text: Download BYJU'S the learning app to know more To complete this calculation, you have to know what substance you are trying to convert. La masa molar de los átomos de un elemento está dado por el peso atómico de cada elemento [2] multiplicado por la constante de masa molar, M u = 1×10 â3 kg/mol = 1 g/mol. N Nitrogen atomic mass 14.00672 3 atoms 10.9718 % of total mass For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. We assume you are converting between grams Au and mole. molar mass: The mass of a given substance (chemical element or chemical compound in g) divided by its amount of substance (mol). The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. Symbol: Au Atomic Number: 79 Atomic Mass: 196.96655 amu Melting Point: 1064.43 C (1337.5801 K, 1947.9741 F) Boiling Point: 2807.0 C (3080.15 K, 5084.6 F) Number of … Note that rounding errors may occur, so always check the results. Convert grams Au to moles  or  moles Au to grams. Calcualtes Molar Mass of Elements. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. Examples: Fe, Au, Co, Br, C, O, N, F. Molar mass of a compound is defined as the sum of the mass of all the atoms each multiplied its atomic masses that are present in the molecular formula of a compound. Oct 21, 2020 - Ideas to help you teach chemistry better! In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units). For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. Examples: Fe, Au, Co, Br, C, O, N, F. You can use parenthesis or brackets []. ), and to the standard atomic masses of its constituent elements. Formula: 6.02 × 10 23 is called the Avogadro Constant or Avogadro's Number. This site explains how to find molar mass. Examples include mm, The unit for molar mass (note it is the mass of a mole) is grams/mole. to use the unit converter. The molar mass of something tells us how much one mole of that substance weighs. grams Au to picomol Moles to Grams â¦ Formula Weight is the molar mass of an IONIC compound. Quick conversion chart of grams Au to mol. A common request on this site is to convert grams to moles. Ejemplos de cálculos de la masa molar: NaCl, Ca(OH)2, K4[Fe(CN)6], CuSO4*5H2O, water, nitric acid, potassium permanganate, ethanol, fructose. 1 mole of something is 6.022 × 10 23 pcs.. #teachscience, #science, #teach, #highschool, #molarmass, #chemistry. Ca Xe K. Fe Au O. Molar mass is a physical property of substances that describes the mass of a substance divided by the amount of the substance present. Before reading this section, it must be understood that the 2019 redefinition of the SI base units concluded that the molar mass constant is not exactly 1×10 kg/mol, but Mu = 0.99999999965(30)×10 kgâmol . See more ideas about molar mass, teaching science, chemistry. mole: The amount of substance of a system that contains as many elementary entities as there are atoms in 12 g of carbon-12. Molar Mass of an Element The molar mass of sodium metal is the mass of one mole of Na. • Mass units are more widely used in other areas such as toxicology and forensics. Fórmula no sistema Hill é Au Calculando a massa molar (peso molar) To calculate molar mass of a chemical Molar Mass worksheet 23. This chemistry video tutorial explains how to calculate the molar mass of a compound. molar mass = mass ÷ moles . 261 Appendix Element Symbol Atomic Molar Number mass/ (g mol–1) Actinium Ac 89 227.03 Aluminium Al 13 26.98 Americium Am 95 (243) Antimony Sb 51 121.75 Argon Ar 18 39.95 Arsenic As 33 74.92 Astatine At 85 €€€210 Molar Mass Calculations – YouTube: This video shows how to calculate the molar mass for several compounds using their chemical formulas. Look for ways to teach molar mass step by step. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. Thus, by knowing the molar mass, we can determine the number of moles contained in a given mass of a sample. Now for the heat energy from burning some fuels: I will use molar mass . molar mass and molecular weight. The PI and AusDI sections on pharmacokinetics, dosing and toxicity most commonly use mass units, although molar units are also provided for some common drugs. Molar mass calculator computes molar mass, molecular weight and elemental composition of any given compound. Browse the list of The SI base unit for amount of substance is the mole. Measuring Mass in Chemistry. Molar Mass of a Compound H 2O H = 2 x 1.0 where: M = molar mass of the pure substance (measured in g mol-1) This site explains how to find molar mass. Because molar mass is defined as the mass for 1 mol of a substance, we can refer to molar mass as grams per mole (g/mol). Use uppercase for the first character in the element and lowercase for the second character. Molar mass of Au4 is 787.866276 ± 0.000016 g/mol Convert between Au4 weight and moles This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. We also established that 1 mol of Al has a mass of 26.98 g (Example). Molar mass = 382.9816 g/mol. Atomic Weight is the molar mass of an element. Element Symbol Atomic Molar Number mass/ (g mol–1) Actinium Ac 89 227.03 Aluminium Al 13 26.98 Americium Am 95 (243) Antimony Sb 51 121.75 Argon Ar 18 39.95 Arsenic As 33 74.92 Astatine At 85 210 Barium Ba 56 The values given are molar mass, heat energy, kJ/mol and MJ/kg H2 = 2g/mol = 286 - 121 CH4 16g/mol - 889 - 50.0 C2H6 = 30g/mol = 1560 - 47.8 C3H8 The reason is that the molar mass of the substance affects the conversion. mol −1 (debug:usquare=4.36E-16; u=2.0880613017821E-8; smart_round=8.6802567553657); value=41 molecular weight of Au or 12 grams is equal to 1 mole of carbon-12, which has 6.02×10 23 atoms. The molar mass, also known as molecular weight, is the sum of the total mass in grams of all the atoms that make up a mole of a particular molecule. You can look up that answer from the table: 22.99 g. You may be wondering why the molar mass of sodium isn't just twice its , the sum Molar Mass Calculator Chemical Calculator Molar Mass Empirical Formula Degree of Unsaturation Unit Converter Lattice Energy d-d Spectrum d-d Spectrum (Jahn-Teller effect) Quadratic Equation Cubic Equation Quartic Equation Quintic Equation Radiation Dose Food Energy Requirements Crystal Structures Colour and Light Molecular Orbital Diagrams The mass of 1 mole of atoms of an element is known as the molar mass of the element and has the units grams per mole, g mol-1 Below is a table of some of the elements you will encounter during your chemistry course: A periodic table is necessary to complete the questions. The molar mass of a substance is the mass of one mole of the substance. Elementos. Type in unit We use the most common isotopes. grams Au to decimol To find this number, multiply the atomic mass by the molar mass constant, which is one gram per one mole because gold in its elemental form is a single atom. The molar mass is written $$M$$ and has the unit $$\text{g/mol}$$. Please enable Javascript area, mass, pressure, and other types. Molecular weight, mass calculations, Avogadro's number, and mole day teaching. Finding Molar Mass. La masa molar de un compuesto viene dada por la suma del peso atómico estándar (es decir, la masa atómica relativa estándar) de los átomos que forman el compuesto multiplicado por la constante de masa molar. Let me make it more clear with an example of sodium chloride. The major difference between molar mass and molecular mass is molar mass refers to mass of a mole of a substance whereas molecular mass refers to the mass of molecules. Thanks for the feedback, my program works for that now. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. The molar amount in question is approximately one-one thousandth (~10 −3) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g): In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass … The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. Help us build an awesome resource for HSC students during the Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. grams Au to nanomol symbols, abbreviations, or full names for units of length, Topic: Mole Concept Enter to win! … 0313 g/mol One mole of a substance is equal to a very large number, 6.023 x 10 23 of atoms (or molecules) which the substance is made of. A common request on this site is to convert grams to moles. You can do the reverse unit conversion from This is known as the molar mass, M, and has the units g mol-1 (grams per mole of substance) The relationship between molar mass, mass and moles can be expressed as a mathematical equation as shown below: g mol-1 = g ÷ mol . Molar mass calculator also displays common compound name, Hill formula, elemental composition, mass percent composition, atomic percent compositions and allows to convert from weight to number of moles and vice versa. The division sign (/) implies “per,” and “1” is implied in the denominator. inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, These relative weights computed from the chemical equation are sometimes called equation weights. It is recognized that there are advantages to the use of molar units such as the clear definition of the molecular entity for assay standardization and understanding of the molecular interactions Molar mass of Au = 196.96655 g/mol. Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. The RCPA has membership on the Working Party. conversion calculator for all types of measurement units. What is the molar mass Molar mass is closely related to the relative molar mass (M r) of a compound, to the older term formula weight (F.W. This collection of ten chemistry test questions deals with calculating and using molar masses. To complete this calculation, you have to know what substance you are trying to convert. Did you mean Au? grams Au to atom Finding molar mass starts with units of grams per mole (g/mol). The division sign (/) implies âper,â and â1â is implied in the denominator. grams Au to kilomol The reason is that the molar mass of the substance affects the conversion. Weight average molecular weights are often calculated from low angle light scattering analysis according to the relation KC/ÎRÎ¸ = 1/Mw + 2A2C, where A2 is the second virial coefficient, K the polymer optical constant, and ÎRÎ¸ the excess Rayleigh's ratio. molar not mass units for therapeutic drugs. La masa molar de los átomos de un elemento viene dada por la masa atómica relativa estándar del elemento multiplicada por la constante de masa molar, 1 × 10â3 kg/mol = 1 g/mol. • Australian Society of Clinical and Experimental Pharmacologists These relative weights computed from the chemical equation are sometimes called equation weights. Mass Average Molar Mass. Molar mass. The molar mass is â¦ The molar mass of sodium chloride is known; it is 58.44 g mol â1. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together. Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The molar mass of gold is 196.97 g/mol. as English units, currency, and other data. If we look at 1 atom of carbon-1, it has a mass of 12 amu. Molar Mass worksheet KEY Determine the molar mass for the following elements: Ca Xe K 40.08 g Ca 131.29 g Xe 39.10 g K Fe Au O 55.85 g Fe 196.97 g Au 16.00 g O Determine the molar mass for the following molecules/compounds La masa molar de un compuesto es calculada por la suma de los pesos atómicos estándar de los átomos que forman el mismo multiplicado por la constante de la masa molar (6,002×10 23). The SI base unit for amount of substance is the mole. Molar Mass is the mass of one mole of a substance (6.02 x 10 23 formula units). You can view more details on each measurement unit: Because molar mass is defined as the mass for 1 mol of a substance, we can refer to molar mass as grams per mole (g/mol). How many grams Au in 1 mol? This determines the molar mass for the entire compound. In the example we've been using, this tells us that because the FW of Example Reactions: â¢ Fe + Au(NO3)3 = Au + Fe(NO3)3. :: Chemistry Applications:: Chemical Elements, Periodic Table. The molar mass is the mass in grams of 1 mole of the substance. The unit used to measure is grams per mole. Use â¦ In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together. Determine the molar mass for the following elements: 1 mole of an element = the atomic mass of the element in grams. Molar mass is used to convert moles to grams. Finding molar mass starts with units of grams per mole (g/mol). 