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# Positive Definite

URGENT: Can you prove or disprove:

Let A and B be (complex matrices) positive definite with trace 1.

Given A < B, (B-A is pos def )

then

A^2 < AB (AB-A^2 is pos def)

CompuChip
Homework Helper
Maybe you can use that
A B - A^2 = A(B - A)
and from the result that
One thing that is true is this: if A and B are hermitian (or real
symmetric) with all their eigenvalues in [0, a] and [0, b]
respectively, then A B has all its eigenvalues in [0, a b].

A or B might not be real symmetric.

This is not homework! But I guess this section will get more viewers.

morphism
Homework Helper
Usually self-adjointness is included in any notion of positivity for complex operators. How are you defining "positive definite" for a complex matrix A?

A matrix M such that for all vectors v, <v, Mv> (inner product, the usual one for complex vector spaces) is a real, positive number.

morphism


text:
Carbon-12

Carbon-12 (12C) is the more abundant of the two stable isotopes of carbon (Carbon-13 being the other), amounting to 98.93% of the element carbon;[1] its abundance is due to the triple-alpha process by which it is created in stars. Carbon-12 is of particular importance in its use as the standard from which atomic masses of all nuclides are measured, thus, its atomic mass is exactly 12 daltons by definition. Carbon-12 is composed of 6 protons 6 neutrons and 6 electrons.

Carbon-12,  12C
General
Name, symbolCarbon,12C
Neutrons6
Protons6
Nuclide data
Natural abundance98.93%
Parent isotopes12N
12B
Isotope mass12 u
Spin0
Excess energy0± 0 keV
Binding energy92161.753± 0.014 keV
Complete table of nuclides

History

Before 1959 both the IUPAP and IUPAC used oxygen to define the mole; the chemists defining the mole as the number of atoms of oxygen which had mass 16 g, the physicists using a similar definition but with the oxygen-16 isotope only. The two organizations agreed in 1959/60 to define the mole as follows.

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 12 gram of carbon 12; its symbol is "mol".

This was adopted by the CIPM (International Committee for Weights and Measures) in 1967, and in 1971 it was adopted by the 14th CGPM (General Conference on Weights and Measures).

In 1961 the isotope carbon-12 was selected to replace oxygen as the standard relative to which the atomic weights of all the other elements are measured.[2]

In 1980 the CIPM clarified the above definition, defining that the carbon-12 atoms are unbound and in their ground state.

Hoyle state

The Hoyle state is an excited, spinless, resonant state of carbon-12. It is produced via the triple-alpha process, and was predicted to exist by Fred Hoyle in 1954.[3] The existence of the 7.7 MeV resonance Hoyle state is essential for the nucleosynthesis of carbon in helium-burning red giant stars, and predicts an amount of carbon production in a stellar environment which matches observations. The existence of the Hoyle state has been confirmed experimentally, but its precise properties are still being investigated.[4] In 2011, an ab initio calculation of the low-lying states of carbon-12 found (in addition to the ground and excited spin-2 state) a resonance with all of the properties of the Hoyle state.[5][6]

Isotopic purification

The isotopes of carbon can be separated in the form of carbon dioxide gas by cascaded chemical exchange reactions with amine carbamate.[7]

References

1. ^ "Table of Isotopic Masses and Natural Abundances" (PDF). 1999.
2. ^
3. ^ Hoyle, F. (1954). "On Nuclear Reactions Occurring in Very Hot Stars. I. the Synthesis of Elements from Carbon to Nickel". The Astrophysical Journal Supplement Series. 1: 121. Bibcode:1954ApJS....1..121H. doi:10.1086/190005. ISSN 0067-0049.
4. ^ Chernykh, M.; Feldmeier, H.; Neff, T.; Von Neumann-Cosel, P.; Richter, A. (2007). "Structure of the Hoyle State in C12" (PDF). Physical Review Letters. 98 (3): 032501. Bibcode:2007PhRvL..98c2501C. doi:10.1103/PhysRevLett.98.032501. PMID 17358679.
5. ^ Epelbaum, E.; Krebs, H.; Lee, D.; Meißner, U.-G. (2011). "Ab Initio Calculation of the Hoyle State" (PDF). Physical Review Letters. 106 (19): 192501. arXiv:1101.2547. Bibcode:2011PhRvL.106s2501E. doi:10.1103/PhysRevLett.106.192501. PMID 21668146.
6. ^ Hjorth-Jensen, M. (2011). "Viewpoint: The carbon challenge". Physics. 4: 38. Bibcode:2011PhyOJ...4...38H. doi:10.1103/Physics.4.38.
7. ^ Kenji Takeshita and Masaru Ishidaa (December 2006). "Optimum design of multi-stage isotope separation process by exergy analysis". ECOS 2004 - 17th International Conference on Efficiency, Costs, Optimization, Simulation, and Environmental Impact of Energy on Process Systems. 31 (15): 3097–3107. doi:10.1016/j.energy.2006.04.002.
 Lighter: carbon-11 Carbon-12 is an isotope of carbon Heavier: carbon-13 Decay product of: boron-12, nitrogen-12 Decay chain of carbon-12 Decays to: stable
Art in Dubai

Art in Dubai is an emerging activity in Dubai, United Arab Emirates. New galleries such as Carbon 12 Dubai, art fairs, artists, art patrons and collectors have grown in number.

Atom

An atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers (a ten-billionth of a meter, in the short scale).

Atoms are small enough that attempting to predict their behavior using classical physics – as if they were billiard balls, for example – gives noticeably incorrect predictions due to quantum effects. Through the development of physics, atomic models have incorporated quantum principles to better explain and predict this behavior.

Every atom is composed of a nucleus and one or more electrons bound to the nucleus. The nucleus is made of one or more protons and typically a similar number of neutrons. Protons and neutrons are called nucleons. More than 99.94% of an atom's mass is in the nucleus. The protons have a positive electric charge, the electrons have a negative electric charge, and the neutrons have no electric charge. If the number of protons and electrons are equal, that atom is electrically neutral. If an atom has more or fewer electrons than protons, then it has an overall negative or positive charge, respectively, and it is called an ion.

The electrons of an atom are attracted to the protons in an atomic nucleus by this electromagnetic force. The protons and neutrons in the nucleus are attracted to each other by a different force, the nuclear force, which is usually stronger than the electromagnetic force repelling the positively charged protons from one another. Under certain circumstances, the repelling electromagnetic force becomes stronger than the nuclear force, and nucleons can be ejected from the nucleus, leaving behind a different element: nuclear decay resulting in nuclear transmutation.

The number of protons in the nucleus defines to what chemical element the atom belongs: for example, all copper atoms contain 29 protons. The number of neutrons defines the isotope of the element. The number of electrons influences the magnetic properties of an atom. Atoms can attach to one or more other atoms by chemical bonds to form chemical compounds such as molecules. The ability of atoms to associate and dissociate is responsible for most of the physical changes observed in nature and is the subject of the discipline of chemistry.

Atomic mass

The atomic mass (ma) is the mass of an atom. Its unit is the unified atomic mass units (abbr. u) where 1 unified atomic mass unit is defined as 1⁄12 of the mass of a single carbon-12 atom, at rest. For atoms, the protons and neutrons of the nucleus account for nearly all of the total mass, and the atomic mass measured in u has nearly the same value as the mass number.

When divided by unified atomic mass units, or daltons (abbr. Da), to form a pure numeric ratio, the atomic mass of an atom becomes a dimensionless value called the relative isotopic mass (see section below). Thus, the atomic mass of a carbon-12 atom is 12 u (or 12 Da), but the relative isotopic mass of a carbon-12 atom is simply 12.

The atomic mass or relative isotopic mass refers to the mass of a single particle, and therefore is tied to a certain specific isotope of an element. The dimensionless standard atomic weight instead refers to the average (mathematical mean) of atomic mass values of a typical naturally-occurring mixture of isotopes for a sample of an element. Atomic mass values are thus commonly reported to many more significant figures than atomic weights. Standard atomic weight is related to atomic mass by the abundance ranking of isotopes for each element. It is usually about the same value as the atomic mass of the most abundant isotope, other than what looks like (but is not actually) a rounding difference.

The atomic mass of atoms, ions, or atomic nuclei is slightly less than the sum of the masses of their constituent protons, neutrons, and electrons, due to binding energy mass loss (as per E = mc2).

Atomic mass unit

The unified atomic mass unit or dalton (symbol: u, or Da or AMU) is a standard unit of mass that quantifies mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1 g/mol. It is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state and at rest, and has a value of 1.660539040(20)×10−27 kg, or approximately 1.66 yoctograms. The CIPM has categorised it as a non-SI unit accepted for use with the SI, and whose value in SI units must be obtained experimentally.The atomic mass unit (amu) without the "unified" prefix is technically an obsolete unit based on oxygen, which was replaced in 1961. However, many sources still use the term amu but now define it in the same way as u (i.e., based on carbon-12). In this sense, most uses of the terms atomic mass units and amu, today, actually refer to unified atomic mass unit. For standardization, a specific atomic nucleus (carbon-12 vs. oxygen-16) had to be chosen because the average mass of a nucleon depends on the count of the nucleons in the atomic nucleus due to mass defect. This is also why the mass of a proton or neutron by itself is more than (and not equal to) 1 u.

The atomic mass unit is not the unit of mass in the atomic units system, which is rather the electron rest mass (me).

Until the 2019 redefinition of SI base units, the number of daltons in a gram is exactly the Avogadro number by definition, or equivalently, a dalton is exactly equivalent to 1 gram/mol. Thereafter, these relationships will no longer be exact, but they will still be extremely accurate approximations.

The Avogadro constant, named after scientist Amedeo Avogadro, is the number of constituent particles, usually molecules, atoms or ions that are contained in the amount of substance given by one mole. It is the proportionality factor that relates the molar mass of a substance to the mass of a sample, is designated with the symbol NA or L, and has the value 6.022140857(74)×1023 mol−1 in the International System of Units (SI).Previous definitions of chemical quantity involved the Avogadro number, a historical term closely related to the Avogadro constant, but defined differently: the Avogadro number was initially defined by Jean Baptiste Perrin as the number of atoms in one gram-molecule of atomic hydrogen, meaning one gram of hydrogen. This number is also known as Loschmidt constant in German literature. The constant was later redefined as the number of atoms in 12 grams of the isotope carbon-12 (12C), and still later generalized to relate amounts of a substance to their molecular weight. For instance, the number of nucleons (protons and neutrons) in one mole of any sample of ordinary matter is, to a first approximation, 6×1023 times its molecular weight. Similarly, 12 grams of 12C, with the mass number 12 (6 protons, 6 neutrons), has a similar number of carbon atoms, 6.022×1023. The Avogadro number is a dimensionless quantity, and has the same numerical value of the Avogadro constant when given in base units. In contrast, the Avogadro constant has the dimension of reciprocal amount of substance. The Avogadro constant can also be expressed as 0.6023... mL⋅mol−1⋅Å−3, which can be used to convert from volume per molecule in cubic ångströms to molar volume in millilitres per mole.

