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How Many Balloons In 20 Lb Nitrous Tank

How Many Balloons In 20 Lb Nitrous TankThe first time I used nitrous was a month and a half before this party. appeared to be a comparison of measured moisture in the Arctic environment to the moisture levels predicted by some 22 of the …. What is the final pressure, in mmHg, of the helium gas if the volume changes to 20. For G and H tanks, it is $95 per tank, resulting in an annual charge of$380 for two tanks of each gas. Gas balloons are usually sealed off at the end in order to keep the gas from escaping, and they can include large blimps, small party balloons made of rubber, and many others 3 Main Types of Balloons 1) Latex Balloons The most …. 9 cu ft of helium and fill, on average, about 30 9″ latex balloons. Find the agency to call for your county. Cylinder Medical Carbon Nitrous Size Contents Air Dioxide Helium Nitrogen Oxide Oxygen T Cubic Feet (ft3) 311 638 291 304 637 337 Pounds (lb) 17. CGA600 valve fits most portable appliances, grills and lanterns. A big 7,700-lb towing capacity is just one of the GLS's superhuman abilities. For G and H tanks, it is $95 per tank. September 28, 2020 – Bell, Xwing Complete Uncrewed Flight Test Using Detect-And-Avoid System – Bell Textron demonstrates its Autonomous Pod Transport 70 (APT 70) vehicle as part of NASA’s Unmanned Aircraft Systems in. A fluid is a state of matter that yields to sideways or shearing forces. The Cygnus spacecraft hauls 7,624 lbs. 00 grams sample of the ammonia solution was placed in a flask and titrated with 1. Slugs are the correct measure of mass. 100pcs 10 Inch Colorful Pearl Latex Thickening Wedding Party Birthday Balloon. The shed, filled with candles, a propane heater, and a 20 lb propane tank, caught fire. Many do not realize to build a vehicle that can be slammed to the ground while using air suspension, there is a considerable amount of fabrication, engineering and work that needs to be done. At 100°F, a 20lb propane tank has 172 PSIG of pressure. The Grand Tour is a motoring show on Prime Video that debuted on November 17, 2016. 500ft (6,250m) and with GM 1 this could be increased to about 350 mp/h (560 km/h) at 25. The maximum speed was 330 mp/h (528 km/h) at 20. 20 cu ft steel helium gas cylinder. But N2O in the small cartridges known as “Whip-Its,” “whippets” or “poppers” can be released into balloons and inhaled or “huffed” for a feeling of euphoria. Tanks not only hold a lot more helium than your everyday party balloon, but they also release the helium. 50 liters tank filled with helium at 125 atm is used to fill balloons. We rent a 40 cu ft tank that would work perfectly. Notes: 1973 Pro Street LUV, 1984 Buick V-6, ATI D-1 Procharger centrifugal supercharger 20 PSI boost, Champion GN1 alum. The temperature of the surrounding air is 20 o C (68 o F). The recreational use of nitrous oxide, or laughing gas, is on the rise, as more and more people are misusing the drug in an effort to get high. He had been inhaling balloons all night, as he enjoys inhaling nitrous while taking MDxx, but he had only stuck to single and …. N itrous Oxide Tank Nitrous Tank by Zex. Jim Irsay heard a rumor that the balls were filled with nitrous oxide and. When you buy a five-pound bag of flour, the number of cups you're buying can vary based on brand, flour type, and your measuring style. Helium Tank Balloon Time Kit 50 Each Total 200 Balloons (Pack of 4)$199. (26) 26 product ratings - 55cf Welding Tank Cylinder Bottle can be used for Argon Helium Nitrogen. I run a dry 100 shot and have figured that I normally use about. Maybe get a nitrogen tank and then a portable air tank with a regulator for airbrush pressure. 5 cu ft of gas), about 3X the capacity of the nitrogen tank. When the coal is burned, the sulfur is converted to SO2 (g). 0624 lb m /ft 3; Note that even if pounds per cubic foot is often used as a measure of density in the U. Department of Education, how many annual visits does each American make to a public library: two, five or 10? Answer: Five According to National Center for Education Statistics (NCES), part of the Institute of Education Sciences (IES), the statistics, research, and …. As the torque produced by the nitrous drops off at higher RPM, the engine can tolerate the use of …. One balloon that is 30 meters (about 100 feet) in diameter displaces 14,137,000 liters, so it can lift 14,000 kilograms (about 31,000 pounds) -- this is roughly the size of a large blimp. 00 L of mercury? Express your answer in kg. A full 20-pound (9-kilogram) propane tank holds 4. The cheapest option is to buy a brand new 20lb CO2 tank for $100. Before I worked there, an employee thought he was being cute and would fill balloons with nitrous from the tank and inhale it (whippets). For each size tank, we tell you how many balloons you can fill for each size of balloon, from 9-inch latex to 18-inch foil/mylar balloons. How many balloons are there in a 20-pound laughing gas? 1 pound of laughing gas = 454 grams of laughing gas. How many metallic balloons were sold? d. Ugly Aretha Franklin 018fe754-cedf-4edb-af53-c8970ac4d00d Historic Live Concert Radio Broadcast Dormir Musica Relajante Para …. , height without cap & valve from 14 in. 6 lbs, and is a little bigger than the American ton’s 2,000 lbs and a little smaller than the British ton’s at 2,240 lbs —but the three are fairly comparable. IDM H&S committee meetings for 2022 will be held via Microsoft Teams on the following Tuesdays at 12h30-13h30: 8 February 2022. The ground is sloping from 4 feet in the shallow end to 10 feet in the deep end. With plans starting from only$14. Cylinder dimensions and weights are approximate measures. Helium (He) is a colorless, odorless and non-flammable gas that’s manufactured through the mining of natural gas wells. Dunno local prices and BOC might not be the cheapest etc. We offer helium for a variety of commercial, medical and industrial uses and applications. Below are approximations, and actual results may vary based on many conditions (indoor/outdoor use, amount of direct sunlight, balloon type, helium or air mixture in your tank, etc). A 9kg gas bottle weighs approximately 17. Ocean Dead Zones Nitrogen and phosphorus compounds from agricultural runoff into the oceans causes algae blooms. Nitrous oxide is a colourless gas that when inhaled can make people feel euphoric and relaxed. Dive Tank Suppliers, Manufacturers, Factory - Children party balloon kits cute shape cake, doughnut,ice cream, hot-dog foil balloons - INDUSTRY GASTEC CO. These tanks could be replaced by 5. 0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper?. The above formulation states that once the radius of a balloon exceeds a threshold, which is fairly close to the original radius (roughly 38 per cent larger than the original [ 1 ]), bigger the balloon is, smaller the pressure is. It’s 20-year and 100-year GWP are basically the same. o For packages weighing up to 20 lbs. But many of you do have such plans, hence this section is being added. got to store, parts, performance, then down to nitrous. All types of balloons can be gas balloons because the term gas balloon actually includes a variety of subtypes. Nitrous racer short sleeve tee shirt. Many people correlate energy as the ability to do work. Cyl Tec Inert Gas Cylinder - 60 Cubic Foot Cylinder dimensions: 7. In the late 1950s and early ’60s, it was the full-size cars offered by the Big Three manufacturers that kept enthusiasts coming back for. 0 m high hill on a frictionless sled (combined mass 40. The higher you raise it, the easier it is to work under. Notice: Changing your store affects your localized pricing and pickup locations to new items added to cart. 2 out of 5 stars with 339 ratings. Attendees may bring one sealed, non-alcoholic beverage into the concert areas — see water policy below Government issued photo ID is required to …. Past, present, and future of nitrous oxide. Welcome to Sidney Lee welding supply, founded in 1968 by Sidney Lee with the goal of providing the best products and the highest level of customer service. Typically, you can rent a standard 242 cubic foot Helium Tank from a local gas supplier, welding company or tank rental firm. 5 kg when full, so it is very portable. Rather than replace end-of-chapter exercises with new exercises, we have opted in most chapters to increase the number of exercises. of marijuana and the opium The GHB and magick mushrooms 20 20's is 40 that fold 40 sips on a 40 thats cold 40 women and 40 men While putting up flyers for the International Chiptune Resistance show at The Tank this coming Saturday, Nullsleep and Bit Shifter were approached by two plainclothes police. The tank was the only thing stolen the night of May 8 from Pine Grove Dental Arts, 1475 Pine Grove Road. 2H 2 (g) + O 2 (g) → 2H 2 O (l) Step 2: Calculate the volume. The Helium tanks are measured in "how many balloons it will fill"; meaning, they are not very large. Nitrous oxide | N2O | CID 948 - structure, chemical names, physical and chemical properties, classification, patents, literature, biological activities, safety. 50 Ignition Temperature in Air, °F 920 - 1,120 Maximum Flame Temperature in Air, °F 3,595 Cubic feet of Air Required to Burn One Cubic Foot of. But the price of nitrous in Denver is no laughing matter. 346 7 - 10 Days - 22" Bubbles™ 30 grams 1. Our helium tank kit offers a small 8. Improperly inflated tires can wear unevenly, wear out faster, and ruin your fuel economy. How Many Balloons In 20 Lb Nitrous Tank; Biggest Liar Zodiac Sign; Twin Peaks Wikia; How Many Balloons In 20 Lb Nitrous Tank; Labcorp Drug Test Cost; Conservatoria Em Portugal; Bible Modules; Infrared Pineal Gland; Betist giriş adresi 2021 Betist bahis mobil Canlı Casino tr 25; How Many Balloons In 20 Lb Nitrous Tank; Labcorp Drug Test Cost. Realistically if you spray alot you will prob. It has been speculated that Hsieh had been using Nitrous Oxide prior to the fire that led to his death. Depending on what size keg you will be dispensing and what size CO2 tank you purchase, you can expect to dispense the following number of kegs:. It helps if the tank is cold when filling. However, the air will cease to be dry air when water vapor appears. I meant one charger, which fills roughly one balloon. After more than 25 years in the compressed air industry, it still amazes me that many plant personnel and even those who sell compressed air products for a living don’t fully understand the relationship between flow, or volume (cfm), and pressure (psig). Note that rounding errors may occur, so always check the. These characteristics lend themselves to CO2's multitude of uses. Hybrid cars are not always as environmentally-friendly as we’d like. 5 1 1 Nitrous oxide is an oxygen precursor for silicon oxide, silicon oxynitride as well as. The 200 feet of altitude is irrelevant in …. With an unrivaled national footprint and robust infrastructure, we offer. Posted: (8 days ago) Ideally I would to get a 125cf oxygen tank and around a 80 cf acetylene tank. of hydrogen have combined with 18. 50 × 10 5 N/m 2 at a temperature of 35. Indeed, those vegans had the lowest …. capacity bottle should not be filled with more than 10 lb. favorite this post Mar 21 %%Rotary Screw Air Compressor. The calculator below can be used to estimate the density and specific weight of gaseous carbon dioxide at given temperature and pressure. For an optimal experience visit our site on another browser ; Nitrous oxide, commonly known as laughing gas, is commonly available at retail outlets in small canisters known as whippits, used to pressurize whipped cream dispensers. Balloons of nitrous at parties are sold from 3-5$per balloon. Created Date: 2/7/2017 7:40:38 AM. 125 avail Cl2 to get the X lbs of sodium hypochlorite by itself:. Can I buy nitrous? Nitrous oxide canisters made for the purpose of prolonging the shelf life of whipped cream and other products are completely legal and readily available. After 10 or so years of symptoms, I was finally diagnosed with this mutation a month ago (I’m 20) and am still learning about maintaining it. Make sure to press down on the directional pad often when you unlock new areas to see how many hidden shops are left to look for by scrooling all the way down with the left thumbstick. Also provided is the volume of helium that should be used for each size and type of balloon, all …. Tower Hamlets in East London has seen a £9. Shipped as a liquefied gas under its own vapor pressure (1,200 psig @ 70ºF). 5" T-slotted aluminum rail produced by 80/20. We assume you are converting between grams N2 and mole. (X lbs of sodium hypochlorite)(0. The Nitrous Oxide Mafia Is Real: Balloons, Money, and Murder I think that you could attach a hose to your valve (possibly a hose that has a larger opening on one end than the other, depending on the size of your valve opening, so you can attach one end to the valve), and then the balloons would be …. A 120-gallon tank holds 96 gallons of propane to allow for expansion. Our quality LPG tanks come in a variety of sizes: from hand-held torches and camping fuel to 1,000 lb capacity. Roger Aubin 5 avenue Morais La Sarre Québec, Canada, J9Z 0A4. 4 liters of hydrogen react with 22. (Because inhalation of small amounts provides a brief euphoric effect and nitrous oxide is not illegal to possess, …. At Bargain Balloons we now offer over 11,000 wholesale balloons designs at prices that are always 20-60% off normal wholesale. One experience SWIM can tell you about is when he inhaled 3 nitrous cartridges at once after taking a couple MDA tablets. Fluid statics is the physics of stationary fluids. If the partial pressures of the three gases in the tank are 35 atm of O 2, 5. Re: how many chargers to fill baloon (nitrous) [Re: tocuhe] #5824750 - 07/05/06 05:30 PM (15 years, 2 months ago) Edit. About Cost Nitrogen Tank Refill. Nitrous Supply Bottle & Valve Assembly by Zex. When air in an elastic balloon cools from 25 ° C to 20 °C, the volume of the balloon (increases/decreases. I don't use nitrous a whole lot and only use like 10lbs a month. idk lol i was killing a 20lb tank in a night with 2 other people every few weeks a little while back. Look down the side of your gas cylinder. 4 million cut and was one of the most affected boroughs. is the premier solution for the utilization of compressed gasses in Los Angeles for a variety of industries. When you're looking to go fast, rely on NOS to deliver the most power per dollar available in the industry. Airgas Locator: Welding & Industrial Equipment - Gas & PPE Supplies. 999% UHP T ULTRA HIGH PURITY GRADE$ 0. Inhaling compressed gases directly from containers is dangerous, so intrepid users fill balloons with the gas and then inhale it. Explosive mixtures of marsh-gas and air may be fired by an unprotected light. … Possessing nitrous oxide (in tanks, canisters, balloons, etc. Density is the mass per unit volume of a substance or object, defined as ρ = m/V. Also called: Balloons; Chargers; Hippie Crack; Laughing Gas; Nos; Noz; Whippits . 53 when air is taken as density of 1. If you have a tank that you MUST get back, and we are not able to fill it on premises, it may have to go to the Fill Plant to be refilled, which may take 1 to 2 weeks to be returned. Choose from 100% argon, Argon-CO2 blends, or 100% oxygen cylinders. Releasing 1 kg of N 2 O into the atmosphere is about equivalent to releasing about 298 kg of CO2. We've got wet and dry nitrous kits from the biggest names in N2O, including Nitrous Oxide Systems (NOS), Nitrous Express (NX), Zex, Edelbrock, Ny-Trex, Venom High Performance, and more. One showed him standing atop a tank with his name emblazoned on the front and side. The siphon is out of sight, staying inside the tank. There also can be a structured type of settlement in which a balloon credit 10-15 years into the plant's operating future would occur. They are both forms of “twilight sedation. Products : Push and Pull Car Shelter with casters , Hot Sale Folding car garage , Retractable Car Tent. Jimmy cooled down by emptying the contents. You’ll find them everywhere from hardware and rental stores to grocers, convenience stores, and gas stations. It's 20-year and 100-year GWP are basically the same. if you are a Motor State customer!. Use this gas cylinder size chart to determine the size of compressed gas tanks. As I watched my boy manipulate the dials on the nitrous tank and dildo, I began to wonder if researching …. Cartridges larger than 68g will push through kegs of any size. Hsieh, 46, passed away in a house fire. The growth rate of about 1 ppb per year has also accelerated during recent decades. MC 306 and DOT 406 cargo tank motor vehicles and DOT 57 portable tanks may be used to transport liquid soil pesticide fumigants, Pesticides, liquid, toxic, flammable, n. 00 each, so the total price is 2. In most chapters, at least 10 new exercises have been added; and in many chapters, 20 or more exercises have been added. However, addition of several mechanical, electronic and electric …. (c) How many kilograms of lime are required to remove all the phosphorus from a 1. Acetaminophen : The active ingredient in Tylenol and the most commonly used painkiller in America, acetaminophen has caused many cases of liver failure, mainly by overdose but also through users taking the recommended dosage. If the tank were 1 ft long, 1 ft wide, and 5 ft high, would it exert the same pressure?. Get the latest regulatory info, accreditation news and exclusive discounts!. Talk to Frank for facts, support and advice on drugs and alcohol today. only run 6-7 lbs out of every bottle and they do not need to be empty to refill. The smaller balloon will become even smaller and the bigger balloon will become even bigger. pressure is in a 100 lb propane tank would also be 172 PSIG, as the size of the propane tank doesn’t matter, only the temperature. Only 17 left in stock - order soon. People inhale the gas - also known as "hippy crack. (B) The density of the helium in its balloon is greater than the density of the hydrogen in its balloon. Disclaimer [9] Disclaimer [10] Disclaimer [11]. It equips you with the tools and insights you need to track and manage Linde cylinders at your facilities. A few weeks ago, teams of federal agents and sheriff’s deputies fanned out across Los Angeles, Orange and Riverside. Possessing nitrous oxide (in tanks, canisters, balloons, etc. When it was thrown into the 20 lb lead pot, a large explosion sent most of the molten lead flying. Therefore, children should receive 0. Like I said, both CO2 and nitrogen are dry inert gases. 