text:
Durham v. Hallmark County Mutual Insurance Co. 2018 WL3469257
Posted on July 30, 2018 Author Categories Volume 21, Edition 7, Cases
SEE TX R RAP RULE 47.2 FOR DESIGNATION AND SIGNING OF OPINIONS.
Court of Appeals of Texas, Eastland.
RANDY DURHAM, Appellant
HALLMARK COUNTY MUTUAL INSURANCE COMPANY, Appellee
Opinion filed July 19, 2018
On Appeal from the 358th District Court
Ector County, Texas
Trial Court Cause No. D-137,194-A
Panel consists of: Willson, J., Bailey, J., and Wright, S.C.J.2
MEMORANDUM OPINION
JIM R. WRIGHT SENIOR CHIEF JUSTICE
*1 This is an appeal from a judgment in which the trial court granted Appellee's combined motion to dismiss for lack of jurisdiction and motion for summary judgment. We affirm.
Appellant originally sued Bobby Burl Straley; L&L Trucking; and Larry Eilers, individually and d/b/a L&L Trucking, after Appellant was injured in a vehicle accident. It was alleged that Straley was the driver of the truck that was involved in the accident; that L&L Trucking owned the truck; and that Eilers, individually and d/b/a L&L Trucking, was the owner of the trucking company. In his first amended petition, Appellant added Hallmark County Mutual Insurance Company, the company that insured "Larry Eilers DBA L&L Trucking Co.," as a party to the suit. Appellant sought a declaratory judgment that Appellee owed its insured a duty to defend. Appellant also sought a declaratory judgment that the incident was "covered by the policy [written by Appellee] and not subject to any exclusions."
Appellee filed an answer in which it denied that Appellant was an insured or third-party beneficiary under the policy or was a judgment creditor of the insured. Appellee then filed a single motion denominated as "Motion for Summary Judgment and Motion to Dismiss for Lack of Jurisdiction." This motion rested on several grounds: first, that Appellant was not a named or additional insured; second, that Appellant was not an intended third-party beneficiary; and third, that Texas is not a "direct action" state and, therefore, that Appellant could not sue Appellee until he procured a judgment against the Appellee's insured. The trial court granted both parts of the motion for summary judgment and the plea to the jurisdiction; it severed and dismissed Appellee from the primary suit, and Appellant's claims against Appellee became final and appealable.
Whether a court has subject-matter jurisdiction is a question of law and is reviewed de novo. Tex. Nat. Res. Conservation Comm'n v. IT-Davy, 74 S.W.3d 849, 855 (Tex. 2002). We note that a review of a plea to the jurisdiction challenging the existence of jurisdictional facts mirrors that of a motion for summary judgment. Tex. Dep't of Parks & Wildlife v. Miranda, 133 S.W.3d 217, 228 (Tex. 2002).
In his first issue, Appellant contends that the trial court erred when it granted summary judgment in favor of Appellee, because there existed a question of material fact as to whether Appellee had a duty to defend its insured. Appellant's first issue requires that this court first decide a preliminary issue: whether the trial court could issue a declaratory judgment regarding Appellee's duty to defend before the insured was found liable as a result of the accident.
Appellee argues that Appellant cannot sue it unless Appellant first obtains a judgment that reflects the tortfeasor's liability. We agree. Texas is not a direct action state; rather, "the general rule … is that an injured party cannot sue the tortfeasor's insurer directly until the tortfeasor's liability has been finally determined by agreement or judgment." Angus Chem. Co. v. IMC Fertilizer, Inc., 939 S.W.2d 138, 138 (Tex. 1997) (citing Great Am. Ins. Co. v. Murray, 437 S.W.2d 264, 265 (Tex. 1969)). With limited exceptions, not applicable here, this rule applies equally in instances where a plaintiff seeks a declaratory judgment and where a plaintiff seeks money damages. See In re Essex Ins. Co., 450 S.W.3d 524, 526 (Tex. 2014).
*2 Appellant argues that the Texas Supreme Court has recognized a third-party claimant's ability to participate in a declaratory judgment action. See Farmers Tex. Cty. Mut. Ins. Co. v. Griffin, 955 S.W.2d 81, 84 (Tex. 1997) (explaining that an insurer must "either accept coverage or make a good faith effort to resolve coverage before adjudication of the plaintiff's claim" and stating that "the plaintiff may wish to participate in that litigation"); see also State Farm Fire & Cas. Co. v. Gandy, 925 S.W.2d 696, 714 (Tex. 1996) ("A plaintiff who thinks a defendant should be covered by insurance may be willing to … assist in obtaining an adjudication of the insurer's responsibility.").
In Griffin, an automobile insurer brought a declaratory judgment suit in which it asked the trial court to determine that it had no duty to defend or indemnify its insured. 955 S.W.2d at 82. The insured had invoked the insurer's duty to defend him. Id. The court held that, under the facts in that case, the insurer's duty to indemnify was properly justiciable by declaratory judgment, even before the trial court rendered a judgment in the underlying suit. Id. at 83–84.
However, in 2014, the Texas Supreme Court declined to extend its holding in Griffin to those instances in which a plaintiff in the underlying lawsuit seeks to ensure that the insureds' coverage dispute is resolved prior to the adjudication of the insureds' negligence. Essex, 450 S.W.3d at 527. In Essex, the plaintiff in the underlying lawsuit sued Essex's insured for personal injuries, then added a declaratory judgment claim against Essex in which that plaintiff sought a declaration that the insurer must indemnify its insured. Id. at 525. The plaintiff argued that he "merely [sought] a declaration that the … policy cover[ed] the [insured's] liability to [the plaintiff], as opposed to a money judgment" and, therefore, that the declaratory judgment constituted an exception to the "no direct action" rule. Id. at 526. The court explained that allowing the plaintiff to pursue claims simultaneously against the insured—for liability—and the insurer—for coverage of that liability—would "prejudice" both parties because it would create a conflict of interest for the insurer and would require the admission of evidence of liability insurance in violation of Rule 411 of the Texas Rules of Evidence. Id. at 526–27. Therefore, the policy of the "no direct action" rule applied. Id.
We find no facts in the procedural history of, or allegations in, this case that distinguish it from Essex. Because a third-party claimant's claims regarding an insurer's coverage are not ripe for adjudication until a judgment is obtained establishing the insured's liability, the trial court did not err when it granted Appellee's motion for summary judgment.1 See id. We overrule Appellant's first issue.
In his second issue, Appellant argues that the trial court erred in dismissing the case for want of jurisdiction because he is an intended third-party beneficiary as to the issue of duty to defend. In other words, Appellant argues that he had standing, as a third-party beneficiary, to bring the declaratory judgment suit.
*3 The "no direct action" rule pertains to standing because there is no justiciable controversy until the liability of the insured has been established. See Farmers Ins. Exch. v. Rodriguez, 366 S.W.3d 216, 223 (Tex. App.—Houston [14th Dist.] 2012, pet. denied). "But the need for a determination of liability before bringing a direct action against an insurer, while often referred to as a standing issue, is more appropriately characterized and analyzed as ripeness." Auzenne v. Great Lakes Reinsurance, PLC, 497 S.W.3d 35, 37–38 (Tex. App.—Houston [14th Dist.] 2016, no pet.). We refer to our holding that Appellant's declaratory judgment was not ripe until the insured was finally determined to be liable to Appellee. Furthermore, if Appellee owed a duty to defend, that particular duty was owed to its insured, not to Appellant, a nonparty to the insurance contract. In any event, under the "no direct action" rule, a plaintiff in an underlying lawsuit cannot maintain a lawsuit against an alleged tortfeasor's insurer until that party has been found to be liable for damages that resulted from the activity sued upon. See Essex, 450 S.W.3d at 527. Accordingly, we overrule Appellant's second issue.
We affirm the judgment of the trial court.
Not Reported in S.W. Rptr., 2018 WL 3469257
Jim R. Wright, Senior Chief Justice (Retired), Court of Appeals, 11th District of Texas at Eastland, sitting by assignment.
Appellant argues that the trial court's judgment, which granted both the motion to dismiss for lack of jurisdiction and the motion for summary judgment, was erroneous because it was "paradoxical." However, matters concerning subject-matter jurisdiction, such as ripeness and standing, may be raised by a plea to the jurisdiction, as well as by other procedural vehicles such as a motion for summary judgment. Bland Indep. Sch. Dist. v. Blue, 34 S.W.3d 547, 553–54 (Tex. 2000).
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text:
Farewell, Crimea
Why Ukrainians Don't Mind Losing the Territory to Russia
By Taras Kuzio
Ukraine's Prime Minister Arseniy Yatsenyuk waves as he leaves a European Union summit in Brussels on March 6, 2014.
When I recently asked one member of the Lvivska Sotnia -- an ardently nationalist self-defense brigade that provided security for the months-long protests at Kiev's Independence Square -- how he wanted Ukraine to respond to Russia's seizure of Crimea, one might have expected a fiercely jingoistic response. This was, after all, one of those Ukrainians whom Russian President Vladimir Putin used as a pretext for his invasion in the first place. But it turns out the last thing he wanted was for Kiev to try to retake Crimea by force -- that would only risk starting World War III, he said.
Lost in the recent discussion in the West about Russian aggression in Crimea has been the question of whether Ukrainians believe that Crimea is worth fighting for in the first place. Although Westerners (and the Ukrainian government) profess the importance of defending Ukrainian territorial integrity, most Ukrainians wouldn't seem to mind letting Crimea go. For them, the issue is much more a matter of prudence than principle.
There are three main factors that inform Ukraine's current passivity toward Russia's aggression in Ukraine. First, there is the fragility and disunity of the current political leadership. Until the elections scheduled for May 25, Ukraine has only an interim government -- and a weak one, at that. Arseniy Yatsenyuk, who is acting prime minister, is an ally of Yulia Tymoshenko, a former prime minister and head of the Fatherland party who was imprisoned from 2011 to 2014. Moreover, only two of the four main opposition forces that toppled the previous government -- Fatherland and the nationalist Svoboda (Freedom) party -- have joined the current governing coalition. The former heavyweight boxer Vitaly Klitschko's moderate UDAR (Ukrainian Democratic Alliance for Reforms) has declined to participate, as has the group of businessmen affiliated with Petro Poroshenko, who was foreign minister under former president Viktor Yushchenko. The government has also failed to reach out to eastern Ukrainian moderates affiliated with ousted Prime Minister Viktor Yanukovych's Party of Regions, including Sergei Tigipko, who came in third place in the 2010 presidential elections.
The government has also been forced to devote most of its time and attention to stabilizing the country's precarious economic and financial situation, rather than to developing an effective military response to Russian aggression. Even after Russia's invasion, acting President Oleksandr Turchynov and Yatsenyuk -- who was preoccupied trying to secure financial assistance from the IMF and United States, and preparing the European Union Association Agreement that will be signed next week -- failed to give clear directions to Ukrainian military forces in Crimea. The troops were ordered neither to withdraw nor to defend themselves and their bases. As a result, they have been stranded in Crimea, with many of their weapons forcibly transferred to local paramilitaries.
A second factor accounting for Ukrainians' passivity is their awareness of how difficult it can be to control Crimea. There is a deep strain of xenophobia among the region's Russian population. In 1991, I was physically attacked for speaking Ukrainian outside the Crimean parliament building, and was refused the purchase of a train ticket unless I spoke in Russian. The acting leader of Crimea, Sergei Aksyonov, has, like other Russian nationalists, openly praised Soviet leader Joseph Stalin's 1944 ethnic cleansing of the region's Tatars, which resulted in the death of half of their entire population. (Tatars have since returned and now account for nearly 20 percent of the area's population.) And in 2006 and 2009, the local Russian nationalists -- under the direction of Yanukovych's Party of Regions -- successfully agitated against joint military exercises between Ukrainian and NATO military forces.
Putin has successfully mobilized these xenophobic sentiments to tie the region to Russia -- the many pro-Russian nationalist demonstrations of recent weeks offer plenty of evidence. Any window that the Ukrainian government had to offer a counternarrative -- it could have pointed out, for instance, that ethnic Russians are only 58 percent of the total population in Crimea, and that they have been guaranteed political and cultural autonomy -- has since shut.
The third and most important factor, however, is that most Ukrainians have always been ambivalent about Crimea. The region was transferred from Russian to Ukrainian control in 1954, and Russians still feel a much stronger sense of attachment to the region, and specifically to the port city of Sevastopol, than do Ukrainians. Some Ukrainians even believe that Crimea has a right to secede, although they may have wished that Crimea had done so in more orderly fashion. "If there is a clear majority of the people on a certain territory that do not want to accommodate themselves with the state they live in," Ukrainian writer Mykola Riabchuk said before the Russian incursion, "they should have a right to secede, which would require some negotiations, international mediation, referendums, and post-separation settlements for national minority rights." Russians, on the other hand, are willing to fight for it.
Should Crimea join Russia, many Ukrainians would mourn its loss as a blow to national self-esteem. But they would soon recognize that they have been freed of many of the burdens associated with the territory. First, Ukrainians would no longer be responsible for providing Crimea with enormous agricultural subsidies, or for propping up its ailing tourism sector. Second, Ukraine could more easily pursue clear, pro-European policies without having to consider blowback from an ardently pro-Russian domestic minority. Third, the government in Kiev would no longer have to invest resources to ward off a festering Islamist insurgency among Crimea's underprivileged and repressed Tatar population (a task to which Kiev has consistently failed to offer adequate resources).
Finally, Kiev's foreign policy would no longer be hostage to debates over the fate of Russia's Black Sea Fleet. In 1997, the Black Sea Fleet was granted basing rights for a term of 20 years, after which Ukraine would become a neutral "non-bloc" country associated with neither NATO nor a Russian military alliance. But in April 2010, Yanukovych railroaded the Kharkiv Accords through parliament, which extended the fleet's rights through 2042 -- although Ukraine's official foreign policy strategy released in July 2010 still maintained that the country intended to pursue non-bloc status. Yanukovych never clarified the contradiction between accepting a de facto permanent foreign base and pursuing military neutrality.
Putin would be wrong to mistake Ukrainians' ambivalence about Crimea for a broader ambivalence about eastern Ukraine. Ukrainians would not hesitate to fight over the other territories there. Although they have sizeable Russian-language populations, Ukrainians think of these regions as an integral part of the country, historically and culturally. Most would feel that the loss of any territory beyond Crimea would represent an existential threat to Ukrainian independence.
And if Russia does end up annexing Crimea, Ukraine is sure to offer some diplomatic resistance, if only to save face. It's likely that Kiev will continue to argue that Crimea should be returned to Ukraine, much as Dublin argued for decades that Northern Ireland should belong to Ireland. Over time, however, Ukrainians will probably learn to thank Putin for taking Crimea off of their hands, thus clearing the way for them to join the European Union -- and, eventually, NATO.
TARAS KUZIO is a research associate at the Centre for Political and Regional Studies at the University of Alberta.
More By Taras Kuzio
Russia & FSU Ukraine Politics & Society

