(x)=(x-1)M_{a_0}(x)$ holds. The theorem is proved if for each $\varepsilon > 0$ the polynomial $P(x)$ has a root $\zeta$ in the open ball $B_\varepsilon(\varphi_{a_0})$ for all sufficiently large $a_1$. We prove this by using Rouch\'e's Theorem. Let $\varepsilon > 0$ be sufficiently small and set $C_\varepsilon : =\partial B_\varepsilon(\varphi_{a_0})$. Then $\delta := \min\{|M_{a_0}(x)|\,:\, x \in C_\varepsilon \} >0$. Thus, on $C_\varepsilon$ we have the following estimations.
\begin{align}
\label{est}
|x^{a_1+1}R(x)+S(x)| &< (\varphi_{a_0}+\varepsilon)^{a_1+1}4(\varphi_{a_0}+\varepsilon)^{\eta+a_0-1} + 4(\varphi_{a_0}+\varepsilon)^{a_0+1},
\\
|x^{a_1+a_2}Q(x)| &> (\varphi_{a_0}-\varepsilon)^{2a_1+\eta}(\varphi_{a_0}-1-\varepsilon)\delta.
\nonumber
\end{align}
Combining these two inequalities yields
\begin{equation}\label{P_est}
|P(x)| > (\varphi_{a_0}-\varepsilon)^{2a_1+\eta}(\varphi_{a_0}-1-\varepsilon)\delta -
4(\varphi_{a_0}+\varepsilon)^{a_1+a_0+\eta} - 4(\varphi_{a_0}+\varepsilon)^{a_0+1}.
\end{equation}
Since \eqref{est} and \eqref{P_est} imply that for sufficiently large $a_1$ we have
\[
|P(x)| > |x^{a_1+1}R(x)+S(x)| \qquad (x\in C_\varepsilon),
\]
Rouch\'e's Theorem yields that $P(x)$ and $x^{a_1+a_2}(x-1)M_{a_0}(x)$ have the same number of roots in $B_\varepsilon(\varphi_{a_0})$. Thus our assertion is proved.
\end{proof}

To prove Theorem~\ref{th:general} we need the following auxiliary lemma.

\begin{lemma}\label{lem:aux}
Let $r\ge 1$, $a_r>  \cdots > a_1 > a_0 \ge 2$, and choose $k\in\{1,\ldots,r-1\}$. If $P$ is as in \eqref{pzdef} and $Q$ as in \eqref{eq:qz} then, for fixed $a_0,\ldots,a_k$ and $a_{k+1},\ldots, a_r$ sufficiently large, we have
\[
\#\{\xi \in \mathbb{C}\;:\, P(\xi)=0,\, |\xi| > 1\} = \#\{\xi \in \mathbb{C}\;:\, Q(\xi)=0,\, |\xi| > 1\} = 1.
\]
\end{lemma}

\begin{proof}
First observe that
\begin{equation}\label{eq:PQesti}
P(z)= z^{a_{k+1} + \cdots + a_r}Q(z) + O(z^{a_{k+2} + \cdots + a_r+\eta})
\end{equation}
for some fixed constant $\eta \in \mathbb{N}$.Since by ~\eqref{eq:decomp} the polynomial $P$ has exactly one (Salem) root outside the unit disk, it is sufficient to prove the first equality in the statement of the lemma.

We first show that $Q$ has at least one root $\xi$ with $|\xi| > 1$. This is certainly true for $k < r-2$ as in this case we have $|Q(0)|>1$. For $k\in \{r-2,r-1\}$ we see that
\[
Q^{(\ell)}(1)=0,\;\hbox{for} \;  0\le \ell < a_0+\cdots+a_k - 1, \quad Q^{(a_0+\cdots+a_k - 1)}(1)<0.
\]
As the leading coefficient of $Q$ is positive, this implies that $Q(\xi)=0$ for some $\xi >1$.

Now we show that $Q$ has at most one root $\xi$ with $|\xi| > 1$. Assume on the contrary that there exist two distinct roots $\xi_1,\xi_2\in\mathbb{C}$ of $Q$ outside the closed unit circle. Applying Rouch\'e's Theorem to $P(z)$ and $z^{a_{k+1}+\cdots+a_r}Q(z)$ shows by using \eqref{eq:PQesti} that also $P$ has two zeroes outside the closed unit circle which contradicts the fact that $P$ is a product of a Salem polynomial and cyclotomic polynomials, see~\eqref{eq:decomp}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{th:general}]
From Lemma~\ref{lem:aux} and \eqref{eq:qz} we derive that
\[
Q(z)=C(z)T(z)z^s,
\]
where $s\in\{0,1\}$ and $T$ is a Pisot or a Salem polynomial. To show the theorem we have to prove that $T$ is a Pisot polynomial. It suffices to show that $C(z)T(z)$ is not self-reciprocal.

We distinguish three cases. If $k < r-2$ then $|C(0)T(0)|>1$, hence, as this polynomial has leading coefficient $1$ it cannot be self-reciprocal.

Denote the $\ell$-th coefficient of the polynomial $f$ by $[z^\ell]f(z)$. If $k=r-2$ we have $[z]C(z)T(z)=2(-1)^{r-1}$ and $[z^{a_0+\cdots+a_{r-2}}]C(z)T(z)=-r$. As $k>0$ we have $r>2$ and again the polynomial cannot be self-reciprocal.