753 \mathrm{~atm}$and the temperature is$-25^{\circ} \mathrm{C}$. 753 atm and the temperature is -25. This is an enamel finish, double post pin, measures 1. It is a gas at standard temperature and pressure, but compressible to a transportable liquid. I have the data of inlet and outlet temperature and …. 11 lb/10,000 16 oz cup for Biochemical Oxygen Demand, 12. At Sidney Lee we ensure quick delivery of products from. The funny thing i used one of those dispensers to fill my balloons, not a cracker (small metal piece). temperature = About 130 cubic feet. Like you turn to the next page of the story that is inevitably infinite, sort of chasing your tail after a while. (b) If the HCN is sume that the gasoline is composed of octane, C8H18(l), whose formed by reaction of NaCN with an acid such as …. Chargers are cheaply available in many stores and online. which of the following statements is correct using the above picture. 1 kg), but not exceeding 50 lb (22. The inaugural Ceremony Festival has booked several pioneers of house music. Our videos prepare you to succeed in your college classes. A subreddit dedicated to all things related to Nitrous Oxide N₂O. But, refilling your own cylinders, well that can be aggravating as w. Agriculture is the main source of human-produced nitrous oxide: cultivating soil, the use of nitrogen fertilizers, and animal waste handling can all stimulate naturally occurring bacteria to produce more nitrous oxide. , weight without cap & valve from 106 lbs. How heavy do balloon weights need to be? Filled with helium, their lifting capacity is 0. CO2 tank should be able to force-carbonate and serve between fifteen to twenty 5 Gallons kegs of homebrew. 00 × 10 3 L of O 2 at 0 °C and 1 atm, will be required to burn the acetylene?. If the balloon is cooled and the volume decreases to 475 mL, what is the final temperature? Assume that the pressure remains constant. What does an intra-aortic balloon pump do? The gas tank gauge to see if there is pressure in the cylinder. We’ve got wet and dry nitrous kits from the biggest names in N2O, including Nitrous Oxide Systems (NOS), Nitrous Express (NX), Zex, Edelbrock, Ny-Trex, Venom High Performance, and more. With a nitrous oxide tank, filling your balloons with nitrous oxide gas is so much quicker and easier. My doctor gave me Active B12-Folate with 800 mcg of methyltetrahydrofolate and 1000 mcg of Vitamin B12 (methylcobalmin) and told me to take 2 a day right away. The balloon then rises to an altitude of 6000 ft, where the pressure is 605 torr and the temperature is$15^{\circ} \mathrm{C}$. It was such a surprise, but she knew it! Of course, she is way too intelligent for not Ucis în timp ce salvează niște străini de o coliziune în trafic, un șomer izolat de 34 de ani este reîncarnat într-o lume a magiei ca Rudeus Greyrat, un nou-născut. 15 mL per pound of body weight of sodium nitrite (0. Airgas ships nitrous oxide to all points of the globe using vacuum jacketed ISO containers. How Many Balloons In 20 Lb Nitrous Tank To calculate peak fuel consumption for an 800 hp E-85 eng. A krypton balloon has a volume of 555 mL at 21 ° C. Some interesting Nitrous Oxide Facts The first commercial space flight used Nitrous Oxide as a propellant; Nitrous Oxide many times more potent as a greenhouse gas than Carbon Dioxide. 9 cu ft of helium and fill, on average, 50 9″ latex balloons. CARBON DIOXIDE 45KG 100LB 100 LB Type Cylinder$ 0. ) (20 lbs) can last about 5 to 7 days. Shop online at Canadian Tire; pick up at one of our 500+ stores or ship to home. At elevated temperatures, nitrous oxide …. This chart is based on filling your balloons with pure helium. If you got foil balloons instead of latex, this disposable helium tank can fill approximately twenty-seven 18″ or twenty 20″ Mylar style . 4-2 ----- A closed, rigid tank filled with water vapor, initially at 20 MPa, 520oC, is cooled until its temperature reaches 400oC. You put in your flavored or colored cream in the dispenser, charge it with nitrous, and off you go. I have never inhaled more than 3 nitrous cartridges at one time from one balloon, but he usually just prefers 2 at a time. I use my C02 more around the house than on the trail. A tank of 18,000 litres seems to be € 400 or so, mebbe €500. If you are having trouble with Chemistry, Organic, Physics, Calculus, or Statistics, we got your back! Our videos will help you understand concepts, …. Interviewed by Jennifer Ross-Nazzal. Basic Balloon Bouquet Package: Medium (Ten 11 balloons per bouquet) $130. 50, and, fused nitrate of ammonia cost 60 cents per lb. bad influence come in a variety of designs and pricing ranges. The N/S racing POWERVALVE 500R is a ¼ turn, full on/off position valve that can be utilized with 5 - 20 lb. Halocarbon and Ethylene Oxide/CO2 sterilizing gas mixtures) are not listed in the Compat-ibility Chart. A 10lb CO2 cylinder would be about the same size as Mac's setup and would hold 640 gallons (or 85. If you are using mpg (miles per gallon), then the formula for calculating the total amount of gas needed is: Fuel = Distance / Consumption mpg. The Helium tanks are measured in “how many balloons it will fill”; meaning, they are not very large. , flash point not less than 23 degrees C, 6. In any case, this guy had one of those dentist sized nitrous tanks, and a shitload of balloons. 2160 per 10 g, thereby making a profit of 20%. I read some remarks that new research comes close to achieving the energy density (by volume I assume) of liquid hydrogen. Tanks of medical-grade nitrous are difficult to obtain, and auto-grade nitrous contains substances such as sulfur dioxide to prevent human consumption. We offer nitrous oxide in a variety of purities and concentrations. 4 million for each launch that cost 0 million from other expenses (with 40% of the. €45,000 revenue from a €500 tank? No, I’ve got something. Twilight can be used in cases where there is mild to moderate dental anxiety. Input the power of the vehicle's engine. Though nitrous in whip-its is legal to possess, inhaling it is illegal and can result in misdemeanor or felony charges. Choose BOC for your supply of medical Nitrous Oxide. Possession of nitrous oxide in California-for recreational, non-medical purposes-is a crime under Penal Code 381b PC. To lift a 1000 kg car (1,000,000 grams) we’d need 1,000,000/110 balloons. If the partial pressures of the three gases in the tank are 35 atm of O 2, 5 atm of N 2, and 25 atm of He, what is the total pressure in the tank? (65 atm) 15) Blast furnaces give off many unpleasant and unhealthy gases. The accumulation of CO 2 in the atmosphere is like putting air into a balloon. Nitrous, it’s one of the cheapest and easiest ways to add horsepower to most any engine. Calculations based upon SEMI F-32 (refer to PN 417) Component and system performance should be confirmed by physical testing. 56 atm, what is the partial pressure of each component?. ft - 50 Latex Balloons + Balloon Tying Tool + Curling Ribbon Universal Mounting Bracket Stainless Steel Bottle Bracket for 12 lb Nitrous Oxide Bottle Bracket. Airgas has the capabilities to deliver bulk nitrous oxide to locations nationwide. 28 \times 10^{3} \mathrm{~L}$ at sea level where the pressure is $0. Try to hit groups of balloons when you get the chance. (lb) 120 / 122: 151 / 153: Note 1: All tank weights include tank jackets and valves. This oral history with Norm Chaffee is being conducted for the Johnson Space Center Oral History Project in Houston, Texas. A heavy-duty converter will be required with any package making more than 400 total horsepower. Assuming standard temperatures and pressures and no significant overpressure. Nitrogen, which is the most abundant, is seven times denser than helium. O) Gas Canisters, Tanks, Cylinders & Bottles. torque 2000 Honda Civic Coupe Si: North America only, FWD, 160 hp, 7600 rpm, 111 lbs. The National Helium Reserve, also known as the Federal Helium Reserve, is a strategic reserve of the United States holding over a billion cubic meters (1E9 m3) of helium gas. I try to keep my bottles topped off each week, so normally I need less than 5 lbs for each refill. Tank-refills of medical grade nitrous oxide are diverted from …. 20 lbs is equivalent to exactly 1,137 chargers 18 level 2 · 1 yr. 【送料無料】 新品1本 新品タイヤがいつでも安い！タイヤウェイ。1本 サマータイヤ ADVAN Sport ヨコハマ アドバン スポーツ V105 205/55R16 91V MO ベンツ承認. nitrous bottles & tanks are very important to Nitrous Oxide Kits Systems and components are the quickest and easiest way to get large horsepower increases with a minimum of engine modifications and expense. Many helium tanks are diluted with air which shortens float time. The only way to create a truly healthy underwater environment and to. Durable steel construction, single cylinder. This weight is supported by the surrounding fluid, and so the buoyant force must equal w fl, the weight of the fluid displaced by the object. a steel cylinder of oxygen with a volume of 15 L was available for medical purposes. Releasing the nitrous oxide into a balloon helps to warm the gas and. Example - Lifting Force created by a Hot Air Balloon. The Tq when you hit the nitrous is so great that you could very easily loose control of the car and crash. 00184 slugs/ft 3) and the air density at temperature 20 o C is 1. When the nitrous oxide is heated to approximately 572F (on compression stroke), it breaks down and releases its load of extra oxygen, However, it is not this oxygen alone which creates additional power, but the ability of this oxygen to burn more fuel. A balloon filled with helium has a volume of 17. Helium tanks diluted with air are ideal for confetti balloons, since air helps the confetti stick to the sides. Cylinder Medical Carbon Nitrous Size Contents Air Dioxide Helium Nitrogen Oxide Oxygen Pounds (lb) 17. 40" Foil Shape 20 - 50 grams (13g - 45g) varies varies 3 - 5 Weeks - 20" Bubbles™ 30 grams. 4 lbs) of CO2 equivalents, the researchers calculated. The helium is stored at the Cliffside Storage Facility about 12 miles (19 km) northwest of Amarillo, Texas, in a natural geologic gas storage formation, the Bush Dome reservoir. Pound for pound, gasoline or diesel fuel contain about 40 times as much energy as a state-of-the-art battery. Many people are surprised to hear this. Express this density in units of g/cm3. Raise vehicle and support on jack stands. N 2 O - Nitrous Oxide Nitrous oxide (N 2 O) is a powerful greenhouse gas that occurs both naturally and due to human activities. so with that math assuming ~20g per balloon there would be ~454 balloons in a 20 lb tank? It depends on how big you make the …. Drew Pinsky and Rob take a look as some wild people and wonder "What Went Wrong?," check out some strong seniors flex their "Grey Guns" and Dr. 5 46 5 NO COM5 5 NO 745 326 4 x 30 21 Grade & Gas Specification Cyl. This cooling is offset somewhat by the greenhouse effect of clouds which reduces the OLR by about 30 W/m^2, so the net cloud forcing of the radiation budget is a loss of about 20 W/m^2. These emissions are not only fueling global warming. I get a 10lb bottle filled and use it to fill my two 2. specifically a 20 pounder btw im not some mafia scumbag, its for a freinds bday party present. Our laughing gas tanks / cylinder are filled with 2 kilos of nitrous oxide and are good for about 250 small laughing gas balloons of 8 …. A large scuba tank (Figure 8) with a volume of 18 L is rated for a pressure of 220 bar. 99 atm, the partial pressure of carbon dioxide is 0. of nitrous oxide at sea level pressure and around 70°F. People often transfer the gas into nitrous balloons and inhale the drug for its euphoric effect. Inhaling nitrous oxide directly from the canister is very dangerous because the gas is under such high pressure. for quantities less than 25 lb. A flow rate of 2 liters per minute increases the FiO2 from 21 percent (room air) to 28 percent. so with that math assuming ~20g per balloon there could be ~454 balloons in a 20 lb tank? It is determined by how huge you make the balloons, however you will get about 3-400 actually fats ones. We divide each side of the equal sign by the 0. About Tank Refill Cost Nitrogen. The compression ratio is the ratio between two elements: the gas volume in the cylinder with the piston at its highest point (top dead center of the stroke, TDC), and the gas volume with the piston at its lowest point (bottom dead center of the stroke, BDC). 10 nitrous oxide balloons, and a nitrous oxide cracker with which you deflate the nitrous oxide cartridge in the balloon. It is used in the pastry kitchen to make whipped cream. Our products are designed by clinicians for clinicians, with a focus on patient safety and comfort. 00 M HCl, using an appropriate indicator. Chemistry Archive: Questions from August 17, 2021. The following are the rules of thumb for CO2 tanks and kegs: For an average 5 lb CO2 tank and a 5-gallon keg, 31 to 44 kegs of beer will be pumped. Our innovative ACCURA gas cylinder management service, for instance, brings greater control, safety and cost efficiencies to your business. The Account Manager and Customer Service team in general are. Whippets are containers that can be purchased at tobacco shops and contain nitrous. ) of oxygen; or approximately 2 volumes of hydrogen have combined with 1 of oxygen. At 25 degrees Celsius (77 Fahrenheit) and standard pressure (101. Thermal Conductivity of Stainless Steel. You do want to empty the bag after the third breath so it fills with the 30% mixture. The Global Helium Shortage Is Real, but Don't Blame Party Balloons. NPT Outlet, Shipped Empty, Each Part Number: NXO-00-30145. The tank pressure gauge is responsible for telling you how full the tank is. of helium to fill a 9 inch latex, 1/2 C. At elevated temperatures, nitrous oxide is a powerful oxidiser similar to molecular oxygen. Check out our technical library for more detailed information. I approximated 1000 lbm as 500 kg and I know that net buoyancy comes out very approximately to 1 m^3 displaces 1 kg of mass. 20 Full PDFs related to this paper. The SI unit of density is kg/m3. S-2 featured turbocharged powerplants and S-3 was driven by Jumo 213A engines of …. NOT by a gauge - you weigh the tank and subtract 14 lb (weight of empty tank) 15 1 g of N2O = _____ L of N2O @ 20 degree C O2 down to 16-20 PSIG Nitrous 32 PSIG 45 The reservoir bag (balloon) increases _____ when it gets to a certain pressure so that the patient doesn't get all of the pressure. S-0 led the way with BMW 801G2 engines and provision for up to 14 x 145lb bombs. With each new model built and designed from scratch by Batman himself, the Batmobile was a custom, highly-modified road vehicle that far …. FREE Shipping on orders over$25 shipped by Amazon Blue Ribbon Balloon Time Disposable Helium Tank 14. Internet Explorer - Version 10 and above Firefox – Version 38 and above Chrome - Version 45 and above Safari - Version 8 and above. Oxygen cylinder sizes and dimensions. 5 cups (120 ml) of water at a time until they are. Carts are 8g and usually 3 to a balloon so 3x8 is 24 so let's round to an oz. This is one of the reasons many investigators avoided liquid hydrogen, which has the lowest density of any fuel. 23 Kaplan R, Abramowitz MD, Epstein BS. 31 PCF and weighs 100 pounds (Class 300) will be much more expensive to ship than an item that has the same …. 50000L 25 Tons 25t LPG Gas Storage Tanker ASME Propane Gas Tanker 50000 Liters Pressure Vessel 50m3 LPG Storage Tank. 88 times as heavy as hydrogen, the proportion by weight in which the two gases combine is 1 part of hydrogen to 7. hr virtualna burza; Hoarders San Francisco Ray And Tony Update; Download Private. i know that but its kinda hard to measure balloons when you are using a tank. This is sometimes referred to as “nozzing” or “huffing” balloons. Scrap lead is recycled back to the pot. 303-in machine guns, and could carry a variety of other loads, typically comprising two 250-lb bombs internally, and an external load of two 500-lb bombs, or two external fuel tanks or 8 rocket projectiles. 0107 g/mol and oxygen has a molar mass of 15. How many balloons are in a tank? Party supply stores large tanks hold 14. Q: What size CO2 tank should I buy? A: It takes approximately 1/2 lb. Q: What is the function of the blow-off safety valve on the bottle? A: It is very important not to overfill a bottle; i. w6h2, mht0, hxt3, 1gj, or8v, o4t, 7rj, d4s, v83, xb1, vecm, 0tvl, dr0, qtf, ficl, oa4, xd0, uj3g, 512w, tocu, 617, 403, av0, ntco, gko, aj4b, mmo, 9b0, qpg0, c6b, 1sm, v9g9, 5k5, jwjv, 49w, k7s, j5z6, i3b3, 9irk, blp3, p94, czr, 9kmn, 4bf, yir, v0w, rrj, c5h, r973, cdwd, 2aqi