Pending revisions in the base set of SI units necessitated redefinitions of the concepts of chemical quantity. The Avogadro number, and its definition, was deprecated in favor of the Avogadro constant and its definition. Based on measurements made through the middle of 2017 which calculated a value for the Avogadro constant of NA = 6.022140758(62)×1023 mol−1, the redefinition of SI units is planned to take effect on 20 May 2019. The value of the constant will be fixed to exactly 6.02214076×1023 mol−1.

CNO cycle

The CNO cycle (for carbon–nitrogen–oxygen) is one of the two known sets of fusion reactions by which stars convert hydrogen to helium, the other being the proton–proton chain reaction (pp-chain reaction). Unlike the latter, the CNO cycle is a catalytic cycle. It is dominant in stars that are more than 1.3 times as massive as the Sun.In the CNO cycle, four protons fuse, using carbon, nitrogen, and oxygen isotopes as catalysts, to produce one alpha particle, two positrons and two electron neutrinos. Although there are various paths and catalysts involved in the CNO cycles, all these cycles have the same net result:

4 11H + 2 e− → 42He + 2 e+ + 2 e− + 2 νe + 3 γ + 24.7 MeV → 42He + 2 νe + 3 γ + 26.7 MeVThe positrons will almost instantly annihilate with electrons, releasing energy in the form of gamma rays. The neutrinos escape from the star carrying away some energy. One nucleus goes on to become carbon, nitrogen, and oxygen isotopes through a number of transformations in an endless loop.

The proton–proton chain is more prominent in stars the mass of the Sun or less. This difference stems from temperature dependency differences between the two reactions; pp-chain reaction starts at temperatures around 4×106 K (4 megakelvin), making it the dominant energy source in smaller stars. A self-maintaining CNO chain starts at approximately 15×106 K, but its energy output rises much more rapidly with increasing temperatures so that it becomes the dominant source of energy at approximately 17×106 K.

The Sun has a core temperature of around 15.7×106 K, and only 1.7% of 4He nuclei produced in the Sun are

born in the CNO cycle. The CNO-I process was independently proposed by Carl von Weizsäcker and Hans Bethe in the late 1930s.

Carbon-13

Carbon-13 (13C) is a natural, stable isotope of carbon with a nucleus containing six protons and seven neutrons. As one of the environmental isotopes, it makes up about 1.1% of all natural carbon on Earth.

Carbon-14

Carbon-14, (14C), or radiocarbon, is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons. Its presence in organic materials is the basis of the radiocarbon dating method pioneered by Willard Libby and colleagues (1949) to date archaeological, geological and hydrogeological samples. Carbon-14 was discovered on February 27, 1940, by Martin Kamen and Sam Ruben at the University of California Radiation Laboratory in Berkeley, California. Its existence had been suggested by Franz Kurie in 1934.There are three naturally occurring isotopes of carbon on Earth: carbon-12, which makes up 99% of all carbon on Earth; carbon-13, which makes up 1%; and carbon-14, which occurs in trace amounts, making up about 1 or 1.5 atoms per 1012 atoms of carbon in the atmosphere. Carbon-12 and carbon-13 are both stable, while carbon-14 is unstable and has a half-life of 5,730±40 years. Carbon-14 decays into nitrogen-14 through beta decay. A gram of carbon containing 1 atom of carbon-14 per 1012 atoms will emit ~0.2 beta particles per second. The primary natural source of carbon-14 on Earth is cosmic ray action on nitrogen in the atmosphere, and it is therefore a cosmogenic nuclide. However, open-air nuclear testing between 1955–1980 contributed to this pool.

The different isotopes of carbon do not differ appreciably in their chemical properties. This resemblance is used in chemical and biological research, in a technique called carbon labeling: carbon-14 atoms can be used to replace nonradioactive carbon, in order to trace chemical and biochemical reactions involving carbon atoms from any given organic compound.

Isotope

Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number. All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.The term isotope is formed from the Greek roots isos (ἴσος "equal") and topos (τόπος "place"), meaning "the same place"; thus, the meaning behind the name is that different isotopes of a single element occupy the same position on the periodic table. It was coined by a Scottish doctor and writer Margaret Todd in 1913 in a suggestion to chemist Frederick Soddy.

The number of protons within the atom's nucleus is called atomic number and is equal to the number of electrons in the neutral (non-ionized) atom. Each atomic number identifies a specific element, but not the isotope; an atom of a given element may have a wide range in its number of neutrons. The number of nucleons (both protons and neutrons) in the nucleus is the atom's mass number, and each isotope of a given element has a different mass number.

For example, carbon-12, carbon-13 and carbon-14 are three isotopes of the element carbon with mass numbers 12, 13 and 14 respectively. The atomic number of carbon is 6, which means that every carbon atom has 6 protons, so that the neutron numbers of these isotopes are 6, 7 and 8 respectively.

Isotopes of carbon

Carbon (6C) has 15 known isotopes, from 8C to 22C, of which 12C and 13C are stable. The longest-lived radioisotope is 14C, with a half-life of 5,700 years. This is also the only carbon radioisotope found in nature—trace quantities are formed cosmogenically by the reaction 14N + 1n → 14C + 1H. The most stable artificial radioisotope is 11C, which has a half-life of 20.334 minutes. All other radioisotopes have half-lives under 20 seconds, most less than 200 milliseconds. The least stable isotope is 8C, with a half-life of 2.0 x 10−21 s.

Isotopes of sulfur

Sulfur (16S) has 24 known isotopes with mass numbers ranging from 26 to 49, four of which are stable: 32S (95.02%), 33S (0.75%), 34S (4.21%), and 36S (0.02%). The preponderance of sulfur-32 is explained by its production from carbon-12 plus successive fusion capture of five helium nuclei, in the so-called alpha process of exploding type II supernovas (see silicon burning).

Other than 35S, the radioactive isotopes of sulfur are all comparatively short-lived. 35S is formed from cosmic ray spallation of 40Ar in the atmosphere. It has a half-life of 87 days. The next longest-lived radioisotope is sulfur-38, with a half-life of 17 minutes. The shortest-lived is 49S, with a half-life shorter than 200 nanoseconds.

When sulfide minerals are precipitated, isotopic equilibration among solids and liquid may cause small differences in the δS-34 values of co-genetic minerals. The differences between minerals can be used to estimate the temperature of equilibration. The δC-13 and δS-34 of coexisting carbonates and sulfides can be used to determine the pH and oxygen fugacity of the ore-bearing fluid during ore formation.

In most forest ecosystems, sulfate is derived mostly from the atmosphere; weathering of ore minerals and evaporites also contribute some sulfur. Sulfur with a distinctive isotopic composition has been used to identify pollution sources, and enriched sulfur has been added as a tracer in hydrologic studies. Differences in the natural abundances can also be used in systems where there is sufficient variation in the 34S of ecosystem components. Rocky Mountain lakes thought to be dominated by atmospheric sources of sulfate have been found to have different δS-34 values from oceans believed to be dominated by watershed sources of sulfate.

Kilogram

The kilogram or kilogramme (symbol: kg) is the base unit of mass in the International System of Units (SI). Until 20 May 2019, it remains defined by a platinum alloy cylinder, the International Prototype Kilogram (informally Le Grand K or IPK), manufactured in 1889, and carefully stored in Saint-Cloud, a suburb of Paris. After 20 May, it will be defined in terms of fundamental physical constants.

The kilogram was originally defined as the mass of a litre (cubic decimetre) of water. That was an inconvenient quantity to precisely replicate, so in 1799 a platinum artefact was fashioned to define the kilogram. That artefact, and the later IPK, have been the standard of the unit of mass for the metric system ever since.

In spite of best efforts to maintain it, the IPK has diverged from its replicas by approximately 50 micrograms since their manufacture late in the 19th century. This led to efforts to develop measurement technology precise enough to allow replacing the kilogram artifact with a definition based directly on physical phenomena, a process which is scheduled to finally take place in 2019.

The new definition is based on invariant constants of nature, in particular the Planck constant which will change to being defined rather than measured, thereby fixing the value of the kilogram in terms of the second and the metre, and eliminating the need for the IPK. The new definition was approved by the General Conference on Weights and Measures (CGPM) on 16 November 2018. The Planck constant relates a light particle’s energy, and hence mass, to its frequency. The new definition only became possible when instruments were devised to measure the Planck constant with sufficient accuracy based on the IPK definition of the kilogram.

List of museums in the United Arab Emirates

This is a list of museums in the United Arab Emirates.

Al Ain National Museum

Al Eslah School Museum

Al Hisn Fort Museum

Dubai Museum

Fujairah Heritage Village

Fujairah Museum

Guggenheim Abu Dhabi

Louvre Abu Dhabi

Saeed Al Maktoum House

Salsali Private Museum

Salwa Zeidan Gallery

Sharjah Art Museum

Sharjah Museum of Islamic Civilization

Sharjah Maritime Museum

Sharjah Heritage Museum

Sharjah Classic Cars Museum

Sharjah Calligraphy Museum

Sharjah Archaeology Museum

Sheikh Obaid bin Thani House

Sheikh Zayed Palace Museum

Zayed National Museum

Tanki Online Museum

Carbon 12 Dubai

Mass excess

The mass excess of a nuclide is the difference between its actual mass and its mass number in atomic mass units. It is one of the predominant methods for tabulating nuclear mass. The mass of an atomic nucleus is well approximated (less than 0.1% difference for most nuclides) by its mass number, which indicates that most of the mass of a nucleus arises from mass of its constituent protons and neutrons. Thus, the mass excess is an expression of the nuclear binding energy, relative to the binding energy per nucleon of carbon-12 (which defines the atomic mass unit). If the mass excess is negative, the nucleus has more binding energy than 12C, and vice versa. If a nucleus has a large excess of mass compared to a nearby nuclear species, it can radioactively decay, releasing energy.