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: The http status code of BadRequestKeyError I recently used postman to test a website and it returns BadRequestKeyError with http status code 500
but its return webpage's title shows
<title>werkzeug.exceptions.BadRequestKeyError: 400 Bad Request: The browser (or proxy) sent a request that this
        server could not understand.
        KeyError: 'user'
        // Werkzeug Debugger</title>

so what is its http status code 500 or 400 ?
The http status code of BadRequestKeyError

redpajama_set_name: RedPajamaStackExchange


text:
Some Car Dealers Refuse to Recalibrate Aftermarket Glass
Some Car Dealers Refuse to Recalibrate Aftermarket Glass | Shared from GlassBytes.com
January 19, 2017 by Phillip Thompson
pthompson@glass.com
Is OE glass now a requirement for recalibration after a windshield replacement?
That seems to be a growing refrain in the auto glass industry.
Several shops have reported that several car dealerships have refused to recalibrate a windshield replaced with aftermarket glass.
One post in the glassBYTEs.com forum asked: "What do you do when you take an Acura to a dealer to have it recalibrated and the dealer says they won't do it because it was not an OE [windshield]?"
Calls to several glass shops confirmed a recent trend by dealers to refuse recalibration on aftermarket glass. And while several also said they have not encountered resistance from dealerships, those that have say it's a growing problem.
"We've seen it several times in the last month or so," said Natalie Schmidtmann of Ray Sands Glass in Rochester, N.Y. "Especially in forward collisions. If it's an aftermarket piece of glass, the dealer won't do it."
Jeff Olive, training manager for Glasspro in Mt. Pleasant S.C., said he has had similar problems with dealerships.
"This is a major problem in our industry right now, getting these dealers on board," he said January 19. "They need to realize that we're the customers in this situation and we're referring business to them. It's an income opportunity."
Olive recounted a recent incident in which his shop replaced a windshield in a Volvo. The shop got approval from an insurance to do the replacement with the aftermarket glass made by Pilkington, which was the same brand as the Volvo OE glass. After the replacement, Glasspro sent the customer to the Volvo dealership to have the glass recalibrated. The dealer refused.
"He won't listen to a word I'm saying, even though I explained it's the same manufacturer that makes Volvo glass," Olive said. "He said if it doesn't say 'Volvo' on the glass it won't work."
Olive pointed out that his company has done other replacements with aftermarket glass, on Honda vehicles, and recalibrated the glass without a problem.
The fact that the glass is made by the same manufacturer as the OE glass should suffice, said Jason Lueders, manager and Lee and Cates Glass Inc. in Jacksonville, Fla.
"I'm not sure why it would matter, since Pilkington makes the glass," he said, after noting that while he hasn't dealt with dealer refusals personally, he's heard of several.
The end result, Olive said, is a black eye for his business.
"It makes us look bad," he said. "The insurance company eventually pays for the OE windshield, but I now have to take out a windshield I can't return and replace it – and spend the time and techs to do it. To the customer, it looks like we tried to install some cheap glass and got told no by the insurance company and the dealer."
This latest trend comes after instances last year of insurance companies, rather than dealerships, refusing to pay for recalibration.
Insurance company Arbella denied payment of a recalibration for a customer of Peter Brown of Tiny & Sons Auto Glass in Pembroke, Mass., claiming that work was part of the customer's deductible and not the comprehensive policy.
Mercedes Calls for OEM Glass in Windshield Replacement
Mercedes Calls for OEM Glass in Windshield Replacement | Shared from GlassBytes.com
"To promote and maintain its rigorous standards of quality and safety, if a collision repair is necessary, MBUSA strongly recommends that all repairs be performed by a certified technician using only Genuine Mercedes-Benz Parts," reads the six-page statement.
Aftermarket variants do not meet the exacting specifications of Mercedes-Benz glass, according to the statement, which points out that aftermarket glass often do not contain such technology as solar glass coatings, which provide UV protection and heat-load reduction, optimize air-conditioning performance and can improve fuel economy.
The position statement also provides guidance for recalibration of on-board systems and Advanced Driver Assistance Systems (ADAS), such as cameras, rain sensors, antennaeand heating elements. "Aftermarket glass often does not account for these complex electrical components and may interfere with your vehicles electronic systems, or cause these electronic systems to not function properly," the company said.
Additionally, post-collision repair evaluations of model year 1996 (OBDII) vehicles " … should have these systems evaluated using an up-to-date Xentry Kit diagnostics computer," the statement reads.
"Post-repair scanning and diagnosis of the vehicle is necessary to ensure that the vehicle's safety and
driver-assist systems are operable and fully functioning," according to the statement. "Many of the safety and driver-assist systems that may have been activated during a collision require vehicle calibration, normalizationor coding. The post-repair scan will also help to ensure that a comprehensive repair has been performed.
Some examples given by Mercedes include:
Vehicle collisions, regardless of the appearance of damage;
Windshield replacement for vehicles with driver-assist sensors (including rain/light sensors) located in the windshield;
And, removal and/or replacement of exterior components, bumpers, SRS sensors, parking sensors, driver-assist system sensors and cameras, wiring harnesses, vehicle control units, seatsor interior trim panels.


GRAYSON COUNTY, Tex. (KXII) - Grayson County commissioners took an unexpected path to replace District Attorney Joe Brown after accepting his resignation Tuesday following the announcement that he'll be a United States attorney for the Eastern District.
Many people expected his top assistant to take over until the election. But commissioners said they're taking applications to appoint the interim district attorney.
"Grayson County is unique in that we're the only county in Texas where commissioners decide the district attorney. It's normally gubernatorial, but in our case, it's the county," said county judge Bill Magers. "All we're doing is asking for applications for a week."
That request caused uproar from several people during public comment.
"They had said all along they didn't want to get into the politics of this," said first assistant district attorney Kerye Ashmore, "and appoint someone and give the appearance of preference."
This allows any attorney to apply for the DA position and carry out the job until the election. Ashmore said that could jeopardize the current prosecutions they're working like the 13 murder cases they have pending.
"I wish it was somebody from this office," said district attorney Joe Brown. "We have great prosecutors who know the courthouse, and judges and policies."
Sheriff Tom Watt and others even said it could stir up the election as well.
"Very few things decimate an organization more than instability," Watt said.
Commissioners can appoint anyone. That could include announced candidates for the position.
"If they do that, they're appearing to give preference to someone, which I wouldn't encourage them to do," Ashmore said.
"This is unchartered territory," Magers said. "There really is no protocol. The commissioners wanted to accept applications so they did."
"My anticipation was Kerye Ashmore would serve during transition time period and keep some consistency until the parties had been able to choose a nominee," Brown said.
Brown's position was just approved by the senate last week, and he's waiting on President Trump to sign off. He expects his last day will be March 2.
Commissioners will take applications for the next week. After that they'll appoint someone, and after the precinct chairs are elected for both the Republican and Democratic parties next month, they'll pick the nominees for the November ballot.


City of Yuma Heritage Festivals invites you to join the fun at the annual Yuma Medjool Date Festival. This festival celebrates the wonderful and tasty Date fruit, grown right here in the desert Southwest! Enjoy a fabulous array of delicious food and drink, live music, celbrity chef cooking demonstrations and contest, plus more!
Featuring Local Date Fruit Companies, Live Celebrity Chef Food Demonstrations, Community Performances, Tasty Treats & Eats, Unique Business & Craft Vendors, Giveaways, our $5 Wow Factor Kids Zone, and a live concert featuring Tommy & Drifters!
THANK YOU TO OUR AMAZINGLY GENEROUS EVENT SPONSORS!
Naked Dates, South West Medjools LLC, Martha's Garden, Imperial Date Gardens, Inc., Berryman Farms, Brocket Farms, A Pop Above Kettle Corn, Cricket Wireless, No Hassle Wireless, Lutes Casino, Tronix Sound, z93 Monster Media Yuma, El Dorado Broadcasters, Power 98, La Campesina, Print Zoom, Imaginary Friends, and B&E Services!


text:
package org.spoofax.modelware.gmf;

import org.eclipse.core.runtime.IConfigurationElement;
import org.eclipse.core.runtime.Platform;
import org.eclipse.ui.IStartup;
import org.eclipse.ui.plugin.AbstractUIPlugin;
import org.osgi.framework.BundleContext;
import org.spoofax.modelware.emf.Language;
import org.spoofax.modelware.emf.LanguageRegistry;

/**
 * Initializes the language registry--containing the languages for which both a textual and
 * graphical editor is provided--by evaluating the 'synchronizer' extensions.
 *
 * @author oskarvanrest
 */
public class RuntimeActivator extends AbstractUIPlugin implements IStartup {

	private static final String SYNCHRONIZER_ID = "org.spoofax.modelware.gmf.synchronizer";

	/**
	 * Like Eclipse IMP, this plug-in is activated on startup of the workbench
	 */
	@Override
	public void earlyStartup() {
	}

	/**
	 * This method is called upon plug-in activation
	 */
	public void start(BundleContext context) throws Exception {
		super.start(context);

		evaluateSynchronizerExtensions();
		EditorPairRegistry.getInstance(); // initialize
	}

	/**
	 * This method is called when the plug-in is stopped
	 */
	public void stop(BundleContext context) throws Exception {
		super.stop(context);
	}

	private void evaluateSynchronizerExtensions() {
		IConfigurationElement[] config = Platform.getExtensionRegistry().getConfigurationElementsFor(SYNCHRONIZER_ID);
		for (IConfigurationElement e : config) {
			String textFileExtension = e.getAttribute("textFileExtension");
			String domainFileExtension = e.getAttribute("domainFileExtension");
			String diagramFileExtension = e.getAttribute("diagramFileExtension");
			String nsURI = e.getAttribute("nsURI");

			if (textFileExtension != null && domainFileExtension != null && diagramFileExtension != null && nsURI != null) {
				Language language = new Language(textFileExtension, domainFileExtension, diagramFileExtension, nsURI);
				LanguageRegistry.getInstance().add(language);
			}
		}
	}
}

redpajama_set_name: RedPajamaGithub


And here's what you can do to ease it.
Do you often get up during the night to go to the bathroom? If so, you'll know that it's very annoying. Not only does it wake you from your warm cocoon of slumber, but it might also disturb your partner and you run the risk of not being able to get back to sleep once you've returned.
This condition is called nocturia, but what causes it and how can you avoid it? Good Housekeeping UK spoke to the guys over at NetDoctor to bring you everything you need to know.