1 grams Au is equal to 0.0050770041918285 mole. Molar Mass Calculator Chemical Calculator Molar Mass Empirical Formula Degree of Unsaturation Unit Converter Lattice Energy d-d Spectrum d-d Spectrum (Jahn-Teller effect) Quadratic Equation Cubic Equation Quartic Equation Quintic Equation Radiation Dose Food Energy Requirements Crystal Structures Colour and Light Molecular Orbital Diagrams If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. Molar mass of Au is 196.9665690 ± 0.0000040 g/mol Compound name is gold Convert between Au weight and moles 1. Calculando el peso molecular (masa molecular) The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. For hydrogen chloride, the molar mass is 1.007 + 35.453 = 36.460 g/mol. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. Compound Name Formula Search. 1 grams Au is equal to 0.0050770041918285 mole. Convert grams Au to moles or moles Au to grams âºâº Percent composition by element Molar Mass: The mass of 1 mole of a substance in grams Also known as: formula mass (formula weight) or gram formula mass or molecular mass (molecular weight) 18.0 g/mol 2. Take the products you obtained in the previous step and add them all together to calculate the molar mass of the compound. grams Au to micromol. Molar mass calculator also displays common compound name, Hill formula, elemental composition, mass percent composition, atomic percent compositions and allows to convert from weight to number of moles and vice versa. of a compound Molar mass is an important physical property of substances. The answers appear after the final question. 36.46 grams is the mass of one mole of hydrogen chloride. the questions is my molality is 1.5 mol/1 kg i have a mass of 1.01 g how do i use that to find the molar mass (g/mol) The molar mass of something tells us how much one mole of that substance weighs. For carbon, we multiply its molar mass of 12.0107 grams per mol by two because we have two carbon atoms. This equals 24.0214 grams per mole. Molar masses (in grams) are therefore numerically equal to formula weights (in amu). GitHub Gist: instantly share code, notes, and snippets. Determine the molar mass for the following molecules/compounds: 1 mole of a compound = to the masses of each element times the number of atoms of each element combined. PLZ EXPLAIN HOW YOU GET THE ANSWER the molecular formula of aspartame , the artificial sweetener marketed as nutrasweet is C14H18N2O5 A) what is the molar mass of aspartame B) how many moles of aspartame are present in 1.00 mg of aspartame C) how many molecules of aspartame are present in1.00 mg of aspartame D) how many hydrogen atoms are present in 1.00 mg … This was predicated on drug prescribing becoming molar units (3). grams Au to millimol Note that rounding errors may occur, so always check the results. moles Au to grams, or enter other units to convert below: grams Au to centimol We have 6 carbon atoms, carbon has a mass of approximately 12 grams per mol from our periodic table we can note that, we have 12 hydrogen atoms hydrogen has approximately a mass of 1 gram for every mol of hydrogen, I'm rounding a lot here just bear with me and then I have 6 oxygen atoms and oxygen has a molar mass of approximately 16 grams per every mol. You can find metric conversion tables for SI units, as well Molar mass 1 mole of something is 6.022 × 10 23 pcs. 2 id. Molecular Weight is the molar mass of a COVALENT compound. The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. Au Gold atomic mass 196.9665694 1 atom 51.4298 % of total mass. of a chemical compound, More information on Type in your own numbers in the form to convert the units! Au(NO3)3. grams Au to molecule What is the mass of hydrogen gas that has twice the number of molecules as in 1.6 g of oxygen gas? This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. M = m ÷ n . Use this page to learn how to convert between grams Au and mole. name molecular formula relative molecular mass molar mass (g mol-1) mass of 1 mole (g) helium gas He 4.003 4.003 g mol-1 4.003 g oxygen gas O 2 2 × 16.00 = 32.00 32.00 g mol-1 32.00 g carbon dioxide gas CO 2 12.01 + (2 × 16 Converting between Mass and Number of Moles A substance’s molar mass can be used to convert between the mass of the substance and the number of moles in that substance. Therefore, the mass of gold, Au = number of moles of Au x molar mass of Au = 0.25 x 197 = 49.25 g 5. molecular weight of Au or mol. common chemical compounds. Por ejemplo, para calcular los gramos mol del ácido clorhídrico es necesario conocer los valores de los elementos que conforman este compuesto en la tabla periódica. Mass Average Molar Mass Weight average molecular weights are often calculated from low angle light scattering analysis according to the relation KC/ΔRθ = 1/Mw + 2A2C, where A2 is the second virial coefficient, K the polymer optical constant, and ΔRθ the excess Rayleigh's ratio. If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. Current reporting practice A review of practice in Australia has shown that the concentrations of some drugs are almost always reported in mass units (eg, gentamicin), while those of other drugs are always reported in molar units … Thus, the molar mass of [Relative atomic 2 -1 Molar Mass: 382.9813. Auric Cyanide. Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. The molar mass links the mass of a substance to its moles. metres squared, grams, moles, feet per second, and many more! ConvertUnits.com provides an online We use the most common isotopes. Finding Molar Mass This post will continue from the last post and go into a little more depth about molar mass for Prelim Chemistry. could someone explain how to convert something that is mol/kg to g/mol? Related Topics: More Lessons for Chemistry Math Worksheets Mole, Mass & Avogadro Constant An amount of substance containing 6.02 × 10 23 particles is called a mole (often abbreviated to mol). Gold(III) Cyanide Au(CN)3 Molar Mass, Molecular Weight. mol âºâº Au molecular weight. Thus, 1 mole of any element has a mass in grams that is numerically equivalent to its atomic mass. This is defined as 0.001 kilogram per mole, or 1 gram per mole. destiny cambero chem 111 farnum mtwr 9:40 am 08/15/18 experiment: 12 molar mass of volatile liquid conclusion: in conclusion, my unknown sample was the average Turnitin CHEM 111 separation of mixtures lab report CHEM 111 Carbonate Analysis current Expt 34 - lab report CHEM 111 Experiment 5 - lab report CHEM 111 experiment 8 lab report CHEM 111 expt. The answer is 196.96655. The molar mass (or mass per amount of substance) is easily calculated from the chemical formula of any pure compound from readily available tables of the molar masses of the elements. It will calculate the total mass along with the elemental composition and mass of each element in the compound. Molar Mass = Mass of the Substance (Kg)/Amount of Substance (Mol) Mole or mol is the unit used to measure the amount of a substance. This converts atomic units to grams per mole, making the molar mass of hydrogen 1.007 grams per mole, of carbon 12.0107 grams per mole, of oxygen 15.9994 grams per mole, and of chlorine 35.453 grams per mole. Calculate the molecular weight Multiply the relative atomic mass by the molar mass constant. Complete the questions is not the same as molecular mass, which has 6.02×10 23.. Moles Au to moles or moles Au to grams rounding errors may occur, so always check the results,! Gold atomic mass know more mass average molar mass is the mass a! Mass is a physical property of substances that describes the mass of one mole of that substance weight... The results one mole of the substance present 0.000016 g/mol convert between grams Au and mole of... ) Cyanide Au ( CN ) 3 molar mass is used to convert between Au weight moles... ( III ) Cyanide Au ( CN ) 3 molar mass of a COVALENT compound mole. 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Also established that 1 mol of Al has a mass of one of!, chemistry ( masa molecular ) Did you mean Au using their chemical formulas we assume you are trying convert! This page to learn how to calculate the molar mass ( note it is the mass of hydrogen,. Is numerically equivalent to its atomic mass of an element the molar mass is written (., chemistry mass links the mass of a substance divided by the of! And forensics which may also be called standard atomic weight or average atomic mass convert between grams and!, pressure, and other types to learn how to calculate the molar mass one! The following elements: 1 mole of the substance affects the conversion ± 0.0000040 g/mol compound name is convert. A substance is the mole is known ; it is the mass in grams ) are numerically! Entire compound and Technology with calculating and using molar masses can find metric conversion tables for units! 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The compound substance affects the conversion is how to convert grams to moles or moles Au to.. ± 0.0000040 g/mol compound name is gold convert between grams Au to grams the element and lowercase the! Names for units of grams per mole, or full names for units of all the atoms in a compound. Measure is grams per mole measure is grams per mole, or 1 gram per mole, or full for... Your own numbers in the element and lowercase for the following elements: 1 mole of that weighs... Of ten chemistry test questions deals with calculating and using molar masses ( in amu ) notes, snippets! H = 2 x au molar mass could someone explain how to calculate the molar mass of one mole an! 51.4298 % of total mass gold ( III ) Cyanide Au ( CN ) 3 mass... Sometimes called equation weights area, mass, molecular weight ), and mole calculating molecular is!, currency, and mole the division sign ( / ) implies “,! Isotropically weighted averages for SI units, as well as English units, as well as units... 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Can find metric conversion tables for SI units, as well as units! It tells us how much one mole of something tells us how many grams are in one mole of,! Use uppercase for the feedback, my program works for that now given mass of an element molar. Is au molar mass to g/mol much one mole of that substance weighs Al has a mass of mole. Of one mole of hydrogen gas that has twice the number of au molar mass contained in a compound! Equivalent to its au molar mass rounding errors may occur, so always check the results grams Au and mole teaching... / ) implies âper, â and â1â is implied in the denominator for of. Sometimes called equation weights teachscience, # teach, # molarmass, highschool... Me au molar mass it more clear with an example of sodium chloride is known ; it the... ( in amu ) compounds using their chemical formulas a mass of sodium metal the! Oxygen gas ( average molecular weight, mass, which is the molecular formula, the mass. 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# Do homodesmotic reactions have a closer matching of the hybridization states of the atoms of reactants and products as compared to isodesmic reactions?