Mass number

The mass number (symbol A, from the German word Atomgewicht (atomic weight), also called atomic mass number or nucleon number, is the total number of protons and neutrons (together known as nucleons) in an atomic nucleus. It determines the atomic mass of atoms. Because protons and neutrons both are baryons, the mass number A is identical with the baryon number B as of the nucleus as of the whole atom or ion. The mass number is different for each different isotope of a chemical element. This is not the same as the atomic number (Z) which denotes the number of protons in a nucleus, and thus uniquely identifies an element. Hence, the difference between the mass number and the atomic number gives the number of neutrons (N) in a given nucleus: ${\displaystyle N=A-Z}$.

The mass number is written either after the element name or as a superscript to the left of an element's symbol. For example, the most common isotope of carbon is carbon-12, or 12
C
, which has 6 protons and 6 neutrons. The full isotope symbol would also have the atomic number (Z) as a subscript to the left of the element symbol directly below the mass number: 12
6
C
. This is technically redundant, as each element is defined by its atomic number, so it is often omitted.

Molar mass constant

The molar mass constant, symbol Mu, is a physical constant which relates relative atomic mass and molar mass. Its value is defined to be 1 g/mol in SI units.

The molar mass constant is important in writing dimensionally correct equations. It is common to see phrases such as

The molar mass of an element is the atomic weight in grams per mole.

However, atomic weight, i.e., relative atomic mass, is a dimensionless quantity, and cannot take the units of grams per mole. Formally, the operation is the multiplication by a constant which has the value 1 g/mol, that is the molar mass constant.

The molar mass constant is unusual (but not unique) among physical constants by having an exactly defined value rather than being measured experimentally. It is fixed by the definitions of the mole and of relative atomic mass. From the definition of the mole, the molar mass of carbon 12 is exactly 12 g/mol. From the definition of relative atomic mass, the relative atomic mass of carbon 12, that is the atomic weight of a sample of pure carbon 12, is exactly 12. The molar mass constant is given by

${\displaystyle M_{\rm {u}}={{M({}^{12}{\rm {C}})} \over {A_{\rm {r}}({}^{12}{\rm {C}})}}={1\ {\rm {g/mol}}}}$

The speed of light, the electric constant and the magnetic constant are other examples of physical constants whose values are fixed by the definitions of the International System of Units (SI), in these cases by the definitions of the metre and the ampere.

The molar mass constant is also related to the mass of a carbon-12 atom in grams:

${\displaystyle m({}^{12}{\rm {C}})={\frac {12M_{\rm {u}}}{N_{\rm {A}}}}}$

Hence the uncertainty in the value of the mass of a carbon-12 atom in SI units is governed by the uncertainty in the Avogadro constant: the CODATA 2006 recommended value is 1.992 646 54(10)×10−26 kg (ur = 5×10−8).

The relatively simple value of the molar mass constant in SI units is also a consequence of the way in which the International System of Units is defined. It is possible to quote the value of the molar mass constant in other units: for example, it is equal to (1/453.592 37) lb/mol ~ 2.204 623 262 × 10−3 lb/mol.

Mole (unit)

The mole is the base unit of amount of substance in the International System of Units (SI). Effective 20 May 2019, the mole is defined as the amount of a chemical substance that contains exactly 6.02214076×1023 (Avogadro constant) constitutive particles, e.g., atoms, molecules, ions or electrons.This definition was adopted in November 2018, revising its old definition based on the number of atoms in 12 grams of carbon-12 (12C) (the isotope of carbon with relative atomic mass 12 Da by definition). The mole is an SI base unit, with the unit symbol mol.

The mole is widely used in chemistry as a convenient way to express amounts of reactants and products of chemical reactions. For example, the chemical equation 2H2 + O2 → 2H2O can be interpreted to mean that 2 mol dihydrogen (H2) and 1 mol dioxygen (O2) react to form 2 mol water (H2O). The mole may also be used to represent the number of atoms, ions, or other entities in a given sample of a substance. The concentration of a solution is commonly expressed by its molarity, defined as the amount of dissolved substance per unit volume of solution, for which the unit typically used is moles per litre (mol/l), commonly abbreviated M.

The term gram-molecule was formerly used for essentially the same concept. The term gram-atom has been used for a related but distinct concept, namely a quantity of a substance that contains an Avogadro's number of atoms, whether isolated or combined in molecules. Thus, for example, 1 mole of MgBr2 is 1 gram-molecule of MgBr2 but 3 gram-atoms of MgBr2.

Relative atomic mass

Relative atomic mass (symbol: Ar) or atomic weight is a dimensionless physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to one unified atomic mass unit. The unified atomic mass unit (symbol: u or Da) is defined as being 1⁄12 of the atomic mass of a carbon-12 atom. Since both values in the ratio are expressed in the same unit (u), the resulting value is dimensionless; hence the value is said to be relative.

For a single given sample, the relative atomic mass of a given element is the weighted arithmetic mean of the masses of the individual atoms (including their isotopes) that are present in the sample. This quantity can vary substantially between samples because the sample's origin (and therefore its radioactive history or diffusion history) may have produced unique combinations of isotopic abundances. For example, due to a different mixture of stable carbon-12 and carbon-13 isotopes, a sample of elemental carbon from volcanic methane will have a different relative atomic mass than one collected from plant or animal tissues.

The more common, and more specific quantity known as standard atomic weight (Ar, standard) is an application of the relative atomic mass values obtained from multiple different samples. It is sometimes interpreted as the expected range of the relative atomic mass values for the atoms of a given element from all terrestrial sources, with the various sources being taken from Earth. "Atomic weight" is often loosely and incorrectly used as a synonym for standard atomic weight (incorrectly because standard atomic weights are not from a single sample). Standard atomic weight is nevertheless the most widely published variant of relative atomic mass.

Additionally, the continued use of the term "atomic weight" (for any element) as opposed to "relative atomic mass" has attracted considerable controversy since at least the 1960s, mainly due to the technical difference between weight and mass in physics. Still, both terms are officially sanctioned by the IUPAC. The term "relative atomic mass" now seems to be replacing "atomic weight" as the preferred term, although the term "standard atomic weight" (as opposed to the more correct "standard relative atomic mass") continues to be used.

Triple-alpha process

The triple-alpha process is a set of nuclear fusion reactions by which three helium-4 nuclei (alpha particles) are transformed into carbon.

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# Audio Jack Short Circuit Detect

In our current design we are using a 3.5 mm headset jack with a normally open switch. The jack is waterproof, now in the scenario when the jack is filled with water, assuming salty water the switch will close (contact established by water). In that case, we will need to detect it as fault and not enable the headset left and right output.

How do I detect that kind of scenario?

• it's not very waterproof if it's filled with water :) Make sure you are not overthinking this design. Do you really need to ensure the headset jack is not powered when underwater (faulted) as you describe it? What harm will it cause? I would think that the water, even salt water, is less of a load then the actual speakers are with copper wire. – Brian Onn Feb 19 '14 at 18:18
• I meant the Jack won't allow the water to go onto the PCB. – Ash Feb 20 '14 at 5:38

Let's see if we can tell the difference between an open and a closed switch in salty water.

A hard closed switch in free air will have just a few milliohms of contact resistance, which could range from 1 m$\Omega$ to 25m$\Omega$, depending on the switch quality. So what will be the resistance between the contacts of an open-switch in salty water? For the purpose of this exercise, let's assume sea-water, but even sea-water has differing levels of salinity, so just be aware of that for now.

Let's assume that the switch contact area is $1$ mm2 and the separation between the open contacts is just $1$ mm. According to Wikipedia, the resistivity $\rho$ of sea water is $\rho = 2 \times 10^{-1} \Omega\cdot m$. So an area $1$ mm2 across a distance of $1$ mm has a resistance of:

$$R = {\rho {l\over A}}$$

where $l$ is the length (or distance) across the resistive region and $A$ is the cross-sectional area of the region. So, perhaps your contact, when open in salt water, has a resistance like this:

$$R = {\rho {l\over A}} = {(2\times 10^{-1})\Omega\cdot m {1\times 10^{-3}m\over {1\times 10^{-3}m^2}}} = 2\times 10^{-1} \Omega \text { or 200}m\Omega$$

So it would seem that a small open switch in sea water, depending on the assumptions I made about contact area and separation, is very close to a closed switch and might be indistinguishable from a closed switch. In this case, given my assumptions, the open switch has a resistance of $200 m\Omega$ whereas the closed switch might have a resistance of $10 m\Omega \text{ to } 20 m\Omega$. You can use this fact with a current source to detect the voltage difference between a closed switch and an open switch in salt water with leakage resistance like this. You basically create a specific milliohmmeter that can detect above and below a threshold, perhaps anything above $50 m\Omega$ is considered an open switch and when below $50 m\Omega$ is a closed switch.

Next, what about capacitance? The capacitance equation for two plates separated by a gap filled with a dielectric is:

$$C = {k \epsilon_0 A \over d }$$

where $k$ is the relative permittivity of the dialectric (sea water) between the plates, $\epsilon_0 = 8.854 \times 10^{-12} F/m, A = 1\times 10^{-3}m^2 \text{ and } d = 1\times 10^{-3}m$

Using the Klein-Swift model for relative permittivity of sea water, and some matlab code from Professor Joel T. Johnson at Ohio State University, and some help from GNU Octave, we can come up with a value of approx $k = 79$, at freq $f = 1$ kHz, $20\,^{\circ}{\rm C}$ and $4.5$ g/kg salinity

octave:3> sea_eps(1000,20, 4.5)
ans =  7.89719e+01 - 1.28469e+07i

$$C = {k \epsilon_0 A \over d } = {(79)\cdot (8.854 \times 10^{-12}) {F\over m}\cdot (1\times 10^{-3}m^2)}\cdot {1\over {1\times 10^{-3}m}} \approx 700 pF$$

So the theoretical capacitance is $700$ pF, but the switch is not a very good capacitor, since it has such a low parallel resistance, we calculated above at about 200m$\Omega$

So, detecting the open or closed switch using the capacitance of the contacts is not going to be easy, because the extremely low parallel resistance of 200m$\Omega$ overwhelms any useful capacitance there might be.

