text:
Coronavirus Overview
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CORONA MEASURES PER DESTINATION
Lift pass included
Ski Guarantee
Board-basis
Catered chalets
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Half-term
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COVID-19 Travel Info
Cancelled Holidays
Info per destination
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Skiing in Madonna di Campiglio
Select a country... Andorra Austria France Italy Switzerland
Select a region... Adamello Ski Bardonecchia Bormio Breuil-Cervinia Dolomiti Superski Folgarida-Marilleva La Via Lattea Livigno Monterosa Val di Fiemme Val di Sole
Select a city... Commezzadura Folgarida Madonna di Campiglio Pejo Pellizzano Pinzolo
Region Val di Sole
Total 173Km
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Travel information relating to the Coronavirus
This information has been carefully compiled by Sunweb based on information provided by the local authorities. Since the measures mentioned are subject to constant change, no rights can be derived from this and Sunweb accepts no liability for any damage resulting from the use of this information.
Regulations per resort
Distance in public areas (in metres)
1 metre, 2 metres while sporting
2m distance in South-Tirol. A mask is mandatory both indoor & outdoors from the age of 6.
All over Italy, masks must be worn in in- and outdoor public spaces from the age of 6.
Covid-19 test before departure or wait in quarantine
A negative COVID PCR test is mandatory. The test must be taken within 48 hours of arrival. This is mandatory for all arrivals no matter the transportation type.
Covid-19 test upon arrival in destination
If you did not complete a test before arrival you will need to complete a test at a local GGD and wait in quarantine for the results.
Health declaration
It is mandatory for all passengers to fill in a passenger locator form 48 before arriving/returning to the UK. The form can be found here.
From December 21 to January 6, everyone must undergo a mandatory quarantine of 14 days. Foreign tourists arriving in Italy during the same period will also have to undergo a mandatory quarantine of 14 days.
Maximum persons together outdoors
Groups are prohibited on the street, in parks and in front of bars / restaurants
Allowed to share a room with someone outside your household?
In all regions from 22:00 to 05:00. On New Year's Eve it will be extended until 7am, January 1st, 2021.
Closed untill 18/01
Skilifts
Access to the ski lifts is organized in such a way that social distancing between skiers can be guaranteed.
A mask is mandatory
Ski lessons
Public transport / taxi
A mask is mandatory. Regional authorities can set their own rules to combat crowds. This includes restrictions on the maximum number of people per mode of transport, etc. For children aged 3-6 years old a mask is mandatory when travelling on all forms of public transport in Turion region.
Only open in areas coded yellow until 18:00. Mandatory to wear a mask when inside and not seated, you will have fixed seats with a maximum of 4 people per table. From 6 p.m. on December 31, 2020 until 7 a.m. on January 1st, 2020, any catering service in hotels and other accommodation facilities, will only be permitted through room service.
Bars/ Apres-ski
A mask is mandatory to wear when inside and not seated, you will have fixed seats. Closed at 6pm and a max of 4 people per table
Club/Nightlife
Events (indoor and outdoor)
Partially Open
A mask is mandatory. Until 6 January 2021, retail business will be permitted until 9 p.m. all areas holidays and pre-holidays days. All stores located in markets and shopping centers, shopping malls, shopping parks, business combinations and other similar structures will be closed.
Supermarkets and pharmacies
Not allowed to sell alcohol in certain hours
Swimming pools and Gym are closed
Additional airline or transport rules
Transfer regulation
A mask is mandatory indoors
Other formailities needed to enter the country
Everyone must complete a PLF form upon arrival in Italy. The form can be found here.
Please always double check the GOV UK website before travelling.
Additional measures per ski region
Other excursions
Starting December 21, 2020, until January 6, 2021 it is forbidden, within the national territory, to move in and out between the territories of different Regions or Autonomous Provinces.
Increased restrictions for international, travel due to identification of a new variant of the virus Covid-19. The rules differ between England, Scotland, Wales and Northern Ireland. Please visit GOV.UK for more information.
Madonna di Campiglio is one of the most internationally renowned places in Trentino and a true paradise for skiers and winter sports enthusiasts. This popular, fashionable ski village is located at an altitude of 1550 meters in the sunny ski resort of Skirama Brenta. With almost all lifts within walking distance of the village, many blue slopes and the many excellent restaurants, Madonna di Campiglio is one of the places in Italy where Italians themselves also like to go on holiday.
Madonna di Campiglio, a beloved residence of the Austrian and Middle-European nobility and wealthy, was discovered by tourism in the second half of the last century. Princess Sissi and Emperor Franz-Joseph stayed here, and the residents here are only proud of it. Madonna is still popular with the international jet set, and you can marvel at chic Italian ladies in fur coats, flashy ski suits and large moon boots.
In addition to skiing, people also like to come to Madonno to shop, enjoy and stroll. The centre of Madonna di Campiglio consists of two main streets with attractive squares, full of restaurants, espresso bars and shops. You will find many shops with artisanal products and also high-quality shops, so there is something for everyone. Buy cheese from the region, honey, jam and delicious local garlic salami.
Una cucina raffinata
The traditional gastronomy is based on simple dishes with natural, organic products from the area. After eight o'clock in the evening, the mainly Italian guests go to the table at one of the luxury restaurants. In addition to luxury and trendy restaurants, Madonna di Campiglio is also rich in clubs; here you can enjoy dancing and dancing until the early hours.
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Any of you are using photoshop to resize image? Or other software? Do you find that it is kind of troubles when you are going to resize a 100px X100px into a 500px X 100px without losing any quality of the image, and, also without cropping any of it? Let say you are really good in photo edit. But, how about make something disappear from a photo and resize it and with all the condition I mentioned before? And, make it just a seconds? I can't do that. But I found this video from youtube today and I just can't believe what I am watching! There is only one thing in my mind, how if this technique can be implemented in web design?
If we can use this in web design. So, we no need to border much with the complicated layout and we can all go for fluid layout with all kind of design! Have a nice day! By the way, the video is about 1 year ago. Anyone of you know about this software? Is it in the market right now?
Here is a link to the demo vidéo of the last Photoshop World.
That techno is called "Seam Carving" and is going to be integrated in Photoshop Cs4 (extended or not, i don't know).
Yeah, I watch this before. It was amazing, Perhaps they convert it to vector format?
http://rsizr.com/ is the web service.
This is fairly old news, this technology came out last year.
There are multiple pieces of software that do this, as well as a few online services.
Just look up the term "content aware resizing". There is a free site online that will do content aware resizing but I don't remember the address…it was some web 2.0 type name like resizr or something like that.
I believe Adobe acquired this technology and showed it off at a MAX conference last year.


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UFABET reports that Manchester United have emerged as another potential interest in Stuttgart striker Zaza Caladjic . during this summer With Darwin Núñez believed to be their number one target , Benfica are expect to sign Caladjcic as Nunez's replacement. Despite the 24-year-old facing injury this season. This
May 30, 2022 by admin in Sportnews and football
Thomas Frank, manager of Brentford Christian Eriksen is believed to be ready to return to the Danish national team. The 30-year-old midfielder has not played for the "Dairy Cow" team since falling due to an arrhythmia in a game that helped the team. At Euro 2020
Mar 15, 2022 by admin in Sportnews and football
Newcastle United fans may have a new favorite after red sign midfielder Bruno Guimares. Move to join the army along with a beautiful girlfriend to accompany Dietitian Dee Dre samba soccer star Decided to move from Olympique Lyon to stay at St James' Park for a huge fee
Feb 05, 2022 by admin in Sportnews and football
Newcastle ready to throw in the "Dzeko" to help escape relegation
The media celebrated the good the Daily Mail revealed that Newcastle United are ready to spend grab the ground Dzeko Striker veteran Inter Milan to help the team win escape relegation on the market this winter. racing age. 35 years, recently moved to the army "Python" as
'Pep' plays the Man City in the first half, the form is not good
Pep Guardiola says Manchester City's 4-0 win over Newcastle United was a positive result. But not in good form especially in the first half. Manchester City took the lead quickly from a mistake by the Newcastle defensive line before Cancelo dribbled in great distance . There
"Conte" says Liverpool are one of the best teams in the world.
Tottenham Hotspur boss Antonio Conte has admitted his rivals Liverpool are among the best in the world. and is a game that is not easy, so satisfied that the points can be divided back successfully The Italian has just led his "Golden Spikes" army to clear
Kane should be Red, pointing out the position of the lion captain's
Gary Neville, Jamie Redknapp and guru Jamie Carragher agreed that Harry Kane should have been sent off for a red card for the tackle. Sandy Robertson. The game where the Liverpool army drew 2-2 with Tottenham Hotspur, there was a dramatic incident in which Robertson
'Aubameyang' from the first team See you again after Afcon
The Daily Mail has revealed that Pierre-Emerick Aubameyang has been banned from Arsenal's first team. And will not be in the team's plans for Mikel Arteta until the end of the Africa Cup of Nations striker Gabon. He was sacked as captain earlier this month for disciplinary

redpajama_set_name: RedPajamaCommonCrawl


Wherever recovery takes you, you can always call Pax House home.
Our treatment program close to Pasadena is designed to treat the mind, body and spirit of those suffering from addiction. From residential detox to outpatient treatment, Pax House is here to help guide you through early recovery so you can continue your new life, free from alcohol and drugs for years to come.
Detox is typically the first step in the addiction treatment process and often times necessary depending on the substance of abuse. Learn more about our program, and how we assist in easy, medical withdrawal.
Our inpatient drug rehab program was designed to help those suffering from addiction have an individualized plan for overcoming addiction. Learn more about the Pax House addiction treatment program.
What you do after leaving treatment can be the difference between sobriety and relapse. Our skilled patient advocates can assist you in planning out your aftercare including locating sober living, scheduling therapy appointments, and attending outpatient treatment.
Pax House provides a uniquely serene and embracing atmosphere for those learning how to overcome drug addiction or substance abuse. Our facility offers a comfortable home-like environment where we are able to facilitate drug & alcohol treatment programs varying in length (30, 60, or 90 days), tailored to the needs of each client and utilizing proven treatment modalities to impart the deeper change necessary for lasting recovery. We have experienced, licensed, and certified clinicians and staff members that focus their attention on meaningfully interacting with each client.
The entire staff works together to not just provide the highest level of pasadena drug & alcohol treatment, but also to serve as personal role models for living happy, fun, healthy, and sober lives. Early recovery is difficult. But by instilling positive and lasting change reinforced through close individualized and personal attention, we help our clients minimize the risk of relapse. We are a beacon of peace in the saturated landscape of residential treatment facilities in California. Pax House Recovery clients come from throughout the United States, Canada and international countries to receive high quality care.
"The message I would give to someone who is struggling with alcohol or drug abuse and searching for help is to enter a substance abuse treatment program in a place where you will feel comfortable. Find a place that will offer the things that you feel are going to help you during early recovery, such as counseling, psychiatrists on staff, nurses and therapists, and people on the recovery team that will actually help you and support you during this time."
"The message I would give to someone who is struggling with alcohol or drug abuse and searching for help is to enter a substance abuse treatment program in a place where you will feel comfortable. Find a place that will offer the things that you feel are going to help you during early recovery, such as counseling, psychiatrists on staff, nurses and therapists, and people on the recovery team that will actually help you and support you during this time. Additionally, you should search for a clean, fun facility."
"Today marks 6 years of sobriety for me. I am extremely humbled by how much I've learned about myself and the world, and how little I truly know. Sobriety for me is not difficult. However, I find that life in itself can be difficult. Regardless of my challenges and my triumphs, I remain vigilant on living to my fullest potential. Living a sober life has taught me to be kind, loving, and understanding. To see the innocent in people that harm me and others. I love my job, my friends, my family. I could not possibly imagine a better life."
"Before discovering Pax House my spirit was broken. I was a shell of a human, and I could not stop using drugs. I traveled from Oklahoma and began a new life at Pax House in August of 2014. They shared with me a different way of life, a beautiful way of life. I felt like I finally belonged and was understood. The amount of love and support I received was immense! Today I have 2 years and 5 months clean and I wouldn't trade my life in recovery for anything. I now work for Pax House and I get to pay it forward every day. It's a privilege to be a source of strength and support for another suffering addict. My life has purpose if I can make a difference in the life of just one addict."
"It's all about God, recovery and everything else is a blessing. When i get that confused is when I start getting in my own head. I'm thankful for the team that I have around me that holds me down. I'm thankful for everyone in my life that's made this possible for me to live free on a day to day basis. I love my family, I love my Pax house family, and my recovery family. I love you all."
"I would recommend the substance abuse treatment program at Pax House to other people with substance abuse or addiction problems. The people at Pax are really helpful and display empathy, not sympathy. They truly understand what you are going through, and what you've been through. There are people here that have gone through recovery before you, which provides the opportunity to learn from others who know the tough path of recovery."
How do you know you are finding what you need when you search for drug and alcohol rehab centers?
It can be difficult to discern the best fit for yourself or a loved one when scrolling through the endless listings of outpatient rehabs, inpatient drug recovery programs, intervention services, life coaches and sober companions, and the list goes on and on. You see pictures of lavish rehab centers and generic stock photos of the beach or mountain top retreats. Hopping from one website to another, seeing the same informational tabs to "contact us" or "meet our staff". It is easy to get caught up in the flashy amenities, mock action shots of group counseling sessions, and promises of holistic approaches. You will most likely see quick paragraphs on what Cognitive Behavioral Therapy is or descriptions of different levels of care. We at Pax House love to show off these talking points just as much as the next one. What we find though is that all of these things are just the icing on the cake. The real power, the real truth of the matter, is that we are far more interested in knowing who you are, what you are struggling with, and how we can help. From our chef and group facilitators, to our administrative staff and counselors, all the way to our owners; we are all people who have been where you are and have found a better way to live. Every single one of our staff members are in recovery themselves from one year all the way to 27 years. We come from all walks of life be it gender, race, ethnicity, socio-economic background, geographically; we as staff have come together as our own family to pass on the solutions we have found to the next still suffering alcoholic and addict. That is our primary purpose.
We believe in human to human, heart to heart connections that foster a sense of hope and purpose in each of our client's lives.
We have transformed our deepest bottoms into incredible, beautiful lives of our own; and we believe with all of our experience that any client who comes in through our doors, who becomes part of the Pax family, can have the same healing journey we have gone on ourselves. Individual and group counseling, psychotherapy, Cognitive Behavioral Therapy, Dialectal Behavioral Therapy, 12-step recovery, and so much more are just the context through which we speak the language of the heart. We just happen to have a beautiful facility, serene location, and customized treatment in order to better facilitate this process. Between our inpatient drug and alcohol rehab center, our outpatient rehab program, and in connection with several community-based recovery programs; we are able to offer a full continuum of care to ensure that when you are ready to take the world by storm, you feel supported, loved, and encouraged to move forward with a new found freedom. So the only question we have left for you, is how free do you want to be?