Finally for $k=r-1$ we have $[z]C(z)T(z)=0$ and $[z^{a_0+\cdots+a_{r-1}-1}]C(z)T(z)=-r$ which again excludes self-reciprocity of $C(z)T(z)$.
\end{proof}

\comment{
\rmus{Finding cyclotomic parts in polynomials:   \cite[Section 2]{beukerssmyth}}

\rmus{Modifying and applying techniques developped in  \cite[Section~7]{grossmcmullen}?}
}

\section{Concluding remarks}

In this note we have studied Coxeter polynomials of star-like trees with special emphasis on star-like trees with three arms. It would be nice to extend Theorem~\ref{th:m} to star-like trees $T(a_0,\ldots, a_r)$ with four and more arms in order to get lower estimates on the degrees of the Salem polynomials involved in Theorem~\ref{th:general}. In fact, the estimate on the maximal multiplicity of the irreducible factors of $R_T$ contained in Lemma~\ref{lem:mult} can be carried over to star-like trees with larger values of $r$. Concerning Proposition~\ref{farc}, the argument based on Mann's Theorem used in order to settle Case~1 of its proof can be extended to $T(a_0,\ldots, a_r)$, however, in the situation of Case~2 we were not able to prove that the orders of the occurring roots of unity are bounded by a reasonable bound. We expect that a generalization of this case requires new ideas.

\bigskip

{\bf Acknowledgment}.
The authors are indebted to the referee for insightful comments on the first version
of this paper.

\def$'${$'$}

redpajama_set_name: RedPajamaArXiv


With the disappointment you've just had, today love will be the last topic you want to talk about. To top it off, you will be forced to act as a loving counselor with that best friend you have.
Giving advice has always been your thing, even though the reality of oneself is completely different from others, you have the ability of empathizing with others, and a lot of people would like to know how to do it.
Virgo in a relationship will also have to make an effort to please the person who they share their life with, they will have to do the activity that they don't like doing.
When you finally get to concentrate with what you have in front of you, you can do whatever you want and more. Even if it was with a lot of effort, yesterday you were able to progress more than you expected in that project you have been told to do and today you will have more energy and a better mood.
It could have something to do that it is Tuesday and you are starting to get back in tune with the rhythm of working days. Virgos who are unemployed should have their telephones nearby today since they may receive a very interesting call.
It could be a temporary job offer or collaboration, but they shouldn't reject it. They might find the opportunity they have been longing for.
Don't let anybody fool you, it has been you and your ability to attract good things to your life what has made you get that chance and good energy professionally speaking. Now that everything seems to go better, it's time for you to go and have a check-up to check your health.
Ask for an appointment for the doctor's as soon as possible, the position of the stars says it's the best time for it. Don't forget to tell your doctor about all your discomforts that you've had these days, they could prescribe you a treatment that will help you so it doesn't happen again. Don't take it as something bad, after all these things are to take care of your health.


We left Jason's sister's house and got 40 miles away when we realized we forgot our hard drive. Oops. So we pulled off at a truck stop, unhitched the car and Jason drove back to retrieve it. Richie was nice enough to meet him. Unfortunately, it added another hour and a half to our already delayed trip.
We continued on towards central Oregon, excited to see his brother and his family. We drove past Mount Hood and saw some beautiful scenery through the passes. But mountain passes always set me on edge and today was no exception. I kept asking Jason how the temperature gauge was doing and it was running high, but still coming back down when we coasted…until it didn't.
We heard a terrible screeching sound that sounded like a belt was dying so immediately he pulled over in the middle of the Willamette Forest. At dusk. On a rainy day. Awesome. Did I mention how cold it was? Brrr.
While driving towards Sisters I was calling every motel I could Google. And they either couldn't accommodate a family of 5 or they were sold out. Around 8:30p, we continued on to Bend and finally found a Quality Inn that had a room for the night. The desk clerk gave us a discount due to the situation we were in. Meanwhile I had been leaving messages with the RV Park about what had happened because we had already paid for our stay.
After getting checked in for the night, we headed to Shari's for a late dinner around 9pm. It happened to be the only place still open on a Sunday night. It wasn't ideal, but we were supposed to have been at the park by 4pm with plenty of time for dinner.
The next morning Jason called Good Sam for a mechanic and a tow. They sent a big orange beast to carry Edelweiss to safety. Jason got to ride in the tow truck – woohoo man stuff!
The mechanic was great and was able to get to it late that day, while we hung out with Erick and Monica and their boys. They were generous enough to let us stay the night and the following morning we picked up our home and pulled into the park in Sisters.