So your switch looks like this underwater:

What can we do with a parallel RC? According to Wikipedia:

The parallel RC circuit is generally of less interest than the series circuit. This is largely because the output voltage $V_{out}$ is equal to the input voltage $V_{in}$ — as a result, this circuit does not act as a filter on the input signal unless fed by a current source

That page indicates that the transfer function from an input current to an output voltage is:

$$\frac{V_{out}}{I_{in}} = \frac{R}{1+sRC}$$

or

$$V_{out} = \frac{R}{1+sRC} \cdot {I_{in}}$$

So this tells me that the voltage out $V_{out}$ is inversely proportional to the capacitance, and so if you fed your switch with an AC current source above audible ranges, you could get an AC voltage out that would be filtered as the capacitance of the switch changes. In this case, the capacitance of the switch starts high, perhaps above $700$pF even, and decreases to 0, which basically means the voltage starts out low and increases to max voltage developed across the R according to Ohms law, $V_{out} = I_{in} \times R$

This is can work but provides no additional benefit above the milliohmeter style of detection I suggested up first above. So I don't suggest using the switch capacitance as a detection method like this, but the exercise above was academic, nonetheless.

So perhaps the best way is just to pass a small DC current through the switch and measure the voltage drop, as stated first. If you rely on the changing DC resistance of the switch, and just allow that anytime the DC switch resistance is less than perhaps $50 m\Omega$ then consider it on. If you KNOW that your specific brand of audio jack has switch contact resistance of less than $2 m \Omega$ then you can make your resistance threshold even tighter.

Finally, the inherent problem with this method is that the resistances involved are so low that you need a high current in order to have a detectable voltage; If your device is battery powered, then high currents are not feasible, and with battery-saving low currents you'll end up with just $10$'s of microvolts. This is fine, however you will need to add an op-amp gain stage to make it measurable. One option here is to pulse a higher current and synchronize your micro with the pulse to detect the voltage drop and determine if the switch is < $50 m\Omega$ or more than that threshold. Unfortunately, it's not going to help, for if you pulse a $10$ mA pulse every second through the switch you still only get $500\mu$V from it, and you still need to gain it up and you have a $1$ second max delay to detecting an audio jack insertion event.Plus the energy consumed per pulse at $10$mA is far greater than the energy consumed by a steady $500\mu$A current across $200{\rm m} \Omega$ It's probably better to just have a constant $500\mu{\rm A}$ current and gain up the voltage drop detect when it is above or below a voltage threshold.

references used
1. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html
2. http://en.wikipedia.org/wiki/Resistivity
3. http://en.wikipedia.org/wiki/RC_circuit#Parallel_circuit
4. http://rime.aos.wisc.edu/MW/models/kleinswift.shtml
5. http://www2.ece.ohio-state.edu/~johnson/5010/progs/sea_eps.m


text:
0
TECHNICAL PAPERS

# Radiative Properties of Patterned Wafers With Nanoscale Linewidth

[+] Author and Article Information
Y.-B. Chen

George W. Woodruff School of Mechanical Engineering, Georgia Institute of Technology, Atlanta, GA 30332

Z. M. Zhang1

George W. Woodruff School of Mechanical Engineering, Georgia Institute of Technology, Atlanta, GA 30332zzhang@me.gatech.edu

P. J. Timans

Mattson Technology, Fremont, CA 94538

1

Corresponding author.

J. Heat Transfer 129(1), 79-90 (Jun 08, 2006) (12 pages) doi:10.1115/1.2401201 History: Received January 31, 2006; Revised June 08, 2006

## Abstract

Temperature nonuniformity is a critical problem in rapid thermal processing (RTP) of wafers because it leads to uneven diffusion of implanted dopants and introduces thermal stress. One cause of the problem is nonuniform absorption of thermal radiation, especially in patterned wafers, where the optical properties vary across the wafer surface. Recent developments in RTP have led to the use of millisecond-duration heating cycle, which is too short for thermal diffusion to even out the temperature distribution. The feature size is already below $100nm$ and is smaller than the wavelength $(200–1000nm)$ of the flash-lamp radiation. Little is known to the spectral distribution of the absorbed energy for different patterning structures. This paper presents a parametric study of the radiative properties of patterned wafers with the smallest feature dimension down to $30nm$, considering the effects of temperature, wavelength, polarization, and angle of incidence. The rigorous coupled wave analysis is employed to obtain numerical solutions of the Maxwell equations and to assess the applicability of the method of homogenization based on effective medium formulations.

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## Figures

Figure 1

Parameters of selected cases, where dG and dT are the depths of gate and trench, respectively, and lG, lP, and lT are the lengths of gate, pitch, and trench

Figure 2

Optical constants of silicon at 25, 700, and 910°C: (a) refractive index, and (b) extinction coefficient

Figure 3

Schematic drawing for the TE wave incidence on a grating layer, showing the reflected diffraction orders j=−2, −1, 0, and 1

Figure 4

Calculated normal, spectral absorptance for plain Si(A‐1) and SiO2-coated Si(A‐2) at 25, 700, and 910°C

Figure 5

Spectral absorptance predicted using RCWA for cases with gratings at 910°C at θ=0 and 45deg: (a) case A-3; (b) case B-1; (c) case B-2; and (d) case B-3

Figure 6

Comparison of the absorptance predicted by different methods at 910°C: (a) TE wave for case A-3; (b) TM wave for case A-3; (c) TE wave for case B-1; (d) TM wave for case B-1; (e) TE wave for case B-3; and (f) TM wave for case B-3

Figure 7

Effect of incidence angle on the absorptance at 910°C for cases without gates: (a) TE wave for case A-1; (b) TM wave for case A-1; (c) TE wave for case A-2; (d) TM wave for case A-2; (e) TE wave for case A-3; and (f) TM wave for case A-3

Figure 8

Effect of incidence angle on the absorptance at 910°C for cases with gates: (a) TE wave for case B-1; (b) TM wave for case B-1; (c) TE wave for case B-2; (d) TM wave for case B-2; (e) TE wave for case B-3; and (f) TM wave for case B-3

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text:
## Thinking Mathematically (6th Edition)

The number is $140$.
step 1 Let x represent the unknown number in the problem. step 2 "$80\%$ of a number" = $\displaystyle \frac{80}{100}\cdot x=0.8x$ " ... $80\%$ of the number is added to the number" = $x+0.8x$ " the result is $252$" is written as " = $252$" step 3 The equation is $x+0.8x=252$ step 4 $x+0.8x=252\qquad$/simplify LHS $1.8x=252\qquad/\div 1.8$ $x=\displaystyle \frac{252}{1.8}=\frac{2520}{18}=140$ The number is $140$. step 5 Check the solution in the original wording of the problem. When 80% of 140, ($\displaystyle \frac{80}{100}\cdot 140=112),$ is added to the number (112+140), the result is 252. OK.


text:
Indets - Maple Help

Units

 Indets
 a units-safe frontend to indets

 Calling Sequence Indets(expresion) Indets(expresion, type)

Parameters

 expression - any expression type - a valid type with which to filter the output

Description

 • This command is a simple frontend to the global indets command. It still returns all the indeterminant subexpressions of the input of the given type but it removes the names of units and the unit data structures from the output.

Examples

 > $\mathrm{with}\left(\mathrm{Units}\right):$
 > $\mathrm{expr}≔{\left(x-30\mathrm{Unit}\left('\mathrm{fathom}'\right)\right)}^{2}+\mathrm{Unit}\left({'m'}^{2}\right)+{m}^{2}$
 ${\mathrm{expr}}{≔}{\left({x}{-}{30}{}⟦{\mathrm{fathom}}⟧\right)}^{{2}}{+}⟦{{m}}^{{2}}⟧{+}{{m}}^{{2}}$ (1)
 > $:-\mathrm{indets}\left(\mathrm{expr}\right)$
 $\left\{{\mathrm{fathom}}{,}{m}{,}{m}{,}{x}{,}⟦{\mathrm{fathom}}⟧{,}⟦{{m}}^{{2}}⟧\right\}$ (2)
 > $\mathrm{Indets}\left(\mathrm{expr}\right)$
 $\left\{{m}{,}{x}\right\}$ (3)


text:
# 13.6 Binomial theorem  (Page 3/6)

 Page 3 / 6

## Key equations

 Binomial Theorem ${\left(x+y\right)}^{n}=\sum _{k-0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}$ $\left(r+1\right)th\text{\hspace{0.17em}}$ term of a binomial expansion $\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$

## Key concepts

• $\left(\begin{array}{c}n\\ r\end{array}\right)\text{\hspace{0.17em}}$ is called a binomial coefficient and is equal to $C\left(n,r\right).\text{\hspace{0.17em}}$ See [link] .
• The Binomial Theorem allows us to expand binomials without multiplying. See [link] .
• We can find a given term of a binomial expansion without fully expanding the binomial. See [link] .

## Verbal

What is a binomial coefficient, and how it is calculated?

A binomial coefficient is an alternative way of denoting the combination $\text{\hspace{0.17em}}C\left(n,r\right).\text{\hspace{0.17em}}$ It is defined as $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ r\end{array}\right)=\text{\hspace{0.17em}}C\left(n,r\right)\text{\hspace{0.17em}}=\frac{n!}{r!\left(n-r\right)!}.$

What role do binomial coefficients play in a binomial expansion? Are they restricted to any type of number?

What is the Binomial Theorem and what is its use?

The Binomial Theorem is defined as $\text{\hspace{0.17em}}{\left(x+y\right)}^{n}=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}\text{\hspace{0.17em}}$ and can be used to expand any binomial.

When is it an advantage to use the Binomial Theorem? Explain.

## Algebraic

For the following exercises, evaluate the binomial coefficient.

$\left(\begin{array}{c}6\\ 2\end{array}\right)$

15

$\left(\begin{array}{c}5\\ 3\end{array}\right)$

$\left(\begin{array}{c}7\\ 4\end{array}\right)$

35

$\left(\begin{array}{c}9\\ 7\end{array}\right)$

$\left(\begin{array}{c}10\\ 9\end{array}\right)$

10

$\left(\begin{array}{c}25\\ 11\end{array}\right)$

$\left(\begin{array}{c}17\\ 6\end{array}\right)$

12,376

$\left(\begin{array}{c}200\\ 199\end{array}\right)$

For the following exercises, use the Binomial Theorem to expand each binomial.