Our company has a highly motivated team in sales & marketing, project management, implementation and customer support. Thanks to over twenty years of practical experience and excellent references in the industry we assure your success.
HIM GmbH is privately owned, the founder and one of the stakeholders is the CEO.


Home » Blog » Why Get a House Inspection?
Why Get a House Inspection?
You've been searching for "the house" for a long time, and you finally found a Houston property that meets your qualifications. You're tempted to forego the house inspection in order to set your offer apart—or perhaps to speed up the real estate process. But we at Bryan & Bryan Inspections want to shout a united, "Don't do it!" We realize that inspecting homes is our livelihood. However, we believe in what we do because of its value to each of our customers. Buying a Houston home is an investment, and even though you're antsy to move, you need to take the necessary steps to protect your hard-earned money. If you decide not to get a home inspection, you open yourself up to liability down the road. You never know what unseen issues might exist in (or under) the home. Most homebuyers cannot spot these concerns—like a faulty foundation—just from a walk-through with the realtor. The house inspection offers protection by arming you with as much knowledge as possible about the condition of the property. From there, you can decide whether the issues are worth continuing with your offer or whether you need to adjust your offer or walk away. The idea of abandoning the house inspection may sound like a good idea, but don't let the temptation get the better of you. Instead, choose a professional and thorough Houston home inspector who can help you know all you can about your potential home. To find out more about our home inspection services or to schedule a house inspection, contact Bryan & Bryan Inspections at (281) 484-8318.


text:
enum
{
	ed25519_seed_size = 32,
	ed25519_private_key_size = 64,
	ed25519_public_key_size = 32,
	ed25519_signature_size = 64,
	ed25519_scalar_size = 32,
	ed25519_shared_secret_size = 32
};

extern "C" {

#ifndef ED25519_NO_SEED
int TORRENT_EXPORT ed25519_create_seed(unsigned char *seed);
#endif

void TORRENT_EXPORT ed25519_create_keypair(unsigned char *public_key, unsigned char *private_key, const unsigned char *seed);
void TORRENT_EXPORT ed25519_sign(unsigned char *signature, const unsigned char *message, size_t message_len, const unsigned char *public_key, const unsigned char *private_key);
int TORRENT_EXPORT ed25519_verify(const unsigned char *signature, const unsigned char *message, size_t message_len, const unsigned char *private_key);
void TORRENT_EXPORT ed25519_add_scalar(unsigned char *public_key, unsigned char *private_key, const unsigned char *scalar);
void TORRENT_EXPORT ed25519_key_exchange(unsigned char *shared_secret, const unsigned char *public_key, const unsigned char *private_key);

}

#endif // ED25519_HPP


text:
Come Back, Mr. Chips
on November 16th, 2007 Posted By BG in Articles
by Julie Scelfo - Newsweek
Stereotyping, low pay, lack of role models. Why the number of men teaching in schools is at a 40-year low.
When numerous fellow teachers asked Josh Holt to mentor their students last year at Heber Hunt Elementary in Sedalia, Mo., Holt initially felt a burst of pride. It was his first year at the school teaching health and PE, and he hoped that he had impressed his senior colleagues. But looking around the school, Holt realized there was a simpler reason teachers and even parents asked him to be a "buddy" in the school's program for at-risk kids: he was born with a Y chromosome. "The principal is the only other guy in the school," says Holt, 24. "Some of these kids don't have any men in their lives, and they really need a male role model."
That is increasingly difficult to find in American classrooms. According to the National Education Association, the number of male schoolteachers is hovering at a 40-year low. Only one quarter of our 3 million teachers are men. In elementary schools, the problem is more acute-just 9 percent are men, down from 18 percent in 1981. "If kids do not see males in the classroom, they begin to believe teaching is only for females," says Reg Weaver, president of the NEA. Unless more men become teachers, says Weaver, the shortage will continue to be a self-perpetuating problem.Although the feminization of the teaching profession has been underway since the 1890s, school administrators say it's becoming a more salient issue as boys fall behind girls in graduation rates and demonstrate more difficulties with reading and writing.
Fourth-grade teacher Steve Swider
There are several reasons many men find it difficult to enter, and stay in, the teaching profession: the starting salary for teachers is about $30,000, and less in early education. "Right now I don't have a wife, I don't have kids," says Bart Tittle, 24, a preschool teacher in Independence, Mo., who earns about $25,000 a year. "Later in life it's going to be much more challenging."
Another reason men fail to consider teaching is a widely held belief that they lack nurturing skills. Conversely, if a man expresses tenderness or too many traits associated with being female, some parents assume the male teacher is gay. Last fall Justin Smith, a first-grade teacher in North Kansas City, Mo., told the mother of one of his students that he'd be happy to tie a scarf on the child's Halloween costume. A few hours later, the principal notified Smith that the mother had requested her son be pulled from the class because she didn't want her child "taught by a homosexual." "At first I was kind of offended," says Smith, who is straight and married, "but then I realized this was a common stereotype and that I'd just have to let my teaching do the talking for me."
Another problem is that grown men who express physical affection for small children can be accused of being pedophiles. Steve Weber, a preschool teacher in Onamia, Minn., who has been working with special-needs children for nearly 30 years, still remembers the pit that formed in his stomach 20 years ago when a student's mother accused him of molesting her profoundly handicapped daughter. During a meeting to discuss the allegations, the mother burst into tears and blurted out that the child's molester was actually her boyfriend. "I was, like, 'Holy crap!' " recalls Weber, both relieved and horrified. Smith, whose first-grade students often need help undoing buckles or rebuttoning pants when using the toilet, insists they step into the public hallway before he will help them. "It's awkward, but I just have to protect myself," he says.
Educators are taking new approaches to address the shortage. This summer, Indiana University offered a seminal course on men and education. In July, the Pennsylvania state legislature appropriated $1 million to create a statewide version of Call Me MISTER, a recruitment program that began at Clemson University aimed at attracting African-American men. The Borough of Manhattan Community College now runs a mentorship program to help male teachers-in-training find jobs. Roy Fox, an education specialist at the University of Missouri-Columbia, is expanding MEET, or Men for Excellence in Elementary Teaching, a group that gathers on a monthly basis to discuss research and serve as a sounding board to deal with one another's male-specific teaching problems.
Bryan Nelson, the founder of MenTeach.org, says the absence of men in America's classrooms deprives kids. Children need good teachers, male and female, says Nelson. With support from the community, that's exactly what men like Josh Holt hope to be.


text:
\section{Introduction}

The one-dimensional Schr\"{o}dinger operator
\begin{equation}
\label{i1} L(y)=-y''+v(x)y,
\end{equation}
considered on $\mathbb{R}$ with $\pi$-periodic real-valued potential
$v\in L^2_{loc}(\mathbb{R})$, is self-adjoint, and its spectrum has a
gap-band structure (see Thm 2.3.1 in  \cite{E}, or Thm 2.1 in
\cite{MW}); namely, there are points
\begin{equation*}
\lambda_0^+ < \lambda_1^- \leq \lambda_1^+ < \lambda_2^- \leq
\lambda_2^+ < \lambda_3^- \leq \lambda_3^+ < \lambda_4^- \leq \lambda_4^+ <\cdots
\end{equation*}
such that
\begin{equation*}
Sp(L)= \bigcup_{n=1}^\infty [\lambda_{n-1}^{+},\lambda_n^{-}]
\end{equation*}
and the intervals of the spectrum are separated by the spectral gaps
\begin{equation*}
(-\infty,\lambda_0^+), (\lambda_1^-,\lambda_1^+),
\ldots,(\lambda_n^-,\lambda_n^+),\ldots.
\end{equation*}

The points $\lambda^-_n, \lambda^+_n $ could be determined as
eigenvalues of the Hill equation
\begin{equation}
\label{i02}
 - y^{\prime \prime} + v(x) y
= \lambda y,
\end{equation}
considered on $[0,\pi],$ respectively,  with periodic boundary
conditions
\begin{equation}
\label{i03}   y(0) = y(\pi), \quad y^\prime (0) = y^\prime (\pi),
\end{equation}
for even $n,$ and with anti-periodic boundary conditions
\begin{equation}
\label{i04}    y(0) =- y(\pi), \quad y^\prime (0) = - y^\prime (\pi),
\end{equation}
for odd $n.$  See basics and details in \cite{E,MW,Mar}.\\

The rate of decay of the size of the spectral gap $\gamma_n =
\lambda_n^+ - \lambda_n^- $ is closely related to the smoothness of
the potential $v.$ Here we mention only Hochstadt's result
\cite{Hoch63} that an $L^2 ([0,\pi])$-potential $v$ is in $C^\infty $
if and only if $(\gamma_n )$ decays faster than any power of $1/n, $
and Trubowitz's result \cite{Tr} that an $L^2 ([0,\pi])$-potential
$v$ is analytic if and only if $(\gamma_n )$ decays exponentially.

If $v$ is a complex-valued potential then the operator (\ref{i1}) is
non-self-adjoint, so one cannot talk about spectral gaps. Moreover,
the periodic and anti-periodic eigenvalues $\lambda^\pm_n $ are
well-defined for large $n$ (see Lemma~\ref{loc} below)  but the
asymptotics of $|\lambda_n^+ - \lambda_n^-|$ does not determine the
smoothness of $v.$ In \cite{Tk92} V. Tkachenko suggested to consider
also the Dirichlet b.v.p. $y(\pi)=y(0)=0. $  For large enough $n, $
close to $n^2 $ there is exactly one Dirichlet eigenvalue $\mu_n, $
so the {\em deviation}
\begin{equation}
\label{0.2}  \delta_n = |\mu_n - \frac{1}{2}(\lambda^+_n +
\lambda^-_n)|
\end{equation}
is well defined. Using an adequate parametrization of potentials in
spectral terms similar to Marchenko--Ostrovskii's ones
\cite{Mar,MO75} for self-adjoint operators, V. Tkachenko
\cite{Tk92,Tk94}  (see also \cite{ST}) characterized
$C^\infty$-smoothness and analyticity in terms of $\delta_n$ and
differences between critical values of Lyapunov functions and
$(-1)^n$. See further references and later results in \cite{DM3, DM5, DM15}.\\

In the case of specific potentials, like the Mathieu potential
\begin{equation}
\label{i05} v(x) = 2a \cos 2x, \quad a\neq 0, \;\text{real},
\end{equation}
or more general trigonometric polynomials
\begin{equation}
\label{i06} v(x) = \sum_{-N}^N c_k \exp (2ikx), \quad c_k =
\overline{c_{-k}}, \quad 0 \leq k \leq N < \infty,
\end{equation}
one comes to two classes of questions:

(i) Is the $n$-th spectral gap closed, i.e.,
\begin{equation}
\label{i07} \gamma_n = \lambda_n^+ - \lambda_n^- = 0,
\end{equation}
or, equivalently, is the multiplicity of $\lambda_n^+ $ equal to 2?

(ii) If $ \gamma_n \neq 0, $ could we tell more about the size of
this gap, or, for large enough $n,$ what is the asymptotic behavior
of $\gamma_n = \gamma_n (v)?$

E. L. Ince \cite{I} proved that the Mathieu-Hill operator has only
{\em simple } eigenvalues both for periodic and anti-periodic
boundary conditions, i.e., $\gamma_n \neq 0 $ for every $n \in
\mathbb{N}. $ His proof is presented in \cite{E}; see other proofs of
this fact in \cite{Hil,Mark,McL}, and further references in
\cite{E,MW58}.\\

For fixed $n$ and $a \to 0, $ D. Levy and J. Keller \cite{LK} gave an
asymptotics of the spectral gap $ \gamma_n = \gamma_n (a),\; v \in
(\ref{i05});$ namely
\begin{equation}
\label{i08} \gamma_n = \lambda_n^+ - \lambda_n^- =
\frac{8(|a|/4)^n}{[(n-1)!]^2} \left ( 1+ O(a) \right ), \quad a \to
0.
\end{equation}
Almost 20 years later, E. Harrell \cite{Har} found, up to a constant
factor, the asymptotics of the spectral gaps of the Mathieu operator
for fixed $a$ as  $ n\to \infty. $  J. Avron and B. Simon \cite{AS}
gave an alternative proof of E. Harrell's asymptotics and found the
exact value of the constant factor, which led to the formula
\begin{equation}
\label{i09} \gamma_n = \lambda_n^+ - \lambda_n^- =
\frac{8(|a|/4)^n}{[(n-1)!]^2} \left ( 1+ O\left (\frac{1}{n^2}\right
) \right ) \quad n \to \infty.
\end{equation}
Later, another proof of (\ref{i09}) was given by H. Hochstadt
\cite{Hoch84}. For general trigonometric polynomial potentials,
A. Grigis \cite{Gr} obtained a generic form of the main
term in the gap asymptotics.\\

In this paper we extend the result of Harrell-Avron-Simon and give
the following more precise asymptotics of the size of spectral gap
for the Mathieu operator (even in the case when the parameter $a$ is
a complex number):
\begin{equation}
\label{BF} \lambda_n^+ - \lambda_n^-= \pm \frac{8(a/4)^n}{[(n-1)!]^2}
\left[1 - \frac{a^2}{4n^3}+ O \left( \frac{1}{n^4}\right)\right],
\quad n\rightarrow\infty.
\end{equation}

Our approach is based on the methods developed in \cite{DM11,DM10},
where the gap asymptotics of the Hill operator with two term
potential of the form
$$ v(x) = A \cos 2x + B \cos 4x, \quad A\neq 0, \; B \neq 0
$$
was found. The same methods and similar asymptotic estimates play a
crucial role in the study of Riesz basis property of the root system
of Hill operators with trigonometric polynomial potentials (see
\cite{DM25}).\\

Let us mention  that the paper \cite{Kirac} claims to provide "the
formula which states the isolated terms of arbitrary number in the
asymptotics of the sequence $\gamma_n.$"  However, this claim is
false due to a unavoidable technical mistake (in \cite{Kirac},
formula (5) does not imply (4) for $m=k+1$ since, by Stirling's
formula, the remainder in (5) is much larger than the main term in
(4)).