Oct 4, 2015

Quick answer: $\textcolor{b l u e}{\text{yes}}$

#### Explanation:

An isodesmic reaction is a reaction that has the same number and type of bonds in the reactants and in the products.

An example of an isodesmic reaction is

${\text{CH"_3"CH"_2"CH"_2"CH"_3 + 2"CH"_4 → "3CH"_3"CH}}_{3}$

Each side has 3 $\text{C-C}$ bonds and 18 $\text{C-H}$ bonds.

Homodesmotic reactions are a subset of isodesmic reactions.

Here, the reactants and products must share even larger similarity.

Homodesmotic reactions are reactions in which reactants and products have

• equal numbers of each type of carbon-carbon bond ($s {p}^{3} - s {p}^{3} , s {p}^{3} - s {p}^{2} , \text{etc.}$)
• equal numbers of carbon atoms with the same hybridization and the same number of hydrogen atoms

The above reaction is not homodesmotic because the reactants have 3 $\text{C-C } s {p}^{3}$ bonds, 2 ${\text{CH}}_{4} \left(s {p}^{3}\right)$, 2 ${\text{CH}}_{3} \left(s {p}^{3}\right)$, and 2 ${\text{CH}}_{2} \left(s {p}^{3}\right)$ groups, while the products have 6 ${\text{CH}}_{3} \left(s {p}^{3}\right)$ groups.

The following reaction is both isodesmic and homodesmotic:

${\text{CH"_3"CH"_2"CH"_2"CH"_3 + "CH"_3"CH"_3 → 2"CH"_3"CH"_2"CH}}_{3}$

It is isodesmic because both reactants and products have 4 $\text{C-C}$ bonds and 16 $\text{C-H}$ bonds.

It is homodesmotic because both reactants and products have 4 $\text{C-C } s {p}^{3}$ bonds, 4 ${\text{CH}}_{3} \left(s {p}^{3}\right)$, and 2 ${\text{CH}}_{2} \left(s {p}^{3}\right)$ groups.

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Number - Subtraction

a - answer    s - solution    v - video    d - discussion

Question 1

Find the answer to the following questions:

a) $2-1$a s v d
b) $5-1$a s v d
c) $6-3$a s v d
d) $8-3$a s v d
e) $9-7$a s v d
f) $17-4$a s v d
g) $19-4$a s v d
h) $12-3$a s v d
i) $15-7$a s v d
j) $20-7$a s v d
k) $37-4$a s v d
l) $35-7$a s v d
m) $38-9$a s v d
n) $142-1$a s v d
o) $122-6$a s v d
p) $176-3$a s v d
q) $199-9$a s v d
r) $217-8$a s v d
s) $125-9$a s v d
t) $215632-8$a s v d

Question 2

Find the answer to the following questions:

a) $25-11$a s v d
b) $39-14$a s v d
c) $315-12$a s v d
d) $632-421$a s v d
e) $32-13$a s v d
f) $45-16$a s v d
g) $85-27$a s v d
h) $78-19$a s v d
i) $156-18$a s v d
j) $56-38$a s v d
k) $63-47$a s v d
l) $176-58$a s v d
m) $275-56$a s v d
n) $463-185$a s v d
o) $312-185$a s v d
p) $588-289$a s v d
q) $732-679$a s v d
r) $1890-897$a s v d
s) $204-96$a s v d
t) $1001-125$a s v d

Question 3

Find the answer to the following questions:

a) $8-3-2$a s v d
b) $15-5-3$a s v d
c) $8-0-7$a s v d
d) $17-7-4-6$a s v d
e) $59-14-23$a s v d
f) $63-21-35$a s v d
g) $94-29-38$a s v d
h) $158-32-17$a s v d
i) $275-88-16$a s v d
j) $532-123-11$a s v d
k) $912-125-622$a s v d
l) $124-32-19-22$a s v d
m) $184-55-55-55$a s v d
n) $1612-312-55$a s v d
o) $6123-1252-628$a s v d
p) $18523-531-621$a s v d
q) $23675-5125-124$a s v d
r) $12478-5123-632-231$a s v d
s) $21985-2531-5632-12-0$a s v d
t) $128532-0-5123-6312-12352$a s v d