${\left(4a-b\right)}^{3}$

$64{a}^{3}-48{a}^{2}b+12a{b}^{2}-{b}^{3}$

${\left(5a+2\right)}^{3}$

${\left(3a+2b\right)}^{3}$

$27{a}^{3}+54{a}^{2}b+36a{b}^{2}+8{b}^{3}$

${\left(2x+3y\right)}^{4}$

${\left(4x+2y\right)}^{5}$

$1024{x}^{5}+2560{x}^{4}y+2560{x}^{3}{y}^{2}+1280{x}^{2}{y}^{3}+320x{y}^{4}+32{y}^{5}$

${\left(3x-2y\right)}^{4}$

${\left(4x-3y\right)}^{5}$

$1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}$

${\left(\frac{1}{x}+3y\right)}^{5}$

${\left({x}^{-1}+2{y}^{-1}\right)}^{4}$

$\frac{1}{{x}^{4}}+\frac{8}{{x}^{3}y}+\frac{24}{{x}^{2}{y}^{2}}+\frac{32}{x{y}^{3}}+\frac{16}{{y}^{4}}$

${\left(\sqrt{x}-\sqrt{y}\right)}^{5}$

For the following exercises, use the Binomial Theorem to write the first three terms of each binomial.

${\left(a+b\right)}^{17}$

${a}^{17}+17{a}^{16}b+136{a}^{15}{b}^{2}$

${\left(x-1\right)}^{18}$

${\left(a-2b\right)}^{15}$

${a}^{15}-30{a}^{14}b+420{a}^{13}{b}^{2}$

${\left(x-2y\right)}^{8}$

${\left(3a+b\right)}^{20}$

$3,486,784,401{a}^{20}+23,245,229,340{a}^{19}b+73,609,892,910{a}^{18}{b}^{2}$

${\left(2a+4b\right)}^{7}$

${\left({x}^{3}-\sqrt{y}\right)}^{8}$

${x}^{24}-8{x}^{21}\sqrt{y}+28{x}^{18}y$

For the following exercises, find the indicated term of each binomial without fully expanding the binomial.

The fourth term of $\text{\hspace{0.17em}}{\left(2x-3y\right)}^{4}$

The fourth term of $\text{\hspace{0.17em}}{\left(3x-2y\right)}^{5}$

$-720{x}^{2}{y}^{3}$

The third term of $\text{\hspace{0.17em}}{\left(6x-3y\right)}^{7}$

The eighth term of $\text{\hspace{0.17em}}{\left(7+5y\right)}^{14}$

$220,812,466,875,000{y}^{7}$

The seventh term of $\text{\hspace{0.17em}}{\left(a+b\right)}^{11}$

The fifth term of $\text{\hspace{0.17em}}{\left(x-y\right)}^{7}$

$35{x}^{3}{y}^{4}$

The tenth term of $\text{\hspace{0.17em}}{\left(x-1\right)}^{12}$

The ninth term of $\text{\hspace{0.17em}}{\left(a-3{b}^{2}\right)}^{11}$

$1,082,565{a}^{3}{b}^{16}$

The fourth term of $\text{\hspace{0.17em}}{\left({x}^{3}-\frac{1}{2}\right)}^{10}$

The eighth term of $\text{\hspace{0.17em}}{\left(\frac{y}{2}+\frac{2}{x}\right)}^{9}$

$\frac{1152{y}^{2}}{{x}^{7}}$

## Graphical

For the following exercises, use the Binomial Theorem to expand the binomial $f\left(x\right)={\left(x+3\right)}^{4}.$ Then find and graph each indicated sum on one set of axes.

Find and graph $\text{\hspace{0.17em}}{f}_{1}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{1}\left(x\right)\text{\hspace{0.17em}}$ is the first term of the expansion.

Find and graph $\text{\hspace{0.17em}}{f}_{2}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{2}\left(x\right)\text{\hspace{0.17em}}$ is the sum of the first two terms of the expansion.

${f}_{2}\left(x\right)={x}^{4}+12{x}^{3}$

Find and graph $\text{\hspace{0.17em}}{f}_{3}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{3}\left(x\right)\text{\hspace{0.17em}}$ is the sum of the first three terms of the expansion.

Find and graph $\text{\hspace{0.17em}}{f}_{4}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{4}\left(x\right)\text{\hspace{0.17em}}$ is the sum of the first four terms of the expansion.

${f}_{4}\left(x\right)={x}^{4}+12{x}^{3}+54{x}^{2}+108x$

Find and graph $\text{\hspace{0.17em}}{f}_{5}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{5}\left(x\right)\text{\hspace{0.17em}}$ is the sum of the first five terms of the expansion.

## Extensions

In the expansion of $\text{\hspace{0.17em}}{\left(5x+3y\right)}^{n},\text{\hspace{0.17em}}$ each term has the form successively takes on the value $\text{\hspace{0.17em}}0,1,2,\text{\hspace{0.17em}}...,\text{\hspace{0.17em}}n.$ If $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k\end{array}\right)=\left(\begin{array}{c}7\\ 2\end{array}\right),\text{\hspace{0.17em}}$ what is the corresponding term?

$590,625{x}^{5}{y}^{2}$

In the expansion of $\text{\hspace{0.17em}}{\left(a+b\right)}^{n},\text{\hspace{0.17em}}$ the coefficient of $\text{\hspace{0.17em}}{a}^{n-k}{b}^{k}\text{\hspace{0.17em}}$ is the same as the coefficient of which other term?

Consider the expansion of $\text{\hspace{0.17em}}{\left(x+b\right)}^{40}.\text{\hspace{0.17em}}$ What is the exponent of $b$ in the $k\text{th}$ term?

$k-1$

Find $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{c}n\\ k\end{array}\right)\text{\hspace{0.17em}}$ and write the answer as a binomial coefficient in the form $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k\end{array}\right).\text{\hspace{0.17em}}$ Prove it. Hint: Use the fact that, for any integer $\text{\hspace{0.17em}}p,\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}p\ge 1,\text{\hspace{0.17em}}p!=p\left(p-1\right)!\text{.}$

$\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{l}n\\ k\end{array}\right)=\left(\begin{array}{c}n+1\\ k\end{array}\right);\text{\hspace{0.17em}}$ Proof:

$\begin{array}{}\\ \\ \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{l}n\\ k\end{array}\right)\\ =\frac{n!}{k!\left(n-k\right)!}+\frac{n!}{\left(k-1\right)!\left(n-\left(k-1\right)\right)!}\\ =\frac{n!}{k!\left(n-k\right)!}+\frac{n!}{\left(k-1\right)!\left(n-k+1\right)!}\\ =\frac{\left(n-k+1\right)n!}{\left(n-k+1\right)k!\left(n-k\right)!}+\frac{kn!}{k\left(k-1\right)!\left(n-k+1\right)!}\\ =\frac{\left(n-k+1\right)n!+kn!}{k!\left(n-k+1\right)!}\\ =\frac{\left(n+1\right)n!}{k!\left(\left(n+1\right)-k\right)!}\\ =\frac{\left(n+1\right)!}{k!\left(\left(n+1\right)-k\right)!}\\ =\left(\begin{array}{c}n+1\\ k\end{array}\right)\end{array}$

Which expression cannot be expanded using the Binomial Theorem? Explain.

• $\left({x}^{2}-2x+1\right)$
• ${\left(\sqrt{a}+4\sqrt{a}-5\right)}^{8}$
• ${\left({x}^{3}+2{y}^{2}-z\right)}^{5}$
• ${\left(3{x}^{2}-\sqrt{2{y}^{3}}\right)}^{12}$

The expression $\text{\hspace{0.17em}}{\left({x}^{3}+2{y}^{2}-z\right)}^{5}\text{\hspace{0.17em}}$ cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.

sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
I want to know trigonometry but I can't understand it anyone who can help
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
f(x)= 1 x    f(x)=1x  is shifted down 4 units and to the right 3 units.
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
I want to know partial fraction Decomposition.
classes of function in mathematics
divide y2_8y2+5y2/y2
wish i knew calculus to understand what's going on 🙂
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn
what's the derivative of 4x^6
24x^5
James
10x
Axmed
24X^5
Taieb
secA+tanA=2√5,sinA=?
tan2a+tan2a=√3
Rahulkumar
classes of function
Yazidu
if sinx°=sin@, then @ is - ?
the value of tan15°•tan20°•tan70°•tan75° -
NAVJIT
0.037 than find sin and tan?
cos24/25 then find sin and tan


text:
# Proving the Product rule

1. Dec 17, 2004

### Patrix0

Hi all!
I've been working with differentiation for quite some time now, and I know how to use the product rule.
f(x) = u(x)*t(x)
f'(x) = u'(x)*t(x) + t'(x)u(x)
But what I don't understand is how one derives it...

Oh, and one more thing.
How do we know a number is irrational, like pi and e ?
Like, how do we know that it has an infinite number of decimals?
Is it because when you do the decimal operation you will eventually see that there's a sequence of decimals which repeats itself, or not? Something else perhaps?

Thanks in advance.

Last edited: Dec 17, 2004
2. Dec 17, 2004

### shmoe

Hi, it can be derived using the definition of the derivative. Start with:

$$f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$

Now use the fact that f(x)=u(x)*t(x). Do you want to try to proceed on your own?

Proving a number like pi or e is irrational can be tricky. I think there's a post somewhere here where HallsOfIvy posted one of the easier proofs, I'll see if I can find it.

edit-aha! https://www.physicsforums.com/showthread.php?t=8193 The method was originally by I. Niven though Halls is probably refering to an article of A. Parks, who generalized it. It's not short, but it's accessible after first year calculus.

Last edited: Dec 17, 2004
3. Dec 17, 2004

### matt grime

That seems a remarkably uneasy* proof of the fact that e is irrational, which can be done from first principles merely from the definition of e.

let e=p/q, where p and q have no common factors

consider p(q-1)! - eq!

the resulting number can be shown to be both integer and between 0,1 and not equal to either. contradiction, hence e is irrational. It's a proof in the first year of any maths course.

irrationality of pi is harder.

* in the sense of one has to prove more complicated theorems to use it

Last edited: Dec 17, 2004
4. Dec 17, 2004

### Hurkyl

Staff Emeritus
Pictures sometimes help (of course, sometimes they don't). Suppose you have a rectangle with a width of u and a height of t. How does the area change if you adjust u and t by small amounts? (Say, by an amount of Δu and Δt respectively)

5. Dec 17, 2004

### shmoe

Absolutely, but for pi it's decent. Getting e is a nice bonus, though it also gets e to the power of any non-zero rational for free too.

I should have pointed out the simpler method for e, thank you. Though it's not p(q-1)! - eq! that will be an integer between 0 and 1, but q! times the terms in the infinite series for e after the qth term (maybe I'm misreading).