My first real attempt at baking started with a classic coffee cake. I have always been fascinated by the crumbs that sit atop coffee cake or crumb cake. In fact, my mother taught me how to choose the perfect crumb cake the same way mothers teach their daughters how to pick out the perfect pair of shoes.
My passion for buttery crumbs led me on a career journey to bake the perfect crumb cake. In order to bake a delicious cake, you need to start off with great ingredients. After all, there are just a handful of ingredients in any one cake.
Flour is one of the main ingredients, and it's an important one! I prefer cake flour to all-purpose flour.
Cake flour yields a more delicate crumb texture that's entirely different from all-purpose flour. It has a lower protein content, about 7 percent, as compared with 10 or 12 percent in all-purpose flour. The lower the protein, the lower the gluten content, and the lighter the cake. And who doesn't like a lovely light and airy cake instead of something that's dead weight?
Cake flour is also ground finer than other flours. When a recipe calls for a specific flour, there is a good reason. Purchase a good-quality cake flour (I like Bob's Red Mill or King Arthur).
Eggs are hugely important. I try always to use large eggs, because they are the perfect size for my recipes. Whatever size the egg, it's important to buy the size eggs that a recipe calls for; otherwise you may end up with a dry cake.
I prefer organic eggs, but they're expensive. When baking, always start with room-temperature eggs. It helps the mixing process if all of the ingredients are at the same temperature. They will take less time to mix into the cake, which means less gluten development. OK, let's just say your cake will have a much better consistency.
Butter is super important when it comes to baking. Here's the dealio: All butters contains between 15 and 20 percent water. This is significant. I prefer butter with a butterfat content between 82 and 84 percent. The higher the fat percentage, the lower the percent of water. Excess water is not helpful in making butter crumbs. Buy the best quality you can afford.
Spices are equally important, especially if the spice you plan on using is as old as your youngest child or has been in your cabinet since you moved into your house. When spices sit and sit and sit, the delicate oils become exposed to air and light. Four-year-old ground cinnamon will not taste like a fresh jar of cinnamon.
Before you throw them away, repurpose spices by throwing them into a pot of water and making potpourri. When you purchase new ones, write the year on the jar and replace it after a year; this will keep your spices fresh and flavorful. I have several friends who pack up their spice cabinet each time they move. They would be better off leaving the old jars behind.
See how to make Hedy's crumb cake, step by step.
Pastry chef and cookbook author Hedy Goldsmith is a James Beard Award finalist for the nationally contested Outstanding Pastry Chef category. Her creations grace the menus of The Genuine Hospitality Group of restaurants, including flagship Michael's Genuine Food & Drink in Miami. Her first cookbook, Baking Out Loud, was published by Clarkson Potter in 2012. Hedy has been featured in multiple national media outlets, including NBC's Today show, The New York Times and People magazine.
Why are the sweets at most airports dry, flavorless, high in fat and sugar, and oddly very appealing? Skip them in favor of pastry chef Hedy Goldsmith's homemade versions of these treats.
These tried-and-true wine and food combinations were practically made for each other.
On the hottest of days, icebox cake is your secret weapon. Pick from Food Network's best recipes.
The Environmental Working Group has updated its list of which fruits and veggies you should buy organic; here's a review.
We've got a dozen delicious ways to celebrate the season.


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Jerry Cantrell: Why Metallica's James Hetfield Is 'The Godfather'
Philip Trapp Published: August 3, 2020
Paul Buck / Ethan Miller, Getty Images
Alice in Chains guitarist Jerry Cantrell called Metallica's James Hetfield a "godfather" among rock icons in an interview that emerged on Monday (Aug. 3), a day that also happens to be Hetfield's 57th birthday.
The Alice in Chains member's commendation came as he shared a fond memory of working with the Metallica frontman. Namely, an occasion when Cantrell teamed up with Hetfield alongside Metallica's Robert Trujillo and Alice in Chains' Sean Kinney for a benefit concert in Los Angeles.
Looking back on that time, Cantrell saluted Hetfield as a revolutionary figure in rock.
"What makes James such a great frontman is his physical presence," the Alice in Chains guitarist recently told Louder. "Nobody else commands the same kind of respect and attention without it being self-seeking or egocentric."
Cantrell continued, "He's all about the music; all about the fun and the celebration and the connection with the fans. And that to me is really inspiring. Especially because he's done this for so long, and he's been at the top of his game for so long and he continues to search for a deeper meaning."
Saying that he "can't wait to see what [Hetfield] does next," the Alice in Chains guitarist claimed that jamming with Hetfield during the aforementioned benefit gig "was one of the coolest things I've ever done in my whole life." The group played a version of Metallica's The Black Album single "Nothing Else Matters."
That concert also saw Cantrell sing Alice in Chains' "Them Bones" and "Would?" with the Hetfield-backed benefit band. And it certainly cemented the Metallica rocker's status as a rock and metal legend in the guitarist's mind.
"He took an underground thing and took it worldwide," Cantrell added. "Metallica is a huge thing for me and for any hard rock or metal band. … Will he go down in history as a rock icon? He already has — he's the godfather, man."
See James Hetfield in a Photo Timeline of Metallica's Remarkable Career
Source: Jerry Cantrell: Why Metallica's James Hetfield Is 'The Godfather'
Filed Under: alice in chains, james hetfield, jerry cantrell, Metallica
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An invalid regex in the allow parameter leads to raising of a misleading exception: The property "allow" is not defined for filters of type "org.apache.catalina.filters.RemoteAddrFilter".
javax.servlet.ServletException: The property "allow" is not defined for filters of type "org.apache.catalina.filters.RemoteAddrFilter"
It seems that IntrospectionUtils.setProperty catches things like InvocationTargetException and logs them at the DEBUG level. I assert that such things should be logged at the WARN level.
Without DEBUG logging enabled, a user cannot see the underlying error.
I looked into changing the current error message to better describe the problem and to mention enabling debug logs for more detail. The issue with that approach is there are a lot of 'normal' debug logs and a single error.
Logging these errors as warnings seems appropriate to me.
I have fixed this in 8.0.x for 8.0.6 onwards and in 7.0.x for 7.0.54 onwards.


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You may remember that I went to the first FashionBloggerCafé shoedition back in summer and as I was so flashed by this gorgeous event I just had to go there again this season. Unfortunately it's not that easy to get there when you don't live in Germany anymore but I managed to stop by for three hours and meet some bloggers I still knew from last year.
Yet again Styleranking, the organisers of the FashionBloggerCafé, have come up with some amazing cooperation partners and funny workshops. Just like last year the brand Clarks was present and gifted every blogger a brand new pair of shoes (you could already get a hint at mine on Instagram but I will surely post them on the blog these days). Also Peter Kaiser was present again, calling for our creativity. While last time we had to design a pair of shoes, this time it was a bag. The winner will get her design realised and send over as her very own unique bag so it would mean the world to me if you could vote for my design!
And finally another creative workshop: Ankle bracelet designing with Compeed.
One of the highlights of the FashionBloggerCafé was the blogger runway show. The models were styled by five bloggers, one of them Alice from I heart Alice, who is also the testimonial of the GDS (the shoe fair on which the event took place). Unfortunately I missed the two speeches about media kit and blog monetization but the FashionBloggerCafé was yet again great fun and I hope to be able to go there again next season.
Done! Voted! :) Looks like it was a very nice event!!!
you attend amazing events! so many yummy treats!
That is just so cool! I will definitely vote for you! That fashion show is just so awesome! The fashions are so unique!
Wow, this looks like a fantastic event! Thanks so much for sharing :) You've got my vote for sure!
Looks like great fun and the photos are amazing.


Memorial Cards were scanned into history at a special 2-day event hosted by the Western Family History Association in Lackagh Parish Centre on Saturday November 2nd and on Sunday November 3rd. The scanning sessions were held from 2pm to 5pm on Saturday and from 3pm to 5pm on Sunday.
Members of the association and the public were invited to bring along their memorial cards which were scanned while they waited and returned immediately to each person. Members were invited to bring their friends.
Memorial cards hold a wealth of information that is of huge interest for people tracing their ancestors and to family historians and genealogists. But, as generations move on and as families die out in a locality, there is a risk that memorial cards may be discarded or destroyed. If so, the associated family details of deceased persons are lost forever to their descendants and to family historians.
The purpose of the 2-day event at Lackagh was to preserve, secure and store this information for people tracing their ancestry. It was a continuation of the project commenced during Heritage Week in August 2012. Memorial cards were scanned and recorded by WFHA personnel. The donors of the cards were asked to sign a consent form giving permission to the WFHA for the use of the recorded information in family history research into the future.


text:
Antigonish—Guysborough fut une circonscription électorale fédérale de Nouvelle-Écosse. La circonscription fut représentée de 1917 à 1968.

La circonscription a été créée d'abord en 1914 avec l'union des circonscriptions d'Antigonish et de Guysborough. Abolie en 1966, elle fut redistribuée parmi Cape Breton Highlands—Canso et Nova-Centre.