text:
Managing risk, Inside Time
The Parole Board welcomes the publication of the Joint Inspectorate Report on Life Sentence Prisoners as it sheds light on the challenging work undertaken with some of the most serious offenders.
The Parole Board will only direct that a Life Sentence Prisoner should be released when it is satisfied that it is no longer necessary for the protection of the public that the prisoner should be confined.
The Parole Board is sometimes asked to make a recommendation on an offender's suitability for a move from closed to open conditions.
In these cases, the Parole Board considers the offender's risk of committing further offences, their risk of absconding, the likelihood that they will comply with the regime in an open prison and whether there is a benefit of a period in an open prison to allow the offender to adequately adjust to life in the community and have access to support, housing etc.
The Parole Board does not have a role in deciding whether an offender can be released on temporary licence. This is a matter for the prison governor.
However, the Parole Board believes that periods of release on temporary licence (ROTL) are an important part of an offender's rehabilitation and a source of information when making a decision about release.
ROTLs provide opportunities to test how an offender's risk can be managed safely in the community and the Parole Board looks for information on this in the reports it is given.
Managing an offender on ROTL is a crucial stage of rehabilitation, as highlighted in the inspection, and as such the Parole Board would expect to see a robust plan for supporting the offender to prepare for life in the community.
In order to reach a decision, the Parole Board considers the assessments prepared by Offender Managers and Offender Supervisors.
It does not simply accept these assessments at face value.
The Parole Board, sitting as a court, seeks a range of additional information, for example from other sources such as the Police.
The Parole Board also consults experts such as psychologists and reports from offending behaviour courses that the offender has completed.
It also has the opportunity to challenge witnesses about their assessments at an oral hearing.
This enables the Parole Board to gain a better understanding of the reasons for the offences, assess the extent to which the offender has reduced their risk and judge whether the plans to manage risk in the community will be adequate to protect the Public.
The quality of the reports provided by prison and probation staff is very important to the Parole Board to enable it to make decisions which take account of the offender's progress and are fair and protect the public.
The Parole Board is encouraged that its work with NOMS to guide report writers to produce good quality reports has improved practice.
More work is needed to support further improvements in reports by Offender Managers and Offender Supervisors.
The Parole Board is currently working with the National Offender Management Service to improve the parole process and ensure a greater focus on quality as well as ensuring reviews are concluded more efficiently and on time.
The Parole Board is pleased to see that the inspectors' report recommends training for Offender Managers in planning and delivering work with lifers to improve on the quality of sentence planning.
Where the Parole Board is not satisfied that an offender's risk can be managed in the community it will identify risk factors, for example alcohol misuse, or anger management which need to be addressed before the offender can be re-considered for release.
In this way, the Parole Board supports sentencing planning work by ensuring that sentence plans tackle the needs of the individual offender.
The Parole Board believes that the recommendations made by the Chief Inspectors are well timed to inform the work to establish the new national Probation Service.
Source: www.insidetime.org/managing-risk
Inside Time
Parole Board
Joint Inspectorate Report on Life Sentence Prisone
ROTL


This is a webcomic. Here's some information about me.
The artist and writer of this comic. Didn't want to post a photo, so here's a picture of him as a fat baby. Please note: he is in all actuality, not a fat baby.
Actually, let's drop the third person thing. I'm kind of an artist, but more than that, I consider myself a creator. I've worked in film, comics, short stories, 2D, 3D, animation, and pretty much every other medium you can imagine. I usually can't decide what I want to do, so I try and do it all. Like most artists, I don't consider myself as talented as I'd like to be, and go through cycles of putting forth immense effort, then wondering why I bothered.


It's funny, a few people have said to me that they think it will be easier to try and focus on one plastic of the four, the thing is the four biggest are just the tip of the iceberg that we can see. I've been thinking a lot this week about 'hidden' plastic. For example, what restaurants use to prepare our meals, the plastic hiding in tea bags, the plastic packaging of so many products we use every day! It's overwhelming… hence focusing on the Top 4!
It is possible to get by in Cambodia without using plastic water bottles, a bit of organisation and know-how and it's quite easy. Remote areas of Cambodia will have large clay/terracotta pots with drinking water in them, Trailblazer Foundation have done a tremendous job at providing remote communities and NGOs with water filters and a lot of restaurants have filtered or 20L bottled water. In Phnom Penh and Sihanoukville a lot of guesthouses have water filters and will refill your drink bottle for 1-2,000 riel (25-50 cents).
I did an experiment over the last week, if I used single use water bottles, I would have consumed around 28-30 500ml bottles – just me, just one week. It would have cost $3.50-$3.75, instead by using the 20L bottle it cost me around $0.50 for the whole week! So… over a year $26 versus $195. Cost effective and better for the environment don't you think?
Myth! A lot of discussion occurs around minerals in water that are filtered out for the bottled water we drink. So I asked my doctor if I was missing out on any essential nutrients because of this. She told me the only mineral I need is fluoride, which I get in my toothpaste (foils my plan to go to plastic-free toothpaste, but there you go).
This is absolutely the easiest thing to avoid – easy!! I always take my own bag shopping and I've easily been able to avoid plastic shopping bags. Lessons learned when market shopping and I now have a container and some jars so I can buy tofu, coconut milk and fresh cut fruit without the need for a bag. Simple. Something new on the horizon will be kasava starch bags…. soon soon.
Okay… this is the hard one! I take a bamboo straw with me everywhere, I ask in Khmer for no straw, I choose drinks I know won't come with a straw and… with a week to go, I have accumulated at least a dozen straws in July 😦 Help!!!!! Please send in suggestions, advice, anything to help combat straws… small but nasty!