Question 4

Find the answer to the following worded questions:

a) I have six apples at the beginning of the week. I ate two during the week. How many do I have left at the end of the week?a s v d

b) Harriet went shopping with £$20$. She bought a top for £$14$. How much does Harried have left?a s v d

c) An electric car has a range of $375$km. After driving $232$km, how much further can it drive before it is recharged?a s v d

d) What number needs to be added to $56$ to get $222$?a s v d

e) Kian bought a discount packet of $1000$ nails. He found that $383$ were bent. How many were not bent?a s v d

f) Michelle bought $60$ free range eggs. She lives in a tall building and after going up the first flght of stairs dropped $14$ of them. On the second flight of stairs she dropped a further $27$ eggs! How many does she have left?a s v d

g) I was paid a salary of £$2000$ this month. Out of my salary I was taxed £$632$ and I spent a further £$525$ on rent. How much do I have left?a s v d

h) Mr Smith sold his house for £$725\,750$. He paid £$6241$ to the estate agent, £$1273$ to the lawyers and bought another house for £$623\,124$. How much does Mr Smith have left?a s v d


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# How does one calculate the energy used depending on the velocity and air resistance of an object [closed]

I'm in need of a equation that can tell me how much energy/work I need to move an object inside a low air-pressure tube. The equation should (if possible) include drag, object velocity and mass. If there isn't such a equation i would also be happy if you could show me a way how to get to the calculate the work/energy needed.

## closed as unclear what you're asking by Aaron Stevens, Gert, ZeroTheHero, Buzz, Kyle KanosJan 24 at 11:04

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• At high Reynolds number, the drag force is usually modeled as: $F_D=\frac12 \rho v^2 C_A A$ – Gert Jan 22 at 17:03

If you know the magnitude of the drag force $$F_{\rm drag}(t)$$ and the speed of the object $$v(t)$$ then rate at which energy is being dissipated at a particular time $$t$$ is $$F_{\rm drag}(t) \,v(t)$$.
To evaluate the total energy dissipation from time $$t_{\rm start}$$ to time $$t_{\rm finish}$$ you will need to do an integration $$\displaystyle \int_{t_{\rm start}}^{t_{\rm finish}} F_{\rm drag}(t) \,v(t) \, dt$$


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# Asset allocation strategies based on penalized quantile regression

Giovanni Bonaccolto (), Massimiliano Caporin () and Sandra Paterlini ()
Additional contact information
Giovanni Bonaccolto: University “Kore” of Enna
Sandra Paterlini: European Business School

Computational Management Science, 2018, vol. 15, issue 1, 1-32

Abstract: Abstract It is well known that the quantile regression model used as an asset allocation tool minimizes the portfolio extreme risk whenever the attention is placed on the lower quantiles of the response variable. By considering the entire conditional distribution of the dependent variable, we show that it is possible to obtain further benefits by optimizing different risk and performance indicators. In particular, we introduce a risk-adjusted profitability measure, useful in evaluating financial portfolios from a ‘cautiously optimistic’ perspective, as the reward contribution is net of the most favorable outcomes. Moreover, as we consider large portfolios, we also cope with the dimensionality issue by introducing an $$\ell _1$$ ℓ 1 -norm penalty on the assets’ weights.

Keywords: Quantile regression; $$\ell _1$$ ℓ 1 -Norm penalty; Asset allocation (search for similar items in EconPapers)
Date: 2018
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Working Paper: Asset Allocation Strategies Based on Penalized Quantile Regression (2015)
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# functional equality of unions/intersections [duplicate]

I came across a statement which said that if we have a function $f:X\rightarrow Y$ where $A\subset X$ and $B\subset X$, then its true that $f(A\cup B)=f(A)\cup f(B)$, however its false that $f(A\cap B)=f(A)\cap f(B)$. I do not understand how this is so. Any help?

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## marked as duplicate by Martin Sleziak, user127.0.0.1, Claude Leibovici, Davide Giraudo, Sami Ben Romdhane Mar 9 '14 at 9:38

Another fact woth knowing is that the second equality $f(A\cap B)=f(A)\cap f(B)$ is true if $f$ is injective. – Martin Sleziak Oct 28 '11 at 10:02
See this question: math.stackexchange.com/questions/239783/… – Martin Sleziak Mar 9 '14 at 7:59

Remember that $A=B$ if and only if $x\in A\iff x\in B$, so to verify an equality of sets we need to see that the elements are on both sets, and to show that two sets are different we can just point out an element in one and not in the other.

To see that $f(A\cup B)=f(A)\cup f(B)$, note that:

\begin{align} f(x)\in f(A\cup B) &\iff x\in A\cup B\\ &\iff x\in A\text{ or } x\in B\\ &\iff f(x)\in f(A)\text{ or } f(x)\in f(B)\\ &\iff f(x)\in f(A)\cup f(B) \end{align}

On the other hand, if $f(x)\in f(A)\cap f(B)$ it just means that for some $a\in A$ we have $f(a)=f(x)$ and for some $b\in B$ we have $f(b)=f(x)$. We can use this to show that this is not the same set.

That is, we can find a function $f\colon X\to Y$, and two sets $A,B\subseteq X$ with some $x\in A$ and $x\notin B$, and $y\in B$ for which $y\notin A$ such that $f(x)=f(y)$. Now try to see why $f(A\cap B)\neq f(A)\cap f(B)$.

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@Srivatsan: Of course! Many thanks for catching that mistake! :-) – Asaf Karagila Oct 24 '11 at 0:36
It is not necessarily the case that $f(x)\in f(A) \implies x\in A$. For example here we could have $x$ coming from some other set $C$ which happens to satisfy $f(C)=f(A)=f(B)$. Therefore to prove $f(x)\in f(A)\cup f(B) \implies f(x)\in f(A\cup B)$ we should argue (wlog) $y\in f(A) \implies$ there exists $x\in A$ with $f(x)=y$. This $x$ certainly is in $A\cup B$. Hence $y=f(x)\in f(A\cup B)$. – massy255 Jul 3 '15 at 4:43

Well, to prove the first note that if $y\in f(A\cup B)$ then there exists $x\in A\cup B$ such that $f(y)=x$, but then it may happen that $x\in A$, and in this case $y\in f(A)$ or $x\in B$, and so $y\in f(B)$, in any case $y\in f(A)\cup f(B)$. Viceversa let $y\in f(A)\cup f(B)$. Then $y$ is in either $f(A)$ or in $f(B)$. Then there exists $x\in A\cup B$ such that $y=f(x)$ and so $y\in f(A\cup B)$.

To prove the other statement, note first that if $y\in f(A\cap B)$, then there exists $x\in A\cap B$, such that $y=f(x).$ Hence $y\in f(a)$ and $y\in f(B)$ so that $f(A\cap B)\subseteq f(A)\cap f(B)$.

To prove that the converse may not hold, consider $A=\{1\}$, $B=\{2\}$ and define $f(1)=f(2)=\{3\}.$ Then $f(A\cap B)=\emptyset$ while $f(A)\cap f(B)=\{3\}.$

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Ha! that's even simpler than the example I tried to get OP to construct in my answer! – Gerry Myerson Oct 24 '11 at 0:42

It should be easy for you to construct an example of sets $A$ and $B$, and a function $f$, such that $f(A\cap B)\ne f(A)\cap f(B)$. You can do it with $A$ and $B$ having 2 elements each, one in common. Try it!

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# Calculate the the displacement

1. Jan 7, 2014

### asymptotically

1. The problem statement, all variables and given/known data

Hi, the problem is to calculate the the displacement current flowing through a surface that lies in between the two plates of a parallel plate capacitor as the capacitor discharges. I think I have a correct solution but I'm not 100% sure on why it's correct.

2. Relevant equations

Gauss Law: $\oint_S E \cdot dA=\frac{Q_{enc}}{\epsilon_0}$
Definition of Displacement Current: $I_d=\epsilon_0\frac{d}{dt}\oint_S E \cdot dA$

3. The attempt at a solution

Constructing a Gaussian surface about one of the plates like this we have $\oint_S E \cdot dA=\frac{Q(t)_C}}{\epsilon_0}$.

Now this is the part I'm slightly confused about, the flux through every surface except the surface
that lies in between the plates has to be zero, why is there no flux through the left surface. Also we also only consider the charge on the capacitor and ignore the charge in the wire? Are these simplifications or is there justification for this.

The answer for $I_d$ is then just derivative of the rate of change of the charge on the capacitor which is the conduction current.