6. Dec 17, 2004

### matt grime

no, you're probably correct, few too many beers not helping my brain

7. Dec 17, 2004

### zeronem

Let there be a y = uv, where u = g(x), v = q(x) and y = f(x)

y = uv,

y+dy = (u +du)(v+dv) = uv + udv + vdu + dudv

then dy = udv +vdu + dudv,

dudv moves toward zero thus,
dy/dx = u(dv/dx) + v(du/dx) which is f'(x) = [g(x)]q'(x) + [q(x)]g'(x)

8. Dec 17, 2004

### fourier jr

i would use the mean-value theorem:
$$f'(x) =\lim_{x\rightarrow y} \frac{f(x)-f(y)}{x-y}$$, $$g'(x) =\lim_{x\rightarrow y} \frac{g(x)-g(y)}{x-y}$$,

for e i think the proof uses power series, for pi,
let $$\pi = a/b$$ where a & b are positive integers, and let

$$f(x) = \frac{x^n(a-bx)^n}{n!}$$

$$F(x) = f(x) - f^(2) (x) + f^(4) (x) - ... + (-1)^nf^(2n)(x)$$

since $$n!f(x)$$ has integer coefficients & terms in x whose degree is at least n, f(x) & its derivatives have integer values @ x=0, and also for $$x= \pi = a/b$$

(note that $$f(x) = \frac{b^n}{n!}(\pi - x)^n$$, so $$f(x) = f(\pi-x)$$ ).

from calculus,
$$\frac{d}{dx} [F'(x)sin(x) - F(x)cos(x)] = F''(x)sin(x) + F(x)sin(x) = f(x)sin(x)$$

and $$\int_{0}^{\pi} f(x)sin(x) dx = [F'(x)sin(x) - F(x)cos(x)]^\pi_0 = F(\pi) - F(0)$$, an integer, since the ith derivative of f evaluated at 0 & at $$\pi$$ are integers. but for $$0 < x < \pi$$,

$$0 < f(x)sin(x) < \frac{\pi^n a^n}{n!}$$

so that the integral is positive, but arbitrarily small for large n, since (another result from calculus) $$\lim_{n\rightarrow \infinity} \frac{x^n}{n!} = 0$$,

contradicting the fact that the integral above is an integer. so $$\pi$$ is irrational (that's ivan niven's excellent proof; i got it as a handout in a course i did last spring)

it seems that most proofs of irrationality are by contradiction. the proofs that e, pi & square roots of a (non-square) numbers are all donce by pretending that they are rational, setting them to a/b & going from there

Last edited: Dec 17, 2004
9. Dec 18, 2004

### Patrix0

Thanks for your help guys! I'll do my best to understand!

10. Jan 4, 2005

### learningphysics

There are definitely more elegant derivations of the product rule. But here's one way:

Let r(x)=p(x)q(x).

Let s(x,h)=[p(x+h)-p(x)]/h, h is not 0
Let t(x,h)=[q(x+h)-q(x)]/h, h is not 0

r'(x)
=limh->0[p(x+h)q(x+h)-p(x)q(x)]/h

write p(x+h) in terms of s(x,h) and q(x+h) in terms of t(x,h)

r'(x)
=limh->0[(hs(x,h)+p(x))(ht(x,h)+q(x))-p(x)q(x)]/h

Multiply out the numerator and simplify

r'(x)
=limh->0 [h^2s(x,h)t(x,h) + p(x)ht(x,h) + hs(x,h)q(x)]/h
=limh->0 h^2s(x,h)t(x,h)/h + limh->0 p(x)ht(x,h)/h + limh->0 hs(x,h)q(x)/h
=limh->0 hs(x,h)t(x,h) + p(x)limh->0 t(x,h) + q(x)limh->0 s(x,h)

Note that limh->0 t(x,h)=q'(x) and limh->0 s(x,h)=p'(x)

r'(x)
=p'(x)q'(x)limh->0 h + p(x)q'(x) + q(x)p'(x)
=0 + p(x)q'(x) + q(x)p'(x)
=p(x)q'(x) + q(x)p'(x)

11. Jan 5, 2005

### Zurtex

Last edited by a moderator: Apr 21, 2017
12. Jan 6, 2005

### fourier jr

what about this:

$$(fg)'(x) = \lim_{y \rightarrow x} \frac{ (fg)(y) - (fg)(x)}{y-x} = \lim_{y \rightarrow x} \frac{ f(y)g(y) - f(x)g(x)}{y-x} = \lim_{y \rightarrow x}\left( f(y) \frac{g(y) - g(x)}{y-x} + g(x) \frac{f(y) - f(x)}{y-x}\right) = f(x)g'(x)+f'(x)g(x)$$

13. Jan 6, 2005

### dextercioby

Sorry to disspoint you,but that's the traditional method of proving it.I was taught this method in school,when in the 11-th grade.
It's nothing spectacular in it,just elegance.

Daniel.

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# What's wrong with this RenameFile?

Bug introduced in 5.2 or earlier and persisting through 11.2.0
The bug appears on Windows only

I want to rename a .txt file.

For example create a file on your Desktop and name it use TEST.TXT, now I want to rename it to test.txt:

RenameFile["C:\\Users\\Administrator\\Desktop\\TEST.TXT",

It seems that RenameFile does not distinguish the UPPER CASE from the lower case in the file extensions. So how can I rename this file?

Ps: I use MATLAB in my homework, but MATLAB can only read *.m in terms of *.M files. So I must rename the *.M files. Therefore I use *.TXT and *.txt to make a example.

• Works like a charm for me on Linux. – Öskå Sep 30 '14 at 13:58
• I suspect this is an OS problem, since Windows, by default, does not utilize a case-sensitive filesystem, although there are some exceptions. – bobthechemist Sep 30 '14 at 14:09
• I am very happy to learn Mathematica as well as English at the same time here.*^_^* – partida Sep 30 '14 at 14:23
• Glad to know there is a workaround, but this ought to be reported to wolfram support. – george2079 Sep 30 '14 at 19:37
• @bobthechemist I don't think this is the problem of windows. windows cmd has ren function which has no problem rename upper case to lower case. I think this is apparently a bug. Latest 11.2 still has this problem. – matheorem Oct 22 '17 at 2:39

winRenameFile[f1_, f2_] := Module[{}, RenameFile[f1, "pre" <> f1];


text:
# transition-matrix's questions - Chinese 1answer

103 transition-matrix questions.

### Ergodic Markov chains and eigenvalues

I just read on wikipedia that a way to check whether a Markov chain is ergodic is to compute the eigenvalues of the transition matrix, and if those are all (except for 1) less than 1, then the chain ...

### Matrix exponential of an LTI system

Is there any direct proof to show that if eigenvalues of an LTI system are negative then the transition matrix ( or matrix exponential with respect to time) e^{At} decays to zero when t goes to ...

### 1 Is this process a Markov chain?

2 answers, 895 views markov-chains transition-matrix
Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state i, i = 0, 1, 2, 3, if the first urn contains i white ...

### How to model a markov chain that randomly switches between two transition matrices?

Let's say I have a discrete time, time-homogeneous Markov chain $X = \{X_{1}, \dots , X_{n}\}$ with state space $S= \{1,2,3\}$ and a transition matrix: \begin{bmatrix} .4 & .3 & .3\\ .3 &...

### Identify integers $1 \leq n /leq 20$ where $p_{0,0}^{(n)} > 0$ by using a transition matrix

As the title says I have to answer the following: Identify integers $1 \leq n /leq 20$ where $p_{0,0}^{(n)} > 0$ by using a transition matrix. I know this is with the states {0,1,2,3,4,5,6,7} and ...

### How to calculate a transition kernel / function?

I m dealing with the Kalman filter and the Signal Process is given by X(n+1) = B*X(n) + C*W(n+1)with W(n) ∈ R^n is i.i.d standard normal distributed, B ...

### Conditional transition probabilites

I am confusing myself with a problem which I give below. Can someone please comment and point out my mistakes. Say I have a two state system $S_1$ and $S_2$ the probability to transition to the two ...

### 1 The Determinant of a Transition Matrix

Suppose we have a linear map $T$ whose matrix is $A$ such that $$A: V \rightarrow V$$ where $A$ is the transition matrix from the basis $v_1,v_2,...,v_n$ to the basis $w_1,w_2,...,w_n$. If the ...

### Repair chain example problem in find the limiting distribution of a Markov Chain

I am working through Durrett's book Essentials of Stochastic Processes, which is quite good. He had a problem in the first chapter on Markov processes, and I had trouble understanding how he obtained ...

### Probability of specific paths in a Markov Chain

I have a transition matrix $$\mathbf P = \begin{bmatrix}0.9 & 0.06 & 0.04 \\ 0.6 & 0.3 & 0.1\\0.7 & 0.25 & 0.05\end{bmatrix}$$ The different states are labeled A, I, O in ...

### Markov transition matrix: $\lim \limits_{n\to \infty} P^n$ and $\lim \limits_{n\to \infty} \frac1n \sum_{k=1}^n P^k$

Given the Markov transition matrix P=\left( \begin{array}{ccccccc} 0 & 0 & 1 & 0 & 0\\ 0 & 0.3 & 0 & 0 & 0.7\\ 1 & 0 & 0 & 0 & 0\\ 0.4 & 0 & ...

### Ergodic Markov Chain [closed]

Suppose there exists an ergodic Markov chain with symmetric transition probabilities. For this Markov chain, why is the stationary distribution uniform?

### transition matrix between bases

1 answers, 157 views inverse transition-matrix


text:
# 12.6: The Reduction of Polar C–X σ Bonds

## Reduction of Halides

Alkyl halides can be reduced to alkanes through reaction with hydride reagents, most commonly $$\ce{LiAlH4}$$

Mechanism


text:
Public Group

# Optimizing Meshes

This topic is 3024 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi guys,

Without BLOWING my mind, can someone gently introduce me to how you go about optimizing a mesh when rendering? I've heard of octrees and quadtrees and all kind of trees bar oak trees and I was wondering if these can be applied to a model 5000 faces in size, as that's how big my main character is in face sizes. When I say main character I mean the one all humanoids will derive from, so there might be a few on screen. Damn I'd only need 5 to be hitting 25000+ faces per frame yikes!

Clearly something else must happen. It musn't be a simple case of chucking out a 5000 face model all the time and I'm not tooooo comfortable modelling LODS although I might just have to learn.