Géographie
En 1914, la circonscription de Antigonish—Guysborough comprenait:

 Le comté de Antigonish
 Le comté de Guysborough

Députés
 1917-1921 — John Howard Sinclair, Libéral
 1921-1925 — Colin Francis McIsaac, Libéral
 1925-1926 — Edward Mortimer MacDonald, Libéral
 1926-1927 — John C. Douglas, Conservateur
 1927¹-1936 — William Duff, Libéral
 1936¹-1958 — James Ralph Kirk, Libéral
 1958-1962 — Clement O'Leary, Progressiste-conservateur
 1962-1968 — John B. Stewart, Libéral

 ¹ = Élections partielles

Lien externe
 Circonscription Antigonish—Guysborough — Chambre des communes du Canada

Ancienne circonscription fédérale en Nouvelle-Écosse

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}
For a~bounded linear operator $A$ on a~Hilbert space, we consider the equation
$Ax=b$ and its GMRES approximations $x_k$ with residuals $r_k^{A,r_0}:=b-Ax_k$,
i.e., $\|r_k^{A,r_0}\| = \min_{p\in\mathcal{P}_k,\,p(0)=1} \|p(A)r_0\|$.
In this paper we are concerned with convergence of
the GMRES method for $A=B+C$ where $B$ is such that it exhibits
at least
linear GMRES convergence and $C$ is compact. The main result, the proof of which
we postpone to Section~\ref{sec:proof}, is as follows.
\begin{theorem} \label{thm:stab}
  Let $H$ be a~complex separable Hilbert space.
  Suppose $A=B+C$ is invertible with
  $B$ an~invertible bounded linear operator on~$H$ and
  $C$ a~compact linear operator on~$H$ with singular values
  $\|C\|=\sigma_1(C)\geq\sigma_2(C)\geq\cdots\geq0$.
  %
  Then for any $r_0\in H$
  the GMRES residuals $r_k^{A,r_0}$ fulfill
  %
  \begin{gather}
    \label{eq:stab}
    \frac{\|r_k^{A,r_0}\|}{\|r_0\|}
    \leq
    \prod_{j=1}^k \Bigl( M_B + (1+M_B)\, \|A^{-1}\|\, \sigma_j(C) \Bigr)
    \qquad
    \text{for all } k\in\mathbb{N}, \\
    \label{eq:limsup}
    \limsup_{k\rightarrow\infty} \frac{\|r_k^{A,r_0}\|}{\|r_{k-1}^{A,r_0}\|}
    \leq M_B,
    \intertext{where $B$'s linear reduction factor $M_B\in[0,1]$ is given by}
    \label{eq:deflinear}
    M_B = \sup_{z\in H} \sup_{k\in\mathbb{N}} \frac{\|r_k^{B,z}\|}{\|r_{k-1}^{B,z}\|}.
    \intertext{Consequently, if there exists $p>0$ such that}
    \nonumber
    \|C\|_{\mathcal{S}_p}
    \coloneqq \Bigl( \sum_{j=1}^\infty \sigma_j(C)^p \Bigr)^\frac{1}{p}
    < \infty,
    \shortintertext{then}
    \label{eq:stabrate}
    \Biggl( \frac{\|r_k^{A,r_0}\|}{\|r_0\|} \Biggr)^{\frac{1}{k}}
    \leq
    M_B + k^{-\frac{1}{p}}\, (1+M_B)\, \|A^{-1}\|\, \|C\|_{\mathcal{S}_p}
    \qquad
    \text{for all } k\in\mathbb{N}.
  \end{gather}
  %
\end{theorem}

A~preliminary version of the result in Theorem~\ref{thm:stab}
has been published
in the thesis \cite[Appendix~III.B]{ble2019}, where it has
been used to study the pressure convection--diffusion
preconditioner~\cite{ElmanSilvesterWathen2014}
for the solution of the Navier--Stokes equations.
We expect the result to be applicable to analysis of
a~broad range of problems comming from the numerical solution
of partial differential equations, where compact perturbations
of ``nice'' operators often appear.

The linear reduction factor $M_B$ as given by~\eqref{eq:deflinear}
is the best constant of the estimate
\begin{align}
  \label{eq:deflinear2}
  \|r_k^{B,r_0}\| \leq M_B \|r_{k-1}^{B,r_0}\|
  \qquad \text{for all } k\in\mathbb{N} \text{ and all } r_0\in H.
\end{align}
Obviously it holds $0\leq M_B\leq1$ but
Theorem~\ref{thm:stab} is non-trivial only when $M_B<1$.
The result can be interpreted as a~type of
stability\footnote{%
  Here the term ``stability'' is \emph{not} used in the sense
  of studying the behavior of GMRES in finite arithmetic precision,
  which itself is a~broad research field
  \cite[section~5.10]{liesen-strakos-2013}.
} of linear GMRES convergence
with respect to compact perturbations. The bound~\eqref{eq:stab}
controls the departure from linear convergence~\eqref{eq:deflinear2}
in terms of the singular values of~$C$. Recall that $\sigma_j(C)\to0$
as $j\to\infty$ for $C$ compact. If additionally $\sum_{j=1}^\infty
\sigma_j(C)^p$ is finite\footnote{%
  Note that $\|\cdot\|_{\mathcal{S}_p}$ in Theorem~\ref{thm:stab}
  is the norm (quasinorm) on the $p$-Schatten--von-Neumann
  ideal~$\mathcal{S}_p(H)$ when $p\geq1$ ($0<p<1$).
  The Hilbert--Schmidt norm $\|\cdot\|_{\mathcal{S}_2}$
  takes the form of the Frobenius matrix norm
  in the finite-dimensional case.
}, then, in view of~\eqref{eq:stabrate},
the mean departure from the linear convergence is
superlinear $\smash{O\bigl(k^{-\frac1p}\bigr)}$.

For the sake of illustration we explicitly give a~variant of
Theorem~\ref{thm:stab} for the finite-dimensional case.
\begin{corollary}
  Let $B\in\mathbb{C}^{N\times N}$ be a~non-singular matrix
  and $C\in\mathbb{C}^{N\times N}$ be such that $A=B+C$ is non-singular.
  %
  Let $M_B\in[0,1]$ be given by~\eqref{eq:deflinear},
  or equivalently,
  let $M_B\in[0,1]$ be the best possible constant
  from the inequality~\eqref{eq:deflinear2}.
  %
  Then \eqref{eq:stab} holds true and \eqref{eq:stabrate},
  in particular, implies that
  %
  \begin{gather*}
    \Biggl( \frac{\|r_k^{A,r_0}\|}{\|r_0\|} \Biggr)^{\frac{1}{k}}
    \leq
    M_B + \frac{1}{\sqrt{k}}\, (1+M_B)\, \|A^{-1}\|\, \|C\|_{\mathcal{S}_2}
    \qquad
    \text{for all } k\in\mathbb{N},
  \end{gather*}
  %
  where $\|C\|_{\mathcal{S}_2}$ is the Frobenius matrix norm of~$C$.
\end{corollary}

We will devote Section~\ref{sec:linear} to obtaining an~intrinsic
characterization of~$M_B$ which does not depend on GMRES.
We will show in Propositions~\ref{thm:linear}
and~\ref{lemma:minimizerlambda} that
\begin{align*}
  M_B
  = \min_{\lambda\in\mathbb{C}} \|I-\lambda B\|
  = \min_{\lambda\in\mathbb{C}} \|I-\lambda B^{-1}\|
\end{align*}
and that the minima are attained on $\fovcl{B^{-1}}$ and $\fovcl{B}$,
respectively,
where $\fovcl{T}$ stands for the closure of~$\fov{T}$,
the numerical range (field of values) of~$T$.
The well-known sufficient condition for $M_B<1$ that $0\notin\fovcl{T}$
is due to Elman's bound~\cite{elman1982}
\begin{align*}
  M_B
  &\leq \sqrt{1-\frac{\nu_B^2}{\|B\|^2}}
\end{align*}
and the improved bound by Starke~\cite{starke1997},
Eiermann, and Ernst~\cite{EiermannErnst2001}
\begin{align*}
  M_B
  &\leq \sqrt{1-\nu_B\nu_{B^{-1}}}
  \leq \sqrt{1-\frac{\nu_B^2}{\|B\|^2}},
\end{align*}
where $\nu_{T}=\inf_{\lambda\in\fov{T}}|\lambda|$.
(Note that $\nu_{T^{-1}}>0$ if and only if $\nu_T>0$ for invertible~$T$.)
If $H$ is finite-dimensional, then the linear convergence~\eqref{eq:deflinear2}
with some $M_B<1$ is equivalent to strictly monotone convergence for any
initial residual, i.e.,
\begin{align}
  \label{eq:defmonotone}
  \|r_k^{B,r_0}\| < \|r_{k-1}^{B,r_0}\|
  \quad\text{or}\quad
  \|r_{k-1}^{B,r_0}\| = 0
  \qquad \text{for every } k\in\mathbb{N} \text{ and every } r_0\in H.
\end{align}
We will also show in Proposition~\ref{thm:linear}
that \eqref{eq:deflinear2} with $M_B<1$ holds
if and only if $0\notin\fovcl{B}$,
that~\eqref{eq:defmonotone} holds
if and only if $0\notin\fov{B}$,
and that the two situations are not, in general, equivalent unless
$H$ is finite-dimensional or $B$ is self-adjoint.
In fact, we will give an~example in Remark~\ref{rem:example}
of a~unitary $B$ with $0\notin\fov{B}$ and
$0\in\fovcl{B}$ so that $B$ exhibits strictly monotone
GMRES convergence~\eqref{eq:defmonotone} but does not exhibit linear
GMRES convergence of any rate, i.e., the best~$M_B$
of~\eqref{eq:deflinear2} is $M_B=1$.

The definition of~$M_B$ that we consider corresponds to the optimal constant
in bounds of the form $ \|r_k^{B,r_0}\| \leq M_B^k\, \|r_0\| $.
On the other hand, one might have a~bound of the type
$ \|r_k^{B,r_0}\| \leq c\,M_{B,c}^k\, \|r_0\| $
with some $c>1$ and $M_{B,c}<1$, which can be a~better bound
in the sense that $M_{B,c}<M_B$. We do not deal
with the latter in the present paper and we point
the interested reader to~\cite{liesen-tichy-2018}.

Theorem~\ref{thm:stab} can be seen as a~generalization Moret's
result~\cite{moret1997}, which only considers $B=\lambda I$,
$\lambda\neq0$. Indeed, $M_{\lambda I}=0$ and hence
\eqref{eq:limsup} gives the \emph{Q-superlinear convergence}
$\lim_{k\rightarrow\infty} \frac{\|r_k\|}{\|r_{k-1}\|}=0$
of~\cite[Theorem 1]{moret1997} and \eqref{eq:stabrate}
gives the rate $\|r_k\|^\frac1k = O\bigl(k^{-\frac1p}\bigr)$
of~\cite[eq.~(1.1)]{moret1997}; only the estimate
\cite[ineq.~(2.7)]{moret1997} is finer than ours:
\begin{align*}
  \frac{\|r_k^{A,r_0}\|}{\|r_0\|}
  \leq
  \prod_{j=1}^k \sigma_j(A^{-1})\, \sigma_j(C)
  \leq
  \prod_{j=1}^k \|A^{-1}\|\, \sigma_j(C);
\end{align*}
the first inequality is due to Moret and the right-most
expression comes from the bound~\eqref{eq:stab}.

Under the conditions of Theorem~\ref{thm:stab},
Hansmann's theorem \cite[Theorem~2.1]{hansmann2011} provides
a~bound for the accumulation rate of discrete eigenvalues of~$A$
at the numerical range of~$B$. A~straightforward application
of the theorem gives the estimate
\begin{align*}
  \sum_{\lambda\in\sigma_\mathrm{disc}(A)}
  \dist\bigl(\lambda,\fovcl{B}\bigr)^p
  \leq
  \|C\|_{\mathcal{S}_p}^p
  \qquad
  \text{for any } p>1,
\end{align*}
where $\sigma_\mathrm{disc}(A)$ is the set of eigenvalues of~$A$
with finite algebraic multiplicity and each eigenvalue is counted
in the sum according to its
algebraic multiplicity. This estimate can be understood as the counterpart
to the perturbation result of Theorem~\ref{thm:stab} concerning
the perturbation of spectra\footnote{%
  Here, recall that
  concerning the finite-dimensional case
  the spectrum itself is not sufficient for obtaining any useful information
  about the behavior of GMRES \cite{greenbaum-strakos,greenbaum-ptak-strakos}.
  On the other hand, in the infinite-dimensional case the spectrum determines,
  in certain sense, the behavior of GMRES at high iteration counts $k\to\infty$;
  see \cite{nevanlinna1993}. The relationship between the finite-dimensional
  and infinite-dimensional cases is not, to our best knowledge, well established.
}.

Nevanlinna~\cite{nevanlinna1996} and Malinen~\cite{malinen1996}
similarly deal with GMRES for compactly perturbed operators
but they use techniques of complex analysis, which we completely
avoid in this paper. As a~consequence, our result is not obscured
by constants that are rather difficult to evaluate.
See also \cite{huhtanen-nevanlinna-2000}, in which splittings
$A=B+C$ with $B$ normal and $C$ compact (or low-rank) are
considered and lower bounds on GMRES convergence speed
are obtained.

The outline of the paper is as follows. In Section~\ref{sec:prelim}
we fix the notation and gather some basic facts. Section~\ref{sec:linear}
contains a~characterization of linear convergence and the
reduction factor~$M_B$. In Section~\ref{sec:proof},
building on Moret's result~\cite{moret1997},
we prove Theorem~\ref{thm:stab}. We briefly conclude
and mention interesting questions this research gave rise to
in Section~\ref{sec:outro}.