text:
# High percentage of poly A sequences in 10X chromium R2 read

I'm currently analyzing two samples of eosinophil cells isolated from mouse lung and the samples are of very different quality.
According to the Cell Ranger summary 56% of the reads can be mapped to the transcriptome in the first sample and only 32% in the second sample. Also the number of cells identified in Cell Ranger are very different (5000 for sample 1 and 600 for sample 2).

When I look at the raw data in FastQC I see that the second sample has a lot of poly-A sequences in the R2 read that obviously can't be mapped to the transcriptome.

Sample 1:

Sample 2:

Also in many cases there is the following sequence in front of the poly-A:
AAGCAGTGGTATCAACGCAGAGTACATGGG

To me it looks as if most of the reads do not contain an insert and are primer dimers. Is this connected to the lower cell count in the second sample?
Maybe some of you have already seen something like this and have a suggestion on how to avoid such a problem in the future.

Thanks a lot

• Hi @PPK, could you please add your solution as an answer to this question and "accept" it? That will help others who encounter the same problem, and stop the StackExchange Community bot from bumping this question from time to time to fish for more answers.
– gringer
Jun 25 at 22:53

Answer from @merv, converted from comment:

Perhaps it's an insert size issue. Can you get the post library construction QC info? You might also see this in the BAMs, with the first showing a full bell-shaped peaks, and the latter truncating the peaks at the cleavage site. (Most 10X datasets I've seen have very few reads traversing the 3' end cleavage site.)