Is there a practical mathematical or code algorithm that can do this? Thanks ;o)

(Please try to keep it simple and always remember I am grateful for every reply).

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Well other than the line.

Mesh->OptimizeInplace( D3DXMESHOPT_ATTRSORT | D3DXMESHOPT_COMPACT | D3DXMESHOPT_VERTEXCACHE, (DWORD*)MeshBuffer->GetBufferPointer(), 0, 0, 0);

My knowledge of how to optimize models is hazy. Having one mesh and lots of animation tracks (per instance) is a decent means of reducing memory consumption but does not lower the CPU to GPU passing involved.

In DX10 I imagine you can pass the model in once and then pass in coordinates and an animation value and make the geometry buffer "build" the instance on the GPU for you cutting down the load between the two devices alot. Although of course I have never done that or I would be more specific or say you can do that.

I am aware of a VCache being a good way to increase rendering speeds down to the DX SDK although I do not know how to use it for all mesh allocation as the DX SDK tutorials haunt my nightmares with their convoluted code. All I can do to help is recommend you give OptimizeInplace a read over on the MSDN. I will watch this thread to see if anyone gives better advice I can follow too :p

As for management trees I think they come down to optimizing calculations for LOD in terrain and shadow lighting calculations. The only other instance you might be thinking of is a management system for ignoring models and objects offscreen in all drawing calculations.

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Nice reply thanks! It triggered off some ideas new and old. I'm using my own format so no access to hard coded DX functions and stuff sadly. I wonder if I can run something on a vertex buffer from hard coded functions? Unlikely probably.

I was thinking of uploading one set of vertices onto the GPU and then transforming these for each model using those vertices. So for example I upload the full set for the humanoid class. Then when animating I do the animation (vertex position changing) on the GPU. But I *somehow* keep an unaltered copy of the vertices on the GPU so that each model applies its own transform to this 'reference' vertex list (am I talking carp again?) so only one set ever has to be copied across.

Am I making enough sense for anyone to understand what I'm talking about?

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I am a sir. You are talking sense but the idea of using the meshes from one models code and drawing multiple instances of it would probably require a geometry shader (DX10+) I am not sure what the OpenGL equivlent version is.

Ok so I have never done this before but I know that if you have data in a vertex buffer of any description you can basically transfer it around so long as your confident of your vertex definition (I dont recommend FVF). What I am trying to do to clarify is take a vertex buffer dump its data into a mesh optimise it and then extract the vertex buffer. This is my first ever attempt at brainstorming a solution so dont hold me to this code because I have never done this before.

	//This by no account is the correct means to do it but if I was trying to do it I would be doing something like.	ID3DXMesh* meshGeneric; //Create a standardised mesh	meshGeneric->LockVertexBuffer(D3DLOCK_DISCARD, *YourVertexBufferData); //Attempt to lock and throw data into it from your vertex buffer.	//Attempt to run the optimisation based on your index buffers data. (might make the above line invalidated by the new data (I dont think so tho)	meshGeneric->OptimizeInplace(D3DXMESHOPT_ATTRSORT | D3DXMESHOPT_COMPACT | D3DXMESHOPT_VERTEXCACHE, (DWORD*)YourIndexBufferData->GetBufferPointer(), 0, 0, 0);	//Extract the optimised code out into your vertex buffer for your own intent.	meshGeneric->GetVertexBuffer(YourIndexBufferData);	//While I cannot be sure this is the right way to do this I can be sure its possible and that you wont be likely to out optimise this process.

Not sure thats how you do it but I am confident you could do it somehow. In retrospect you might need to perform lock and unlock commands during some of those processes.

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I'm sure I won't be the only one who will warn you about premature optimization, but mind the warning signs. If you don't know how much it will help things, than it's probably not worth it at the moment.

5000 faces is not much at all for remotely modern hardware... I've had 100 models(single texture, though for sure I'm not PS bound yet), rendering 650 verts in 400 faces through 4 DIP calls, rendered from 2 cameras... and even with the 800 DIP calls per frame(no instancing) I had a very happy 60 fps with no optimizations at all, and my rig is from the pre-vista days...

According to the research i've seen, which is by no means implied to be correct, a settexture(2500-3500cpu cycles) or worse setstreamsource(3500-6000) call is likely to stall you faster than 1200 - 1400 for DrawIndexedPrimitive. So having your tris ordered according to the texture/shader, will do wonders to help speed things up..

With that being said, the goal is to keep data in the cache as long as possible.
This means drawing nearby faces together. If I have to pull 3 verts for a tri in your nose, and then 3 more for a tri in your toes, and so on, it won't be long before the GPU says enough and kills your Framerate.

Look forward to seeing your work... I'm ready to tackle my own animation, but decided to stick around in GUI hell for a while...

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I can show 16 seconds of boredom in the video below of a repeated arm wave with my own mesh format:

It's just loaded out of a binary file with arrays corresponding to vertices/indices/normals/textures and the like laid out in a pre-defined way.

Every time I ask a question on here a whole new universe opens up. It will take me a while to digest what you have both said I'll reply properly shortly.

**Update**

EnlightenedOne:

Hi, is that only for .x files or can I use it on any vertexbuffer? Maybe a dumb question sorry. Very good reply! Lots of research needed there thanks alot for a great reply ;o)

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Quote:
 Original post by Burnt_FyrI'm sure I won't be the only one who will warn you about premature optimization, but mind the warning signs. If you don't know how much it will help things, than it's probably not worth it at the moment.

Ok thanks mate I'll keep that in mind!

Quote:
 Original post by Burnt_Fyr5000 faces is not much at all for remotely modern hardware... I've had 100 models(single texture, though for sure I'm not PS bound yet), rendering 650 verts in 400 faces through 4 DIP calls, rendered from 2 cameras... and even with the 800 DIP calls per frame(no instancing) I had a very happy 60 fps with no optimizations at all, and my rig is from the pre-vista days...

Sounds good!

Quote:
 Original post by Burnt_FyrAccording to the research i've seen, which is by no means implied to be correct, a settexture(2500-3500cpu cycles) or worse setstreamsource(3500-6000) call is likely to stall you faster than 1200 - 1400 for DrawIndexedPrimitive. So having your tris ordered according to the texture/shader, will do wonders to help speed things up..

So this means that I need to keep setStreamSource calls down right? So can I use one setStreamSource call and draw all my models off it, yet still animate them independently?

Quote:
 Original post by Burnt_FyrWith that being said, the goal is to keep data in the cache as long as possible.This means drawing nearby faces together. If I have to pull 3 verts for a tri in your nose, and then 3 more for a tri in your toes, and so on, it won't be long before the GPU says enough and kills your Framerate.

I didn't understand that bit. N00b time - what is the cache? Why does drawing verts in different places wreck the GPU's output?

Thanks ;o)

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The Cache as far as I understand it is your RAM although he might be refering to the tiny Cache of RAM on the CPU. D3DX calls might look specifically like they are for one purpose which is .x mesh format but you use D3DXMATRIX and D3DX all sorts whenever your using DX. That container I believe is just a generic form of storage for verticies which can hold any elaborate combination of vertex data you tell it too (to my knowledge). If that doesn't serve you true MSDN it.

HOLD IT!

You mesh looks nice but when I realised that was wireframe I was abit worried why on earth do you need such a high poly mesh? If you reduce the poly count to 2000 you could probably get a near identical looking character provided you used a shader to do per pixel lighting calculations so the light got distributed as it should rather than shading being done per vertex. It looks like your using the Fixed Function Pipeline to draw that and you have only upped the poly count to smooth the light/shadow appearance, am I wrong?

If not you need to stop worrying about optimisation and go straight to here and assimilate some critical code. My capabilities would be dead in the water without these invaluable examples.

Go here and click on HLSL and start learning the beauty of what the GPU can do for you.

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When drawing a triangle, the relevant vertices are fetched from the vertex-buffer, and are processed with the vertex shader. The outputs from the vertex shader are stored in the vertex-cache (but it has limited space, so new results will overwrite old results).

If another triangle is drawn soon after, and shares some vertices with the previous triangles, then it won't have to run the vertex shader 3 times - instead it can fetch some of the results from the cache.

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Quote:
 Original post by EnlightenedOneHOLD IT!You mesh looks nice but when I realised that was wireframe I was abit worried why on earth do you need such a high poly mesh? If you reduce the poly count to 2000 you could probably get a near identical looking character provided you used a shader to do per pixel lighting calculations so the light got distributed as it should rather than shading being done per vertex. It looks like your using the Fixed Function Pipeline to draw that and you have only upped the poly count to smooth the light/shadow appearance, am I wrong?

Spot on. I am using the fixed function pipeline. No shaders no HLSL. The model was mainly just to demo my animation explorations. I've found a much more suitable mesh of only 2000 polys for the body on TurboSquid. It will be being integrated in when I have some more cash.

I have used HLSL before but never got too deep into it although doing animation stuff on the GPU is something I consider a must before any release.

Thanks for a great reply time to check out that link. This kind of stuff is just what I wanted to know so I get any nasty stuff out of my head, and my code early on ;o)

Thanks for your reply too Hodgman ;o)

**Edit**

If I seem a bit daft it's because I'm quite n00b about all this to be truthful. I've book marked that page EnlightenedOne thanks alot.

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Groups of the periodic table [duplicate]

The groups go from 1 to 18 in the periodic table. When looking at the transition metals in the periodic table, You see roman numerals labelled with 'B,' in the order of III, IV, V, VI, VII, VIII, VIII, VIII, I, II. Why does III - VIII come before I and II? And why does VIII get repeated three times?

Also, what is the difference in the groups labelled A and B, besides the fact that the B groups are the transition metals?