\section{Preliminaries}
\label{sec:prelim}
Throughout the paper we assume that $H$ is
a~complex Hilbert space with an~inner
product~$(\cdot,\cdot)$ which is linear and antilinear
in the first and second argument, respectively,
and the symbol ${\|\cdot\|}$ stands for
either the norm on~$H$ induced by $(\cdot,\cdot)$ or
the induced operator norm on~$\mathcal{L}(H)$, the space
of bounded linear operators on~$H$.
For operator $A\in\mathcal{L}(H)$, initial residual $r_0\in H$,
and integer $k=0,1,2,\ldots$ we define the Krylov subspace
$\mathcal{K}_k(A,r_0)\coloneqq
\operatorname{span}\{r_0,Ar_0,A^2r_0,\ldots,A^{k-1}r_0\}
\subset H$.
For a~right-hand side $b\in H$ and an~initial guess $x_0\in H$,
the initial residual is given by $r_0=b-Ax_0$.
Then the GMRES algorithm constructs a~sequence
$\{x_k\}_{k=1}^\infty\subset H$ given by minimizing the norm of
residuals $r_k^{A,r_0}=b-Ax_k$ over $x_k\in x_0+\mathcal{K}_k(A,r_0)$, i.e.,
\begin{align}
  \label{eq:gmreskrylov}
  x_k &= \argmin_{\substack{x_k\in H \\x_k-x_0\in\mathcal{K}_k(A,r_0)}} \|b-Ax_k\|.
\end{align}
We will write just $r_k$ instead of~$r_k^{A,r_0}$
if there is no risk of confusion.

\begingroup\setlength\emergencystretch{\hsize}\hbadness=10000
Assume that $t_1,t_2,\ldots,t_k$ is
the ascending orthonormal basis of the spaces $\mathcal{K}_k(A,r_0)$, $k=1,2,\ldots$, and
$z_1,z_2,\ldots,z_k$ is the ascending orthornormal basis of the spaces $A\mathcal{K}_k(A,r_0)$,
$k=1,2,\ldots$. This is well-defined~if
\begin{align*}
  \mathcal{K}_{k+1}(A,r_0) \supsetneq \mathcal{K}_k(A,r_0)
  \qquad \text{for all } k=1,2,\ldots.
\end{align*}
\endgroup
It is well-known that in the converse case, when
$\mathcal{K}_{m+1}(A,r_0)={}\allowbreak \mathcal{K}_{m}(A,r_0)
\supsetneq{}\allowbreak \mathcal{K}_{m-1}(A,r_0)\supsetneq{}\allowbreak \ldots$
for certain $m$, the solution has been reached, i.e., $Ax_m = b$,
provided that $A$ is invertible.
To see this, observe that $A\mathcal{K}_{m}(A,r_0)\subset\mathcal{K}_{m+1}(A,r_0)$
but at the same time
$\dim A\mathcal{K}_{m}(A,r_0)=\dim\mathcal{K}_{m}(A,r_0)=\dim\mathcal{K}_{m+1}(A,r_0)$
by the invertibility of $A$; hence
$A\mathcal{K}_{m}(A,r_0)=\mathcal{K}_{m+1}(A,r_0)\ni r_0$, i.e.,
$r_0=\hat p(A)Ar_0$ with some $\hat p\in\mathcal{P}_{m-1}$
which means that $r_m=r_0-\hat p(A)Ar_0=0$; here and throughout the paper
$\mathcal{P}_n$ stands for the space of polynomials of degree at most~$n$.

Hence, if $A$ is invertible, we can assume that $\{t_j\}_{j=1}^\infty$ and
$\{z_j\}_{j=1}^\infty$ are extended orthonormal systems, i.e.,
either (i) $\{t_j\}_{j=1}^\infty$ and $\{z_j\}_{j=1}^\infty$ are
orthonormal systems, or (ii) $\{t_j\}_{j=1}^\infty$ and $\{z_j\}_{j=1}^\infty$
are such that $\{t_j\}_{j=1}^m$ and $\{z_j\}_{j=1}^m$
are orthonormal systems and $\{t_j\}_{j=m+1}^\infty$ and $\{z_j\}_{j=m+1}^\infty$
are zero sequences. In both cases $t_1,t_2,\ldots,t_k$ and
$z_1,z_2,\ldots,z_k$ are orthonormal bases of $\mathcal{K}_k(A,r_0)$
and $A\mathcal{K}_k(A,r_0)$, respectively, for each $k\in\mathbb{N}$.

The numerical range of a~bounded linear operator~$A$ on~$H$ is
a~subset of~$\mathbb{C}$ given by
\begin{gather*}
  \fov A \coloneqq
  \bigl\{ (Az,z),\, z\in H,\, \|z\|=1 \bigr\}.
  \shortintertext{We set}
  \nu_A \coloneqq
  \inf_{\lambda\in\fov A} |\lambda|.
\end{gather*}
The closure of the numerical range, $\fovcl A$, contains the spectrum
of~$A$. The Toeplitz--Hausdorff theorem
\cite[Theorem 9.3.1]{davies2007}, \cite{gustafson-1970}
says that $\fov A$ is a~convex set. The Lax--Milgram theorem guarantees that
$A$ is invertible provided $0\notin\fovcl A$ and in such a~situation
it holds that $\|A^{-1}\|\leq\nu_A^{-1}$.

\section{Characterization of linear GMRES convergence}
\label{sec:linear}
\begin{proposition}
  \label{thm:linear}
  Let $B$ be a~bounded linear operator on a~complex Hilbert space $H$.
  Further suppose that $B$ is invertible.
  The linear reduction factor~\eqref{eq:deflinear} fulfills
  %
  \begin{align}
     \label{eq:bndM}
     M_B
     = \inf_{\lambda\in\mathbb{C}} \|I-\lambda B\|
     = \inf_{\lambda\in\mathbb{C}} \|I-\lambda B^{-1}\|
     \leq \sqrt{1 - \nu_B \nu_{B^{-1}}}
     \leq \sqrt{1 - \frac{\nu_B^2}{\|B\|^2}}
     .
  \end{align}
  %
  Furthermore, the following assertions are equivalent:
  %
  \begin{enumerate}[\normalfont(L1)]
    \item
      \label{L1}
      the operator $B$ exhibits
      linear GMRES convergence~\eqref{eq:deflinear2} with
      some $M_B<1$;
    \item
      \label{L2}
      $\nu_B > 0$;
    \item
      \label{L3}
      $\nu_{B^{-1}} > 0$.
  \end{enumerate}
  %
  %
  %
  %
  The following assertions are equivalent as well:
  %
  \begin{enumerate}[\normalfont(M1)]
    \item
      \label{M1}
      the operator $B$ exhibits
      strictly monotone GMRES convergence~\eqref{eq:defmonotone};
      %
    \item
      \label{M2}
      $0\notin\fov B$;
    \item
      \label{M3}
      $0\notin\fov B^{-1}$.
  \end{enumerate}
  %
  If $H$ is finite-dimensional or $B$ is self-adjoint then
  \LREF{L1}, \LREF{L2}, \LREF{L3}, \MREF{M1}, \MREF{M2}, and \MREF{M3}
  are equivalent.
  %
\end{proposition}
\begin{proof}
  From~\eqref{eq:gmreskrylov} we have, for all $\lambda\in\mathbb{C}$,
  $r_0\in H$, and $k\in\mathbb{N}$,
  %
  \begin{align*}
    \|r_k\|
    = \min_{\substack{p\in\mathcal{P}_k \\ p(0)=1}} \|p(B)r_0\|
    \leq \|I-\lambda B\|
    \min_{\substack{p\in\mathcal{P}_{k-1} \\ p(0)=1}} \|p(B)r_0\|
    = \|I-\lambda B\|\, \|r_{k-1}\|.
  \end{align*}
  %
  Dividing by $\|r_{k-1}\|$,
  taking the infimum over $\lambda\in\mathbb{C}$ and the supremum
  over $r_0\in H$ and $k\in\mathbb{N}$, we immediately obtain
  $M_B\leq\inf_{\lambda\in\mathbb{C}}\|I-\lambda B\|$.
  We continue by showing the opposite inequality.
  %
  The minimax theorem of Asplund and Pt\'{a}k~\cite{asplund-ptak-1971}
  says that
  %
  \begin{align}
    \label{eq:minimax}
    \inf_{\lambda\in\mathbb{C}} \|I-\lambda B\|
    = \inf_{\lambda\in\mathbb{C}} \sup_{\substack{z\in H \\ \|z\|=1}} \|z-\lambda B z\|
    = \sup_{\substack{z\in H \\ \|z\|=1}} \inf_{\lambda\in\mathbb{C}} \|z-\lambda B z\|
    .
  \end{align}
  %
  The minimization on the right hand-side is nothing other than
  the first GMRES step so that
  $\inf_{\lambda\in\mathbb{C}} \|I-\lambda B\|=
  \sup_{r_0\in H} \frac{\|r_1\|}{\|r_0\|}\leq M_B$
  and the first equality in~\eqref{eq:bndM} is thus proved.
  %
  Furthermore, the minimization on the right-hand side of~\eqref{eq:minimax}
  can be carried out explicitly:
  for any $z\in H$, $z\neq0$ it holds that
  %
  \begin{align}
    \label{eq:stepone}
    \min_{\lambda\in\mathbb{C}} \frac{\|z-\lambda B z\|^2}{\|z\|^2}
    = 1 - \frac{|(B z,z)|^2}{\|B z\|^2 \|z\|^2}
  \end{align}
  %
  and the minimum is attained for $\lambda=\frac{(z,B z)}{\|B z\|^2}$.
  %
  Equations~\eqref{eq:minimax} and~\eqref{eq:stepone} imply
  %
  \begin{subequations}
    \label{eq:stepone2}
    \begin{align}
      \label{eq:stepone2a}
      \inf_{\lambda\in\mathbb{C}} \|I-\lambda B\|
      &= \sup_{z\in H}
         \sqrt{1-\frac{|(B z,z)|^2}{\|B z\|^2 \|z\|^2}},
      \\
      \label{eq:stepone2b}
      \inf_{\lambda\in\mathbb{C}} \|I-\lambda B^{-1}\|
      &= \sup_{z\in H}
         \sqrt{1-\frac{|(B^{-1}z,z)|^2}{\|B^{-1}z\|^2 \|z\|^2}},
    \end{align}
  \end{subequations}
  %
  but by virtue of the invertibility of~$B$,
  the right-hand sides in~\eqref{eq:stepone2} are the same, which shows
  the second equality in~\eqref{eq:bndM}.
  %
  Using the obvious inequalities
  %
  \begin{subequations}
    \label{eq:ineq}
    \begin{gather}
      \label{eq:ineqa}
      \inf_{z\in H} \frac{|(B z,z)|^2}{\|B z\|^2 \|z\|^2}
      \geq \inf_{z\in H} \frac{|(B z,z)|}{\|z\|^2}
           \inf_{z\in H} \frac{|(B z,z)|}{\|B z\|^2}
      %
      %
      = \nu_B \nu_{B^{-1}},
      \\
      \label{eq:ineqb}
      \nu_{B^{-1}} \geq \frac{\nu_B}{\|B\|^2},
      \qquad
      \nu_B \geq \frac{\nu_{B^{-1}}}{\|B^{-1}\|^2},
    \end{gather}
  \end{subequations}
  %
  together with~\eqref{eq:stepone2}
  immediately yields both inequalities in~\eqref{eq:bndM}.
  Thus~\eqref{eq:bndM} is proved.

  The inequalities in~\eqref{eq:ineqb} establish the equivalence
  of~\LREF{L2} and~\LREF{L3}. The assertion~\LREF{L2}
  or~\LREF{L3} implies~\LREF{L1} by~\eqref{eq:bndM}.
  We proceed by showing that~\LREF{L1} implies~\LREF{L2}.
  Assume that~\LREF{L2} is violated, i.e., that $\nu_B=0$.
  Hence there exists $\{z_j\}_{j=1}^\infty\subset H$ such that
  $\|z_j\|=1$ and $(B z_j,z_j)\rightarrow0$ as $j\rightarrow\infty$.
  Therefore $\frac{|(B z_j,z_j)|}{\|B z_j\|} \leq \|B^{-1}\|\, |(B z_j,z_j)|
  \rightarrow0$. Thus by~\eqref{eq:stepone} we obtain
  %
  \begin{gather*}
    \inf_{\lambda\in\mathbb{C}} \|z_j-\lambda B z_j\|^2
    = 1 - \frac{|(B z_j,z_j)|^2}{\|B z_j\|^2}
    \rightarrow 1.
    %
    \shortintertext{Thus}
    %
    \sup_{\substack{r_0\in H \\ \|r_0\|=1}}
    \inf_{\lambda\in\mathbb{C}} \|r_0-\lambda B r_0\|
    = 1
  \end{gather*}
  %
  so that $M_B=1$ and implication \LREF{L1}$\Rightarrow$\LREF{L2}
  is proved.