Here, in my Kitchen, you will find lots of wonderful things to eat!!!
I love food - everything about it. growing it, cooking it, eating it. I especially love bread and you will find some of my favourite bread recipes here, along with other standards that have become family faviourites. This is not a recipe site like some - where there are dozens upon dozens of recipes and you can search for what you want. I have links at the bottom of the page that will take you to sites like that. Here are MY favourites. I hope you enjoy them.
Most of the following recipes came from friends or from newsgroups. The original sender is identified whenever possible.
Great masses of Wild Roses grew over the fence on the lane to our house beneath the pines. We had sweet wild berries, old roses, lilies, wild flowers, daffodils, all growing in a colorful riot alongside the rutted dirt lane. The homestead flock of geese would waddle out of the yard and cross over the driveway to pick berries. At least they did until an elf would jump out from behind a rock or some leafs and frighten them, then they would flap, running and hollering back to the safety of the yard.
Flowers freshly picked by a child's hand sat in a canning jar in the center of the old round dinning table. The house scented with the fragrance of blooming sunshine. We picked the berries, ate them fresh, in syrup, pies, cobblers, and dried. What we couldn't eat, we canned for feasting during the winter snows.
We harvested the roses from the wiry rambling brambles to make pale rose jam and jelly, dried the rose hips for use in cough syrups, simples and teas in the cold winter months. Dried the pastel petals for scenting the water that was always kept in the old black cast iron pot on the back of the wood burning cook stove. We've moved on since then. the geese are long gone. The children too. The old plank house is no longer there, the stand of pines has been cut down for lumber, the well was capped and filled in. the barns gone along with the livestock.
The old lane still has mounds of old fashioned roses cascading from the fence to the bar ditch, daffodils, jonquils, lilies, geese, and children still walk the lane in memory.
Early in the morning, before the dew has vanished, take your basket to the rose bushes, and gather the ripe hips. Take only those that are ripe and only enough for that day's use. Be sure to leave some of the ripe hips on the bush. Cut the hips from the stem with sharp shears, or scissors, laying them in the basket carefully, so as not to bruise them. Inspect the hips, and carefully remove any insects before proceeding.
Soak the hips in cool clean water for a short while. Clean them and rinse well to remove all sticks and dirt.
Add the juice of a small lemon, or orange, to this, and stir all together. Let stand another thirty minutes.
Put the juicy (sweated) hips into a heavy pan, add a small amount of water, and boil until the mixture sheets when dripped from a wooden spoon.
Pour the Rose Hip Jam into jam or jelly jars, and seal.
*sweat = the rose hips and sugar will become moist in the crock, the sugar will coat the fruits, and juice will appear in the bottom of the container.
Here is a recipe for kids. It�s very simple, easy, and delicious, too.
Cut a watermelon meat into small pieces. Put the small pieces into the blender, Place the lid on the blender. Turn the blender on, and blend until the pieces are all turned into juice.
Pour 3 cups of the watermelon juice into a shallow metal pan, or bowl. Add the sugar and lemon juice. Stir them all well, until the sugar is completely dissolved.
Put the juice mixture, still in the metal pan, or bowl, in the freezer. Be careful, the juice is sticky if it's spilled.
Every 20 to 30 minutes, carefully remove the juice from the freezer, and stir it up really well. Break and mash all the frozen pieces into small pieces. This will make the juice look a little like slush, or oatmeal. Do this until all the juice is completely frozen.
Depending on the temperature of your freezer, It will take about 2 hours for the juice to freeze solid , but it needs to be stirred, and mashed up while it is freezing.
Serve in pretty glasses, or bowls. Very good!


Tank Cleaning and issuing of Environmental Manifest.
Supply of Technical materials and spare parts.
Industrial Waste Disposal and issuing of Environmental Manifest.
Hot work, Painting and Certification.
Supply of Safety Materials, Fuel and Food Stores, AB Marine.
Garbage Disposal and issuing of Garbage Certificate and Environmental Manifest.


text:
Q: Gtk3+ - Pygobject and HiDPI screen I'm using Python 2 with matplotlib, pygobjet and glade.
For some reason, In my macbook retina I'm having trouble seeing my application.
When I use the FigureCanvasGTK3Cairo, my plots are being scaled of screen. If I use FigureCanvasGTK3Agg backend, the image is well rendered but using a low resolution.
Here a test code (got from gtk3 python cookbook)
from gi.repository import Gtk

from matplotlib.figure import Figure
from numpy import arange, pi, random, linspace
import matplotlib.cm as cm

#from matplotlib.backends.backend_gtk3agg import FigureCanvasGTK3Agg as FigureCanvas
#from matplotlib.backends.backend_gtk3cairo import FigureCanvasGTK3Cairo as FigureCanvas

myfirstwindow = Gtk.Window()
myfirstwindow.connect("delete-event", Gtk.main_quit)
myfirstwindow.set_default_size(400, 400)

fig = Figure(figsize=(5,5), dpi=100)
ax = fig.add_subplot(111, projection='polar')

N = 20
theta = linspace(0.0, 2 * pi, N, endpoint=False)
radii = 10 * random.rand(N)
width = pi / 4 * random.rand(N)

bars = ax.bar(theta, radii, width=width, bottom=0.0)

for r, bar in zip(radii, bars):
    bar.set_facecolor(cm.jet(r / 10.))
    bar.set_alpha(0.5)

ax.plot()

sw = Gtk.ScrolledWindow()
myfirstwindow.add(sw)

canvas = FigureCanvas(fig)
canvas.set_size_request(400,400)
sw.add_with_viewport(canvas)

myfirstwindow.show_all()
Gtk.main()

The results:
Agg Backend
CairoGtk backend
I have no problem on my other computers.