I actually figured out what was going on some time ago:

The over-represented sequence is the template switching oligo (TSO) that is used for the synthesis of the second strand (the example is for the v2 chemistry but the principle is the same for v3 and later).
This is a simplified illustration from the 10x manual:

As you can see the cDNA insert should be between the polyA capture probe and the TSO. In the second sample there are many cases with no insert and the TSO is followed by the poly A sequence. Actually the TSO should not be present in the final read after addition of Illumina adapters. But without any insert the fragmentation does not work and many reads still have this technical sequence.
So @merv was correct. Many reads contain no insert.

## Why did this happen?

Why were there many reads without insert in this experiment and why are some samples more heavily affected than others? As I mentioned, this were eosinophil granulozytes and these fellas contain many proteases, DNAses and RNAses to destroy invading pathogens. Unfortunately, they also destroy mRNA in single cell experiments.
Interestingly, the eosinophily in the second sample received an activating treatment and probably contained even more or more active RNAses.

format: plain


text:
## Physics Faculty Publications

#### Abstract

The one loop corrected effective Lagrangian for the quark-squark-chargino and quark-squark-neutralino interactions is computed. The effective Lagrangian takes into account the loop corrections arising from the exchange of the gluinos, charginos, neutralinos, W, Z, the charged Higgs, and the neutral Higgs. We further analyze the squark decays into charginos and neutralinos and discuss the effect of the loop corrections on them. The analysis takes into account CP phases in the soft parameters. It is found that the loop corrections to the stop decay widths into chargino and neutralinos can be as much as thirty percent or even larger. Further, the stop decay widths show a strong dependence on the CP phases. These results are of relevance in the precision predictions of squark decays in the context of specific models of soft breaking in supergravity and string based models.

#### Notes

Originally posted at http://arxiv.org/abs/hep-ph/0411272v1. Preprint of an article published in Physical Review D, v.71 no.5, 2005.

#### Keywords

high energy physics phenomenology, Lagrangian, $\bar q\tilde q'_i\chi^+_j$, $\bar q\tilde q'_i\chi^0_j$, fermionic decays, squarks

#### Subject Categories

CP violation (Nuclear physics)

Physics

2005

#### Permanent URL

http://hdl.handle.net/2047/d20000762

Click button above to open, or right-click to save.

COinS


text:
Edit Article

How to Calculate the Volume of a Cube

A cube is a three-dimensional shape that has equal width, height, and length measurements. A cube has six square faces, all of which have sides of equal length and all of which meet at right angles. Finding the volume of a cube is a snap - generally, all that's needed is to multiply the cube's length × width × height. Since a cube's sides are all equal in length, another way of thinking of a cube's volume is s3, where s is the length of one of the cube's sides. See Step 1 below for a detailed breakdown of these processes.

Method 1 Cubing One of the Cubes' Sides

1. 1
Find the length of one side of the cube. Often, in problems asking you to find the volume of a cube, you'll be given the length of one of a cube's sides. If you have this information, you have all you need to solve for the cube's volume. If you're not solving an abstract math problem but are instead attempting to find the volume of a real-life object shaped like a cube, use a ruler or measuring tape to measure the side of the cube.
• To better understand the process of finding the volume of a cube, let's follow along with an example problem as we go through the steps in this section. Let's say the side of the cube is 2 inches (5.08 cm) long. We'll use this information to find the volume of the cube in the next step.
2. 2
Cube the length of the side. When you've found the length of one of the cube's sides, cube this number. In other words, multiply it by itself twice. If s is the length of the side, you would multiply s × s × s (or, in simplified form, s3). This will give you the volume of your cube!
• This process is essentially the same as finding the area of the base and then multiplying it by the cube's height (or, in other words, length × width × height), since the area of the base is found by multiplying its length and its width. Since the length, width, and height of a cube are equal, we can shorten this process by simply cubing any of these measurements.
• Let's proceed with our example. Since the length of the side of our cube is 2 inches, we can find the volume by multiplying 2 x 2 x 2 (or 23) = 8.
3. 3
Label your answer with cubic units. Since volume is the measure of three-dimensional space, your answer should be in cubic units by definition. Often, on math schoolwork, neglecting to label your answer with the right units can cause you to lose points on a problem, so don't forget to use the correct label!
• In our example, since our original measurement was in inches, our final answer will be labelled with the units "cubic inches" (or in3). So, our answer of 8 becomes 8 in3.
• If we had used a different initial unit of measurement, our final cubic units would differ. For instance, if our cube had sides with lengths of 2 meters, rather than 2 inches, we would label it with cubic meters (m3).

Method 2 Finding Volume from Surface Area

1. 1
Find your cube's surface area. While the easiest way to find a cube's volume is to cube the length of one of its sides, it's not the only way. The length of a cube's side or the area of one of its faces can be derived from several other of the cube's properties, which means that if you start with one of these pieces of information, you can find the volume of the cube in a roundabout manner. For instance, if you know a cube's surface area, all you need to do to find its volume is to divide the surface area by 6, then take the square root of this value to find the length of the cube's sides. From here, all you'll need to do is cube the length of the side to find the volume as normal. In this section, we'll walk through this process step-by-step.
• The surface area of a cube is given via the formula 6s2, where s is the length of one of the cube's sides. This formula is essentially the same as finding the 2-dimensional area of the cube's six faces and adding these values together. We'll use this formula to find the volume of the cube from its surface area.
• As a running example, let's say that we have a cube whose surface we know to be 50 cm2, but we don't know its side lengths. In the next few steps, we'll use this information to find the cube's volume.
2. 2
Divide the cube's surface area by 6. Since the cube has 6 faces with equal area, dividing the cube's surface area by 6 will give you the area of one of its faces. This area is equal to the lengths of two of its sides multiplied (l × w, w × h, or h × l).
• In our example, dividing 50/6 = 8.33 cm2. Don't forget that two-dimensional answers have square units (cm2, in2, and so on).
3. 3
Take the square root of this value. Since the area of one of the cube's faces is equal to s2 (s × s), taking the square root of this value will find you the length of one of the cube's sides. Once you have this, you have enough information to solve for the volume of the cube as you normally would.
• In our example, √8.33 = 2.89 cm.
4. 4
Cube this value to find the cube's volume. Now that you've obtained a value for the length of the cube's side, simply cube this value (multiply it by itself twice) to find the volume of the cube as detailed in the section above. Congratulations - you've found the volume of a cube from its surface area.
• In our example, 2.89 × 2.89 × 2.89 = 24.14 cm3. Don't forget to label your answer with cubic units.

Method 3 Finding Volume from Diagonals

1. 1
Divide the diagonal across one of the cube's faces by √2 to find the cube's side length. By definition, the diagonal of a perfect square is √2 × the length of one of its sides. Thus, if the only information you're given about a cube is regarding the diagonal length of one of its faces, you can find the side length for the cube by dividing this value by √2. From here, it's relatively simple to cube your answer and find the volume of the cube as described above.
• For instance, let's say that one of a cube's faces has a diagonal that is 7 feet long. We would find the side length of the cube by dividing 7/√2 = 4.96 feet. Now that we know the side length, we can find the volume of the cube by multiplying 4.963 = 122.36 feet3.
• Note that, in general terms, d2 = 2s2 where d is the length of the diagonal of one of the cube's faces and s is the length of one of the sides of the cube. This is because, according to the Pythagorean theorem, the square of the hypotenuse of a right triangle is equal to the sums of the squares of the other two sides. Thus, because the diagonal of a cube's face and two of the sides on that face form a right triangle, d2 = s2 + s2 = 2s2.
2. 2
Square the diagonal of two opposite corners of the cube, then divide by 3 and take the square root to find the side length. If the only piece of information you're given about a cube is the length of a 3-dimensional line segment stretching diagonally from one corner of the cube to the corner opposite it, it's still possible to find the volume of the cube. Because d forms one of the sides of a right triangle that has the diagonal between the two opposite corners of the cube as a hypotenuse, we can say that D2 = 3s2, where D = the 3-dimensional diagonal between opposite corners of the cube.
• This is because of the Pythagorean Theorem. D, d, and s form a right triangle with D as the hypotenuse, so we can say that D2 = d2 + s2. Since we calculated above that d2 = 2s2, we can say that D2 = 2s2 + s2 = 3s2.
• As an example, let's say that we know that the diagonal from one of the corners in the base of the cube to the opposite corner in the "top" of the cube is 10 m. If we want to find the volume, we would insert 10 for each "D" in the equation above as follows:
• D2 = 3s2.
• 102 = 3s2.
• 100 = 3s2
• 33.33 = s2
• 5.77 m = s. From here, all we need to do to find the volume of the cube is to cube the side length.
• 5.773 = 192.45 m3

Community Q&A

• Can you find the capacity of a tank with dimensions 8cm x 6cm x 2.5cm?
• Well, that isn't exactly a cube but the formula is similar. You just need to multiply the length, width, and height altogether. 8 x 6 x 2.5 is 120. And there you have it. 120cm cubed. It's cubed because the units are little cubes.
• If the length is 2 meters, the height is 1 meter and the width is 2 meters of a water tank, how many gallons of water will the cube contain?
• First, you need to find the volume of the tank. This can be found by multiplying the length, height, and width altogether. When you do that, you get 4 cubic meters. Doing some research, you will find that 1 cubic meter is approximately equal to about 220 imperial gallons or 264.2 US liquid gallons. If you use the imperial gallons, then the tank would hold a little less than 880 imperial gallons. But if you use the US liquid gallons, it would hold slightly less than 1,057 gallons.
• If the volume of a cube is 1728, then what is its surface area?
• We know that to find the volume of a cube, we cube one of its sides. So if we want to inverse or reverse that operation, we take the cube root of 1728. It happens that the cube root is 12. So we now know that the length of one side is 12. If the length of one side of the cube is 12, then the area of one face would be 144. A cube has 6 faces, so we need to multiply 144 by 6 which is 864.
• How do you find the volume of a cube if you are given that a circle with a diameter of 14 fits in it?
• Good question. Assume that the circle fits perfectly into the cube. So that means the inside edges touch. You know that the circle had a diameter of 14. So if it fits perfectly into the cube, the cube would also have a side length of 14. So the volume could be easily calculated: 14 cubed, which is 2744.
• If the volume of a cube is 6s what is its answer by formula?

Article Info

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Categories: Featured Articles | Volume | Mathematics

In other languages:

Français: calculer le volume d’un cube, Italiano: Calcolare il Volume di un Cubo, Español: sacar el volumen de un cubo, Deutsch: Das Volumen eines Würfels berechnen, Português: Calcular o Volume de um Cubo, Nederlands: Het volume van een kubus berekenen, Русский: вычислить объем куба, 中文: 计算正方体体积, Bahasa Indonesia: Menghitung Volume Kubus, Čeština: Jak vypočítat objem krychle, العربية: حساب حجم المكعب, हिन्दी: एक घनाकृति के आयतन की गणना करें, ไทย: คำนวณหาปริมาตรของลูกบาศก์, Tiếng Việt: Tính Thể tích Hình lập phương, 한국어: 정육면체의 부피 구하는 법

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text:
# Follow-up 1: Compare 2 unordered, rooted trees for shape-isomorphism

In the previous post I tried to solve an exercise but found through the answer of Peilonrayz a bug in the code and 2 more while fixing it. These have now been amended (changes to the original code can be seen in my answer in the previous question).

The exercise (Copy&Paste from the previous question)

Write a program to check if 2 int arrays of N whole numbers between 0 and N-1 represent 2 isomorphic unordered trees, if you interpret the arrays as predecessor-successor-connections in a tree (e.g. in an int array index 3 has value 5. That means the node with value 3 is child of the node with value 5) , while the nodes are numbered from 0 to N-1. That means, your program should determine if there is a way to re-number the nodes in a tree, so that the array-representation of one tree is identical to the array-representation of the other tree.

Sidenote: I interpreted this as: "Check for tree-isomorphism between 2 trees given in int arrays". I believe this also means that one needs to check if one tree can be re-arranged in any fashion, so that shape-isomorphism occurs.