2. Jan 7, 2014

### maajdl

The total current is conserved, that's the meaning of the displacement current.
The displacement current outside the capacitor vanishes because the field vanishes.
The displacement current inside the capacitor is therefore the electric current in the wires.

The field outside the capacitor is the superposition of the field produced by opposite charges.
Therefore is is zero (or almost zero).

3. Jan 7, 2014

### asymptotically

Is my solution correct then?, what about the charge in the wire?

4. Jan 8, 2014

### maajdl

Yes the "Relevant equation" is correct, but it only helps you to prove you last sentence.
The wire has no charges, the charges accumulate on the plates (can you explain why?).
You still need to calculate Id=Ic!

5. Jan 8, 2014

### asymptotically

Nope, why do wires need to be covered in an insulator if they don't have a charge?

Na the problem just say's Ic=I at some time, what is the displacement current at this time, to calculate Ic as a function of t I assume I just use one of kirchoff's law and solve a Diff Eq?


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# The Host Galaxy of GRB 970508

### Abstract

We present late-time imaging and spectroscopic observations of the optical transient (OT) and host galaxy of GRB 970508. Imaging observations roughly 200 and 300 days after the burst provide unambiguous evidence for the flattening of the light curve. The spectroscopic observations reveal two persistent features that we identify with [O II] łambda3727 and [Ne III] łambda3869 at a redshift of z=0.835–the same redshift as the absorption system seen when the transient was bright. The OT was coincident with the underlying galaxy to better than 370 mas or a projected radial separation of less than 2.7 kpc. The luminosity of the [O II] line implies a minimum star formation rate of 1 M$_solar$ yr$^-1$. In our assumed cosmology, the implied rest-frame absolute magnitude isM$_B$=-18.55, or L$_B$=0.12 L$_*$. This object, the likely host of GRB 970508, can thus be characterized as an actively star-forming dwarf galaxy. The close spatial connection between this dwarf galaxy and the OT requires that at least some fraction of gamma-ray burst progenitors are not ejected in even the weakest galactic potentials.

Based on the observations obtained at the W. M. Keck Observatory which is operated by the California Association for Research in Astronomy, a scientific partnership among California Institute of Technology, the University of California and the National Aeronautics and Space Administration.

Publication
Astrophysical Journal Letters


text:
Werner's Blog — Opinion, Analysis, Commentary
Electricity conservation tariffs: how do they work?

For a number of years B.C. Hydro has been offering electricity tariffs to businesses that promote energy conservation. The General Service Business Rates that are currently in effect promote energy conservation by using locally-high marginal costs, which is tantamount to a feebate system where increases in electricity consumption are punished with an extra fee and decreases in electricity use are rewarded with a rebate. Businesses pay both a capacity charge per kilowatt (kW) maximum use and an energy charge per kilowatthour (kWh) actual use. The energy charge is modified to provide an incentive, customized to each electricity user, to reduce consumption. Here is how it works, and how to make a cost-benefit analysis for investing into energy-saving devices.

B.C. Hydro calculates a baseline for each consumer for each month as a rolling average of the last three years of consumption. In the illustration below, the averages of May 2013, May 2014, and May 2015 are used to calculate the baseline for May 2016. In part 1 of the tariff, the customer is charged for the baseline consumption regardless of the actual level of electricity consumption. The part 1 tariff consists of a higher rate up to 14,800 kWh (Tier 1), and a lower rate for consumption over 14,800 kWh (Tier 2). For the medium-sized customer, the Tier 1 rate is currently $0.1030/kWh, and the Tier 2 rate is$0.0719/kWh.

It gets more interesting in the next step. Let us assume that a customer has a baseline consumption of 30 MWh (i.e., 30,000 kWh), as shown by the red vertical line in the next diagram. B.C. Hydro now calculates a 20-percent band to each side, which ranges from 24 MWh to 36 MWh for a baseline of 30 MWh. The part 2 Tariff now sets a relatively high marginal rate of $0.1009/kWh for consumption exceeding the baseline and consumption below the baseline. A consumer who uses less than the baseline receives a rebate, while the consumer who uses more than the baseline pays a premium. As the diagram below shows, the marginal electricity rate—what you pay for the next kWh of consumption—jumps up from$71.90/MWh to $100.90/MWh when consumption falls within the 20-percent band around the baseline. What is the benefit of this jump in marginal rate? click on image for high-resolution PDF version Economic theory holds that consumers react to marginal rates. The energy conservation tariff locally raises the marginal rate, which means that the consumer should become more reactive to the price of electricity. Of course, if the electricity consumer does not change behaviour and consumption drifts randomly around the baseline, the locally-high marginal rate accomplishes nothing. However, the power-smart consumer will benefit if she invests in energy-saving equipment, and the higher marginal rate boosts the return on investment. Let us see how this calculation works. Let us assume that a business has indeed a 30 MWh baseline of electricity use based on the last three years of consumption. The business has an opportunity to invest$25,000 into a new heating system that will permanently reduce energy consumption by 10%, a 3 MWh per month reduction from 30 MWh to 27 MWh. In the first twelve months of using this technology, the 10% reduction yields (12 months) times (3 MWh) times ($110.9/MWh) equal$3,992.40. In the second year, the baseline adjusts from 30 MWh to (30+30+27)/3 = 29 MWh. So in the second year, the company gains another $2,661.60. In the third year the baseline drops to (30+27+27)/3= 28 MWh, and the business gains another$1,330.80. The electricity rebate is thus twice the amount of the first year alone, so $7,984.80. But this is not all the money that is saved; this was just the rebate. In perpetuity, the business saves (12 months) times (3 MWh) times ($71.90/MWh)= $2,588.40 per year. And in years 2 and 3, the shift in the baseline saves$826.80 and $1,725.60, respectively. Now we simply need to add together all these savings, and calculate the present-discounted value. For that purpose, a bit of math helps along the way. Let $$\tau_1$$ denote the base tariff per kWh, and let $$\tau_2>\tau_1$$ denote the conservation tariff per kWh. The electricity conservation that results from investing in equipment is valued at $$s$$ MWh. Further let $$\rho$$ denote the discount rate. Then, starting in year zero, the benefit is given by: $\mathcal{B}(\tau_1,\tau_2)=s\left[ \tau_2 +\frac{2\tau_2 +\tau_1}{1+\rho}+\frac{\tau_2+2\tau_1}{(1+\rho)^2} +\sum_{t=3}^\infty \frac{\tau_1}{(1+\rho)^t}\right]$ Simplifying this expression yields $\mathcal{B}(\tau_1,\tau_2)=s\left[\frac{\tau_1(\rho^2+3\rho+3)+\tau_2(3\rho^3+8\rho^2+6\rho)}{3\rho(1+\rho)^2}\right]$ Compare that to the case where $$\tau=\tau_1=\tau_2$$, and thus $\mathcal{B}(\tau_1)=s\cdot\tau_1\cdot\left(\frac{1+\rho}{\rho}\right)$ Here, the gains are simply the present-discounted value of $$\tau_1 s$$. As a rule of thumb, with an 8% discount rate, the present-value multiplier in the round parentheses is 13.5. Multiply the annual savings by that number, and you have the total benefit. With the electricity feebate the savings are larger. We can further calculate the ratio of the two benefits: $\mathcal{C}\equiv\frac{\mathcal{B}(\tau_1,\tau_2)}{\mathcal{B}(\tau_1)}= \frac{\rho^2+3\rho+3}{3(1+\rho)^3}+\frac{3\rho^3+8\rho^2+6\rho}{3(1+\rho)^3}\left(\frac{\tau_2}{\tau_1}\right)$ which turns out to be a function of the ratio $$\tau_2/\tau_1$$ and is independent of $$s$$. We can use the actual ratio$100.90/$71.90=1.4033 and a discount rate of 5% to calculate the increase in benefits. It turns out to be a 3.7% increase in benefits. If the discount rate was higher, say 10%, then the gains would increase to 6.9%. Essentially, the utility "subsidizes" the installation of energy efficiency devices at a rate of roughly 4-7%. Returning to our business, the 3 MWh energy conservation per month equals $$s=36$$ MWh per year. We allow for a discount rate of 8%, considering that a commercial entity has higher profit expectations than a household. Thus, $$\mathcal{B}=$$$36,930. This means that an investment of $25,000 will pay off: the benefits exceed the cost. You may wonder if a$25,000 investment is a realistic figure. The 3 MWh per month is equal to a constant load of 4.167 kW (=3,000 kWh divided by 720 hours per month). This means the capital cost calculates as \$6,000/kW. If you compare that to capital costs reported by the US Energy Information Administration in Updated Capital Cost Estimates for Utility Scale Electricity Generating Plants, April 2013, the number is roughly in the ballpark for new generation capacity (once you allow for the USD exchange rate).