redpajama_set_name: RedPajamaStackExchange


The MileagePlus frequent flier loyalty program of United Airlines has a promotion during which you can earn a bonus of up to 75 percent when you purchase MileagePlus miles — plus you will be entered for a chance to win 1,000,000 MileagePlus award miles and $9,800.00 in cash which may be used to help offset any tax liability — but this offer expires on Friday, March 9, 2018 at 11:59 in the evening Central Standard Time.
No metered mail will be accepted. All mail-in entries must be postmarked by Friday, March 9, 2018 and received by Thursday, March 15, 2018. No copies, facsimiles or mechanical reproductions will be accepted. The sponsor assumes no responsibility for lost, late, incomplete, stolen, misdirected, illegible or postage due entries or mail. Limit one entry per outer mailing envelope. Each page or postcard will receive two entries into the grand prize drawing.
You must be a legal resident of the 50 United States — including the District of Columbia — who is a minimum of 21 years age at the time of entry.
Bonus MileagePlus miles will generally be credited to the MileagePlus membership account of the recipient when the transaction is complete — provided that certain transactions may take up to 48 hours.
Bonus MileagePlus miles count towards the maximum annual limit of 150,000 MileagePlus miles per account. You can purchase up to 150,000 miles per account per calendar year.
Your credit card will be billed immediately upon purchase.
Purchased miles cannot be refunded; and they do not count toward MileagePlus Premier elite level status.
The maximum limit is two entries per person — no matter however you decide to participate in this sweepstakes, which is void where prohibited and is not sponsored or administered by United Airlines.
…and if you plan on participating in this sweepstakes, I hope that you win the lone grand prize — which has an estimated retail value of $29,800.00 in United States dollars.


Ronnie and Arthur are shown enjoying the Cinco de Mayo celebration sponsored by the Volunteer Services Council.
Mikey is enjoying the Washington County Fair thanks to partnerships with the Fair Association and businesses in the area.
If you would like to apply to become a volunteer at Brenham State Supported Living Center please visit the Department of Aging and Disabilities website to apply or contact Kristie Wensel in Community Relations at 979-277-1507.
Your gift to the VSC benefits the residents in unrestricted ways across the campus and throughout the community.
Also check out our Wish List.
Brenham SSLC and it's Volunteer Services Council celebrated it's 40th anniversary in 2014.
FACT: There are approximately 260 people living at Brenham State Supported Living Center.
Ground was recently broken for the NEW Hazel Lawson Adaptive Sports Complex. A field with a rubber surface to allow for athletes of all abilities to play field games like baseball, football, soccer and more. JOIN THE TEAM and help make this field a reality for residents & the community to enjoy the thrill of victory and the agony of defeat.
Art work from BSSLC artists is on sale at Hermann's General Store in downtown Brenham or at the Expression's Bistro & Gift Shop in the Education & Training Bldg 503 Texan Trail on campus.
Community Relations office: 4001 Hwy. 36 South • Brenham, Texas 77833. • ©2013 Volunteer Services Council for BSS. All rights reserved.


Could not boot unless picking "recovery mode"
Thank you for taking the time to report this bug and helping to make Ubuntu better. Unfortunately we can't fix it, because your description does not yet have enough information.
Could you please check if this problem still occurs in the newest hardy kernel.
4. Please run the command "cat /proc/version_signature > version.log" and attach the resulting file "lspci-vvnn.log" to this bug report.
For your reference, the full description of procedures for kernel-related bug reports is available at http://wiki.ubuntu.com/KernelTeamBugPolicies. Thanks in advance!
the information I already provided. Good luck with that.
> does not yet have enough information.
> 1. Please include the output of the command "uname -a" in your next response. It should be one, long line of text which includes the exact kernel version you're running, as well as the CPU architecture.
> 2. Please run the command "dmesg > dmesg.log" after a fresh boot and attach the resulting file "dmesg.log" to this bug report.
> 3. Please run the command "sudo lspci -vvnn > lspci-vvnn.log" and attach the resulting file "lspci-vvnn.log" to this bug report.
> 4. Please run the command "cat /proc/version_signature > version.log" and attach the resulting file "lspci-vvnn.log" to this bug report.
> http://wiki.ubuntu.com/KernelTeamBugPolicies. Thanks in advance!
This is likely hardware specific and since we are unable to reproduce the issue, we will not be able to debug/test any fixes without your help. I'll close this report for now. If you do happen to test Ubuntu in the future, please let us know your results. Thanks.


A lot, but probably not as many as you might think.
Does "kind" mean species? There are millions of species—somewhere between three and thirty million, a rather large range caused in part because taxonomists (biologists who figure out classifications) don't agree on what a species is. The most common definition of a species is a group of organisms that can interbreed and produce fertile offspring.
Or does "kind" mean genus? A group of similar species is a genus. The Boa genus, for instance, includes several snakes, one of which is the Boa constrictor.
Or does "kind" mean family? A group of similar genera (the plural of genus) is a family. Canidae is a family that includes dogs, wolves, foxes, jackals, coyotes, and similar animals. Biological families number in the thousands, not the millions. Did Noah take two dachshunds, two beagles, and two grey wolves? Or did he take a pair of animals representing the Canidae family?
Many animals did not need to be on the ark for survival. Sea animals, for instance, could survive a flood. Many of the one million species of insects could probably survive without being on the ark. Interestingly, the Bible says, "Everything on the dry land in whose nostrils was the breath of life died," but insects don't have nostrils or lungs. They get their oxygen through their cell walls.
While God told Noah to take two of every kind of animal into the ark, He clarified his instructions so that the animals should be two of every unclean animal and "seven of each kind of clean animal" and "seven of each kind of the birds of the heavens." People who study the Bible are equally divided on whether there were seven of each clean animal or seven pairs of each clean animal on the ark. We don't know.
John Woodmorappe, author of the most detailed study of the ark and its animals, says if a "kind" means what we call today a genus, there would have to be just under 16,000 animals (8,000 genera). Others say that a "kind" means a family and so there would be about 2,000 animals (1,000 families).
How many animals were on the ark? There were perhaps as many as sixteen thousand. But there didn't have to be hundreds of thousands. A lot—but not as many as you might think. Check out eight more of the most common questions about the animals on Noah's ark in Noah: The Real Story.