To the best of our knowledge, (\ref{BF}) is the first formula that
gives more asymptotic terms than the formula of Harrell-Avron-Simon.

\section{Preliminaries}

Let $L_{Per^+}(v)$ and $L_{Per^-}(v)$ denote, respectively, the
operator (\ref{i1}) considered with periodic $(Per^+)$ or
antiperiodic $(Per^-)$ boundary conditions. Further we assume that
$v\in L^2 ([0,\pi])$ is a complex-valued potential such that
\begin{equation}
\label{p1} V(0) = \int_0^\pi v(x) dx =0.
\end{equation}
The following assertion is well-known (e.g., \cite[Proposition
1]{DM10}).
\begin{Lemma}
\label{loc} The spectra of $L_{Per^\pm} (v)$ are discrete. There is
an $N_0=N_0 (v)$ such that the union $\cup_{n>N} D_n $ of the discs
$D_n =\{z: \, |z-n^2|< 1 \}$ contains all but finitely many of the
eigenvalues of $L_{Per^\pm}.$

Moreover, for $n>N$ the disc $D_n$ contains exactly two (counted with
algebraic multiplicity) periodic (if $n$ is even) or antiperiodic (if
$n$ is odd) eigenvalues $\lambda_n^-, \lambda_n^+$  (where $Re
\,\lambda_n^- < Re \,\lambda_n^+$  or $Re \,\lambda_n^- = Re
\,\lambda_n^+$ and $Im \,\lambda_n^- \leq Im \,\lambda_n^+).$
\end{Lemma}

In view of Lemma \ref{loc},
\begin{equation}
|\lambda_n^{\pm}-n^2|<1, \quad \text{for } n\geq N_0.
\end{equation}
 Moreover, Lemma \ref{loc} allows
us to apply the Lyapunov--Schmidt projection method and reduce the
eigenvalue equation $Ly = \lambda y $ for  $\lambda \in D_n $ to an
eigenvalue equation in the two-dimensional space $E_n^0 = \{ L^0 y =
n^2 y\}$   (see \cite[Section 2.2]{DM15}).

This leads to the following (see the formulas (2.24)--(2.30) in
\cite{DM15}).
\begin{Lemma}
\label{lem1} In the above notations, $\lambda_n^\pm=n^2 + z$, for
$|z|<1$, is an eigenvalue of $L_{Per^\pm} (v)$ if and only if $z$ is
a root of the equation
\begin{equation}
\label{determinant S-z}
\begin{vmatrix} z-S^{11} & S^{12} \\ S^{21} & z-S^{22} \end{vmatrix} =0,
\end{equation}
where $S^{11},S^{12},S^{21},S^{22}$ can be represented as
\begin{equation}\label{S_ij}
S^{ij}(n,z)= \sum_{k=0}^{\infty}S_k^{ij}(n,z), \quad i,j=1,2,
\end{equation}
with
\begin{equation}\label{S_0}
S_0^{11}= S_0^{22}=0, \quad S_0^{12}=V(-2n), \quad S_0^{21}=V(2n),
\end{equation}
and for each $k=1,2,...,$
\begin{align}
\label{S_11}  S_k^{11}(n,z) &= \sum_{j_1,...,j_k\neq\pm n}
\frac{V(-n+j_1)V(j_2-j_1)\cdots V(j_{k}-j_{k-1})V(n-j_k)}{(n^2-j_1^2+z)\cdots(n^2-j_k^2+z)},\\
\label{S_22}  S_k^{22}(n,z) &= \sum_{j_1,...,j_k\neq\pm n}
\frac{V(n+j_1)V(j_2-j_1)\cdots V(j_k -j_{k-1})V(-n-j_k)}{(n^2-j_1^2+z)\cdots(n^2-j_k^2+z)},\\
\label{S_12}   S_k^{12}(n,z) &= \sum_{j_1,...,j_k\neq\pm n}
\frac{V(-n+j_1)V(j_2-j_1)\cdots V(j_k -j_{k-1})V(-n- j_k)}{(n^2-j_1^2+z)\cdots(n^2-j_k^2+z)},\\
\label{S_21} S_k^{21}(n,z) &= \sum_{j_1,...,j_k\neq\pm n}
\frac{V(n+j_1)V(j_2-j_1)\cdots
V(j_k-j_{k-1})V(n-j_k)}{(n^2-j_1^2+z)\cdots(n^2-j_k^2+z)}.
\end{align}
The above series converge absolutely and uniformly for $|z|\leq 1$.
\end{Lemma}

Moreover, (\ref{S_ij})--(\ref{S_21}) imply the following (see
Lemma~23 in \cite{DM15}).
\begin{Lemma} For any (complex-valued) potential $v$
\begin{equation}
\label{i} S^{11}(n,z)=S^{22}(n,z).
\end{equation}
 Moreover, if $V(-m)=\overline{V(m)}\;\; \forall m,$ then
\begin{equation}\label{ii}
 S^{12}(n,z)=\overline{S^{21}(n,\bar{z})},
\end{equation}
and if $V(-m)=V(m) \;\; \forall m,$ then
\begin{equation} \label{iii}
S^{12}(n,z)=S^{21}(n,z).
\end{equation}
\end{Lemma}

\begin{proof}
For each $k\in \mathbb{N}, $ the change of summation indices $i_s =-
j_{k+1-s}, $ $ s=1, \ldots, k$ proves that $S^{11}_k (n,z) = S^{22}_k
(n,z). $  In view of (\ref{S_ij}) and (\ref{S_0}), (\ref{i}) follows.

In a similar way,  we obtain that (\ref{ii}) and (\ref{iii}) hold by
using for each $k\in \mathbb{N} $ the change of indices $i_s=
j_{k+1-s},$ $ s=1,2,\ldots,k.$
\end{proof}

Further we consider only the Mathieu potential, i.e.,
\begin{equation}
\label{mp} v(x)= 2a \cos 2x = a e^{-2ix} + a e^{2ix}, \quad V(\pm
2)=a, \;\; V(k)=0 \;\;\text{if} \; k \neq \pm 2.
\end{equation}
For convenience, we set
\begin{equation}
\label{ab}
\alpha_n(z):=S^{11}(n,z)=S^{22}(n,z),\quad \beta_n(z):=S^{21}(n,z)=S^{12}(n,z).
\end{equation}
In these notations the basic equation  \eqref{determinant S-z}
becomes
\begin{equation}
\label{be}
(z-\alpha_n(z))^2=(\beta_n(z))^2.
\end{equation}
By Lemmas \ref{loc} and \ref{lem1},  for large enough $n\in
\mathbb{N},$ this equation has in the unit disc exactly the following
two roots (counted with multiplicity):
\begin{equation}
\label{zn}  z_n^- = \lambda_n^- -n^2, \quad z_n^+ = \lambda_n^+ -n^2.
\end{equation}

\section{Asymptotic Estimates for $z_n^\pm$ and $\alpha_n (z).$}

In this section we use the basic equation (\ref{be})  to derive
asymptotic estimates for the deviations $z_n^\pm.$ It turns out that
$|\beta_n (z)|, \; |z|\leq 1, $ is much smaller than $|\alpha_n (z)|,
$  so it is enough to analyze the asymptotics of $\alpha_n (z_n^\pm)
$ in order to find asymptotic estimates for $z_n^\pm.$

The following inequality is well known (e.g., see Lemma~78 in
\cite{DM15}):
\begin{equation}
\label{In1} \sum_{j \neq \pm n} \frac{1}{|n^2-j^2|} < \frac{2\log
6n}{n}, \quad \text{for} \; n \in \mathbb{N}.
\end{equation}

\begin{Lemma}
\label{lem2}
 If $|z|\leq 1$, then
\begin{equation}\label{In2}
 \sum_{j_1,...,j_\nu\neq\pm n} \frac{1}{|n^2-j_1^2+z|\cdots |n^2-j_\nu^2+z|} <
  \left( \frac{4\log 6n}{n} \right)^\nu.
\end{equation}
\end{Lemma}

\begin{proof} If $|z|\leq 1$ and $j\neq \pm n$, then
\begin{equation*}
|n^2-j^2+z| \geq |n^2-j^2|-1 \geq \frac{1}{2} |n^2-j^2|.
\end{equation*}
Therefore,
\begin{align*}
\sum_{j_1,...,j_\nu\neq\pm n} \frac{1}{|n^2-j_1^2+z|\cdots
|n^2-j_\nu^2+z|} \leq 2^\nu \left(\sum_{j \neq \pm n}
\frac{1}{|n^2-j^2|} \right)^\nu,
\end{align*}
so (\ref{In2}) follows from (\ref{In1}).
\end{proof}

The next lemma  gives a rough estimate for $\beta_n(z);$ we improve
this estimate in the next section.
\begin{Lemma}
\label{lem3} For $|z|\leq 1  $ we have
\begin{equation}
\label{BE1} \beta_n(z)=O\left( \left( \frac{\log n}{n} \right)^n
\right).
\end{equation}
\end{Lemma}

\begin{proof}
If $\nu < n-1, $ then all terms of the sum $S^{21}_\nu (n,z) $ in
(\ref{S_21}) vanish. Indeed, each term of the sum $S^{21}_\nu (n,z) $
is a fraction which nominator has the form $V(x_1) V(x_2) \cdots
V(x_{\nu+1}) $ with $ x_1= n+j_1, \; x_2 = j_2-j_1,  \ldots,  x_{\nu
+1}= n- j_{\nu}. $ Therefore, if $\nu < n-1$ then there are no $x_1,
x_2,\ldots, x_{\nu+1} \in \{-2,2\}$ satisfying
$x_1+x_2+\cdots+x_{\nu+1}=2n,$ so every term of the sum $S^{21}_\nu
(n,z) $ vanishes due to (\ref{mp}). Hence, by (\ref{mp}) we have
\begin{equation*}
|\beta_n(z)| \leq \sum_{\nu=n-1}^\infty \sum_{j_1,...,j_\nu \neq\pm
n} \frac{|a|^{\nu+1}}{|n^2-j_1^2+z|\cdots |n^2-j_\nu^2+z|},
\end{equation*}
so (\ref{BE1}) follows from (\ref{In2}).
\end{proof}

\begin{Lemma}
\label{lemA} In the above notations,
\begin{equation}
\label{In0} z_n^\pm = \frac{a^2}{2n^2} + O\left( \frac{1}{n^4}
\right), \quad  \alpha_n (z_n^\pm)=\frac{a^2}{2n^2} + O\left(
\frac{1}{n^4} \right), \quad  n \to \infty.
\end{equation}
\end{Lemma}

\begin{proof}
In view of  (\ref{S_ij}), (\ref{S_11}) and (\ref{ab}), we have
\begin{equation}
\label{A} \alpha_n(z)= \sum_{p=1}^\infty A_p(n,z),
\end{equation}
where
\begin{equation}\label{A_p}
A_p(n,z)= \sum_{j_1,...,j_p\neq\pm n} \frac{V(-n+j_1)V(j_2-j_1)\cdots
V(j_p - j_{p-1})V(n-j_p)}{(n^2-j_1^2+z)\cdots(n^2-j_p^2+z)}.
\end{equation}