The utility's problem is to figure out the optimal ratio $$\tau_2/\tau_1$$. This depends crucially on the the capital cost of bringing new generation capacity online. It would not make sense to pay much more for energy conservation than for providing new capacity. In other words, the ratio $$\mathcal{C}$$ above should not exceed the ratio of new generation cost to average generation cost; let us call that generation cost ratio $$\phi>1$$. If we continue assuming a discount rate of 8%, then we can solve the equation above for the ratio of $$\tau_2/\tau_1$$ as follows: $\tau_2/\tau_1\approx 7.0938\cdot\phi-6.0938$ So if the new generation cost is 20% higher than the average cost, then $$\phi=1.20$$ and thus $$\tau_2/\tau_1=2.42$$. The energy conservation tariff $$\tau_2$$ can be up to 2.42 as high as the base tariff $$\tau_1$$. Unfortunately, I do not know B.C. Hydro's ratio of new-to-old generation cost $$\phi$$, and of course one can argue the size of the discount rate. However, my sense is that B.C. Hydro has still room left to increase the $$\tau_2/\tau_1$$ ratio to make the energy conservation tariff more effective. This said, I applaud B.C. Hydro for using marginal cost pricing to incentivize energy conservation. This tool is particularly effective for businesses because: (i) they are more likely than households to overcome information problems (figuring out costs and benefits for energy conservation investments); (ii) the discount rate for businesses is generally assumed to be higher than for individuals; (iii) the electricity use profile of businesses tends to be much "flatter" than for households, which means that deviations from the baseline are more likely the result of business decisions than random "noise".

The remaining question is a research question: is there empirical evidence that energy conservation tariffs deliver the intended reductions in electricity use? If I had the data from B.C. Hydro, I would love to look into it.

Perhaps the biggest challenge with energy conservation tariffs is overcoming the information problem. Do businesses understand how this system works? Do they pay attention? Perhaps we need more energy consultants advising businesses on opportunities to save money through smart investments in energy conservation.

Posted on Sunday, May 22, 2016 at 12:00 — #Energy | #BC | #Economics

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# A JOINING CHARACTERIZATION OF FACTORS OF SIMPLE MAPS

• Ahn, Young-Ho (Department of Mathematics, Mokpo National University)
• Published : 2007.09.30

#### Abstract

Let T be an ergodic measure preserving transformation on (X, B, ${\mu}$). It is called 2-simple if every 2-fold ergodic joining is either a product measure or an off-diagonal measure. In general, factors of simple maps are not simple. So far, there has been no characterization of the factor of simple maps. In this paper, we give a joining characterization of factors of simple maps.


text:
location:  Publications → journals → CMB
Abstract view

# Exact Filling of Figures with the Derivatives of Smooth Mappings Between Banach Spaces

Published:2005-12-01
Printed: Dec 2005
• D. Azagra
• M. Fabian
• M. Jiménez-Sevilla
 Format: HTML LaTeX MathJax PDF PostScript

## Abstract

We establish sufficient conditions on the shape of a set $A$ included in the space $\mathcal L _s^n(X,Y)$ of the $n$-linear symmetric mappings between Banach spaces $X$ and $Y$, to ensure the existence of a $C^n$\nobreakdash-smooth mapping $f\colon X \rightarrow Y$, with bounded support, and such that $f^{(n)}(X)=A$, provided that $X$ admits a $C^{n}$-smooth bump with bounded $n$-th derivative and $\dens X=\dens \mathcal L ^n(X,Y)$. For instance, when $X$ is infinite-dimensional, every bounded connected and open set $U$ containing the origin is the range of the $n$-th derivative of such a mapping. The same holds true for the closure of $U$, provided that every point in the boundary of $U$ is the end point of a path within $U$. In the finite-dimensional case, more restrictive conditions are required. We also study the Fr\'echet smooth case for mappings from $\mathbb R^n$ to a separable infinite-dimensional Banach space and the G\^ateaux smooth case for mappings defined on a separable infinite-dimensional Banach space and with values in a separable Banach space.
 MSC Classifications: 46B20 - Geometry and structure of normed linear spaces

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Integral model of a reductive group over a prime field

Let $p$ be a rational prime, $\mathbb{Z}_p$ the ring of $p$-adic integers, and $k$ an algebraic closure of the residue field $\mathbb{F}_p$. Suppose $G$ is an affine smooth group scheme over $\mathrm{Spec}~\mathbb{F}_p$ such that $G_k$ is a connected reductive algebraic group.

Then, is there an affine smooth group scheme $\mathbf{G}$ over $\mathrm{Spec}~\mathbb{Z}_p$ such that the geometric fibres of $\mathbf{G}$ are connected reductive algebraic groups and such that $\mathbf{G}_{\mathbb{F}_p}=G$?

I am also interested in the case that $\mathbb{Z}_p$ is replaced by a general complete discrete valuation ring of characteristic zero and $\mathbb{F}_p$ is replaced by the corresponding residue field. Any reference would be appreciated.

(The below statement is not true; see the comments.)

I learnt from Conrad's notes http://math.stanford.edu/~conrad/papers/redgpZsmf.pdf that if $G$ can be defined over $\mathbb{Z}$ then it is necessarily split, and the reason is roughly that $\mathbb{Q}$ admits no non-trivial unramified extension. But here we only concern $\mathbb{Z}_p$.

• The statement at the end is not true (so perhaps you misread something in that link, such as Remark 5.1.5): it has to be assumed that the $\mathbf{Q}$-fiber is split. Indeed, there do exist non-split reductive group schemes over $\mathbf{Z}$, such as special orthogonal groups of unimodular positive-definite quadratic lattices of even rank $\ge 4$ and even more, as discussed in math.stanford.edu/~conrad/papers/redgpZsmf.pdf (see Theorem 1.4 and then Examples 2.9 and 2.10 and the "Summary" at the end of section 2 there for first steps). Nov 21 '16 at 17:32
• @nfdc23 you are right, editted. Nov 21 '16 at 17:49

Yes, for any henselian local (e.g., complete local noetherian) ring $A$ with residue field $\kappa$ and any connected reductive $\kappa$-group $G$, there exists a connected reductive $A$-group scheme $\mathbf{G}$ with special fiber $G$. Note that your hypothesis on the $\mathbf{F}_p$-group just says that this $\mathbf{F}_p$-group is connected reductive (there is no need to state a condition separately on its geometric fiber over $k$).

To prove this, it seems simplest to use the crutch of the existence of Chevalley groups over $\mathbf{Z}$. More specifically, there does exist a split reductive $A$-group $\mathbf{H}$ such that $H := \mathbf{H}_{\kappa}$ is a Galois-twisted form of $G$. Over any field at all, connected reductive groups always admit a (unique up to isomorphism) quasi-split inner form. Let $G'$ denote the quasi-split inner form of $G$, so $G'$ is a quasi-split form of the split $\kappa$-group $H$.

We will first solve the lifting problem for $G'$, and then solve it for $G$. Put another way, we will first solve it when $G$ is quasi-split, and then will have to address the behavior under passage to inner forms. Handling the latter step will be more serious, involving automorphism schemes of reductive groups over rings, so it is worth noting that when $\kappa$ is finite or separably closed then inner forms are always trivial and hence $G$ is always quasi-split. Thus, for such special $\kappa$ the more technically difficult task of going beyond the quasi-split case does not arise.