It's Wednesday, which means it's time for our weekly installment of new arrivals. For today's roundup, we've searched high and low to find the best in designer and fast-fashion pieces—all of which just hit stores—so that you can be the first of your friends to wear them. From the tops we're loving for the season ahead to the accessories we can't take our eyes off of, we have something for every style type and price point. Now your only problems will be narrowing down your favorites and making room in your closet. So what are you waiting for? Shop our picks of the week below, and don't forget to check back next Wednesday for another batch of fresh fashion finds.
If it's not already on your radar, this Australian brand is about to be your new go-to for summer threads.
Another day, another Jacquemus statement accessory to love.
You can never have enough denim jackets—trust me.
Even the most pink-averse girls have to love this dress.
We're so into every type of woven bag for spring.
Who doesn't love a good matching set?
The only thing better than this top is the matching skirt.
Filing this under how to make any outfit instantly cooler.
Available in 70cm to 85cm.
The bustier-top trend is definitely sticking around for at least another season.
No word for this other than obsessed.
Are you ready for the comeback of clogs?
Have we mentioned we're loving woven bags for spring?
You can never have enough lace tops, especially in a color this good.
There's nothing we love more than a good, new blogger brand.
If you want to win at the Western trend for spring, we humbly suggest snagging these shoes.
Anyone in need of a brunch outfit?
Leave it to Proenza Schouler to reinvent a classic.
This easy spring dress will be your new weekend hero piece.
The perfect pop of color for all your outfits.
Knit tanks are an effortless way to step up your outfit beyond a basic tee.
The latest Garrett Leight collab to obsess over and of course, buy.