First we show that
\begin{equation}
\label{A2k} A_{2k}(n,z) \equiv 0 \quad \forall k \in \mathbb{N}.
\end{equation}
Indeed, for $p=2k$ each term of the sum in (\ref{A_p}) is a fraction
which nominator has the form $V(x_1) V(x_2) \cdots V(x_{2k+1}) $ with
\begin{equation*}
x_1= -n+j_1, \quad x_2= j_2-j_1, \quad \ldots, \quad x_{2k+1}= n-
j_{2k}.
\end{equation*}
Since  $x_1+x_2+\cdots+x_{2k+1}=0, $ it follows that there is $i_0$
such that $x_{i_0} \neq \pm 2, $ so $V(x_{i_0}) =0 $ due to
(\ref{mp}). Therefore, every term of the sum $A_{2k} (n,z) $
vanishes, hence (\ref{A2k}) holds.\\

Next we estimate iteratively, in two steps,   $\alpha_n (z)$ and
$z_n^\pm.$ The first step provides rough estimates which we improve
in the second step.\\

{\em Step 1.} By (\ref{A_p}), we have
\begin{equation*}
A_1(n,z)= \sum_{j_1\neq \pm n}\frac{V(-n+j_1)V(n-j_1)}{n^2-j_1^2+z}.
\end{equation*}
 In view of (\ref{mp}), we get a non-zero term in the above sum if and only if
$j_1=n+2,$ or $ j_1=n-2.$ Therefore,
\begin{align}
\label{A_1} A_1 (n,z) = \frac{a^2}{n^2-(n-2)^2+z} +
\frac{a^2}{n^2-(n+2)^2+z} = a^2 \frac{8-2z}{(4n)^2-(4-z)^2},
\end{align}
which implies that
\begin{equation}
\label{In4} A_1(n,z)= O\left(\frac{1}{n^2}\right)  \quad \text{for}
\;\; |z|\leq 1.
\end{equation}

On the other hand, from (\ref{mp}), \eqref{In2} and (\ref{A_p}) it
follows that
\begin{equation}
\label{In5} |A_{2k-1}(n,z)| \leq |a|^{2k}\left( \frac{ 4\log 6n }{n}
\right)^{2k-1}, \quad k=2,3, \ldots,
\end{equation}
which implies
\begin{equation}
\label{In6} \sum_{k=2}^\infty |A_{2k-1}(n,z)| \leq \sum_{k=2}^\infty
|a|^{2k} \left( \frac{4\log 6n}{n} \right)^{2k-1}= o \left(1/n^2
\right).
\end{equation}
Hence, by (\ref{In4}) and (\ref{In6}) we obtain
\begin{equation}
\label{In7} \alpha_n(z)=O\left( \frac{1}{n^2} \right) \quad
\text{for} \;\; |z|\leq 1.
\end{equation}

Furthermore, from (\ref{be}), (\ref{zn}) and (\ref{BE1}) it follows
immediately that
\begin{equation}
\label{In8} z_n^\pm -  \alpha_n(z_n^\pm)=O\left( \frac{1}{n^k}
\right), \quad  \forall k \in \mathbb{N}.
\end{equation}
Therefore, (\ref{In7}) implies that
\begin{equation}
\label{In11} z_n^\pm= O\left(\frac{1}{n^2} \right).
\end{equation}

\emph{Step 2.}    By \eqref{A_1} we have
\begin{equation}\label{In10}
A_1 (n,z)= \frac{a^2}{2n^2}+ O\left(\frac{1}{n^4}\right) \quad
\text{if}  \;\; z=O(1/n^2).
\end{equation}
Let us consider
\begin{equation*}
A_3(n,z)= \sum_{j_1,j_2,j_3\neq \pm n} \frac{V(-n+j_1)V(j_2-j_1)
V(j_3 -j_2) V(n-j_3)}{(n^2-j_1^2+z)(n^2-j_2^2+z)(n^2-j_3^2+z)}.
\end{equation*}
In view of (\ref{mp}), we get a non-zero term in the above sum if and
only if
\begin{equation*}
j_1=n+2; \quad j_2=n+4; \quad j_3=n+2,
\end{equation*}
or
\begin{equation*}
j_1=n-2; \quad j_2=n-4; \quad j_3=n-2.
\end{equation*}
Hence,
\begin{align*}
  A_3(n,z) &=
\frac{a^4}{[n^2-(n+2)^2+z][n^2-(n+4)^2+z][n^2-(n+2)^2+z]}
\\  &+ \frac{a^4}{[n^2-(n-2)^2+z][n^2-(n-4)^2+z][n^2-(n-2)^2+z]},
\end{align*}
so it is easy to see that
\begin{align}\label{In12}
A_3(n,z)=  O\left( \frac{1}{n^4}\right) \quad \text{if} \;\;|z|\leq
1.
\end{align}
On the other hand, by \eqref{In5} we have
\begin{equation}
\label{In13} \sum_{k=3}^\infty |A_{2k-1}(n,z)| \leq \sum_{k=3}^\infty
|a|^{2k} \left( \frac{4\log 6n}{n} \right)^{2k-1}= o \left(1/n^4
\right).
\end{equation}
Therefore, by (\ref{In10}), (\ref{In12}) and (\ref{In13}) imply that
\begin{equation}
\label{In14} \alpha_n(z)=\frac{a^2}{2n^2}+
O\left(\frac{1}{n^4}\right) \quad \text{if} \;\; z=O(1/n^2).
\end{equation}

Hence, from \eqref{In8} it follows that
\begin{equation}
\label{In15}
z_n^\pm = \frac{a^2}{2n^2} + O \left( \frac{1}{n^4} \right).
\end{equation}
\end{proof}

\emph{Remark.}  From \eqref{A_1} and \eqref{In15} it follows that
\begin{equation}
\label{In16} A_1(n,z_n^\pm)= \frac{a^2}{2n^2} + \frac{a^2}{2n^4} -
\frac{a^4}{16n^4} + O\left( \frac{1}{n^6} \right).
\end{equation}
Similarly, it is easily seen that
\begin{equation}
\label{In17} A_3 (n,z_n^\pm)= \frac{a^4}{16n^4} + O\left(
\frac{1}{n^6} \right).
\end{equation}
On the other hand, analyzing $A_5 (n,z) $ one can show that
\begin{equation}
\label{In18} A_5 (n,z)=  O\left( \frac{1}{n^6} \right) \quad
\text{if} \;\; |z|\leq 1.
\end{equation}
Moreover, by (\ref{In5}) we have
\begin{equation}
\label{In19} \sum_{k=4}^\infty |A_{2k-1}(n,z)| = o \left(
\frac{1}{n^6} \right) \quad \text{if} \;\; |z|\leq 1.
\end{equation}
Hence, in view of \eqref{In8}, the estimates
\eqref{In16}--\eqref{In19} lead to
\begin{equation}
\label{BestEstimateZ_n}
z_n^\pm= \frac{a^2}{2n^2} + \frac{a^2}{2n^4} + O\left( \frac{1}{n^6} \right).
\end{equation}
This analysis could be extended in order to obtain more asymptotic
terms of $z_n^\pm, $ and even to explain that the corresponding
asymptotic series along the powers of $1/n$ contains only even
nontrivial terms. However,
in this paper we need only the estimate (\ref{In15}).\\

The following assertion plays an essential role later.
\begin{Lemma} With $\gamma_n = \lambda_n^+ - \lambda_n^- = z_n^+ -
z_n^-,$
\begin{equation}
\label{D1} \alpha_n(z_n^+)-\alpha_n(z_n^-)= \gamma_n \left[
-\frac{a^2}{8n^2} + O \left( \frac{1}{n^4}\right) \right].
\end{equation}
\end{Lemma}

\begin{proof} By \eqref{A} and \eqref{A2k} we obtain
\begin{align}\label{D2}
\alpha_n(z_n^+)-\alpha_n(z_n^-)= A_1(n,z_n^+)- A_1(n,z_n^-) +
\int_{z_n^-}^{z_n^+}  \frac{d}{dz} \tilde{\alpha}_n(z) \, dz,
 \end{align}
where
\begin{align*}
\tilde{\alpha}_n(z)= \alpha_n(z) -A_1(n,z) =A_3(n,z) + A_5(n,z) +
\cdots.
\end{align*}
In view of \eqref{In12} and  \eqref{In13},
\begin{equation*}
\tilde{\alpha}_n(z) = O(1/n^4) \quad \text{for} \;\; |z|\leq 1.
\end{equation*}
By the Cauchy formula for derivatives, this estimate implies that
$$
d\alpha_n/dz =O(1/n^4) \quad \text{for} \;\; |z|\leq 1/2.
$$
Hence, we obtain
\begin{align} \label{D3}
\int_{z_n^-}^{z_n^+}  \frac{d}{dz} \tilde{\alpha}_n(z) \, dz =
\gamma_n O \left( \frac{1}{n^4} \right).
\end{align}
On the other hand, by \eqref{A_1}
\begin{align} \nonumber
A_1(n,z_n^+)-A_1(n,z_n^-) &= \left[ \frac{8-2z_n^+}{(4n)^2-
(4-z_n^+)^2} - \frac{8-2z_n^-}{(4n)^2- (4-z_n^-)^2}  \right] a^2 \\
\nonumber &=\gamma_n \left[\frac{-32n^2  -32 + 8(z_n^+ + z_n^-) -2
z_n^+ z_n^-} {[(4n)^2- (4-z_n^+)^2][(4n)^2- (4-z_n^-)^2]}  \right]
a^2.
\end{align}
Therefore, taking into account \eqref{In0}, we obtain
\begin{equation}
\label{D4} A_1(n,z_n^+)-A_1(n,z_n^-)= \gamma_n \left[
\frac{-a^2}{8n^2} + O \left( \frac{1}{n^4} \right) \right].
\end{equation}
In view of  \eqref{D2}, the estimates \eqref{D3} and \eqref{D4} lead
to \eqref{D1}.
\end{proof}

\section{Asymptotic formulas for $\beta_n(z_n^\pm)$ and $\gamma_n.$}

In this section we find more precise asymptotics of $\beta_n
(z_n^\pm)$. These asymptotics, combined with the results of the
previous section,
lead to an asymptotics for $\gamma_n.$  \\

In view of (\ref{mp}), each nonzero term in \eqref{S_21} corresponds
to a \emph{k}-tuple of indices $(j_1,...,j_k)$ with $j_1, \ldots ,
j_k \neq \pm n$ such that
\begin{equation}
\label{a1} (n+j_1)+(j_2-j_1)+ \cdots +(j_k-j_{k-1})+(n-j_k)=2n
\end{equation}
and
\begin{equation}
\label{a2} n+j_1,  j_2-j_1, \ldots, j_k-j_{k-1},   n-j_k \in \{-2,
2\}.
\end{equation}
By (\ref{a1}) and (\ref{a2}), there is one-to-one correspondence
between the nonzero terms in \eqref{S_21} and the "admissible" walks
$x=(x(t))_{t=1}^{k+1}$ on $\mathbb{Z}$ from $-n$ to $n$ with steps
$x(t)= \pm 2$ and vertices $j_0= -n, \; j_{k+1}=n, $
\begin{equation}
\label{j_k}  j_s=-n+\sum_{t=1}^s x(t) \neq \pm n, \quad s=1,\ldots,
k.
\end{equation}

Let $X_n(p), \; p=0, 1, 2, \ldots $ denote set of all such walks with
$p$ negative steps. It is easy to see that every walk $x \in X_n (p)$
has totally $n+ 2p $  steps because $\sum x(t) = 2n. $ Therefore,
every admissible walk has  at least $n$ steps.

In view of (\ref{S_ij}), (\ref{S_21}), (\ref{mp}) and (\ref{ab}), we
have
\begin{equation}\label{sigma_p}
\beta_n(z)= \sum_{p=0}^\infty \sigma_p(n,z) \quad \text{with} \quad
\sigma_p(n,z)=\sum_{x\in X_n(p)} h(n,z),
\end{equation}
where, for $x=(x(t))_{t=1}^{k+1},$
\begin{equation}
\label{h(x,z)} h(x,z)=\frac{a^{k+1}}{(n^2-j_1^2+z)(n^2-j_2^2+z)\cdots
(n^2-j_k^2+z)}
\end{equation}
with $j_1, \ldots, j_k $ given by (\ref{j_k}).