How does one make quasi-split forms of a split connected reductive group? These are classified (up to isomorphism) by degree-1 Galois cohomology valued in the automorphism group of the associated based root datum $(R, \Delta)$. In other words, the pointed set ${\rm{H}}^1(\kappa, {\rm{Aut}}(R, \Delta))$ of conjugacy classes of continuous maps ${\rm{Gal}}(\kappa_s/\kappa) \rightarrow {\rm{Aut}}(R, \Delta)$ classifies all such possibilities. In concrete terms, the Galois-twisting of $H$ to create the quasi-split $G$ is given through "pinned automorphisms". But the notion of "pinned automorphism" makes sense for any pinned split reductive group over any base, and in particular we can choose a pinning of $\mathbf{H}$ over $A$ to make ${\rm{Aut}}(R, \Delta)$ act on $\mathbf{H}$.

Thus, if we fix a finite Galois extension $\kappa'/\kappa$ so that $G$ is made from $H$ through twisting by a homomorphism $c: {\rm{Gal}}(\kappa'/\kappa) \rightarrow {\rm{Aut}}(R, \Delta)$ and view $c$ as a function on ${\rm{Aut}}(A'/A)$ for the (unique) local finite etale $A$-algebra $A'$ with residue field $\kappa'/\kappa$ then we can carry out the same "Galois-twisting" directly over $A'$ (using "Galois descent" relative to the finite etale cover ${\rm{Spec}}(A') \rightarrow {\rm{Spec}}(A)$) to make an $A$-form $\mathbf{G}$ of $\mathbf{H}$ with special fiber $G$. We are now done if $\kappa$ is finite or separably closed.

For the general case, we now have a reductive $A$-group scheme $\mathbf{G}'$ whose reduction $G'$ is an inner form of $G$, and we seek to make a reductive $A$-group $\mathbf{G}$ with reduction $G$; the split $\mathbf{H}$ and $H$ have done their work above and will no longer be used.

By the very definition of inner forms, $G$ is obtained from $G'$ through twisting against a cohomology class in ${\rm{H}}^1(\kappa, {G'}^{\rm{ad}})$ where ${G'}^{\rm{ad}} := G'/Z_{G'}$ is the adjoint semisimple central quotient of $G'$. The natural map $$f:{\rm{H}}^1(\kappa, {G'}^{\rm{ad}}) \rightarrow {\rm{H}}^1(\kappa, {\rm{Aut}}_{G'/\kappa})$$ into the pointed set of isomorphism classes of forms of $G'$ is generally not injective. Rather, by Proposition 39(ii) in section 5.5 of Chapter I of Serre's book "Galois cohomology" applied to the short exact sequence of $\kappa_s$-point arising from the short exact sequence of smooth $\kappa$-groups $$1 \rightarrow {G'}^{\rm{ad}} \rightarrow {\rm{Aut}}_{G'/\kappa} \rightarrow E_{G'} \rightarrow 1$$ (with $E_{G'}$ the etale component group of the automorphism scheme of $G'$), there is a natural action by $E_{G'}(\kappa)$ on ${\rm{H}}^1(\kappa, {G'}^{\rm{ad}})$ and the $f$-fibers are the $E_{G'}(\kappa)$-orbits. By using more work with the etale topology over rings in place of Galois cohomology over fields, we can define the separated etale $A$-group $E_{\mathbf{G}'}$ similarly using the automorphism scheme ${\rm{Aut}}_{\mathbf{G}'/A}$. This reduces our task to showing that the pointed set of torsors ${\rm{H}}^1(A, {\mathbf{G}'}^{\rm{ad}})$ maps surjectively to ${\rm{H}}^1(\kappa, {G'}^{\rm{ad}})$ and that $E_{\mathbf{G}'}(A) \rightarrow E_{G'}(\kappa)$ is surjective too.

Since $A$ is henselian, for any smooth $A$-scheme $X$ the natural map $X(A) \rightarrow X(k)$ is surjective due to the Zariski-local structure of smooth maps and the various ways of characterizing henselian local rings; see [EGA, IV$_4$, 18.5.17]. Thus, setting $X = E_{\mathbf{G}'}$ settles our second surjectivity task. It remains to prove surjectivity of ${\rm{H}}^1(A, {\mathbf{G}'}^{\rm{ad}}) \rightarrow {\rm{H}}^1(\kappa, {G'}^{\rm{ad}})$. For the smooth affine $A$-group $\mathscr{G} := {\mathbf{G}'}^{\rm{ad}}$, the vector bundle $\mathfrak{g} := {\rm{Lie}}(\mathscr{G})$ over $A$ is a finite free $A$-module (since $A$ is local) and the adjoint representation ${\rm{Ad}}_{\mathscr{G}}: \mathscr{G} \rightarrow {\rm{GL}}(\mathfrak{g}) = {\rm{GL}}_N$ has trivial kernel.

[The difficult Theorem 5.3.5 in the linked article on reductive group schemes in the question posed shows that this monomorphism is a closed immersion. Likewise, further arguments show that the associated coset sheaf for the adjoint representation of any adjoint semisimple group over any scheme is represented by an affine scheme, necessarily smooth over the base -- the key point is to use effective descent for affineness to work etale-locally on the base to reduce to working with split groups and thereby reduce to working over $\mathbf{Z}$ that is Dedekind, to which Theorem C.2.5 in the PhD thesis of Brandon Levin is applicable. We will not need either of these refinements.]

Now comes the trick to turn our H$^1$-surjectivity problem into a more tangible "geometric" lifting problem that makes contact with the henselian property of $A$. Consider the $\mathscr{G}$-torsor $${\rm{GL}}(\mathfrak{g}) \rightarrow Z := {\rm{GL}}(\mathfrak{g})/\mathscr{G}$$ where $Z$ is a sheaf for the fppf (or etale) topology over $A$. This defines a natural map $Z(A) \rightarrow {\rm{H}}^1(A, {\mathbf{G}'}^{\rm{ad}})$ compatible with the natural map $Z(\kappa) \rightarrow {\rm{H}}^1(\kappa, {G'}^{\rm{ad}})$.
The vanishing of ${\rm{H}}^1(\kappa, {\rm{GL}}_N)$ implies that the map $Z(\kappa) \rightarrow {\rm{H}}^1(\kappa, {G'}^{\rm{ad}})$ is surjective (!), so our surjectivity problem for ${\rm{H}}^1$'s reduces to proving the surjectivity of the natural map $Z(A) \rightarrow Z(\kappa)$. If $Z$ were known to be a smooth $A$-scheme then we would be done again by [EGA, IV$_4$, 18.5.17] since $A$ is henselian. But $Z$ seems to be "just" a quotient sheaf, without evident geometric substance, so what can we do?

We saw above that $Z$ is in fact a smooth affine $A$-scheme, but this isn't needed: it would suffice to know that $Z$ is a quasi-separated algebraic space (then necessarily smooth over $A$), as that provides enough "geometric structure" on $Z$ to apply the same EGA result with the aid of the henselian local ring at a given $\kappa$-point of $Z$ (which makes sense for quasi-separated algebraic spaces). The definition of $Z$ as an fppf quotient sheaf associated to the equivalence relation defined by the free action arising from translation by an inclusion between finite type flat $A$-group schemes implies that $Z$ is an algebraic space by a deep theorem of Artin concerning quotients by fppf equivalence relations for schemes. This thereby settles the case of general residue fields $\kappa$.


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# sigma notation algerbra 2

0
139
2

use sigma notation to write the sum. 4,7,10,13,16

Guest Feb 28, 2017
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#1
0

$$\sum_{n=0}^{4}[4 + 3n]=50$$

Guest Feb 28, 2017
#2
+19096
+5

use sigma notation to write the sum. 4,7,10,13,16

$$\begin{array}{|rcll|} \hline \sum \limits_{n=1}^{5} 4+(n-1)\cdot 3 = \sum \limits_{n=1}^{5} 1+3n \\ \hline \end{array}$$

heureka  Mar 1, 2017

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text:
# centre of gravity

• Dec 8th 2008, 02:23 PM
bandy1989
centre of gravity
have managed to work out th centre of gravity by splitting the shape up into a rectagle and triangle
but i dont know how to combine them to get a c.o.g for he full shape.
the informaion is attatched.
• Dec 8th 2008, 03:21 PM
Skalkaz
If we have two mass point than the centre of gravity is the weighted mean:

$\vec r = (m_1*\vec {r_1} + m_2*\vec {r_2}) : (m_1 + m_2)$


text:
# How to do MC integration from gibbs sampling of posterior

I'm a beginner of MCMC. Two questions confused me.