text:
Posts Tagged 'Iain Mackay'
Review – Autumn Glory (Checkmate, Symphonic Variations, Pineapple Poll) – Birmingham Royal Ballet at Sadler's Wells
The Birmingham Royal Ballet opened their fall visit to London with a series of ballets that were a treat of historical information – ballets by the great and the good of years gone by, that I'd mostly only ever known by images ("Ooh, look at those hats!") or reputation ("What, you say there's a ballet done to the music of Gilbert and Sullivan?"). I've seen a fair amount of Ashton (represented in this case by his "Symphonic Variations"), but I was burning with curiosity over Ninette de Valois's "Checkmate," which has some of the strangest costumes I've ever seen (excepting some of the Surrealist designed ones done for Diaghilev). And "Pineapple Poll," well, I'd seen a version of it done by Spectrum Dance in 2004 (with choreography by Donald Byrd), but I anticipated that this one would also be a good time.
In practice, Checkmate won the prize for weird classic ballet of the year. While the movements of the various chess pieces were supposed to be stylized versions of their actual allowed movements in the game, I was not able to see this. Instead, I was caught up in drama and metaphor, as the seductive Black Queen (Victoria Marr) went from terrorizing to enticing the Red Knights (Iain Mackay and Jamie Bond), dropping their guard enough that she was able to pull Iain in for a kill. The queen's rattling of her daggers and bum-shaking was almost insect-like; she was certainly menacing and a most unique (ballet) character as a deadly female. Unfortunately I found myself spending more time thinking about what it all "meant" rather than enjoying the movement. (It seemed to me to be warning of the rise of the forces that would lead to World War II; I assumed the gentle but weak "reds" were supposed to be England.) The aesthetic pleasures were most certainly there, but I hadn't actually come (I promise!) to ogle the very handsome male dancers. The use of poles was fascinating, however, from grills to put dead pieces on to traps (when surrounding the checked Red King (Jonathan Payn)) to simply the linear effect they had on stage (very good with the graphics of the drop) – it was a most unique effect. However, unfortunately, I don't feel this piece reached me either through dance or generated emotion – it seemed very much like an intellectual effort and one that hadn't aged well.
"Symphonic Variations," by Ashton, failed to make almost any impression on me at all. Three men in white, toga-like half-shirts (phoar!) paired three women in white with pleated, short skirts (design Sophie Fedorovitch); and while I loved Cesar Franck's piano music (thank you Jonathan Higgins!) I was only able to think of Balanchine's "Apollo," which I've frequently groused about for being too silly and male ego-centric. However, I felt Balanchine's choreography glowed like a sun, and Ashton's was a pale moon beside it – not Diana so much as Phobos. Chi Cao was a strong lead and great partner to Natasha Oughtred, but … I found the choreography forgettable even if I was having a bit of a Chippendale's experience as I sat blushing in my chair.
Next up was "Pineapple Poll," and as the curtain rose on a cartoony set painted to look like an 1830s port town, my heart sunk a bit; it looked like I was in for 45 minutes of twee. A bevy of ladies came in and danced with young men dressed as sailors; some drama developed as the pub lackey (Tzu-Chao Chou, officially credited as "Jasper the potboy") showed clearly he was in love with Pineapple Poll (Carol-Anne Millar), a "bumboat woman" (this appears to be a person who makes a living selling stuff to people who live on ships). But then Captain Belaye (Robert Parker) showed up, the women started swooning over him … and Birmingham Royal Ballet exploded in a festival of fantastic dancing and expressive acting that made me completely lose track of my critic's notebook. The girls were a series of faints and flutterings, the boat's crewmen were angry and boisterous, the simpering fiancee, Blanche (Arancha Baselga), a hoot … at the time I thought it was just a case of good choreography but in fact it was the cast that took the structure and covered the whole thing with ribbons and fun. Just like in Gilbert and Sullivan operettas, it's not just one or two good leads that make the show, it's everyone in the cast giving it 100% and acting like they, too, could be the full focus of someone's attention at any given moment. Millar was amazing, a real comic genius, so expressive with her body. She owned the stage when she was on it and is now on my top list of ballerinas to arrange my show schedule around. But everyone was just so very good in this show and there's no doubt for me it was the highlight of the night. Thanks, guys, for another marvellous evening out.
(This review is for a performance that took place on Tuesday, October 18th, 2011 at Sadler's Wells. It was repeated on October 19th. For a five star review of the alternate cast, see Clement Crisp; Mark Monahan's less excited review is here.)
Tags:Arancha Baselga, Birmingham Royal Ballet, Carol-Anne Millar, Checkmate, Donald Byrd, Iain Mackay, Jamie Bond, Jonathan Higgins, Jonathan Payn, Pineapple Poll, sadler's wells, Symphonic Variations, Tzu-Chao Chou, Victoria Marr
Review – Romeo and Juliet – Birmingham Royal Ballet at Sadler's Wells
As the first ballet I've been able to see in London since the Bolshoi departed for pastures snowy in August, Birmingham Royal Ballet's Romeo and Juliet was very high on the excitement level. I consider this company very accomplished and had high expectations for their performance of a ballet I love, with its striking Prokofiev score and deeply emotional story.
Unfortunately, BRB failed to deliver, a problem which I'll attribute greatly to the Kenneth Macmillan choreography, which I had not seen before*. It was full of over-dramatic movement: arms held high in grief a la Laurencia; women standing in heavily curved positions meant to look like Renaissance painting; the maneki-neko hands of the Capulet court ladies I found the gestures and posturing grating, unnatural, off-putting, and occasionally comic. The simplistic sword fighting was at least fun and active; but the mannerized, stiff choreography that filled most of the night left me flat and seemed an incredible waste of a great score. The "Dance of the Capulets" was forgettable – an outrage! – and while Romeo (Iain Mackay) and Tybalt (Robert Parker)'s power struggle during the second iteration of this bit of music was interesting, the dances themselves were … overly complex and indigestible.
On the positive side, our lead characters (Jenna Roberts as Juliet) had great chemistry, stage presence, and acting skills. Juliet was fluid and charming; her Romeo was utterly devoted and star-struck. However, Iain Mackay once again displayed a painful inability to manage overhead lifts, even in scene one, when he should have been at his strongest; he needs to be sent for remedial weight training stat.
My complaints end with a sort of sadness about the lack of emotional impact this ballet had on me. I remember seeing the bedroom scene – as done by Kent Stowell of Pacific Northwest Ballet, of all people – leaving me damp-eyed and reminding myself "These are dancers playing fictional characters! There is nothing to be sad about here!" but this Romeo and Juliet left me with none of that. Instead, I grew impatient for this very long ballet to simply end, hoping (at around 10 PM) that Romeo would make his appearance and get on with offing himself. Admittedly, Roberts was the floppiest dead "star-cross'd lover" I'd ever seen – but I felt no pain for her or Romeo in the final scene. And I really want that, a chance for a good weep, not just an opportunity to ooh-aah over lovely costumes; and, unfortunately, Birmingham Royal Ballet's production denied me this pleasure. Bah. On to "Pointes of View" on Friday, which I'm sure I will find far more satisfying.
*Seriously, why the hookers? It's not in the original and it adds an unnecessary layer of smarm to the ballet. Prostitutes dancing with Montagues is one of the least attractive additions to a known story ballet I've ever seen.
(This review is for a performance that took place on Tuesday, October 12, 2010. The show continues through Thursday, October 14th. For a different point of view, see Graham Watt's review on LondonDance.com.)
Tags:Birmingham Royal Ballet, Iain Mackay, Jenna Roberts, Robert Parker, Romeo and Juliet, sadler's wells
Review – Cyrano – Birmingham Royal Ballet at Sadler's Wells
Tonight's Cyrano at Sadler's Wells was the long-sought for Holy Grail of ballet: a new story ballet that was good. Birmingham Royal Ballet managed to take a play about poetry and turn it into a ballet that … well, frequently was poetry. Perhaps I exaggerate a little, but I can't remember the last time Romeo and Juliet gave me the sniffles ("Cmon and die already!"), and Cyrano made me weepy at least twice. The music wasn't really amazing or memorable (though it frequently sounded familiar), but it did the job and was enjoyable; the costumes were good and a nice break from fake renaissance and "modernesque" fluffy skirts and corsets.
But really, it was about the dance, wasn't it? Elisha Willis came back from a sizzling modern turn in E=MC2 to take on the role of Roxanne, and managed to be both adorable, nimble, and graceful – but, most importantly, to be 100% believable as Roxanne all the way though. You could see why three guys were completely in love with her, and when she twirled and leapt her way through the regiment and then did an Irish step-dance turn on top of a drum, you absolutely bought that all the soldiers were now going to go over the top full of patriotic fervor. Passion, joy, fear, anguish – she had me sold every step of the way. And God, she was pretty, and cute as a butterbean in her "sneak into the battle to see my boyfriend" velvet pantsuit. I will have to make a special effort to see her next time BRB are in town.
I, of course, must now turn to Cyrano, as Robert Parker was hardly offstage the whole night. He managed an incredible upper-body expressiveness – vital when this is how you are "speaking" love letters – and also had great leaps and heart-wrenching solos. But even better, he got to fight duels, which were incredibly fun to watch, and do a love duet with someone who was in love with someone else – all while making me feel bad for him (and not just because his nose fell off after the balcony scene and he had to dance around it for about five minutes) – all while wearing a cape, a wig, and yummy custom boots. What a turnaround from Mayerling – I struggled so hard to care about its lead, but Parker had me bought the minute he kissed a market woman while fighting off a man with his blade.
Most of the "heavy lifting" was left to Iain Mackay in the role of Christian, who had to, again and again, do a Macmillan-esque lift in which Roxanne went over and around his shoulders to land en pointe. After the balcony scene, it looked like he almost dropped her, and in his final scene his arms were visibly trembling as he held her overhead. For all that Parker had to act and dance, there's no doubt that Mackay was really, really working hard during this show – perhaps a little beyond his abilities. Still, he was compelling in the role of a simpleton, and I found that, like Roxanne and Cyrano, he too had charmed me before the end of the ballet. Damn these dancers and their mad acting skillz! I'm so used to just watching the dance, since when am I supposed to get caught up in the story?
Really, this was such a fun show, with lots of compelling action on stage (such as the "baguette adagio" in the bakery, and of course the duelling, and the very exciting battle scene), so much so that I never looked at my watch once – well, only to make sure I was coming back on time from intermission. My companions Amy, Alice and J also thought it was great. Doubtlessly there have been ballets in which the choreography was more exciting – I sit here still befuddled about people's enthusiasm for Mayerling – but David Bintley has created something in which we, as the audience, care about the story and the characters. This, then, is a ballet worth watching – and as it's on for two more days, I advise you go buy your tickets ASAP and make sure every night has a full house.
(This review is for a performance that took place on Thursday, November 11th, 2009. Cyrano continues at Sadler's Wells through Saturday, November 13th. If you're skint, there is a half price ticket deal available, but I recommend paying full price if you can afford it, as Sadler's Wells is an institution that deserves to be rewarded for consistently bringing great dance talent to London.)
Tags:Birmingham Royal Ballet, Cyrano, Cyrano de Bergerac the ballet, Cyrano the ballet, David Bintley, Elisha Willis, Iain Mackay, Robert Parker, sadler's wells