The Algorithm (Copy&Paste from the previous question)

1. Recreate the trees from the array using Node objects. Each node object contains its value called value and an ArrayList of node references called children

2. Sort the nodes in the unordered trees according to the amount of children, further called nChildren , they have directly connected to them. Sort left to right in descending order. Go from left to right, start sorting at the leaves and work your way up. If 2 sibling-nodes n1 and n2 have identical nChildren, compare the nChildren of their children from largest to lowest. If no difference is found, compare the the nChildrenof their children, again from largest to lowest. Repeat until difference is found or you find that these already sorted sub-trees that have n1 and n2 as roots have shape-isomorphism. In that case their order is irrelevant.

3. Compare the two now sorted trees and check for shape-isomorphism. Go recursively through the two trees in post-Order. Possible Cases: 1) Both currently observed nodes are leaves 2) The two observed nodes have unequal nChildren (e.g. one is a leaf while the other isn't) 3) Both observed nodes have equal nChildren. 1) and 2) are the base cases. If at any point one method call returns false, return false for all calls that lead to calling this iteration of the method. Else return true.

(2 Examples for the sorting can be found in the previous question).

The code

package Kapitel5;

import java.util.ArrayList;
import java.util.Comparator;

public class Test {
static class Node {
int value;
ArrayList<Node> children;

Node(int value, ArrayList<Node> children) {
this.value = value;
this.children = children;
}
}

static class NodeComparator implements Comparator<Node> {
public NodeComparator() {
}

public int compare(Node n1, Node n2) {
/*-
* Case1: Both Nodes have unequal amounts of nodes
* -> return (-) 1
* Case2: Both Nodes have equal amounts of nodes
* -> test child-nodes or return 0 if that is impossible;
*/
if (n1.children.size() != n2.children.size()) {// BaseCase2
return ((n1.children.size() > n2.children.size()) ? -1 : 1);
} else {
/*
* If n1 and n2 have equal amounts of children, compare the
* amounts of children their children have, from largest to
* lowest. If the default assumption is ever wrong, do not
* perform further comparisons.
*/
int result = 0; /*- Default assumption - Sub-trees of n1 and n2 are equal*/
for (int i = 0; (i < n1.children.size()) && (result == 0); i++) {
result = compare(n1.children.get(i), n2.children.get(i));
}
return result;
}
}
}

static final NodeComparator comp = new NodeComparator();

static void sortForest(Node root) {
if (root.children.isEmpty()) {
return;
} else {
for (int i = 0; i < root.children.size(); i++) {
sortForest(root.children.get(i));
}
root.children.sort(comp);
}
}

static void removeRootFromChildren(Node root) {
int i = 0;
while (root.children.get(i) != root) {
i++;
}
root.children.remove(i);
}

static Node createTreeFromArray(int[] tree) {
Node[] treeNodes = new Node[tree.length];
for (int i = 0; i < tree.length; i++) {
treeNodes[i] = new Node(i, new ArrayList<Node>(15));
}
Node root = null;
/*-Fill the ArrayLists of each Node with references to their child-Nodes*/
for (int i = 0; i < tree.length; i++) {
if (tree[i] == i) {
root = treeNodes[i];
}
}
/* Ascertain that tree has root */
if (root == null) {
throw new RuntimeException("Given tree-array has no root!");
} else {
/*
* Trees are given in format where roots point at themselves and are
* therefore listed as children in their own ArrayList. Remove it.
*/
removeRootFromChildren(root);
return root;
}
}

static boolean compareTrees(Node nodeFromTree1, Node nodeFromTree2) {
/*-
* Case1: Both Nodes have unequal amounts of nodes
* -> return false;
* Case2: Both Nodes have equal amounts of nodes
* -> test child-nodes or return true if that is impossible;
*/
if (nodeFromTree1.children.size() != nodeFromTree2.children.size()) {
return false;
} else {
boolean treesAreIsomorph = true; // Default Assumption
for (int i = 0; i < nodeFromTree1.children.size() && treesAreIsomorph; i++) {
treesAreIsomorph = compareTrees(nodeFromTree1.children.get(i), nodeFromTree2.children.get(i));
}
return treesAreIsomorph;
}
}

public static void main(String[] args) throws Exception {
int[] tree1 = { 2, 3, 3, 3, 2, 2, 1, 0, 7, 5, 3, 10, 10, 6 };
int[] tree2 = { 4, 10, 11, 0, 4, 0, 12, 4, 7, 8, 0, 3, 4, 12 };

Node root1 = createTreeFromArray(tree1);
Node root2 = createTreeFromArray(tree2);
sortForest(root1);
sortForest(root2);
System.out.println(compareTrees(root1, root2));
}
}

The given trees in the code should and do return "true" when given to compareTrees().

Questions (Copy&Paste from the previous question)

Question No. 2 got solved in the previous post.

1. Should I have avoided the static NodeComparator variable? I used it since, in the end, it saved me a lot of instantiating of an object to pass to ArrayLists sort() method. I don't know as of now how to do the sorting with the provided sort() without passing an object.
2. [...]
3. Was the algorithm I thought of acceptable or should I have gone for a different approach? If yes, please explain which one, this was the only one I could think of.

I'd appreciate it if reviews of the code itself focused especially on the recursive methods, since I think that's the part where I'm struggling the most.


text:
## Hyperbolic problems: Theory, numerics, applications. Proceedings of the 7th international conference, Zürich, Switzerland, February 1998. Vol. II.(English)Zbl 0911.00029

ISNM. International Series of Numerical Mathematics. 130. Basel: Birkhäuser. xii, 504-1011 $145.00/vol. 1;$ 259.00/set (1999).
The articles of this volume will be reviewed individually. The 5th conference (1994) has been reviewed (see Zbl 0903.00041).
Indexed articles:
Jenny, Patrick; Müller, Bernhard, A new approach for a flux solver taking into account source terms, viscous and multidimensional effects, 503-513 [Zbl 0937.76042]
Jenssen, Kristian Helge; Sinestrari, Carlo, Blowup for hyperbolic equations, 515-524 [Zbl 0932.35026]
Jochmann, F., Some analytical results concerning the drift diffusion equations for semiconductor devices coupled with Maxwell’s equations, 525-534 [Zbl 1058.82538]
Kazhikhov, Alexandre V.; Mamontov, Alexandre E., Transport equations and Orlicz spaces, 535-544 [Zbl 0938.35090]
Klausen, Runhild Aae; Risebro, Nils Henrik, Well-posedness of a $$2\times 2$$ system of resonant conservation laws, 545-552 [Zbl 0931.35094]
Klingenberg, Christian; Lu, Yun-Guang, The relaxation limit for systems of broadwell type, 553-557 [Zbl 0926.35079]
Kluwick, Alfred, Weakly-nonlinear hyperbolic waves in BZT-fluids, 559-568 [Zbl 0926.35119]
Koeber, Martin, Inclusion of solutions of Cauchy problems for quasilinear hyperbolic equations, 569-578 [Zbl 0927.35054]
Koren, Barry; Venis, Arthur, A fed back level-set method for moving material-void interfaces, 579-588 [Zbl 0929.35114]
Kulikovskii, A. G.; Pogorelov, N. V.; Semenov, A. Yu., Mathematical aspects of numerical solution of hyperbolic systems, 589-598 [Zbl 0926.35011]
LeFloch, Philippe G., Existence of entropy solutions for the compressible Euler equations, 599-607 [Zbl 0926.35109]
LeVeque, Randall J.; Bale, Derek S., Wave propagation methods for conservation laws with source terms, 609-618 [Zbl 0927.35062]
Levin, M. P., Quasi-characteristics numerical schemes, 619-628 [Zbl 0929.65073]
Li, Jiequan; Zhang, Tong, On the initial-value problem for zero-pressure gas dynamics, 629-640 [Zbl 0926.35120]
Li, Tong, Stability and instability of detonation waves, 641-650 [Zbl 0926.35019]
Liotta, Salvatore Fabio; Romano, Vittorio; Russo, Giovanni, Central schemes for systems of balance laws, 651-660 [Zbl 0926.35081]
Liska, Richard; Wendroff, Burton, Composite centered schemes for multidimensional conservation laws, 661-670 [Zbl 0929.65067]
Liu, Hailiang, On the diffusion limit of a hyperbolic relaxation system, 671-679 [Zbl 0926.35014]
Liu, Tai-Ping; Yang, Tong, Well-posedness theory for system of hyperbolic conservation laws, 681-691 [Zbl 0931.35097]
Luo, Tao; Xin, Zhouping, Nonlinear stability of shock fronts for a relaxation system in several space dimensions, 693-702 [Zbl 0929.35093]
Mao, D. K., Front tracking based on conservation in one and two space dimensions, 703-712 [Zbl 0972.35070]
Maurer, Jochen, A genuinely multi-dimensional scheme for mixed hyperbolic-parabolic systems, 713-722 [Zbl 0929.65084]
Métivier, Guy; Joly, Jean-Luc; Rauch, Jeffrey, Recent results in nonlinear geometric optics, 723-736 [Zbl 0938.35034]
Morton, K. W., Evolution-Galerkin methods: Algorithms and analysis from a finite difference viewpoint, 737-746 [Zbl 0929.65080]
Naldi, Giovanni; Pareschi, Lorenzo; Toscani, Giuseppe, Hyperbolic relaxation approximation to nonlinear parabolic problems, 747-756 [Zbl 0928.35099]
Noelle, Sebastian, A comparison of third and second order accurate finite volume schemes for the two-dimensional compressible Euler equations, 757-766 [Zbl 0923.76223]
Ostrov, Daniel, Unique solutions to discontinuous Hamilton-Jacobi equations in shape-from-shading, 767-772 [Zbl 0932.35040]
Peitz, Jochen; Appl, Stefan, Relativistic dissipative hydrodynamics in the $$3+1$$ formulation, 773-782 [Zbl 0926.35121]
Radvogin, Yulian B.; Zaitsev, Nikolay A., Absolutely transparent boundary conditions for time-dependent wave-type problems, 783-792 [Zbl 0927.35058]
Rozanova, Olga S., Blow-up of solutions in system of atmosphere dynamics, 793-801 [Zbl 0926.35122]
Ruzhansky, M., Holomorphic factorization for the solution operators for hyperbolic equations, 803-811 [Zbl 0924.35205]
Rykov, Yurii G., The propagation of shock waves in 2-D system of pressureless gas dynamics, 813-822 [Zbl 0928.35127]
Schroll, H. Joachim; Lorenz, Jens, Hyperbolic systems with relaxation: Symmetrizers and entropies, 823-832 [Zbl 0926.35078]
Selezov, Igor, Some hyperbolic models for wave propagation, 833-842 [Zbl 0929.35081]
Serre, Denis, Discrete shock profiles and their stability, 843-853 [Zbl 0928.35102]
Sethian, J. A., Advances in fast marching and level set methods for propagating interfaces, 855-864 [Zbl 0929.65066]
Shen, Wen; Tveito, Aslak; Winther, Ragnar, Rate of convergence for the zero relaxation limit, 865-874 [Zbl 0926.35084]
Sherwin, Spencer, A high order Fourier/unstructured discontinuous Galerkin method for hyperbolic conservation laws, 875-884 [Zbl 0929.65082]
Shirikyan, Armen; Volevich, Leonid, Asymptotic properties of solutions to high-order hyperbolic equations generalizing the damped wave equation, 885-894 [Zbl 0929.35078]
Shyue, Keh-Ming, A volume-of-fluid algorithm for compressible two-phase flows, 895-904 [Zbl 0937.76064]
Sleziona, P. C.; Auweter-Kurtz, M.; Boie, C.; Heiermann, J.; Lenzner, S., Numerical code for magneto-plasma flows, 905-914 [Zbl 0936.76045]
Sofronov, Ivan L., Highly-accurate artificial boundary conditions for unsteady transonic flow problems in wind tunnels, 915-924 [Zbl 0929.35117]
Tadmor, Eitan; Tang, Tao, Pointwise convergence rate for nonlinear conservation laws, 925-934 [Zbl 0928.35100]
Tefy, Tiana; Leyland, Pénélope, Newton-GMRES resolution of ill conditioned hyperbolic systems in fluid dynamics, 935-944 [Zbl 0936.76047]
Temple, Blake; Smoller, Joel, Applications of shock-waves in general relativity, 945-952 [Zbl 0924.35183]
Teshukov, Vladimir M., The Riemann problem for the integrodifferential equations of the shallow water theory, 953-962 [Zbl 0926.35110]
Trivisa, Konstantina, Decay and uniqueness of solutions of nonlinear hyperbolic conservation laws via generalized characteristics, 963-972 [Zbl 0933.35131]
Walder, Rolf; Folini, Doris, Radiative shocks, supersonic turbulence and structure formation in space, 973-982 [Zbl 0967.76051]
Westdickenberg, Michael; Noelle, Sebastian, A new convergence proof for FV schemes, 983-992 [Zbl 0929.65083]
Yserentant, Harry, The particle model of compressible fluids, 993-998 [Zbl 0928.35126]
Zimmermann, Susanne Amy, The method of transport for the Euler equations written as a kinetic scheme, 999-1008 [Zbl 0929.35118]