  The equivalence of~\MREF{M2} and~\MREF{M3} follows directly
  from the definition of numerical range owing to the invertibility
  of~$B$. We proceed by showing that~\MREF{M2} implies~\MREF{M1}.
  We fix $k\in\mathbb{N}$. If $r_{k-1}=0$ we are done so let us assume
  the converse. We have
  %
  \begin{align*}
    \frac{\|r_k\|}{\|r_{k-1}\|}
    = \min_{\substack{p\in\mathcal{P}_k \\ p(0)=1}}
      \frac{\|p(B)r_0\|}{\|r_{k-1}\|}
    \leq \frac{\|(I-\lambda B)r_{k-1}\|}{\|r_{k-1}\|}
  \end{align*}
  %
  for any $\lambda\in\mathbb{C}$. Hence with $\lambda$ minimizing
  the right-hand side we obtain, by~\eqref{eq:stepone}
  and~\MREF{M2},
  %
  \begin{align*}
    \frac{\|r_k\|}{\|r_{k-1}\|} \leq
    \sqrt{1-\frac{|(B r_{k-1},r_{k-1})|^2}{\|B r_{k-1}\|^2 \|r_{k-1}\|^2}}
    < 1
  \end{align*}
  %
  so that~\MREF{M1} follows.
  %
  On the other hand, if~\MREF{M2} is violated
  then there exists
  $r_0\in H$ with $r_0\neq0$ and $(B r_0,r_0)=0$
  so that~\eqref{eq:stepone} immediately implies
  $\|r_1\|=\|r_0\|$, which contradicts~\MREF{M1}.
  Hence the equivalence of~\MREF{M1} and~\MREF{M2}
  is now clear.

  If $H$ is finite-dimensional then $\fov B$
  is closed and thus \LREF{L2} and \MREF{M2} are equivalent.
  %
  Now assume $B$ is self-adjoint. Clearly \LREF{L2} implies
  \MREF{M2}. To prove the opposite, assume that $\nu_B=0$
  but $0\notin\fov B$. As $\fov B$ is convex, it must be
  of the form $[a,0)$, $(a,0)$, $(0,b)$, or $(0,b]$ with
  some $a<0$ or $b>0$. On the other hand, the endpoints of
  $\fov B$ belong to the spectrum of~$B$, which is always
  closed. Hence the conclusion is that $0$ is in the spectrum,
  which contradicts the invertibility of~$B$.
  %
\end{proof}

\begin{remark}
  \label{rem:example}
  %
  We give an~example of a~unitary operator fulfilling~\MREF{M1}
  but contradicting~\LREF{L1}, thus showing that~\LREF{L1}
  and~\MREF{M1} are not in general equivalent.
  %
  Let $B$ be a~diagonal operator on~$\ell^2$
  given by
  %
  \begin{align*}
    B = \sum_{j\in\mathbb{Z}} \lambda_j e_j (\cdot,e_j),
    \qquad
    \lambda_j=e^{i\arctan j} \text{ for }j\in\mathbb{Z},
  \end{align*}
  %
  where $\{e_j\}_{j\in\mathbb{Z}}$ is the canonical basis in~$\ell^2$,
  which is orthonormal with respect to $(\cdot,\cdot)$.
  The eigenvalues $\{\lambda_j\}_{j\in\mathbb{Z}}$ fulfill
  $|\lambda_j|=1$ and $\real\lambda_j>0$ and,
  indeed, $B$ is unitary and hence invertible. In fact,
  the spectrum of~$B$ is $\{\lambda_j\}_{j\in\mathbb{Z}} \cup \{i,-i\}$.
  Indeed, $\|(B\mp iI)e_k\|=|\lambda_k\mp i|\rightarrow0$
  as $k\rightarrow\pm\infty$, i.e., $\pm i$ is in
  the approximate point spectrum of~$B$.
  %
  For $f_k\coloneqq e_k+e_{-k}$ it holds that
  $\frac{(B f_k,f_k)}{\|f_k\|^2}=\real\lambda_k\rightarrow0$
  as $k\rightarrow\pm\infty$. Hence $\nu_B=0$, which
  is~\LREF{L2} or, equivalently, \LREF{L1}.
  %
  On the other hand, $(B z,z)=\sum_{j\in\mathbb{Z}}\lambda_j|(z,e_j)|^2$
  which is non-zero whenever $z\neq0$, because
  $\real\lambda_j>0$ for all $j\in\mathbb{Z}$. Thus $0\notin\fov B$,
  which is~\MREF{M2} so that~\MREF{M1} holds true.
  %
\end{remark}

The following result characterizes $\lambda$ that minimize~\eqref{eq:bndM}.
\begin{proposition}
  \label{lemma:minimizerlambda}
  Let $H$ be a~complex Hilbert space and $B\in\mathcal{L}(H)$ be invertible.
  %
  There exist $\lambda_B\in\mathbb{C}$ and $\lambda_{B^{-1}}\in\mathbb{C}$ such that the infima
  on the left-hand sides of~\eqref{eq:stepone2a} and~\eqref{eq:stepone2b},
  respectively, are attained.
  %
  Any such $\lambda_B$ and $\lambda_{B^{-1}}$ fulfill
  %
  \begin{align*}
    \lambda_B
    &\in\fovcl B^{-1},
    &
    \|\lambda_B B\|
    &\leq 1 + M_B,
    \\
    \lambda_{B^{-1}}
    &\in\fovcl B,
    &
    \|\lambda_{B^{-1}} B^{-1}\|
    &\leq 1 + M_B.
  \end{align*}
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
  %
\end{proposition}
\begin{proof}
  We will first prove that any minimizer of~\eqref{eq:stepone2a}, if it
  exists, fulfills $\lambda_B\in\fovcl{B^{-1}}$. For the sake of contradiction
  assume that $\lambda_B\in\mathbb{C}\setminus\fovcl{B^{-1}}$. Let $z\in H$ with
  $\|z\|=1$ be arbitrary. The function $\lambda\mapsto\|z-\lambda B z\|^2$
  is minimized by $\lambda_z = \frac{(z,B z)}{\|B z\|^2}\in\fov{B^{-1}}$;
  see \eqref{eq:stepone}. Hence $|\lambda_B-\lambda_z|\geq C>0$,
  where $C=\dist(\lambda_B,\fov{B^{-1}})$ is independent of~$z$.
  %
  Now consider that
  %
  \begin{align*}
    \|z-\lambda_B B z\|^2 &=
    1 - 2\real\bigl( \lambda_B (B z,z) \bigr) + |\lambda_B|^2\,\|B z\|^2,
    \\
    \|z-\lambda_z B z\|^2 &=
    1 - \frac{|(B z,z)|^2}{\|B z\|^2},
  \end{align*}
  %
  where the second equality has been already shown in~\eqref{eq:stepone}.
  Combining the last two equalities we obtain
  %
  \begin{gather*}
    \begin{aligned}
      \|z-\lambda_B B z\|^2 - \|z-\lambda_z B z\|^2
      &= \frac{|(B z,z)|^2}{\|B z\|^2}
      - 2\real\bigl( \lambda_B (B z,z) \bigr) + |\lambda_B|^2\,\|B z\|^2
      \\
      &= \|B z\|^2\, |\lambda_B-\lambda_z|^2
      \geq \|B^{-1}\|^{-2}\, C^2.
    \end{aligned}
    %
    \shortintertext{Hence}
    %
    \begin{aligned}
      \|I-\lambda_B B\|^2 = \sup_{\|z\|=1} \|z-\lambda_B B z\|^2
      &\geq \sup_{\|z\|=1} \|z-\lambda_z B z\|^2 + \|B^{-1}\|^{-2}\, C^2
      \\
      &> \sup_{\|z\|=1} \|z-\lambda_z B z\|^2
      = \inf_{\lambda\in\mathbb{C}} \|I-\lambda B\|^2,
    \end{aligned}
  \end{gather*}
  %
  where we used~\eqref{eq:minimax} in the last equality,
  and this gives the desired contradiction.
  Hence any minimizers of~\eqref{eq:stepone2a} belong
  to the compact set $\fovcl{B^{-1}}$, where  the function
  to be minimized is continuous, so the existence is also proved.
  %
  By the triangle inequality we have
  $\|\lambda_B B\| \leq \|I\| + \|I-\lambda_B B\| = 1 + M_B$.
  %
  The proof for $\lambda_{B^{-1}}$ is analogous.
\end{proof}

\section{Proof of the main result}
\label{sec:proof}
Moret~\cite[Lemma 6]{moret1997}, as well as
Eiermann and Ernst~\cite[Lemma 3.3]{EiermannErnst2001},
proves the following auxiliary result,
which will be crucial for us.
\begingroup\setlength\emergencystretch{\hsize}\hbadness=10000
\begin{lemma}[Moret's formula]
  \label{lemma:moret}
  %
  Let $H$ be a~separable complex Hilbert space,
  $A\in\mathcal{L}(H)$ be invertible,
  and $r_0\in H$ be given.
  For every $k\in\mathbb{N}$ and $\lambda\in\mathbb{C}$ the GMRES residuals
  $r_k \coloneqq r_k^{A,r_0}$ fulfill
  %
  \begin{align}
    \label{eq:moret}
    \|r_k\|
    = |(t_{k+1},z_k)| \|r_{k-1}\|
    = |(t_{k+1},(I-\lambda A^{-1})z_k)| \|r_{k-1}\|,
  \end{align}
  %
  where $\{t_j\}_{j=1}^\infty$ and $\{z_j\}_{j=1}^\infty$
  are the ascending extended orthonormal bases of
  %
  %
  $\{\mathcal{K}_1(A,r_0),\allowbreak \mathcal{K}_2(A,r_0),\allowbreak \ldots\}$
  and
  %
  %
  $\{A\mathcal{K}_1(A,r_0),\allowbreak A\mathcal{K}_2(A,r_0),\allowbreak \ldots\}$,
  respectively.
  %
\end{lemma}
\endgroup
\noindent
Note that the second equality in~\eqref{eq:moret} follows trivially
from the definition of $t_{k+1}$ and $z_k$;
indeed $A^{-1}z_k\in\mathcal{K}_k(A,r_0)$, which is a~space spanned
by $\{t_1,\ldots,t_k\}$, and $(t_{k+1},t_j)=0$ for all $j\leq k$.

The approximation numbers of a~bounded linear operator $T$
on a~Hilbert space~$H$ are defined by
\begin{align}
  \label{eq:apnums}
  \sigma_j(T) = \inf_{\substack{
      M\in\mathcal{L}(H) \\
      \operatorname{rank}(M) < j
    }} \|T-M\|,
    \qquad j=1,2,\ldots
\end{align}
The approximation numbers fulfill
\begin{subequations}
  \label{eq:snum}
  \begin{gather}
    \label{eq:snum1}
    \|T\| = \sigma_1(T) \geq \sigma_2(T) \geq \ldots \geq 0,
    \qquad \qquad \qquad \qquad \quad
    \\
    \label{eq:snum2}
    \sigma_{j+k-1}(S+T) \leq \sigma_j(S) + \sigma_k(T),
    \qquad
    j,k=1,2,\ldots,
    \\
    \label{eq:snum3}
    \sigma_j(WTU) \leq \|W\|\, \sigma_j(T)\, \|U\|,
    \qquad \quad
    j=1,2,\ldots
  \end{gather}
\end{subequations}
with any $S,T,U,W\in\mathcal{L}(H)$;
see~\cite[paragraph~2.2.1, p.~79, Theorem~2.3.3, p.~83]{pietsch1987}.
If $T$ is compact, the numbers $\sigma_j(T)$ are the singular
values of~$T$.