The set $X_n (0) $ has only one element, namely the walk
\begin{equation}
\label{xi} \xi=(\xi(t))_{t=1}^n, \quad \xi(t)=2 \;\; \forall t.
\end{equation}
Therefore,
\begin{equation}
\label{xiz}  \sigma_0 (n,z)= h(\xi, z) =\frac{a^n}{(n^2-j_1^2+z)
\cdots (n^2-j_{n-1}^2+z)}
\end{equation}
with $j_k=-n+2k$, $ k=1,\cdots, n-1$. Moreover, since
\begin{align*}
\prod_{k=1}^{n-1} \left(n^2-(-n+2k)^2 \right) = 4^{n-1} [(n-1)!]^2,
\end{align*}
the following holds.
\begin{Lemma} In the above notations,
\begin{equation}\label{Sigma0}
 \sigma_0(n,0)= h(\xi,0)= \frac{4(a/4)^n}{[(n-1)!]^2}.
\end{equation}
\end{Lemma}

It is well known that
\begin{align}\label{HarmonicSum}
\sum_{k=1}^n \frac{1}{k}= \log n + \gamma + \frac{1}{2n} -
\frac{1}{12n^2} + O\left( \frac{1}{n^4} \right), \quad n\in \mathbb{N},
\end{align}
where $\gamma$ is the Euler's constant.

\begin{Lemma}
\label{lemb2} In the above notations,
\begin{equation}
\label{b2} \sigma_0 (n,z_n^\pm)=\sigma_0(n,0) \left[1- \frac{a^2 \log
n}{4n^3} - \frac{a^2 \gamma}{4n^3} + O\left(
\frac{1}{n^4}\right)\right].
\end{equation}
\end{Lemma}

\begin{proof} By (\ref{xiz}), we have
\begin{equation}\label{b3}
\sigma_0(n,z_n^\pm) =\sigma_0(n,0) \prod_{k=1}^{n-1}
\left(1+\frac{z_n^\pm}{n^2-(-n+2k)^2}\right)^{-1}.
\end{equation}
For simplicity, we set $\displaystyle b_k =
\frac{z_n^\pm}{n^2-(-n+2k)^2}= \frac{z_n^\pm}{4k(n-k)}.$ Then,
\begin{align*}
\log\left( \prod_{k=1}^{n-1} (1+b_k)^{-1} \right) &=
-\sum_{k=1}^{n-1} \log(1+b_k) = -\sum_{k=1}^{n-1} b_k + O \left(
\sum_{k=1}^{n-1} |b_k|^2 \right).
\end{align*}
Using \eqref{In0}, we obtain
\begin{align*}
\sum_{k=1}^{n-1} b_k &= \left( \sum_{k=1}^{n-1} \frac{1}{4k(n-k)} \right)
\left[ \frac{a^2}{2n^2} +  O\left( \frac{1}{n^4} \right) \right] \\
&= \frac{1}{2n} \left(\sum_{k=1}^{n-1} \frac{1}{k} \right) \left[
\frac{a^2}{2n^2} +  O\left( \frac{1}{n^4} \right) \right].
\end{align*}
By \eqref{HarmonicSum}, it follows that
\begin{equation*}
\sum_{k=1}^{n-1} b_k =\frac{a^2 \log n}{4n^3} +
\frac{a^2 \gamma}{4n^3} + O\left( \frac{1}{n^4}\right).
\end{equation*}
On the other hand, by \eqref{In0},
\begin{align*} \sum_{k=1}^{n-1} |b_k|^2 =
\left( \sum_{k=1}^{n-1} \frac{1}{[4k(n-k)]^2} \right) O\left(
\frac{1}{n^4}\right) = O \left( \frac{1}{n^4} \right).
\end{align*}
Hence,
\begin{align*} \log  \left (\prod_{k=1}^{n-1} (1+ b_k)^{-1} \right )  =
 - \frac{a^2 \log n}{4n^3} -\frac{a^2 \gamma}{4n^3}  + O\left(
 \frac{1}{n^4}\right),
\end{align*}
which implies (\ref{b2}).

\end{proof}

Next we study the ratio $\sigma_1(n,z)/\sigma_0(n,z).$
\begin{Lemma}
\label{lemR} We have
\begin{equation}
\label{r1} \sigma_1(n,z)=\sigma_0(n,z) \cdot \Phi (n,z),
\end{equation}
where
\begin{equation}
\label{r2}  \Phi (n,z)=\sum_{k=2}^{n-1} \varphi_k (n,z)
\end{equation}
with
\begin{equation}
\label{r3}  \varphi_k (n,z)=
\frac{a^2}{[n^2-(-n+2k)^2+z][n^2-(-n+2k-2)^2+z]}.
\end{equation}
\end{Lemma}

\begin{proof}
From the definition of $X_n (1) $ and
 \eqref{sigma_p} it follows that
\begin{equation}\label{sigma1}
\sigma_1(n,z)=\sum_{x\in X_n(1)} h(x,z)= \sum_{k=2}^{n-1} h(x_k,z),
\end{equation}
where $x_k$ denotes the walk with $(k+1)$'th step equal to -2, i.e.,
\begin{equation*}
x_k(t)=
\begin{cases}
~~2 & \text{if } t\neq k \\
-2 & \text{if } t=k
\end{cases},
\quad 1\leq t \leq n+2.
\end{equation*}
Now, we figure out the connection between vertices of $\xi$ and $x_k$
as follows:
\begin{equation*}
j_\alpha(x_k)= \begin{cases} j_\alpha(\xi), & 1\leq \alpha\leq k,\\
j_{k-1}(\xi) &   \alpha = k+1,\\
j_{\alpha-2}(\xi) &  k+2  \leq \alpha \leq n=2.
\end{cases}
\end{equation*}
Therefore, by \eqref{h(x,z)}
\begin{equation} \label{b12}
h(x_k,z)=
h(\xi,z)\frac{a^2}{(n^2-[j_{k-1}(\xi)]^2+z)(n^2-[j_{k}(\xi)]^2+z)}.
\end{equation}
Since $j_k(\xi)=-n+2k, \; k=2,\ldots,n-1,$ (\ref{sigma1}) and
(\ref{b12}) imply (\ref{r1}).
\end{proof}

\begin{Lemma}
\label{lemRR} In the above notations, if $z=O(1/n^2)$ then
\begin{equation}
\label{r12} \Phi (n,z) = \Phi (n,0) + O \left( 1/n^4 \right)
\end{equation}
and
\begin{equation}
\label{r13} \Phi^*(n,z) := \sum_{k=2}^{n-1} |\varphi_k (n,z)| = \Phi
(n,0) + O \left( 1/n^4 \right).
\end{equation}
Moreover,
\begin{equation}\label{r11}
\Phi (n,0) = \frac{a^2}{8n^2} + \frac{a^2\log n}{4n^3} +
\frac{a^2(\gamma-1)}{4n^3}  + O \left( 1/n^4 \right).
\end{equation}
\end{Lemma}

\begin{proof}
Since
$$
\frac{\varphi_k (n,z)}{\varphi_k (n,0)} = \left[ 1+
\frac{z}{n^2-(-n+2k)^2}\right]^{-1} \left[ 1+
\frac{z}{n^2-(-n+2k-2)^2}\right]^{-1},
$$
it is easily seen that
$$
\varphi_k (n,z)/\varphi_k (n,0) = 1+  O \left( 1/n^3 \right) \quad
\text{if} \;\; z=O(1/n^2).
$$
On the other hand, $\varphi_k (n,0)=O (1/n^2),$ so it follows that
$$
\varphi_k (n,z)-\varphi_k (n,0) =\varphi_k (n,0)\, O \left( 1/n^3
\right) = O \left( 1/n^5 \right) \;\; \text{if} \;\; z=O \left( 1/n^2
\right).
$$
Therefore, we obtain that
$$
\sum_{k=2}^{n-1} |\varphi_k (n,z)-\varphi_k (n,0)| = O \left( 1/n^4
\right) \quad \text{if} \;\; z=O(1/n^2).
$$
The latter sum dominates both $|\Phi (n,z)- \Phi (n,0)|$ and $|\Phi^*
(n,z)- \Phi (n,0)|. $ Hence,  (\ref{r12}) and (\ref{r13}) hold.\\

Next we prove (\ref{r11}). Since
\begin{equation*}
\Phi (n,0)=\sum_{k=2}^{n-1} \frac{a^2}{16(k-1)k(n-k)(n+1-k)},
\end{equation*}
by using the identities
\begin{equation*}
\frac{1}{k(n-k)}= \frac {1}{n} \left( \frac{1}{k}+\frac{1}{n-k} \right),
\quad \frac{1}{(k-1)(n+1-k)}= \frac {1}{n} \left( \frac{1}{k-1}+\frac{1}{n+1-k} \right)
\end{equation*}
we obtain
\begin{equation}
\label{Di} \Phi (n,0) = \frac{a^2}{16n^2} \sum_{i=1}^4 D_i (n),
\end{equation}
where
\begin{equation*}
D_1(n)= \sum_{k=2}^{n-1} \frac{1}{k(k-1)},
\quad D_2(n)= \sum_{k=2}^{n-1} \frac{1}{(n-k)(n+1-k)},
\end{equation*}
\begin{equation*}
D_3(n)= \sum_{k=2}^{n-1} \frac{1}{k(n+1-k)},
\quad D_4(n)= \sum_{k=2}^{n-1} \frac{1}{(k-1)(n-k)}.
\end{equation*}
The change of summation index  $m=n+1-k$ shows that $D_2(n)=D_1(n),$
and we have
\begin{equation}\label{D_1}
D_1(n)=\sum_{k=2}^{n-1} \left (\frac{1}{k-1}- \frac{1}{k} \right )=1-
\frac{1}{n-1}= 1 - \frac{1}{n} + O \left( \frac{1}{n^2}\right).
\end{equation}
Moreover, since
\begin{equation*}
D_3(n)= \frac{1}{n+1} \left( \sum_{k=2}^{n-1} \frac{1}{k} +
\sum_{k=2}^{n-1} \frac{1}{n+1-k} \right) =\frac{2}{n+1}
\sum_{k=2}^{n-1} \frac{1}{k},
\end{equation*}
by \eqref{HarmonicSum} we obtain that
\begin{equation}\label{D_3}
D_3(n)=\frac{2 \log n}{n} +\frac{2(\gamma -1)}{n}-
\frac{2 \log n}{n^2} + O \left( \frac{1}{n^2}\right).
\end{equation}
Similarly,
\begin{equation*}
D_4(n)= \frac{1}{n-1} \left( \sum_{m=1}^{n-2} \frac{1}{m} +
\sum_{m=1}^{n-2} \frac{1}{n-m-1} \right)= \frac{2}{n-1}
\sum_{m=1}^{n-2} \frac{1}{m},
\end{equation*}
and \eqref{HarmonicSum} leads to
\begin{equation}\label{D_4}
D_4(n)=\frac{2 \log n}{n} + \frac{2\gamma}{n} +
\frac{2\log n}{n^2} +  O \left( \frac{1}{n^2}\right).
\end{equation}
Hence, in view of \eqref{Di}--\eqref{D_4}, we obtain \eqref{r11}.
\end{proof}

\begin{Proposition}
We have
\begin{equation}\label{beta}
\beta_n(z_n^\pm)=\sigma_0(n,0) \left[  1 + \frac{a^2}{8n^2} -
\frac{a^2}{4n^3} + O \left( \frac{1}{n^4}\right) \right] .
\end{equation}
\end{Proposition}

\begin{proof}
From (\ref{b2}), (\ref{r1}), (\ref{r12}) and (\ref{r11}) it follows
immediately that
$$
\sigma_1(n,z_n^\pm)+\sigma_0(n,z_n^\pm) = \sigma_0(n,0) \left[  1 +
\frac{a^2}{8n^2} - \frac{a^2}{4n^3} + O \left( \frac{1}{n^4}\right)
\right] .
$$
Since $\beta_n (z) = \sum_{p=0}^\infty \sigma_p (n,z), $ in view of
(\ref{b2}) to complete the proof it is enough to show that
\begin{equation}
\label{S2} \sum_{p=2}^\infty \sigma_p(n,z_n^\pm)= \sigma_0(n,z_n^\pm)
\, O\left(\frac{1}{n^4}\right).
\end{equation}

Next we prove (\ref{S2}). Recall that $\sigma_p (n,z) = \sum_{x\in
X_n(p)} h(x,z).$ Now we set
$$
\sigma^*_p (n,z) = \sum_{x\in X_n(p)} |h(x,z)|.
$$
We are going to show that there is an absolute constant $C>0$ such
that
\begin{equation}
\label{S3} \sigma^*_p (n,z_n^\pm) \leq \sigma^*_{p-1} (n,z_n^\pm)
\cdot \frac{C}{n^2}, \quad p\in \mathbb{N}, \;\; n \geq N_0.
\end{equation}
Since $\sigma_0 (n,z)$ has one term only, we have $\sigma^*_0
(n,z)=|\sigma_0 (n,z)|.$

Let $p \in \mathbb{N}.$ To every walk $x \in X_n(p)$ we assign a pair
$(\tilde{x},j)$, where $\tilde{x} \in X_n(p-1)$ is the walk that we
obtain after dropping the first cycle $ \{ +2, -2 \}$ from $x,$ and
$j$ is the vertex of $x$ where the first negative step of $x$ is
performed. In other words, we consider the map
\begin{equation*}
\varphi: X_n(p)\longrightarrow X_n(p-1) \times I, \qquad I=\{-n+4, -n+6, \ldots, n-2\},
\end{equation*}
defined by $\varphi(x)=(\tilde{x},j)$, where
\begin{equation*}
\tilde{x}(t)=
\begin{cases}
x(t) & \text{if } 1\leq t\leq k-1 \\
x(t+2) & \text{if } k \leq t \leq n+2p-2
\end{cases},
\end{equation*}
where $k=\min\{t:x(t)=2,~x(t+1)=-2\}$ and $j=-n+2k$.