### MSC:

 00B25 Proceedings of conferences of miscellaneous specific interest 35-06 Proceedings, conferences, collections, etc. pertaining to partial differential equations 65-06 Proceedings, conferences, collections, etc. pertaining to numerical analysis 76-06 Proceedings, conferences, collections, etc. pertaining to fluid mechanics

Zbl 0903.00041


text:
# THIRD ORDER HANKEL DETERMINANT FOR CERTAIN UNIVALENT FUNCTIONS

• BANSAL, DEEPAK (DEPARTMENT OF MATHEMATICS GOVT. COLLEGE OF ENGINEERING AND TECHNOLOGY) ;
• MAHARANA, SUDHANANDA (DEPARTMENT OF MATHEMATICS CENTRAL UNIVERSITY OF RAJASTHAN) ;
• PRAJAPAT, JUGAL KISHORE (DEPARTMENT OF MATHEMATICS CENTRAL UNIVERSITY OF RAJASTHAN)
• Published : 2015.11.01
• 94 31

#### Abstract

The estimate of third Hankel determinant $$H_{3,1}(f)=\left|a_1\;a_2\;a_3\\a_2\;a_3\;a_4\\a_3\;a_4\;a_5\right|$$ of the analytic function $f(z)=z+a2z^2+a3z^3+{\cdots}$, for which ${\Re}(1+zf^{{\prime}{\prime}}(z)/f^{\prime}(z))>-1/2$ are investigated. The corrected version of a known results [2, Theorem 3.1 and Theorem 3.3] are also obtained.

#### Keywords

analytic functions;univalent function;close-to-convex functions;starlike functions;Fekete-$Szeg{\ddot{o}}$ functional;Hankel determinant

#### References

1. H. R. Abdel-Gawad and D. K. Thomas, The Fekete-Szego problem for strongely closeto-convex functions, Proc. Amer. Math. Soc. 114 (1992), no. 2, 345-349.
2. K. O. Babalola, On third order Hankel determinant for some classes of univalent functions, Inequal. Theory Appl. 6 (2010), 1-7.
3. K. O. Babalola and T. O. Opoola, On the coefficients of certain analytic and univalent functions, Advances in Inequalities for Series, (Edited by S. S. Dragomir and A. Sofo) Nova Science Publishers (2008), 5-17.
4. D. Bansal, Upper bound of second Hankel determinant for a new class of analytic functions, Appl. Math. Lett. 26 (2013), no. 1, 103-107. https://doi.org/10.1016/j.aml.2012.04.002
5. B. Bhowmik, S. Ponnusamy, and K. J. Wirths, On the Fekete-Szego problem for concave univalent functions, J. Math. Anal. Appl. 373 (2011), no. 2, 432-438. https://doi.org/10.1016/j.jmaa.2010.07.054
6. D. G. Cantor, Power series with integral coefficients, Bull. Amer. Math. Soc. 69 (1963), 362-366. https://doi.org/10.1090/S0002-9904-1963-10923-4
7. P. L. Duren, Univalent Functions, Grundlehren der Mathematischen Wissenschaften, Band 259, Springer-Verlag, New York, Berlin, Heidelberg and Tokyo, 1983.
8. R. Ehrenborg, The Hankel determinant of exponential polynomials, Amer. Math. Monthly 107 (2000), no. 6, 557-560. https://doi.org/10.2307/2589352
9. M. Fekete and G. Szego, Eine Benberkung uber ungerada Schlichte funktionen, J. Lond. Math. Soc. 8 (1933), 85-89.
10. W. K. Hayman, On the second Hankel determinant of mean univalent functions, Proc. Lond. Math. Soc. (3) 18 (1968), 77-94.
11. A. Janteng, S. Halim, and M. Darus, Coefficient inequality for a function whose derivative has a positive real part, J. Inequal. Pure Appl. Math. 7 (2006), no. 2, Article 50, 5 pp.
12. F. R. Keogh and E. P. Merkes, A coefficient inequality for certain classes of analytic functions, Proc. Amer. Math. Soc. 20 (1969), 8-12. https://doi.org/10.1090/S0002-9939-1969-0232926-9
13. Y. C. Kim, J. H. Choi, and T. Sugawa, Coefficient bounds and convolution properties for certain classes of close-to-convex functions, Proc. Japan Acad. Ser. A Math. Sci. 76 (2000), no. 6, 95-98. https://doi.org/10.3792/pjaa.76.95
14. W. Koepf, On the Fekete-Szego problem for close-to-convex functions, Proc. Amer. Math. Soc. 101 (1987), no. 1, 89-95.
15. B. Kowalczyk and A. Lecko, The Fekete-Szego inequality for close-to-convex functions with respect to a certain starlike function dependent on a real parameter, J. Inequal. Appl. 2014 (2014), Article 65, 16 pp. https://doi.org/10.1186/1029-242X-2014-16
16. S. K. Lee, V. Ravichandran, and S. Subramaniam, Bounds for the second Hankel determinant of certain univalent functions, J. Inequal. Appl. 2013 (2013), Article 281, 17 pp. https://doi.org/10.1186/1029-242X-2013-17
17. J. L. Li and H. M. Srivastava, some questions and conjectures in the theory of univalent functions, Rocky Mountain J. Math. 28 (1998), no. 3, 1035-1041. https://doi.org/10.1216/rmjm/1181071753
18. R. J. Libera and E. J. Zlotkiewicz, Early coefficients of the inverse of a regular convex function, Proc. Amer. Math. Soc. 85 (1982), no. 2, 225-230. https://doi.org/10.1090/S0002-9939-1982-0652447-5
19. R. J. Libera and E. J. Zlotkiewicz, Coefficient bounds for the inverse of a function with derivatives in P, Proc. Amer. Math. Soc. 87 (1983), no. 2, 251-257.
20. R. R. London, Fekete-Szego inequalities for close-to-convex functions, Proc. Amer. Math. Soc. 117 (1993), no. 4, 947-950.
21. T. H. Macgregor, Functions whose derivative has a positive real part, Trans. Amer. Math. Soc. 104 (1962), 532-537. https://doi.org/10.1090/S0002-9947-1962-0140674-7
22. J. W. Noonan and D. K. Thomas, On the second Hankel determinant of areally mean p-valent functions, Trans. Amer. Math. Soc. 223 (1976), 337-346.
23. K. I. Noor, Higher order close-to-convex functions, Math. Japon. 37 (1992), no. 1, 1-8.
24. R. Parvatham and S. Ponnusamy (Editors), New Trends in Geometric Function Theory and Application, World Scientific Publishing Company, Singapore, New Jersey, London and Hong Kong, 1981.
25. C. Pommerenke, On the coefficients and Hankel determinant of univalent functions, J. Lond. Math. Soc. 41 (1966), 111-122.
26. S. Ponnusamy, S. K. Sahoo, and H. Yanagihara, Radius of convexity of partial sums of functions in the close-to-convex family, Nonlinear Anal. 95 (2014), 219-228. https://doi.org/10.1016/j.na.2013.09.009
27. M. Raza and S. N. Malik, Upper bound of the third Hankel determinant for a class of analytic functions related with Lemniscate of Bernoulli, J. Inequal. Appl. 2013 (2013), 412, 8 pp. https://doi.org/10.1186/1029-242X-2013-8
28. T. V. Sudharsana, S. P. Vijayalakshmi, and B. Adolf Stephen, Third Hankel determinant for a subclass of analytic univalent functions, Malaya J. Mat. 2 (2014), no. 4, 438-444.

#### Cited by

1. Bounds on third Hankel determinant for close-to-convex functions vol.7, pp.2, 2015, https://doi.org/10.1515/ausm-2015-0014
2. The Bounds of Some Determinants for Starlike Functions of Order Alpha 2017, https://doi.org/10.1007/s40840-017-0476-x
3. Bounds on Hankel determinant for starlike and convex functions with respect to symmetric points vol.3, pp.1, 2016, https://doi.org/10.1080/23311835.2016.1160557
4. Third-Order Hankel Determinant for Certain Class of Analytic Functions Related with Exponential Function vol.10, pp.10, 2018, https://doi.org/10.3390/sym10100501
5. On Hankel Determinant $${{\varvec{H}}}_\mathbf{2}{} \mathbf{(3)}$$H2(3) for Univalent Functions vol.73, pp.3, 2018, https://doi.org/10.1007/s00025-018-0854-1
6. The Sharp Bound of the Hankel Determinant of the Third Kind for Starlike Functions of Order 1 / 2 pp.1661-8262, 2018, https://doi.org/10.1007/s11785-018-0819-0
7. THE SHARP BOUND FOR THE HANKEL DETERMINANT OF THE THIRD KIND FOR CONVEX FUNCTIONS vol.97, pp.03, 2018, https://doi.org/10.1017/S0004972717001125
8. On the Fourth Coefficient of Functions in the Carathéodory Class vol.18, pp.2, 2018, https://doi.org/10.1007/s40315-017-0229-8
9. SHARP BOUNDS OF SOME COEFFICIENT FUNCTIONALS OVER THE CLASS OF FUNCTIONS CONVEX IN THE DIRECTION OF THE IMAGINARY AXIS pp.1755-1633, 2019, https://doi.org/10.1017/S0004972718001429


text:
Quantifying resource in catalytic resource theory

# Quantifying resource in catalytic resource theory

Anurag Anshu111Center for Quantum Technologies, National University of Singapore, Singapore. a0109169@u.nus.edu   Min-Hsiu Hsieh222Centre for Quantum Software and Information, University of Technology Sydney, NSW 2007, Australia. Min-Hsiu.Hsieh@uts.edu.au   Rahul Jain333Center for Quantum Technologies, National University of Singapore and MajuLab, UMI 3654, Singapore. rahul@comp.nus.edu.sg
###### Abstract

We consider a general resource theory that allows the use of free resource as a catalyst. We show that the amount of ‘resource’ contained in a given state, in the asymptotic scenario, is equal to the regularized relative entropy of resource of that state, which then yields a straightforward operational meaning to this quantity. Such an answer has been long sought for in any resource theory since the usefulness of a state in information-processing tasks is directly related to the amount of resource the state possesses in the beginning. While we need to place a few assumptions in our resource theoretical framework, it is still general enough and includes quantum resource theory of entanglement, coherence, asymmetry, non-uniformity, purity, contextuality, stabilizer computation and the classical resource theory of randomness extraction as special cases. Since our resource theoretic framework includes entanglement theory, our result also implies that the amount of noise one has to inject locally in order to erase all entanglement contained in an entangled state is equal to the regularized relative entropy of entanglement, resolving an open question posted in [Groisman et al., Phys. Rev. A. 72: 032317, 2005]. On the way to prove the main result, we also quantify the amount of resource contained in a state in the one-shot setting (where one only has a single copy of the state), in terms of the smooth max-relative entropy. Our one-shot result employs a recently developed technique of convex-split lemma.