If I know the posterior distribution, and from the Gibbs sampling, I got the sampled parameter, so

1. How to draw the histogram with y axis as marginal density in R ?
2. How to further apply Monte Carlo integration on the resulting Markov Chain, which are sampled from this posterior. It seems I couldn't get a connection between these two.

Could anyone give an example to explain this?

Thank you

First I presume that

If I know the posterior distribution, and from the Gibbs sampling, I got the sampled parameter

simply means that you can compute $\pi(\theta|x)$ up to a constant and derive a Markov chain $\theta¹,\theta²,\ldots,\theta^T$ that converges to $\pi(\theta|x)$.

1. How to draw the histogram with y axis as marginal density in R ?

is unclear. You can draw the histogram of your simulations, one component at a time, but you do not know the marginal density $\pi_1(\theta_1|x)$ in most cases. You can always produce and display an estimate of this marginal density, e.g., by Rao-Blackwellisation:$$\hat{\pi}_1(\theta_1|x)=\frac{1}{T}\sum_{t=1}^T \pi(\theta_1|\theta_{-1}^t,x)$$where the conditional density on the right represents the full conditional of $\theta_1$ given all other components that you used in Gibbs sampling. But this display cannot be used as a proof of convergence as the histogram _and_ the alternative density estimate are based on the same simulation.

In Example 7.2 of our book Introducing Monte Carlo methods with R, we consider the pair of distributions $$X\vert\theta \sim \mathcal{B}\text{in}(n,\theta)\,,\quad \theta \sim {\mathcal{B}}e(a,b)$$ (i.e., the joint distribution) as our target (this is a toy example where there is no observation, no parameter, no posterior, despite the possibly confusing notations of $X$ and $\theta$: both are random variables to be simulated there) and this leads to the joint distribution $$f(x,\theta) = {n \choose x} \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\, \theta^{x+a-1} (1-\theta)^{n-x+b-1}$$ that we can simulate by Gibbs sampling since the corresponding conditional distribution of $X \vert \theta$ is given above, while $\theta \vert x \sim {\cal B}e(x+a,n-x+b)$. The histograms in the associated figure (Fig. 7.1) are constructed by Gibbs sampling, while the overlaid curves are the true marginal densities, because this is a toy problem where the exact marginals are known to be a Beta-Binomial for $X$ and a Beta for $\theta$. This hardly ever happens in practical situations.

[Reproduced from Introducing Monte Carlo methods with R]

And

1. How to further apply Monte Carlo integration on the resulting Markov Chain, which are sampled from this posterior. It seems I couldn't get a connection between these two.

is also unclear. If you mean producing an approximation of a posterior expectation $\mathbb{E}[\mathfrak{h}(\theta)|x]$, you simply apply the regular Monte Carlo formula: $$\mathbb{E}[\mathfrak{h}(\theta)|x]\approx\frac{1}{T}\sum_{t=1}^T \mathfrak{h}(\theta^t)$$

• Thank you @Xi'an . For the first questions, I'm referring the Example of 7.2 in the book of Introducing Monte Carlo Methods with R by Robert and Casella. In that example, they simply using Gibbs Sampler to draw sample from a beta-binomial distribution, and plot a marginal density histogram. Quite confusing about that. – rifle123 Nov 3 '16 at 2:28
• @Xi'an I'm a little puzzled by "you can at best produce an estimate of this marginal density by Rao-Blackwellisation" right after discussing the histogram -- wouldn't the histogram of the sampled parameter itself be an estimate of the marginal density? [It's not as efficient as a Rao-Blackwellized estimate, of course.] Possibly I misunderstood, however -- were you saying that the histogram of the sampled values isn't such an estimate or were you saying that it may not be a good estimate? In either case are there some conditions that hold when it works (or perhaps when it doesn't)? – Glen_b Nov 3 '16 at 5:45
• @Glen_b: sorry to confuse everyone with the book example behind this question and with this answer!!! The example utilises the true marginal distributions to show that Gibbs converges to the right limiting distributions. In realistic situations, this is not possible. Hence, neither is plotting the true target. When drawing the marginal density based on the MCMC sample, the checking aspect disappears, whether one uses histogram, kernel, or RB estimates. – Xi'an Nov 3 '16 at 7:03
• @Xi'an Thank you. Actually, I have one more confusion about the the full conditional equation of Example 7.3 just below the marginal density figure. For the example 7.3, when you are generating the full conditional Eqn. 7.4 from the posterior Eqn.7.3, I'm curious about it seems you have modified the prior distribution of theta in Eqn. 7.3 by multiplying some form of sigma in it. My understanding for full conditional of , e.g. theta, should drop out the second exponential in Eqn. 7.4. – rifle123 Nov 5 '16 at 22:50
• First there is a typo in Eqn. (7.3), the last term should be $e^{\mathbf{-}1/b\sigma^2}$. Second, there is indeed a mistake in the model: we should have used the prior $\theta\sim\mathcal{N}(\theta_0,\tau^2\sigma^2)$ instead of the one we propose at the bottom of page 202, in order to make the example easier [and Eqn. (7.4)]. Apologies for the confusion! If you put this as another question, I can provide the complete correction to Example 7.3. – Xi'an Nov 6 '16 at 11:23


text:
It's possible to solve this problem in $O(n)$?

Suppose given two strings $$s_1$$ and $$s_2$$ each of length $$n$$.

We consider $$s_1$$ and $$s_2$$ as dual, if one of the following conditions satisfied:

1. Strings are equal.

2. Else, if $$n$$ is even then:

Divide $$s_1$$ into two equal partitions $$a_1,a_2$$, also divide $$s_2$$ into $$b_1,b_2$$ then:

• $$a_1$$ with $$b_1$$ and $$a_2$$ with $$b_2$$ are dual.
• Or $$a_1$$ with $$b_2$$ and $$a_2$$ with $$b_1$$ are dual.

The description above implies a recursive algorithm in $$O(n\log n)$$. But my question is that, is it possible to solve the problem in $$O(n)$$?

• "Divide $s_1$ into two equal partition $a_1$, $a_2$" - does it mean that the length of $s_1$ is even and $a_1$ is the first half of $s_1$ and $a_2$ is the second half of $s_1$? How can it be solved in $O(n \log n)$? I see $O(n^2)$ algorithm. where running time is given by recurrence $T(n) = 4 T(\frac n2) + O(1)$. Nov 17 '21 at 9:22
• You can probably solve it in $O(n)$ by using hashes: for string $s$, which consist of two parts $\alpha_1$ and $\alpha_2$, you can recursively compute $h(s) = h'(h(\alpha_1), h(\alpha_2))$, where $h'$ is some symmetric hash-function. If you properly select $h'$, it should be correct with high probability (it can only result if false-positive, and you can always check it). Nov 17 '21 at 10:39

This problem can be formulated as a rooted tree isomorphism problem. Let $$k$$ be the maximum integer such that $$2^k$$ divides $$n$$. Then, each input string represents a full binary tree of height $$k + 1$$. Each leaf $$i$$ has the label $$s[i n/2^k, (i+1)n/2^k-1]$$ while internal nodes are unlabeled. Two trees are isomorphic if and only if the strings are dual.

Assume the alphabets of the strings can be represented as an integer less than $$O(n^c)$$ for constant $$c$$. Then, the bottom-up tree canonization algorithm of Aho-Hopcroft-Ullman solves the tree isomorphism problem in $$\Theta(n)$$ time in the word-RAM model. Explanation of the algorithm can be found in e.g. this lecture slide https://www.cs.uu.nl/docs/vakken/an/an-isomorphism-2016.pdf.

Because the trees are binary in this case, the algorithm can be simplified. Also, the algorithm can be implemented without integer sorting algorithms if a hash table is used, though the runtime bound becomes an expected time theoretically. Pseudocode is:

labels = {} # a hash table

def canonicalize(s, depth, i):
if depth == k:
return s[i] # Assumes n == 2^k for simplicity

l = canonicalize(s, depth + 1, i * 2)
r = canonicalize(s, depth + 1, i * 2 + 1)
if l > r:
l, r = r, l

if (l, r) not in labels:
labels[(l, r)] = labels.len()

return labels[(l, r)]

label1 = canonicalize(s_1, 0, 0)
label2 = canonicalize(s_2, 0, 0)
print(label1 == label2)
$$$$
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