\begin{lemma}[{Pietsch~\cite[Lemma 2.11.13, p. 125]{pietsch1987}}]
  Let $T$ be a~bounded linear operator on a~complex Hilbert space $H$.
  Let $\sigma_1(T) \geq \sigma_2(T) \geq \sigma_3(T) \geq \ldots \geq 0$
  denote the approximation numbers of $T$ as defined by~\eqref{eq:apnums}.
  Then for any pair of orthonormal families
  $\{f_1,f_2,\ldots,f_k\}$, $\{g_1,g_2,\ldots,g_k\}\subset H$
  it holds that
  %
  \begin{align}
    \label{eq:determinant}
    \det\bigl\{ (Tf_i,g_j) \bigr\}_{i,j=1}^{k} \leq \prod_{j=1}^k \sigma_k(T).
  \end{align}
  %
\end{lemma}
Moret \cite[ineq.~(2.7)]{moret1997} proves that if $A-\lambda I$
is compact for some $\lambda\in\mathbb{C}$, then
\begin{align*}
  \frac{\|r_k\|}{\|r_0\|} \leq \prod_{j=1}^k
  \sigma_j(A-\lambda I)\,\sigma_j(A^{-1}).
\end{align*}
We will need a~modification, which follows.
\begin{lemma}
  Suppose that the assumptions of Lemma~\ref{lemma:moret} are fullfilled.
  Then for every $k\in\mathbb{N}$ and $\lambda\in\mathbb{C}$ the GMRES residuals
  $r_k \coloneqq r_k^{A,r_0}$ fulfill
  %
  \begin{align}
    \label{eq:resestapprox}
    \frac{\|r_k\|}{\|r_0\|} \leq \prod_{j=1}^k \sigma_j(I-\lambda A^{-1}),
  \end{align}
  %
  where
  $\sigma_1(I-\lambda A^{-1}) \geq \sigma_2(I-\lambda A^{-1}) \geq \ldots \geq 0$
  are the approximation numbers of $I - \lambda A^{-1}$
  as defined by~\eqref{eq:apnums}.
  %
\end{lemma}
\begin{proof}
  From~\eqref{eq:moret} we have
  %
  \begin{gather*}
    \frac{\|r_k\|}{\|r_0\|}
    = \prod_{j=1}^k |(t_{j+1},(I-\lambda A^{-1})z_j)|.
    %
    \shortintertext{%
      The matrix
    }
    %
    \bigl\{ |(t_{i+1},(I-\lambda A^{-1})z_j)| \bigr\}_{i,j=1}^{k}
  \end{gather*}
  %
  is upper triangular because by construction,
  %
  \begin{align*}
    0 &= (t_{j+2},z_j) = (t_{j+3},z_j) = \ldots, \\
    0 &= (t_{j+1},A^{-1}z_j) = (t_{j+2},A^{-1}z_j) = \ldots.
  \end{align*}
  %
  This implies that $\prod_{j=1}^k |(t_{j+1},(I-\lambda A^{-1})z_j)| =
  \det\bigl\{ |(t_{i+1},(I-\lambda A^{-1})z_j)| \bigr\}_{i,j=1}^{k}$
  which is bounded by
  $\prod_{j=1}^k \sigma_j(I-\lambda A^{-1})$ due to~\eqref{eq:determinant}.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:stab}]
  By virtue of the invertibility of $A=B+C$ and the invertibility
  of~$B$ we have
  $A^{-1}=B^{-1}-B^{-1}CA^{-1}$.
  Hence by Lemma~\ref{lemma:moret} we have
  %
  \begin{align*}
    \frac{\|r_k\|}{\|r_{k-1}\|}
    &\leq |(t_{k+1},(I-\lambda B^{-1})z_k)|
        + |(t_{k+1},\lambda B^{-1}CA^{-1}z_k)|
    \\
    &\leq \|I-\lambda B^{-1}\| + \|\lambda B^{-1}CA^{-1}z_k\|.
  \end{align*}
  %
  The sequence $\{z_k\}_{k=1}^\infty$ is an extended orthonormal system, which is,
  by Bessel's inequality, weakly null. By the compactness
  of $C$, \eqref{eq:limsup} follows.

  Using~\eqref{eq:resestapprox} and~\eqref{eq:snum}
  we obtain, for any $\lambda\in\mathbb{C}$,
  %
  \begin{align*}
    \frac{\|r_k\|}{\|r_0\|}
    \leq \prod_{j=1}^k \sigma_j(I-\lambda A^{-1})
    &\leq \prod_{j=1}^k \Bigl(
        \sigma_1(I-\lambda B^{-1}) + \sigma_j(\lambda B^{-1}CA^{-1})
    \Bigr)
    \\
    &\leq \prod_{j=1}^k \Bigl(
        \|I-\lambda B^{-1}\| + \|\lambda B^{-1}\|\,\|A^{-1}\|\,\sigma_j(C)
    \Bigr).
  \end{align*}
  %
  Taking $\lambda\in\mathbb{C}$ minimizing $\|I-\lambda B^{-1}\|$, we
  obtain~\eqref{eq:stab} using Proposition~\ref{lemma:minimizerlambda}.

  From~\eqref{eq:stab} we obtain,
  using the AM--GM inequality and H\"older's inequality,
  %
  \begin{align*}
    \biggl( \frac{\|r_k\|}{\|r_0\|} \biggr)^\frac1k
    &\leq \tfrac1k \sum_{j=1}^k \Bigl( M_B + (1+M_B)\, \|A^{-1}\|\, \sigma_j(C) \Bigr)
    \\
    &= M_B + \tfrac1k\, (1+M_B)\, \|A^{-1}\|\, \sum_{j=1}^k \sigma_j(C)
    \\
    &\leq M_B + \tfrac1k\, (1+M_B)\, \|A^{-1}\|\,
               \Biggl( \sum_{j=1}^k \sigma_j(C)^p \Biggr)^{\frac1p} k^{\frac{p-1}{p}},
  \end{align*}
  %
  which immediately yields~\eqref{eq:stabrate}.
  %
\end{proof}

\section{Conclusion and outlook}
\label{sec:outro}
We generalized Moret's result~\cite{moret1997} in Theorem~\ref{thm:stab}
and thus obtained a~certain stability of linear GMRES convergence
under compact perturbations. Indeed, the superlinear
convergence of~\cite{moret1997} follows as a~corollary.
In Section~\ref{sec:linear} we obtained a~useful characterization
of linear convergence, which sheds some light on the limits of
the applicability of Theorem~\ref{thm:stab}.

Bound~\eqref{eq:limsup} has an interesting consequence:
\begin{align}
  \label{eq:limsup2}
  \limsup_{k\rightarrow\infty} \frac{\|r_k^{A,r_0}\|}{\|r_{k-1}^{A,r_0}\|}
  \leq \inf_{K\in\mathcal{C}(H)} M_{A+K}
  \eqqcolon
  M_{\pi(A)},
\end{align}
where the infimum is taken over $\mathcal{C}(H)$, the ideal
of all compact operators in $\mathcal{L}(H)$.
Here $\pi$ is the quotient map
$\pi:\mathcal{L}(H)\to\mathcal{L}(H)/\mathcal{C}(H)$
with the quotient space known as the Calkin algebra
and the quotient norm
$\|\pi(T)\|=\inf_{K\in\mathcal{C}(H)}\|T+K\|$.
In particular, every finite-dimensional eigenspace can be ``removed''
from~$A$ when computing the Q-rate~\eqref{eq:limsup2}. Interestingly,
for the R-rate, one has
\begin{align*}
  \limsup_{k\rightarrow\infty}
  \Biggl( \frac{\|r_k^{A,r_0}\|}{\|r_0\|} \Biggr)^{\frac{1}{k}}
  \leq
  \lim_{k\to\infty}
  \inf_{\substack{p\in\mathcal{P}_k\\p(0)=1}}
  \|p(A)\|^{\frac1k}
  =
  \inf_{\substack{k\in\mathbb{N}\\p\in\mathcal{P}_k\\p(0)=1}} \max_{z\in\sigma_0(A)}
  |p(z)|^{\frac1k},
\end{align*}
where the equality is due to Nevanlinna~\cite[Theorem~3.3.4]{nevanlinna1993}
and $\sigma_0(A)$ consists of the spectrum of~$A$ except for
its isolated points.
By this analogy, one could conjecture that $M_{\pi(A)}$ also
``cannot see'' any eigenspaces, including those of infinite dimension.
One might wonder whether there is a~useful characterization of~$M_{\pi(A)}$.

Another interesting question concerns $s$-step linear convergence
$\|r_k\|\leq M\|r_{k-s}\|$ with some $s\in\mathbb{N}$ and $M<1$.
The present result assumes linear reduction in every step, i.e.,
$s=1$. On the other hand, an~important class of problems, e.g.,
self-adjoint operators, exhibit a~reduction in every second step,
i.e., $s=2$ and $M<1$. To this end, we cannot see how the technique
of Section~\ref{sec:proof} generalizes to $s\geq2$. Note that
\cite{huhtanen-nevanlinna-2000} is concerned with related issues but
does not directly provide such stability we obtained for the case $s=1$.

\section*{Acknowledgement}
The author would like to thank
Erin Carson, Oliver Ernst, Zden\v{e}k Strako\v{s}, and Petr Tich\'{y}
for fruitful discussions and useful suggestions, which improved the paper.

redpajama_set_name: RedPajamaArXiv


text:
A batalha de Azaz ocorreu em 11 ou 13 de junho de 1125, entre as forças dos estados cruzados comandadas pelo rei Balduíno II de Jerusalém e o exército seljúcida de Suncur Burcuci, quando a vitória dos cristãos forçou o fim do cerco muçulmano a Azaz.

Antecedentes
Em 1118, o conde Joscelino I de Edessa conquistara a praça de Azaz ao atabei de Alepo, mas no ano seguinte os cruzados liderados por Rogério de Salerno foram chacinados na batalha do Campo de Sangue. Joscelino foi aprisionado em batalha pelos turcos em 1122, e em 1123 Balduíno II de Jerusalém sofreu o mesmo destino durante uma patrulha no Condado de Edessa.

Ambos foram resgatados em 1124, e a 8 de Outubro do mesmo ano Balduíno cercou Alepo. Em Janeiro de 1125 o atabei seljúcida Suncur Burçuci de Moçul marchou para sul em auxílio da cidade que estava prestes a render-se, após três meses de cerco. Apesar de Alepo ser apelidada de «o maior prémio que a guerra poderia oferecer», Balduíno retirou sem dar batalha.

Batalha

Depois de forçar os cristãos a abandonar o cerco e reforçado com contingentes de Toguetequim de Damasco e de outras cidades-estado muçulmanas, Aque Suncur Bursuqui cercou a cidade de Azaz, a norte de Alepo, em territórios do Condado de Edessa. Balduíno II, Joscelino I e Pôncio de Trípoli trouxeram 1100 cavaleiros dos seus domínios, que incluíam o Principado de Antioquia do qual Balduíno assumira a regência, e mais outros 2000 soldados de infantaria.

Repartindo o exército em treze destacamentos, os cruzados enfrentaram o inimigo com três batalhões de cavaleiros: os de Antioquia à direita, Trípoli e Edessa no centro, Jerusalém à esquerda. O rei de Jerusalém usou uma habitual táctica muçulmana: acossado de perto pelo inimigo, simulou uma retirada para atraí-lo para um terreno aberto, onde montara uma emboscada. De seguida virou-se para o inimigo e carregou, tirando vantagem da rapidez do ataque.

Depois de uma batalha longa e sangrenta, os seljúcidas foram derrotados, perdendo quinze emires e milhares de soldados. O campo muçulmano foi tomado por Balduíno, que obteve butim suficiente para resgatar prisioneiros cristãos em poder dos turcos, incluindo o futuro conde Joscelino II de Edessa.

Consequências
Para além de levantar o cerco a Azaz, esta vitória recuperou muita da influência que os cruzados tinham perdido após a derrota na batalha do Campo de Sangue em 1119. O rei de Jerusalém ainda planearia um ataque a Alepo, mas em 1126 Boemundo III de Antioquia atingiu a maioridade e entrou em conflito com Edessa, pelo que a aliança cruzada não se realizou. Em 1128, Zengui obteve o controlo de Alepo e Moçul - a sua forte liderança levaria ao enfraquecimento do poder cruzado no norte da Síria e, em 1144, à reconquista de Edessa pelos muçulmanos.

Azaz, Batalha De
Batalhas do Principado de Antioquia
Batalhas do Condado de Edessa
Batalhas do Condado de Trípoli
Batalhas envolvendo o Reino de Jerusalém
Batalhas envolvendo o Império Seljúcida


A stay at 25hours Hotel Terminus Nord places you in the heart of Paris, within a 15-minute walk of Canal Saint-Martin and Grands Boulevards. This hotel is 0.7 mi (1.1 km) from Folies Bergere and 1.2 mi (1.9 km) from Grevin Museum.
Stay in one of 237 guestrooms featuring LCD televisions. Complimentary wireless Internet access keeps you connected, and satellite programming is available for your entertainment. Private bathrooms with bathtubs or showers feature complimentary toiletries and hair dryers. Conveniences include irons/ironing boards and blackout drapes/curtains.
Enjoy a satisfying meal at Neni serving guests of 25hours Hotel Terminus Nord. Quench your thirst with your favorite drink at the bar/lounge. Buffet breakfasts are available daily from 6:30 AM to 10:30 AM for a fee.