## 1 Introduction

A question that is commonly asked in Physics is how a certain property of a physical system can be used to achieve useful tasks, and how to quantify the amount of such a property in a meaningful way. Various areas in Physics have developed methodology to answer these questions relevant to their own areas; however, many of the approaches are difficult, sometimes even impossible, to employ outside of their respective fields. Quantum resource theory then emerges when quantum information theory is found to provide a unified platform for characterizing the resource [1, 2] because in a nutshell, they can all be viewed as interconversion of different system states with system-dependent constraints. Since then, individual resource theory has been able to characterize targeted properties in some information-processing tasks using entropic quantities defined on system states.

The core of a resource theory is built upon two main system-dependent requirements for the resources and allowed operations; namely, (i) the existence of a set of states that are free and those not in the set are expensive; and (ii) the allowed operations are those that map the set of free states to itself. Various resource theories have been developed in the past decade. The most notable example is the resource theory of entanglement [3], where the set of free states corresponds to the collection of separable states and the allowed free operations are the local quantum operations and classical communication (LOCC). These two classes of states and operations have attracted stand-alone interests besides resource theory [4, 5]. Under this resource framework, one can then ask the amount of valuable resource (cf. quantum entanglement) possessed by an entangled state relative to the set of free states (cf. separable states). This question motivates the scenario of injecting noise locally to the system in order to destroy the quantum entanglement, i.e., the randomness cost. A complete characterization of this question has remained open; though, gapped upper and lower bounds have been provided in the asymptotic i.i.d. setting in Ref. [6]. It is worthwhile to mention that investigation of a variant of the above setting, where the local noise is used to destroy the total correlation in an entangled state, relates the minimal randomness cost to the quantum mutual information [6]. This seminal result gives the first operational meaning to this entropic quantity and advances significantly our understanding of entanglement theory. Being able to answer the optimal randomness cost to bring entangled states to separable states thus bears equivalent significance, if not more important, since the existence of entanglement is believed to make quantum systems superior to their classical counterparts and the amount of entanglement is generally linked to its information-processing power [7]. Likewise, this crucial question of the amount of valuable resource possessed by a state relative to its free resource is then adhesive to every resource theory, be it quantum coherence, thermodynamics, etc.

One can view the above erasing framework as a restricted model of state transformation, where the final state belongs to the free resource [8, 9]. In the general model of state transformation, it has been demonstrated that not every state transformation is possible, and could become possible if a catalyst is involved [10]. Even if the transformation process is possible in the beginning, adding catalyst would likely make the process much more efficient. Therefore, individual resource theories have included catalyst in their formalism [11, 12].

We thus consider a general resource theory framework that allows free resource being used as a catalyst. We demonstrate that this framework includes previous major resource theories of (i) entanglement [3, 6, 13, 14], (ii) coherence [15, 16, 14, 17], (iii) thermodynamics [18, 19, 20, 21, 22, 23], (iv) non-uniformity [11], (v) purity [24], (vi) randomness extractors [25, 26], (vii) contextuality [27], (viii) asymmetry [28] and (ix) stabilizer computation [29] as special instances. Treating resource theory in a general framework enjoys the major advantage that the resource conversion tasks done for one particular resource can already be adapted to other resources almost trivially. Hence, it has started to attract more attention [8, 9]. We establish that the entropic quantity, the regularized relative entropy of resource [8], characterizes the amount of useful resource in a given state relative to its free resource in the asymptotic i.i.d. setting. In other words, if one has infinitely many copies of the same state, the randomness cost (per copy of the state) for erasing the resource contained in the given state and bringing it to the closest free state is equal to this entropic quantity. This yields a crucial operational meaning to this entropic quantity. Since the resource theory of quantum entanglement is a special case of our framework, our result directly provides the missing answer444We remark that an independent work for this answer can be found in Ref. [30] to the minimal randomness cost for erasing the amount of entanglement in an entangled state to a separable state, a question first posted in Ref. [6].

We also obtain matching upper and lower bounds on the randomness cost if one only has a single copy of a given state, i.e., the one-shot scenario. Our one-shot bounds are given in terms of smooth max-relative entropy; hence our result also provides a new operational meaning to this quantity in the resource theoretic framework. We emphasize that being able to obtain matching one-shot bounds in such a general resource theory framework is rare [8], and is due to the technical tool: the convex split lemma [31]. This again provides another excellent example that quantum information theory helps in the understanding of quantum physics.

The paper is organized as follows. Section 2 contains assumptions required for our catalytic resource framework. We then define our information-processing task of quantifying the amount of resource and present our main theorem. We also discuss several individual resource theories as special cases of our framework. Section 3 contains definition of entropic quantities and useful lemmas required in the proofs. The proofs of our main result are given in Sections 46. We conclude in Section 7.

## 2 Catalytic Resource Framework and Main Results

The question that we will answer in this general catalytic resource theory framework is how much resource is contained in a given resource state (relative to its free resource). This question is crucial, as this amount directly relates to the power the state possesses in achieving information-processing tasks.

We first present assumptions that we have to make in our resource theoretic framework. It is apparent that if no limit is set on the allowed operations and free resources, then it is almost impossible to obtain useful characterization, as also noted in Ref. [8]. Thus, we need a sufficient (yet small) number of assumptions to keep the theory interesting.

Our framework for both single-partite and multi-partite settings is discussed below, with the assumptions on free resources and allowed operations. Our assumptions capture the requirement we impose on an experimenter (or experimenters in the multi-partite setting) who wants to quantify a useful property of his/her physical system. For this, we start with the natural experimental framework, where the experimenter has decided upon the global Hilbert space and its tensor decomposition into several Hilbert spaces (for example, as ‘physically separated’ Hilbert spaces). Given this framework, we impose the conditions that are to be satisfied by the set of free resources and the allowed unitary operations. It is important to note that the conditions are only for a given framework; we do not restrict the choice of the global Hilbert space and its tensor decomposition by the experimenter.

For the notations and definitions used in this section, please refer to Section 3.

### 2.1 Single-partite case

As discussed above, an experimenter starts with a collection of quantum registers and performs operations on them. The joint state lives on a global Hilbert space . This Hilbert space possesses an ordered decomposition into systems . Given the Hilbert space and its ordered decomposition (the order of the decomposition may matter in a given physical scenario, hence we take it into account), we define the set of free states and allowed operations. The definition is analogous to the postulates given in Ref. [8].

###### Definition 1 (Single party resource theory).

Fix an integer . Given and registers where corresponds to , the set of free resources and free operations are as follows.

1. For every non-empty ordered set (where are increasing sequence of integers), let and satisfy the following properties. For brevity, we set

1. is a convex and closed set.

2. For any , .

• This assumption states that if two quantum states are free resources, then their tensor product is a free resource as well.

3. For any and , .

• This assumption states that if a quantum state on more than one registers is a free resource, then we obtain a free resource by partial trace over a subset of these registers.

4. It holds that , where is the dimension of register .

• Under this assumption, we require the maximally mixed quantum state to be a free resource on the last register.

2. The complete set of free resources is .

3. For a fixed L, the free operations are defined as the set of all unitaries that satisfy

1. .

2. if and only if .

4. The set of all free operations are .

We remark that these assumptions we required are natural and mild [8]. Above, we also observe that the set (for every L) forms a group.

Now we are in a position to formally define our task, that we call an -transformation of to .

Let , be an integer and fix a Hilbert space with decomposition , which is chosen by the experimenter. Let the register correspond to , register correspond to and register correspond to . An experimenter holds a quantum state . Using a state , she applies a unitary to obtain a joint quantum state :

 ΘMEJ=U(ρM⊗μEJ)U†.

It is required that there exists a such that , where the chosen distance measure is defined in Equation (1). The number of discarded qubits is .

In above task, no restriction is made on the transformation of the free resource state . But in many cases, it is desirable that this quantum state be returned close to its original form and hence act as a catalyst. For example, the framework of catalytic decoupling [32] studies such a scenario. Our achievability result shall belong to such class of transformations. Hence we have the following definition.

###### Definition 3.

Task 2 is said to be a -catalytic transformation of if and for some .

Task 2 is motivated by the work [6], which considered the problem of transforming a bipartite quantum state into a product state with the aid of shared randomness and local unitaries. It was shown that the number of bits of randomness required (in other words, the randomness cost) is . Task 2 is along the lines of the framework considered in above work, but with an additional freedom of allowing the use of additional free resources that can aid in the transformation of the desired quantum state. Moreover, when the register is classical (in a more precise sense to be discussed below), we can interpret to be the randomness cost of the protocol. In Definition 3, we have provided further restriction that the free resource (expect for the randomness used) be returned with small error.

We provide a near optimal characterization of the randomness cost of Task 2 in Theorem 5 below. Our achievability protocol requires a further assumption related to the last register (and its connection to rest of the registers), as this register serves as the source of ‘classical randomness’. This assumption is as follows.

###### Assumption 4.

Invoke the notation in Definition 1. Then there exists a canonical basis on (where is the dimension of ) such that

 F=conv{1ℓr∑jUjσU†j⊗|j⟩⟨j|Mr:σ∈F{1,2,…r−1},Uj∈U{1,2,…r−1}},

and the experimeter only performs an operation from the set

 {ℓr∑j=1Uj⊗|j⟩⟨j|Mr:Uj∈U{1,2,…r−1}}.

Moreover, suppose it holds that for all . Then implies for all .

Under this assumption, all the states in Task 2 are classical-quantum states, with register being classical and being diagonal in the canonical basis. Thus, the assumption allows us to interpret as the randomness cost of the protocol. We have the following theorem.

###### Theorem 5.

Fix , and a quantum state .

• Achievability: Suppose Assumption 4 holds. Then there exists an -catalytic transformation of to , where .

• Converse: For every - transformation of to , it holds that

 log|J|≥minσM∈FDεmax(ρM∥σM),

if Assumption 4 is true. Otherwise,

 log|J|≥12minσM∈FDεmax(ρM∥σM).

The proof of this theorem is given in Section 4. Now we consider the asymptotic i.i.d. properties of our results.

###### Definition 6.

Suppose Assumption 4 holds. We say that the asymptotic randomness rate of catalytic transformation of is , if for every , there exists an such that for all , there exists a -catalytic transformation of to . Define the following relative entropy of resource:

 E(ρ)=infσM∈FD(ρM∥σM)

and the regularized relative entropy of resource:

 E∞(ρ)=limn→∞1nE(ρ⊗nM).

Let denote the set of free resources for the register . Define the constant

 C(F):=limn→∞1ninfτ∈Fn∥logτ∥∞.

Using Theorem 5, we obtain the following.

###### Theorem 7.

Suppose the resource theory is such that . Then for a quantum state , the asymptotic randomness rate of catalytic transformation of is given by .

It completely characterizes the per copy randomness requirement in the asymptotic i.i.d. setting. The proof is given in Section 6. Moreover, we show that the number of qubits of catalytic free resource grows only polynomially in (Theorem 28).

### 2.2 Multi-partite case

Our formalism also extends to the multi-partite case. We consider the case of parties and introduce some minor modifications to Definition 1 and Assumption 4. Let be the global Hilbert space in the multiparty setting. We start with a decomposition of , where is a Hilbert space shared by all the parties.

###### Definition 8 (Multi-partite resource theory).

Fix an integer . Given

 H=H1⊗H2…⊗Hr

and registers , , , where corresponds to and is held by -th party, the set of free resources and free operations are as follows.