Stranded - Try to survive for 1 minute while monsters from the sea try to eat you!
I was kinda in a hurry and forgot to implement the score system.
That was fun. I like it. When I had ten seconds left I stopped killing the monsters and started to run away from them.... I have like 50+ on the screen :D Still won though!
Simple To Understand, But Difficult To Accomplish. i like It. my Only Gripe Is The Weird Hitboxes For Using The Axe.
Pretty nice! Almost more fun to dodge the monsters than kill them.
xD The sea monster ate me, this was interesting to play.
Really nice gameplay! It also had a really cool concept! I was able to win when I got to the end and I used my campfire at about 10 seconds.
First of all, a try again button would be very good since it asked me many permissions each time I refreshed the page.
I really like this game concept. Depending on what you choose, the strategy and the game itself is completly different ! Good job here.
However, I don't understand what to do when I choose the lighter, I try to press C in front of the wood, nothing. Rock nothing. So any clue ?
Overall, pretty strong entry !
@lordjosephdeburg I think you have an old version of Java.
@Masadow Yeah, I could've told that you can press esc in the end screen to restart :'D Also, the lighter lights the tree up when you press space.
"Server not found, Firefox can't find the server at www.jeasonfire.tk."
Currently, jeasonfire.tk can't be found.
Sorry guys, my domain name expired (I think) so now I don't have a website. I'll make a new one but it'll take a day or two. Meanwhile, you can download the html & jar so you can play it.
Funny small goodie. Even though I don't have much to do, it still deserves a solid 3 overall!
Interesting game, pretty good, especially considering you made it in java, which doesn't have high level support for games, only low level stuff like lwjgl and such.


The technic called Infrared thermography (IRT) or thermal imaging is based on the Infra Red radiation of an object or a place in relation with it's temperature. The picture obtained this way is called a thermogram.
Several applications are directly related to thermography like the thermal building diagnostics, sensing the heat losses through the roofs, the detection of water scarcity in agriculture, the temperature measurements used to monitor industrial facilities and others. The aerial thermography is the method that allows to have an overview of a large space in a minimum of time.
See our FOX-C8 THERMO drone especially designed for thermography, equipped with an OPTRIS thermal camera.