Everyone pictures fair day as a day off of school to go on fun rides, eat funnel cakes, and hang out with friends. However, in reality, Fair Day may not be all it's cracked up to be. Students do get the advantage of a free ticket, but is it worth going? The lines are excessively long for rides that only last two to three minutes and it is overcrowded with students.
Fair Day is Friday, Feb. 9 and Robinson, along with many other local schools, get the day off. But there could be better things to do with your day then go to the fair, and here are five suggestions for you.
Whichever beach you choose to go to will probably be less packed then the fair and you'll also get some sun while you're at it. It's supposed to be around 80 degrees on Friday, so if you're at the beach you'll be able to cool down in the water instead of being obnoxiously hot and sweaty waiting in fair lines.
This might be a little crowded because it is a popular park, but since not everyone has a pass (or a free ticket), the lines might be shorter than the ones at the fair. If you're looking for a place to ride fun rides, Busch Gardens beats the fair.
This is a relatively new place, but it's still seems more entertaining than any ride at the fair. Altitude is the largest trampoline park in Tampa Bay and has many different activities to do. They have trampolines, a foam pit, a rock climbing wall, and more. You can bounce around or play dodge ball with your friends, with probably no lines at all. And although you don't get in for free, it is only $15 to jump for one hour and $22 for two hours.
Since most people will be at the fair, the mall will probably be pretty empty, giving you a lot of time and space to shop. And if you go to Westshore Mall, you'll be able to watch a movie along with it, in a pretty empty theater as well. Just a few of the many movies out on Friday you could watch are Jumaji: Welcome to the Jungle, Fifty Shades Freed, or The Maze Runner: The Death Cure. The Death Cure is the third movie in the Maze Runner trilogy and is a great choice if you're looking for an exciting action movie. You could also see The Post, a movie about the media exposing government lies, which is an Academy Award nominated film for Best Picture.
If you ever get the opportunity to not go to school on a Friday, you should sleep. Instead of getting up and waiting in lines, take a nap instead. Students often don't get enough sleep during the weekdays, so it could be refreshing to take advantage of the day off. Why wake up early and wear yourself out when you could turn off your alarm and snooze the day away.


text:
Q: How to get "Project Id" to create a Direct Link? I have my project name, but not the numeric Project Id. The latter is needed to use HTML Direct Links.I'm using JIRA 5.0.1
How do I get the numeric Project Id for a given project name?
I've searched the Project Administration area, several other places, the documentation, Google, etc but still can't find a way to get that value.
Thanks.

A: This solution does not require admin rights:
Navigate to https://jira.YOURDOMAIN.TLD/rest/api/2/project/YOURPROJECTNAME and read the id in the JSON response:
{
    "self":"https://jira.YOURDOMAIN.TLD/rest/api/2/project/YOURPROJECTNAME",
    "id":"12345",  ☜ Project Id
    "key":"YOURPROJECTNAME",
    "description":..
    :
}

Navigate to https://jira.YOURDOMAIN.TLD/rest/api/2/project to get a JSON list of projects.
Bonus: here's a one-liner in Groovy to get the ID:
groovy -e "println new groovy.json.JsonSlurper().parseText("https://jira.YOURDOMAIN.TLD/rest/api/2/project/YOURPROJECTNAME".toURL().text)?.id ?: 'not found'"

A java.io.FileNotFoundException probably means that your JIRA server requires authentication.
Here's a one-liner to list all the visible projects and their ID:
groovy -e "new groovy.json.JsonSlurper().parseText('https://jira.YOURDOMAIN.TLD/rest/api/2/project'.toURL().text)?.each{println it.key+' = '+it.id}"

A: The easiest way is to do it from the web browser:

*

*Go to the Administration page.

*Select the Project from the menu.

*Hover over 'Edit Project' link and check the link href (in the status bar).
It should be something like http://servername:8080/secure/project/EditProject!default.jspa?pid=10040
Where pid is the id you are looking for.
For Jira 6.x:

*

*place the cursor on EDIT Project button and

*look at the url being redirected at bottom left of the screen

A: This solution doesn't require admin rights and shows you all of the projects the current user can view.
https://example.com/rest/api/2/project
Responses found here.
https://docs.atlassian.com/jira/REST/latest/#d2e4972
returns a json array.
[
    {
        "self": "http://www.example.com/jira/rest/api/2/project/EX",
        "id": "10000",
        "key": "EX",
        "name": "Example",
        "avatarUrls": {
            "24x24": "http://www.example.com/jira/secure/projectavatar?size=small&pid=10000",
            "16x16": "http://www.example.com/jira/secure/projectavatar?size=xsmall&pid=10000",
            "32x32": "http://www.example.com/jira/secure/projectavatar?size=medium&pid=10000",
            "48x48": "http://www.example.com/jira/secure/projectavatar?size=large&pid=10000"
        },
        "projectCategory": {
            "self": "http://www.example.com/jira/rest/api/2/projectCategory/10000",
            "id": "10000",
            "name": "FIRST",
            "description": "First Project Category"
        }
    },
    {
        "self": "http://www.example.com/jira/rest/api/2/project/ABC",
        "id": "10001",
        "key": "ABC",
        "name": "Alphabetical",
        "avatarUrls": {
            "24x24": "http://www.example.com/jira/secure/projectavatar?size=small&pid=10001",
            "16x16": "http://www.example.com/jira/secure/projectavatar?size=xsmall&pid=10001",
            "32x32": "http://www.example.com/jira/secure/projectavatar?size=medium&pid=10001",
            "48x48": "http://www.example.com/jira/secure/projectavatar?size=large&pid=10001"
        },
        "projectCategory": {
            "self": "http://www.example.com/jira/rest/api/2/projectCategory/10000",
            "id": "10000",
            "name": "FIRST",
            "description": "First Project Category"
        }
    }
]

A: Exporting a ticket in XML reveals the project ID for me. I am not admin, so can't access the admin page. The rest/json trick didn't work for me, either. The XML of an issue has the following,
<project id="1234" key="test">TEST Project</project>

redpajama_set_name: RedPajamaStackExchange


text:
Session Type: Convention, Management Workshop
Location: Convention CenterRoom Number: 157B + 157C
Date: Monday, October 10, 2016Time: 8:00 AM - 8:50 AMTrack Number: 34
To compete in a changing world, leaders must set the strategic priorities for their companies and engage their teams to move the strategy forward. This program will provide practical insights for leading strategic change by creating a sense of urgency, communicating consistently, focusing on behavior, and leveraging the company's culture to achieve results.
Kathy Mazzarella
Graybar Kathleen M. Mazzarella is Chairman, President and CEO of Graybar, an employee-owned, FORTUNE 500 distributor of electrical, communications and data networking products. With more than 8,000 employees, 260 locations across North America and $6 billion in annual revenue (2014), Graybar offers its customers convenient local service, backed by the strength and supply chain capabilities of a large company. Kathy joined Graybar in 1980 in customer service and soon became one of Graybar's top salespeople. Kathy advanced into field sales management and later was promoted into corporate product management. At Graybar's corporate headquarters, she held executive-level positions in strategic accounts, human resources, strategic planning, sales and marketing. She was named executive vice president and chief operating officer in 2010. Kathy was appointed Graybar's 11th president and CEO in 2012, and she became chairman of Graybar's board of directors the following year. Kathy holds an associate degree in telecommunications engineering, a bachelor's degree in applied behavioral sciences and a master's degree in business administration. She serves on several boards, including the Federal Reserve Bank of St. Louis, Waste Management, Inc., St. Louis Regional Chamber, United Way of Greater St. Louis, Civic Progress, the Saint Louis Club and the NAW Institute. She also serves on the NAED Channel Advantage Partnership Council and the Webster University School of Business and Technology's Advisory Board. In addition, she is a member of The Committee of 200 and a member of the International Women's Forum. Kathy was recognized by the St. Louis Business Journal with the "40 Under 40" award in 2000 and as one of St. Louis' "Most Influential Businesswomen" in 2004. In 2011, she received the Outstanding Alumna Award from the George Herbert Walker School of Business & Technology at Webster University. In 2014, Kathy was recognized by the Urban League of Metropolitan St. Louis as one of its distinguished women in leadership. In 2015, she earned the National Association of Electrical Distributors' Trailblazer Award and was recognized as a Leader of Distinction by the YWCA of Metro St. Louis.


text:
Tag: qhubeka
My Blog, Pro Cycling., Tour de of Britain 2017
Tour de of Britain, Stage 6. Newmarket to Aldeburgh.
Today's stage started in Newmarket which is more known for its horse racing than cycling, though Tony Gallopin or Laura Trott would feel at home there! Also, this is a sprint stage so there will be no horse category climbs! Seriously though, it has been lovely talking to so many new riders this year. At my last Tour de of Britain I spent most of … Continue reading Tour de of Britain, Stage 6. Newmarket to Aldeburgh.
jamesrannochcc Sep 8, 2017 Apr 24, 2018 1 Comment
My Blog, Pro Cycling., Tour de Past
Tour de Past, Stage 14. 2015, Qhubeka Get Their First Tour Stage.
Over the 25 years or so of being a cycling fan I have had many favourite riders who I have willed on during many different races. I have shouted with excitement when they've won and felt their pain in defeat. I have never been an avid follower of any cycling teams though. I liked ONCE and T-Mobile back in the day because they had a few of my types of riders but I have never been like the avid Belgium fans of either Quick-Step or Lotto or the Team Sky diehards.
When MTN-Qhubeka came onto my radar though I instantly thought that they were a team I could really get behind. That was in 2013 when they took a step up to Pro-Continental level and made an immediate impact when Gerald Ciolek won the snow effected Milan-San Remo that spring. What most impressed me about them was that they chose to have the name of a charity on their jersey's, promoting the work of Qhubeka instead of receiving money from some corporate entity. Here is some information on how providing bikes for schoolchildren can help improve their lives as well as the education of third world countries.
In 2014 the team gained its first entry to a Grand Tour after getting a Wild Card for that years Vuelta. They put in a solid performance with all their riders finishing and Sergio Pardilla their top rider in 17th place to show they belonged in the biggest races on the calendar. At the end of the season their new General Manager Brian Smith announced some big name signings for 2015 such as Edvald Boasson Hagen and Matt Goss and that along with their showing in the Vuelta convinced ASO to give them a wildcard berth for the 2015 Tour de France.
Brian Smith introduced another aspect to the team for 2015 with made them more supportable for me, the distinctive black and white jerseys. I had hoped that fellow Paisley Buddie had got inspiration for the design from our local football team St Mirren but he admitted in an interview that he was thinking more along the lines of Juventus.
Continue reading "Tour de Past, Stage 14. 2015, Qhubeka Get Their First Tour Stage."
jamesrannochcc Jul 15, 2017 Jul 21, 2017 Leave a comment


text:
Best High Volatility...
New Award for Push Gaming
Push Gaming appeared unusually joyful for the last days of November, as the company proudly received the 'Best High Volatility Game' award for one of its outstanding works. The nominee was one of the most famous slots of the company – Jammin 'Jars.
The award ceremony took place in the Heart Club lounge, hosting dozens of strong participants, among whom it was easy to spot more than one leader in the industry. The competition turned out to be at a rather high level, but still Push Gaming was honored with the prestigious award to be extremely proud of. In addition, this was not the only nomination in which the operator was declared. The name Push Gaming sounded in relation to "Mobile Game of the Year", "Provider of the Year" and "Rising Star", which testifies to the hard work, effort and patience with which the company is expanding its exclusive range.
The CEO of Push Gaming, James Marshall, was quick to note that the company is sincerely happy with this kind of recognition, because this is what further serves to assert the operator in the field of online entertainment. Jammin 'Jars was well received by fans of online gambling, which is good news, but still, Push Gaming does not stand still and is ready to continue its fruitful work for the benefit of avid fans of computer games.
nineteen − 13 =
Gambling Horoscope 2020: Is Today My Lucky Day?
Grande Vegas Is About to Give Out Immeasurable Treasure ...
BGaming Goes to China with Sic Bo!
Gambling Horoscope 2021: Is Today My Lucky Day to Gambl ...
Yggdrasil Starts Collaboration with SPER


Sanjeev Padmanabhan R favorited Aatmn's blog post How can you receive with ease?
Sanjeev Padmanabhan R favorited Aatmn's blog post Contribute by being the boss in you set up.
Sanjeev Padmanabhan R favorited Aatmn's blog post Restrictive factors can inspire you to fly high!
Sanjeev Padmanabhan R favorited Aatmn's blog post Let there be respect and honour from one and all.
Sanjeev Padmanabhan R favorited Aatmn's blog post How can you remain cheerful amidst adversities?
Sanjeev Padmanabhan R favorited Aatmn's blog post Who is the author of your life?
Sanjeev Padmanabhan R favorited Aatmn's blog post How to stop being concerned or worried about the future?
I'm a person with a spiritual mindset. And I like to understand things at a deeper perspective. Being a medical student I use my spiritual learnings from workshops to understand the body and diseases better.
I enjoy playing basketball, going on treks. Dancing and meeting new people.