The map $\varphi$ is clearly injective, and moreover, we have
\begin{equation} \label{S5}
h(x,z)= h(\tilde{x},z)\frac{a^2}{(n^2-j^2+z)(n^2-(j-2)^2+z)}.
\end{equation}
Since the mapping $\varphi$ is injective, from (\ref{r3}),
(\ref{r13}) and \eqref{S5} it follows that
\begin{equation}\label{S6}
\sigma^*_p(n,z) \leq \sigma^*_{p-1}(n,z)\cdot \Phi^*(n,z).
\end{equation}
Hence, by \eqref{r13} and (\ref{r11}), we obtain that (\ref{S3})
holds.

From (\ref{S3}) it follows (since $\sigma^*_0 (n,z_n^\pm)=|\sigma_0
(n,z_n^\pm)|$) that
$$
\sigma^*_p(n,z_n^\pm) \leq |\sigma_0 (n,z_n^\pm)| \cdot \left
(\frac{C}{n^2}\right)^p.
$$
Hence, (\ref{S2}) holds, which completes the proof.
\end{proof}

\begin{Theorem}
The  Mathieu operator
\begin{equation*}
L(y)=-y''+2a \cos{(2x)}y, \quad a\in \mathbb{C}, \;\; a\neq 0,
\end{equation*}
considered with periodic or anti-periodic boundary conditions has,
close to $n^2$ for large enough $n$, two periodic (if $n$ is even) or
anti-periodic (if $n$ is odd) eigenvalues $\lambda_n^-$,
$\lambda_n^+$. For fixed nonzero $a\in \mathbb{C}$,
\begin{equation}\label{G}
\lambda_n^+ -\lambda_n^-  = \pm \frac{8(a/4)^n}{[(n-1)!]^2} \left[1 -
\frac{a^2}{4n^3}+ O \left( \frac{1}{n^4}\right)\right], \quad n \to
\infty.
\end{equation}
\end{Theorem}

\begin{proof} The basic equation \eqref{be}
splits into two equations
\begin{align}\label{E1}
z- \alpha_n(z)- \beta_n(z)&=0, \\ \label{E2} z- \alpha_n(z)+
\beta_n(z)&=0.
\end{align}
In view of  (\ref{BE1}) and  \eqref{In7}, it follows that  for large
enough $n$
\begin{equation*}
|z- \alpha_n(z) \pm \beta_n(z)| < |z| \quad \text{if} \;\;  |z|=1.
\end{equation*}
Hence, for large enough $n,$ each of the equations \eqref{E1} and
\eqref{E2} has only one root in the unit disc due to Rouche's
theorem.

On the other hand, by Lemmas~\ref{loc} and \ref{lem1}, for large
enough $n$ the basic equation has exactly two roots $z_n^-, z_n^+$ in
the unit disc, so either $z_n^-$ is the root of (\ref{E1}) and $
z_n^+$ is the root of (\ref{E2}), or $z_n^+$ is the root of
(\ref{E1}) and $ z_n^-$ is the root of (\ref{E2}). Therefore,  we
obtain
\begin{equation*}
z_n^+ - z_n^- -[\alpha_n(z_n^+)- \alpha_n(z_n^-)]  = \pm [
\beta_n(z_n^+) + \beta_n(z_n^-)].
\end{equation*}
Now, \eqref{D1} and \eqref{beta} imply, with $\gamma_n= \lambda_n^+ -
\lambda_n^-,$
\begin{equation*}
\gamma_n \left[1 +\frac{a^2}{8n^2} + O \left( \frac{1}{n^4}\right)
\right] = \pm 2\sigma_0(n,0) \left[1+ \frac{a^2}{8n^2} -
\frac{a^2}{4n^3} + O \left( \frac{1}{n^4} \right)  \right].
\end{equation*}
Therefore,
\begin{align*}
\gamma_n &= \pm 2\sigma_0(n,0) \left[1+ \frac{a^2}{8n^2} -
\frac{a^2}{4n^3} + O \left( \frac{1}{n^4} \right)  \right]
\left[1 -\frac{a^2}{8n^2} + O \left( \frac{1}{n^4}\right) \right] \\
&= \pm 2\sigma_0(n,0) \left[1 - \frac{a^2}{4n^3} + O \left(
\frac{1}{n^4}\right)\right].
\end{align*}
Hence, in view of \eqref{Sigma0}, \eqref{G} holds.
\end{proof}

redpajama_set_name: RedPajamaArXiv


Design Manual for Urban Roads and Streets – are we culturally ready?
Last month the Minister for Transport, Tourism and Sport, Leo Varadkar, launched the 'Design Manual for Urban Roads and Streets'. Produced jointly by the Department of Transport, Tourism and Sport and the Department of the Environment, Community and Local Government with a multi-disciplinary local authority led team; it illustrates the Irish Government's commitment to the creation of sustainable and inclusive public spaces.
The Manual notes that the way forward requires "a shift away from conventional design solutions toward those which prioritise sustainable modes of transport, safeguard vulnerable users and promote a sense of place." It provides a revision of the way we think about, design and plan our local spaces, promoting our streets to something more than just car corridors.
The principles of integrated street design and modal reprioritisation, if followed, have the potential to enhance local liveability: that is to support a sense of place and identity; connectivity and community; movement and rest. The design guidance relates to new build and retrofit schemes, and is usefully illustrated with photographed examples of best practice. Many of these can already be found around the country such as in Ballina, Co. Mayo, Drogheda, Co. Louth and Adamstown, Co. Dublin.
The principles promoted by the manual follow contemporary European standards, which have already been championed globally: a testament to the perceived value in this revised approach. In addition to the much referenced UK 'Manual for Streets' published in 2007, the past decade has seen a raft of similar guidelines across Europe, Australia, Emirates (Abu Dhabi) and the USA (Seattle, 2005; Los Angeles 2008; New York, 2009; San Francisco, Draft, 2008). A quiet international planning revolution is evidently underway with modal reprioritisation and street design supporting a new emphasis on 'place-making'. For example, the transformation of London's Exhibition Road into a 'shared space' has unequivocally given it the stamp of a cultural heartland.
The challenge now is to close the gap between evolving policy thinking and practical implementation. A cultural readiness, which entails both vision and purpose, is required to see such a project through. Are we in Ireland ready to join this quiet revolution? What is needed is strong local leadership and clarity of vision to translate this momentum into consistent and sustainable gains. As professionals, can we successfully engage with local communities to identify priorities, and engender support for innovative and effective neighbourhood schemes?
Only time will tell if we can rise to the challenge of DMURS implementation. Why not share your views!
Previous: Previous post: Big and bold versus small and beautiful?
Next: Next post: Transport Insights completes first project!


text:
Martha Bakes Returns for its Tenth Season on PBS Beginning July 7, 2018
NEW YORK, July 03, 2018 (GLOBE NEWSWIRE) -- The tenth season of Martha Stewart's Emmy®-nominated baking show, Martha Bakes, premieres on PBS stations nationwide on Saturday, July 7th, 2018. The popular program is presented in partnership with Domino® Sugar and Kerrygold® Butter.
In the tenth season of Martha Bakes, Martha responds to the growing interest in innovative and show stopping decorating techniques that are becoming more prevalent on social media. Martha's tricks and tips make it possible for the home baker to create recipes with pristine presentation. Some highlights from this season include mirror glazed cakes, intricate lattice designs, decorative breads and over the top desserts.
On each of the season's thirteen episodes, a special culinary guest will join Martha to share some of their secrets. Season ten's guests are some of the top pastry chefs and bakers from across the country, including Dominique Ansel of Dominique Ansel Bakery, Ron Ben Israel of Ron Ben Israel Cakes, Daniel Alvarez of Daily Provisions and Union Square Café, and famed chocolatier Jacques Torres.
"I'm thrilled to present our tenth season of Martha Bakes. This success could not be possible without the six seasons of partnership with Domino® Sugar and the two seasons with Kerrygold®," said Martha. "There is no better place to teach engaged viewers than on PBS. PBS continues to support informative, beautiful and quality programming."
"Domino® Sugar is proudly sponsoring another new season of Martha Bakes, Martha Stewart is a reflection of expertise in baking and she only bakes with quality ingredients that provide the best results. Domino Foods Inc. understands how important it is to have the best results when you bake for family and friends, which is why we have been crafting premium quality sugar for over a century," said Brian O'Malley, President and CEO of Domino Foods, Inc.
"We found a natural fit partnering with Martha Stewart. Like Martha, Kerrygold® dairy products also embrace the highest standards - beginning in the pastures of Ireland with our grass-fed cows," said Lisa Miller, Marketing Director, Ornua North America.
Martha Bakes reaches 98% of the United States via PBS' nationwide stations. Check local listings for season ten episode airings. For more information, visit marthastewart.com and pbs.org/show/martha-bakes.
About Martha Stewart
Martha Stewart is an Emmy Award-winning television show host, entrepreneur, bestselling author of 90 books, and America's most trusted lifestyle expert and teacher. Millions of people rely on Martha Stewart as a source of useful "how-to" information for all aspects of everyday living - cooking, entertaining, gardening, home renovating, collecting, organizing, crafting, healthy living, holidays, weddings, and pet care. The Martha Stewart brand reaches approximately 100 million consumers across all media and merchandising platforms each month. Her branded products can be found in over 70 million households and have a growing retail presence in thousands of locations.
Sequential Brands Group, Inc. (Nasdaq:SQBG) owns, promotes, markets, and licenses a portfolio of consumer brands in the fashion, active, and home categories, which includes the Martha Stewart media and merchandising properties. Sequential seeks to ensure that its brands continue to thrive and grow by employing strong brand management, design and marketing teams. Sequential has licensed and intends to license its brands in a variety of consumer categories to retailers, wholesalers and distributors in the United States and around the world. For more information, please visit Sequential's website at: www.sequentialbrandsgroup.com.
About Domino® Sugar and C&H® Sugar
For more than 100 years, Domino® and C&H® Sugar have joined millions of Americans in their kitchens to help bake something special and create memories that last a lifetime. Share the joy of baking with family and friends by using 100% pure cane sugar from Domino and C&H. For classic recipes, baking tips, product information and so much more, visit www.dominosugar.com and www.chsugar.com.
About Kerrygold®
Kerrygold butters and cheeses are made from the milk of grass-fed cows that graze on Ireland's lush, green pastures for up to 300 days a year. The traditional practices upheld by generations of small, Irish dairy farms yield products that have a rich golden color from the naturally-occurring beta-carotene in the grass, and creamy flavor that is unparalleled.
Magrino PR
Jessie Lison
E: Jessie.lison@magrinopr.com


text:
Hillingdon councillor tells of his persecution for being a Muslim
Unhappy experiences: Jazz Dhillon
HILLINGDON Labour councillor Jazz Dhillon spoke of his unhappy personal experiences when proposing a motion on Islamophobia at a council meeting last week.
"I know what it is like to be called a 'paki', or a 'wog', to be strung up on my first day of senior school because I was a different colour," he told colleagues and opponents.
The motion called on the council to adopt a definition of Islamophobia put forward by the All Party Parliamentary Group on British Muslims, that "Islamophobia is rooted in racism and is a type of racism that targets expressions of Muslimness or perceived Muslimness".
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It urged that any person who has expressed Islamophobic views should have no place in public life in Hillingdon.
The motion was amended by the ruling Conservative Group to say the council would consider the all-party recommendations as part of its regular review of council policy.
Cllr Dhillon said later: "It is still something to recognise that the council has taken action to condemn attacks and regularly review our own policies."