• If you don't get old numeration than just uses new Table. – Mithoron May 5 at 19:36
• It doesn't. I see that there are 18 groups but why are they separated into A and B? Is there more significance to that than "transition vs non-transition?" And, within the B transition metals, why are the numerals ordered the way that they are? why does the numeral VIII appear 3 times whereas the other numerals appear only once? And why does III - VIII comes before I and II? These are my questions. – Kameron Shrum May 5 at 19:53
• The numbering with Roman numerals is outdated. Modern tables just label them 1-18. For a bit about why there are three columns of VIII, see the comments here – Tyberius May 6 at 15:14

Even though your question get downvoted, I think it is an legitimate question for people to understand the trend. I believe I-VIII Roman numeral nomenclature belong to CAS system while 1-18 Arabic numeral groups are recommended by IUPAC (Ref.1). The IUPAC recommended Periodic Table is given below (ignore the Roman numerals, which I put in there to explain):

According to Ref.1, the confusion of Periodic Table of Elements has been addressed as follows:

H.G. Deming used the long periodic table in his textbook General Chemistry (See following diagram with Roman numerals only; Ref.2), which appeared in the USA for the first time in 1923, and designated the first two and the last five Main Groups with the notation "A", and the intervening Transition Groups with the notation "B". The numeration was chosen so that the characteristic oxides of the B groups would correspond to those of the A groups. The iron, cobalt, and nickel groups were designated neither A nor B. The Noble Gas Group was originally attached by Ueming to the left side of the periodic table. The group was later switched to the right side and usually labeled as Group VlllA. This version of the periodic table was distributed for many years by the Sargent-Welch Scientific Company, Skokie, Illinois, USA. $$[\cdot\cdot\cdot]$$

The designations A and B have been extensively and rather arbitrarily used in the meantime in textbooks and in publications. An investigation of the application of the subgroup designations A and B in all articles, which appeared between 1972 and 1981 and covered by Chemical Abstracts, revealed a completely arbitrary use of the designation. Moreover, more than 10% of the articles it was nearly impossible, from the wording of the text, to recognize which elements were being discussed without further information (Ref.3).

The IUPAC Commission on the Nomenclature of Inorganic Chemistry (CNIC) proposed the designations of groups by Arabic numerals in 1984 (Ref.4), which was approved by the Nomenclature Commission of the American Chemical Society (ACS) and rest is history.

Thus, it is safe to say that this nomenclature is now outdated and not worth discussing.

References:

1. E. Fluck, “New notations in the periodic table,” Pure and Applied Chemistry 1988, 60(3), 431-436 (https://dx.doi.org/10.1351/pac198860030431).
2. Horace Grove Deming, General Chemistry; J. Wiley & Sons, Inc.: New York, NY, 1923.
3. W. C. Fernelius, W. H. Powell, “Confusion in the periodic table of the elements,” J. Chem. Educ. 1982, 59(6), 504-504 (https://doi.org/10.1021/ed059p504).
4. K. L. Loening, “Recommended Format for the Periodic Table of the Elements,” J. Chem. Educ. 1984, 61(2), 136-136 (https://doi.org/10.1021/ed061p136).


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# So little information!!!

Algebra Level 5

An arithmetic progression of $$n$$ terms has the following characteristics:

1. The first term $$a$$ is a positive odd integer
2. The common difference $$d$$ is a positive integer
3. The terms of the progression alternate between odd and even integers.

Now, it is given that

$\left(T_2 + T_4 + T_6 + T_8 + \ldots\right) - \left(T_1 + T_3 + T_5 + T_7 + \ldots\right) = 501$

Find the value of $$p$$ if $$n \equiv p \pmod{4}$$

Clarification: $$T_k$$ represents the $$k$$-th term in the progression

Inspiration (Recommended to solve this first)

×


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# Name for vectors in a Krylov space but not the preceding one

It seems to me that a useful concept to define when studying Krylov subspace methods is the idea of a vector that belongs to a Krylov subspace $\mathcal{K}_{n+1}(A,b)$ but not to the preceding one $\mathcal{K}_{n}(A,b)$, i.e., they can be written as a polynomial $p(A)b$ of degree exactly $n$, with nonsingular leading term.

Is there a standard name or notation for vectors that belong to $\mathcal{K}_{n+1}(A,b) \setminus \mathcal{K}_{n}(A,b)$? Or, how can I describe or denote them? Something like "$v$ is a proper vector in $\mathcal{K}_{n+1}(A,b)$" sounds good, but I don't want to invent a new term when there already exists a perfectly cromulent word for it1.


text:
Posted on

Post-Quantum Cryptography: What and Why Needs to be Changed Soon?

This article was prepared in cooperation with Microsec Ltd. Check out their blog for more interesting readings!

In this post, we explain the impact of prospective quantum computers on current implementations of cryptographic primitives, including digital signatures.

What is Post-Quantum Cryptography?

The goal of cryptography is to make any unwanted action (like reading a secret message or forging a signature) far more expensive than the benefit of such actions. Naturally, the cost of an attack largely depends on the cost of the tools necessary for the attack, meaning that the possible tools of an adversary must be taken into account in the design phase of every cryptosystem. For a long time, attacks that require a quantum computer were considered to have infinite cost since it is still an active research area being far from practical applications (e.g. for attacking cryptosystems ). At the same time, we cannot ignore the advancement of this field, as by the time the cost of using a quantum computer drops (i.e. it will be available and useful), we must be prepared for attacks that utilize this new tool. This is exactly what post-quantum cryptography (PQC) is dealing with. More precisely, the goal of PQC is to design classical cryptographic algorithms (for traditional computers) that resist both traditional and quantum attacks. It is important to note that PQC stands in contrast to the so-called quantum cryptography (e.g. quantum key distribution) that aims to build cryptosystems utilizing the principles of quantum mechanics. As these latter methods themselves require such infrastructure that is not widely available yet, they cannot be alternatives for the current techniques soon.

Next, we explore the challenges posed by the possibility of the two known quantum attacks.

Symmetric key Cryptography and Quantum Database Search

In cryptography, one-way functions (OWF) play a crucial role. Practically, a function is called one-way if it is easy to evaluate, but hard to invert, i.e. given an output of the function to find an input that is mapped to the same output by the function. Obviously, no encryption function can be secure without this property and the level of security is closely connected to the hardness of inversion. More precisely, a crypto primitive, constructed from a OWF, is said to offer $k$-bit security if inverting the OWF is expected to take $N=2^k$ evaluations of the function. When the structure of a function does not facilitate its inversion, the exhaustive search for the preimage is the best one can do. This is expected to take roughly $N/2$ invocation of the function, where $N$ is the size of the domain.

Perhaps surprisingly, we are not even sure about the existence of OWFs. We only conjecture that several functions have this important property based on the experience that no one could come up with efficient algorithms to invert them (or at least no one has revealed his or her ability to do so). Complexity theorists call “Minicrypt” the imaginary world, where OWFs provably exist; thus a significant portion of cryptography, roughly symmetric key crypto, can be securely realized regardless of the actual state of the art of algorithm design. The question arises: can we still hope that we are living in Minicrypt after the advent of quantum computers?

Fortunately, the answer seems to be positive. According to a lower bound, any generic brute force quantum attack to invert a function needs to evaluate the function at least roughly $\sqrt{N}$ times. This quadratic speedup over classical brute force is achievable. In 1996, Lov Grover proposed a quantum search algorithm that needs around $\sqrt{N}$ steps to find an element in an unordered database. This method naturally translates to a generic function inversion method with roughly $\sqrt{N}$ function evaluations. Regarding the security loss, any OWF based scheme with $k$ bit classical security will have at least ${k/2}$ bit security if we consider the possibility of quantum search. This is the reason why doubling the key length of a symmetric key cipher (like AES, the workhorse of internet security) or the output length of a hash function is considered as a simple but effective defense. However, it is worth noting that even a quadratic speedup of the attack seems to be too pessimistic for some schemes like the SHA3 hash function due to implementation issues. At the same time, it is important to keep in mind that the lower bound holds only for the generic case. So it does not rule out the existence of other quantum algorithms that invert only specific OWFs but faster than Grover’s method by utilizing their inner structure. This leads us to the second quantum algorithm, that turns out to be a more devastating attack for another branch of cryptography.

Public key Cryptography and the Shor Algorithm

For several cryptographic tasks, OWFs provably do not suffice. These include non-interactive key exchange, public key encryption, but currently used digital signatures are also built on stronger assumptions (however, this is not inherent in the last case as in the other cases). These primitives can be realized using a very specially structured subset of OWFs, called trapdoor functions that are also easy to evaluate and hard to invert, but the knowledge of a “trapdoor” (or “backdoor”) makes inversion easy as well. (Of course, the trapdoor must be hard to find knowing the function. Otherwise, the inversion would become easy.) The corresponding imaginary world, where public key cryptography is possible, is called Cryptomania. While quantum computing also does not seem to rule out this much more appealing world, by now, we know that public key crypto must be drastically different in “Post-quantum Cryptomania” than what we use today.

The (oversimplified) recipe for constructing public key cryptosystems is the following. Choose a -preferably long-standing- mathematical problem with sufficiently rich structure (not as easy as it may sound) and build a cryptosystem around it (this is at least as hard as it sounds). Then prove that any adversary, who can break it, must also be able to solve the underlying mathematical problem. This is a win-win situation as either your cryptosystem is secure or the open problem is solved. The only issue that causes concerns is that the mathematical problems we use today are not versatile enough. Namely, all ubiquitously used public key cryptosystems depend on variants of the following two assumptions.

1. Factorization, i.e. the decomposition of a sufficiently large composite number into a product of primes, is hard. Neither RSA encryption nor the RSA signature are secure if factorization is solvable efficiently.
2. The computation of the discrete logarithm in several algebraic structures is hard (e.g. finding the exponent $x$ that satisfies $g^x = y\mod p$ for given $g,y$ and certain $p$). The Diffie-Hellman key-exchange used, e.g. in TLS, and the entire elliptic curve cryptography, including ElGamal encryption or the ECDSA signature scheme, rely on this assumption.

Moreover, the above two problems are not even independent of each other. Both of them are encompassed by the so-called finite Abelian hidden subgroup problem (FA-HSP). This huge dependence on a single problem causes that cryptography has to fear of any new opportunities in computing. Indeed, in 1994 Peter Shor proposed a fast quantum algorithm for factorization. Furthermore, this can be extended to solve the more general FA-HSP and thus the discrete logarithm problem as well. Accordingly, the first quantum computer capable of running Shor’s algorithm will undermine the security of the whole public key infrastructure.

Fortunately, there is no fundamental reason to stick to these assumptions. The main cause why cryptosystems based on them became so widespread is their outstanding efficiency compared to other solutions. Because other approaches also exist, they were just suppressed by the success of RSA and elliptic curve crypto. But quantum computers will initiate a new race that hopefully leads to more versatile solutions in applied cryptography. If in the meanwhile no one refutes our hope that we are in Cryptomania…


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## Essential University Physics: Volume 1 (3rd Edition)

Recall, the cabin is pressurized, so the pressure inside of the plane is actually the pressure we feel. However, the question asks why we don't feel the $\underline{changing}$ pressure. We do not feel this because it impacts objects evenly from all different sides.