1. I agree to satisfy and conduct myself in accordance with all IAPM certification programme policies and requirements.
2. I shall maintain confidentiality of IAPM examination questions and content. Furthermore, I agree not to copy, discuss, debrief or disclose, in any manner, the specific content of IAPM examination questions and answers to any individual.
3. I certify and swear that I have performed the exam entirely alone and without the help of any other individual, literature or any other means of assistance.
4. I agree that I shall at all times act in a truthful and honest manner and provide truthful and accurate information to the IAPM. I agree that any intentional or unintentional failure to provide true, timely and complete responses to questions in this application may lead to further investigation and/or sanctions by the IAPM.
5. I agree that all materials that I submit to the IAPM certification department become the property of the IAPM certification department, and that the IAPM certification department is not required to return any of these materials to me.
6. I agree that information related to my participation in the IAPM certification process may be used in an anonymous manner for research purposes only.
7. I agree that all disputes relating in any way to my application for an IAPM certificate and/or my involvement generally in an IAPM certification program, will be resolved solely and exclusively by means of IAPM certification department policies, procedures and rules. Any disputes or lawsuits are excluded.
8. The IAPM reserves the right to suspend or revoke the certificate of any individual who is determined to have failed to up - hold, or otherwise breached this agree - ment. In this case, the IAPM reserves the right to pursue criminal proceedings.
9. I release and indemnify the IAPM and the IAPM certification department from all liability and claims that may arise out of, or be related to, my project manage - ment and related activities.
10. I hereby release, discharge and in - demnify the IAPM, its directors, offic - ers, members, examiners, employees, attorneys, representatives, agents and the IAPM certification department from any actions, suits, obligations, damages, claims or demands arising of or in con - nection with this application, the scores given with respect to the examination or any other action taken by the IAPM with regard to certifying, testing and professional development including, but not limited to, all actions related to ethics matters and cases.
11. I understand and agree that any de - cision concerning my qualification for any certificate, as well as any decisions regarding my continuing qualification for any certificate rest within the sole and ex - clusive discretion of the IAPM, and that these decisions are final.


I was thrilled to be a part of The Allan House Art Nouveau open house at the start of this year. We filled the evening with beautiful flowers, this gorgeous Anna Campbell wedding dress that I absolutely adore (!!), and delicious food and drinks. Always a treat working with a great team in a beautiful space.


"I have heard the same .... it must be so incredibly beautiful there. I've always wanted to go ... perhaps I can convince you to post some pictures ... sacred beautiful places of our Mother Earth"
"In my early 20s, Nienna. I had a very hard physical 3D life up until the last few years. 20 to 25, I rested in Nienna, and wept with her. Life is better now, but I might not have survived if not for the strength I found in Nienna, knowing that I was…"
"YES! Pretty much accurate :D "
T.E. Pelton liked Ryna de Liane's blog post Holy Homemaking.
The ORIGINAL Pleiades Elf, now with 100% more Kari'El!
I am now blessed beyond words, as I am handfasted, happily, to my soulmate, who is as real as you or I.
I am a personal assistant for my Selkie of a wonderful mother in law, with whom I love to swim in the summer. It's a beautiful setting for low stress work that pays decently. Much of my work puts me close to nature--gardening three out of four seasons, shovelling snow in winter. I avoid cities and don't have to deal with office politics, and I get to grow my own food in season.
I no longer have a cell phone and prefer writing letters to those I love. In fact I tend to be a bit wary of technology, only using it to connect to/seek out like minds (I do not even use the Internet or any kind of computers for work!). A person is so much more important than a screen or a machine.
I am here because I'm grateful to have created a life in which I have enough time to give myself over to the spiritual pursuits I ignored for over a decade.
Tie Eldalieva, the Elven Path.
Cash only lifestyle. Upcycling. Kawaii/Harajuku fashion is my high tech weakness. I'm also a homemaker and homesteader, and ovo-lacto-pescatarian moving towards ovo-lacto-vegetarian.
I am grateful to be able to have a job that also allows me to be a full time homemaker, as my work largely keeps me at-home.
There are several lives I recall.
So I'm of direct Azorean maternal mitochondrial descent--my mother's mother is full blood Azorean from Sao Miguel. I grew up with her telling me that the Azores was all that remains of Atlantis.
Eshakal em te laktu kala,Shek-na-Krena Nekomei ..te moi..
Wonderful to meet you Here!


Zeroconf is a collection of tools and protocols to allow networks to configure themselves. It is called Bonjour (formerly Rendezvous) by Apple, and used extensively on Mac OS X.
Traditionally most of its work is done by DHCP and DNS. But the decentralised zeroconf method is appropriate in some situations, for example ad-hoc networks. It also requires no configuration (apart from the installation).
It works happily along side traditional tools. For example you may have an IP address allocated by DHCP and use DNS to resolve address on the web, but still use the hostname.local from MDNS (Multicast DNS) to resolve the addresses of other computers on the LAN.
Note, if you are not doing this on your own network you may want to talk to the Network Admin first. Although zeroconf is a fairly efficient protocol the admin may not want the extra traffic. Also be aware that it is trivially easy to spoof an mdns .local name.
Go to System > Preferences > Remote Desktop. Check the first two boxes under Sharing and configure the Security settings as desired.
From another computer, launch a VNC client and enter myhost.local as the host name, replacing the "myhost" portion with the host name deduced from step 1 above.
Assuming the remote Ubuntu computer responded to the VNC connection request, then Zeroconf is properly resolving the local host name.
then it is possible that your firewall is blocking the zeroconf communication. If you can turn your firewall off and this fixes the problem, then it is definitely the firewall.
Now you should be able to resolve .local names.
PING flute.local (192.168.1.101) 56(84) bytes of data.
If you are using [Guarddog] as a firewall, you'll need to create a new zone for multicast traffic and add an entry to the networkprotocoldb.xml file.
Call the new zone "multicast" and set its Zone Addresses to 224.0.0.0/255.0.0.0 Add the XML below into /usr/share/apps/guarddog/networkprotocoldb.xml just above where it says "<protocol name="domain">". Restart [Guarddog] & then check the boxes to allow mDns traffic between your zones.


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text:
Erika Sulzmann (* 7. Januar 1911 in Mainz; † 17. Juni 1989) war eine deutsche Ethnologin. Erika Sulzmann leitete 1951 bis 1954 die erste größere Feldforschungsreise der Nachkriegszeit, die Mainzer Kongo-Expedition. Dieser ersten ethnographischen Forschungsreise folgten in den Jahren 1956 bis 1980 noch acht weitere Reisen zu den südwestlichen Mongo in Zaïre. Erika Sulzmann errichtete 1984 zusammen mit Irma und Rosemarie Sulzmann die Sulzmann-Stiftung, die ethnologische, sozial-, sprach- und literaturwissenschaftliche Forschungen in und über Afrika fördert.

Leben
Erika Sulzmann gelangte auf einem kleinen Umweg zur Ethnologie. Nach ihrem Abitur arbeitete sie als Bibliothekarin und Fotografin am Institut für Kulturmorphologie in Frankfurt (heute Frobenius-Institut). Durch ihr wachsendes Interesse an Anthropologie und vor allem an afrikanischen Kulturen begann Sulzmann 1940 bei Hermann Baumann in Wien Ethnologie zu studieren. Forschungen zur Wiener Völkerkunde während der NS-Zeit haben ergeben, dass sie eine Befürworterin der NS-Ideologie war und diesbezüglich Druck auf die Studierenden ausgeübt hat.
1947 promovierte sie zum Thema Die Mongo – Studien zu einer regionalen Monographie bei Wilhelm Koppers.
Im Sommer 1948 kam Erika Sulzmann als Assistentin an das neueingerichtete Institut für Völkerkunde der Johannes-Gutenberg-Universität Mainz. 1960 wurde sie zunächst Kustodin der ethnographischen Studiensammlung, dann akademische Oberrätin und schließlich Akademische Direktorin. Sie war bis zu ihrem Tod im Juni 1989 eine aktive Mitarbeiterin des Institutes.

Schriften (Auswahl)
 Quellen zur Geschichte und Sozialstruktur der Mbóle und Imoma (Ethnie Móngo, Zaire). Berlin, Reimer, 1986.
 La soumission des Ekonda par les Bombomba. In: Annales aequatoria (1985), Vol. 6, S. 1–17.
 Orale Tradition und Chronologie : der Fall Baboma – Bolia (Northwest Zaire). In: Mélanges de culture et de linguistique africaines; publiés à la mémoire de Leo Stappers / éd. par C.M. Faïk-Nzuji et E. Sulzmann, 1983, S. 525–586.
 Mélanges de culture et de linguistique africaines : publiés à la mémoire de Leo Stappers. Berlin, Reimer, cop. 1983 (mit Clémentine Madiya Faïk-Nzuji).
 Ein Jagdbericht im Dialekt der Batwa von Ebungu (Ekonda). In: Annales Aequatoria; recueil d'études offert au R.P.G. Hulstaert à l'occasion de son 80e anniversaire; tome I, vol. 2 / A. Claessens, et H. Vinck / 1980.

Literatur
 Beer, Bettina: Sulzmann, Erika, in: Bettina Beer: Frauen in der deutschsprachigen Ethnologie. Ein Handbuch. Köln: Böhlau 2007, ISBN 978-3-412-11206-6, S. 218–222
 Brandstetter, Anne-Maria: In memoriam Erika Sulzmann, in: Forschungen in Zaire. Münster, Hamburg: Lit 1992, S. VII-X
 Brandstetter, Anne-Maria: Dr Erika Sulzmann (1911–1989), in: Hedwig Brüchert (Ed.): Rheinland-Pfälzerinnen : Frauen in Politik, Gesellschaft, Wirtschaft und Kultur in den Anfangsjahren des Landes Rheinland-Pfalz. Mainz: Hase und Koehler 2001, S. 414–417
 Geisenhainer, Katja: Erika Sulzmann und die "Stammeskarte von Afrika": Der frühe Werdegang einer Ethnologin im NS-Kontext, in: Andre Gingrich; Peter Rohrbacher (Hg.), Völkerkunde zur NS-Zeit aus Wien (1938–1945): Institutionen, Biographien und Praktiken in Netzwerken (Phil.-hist. Kl., Sitzungsberichte 913; Veröffentlichungen zur Sozialanthropologie 27/1). Wien: Verlag der ÖAW 2021, S. 493–520 doi:10.1553/978OEAW86700
 Schröter, Suzanne: Die Sammlerin der Dinge: Erika Sulzmann, Forscherin in Afrika, in: Karl-Heinz Kohl (Hg.): Das exotische Ding. Geschichten einer Sammlung. Mainz: Institut für Ethnologie und Afrika-Studien 1996, S. 23–34
 Schröter, Suzanne: Noch ein Fahrrad im Kongo : Erika Sulzmann, Forscherin in Afrika, Videofilm 1999, 60 min.

Weblinks
 Interviews with German anthropologists: Erika Sulzmann
 Erika Sulzmann (1911-1989): notice biographique / Honoré Vinck, in: Annales Aequatoria 11(1990), S. 458–459

Einzelnachweise

Anthropologe (20. Jahrhundert)
Ethnologe
Person (Johannes Gutenberg-Universität Mainz)
Deutscher
Geboren 1911
Gestorben 1989
Frau


You can learn the Alexander Technique by taking a series of lessons with a qualified teacher of the Alexander Technique. You can take lessons either on an individual basis or in a group with other students.
The principles of FM Alexander's work are learnt with the help of a teacher who uses verbal instruction together with gentle, specialised manual skills.
The lessons that I teach are designed to suit your individual needs and are taught in a very interactive way, encouraging questions, originality and diversity of ideas.
While it is usual in early lessons to work on simple practical activities such as sitting and walking, you can learn the work in relation to any activities that are of particular interest or focus for you in your life, work or recreation.
While you can learn something of great value to yourself in a relatively short series of lessons, the value of the work for you and your mastery of the principles is increased considerably by a more long term commitment.