Whole Foods Market and Trader Joe's, two of the largest US grocery store chains, have pledged to take action to eliminate per- and polyfluoroalkyl substances (PFASs) from some food packaging – commitmentsthat could lead to a domino effect across the entire grocery sector.
The actions taken by these two companies signal that toxic PFAS chemicals are increasingly becoming a sticky issue for grocery retailers, restaurant and fast food chains, food packaging companies and chemical suppliers.
The commitments made by Whole Foods and Trader Joe's were in response to the release of our report, Take out toxics: PFAS chemicals in food packaging, which found PFAS chemicals are likely present in takeout containers and other food packaging sold at top grocery retailers.
PFASs have been linked to liver damage, harm to the immune system, and cancer, and can stay in people and the environment for a long time.
Harvard researchers found that water sources for an estimated six million US residents are contaminated at levels exceeding federal standards.
Some PFASs are used for grease and water resistance in food-contact materials (FCMs). The chemicals can leach into the food, where consumers are exposed as they eat.
PFASs have also been detected in compost that included food contact materials, and these mobile chemicals have been shown to be taken up by crop plants.
Last summer, our Mind the Store campaign, Toxic-Free Future, and the Environmental Health Strategy Center wrote to almost 80 grocery, restaurant and fast food chains demanding action on both PFASs and phthalates in food packaging. We also investigated places where PFASs might be hiding in common food packaging and FCMs used or sold at leading US grocery stores.
We collected 78 samples across 12 states and sent them to a laboratory for total fluorine analysis to determine likely PFAS treatment. Items tested included paper takeout containers, bakery and deli papers, microwavable trays, and baking supplies like muffin cups.
Samples were collected from five of the nation's largest grocery stores: Ahold Delhaize (parent of Food Lion, Stop and Shop, and Hannaford), Albertsons, Kroger, Trader Joe's, and Whole Foods Market (Amazon).
four out of 38 deli and bakery papers tested were likely treated with PFASs.
Here's the good news: in many cases, retailers use or sell packaging that is free of PFAS treatment, indicating that PFAS-free alternatives are widely available and competitively priced.
In response to the report, Whole Foods Market removed all prepared foods and bakery packaging that tested positive for likely PFAS treatment from its store shelves, a massive undertaking. This is a step in the right direction, and we are now calling on Whole Foods to take the next step and ban PFAS in all food contact materials.
Trader Joe's also pledged action, telling the media that the company has asked vendors to avoid the use of PFASs in packaging its products.
Our Mind the Store campaign plans to leverage the commitments we won from Whole Foods and Trader Joe's to increase pressure on other leading grocery, restaurant and fast food companies to follow suit in the months ahead.
PFASs are not only facing "retail regulation" but also increased action by policy makers. Our report was released on the heels of policies adopted by Washington state and San Francisco restricting PFASs in certain food packaging. Since our report was published, Berkeley, California passed an ordinance that requires all disposable takeout containers to be both compostable and PFAS-free.
This is only the beginning. Other cities and states are expected to follow, underscoring the need for food retailers to get out in front of the regulatory curve. On 6 February, SAFER States announced at least eight states (Connecticut, Maryland, Maine, Minnesota, New Jersey, New York, Rhode Island and Vermont) will consider policies to eliminate or reduce PFASs in food packaging during this year's legislative sessions. Other states are considering action on PFASs in firefighting foam and drinking water.
These chemicals also face greater scrutiny at the federal level by the newly formed bipartisan Congressional PFAS task force and the EPA.
Additionally, in October, Congress took bipartisan action to protect drinking water from contamination by passing legislation that directs the Federal Aviation Administration to allow airports to use PFAS-free firefighting foam. And in Europe, the European Food Safety Authority (Efsa) is proposing to lower the tolerable intake of the PFAS chemicals PFOS and PFOA in food.
Retailers have an important role to play when it comes to eliminating consumer exposure to PFAS chemicals. They have both the power and the moral responsibility to eliminate and safely replace toxic chemicals such as PFASs in order to "mind the store".
develop comprehensive safer chemicals policies to reduce and eliminate other toxic chemicals, such as hormone-disrupting phthalates, in food contact materials.
We trust retailers to provide the best for our families. It's only fair that they enact sensible policies to protect our food, health, and environment.
The views expressed in contributed articles are those of the expert authors and are not necessarily shared by Chemical Watch.


Originally launched in 1975, the Yamaha Recording Custom drum kit, or the Yamaha 9000 as it was known back then, has become an industry standard in the studio.
Delivering an incredible, studio-ready sound, it is no wonder why the Recording Custom series is widely regarded as being the most recorded drum kit in history.
Directly collaborating with legendary drummer, Steve Gadd, Yamaha have re-designed the Recording Custom drum series and improving it in almost every way, upgrading it to ensure it perfectly suits modern drumming styles and recording technology.
Precision engineered and designed to deliver an incredible snare sound that is perfect for recording in the studio and delivering incredible, versatile snare sound as part of your live drum kit.
For three years, Yamaha worked closely with legendary Steve Gadd to improve on the classic Recording Custom drum series. This unique collaboration resulted in an all-around improvement of the Yamaha Recording Custom drums series.
One major addition to the Recording Custom series that Gadd requested was metal snare drums and so stainless steel, aluminium and brass snare drums have all entered the Recording Custom series with a bang!
Steve Gadd personally requested that the aluminium snare drum be seamless due to his preference for aluminium die-cast hoops when playing with this particular snare, while the other snare drums in this line come with 2.3mm steel triple-flanged dyna hoops.
Allowing you to get the perfect sound to suit your needs, the Yamaha Recording Custom Snare has a really wide tuning range.
Being aluminium, it has a very dry tone providing a really precise reflection of your performance; perfect for intricate playing. It also allows the snare sound to cut through the mix, breaking through with a powerful presence even in loud and heavy music.
You'll also enjoy its wide dynamic range, making it a versatile snare drum suitable for all genres of music and drumming styles.
A heavy-duty Q-type strainer has been used, bolstering the snare drum and ensuring a solid, smooth operation while allowing for easy adjustment and replacement of the snare.
Traditionally, Yamaha snare drums usually have a metal centre bead on the interior drum but on the newly designed Recording Custom snare drum, Yamaha have fixed the centre bead to the outside of the drum.
This new design element allowing the Recording Custom snare drum to produce a more open sound than their conventional snares, giving you a more crisp, articulate performance.


The Lamy Safari fountain pen has the classic Lamy design. The pen, first released in 1980, made a strong construction of colorful ABS plastic with a solid clip and ergonomically designed grip. The Lamy Safari Fountain Pen is robust and very popular at school as successor to the Lamy ABC, but due to its quality and highly regarded writing experience, this pen is also very popular with experienced writers and enthusiasts.


text:
Evaluating the role of viral proteins in HIV-mediated neurotoxicity using primary human neuronal cultures
Vasudev R. Rao, Eliseo A. Eugenin, Vinayaka R. Prasad
Despite the inability of HIV-1 to infect neurons, over half of the HIV-1-infected population in the USA suffers from neurocognitive dysfunction. HIV-infected immune cells in the periphery enter the central nervous system by causing a breach in the blood-brain barrier. The damage to the neurons is mediated by viral and host toxic products released by activated and infected immune and glial cells. To evaluate the toxicity of any viral isolate, viral protein, or host inflammatory protein, we describe a protocol to assess the neuronal apoptosis and synaptic compromise in primary cultures of human neurons and astrocytes.
Methods in Molecular Biology
Humana Press Inc.
https://doi.org/10.1007/978-1-4939-3046-3_25
Viral Proteins
HIV-associated dementia
HIV-associated neurocognitive dysfunction
Neuronal apoptosis
Neuronal damage
Rao, V. R., Eugenin, E. A., & Prasad, V. R. (2016). Evaluating the role of viral proteins in HIV-mediated neurotoxicity using primary human neuronal cultures. In Methods in Molecular Biology (Vol. 1354, pp. 367-376). (Methods in Molecular Biology; Vol. 1354). Humana Press Inc.. https://doi.org/10.1007/978-1-4939-3046-3_25
Evaluating the role of viral proteins in HIV-mediated neurotoxicity using primary human neuronal cultures. / Rao, Vasudev R.; Eugenin, Eliseo A.; Prasad, Vinayaka R.
Methods in Molecular Biology. Vol. 1354 Humana Press Inc., 2016. p. 367-376 (Methods in Molecular Biology; Vol. 1354).
Rao, VR, Eugenin, EA & Prasad, VR 2016, Evaluating the role of viral proteins in HIV-mediated neurotoxicity using primary human neuronal cultures. in Methods in Molecular Biology. vol. 1354, Methods in Molecular Biology, vol. 1354, Humana Press Inc., pp. 367-376. https://doi.org/10.1007/978-1-4939-3046-3_25
Rao VR, Eugenin EA, Prasad VR. Evaluating the role of viral proteins in HIV-mediated neurotoxicity using primary human neuronal cultures. In Methods in Molecular Biology. Vol. 1354. Humana Press Inc. 2016. p. 367-376. (Methods in Molecular Biology). https://doi.org/10.1007/978-1-4939-3046-3_25
Rao, Vasudev R. ; Eugenin, Eliseo A. ; Prasad, Vinayaka R. / Evaluating the role of viral proteins in HIV-mediated neurotoxicity using primary human neuronal cultures. Methods in Molecular Biology. Vol. 1354 Humana Press Inc., 2016. pp. 367-376 (Methods in Molecular Biology).
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abstract = "Despite the inability of HIV-1 to infect neurons, over half of the HIV-1-infected population in the USA suffers from neurocognitive dysfunction. HIV-infected immune cells in the periphery enter the central nervous system by causing a breach in the blood-brain barrier. The damage to the neurons is mediated by viral and host toxic products released by activated and infected immune and glial cells. To evaluate the toxicity of any viral isolate, viral protein, or host inflammatory protein, we describe a protocol to assess the neuronal apoptosis and synaptic compromise in primary cultures of human neurons and astrocytes.",
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AB - Despite the inability of HIV-1 to infect neurons, over half of the HIV-1-infected population in the USA suffers from neurocognitive dysfunction. HIV-infected immune cells in the periphery enter the central nervous system by causing a breach in the blood-brain barrier. The damage to the neurons is mediated by viral and host toxic products released by activated and infected immune and glial cells. To evaluate the toxicity of any viral isolate, viral protein, or host inflammatory protein, we describe a protocol to assess the neuronal apoptosis and synaptic compromise in primary cultures of human neurons and astrocytes.
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text:
Council Strategy
The Council Strategy shapes the work of Mole Valley District Council (MVDC) and sets out its vision and priorities.
Council Strategy 2019-24
MVDC's vision and priorities are set out in the Council Strategy 2019-2024 (see 'Downloads'). The strategy also includes four guiding principles, which guide our approach to decision-making.
The strategy was developed alongside the budget setting process in 2018-19. As part of the development we conducted a consultation to gain feedback on the draft strategy. The Strategy was approved by Council in February 2019 and came into effect on 1 April 2019. The Strategy was revised in September 2019 and the revisions were approved by Council in October 2019.
Our strategy includes five core values (see 'Downloads'):
We respect
The Annual Plan sets out the key areas of work on which MVDC will have a particular focus over the financial year and demonstrates how it will deliver the commitments for each of the Council Strategy priorities. Progress made on these are reported in the Business and Budget monitoring reports to the Cabinet.
MVDC - Corporate Values


In an environment with people, animals and nature, there is always a high potential for conflict. Many of the approved mountain bike routes will take you through Alpine pastures and along forest paths, where there is always a possibility that you will come across animals in summer. Therefore, if you see signs prohibiting access as well as other warning signs, please make sure you adhere to them. Hikers and bikers should have respect for each other, as well as consideration for animals and nature, we would ask you to observe the following rules.
Please consider nature and wildlife!
Ride only on the official MTB routes. Respect signs indicating road closure and do not drive across country!
Ride with consideration for others and with foresight. Make sure you let others know your intention to overtake in advance, as not to frighten other riders. A friendly greeting often works wonders!
Take care not to leave tracks. Be careful how you use your brakes and avoid using trails after rainfall!
Always put rubbish in your rucksack to dispose of later, don't just simply throw away whilst riding!
Always be prepared for the unexpectable. Make sure you adjust your speed and choice of trail to your own level of fitness and ability. Ride with foresight, other bikers and hikers can turn up at any time!
Always ride with a helmet!
Check your equipment before you start off!
Inform yourself about the weather and surroundings!
Don't forget tools, first-aid kit and mobile telephone!
In many regions, mountain biking is only permitted during the day in the summer months. That being, two hours after sunrise and one hour before the sun going down. If you intend to make a tour by yourself, please inform a relative, friend or host about the details of tour and estimated time you will return!