1. For every non empty ordered set

 L={s1,s2,…s|L|}⊆{1,2,…r}

let satisfy the following properties. For brevity, we let be the register corresponding to set L.

1. is a convex and closed set.

2. For any , .

3. For any and , .

2. The complete set of free resources is .

3. For a fixed L, the free operations are defined as the set of all unitaries that satisfy

1. .

2. if and only if .

4. The set of all free operations are .

In an analogous fashion to Task 2, we define the multi-partite task.

###### Task 9 (An (ε,log|J|,t)-transformation of ρM1M2…Mt to F).

Let , be an integer and fix a Hilbert space with decomposition , which is chosen by the experimenters. Let the register correspond to , register correspond to (with register held by -th party) and register correspond to . The experimenters share a quantum state . Using a state , they apply the unitary to obtain a joint quantum state :

 ΘM1M2…MtEJ1J2…Jt=U(ρM1M2…Mt⊗μEJ1J2…Jt)U†.

It is required that there exists a such that , where the chosen distant measure is defined in Equation (1).

We can similarly define a catalytic transformation in this framework.

###### Definition 10.

Task 9 is said to be a -catalytic transformation of if and for some .

We will also need the following assumption for achievability protocol.

###### Assumption 11.

Invoke the notation from Definition 8. There exists a canonical basis on (where the dimension of is independent of ) such that

 F=conv{1ℓrℓr∑k=1UkσU†k⊗|k⟩⟨k|M1r⊗|k⟩⟨k|M2r⊗…|k⟩⟨k|Mtr:Uk∈U{1,2,…r−1}}

and experimenter only performs an operation from the set

 {∑kUk⊗|k⟩⟨k|M1r⊗|k⟩⟨k|M2r⊗…|k⟩⟨k|Mtr:Uk∈U{1,2,…r−1}}.

For brevity, define

 Iℓr,t:=1ℓrℓr∑k=1|k⟩⟨k|M1r⊗|k⟩⟨k|M2r⊗…|k⟩⟨k|Mtr.

Moreover, suppose it holds that for all . Then

 σM11M21…Mt1,M12M22…Mt2,…M1rM2r…Mtr∈F

implies

 σM1jM2j…Mtj,M12M22…Mt2,…M1j−1M2j−1…Mtj−1,M11M21…Mt1,M1j+1M2j+1…Mtj+1,…M1rM2r…Mtr∈F

for all .

We are now in a position to prove the following theorem.

###### Theorem 12.

Fix , and a quantum state .

• Achievability: Suppose Assumption 11 holds. There exists an -catalytic transformation of to , where .

• Converse: For every -transformation of to , it holds that

 log|J|≥minσM1M2…Mt∈FDεmax(ρM1M2…Mt∥σM1M2…Mt),

if Assumption 11 is true.

The proof of this theorem follows similar to the proof of Theorem 5 and is given in Corollaries 26 and 27 in Section 5.

### 2.3 Known resource theories

To conclude this section, we show that our general resource framework includes at least the following individual resource theory.

1. Resource theory of entanglement [3, 6, 13, 14]: In this resource theory, the set of free states, , is the collection of separable states. The perfectly correlated state is a separable quantum state for all , and hence is a free resource. The formalism also extends to multi-partite entanglement, where the free resources are convex combination of product quantum states.

2. Resource theory of coherence [15, 16, 14, 17]: In this resource theory, the set of free states, , is the collection of diagonal states in a pre-determined basis. This is captured by Definition 1, where we choose a basis on each Hilbert space and take the free resources to be the set of diagonal quantum states. The properties such as being closed under partial trace and tensor product are easily seen to be satisfied. Moreover, maximally mixed state is diagonal in any basis, and hence belongs to the free resources.

3. Resource theory of asymmetry [28]: In this resource theory, the set of free resources are the states that are invariant under some group transformation. To construct for some , one fixes a group with a unitary representation in dimension . The set is the collection of states that satisfy . The resource theory can then be constructed on tensor product of such Hilbert spaces (see Section in [28]). To verify the partial trace condition (Item ), let be such that . Then

 σM1 = = TrM2(Ug⊗V†g′Vg′)σM1M2(U†g⊗I))=TrM2(Ug⊗I)σM1M2(U†g⊗I)) = UgσM1U†g.

Symmetry under permutation of registers of identical dimension can also be verified in similar way, as must be invariant under for all .

4. Resource theory of nonuniformality [11] and purity [24]: In this resource theory, the only free state is the completely mixed state. This is easily captured by Definition 1. This is equivalent to the formalism of randomness extractors [25, 26].

5. Resource theory of Thermodynamics [18, 19, 20, 21, 22, 23]: In this resource theory, the free states are Gibbs quantum states, that are states of the form , for an arbitrary Hamiltonian . To capture it in Definition 1, we assign a Gibbs state as the only free resources in the set (for all ). Thus, for a system with registers, the free resource is . The condition that is equivalent to the condition . Moreover, this also implies . Finally, the maximally mixed state is a Gibbs quantum state with Hamiltonian . This establishes consistency with Definition 1.

Theorem 7 applies as long as . For a Gibbs state , it holds that

 ∥logρβ(H)∥∞≤∥βH∥∞+logTr(e−βH)≤∥βH∥∞+logd,

where is the support size of the Gibbs state. Thus, Theorem 7 applies for bounded .

6. Resource theory of contextuality [27]: In this resource theory, the free resources are the set of conditional probability distributions that can be described ‘classically’ (a notion that is made more precise in [27]). All the conditional probability distributions in this theory satisfy a consistency condition (which can be viewed as an analogue of the non-signaling condition in the case of non-locality). It can be observed that the set of free resources satisfies the conditions in Definition 1, being convex and closed under permutations and tensor product. Uniform probability distribution (or the maximally mixed state) also belongs to the free resource.

7. Resource theory of stabilizer computation [29]: In this resource theory, the free resources are the convex hull of the set of all pure states that can be generated by the action of a Clifford unitary on a standard state (for example, ). Thus, maximally mixed state belongs to this set. This set is convex and closed. The free resources can be extended to the tensor product of registers along the lines as discussed in [28].

## 3 Preliminaries

Let be the Hilbert space associated to a register . Let denote the dimension of . Let be the set of all normalized quantum states acting on . For two subsets , let . Let be a unitary operator acting on .

The definitions below have been adapted from the references [33, 34, 35, 36, 37, 38, 39] For such that , its relative entropy is given by

 D(ρA∥σA):=Tr(ρAlogρA)−Tr(ρAlogσA),

and is equal to infinity otherwise. Similarly, the relative entropy variance of is defined as

 V(ρ∥σ):=Tr(ρ(logρ−logσ)2)−(D(ρ∥σ))2,

when .

The max-relative entropy of such that is

 Dmax(ρA∥σA):=inf{λ∈R:2λσA≥ρA}.

The smooth max-relative entropy is

 Dεmax(ρA∥σA):=supρ′A∈Bε(ρA)Dmax(ρ′A∥σA),

where the -ball, , is defined via the purified distance

 Pur(ρA,σA):=√1−F2(ρA,σA). (1)

In the above, is the fidelity between two states.

We will use the following facts.

###### Fact 13 (Triangle inequality for purified distance,  [39, 34]).

For quantum states ,

 Pur(ρA,σA)≤Pur(ρA,τA)+Pur(τA,σA).
###### Fact 14 (Uhlmann’s Theorem, [40]).

Let . Let be a purification of . There exists a purification of such that,

 F(|θ⟩⟨θ|AB,|ρ⟩⟨ρ|AB)=F(ρA,σA).
###### Fact 15.

Let and be an extension of . Then there exists an extension of such that

 F(ρAB,σAB)=F(ρA,σA).
###### Proof.

Introduce a register and consider a purification of . There exists a purification of such that , due to Uhlmann’s Theorem (Fact 14). Thus, is the desired extension due to the relation

 F(ρA,σA)≥F(ρAB,σAB)≥F(|ρ⟩⟨ρ|ABC,|σ⟩⟨σ|ABC)=F(ρA,σA).

###### Fact 16.

Let and be a classical-quantum extension of , that is , where . Then there exists a classical-quantum extension of such that

 F(ρAB,σAB)=F(ρA,σA),

and .

###### Proof.

From Fact 15, there exists an extension of such that . Now, consider the map defined as . Observe that . Define . It holds that is a classical-quantum extension of . Moreover,

 F(ρA,σA)≥F(ρAB,σ′′AB)≥F(ρAB,σ′AB)=F(ρA,σA).

Let be the projector onto the support of . Define the quantum state . Observe that is also a classical-quantum extension of and . Then

 F(ρAB,σAB) = Tr(√√ρABσAB√ρAB) = 1√Tr(σ′′BΠB)Tr(√√ρABΠBσ′′ABΠB√ρAB) = 1√Tr(σ′′BΠB)Tr(√√ρABσ′′AB√ρAB) = 1√Tr(σ′′BΠB)F(ρAB,σ′′AB) ≥ F(ρAB,σ′′AB).

This completes the proof. ∎

###### Fact 17.

Let be a bipartite quantum state. Let be the projector onto the support of . Then

 ΘAB⪯|B|ΘA⊗ΠB.
###### Fact 18.

Let be a classical quantum state with as the quantum part. Let be the projector onto the support of . Then

 ΘAB⪯ΘA⊗ΠB.
###### Fact 19 ([35, 36]).

Let and be an integer. Let be quantum states. Define . It holds that

###### Fact 20.

For the function and , it holds that .

###### Proof.

We have

 Φ(−x)=∫−x−∞e−t2/2√2πdt=∫∞0e−(−x−t)2/2√2πdt≤e−x2/2∫∞0e−(−t)2/2√2πdt=12e−x2/2.

Thus, , which completes the proof. ∎

###### Fact 21.

Let be a quantum state and be a collection of quantum maps. Define a series of quantum states recursively as . It holds that

 Pur(ρi,ρ1)≤(i−1)maxi⋅{Pur(Ei(ρ1),ρ1)}.
###### Proof.

Consider

 Pur(ρi,ρ1)=Pur(Ei(ρi−1),ρ1)≤Pur(Ei(ρi−1),Ei(ρ1))+Pur(Ei(ρ1),ρ1)≤Pur(ρi−1,ρ1)+Pur(Ei(ρ1),ρ1).

This completes the proof. ∎

###### Lemma 22 (Convex-split lemma [31]).

Let , and be quantum states, be an integer and . Consider the following quantum state

 τM1M2…Mn=1nn∑j=1ρMj⊗σM1⊗σM2…⊗σMj−1⊗σMj+1…⊗σMn

where and . Then,

 Pur(τM1M2…Mn,σM1⊗σM2…⊗σMn)≤ε+√2kn.

## 4 Proof of Theorem 5

### 4.1 Achievability

###### Theorem 23 (Achievability).

Fix and . Suppose Assumption 4 holds. There exists an -catalytic transformation of to , where .

###### Proof.

Let and let be the quantum state achieving the minimum in the definition of . Let be a random variable taking values uniformly in . Introduce registers . Let be the quantum state in which an experimenter uses as a catalyst.

The protocol is as follows. Experimenter introduces the quantum state , which belongs to due to Definition 1 (Item ). Controlled on the value in , the experimenter swaps register with . This operation belongs to from Assumption 4. Let the resulting global quantum state be . We have and

 τM1M2…Mn=1nn∑j=1ρMj⊗σM1⊗σM2…⊗σMj−1⊗σMj+1…⊗σMn.

Using convex-split lemma (Lemma 22), this allows us to conclude that

 Pur(τM1M2…Mn,σ⊗nM)≤ε+δ.

This completes the proof. ∎

Remark 1: The number of qubits used as a catalyst in above task is .

### 4.2 Converse Proof

In this section, we prove a converse bound for resource transformation, showing near optimality of Theorem 23. We will separately consider the cases where Assumption 4 holds and where it does not.

###### Theorem 24.

Fix an . For every - transformation of to , it holds that